About the Authors
S P Ghosh obtained his BE (Hons) in Electrical Engineering from National Institute of Technology, Durgapur, and received a Master of Electrical Engineering
degree from Jadavpur University with specialization in High Voltage Engineering. He joined College of Engineering and Management, Kolaghat, as a lecturer
in 2002. Presently, he is working as an Assistant Professor in the department of
Electrical Engineering at College of Engineering and Management, Kolaghat.
He has published several papers in national and international conferences. He is
also pursuing his PhD at Bengal Engineering and Science University, Shibpur.
His areas of interest include Power Systems, Electrical Machines, and Artificial
Neural Networks.
A K Chakraborty received his BEE (Hons) from Jadavpur University, MTech
in Power System Engineering from IIT Kharagpur and PhD (Engineering)
from Jadavpur University. He joined College of Engineering and Management,
Kolaghat, in 1998 as Assistant Professor and was elevated to the rank of Professor in the Electrical Engineering Department. He served as HOD from 2002 to
2005 in the same department. Presently, he is working as Professor and Head of
the Department of Electrical Engineering. He also worked as a Lecturer in NIT
Silchar for five years. He served industries, namely, CESC Ltd. and Tinplate Company of India Ltd (a Tata Enterprise) for over fourteen years before joining this
institute. He is a Fellow of Institute of Engineers (India), Chartered Engineer,
Member IET (UK) and Life Member of ISTE. He has published several technical
papers in national and international conferences and also in reputed journals. He has guided several MTech
and PhD scholars. His research interests are in the field of Power System Protection, Economic Operation of
Power Systems, Deregulated Power System, and HVDC.
S P Ghosh
Assistant Professor
Department of Electrical Engineering
College of Engineering and Management
Kolaghat, West Bengal
A K Chakraborty
Professor and Head
Department of Electrical Engineering
College of Engineering and Management
Kolaghat, West Bengal
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To my family
My Wife, Lipika
Daughter, Adrita
S P Ghosh
To my family
Wife, Indira
Daughters, Amrita and Ananya
A K Chakraborty
Contents
Foreword
Preface
xv
xvii
1.
1–22
Introduction to Different Types of Systems
Introduction 1
1.1 Concepts of Signals and Systems 1
1.2 Different Types of Signals 2
1.3 Different Types of Systems 6
1.4 Interconnection of Systems 10
Solved Problems 11
Summary 17
Short-Answer Questions 18
Exercises 20
Questions 20
Multiple-Choice Questions 20
Answers 22
2
Introduction to Circuit-Theory Concepts
23–102
Introduction 23
2.1 Some Basic Terminologies of Electric Circuits 23
2.2 Different Notations 26
2.3 Basic Circuit Elements 27
2.4 Passive Circuit Elements 28
2.5 Types of Electrical Energy Sources 39
2.6 Fundamental Laws 41
2.7 Source Transformation 43
2.8 Network Analysis Techniques 48
2.9 Duality 50
2.10 Star-Delta Conversion Technique 52
Solved Problems 55
Summary 79
Short-Answer Questions 81
Exercises 89
Questions 92
Multiple-Choice Questions 92
Answers 102
3
Network Topology (Graph Theory)
Introduction 103
3.1 Graph of a Network 103
3.2 Terminology 104
3.3 Concept of a Tree 105
103–154
viii
Contents
3.4 Incidence Matrix [Aa] 107
3.5 Tie-Set Matrix and Loop Currents 110
3.6 Cut-Set Matrix and Node-Pair Potential 112
3.7 Formulation of Network Equilibrium Equations 115
3.8 Generalized Equations in Matrix Forms for Circuits having Sources 116
Solved Problems 118
Summary 147
Short-Answer Questions 147
Exercises 150
Questions 151
Multiple-Choice Questions 152
Answers 154
4
Network Theorems
155–230
Introduction 155
4.1 Network Theorems 155
4.2 Substitution Theorem 156
4.3 Superposition Theorem 156
4.4 Reciprocity Theorem 159
4.5 Thevenin’s Theorem 160
4.6 Norton’s Theorem 161
4.7 Maximum Power Transfer Theorem 166
4.8 Tellegen’s Theorem 170
4.9 Millman’s Theorem 172
4.10 Compensation Theorem 175
Solved Problems 177
Summary 217
Short-Answer Questions 218
Exercises 220
Questions 224
Multiple-Choice Questions 225
Answers 230
5
Laplace Transform and Its Applications
Introduction 231
5.1 Advantages of Laplace-Transform Method 231
5.2 Definition of Laplace Transform 232
5.3 Concept of Complex Frequency 232
5.4 Basic Theorems of Laplace Transform 233
5.5 Region of Convergence (ROC) 237
5.6 Laplace Transform of some Basic Functions 238
5.7 Laplace Transform Table 242
5.8 Other Important Laplace Transforms 243
5.9 Laplace Transform of Periodic Functions 244
5.10 Inverse Laplace Transform 244
5.11 Applications of Laplace Transform 248
5.12 Transient Analysis of Electric Circuits using Laplace Transform 250
5.13 Response with Pulse Input Voltage 268
5.14 Steps for Circuit Analysis using Laplace Transform Method 271
5.15 Concept of Convolution Theorem 271
231–326
ix
Contents
Solved Problems 273
Summary 303
Short-Answer Questions 304
Exercises 309
Questions 312
Multiple-Choice Questions 313
Answers 325
6
Two-Port Network
327–412
Introduction 327
6.1 Relationships of Two-Port Variables 327
6.2 Conditions for Reciprocity and Symmetry 334
6.3 Interrelationships between Two-Port Parameters 338
6.4 Interconnection of Two-Port Networks 339
6.5 Two-Port Network Functions 344
6.6 Transfer Functions of Terminated Two-Port Networks 345
6.7 Application of Network Parameters to the Analysis of Typical Two-Port Networks 348
6.8 Some Special Two-Port Networks 351
6.9 Image Parameters of a Two-Port Network 354
Solved Problems 359
Summary 398
Short-Answer Questions 398
Exercises 402
Questions 405
Multiple-Choice Questions 406
Answers 412
7
Fourier Series and Fourier Transform
Part I: Fourier Series 413
Introduction 413
7.1 Definition of Fourier Series 414
7.2 Dirichlet’s Conditions 414
7.3 Convergence of Fourier Series 414
7.4 Fourier Analysis 415
7.5 Waveform Symmetry 419
7.6 Truncating Fourier Series 422
7.7 Steady-State Response of Network to Periodic Signals 424
7.8 Steps for Application of Fourier Series to Circuit Analysis 424
7.9 Power Spectrum 425
Part II: Fourier Transform 425
Introduction 425
7.10 Definition of Fourier Transform 425
7.11 Convergence of Fourier Transform 426
7.12 Fourier Transform of Some Functions 427
7.13 Properties of Fourier Transforms 429
7.14 Energy Density and Parseval’s Theorem 432
7.15 Comparison between Fourier Transform and Laplace Transform 433
7.16 Steps for Application of Fourier Transform to Circuit Analysis 434
Solved Problems 434
413–472
x
Contents
Summary 460
Short-Answer Questions 460
Exercises 467
Questions 469
Multiple-Choice Questions 470
Answers 472
8
Sinusoidal Steady State Analysis
473–542
Introduction 473
8.1 Advantages of using Alternating Currents in Electrical Engineering 473
8.2 Basics of Sinusoids 474
8.3 Terminologies 474
8.4 Some Values of Alternating Quantities 476
8.5 Complex Number Systems 479
8.6 Phasor Representation 480
8.7 The j Operator 484
8.8 Phasor Diagrams 484
8.9 Circuit Response to Sinusoids 484
8.10 Kirchhoff’s Laws in Phasor Domain 485
8.11 Voltage and Current Phasors in Single-Element Circuits 485
8.12 Phasor Analysis of R-L Series Circuit 488
8.13 Phasor Analysis of RC Series Circuit 490
8.14 Phasor Analysis of RLC Series Circuit 492
8.15 Steps for Sinusoidal Steady-State Analysis (Phasor Approach to Circuit Analysis) 494
8.16 Concept of Reactance, Impedance, Susceptance and Admittance as Phasors 494
8.17 AC Power Analysis 496
8.18 Power Calculations in Different Electrical Elements 498
8.19 Sinusoidal Steady-State Response of Parallel AC Circuits 503
8.20 Sinusoidal Steady-State Response of Series–Parallel AC Circuits 507
Solved Problems 507
Summary 527
Short-Answer Questions 528
Questions 534
Exercises 535
Multiple-Choice Questions 537
Answers 542
9
Magnetically Coupled Circuits
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
543–590
Introduction 543
Self-Inductance 543
Coupled Inductor 544
Mutual Inductance 544
Mutual Inductance between Two Coupled Inductors 545
Dot Convention 546
Determination of Coefficient of Coupling from Energy Calculations in Coupled Circuits 548
Inductive Coupling 549
Linear Transformer 551
(conductively Equivalent Circuit of a Magnetically Coupled Circuit) 552
9.10 Ideal Transformer 553
xi
Contents
9.11 Tuned Coupled Circuits 555
Solved Problems 560
Summary 579
Short-Answer Questions 579
Exercises 585
Questions 587
Multiple Choice Questions 588
Answers 590
10 Three Phase Circuits
591–633
Introduction 591
10.1 Advantages of Polyphase Systems 591
10.2 Some Terminologies 592
10.3 Generation of Balanced Three-Phase Supply 593
10.4 Phase Sequence 594
10.5 Interconnection of Three-Phase Systems 595
10.6 Measurement of Power in Three-Phase Circuits 599
10.7 Conversion of Balanced Three-Phase System from Star to Delta 604
10.8 Analysis of Balanced Parallel Load 605
10.9 Analysis of Unbalanced Load Circuits 606
Solved Problems 610
Summary 625
Short-Answer Questions 626
Exercises 628
Questions 629
Multiple-Choice Questions 630
Answers 633
11 Resonance
634–685
Introduction 634
11.1 Series Resonance or Voltage Resonance 634
11.2 Parallel Resonance or Current Resonance or Anti-Resonance 641
11.3 Relation between Damping Ratio and Quality Factor 645
11.4 A More Realistic Parallel Resonant Circuit 646
11.5 Universal Resonance Curve 652
11.6 Applications of Resonance 653
Solved Problems 654
Summary
671
Short-Answer Questions 671
Exercise 678
Questions 679
Multiple-Choice Questions 681
Answers 685
12 Network Functions and Their Time-Domain and
Frequency-Domain Response
Introduction 656
12.1 Terminal and Terminal Pairs 686
12.2 Network Functions for a One-Port Network 687
12.3 Network Functions for Two-Port Networks 687
12.4 Poles and Zeros of Network Functions 688
686–754
xii
Contents
12.5 Pole Zero Diagram 689
12.6 Significance of Poles and Zeros 689
12.7 Natural Response and Natural Frequencies 690
12.8 Relation between Pole Position, Natural Response and Stability 691
12.9 Restriction on the Location of the Poles and Zeros in the s-Plane 692
12.10 Necessary Conditions for Driving Point Functions
(Restriction on Pole-Zero Locations in the s-Plane for Driving Point Functions) 693
12.11 Necessary Conditions for Transfer Functions
(Restriction on Pole-Zero Locations in the s-Plane for Transfer Functions) 697
12.12 Time Domain Behaviour from Pole–Zero Plot 697
12.13 Frequency Domain Behaviour from Pole–Zero Plot 699
Solved Problems 715
Summary 743
Short-Answer Questions 744
Exercises 747
Questions 749
Multiple-Choice Questions 749
Answers 754
13 Elements of Realizability and Network Synthesis
755–860
Part I: Elements of Realizability 755
Introduction 755
13.1 Elements of Realizability Theory 755
13.2 Hurwitz Polynomial 757
13.3 Positive Real Functions 759
Part II: Synthesis of Driving Point Functions 767
Introduction 767
13.4 Basic Synthesis Procedure 767
13.5 Methods of Synthesis 770
13.6 Driving Point Synthesis of One-Port Networks with Two Types of Elements 771
13.7 Synthesis of RLC Driving point Functions 792
Solved Problems 803
Summary 851
Short-Answer Questions 852
Exercises 855
Questions 857
Multiple-Choice Questions 858
Answers 860
14 Operational Amplifier and Active Filter
Introduction 861
14.1 Operational Amplifier (Op-Amp) 861
14.2 Filter 862
14.3 Advantages of Active Filters over Passive Filters
14.4 Application of Active Filters 863
14.5 Types of Active Filters 863
14.6 Low-Pass-Active Filter 864
14.7 High-Pass-Active-Filter 867
14.8 Band-Pass Active Filter 869
14.9 Band-Reject (Notch) Active Filter 876
14.10 Filter Approximation 878
14.11 All-Pass Active Filter 884
861–892
862
xiii
Contents
Summary 885
Short-Answer Questions 887
Exercises 889
Questions 889
Multiple-Choice Questions 889
Answers 892
15 Introduction To Software SPICE
Introduction 893
15.1 Types of Spice 893
15.2 Execution of SPICE (How Spice Works) 894
15.3 Types of Analysis 894
15.4 Model Statements 895
15.5 DC Circuit Analysis 903
15.6 Transient Analysis 903
15.7 AC Circuit Analysis 905
15.8 Fourier Analysis and Harmonic Decomposition using SPICE
15.9 Harmonic Recomposition 906
15.10 DC Sensitivity Analysis 906
Solved Problems 906
Summary 921
Questions 922
895–922
905
16 Indefinite Admittance Matrix (IAM)
923–930
16.1 Definition of Indefinite Admittance Matrix (IAM) 923
16.2 Properties of IAM 924
16.3 Applications of IAM 927
Exercises 931
17 Symmetrical Components
Introduction 933
17.1 Advantages of Symmetrical Component Method 933
17.2 a Operator 933
17.3 Symmetrical Components of an Unbalanced Three-Phase System 934
17.4 Component Synthesis (Evaluation of the Components) 935
17.5 Component Analysis 935
17.6 Graphical Method of Determining Sequence Components 935
17.7 Symmetrical Components of Current Phasors 936
17.8 Absence of Zero Sequence Components of Voltage and Current 936
17.9 Three-Phase Power in terms of Symmetrical Components 937
17.10 Sequence Impedances and Sequence Networks 938
17.11 Solution of 3-Phase Unbalanced Loads supplied from Unbalanced Supply 939
17.12 Solution of 3-Phase Unbalanced Loads supplied from Balanced Supply 941
Solved Problems 941
Summary 947
Exercises 948
Questions 949
Multiple-Choice Questions 949
Answers 950
Appendix A, B, C
931–948
Foreword
There is no necessity to emphasize that all engineering systems use electric circuits as components. Again,
the knowledge of circuit theory is very much essential to understand the operation of these systems. Circuit
Theory and Networks is an important subject which is common to almost all core and modern engineering
branches. Having a clear idea of the basic concepts is very much essential to both students, who are pursuing
their engineering courses, and the practicing engineers, who run plants and systems in a day-to-day way.
The book Network Analysis and Synthesis written by S P Ghosh and A K Chakraborty is the result of their
long association with teaching and plant-operation experiences. The book consists of 17 chapters which are
nicely written, starting from the fundamentals. Every chapter contains live examples and worked-out problems of standard universities, UPSC, IETE, AMIE and GATE examinations. The book has been written with
an up-to-date approach to accommodate the students of present standards and also to overcome the difficulties of teachers.
I am sure this book will be an asset not only to the teachers and students but also to practicing engineers
and technicians who are engaged in system operations.
I wish the publication all success.
S K Sen
BE Cal, Ph D (London), FIE, FNAE, DIC
FELLOW IMPERIAL COLLEGE (LONDON)
Former Minister-in-Charge, Power, Science Technology and
Non-Conventional Energy Sources, Govt. of West Bengal
Ex-Vice-Chancellor, Jadavpur University (Kolkata)
Ex-Prof. and Head, Electrical Engineering, B E College (Shibpur)
Hon. Member, Sikkim State Planning Commission
Hon. Advisor to the Chief Minister, Govt. of Sikkim, India
Preface
Brief Introduction to the Subject
Network analysis and Synthesis is a gateway course to all engineering subjects; Electrical Engineering, Electronics and
Communication Engineering, Computer Science and Engineering, Information Technology, Instrumentation Engineering in particular. Almost all engineering systems use electric circuits as components. Knowledge of Network Analysis
is very essential to understand the operation of these systems. Also, the subject of Network Analysis provides the background for understanding the behaviour of many other electrical and electronic devices. The present book has been written to provide knowledge of network analysis and synthesis, starting from the fundamentals.
Objectives
This book has been written as per the syllabi of Network Analysis and Synthesis, or Circuit Theory and Networks, as it is
taught under different universities in India. This text works well in our self-paced course, where students must rely on it as
their primary learning resource. Nonetheless, completeness and clarity are equally advantageous when the book is used in
a more traditional classroom setting. Cognizance of the present standard of students and the difficulties of the teachers has
been given due thought. The conceptual examples and practice problems and a variety of conceptual and multiple choice
questions at the end of each chapter give students a chance to check and to enhance their conceptual understanding.
Scope
This book is mainly written for the engineering students of different universities all over India for the subject of network
analysis and synthesis. However, as this course mitigates a definite percentage in every competitive examination of engineering professionals, viz., IES, UPSC, GATE, etc., we have written this book to help students see that a relatively small
number of basic concepts are applied to a wide variety of situations.
Salient Features
Some salient features of this book are
• Covers both network analysis and synthesis
• Rich pedagogy with large number of examples, solved problems, unsolved problems, MCQ's and short-answer type
questions and answers
• Contains large number of problems and questions from Indian universities, GATE, UPSC, AMIE, IETE and other
competitive examinations
• Discussion of the software packages PSPICE and MATLAB for solving network analysis problems
• Detailed coverage of different types of systems and networks
• Simple and student-friendly approach of writing
Organization
This book has a total of seventeen chapters. The first chapter provides information about the basic characteristics of different types of systems. The second chapter deals with the basic circuit components, laws and techniques for circuit analysis. Chapter 3 discusses the application of graph-theory concepts in circuit analysis. In this chapter, the application of a
xviii
Preface
mathematical tool like graph theory has been presented with the help of a large number of practical examples. Chapter 4 is
devoted to various network theorems necessary for simplified analysis of electrical problems. For examination purposes,
this chapter is very important as several questions frequently are set from this chapter. Chapter 5 introduces a new method
of circuit analysis—Laplace Transform method. Starting from the very fundamental concept of Laplace transform, its
applications in various complicated circuit problems has been discussed in detail in this chapter. The sixth chapter deals
with the concept of two-port network which has a vast application in many fields like transmission lines, filters and attenuators, and so on. Chapter 7 is divided into two parts. Part I presents the fundamentals of Fourier series and its application
for circuit analysis. Part II discusses Fourier transform and its applications. Chapter 8 discusses the method of studying
steady-state behaviour of electrical networks when sinusoidal excitations are applied. Chapter 9 deals with mutual inductance and magnetically coupled circuits. The tenth chapter explains the different aspects of three-phase circuits. Chapter 11
discusses a very important practical phenomenon of electrical engineering, called resonance. In this chapter, starting from
the basic concept, the conditions of different circuits under resonance and its application have been discussed. In Chapter
12, the relations between the various voltages and currents in a circuit have been discussed in terms of network functions
and responses. Chapter 13 discusses a new concept in the subject, known as network synthesis which aims at determining a
suitable electrical network given some operating characteristics. Chapter 14 is devoted to operational amplifiers and active
filters. Chapter 15 deals with a software package for circuit analysis, termed as SPICE. Chapter 16 explains circuit analysis
with the help of a tool, called indefinite admittance matrix. However, it should be mentioned that this method was very
useful earlier; with the advancement of digital computers, this method is becoming obsolete. The last chapter, Chapter 17
aims at the discussion of symmetrical component method of unbalanced three-phase circuits.
Acknowledgements
Authors are indebted to Prof. Nirmal Chatterjee, Prof. Kalyan Dutta, Prof. C K Roy, Prof. A N Sanyal of Jadavpur
University for their encouragement. We are grateful to Prof. S N Bhadra, HOD, Department of Electrical Engineering,
College of Engineering and Management, Kolaghat, and Ex-Professor of IIT, Kharagpur, Prof. P B Duttagupta, IIT
Kharagpur, Prof. H P Bhowmik, Ex-Principal, Institute of Leather Technology and Dr Abhinandan De, Bengal Engineering and Science University, Shibpur, for their constant inspiration and encouragement in the filed of academics. We
would also like to thank the reviewers for taking out time to review the book. Their names are given below.
Urmila Kar
Tirtha Shankar Das
T L Singhal
Ashutosh Marathe
Arvind Pachorie
A A Ansari
Shashi Gandhar
Vinay Pathak
Mahavir Singh
NSEC (Netaji Subhash Engineering College)
Garia, West Bengal
Guru Nanak Institute of Technology
Kolkata, West Bengal
Chitkara Institute of Enginering
Punjab
Pune University
Pune, Maharashtra
Government Engineering College
Jabalpur, Madhya Pradesh
Sagar Institute of Research and Technology
Bhopal, Madhya Pradesh
Bharati Vidyapeeth College of Engineering,
GGSIP University, New Delhi
Bhopal Institute of Technology
Bhopal, Madhya Pradesh
Accurate Institute
Greater Noida, Uttar Pradesh
xix
Preface
Pranita Joshi
A Subramanian
R Joseph Xavier
Vishwanath Hegde
K Amaresh
B Venkata Prasanth
Mumbai University
Mumbai, Maharashtra
V R S Engineering College
Villupuram, Tamil Nadu
Ramakrishna Institute of Technology
Coimbatore, Tamil Nadu
Malnad Engineering College
Hasan, Karnataka
KSRM Engineering College
Kadapa, Andhra Pradesh
NBKR Institute of Technology
Nellore, Andhra Pradesh
We are also thankful to the editorial and production staff of Tata McGraw Hill Education Private Limited for taking
interest in publishing this edition. Last but not the least, we acknowledge the support offered by our respective wives and
children without which this work would have not been successful.
Feedback
Criticism and suggestions for improvement shall be gratefully acknowledged. Readers may contact S P Ghosh at
ghosh_shankar@rediffmail.com and Dr A K Chakraborty at akcalll@yahoo.co.in.
S P Ghosh
A K Chakraborty
Visual Walkthrough
5
Each
c
that g hapter be
g
i
the c ves an id ins with
hapte
e
a abo an Intr
r.
o
ut th
e con duction
tents
of
Laplace Transform
and Its Applications
Introduction
Classical methods of solving differential equations become quite cumbersome when used for networks
involving higher order differential equations. In such cases, the Laplace transform method is used.
The classical methods consist of three steps:
(i) determination of complementary function,
(ii) determination of particular integral, and
iii) determination of arbitrary constants.
But, these methods become difficult for the equations containing derivatives; and transform methods
prove to be superior.
The Laplace transform is an integral that transforms a time function into a new function of a
complex variable. The term Laplace comes from the name of the French mathematician Pierre Simon
Laplace (1749–1827). The transformation method is a very effective tool for solving integro-differential
equations.
Laplace transformation is also a very powerful tool for network analysis. Any linear circuit consisting
of linear circuit elements can be solved by the knowledge of Laplace transformation.
In this chapter, we will first discuss the basics of Laplace transformation and then apply this transform
method to study the transient behaviour of electric circuits.
603
Three Phase Circuits
) = V I cos( 30° + )
(
W1 = V12 I1 cos 30° +
1. It gives complete solution.
2. Initial conditions are automatically considered in the transformed equations.
3. Much less time is involved in solving differential equations.
4. It gives systematic and routine solutions for differential equations.
L L
For a switch in the position 2, the wattmeter reading,
(
) = V I cos( 30° − )
) +V I cos( 30 − ) = 3V I cos = total power of the load
W2 = V13 I1 cos 30° −
)
(
(
∴ W1 + W2 = VL I L cos 30 +
5.1 ADVANTAGES OF LAPLACE-TRANSFORM METHOD
Laplace-transform methods offer the following advantages over the classical methods:
For a switch in the position 1, the wattmeter reading
L L
L L
L L
Thus, the sum of the wattmeter readings gives the load power, same as in a twowattmeter method. Here, also, if the current coil is to be reversed to obtain one
of the wattmeter readings then that reading should be treated as negative.
In case of a balanced delta-connected load, for a three-phase power measurement by one-wattmeter method, the resistance (say, R) of value equal to that of the
pressure coil of the wattmeter is connected in each of the remaining two phases,
as shown in Fig. 10.19. The pressure coil and the resistances form a balanced
star-connection.
W
1
i1
v3
i3
R
v1
i2
3
in
ded tter
i
v
o
pr
be
are pic for
s
o
e
.
Example 10.5 The power input to a three-phase induction motor is read by two wattmeters. The readrial
mpl
ch t
ings are 1000 W and 500 W. Find out the pf of the motor. If the line voltage is 400 V, find the line current.
Exa fter ea xt mate
Solution Here, W
1000 W; W
500 W, V
400 V
d
a
e
e
t
r
k
⬖ power factor of the motor,
Wor chapte g of the
⎡
⎡
n
3 (W − W ) ⎤
3 (1000 − 500 ) ⎤
h
⎡
i
1 ⎤
c
d
⎥
⎥
⎢
⎢
a
= cos tan
= cos ⎢ tan
cos = cos tan
n
⎥ = 0.5
e
W +W
1000 + 1500 ⎥
⎥
⎢
⎢
3⎦
⎣
⎣
⎣
⎦
⎦
ersta
und
1
2
−1
v2
R
2
Fig. 10.19 One-wattmeter
method for a balanced deltaconnected load
L
1
−1
2
1
−1
2
Line current is
IL =
P
3VL cos
W1 + W2
=
1000 + 500
=
3VL cos
3 × 400 × 0.5
= 4.33 A
W
10.6.2 Measurement of Reactive Power
In case of a balanced three-phase load, the reactive
power can be measured using one wattmeter.
The connection is shown in Fig. 10.20. Here, the
current coil of the wattmeter is connected in one line
and the pressure coil is connected across the other two
lines.
The phasor diagram is shown in Fig. 10.21.
The wattmeter reading is
(
W = V32 I1 cos 90° +
1
i1
N
3
2
i3
V3
i2
p
⇒ W = − 3V p I p sin
V2
V32
V2
Fig. 10.20 Measurement
of reactive power for a
3-phase balanced starconnected load
) = V I cos( 90° + )
L
V1
I3
(90
␾) V1
I
␾ 1
␾
V3
I2
␾
V2
Fig. 10.21 Phasor
diagram for reactive
power measurement for
balanced 3-phase starconnected load
560
Network Analysis and Synthesis
When k ⴝ kC
Each
c
taken hapter co
n
f
all ov rom the tains a lar
ge nu
quest
er Ind
m
io
ia an
d oth n papers ber of solv
of dif
er co
ed pr
mpet
itive ferent uni oblems
exam
v
inatio ersities
ns.
• In this condition, the resistance which the secondary circuit couples into the primary at resonance is
equal to the primary resistance.
• The secondary current will be maximum.
• The curve of the secondary current will be broader and flat-topped.
• The curve of the primary current will have two peaks.
When k > kC
• The double peaks of the primary current become more prominent; the peaks being separated from each
other.
• The magnitude of the primary current at peaks becomes smaller as the value of k is increased.
• The curve of the secondary current will also have two peaks.
Solved Problems
Problem 9.1 Find the effective value of the inductance for the following connections:
(a)
(b)
2H
i
i
i
5H
(c)
i
10 H
1H
2H
4H
1H
i
5H
2H
1H
i
3H
2H
Fig. 9.33
Solution
(a) This is a series-aiding connection. The effective inductance is,
)
(
⬗ Leq = L1 + L2 + 2 M = 5 + 10 + 2 × 2 = 19 H
(b) This is a series-opposing connection. The effective inductance is,
)
(
⬗ Leq = L1 + L2 − 2 M = 2 + 4 − 2 × 1 = 4 H
(c) Since the coils are magnetically coupled in series aiding or they assist each other, therefore,
(
) (
)
(
) (
)
= ( L + M + M ) = ( 5 + 2 + 1) = 8 H
effective inductance for the coil 1 is L1eff = L1 + M12 + M13 = 2 + 1 + 2 = 5 H
effective inductance for the coil 2 is L2 eff = L2 + M12 + M 23 = 3 + 1 + 1 = 5 H
effective inductance for the coil 3 is L3eff
Total effective inductance is
(
3
)
13
23
(
)
Leff = L1eff + L2 eff + L3eff = L1 + L2 + L3 + 2 M12 + M 23 + M13 = 18 H
527
Sinusoidal Steady State Analysis
Summary
1. For transmission and distribution, alternating current
has a number of advantages over direct current.
2. A sinusoid is a signal that has the form of a sine or
cosine function and in general can be written as
)
)
v (t = V m sin t . A shifted sinusoid can be written as
)
v (t = V m sin( t +
where, Vm is the amplitude, ␻ is
2
, T is the time period
2␲f
T
of the sinusoid and ␾ is the phase of the sinusoid.
Use of sinusoids has several advantages like minimum
disturbance in electrical circuits, less interference to
nearby communication lines and less iron and copper
losses.
The value of an alternating quantity at any instant of
time is known as the instantaneous value.
The maximum value of an alternating quantity attained
in each cycle is known as the peak or maximum or crest
value.
The average value of an alternating quantity over a
given time interval is the summation of all instantaneous values divided by the number of values taken
T
1
over that interval. Mathematically, V av = ∫vdt , where
T 0
T is the time period of the quantity.
the angular frequency
3.
4.
5.
6.
7. The rms or effective value of a continuous periodic
t
T2 is
function f(t) defined over the interval T1
T
f rms =
T
2
2
1 2
⎡f t ⎤ dt or, f = 1 ⎡f t ⎤ dt
⎦
rms
⎦
T2 −T1 T∫ ⎣
T ∫0 ⎣
1
()
()
8. Form factor is the ratio of the rms value to the average
value for an alternating wave.
rms Value
( ) average
value
∴ form factor K f =
For a sinusoidal wave, its value is 1.11.
9. Peak factor is the ratio of the peak value to the rms
value for an alternating wave.
( )
peak value maximum value
=
rms value
rms value
For a sinusoidal wave its value is 1.414.
10. A phasor is a complex quantity that represents both
the magnitude and phase angle of a sinusoid. For a
sinusoid given as v t = V m cos t + , the corre∴ peak factor K p =
()
(
)
sponding phasor is written as, v (t ) = V cos ( t + ) .
m
11. The graphical representation of the phasors of sinusoidal quantities taken all at the same frequency and with
proper phase relationships with respect to each other
is called a phasor diagram.
12. Both KVL and KCL hold good in phasor domain, i.e.,
V1 +V 2 +V 3 + ⋅⋅⋅ +V n = 0 and I 1 + I 2 + I 3 + ⋅⋅⋅ + I n = 0 .
13. The voltage and current in different circuit elements
have definite phase relations. For a resistor, the voltage
and current are always in phase, i.e., the phase angle is
zero. In a pure inductor, the current lags behind the
voltage by 900 and in a pure capacitor, the current
leads the voltage by 900.
14. Impedance (Z) of any two-terminal network is the
ratio of the phasor voltage (V) to the phasor current (I )
V
j ∠Z
= Z ∠Z . The real part
i.e. Z = = R + jX = Z e
I
of impedance Re[Z]
R is called the resistance. The
(
)
imaginary part of impedance Im[Z]
X is called the
reactance. Impedance, resistance and reactance are all
measured by the same unit, ohm ( ).
Z = R ; for aresistor
= j L ; for aninductor
1
=
; for a capacitor
j C
15. The reciprocal of the impedance Z is called admittance. So, it is the ratio of the phasor current to the
phasor voltage, i.e. . The real part of admittance
R
is called conductance, G = Re ⎡⎣Y ⎤⎦ = 2
. The
R +X 2
imaginary part of admittance is called susceptance,
X
B = Im ⎡⎣Y ⎤⎦ = 2
. Admittance, conductance and
R +X 2
susceptance are all measured by the same unit, siemen (S).
16. Instantaneous power absorbed by an element is the
product of the instantaneous voltage v(t) and the
instantaneous current i(t), i.e., p(t) v(t) i(t) (in watts).
17. Average or real or active power (in watts) is the
average of the instantaneous power over a time
T
1
interval, i.e., P = ∫ p t dt . For the sinusoidal voltT 0
age and current given as v t = V m cos t + v and
()
()
)
)
(
)
i (t = I m cos ( t + i , the average power is given as
1
P = V m I m cos ( v − i = V rms I rms cos ( v − i .
2
)
)
that
ary the
m
sum d in
ns a covere
i
a
t
on opics
t
er c
hapt portant
c
h
c
m
Ea
ei
s th
give ter.
chap
744
Network Analysis and Synthesis
N(s)
( s − z )( s − z ) ⋅⋅⋅( s − z )
( ) D s = K s − p s − p ⋅⋅⋅ s − p
( ) ( )( ) (
)
j − z )( j − z )( j − z ) ⋅⋅⋅( j − z )
(
=K
( j − p )( j − p )( j − p ) ⋅⋅⋅( j − p )
F s =
1
If, j − z i =
Each
with chapter c
o
a
writi nswers; w ntains a s
ng br
ief an hich acts et of shor
ta
d to-t
he-po s a guide answer q
uesti
to the
int an
on
swer
s in e students s
for
xami
natio
ns.
(
(
)
i
s= j
3
n
1
2
3
m
of all zeros lines to j
( ) = K Product
Productt of all poles lines to j
F j
n
∏(j −z )
2
zi
∏(j −p )
i
2
(
+ j −
) =q
⬔F( j␻)
2
i
⎛ − pi ⎞
∠ j − pi = tan ⎜
⎟= i
⎝ − pi ⎠
(
)
r1r2 ⋅⋅⋅ rn
q1q2 ⋅⋅⋅ qm
and angle
i
pi
K
i
i =1
−1
pi
i
= K im=1
) =r
⎛ − zi ⎞
∠ j − z i = tan ⎜
⎟=
⎝ − zi ⎠
and, j − pi =
m
2
+ j −
zi
2
Hence, the magnitude and phase angle of the complete frequency response may be written as
n
2
2
1
1
(summation of angles of the vectors from
zeros to j␻-point)
(summation of angles
of the vectors from poles to j␻-point)
n
)
(
m
(
= ∑ ∠ j − z i − ∑ ∠ j − pi
−1
i =1
)=( +
then the network function may be written as
1
2
+
3
i =1
+ ⋅⋅⋅ +
n
)
) − ( + + + ⋅⋅⋅+ )
1
2
3
n
15. The variation of magnitude and phase of a network
function with frequency in logarithmic scale is known
as Bode plot.
of all zeros lines to j
( ) = K Product
Productt of all poles lines to j
F j
Short-Answer Questions
1. What are the poles and zeros? What information
do they provide in respect of the network to which
they relate?
We consider a network function given by the ratio of
two polynomials as
an s n + an−1s n−1 + ⋅⋅⋅+ a2 s 2 + a1s + a0
( ) b s +b s
F s =
m
m
m−1
m −1
+ ⋅⋅⋅+ b2 s 2 + b1s + b0
(1)
It is often convenient to factor the polynomials in the
numerator and denominator, and to write the transfer
function in terms of those factors:
N(s)
( s − z )( s − z ) ⋅⋅⋅( s − z )
( ) D s = K s − p s − p ⋅⋅⋅ s − p
( ) ( )( ) (
)
F s =
1
2
n
1
2
m
(2)
where, the numerator and denominator polynomials,
N(s)and D(s), have real coefficients defined by
a
the system’s differential equation and K = n
bm
is a positive constant, known as scale factor.
From Eq. (2), we can observe the following:
At s zi,i 1,2,3....,n, the numerator polynomial N(s) 0;
these complex frequencies are known as the zeros of
the network function F(s). At zeros, the value of the network function is zero, i.e., Lim F(s) 0.
s →Z i
At s pi, i 1,2,3...., m, the denominator polynomial
D(s) 0; these complex frequencies are known as the
poles of the network function F(s). At poles, the value of
.
the network function is infinity, i.e., Lim F(s)
s → Pi
• Significance of poles and zeros The values of poles
and zeros of F(s) and their locations in the s-plane completely specify a network function. All the coefficients
of polynomials N(s) and D(s) are real, therefore the
poles and zeros must be either purely real, or appear
in complex conjugate pairs. In general for the poles,
either pi ␴i , or else pi , pi 1 ␴ j␻i. The existence of
a single complex pole without a corresponding conjugate pole would generate complex coefficients in the
polynomial D(s). Similarly, the system zeros are either
real or appear in complex conjugate pairs.
The poles and zeros are properties of the transfer function, and therefore of the differential equation describing the input–output system dynamics.
Together with the gain constant K they completely
characterize the differential equation, and provide a
complete description of the system.
2. What do you understand by driving point impedance of a two-port network? Enumerate important
properties of driving point impedance functions of
a two-port passive network.
467
Fourier Series and Fourier Transform
Exercises
Fourier Series
1F
1. Find the Fourier series expansion for the following
functions and sketch the frequency spectrum.
(a)
f (t )
t
0
T
2T
T
f (t)
T/2
T T/2 0 T/2 T
2T
t
2
2␲
␲
0
␲␻
2
v
␲
1
2␲ ␻
20
3␲
2
f (t)
(c)
2␲
␲
2H
v(t )
t
0
A
(b)
hat
ter t ut
p
a
h
c
,b
each roblems ents,
Fig. 7.63 (a)
Fig. 7.63 (b)
f
o
end
4. Find the Fourier series expansion for the waveforms
ng p ignm
shown in Fig. 7.64.
t the practici ks, ass
a
n
⎡
⎤
1
1
1
wor
for
give
[(a) v = −2 ⎢ sin x + sin2 x + sin3 x + sin 4 x + ⋅⋅⋅⎥
2
3
4
⎣
⎦
s is udents tutorial
e
s
i
c
t
V 4V
4V
4V
cos x +
cos 3 x +
cos 5 x + ⋅⋅⋅ ]
exer
ting
he s
(b) v = 2 +
ns.
(3 )
(5 )
t of only t s in set questio
e
s
t
r
A s no
(a)
tion
ache
help the te xamina
e
s
help es and
z
z
i
u
(b)
q
v (t )
10
␲␻
0
2␲ ␻
t
Fig. 7.56
4␲
x
3␲
␲
f (x )
∞
( ) A2 + ∑ nA sinn t
[Ans: (a) f t =
V
n =1
⎤
T 2T ⎡
1
(b) f (t = − 2 ⎢cos t + 2 cos 3 t + ⋅⋅⋅⎥
4
3
⎣
⎦
)
()
(c) f t =
1
1
2 ∞
1
+ sin t − ∑ 2 cos 2 n t ]
2
n =1 4 n − 1
2. A periodic waveform as shown in Fig. 7.62 feeds an RL
1
H. Calculate the power
load with R 10 ohm and L
2
␲
[v =
at the fundamental frequency supplied to the load.
x
V m 2V m ⎛
⎞
1
− 2 ⎜ cos x + cos 5 x + ⋅⋅⋅⎟ +
4
25
⎝
⎠
⎤
Vm ⎛
⎞
1
1
1
⎜⎝ sin x − 2 sin2 x + 3 sin3 x − 4 sin 4 x + ⋅⋅⋅⎟⎠ ]
⎦
f (t )
A
t
0
0 ␲ 2␲ 3␲
Fig. 7.64
5. A triangular wave increases linearly from 0 to Vm during
the interval 0 to ␲. The wave has zero value during the
interval ␲ to 2␲ and this cycle is repeated. Find the
Fourier series representation of the wave.
T
2T
Fig. 7.62
3. A waveform of the shape shown in Fig. 7.63 (a) is
applied to the network shown in Fig. 7.63 (b). Calculate the power dissipated in a 20- resistor. Take
␻ 1 rad/s.
[1.17 W]
6. A wave has a constant value Im during the interval −
2
3
and Im during the interval
to
. This cycle
2
2
2
is repeated in the next intervals. Find the Fourier series
for the wave.
to
⎡ 4I m ⎛
⎞⎤
1
1
1
cos − cos 3 + cos 5 − cos 7 + ⋅⋅⋅⎟ ⎥
⎢i =
3
5
7
⎝⎜
⎠⎦
⎣
405
Two-Port Network
I1
40
2
1
Ques
t
each ions are g
i
c
prepa hapter to ven at th
ee
re the
h
topic elp the nd of
reade
.
rs
40
24. A two-port network has
I2
V1
20
1
V2
2
Fig. 6.147
(i) at Port 1, driving point impedances of 60 and
50 with Port 2 open circuited and short circuited
respectively.
(ii) at Port 2, driving point impedances of 80 and
70 with Port 1 open circuited and short circuited
respectively.
⎤
⎡ z 11 = z 22 = 60 ; z 12 = z 21 = 20 ; A = D = 3 ;
⎥
⎢
⎢⎣ B = 160 ; C = 0.05 mho; Z 0 = 56.57 ; = 1.762 ⎥⎦
Find the image parameters of the network. Derive the
expressions used.
[54.77 , 74.83 ]
1. (a) Consider a linear passive two-port network and
explain what are meant by i) open-circuit impedance parameters, and ii) short-circuit admittance
parameters.
4. What are transmission parameters? Where are they
most effectively used? Establish, for two-port networks,
the relationship between the transmission parameters
and the open-circuit impedance parameters.
(b) What are the open-circuit impedance parameters
of a two-port network? How can the transmission
parameters be obtained from open-circuit impedance parameters?
5. (a) Two two-port networks are connected in parallel.
Prove that the overall y-parameters are the sum of
corresponding individual y-parameters.
Questions
(c) Establish for two-port networks, the relationship
between the transmission parameters and the
open-circuit parameters.
(b) Two two-port networks are connected in cascade.
Prove that the overall transmission parameter
matrix is the product of individual transmission
parameter matrices.
(d) Define z and y parameters of a typical four-terminal network. Determine the relationship between
the z and y parameters.
(c) Two two-port networks are connected in series.
Prove that the overall z-parameters are the sum of
corresponding individual z-parameters.
(e) Express h-parameters in terms of z-parameters for
a two-port network.
6. (a) Define ‘transfer function’ and ‘driving point function’ of a two-port network.
(f ) Derive expressions for the y-parameters in terms of
ABCD parameters of a two-port network.
(b) Derive the expression of input impedance of a
two-port network terminated with a load-impedance ZL, in terms of its -parameters.
2. (a) What do you understand by a reciprocal network?
What is a symmetrical network?
(b) Write technical note on derivation of short-circuit
admittance parameter y12 of a symmetrical and
reciprocal two-port lattice network.
(c) How will you find the ␲-equivalent of a given network when its y-parameters are known?
3. (a) Explain what are meant by the transmission (ABCD)
parameters of a two-port network. Derive the conditions necessary to be satisfied for the network to
be i) reciprocal, and ii) symmetrical.
Or,
Prove that for a reciprocal two-port network, T
(AD BC ) 1
(b) Prove that for a symmetrical two-port network,
h (h11h22 h12h21) 1
(c) Derive the expression of output impedance of a
two-port network terminated with a load-impedance ZL, in terms of its transmission parameters.
7. What is a gyrator? Mention some properties of an
ideal gyrator. Show that a gyrator is a non-reciprocal
device.
8. What is negative impedance converter (NIC)? Show
that an NIC is a non-reciprocal device.
9. What are image parameters? Derive expression of
image parameters in terms of (i) ABCD parameters
(ii) open-circuit and short-circuit impedances.
10. What is a symmetrical network? Derive expressions for
characteristic impedance and propagation constant of
a symmetrical networks in terms of short-circuit and
open-circuit impedances.
225
Network Theorems
Prove that the load impedance which absorbs the
maximum power from a source is the conjugate of the
impedance of the source.
14. Derive the condition for maximum power transfer for
(a) Load impedance with variable resistance and variable reactance
11. Prove the condition for maximum power transfer for
an ac circuit.
(b) Load impedance with variable resistance and fixed
reactance
12. A source with internal impedance RS
jXS delivers
power to a variable load impedance RL
j0. Show
that the condition for maximum power in the load is
15. State and clearly prove with the help of a suitable
example the maximum power transfer theorem as
applicable to RLC circuits excited from the sinusoidal
energy source. Hence explain clearly the concept and
its significance in impedance matching.
RL 2 = RS 2 + X S 2 .
13. State the maximum power transfer theorem and
verify that only 50% of the total power supplied by the
source can be transferred to the load.
16. State and prove the following theorems:
( i) Tellegen’s theorem
Or,
(ii) Millman’ theorems
State and explain the maximum power transfer theorem. Derive the expression for efficiency for maximum
power transfer.
(iii) Compensation theorem
Multiple-Choice Questions
1. Which one of the following theorems is a manifestation of the law of conservation of energy?
(i) Tellegen’s Theorem
(ii) Reciprocity Theorem
(iii) Thevenin’s Theorem
(iv) Norton’s Theorem
2. Tellegen’s theorem is applicable to
(i) circuits having passive elements
(ii) circuits having time-invariant elements only
(iii) circuits with linear elements only
(iv) circuits with active or passive, linear or non-linear
and time-invariant or time-varying elements
7.
8.
9.
3. In any lumped network with elements in b branches,
b
∑
k =1
k
(t ) ⋅ i k (t ) = 0 , for all t, holds good according to
(i) Norton’s theorem
(ii) Thevenin’s theorem
(iii) Millman’s theorem (iv) Tellegen’s theorem
4. Millman’s theorem yields
(i) equivalent voltage source
(ii) equivalent voltage or current source
(iii) equivalent resistance
(iv) equivalent impedance
5. The superposition theorem is applicable to
(i) current only
(ii) voltage only
(iii) both current and voltage
(iv) current, voltage and power
6. Superposition theorem is not applicable for
10.
11.
(i) voltage calculations (ii) bilateral elements
(iii) power calculations (iv) passive elements
Thevenin’s theorem can be applied to calculate the
current in
(i) any load
(ii) a passive load only
(iii) a linear load only
(iv) a bilateral load only
Norton’s equivalent circuit consists of a
(i) voltage source in parallel with impedance
(ii) voltage source in series with impedance
(iii) current source in parallel with impedance
(iv) current source in series with impedance
The superposition theorem is applicable to
(i) linear responses only
(ii) linear and non-linear responses
(iii) linear, non-linear and time-variant responses
When a source is delivering maximum power to a
load, the efficiency of the circuit
(i) is always 50%
(ii) depends on the circuit parameters
(iii) is always 75%
(iv) none of these.
Maximum power transfer occurs at a
(i) 100% efficiency
(ii) 50% efficiency
(iii) 25% efficiency
(iv) 75% efficiency
12. Which of the following statements is true?
(i) A Norton’s equivalent is a series circuit.
(ii) A Thevenin’s equivalent circuit is a parallel circuit.
(iii) R-L circuit is a dual pair.
(iv) L-C circuit is a dual pair.
ns
estio re
u
q
a
e
hoic
ions
le c e quest d help
p
i
t
l
n
s
u
The ons a
ject
r, m
apte ovided. minati the sub
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c
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ach
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of e wers ar petitive hensio
d
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ans
com
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mpr
At t Q) with ifferent lear co
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(MC from d have a
n
take eader to
r
the atter.
m
1
Introduction to Different
Types of Systems
Introduction
An electrical network is one of the many important physical systems. In order to understand the basic
characteristics of an electric network, we must first know the different concepts of systems. In this
chapter, different types of systems have been discussed.
1.1
CONCEPTS OF SIGNALS AND SYSTEMS
1.1.1 Signal
A signal is defined as a function of one or more variables, which provides information on the nature of a
physical phenomenon.
When the function depends on a single variable, the signal is said to be one-dimensional. Example A
speech signal whose amplitude varies with time, depending on the spoken word and who speaks it.
When the function depends on two or more variables, the signal is said to be multidimensional. Example An
image (2-D signal).
1.1.2 Systems
A system is an entity that takes an input signal and produces an output signal. It is a combination and interconnection of several components to perform a desired task.
Input signals
x1(t )
x2(t)
Output signals
y1(t )
System
xn(t)
Fig. 1.1
y2(t )
yn(t )
Block-diagram representation of a system
2
Network Analysis and Synthesis
The system responds to one or more input quantities, called input signals or excitation, to produce one or
more output quantities, called output signals or response.
1.2
DIFFERENT TYPES OF SIGNALS
Signals can be classified into different categories, as given below.
1. Continuous-time and discrete-time signals
2. Periodic and non-periodic signals
3. Odd and even signals
1.2.1 Continuous-Time and Discrete-Time Signals
x(t )
Signals are represented mathematically as functions of one or more independent variables. We classify signals as being either continuous-time
(functions of a real-valued variable) or discrete-time (functions of an integer-valued variable).
In other words, a continuous-time signal has a value defined for each point in
time and a discrete-time signal is defined only at discrete points in time.
To signify the difference, we (usually) use round parenthesis around the
argument for continuous time signals, e.g., x(t) and square brackets for discrete-time signals, e.g., x[n]. We will also use the notation xn for discretetime signals.
The sequences of values of the discrete-time signal shown
in Fig. 1.2 (b) defined at discrete points in time are called samples, and the spacing between them is called the sample spacing. For equal sample spacing, the sequences of values are
expressed as a function of the signed integer n as x[n], where n
X [ 4]
is termed as a sequence of samples or sequence, in short.
time (t)
Fig. 1.2(a) Continuous-time
signal
X [n]
X [3]
X [ 2]
1
4
A signal f (t) is said to be periodic if
f (t ) = f (t ± nT )
X [ 3]
(1.1)
X [2]
X [0]
3
1.2.2 Periodic and Non-Periodic Signals
X [1]
2
0
X [ 1]
1
2
3
time
[n]
Fig. 1.2(b) Discrete-time signal
where n is a positive integer and ‘T’ is the period. Thus, a
periodic signal repeats itself every T seconds. Some periodic
signals are shown in Fig. 1.3.
v(t )
V
2T
(a)
Fig. 1.3 Periodic signals
T
0
T
(b)
2T
3T
4T
t
3
Introduction to Different Types of Systems
A signal not satisfying the above condition of Eq. (1.1) is called a non-periodic signal. Examples of some
non-periodic signals are et, t, etc.
1.2.3 Odd and Even Signals
A signal f (t) is said to be odd if
()
( )
f t = − f −t
(1.2)
Some examples of odd signals are sine functions, triangular functions and square function, as shown in Fig. 1.4.
v(t)
f (t )
V
T/2
0 T/4
−V
− T/2
−T
Fig. 1.4
t
T
0
ωt
Odd signals
A signal f (t) is said to be even if
() ( )
f t = f −t
(1.3)
Some examples of even signals are shown in Fig. 1.5.
f(t )
f (t)
V
− T/2
0
T/2
t
0
ωt
−V
Fig. 1.5 Even signals
Decomposition of a signal into odd and even components For any function f(t), let the odd component be denoted by f0(t) and the even component by fe(t), so that,
() () ()
∴ f ( −t ) = f ( −t ) + f ( −t ) = − f ( t ) + f ( t )
f t = f0 t + f e t
0
e
0
(1.4)
(1.5)
e
[by Eq. (1.2) and (1.3)]
By addition and subtraction of Eqs (1.4) and (1.5), we get,
1
f e t = ⎡⎣ f t + f −t ⎤⎦
2
()
() ( )
(1.6)
()
() ( )
(1.7)
1
f 0 t = ⎡⎣ f t − f −t ⎤⎦
2
By these two equations, we can decompose a signal into its odd and even components.
4
Network Analysis and Synthesis
Example 1.1 Decompose the following signal into its odd and even components.
Solution To find the even and odd components we need the folded signal, i.e. f ( t), as shown in Fig. 1.6 (b).
By point-by-point addition and subtraction, we get the even and odd components as shown in Fig. 1.6 (c)
and Fig. 1.6 (d).
f0(t )
f (t )
1
1/2
fe(t )
f( t )
1
1
1/2
1
0
0
1
Fig. 1.6 (a) Signal of
Ex .1.1
1/2
−1
t
t
0
t
1
0
1
t
Fig. 1.6 (c) Even component
of signal of Fig. 1.6 (a)
Fig. 1.6 (b) Folded
signal of Fig. 1.6 (a)
1.2.4 Some Standard Signals
Fig. 1.6 (d) Odd component
of signal of Fig 1.6 (a)
f (t)
There are some standard signals which can be generated easily in the
laboratory. Some of these standard signals are discussed below.
Sinusoidal Signal A sinusoid is a signal that has the form of a
sine or cosine function.
t
()
We consider a sinusoidal voltage, v t = Vm sin t
where Vm is the amplitude,
t is the argument of the sinusoid,
is the angular frequency of the sinusoid in rad/s 2 f
and T is the time period of the sinusoid.
As the sinusoid is periodic, it repeats itself; such that
⎛ 2 ⎞
v t = v t + T = Vm sin ⎜ t +
⎟⎠ = Vm sin t + = Vm sin t
⎝
() (
)
(
()
A shifted sinusoid can be written as, v t = Vm sin
where
is the phase of the sinusoid.
Thus, we see that,
( t ± 180 )
− cos t = cos( t ± 180 )
± cos t = sin ( t ± 90 )
sin t = cos( t ± 90 )
− sin t = sin
Fig. 1.7(a) sin t
2
,
T
f (t )
)
( t+ )
t
Fig. 1.7 (b) cos t
f (t )
Ke
τ
1/a
at
K
±
Exponential Signal An exponential signal is a function of time defined as
()
f t = 0,
t <0
= Ke − at , t ≥ 0
0.37 K
0
Fig. 1.8 Exponential
signal
t
5
Introduction to Different Types of Systems
where
K and a are some real constants. The reciprocal of a has the dimension
⎛
1⎞
of time and is known as time constant, ⎜ = ⎟ . This is the time to reach
a⎠
⎝
63.2% of the total change from the initial to final value.
u(t )
1
0
t
Fig. 1.9 (a) Unit step
function
Singularity Signals There are three singularity signals, namely,
a. Step signal,
b. Ramp signal, and
c. Impulse signal.
Ku(t)
K
Step Signal This function is also known as Heaviside unit function. It is defined
as given below.
0
f (t ) = u(t ) = 1 for t > 0
= 0 for t < 0
and is undefined at t 0.
A step function of magnitude K is defined as
u(t − T )
1
f (t ) = Ku(t ) = K for t > 0
= 0 for t < 0
and is undefined at t 0.
A shifted or delayed unit step function is defined as
0
Ramp Signal A unit ramp function is defined as
T
t
Fig. 1.9 (c) Shifted unit step
function
f (t ) = u(t − T ) = 1 for t > T
= 0 for t < T
and is undefined at t T.
Another function, called gate function, can be obtained from step
function as follows.
Therefore, g(t) Ku(t − a) Ku(t − b)
t
Fig. 1.9 (b) Step function of magnitude K
K
0
a
b
Fig. 1.9 (d) Gate function
f (t ) = r (t ) = t for t ≥ 0
= 0 for t < 0
A ramp function of any slope K is defined as
f (t ) = Kr (t ) = Kt for t ≥ 0
= 0 for t < 0
A shifted unit ramp function is defined as
f (t ) = r (t − T ) = t for t ≥ T
= 0 for t < T
Impulse Signal This function is also known as Dirac Delta function, denoted by d(t). This is a function of a
real variable t, such that the function is zero everywhere except at the instant t 0. Physically, it is a very sharp
pulse of infinitesimally small width and very large magnitude, the area under the curve being unity.
6
Network Analysis and Synthesis
Kr(t)
r(t)
r(t − T )
K
1
1
1
1
t
0
Fig. 1.10 (a) Unit ramp function
1
0
t
Fig. 1.10 (b) Ramp function
Consider a gate function as shown in Fig. 1.11.
The function is compressed along the time-axis
and stretched along the y-axis, keeping the area under
1
the pulse as unity. As a
0, the value of
a
and the resulting function is known as impulse.
It is defined as (t ) = 0 for t ≠ 0
∞
and ∫ (t )dt = 1
−∞
0
T
t
Fig. 1.10 (c) Shifted unit ramp function
f (t )
δ (t )
3/a
∞
2/a
1/a
0
a/3
a/2
a
Fig. 1.11(a) Generation of
impulse function from gate
function
t
0
t
Fig. 1.11(b) Impulse
signal
1
d
Also, (t ) = Lim ⎡⎣ u(t ) − u(t − a ) ⎤⎦ = ⎡⎣ u(t ) ⎤⎦
dt
a ⎯⎯
→0 a
1.3
DIFFERENT TYPES OF SYSTEMS
Systems can be classified from different points of view as given below.
1. Continuous and discrete-time systems
2. Fixed and time-varying systems
3. Linear and non-linear systems
4. Lumped and distributed systems
5. Instantaneous and dynamic systems
6. Active and passive systems
7. Causal and non-causal systems
8. Stable and unstable systems
9. Invertible and non-invertible systems
1.3.1 Continuous- and Discrete-Time Systems
A continuous-time system is a system which accepts only continuous-time signals to produce continuoustime internal and output signals. On the other hand, a discrete-time system is a system that transforms discrete-time input(s) into discrete-time output(s).
Examples
Continuous-Time Systems
(i) Atmospheric pressure as a function of altitude
(ii) Electric circuits composed of resistors, inductors, capacitors driven by continuous-time sources
7
Introduction to Different Types of Systems
Discrete-Time Systems
(i) Weekly stock market index
(ii) Balance in a bank account from month to month
1.3.2 Time-Invariant (Fixed) and Time-Varying Systems
A system is time-invariant or fixed if the behavInput x(t)
Input x(t
T)
iour and characteristics of the system do not
change with time. Otherwise, the system is
time-varying.
time
time
0
0
T
Mathematically, if the input x(t) gives the output
Output y(t)
Output y(t T )
y(t) then the system is time-invariant if the input
x(t − T ) gives the output y(t − T ) for any delay
T. Hence, a time-shift of the input gives the same
time
time
T
0
0
time-shift of the output.
Fig. 1.12 Time-invariant system
Whether a system is time-invariant or timevarying can be seen in the differential equation (or
difference equation) describing it. Time-invariant systems are modeled with constant-coefficient equations.
A constant-coefficient differential (or difference) equation means that the parameters of the system are not
changing over time and an input now will give the same result as the input later.
Example 1.2 A continuous-time system is modeled by the equation y(t) ⴝ tx(t) ⴙ 4, and a discrete-time
system is modeled by y(n) ⴝ x 2[n]. Are these systems time-invariant?
Solution For continuous-time system
For input x(t) x1(t), output y1(t) tx1(t) 4
For input x(t) x1(t − T ), output, y2(t) tx1(t − T )
(i)
(ii)
4
From the condition of time-invariance, the output should be
y1(t − T ) (t − T) x1 (t − T)
From equations (ii) and (iii), y2(t) y1(t − T)
Hence, the system is not time-invariant.
4
For discrete-time system
For input x1[n ], output y1[n] x12[n]
For input x1[n − n0], output x12[n − n0]
From the condition of time-invariance, the shifted output y1[n − n0]
Hence, the system is time-invariant.
(iii)
x12[n − n0]
1.3.3 Linear and Non-Linear Systems
A system, in continuous-time or discrete-time, is said to be linear, if it obeys the properties of superposition,
i.e. additivity and homogeneity (or scaling); while a system is non-linear that does not obey at least any one
of these properties.
8
Network Analysis and Synthesis
The superposition principle says that the output to a linear combination of input signals is the same linear
combination of the corresponding output signals. Mathematically, the linearity condition is based on two
properties:
Additivity If the input signals x1(t) and x2(t) correspond to the output signals y1(t) and y2(t), respectively
then the input signal {x1(t) x2(t)} should correspond to the output signal {y1(t) y2(t)}.
Homogeneity If the input signal x1(t) corresponds to the output signal y1(t), then the input signal a1x1(t)
should correspond to the output signal a1y1(t) for any constants a1.
Combining these two properties, the condition for a linear system can be written as, if the input signals
x1(t) and x2(t) correspond to the output signals y1(t) and y2(t), respectively then the input signal a1x1(t) a2x2(t)
should correspond to the output signal a1y1(t) a2y2(t) for any constants a1 and a2.
Example 1.3 Check whether the systems with the input–output relationship given below are linear:
(a) y(t) ⴝ mx(t) ⴙ c,
(b) y(t) ⴝ tx(t)
Solution
(a) For an input x1(t), output, y1(t) mx1(t) c
For an input x2(t), output, y2(t) mx2(t) c
For an input {x1(t) x2(t)}, output, y3(t) m{x1(t) x2(t)} c
From the condition of linearity, the output should be
{ y1(t) y2(t)} m{x1(t) x2(t)} 2c
From equations (i) and (ii), we conclude that the system is non-linear.
(b) For an input x1(t), output, y1(t) tx1(t)
For an input x2(t), output, y2(t) tx2(t)
For an input {k1x1(t) k2x2(t)}, output, y3(t) t{k1x1(t) k2x2(t)}
where k1 and k2 are any arbitrary constants.
From the condition of linearity, the output should be
{k1y1(t) k2y2(t)} k1tx1(t) k2tx2(t) t{k1x1(t)
From equations (i) and (ii), we conclude that the system is linear.
(i)
(ii)
(i)
k2x2(t)}
(ii)
1.3.4 Lumped and Distributed Systems
All physical systems contain distributed parameters because of the physical size of the system components.
For example, the resistance of a resistor is distributed throughout its volume.
However, if the size of the system components is very small with respect to the wavelength of the highest
frequency present in the signals associated with it then the system components behave as if it all were occurring at a point. This system is said to be a lumped-parameter system.
Distributed parameter systems are modeled
•
•
by partial differential equations if they are continuous-time systems, and
by partial difference equations if they are discrete-time systems.
Lumped parameter systems are modeled with ordinary differential or difference equations.
9
Introduction to Different Types of Systems
Example Consider an electric power system of frequency 50 Hz. The wavelength of the signal is
obtained as,
C 3 × 105
n =C ⇒ = =
km = 6000 km
n
50
Thus, the electrical system inside a room can be treated as a lumped-parameter system, but will be treated
as distributed system for long-distance transmission lines.
1.3.5 Instantaneous (Static or Memoryless) and Dynamic Systems
An instantaneous or static or memoryless system is a system where the output at any specific time depends
on the input at that time only. On the other hand, a dynamic system is one whose output depends on the past
or future values of the input in addition to the present time.
A static system has no memory. Physically, it contains no energy-storage elements; while a dynamic
system has one or more energy-storage element(s).
Example An electrical circuit containing resistance R, has the v i relationship as, v(t)
Ri(t), and so the
t
1
system is static. But an electrical circuit containing the capacitor C has the v i relationship as v (t ) = ∫ i(t )dt ,
C0
and so, the system is a dynamic system.
1.3.6 Active and Passive Systems
A system having no source of energy is known as a passive system. Examples of passive systems are electric
circuits containing resistance, capacitance, inductance, diodes, etc.
A system having a source of energy together with other passive elements is known as an active system.
Examples of active systems are electric circuits containing voltage sources or current sources or op-amps.
1.3.7 Causal and Non-Causal Systems
A system is said to be causal if the output of the system depends only on
the input at the present time and/or in the past, but not the future value
of the input. Thus, a causal system is non-anticipative, i.e. output cannot
come before the input.
On the other hand, the output of a non-causal system depends on the
future values of the input.
Input x(t)
0
time
Output y (t)
Example The moving-average system described by
1
y[ n] = {x[ n] + x[ n − 1] + x[ n − 2 ]}
3
is causal; but the moving-average system described by
1
y[ n] = {x[ n + 1] + x[ n] + x[ n − 1]}
3
is non-causal since the output depends on the future value of the input
x[n 1].
It is obvious that the idea of future inputs does not have any physical
meaning if we take time as our independent variable and for that reason all
0
time
Output y (t)
0
time
Fig. 1.13 (a) Causal systems
10
Network Analysis and Synthesis
real-time systems are causal. However, for the case of image processing, the dependent variable may by the pixels
to the left and right (the ‘future’) of the current position on the image, and thus, we can have a non-causal system.
1.3.8 Stable and Unstable Systems
A stable system is one where the output does not diverge as long as the
input does not diverge. A bounded input produces a bounded output. For
this reason, this type of system is known as a bounded input–bounded
output (BIBO) stable system.
Mathematically, a stable system must have the following property:
If x(t) be the input and y(t) be the output then the output must satisfy
the condition
y (t ) ≤ M y < ∝;
Input x(t )
Output y (t)
for all t
whenever the input satisfies the condition
x (t ) ≤ M x < ∝;
time
0
time
0
for all t
Fig. 1.13 (b) Non-causal system
where Mx and My both represent a set of finite positive numbers.
If these conditions are not met, i.e. the output of the system grows without limit (diverges) from a bounded
input then the system is unstable.
1.3.9 Invertible and Non-Invertible Systems
A system is referred as an invertible system if
x(t )
y(t )
System
(i) distinct inputs lead to distinct outputs, and
(ii) the input can be recovered from the output.
w (t) = x (t)
Inverse system
Fig. 1.14 Invertible system
The property of invertibility is important in the design of communication systems. When a transmitted
signal propagates through a communication channel, it becomes distorted due to the physical characteristics
of the channel. An equalizer is connected in cascade with the channel in the receiver to compensate this distortion. By designing the equalizer to be inverse of the channel, the transmitted signal is restored.
1.4
INTERCONNECTION OF SYSTEMS
Most of the physical systems are built as interconnections of several subsystems. Different types of interconnections are shown below.
Series or Cascade Interconnection The output of
the system 1 is the input to the system 2.
Parallel Interconnection The same input signal is
applied to systems 1 and 2.
Input
System 1
Fig. 1.15
System 1
Input
+
System 2
Fig. 1.16
Output
System 2
Output
11
Introduction to Different Types of Systems
Combination of Both Cascade and Parallel Interconnections
System 1
System 2
Input
System 4
Output
System 3
Fig. 1.17
Feedback Interconnection The output of the system 2 is fed back and added to the external input to
produce the actual input to the system 1.
Input
System 1
Output
System 2
Fig. 1.18
Solved Problems
Problem 1.1 Check whether the system defined by y(t) ⴝ sin[x( t)] is time-invariant.
Solution For input x(t) x1(t), output y1(t) sin[x1(t)]
For input x(t) x1(t − T), output, y 2 (t) sin[x1(t − T )]
From the condition of time-invariance, the output should be
y1(t − T ) sin[x1(t − T )]
From equations (ii) and (iii), y2(t) y1(t − T )
Hence, the system is time-invariant.
(i)
(ii)
(iii)
Problem 1.2 Consider a system S with input x[n] and output y[n] related by,
y [n] ⴝ x[n]{g[n] ⴙ g[n − 1]}
(a) If g[n] 1, for all n, show that S is time-invariant.
(b) If g[n] n, show that S is not time-invariant.
(c) If g[n] 1 (−1)n, show that S is time-invariant.
Solution
(a) If g[n] 1, for all n then y[n] x[n]{1 1 1} 2x[n]
For input x[n] x1[n], output y1[n] 2x1[n]
For input x[n] x1[n − n0], output, y2[n] 2x1[n − n0]
From the condition of time-invariance, the output should be
y1[n − n0] 2x1[n − n0]
From equations (ii) and (iii), y2[n] y1[n − n0]
Hence, the system is time-invariant.
(i)
(ii)
(iii)
12
Network Analysis and Synthesis
(b) If g[n] n, then y[n] x[n]{n n − 1} (2n − 1) x[n]
For input x[n] x1[n], output y1[n] (2n − 1)x1[n]
For input x[n] x1[n − n0], output, y2[n] (2n − 1)x1[n − n0]
From the condition of time-invariance, the output should be
y1[n − n0] {2(n − n0) − 1}x1[n − n0]
From equations (ii) and (iii), y2[n] y1[n − n0]
Hence, the system is not time-invariant.
(i)
(ii)
(iii)
(c) If g[n] 1 (−1)n, then y[n] x[n]{1 (−1)n 1 (−1)n−1} 2x[n]
This relation is same as that of Part (a). Hence the system is time-invariant.
Problem 1.3 Consider the systems S whose input and output are related by
y(t) ⴝ x 2(t)
Check whether S is linear.
Solution For an input x1(t), output, y1(t) x12(t)
For an input x2(t), output, y2(t) x22(t)
For an input {k1x1(t) k2x2(t)}, output, y3(t) [k1x1(t) k2 x2(t)]2
where, k1 and k2 are any arbitrary constants.
From the condition of linearity, the output should be
{k1y1(t) k2y2(t)} k1x12(t) k2x22(t)
From equations (i) and (ii), we conclude that the system is not linear.
(i)
(ii)
Problem 1.4 Consider the following discrete-time systems with input-output relationships as given,
y[n] ⴝ Re{x[n]}
Check whether the system is linear.
Solution Let, the input be, x1[n] r[n] js[n]
Therefore, the output is, y1[n] Re{x1[n]} Re {r[n] js[n]} r[n]
Now we consider scaling of the input x1[n] by a complex number, say, (a
x2[n]
(a
jb)x1[n]
Corresponding output is, y2[n]
(a
jb){r[n]
Re{x2[n]}
js[n]}
jb) , i.e. the input is,
{ar[n] − bs[n]}
Re{ar[n] − bs[n]}
But the scaled output for linear system is, (a jb)y1[n] ar[n]
As the two outputs are not the same, the system is not linear.
j{br[n]
j{br[n] as[n]}
as[n]}
ar[n] − bs[n]
jbr[n]
Problem 1.5 Consider a discrete-time system whose output y[n] is the average of the three most recent
values of the input signal, x[n], given as
1
y [n] = x [n] + x [n −1] + x [n − 2 ]
2
Show that the system is BIBO stable.
{
Solution Let us assume that, x[n] < Mx <
∴ y ⎡⎣ n ⎤⎦ =
}
for all n,
{
}
1
1
1
x ⎡ n ⎤ + x ⎡ n − 1⎤⎦ + x ⎡⎣ n − 2 ⎤⎦ ≤ x ⎡⎣ n ⎤⎦ + x ⎡⎣ n − 1⎤⎦ + x ⎡⎣ n − 2 ⎤⎦ ≤ ⎡⎣ M x + M x + M x ⎤⎦ ≤ M x
3 ⎣ ⎦ ⎣
3
3
13
Introduction to Different Types of Systems
Hence, the absolute value of the output signal y[n] is always less than the maximum absolute value of the
input signal x[n] for all n; which shows that the system is stable.
Problem 1.6 Determine whether the following continuous-time systems are stable:
(a) y(t) tx(t)
(b) y(t) x (t) sin 100 t
Solution Here, let the input be bounded.
(a) y(t) tx(t)
As t
, y(t)
[since x(t) is multiplied by t]
Hence the system is unstable system.
(b) y(t) x(t) sin 100 t
Here x(t) is multiplied by sin 100 t. We know that the value of sine varies between −1 and 1.
Hence y(t) is bounded as long as x(t) is bounded. Hence the system is stable.
Problem 1.7 Determine whether the following continuous-time systems are causal or non-causal:
(a) y(t)
(d)
x(t) cos(t
1)
dy ( t )
+10 y ( t ) + 5 = x ( t )
dt
(b) y(t)
x( t)
(c) y(t)
x( t)
t
(e) y ( t ) = ∫ x ( t )dt
−∞
Solution (a) y(t) x(t) cos(t 1)
Here, y(t) depends on the present input x(t). A cosine function can be evaluated at (t 1). Therefore, the
system is causal.
(b) y(t) x(2t)
Here, if t 5 then y(5) x(10)
Thus, the output y(t) depends on the future input. Therefore, the system is non-causal.
(c) y(t) x(−t)
Here, if t −3, then y(−3) x(3)
Thus, the output y(t) depends on the future input. Therefore, the system is non-causal.
(d)
( ) + 10 y t + 5 = x t
()
()
dt
dy t
Here, y(t) depends upon the present value of x(t). Therefore, the system is causal.
()
t
()
(e) y t = ∫ x t dt
−∞
Here, y(t) depends upon the present and the past values of x(t), but not on the future value. Therefore,
the system is causal.
Problem 1.8 Determine whether the following systems are invertible:
10 x(t)
(b) y(t) x2(t)
c) y(t) x(t n)
(d) y(t) x(2t)
(a) y(t)
Solution
(a) y(t) 10x(t)
( ) 101 y (t )
For this system, the inverse system will be w t =
14
Network Analysis and Synthesis
x (t)
w(t ) =
y (t ) = 10x (t)
System
Inverse system
1
y(t) = x(t )
10
Therefore, the system is an invertible system.
(b) y(t) x2(t)
2
Inverse system would be w t = y t = x t = ± x t
()
()
()
()
Here, two outputs are possible: x(t) or −x(t). This implies that there is no unique output for unique input.
Therefore, the system is a non-invertible system.
(c) y(t) x(t − n)
Here, output is the delayed input, by ‘n’ samples. Clearly, the system is invertible. There can be another
system for which the output is the advanced input by ‘n’ samples. The inverse system is w(t) y(t n).
(d) y(t) x(2t)
Here, the input is compressed by a factor 2. Hence, there can be another system which will expand the input
⎛ 1⎞
by the same factor. Hence the system is invertible. The inverse system is w (t ) = y ⎜ ⎟ .
⎝ 2⎠
Problem 1.9. Determine whether the following systems are static or dynamic:
(b) y ( t ) =
ex(t)
(a) y(t)
d
x (t)
dt
Solution (a) y(t) ex(t)
Here, the output depends on present input only. Hence the system is a static system.
d
(b) y t = x t
dt
Here, the output depends on differentiation of the input. Calculation of differentiation depends on the
present as well as past values. Therefore, the system is a dynamic system.
()
()
Problem 1.10 Express the following signals in terms of the standard signals:
a)
f (t )
Vm
0
π
b) f(t )
c) f(t)
K
1
t
0
1
t
0
f (t)
Vm
1
2
3
t
0
π
t
Fig. 1.21
Fig. 1.20
Solution
(a) Here, the signal can be expressed in terms of step signal as
⎛ T⎞ ⎛ T⎞
f t = Vm sin t u t + Vm sin ⎜ t − ⎟ u ⎜ t − ⎟
⎝ 2⎠ ⎝ 2⎠
()
()
()
= Vm sin t u t + Vm sin
(t − ) u(t − )
(b) Here, the signal starts with a straight line of slope K passing through the origin
and then comes to zero at t 1. Hence the signal can be expressed in terms of
ramp and step signals as f (t) Kr(t) − Kr(t −1) − Ku(t −1)
f (t )
K
0
Fig. 1.22
1
t
15
Introduction to Different Types of Systems
(c) Here, the signal starts from the origin with a slope of 1. At
t
1, the slope becomes zero (parallel to time axis). At
t
2, the slope becomes −1 and again at t
3, the slope
becomes zero. Therefore, the signal can be written in terms
of the ramp signals as given below.
f (t) r(t) − r(t − 1) − r(t − 2) r(t − 3)
Problem 1.11 Determine the even and odd components of the
following signals:
f(t )
1
0
1
2
t
3
Fig. 1.23
(a) Unit step signal;
f (t )
1
f(t )
f(t)
1
f (t )
1/2
1
1
1
0
0
1
Fig. 1.24 (a)
(f ) f(t)
1
t
1
1
t
1/2
1
1/2
Fig. 1.24 (b)
u(t) − r(t − 1)
t
0
Fig. 1.24 (c)
2r(t − 2) − r(t − 3)
0
t
1
Fig. 1.24 (d)
u(t − 4) − 2u(t − 5)
Solution
(a) To find the even and odd components of a unit step signal, we need to find the folded signal, i.e. u(−t), as
shown in the figure below.
u(t )
u( t)
1
1
0
t
0
Fig. 1.25 (a) Unit step signal
t
Fig. 1.25 (b) Folded signal
Now,
()
() ( )
()
1
f e t = ⎡⎣ f t + f −t ⎤⎦
2
() ( )
1
f 0 t = ⎡⎣ f t − f −t ⎤⎦
2
By point-by-point addition and subtraction of the signals of Fig. 1.25 (a) and Fig. 1.25 (b), we get the
even and odd components, respectively, as shown in Fig. 1.25 (c) and Fig. 1.25 (d) below.
f0(t)
fe(t)
1/2
1/2
0
t
1/2
0
Fig. 1.25 (c) Even component of unit step signal
t
Fig. 1.25 (d) Odd component of unit step singal
16
Network Analysis and Synthesis
(b) To find the even and odd components, we need the folded signal, i.e. f (−t), as shown in the
Fig. 1.26 (b)
f0(t)
fe(t )
f ( t)
f (t)
12
12
1
1
1
0
0
t
1
1
Fig. 1.26 (a) Signal
t
0
1
Fig. 1.26 (b) Folded
signal
0
12
t
1
t
1
Fig. 1.26 (c) Even
component of signal
Fig. 1.26 (d) Odd component
of signal
By point-by-point addition and subtraction, we get the even and odd components as shown in
Fig. 1.26 (c) and Fig. 1.26 (d).
(c) The procedure is followed as mentioned below.
By point-by-point addition and subtraction, we get the even and odd components as shown in
Fig. 1.27 (c) and Fig. 1.27 (d).
f0(t)
1
f (t )
f( t )
fe(t)
1
1/2
1
1
1
0
1/2
1
1
t
1
1
t
0
1/2
1
1
Fig. 1.27 (a) Signal
t
1
Fig. 1.27 (b) Folded
signal
t
1
0
1
Fig. 1.27 (c) Even
component of the signal
Fig. 1.27 (d) Odd
component of the signal
(d) To find the even and odd components we need the folded signal, i.e. f (−t), as shown in Fig. 1.28 (b).
By point-by-point addition and subtraction, we get the even and odd components as shown in
Fig. 1.28 (c) and Fig. 1.28 (d).
1
f(t )
f( t )
1
1
1/2
1/2
1
1
0
1/2
Fig. 1.28 (a) Signal
t
f0(t )
fe(t )
1/2
1
1
0
1
1/2
Fig. 1.28 (b) Folded
signal
1
1/2
t
1
0
1
t
Fig. 1.28 (c) Even
component of the signal
0
t
1/2
Fig. 1.28 (d) Odd
component of the signal
(e) To find the even and odd components the signal and the folded signal, i.e. f (−t) are shown in
Fig. 1.29 (a) and Fig.1.29 (b), respectively.
17
Introduction to Different Types of Systems
f(t)
1/2
1/2
1/2
1
0
t
1
1
Fig. 1.29 (a) Signal
0
f0(t )
fe(t )
f( t)
t
1
1
Fig. 1.29 (b) Folded
signal
1/4
1/4
0
1
t
1
Fig. 1.29 (c) Even
component of the signal
0
1/4
1
t
Fig. 1.29 (d) Odd
component of the signal
By point-by-point addition and subtraction, we get the even and odd components as shown in
Fig. 1.29 (c) and Fig. 1.29 (d).
(f) f (t) u(t) − r(t − 1) 2r(t − 2) − r(t − 3) u(t − 4) − 2u(t − 5)
fe(t)
f (t)
1
2
1/2
1
0
1
2
3
4
t
5
Fig. 1.30 (a) Signal
5
4
3
2
1
5 4 3 2 1
0
1
3
2
2
3
4
5
t
f0(t)
t)
2
4
1
Fig. 1.30 (c) Even component of the signal
f(
5
1 0
1
0
Fig. 1.30 (b) Folded signal
t
1/2
1 2 3 4 5
1/2
1
t
Fig. 1.30 (d) Odd component of the
signal
Here, the signal is drawn as shown in Fig. 1.30 (a). The folded signal is shown in Fig. 1.30 (b).
The even component and the odd components of the signal are obtained by point-by-point addition and subtraction of signals of Fig. 1.30 (a) and Fig. 1.30 (b), respectively. These are shown in
Fig. 1.30 (c) and Fig. 1.30 (d).
Summary
1. A signal is defined as a function of one or more variables, which provides information on the nature of a
physical phenomenon.
2. A system is an entity that takes an input signal and
produces an output signal. It is a combination and
interconnection of several components to perform a
desired task.
3. Signals can be classified into different categories, such
as continuous-time and discrete-time signals, periodic
and non-periodic signals, and odd and even signals.
4. Systems can be classified from different points of view,
such as continuous and discrete-time systems, fixed and
time-varying systems, linear and non-linear systems,
lumped and distributed systems, instantaneous and
dynamic systems, active and passive systems, causal
and non-causal systems, stable and unstable systems, and
invertible and non-invertible systems.
5. The principle of superposition is based upon two conditions—additivity and homogeneity. The superposition principle is applicable only for linear systems.
18
Network Analysis and Synthesis
6. A causal system is a non-anticipative system where
output cannot come before the input. Almost all physical systems are causal.
7. A system where a bounded input produces a bounded
output is known as bounded input-bounded output
(BIBO) stable system.
Short-Answer Questions
1. What is a system? What are the different types of
systems?
A system is an entity that takes an input signal and produces an output signal. It is a combination and interconnection of several components to perform a desired task.
Input signals
x1(t )
A function f(t) is said to be even if
f(t) f(−t)
(2)
Some examples of even functions are shown in Fig. 1.33.
V
Output signals
y1(t )
System
x2(t )
xn(t )
f(t )
f (t )
y2(t)
T/2 0 T/2
−V
yn(t)
Fig. 1.33 Even functions
Fig. 1.31 Block-diagram representation of a system
The system responds to one or more input quantities,
called input signals or excitation, to produce one or more
output quantities, called output signals or response.
Systems can be classified from different points of
view as given below.
1. Continuous and discrete-time systems
2. Fixed and time-varying systems
3. Linear and non-linear systems
4. Lumped and distributed systems
5. Instantaneous and dynamic systems
6. Active and passive systems
7. Causal and non-causal systems
8. Stable and unstable systems
9. Invertible and non-invertible systems
2. Define an odd and an even function. How can you
decompose a general function into its odd and
even components?
A function f(t) is said to be odd if
f(t) −f(−t)
(1)
Some examples of odd functions are sine functions, triangular functions, and square functions, as shown in Fig. 1.32.
v (t )
t
0
ωt
For any function f(t), let the odd component be
denoted by f0(t) and even component by fe(t), so that,
⬖ f (−t)
f (t)
f0(t)
f0(−t)
fe( t)
fe(t)
(3)
−f0(t)
fe(t)
(4)
[by Eqs (1) and (2)]
By addition and subtraction of Eqs (3) and (4),
we get
)
)
)
)
)
)
1
(5)
f e (t = ⎡⎣f (t + f ( −t ⎤⎦
2
1
f 0 (t = ⎡⎣f (t − f ( −t ⎤⎦
(6)
2
By these two equations, we can decompose a signal
into its odd and even components.
3. Define the following functions:
(a) Step function
(c) Ramp function
(b) Gate function
(d) Impulse function
(a) Step function
A step function is defined as given below.
f (t)
u(t)
1 for t > 0
1 for t > 0
and is undefined at t 0.
u(t)
Ku(t )
1
K
f (t)
T/2
V
T
T/2
0 T/4
V
Fig. 1.32 Odd functions
T
t 0
ωt
0
t
Fig. 1.33 (a) Unit
step function
0
t
Fig. 1.33 (b) Step
function of magnitude K
19
Introduction to Different Types of Systems
A step function of magnitude K is defined as
f (t)
Ku(t)
and is undefined at t
f (t )
K for t > 0
1 for t > 0
3/a
0.
(t )
2/a
1/a
K
0
0
a
a/3
a/2
a
0
t
Fig. 1.34 (b)
Impulse Signal
Fig. 1.34 (a) Generation of
impulse function from gate
function
b
Fig. 1.33 (c) Gate
function
t
It is defined as,
∞
(b) Gate function
A gate function can be obtained from a step function as shown in Fig. 1.33 (c).
Therefore, g(t)
Ku(t−a)−Ku(t−b).
r(t)
t for t
0 for t
Kr(t)
Kt for t 0
0 for t < 0
K
1
1
1
0
t
Fig. 1.33 (d) Unit
ramp function
0
0 and ∫ (t )dt = 1
−∞
Also, (t ) = Lim 1 [ u(t ) − u(t − a )] = d [ u(t )]
dt
a ⎯→
⎯oa
The superposition principle says that the output to a
linear combination of input signals is the same linear combination of the corresponding output signals. Mathematically, the linearity condition is based on two properties:
Kr (t)
r (t )
for t
A system in continuous-time or discrete-time, is said
to be linear if it obeys the properties of superposition,
i.e. additivity and homogeneity (or scaling); while a
system is non-linear that does not obey at least any
one of these properties.
0
0
A ramp function of any slope K is defined as
f(t)
0
4. What are the conditions for a system to be a linear
system?
(c) Ramp function
A unit ramp function is defined as
f(t)
(t)
t
Fig. 1.33 (e) Ramp
funtion
(d) Impulse function
This function is also known as Dirac Delta function,
denoted by (t). This is a function of a real variable t,
such that the function is zero everywhere except at
the instant t 0. Physically, it is a very sharp pulse of
infinitesimally small width and very large magnitude,
the area under the curve being unity.
Consider a gate function as shown in Fig. 1.34 (a).
The function is compressed along the timeaxis and stretched along the y-axis, keeping
0, the
area under the pulse unity. As a
value of [1/a]
and the resulting function
is known as impulse.
1. Additivity If the input signals x1(t) and x2(t) correspond to the output signals y1(t) and y2(t), respectively then the input signal {x1(t)
x2(t)} should
correspond to the output signal {y1(t) y2(t)}.
2. Homogeneity If the input signal x1(t) corresponds
to the output signal y1(t) then the input signal a1x1(t)
should correspond to the output signal a1y1(t) for
any constants a1.
Combining these two properties, the condition for
a linear system can be written as if the input signals
x1(t) and x2(t) correspond to the output signals y1(t)
and y2(t), respectively then the input signal a1x1(t)
a2x2(t) should correspond to the output signal a1y1(t)
a2y2(t) for any constants a1 and a2.
5. Give the conditions for a BIBO stability of a
system.
A stable system is one where the output does not
diverge as long as the input does not diverge. A
bounded input produces a bounded output. For this
20
Network Analysis and Synthesis
reason, this type of system is known as bounded inputbounded output (BIBO) stable system.
Mathematically, a stable system must have the following property:
If x(t) be the input and y(t) be the output then the
output must satisfy the condition
y (t ) ≤ M y <∝ ;
for all t
whenever the input satisfy the condition
y (t ) ≤ M x <∝ ;
for all t
where, Mx and My both represent a set of finite positive
numbers.
A system is referred to as an invertible system if
i) distinct inputs lead to distinct outputs, and
ii) the input can be recovered from the output.
The property of invertibility is important in the design
of communication systems. When a transmitted signal
propagates through a communication channel, it
becomes distorted due to the physical characteristics
of the channel. An equalizer is connected in cascade
with the channel in the receiver to compensate this
distortion. By designing the equalizer to be inverse of
the channel, the transmitted signal is restored.
y(t )
x(t )
System
6. Define invertible systems. Why is it important to
have an inverse system of a system?
Fig. 1.35
w (t) = x(t)
Inverse system
Invertible system
Exercises
1. A discrete-time system is modeled by
Y[n]
2
X [n]
2. Consider the systems S whose input and output are
related by
tx(t)
(b) y(t)
x2(t)
(d) y
mx
c
Check whether S is linear.
Is this system time-invariant?
(a) y(t)
(c) y(t)
x(t)x(t − 1)
3. Consider the following discrete-time systems with
input–output relationships as given:
(a) y[n]
2x[n]
3
(b) y[n]
nx[n]
Check whether the systems are linear.
Questions
1. What is a system? What are the different types of
systems? Give their definitions.
2. Define the following and give examples:
(a) Continuous and discrete signals
(b) Periodic and non-periodic signals
(c) Odd and even signals
(d) Step, ramp and impulse signals
3. Define the following and give examples:
(c) Lumped and distributed system
(d) Instantaneous (static or memoryless) and dynamic
system
(e) Causal and non-causal system
(f) Active and passive system
4. (a) What are the conditions for a system to be a linear
system?
(b) Give the conditions for BIBO stability of a system.
(a) Continuous and discrete system
(b) Time-invariant and Time-varying system
Multiple-Choice Questions
1. The output y(t) and the input x(t) of a system are
related by the equation y(t) mx(t) c, where m and
c are constants. The system is
(i) linear
(ii) non-linear
(iii) may be linear or non-linear depending on y(t)
and x(t)
(iv) none of the above
21
Introduction to Different Types of Systems
2. If the impulse response is realizable by delaying it
appropriately and is bounded for bounded excitation
then the system is said to be
(i) causal and stable
(ii) causal but not stable
(iii) non-causal but stable
(iv) non-causal, not stable
3. In a linear circuit, when the ac input is doubled, the ac
output becomes
(i) one-fourth
(ii) half
(iii) two times
(iv) four times
4. A circuit having an emf source or any energy source is
a/an
(i) active circuit
(ii) passive circuit
(iii) unilateral circuit
(iv) bilateral circuit
9. What is the input–output relation of the causal movingaverage system (discrete time)?
{
}
{
}
(i)
y [n ] =
1
x [ n ] + x [ n − 1] + x [ n − 2 ]
3
(ii)
y [n ] =
1
x [ n − 1] + x [ n ] + x [ n + 1]
3
(iii)
y [n ] =
1
2
1
x [n ] + x [n ] + x [n ] 2
3
{
{
)
}
}
1
x [ n ] + x [ n + 1] + x [ n + 2 ]
3
10. The v−i characteristic of an element is shown in
Fig. 1.36. The element is
(i) non-linear, active, non-bilateral
(ii) linear, active, non-bilateral
(iii) non-linear, passive, non-bilateral
(iv) non-linear, active, bilateral
(iv)
y [n ] =
v
5. A network is said to be linear if and only if
(i) a response is proportional to the excitation
function
(ii) the principle of superposition applies
(iii) the principle of homogeneity applied
(iv) both the principles (ii) and (iii) apply
6. Consider the following data
1. Input applied for t t0
2. Input applied for t t0
3. State of the network at t
4. State of the network at t
) (
(
0
i
Fig. 1.36
t0
t0
Among these, those needed for determining
the response of a linear network for t > t0 would
include
(i) 1, 3 and 4
(ii) 2, 3 and 4
(iii) 2 and 3
(iv) 2 and 4
7. An excitation is applied to a system at t T and its
response is zero for
< t < T. Such a system is a/an
(i) non-causal system
(ii) stable system
(iii) causal system
(iv) unstable system
8. The elements which are not capable of delivering
energy by their own are known as
(i) unilateral elements
(ii) non-linear elements
(iii) passive elements
(iv) active elements
11. Which one of the following is a linear system?
(i) y(t) 2u(t)
(ii) y(t) 2u(t) 5
(iii) y(t) 2u2(t)
(iv) y(t) 2u2(t) 5
12. A function f( . ) is linear under the following conditions
(i) f (x1 x2) f (x1) f (x2) only
(ii) f (kx) kf (x) only
(iii) f (x1 x2) f (x1) f (x2) and f (kx) kf (x)
(iv) f (x1 x2) f (x1) f (x2) or f (kx) kf (x)
13. The v−i characteristic of a resistor is i 2v 2. The resistor is
(i) linear, passive, bilateral
(ii) non-linear, passive, bilateral
(iii) non-linear, active, bilateral
(iv) non-linear, active, unilateral
14. The system y(t) tx(t) 4 is
(i) non-linear, time-varying and unstable
(ii) linear, time-varying and unstable
(iii) non-linear, time-invariant and unstable
(iv) non-linear, time-varying and stable
15. The following is true.
(i) A finite signal is always bounded.
(ii) A bounded signal always possesses finite energy.
22
Network Analysis and Synthesis
(iii) A bounded signal is always zero outside the
interval [−t0, t0] for some t0.
(iv) A bounded signal is always finite.
16. The function x(t) is shown in Fig. 1.37. The even and
odd parts of a unit-step function u(t) are respectively
1 1
1 1
(ii) , x (t )
(i)
, x (t )
2 2
2 2
(iii)
1 1
, − x (t )
2 2
(iv)
1 1
, − x (t )
2 2
x (t)
1
(t
(t
4)x(t
5)x(t
1)
5)
18. The impulse response h(t) of a linear time–invariant continuous time system is described by h(t) exp( t)u(t)
exp( t)u( t), where, u(t) denotes the unit step function,
and and are real constants. The system is stable if
(i)
is positive and is positive
(ii)
is negative and is negative
(iii)
is positive and is negative
(iv)
is negative and is positive
19. Which of the following represent a stable system?
1. Impulse response of the system decreases exponentially.
2. Area within the impulse response is finite.
3. Eigen values of the system are positive and real.
4. Roots of the characteristic equation of the
system are real and negative.
0
1
(iii) y(t)
(iv) y(t)
t
Fig. 1.37
17. The input and output of a continuous-time system are
respectively denoted by x(t) and y(t). Which of the following description corresponds to a causal system?
(i) y(t) x(t 2) x(t 4)
(ii) y(t) (t 4)x(t 1)
Select the correct answer using the codes given below.
(i) 1 and 4
(ii) 1 and 3
(iii) 2, 3 and 4
(iv) 1, 2 and 4
Answers
1.
2.
3.
4.
(ii)
(i)
(iii)
(i)
5.
6.
7.
8.
(iv)
(iii)
(iii)
(iii)
9. (i)
10. (ii)
11. (i)
12. (iii)
13. (ii)
14. (i)
15. (ii)
16. (i)
17. (i)
18. (iv)
19. (ii)
2
Introduction to CircuitTheory Concepts
Introduction
The most fundamental branch of electrical engineering is electric circuit theory. All other branches of electrical engineering, such as electric power, electric machines, control, electronics, computers, communications and instrumentation are built on the electric circuit theory. Thus, it is very essential to have a proper
grounding with electric circuit theory as the base. In this chapter, we will discuss about the basic terms
related to electric circuit theory, basic circuit elements and their properties. We will also discuss the different
laws which are required to analyze an electric circuit where many circuit elements are interconnected.
2.1
SOME BASIC TERMINOLOGIES OF ELECTRIC CIRCUITS
2.1.1 Concept of Electric Charge
The most basic quantity in an electric circuit is the electric charge q. Electric charge is a fundamental conserved property of some subatomic particles, which determine their electromagnetic interaction. Electrically
charged matter is influenced by, and produces, electromagnetic fields.
It is known that an atom consists of a positively charged nucleus surrounded by negatively charged electrons. In a neutral atom, the total charge of the nucleus is equal to the total charge of the electrons. When
electrons are removed from a substance, the substance becomes positively charged and if excess electrons are
given to a substance, it becomes negatively charged.
The SI unit of charge is coulomb (C). The charge of an electron is 1.602 10 19 C. Thus, one coulomb
⎛
⎞
1
charge is defined as the charge possessed by ⎜
electrons.
⎝ 1.602 × 10−19 ⎟⎠
1 coulomb charge
charge of 6.24
1018 electrons
The total electric charge of an isolated system remains constant regardless of changes within the system
itself. This is known as the law of conservation of charge. The law of conservation of charge states that
charge can neither be created nor destroyed.
24
Network Analysis and Synthesis
The electric charge of a macroscopic object is the sum of the electric charges of its constituent particles.
Often, the net electric charge is zero, because it is favorable for the number of electrons in every atom to
equal the number of protons (or, more generally, for the number of anions, or negatively charged atoms, in
every molecule to equal the number of cations, or positively charged atoms). When the net electric charge is
non-zero and motionless, the phenomenon is known as static electricity. Even when the net charge is zero, it
can be distributed non-uniformly due to an external electric field, or due to molecular motion; in such cases
the material is said to be polarized. The charge due to the polarization is known as bound charge, while the
excess charge brought from outside is called free charge. The motion of charged particles (e.g., of electrons
in metals) in a particular direction is said to constitute an electric current.
2.1.2 Conductor, Insulators and Semiconductors
In some materials, there is a large number of free electrons or loosely bound valence-band electrons present.
These electrons are easily knocked out of their orbit and easily constitute a large current. Such materials are
known as conductors. Almost all metals and some liquids are good conductors.
In some materials, no free electrons are available; the valence-band electrons are tightly bound to the nucleus.
Such materials are known as insulators. Examples of some insulators include glass, mica, plastics, etc.
In between the limits of these two major categories is a third general class of materials called semiconductors; where there are no such free electrons present, but free electrons can easily be created by adding
some impurities. Examples of some insulators include germanium and silicon. For example, germanium, a
semiconductor, has approximately one trillion times (1 1012) the conductivity of glass, an insulator, but has
only about one thirty-millionth (3 10 8) part of the conductivity of copper, a conductor.
2.1.3 Concept of Electric Current
The phenomenon of transferring electric charge from one point in a circuit to another is described by the term
electric current. Electric current is defined as the rate of flow of electric charges or electrons through a crosssectional area. By convention, the electric current flows in the opposite direction to the electrons.
If Q amount of charges flow through an area in time t, then the current is given as,
I=
or in differential form,
i=
dq
dt
Q
t
(2.1)
(2.2)
t
and the charge transferred between time t0 and t is given by
q = ∫ idt
(2.3)
t0
As Q is expressed in coulombs, the unit of electric current is coulomb per second and it is given the name
ampere (A).
Thus,
1A current flow of 6.24 1018 electrons per second through an area
2.1.4 Current Density
Current density at any point is a vector whose magnitude is the electric current per unit cross-sectional area and
I
whose direction is normal to the cross-sectional are a, i.e. J = nˆ. Its unit is ampere per square metre (A/m2).
A
25
Introduction to Circuit-Theory Concepts
2.1.5 Concept of Electric Potential and Potential Difference
To move an electron in a conductor in a particular direction, or to create a current, requires some work or
energy. This work is done by the potential or the potential difference. This is also known as voltage difference or voltage (with reference to a selected point such as earth). The unit of potential is volt.
The potential of a point is 1volt if 1joule of work is done in bringing a 1-coulmb charge from infinity to
that point.
The voltage Vab between two points a and b is the energy (or work) w required to move a unit positive
charge from a to b. [Unit of voltage is volt (V ).]
dw
(2.4)
dq
The potential difference between two points is 1volt if 1joule of work is done to displace 1coulomb of
charge from one point to another.
Vab =
2.1.6 Drift Velocity
Electric current is the number of coulombs of charge which pass a point in the circuit per unit time. Because
of its definition, it is often confused with the quantity ‘drift velocity.’ Drift velocity refers to the average
distance traveled by a charge carrier per unit time. Like the velocity of any object, the drift velocity of an
electron is the distance-to-time ratio. The path of a typical electron through a wire could be described as a
rather chaotic, zigzag path characterized by collisions with fixed atoms. Each collision results in a change in
direction of the electron.
The net effect of these collisions results in slow drifting of the electrons with a constant average drift
velocity. The drift velocity is defined as the vector average velocity of the charge carriers moving under the
influence of an electric field.
Mathematically, if n number of charge carriers (electrons) with charge Q each passes through an area A
with drift velocity v, then the current is given by, I nQvA.
2.1.7 Concept of Electromotive Force (emf)
The phenomenon of electric current depends on the presence of free
electrons. If a material has a large number of free electrons, these electrons will always move in random directions as shown in Fig. 2.1 (a). If
an external effort is applied to the material, it is possible to drift all the
electrons in a definite direction as shown in Fig. 2.1 (b). Such an external factor is known as electromotive force (emf). In other words, the
voltage or potential of an electrical energy source is known as emf.
Fig. 2.1 (a) Typical path of an electron
When we say something as ‘electrical energy source’, we mean
that the energy is converted from non-electrical form (such as,
mechanical, chemical, tidal, etc.) into an electrical form. Please note that emf is not a force, but it is the
energy or work done.
2.1.8 Electric Circuits and Networks
Any combination and interconnection of network elements like resistors or inductors or capacitors or electrical energy sources are known as a ‘networks’. However, a closed energized network is known as a ‘circuit’.
26
Network Analysis and Synthesis
A network need not contain an energy source; but a circuit must contain an
energy source. Therefore, it can be stated that all circuits are networks, but all
networks are not circuits.
2.1.9 Loop and Mesh
A loop or mesh denotes a closed path obtained by starting at a node and returning
back to the same node through a set of connected circuit elements without passing
through any intermediate node more than once. However, the difference between
mesh and loop is that a mesh does not contain any other loop within it, i.e. a mesh
is the smallest loop. In Fig. 2.2, some loops are: a-b-e-d-c-a, a-b-e-g-f-c-a, c-d-eb-g-f-c, etc; and some meshes are: a-b-e-d-c-a, c-d-e-g-f-c, g-e-b-g (through R7)
and g-e-b-g (through I ).
2.1.10
a
Node and Branch
R1
A high current results from
many charge carriers passing
through a given cross-section
of wire on a circuit.
Fig. 2.1 (b) Current is
constituted by flow of
many charge carriers
through a cross-section.
b
A node is a point in a circuit where two or more circuit eleV1 +
R5
_
ments join. A node is said to be an essential node if it joins
R2
R
3
d
R7
e
c
I
three or more elements. Examples of nodes for Fig. 2.2 are
a, b, c, d, e, f and g and examples of some essential node of
V2 +
R6
_
Fig. 2.2 are b, c, e and g.
R4
A branch is a path that connects two nodes. Those paths
f
g
that connect essential nodes without passing through an
essential node are known as essential branches. Examples of
Fig. 2.2 Circuit illustrating terminologies
branches of Fig. 2.2 are V1, R1, R2, R3, V2, R4, R5, R6, R7 and I
and some essential branches of Fig. 2.2 are c-a-b, c-d-e, c-f-g, b-e, e-g, b-g (through R7 ), and b-g (through I ).
2.2
DIFFERENT NOTATIONS
Notations
C
E
e
G
Name
capacitance
voltage source
instantaneous value of E
conductance
Unit
farad, F
volt, V
volt, V
siemens, S
I
i
k
L
M
N
P
Q
q
current
instantaneous current
coefficient
inductance
mutual inductance
number of turns
power
charge
instantaneous charge
ampere, A
ampere, A
unit less
henry, H
henry, H
unit less
watt, W
coulomb, C
coulomb, C
27
Introduction to Circuit-Theory Concepts
R
2.3
resistance
time constant
ohm,
second
t
V
v
instantaneous time
voltage drop
instantaneous V
second
volt, V
volt, V
W
energy
magnetic flux
joule, J
weber, Wb
magnetic linkage
weber, Wb
instantaneous
weber, Wb
BASIC CIRCUIT ELEMENTS
Active and Passive Elements Electric circuits consist of two basic types of elements. These are the
active elements and the passive elements.
An active element is capable of generating electrical energy. In electrical engineering, generating or producing electrical energy actually refers to conversion of electrical energy from a non-electrical form to an
electrical form. Similarly, energy loss would mean that electrical energy is converted to a non-useful form of
energy and not actually lost.
Examples of active elements are voltage source (such as a battery or generator) and current source. Most
sources are independent of other circuit variables, but some elements are dependent (modeling elements such
as transistors and operational amplifiers would require dependent sources).
Active elements may be ideal voltage sources or current sources. In such cases, the particular generated
voltage (or current) would be independent of the connected circuit.
A passive element is one which does not generate electricity but either consumes it or stores it. Resistors,
inductors and capacitors are simple passive elements. Diodes and transistors are also passive elements.
Passive elements may either be linear or non-linear. Linear elements obey a straight-line law. For example, a linear resistor has a linear voltage vs. current relationship which passes through the origin (V RI ).
A linear inductor has a linear flux vs. current relationship which passes through the origin (
kI ) and
a linear capacitor has a linear charge vs. voltage relationship which passes through the origin (q CV ).
[R, k and C are constants.]
Resistors, inductors and capacitors may be linear or non-linear, while diodes and transistors are always non-linear.
Linear Element A circuit/network element is linear if the relation between the current and voltage involves
a constant coefficient.
Examples Voltage–current relationship of a resistor, inductor and capacitor (both with zero initial condidi
1
tions) are linear (v ri, v = L , v = ∫ idt ). Hence, the elements are linear.
dt
C
Diode and transistors are non-linear devices having non-linear characteristics.
Bilateral System In a bilateral system, the same relationship between current and voltage exists for current flowing in either direction. On the other hand, a unilateral system has different current–voltage relationships for the two possible directions of current, as in diodes and transistors.
28
Network Analysis and Synthesis
2.4
PASSIVE CIRCUIT ELEMENTS
We will consider three basic passive elements, namely,
1. Resistor,
2. Inductor, and
3. Capacitor.
Name of Passive Element
Resistor
Symbol
Inductor
Capacitor
2.4.1 Electrical Resistance
Electrical resistance is a measure of the degree to which an object opposes an electric current through it.
The SI unit of electrical resistance is the ohm. Its reciprocal quantity is electrical conductance measured
in siemen. Electrical resistance shares some conceptual parallels with the mechanical notion of friction.
The resistance of an object determines the amount of current through the object for a given voltage across
the object.
V
I=
(2.5)
R
where, R is the resistance of the object, measured in ohm, equivalent to J-s/C 2,
V is the voltage across the object, measured in volt, and
I is the current through the object, measured in ampere.
For a wide variety of materials and conditions, the electrical resistance does not depend on the amount of
current through or the amount of voltage across the object, meaning that the resistance R is constant.
Factors Affecting the Resistance
1. Length of the Material The resistance of a material is directly proportional to the length of the
material.
2. Cross-sectional Area The resistance of a material is inversely proportional to the cross-sectional area of
the material.
3. Type and Nature of the Material The resistance of a material is dependent upon the nature of the material in the sense that it depends upon the number of free electrons present in the materials. For example,
for a conductor with plenty of free electrons, the resistance is least and for insulators with no free electrons, the resistance is the largest.
4. Temperature The resistance of a material is affected by the temperature of the material. Near
room temperature, the electric resistance of a typical metal conductor increases linearly with the
temperature:
T)
(2.6)
R R0(1
where,
is the thermal resistance coefficient.
29
Introduction to Circuit-Theory Concepts
The electric resistance of a typical intrinsic (non-doped) semiconductor decreases exponentially with the
temperature:
R R0e T
(2.7)
Extrinsic (doped) semiconductors have a far more complicated temperature profile. As temperature
increases starting from absolute zero, they first decrease steeply in resistance as the carriers leave the donors
or acceptors. After most of the donors or acceptors have lost their carriers the resistance starts to increase
again slightly due to the reducing mobility of carriers (much as in a metal). At higher temperatures it will
behave like intrinsic semiconductors as the carriers from the donors or acceptors become insignificant compared to the thermally generated carriers.
The electric resistance of electrolytes and insulators is highly non-linear, and case-dependent, therefore no
generalized equations are given.
Resistance of a Conductor
dc Resistance As long as the current density is totally uniform in the conductor, the dc resistance R of a
conductor of regular cross-section can be computed as
l
R=
(2.8)
A
where, l is the length of the conductor, measured in metres
A is the cross-sectional area, measured in square metres
(Greek: rho) is the electrical resistivity (also called specific electrical resistance) of the material,
measured in ohm-metre. Resistivity is a measure of the ability of the material to oppose the flow of
electric current.
For practical reasons, any connections to a real conductor will almost certainly mean the current density is not
totally uniform. However, this formula still provides a good approximation for long thin conductors such as wires.
AC Resistance If a wire conducts high-frequency alternating current then the effective cross-sectional
area of the wire is reduced. This is because of the skin effect.
This formula applies to isolated conductors. In a conductor close to others, the actual resistance is higher
because of the proximity effect.
Differential Resistance When resistance may depend on voltage and current, differential resistance,
incremental resistance or slope resistance is defined as the slope of the U-I graph at a particular point.
dU
R=
(2.9)
Thus:
dI
This quantity is sometimes simply called resistance, although the two definitions are equivalent only for an
ohmic component such as an ideal resistor. If the U-I graph is not monotonic (i.e., it has a peak or a trough), the
differential resistance will be negative for some values of voltage and current. This property is often known as
negative resistance, although it is more correctly called negative differential resistance, since the absolute resistance U/I is still positive.
Resistor A resistor is a two-terminal electrical or electronic component that resists an electric current by
producing a voltage drop between its terminals in accordance with Ohm’s law:
V
(2.10)
R=
I
30
Network Analysis and Synthesis
The electrical resistance is equal to the voltage drop across the resistor divided by the current through the
resistor. Resistors are used as part of electrical networks and electronic circuits.
Energy in a Resistor Instantaneous power absorbed in the resistor is
p
vi
iR
i
i2R
(in Watt)
(2.11)
Therefore, the energy converted into heat energy is given by
t
t
0
0
W = ∫ pdt = ∫ i 2 Rdt = i 2 Rt (in joules)
(2.12)
Four-Band Axial Resistors Four-band identification is the most commonly used color-coding scheme
on all resistors. It consists of four colored bands that are painted around the body of the resistor. The scheme
is simple. The first two numbers are the first two significant digits of the resistance value, the third is a multiplier, and the fourth is the tolerance of the value. Each color corresponds to a certain number, shown in the
chart below. The tolerance for a 4-band resistor will be 2%, 5%, or 10%.
The Standard EIA Color Code Table per EIA-RS-279 is as follows:
Color
Black
Brown
Red
Orange
Yellow
Green
Blue
Violet
Grey
1 band
2 band
3rd band
(multiplier)
0
0
100
—
—
1
10
1
1% (F)
100 ppm
10
2
10
2% (G)
—
50 ppm
3
10
4
—
25 ppm
10
5
0.5% (D)
—
10
6
0.25% (C)
—
10
7
0.1% (B)
—
10
8
0.05% (A)
—
—
9
st
1
2
3
4
5
6
7
8
nd
2
3
4
5
6
7
8
4th band
(tolerance)
Temp.
Coefficient
15 ppm
—
White
9
9
10
Gold
—
—
0.1
5% (J)
—
Silver
—
—
10% (K)
—
None
—
—
0.01
—
20% (M)
—
Note: Red to violet are the colors of the rainbow where red is low energy and violet is high energy.
As an example, let us take a resistor which (read left to right) displays the colors yellow, violet, yellow,
brown. We take the first two bands as the value, giving us 4, 7. Then the third band, another yellow, gives
us the multiplier 104. Our total value is then 47 104 , totaling 470,000 or 470 k . Our brown is then a
tolerance of 1%.
Resistors use specific values, which are determined by their tolerance. These values repeat for every exponent; 6.8, 68, 680, and so forth. This is useful because the digits, and hence the first two or three stripes, will
always be similar patterns of colors, which make them easier to recognize.
31
Introduction to Circuit-Theory Concepts
5-Band Axial Resistors 5-band identification is used for higher tolerance resistors (1%, 0.5%, 0.25%,
0.1%), to notate the extra digit. The first three bands represent the significant digits, the fourth is the multiplier, and the fifth is the tolerance. 5-band standard tolerance resistors are sometimes encountered, generally
on older or specialized resistors. They can be identified by noting a standard tolerance color in the 4th band.
The 5th band in this case is the temperature coefficient.
Series and Parallel Arrangements of Resistors Resistors
in a parallel configuration each have the same potential difference
(voltage). To find their total equivalent resistance (Req):
1
1 1
1
= + + ⋅⋅⋅+
Req R1 R2
Rn
The parallel, property can be represented in equations by two
vertical lines “||” (as in geometry) to simplify equations. For two
resistors,
R1 R2
(2.14)
R1 + R2
The current through resistors in series stays the same,
but the voltage across each resistor can be different. The
sum of the potential differences (voltage) is equal to the
total voltage. To find their total resistance:
Req = R1
Req
R1
(2.13)
Rn
Fig. 2.3 Parallel arrangement
of resistors
R2 =
R1
R2
Rn
R1
R2
R2
Fig. 2.4 Series arrangement of resistors
R3
(2.15)
A resistor network that is a combination of parallel and series can sometimes
be broken up into smaller parts that are either one or the other. For instance,
RR
Req = R1 R2 + R3 = 1 2 + R3
(2.16)
R1 + R2
(
R2
)
R2
R1
Fig. 2.5 Series–parallel
arrangement of resistors
Characteristics of Series Circuits
1. The same current flows through each resistance.
2. The supply voltage V is the sum of the voltage drops across each resistance, i.e. V
3. The equivalent resistance is equal to the sum of the individual resistances.
V1
V2
V3
Vn.
Current Division by Parallel Resistances When a total current IP is passed through parallel-connected
resistances R1 and R2, the voltage VP which appears across the parallel circuit is
VP IPRP IPR1R2/(R1 R2)
The currents I1 and I2 which pass through the respective resistances R1 and R2 are
I1 VP /R1 IPRP/R1 IPR2/(R1 R2)
I2 VP/R2 IPRP/R2 IPR1/(R1 R2)
In general terms, for resistances R1, R2, R3, . . . (with conductances G1, G2, G3, . . . ) connected in parallel:
VP IPRP IP/GP IP/(G1 G2 G3
)
In VP/Rn VPGn IPGn/GP IPGn/(G1 G2 G3
)
where Gn 1/Rn and In is the current through the nth resistance Rn.
32
Network Analysis and Synthesis
Characteristics of Parallel Circuits
1. The voltage across all the resistances is the same.
2. The total current is the sum of the currents flowing through the parallel resistances.
3. The reciprocal of the equivalent resistance of a parallel circuit is equal to the sum of the reciprocal of
the individual resistances.
4. The highest current passes through the highest conductance (with the lowest resistance).
Example 2.1 Find the equivalent resistance between the terminals A and B.
A
15
10
B
C
D
6
Fig. 2.6
4
Circuit of Example 2.1
Solution The circuit is redrawn as shown in Fig. 2.7 (a).
Here, 15 and 10 are in parallel and 6 and 4 are also in parallel. These
two resisiances are then connected in series. Therefore, the equivalent resistance
between terminals A and B is
{
} {
× 10 6 × 4
+
= 6 + 2.4 = 8.4
} 15
15 + 10 6 + 4
A
15
C
6
10
D
4
B
Fig. 2.7 (a)
Req = 15 10 + 6 4 =
2.4.2
Capacitance
Capacitance is a measure of the amount of electric charge stored for a given electric potential. The most
common form of charge storage device is a two-plate capacitor. If the charges on the plates are Q and −Q,
and V gives the voltage difference between the plates then the capacitance is given by
Q
C=
(2.17)
V
The SI unit of capacitance is farad (F); 1 farad 1 coulomb per volt.
The capacitance can be calculated if the geometry of the conductors and the dielectric properties of the
insulator between the conductors are known. For example, the capacitance of a parallel-plate capacitor constructed of two parallel plates of area A separated by a distance d is approximately equal to the following:
A
C=
(2.18)
d
where C is the capacitance in farad, F,
is the permittivity of the insulator used (or 0 for a vacuum),
A is the area of each plate, measured in square metre and,
d is the separation between the plates, measured in metre.
The equation is a good approximation if d is small compared to the other dimensions of the plates.
The dielectric constant for a number of very useful dielectric changes as a function of the applied electrical
field, e.g., for ferroelectric materials, so the capacitance for these devices is no longer purely a function of
33
Introduction to Circuit-Theory Concepts
device geometry. If a capacitor is driven with a sinusoidal voltage, the dielectric constant, or more accurately
the dielectric permittivity, is a function of frequency. A changing dielectric constant with frequency is referred
as a dielectric dispersion, and is governed by dielectric relaxation processes, such as Debye relaxation.
Capacitor A capacitor is an electrical device that can store energy in the electric field between a pair of
closely spaced conductors. When a current is applied to the capacitor, electric charges of equal magnitude, but
of opposite polarity, build up on each plate.
Capacitors are used in electrical circuits as energy-storage devices. They can also be used to differentiate
between high-frequency and low-frequency signals and this makes them useful in electronic filters.
Capacitors are occasionally referred as condensers. This is now considered an antiquated term.
Properties of Capacitance The relation between charge and voltage in a capacitor is written as
Q CV
(2.19)
dQ
dV
dV
dC
=C
=C
+V
dt
dt
dt
dt
In most physical cases, the capacitance is constant with time.
dV
∴ i =C
dt
1
∴ dV = i dt
C
v
t
1
Taking integration on both sides, ∫ dV = ∫ i dt
C0
0
The current,
i=
(2.20)
t
1
i(t )dt + vc (0)
(2.21)
C ∫0
where, vc(0) is the initial voltage across the capacitor. For zero initial voltage,
t
1
vc = ∫ i dt
(2.22)
C0
From Eq. (2.20), it is clear that for an abrupt change of voltage across the capacitor, the current becomes infinite.
Also, from Eq. (2.22), it is observed that for a finite change of current in zero time, the integral must be zero.
Therefore, the voltage across a capacitor cannot change instantaneously.
Let us explain the meaning of the initial voltage vC (0). It is possible that this capacitor might have been used
in some other circuit earlier, where it absorbed some energy and then it was disconnected. Because of its nondissipative nature, the energy was stored within the capacitor. Now, as this capacitor is connected to a circuit, it
gets some path to release its stored energy. Here, this stored energy is represented by the initial voltage vC(0).
or,
vc (t ) =
Energy Stored in Capacitors The energy (measured in joules) stored in a capacitor is equal to the work done
to charge it. Consider a capacitance C, holding a charge q on one plate and q on the other. Moving a small
element of charge dq from one plate to the other against the potential difference V q/C requires the work dW:
q
dW = dq
(2.23)
C
where, W is the work measured in joule,
q is the charge measured in coulomb, and
C is the capacitance, measured in farad.
34
Network Analysis and Synthesis
We can find the energy stored in a capacitance by integrating this equation. Starting with an uncharged
capacitance (q 0) and moving the charge from one plate to the other until the plates have charge Q and
Q requires the work W:
Q
q
1 Q2 1 2
Wcharging = ∫ dq =
= CV = Wstored
(2.24)
C
2 C 2
0
Combining this with the above equation for the capacitance of a flat-plate capacitor, we get
1
1 A 2
Wstored = CV 2 =
V
2
2 d
where, W is the energy measured in joule,
C is the capacitance, measured in farad, and
V is the voltage measured in volt.
(2.25)
Series or Parallel Arrangements of Capacitors
Capacitors in a parallel configuration each have the same potential difference (voltage). Their total capacitance (Ceq) is given by
Ceq
C1
C2
Cn
(2.26[a])
The reason for putting capacitors in parallel is to increase the total
amount of charge stored. In other words, increasing the capacitance also
increases the amount of energy that can be stored.
The current through capacitors in series stays the same, but the voltage
across each capacitor can be different. The sum of the potential differences (voltage) is equal to the total voltage. Their total capacitance is
given by
1
1 1
1
= + + ⋅⋅⋅+
Ceq C1 C2
Cn
C1
In parallel the effective area of the combined capacitor has increased,
increasing the overall capacitance. While in series, the distance between
the plates has effectively been increased, reducing the overall capacitance.
Cn
Fig. 2.8 Parallel arrangement of
capacitors
C1
(2.26[b])
C2
C2
Cn
Fig. 2.9 Series arrangement of
capacitors
Voltage Division by Capacitances
In Series Connection When a total voltage ES is applied to series-connected capacitances C1 and C2, the
charge QS which accumulates in the series circuit is
QS iSdt ESCS ESC1C2/(C1 C2)
The voltages V1 and V2 which appear across the respective capacitances C1 and C2 are
V1 iS dt/C1 ESCS/C1 ESC2/(C1 C2)
V2 iSdt/C2 ESCS/C2 ESC1/(C1 C2)
In general terms, for capacitances C1, C2, C3, . . . connected in series:
QS iSdt ESCS ES/(1/CS) ES/(1/C1 1/C2 1/C3
/
Vn iS dt/Cn ESCS/Cn ES/Cn(1/CS) ES/Cn(1/C1 1/C2 1/C3
)
Note that the highest voltage appears across the lowest capacitance.
35
Introduction to Circuit-Theory Concepts
In Parallel Connection When a voltage EP is applied to parallel-connected capacitances C1 and C2, the
charge QP which accumulates in the parallel circuit is
QP iPdt EPCP EP(C1 C2)
The charges Q1 and Q2 which accumulate in the respective capacitances C1 and C2 are
Q1 i1dt EPC1 QPC1/CP QPC1/(C1 C2)
Q2 i2dt EPC2 QPC2/CP QPC2/(C1 C2)
In general terms, for capacitances C1, C2, C3, . . . connected in parallel:
QP iPdt EPCP EP(C1 C2 C3
)
Qn indt EPCn QPCn /CP QPCn /(C1 C2 C3
)
Note that the highest charge accumulates in the highest capacitance.
2.4.3 Inductance
An electric current i flowing round a circuit produces a magnetic field and hence a magnetic flux through
the circuit. The ratio of the magnetic flux to the current is called the inductance, or more accurately selfinductance of the circuit. The term was coined by Oliver Heaviside in February 1886. Inductance is denoted
by L, in honour of the physicist Heinrich Lenz. The quantitative definition of the inductance is therefore
(2.27)
L=
i
It follows that the SI unit for inductance is weber per ampere. In honour of Joseph Henry, the unit of
inductance has been given the name henry (H): 1H 1Wb/A.
Properties of Inductance The equation relating inductance and flux linkages can be rearranged as
follows:
Li
(2.28)
Taking the time derivative of both sides of the equation yields
d
di dL
= L +i
dt
dt
dt
In most physical cases, the inductance is constant with time and so
di
d
=L
(2.29)
dt
dt
By Faraday’s law of induction we have
d
= −E = v
(2.30)
dt
where E is the electromotive force (emf) and v is the induced voltage. Note that the emf is opposite to the
induced voltage. Thus
di
v=L
(2.31)
dt
t
1
i(t ) = ∫ v (t ) dt + i(0)
or
(2.32)
L0
where i(0) is the initial current. When initial current is zero,
t
1
i(t ) = ∫ v (t )dt
(2.33)
L0
36
Network Analysis and Synthesis
These equations together state that for a steady applied voltage v, the current changes in a linear
manner, at a rate proportional to the applied voltage, but inversely proportional to the inductance.
Conversely, if the current through the inductor is changing at a constant rate, the induced voltage is
constant.
From Eq. (2.31), it is clear that for an abrupt change in current, the voltage across the inductor becomes
infinite. Also, from Eq. (2.33), it is observed that for a finite change in voltage in zero time the integral must
be zero.
Therefore, the current through an inductor cannot change instantaneously.
Let us explain the meaning of the initial current i(0). It is possible that this inductor might have been
used in some other circuit earlier, where it absorbed some energy and then it was disconnected. Because
of its non-dissipative nature, the energy was stored within the inductor core. Now, as this inductor is connected to a circuit, it gets some path to release its stored energy. Here, this stored energy is represented by
the initial current i(0).
The effect of inductance can be understood using a single loop of wire as an example. If a voltage is suddenly applied between the ends of the loop of wire, the current must change from zero to non-zero. However,
a non-zero current induces a magnetic field by Ampère’s law. This change in the magnetic field induces
an emf that is in the opposite direction of the change in current. The strength of this emf is proportional to
the change in current and the inductance. When these opposing forces are in balance, the result is a current
that increases linearly with time where the rate of this change is determined by the applied voltage and the
inductance.
Inductor An inductor is a passive electrical device employed in electrical
circuits for its property of inductance. An inductor can take many forms.
Series and Parallel Arrangement of Inductors Inductors in a parallel
configuration each have the same potential difference (voltage). To find their
total equivalent inductance (Leq),
1
1 1
1
= + + ⋅⋅⋅+
Leq L1 L2
Ln
(2.34)
The current through inductors in series stays the same, but the voltage
across each inductor can be different. The sum of the potential differences
(voltage) is equal to the total voltage. To find their total inductance.
Leq L1 L2
Ln
(2.35)
These simple relationships hold true only when there is no mutual coupling
of magnetic fields between individual inductors.
L1
Ln
L2
Fig. 2.10 Parallel arrangement of inductors
L1
L2
Ln
Fig. 2.11 Series arrangement of inductors
Energy Stored in Inductors When an electric current is flowing in an inductor, there is energy stored
in the magnetic field.
Suppose that an inductor of inductance L is connected to a variable dc voltage supply. The supply is
adjusted so as to increase the current i flowing through the inductor from zero to some final value I.
As the current through the inductor is increasing, the emf generated is
di
(2.36)
E = −L
dt
37
Introduction to Circuit-Theory Concepts
and this emf acts to oppose the increase in the current. Clearly, work must be done against this emf by the voltage source
in order to establish the current in the inductor. The work done by the voltage source during a time interval dt is
di
dW = Pdt = − Eidt = i L dt = Lidi
dt
Here, P
Ei is the instantaneous rate at which the voltage source performs work. To find the total work
W done in establishing the final current I in the inductor, we must integrate the above expression. Thus,
I
1
W = L ∫ idi = LI 2
2
0
(2.37)
This energy is actually stored in the magnetic field generated by the current flowing through the inductor.
In a pure inductor, the energy is stored without loss, and is returned to the rest of the circuit when the current
through the inductor is ramped down, and its associated magnetic field collapses.
2.4.4 Coupled Inductors
When the magnetic flux produced by an inductor links another inductor, these inductors are said to be coupled. Coupling is often undesired but in many cases, this coupling is intentional and is the basis of the transformer. When inductors are coupled, there exists a mutual inductance that relates the current in one inductor
to the flux linkage in the other inductor. Thus, there are three inductances defined for coupled inductors:
L11—the self-inductance of the inductor 1
L22—the self-inductance of the inductor 2
L12 L21—the mutual inductance associated with both inductors
When either side of the transformer is a tuned circuit, the amount of mutual inductance between the two
windings determines the shape of the frequency response curve. Although no boundaries are defined, this is
often referred as loose-, critical-, and over-coupling. When two tuned circuits are loosely coupled through
mutual inductance, the bandwidth will be narrow. As the amount of mutual inductance increases, the bandwidth continues to grow. When the mutual inductance is increased beyond a critical point, the peak in the
response curve begins to drop, and the centre frequency will be attenuated more strongly than its direct sidebands. This is known as over-coupling.
Mutual Inductance The two vertical lines between the inductors indicate a solid core that the wires
of the inductor are wrapped around. n:m shows the ratio between the number of windings of the left
inductor to windings of the right inductor. This picture also shows the dot
M
convention.
I2
I
1
Mutual inductance is the concept that the current through one inductor
can induce a voltage in another nearby inductor. It is important as the mechanism by which transformers work, but it can also cause unwanted coupling
between conductors in a circuit.
The mutual inductance, M, is also a measure of the coupling between two
inductors. The mutual inductance by the circuit i on the circuit j is given by
n:m
the double integral Neumann formula:
ds
ds
Fig. 2.12 Circuit diagram
M ij = 0 ∫ ∫ i j
(2.38)
representation of mutually
4 Ci C j Rij
inducting inductors
38
Network Analysis and Synthesis
The mutual inductance also has the relationship:
M21 N1N2P21
where M21 is the mutual inductance, and the subscript specifies the relationship of the voltage induced in
the coil 2 to the current in the coil 1.
N1 is the number of turns in the coil 1,
N2 is the number of turns in the coil 2,
P21 is the permeance of the space occupied by the flux.
The mutual inductance also has a relationship with the coefficient of coupling. The coefficient of coupling is always between 1 and 0, and is a convenient way to specify the relationship between a certain orientation of the inductor with arbitrary inductance:
M = k L1 L2
(2.39)
where k is the coefficient of coupling and 0 k 1,
L1 is the inductance of the first coil, and
L2 is the inductance of the second coil.
Once this mutual inductance factor M is determined, it can be used to predict the behavior of a circuit:
dI
dI
V = L1 1 + M 2
(2.40)
dt
dt
where V is the voltage across the inductor of interest,
L1 is the inductance of the inductor of interest,
dI1 /dt is the derivative, with respect to time, of the current through the inductor of interest,
M is the mutual inductance, and
dI2 /dt is the derivative, with respect to time, of the current through the inductor that is coupled to the
first inductor.
When one inductor is closely coupled to another inductor through mutual inductance, such as in a transformer, the voltages, currents, and number of turns can be related in the following way:
N
(2.41)
Vs = I p s
Np
where Vs is the voltage across the secondary inductor,
Vp is the voltage across the primary inductor (the one connected to a power source),
Ns is the number of turns in the secondary inductor, and
Np is the number of turns in the primary inductor.
Conversely the current is
Np
Is = I p
(2.42)
Ns
where
Is is the current through the secondary inductor,
Ip is the current through the primary inductor (the one connected to a power source),
Ns is the number of turns in the secondary inductor, and
Np is the number of turns in the primary inductor.
Note that the power through one inductor is the same as the power through the other. Also note that these
equations don’t work if both transformers are forced (with power sources).
39
Introduction to Circuit-Theory Concepts
Inductance and Capacitance as Linear Circuit Elements We consider that an alternating voltage v(t )
is applied to an inductor L at a reference time t 0. Then the current carried by the inductor is given by
t
1
i(t ) = ∫ v (t ) dt + i(0)
(2.32)
L0
and the relation between flux linkage and current is given by
(t ) Li(t )
(2.28)
The properties of an inductor can be explained by plotting the characteristics in the i– plane. If the characteristic is a straight line passing through the origin, the inductor will be considered as a linear element.
But if the i– characteristic is not a straight line and/or does not pass through the origin (e.g., Hysteresis
curve), the inductor will behave as a non-linear element.
λ
q
λ
Slope = L
Slope = C
i
v
i
Fig. 2.13 (a) Characteristic
of a linear inductor
Fig. 2.13 (b) Characteristic of
a non-linear inductor
Fig. 2.13 (c) Characteristic
of a linear capacitor
Similarly, for a capacitor the voltage is given by
t
1
v t = ∫ i t dt + v 0
C0
()
()
()
(2.21)
and the relation between charge and voltage is given by
q(t ) Cv(t )
(2.19)
The properties of a capacitor can be explained by plotting the characteristics in the q–v plane. If the characteristic is a straight line passing through the origin, the capacitor will be considered as a linear element.
But if the q–v characteristic is not a straight line and/or does not pass through the origin (e.g., space-charge
capacitance of a diode), the capacitor will behave as a non-linear element.
2.5
TYPES OF ELECTRICAL ENERGY SOURCES
Energy source is defined as the device that generates electrical energy. They are classified according to the
current–voltage characteristics. The classification is given below.
Electrical energy source
Independent sources
Dependent sources
Voltage-controlled
voltage source (VCVS)
Voltage source
Current source
Voltage-controlled
current source(VCCS)
Current-controlled
curren source(CCCS)
Current-controlled
voltage source(CCVS)
40
Network Analysis and Synthesis
Independent Voltage Source An ideal voltage source has the following features:
(i) It is a voltage generator whose output voltage remains absolutely constant whatever be the value of the
output current.
(ii) It has zero internal resistance so that voltage drop in the source is zero.
(iii) The power drawn by the source is zero.
In practical, the voltage does not remain constant, but falls slightly; this is taken care of by connecting a
small resistance (r ) in series with the ideal source. In this case, the terminal voltage will be
v1 (t ) = v (t ) − ir
i.e. it will decrease with increase in the current i.
An ideal voltage source is not practically possible. No voltage source can maintain its terminal voltage constant even when its terminals are short-circuited. The terminal voltage of a practical voltage source
decreases as the load current increases. The v–i characteristics of an ideal and practical voltage source are
shown in Fig. 2.14. A dc or ac generator or batteries are some examples of independent voltage sources. A
lead–acid battery and a dry-cell are some examples of constant voltage source which can produce constant
terminal voltage within a specified range of output current.
i
V
v (t)
v (t)
r
v (t)
v1(t )
Ideal
Practical
i
Fig. 2.14
Independent voltage sources and their characteristics
Independent Current Source An ideal current source has the following features.
(i) It produces a constant current irrespective of the value of the voltage across it.
(ii) It has infinity resistance.
(iii) It is capable of supplying infinity power.
In practical, the output current does not remain constant but decreases with increase in voltage. So, a
practical current source is represented by an ideal current source in parallel with a high resistance (R) and the
output current becomes
v (t )
i1 (t ) = i(t ) −
R
Similar to voltage sources, an ideal current source is not practically possible. No current source can maintain constant current even when its terminals are open-circuited. The output current of a practical current
source decreases as the output voltage increases. The v–i characteristics of an ideal and practical current
source are shown in Fig. 2.15. A solar cell, which can produce constant current within a specified range of
output voltage, is an example of an independent current source. A natural lightning can be considered to be
an ideal current source. When a natural lightning strikes the top of a conductor, the resistance to the ground
path is ideally zero. But, when the lightning strikes a non-conducting element (like the top of a tree), a large
voltage is developed across the element which is flashed out immediately.
41
Introduction to Circuit-Theory Concepts
i1
v (t)
Practical
Ideal
i(t)
I
i(t)
R
v (t )
i
Fig. 2.15 Independent current sources and their characteristics
Dependent Sources In dependent sources (also referred as controlled sources), the source voltage or
current is not fixed, but is dependent on a voltage or current at some other location in the circuit. Thus, there
are four types of dependent sources.
(a) Voltage-controlled voltage source (VCVS)
(b) Current-controlled voltage source (CCVS)
(c) Voltage-controlled current source (VCCS)
(d) Current-controlled current source (CCCS)
K1VX
K2Ix
K3Vx
K4Ix
VCVS
CCVS
VCCS
CCCS
Fig. 2.16 Symbols of dependent sources
Dependent sources are unilateral, because for a voltage-controlled voltage source, say, v2 kv1, the output
voltage v2 is controlled by the input voltage v1, but the output current i2 has no influence on the input v1.
Application in electronic systems that uses either transistors or vacuum tubes needs dependent sources.
2.6
FUNDAMENTAL LAWS
The fundamental laws that govern electric circuits are Ohm’s law and Kirchhoff’s laws.
2.6.1 Ohm’s Law
Ohm’s law states that the voltage v(t ) across a resistor R is directly proportional to the current i(t ) flowing
through it.
v(t ) i(t ) or v(t ) R i(t )
This general statement of Ohm’s law can be extended to cover inductances and capacitors as well alternating current conditions and transient conditions. This is then known as the generalized Ohm’s law. This may be stated as
v(t ) Z(p) i(t ), where p d/dt differential operator
Z(p) is known as the impedance function of the circuit, and the above equation is the differential equation
governing the behaviour of the circuit.
Z(p) R for a resistor
Lp for an inductor
1
for a capacitor
Cp
42
Network Analysis and Synthesis
In the particular case of alternating current, p j , so that the equation governing circuit behaviour may
be written as
V Z( j )I
Z( j )
R for a resistor
j L for an inductor
1
for a capacitor
jω C
Definition of Ohm’s Law Physical states (temperature, material, etc.) of a conductor remaining constant,
the current flowing through a capacitor is directly proportional to the potential difference across the two ends
of the conductor.
2.6.2 Kirchhoff’s Current Law (KCL)
Kirchhoff’s current law is based on the principle of conservation of charge. This requires that the algebraic
sum of the charges within a system cannot change. Thus the total rate of change of charge must add up to
zero. The rate of change of charge is the current.
i1
i4
id
i5
ie
ia
i2
i3
ic
ib
Fig. 2.17
Illustration of KCL
This gives us our basic Kirchhoff’s current law as the algebraic sum of the currents meeting at a point is
zero, i.e. at a node, .In 0, where In are the currents in the branches meeting at the node.
This is also sometimes stated as that the sum of the currents entering a node is equal to the sum of the
current leaving the node.
The theorem is applicable not only to a node, but to a closed system.
i1 i2 i3 i4 i5 0; Also, for the closed boundary, ia ib ic id ie 0.
2.6.3 Kirchhoff’s Voltage Law (KVL)
Kirchhoff’s voltage law is based on the principle of conservation of
energy. This requires that the total work done in taking a unit positive
charge around a closed path and ending up at the original point is zero.
This gives us our basic Kirchhoff’s law as the algebraic sum of the
potential differences taken round a closed loop is zero, i.e. around a loop,
.Vn 0, where Vn are the voltages across the branches in the loop.
va vb vc vd ve 0
Vd
Vc
Ve
Loop
Va
Vb
Fig. 2.18 Illustration of KVL
43
Introduction to Circuit-Theory Concepts
This is also sometimes stated as the sum of the emfs taken around a closed loop is equal to the sum of the
voltage drops around the loop.
Although all circuits could be solved using only Ohm’s law and Kirchhoff’s laws, the calculations would
be tedious. Various network theorems have been formulated to simplify these calculations.
Sign Conventions for Applying Kirchhoff’s Laws
(i) When tracing through a voltage source from a positive to a negative terminal, the voltage should be
given a positive sign.
(ii) When tracing through a voltage source from a negative to a positive terminal, the voltage should be
given a negative sign.
(iii) When tracing through a resistance in the direction of current flow, the voltage should be given a positive sign.
(iv) When tracing through a resistance in a direction opposite to the direction of current flow, the voltage
should be given a negative sign.
2.7
SOURCE TRANSFORMATION
Transformation of several voltage (or current) sources into a single voltage (or current) source and a voltage
source into a current source or vice-versa is known as source transformation. This makes circuit analysis easier.
There are some rules of source transformation.
Rule (1) Several voltage sources {V1(t ), V2(t ), . . . , Vn(t )} connected in series will be replaced by a single
voltage source of value V V1(t ) V2(t )
Vn(t ). Similarly, a number of current sources {I1(t ), I2(t ), . . . ,
In(t ).
In(t )} connected in parallel is replaced by a single current source of value I(t ) I1(t ) I2(t )
V1(t )
V2(t )
⬅
V(t) {V1(t) V2(t)
... V (t)}
n
I1
I2
... I
n
⬅
I
I1 I2 ... In
Vn(t )
Fig. 2.19 Source transformation technique: Rule (1)
Rule (2) A number of voltage sources V1(t ), V2(t ), . . . , Vn(t ) in parallel will result in a single voltage source,
V(t ) V1(t ) V2(t ) . . . Vn(t ).
Therefore, voltage sources should not be connected in parallel unless they have identical potential, as
paralleling of sources with non-similar potential waveforms will result in heavy current, which may damage
the equipment.
Similarly, a number of current sources I1(t ), I2(t ), . . . , In(t ) in series will result in a single current source
of value I(t ) I1(t ) I2(t ) . . . In(t ) and thus, current sources cannot be connected in series if they are
not identical.
44
Network Analysis and Synthesis
I1(t )
V1
V2
Vn
V
⬅
V1
Vn
V2
I2(t )
⬅
I(t)
I1(t )
I2(t )
In(t )
In(t )
Fig. 2.20
Source transformation technique: Rule (2)
Rule (3) As far as the computations in the remainder of the network are concerned, a resistor in parallel
with an ideal voltage source and a resistor in series with an ideal current source may be ignored.
R
R
v (t)
⬅
⬅ v (t)
I(t)
I(t )
Fig. 2.21 Source transformation technique: Rule (3)
Rule (4) A voltage source V(t ) in series with a resistor R can be converted into a current source I(t ) in parallel with the same resistor R, where, I t = V (t ) .
R
()
R
V(t)
Fig. 2.22
I1(t)
I1(t)
V1(t)
⬅
I(t)
R
V1(t)
Source transformation technique: Rule (4)
Similarly, a voltage source V(t ) in series with a capacitor C may converted into a current source I(t ) in
dV (t )
; and a voltage source V(t ) in series with an inductor L may converted
parallel with C, where, I (t ) = C
dt
1
into a current source I(t ) in parallel with L, where, I t = ∫V (t )dt .
L
()
2.7.1
V-Shift and I-Shift in Source Transformation
This method of shifting a voltage source or a current source is useful for a voltage source without any
series passive element and a current source without any parallel passive element. For source transformation,
45
Introduction to Circuit-Theory Concepts
i.e. transforming a voltage source into a current source and vice-versa, it is first necessary to shift the sources
within the network. This is done by V or I shifting.
The methods are explained below.
For voltage source shifting, we consider a network as shown in Fig. 2.23 (a). We can shift the voltage
source within the network as shown in Fig. 2.23 (b) and Fig. 2.23 (c).
R3
R3
R1
R2
R3
R1
R2
R1
V
V
(a)
V
V
V V
V
(c)
(b)
Fig. 2.23
R2
V-Shift in source transformation
Similarly, a current source can be shifted within a network as explained in Fig. 2.24 (a) to Fig. 2.24 (b).
I
R1
R1 R
2
R3
R2
I
I
I
R3
(a)
Fig. 2.24
(b)
I-Shift in source transformation
Example 2.2 Using source transformation, find the current through
the 3- resistor shown in Fig. 2.25.
Solution We convert the four voltage sources in series with the resistances into equivalent currents in parallel with the same resistances.
The simplified circuit is shown in Fig. 2.26 (a) and Fig. 2.26 (b).
6V 2
8V
6V 3V
2
1
1
3
2
Fig. 2.25 Circuit of Example 2.2
3A
2/3
1
4A
2
6A
1
(a)
Fig. 2.26
3A
3
10 A
2/3
6A
(b)
3
46
Network Analysis and Synthesis
Now we convert back the current sources into their equivalent voltage
sources. The simplified circuit is shown in Fig. 2.26 (c).
From Fig. 2.26 (c), we get the current through the 3- resistor as
20
4+
3 = 32 = 2.46 A
I=
2 2
13
+ +3
3 3
2.8
2/3
20
V
3
2
3
4V
I
3
Fig. 2.26 (c)
NETWORK ANALYSIS TECHNIQUES
Network analysis is the determination of the response output of a network when the input excitation is given.
There are two techniques of network analysis:
1. Nodal analysis
2. Loop or mesh analysis
2.8.1 Nodal Analysis
It is based on Kirchhoff’s current law (KCL). In this method, the unknown variables are the node voltages. It
is generally used when the circuit contains several current sources.
Steps
(i) If there are ‘N’ number of nodes in a network, all nodes are labeled. One node is treated as the datum or
reference node (zero potential) and the other node voltages are treated as unknowns to be determined
with respect to this reference.
(ii) KCL is written at each node in terms of node voltages.
(a) KCL is applied at N 1 of the N nodes of the circuit using assumed current directions, as necessary. This will create N 1 linearly independent equations, known as node equations.
(b) In a circuit with independent voltage sources, if two nodes of interest are separated by a voltage
source instead of a resistor or current source then the concept of supernode is used that creates
constraint equations.
(c) The current is computed based on voltage difference between two nodes. The current in any branch
is obtained via Ohm’ law as
i=
I=
where, Vm
Vmn
R
Vmn
Z
=
Vm − Vn
;
R
for dc
=
Vm − Vn
;
Z
for ac
Vn and current flows from the node m to n.
(iii) Solution of the N 1 simultaneous equations (by Gaussian elimination or matrix method) gives the
unknown node voltages.
47
Introduction to Circuit-Theory Concepts
Example 2.3 For the network shown in Fig. 2.27, apply Kirchhoff’s current law and write the node equations.
R2
E1
I3 I4
I1
I5 I6
R1
Fig. 2.27
R4
E2
E3
I7
R3
I8
R5
R6
I2
Network of Example 2.3
Solution Let node voltages be E1, E2 and E3 at nodes 1, 2 and 3 respectively.
At the node 1,
I1 = I 3 + I 4
I1 =
E1 (E1 − E2 )
+
R1
R2
⎛ 1 1⎞
E
I1 = ⎜ + ⎟ E1 − 2
R
R
R2
⎝ 1
2⎠
(i)
At the node 2,
I4 = I5 + I6
(E1 − E2 ) (E1 − E3 ) E2
=
+
R2
R4
R3
0=−
⎛ 1 1 1⎞ E
E1
+ E2 ⎜ + + ⎟ − 3
R2
⎝ R2 R3 R4 ⎠ R4
(ii)
At the node 3,
I 6 = I 7 + I8 − I 2
(E2 − E3 ) E3 E3
= + − I2
R4
R5 R6
I2 = −
⎛ 1 1 1⎞
E2
+ E3 ⎜ + + ⎟
R4
⎝ R4 R5 R6 ⎠
(iii)
Given the other values, solution of equations (i), (ii), and (iii) gives the values of E1, E2 and E3.
Concept of Supernode This concept is used when a circuit contains voltage sources. A supernode
is formed by enclosing a dependent or independent voltage source connected between two non-reference
nodes and any elements connected in parallel with it. This concept is necessary for nodal analysis with
voltage source, because the current through a voltage source is unknown. We consider the following
example.
48
Network Analysis and Synthesis
Example 2.4 Determine the node voltages V1 , V2 , and V3 for the circuit of Fig. 2.28 (a).
V1
5V
V2
5
i1
10V
10V
V3 Supernode
i3
i2 10
Fig. 2.28 (a)
20
Circuit of Example 2.4
V2
V3
Fig. 2.28 (b) KVL with supernode
Solution For this problem we have two cases:
Case-1 When a voltage source is connected between the reference node and a non-reference node
In this case, the voltage of the non-reference node is taken equal to the voltage of the voltage source. For the
circuit shown in Fig. 2.28 (a),
V1 5 V
(1)
Case-2 When a voltage source is connected between two non-reference nodes
In this case, a supernode is considered enclosing the non-reference nodes. Both KCL and KVL is written for
the supernode.
For this example, nodes 2 and 3 are forming the supernode.
By KCL at the supernode, i1 i2 i3
V1 − V2 V2 − 0 V3 − 0
or,
(2)
=
+
5
10
20
To apply KVL to the supernode, the circuit is drawn as shown in Fig. 2.28 (b). By KVL,
10 V3 V2 0
(3)
5 V, V2
4.2857 V, and
Solving equations (1), (2) and (3), the node voltages are obtained as V1
V3
5.7143 V.
Properties of a Supernode
(i) It provides the constraint equations.
(ii) Both KCL and KVL are written for a supernode.
(iii) A supernode does not have any voltage of its own.
2.8.2 Loop or Mesh Analysis
It is based on Kirchhoff’s voltage law (KVL). In this method, the unknown variables are the loop currents. It
is generally used when the circuit contains several voltage sources.
Steps
(i) If there are ‘N’ number of loops/meshes in a network, all loops are labeled.
(ii) KVL is written at each loop/mesh in terms of loop/mesh currents. Loop currents are those currents
flowing in a loop; they are used to define branch currents.
(a) For N independent loops, a total of N equations are written using KVL around each loop. These
equations are known as loop/mesh equations.
49
Introduction to Circuit-Theory Concepts
(b) The concept of supermesh is used in case a circuit contains current source that provides the constraint equations.
(iii) Solution of the N simultaneous equations gives the required loop/mesh currents.
Example 2.5 Write the mesh equations for the circuit shown in Fig. 2.29.
R1
Solution Two meshes are labeled as meshes 1 and 2.
Applying KVL for the mesh 1,
Vs
R2(I1
R1I1
I2)
(1)
Vs
R3
R2
I1
R4
I2
Applying KVL for the mesh 2,
0
I1R2
I2(R2
R3
R4)
(2)
Fig. 2.29 Circuit of Example 2.5
Solving the equations, we get I1 and I2.
Concept of Supermesh This concept is used when a circuit contains current sources. A supermesh
is formed by excluding the branch containing a dependent or independent current source connected in
common to two meshes and any elements connected in series with it. This concept is necessary for loop
analysis with a current source, because the voltage drop across a current source is unknown. We consider
two examples.
Example 2.6 Find the mesh currents in the circuit of Fig. 2.30.
Solution Here, a current source is in one mesh.
In this case, the mesh current is taken equal to the current of the
current source. For example, for the circuit shown in Fig. 2.30,
i2
10 A
By KVL for the mesh 1, we get
5
i1
10
(i1
10)
5 ⇒ i1
6.33A
5
5V
20
i1
10
10 A
i2
Fig. 2.30 Circuit of Example 2.6
Example 2.7 Find the mesh currents in the circuit of Fig. 2.31.
Solution Here, a current source is connected between two meshes.
In this case, a supermesh is considered excluding the branch
with the current source and any elements connected in series with
it. Both KCL and KVL is written for the supermesh. For example,
consider the circuit shown in Fig. 2.31. A supermesh is formed by
excluding the branch with the 3-A current source.
By KVL for the supermesh,
2(i1 i2) 4(i3 i2) 8i3 5
(i)
By KCL at any one node of the omitted branch (say, X ),
(ii)
i1 3 i3
Also by KVL for the second mesh, 2i2 4(i2
i3) 2(i2 i1) 0
i2
2
4
Supermesh
3A
i3
8
X
i3
2
6V
i1
1
i1
Fig. 2.31 Current source connected
between two meshes
(iii)
50
Network Analysis and Synthesis
Solving equations (i), (ii) and (iii), the mesh currents are obtained as i1
and i3 0.4737 A.
3.437 A, i2
1.1052 A
Properties of a Supermesh
(i) It provides the constraint equations.
(ii) Both KCL and KVL are written for supermesh.
(iii) A supermesh does not have any current of its own.
2.8.3 Comparison of Loop and Node Analysis
In any network having ‘N’ nodes and ‘B’ branches, there are 2B unknowns, i.e. ‘B’ branch currents and ‘B’
branch voltages. These unknowns can be determined either by loop analysis or nodal analysis.
The choice of the method depends on two factors:
Nature of the Network The mesh-method is generally used for circuits having many series-connected
elements, voltage sources, or supermeshes. On the other hand, nodal analysis is more suitable for networks
for circuits having many parallel-connected elements, current sources, or supernodes.
The main factor for selecting any one method is the minimum number of equations. If a circuit has fewer nodes than
meshes then nodal analysis is used, while if a circuit has fewer meshes than nodes then the loop method is used.
Requirement of the Problem If node voltages are required, nodal analysis is used; if branch/mesh currents are required, loop analysis is used.
However, there are some particular circuits, where only one method can be applied. For example, in
analyzing transistor circuits, the mesh method is the only possible method; while for op-amp circuits and for
non-planar networks, the node method is the only possible method.
2.9
DUALITY
Duality is a transformation in which currents and voltages are interchanged. Two phenomena are said to be dual if they are described by
equations of the same mathematical form.
There are a number of similarities and analogies between the two
circuit analysis techniques based on loop-current method and node-voltage method. The principal quantities and concepts involved in these two
methods based on KVL and KCL are dual of each other with voltage variables substituted by current variables, independent loop by independent
node-pair, etc.
This similarity is termed as ‘principle of duality’.
Some dual relations are
v = Ri
di
v=L
dt
1
v = ∫ idt
C
R
L
C
v
i(t)
Fig. 2.32 (a)
Series RLC circuit
i = Gv
i =C
i=
dv
dt
1
vdt
L∫
L
i
R
Fig. 2.32 (b)
Parallel RLC circuit
C
v(t )
51
Introduction to Circuit-Theory Concepts
Thus, the circuit elements (R, L, C) have some dual relationship. Duality also appears as a relation between
two networks. For example, an RLC series circuit with voltage excitation is the dual of an RLC parallel circuit
with current excitation.
di 1
For a series circuit,
v = Ri + L + ∫ idt
dt C
dv 1
For a parallel circuit,
i = Gv + C + ∫ vdt
dt L
Dual Quantities and Concepts
Sl. No.
Quantity/Concept
Dual
1
Current
Voltage
2
3
4
5
6
Resistance
Inductance
Impedance
Reactance
Branch current
Conductance
Capacitance
Admittance
Susceptance
Branch voltage
7
8
9
10
11
12
13
Mesh or loop
Mesh current or loop current
Link
Link current
Tree branch current
Tie-set
Short-circuit
Node or node-pair
Node voltage or node-pair voltage
Tree branch
Tree branch voltage
Link voltage
Cut-set
Open-circuit
14
Parallel paths
Series paths
Construction of Dual of a Network
1. A dot is placed inside each independent loop of the given network; these dots correspond to the nonreference nodes of the dual network.
2. A dot is placed outside the network; this dot corresponds to the datum node.
3. All internal dots are connected by dashed lines crossing the common branches and placing the elements which are duals of the elements of the original network.
4. All internal dots are connected to the external dot by dashed lines crossing all external branches and
placing dual elements of the external branch.
Conventions for Reference Polarities of Voltage Source and Reference Directions of the Current Source
(i) A clockwise current in a loop corresponds to a positive polarity (with respect to reference node) at the
dual independent node.
(ii) A voltage rise in the direction of a clockwise loop current corresponds to a current flowing towards the
dual independent nodes.
Finally, the dual construction can be checked by writing mesh equations and node equations of two networks.
52
Network Analysis and Synthesis
Example 2.8 Draw the dual of the network shown in Fig. 2.33.
5A
3
5
6
4
100 V
Fig. 2.33
Circuit of Example 2.8
Solution Following the steps, a dual network is drawn as shown in Fig. 2.34.
5V
5A
1/3
1/5
3
100V
4
5
1/4
6
1/5
100A
1/4
100 A
1/6
1/6
1/3
5V
Fig. 2.34 Figure explaining drawing dual of
network of Fig. 2.33
Fig. 2.35 Dual of network of Fig. 2.32
Therefore, the dual network becomes as shown in Fig. 2.35.
By KVL to the original network,
I1(3 4) I2(4) 100
I1(4)
I2(4
5
6)
5I3
0
I3
5
5A
The dual equations will be,
V1(3
V1(4)
V2(4
5
4)
V2(4)
100
6)
5V3
0
V3
I
4
1
3
I
6
2
5
These equations satisfy the dual network.
2.10
100 V
I
5
3
Fig. 2.36 Labeled circuit of Fig. 2.33
STAR-DELTA CONVERSION TECHNIQUE
The Y- transform, also written Y-delta, Wye-delta, Kennelly’s delta-star transformation, star-mesh
transformation, T- or T-pi transform, is a mathematical technique to simplify the analysis of an electrical
network. The name derives from the shapes of the circuit diagrams, which look respectively like the letter Y
and the Greek capital letter .
The transformation is used to establish equivalence for networks with three terminals. For equivalence,
the impedance between any pair of terminals must be the same for both networks.
53
Introduction to Circuit-Theory Concepts
For the star connection, the impedance between terminals 1
and 2 is Z1 Z2.
For the delta connection, the the impedance between terminals 1 and 2 is
Z12
1
2
Z2
Z1
23
12
23
31
12
23
31
As the impedance between terminals 1 and 2 should be
same, therefore,
(
Z12 Z 23 + Z 31
3
(b)
Fig. 2.37 (a)
connection
Star connection (b) Delta
)
2
Z23
3
(a)
31
Z1 + Z 2 =
Z31
Z3
Z (Z + Z )
(Z + Z ) = Z + Z + Z .
Z12
1
(i)
Z12 + Z 23 + Z 31
Similarly, for terminals 2 and 3 we get,
(
Z 23 Z 31 + Z12
Z2 + Z3 =
(ii)
Z 23 + Z 31 + Z12
Z 3 + Z1 =
2.10.1
)
(
Z 31 Z12 + Z 23
)
(iii)
Z 31 + Z12 + Z 23
Delta to Star Conversion
In this case, Z1, Z2, and Z3 are to be written in terms of Z12, Z23, and Z31.
Z12 Z 31
By, (i) (ii) (iii);
Z1 =
Z12 + Z 23 + Z 31
(iv)
Similarly, we get,
Z2 =
Z 23 Z12
Z12 + Z 23 + Z 31
(v)
and
Z3 =
Z 31 Z 23
Z12 + Z 23 + Z 31
(vi)
2.10.2
Star to Delta Conversion
In this case, Z12, Z23, and Z31 are to be written in terms of Z1, Z2, and Z3.
Let Z Z1Z2 Z2Z3 Z3Z1. Then from (iv) to (vi), we get
Z=
Z12 Z 232 Z 31
+
2
Z12 Z 23 Z 312
Z12 2 Z 23 Z 31
(Z + Z + Z ) (Z + Z + Z ) (Z + Z + Z )
12
23
31
12
23
3
From (vii) and (iv), we get Z = Z12 Z 3 ⇒ Z12 =
Therefore,
+
2
Z12 =
31
12
23
2
=
31
Z
Z3
Z1 Z 2 + Z 2 Z 3 + Z 3 Z1
ZZ
= Z1 + Z 2 + 1 2
Z3
Z3
Z12 Z 23 Z 31
Z12 + Z 23 + Z 31
(viii)
54
Network Analysis and Synthesis
Similarly,
Z 23 =
Z1 Z 2 + Z 2 Z 3 + Z 3 Z1
Z Z
= Z2 + Z3 + 2 3
Z1
Z1
and
Z 31 =
Z1 Z 2 + Z 2 Z 3 + Z 3 Z1
ZZ
= Z 3 + Z1 + 3 1
Z2
Z2
Example 2.9 Find the equivalent resistance between the terminals A and B of the circuit shown below.
A
6Ω
4Ω
5Ω
8Ω
3Ω
5Ω
4Ω
B
A
6
Fig. 2.38 Circuit of Example 2.9
4
Solution Converting star into delta,
⎛
rr ⎞
15
r12 = ⎜ r1 + r2 + 1 2 ⎟ = 8 + = 9.875 Ω
r
8
⎝
3 ⎠
9.875
5
B
Fig. 2.39 (a)
⎛
rr ⎞
40
r23 = ⎜ r2 + r3 + 2 3 ⎟ = 13 + = 26 ⋅ 33 Ω
3
r1 ⎠
⎝
⎛
rr ⎞
24
r31 = ⎜ r3 + r1 + 3 1 ⎟ = 11 + = 15 ⋅ 8 Ω
r
5
⎝
2 ⎠
Combining the parallel connections of 5 and 15.8 and 4 and 26.33 ,
we have the reduced circuit.
Again, converting the delta made of 6 , 4 and 9.875 into equivalent
star,
r r
6×4
= 1.2075 Ω
r1 = 12 31 =
r1 + r2 + r3 19 ⋅ 875
r2 =
4 × 9.875
= 1.987 Ω
19.875
r3 =
6 × 9.875
= 2.981 Ω
19.875
So, the given circuit becomes as shown in Fig. 2.39 (d).
∴ RAB = 1.2075 +
6.779 × 5.459
= 4.23 Ω
6.779 + 5.459
4
26.33
15.8
A
6
4
9.875
3.798
3.472
B
Fig. 2.39 (b)
A
1.2075
2.981
1.987
3.798
3.472
B
Fig. 2.39 (c)
A
1.2075
6.779
5.459
B
Fig. 2.39 (d)
55
Introduction to Circuit-Theory Concepts
Solved Problems
Problem 2.1 Find the values of V, Vab and the power delivered by the 5-V source.
20i
20
60
V
5V
70i 0
1
v = −7 − 90i = −7 − 90 × = −10 V
30
vab 20i v 30i 50i 10
1
= 50 × − 10 = −8.33 V
30
Power drawn by the 5-V source
2
5
v
b
40
a
2V
30
Fig. 2.40
60
20
V
i
5V
− (power taken the source)
= −5 ×
30
1
= −0.166 W
30
A
4
5
6
5
25
C
Fig. 2.42
A
⎛
rr ⎞
5×6
= 18.5 Ω
r23 = ⎜ r2 + r3 + 2 3 ⎟ = 5 + 6 +
r1 ⎠
4
⎝
10
5
14.8
12.3
18.5
⎛
rr ⎞
6×4
r31 = ⎜ r3 + r1 + 3 1 ⎟ = 6 + 4 +
= 14.8 Ω
r2 ⎠
5
⎝
The circuit becomes as shown below in Fig. 2.44
B
(i) Equivalent resistance between A and B,
Fig. 2.43
C
25
A
RAB =
5.52 × (10.06 + 3.73)
= 3.94 Ω
5.52 + 10.06 + 3.73
10
N
B
⎛
rr ⎞
5×4
= 12 ⋅ 33 Ω
r12 = ⎜ r1 + r2 + 1 2 ⎟ = 4 + 5 +
6
r3 ⎠
⎝
3.73 × (10.06 + 5.52 )
= 3⋅ 035 Ω
3.73 + 10.06 + 5.52
10.06 × ( 3.73 + 5.52 )
4.82 Ω
(ii) RBC =
10.06 + 3.73 + 5.52
b 40
Fig. 2.41
Problem 2.2 Find the equivalent resistance between
(i) A and B,
(ii) B and C,
(iii) C and A, and
(iv) A and N of the circuit shown.
Solution Converting the star into delta,
(iii) RCA =
2V
a
2
1
Solution Current, i = = A
60 30
By KVL,
5.52
3.73
B
10.063
Fig. 2.44
C
56
Network Analysis and Synthesis
(iv) Converting the delta into equivalent star,
5×6
r1 =
= 0.83 Ω
5 + 6 + 25
25 × 6
= 4.167 Ω
r2 =
5 + 6 + 25
25 × 5
= 3.472 Ω
r3 =
5 + 6 + 25
The circuit becomes as shown in Fig. 2.46.
A
4
5
0.83
B
C
3.472
4.167
Fig. 2.45
A
A
4
9.167
10
N
4
13.472
N
5.455
0.83
N
0.83
B
B
Fig. 2.46
∴ RAN =
4 × 6.288
= 2.4448 Ω
4 + 6.288
Problem 2.3 Find the current through the galvanometer using
delta–star conversion.
Solution Converting the delta consisting of 20
we get,
RAC
, 30
and 50
B
20
,
10
A
r1 =
20 × 30
=6Ω
20 + 30 + 50
r2 =
20 × 50
= 10 Ω
20 + 30 + 50
r3 =
30 × 50
= 15Ω
20 + 30 + 50
5
30
D
8V
Fig. 2.47
B
16
20
8
Main current i = = 0 ⋅ 5 A
16
Now, to calculate potential difference between the points B and D;
Vxc 10 0.5 5 V
VBD
(10
0.25
5
0.25)
current through the galvanometer, (50
iG =
1.25
= 0.025A
50
C
50
A
10
r2
r1
r3
5
30
D
1.25 V
8V
)
Fig. 2.48
C
50
57
Introduction to Circuit-Theory Concepts
10
6
A
B
10
C
X
15
6
A
5
10
C
X
D
8V
8V
Fig. 2.49
Problem 2.4 The current and voltage profile of an element vs. time has been shown in Fig. 2.50. Determine the element and find its value.
Voltage (V)
Current (A)
1A
5
0
5
Time (ms)
5
0
Time (ms)
Fig. 2.50
Solution Here, the voltage is not proportional to the current; therefore, the element is not a resistance.
Also, at t 5 ms, i 0, but the voltage suddenly drops to zero value, i.e. the element acts as a short circuit.
As the voltage across a capacitor cannot change instantaneously, the element is not a capacitor.
Now, the current is zero at t 0 and the voltage is zero at t 5 ms. Therefore, we conclude that the element is an inductor.
di
1
From the figure,
=
= 200 A/s and v 5 V.
dt 5 × 10−3
di
v
5
⇒ L=
=
= 25 mH
v=L
di
200
dt
dt
Problem 2.5 The voltage across a capacitor of value C
waveform.
Solution Here, the voltage can be expressed as
v(t )
0
t 0
10t
0 t 1
20 10t
1 t 2
0
t 2
()
Since, i t = C
i(t )
0
10 2
( ) , we have the current given as
dv t
dt
t
0
0
t
1
1 mF is shown in Fig. 2.51. Find the current
v(t)(V )
10
0
Fig. 2.51
1
2
t (s)
58
Network Analysis and Synthesis
10 2
1 t 2
0
t 2
This means that the current waveform consists of two sharp positive
and negative pulses of magnitude 10 mA as shown in Fig. 2.52.
i (t) (mA)
10
I
I/3 R
I/6
R
I/6
I/6
R
R
R
R
2R
2R
2R
A
B
2R
R
2R 2R
R
R
Fig. 2.54
R/2
2R
4R
R
A
B
4R
I/3 R
R
R
R
I/3
R
I/3
Fig. 2.53
Y
R
I/3
I/6 R
A
Solution The hexagon can be redrawn as shown in
Fig. 2.55.
X
I/6
R
Problem 2.7 A regular hexagon is formed from
6 wires of R ohms each. The corners are joined to the
centre by six more wires of 2R ohms each. Calculate the
resistance of the hexagon between any two nodes diametrically opposite.
4R
I
B
I/6
Equivalent resistance,
R
I/3
R
R
I
I
I
5
+ R⋅ + R⋅ = R× I
3
6
3
6
V
5
RAB = AB = R
I
6
2
Fig. 2.52 Current waveform of
the capacitor
Solution The configuration is shown in Fig. 2.53.
The current distribution is shown.
So, the total voltage drop between two opposite corners A and
B for a total current of I is
R/2
1
10
Problem 2.6 Twelve similar conductors each of ‘R’ resistance form a
cubical frame. Find the resistance across two opposite corners of the cube.
VAB = R ⋅
t (s)
0
4R
2R 2R
R/2
X'
R
R/2
Y'
Fig. 2.55
The hexagon is symmetrical about XX´.
Equivalent resistance of the second quadrant,
28
R1 = ( 2 R R / 2 + R) 4 R = R
27
So, the figure is modified as shown in Fig. 2.56.
28
∴ RAB = ( R1 R1 ) + ( R1 R1 ) = R1 = R
27
R1
R1
A
B
R1
Fig. 2.56
R1
59
Introduction to Circuit-Theory Concepts
Problem 2.8 Find the input resistance of the infinite section resistive network shown below.
R
R
Rin
R
A
R
R
R
R
B’
A
R
R
Infinity
B
Fig. 2.57
Solution Let the equivalent resistance be Rin.
The network can be terminated at A’B’ instead of AB.
RA′B ' = [ R + ( Rin ) ( R)]
By assumption,
Rin = R +
Rin R
R + Rin
=
2 Rin R + R 2
R + Rin
⇒ RRin + Rin 2 = 2 RRin + R 2 ⇒ Rin 2 − RRin − R 2 = 0
⇒ Rin =
R ± R2 + 4 R2 R
= [1 ± 5 ]
2
2
⎛ 5 + 1⎞
Taking positive sign, Rin = ⎜
⎟R.
⎝ 2 ⎠
Problem 2.9 In the network shown, calculate the power input to each of the following elements when it is connected across A and B:
2100V
(a) a resistance RAB of 59
(b) a voltage source of 160 V
7
18
Solution (a) Converting the two deltas into star, the circuit is shown
in Fig. 2.59.
r1 =
18 × 6
=3
18 + 12 + 6
r2 =
6 × 12
=2Ω
36
r3 =
18 × 12
=6
36
14 × 7
28 × 14
r =
= 2 , r21 =
= 8 , r31 = 4
49
49
1
1
⎛ 69 × 20 ⎞
∴ Req = ⎜ 3 +
+ 2 = 20.5
⎝ 69 + 20 ⎟⎠
Main current, i =
2100
= 102.41 A
20.5
A
6
B
12
28
10
Fig. 2.58
14
60
Network Analysis and Synthesis
2100 V
2100 V
2
3
3
2
69
6
59
4
2
10
8
20
Fig. 2.59
Current in the 59
Power input, Pi
resistor, i59 = 102.41 ×
2
20
= 23.01A
89
2100 V
2
(i59)
59 (23 01)
59
31248 189 W 31 kW
(b) By KVL, for circuit of Fig. 2.60,
15i1 10i2 2260 0 and 30i2 10i1 160 0
Solving, i1
206.285A
i2
63.43A
power input, Pi v i
160 (i1 i2)
160 ( 206 285 63 43)
17.37 kW
Problem 2.10 The two-dimensional network of the
Fig. 2.61 consists of an infinite number of square meshes,
each side of which has a resistance of R. Find the effective resistance between two adjacent nodes such as
X and Y.
i1
3
6
2
2
160 V
4
i2
10
8
Fig. 2.60
P
X
T
Y
S
Q
R
Solution Let the current flowing into the circuit at the node
X be I. Since the infinite network is symmetrical about X, the
T
current ‘I’ in going from X to infinity, is divided equally along
the branches XQ, XT, XP and XY.
Fig. 2.61
The current ‘I’ then returns from infinity and is taken from
the network at the node Y.
Again, by symmetry, the currents flowing along RY, XY, SY and TY are each I/4.
Hence, the total current flowing along XY is I兾4 I兾4 I兾2. So, the voltage between X and Y.
VXY I兾2 R
So, the effective resistance between X and Y, RXY VXY 兾I R兾2
Problem 2.11 Use loop current analysis to find currents in all branches of the network of Fig. 2.62. Also,
find the power delivered by the 5-A current source.
61
Introduction to Circuit-Theory Concepts
5
Solution By KVL, 5i1 + 10i2 + 5(i2 − i4 ) + 15(i1 − i3 ) = 50
or,
20i1 + 15i2 − 15i3 − 5i4 = 50
and,
5(i4 − i2 ) + 30 + 10i4 + 20(i4 − i3 ) = 0
(i)
−5i2 − 20i3 + 35i4 = −30
or,
By constraint equations,
(ii)
50V
10A
I2
I1 5 A
15
5
I3 20
I4
10
30 V
10
(i2
and
From (i) and (ii),
i1) 5
i3 10
(iii)
(iv)
Fig. 2.62
20(i2 − 5) + 15i2 − 15 × 10 − 5i4 = 50
or,
35i2 − 5i4 = 300 ⇒ 7i2 − i4 = 60
and,
−5i2 + 35i4 = 170 ⇒ − i2 + 7i4 = 34
Solving i4
i2
6.02083A
4.4583A
i1 4.4583A and i3 10A
Power delivered by the 5-A current source
v
110.83
i
5
554.16 W
[To calculate the voltage across the 5-A current source, v, writing KVL for mesh (1),
5i1 + v + 15(i1 − i3 ) = 50 ⇒ v = 50 − 20i1 + 15i3 = 200 − 20 × 4.4
4583 = 110.83 V ]
Problem 2.12 For the circuit of Fig. 2.63 (a), find the voltage Vx using nodal analysis.
Iy
0.6 A
40
100
50
0.2 Vx
Vx
25 Iy
Fig. 2.63 (a)
Solution By KCL at the node (1),
− 0.6 + I y +
Vx
v −v
− 25 I y + 1 2 = 0
50
40
(i)
1
2
Iy
0.6A
40
100
50
25Iy
Fig. 2.63 (b)
Vx
0.2Vx
62
Network Analysis and Synthesis
By KCL at the node (2),
v2
0.2Vx
(ii)
and other constraint equation,
Vx
and v1 = Vx
100
V
V
V
v −v
− 0 ⋅ 6 + x + x − 25 x + 1 2 = 0
100 50
100
40
Iy =
From (1),
(iii)
⇒ −120 + 2Vx + 4Vx − 50Vx + 5Vx − 5 × 0.2Vx = 0
⇒ Vx =
120
= −3 V
−40
Problem 2.13 Use nodal analysis to find the voltages VA , VB and Vx in the circuit of Fig. 2.64, in which I1
Solution By KCL at the node (A),
VA VA − VB
+
+ 0.03Vx = 0
100
20
(1)
By KCL at the node (B),
VB − VA VB VB − VC
+ +
=0
20
40
40
(2)
Constraint equations,
Iy =
and
VC = 80 I y
− 0.4 +
VB
40
I1
(3)
Vx
VB
20
Iy 40
80Iy
40
VA
100
C
Fig. 2.64
(4)
(VA − VB ) = Vx
and
0.4 A.
0.03 Vx
(5)
V
V V
From (1), − 0.4 + A + A − B + 0 ⋅ 03VA − 0 ⋅ 03VB = 0 ⇒ ( 9VA − 8VB ) = 40
100 20 20
(6)
From (2), VA VB [by (3) and (4)]
Thus, solving (6) VA VB 40V
Vx (VA VB) 0
I1
300
Problem 2.14 For the circuit, use loop analysis to find I1 and the power
absorbed by the 500- resistor.
50V
Solution Converting the dependent current source into dependent voltage
source,
Fig. 2.65
By KVL, 800 I1 − 200 I1 = 50 ⇒ I1 =
Power absorbed by the 500-
I1
50
= 0.083A
600
resistor
200I1
300
50V
2
⎛ 50 ⎞
500
× 500 =
= 3 ⋅ 47W
= I12 R = ⎜
⎟
144
⎝ 600 ⎠
0.4I1
500
Fig. 2.66
500
63
Introduction to Circuit-Theory Concepts
Problem 2.15 Determine the currents in all the branches
of the network.
I1
I2
5
5 I1 + 10 I 2 + 10( I1 − I 2 ) + 5 I1 = 5
(I2 I1) 10
(3.75 10)
10
10V
5V
5I1
⇒ 20 I1 = 5 ⇒ I1 = 0 ⋅ 25A
By KVL for the mesh (2),
5I2 10 5I1
15I2 15I1 10
0.416 A
I2
5
10I2
Solution By KVL, for the mesh (1)
Fig. 2.67
0
6.25
Problem 2.16 For the circuit shown in Fig. 2.68,
(a) determine the KVL equations
(b) find the two loop currents I1 and I2
(c) find the power supplied by the source and the power dissipated in each resistor
j2
2
j5
3
j2
10 00 V
1
Fig. 2.68
Solution
(a) KVL equations:
2 I1 − j 2 I 2 = 10 ⎫⎪
⎬
and − j 2 I1 + 4 − j 3 I 2 = 0 ⎭⎪
(
(b) Solving for the currents, I1 =
10
0
(
2
− j2
2
and
I2 =
2
− j2
− j2
4 − j3
) = 40 − j 30 = 3.773∠ − 10.3 A
(
− j2
(
)
12 − j 6
− j2
4 − j3
)
10
0
− j2
4 − j3
)
=
j 20
= 1.5∠116.6 A
12 − j 6
(
)
(c) Power supplied by the source, Ps = VI1 cos 1 = 10 × 6.933cos 19.44 = 65.28 W
64
Network Analysis and Synthesis
Power dissipated in resistors,
P2 = I1 × 2 = 96.15 W ⎫
⎪
2
⎪
P3 = I 2 × 3 = 23.08 W ⎬
⎪
2
P1 = I 2 × 1 = 7.69 W ⎪
⎭
2
8
Problem 2.17 For the circuit shown in Fig. 2.69, determine
the voltage ‘v’ using nodal analysis.
Solution Let the node voltages be V1 and V2. Here, V2
By KCL,
V1 − 100 V1 V1 − V2
+ +
= 0 ⇒ 17V1 − 12 v = 300
8
12
2
V2 − V1 V2
+ − 10 = 0 ⇒ − 3V1 + 4 v = 60
2
6
and
Solving equations (i) and (ii), we get,
v=
17 300
−3 60
17 −12
−3
=
v
12
100V
v
2
6
10A
6
10A
Fig. 2.69
8
(i)
2
1
2
v
12
100V
(ii)
Fig. 2.70
1920
= 60 V
32
4
Problem 2.18 Determine the voltage v in the network
in Fig. 2.71 using nodal analysis.
Solution Converting the current source into equivalent
voltage source, we get the following circuit.
By KVL,
14 I1 − 12 I 2 = 100
2
2
12
100 V
v
6
10 A
Fig. 2.71
−12 I1 + 20 I 2 = −60
I2 =
v
(6I2
2
2
Solving for I2,
14 100
−12 − 60
14 −12
−12 20
60)
=
−840 + 1200 360
= 2.64 A
=
280 − 144 136
75.88 V
12
6
100V
v
I1
I2
60 v
Fig. 2.72
65
Introduction to Circuit-Theory Concepts
Problem 2.19 Determine the voltage V using source transformation and
simplification in Fig. 2.73.
3
4
8V
6A
Solution By KVL,
3
)
(
32
A
7
28
and 9i2 + 36 − 8 = 0 ⇒ i2 = − A
9
4 i1 + 6 + 8 + 31 = 0 ⇒ i1 = −
6
Fig. 2.73
i1
3
Thus, the voltage is,
) (
)
3
∴i = −
20
10
=− A
6
3
i2
6
2
5A V0
2
5A
Fig. 2.75
5
2
j5
30 V
10 V
2
2
2
Problem 2.21 In the network shown in Fig. 2.77, determine the voltage Vb
which results in a zero current through the (2 j3) impedance branch.
j3
i
10 V
Fig. 2.76
4
Vb
6
Fig. 2.77
Solution When the 30-V source is acting alone, let the current through the branch (2
j5
30V
2
j3
I1
6
(a)
Fig. 2.78
Impedance, Z = 5 +
(
) = ⎛ 7 + j 62 ⎞
j 5 × 4.4 + j 3
4.4 + j8
i3
2
⎛ 10
⎞ 10
∴V0 = 2i + 10 = 2 × ⎜ − + 10⎟ = = 3.33 V
⎠ 3
⎝ 3
5
V
Fig. 2.74
Problem 2.20 Convert the current sources into the equivalent voltage source given in Fig. 2.75 and hence find the voltage V0.
Solution Converting the current sources into voltage sources, we get
the following circuit.
4
8V
6A
⎛ 32 ⎞
⎛ 28 ⎞
V = 4 i1 + 6 + 6 i2 + 6 = 4 ⎜ − + 6⎟ + 6 ⎜ − + 6⎟ = 23.05 V
⎠
⎝ 7
⎠
⎝ 9
(
V
⎜⎝ 4.4 + j8 ⎟⎠
4
5
2
j3
30V
j 5 I1
2.4
(b)
j3)
be I1.
V0
66
Network Analysis and Synthesis
∴I =
(
30 30 4.4 + j8
=
Z
7 + j 62
∴ I1 = I ×
)
)
(
30 4.4 + j8 ⎛ j 5 ⎞
j150
j5
=
A
×⎜
=
⎟
4.4 + j8
7 + j 62
⎝ 4.4 + j8 ⎠ 7 + j 62
When the Vb source is acting alone, let the current through the
branch (2 j3) be I2.
Impedance, Z = 4 +
(
6 × 4.5 + j 5.5
10.5 + j 5.5
∴I′=
) = ⎛ 69 + j 55 ⎞
(
Vb Vb 10.5 + j 5.5
=
69 + j 55
Z
∴ I2 = I ′ ×
Current through the branch (2
I1 = I 2 ⇒
j5
5
⎜⎝ 10.5 + j 5.5 ⎟⎠
)
j3
2
4
Vb
6
Fig. 2.79
)
(
V 10.5 + j 5.5 ⎛
6Vb
⎞
6
6
×⎜
=
A
= b
10.5 + j 5.5
69 + j 55
⎝ 10.5 + j 5.5 ⎟⎠ 69 + j 55
will be zero, if
j3)
6Vb
j150
=
⇒ Vb = 25 + j 25 V = 35.35∠45 V
7 + j 62 69 + j 55
)
(
Problem 2.22 Determine the current through the impedance (2 + j3)
where, Vb 20 0 (V ).
5
2
j3
j5
30 V
in the circuit shown in Fig. 2.80,
4
Vb
6
Fig. 2.80
Solution When the 30-V source is acting alone, let the current through the branch (2
5
2
j5
30V
j3
I1
6
(a)
Fig. 2.81
Impedance, Z = 5 +
∴I =
(
) = ⎛ 7 + j 62 ⎞
j 5 × 4.4 + j 3
4.4 + j8
(
30 30 4.4 + j8
=
Z
7 + j 62
)
⎜⎝ 4.4 + j8 ⎟⎠
4
j3)
5
2
j3
30V
j 5 I1
2.4
(b)
be I1.
67
Introduction to Circuit-Theory Concepts
∴ I1 = I ×
)
(
30 4.4 + j8 ⎛ j 5 ⎞
j5
j150
=
=
= 2.4 ∠6.44 = 2.38 + j 0.27 A
×⎜
4.4 + j8
7 + j 62
⎝ 4.4 + j8 ⎟⎠ 7 + j 62
When the 20-V (Vb) source is acting alone, let the current through the branch (2
Impedance,
Z =4+
(
6 × 4.5 + j 5.5
10.5 + j 5.5
∴I′=
∴ I2 = I ′ ×
) = ⎛ 69 + j 55 ⎞
⎜⎝ 10.5 + j 5.5 ⎟⎠
(
Vb 20 10.5 + j 5.5
=
Z
69 + j 55
)
(
j3)
j3
2
j5
5
be I2.
I1
4
Vb
6
)
Fig. 2.82
)
(
20 10.5 + j 5.5 ⎛
⎞
120
6
6
=
×⎜
=
= 1.36 ∠ − 38.56 = 1.06 − j 0.85 A
⎟
10.5 + j 5.5
69 + j 55
⎝ 10.5 + j 5.5 ⎠ 69 + j 55
Total current through the branch (2
(
) (
j3)
(
is
) (
) (
)
)
( )
I = I1 − I 2 = 2.38 + j 0.27 − 1.06 − j 0.85 = 1.32 + j1.12 = 1.73∠40.31 A
Problem 2.23 Write the loop equations of the circuit and find the voltage Vx.
2Ω
5 300 (V)
j2 Ω
j5 Ω
10 00 (V)
5Ω
2Ω
10 Ω
j2 Ω
10 Ω
Vx
Fig. 2.83
Solution By KVL for the three meshes, we get,
2Ω
10 00 (V )
5 300 (V )
j2 Ω
j5 ΩI
I1
5Ω
2Ω
I3
10 Ω
2
j2 Ω
10 Ω
Vx
Fig. 2.84
(7 + j 3) I − j 5 I − 5 I = 10
1
2
3
(i)
68
Network Analysis and Synthesis
j5I1
5I1
(12
j3) 5I2
(2
j2)I2
(2
(17
(4.33
j2) I3
j2) I3
(ii)
j2.5)
0
(iii)
Solving for I3 from equations (i), (ii) and (iii), we get
(7 + j 3)
− j5
10
(12 + j 3) − ( 4.33 + j 2.5)
−5
−(2 − j 2)
0
= 0.435∠ − 194.15 ( A )
−5
(7 + j 3) − j 5
− j5
(12 + j 3) − ( 2 − j 2 )
−5
− ( 2 − j 2 ) (17 − j2 )
− j5
I3 =
Therefore, the required voltage is
( )
Vx = 10 × I 3 = 10 × 0.435∠ − 194.15 = 4.35∠ − 194.15 V
Problem 2.24 For the network shown, find the value of the voltage V which results in the output voltage
V0 5 V.
5
j2
j5
2
3
V
V0
5
j2
Fig. 2.85
Solution For V0
5 V, the current in the (2
I5 =
j 2)
⎛
5 ⎞ ⎛ 7 − j2 ⎞
∴ I3 = I4 + I5 = ⎜1+
A
=
2
−
j 2 ⎟⎠ ⎜⎝ 2 − j 2 ⎟⎠
⎝
(
I1 5
5
2 − j2
5
I4 = = 1 A
5
Also,
branch is
)
j2
x I3 j5
I4
I2
V
3
Fig. 2.86
⎛ 7 − j2 ⎞
⎛ 20 + j 25 ⎞
Vx = 5 + I 3 × j 5 = 5 + ⎜
× j5 = ⎜
⎟
−
j
2
2
⎝
⎠
⎝ 2 − j 2 ⎟⎠
⎛ 20
25 ⎞
Vx ⎜ 3 + j 3 ⎟
=
3 ⎜ 2 − j2 ⎟
⎝
⎠
2
V0
5
j2
Voltage at the node x is
∴ I2 =
I5
69
Introduction to Circuit-Theory Concepts
⎛ 20 + j 25 ⎞
3 ⎟ + ⎛ 7 − j 2 ⎞ = ⎛ 13.67 + j 6.333 ⎞
∴ I1 = I 2 + I 3 = ⎜ 3
⎜ 2 − j 2 ⎟ ⎜⎝ 2 − j 2 ⎟⎠ ⎜⎝
2 − j 2 ⎟⎠
⎝
⎠
)
(
Now, by KVL for the left mesh, we get,
⎛ 20 + j 25 ⎞ ⎛ 13.67 + j 6.33 ⎞
× 5 − j2
V = Vx + I1 5 − j 2 = ⎜
+
2 − j 2 ⎟⎠
⎝ 2 − j 2 ⎟⎠ ⎜⎝
)
(
=
(
)
101 + j 29.33 105.17∠16.119
=
= 37.18∠61.19 V
2.83∠ − 45
2 − j2
( )
Problem 2.25 (a) Determine the voltages of
node ‘m’ and ‘n’ with respect to the reference in the
circuit shown.
(b) Find the current ‘I’ using node voltage method.
5
4
m
I
j2
n
2
j2
50 0 (V )
50 90 (V )
Fig. 2.87
Solution (a) By KCL at node (m), we get,
Vm − 50 Vm Vm − Vn
+ +
= 0 ⇒ 10 + j 9 Vm − j 5Vn = j 200
j2
5
4
By KCL at node (n), we get,
(
)
Vn − Vm Vn Vn − j 50
+
+
= 0 ⇒ j1Vm + 2 − j 3 Vn = 100
− j2
4
2
(
)
Solving for Vm and Vn from equations (i) and (ii), we get,
Vm =
j 200
100
(10 + j 9)
j1
(
− j5
2 − j3
) = 600 − j 900 = 24.76∠ − 40.36 V
( )
42 − j12
− j5
( 2 − j 3)
(10 + j 9) j 200
Vn =
j1
(10 + j 9)
j1
100
− j5
=
1200 + j 900
= 34.34 ∠52.81 V
42 − j12
( )
( 2 − j 3)
(b) Therefore, the required current is,
V
24.76 ∠ − 40.36
I= m =
= 12.38∠ − 130.36 A
j2
2 ∠90
( )
Problem 2.26 Use node voltage method to find V in the circuit.
Solution Converting the voltage source into current source, we get the circuit shown below.
(i)
(ii)
70
Network Analysis and Synthesis
40
j20
V
V
j30
6 30
120
2.68
50
40
6 30
j 20
41.56
j30
50
15
Fig. 2.88
Fig. 2.89
By KCL,
V
V
V
+
+ = 2.68∠ − 41.56 − 6 ∠30
40 + j 20 − j 30 50
⇒ V ⎡⎣0.022 ∠26.56 + j 0.033 + 0.02 ⎤⎦ = 2 − j1.78 − 5.196 − j 3
⇒ V=
−3.196 − j 4.78
−3.196 − j 4.78
=
⇒ V = −97.62 ∠8.94 V
0.02 + j 0.01 + j 0.033 + 0.02 0.04 + j 0.043
Problem 2.27 Using source transformation and simplification,
determine the voltage between the points, P and Q shown in Fig. 2.90.
2A
Solution By KCL,
P
V − 10 VP
At the node-1, P
+ + 2 = 0 ⇒ VP = 4.8 V
2
8
VQ − 10 VQ
At the node-2,
+ − 2 = 0 ⇒ VQ = 10.8 V
4
6
Therefore, the voltage between the points P and Q is
(Vp VQ ) (4.8 10.8)
6V
Problem 2.28 Find the voltage across the resistor R
in Fig. 2.91.
2Ω
10 V
2-
10 A
2Ω
R = 2Ω
2
10 A
2
10V
2
i
R=2
R=2
(a)
(b)
Fig. 2.92
Voltage across R
6
Fig. 2.91
−10
5
=− A
6
3
2
8
2
2
10 V
4
10 V
Fig. 2.90
Solution Since the 2- resistor is in parallel with the 10-V
voltage source, it may be ignored. Also, converting the current source into equivalent voltage source, we get the simplified circuit as shown in Fig. 2.92.
∴i =
Q
2
5
10
resistor, is, V = i × 2 = − × 2 = − = −3.33 V
3
3
20 V
2Ω
71
Introduction to Circuit-Theory Concepts
Problem 2.29 Find the current through the 5Fig. 2.93 using mesh analysis.
resistor in
10
50 V
Solution By KVL for the first mesh,
15i1 − 10i2 − 5i3 = 50 ⇒ 3i1 − 2i2 − i3 = 10
By KVL for the supermesh,
)
(
(i)
2i2 + i3 + 5 i3 − i1 + 10 i2 − i1 = 0 ⇒ − 15i1 + 12i2 + 6i3 = 0
(i − i ) = 2 ⇒ i = ( 2 + i )
3
2
3
50 V
Putting this value of i2 in equations (i) and (ii), we get
3i1 3i3 14
15i1 18i3
24
46
Solving these two equations, i1 = 20 A and i2 =
A = 15.33 A
3
current through the 5-
2A
10
I2
2
3
I1
2 A I3
5
1
Fig. 2.94
) 143 = 4.67 A
(
resistor is, i = i1 − i3 =
Problem 2.30 Obtain the current ‘I’ in the network shown in Fig. 2.95.
Solution By KVL for the second mesh
3VR 5I 4 VR 0
or,
2VR 5I 4 0
Also,
VR 2 (I 2) putting this in (i),
2 2 (I 2) 5I 4 0
⇒ −4 I + 8 + 5 I + 4 = 0 ⇒ I = −12A
5
2A
VR
(i)
10
3VR
2
I
Fig. 2.95
20
Solution We convert the 5-A current source into its equivalent voltage source.
From the first loop, we get, i1 2A
i1
4V
3
Problem 2.31 Use mesh analysis to find the current ix .
2A
1
(ii)
Also, the constraint equation is that
2
5
Fig. 2.93
)
(
2
3
20
25
i2
1.5ix
10
ix
2A
25
1.5ix 5
5A
Fig. 2.96
5
i3
25 V
ix
Fig. 2.97
By KVL for the supermesh as shown by the dotted line, we get
20i2 30i3 25 10 (i2
(
)
i1)
0
Putting the value of i1, 20i2 + 30i3 + 25 + 10 × i2 − 2 = 0 ⇒ 6i2 + 6i3 + 1 = 0
(i)
72
Network Analysis and Synthesis
Also by KCL we get the following constraint equations.
) (
) ⇒ i = (2 − i )
and 1.5i = ( i − i ) ⇒ i = (1.5i + i ) = (1.5i + 2 − i ) = ( 2 + 0.5i )
(
ix = i1 − i2 = 2 − i2
x
3
2
3
x
2
x
2
x
x
x
Putting the values of i2 and i3 in equation (i), we get
)
(
)
(
6i2 + 6i3 + 1 = 0 ⇒ 6 × 2 − ix + 6 × 2 + 0.5ix = −1 ⇒ ix =
25
= 8.33 A
3
Problem 2.32 Calculate the effective resistance between the points A and B in the circuit shown in Fig. 2.98.
2Ω
3Ω
4Ω
A
6Ω
6Ω
2Ω
2Ω
2Ω
5Ω
5Ω
3Ω
B
Fig. 2.98
Solution The 2- , 2- and 3- resistances are in series and the 4- , 2series. The reduced circuit is shown in Fig. 2.99.
2Ω
and 5-
resistances are also in
3Ω
A
r1 r2
6Ω
6Ω
7Ω
r3
11 Ω
5Ω
B
Fig. 2.99
Converting the delta consisting of the resistances of 6
reduced as shown in Fig. 2.100.
,6
and 3
into equivalent star, the circuit is
3.2 Ω
6×3
= 1.2
r1 =
6 + 3+ 6
6×3
= 1.2
r2 =
6 + 3+ 6
6×6
= 2.4
r3 =
6 + 3+ 6
A
7
7.4
12.2
B
Fig. 2.100
73
Introduction to Circuit-Theory Concepts
The equivalent resistance between terminals A and B is given as
RAB = 7
⎡
7.4 × 12.2 ⎤
⎢ 3.2 + 7.4 + 12.2 ⎥ = 7
⎣
⎦
7 × 7.8061
⎡⎣ 3.2 + 4.6061⎤⎦ =
= 3.69
7 + 7.8061
Problem 2.33 Find the currents i1, i2 and i3 and powers delivered by the sources of the network shown in
Fig. 2.101.
i3
6
A
12
i1
E
C
4
D
B
F
4V
12 V
i2
4
Fig. 2.101
Solution We consider the four meshes and the mesh currents as shown in Fig. 2.102.
6Ω
12 Ω
A
i4
i3
C
E
B
4Ω
F
D
12 V
i1
i2
4V
4Ω
Fig. 2.102
By KVL for the meshes, we get,
3
18i1 − 12i4 = 0 ⇒ 3i1 = 2i4 ⇒ i4 = i1
2
⎛3 ⎞
−12i1 + 12i4 = 12 ⇒ 12i1 = 12i4 − 12 = 12 ⎜ i1 ⎟ − 12 = 18i1 − 12 ⇒ i1 = 2 A
⎝2 ⎠
∴ i4 = 3 A
4i2 = 16 ⇒ i2 = 4 A
4i3 = 4 ⇒ i3 = 1A
Therefore, the required currents are
i1
2A;
i2
4A;
Power delivered by the 12-V source 12 (i4 i2) 12
Power delivered by the 4-V source 4 (i2 i3) 4 5
i3
1A
7 84W
20W
74
Network Analysis and Synthesis
Problem 2.34 Determine the current through 10resistance in the network shown in Fig. 2.103 by using
star–delta conversion.
24
4
13
30
12
8
17
10
Solution The resistances of 8 and 4 are in series
and the resistances of 13 and 17 are also in series.
12
The reduced circuit is shown in Fig. 2.104 (a).
There are two deltas in the circuit, one consisting of
the resistances of 12 each and the other consisting of
Fig. 2.103
the resistances of 30 each. We convert the deltas into
their equivalent star and the reduced circuit is shown in Fig. 2.104 (b).
30
180 V
24
24
180 V
12
4
30
12
30
10
12
10
10
4
4
10
10
30
180 V
180 V
Fig. 2.104(a)
Fig. 2.104 (b)
Equivalent resistances in star are,
12 × 12
=4
12 + 12 + 12
30 × 30
R' =
= 10
30 + 30 + 30
From Fig. 2.104 (b), the further modified circuit is shown in Fig. 2.104 (c).
Therefore, the current through the 10- resistance is the current through the 24Fig. 2.104 (c). This is given as total current
R=
I=
38
180
= 6.27 A
38 × 24
4+
+ 10
38 + 24
38
∴ I10 = I ×
38 + 24
38
= 6.27 ×
= 3.8426 A
38 + 24
Problem 2.35 Find the equivalent
resistance branch in
4
10
I10
24
180 V
Fig. 2.104(c)
network for the circuit shown in
R
Fig. 2.105.
Solution We convert the outer star into its equivalent delta with each
R×R
resistance equal to R ' = R + R +
= 3R . The reduced circuit is shown in
R
Fig. 2.106 (a).
R
R
Fig. 2.105
R
R
R
75
Introduction to Circuit-Theory Concepts
Now, we convert the inner star into its equivalent delta with each resistance equal to R ′′ = R + R +
The reduced circuit is shown in Fig. 2.106 (b).
Combining all the parallel resistances the equivalent
3R
R
R
3R
3R
3R
network is shown in Fig. 2.106 (c).
3R
3R
R
R×R
= 3R .
R
1.5 R
R
3R
3R
3R
1.5 R
Fig. 2.106 (c)
Fig. 2.106 (b)
Fig. 2.106 (a)
1.5 R
Problem 2.36 The element of a 500-watt electric iron is designed for use on a 200-V supply. What value of
resistance is needed to be connected in series in order that the iron can be operated from a 240-V supply?
Solution Since the iron is rated for 500 W, 200 V, the resistance of the iron coil is
2002
R=
= 80
500
When an external resistance Rx is connected in series with the iron, the total resistance in the circuit is RT
(R Rx). If this is connected to a 240-V supply, the power equation becomes
V2
2402
P=
⇒ 500 =
⇒ Rx = 35.2
RT
80 + Rx
Problem 2.37 Find the value of the constant ‘K’ in the circuit shown in Fig. 2.107, such that the power
dissipated in 2- resistor does not exceed 50-W.
4
KI
I
6A
8
2
16 V
Fig. 2.107
Solution Here, the 8- resistance in parallel with the 16-V source can be ignored. Converting the dependent voltage source into its equivalent current source, we get the following circuit.
KI
4
I
6A
I1
2
Fig. 2.108
⎛ KI − 4 ⎞
By KVL for the right mesh, we get 4 I1 − KI + 16 + 2 × I1 − 6 = 0 ⇒ I1 = ⎜
⎝ 6 ⎟⎠
(
)
16 V
76
Network Analysis and Synthesis
⎛ KI − 4 ⎞ 40 − KI
40
=
⇒ I=
Also, I = 6 − I1 = 6 − ⎜
6
6+ K
⎝ 6 ⎟⎠
)
(
Now, the power dissipated in the 2-
resistance is 50 W.
2
⎛ 40 ⎞
40
×2 ⇒
=5 ⇒ K =2
50 = ⎜
6+ K
⎝ 6 + K ⎟⎠
∴ P2 = I 2 × 2
Problem 2.38 Use nodal analysis to determine v1 and power
being supplied by the dependent current source in the circuit
shown in Fig. 2.109.
Solution We first label the circuit as shown in Fig. 2.110 below.
By KCL at the node 1,
By KCL at the node 3,
v1 − v3 v1 − v2
+
=5
50
20
7v1 − 5v2 − 2 v3 = 500
)
(
)
9v1 + 10 0.4 v1 − 16 v3 = 0 ⇒ 13v1 = 16 v3
5A
30
0.4 v1
v1
0.01v1
Fig. 2.109
50
v1
20
v2
(ii)
5A
Also, by constraint equation, v2 0.4v1
Putting this value in (i) and (ii), we get,
(
20
(i)
v3 − v1 v3 − v2
+
= 0.01v1
50
30
9v1 + 10v2 − 16 v3 = 0
7v1 − 5 0.4 v1 − 2 v3 = 500 ⇒ 5v1 − 2 v3 = 500
50
(iii)
30
v3
0.4 v1
v1
0.01v1
Fig. 2.110
(iv)
⎛ 13 ⎞
500 × 16
From (iii), putting the value of v3, we get, 5v1 − 2 ⎜ ⎟ v1 = 500 ⇒ v1 =
= 148.148 V
54
⎝ 16 ⎠
13
13
× v1 = × 148.148 = 120.37 V
16
16
power supplied by the dependent current source is
P v3 0.01v1 120.37 0.01 148.148 178.32 W
∴ v3 =
Problem 2.39 Calculate the node voltages in the circuit shown in
Fig. 2.111.
Solution By KCL for the two nodes, we get
At the node 1,
V1
V −V
− 0.8 I + 12 × 10−3 + 1 2 3 = 0
3
10 × 10
20 × 10
⇒ 3V1 − V2 = 16 × 103 I − 240
20 k
V2
V1
I
10 k
(i)
12 mA
0.8I
30 k
Fig. 2.111
At the node 2,
−12 × 10−3 +
V2 − V1
20 × 10
3
+
V2
30 × 103
= 0 ⇒ − 3V1 + 5V2 = 720
(ii)
77
Introduction to Circuit-Theory Concepts
I =−
Also,
V2
30 × 103
⎛
V2 ⎞
8
Putting this in (i), we get, 3V1 − V2 = 16 × 103 ⎜ −
− 240 = − V2 − 240 ⇒ − 45V1 + 7V2 = 3600
3⎟
15
⎝ 30 × 10 ⎠
720 5
3600 7
Solving (ii) and (iii), we get, V1 =
−3 5
−45 7
−3
=−
12960
= −63.53 V
204
720
−45 3600
21600
=−
= 105.88 V
204
−3 5
−45 7
Problem 2.40 Draw a circuit and its dual if the mesh equations of the circuit are given as
8i1 2i2 4i3 5
14i2 6i3 3
4i1 6i2 15i3 6
Solution The circuit satisfying the mesh equations is shown in Fig. 2.112 below.
The dual equations will be
i1 2
6
8v1 2v2 4v3 5
14v2 6v3 3
i2
2
i1
5V
4v1 6v2 15v3 6
Here, v1, v2, and v3 are the node voltages. In the dual circuit, resis4
6
tances will be replaced by conductances and voltage sources by
i3
the current sources.
V2 =
(iii)
5
Following the procedure mentioned in section 2.9, we construct the dual circuit as shown below.
1/2
5A
Fig. 2.112
2v1
1/6
i1 2
2i1
6
3A
5V
v1 2
4
1/2
1/6
6A
6
1/4
v3
1/5
Fig. 2.113
3V
v2
6V
5
2i1
3V
6V
78
Network Analysis and Synthesis
Therefore, the dual circuit is shown below.
1/4
1/2
v1
2v 1
1/2
5A
1/6
v2
3A
v3
1/5
1/6
Fig. 2.114
Problem 2.41 Draw the dual of the circuit shown in Fig. 2.115.
(a)
L2
C
vg
R2
R3
L1
R1
i0
(b)
2
10 V
1F
3H
1
2
1H
1
2F
Fig. 2.115
Solution (a) The dual network is drawn as shown below.
2
1
C
R3
vg
R2
R1
L1
3
i0
Fig. 2.116
L2
6A
79
Introduction to Circuit-Theory Concepts
The final dual circuit becomes as shown below.
C1 = L1
1
G1 = R1 mho
L=C 2
G2
G3
C2 = L2
3
G2 = R2 mho
ig= vg
G1
v0= i0
G3 = R3 mho
Fig. 2.117
(b) The dual network is drawn as shown below.
2
1
10 V
1F
3H
2
2
1H
3
4
1
2F
Fig. 2.118
The final dual network is shown below.
1
1F
1/2
2H
1/2
10A
3F
1H
1
Fig. 2.119
Summary
1. Electric charge is a fundamental conserved property of
some subatomic particles, which determines their electromagnetic interaction. The SI unit of charge is coulomb
(C). The charge of an electron is 1.602 10 19C. Hence,
1-coulomb charge charge of 6.24 1018 electrons.
2. Electric current is defined as the rate of flow of electric charges or electrons through a cross-sectional area
i dq兾dt.
3. The work done to move an electron in a conductor in
a particular direction or to create a current is known
80
Network Analysis and Synthesis
as the potential of a point. V dw兾dq The potential
of a point is 1 volt if 1 joule of work is done in bringing a 1-coulomb charge from infinity to that point.
4. Any combination and interconnection of network
elements like a resistor or inductor or capacitor or
electrical energy sources are known as ‘networks’.
5. A closed energized network is known as a ‘circuit’.
6. A loop or mesh denotes a closed path obtained by
starting at a node and returning back to the same
node through a set of connected circuit elements
without passing through any intermediate node more
than once. A mesh does not contain any other loop
within it.
7. A node is a point in a circuit where two or more circuit
elements join. A node is said to be an essential node if
it joins three or more elements.
8. A branch is a path that connects two nodes. Those
paths that connect essential nodes without passing
through an essential node are known as essential
branches.
9. An active element is capable of generating electrical
energy. Examples of active elements are voltage source
(such as a battery or generator) and current source.
10. A passive element is one which does not generate
electricity but either consumes it or stores it. Resistors,
inductors and capacitors are simple passive elements.
11. Electrical resistance is a measure of the degree to
which an object opposes an electric current through
it. The voltage–current relationship of a resistance is
v Ri. The SI unit of electrical resistance is the ohm
( ). Resistance of a conductor depends on several
factors like length, cross-section, temperature, etc.
12. For interconnection of several resistances, the equivalent resistance is given as
when they are connected in
Req R1 R2 R3
series
When they are connected in
1
=
parallel
1 1 1
+ + + ⋅⋅⋅
R1 R 2 R 3
13. Capacitance (C ) is a measure of the amount of electric
charge stored for a given electric potential; C Q兾V.
The voltage–current relationship of a capacitor is given
as i C dv兾dt and the energy stored in a capacitor is
W 1兾2 CV 2.
14. For interconnection of several capacitances, the equivalent resistance is given as
when they are connected in
Ceq C1 C2 C3
parallel
=
1
1 1 1
+ + + ⋅⋅⋅
C1 C2 C 3
when they are connected in
series
15. When an electric current i flows round a circuit and
produces a magnetic field through the circuit, the
ratio of the magnetic flux to the current is called the
inductance, or more accurately self-inductance of
the circuit; L
兾i. The voltage–current relationship of an inductor is given as v
L di兾dt and the
energy stored in a inductor is W 1兾2 LI2.
16. For interconnection of several inductances, the equivalent inductance is given as
Leq L1 L2 L3
when they are connected in
series
1
When they are connected in
=
1 1 1
+ + + ⋅⋅⋅ parallel
L1 L2 L3
17. When the magnetic flux produced by one inductor
links another inductor, these inductors are said to be
magnetically coupled. The mutual inductance, M, is
a measure of the coupling between two inductors L1
and L2 and it is given as M = k L1L2
where, k is
called the coefficient of coupling.
18. An ideal voltage source can produce constant output
voltage, whatever be the value of the output current.
However, in most of the practical voltage sources, the
output voltage reduces as the load current increases.
19. An ideal current source can produce constant output
current, whatever be the value of the output voltage.
However, in most of the practical current sources, the
output current reduces as the load voltage increases.
20. Dependent sources or controlled sources are the
sources where the source voltage or current is not
fixed, but is dependent on a voltage or current at some
other location in the circuit. There are four types of
dependent sources, voltage-controlled voltage source
(VCVS), current-controlled voltage source (CCVS), voltage-controlled current source (VCCS), and currentcontrolled current source (CCCS).
21. Ohm’s law states that physical states (temperature,
material, etc.) of a conductor remaining constant, the
current flowing through a conductor is directly proportional to the potential difference across the two
ends of the conductor; i.e., V RI.
22. Kirchhoff’s current law (KCL) states that the algebraic
sum of the currents meeting at a node is zero.
81
Introduction to Circuit-Theory Concepts
23. Kirchhoff’s voltage law (KVL) states that the algebraic
sum of the potential differences taken round a closed
loop is zero.
24. Nodal analysis is based on Kirchhoff’s current law. In
this method, the solution of the KCL equations gives
the unknown node voltages. It is generally used when
the circuit contains several current sources.
25. The concept of supernode is used when a circuit contains independent voltage sources between two nonreference nodes.
26. Mesh analysis is based on Kirchhoff’s current law. In
this method, the solution of the KCL equations gives
the unknown node voltages. It is generally used when
the circuit contains several current sources.
27. The concept of supermesh is used when an independent current source is connected in common to two
meshes.
28. Two phenomena are said to be dual if they are
described by equations of the same mathematical
form. There are a number of dualities between two
circuit quantities, like resistance and conductance,
inductance and capacitance, series circuit and parallel circuit, and so on.
29. A star-connected circuit can be converted into a delta-connected circuit and vice versa. The formulas are as given.
Delta to star conversion
Z1 =
Z12 Z 31
Z12 + Z 23 + Z 31
and
Z3 =
Z2 =
Z 23 Z12
Z12 + Z 23 + Z 31
Z 31 Z 23
Z12 + Z 23 + Z 31
Star to delta conversion
Z12 =
Z1 Z 2 + Z 2 Z 3 + Z 3 Z1
ZZ
= Z1 + Z 2 + 1 2
Z3
Z3
Z 23 =
Z Z
Z 1Z 2 + Z 2 Z 3 + Z 3 Z 1
= Z2 + Z3 + 2 3
Z1
Z1
and
Z 31 =
Z 1Z 2 + Z 2 Z 3 + Z 3 Z 1
Z Z
= Z3 + Z1 + 3 1
Z2
Z2
Short-Answer Questions
1. Define the following terms:
(a) Electric charge
(b) Electric current
(c) Current density
(d) Electric potential and potential difference
(e) Drift velocity
(f) EMF
(a) Electric charge The most basic quantity in an
electric circuit is the electric charge q. Electric charge
is a fundamental conserved property of some subatomic particles, which determines their electromagnetic interaction. Electrically charged matter is
influenced by, and produces, electromagnetic fields.
It is known that an atom consists of a
positively charged nucleus surrounded by negatively charged electrons. In a neutral atom, the total
charge of the nucleus is equal to the total charge
of the electrons. When electrons are removed from
a substance, the substance becomes positively
charged and if excess electrons are given to a substance, it becomes negatively charged.
The SI unit of charge is coulomb (C). The charge
of an electron is 1.602
10 19C. Thus, one coulomb charge is defined as the charge possessed by
⎛
⎞
1
⎝⎜ 1.602 × 10 −19 ⎠⎟
electrons.
1 coulomb charge
charge of 6.24
1018 electrons
(b) Electric current The phenomenon of transferring electric charge from one point in a circuit
to another is described by the term electric current. Electric current is defined as the rate of
flow of electric charges or electrons through a
cross-sectional area. By convention, the electric
current flows in the opposite direction to the
electrons.
If Q amount of charges flow through an area in
time t then the current is given as I Qyt or in differential form, i dqydt and the charge transferred
t
between time t0 and t is given by q = ∫ idt .
t0
82
Network Analysis and Synthesis
As Q is expressed in coulomb, the unit of electric
current is Coulomb per second and it is given the
name ampere (A).
Thus, 1A current flow of 6.24 1018 electrons
per second through an area
(c) Current Density Current density at any point
is a vector whose magnitude is the electric current per unit cross-sectional area and whose direction is normal to the cross-sectional are a, i.e.,
J I兾A n̂. Its unit is ampere per square metre (A/m2).
(d) Electric potential and potential difference To
move an electron in a conductor in a particular
direction, or to create a current, requires some work
or energy. This work is done by the potential or the
potential difference. This is also known as voltage
difference or voltage (with reference to a selected
point such as earth). The unit of potential is volt.
The potential of a point is 1volt if 1joule of work
is done in bringing 1 coulmb of charge from infinity
to that point.
The voltage Vab between two points a and b
is the energy (or work) w required to move a unit
positive charge from a to b. [Unit of voltage is
volt (V).]
Vab dw兾dq
(2.4)
The potential difference between two points
is 1 volt if 1 joule of work is done to displace
1 coulomb of charge from one point to the other.
(e) Drift velocity Drift velocity refers to the average distance traveled by a charge carrier per
unit time. Like the velocity of any object, the drift
velocity of an electron is the distance-to-time
ratio. The path of a typical electron through a wire
could be described as a rather chaotic, zigzag path
characterized by collisions with fixed atoms. Each
collision results in a change in direction of the
electron.
The net effect of these collisions results in slow
drifting of the electrons with a constant average drift velocity. The drift velocity is defined as
the vector average velocity of the charge carriers
moving under the influence of electric field.
Mathematically, if n number of charge carriers
(electrons) with charge Q each passes through an
area A with drift velocity v then the current is given
by I nQvA.
(f) Electromotive force (emf) The phenomenon
of electric current depends on the presence of
free electrons. If a material has a large number of
free electrons, these electrons will always move in
random directions as shown in Fig. 2.120 (a). If an
external effort is applied to the material, it is possible to drift all the electrons in a definite direction
as shown in Fig. 2.120 (b). Such an external factor is
known as electromotive force (emf ). In other words,
the voltage or potential of an electrical energy
source is known as emf.
Fig. 2.120(a) Typical path
of an electron
A high current results from
many charge carriers passing
through a given cross-section
of wire on a circuit.
Fig. 2.120 (b) Current
is constituted by flow of
many charge carriers
through a cross section
When we say something as electrical energy
source, we mean that the energy is converted from
a non-electrical form (such as, mechanical, chemical,
tidal, etc.) into electrical form. Please note that emf is
not a force, but it is the energy or work done.
2. Why should the current in different cross sections
of a cable be constant even though the crosssectional area is different at different places? Is
current a scalar or vector quantity?
An electric current is defined as the time rate of flow of
electric charge across a cross-sectional area.
By convention, the electric current flows in the
opposite direction to the electrons.
If Q amount of charges flow through an area in time
t then the current is given as I Qyt or in differential
form, i dqydt.
The current is the same for all cross-sections of
a conductor even though the cross-sectional areas
83
Introduction to Circuit-Theory Concepts
are different at different places of the conductor. This
is because of the fact that electric charge cannot be
accumulated at any point in a conductor.
Although the current in a conductor has some magnitude in a certain direction, it is not a vector quantity;
it is a scalar quantity that indicates the rate of charge
flow. In case of alternating currents, it may be represented as a phasor quantity.
3. What is the difference between circuits and networks?
or,
“All circuits are networks, but all networks are not
circuits.”-Justify this statement.
Any combination and interconnection of network elements like resistor or inductor or capacitor or electrical energy sources are known as
‘networks’. However, a closed energized network
is known as ‘circuit’. A network need not contain
an energy source; but a circuit must contain
energy source.
4. What is the difference between loop and mesh?
A loop or mesh denotes a closed path obtained by
starting at a node and returning back to the same node
through a set of connected circuit elements without
passing through any intermediate node more than
once. However, the difference between mesh and loop
is that a mesh does not contain any other loop within
it, i.e., a mesh is the smallest loop.
5. Explain the limitations of ohm’s law.
(a) It is not applicable to non-linear circuits like circuits
with powdered carbon, thyrite, etc.
(b) It is not applicable to unilateral circuits, like circuits
with electron tubes, transistors, etc.
6. Explain linearity conditions of elements in detail.
We consider that an alternating voltage v(t ) is
applied to an inductor L at a reference time t 0. Then
the current carried by the inductor is given by
t
i (t ) =
1
v (t )dt + i (0)
L ∫0
(1)
and the relation between flux linkage and current is
given by
(t)
Li(t)
(2)
The properties of an inductor can be explained by plotting the characteristics in the i– plane. If the characteristic is a straight line passing through the origin, the inductor
will be considered as a linear element. But if the i– characteristic is not a straight line and /or does not pass through
the origin (e.g., hysteresis curve), the inductor will behave as
a non-linear element.
λ
i
Fig. 2.121 Characteristic
of a non-linear inductor
q
Slope = C
Slope = L
v
i
Fig. 2.122 (a)
Characteristic of a
linear inductor
Fig. 2.122 (b)
Characteritic of a
linear capcitor
Similarly, for a capacitor the voltage is given by
t
) C1 ∫ i (t )dt +v (0 )
v (t =
(3)
0
A circuit/network element is linear if the relation
between the current and voltage involves a constant
coefficient.
and the relation between charge and voltage is
given by
For example, voltage–current relationship of a resistor, inductor and capacitor (both with zero initial condi
ditions) are linear (v = Ri , v = L , v = C1 ∫ idt ) . Hence
dt
these elements are linear.
The properties of a capacitor can be explained by plotting the characteristics in the q–v plane. If the characteristic is a straight line passing through the origin, the
capacitor will be considered as a linear element. But if the
q–v characteristic is not a straight line and/or does not
pass through the origin (e.g., space-charge capacitance
of a diode), the capacitor will behave as a non-linear
element.
However, if the initial conditions in inductors and
capacitors are non-zero, these elements will become
non-linear as explained below.
q(t )
Cv(t )
(4)
84
Network Analysis and Synthesis
7. Differentiate between unilateral and bilateral elements. Give examples.
In practical voltage sources, the voltage does not
remain constant, but falls slightly; this is taken care of
by connecting a small resistance (r) in series with the
ideal source. In this case, the terminal voltage will be
A system where the voltage–current relationship is
different for two possible directions of current flow is
known as a unilateral system.
v1 (t )
An ideal current source has the following characteristics:
For example, the v–i relationships in a resistor,
inductor and capacitor are same for any direction of
current flow. So, these are bilateral elements. But, in
a diode transistors, the v–i relationships change for
change in the direction of current flow. These elements
are unilateral.
(i)
It produces a constant current irrespective of the
value of the voltage across it.
(ii) It has infinity resistance.
(iii) It is capable of supplying infinity power.
i1
Practical
Ideal
i(t )
R v(t)
i(t)
i
0
Fig. 2.123 (a) v–i
Fig. 2.125
Fig. 2.123 (b)
relationship for a bilateral
element
8. Discuss the characteristics of ideal and practical
sources (voltage and current). What is loading of
sources? Explain.
Or,
Draw the V–I characteristics for voltage and current source for ideal and actual cases.
Or,
Draw the symbol and characteristics of ideal and
practical voltage and current sources.
An ideal voltage source has the following characteristics:
(i) It is a voltage generator whose output voltage
remains absolutely constant whatever be the
value of the output current.
(ii) It has zero internal resistance so that voltage drop
in the source is zero.
(iii) The power drawn by the source is zero.
i
v (t )
r
i
i
0
v (t )
v (t)
v
I
V
ir
i.e., it will decrease with increase in current i.
On the other hand, in a bilateral system, the same
relationship between current and voltage exists for the
current flowing in either direction.
v
v (t )
In practical current sources, the output current does
not remain constant but decreases with increase in
voltage. So, a practical current source is represented
by an ideal current source in parallel with a high resistance (R) and the output current becomes
i 1(t ) = i (t ) −
v (t )
R
Loading of sources It has been mentioned that the
output voltage of a voltage source decreases as the
load current increases. If the source is loaded in such
a way that the output (or load) voltage falls below
a specified full load value, then the source is said to
be loaded and the situation is known as loading of
source.
For example, we consider a voltage source of
100 V as shown in Fig. 2.126 with an internal resistance
of 1 .
1
IL(t )
v (t)
v1(t)
Ideal
Practical
100 V
VL(t )
t
Fig. 2.124 Independent Voltage sources and their
characteristics
Fig. 2.126 (a) Loading of source
Load, RL
85
Introduction to Circuit-Theory Concepts
Thus, we cannot convert a voltage source V with zero
internal resistance to a corresponding current source.
v(t )
VNL
10. Give one practical example each of an ideal voltage source and an ideal current source.
Ideal
Example of ideal voltage source An ideal voltage
source is not practically possible. No voltage source
can maintain its terminal voltage constant even when
its terminals are short-circuited. The terminal voltage of
a practical voltage source decreases as the load current
increases. A dc or ac generator or batteries are some
examples of independent voltage sources. A lead–
acid battery and a dry-cell are some examples of
constant voltage source which can produce constant
terminal voltage within a specified range of output
current.
Loading of source
i
IFL
0
Fig. 2.126 (b) Loading of source
Here, load current, I L =
100
RL + 1
No load voltage is, VNL
100 V
If the specified full load current is 10 A then the load
⎛ 100 ⎞
resistance on full load is R L = ⎜
−1 = 9 Ω
⎝ 10 ⎟⎠
Then, the full load voltage is,
⎛ 100 ⎞
V FL = I FL × R L = 10 × ⎜
− 1 = 90 V
⎝ 10 ⎟⎠
If the load resistance is increased beyond 9 , the
load voltage falls below the specified full load voltage of 90 V. In that case, the source is said to be
loaded.
9. How can ideal voltage sources be converted into
ideal current sources and vice-versa?
A voltage source V(t ) with an internal resistance R can
be converted into a current source I(t ) in parallel with
the same resistance R, where, I(t ) V(t) yR.
R
V (t )
Fig. 2.127
I1(t )
V1(t ) ⬅ I (t )
I1(t )
R
V1(t )
Conversion of voltae source into current
source
A voltage source can be converted into a current
source and vice-versa if and only if their respective
open-circuit voltage and short-circuit current are same.
However, an ideal voltage source can never be opencircuited and an ideal current source can never be
short-circuited, as this is in contrary to the definitions
of ideal voltage and current sources.
Example of ideal current source Similar to voltage
sources, an ideal current source is not practically possible.
No current source can maintain constant current even
when its terminals are open-circuited. The output current
of a practical current source decreases as the output voltage increases. A solar cell, which can produce constant
current within a specified range of output voltage, is an
example of independent current source. A natural lightning can be considered to be an ideal current source.
When a natural lightning strikes the top of a conductor,
the resistance to the ground path is ideally zero. But,
when the lightning strikes a non-conducting element
(like the top of a tree) a large voltage is developed across
the element which is flashed out immediately.
11. Explain why a capacitor is considered as a linear
circuit element.
Let VC1 and VC2 individually excite a relaxed capacitor,
producing the respective currents,
dV
dVC
1
and i C = C C 2
iC = C
2
1
dt
dt
Let iC be the current induced by a voltage (VC1
(
VC2)
)
dVC
dVC
d
1
2
V +V = C
+C
(ic1 ic2)
dt C 1 C 2
dt
dt
This shows that the v–i characteristic of a capacitor
obeys the superposition principles. Therefore, a capacitor is considered as a liner element.
∴i C = C
12. Explain why an inductor is considered as a linear
circuit element.
Let VL1 and VL2 individually excite a relaxed capacitor,
producing the respective currents,
iL =
1
1
V dt
L ∫ L1
and i L =
2
1
V dt
L ∫ L2
86
Network Analysis and Synthesis
Let iL be the current induced by a voltage (VL1
) (
(
1
1
∴ iL = ∫VL dt = ∫ VL + VL dt = iL + iL
1
2
1
2
L
L
VL2)
)
This shows that the v–i characteristic of an inductor obeys the superposition principles. Therefore, an
inductor is considered as a linear element.
13. Explain the following:
(a) The current through an inductor cannot
change instantaneously.
(b) The voltage across a capacitor cannot change
instantaneously.
(a) The equation relating inductance and flux linkages
can be rearranged as follows:
Li
(1)
Taking the time derivative of both sides of the
equation yields
d
di dL
= L +i
dt
dt
dt
From the equation (2), it is clear that for an
abrupt change in current, the voltage across the
inductor becomes infinite. Also, from the equation
(3), it is observed that for a finite change in voltage
in zero time, the integral must be zero.
Therefore, the current through an inductor cannot
change instantaneously.
(b) The relation between charge and voltage in a
capacitor is written as
Q
t
1
idt
C ∫0
1
i (t )dt + v c (0)
C ∫0
where, vc(0) is the initial voltage across the capacitor. For zero initial voltage,
where E is the electromotive force (emf ) and v is the
induced voltage. Note that the emf is opposite to
the induced voltage. Thus
(2)
t
1
v (t )dt + i (0)
L ∫0
where i(0) is the initial curent. When initial current
is zero,
t
1
i (t ) = ∫v (t )dt
L0
∫ dV =
t
d
= −E = v
dt
i (t ) =
vc
or, v c (t ) =
By Faraday’s law of induction we have
or
i=
0
d
di
=L
dt
dt
(4)
dQ
dV
dV
dC
=C
=C
+V
dt
dt
dt
dt
In most physical cases, the capacitance is constant with time
(5)
dV
∴ i =C
dt
1
∴ dV = idt
C
Taking integration on both sides,
The current,
In most physical cases, the inductance is constant
with time and so
di
v =L
dt
CV
(3)
These equations together state that for a steady
applied voltage v, the current changes in a linear
manner at a rate proportional to the applied voltage,
but inversely proportional to the inductance. Conversely, if the current through the inductor is changing
at a constant rate, the induced voltage is constant.
t
vc =
1
idt
C ∫0
(6)
From the equation (5), it is clear that for an abrupt
change of voltage across the capacitor, the current
becomes infinite. Also, from the equation (6), it is
observed that for a finite change of current in zero
time, the integral must be zero.
Therefore, the voltage across a capacitor cannot
change instantaneously.
14. Elaborate the statement: “A voltage impulse causes
a current to be established in an inductance in
zero time.” What is the value of this current? Is it a
violation of the fact that current in an inductance
cannot change instantaneously?
The voltage–current relationship of an inductor is
t
iL =
()
1
∫ v t dt
L −∞
87
Introduction to Circuit-Theory Concepts
If an impulse voltage is applied to an inductor then the
resulting current is given by
t
iL =
)
(
(
1
1
∫ t −T dt = L u t −T
L −∞
)
Thus, an impulse voltage applied to an inductor L
results instantaneously in a current of 1兾L.
However, we know that the current in an inductor
cannot change instantaneously. Here, the instantaneous current generated is an unusual behaviour of
an inductor and this happens because of the fact that
the driving voltage in the form of an impulse is also an
unusual voltage.
15. If a unit impulse current is applied to a capacitor
what will be the result?
The voltage–current relationship of a capacitor
t
is v C =
()
1
∫ i t dt .
C −∞
If an impulse current is applied to a capacitor then
the resulting voltage across the capacitor is given by
t
vC =
(
)
(
1
1
∫ t −T dt = C u t −T
C −∞
)
Thus, an impulse current applied to a capacitor C
results in an instantaneous voltage of 1兾C.
16. Derive an expression of the energy stored in a
capacitor.
Energy stored in a capacitor The energy (measured
in joules) stored in a capacitor is equal to the work
done to charge it. Consider a capacitance C, holding a
charge q on one plate and q on the other. Moving a
small element of charge dq from one plate to the other
against the potential difference V q/C requires the
work dW,
q
dW = dq
C
where, W is the work measured in joules
q is the charge measured in coulombs
C is the capacitance, measured in farads
We can find the energy stored in a capacitance by
integrating this equation. Starting with an uncharged
capacitance (q 0) and moving the charge from one
plate to the other until the plates have charge Q and
Q requires the work W:
Q
q
1Q 2 1 2
W = ∫ dq =
= CV
C
2C 2
0
Combining this with the above equation for the capacitance of a flat-plate capacitor, we get the energy stored
in a capacitor as
1
1 A 2
W = CV 2 =
V
2
2 d
where, W is the energy measured in joules
C is the capacitance, measured in farads
V is the voltage measured in volts
17. Derive an expression of the energy stored in an
inductor.
When an electric current is flowing in an inductor, there
is energy stored in the magnetic field.
Suppose that an inductor of inductance L is connected to a variable dc voltage supply. The supply is
adjusted so as to increase the current i flowing through
the inductor from zero to some final value I.
As the current through the inductor is increasing,
the emf generated is E
L di兾dt and this emf acts
to oppose the increase in the current. Clearly, work
must be done against this emf by the voltage source
in order to establish the current in the inductor. The
work done by the voltage source during a time interval dt is
dW = Pdt = − Eidt = iL
di
dt = Lidi
dt
Here, P
Ei is the instantaneous rate at which the
voltage source performs work. To find the total work W
done in establishing the final current I in the inductor,
we must integrate the above expression. Thus,
I
1
W = L ∫ idi = LI 2
2
0
18. Define V-shift and I-shift in the source transformation.
For source transformation, i.e., transforming a voltage
source into a current source and vice-versa, it is first
necessary to shift the sources within the network. This
is done by V or I shifting.
The methods are explained below.
For voltage source shifting, we consider a network as shown in Fig. 2.128 (a). We can shift the
voltage source within the network as shown in
Fig. 2.128 (b) and Fig. 2.128 (c).
Similarly, a current source can be shifted within a
network as explained in Fig. 2.128 (d) to Fig. 2.128 (e).
88
Network Analysis and Synthesis
R3
R3
R2
However, there are some particular circuits, where
only one method can be applied. For example, in analyzing transistor circuits, mesh method is the only possible method; while for op-amp circuits and for nonplanar networks, node method is the only possible
method.
R1
R2
R1
2. Requirement of the problem If node voltages are
required, nodal analysis is used; if branch/mesh
currents are required, loop analysis is used.
V
(a)
V
V
(b)
R3
20. Explain ‘duality’ in electrical engineering. State the
steps followed in finding the dual of a network.
R2
Two phenomena are said to be dual if they are described
by equations of the same mathematical form.
R1
V
V
There are a number of similarities and analogies
between the two circuit analysis techniques based
on loop-current method and node voltage method.
The principal quantities and concepts involved in
these two methods based on KVL and KCL are dual
of each other with voltage variables substituted by
current variables, independent loop by independent
node-pair, etc.
V
(c)
I
R1
R1 R
2
R3
R2
I
I
This similarity is termed as ‘principle of duality’.
Some dual relations are
I
R3
(d)
Fig. 2.128
(e)
V-Shift and I-Shift in source transformation
19. Comment briefly on the choice between loop and
node methods of analyzing a network
In any network having ‘N’ nodes and ‘B’ branches,
there are 2B unknowns, i.e., ‘B’ branch currents and
‘B’ branch voltages. These unknowns can be determined either by loop analysis or nodal analysis.
The choice of the method depends on two factors:
1. Nature of the network The mesh-method is generally used for circuits having many series-connected elements, voltage sources, or supermeshes. On the other hand, nodal analysis is more
suitable for networks for circuits having many
parallel-connected elements, current sources, or
supernodes.
The main factor for selecting any one method
is the minimum number of equations. If a circuit has
fewer nodes than meshes then nodal analysis is
used, while if a circuit has fewer meshes than nodes
then loop method is used.
v = Ri
i = Gv
di
dv
v =L
i =C
dt
dt
1
1
v = ∫ idt i = ∫vdt
C
L
Thus, the circuit elements (R, L, C) have some dual relationship. Duality also appears as a relation between
two networks. For example, an RLC series circuit with
voltage excitation is the dual of an RLC parallel circuit
with current excitation.
Steps for construction of the dual of a network
1. A dot is placed inside each independent loop of the
given network; these dots correspond to the nonreference nodes of the dual network.
2. A dot is placed outside the network; this dot corresponds to the datum node.
3. All internal dots are connected by dashed lines
crossing the common branches and placing the
elements which are duals of the elements of the
original network.
4. All internal dots are connected to the external dot
by dashed lines crossing all external branches and
placing dual elements of the external branch.
89
Introduction to Circuit-Theory Concepts
Conventions for reference polarities of voltage source
and reference directions of current source
1. A clockwise current in a loop corresponds to a positive polarity (with respect to reference node) at the
dual independent node.
2. A voltage rise in the direction of a clockwise loop
current corresponds to a current flowing towards
the dual independent nodes.
Finally, the dual construction can be checked by writing
mesh equations and node equations of the two networks.
Exercises
1. Find RAB in the network shown below. All resistance
values are in ohms.
[23.52 ]
A
10
20
5
15
15V
5
8
10
25
2
8A
4
10 V
12A
6
(a)
B
(b)
Fig. 2.132
30
5
Fig. 2.129
2. Use loop current analysis to find the current in each
battery in the network shown. All resistance values are
in ohm.
[0.793 A, 0.408 A, 0.295 A]
60
20
40
4
25
5. Convert the circuit shown in
Fig. 2.133 to a single current
source in parallel in with a
single resistor.
[I
1 A, R
10V
18V
6
2.73 ]
5
Fig. 2.133
6. Determine the voltage V in the circuit, using the source
transformation technique and/ or any other method.
[V 56.25 V]
50
5A
120 V
60V 40 V
3
Fig. 2.130
2
5
3. Find the current through the 2- resistance in the network shown below. Use loop current method.
[ 0.841 A]
60
2
1
3
8
20A
10
V
Fig. 2.134
7. Find the current flowing through the 5- resistor using
⎡ 11 ⎤
source transformation technique.
⎢ A⎥
⎣ 27 ⎦
10
10 V
3
20V
1
Fig. 2.131
5V
4. Convert the circuits shown in Fig. 2.132 to a single voltage source in series with a single resistor.
[V
5兾3 V, R
8兾3 ]
[V
104 V, R
10 ]
Fig. 2.135
3
1
2A
5
90
Network Analysis and Synthesis
8. Reduce the network shown in Fig. 2.136 (a) to a form
shown in Fig. 2.136 (b) using successive source transformations.
[I 2.14 A, R 1.75 ]
12. Use mesh analysis to find the current ix in the circuit
shown in Fig. 2.140.
[8.33 A]
20
25
a
1A
1.5ix 5
10
ix
2A
2
2
5A
4V
Fig. 2.140
a
3
1
6V
3V
13. Use mesh analysis to find the current ix in the circuit
shown in Fig. 2.141.
[2.79 A]
R
I
b
(a)
b
8
Fig. 2.136
5
0.5V1
V1
10
_
40 V
_
20_V
3
10
5
Fig. 2.141
20 V
Fig. 2.137
3
j4
j3
j5
4
V2
4A
2
_
4
14. Determine the value of V2, such that the current
through (3 j4) impedance is zero.
[80.43 119.55 (V)]
_
20
2
100V
9. For the circuit of Fig. 2.137, apply source transformation and then find V1 and V2 by nodal analysis.
[V1 40 V, V2 15 V]
2A
8A
ix
(b)
4
V2
Fig. 2.142
10. In the circuit shown in Fig. 2.138 if I1
and the power delivered in it.
2 A, determine RL
[2 ; 18 W]
15. Find the current ix in the circuit shown in Fig. 2.143.
[0.571 mA]
I1
12 A
6
3
3I1
5k
RL
4 mA
i1
20 k
ix
3i1
Fig. 2.143
Fig. 2.138
11. Find the node voltages Va, Vb and Vc using nodal analysis.
[4.3 V; 3.9 V; 3.3 V]
16. Find the current i1 in the circuit shown in Fig. 2.144.
[ 1 A]
2
Va
4A
Fig. 2.139
2
1
2V
10
Vb
i
3
2i
i1
Vc
5
20V
Fig. 2.144
90 V
40
v2
2v2
91
Introduction to Circuit-Theory Concepts
17. Find the equivalent resistance between the terminals
A and B for the circuit shown in Fig. 2.145.
[60 ]
80
10
A
Req
22. Construct the dual of the networks shown below.
L
R
40
v (t )
30
100
C
20
(a)
B
5A
Fig. 2.145
18. Using mesh analysis, find the current ix in the circuit
shown in Fig. 2.146.
[2.79 A]
3
5
8A
100V
ix
10
8
2
4
100 V
6
4
3
(b)
5
Fig. 2.146
5
4H
19. For the circuit shown in Fig. 2.147, find the currents iA, iB,
[3 A, 5.4 A, 6 A]
and iC.
0.2F
2A
iB
iA
Vx
5.6 A
18
3
20V
iC
0.1Vx
9
2A
(c)
3F
Fig. 2.147
4H
50 mA
10
20. Use nodal analysis to find the voltage Vxy in the circuit
shown in Fig. 2.148 below.
[ 0.257 V]
(d)
X
10
2
V1
30
0.55 V
40
12
6
3 V1
5H
10 V
20
2F
3A
Y
Fig. 2.148
(e)
21. Determine Va and Vb in the circuit shown in Fig. 2.149.
[5.17
75 V;1.33 V]
j6
10 0 (V )
Va
3
j6
Vb
j4
2
t=0
j5
6V
2H
j4
(f)
Fig. 2.149
Fig. 2.150
10 mF
92
Network Analysis and Synthesis
23. Draw a circuit and its dual if the mesh equations of the
circuit are
3; 4i1 5i2
(a) 8i1 2i2 4i3 6; 7i2 5i3
9i3 5
4; i1 6i2 5i3 6; i1 5i2
(b) 4i1 i2 i3
8i3 2
24. Draw a circuit and its dual if the node equations of
the circuit are
(a) 4v1 v2 v3
4; v1 6v2 5v3 6; 5v2 8v3
3; v1 v2 2v3
(b) 4v1 v2 v3 5; 3v2 v3
2v2
v3
5; 2v1
8v2 3v3
(c) 6v1
v1 3v2 9v3 0
2
6
4;
Questions
1. Define an electrical network. “All circuits are networks,
but all networks are not circuits.”-Justify this statement.
2. Explain linearity conditions of elements in detail.
3. Differentiate between unilateral and bilateral elements. Give examples.
4. (a) State the basic assumptions for circuit analysis.
(b) Briefly mention the different source transformation techniques.
(c) Discuss the properties of an ideal current source
and ideal voltage source.
(d) Explain how a voltage source can be converted
into an equivalent current source and vice-versa.
5. Explain the properties of basic elements R, L and C in
the network.
6. What is electrical resistance? Explain the factors that
affect the resistance.
7. Define capacitance. Derive an expression of the energy
stored in a capacitor.
8. Define self-inductance of a coil. Derive an expression
of the energy stored in an inductor.
9. What is mutual inductance? Explain coefficient of coupling of two mutually coupled coils.
10. (a) Explain why a capacitor is considered as a linear circuit element.
(b) Explain why an inductor is considered as a linear
circuit element.
11. Explain why
(a) the current through an inductor cannot change
instantaneously
(b) the voltage across a capacitor cannot change
instantaneously
12. Discuss the characteristics of ideal and practical
sources (voltage and current). What is loading of
sources? Explain.
Or,
Draw the V–I characteristics for voltage and current
source for ideal and actual cases.
Or,
Draw the symbol and characteristics of ideal and practical voltage and current sources.
13. Explain voltage source to current source transformation. Define V-shift in the source transformation.
14. Establish the conditions for equivalence of practical
voltage and current sources.
15. Give a brief introduction to the dependent (controlled)
sources.
16. (a) State Kirchhoff’s voltage and current laws.
(b) Give a brief comparison of the loop method and
node method of circuit analysis.
(c) Comment briefly on the choice between loop and
nodal methods of analyzing a network.
17. Explain ‘duality’ in electrical engineering. How can you
draw the dual of a network?
18. State the steps followed in finding the dual of a network.
19. Elaborate the statement: “A voltage impulse causes
a current to be established in an inductance in zero
time.” What is the value of this current? Is it a violation of the fact that current in an inductance cannot
change instantaneously?
Multiple-Choice Questions
1. Find the odd one from the following elements:
(i) Inductor
(ii) Capacitor
(iii) Resistor
(iv) Transistor
2. Kirchhoff’s laws are valid for
(i) linear circuits only
(ii) passive time-invariant circuits
93
Introduction to Circuit-Theory Concepts
(iii) non-linear circuits only
(iv) both linear and non-linear circuits
12. The voltage across the 5-A current source in the circuit
shown in Fig. 2.151 is
3. Kirchhoff’s laws are applicable to
(i) dc circuits
(ii) circuits with sinusoidal excitation only
(iii) circuits with dc and sinusoidal excitation only
(iv) circuits with any excitation.
5
5V
4. Kirchhoff’s law fails in case of
(i) linear networks
(ii) non-linear networks
(iii) dual networks
(iv) distributed parameter networks
5. KCL is a consequence of law of conservation of
(i) energy
(ii) charge
(iii) flux
(iv) all of the above
6. A component that opposes the change in circuit current is
(i) resistance
(ii) capacitance
(iii) inductance
iv) conductance
9. A network N’ is a dual of a network N if
(i) both of them have same mesh equations
(ii) both of them have same node equations
(iii) mesh equations of one of them are node equations of the other
(iv) none of the above
10. A connected planar network has 4 nodes and
5 elements. The number of meshes in its dual network is
(i) 4
(ii) 3
(iii) 2
(iv) 1
11. Two networks can be dual when
(i) their nodal equations are the same
(ii) the loop equations of one network are the nodal
equations of the other
(iii) their loop equations are the same
(iv) none of these
5
Fig. 2.151
ix
6
3V
12A
9
Fig. 2.152
(i) 25 V
(ii) 15 V
(iii) 17.5 V
(iv) 20 V
13. The current ix in the network of Fig. 2.152 is,
(i) 1A
(ii) 1兾2A
(iii) 1兾3A
(iv) 4兾5A
14. The equivalent circuit of the capacitor shown is
C
7. A component that opposes the change in circuit voltage is
(i) resistance
(ii) capacitance
(iii) inductance
(iv) conductance
8. For a dc voltage an inductor
(i) is virtually a short circuit.
(ii) is an open circuit
(iii) depends on polarity
(iv) depends on voltage value
5A
V0 = q0/C
Fig. 2.153
(i)
(iii)
C
V0
(ii)
C
(iv)
C
C
V0
15. A network has seven nodes and five independent
loops. The number of branches in the network is
(i) 7
(ii) 5
(iii) 11
(iv) 12
16. An electric circuit with 10 branches and 7 nodes will
have
(i) 3 loop equations
(ii) 4 loop equations
(iii) 7 loop equations
(iv) 10 loop equations.
17. A circuit having an emf source or any energy source is
(i) active circuit
(ii) passive circuit
(iii) unilateral circuit
(iv) bilateral circuit
94
Network Analysis and Synthesis
18. The internal impedance of an ideal current source is
(i) zero
(ii) infinite
(iii) both (i) and (ii)
(iv) none of these
19. The internal impedance of an ideal voltage source is
(i) zero
(ii) infinite
(iii) both (i) and (ii)
(iv) none of these
20. The internal impedance of a dependent voltage
source is
(i) zero
(ii) infinity
(iii) fraction of ohm
(iv) any unknown value
21. An ideal voltage source will charge an ideal capacitor
(i) in infinite time
(ii) exponentially
(iii) instantaneously
(iv) none of the above
22. A practical current source is usually represented by
(i) a resistance in series with an ideal current source
(ii) a resistance in parallel with an ideal current source
(iii) a resistance in series with an ideal voltage source
(iv) none of the above
23. Energy stored in a capacitor is
∞
(i) 1兾4CV 2
(ii) 1兾2CV 2
(iii)
1
∫2C
(iv) 0
0
24. The node method of circuit analysis is based on
(i) KVL and Ohm’s law
(ii) KCL and KVL
(iii) KCL, KVL and Ohm’s law
(iv) KCL and Ohm’s law
25. The loop method of circuit analysis is based on
(i) KVL and Ohm’s law
(ii) KCL and KVL
(iii) KCL, KVL and Ohm’s law
(iv) KCL and Ohm’s law.
26. If there are b branches and n nodes, the number of
KVL equations required will be
(i) b
(ii) b n
(iii) n 1
(iv) b n 1
27. If the number of branches is ‘B’, the number of nodes is
‘N’ and the number of dependent loops is ‘L’ then the
number of independent node equations will be
(i) N L 1
(ii) B 1
(iii) B N
(iv) N 1
28. A network has 10 nodes and 17 branches in all. The
number of different node pair voltages would be
(i) 7
(ii) 9
(iii) 10
(iv) 45
29. Two wires A and B of the same material and lengths
L and 2L have radii r and 2r, respectively. The ratio of
their specific resistance will be
(i) 1 : 1 (ii) 1 : 2
(ii) 1 : 4
(iv) 1 : 8
30. There are two wires A and B. A is 20 times longer than
B and has half the cross section of that of B. If the
resistance of B is 1 , the resistance of A will be
(i) 40
(ii) 1兾40
(iii) 20
(iv) 10
31. The resistance between the opposite faces of a
1-m cube is found to be 1 . If its length is increased
to 2 m, with its volume remaining the same then
its resistance between the opposite faces along its
length is
(i) 2
(ii) 4
(iii) 1
(iv) 8
(v) ½
32. A wire of length l and of circular cross section of radius
r has a resistance of R ohms. Another wire of the same
material and cross-sectional radius 2r will have the
same resistance R if the length is
(i) 2l
(ii) l兾2
(iii) 4l
(iv) l2
33. Two resistances of equal value, when connected in
parallel, give an equivalent resistance of R. If these
resistances are connected in series, the equivalent
resistance will be
(i) R
(ii) 4R
(iii) 2R
(iv) R兾2
34. A series arrangement of ‘n’ identical resistances is
changed into a parallel arrangement. The new total
resistance will become…times the original resistance.
(i) 1兾n
(ii) 1兾n 2
(iii) 1兾n 3
(iv) 1兾n 4
35. If a two-terminal network element in a circuit has voltage and current variables that follow the associated
reference directions and its power is negative, which
of the following is true?
(i) The element is supplying energy to the rest of
the circuit.
(ii) The element is receiving energy from the rest of
the circuit.
(iii) Either (i) or (ii) could be true.
36. If an ideal voltage source and an ideal current source
are the connected in parallel, what are the properties
of the combination?
(i) The same as a voltage source
(ii) The same as a current source
(iii) Different from either a voltage source or a current source
37. If an ideal voltage source and an ideal current source
are connected in series, what are the properties of the
combination?
(i) The same as a voltage source
(ii) The same as a current source
(iii) Different from either a voltage source or a current source
95
Introduction to Circuit-Theory Concepts
38. When ideal voltage sources are connected in series,
which of the following is true?
(i) The voltages add, independent of whether the
individual sources are constant valued or have
outputs that are functions of time.
(ii) The connection violates KVL; thus it is not permitted.
(iii) Neither is true.
39. When ideal arbitrary voltage sources are connected in
parallel, which of the following is true?
(i) The voltages add, independent of whether the
individual sources are constant valued or have
outputs that are functions of time.
(ii) The connection violates KVL; thus it is not permitted.
(iii) Neither is true.
40. When ideal arbitrary current sources are connected in
series, which of the following is true?
(i) The currents add, independent of whether the
individual sources are constant valued or have
outputs that are functions of time.
(ii) The connection violates KCL; thus it is not permitted.
(iii) Neither is true.
41. When ideal current sources are connected in parallel,
which of the following is true?
(i) The currents add, independent of whether the
individual sources are constant valued or have
outputs that are functions of time.
(ii) The connection violates KCL; thus it is not permitted.
(iii) Neither is true.
42. In a network containing only independent current
sources and resistors, if the values of all resistors are
doubled, the values of the node voltages
(i) are doubled
(ii) remain the same
(iii) are halved
(iv) change in some other way
43. In a network containing only independent current
sources and resistors, if the values of all the current
sources are doubled, the values of the node voltages
(i) are doubled
(ii) remain the same
(iii) are halved
(iv) change in some other way
44. In a network containing only independent voltage
sources and resistors, if the values of all the voltage sources are doubled, the values of the mesh
currents
(i) are doubled
(ii) remain the same
(iii) are halved
(iv) change in some other way
45. In a network containing only independent voltage
sources and resistors, if the values of all the resistors
are doubled, the values of the mesh currents
(i) are doubled
(ii) remain the same
(iii) are halved
(iv) change in some other way
46. If the same constant value of current is added to all
the independent current sources in a network, the
node voltages
(i) will all have a constant value added
(ii) will remain the same
(iii) will all have a constant value subtracted
(iv) will change in some other way
47. If the same constant value of voltage is added to each
of the independent voltage sources in an arbitrary
network containing only resistors and independent
voltage sources, the mesh currents
(i) will all have a constant value added
(ii) will remain the same
(iii) will all have a constant value subtracted
(iv) will change in some other way
48. Two resistors R1 and R2 give combined resistance of
4.5 when in series and 1 when in parallel. The
resistances are
(i) 2 and 2.5
(ii) 1 and 3.5
(iii) 1.5 and 3.5
(iv) 4 and 0.5
49. When all the resistance in the circuit are of 1 each,
the equivalent resistance across the points A and B
will be
B
A
Fig. 2.154
(i) 1
(ii) 0.5
(iii) 2
(iv) 1.5
50. The energy expanded or heat generated in joules
when a current of ‘I’ flows through a conductor ‘R’ for ‘t’
seconds is given by
(i) I 2Rt
(ii) IRt
(iii) IR 2t
(iv) IRt 2
51. A 2- resistance having a current of 2 A will dissipate
a power of
(i) 2 W
(ii) 4 W
(iii) 8 W
(iv) 8 J
52. The ratio of resistances of a 100-W, 220-V lamp to that
of a 100-W, 110-V lamp will be, at the respective voltages
(i) 4
(ii) 2
(iii) 1兾2
(iv) 1兾4
96
Network Analysis and Synthesis
53. The elements which are not capable of delivering
energy by their own are known as
(i) unilateral elements (ii) non-linear elements
(iii) passive elements
(iv) active elements
54. For the circuit shown in Fig. 2.155, the value of current I is
i 3
2
(ii) 0.5 A
2
2
6
100 V
(i) 1 A
(iv) none of these
5
A
12
6
(iii) 1.5 A
60. If the current in the 7- resistor branch is 0.5 A as
shown in Fig. 2.160 and now if the source is connected
in series with the 7- branch and the terminals AB are
shorted, the current in the 5- resistor is
3
4
10
10 V
7
Fig. 2.155
(i) 10 A
(ii) 15 A
(iii) 20 A
55. The current in the 1- resistor is
(ii) 0.5 A
(iii) 9.75 A
(iv) none of these.
61. The voltage across the 5-A source in the given circuit is
5V
1
Fig. 2.160
(i) 1 A
A
10 V
B
(iv) 25 A
5
B
Fig. 2.156
(i) 5 A
(ii) 10 A
(iii) 15 A
56. The current in a 5- resistor branch in a linear network is 5 A. If this branch is replaced by a resistor of
10 , the current in this branch will be
(i) 5 A
(ii) 10 A
(iii) less than 4 A
(iv) none of these
57. The potential of the point A in the given network is
1/2
A 1/3
5V
10 V
1
5
10 V
(iv) zero
5A
Fig. 2.161
(i) 25 V
(ii) 15 V
(iii) 17.5 V
(iv) 20 V
62. In the circuit shown in Fig. 2.162 current I flows
through the resistance R. If a battery with an emf of
2 V and an internal resistance of 1
is connected
between the terminals A and A’ with the positive terminal connected to A’, the current through R would be
1
A
1
R=2
B
Fig. 2.157
(i) 6 V
(ii) 7 V
I amp
(iii) 8 V
(iv) none of these
A
58. The current through the 30- branch in the given circuit is
Fig. 2.162
10
(i) 2 A
5
10
(ii) 1.66 A
30
10 A
Fig. 2.158
(i) 2.5 A
I
(ii) 2.25 A
(iii) 2 A
10 V
Fig. 2.159
5
2
10
10
(iv) 1.5 A
5
(iv) 10 A
5 volts
59. The current through the 8- branch is
(iii) 1 A
63. The circuit shown in Fig. 2.163 is linear and timeinvariant. The sources are ideal. The voltage across
the 1- resistor and the current through it will be
1A
1
v(t )
8
Fig. 2.163
(i) −5 V and −5 A
(iii) 1 V and 6 A
(ii) 1 V and 1 A
(iv) 5 V and 5 A.
97
Introduction to Circuit-Theory Concepts
64. The number of 2-μF, 400-V capacitors needed to obtain
a capacitance value of 1.5 μF rated for 1600 V is
(i) 12
(ii) 8
(iii) 6
(iv) 4
(i)
A
A
(ii)
R
R
I
V
65. The value of the current I flowing in the 1- resistor in
the circuit, shown in Fig. 2.164 will be
B
B
(iii)
(iv)
A
A
I
5V
5A
R
1
V
I
B
Fig. 2.164
(i) 10 A
(ii) 6 A
(iii) 5 A
(iv) zero.
66. In the circuit shown in Fig. 2.165, the current I through
RL is
60
120
420 V
RL = 30
69. Two condensers of 20-μF and 40-μF capacitances are
connected in series across a 90-V supply. After charging,
they are removed from the supply and are connected
in parallel with positive terminals connected together.
Similar is done to the negative terminals. Then the voltage across them will be
(i) 90 V
(ii) 60 V
(iii) 40 V
(iv) 20 V
70. The current read by the ammeter A in the ac circuit
shown in Fig. 2.167 is
420 V
Fig. 2.165
(i) 2 A
A
(ii) zero
(iii)
2A
(iv)
(ii) 5 A
(iii) 3 A
RL
I
3
(iv)
P
4
10 V
Fig. 2.168
RL
RL
68. A simple equivalent circuit of the 2-terminal network
shown in Fig. 2.166 is
(i)
2兾5 A
(iii) 18兾5 A
(ii) 24兾5 A
(iv) 2兾5 A
72. For the circuit shown in Fig. 2.169, the voltage VAB is
B
A
10 V
R
5
A
V
Fig. 2.169
B
Fig. 2.166
50 V
10
5
I
(iv) 1 A
2
1
RL
P
5A
71. In the circuit shown in Fig. 2.168, current I is
P
(iii)
3A
Fig. 2.167
(i) 9 A
(ii)
P
1A
6A
67. A voltage source with an internal resistance RS, supplies power to a load RL. The power delivered to the
load varies with RL as
(i)
B
(i) 6 V
(ii) 10 V
(iii) 25 V
(iv) 40 V
98
Network Analysis and Synthesis
73. The equivalent resistance between the terminal
points X and Y in the circuit shown is
15
Y
30
15
30
15
30
X
15
79. For the circuit shown in Fig. 2.176, the current I is
given by
Fig. 2.170
(i) 15
(ii) 45
(iii) 55
(iv) 30
74. In the circuit shown in Fig. 2.171, if I = 2 then the value
of the battery voltage V will be
I
78. For the circuit shown
A
C
Linear
in Fig. 2.175, when the
i
E
passive
voltage E is 10 V, the
network
D
B
current i is 1 A. If the
applied voltage across Fig. 2.175
the terminal C–D is 100
V, the short-circuit current flowing through the terminals A–B will be
(i) 0.1 A
(ii) 1 A
(iii) 10 A
(iv) 100 A
2
4
6A
I
1
3V
3
0.5
1
1
Fig. 2.176
V
1
(i) 3 A
(ii) 2 A
(iii) 1 A
(iv) zero
80. The value of V in the circuit shown in Fig. 2.177 is
Fig. 2.171
(i) 5 V
(ii) 3 V
(iii) 2 V
(iv) 1 V
75. The effective resistance between the terminals A and
B in the circuit shown in Fig. 2.172 is
3V
1
V
1
1
3A
A
R R
O
R
B
Fig. 2.177
R
(i) 1 V
R
C
R
(ii) 2 V
(iii) 3 V
81. For the circuit given in figure, the power delivered by
the 2 volt source is given by:
(i) 4 W
(ii) 2 W
(iii) 2 W
(iv) 4 W
Fig. 2.172
(i) R
(ii) R 1
(iii) R兾2
(iv) 6兾11 R
76. The current in the given circuit with a dependent source is
3
2 Vb
1
2A
Vb
24 V
3
(i) 4 W
(ii) 2 W
6
4A
2 W (iv)
4W
resistor in the circuit
4/7 A
25/7 A
120
R
Fig. 2.179
Fig. 2.174
(ii) 2.5
(iii)
5V
(i) 10 A
(ii) 12 A
(iii) 14 A
(iv) 16 A
77. The value of the resistance ‘R’ shown in Fig. 2.174 is
(i) 3.5
1V
82. The current through 120shown in the Fig. 2.178 is
Fig. 2.173
7
1
Fig. 2.178
4
50 V
(iv) 4 V
(iii) 1
(iv) 4.5
(i) 1 A
(ii) 2 A
(iii) 3 A
(iv) 4 A
99
Introduction to Circuit-Theory Concepts
83. Four resistors of equal value when connected in series
across a supply dissipate 25 W. If the same resistors
are now connected in parallel across the same supply,
what is the power dissipated?
(i) 75 W
(ii) 100 W
(iii) 200 W (iv) 400 W
(i) 4 A
20 A
8
2
10
I=2
(i) 3.0 A
(ii) 115 V
(iii) 85 V
(iv) 55 V
I3
10 A
10
(iv) 63兾17
3
3
6V
3
1
3
I
Fig. 2.182
(ii) 1 A
(iii) 2 A
(iv) 3 A
87. In the circuit given when R is infinite, V 4 V and when
R 0, the current through R is 4 A. If R 3 , what is
the current through it?
Sources
and
resistors
Fig. 2.183
12 V
(i) 0
86. What is the current I in the circuit given in Fig. 2.182?
2
L2
t=0
C
Fig. 2.185
Fig. 2.181
(i) 0
1
L1
6
(iii) 65
(iv) 0.0 A
5
R3
30 V
(ii) 5
(iii) 1.0 A
65 V
I2
I1
(ii) 2.0 A
89. The circuit shown in Fig. 2.185 is in steady state with
the switch open. At t = 0, the switch is closed. What
is the current through the 1- resistor, i(0 )?
85. A part of an electrical network has the configuration
shown in Fig. 2.180. The voltage drops across the resistances are 20 V, 30 V and 65 V with respective polarities
shown. Which one of the following gives the correct
value of the resistance R3?
(i) 13
2
1
Fig. 2.184
V
Fig. 2.180
20 V
(iv) 1 A
5A
3
(i) 185 V
(iii) 2 A
9
84. What is the voltage V in the circuit shown in Fig. 2.180?
5
(ii) 3 A
88. For the circuit given, what is the current delivered by
the battery?
(ii) 1.33 A
(iii) 1.66 A
(iv) 2 A
90. A 2-terminal network is one of the R-L-C elements.
The element is connected to an ac supply. The current
through the element is I. When a capacitor is inserted
in series between the source and the element then current through the element becomes 2I. The element
(i) is a resistor
(ii) is an inductor
(iii) is a capacitor (iv) cannot be a single element
91. For the circuit shown in
R
Fig. 2.186, if the current I = 3 A
and 1.5 A for RL = 0 and 2
respectively, then what is the V
value of I for RL = 1 ?
(i) 0.5 A
(ii) 1.0 A
Fig. 2.186
(iii) 2.0 A (iv) 3.0 A
I
92. What is the value of current I in the circuit shown in
Fig. 2.187?
2
1
6V
30 V
1
2
V
RL
R
I
Fig. 2.187
(i) 1 A
(ii)
3A
(iii)
6A
(iv) 9 A
100
Network Analysis and Synthesis
93. In the circuit shown, the current through R is
I 10
10 V
I1
5
5
10
(iii) 2.5 A
(iv) 3.33 A
94. Referring to the circuit shown in Fig. 2.189, the current
in the 18- resistor is
13
11
14
18
44 V
5
9
22
Fig. 2.189
(i) 2 A
(ii) 1.5 A
(iii) 1 A
(iv) 0.5 A
95. An ideal ammeter is connected between terminals
A and B of the network shown above. The current
through the ammeter is
3
6
9.6 V
A
6
5
6
(i) 8.58 A (ii) 7.54 A
(iii) 11.66 A (iv) 15 A
99. A lamp rated at 10 W, 50 V is proposed to be used in a
110-V system. The wattage and resistance of the resistor to be connected in series with the lamp should be
(i) 15 watts, 350 ohms
(ii) 10 watts, 250 ohms
(iii) 12 watts, 300 ohms
(iv) 15 watts, 250 ohms
100. Fig. 2.194 shows the waveform of the current passing
through an inductor of 16A
resistance and 2-H induct
tance. The energy absorbed
0
2s
4s
by the inductor in the first
Fig. 2.194
four seconds is
(i) 144 J
(ii) 98 J
(iii) 132 J
(iv) 168J
101. A segment of a circuit is shown in Fig. 2.195. VR 5 V,
VC = 4sin2t. The voltage VL is given by
Q
B
1A
Fig. 2.190
(i) 0.8 A
100 V
Fig. 2.193
Fig. 2.188
(ii) 0.5 A
10
20
5A
5I
R(= 10 )
(i) 0
98. The current I1 through the 5- resistor in the network
shown in Fig. 2.193, is
(ii) 1.6 A
(iii) 0 A
(iv) 3.2 A
96. For the network shown in Fig. 2.191, the current in the
2- resistor would be
VR
2A 5
1F
P
R
VL
2x
VC
iC
10
S
5A
2
10
25 A
Fig. 2.195
Fig. 2.191
(i) 5 A
(ii) 20 A
(iii) 25 A
(iv) 30 A
97. The branch voltages are marked with proper polarity
for the network shown in Fig. 2.192. The value of V5 is
(i) 3 8 cos 2t
(ii) 32 sin 2t
(iii) 16 sin 2t
(iv) 16 cos 2t
102. In the circuit of Fig. 2.196, the magnitudes of VL and VC
are twice that of VR. The inductance of the coil is
VR
V1 = 1V
A
V2 = 2 V B
V4
V5
5
V3
VC
C
C
5 00
V6
L
VL
D
Fig. 2.192
(i) 3 V
(ii) 2 V
Fig. 2.196
(iii) 1 V
(iv) 0 V
(i) 2.14 mH (ii) 5.30 H (iii) 3.18 mH
(iv) 1.32 H
101
Introduction to Circuit-Theory Concepts
103. In Fig. 2.197, the value of the source voltage is
10
Fig. 2.197
(iii) 30 V
(iv) 44 V
104. In Fig. 2.198, Ra, Rb and Rc are 20 , 10
and 10
respectively. The resistances R1, R2 and R3 in of an
equivalent star-connection are
a
R1
Rc
Ra
c
(ii) 5, 1
Vab
1A
Fig. 2.198
(ii) 5, 2.5, 5
(iv) 2.5, 5, 2.5
105. In Fig. 2.199, the value of resistance R in
10
5V
i
b
is
Fig. 2.202
2A
10
(i)
R
3V
(ii) 0 V
(iii) 3 V
Vab b 3
Fig. 2.199
(iii) 30
8A
Fig. 2.203
R
(i) 0.31 A
10
10
100 V
(ii) 1.25 A
(iii) 1.75 A
Fig. 2.200
C
R
(ii) 5.0
(iii) 7.5
R
R
(iv) 10.0
107. In the circuit shown in Fig. 2.201, the current source
I = 1 A, voltage source V = 5 V, R1 = R2 = R3 = 1 ,
L1 = L2 = L3 = 1 H, C1 = C2 = 1 F. The current (in A) through
R3 and the voltage source V respectively will be
(iv) 2.5 A
111. The minimum number of equations required to analyze circuit shown in Fig. 2.204 is
C
(i) 2.5
i
(iv) 40
106. In the Fig. 2.200, the value of R is
4Vab
1
1
5V
(ii) 20
(iv) 5 V
110. In the circuit shown in Fig. 2.203, the value of the current i will be given by
1
(i) 10
(iv) 5, 4
2
b
100 V
(iii) 5, 2
R2
b c
(i) 2.5, 5, 5
(iii) 5, 5, 2.5
C2
108. A 3-V dc supply with an internal resistance of 2- supplies a passive non-linear resistance characterized by
the relation VNL = I2NL. The power dissipated in the nonlinear resistance is
(i) 1.0 W
(ii) 1.5 W
(iii) 2.5 W
(iv) 3.0 W
a
R3
V
R3
109. Assuming ideal elements in the circuit shown below,
the voltage Vab will be
a
Rb
L3
L2
(i) 1, 4
(ii) 24 V
R2
Fig. 2.201
Q
(i) 12 V
C1
I
6
1A
R1
L1
P 2A
6
C
R
Fig. 2.204
(i) 3
(ii) 4
(iii) 6
(iv) 7
102
Network Analysis and Synthesis
112. Twelve 1- resistances are used as edges to form a
cube. The resistance between two diagonally opposite corners of the cube is
(i) 5兾6
(ii) 1
(iii) 6兾5
(iv) 3兾2
113. A two-terminal black box contains one of the R, L, C
elements. The black
box is connected to a 220-V
ac supply. The current through the source is I. When
a capacitance of 0.1F is inserted in series between
the source and the box then the current through the
source is 2I. The element is
(i) a resistance
(ii) an inductance
(iii) a capacitance of 0.5 F
(iv) not readily identifiable from the given data
Answers
1. (iv)
2. (iv)
3. (iv)
4. (iv)
5. (ii)
6. (iii)
7. (ii)
8. (i)
9. (iii)
10. (ii)
11. (ii)
12. (ii)
13. (i)
14. (i)
15. (iii)
16. (ii)
17. (i)
18. (ii)
19. (i)
20. (iv)
21. (iii)
22. (ii)
23. (ii)
24. (iv)
25. (i)
26. (iv)
27. (iv)
28. (iv)
29. (i)
30. (i)
31. (ii)
32. (iii)
33. (ii)
34. (ii)
35. (i)
36. (i)
37. (ii)
38. (i)
39. (ii)
40. (ii)
41. (i)
42. (i)
43. (i)
44. (i)
45. (iii)
46. (iv)
47. (iv)
48. (iii)
49. (ii)
50. (i)
51. (iii)
52. (i)
53. (iii)
54. (iv)
55. (iv)
56. (iii)
57. (iii)
58. (iii)
59. (ii)
60. (ii)
61. (iii)
62. (iv)
63. (iv)
64. (i)
65. (iii)
66. (iii)
67. (iii)
68. (i)
69. (iii)
70. (ii)
71. (ii)
72. (i)
73. (iv)
74. (iii)
75. (iii)
76. (ii)
77. (i)
78. (iii)
79. (iii)
80. (iii)
81. (ii)
82. (iii)
83. (iv)
84. (iv)
85. (ii)
86. (ii)
87. (iv)
88. (iv)
89. (i)
90. (ii)
91. (iii)
92. (iii)
93. (i)
94. (iii)
95. (i)
96. (iii)
97. (iv)
98. (i)
99. (iii)
100. (i)
101. (ii)
102. (iii)
103. (iii)
104. (i)
105. (ii)
106. (iii)
107. (iv)
108. (i)
109. (i)
110. (ii)
111. (ii)
112. (i)
113. (ii)
3
Network Topology
(Graph Theory)
Introduction
The word topology refers to the science of place. In mathematics, topology is a branch of geometry in
which figures are considered perfectly elastic.
Network topology refers to the properties that relate to the geometry of a network (circuit). These
properties remain unchanged even if the circuit is bent into any other shape provided that no parts are
cut and no new connections are made.
In electrical engineering, solution of network analysis problems involves finding the current through
and voltage across different circuit elements. Different laws (like Ohm’s law, Kirchhoff’s laws, etc.) have
been postulated for simplifying the solution method. However, it is sometimes found that the algebraic
equations written by different laws are not independent. On the other hand, the equations formed by
network topology method are all independent.
The network topology method has many other merits and can be listed as follows.
1. The graph theory or network topology deals with those properties of networks which do not
change with the change in the shape of the networks.
2. All the equations (KCL and KVL) formed by graph theory concept are independent equations.
3. The graph theory concept eases the solution method for solving networks with a large number of
nodes and branches.
In this chapter, we will discuss the fundamentals of graph theory (network topology) and their
applications for solving network-analysis problems.
3.1
GRAPH OF A NETWORK
A linear graph (or simply a graph) is defined as a collection of points called nodes, and line segment called
branches, the nodes being joined together by the branches.
104
Network Analysis and Synthesis
6
4
b
a
1
6
b
4
a
5
5
c
c
3
2
2
1
3
d
d
Fig. 3.1 (a) Circuit
Fig 3.1 (b)
Graph of the circuit
While Drawing the Graph of a Given Network
(i) All passive elements between the nodes are represented by lines.
(ii) The independent current sources and voltage sources are represented by their internal impedances
(i.e., current sources by an open circuit and voltage sources by a short circuit) if they are accompanied
by a passive element, viz, a shunt admittance in a current source and a series impedance in a voltage
source.
(iii) If the sources are not accompanied by passive elements, an arbitrary impedance (say resistance R) or
admittance is assumed to accompany the sources and finally, we find the results by letting the impedance R → 0 or R → as the case may be for the current or voltage sources.
3.2
TERMINOLOGY
In order to discuss the more involved methods of circuit analysis, we must define a few basic terms necessary
for a clear, concise description of important circuit features.
R1
a
Node A node is a point in a circuit where
two or more circuit elements join.
i1
v1
Example a, b, c, d, e, f and g
Essential Nodes A node that joins three
or more elements.
c
R2
d
R5
i2
R3
i3
v2
e
i4
i6
R7
I
R6
R4
Example b, c, e and g
Branc A branch is a path that connects
two nodes.
b
f
i5
g
Fig. 3.2 Circuit illustrating terminologies
Example v1, R1, R2, R3, v2, R4, R5, R6, R7 and I
Essential Branch Those paths that connect essential nodes without passing through an essential node.
Example c-a-b, c-d-e, c-f-g, b-e, e-g, b-g (through R7), and b-g (through I)
Loop A loop is a complete path, i.e., its starting at a selected node, tracing a set of connected basic-circuit elements
and returning to the original starting node without passing through any intermediate node more than once.
105
Network Topology (Graph Theory)
Example abedca, abegfca, cdebgfc, etc.
Mesh A mesh is a special type of loop, i.e., it does not contain any other loops within it.
Example abedca, cdegfc, gebg (through R7) and gebg (through I)
Oriented Graph A graph whose branches are oriented is called a directed or oriented graph.
Rank of a Graph The rank of a graph is (n
1) where n is the number of nodes or vertices of the graph.
Planar and Non-Planar Graph A graph is planar if it can be drawn in a plane such that no two branches
intersect at a point which is not a node.
6
4
a
b
2
5
c
3
1
d
Fig. 3.3 (b) Non-planar graph
Fig. 3.3 (a) Planar graph
Fig. 3.3 (c)
Subgraph A subgraph is a subset of the branches and nodes of a graph. The subgraph is said to be proper
if it consists of strictly less than all the branches and nodes of the graph.
Path A path is a particular sub graph where only two branches are incident at every node except the
terminal nodes (i.e., starting and finishing nodes). At the terminal nodes, only one branch is incident.
In the example in the Fig. 3.3 (c), branches 2, 3, and 4, together with all the four nodes, constitute a path. A
graph is connected if there exists a path between any pair of vertices. Otherwise, the graph is disconnected.
3.3
CONCEPT OF A TREE
For a given connected graph of a network, a connected subgraph is known as a tree of the graph if the subgraph has all the nodes of the graph without containing any loop.
R1
R2
1
v1
R4
2
3
R3
is
R5
4
Fig. 3.4 (a) Circuit
Fig. 3.4 (b) Trees and links of the circuit of Fig. 3.4 (a)
106
Network Analysis and Synthesis
Twigs The branches of a tree are called twigs or tree-branches. The number of branches or twigs, in any
selected tree is always one less than the number of nodes, i.e.,
twigs (n 1), where n is the number of nodes of the graph
For the graph shown in Fig. 3.3 (c), twigs
(4
1)
3 twigs. These are shown by solid lines in
Fig.3. 4 (b).
Links and co-tree If a graph for a network is known and a particular tree is specified, the remaining
branches are referred as the links. The collection of links is called a co-tree. So, a co-tree is the complement
of a tree. These are shown by dotted lines in Fig. 3.4 (b).
The branches of a co-tree may or may not be connected, whereas the branches of a tree are always connected.
To Summarize Number of nodes in a graph n
Number of independent voltages n 1
Number of tree-branches n 1
Number of links L (Total number of branches)
b (n 1)
Total number of branches b L (n 1)
(Number of tree-branches)
Properties of a tree
1. In a tree, there exists one and only one path between any pairs of nodes.
2. Every connected graph has at least one tree.
3. A tree contains all the nodes of the graph.
4. There is no closed path in a tree and hence, a tree is circuitless.
5. The rank of a tree is (n 1).
Example 3.1 For the network shown in Fig. 3.5, draw the graph and show some possible trees.
Solution Before drawing the graph we first label the nodes and
branches of the network as shown in Fig. 3.6 (a). Since the voltage
source is accompanied by a series resistance and the current source
by a parallel resistance, while drawing the graph they will be opencircuited and short-circuited, respectively.
(5)
(5)
(4)
L1
(2)
I
C
B
(3)
(2)
R3
(4)
(1)
Fig. 3.6 (a)
I
Circuit of Example 3.1
A
(3)
(1)
V
Fig. 3.5
C1
B
R2
A
L1
V
The graph of the network is shown in Fig. 3.6 (b) and some trees
are shown in Fig. 3.6 (c) to Fig. 3.6 (e).
R1
R2 C1
R1
C
Fig. 3.6 (b) Graph of the
circuit of Fig. 3.5
R3
107
Network Topology (Graph Theory)
The twigs are shown by solid lines and the links by dashed lines.
(5)
(5)
A
B
A
B
B
(3)
(4)
(2)
(1)
(4)
(1)
(4)
(2)
(2)
C
C
C
Fig. 3.6 (c)
3.4
A
(3)
(3)
(1)
(5)
Fig. 3.6 (d)
Fig. 3.6 (e)
INCIDENCE MATRIX [Aa]
The incidence matrix symbolically describes a network. It also facilitates the testing and identification of the
independent variables. The incidence matrix is a matrix which represents a graph uniquely.
For a given graph with ‘n’ nodes and ‘b’ branches, the complete incidence matrix Aa is a rectangular matrix
of order n b, whose elements have the following values:
Number of columns in [A] Number of branches b
Number of rows in [A] Number of nodes n
Aij 1, if the branch j is associated with the node i and oriented away from the node j.
1, if the branch j is associated with the node i and oriented towards the node j.
0, if the branch j is not associated with the node i.
This matrix tells us which branches are incident at which nodes and what the orientations relative to the
nodes are.
Example 3.2 Draw the graph of the network shown in Fig. 3.7 (a) and write the incidence matrix.
a
4
b
5
c
a
4
2
6
1
2
3
d
Fig. 3.7 (a) Network
b
5
c
3
1
6
d
Fig. 3.7 (b) Graph of the network
108
Network Analysis and Synthesis
Solution
The graph of the network is shown in Fig. 3.7 (b). The incidence matrix Aa is given as
Branches
Aa = Nodes
Reference node
1 2
3
4
5
6
a
1
0
0
1 0
0
Reduced
b
0
1
0
1
1 0
incidence
c
0
0
1
0
1
1
matrix AI
d
1
1
1
0
0
1
3.4.1 Incidence Matrix and KCL
For the graph shown in Fig. 3.8, Kirchhoff’s current law for the branch currents (i1, i2,
i1
i2
i6
0
i1
i2
i4
i3
i3
i5
i5
i4
i6
0
0
0
, i6) gives the equations
1
(1)
(6)
In matrix form, these equations can be represented as
(2)
(3)
2
3
(5)
⎡ 1 1 0 0 0
⎢
⎢ −1 0 1 0 −1
⎢ 0 −1 −1 1 0
⎢
⎣ 0 0 0 −1 1
Or,
where,
⎡ i1 ⎤
⎢ ⎥
1 ⎤ ⎢i2 ⎥
⎥⎢ ⎥
0 ⎥ ⎢ i3 ⎥
=0
0 ⎥ ⎢i4 ⎥
⎥⎢ ⎥
−1⎦ ⎢i5 ⎥
⎢ ⎥
⎣i6 ⎦
(4)
4
Fig. 3.8 Graph illustrating
incidence matrix and KCL
Aa I b = 0
Aa is the complete incidence matrix of the graph.
Reduced Incidence Matrix [A] The matrix obtained from Aa by eliminating one of the rows is called the
reduced incidence matrix. In other words, suppression of the datum node (reference node) from the incidence
matrix results in a reduced incidence matrix.
3.4.2 Incidence Matrix and KVL
For the graph shown in Fig. 3.8, the branch voltages (vb1, vb2,
voltages (vn1, vn2, vn3, vn4) as
vb1
(vn1 – vn2), vb2
(vn1 – vn3), vb3
(vn2 – vn3), vb4
vb6) can be represented in terms of the node
(vn3 – vn4),
vb5
( vn1
vn4), vb6
(vn1 – vn4)
109
Network Topology (Graph Theory)
Thus, the Kirchhoff’s voltage law in matrix form can be written as
⎡ vb1 ⎤
⎡1 −1 0 0 ⎤
⎢
⎥ ⎡ vn1 ⎤ ⎢ v ⎥
⎢1 0 −1 0 ⎥ ⎢ ⎥ ⎢ b 2 ⎥
⎢0 1 −1 0 ⎥ ⎢ vn2 ⎥ ⎢ vb 3 ⎥
⎢
⎥⎢ ⎥ = ⎢ ⎥
0
0
1
−
1
⎢
⎥ ⎢ vn3 ⎥ ⎢ vb 4 ⎥
⎢0 −1 0 1 ⎥ ⎢ ⎥ ⎢ v ⎥
⎢
⎥ ⎢⎣ vn4 ⎥⎦ ⎢ b 5 ⎥
⎢ ⎥
⎢⎣1 0 0 −1⎥⎦
⎢⎣ vb 6 ⎦⎥
AaT Vn = Vb
Or,
Properties of complete incidence matrix
(i) The sum of the entries in any column is zero.
(ii) The determinant of the incidence matrix of a closed loop is zero.
(iii) The rank of the incidence matrix of a connected graph is (n 1).
3.4.3 Number of Possible Trees of a Graph
The number of possible trees of a graph, det {[A] [A]T}
where, A is the reduced incidence matrix obtained by eliminating any one row of the complete incidence
matrix Aa, and [A]T is the transpose of the matrix [A].
Example 3.3 For the graph shown in Fig. 3.8, find the number of possible trees.
Solution The complete incidence matrix is
So, the reduced incidence matrix is
⎡ 1 1 0 0 0 1⎤
⎢
⎥
−1 0 1 0 −1 0 ⎥
Aa = ⎢
⎢ 0 −1 −1 1 0 0 ⎥
⎢
⎥
⎣ 0 0 0 −1 1 −1⎦
⎡ 1 1 0 0 0 1⎤
⎢
⎥
A = ⎢ −1 0 1 0 −1 0 ⎥
⎢ 0 −1 −1 1 0 0 ⎥
⎣
⎦
Thus, the number of possible trees of the graph of Fig. 3.8
⎧
⎡1 −1 0 ⎤ ⎫
⎪
⎢
⎥⎪
⎪ ⎡ 1 1 0 0 0 1 ⎤ ⎢1 0 −1⎥ ⎪ 3 −1 −1
⎪⎪ ⎢
⎥ ⎢0 1 −1⎥ ⎪⎪
= det ⎨ ⎢ −1 0 1 0 −1 0 ⎥ ⎢
⎥ ⎬ = −1 3 −1 = 16
⎪ ⎢ 0 −1 −1 1 0 0 ⎥ ⎢0 0 1 ⎥ ⎪ −1 −1 3
⎦ ⎢0 −1 0 ⎥ ⎪
⎪⎣
⎢
⎥⎪
⎪
⎢⎣1 0 0 ⎥⎦ ⎪⎭
⎪⎩
110
Network Analysis and Synthesis
3.5
TIE-SET MATRIX AND LOOP CURRENTS
Tie-Set A tie-set is a set of branches contained in a loop such that each loop contains one link or chord and
the remainder are tree branches.
Consider the graph and the tree as shown in Fig. 3.9. This selected tree will result in three fundamental
loops as we connect each link, in turn to the tree.
1
1
2
2
5
3
3
4
FL 1
4
6
2
Fig. 3.9 (b) Tree of the graph Fig. 3.9 (c) Loop-1
Fig. 3.9 (a) Graph
3
2
FL 3
FL 2
5
3
4
Fig. 3.9 (d) Loop-2
4
6
Fig. 3.9 (e) Loop-3
Fundamental Loop 1 (FL1): Connecting link 1 to the tree
Fundamental Loop 2 (FL2): Connecting link 5 to the tree
Fundamental Loop 3 (FL3): Connecting link 6 to the tree
These sets of branches (1, 2, 3), (2, 4, 5) and (3, 4, 6) form three tie-sets.
3.5.1 Tie-Set Matrix or Loop Incidence Matrix or Circuit Matrix (Ba)
For a given graph having ‘n’ nodes and ‘b’ branches, the tie-set matrix is a rectangular matrix with ‘b’ columns and as many rows as there are loops. Its elements have the following values:
Bij 1, if the branch j is in the loop i and their orientations coincide (i.e., the loop current and branch current flows in the same direction)
1, if the branch j is in the loop i and their orientations do not coincide
0, if the branch j is not in the loop i
Example 3.4 For the graph shown in Fig. 3.10 (a), select a tree, identify the
tie-sets and write the tie-set matrix.
Solution The tree is shown in Fig. 3.10 (b) and three tie-sets are identified and
shown in Fig. 3.10 (b). The tie-set matrix is written as follows. The entries in the
tie-set schedule are given as 1 or 1 depending on whether the branch current
is in the same direction as the link current or not. If the branch current does not
depend on the link current then the entry is zero
a
4
b 5
c
2
1
3
d
Fig. 3.10 (a)
Graph
6
111
Network Topology (Graph Theory)
Links ( j) 1
Tie-set Matrix, Ba
4
5
6
j4
Branches no (i)
2 3 4 5 6
1
1
0
1
0
0
0
1
1 0
1
0
0
0
1 0
0
1
j5
b
a
c
i4 2
i5
i6
1
j6
3
d
Fig. 3.10 (b)
Formation of loops
3.5.2 Tie-Set Matrix and KVL
For the graph shown in Fig. 3.9 (a) and three loops shown in Fig. 3.9 (c), (d) and (e), three fundamental mesh
KVL equations can be written as follows:
For Fundamental Loop 1 (FL1): vb1 vb3 vb2 0
For Fundamental Loop 2 (FL2): vb2 vb4 vb5 0
For Fundamental Loop 3 (FL3): vb3 vb6 vb4 0
These equations in matrix form is written as
⎡ vb1 ⎤
⎡1 1 −1 0 0 0 ⎤ ⎢ ⎥
⎢
⎥ ⎢ vb 2 ⎥
⎢0 −1 0 −1 −1 0 ⎥ ⎢ v ⎥ = 0
⎢0 0 1 1 0 1 ⎥ ⎢ b 3 ⎥
⎣
⎦⎢ ⎥
⎣ vb44 ⎦
BaVb = 0
Or,
3.5.3 Tie-Set Matrix and KCL
For the graph shown in Fig. 3.9 (a) and three loops shown in Fig. 3.9 (c), (d) and (e), the branch currents
(ib1, ib2, ,ib6) can be represented in terms of the loop currents (IL1, IL2, IL3) as
ib1 IL1,
ib2 (IL1 – IL2),
ib3 (−IL1 IL3),
In matrix form, these equations can be written as
⎡ ib1 ⎤ ⎡ 1 0
⎢ ⎥ ⎢
⎢ib 2 ⎥ ⎢ 1 −1
⎢i ⎥ ⎢
⎢ b 3 ⎥ = ⎢ −1 0
⎢i ⎥ ⎢ 0 −1
⎢ b4 ⎥ ⎢
⎢ ib 5 ⎥ ⎢ 0 1
⎢ ⎥ ⎢ 0 0
⎢⎣ ib 6 ⎥⎦ ⎣
Or,
I b = Ba T I L
ib4
(−IL2
0⎤
⎥
0⎥
⎡ IL1 ⎤
1⎥ ⎢ ⎥
⎥⎢I ⎥
1⎥ ⎢ L 2 ⎥
I
0⎥ ⎣ L 3 ⎦
⎥
1 ⎥⎦
IL3),
ib5
IL2,
ib6
IL3
112
Network Analysis and Synthesis
3.6
CUT-SET MATRIX AND NODE-PAIR POTENTIAL
Cut-set A cut-set is a minimum set of elements that when cut, or removed, separates the graph into two
groups of nodes. A cut-set is a minimum set of branches of a connected graph, such that the removal of
these branches from the graph reduces the rank of the graph by one.
In other words, for a given connected graph (G), a set of branches (C) is defined as a cut-set if and only if
(i) the removal of all the branches of C results in an unconnected graph
(ii) the removal of all but one of the branches of C leaves the graph still connected
Example Consider the graph shown in Fig. 3.11 (a). The rank of the graph is 3.
The removal of branches 1 and 3 reduces the graph into two connected subgraphs as shown in Fig. 3.11 (b).
The rank of the graph of Fig. 3.11 (a) (4 1) 3
The rank of the graph of Fig. 3.11 (b) addition of the ranks of the subgraphs (1 1) 2
So, branches [1, 3] may be a cut-set.
(5)
(5)
(2)
(2)
(2)
(3)
(1)
(4)
Fig. 3.11 (a) Graph
(4)
Fig. 3.11 (b) Subgraphs
with removal of 1 and 3
(4)
Fig. 3.11 (c) Subgraphs
with removal of 1, 3 and 5
Also, removal of the branches 1, 3 and 5 reduces the graph into two connected subgraphs as shown in
Fig. 3.11 (c) and the rank becomes 2. So, [1, 3, 5] may also be a cut-set.
As a cut-set is the minimum set of branches and [1, 3] is a subset of [1, 3, 5], so, [1, 3] is the cut-set, and
[1, 3, 5] is not a cut-set.
1
Fundamental Cut-Set A fundamental cut-set (FCS) is a cut-set
that cuts or contains one and only one tree branch. Therefore, for
a given tree, the number of fundamental cut-sets will be equal to the
number of twigs.
(1)
(6)
The Procedure for Finding the Fundamental Cut-Sets
1. First, select a tree of the given graph.
2. Focus on a tree branch (bk).
3. Check whether removing this tree branch (bk) from the tree disconnects the tree into two separate parts.
4. All the links which go from one part of this disconnected tree
to the other, together with the tree branch (bk) forms a fundamental cut-set.
C1
C2
(2)
(bk)
3
2
(3)
(5)
(4)
4
C3
Fig. 3.12 Graph illustrating
fundamental cut-set
113
Network Topology (Graph Theory)
Following this procedure, the fundamental cut-sets for the above graphs will be
f-cut-set – 1: [1, 2, 6]
f-cut-set – 2: [2, 3, 5, 6]
f-cut-set – 3: [4, 5, 6]
Properties of a Cut-Set
1. A cut-set divides the set of nodes into two subsets.
2. Each fundamental cut-set contains one tree-branch, the remaining elements being links.
3. Each branch of the cut-set has one of its terminals incident at a node in one subset and its other terminal
at a node in the other subset.
4. A cut-set is oriented by selecting an orientation from one of the two parts to the other. Generally, the
direction of a cut-set is chosen same as the direction of the tree branch.
3.6.1
Cut-Set Matrix (QC)
For a given graph, a cut-set matrix (QC) is defined as a rectangular matrix whose rows correspond to cut-sets
and columns correspond to the branches of the graph. Its elements have the following values:
Qij 1, if the branch j is in the cut-set i and the orientations coincide
1, if the branch j is in the cut-set i and the orientations do not coincide
0, if the branch j is not in the cut-set i
Example 3.5 For the graph shown in Fig. 3.12, write the fundamental cut-set matrix.
Solution The fundamental cut-sets have been identified as
f-cut-set – 1: [1, 2, 6]
f-cut-set – 2: [2, 3, 5, 6]
f-cut-set – 3: [4, 5, 6]
So, the cut-set matrix is written as
Branch no.
2 3 4 5
6
1
1 0
0
0
1
0
1 1
0
1
1
0
0 0
1
1
1
f-cut-sets 1
1
2
3
3.6.2 Cut-Set Matrix and KVL
By cut-set schedule, the branch voltages can be expressed in terms of the
tree-branch voltages.
A cut-set consists of one and only one branch of the tree together with
any links which must be cut to divide the network into two parts. A set of
fundamental cut-sets includes those cut-sets which are obtained by applying a cut-set division for each of the branches of the network tree.
Consider the following graph shown in Fig. 3.13.
FCS-2
1 6
FCS-1
2
7
5
FCS-3
8
3
4
FCS-4
Fig. 3.13 (a) Graph
114
Network Analysis and Synthesis
Applying cut-sets at nodes a, b, c, d, which are the fundamental cut-sets (FCS),
we can write the cut-set schedule as
b
6
FCS-1→
FCS-2→
FCS-3→
FCS-4→
a
b
c
d
1
1
2
0
3
0
4
1
5
1
6
0
7
0
8
0
1
1
0
0
0
1
0
0
0
0
1
0
1
1
0
1
0
0
0
0
1
0
0
1
a
7
5
c
8
d
Fig. 3.13 (b) Tree
The tree-branch voltages are [vt 5, vt 6, vt 7, vt 8], the branch voltages are [Vb1, Vb2, … Vb8] and the relationship
between tree-branch voltages and branch voltages are
Vb1
−vt5
vt6
Vb5
vt5
Vb2
−vt6
vt7
Vb6
vt6
Vb3
vt7 − vt8
Vb7
vt7
Vb4
vt5 − vt8
Vb8
vt8
The above equations can be related by using the cut-set schedule as
⎡Vb1 ⎤
⎢ ⎥ ⎡ −1 1
⎢Vb 2 ⎥ ⎢ 0 −1
⎢V ⎥ ⎢
⎢ b3 ⎥ ⎢ 0 0
⎢V ⎥ ⎢ 1 0
⎢ b4 ⎥ = ⎢
⎢Vb 5 ⎥ ⎢ 1 0
⎢ ⎥ ⎢
⎢Vb 6 ⎥ ⎢ 0 1
⎢V ⎥ ⎢ 0 0
⎢ b7 ⎥ ⎢
⎢⎣Vb8 ⎥⎦ ⎢⎣ 0 0
Or,
0
1
1
0
0
0
1
0
0⎤
⎥
0⎥
−1⎥ ⎡ vt 5 ⎤
⎥⎢ ⎥
−1⎥ ⎢ vt 6 ⎥
0 ⎥ ⎢⎢ vt 7 ⎥⎥
⎥
0 ⎥ ⎢⎣ vt 8 ⎥⎦
⎥
0⎥
1 ⎥⎦
Vb = QC T Vt
3.6.3 Cut-Set Matrix and KCL
For the graph of Fig. 3.13, writing Kirchhoff ’s current laws for the nodes, the branch currents can be expressed as
Node a:
ib1
ib4
ib5
0
Node b:
ib1
ib2
ib6
0
Node c:
ib2
ib3
ib7
0
Node d:
ib3
ib4
ib8
0
115
Network Topology (Graph Theory)
In matrix form they can be written as
⎡ −1 0 0 1
⎢
⎢ 1 −1 0 0
⎢ 0 1 1 0
⎢
⎣ 0 0 −1 −1
1 0 0
0 1 0
0 0 1
0 0 0
⎡ ib1 ⎤
⎢ ⎥
⎢ib 2 ⎥
⎢ ⎥
0 ⎤ ⎢ ib 3 ⎥
⎥
0 ⎥ ⎢ib 4 ⎥
⎢ ⎥=0
0 ⎥ ⎢ ib 5 ⎥
⎥⎢ ⎥
1 ⎦ ⎢ ib 6 ⎥
⎢i ⎥
⎢ b7 ⎥
⎢⎣ ib8 ⎥⎦
QC I b = 0
Or,
There is a cut-set matrix for a given tree. If a graph contains more than one tree, there will be as many
numbers of cut-set matrices as the number of trees of the graph.
To summarize, KVL and KCL equations in three matrix forms are given below.
Matrix
Incidence matrix (Aa)
Aa Ib
Tie-set matrix (Ba)
Ib
Cut-set matrix (QC)
3.7
KCL
0
BaT IL
QC Ib
0
KVL
Vb
AaT Vn
Ba Vb
Vb
QC
0
T
Vt
FORMULATION OF NETWORK EQUILIBRIUM EQUATIONS
The network equilibrium equations are a set of equations that completely and uniquely determine the state
of a network at any instant of time. These equations are written in terms of suitably chosen current variables
or voltage variables.
These equations will be unique if the number of independent variables are equal to the number of independent equations.
Number of Independent variables or equations b (n 1); for loop method of analysis
(n 1); for node method of analysis
The equations for a network can be formed in either of the two methods as given below:
1. Through a set of voltage law equations in which the currents are the independent variables (loop-basis
method)
2. Through a set of current law equations in which the node-pair voltages are the independent variables
(node-basis method)
3.7.1 Formulation of Network Equations on Loop Basis
Steps
1. Draw the directed graph of the network selecting the direction of assumed current flow to coincide for
current sources.
2. Select a tree of the graph.
116
Network Analysis and Synthesis
3. Place all voltage sources in the tree and all current sources in the co-tree.
4. Place all control-voltage branches for voltage-controlled dependent sources in the tree and all controlcurrent branches for current-controlled dependent sources in the co-tree, if possible.
5. Add one link to the tree, creating a fundamental loop, and write a KVL equation for this fundamental
loop (FL). Repeat for each additional link until L ( b n 1) mesh equations are obtained in the
form Ba Vb 0.
6. The current sources in the co-tree, if present, will provide the constraint equations.
7. The KCL equations are obtained by representing the branch currents in terms of loop currents in the
form Ib BaT IL.
8. For each branch, the relationship between the voltage and current is obtained from Ohm’s law (V ⴝ RI).
9. Finally, the equilibrium equations are obtained in terms of loop currents by suitable substitution of the
equations obtained in steps 5 to 8.
3.7.2 Formulation of Network Equations on Node Basis
Steps
1. Draw a directed graph of the circuit under considerations, selecting the directions of assumed current
flow to coincide for current sources.
2. Select the tree of the graph so that current sources are in the co-tree and the voltage sources are
within the tree, if possible. Also, if possible, select the tree so that at least two branches of the tree are
incident at the reference node.
3. Identify (n 1) fundamental cut-sets (FCS) and draw the FCS lines.
4. Write the (n 1) FCS KCL equations in the form Aa Ib 0 or QC Ib ⴝ 0.
5. Obtain each of the branch currents in terms of node voltages in the form Vb
AaT Vn or, Vbⴝ
T
QC ×Vt.
6. For each branch, the relationship between the voltage and current is obtained from Ohm’s law
(V ⴝ RI).
7. Substitute the equations of the step 6 into the KVL equations of the step 5 and finally into the KCL
equations of the step 4, thus obtaining the (n 1) independent node voltage equations.
3.8
GENERALIZED EQUATIONS IN MATRIX FORMS FOR CIRCUITS
HAVING SOURCES
A general branch consisting of a voltage source Vs and a current source Is
is shown in Fig. 3.14.
Here, the branch current is (Ib Is) and the branch voltage is (Vb
Without sources, the KCL and KVL equations are
Aa Ib 0
(3.1)
and
Ib ⴝ BaT IL
(3.2)
QC Ib ⴝ 0
(3.3)
Vb ⴝ AaT Vn
(3.4)
Ba Vb ⴝ 0
(3.5)
T
C
(3.6)
Vb ⴝ Q
Vt
Ib
Vs
Zb
Vs)
Is
KCL
KVL
Fig. 3.14
117
Network Topology (Graph Theory)
With the sources, the KCL and KVL equations are modified as
Aa Ib Aa Is 0
Ib
Is
Ba IL
Qc Ib
and
Vb
Qc Is
Ba Vb
(3.8)
0
(3.9)
Aa Vn
(3.10)
T
Vs
Vb
(3.7)
T
Ba Vs
0
(3.11)
Q Vt
(3.12)
T
c
Vs
The branch voltage–current relations for the passive network elements are written in matrix form as
Vb Zb Ib
(3.13)
and
where,
Ib
Yb Vb
(3.14)
Zb is the branch impedance matrix and Yb is the branch admittance matrix, both of the order b × b.
On the basis of these equations, the general equations can be written in terms of three matrices
as follows.
Node Equations From Eq. (3.7),
Aa Is
Or,
T
Aa Yb Aa Vn
Aa Ib
Aa Yb Vb
Aa Yb Vs
Aa Is
Aa Yb (AaT Vn
Vs)
{by Eq. (3.10)}
Aa [Yb Vs – Is]
YVn =Aa [Yb Vs − I s ]
Or,
In case of node analysis, one node is taken as the datum node and the potential of that node is zero. Consequently, the complete incident matrix becomes the reduced incidence matrix. Thus, the node equations become
Y Vn =A [Yb Vs − I s ]
where, Y ⴝ AYb AT is called the nodal admittance matrix of the order of (n – 1) (n – 1). The above equation represents a set of (n – 1) number of equations, known as node equations.
Mesh Equations
From Eq. (3.11), Ba Vs
Ba Vb
Ba Zb Ib
Or,
Ba Zb BaT IL Ba [Zb Is Vs]
Or,
Ba Zb (BaT IL
Is)
{by Eq. (3.8)}
Z I L =Ba [ Z b I s −Vs ]
where, Z Ba Zb BaT is the loop-impedance matrix of the order of (b n 1) (b n 1). The above
equation represents a set of (b n 1) number of equations, known as mesh or loop equations.
Cut-Set Equations From Eq. (3.8),
Qc Is
Or,
Qc Yb QcT Vt
Qc Ib
Qc [Yb Vs
Qc Yb Vb
Is]
Qc Yb (QcT Vt
Vs)
{by Eq. (3.12)}
118
Network Analysis and Synthesis
Or,
where, Yc
(n
YcVt =Qc [YbVs − I s ]
Qc Yb QcT is the cut-set admittance matrix of the order of (n
1)
(n
1) and the set of
1) equations represented by the above equation is known as cut-set equations.
Solution of Equilibrium Equations
There are two methods of solving equilibrium equations:
Elimination Method By eliminating variables until an equation with a single variable is achieved, and
then by the method of substitution.
Determinant Method By the method known as Cramer’s rule.
Solved Problems
Problem 3.1 Draw the graph of the network shown in Fig. 3.15 (a)
7
3
2
1
5
4
6
Fig. 3.15 (a)
Solution The graph of the network is shown below.
1
(2)
2
(7)
3
(5)
(3)
(1)
4
(4)
(6)
5
Fig. 315 (b)
Problem 3.2 From Fig. 3.16, make the graph and find one tree. How many mesh currents are required for
solving the network? Find the number of possible trees.
Fig. 3.16
119
Network Topology (Graph Theory)
Solution The graph of the network is shown below. One tree of the graph is shown.
(2)
(1)
(2)
2
(5)
3
1
(1)
4
(3)
2
(7)
(9)
(5)
3
1
4
(3)
(6)
(7)
(6)
(4)
(4)
7
(8)
(9)
5
5
6
7
(10)
Fig. 3.17 (a) Graph of the network
(8)
6
(10)
Fig. 3.17 (b) Tree of the graph
The complete incidence matrix is obtained as
Branches
Aa
Nodes
1
2
3
4
5
6
7
8
9
10
1
1
0
0
0
0
0
0
0
1
0
2
1
1
1
0
0
0
0
0
0
0
3
0
1
1
1
1
0
0
0
0
0
4
0
0
0
0
1
1
1
0
0
0
5
0
0
0
0
0
1
0
0
0
1
6
0
0
0
0
0
0
1
1
0
0
7
0
0
0
1
0
0
0
1
1
1
The reduced incidence matrix becomes
Branches
A
Nodes
1
2
3
4
5
6
7
8
9
10
1
1
0
0
0
0
0
0
0
1
0
2
1
1
1
0
0
0
0
0
0
0
3
0
1
1
1
1
0
0
0
0
0
4
0
0
0
0
1
1
1
0
0
0
5
0
0
0
0
0
1
0
0
0
1
6
0
0
0
0
0
0
1
1
0
0
120
Network Analysis and Synthesis
Hence the number of possible trees is
⎧
⎪
⎪
⎪⎡ 1 0 0
⎪⎢
⎪ ⎢ −1 1 1
⎪⎪ ⎢ 0 −1 −1
n = det ⎨ ⎢
⎪⎢ 0 0 0
⎪⎢ 0 0 0
⎪⎢
⎪ ⎢⎣ 0 0 0
⎪
⎪
⎪⎩
0
0
0
0 0 0
1 1 0
0 −1 1
0 0 −1
0
0
0 0 −1
0 0
0 0
0
0
1 0
0 0
0 −1 1
0
0
0
⎡ 1 −1 0 0 0 0 ⎤ ⎫
⎢
⎥⎪
⎢ 0 1 −1 0 0 0 ⎥ ⎪
0 ⎤ ⎢ 0 1 −1 0 0 0 ⎥ ⎪
⎥⎪
⎥⎢
0⎥ ⎢ 0 0 1 0 0 0 ⎥⎪
0 ⎥ ⎢ 0 0 1 −1 0 0 ⎥ ⎪⎪
⎥⎬
⎥⎢
0 ⎥ ⎢ 0 0 0 1 −1 0 ⎥ ⎪
⎢
⎥
1 ⎥ ⎢ 0 0 0 1 0 −1⎥ ⎪
⎥
⎪
0 ⎥⎦ ⎢ 0 0 0 0 0 1 ⎥ ⎪
⎢
⎥⎪
⎢ −1 0 0 0 0 0 ⎥ ⎪
⎢⎣ 0 0 0 0 1 0 ⎥⎦ ⎪
⎭
⎡ 2 −1 0 0 0 0 ⎤
⎢
⎥
⎢ −1 3 −2 0 0 0 ⎥
⎢ 0 −2 4 0 0 0 ⎥
= det ⎢
⎥ ⇒ n = 12
⎢ 0 0 −1 3 −1 −1⎥
⎢ 0 0 0 −1 2 0 ⎥
⎢
⎥
0 −1 0 2 ⎥⎦
⎢⎣ 0 0
Problem 3.3 Branch current and loop current relations are expressed in matrix form as,
⎡ i1 ⎤
⎢ ⎥ ⎡ 1 0 0 −1⎤
⎢i2 ⎥ ⎢ 0 1 0 −1⎥
⎥
⎢i ⎥ ⎢
⎡ ⎤
⎢ 3 ⎥ ⎢ 0 1 1 0 ⎥ ⎢ I1 ⎥
⎢
⎥
⎢i ⎥
0 1 1 0 ⎥ ⎢ I2 ⎥
⎢ 4⎥=⎢
⎢
1 −1 0 0 ⎥ ⎢⎢ I 3 ⎥⎥
⎢ i5 ⎥
⎢
⎥
⎢ ⎥
⎢ ⎥
⎢ i6 ⎥ ⎢ 0 0 −1 0 ⎥ ⎣ I 4 ⎦
⎢
⎥
⎢ i ⎥ −1 0 0 0
⎥
⎢ 7⎥ ⎢
⎢⎣ i8 ⎥⎦ ⎢⎣ 0 0 0 1 ⎥⎦
Draw the oriented graph.
Solution We know that, [Ib]
[Ba]T [IL]. So, the tie-set matrix, here, is
121
Network Topology (Graph Theory)
Ba
Branches
Loop or Link
Currents
1
2
3
4
5
6
7
8
1
1
0
0
0
1
0
1
0
2
0
1
1
1
1
0
0
0
3
0
0
1
1
0
1
0
0
4
1
1
0
0
0
0
0
1
So, the graph consists of four loops and eight branches. Loop1 consists of branches 1, 5 and 7. The orientations are given following the sign 1 or 1. Following the procedure, the complete oriented graph is
shown below.
(8)
I4
(1)
(2)
I1
(7)
(5)
(6)
I2
I3
(3)
(4)
Fig. 3.18
Problem 3.4 The fundamental cut-set matrix is given as
Twigs
Links
1
2
3
4
5
6
7
1
0
0
0
1
0
0
0
1
0
0
1
0
1
0
0
1
0
0
1
1
0
0
0
1
0
1
0
Draw the oriented graph of the network.
Solution The graph has seven branches and three fundamental cut-sets:
Cut-set-1: [1, 5]
Cut-set-2: [2, 5, 7]
Cut-set-3: [3, 6, 7]
Cut-set-4: [4, 6]
122
Network Analysis and Synthesis
So, the oriented graph is as shown in Fig. 3.19 (a), (b), (c).
C1
i3
(2)
C4
(6)
(1)
(3)
(4)
(3)
(4)
(3)
C2
(4)
(6)
(5)
(5)
(5)
(1)
(2)
(1)
i1
(6)
i2
(2)
(7)
C3
Fig. 3.19 (b)
Fig. 3.19 (a)
Fig. 3.19 (c)
Problem 3.5 Write the complete incidence matrix for the graph shown in Fig. 3.20 (a).
2
1
4
3
7
6
5
Fig. 3.20 (a)
Solution We first label the nodes as shown in Fig. 3.20 (b)
A
2
1
B
C
4
3
7
6
D
5
E
Fig. 3.20 (b)
The complete incidence matrix is given as
Aa
A
1
1
2
1
3
0
4
1
5
0
6
0
7
0
B
C
D
E
1
0
0
0
0
1
0
0
0
1
1
0
0
0
1
0
0
1
0
1
1
0
0
1
1
0
1
0
123
Network Topology (Graph Theory)
Problem 3.6 Write down the incidence matrix and cut-set matrices for the network shown.
Solution The graph and a suitable tree for the network are shown in Fig.3.21 (b).
A
C3
5
1
5
2
4
5
4
B
5
4
3
C
4
10 V
6
C2
C1
Fig. 3.21 (a)
D
Fig. 3.21 (b)
The complete incidence matrix is given as
Aa
1
2
3
4
5
6
A
1
1
1
0
0
0
B
1
0
0
1
0
1
C
0
1
0
1
1
0
D
0
0
1
0
1
1
The fundamental cut-sets are identified as
f-cutset-1: [1, 4, 6]
f-cutset-2: [3, 5, 6]
f-cutset-3: [1, 2, 3]
The fundamental cutset matrix is given as
Q
Problem 3.7
1
2
3
4
5
6
C1
1
0
0
1
0
1
C2
0
0
1
0
1
1
C3
1
1
1
0
0
0
For the network shown in Fig. 3.22 (a), give fundamental cut-set matrix and hence find KCL
equations.
1
1A
Fig.3.22 (a)
2
2
1
124
Network Analysis and Synthesis
Solution The graph and one tree are shown for the network.
The fundamental cutsets are identified as
f-cutset-1: [1, 2]
f-cutset-2: [2, 3, 4]
(2)
C1
C2
1
2
3
4
1
1
0
0
0
(3)
(1)
The fundamental cut-set matrix is given as
Qa
B
A
1
1
(4)
C
Fig. 3.22 (b)
1
The KCL equations in terms of cut-set matrix is given as
[Q] [Yb][QT][Vt]
[Q] [IS]
Here,
⎡2
⎢
⎡1 1 0 0 ⎤ ⎢ 0
⎡⎣Q ⎤⎦ ⎡⎣Yb ⎤⎦ ⎡⎣Q T ⎤⎦ = ⎢
⎥⎢
⎣0 −1 1 1 ⎦ 0
⎢
⎣0
⎡1 0 ⎤
0 0 0 ⎤ ⎡1 0 ⎤
⎥⎢
⎥
⎢
⎥
1 0 0 ⎥ ⎢1 −1⎥ ⎡ 2 1 0 0 ⎤ ⎢1 −1⎥
=⎢
⎥
0 2 0 ⎥ ⎢0 1 ⎥ ⎣ 0 −1 2 1 ⎦ ⎢0 1 ⎥
⎥⎢
⎥
⎢
⎥
0 0 1 ⎦ ⎣0 1 ⎦
⎣0 1 ⎦
⎡ 3 −1⎤
=⎢
⎥
⎣ −1 4 ⎦
⎡ −1⎤
⎢ ⎥
⎡1 1 0 0 ⎤ ⎢ 0 ⎥ ⎡1 ⎤
− ⎡⎣Q ⎤⎦ ⎡⎣ I S ⎤⎦ = − ⎢
=⎢ ⎥
⎥
⎣0 −1 1 1 ⎦ ⎢ 0 ⎥ ⎣0 ⎦
⎢ ⎥
⎣ 0⎦
Thus, the KCL equations are
⎡ 3 −1⎤ ⎡Vt 1 ⎤ ⎡1 ⎤
⎢
⎥⎢ ⎥ = ⎢ ⎥
⎣ −1 4 ⎦ ⎢⎣Vt 3 ⎦⎥ ⎣0 ⎦
Problem 3.8 For the network shown in Fig. 3.23 (a), draw the oriented graph, select a suitable tree and obtain the fundamental cut-set
matrix. Determine the node equations and find v.
Solution The oriented graph of the network is shown in Fig. 3.23 (b).
Since we have to find v, we take the branch (2) in the twig and a possible tree is selected.
The fundamental cutsets are identified as
f-cut-set-1: [1, 2, 3]
f-cut-set-2: [3, 4]
2v
2
2
2V
Fig. 3.23 (a)
v
2
2
125
Network Topology (Graph Theory)
The fundamental cut-set matrix is given as
Qa
A
1
2
3
4
C1
1
1
1
0
C2
0
0
1
1
The node equations are given as
[Q][Yb][QT][Vt ]
(4)
C3
B
(2)
(3)
(1)
C
C1
Fig. 3.23 (b)
[Q]
{[Yb][Vs]
[IS]}
Here,
⎡1
⎢ 2
⎢
⎡ −1 1 1 0 ⎤ ⎢ 0
T
⎡⎣Q ⎤⎦ ⎡⎣Yb ⎤⎦ ⎡⎣Q ⎤⎦ = ⎢
⎥⎢
⎣ 0 0 −1 1 ⎦ ⎢ 0
⎢
⎢ 0
⎣
0
0
1
0
2
0
1
0
0
⎧⎡ 1
⎪⎢ 2
⎪⎢
⎡1 1 0 0 ⎤ ⎪ ⎢ 0
⎡⎣Q ⎤⎦ × ⎡⎣Yb ⎤⎦ ⎡⎣Vs ⎤⎦ − ⎡⎣ I S ⎤⎦ = ⎢
⎥ ⎨⎢
⎣0 −1 1 1 ⎦ ⎪ ⎢ 0
⎪⎢
⎪⎢ 0
⎩⎣
{
}
0 ⎤
⎥ ⎡ −1 0 ⎤
⎥
0 ⎥⎢ 1 0⎥ ⎡ 3
⎢
⎥=⎢ 2
⎥⎢
⎥ ⎢− 1
1
1
−
0 ⎥⎢
⎥ ⎢
⎥ ⎣ 0 1⎦ ⎣ 2
1 ⎥
2⎦
2
0
0
1
0
2
0
1
0
0
2
−1 ⎤
2⎥
⎥
1 ⎥
⎦
⎫
0 ⎤
⎥ ⎡2 ⎤ ⎡ 0 ⎤ ⎪
⎪
⎥
0 ⎥ ⎢0 ⎥ ⎢ 0 ⎥ ⎪ ⎡ 1 ⎤
⎢ ⎥ − ⎢ ⎥⎬ =
⎥
⎢
⎥
0 ⎥ ⎢⎢ 0 ⎥⎥ ⎢⎢ 0 ⎥⎥ ⎪ ⎣ −2v ⎦
⎥ 0
2v ⎪
1 ⎥ ⎣ ⎦ ⎣ ⎦⎪
2⎦
⎭
Thus, the KCL equations are
⎡3
⎢ 2
⎢ −1
⎢⎣ 2
Here, Vt2
−1 ⎤
2 ⎥ ⎡Vt 2 ⎤ = ⎡ 1 ⎤
⎥
⎥⎢ ⎥ ⎢
1 ⎥ ⎣⎢Vt 4 ⎦⎥ ⎣ −2 v ⎦
⎦
v. Putting this in the KCL equations and solving we get, v
4
V
9
Problem 3.9 For the resistive network, write a cut-set schedule and equilibrium equations on voltage
basis. Hence obtain values of branch voltages and branch currents.
2
5
5
10
10
910 V
Fig. 3.24
5
126
Network Analysis and Synthesis
Solution The graph of the network is shown in Fig. 3.25. A suitable tree is shown.
6
6
2
2
3
4
1
C2
3
C1
4
1
5
5
C3
(a)
(b)
Fig. 3.25
The fundamental cut-sets are identified as
f-cut-set-1: [1, 2, 6]
f-cut-set-2: [3, 5, 6]
f-cut-set-3: [1, 4, 5]
The fundamental cutset matrix is given as
Q
1
2
3
4
5
6
C1
1
1
0
0
0
1
C2
0
0
1
0
1
1
C3
1
0
0
1
1
0
The node equations are given as
[Q][Yb][QT][Vt] [Q] {[Yb][VS]
Here,
⎡ −1
⎢
T
⎡
⎤
⎡⎣Q ⎤⎦ ⎡⎣Yb ⎤⎦ ⎣Q ⎦ = ⎢ 0
⎢ 1
⎣
⎡0.9
⎢
= ⎢ 0.5
⎢ 0.5
⎣
[IS]}
[Q] [Yb][VS]
⎡1
⎢ 5
⎢
⎢ 0
⎢
1 0 0 0 1⎤ ⎢ 0
⎥⎢
0 1 0 −1 1 ⎥
⎢
0 0 1 −1 0 ⎥⎦ ⎢ 0
⎢
⎢ 0
⎢
⎢ 0
⎣
0.5 −0.2 ⎤
⎥
0.8 0.2 ⎥
0.2
0.3 ⎥⎦
{since IS
0 here}
0
0
0
0
1
0
0
0
0
1
10
0
0
0
0
1
10
0
0
0
0
1
0
0
0
0
5
5
0 ⎤
⎥
⎥
0 ⎥ ⎡ −1 0 1 ⎤
⎢
⎥
⎥⎢ 1 0 0⎥
0 ⎥⎢ 0 1 0⎥
⎥⎢
⎥
0 ⎥ ⎢ 0 0 1⎥
⎥⎢
⎥
⎥ 0 −1 −1⎥
0 ⎥⎢
⎢⎣ 1 1 0 ⎥⎦
⎥
1 ⎥
2⎦
127
Network Topology (Graph Theory)
⎡1
⎢ 5
⎢
⎢ 0
⎡ −1 1 0 0 0 1 ⎤ ⎢⎢ 0
⎢
⎥
⎡⎣Q ⎤⎦ ⎡⎣Yb ⎤⎦ ⎡⎣Vs ⎤⎦ = ⎢ 0 0 1 0 −1 1 ⎥ ⎢
⎢ 1 0 0 1 −1 0 ⎥ ⎢⎢ 0
⎣
⎦
⎢
⎢ 0
⎢
⎢ 0
⎣
0
0
0
0
1
0
0
0
0
1
10
0
0
0
0
1
10
0
0
0
0
1
0
0
0
0
5
5
0 ⎤
⎥
⎥
0 ⎥ ⎡ −910 ⎤
⎢
⎥
⎥⎢ 0 ⎥ ⎡
0 ⎥ ⎢ 0 ⎥ 182 ⎤
⎢
⎥
⎥⎢
⎥=⎢ 0 ⎥
0 ⎥⎢ 0 ⎥ ⎢ 0 ⎥
⎥⎢
⎦
⎥ ⎣
⎥⎢ 0 ⎥
0 ⎥
⎢⎣ 0 ⎥⎦
⎥
1 ⎥
2⎦
Thus, the KCL equations are
⎡0.9 0.5 −0.2 ⎤ ⎡Vt 2 ⎤ ⎡182 ⎤
⎢
⎥⎢ ⎥ ⎢
⎥
= ⎢ 0.5 0.8 0.2 ⎥ ⎢Vt 3 ⎥ = ⎢ 0 ⎥
⎢ 0.5 0.2 0.3 ⎥ ⎢V ⎥ ⎢ 0 ⎥
⎣
⎦ ⎣ t4 ⎦ ⎣
⎦
Solving by Cramer’s rule, we get the tree-branch voltages as
Vt2
143 V;
Vt3
14.3 V; Vt4
300 V
Problem 3.10 Using topological method, obtain node equations and node voltages in the s domain
for the network shown in Fig. 3.26 (a), when L1 ⴝ L2 ⴝ 1 H, C5 ⴝ 1 F, G3 ⴝ G4 ⴝ 1 ⍀, Vgt (t) ⴝ 2u (t) and
ig4 (t) ⴝ 2 (t), where, u(t) is the unit step function and ␦(t) is the unit impulse function.
L1
L2
1
Vg1(t )
2
C5
G3
ig4(t)
G4
3
Fig. 3.26 (a)
Solution
The graph of the network is shown in Fig. 3.26 (b).
(3)
1
2
(4)
(1)
(2)
(5)
3
Fig. 3.26 (b)
128
Network Analysis and Synthesis
The incidence matrix is given as
Aa
1
2
3
4
5
1
1
1
1
0
0
2
0
0
1
1
1
3
1
1
0
1
1
The reduced Incidence matrix is
⎡ −1 1 1 0 0 ⎤
A= ⎢
⎥
⎣ 0 0 −1 1 1 ⎦
The branch admittance matrix is
⎡1
⎢ s
⎢ 0
⎢
Yb = ⎢ 0
⎢
⎢ 0
⎢
⎢⎣ 0
⎡1
⎢ s
⎢ 0
⎡ −1 1 1 0 0 ⎤ ⎢
∴ AYb = ⎢
⎥⎢ 0
⎣ 0 0 −1 1 1 ⎦ ⎢
⎢ 0
⎢
⎢⎣ 0
⎡− 1
∴ AYb A = ⎢⎢ s
⎢⎣ 0
1
1
s
1
0 −
s
T
0
0
1
0
0
1
s
0
0
0
0
0
0
1
0
1
s
0
0
0
0
0
0 0⎤
⎥
0 0⎥
⎥
0 0⎥
⎥
s 0⎥
⎥
0 1 ⎥⎦
0 0⎤
⎥
0 0⎥ ⎡− 1
⎥ ⎢ s
0 0⎥ = ⎢
⎥ ⎢ 0
⎣
s 0⎥
⎥
0 1 ⎥⎦
1
0 0⎤
⎥
⎥
s 1⎥
⎦
−1
⎤
⎥
⎥
⎥
s ⎦
s
1
0 −
s
⎡ −1 0 ⎤
⎢
⎥
0 0 ⎤ ⎢ 1 0 ⎥ ⎡⎢ 2 + 1
s
⎥⎢
⎥
⎥ ⎢ 1 −1⎥ = ⎢
s 1⎥ 0 1
⎢ −1
⎦⎢
⎥ ⎣
s
⎢0 1⎥
⎣
⎦
(
1
)
(
s
s +1+ 1
)
Now,
⎡− 1
AYbVs − AI s = ⎢⎢ s
⎢⎣ 0
1
1
s
1
0 −
s
⎡2 ⎤
⎡0⎤
⎢ s2 ⎥
⎢ ⎥
0 0 ⎤ ⎢ 0 ⎥ ⎡ −1 1 1 0 0 ⎤ ⎢ 0 ⎥ ⎡ 2 ⎤
⎥
⎥⎢
⎢ ⎥ ⎢ s2 ⎥
⎥ ⎢ 0 ⎥ − ⎢ 0 0 −1 1 1 ⎥ ⎢ 0 ⎥ = ⎢
⎣
⎦
s 1⎥ ⎢
2 ⎥⎦
⎦ 0 ⎥
⎢0⎥ ⎣
⎢
⎥
⎢ −2 ⎥
⎣ ⎦
⎢⎣ 0 ⎥⎦
129
Network Topology (Graph Theory)
Thus, node equations are
)
(
⎡ 2 +1
⎢ s
⎢
⎢ −1
s
⎣
−1
(
s
s +1+ 1
⎤
⎥ ⎡V1 ⎤ ⎡⎢ 2 2 ⎤⎥
⎥⎢ ⎥= ⎢ s ⎥
⎥ ⎢⎣V2 ⎥⎦ ⎣ 2 ⎦
s ⎦
)
Solving by Cramer’s rule, we get the voltages as
V1 =
(
)
2 2s2 + s + 1
(
)(
)
s s + 1 s + 2s + 1
2
V2 =
and
(
)
2 s3 + s2 + 1
(
)(
)
s s + 1 s + 2s + 1
2
Problem 3.11 For the network of Fig. 3.27, draw the graph and write a tie-set schedule. Using the tie-set
schedule obtain the loop equations and find the currents in all branches.
0.2
1
1
0.5
1
0.5
9V
Fig. 3.27
Solution The graph and one tree are shown in Fig. 3.28.
(5)
(5)
i3
(4)
(3)
(6)
(3)
(4)
i1
(2)
(1)
(6)
(1)
(a)
(b)
Fig. 3.28
The tie-set matrix
⎡1 0 1 0 0 −1⎤
⎢
⎥
Ba = ⎢0 1 0 1 0 1 ⎥
⎢0 0 −1 −1 1 0 ⎥
⎣
⎦
i2
(2)
130
Network Analysis and Synthesis
Branch impedance matrix is
⎡0.5 0
⎢
⎢ 0 0.5
⎢0
0
Zb = ⎢
0
⎢0
⎢0
0
⎢
0
⎢⎣ 0
0 0
0 0
⎡0.5 0
⎢
0 0.5
⎡1 0 1 0 0 −1⎤ ⎢
⎢
0
⎢
⎥ 0
⎡⎣ Ba ⎤⎦ ⎡⎣ Z b ⎤⎦ = ⎢0 1 0 1 0 1 ⎥ ⎢
0
⎢0 0 −1 −1 1 0 ⎥ ⎢ 0
⎣
⎦⎢ 0
0
⎢
0
⎢⎣ 0
0 0
0 0
1 0
0 1
0 0
0 0
0⎤
⎥
0⎥
0 0⎥
⎥
0 0⎥
0.2 0 ⎥
⎥
0 1 ⎥⎦
0
0
Thus,
1 0
0 1
0 0
0 0
0⎤
⎥
0⎥
⎡0.5 0
1 0 0 −1⎤
0 0⎥ ⎢
⎥
0
0
5
0
1 0
1⎥
=
.
⎥ ⎢
0 0⎥ ⎢
0
0 −1 −1 0.2 0 ⎥⎦
0.2 0 ⎥ ⎣
⎥
0 1 ⎥⎦
0
0
⎡ 1
⎢
0
⎡0.5 0
1 0 0 −1⎤ ⎢
⎢
T
⎢
⎥ 1
∴⎡⎣ Ba ⎤⎦ ⎡⎣ Z b ⎤⎦ ⎡⎣ Ba ⎤⎦ = ⎢ 0 0.5 0 1 0
1⎥ ⎢
0
⎢0
0 −1 −1 0.2 0 ⎥⎦ ⎢⎢
⎣
0
⎢
⎢⎣ −1
0
1
0
1
0
1
0⎤
⎥
0⎥
⎡ 2.5 −1 −1 ⎤
−1⎥
⎢
⎥
⎥ = ⎢ −1 2.5 −1 ⎥
−1⎥
⎢ −1 −1 2.2 ⎥
⎣
⎦
1⎥
⎥
0 ⎥⎦
⎡ −9 ⎤
⎢ ⎥
0
⎡1 0 1 0 0 −1⎤ ⎢ ⎥ ⎡ 9 ⎤
⎢
⎥⎢ 0 ⎥ ⎢ ⎥
Now, − ⎡⎣ Ba ⎤⎦ ⎡⎣Vs ⎤⎦ = − ⎢0 1 0 1 0 1 ⎥ ⎢ ⎥ = ⎢ 0 ⎥
⎢
⎥⎢ 0 ⎥ ⎢ ⎥
⎣0 0 −1 −1 1 0 ⎦ ⎢ 0 ⎥ ⎣ 0 ⎦
⎢ ⎥
⎢⎣ 0 ⎥⎦
So, the loop equations are
⎡ 2.5 −1 −1 ⎤ ⎡ i1 ⎤ ⎡ 9 ⎤
⎢
⎥ ⎢ ⎥ ⎢ ⎥
⎢ −1 2.5 −1 ⎥ × ⎢i2 ⎥ = ⎢ 0 ⎥
⎢ −1 −1 2.2 ⎥ ⎢ i ⎥ ⎢ 0 ⎥
⎣
⎦ ⎣ 3⎦ ⎣ ⎦
131
Network Topology (Graph Theory)
Solving the three equations,
i1
8.9 A, i2
6.33 A,
i3
6.92 A
Problem 3.12 Figure 3.29 (a) shows a dc network. (a) Draw a graph of the network. Which elements are
not included in the graph and why? (b) Write a loop incidence matrix and use it to obtain loop equations. (c)
Find branch currents.
2
2
2A
2
2
5V
2
Fig. 3.29 (a)
Solution (a) The graph is shown below.
A
(3)
(4)
(5)
i2
i1
B
(2)
(1)
C
Fig. 3.29 (b)
The 2-V resistor in parallel with the voltage source and the 2-A current source have not been included in the
graph. This is because of the reason that passive elements in parallel with a voltage source are not included in a
graph and the current source in parallel with a passive element is open-circuited while drawing a graph.
(b) The tie-set matrix for the tree chosen is
⎡1 0 0 −1 1 ⎤
Ba = ⎢
⎥
⎣0 1 −1 0 −1⎦
Branch impedance matrix is
⎡2
⎢
⎢0
Zb = ⎢ 0
⎢
⎢0
⎢0
⎣
0
2
0
0
0
0
0
0
0
0
0
0
0
2
0
0⎤
⎥
0⎥
0⎥
⎥
0⎥
2 ⎥⎦
132
Network Analysis and Synthesis
⎡2
⎢
0
⎡1 0 0 −1 1 ⎤ ⎢
T
⎢
Ba Z b Ba = ⎢
⎥ 0
⎣0 1 −1 0 −1⎦ ⎢
⎢0
⎢0
⎣
0 0 0 0⎤⎡ 1 0 ⎤
⎥⎢
⎥
2 0 0 0 ⎥ ⎢ 0 1⎥
0 0 0 0 ⎥ ⎢ 0 −1⎥
⎥⎢
⎥
0 0 2 0 ⎥ ⎢ −1 0 ⎥
0 0 0 2 ⎥⎦ ⎢⎣ 1 −1⎥⎦
⎡ 1 0⎤
⎢
⎥
0 1⎥
⎡ 2 0 0 −2 2 ⎤ ⎢
⎡ 6 −2 ⎤
=⎢
⎥ ⎢ 0 −1⎥ = ⎢
⎥
⎥ ⎣ −2 4 ⎦
⎣ 0 2 0 0 −2 ⎦ ⎢
⎢ −1 0 ⎥
⎢ 1 −1⎥
⎣
⎦
Now,
⎡2 ⎤
⎡ 0⎤
⎢ ⎥
⎢ ⎥
0
0
⎡ 2 0 0 −2 2 ⎤ ⎢ ⎥ ⎡1 0 0 −1 1 ⎤ ⎢ ⎥ ⎡ 4 ⎤ ⎡0 ⎤ ⎡ 4 ⎤
⎢
⎥
⎢
Ba Z b I s − BaVs = ⎢
−
0
−
⎥
⎢
⎥ 5⎥ = ⎢ ⎥ − ⎢ ⎥ = ⎢ ⎥
⎣ 0 2 0 0 −2 ⎦ ⎢ ⎥ ⎣0 1 −1 0 −1⎦ ⎢ ⎥ ⎣ 0 ⎦ ⎣ 5 ⎦ ⎣ −5 ⎦
⎢0 ⎥
⎢ 0⎥
⎢0 ⎥
⎢ 0⎥
⎣ ⎦
⎣ ⎦
So, the loop equations are
⎡ 6 −2 ⎤ ⎡ i1 ⎤ ⎡ 4 ⎤
⎢
⎥⎢ ⎥ = ⎢ ⎥
⎣ −2 4 ⎦ ⎢⎣i2 ⎦⎥ ⎣ −5 ⎦
Solving these equations, i1 0.3A, i2
1.1 A
(c) Putting these values, the branch voltages are
V1 2 i1 0.6 V, V2 2 i2
Thus, the branch currents are
I AB =
2.2 V, V3
5 V, V4
2
i1
4
3.4 V, V5
2.8 V
5
2.8
2.2
3.4
0.6
= 1.7 A, I AD =
= 1.4 A, I AC = = 2.5 A, I DB =
= 0.3A, I DC =
= 1.1A
2
2
2
2
2
So, the current supplied by the battery
(1.7
1.4
2.5
2)
3.6 A
Problem 3.13 For the network shown in Fig. 3.30, draw the oriented graph and obtain the tie-set matrix.
Use this matrix to calculate i.
1
2
2
i
1
1V
2V
3
Fig. 3.30
1
133
Network Topology (Graph Theory)
Solution The oriented graph and any one tree are shown.
The tie-set matrix is given as
⎡1 1 0 0 1 0 ⎤
⎢
⎥
Ba = ⎢0 −1 1 −1 0 0 ⎥
⎢0 0 0 1 −1 1 ⎥
⎣
⎦
1
⎡1
⎢
⎢0
⎢0
Zb = ⎢
⎢0
⎢0
⎢
⎢⎣0
0 0 0 0 0⎤
⎥
2 0 0 0 0⎥
0 2 0 0 0⎥
⎥
0 0 1 0 0⎥
0 0 0 3 0⎥
⎥
0 0 0 0 1 ⎥⎦
⎡1
⎢
0
⎡1 1 0 0 1 0 ⎤ ⎢
⎢
⎢
⎥ 0
∴ Ba Z b = ⎢0 −1 1 −1 0 0 ⎥ ⎢
⎢0 0 0 1 −1 1 ⎥ ⎢0
⎣
⎦ ⎢0
⎢
⎢⎣0
(3)
(3)
(2)
(2)
(4)
2
The branch impedance matrix
1
(1)
(5)
3
(1)
(6)
4
(a)
I1
I2
3
(5)
I3
(6)
4
(b)
Fig. 3.31
0 0 0 0 0⎤
⎥
2 0 0 0 0⎥
⎡1 2 0 0 3 0 ⎤
0 2 0 0 0⎥ ⎢
⎥
=
⎥ ⎢0 −2 2 −1 0 0 ⎥
0 0 1 0 0⎥ ⎢
0 0 0 1 −3 1 ⎥⎦
0 0 0 3 0⎥ ⎣
⎥
0 0 0 0 1 ⎥⎦
⎡1 0 0 ⎤
⎢
⎥
1 −1 0 ⎥
⎡1 2 0 0 3 0 ⎤ ⎢
⎡ 6 −2 −3⎤
⎢
⎥ ⎢0 1 0 ⎥ ⎢
⎥
T
∴ Ba Z b Ba = ⎢0 −2 2 −1 0 0 ⎥ ⎢
⎥ = ⎢ −2 5 −1⎥
0
−
1
1
⎥ ⎢
⎢0 0 0 1 −3 1 ⎥ ⎢
⎥
⎣
⎦ ⎢1 0 −1⎥ ⎣ −3 −1 5 ⎦
⎢
⎥
⎢⎣0 0 1 ⎥⎦
Now,
⎡ −2 ⎤
⎢ ⎥
0
⎡1 1 0 0 1 0 ⎤ ⎢ ⎥
⎡ −2 ⎤ ⎡ 2 ⎤
⎢
⎥ ⎢ −1 ⎥
⎢ ⎥ ⎢ ⎥
− BaVs = − ⎢0 −1 1 −1 0 0 ⎥ ⎢ ⎥ = − ⎢ −1 ⎥ = ⎢ 1 ⎥
⎢0 0 0 1 −1 1 ⎥ ⎢ 0 ⎥
⎢ 0 ⎥ ⎢0 ⎥
⎣
⎦⎢ 0 ⎥
⎣ ⎦ ⎣ ⎦
⎢ ⎥
⎢⎣ 0 ⎥⎦
(4)
2
134
Network Analysis and Synthesis
So, the loop equations become
⎡ 6 −2 −3⎤ ⎡ I1 ⎤ ⎡ 2 ⎤
⎢
⎥⎢ ⎥ ⎢ ⎥
⎢ −2 5 −1⎥ ⎢ I 2 ⎥ = ⎢ 1 ⎥
⎢ −3 −1 5 ⎥ ⎢ I ⎥ ⎢ 0 ⎥
⎣
⎦⎣ 3 ⎦ ⎣ ⎦
Solving for I1
I1 =
I1
2 −2 −3
1 5 −1
0 −1 5
6 −2 −3
−2 5 −1
−3 −1 5
= 0.91A
0.91 A
Problem 3.14 Determine the currents in all branches of the network shown in Fig. 3.32 using the node
analysis method. Use the graph theory method.
1
2
1
2
1A
2V
Fig. 3.32
Solution Here, the 1- resistance in parallel with the 2-V voltage source can be ignored. Also, there is no passive element in parallel with the 1-A current source. We assume a resistance R in parallel with the 1-A current
source and finally let R → . Therefore, the graph of the network is shown in Fig. 3.33.
1
1A
R
1
2
2
2
(1)
1
2V
2
(3)
(2)
(4)
3
(a)
Fig. 3.33
The complete incidence matrix is
⎡ 1 1 0 0⎤
⎢
⎥
Aa = ⎢ −1 0 1 1 ⎥
⎢ 0 −1 −1 −1⎥
⎣
⎦
(b)
135
Network Topology (Graph Theory)
Reduced incidence matrix is
⎡ 1 1 0 0⎤
A= ⎢
⎥
⎣ −1 0 1 1 ⎦
Branch admittance matrix is
⎡1
⎢
⎢0
Yb = ⎢
⎢0
⎢
⎢0
⎢⎣
⎡1
⎢
⎢0
⎡ 1 1 0 0⎤ ⎢
∴ AYb = ⎢
⎥
⎣ −1 0 1 1 ⎦ ⎢⎢0
⎢0
⎣⎢
⎡ 1
∴ AYb AT = ⎢⎢
⎢⎣ −1
1
R
0
0
0
1
0
R
0
1
0
0
0
1
R
0 ⎤
⎥
0 ⎥ ⎡ 1
⎥=⎢
0 ⎥ ⎢ −1
⎥ ⎢⎣
1 ⎥
2 ⎥⎦
0
1
0
0
2
1
R
0
⎡1 −1⎤
0 ⎤ ⎢1 0 ⎥ ⎡ 1 + 1
⎥⎢
⎥=⎢
R
1 ⎥ ⎢0 1 ⎥ ⎢ −1
⎥ ⎣
2 ⎥⎦ ⎢
⎣0 1 ⎦
(
0
1
2
0
0
0 ⎤
⎥
0 ⎥
⎥
0 ⎥
⎥
1 ⎥
2 ⎥⎦
2
0
1
2
0 ⎤
⎥
1 ⎥
2 ⎥⎦
) −1⎤⎥
2 ⎥⎦
Now,
⎡ 1
AYbVs − AI s = ⎢⎢
⎢⎣ −1
1
R
0
0
1
2
⎡0 ⎤
⎡ 0⎤
0 ⎤ ⎢ 0 ⎥ ⎡ 1 1 0 0 ⎤ ⎢ −1⎥ ⎡0 ⎤ ⎡ −1⎤ ⎡1⎤
⎥⎢ ⎥ −
⎢ ⎥=
⎢
⎥
⎢ ⎥− ⎢ ⎥= ⎢ ⎥
1 ⎥ ⎢ 0 ⎥ ⎣ −1 0 1 1 ⎦ ⎢ 0 ⎥ ⎣1 ⎦ ⎣ 0 ⎦ ⎣1⎦
⎢ ⎥
2 ⎥⎦ ⎢ ⎥
⎣2 ⎦
⎣ 0⎦
Thus, node equations are
(
⎡ 1+ 1
⎢
R
⎢ −1
⎣
) −1⎤⎥ ⎡⎢V ⎤⎥ = ⎡⎢1⎤⎥
1
2 ⎥⎦ ⎢⎣V2 ⎥⎦ ⎣1⎦
With R → , the equations become
V1 − V2 = 1
−V1 + 2V2 = 1
Solving equations, we get
V1 = 3 V,
V2 = 2 V
136
Network Analysis and Synthesis
Hence, the currents in different branches are shown in Fig. 3.34.
1
V1
1A
1A
2
V2
2
2A
0A
2A
1A
1
2V
Fig. 3.34
Problem 3.15 Consider the network shown in Fig. 3.35. Using
loop method of analysis, determine currents in all the branches
indicating their directions. Use graph theory method.
i2
2
1
1
4V
Solution The graph of the network is shown below. Also the tree
is selected as shown.
For the selected tree, the tie-set matrix is given as
3V
3i2
Fig. 3.35
⎡1 0 −1⎤
Ba = ⎢
⎥
⎣0 1 1 ⎦
⎡1 0 0 ⎤
⎢
⎥
The branch impedance matrix is Z b = ⎢0 2 0 ⎥
⎢0 0 1 ⎥
⎣
⎦
i2
2
1
1
4V
⎡1 0 0 ⎤
⎡1 0 −1⎤ ⎢
⎥ ⎡1 0 −1⎤
∴ Ba Z b = ⎢
⎥ ⎢0 2 0 ⎥ = ⎢
⎥
⎣0 1 1 ⎦ ⎢0 0 1 ⎥ ⎣0 2 1 ⎦
⎣
⎦
i1
3i2
3V
i2
Fig. 3.36
⎡ 1 0⎤
⎡1 0 −1⎤ ⎢
⎥ ⎡ 2 −1⎤
∴ Ba Z b B = ⎢
⎥ ⎢ 0 1⎥ = ⎢
⎥
⎣0 2 1 ⎦ ⎢ −1 1 ⎥ ⎣ −1 3 ⎦
⎣
⎦
T
a
⎡ −4 ⎤
⎡1 0 −1⎤ ⎢
⎥ ⎡ 4 − 3i2 ⎤
Now, − BaVs = − ⎢
⎥
⎥⎢ 3 ⎥ = ⎢
⎣0 1 1 ⎦ ⎢ −3i ⎥ ⎢⎣ −3 + 3i2 ⎥⎦
⎣ 2⎦
1
⎡ 2 −1⎤ ⎡ i1 ⎤ ⎡ 4 − 3i2 ⎤
So, the loop equations become ⎢
⎥
⎥⎢ ⎥ = ⎢
⎣ −1 3 ⎦ ⎢⎣i2 ⎥⎦ ⎢⎣ −3 + 3i2 ⎥⎦
Fig. 3.37
These equations reduce to,
2i1 − i2 = 4 − 3i2
3
⇒ i1 + i2 = 2
(b)
2
1
1
3V
3i2
Fig. 3.38
1A
4A
4V
∴ i2 = −1A
Thus, the branch currents are shown with their directions.
2
(a)
3A
−i1 + 3i2 = −3 + 3i2 ⇒ i1 = 3A
1
137
Network Topology (Graph Theory)
Problem 3.16 For the circuit shown in Fig. 3.39 construct a tree
in which 10 ⍀ and 20 ⍀ are in tree branches. Using node analysis,
solve for V1 and V2.
2A
5
Solution Here, we have one current source without parallel resistance
and one voltage source without series resistance. Therefore, we connect
a parallel resistance R1 in parallel with the 2-A current source and a
series resistance R2 in series with the 20-V voltage source. Finally, we
will let R1 → and R2 → 0.
Now, we construct the graph of the network as
shown below. A tree, in which 10 and 20 are in tree
branches, is selected.
The complete incidence matrix is
⎡ −1 1 0 0 0 1 ⎤
⎢
⎥
0 −1 1 1 0 0 ⎥
Aa = ⎢
⎢ 0 0 −1 0 −1 −1⎥
⎢
⎥
⎣ 1 0 0 −1 1 0 ⎦
10
20
V1
V2
50
80V
Fig. 3.39
R1
2A
5
A
10
20
B
V1
C
V2
20V
50
80V
R2
Fig. 3.40
(6)
⎡ −1 1 0 0 0 1 ⎤
⎢
⎥
The reduced Incidence matrix is A = ⎢ 0 −1 1 1 0 0 ⎥
⎢ 0 0 −1 0 −1 −1⎥
⎦
⎣
⎡0.2 0
0
0
⎢
0
⎢ 0 0.1 0
⎢0
0 0.05
0
⎢
0
0
0
0.02
The branch admittance matrix is Yb = ⎢
⎢
⎢0
0
0
0
⎢
⎢
⎢0
0
0
0
⎢⎣
VB
VA
(2)
(4)
(1)
0⎤
⎥
0⎥
0⎥
⎥
0⎥
⎥
0⎥
⎥
⎥
1⎥
R1 ⎥⎦
0
0
0
0
1
R2
0
0
0
0
0
1
R2
0
0⎤
⎥
0⎥
0⎥
⎥
0⎥
⎥
0⎥
⎥
⎥
1⎥
R1 ⎥⎦
(5)
Fig. 3.41 (a)
(6)
VB
VA
VC
(2)
(1)
Fig. 3.41 (b)
⎡0.2 0
0
0
⎢
0
⎢ 0 0.1 0
⎢0
0 0.05
0
⎡ −1 1 0 0 0 1 ⎤ ⎢
0
0 0.02
⎢
⎥⎢ 0
∴ AYb = ⎢ 0 −1 1 1 0 0 ⎥ ⎢
⎢ 0 0 −1 0 −1 −1⎥ ⎢ 0
0
0
0
⎣
⎦⎢
⎢
⎢0
0
0
0
⎢⎣
VC
(3)
(3)
(4)
(5)
138
Network Analysis and Synthesis
⎡
0
0
0
⎢ − 0.2 0.1
⎢
0
= ⎢ 0 −0.1 0.05 0.02
⎢
1
⎢ 0
0
−0.05
0
−
⎢
R
2
⎣
1 ⎤
R1 ⎥⎥
0 ⎥
⎥
1
− ⎥⎥
R1 ⎦
⎡ −1 0 0 ⎤
1 ⎤⎢
⎥
1 −1 0 ⎥
R1 ⎥⎥ ⎢
⎢ 0 1 −1⎥
0 ⎥⎢
⎥
⎥ 0 1 0⎥
1 ⎥ ⎢⎢
− ⎥ 0 0 −1⎥
R1 ⎦ ⎢
⎥
⎢⎣ 1 0 −1⎥⎦
⎤
1
−
−0.1
⎥
R1
⎥
⎥
0.17
−0.05
⎥
⎛
1 1 ⎞⎥
−0.05 ⎜ 0.05 + + ⎟ ⎥
R2 R1 ⎠ ⎥
⎝
⎦
⎡
0
0
0
⎢ −0.2 0.1
⎢
∴ AYb AT = ⎢ 0
−0.1 0.05 0.02
0
⎢
1
⎢ 0
−0.05
0
0
−
⎢
R2
⎣
⎡⎛
1⎞
⎢⎜ 0.3 + ⎟
R1 ⎠
⎢⎝
⎢
= ⎢ −0.1
⎢
1
⎢ −
R
⎢⎣
1
Now,
⎡
0
0
0
⎢ −0.2 0.1
⎢
AYbVs − AI s = ⎢ 0
−0.1 0.05 0.02
0
⎢
1
⎢ 0
0
−0.0
05
0
−
⎢
R2
⎣
⎡⎛
1⎞
⎢⎜ 0.3 + ⎟
R1 ⎠
⎢⎝
⎢
Thus, node equations are
⎢ −0.1
⎢
1
⎢ −
R1
⎢⎣
⎡ −80 ⎤
⎡0 ⎤
1 ⎤⎢
⎥
⎢ ⎥
0 ⎥
0
R1 ⎥⎥ ⎢
⎡ −1 1 0 0 0 1 ⎤ ⎢ ⎥
14
⎢ 0 ⎥ ⎢
⎢
⎥ 0⎥
0 ⎥⎢
−
0
−
1
1
1
0
0
=
0
⎥
⎥ ⎢0 ⎥
⎥⎢ 0 ⎥ ⎢
⎢
⎥
⎢ 0 0 −1 0 −1 −1⎥
1
⎦ ⎢ 0 ⎥ ⎛ 20 + 2⎞
− ⎥⎥ ⎢ −20 ⎥ ⎣
⎟
R1 ⎦ ⎢
⎥
⎢ ⎥ ⎜⎝ R2
⎠
⎢⎣ 0 ⎥⎦
⎢⎣ 2 ⎥⎦
⎤
⎥
14
⎥ ⎡VA ⎤
⎢ ⎥
⎥
0.17
−0.05
0
⎥ ⎢VB ⎥ =
⎢
⎥
⎛
⎛ 20 ⎞
1 1 ⎞⎥ V
−0.05 ⎜ 0.05 + + ⎟ ⎥ ⎣ C ⎦ ⎜ + 2⎟
R2 R1 ⎠ ⎥
⎝
⎝ R2
⎠
⎦
−0.1
−
1
R1
0.3VA − 0.1VB = 14
With R1 → , the equations become
−0.1VA + 0.17VB − 0.05VC = 0
⎛
⎛ 20 ⎞
1⎞
−0.05VB + ⎜ 0.05 + ⎟ VC = ⎜ + 2⎟
R2 ⎠
⎝
⎝ R2
⎠
139
Network Topology (Graph Theory)
Solving equations, we get
− 0.1
0.17
14
0
0
− 0.05
⎛ 20 ⎞
⎛
1⎞
⎜ R + 2⎟ −0.05 ⎜ 0.05 + R ⎟
14 ⎡0.17 0.05 R2 + 1 − 0.0025 R2 ⎤⎦ + 0.005 20 + 2 R2
⎝ 2
⎝
⎠
2⎠
VA =
= ⎣
0.3 −0.1
0
0.3 ⎡⎣0.17 0.05 R2 + 1 − 0.0025 R2 ⎤⎦ − 0.001 0.05 R2 + 1
−0.1 0.17
−0.05
)
)
(
(
(
)
)
(
⎛
1⎞
−0.05 ⎜ 0.05 + ⎟
R2 ⎠
⎝
0
2.48
= 60.49 V
0.041
Similarly, with R2 0, we get,
With R2
0, VA =
VB = 41.47 V
VC = 20 V
(
) (
)
V = (V − V ) = ( 41.420 ) = 21.47 V
V1 = VA − VB = 60.49 − 41.47 = 19.02 V
and
2
B
C
Problem 3.17 The circuit of Fig. 3.42 contains a voltage-controlled voltage source. For this circuit, draw the oriented graph. By selecting a proper
tree obtain the tie-set matrix and hence calculate the voltage, Vx .
Vx
5
Solution Since the controlled voltage source is not accompanied by
any passive element, we will consider a resistance R1 in series with the
controlled voltage source, and finally let R1 → 0.
5
Fig. 3.42
⎡1 1 −1 0 0 0 ⎤
⎢
⎥
The tie-set matrix is Ba = ⎢0 0 1 −1 1 0 ⎥
⎢0 −1 0 1 0 1 ⎥
⎣
⎦
The branch impedance matrix
0
5
0
0
0
0
0
0
5
0
0
0
0
0
0
5
0
0
0 0⎤
⎥
0 0⎥
0 0⎥
⎥
0 0⎥
4 0⎥
⎥
0 R1 ⎥⎦
Vx
4
1V
The graph of the network is shown with one tree.
⎡5
⎢
⎢0
⎢0
Zb = ⎢
⎢0
⎢0
⎢
⎢⎣0
5
5
(6)
R1
1
Vx
1
5
2
5
5
(1)
5
1V
(a)
Fig. 3.43
3
4
Vx
I3
2
3
(4)
(2)
I1
I2
(3)
4
(b)
(5)
140
Network Analysis and Synthesis
⎡5
⎢
0
⎡1 1 −1 0 0 0 ⎤ ⎢
⎢
⎢
⎥ 0
∴ Ba Z b = ⎢0 0 1 −1 1 0 ⎥ ⎢
0
⎢0 −1 0 1 0 1 ⎥ ⎢
⎣
⎦ ⎢0
⎢
⎢⎣0
0 0 0 0
5 0 0 0
0 5 0 0
0 0 5 0
0 0 0 4
0 0 0 0
0⎤
⎥
0⎥
⎡ 5 5 −5 0 0 0 ⎤
0⎥ ⎢
⎥
5 −5 4 0 ⎥
⎥ = ⎢0 0
0⎥
⎢0 −5 0
5 0 R1 ⎥⎦
0⎥ ⎣
⎥
R1 ⎥⎦
⎡ 1 0 0⎤
⎢
⎥
1 0 −1⎥
⎡ 15
⎡ 5 5 −5 0 0 0 ⎤ ⎢
⎢
⎥ ⎢ −1 1 0 ⎥ ⎢
T
∴ Ba Z b Ba = ⎢0 0
5 −5 4 0 ⎥ ⎢
⎥ = ⎢ −5
0 −1 1 ⎥ ⎢
⎢
⎢0 −5 0
⎥
5 0 R1 ⎦ ⎢
−5
⎣
0 1 0⎥ ⎣
⎢
⎥
⎢⎣ 0 0 1 ⎥⎦
−5
14
−5
⎡ 0⎤
⎢
⎥
0⎥
⎡1 1 −1 0 0 0 ⎤ ⎢
⎡ 1 ⎤ ⎡ −1⎤
⎢
⎥ ⎢ −1 ⎥
⎢
⎥ ⎢ ⎥
Now, − BaVs = − ⎢0 0 1 −1 1 0 ⎥ ⎢
⎥ = − ⎢ −1 ⎥ = ⎢ 1 ⎥
0⎥
⎢
⎢0 −1 0 1 0 1 ⎥
⎢ −V ⎥ ⎢V ⎥
⎣
⎦⎢ 0 ⎥
⎣ x⎦ ⎣ x⎦
⎢
⎥
⎢⎣ −Vx ⎥⎦
⎡15 −5
⎢
So, the loop equations become ⎢ −5 14
⎢ −5 −5
⎣
With R1 → 0 and Vx
⎤ ⎡ I1 ⎤ ⎡ −1⎤
⎥⎢ ⎥ ⎢ ⎥
−5 ⎥ ⎢ I 2 ⎥ = ⎢ 1 ⎥
10 + R1 ⎥⎦ ⎢⎣ I 3 ⎥⎦ ⎢⎣Vx ⎥⎦
−5
(
)
⎡15 −5 −5 ⎤ ⎡ I1 ⎤ ⎡ −1⎤
⎢
⎥⎢ ⎥ ⎢ ⎥
4I2, the equations reduce to, ⎢ −5 14 −5 ⎥ ⎢ I 2 ⎥ = ⎢ 1 ⎥
⎢ −5 −9 10 ⎥ ⎢ I ⎥ ⎢ 0 ⎥
⎣
⎦⎣ 3 ⎦ ⎣ ⎦
Solving for I2,
I2 =
15 −1 −5
−5 1 −5
−5 0 10
15 −5 −5
−5 14 −5
−5 −9 10
Vx = 4 × I 2 = 4 ×
=
1
A
19
1 4
= V
19 19
⎤
⎥
⎥
10 + R1 ⎥⎦
(
−5
−5
)
141
Network Topology (Graph Theory)
Problem 3.18 In the following circuit of Fig. 3.44, determine the voltages V2 and V3 using cut-set analysis. Select the circuit elements (1), (2) and
(3) in the tree.
Solution The graph and tree are shown in Fig. 3.45 Hence, there is
no series impedance with voltage source and parallel admittance with
current source. We consider two resistances R1 and R2 in series with
the voltage source and in parallel with the current source, respectively.
Finally, we will let R1 → 0, R2 → .
(5)
8V
(2) 1
1A
2
V2
(6)
(4) 2
(1)
V3
1
(3)
Fig. 3.44
(5)
Three fundamental cut-sets are
C1
f-cutset-1: [1, 4, 5, 6]
f-cutset-2: [2, 4, 6]
f-cutset-3: [3, 5, 6]
C2
(6)
The fundamental cut-set matrix is given as
Q
(1)
1
2
3
4
5
6
C1
1
0
0
1
1
1
C2
0
1
0
1
0
1
C3
0
0
1
0
1
1
Fig. 3.45
The node equations are given as
[Q][Yb][QT][Vt]
[Q] × {[Yb][VS] −[IS]}
Here,
⎡ 1
⎢ R1
⎢
0
⎡1 0 0 −1 −1 1 ⎤ ⎢
⎢
⎢
⎥ 0
⎡⎣Q ⎤⎦ ⎡⎣Yb ⎤⎦ ⎡⎣Q T ⎤⎦ = ⎢0 1 0 1 0 −1⎥ ⎢
⎢0 0 1 0 1 1 ⎥ ⎢ 0
⎣
⎦⎢ 0
⎢
⎢ 0
⎢⎣
⎡⎛
1
1 ⎞
⎢⎜ 4 + R + R ⎟
1
2⎠
⎢⎝
⎢
⎛
⎞
= ⎢ −⎜ 2 + 1 ⎟
R
⎝
2⎠
⎢
⎢
⎛
⎞
⎢ −⎜ 2 + 1 ⎟
R2 ⎠
⎝
⎢⎣
C3
(2)
0 0 0 0
1
0
0
0
1
R2
0
0
2
0
0
0
0
2
0 0 0 0
⎛
⎞
−⎜ 2 + 1 ⎟
R2 ⎠
⎝
⎛
⎞
−⎜ 3+ 1 ⎟
R2 ⎠
⎝
0
1
0
0
0 ⎤
⎥⎡ 1 0
⎥⎢
0 ⎥⎢ 0 1
0 ⎥⎢ 0 0
⎥⎢
0 ⎥ ⎢ −1 1
0 ⎥ ⎢ −11 0
⎥⎢
1 ⎥ ⎢⎣ 1 −1
R2 ⎥⎦
⎛
⎞⎤
−⎜ 2 + 1 ⎟ ⎥
R2 ⎠
⎝
⎥
⎥
1
⎥
R2
⎥
⎥
⎛
⎞
1
⎥
3
+
⎜⎝
R2 ⎟⎠ ⎥⎦
1⎤
⎥
0⎥
1⎥
⎥
0⎥
1⎥
⎥
1 ⎥⎦
(4)
(3)
142
Network Analysis and Synthesis
⎡ 1
⎢ R1
⎢
0
⎡1 0 0 −1 −1 1 ⎤ ⎢
⎢
⎢
⎥ 0
⎡⎣Q ⎤⎦ ⎡⎣Yb ⎤⎦ ⎡⎣Vs ⎤⎦ = ⎢0 1 0 1 0 −1⎥ ⎢
⎢0 0 1 0 1 1 ⎥ ⎢ 0
⎣
⎦⎢ 0
⎢
⎢ 0
⎢⎣
0 ⎤
⎥ ⎡ −8 ⎤
⎥⎢ ⎥
0 ⎥⎢ 0 ⎥ − 8
R1
0 ⎥⎢ 0 ⎥
0
⎥⎢ ⎥ =
0 ⎥⎢ 0 ⎥
0
0 ⎥⎢ 0 ⎥
⎥⎢ ⎥
1 ⎥ ⎢⎣ 0 ⎥⎦
R2 ⎥⎦
0 0 0 0
1 0 0 0
0 1 0 0
0 0 2 0
0 0 0 2
0 0 0 0
⎡0 ⎤
⎢ ⎥
0
⎡1 0 0 −1 −1 1 ⎤ ⎢ ⎥ ⎡ −1⎤
⎢
⎥ ⎢0 ⎥ ⎢ ⎥
− ⎡⎣Q ⎤⎦ ⎡⎣ I s ⎤⎦ = − ⎢0 1 0 1 0 −1⎥ ⎢ ⎥ = ⎢ 1 ⎥
⎢0 0 1 0 1 1 ⎥ ⎢0 ⎥ ⎢ 1 ⎥
⎣
⎦ ⎢0 ⎥ ⎣ ⎦
⎢ ⎥
⎢⎣1 ⎥⎦
Thus, the KCL equations are
⎡⎛
⎞
⎢⎜ 4 + 1 R + 1 R ⎟
1
2⎠
⎢⎝
⎢
⎛
⎞
= ⎢ −⎜ 2 + 1 ⎟
R
⎝
2⎠
⎢
⎢
⎛
⎞
⎢ −⎜ 2 + 1 ⎟
R
⎝
⎢⎣
2⎠
⎛
⎞
−⎜ 2 + 1 ⎟
R
⎝
2⎠
⎛
⎞
−⎜ 3+ 1 ⎟
R2 ⎠
⎝
1
R2
⎛
⎞⎤
−⎜ 2 + 1 ⎟ ⎥
⎛
⎞
R
− ⎜ 8 + 1⎟
⎝
2⎠ ⎥
⎡Vt 1 ⎤
⎝ R1 ⎠
⎥⎢ ⎥
1
1
⎥ ⎢Vt 2 ⎥ =
R2
⎥⎢ ⎥
1
V
⎛
⎞ ⎥⎣ t3 ⎦
1
⎜⎝ 3 + R ⎟⎠ ⎥
⎥⎦
2
When R1 → 0, R2 → , the equations reduce to the form as given below.
Vt1
8V
2Vt1 3Vt2 1
Solving the last two equations, Vt2
Therefore,
5V; Vt3
V2
V3
2Vt1 3Vt3
5V
1
Vt2 5V
Vt3 5V
Problem 3.19 For the network shown in Fig. 3.46, write
the tie-set matrix and determine the loop currents and
branch currents.
Solution The graph and a suitable tree for the network are
shown in Fig. 3.47.
A
(1)
5
(5)
C
10
10V
10
5
Fig. 3.46
(4)(6)
5
5
(2)
B
(3)
Fig. 3.47
D
143
Network Topology (Graph Theory)
The tie-set matrix is given as
⎡1 0 0 1 −1 0 ⎤
⎢
⎥
Ba = ⎢0 1 0 0 1 −1⎥
⎢0 0 1 −1 0 1 ⎥
⎣
⎦
The branch impedance matrix is given as
⎡5 0 0 0 0 0 ⎤
⎢
⎥
⎢0 10 0 0 0 0 ⎥
⎢0 0 5 0 0 0 ⎥
Zb = ⎢
⎥
⎢0 0 0 10 0 0 ⎥
⎢0 0 0 0 5 0 ⎥
⎢
⎥
⎢⎣0 0 0 0 0 5 ⎥⎦
⎡5 0 0 0 0 0 ⎤
⎢
⎥
0 10 0 0 0 0 ⎥
⎡1 0 0 1 −1 0 ⎤ ⎢
⎡ 5 0 0 10 −5 0 ⎤
⎢
⎥ ⎢0 0 5 0 0 0 ⎥ ⎢
⎥
∴ Ba Z b = ⎢0 1 0 0 1 −1⎥ ⎢
0
5 −5 ⎥
⎥ = ⎢0 10 0
0
0
0
10
0
0
⎥ ⎢
⎢0 0 1 −1 0 1 ⎥ ⎢
⎥
⎣
⎦ ⎢0 0 0 0 5 0 ⎥ ⎣0 0 5 −10 0 5 ⎦
⎢
⎥
⎢⎣0 0 0 0 0 5 ⎥⎦
⎡ 1 0 0⎤
⎢
⎥
0 1 0⎥
⎡ 20 −5 −10 ⎤
⎡ 5 0 0 10 −5 0 ⎤ ⎢
⎥ ⎢ 0 0 1⎥ ⎢
⎢
⎥
T
0
5 −5 ⎥ ⎢
∴ Ba Z b Ba = ⎢0 10 0
⎥ = ⎢ −5 20 −5 ⎥
−
1
0
1
⎥ ⎢
⎢0 0 5 −10 0 5 ⎥ ⎢
⎥
⎦ ⎢ −1 1 0 ⎥ ⎣ −10 −5 15 ⎦
⎣
⎢
⎥
⎢⎣ 0 −1 1 ⎥⎦
⎡ −10 ⎤
⎢
⎥
0⎥
⎡1 0 0 1 −1 0 ⎤ ⎢
⎡10 ⎤
⎢
⎥⎢ 0 ⎥ ⎢ ⎥
− BaVs = − ⎢0 1 0 0 1 −1⎥ ⎢
⎥ = ⎢ 0⎥
⎢0 0 1 −1 0 1 ⎥ ⎢ 0 ⎥ ⎢ 0 ⎥
⎣
⎦⎢ 0 ⎥ ⎣ ⎦
⎢
⎥
⎢⎣ 0 ⎥⎦
Thus, the lop equations are given as
⎡ 20 −5 −10 ⎤ ⎡ I1 ⎤ ⎡10 ⎤
⎢
⎥⎢ ⎥ ⎢ ⎥
⎢ −5 20 −5 ⎥ ⎢ I 2 ⎥ = ⎢ 0 ⎥
⎢ −10 −5 15 ⎥ ⎢ I ⎥ ⎢ 0 ⎥
⎣
⎦⎣ 3 ⎦ ⎣ ⎦
144
Network Analysis and Synthesis
Solving by Cramer’s rule, we get the loop currents as
10 −5 −10
0 20
0 −5
I1 =
I2 =
−5
15
20 −5 −10
− 5 20
−5
−10 −5
15
20 10 −10
−5 0 −5
15
−10 0
20 −5 −10
− 5 20
−10 −5
=
2750
= 1.047 A
2625
=
1250
= 0.476 A
2625
=
2250
= 0.857 A
2625
−5
15
20 −5 10
− 5 20 0
−10 −5
I3 =
0
20 −5 −10
− 5 20
−10 −5
Also, the branch currents are given as, Ib
−5
15
BaT IL
⎡ I b1 ⎤ ⎡ 1 0 0 ⎤
⎡ 1 0 0⎤
⎢ ⎥ ⎢
⎢
⎥
⎥
⎢ Ib2 ⎥ ⎢ 0 1 0 ⎥
⎢ 0 1 0 ⎥ ⎡1.047 ⎤
⎡
⎤
I
1
⎢I ⎥ ⎢
0 0 1⎥ ⎢ ⎥ ⎢ 0 0 1⎥ ⎢
⎥
∴⎢ b3 ⎥ = ⎢
⎥ ⎢ I2 ⎥ = ⎢
⎥ ⎢0.476 ⎥
⎢ I ⎥ ⎢ 1 0 −1⎥
1
0
−
1
⎥⎢
⎢ ⎥ ⎢
⎥
⎢ b4 ⎥ ⎢
⎥ ⎣ I 3 ⎦ ⎢ −1 1 0 ⎥ ⎣ 0.857 ⎦
⎢ I b 5 ⎥ ⎢ −1 1 0 ⎥
⎢
⎥
⎢ ⎥ ⎢ 0 −1 1 ⎥
0
−
1
1
⎢
⎥⎦
⎣
⎣
⎦
I
⎢⎣ b 6 ⎥⎦
⎫
(
)
⎪⎪
∴ I = ( −1.047 + 0.476 ) = −0.571 A ⎬
⎪
∴ I = ( −0.476 + 0.857 ) = 0.381 A ⎪⎭
∴ I b 4 = 1.047 − 0.857 = 0.19 A
b5
b6
Problem 3.20 Determine the current i1 in the circuit of Fig. 3.48
using nodal analysis method and graph theory concepts.
Solution By source transformation technique, we convert the
19-V and 25-V voltage sources into current sources.
30 V
5
19 V 2
4
i1
4A
1.5i1
Fig. 3.48
25 V
145
Network Topology (Graph Theory)
Since the 30-V voltage source, the 4-A current
source, and the controlled current source are not
accompanied by the passive elements, we consider
three resistors R1, R2 and R3 and finally let, R1 → 0, R2
→ , and R3 → 0.
The graph of the network is shown.
9.5 A
5
2
i1
4A
30 V
1.5i1
4
6.25 A
4
6.25 A
Fig. 3.49
The complete incidence matrix is
9.5 A
⎡1 −1 0 0 0 0 ⎤
⎢
⎥
0 1 1 1 0 0⎥
Aa = ⎢
⎢0 0 −1 −1 1 1 ⎥
⎢
⎥
⎣1 0 0 0 −1 −1⎦
1
R1
5
2
3
2
R2 1.5i1
i1
4A
30 V
R3
Fig. 3.50
The reduced incidence matrix is
1
⎡1 −1 0 0 0 0 ⎤
⎢
⎥
A = ⎢0 1 1 1 0 0 ⎥
⎢0 0 −1 −1 1 1 ⎥
⎣
⎦
2
(2)
(4)
(3)
(1)
3
(5)
(6)
4
The branch admittance matrix is
⎡G1
⎢
⎢0
⎢
⎢0
Yb = ⎢
⎢0
⎢
⎢0
⎢
⎢0
⎣
Fig. 3.51
0
0
0
0
1
5
0
0
0
0
G2
0
0
0
0
1
0
0
0
0
2
0
G3
0
0
0
0 ⎤
⎥
0 ⎥
⎥
0 ⎥
⎥ where, G = 1 , G = 1 , G = 1
1
2
R2 3 R3
R1
0 ⎥
⎥
0 ⎥
⎥
1 ⎥
4⎦
⎡G1
⎢
⎢0
⎢
⎡1 −1 0 0 0 0 ⎤ ⎢ 0
⎢
⎥
∴ AYb = ⎢0 1 1 1 0 0 ⎥ ⎢
⎢
⎢0 0 −1 −1 1 1 ⎥ ⎢ 0
⎣
⎦
⎢0
⎢
⎢0
⎣
0
0
0
0
1
0
0
0
5
0
G2
0
0
0
0
1
0
0
2
0
0
G3
0
0
0
0
0 ⎤
⎥
0 ⎥
⎥ ⎡G1
0 ⎥ ⎢⎢
⎥= 0
0 ⎥ ⎢⎢
⎥
⎢0
0 ⎥ ⎣
⎥
1 ⎥
4⎦
−1
1
5
5
0
0
0
0
G2
1
0
2
1
−1 −
2
G3
0 ⎤
⎥
⎥
0 ⎥
⎥
1 ⎥
4⎦
146
Network Analysis and Synthesis
⎡G
⎢ 1
⎢
T
∴ AYb A = ⎢ 0
⎢
⎢0
⎣
−1
1
0
0
0
G2
1
0
5
5
2
1
−1 −
2
0
(
⎡G +1
⎢ 1
5
⎢
1
=⎢ −
5
⎢
⎢
0
⎢⎣
)
⎡ 1
⎢
⎤
0 ⎢ −1
⎥
⎥⎢ 0
0 ⎥⎢
⎥⎢ 0
1 ⎥⎢ 0
4 ⎦⎢
⎢⎣ 0
G3
−1
(
5
1
G1 +
+1
5
2
1
−
2
)
(
0⎤
⎥
1 0⎥
1 −1⎥
⎥
1 −1⎥
0 1⎥
⎥
0 1 ⎥⎦
⎤
0
⎥
⎥
−1
⎥
2
⎥
⎥
1
1
G3 +
+
2
4 ⎥⎦
0
)
Now,
⎡ −30G1 ⎤
⎢
⎥
0 ⎥
⎡1 −1 0 0 0 0 ⎤ ⎢
⎢
⎥⎢ 4 ⎥
⎥
AYbVs − AI s = − AI s = − ⎢0 1 1 1 0 0 ⎥ ⎢
⎢0 0 −1 −1 1 1 ⎥ ⎢ −9.5 ⎥
⎣
⎦ ⎢ −1.5i ⎥
1 ⎥
⎢
⎢⎣ 6.25 ⎥⎦
⎡ −30G1 ⎤
⎥
⎢
= −⎢
−5.5
⎥
⎢ 15.75 − 1.5i ⎥
1 ⎦
⎣
)
(
Thus, node equations are
(
⎡G +1
⎢ 1
5
⎢
⎢ −1
5
⎢
⎢
0
⎢⎣
)
−1
(
5
G2 + 7
10
−1
2
⎤
⎥ ⎡V ⎤ ⎡ 30G
⎤
1
⎥⎢ 1 ⎥ ⎢
⎥
−1
5.5
⎥ ⎢V2 ⎥ = ⎢
⎥
2 ⎥
⎥
⎢V ⎥ ⎢1.5i − 15.7
7
5
1
⎦
G3 + 3 ⎥ ⎣ 3 ⎦ ⎣
4 ⎦⎥
0
)
(
)
With R1 → 0, G1 → , R2 → , G2 → 0, the equations become
⎛
1⎞
1
⎜⎝ G1 + 5 ⎟⎠ V1 − 5V2 = 30G1
⎛
1
7⎞
1
− V1 + ⎜ G2 + ⎟ V2 − V3 = 5.5
2
5
10 ⎠
⎝
(
1
3
− V2 + V3 = 1.5i1 − 15.75
2
4
)
{ we made, Vs
0}
147
Network Topology (Graph Theory)
or,
V1 = 30
(i)
1
7
1
− V1 + V2 − V3 = 5.5 ⇒ 7V2 − 5V3 = 115
5
10
2
⎡
⎤
⎛ V −V ⎞
1
3
− V2 + V3 = ⎢1.5⎜ 2 1 ⎟ − 15.75 ⎥ ⇒ 16V2 − 15V3 = 495
2
4
⎢⎣ ⎝ 5 ⎠
⎥⎦
Solving equations (i), (ii), and (iii), we get
V2 = −30 V,
(ii)
(iii)
V3 = 65 V
⎛ V − V ⎞ −30 − 30
Hence, the current, i1 = ⎜ 2 1 ⎟ =
= −12 A
5
⎝ 5 ⎠
Summary
1. Network Topology refers to the properties that relate
to the geometry of the network (circuit).These properties remain unchanged even if the circuit is bent into
any other shape provided that no parts are cut and no
new connections are made.
2. A graph is defined as a collection of points called
nodes, and line segments called branches, the nodes
being joined together by the branches. A subgraph is
subset of the branches and nodes of a graph.
3. For a given connected graph of a network, a connected subgraph is known as a tree of the graph if the
subgraph has all the nodes of the graph without containing any loop.
4. The branches of tree are called twigs. If a graph for a
network is known and a particular tree is specified, the
remaining branches are referred as the links. The collection of links is called a co-tree.
5. Number of nodes in a graph n
Number of independent voltages n 1
Number of tree-branches n 1
Number of links L (Total number of branches)
(Number of tree-branches)
b (n 1)
Total number of branches b L (n 1)
6. Network analysis by the topological method involves
writing KCL and KVL equations with the help of any
one of the three matrices i.e. incidence matrix, tie-set
matrix and cut-set matrix.
Short-Answer Questions
1. Explain ‘network topology’.
Network Topology The word topology refers to the science of place. In mathematics, topology is a branch
of geometry in which figures are considered perfectly
elastic.
Network topology refers to the properties that
relate to the geometry of the network (circuit). These
properties remain unchanged even if the circuit is bent
into any other shape provided that no parts are cut and
no new connections are made.
2. State the advantages offered by the graph theory
as applied to electric circuit problems.
1. Graph theory or network topology deals with
those properties of networks which do not change
with the change in the shape of the networks.
2. All the equations (KCL and KVL) formed by graphtheory concept are independent equations.
3. The graph-theory concept eases the solution
method for solving networks with a large number
of nodes and branches.
148
Network Analysis and Synthesis
3. Define the following terms:
(i) Graph of a network
(ii) Oriented graph
(iii) Rank of a graph
(iv) Planar and non-planar graph
(v) Subgraph
(vi) Path
The subgraph is said to be proper if it consists of strictly less than all the branches and
nodes of the graph.
1
(1)
6
b
6
a
1
b
a
5
4
5
3
d
d
Fig. 3.52 (b)
Graph of the Circuit
(ii) Oriented graph A graph
whose branches are oriented is called a directed
or oriented graph. The
orientation is indicated by
an arrow head in each of
the branches representing the direction of current flow in the branch.
3
3
2
(5)
(5)
(4)
4
Subgraph 1
4
Subgraph 2
3
1
In the example in
Fig. 3.56, branches 2,
d
3, and 4, together with
Fig. 3.56
all the four nodes, constitute a path. A graph
is connected if there
exists a path between any pair of vertices. Otherwise, the graph is disconnected.
1
(1)
(6)
(2)
(3)
2
(5)
3
(4)
4
Fig. 3.53
(iii) Rank of a graph The rank of a graph is (n −1)
where n is the number of nodes or vertices of
the graph.
(iv)
2
(vi) Path A path is a particular subgraph where
only two branches are incident at every node
except the internal
nodes (i.e., starting and
6
finishing nodes). At the
b 5
4
a
c
internal nodes, only
2
one branch is incident.
2
1
Fig. 3.52 (a) Circuit
(2)
Fig. 3.55
c
c
3
2
(2)
(6)
(i) Graph of a network A linear graph (or simply a
graph) is defined as a collection of points called
nodes, and line segments called branches, the
nodes being joined together by the branches.
4
1
Planar and non-planar graph A graph is planar
if it can be drawn in a plane such that no two
branches intersect at a point which is not a node.
4. Show that the number of links for a graph having n
nodes and b branches is b ⴚ n 1.
Let, n
Number of nodes in a graph
number of independent voltages
number of tree-branches
number of links, L
n
n
1
1
(total number of branches)
(number of tree-branches)
b (n 1) b n 1
5. Enlist the properties of a tree.
Fig. 3.54 (a)
Planar graph
Fig. 3.54 (b)
non-planar Graph
(v) Subgraph A subgraph is a subset of the
branches and nodes of a graph. For example,
for the graph shown in Fig. 3.53, some subgraphs are shown below, in Fig. 3.55.
1. In a tree, there exists one and only one path between
any pairs of nodes.
2. Every connected graph has at least one tree.
3. A tree contains all the nodes of the graph.
4. There is no closed path in a tree and hence, tree is
circuitless.
5. The rank of a tree is (n 1).
149
Network Topology (Graph Theory)
6. List the properties of an incidence matrix.
1. The sum of the entries in any column is zero.
2. The determinant of the incidence matrix of a closed
loop is zero.
3. The rank of incidence matrix of a connected graph is
(n 1).
7. Show that the determinant of the complete incidence matrix of a closed loop is zero.
a
4
1
b
5
6
3
2
b 5
4
a
c
2
1
c
3
4. A cut-set is oriented by selecting an orientation
from one of the two parts to the other. Generally,
the direction of a cut-set is chosen to be same as the
direction of the tree branch.
9. Show that for a network graph with P separate
parts, n nodes and b branches, the number of
chords C is given as C b n P.
We know that if a connected graph of a network has n
nodes and b branches then number of tree branches or
twigs, bt (n 1)
number of links or chords, C
6
(b bt )
n 1
b
d
d
Fig. 3.57 (a)
Network
Fig. 3.57 (b)
Graph of network
The complete incidence matrix of the graph Aa is given
as Fig. 3.57
2
3
5
1 0
6
0 Reduced
1
0
Aa = Nodes b
0
1 0
c
0
0
1
0
1
1 matrix AI
d
1
1
1
0
0
1
1
1 0
incidence
However, for the closed loop consisting of the branches
1, 2 and 4, the complete incidence matrix is given as
⎡
1 2 4⎤
⎢
⎥
a −1 0 1 ⎥
Aa = ⎢
⎢ b 0 −1 −1⎥
⎢
⎥
⎣d 1 1 0 ⎦
The determinant value of this matrix comes to be zero.
Therefore, we can conclude that the complete incidence matrix of a closed loop is zero.
8. Mention some properties of a cut-set.
n1
n2
n3
np
Again, number of twigs for the first part, bt1
(n1
1)
Number of twigs for the second part, bt2
(n2
1)
Number of twigs for the third part, bt3
Number of twigs for the pth part, btp
(n3
(np
1)
1)
Hence, total number of twigs is,
bt
(n1
1) + (n2
1)
(n3
number of chords, C
(n1
n2
(b
n3
1)
np)
(np
1)
1
P
(n
P)
bt )
b
(n
P)
b
n
P
10. Prove that in a linear graph, every cut-set has
an even number of branches in common with every
loop.
A cut-set is a minimum number of branches of a connected graph that when cut, or removed from the
graph, separates the graph into two groups of nodes. A
cut-set is said to be a fundamental cut-set if it contains
only one tree branch.
1. A cut-set divides the set of nodes into two subsets.
We consider the graph as shown in Fig. 3.58.
2. Each fundamental cut-set contains one tree-branch,
the remaining elements being links.
The fundamental cut-sets are
3. Each branch of the cut-set has one of its terminals
incident at a node in one subset and its other terminal at a node in the other subset.
1)
Let, n1, n2, n3, …nP be the number of nodes of the
first, second, third, … pth part of the graph, respectively,
so that the total number of nodes of the graph is
n
a
Reference
node
0
4
(n
Now, instead of a connected graph, if we have a network graph with P separate parts then the number of
chords is calculated as explained below.
Branches
1
b
f-cut-set – 1: [1, 2, 6]
f-cut-set – 2: [2, 3, 5, 6]
f-cut-set – 3: [4, 5, 6]
150
Network Analysis and Synthesis
C1
1
1
C2
(1)
(1)
(2)
(6)
(6)
Loop 2
(5)
(3)
(5)
3
2
3
2
(2)
Loop 1
(3)
Loop 3
(4)
(4)
4
Fig. 3.59
Graph Illustrating Loops
C3
4
Fig. 3.58 Graph Illustrating
Fundamental cut-set
Therefore, a fundamental cut-set contains only one tree
branch and the other branches being the links of the graph.
On the other hand, a loop always consists of one link and
the other branches being the tree branches of the graph.
It is shown in Fig. 3.59.
The loops are
Loop 1: [1, 2, 3] Loop2: [3, 4, 5] Loop 3: [1, 2, 4, 6]
As every fundamental cut-set must contain one tree
branch and at least one link, and every loop also must
contain one link and at least one tree branch, we can
say that every fundamental cut-set has two branches
in common with every loop.
In the similar way, considering a cut-set for an unconnected graph, which contains more than one tree
branch, we can show that every fundamental cut-set
has an even number of branches in common with
every loop.
Exercises
1. For the network shown in Fig. 3.60, draw the graph
and a possible tree. Show the links and write the tieset matrix. Write the equations of the branch currents
in terms of loop currents.
4
2
5
1
tions containing branch currents and loop currents.
All the values are in ohms.
A
3
4
6
1
6
Fig. 3.60
6
8V
4
2
3
5
B
2. Find out the currents through and voltage across all
branches of the network shown in fig. 3.61 with the
help of its tie-set schedule.
2
8
D
7
C
9
Fig. 3.62
4. Draw the graph of the circuit shown in Fig. 3.63 and
select a suitable tree to write tie-set matrix.
6
10
10
4
2
12 V
6V
Fig. 3.61
3. Find a tree from the graph of the network shown in
Fig. 3.62 Make the tie-set matrix and write the equa-
5V
10
20
20
1 A, i3
0.5 A]
Fig. 3.63
[i1
3 A, i2
151
Network Topology (Graph Theory)
5. For the given network of
Fig. 3.64 draw the graph
and a tree. Write the cutsets and the cut-set matrix
of the tree. Write the
equations of link branch
voltages in terms of tree
branch voltages.
A
8
E
VX
F
D
B
15
4
100 V
Vx /14
C
Fig. 3.67
Fig. 3.64
6. For the given network of
Fig. 3.65 draw the graph and a tree. Write the cut-sets
and the cut-set matrix of the tree. Write the equations of
link branch voltages in terms of tree branch voltages. All
the values are in ohms.
A
2
3
3
1
2
1
1
C
3
Fig. 3.65
7. The linear oriented
graph is given in
Fig. 3.66 Considering
a tree, mark all the
fundamental cut-sets
and form the cut-set
matrix.
8
9
6
7
3
1
2
Fig. 3.66
8. For the network shown in Fig. 3.67, determine
(a) tie-set matrix,
(b) loop impedance matrix, and
(c) loop currents.
9. Select the (i) fundamental
1
2
loops, and (ii) fundamental
5
4
7
3
6
cut-sets corresponding to
a tree of the network graph
Fig. 3.68
which is shown by solid lines
in Fig. 3.68. Hence write
the KCL and KVL equations for the network in matrix
form.
10. Draw the graph of the network in Fig. 3.69. Select a
tree with tree branches of
elements (1) and (2) and
write the equilibrium equation taking tree branch
voltages as variables.
2
B
[7 A, 4 A]
(4) 3
2A
(1)
4
2
(2)
5
(3)
Fig. 3.69
4
5
11. The incidence matrix of
a network graph is given below. Draw the oriented
graph.
⎡1
⎢
0
A= ⎢
⎢0
⎢
⎣0
0 0 1⎤
⎥
1 0 0⎥
0 −1 1 −1⎥
⎥
0 0 −1 0 ⎦
0 0 0 1
1 0 0 −1
0 1 0
0 0 1
Questions
1. Give the topological description of networks.
2. (a) Define the following terms:
(i) Graph of a network
(ii) Oriented graph
(iii) Rank of a graph.
(iv) Planar and non-planar graph
(vi) Subgraph
(v) Path
(b) State the advantages offered by graph theory as
applied to electric circuit problems.
3. What is meant by a graph? How does a graph help in
circuit analysis?
4. (a) Define a tree of a graph of a network. Mention some
basic properties of a ‘tree’. How can you calculate
the number of possible trees of a given graph?
(b) Define the followings:
(i) Twigs
(ii) Co-tree
(iii) Links or chords
152
Network Analysis and Synthesis
5. Show that the number of links for a graph having n
nodes and b brances is b n 1.
(c) Write a tie-set schedule and formulate the equilibrium equation on loop current basis.
6. Show that for a network graph with P separate parts, n
nodes and b branches, the number of chords C is given
as C b n P.
11. (a) Define a cut-set in a network graph. How can you
find out a fundamental cut-set? Mention some
properties of a cut-set.
(b) Define a cut-set matrix with an illustrative example
and show that the matrix equation QIb 0, where
Q is the cut-set matrix and Ib represents the branch
current matrix of the graph.
(c) Briefly discuss the relation between branch voltage matrix and node voltage matrix in terms of
cut-sets.
7. Explain with illustrative examples the meaning of the
following terms:
(a) Incidence matrix
(b) Tie-set matrix
(c) Cut-set matrix
8. (a) Explain what is meant by incidence matrix of a
graph and indicate how the values of the incidence
matrix elements are obtained.
13. (a) Write notes on network equilibrium equation.
(b) List the properties of an incidence matrix.
(c) How can you determine the number of possible
trees of a graph with this matrix?
9. Show that the determinant of the complete incidence
matrix of a closed loop is zero.
10. (a) Explain the term ‘tie-set’ and ‘tie-set matrix’ of a
network with an illustrative example.
(b) Show that the matrix equation,
Ib
12. Prove that in a linear graph, every cut-set has an even
number of branches in common with every loop.
BTIL
where, B is the tie-set matrix and Ib and IL represent the branch current and loop current matrix respectively.
(b) Establish that the independent loop equations of a
network can be formulated from the tie-set matrix
of its graph, with illustrative examples.
(c) Establish the formulation of node equations of a
network from the cut-set matrix.
14. Using the topological properties of a network graph,
describe the step-by-step procedure of analyzing a
network by the node voltage method.
15. Using the topological properties of a network graph,
describe the step-by-step procedure of analyzing a
network by the loop current method.
Multiple-Choice Questions
1. The number of links for a graph having ‘n’ nodes and
‘b’ branches are
(i) b – n 1
(iii) b n – 1
(ii) n – b 1
(iv) b n
2. The tree branches of a graph are called
(i) chords
(iii) twigs
(ii) links
(iv) co-tree
5. For a connected planar graph of v vertices and e
edges, the number of meshes is
(i) (e
v
1)
(ii) (e
v
1)
(iii) (e
v
1)
6. The number of chords of a tree of a connected graph
G of v vertices and e edges is
(i) (v
1)
(ii) (e
v
1)
(iii) (e
v
1)
7. The table meant for the oriented graph represents a/an
3. The tie-set matrix gives the relation between
(i) branch currents and link currents
(ii) branch voltages and link currents
(iii) branch currents and link voltages
(iv) none of these
3
4. The graph of a network has six branches with three
tree branches. The minimum number of equations
required for the solution of the network is
(i) 2
(ii) 3
(iii) 4
(iv) 5
Fig. 3.70
I3
2
I1
1
153
Network Topology (Graph Theory)
The parallel branches in the graph are
← Branch →
Link or loop current
1
2
3
i1
1
−1
0
i2
0
1
1
(i) tie-set matrix
(iii) incidence matrix
(i) 1 and 2
(ii) 2 and 3
(iii) 6 and 7
(iv) none of the above
12. For a given network, the incidence matrix is given as
1
(ii) cut-set matrix
(iv) none of the above
1
2
3
4
5
4
5
6
⎡1 0 0 1 −1 0 ⎤
⎢
⎥
⎢0 1 0 −1 1 −1⎥
⎢0 0 1 0 0 1 ⎥
⎣
⎦
8. The reduced incidence matrix of a circuit is given by
6
a ⎡1 −1 −1 −1 0 0 ⎤
⎢
⎥
Ai = b ⎢0 1 0 0 −1 1 ⎥
c ⎢⎣0 0 1 0 1 0 ⎥⎦
2 3
The series branches in the graph are
(i) 3 and 4
(ii) 3 and 5
(iii) 3 and 6
(iv) none of the above
13. For a given network the incidence matrix is given as
The set of branches forming a tree are
(i) 1, 2 and 3
(ii) 2, 3 and 5
1
(iii) 1, 2 and 4
(iv) 1, 2 and 6.
⎡1 0 0 1 −1 0 ⎤
⎢
⎥
⎢0 1 0 −1 1 −1⎥
⎢0 0 1 0 0 1 ⎥
⎣
⎦
9. Relative to a given fixed tree of a network,
1. link currents form an independent set
2. branch currents form an independent set
3. link voltages form an independent set
4. branch voltages form an independent set of these
statements,
(i)
(ii)
(iii)
(iv)
1, 2, 3 and 4 are correct
1, 2 and 3 are correct
2, 3 and 4 are correct
1, 3 and 4 are correct
2 3
4
5
The parallel branches in the graph are
(i) 3 and 5
(ii) 4 and 5
(iii) 3 and 6
(iv) none of the above
14. Which one of the following represents the total
number of trees in the graph given in Fig. 3.71?
1
2
2
3
10. For a given network, the incidence matrix is given as
1
2
3
4
5
6
6
4
7
⎡ 1 0 0 1 0 −1 1 ⎤
⎢
⎥
⎢ −1 −1 1 0 0 0 0 ⎥
⎢ 0 1 0 −1 1 0 0 ⎥
⎣
⎦
1
5
3
4
Fig. 3.71
The series branches in the graph are
(i) 4
(i) 3 and 4
(ii) 6 and 7
(iii) 2 and 3
(iv) none of the above
11. For a given network, the incidence matrix is given as
1
2
3
4
5
6
7
⎡ 1 0 0 1 0 −1 1 ⎤
⎢
⎥
⎢ −1 −1 1 0 0 0 0 ⎥
⎢ 0 1 0 −1 1 0 0 ⎥
⎣
⎦
(ii) 5
(iii) 6
15. In the graph and the tree
shown in Fig. 3.72 the fundamental cut-set for the
branch 2 is
(i) 2, 1, 5
(ii) 2, 6, 7, 8
(iii) 2, 1, 3, 4, 5
(iv) 2, 3, 4
(iv) 8
7
6
4
5
8
3
1
2
Fig. 3.72
154
Network Analysis and Synthesis
16. In the graph shown in Fig. 3.73, one possible tree is
formed by the branches 4, 5, 6, 7. Then one possible
fundamental cut set is
19. The number of chords in the graph of the given circuit
will be
8
6
7
1
3
2
4
5
Fig. 3.74
Fig. 3.73
(i) 1, 2, 3, 8
(ii) 1, 2, 5, 6
(iii) 1, 5, 6, 8
(iv) 1, 2, 3, 7, 8
(i) 3
17. Which one of the following statements is correct?
A tree in a network is a connected graph containing
(ii) 4
(iii) 5
20. Consider the network graph
shown in Fig. 3.75. Which one of
the following is NOT a ‘tree’ of
this graph?
(iv) 6
Fig. 3.75
(i) all the nodes only
(ii) all the branches only
(i)
(ii)
(iii)
(iv)
(iii) all the branches and nodes
(iv) all the nodes but no close path
1
2
3
4
5 6
⎡1 −1 −1 −1 0 0 ⎤
⎢
⎥
18. A = 0
1 0 0 −1 1 ⎥
⎢
⎢0 0 1 0 1 0 ⎥
⎣
⎦
21. In the following graph, the
number of trees (P) and the
number of cut-sets (Q) are
For the reduced incidence matrix given, which is the
set of branches forming a tree?
(i) 1, 2, 3
(ii) 2, 4, 6
(iii) 2, 3, 5
(iv) 1, 4, 6
(i) P
(ii) P
(iii) P
(iv) P
2, Q
2, Q
4, Q
4, Q
2
6
6
10
(1)
(2)
(3)
(4)
Fig. 3.76
Answers
1.
2.
3.
4.
5.
(i)
(iii)
(i)
(ii)
(i)
6.
7.
8.
9.
(ii)
(i)
(i)
(ii)
10.
11.
12.
13.
(iv)
(iii)
(iii)
(ii)
14.
15.
16.
17.
(iv)
(ii)
(iv)
(iv)
18.
19.
20.
21.
(i)
(i)
(ii)
(iii)
4
Network Theorems
Introduction
A theorem is a relatively simple rule used to solve a problem, derived from a more intensive analysis
using fundamental rules of mathematics. At least hypothetically, any problem in mathematics
can be solved just by using the simple rules of arithmetic, but human beings are not as consistent
or as fast as a digital computer. We need some shortcut methods in order to avoid procedural
errors.
In electric network analysis, the fundamental rules are Ohm’s law and Kirchhoff’s laws. While these
humble laws may be applied to analyze any circuit configuration, for complex circuits, it is sometimes
necessary to simplify the network to find current or voltage in a particular branch without solving the
entire circuit. For this purpose, there are some ‘shortcut’ methods of analysis, known as network theorems. As with any theorem of geometry or algebra, the network theorems are also derived from fundamental rules.
4.1
NETWORK THEOREMS
In this chapter, we will discuss the following network theorems:
1. Substitution theorem
2. Superposition theorem
3. Reciprocity theorem
4. Thevenin’s theorem
5. Norton’s theorem
6. Maximum power transfer theorem
7. Tellegen’s theorem
8. Millman’s theorem
9. Compensation theorem
156
Network Analysis and Synthesis
4.2
SUBSTITUTION THEOREM
Statement
Any branch in a network may be substituted by a different branch without disturbing the voltages and currents in the
entire network, provided the new branch has the same set of terminal voltage and current as the original branch.
Proof In a network N, let the number of branches be ‘b’. The branch method requires the solution of 2b
equations. Now, after substitution, (2b 1) branch equations remain unaltered. However, as the branch voltage and current of the replaced branch remain unaltered, it implies that the set of 2b simultaneous equations
will still be satisfied with the same voltage and currents as before. This proves the substitution theorem.
Points to be Noted
(i) The substitution theorem is a general theorem and is applicable for any arbitrary network.
(ii) This theorem is used to replace an impedance branch by either a voltage source or a current source or
a voltage source with a series impedance without altering the branch voltage and current. The restrictions imposed are that the branch should not be coupled to other branches in the circuit and the modified network must have a unique solution.
(iii) This theorem is very useful in circuit analysis of network having non-linear elements.
(iv) This theorem cannot be applied to a branch which is coupled to other branches in the circuit.
(v) This theorem cannot be applied if the branch voltage and current are not known.
Example 4.1 We consider the branch xy of the circuit shown in Fig. 4.1. The branch voltage Vxy ⴝ 50 V,
and branch current Ixy ⴝ 5 A. This branch can be substituted by any other branch as shown without altering
the voltage and current in the branch.
10
10
100V
10
x
10
x
10
10
x
x
5A
5
5A
50V
25V
y
y
y
(a)
(b)
(c)
(d)
y
Fig. 4.1 Illustration of substitution theorem
The branch can be substituted using the relation as given below.
Vxy ZxyIxy E, before substitution
Vxy Zxy Ixy E , after substitution
4.3
SUPERPOSITION THEOREM
Statement
This theorem states that in a linear bilateral network, the current at any point (or voltage between any two
points) due to the simultaneous action of a number of independent sources in the network is equal to the
summation of the component currents (or voltages). A component current (or voltage) is defined as that due
to one source acting alone in the network with all the remaining sources removed.
157
Network Theorems
r1(t )
r (t)
Linear
passive
bilateral
network
e1(t )
r2(t)
Linear
passive
bilateral
network
e1(t)
Linear
passive
bilateral
network
e2(t)
e2(t )
Fig. 4.2 Illustration of superposition theorem
Proof
Z1
E1
I1
Z2
Z3
I2
E1
E2
Z1
I1
Z3
Z2
Z1
Z2
I2
I 1 Z3
I 2
(b)
(a)
(c)
Fig. 4.3 Proof of superposition theorem
Using KVLfor the above network, as shown in Fig. 4.3 (a),
E1 = I1 ( Z1 + Z 3 ) + I 2 Z 3
and
E2 = I1 Z 3 + I 2 ( Z 2 + Z 3 )
Solving the above two equations,
I1 =
Z2 + Z3
Z3
E1 −
E
Z1 Z 2 + Z 2 Z 3 + Z 3 Z1
Z1 Z 2 + Z 2 Z 3 + Z 3 Z1 2
I1 =
Z1 + Z 3
− Z3
E
E1 +
Z1 Z 2 + Z 2 Z 3 + Z 3 Z1 2
Z1 Z 2 + Z 2 Z 3 + Z 3 Z1
Making E2 inoperative, the circuit diagram becomes as shown in Fig. 4.3 (b),
Then the KVL equations are
E1 = I1′ ( Z1 + Z 3 ) + I 2′ Z 3
and
0 = I1′ Z 3 + I 2′ ( Z 2 + Z 3 )
and
I 2′ =
Solving the above two equations,
I1′ =
Z2 + Z3
E
Z1 Z 2 + Z 2 Z 3 + Z 3 Z1 1
− Z3
E
Z1 Z 2 + Z 2 Z 3 + Z 3 Z1 1
Making E1 inoperative, the circuit diagram becomes as shown in Fig. 4.3 (c),
Then the KVL equations are
0 = I 2′′( Z1 + Z 3 ) + I 2′′Z 3 and
E2 = I 2′′Z 3 + I 2′′( Z 2 + Z 3 )
Solving the above two equations,
I1′′=
− Z3
E
Z1 Z 2 + Z 2 Z 3 + Z 3 Z1 2
and
I 2′′ =
Z2 + Z3
E
Z1 Z 2 + Z 2 Z 3 + Z 3 Z1 2
E2
158
Network Analysis and Synthesis
I1 = I1′ + I1′′, I 2 = I 2′ + I 2′′
So,
If an excitation e1(t) alone gives a response r1(t), and an excitation e2(t) alone gives a response r2(t) then,
by the superposition theorem, the excitation e1(t) and the excitation e2(t) together would give a response r(t)
r1(t) r2(t)
The superposition theorem can even be stated in a more general manner, where the superposition occurs
with scaling.
Thus an excitation of k1e1(t) and an excitation of k2e2(t) occurring together would give a response of k1r1(t)
k2 r2(t).
Steps to Apply Superposition Theorem
1. Only one source is considered to act alone. The other sources are replaced by their internal impedances,
i. e., ideal independent voltage sources are short-circuited and ideal independent current sources are
open-circuited. All dependent sources will act normally.
2. Using any suitable network analysis technique, the current through or the voltage across the desired
element is found out due to the source under consideration.
3. The above steps are repeated considering all the independent sources one by one.
4. The total response (current or voltage) is obtained by taking the algebraic sum of all the responses.
Points to be Noted
(i) This theorem is valid for all types of linear circuits having time-varying or time-invariant elements.
(ii) This theorem is used to find the current or voltage in a branch when the circuit has a large number of
independent sources.
(iii) This theorem is not valid for power relationship.
(iv) This theorem is not applicable to circuits containing only dependent sources. With dependent sources,
superposition can be used only when the controlling functions are external to the network containing
sources, so that the controls are unchanged when the sources act at a time.
(v) This theorem is not applicable for circuits with non-linear elements.
(vi) This theorem is not useful for circuits with only one independent source.
Example 4.2 Find the current ‘I’ in the circuit shown in Fig. 4.4
using the superposition theorem.
1
I
2
Solution We consider two cases:
1A
1
3
Case (1) When the 1-V voltage source is acting alone
For Fig. 4.5 (a), the current through the 2- resistance in this case is
Fig. 4.4 Circuit of Example 4.2
1
I′=− A
3
Case (2) When the 1-A current source is acting alone
1
1
For Fig. 4.5(b), the current through the 2- resistance in this case is I ′′ = 1 ×
= A
1+ 2 3
By superposition theorem, the current when both the sources are acting simultaneously is
1 1
I = ( I ′ + I ′′ ) = − + = 0
3 3
1V
159
Network Theorems
I
I
1
1
2
2
1
3
Fig. 4.5 (a) Voltage source acting alone
4.4
1A
1V
1
Fig. 4.5 (b)
3
Current source acting alone
RECIPROCITY THEOREM
Statement
In any linear, bilateral and time-invariant network, the ratio of response to excitation remains same for an
interchange of the position of excitation and response in the network.
Proof Let us consider a network ‘N’ having only one driving voltage source E Ei in the loop ‘i’ and the
current source Ij in the loop j, Then Ij YjiEi.
Next, interchanging the positions of cause and effect, i.e., placing the same voltage source E Ej in the
loop j, we get the current response Ii in the loop ‘i’ as Ii Yij Ej
Then Ii will be equal to Ij provided, Yij Yji
This is the condition for reciprocity, Yij Yji for all j and i signifies that the admittance matrix Y is symmetric.
Points to be Noted
(i) This theorem is applicable to the networks comprising of linear, time-invariant, bilateral, passive elements, such as ordinary resistors, inductors, capacitors and transformers.
(ii) This theorem is inapplicable to unilateral networks, such as networks comprising of electron tubes or
other control devices.
(iii) This theorem is inapplicable to circuits with time-varying elements.
(iv) This theorem is inapplicable to circuits with dependent sources.
(v) To apply this theorem, we have to consider only the zero-state response by taking all the initial conditions to be zero.
Example 4.3 Verify the reciprocity theorem for the network
shown in Fig. 4.6 using a current source and a voltmeter.
2
Solution Using a current source and a voltmeter, the circuit is
shown in Fig. 4.7 (a).
By KCL,
At the node (1), ⇒ 3e1 − e2 − 2i1 = 0
(i)
At the node (2)
⇒ − 6 e1 + 13e2 − 3v1 = 0
At the node (3)
9v1 = 5e2
From (ii)
9
⇒ − 6 e1 + 13 × v1 − 3v1 = 0
5
117
⇒ − 6 e1 + (
− 3)v1 = 0
5
17
102
⇒ 6 e1 +
v ⇒ e1 = v1
5 1
5
1
3
Fig. 4.6
(ii)
(iii)
1
i1
4
1
5
Circuit of Example 4.3
2
2
3
4
3
5
Fig. 4.7 (a) Circuit of Example 4.3 with current
source and voltmeter
V
160
Network Analysis and Synthesis
⇒ 3×
From (i)
⎛ i ⎞ ⎛ 21⎞
17
9
v1 − v1 = 2i ⇒ ⎜ 1 ⎟ = ⎜ ⎟ (A)
5
5
⎝ v1 ⎠ ⎝ 5 ⎠
Interchanging the positions of the current source and the
voltmeter, the circuit is shown in Fig. 4.7 (b).
By KCL,
At the node (1)
⇒ 3v2 = e2
(iv)
At the node (2)
V
⇒ − 6 v2 + 13e2 − 3e3 = 0
At the node (3)
2
4
3
1
3
i2
5
Fig. 4.7 (b) Circuit of Fig. 4.7 (a) interchanging
the position of source and excitation
⇒ − 6 v2 + 13 × 3v2 − 3e3 = 0
⇒ e3 = 11v2
2
1
(v)
⇒ 5e3 − 5e2 + 4 e3 − 20i2 = 0
⎛ i ⎞ ⎛ 21⎞
⇒ 20i2 = 9e3 − 5e2 = 9 × 11v2 − 5 × 3v2 = 84 v2 ⇒ ⎜ 2 ⎟ = ⎜ ⎟ (B)
⎝ v2 ⎠ ⎝ 5 ⎠
From equations (A) and (B), reciprocity theorem is proved.
4.5
THEVENIN’S THEOREM
Statement
A linear active bilateral network can be replaced at any
two of its terminals by an equivalent voltage source
(Thevenin’s voltage source), Voc, in series with an
equivalent Impedance (Thevenin’s impedance), Zth.
Here, Voc is the open circuit voltage between the two
terminals under the action of all sources and initial conditions, and Zth is the impedance obtained across the
terminals with all sources removed by their internal
impedance and initial conditions reduced to zero.
Linear
active
bilateral
network
Fig. 4.8
Z Thevenin
Z in
where, K0, K1, K2, K3, K4 are constants.
E Thevenin
Illustration of Thevenin’s theorem
a
Proof We consider a linear active circuit of Fig. 4.9 (a). An external current
source is applied through the terminals a–b where we have access to the circuit.
We have to prove that the v–i relation at terminals a–b of Fig. 4.9 (a) is
identical with that of the Thevenin’s equivalent circuit of Fig. 4.9 (b).
For simplicity, we assume that the circuit contains two independent
voltage sources Vs1 and Vs2 and two independent current sources
Is1 and Is2.
Considering the contribution due to each independent source
I
including the external one, the voltage at a–b, V, is, by superposition
theorem,
V = K0 I + K1Vs1 + K 2Vs 2 + K 3 I s1 + K 4 I s 2
Voc
I
Linear
circuit
V
b
Fig. 4.9 (a) A currentdriven circuit
a
Zth
V
b
Fig. 4.9 (b) Thevenin’s Equivalent
Circuit
Vth
161
Network Theorems
V = K0 I + P0
or,
where,
P0 = K1Vs1 + K 2Vs 2 + K 3 I s1 + K 4 I s 2
(4.1)
total contribution due to internal independent sources
To evaluate the constants K0 and P0 of Eq. (4.1), two conditions are
(i) When the terminals a and b are open-circuited
I 0, and V Voc Vth
From Eq. (4.1), Vth Voc P0 ⇒ Vth = P0
(ii) When all the internal sources are turned off
P0 0 and the equivalent impedance is Zth
From Eq. (4.1), V K0I
V
= K0 = Z th ⇒ K0 = Z th
I
Thus, substituting the values of K0 and P0, the v–i relation becomes,
Or,
V = Z th I + Vth
This represents the v–i relationship of Fig. (b). So, Thevenin’s theorem is proved.
Points to be Noted
(i) This theorem is very useful for replacement of a large portion of a network with a small equivalent
circuit. This is useful for calculating the load resistance in impedance-matching problems.
(ii) This theorem is applicable to any linear, bilateral, active network.
(iii) To apply this theorem, the load branch should not be magnetically coupled to any other branch in the
circuit and the load should not contain any dependent source.
(iv) This theorem is inapplicable to non-linear and unilateral networks.
4.6
NORTON’S THEOREM
Statement
A linear active bilateral network can be replaced at any
Linear
two of its terminals, by an equivalent current source
active
YN
Isc
(Norton’s current source), Isc, in parallel with an equivaIsc
YN
bilateral
lent admittance (Norton’s admittance), YN.
network
Here, Isc is the short-circuit current flowing from one
terminal to the other under the action of all sources and
Fig. 4.10 Illustration of Norton’s theorem
initial conditions, and YN is the admittance obtained across
the terminals with all sources removed by their internal impedance and initial conditions reduced to zero.
Proof We consider a linear active circuit of Fig. 4.11 (a). An external voltage source is applied through
the terminals a–b where we have access to the circuit.
We have to prove that the v–i relation at terminals a–b of Fig. 4.11 (a) is identical with that of the Norton’s
equivalent circuit of Fig. 4.11 (b).
162
Network Analysis and Synthesis
For simplicity, we assume that the circuit contains
two independent voltage sources Vs1 and Vs2 and two
independent current sources Is1 and Is2.
Considering the contribution due to each independent source including the external one, the entering at
a, I, is, by superposition theorem,
I = K0V + K1Vs1 + K 2Vs 2 + K 3 I s1 + K 4 I s 2
where, K0, K1, K2, K3, K4 are constants.
I = K0V + P0
Or,
I
I
a
Linear
circuit
V
b
Fig. 4.11 (a) A voltagedriven circuit
a
YN
V
Is
b
Fig. 4.11 (b) Norton’s
equivalent circuit
(4.2)
where, P0 = K1Vs1 + K 2Vs 2 + K 3 I s1 + K 4 I s 2 total contribution due to internal independent sources
To evaluate the constants K0 and P0 of Eq. (4.2), two conditions are
When the terminals a and b are short-circuited,
V 0, and I
Isc IN
From Eq. (4.2), Isc P0 ⇒ I sc = − P0
When all the internal sources are turned off
P0 0 and the equivalent admittance is YN.
From Eq. (4.2), I K0V
Or,
I
= K0 = YN
V
⇒
K0 = YN
Thus, substituting the values of K0 and P0, the v–i relation becomes,
I = VYN − I N
(4.3)
This represents the v–i relationship of Fig. (b). So, Norton’s theorem is proved.
Points to be Noted
(i) This theorem is very useful for replacement of a large portion of a network with a small equivalent
circuit. This is useful for calculating the load resistance in impedance-matching problems.
(ii) This theorem is applicable to any linear, bilateral, active network.
(iii) To apply this theorem, the load branch should not be magnetically coupled to any other branch in the
circuit and the load should not contain any dependent source.
(iv) This theorem is inapplicable to non-linear and unilateral networks.
(v) This theorem is inapplicable for active load.
4.6.1 Steps for Determination of Thevenin’s/Norton’s Equivalent Circuit
1. The portion of the network across which the Thevenin’s or Norton’s equivalent circuit is to be found
out is removed from the network.
2. (a) The open-circuit voltage (Voc or Vth) is calculated keeping all the sources at their normal values.
(b) The short-circuit current (Isc or IN) flowing from one terminal to the other is calculated keeping all
the sources at their normal values.
163
Network Theorems
Calculation of Zth or YN
When the Circuit Contains Only Independent Sources
• All voltage sources are short-circuited.
• All current sources are open-circuited.
• Equivalent impedance or admittance is calculated looking back to the circuit with respect to the two terminals.
When the Circuit Contains Both Dependent and Independent Sources
• Open-circuit voltage (Voc) is calculated with all sources alive.
• Short-circuit current (Isc) is calculated with all sources alive.
V
I sc
• Thevenin’s impedance is obtained as, Z th = oc =
1
YN
When the Circuit Contains Only Dependent Sources
• In this case, Voc 0.
• We connect a test voltage (or current) source at the terminals a and b and the current flowing through
a–b (voltage drop between the terminals a–b) is calculated.
V
• Thevenin’s impedance is obtained as, Z th = test = 1
I test YN
Finally, Thevenin’s equivalent circuit is obtained by placing Voc in series with Zth and Norton’s equivalent
is obtained by placing Isc in parallel with YN.
Example 4.4 Find both Thevenin’s and Norton’s equivalent circuit for the network shown in Fig. 4.12. All
resistance values are in ohm.
2
5V
2
2
1
1
2A
Fig. 4.12 Circuit of Example 4.4
Solution The circuit has only independent sources. We find the Thevenin equivalent impedance by removing
the sources.
⎛2 ⎞ 5
⇒ RN = Rth = ⎜ + 1⎟ =
⎝3 ⎠ 3
2
2
Fig. 4.13 (a)
2
2
1
1
R th
1
1
Fig. 4.13 (b)
R th
164
Network Analysis and Synthesis
Short-circuiting the terminals,
1
2
1
2
2
5V
2A
Isc
Fig. 4.13 (c)
By superposition theorem, when the 5-V source is acting alone,
5
= 7A
5
7
I1 4.5 A
I2 2 A
Isc 1 A
and when the 2-A source is acting alone,
I=
2
2
2
5V
1
2
I2
1
1
2
Isc
Fig. 4.13 (d)
1
2
I1
I
2
1
2A Isc
2A Isc
1
Fig. 4.13 (e)
Fig. 4.13 (f)
3 = 4A
∴ I sc′′ = 2 ×
2 +1 5
3
⎛ 4⎞ 9
∴ total I sc = ( I sc′ + I sc′′) = ⎜ 1 + ⎟ = A
⎝ 5⎠ 5
5/3
A
2
A
or, 3 V
5/3
9/5 A
B
Fig. 4.13 (g)
B
Fig. 4.13 (h)
9 5
∴Vth = I sc × Rth = × = 3 V
5 3
i0
So, the circuits are shown.
10
A
Example 4.5 Find Thevenin’s equivalent circuit across the terminals A and B for the network
shown in Fig. 4.14.
12V
Solution The circuit has both dependent and
independent sources. We find Vth and Isc and then
taking the ratio we get Zth.
Fig. 4.14
2i0
5
B
165
Network Theorems
To find Vth
i0
10
A
By KVL for the supermesh shown,
10i0 + Vth − 12 = 0 ⇒ Vth = 12 − 10i0
(i)
5
2i0
12V
Vth
By KCL at the node A,
−i0 − 2i0 +
Vth
= 0 ⇒ Vth = 15i0
5
From (i) and (ii) we get, Vth 7.2 V
B
(ii)
Fig. 4.15 (a)
i0
To find Isc
When the terminals A and B are shorted, no current
flows through the 5- resistance. The circuit is shown in
Fig. 4.15 (b).
By KVL for the supermesh,
10i0 12 ⇒ i0 1.2 A
By KCL at the node A, Isc 3i0 3.6 A
Therefore, the Thevenin impedance is given as
Z th =
Vth
I sc
=
10
A
2i0
12 V
ISC
B
Fig. 4.15 (b)
2
A
7.2 V
7.2
=2Ω
3.6
B
Fig. 4.15 (c)
Example 4.6 Find Thevenin’s equivalent circuit for the network shown
10
in Fig. 4.16.
Solution This circuit does not have any independent source; it has only a
dependent current source. Therefore, the Thevenin equivalent voltage will
be zero.
Vth 0
To find Thevenin’s equivalent impedance, we connect a test current source
of value I. Let the voltage across this test source be V.
20
Fig. 4.16
10
By KCL at the node x,
V − v0
− 0.5v0 − I = 0
10
⇒ V
x
I
10I
)
(
0.5v0
v0
Also, v0 = 20 × 0.5v0 + I = 10v0 + 20 I ⇒ v0 = −
6v0
20
I
9
Putting the value of v0 in (i) from (ii),
(i)
(ii)
20
v0
0.5v0
Fig. 4.17 (a)
3.33
⎛ 20 ⎞
10
V = 10 I + 6 × ⎜ − I ⎟ = − I
3
⎝ 9 ⎠
V
10
= − = −3.33
I
3
The Thevenin’s equivalent circuit is shown in Fig. 4.17 (b).
V
A
Z th =
B
Fig. 4.17 (b)
166
Network Analysis and Synthesis
4.7
MAXIMUM POWER TRANSFER THEOREM
As we are probably aware, a normal car battery is rated at 12 V and generally has an open circuit voltage of
around 13.5 V. Similarly, if we take 9 pen-torch batteries, they too will have a terminal voltage of 9 1.5
13.5 V. However, we know that if our car battery is dead, we cannot start our car with 9 pen-torch batteries.
The reason behind that is that a pen-torch batteries, although having the same open-circuit voltage do not
have necessary power (or current capacity) and hence the required current cannot be given. Or if stated in
different terms, it has too high an internal resistance so that the voltage would drop without giving the necessary current.
This means that a given battery (or any other energy supply, such as the mains) can only give a limited
amount of power to a load. The maximum power transfer theorem defines this power, and tells us the condition at which this occurs.
Statement Maximum power is absorbed by one network from another connected to it at two terminals,
when the impedance of one is the complex conjugate of the other.
This means that for maximum active power to be delivered to the load, the load impedance must correspond to the conjugate of the source impedance (or in the case of direct quantities, be equal to the source
impedance).
The statement and proof of this theorem are discussed for four different cases:
1. Purely resistive circuit with variable load resistance
In this case, the statement of this theorem is given as ‘Maximum power will be delivered from a source to a
load when the load resistance is equal to the source resistance.’
Proof Let V be the voltage source, RS the internal resistance of the source and
RL the load resistance.
power delivered to the load is,
V 2 RL
2
P = I RL =
(R + R )
S
RS
(4.4)
2
RL
V
L
For maximum power,
(
)
(
)
Fig. 4.18 (a) Purely
resistive circuit with
variable load resistance
)
⎡ R + R − 2R R + R ⎤
∂P
S
L
S
L ⎥
=0 ⇒ V2 ⎢ L
=0
4
⎢
⎥
∂RL
+
R
R
S
L
⎣
⎦
(
RS = RL
i.e., the load resistance is equal to the source resistance.
(V )
V
R in Eq. (4.4), the maximum power transferred will be P =
= 2 and thus, the
2
2
Putting RL
S
efficiency will be 50%. This case arises in a purely dc circuit.
max
4 RL
RL
2. Load impedance with variable resistance and variable reactance
In this case, the statement of the theorem is given as ‘Maximum power will be delivered from a source to a
load when the load impedance is the complex conjugate of the source impedance.’
167
Network Theorems
Proof Let V be the voltage source, (RS jXS) the internal impedance of the source and (RL jXL) the load impedance.
V
V
=
(4.5)
current, I =
ZS + ZL
RS + RL + j X S + X L
) (
(
(RS
)
jXS)
(RL
V
jXL)
Power delivered to the load is
V 2 RL
2
P = I RL =
where,
( RS + RL )2 + ( X S + X L )2
Z S = RS + jX S , Z L = RL + jX L
For maximum power,
Now,
Fig. 4.18 (b) Load impedance with
variable resistance and variable reactance
(4.6)
∂P
must be zero.
∂X L
( )
2
−2 V RL ( X L + X S )
∂P
=
=0
∂X L ⎡( R + R )2 + ( X + X )2 ⎤ 2
S
L
S
⎣ L
⎦
From which, XL
XS
0
or
XL = −XS
i.e., the reactance of the load impedance is of opposite sign to the reactance of the source impedance.
Putting XL
XS in Eq. (4.6) P =
V 2 RL
( RL + RS )2
∂P V ( RL + RS ) − 2V RL ( RL + RS )
=
=0
∂RL
( RL + RS )4
2
For maximum power,
V 2 ( RL + RS ) − 2V 2 RL = 0 or
or,
2
2
RL = RS
(V 2 ) and thus, the efficiency will be 50%.
2
V2
The maximum power transferred will be Pmax =
=
4 RL
RL
3. Load impedance with variable resistance and fixed reactance
Maximum power transfer in this case takes place under certain conditions as obtained below.
Here, the current,
V
V
I=
=
Z S + Z L ( RS + RL ) + j ( X S + X L )
(RS
(4.7)
RL
V
Power delivered to the load is
where,
j XL
V 2 RL
2
P = I RL =
jXS)
( RS + RL )2 + ( X S + X L )2
Z S = RS + jX S , Z L = RL + jX L
(4.8)
Fig. 4.18 (c) Load impedance with variable resistance
and fixed reactance
168
Network Analysis and Synthesis
For maximum power,
∂P
=0
∂RL
2
2
⎡
⎤
RS + RL + X S + X L − 2 RL ( RS + RL ) ⎥
⎢
⇒ V
=0
2
⎢
⎥
2
2
⎡
⎤
+
X
)
(
R
+
R
)
+
(
X
⎥⎦
L
S
L
⎣ S
⎦
⎣⎢
2
(
) (
)
⇒ RL 2 = RS 2 + ( X S + X L )2 ⇒ RL = RS 2 + ( X S + X L )2
Case (a) If the source impedance is purely resistive, i. e., XS
transfer becomes
0, the condition for maximum power
RL = RS 2 + X L 2
Case (b)
becomes
if the load impedance is purely resistive, i. e., XL
0, the condition for maximum power transfer
RL = RS 2 + X S 2 = Z S
i. e., the load resistance is equal to the source impedance.
4. Load impedance with fixed ratio, i. e., with variable magnitude but fixed angle
In this case, the statement of the theorem is given as ‘Maximum power is delivered from a source to a load
when the magnitude of the load impedance is equal to the magnitude of the source impedance.’
Proof
Let the angle of the load impedance be f.
∴ Z L = Z L cos + j Z L sin
power delivered to the load is
P=
V 2 Z L cos
( RS + Z L cos )2 + ( X S + Z L sin )2
For maximum power transfer
⎤
V 2 Z L cos
dP
d ⎡
=0 ⇒
⎢
⎥=0
2
2
d ZL
d Z L ⎢⎣ ( RS + Z L cos ) + ( X S + Z L sin ) ⎥⎦
Simplifying we get
2
Z L = RS 2 + X S 2 = Z S
2
ZL = ZS
This case arises in a transformer where the turns ratio is varied for maximum power transfer.
Points to be Noted
(i) It is to be noted that when maximum power is being transferred, only half the applied voltage is available to the load and the other half drops across the source. Also, under these conditions, half the power
supplied is wasted as dissipation in the source.
169
Network Theorems
Thus, the useful maximum power will be less than the theoretical maximum power derived and will
depend on the voltage required to be maintained at the load.
(ii) For circuits having a resistive load being supplied from a source with only an internal resistance (the
case for dc), the maximum power will be transferred to the load when the load resistance is equal to the
source resistance.
Example 4.7 Find the value of R in the circuit of Fig. 4.19
such that maximum power transfer takes place. What is the
amount of this power?
1
4V
Solution Removing the résistance R,
5
2
R
1
6V
∴ 3i1 − 2i1 = 4
and
−2i1 + 8i2 = 0
Solving,
Figure 4.19 Circuit of Example 4.7
2
i2 = A
5
1
⎛
2 ⎞ 32
∴ 1 × i2 + 6 = Voc ⇒ Voc = ⎜ 6 + ⎟ =
V
5⎠ 5
⎝
Also, to find the Rth,
4V
2
Voc
1
i2
i1
17
×1
⎡⎛ 1 × 2 ⎞ ⎤
⎛2 ⎞
17
Rth = ⎢⎜
=
+ 5 ⎥ [1] = ⎜ + 5⎟ 1 = 3
⎟
17
20
1
+
2
3
⎝
⎠
⎝
⎠
⎦
⎣
+1
3
for maximum power transfer, R = Rth =
5
Fig. 4.20 (a)
5
1
17
= 0 ⋅ 85
20
2
1
Rth
2
maximum power Pmax =
Voc
= 12 W
4R
Fig. 4.20 (b)
4.7.1 Concept of Internal Resistance of Voltage and Current Sources
A voltage source is any device or system that produces an electromotive force
between its terminals. An example of a primary source is a common battery.
Similarly, a current source is an electrical or electronic device that delivers electric current. Examples of current sources are a large voltage source in series with
a large resistor (however, this type of current source has very poor efficiency), an
active current source involving transistors, high-voltage current source like Van
de Graff generator, etc. A current source is the dual of a voltage source.
In circuit theory, an ideal voltage source is a circuit element where the voltage
across it is independent of the current through it. It only exists in mathematical
models of circuits. The internal resistance of an ideal voltage source is zero; it is
able to supply any amount of current. The current through an ideal voltage source
is completely determined by the external circuit. When connected to an open cir-
Battery
B
r
A
R
I
Fig. 4.21 A battery of emf
E and internal resistance r
connected to a load resistor
of resistance R
170
Network Analysis and Synthesis
cuit, there is zero current and thus zero power. When connected to a load resistance, the current through the source
approaches infinity as the load resistance approaches zero (a short circuit). Thus, an ideal voltage source can supply
unlimited power.
Similarly, an independent current source with zero current is identical to an ideal open circuit. For this
reason, the internal resistance of an ideal current source is infinite. The voltage across an ideal current source
is completely determined by the circuit it is connected to. When connected to a short circuit, there is zero
voltage and thus zero power delivered. When connected to a load resistance, the voltage across the source
approaches infinity as the load resistance approaches infinity (an open circuit). Thus, an ideal current source
can supply unlimited power forever and so represents an unlimited source of energy. Connecting an ideal
open circuit to an ideal non-zero current source is not valid in circuit analysis as the circuit equation would
be paradoxical, e.g., 3 0.
Now, real batteries are constructed from materials which possess non-zero resistivities. They possess
internal resistances. Incidentally, a pure voltage source is usually referred as an emf. A battery can be modeled as an emf E connected in series with a resistor r, which represents its internal resistance as shown in the
Fig. 4.21.
The voltage V of the battery is defined as the difference in electric potential between its positive and negative terminals, i.e., the points A and B, respectively. Thus, the voltage V of the battery is related to its emf E
and internal resistance r via V E Ir
Now, the emf of a battery is essentially constant; so we must conclude that the voltage of a battery actually decreases as the current drawn from it increases. In fact, the voltage only equals the emf when the current
is negligibly small. The maximum current drawn from the battery is I0 E/r (since for I I0 the voltage V
becomes negative which can only happen if the load resistor R is also negative; that is not feasible.). It follows
that if we short-circuit a battery, by connecting its positive and negative terminals together using a conducting
wire of negligible resistance, the current drawn from the battery is limited to I0 by its internal resistance.
A real battery is usually characterized in terms of its emf E (i.e., its voltage at zero current), and the maximum current I0 which it can supply.
Therefore, we conclude that, no real voltage source is ideal; all have a non-zero effective internal resistance, and none can supply unlimited current. However, the internal resistance of a real voltage source is
effectively modeled in linear circuit analysis by combining a non-zero resistance in series with an ideal
voltage source. Similarly, no real current source is ideal (no unlimited energy sources exist) and all have a
finite internal resistance (none can supply unlimited voltage). The internal resistance of a physical current
source is effectively modeled in circuit analysis by combining a non-zero resistance in parallel with an
ideal current source.
4.8
TELLEGEN’S THEOREM
Statement
Consider an arbitrary lumped network whose graph ‘G’ has ‘b’ branches and ‘n’ nodes. Let the associated
reference polarities and directions be chosen for the branch voltages v1 , v2 , v3 ,...vb and the branch currents
i1 , i2 , i3 ,... ib, which satisfy all the constraints imposed by KVL and KCL respectively.
Then, the summation of instantaneous power delivered to all branches is zero
b
i.e.,
∑v i =0
k =1
k k
171
Network Theorems
Proof
We have to prove that
b
∑v i = 0
k k
k =1
⎡⎣ v1
or,
⎡ i1 ⎤
⎢ ⎥
⎢i ⎥
... vb ⎤⎦ ⎢ 2 ⎥ = 0
⎢.⎥
⎢⎣ ib ⎥⎦
v2
or,
Vb]T [Ib] 0
Now, by KCL and KVL using complete incidence matrix, we have
[Aa] [Ib] 0
and
[Vb] [Aa]T [Vn]
So,
[Vb]T [Ib] [[Aa]T [Vn]]T [Ib]
[Aa] [Ib] [Vn]T
0. [Vn]T
(4.9)
(4.10)
(4.11)
{by Eq. (4.11)}
{by Eq. (4.10)}
[Vb ] × [ I b ] = 0
⇒
T
Thus, Tellegen’s theorem is proved.
Points to be Noted
(i) This theorem is applicable for any lumped network having elements which are linear or non-linear,
active or passive, time-varying or time-invariant.
(ii) This theorem is completely independent of the nature of the elements and is only concerned with the
graph of the network.
(iii) This theorem is based on two Kirchhoff’s laws, i.e., KVL and KCL.
(iv) This theorem implies that the power delivered by independent sources of the network must be equal to
the sum of the power absorbed (dissipated or stored) in all other elements in the network.
(v) If the network is in sinusoidally steady-state (ac circuits) then Tellegen’s theorem is given as
b
∑V I = 0
*
k =1
k k
where, Vk are the phasor voltages, Ik are the phasor currents and Ik* is the complex conjugate of Ik.
(vi) If t1 and t2 refer to two different instants of observations, it still follows from Tellegen’s theorem that
b
∑ v (t ) ⋅ i (t ) = 0
k =1
k
1
k
2
(vii) If N1 and N2 refer to two different circuits having the same graph, with the same reference directions
assigned to the branches in the two circuits then by Tellegen’s theorem,
b
∑ v ⋅i = 0
k =1
1k
2k
b
and
∑ v ⋅i = 0
k =1
2k
1k
where, v1k and i1k are the voltages and currents in N1 and v2k and i2k are the voltages and currents in N2,
all satisfying the Kirchhoff’s laws.
172
Network Analysis and Synthesis
Example 4.8 Verify Tellegen’s theorem for the network shown in
V2
Fig. 4.22.
It is given that V1 4 V, V2
2 V, V3
I2 2 A, I3
6 A, I4 4 A, I5 4 A.
2 V, V4
8 V, V5
6 V, and I1
2 A,
I1
I2
I4
V4
V3
V1
Solution Before verifying Tellegen’s theorem, we have to check
whether the voltage and current values satisfy the KVL and KCL,
respectively.
At the node (A), (i1 − i2 ) = ( 2 − 2 ) = 0
I3
V5
Fig. 4.22
I5
Circuit of Example 4.8
At the node (B), (i2 + i3 + i4 ) = ( 2 + 2 − 4 ) = 0
At the node (C), (i5 − i4 ) = ( 4 − 4 ) = 0
At the node (D), ( −i1 − i3 − i5 ) = ( −2 + 6 − 4 ) = 0
Thus, the currents satisfy KCL.
For the loop ABDA,
( − v2 + v3 − v1 ) = ( 2 + 2 − 4 ) = 0
For the loop ABCDA,
( − v2 + v4 + v5 − v1 ) = ( 2 + 8 − 6 − 4 ) = 0
For the loop BCDB,
( v4 + v5 − v3 ) = (8 − 6 − 2 ) = 0
Thus, the voltages satisfy KVL.
5
So, by Tellegen’s theorem, ∑Vk ik = ( 4 × 2 ) + ( −2 × 2 ) + ( 2 × −6 ) + (8 × 4 ) + ( −6 × 4 ) = 0
k =1
4.9
MILLMAN’S THEOREM
Consider a number of admittances Y1, Y2, Y3 …Yp … Yq, …Yn be connected together at a common point S. If
the voltages of the free ends of the admittances with respect to a common reference N are known to be V1N,
V2N, V3N …VpN… VqN, …VnN, then Millman’s theorem gives the voltage of the common point S with respect to
the reference N as follows.
Y1
Applying Kirchhoff’s current law at the node S,
S
∑ I = 0, I = Y (V −V )
p
p =1
p
p
pN
1
Yn
n
Y2
sN
n
n
or,
∑Y (V −V ) = 0
p =1
p
pN
Yq
sN
Y3
2
3
n
or,
Yp
p =1
q
n
∑Y V = V ∑Y
p
pN
sN
p =1
n
⇒ VsN =
p
∑YpVpN
p
N reference
Fig. 4.23 Illustration of Millman’s theorem
p =1
n
∑Y
p =1
p
An extension of the Millman’s theorem is the equivalent generator theorem.
173
Network Theorems
Statement
(I) This theorem states that if several ideal voltage sources (V1, V2, …) in series with impedances
(Z1, Z2,…) are connected in parallel , then the circuit may be replaced by a single ideal voltage source
(V) in series with an impedance (Z );
n
∑V Y
where,
i i
V = i =1n
∑Y
i =1
A
A
i
1
and, Z = n
∑Yi
i =1
Z1
Z2
V1
V2
…..
Zn
Z
Vn
V
B
(II) If several ideal current sources (I1, I2, …) in parallel
with impedances (Z1, Z2, …) are connected in series,
they may be replaced by a single ideal current source
(I ) in parallel with an impedance (Z);
n
where,
I=
Fig. 4.24 Voltage source equivalent using
Millman’s theorem
Ii
∑ Y
i =1
n
i
i =1
i
∑ 1Y
I1
n
1
and, Y = n
∑ 1Y
i =1
or Z = ∑ Z i
i =1
i
I2
In
I
A
B
Z1
B
Z2
A
B
Z
Zn
Fig. 4.25 Current source equivalent using Millman’s theorem
Proof
(I) Using the superposition theorem, the short-circuit current through A–B considering only one source acting
alone and replacing other sources by their internal impedances, (i.e., short circuit for ideal voltage sources),
I sc1 = V1Y1
Total short-circuit current through A–B, Isc
I sc 2 = V2Y2
(Isc1
I scn = VnYn
Isc2
…
Iscn)
= V1Y1 + V2Y2 + ⋅⋅⋅+ VnYn
n
= ∑ViYi
(4.12)
i =1
Impedance looking back from A–B with all the sources removed
Z=
1
1
= n
Y1 + Y2 + ⋅⋅⋅+ Yn
∑Yi
i =1
(4.13)
174
Network Analysis and Synthesis
Thus, by Thevenin’s theorem, the equivalent voltage is,
n
∑V Y
V = I sc ⋅ Z = i =1n
i i
(4.14)
∑Yi
i =1
Form Eq. (4.12), (4.13) and (4.14), Millman’s theorem is proved.
(II) Using the superposition theorem, the short-circuit current through A–B considering only one source acting
alone and replacing other sources by their internal impedances, (i.e., open circuit for ideal current sources),
I Z
IZ
I Z
I sc1 = n1 1 ; I sc2 = n2 2 ; ⋅⋅⋅ I scn = nn n
∑ Zi
∑ Zi
∑ Zi
i =1
Total short-circuit current, Isc
i =1
I
(Isc1
Isc2
i =1
…
Iscn)
n
∑I Z
I = i =1n
or,
i
∑Z
i =1
i
(4.15)
i
Impedance looking back from A–B with all the sources removed
n
Z = ∑ Zi
(4.16)
i =1
From Eq. (4.15) and (4.16), Millman’s theorem is proved.
Points to be Noted
(i) This theorem provides the equivalent circuits which are either Thevenin or Norton equivalent circuits.
(ii) This theorem is applicable only to independent voltage sources with their internal series impedances
connected directly in parallel, or independent current sources with their internal series admittances
connected directly in series.
(iii) This theorem is not applicable to circuits where impedances or dependent sources are present between
the independent sources.
(iv) This theorem is not useful for circuits with less than two independent sources.
Example 4.9 Find the load current using Millman’s theorem.
Solution Here, E1
Z1
1V,
1 ,
2V,
E3
2 ,
Z3
1
mho
Y1 1 mho, Y2 0.5 mho, Y3
3
By Millman’s theorem, the equivalent voltage is
3
∴E =
∑EY
i =1
3
i i
∑Y
i =1
i
=
1 ×1+ 2 × 0 ⋅5 + 3 ×
1+ 0 ⋅5 +
1
3
E2
Z2
1
3 = 3 = 18 V
11
11
6
3V
3 ,
I
1
2
3
10
1V
2V
3V
Fig. 4.26 Circuit of Example 4.9
175
Network Theorems
and the equivalent impedance is
1
6
Z= 3 =
11
∑Yi
i =1
Therefore, the current through the resistor is
E
=
Z + 10 6
∴I =
4.10
18
11
11 = 18 = 9 A
+ 10 116 58
COMPENSATION THEOREM
In many circuits, after the circuit is analyzed, it is realized that only a small change needs to be made to a
component to get a desired result. In such a case, we would normally have to recalculate. The compensation
theorem allows us to compensate properly for such changes without sacrificing accuracy.
Statement
In any linear bilateral active network, if any branch carrying a current I has its impedance Z changed by an
amount ␦Z, the resulting changes that occur in the other branches are the same as those which would have
been caused by the injection of a voltage source of ( I␦Z) in the modified branch.
In other words, in a linear network N, if the current in a branch is I and the impedance Z of the branch is
increased by ␦Z then the increment of voltage and current in each branch of the network is that voltage or
current that would be produced by an opposite voltage source of value vc( I␦Z) introduced into the altered
branch after the modification.
Zth
Proof Consider the network N, having branch impedance Z.
Let the current through Z be I and its voltage be V.
Let ␦Z be the change in Z.
Then, I (the new current) can be written as
I′=
Voc
;
Z + δ Z + Z th
δI = I′− I =
=−
⎛ V
⎞⎛
⎞
Voc
V
δZ
− oc = − ⎜ oc ⎟ ⎜
⎟
Z + δ Z + Z th Z + Z th
⎝ Z + Z th ⎠ ⎝ Z + δ Z + Z th ⎠
Vc
Iδ Z
=−
where Vc = I δ Z
Z + δ Z + Z th
Z + δ Z + Z th
How to Find ␦I?
(i) Find the product I␦Z, where I is the current through the branch
before changing the impedance.
(ii) Remove all the independent sources.
Voc
Z
I
Fig. 4.27 (a) Circuit for
explaining compensation
theorem
Zth
I
Z
Vc = I
Fig. 4.27 (b) Equivalent circuit
using compensation theorem
176
Network Analysis and Synthesis
(iii) Connect a voltage source of magnitude Vc I␦Z, in series with the branch. The polarity of Vc is such
as to oppose the direction of the current I.
(iv) Solve the network assuming current flowing to be ␦I and thus the value of ␦I.
Points to be Noted
(i) This theorem is used to calculate the incremental changes in the voltages and currents in the branches
of a circuit due to a change of impedance in one branch.
(ii) This theorem is not applicable to circuits with only dependent sources.
(iii) This theorem is not applicable to circuits with non-linear elements.
Example 4.10 In the network shown in Fig. 4.28, the resistance
R is changed from 4 ⍀ to 2 ⍀. Verify the compensation theorem.
Solution By KCL,
5i1 − 4i2 = 1 and − 4i1 + 12i2 = 0
Solving
⇒ i1 =
∴ I1 = (i1 − i2 ) =
3
A
11
and
i2 =
1
1V
1
A
11
8
R
Fig. 4.28 Circuit of Example 4.10
1
2
A
11
After changing the value of the resistance from 4
KCL
3i1′ − 2i2′ = 1
and
−2i1′ + 10i2 = 0
Solving
5
⇒ i1′ = A
13
to 2 , by
I1
1V
4
I2
8
2
I2
8
Fig. 4.29 (a)
1
and
1
i2′ = A
13
I1
1V
4
∴ I1′ = A
13
change in current,
Fig. 4.29 (b)
⎛ 4 2⎞ 8
A
δ I = ( I1 − I1 ) = ⎜ − ⎟ =
⎝ 13 11⎠ 143
1
(I)
2
Using the compensation theorem,
Vc = I1 × δ Z =
2
4
( −2 ) = − V.
11
11
4/11
4
11 = 8 A
143
2+ 8
9
From (I) and (II), the compensation theorem is proved.
δI =
8
I
(II)
Fig. 4.29 (c)
177
Network Theorems
Solved Exercises
Superposition Theorem
Problem 4.1 Calculate the voltage V across the resistor R by using the superposition theorem.
j1
j5
R=1
j4
1A
1V
Fig. 4.30
Solution We consider two cases:
Case (1) When the 1-A current source is acting alone
For Fig. 4.31(a), the voltage across the resistor R
1
is, V ′ =
j1
j
.
1+ j
Case (2) When the 1-V voltage source is acting alone
1
For Fig. 4.31(b), the current through the resistor I ′′ =
1+ j
1A
Fig. 4.31 (a) Circuit with current
source acting alone
j1
1
.
1+ j
So, by the superposition theorem, total voltage across the resistor when
both the sources are acting simultaneously is,
and hence, the voltage across the resistor R
V = (V ′ + V ′′ ) =
1
V
1
is V ′′ = I ′′ × 1 =
j5
V
1V
j4
j
1
+
= 1V
1+ j 1+ j
Fig. 4.31 (b) Circuit with
voltage source acting alone
Problem 4.2 Use the superposition theorem on the circuit shown in
Fig. 4.32 to find ‘I’.
Solution We consider two cases:
Case(1) When the 10-V voltage source is acting alone
1
I
2Vx
5
2
10 V
Vx
2A
For Fig. 4.33(a), by KVL, 5i ′ − 2 v x′ + 2i ′ = 10 with v x′ = −2i ′
⇒ 7i ′ + 4i ′ = 10 ⇒ i ′ = 10
11
A
Fig. 4.32
Case (2) When 1-V voltage source is acting alone
For Fig 4.33(b), by KCL at the node (x)
v′
2 = ix + i ′′ = − x + i ′′
2
But loop analysis in the left loop gives
3
5i ′′ + 3v x ′′ = 0 or, i ′′ = − v x′′
5
2 Vx
5
(i)
10 V
i
2
Vx
Fig. 4.33 (a) Voltage source
acting alone
178
Network Analysis and Synthesis
v ′′ 3
20
From (i), 2 = − x − v x′′ ⇒ v x′′= −
2 5
11
2 Vx
5
3 ⎛ 20 ⎞ 12
∴ i ′′ = − × ⎜ − ⎟ = A
5 ⎝ 11 ⎠ 11
i
X
Vx
2
2A
ix
So, by the superposition theorem total current, when both the sources are
acting simultaneously, is,
Fig. 4.33 (b) Current source
acting alone
⎛ 10 12 ⎞
2
I = (i ′ − i ′′ ) = ⎜ − ⎟ = − A
11
⎝ 11 11 ⎠
Problem 4.3 Determine the current in the capacitor branch by the superposition
theorem.
j4
3
4 ∠0°
2
I′=
= ∠0° A
3+ j4 + 3− j4 3
) (
(
2 90 A
Fig. 4.34
⎛ 4
⎞
(3 + j 4)
= ⎜ − + j1⎟ A
(3 + j 4) + (3 − j 4) ⎝ 3
⎠
3
total current when both the sources are acting simultaneously is
j4
3
4 0° V
3
⎛2 4
⎞ ⎛ 2
⎞
I = ( I ′ + I ′′ ) = ⎜ − + j1⎟ = ⎜ − + j1⎟
⎝3 3
⎠ ⎝ 3
⎠
j4
3
2 90° A
Fig. 4.35 (a)
When voltage source
acting alone
Problem 4.4 Find the current i0 using superposition
theorem.
(a)
(c)
4
j5
Fig. 4.35 (b)
When current source
acting alone
2 0 (A)
i0
j2
j2
2 0 A
6
i0
4
8
2
i0
(b) 10 cos 4t (V)
Fig. 4.36
1H
j4
j4
= 1.2 ∠123.7° A
5 0 V
j4
)
When the current source is acting alone
Here, the current in the capacitor branch is
I ′′ = 2 ∠90° ×
4 0 V
3
Solution When the voltage source is acting alone
Here, the current in the capacitor branch is
8V
j4
10 30 (V)
179
Network Theorems
Solution (a) When the voltage source is acting alone
The current in this case is
i0
⎛
5
1⎞
i0′ =
= ⎜1+ j ⎟ A
4 − j2 ⎝
2⎠
When the current source is acting alone
In this case, the current is,
i0′′= 2 ∠0° ×
5 0 V
j2
Fig. 4.37 (a)
alone
Voltage source acting
⎛ 8 14 ⎞
4
=⎜ + j ⎟ A
4 − j2 ⎝ 5
5⎠
by the superposition theorem, total current is
j5
4
⎛ 8⎞ ⎛ 1 4⎞
i0 = i0′ + i0′′ = ⎜ 1 + ⎟ + j ⎜ + ⎟ = 2.9∠26.56° A
⎝ 5⎠ ⎝ 2 5 ⎠
)
(
j5
4
i0
2 0 A
j2
(b) When the dc source is acting alone
⎛ j 4 × 4 ⎞ 2 + j6
Equivalent impedance, Z = ⎜
+2 =
⎝ 4 + j 4 ⎟⎠ 1 + j
main current, I =
(
) (
8 8 1+ j 4 1+ j
=
=
1+ j3
Z 2 + j6
the current, i0′ = I ×
Fig. 4.37 (b) Current source acting
alone
)
4
i0
j4
⎛2
4
4(1 + j )
4
6⎞
=
×
=⎜ − j ⎟ A
4 + j 4 1+ j 3 4 + j 4 ⎝ 5
5⎠
When the ac source is acting alone
2
8V
Fig. 4.38 (a) dc source acting alone
⎛ j 4 × 2 ⎞ 4 + j6
Equivalent impedance, Z = 4 + ⎜
=
⎝ 2 + j 4 ⎟⎠ 1 + j 2
4
2
i0
main current,
I=
10∠0°
(1 + j 2 ) 10 + j 20
= 10∠0°
=
Z
4 + j6
4 + j6
the current,
⎛ 10 15 ⎞
2
10(1 + j 2 )
1
i0 ′′ = I ×
=
×
=
−j
A
2 + j4
4 + j6
1 + j 2 ⎜⎝ 13 13 ⎟⎠
10 cost 4t(V)
j4
Fig. 4.38 (b) ac source acting alone
⎛ 2 10 ⎞ ⎛ 6 15 ⎞
by the superposition theorem, total current is, i0 = (i0′ + i0′′ ) = ⎜ + ⎟ − j ⎜ + ⎟ = 2.63∠ − 63.58° A
⎝ 5 13 ⎠ ⎝ 5 13 ⎠
(c) When the voltage source is acting alone
Equivalent impedance, Z =
j 4(8 − j 2 )
28 + j 22
+6=
8 + j2
4+ j
180
Network Analysis and Synthesis
main current, I =
j2
10∠30° ( 4 + j ) (8.66 + j 5)( 4 + j )
=
28 + j 22
28 + j 22
i0
j4
8
the current,
i0′ = I ×
6
8 − j 2 8.66 + j 5
=
= 0.14 ∠ − 8.16° A
8 + j 2 56 + j 44
10 30 (V)
Fig. 4.39 Voltage source acting alone
When the current source is acting alone
2 0 (A)
j2
j2
6
i0
i0
j4
8
j4
8
6
2 0 (A)
Fig. 4.40
j2
j4 × 6
j12
Z=
=
6 + j4 3 + j2
where,
i0
8
the current,
i0′′ = 2 ∠0 ×
Z
j12
=
= 0.73∠47.49 A
8 − j 2 + Z 12 + j11
Z
Fig. 4.41
by the superposition theorem, total current is
(
) (
) (
)
i0 = i0′ + i0′′ = 0.14 ∠ − 8.16° + 0.73∠47.49° = 0.631 + j 0.518 = 0.81∠39.38° A
Problem 4.5 Find v0 using the superposition theorem.
8
30 sin 5t (V)
v0
0.2 F
1H
2 cos 10t (A)
Fig. 4.42
Solution (a) When the voltage source is acting alone
Here, X C =
−j
= − j1
5 × 0.2
and
X L = j × 5 ×1 = j5
By KCL,
−
30 − v0′ v0′ v0′
30
+
+ = 0 ⇒ v0′ =
= 4.631∠ − 81.12° ( V )
8
− j1 j 5
8 0.125 + j 0.8
(
)
2 0 (A)
181
Network Theorems
8
8
v0
30 0 (V)
j1
v0
j5
Fig. 4.43 (a) Voltage source acting alone
j10
j0.5
2 0 (A)
Fig. 4.43 (b) Current source acting alone
When the current source is acting alone
Here, X C =
−j
= − j 0.5
10 × 0.2
and
X L = j × 10 × 1 = j10
⎛1 1
1 ⎞
2
⇒ v0 ′′ =
= 1.051∠ − 86.24° ( V )
By KCL, 2 = v0 ′′ ⎜ +
+
⎟
0.125 + j1.9
⎝ 8 j10 − j 0.5 ⎠
By the superposition theorem, when both the sources are acting simultaneously, the voltage is
v0 = ( v0′ + v0′′) = 4.631 sin(5t − 81.12° ) + 1.051 cos(110t − 86.24° ) ( V )
Problem 4.6 Find i0 and i from the circuit of Fig. 4.44 using
superposition theorem.
i0
Solution When the 6-V source is acting alone
The circuit is shown.
1
5
1A
6V
Here, i0′ = i ′
i
2i0
Fig. 4.44
6 3
By KVL, 6i ′ + 2i ′ = 6 ⇒ i ′ = i0′ = = A = 0.75 A
8 4
i0
1
5
i
When the 1-A source is acting alone
By KCL, we get, 1 = i ′′ − i0′′ ⇒ i ′′ = 1 + i0′′
6V
2i0
By KVL for the supermesh,
1 × i0′′+ 5i ′′ + 2i0′′= 0 or, 3i0′′+ 5i ′′ = 0
(
Fig. 4.45 (a) 6-V Source acting alone
)
5
or, 3i0′′+ 5 1 + i0′′ = 0 or, i0′′= − = −1.25 A
4
∴ i ′′ = 1 − 1.25 = −0.25 A
By the superposition theorem, the total currents when both the
sources are acting simultaneously is given as
i = (i ′ + i ′′ ) = (0.75 − 0.25) = 0.5 A ⎫⎪
⎬
i0 = (i0′ + i0′′) = (0.75 − 1.25) = − 0.5 A ⎭⎪
i0
1
5
1A
i
2i0
Fig. 4.45 (b) 1-A source acting alone
182
Network Analysis and Synthesis
Problem 4.7 Using the superposition theorem, calculate the current through the (2 ⴙ j3)⍀ impedance
branch of the circuit shown in Fig. 4.46.
5
2
j3
j5
30 V
4
6
20 V
Fig. 4.46
Solution
Case (I) When the 30-V source is acting alone
Impedance, Z = 5 +
∴I ′ =
( 4.4 + j 3) × j 5
4.4 + j 3 + j 5
5
= ( 6.32 + j 2.6 ) Ω
I
30 V
)
(
30
30
=
= 4.06 − j1.67 A
Z 6.32 + j 2.6
i′ = I ′ ×
2
j3
4
j5
i
6
2
j3
4
Fig. 4.47
j5
= ( 2.39 + j 0.27) A
4.4 + j 3 + j 5
Case (II) when the 20-V source is acting alone
Impedance, Z = 4 +
( 4.5 + j 5.5) × 6
4.5 + j 5.5 + 6
= ( 7.31 + j1.41) Ω
5
j5
20
20
∴ I ′′ = =
= ( 2.64 − 0.509) A
Z 7.31 + j1.41
i
I
6
20 V
Fig. 4.48
6
i ′′ = − I ′′ ×
= −(1.064 − j 0.848) A
4.5 + j 5.5 + 6
By the superposition theorem, total current flowing through the (2
j3) impedance is
i = (i ′ + i ′′ ) = ( 2.39 + j 0.27) − (1.064 − j 0.848) = (1.32
25 + j1.117) A = 1.733∠40.14° A
6
Problem 4.8 Using the superposition theorem, find VAB.
4V
Solution We consider three cases:
Case (I) When the 2-V source is acting alone
The circuit is shown Fig. 4.50.
A
Fig. 4.49
6
I
A
Fig. 4.50
4
2V
2
2V
2A
B
4
2
B
183
Network Theorems
For this circuit, the current in the loop is obtained as I ′ =
2 1
= A
12 6
1
the voltage between A and B is VAB
′ = I ′ × 6 = × 6 = 1V
6
Case (II) When the 4-V source is acting alone
The circuit is shown in Fig. 4.51.
In this circuit, the loop current is obtained as
4V
6
I
4 1
I ′′ = = A
12 3
A
4
B
2
Fig. 4.51
voltage between A and B is,
6
1
VAB
′′ = − I ′′ × 6 = − × 6 = −2 V
3
Case (III) When the 2-A source is acting alone
The circuit is shown in Fig. 4.52.
2A
A
We convert the current source into its equivalent voltage source as
shown in Fig. 4.53.
8 2
The loop current is I ′′′ = = A
12 3
voltage between A and B is
2
VAB
′′′= − I ′′′ × 6 = − × 6 = − 4 V
3
voltage between A and B when all the sources are acting
simultaneously is given by superposition theorem as
4
B
2
Fig. 4.52
6
8V
I
A
4
2
B
3
8A
Fig. 4.53
)
(
VAB = VAB
′ + VAB
′′ + VAB
′′′= 1 − 2 − 4 = −5 V
4i
Problem 4.9 Find the current i in the circuit shown in the Fig. 4.54 using the
superposition theorem.
2
Solution We consider the three cases:
Case (I) When the 10-V source is acting alone
The circuit is shown in Fig. 4.55.
2Ai
10V
4i
Fig. 4.54
4i
2
3
i
2
3
2A
2i
i
Fig. 4.56
3
2A
i
10 V
Fig. 4.55
2
Fig. 4.57
184
Network Analysis and Synthesis
By KVL for the loop, we get, − 4i ′ + 3i ′ − 10 + 2i ′ = 0 ⇒ i ′ = 10 A
Case (II) When the 2-A source is acting alone
The circuit is shown in Fig. 4.56.
We convert the dependent voltage source into its equivalent dependent current source as shown in Fig. 4.57.
The total current (2 2i ) is divided into two paths, resistors 2 and 3 .
by current divider rule, current through the 3- resistor is
⎛ 2 ⎞
i ′′ = ⎜
× ( 2 + 2i ′′ ) ⇒ i ′′ = 4 A
⎝ 2 + 3 ⎟⎠
Case (III) When the 8-A source is acting alone
The circuit is shown in Fig. 4.58.
By KVL for the loop, we get,
−4i ′′′ + 3( I − 8) + 2 I = 0
where, i ′′′ = ( I − 8) or, I = (i ′′′ + 8)
4i
2
8A
3
I
i
⇒ − 4i ′′′ + 3i ′′ + 2(i ′′′ + 8) = 0 ⇒ i = −16 A
Fig. 4.58
current when all the sources are acting simultaneously is given by the
superposition theorem as
i = (i ′ + i ′′ + i ′′′ ) = (10 + 4 − 16 ) = −2 A
Problem 4.10 Using the superposition theorem determine V1, the
voltage across the 3-ohm resistor in Fig. 4.59.
V1
4i
3
8A
Solution Case (I) When the 8-A current source is acting alone
i
2A
1
By KVL for the supermesh, 3i ′ + 2i1 − 4i ′ = 0 ⇒ i1 = i ′
2
2
10V
By KCL at the node x,
i1 = (8 + i ′ ) ⇒
1
i ′ = 8 + i ′ ⇒ i ′ = −16 A
2
Fig. 4.59
∴V1′= 3i ′ = 3 × ( −16 ) = −48 V
Case (II) When the 2-A current source is acting
alone
By KVL,
V1
4i
i
x
3(i2 + 2 ) + 2i2 − 4i ′′ = 0 ⇒ 5i2 + 6 − 4i ′′ = 0
Now, i ′′ = (i2 + 2 )
∴ 5i2 + 6 − 4(i2 + 2 ) = 0 ⇒ i2 = 2 A
∴ i ′′ = (i2 + 2 ) = ( 2 + 2 ) = 4 A
∴V1′′= 3i ′′ = 3 × 4 = 12 V
3
8A
V1
4i
i2
2
i1
Fig. 4.60 (a)
2
Fig. 4.60 (b)
3
2A
i
2A
185
Network Theorems
Case (III) When the 10-V voltage source is acting alone
By KVL, 3i ′′′ − 10 + 2i ′′′ − 4i ′′′ = 0 ⇒ i ′′′ = 10 A
V1
4i
3
i
∴V1′′′= 10 × 3 = 30 V
When all the sources are acting simultaneously, by the superposition theorem the
voltage is given as
V1 = (V1′+ V1′′+ V1′′′) = ( − 48 + 12 + 30) = − 6 V
Fig. 4.60 (c)
Problem 4.11 For the network shown in Fig. 4.61
calculate the current throughout the impedance (3 ⴙ j4)
using superposition theorem.
Main current, I =
−10
j5
j10 × j 5
=
=
3 + j 9 − 5 + j 60 −1 + j12
5
10
10∠0°
10(8 + j 4 )
=
( 3 + j 4 )5 − 5 + j 60
j5 +
3+ j4 + 5
∴ I ′′ = I ×
0V
j5
3
90 V
I
j4
Fig. 4.62
5
10 × 5
10
=
=
8 + j 4 −5 + j 60 −1 + j12
j5
5
When both the sources are acting simultaneously, by the superposition
theorem, the total current flowing through the impedance (3 j4) is
I = ( I ′ + I ′′ ) =
10
Fig. 4.61
When the 10 0 V is acting alone
Main current, I =
3
90 V
j4
j10( 3 + j 9)
10∠90°
=
(3 + j 4) j 5
− 5 + j 60
5+
3 + j 4 + j5
∴I ′ = I ×
j5
5
10
Solution When the 10 90ⴗ V is acting alone
10 V
2
3
−10
10
+
=0A
−1 + j12 −1 + j12
10 0 V
j4
I
Fig. 4.63
Problem 4.12 Using the superposition theorem, determine the current in the 4-⍀ resistor in the network
shown in Fig. 4.64.
4
20
0A
Fig. 4.64
5
j2
2
j2
100
90 V
186
Network Analysis and Synthesis
Solution Case (I) When the 20 0 A source is acting alone
The circuit is shown in Fig. 4.65.
4
20
j2
5
0A
2
I1
j2
Fig. 4.65
Reducing the parallel combination, the simplified circuit is
shown in Fig. 4.66.
Z1 =
5 × j2
= 1.857∠68.2° = 0.69 + j1.72
5+ j2
Z2 =
2 × (− j 2)
= (1 − j1) = 1.414 ∠ − 45°
2 − j2
(
)
By current division rule, the current through the 4I1 = 20∠0 ×
4
Z1
20 0 A
I1
Z2
Fig. 4.66
resistor is
Z1
1.857∠68.2
= 20∠0 ×
= 6.48∠61 = 3.14 + j 5.66 A
0.69 + j1.72 + 4 + 1 − j1
Z1 + 4 + Z 2
Case (II) When the 100 90 V source is acting alone
Here, the current source is open-circuited. Combining the parallel connection of 5 and j 2 the simplified circuit is shown
in Fig. 4.67.
By KVL for the two loops, we get,
( 4 + 0.69 + j1.72 − j 2 ) I 2 + j 2 I = 0
⇒
)
(
4
Z1
I2
2
j2
20 90 A
I
Fig. 4.67
( 4.69 − j 0.28) I 2 + j 2 I = 0
j 2 I 2 + ( 2 − j 2 ) I = 100∠90° = j100
and,
(i)
(ii)
Solving (i) and (ii), we get
I2 =
0
j2
j100 ( 2 − j 2 )
j2
( 4.69 − 0.28)
j2
(2 − j 2)
=
200
= 12.33∠37.75° ( A ) = ( 9.75 + j 7.55) A
−12.83 + j 9.93
By superposition theorem, when both the sources are acting simultaneously, the current through the 4resistor is
I = I1 − I 2 = ( 3.14 + j 5.66 ) − ( 9.75 + j 7.55) = ( − 6.61 − j1.9) = 6.89∠ − 163.67° A
The direction of the current is from right to left.
187
Network Theorems
Problem 4.13 Find I in the Fig. 4.68 using the superposition theorem.
4V
Solution When the 4-V voltage source is acting alone
The circuit is shown in Fig. 4.69.
Here, by KVL,
3
1
VX
2
I
− 4 + 3 I ′ + 5Vx′− Vx′= 0
3I ′ + 4 × (− 2 I ′ ) = 4
or,
Fig. 4.68
[ Vx′= −2 I ′ ]
4V
4
I ′ = − A = − 0.8 A
5
When the 2-A current source is acting alone
The circuit is shown in Fig. 4.70.
or,
By KCL, 2 =
VX
3
5 VX
2
I
Vx′′ Vx′′− 5Vx′′
12
+
⇒ Vx′′= − = −2.4 V
2
3
5
Fig. 4.69
3
⎛ 12 ⎞
12
− −5×⎜− ⎟
V ′′− 5Vx′′
5
⎝ 5 ⎠ 16
∴ I ′′ = x
=
= = 3.2 A
3
3
5
When both the sources are acting simultaneously, the current by superposition theorem is given as I = ( I '+ I ") = ( − 0.8 + 3.2 ) = 2.4 A
1
VX
2
Fig. 4.70
1
1
i
Problem 4.14 Draw the Thevenin’s equivalent of the circuit in Fig. 4.71 and find the load current, i.
Solution Open-circuiting the terminals, by KVL for two
meshes,
10 V
5V
1
R =2
1
2
Fig. 4.71
3i1 − i2 = 10 and − i1 + 4i2 = − 5
Solving, i1 = 5 , and i2 = − 5
11
11
1
1
i1 1
i2
10 V
⎛ 10 ⎞ 45
∴Voc = (5 + 2i2 ) = ⎜ 5 − ⎟ = V
⎝ 11 ⎠ 11
Voc
2
1
Fig. 4.72 (a)
1
Fig. 4.72 (b)
1
1
5 VX
2A
I
Thevenin’s and Norton’s Theorem
1
5 VX
2A
1
2
Rth
2/3
2
Rth
188
Network Analysis and Synthesis
R th
5
×2
10
3
=
Equivalent resistance, Rth =
5 + 2 11
3
So, the load current is, i =
i
45
Voc
11 = 45 = 1.40625 A
=
Rth + 2 10 + 2 32
11
Fig. 4.73
3V0
V0
Problem 4.15 Find I, in the given figure, using Thevenin’s theorem.
Solution Removing the 2-
2
Voc
I
1
resistor,
By KVL for the supermesh, −10 − v0 + 3v0 + v0 c = 0 ⇒ v0 c = 10 − 2 v0
But, due to open-circuit, the 1-A source will circulate through the1resistor.
10 V
∴ v0 = 1 × 1 = 1 V
1A
2
Fig. 4.74
∴V0 c = (10 − 2 ) = 8 V
3V0
V0
Let’s short circuit the terminals x–y,
By KVL,
1
−10 − v0 + 3v0 = 0 or, v0 = 5
1A
10V
VOC
But, by KCL at the node (a),
v0
= 1 − I sc
1
⇒ I sc = (1 − v0 ) = − 4 A ( e.g., current is f lowing f rom y to x )
∴ Rth =
Fig. 4.75 (a)
V0
Voc
8
= =2
I sc 4
So, the current through the 2-
a
3V0
x
1
1A
10V
Isc
8
resistor, I =
=2A
2+2
y
Problem 4.16 By the iterative use of Thevenin’s theorem, reduce the
circuit shown in Fig. 4.76 to a single emf acting in series with a single resistor. Hence, calculate the current in the 10-⍀ resistor connected across XY.
10
100
100
1000
Fig. 4.75 (b)
X
90
1000
10
100
10
100 V
Y
Fig. 4.76
189
Network Theorems
Solution Consider the section of the network to the left of A–B: By use of Theremin’s
theorem, this portion is reduced to the form of
Fig. 4.77 (b).
1000 × 100 1000
=
1000 + 100 11
100 × 1000 1000
∴Vth =
=
V
1100
11
∴ Rth =
10
90
1000
th
1000
X
10
100
10
100V
Y
B
Fig. 4.77 (a)
Applying Thevenin’s theorem to the section left
of CD of Fig. 4.77 (b),
(100011) ×10 = 2100
∴R =
(210011) +10 221
1000 × 10
(
1000
11)
=
V
∴V =
(210011) +10 221
100
A 100
A 100
1000
C 100
X
1000/11
10
100
10
1000/11 V
B
Y
D
Fig. 4.77 (b)
th
Applying Theremin’s theorem to the section left
of EF of Fig. 4.77 (c),
∴ Rth =
∴Vth =
(24200 221) ×100 = 24200
(24200 221) +100 463
C
(
)
E 1000
X
2100/221
10
100
100 0/221V
Y
(1000
) × 100 1000
221
V
=
24200
+ 100 463
221
100
D
F
Fig. 4.77 (c )
Section left to XY is put as in Fig. 4.77 (d).
487200
24200
+ 1000 =
463
463
1000
× 1000 1000
463
Vth =
=
V
24200
+ 1000 4872
463
E
∴ Rth =
(
(
)
)
Hence, the current in the 10-
X
24200/463
10
1000/463 V
Y
resistor is
(1000 487.2) = 0.0193 A
I=
(487200 436) +10
100 0
F
Fig. 4.77 (d)
190
Network Analysis and Synthesis
Problem 4.17 In the operational-amplifier circuit shown
in Fig. 4.78 find I, in the R ⴝ 4-k⍀ resistor, using Thevenin’s
theorem.
4k
Solution Open- circuiting the 4-k
e2 = 0, e3 = V0
Here,
e1 − 12
2 × 103
+
e1 − V0
4 × 103
0 − e1
8 × 10
+
3
+
e1
8 × 103
12k
2k
12 V
8k
resistor,
R =4k
OPAMP
Vo
= 0 ⇒ 7e1 = ( 48 + 2V0 )
(i)
Fig. 4.78
0 − V0
3
= 0 ⇒ V0 = − e1
2
12 × 103
(ii)
From (i) and (ii), ⇒ e1 4 8 V eoc
Now, we connect a 1-A current source at the place of the
4-k resistor.
By KCL at the node (1),
e −V
e1
e1
+ 1 0 +
= 1 ⇒ 7e1 = 8000 + 2V0
2 × 103 4 × 103 8 × 103
By KCL at the node (2),
⎛ 3 ⎞
3
V0 = − e1 ⇒ 7e1 = 8000 + 2 ⎜ − e1 ⎟ ⇒ e1 = 800 V
2
⎝ 2 ⎠
4k e3
2k e1
12 V
12k
8k
e2
1A
V0
Fig. 4.79
e
∴ Rth = 1 = 800
1
4 ⋅8
4 ⋅8
= 1 mA
∴i =
=
4000 + 800 4 ⋅8 × 103
VS
4V S
B
Fig. 4.80
VS
A
2
V1
4
A
2
2
2
V oc
10V
Isc
10V
4V S
4VS
B
Fig. 4.81
4
10 V
resistor by KCL,
4
A
2
Voc − 10
= 4 vs = 4(10 − Voc ) ⇒ Voc = 10 V
2
VS
4
2
Problem 4.18 Find Thevenin’s equivalent about AB for the circuit
shown in Fig. 4.80.
Solution Open-circuiting the 4-
OPAMP
B
191
Network Theorems
Short-circuiting the terminals AB, by KCL
V1 − 10 V1
+ = 4 vs = 4(10 − V1 )
2
4
180
V1 =
= 9 ⋅ 47 V
19
9 ⋅ 47
∴
I sc =
= 2 ⋅ 368 A
4
Vth
Rth =
∴
= 4 ⋅ 22
I sc
Problem 4.19 In the network, determine the steady current in the 8-⍀ inductor using Thevenin’s theorem.
j4
100
a
0 ° (V)
j8
j4
b
j8
100
j6
60° (V)
Fig. 4.82
Solution With a-b open-circuited,
j4
a
100 0 ° (V)
j4
b
j8
j6
Fig. 4.83
100∠0
( − j8) = 200∠0 V
j 4 − j8
100∠60
Vb =
( − j 6 ) = 300∠60 V
j 4 − j6
∴ Vth
= (Va − Vb ) = 200∠0 − 300∠60 = (50 − j 259.81) V
Va =
∴ Z th =
current in the 8-
( j 4 )( − j8) ( j 4 )( − j 6 )
+
= j 20
j 4 − j8
j 4 − j6
inductor, i =
Vth
Z th + Z L
=
(50 − j 259.81)
= 9.45∠ − 169.1 A
j 20 + j8
100 60 ° (V)
192
Network Analysis and Synthesis
Problem 4.20 Obtain Thevenin’s equivalent circuit with respect to terminals A–B in the networks shown
below.
(a)
(b)
10
0 (A)
j15
5
j 10
A
2
3
20
A
90 ° (V)
j 15
3
j4
B
B
(c)
10
(d)
5
A
100
0 (V)
j5
8
1
j6
2cos 2tu(t)
B
1/2 H
1/4 F
1
4
B
(e)
A
1/4 F
5I
I 100
A
j5
j10
10 0 (V)
B
Fig. 4.84
Solution
(a) With A–B open, the current is
10∠0
150∠90
I=
× j15 =
5 − j 5 + j15
5 + j10
I
10
0 ° (A)
2
j 15
Thevenin voltage
A
150∠90
× (5∠ − 90 ) = 67.08∠ − 63.4 V
Vth = VAB = I × − j 5 =
5 + j10
(
)
j5
B
Thevenin impedance,
− j 5 × (5 + j15)
= 7.07∠ − 81.86°
Z th = Z AB =
− j 5 + 5 + j15
Fig. 4.85 (a)
Z th = 7.07
81.86° ( )
A
Thus, the Thevenin’s equivalent circuit is shown in Fig. 4.85 (b).
(b) Here, Thevenin voltage,
20∠90°
j120( 3 − j 4 )
× (3 − j 4) =
5 + j10 + 3 − j 4
8 + j6
50∠36.87°
= 10∠0° ( V )
Vth =
5∠36.87°
V th = 67.08
Vth =
63.4 ° (V)
B
Fig. 4.85 (b)
193
Network Theorems
Thevenin impedance,
Z th =
(5 + j10) × ( 3 − j 4 ) 11.8∠63.43 × 5∠ − 53.13
= 5.59∠ − 26.56 ( )
=
10∠36.87
(5 + j10) + ( 3 − j 4 )
Thus, the Thevenin’s equivalent circuit is shown in Fig. 4.86 (b).
Z th = 7.07
j 10
5
81.86°( )
A
A
20
3
90 (V)
V th = 6 7.08
63.4 ° (V)
5
10
A
100 0 (V)
j4
Fig. 4.86 (a)
Fig. 4.86 (b)
j6
B
B
B
8
j5
Fig. 4.87
(c) Here, with A–B open, the equivalent impedance,
Z = 10 +
main current,
I=
− j 5 × (13 + j 6 ) 160 − j 55
=
− j 5 + (13 + j 6 )
13 + j1
= 12.98∠ − 23.37°
( )
100∠0°
100∠0°
=
= 7.7∠23.37° ( A )
Z
12.98∠ − 23.37°
Thevenin voltage,
⎛
⎛ − j5 ⎞
⎞
− j5
Vth = I × ⎜
× (8 + j 6 ) = 7.7∠23.37° × ⎜
× (8 + j 6 ) = 29.553∠ − 34.16° ( V )
⎟
⎝ − j5 + 5 + 8 + j6 ⎠
⎝ 13 + j1⎟⎠
⎡ 10 × ( − j 5) ⎤
Z th = ⎢
+ 5 ⎥ (8 + j 6 ) = 5.33∠ − 0.5° ( )
⎣ 10 − j 5
⎦
Thevenin impedance,
(d) The circuit is redrawn as shown in Fig. 4.88, considering two capacitors in parallel.
⎛ 1 1⎞ 1
Ceq = (C1 + C2 ) = ⎜ + ⎟ = F
⎝ 4 4⎠ 2
Thevenin voltage is given as
(1+ 2 s )
(
Thevenin impedance, Z th ( s ) = 1 + 2
s
Fig. 4.88
) (1+ s 2 ) = 1
10∠0°
= 0.09995∠ − 5.7° (A)
100 + j10
2/ s
B
)
(e) To find Vth
With A–B open, the current of the dependent source can
flow through the capacitor only.
∴I =
1
2s /s 2 4
2s
4s
(V)
Vth ( s ) = 2
×
= 2
s + 4 1+ 2 +1+ s
( s + 4 )( s + 2 )
s
2
(
A
s/ 2
1
5I
I 100
A
j5
10
0 0 (V)
j 10
Vth
B
Fig. 4.89
194
Network Analysis and Synthesis
Thevenin voltage,
Vth = VAB = ( I × j10) − {5 I × ( − j 5)} = j 35 I
= j 35 × 0.09995
5∠ − 5.7° = 3.48∠84.3° ( V )
To find IN
I 100
Converting the dependent current source into the
voltage source, by KVL,
10∠0 = (100 + j10) I − j10 I N
A
j5
j 2 5I
10
0
IN
j10
0 0 (V)
and −( − j 25 I ) = − j10 I + I N ( j10 − j5)
B
Fig. 4.90
Solving for IN, I N = 0.6 ∠31° ( A )
Thevenin impedance,
Z th =
Vth
IN
=
3.48∠84.3
= 5.8∠53.3 ( )
0.6 ∠31
Problem 4.21 Find V0 using Thevenin’s theorem
Solution To find Vth
Removing the 2- resistor and open circuiting the terminals and then converting the dependent current source into
dependent voltage source, we redraw the circuit as follows.
By KVL for the two loops, (here, i0
I1)
3i 0
i0 4
2H
12cos t (V)
1/4 F
1/4 F
2
V0
( 4 − j 4 ) I1 + j 4 I 2 = −12 and − j 2 I1 + ( − j 6 ) I 2 = 0
Fig. 4.91
Solving for I2,
I2 =
( 4 − j 4 ) −12
− j2
0
(4 − j 4) j 4
− j2
− j6
3i 0
=
− j 24
− j 24 − 24 − 8
i0 4
2H
12 cos t (V)
1/4 F
1/4 F
V th
j3
=
= 0.6 ∠53.13° ( A )
4 + j3
Therefore, Thevenin voltage is
Vth = I 2 × ( − j8) =
24
= 4.8∠ − 36.87° ( V )
4 + j3
i0 4
12
0 (V)
I1
I2
j4
To find IN
Fig. 4.92
Removing the 2- resistor and short-circuiting the terminals and then converting the dependent current source into
dependent voltage source, we redraw the circuit as shown in Fig. 4.92 (b)
j 6i 0
j2
vth
j4
195
Network Theorems
By KVL for the two loops,
j 6i 0
j2
i0 4
( 4 − j 4 ) I1 + j 4 I 2 = −12
− j 2 I1 + ( − j 2 ) I 2 = 0
12
0 ° (V)
j4
I1
IN
I2
Solving for I2,
I2 = IN =
( 4 − j 4 ) −12
− j2
0
(4 − j 4) j 4
− j2
− j2
Fig. 4.92 (b)
=
j3
− j 24
=
= 1.341∠63.435° ( A )
−8 − j8 − 8 2 + j
Z th = 3.58
100.3 °( )
A
Therefore, Thevenin impedance is,
Z th =
Vth
IN
=
4.8∠ − 36.87°
3.58∠ − 100.3° ( Ω )
1.341∠63.435°
Vth =4.8
v0
36.87° (V)
2
B
Thus, Thevenin’s equivalent circuit becomes as shown
in Fig. 4.93.
Thus, the required voltage,
Fig. 4.93
⎛ Vth ⎞
⎛ 4.8∠ − 36.87° ⎞
× 2 = 1.27∠32° ( V )
v0 = ⎜
⎟ ×2=⎜
⎝ 3.58∠ − 100.3° + 2 ⎟⎠
⎝ Z th + 2 ⎠
5
Problem 4.22 Obtain the Norton’s equivalent circuit
with respect to the terminals AB for the network shown
in Fig. 4.94.
∴ Z eq =
20 V
B
5 × 15 75
= = 3⋅ 75
5 + 15 20
5
A
10V
Solution Removing the source,
15
Fig. 4.94
5
15
A
10 V
B
15
A
20 V
B
Fig. 4.94 (b)
Fig. 4.95
A
B
3.75
Short-circuiting AB,
I sc =
10 20
+ = 3.33 A
5 15
So, Norton’s equivalent circuit is shown in Fig.
3.33 A
Fig. 4.96
196
Network Analysis and Synthesis
Problem 4.23 Replace the circuit in Fig. 4.97 with the Thevenin’s equivalent circuit across A and B.
1k
I
A
10 mV
V0 /104 V
30 k
V0
75I
B
Fig. 4.97
Solution By KVL for the left-hand side loop,
1 × 103 × I +
V0
104
= 10 × 10−3
(i)
In the right-hand side loop, the dependent current source current will circulate in the resistor. By KVL,
)
(
V0 = 30 × 103 × −75 I = −225 × 104 I
(ii)
Substituting the value of I from (ii) in (i), we get,
⎛
⎞ V0
V0
+ 4 = 10 × 10−3
⇒ 1 × 103 × ⎜ −
4⎟
⎝ 225 × 10 ⎠ 10
⇒ − 4.44 × 10− 4 V0 + 1 × 10− 4 V0 = 10 × 10−3
⇒ V0 = −
10 × 10−3
= −29 V
3.44 × 10−4
1 × 10 × I + 0 = 10 × 10
−3
A
V0 = 0
Now, short circuiting the terminals A and B, we get by
KVL to left-hand-side loop,
3
1k
I
30 k
10 mV
Isc
75I
−5
⇒ I = 1 × 10 A
B
Fig. 4.98
Also, from right-hand side loop on the short circuit,
38.67 k
I sc = −75 I = −75 × 1 × 10−5 = −75 × 10−5 A
A
Thus, the Thevenin equivalent impedance is given as
29 V
V
−29
Z th = oc =
= 38.67 k
I sc −75 × 10−5
B
Fig. 4.99
Thevenin’s equivalent circuit is shown in the Fig. 4.99.
I0
Problem 24 Find the Thevenin’s equivalent between
terminals a and b of the circuit shown in Fig. 4.100.
Solution By KVL for the right-hand side mesh,
Voc = Vx = ( − 40 I 0 ) × 50 = −2000 I 0
From the left-hand side loop,
3V
1k
a
2Vx
5
40I0
(i)
Fig. 4.100
Vx
b
197
Network Theorems
I0 =
I0
3 − 2Vx 3 − 2Voc
=
1000
1000
1k
a
(ii)
3V
From (i) and (ii), we get,
Isc
50
40I0
⎛ 3 − 2Voc ⎞
Voc = −2000 ⎜
⎟ ⇒ Voc = 2 V
⎝ 1000 ⎠
b
Fig. 4.101
To determine the Thevenin’s impedance, we short circuit the terminals a and b.
Here,
16.67
a
⎛ 3 ⎞
I sc = −40 I 0 = −40 × ⎜
= −0.12 A
⎝ 1000 ⎟⎠
∴ Rth =
2V
Voc
2
=
= 16.67
I sc 0.12
b
Fig. 4.102
Thevenin’s equivalent circuit is shown in Fig. 4.102.
R1 = 2
Problem 25 In the network shown in Fig. 4.103
the switch is closed at time t ⴝ 0. Assuming all the
initial currents and voltages as zero, find the current through the inductor L2 by the use of Norton’s
theorem.
2
3
s
L2 = 1H
C= 2F
3V
Fig. 4.103
s
s
R2 = 2
t=0
Solution The network for t
0 in Laplace
domain is shown in Fig. 4.104.
The equivalent network reduces to one as shown
in Fig. 4.105.
2
L1 = 1 H
2
A
s
3
s
1
s
1
s
s
2
B
Fig. 4.105
Fig. 4.104
To find the current in L2, we have to find Thevenin’s equivalent circuit across the terminals A and B. The
impedance between terminals A and B is given as
Z th = Z AB =
( s + 2 ) × 1s
s+2+
1
s
=
( s + 2) = ( s + 2)
s + 2 s + 1 ( s + 1)
2
2
198
Network Analysis and Synthesis
Short circuit current flowing from A to B is given as
3
3
I sc = s =
s+2 s s+2
)
(
A
IL
3
s(s 2)
(s 2)
(s 1)2
Therefore, the Norton’s equivalent circuit is shown in Fig.
4.106.
Hence the current,
Fig. 4.106
s+2
3
3
1
IL =
×
×
=
2
s s+2
s+2
s s + 2 s2 + 2s + 2
s +1
+
+
s
2
( s + 1)2
(
( )
) ( ) (
) (
)(
) (
s 2
B
)
By partial fraction expansion,
IL =
where,
k3
k3 *
k
k
= 1+ 2 +
+
s
s
+
s
+
+
j
s
+
1 − j1
2
1
1
s s + 2 s + 2s + 2
(
)(
3
)
2
k1 =
3
3
=
( s + 2 )( s + 2 s + 2 ) s =0 4
k2 =
3
3
=−
4
s( s + 2 s + 2 ) s =−2
k3 =
3
3
=j
s( s + 2 )( s + 1 − j1) s =−1− j1
4
2
2
k3 * = − j
3
4
3
3
3
3
3
j3
j3
2
4 −
4 = 4− 4 −
∴IL = 4 − 4 +
s s + 2 s + 1 + j1 s + 1 − j1 s s + 2 ( s + 1)2 + 1
Taking inverse Laplace transform we get the required current as
3 3
3
i(t ) = − e −2 t − e − t sin t
4 4
2
Problem 4.26 The following circuit of Fig. 4.107 has a dependent
current source and an independent voltage source. Find the Thevenin
equivalent network of the circuit across the terminals a and b.
Solution
100 V
100 V
a
v1
100
100 V
20
v1
b
v1
100
Fig. 4.108
20
v1
20
Isc v1 = 0
Fig. 4.107
199
Network Theorems
With open circuit, v1
−
voc. By KCL,
voc 100 + voc
+
= 0 ⇒ − voc + 500 + 5voc = 0 ⇒ voc = −125
5V
100
20
With short-circuit, v1
5A
that, Isc
25
a
0 and the dependent current source is open, so
v
−125
Thus, Thevenin impedance, Rth = oc =
= 25
I sc
−5
125 V
So, the Thevenin’s equivalent circuit is shown in Fig. 4.109.
Fig. 4.109
b
Problem 4.27 In the network of Fig. 4.110, the switch
K is closed at time t ⴝ 0, a steady state having previously existed. Obtain the current in the resistor R
using Thevenin’s theorem.
K
10
100 V
Solution When the switch K is opened, under steady
state condition, two inductors behave as short circuits.
Therefore, the initial currents flowing through the inductors can be found out by writing the KVL equations for
the circuit at t 0 .
By KVL for the two meshes,
s
L1
L2
10
1H
R3
1H
R
10
10
10
10
100 V
Solving, I1 = 4 A, I 2 = 2 A
Hence, the transform network for t
0 is shown in
Fig. 4.111 (b).
Thevenin equivalent impedance with respect to the
terminals a and b is given as
10
R2
Fig. 4.110
30 I1 − 10 I 2 = 100 and − 10 I1 + 20 I 2 = 0
Z th =
R1
I1
Fig. 4.111 (a) Circuit at t
10
I2
10
0
( s + 10) × 10 10( s + 10)
=
s + 10 + 10
( s + 20)
L 1 I 1=4 V
L 2I 2 =2 V
s
a
100
s
10
10
ZL
b
Fig. 4.111 (b) Transform network for t
0
To find the open-circuit voltage across the terminals a and b, we have the current flowing in the left mesh
100 + 4
4 s + 100
s
I (s) =
=
s + 10 + 10 s( s + 20)
200
Network Analysis and Synthesis
4 s + 100
2 s 2 + 80 s + 1000
× 10 + 2 =
s( s + 20)
s( s + 20)
()
∴Voc ( s ) = I s × 10 + 2 =
Therefore, the Thevenin’s equivalent circuit is shown in
Fig. 4.111 (c).
Hence, the current through the resistor R 10 is given as,
I L (s) =
Voc ( s )
=
Z th + R
Z th=10(s+10) /(s+20)
2 s 2 + 80 s + 1000
⎡ 10( s + 10)
⎤
+ ( s + 10) ⎥
s( s + 20) ⎢
⎣ ( s + 20)
⎦
a
s
10
V OC =2s 2+80s+1000
s(s+20)
b
2 s 2 + 80 s + 1000
=
s( s + 10)( s + 30)
Fig. 4.111 (c) Thevenin’s equivalent circuit
By partial fraction expansion, let
I L (s) =
K3
K
2 s 2 + 80 s + 1000 K1
=
+ 2 +
+
30
s
s
s
+
10
s s + 10 s + 30
)(
(
)
⎡ 2 s 2 + 80 s + 1000 ⎤
100
∴ K1 = s ⎢
⎥ =
s
s
+
s
10
+
30
⎢⎣
⎥⎦ s =0 3
⎡ 2 s 2 + 80 s + 1000 ⎤
∴ K 2 = s + 10 ⎢
= −2
⎥
⎢⎣ s s + 10 s + 30 ⎥⎦ s =−10
⎡ 2 s 2 + 80 s + 1000 ⎤
2
=
∴ K 3 = s + 30 ⎢
⎥
⎢⎣ s s + 10 s + 30 ⎥⎦ s =−30 3
)(
)
(
) (
)(
)
(
) (
)(
)
(
10
I L (s) =
s
3−
2
2
+ 3
s + 10 s + 30
Taking inverse Laplace transform, we get
( ) 103 − 2e
iL t =
−10 t
2
+ e −30t = 3.33 − 2 e −10t + 0.67e −30t
3
Problem 4.28 For the network shown in Fig. 4.112,
show that the Thevenin equivalent at the terminals a–b is
represented by,
Vth =
V1
1+ a + b − ab
2
(
) and Z = 3 −2 b
th
Solution When the terminals a–b are open-circuited no current will flow through the right side of the
1 resistor. By KVL for the left mesh,
V
2 I1 + aV1 = V1 ⇒ I1 = 1 (1 − a )
2
1
a
1
I1
V1
bI1
1
aV1
b
Fig. 4.112
201
Network Theorems
V
V
V
∴ Vth = 1 × I1 + aV1 + bI1 = 1 × 1 (1 − a ) + aV1 + b 1 (1 − a ) = 1 (1 − a + 2 a + b − ab )
2
2
2
V1
∴ Vth = (1 + a + b − ab ) ( Proved )
2
To find the Thevenin impedance, we have to find the short-circuit current flowing through the terminals a–b.
bI1
By KVL for the two meshes, we get,
1
1
I1
2 I1 − I sc = V1 (1 − a )
(i)
and, 1 × ( I sc − I1 ) − bI1 + 1 × I sc = aV1 ⇒ − (1 + b ) I1 + 2 I sc = aV1
(ii)
V1
1
a
Isc
aV 1
b
Solving (i) and (ii), we get
I sc =
2
V1 (1 − a )
−(1 + b )
aV1
2
− 1+ b
(
)
Fig. 4.113
=
−1
2
2 aV1 + V1 (1 − a + b − ab ) V1 (1 + a + b − ab )
=
4 −1− b
3− b
Therefore, the Thevenin impedance is, (Proved)
V1
Vth 2 (1 + a + b − ab ) 3 − b
Z th =
=
=
I sc V1 (1 + a + b − ab )
2
3− b
I =5 30 A
Problem 4.29 Find the Thevenin equivalent circuit for the network
shown in Fig. 4.114 at terminals A–B.
Solution When the terminals A and B are open-circuited, the current
flowing through the right branch (50 j50) is
50 + j 50
I = 5∠30° ×
100 + 50 + j 50 + 50 + j 50
⎛ 1+ j ⎞
⎛ 50 + j 50
0 ⎞
= 5∠30° × ⎜
= 5∠30° × ⎜
⎟
⎝ 4 + j 2 ⎟⎠
⎝ 200 + j100 ⎠
Therefore, the Thevenin voltage is,
⎛ 1+ j ⎞
Vth = I × (50 + j 50) = 5∠30° × ⎜
× (50 + j 50) = 111.8∠93.43° V
⎝ 4 + j 2 ⎟⎠
100
50
50
j 50
j50
B
Fig. 4.114
100
(
(150 + j 50 × (50 + j 50
) (50 + j 50) = (150 + j 50)) + (50 + j 50))
50
j 50
j 50
Zth
B
Fig. 4.115
50 36.87
a
111.8 93.43 V
= 50∠36.87°
Thus, Thevenin equivalent circuit is shown in Fig. 4.116.
A
50
Thevenin impedance is given as
Z th = 150 + j 50
A
Vth
b
Fig. 4.116
202
Network Analysis and Synthesis
Problem 4.30 For the one port shown in Fig. 4.117,
determine the Norton’s equivalent at the terminals AB, if
the v–i characteristic is given by, 16v ⴝ 80 ⴚ 2i.
Solution The v–i characteristic is given as,
V
i
16 v = 80 − 2i ⇒
+ =1
5 40
Thus, short-circuit current, Isc 40 A (where v
and open-circuit voltage, Voc 5 V (where i
∴ RN =
i
A
i
N
40
v
0
B
Fig. 4.117
5
v
Fig. 4.118
0)
A
0)
Voc 5 1
= =
I sc 40 8
1
8
40A
B
Norton, equivalent circuit is shown in Fig. 4.119.
Fig. 4.119
Maximum Power Transfer Theorem
Problem 4.31 Find the Thevenin’s equivalent between
the points a and b for the circuit given in Fig. 4.120. What
should be the value of impedance connected between
a and b for maximum power to be transferred from the
sources? Obtain the amount of the maximum power.
2
j6
a
(3 j5)
100 V
Solution Here the current
b
100
100
I=
=
= (10 − j10) A
2 + 3 + j5 5 + j5
Fig. 4.120
∴Vth = I × ( 3 + j 5) = (10 − j10) × ( 3 + j 5) = (80 + j 20) = 82.46
6 ∠14° V
(1.6 j 6. 4)
a
2 × ( 3 + j 5)
= (1.6 + j 6.4 )
∴ Z th = j 6 +
2 + 3 + j5
Thevenin’s equivalent circuit is shown in Fig. 4.121.
For maximum power transfer, the impedance should be complex conjugate
of Thevenin impedance.
∴ Z L = 1.6 − j 6.4
)
(
82.46 14 V
b
Fig. 4.121
2
Amount of the maximum power is, Pmax =
Vth (82.46 )2
=
= 1062.5 W
4R
4 × 1.6
5
Problem 4.32 In the network in Fig. 4.122, two voltage
sources act on the load impedance connected to the terminals
A and B. If the load is variable in both reactance and resistance,
for what load, will ZL receive maximum power? What is the value
of maximum power?
Solution Here, V1 = 50∠0 = 50 V; and V2 = 25∠90 = j 25 V
50
0 (V)
j5
A
ZL
3
j4
25 90 (V)
B
Fig. 4.122
vth
203
Network Theorems
Current in the circuit, I =
Z th
50 − j 25
50 − j 25
=
A
5 + j5 + 3 − j 4
8 + j1
a
Thevenin voltage,
V th
⎛ 50 − j 25 ⎞
25 − j 75
Vth = 50 − I × (5 + j 5) = 50 − ⎜
× (5 + j 5) =
8 + j1
⎝ 8 + j1 ⎟⎠
= 9.8∠ − 78.7° = (1.923 − j 9.615) V
b
Fig. 4.123
Thevenin impedance,
Z th =
(5 + j 5) × ( 3 − j 4 ) 35 − j 5
=
= ( 4..23 − j1.154 )
(5 + j 5) + ( 3 − j 4 ) 8 + j1
Thus, the Thevenin’s equivalent circuit is shown in Fig. 4.123.
For maximum power transfer to the load, Z L = Z m * = ( 4.23 + j1.154 )
The value of the maximum power is, Pmax =
Vth 2
9.82
=
= 5.676 W
4 R 4 × 4.23
Problem 4.33 In the network shown, the power dissipated in R when E1,
E2 or E3 acting alone is
(a) 20 W, 80 W, and 5 W respectively
(b) 30 W, 270 W, and 120 W respectively
Calculate the maximum power that R can dissipate due to the simultaneous
action of all the sources. Calculate both for (a) and (b).
What will be the minimum power dissipated in R when all the sources are
acting simultaneously?
Solution Current for E1 at R, i1 = ±
Current for E2 at R, i2 = ±
P2
R
Current for E3 at R, i3 = ±
P3
R
E1
E2
E3
Fig. 4.124
P1
R
total current flow for simultaneous action of all the three sources is
i = ±i1 ± i2 ± i3 = ±
P
P1
P
± 2± 3
R
R
R
2
⎡ P
2
P ⎤
P
∴ power, P = i R = ⎢ ± 1 ± 2 ± 3 ⎥ R = ⎡ ± P1 ± P2 ± P3 ⎤
⎣
⎦
R
R⎥
R
⎢⎣
⎦
2
Resistive
network
R
204
Network Analysis and Synthesis
• For maximum power,
Pmax = ⎡ P1 + P2 + P3 ⎤
⎣
⎦
2
2
2
(a) Pmax = ⎡ 20 + 80 + 5 ⎤ = ⎡ 2 5 + 4 5 + 5 ⎤ = 49 × 5 = 245 W
⎣
⎦ ⎣
⎦
2
2
(b) Pmax = ⎡ 30 + 270 + 120 ⎤ = ⎡ 4 5 − 3 5 ⎤ = 1080 W
⎣
⎦ ⎣
⎦
• For minimum power,
2
2
(a) Pmin = ⎡ − 20 + 80 − 5 ⎤ = ⎡ 4 5 − 3 5 ⎤ = 5 W
⎣
⎦ ⎣
⎦
2
2
(b) Pmin = ⎡ − 30 + 270 − 120 ⎤ = ⎡ − 30 + 3 30 − 2 3 ⎤ = 0 W
⎣
⎦ ⎣
⎦
Problem 4.34 Find the value of R in the circuit of Fig. 4.124 such that maximum power transfer takes
place. What is the amount of this power?
3
(a)
5V
2
(b)
1
2A
1
10
R
5A
R
2
24 V
5
Fig. 4.124
Solution (a) In the network, the 2- resistor is connected in parallel with an ideal voltage source of 5 V;
hence this resistance can be removed without affecting the current flows in the other branches.
1
3
5V
1
2A
1
R
5 /3 A
Fig. 4.125
Converting the voltage source into current source,
⎛5 ⎞
11
⎜⎝ 3 + 2⎟⎠ A = 3 A
3
1
2A
R
205
Network Theorems
7
4
For maximum power transfer, R =
2
⎛ 11⎞
⎜⎝ 4 ⎟⎠
= 1⋅ 08 W
Maximum Power, Pmax =
4× 7
4
1
1
7/4
3/4
3/4
11/3 A
R
Fig. 4.126
R
11/4 V
R
11/4 V
Rth
2
10
5
Fig. 4.127 (a)
Rth =
(b) To find Rth
10 × 5
+ 2 = 5.33
10 + 5
2
10
Vo c
To find VOC
24
= −1⋅ 6 A
15
∴ Voc = 5i + 10 = −8 + 10 = 2 V
∴ Pmax =
5
10V
i =−
2V
i
Fig. 4.127 (b)
4
= 0 ⋅188 W
4 × 5 ⋅ 33
3
3
A
Problem 4.35 In the network shown, find the
value of ZL to which the maximum power can be
delivered. Hence, find the value of the maximum
power.
Solution With respect to terminals A and B, the
Thevenin voltage is
Vth =
5 0 (V)
ZL
j3
j3
B
Fig. 4.128
⎛
⎞ 45∠0°
j3
5∠0°
= 2.236 ∠ − 26.56° ( V )
×⎜
=
j 3( 3 − j 3) ⎝ 3 + j 3 − j 3 ⎟⎠ 18 + j 9
3+
3− j3+ j3
⎛ 3 × j 3⎞
⎜⎝ 3 + 3 + j 3 ⎟⎠ × ( − j 3)
and Thevenin impedance, Z th =
= 3∠ − 53.12°
3× j3
3+
− j3
3+ j3
= (1.8 − j 2.4 )
206
Network Analysis and Synthesis
For maximum power transfer, Z L = Z th * = (1.8 + j 2.4 )
current, I =
2.236 ∠ − 26.56°
= 0.621∠ − 26.56° A
1.8 × 2
The value of the maximum power is, P
max
=
(Vth )2
4R
=
( 2.236 )2
= 0.694 W
4 × 1.8
Problem 4.36 A loudspeaker is connected across terminals A and B of the network. What should its
impedance be to obtain maximum power dissipation in it?
(b)
j5
4
j6
A
(a)
(3
j4)
10
A
10
30 (V)
120
15 (V)
10
5 0 (A)
j8
j5
B
B
Fig. 4.129
Solution (a) Equivalent impedance with respect to the terminals A and B is
Z th =
( 3 + j 4 )( − j 5)
= 7.9∠ − 18.43°
3 + j 4 − j5
= (7.5 − j 2.5)
For maximum power transfer, Z L = Z th * = (7.5 + j 2.4 )
(b) Equivalent impedance with respect to the terminals A and B is
⎛ − 40 + j 50 + 40 + j 52 + j 60 − 78 ⎞
⎡ (10 + j8) j 5
⎤
Z th = ⎢
+ 4 + j 6 ⎥ 10 = ⎜
⎟⎠
10 + j13
⎝
⎣ 10 + j8 + j 5
⎦
= 6.14 ∠30° = (5.316 + j 3.07)
For maximum power transfer, Z L = Z th * = 6.14 ∠ − 30°
10
= (5.316 − j 3.07)
Problem 4.37 Two inductors each of 1-⍀ reactance and negligible resistance are connected in series
across a 2-V ac source. Find the value of resistance which should be connected across one of the inductors
for maximum power dissipation. Also, find the maximum power.
Solution Here, Z =
current I =
R × j1
−1 + j 2 R
+ j1 =
R + j1
R + j1
2 ∠0° 2 ∠0° × ( R + j1)
=
−1 + j 2 R
Z
current through the resistance, I R = I ×
j1
j2
=
R + j1 −1 + j 2 R
207
Network Theorems
2
power, P = I R =
4R
1 + 4 R2
For maximum power,
dP
(1 + 4 R 2 ) × 4 − 4 R × 8 R
=0 ⇒
= 0 ⇒ R = 0.5
dR
(1 + 4 R 2 )2
maximum power, Pmax =
4 × 0.4
=1W
1 + 4 × (0.5)2
Problem 4.38 A network has two output terminals. The open-circuit voltage at these terminals is 260 V.
The current flowing through the terminals is 20 A when the terminals are short circuited. Also the current is
13 A when a coil of 11-ohm reactance and negligible resistance is connected across the terminals. Find the
impedance components of the equivalent circuit feeding the terminals. What value of load impedance will
give maximum power transfer and what is the value of this power?
Solution Here, Vth = 260 V; Isc 20 A
Let the Thevenin impedance across the terminals is Z (R
Vth 260
∴Z =
=
= 13
I sc 20
jX)
∴ R 2 + X 2 = 169
When the 11-ohm reactance is connected across the terminals, the current is 13 A.
∴
(
(i)
)
260
260
= 13 ⇒ R + j X + 11 =
= 20
13
R + j X + 11
)
(
(
)
2
∴ R 2 + X + 11 = 400
(ii)
Solving (i) and (ii), we get, R = 12
X =5
Therefore, Thevenin impedance, Z th = (12 + j 5)
For maximum power transfer, Z L = Z th * = (12 − j 5)
Value of maximum power, Pmax =
Vth 2 ( 260)2
=
= 1408.33 W
4 R 4 × 12
Problem 4.39 What should be the value of ZL for
maximum power to be delivered in the circuit shown in
Fig. 4.130?
3
5cos (␻t
4
j2
30 )
Solution In this circuit, when the voltage sources are
replaced by their internal impedances; ie., when they are
short-circuited, the equivalent Thevenin impedance with
Fig. 4.130
respect to the load terminals is given by
11 ⎞
( 3 + j 2 ) × ( 4 − j 3) 18 − j1 ⎛ 127
Z th = ( 3 + j 2 ) ( 4 − j 3) =
=
=⎜
+j ⎟
50 ⎠
( 3 + j 2 ) + ( 4 − j 3) 7 − j1 ⎝ 50
= ( 2.54 + j 0.22 )
= 2.55∠4.95° ( )
ZL
j3
2cos ␻t
208
Network Analysis and Synthesis
For maximum power to be delivered, the load impedance should be complex conjugate of the Thevenin impedance, so that, Z L = Z th * = ( 2.54 − j 0.22 ) = 2.55∠ − 4.95° ( )
2I1
Problem 4.40 In the network shown, calculate the maximum
power that may be dissipated in the external resistor R.
I1
10 A
4
3
6
R
Solution Transforming the current source into voltage source,
Fig. 4.131
By KVL, 6i1 + 4i1 − 40 − 2i1 = 0 ⇒ i1 = 5 A
2i1
∴eoc = 6i1 = 30 V
i1
For maximum power, R Req
Shorting the terminals and solving by loop method,
40 V
3
eoc
6
4
I sc = 5 A
Fig. 4.132
30
=6
5
( 30)2 900
∴ Pmax =
=
= 37.5 W
4 × 6 24
∴ Rth =
2i1
i1
40 V
3
Isc
6
4
Reciprocity Theorem
Problem 4.41 Solve the network shown in
Fig. 4.134 (a) and hence find the zcurrent in the 2-⍀
resistor in Fig. 4.134 (b) when an emf of 36 V is added
in the branch BD as shown in Fig. 4.134 (b). All values
are in ohm.
Solution
Fig. 4.133
B
72 36
=
⇒ I = 0 ⋅5 A
1
I
current in the 2action of two sources
resistor for simultaneous
I = (6 − 0.5) = 5.5 A
21
6
A
C
18
21
18
A
C
36 V
• Solve by any method of network
analysis.
• We consider the 36-V source acting alone.
When the 72-V source is acting alone,
The current in 2- resistor 6 A
By the reciprocity theorem,
B
6
6
12
12
D
2
72V
72V
Fig. 4.134 (a)
B
6
3A
A
Fig. 4.134(b)
6A
12
B
2A
1A
18
3A
D
6
21
C
6
4A
2
6
D
2
21
18
A
I
C
36V
12
2
6
D
72V
Fig. 4.135 (a)
Fig. 4.135(b)
Problem 4.42 An emf source E, having negligible internal impedance is connected in series with
an impedance Z1 to the input terminals 1–2 of a linear, bilateral four terminal network. It produces a current
I2 in impedance ZL connected across the output terminals 3–4. The emf source is now transferred so as to
209
Network Theorems
act, in series with Z2, between terminal 3–4. Z1 is disconnected and the input
terminals 1–2 are short-circuited. The short-circuited current traversing terminals 1–2 is then I1. Prove that the impedance looking into terminals 1–2
under the first condition is, Z 12 =
Z 1I 2
.
I1 − I2
Solution Let the impedance looking into terminals 1–2 be Z12.
Thus the network becomes as shown in Fig. 4.136.
∴I =
voltage across 1–2, V12 =
Z1
I
E
Z 12
2
Fig. 4.136
E
Z1 + Z12
1
E × Z12
Z1 + Z12
V12
So, the circuit becomes as shown in Fig. 4.137.
The given network is linear and bilateral and according to the reciprocity
theorem, if the source E is put across terminals 1–2, the response current flowing through Z2 will be I1 as shown in Fig. 4.138.
Now, if a voltage equal to V12 is applied instead of E, the current flowing
through Z2 will be,
N
2
E
N
2
Problem 4.43 Verify the reciprocity theorem for the ladder network shown in Fig. 4.139.
20
10
j10
Fig. 4.139
Solution Let the three loop currents be I1, I2, and I3. By KVL for the three loops,
20
20
I2
200 45 (V)
I1
j 10
I3
j10
Fig. 4.140
3
I1
Fig. 4.138
j 10
Z 12
4
1
But, this current is equal to I2.
⎛ ZI ⎞
Z1− 2
∴ I 2 = I1
⇒ Z12 = ⎜ 1 2 ⎟ (Proved
d)
Z1 + Z1− 2
⎝ I1 − I 2 ⎠
200 45 (V)
3
Fig. 4.137
I1
I
E × Z12
Z12
× V12 = 1 ×
= I1 ×
E
E Z1 + Z12
Z1 + Z12
20
1
10
Z 12
4
210
Network Analysis and Synthesis
( 20 + j10) I1 − j10 I 2 = 200∠45°
− j10 I1 + 20 I 2 + j10 I 3 = 0
j10 I 2 + (10 − j10) I 3 = 0
Solving for I3,
( 20 + j10) − j10 200∠45°
I3 =
− j10
0
(
20 + j10
)
− j10
0
20
j10
0
0
− j10
0
20
j10
j10
10 − j10
(
=
200∠45° × 100
( 20 + j10)( 200 − j 200 + 100) − j10( j100 + 100)
)
Now by interchanging the positions of the voltage source and the response current, we get,
By KVL,
20
I2
200 45 (V)
I3
j10
j 10
2
− j10 I1 + 20 I 2 + j10 I 3 = 0
(
20
I1
( 20 + j10) I − j10 I = 0
1
( )
= 2.169∠57.53° A
)
10
Fig. 4.141
j10 I 2 + 10 − j10 I 3 = 200∠45°
Solving for I1,
I1 =
0
0
− j10 0
20 0
200∠45°
j10
0
0
( 20 + j10) − j10
j10
20
− j10
0
j10 (10 − j10)
= 2.169∠57.53° ( A )
Since the currents in both the cases are the same, reciprocity theorem is verified.
Problem 4.44 In the given circuit of Fig. 4.142, find the reading of
the voltmeter V. Interchange the current source and voltmeter and
verify the reciprocity theorem.
1 + j1
= 0.707∠45° ( A )
1 + j1 + 1 − j1
The voltage, V = I 2 × ZC = 0.707∠45° × (1) = 0.707∠45° ( V )
1 /j
1
1
1 0 (A)
Solution Here, the current
I 2 = 1∠0° ×
j1
Fig. 4.142
V
211
Network Theorems
Now, interchanging the positions of the current source and the finding
the resulting voltage, we get
I1 = 1∠0° ×
I1
I2
j1
1 /j
1
1
1 0 (A)
1
= 0.5∠0° ( A )
1 − j1 + j1 + 1
V
the voltage,
V = 0.5∠0° × (1 + j1) = 0.5∠ − 23.2° × 2 ∠45° = 0.707∠45° (V
V)
As ‘V’ is same as obtained before interchanging the position of the current source, the reciprocity theorem is verified.
Fig. 4.143 (a)
I1
I2
j1
1 /j
1
1
V
Problem 4.45 In this circuit of Fig. 4.144, find voltage V. Interchange
the current source and resulting voltage V and show that the reciprocity theorem is verified.
1 0 (A)
Fig. 4.143 (b)
Solution Here, the current
I 2 = 5∠90° ×
I1
5 + j5
= 4.64 ∠111.8° ( A )
5 + j5 + 2 − j 2
I2
j5
2
5 90 (A)
the voltage,
5
j2
V
V = I 2 × ZC = 4.64 ∠111.8° × ( − j 2 ) = 9.28∠21.8° ( V )
Now, interchanging the positions of the current source and the finding
the resulting voltage, we get,
− j2
I1 = 5∠90° ×
= 1.31∠ − 23.2° ( A )
− j 2 + 5 + 2 + j5
the voltage,
V = 1.31∠ − 23.2 × (5 + j 5)
= 1.31∠ − 23.2 × 7.075∠45 = 9.28∠21.8 ( V )
Fig. 4.144
I1
I2
j5
2
5
j2
V
5 90 (A)
Fig. 4.145
As ‘V’ is same as obtained before interchanging the position of the
current source, reciprocity theorem is verified.
Compensation Theorem
Problem 4.46 Find the current flowing in the resistor R4 of the
network shown in Fig. 4.146. If a resistance of 0.5 ⍀ is inserted in
series with R4, find, using the compensation theorem, the current
that will flow through R4. All values are in ohms.
Solution Solving the network by any method of network analysis, I 0.5 A
Now z = 0.5
0.25
A = 0.01269 A
19.7
∴ I ′ = ( I − I ) = (0.5 − 0.01269) A = 0.4873 A
∴ Vc = I . Z = 0.5 × 0.5 = 0.25 V ⇒
I=
R 1 =4
12 V
Fig. 4.146
R 2 =16
R 3= 8
R 4 =8
212
Network Analysis and Synthesis
4
4
8
16.5
I
8
I
8
16
12 V
8
0.025 V
Fig. 4.147
10
Problem 4.47 Find the current through the 10-ohm resistance
in the circuit shown in Fig. 4.148. If the impedance (3 ⴙ j4) ohms
is changed to (4 ⴙ j4) ohms, find the new current in the 10-ohms
resistance using compensation theorem.
3
j5
50 0 (V)
j4
Solution Before changing the impedance the current through
the 10- resistance is given as
Fig. 4.148
10
10
I2
I1
3
50 0 (V)
Vc
I2
j5
j5
j4
4
j4
(a)
(b)
Fig. 4.149
I1 =
50
50
50( 3 + j 9)
= 4.5∠ − 13° ( A )
=
=
( j 5) × ( 3 + j 4 ) 10 + j105 11.1∠130
10 +
j5 + 3 + j 4
Now before changing the impedance,. The current through the ( 3 + j 4 ) Ω branch is
⎛
⎞ 50( 3 + j 9) ⎛ j 5 ⎞
j 250
j5
=
= 2.37∠5.44° ( A )
I 2 = I1 × ⎜
=
×⎜
⎟
⎟
⎝ j 5 + 3 + j 4 ⎠ 10 + j105 ⎝ 3 + j 9 ⎠ 10 + j105
Now Z = ( 4 + j 4 ) − ( 3 + j 4 ) = 1
∴Vc = I ⋅ Z = 2.37∠5.44° × 1 = 2.37∠5.44° V
The compensating circuit is shown in Fig. 4.149 (b).
⎛
⎞
⎜ 2.37∠5.44° ⎟ ⎛ j 5 ⎞ 11.85∠95.44°
⇒ I1 = ⎜
⎟ ×⎜
⎟ = 20 + j110 = 0.106 ∠15.74° A
⎜ ( 4 + j 4 ) + 10 × j 5 ⎟ ⎝ 10 + j 5 ⎠
⎜⎝
10 + j 5 ⎠⎟
∴ I1′ = ( I1 − I1 ) = ( 4.5∠ − 13° − 0.106 ∠15.74° ) A = 4.39∠ − 12.93° A
213
Network Theorems
Millman’s Theorem
Problem 4.48 Calculate the load current I in the circuit in Fig. 4.150
by Millman’s theorem.
2
2
I
5
15
+
Solution By Millman’s Theorem, equivalent voltage,
2 3 5
EY 2 + 2 + 5 35
∑
V=
=
= = 2.91667 V
∑ Y 1 + 1 + 1 12
2 2 5
2V
3V
-
5V
Fig. 4.150
and equivalent impedance,
1
1
10
=
= = 0.833
1
1
1
12
Y
∑
+ +
2 2 5
Therefore the current through the load resistance,
V
2.91667
I=
=
= 0.184 A
Z + 15 0.833 + 15
Z=
Problem 4.49 Obtain the potential of the node F with respect to the node G in the circuit of Fig. 4.151. All
values are in ohms.
1
2
3
4
5
1V
2V
3V
4V
5V
F
6
G
Fig. 4.151
Solution By Millman’s Theorem, equivalent voltage,
5
∑EY
V = i =15
i i
∑Yi
i =1
=
1 ×1− 2 × 1 + 3 × 1 − 4 × 1 + 5 × 1
2
3
4
5 = 60 V
137
1+
+1 +1 +1 +1
2
3
4
5
1
1
60
=
=
Y 1+ 1 + 1 + 1 + 1
137
2
3
4
5
Therefore the current through the 6 resistance,
60
V
137 + 60 A
I=
=
Z + 6 60
+ 6 882
137
Z
F
V
I
6
G
Equivalent impedance, Z =
Fig. 151 (a)
214
Network Analysis and Synthesis
Hence the voltage between the points F and G is, VFG = 6 × I = 6 ×
60 60
=
V
882 147
Problem 4.50 Use Millman’s theorem to obtain an equivalent current source for the circuit shown in
Fig. 4.152. Also, obtain the equivalent voltage source.
20
10
15
j30
100 0 mA
20 0 mA
j20
4 30 V
Fig. 4.152
Solution We convert the voltage source into equivalent current source as
I=
V 4 ∠30°
=
= 0.1108∠ − 26.3° ( A )
Z 20 + j 30
100 0 mA
j20
110.8
10
26.3 mA
The modified circuit is shown in Fig. 4.153.
Total equivalent current source is
I = 100∠0° + 110.8∠ − 26.3° − 20∠0° = (89.33 − j 49.1) = 101.93∠ − 28.8° ( mA )
20
15
j30
j20
20 0 mA
Fig. 4.153
Total equivalent impedance is obtained as
1
1
1
1
=
+
+
= 0.059 − j 0.095 ⇒ Z = 4.73 + j 7.57
Z 10 + j 20 20 + j 30 15 + j 20
(
)
(
)
Equivalent voltage source is obtained as, V = 101.93∠ − 28.8° × 10−3 × ( 4.73 + j 7.57) = 0.9∠29.2° ( V )
Problem 4.51 In the network, two voltage sources
act on the load impedance connected to terminals
a, b. If the load is variable in both reactance and
resistance, what load ZL will receive the maximum
power? What is the value of the maximum power?
Use Millman’s theorem.
Solution Here,
V1 = 50∠0° = 50 V ; Z1 = (5 + j 5) ;
Y1 =
1
1
=
= (0.1 − j0.1) mho
Z1 (5 + j 5)
5
j5
ZL
50 0 (V)
b
Fig. 4.154
3
a
j4
25 90 (V)
215
Network Theorems
V2 = 25∠90° = j 25V ; Z 2 = ( 3 − j 4 ) ; Y2 =
1
1
=
= (0.12 + j 0.16 ) mho
Z2 (3 − j 4)
Millman voltage source,
V Y + V Y 50(0.1 − j 0.1) + j 25(0.12 + j 0.116 )
= 9.807∠ − 78.65° ( V))
Vm = 1 1 2 2 =
(0.1 − j 0.1) + (0.12 + j 0.16 )
Y1 + Y2
Millman impedance,
Zm =
1
1
=
= 4.385∠ − 15.25° = ( 4.23 − j1.15)
Y1 + Y2 0.22 − j 0.06
(
For maximum power transfer to the load, Z L = Z m * = 4.23 + j1.15
Maximum power, Pmax =
)
Vm 2 ( 9.807)2
=
= 5.68 W
4 RL 4 × 4.23
Tellegen’s Theorem
Problem 4.52 Find the current through the 1-⍀ resistor in the circuit in Fig. 4.155 using Tellegen’s theorem.
2
1A
Solution To find the current, using Tellegen’s theorem, we first find
the Thevenin’s equivalent circuit with respect to terminals a and b.
Thevenin voltage,
3
1
2
2
3
4
Vth =
×4= V
3+ 4 + 2
3
Fig. 4.155
a
a
a
3
2
3V
1A
1A
3
2
3
b
2
2
V th
4
4
b
b
2
Fig. 4.156
Thevenin impedance,
Z th =
5 × 4 20
=
5+ 4 9
Thus, the equivalent circuit is shown in Fig. 4.157.
Now, applying Tellegen’s Theorem,
4
20
− × I + × I × I + 1 × I × I = 0 ⇒ I = 0.414 A
3
9
20
9
i
4
V
3
Fig. 4.157
1
216
Network Analysis and Synthesis
Problem 4.53 Find the value of source E2 using Tellegen’s
theorem if the power absorbed by E2 is 20 W.
Solution We have to find out the Thevenin’s equivalent
across XY.
100
Here,
∴ i=
=5A
20
∴ Voc = i × 10 = 50 V
)
(
5
10
100 V
E2
Rth = 10 10 + 10 = (5 + 10) =15
Fig. 4.158
5
10
10
5
10
5
5
100 V
10
5
X
100 V
10
X
E2
Voc
i
Y
Y
Fig. 4.159
10
5
Now Applying Tellegen’s Theorem to the equivalent circuit,
5
−50 I + 15 I 2 + E2 I = 0
10
X
But it is given that
Rth
Y
E2 I = 20 ⇒ 15 I − 50 I + 20 = 0
2
Fig. 4.160
⇒ 3 I − 10 I + 4 = 0
2
Rth = 15
10 ± 100 − 4 × 3 × 4 10 ± 52 5 ± 13
=
=
6
6
3
= 2.8685 A or 0.4648 A
∴I =
So, the value of E2 =
50 V
20
= 6.97 V or, 43.03 V
I
i
E2
Fig. 4.161
4A
Problem 4.54 A set of measurements is made on a linear
time-invariant resistive circuit as shown in Fig. 4.162 a. The
circuit is then reconnected as shown in Fig. 4.162 b. Find the
current through the 5-⍀ resistance.
10 V
Solution By Tellegen’s theorem, if the set of voltages and
currents is taken corresponding to two different instants of time t1
and t2, then
b
b =1
(a)
5
i
b
∑ v (t )i (t ) = ∑ v (t )i (t ) = 0
b
1
b
2
b =1
b
2
b
1
4V
N
N
(b)
Fig. 4.162
6A
217
Network Theorems
Here, the circuits for two different instants of time are as shown below:
4A
10 V
4V
N
5
i
6A
N
(a)
(b)
Fig. 4.163
By Tellegen’s theorem,
2
2
∑ v (t )i (t ) = ∑ v (t )i (t ) ⇒ v (t )i (t ) + v (t )i (t ) = v (t )i (t ) + v (t )i (t )
b =1
k
1
Here, v1 (t1 ) = 10V ; i1 (t1 ) = − 4 A;
v1 (t2 ) = 5i; i1 (t2 ) = i
and
k
2
b =1
and
k
2
k
1
1
1
1
2
2
1
2
2
1
2
1
1
2
2
2
1
v2 (t1 ) = 4 V; i2 (t1 ) = 0
v2 (t2 ) = 0; i2 (t2 ) = 6 A
So, from (1); (10 × i ) + ( 4 × 6 ) = (5i × −4 ) + (0 × 0)
10 i + 24 = −20 i ⇒ i = −
24
= −0.8 A
30
Problem 4.55 Two sets of measurements are taken on
a resistive network shown in Fig. 4.162. Find V2.
(a) R2 ⴝ 1 , V1 ⴝ 5 V, I1 ⴝ 2 A, V2 ⴝ 1 V
(b) R2 ⴝ 10 , V1 ⴝ 6 V, I1 ⴝ 6 A
V2
I2
I1
V1
N
V2
R2
Fig. 4.162
Solution Here,
⎡ v (t ) ⎤
⎛ 1⎞
v1 (t1 )i1 (t2 ) + v2 (t1 )i2 (t2 ) = v1 (t2 )i1 (t1 ) + v2 (t2 )i2 (t2 ) ⇒ (5 × 6 ) + 1 × ⎢ − 2 2 ⎥ = (6 × 2 ) + v2 (t2 ) × ⎜ − ⎟
10
⎝ 1⎠
⎣
⎦
v (t )
v (t )
v (t ) ⎫⎪
⎪⎧
⇒ 30 − 2 2 = 12 − v2 (t2 ) ⎨ i2 (t2 ) = − 2 2 and i2 (t1 ) = − 2 1 ⎬
10
R
t
(
)
R
(t ) ⎪
2 2
2 1 ⎭
⎩⎪
⇒ v2 (t2 ) = −
18
= −20 V
9
10
Summary
1. Network theorems are used to simplify a complex circuit to a simpler circuit and to thereby make the circuit
analysis much easier.
2. The superposition theorem states that in a linear
bilateral network, the current at any point (or voltage
between any two points) due to the simultaneous
action of a number of independent sources in the net-
work is equal to the summation of the component currents (or voltages) due to one source acting alone in
the network with all the remaining sources removed.
3. As per the reciprocity theorem, in any linear timeinvariant, bilateral network, the ratio of response to
excitation remains same for an interchange of the
position of excitation and response in the network.
218
Network Analysis and Synthesis
4. Thevenin’s and Norton’s theorems state that a linear
active bilateral network can be replaced at any two
of its terminals by the Thevenin equivalent voltage
source, Vth, in series with an equivalent impedance, Zth,
or by the Norton equivalent current source, IN, in parallel with an equivalent impedance, ZN. The relations are
Z th = Z N
IN =
V th
Zt
5. The maximum power is absorbed by one network
from another connected to it at two terminals, when
the impedance of one is the complex conjugate of
the other. For dc circuits, the condition for maximum
power transfer is RL RS; and for ac circuits, the condition is ZL ZS*.
6. The superposition, Thevenin’s, Norton’s and maximum
power transfer theorems are all valid for linear circuits
only.
7. Tellegen’s theorem states that the summation of
instantaneous powers delivered to all branches is zero.
8. The voltage source equivalent Millman’s theorem
states that if several ideal voltage sources (V1, V2, …)
in series with impedances (Z1, Z2,…) are connected in
parallel , then the circuit may be replaced by a single
ideal voltage source (V) in series with an impedance (Z);
n
∑V Y
where V = i =1n
i
∑Y
i =1
i
i
and,
Z= n
1
∑Y
i =1
.
i
9. The current source equivalent Millman’s theorem
states that if several ideal current sources (I1, I2,…) in
parallel with impedances (Z1, Z2, …) are connected in
series, they may be replaced by a single ideal current
source (I) in parallel with an impedance (Z); where,
n
Ii
∑
n
Yi
1
i =1
I= n
and, Y = n
or, Z = ∑ Z i .
i =1
1
1
∑
∑
Yi
Yi
i =1
i =1
10. According to the compensation theorem, in any linear
bilateral active network, if any branch carrying a current I has its impedance Z changed by an amount ␦Z,
the resulting changes that occur in the other branches
are the same as those which would have been caused
by the injection of a voltage source of ( I␦Z) in the
modified branch.
Short-Answer Questions
1. Under what conditions are network theorems preferred over Kirchhoff’s laws in analyzing electric
circuits?
In electric network analysis, the fundamental rules are
Ohm’s law and Kirchhoff’s laws. While these humble laws
may be applied to analyze any circuit configuration, for
complex circuits, it is necessary to simplify the network
to find current or voltage in a particular branch without
solving the entire circuit. For these complex networks,
network theorems are preferred over Kirchhoff’s laws.
2. Mention some examples where the reciprocity theorem is not applicable.
(i)
(ii)
(iii)
(iv)
This theorem is inapplicable to unilateral networks, such as networks comprising of electron
tubes or other control devices.
This theorem is inapplicable to circuits with timevarying elements.
This theorem is inapplicable to circuits with
dependent sources.
To apply this theorem, we have to consider only
the zero-state response by taking all the initial
conditions to be zero.
3. Mention some limitations of the superposition
theorem.
(i)
(ii)
(iii)
This theorem is not valid for power relationship.
This theorem is not applicable to circuits containing only dependent sources. With dependent
sources, superposition can be used only when
the controlling functions are external to the network-containing sources, so that the controls are
unchanged when the sources act at a time.
This theorem is not applicable for circuits with
non-linear elements.
4. What is the use of superposition theorem?
The superposition theorem is used to find the current or voltage in a branch when the circuit has a
large number of independent voltage and/or current
sources.
5. Mention some examples where Thevenin’s theorem cannot be applied.
(i)
This theorem is inapplicable to loads which
are magnetically coupled to other parts of the
circuit.
219
Network Theorems
(ii)
(iii)
(iv)
This theorem is inapplicable for non-linear and
unilateral networks.
This theorem is inapplicable for active load.
To apply this theorem, the load should not contain any dependent source.
Power delivered to the load is
2
P = I RL =
E 2RL
( R + R L )2 + ( X + X L )2
Z = R + jX , Z L = R L + jX L
where,
6. Explain the use of Thevenin’s theorem.
Thevenin’s theorem is very useful for replacement of
a large portion of a network with a small equivalent
circuit. This theorem is used to find the current in a
particular passive element in a linear bilateral network.
This theorem is also useful for calculating the load
resistance in impedance-matching problems.
7. Show that Thevenin’s and Norton’s theorems are
dual to each other.
The Thevenin’s equivalent circuit with respect to two
terminals a–b is shown in Fig. 4.163 (a).
Zth
For maximum power,
Now,
−2( E )2 R L ( X L + X )
∂P
=
=0
∂X L [( R L + R )2 + ( X L + X )2 ]2
i.e., the reactance of the load impedance is of opposite
sign to the reactance of the source impedance.
X in the equation (2) P =
Putting XL
E 2RL
( R L + R )2
For maximum power,
∂P E ( R L + R ) − 2 E R L ( R L + R )
=
=0
∂R L
( R L + R )4
2
IN
ZL
ZL ZL
V
b
Fig. 4.163 (a)
Thevenin’s
equivalent circuit
E2
Pmax =
=
4RL
E
9. Explain the application and limitations of Millman’s
theorem.
Applications
(i)
RL
jXL
Let E be the voltage
source, (R
jX) the
Fig. 4.164
internal impedance of
the source and (RL jXL) the load impedance.
E
E
=
Z + Z L (R + R L ) + j ( X + X L )
RL
Thus, the efficiency of the circuit is 50%.
(ii)
jX
(E 2 )
2
This equation is identical with the KCL equation of
Norton’s equivalent circuit as shown in Fig. 4.163 (b).
Therefore, we conclude that Thevenin’s and Norton’s
theorems are dual to each other.
R
2
The maximum power transferred will be
Replacing the voltage by current, the impedances by conductances, the equation becomes
I N = V (Y N +Y L ) = VY N +VY L
8. Show that under the
condition of maximum power transfer,
the efficiency of a circuit is 50%.
2
E 2 ( R L + R ) − 2 E 2 R L = 0 or R L = R
Or,
Fig. 4.163 (b)
Norton’s equivalent
circuit
The KVL equation for the Thevenin’s equivalent
circuit can be written as V th = I ( Z th + Z L ) = IZ th + IZ L
E=
∂P
must be zero.
∂X L
From which, X L + X = 0 or X L = − X
a
Vth I
(2)
This theorem provides the equivalent circuits
which are either Thevenin or Norton equivalent
circuits.
This theorem is applicable only to independent voltage sources with their internal series
impedances connected directly in parallel, or
independent current sources with their internal series admittances connected directly in
series.
Limitations
(i)
(1)
(ii)
This theorem is not applicable to circuits where
impedances or dependent sources are present
between the independent sources.
This theorem is not useful for circuits with less
than two independent sources.
220
Network Analysis and Synthesis
10. Mention some salient features of Tellegen’s theorem.
(i)
(ii)
(iii)
(iv)
(v)
This theorem is applicable for any lumped
network having elements which are linear or
non-linear, active or passive, time-varying or
time-invariant.
This theorem is completely independent of the
nature of the elements and is only concerned
with the graph of the network.
This theorem is based on two Kirchhoff’s laws,
i.e., KVL and KCL.
This theorem implies that the power delivered
by independent sources of the network must be
equal to the sum of the power absorbed (dissipated or stored) in all other elements in the network.
If the network is in sinusoidally steady state (ac
circuits) then Tellegen’s theorem is given as
b
∑V I * = 0
k =1
k k
(vi)
where Vk are the phasor voltages, Ik are the
phasor currents and Ik* is the complex
conjugate of Ik.
If t1 and t2 refer to two different instants of
observations, it still follows from Tellegen’s theorem that
b
∑v (t ) ⋅ i (t ) = 0
k
k =1
(vii)
k
1
2
If N1 and N2 refer to two different circuits having
the same graph, with the same reference directions assigned to the branches in the two circuits
then by Tellegen’s theorem,
b
∑v ⋅ i
k =1
1k
b
2k
= 0 and ∑v 2 k ⋅ i 1k = 0
k =1
where, v1k and i1k are the voltages and currents
in N1 and v2k and i2k are the voltages
and currents in N2, all satisfying the
Kirchhoff’s laws.
Exercises
TELLEGEN’S THEOREM
I1
1. The circuit of Fig. 165 (a) is reconnected as of
Fig. 4.165 (b).
R3
R1
V1
V3
NETWORK
IL = 2 A
R2 VL= 2 V
R =1
V1 = 5e j 5°
Fig. 4.165 (a)
V 2 = 15e
I 2 = 8e j 10°
I 3 = 10e j 15°
At a frequency of 100 Hz, the readings are
R1
V 1 ' = 10e j 20°
R3
R2
I 1 ' = 2 j 25°
V 2 ' = 12e j 35 I 2 ' = 10e − j 10°
V 3 ' = 5e j 15°
Vs =3 V
V2
I 1 = 12e j 40°
− j 20°
V3 = ?
Is
I2
Fig. 4.166
Is = 1 A
Vs
I3
VL
I 3 ' = 14.93e j 68°
R =2
Fig. 4.165 (b)
If Vs
2 V and Is
1 A, find the voltage VL of
Fig. 4.165 (b). Use Tellegen’s Theorem.
[1V]
2. The following readings were taken at a frequency of 50
Hz in a linear RLC network shown in figure.
The reading of V3 was missed. Calculate V3 using
Tellegen’s Theorem.
[ 18e j 15° ]
Reciprocity Theorem
1. In the network shown in Fig. 4.167, verify the reciprocity
theorem using a voltage source and an ammeter. What
are the methods of verifying the Reciprocity Theorem? All
values are in ohms.
221
Network Theorems
2. Find the current
3
4
in the 6- resistor
and the source cur1
2
rent in Fig. 4.168 a.
Hence, determine
the current in the 3Fig. 4.167
resistor when an
emf of 72 V is added
in series with the 6- resistor as shown in Fig. 4.168 b.
[ 0.5 A, 6 A]
3
5
7
A
C
2.5
3
2
5
10 V
5
Fig. 4.172
C
7
3
24 V
24 V
2
2. If the 5- resistance increases to 6 , determine the
compensation source and find the current through the
20
6 resistance.
[1 V;
A]
23
B
B
A
change in current through the 3- resistor.
[4.74
23.23 V; 0.271 159.5 A]
D
D
2
16
6
Millman’s Theorem
16
6
1. Find the load current using Millman’s theorem. All
resistance values are in ohms.
[1.176 A]
72 V
(a)
Fig. 4.168
(b)
I
4
3. In this circuit, find the voltage V. Interchange the current source and resulting voltage V and show that the
reciprocity theorem is verified.
[9.28 21.8 (V)]
5 90 A
5
4
10
4V
2V
10V
Fig. 4.173
2
j2
j5
4
2. Using Millman’s theorem, find the current in the load
[1.06
58.46 (A)]
impedance, ZL (2 j4) .
V
5 0V
Fig. 4.169
1 0V
4. Two sets of measurements are made on a linear passive resistive network in Fig. 4.170 a and Fig. b. Find the
current through the 2- resistor.
1
1
ZL
5 0A
2A
5A
20 V
5
N
2
I
(a)
30 V
N
Fig. 4.174
3. Determine the current through the branch AB using
⎡ 36 ⎤
Millman’s theorem.
⎢ A⎥
⎣ 67 ⎦
(b)
Fig. 4.170
A
Compensation Theorem
1. The 5- resistor has
been changed to an
8- resistor in the circuit. Determine the
compensation source
VC and calculate the
4
4
4
5
5
2
10 0 V
j5
j4
2V
4V
6V
B
Fig. 4.171
Fig. 4.175
222
Network Analysis and Synthesis
Thevenin’s and Norton’s Theorems
1. Determine the Thevenin equivalent circuit with
respect to the terminals A and B for the circuit shown
in Fig. 4.176 and hence the current flowing through
10- resistor.
[0.193 A]
50
4V
2
30
50
B
10 V
3
3
10
1
Fig. 4.181
6. Find the Thevenin equivalent circuit with respect to
the terminals A and B.
Fig. 4.176
[a) Vth
2. Find the Thevenin equivalent circuit for the following
networks:
(i)
10
10
10 V
10
A
5. Use Thevenin’s theorem to find the current supplied
by the battery.
[Rth 33.34 ; Vth 10 V; i 0.3 A]
c) Vth
d) Vth
3k
2k
(a)
V x /4000 V x
4V
50 mA
5.59
v1
200
Fig. 4.182
A
4
(b)
1
A
ix
10 ix
Fig. 4.178
2
[(i) 8 V; 10 k (ii) 25V; 350 ]
3. Determine the current in the branch AB for the circuit shown in Fig. 4.179 by using Thevenin’s theorem.
[1.818 A]
B
Fig. 4.183
(c)
5
j10
A
A
5
100 V
10
26.6 ( );
5 36.87 ( )]
0.1v1
B
15
1 ;
B
10
ix
1k
1k
10 0 (V); Zth
0; Rth
50
A
Fig. 4.177
500ix
10.64 ; b) Vth
11.18 93.44 (V); Zth
100
(ii)
0; Rth
20 90 (V)
10 A
3
j4
B
B
Fig. 4.184
Fig. 4.179
4. Find Norton’s equivalent at terminals a–b.
[0 A; 10.64 ]
(d)
10
5 30 (A)
50
a
v1
200
100
b
Fig. 4.180
A
5
5
j5
j5
B
0.1v1
Fig. 4.185
7. Compute I0 using Norton’s theorem.
[I0 0.542 cos(2t
77.47 ) (A)]
223
Network Theorems
2
i1
c)
1/4 F
4H
I0
20
1/2 F
40
R
Maximum Power Transfer Theorem
1. Determine the value of
the resistor RL that will
draw maximum power
from the rest of the circuit. What is the maximum power?
[4.22 , 2.901 W]
2
4
VX
1
9V
3 VX
Fig. 4.191
4. Determine ZL so that the maximum power is absorbed
by it.
[40 0 V; (8 j20) , 50 W]
VR
Fig. 4.192
5. Determine the value of R such that the 6consumes the maximum power.
[R
1k
R
40
60
100 V
30
3
Fig. 4.193
1. Apply the superposition theorem to the circuit to
[ 0.75 A]
find I3.
30
30
I3
50
8A
60
100 V
60 V
2. Use the superposition theorem to find the voltage Vx.
[12.5 V]
2
20
1
Vx
R
10 V
Fig. 4.190
6
Fig. 4.194
1
10 V
3
3
ZL
Fig. 4.189
b)
18 ]
Superposition Theorem
50 V
60 V
resistor
R
10V
3. Find the value of the resistance R for maximum power
to be transferred to it. Also, find the maximum power.
[a) 44 , 0.568 W; b) 4.5 , 1.39 W; c) 16 ]
60
j 20
Fig. 4.187
Fig. 4.188
a)
j10
40
50 0 V
2 VR 0.5 H
1 F
ZL
10
RL
2. The circuit operates in
the sinusoidal steady state with ␻ 1000 rad/s and Is
I 0 A(rms). Find the value of the load impedance for
maximum average power transfer. Also, find the average power absorbed by the load under this condition.
[(1500 j1000) V; 83.33 W]
IS
50 V
10i1
Fig. 4.186
Fig. 4.195
2A
4
0.1Vx
224
Network Analysis and Synthesis
3. Determine the voltage vx in the circuit using the superposition theorem.
[ 38.5 V]
2
ix 5
[5 A]
1
4
30 A
20 V
VX
50 V
6. Find the current ix by the superposition theorem.
0.1VX
4i x
100 V
Fig. 4.199
Fig. 4.196
4. Use the superposition theorem to find the voltage vx.
[5 2.56 sin(500t 39.8 ) (V)]
5
7. Find v0 using the superposition theorem.
1 2.498 cos(2t 30.79 )
[v0
2.328 sin (5t 10 ) V]
1
vo
2H
4
5V
1
20sin 50t (V)
2 mF
vx
6V
10cos2t (V)
0.1 F
2sin 5t (A)
Fig. 4.200
Fig. 4.197
5. Find the current through the capacitor using the
superposition theorem.
[4.86 80.8 (A)]
j1
5
20 0 (V)
8. Find the current i0 using
the superposition theorem.
[ 0.4706 A]
3
4A
5
2
1
i0
5i0
20V 4
j5
Fig. 4.201
10 0 (V)
Fig. 4.198
Questions
1. State and explain the substitution theorem.
2. State and explain the superposition theorem. Give
a proof for a general n-mesh network indicating the
conditions under which it is applicable.
3. State the reciprocity theorem as applied to a network
and give a proof of the same for a general network.
Mention two networks where this theorem is not
applicable.
4. State Thevenin’s theorem and give a proof of the
same. Mention one example of a network where this
theorem is not applicable.
5. a) State Norton’s theorem as applied to a network and
give a proof of the same.
b) What is ‘dual network’? Mention the procedure for
drawing the dual of a given network.
6. State and explain clearly Thevenin’s theorem as
applied in ac circuits.
7. State and explain Thevenin’s theorem, and specify
the types of networks to which it is applicable. Also,
state the theorem which is the dual of the above
theorem.
8. State the maximum power transfer theorem for all the
various kinds of networks and loads.
9. State the maximum power transfer theorem. Derive
conditions for maximum power transfer for a resistive
network and resistive load.
10. State and prove the maximum power transfer
theorem.
Or,
In the circuit, the
source emf ES, resisRs
RL
jX s
tance RS and reactance
j XL
ES
jXS are fixed but both
the load resistance RL
and reactance jXL are
variable. Show that
Fig. 4.202
maximum power is
XS and RL RS.
consumed in the load when XL
225
Network Theorems
Prove that the load impedance which absorbs the
maximum power from a source is the conjugate of the
impedance of the source.
14. Derive the condition for maximum power transfer for
11. Prove the condition for maximum power transfer for
an ac circuit.
(b) Load impedance with variable resistance and fixed
reactance
12. A source with internal impedance RS
jXS delivers
j0. Show
power to a variable load impedance RL
that the condition for maximum power in the load is
15. State and clearly prove with the help of a suitable
example the maximum power transfer theorem as
applicable to RLC circuits excited from the sinusoidal
energy source. Hence explain clearly the concept and
its significance in impedance matching.
RL 2 = RS 2 + X S 2 .
13. State the maximum power transfer theorem and
verify that only 50% of the total power supplied by the
source can be transferred to the load.
(a) Load impedance with variable resistance and variable reactance
16. State and prove the following theorems:
( i) Tellegen’s theorem
Or,
(ii) Millman’ theorems
State and explain the maximum power transfer theorem. Derive the expression for efficiency for maximum
power transfer.
(iii) Compensation theorem
Multiple-Choice Questions
1. Which one of the following theorems is a manifestation of the law of conservation of energy?
(i) Tellegen’s Theorem
(ii) Reciprocity Theorem
(iii) Thevenin’s Theorem
(iv) Norton’s Theorem
2. Tellegen’s theorem is applicable to
(i) circuits having passive elements
(ii) circuits having time-invariant elements only
(iii) circuits with linear elements only
(iv) circuits with active or passive, linear or non-linear
and time-invariant or time-varying elements
7.
8.
9.
3. In any lumped network with elements in b branches,
b
∑
k =1
k
(t ) ⋅ i k (t ) = 0 , for all t, holds good according to
(i) Norton’s theorem
(ii) Thevenin’s theorem
(iii) Millman’s theorem (iv) Tellegen’s theorem
4. Millman’s theorem yields
(i) equivalent voltage source
(ii) equivalent voltage or current source
(iii) equivalent resistance
(iv) equivalent impedance
5. The superposition theorem is applicable to
(i) current only
(ii) voltage only
(iii) both current and voltage
(iv) current, voltage and power
6. Superposition theorem is not applicable for
10.
11.
(i) voltage calculations (ii) bilateral elements
(iii) power calculations (iv) passive elements
Thevenin’s theorem can be applied to calculate the
current in
(i) any load
(ii) a passive load only
(iii) a linear load only
(iv) a bilateral load only
Norton’s equivalent circuit consists of a
(i) voltage source in parallel with impedance
(ii) voltage source in series with impedance
(iii) current source in parallel with impedance
(iv) current source in series with impedance
The superposition theorem is applicable to
(i) linear responses only
(ii) linear and non-linear responses
(iii) linear, non-linear and time-variant responses
When a source is delivering maximum power to a
load, the efficiency of the circuit
(i) is always 50%
(ii) depends on the circuit parameters
(iii) is always 75%
(iv) none of these.
Maximum power transfer occurs at a
(i) 100% efficiency
(ii) 50% efficiency
(iii) 25% efficiency
(iv) 75% efficiency
12. Which of the following statements is true?
(i) A Norton’s equivalent is a series circuit.
(ii) A Thevenin’s equivalent circuit is a parallel circuit.
(iii) R-L circuit is a dual pair.
(iv) L-C circuit is a dual pair.
226
Network Analysis and Synthesis
13. For a linear network containing generators and impedances, the ratio of the voltage to the current produced
in the other loop is the same as the ratio of voltage
and current obtained if the position of the voltage
source and the ammeter measuring the current are
interchanged. This network theorem is known as
(i) Millman’s theorem (ii) Norton’s theorem
(iii) Tellegen’s theorem (iv) Reciprocity theorem
14. Under conditions of maximum power transfer from
an ac source to a variable load,
(i) the load impedance must also be inductive, if
the generator impedance is inductive
(ii) the sum of the source and load impedance is zero
(iii) the sum of the source reactance and load reactance is zero
(iv) the load impedance has the same phase angle
as the generator impedance
2. XL
(
Of these statements,
(i) 1 and 2 are correct
(ii) 1, 3 and 4 are correct
(iii) 2 and 4 are correct
(iv) 1, 2, 3 and 4 are correct
16. In a linear network, the ratio of voltage excitation
to current response is unaltered when the position
of excitation and response are interchanged. This
assumption stems from the
(i) principle of duality
(ii) reciprocity theorem
(iii) principle of superposition
(iv) equivalence theorem
17. If all the elements in a particular network are linear
then the superposition theorem holds when the excitation is
(i) dc only
(ii) ac only
(iii) either ac or dc (iv) an impulse.
18. An ac source of voltage Es and an internal impedance
ZS (RS jXS) is connected to a load of impedance
ZL (RL jXL). Consider the following conditions in
this regard:
1. XL XS , if only XL is varied
)
2
3. R L = R S 2 + X S + X L , if only RL is varied.
19.
15. Consider the following statements:
The transfer impedances and admittances of a network remain constant when the position of excitation
and response are interchanged if the network
1. is linear
2. consists of bilateral elements
3. has high impedance or admittance as the case
may be
4. is resonant
XS , if only XS is varied
20.
21.
22.
23.
24.
4. Z L = Z S if the magnitude of ZL is varied, keeping
the phase angle fixed.
Among these conditions, those which are to be satisfied for maximum power transfer from the source to
the load would include
(i) 2 and 3
(ii) 1 and 3
(iii) 1, 2 and 4
(iv) 2, 3 and 4
The reciprocity theorem is applicable to a network
1. which contains R, L and C as elements
2. which is initially relaxed system
3. which has both independent and dependent
sources
Tick out the correct combination:
(i) 1 and 2
(ii) 1 and 3
(iii) 2 and 3
(iv) 1, 2 and 3.
The reciprocity theorem is applicable to
(i) circuits with one independent source
(ii) circuits with only one independent source and
no dependent source
(iii) circuits with any number of independent
sources
(iv) circuits with any number of sources
The substitution theorem is applicable for a network
which has
1. a unique solution
2. one or two non-linear elements
3. one non-linear or time-varying element
Choose the correct combination:
(i) 1 and 2
(ii) 1 and 3
(iii) 2 and 3
(iv) 1, 2 and 3.
The substitution theorem applies to
(i) linear networks
(ii) non-linear networks
(iii) linear time-invariant networks
(iv) any network
Which of the following theorems is applicable for
both linear and non-linear circuits?
(i) Superposition (ii) Thevenin
(iii) Norton
(iv) None of these
A network is composed of two sub-networks N1 and
N2 as shown in Fig. 4.203.
Sub-network
Sub-network
N1
N2
Fig. 4.203
227
Network Theorems
25.
26.
27.
28.
If the sub-network N1 contains only linear, bilateral,
time-invariant elements then it can be replaced by
its Thevenin equivalent even if the sub-network N2
contains
(i) a two-terminal element which is non-linear
(ii) a non-linear inductance mutually coupled to an
element in N1
(iii) an element which is linear, but mutually coupled
to some element in N1
(iv) a dependent source, the value of which
depends upon the voltage or current in some
element in N1
A certain network consists of two ideal identical voltage sources and a large number of ideal resistors. The
power consumed in one of the resistors is 4 W when
either of the two sources is active and the other is
replaced by a short-circuit. The power consumed by
the same resistor when both the sources are active
would be
(i) zero or 16 W
(ii) 4 W or 8 W
(iii) zero or 8 W
(iv) 8 W or 16W
If a network has all linear elements except for a few
non-linear ones then superposition theorem:
(i) cannot hold at all
(ii) always holds
(iii) may hold on careful selection of element values,
source waveform and response
(iv) holds in case of direct current excitations
The maximum power
that can be dissipated
1
3
in the load in the circuit
RL
shown in Fig. 4.204 is
9V
6
(i) 3 W
(ii) 6 W
(iii) 6.75 W (iv) 13.5 W Fig. 4.204
If Rg in the circuit shown in Fig. 4.205 is variable
between 20 and 80 then the maximum power
transferred to the load RL will be
(i) 15 W
(ii) 13.33 W
(iii) 6.67 W
(iv) 2.4 W
RL = 60
Fig. 4.205
29. Thevenin impedance across the terminals AB of the
given network is
(i) 10
3
(ii) 20
9
3
1A
1V
2
2
Fig. 4.206
30. The V-I relation for the network shown in the given
box is V 4I 9.
If now a resistor R
2 is connected across it,
then the value of I will be
(i)
4.5 A
(ii)
1.5 A
(iii) 1.5 A
(iv) 4.5 A
I
N
R=2
V
Fig. 4.207
i/4
31. In the network shown
i
in Fig. 4.208, the effective resistance faced by
the voltage source is
4
V
(i) 4
(ii) 3
(iii) 2
(iv) 1
32. For the network Fig. 4.208
shown in Fig. 4.209, if
5 A and if Vs 0, then I
Vs V1 and V 0 then I
½ A. The values of Isc and R1 of the Norton’s equivalent
across AB would be respectively
(i)
5 A and 2
(ii) 10 A and 0.5
(iii) 5 A and 2
(iv) 2.5 A and 5
A
Vs
Resistive
circuit
I
V
B
Fig. 4.209
Rg
40 V
A
2
(iii) 13
4
(iv)
11
5
33. In the network shown
in Fig. 4.210, the
Thevenin source and
the impedance across
terminals A–B will be
respectively
(i) 15 V and 13.33
(ii) 50 V and 15
(iii) 115 V and 20
(iv) 100 V and 25
10
15V
5
10
10A
Fig. 4.210
A
B
228
Network Analysis and Synthesis
34. Which one of the fol1k
lowing combination
of open-circuit voltage
I1
99I1
and Thevenin’s equiva- 1 V
lent resistance represents the Thevenin’s
equivalent of the circuit Fig. 4.211
shown in Fig. 4.211?
(i) 1 V, 10
(ii) 1 V, 1k
(iii) 1m V, 1k
(iv) 1m V, 10
a
b
35. For the circuit shown in Fig. 4.212, the current
through R, when VA 0 and VB 15 V is I amperes.
Now, if both VA and VB are increased by 15 V then the
current through R will be
3
R
(i)
(ii)
(iii)
(iv)
10 V in series with the 1.2- resistance
6 V in series with the 1.2- resistance
10 V in series with the 5- resistance
6 V in series with the 5- resistance
38. A dc current source
is connected as shown
in Fig. 4.215:
The Thevenin’s equivalent of the network at
terminals a b
(i) will be
4V
I
(iii) 3I amperes
(iv)
I
36. Thevenin’s equivalent circuit of the
network shown in
Fig. 4.213, between
terminals T1 and
T2 is
a
2
2
3
39. Which one of the following impedance values of the
load will cause maximum power to be transferred to
the load for the network shown in Fig. 4.216?
amperes
amperes
15 V
(iv) T1
20 V
j2
6
I1
0.8I1
40 V
T1
Vs
T2
Fig. 4.216
10
24
T2
16
T2
16
T2
24
T2
A
Net work
(ii) (2
(iv) 2
40. The Thevenin’s equivalent resistance Rth for
the given network is
(i) 1
(ii) 2
(iii) 4
(iv) infinity
A
Net work
zL
j2
(i) (2 j2)
(iii)
j2
37. The Thevenin equivalent of the network shown in
Fig. 4.214 (a) is 10 V in series with a resistance of 2
. If now, resistance of 3 is connected across AB in
Fig. 4.214 (b), the Thevenin equivalent of the modified network across AB will be
3
B
B
Fig. 4.214 (a)
j2
2
Fig. 4.213
(iii) T1
b
b
(ii)
40 V
a
(iv) is NOT feasible
VB
3
(i) I amperes
(ii) T1
(ii) will be
4V
Fig. 4.212
40 V
b
Fig. 4.215
b
2V
(i) T1
2A
2
3
3
a
a
(iii) will be
VA
2
Fig. 4.214 (b)
j2)
2
2
1A
Rth
1V
41. The Norton’s equiva- Fig. 4.217
lent of circuit shown in
Fig. 4.218 (a) is drawn in the circuit shown in
Fig. 4.218 (b). The values of Isc and Req in Fig. 4.218 (b)
are respectively
3
2V
4/5
1
1
Req
Isc
2
(a)
Fig. 4.218 (a)
2
(b)
Fig. 4.218 (b)
2
229
Network Theorems
(i)
(iii)
5
A and 2
2
(ii)
4
12
A and
V
5
5
(iv)
42. For the circuit shown
in Fig. 4.219, the current flowing through
the 1resistor is
adjusted to zero by
varying the value of R.
What is the value of R?
(i) 2
(ii) 3
(iii) 4
(iv) 6
Thevenin’s equivalent of the
network shown in Fig. 4.222
(a) would correspond to the
network shown in Fig. 4.222
(b), if one or more of the following conditions are met:
2
A and 1V
5
2
A and 2
5
4
6
1. IL
R
2
10
10V
Fig. 4.219
A
1
1
1
8A
B
Fig. 4.220
(i)
(iii) 16 V, 2 Ω
3
3
(iv) 16 V , 3 Ω
3
44. If Thevenin’s equivalent resistance of the circuit
shown in Fig. 4.221 seen from the open terminals is
2 then the value of ‘R’ will be
Fig. 4.222 (b)
IL
3. IL
2IL, if the voltage Vth is doubled.
The correct set of conditions would include
(i) 1, 2 and 3
(ii) 1 and 2
(iii) 2 and 3
(iv) 1 and 3
46. In a given network of generators and impedances, the
impedance ‘Z’ of the branch with current ‘I’ flowing
through it, has increased from Z to Z
Z. The solution of the network will remain the same if
(i) an emf of I Z is introduced in series with the
branch
(ii) the impedance of all the other branches are
reduced by the same amount
(iii) the voltages of the generators in all the other
branches are increased proportionately
(iv) a negative resistance device is introduced in the
network
10
60 , Z2
10 60 , Z3
48. In Fig. 4.223, Z1
50 53.13 . Thevenin impedance seen from X–Y is
X
Z1
100 0
R
2
RL
47. Thevenin’s theorem is not applicable for circuits with
(i) passive load
(ii) active load
(iii) bilateral load
(iv) none of these
(ii) 4 V, 3 Ω
2
12 V, 3 Ω
2
IL
Vth
2. The equivalence is valid only if the frequency of Vth
is maintained at 50 Hz.
1
43. What is the Thevenin’s equivalent between A and B
for the circuit shown in Fig. 4.220?
4V
Zth
Z2
Z3
Y
1A
5V
2
Fig. 4.223
(i) 56 45
(iii) 70/300
Fig. 4.220
(i) 4
(ii) 2
(iii) 1
(iv) zero.
45.
L1
240 V,
50 Hz
Fig. 4.222 (a)
49. Two ac sources feed a common variable resistive load
as shown in Fig. 4.224 . Under the maximum power
transfer condition, the power absorbed by the load
resistance RL is
R2
R1
C1
L2
C2
(ii) 60/300
(iv) 34.4/65
IL
RL
6
V
110 0
Fig. 4.224
j8
6
RL
j8
90 0
230
Network Analysis and Synthesis
(i) 2200 W
(iii) 1000 W
(ii) 1250 W
(iv) 625 W
52. In Fig. 4.227 the current source is 1 0 A, R 1 ,
the impedances are ZC
j , and ZL
2j . The
Thevenin equivalent looking into the circuit across
X–Y is
50. In Fig. 4.225, the value of R is
R
14
X
1
5A
10A
2
100 V
40V
Y
Fig. 4.225
(i) 10
(iii) 24
Fig. 4.227
(ii) 18
(iv) 12
(i)
2 0 V, (1 2j )
(ii) 2 45 V, (1 2j )
(iii) 2 45 V, (1 j )
51. In Fig. 4.226, the Thevenin’s equivalent pair (voltage,
impedance), as seen at the terminals P–Q, is given by
(iv)
10
P
20
Unknown
network
4 V 10
Q
Fig. 4.226
(i) (2 V, 5 )
(iii) (4 V, 5 )
2
45 V, (1
j)
53. A source of angular frequency of 1 rad/s has a source
impedance consisting of a 1- resistance in series
with 1-H inductance. The load that will obtain the
maximum power transfer is
(i) 1 resistance
(ii) 1- resistance in parallel with 1-H inductance
(iii) 1- resistance in series with 1-F capacitor
(iv) 1- resistance in parallel with 1-F capacitor
(ii) (2 V, 7.5 )
(iv) (4 V, 7.5 )
Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
(i)
(iv)
(iv)
(ii)
(iii)
(iii)
(i)
(iii)
(i)
(i)
(ii)
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
(iv)
(iv)
(iii)
(i)
(ii)
(iii)
(iv)
(i)
(ii)
(ii)
(iv)
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
(iv)
(i)
(i)
(i)
(i)
(iii)
(iv)
(ii)
(iv)
(iii)
(iii)
34.
35.
36.
37.
38.
39.
40.
41.
42.
43.
44.
(ii)
(i)
(i)
(ii)
(iv)
(iv)
(ii)
(iv)
(ii)
(iii)
(iii)
45.
46.
47.
48.
49.
50.
51.
52.
53.
(i)
(i)
(ii)
(i)
(iv)
(iv)
(i)
(iv)
(iii)
5
Laplace Transform
and Its Applications
Introduction
Classical methods of solving differential equations become quite cumbersome when used for networks
involving higher order differential equations. In such cases, the Laplace transform method is used.
The classical methods consist of three steps:
(i) determination of complementary function,
(ii) determination of particular integral, and
iii) determination of arbitrary constants.
But, these methods become difficult for the equations containing derivatives; and transform methods
prove to be superior.
The Laplace transform is an integral that transforms a time function into a new function of a
complex variable. The term Laplace comes from the name of the French mathematician Pierre Simon
Laplace (1749–1827). The transformation method is a very effective tool for solving integro-differential
equations.
Laplace transformation is also a very powerful tool for network analysis. Any linear circuit consisting
of linear circuit elements can be solved by the knowledge of Laplace transformation.
In this chapter, we will first discuss the basics of Laplace transformation and then apply this transform
method to study the transient behaviour of electric circuits.
5.1
ADVANTAGES OF LAPLACE-TRANSFORM METHOD
Laplace-transform methods offer the following advantages over the classical methods:
1. It gives complete solution.
2. Initial conditions are automatically considered in the transformed equations.
3. Much less time is involved in solving differential equations.
4. It gives systematic and routine solutions for differential equations.
232
Network Analysis and Synthesis
5.2
DEFINITION OF LAPLACE TRANSFORM
Let f (t ) be a function of time which is zero for t 0 and which is arbitrarily defined for t 0 subject to some
mild conditions. Then the Laplace transform of the function f (t ), denoted by F (s ) is defined as
∞
L ⎡⎣ f (t ) ⎤⎦ = F ( s ) = ∫ f (t )e − st dt
0_
Thus, the operator L[ ] transforms f (t ), which is in the time domain, into F (s), which is in the complex frequency
domain, or simply the s-domain, where,
s Complex frequency (␴ j␻)
where, ␴ real part of s neper frequency
␻ Imaginary part of s radian frequency
Note
The lower limit of the integration should be 0 instead of 0 or simply 0. If f (t ) is continuous at t 0,
then the value of f (0) is well-defined. But, if f (t ) is not continuous at t 0 then the meaning of f (0) becomes
ambiguous. To consider the effect of ‘instantaneous energy transfer’ we must use 0 as the lower limit to
include the impulses at t 0. The use of 0 will exclude the existence of any impulses at the origin.
So, we use 0− as the lower limit.
5.3
CONCEPT OF COMPLEX FREQUENCY
The complex frequency (s) is the sum of two frequencies the real and imaginary.
s Complex frequency (␴ j␻)
where, ␴ real part of s neper frequency
␻ imaginary part of s radian frequency
The general solution of the differential equation in time-domain is
i (t) I0est, where s (␴ j␻)
Since est is a dimensionless quantity and so, also, the product ‘st’ is a dimensionless quantity, the unit of ‘s’ must
be (time) 1 or Hz.
Here, is interpreted as radian frequency; as a radian is a ratio of two lengths, ‘␻’ is effectively (time) 1, i.e.,
frequency in Hz.
Also, as ␴ and ␻ must have the same dimension, i.e., the dimension of should be (time) 1. Also, with ␻ 0,
1 ⎡ i (t ) ⎤
I0e␴t ⇒ σ = ln ⎢
⎥
t ⎣ I0 ⎦
Since the unit of ln of some number is neper, the unit of ␴ is neper per second.
i(t)
Physical Significance of Complex Frequency
We have,
i(t)
I0est
I0e(␴
j␻)t
I0e␴t [cos ␻t
j sin ␻t]
If ␴ 0 then the variation of the real and imaginary parts
of the function is shown in Fig. 5.1.
vt
sin vt
vt
cos vt
Fig. 5.1 Variation of real and imaginary parts
with ␴ 0
233
Laplace Transform and Its Applications
0 then the variation of the real and imaginary parts of the function is shown in Fig. 5.2.
Damped Sinusoidal
1
0.8
0.6
0.4
0.2
0
0.2
0.4
Damped Cosinusoidal
1
0.5
f(t)
f(t)
If ␴
0
0.5
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (second)
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (second)
Fig. 5.2 Variation of real and imaginary parts with ␴
0 then the variation of the real and imaginary parts of the function is shown in Fig. 5.3.
8
6
4
2
0
2
4
6
8
Damped Sinusoidal
f(t)
f(t)
If ␴
0
0 1 2 3 4 5 6 7 8 9 10
Time (second)
8
6
4
2
0
2
4
6
8
Damped Cosinusoidal
0 1 2 3 4 5 6 7 8 9 10
Time (second)
Fig. 5.3 Variation of real and imaginary parts with ␴
0
From these figures, it is clear that
␻ decides the number of oscillations per second
␴ decides the magnitude of these oscillations
5.4
BASIC THEOREMS OF LAPLACE TRANSFORM
Linearity Theorem If Laplace transform of the functions f1(t) and f2(t) are F1(s) and F2(s) respectively then
Laplace transform of the functions [K1 f1(t) K2 f2(t)] will be [K1 F1(s) K2 F2(s)].
L [K1 f1(t)
K2 f2(t)]
[K1 F1(s)
K2 F2(s)]
where, K1 and K2 are constants.
Scaling Theorem
If Laplace transform of f (t ) is F (s ) then
L [ f (Kt )]
∞
1
s
F ( ) , where
K K
∞
− st
∫ f ( Kt )e dt = ∫ f ( x )e
Proof L [ f (Kt )]
0_
K
)
0_
∞
=
− x( s
− x( s )
1
1
s
f ( x )e K dx = F ( )
∫
K 0_
K K
dx
K
K is a constant and K
Taking, Kt
x, dx
0.
Kdt
234
Network Analysis and Synthesis
Time Differentiation Theorem
If Laplace transform of f (t ) is F (s ) then,
L[
df (t )
] = sF ( s ) − f (0− )
dt
∞
∞
Proof
L[
df (t )
df (t ) − st
]= ∫
e dt = [ e − st f (t )]0∞_ + s ∫ f (t )e − st dt , by integration by parts
dt
dt
0_
0_
sF (s ) – f (0_)
In general, for n-th order differentiation,
L[
df n (t )
] = s n F s − s n−1 f (0 _) − s n− 2 f ′ (0 _) − ⋅⋅⋅ − s f n− 2 (0 _) − f n−1 (0 _)
dt n
()
Frequency Differentiation Theorem
If Laplace transform of f (t ) is F (s ) then,
L ⎡⎣tf (t ) ⎤⎦ = −
dF ( s )
ds
∞
F ( s ) = ∫ f (t )e − st dt
Proof
0_
Taking derivative with respect to s,
∞
∞
)
(
(
)
L[ ∫ f (t )dt ] =
F (s)
s
dF ( s )
= ∫ f (t ) −te − st dt = ∫ −tf (t ) e − s t dt = L ⎡⎣ −tf (t ) ⎤⎦
ds
0
0
−
−
Time Integration Theorem
If Laplace transform of f (t ) is F (s ) then,
t
0
Proof
∞
∞
⎡⎛ t
⎞ e − st ⎤
1
1
1
L[ ∫ f (t )dt ] = ∫ {[ ∫ f (t )dt ]e }dt = ⎢⎜ ∫ f (t )dt ⎟
⎥ + ∫ f (t )e − st dt = 0 + F ( s ) = F ( s )
−
s
s
s
s
⎠
⎢⎣⎝ 0
⎥⎦0 _
0
0_
0
0_
t
∞
t
− st
In general, for nth order integration,
t1 t2
tn
0 0
0
L[ ∫ ∫ ⋅⋅⋅∫ f (t )dt1dt2 ⋅⋅⋅ dtn ] =
F (s)
sn
Shifting Theorem The shifting may be done with respect to time or frequency.
Time Shifting Theorem
If Laplace transform of f (t ) is F (s ), then
L[ f (t
a)]
e
as
F (s )
235
Laplace Transform and Its Applications
Proof Let, (t a) x,
dt dx
As, t → a, x → 0 and as t → , x →
and t
(x
a)
∞
∞
∞
∞
0−
a
0−
0−
∴ L[ f (t − a )] = ∫ f (t − a )e − st dt = ∫ f (t − a )e − st dt = ∫ f ( x )e − s ( x + a )t dt = e − as ∫ f ( x )e − sx dx =e − as F ( s )
Frequency Shifting Theorem
If Laplace transform of f (t ) is F (s ), then
L[e at f (t)]
F (s
a)
∞
∞
0−
0−
L ⎡⎣ e − at f (t ) ⎤⎦ = ∫ e − at f (t )e − st dt = ∫ f (t )e − ( s + a )t dt = F ( s + a )
Proof
Initial Value Theorem
If the Laplace transform of f (t ) is F (s ) and the first derivative of f (t ) is Laplace transformable, then, the initial
value of f (t ) is
f (0+ ) = Lt f (t ) = Lt [ sF ( s )]
t →0
s →∞
⎡d
⎤ ∞ ⎡ df (t ) ⎤ − st
L ⎢ f (t ) ⎥ = ∫ ⎢
⎥ e dt
⎣ dt
⎦ 0− ⎣ dt ⎦
Proof
∞
⎡ df (t ) ⎤ − st
sF ( s ) − f (0− ) = ∫ ⎢
e dt
dt ⎥⎦
0− ⎣
Or,
Taking limit s
[by time differentiation theorem]
,
∞
⎡ df (t ) ⎤ − st
Lt ⎡ sF ( s ) − f (0− ) ⎤⎦ = Lt ∫ ⎢
e dt
s →∞ ⎣
s →∞
dt ⎥⎦
0− ⎣
or,
∞
⎡ 0+ df (t )
df (t ) ⎤
dt ⎥
Lt ⎡⎣ sF ( s ) ⎤⎦ − f (0− ) = Lt ⎢ ∫ e 0
dt + ∫ e − st
s →∞
s →∞
dt
dt
⎢⎣ 0−
⎥⎦
0+
or,
⎡ 0+ df (t ) ⎤
dt ⎥
Lt ⎡⎣ sF ( s ) ⎤⎦ − f (0− ) = Lt ⎢ ∫ e 0
s →∞
s →∞
dt
⎢⎣ 0−
⎥⎦
[as s is not a function of time t]
0+
or,
Lt ⎡⎣ sF ( s ) ⎤⎦ − f (0− ) = Lt ∫ df (t ) = f (0+ ) − f (0− )
s →∞
or,
s →∞
0−
f (0+ ) = Lt ⎡⎣ sF ( s ) ⎤⎦
s →∞
Final Value Theorem
If a function f (t ) and its derivatives are Laplace transformable, then the final value of f (t ) is,
f (∞) = Lt f (t ) = Lt ⎡⎣ sF ( s ) ⎤⎦
t →∞
s →0
236
Network Analysis and Synthesis
∞
⎡d
⎤
⎡ df (t ) ⎤ − st
L ⎢ f (t ) ⎥ = ∫ ⎢
⎥ e dt
⎣ dt
⎦ 0− ⎣ dt ⎦
Proof
∞
⎡ df (t ) ⎤ − st
sF ( s ) − f (0− ) = ∫ ⎢
e dt
dt ⎥⎦
0− ⎣
Or,
[by time differentiation theorem]
Taking limit s → 0,
∞
t
∞
⎛ df (t ) ⎞
⎡ df (t ) ⎤ − st
⎡ df (t ) ⎤
Lt ⎡⎣ sF ( s ) − f (0− ) ⎤⎦ = Lt ∫ ⎢
e
dt
=
dt
Lt
dt
=
⎥
⎢ dt ⎥
∫
s →0
s →0
t →∞ ∫ ⎜
dt
dt ⎟⎠
⎦
⎦
0− ⎣
0− ⎝
0− ⎣
Lt ⎡⎣ sF ( s ) − f (0− ) ⎤⎦ = Lt ⎡⎣ f (t ) − f (0− ) ⎤⎦
or,
s →0
t →∞
Lt ⎡⎣ sF ( s ) ⎤⎦ − f (0− ) = Lt ⎡⎣ f (t ) ⎤⎦ − f (0− )
or,
s →0
t →∞
Lt ⎡⎣ f (t ) ⎤⎦ = Lt ⎡⎣ sF ( s ) ⎤⎦
or,
t →∞
s →0
This theorem is only applicable if the value of the function f (t ) is finite as t becomes infinity, i.e., F (s ) has all
poles lying in the left half of the s-plane or at most one simple pole at the origin.
Example 5.1 Find the initial and final value of the functions given as
(
)
4 s +1
( ) s + 4s + 6
(a) F s =
2
5s − 1600
( ) s s + 18
s + 90 s + 800
3
(b) F s =
(
3
2
)
Solution a) By initial-value theorem, the initial value of the function is given as
⎡
4 ⎤
⎢ 4+
⎥
⎡
4 s +1 ⎤
s ⎥=4
f 0 + = Lim ⎡⎣ sF s ⎤⎦ = Lim ⎢ s × 2
⎥ = Lim ⎢
s →∞
s →∞ ⎢
⎣ s + 4 s + 6 ⎥⎦ s→∞ ⎢ 1 + 4 + 62 ⎥
⎢⎣ s s ⎥⎦
( )
(
()
)
By final-value theorem, the final value of the function is given as
(
)
⎡
4 s +1 ⎤
f ∞ = Lim ⎡⎣ sF s ⎤⎦ = Lim ⎢ s × 2
⎥=0
s →0
s →0 ⎢
⎣ s + 4 s + 6 ⎥⎦
( )
()
(b) By initial-value theorem, the initial value of the function is given as
⎡
⎤
⎡
⎤
5s 3 − 1600
5s 3 − 1600
⎥ = Lim ⎢
⎥
f 0 + = Lim ⎡⎣ sF s ⎤⎦ = Lim ⎢ s ×
2
3
2
3
s →∞
s →∞ ⎢
s →∞ ⎢ s + 18 s + 90 s + 800 ⎥
⎥
⎣ s s + 18 s + 90 s + 800 ⎦
⎣
⎦
( )
()
(
⎡
⎤
1600
5− 3
⎢
⎥
s
= Lim ⎢
⎥=5
18 90 800
s →∞ ⎢
1+ + 2 + 3 ⎥
⎢⎣
s s
s ⎥⎦
)
(
)
237
Laplace Transform and Its Applications
By final-value theorem, the final value of the function is given as
⎡
⎡
⎤
⎤ −1600
5s 3 − 1600
5s 3 − 1600
⎥ = Lim ⎢
⎥=
f ∞ = Lim ⎡⎣ sF s ⎤⎦ = Lim ⎢ s ×
= −2
2
3
s →0
s →0 ⎢
⎥ s→0 ⎢ s 3 + 18 s 2 + 90 s + 800 ⎥ 800
s
+
90
s
+
800
+
18s
s
s
⎣
⎣
⎦
⎦
( )
5.5
()
)
(
)
(
REGION OF CONVERGENCE (ROC)
The existence of Laplace transform X(s ) of a given x(t ) depends on whether the transform integral converges
∞
∞
−∞
−∞
X ( s ) = ∫ x (t )e − s t dt = ∫ x (t )e − t e − jwt dt < ∞
which in turn depends on the duration and magnitude of x(t) as well as the real part of s, Re[s] ␴ (the imaginary part of s Im[s] j␻ determines the frequency of a sinusoid which is bounded and has no effect on the
convergence of the integral).
This limits the variable s (␴ j␻) to a part of the complex plane. The subset of values of s for which the
Laplace transform exists is called the region of convergence (ROC) or the domain of convergence.
Thus, the Laplace transform F (s ) typically exists for all complex numbers such that Re{s} a, where a is
a real constant which depends on the growth behavior of f (t ), whereas the two-sided transform is defined in a
range a Re{s} b. In the two-sided case, it is sometimes called the strip of convergence.
Causal Signals When x(t) is right sided (i.e., x(t) 0 for t
a positive ␴ 0 tends to attenuate x (t )e ␴t as t → .
t0), it may have infinite duration for t
0, and
Non-Causal Signals When x(t ) is left sided (i.e., x(t) 0 for t t0), it may have infinite duration for t 0,
and a negative ␴ 0 tends to attenuate x (t )e ␴t as t → .
Based on these observations, we can get the following properties for the ROC:
• If x(t) is absolutely integrable and of finite duration then the ROC is the entire s-plane (the Laplace transform integral is finite, i.e., X(s) exists, for any s).
• The ROC of X(s) consists of strips parallel to the j␻-axis in the s-plane.
• If x(t) is right sided and Re[s] ␴0 is in the ROC, then any s to the right of ␴0 (i.e., Re[s] ␴0) is also in
the ROC, i.e., ROC is a right-sided half plane.
• If x(t) is left sided and Re[s]
␴0) is also in the
0 is in the ROC then any s to the left of ␴0 (i.e., Re[s]
ROC, i.e., ROC is a left-sided half plane.
• If x(t) is two-sided then the ROC is the intersection of the two one-sided ROCs corresponding to the two
one-sided components of x(t). This intersection can be either a vertical strip or an empty set.
()
• If X(s) is rational then its ROC does not contain any poles (by definition X s
s= sp
= ∞ dose not exist).
The ROC is bounded by the poles or extends to infinity.
• If X(s) is a rational Laplace transform of a right-sided function x(t) then the ROC is the half plane to the
right of the rightmost pole; if X(s) is a rational Laplace transform of a left-sided function x(t), then the
ROC is the half plane to the left of the leftmost pole.
• A signal x(t) is absolutely integrable, i.e., its Fourier transform X( j␻) exists (first Dirichlet condition,
assuming the other two are satisfied), if and only if the ROC of the corresponding Laplace transform X(s)
contains the imaginary axis Re[s] 0 or s j␻.
238
Network Analysis and Synthesis
3e2t
Example 5.2 Find the ROC of the function x(t)
jv
ROC
Solution Laplace transform of the function,
()
∞
X s = ∫ 3e 2 t e − st dt =
0
provided Re[s]
2
0
3 ,
s−2
s
Fig. 5.4 Schematic
of Example 5.2
2, which defines the ROC.
Example 5.3 Given the following Laplace transform, find the corresponding signal:
) ( s + 1)1( s + 2 ) = s 1+ 1 − s +1 2
X (s =
and the region of convergence.
Solution There are three possible ROCs determined by the two poles sp1
−1 and sp2
−2:
• The half plane to the right of the rightmost pole sp2 −1, with the corresponding right sided time function
x(t ) [e t − e 2t ] u(t )
• The half plane to the left of the leftmost pole sp1 −2, with the corresponding left sided time function
x(t ) [−e t e 2t] u(−t )
• The vertical strip between the two poles −2 Re[s] −1, with the corresponding two-sided time function
x(t ) − e−tu(−t ) − e 2tu(t )
5.6
LAPLACE TRANSFORM OF SOME BASIC FUNCTIONS
Exponential Function
f (t )
eat
By definition of Laplace transform,
∞
F ( s ) = L ⎡⎣ f (t ) ⎤⎦ = ∫ e ⋅ e dt = ∫ e
0−
Similarly, for f (t ) = e
Sine Function
− at
, F (s) =
∞
∞
at
− st
0−
( a− s )t
⎡ e ( a− s )t ⎤
⎛
1 ⎞
1
=
dt = ⎢
⎥ = ⎜0−
⎟
⎣ ( a − s ) ⎦ 0− ⎝ ( a − s ) ⎠ ( s − a )
1
s+a
f (t ) = sin t =
1 j t −j t
⎡e − e ⎤
⎦
2j⎣
∞
∞
⎡1
⎤
1 ⎡ ( j − s )t − ( j + s )t ⎤
1 ⎡ 1
1 ⎤
F ( s ) = L ⎡⎣ f (t ) ⎤⎦ = ∫ ⎢ ⎡⎣ e j t − e − j t ⎤⎦ ⎥ ⋅ e − st dt =
e
−e
⋅dt = ⎢
−
=
∫
⎣
⎦
2j
2 j 0−
2 j ⎣s− j
s + j ⎥⎦ s 2 +
⎦
0− ⎣
Cosine Function
2
1
f (t ) = cos t = ⎡⎣ e j t + e − j t ⎤⎦
2
∞
∞
⎡1
⎤
1
1⎡ 1
1 ⎤
s
j −s t
+
= 2
F ( s ) = L ⎡⎣ f (t ) ⎤⎦ = ∫ ⎢ ⎡⎣ e j t + e − j t ⎤⎦ ⎥ .e − st dt = ∫ ⎡ e ( ) + e − ( j + s )t ⎤ ⋅dt = ⎢
⎥
⎣
⎦
2
2 0−
2⎣s− j
s+ j ⎦ s +
⎦
0− ⎣
2
239
Laplace Transform and Its Applications
Hyperbolic Sine Function
1
f (t ) = sinh at = ⎡⎣ e at − e − at ⎤⎦
2
∞
∞
⎡1
⎤
1
1⎡ 1
1 ⎤
a
a− s t
F ( s ) = L ⎡⎣ f (t ) ⎤⎦ = ∫ ⎢ ⎡⎣ e at − e − at ⎤⎦ ⎥ ⋅ e − st dt = ∫ ⎡ e ( ) − e − ( a+ s )t ⎤ ⋅dt = ⎢
−
= 2 2
⎥
⎣
⎦
2
2 0−
2⎣s−a s+a⎦ s −a
⎦
0− ⎣
Hyperbolic cosine function
1
f (t ) = cosh at = ⎡⎣ e at + e − at ⎤⎦
2
∞
∞
⎡1
⎤
1
1⎡ 1
1 ⎤
s
a− s t
F ( s ) = L ⎡⎣ f (t ) ⎤⎦ = ∫ ⎢ ⎡⎣ e at + e − at ⎤⎦ ⎥ ⋅ e − st dt = ∫ ⎡ e ( ) + e − ( a+ s )t ⎤ ⋅dt = ⎢
+
⎥= 2 2
⎣
⎦
s
−
a
2
2
2
s
+
a
⎦
⎣
⎦ s −a
0− ⎣
0−
Damped sinusoidal function
⎫ ⎧1
⎫
⎧1
f (t ) = e − at sin t = e − at ⋅ ⎨ ⎡⎣ e j t − e − j t ⎤⎦ ⎬ = ⎨ ⎡⎣ e − ( a− j )t − e − ( a+ j )t ⎤⎦ ⎬
⎭ ⎩2 j
⎭
⎩2 j
∞
∞
⎡1
⎤
1
⎡ e − ( s + a− j )t − e − ( s + a+ j )t ⎤ ⋅ dt
F ( s ) = L ⎡⎣ f (t ) ⎤⎦ = ∫ ⎢ ⎡⎣ e − ( a− j )t − e − ( a+ j )t ⎤⎦ ⎥ ⋅ e − st dt =
⎦
2j
2 j 0∫− ⎣
⎦
0− ⎣
⎤
1 ⎡
1
1
= ⎢
−
=
2 j ⎣ {( s + a ) − j } {( s + a ) + j } ⎥⎦ ( s + a )2 + 2
Damped cosine function
⎫ ⎧1
⎫
⎧1
f (t ) = e − at cos t = e − at . ⎨ ⎡⎣ e j t + e − j t ⎤⎦ ⎬ = ⎨ ⎡⎣ e − ( a− j )t + e − ( a+ j )t ⎤⎦ ⎬
⎭ ⎩2
⎭
⎩2
∞
∞
⎡1
⎤
1
F ( s ) = L ⎡⎣ f (t ) ⎤⎦ = ∫ ⎢ ⎡⎣ e − ( a− j )t + e − ( a+ j )t ⎤⎦ ⎥ ⋅ e − st dt = ∫ ⎡⎣ e − ( s + a− j )t + e − ( s + a+ j )t ⎤⎦ ⋅ dt
2
2 0−
⎦
0− ⎣
⎤
1⎡
1
1
( s + a)
= ⎢
+
=
2 ⎣ {( s + a ) − j } {( s + a ) + j } ⎥⎦ ( s + a )2 + 2
5.6.1 Singularity Functions and Waveform Synthesis
In order to synthesize any signal, there are some standard or singularity functions which can be realized in
the laboratory. Other signals can be written in terms of these singularity functions. Those singularity functions are
1. Step function,
2. Ramp function,
3. Impulse function, and
4. Unit doublet function.
Step function
This function is also known as Heaviside unit function. It is defined as given below.
f (t ) u(t) 1 for t > 0
0 for t < 0
and is undefined at t 0.
u(t )
1
t
0
Fig. 5.5 (a) Unit step
function
240
Network Analysis and Synthesis
A step function of magnitude K is defined as,
f (t ) Ku(t) K for t > 0
0 for t < 0
and is undefined at t 0.
A shifted or delayed unit step function is defined as
f (t ) u(t T) 1 for t > T
0 for t < T
and is undefined at t T.
The Laplace transform of a unit step function is given as
∞
Ku(t)
K
t
0
Fig. 5.5 (b) Step function
of magnitude K
u(t
T)
1
∞
∞
⎡ e − st ⎤
1 1
F ( s ) = L ⎡⎣ f (t ) ⎤⎦ = ∫ u(t ) ⋅ e − st dt = ∫ 1⋅ e − st dt = ⎢
⎥ = 0− =
−
−
s
s s
⎦ 0−
⎣
0−
0−
K
s
Similarly, the Laplace transform of the shifted unit step function u(t − T) is,
Also, the Laplace transform of step function of magnitude K is L[ Ku(t )]=
0
T
t
Fig. 5.5 (c) Shifted unit
step function
e − st
{by differentiation theorem}
s
g(t)
Another function, called gate function can be obtained from step function as follows.
K
Therefore, g(t) Ku(t a) Ku(t b)
L[ u(t − T )] =
(
K − as − bs
e −e
s
L ⎡⎣ g (t ) ⎤⎦ =
)
a
0
Fig. 5.6
b
Gate function
Ramp Function
r (t)
A unit ramp function is defined as
f (t )
r(t)
t for t
0 for t
A ramp function of any slope K is defined as
f (t ) Kr(t)
0
0
1
0
0
t for t
0 for t
T
T
Fig. 5.7 (a) Unit
ramp function
Kr (t)
K
1
()
∞
0−
0−
L ⎡⎣ r t ⎤⎦ = ∫ r (t ) ⋅ e − s t dt = ∫ te − s t dt
Fig. 5.7 (b)
function
r (t
1
1
u
du
Ramp
T)
Integrating by parts, let,
then
t
0
The Laplace transform of a unit ramp function is
∞
t
0
Kt for t
0
for t
A shifted unit ramp function is defined as
f (t ) r(t T)
1
t and dv
dt and v = ∫ e − s t dt = −
e
−st
s
e
st
dt
0
Fig. 5.7 (c) Shifted unit
ramp function
t
241
Laplace Transform and Its Applications
Now,
∞
∞
∞
∞
∞
⎡ t
⎤
1
1
1
L[ r (t )] = ∫ udv = uv |0∞− − ∫ vdu = ⎢ − ( e − st ) ⎥ + ∫ e − st dt = ∫ e − st dt = 2
s 0−
s
⎣ s
⎦ 0− s 0−
0−
0−
Similarly, Laplace transform of a ramp of slope K is L ⎡⎣ Kr (t ) ⎤⎦ =
K
s2
Ke − Ts
s2
Impulse function This function is also known as Dirac Delta function, denoted by (t). This is a function
of a real variable t, such that the function is zero everywhere except at the instant t 0. Physically, it is a very
sharp pulse of infinitesimally small width and very large magnitude, the
f(t )
area under the curve being unity.
and Laplace transform of a shifted ramp function is L ⎡⎣ Kr (t − T ) ⎤⎦ =
3/a
Consider a gate function as shown in Fig. 5.8.
The function is compressed along the time-axis and stretched along the
y-axis, keeping area under the pulse unity. As a → 0, the value of 1 → ∞
a
and the resulting function is known as impulse.
2/a
1/a
0
∞
It is defined as
(t ) = 0 for t ≠ 0 and ∫ (t )dt = 1
a/2
t
a
Fig. 5.8 Generation of impulse
function from gate function
−∞
Also,
a /3
(t ) = Lim a1 ⎡⎣ u(t ) − u(t − a) ⎤⎦
a →0
The Laplace transform of the impulse function is obtained as
⎫
⎧1
1 ⎡ 1 e − as ⎤
1 − e − as
se − as
=
=
L ⎡⎣δ t ⎤⎦ = Lim L ⎨ ⎡⎣ u(t ) − u(t − a ) ⎤⎦ ⎬ = Lim ⎢ −
Lim
Lim
=1
⎥
s ⎦ a→0 as
s
a →0
a →0
⎭ a →0 a ⎣ s
⎩a
()
f (t)
Unit Doublet Function The derivative of unit
impulse function with respect to time at any instant
of time is known as unit doublet function. It is
defined as
(
)
(
1
2
a
a
)
t
1
a
The name of the function is given as doublet because
it can be obtained from the function shown in
Fig. 5. 9 (a) with a → 0.
The Laplace transform of a unit doublet function is obtained as
2
Fig. 5.9 (a) Generation
of unit doublet function
with a → 0
⎡d
⎤
L ⎡⎣δ ′ t − T ⎤⎦ == L ⎢ δ t − T ⎥ = sL ⎡⎣δ t − T ⎤⎦ = se − Ts
⎣ dt
⎦
)
f(t)
t
a
d
⎡δ t − T ⎤ = δ ′ t − T = 0
for t ≠ 0
⎦
dt ⎣
= +∞ and − ∞ for t = T
(
[by L’Hospital’s rule]
(
)
(
)
Fig. 5.9 (b) Unit
doublet function
242
Network Analysis and Synthesis
Example 5.4 Express the function in terms of the standard signals and find its Laplace transform.
Solution The function can be written as the summation of
some ramp functions as given below.
f (t ) r(t ) r (t 1) r (t 2) r (t 3)
1 e − s e −2 s e −3s
∴F s = 2 − 2 − 2 + 2
s
s
s
s
()
5.7
f(t)
1
0
1
2
t
3
Fig. 5.10 Waveform of Example 5.4
LAPLACE TRANSFORM TABLE
Table 5.1
Standard Laplace Transforms
Sl. No.
Functions [ f (t )]
Laplace Transform [F (s )]
In Time (t ) Domain
In Frequency (s) Domain
∞
Definition
If f (t ) is Laplace transformable
Then L[ f (t )] = F (s )= ∫ f (t )e − st dt
0−
1
U(t ) (unit step function)
1
s
2
U(t
e − sT
s
3
(t ) (unit impulse)
4
e at (exponential function)
1
s−a
5
e at (exponential function)
1
s+a
6
sin t (sine function)
7
cos t (cosine function)
s
s +
8
t n (n =1, 2, 3, …) (ramp function)
n!
s n+1
9
t (unit ramp function)
1
s2
10
e at sin t (damped sine function)
T ) (unit step function shifted/delayed by T )
1
s2 +
2
2
2
( s + a )2 +
2
243
Laplace Transform and Its Applications
11
e at cos t (damped cosine function)
( s + a)
( s + a )2 +
12
e at t n (damped ramp function)
n!
( s + a )n+1
13
d
f (t ) (differentiation theorem)
dt
sF ( s ) − f (0− )
F ( s ) f (0− )
+
s
s
t
14
∫ f (t )dt (integration theorem)
0
15
sinh t (hyperbolic Sine function)
16
cosh t (hyperbolic Cosine function)
17
e at sinh t (damped hyperbolic Sine function)
18
e at cosh t (damped hyperbolic Cosine function)
19
Initial-value theorem
20
Final-value theorem
21
5.8
2
s2 −
2
s
s −
2
2
( s + a )2 −
2
( s + a)
( s + a )2 −
2
Lt f (t ) = Lt sF ( s )
t →0
s →∞
Lt f (t ) = Lt sF ( s )
t →∞
Shifting theorem f (t
a)
e
± as
s →0
F (s)
OTHER IMPORTANT LAPLACE TRANSFORMS
1
(t)
2
(t
a)
e as
3
(t
a) g(t)
e as g(a)
Note: g(a) Not G(a)
4
5
1
n
1−
1−
2
e
−
1−
where
e
−
nt
2
nt
sin
n
1−
sin( n 1 −
cos 1
2
2
t
t + ),
2
n
s2 + 2
n
s+
2
n
( < 1)
2
n
s( s 2 + 2
n
s+
2
n
( < 1)
244
Network Analysis and Synthesis
5.9
LAPLACE TRANSFORM OF PERIODIC FUNCTIONS
If f (t ) is periodic with time period T ( 0), so that f (t T) f (t ) then the Laplace transform of the function
⎛ 1 ⎞
times the Laplace transform of the first cycle.
is equal to ⎜
⎝ 1− e − Ts ⎟⎠
Proof
⎡ 1 ⎤
∴ L ⎡⎣ f (t ) ⎤⎦ = F ( s ) = F1 ( s ) ⎢
− Ts ⎥
⎣1− e ⎦
Let
f (t ) − be the periodic function,
T − the time period,
f1(t ), f2(t ), . . . , f n(t ) − the functions representing the first, second, . . . , nth cycle, respectively
fn(t )
f (t ) f1(t) f2(t )
f1(t) f1 (t T ) f1 (t 2T )
Taking Laplace transform,
L ⎡⎣ f (t ) ⎤⎦ = F ( s ) = L ⎡⎣ f1 (t ) ⎤⎦ + L ⎡⎣ f1 (t − T ) ⎤⎦ + L ⎡⎣ f1 (t − 2T ) ⎤⎦ + ⋅⋅⋅
= F1 ( s ) + e − Ts F1 ( s ) + e −2Ts F1 ( s ) + ⋅⋅⋅= F1 ( s ) ⎡⎣1 + e − Ts + e −2Ts + e −3Ts + ⋅⋅⋅⎤⎦
⎡ 1 ⎤
F ( s ) = F1 ( s ) ⎢
− Ts ⎥
⎣1− e ⎦
Therefore,
Example 5.5 Find the Laplace transform of the square wave.
Solution The first cycle is shown below.
It can be written as
f1(t ) u(t ) − 2 u(t − T )
Taking Laplace transform of the first cycle,
u (t − 2T )
f(t )
1
0
1
T
Fig. 5.11 (a)
2
1 2 e − Ts e −2Ts 1
Example 5.5
−
+
= 1 − e − Ts
s
s
s
s
f1(t )
By the theory of time periodicity, the Laplace transform of the square wave is given
1
2
1
1
0
as, F ( s ) = 1 − e − Ts ×
(since
time
period
of
the
square
wave
is
2T)
s
1 − e −2Ts
1
F1(s)
(
(
)
⎛ Ts ⎞
1 ⎛ 1 − e − Ts ⎞ 1
= ⎜
= tanh ⎜ ⎟
s ⎝ 1 + e − Ts ⎟⎠ s
⎝ 2⎠
5.10
)
time
2T 3T
Square wave of
2T
T
time
Fig. 5.11 (b) First cycle of
square wave of Fig 5.12 (a)
INVERSE LAPLACE TRANSFORM
N (s)
D( s )
where, N(s) is the numerator polynomial and D(s) is the denominator polynomial. The roots of N(s)
called the zeros of F (s ) while the roots of D(s) 0 are the poles of F (s ).
Let F (s ) have the general form of F ( s ) =
0 are
245
Laplace Transform and Its Applications
For example, for the function, F ( s ) =
s −1
, the zero is at s
s s−2 s−3
)(
(
)
1 and the poles are at s
0, 2 and 3.
We use partial fraction expansion to break F (s ) down into simple terms. Thus, there are two steps to find
inverse Laplace transform:
I. Decomposition of F (s ) into simple terms using partial fraction expansion.
II. Evaluation of the inverse of each term comparing with the standard forms of Laplace transforms.
We consider the following three cases:
F (s) =
Simple poles Let
(
N (s)
s + p1 s + p2 s + p3 ⋅⋅⋅ s + pn
)(
)(
) (
)
where, s −p1, −p2 −p3, …, −pn are the simple poles, and pi pj for all i
Assuming that the degree of N(s) is less than the degree of D(s)
F (s) =
j (i.e., poles are distinct)
k
k
k1
k
+ 2 + 3 + ⋅⋅⋅+ n
s + p1 s + p2 s + p3
s + pn
(5.1)
where, expansion coefficients k1, k2, k3, …, kn are known as the residues of F (s ). These can be found out by
residue method explained below.
Multiplying both sides of Eq. (5.1), by (s p1),
( s + p )k ( s + p )k
( s + p )k
( s + p ) F ( s) = k + s + p + s + p + ⋅⋅⋅ + s + p
1
1
2
1
2
Putting
( s + p ) F (s)
s = − p1 ⇒
(
In general, ki = s + pi
1
)
s =− pi
n
3
s = pi
3
n
= k1
This is known as Heaviside’s theorem.
Once the values of ki are known, the inverse Laplace is obtained as
(
f (t ) = k1e
− p1t
+ k2 e
− p2 t
+ k3e
− p3t
+ ⋅⋅⋅+ k n e
− pnt
Example 5.6 Find the inverse Laplace transform of the function, F ( s ) =
Solution
Let
F (s) =
(
k
k
k
2s +1
= 1 + 2 + 3
s
+
1
s
+
2
s
+3
s +1 s + 2 s + 3
)(
)(
)
(
2s +1
s+2 s+3
∴k 2 = s + 2 F ( s ) s =−2 =
2s +1
s +1 s + 3
(
)
∴k1 = s + 1 F ( s ) s =−1 =
(
n
1
)
(
)(
)(
)
)
=−
s =−1
=3
s =−2
1
2
) u(t )
2s + 1
.
( s + 1)( s + 2)( s + 3)
246
Network Analysis and Synthesis
)
(
∴k 3 = s + 3 F ( s ) s =−3 =
∴ F (s) = −
(
2s +1
s +1 s + 2
)(
)
=−
s =−3
5
2
1
3
5
+
−
s
+
2
2 s +1
2 s+3
)
(
)
(
1
5
Thus, the inverse Laplace transform is given as f (t ) = − e − t + 3e −2 t − e 3t
2
2
Repeated poles Suppose, F (s ) has ‘n’ repeated poles at s
∴ F (s) =
kn
k n−1
+
( s + p) ( s + p)
n
n−1
+
k n− 2
( s + p)
n− 2
− p.
+ ⋅⋅⋅ +
k2
+
where, F1(s ) is the remaining part of F (s ) that does not have a pole at s
)
(
k1
( s + p) ( s + p)
2
+ F1 ( s )
− p.
n
We find, k n = s + p F ( s ) s =− p
To find kn−1, kn 2,…, kn m, the procedure is
In general, k n− m =
)
(
k n−1 =
n
d ⎡
s + p F (s)⎤
⎢
⎣
⎦⎥ s =− p
ds
k n− 2 =
n
1 d2 ⎡
s + p F (s)⎤
2 ⎢
⎣
⎦⎥
2! ds
s =− p
)
(
n
1 dm ⎡
s + p F (s)⎤
, where, m
m ⎢
⎣
⎦⎥
m! ds
s =− p
(
)
1, 2, …, (n − 1).
Once the values of k1, k2, …, kn are known, the inverse Laplace is obtained as
⎛
k
k n n−1 − pt ⎞
t e ⎟ u ( t ) + f1 ( t )
f (t ) = ⎜ k1e − pt + k 2 te − pt + 3 t 2 e − pt + ⋅⋅⋅+
2!
n −1 !
⎝
⎠
)
(
Example 5.7 Find the inverse Laplace transform of the function F ( s ) =
Solution
Let
F (s) =
12
k1
=
( s + 2) ( s + 4) ( s + 2)
2
2
+
k
k2
+ 3
s+2 s+4
12
(s + 2) (s + 4 )
By residue method,
)
(
2
k1 = s + 2 F ( s ) s =−2 =
∴k2 =
(
12
s+4
)
=6
s =−2
2
d ⎡
d ⎡ 12 ⎤
s + 2 F (s) ⎤
= ⎢
⎥
⎢
⎥
⎦ s =−2 ds ⎢ s + 4 ⎥
ds ⎣
⎣
⎦
(
)
(
)
= −3
s =−2
2
.
247
Laplace Transform and Its Applications
)
(
k 3 = s + 4 F ( s ) s =−4 =
F (s) =
Thus,
6
( s + 2)
2
12
( s + 2)
=3
2
s =−4
3
3
+
s+2 s+4
−
3e 4t
Taking inverse Laplace transform, f (t )
3e 2t
6te 2t
Complex Poles Since N(s) and D(s) always have real coefficients and as the complex roots of polynomials
with real co-efficients occur in conjugate form, F (s ) may have the general form
F (s) =
A1 s + A2
s + as + b
2
+ F1 ( s ) =
k1
k2
+
+ F1 ( s )
s+ − j
s+ + j
where, F1(s ) is the remaining part of F (s ) that does not have this pair of complex poles.
Let
( s + as + b ) = ( s + 2 s +
2
2
2
+
2
) = (s + ) +
2
2
2
(
s1,2 = − ± j
) = − a2 ± j b − a4
Thus, the coefficients are
)
(
k1 = s − s1 F ( s ) s = s
and k2
k1
Complex conjugate of k1
1
Example 5.8 Find the inverse Laplace transform of the function F ( s ) =
Solution
Let
F (s) =
2s +1
( s + 1)( s + 2 s + 5)
2
(
)
∴ A = s + 1 F ( s ) s =−1 =
=
)
)(
k1
k2
A
+
+
s + 1 s + 1 − j 2 s + 1 + j2
2s +1
1
=−
4
s + 2 s + 5 s =−1
2
2s +1
k1 = s + 1 − j 2 F ( s ) s = −1+ j 2 =
(
) s +1 s +1+ j 2
(
2s + 1
.
1
s
+
s 2 + 2s + 5
(
(
)(
⎛1
1⎞
∴k 2 = k1* = ⎜ + j ⎟
2⎠
⎝8
1
1
1
1
−j
+j
1⎛ 1 ⎞ 8
2 + 8
2
∴ F (s) = − ⎜
+
4 ⎝ s + 1⎟⎠ s + 1 − j 2 s + 1 + j 2
) (
s = −1+ j 2
)
⎛1
1⎞
=⎜ − j ⎟
2⎠
⎝8
)
248
Network Analysis and Synthesis
Taking inverse Laplace transform,
1
1
f (t ) = − ⎡⎣e −t − e −t cos 2t ⎤⎦ + e −t sin 2t = − e −t sin 2 t + e −t sin 2t
4
2
5.11
APPLICATIONS OF LAPLACE TRANSFORM
1. Solving integro-differential equations and simultaneous differential equations
2. Transient analysis of electrical circuits
5.11.1 Solving integro differential equations and simultaneous differential equations
An integro-differential equation is an integral equation in which various derivatives of the unknown function
can also be present. A standard form of an integro-differential equation is
()
()
()
()
t
()
()
an x n t + an−1 x n−1 t + an− 2 x n− 2 t + ⋅⋅⋅+ a0 x t + a−1 ∫ x t dt = f t
0
where all the coefficients (an, an 1,..., a0 , a 1) are constants.
Another type of differential equations applicable for more than one unknown variables is known as simultaneous differential equation. Considering two unknowns, x(t) and y(t), the equations take the form as given below.
)
)
)
)
γ x ′(t ) + γ x (t ) + δ y ′(t ) + δ y (t ) = 0
α1 x ′ ( t + α 0 x ( t + β1 y ′ ( t + β0 y ( t = 0
1
0
1
0
where,
i, i, i, and i are arbitrary constants.
Using the Laplace transform of integrals and derivatives, an integro-differential equation can be solved.
Similarly, it is easier with the Laplace transform method to solve simultaneous differential equations by transforming both equations and then solve the two equations in the s-domain and finally obtain the inverse to get
the solution in the time domain.
Example 5.9 (Integro-differential equation)
t
Solve the equation for the response i(t), given that
di
+ 2 i + 5 ∫ idt = u (t ) and i(0)
dt
0
⎡ di ⎤
∴L ⎢ ⎥ = sI ( s ) − i(0) = sI ( s ) − 0 = sI ( s )
⎣ dt ⎦
Taking Laplace transform on both sides of the given equation,
Solution Let L[i(t)]
I(s)
I (s) 1
=
s
s
1
1
2
I (s) = 2
=
2
s + 2s + 5 2 s +1 + 2 2
sI ( s ) + 2 I ( s ) + 5
or,
(
) ()
1
Taking inverse Laplace transform, we get i(t ) = e − t sin 2t ( A ), t > 0
2
0.
249
Laplace Transform and Its Applications
Example 5.10 (Integro-differential equation)
Solve the initial-value problem for y(t) when
d 2y
+ y (t ) = 3 sin2t and y (0) = 1, y ′(0) = −2.
dt 2
Solution Let L[y(t )]
⎡d2 y ⎤ 2
2
Y(s). ∴L ⎢ 2 ⎥ = s Y ( s ) − sy (0) − y ′(0) = s Y ( s ) − s + 2
dt
⎣
⎦
s
2
− 2
s +1 s + 4
Taking inverse Laplace transform, we get, y (t ) (cos t
Y (s) =
Or,
2
sin 2t )
Example 5.11 (Simultaneous differential equations)
Find the solution of the system
x(0)
1, y(0)
Solution
dx
− 6 x + 3 y = 8e t
dt
and
dy
− 2 x − y = 4e t
dt
with initial conditions
0.
Taking Laplace transform,
( s − 6 ) X + 3Y = −ss−+19
(i)
)
(ii)
(
−2 X + s − 1 Y =
4
s −1
Solving for X and Y,
X=
Y=
(
2
1
−s + 7
=−
+
1
4
s
−
s
−
s −1 s − 4
)(
)
2
=
( s − 1)( s − 4 )
−2
2
3+ 3
s −1 s − 4
Taking inverse Laplace transform,
2
2
x (t ) = −2 e t + e 4 t and y (t ) = − e t + e 4 t
3
3
Example 5.12 (Simultaneous differential equations)
Solve for x(t) and y(t), given that x(0)
Solution
4, y(0)
3 and
dx
dy
+ x + 4 y = 10 and x −
− y =0
dt
dt
Following the same procedures, as in Ex. 5.11, we get,
4 s 2 + 2 s + 10
3s 2 + s + 10
X=
and
Y
=
s s2 + 3
s s2 + 3
(
)
Taking inverse Laplace transform, we get the desired results.
(
)
250
Network Analysis and Synthesis
5.11.2
Application of Laplace Transform Method to Circuit Analysis
We now apply the mathematical tool for the analysis of electric circuits.
Transform Impedance of Network Elements
Element
1. Resistor (R)
2 Inductor (L)
Time Domain
v(t ) Ri(t )
V(s)
v (t) i(t)
R
v (s) I(s)
v (t ) = L
di(t )
dt
V(s)
L[sI (s)
I (s) =
1 ⎡ V ( s ) i (0− ) ⎤
+
L ⎢⎣ s
s ⎥⎦
t
i (t ) =
v (t)
1
∫ v(t )dt
L −∞
i(t)
L
s-Domain
RI(s)
R
I(s)
i(0 )]
sL
V(s)
Li(0 )
3 Capacitor (C )
dv (t )
dt
t
1
v (t ) = ∫ i(t )dt
C −∞
i (t ) = C
v (t)
i(t)
C
I(s)
sCV (s)
Cv (0 )
V (s) =
I ( s ) v (0− )
+
Cs
s
I (s)
V (s)
Z (s)
1/sC
v (0 )/s
Advantages of analyzing the circuits using frequency domain rather than time domain
The following are some advantages of analyzing an electrical network in s-domain rather that in t-domain:
1. Each element can easily be replaced by a transform impedance.
2. No integration or differentiation is involved in the transform equations.
3. The response obtained after solution is a complete response, i.e., both the steady state and transient
responses are obtained.
5.12 TRANSIENT ANALYSIS OF ELECTRIC CIRCUITS USING LAPLACE TRANSFORM
In electrical engineering, a transient response or natural response is the electrical response of a system to a
change from equilibrium.
The condition prevailing in an electric circuit between two steady-state conditions is known as the transient
state; it lasts for a very short time. The currents and voltages during the transient state are called transients.
In general, transient phenomena occur whenever
(i) a circuit is suddenly connected or disconnected to/from the supply,
(ii) there is a sudden change in the applied voltage from one finite value to another,
(iii) a circuit is short-circuited.
251
Laplace Transform and Its Applications
A simple example would be the output of a 5-volt dc power supply when it is turned on. The transient
response is from the time the switch is turned on and the output is a steady 5V. At this point, the power supply
reaches its steady-state response of a constant 5V.
The transient response is not necessarily tied to on–off events but to any event that affects the equilibrium of the system. If in an RC circuit, the resistor or capacitor is replaced with a variable resistor
or variable capacitor (or both) then the transient response is the response to a change in the resistor or
capacitor.
The transient currents are not caused by any part of the supply voltage, but are entirely associted with the
changes in the stored energy in capacitors and inductors. As there is no energy stored in resistors, there are no
transients in purely resistive circuits.
Although transients last for a very short time, their study is very important because
• they indicate what dangerous rises in voltage or current may happen in individual sections of a circuit
• they indicate how signals are distored in waveform or amplitude as they pass through amplifiers, filters,
or other circuit elements
We consider the transient analysis for the following circuits subject to step input, impulse input and sinusoidal
input:
1. RL series circuit,
2. RC series circuit,
3. RLC series circuit, and
4. RLC parallel circuit.
5.12.1
RL Series Circuit
RL series circuit with step input
Inductors store energy in a magnetic field (produced by the current through the wire). Thus, the stored energy in
an inductor tries to maintain a constant current through its windings. Because of this, inductors oppose changes
in current, and act precisely the opposite of capacitors, which oppose changes in voltage.
A fully discharged inductor, having zero current through it, will initially act as an open-circuit when attached
to a source of voltage, dropping maximum voltage across its leads. Over time, the inductor current rises to the
maximum value allowed by the circuit, and the terminal voltage decreases correspondingly. Once the inductor
terminal voltage has decreased to a minimum (zero for an ideal inductor), the current will stay at a maximum
level, and it will behave essentially as a short-circuit.
If the switch is closed at time t 0, the voltage across the RL combiR
Switch
nation would be v(t ) which is a step of magnitude V [or Vu(t )] and not a
constant as is the supply voltage V.
v(t ) 0, for t 0
L
i(t)
V, for t 0
Thus the differential equation governing the behaviour of the circuit would be
Ri(t ) + L
Fig. 5.12
di(t )
= Vu(t )
dt
Taking Laplace transform, we get
RI ( s ) + L ⎡⎣ sI ( s ) − i(0− ) ⎤⎦ =
V
s
RL series circuit
252
Network Analysis and Synthesis
V
or
L
I (s) =
s s+ R
(
⎛
⎞
i (0− ) V ⎜ 1
1 ⎟ i (0− )
+
+
=
−
R⎜ s s+ R ⎟ s+ R
s+ R
⎝
⎠
L
L
L
L
)
Taking inverse Laplace transform,
i (t ) =
−( R )t ⎞
−( R )t
−( R )t ⎞
V⎛
V⎛
1 − e L ⎟ + i (0− )e L = ⎜ 1 − e L ⎟
⎜
⎠
⎠
R⎝
R⎝
with i(0− ) = 0.
V − Rt
The transient part of the current response, itr = ⎡⎣i(t ) − is ⎤⎦ = − e L
R
)
(
L
V
V
, i = 1 − e −1 = 0.63 = 0.63is
R
R
R
When the switch is first closed, the voltage across the inductor will immediately jump to battery voltage
(acting as though it were an open-circuit) and decay down to zero over time (eventually acting as though it
were a short-circuit). The voltage across the inductor is determined by calculating how much voltage is being
dropped across R, given the current through the inductor, and subtracting that voltage value from the battery
voltage. When the switch is first closed, the current is zero, then it increases over time until it is equal to the
battery voltage divided by the series resistance. This behavior is precisely opposite that of the series resistor–
capacitor circuit, where current started at a maximum and capacitor voltage at zero.
From the current equation at t = =
The steady state part of the current response, is =
V
R
The variation of the current is shown in Fig. 5.13.
1.0
Current
0.63
0.5
0.0
0 s 0. 5s 1.0 s 1 .5 s 2 .0s 2. 5s 3. 0s 3 .5s 4. 0s 4.5s 5 .0 s
Time
t
L/R
Fig. 5.13
The quantity
=
Variation of current with time in R-L series circuit with step input
L
is known as the time-constant of the circuit and is defined as follows.
R
Definitions of time-constant ( )
1. It is the time taken for the current to reach 63% of its final value. Thus, it is a measure of the rapidity with
which the steady state is reached.
Also, at t 5 , i 0.993is; the transient is therefore, said to be practically disappeared in five time constants.
253
Laplace Transform and Its Applications
2. The tangent to the equation i =
R
− t⎞
V⎛
1 − e L ⎟ at t
⎜
R⎝
⎠
0, intersects the straight line, i =
V
L
at t = = . Thus,
R
R
time-constant is the time in which steady state would be reached if the current increases at the initial rate.
Physically, time-constant represents the speed of the response of a circuit. A low value of time-constant
represents a fast response and a high value of time-constant represents a sluggish response.
R
⎛
− t⎞
Calculations of the Voltage Across Elements Voltage across the resistor, VR = Ri(t ) = V ⎜ 1 − e L ⎟
⎝
⎠
Voltage across the inductor, VL = L
R
R
− t⎞⎤
− t
di(t )
d ⎡V ⎛
= L ⎢ ⎜ 1 − e L ⎟ ⎥ = Ve L
dt
dt ⎢⎣ R ⎝
⎠ ⎥⎦
RL Series Circuit with impulse input By KVL, the mesh equation becomes
Ri(t ) + L
Taking Laplace transform,
RI(s) sLI(s) V with i(0 )
or,
di(t )
= V (t )
dt
VR/L
0
Voltage across resistor
Voltage
across
R and L
⎛
⎞
V⎜ 1 ⎟
I (s) =
L⎜ s+ R ⎟
⎝
L⎠
Taking inverse Laplace transform, i(t ) =
Voltage across the resistor, VR = Ri(t ) =
Voltage across inductor
V − RL t
e
L
VR
e
L
VR/L
time
Fig. 5.14 Variation of voltages with time in R-L series
circuit with impulse input
R
− t
L
di(t )
d ⎛ V − Rt ⎞
VR − R t
=L ⎜ e L ⎟ =− e L
dt
dt ⎝ L
L
⎠
The plots of the voltages are shown in Fig. 5.14.
Voltage across the inductor, VL = L
RL series circuit with sinusoidal input Here, the input voltage is given as, v(t)= V sin t
By KVL,
Ri(t ) + L
I ( s ) ⎡⎣ R + sL ⎤⎦ =
or,
V
or,
di(t )
= V sin t with i(0− ) = 0
dt
I (s) =
( s + )(
2
2
L
s+ R
L
)
=
⎧
V ⎪
⎨
L ⎪ s+ j
⎩
(
V
s2 +
2
1
)( s − j )(
⎫
⎡
⎤
A3 ⎥
A2
⎪ V ⎢ A1
=
+
+
⎬
⎢
s+ j
s + R ⎥⎥
s+ R ⎪ L ⎢s− j
⎣
L⎦
L ⎭
)
254
Network Analysis and Synthesis
where,
⎧
⎪
A1 = ⎨ s − j
⎪
⎩
(
)
⎧
⎪
A2 = ⎨ s + j
⎪
⎩
)
(
1
( s + j )( s − j )(
1
( s + j )( s − j )(
⎧
⎪
A3 = ⎨ s + R
L s+ j
⎪
⎩
)(
(
and
⎡
V ⎢
∴ I (s) =
L ⎢2 j
⎢⎣
1
⎫
⎪
=
⎬
2j
R
s+
⎪
L ⎭s = j
)
⎫
⎪
=−
⎬
2j
R
s+
⎪
L ⎭s =− j
)( s − j )(
L
(
L
R+ j L
)
(
L
R− j L
⎫
L2
⎪
= 2
⎬
R + 2 L2
s+ R ⎪
L ⎭s =− R
L
)
(
L
−
)
+
)
)
( R + j L )( s − j ) 2 j ( R − j L )( s + j ) ( R +
2
L2
2
L2
)(
⎤
⎥
⎥
s+ R ⎥
L ⎦
)
Taking inverse Laplace transform,
R
R
⎡
⎤
− t
− t
V ⎢
Le j t
Le − j t
L2 e L ⎥ V ⎡ e j t
e− j t ⎤
e L
i (t ) =
−
+
= ⎢
−
⎥ +V L 2
L ⎢ 2 j R + j L 2 j R − j L R 2 + 2 L2 ⎥ 2 j ⎣ R + j L R − j L ⎦
R + 2 L2
⎢⎣
⎥⎦
)
(
(
)
(
)
(
)
Let, R + j L = Ze j and R − j L = Ze − j so t hat, Z =
( R + L ) and = tan ⎛⎜⎝ RL ⎞⎟⎠
2
2
−1
2
Putting these values,
R
− t
j t−
− j t−
e L V ⎡ e ( ) − e ( ) ⎤ V L − RL t
V ⎡ e j t e− j t ⎤
i (t ) = ⎢ j − − j ⎥ + V L 2 = ⎢
⎥+ 2 e
Z ⎢⎣
2 j ⎣ Ze
2j
Z
Ze ⎦
⎥⎦ Z
or, finally, the current is,
i (t ) =
V
sin
Z
( t − ) + VZ L e
R
− t
L
2
From this result, it is clear that the current in an RL series circuit lags behind the voltage by an angle,
⎛ L⎞
= tan −1 ⎜
If the resistance R 0 then
90 as is the case for a perfect inductor.
⎝ R ⎟⎠
Example 5.13 The series RL circuit shown in Fig. 5.15 is excited by adc
voltage of 50 V. Assume the initial current flowing through the inductor to
be 5 A and find the current i(t) for t > 0. Use Laplace transform method.
Solution Applying KVL for the loop, we get,
Ri(t ) + L
di(t )
= 50
dt
Switch
50 V
Fig. 5.15
5
i(t)
1H
Circuit of Example 5.13
255
Laplace Transform and Its Applications
Taking Laplace transform,
()
() ( )
RI s + L ⎡⎣ sI s − i 0 − ⎤⎦ =
50
s
( ) ( ) 50s + 5
⇒ 5 I s + sI s =
⇒
( s + 5) I ( s ) = 50s + 5
( ) s s50+ 5 + s +5 5 = 50 ⎡⎢ 1s − s +1 5 ⎤⎥ + s +5 5
( ) ( ) ⎣
)
⎦ (
⇒ I s =
Taking inverse Laplace transform, we get,
()
(
)
(
i t = 50 1 − e −5t + 5e −5t = 50 − 45e −5t
5.12.2
)(A)
t >0
RC Series Circuit
RC series circuit with step input As capacitors store energy in the form of an electric field, they tend
to act like small secondary-cell batteries, being able to store and release electrical energy. A fully discharged
capacitor maintains zero volts across its terminals, and a charged capacitor maintains a steady quantity of voltage across its terminals, just like a battery.
When capacitors are placed in a circuit with other sources of voltage,
Switch
R
they will absorb energy from those sources, just as a secondary-cell battery
will become charged as a result of being connected to a generator. A fully
discharged capacitor, having a terminal voltage of zero, will initially act as a
v(t)
C
i(t )
short-circuit when attached to a source of voltage, drawing maximum current as it begins to build a charge. Over time, the capacitor terminal voltage
Fig. 5.16 R-C series circuit
rises to meet the applied voltage from the source, and the current through
the capacitor decreases correspondingly. Once the capacitor has reached
the full voltage of the source, it will stop drawing current from it, and behave essentially as an open-circuit.
t
By KVL,
Ri(t ) +
1
i(t )dt = Vu(t )
C ∫0
Taking Laplace transform,
1 ⎡ I ( s ) q(0− ) ⎤ V
RI ( s ) + ⎢
+
=
C⎣ s
s ⎥⎦ s
⎡
1 ⎤ V q(0− )
or, I ( s ) ⎢ R + ⎥ = −
Cs
Cs
⎣
⎦ s
q(0− )
1V −
C
C
=
or, I ( s ) =
R
1
1
s R+
s+
Cs
RC
Taking inverse Laplace transform,
⎡V q(0− ) ⎤ − t RC
i (t ) = ⎢ −
, for t ≥ 0
⎥e
⎣ R RC ⎦
V −t
= e RC ,
if q(0− ) = 0
R
The steady state part of the current response, is = 0
V − q(0− )
(
) (
)
256
Network Analysis and Synthesis
The transient part of the current response, itr = ⎡⎣i(t ) − is ⎤⎦ =
From the current equation at, t = = RC , i =
V − t RC
e
R
V −1
V
e = 0.37
R
R
When the switch is first closed, the voltage across the capacitor is zero; thus, it first behaves as though it
were a short-circuit. Over time, the capacitor voltage will rise to equal battery voltage, ending in a condition
where the capacitor behaves as an open circuit. Current through the circuit is determined by the difference in
voltage between the battery and the capacitor, divided by the resistance. As the capacitor voltage approaches the
battery voltage (V), the current approaches zero. Once the capacitor voltage has reached V, the current will be
exactly zero.
The variation of current in the circuit is shown in Fig. 5.17.
1.0
Current
0.5
0.37
0 .0
0 s 0. 5s 1 .0s 1 .5s 2 .0s 2 .5s 3 .0 s 3.5 s 4. 0s 4. 5s 5 .0s
t
Fig. 5.17
The quantity
Time
RC
Variation of current with time in R-C series circuit with step input
RC is known as the time-constant of the circuit and it is defined as follows.
Definitions of Time-Constant ( )
1. It is the time in which the current decays to 37% of its initial value.
V
Also, at t 5 , i = 0.07 ; the transient is therefore, said to be practically disappeared in five time
R
constants.
V − t RC
e
at t 0, intersects the time axis at t
RC.
R
Thus, time-constant is the time in which the current would reach the steady-state zero value if the current
decays at the initial rate.
Physically, time-constant represents the speed of the response of a circuit. A low value of time-constant represents a fast response and a high value of time-constant represents a sluggish response.
2. The tangent to the equation i =
Calculations of the Voltage Across Elements Voltage across the resistor, VR
t
Voltage across the capacitor, VC =
t
−t
1
1 V −t
⎛
i(t )dt = ∫ e RC dt = V ⎜ 1 − e RC ⎟⎞
∫
⎠
⎝
C0
C0R
Ri(t)
Ve t/RC
257
Laplace Transform and Its Applications
RC Series Circuit with Impulse Input With zero initial condition, q(0 )
0, KVL equation becomes,
t
Ri(t ) +
1
i(t )dt = V (t )
C ∫0
RI ( s ) +
I (s)
=V
Cs
V
I (s) =
R+ 1
or,
Taking inverse Laplace transform, i(t ) =
⎡
⎤
⎛
⎞
1
⎢
⎥
V⎜
s
V
RC
⎟ = ⎢1 −
=
⎥
1 ⎥
R⎜ s+ 1 ⎟ R ⎢
s+
⎝
Cs
RC ⎠
⎢⎣
RC ⎥⎦
1 − t RC ⎤
V⎡
(t ) −
e
⎢
⎥ , for t ≥ 0
R⎣
RC
⎦
Voltage across the resistor,
⎡
1 − t RC ⎤
VR = Ri(t ) = V ⎢ (t ) −
e
⎥
RC
⎣
⎦
Voltage across the capacitor,
VC = V (t ) − VR =
{
V
e
} RC
These variations of the voltages are shown in
Fig. 5.18.
RC Series Circuit with Sinusoidal Input
Here, the input voltage is given as v (t) = V sin t
By KVL,
−t
RC
V/RC
Voltage across C
Voltage
across
R and C
t
Ri(t ) +
or,
or,
where,
1
i(t )dt = V sin t , with q(0− ) = 0
C ∫0
⎡
1⎤
V
I (s) ⎢ R + ⎥ = 2
Cs
⎣
⎦ s +
I (s) =
V Cs
( s + )(1+ sRC )
2
2
⎧
⎪
A1 = ⎨ s − j
⎪
⎩
(
)
⎧
⎪
A2 = ⎨ s + j
⎪
⎩
)
(
Time
Fig. 5.18 Variation of voltages with time in R-C series
circuit with impulse input
2
=
Voltage across R
V/RC
⎧
V ⎪
⎨
R ⎪ s+ j
⎩
(
1
( s + j )( s − j )(
1
( s + j )( s − j )(
s
)( s − j )(
⎫
⎡
⎤
A3 ⎥
A2
⎪ V ⎢ A1
=
+
+
⎬
⎢
s+ j
s + R ⎥⎥
s+ 1
⎪ L ⎢s− j
⎣
L⎦
RC ⎭
)
⎫
RC
⎪
=
⎬
2 1 + j RC
s+ 1
⎪
RC ⎭s = j
)
(
)
⎫
RC
⎪
=−
⎬
2
1
j RC
−
1
s+
⎪
RC ⎭s =− j
)
(
)
258
Network Analysis and Synthesis
and
⎧
⎪
A3 = ⎨ s + 1
RC s + j
⎪
⎩
)(
(
1
)( s − j )(
⎡
⎢
V ⎢
RC
∴ I (s) =
R ⎢ 2 1 + j RC s − j
⎢
⎣
)(
(
−
⎫
⎪
=
⎬
⎛
s+ 1
⎪
RC ⎭s =− 1
⎜⎝
RC
)
RC
2 1 − j RC s + j
) (
)(
1
RC
1 ⎞
2
+ 2 2⎟
RC ⎠
−
⎤
1
⎥
RC
⎥
+
⎥
⎛ 2
1 ⎞
1
⎜⎝ + R 2C 2 ⎟⎠ s + RC ⎥
⎦
−
)
)
(
Taking inverse Laplace transform,
t
−
⎡
⎤
Ve RC
ej t
e− j t
i (t ) = V C ⎢
−
⎥−
⎛
1 ⎞
⎢⎣ 2 1 + j RC 2 1 − j RC ⎦⎥
RC ⎜ R 2 + 2 2 ⎟
⎝
C ⎠
) (
(
)
⎡
⎤
−t
V ⎢ ej t
e− j t ⎥
Ve RC
= ⎢
−
−
⎥
2 j R+ 1
⎛
1 ⎞
R− 1
⎢
⎥
RC ⎜ R 2 + 2 2 ⎟
j
C
j
C
⎣
⎦
⎝
C ⎠
⎛
⎛
1 ⎞ ⎛
j ⎞
1 ⎞ ⎛
j ⎞
−j
j
⎜⎝ R + j C ⎟⎠ = ⎜⎝ R − C ⎟⎠ = Ze and ⎜⎝ R − j C ⎟⎠ = ⎜⎝ R + C ⎟⎠ = Ze so that,
Let,
⎛
Z = ⎜ R2 +
⎝
⎛ 1 ⎞
1 ⎞
and = tan −1 ⎜
2⎟
⎠
⎝ RC ⎟⎠
C
2
Putting these values,
i (t ) =
− j t+
j t+
−t
−t
V ⎡e ( ) −e ( ) ⎤
V
V ⎡ e j t e− j t ⎤
V
RC
−
=
e RC
−
e
⎢
⎥−
⎢ −j
j ⎥
2
2
2 j ⎣ Ze
Z ⎢⎣
2j
Ze ⎦ CZ
⎥⎦ CZ
or, finally, the current is,
i (t ) =
V
sin
Z
V
e
( t + ) − CZ
−t
RC
2
⎛ 1 ⎞
From this result, it is clear that the current in RC series circuit leads the voltage by an angle, = tan −1 ⎜
.
⎝ RC ⎟⎠
If the resistance R 0, then = 90 as is the case for a perfect capacitor.
Example 5.14 Find the current i(t) for the circuit shown in Fig. 5.19, if the
voltage source is v(t) 5e 2t u(t) and vc(0 ) 0.
t
Solution
()
Switch
1
Applying KVL for the loop, Ri(t ) +
1
i(t )dt = v (t ) = 5e −2 t u t
C ∫0
v( t )
RI ( s ) +
1 ⎡ I ( s ) q(0− ) ⎤
5
=
+
⎢
⎥
C⎣ s
s ⎦ s+2
Fig. 5.19 Circuit of
Example 5.14
Taking Laplace transform,
i(t )
1F
259
Laplace Transform and Its Applications
⎡ 1⎤
5
I ( s ) ⎢1 + ⎥ =
⎣ s⎦ s+2
(since vc(0 )
0)
I (s) =
(
5s
10
5
=
−
s +1 s + 2 s + 2 s +1
)(
)
Taking inverse Laplace transform,
10e 2t
i(t )
5e t ; for t
0
RLC Series Circuit
RLC Series Circuit with Step Input
R
With zero initial conditions, the Kirchhoff’s voltage law equation becomes,
L
C
t
Ri(t ) + L
or,
di(t ) 1
+ ∫ i(t )dt = Vu(t )
dt
C0
RI ( s ) + sLI ( s ) +
Vu(t)
V
1
I (s) =
Cs
s
Fig. 5.20
i(t)
RLC series circuit
V
L
R
1
s + s+
L
LC
The roots of the denominator polynomial of equation are,
I (s) =
or,
(5.2)
2
s2 +
or,
s1 = −
Let
0
Then,
1
R
R2
+
−
2
2L
4 L LC
1
=
and
0
LC
s1 = −
R
1
s+
=0
L
LC
0
+
2
0
=
and,
s2 = −
i.e.
=
R
2L
−1
R C
2 L
s2 = −
and
1
R
R2
−
−
2
2L
4 L LC
0
−
damping ratio
2
0
−1
V
So,
I (s) =
(
A
B
L
=
+
s − s1 s − s2
s − s1 s − s2
(
A = s − s1
and
(
B = s − s2
)(
)(
)
V
L
s − s1 s − s2
)(
)(
V
)
=
s = s1
(
L =
s1 − s2 2
V
L
s − s1 s − s2
)(
)
V
0
V
)
=
s = s2
(
L =−
s2 − s1
2
)
2
L
−1
V
0
L
2
−1
260
Network Analysis and Synthesis
Putting these values of A and B, we get,
V
I (s) =
2 0L
⎡ 1
1 ⎤
−
⎢
⎥
2
− 1 ⎣ s − s1 s − s2 ⎦
Taking inverse Laplace transform,
V
i (t ) =
2 0L
2
⎡ e s1t − e s2t ⎤ =
⎣
⎦
−1
V
2
2 0L
−1
e
−
0t
⎡ ⎛⎜⎝ 0
⎢e
⎣
2
−1⎞⎟ t
⎠
− ⎛⎜
−e ⎝
2
0
−1⎞⎟ t
⎠
⎤
⎥
⎦
Depending upon the values of R, L and C, three cases may appear:
R
1
>
(overdamped condition)
2L
LC
R
1
<
(underdamped condition)
(b)
2L
LC
(a)
0
L
1
0
)
and 0 =
R
LC
Under this condition, the current becomes
2 0L
2
−1
−
0t
⎡ ⎛⎝⎜ 0
⎢e
⎣
2
−1⎞⎟ t
⎠
0
1
2
3
4 5
6 7
Time (seconds)
− ⎛⎜
−e ⎝
2
0
−1⎞⎟ t
⎠
⎤
⎥=
⎦
V
L
0
2
−1
e
−
0t
sinh
The graphical plot for the current is shown in Fig. 5.21.
R
1
=
or
Critically Damped Condition The condition is
2L
LC
From the equation (5.2),
V
I (s) =
Taking inverse Laplace transform i(t ) =
L
s2 + 2 0 s +
=
2
0
= 1 or Q =
⎛
⎞
V
1
⎜
⎟
2
L⎜ s+
⎟⎠
⎝
0
(
)
V − 0t
te
L
The graphical plot for the current is shown Fig. 5.21.
Underdamped Condition The condition is
8
9
10
Fig. 5.21 Current response in RLC series circuit for
three different damping conditions
(since, quality factor, Q =
e
Overdamped condition
0.2
0.1
V
Critically damped condition
0.3
0.1
R
1
>
2L
LC
1
> 1 or Q <
2
i (t ) =
Underdamped condition
0.4
R
1
(critically damped condition)
=
2L
LC
Overdamped Condition The condition is
or
0.5
Amplitude
(c)
0.6
1
R
<
or
2L
LC
< 1 or Q >
1
2
1
2
(
2
0
)
−1 t
261
Laplace Transform and Its Applications
So, the current becomes
⎡ ⎛⎜ j 0 1− 2 ⎞⎟ t −⎛⎜ j 0 1− 2 ⎞⎟ t ⎤
⎠
⎠ ⎥
⎢ ⎝
V
−e ⎝
t e
−
e 0 ⎢
⎥
2j
⎢
⎥
L 1− 2
0
⎢⎣
⎥⎦
⎡ ⎛⎜ 0 2 −1 ⎞⎟ t −⎛⎜ 0 2 −1 ⎞⎟ t ⎤
V
− 0t
⎠ ⎥
⎠
⎢e⎝
i (t ) =
−e ⎝
=
e
⎢
⎥
2 0 L 2 −1
⎣
⎦
V
=
0
L 1−
2
e
−
0t
sin
(
)
1− 2 t
0
So, the circuit is oscillatory. When R 0,
0, the oscillations are undamped or sustained. The frequency of
the undamped oscillation ( 0) is known as undamped natural frequency.
RLC Series Circuit with Impulse Input
With zero initial conditions, the Kirchhoff’s voltage law equation becomes
t
Ri(t ) + L
di(t ) 1
+ ∫ i(t )dt = V (t )
dt
C0
RI ( s ) + sLI ( s ) +
or,
V s
(
L)
I (s) =
or,
s2 +
R
1
s+
L
LC
The roots of the denominator polynomial of equation are
s2 +
or,
s1 = −
Let
0
Then,
So,
1
=
and
0
LC
s1 = −
I (s) =
1
R
R2
+
−
2
2L
4 L LC
0
+
2
0
(V L )s
( s − s )( s − s )
1
2
=
s2 = −
i.e.
=
R
2L
−1
=
and,
0
−
V s
(
)
= L =
(s − s ) 2 L
2
1
2
(
(V )s
) ( s − s )L( s − s )
2
1
=
2
(V L )s = −
2
(s − s )
2
s = s2
2
0
s = s1
B = s − s2
−1
Vs1
1
( s − s )( s − s )
1
and
2
0
A
B
+
s − s1 s − s2
V s
(
L)
A = (s − s )
1
1
R
R2
−
−
2
2L
4 L LC
R C
= damping ratio
2 L
s2 = −
and
1
R
=0
s+
L
LC
1
−1
Vs2
2 0L
2
−1
1
I (s) = V
Cs
(5.3)
262
Network Analysis and Synthesis
Putting these values of A and B, we get,
⎡ s1
s ⎤
− 2 ⎥
⎢
2
− 1 ⎣ s − s1 s − s2 ⎦
V
I (s) =
2 0L
Taking inverse Laplace transform,
V
i (t ) =
2
2 0L
−1
V
=
2
2 0L
−1
⎡ s1e s1t − s2 e s2t ⎤ =
⎣
⎦
e
−
0t
(
⎡
⎢ −
⎣
+
0
V
2 0L
2
0
2
−1
)
⎛
⎜
e
−
2
−1 e⎝
0
0t
−1⎞⎟ t
⎠
⎡ ⎛⎜⎝ 0
⎢ s1e
⎣
(
− −
2
−1⎞⎟ t
⎠
−
0
− ⎛⎜
− s2 e ⎝
2
0
2
0
)
−1⎞⎟ t
⎠
− ⎛⎜
⎤
⎥
⎦
2
−1 e ⎝
0
−1⎞⎟ t
⎠
Three cases are considered:
R
1
(a)
(overdamped condition)
>
2L
LC
(b)
R
1
<
(underdamped condition)
2L
LC
(c)
R
1
=
(critically damped condition)
2L
LC
Overdamped Condition Here,
The current becomes
V
i (t ) =
2
2 0L
V
=
2
L
−1
e
⎡
− 1 ⎣⎢
−
2
1
0t
(
⎡
⎢ −
⎣
− 1 cosh
+
0
(
2
0
2
0
)
I (s) =
where,
(
A= s+
0
)
s
2
(s + )
B=
⎡
d ⎢
s+
ds ⎢
⎣
(
0
)
=− 0
2
0
and,
s2 + 2 0 s +
s =−
⎤
⎥
=1
2
s + 0 ⎥⎦
s =− 0
s
2
(
0
)
=
2
0
2
0
)
− 1 t − sinh
Critically damped condition The condition is
From the equation (5.3),
(V L )s
⎛
⎜
−1 e⎝
(
−1⎞⎟ t
⎠
(
− −
2
0
−
0
2
0
)
)
−1 t ⎤
⎦⎥
1
⎛
⎞
⎡
⎤
V
s
V⎢
A
B ⎥
⎜
⎟
=
+
2
2
L⎜ s+
s+ 0 ⎥
⎟⎠ L ⎢ s +
⎝
0
0
⎣
⎦
(
)
(
)
− ⎛⎜
−1 e ⎝
2
0
−1⎞⎟ t
⎠
⎤
⎥
⎦
⎤
⎥
⎦
263
Laplace Transform and Its Applications
⎡
( ) VL ⎢⎢ s +1 −
I s =
So,
⎣
V
⎡1 −
L⎣
Taking inverse Laplace transform, i(t ) =
0
0
t ⎤⎦ e
−
Underdamped Condition The condition is,
So, the current becomes
V
i (t ) =
2
2 0L
=
−1
V
2 0 Lj 1 − 2
i (t ) =
=
V
L 1−
2
V
L 1− 2
e
e
−
−
e
−
e
t
t
⎡
⎢
⎢⎣
0t
−
(
2
0
(s + )
2
0
⎤
⎥
⎥
⎦
0t
1
)
⎧ ⎛⎜ 0
− 1 ⎨e ⎝
⎩
2
−1⎞⎟ t
⎠
− ⎛⎜
2
+e ⎝
0
−1⎞⎟ t
⎠
⎪⎫
⎬−
⎭
0
⎧ ⎛⎝⎜ 0
⎨e
⎩
)
(
⎡
⎧ ⎛⎝⎜ j 0 1− 2 ⎞⎠⎟ t −⎛⎝⎜ j 0 1− 2 ⎞⎠⎟ t ⎪⎫
2
+e
⎢ j 0 1−
⎬−
⎨e
⎢⎣
⎭
⎩
0t
⎡ 1 − 2 cos
⎢⎣
cos
0
{(
(
)
1 − 2 t − sin
0
(
0
0
2
−1⎞⎟ t
⎠
− ⎛⎜
−e ⎝
2
0
−1⎞⎟ t
⎠
⎪⎫ ⎤
⎬⎥
⎭ ⎥⎦
⎧ ⎛⎝⎜ j 0 1− 2 ⎞⎠⎟ t −⎛⎝⎜ j 0 1− 2 ⎞⎠⎟ t ⎪⎫ ⎤
−e
⎬⎥
⎨e
⎭ ⎥⎦
⎩
)
1− 2 t ⎤
⎥⎦
) }
1− 2 t +
0
⎛ 1− 2 ⎞
, wheree = tan −1 ⎜
⎟
⎟⎠
⎜⎝
RLC Series Circuit with Sinusoidal Input Sinusoidal
voltage v(t ) Vm sin( t
) is applied to a series RLC circuit at time t 0. We want to find the complete solution for
the current i(t ) using Laplace transform method.
L
R
v(t ) V msin(vt
u)
C
i(t )
t
By KVL, Ri(t ) + L
di(t ) 1
+ ∫ i(t )dt = Vm sin( t + )
dt
C −∞
Fig. 5.22 RLC series circuit with sinusoidal input
Taking Laplace transform with zero initial conditions,
(
s sin + cos
⎡
1⎤
I ( s ) ⎢ R + sL + ⎥ = Vm
Cs ⎦
s2 + 2
⎣
or,
I (s) =
(
Vm s s sin + cos
(
L s2 +
2
)
)⎛⎜⎝ s + RL s + LC1 ⎞⎟⎠
2
=
(
⎛ 2 R
1 ⎞
⎜⎝ s + L s + LC ⎟⎠ = 0
1
R
R2
+
−
2
2L
LC
4L
and,
s2 = −
)
)(
s s sin + cos
Vm
L s + j s − j s − s1 s − s2
where, s1, s2 are the roots of the quadratic equation:
Thus, s1 = −
(
)(
)
1
R
R2
−
−
2
2L
LC
4L
)(
)
264
Network Analysis and Synthesis
Now, let
K
K
K
K
(
)
=
+
+
+
( s + j )( s − j )( s − s )( s − s ) s − s s − s s + j s − j
s s sin + cos
1
1
2
3
2
1
4
2
So, by residue method, multiplying by (s – s1) and putting s s1,
s1 s1 sin + cos
s2 s2 sin + cos
K1 =
and K 2 =
s1 + j s1 − j s1 − s2
s2 + j s2 − j s2 − s1
(
(
Similarly, multiplying by (s
)(
)
)(
j ) and putting s
)
(
(
–j ,
)(
)
(cos − j sin )
( − j − j )( − j − s )( − j − s ) 2( s + j )( s + j )
j ( − sin + cos )
(coos + j sin )
K =
=
( j + j )( j − s )( j − s ) 2( s − j )( s − j )
(
−j
K3 =
)
)(
)
− j sin + cos
=
1
and,
2
1
2
4
1
2
1
2
Hence the current response becomes,
V
V
st
st
i(t ) = m ⎡⎣ K1e 1 + K 2 e 2 ⎤⎦ + ⎡⎣ K 3 e − j t + K 4 e j t ⎤⎦ = I tr + I ss
L
L
Thus, the transient part of the total current is
⎤
⎡
⎥
⎢ s s sin + cos
s2 s2 sin + cos
Vm ⎢ 1 1
s1t
s2 t ⎥
I tr =
e −
e
⎥
L⎢ 2
R2 4
R2 4
2
2
⎢ s1 + 2
⎥
−
+
−
s
2
⎢⎣
⎥⎦
L2 LC
L2 LC
)
(
)
(
)
(
)
(
The steady-state part of the total current is obtained as follows.
− j t+
⎤ V ⎡
V ⎡
e− j e− j t
ej ej t
e ( )
I ss = m ⎢
+
⎥= m ⎢
2 L ⎢⎣ s1 + j s2 + j
s1 − j s2 − j ⎥⎦ 2 L ⎢⎣ s1 + j s2 + j
)(
(
I ss =
or,
=
=
or,
Vm
(
2
(
2
2 L s1 +
Vm
2 L s1 +
Vm
L
I ss =
Vm
L
V
= m
L
2
2
)(
) (
s2 2 +
2
)
(
⎡ ss −
2
( s + )( s + )
2
2
1
2
2
⎡ R
sin
⎢
⎣ L
(
( s + )(
2
2
⎡⎛ 1
−
⎢⎜
⎣⎝ LC
( t + ) − ⎛⎜⎝
1
1
1 2
2
2
)(
(
) (
j t+
e( )
s1 − j s2 − j
)
− j s1 − j s2 ⎤
⎦
)2 cos( t + ) − ( s + s )2 sin ( t + )⎤⎦
1
2
2
1
2
)
⎡ e − j( t + ) s s −
1 2
⎣
)( s + ) ⎣(
2
)(
+
s12 +
2
2
)(
−
2
⎞
⎟⎠ cos
1 ⎞
cos
LC ⎟⎠
s2 2 +
2
)
⎛
⎤
( t + ) − ⎜⎝ − LR ⎞⎟⎠ sin ( t + )⎥
⎦
⎤
( t + )⎥
⎦
⎧
⎛
1 ⎞⎫
2
L−
⎪⎪
⎜
⎟ ⎪⎪
⎛
1 ⎞
−1
2
C
R +⎜ L−
sin ⎨ t + − tan ⎜
⎟ ⎬×
R
C ⎟⎠
⎝
s2 2 + 2
⎪
⎜
⎟⎪ L
⎝
⎠
⎪⎩
⎪⎭
)
)(
)
⎤
⎥
⎥⎦
265
Laplace Transform and Its Applications
⎧
⎛
1 ⎞⎫
L−
⎪
⎪⎪
⎜
C⎟⎪
I ss =
sin ⎨ t + − tan −1 ⎜
⎬
⎟
2
R
⎪
⎪
⎜
⎟
⎛
⎞
1
⎝
⎠ ⎪⎭
R2 + ⎜ L −
⎪⎩
⎟
C⎠
⎝
Vm
or,
This gives the steady-state current of the series RLC circuit to a sinusoidal voltage.
Example 5.15 Determine the current i(t) in a series RLC circuit consisting of R 5 , L 1 H and C ¼ F
when the source voltage is given as (a) ramp voltage 12r(t 2), and (b) step voltage 3u(t − 3). Assume that
the circuit is initially relaxed.
Solution Applying KVL for the series RLC circuit we get,
di t
()
di t
( )+ 1
⇒ 5i t +
(a) When v(t )
12r(t
( ) + 1 i t dt = v t
() ()
dt
C∫
()
Ri t + L
2)
()
5i t +
( )+ 1
di t
1
dt
( ) s s + 5s + 4 = 12e
⇒ I s =
(
)
2
3u (t
4
∫ i (t )dt = v (t )
4
()
−2 s
⎡1
1
1 ⎤
⎤
⎡
3e −2 s 4 e −2 s e −2 s
1
−2 s ⎢
3
4
−
+ 12 ⎥ =
−
+
⎢
⎥ = 12 e
⎢ s s +1 s + 4 ⎥
s
s +1 s + 4
⎢⎣ s s + 1 s + 4 ⎥⎦
⎢⎣
⎥⎦
)(
(
)
Taking inverse Laplace transform, we get
i(t ) 3u(t
(b) When v (t )
∫ i (t )dt = v (t )
⎛
4⎞
12 −2 s
⎜⎝ 5 + s + s ⎟⎠ I s = s 2 e
Taking Laplace transform,
12 e −2 s
1
dt
3)
()
5i t +
4e (t
2)
( )+ 1
di t
1
dt
2)
e 4(t
2)
∫ i (t )dt = v (t )
4
⎛
4⎞
3 −3s
Taking Laplace transform, ⎜ 5 + s + ⎟ I s = e
s⎠
s
⎝
()
()
⇒ I s =
(
⎡ 1
1 ⎤
−3 s
−3 s
⎡
⎤
1
−3 s
−3 s ⎢
3 − 3 ⎥= e − e
e
3
=
3
e
=
⎥
⎢
⎢ s + 4 s + 1 ⎥ s + 4 s +1
s 2 + 5s + 4
⎢⎣ s + 1 s + 4 ⎦⎥
⎢⎣
⎥⎦
3e −3s
)
(
)(
)
Taking inverse Laplace transform, we get
i(t )
e (t
3)
e 4(t
3)
266
Network Analysis and Synthesis
5.12.4
RLC Parallel Circuit
RLC Parallel Circuit with Step Current Input
R
Iu(t)
With zero initial conditions, the Kirchhoff’s current law equat
v (t )
dv (t ) 1
+C
+ ∫ v (t )dt = Iu(t )
tion becomes
R
dt
L0
L
C
v(t)
Fig. 5.23 RLC parallel circuit
V (s)
1
I
+ sCV ( s ) + V ( s ) =
R
sL
s
or,
I
C
1
1
s +
s+
RC
LC
The roots of the denominator polynomial of equation are
V (s) =
or,
(5.4)
2
s2 +
or,
s1 = −
Let
0
Then,
1
1
1
+
−
2 2
2 RC
LC
4R C
1
=
s1 = −
+
0
V (s) =
2
0
−1
and
s2 = −
=
A
C
+
( s − s )( s − s ) s − s
1
(
∴ A = s − s1
and
i.e.
LC
I
So,
1
2 RC
0
(
∴ B = s − s2
1
2
)(
)(
1 L
2R C
0
−
damping ratio
2
0
−1
I
=
)
s = s1
(
I
C =
s1 − s2 2 C
0
I
) ( s − s )(Cs − s )
1
=
1
1
1
−
−
2 2
2 RC
LC
4R C
B
s − s2
I
C
s − s1 s − s2
s2 = −
and,
=
and
1
1
=0
s+
RC
LC
2
)
2
−1
I
=
s = s2
(
I
C =−
s2 − s1
2 0C
)
2
−1
Putting these values of A and B, we get,
I
V (s) =
2 0C
2
⎡ 1
1 ⎤
−
⎢
⎥
− 1 ⎣ s − s1 s − s2 ⎦
Taking inverse Laplace transform,
I
v (t ) =
2 0C
2
−1
⎡ e s1t − e s2t ⎤ =
⎣
⎦
I
2 0C
2
−1
e
−
0t
⎡ ⎛⎝⎜ 0
⎢e
⎣
2
−1⎞⎟ t
⎠
− ⎛⎜
−e ⎝
2
0
−1⎞⎟ t
⎠
⎤
⎥
⎦
267
Laplace Transform and Its Applications
Depending upon the values of R, L and C, three cases may appear:
0.6
1
1
>
(a)
(overdamped condition)
2 RC
0.5
LC
(c)
0.4
1
1
<
(underdamped condition)
2 RC
LC
Critically damped condition
Amplitude
(b)
1
1
=
(critically damped condition)
2 RC
LC
0.3
Overdamped condition
0.2
0.1
0
0.1
0
Overdamped Condition The condition is,
1
1
>
or,
2 RC
LC
1
> 1 or Q <
2
1
1
and 0 =
)
RC
LC
0
Under this condition, the current becomes
I
2 0C
2
e
−1
−
0t
⎡ ⎛⎝⎜ 0
⎢e
⎣
2
−1⎞⎟ t
⎠
1
2
3
4 5
6 7
Time (seconds)
− ⎛⎜
2
−e ⎝
0
−1⎞⎟ t
⎠
⎤
⎥=
⎦
I
C
0
2
−1
e
−
0t
sinh
(
2
0
)
−1 t
The graphical plot for the voltage is shown in Fig. 5.24.
Critically Damped Condition The condition is
I
From the equation (5.4), V ( s ) =
C
s2 + 2 0 s +
=
2
0
1
1
=
or
2 RC
LC
= 1 or, Q =
1
2
⎛
⎞
I
1
⎜
⎟
2
C⎜ s+
⎟⎠
⎝
0
(
)
I − 0t
te
C
The graphical plot for the voltage is shown in Fig. 5.24.
Taking inverse Laplace transform, v (t ) =
Underdamped Condition The condition is
1
1
<
or
2 RC
LC
< 1 or Q >
1
2
So, the voltage becomes,
I
v (t ) =
2 0C
2
−1
I
=
0
C 1−
2
e
e
−
−
0t
8
9
Fig. 5.24 Voltage response in RLC parallel circuit
for three different damping conditions
(since, quality factor, Q =
v (t ) =
Underdamped condition
⎡ ⎛⎜ 0 2 −1 ⎞⎟ t −⎛⎜ 0 2 −1 ⎞⎟ t ⎤
⎠
⎠ ⎥
⎢e⎝
−e ⎝
⎢
⎥
⎣
⎦
⎡ ⎛⎜ j 0 1− 2 ⎞⎟ t −⎛⎜ j 0 1− 2 ⎞⎟ t ⎤
⎠ ⎥
⎠
⎢ e⎝
−e ⎝
0t
⎢
⎥=
2j
⎢
⎥
⎢⎣
⎥⎦
I
0
C 1−
2
e
−
0t
sin
(
0
)
1− 2 t
10
268
Network Analysis and Synthesis
Similarly, we can find out the impulse response and sinusoidal response of a parallel RLC circuit using Laplace
transform method as for the series RLC circuit.
Example 5.16 For the RC parallel circuit shown in Fig. 5.25, determine the voltage across the capacitor using Laplace transform method.
Assume the capacitor to be initially relaxed.
( ) + C dv (t ) = i t =10
()
R
dt
10 A
1F
5
v t
Solution Applying KCL at the upper node,
Fig. 5.25 Circuit of Example 5.16
Taking Laplace transform and putting the values of R and C,
( ) + sV s = 10
() s
5
V s
⎡
⎤
⎢1
1 ⎥
= 50 ⎢ −
⇒ V s =
⎥
⎛ 1⎞
⎢s s+ 1 ⎥
s⎜ s + ⎟
⎢⎣
5 ⎥⎦
⎝ 5⎠
()
10
Taking inverse Laplace transform, we get
−t
⎛
v t = 50 ⎜ 1 − e 5 ⎞⎟ V
⎠
⎝
()
5.13
( )
RESPONSE WITH PULSE INPUT VOLTAGE
5.13.1
RC Series Circuit
v( t )
If a voltage pulse of width T as shown in Fig. 5.26 is applied to an RC series circuit
1
then by KVL, Ri(t ) + ∫ i(t )dt = v(t )
C
Taking Laplace transform with zero initial condition,
RI ( s ) +
1
V Ve − sT
I (s) = −
Cs
s
s
Taking inverse Laplace transform,
or, I ( s ) =
i (t ) =
V 1 − e − sT
R s+ 1
RC
V
0
T
t
Fig. 5.26 Pulse Voltage
V ⎡ − t RC −(t −T ) RC ⎤
−e
⎥
⎢e
R⎣
⎦
( t −T ) ⎤
⎡ −t
−
RC
Hence the voltage across the resistance is given as vR (t ) = Ri(t ) = V ⎢ e RC − e
⎥
⎦
⎣
( t −T ) ⎤
⎡
−t
−
RC
and the voltage across the capacitor is given as vc (t ) = V − vR (t ) = V ⎢1 − e RC + e
⎥
⎦
⎣
To plot the two voltages with varying time, we have the following observations:
i. At t
0, all the voltage appears across the resistance R and thus,
vR V and vC 0
ii. As the time increases, the voltage vC grows and the voltage vR decays exponentially, with time-constant
RC.
269
Laplace Transform and Its Applications
Voltage
across
R and C
Vo l t a g e a c r o s s C
Vo l t a g e a c r o s s R
Time
Fig. 5.27
Voltage response of RC series circuit with pulse input
iii. At t T, voltage across the network drops abruptly to zero from V. Again this entire drop is instantaneously felt across the resistance R.
iv. For time t T, total voltage across the circuit is zero. So, at any instant of time t, vR (t) vC (t) 0 and
both vR and vC asymptotically approach zero.
Case (1): If time-constant (
RC) << pulse-width ( T ) The voltage across the resistance vR will consist of two trigger pulses, one positive and the other negative, of height V at the points where the voltage across
the network changes abruptly (i.e., t 0 and T ).
In this case, the voltage across capacitor attains the steady state very quickly, i.e., vc V.
vR = Ri = RC
dvC
dV
≈ RC
dt
dt
or,
vR = RC
dV
dt
Thus, the voltage vR is the differentiation of the input voltage and hence the circuit acts as a differentiator.
1.0V
Voltage
across C
0.5V
Voltages
0V
Voltage
across R
0.5V
1.0V
0s
2s
4s
6s
8s
10s 12s 14s 16s 18s 20s
Time
Fig. 5.28 Voltage response of RC series circuit ( RC
pulse input
T) with
Case (2): If time-constant (
RC) >> pulse-width (T ) In this case, the voltage across the capacitor
varies with time almost linearly and the value is far from the steady state value V; i.e., vR V.
t
vC =
t
t
1
1 v
1
idt = ∫ R dt ≈
Vdt
∫
C0
C0 R
RC ∫0
t
or,
vC ≈
1
Vdt
RC ∫0
Thus, the voltage vC is the integration of the input voltage and hence the circuit acts as an integrator.
270
Network Analysis and Synthesis
1.0V
0.8V
Voltage across R
0.6V
Voltages
0.4V
across
resistor and 0.2V
capacitor
0V
0.2V
0s
Voltage across C
2s
4s
6s
8s 10s 12s 14s 16s 18s
Time
Fig. 5.29 Voltage response of RC series circuit ( RC
5.13.2
20S
T) with pulse input
RL series circuit
di
If a similar pulse voltage is applied to an RL series circuit then the KVL equation will be, Ri(t ) + L = v (t )
dt
Taking Laplace transform with zero initial condition,
V Ve − sT
RI ( s ) + sLI ( s ) = −
s
s
Taking inverse Laplace transform, i(t ) =
or,
⎡
⎤
V⎢
e − sT ⎥
1
I (s) = ⎢
−
⎥
L s s+ R
s s+ R ⎥
⎢⎣
L
L ⎦
) (
(
)
R
R
⎤
⎛
− t⎞
− ( t −T ) ⎞
V ⎡⎛
u(t − T ) ⎥
⎢ ⎜ 1 − e L ⎟ u(t ) − ⎜ 1 − e L
⎟
R ⎢⎣⎝
⎠
⎠
⎝
⎥⎦
The variation of the two voltages is shown in Fig. 5.30.
1.0V
Voltage across R
Voltage
across
resistor
and
inductor
0.5V
0V
0.5V Voltage across L
1.0V
0s 0.5s 1.0s 1.5s 2.0s 2.5s 3.0s 3.5s 4.0s 4.5s 5.0s
Time
Fig. 5.30
Voltage response of RL series circuit with pulse input
Case (1): If time-constant (
L /R ) << pulse-width ( T ) In this case, the voltage across resistor attains
the steady state very quickly, i.e., vR V.
vL = L
di
d ⎛v ⎞
d ⎛ V ⎞ L dV
=L ⎜ R⎟=L ⎜ ⎟≈
dt
dt ⎝ R ⎠
dt ⎝ R ⎠ R dt
or,
vL =
L dV
R dt
Thus, the voltage v L is the differentiation of the input voltage and hence the circuit acts as a
differentiator.
271
Laplace Transform and Its Applications
1.0V
0.5V
voltage
across
0V
resistors
inductor 0.5V
1.0V
0s
Fig. 5.31
Voltage across R
Voltage across L
2s
4s
6s
8s 10s 12s
Time
14s 16s 18s 20s
Voltage response of RL series circuit ( L/R
T) with pulse input
Case (2): If time-constant (
L /R ) >> pulse-width ( T ) In this case, the voltage across the resistor
varies with time almost linearly and the value is far from the steady-state value V; i.e., vL V.
t
t
t
R
1
vL dt ≈ ∫Vdt
∫
L0
L0
vR = Ri = R
or,
∴ vR ≈
R
Vdt
L ∫0
Thus, the voltage vR is the integration of the input voltage and hence the circuit acts as an integrator.
1.0V
Voltages 0.8V
0.6V
Voltage across L
Voltage across R
0.4V
0.2V
0V
0.2V
0s 2s
Fig. 5.32
5.14
4s
6s
8s 10s 12s
Time
14s 16s 18s 20s
Voltage response of RL series circuit ( L/R
T) with pulse input
STEPS FOR CIRCUIT ANALYSIS USING LAPLACE TRANSFORM METHOD
1. All circuit elements are transformed from time-domain to Laplace domain with initial conditions.
2. Excitation function is transformed into Laplace domain.
3. The circuit is solved using different circuit analysis techniques, such as mesh analysis, node analysis, etc.
4. Time domain solution is obtained by taking inverse Laplace transform of the solution.
5.15
CONCEPT OF CONVOLUTION THEOREM
Convolution Integral If h(t ) is the impulse response of a linear network then the response of the same
network y (t ) subject to any arbitrary input w (t ) is given by the convolution integral as
∞
∞
−∞
−∞
y (t ) = ∫ h( )w (t − )d = ∫ w ( ) h(t − )d
Thus, if the impulse response of any linear time-invariant system is known, we can obtain the zero-state response
of the system to any other type of input.
272
Network Analysis and Synthesis
Convolution Theorem
If f1(t ) and f2(t ) are two functions of time which are zero for t 0, and if their Laplace transforms are F1(s) and
F2(s), respectively then the convolution theorem states that the Laplace transform of the convolution of f1(t ) and
f2(t ) is given by the product F1(s) F2(s).
Mathematically, the convolution of f1(t ) and f2(t ) is written as
()
t
()
f1 t * f 2 t = ∫ f1
() (
)
f2 t −
0
Where
Proof
t
(
d = ∫ f1 t −
) f ( )d = f (t )* f (t )
2
2
1
0
()
()
() ()
is a dummy variable for time t, the convolution theorem is written as, L ⎡⎣ f1 t * f 2 t ⎤⎦ = F1 s F2 s
By the definition of convolution,
⎡t
⎤ ∞⎡ t
⎤
L ⎡⎣ f1 t * f 2 t ⎤⎦ = L ⎢ ∫ f1 f 2 t − d ⎥ = ∫ ⎢ ∫ f1 t − f 2 d ⎥ e − st dt
⎢⎣ 0
⎥⎦ 0 ⎢⎣ 0
⎥⎦
Also, by the definition of a shifted unit step function, using dummy variable,
u(t ) 1; for
t
1; for
t
()
()
t
(
∫ f1 t −
() (
)
) ()
d = ∫ f1 t −
f2
0
∞
(5.5)
)u ( t − ) f ( ) d
(
2
0
∞ ∞
⎡
Putting this in (5.5), we get, L ⎡⎣ f1 t * f 2 t ⎤⎦ = ∫ ⎢ ∫ f1 t −
0⎢
⎣0
Now, let (t
) x
dt dx,
()
) ()
(
()
(
t
⎤ − st
)u(t − ) f ( )d ⎥ e dt
2
⎥⎦
(5.6)
0
x
From (5.6), we get,
∞ ∞
⎡
L ⎡⎣ f1 t * f 2 t ⎤⎦ = ∫ ⎢ ∫ f1 x u x f 2
0⎢
⎣−
()
()
⎤ − s x+
( ) ( ) ( )d ⎥ e ( )dx
∞
()() ()
= ∫ f1 x u x f 2
−
⎥⎦
∞
e − sx dx ∫ f 2
()
∞
()
0
0
()
∞
d e − s d = ∫ f1 x e − sx dx ∫ f 2
()
0
( )e d { u( x ) = 0 for x < 0}
−s
() ()
∴ L ⎡⎣ f1 t * f 2 t ⎤⎦ = F1 s F2 s
Thus, the convolution in time domain becomes multiplication in the frequency domain, and vice-versa.
Application of Convolution Theorem The convolution theorem is used to find the response of a linear
system to any arbitrary excitation if the impulse response of the system is known.
We know that the transfer function is defined as the ratio of response transform to excitation transform with
zero initial conditions. Thus,
Laplace transform of response
Transfer function
Laplace transform of Excitation all initial conditions reduced to zero
273
Laplace Transform and Its Applications
Y (s)
() Ws
()
H s =
or,
IC =0
Thus, Y(s) H(s)W(s)
Here, W(s) L[w(t )], is the input Laplace transform and Y(s) L[y(t )], is the output Laplace transform.
Now, if the input is an impulse function then w(t )
(t ) or W(s) 1
Y(s) H(s)W(s) H(s)
Taking inverse Laplace transform,
y(t )
h(t )
Thus, h(t ) is the impulse response of the system. If this impulse response of the system is known, we can find
out the response of the system due to any arbitrary input w(t ) from the following relation:
()
t
t
0
0
( ) ( ) or y (t ) = h(t )* w (t ) = ∫ h( )w (t − )d = ∫ h(t − )w ( )d
Y s =H s W s
Example 5.17 Find the convolution integral when f1( t )
e at
and
f2 ( t )
t.
Solution Here, the convolution integral is given as
()
()
t
f1 t * f 2 t = ∫ e
( )
−a t−
t
t
d =e
0
− at
∫ e d =e
a
0
− at
t
⎡ ea
⎤
⎡ ea
ea
ea ⎤
− ∫ 1⋅ d ⎥ = e − at ⎢
− 2 ⎥
⎢
a
a ⎦0
⎣ a
⎦0
⎣ a
⎡ te at e at 1 ⎤ 1
= e − at ⎢
− 2 + 2 ⎥ = 2 ⎡⎣ at − 1 + e − at ⎤⎦ .
a
a
a ⎦ a
⎣
Solved Problems
Problem 5.1 (a) Find the initial value of the function whose Laplace transform is
V (s ) = A.
( s + a ) sin + b cos
( s + a )2 + b 2
Check the result by solving it for v(t).
I (s ) =
(b) Find the final value of the function whose Laplace Transform is
s +6
s (s + 3 )
Solution
(a) By initial value theorem,
⎛
a⎞
b
⎜⎝ 1 + s ⎟⎠ sin + s cos
s + a )sin + b cos
(
= Asin
= Lim A
V (0+ ) = Lim sV ( s ) = Lim sA
⎛ a⎞ ⎛ b⎞
( s + a) + b
+
1+
s →∞
s →∞
2
2
2
s →∞
⎝⎜
s ⎟⎠
2
⎜⎝ s ⎟⎠
274
Network Analysis and Synthesis
In order to check this result, we find v(t) and then put t 0.
⎡ s + a sin + b cos ⎤
⎡ s + a sin
⎤
b cos
⎥
⎥ = AL−1 ⎢
v (t ) = L−1 ⎢ A
+
2
⎢
⎢ s + a 2 + b2 s + a 2 + b2 ⎥
⎥
s + s + b2
⎣
⎣
⎦
⎦
(
(
= A ⎡⎣sin e
At t 0, v(0 ) Ae sin (0
(b) By final-value theorem,
− at
)
(
(
)
cos bt + cos e
− at
s →0
(
(
sin bt +
)
)
s →0
s+6
s+6
= Lim
=2
s
→
0
s s+3
s+3
(
)
(
)
⎡ s + 6 ⎤ −1 ⎡ 2
1 ⎤
−3t
i(t ) = L−1 ⎢
⎥= L ⎢ −
⎥=2−e
s
s
3
+
3
s
s
+
⎢⎣
⎥⎦
⎣
⎦
For checking it,
, i( )
− at
) = A sin
I (∞) = Lim sI ( s ) = Lim s
At t
sin bt ⎤⎦ = Ae
)
)
(
)
2 e =2
Problem 5.2 a) Obtain the Laplace transform of a square wave of unit amplitude and periodic time 2T,
as shown in Fig. 5.33 (a).
f(t )
f(t )
1
1
0
T
2T
3T
time
1
0
Fig. 5.33(a)
1/2
1
Fig. 5.33(b)
b) Find the Laplace Transform of the function, shown in Fig. 5.33 (b).
Solution
(a) The equation of the square wave is
f (t ) = u(t ) − u(t − T ) − u(t − T ) + u(t − 2T ) + u(t − 2T ) − u(t − 3T ) − ⋅⋅⋅
= u(t ) − 2 u(t − T ) + 2 u(t − 2T ) − 2 u(t − 3T ) + ⋅⋅⋅
Taking Laplace transform,
1 2 e − Ts 2 e −2Ts 2 e −3Ts
1
F (s) = −
+
−
+ ⋅⋅⋅= ⎡⎣1 − 2 e − Ts 1 − e − Ts + e −2Ts − e −3Ts + ⋅⋅⋅ ⎤⎦
s
s
s
s
s
− Ts
− Ts
⎤
⎡
⎤
⎡
⎧
1
2e
1 1− e
1 ⎫
= ⎢1 −
⎬
⎨ sum of G.P. series =
⎥
⎥= ⎢
s ⎣ 1 + e − Ts ⎦ s ⎣ 1 + e − Ts ⎦
1
+
e − Ts ⎭
⎩
(
⎛ Ts ⎞
1
F ( s ) = tanh ⎜ ⎟
s
⎝ 2⎠
1
(b) The equation can be written as f (t ) = 2 r (t ) − 4 r (t − ) + 2 r (t − 1)
2
)
t
275
Laplace Transform and Its Applications
1
− s
−s
−s
2
1 4e 2 2e − s 2
Taking Laplace transform, F ( s ) = 2 2 − 2 + 2 = 2 ⎡⎢1 − 2 e 2 + e − s ⎤⎥ = 2 ⎡⎢1 − e 2 ⎤⎥
⎦
⎦ s ⎣
s
s
s
s ⎣
Problem 5.3 Find the current i( t) flowing through the circuit
if the circuit is initially relaxed. Find the voltage across the capacitor vc( t) also. What is the value of the steady-state current?
5
( )(
1
10V
⎛
⎞
1 ⎟
10
Solution By KVL, ⎜ 5 +
I s =
⇒ I s 5s + 2 = 10
⎜
s ⎟
s
⎝
2⎠
10
2
I s =
=
5s + 2 s + 2
()
2
)
2
F
V c ( t)
Fig. 5.34
()
5 2t
−
Taking inverse Laplace transform, the current in the circuit i t = 2 e 5 A
()
( )
⎛
⎞
1 2
2
4
1
1 ⎟
⎜
Voltage across the capacitor is VC s = I s ×
= ×
=
= 10 −
⎜ s s+ 2 ⎟
1
s s+ 2
s s+ 2
s
⎝
5⎠
5
5
2
() ()
)
(
−2t
Taking inverse Laplace transform, VC t = 10 ⎡⎢1 − e 5 ⎤⎥ V
⎣
⎦
()
( )
From the current expression, as t → , i(t) → 0. So, the steady state value of the current is, zero.
t Close 0
2
1
Problem 5.4 A sinusoidal voltage 25sin10t is applied at time t 0
to a circuit as shown in Fig. 5.35. Find the current i(t) by Laplace transform method. R 5 and L 1 H.
25sin10t
10
Solution By KVL, RI ( s ) + sLI ( s ) = 25 2
with zero initial
s + 100
condition.
I (s) =
250
( s + 5)( s + 100) (
(
A1 = s + 5
where,
=
2
(
(
⎡ A
A3 ⎤
A2
250
= 250 ⎢ 1 +
+
⎥
s
s
j
s
j10 ⎦
+
+
−
5
10
s + 5 s + j10 s − j10
⎣
) s + 5 s1 + 100
( )(
)
A2 = s + j10
A3 = s − j10
Fig. 5.35
)(
)(
=
2
)
1
125
s =−5
) ( s + 5)( s + j110)( s − j10)
) ( s + 5)( s + j110)( s − j10)
=−
s =− j 10
=
s = j 10
1
1
=−
100( 2 + j )
j 20 5 − j10
(
1
100( −2 + j )
)
i(t)
R
L
276
Network Analysis and Synthesis
⎡ A
A3 ⎤
A2
Substituting these, I ( s ) = 250 ⎢ 1 +
+
⎥
⎣ s + 5 s + j10 s − j10 ⎦
Taking inverse Laplace transform,
⎫
⎧
1
1
i(t ) = 250 ⎡⎣ A1e −5t + A2 e − j10t + A3 e j10t ⎤⎦ = 2 e −5t + 250 ⎨−
e − j10t +
e j10t ⎬
+
−
+
100
2
100
2
(
j
)
(
j
)
⎭
⎩
− j 10 t
j 10 t
⎫
⎧
−2 − j e ⎪
5⎪ 2− j e
1
−5t
− j 10 t
= 2 e −5t − ⎨
−
− je − j10t + 2 e j10t + je j10t
⎬ = 2e − 2e
2⎪
5
5
2
⎪
⎭
⎩
(
or,
i(t)
2e 5t
2cos10t
)
)
(
{
}
sin 10t (A)
Problem 5.5 The circuit was in steady state with the switch in the
position 1. Find the current i(t) for t 0 if the switch is moved from the
position 1 to 2 at t 0.
1
2
10
50 V
10 V
When the switch is in the position 1, steady state exists
10
and the initial current through the inductor is, i(0− ) = = 1 A
Fig. 5.36
10
After the switch is moved to the position 2, the KVL gives in Laplace transform,
⎡1
50
100
1
1 ⎤
1
10 I ( s ) + 0.5sI ( s ) − 0.5 × 1 =
or, I ( s ) =
+
= 5⎢ −
⎥+
s
s s + 20 s + 20 ⎣ s s + 20 ⎦ s + 20
Solution
Taking inverse Laplace transform, i(t)
5
(
)
4e 20t (A); t
0;
0.5 H
Problem 5.6 (a) In the circuit shown in Fig. 5.37, the switch S has been thrown to the position 1 for a long
R1
period of time. Find the complete expression for the current after
1
throwing the switch S to 2 which removes R1 from the circuit.
R2
S
(b) If the values of V, R1, R2 and L be 10 V, 1 ohm, 2 ohm and 1 H
respectively, calculate
(i) the steady-state current
(ii) the energy stored in the inductance at steady-state period
(iii) time constant of the circuit for both the positions of the
switch S
t
2
V
0
L
Fig. 5.37
Also, calculate the voltage across the resistor R2 and inductor L, at 0.05 seconds after the switch S has been
thrown to the position 2.
Solution
(a) For t 0, as the circuit was in steady state with the switch in the position 1, the circuit becomes as shown in
Fig 5.38 (a)
R1
V
V
i (0 )
Circuit for t
Fig. 5.38 (a)
R2
R2
0
L
i(t)
Circuit for t
Fig. 5.38 (b)
0
277
Laplace Transform and Its Applications
( ) R V+ R
i 0− =
1
For t
2
0, the circuit becomes as shown, in Fig. 5.38 (b).
()
()
( ) Vs ⇒ ⎡⎣ R + sL ⎤⎦ I ( s ) = Vs + R VL
+R
R2 I s + sLI s − Li 0 − =
By KVL,
2
1
2
⎛
⎡
⎤
⎞
⎢
⎥
⎜
⎟
1
1
V ⎢
V
⎥+
⎜
⎟
⇒ I s =
R2 ⎢ ⎛ R2 ⎞ ⎥ R1 + R2 ⎜ ⎛ R2 ⎞ ⎟
⎢ s⎜ s + L ⎟ ⎥
⎜⎝ ⎜⎝ s + L ⎟⎠ ⎟⎠
⎠⎦
⎣ ⎝
()
⎛R ⎞
⎛R ⎞
−⎜ 2 ⎟ t ⎞
−⎜ 2 ⎟ t
V ⎛
V
L
⎝ L⎠
Taking inverse Laplace transform, i t = ⎜ 1 − e
e ⎝ ⎠
⎟+
+
R2 ⎝
R
R
⎠
1
2
()
(b) V
10 V, R1
( A ), t > 0
1 ohm, R2
2 ohm and L 1H
V 10
(i) Steady-state current, I ss = = = 5 A
R2 2
1
1
(ii) Energy stored in the inductance at steady-state period, W = LI 2 = × 1 × 52 = 12.5 W
2
2
(iii) Time constant of the circuit for switch in the position 1 is,
1
=
L
1
=
= 0.33 second
R1 + R2 1 + 2
Time constant of the circuit for switch in the position 2 is,
2
=
L 1
= = 0.5 second
R2 2
⎤
⎡
20
×2=7 V
0.05, voltage across the resistor, VR = i × R2 = ⎢5 1 − e −2 t + e −2 t ⎥
2
3
⎦t =0.05
⎣
and voltage across the inductor, VL (10 7) 3 V
R
(
For t
)
Problem 5.7 The circuit of Fig. 5.39 is initially in the steady state.
The switch S is closed at t 0.
R
C
Vc(t )
1
(a) Find VC ( t).
R
2
(b) Determine the final value of VC ( t) and verify it from the final-value
theorem of laplace transform.
Fig. 5.39
Solution At steady-state before closing the switch, the capacitor becomes
2
open-circuited. So, the circuit becomes as shown in Fig. 5.40. v(0+ ) = V
3
V
V
For t 0, by KVL,
RI1 + R I1 − I 2 =
⇒ 2 RI1 − RI 2 =
(i)
V
s
s
(
and
S
V
)
⎛
1
2V
1⎞
2V
⇒ − RI1 + ⎜ R + ⎟ I 2 = −
I + R I 2 − I1 = −
Cs 2
3s
Cs ⎠
3s
⎝
(
)
R
R
v(0 )
R
(ii)
Fig. 5.40
278
Network Analysis and Synthesis
Solving equations (i) and (ii),
V
2R
I2 =
s
2
V
−R −
2R
−R
∴Vc ( s ) = I 2 ×
(
− 4VR + VR
3s =
s = − V ⎛ Cs ⎞
3s
3s ⎜⎝ 2 + RCs ⎟⎠
−R
2 R R + 1 − R2
Cs
R+ 1
Cs
)
(
)
⎛
⎞
1 2V
V
1
2V V ⎡
1 ⎤ V V⎜
⎟
=−
+
+
= ⎢2 −
=
+
⎥
⎟
Cs 3s
3s 2 + RCs 3s 3s ⎣ RCs + 2 ⎦ 2 s 6 ⎜ s + 2
⎝
RC ⎠
)
(
Taking inverse Laplace transform, vc (t ) =
V V − 2 t RC
+ e
( V ), t > 0
2 6
Thus, the final value of the voltage, vc (∞) = Lim vc (t ) =
t →∞
V
2
⎛
⎞
V
Vs
⎜
⎟ =V
SVc ( s ) = Lim
+
By final-value theorem, vc (∞) = Lim
⎟ 2
s →0
s →0 ⎜ 2
2
6 s+
⎜⎝
⎟
RC ⎠
)
(
Problem 5.8 The circuit given in Fig. 5.41 is initially at steady state with
the switch ‘ K’ open. If the switch is closed at time t 0, find the voltage ‘VC( t)’
across the capacitor.
Solution At steady-state before closing the switch, the capacitor becomes open2
circuited. So, the circuit becomes as shown in Fig. 5.42. v(0− ) = × 6 = 4 V
3
For t 0, by KVL,
6
6
1 × 103 × I1 + 1 × 103 × I1 − I 2 = ⇒ 2000 I1 − 1000 I 2 =
(i)
s
s
)
(
and
⎛
106
4
106 ⎞
4
I 2 + 1 × 103 × I 2 − I1 = − ⇒ − 1000 I1 + ⎜ 1000 +
I =−
s
s
s ⎟⎠ 2
s
⎝
(
)
2000
Taking inverse Laplace transform,
vc(t)
3
=−
(1000 +10 Cs)
Fig. 5.41
1k
1k
)
(
e
(V), t
0
v (0 )
6V
1k
Fig. 5.42
2 ⎛
1 ⎞
⎜
1000 ⎝ s + 2000 ⎟⎠
106 4
2000
4 3
1
+ =−
+ = +
s
s
s s + 2000 s s s + 2000
2000t
1 F
Vc( t)
1k
6
−1000
∴Vc ( s ) = I 2 ×
1k
6V
6
s
4
−1000 −
s
I2 =
2000
−1000
Solving equations (i) and (ii),
(ii)
1k
279
Laplace Transform and Its Applications
Problem 5.9 In the circuit in Fig. 5.43, the steady state exists when
the switch S is in the position a for a considerable period of time. Find the
current response after throwing the switch from the position a to b. What
will be the steady-state value of the current?
Solution When the switch is in the position a, steady state exists and the
20
initial current through the inductor is i(0− ) = = 2 A
10
After the switch is moved to the position b, the KVL gives, in Laplace
transform,
1
I ( s ) + 1sI ( s ) − 1 × 2 = 0 or, I ( s ) =
100 × 10−6 s
2
2s
= 2
104
s + 104
s+
s
(
)
10
a
b
20V
1H
100 F
Fig. 5.43
10
i
20V
104
s
Fig. 5.44
Fig. 5.45
s
2V
Taking inverse Laplace transform, i(t) 2cos100t
(A); t 0
The steady state current will oscillate sinusoidally following the relation i(t) 2 cos100t with a peak magnitude of 2 A and frequency of 100 rad/s or 15.9 Hz.
Problem 5.10 In the network shown in Fig. 5.46, the switch S is
closed and a steady state is attained. At t 0, the switch is opened.
Determine the current through the inductor for t > 0.
Solution When the switch S is closed and the steady-state exists,
V
5
the current through the inductor is, i(0− ) = =
=2 A
R 2.5
The voltage across the capacitor, Vc(t)
1
5V
2
S
R = 2.50hm
L = 0.5 H
C = 200 uF
Fig. 5.46
0 as it is shorted.
t
For t
0, the switch is opened. By KVL, L
di 1
+
idt = 0
dt C ∫0
Taking Laplace transform, L ⎡⎣ sI ( s ) − i(0− ) ⎤⎦ +
s
Putting the values, I ( s ) = 2 2
s + 104
Taking inverse Laplace transform, i(t)
⎡
I (s)
1⎤
= 0 or, I ( s ) ⎢ sL + ⎥ = Li(0− )
Cs
Cs
⎣
⎦
2 cos100t (A); t
0
Problem 5.11 The circuit shown in Fig. 5.47 is initially in the steady state with
the switch S open. At t 0, the switch S is closed. Obtain the current through the
inductor for t 0. Take R1 R2 R4 1- and R3 2- and L 1-H.
(
)
1
Solution When the switch S is open and steady state exists, the current through the
1
inductor is i2 (0− ) =
=1 A
R1 + R2
R3
R3 + R1 + R2
R3
R1 S
R2
2 L
1V
R4
Fig. 5.47
280
Network Analysis and Synthesis
After S is closed, for t
2i1 − i2 − i3 = 1
0, by KVL,
di2
−i =0
dt 3
−i1 − i2 + 4i3 = 0
−i1 + 2i2 +
1
s
− I1 ( s ) + I 2 ( s ) ⎡⎣ s + 2 ⎤⎦ − I 3 ( s ) = i2 (0− ) = 1
2 I1 ( s ) − I 2 ( s ) − I 3 ( s ) =
Taking Laplace transform,
− I1 ( s ) − I 2 ( s ) + 4 I 3 ( s ) = 0
1
2
By Cramer’s rule,
I2 (s) =
−1
s
1
−1
0
−1
−1
4
5
1
6
= 6+
s s+ 6
2
−1
−1
7
−1 ( s + 2 ) −1
−1
−1
4
Taking inverse Laplace transform,
5 1 −6 t
i2 (t ) = + e 7 ( A ); t > 0
6 6
Problem 5.12 A series R-L-C circuit with R 3 , L 1 H and C 0.5 F is excited with a unit step voltage.
Obtain an expression for the current using Laplace transform. Assume that the circuit is relaxed initially.
Solution By KVL,
RI ( s ) + sLI ( s ) − Li(0− ) +
1
Q (0− ) 1
I (s) +
=
sC
sC
s
Since the circuit is initially relaxed,
i(0 ) 0 and Q(0 ) 0
Putting the values,
or,
⎡
2⎤ 1
I (s) ⎢3 + s + ⎥ =
s⎦ s
⎣
I (s) =
A
A
1
1
= 1 + 2
=
s + 3s + 2 s + 1 s + 2 s + 1 s + 2
A1 =
1
1
= 1 and A2 =
= −1
s + 2 s =−1
s + 1 s =−2
where,
I (s) =
2
(
)(
)
1
1
−
s +1 s + 2
Taking inverse Laplace transform,
i(t ) = e − t + e −2 t ( A ) = 2 e
− 3t
2
( 2 ) ( A)
sinh t
281
Laplace Transform and Its Applications
S
I = 2A
Fig. 5.48
L = 1H
C = 0.5 F
R=1
S
L = 0.5 H
I = 2A
R = 0.5
Problem 5.13 The switch S in the figure is opened at t 0. Determine the voltage v(t), for t
the nature of the response?
(a)
(b)
1
2
0. What is
C = 1F
Fig. 5.49
Solution
t
(a) By KCL,
v (t )
dv
1
+ i(0− ) + ∫ vdt + C = I
R
L0
dt
⎡1 1
⎤ I
Taking Laplace transform, V ( s ) ⎢ + + sC ⎥ =
R
sL
⎣
⎦ s
⎡ 2 s⎤ 2
4
4
Putting the values, V ( s ) ⎢ 2 + + ⎥ =
or, V ( s ) = 2
=
2
s
s
2
s
4
s
4
+
+
⎣
⎦
s+2
(
Taking inverse Laplace transform, v(t ) 4te
The response is critically damped (
1)
2t
(V ), t
)
0
(b) Proceeding in the same way as Prob. 5.13(a),
⎛ 3 ⎞
⎜⎝ 2 ⎟⎠
2
1
V (s) = 2
=
×
2
2
s + s +1 ⎛ 1 ⎞ ⎛ 3 ⎞
3
⎜⎝ s + 2 ⎟⎠ + ⎜ 2 ⎟
⎝
⎠
⇒ v (t ) =
The response is under-damped (
2
e
3
−t
2
⎛ 3 ⎞
sin ⎜
t ⎟ ( V ); t > 0
⎝ 2 ⎠
1)
Problem 5.14 In the R-C series circuit of Fig. 5.50, the capacitor has
an initial charge of 2.5 mC. At t 0, the switch is closed and a constant
voltage source of V
100 V is applied. Use the Laplace transform
method to find the current i( t) in the circuit.
S
100 V
Solution By KVL, after the switch is closed,
Ri(t ) +
t
⎤
1⎡
⎢Q(0− ) + ∫ i(t )dt ⎥ = V
C ⎢⎣
⎥⎦
0
Taking Laplace transform, 10 I ( s ) +
Fig. 5.50
I (s)
2.5 × 10−3 100
15
−
=
or,, I ( s ) =
−6
s
50 × 10 s 50 × 10−6 s
s + 2 × 103
Taking inverse Laplace transform, i(t )
3
15e 2 10 t (A);
t
0
10
i(t)
50 uF
Q0
282
Network Analysis and Synthesis
Problem 5.15 In the R-L circuit as shown, in Fig. 5.51, the switch is in the position 1 long enough to establish steady-state condition and at t 0 it is switched
to the position 2. Find the resulting current, i( t).
Solution When the switch is in the position 1, steady-state exists and the initial
50
current through the inductor is i(0− ) = = 2 A
25
After the switch is moved to the position 2, the KVL gives in Laplace transform,
25 I ( s ) + 0.01sI ( s ) − 0.01 × 2 =
50 V
100 V
25
0.01 H
Fig. 5.51
100
s
or, I ( s ) =
A
A2
104
2
2
−
= 1+
−
s s + 2500 s + 2500 s s + 2500 s + 2500
where, A1 =
104
s + 2500
)
(
(
2
1
)
= 4 and A2 =
s =0
104
= −4
s s =−2500
4
4
2
4
6
I (s) = −
−
= −
s s + 2500 s + 2500 s s + 2500
Taking inverse Laplace transform, i(t )
6e 2500t (A);
4
t
0
Problem 5.16 In the series R-L-C circuit as shown, there
is no initial charge on the capacitor. If the switch is closed
at t 0, determine the resulting current at i( t).
Solution By KVL,
for t
1
0,
Fig. 5.52
di 1
idt = V ⎡⎣ i(0− ) = 0 ⎤⎦
+
dt C ∫0
Taking Laplace transform,
Putting the values,
1H
0.5 F
2
50V
t
Ri + L
2
S
RI ( s ) + sLI ( s ) +
2 I ( s ) + sI ( s ) + 2
By partial fraction expansion,
I (s) V
=
Cs s
I ( s ) 50
=
s
s
I (s) =
or,
I (s) =
50
50
50
=
=
2
s + 2s + s s +1+ j s +1− j
s +1 +1
(
)(
50e t sin t (A);
t
2
) (
)
j 25
j 25
−
s +1+ j s +1− j
Taking inverse Laplace transform, i(t )
j25 e( 1 j ) t
e( 1 j ) t
Problem 5. 17 In the two-mesh network shown in Fig. 5.53,
there is no initial charge on the capacitor. Find the loop currents
i1( t) and i2( t) which result when the switch is closed at t 0.
Solution Writing the two mesh equations,
t
1
10i1 (t ) +
i (t )dt + 10i2 (t ) = 50 and 50i2 (t ) + 10i1 (t ) = 50
0.2 0∫− 1
S
Fig. 5.53
10
i2(t)
i1(t ) 0.2 F
40
2
1
50 V
0
283
Laplace Transform and Its Applications
I (s)
⎡
50
5⎤
50
10 I1 ( s ) + 1 + 10 I 2 ( s ) =
⇒ I1 ( s ) ⎢10 + ⎥ + 10 I 2 ( s ) =
0.2 s
s
s
s
⎦
⎣
Taking Laplace transform,
and
10 I1 ( s ) + 50 I 2 ( s ) =
Solving,
I1 ( s ) =
50
s
5
1
1
and I 2 ( s ) = −
s + 0.625
s s + 0.625
i1 (t ) = 5e − 0.625t ( A ) and i2 (t ) = 1 − e − 0.625t ( A ), t > 0
Taking inverse Laplace transform,
Solution Circuit for t
I1 ( s ) ⎡⎣ 2 + 2 + 0.5s ⎤⎦ − ⎡⎣ 2 + 0.5s ⎤⎦ I 3 ( s ) =
I1 ( s ) ⎡⎣ s + 8 ⎤⎦ − ⎡⎣ s + 4 ⎤⎦ I 3 ( s ) =
48
s
and
− I1 ( s ) ⎡⎣ 2 + 0.5s ⎤⎦ + ⎡⎣ 4 + 0.5s ⎤⎦ I 3 ( s ) = 0
or,
− I1 ( s ) ⎡⎣ s + 4 ⎤⎦ + ⎡⎣ s + 8 ⎤⎦ I 3 ( s ) = 0
(
48
Solving (i) and (ii),
(
− s+4
)
and
(
s +8
(
− s+4
I 2 ( s ) = I1 ( s ) − I 3 ( s ) =
)
)
=
6 s +8
s
( )
−( s + 4) s( s + 6)
0
=
6 s+4
s +8
( ) − 6( s + 4 ) = 24
s( s + 6) s( s + 6) s( s + 6)
24
=4 A
s →0 s + 6
i2 (∞) = Lim sI 2 ( s ) = Lim
s →0
1
i3
i1
0.5 H
6 s +8
final value of the current,
Fig. 5.54
2
24 V
)
48
− s+4
I3 (s) =
(ii)
s +8
s +8
24
s
2
( )
−( s + 4) s( s + 6)
s +8
2
(i)
s +8
0
I1 ( s ) =
2
i2(t)
− s+4
s
S 2
i 2(t)
24 V
0.5 H
0 is shown in Fig. 5.55.
By KVL, in Laplace transform,
or,
2
Problem 5. 18 Find using final-value theorem, the steady-state value of i2( t )
in the circuit shown in Fig. 5.54. Switch S is closed at t 0. The inductor is initially
de-energized.
Fig. 5.55
2
284
Network Analysis and Synthesis
Problem 5.19 In a series LC circuit, the supply voltage being v
ditions. Assume L 1H, C 1F.
Solution By KVL, for t
or,
Vmcos( t ), find i( t ) with zero initial con-
⎡
1 ⎤ sV
I ( s ) ⎢ sL + ⎥ = 2 m
Cs ⎦ s + 1
⎣
0,
⎡
⎤
⎡
⎤
s2
s2 ⎥
⎢
V
= Vm
I (s) =
=
⎢
⎥
m
2 ⎥
⎢ 2
⎛ 1⎞
⎢⎣ s + j s − j s + j s − j ⎥⎦
s2 +1 ⎜ s + ⎟
⎢⎣ s + 1 ⎥⎦
⎝ s⎠
sVm
(
)
(
)(
(
)
)(
)(
)
⎡
⎡
⎤
*
* ⎤
s2
⎥ = V ⎢ K1 + K1 + K 2 + K 2 ⎥
= Vm ⎢
⎢ s+ j 2 s− j 2 ⎥ m ⎢ s− j 2 s+ j 2 s− j
s+ j ⎥
⎣
⎣
⎦
⎦
)(
(
where,
(
K1 = I ( s ) × s − j
)
)
2
s= j
=
(
) (
) (
(
)
(
) (
)
1
4
2
)
s + j 2s − s2 × 2 s + j
2
1 d
j
=−
K2 =
s − j I (s) =
4
4
ds
2 −1 !
s+ j
s= j
)
(
Thus,
)
(
)
(
1
j
K1* = ; and K 2* =
4
4
⎤
V ⎡ 1
j
j ⎥
1
I (s) = m ⎢
+
−
+
4 ⎢ s− j 2 s+ j 2 s− j
s+ j ⎥
⎣
⎦
(
) (
) (
Taking inverse Laplace transform,
i (t ) =
) (
)
Vm
V
⎡te jt + te − jt − je jt + je jt ⎤ = m ⎡⎣t cos t + sin t ⎤⎦ ( A ); t > 0
⎦ 4
4 ⎣
Problem 5.20 The series RC circuit of Fig. 5.56 has a sinusoidal voltage
source, v 180sin(2000 t
)( V) and an initial charge on the capacitor
Q 0 1.25 mC with polarity as shown. Determine the current if the switch
is closed at a time corresponding to
90 . What is the current at time
t 0?
Solution By KVL, for t
40i(t ) +
v(t)
25 F
Fig. 5.56
0,
t
⎡
⎤
1
−3
1
25
10
.
×
+
i(t )dt ⎥ = 180 cos 2000t
⎢
∫
−6
25 × 10 ⎢⎣
⎥⎦
0
Taking Laplace transform, 40 I ( s ) +
1.25 × 10−3 4 × 104
180 s
+
I (s) = 2
−6
s
25 × 10 s
s + 4 × 106
⇒ I (s) =
4.5s 2
( s + 4 × 10 )( s + 10 )
2
6
40
i(t)
3
−
1.25
s + 103
Q0
285
Laplace Transform and Its Applications
Applying Heaviside expansion formula to find the first term on the right-hand side, we have,
P( s ) = 4.5s 2 , Q( s ) = s 3 + 103 s 2 + 4 × 106 s + 4 × 109 , Q ′( s ) = 3s 2 + 2 × 103 s + 4 × 106 ,
a1 = − j 2 × 103 ; a2 = j 2 × 103 and a3 = −103
Then,
i (t ) =
(
)e
Q ′ ( − j 2 × 10 )
P − j 2 × 103
− j 2 ×103 t
3
(
)
(
3
+
(
)e
Q ′ ( j 2 × 10 )
P j 2 × 103
j 2 ×103 t
3
)
+
( )e
Q ′ ( −10 )
P −103
3
−103 t
3
3
− 1.25e −10 t
3
= 1.8 − j 0.9 e − j 2 ×10 t + 1.8 + j 0.9 e j 2 ×10 t − 0.35e −10 t
)
(
3
3
= −1.8 sin 2000t + 3.6 cos 2000t − 0.35e −10 t = 4.02 sin 2000t + 116.6° − 0.35e −10 t ( A ); t > 0
Problem 5.21 In the RL circuit of Fig.5.57, the source is v 100sin(500 t
).
Determine the resulting current if the switch is closed at a time corresponding to
0.
Solution By KVL,
v(t )
RI ( s ) + sLI ( s ) − Li(0− ) = V ( s )
or,
5 I ( s ) + 0.01sI ( s ) =
or,
I (s) =
5
i(t)
0.0 H
100 × 500
[ i(0 ) 0]
s + 25 × 104
Fig. 5.57
2
5 × 106
( s + 25 × 10 )( s + 500)
2
4
⎛ −1 + j ⎞ ⎛ −1 − j ⎞
10
+
+ 5⎜
I ( s ) = 5⎜
⎟
⎟
⎝ s + j 500 ⎠ ⎝ s − j 500 ⎠ s + 500
By partial fraction expansion,
Taking inverse Laplace transform,
)
(
i(t ) = 10 sin 500t − 10 cos 500t + 10e −500t = 14.14 sin 500t − 45° + 10e − 500t ( A ); t > 0
Problem 5.22 A dc voltage applied to a coil of inductance L and resistance R is suddenly changed from
V1 to V2. (a) Find an expression for current in the circuit. (b) If R 10 , L 1 H, V1 100 V, and V2 200 V,
find current at t 0.5 s. (c) If R 10 , L 1 H, V1 200 V, and V2 100 V, find current at t 0.5 s.
( )
V
i 0− = 1
R
Solution Here, initial current in the circuit,
()
di t
(a) After changing the voltage, the KVL equations is
Ri t + L
= V2 u t
dt
Taking Laplace transform,
V
V VL
RI s + L ⎡⎣ sI s − i 0 − ⎤⎦ = 2 ⇒ I s ⎡⎣ R + sL ⎤⎦ = 2 + 1
s
s
R
⎛ V1
⎞
V2
V2
V1 L ⎛ 1 ⎞
⎜
⎟
L
R
⇒ I s =
+
=
+⎜
⎟
R
R ⎜⎝ R + sL ⎟⎠ s s + R
s R + sL
s+
L ⎟⎠
L ⎜⎝
()
()
() ( )
()
()
(
)
(
)
()
286
Network Analysis and Synthesis
Taking inverse Laplace transform,
( ) R ⎡⎢⎣1 − e ( ) ⎤⎥⎦ + R e ( ) = R + ⎜⎝ R − R ⎟⎠ e ( )
i t =
(b) If R
10
,L
1 H, V1
V2
− R
L
V1
t
100 V, and V2
− R
L
(c) If R
10
,L
1 H, V1
⎛
− 10
200 V, and V2
1
× 0.5
⎛
− 10
1
× 0.5
0.5 s as
= 20 − 10e −5 = 19.93 A
100 V, we get the current at t
200 100 ⎞ ( )
+⎜
−
e
( ) 200
10 ⎝ 10 10 ⎟⎠
i t =
V2 ⎛ V1 V2 ⎞ − R L t
200 V, we get the current at t
100 200 ⎞ ( )
+⎜
−
e
( ) 200
10 ⎝ 10 10 ⎟⎠
i t =
t
0.5 s as
= 10 + 10e −5 = 10.07 A
Problem 5.23 A 50 F capacitor and 20000- resistor are connected in series across a 100-V battery at t 0. At t 0.5 s, the battery voltage is suddenly increased to 150 V. Find the charge on capacitor
at t 0.75 s.
Solution When the circuit is connected to a 100-V supply, the equation of voltage across the capacitor is
− t
⎛
−6 ⎞
−t
⎛
vC = E ⎜ 1 − e RC ⎞⎟ = 100 ⎜ 1 − e 20000 × 50 × 10 ⎟ = 100 1 − e − t
⎠
⎝
⎝
⎠
(
At t
At t
0.5 s, the voltage across the capacitor is vC
100(1
e 0.5)
)
39.347 V
0.5 s, charge on the capacitor is, q = CvC = 50 × 10−6 × 39.347 = 1967.35 × 10−6 C
This charge is the initial charge q0 when the battery voltage is suddenly increased to 150 V.
When the circuit is connected to 150 V, the KVL equation becomes,
t
( ) C1 ∫ i (t )dt = Vu(t )
Ri t +
0
Taking Laplace transform,
V q0
−
1 ⎡ I s q0 ⎤ V
RI s + ⎢
+ ⎥=
⇒ I s = R RC
1
C ⎢⎣ s
s ⎥⎦ s
s+
RC
()
()
()
Taking inverse Laplace transform,
⎡V q ⎤ − t
i(t ) = ⎢ − 0 ⎥ e RC
⎣ R RC ⎦
Therefore, the voltage across the capacitor,
Vc =
t
t
−t
q −t
1
1 ⎛ V q0 ⎞ − t RC
⎛
i
(
t
)
dt
=
−
e
dt = V ⎜ 1 − e RC ⎞⎟ + 0 e RC
⎜
⎟
∫
∫
⎝
⎠ RC
C0
C 0 ⎝ R RC ⎠
287
Laplace Transform and Its Applications
Substituting the values, the voltage across the capacitor at t
voltage,
0.75 s i.e., 0.25 second after changing the battery
q −t
−t
1967.35 × 10−6 − 0.25
⎛
= 63.82 V
VC = V ⎜ 1 − e RC ⎞⎟ + 0 e RC = 150 1 − e − 0.25 +
e
⎝
⎠ RC
1
)
(
charge on the capacitor is, q = CvC = 50 × 10− 6 × 63.82 = 3.19 × 10−3 C
Problem 5.24 For the circuit shown in figure, find an expression for the current supplied by the source. How much time it will
take for the current to reach 25 mA? Assume the circuit to be initially relaxed.
t=0
10 V
500
Solution Applying KVL for the two meshes, we get
Fig. 5.58
t
1
i dt = 0
100 × 10−6 ∫0 2
Taking Laplace transform,
()
( ) 10s
500 I1 s − 500 I 2 s =
⎛
104 ⎞
−500 I1 s + ⎜ 1200 +
I s =0
s ⎟⎠ 2
⎝
()
()
Solving for I1(s), we get
10
−500
s
()
I1 s =
0
⎛
104 ⎞
1200
+
⎜⎝
s ⎟⎠
500
−500
=
⎛
104 ⎞
−500 ⎜ 1200 +
s ⎟⎠
⎝
24 s + 200
(
s 700 s + 104
)
=
24
200
+
700 s + 104 s 700 s + 104
(
⎛
⎞
⎛
⎛
⎞
⎞
24 ⎜
1
2
1
1
⎜
⎟ = 1 ⎛ 1⎞ + 1 ⎜
⎟+
⎟
=
700 ⎜ s + 100 ⎟ 7 ⎜⎜ s s + 100 ⎟⎟ 50 ⎜⎝ s ⎟⎠ 70 ⎜ s + 100 ⎟
⎝
⎝
⎝
7⎠
7⎠
7 ⎠
(
)
Taking inverse Laplace transform,
( ) 501 + 701 e
i1 t =
−100 t
7
(A)
For the current to be 25mA, we get,
25 × 10−3 =
1 1 −100t 7
+ e
⇒ t = 0.0735 second
50 70
700
100 F
i1
500i1 − 500i2 = 10
−500i1 + 1200i2 +
i2
)
288
Network Analysis and Synthesis
Problem 5.25 Figure 5.59 shows a parallel RLC circuit fed from a dc current source through a switch. The
circuit elements are R 400 , L 25 mH, C 25 nF.
The source current is 24 mA. The switch which has been
in the closed position for a long time is opened at t 0.
S
I
L
R
C
v( t )
Fig. 5.59
(a) What is the initial value of current iL (i. e., at t 0)?
(b) What is the initial value of voltage across L at t 0?
(c) What is the expression for current through inductance, capacitance and resistance?
(d) What is the final value of iL?
(e) What happens to iL(t) if R is increased from 400 to 625 ? Assume that initial energy is zero.
t
Solution Applying KCL for the node, we get
v (t ) 1
dv
+
vdt + C = I
R L ∫0
dt
⎡1 1
⎤ I
I
V ( s ) ⎢ + + sC ⎥ =
⇒ V s =
⎛
s
1 ⎞
⎣ R sL
⎦ s
C ⎜ s2 +
+
RC LC ⎟⎠
⎝
()
Taking Laplace transform,
Substituting the values,
()
V s =
=
24 × 10−3
24 × 106
=
⎛
⎞ 25 s 2 + 105 s + 16 × 108
1
s
25 × 10−9 ⎜ s 2 +
+
−9 ⎟
−9
−3
⎝
400 × 25 × 10
25 × 10 × 25 × 10 ⎠
(
(
24 × 106
)(
25 s + 2 × 104 s + 8 × 104
Taking inverse Laplace transform,
)
=
16
16
−
s + 2 × 104 s + 8 × 104
4
4
v (t ) = 16 e −2 ×10 t − 16 e −8×10 t ( V )
()
IL s =
Also, the current through the inductor,
=
=
Taking inverse Laplace transform,
)
()
( )=
()
V s
V s
sL
25 × 10−3 s
(
384 × 10
=
24 × 106
)
5
)(
s s + 2 × 104 s + 8 × 104
)
24
32
8
−
+
4
1000 s 1000 s + 2 × 10
1000 s + 8 × 104
)
(
4
)(
(
25 × 10−3 × 25s s + 2 × 104 s + 8 × 104
4
(
)
( )
iL t = 24 − 32 e −2 ×10 t + 8e −8×10 t mA
(a) At t 0, we get, iL(0) 0
(b) At t 0, we get, v(0) 0
(c) Current through inductance
()
4
4
( )
iL t = 24 − 32 e −2 ×10 t + 8e −8×10 t mA
Current through capacitance,
iC (t ) = C
4
4
4
4
dv (t )
d
= 25 × 10−9 [16 e −2 ×10 t − 16 e −8×10 t ] = 32 e −8×10 t − 8e −2 ×10 t ( mA )
dt
dt
289
Laplace Transform and Its Applications
()
iR t =
Current through resistance,
(d) At t
( ) = 1 × ⎡16e
v t
, the final value of iL is,
(e) Putting the value of resistance R
()
IL s =
=
( )=
400 ⎣
R
V s
I
sL
⎛
s
1 ⎞
sLC ⎜ s 2 +
+
RC LC ⎟⎠
⎝
−2 ×104 t
( )
4
4
4
1
− 16 e −8×10 t ⎤ = ⎡ e −2 ×10 t − 16 e −8×10 t ⎤ A
⎦ 25 ⎣
⎦
( )
( )
iL ∞ = 24 − 32 e −∞ + 8e −∞ = 24 mA
625
in the expression of iL, we get,
24 × 10−3
384 × 105
=
⎛
⎞ s s 2 + 64 × 103 s + 16 × 108
s
1
s × 25 × 10−3 × 25 × 10−9 ⎜ s 2 +
+
⎝
625 × 25 × 10−9 25 × 10−3 × 25 × 10−9 ⎟⎠
(
()
) ( )
(
3
)
Taking inverse Laplace transform and simplifying, we get, iL t = 106.67 × 109 e −32 ×10 t sin 14.4 × 103 t A
Here, with R 400 , the circuit was in overdamped condition. As the value of the resistance is increased to
625 , the circuit becomes underdamped.
1000
Problem 5.26 In the network of Fig. 5.60, the switch S has been closed
for a long time. The switch is suddenly opened at t 0 and reclosed at t
20 s. Find the expression for the voltage V0 for t 20 s and t > 20 s. 120V
3000
S
V0
2000
Solution With the switch closed, the initial voltage across the capacitor is
120
vC 0 − =
× 2000 = 80 V
Fig. 5.60
1000 + 2000
1000
After the switch is opened, the transformed circuit is shown in Fig. 5.62
below.
120
80
2000
120V
Applying KCL at node X, we get
VX s −
VX −
V
s + X +
s =0
1
4000
2000
Fig. 5.61
10−8 s
0.001 F
( )
()
⎡ 1
⎤ 120
1
0.03
+ 80 × 100−8
⇒ VX ⎢
+
+ 18−8 s ⎥ =
+ 80 × 100−8 =
4000
2000
4000
s
s
⎣
⎦
0.03
80 × 100−8
⇒ VX =
+
−8
s 0.00075 + 10−8 s 0.00075 + 10 s
)
(
=
40
40
80
40
40
−
+
= +
s s + 0.075 × 106 s + 0.075 × 106
s s + 0.075 × 106
Taking inverse Laplace transform,
Therefore, the desired voltage is
At t
20
()
VC(0 )
1000
3000
X
120/s
V0(s)
1/10 8s
80/s
2000
Fig. 5.62
6
VX t = 40 + 40e − 0.075 × 10 t
V0 (t ) = VX (t ) +
s, the voltage of node X is
120 − VX (t )
6
× 3000 = 100 + 10e − 0.075 × 10 t
4000
6
−6
VX = 40 + 40e −0.075×10 × 20×10 = 48.925 V
for 0
t
20
s
290
Network Analysis and Synthesis
When the switch is reclosed at t 20 s, the voltage across the capacitor
will be 48.925 V. After reclosing the switch, the transformed circuit is shown
in Fig. 5.63.
Now let the voltage of node X be VX . Applying KCL at node X, we get
120
48.925
V′ s −
VX′ s −
s =0
s + VX′ s + X
1
1000
2000
10−8 s
⎡ 1
⎤ 120
1
⇒ VX′ s ⎢
+
+ 10−8 s ⎥ =
+ 48.925 × 10−8
⎣ 1000 2000
⎦ 1000 s
()
1000
X Vx
120/s
()
()
2000
1 V0(s)
10 8s
48.925/s
Fig. 5.63
()
()
0.12
+ 48.925 × 10−8
s
0.12
48.925 × 10−8
0.12
48.925
=
⇒ VX′ s =
+
+
−8
6
−8
s + 0.15 × 106
s 0.0015 + 10 s 0.0015 + 10 s s s + 0.15 × 10
⇒ VX′ s ⎡⎣0.0015 + 10−8 s ⎤⎦ =
()
)
(
.075
( ) 80s − s + 31
0.15 × 10
⇒ VX ′ s =
)
(
6
()
6
Taking inverse Laplace transform, we get VX′ t = 80 − 31.075e − 0.15 × 10 t
In this case, the output voltage V0 is equal to VX (t). Since time t is to be counted from the instant the switch
is reclosed, t is replaced by (t 20 10 6 ).
()
−6
6
∴V0 t = 80 − 31.075e − 0.15 × 10 ( t − 20 × 10 )
for t
20
s
Problem 5.27 In the circuit of Fig. 5.64 the switch S is closed at t 0 and opened
again at t
seconds. Prior to closing the switch at t 0, vC 10 V while L and
seconds.
C2 do not have any stored energy. Find the voltages vC1 and vC2 at t
C1 C2 1F; C 2H
S
C1
vc
1
Solution After closing the switch, applying KVL in the circuit, we get
L
( ) + 1 i t dt + 1 i t dt = 0
() C ∫()
dt
C ∫
di t
t
t
1 −∞
2 −∞
Since initial voltage across C1 is 10 V, we get,
()
2 sI s +
Fig. 5.64
( ) − 10 + I ( s ) = 0
I s
s
s
⎡
2 ⎤ 10
⇒ I s ⎢2s + ⎥ =
⇒
s⎦ s
⎣
()
s
( ) s 5+ 1
I s =
Taking inverse Laplace transform, we get i(t)
2
5sint
1
∴ vC t = −10 + ∫ 5sin tdt = −10 + −5csot 0 = 0
1
C1 0
()
( ) C1 ∫ 5sin tdt = −5csot = 10 V
∴ vC t =
2
0
2 0
L
C2
vc
2
291
Laplace Transform and Its Applications
Problem 5.28 The network shown in Fig. 5.65 is in
steady state with S1 closed and S2 open. At t t1, S1 is
opened and S2 is closed. Find current through capacitor for t t1.
2H
2
S1
S2
3H
10 V
1 F
Solution When switch S1 is closed and S2 is
opened, the initial current through the 3-H inductor is
Fig. 5.65
10
i 0 − = = 5 A . Initial voltage across the capacitor is
2
zero.
When the switch S1 is opened and S2 is closed, the current through the capacitor is given by the KVL equa-
( )
tion as 3
( )+
di t
dt
t
()
1
i t dt = 0
1 × 10−6 ∫0
Taking Laplace transform,
()
3sI s − 3 × 5 +
()
1 I s
15s
=0 ⇒ I s = 2
=
−6
s
10
3s + 106
()
5s
106
s2 +
3
⎡⎛ 106 ⎞ ⎤
i t = 5cos ⎢⎜
⎟ t ⎥ = 5cos 577.35t
⎢⎝ 3 ⎠ ⎥
⎣
⎦
Since the switch is closed at t t1, the time will be shifted by (t t1) so that the current through the capacitor
is given as i(t) 5 cos[577.35(t t1)] for t t1
Taking inverse Laplace transform we get
()
(
500 x 103
Problem 5.29 The switch in Fig. 5.66 has been in the
position A for a long time. At t 0 it is moved to B and at t
1 second it is moved to A again. Find the voltage across the
capacitor after a further lapse of 1 millisecond.
)
B
A
10V
1500
1 F
vc
Solution As the switch is in the position A for a long time,
the initial charge across the capacitor is zero.
Fig. 5.66
When the switch is moved to the position B, the current in the circuit is obtained from the KVL equation as
⎛
1 ⎞ 10
10
1 ⎛ 1 ⎞
I s ⎜ 500 × 103 +
=
⇒ I s =
=
⎝
1 × 10−6 s ⎟⎠ s
500 × 103 s + 2 5 × 104 ⎜⎝ s + 2 ⎟⎠
()
()
Taking inverse Laplace transform,
)
(
( ) 5 ×110 e ( A )
i t =
−2 t
4
Therefore, voltage across the capacitor is
()
(
)()
(
vC t = 10 − 500 × 103 i t = 10 − 500 × 103
) 5 ×110 e = 10(1− e ) ( V )
−2 t
−2 t
4
Therefore, voltage across the capacitor at t 1second is vc(t) 10(1 e 2) 8.65 ( V )
At t 1second, the switch is moved to position A, so that the KVL equation becomes,
t
1
i t dt + 1500i t = 0
1 × 10−6 ∫0
()
()
292
Network Analysis and Synthesis
⎡
⎤ 8.65
1
Taking Laplace transform, I s ⎢1500 +
=
since initial voltage across the capacitor is 8.65 V.
−6 ⎥
s
1 × 10 s ⎦
⎣
⎞
8.65 ⎛
1
⇒ I s =
⎜
1500 ⎝ s + 666.67 ⎟⎠
()
()
8.65
e
( ) 1500
i t =
Taking inverse Laplace transform,
(A)
− 666.67 t
()
()
Hence, the voltage across the capacitor is vC t = 1500i t = 1500 ×
At t
8.65 − 666.67t
e
= 8.65e − 666.67t
1500
1ms, the voltage is, vC = 8.65e − 666.67 × 10 × − 3 = 4.44 V
Problem 5.30 Determine the Laplace transform of f (t ) =
−t
2 − 2 e −t
.
t
⎛
⎞
( ) 2 − 2t e = 2t (1 − e ) = te2 ( e − 1) = te2 ⎜⎝ 1 + t + t2! + t3! + t4! + t5! + ⋅⋅⋅− 1⎟⎠
f t =
Solution
−t
t
t
⎛
2
3
4
2
3
4
5
(expanding et )
t
⎞
5
( ) te2 ⎜⎝ t + t2! + t3! + t4! + t5! + ⋅⋅⋅⎟⎠
∴f t =
t
⎛
⎞
1
1
1
1
= 2 ⎜ e − t + te − t + t 2 e − t + t 3 e − t + t 4 e − t + ⋅⋅⋅⎟
2!
3!
4!
5!
⎠
⎝
Taking Laplace transform of each term, we get
⎡
⎤
1
1 1!
1 2!
1 3!
⎥
F s = 2⎢
+
+
+
+
⋅⋅⋅
⎢ s + 1 2! s + 1 2 3! s + 1 3 4! s + 1 4
⎥
⎣
⎦
⎡
⎤
⎛
⎞ ⎛ ⎛
⎞ ⎛ ⎛
⎞
1⎞
1
1⎞
1
1 ⎛ 1⎞
1
⎥
⎟
⎜
⎟
⎜
⎟
+
= 2⎢
+⎜ ⎟⎜
+
+
⋅⋅⋅
⎢ s + 1 ⎝ 2 ⎠ ⎜ s + 1 2 ⎟ ⎜⎝ 3 ⎟⎠ ⎜ s + 1 3 ⎟ ⎜⎝ 4 ⎟⎠ ⎜ s + 1 4 ⎟
⎥
⎠
⎝
⎝
⎠
⎝
⎠
⎣
⎦
()
)
(
(
(
)
)
)
(
)
(
(
)
Problem 5.31 Express the following functions in terms of singularity functions and find their Laplace transform:
(a)
(b) f (t )
f(t )
Vm
0
(c) f (t )
K
t
0
(d)
e t /2
1
t )2
K(2
Kt 2
t
t
0
1
t
0 1 23 4 56
f (t )
2
Fig. 5.67
()
f (t )
K
Solution
(a) Here, the signal can be expressed in terms of step signal as,
⎛ T⎞ ⎛ T⎞
f t = Vm sin t u t + Vm sin ⎜ t − ⎟ u ⎜ t − ⎟ = Vm sin t u t + Vm sin
⎝ 2⎠ ⎝ 2⎠
()
1
()
Vm
(
t−
)(
u t−
)
0
Fig. 5.68
t
293
Laplace Transform and Its Applications
Taking Laplace transform of individual terms, we get the Laplace transform of the functions as,
V
V e− s
V
F s = 2 m + m2
= 2 m 1+ e− s
s +1 s +1 s +1
)
(
()
(b) Here, the signal starts with a straight line of slope K passing through origin and then
comes to zero at t 1. Hence the signal can be expressed in terms of ramp and step signals
as f t = Kr t − Kr t − 1 − Ku t − 1
f(t )
K
Taking Laplace transform of individual terms, we get the Laplace transform of the funcK Ke − s Ke − s K
= 2 ⎡⎣1 − 1 + s e − s ⎤⎦
tions as, F s = 2 − 2 −
s
s
s
s
f(t )
(c) The function can be written as
0
()
()
( )
( )
()
()
)
(
()
( ) ( ) ( ) ( ) (
= Kt u ( t ) − K ⎡t − ( 2 − t ) ⎤ u ( t − 1) − K ( 2 − t ) u ( t − 2 )
⎢⎣
⎥⎦
= Kt u ( t ) − K 4 ( t − 1) u ( t − 1) − K ( 2 − t ) u ( t − 2 )
2
2
f t = Kt u t − Kt u t − 1 + K 2 − t u t − 1 − K 2 − t u t − 2
2
2
2
2
2
)
1
Fig. 5.69
K
Kt 2
t )2
K(2
t
2
0
2
2
t
2
1
Fig. 5.70
Taking Laplace transform of individual terms, we get the Laplace transform
⎡ 2 4 e − s 2 e −2 s ⎤
2
of the functions as F s = K ⎢ 3 − 2 − 3 ⎥ = K 3 ⎡⎣1 − 2 se − s − e −2 s ⎤⎦
s
s ⎦
s
⎣s
()
f (t )
1
e i2
(d) The function can be written as
()
f t =e
=e
−t
−t
2
2
⎡ u ( t − 2 ) − u ( t − 3) ⎤ + e ⎡ u ( t − 4 ) − u ( t − 5) ⎤ + ⋅⋅⋅
() ( )
⎣
⎦
⎣
⎦
⎡ ⎡ u ( t ) − u ( t − 1) ⎤ + u ( t − 2 ) − u ( t − 3) + u ( t − 4 ) − u ( t − 5) + ⋅⋅⋅⎤
⎦
⎣⎣
⎦
⎡u t − u t −1 ⎤ + e
⎣
⎦
−t
−t
2
2
Taking Laplace transform of individual terms, we get the Laplace transform
of the functions as
()
(
− s+ 1
)
(
−2 s + 1
)
(
−3 s + 1
t
0 1 2 3 4 5 6
Fig. 5.71
)
2
2
2
e
e
e
1
−
+
−
+ ⋅⋅⋅
s+ 1
s+ 1
s+ 1
s+ 1
2
2
2
2
⎛
⎛
⎞
⎞
⎞
−( s + 1 )
−2( s + 1 )
−3( s + 1 )
1 ⎟⎡
1
⎤ ⎜ 1 ⎟⎛
2
2
2
1
−
e
+
⋅⋅⋅
=
=⎜
−
e
+
e
⎜
⎟
⎢
⎥
1
⎜ s+ 1 ⎟ ⎣
⎦ ⎜ s + 1 ⎟ ⎜⎝ 1 + e −( s + 2 ) ⎟⎠
⎝
⎝
2⎠
2⎠
F s =
Problem 5.32 Determine the Laplace transform of the following periodic waveform.
Solution Let for the first half sine wave, the transform is F1(s).
Now, f1(t) sin tu(t) sin(t
) u (t
)
1
e− s 1+ e− s
=
Taking Laplace transform, F1 ( s ) = 2 + 2
s +1 s +1 s2 +1
f (t )
1
0
Fig. 5.72
2
3
t(second )
4
294
Network Analysis and Synthesis
By the theory of periodicity of Laplace transform, the Laplace transform of the full periodic waveform will be,
1
1+ e− s
1
=
×
[ T
1 − e − Ts
s2 +1 1− e− s
⎛ 1+ e− s ⎞ 1
⎛ s⎞
1
= 2 coth ⎜ ⎟
=⎜
− s⎟ 2
⎝ 2⎠
⎝ 1− e ⎠ s +1 s +1
F ( s ) = F1 ( s ) ×
for the waveform given]
f(t)
Problem 5.33 Determine the Laplace transform of the sawtooth
waveform as shown in Fig. 5.73.
Solution For the first cycle,
f1 ( t ) =
1
1
1
r (t ) − u(t − T ) − r (t − T )
T
T
T
0
2T
3T
4T
t
Taking Laplace transform,
Fig. 5.73
)
(
1 1 1 − Ts 1 1 − Ts 1
− e −
e = 2 ⎡⎣1 − 1 + Ts e −TTs ⎤⎦
T s2 s
T s2
Ts
By Scalling theorem (the theory of periodicity), the Laplace transform of the given periodic function is
F1 ( s ) =
F ( s ) = F1 ( s ) ×
1
1
1
1
e − Ts
= 2 ⎡⎣1 − 1 + Ts e − Ts ⎤⎦ ×
= 2−
− Ts
− Ts
1− e
1− e
Ts
Ts s 1 − e − Ts
(
)
(
Problem 5.34 Find the Laplace transform of the waveform
shown in Fig. 5.74.
Solution Here,
V (t )
1
2
4
2
v1 (t ) = r (t ) − r (t − a ) + r (t − a )
2
a
a
a
0
− as
Taking Laplace transform,
V1 ( s ) =
)
2 1 4 e 2 2 e − as
−
+
a s2 a s2
a s2
a/2
a
2a
t
Fig. 5.74
2
− as
− as
2 ⎛
2 ⎛
1 − 2 e 2 + e − as ⎞⎟ = 2 ⎜ 1 − e 2 ⎞⎟
2 ⎜
⎠ as ⎝
⎠
as ⎝
By Scalling theorem (the theory of periodicity), the Laplace transform of the given periodic function is,
=
− as
2
⎛
⎛ as ⎞
− as ⎞
1
2 ⎛
1
2 1− e 2 ⎞
2
2
⎟ = 2 tanh ⎜ ⎟
V ( s ) = V1 ( s ) ×
= 2 ⎜1− e ⎟ ×
= 2⎜
− Ts
− as
as
−
⎝
⎠
⎝ 4⎠
1− e
1− e
as
as ⎜⎝ 1 + e 2 ⎟⎠ as
Problem 5.35 A pulse voltage of width a and magnitude 10 V is applied at time t 0 to a series RL circuit
consisting of a resistance R 4 and an inductor L 2 H. Find the current i( t). Assume zero current through
the inductor L before application of the voltage pulse.
Solution The pulse voltage can be written as v(t)
10u(t)
Applying KVL for the RL series circuit with the pulse voltage,
10u(t
()
a)
Ri t + L
( )=v t
()
di t
dt
295
Laplace Transform and Its Applications
Taking Laplace transform, RI(s)
L[sI(s)
i(0 )]
V(s)
( ) 10s (1 − e )
()
− as
With zero initial current, substituting the values we get 4 I s + 2 sI s =
⎛
− as
⎞
(
)
− as
()
(
) () (
5 1 − e − as
⎡
− as
− ass
⎤
( ) 10s ⎜⎝ 12−se+ 4 ⎟⎠ = s s + 2 = 25 (1 − e ) ⎡⎢ 1s − s +1 2 ⎤⎥ = 25 ⎢ 1s − s +1 2 − e s + se+ 2 ⎥
( )
⎣
⎦
⎦
⎣
⇒ I s =
Taking inverse Laplace transform,
)(
)
5
−2 t − a
i t = ⎡⎢ 1 − e −2 t u t − 1 − e ( ) u t − a ⎤⎥
⎦
2⎣
Problem 5.36 A voltage pulse of width b and magnitude 10 V is applied at time t 0 to a series RC circuit
1
consisting of a resistor R 1 and a capacitor C = F . Find the current i(t). Assume zero charge across
4
the capacitor C before application of the voltage pulse.
Solution The pulse voltage can be written as v(t)
10u(t)
10u(t
b)
t
1
Ri t + ∫ i t dt = v t
C −∞
()
Applying KVL for the RC series circuit with the pulse voltage,
()
()
⎡ I ( s ) v (0 − ) ⎤
( ) C1 ⎢ s + s ⎥ = V ( s )
RI s +
Taking Laplace transform,
⎢⎣
⎥⎦
()
With zero initial voltage, substituting the values we get 4 I s +
(
− bs
⎛ 1 − e − bs ⎞ 10 1 − e
⇒ I s = 10 ⎜
⎟=
s+4
⎝ s+4 ⎠
()
( ) = 10 1 − e
I s
s1
4
()
(
− bs
)
) = ⎡ 10 − 10e ⎤
− bs
⎢
⎣s+4
− 4 t −b
Taking inverse Laplace transform, i t = 10 ⎡ e − 4 t u t − e ( ) u t − b ⎤
⎣
⎦
()
s
⎥
s+4 ⎦
)
(
Problem 5.37 Find the response current of a series RL circuit consisting of a resistor R
inductor L 1 H when each of the following driving force voltage is applied:
(a) unit ramp voltage , r(t 2)
(b) unit impulse voltage (t 2)
(c) unit step voltage u(t 2)
(d) unit doublet voltage (t 2)
(e) pulse of width a and magnitude 1 V beginning at time t
2 second
Solution
(a) Unit ramp voltage
Taking Laplace transform,
(
() (
)
di
=v t =r t−2
dt
e −2 s
1
R + sL I s = 2 e −2 s ⇒ I s = 2
s sL + R
s
Applying KVL to RL series circuit,
Ri + L
)()
()
(
)
3
and an
296
Network Analysis and Synthesis
−2 s
( ) s es + 3 = e
( )
I s =
Substituting the values,
−2 s
2
⎡ K1 K 2 K 3 ⎤
+
⎢ 2 +
⎥
s s + 3⎦
⎣s
∴ K1 =
1
1
=
s + 3 s =0 3
∴ K2 =
d ⎡ 1 ⎤
1
=−
2
ds ⎢⎣ s + 3 ⎥⎦ s =0
s+3
)
(
=−
1
9
s =0
1
1
∴ K3 = 2
=
s s =−3 9
⎡1
1 ⎤
−1
9+ 9 ⎥
I s = e −2 s ⎢ 23 +
⎢s
s
s + 3⎥
⎥⎦
⎣⎢
()
()
)
(
)
(
(
1
1
1 −3 t − 2
i t = − u t − 2 + r t − 2 + e ( )u t − 2
9
3
9
Taking inverse Laplace transform,
(b) Unit impulse voltage (t 2)
di
In this case, Ri + L = v t =
dt
)
( ) (t − 2 )
Taking Laplace transform,
( R + sL ) I ( s ) = e
−2 s
()
()
⇒ I s =
(
−3 t − 2
i t = e ( )u t − 2
Taking inverse Laplace transform,
)
(
e −2 s
e −2 s
=
sL + R
s+3
) (
)
(c) Unit step voltage u(t 2)
di
In this case, Ri + L = v (t ) = u(t − 2 )
dt
−2 s
Taking Laplace transform,
( R + sL ) I ( s ) = e s
()
In this case,
Ri + L
(t
)
) (
(
)
)
(
)
2)
()
(
di
=v t = ′ t−2
dt
Taking Laplace transform,
(
⎡1
e −2 s
e −2 s
1
1 ⎤
=
= e −2 s ⎢ −
⎥
sL + R s s + 3 3
⎢⎣ s s + 3 ⎥⎦
1
1 −3 t − 2
i t = u t − 2 − e ( )u t − 2
3
3
Taking inverse Laplace transform,
(d) Unit doublet voltage
(
()
⇒ I s =
)
⎡
e −2 s
se −2 s
3 ⎤
=
= e −2 s ⎢1 −
⎥
sL + R
s+3
s + 3 ⎥⎦
⎢⎣
−3 t − 2
t − 2 u t − 2 − 3e ( ) u t − 2
( R + sL ) I ( s ) = se
Taking inverse Laplace transform,
() (
i t =
−2 s
()
⇒ I s =
)(
(e) Pulse of width a and magnitude 1 V beginning at time t
di
In this case, Ri + L = v t = u t − 2 − u t − 2 − a
dt
() (
) (
)
)
(
) ( )
( )
2 seconds
(
)
297
Laplace Transform and Its Applications
Taking Laplace transform,
( R + sL ) I ( s ) = 1s e
)
(
−2 s
1 − 2+ a s
− e( )
s
⎡
1 ⎤
− 2+ a s 1 1
) ⎤ 1
= ⎡ e −2 s − e ( ) ⎤ ⎢ −
⎥
⎦s s+3 ⎣
⎦3⎢s s+3 ⎥
⎣
⎦
−( 2 + a )s
−( 2 + a )s
−
2
s
−
2
s
⎤
1⎡e
e
e
e
= ⎢
−
−
+
⎥
3 ⎢⎣ s
s
s + 3 s + 3 ⎥⎦
−2 s
− 2+ a s
( ) e s sL− e+ R = ⎡⎣ e
(
)
I s =
Taking inverse Laplace transform,
−e (
− 2+ a s
−2 s
()
(
(
)
(
) (
)
)
)
(
Problem 5.38 Find the response current of a series RC circuit consisting of a resistor R
1
itor C = F when each of the following driving force voltage is applied:
4
(a) ramp voltage 2r( t 3)
(b) impulse voltage 2 ( t 3)
(c) step voltage 2u( t 3)
(d) doublet voltage 2 ( t 3)
Solution
(a) Ramp voltage 2r(t
3)
t
()
)
(
1
∫ idt = v t = 2r t − 3
C −∞
Ri +
Applying KVL to RC series circuit,
Taking Laplace transform,
⎛
1⎞
2 −3s
⎜⎝ R + Cs ⎟⎠ I s = s 2 e
()
()
I s =
Taking inverse Laplace transform,
⎡1
2 e −3s
e −3s
1
2 e −3s
1 ⎤
=
=
= e −3s ⎢ −
2
s
s
+
2 ⎥⎦
⎛
⎛
⎞
⎞
s s+2
4
1
⎣
s2 ⎜ R + ⎟ s2 ⎜ 2 + ⎟
Cs ⎠
s⎠
⎝
⎝
)
(
()
(
)
(
)
1
1 −2 t − 3
i t = u t − 3 − e ( )u t − 3
2
2
(b) Impulse voltage 2 (t 3)
t
1
In this case, Ri + ∫ idt = v t = 2 t − 3
C −∞
()
Taking Laplace transform,
(
)
⎛
1⎞
−3 s
⎜⎝ R + Cs ⎟⎠ I s = 2 e
()
−3 s
−3 s
−3 s
( ) ⎛ 2e 1 ⎞ = ⎛ 2e 4 ⎞ = ses+ 2 = e
( )
R+
2+
I s =
(
1
−3 t − 2
−3 t − 2 − a )
i t = u t − 2 − u t − 2 − a − e ( )u t − 2 + e (
u t −2−a
3
⎜⎝
Cs ⎟⎠ ⎜⎝
s ⎟⎠
Taking inverse Laplace transform, i(t)
(t 3) u (t 3)
2e 2(t
−3 s
3)
⎡
2 ⎤
⎢1 − s + 2 ⎥
⎣
⎦
u (t
3)
2
)
and a capac-
298
Network Analysis and Synthesis
(c) Step voltage 2u(t
3)
t
In this case,
Ri +
()
)
(
1
∫ idt = v t = 2u t − 3
C −∞
Taking Laplace transform,
⎛
1⎞
−3 s 1
⎜⎝ R + Cs ⎟⎠ I s = 2 e s
()
()
I s =
2 e −3s
2 e −3s
e −3s
=
=
⎛
⎛
s+2
1⎞
4⎞
s⎜ R + ⎟ s⎜ 2 + ⎟
s⎠
Cs ⎠
⎝
⎝
(
e 2(t
Taking inverse Laplace transform, i(t)
(d) Doublet voltage 2 ( t
Ri +
u (t
3)
3)
t
In this case,
3)
)
()
)
(
1
idt = v t = 2 ′ t − 3
C −∫∞
Taking Laplace transform,
⎛
1⎞
−3 s
⎜⎝ R + Cs ⎟⎠ I s = 2 se
()
−3 s
−3 s
2 −3 s
( ) ⎛ 2 se 1 ⎞ = ⎛2 se 4 ⎞ = ss e+ 2 = e
( )
R+
2+
I s =
Cs ⎟⎠
⎜⎝
Taking inverse Laplace transform,
()
⎜⎝
s ⎟⎠
)
(
−3 s
(
)
⎡
4 ⎤
⎢s − 2 + s + 2 ⎥
⎣
⎦
(
)
−2 t − 3
i t = ′ t − 3 − 2 t − 3 + 4e ( )u t − 3 t ≥ 3
Problem 5.39 Find the response current of a series RLC circuit consisting of a resistor R 2 , an inductor L 1 H and a capacitor C = 1 F when each of the following driving force voltage is applied:
4
(a) ramp voltage 12r(t 2)
(b) step voltage 3u(t 3)
(c) impulse voltage 3 (t 1)
(d) doublet voltage 2 (t 3)
Solution Applying KVL for the series RLC circuit we get
()
⇒ 5i t +
(a) Ramp voltage 12r(t 2)
When v(t) 12r(t 2)
()
5i t +
( )+ 1
di t
dt
( )+ 1
dt
1
∫ i (t )dt = v (t )
4
di t
⎛
4⎞
12
Taking Laplace transform, ⎜ 5 + s + ⎟ I s = 2 e −2 s
s⎠
⎝
s
()
1
()
Ri t + L
4
∫ i (t )dt = v (t )
( ) + 1 i t dt = v t
() ()
dt
C∫
di t
299
Laplace Transform and Its Applications
12 e −2 s
( ) s s + 5s + 4 = 12e
⇒ I s =
(
)
2
⎡1
1
1 ⎤
⎤
⎡
3e −2 s 4 e −2 s e −2 s
1
−2 s ⎢
3
4
−
+
−
+ 12 ⎥ =
⎢
⎥ = 12 e
⎢ s s +1 s + 4 ⎥
s
s +1 s + 4
⎢⎣ s s + 1 s + 4 ⎥⎦
⎢⎣
⎥⎦
−2 s
(
)(
)
2)
4e (t
Taking inverse Laplace transform, we get,
i(t) 3u(t
(b) Step voltage 3u(t 3)
When v(t) 3u(t 3)
()
5i t +
Taking Laplace transform,
()
⇒ I s =
( )+ 1
di t
e 4(t
2)
t
∫ i (t )dt = v (t )
4
()
( s + 5s + 4 )
2
= 3e
−2 s
⎡ 1
1 ⎤
⎡
⎤
e −2 s e −2 s
1
−2 s ⎢
3
− 3 ⎥=
−
⎢
⎥ = 3e
⎢ s + 4 s + 1 ⎥ s + 4 s +1
⎢⎣ s + 1 s + 4 ⎥⎦
⎢⎣
⎥⎦
)(
(
)
()
− t −3
−4 t − 3
i t =e ( ) +e ( ) t ≥3
Taking inverse Laplace transform, we get
(c) Impulse voltage 3 (t 1)
When v(t) 3 (t 1)
()
5i t +
( )+ 1
di t
dt
1
∫ i (t )dt = v (t )
4
⎛
4⎞
−s
⎜⎝ 5 + s + s ⎟⎠ I s = 3e
()
⇒ I s =
()
(
⎡
⎤
⎡ K
K ⎤
s
−s
=
3
e
⎢
⎥ = 3e − s ⎢ 1 + 2 ⎥
2
s + 5s + 4
⎢⎣ s + 1 s + 4 ⎥⎦
⎣ s +1 s + 4 ⎦
3se − s
)
(
)(
)
⎡ s ⎤
−1
∴ K1 = ⎢
=
⎥
⎣ s + 4 ⎦ s =−1 3
⎡ s ⎤
4
∴ K2 = ⎢
=
⎥
⎣ s + 1 ⎦ s =− 4 3
⎡ 1
4 ⎤
⎢ −3
⎥ 4e − s e − s
∴ I s = 3e ⎢
+ 3 ⎥=
−
⎢ s + 1 s + 4 ⎥ s + 4 s +1
⎢⎣
⎥⎦
−4( t −1)
− t −1
Taking inverse Laplace transform, we get, i t = 4 e
− e ( ) t ≥1
()
−s
()
(d) Doublet voltage 2 (t 3)
When v(t) 2 (t 3)
2
⎛
4⎞
3 −3s
⎜⎝ 5 + s + s ⎟⎠ I s = s e
3e −2 s
Taking Laplace transform,
1
dt
2)
300
Network Analysis and Synthesis
( )+ 1
di t
()
5i t +
Taking Laplace transform,
∫ i (t )dt = v (t )
4
⎛
4⎞
−3 s
⎜⎝ 5 + s + s ⎟⎠ I s = 2 se
()
()
⇒ I s =
Let,
1
dt
(
2 s 2 e −3s
⎡
10 s + 8 ⎤
= e −3s ⎢ 2 − 2
⎥
s + 5s + 4
⎣ s + 5s + 4 ⎦
2
)
K
K
10 s + 8
= 1 + 2
s
+
1
s
+4
s + 5s + 4
2
⎡ 10 s + 8 ⎤
−2
∴ K1 = ⎢
⎥ = 3
4
s
+
⎣
⎦ s =−1
⎡ 10 s + 8 ⎤
32
=
∴ K2 = ⎢
⎥
⎣ s + 1 ⎦ s =−4 3
⎡ 2
32 ⎤
−
⎢
⎥
2 e −3s 32 e −3s
∴ I s = 2 e −3s − e −3s ⎢ 3 + 3 ⎥ = 2 e −3s +
−
3 s +1 3 s + 4
⎢ s +1 s + 4 ⎥
⎢⎣
⎥⎦
()
()
)
(
2 − t − 3 32 −4 t − 3
i t =2 t−3 + e ( ) − e ( ) t ≥3
3
3
Taking inverse Laplace transform, we get
Problem 5.40 A voltage pulse of magnitude 6 V and duration 3 seconds to 6 seconds is applied to a
series RL circuit consisting of R 6 and L 2 H. Obtain the current i(t). Also calculate the voltage across
L and R.
Solution Applying KVL for the series RL circuit,
( R + sL ) I ( s ) = 6s ⎡⎣ e
Taking Laplace transform,
−3 s
Ri + L
) (
(
)
di
= v (t ) = 6 ⎡⎣ u t − 3 − u t − 6 ⎤⎦
dt
− e −6 s ⎤⎦
6 ⎡ e −3s − e −6 s ⎤ 6 ⎡ e −3s − e −6 s ⎤ 3 ⎡ e −3s − e −6 s ⎤
1 ⎤
−3 s
−6 s ⎡ 1
⇒ I s = ⎢
⎥= ⎢
⎥ = ⎡⎣ e − e ⎤⎦ ⎢ −
⎥= ⎢
⎥
s ⎣ R + sL ⎦ s ⎣ 6 + 2 s ⎦ s ⎣ s + 3 ⎦
⎣ s s + 3⎦
()
)( )(
)( )
di
Voltage across inductor, v = L = 2 ⎡{− ( −3) e ( ) } u ( t − 3) − {− ( −3) e ( ) } u ( t − 6 ) ⎤
⎣⎢
⎦⎥
dt
Taking inverse Laplace transform,
() (
−3 t − 3
−3 t − 6
i t = 1− e ( ) u t − 3 − 1− e ( ) u t − 6
−3 t − 3
−3 t − 6
L
(
)
(
−3 t − 3
−3 t − 6
= 6e ( )u t − 3 − 6e ( )u t − 6
Voltage across resistor,
(
)(
)
) (
)(
)
−3 t − 3
−3 t − 6
vR = Ri = 6i = 6 ⎡⎢ 1 − e ( ) u t − 3 − 1 − e ( ) u t − 6 ⎤⎥
⎣
⎦
301
Laplace Transform and Its Applications
Problem 5.41 Voltage having waveform of truncated ramp as shown
in Fig. 5.75 is applied to an RL series circuit consisting of a resistor R 3
and inductor L 1 H. The rise time t0 2 s. Find the current i(t).
Solution The applied voltage can be synthesized in terms of two ramp
functions as v t = 1 r t − 1 r t − t
0
t0
t0
Applying KVL for the series RL circuit,
1
1
di
Ri + L = v t = r t − r t − t0
dt
t0
t0
()
()
()
( R + sL ) I ( s ) = t1 ⎡⎢ s1 − s1 e
⎣
2
0
⎣
2
− t0 s
2
( ) t1 ⎡⎢ s1 − s1 e
⇒ I s =
t0
0
t
Fig. 5.75
)
(
0
Let
1
)
(
()
Taking Laplace transform,
v(t )
− t0 s
2
⎤
⎥
⎦
) ( )
⎤⎛ 1 ⎞ 1
1
− t0 s
⎥ ⎜ R + sL ⎟ = t 1 − e
2
⎠
⎝
s
s
+3
⎦
0
(
K
K K
1
= 1+ 2+ 3
s s+3
s2 s + 3 s2
⎡ 1 ⎤
1
∴ K1 = ⎢
=
⎥
⎣ s + 3 ⎦ s =0 3
)
(
⎡d 1 ⎤
1
∴ K2 = ⎢
=−
⎥
9
⎣ ds s + 3 ⎦ s =0
⎡1⎤
1
∴ K3 = ⎢ 2 ⎥ =
⎣ s ⎦ s =−3 9
( ) t1 (1 − e
∴I s =
− t0 s
0
) ⎡⎢⎣ 13 ⎛⎜⎝ s1 ⎞⎟⎠ − 19 ⎛⎜⎝ 1s ⎞⎟⎠ + 19 ⎛⎜⎝ s +1 3⎞⎟⎠ ⎤⎥⎦ = t1 (1− e ) ⎡⎢⎣− 19 ⎛⎜⎝ 1s ⎞⎟⎠ + 13 ⎛⎜⎝ s1 ⎞⎟⎠ + 19 ⎛⎜⎝ s +1 3⎞⎟⎠ ⎤⎥⎦
− t0 s
2
2
0
⎤
1⎡ 1 1
1
1⎡ 1 1
1 −3 t −t ⎤
Taking inverse Laplace transform, i t = ⎢ − + r t + e −3t ⎥ u t − ⎢ − + r t − t0 + e ( 0 ) ⎥ u t − t0
9
9
t0 ⎣ 9 3
t0 ⎣ 9 3
⎦
⎦
where, t0 2 s.
()
()
()
Problem 5.42 Figure 5.76 shows a staircase voltage
waveform. Assuming that the staircase is not repeated,
express its equation in terms of step functions. If this voltage
is applied to a series RL circuit with R 2 ohm and L 1 H,
find an expression for the resulting current i(t); i(0 ) 0.
() (
) (
) (
) (
) (
)
(
Taking Laplace transform,
1
V s = ⎡⎣ e −2 s + e −4 s + e −6 s + e −8 s + e −10 s − 5e −12 s ⎤⎦
s
()
)
4
6
(
Voltage in volt
5
Solution Here, the applied voltage is a combination of several shifted step functions and can be written as
v t = u t − 2 + u t − 4 + u t − 6 + u t − 8 + u t − 10 − 5u t − 12
(
)
4
3
2
1
0
2
8
10
Time t in seconds
Fig. 5.76
12
)
302
Network Analysis and Synthesis
If this voltage is applied to RL series circuit, applying KVL we get
( R + sL ) I ( s ) = V ( s ) = 1s ⎡⎣ e
Taking Laplace transform,
()
⇒
I s =
−2 s
()
Ri t + L
( )=v t
()
dt
di t
+ e −4 s + e −6 s + e −8 s + e −10 s − 5e −12 s ⎤⎦
1
⎡ e −2 s + e −4 s + e −6 s + e −8 s + e −10 s − 5e −12 s ⎤
⎦
s s+2 ⎣
(
)
1 ⎡1
1 ⎤ −2 s −4 s −6 s −8 s −10 s
⎡ e + e + e + e + e − 5e −12 s ⎤
= ⎢ −
⎦
2 ⎣ s s + 2 ⎥⎦ ⎣
Taking inverse Laplace transform,
()
)
(
)
(
(
)
(
1
1
1
1
−2 t − 2
−2 t − 4
−2 t − 6
−2 t −8
i t = ⎡1 − e ( ) ⎤ u t − 2 + ⎡1 − e ( ) ⎤ u t − 4 + ⎡1 − e ( ) ⎤ u t − 6 + ⎡1 − e ( ) ⎤ u t − 8
⎦
⎦
⎦
⎦
2⎣
2⎣
2⎣
2⎣
1⎡
5
−2 t −10
−2 t −12
+ 1 − e ( ) ⎤ u t − 10 − ⎡1 − e ( ) ⎤ u t − 12
⎦
⎦
2⎣
2⎣
)
(
(
)
Problem 5.43 The unit step response of a network is given by (1
response h(t) of this network.
Solution Here, the input is,
() (
and the output is y t = 1 − e − bt
By convolution theorem,
)
e bt). Determine the unit impulse
( ) ( ) ⇒ W ( s ) = 1s
w t =u t
) ⇒ Y ( s ) = 1s − s +1 b = s( sb+ b )
()
() ()
1
b
b
⇒
= H (s) ⇒ H (s) =
s
+
s( s + b)
s
( b)
Y s =H s W s
Taking inverse Laplace transform, the impulse response is, h(t)
be bt
Problem 5.44 The unit impulse response of current of a circuit having R 1 ohm and C 1 F in series is
given by [ (t) exp( t) u (t)]. Find the current expression when the circuit is driven by the voltage given as
[1 exp( 2t)] u (t).
()
()
( ) ()
()
Solution Here, the impulse response is h t = ⎡⎣ t − exp −t u t ⎤⎦ ⇒ H s = 1 −
()
( ) ( ) ⇒ W ( s ) = 1s − s +1 2 = s s2+ 2
( )
The input is, w t = ⎡⎣1 − exp −2t ⎤⎦ u t
By convolution theorem, the output is given by
()
( ) ( ) s +s 1 × s s2+ 2 = s + 1 2 s + 2 = s 2+ 1 − s +2 2
( ) ( )( )
taking inverse Laplace transform, y ( t ) = ( 2 e − 2 e )
Y s =H s W s =
−t
−2 t
1
s .
=
s +1 s +1
303
Laplace Transform and Its Applications
)
(
Problem 5.45 The response of a network to an impulse is h ( t ) = 0.18 e − 0.32 t − e −2.1 t . Find the response of
the network to a step function using the convolution theorem.
Solution By convolution theorem,
A3
A
A2
⎡ 1
1 ⎤ 1
0.32
Y s = H s W s = 0.18 ⎢
−
× =
= 1+
+
⎥
s s + 0.32 s + 2.1
⎣ s + 0.32 s + 2.1 ⎦ s s s + 0.32 s + 2.1
()
() ()
)(
(
∴ A1 =
∴ A2 =
∴ A3 =
(
0.32
s + 0.32 s + 2.1
)(
0.32
s s + 2.1
(
)
0.32
s s + 0.32
(
)
)
= 0.477
s =0
= −0.562
s =−0.32
)
= 0.0856
s =−2.1
0.562 0.0856
−
+
( ) 0.477
s
s + 0.32 s + 2.1
Taking inverse Laplace transform, y ( t ) = 0.477 − 0.562 e
Putting these values, Y s =
− 0.32 t
+ 0.0856 e −2.1t
Summary
1. Laplace transform is defined as
∞
L [f (t )] = F ( s ) = ∫ f (t )e − st dt
0−
where s
complex frequency
(
j ), with,
Real part of s neper frequency and =
Imaginary part of s radian frequency.
2. Laplace transform of some functions are listed in
Table 5.1.
3. The Laplace transform of a periodic function is equal
⎛ 1 ⎞
times the Laplace transform of the first
to ⎜
⎝ 1− e −Ts ⎟⎠
cycle where T is the time period of the function.
4. Inverse Laplace transform can be found by using
partial fraction expansion method and using Laplace
transform pairs as listed in Table 5.1.
5. In Laplace transform domain, the passive circuit elements are replaced as follows.
Resistor vR Ri → VR RI
di
Inductor v L = L → VL sL Li(0 )
dt
t
I v (0 − )
1
−
Capacitor v C = ∫ idt → v C =
sC
s
C −∞
6. Laplace transform is a powerful transform method
for solving network analysis problems. This method
is generally used to find the complete response (both
transient and steady state) of a circuit.
7. If h(t ) is the impulse response of a linear network
then the response of the same network y (t ) subject
to any arbitrary input w(t ) is given by the convolution
integral as
∞
∞
−∞
−∞
y (t ) = ∫ h ( )w (t − )d = ∫ w ( )h (t − )d
8. If f1(t ) and f2(t ) are two functions of time which are
zero for t
0, and if their Laplace transforms are
F1(s) and F2(s ), respectively then the convolution
theorem states that the Laplace transform of the
convolution of f1(t ) and f2(t ) is given by the product
F1(s) F2(s).
Mathematically, the convolution of f1(t ) and f2(t ) is
written as
)
)
t
)
)
t
) )
f 1 (t * f 2 (t = ∫ f 1 ( f 2 (t − d = ∫ f 1 (t − f 2 ( d
0
0
() ()
= f2 t * f1 t
304
Network Analysis and Synthesis
Short-Answer Questions
1. Discuss the advantages of the Laplace transform
method over the conventional classical methods of
solving the linear differential equations with constant coefficients.
If
0 then the variation of the real and imaginary
parts of the function is shown below.
1. It gives complete solution.
2. Initial conditions are automatically considered in
the transformed equations.
3. Much less time is involved in solving differential
equations.
4. It gives systematic and routine solutions for differential equations.
f(t)
Advantages of Laplace Transform Method
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (second)
2. What do you understand by ‘complex frequency’?
Give its physical significance.
i(t )
I0e t ⇒
Damped cosinusoidal
1
f(t)
0.5
0
0.5
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (second)
Fig. 5.78
If
0 then the variation of the real and imaginary
parts of the function is shown below.
f(t)
Complex frequency The complex frequency (s) is the
sum of two frequencies, the real and imaginary.
s complex frequency (
j )
where,
Real part of s neper frequency
Imaginary part of s radian frequency
The general solution of the differential equation in
time-domain is
j )
i(t ) I0e st, where s (
st
Since e is a dimensionless quantity and so, also, the
product ‘st ’ a dimensionless quantity, the unit of ‘s’
must be (time) 1 or Hz.
Here,
is interpreted as radian frequency; as
radian is a ratio of two lengths, ‘ ’ is effectively (time) 1,
i.e. , frequency in Hz.
Also, as and must have the same dimension, i.e.,
0,
the dimension of should be (time) 1. Also, with
Damped sinusoidal
1
0.8
0.6
0.4
0.2
0
0.2
0.4
1 ⎡ i (t ) ⎤
= ln ⎢
⎥
t ⎣ I0 ⎦
8
6
4
2
0
2
4
6
8
Damped sinusoidal
0 1 2 3 4 5 6 7 8 9 10
Time (second)
Physical significance of complex frequency
We have,
i(t ) I0e st I0e ( j )t I0e t[cos t jsin t ]
If
0, then the variation of the real and imaginary
parts of the function is shown below.
Damped cosinusoidal
8
6
4
2
0
2
4
6
8
0 1 2 3 4 5 6 7 8 9 10
Time (second)
sin t
Fig. 5.77
t
cos t t
f(t )
Since the unit of ln of some number is neper, the unit
of is neper per second.
Fig. 5.79
305
Laplace Transform and Its Applications
From these figures, it is clear that
decides the number of oscillations per second
decides the magnitude of these oscillations
3. Explain why the lower limit of the Laplace trans⎛∞
⎞
form integral ⎜ ∫ f (t )e − st dt ⎟ is taken as 0 instead
⎝ 0−
⎠
of 0 .
The lower limit of the integration should be 0 instead
of 0 or simply 0. If f (t ) is continuous at t 0 then the
value of f (0) is well-defined. But, if f (t ) is not continuous at t 0, then the meaning of f (0) becomes ambiguous. To consider the effect of ‘instantaneous energy
transfer’, we must use 0 as the lower limit to include
the impulses at t 0. The use of 0 will exclude the
existence of any impulses at the origin.
So, we use 0 as the lower limit.
4. What is the Laplace transform of a function which
is non-zero for t < 0?
As the lower limit of integration of Laplace transform
is 0 , the Laplace transform does not distinguish
between functions that are different for t 0 bur identical for t 0.
For example, the Laplace transforms of u(t ) and
u(t 1) will be same.
However, t 0 is physically the starting time of a
circuit or system and all the signals considered are usually zero for t 0. For this reason, all will have unique
(one-sided) Laplace transform. Conversely, all Laplace
transform F(s ) will have a unique time function such
that f (t ) 0 for t 0.
5. Does every signal f (t ), such that f (t ) 0 for t < 0,
have a Laplace transform?
The existence of Laplace transform X(s) of a given x(t )
depends on whether the transform integral converges
)
∞
)
∞
)
X ( s = ∫ x (t dte − st dt = ∫ x (t dte − t e − j t dt < ∞
−∞
−∞
which in turn depends on the duration and magnitude
of x(t ) as well as the real part of s, Re s
(the imaginary part of s Im s
j determines the frequency of
a sinusoid which is bounded and has no effect on the
convergence of the integral).
This limits the variable s
(
j ) to a part
of the complex plane. The subset of values of s for
which the Laplace transform exists is called the
region of convergence (ROC) or the domain of convergence.
Thus, the Laplace transform F(s) typically exists for
all complex numbers such that Re{s} a, where a is a
real constant which depends on the growth behavior
of f(t), or precisely the condition is given as
)
f (t = k 1e k 2t
where, k1 and k2 are some constants
2
For example, for the function, f (t ) = e t u (t ), Laplace
transform integral becomes
∞
∞
∞
− st
t
t − st
t − t−j t
dt
∫ e e dt = ∫ e dt = ∫ e
2
2
0−
2
0−
0−
As t approaches infinity, the area under the curve
t) goes to infinity. Thus, the Laplace transform of
(t2
this function does not exist.
6. Define ROC of Laplace transform and mention its
properties.
Region of Convergence (ROC ) The existence of
Laplace transform X(s) of a given x(t ) depends on
whether the transform integral converges
∞
)
∞
)
X ( s ) = ∫ x (t e − st dt = ∫ x (t e − t e − j t dt < ∞
−∞
−∞
which in turn depends on the duration and magnitude
of x(t ) as well as the real part of s, Re s
(the imaginary part of s Im s
j determines the frequency of
a sinusoid which is bounded and has no effect on the
convergence of the integral).
This limits the variable s (
j ) to a part of the
complex plane. The subset of values of s for which the
Laplace transform exists is called the region of convergence (ROC) or the domain of convergence.
Thus, the Laplace transform F(s) typically exists for
all complex numbers such that Re{s} a, where a is a
real constant which depends on the growth behavior
of f (t ), whereas the two-sided transform is defined in a
range a Re{s} b. In the two-sided case, it is sometimes called the strip of convergence.
Properties of region of convergence
1. If x(t ) is absolutely integrable and of finite duration then the ROC is the entire s-plane (the Laplace
transform integral is finite, i.e., X(s) exists, for any s).
2. The ROC of X(s) consists of strips parallel to the
j -axis in the s-plane.
3. If x(t ) is right sided and Re s
0 is in the ROC then
any s to the right of 0 (i.e., Re s
0 ) is also in the
ROC, i.e., ROC is a right-sided half plane.
4. If x(t ) is left sided and Re s
0 is in the ROC then
any s to the left of 0 (i.e., Re s
0 ) is also in the
ROC, i.e., ROC is a left-sided half plane.
306
Network Analysis and Synthesis
5. If x(t ) is two-sided then the ROC is the intersection
of the two one-sided ROCs corresponding to the
two one-sided components of x(t ). This intersection
can be either a vertical strip or an empty set.
6. If X(s) is rational then its ROC does not contain any
= ∞ dose not exist). The
poles (by definition X s
()
s =s p
ROC is bounded by the poles or extends to infinity.
7. If X(s) is a rational Laplace transform of a right-sided
function x(t ) then the ROC is the half plane to the
right of the rightmost pole; if X(s) is a rational Laplace
transform of a left-sided function x(t ) then the ROC
is the half plane to the left of the leftmost pole.
8. A signal x(t ) is absolutely integrable, i.e., its Fourier transform X( j ) exists (first Dirichlet condition,
assuming the other two are satisfied), if and only if
the ROC of the corresponding Laplace transform
X(s) contains the imaginary axis Re s
0 or s j .
7. Derive from the first principle the Laplace transform of a unit step function. Hence or otherwise,
determine the Laplace transform of unit ramp
function and unit impulse function.
A unit step function is defined as given below.
u(t)
f (t ) u(t ) 1 for t 0
1
0 for t 0
t
0
and is undefined at t 0.
The Laplace transform of a unit Fig. 5.80 (a)
step function is given as,
Unit step
)
∞
function
∞
L ⎡⎣u (t ⎤⎦ = ∫ u (t ).e − st dt = ∫ 1.e − st dt
0−
0−
∞
⎡ e − st ⎤
1 1
=⎢
=
⎥ =0−
−
s
−
s s
⎣
⎦0 −
A unit ramp function is defined as
f (t ) r(t ) t for t 0
0 for t 0
We see that unit ramp function is the integrations of unit
step function. Hence, by integration property of Laplace
transform, the Laplace transform of unit ramp function is
obtained as
)
)
r(t)
1
1
0
Fig. 5.80 (b)
Unit ramp
function
)
1
1 1 1
L ⎡⎣ r (t ⎤⎦ = L ⎡⎣ ∫ u (t dt ⎤⎦ = L ⎡⎣u (t ⎤⎦ = × = 2
s
s s s
A unit impulse function is defined as
t
∞
(t )
0 for
t
∫ (t )dt = 1
0 and
−∞
We see that an impulse function is the derivative of
a step function. Hence, by differentiation property
of Laplace transform, the Laplace transform of unit
impulse function is obtained as
⎡d
⎤
1
L ⎡⎣ (t ⎤⎦ = L ⎢ u (t ⎥ = sL ⎡⎣u (t ⎤⎦ = s × = 1
s
⎣d
⎦
)
)
8. Explain gate function.
Obtain the equation of a
gate function starting at
origin and duration T.
)
g(t)
K
a
0
b
Gate function
A gate
function is shown in figure. Fig. 5.81 Gate function
It can be obtained from step function as follows.
Therefore, g(t ) Ku(t a) Ku(t b)
The Laplace transform of the gate function is
K − as
e − e − bs
obtained as L ⎡⎣ g (t ) ⎤⎦ =
s
(
)
9. What do you understand by transient and steadystate response? How can they be identified in a
general solution?
In electrical engineering, a transient response or natural response is the electrical response of a system to
a change from equilibrium.
The condition prevailing in an electric circuit between
two steady-state conditions is known as the transient
state; it lasts for a very short time. The currents and voltages during the transient state are called transients.
In general, transient phenomena occur whenever
(i) a circuit is suddenly connected or disconnected
to/from the supply,
(ii) there is a sudden change in the applied voltage
from one finite value to another, or
(iii) a circuit is short-circuited.
The transient currents are not caused by any part of
the supply voltage, but are entirely associted with the
changes in the stored energy in capacitors and inductors. As there is no energy stored in resistors, there are
no transients in purely resistive circuits.
When the transient phenomena die out, the circuit
becomes steady and the state of the circuit is called
‘steady state’.
In electrical engineering, a simple example would
be the output of a 5-volt dc power supply. when it is
turned on; the transient response is from the time the
307
Laplace Transform and Its Applications
1.0 A
0.5 A
0A
0s
0.5 s
I( R1)
1. 0s
1.5s
2.0s
2. 5s
Time
3 .0s
3.5 s
4.0s
4.5s
5 .0 s
Fig. 5.82
switch is turned on and the output is a steady 5 volts.
At this point, the power supply reaches its steady-state
response of a constant 5 volts.
Another practical example will be an RC series circuit. When it is suddenly switched to a dc supply, the
transient current through the circuit is the maximum
and it gradually decreases so that the steady state current in the circuit becomes zero.
In a general solution, the part of the solution that
diminishes with time is identified as the transient part,
and the part that exists with time is identified as the
steady-state part. For example, for the general solution, f (t ) A Be t, the transient response is Be t and
steady state response is A.
10. What do you understand by initial conditions
before and after switching?
It is possible that a capacitor or an inductor might have
been used in some other circuit earlier, where it absorbed
some energy and then it was disconnected. Because of
its non-dissipative nature, the energy was stored within
the capacitor (or the inductor). Now, as this capacitor
(or inductor) is connected to a circuit, it gets some path
to release its stored energy. This stored energy is represented by the initial voltage VC(0) or initial current IL(0).
11. Discuss the advantages of analyzing the circuits using
frequency domain rather than the time domain.
The following are some advantages of analyzing an
electrical network in s-domain rather that in t- domain:
1. Each element can easily be replaced by a transform
impedance.
2. No integration or differentiation is involved in the
transform equations.
3. The response obtained after solution is a complete
response, i.e., both the steady state and transient
responses are obtained.
12. Define and distinguish between zero Input
Response (ZIR) and Zero State Response (ZSR).
Zero Input Response (ZIR) In circuit thory, the
Zero Input Response or ZIR is the behavior or
response of a circuit with zero inputs. The ZIR
results only from the initial state of the circuit and
not from any external source or forcing function.
The ZIR is also called the natural response, and the
resonant frequencies of the ZIR are called the natural frequencies.
Zero State Response (ZSR) In electrical circuit theory,
the Zero State Response or ZSR is the behavior or
response of a circuit with zero initial conditions. The
ZSR results only from the external inputs or driving
functions of the circuit and not from the initial conditions. Such a network is said to be an initially relaxed
network. The ZSR is also called the forced or driven
response of the circuit.
The total response of the circuit is the superposition of the ZSR and the ZIR.
13. Explain under what condition, an RC series circuit
behaves as
a) Differentiator
c) Coupling network
b) Low-pass filter
d) Integrator
We consider the RC series circuit.
308
Network Analysis and Synthesis
(a) RC series cirR
cuit as differentiator We have an ac
Vi(t)
C
i
source with voltage
vin(t), input to an RC
series circuit. This Fig. 5.83
time the output is the
voltage across the resistor.
We consider only low frequencies << 1/RC, so
that the capacitor has time to charge up until its voltage almost equals that of the source.
⎛ 1 ⎞
V in = IZ = I R 2 + ⎜
⎝ C ⎟⎠
R <<
But,
2
I
1
, so V in =
C
C
C
i
i
Vin
VC
vOut
R
RC dvin /dt
( << 1/RC)
Fig. 5.84
1
, V ≅V
RC in C
d
dq
∴V out = V R = iR = R
= R CVC
dt
dt
For frequencies,
∴g =
Gain
V0
Vi 1
0.707
0
(b) RC series circuit as low pass filter If the RC series
circuit is supplied with a frequency-varying source
then it will act as a low-pass filter if the output is taken
as the voltage across the capacitor.
The voltage across the capacitor is IX C =
I
C .
The voltage across the series combination is:
2
⎛ 1 ⎞
IZ = I R 2 + ⎜
, so the gain is
⎝ C ⎟⎠
R i
i
VOut
( C)
2
1
1+ ( RC )2
fc
Stop-band
Frequency
Fig. 5.86
d
V
dt in
Vseries
C
C
R + 1
2
Actual characteristics
Pass-band
Thus, the output is the differentiation of the input
and the RC series circuit acts as differentiator.
Vin
1
Here, at low frequencies, the capacitive reactance
⎛
1 ⎞
⎜⎝ X C = j 2 fC ⎟⎠ is very high and therefore the circuit
can be considered as an open circuit. Under these
conditions, the input signal is equal to output signal.
At very high frequencies, the capacitive reactance
⎛
1 ⎞
⎜⎝ X C = j 2 fC ⎟⎠ is very low and therefore the output
signal is very small as compared with the input signal.
Thus, the circuit acts as low pass filter with the frequency characteristics as shown in Fig. 5.86.
<<
∴V out ≅ RC
Fig. 5.85
V
IX
g ≡ out = C =
V in IZ
VC
c) RC series circuit as coupling network A coupling network is used for coupling a signal at a frequency from a voltage source to a load. The voltage
source has a source resistance. The load has a load
resistance and a load reactance. The ratio of the load
reactance to the load resistance is greater than 100.
The coupling network includes a reactive element
and a delay circuit. The reactive element is arranged
in series with the load to resonate with the load reactance at the frequency. The delay circuit is between
the reactive element and the source, has a delay
equivalent to a quarter wavelength transmission line
at the frequency and has a characteristic impedance
equal to the square root of the product of the values
of the load resistance and the source-required resistance.
Thus, an RC series circuit will act as coupling network only when the ratio of load resistance to load
reactance is greater than 100 and the suply frequency is such that the capacitor resonates at that
frequency.
309
Laplace Transform and Its Applications
d) RC series circuit as integrator We have an ac
source with voltage vin(t), input to an RC series circuit.
The output is the voltage across the capacitor.
We consider only high frequencies
1/RC, so
that the capacitor has insufficient time to charge up,
its voltage is small, so the input voltage approximately
equals the voltage across the resistor.
⎛ 1 ⎞
V in = IZ = I R + ⎜
⎝ C ⎟⎠
V out = VC =
R i
i
2
Vin
>>
iR
VOut
C
C >> 1 , so V in ≅ IR
R
For frequencies,
1
V dt
RC ∫ in
Vout ≅
2
But,
1
1 V
idt ≅ ∫ in dt
C∫
C R
1/RC Vindt
(v >> 1/RC)
Fig. 5.87
Thus, the voltage vC is the integration of the input
voltage and hence the RC series circuit acts as an
integrator.
1
, V ≅V
RC in R
Exercises
1. (a) Find the initial values of the functions:
()
(i) f t = e
− at
()
10
[(i) 1, (ii) 2]
cos tu t
2 ( s + 1)
( ) s + 2s + 5
(ii) F s =
i2(t)
100 V
(b) Find the final value of the functions:
7
7
[(i)
, (ii) 0]
(ii) F s =
2
9
s s +3
()
)
(
s −1
(ii) F ( s =
(s + 1 (s + 2
)
)
)
Fig. 5.89
2
()
(i) f(t )
(ii) f(t)
V
A
Fig. 5.88(a)
1
2
S
10 V
()
V
100␮F
4. Find for the circuit shown, the current through C using
Laplace transform. The switch is closed at t 0 and the
initial charge in the capacitor, i.e., at t 0 is zero.
[10sin 100t (A)]
2. Obtain the Laplace transform of the following functions:
V
[(i) F s = 2 (1− e −Ts − se −Ts )
Ts
A
(ii) F s = 2 (1− e −Ts −Te −Ts ) ]
Ts
0
1H
2
100␮F
Fig. 5.90
5. The circuit of Fig. 5.91 was initially in the steady state with
the switch S in the position a. At t 0, the switch goes from
a to b. Find an expression for the voltage v0(t) for t 0.
Take the initial current in the inductor L2 to be zero.
1 − 3t
[ v 0 (t ) = e 2 ( V ) ]
2
t
0
T
a
t
R1
Fig. 5.88(b)
3. In the network shown, the switch is closed and a steady
state is reached in the network. At time t 0, the switch
is opened. Find an expression for the current through
[10 cos 100t (A)]
the inductor i2(t).
2V
Fig. 5.91
b
2
L1
2H
R2
1
L2
1H V
0
310
Network Analysis and Synthesis
6. In the circuit of Fig. 5.92, the applied voltage is v(t)
10sin(10t
␲/6), R
1 ,C
1 F. Using Laplace
Transformation, find complete solution for current i(t).
Switch K is closed at time t
0. Assume zero charge
across the capacitor before switching.
5
100
(1− 10 3 )e −t +
cos(10t − 54° 8 ')( A ) ]
[ i (t ) =
101
101
(b) i2(t)
5 16.3375e 0.707t
i1
1H
1
2
1
1
5u(t)
v1(t)
1
K
V
(b) determine i2(t), using the Laplace transform
method if k1 3.
(a) va(t) 4 e 0.75t (1.5cos0.25t 0.5sin0.25t)
Va
1
1.3375e 0.707t (A)]
k1i1
1
1F
1F
i2(t)
i(t )
Fig. 5.95
Fig. 5.92
7. A series RLC circuit, with R 5⍀, L 0.1 H and C
500 μF, has a sinusoidal voltage source, v
1000 sin250t. Find the resulting current if the switch
is closed at t 0.
[i(t) e 25t (5.42cos139t 1.89sin 139t) 5.65sin(250t
73.6 ) (A)]
8. The two-mesh network shown in Fig. 5.93 contains a
sinusoidal voltage source, v
100 sin(200t
␾)(V).
The switch is closed at an instant when the voltage
is increasing at its maximum rate. Find the resulting
mesh currents, with directions as shown in the figure.
[i1(t) 3.01e 100t 8.96 sin(200t 63.4 )
i2(t) 1.505e 100t 4.48 sin(200t 63.4 )]
50 mH
V
i1
1
−t
b
a
2
⎛ 3 ⎞ 20 − t
⎛ 3 ⎞
cos ⎜
t⎟−
e 2 sin⎜ t ⎟ (V))]
3
⎝ 2 ⎠
⎝ 2 ⎠
s
10
1H
iL(t ) 1F
1
Fig. 5.96
i2
100
0; assume the all initial conditions to be
10
1
100 V
Vc(t)
12. Find the source current after the switch is closed at
t 0. Take initial current to be zero. [(3 e 25t)(A)]
[ i 2 (t ) =
2
[(i) 5 A, 5 V; (ii) 15 − 10e
5V
Fig. 5.93
9. Find i2(t) for t
zero.
(i) Determine initial values for iL(t) and Vc(t) with switch
in the position b.
(ii) Determine Vc(t) for t 0. Sketch Vc(t) as a function
of time.
(iii) Determine damping ratio, undamped and damped
natural frequencies.
10 V
2
10
11. The network shown in Fig. 5.96, has reached steady
state when the switch S moves from a to b.
10 5 −30t
+ e
− 5e −10t , for t
3 3
1H
1H
10
10
t
0]
i2(t)
Fig. 5.94
10. In the network shown, in Fig. 5.95,
(a) determine Va(t), using the Laplace transform
3.
method if k1
100 V
0
50
4H
Fig. 5.97
13. Find an expression for the current in the inductor at time
t after the switch is closed. What is the final value of the
current and how long will it take for the inductor current
to reach 95% of its final value?
− 50 t
[ 2 ⎛⎜ 1− e 6 ⎞⎟ A ; 2 A, 0.36 second]
⎝
⎠
( )
311
Laplace Transform and Its Applications
t
50
0
4H
100
)
(
18. Show that the Laplace transform of the square wave is
100 V
F (s ) =
Fig. 5.98
1
s (1+ e − as )
f(t)
14. In the circuit, find the initial and final values of currents
i1 and i2 when the switch is closed at t 0. Use initialvalue and final-value theorems.
[i1(0)
)
(
1 −2 t − 3
1
u t − 3 − e ( )u t − 3 ]
2
2
[(a) e 2(t 3) u(t 3) ; (b)
iL
7.14 A, i1 ( )
10 A; i2 (0)
7.14 A, i2( )
1
0 A]
0
15
2a
a
3a
4a
t (second)
Fig. 5.102
t
0
30 mH
150 V
6
i2
i1
19. Determine the current response of a series RL circuit
with R 6 and L 3 H for each of the following driving voltages:
(a) a step voltage 2u(t 2)
(b) a ramp voltage 2r(t 3)
Fig. 5.99
15. In the network shown in Fig. 5.100, the switch is closed at
t 0, prior to which the circuit is in the zero state. Using
Thevenin’s theorem, transform the circuit to the left of
points A and B into its Thevenin equivalent in frequency
domain and find the current in the 30- resistance. Convert the expression for current in the time domain.
t
[0.1818
0.265e 13.14t
10
1H
0.083e 41.86t (A)]
A
2H
0
20
10 V
B
2
103(t
t1)} for t
t1]
2 H
S1
10 V
(b)
3H
S2
1 F
Fig. 5.101
17. An RC series circuit has R 2 , C 0.25 F. Find the current response if the driving voltage is (a) step voltage
2u(t 3), and (b) ramp voltage 2r(t 3).
)
1 −2(t − 3)u (t − 3) ⎤
1
2⎡
⎢2 r t − 3 − u t − 3 + e
⎥]
4
4
3⎣
⎦
(
)
)
(
20. A series RL circuit has a resistor R 4 and an inductor L 2 H. A pulse of magnitude 10 V and duration
5 ms is applied to the circuit at t
3 ms. Find i(t).
Assume that the circuit was initially relaxed.
⎡5
⎢ ⎡⎣u t − 0.003 − u t − 0.008 ⎤⎦ −
⎣2
) (
(
)
⎤
5 −2 t − 0.003)
−2 t − 0.008 )
− ⎡e (
u t − 0.003 − e (
u t − 0.008 ⎤ ⎥
⎦⎦
2⎣
(
16. The network shown in Fig. 5.101 is in steady state with
switches S1 and S2 open. At t t1, S1 is opened and S2 is
closed. Find the current through the capacitor for t t1.
)(
(
30
Fig. 5.100
[i(t) 5cos{0.577
Assume that the circuit is initially relaxed.
1
−2 t − 2
1− e ( ) u t − 2 ;
[(a);
3
)
)
(
21. A voltage pulse of 20-V magnitude and 10-μs duration is applied to an RC circuit. Determine the current.
Assume that the circuit was initially relaxed. Take R
10- and C 10 μF.
−10 −4 (t −10 −5 ) ⎤
⎡ ⎡ −10−4 t
⎤
−e
u t − 10 −5 ⎥
⎢⎣2 ⎣⎢e
⎦⎥
⎦
)
(
22. A unit doublet voltage ␦’(t 5)is applied at t 0 to a
series RLC circuit consisting of a resistor R 4 , L 1 H
1
and C
F. Determine i(t). Assume that the circuit was
3
initially relaxed.
⎡
⎤
1 −(t − 5)
9 −3 t − 5
u t − 5 − e ( )u t − 5 ⎥
⎢ t −5 + e
2
2
⎦
⎣
(
)
(
)
(
)
312
Network Analysis and Synthesis
23. Figure 5.103 shows a staircase voltage waveform.
Assuming that the staircase is not repeated, express its
equation in terms of step functions. If this voltage is
applied to a series RL circuit with R 4 ohms and L 2
H, find an expression for the resulting current i(t);
i(0 ) 0
1
farad when each of the following driving volt3
ages is applied:
C
(a) ramp voltage 9r (t 2) (b) step voltage 4u (t 3)
(c) impulse voltage 9␦ (t 1).
⎡ ⎡ 9 −(t −2 ) 3 −3(t −2 ) ⎤
− e
⎢(a ) ⎢3 − e
⎥u t − 2 ;
2
⎦
⎢ ⎣ 2
⎢
⎡e −(t − 3) − e −3(t − 3) ⎤u t − 3 ;
2
(
b
)
⎢
⎣
⎦
⎢
−3(t −1)
− t −1
u t − 1 − e ( )u t − 1
⎢(c ) 3e
⎢
⎣
4
Voltage in volts
)
(
)
(
3
2
2
4
6
8
10
Time t in seconds
26. Find the response of the network shown in Fig. 5.104 when
the input voltage is (a) unit impulse, and (b) vi(t) e 2t.
Fig. 5.103
()
)
(
)
(
(
[(a) e t; (b) (e t e 2t)]
)
(
⎡ i t = ⎡1− e −2(t −2 ) ⎤u t − 2 + ⎡1− e −2(t − 4 ) ⎤u t − 4 ⎤
⎣
⎦
⎣
⎦
⎥
⎢
⎥
⎢ ⎡
−2(t − 6 )
−2(t − 8 )
⎤
⎡
⎤
1
6
1
+
−
e
u
t
−
+
−
e
u
t
−
8
⎥
⎢ ⎣
⎦
⎣
⎦
⎥
⎢
−2(t −10 )
⎡
⎤
⎥
⎢ − 4 1− e
u t − 10
⎦
⎦
⎣ ⎣
(
⎥
⎥
⎥
⎥
⎥
⎥
⎦
25. Verify that the convolution between two functions
2
f1(t) 2u(t) and f2(t) exp ( 3t)u(t) is [1 exp( 3t)];t > 0
3
where u(t) is the unit step function.
1
0
)
(
⎤
) ⎥
(
R 1
)
)
24. Find the current i(t) in a series RLC circuit comprising
resistor R 4 , inductor L 1 henry and capacitor
Vi(t )
C
1F
V0(t)
Fig. 5.104
Questions
1. (a) What do you understand by complex frequency?
Give its physical significance.
(b) Define Laplace transform of a function f (t ). What
are the advantages of Laplace transform?
Or,
Discuss the advantages of Laplace transform method
over the conventional classical methods of solving differential equations with constant coefficient.
4. Explain why the lower limit of the Laplace transform
⎤
⎡∞
− st
integral ⎢ ∫ f (t )e dt ⎥ is taken as 0 instead of 0 .
⎥⎦
⎢⎣ 0 −
5. What is the Laplace transform of a function which is
non-zero for t 0?
6. Does every signal f (t ), such f (t )
Laplace transform?
0 for t
0, have a
(c) State and deduce initial-value and final-value theorems.
7. (a) Define unit-step, unit ramp and unit impulse functions
and derive their Laplace transform from first principles.
(d) Write notes on application of Laplace transform to
network analysis.
(b) Define ROC of Laplace transform and mention its
properties.
2. What is Laplace transformation? Give reasons for its
wide use in the electric circuit analysis.
8. Define and sketch ramp, unit step and unit impulse
functions.
3. Discuss the advantages of analyzing circuits using
frequency domain rather than time domain. How can
the initial conditions of a circuit be incorporated using
Laplace transform?
9. Derive from the first principle the Laplace transform
of a unit step function. Hence or otherwise, determine
the Laplace transform of a unit ramp function and a
unit impulse function.
313
Laplace Transform and Its Applications
10. Explain gate function. Obtain the equation of a gate
function starting at origin and duration T.
11. (a) Find the current i(t) if unit step voltage is applied to
an RL circuit.
(b) Derive an expression for the current i(t) flowing
through an RLC series circuit. Explain with suitable
sketches the variation of current with time under
three conditions:
I. Underdamped
Or,
Derive an expression for the current response in
an R-L series circuit excited with constant voltage
source.
(b) Define the term ‘time-constant’ of a circuit. What is
the physical significance of time-constant of a circuit? Find its value for an R-L series circuit.
12. (a) Derive an expression for the decay current in an RC
circuit excited by a unit step voltage. What is the
time-constant of the circuit?
Also, determine the nature of the voltage response
across the capacitor.
II. Critically damped
III. Overdamped
14. What do you understand by the impulse response of
a network? Briefly explain its importance in network
analysis.
15. What do you understand by transient and steady-state
parts of response? How can they be identified in a general solution?
Or,
Discuss the natural and steady-state response of an
electrical circuit with illustrative examples.
Or,
(b) Under what conditions an RC series circuit will act
as: i) a differentiator? ii) an integrator?
Write notes on (a) Transient and steady state response
13. (a) Explain the terms ‘critical resistance’, ‘damping ratio’
and ‘frequency’ as applied to the study of RLC series
circuit. How do they help in simplifying the analysis
of the circuit?
16. State and prove convolution theorem.What is the necessity of the convolution theorem in circuit analysis?
(b) Free and forced response
Multiple-Choice Questions
1. The condition for over-damped response of an RLC
series circuit is
1
R2
R2
1
i)
ii)
>
=
2
2
LC
LC
4L
4L
R2
1
R2
1
iii)
iv)
≤
<
2
4 L2 LC
4 L LC
2. Transient current in an RLC circuit is oscillatory when
(i) R = 2
L
C
(ii) R > 2
L
C
L
(iv) R = 0.
C
3. Laplace transform analysis gives
(i) time domain response only
(ii) frequency domain response only
(iii) both (i) and (ii)
(iv) none of these
(iii) R < 2
4. A function f(t) is shifted by a then it is correctly represented as
(i) f(t a)u(t)
(ii) f(t)u(t a)
(iii) f(t a)u(t a)
(iv) f(t a)(t a)
5. Laplace transform of a delayed unit impulse function
␦s(t) ␦(t 1) is
(i) unity
(ii) zero
(iv) s
(iii) e s
6. The condition for under-damped response of an RLC
series circuit is
i)
R2
1
=
4 L2 LC
ii)
R2
1
>
4 L2 LC
2
R2
1
iii) R < 1
iv)
≤
2
2
LC
4
L
LC
4L
7. The value of the impulse function ␦(t) at t 0 is
(i) 0
(ii)
(iii) 1
(iv) indeterminate
8. The value of the ramp function at t
is
(i) infinity (ii) unity (iii) zero (iv) indeterminate
9. The value of the ramp function at t
(i) 0
(ii)
(iii)
(iv) 1
is
10. The value of the impulse function ␦(t) for t 0 is
(i) zero
(ii) unity
(iii) k, where k is a constant
(iv) infinity
314
Network Analysis and Synthesis
11. The free response of RL and RC series networks having
a time constant is of the form
(i) A + Be
(iii) Ae
−
−
t
t
+ Be
(ii) Ae
−
t
(iv)
−
t
( A + Bt )e
(i)
−
t
12. In the complex frequency s ␴ j␻, ␻ has the units
of rad/s and ␴ has the units of
(i) Hz (ii) neper/s (iii) rad/s (iv) rad
13. Time constant of a series RC circuit is
(i) C/R
(ii) R/C
(iii) RC
(iv) 1/RC
14. Time constant of a series RL circuit is
(i) L/R
(ii) R/L
(iii) LR
(iv) 1/LR
15. A coil with a certain number of turns has a specified
time constant. If the number of turns is doubled, its
time constant would
(i) remain unaffected
(ii) become doubled
(iii) become four-fold
(iv) get halved
16. An RLC series circuit has R 1 , L 1 H and C 1 F.
Damping ratio of the circuit will be
(i) more than unity (ii) unity (iii) 0.5 (iv) zero
17. A step-function voltage is applied to an RLC series circuit having R 2 , L 1 H and C 1 F. The transient
current response of the circuit would be
(i) over-damped
(ii) critically damped
(iii) under damped
(iv) over, under or critically damped depending upon
magnitude of the step voltage
18. For an RC circuit comprising a capacitor C 2 μF in
series with a resistance R 1 M , the period 6 seconds
will be equal to
(i) one time constant
(ii) two time constants
(iii) three time constants (iv) four time constants
19. A series RL circuit with R 100 ohms; L 50 H, is supplied to a dc source of 100 V. The time taken for the
current to rise 70% of its steady-state value is
(i) 0.3 s
(ii) 0.6 s
(iii) 2.4 s
(iv) 70% of time required to reach steady state
20. If f(t) and its first derivative are Laplace transformable
then the initial value of f(t) is given by
(i)
Lt f (t ) = Lt sF ( s )
t →0
s →0
F (s )
s
F (s )
(iii) Lt f (t ) = Lt
t →0
s →0 s
(ii)
(iv)
Lt f (t ) = Lt
t →0
s →∞
Lt f (t ) = Lt sF ( s )
t →0
s →∞
21. If f(t) and its first derivative are Laplace transformable
then the final value of f(t) is given by
(ii)
(iii)
(iv)
Lt f (t ) = Lt sF ( s )
t →∞
s →0
Lt f (t ) = Lt
F (s )
s
Lt f (t ) = Lt
F (s )
s
t →∞
t →∞
s →∞
s →0
Lt f (t ) = Lt sF ( s )
t →∞
s →∞
22. At t 0 with zero initial condition which of the following will act as a short circuit?
(i) Inductor
(ii) Capacitor
(iii) Resistor
(iv) None of these
23. At t 0 with zero initial condition which of the following will act as an open circuit?
(i) Inductor
(ii) Capacitor
(iii) Resistor
(iv) None of these
24. A capacitor at time t
as a
(i) short circuit
(iii) current source
0 with zero initial charge acts
(ii) open circuit
(iv) voltage source
25. A series RC circuit is suddenly connected to a dc voltage of V volts. The current in the series circuit just after
the switch is closed is equal to
V
VC
V
(i) zero
(ii)
(iii)
(iv)
RC
R
R
26. A series LC circuit is suddenly connected to a dc voltage of V volts. The current in the series circuit just after
the switch is closed is equal to
V
V
(ii)
(iii) zero
(iv) V
L
C
LC
27. The steady-state current in the RC series circuit, on
the application of a step voltage of magnitude E
will be
E
(i) zero
(ii)
R
E − t CR
E −t
(iii)
(iv)
e
e
R
RC
28. A 10- resistor, a 1-H inductor and a 1-F capacitor are
connected in parallel. The combination is driven by a
unit step current. Under steady-state conditions, the
source current flows through the
(i) resistor
(ii) inductor
(iii) capacitor only
(iv) all the three elements
(i)
29. When a unit impulse voltage is applied to an inductor
of 1 H, the energy supplied by the source is
(i)
(ii) 1 Joule
(iii) 1 Joule
(iv) 0
2
315
Laplace Transform and Its Applications
30. Which of the following conditions are necessary for
validity of the initial value theorem: Lim sF ( s ) = Limf (t )?
s →∞
t →0
(i) f(t) and its derivative f’(t) must have Laplace
transform.
(ii) If the Laplace transform of f(t) is F(s) then Lim
sF(s) must exist.
(iii) Only f(t) must have Laplace transform.
(iv) (i) and (ii) both.
1
31. Inverse Laplace transform of
is
s −a
at
(iv) e at
(i) sin at (ii) cos at (iii) e
32. The impulse response of an RL circuit is a
(i) rising exponential function
(ii) decaying exponential function
(iii) step function
(iv) parabolic function
∞
2.
3.
( s + 2)
, the initial and final values of v(t)
s ( s + 1)
will be respectively
35. For V ( s ) =
(i) 1 and 1 (ii) 2 and 2 (iii) 2 and 1 (iv) 1 and 2
36. The Laplace transform of the function i(t) is
10 s + 4
I s =
. Its final value will be:
s s + 1 s 2 + 4s + 5
(
)(
(ii) 5 4
)
(iii) 4
(iv) 5
37. An initially relaxed 100-mH inductor is switched ‘ON’
at t 1 second to an ideal 2-A dc current source. The
voltage across the inductor would be
(i) zero
(ii) 0.2␦(t) V
(iii) 0.2␦(t 1) V
(iv) 0.2tu (t 1) V
38. If the unit step response of a network is (1
its unit impulse response will be
1 −t
(i) e t
(ii) e
(iii)
1
e
−t
(iv) (1
)e
∫ (t )dt = 1
−∞
34. An initially relaxed RC series network with R 2 M ,
and C 1 ␮F is switched on to a 10-V step input. The
voltage across the capacitor after 2 seconds will be
(i) zero (ii) 3.68 V (iii) 6.32 V (iv) 10 V
(i) 4 5
∫ (t )dt = 1
0+
∞
33. Laplace transform of the output response of a linear
system is the system transfer function when the input is
(i) a step signal
(ii) a ramp signal
(iii) an impulse signal
(iv) a sinusoidal signal
()
40. A series circuit containing R, L and C is excited by a
step voltage input. The voltage across the capacitance
exhibits oscillations. The damping coefficient (ratio) of
this circuit is given by
R
R
(i) =
(ii) =
LC
2 LC
R
R
(iii) =
(iv) =
2 L
2 C
C
L
41. Consider the following statements:
A unit impulse ␦(t) is mathematically defined as
1. ␦(t) 0, t 0
t
e ) then
t
39. The response of an initially relaxed system to a
unit ramp excitation is (1 e t) . Its step response
will be
1 2 −t
(ii) 1 e t
(iii) e t
(iv) t
(i)
t −e
2
Of these statements,
(i) 1, 2 and 3 are correct
(iii) 2 and 3 are correct
(ii) 1 and 2 are correct
(iv) 1 and 3 are correct
42. With symbols having their usual meanings, the
Laplace transform of u(t a) is
− as
as
1
1
(ii)
(iii) e
(iv) e
(i)
s
s −a
s
s
43. Two coils having equal resistances but different inductances are connected in series. The time constant of
the series combination is the
(i) sum of the time constants of the individual coils
(ii) average of the time constants of the individual
coils
(iii) geometric mean of the time constants of the
individual coils
(iv) product of the time constants of the individual coils
44. If the step response of an initially relaxed circuit is
known then the ramp response can be obtained by
(i) integrating the step response
(ii) differentiating the step response
(iii) integrating the step response twice
(iv) differentiating the step response twice
45. If a capacitor is energized by a symmetrical squarewave current source then the steady state voltage
across the capacitor will be a
(i) square wave
(ii) triangular wave
(iii) step function
(iv) impulse function
46. A square wave is fed to an RC circuit. Then
(i) voltage across R is square and across C is not
square
(ii) voltage across C is not square and across R is not
square
(iii) voltage across both R and C is square
(iv) voltage across both R and C is not square
316
Network Analysis and Synthesis
47. A step voltage is applied to an under-damped series
RLC circuit with variable R. Which of the following
statements correctly describes the behaviour of the
circuit?
1. If R is increased, the steady-state voltage across C
will be reduced
2. If R is increased, the frequency of transient oscillation across C will be reduced.
3. If R is reduced, the transient oscillation will die
down faster.
4. If R is reduced to zero, the peak amplitude of the voltage across C will be double the input step voltage.
Select the correct answer using the codes given
below:
Codes: (i) 1 and 2
(ii) 2 and 3
(iii) 2 and 4
(iv) 1, 3 and 4
48. The number of turns of a coil having a time constant T
is doubled. Then the new time constant will be
(i) T
(ii) 2T
(iii) 4T
(iv) T/2
st
49. The response of a network is of the form ke ,where
s ␴ j␻. Then ␴ is known as
(i) radian frequency
(ii) neper frequency
(iii) complex frequency (iv) none of these
50. In a Laplace transform the variable ‘s’ equals (␴ j␻).
Which of the following represent the true nature of ␴ ?
1. ␴ has a damping effect.
2. ␴ is responsible for convergence of integral
∞
∫ f (t )e dt .
− st
0
3. ␴ has a value less than zero.
Select the correct answer using the coeds given below:
Codes:
(i) 1, 2 and 3
(ii) 1 and 2
(iii) 2 and 3
(iv) 1 and 3
51. Laplace transform of tn e at is
n
(i)
(ii)
n +1
s −a
)
(
(iii)
52.
(
s
s2 +
n!
(s − a
2
)
(iv)
n
n +1
56. The dc gain of a system represented by the transfer
25
is
function
s +2 s +3
)
(i) 25
(ii) 25/6
(i) 1 and 2
(iii) 3 and 4
(iii) 5
(iv) 10
(ii) 2 and 3
(iv) 1 and 4
58. Double integration of a unit step function would lead to
(i) an impulse
(ii) a parabola
(iii) a ramp
(iv) zero
n +1
(iii) cosh␻t
)
57. Consider the following statements:
The impulse response of a linear network can be
used to determine the
1. step response
2. response of the sinusoidal input
3. elements of the network uniquely
4. interconnection of network elements
Which of these statements are correct?
∫ f (t )f ( − t )d
1
2
t
(ii)
0
t
(ii) cos␻t
)(
(
t
(iii)
) is the Laplace transform of
(i) sin␻t
55. The convolution of a function f(t) with the unit impulse
function ␦(t) is
(i) ␦(t)
(ii) f(t)␦(t) (iii) f(t)
(iv) f( )␦(t)
(i)
n!
(s − a
54. The Laplace transform method enables one to find
the response in
(i) the transient state only
(ii) the steady state only
(iii) both transient and steady states
(iv) the transient state provided sinusoidal forcing
functions do not exist.
59. Which of the following integrals represents the convolution of two functions f1(t) and f2(t)?
n!
(s + a )
2. The output is a ramp for rectangular input
pulse.
3. The output has zero average for all inputs.
Of these statements,
(i) 1, 2 and 3 are correct
(ii) 1 and 2 are correct
(iii) 2 and 3 are correct
(iv) 1 and 3 are correct
)
)
∫ f 1 (t − f 2 (t dt
(iv) sinh␻t
1
2
0
t
(iv)
∫ f ( − t )f ( )dt
1
2
0
0
53. Consider the following statements regarding an RC
differentiating network:
1. For an applied rectangular pulse, the output is
spiky in nature for RC << pulse duration.
∫ f (t − )f ( )d
1
( s + 1)
( ) s1 s + k and f(t) as t → is 2 then the value
( )
60. If F s =
of K is
(i) ½
(ii) 1
(iii) 2
(v)
317
Laplace Transform and Its Applications
61. The transient response of the initially relaxed network
shown in Fig. 5.105 is
Switch
66. The time constant of the network shown in Fig. 5.106 is
R
C
R
C
R
V
V
i(t)
C
Fig. 5.106
(iii) CR
4
(ii) 2CR
(iv) CR
2
67. A non-linear system cannot be analyzed by Laplace
transform because
(i) it has no zero initial conditions
(ii) superposition law cannot be applied
(iii) non-linearity is generally not well defined
(iv) all of the above
(i) CR
Fig. 5.105
( ) VR e
V
(ii) i (t ) = e
R
(i) i t =
−t
t
RC
RC
( ) VR ⎛⎝⎜ 1− e
−t
( ) VR ⎛⎜⎝ 1+ e
−t
(iii) i t =
(iv) i t =
RC
⎞
⎠⎟
RC
⎞
⎟⎠
68. In the circuit shown in Fig. 5.107, the response i(t) is
Switch
62. A first-order linear system is initially relaxed. For a unit
step signal u(t), the response is v1(t) (1 e 3t) for
t 0. If a signal 3u(t)
␦ (t) is applied to the same
initially relaxed systems, the response will be
(ii) (3 3e 3t)u(t)
(i) (3 6e 3t)u(t)
(iii) 3u(t)
(iv) (3 3e 3t)u(t)
63. A unit impulse input to a linear network has a response
R(t) and a unit step input to the same network has
response S(t). The response R(t)
)
dS
(t
dt
(ii) equals the integral of S(t)
(iii) is the reciprocal of S(t)
(iv) has no relation with S(t)
(i) equals
()
()
Fig. 5.107
⎛ t ⎞
V
exp ⎜ −
R
⎝ RC ⎟⎠
V
(ii)
(t
R
⎛ t ⎞⎤
1
V ⎡
(iii)
exp ⎜ −
⎢ (t −
⎥
R⎣
RC
⎝ RC ⎠⎟ ⎦
⎛ t ⎞⎤
V ⎡
(iv)
⎢ (t − exp ⎜ −
⎥
R⎣
⎝ RC ⎟⎠ ⎦
(i)
)
65. The impulse response of a circuit is given by
1 −R t
h t = e L u t . Its step response is given as
L
R
R
⎛
− t⎞
− t⎞
1⎛
(ii)
(i) ⎜ 1− e L ⎟ u t
1− e L ⎟ u t
⎜
R⎝
⎠
⎝
⎠
R
⎛
⎞
t
−
L
(iii)
(iv) none of these
1− e L ⎟ u t
R ⎜⎝
⎠
()
C
i(t)
V (t )
)
64. The response of an initially relaxed linear circuit
to a signal VS is e 2t u(t). If the signal is changed to
⎛
dV S ⎞ , the response would be
⎜V S + 2 dt ⎟
⎝
⎠
(i)
4e 2tu(t)
(ii)
3e 2tu(t)
2t
(iv) 5e 2tu(t)
(iii) 4e u(t)
()
R
()
)
69. A voltage v(t) 6e 2t is applied at t 0 to a series RL
circuit with L 1 H. If i(t) 6[e 2t e 3t] then R will
have a value of
1
(i) 2
(ii) 1
(iii) 3
(iv)
Ω
3
3
70. The Laplace transform of the signal described in
Fig. 5.108 is
f( t)
a
Fig. 5.108
b
t
318
Network Analysis and Synthesis
− bs
(ii) e
(i) e − as
(iii)
(
s
e − as + e − bs
)
(iv)
s
71. If a pulse voltage v(t) of
4-V magnitude and 2-second
duration is applied to a pure
inductor of 1 H, with zero initial current, the current (in A)
drawn at t 3 seconds, will be
(i) zero
(ii) 2
(iii) 4
(iv) 8
(e
S
s2
− as
− e − bs
)
1
s
1F
1
1
1H
v (t)
Fig. 5.109
72. At a certain current, the energy stored in an iron-cored
coil is 1000 J and its copper loss is 2000 W. The time
constant (in second) of the coil is
(i) 0.25
(ii) 0.5
(iii) 1.0
(iv) 2.0
Fig. 5.112
76. The circuit shown in Fig. 5.113 is in steady state with
the switch ‘S’ open. The switch is closed at t 0. The
values of VC(0 ) and VC( ) will be respectively
(i) 2 V, 0 V
(ii) 0 V, 2 V
(iii) 2 V, 2 V
(iv) 0 V, 0 V
1/2 F
Vc
S
2A
v (t)
Fig. 5.113
3
77. In the circuit shown, the switch is opened at t
Prior to that switch was closed, i(t) at t 0 is
(ii) 3 A
(i) 2 A
2
3
1
A
(iv) 1 A.
(iii)
3
2
1
0
1 2 3 4 5 t sec
Fig. 5.110
The equation for v(t) is
(i) u t − 1 + u t − 2 + u t − 3
( ) ( ) ( )
(ii) u (t − 1) + 2u (t − 2 ) + 3u (t − 3 )
(iii) u (t ) + u (t − 1) + u (t − 2 ) + u (t − 4 )
(iv) u (t − 1) + u (t − 2 ) + u (t − 3 ) − 3u (t − 4 )
i (t)
2
1
2
4V
1F
Fig. 5.114
()
∞
()
78. Given the Laplace transform L ⎡⎣v t ⎤⎦ = ∫ e − st v t dt ,
0
the inverse transform v(t) is
S
V
C
0.25 H
0.25
+ j∞
(i)
(ii) 2 V
(iv) 0.25 V
75. In the circuit shown in Fig. 5.112, S is open for a long
time and steady state is reached. S is closed at t 0.
The current I at t 0 is
(i) 4 A
(ii) 3 A
(iii) 2 A
(iv) 2 A
∫ e V ( s )ds
st
(ii)
1
∞
2 j ∫0
)
e V ( s ds
st
1
2 j
− j∞
(iii)
Fig. 5.111
(i) 3 V
(iii) 0.5 V
0.
Switch
74. For the circuit given in Fig. 5.111 V0 2 V and inductor
is initially relaxed. The switch S is closed at t 0. The
value of v at t 0 is
V0 = 2 V
1
1
73. Consider the voltage waveform shown in Fig. 5.110.
0
8A
3
I
(iv)
1
2 j
+ j∞
∫ e V ( s )ds
st
− j∞
+ j∞
∫ e V ( s )ds
− st
− j∞
79. In the circuit shown in Fig. 5.115, switch ‘S’ is closed at
t 0. After some time when the current in the inductor was 6 A, the rate of change of current through it
was 4 A/s. The value of the inductor is
(i) indeterminate
(ii) 1.5 H
(iii) 1.0 H
(iv) 0.5 H
319
Laplace Transform and Its Applications
t= 0
2F
5
S
L
20 V
Vs(t)
10 V
3
10
1H
Fig. 5.119
Fig. 5.115
80. A circuit consisting of a 1- resistor and a 2-F capacitor in series is excited from a voltage source with the
voltage expressed as 3e t, as shown in Fig. 5.116. If
i(0 ) and vc(0 ) are both zero then the values of i(0 )
and i( ) will be respectively
(i) 3 A and 1.5 A
(ii) 1.5 A and zero
(iii) 3 A and zero
(iv) 1.5 A and 3 A
84. The steady state in the circuit, shown in Fig. 5.120 is
reached with S open. S is closed at t 0. The current I
at t 0 is
(i) 1 A
(ii) 2 A
(iii) 3 A
(iv) 4 A
2V
1
S
1
C
2
2
I
3e
t
i(t)
2F
Vc(t)
Fig. 5.120
Fig. 5.116
81. The time constant associated with the capacitor
charging in the circuit shown in Fig. 5.117 is
(i) 6 ␮s
(ii) 10 ␮s
(iii) 15 ␮s (iv) 25 ␮s
85. For the circuit shown in Fig. 5.121, the current through
L and the voltage across C2 are respectively
(i) zero and RI
(ii) I and zero
(iii) Zero and zero
(iv) I and RI
L
2
3
Vdc
I
C1
C2
R
5 F
Fig. 5.121
Fig. 5.117
82. In the network shown in Fig. 5.118, the switch ‘S’ is
closed and a steady state is attained. If the switch
is opened at t 0, then the current i(t) through the
inductor will be
(i) cos 50t A
(ii) 2 A
(iii) 2 cos 100t A
(iv) 2sin50t A
S
2.5
86. In the circuit shown in Fig. 5.122, the switch is closed at
t 0. The current through the capacitor will decrease
exponentially with a time constant
t=0
1
1F
1
10 V
i(t)
Fig. 5.122
5V
0.5H
2 00 F
(i) 0.5 s
(ii) 1 s
(iii) 2 s
(iv) 10 s
87. The Laplace transformation of f(t) is F(s). Given
Fig. 5.118
83. In the network shown, the switch is opened at t 0.
Prior to that, the network was in the steady state. Vs(t)
at t 0 is
(i) 0
(ii) 5 V
(iii) 10 V
(iv) 15 V.
) s +
F (s =
2
(i) infinity
(iii) one
2
, the final value of f(t) is
(ii) zero
(iv) none of the above
88. The v–i characteristics as seen from the terminal-pair
(A, B) of the network of Fig. 5.123 (a) is shown in
320
Network Analysis and Synthesis
Fig. 5.123 (b). If an inductance of 6-mH value is connected across the terminal-pair (A, B), the time constant of the system will be
(i) 3 ␮s
(ii) 12 s
(iii) 32 s
iv) unknown, unless the actual network is specified
(i) 3 and 4 are correct
(iii) 1 and 2 are correct
4 mA
L
I0
Fig. 5.125
I (s)
A
Network of
linear resistors
and independent
sources
I (t)
94. An inductor with inductance
L and initial current I0 is
shown as Fig. 5.125
The correct admittance diagram for it is
i
i
(ii) 1 and 4 are correct
(iv) 2 and 3 are correct
(a) 1/Ls
v
I0/s
I1(s)
v
B
(0,0)
(a)
I0
8V
(b)
Fig. 5.123
(b)
89. In the circuit shown in Fig. 5.124, it is desired to have a
constant direct current i(t) through the ideal inductor L.
The nature of the voltage source v(t) must be
(i) constant voltage
(ii) linearly increasing voltage
(iii) an ideal impulse
(iv) exponentially increasing voltage
I0
1/Ls
(c) Ls
I0
I(s)
I(t)
(d) Ls
I0/s
I1(s)
I(t)
L
v (t)
95. An inductor with inductance L and
initial current I0 is shown as Fig. 5.126.
The correct impedance diagram for it is
Fig. 5.124
(a)
∞
90. The value of the integral ∫ e
−∞
(i) 1
(ii) (e5
5t
I(s)
(t − 5)dt is
1 (iii) e25
Ls
(iv) zero
91. An inductor at t 0 with initial current I0 acts as a/an
(i) voltage source
(ii) current source
(iii) open circuit
(iv) short circuit
92. A capacitor at t 0 with initial charge Q0 acts as a/an
(i) voltage source
(ii) current source
(iii) open circuit
(iv) short circuit
93. Consider the following statements:
1. Current through an inductor cannot change abruptly.
2. Voltage across the capacitor cannot change abruptly.
3. Initial value of a function f(t) is Lim sF ( s )
s →0
4. Final value of a function f(t) is Lim sF ( s )
s →∞
Of these statements,
(c)
I0
L
Fig. 5.126
(b)
I0
I (s)
I0/s
1/Ls
sL I(s)
(d)
LI0
96. A capacitor with
capacitance C and
initial voltage vc(t) is
shown here.
I(s)
I0/s
sL
C
i(t )
Fig. 5.127
The correct admittance diagram for this circuit is
vc(t )
321
Laplace Transform and Its Applications
(a)
99. Consider the following functions for the rectangular
voltage pulse shown in Fig. 5.130
I (s)
v(t )
sC
Cvc (0)
1
I(s)
(b)
vc (0)
1/Cs
a
b
Fig. 5.130
(c)
I(s)
vc (0)/s
sC
(d)
I(s)
(i) v(t)
u(t a)
u(t b)
(ii) v(t)
u(b t)
u(a t)
(iii) v(t)
u(b t)·u(t a)
(iv) v(t)
u(a t)·u(t b)
( ) s +1 3 , F ( s ) = s 2+ 4 ; what is the Laplace
100. If F1 s =
sC
2
2
transform of the product F1(s) F2(s)?
Cvc (0)/s
()
1
(i) f t = ⎡⎣e −t + 3 cos 2t − 2 sint ⎤⎦
5
97. Laplace transform of f(t) shown in Fig. 5.128 is
( ) 131 ⎡⎣2e
(ii) f t =
2.0
1.0
f (t)
−3t
+ 3 sin2t − 2 cos 2t ⎤⎦
()
1
2
3
1
(iii) f t = ⎡⎣e −2t + 2 sin2t − cos 2t ⎤⎦
7
t
( ) 111 ⎡⎣e
1.0
(iv) f t =
Fig. 5.128
−2 t
+ sint − 2 sin2t ⎤⎦
101. The impulse response of a linear network is given
by e 2t. Which one of the following gives its unit step
response?
(ii) e t e 2t
(i) 1 e 2t
()
1 2
3
(i) F s = − e − s + e − s
s s
s
1 2 − s 3 −2 s 2 −3 s
(ii) F s = − e + e − e
s s
s
s
−s
1 e
2
2
(iii) F s = −
+ e −2 s − e −3 s
s s s
s
1 2 −s 3 −s
(iv) F s = + e − e
s s
s
()
()
(iii)
(
1
1− e −2t
2
102.
()
)
2
(iv)
a
K
1 −t
e − e −2t
2
(
)
1/2
2H
b
98. The time constant of the circuit shown in Fig. 5.129 is
2R
C
5V
R
Fig. 5.131
2R
t=0
Fig. 5.129
(i) RC
(ii) 2 RC
(iii) 3 RC
1H
(iv) 5 RC
The network shown in Fig. 5.131 reaches a steady state
with the switch K in the position a. At t 0, the switch
is moved from a to b by a make-before-break mechanism. Assume the initial current in the 2-H inductor as
322
Network Analysis and Synthesis
zero. What is the current in the 1-H inductor at t 0
and t
, respectively?
(i) 1 A and 0 A
(ii) 2.5 A and 0 A
(iii) 1 A and 2.5 A
(iv) 2.5 A and 2.5 A
2 ( s + 1)
( ) s + 2s + 5 , then what are the values of f(0 )
103. If F s =
2
and f( ) respectively?
(i) 0, 2
(ii) 2, 0
(iii) 0, 1
(iv) 2/5, 0
107. If f1(t ) and f2(t ) have the widths (duration) T1 and T2
respectively then what is the width (duration) of
f1(t)* f2(t ) (where * denotes convolution)?
(i) The larger of T1 and T2
(ii) The smaller of T1 and T2.
(iii) T1 T2
(iv) T1 T2
108. The Laplace transform of v(t) shown in Fig. 5.135 is
v(t )
104. In the circuit shown in Fig. 5.132, the switch S is closed
at t 0. Which one of the following gives the expression for the voltage across the inductance as a function of time?
1
1
S
0
1
t
2
Fig. 5.135
1V
1H
Fig. 5.132
1− e )
(ii) (
−t
t/2
(i) e
(
)
(
)
(i)
1
1
1− e − s − e −2 s
s
s2
(
)
(ii)
1
1
1− e s − e 2 s
s
s2
(iii)
1
1
1+ e − s + e −2 s
2
s
s
(
)
(iv)
1
1
1+ e s + e 2 s
2
s
s
2
(iii) (1
e t)
(iv) e t
105. For the circuit shown in Fig. 5.133, the initial capacitor
voltage is 2 V and I is a unit step function. Then, what
is the expression for v(t) for t 0?
109. If f(t) and F(s) form the Laplace transform pair then
⎛
⎞
what is the Laplace transform of f ⎜ t ⎟ ?
⎝ t0 ⎠
1
I
1
0.25 F
v (t)
e t (ii) 2
(ii)
1
F (t 0 s
t0
⎛1 ⎞
(iii) t 0 F ⎜ s ⎟
⎝ t0 ⎠
(iv)
1 ⎛1 ⎞
F
s
t 0 ⎜⎝ t 0 ⎟⎠
110. The switch in the circuit is closed at t 0. The current
through the battery at t 0 and t
is, respectively
Fig. 5.133
(i) 2
)
(i) t0F(t0s)
1
e 2t (iii) 1
e 2t
(iv) 1
e 2t
106. In the circuit shown in Fig. 5.134, steady state is reached
with the switch S open. Switch S is then closed at t 0.
What is the value of voltage V under steady state (when
t
)?
1H
1V
1F
5
S
5A
V
10
5
Fig. 5.134
(i) 50 V
(ii) 12.5 V
(iii) 25 V
Fig. 5.136
(iv) 0 V
(i) 10 A and 10 A
(iii) 10 A and 0 A
(ii) 0 A and 10 A
(iv) 0 A and 0 A
111. The Laplace transform of the voltage across the
s +1
capacitor of 0.5 F is V s = 3 2
s + s + s +1
()
323
Laplace Transform and Its Applications
Then the value of the current through the capacitor at
t 0 is given by
(i) 0 A
(ii) 0.5 A
(iii) 1.0 A
(iv) 1.5 A
118. In the given circuit, if the inductor is initially relaxed,
then the current in the circuit will be
R
112. If u(t) and ␦(t ) are the step function and the impulse
function respectively at t = 0, then the Laplace transform of the function f(t ) u(t 1) ␦(t ) is equal to
1
1
(iii) 0
(iv)
(i) 1
(ii)
s
s +1
113. The step response of a system is C(t )
1 5e t
2t
3t
6e . The impulse response of the system is
10e
(i) 5e t 20e 2t 18e 3t
(ii) 5et 20e2t 18e 3t
(iii) 5e t 20e 2t 18e 3t
(iv) 5e t 20e 2t 18e3t
Fig. 5.138
(iii)
(s + )
(s + ) +
(s + )
(iii)
(s − ) +
(ii)
2
2
2
2
(iv)
(s − )
(s − ) +
2
(s − )
(s + ) +
2
2
)
L
(t
R
Rt
− ⎞
1⎛
(iv)
1− e L ⎟
⎜
L⎝
⎠
(i) zero
114. The Laplace transform of e t cos t is
(i)
L
d(t)
(ii)
1 − RtL
e
L
119. For the circuit shown in Fig. 5.139, the switch ‘K’ was
closed for a long time till steady-state conditions
reached. At time t 0, the switch ‘K’ is opened then
the current through inductor will be
K
2
2
115. For the circuit shown in Fig. 5.137, the initial inductor
current is 2 A. The value of i(t ) for t 0 is
1
1H
10 V
1 F
i
2
u(t )
Fig. 5.139
2
(i) 5cos10t
(iii) 5cos1000t
Fig. 5.137
(i) 0.5
(iii) 0.5
0.75e t
0.25e t
(ii) 1 e t
(iv) 0.5 0.75e t
116. Consider a system described by the transfer function
2 s + 3 . It is subjected to an input f (t)
G s = 2
s + 2s + 5
10u(t). The initial and final values of the response are
given by
(i) 0, 2 3
(ii) 1, 4
(iii) 0, 6
(iv) 0, 4
()
117. The impulse response of a linear time invariant system
is given by h(t) 2e t u(t)
The unit step response is given by
(ii) y(t)
(i) y(t) 2(1 e t )u(t)
(iv) y(t)
(iii) y(t) 2(1 e 2t )u(t)
2(e t 1)u(t)
2(2 e 2t )u(t)
(ii) 5cos100t
(iv) 5cos 10000t
120. The response of a system to a unit ramp input is
1 t 1 u(t) 1 e 4t. Which one of the following is
8
8
2
the unit impulse response of the system?
(i) 1 e 4t
(ii) 2(1 e 4t)
4t
(iv) 2 e 4t
(iii) e
121. The Laplace transform of current in an RLC series circuit
1
1
F is I s = 2
.
with R 2 , L 1 H and C
2
s + 2s + 2
The voltage across the inductor ‘L’ will be
(ii) e t costu(t)
(i) e t sintu(t)
t
(iv) e t (cost sint)u(t)
(iii) e (sint cost)u(t)
()
122. For the network shown in Fig. 5.140, the initial position of switch ‘S’ is ‘1’. After reaching steady state, if the
324
Network Analysis and Synthesis
position of the switch is changed over to ‘2’, the current ‘i’ for t 0 will be equal to
2R
1
L
126. In Fig. 5.143, the capacitor initially has a charge of
10 couloms. The current in the circuit one second after
the switch S is closed will be
2
V
i
R
Fig. 5.140
(i)
0.5 F
Fig. 5.143
⎛ Rt ⎞
V
exp ⎜ − ⎟
2R
⎝ L⎠
(ii)
⎛ 3Rt ⎞
V
(iii)
exp ⎜ −
R
⎝ L ⎟⎠
⎛ 2 Rt ⎞
V
exp ⎜ −
R
⎝ L ⎟⎠
(iv) V exp ⎛ − 3Rt ⎞
⎜⎝ L ⎟⎠
2R
(i) 14.7 A
T=0
E
i(t)
(iii) 40.0 A
(iv) 50.0 A
10
10 ␮F
20 V
R
K
(ii) 18.5 A
127. In Fig. 5.144, the initial capacitor voltage is zero. The
switch is closed at t 0. The final steady-state voltage
across the capacitor is
123. The correct value of the current i(t ) at any instant
when K is switched on at t = 0 in the network shown
in Fig. 5.141 is
10
Fig. 5.144
L
(i) 20 V
R
(i) E + E e ( L )t
R R
(ii)
E E ( R L )t
− e
R R
E E −( R L )t
+ e
R R
(iv)
E E −( R L )t
− e
R R
(ii) 10 V
(iii) 5 V
(iv) 0 V
128. The circuit shown in Fig. 5.145 is in steady state, when
the switch is closed at t 0. Assuming that the inductance is ideal, the current through the inductor at
t 0 equals
Fig. 5.141
(iii)
2
S
100 V
10
10 mH
10 V
t=0
124. In the circuit shown in Fig. 5.142, the switch S is closed
at t 0. The voltage across the inductance at t 0 is
Fig. 5.145
3
4F
S
10 V
4
4
4H
(i) 0 A
(iii) 1 A
(ii) 0.5 A
(iv) 2 A
129. If, at t 0 , the voltage across the coil is 120 V, the
value of resistance R is
1
Fig. 5.142
(i) 2 V
(ii) 4 V
(iii)
6V
( ) s s +53s + 2
125. Consider the function, F s =
(
2
120 V
)
where
F(s) is the Laplace transform of the function f(t). The
initial value of f(t) is equal to
(i) 5
(ii)
5
2
(iii)
5
3
S
(iv) 8 V
(iv) 0
20
2
10 H
R
40
Fig. 5.146
(i) 0
(iii) 40
(ii) 20
(iv) 60
325
Laplace Transform and Its Applications
130. For the value obtained in Q. 129, the time taken for
95% of the stored energy to be dissipated is close to
(i) 0.10 second
(ii) 0.15 second
(iii) 0.50 second
(iv) 1.0 second
131. An ideal capacitor is charged to a voltage V0 and connected at t 0 across an ideal inductor L. (The circuit
now consists of a capacitor and inductor alone.) If we
1
let 0 =
, the voltage across the capacitor at time
LC
t > 0 is given by
(i) V0
(ii) V0 cos( 0t)
(iv) V0e 0t cos( 0t)
(iii) V0 sin( 0t)
132. In the circuit shown in Fig. 5.147, the switch SW1 is initially CLOSED and SW2 is OPEN. The inductor L carries
a current of 10 A and the capacitor is charged to 10 V
with polarities as indicated. SW2 is initially CLOSED at
0. The current
t
0 and SW1 is OPENED at t
through C and the voltage across L at t 0 is
SW1
R110
1F
1F
1F
3
3A
Fig. 5.148
(i)
1
s
9
(ii)
1
s
4
(iii) 4 s
(iv) 9 s
( ) s 12+ s
( )
134. The Laplace transform of i(t) is given by I s =
As t → , the value of i(t) tends to
(i) 0
(ii) 1
(iii) 2
(iv)
135. In what range should Re(s) remain so that the Laplace
transform of the function e(a 2)t 5 exits?
(i) Re(s) a 2
(ii) Re(s) a 7
(iii) Re(s) 2
(iv) Re(s) a 5
10 A
j2
0.1mF
C
L
10 V
(b)
(a) V
i
3V
Fig. 5.147
(i) 55 A, 4.5 V
(iii) 45 A, 5.5 V
3
136. A square pulse of 3-V amplitude is applied to C-R circuit
shown in Fig. 5.149. The capacitor is initially uncharged.
The output voltage V0 at time t = 2 seconds is
R210
SW2
133. The time constant for the given circuit will be
2 seconds
Vi
t
V0
1k
Fig. 5.149
(ii) 5.5 A, 45 V
(iv) 4.5 A, 5.5 V
(i) 3 V
(ii)
3V
(iii) 4 V
(iv)
Answers
1. (ii)
2. (iii)
3. (i)
4. (iii)
5. (iii)
6. (iii)
7. (iii)
8. (i)
9. (i)
10. (i)
11. (i)
12. (ii)
13. (iii)
14. (i)
15. (ii)
16. (iii)
17. (ii)
18. (iii)
19. (ii)
20. (iv)
21. (i)
22. (ii)
23. (i)
24. (i)
25. (iv)
26. (iii)
27. (i)
28. (ii)
29. (iii)
30. (iv)
31. (iii)
32. (ii)
33. (iii)
34. (iii)
35. (iv)
36. (i)
37. (i)
38. (i)
39. (iii)
40. (iv)
41. (iv)
42. (iv)
43. (ii)
44. (i)
45. (ii)
46. (iv)
47. (iii)
48. (ii)
49. (ii)
50. (ii)
51. (ii)
52. (ii)
53. (iv)
54. (iii)
55. (iii)
56. (ii)
57. (i)
58. (ii)
59. (ii)
60. (iii)
61. (i)
62. (iii)
63. (i)
64. (ii)
65. (ii)
66. (i)
67. (i)
68. (iii)
69. (iii)
70. (iv)
71. (iv)
72. (ii)
73. (iv)
74. (ii)
75. (i)
76. (ii)
77. (iv)
78. (ii)
79. (iv)
80. (iii)
4V
326
Network Analysis and Synthesis
81. (i)
82. (iii)
83. (ii)
84. (ii)
85. (iv)
86. (ii)
87. (iv)
88. (i)
89. (iii)
90. (iii)
91. (ii)
92. (i)
93. (iii)
94. (i)
95. (iii)
96. (i)
97. (ii)
98. (iii)
99. (i)
100. (ii)
101. (iii)
102. (ii)
103. (ii)
104. (iv)
105. (iii)
106. (ii)
107. (iv)
108. (i)
109. (i)
110. (i)
111. (i)
112. (iii)
113. (i)
114. (ii)
115. (iv)
116. (iii)
117. (i)
118. (iii)
119. (iii)
120. (iv)
121. (iv)
122. (iv)
123. (iv)
124. (ii)
125. (iv)
126. (i)
127. (ii)
128. (iii)
129. (i)
130. (iii)
131. (ii)
132. (iv)
133. (iii)
134. (iii)
135. (i)
136. (ii)
6
Two-Port Network
Introduction
A port is a pair of nodes across which a device can be connected. The voltage is measured across the pair
of nodes and the current going into one node is the same as the current coming out of the other node in
the pair. These pairs are entry (or exit) points of the network.
So, a network with two input terminals and two output terminals is called a four-terminal network
or a two-port network.
It is convenient to develop special methods for the systematic treatment of networks. In the case
of a single port linear active network, we
I1
I2
obtained the Thevenin’s equivalent circuit and
Linear
the Norton’s equivalent circuit. When a linear
passive
Port 2 V
V1 Port 1
2
passive network is considered, it is convenient
network
to study its behaviour relative to a pair of designated nodes.
Fig. 6.1
Block diagram of a two-port network
In a two-port network, there are two voltage
variables and two current variables. According to the choice of input and output ports, these voltage and
current variables can be arranged in different equations, giving rise to different port parameters.
In this chapter, we will discuss the behaviours of two-port networks and then will learn about some
special two-port networks.
6.1
RELATIONSHIPS OF TWO-PORT VARIABLES
In order to describe the relationships among the port voltages and currents of an n-port network, ‘n’ number
of linear equations is required. However, the choice of two independent and two dependent variables is
dependent on the particular application.
328
Network Analysis and Synthesis
For an n-port network, the number of voltages and current variables is 2n. The number of ways in which
2 n!
2 n!
2 n!
these 2n variables can be arranged in two groups of n each is
=
. So, there will be
types of
2
n
!×
n
!
(
n
!
)
(
n!)2
port parameters.
For a two-port network (n
2), there are six types of parameters as mentioned below:
1. Open-circuit impedance parameters (z-parameters)
2. Short-circuit admittance parameters (y-parameters)
3. Transmission or chain parameters (T-parameters or ABCD–parameters)
4. Inverse transmission parameters (T -parameters)
5. Hybrid parameters (h-parameters)
6. Inverse hybrid parameters (g-parameters)
6.1.1 Open-Circuit Impedance Parameters (z-Parameters)
The impedance parameters represent the relation between the voltages and the currents in the two-port network.
The impedance parameter matrix may be written as
⎡V1 ⎤ ⎡ z11
⎢ ⎥=⎢
⎢⎣V2 ⎥⎦ ⎢⎣ z21
V1 = z11 I1 + z12 I 2
z12 ⎤ ⎡ I1 ⎤
⎥ ⎢ ⎥ or,
V2 = z21 I1 + z22 I 2
z22 ⎥⎦ ⎢⎣ I 2 ⎥⎦
In this matrix equation, it is easily seen without even expanding the individual equations, that
V
z11 = 1
I1 I = 0
Driving point impedance at Port-1
2
z12 =
V1
I 2 I =0
Transfer impedance
V2
I1 I = 0
Transfer impedance
1
z21 =
2
V
z22 = 2
I 2 I =0
Driving point impedance at Port-2
1
It can be seen that the z-parameters correspond to the driving point and transfer impedances at each port with the other
port having zero current (i.e., open circuit). Thus these parameters are also referred as the open-circuit parameters.
Example 6.1 Determine the z-parameters for the network shown in Fig. 6.2.
Solution We consider two situations:
(a) When I1 0, i.e., Port-1 is open-circuited
In this case no current will flow through the 5- resistor.
By KVL in the right mesh, we get 10 I 2 + 20 I 2 − V2 = 0
V
∴ z22 = 2
= 30
I 2 I =0
1
I1
I2
1
5
V1
1
Fig. 6.2
20
10
2
V2
2
Network of Example 6.1
329
Two-Port Network
From Fig. 6.3 (a), we get V2
20I1
∴ z12 =
I1 = 0
I2
1
V1
= 20
I 2 I =0
2
5
V1
10
V2
20
1
(b) When I2 0, i.e., Port-2 is open-circuited
In this case no current will flow through the 10- resistor.
By KVL in the left mesh, we get 5 I1 + 20 I1 − V1 = 0
V
∴ z11 = 1
= 25
I1 I = 0
2
From Fig. 6.3 (b), we get V2
∴ z21 =
20I1
1
2
Fig. 6.3 (a) When I1
0
I2 = 0
I1
1
2
5
V1
10
V2
20
1
Fig. 6.3 (b)
2
When I2
0
V2
= 20
I1 I = 0
2
⎡ 25 20 ⎤
Therefore, the z-parameters of the network are ⎡⎣ z ⎤⎦ = ⎢
⎥( )
⎣ 20 30 ⎦
6.1.2 Short-Circuit Admittance Parameters ( y-Parameters)
The admittance parameters represent the relation between the currents and the voltages in the two-port
network.
The admittance parameter matrix may be written as
⎡ I1 ⎤ ⎡ y11
⎢ ⎥=⎢
⎢⎣ I 2 ⎥⎦ ⎢⎣ y21
I1 = y11V1 + y12V2
y12 ⎤ ⎡V1 ⎤
⎥ ⎢ ⎥ or,
I 2 = y21V1 + y22V2
y22 ⎥⎦ ⎢⎣V2 ⎥⎦
The parameters y11, y12, y21, y22 can be defined in a similar manner, with either V1 or V2 on short circuit.
I
y11 = 1
V1 V =0
Driving point admittance at Port-1
2
y12 =
I1
V2 V =0
Transfer admittance
I2
V1 V =0
Transfer admittance
I2
V2 V =0
Driving point admittance at Port-2
1
y21 =
2
y22 =
1
It can be seen that the y-parameters correspond to the driving point and transfer admittances at each port
with the other port having zero voltage (i.e., short circuit). Thus these parameters are also referred as the short
circuit parameters.
330
Network Analysis and Synthesis
Example 6.2 Find the y-parameters for the network shown in Fig. 6.4.
Solution We consider two situations:
When V1 0, i.e., Port-1 is short-circuited
In this case, no current will flow through the 20circuit is shown in Fig. 6.5 (a).
By KCL at the node 2,
50
20
resistor. The modified
V2 − 0 V2 − 0
+
= I2
10
50
∴ y22 =
I2
1 1
= + = 0.12
V2 V =0 10 50
10
Fig. 6.4
Network of Example 6.2
I1
50
2
I2
1
∴ y12 =
V1 = 0
0 − V2
50
Also, from Fig. 6.5 (a) we get I1 =
Fig. 6.5 (a)
I1
1
= = 0.02
V2 V =0 50
V2
10
When V1
0
1
When V2 0, i.e., Port-2 is short-circuited
In this case, no current will flow through the 10By KCL at the node 1,
resistor. The modified circuit is shown in Fig. 6.5 (b).
V1 − 0 V1 − 0
+
= I1
20
50
I
1 1
∴ y11 = 1
= + = 0.07
V1 V =0 20 50
I1
1
V1
50
2
I2
V2 = 0
20
2
Also, from Fig. 6.5 (b) we get I 2 =
∴ y21 =
0 − V1
50
Fig. 6.5 (b)
When V2
0
I2
1
= = 0.02
V1 V =0 50
2
⎡ 0.07 0.02 ⎤
Therefore, the y-parameters of the network are ⎡⎣ y ⎤⎦ = ⎢
⎥
⎣0.02 0.12 ⎦
6.1.3 Transmission Parameters (ABCD-Parameters)
The ABCD parameters represent the relation between the input quantities and the output quantities in the
two-port network. They are thus voltage–current pairs.
However, as the quantities are defined as an input–output relation, the output current is marked as going
out rather than as coming into the port.
The transmission parameter matrix may be written as
331
Two-Port Network
⎡V1 ⎤ ⎡ A B ⎤ ⎡ V2 ⎤
V1 = AV2 − BI 2
⎢ ⎥=⎢
⎥ or,
⎥⎢
I1 = CV2 − DI 2
⎢⎣ I1 ⎥⎦ ⎣C D ⎦ ⎢⎣ − I 2 ⎥⎦
I1
Port 1
V1
The parameters A, B, C, D can be defined in a similar manner
with either Port 2 on short circuit or Port 2 on open circuit.
Port 2
V2
Fig. 6.6 Two-port current and voltage variables
for calculation of transmission line parameters
V1
V2 I =0
Open-circuit reverse voltage gain
V
B=− 1
I 2 V =0
Short-circuit transfer impedance
I1
V2 I =0
Open-circuit transfer admittance
I
D=− 1
I 2 V =0
Short-circuit reverse current gain
A=
I2
Linear
passive
network
2
2
C=
2
2
These parameters are known as transmission parameters as in a transmission line, the currents enter at one
end and leave at the other end, and we need to know a relation between the sending-end quantities and the
receiving-end quantities.
Example 6.3 For the network shown Fig. 6.7, determine the
ABCD parameters.
Solution The ABCD-parameter equations are
V1
AV2
BI2
I1
CV2
DI2
I1
1
V1
2
1
I2
I1
V2
2
Network of Example 6.3
2/3
1
1
2/3
V1
2×2
2
=
2+2+2 3
2/3
1
To find the ABCD parameters, we consider two situations:
When V2 0, i.e., port-2 is short-circuited
As shown in Fig. 6.8 (c), by KVL we get,
(
)
1.67 I1 + 0.67 I1 + I 2 = V1
2.33 I1 + 0.67 I 2 = V1
2
2
Fig. 6.7
or,
1
1
For the network shown in Fig. 6.7, we convert the delta
consisting of the resistances of 2 each into its equivalent
star so that the circuit becomes as shown in Fig. 6.8 (a)
and Fig. 6.8 (b).
r1 = r2 = r3 =
2
Fig. 6.8 (a)
1
I2
2
+
V2
2
Modified network of Fig. 6.7
332
Network Analysis and Synthesis
I1
)
(
0.67 I1 + I 2 + 1.67 I 2 = 0
and,
I2
1.67
2
1
2.33
I1 = −
I = −3.5 I 2
0.67 2
or,
1.67
V1
I
∴D= − 1
= 3.5
I 2 V =0
V2
0.67
1
2
Fig. 6.8 (b)
2
I1
Putting this value in the first equation, we get,
)
V
2.33 × −3.5 I 2 + 0.67 I 2 = V1 ⇒ B = − 1
= 7.5
I 2 V =0
(
1.67
1.67
I2
2
1
V1
V2 = 0
0.67
2
When I2 0, i.e., Port-2 is open-circuited
Here, no current will flow through the right side of the 1.67 resistance.
By KVL, we get, V1 (1.67 0.67)I1 2.33I1 and, V2 0.67I1
∴C =
I1
V2
=
1
I1
1
= 1.5
0.67
1.67
1.67
V1
2.33 I1
V
= 3.5
∴A= 1
=
V2 I =0 0.67 I1
0.67
1
2
I2 = 0
2
+
V2
1
I 2 =0
Therefore, the ABCD parameters of the network are A
2
Fig. 6.8 (c)
2
Fig. 6.8 (d)
3.5; B
7.5
;C
1.5
; and D
3.5
6.1.4 Inverse Transmission Parameters (A B C D -Parameters)
The inverse A B C D parameters represent the inverse relation between the input quantities and the output
quantities in the two-port network. They are also voltage–current pairs.
Here also, the output current is marked as going out rather than as coming into the port as shown in Fig. 6.6.
The inverse transmission parameter matrix may be written as
⎡V2 ⎤ ⎡ A′ B ′ ⎤ ⎡ V1 ⎤
V2 = A′V1 − B ′I1
⎢ ⎥=⎢
⎥ ⎢ ⎥ or,
I 2 = C ′ V1 − D ′I1
⎢⎣ I 2 ⎥⎦ ⎣C ′ D ′ ⎦ ⎢⎣ − I1 ⎥⎦
The parameters A , B , C , D can be defined in a similar manner with either Port 1 on short circuit or
Port 1 on open circuit.
A′ =
V2
V1 I =0
Open-circuit voltage gain
1
V
B′ = − 2
I1 V = 0
Short-circuit transfer impedance
I2
V1 I =0
Open-circuit transfer admittance
1
C′ =
1
333
Two-Port Network
I
D′ = − 2
I1 V = 0
Short-circuit current gain
1
6.1.5 Hybrid Parameters (h-Parameters)
The hybrid parameters represent a mixed or hybrid relation between the voltages and the currents in the twoport network.
The hybrid parameter matrix may be written as
⎡V1 ⎤ ⎡ h11
⎢ ⎥=⎢
⎢⎣ I 2 ⎥⎦ ⎢⎣ h21
V1 = h11 I1 + h12V2
h12 ⎤ ⎡ I1 ⎤
⎥ ⎢ ⎥ or,
I 2 = h21 I1 + h22V2
h22 ⎥⎦ ⎢⎣V2 ⎥⎦
The h-parameters can be defined in a similar manner and are commonly used in some electronic circuit analysis.
V
h11 = 1
I1 V = 0
Short-circuit impedance input impedance
2
h12 =
V1
V2 I =0
Open-circuit reverse voltage gain
I2
I1 V = 0
Short-circuit current gain
I2
V2 I =0
Open-circuit output admittance
1
h21 =
2
h22 =
1
As the h-parameters are dimensionally mixed, they are also named mixed parameters. Transistor circuit
models are generally represented by these parameters, as the input impedance (h11) and the short-circuit current gain (h21) can be easily measured by making the output short-circuited.
Example 6.4 Find the hybrid parameters for the network shown in
Fig. 6.9.
I1
I2
1
Solution By KVL, 15 I1 + 5 I 2 = V1
(i)
5 I1 + 20 I 2 = V2
(ii)
Thus, the z-parameters are z11 (5 j10)
Z22 (5 j15)
The hybrid parameter equations are, V1
From Eq. (ii), we get,
I2 = −
2
10
15
5
z12
h11I1
z21
5
h12V2 and
V
5
1
1
I1 + 2 = − I1 + V2
20
20
4
20
1
2
Fig. 6.9 Network of Example 6.4
I2
h21I1
h22V2
(iii)
334
Network Analysis and Synthesis
⎡ 1
V ⎤
55
1
15 I1 + 5 ⎢ − I1 + 2 ⎥ = V1 ⇒ V1 = I1 + V2
20 ⎦
4
4
⎣ 4
Comparing Eq. (iii) and (iv) with the standard equations of h-parameters, we get,
Putting this value of I2 in Eq. (i), we get,
h11 =
55
4
(iv)
1
1
1
; h12 = ; h21 = − ; h22 =
4
4
20
6.1.6 Inverse Hybrid Parameters (g-Parameters)
The inverse hybrid parameters also represent a mixed or hybrid relation between the voltages and the currents
in the two-port network.
The inverse hybrid parameter matrix may be written as
⎡ I1 ⎤ ⎡ g11
⎢ ⎥=⎢
⎢⎣V2 ⎥⎦ ⎢⎣ g21
I1 = g11V1 + g12 I 2
g12 ⎤ ⎡V1 ⎤
⎥ ⎢ ⎥ or,
V2 = g21V1 + g22 I 2
g22 ⎥⎦ ⎢⎣ I 2 ⎥⎦
The g-parameters can be defined in a similar manner and are commonly used in some electronic circuit
analysis.
I
g11 = 1
Open-circuit input admittance
V1 I =0
2
g12 =
I1
I 2 V =0
Short-circuit reverse current gain
V2
V1 I =0
Open-circuit voltage gain
V2
I 2 V =0
Short-circuit output impedance
1
g21 =
2
g22 =
1
6.2 CONDITIONS FOR RECIPROCITY AND SYMMETRY
A network is said to be reciprocal if the ratio of the response transform to the excitation transform is invariant
to an interchange of the positions of the excitation and response of the network.
A two-port network will be reciprocal if the interchange of an ideal voltage source at one port with an ideal
current source at the other port does not alter the ammeter reading.
A two-port network is said to be symmetrical if the input and output ports can be interchanged without
altering the port voltages and currents.
1
I2
I1
2
Conditions in terms of z-parameters
Condition for Reciprocity We short-circuit Port 2 – 2 and
apply a voltage source Vs at Port 1 – 1 .
Therefore, V1 Vs, V2 0, I2 – I2
N
VS
1
Fig. 6.10 (a) Reciprocal network
I2
2
335
Two-Port Network
Writing the equations of z-parameters,
Vs z11 I1 z12 I2 and 0 z21 I1
Solving these two equations for I2 ,
z21
I 2′ = Vs
z11 z22 − z12 z21
I1
1
z22 I2
I1
(6.1)
1
I2
2
VS
N
2
Fig. 6.10 (b) Reciprocal network
Now, we interchange the positions of response and excitations, i.e., short
Port 1 – 1 and apply Vs at Port 2 – 2 ; V1 0, V2 Vs, I1 I1 .
Writing the equations of z-parameters,
z11 I1
z12 I2 and V5
z21 I1
z22 I2
0
Solving these two equations for I1 ,
I1′ = Vs
z12
z11 z22 − z12 z21
(6.2)
For the two-port network to be reciprocal, from Eq. (6.1) and Eq. (6.2), we have the condition as z12 = z21
Condition for symmetry Applying a voltage Vs at Port 1 – 1 with Port 2 – 2 open, we have the equation,
V
Vs = z11 I1 − z12 ⋅ 0 = z11 I1 ⇒ s
=z
(6.3)
I1 I =0 11
2
Now, applying a voltage Vs at Port 2 – 2’ with Port 1 – 1’ open, we have the equation,
V
Vs = z21 ⋅ 0 + z22 I 2 = z22 I 2 ⇒ s
=z
I 2 I =0 22
(6.4)
1
For the network to be symmetrical, the voltages and currents should be same. From Eq. (6.3) and Eq. (6.4),
we have the condition for symmetry as z11 = z22
Conditions in terms of y-parameters
Condition for reciprocity From Fig. 6.10 (a), writing the y-parameter equations,
I1 = y11V
I′
⇒ − 2 = y21
Vs
− I 2′ = y21Vs
From Fig. 6.10 (b), writing the y-parameter equations,
− I1′ = y12Vs
I′
⇒ − 1 = y12
Vs
I 2 = y22Vs
(6.5)
(6.6)
From the principle of reciprocity, the condition for reciprocity is y11 = y21
Condition for symmetry As already stated, a two-port network is said to be symmetric if the ports can be
interchanged without changing the port voltages and currents, and thus the condition of symmetry becomes,
y11 = y22
336
Network Analysis and Synthesis
Conditions in terms of ABCD-parameters
Condition for Reciprocity From Fig. 6.10 (a), writing the ABCD-parameter equations,
Vs = A ⋅ 0 − B( − I 2′ ) = BI 2′
I′ 1
⇒ 2=
Vs B
I1 = C ⋅ 0 − D( − I 2′ ) = DI 2′
From Fig. 6.10 (b), writing the ABCD-parameter equations,
0 = AVs − BI 2
I ′ AD − BC
⇒ 1=
Vs
B
− I1′ = CVs − DI 2
From the principle of reciprocity, the condition for reciprocity is
(6.7)
(6.8)
1 ( AD − BC )
=
B
B
( AD − BC ) =1
V
I1 = DI 2′ = D s
B
Condition for symmetry From Eq. (6.7),
From Eq. (6.8),
⎫⎪
I ′+ CVs 1 ⎧ ⎛ AD − BC ⎞
A
= ⎨Vs ⎜
I2 = 1
⎟⎠ + CVs ⎬ = Vs B
D
D ⎩⎪ ⎝
B
⎭
(6.9)
(6.10)
From Eq. (6.9) and Eq. (6.10), we have the condition for symmetry as A = D
Conditions in terms of h-parameters
Condition for Reciprocity From Fig. 6.10 (a), writing the h-parameter equations,
Vs = h11 I1 + h12 ⋅ 0 = h11 I1
I′
h
⇒ 2 = − 21
V
h11
− I 2′ = h21 I1 + h22 ⋅ 0 = h21 I1
s
From Fig. 6.10 (b), writing the h-parameter equations,
0 = − h11 I1′+ h12Vs
I′ h
⇒ 1 = 12
Vs h11
I 2 = − h21 I1′+ h22Vs
(6.11)
(6.12)
From the principle of reciprocity, the condition for reciprocity is h12 = − h21
I1 =
Condition for symmetry From Eq. (6.11),
Vs
h11
⎛h
⎞
h h −h h
From Eq. (6.12), I 2 = − h21 ⎜ 12 Vs ⎟ + h22Vs = Vs 11 22 12 21
h11
⎝ h11 ⎠
(6.13)
(6.14)
From Eq. (6.13) and Eq. (6.14), we have the condition for symmetry as ( h11 h22 − h12 h21 ) = 1
Conditions in terms of inverse T-parameters
Condition for Reciprocity From Fig. 6.10 (a), writing the T -parameter equations,
0 = A′Vs − B ′I1
− I 2′ = C ′Vs − DI1
⇒
I 2′ A′D ′ − B ′C ′
=
Vs
B′
(6.15)
337
Two-Port Network
From Fig. 6.10 (b), writing the T -parameter equations,
Vs = 0 ′ − B ′ − I1′ = B ′I1′
( )
I′ 1
=
⇒
V B′
I = 0 ′ − D ′ ( − I ′) = D ′ I ′
2
1
1
(6.16)
s
1
From the principle of reciprocity, the condition for reciprocity is ( A′D ′ − B ′C ′ ) = 1
Condition for Symmetry From Eq. (6.15),
From Eq. (6.16), I 2 =
I1 =
A′
V
B′ s
(6.17)
D′
V
B′ s
(6.18)
From Eq. (6.17) and Eq. (6.18), we have the condition for symmetry as A′ = D ′
Conditions in terms of inverse hybrid (g)-parameters
Condition for reciprocity From Fig. 6.10 (a), writing the g-parameter equations,
I1 = g11Vs − g12 I 2′
I′ g
⇒ 2 = 21
V
g22
0 = g21Vs − g22 I 2′
s
From Fig. 6.10 (b), writing the g-parameter equations,
− I1′ = g11 0 + g12 I 2 = g12 I 2
I′
g
⇒ 1 = − 12
Vs
g22
Vs = g21 0 + g22 I 2 = g22 I 2
(6.19)
(6.20)
From the principle of reciprocity, the condition for reciprocity is g12 = − g21
⎛ g g −g g ⎞
I1 = ⎜ 11 22 12 21 ⎟ Vs
g22
⎝
⎠
Condition for symmetry From Eq. (6.19),
From Eq. (6.20), I 2 =
(6.21)
1
V
g22 s
(6.22)
From Eq. (6.21) and Eq. (6.22), we have the condition for symmetry as ( g11 g22 − g12 g 21 ) = 1
Table 6.1 Conditions of Reciprocity and Symmetry in Terms of Different
Two-Port Parameters
Parameter
Condition of Reciprocity
Condition of Symmetry
z
z12
z21
z11
z22
y
y12
y21
y11
y22
T (ABCD)
(AD
BC )
A
D
T (A B C D )
(A D
BC)
A
D
h
h12
h21
(h11h22
h12 h21)
1
g
g12
g21
(g11 g22
g12 g21)
1
1
1
338
Network Analysis and Synthesis
Example 6.5 Find the z-parameters for the network shown in Fig. 6.11 and
state whether the network is reciprocal and symmetrical.
Solution By writing the KVL for the two meshes of Fig. 6.11, we get,
(
)
2 + 2 + 2 I1 + 2 I 2 = V1 ⇒ V1 = 6 I1 + 2 I 2
(
)
2 I1 + 2 + 2 + 2 I 2 = V2 ⇒ V2 = 2 I1 + 6 I 2
Form these two equations; we get the z-parameters of the network as,
z11
Since z11
Since z11
6.3
6
;
z12
z21
2
;
z22
2
2
2
I1
2
I2
2
Fig. 6.11 Network of
Example 6.5
6
z22, for this network, the network is symmetrical.
z21, for this network, the network is reciprocal.
INTERRELATIONSHIPS BETWEEN TWO-PORT PARAMETERS
Each type of two-port parameter has its own utility and is suited for certain specific applications. However, it
is sometimes necessary to convert one set of parameters to another. It is possible through simple mathematical manipulations to convert one set to any of the remaining sets. It is discussed below.
6.3.1
z-Parameters in Terms of Other Parameters
In Terms of y-parameters The z-parameter equations are
V1 = z11 I1 + z 12 I 2
(6.23)
V2 = z21 I1 + z22 I 2
The y-parameter equations are
I1 = y11V1 + y12V2
(6.24)
I 2 = y21V1 + y 22V2
From Eq. (6.24), V2 =
I 2 y21
−
V ; substituting this in the first equation,
y22 y22 1
⎛ I
⎞
y
I1 = y11V1 + y12 ⎜ 2 − 21 V1 ⎟
y
y
⎝ 22
⎠
22
or, V1 =
y22
y
I − 12 I 2
y 1
y
(6.25)
where,
y (y11y22 y12y21)
Substituting this value in the second equation of Eq. (6.24)
⎛y
⎞
y
I 2 = y21 ⎜ 22 I1 − 12 I 2 ⎟ + y22V2
y ⎠
⎝ y
or
V2 = −
Comparing Eq. (6.23), (6.25) and (6.26), we get
y21
y
I1 + 11 I 2
y
y
z11 =
y22
y
y
y
; z12 = − 12 ; z21 = − 21 ; z22 = 11
y
y
y
y
(6.26)
339
Two-Port Network
In Terms of transmission parameters The transmission parameter equations are,
V1 = AV2 − BI 2
(6.27)
I1 = CV2 − DI 2
From the second equation of Eq. (6.27),
From the first equation of Eq. (6.27),
⎛ D⎞
⎛ 1⎞
V2 = ⎜ ⎟ I1 + ⎜ ⎟ I 2
⎝C⎠
⎝C⎠
(6.28)
⎡⎛ 1 ⎞
⎛ D⎞ ⎤
⎛ AD − BC ⎞
⎛ A⎞
V1 = A ⎢⎜ ⎟ I1 + ⎜ ⎟ I 2 ⎥ − BI 2 = ⎜ ⎟ I1 + ⎜
⎟⎠ I 2
C
⎝C⎠ ⎦
⎝
⎝C⎠
⎣⎝ C ⎠
Comparing Eq. (6.28) and (6.29) with Eq. (6.23), we get
z11 =
(6.29)
A
AD − BC
T
D
1
; z12 =
=
; z21 = ; z22 =
C
C
C
C
C
In terms of hybrid parameters The hybrid parameter equations are
V1 = h11 I1 + h12V2
(6.30)
I 2 = h21 I1 + h22V2
From the second equation of Eq. (6.30),
⎛ h ⎞
⎛ 1 ⎞
V2 = ⎜ − 21 ⎟ I1 + ⎜ ⎟ I 2
⎝ h22 ⎠
⎝ h22 ⎠
(6.31)
From the first equation of Eq. (6.30),
⎡⎛ h ⎞
⎛ 1 ⎞ ⎤ ⎛ h h −h h ⎞
⎛h ⎞
V1 = h11 I1 + h12 ⎢⎜ − 21 ⎟ I1 + ⎜ ⎟ I 2 ⎥ = ⎜ 11 22 12 21 ⎟ I1 + ⎜ 12 ⎟ I 2
h22
⎝ h22 ⎠ ⎥⎦ ⎝
⎠
⎝ h22 ⎠
⎢⎣⎝ h22 ⎠
(6.32)
Comparing Eq. (6.31) and Eq. (6.32) with Eq. (6.23), we get
h h −h h
h
h
1
h
z11 = 11 22 12 21 =
; z12 = 12 ; z21 = 21 ; z22 =
h22
h22
h22
h22
h22
Similarly, the inter-relation of the other parameter in terms of the remaining parameters are obtained by writing the remaining parameter equations in the same format as those of the other parameter; and comparing the
coefficients of the two sets of equations, a relation is obtained.
A summary of the relationships between impedance z-parameters, admittance y-parameters,
hybrid h-parameters, and transmission ABCD-parameters is shown in Table where z
(z11z22
z12z21),
h (h11h22 h12h21), T (AD BC), T
(A D
B C ), and
g (g11g22 g12g21).
6.4
INTERCONNECTION OF TWO-PORT NETWORKS
In certain applications, it becomes necessary to connect the two-port networks together.
The common connections are (a) series connection, (b) parallel connection, (c) cascade connection
(d) series–parallel connection, and (e) parallel–series connection.
340
Network Analysis and Synthesis
Table 6.2 Interrelationships Between Two-Port Parameters
[z]
[y]
y22
y
y
− 21
y
[z]
[ y]
[ABCD]
[A B C D ]
z11
z12
z21
z22
z22
z
z
− 21
z
z
− 12
z
z11
z
z11
z21
z
z21
1
z21
y
− 12
y
y11
y
y11
y12
y21
y22
−
y22
y21
−
z22
z21
−
y
y
− 11
y21
z22
z12
z
z12
y
− 11
y12
−
1
z12
z11
z12
−
y
y12
y
− 22
y12
z22
z12
z22
1
y11
y
− 12
y11
z
− 21
z22
1
z22
y21
y11
y
1
z11
z
− 12
z11
y
z21
z11
z
z
[h]
[g]
[ABCD]
z11
y21
y22
−
y21
y22
1
y21
1
y12
y11
y12
y22
−
1
y22
A
C
1
C
T
C
D
C
D
B
1
−
B
−
[A B C D ]
D′
C′
T′
C′
T
B
A
B
1
C′
A′
C′
A′
B′
T′
−
B′
1
B′
D′
B′
−
D′
T′
C′
T′
A B
C D
B′
T′
A′
T′
D
T
C
T
B
T
A
T
A′ B ′
C ′ D′
B
D
1
−
D
T
D
C
D
B′
A′
T′
−
A′
C
T
−
A
A
B
1
A
A
C′
D′
T′
D′
1
A′
D′
B′
1
D′
B′
D′
−
[h]
[g]
g12
g11
h22
h12
h22
1
g11
−
h
− 21
h22
1
h22
g21
g11
g
g11
1
h11
h
− 12
h11
g
g22
g12
g22
h21
h11
h
g21
g22
1
g22
h
−
h11
h
h21
−
h
− 11
h21
1
g21
−
g22
g21
1
h21
g11
g21
g
g21
h
− 22
h21
−
1
h22
h11
h12
−
g
g12
−
g22
g12
h22
h12
h
−
g11
g12
−
1
g12
g22
g
g
− 21
g
−
g11
g12
g21
g22
h12
h11
h12
h21
h22
h22
h
h
− 21
h
h
− 12
h
h11
h
g12
g
g11
g
6.4.1 Series Connection of Two-Port Networks
As in the case of elements, a series connection is defined when the currents in the series elements are equal
and the voltages add up to give the resultant voltage.
In the case of two-port networks, this property must be applied individually to each of the ports. Thus, if
we consider two networks r and s connected in series, at Port 1, Ir1 Is1 I1, and Vr1 Vs1 V1
Similarly, at Port 2, Ir2 Is2 I2 and Vr2 Vs2 V2
The two networks, r and s can be connected in the following manner to be in series with each other.
341
Two-Port Network
Under these conditions,
V1 = (Vr 1 + Vs1 ) = ( z11r + z11s ) I1 + ( z12 r + z12 s ) I 2
Ir 1
Vr 1
V2 = (Vr 2 + Vs 2 ) = ( z21r + z21s ) I1 + ( z22 r + z22 s ) I 2
It is seen that the resultant impedance parameter matrix
for the series connection is the addition of the two individual impedance matrices.
[z] [zr] [zs]
)
(
)
(
)
(
V1
Vs1
Port r 1
+
Is1
Vb
Port s1
Liner
passive
network
s
Ir 2
Vr 2
Port r 2
Is2
Port s 2
+
Va
-
V2
Vs2
Fig. 6.12 Series connection of two-port networks
)
(
∴ z11 = z11r + z11s ; z12 = z12 r + z12 s ; z21 = z21r + z21s ; z22 = z22 r + z22 s
Note
Liner
passive
network
r
In the interconnection of series networks, there is a strong requirement of isolation, since the ground node
of upper network forms the non-ground node of the lower network. For the port properties to be valid, the
voltages Va and Vb must be identically zero for the two networks r and s to be connected in series. If Va and Vb
are not zero, then by connecting the two ports there will be a circulating current and the port property of the
individual networks r and s will be violated.
6.4.2 Parallel Connection of Two-Port Networks
As in the case of elements, a parallel connection is defined when the voltages in the parallel elements are equal
and the currents add up to give the resultant current.
In the case of two-port networks, this property must be applied individually to each of the ports.
Thus, if we consider two networks r and s connected in parallel, at Port 1, Ir1 Is1 I1 and Vr1 Vs1 V1
Similarly, at Port 2, Ir2 Is2 I2 and Vr2 Vs2 V2
Ir 1
I1
I2
Ir2
Linear
The two networks, r and s can be connected in the following manner
passive
to be in parallel with each other.
Vr1
Vr 2
network
Under these conditions,
r
V1
V2
Is 2
Is1
I1 = ( I r 1 + I s1 ) = ( y11r + y11s )V1 + ( y12 r + y12 s )V2
Linear
I 2 = ( I r 2 + I s 2 ) = ( y21r + y21s )V1 + ( y22 r + y22 s )V2
Vs1
It is seen that the resultant admittance parameter matrix for the parallel
connection is the addition of the two individual admittance matrices.
[Y ] [Yr] [Ys]
y11
(y11r
y11s);
y12
Ir 1
Vr 1
Vb Is1
Vs1
(y12r
y12s);
Linear
passive
network
r
Linear
passive
network
s
y21
Ir 2
(y21r
y21s);
I2
Is2
V2
Vs 2
Fig. 6.14 (a) Condition of parallel
connection: Vb 0
V1
(y22r
I1
Ir1
Is1
Vs1
Vs2
Fig. 6.13 Parallel connection of
two-port networks
y22
Vr1
Vr 2
passive
network
s
y22s)
Linear
passive
network
r
Linear
passive
network
s
Ir 2
Vr 2
Is2 Va
Vs2
Fig. 6.14 (b) Condition of parallel
connection: Va 0
342
Network Analysis and Synthesis
Note
As in series connection, parallel connection is also possible under the condition that Va
cannot be connected in parallel as that will violate the port properties.
Vb
0; otherwise they
6.4.3 Cascade Connection of Two-Port Networks
A cascade connection is defined when the output of one network becomes the input to the next network.
Ir 1
Port rI
Vr 1
Ir 2
Linear
passive
network
Is1
Port r2
Vr2 Vs1
Port s1
Is2
Linear
passive
network
Port s2
Vs 2
Fig. 6.15 Cascade connection of two-port networks
It can be easily seen that Ir2 Is1 and Vr2 Vs1
Therefore, it can easily be seen that the ABCD parameters are the most suitable to be used for this
connection.
⎡Vr 1 ⎤ ⎡ Ar
⎢ ⎥=⎢
⎢⎣ I r 1 ⎥⎦ ⎢⎣Cr
Br ⎤ ⎡Vr 2 ⎤ ⎡Vs1 ⎤ ⎡ As
⎥ ⎢ ⎥, ⎢ ⎥ = ⎢
Dr ⎥⎦ ⎢⎣ I r 2 ⎥⎦ ⎢⎣ I s1 ⎦⎥ ⎢⎣Cs
Bs ⎤ ⎡Vs 2 ⎤
⎥⎢ ⎥
Ds ⎥⎦ ⎢⎣ I s 2 ⎥⎦
Br ⎤ ⎡ As Bs ⎤ ⎡Vs 2 ⎤ ⎡ Ar Br ⎤ ⎡ As Bs ⎤ ⎡V2 ⎤
⎥⎢ ⎥
⎥⎢
⎥⎢ ⎥ = ⎢
⎥⎢
Dr ⎥⎦ ⎢⎣Cs Ds ⎥⎦ ⎢⎣ I s 2 ⎥⎦ ⎢⎣Cr Dr ⎥⎦ ⎢⎣Cs Ds ⎥⎦ ⎢⎣ I 2 ⎥⎦
Thus it is seen that the (overall ABCD matrix is the product of the two individual ABCD matrices). This is a
very useful property in practice, especially when analyzing transmission lines.
⎡ A B ⎤ ⎡ Ar Br ⎤ ⎡ As Bs ⎤
⎥
⎥⎢
⎢
⎥=⎢
⎣C D ⎦ ⎢⎣Cr Dr ⎥⎦ ⎢⎣Cs Ds ⎥⎦
⎡V1 ⎤ ⎡Vr 1 ⎤ ⎡ Ar
⎢ ⎥= ⎢ ⎥= ⎢
⎢⎣ I1 ⎥⎦ ⎢⎣ I r 1 ⎥⎦ ⎢⎣Cr
Br ⎤ ⎡Vr 2 ⎤ ⎡ Ar
⎥⎢ ⎥ = ⎢
Dr ⎥⎦ ⎢⎣ I r 2 ⎥⎦ ⎢⎣Cr
Br ⎤ ⎡Vs1 ⎤ ⎡ Ar
⎥⎢ ⎥ = ⎢
Dr ⎥⎦ ⎢⎣ I s1 ⎥⎦ ⎢⎣Cr
6.4.4 Series–Parallel Connection of Two-Port Networks
Two two-port networks are said to be connected in series–parallel if the input ports are connected in series
and the output ports in parallel as shown in Fig. 6.16.
I1
Ir 1
Ir2
I2
V1 = (Vr 1 + Vs1 )
V =V =V
Linear
and 2 r 2 s 2
Under these conditions,
Vr1 passive Vr 2
I1 = I r 1 = I s1
I 2 = ( I r 2 + I 21 )
network
For the network r,
For the network s,
Now,
⎡Vr 1 ⎤ ⎡ h11r
⎢ ⎥=⎢
⎢⎣ I r 2 ⎥⎦ ⎢⎣ h21r
h12 r ⎤ ⎡ I r 1 ⎤
⎥⎢ ⎥
h22 r ⎥⎦ ⎢⎣Vr 2 ⎥⎦
⎡Vs1 ⎤ ⎡ h11s
⎢ ⎥=⎢
⎢⎣ I s 2 ⎥⎦ ⎢⎣ h21s
h12 s ⎤ ⎡ I s1 ⎤
⎥⎢ ⎥
h22 s ⎥⎦ ⎢⎣Vs 2 ⎥⎦
⎡V1 ⎤ ⎡Vr 1 ⎤ ⎡Vs1 ⎤ ⎡ h11r
⎢ ⎥= ⎢ ⎥+ ⎢ ⎥= ⎢
⎢⎣ I 2 ⎥⎦ ⎢⎣ I r 2 ⎥⎦ ⎢⎣ I s 2 ⎥⎦ ⎢⎣ h21r
⎡h
= ⎢ 11r
⎢⎣ h21r
r
V2
V1
Is1
Is 2
Linear
Vs1 passive Vs2
network
s
Fig. 6.16 Series–parallel
h12 r ⎤ ⎡ I r 1 ⎤ ⎡ h11s h12 s ⎤ ⎡ I s1 ⎤
connection of two-port networks
⎥⎢ ⎥
⎥⎢ ⎥ + ⎢
h22 r ⎥⎦ ⎢⎣Vr 2 ⎥⎦ ⎢⎣ h21s h22 s ⎥⎦ ⎢⎣Vs 2 ⎥⎦
h12 r + h12 s ⎤ ⎡ I1 ⎤
h12 r ⎤ ⎡ I1 ⎤ ⎡ h11s h12 s ⎤ ⎡ I1 ⎤ ⎡ h11r + h11s
⎥⎢ ⎥
⎥⎢ ⎥ = ⎢
⎥⎢ ⎥ + ⎢
h22 r + h22 s ⎥⎦ ⎢⎣V2 ⎥⎦
h22 r ⎥⎦ ⎢⎣V2 ⎥⎦ ⎢⎣ h21s h22 s ⎥⎦ ⎢⎣V2 ⎥⎦ ⎢⎣ h21r + h21s
(
(
) (
) (
)
)
343
Two-Port Network
Thus, it is seen that the resultant hybrid parameter matrix for the series–parallel connection is the addition
of the two individual hybrid parameter matrices.
[h] [hr] [hs]
h11 (h11r h11s); h12 (h12r h12s); h21 (h21r h21s); h22 (h22r h22s)
6.4.5 Parallel–Series Connection of Two-Port Networks
Two two-port networks are said to be connected in parallel–series if
the input ports are connected in parallel and the output ports in series
as shown in Fig. 6.17.
V1 = Vr 1 = Vs1
V = (Vr 2 + Vs 2 )
and 2
Under these conditions,
I1 = ( I r 1 + I s1 )
I 2 = I r 2 = I 21
In a similar way in series–parallel connection, it can be shown that the resultant inverse hybrid parameter matrix for the parallel–series connection is
the addition of the two individual inverse hybrid parameter matrices.
[g] [gr] [gs]
g11 (g11r g11s); g12 (g12r g12s);
g21 (g21r g21s); g22 (g22r g22s)
Example 6.6 Find the transmission parameters for the network
shown in Fig. 6.18 considering two networks connected in cascade.
I1
Ir1
Ir 2
Linear
Vr 1 passive Vr2
network
r
I2
V2
V1
Is1
Is2
Linear
Vs1 passive Vs2
network
s
Fig. 6.17 Parallel–series connection
of two-port network
1
2
1
Solution The network of Fig. 6.18 can be considered to be the casV2
V2
2
cade connection of two two-port networks as shown in Fig. 6.19.
We know that for cascade connection, the overall transmission
Fig. 6.18 Network of Example 6.6
parameter matrix is the product of the individual transmission parameter
matrices.
We find the transmission parameter matrix of the individual sections.
For each section, the z-parameters are given as
Network 1
Network 2
z11 (1 2) 3
z12 z21 2
z22 (1 2) 3
1
1
1
1
By the interrelationship between z-parameters and transmission parameters, we get
V2
V2
2
2
⎡ z11
z⎤
−
−
⎢
⎥
⎡ A1 B1 ⎤ ⎡ A2 B2 ⎤ ⎢ z21 z21 ⎥
Fig. 6.19 Cascade connection of two-port
⎢
⎥=⎢
⎥=⎢
networks
⎥
C
D
C
D
z
⎥ ⎣⎢ 2
⎥ ⎢ 1
⎣⎢ 1
1⎦
2⎦
22
⎥
⎢⎣ z21 z21 ⎥⎦
z
3
z 3 × 3 − 22 5
∴ A = 11 = ; B =
=
=
2
2
z21 2
z21
; C=
z
1 1
3
= mho; D = 22 =
z21 2
z21 2
Therefore, the transmission parameter matrix of the network of Fig. 6.18, is
⎡3
5 ⎤ ⎡3
5 ⎤ ⎡7
15 ⎤
⎡A B⎤ ⎢ 2
⎥
⎥
⎢
⎢
2
2
2
2
2⎥
=⎢
×⎢
⎢
⎥=⎢
⎥
⎥
3
3
3
7 ⎥
1
⎣C D ⎦ ⎢ 1
⎣ 2
2 ⎥⎦ ⎢⎣ 2
2 ⎥⎦ ⎢⎣ 2
2 ⎥⎦
344
Network Analysis and Synthesis
6.5
TWO-PORT NETWORK FUNCTIONS
Two-port network functions are broadly divided into two groups:
(I) Transfer function, and
(II) Driving point function.
6.5.1 Transfer Function
It is defined as the ratio of an output transform to an input transform, with zero initial condition and with no
internal energy sources except the controlled sources.
For a two-port network, having the variables I1(s), I2(s), V1(s), and V2(s), the transfer function can take the
following four forms:
Voltage Transfer Function
G12 ( s ) =
Current Transfer Function
Transfer Impedance Function
Transfer Admittance Function
Note
12
V1 ( s )
V (s)
; G21 ( s ) = 2
V2 ( s )
V1 ( s )
I (s)
I (s)
( s ) = 1 ; 21 ( s ) = 2
I2 (s)
I1 ( s )
V1 ( s )
V (s)
Z12 ( s ) =
; Z 21 ( s ) = 2
I2 (s)
I1 ( s )
Y12 ( s ) =
I1 ( s )
I (s)
; Y21 ( s ) = 2
V2 ( s )
V1 ( s )
(i) For a one-port network, Z(s) 1/ Y(s); but for a two-port network, in general Z12 ⬆ 1/Y12; G12 ⬆ 1/␣12.
(ii) Z and Y functions will become z and y parameters under the conditions of open-circuits or short-circuits,
respectively.
6.5.2 Driving Point Function
It takes two forms:
Driving Point Impedance [Z(s)] For a two-port network in zero state with no internal energy sources, the
driving point impedance is the ratio of transform voltage at any port to the transform current at the same port.
V (s)
V (s)
Z11 ( s ) = 1 ; Z 22 ( s ) = 2
I1 ( s )
I2 (s)
Driving point admittance [Y(s)] For a two-port network in zero state with no internal energy sources, the
driving point admittance is the ratio of transform current at any port to the transform voltage at the same port.
I (s)
I (s)
Y11 ( s ) = 1 ; Y22 ( s ) = 2
V1 ( s )
V2 ( s )
Note
(i) Driving point impedance and admittance functions together are known as immittance function.
(ii) Z and Y functions will become z and y parameters under the conditions of open circuits or short circuits,
respectively.
345
Two-Port Network
6.6
TRANSFER FUNCTIONS OF TERMINATED TWO-PORT NETWORKS
A two-port network may be terminated by impedance. The impedance may be connected either in input port
or in the output port as shown in figures.
I2
I1
1
N
V1
V2
1
I2
I1
2
1
ZL
ZL
2
1
Fig. 6.20 (a) Two-port network with
terminted output
2
V2
N
V1
2
Fig. 6.20 (b) Two-port network with
terminated input
6.6.1 Determination of Input Impedance (Zin ) in Terms of Network Parameters and
Terminated Impedance
In terms of z-parameters In this case (Fig. 6.20 (a)), V2
So,
V2 = − I 2 Z L = z21 I1 + z22 I 2 ⇒ I 2 = −
Putting this value in the first equation of z-parameters,
or,
Note
I2ZL
⎛
z21 ⎞
V1 = z11 I1 + z12 I 2 = ⎜ z11 −
I
z22 + Z L ⎟⎠ 1
⎝
)
(
z z − z z + z11 z L
V
Z in = 1 = 11 22 12 21
I1
z22 + Z L
(i) For an open-circuited output, ZL → ; then Zin
(ii) For a short-circuited output, ZL 0; then Zin
In Terms of y-parameters Putting the value V2
y-parameters, we get
z11
z/z22
1/y11
I2ZL ⇒ , I2
I 2 = −YLV2 = y21V1 + y22V2 ⇒ V2 = −
I1 = y11V1 + y12V2 = y11V1 −
Note
y12 y21
V
y22 + YL 1
V
y22 + YL
Z in = 1 =
I1 y11 y22 − y12 y21 + y11YL
(i) For an open-circuited output, YL 0; then Zin y22/ y
(ii) For a short-circuited output, YL → ; then Zin 1/y11
YLV2 in the second equation of
y21
V
y22 + YL 1
Putting this value in the first equation of y-parameters, we get
or,
z21
I
z22 + Z L 1
z11
346
Network Analysis and Synthesis
In terms of ABCD-parameters Putting the value V2
we get
I2ZL in the second equation of ABCD-parameters,
I1 = CV2 − DI 2 = −CI 2 Z L − DI 2 ⇒ I 2 = −
I1
CZ L + D
Putting this value in the first equation of ABCD-parameters, we get
)
(
V1 = AV2 − BI 2 = A( − I 2 Z L ) − BI 2 = − AZ L + B I 2 =
V AZ L + B
Z in = 1 =
I1 CZ L + D
or,
Note
AZ L + B
I
CZ L + D 1
(i) For an open-circuited output, ZL → ; then Zin A/C z11
(ii) For a short-circuited output, ZL 0; then Zin B/D 1/y11
In Terms of h-parameters From the second equation of h-parameters, putting the value, V2
I 2 = h21 I1 + h22V2 = h21 I1 + h22 ( − I 2 Z L ) ⇒ I 2 =
I2 ZL, we get
h21
I
1 + h22 Z L 1
Putting this value in the first equation of h-parameters,
V1 = h11 I1 + h12V2 = h11 I1 + h12 ( − I 2 Z L ) = h11 I1 − h12 Z L
or,
Note
)
(
h21
I
1+ h22 Z L 1
h h −h h Z +h
V
Z in = 1 = 11 22 12 21 L 11
I1
1 + h22 Z L
(i) For an open-circuited output, ZL → ; then Zin
h/h22 z11
(ii) For a short-circuited output, ZL 0; then Zin h11 1/y11
6.6.2 Determination of Output Impedance (Zout) in terms of Network Parameters
and Terminated Impedance
I1Z1 ⇒ I1
Y1V1
z12
V1 = − I1 Z1 = z11 I1 + z12 I 2 ⇒ I1 = −
I
z11 + Z1 2
In terms of z-parameters In this case (Fig. 6.20 (b)), V1
So,
Putting this value in the second equation of z-parameters,
or
Note
Z out =
⎛
⎞
z
V2 = z21 I1 + z22 I 2 = z21 ⎜ − 12 ⎟ I 2 + z22 I 2
⎝ Z1 + z11 ⎠
)
(
z z − z z + z22 Z1
V2
= 11 22 12 21
I2
z11 + Z1
(i) For an open-circuited output, Z1 → ; then
Zout
z22
(ii) For a short-circuited output, Z1
Zout
z/z11
0; then
1/y22
347
Two-Port Network
In terms of y-parameters Putting the value V1
y-parameters, we get
I1ZL
I1 = −Y1V1 = y11V1 + y12V2 ⇒ V1 = −
⇒
I1
Y1I2 in the first equation of
y12
V
y11 + Y1 2
Putting this value in the second equation of y-parameters, we get
⎛
y ⎞
I 2 = y21V1 + y22V2 = y21 ⎜ − 12 ⎟ V2 + y22V2
⎝ y11 + Y1 ⎠
or,
Note
Z out =
V2
y11 + Y1
=
I 2 y11 y22 − y12 y21 + y22Y1
(i) For an open-circuited output, Y1 0; then Zout
(ii) For a short-circuited output, Y1 → ; then Zout
y11/ y
1/y22
In terms of ABCD-parameters Putting the value V1
z22
I1Z1, we get
V1 = AV2 − VI 2
I1 = CV2 − DI 2
V1
AV − BI 2
= − Z1 = 2
I1
CV2 − DI 2
−CZ1V2 + DZ1 I 2 = AV2 − BI 2
I 2 ( B + DZ1 ) = V2 ( A + CZ1 )
Z out =
or,
Note
V2 B + DZ1
=
I 2 A + CZ1
(i) For an open-circuited output, Z1 → ; then Zout D/C z22
(ii) For a short-circuited output, Z1 0; then Zout B/A 1/y22
In terms of h-parameters From the equation, putting the value V1
h-parameters, we get
h
V1 = − I1 Z1 = h11 I1 + h12V2 ⇒ I1 = − 12 V2
h11 + Z1
Putting this value in the second equation of h-parameters,
or,
Note
Z out =
I1Z1 in the first equation of
⎛
⎞
h
I 2 = h21 I1 + h22V2 = h21 ⎜ − 12 ⎟ V2 + h22V2
⎝ h11 + Z1 ⎠
V2
h11 + Z1
=
I 2 h11 h22 − h12 h21 + h22 Z1
(i) For an open-circuited output, Z1 → ; then Zout 1/ h22
(ii) For a short-circuited output, Z1 0; then Zout h11 / h
z22
1/ y22
348
Network Analysis and Synthesis
Example 6.7 The currents I1 and I2 at input and output ports respectively of a two-port network are
expressed as I1
5V1
V2
and
I2
V1
Find the y-parameters. If a load impedance of (3
V2
j5)
is connected across the output port, find the input impedance.
Solution Comparing the equations with the standard y-parameter equations, we get
1 ; and y22 1
y11 5 ; y12 y21
Here, load impedance is ZL (3 j5)
⎛ 3
1
1
5⎞
=
=⎜ − j ⎟
load admittance, YL =
ZL
34 ⎠
3 + j 5 ⎝ 34
)
(
The input impedance is
V
y22 + YL
Z in = 1 =
=
I1 y11 y22 − y12 y21 + y11YL
6.7
3
5
−j
34
34
= 0.248∠1.89° ( )
2
⎛ 3
5⎞
5 × 1 − −1 + 5 × ⎜ − j ⎟
34 ⎠
⎝ 34
1+
( )
APPLICATION OF NETWORK PARAMETERS TO THE ANALYSIS OF TYPICAL
TWO-PORT NETWORKS
We consider six typical two-port networks
1. T network
2. ␲ network
3. Ladder network
4. Lattice network
5. Bridge-T network
6. Parallel-T or twin-T or notch-filter network
6.7.1
T or Star or Y Network
The configuration of a typical T-network is shown in Fig. 6.21.
I1
By KVL equations for the two meshes, we get
Zb
Za
1
(Za Zb)I1 ZcI2 V1 and ZcI1 (Zb Zc)I2 V2
Thus, the z-parameters are Z11 (Za Zb); z12 z21 Zc;
Zc
V1
z22 (Zb Zc)
Rearranging, Za (z11 z12); Zb (z22 z12) Zc z12 z21
1
From the inter-relationship, we get for T-network, the transmisFig. 6.21 T-network
sion parameters as
⎛ Z ⎞
⎛ Z ⎞
Z Z ⎞
z
z
z ⎛
1
1
A = 11 = ⎜ 1 + a ⎟ ; B =
= ⎜ Z a + Zb + a b ⎟ ; C =
= ; D = 22 = ⎜ 1 + b ⎟
z21 ⎝
Zc ⎠
z21 ⎝ Z c ⎠
z12 Z c
z12 ⎝ Z c ⎠
Conversely, Z a =
( A − 1) ; Z = ( D − 1) ; and Z = 1
C
and the y-parameters,
b
C
c
C
I2
2
V2
2
349
Two-Port Network
y11 =
Za + Zc
Zb + Z c
− Zc
z
z22
z
=
; y12 = y21 = − 12 =
; y22 = 11 =
z Z a Zb + Zb Z c + Z c Z a
z Z a Zb + Zb Z c + Z c Z a
z Z a Zb + Zb Z c + Z c Z a
6.7.2
or Delta Network
The configuration of a typical ␲-network is shown in Fig. 6.22.
By KCL equations at the two nodes, we get
(V −V )Y +V Y = I ⇒ (Y + Y )V − Y V = I
V Y + (V −V
V )Y = I ⇒ − Y V + (Y + Y )V = I
1
2
c
2 b
1 a
2
1
1
c
a
2
c
c 1
1
c 2
b
1
1
c
2
)
(
(
)
(
(
2
Yc
2
V1
Thus, the y-parameters are
y11 = Ya + Yb ; y12 = y21 = −Yc ; y22 = Yb + Yc
I2
I1
)
)
Rearranging, Ya = y11 + y12 ; Yb = y22 + y12 ; Yc = − y12 = − y21
Ya
V2
Yb
2
1
Fig. 6.22 ␲-network
From the inter-relationship we get for ␲-network, the transmission parameters as
A= −
Conversely, Ya =
⎛ Y ⎞
YY ⎞
y
y22 ⎛ Yb ⎞
1
1
y ⎛
= ; C=−
= 1+
; B=−
= Y + Y + a b ; D = − 11 = ⎜ 1 + a ⎟
y21 ⎝ Yc ⎠
y21 ⎜⎝ Yc ⎟⎠
y21 Yc
y21 ⎜⎝ a b Yc ⎟⎠
( D − 1) ; Y = ( A − 1) ; and Y = 1
B
b
B
c
B
6.7.3 Ladder Network
The configuration of a typical ladder-network is
Z1
Z3
1
shown in Fig. 6.23.
The series arms are indicated by their impedances
Y2
Y4
Z1, Z3, Z5, … and the shunt arms are indicated by their
admittances Y2, Y4, Y6, …
1
In order to find the driving point impedance at Port- 1,
Fig. 6.23 Ladder network
we start computation at Port- 2 with Y6; i.e., inverting Y6,
combining with Z5, inverting the sum, and so on.
Thus, the driving point impedance at Port 1 1 is given as
1
Z11 = Z1 +
1
Y2 +
1
Z3 +
1
Y4 +
1
Z5 +
Y6
Note
Z5
2
Y6
2
This equation is known as continued fraction.
In order to find the transfer function, we again start at the output port and then proceed backward, applying
KCL and KVL where necessary.
350
Network Analysis and Synthesis
6.7.4 Lattice Network
I1
A Lattice network forms the basis of the design of most four-terminal networks
like attenuators, filters etc.
It consists of two identical impedances in series arm and two identical
impedances in shunt arm as shown in Fig. 6.24 (a).
Here, Za are the series arms and Zb are the diagonal or shunt arms.
To find the z-parameters, we redraw the network as shown in Fig. 6.24 (b).
0, the current I1 enters the bridge at the point A and
Assuming I2
divides equally between the two arms.
Zb
V1
Zb
V2
Za
Fig. 6.24 (a)
Lattic network
A
I1
1
⎛ Z − Za ⎞
I
I
∴ 1 Z a + V2 = 1 Z b ⇒ V2 = I1 ⎜ b
⎟
2
2
⎝ 2 ⎠
∴ z21 =
I2
Za
Zb
Za
⎛ Z − Za ⎞
V2
=⎜ b
I1 I =0 ⎝ 2 ⎟⎠
V1
I2
V2
2
Zb
2
Za
2
Also,
I
V1 = 1 Z a + Z b
2
(
)
1
⎛ Z + Za ⎞
V
⇒ z11 = 1
=⎜ b
I1 I =0 ⎝ 2 ⎟⎠
Fig. 6.24 (b) Equivalent
network
2
As the network is reciprocal and symmetrical,
⎛ Z + Za ⎞
⎛ Z − Za ⎞
∴ z21 = z12 = ⎜ b
and z11 = z22 = ⎜ b
⎟
⎟
⎝ 2 ⎠
⎝ 2 ⎠
(
∴ Z a = z11 − z12
) and Z = ( z + z )
b
11
12
6.7.5 Bridge-T Network
The configuration of a typical bridge-T network is shown in
Fig. 6.25.
In this case, the current in the output depends on the voltages
and a number of nodes instead of one; thus, resulting in number of
simultaneous equation.
Therefore, for such networks, analyzing either on node basis or
on loop basis, the network function is expressed as a quotient of
determinants.
For loop basis, the admittance function, Y =
jk
kj
, where
Z4
I1
1
V2
Z1
Z2
I2
Z3
2
V2
1
Fig. 6.25
2
Bridged T- network
is the loop basis system determinant and
is the cofactor. Here, Y s will be y-parameters if the output is shorted.
′
For node basis, the impedance function, Z jk = kj , where is the loop basis system determinant and
′
is the cofactor. Here, Z s will be z-parameters if the output is open-circuited.
V
I
′
Also, G21 = 2 = 21 21 = 2 = 21
V1
I1
′11
11
kj
kj
351
Two-Port Network
Example 6.8
For the bridge-T RC network, find the transfer admit-
0.5 F
tance Y21.
Solution By KVL,
⎛ 2⎞
2
⎜⎝ 1 + s ⎟⎠ I1 + s I 2 − I 3 = V1
1
⎛ 1 2⎞
2
1
I1 + ⎜ + ⎟ I 2 + I 3 = V2
s
2
⎝ 2 s⎠
1
I3 0.5
1
2
0.5 F
I2
I1
2
Fig. 6.26 Network of
Example 6.8
⎛ 3 2⎞
1
− I1 + I 2 + ⎜ + ⎟ I 3 = 0
2
⎝ 2 s⎠
⎛ 2⎞
⎜⎝ 1 + s ⎟⎠
2
s
−1
2
s
⎛ 1 2⎞
⎜⎝ 2 + s ⎟⎠
1
2
−1
1
2
⎛ 3 2⎞
+
⎝⎜ 2 s ⎟⎠
∴ =
2
1+ 2 s
( )
= −1
12
∴ Y21 =
12
=−
1
2
⎛ 3 2⎞
−1 ⎜ + ⎟
⎝ 2 s⎠
=−
=
s+6
s2
s2 + 6s + 8
2s2
s2 + 6s + 8 s2
s2 + 6s + 8
×
=
−
s+6
2s2
2 s+6
(
)
6.7.6 Parallel-T or Twin-T or Notch-Filter Network
The configuration of typical twin-T network is shown in Fig. 6.27.
In this case also, the current in the output depends on the voltages and
a number of nodes instead of one; thus, resulting in a number of simultaneous equations. We solve the network in the similar process as in the
bridge-T network.
6.8
2F
2
1
V1
1F
2F
2
2
1
V2
1
Fig. 6.27
2
Parallel T-network
SOME SPECIAL TWO-PORT NETWORKS
6.8.1 Gyrator
The gyrator is a two-port network that is designed to transform a load impedance into an input impedance
where the input impedance is proportional to the inverse of the load impedance. It is characterized by a single
resistance, R, known as the gyration resistance.
It can be shown that a gyrator is a non-reciprocal device.
The symbol of a gyrator is shown in Fig. 6.28 (a) and Fig. 6.28 (b). The arrow head indicates the direction
of gyration.
352
Network Analysis and Synthesis
The v–i relationships for the gyrators of Fig. 6.28 (a) and (b) are given below:
For Fig. 6.28 (a),
⎡ v1 ⎤ ⎡ 0 − R ⎤ ⎡ i1 ⎤
⎢ ⎥=⎢
⎥ ⎢ ⎥ or, v1 = − Ri2 and v2 = Ri1
⎢⎣ v2 ⎥⎦ ⎣ R 0 ⎦ ⎢⎣i2 ⎥⎦
For Fig. 6.28 (b),
⎡ v1 ⎤ ⎡ 0 R ⎤ ⎡ i1 ⎤
⎢ ⎥=⎢
⎥ ⎢ ⎥ or, v1 = Ri2 and v2 = − Ri1
⎢⎣ v2 ⎥⎦ ⎣ − R 0 ⎦ ⎢⎣i2 ⎥⎦
i1
R
i2
R
i1
v2
v1
v2
v1
(a)
Fig. 6.28
i2
(b)
Symbol of gyrator
From the v–i relationships, it is clear that for a gyrator, z12
z21 and hence it is non-reciprocal.
Properties of a gyrator
Non-energic (passive) element A gyrator is a non-energic or passive element, i.e., at all times the power
delivered to the two-port is identically zero.
Total instantaneous power entering a gyrator is p(t) v1i1 v2i2
Ri2i1 Ri1i2 0
Hence it is a passive element.
Resistance gyration property When a gyrator is terminated at the output port with a linear resistance RL, as shown
in Fig. 6.29, the input port behaves as a linear resistor with
⎛ R2 ⎞
resistance ⎜ ⎟ . This is explained below.
⎝ RL ⎠
Here, v2
i2RL
R
R ⎛ R2 ⎞
v1 = − Ri2 = v2
= Ri1
=
i
RL
RL ⎜⎝ RL ⎟⎠ 1
i1
v1
R
i2
v2
RL
Fig. 6.29 Gyrator terminated with resistance
input resistance,
v ⎛ R2 ⎞
Ri = 1 = ⎜ ⎟
i1 ⎝ RL ⎠
Since the input resistance is inversely proportional to the load resistance, the gyrator has the property of
resistance inversion or gyration.
353
Two-Port Network
Capacitor-to-inductor mutation property If the output port
of an ideal gyrator is terminated with a capacitor C as shown in
Fig. 6.30, the input port behaves like an inductor. This is explained
below.
dv
Here, i2 = −C 2
dt
i1
i2
R
v2
v1
C
( )
dv2
di
di
d
Fig. 6.30 Gyrator terminated with
= RC
Ri1 = R 2C 1 = Leff 1
capacitor
dt
dt
dt
dt
Thus, at the input port the capacitor behaves as an inductor of
value Le f f R2C. This property is very useful in the design of electronic circuits where it is very difficult to
have inductances of suitable values; the inductor is simulated by using a gyrator terminated with a suitable
capacitor.
∴ v1 = − Ri2 = RC
Inductor-to-capacitor mutation property If the output port
of an ideal gyrator is terminated with an inductor L as shown in
Fig. 6.31, the input port behaves like a capacitor. This is explained
below.
di
Here, v2 = − L 2
dt
v
dv
L di2
L d ⎛ v1 ⎞ L dv1
∴ i1 = 2 = −
=−
−
=
= Ceff 1
R
R dt
R dt ⎜⎝ R ⎟⎠ R 2 dt
dt
R
i1
i2
v2
v1
Fig. 6.31
inductor
L
Gyrator terminated with
Thus, at the input port the inductor behaves as a capacitor of
L
value Ceff = 2 .
R
Current-source-to-voltage source mutation property If
the output port of an ideal gyrator is terminated with a voltage
source as shown in Fig. 6.32, the input port behaves like a current
source. Similarly, connecting a current source across the output
port of a gyrator we get a voltage source at the input port.
v E
Here, i1 = 2 = = I eff .
R R
i1
v1
Fig. 6.32
R
i2
v1
E
Gyrator terminated with voltage
Driving-point-characteristic reflection property If the output source
port of a gyrator is connected across a current-controlled two-terminal resistor, i.e., v2 f ( i2) then the input port becomes a voltagecontrolled resistor as shown in Fig. 6.33. The resulting voltage-controlled resistor is then the dual of the original
current-controlled resistor.
Similarly, if a voltage-controlled resistor is connected at the output port, we get its dual current-controlled
resistor at the input port.
A gyrator is a hypothetical device used for physical systems where the reciprocity condition does not hold
good.
354
Network Analysis and Synthesis
i1
R
i2
i1
i2 f ( i )
2
i1
f (v1)
⬅
v2
v2 = f ( i2) ⬅ V1
v1
v2
Fig. 6.33 A gyrator terminated at the output port with a current controlled resistor behaves like a voltage-controlled resistor.
6.8.2 Negative Impedance Converter (NIC)
A negative impedance converter (NIC) is a two-port device that offers negative impedance, i.e., the impedance
seen at the input port is equal to the negative of the load impedance with some conversion ratio.
It is characterized by the following v –i relationships:
v1 = kv2
i2 = ki1
⎡ v ⎤ ⎡ 0 k ⎤ ⎡ i1 ⎤
or, ⎢ 1 ⎥ = ⎢
⎥⎢ ⎥
⎢⎣ i2 ⎥⎦ ⎣ k 0 ⎦ ⎣⎢ v2 ⎥⎦
(6.33)
⎡v ⎤ ⎡ 0
or, ⎢ 1 ⎥ = ⎢
⎢⎣ i2 ⎥⎦ ⎣ − k
(6.34)
i2
i1
v1
NIC
v2
ZL
or,
v1 = − kv2
i2 = − ki1
− k ⎤ ⎡ i1 ⎤
⎥⎢ ⎥
0 ⎦ ⎢⎣ v2 ⎥⎦
Fig. 6.34 Negative impedance
converter (NIC)
where k is the conversion ratio.
From equations (6.33) and (6.34), it is seen that h12 h21; hence NIC is a non-reciprocal device.
For Eq. (6.33), it is seen that when i1 is in the reference direction, i2 is also in the reference direction and
hence current is said to be inverted. However, the voltage is not inverted. This set of equations characterizes
a current negative impedance converter (CNIC).
For Eq. (6.34), it is seen that the voltage is inverted but the current is not and hence, it characterizes a voltage negative impedance converter (VNIC).
Now, we study the behaviour of an NIC when terminated with a passive element. When it is terminated
di
with an inductor L, we have v2 = − L 2
dt
Putting the value of v2 in the v–i relations of equations (6.33 and (6.34), we get,
⎛ di ⎞
di
di
⎡
⎤
d
v1 = ± kv2 = ± k ⎜ − L 2 ⎟ = ± k ⎢ − Lk
±i1 ⎥ = − k 2 L 1 = Leff 1
dt ⎠
dt
dt
dt
⎝
⎣
⎦
( )
Thus, at the input port, the equivalent inductance is Leff
k2L, i.e., negative of k2L.
Similar conclusions can be obtained when an NIC is terminated with a capacitor or a resistor.
6.9
IMAGE PARAMETERS OF A TWO-PORT NETWORK
We consider a two-port network. Let,
Zi 1 driving point impedance at Port 1 with impedance Zi 2 connected across Port 2, and
Zi 2 driving point impedance at Port 2 with impedance Zi 1 connected across Port 1
Then the impedances Zi 1 and Zi 2 are known as the image impedances of the two-port network.
355
Two-Port Network
From Fig. 6.35 (a), we get the input impedance, Z i1 =
But, V2
I2
I1
AV2 − BI 2
CV2 − DI 2
V1
Zi 1
V2
Zi 2
V2
Zi 2
N
I2Zi 2
AZ i 2 + B
CZ i 2 + D
Similarly, from Fig. 6.35 (b), we get the input impedance,
DZ i1 + B
∴ Zi 2 =
CZ i1 + D
∴ Z i1 =
(a)
(6.35)
I2
I1
Zi 1
V1
N
(6.36)
(b)
AB
From Equations (6.35) and (6.36), we get, Z i1 =
CD
Fig. 6.35 Image parameters of a
two-port network
BD
AC
These two expressions represent the image impedances in terms of the ABCD parameters. However, these
two image impedances do not completely define a two-port network; a third parameter, called image transfer
parameter, is needed. It is obtained as follows.
Zi 2 =
⎡
B ⎤
V1 = AV2 − BI 2 = ⎢ A +
⎥V
Zi 2 ⎦ 2
⎣
From Fig. 6.35 (a),
(6.37)
I1 = CV2 − DI 2 = − ⎡⎣C Z i 2 + D ⎤⎦ I 2
(6.38)
From Eq. (6.37),
V1 ⎛
B⎞ ⎛
AC ⎞ ⎛
ABCD ⎞
A
B
= ⎜ A+
=
+
⎟
⎟ = ⎜ A+
⎜
⎟
V2 ⎝
Zi 2 ⎠ ⎝
BD ⎠ ⎝
D ⎠
(6.39)
From Eq. (6.38),
⎛
I
BD ⎞ ⎛
ABCD ⎞
− 1 = D + CZ i 2 = ⎜ D + C
⎟ =⎜D+
⎟
I2
AC ⎠ ⎝
A ⎠
⎝
(6.40)
)
(
Multiplying equations (6.39) and (6.40),
(
AD + ABCD
V I
− 1× 1=
V2 I 2
AD
V I
− 1 × 1 = AD + BC = AD + AD − 1
V2 I 2
Let,
AD = cosh ,
) = ( AD + BC )
2
( AD
BC
2
1)
AD − 1 = sinh
VI
∴ − 1 1 = cosh + sinh = e
V2 I 2
VI
= loge − 1 1
V2 I 2
where, ␪ is called the image transfer parameter.
(6.41)
356
Network Analysis and Synthesis
Here, the sign of ␪ is ambiguous, because either sign of ␪ will satisfy the equations
AD = ccosh ␪, and
BC = sinh . For the direction of propagation from Port 1 to Port 2, the magnitude of V1I1 exceeds the magnitude V2I2 so that the real part of ␪ is positive.
Zi2I2
Also, V1 Zi1I1 and V2
⎛I ⎞
1 ⎛Z ⎞
= ln ⎜ i1 ⎟ + ln ⎜ 1 ⎟
2 ⎝ Zi 2 ⎠
⎝ I2 ⎠
Hence, the image transfer parameter can be written as,
⎛ BC ⎞
= cosh −1 AD = sinh −1 BC = tanh −1 ⎜
⎟
⎝ AD ⎠
In the other way, may be written as,
(6.42)
(6.43)
6.9.1 Image Parameters in Terms of Short-Circuit and Open-Circuit Impedances
The general equations of ABCD parameters are, V1 AV2 BI2
I1 CV2 DI2
When Port 2 is opened, I2 0
V1 AV2 and I1
V A
Z io = 1 =
I1 C
open-circuit impedance at Port 1 is
When Port 2 is shorted, V2
0
V1
BI2
0
CV2
0
AV2
Similarly,
Also,
Z i1 =
(6.45)
V2 D
=
I2 C
(6.46)
V2 B
=
I2 A
(6.47)
BI2
Z os =
short-circuit impedance at Port 2 is
Now, image impedances are,
DI2
DI2
Z oo =
open-circuit impedance at Port 2 is
When Port 1 is shorted, V1
and I1
(6.44)
V B
Z is = 1 =
I1 D
short-circuit impedance at Port 1 is
When Port 1 is opened, I1
CV2
AB
A
B
=
×
= Z io × Z is
CD
C
D
Z i1 = Z io × Z is
(6.48)
Z i 2 = Z oo × Z os
(6.49)
= tanh −1
In the other way, ␪ can be evaluated as follows.
Z is
Z os
BC
= tanh −1
= tanh −1
AD
Z io
Z oo
tanh =
BC
AD
(6.50)
357
Two-Port Network
BC
⇒
AD
=
e − e− e2 −1
=
e + e− e2 +1
AD + BC ( AD + BC )
= ( AD + BC
=
⇒ e =
2
2
AD − BC
= ln
1
) ⇒ = 12 ln( AD + BC ) { AD BC 1}
2
2
( AD + BC )
Alternatively, tanh =
(6.51)
Z is
Z os
BC
=
=
=k
AD
Z io
Z oo
⇒
(say ) [by equations (6.44), (6.45), (6.46) and (6.47)]
e2 −1
e − e−
1+ k
= k ⇒ e2 =
=
⇒
k
−
2
1− k
e +1
e +e
1 ⎛ 1+ k ⎞
= ln ⎜
2 ⎝ 1 − k ⎟⎠
where,
k=
Z is
Z io
=
Z os
Z oo
(6.52)
Equations (6.48), (6.49), (6.50) and (6.51) are used to find the image parameters Zi1, Zi2 and ␪ from physical
measurements of the open-circuit and short-circuit impedances.
In general, ␪ is a complex quantity. The real part of ␪ is called the image attenuation constant and the
imaginary part of ␪ is called the image phase constant.
Example 6.9 For the two-port network, calculate the z-parameters,
ABCD parameters, open-circuit and short-circuit impedances and image
parameters.
Solution The z-parameters for the T network are,
z11 = 15 ; z12 = z21 = 5 ; z22 = 25
B=
z 15 × 25 − 52
=
= 70
5
z21
Open-circuit impedance at Port 1 is, Zio
z11
V1 5
1
C=
z
1 1
25
= = 0.2 mho D = 22 = = 5
z21 5
z21 5
15
B
Short-circuit impedance at Port 1 is, Z is = =14
D
Open-circuit impedance at Port 2 is, Zoo z22 25
Short-circuit impedance at Port 2 is, Z os =
Image parameters are,
Z i1 =
B
= 23.33
A
AB
3 × 70
=
= 14.49
0.2 × 5
CD
= ln
20
I2
2
V2
2
Fig. 6.36 Network of
Example 6.9
The ABCD parameters are obtained as
z
15
A = 11 = = 3
z21 5
I1 10
1
Zi 2 =
BD
70 × 5
=
= 24.15
3 × 0.2
AC
( AD + BC ) = ln( 15 + 14 ) = 2.03
358
Network Analysis and Synthesis
6.9.2 Symmetrical Networks
A two-port network which is symmetrical with respect to the two ports is termed as a symmetrical network.
For a symmetrical network, the image impedance is referred as the characteristic impedance or iterative
impedance, denoted by Z0. The image transfer parameter of a symmetrical network is termed as the propagation constant, denoted by ␥.
For a symmetrical network, z11 z22; y11 y22; A D; zis zos; zio zoo;
B
C
Z i1 = Z i 2 = Z 0 =
Also,
⎛I ⎞
Z 0 I12
VI
= = ln − 1 1 = ln
= ln ⎜ 1 ⎟
2
V2 I 2
Z0 I 2
⎝ I2 ⎠
And,
In general, ␥ is a complex quantity, expressed as, ␥ ␣ j␤
where, ␣ is known as the attenuation constant, in neper and
␤ is known as the phase constant, in radian
∴ = cosh −1 AD = cosh −1 A = sinh −1 BC
Z0 and ␥ in terms of open-circuit and short-circuit impedances
= tanh −1
Z is
Z io
= tan −1
Z os
or
Z oo
ABCD parameters in terms of Z0 and
Also,
Z 0 = Z is × Z io = Z os × Z oo
⎛ Z + Z ⎞
1 ⎛ Z os + Z oo ⎞
1
is
io
⎟ = ln ⎜
= ln ⎜
⎟
2 ⎜ Z − Z ⎟ 2 ⎜⎝ Z − Z ⎟⎠
⎝ is
io ⎠
os
oo
A = D = cosh ,
B
= Z0
C
∴ B = Z 0 sinh , C =
BC = sinh ␥
sinh
Z0
Example 6.10 For the symmetrical two-port network, calculate the z-parameters and ABCD parameters. Hence or otherwise, find the characteristic impedance
and propagation constant for this network.
Solution The z-parameters for the T network are z11 z22 40 ; z12 z21 10
The ABCD parameters are obtained as,
z
40
z 402 − 102
1
1
=
= 150
C=
A = D = 11 = = 4
B=
= = 0.1 mho
10
z21
z21 10
z21 10
Characteristic impedance is Z 0 =
Propagation constant is
= ln
B
150
=
= 38.73
C
0.1
( AD + BC ) = ln(4 + 15 ) = 2.063
I1 30
30
I2
1
V1 10
2
V2
1
Fig. 6.37 Network
of Example 6.10
2
359
Two-Port Network
Solved Problems
Problem 6.1 Find the z and y parameter for the networks shown in Fig. 6.38.
(a) 1
(c)
(d)
2
Z
2
(b) 1
Za
Zc
2
1
Y
Zb
2
1
2
1
Yc
Ya
Yb
1
2
Fig. 6.38
1
Solution
(a) By KVL,
2
Za
(Z + Z ) I + Z I =V
Z I + (Z + Z ) I =V
a
and
c
c 1
1
b
c 2
1
2
2
c
Za
Zc
I1
I2
1
2
Fig. 6.39
Thus, the z-parameters are
(
z11 = Z a + Z c
) z = z = Z Z = (Z + Z )
12
21
c
22
b
c
(b) By KCL,
1
V −V 1
1
I1 = 1 2 = V1 − V2
Z
Z
Z
V2 − V1
1
1
= − V1 + V2
I2 =
Z
Z
Z
and
Z
2
1
2
Fig. 6.40
Thus, the y-parameters are
y11 =
Since,
1
1
= y22 y12 = y21 = −
Z
Z
y = y11 y22 − y12 y21 = 0 , the z-parameters do not exist for this network.
(c) By KVL,
I +I
⎛ 1⎞
⎛ 1⎞
⎛ 1⎞
⎛ 1⎞
V1 = 1 2 = V2 or, V1 = ⎜ ⎟ I1 + ⎜ ⎟ I 2 and V2 = ⎜ ⎟ I1 + ⎜ ⎟ I 2
Y
⎝Y ⎠
⎝Y ⎠
⎝Y ⎠
⎝Y ⎠
1
z11 = z22 =
Since,
1
= z12 = z21
Y
(
(
)
)
I = Y V + (V −V
V )Y = −V Y + V (Y + Y )
I1 = YaV1 + V1 − V2 Yc = V1 Ya + Yc − V2Yc
2
b 2
2
1
c
1 c
2
b
Y
1
z = z11 z22 − z12 z21 = 0 , the y-parameters do not exist for this network.
(d) By KCL,
2
I2
I1
Thus, the z-parameters are
c
Thus, the y-parameters are
Fig. 6.41
1
2
Ya
1
Fig. 6.42
y11 = Ya + Yc ; y12 = y21 = −Yc ; y22 = Yb + Yc
2
Yc
Yb
2
360
Network Analysis and Synthesis
Problem 6.2 Obtain the z-parameters for the circuit shown in Fig. 6.43.
I1 1
(b)
(a)
I1 2
2
1 I2
2
1
V1
2
2
2
1
V1
V1
2
1
I2
2
1
4
V2
1
2
Fig. 6.43
Solution
(a) The given circuit can be considered as the cascade connection of the following two networks:
2
1
1
1
2
1
1
2
1
2
1
2
1
2
(a)
(b)
Fig. 6.44
From Prob. 6.1 (a),
z11a = z11b = z22 a = z22 b = 3
z12 a = z21a = z12 b = z21b = 2
So, the transmission parameters are
z
3
∴ Aa = Ab = 11 =
z21 2
∴ Ca = Cb =
∴ Ba = Bb =
z 9− 4 5
=
=
2
2
z21
z
1 1
3
= mho ∴ Da = Db = 22 =
z21 2
z21 2
So, the transmission parameters of the resulting network are
⎡3
15 ⎤
5 ⎤⎡ 3
5 ⎤ ⎡7
2⎥
2 ⎥⎢ 2
2⎥=⎢ 2
T = Ta × Tb = ⎢⎢ 2
7 ⎥
3 ⎥⎢ 1
3 ⎥ ⎢3
1
⎣⎢ 2
2 ⎥⎦
2 ⎦⎥ ⎣⎢ 2
2 ⎦⎥ ⎣⎢ 2
So, the z-parameters are
⎫
A 7
z11 = =
⎪
C 3
⎪
T 2 ⎪
z12 =
=
C 3 ⎪
⎬
1 2
⎪
z21 = =
⎪
C 3
⎪
D 7
⎪
z22 = =
C 3
⎭
b) By KVL,
and
V1 = 2 I1 + 4 I 3
(
)
V2 = I1 + I 2 − I 3
2 I1 − I 3 + I1 + I 2 − I 3 − 4 I 3 = 0
361
Two-Port Network
Eliminating I3 from above equations,
I1
26
4
4
6
V1 = I1 + I 2 and V2 = I1 + I 2
7
7
7
7
Thus, the z-parameters are
⎡ 26
4 ⎤
7⎥
⎡⎣ z ⎤⎦ = ⎢⎢ 7
6 ⎥
4
⎢⎣ 7
7 ⎥⎦
I2
2
2
1
2
I3
V1
4
V2
1
1
2
Fig. 6.45
Problem 6.3 Find the open-circuit impedance parameters for the
two-port network shown in Fig. 6.46.
Solution For this ␲-network, the y-parameters are given as
⎛1
1 ⎞ ⎛
100 ⎞
;
= ⎜ 0.2 +
y11 = ⎜ +
⎟
s ⎟⎠
⎝ 5 0.01s ⎠ ⎝
10 mH
1
2
5
10
1
2
Fig. 6.46
1
100
;
=−
0.01s
s
⎛ 1
1 ⎞ ⎛
100 ⎞
= 0.1 +
y22 = ⎜ +
s ⎟⎠
⎝ 10 0.01s ⎟⎠ ⎜⎝
y12 = y21 = −
2
2
2
⎛
100 ⎞ ⎛
100 ⎞ ⎛ 100 ⎞ = 0.02 + 30 + ⎛ 100 ⎞ − ⎛ − 100 ⎞ = ⎛ 0.02 + 30 ⎞
×
+
−
−
∴ y = y11 y22 − y12 y21 = ⎜ 0.2 +
0
.
1
s ⎜⎝ s ⎟⎠ ⎜⎝ s ⎟⎠ ⎜⎝
s ⎟⎠
s ⎟⎠ ⎜⎝
s ⎟⎠ ⎜⎝ s ⎟⎠
⎝
(
)
Thus, the z-parameters are,
0.1 + 100
y
s = 0.1s + 100 = 5s + 5000
z11 = 22 =
0.02 s + 30 s + 1500
y 0.02 + 30
s
−100
y12
s = 100 = 5000
z12 = z21 = −
=−
y
0.02 s + 30 s + 1500
0.02 + 30
s
0.2 + 100
y
s = 0.2 s + 100 = 10 s + 5000
z22 = 11 =
y 0.02 + 30
0.02 s + 30
s + 1500
s
⎫
⎪
⎪
⎪
⎪
⎪⎪
⎬
⎪
⎪
⎪
⎪
⎪
⎪⎭
Problem 6.4 Find the open-circuit impedance parameters of the circuit
given in Fig. 6.47. Also, find the h-parameters of the circuit.
1
Solution By KVL,
I2
2
j15
j 10
( j10 + 5) I + 5 I = V
5 I + ( j15 + 5) I = V
1
and
I1
Thus, the z-parameters are
1
(
2
1
2
2
z11 = 5 + j10
)
5
(i)
(ii)
z12 = z21 = 5
(
Z 22 = 5 + j15
)
1
Fig. 6.47
2
362
Network Analysis and Synthesis
The hybrid parameter matrix may be written as
h12 ⎤ ⎡ I1 ⎤
⎥⎢ ⎥
h22 ⎥⎦ ⎢⎣V2 ⎥⎦
⎡V1 ⎤ ⎡ h11
⎢ ⎥=⎢
⎢⎣ I 2 ⎥⎦ ⎢⎣ h21
From Eq. (ii), we get,
I2 = −
1
1
V2
5
I +
V
=−
I +
1 + j 3 1 5 + j15 2
5 + j15 1 5 + j15
(iii)
Putting this value of I2 in Eq. (i), we get,
⎡
V2
⎤
⎣
5 + j10 × 5 + j15 − 25
⎦
(5 + j10) I + 5 ⎢ − 5 +5j15 I + 5 + j15 ⎥ = V
1
⇒ V1 =
(
1
) (
)
(5 + j15)
I1 +
1
5
30 + j 25
1
V =
I +
V
5 + j15 2
1+ j 3 1 1+ j 3 2
(iv)
Comparing Eq. (iii) and (iv) with the standard equations of h-parameters, we get,
h11 =
30 + j 25
1
1
1
; h12 =
; h =−
; h =
1+ j 3
1 + j 3 21
1 + j 3 22 5 + j15
Problem 6.5 For the network shown in Fig. 6.48, find z and y-parameters.
Solution From Fig. 6.49, we can write the KVL equations,
and
I1
V1 = I 3
(i)
V2 = 2 I 2 − 4 I1 − 2 I 3
(ii)
V1
2 I1 − 2 I 3 + 2 I 2 − 4 I1 − 2 I 3 − I 3 = 0
(
2
⇒ I 3 = I 2 − I1
5
2
1
3I1
V2
Fig. 6.48
)
I1 (I1 I3)
I3
2
2
From (i), V1 = − I1 + I 2 = −0.4 I1 + 0.4 I 2
5
5
V1
4
4
From (ii), V2 = 2 I 2 − 4 I1 − I 2 + I1 = −3.2 I1 + 1.2 I 2
5
5
Fig. 6.49
⎡ − 0.4 0.4 ⎤
∴ ⎡⎣ z ⎤⎦ = ⎢
⎥
⎣ −3.2 1.2 ⎦
I2
2
(I2 3I1)
2
1
) ( ) (
)
z = − 0.4 × 1.2 − 0.4 × −3.2 = 0.8
⎡1.2
− 0.4 ⎤
⎢
0
.
8
0.8 ⎥ mho = ⎡1.5 −0.5 ⎤ mho
∴ ⎡⎣ y ⎤⎦ = ⎢
⎢
⎥
3.2
0.4 ⎥
⎣ 4 −0.5 ⎦
⎢⎣ 0.8 − 0.8 ⎥⎦
V2
2
(I2 2I1 I3)
(
I2
3I1
363
Two-Port Network
Problem 6.6 Find the y parameters for the network shown in Fig. 6.50.
20
Solution This two-port network can be considered as the parallel connection of two two-port networks as shown in Fig. 6.51 (a) & (b).
V1
10
20
5
10
V2
40
V2
V1
(a)
5
40
V1
(b)
V2
Fig. 6.50
Fig. 6.51
For network 6.51 (a), the z-parameters are
)
(
z11a = 50 ; z12 a = z21a = 40 ; z22 a = 45 ; ∴ z = 50 × 45 − 402 = 650
Thus, the y-parameters are
z22 a 45
9
=
=
mho
z 650 130
z
40
4
y12 a = y21a = − 12 = −
=−
mho
650
65
z
y11a =
z
50 1
y22 a = 11a =
=
mho
z 650 13
For the network 6.51 (b), the y-parameters are
1
1
y11b = y22 b =
mho; y12 b = y21b = −
mho
20
20
We know that for parallel connection of two two-port networks, the overall y-parameters are the summation
of individual y-parameters. Thus,
⎛ 9
1⎞
y11 = y11a + y11b = ⎜
+ ⎟ = 0.119 mho
⎝ 130 20 ⎠
)
(
⎛ 4 1⎞
y12 = y21 = y12 a + y12 b = ⎜ − − ⎟ = − 0.111 mho
⎝ 65 20 ⎠
)
(
⎛1 1⎞
y22 = y22 a + y22 b = ⎜ + ⎟ = 0.127 mho
⎝ 13 20 ⎠
)
(
10
Problem 6.7 Obtain the ABCD parameters for the network
Input
shown in Fig. 6.52.
Solution This two-port network can be considered as the casFig. 6.52
cade connection of two two-port networks as shown below.
10
20
50
50
20
Network (a)
Fig. 6.53
10
Network (b)
20
50
50
20
10
Output
364
Network Analysis and Synthesis
For the network (a), as this is a T-network, the z-parameters are given as
(
) (
Ba =
z 1700
=
= 34
50
z21
)
z11 = 60 ; z12 = 50 ; z22 = 70 ; ∴ z = z11 z22 − z12 z21 = 60 × 70 − 502 = 1700
z
60 6
∴ Aa = 11 = =
z21 50 5
Ca =
z
1
1
70 7
=
mho Da = 22 = =
z21 50
z21 50 5
For the network (b), as this is a ␲-network, the y-parameters are given as
⎛ 1 1⎞
⎛ 1 1⎞ 3
7
1
y11 = ⎜ + ⎟ =
mho
mho; y12 = y21 = − mho; y22 = ⎜ + ⎟ =
50
⎝ 50 20 ⎠ 100
⎝ 50 10 ⎠ 25
2
7
3 ⎛ 1⎞
1
× −⎜− ⎟ =
) 100
25 ⎝ 50 ⎠ 125
(
∴ y = y11 y22 − y12 y21 =
3
y22
1
1
= − 25 = 6 Bb = −
=−
= 50
y21
y21
−1
−1
50
50
7
1
y11
2
7
y
125
Cb = −
=−
= mho Db = −
= − 100 =
1
1
−
5
2
y21
y
−
21
50
50
∴ Ab = −
For the entire network, the ABCD parameters are given as
⎡ A B ⎤ ⎡ Aa
⎢
⎥=⎢
⎣C D ⎦ ⎢⎣Ca
Ba ⎤ ⎡ Ab
⎥×⎢
Da ⎥⎦ ⎢⎣Cb
⎡6
Bb ⎤ ⎢ 5
⎥=
Db ⎦⎥ ⎢ 1
⎢⎣ 50
34 ⎤ ⎡ 6
⎥×⎢
7 ⎥ ⎢2
5 ⎥⎦ ⎣ 5
50 ⎤ ⎡ 20.8 179 ⎤
⎥
7 ⎥ = ⎢ 0.68 5.9 ⎥
⎦
2⎦ ⎣
Problem 6.8 Calculate the ABCD parameters of the network shown in Fig. 6.54.
j 20
j 20
2
1
30
2
1
Fig. 6.54
Solution For this T-circuit, the z-parameters are given as
(
z11 = z22 = 30 + j 20
)
z12 = z21 = 30
(
) (
)
2
(
)
(
∴ z = z11 z22 − z12 z21 = 30 + j 20 − 302 = 60 + j 20 j 20 = − 400 + j1200
)
365
Two-Port Network
⎫
⎪
⎪
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎪
⎪
⎭
z
⎛
30 + j 20
2⎞
∴ A = 11 =
= ⎜1+ j ⎟
3⎠
z 60 + j 20 j 20 ⎝
(
)
⎞
z ( 60 + j 20 ) j 20 ⎛ 40
=
∴B=
= − + j 40
⎜⎝
30
z21
∴C =
1
1
=
mho
z12 30
∴D=
z22 30 + j 20 ⎛
2⎞
=
= ⎜1+ j ⎟
30
z12
3⎠
⎝
⎟⎠
3
Problem 6.9 Determine the hybrid parameters for the network in Fig. 6.55.
Solution For this ␲-network, the y-parameters are given as
r1
V1
⎛ 1 1⎞ ⎛ r +r ⎞
⎛ 1 1⎞ ⎛ r +r ⎞
1
y11 = ⎜ + ⎟ = ⎜ 1 2 ⎟ ; y12 = y21 = − ; y22 = ⎜ + ⎟ = ⎜ 2 3 ⎟
r2
⎝ r1 r2 ⎠ ⎝ r1r2 ⎠
⎝ r2 r3 ⎠ ⎝ r2 r3 ⎠
I2
r2
I1
V2
r3
Fig. 6.55
By inter-relationship, the h-parameters are obtained as
h11 =
1 ⎛ r1r2 ⎞
=
y11 ⎜⎝ r1 + r2 ⎟⎠
1
−
y12
r2
r
h12 = −
=−
= 1
y11
⎛ r1 + r2 ⎞ r1 + r2
⎜ rr ⎟
⎝ 12 ⎠
1
−
y21
r2
r
h21 =
=
=− 1
y11 ⎛ r1 + r2 ⎞
r1 + r2
⎜ rr ⎟
⎝ 12 ⎠
)(
(
)
)
Problem 6.10 Find the hybrid parameters of the circuit given in Fig. 6.56.
I1
h22 =
(
)(
)
(
y ⎧⎪ r1 + r2 r2 + r3 − r1r3 ⎫⎪ ⎛ r1r2 ⎞ r1 + r2 r2 + r3 − r1r3
=⎨
⎬×⎜
⎟=
y11 ⎪
r1r2 2 r3
r2 r1 + r2
⎭⎪ ⎝ r1 + r2 ⎠
⎩
Solution For this ␲-network, the y-parameters are given as
⎛ 1 1⎞ 5
⎛ 1 1⎞ 3
1
y11 = ⎜ + ⎟ = ; y12 = y21 = − ; y22 = ⎜ + ⎟ =
2
⎝ 2 3⎠ 6
⎝ 1 2⎠ 2
2
3 5 ⎛ 1⎞
∴ y = y11 y22 − y12 y21 = × − ⎜ − ⎟ = 1
2 6 ⎝ 2⎠
V1
Fig. 6.56
2
1
I2
3
V2
366
Network Analysis and Synthesis
By inter-relationship, the h-parameters are obtained as
h11 =
−1
y
2 =1
h12 = − 12 = −
3
3
y11
2
1 2
=
y11 3
1
y21 − 2
y
3 3
1
=1× =
=
= − h22 =
h21 =
y11
2 2
3
y11 3
2
Problem 6.11 Find the y-parameters for the 2-port networks shown.
(a)
I1 5
V1
0.2V2
I2
20
0.4I1 V2
I1
(b)
I2
12
0.25V2 3 V2
V112
I1
1mho
1mho
V1 1mho
3V1
2mho
(c)
Fig. 6.57
Solution
(a) We consider two cases to find out the y-parameters:
Case (I): Making Port-2 shorted and applying a voltage of V1 at Port- 1
5
I1 5
V1
20
I2
20
0.4I1
V1
V2 = 0
+ I
− 1
−
8I1 +
0.4I1
I2
Fig. 6.58
17 I1 + 20 I 2 = V1
By KVL,
Solving,
I1 =
V1
20
0
20
17 20
12 20
= 0.2V1
and
I
⇒ y11 = 1
= 0.2 mho
V1 V =0
2
17 V1
I2 =
12
12 I1 + 20 I 2 = 0
y21 =
0
17 20
12 20
= − 0.12V1
I2
= − 0.12 mho
V1 V =0
2
Case (II): Making Port-1 shorted and applying a voltage of V2 at Port-2
I1
V1 = 0
Fig. 6.59
0.2V2
20
5
I2
5
0.4I1
V2
I1 0.2 V2
20
8I1
I2
V2
I2
V2
367
Two-Port Network
17 I1 + 20 I 2 = − 0.2V2
By KVL,
Solving,
− 0.2V2
20
V2
20
I1 =
17 20
12 20
= −0.24V2
12
V2
⇒ y12 =
⇒
17 − 0.2V2
I2 =
12 I1 + 20 I 2 = V2
and
= 0.194V2
17 20
12 20
I1
= − 0.24 mho
V2 V =0
1
I
= 0.194 mhho
y22 = 2
V2 V =0
1
⎡ 0.2
− 0.24 ⎤
⎡⎣ y ⎤⎦ = ⎢
⎥ mho
⎣ − 0.12 0.194 ⎦
Thus,
(b) We consider two cases:
Case (I): V1 0
Case (II): V2 0
I1
I1
V1 12
12
0.25V2
I2
12
3
V1= 0
V2
3
V2
0.25V2
Case (I):V1
Fig. 6.60
0
Fig. 6.61
By KCL,
⎫
V2 ⎛ 0 − V2 ⎞
1
+⎜
⇒ y12 = mho ⎪
⎟
4 ⎝ 12 ⎠
6
⎪
⎪
V V
5
⎪
mho
I 2 = y22V2 V =0 = 2 + 2 ⇒ y22 =
⎪
1
3 12
12
⎬
⎛ 1 1⎞
1
⎪
I1 = y11V1 V =0 = ⎜ + ⎟ V1 ⇒ y11 = mho ⎪
2
6
⎝ 12 12 ⎠
⎪
V1
⎪
1
I 2 = y21V1 V =0 = −
⇒ y21 = − mho
⎪⎭
2
12
12
I1 = y12V2 V =0 =
1
(c) For V1
0, the circuit becomes as shown in Fig. 6.62.
(
)
∴ I 2 = y22V2 = 1 + 2 V2 = 3V2 ⇒ y22 = 3 mho
Also,
I1
I2
I
− 1 = V2 ⇒ y12 = −1 mho
1
V1
12
12
I2
V2 = 0
Case (II):V2 = 0
368
Network Analysis and Synthesis
I1
1 mho
1 mho
I2
I1 1mho
I2
I1
V1 = 0
1 mho
2 mho
2 mho
V2
V2
V1
Fig. 6.62
For V2
1mho I3
1mho I2
1mho
3V1
V2 = 0
Fig. 6.63
0, the circuit becomes as shown in Fig. 6.63.
I
∴ − 2 = 3V1
1
I3
+ 3V1 = V1 ⇒ 2V1 = − I 3
1
and
(i)
(ii)
I1 = I 3 + I 4
(iii)
I4
1
(iv)
V1 =
From (i) to (iv), I1 = V1 + I 3 = V1 − 2V1 = −V1 ⇒ y11 = −1 mho
From (i),
y21
3 mho
Thus, the y-parameters are:
⎡ −1 −1⎤
⎡⎣ y ⎤⎦ = ⎢
⎥ mho
⎣ −3 3 ⎦
From the inter-relationship, we get the z-parameters as:
⎡ −1
⎡⎣ z ⎤⎦ = ⎢
⎢ −1
⎣
0 ⎤
⎥
1 ⎥
3⎦
( )
Problem 6.12 Measurements were made on a two-port network shown in Fig. 6.64.
I1
V1
I2
V2
RL = 10
( i) With Port-2 open, a voltage of 100∠0 volt is applied to Port-1,
resulting in I1 10∠0 amp and V2 50∠0 volt.
Fig. 6.64
(ii) With Port-1 open, a voltage of 100∠0 volt is applied to Port-1,
resulting in I2 20∠0 amp and V1 50∠0 volt.
(a) Write the loop equations for the network and also find the driving point and transfer impedance.
(b) What will be the voltage across a 10- resistor connected across Port-2 if a 100∠0 volt source is connected across Port-1.
Solution
(a) From the given data, we get the z-parameters as
V
100∠0°
z11 = 1
=
= 10
I1 I =0 10∠0°
2
369
Two-Port Network
z21 =
V2
25∠0°
=
= 2.5
I1 I =0 10∠0°
2
z12 =
V1
50∠0°
=
= 2.5
I 2 I =0 20∠0°
1
z22 =
V2
100∠0°
=
=5
I 2 I =0 20∠0°
1
So, the loop equations are
(b) Here, V1 = 100∠0°
V1 = 10 I1 + 2.5 I 2 ⎫⎪
⎬
V2 = 2.5 I1 + 5 I 2 ⎭⎪
and V2 = − I 2 RL = −10 I 2
Putting these values in loop equations, 100 = 10 I1 + 2.5 I 2 ⇒ I1 = 10 − 0.25 I 2
and
−10 I 2 = 2.5 I1 + 5 I 2
or,
−110 I 2 = 2.5 10 − 0.25 I 2 + 5 I 2
or,
−15 I 2 = 25 − 0.625 I 2
−25
I2 =
= −1.74 A
14.375
(
or,
)
voltage across the resistor = − I 2 RL = 17.4 V
Problem 6.13 Determine the h-parameter with the following data:
(i) with the output terminals short-circuited, V1 25 V, I1
(ii) with the input terminals open-circuited, V1 10 V, V2
1 A, I2 2 A
50 V, I2 2 A
Solution The h-parameter equations are,
V1 = h11 I1 + h12V2
I 2 = h21 I1 + h22V2
With output short-circuited
V2
0,
given: V1
25 V, I1
1 A and I2
2A
∴ 25 = h11 × 1 ⎫⎪
⎬ ⇒ h11 = 25 , and h21 = 2
2 = h21 × 1⎭⎪
and
With input open-circuited
I1
and
0,
given: V1
10 V, V2
50 V and I2
2A
∴ 10 = h12 × 50 ⎪⎫
1
1
⎬ ⇒ h12 = = 0.2 and h22 = mho = 0.04 mho
5
25
2 = h22 × 50 ⎪⎭
⎡ 25
Thus, the h-parameters are ⎡⎣ h ⎤⎦ = ⎢
⎣ 2
0.2 ⎤
⎥
0.04 −1 ⎦
370
Network Analysis and Synthesis
Problem 6.14 (a) The following equations give the voltages V1 and V2 at the two ports of a two-port
network,
V1
5I1
2I2
V2
2I1
I2
load resistance of 3 is connected across Port-2. Calculate the input impedance.
(b) The z-parameters of a two-port network are z11 5 , z22 2 ⍀, z12 z21 3 . A Load resistance of 4
nected across the output port. Calculate the input impedance.
is con-
Solution
(a) From the given equations,
V1 = 5 I1 + 2 I 2
(i)
V2 = 2 I1 + I 2
(ii)
At the output, V2 = − I 2 RL = −3 I 2
I
Putting this value in (ii), −3 I 2 = 2 I1 + I 2 ⇒ I 2 = − 1
2
⎛−I ⎞
Putting in (i), V1 = 5 I1 + 2 ⎜ 1 ⎟ = 4 I1
2⎠
⎝
V
Input impedance, Z in = 1 = 4
I1
V
(b) [Same as Prob. (a)] Z in = 1 = 3.5
I1
1
Problem 6.15 The y-parameters for a two-port network N are given as,
y11 4 mho, y22 5 mho, y12 y21 4 mho
z z −z z
Solution Output impedance is given as Z out = 11 22 12 21
z11 + Z L
−1
, y12 = y21 = 4
−1
, y22 = 5
−1
y22
5
5
=
=
y 20 − 16 4
y
4
z12 = z21 = − 12 = − = −1
4
y
y
4
z22 = 11 = = 1
y 4
∴ z11 =
and
( ) ( )
5
z11 z22 − z12 z21 4 × 1 − −1 × −1 + 1 × 1 5
Putting these values, Z =
=
=
out
5 +1
z11 + Z L
9
4
V1
1
Fig. 6.65
If a resistor of 1 ohm is connected across Port-1 of N then find out the output impedance.
Here, y11 = 4
I2
1 I1
N
2
V2
2
371
Two-Port Network
Problem 6.16 (a) The h-parameters of a two-port network shown in Fig. 6.66, are h11
20 and h22 1milimho. Find V2/V1.
10
, h12
0.0025, h21
(b) The h-parameters of a two-port network are h11 1 , h12
h21 2, h22 1 mho. The power absorbed by a load resistance of Vi
1 connected across Port-2 is 100 W. The network is excited by a
voltage source of generated voltage Vs and an internal resistance
Fig. 6.66
of 2 . Calculate the value of Vs.
1k
N
V1
V2
RL = 2 k
Solution
(a) The h-parameter equations are
V1 = 100 I1 + 0.0025V2
I 2 = 20 I1 + 0.001V2
(i)
(ii)
By KVL at the output mesh, V2 = −2000 I 2
From (i),
or,
(iii)
⎛ V2 ⎞
⎡ I − 0.001V2 ⎤
V1 = 100 ⎢ 2
⎥ + 0.0025V2 = 5⎜ −
⎟ − 0.005V2 + 0.0025V2
20
⎝ 2000 ⎠
⎣
⎦
V2
= −200
V1
(b) The h-parameter equations are
V1 = I1 + 2V2
I 2 = −2 I1 + V2
Since the load resistance of 1
(i)
(ii)
is connected across Port-2,
V2
∴ 2 = 100 ⇒ V2 = 10 V
1
By KVL, V2 = − I 2 RL = − I 2 ⇒ I 2 = −10 A
2 I1 + V1 = Vs
and
(iii)
From (ii), putting the values of I2 and V2, −10 = −2 I1 + 10 ⇒ I1 = 10 A
From (iii), Vs = 2 × 10 + V1 = 20 + I1 + 2V2
or,
{by (i)}
= 20 + 10 + 2 × 10
Vs = 50 V
Problem 6.17 The z-parameters for a network N are
⎡2 1⎤
⎢
⎥
⎣ 2 5⎦
The terminal connections for the network are shown in Fig. 6.67. Calculate the
voltage ratio V2 / Vs , current ratio –I2 / I1 and input resistance V1 / I1.
1
VS
Fig. 6.67
I1
V1
I2
N
V2
5
372
Network Analysis and Synthesis
Solution The z-parameter equations are
V1 = 2 I1 + I 2
(i)
V2 = 2 I1 + 5 I 2
(ii)
By KVL at the input and output circuits,
I1 + V1 = Vs ⇒
5 I 2 + V2 = 0
and
3 I1 + I 2 = Vs
{by (i)}
(iii)
⇒ 2 I1 + 10 I 2 = 0
{by (ii)}
(iv)
Solving (iii) and (iv),
I1 =
Vs
1
0
10
3 1
2 10
3 Vs
=
2
10
Vs and I 2 =
28
3
0
1
2 10
=−
2
V
28 s
I 1
∴− 2 =
I1 5
Now,
⎛ 20 10 ⎞
10
V2 = 2 I1 + 5 I 2 = ⎜ − ⎟ Vs = Vs
28
⎝ 28 28 ⎠
)
(
∴
V2 5
=
Vs 14
⎛ 20 2 ⎞
18
V1 = 2 I1 + I 2 = ⎜ − ⎟ Vs = Vs
28
⎝ 28 28 ⎠
(
Again,
)
V 9
∴ 1=
I1 14
I2
Problem 6.18 For the two-port network in Fig. 6.68, terminated in a
1-ohm resistance, show that
I1
V2
z
= 21
I1 1+ z 22
and
V1 z 11 + z
=
I1
1+ z 22
V1
N
1
V2
Fig. 6.68
Solution The z-parameter equations are
V1 = z11 I1 + z12 I 2
(i)
V2 = z21 I1 + z22 I 2
(ii)
By KVL at the output, V2 = − I 2 × 1 ⇒ I 2 = −V2
From (ii),
(
( )
V2 = z21 I1 + z22 I 2 = z21 I1 + z22 −V2
)
or,
V2 1 + z22 = z21 I1
or,
V2
z
= 21
I1 1 + z22
(iii)
(Proved)
373
Two-Port Network
From (i),
)
(
⎡V 1 + z22 ⎤
V1 = z11 ⎢ 2
by ( iii )
⎥ + z12 −V2
z21
⎢⎣
⎥⎦
⎡z + z⎤
⎡z +z z −z z ⎤
= V2 ⎢ 11 11 22 12 21 ⎥ = V2 ⎢ 11
⎥
z
⎣ z21 ⎦
⎦
⎣
21
( ) {
}
V V V z + z
z
z + z
(Proved)
∴ 1 = 1 × 2 = 11
× 21 = 11
1 + z22 1 + z22
I1 V2 I1
z21
Problem 6.19 Calculate the T-parameters for the blocks A and B separately and then using these results,
calculate the T-parameters of the whole circuit shown in Fig. 6.69 Prove any formula used.
(a)
A
B
2
1F
2
1F
1
1
4
2
(b)
5
3
A
B
Fig. 6.69
Solution
(a) We consider the given network as a cascade connection of two networks as shown in Fig. 6.69.
For Block A
Opening Port-2,
⎛ 1 1⎞
1
⎛
⎞
By KCL, ⎜ + ⎟ V1 − V2 = I1 and − 1V1 + 1 + s V2 = 0
⎜
3
⎝ 2 3⎠
3
⎝ 3 ⎟⎠
2 I 1 + 3s
2 I1
Solving for V1 and V2, V1 = 1
and V2 =
1 + 5s
1 + 5s
I1
3
(
(
)
)
(
)
⎫
V1
= 1 + 3s ⎪
V2 I =0
⎪⎪
2
⎬
1 + 5s ⎪
I1
=
Ca =
2 ⎪
V2 I =0
⎪⎭
2
∴ Aa =
and
(
)
(
)
Short-circuiting Port-2,
V V 5
∴ I1 = 1 + 1 = V1
2 3 6
and
V
⇒ Ba = − 1
=3
I 2 V =0
V1 = −3 I 2
2
and
I
5V 3 5
Da = − 1
= 1× =
I 2 V =0 6 V1 2
2
V1
I2
1F
2
V2
Block A
Fig. 6.70
I1
V1
3
2
Fig. 6.71
I2
V2 = 0
374
Network Analysis and Synthesis
I1
For Block B
Opening Port-2,
⎛1 ⎞
1
By KCL, ⎜ + s⎟ V1 − V2 = I1
5
⎝5 ⎠
⎛ 1 1⎞
1
− V1 + ⎜ + ⎟ V2 = 0
5
⎝ 5 4⎠
and
I2
5
1F
V1
V2
4
Block B
Solving for V1 and V2,
V1 =
9 I1
(1 + 9s )
and V2 =
Fig. 6.72
4 I1
(1 + 9s )
⎫
⎪
⎪⎪
2
⎬
1 + 9s ⎪
I1
=
Cb =
4 ⎪
V2 I =0
⎪⎭
2
∴ Ab =
V1
9
=
V2 I =0 4
)
(
and
I1
5
V1
1F
I2
V2 = 0
Fig. 6.73
Short-circuiting Port-2,
⎛1 ⎞
∴ I1 = ⎜ + s⎟ V1
⎝5 ⎠
V
=5
and V1 = −5 I 2 ⇒ Bb = − 1
I 2 V =0
2
)
I
Db = − 1
= 5s + 1
I 2 V =0
(
and
2
Since the two networks are connected in cascade, the overall transmission parameter matrix is obtained as
(
)
⎡ 3s + 1
⎢
⎡⎣T ⎤⎦ = ⎡⎣Ta ⎤⎦ × ⎡⎣Tb ⎤⎦ = ⎢⎛ 5s + 1⎞
⎢⎜⎝ 2 ⎟⎠
⎣
3 ⎤ ⎡ 9
4
⎥ ⎢
× ⎢⎛ 1 + 9s ⎞
⎥
5
⎢
2 ⎥⎦ ⎢⎜⎝ 4 ⎟⎠
⎣
⎤
⎥ ⎡ 13.5s + 3
⎥=⎢
5s + 1 ⎥ ⎢⎣ 11.25s + 1.75
⎥⎦
5
(
(
)
(
)
( 30s + 8)⎤⎥
) ( 25s + 5)⎥⎦
(b) [Same as Prob. (a)]
⎡3
1⎤
1 ⎤
⎡3 2 ⎤
⎥ and ⎡⎣Tb ⎤⎦ = ⎢ 2 ⎥ ∴ ⎡⎣T ⎤⎦ = ⎡⎣Ta ⎤⎦ × ⎡⎣Tb ⎤⎦ = ⎢
⎥
3 ⎥
⎢3
⎥
⎣3 2 ⎦
2⎦
⎢⎣ 2 1⎦⎥
Problem 6.20 Two identical sections of the network shown in Fig. 6.74 are connected in parallel. Obtain
the y-parameters of the resulting network and verify by direct calculation.
⎡ 1
Here, ⎡⎣Ta ⎤⎦ = ⎢
⎢1
⎣ 2
Solution For the circuit of Fig. 6.74, y11 = 3
The y-parameters for the combination will be
(
−1
, y12 = y21 = −2
)
−1
⎫
⎪⎪
y12 = y21 = y12′ + y12′′ = −4 −1 ⎬
⎪
y22 = y22
′ + y22
′′ = 6 −1
⎪⎭
y11 = y11′ + y11′′ = 6
(
(
)
−1
and y22 = 3
−1
2
1
)
Fig. 6.74
1
375
Two-Port Network
To find the y-parameters by direct calculation, we consider the resulting network as shown in Fig. 6.75.
I1
I2
2
V1 1
1
2
1
I1
4
I2
V1
2
2
V2
V2
1
Fig. 6.75
For the entire network, y11 = 4 + 2 = 6
−1
; y12 = y21 = −4
−1
−1
; y22 = 4 + 2 = 6
Problem 6.21 Two networks have general ABCD parameters as shown:
Parameter
A
B
C
D
Network-1
1.50
11
0.25 siemens
2.5
Network-2
5/3
4
1 siemens
3.0
If the two networks are connected with their inputs and outputs in parallel, obtain the admittance matrix of the
resulting network.
Solution For the network-1
D 2.5 5 −1
=
=
B 11 22
AD − BC
1.5 × 2.5 − 11 × 0.25
1
y12 = −
=−
=−
11
11
B
1
1 −1
y21 = − = −
B
11
A 1.5 3 −1
y22 = =
=
B 11 22
y11 =
−1
For the network-2
D 3 −1
=
B 4
AD − BC
1
y12 = −
=−
B
4
y11 =
1
1 −1
=−
4
B
5
5
A
=
y22 = =
B 3 × 4 12
So, the admittance matrix of the resulting network is
−1
y21 = −
⎡ 5
−1 ⎤ ⎡ 3
11⎥ + ⎢ 4
⎡⎣ y ⎤⎦ = ⎢ 22
⎢ 1
⎥ ⎢ 1
3
−
⎢⎣ − 11
22 ⎥⎦ ⎢⎣ 4
−1
−15 ⎤
− 1 ⎤ ⎡ 43
44 ⎥
4 ⎥ = ⎢ 44
⎥
73
5 ⎥ ⎢ −15
44
132 ⎥⎦
12 ⎥⎦ ⎢⎣
−1
376
Network Analysis and Synthesis
Problem 6.22 Two identical sections of Fig. 6.76 are connected in series. Obtain
the z-parameters of the resulting network and verify by direct calculation.
1
Solution The z-parameters of each section:
z11 = 3 , z12 = z21 = 1 , z22 = 3
So, the z-parameters of the combined series network are
)
(
(
2
2
Fig. 6.76
)
)
(
z11 = 3 + 3 = 6 , z12 = z21 = 1 + 1 = 2 , z22 = 3 + 3 = 6
To find the z-parameters by direct calculation, we consider the resulting network as shown in Fig. 6.77.
I1
2
I2
2
I1 2
1
2
I2
1
V1
2
V1
V2
2
V2
1
1
2
2
Fig. 6.77
For the resulting network,
V
z11 = 1
=6
I1 I = 0
z21 =
V
=6
z22 = 2
I 2 I =0
V
z12 = 1
=2
I 2 I =0
2
1
V2
=2
I1 I = 0
2
1
⎫
⎪
⎪⎪
⎬
⎪
⎪
⎭
Problem 6.23 (a) Find out the z- and h-parameters for the circuit shown in Fig. 6.78 (a).
1
(b) Hence, obtain the hybrid parameters for the two-port network of
1
Fig. 6.78 (b).
1
1
2
1
V
V
z11 = 1
= 4 , z12 = z21 = 2 , z11 = 2
=4
I1 I = 0
I 2 I =0
⎫
z 16 − 4
∴ h11 =
=
=3 ⎪
4
z12
⎪
⎪
z12 2
h12 =
= = 0.5
⎪
z22 4
⎪
⎬
z21
2
= − = − 0.5 ⎪
h21 = −
⎪
z22
4
⎪
1 1
⎪
h22 =
= = 0.25 −1 ⎪
z12 4
⎭
2
2
Solution
(a) For Fig. 6.78 (a), the z-parameters are
2
1
1
(a)
1
1
1
2
2
1
1
1
1
2
1
(b)
Fig. 6.78
1
2
1
377
Two-Port Network
(b) The connection is series–parallel connection. For this connection, the overall h-parameters will be the
sum of individual h-parameters.
( )
h = ( 0.5 + 0.5) = 1
h = ( −0.5 − 0.5) = −1
h = ( 0.25 + 0.25) = 0.5
∴ h11 = 3 + 3 = 6
12
21
22
⎫
⎪
⎪⎪
⎬
⎪
−1 ⎪
⎪⎭
Problem 6.24 (a) Find the equivalent -network for the T-network shown in Fig. 6.79 (a).
1
(b) Find the equivalent T -network for the -network shown in Fig. 6.79 (b).
Solution
(a) Let the equivalent -network have YC as the series admittance and YA
and YB as the shunt admittances at Port-1 and Port-2, respectively.
Now, the z-parameters are given as
(
)
(
)
z11 = Z A + ZC = 7 , z12 = z21 = ZC = 5 , z22 = Z B + ZC = 7.5
5
)
(
∴ z = 7 × 7.5 − 5 × 5 = 27.5
2
Zc = 5
2
1
(a)
Y3 = 1 mho
1
2
Y2 = 0.5 mho
Y1 = 0.2 mho
2
z22 7.5
=
mho
z 27.5
z
5
o
y12 = y21 = − C = −
mho
z
27.5
z
7
y22 = 11 =
mho
z 27.5
∴ y11 =
(
Zb = 2.5
Za = 2
) 272.5.5 = 111 mho
2
1
(b)
Fig. 6.79
I1
V2
I2
YC
YA
V2
YB
∴YA = y11 + y12 =
(
)
2
mho
27.5
5
2
and
YC = − y21 =
= mho
27.5 11
Thus, the impedances of the equivalent -networks are
Fig. 6.80
∴YB = y22 + y12 =
⎫
⎪
⎪
⎪⎪
1
Z B = = 13.73 , ⎬
YB
⎪
⎪
1
ZC = = 5.5
⎪
YC
⎪⎭
I1
1
Z A = = 11 ,
YA
V1
Fig. 6.81
network
5.5
11
13.75
Equivalent ␲
I2
V1
378
Network Analysis and Synthesis
2
2
1
Y2 0.5 mho
Y1 0.2 mho
Zb 0.25
Za 0.625
Y3 1 mho
1
2
1
Zc 1.25
1
Fig (a) -network
2
Fig (b) Equivalent T-network
Fig. 6.82
(b) The y-parameters,
y11 = 1.2 mho, y12 = y21 = −1 mho, and y22 = 1.5 mho
(
)
y22 1.5
=
y 0.8
, z12 = z21 = −
∴ y = 1.2 × 1.5 − 1 = 0.8
∴ z11 =
y12
1
=
y 0.8
(
) 00..85 = 0.625 ⎫⎪
(
)
ZC = z12 =
1
= 1.25
0.8
∴ Z A = z11 − z12 =
Z B = z22 − z12 =
0.2
= 0.25
0.8
, z22 =
y11 1.2
=
y 0.8
⎪
⎪
⎬
⎪
⎪
⎪
⎭
Problem 6.25 The z-parameter of a 2-port network are z11
20 , z12 z21 5 .
10
, z22
1
ZA 5
ZB 15
2
Find the ABCD-parameters. Also find the equivalent T-network.
ZC 5
Solution From the inter-relationship, we get the ABCD parameters as
z
10
A = 11 = = 2
z21 5
Fig. 6.83
network
z Z − Z12 Z 21 10 × 20 − 5 × 5
= 35
B = 11 22
=
5
z21
C=
1 1
= = 0.2 mho
z21 5
D=
z22 20
= =4
z21 5
To find the equivalent T-network, we have the relations,
(
)
⎫
⎪⎪
z12 = z21 = ZC = 5
⎬ ⇒ ZA = 5 ,
⎪
z22 = Z B + ZC = 20 ⎪⎭
z11 = Z A + ZC = 10
and
(
)
2
1
Z B = 15 ,
ZC = 5
Equivalent T-
379
Two-Port Network
Problem 6.26 The z-parameters of the two-port network N in Fig. 6.84 are z11 4s, z12 z21 3s, z22 9s.
(a) Replace N by its T-equivalent.
(b) Use part a) to find the input current I1 for
Vs cos1000t.
I2
I1
V1
N
ZA
V2
VS
2
ZC
12k
2
1
6k
Solution
ZB
1
Fig. 6.85 Equivalent
T-network
Fig. 6.84
⎡ 4 s 3s ⎤
(a) The z-parameters are ⎡⎣ z ⎤⎦ = ⎢
⎥
⎣ 3s 9 s ⎦
Since the network is reciprocal, its T-equivalent exists. Its elements are
( )
)
(
)
(
Z A = z11 − z12 = s, Z B = z22 − z21 = 6 s, and ZC = z21 = z12 = 3s
So, the equivalent circuit is shown in Fig.6.85.
(b) We repeatedly combine the series and parallel elements of Fig. 6.86, with resistors in k and s in krad/s
to find the input impedance, Zin in k .
I1
I2
6 s + 12 3s + 6
Vs
s
Zin
∴ Z in = = s +
= 3s + 4 k
3s 6s
I1
6 s + 12 + 3s + 6
12k
or
(
)
(
(
)(
) (
)
) (
)
VS
Z in ( j ) = 3 j + 4 = 5∠36.9° k
So, the current,
6k
i1 (t ) =
vs ( t ) 1
= cos 1000t − 36.9° ( mA )
Z in ( j ) 5
)
(
Fig. 6.86
Problem 6.27 For the bridge-T RC network, find the y-parameters and its equivalent -network.
Solution The given network is the parallel combination of the two networks:
⎡ s
−s ⎤
2 ⎥ mho
For the network (a), the y-parameters are ⎡⎣ ya ⎤⎦ = ⎢⎢ 2
s
s ⎥
⎢⎣ − 2
2 ⎥⎦
(
⎡ 1+ 2
⎢
s
For the network (b), the z-parameters are ⎡⎣ zb ⎤⎦ = ⎢
2
⎢
s
⎣
∴ y11b =
)
(
(
(
)
(
)
=
)
4
s+6
0.5
2
0.5 F
2
1
⎤
⎥
s
⎥
1 +2 ⎥
s ⎦
2
Fig. 6.87
)
1 2F
2
2
z
s
∴ y12 b = y21b = − 12 b =
zb
s+6
1
1
2
( 12 + 2 s ) = s + 4
( s )( 12 + 2 s ) − 4 s s + 6
z22 b
=
zb 1 + 2
0.5 F
1
2
1
2
Fig. 6.88 (a) Network (a)
1
12
2
1
2s
s+2
z11b
2 = 2 s+2
∴ y22 b =
=
zb
s+6
s+6
2s
(
)
1 2F
1
Fig. 6.88 (b) Network (b)
2
380
Network Analysis and Synthesis
⎡s+4
⎢s+6
For the network (b), the y-parameters are ⎡⎣ yb ⎤⎦ = ⎢
⎢ 4
⎢
⎣s+6
4 ⎤
s + 6 ⎥⎥
2 s+2 ⎥
⎥
s+6 ⎦
)
(
I2
I1
1
Yc
Ya
V1
Yb
2
V2
1
2
Fig. 6.89 Equivalent π
Thus, the overall y-parameters are
network
⎡ s
⎡⎣ y ⎤⎦ = ⎡⎣ ya ⎤⎦ + ⎡⎣ yb ⎤⎦ = ⎢ 2
⎢ s
⎢⎣ − 2
s2 + 6s + 8 ⎤
⎥
2 s+6 ⎥
⎥
s 2 + 10 s + 8 ⎥
2 s + 6 ⎥⎦
2
4 ⎤ ⎡⎢ s + 8 s + 8
s + 6 ⎥⎥ ⎢ 2 s + 6
=
2 s + 2 ⎥ ⎢ s2 + 6s + 8
⎥ ⎢−
s + 6 ⎦ ⎢⎣ 2 s + 6
⎡s+4
− s ⎤ ⎢s+6
2 ⎥+ ⎢
s ⎥ ⎢ 4
2 ⎥⎦ ⎢
⎣s+6
)
(
(
)
(
)
−
(
)
(
)
Equivalent π network can be found out from the following relations:
s
s2 + 6s + 8
2s
Ya = y11 + y12 =
; Yb = y22 + y12 =
; Yc = − y12 = − y21 =
2 s+6
s+6
s+6
) (
(
) (
(
)
)
)
(
2F
Problem 6.28 For the notch-filter network, determine the y-parameters.
Solution The given network is the parallel combination of the two networks:
For the network (a),
1+ 2s
1+ 2s
z11a = 1 + 1 =
; z12 a = z21a = 1; z22 a = 1 + 1 =
2s
2
s
2s
2s
1+ 4s
∴ za =
4s2
(
)
(
)
(
)
(
)
(
(
)
2
)
(
)
)
(
(
)
(
Thus, the overall y-parameters are,
and
) + (1 + 2 s ) = (1 + 2 s )(8s + 12 s + 1)
1+ 4s
4 + 4s
4 ( s + 1)( 4 s + 1)
(
(
)
2s 1+ 2s
(
)
4s2
1
16 s 3 + 16 s 2 + 4 s + 1
−
=−
1+ 4s 4 s +1
4 4ss + 1 s + 1
y11 = y22 = y11a + y11b =
y12 = y21 = y12 a + y12 b = −
2
(
2
Fig. 6.91 (a) Network (a)
2
)
)
1+ 2s
1+ 2s
z
z
z
1
; y12 b = y21b = − 12 b = −
∴ y11b = 22 b =
; y22 b = 11b =
zb 4 s + 1
zb 4 s + 1
zb
4 s +1
(
2
1
)
s
)
(
)(
)
2F
1
4 s +1
b
2F
) 1
1+ 2s
1
1+ 2s
For network (b), z11b = 1 + 2 =
; z12 b = z21b = ; z22 b = 1 + 2 =
s
s
s
s
s
(
∴ z =
1
Fig. 6.90
)
(
2
V2
1
2s 1+ 2s
2s 1+ 2s
z
z
z
4s2
; y12 a = y21a = − 12 a = −
∴ y11a = 22 a =
; y22 a = 11a =
za
za
za
1+ 4s
1+ 4s
1+ 4s
(
2
V1 1 F
)
(
2
1
2F
2
1
2
1F
1
Fig. 6.91 (b) Network (b)
2
381
Two-Port Network
Problem 6.29 A network has two input terminals a, b and two output terminals c, d. The input impedance with c–d open-circuited is (250
j100) ohms and with c–d short-circuited is (400
j300) ohms.
The impedance across c–d with a–b open-circuited is 200 ohms. Determine the equivalent T-network
parameters.
Solution We consider Fig. 6.21. For c–d terminals opened,
(
ZA +
Again, with a–b terminals opened,
( Z + Z ) = 200
B
(i)
)
(ii)
A
Z B ZC
= 400 + j 300
Z B + ZC
But, for c–d terminals shorted,
( Z + Z ) = ( 250 + j100)
(iii)
C
From (ii) and (i), we get
Z B ZC
− Z B = 150 + j 200
Z B + ZC
or,
Z B ZC − Z B 2 − Z B ZC = 200 150 + j 200
(
(
)
Z B 2 = 200( −150 − j 200) = 104 1 − j 2
or,
B
{by (iii)}
)
2
(
) ⎫⎪
⎪
∴ Z = (150 + j 300 ) ⎬
⎪
Z = (100 + j 200 ) ⎪⎭
∴ Z B = 100 − j 200
A
and
C
3
Problem 6.30 The z-parameters of a two-port network
N are given by
z11
(2s
1/s), z12
z21
2s, z22
(2s
4).
I1
V1
VS 12cost
I2
N
1
V2
1H
(a) Find the T-equivalent of N.
Fig. 6.92
(b) The network N is connected to a source and a load as
shown in Fig.6.92. Replace N by its T-equivalent and then find I1, I2, V1, and V2.
Solution
(a) To find the equivalent T-network, we have the relations,
⎛
1⎞
z11 = Z A + ZC = ⎜ 2 s + ⎟
s⎠
⎝
(
)
z12 = z21 = ZC = 2 s
and
(
) (
z22 = Z B + ZC = 2 s + 4
⇒ ZA =
)
⎫
⎪
⎪
⎪
⎬
⎪
⎪
⎪⎭
1
, Z B = 4 , ZC = 2 s
s
zA 1/s
1
zB 4
2
zC 2s
1
Fig. 6.93 Equivalent
T-network
2
382
Network Analysis and Synthesis
(b) The equivalent circuit is shown in Fig. 6.94.
)
( )
I ( j 2 ) + I ( 5 + j 3) = 0
(
By KVL, I 3 + j + I j 2 = 12 ∠0°
1
2
1
Solving,
I1 =
2
12 ∠0°
0
(
(3+ j )
j2
5 + j3
)=
j2
12 ∠0°
0
2 ∠90°
5.831∠30.96°
= 3.29∠ − 10.22° A
16 + j14
( )
(5 + j 3)
j2
1/s
3
( 3 + j ) 12∠0°
I2 =
and
j2
1
= 1.13∠ − 131.19° A
j2
(5 + j 3)
j2
V1
Vs 12 0
( )
0
(3+ j )
4
2s
V2
s
Fig. 6.94
( )
∴V1 = 12 ∠0° − I1 × 3 = 12 − 3.29 × 3∠ − 10.22° = 2.28 + j1.75 = 2.88∠37.504° V
and
)
(
)
(
V2 = − I 2 1 + j = −1.13 1 + j ∠ − 131.186° = 1.59∠93.81°
So, the currents and voltages are
(
) ( A ) ⎫⎪
i (t ) = 1.13cos( t − 131.2° ) ( A ) ⎪⎪
⎬
v (t ) = 2.88 cos( t + 37.5° ) ( A ) ⎪
⎪
v (t ) = 1.6 cos( t + 93.8° ) ( A ) ⎪⎭
i1 (t ) = 3.29 cos t − 10.2°
2
1
2
Problem 6.31 For the network shown in Fig. 6.95, determine
the z and y parameters.
2V3
I 1 10
2
I2
Solution By KVL for the three meshes, we get
(i)
) ⇒ 12 I + 5 I = V
3I 2
V
V = 2 ( I − 2V ) + 2 ( I + I ) ⇒ 2 I + 4 I − 4V = V
(ii)
V = 2( I + I )
(iii) Fig. 6.95
From (ii) and (iii),
V = 2 I + 4 I − 4 ( 2 I + 2 I ) ⇒ V = −6 I − 4 I
(
V1 = 10 I1 + 3 I 2 + 2 I1 + I 2
1
2
1
1
2
3
1
3
1
2
1
2
3
V3
V2
2
2
2
From (i) and (iv), we get,
1
2
1
2
⎡12 5 ⎤
z=⎢
⎥
⎣ −6 −4 ⎦
⎡ 2
−1
∴ y = ⎡⎣ z ⎤⎦ = ⎢⎢ 9
1
⎢⎣ − 3
2
( )( )
⎤
18 ⎥
⎥
−2 ⎥
3⎦
5
( )
2
2
1
2
(iv)
383
Two-Port Network
Problem 6.32 The h-parameters of a two-port network
shown in Fig. 6.96 are h11 1000 , h12 0.003, h21 100,
and h22 50 10 6 mho. Find V2 and z-parameters of the network if Vs 10 2 0 ( V).
Solution The h-parameter equations are
I2
I1
500
VS
V2
2000
Fig. 6.96
V1 = h11 I1 + h12V2 = 1000 I1 + 0.003V2
(i)
I 2 = h21 I1 + h22V2 = 100 I1 + 50 × 10−6 V2
(ii)
V1 = Vs − 500 I1
(iii)
V2 = −200 I 2
(iv)
By KVL for the two meshes,
Vs − 500 I1 = 1000 I1 + 0.003V2
From (i) and (iii),
or,
10−2 − 1500 I1 = 0.003V2
−
From (ii) and (iv),
(v)
V2
= 100 I1 + 50 × 10− 6 V2
2000
I1 = −5.5 × 10− 6 V2
or,
(vi)
(
0.003V2 = 10−2 + 1500 −5.5 × 10−6 V2
From (v) and (i),
)
⇒ V2 = −1.905 V
The z-parameters are calculated as follows.
z11 =
h
h22
= −500
h
z12 = 12 = 60
h22
h
z21 = − 21 = −2 × 106
h22
z22 =
1
= 20 × 103
h22
Problem 6.33 For the two-port network shown in Fig. 6.97, find the z-parameters.
Solution We consider two cases:
When I2 0
I1
1
1
V1
2
2V1
I1 1
I2
1
V1 (I 1 I )
V1
2
Fig. 6.97
1
2V1
I
2
2
I2 0
V2
Fig. 6.98 (a)
Here, as the output port is open-circuited, no current will flow through the 1The modified circuit is shown in Fig. 6.98 (a).
resistor connected at Port 2.
⎛2
2 ⎞
By KVL for the middle mesh, we get I + 2V1 + 2 I − 2 × I1 − I = 0 ⇒ I = ⎜ I1 − V1 ⎟
5 ⎠
⎝5
(i)
⎛2
2 ⎞
By KVL for the left mesh, we get V1 = I1 + 2 × I1 − I = 3 I1 − 2 I = 3 I1 − 2 × ⎜ I1 − V1 ⎟
5 ⎠
⎝5
{by equation (i)}
(
(
)
)
384
Network Analysis and Synthesis
V1 = 11I1
or,
V
∴ z11 = 1
= 11
I1 I = 0
2
Also, by KVL for the right mesh, we get
⎛2
2 ⎞ 4
4
4
4
V2 = 2 I = 2 × ⎜ I1 − V1 ⎟ = I1 − V1 = I1 − × 11 × I1 = −8 I1
5 ⎠ 5
5
5
5
⎝5
∴ z21 =
2
When I1 0
Here, as the output port is open-circuited, no current will flow through the 1The modified circuit is shown in Fig. 6.98 (b).
I1 0
2V1
1
V1
2
I
1
(I 2 I)
2
V2
= −8
I1 I = 0
resistor connected at Port 1.
I2
V2
Fig. 6.98(b)
⎛2
2 ⎞
By KVL for the middle mesh, we get I − 2V1 + 2 I − 2 × I 2 − I = 0 ⇒ I = ⎜ I 2 + V1 ⎟
5 ⎠
⎝5
(
)
⎛2
2 ⎞ 4
4
By KVL for the left mesh, we get V1 = 2 I = 2 × ⎜ I 2 + V1 ⎟ = I 2 + V1 ⇒ V1 = 4 I 2
5 ⎠ 5
5
⎝5
(ii)
∴ z12 =
V1
=4
I2 I =
1
Also, by KVL for the right mesh, we get
⎛2
2 ⎞
V2 = I 2 + 2 × I 2 − I = 3 I 2 − 2 I = 3 I 2 − 2 × ⎜ I 2 + V1 ⎟
5 ⎠
⎝5
(
=
∴ z22 =
)
{by equation (ii)}
11
4
11
4
I − V = I − × 4 I2 = − I2
5 2 5 1 5 2 5
V2
= −1
I 2 I =0
1
Therefore, the z-parameters of the network are
⎡ 11 4 ⎤
⎡⎣ z ⎤⎦ = ⎢
⎥
⎣ −8 −1⎦
( )
0.9I 1
Problem 6.34 Find the z and y parameters of the network shown
I1
1
10
I2
in Fig. 6.99.
Solution We convert the dependent current source into its equivalent
voltage source as shown in fig. 6.100 below.
By KVL for the two meshes, we get
(
)
I1 + 1 × I1 + I 2 = V1 ⇒ V1 = 2 I1 + I 2
V1
1
V2
Fig. 6.99
(i)
385
Two-Port Network
and,
)
(
10 I 2 + 9 I1 + 1 × I1 + I 2 = V2 ⇒ V2 = 10 I1 + 11I 2
9I1
I1 1
From (i) and (ii), we get the z-parameters as
⎡2 1⎤
⎡⎣ z ⎤⎦ = ⎢
⎥
⎣10 11⎦
V1
Therefore, the y-parameters are
Fig. 6.100
( )
10
I2
V2
1
⎡ 11
−1
−1 ⎤
−1
⎡2 1⎤
12 ⎥
⎢ 12
⎡⎣ y ⎤⎦ = ⎡⎣ z ⎤⎦ = ⎢
=
⎥
⎢ 10
⎥
10
11
2
⎣
⎦
⎢⎣ − 12
12 ⎥⎦
Problem 6.35 The network shown in Fig. 6.101 contains both
dependent current source and a dependent voltage source. For
this circuit, determine the y and z parameters.
2V1
I1
1
V1 1
2V2
Solution We first find out the y parameters. To find the y paramFig. 6.101
eters, we consider two situations:
I1
When V1 0
A
Here, Port 1 is shorted and hence, the dependent voltage source is zero, i. e.,
short-circuited. The 1- resistance in Port 1 becomes redundant. The circuit V1 0 2V2
is shown in Fig. 6.102 (a).
By KCL at the node (A), we get
Fig. 6.102 (a)
I2
V2
2
1
I2
[I 2 V2 2]
2
⎛
V ⎞
3V
− I1 − 2V2 − ⎜ I 2 − 2 ⎟ = 0 ⇒ I1 + I 2 = − 2
2⎠
2
⎝
V2
(i)
By KVL for the outer loop, we get
⎛
V ⎞
V
V2 = 1 × ⎜ I 2 − 2 ⎟ = I 2 − 2
2
2
⎝
⎠
∴ y22 =
⇒
3
V =I
2 2 2
I2
3
=
V2 V =0 2
1
Substituting the value of I2 in (i), we get
3
3
I1 + V2 = − V2 ⇒ I1 = −3V2
2
2
∴ y12 =
I1
= −3
V2 V =0
1
When V2 0
Here, Port 2 is shorted and hence, the dependent current source is zero, i. e.,
open-circuited. The 2- resistance in Port 2 becomes redundant. The circuit
is shown in Fig. 6.102 (b).
By KVL for the left loop, we get V1 = I1 + I 2
(
)
I1
V1 1
1
(I 1 I2 )
2V1
I2
V2 0
Fig. 6.102 (b)
(ii)
386
Network Analysis and Synthesis
By KVL for the outer loop, we get
2V1 + I 2 + V1 = 0 ⇒ I 2 = −3V1
∴ y21 =
I2
= −3
V1 V =0
2
From (ii),
V1 = I1 − 3V1 ⇒ I1 = 4V1
I
∴ y11 = 1
=4
V1 V =0
2
Therefore, the y parameters of the network is given as
⎡ 4 −3⎤
⎥
⎡⎣ y ⎤⎦ = ⎢
⎢ −3 3 ⎥
⎢⎣
2 ⎥⎦
Hence, the z parameters are given as
−1
⎡ −1
⎤
⎡ 4 −3⎤
⎢ 2 −1 ⎥
⎥ =⎢
⎡⎣ z ⎤⎦ = ⎡⎣ y ⎤⎦ = ⎢
⎥
⎢ −3 3 ⎥
4⎥
⎢
⎢⎣
2 ⎥⎦
⎢⎣ −1 − 3 ⎥⎦
−1
Problem 6.36 The model of a transistor in CE mode is shown
in Fig. 6.103. Determine the h parameters of the model.
( )
I1
rb
I
V1
V1
=
V2 I =0
re
V2
V2
bc
To find h parameters, we consider two cases:
Fig. 6.103
When I1 0
Here, the dependent current source is open-circuited. The modified circuit is
shown in Fig. 6.104 (a).
V ⇒ h12 =
bc 2
I2
cb 1
Solution The equations of h parameters are
V1 = h11 I1 + h12V2 and I 2 = h21 I1 + h22V2
∴ V1 =
re
rd
I1 = 0 rb
re
V1
mbcV2
I2
re
V2
rd
bc
1
Fig. 6.104 (a)
Also,
(
V2 = I 2 re + rd
)
I
1
⇒ h22 = 2
=
V2 I =0 re + rd
I1
1
When V2 0
Here, the dependent voltage source is short-circuited.
The modified circuit is shown in Fig. 6.104 (b).
(
∴V1 = I1 rb + re
) ⇒ h =I
V1
11
1 V =0
2
(
= rb + re
)
rb
re
I2
␣cbI1
V1
Fig. 6.104 (b)
V2
V2 = 0
387
Two-Port Network
Also,
I2 =
I
cb 1
⇒ h21 =
I2
=
I1 V = 0
cb
2
Therefore, the h parameters for the transistor model is given as
)
(
⎡ rb + re
⎢
⎡⎣ h ⎤⎦ = ⎢
⎢ cb
⎣
⎤
⎥
1 ⎥
re + rd ⎥⎦
bc
Problem 6.37 Find the hybrid parameters for the network of
Fig. 6.105 (which represents a transistor).
I 1 R1
Solution Case (1): When V2 0
The circuit is modified as shown in Fig. 6.106.
By KCL at the node x,
Vx Vx
+ + I1 = I1 ⇒ Vx = 1 −
R2 R3
(
I1
i2
R2
V1
R2 R3
)R +R I
2
R3
V2
Fig. 6.105
1
3
␣I 1
By KVL,
(
V1 = I1 R1 + Vx = I1 R1 + 1 −
(
)
⎛ R2 R3 ⎞
)⎜⎝ R + R ⎟⎠ I
2
I1
R1
I2
x
y
1
3
R3
⎡
1 − R2 R3 ⎤
V
∴ h11 = 1
= ⎢ R1 +
⎥
I1 V =0 ⎢⎣
R2 + R3 ⎥⎦
2
R2
V1
V2 0
Fig. 6.106
By KCL at the node y,
0 − Vx
= I 2 + I1 ⇒ I 2 = − I1 − 1 −
R3
(
∴ h21 =
⎛ R2 + R3 ⎞
⎛ R2 R3 ⎞
)⎜⎝ R + R ⎟⎠ I = − I ⎜⎝ R + R ⎟⎠
1
2
3
1
2
3
⎛ R + R3 ⎞
I2
= −⎜ 2
⎟
I1 V = 0
⎝ R2 + R3 ⎠
2
Case (2): When I1 0
Here, the dependent current source is to be opened (since I1
The circuit is modified as shown in Fig. 6.107.
(
∴V2 = I 2 R2 + R3
∴ h12 =
0).
) and V = I R
1
2
2
V1
R2
I
1
=
and h22 = 2
=
V2 I =0 R2 + R3
V2 I =0 R2 + R3
1
I2
I1 0
R1
V1
Fig. 6.107
1
Therefore, the hybrid parameters are
(
)
⎡
1 − R2 R3 ⎤
⎛ R + R3 ⎞
R2
1
h11 = ⎢ R1 +
; h21 = − ⎜ 2
; h22 =
⎥ ; h12 =
⎟
R2 + R3
R2 + R3 ⎥⎦
R2 + R3
⎝ R2 + R3 ⎠
⎢⎣
R3
R2
V2
388
Network Analysis and Synthesis
Problem 6.38 Determine the y and z parameters for the network
shown in Fig. 6.108.
1
Solution We convert the dependent current source into equivalent
dependent voltage source. The modified network is shown in Fig. 6.109.
I1
2V1
1
1
I3
V1
V1 1
I2
2
2V1
2
2V2
V2
Fig. 6.108
V2
2V2
Fig. 6.109
By KVL for three meshes, we get
(
)
V1 = 1 × I1 − I 3 + 2V2 ⇒ I 3 = I1 + 2V2 − V1
)
(
)
(
(i)
and
1 × I 3 − 2V1 + 2 I 2 + I 3 − 2V2 + 1 × I 3 − I1 = 0 ⇒ 2V1 + 2V2 = − I1 + 2 I 2 + 4 I 3
and,
V2 = 2 × I 2 + I 3
(
Substituting the value of I3 from (i) into (ii) and (iii), we get
(
2V1 + 2V2 = − I1 + 2 I 2 + 4 I1 + 2V2 − V1
(
V2 = 2 I 2 + I1 + 2V2 − V1
and,
By (iv)
)
(iii)
) ⇒ 6V − 6V = 3I + 2 I
1
2
) ⇒ 2V − 3V = 2 I + 2 I
1
2
1
1
(iv)
2
(v)
2
I1 = 4V1 − 3V2
(v), we get
Also, from (v) and (vi), we get
(ii)
(
(vi)
)
3
2V1 − 3V2 = 2 4V1 − 3V2 + 2 I 2 ⇒ I 2 = −3V1 + V2
2
(vii)
From (vi) and (vii), we get
⎡ 4 −3⎤
⎥ mho
y=⎢
⎢ −3 3 ⎥
⎢⎣
2 ⎥⎦
(
)
−1
⎡ 1
⎡ 4 −3⎤
⎢−
⎥ =⎢ 2
∴ z = ⎡⎣ y ⎤⎦ = ⎢
⎢ −3 3 ⎥
⎢
⎢⎣
2 ⎥⎦
⎢⎣ −1
−1
Note
⎤
−1 ⎥
⎥
4⎥
− ⎥
3⎦
( )
See Problem 6.35 and compare the two methods of solution.
Solution To find h parameters, we consider two cases:
When I1 0
Here, no current will flow through the 3- resistance.
0.5 V1
I1 3
Problem 6.39 Find the h-parameters for the two-port network shown
in Fig. 6.110.
I2
4
V1
3I2
Fig. 6.110
1
V2
389
Two-Port Network
)
(
By KVL at the left mesh, we get V1 = 4 × 0.5V1 + 3 I 2 = 2V1 + 3 I 2
⇒ V1 = −3 I 2
I2
X
4
V1
Also, by KCL at the node (X), we get
I2 =
0.5V1
I1 0 3
V2
+ 0.5V1 = V2 + 0.5V1 = V2 + 0.5 × −3 I 2
1
(
∴ h22 =
) ⇒ 2.5 I = V
2
2
1
3I2
V2
Fig. 6.111 (a)
I2
1
=
= 0.4
V2 I =0 2.5
I1
0.5V1
3
I2
1
⎛V ⎞
∴V1 = −3 I 2 = −3 × ⎜ 2 ⎟ = −1.2V2
⎝ 2.5 ⎠
∴ h12 =
V1
= −1.2
V2 I =0
4
V2 = 0
V1
3I2
Fig. 6.111 (b)
1
When V2 0
Here, Port 2 is short circuited. The 1Fig. 6.111 (b).
resistance becomes redundant. The modified circuit is shown in
∴ I 2 = 0.5V1 = 0.5 × ⎡⎣ 3 I1 + 4 I1 + 4 I 2 + 3 I 2 ⎤⎦ = 3.5 I1 + 3.5 I 2 ⇒ 2.5 I 2 = −3.5 I1
∴ h21 =
I2
3.5
=−
= −1.4
2.5
I1 V = 0
2
Also,
(
)
V1 = 3 I1 + 4 I1 + 4 I 2 + 3 I 2 = 7 I1 + 7 I 2 = 7 I1 + 7 × −1.4 I1 = −2.8 I1
V
∴ h11 = 1
= −2.8
I1 V = 0
2
⎡ −2.8 −1.2 ⎤
Therefore, the h parameters of the network are given as ⎡⎣ h ⎤⎦ = ⎢
⎥
⎣ −1.4 0.4 ⎦
Problem 6.40 Find the driving point impedance at the terminals 1 1’ of the ladder network shown in
Fig. 6.112.
(b)
(a)
1H
1H
1H
1H
1H
1
1
1
2
1
1F
1
Fig. 6.112
2
1
1F
1F
2
1
1F
1F
2
390
Network Analysis and Synthesis
Solution (a) The driving point impedance at 1 1’ is
)
(
Z11 = s + 1 +
1
s+
1
( s + 1) +
s 4 + 3s 2 + 1
s 2 + 2s
=
1
s+
1
( s + 1) + 1s
(b) The driving point impedance at 1 1’ is
(
)
Z11 = s + 1 +
1
s+
=
1
( s + 1) +
s 6 + 3s 5 + 8 s 4 + 11s 3 + 11s 2 + 6 s + 1
s 5 + 2 s 4 + 5s 3 + 4 s 2 + 3s
1
s+
1
( s + 1) + 1s
Solution Writing two mesh equations,
)
⎡1
⇒ ⎢
⎢⎣ −1
∴ I2 =
1
−1
∴ Y21 =
−1
0
−1
(2s + 2)
=
V2
(
(ii)
⎤ ⎡ I1 ⎤ ⎡ sV ⎤
⎥⎢ ⎥ = ⎢ 1 ⎥
2 s + 2 ⎥⎦ ⎢⎣ I 2 ⎥⎦ ⎣ 0 ⎦
−1
2
)
sV1
2s2 +1
2
I2
s
=
V1 2 s 2 + 1
sV1
0
V
1
∴V2 = I 2 = 21
and I1 =
s
2s +1
1
−1
∴ Z 21 =
1F
(i)
)
(
sV1
V1 1F
Fig. 6.113
1
I1 − I 2 = V1 ⇒ I1 − I 2 = sV1
s
⎛
1
2⎞
− I1 + ⎜ 2 s + ⎟ I 2 = 0 ⇒ − I1 + 2 s 2 + 2 I 2 = 0
s
s⎠
⎝
1
2H
I2
Problem 6.41 Determine the network functions Y21 and Z21 for the network shown.
(
I1
−1
( 2 s + 2 ) = 2( s + 1) sV
2
2
−1
2s2 +1
1
(2s + 2)
2
V2
V
1
2s2 +1
= 21 ×
=
2
I1 2 s + 1 2 s + 1 sV1 2 s s 2 + 1
(
)
(
)
391
Two-Port Network
Problem 6.42 Determine driving point impedance Z11, transfer impedance Z21
and voltage transfer function G21 for the network shown.
1H
2
(
)
1
( s + 1) +
2
1
s 4 + 3s 2 + 1
= 2
s + 2s
1
s+
1F
1F
Solution The driving point impedance at 1 1’ is
Z11 = s + 1 +
1H
1
Fig. 6.114
1
s+
1
( s + 1) + 1s
To find the transfer impedance, Z21, we start from the right end,
∴ I 2 = V2 × s
(
)
1
∴V ′ = I 2 × s + V2 = s 2 + 1 V2
(
) (
V1
)
3
(
) (
1F
1H
V 1F
I2
2
V2
2
Fig. 6.115
V
1
∴ Z 21 = 2 = 3
I1 s + 2 s
Also,
1H
1
∴ I1 = I 2 + V ′s = V2 s + s + s V2 = s + 2 s V2
3
I1
) (
)
V1 = I1 × s + V ′ = s 4 + 2 s 2 V2 + s 2 + 1 V2 = s 4 + 3s 2 + 1 V2
∴ G21 =
V2
1
= 4
V1 s + 3s 2 + 1
1
Problem 6.43 Determine the current-transfer ratio
point impedance Z21 for the circuit shown.
21
and driving
I1
R1 1
C1 1F
2
C2 2F V2
Fig. 6.116
Solution By KCL at the the node 1,
V1 − V2
+ sV1 = I1 or, V1 1 + s − V2 = I1
1
(i)
V2 − V1
+ 2 sV2 = 0 or, − V1 + V2 1 + 2 s = 0
1
(ii)
)
(
By KCL at the node 2,
)
(
Solving for V2,
(1 + s ) I
V2 =
−1
(1 + s )
−1
∴ Z 21 =
1
0
=
−1
I1
2 s + 3s
2
(1 + 2 s )
V2
1
0.5
= 2
=
I1 2 s + 3s s s + 1.5
(
)
392
Network Analysis and Synthesis
␣21 =
I 2 V2 × 2 s 0.5 × 2 s
1
=
=
=
I1
I1
s s + 1.5 s + 1.5
)
(
R
R
Problem 6.44 For the notch-filter (Twin-T) network, determine
C
(a) y-parameters,
(b) the voltage ratio transfer function V2 / V1 when no-load impedance is present, and
(c) the value of the frequency at which the output voltage is zero.
V1
Solution
(a) The given network is the parallel combination of the two networks:
C
C
1
2
R
2
R 2
2C
1
2
Fig. 6.118
V2
R 2
Fig. 6.117
R
1
2C
C
1
Network (a)
2
Fig. 6.118
Network (b)
For the network (a),
( Cs + R 2 ) = 2 +2CsRCs ; z = z = R 2 ; z = ( 1Cs + R 2 ) = 2 +2CsRCs
z11a = 1
∴ za =
12 a
21a
22 a
1 + RCs
C 2 s2
)
)
(
(
RCs 2 + RCs
z
z
z
R 2C 2 s 2
; y12 a = y21a = − 12 a = −
; y22 a = 11a =
∴ y11a = 22 a =
za
za 2 R 1 + RCs
za
2 R 1 + RCs
For the network (b), z
11b
∴ y11b =
12 b
21b
22 b
)
(
RCs RCs + 1
C 2 s2
(
)
(
)
1 + 2 RCs
1 + 2 RCs
z
z
z22 b
1
; y12 b = y21b = − 12 b = −
=
; y22 b = 11b =
zb 2 R RCs + 1
zb 2 R RCs + 1
zb
2 R RCs + 1
(
Thus, the overall y-parameters are
)
)
(
(
(
(1 + 2 RCs ( R C s + 4 RCs + 1)
) 2 R(1 + RCs )) + 2 R( RCs + 1)) = 2 R( RCs + 1)
(
)
y11 = y22 = y11a + y11b =
and
(1 + RCs )
( 2Cs + R) = 1+22CsRCs ; z = z = 21Cs ; z = ( 1 s + 2) = 1+22CsRCs
= 1
∴ zb =
)
(
⎛ 1 ⎞
Cs ⎜ 1 + Cs⎟
⎝ 2 ⎠
RCs 2 + RCs
y12 = y21 = y12 a + y12 b = −
2
2 2
R 2C 2 s 2
1
R 2C 2 s 2 + 1
−
=−
2 R 1 + RCs 2 R RCs + 1
2 R RCs + 1
)
(
(
)
(b) Now,
I1 = y11V1 + y12V2
and
I 2 = y21V1 + y22V2
(
)
(
)
393
Two-Port Network
When no-load impedance is present, I2
0,
)
(
2 R RCs + 1
V
y
R 2C 2 s 2 + 1
R 2C 2 s 2 + 1
∴ 2 = − 21 =
× 2 2 2
= 2 2 2
V1
y22 2 R RCs + 1
R C s + 4 RCs + 1
R C s + 4 RCs + 1
) (
(
(c)
) (
)
For V2 = 0, ⇒ 1 + R 2C 2 s 2 = 0
putting s = j , 1 −
2
R 2C 2 = 0
∴ =
Thus, the notch frequency is given by, f N =
1
RC
1
2 RC
Problem 6.45 For the given bridged T network, find the driving
point admittance Y11 and the transfer admittance Y21 with a 2- load
resistor connected across Port 2.
Solution By KVL,
1F
1
1
V 1 I1
⎛ 2⎞
2
⎜⎝ 1 + s ⎟⎠ I1 + s I 2 − I 3 = V1
1
Fig. 6.119
⎛ 1 2⎞
2
1
I1 + ⎜ + ⎟ I 2 + I 3 = V2
s
2
⎝ 2 s⎠
⎛ 3 2⎞
1
− I1 + I 2 + ⎜ + ⎟ I 3 = 0
2
⎝ 2 s⎠
∴ =
11
⎛ 1⎞
⎜⎝ 1 + s ⎟⎠
1
s
−1
1
s
⎛ 1⎞
⎜⎝ 1 + s ⎟⎠
1
−1
1
⎛
1⎞
⎜⎝ 2 + s ⎟⎠
⎛ 1⎞
⎜ 1 + s ⎟⎠
1+1 ⎝
1
1
⎛
1⎞
⎜⎝ 2 + s ⎟⎠
( )
= −1
1
1+ 2 s
( )
= −1
12
1
⎛
1⎞
1 ⎜2+ ⎟
s⎠
⎝
=−
=
=
s+2
s2
s 2 + 3s + 1
s2
s2 + 2s +1
s2
1
2
2
2 F I2
2
I3
2
394
Network Analysis and Synthesis
∴Y11 =
11
=−
1F
s 2 + 3s + 1 s 2
s 2 + 3s + 1
×
=
2
s+2
s+2
s
A
1H
1F
s2 + 2s +1 s2
s2 + 2s +1
∴ Y21 = 21 = −
×
=−
2
s+2
s+2
s
1F
1H
1H
1
V1
V2
1F
Problem 6.46 Determine the voltage transfer function of the
symmetrical lattice network shown in Fig. 6.120.
1
s×
s= s
Solution Let Z1 Series Arm Impedance =
1 s2 +1
s+
s
1 s2 +1
Z2 Shunt Arm Impedance = s + =
s
s
Rearranging the figure, we have Fig. 6.121 as shown.
Applying KVL to the mesh 1ABDC1’, we get
D
C
1H
Fig. 6.120
V1 ( s ) = I ′( s ) Z1 ( s ) + I 2 ( s ) × 1 + Z1 ( s ) ⎡⎣ I1 ( s ) − I ′( s ) + I 2 ( s ) ⎤⎦
V1 ( s ) = I 2 ( s ) ⎡⎣ Z1 ( s ) + 1⎤⎦ + Z1 ( s ) I1 ( s )
or,
B
1
I1
A
Z1
V1 B
I1 I
I
V2
I2 2 1
2
Z2
1
Z2
D
Z1
C
Fig. 6.121 Equivalent
network
(i)
Applying KVL to the mesh 1ADBC1’, we get
V1 ( s ) = Z 2 ( s ) ⎡⎣ I1 ( s ) − I ′( s ) ⎤⎦ − I 2 ( s ) × 1 + Z 2 ( s ) ⎡⎣ I ′( s ) − I 2 ( s ) ⎤⎦
or,
V1 ( s ) = Z 2 ( s ) I1 ( s ) − ⎡⎣ Z 2 ( s ) + 1⎤⎦ I 2 ( s )
(ii)
Multiplying (i) by Z2(s) and (ii) by Z1(s) and subtracting (ii) from (i),
V1 ( s ) ⎡⎣ Z 2 ( s ) − Z1 ( s ) ⎤⎦ = I 2 ( s ) ⎡⎣ Z1 ( s ) + Z 2 ( s ) + 2 Z1 ( s ) Z 2 ( s ) ⎤⎦
I2 (s)
Z 2 ( s ) − Z1 ( s )
=
V1 ( s ) Z1 ( s ) + Z 2 ( s ) + 2 Z1 ( s ) Z 2 ( s )
or,
Now the output voltage V2 ( s ) = I 2 ( s ) × 1
s2 +1
s
− 2
V2 ( s )
Z 2 ( s ) − Z1 ( s )
s
s +1
=
=
V1 ( s ) Z1 ( s ) + Z 2 ( s ) + 2 Z1 ( s ) Z 2 ( s )
s
s2 +1
s2 +1
s
+
+2×
× 2
2
s
s
s +1
s +1
( s + 1) − s
( s + 1+ s )( s + 1− s )
=
=
s + ( s + 1) + 2 s ( s + 1)
( s +1+ s )
V ( s ) ( s − s + 1)
⇒
=
V ( s ) ( s + s + 1)
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
Note
This symmetrical lattice network is used as all-pass network because it has the property for sinusoidal inputs
that everything which comes in goes out without any change in magnitude but distortion in phase.
395
Two-Port Network
Problem 6.47
Z02and R0
(a) If ZaZb
Z0, show that the voltage transfer
Za
1
function of the network is given by V2 =
V1 1+ Za / R 0
(b) Under the condition ZaZb
network of Fig.6.122 is R0.
2
0
R and R0
R0
R0
Fig. 6.122
A
and
B=
R
2 R0 + Z a
B
V1
Fig. 6.123
∴V1 = I1 Z in = I1 Z 0
V2 = I 2 Z0 = I1
So, the voltage transfer function is
=
R0 Z a
+ Z0
R0 + Z a
2
R0
+ Zb
2 R0 + Z a
=
(
)
2Z ( Z + Z )
2 R0 Z 0 + Z a
0
b
B + Zb
Z
A + B + Z0 + Zb 0
B + Zb
V2
1
= I1
=
Z0 ×
V1
A + B + Z0 + Zb
I1 Z 0
R0 Z a + 2 R0 Z 0 + Z a Z 0
R0 + 2 R0 Z b + Z a Z b
2
=
1
A + Z0
1+
B + Zb
R0 Z a + 2 R0 Z 0 + Z a R0
Z a Zb + 2 Z0 Zb + Z a Zb
[putting the co
onditions, R0 = Z 0 , Z a Z b = Z 0 2 ]
a
Z
Z
Z
= 0 = 20 = a
Zb Z0
Z0
Za
V
∴ 2=
V1
1
1
=
A + Z0
Z
1+
1+ a
B + Zb
Z0
(b) So, the input impedance,
⎧⎪
A + Z0 B + Zb
A A + B + Z0 + Zb + A + Z0 B + Zb
=
= A+
⎨ Z in
A + Z0 + B + Zb
A + B + Z0 + Zb
⎩⎪
(
=
(
)(
)
)
(
(
)
Z 0 A + B + Z 0 + Z b + 2 A B + Z b + A2 − Z 0 2
A + B + Z0 + Zb
Z0
Zb
(a) Let I1 Input current,
I2 current through Z0.
and
A
2
RZ
A= 0 a
R0 + Z a
A + Z0
=
B + Zb
2
Z0 , show that the input impedance of the
Solution Replacing the R0, R0, and Za delta by equivalent star, we have
Now,
Z0 V
Zb
V1
) (
= Z0 +
)(
(
)
)
2 A B + Z b + A2 − Z 0 2
A + B + Z0 + Zb
V2
396
Network Analysis and Synthesis
= Z0 +
⎞
2 R0 3 Z a
2 R0 Z a Z b
2 R0 Z a ⎛ R0 2
R0 2 Z a 2
R0 2 Z a 2
2
Z
Z
+ Zb ⎟ +
−
=
+
+
− Z0 2
+
⎜
0
0
2
2
2 R0 + Z a ⎝ 2 R0 + Z a
2 R0 + Z a 2 R + Z 2
⎠ 2R + Z
2R + Z
(
3
= Z0 +
2
2 R0 Z a
R0 Z a
+
(2R + Z ) (2R + Z )
2
0
0
a
0
a
)
(
2
+
a
0
(
a
)
(
)
a
0
)
R Z 2 R0 + Z a 2 R0 Z a Z b
2 R0 Z a Z b
− Z0 2 = Z0 + 0 a
+
− Z0 2
2
R
+
Z
2 R0 + Z a
2
2R + Z
a
0
2
2
(
0
a
)
⎫⎪
R0 Z a
2R Z Z
+ 0 a b − Z0 2 ⎬
2 R0 + Z a 2 R0 + Z a
⎭⎪
2
= Z0 +
Z in = Z 0 +
= Z0 +
2 Z0 Z0 2
Z0 2 Z a
+
− Z 0 2 [ putting t he conditions, R0 = Z 0 , Z a Z b = Z 0 2 ]
2 Z0 + Z a 2 Z0 + Z a
(
Z0 2 2 Z0 + Z a
2 Z0 + Z a
)−Z
2
0
= Z0 + Z0 2 − Z0 2
⇒ Z in = Z 0
Problem 6.48 For the given two-port network, calculate the
z-parameters and the image parameters.
1
Solution The z-parameters for the T network are,
V1
z11 = 30 ; z12 = z21 = 10 ; z22 = 40
The ABCD parameters are obtained as
1
Fig. 6.124
z
30
A = 11 = = 3
z21 10
B=
z
z21
=
30 × 40 − 102
= 110
10
C=
1
1
= = 0.1 mho
z21 10
D=
z22 40
= =4
z21 10
Image parameters are
Z i1 =
AB
3 × 110
=
= 28.72
CD
0.1 × 4
Zi 2 =
BD
110 × 4
=
= 38.3
AC
3 × 0.1
= ln
I1 20
( AD + BC ) = ln( 12 + 11) = 1.914
30
10
I2
2
V2
2
397
Two-Port Network
Problem 6.49 A two-port network has
(i) at Port 1, driving point impedances of 60 and 55 with Port 2 open circuited and short-circuited respectively.
(ii) at Port 2, driving point impedances of 80 and 73.33 with Port 1 open circuited and short-circuited respectively.
Find the image parameters of the network.
Solution It is given that Z io = 60 ; Z is = 55
Z oo = 80 ; Z os = 73.33
Hence, image parameters are given as
Z i1 = Z io × Z is = 60 × 55 = 57.45
Z i 2 = Z io × Z is = 80 × 73.33 = 76.59
1 ⎛ 1+ k ⎞
= ln ⎜
2 ⎝ 1 − k ⎟⎠
where,
k=
Z is
Z io
=
Z os
= 0.957
Z oo
Za
1 ⎛ 1 + 0.957 ⎞
∴ = ln ⎜
= 1.194
2 ⎝ 1 − 0.957 ⎟⎠
Short-circuit input impedance, Z is =
(
)
)
)
Z a + 2 Zb
Z a Zb
Z a + Zb
characteristic impedance is
Z 0 = Z is × Z io = Z b
Za
Zb
Fig. 6.125 (b)
(b) For the symmetrical ␲ network shown in Fig.6. 125 (b),
(
2
1
1
Z Z + 2 Zb
Z a Zb
= a a
Z a + Zb
Z a + Zb
Zb Z a + Zb
2
Zb
characteristic impedance is Z 0 = Z is × Z io = Z a Z a + 2 Z b
Open-circuit input impedance, Z io =
Za
Fig. 6.125 (a)
Solution
(a) For the symmetrical T network shown in Fig.6.125 (a),
Open-circuit input impedance, Z io = Z a + Z b
(
Zb
1
Problem 6.50 Find the characteristic impedance for (a) the symmetrical
T network, and (b) the symmetrical network.
Short-circuit input impedance, Z is = Z a +
2
1
Za
Z a + 2 Zb
2
398
Network Analysis and Synthesis
Summary
1. A network with two input terminals and two output
terminals is called a four-terminal network or a twoport network.
2. There are six types of parameters in a two-port
network, as open-circuit impedance parameters
( z-parameters), short-circuit admittance parameters ( y-parameters), transmission or chain parameters (T-parameters or ABCD-parameters), inverse
transmission parameters (T ’-parameters), hybrid
parameters (h-parameters), and inverse hybrid
parameters (g-parameters).
3. The equations of different parameters are as given
below.
⎡V1 ⎤ ⎡ z 11
⎢ ⎥=⎢
⎢⎣V 2 ⎥⎦ ⎢⎣ z 21
z 12 ⎤ ⎡ I 1 ⎤
⎥⎢ ⎥
z 22 ⎥⎦ ⎢⎣ I 2 ⎥⎦
⎡V 2 ⎤ ⎡ A ′ B ′ ⎤ ⎡ V1 ⎤
⎢ ⎥=⎢
⎥⎢ ⎥
⎢⎣ I 2 ⎥⎦ ⎣C ′ D ′ ⎦ ⎢⎣ − I 1 ⎥⎦
⎡ I 1 ⎤ ⎡ y 11
⎢ ⎥=⎢
⎢⎣ I 2 ⎥⎦ ⎢⎣ y 21
y 12 ⎤ ⎡V1 ⎤
⎥⎢ ⎥
y 22 ⎥⎦ ⎢⎣V 2 ⎥⎦
⎡V1 ⎤ ⎡ h11 h12 ⎤ ⎡ I 1 ⎤
⎢ ⎥=⎢
⎥⎢ ⎥
⎢⎣ I 2 ⎥⎦ ⎢⎣ h21 h22 ⎥⎦ ⎢⎣V 2 ⎦⎥
⎡V1 ⎤ ⎡ A
⎢ ⎥=⎢
⎢⎣ I 1 ⎥⎦ ⎣C
B ⎤ ⎡ V2 ⎤
⎥⎢ ⎥
D ⎦ ⎢⎣ − I 2 ⎥⎦
⎡ I 1 ⎤ ⎡ g11
⎢ ⎥=⎢
⎢⎣V 2 ⎥⎦ ⎢⎣ g 21
g12 ⎤ ⎡V1 ⎤
⎥⎢ ⎥
g 22 ⎥⎦ ⎢⎣ I 2 ⎥⎦
4. A two-port network is said to be reciprocal if, z12 z21,
or y12 y21, or (AD BC ) 1, or (A’D’ B’C ’) 1, or
h21, or g12
g21.
h12
5. A two-port network is said to be symmetrical if, z11 z22,
or y11 y22, A D, or A’ D’, or (h11h22 h12h21) 1, or
(g11g22 g12g21) 1.
6. The inter-relationships between six types of parameters are given in Table 6.2.
7. Two-port networks can be connected in series, parallel, or in cascade. For different types of interconnection, the parameter calculation will be as given below.
For series connection z-parameters are added
For parallel connection y-parameters are added
For cascade connection transmission parameter matrices are multiplied
For series–parallel connection h-parameters are added
For parallel–series connection g-parameters are added
8. Some typical two-port networks are ladder network,
lattice network, etc.
9. A gyrator is a non-reciprocal two-port network that is
designed to transform a load impedance into an input
impedance where the input impedance is proportional to
the inverse of the load impedance. It is characterized by a
single resistance, R, known as the gyration resistance.
10. A negative impedance converter (NIC) is a non-reciprocal
two-port device that offers negative impedance, i.e., the
impedance seen at the input port is equal to the negative
of the load impedance with some conversion ratio.
Short-Answer Questions
1. What are the open-circuit impedance parameters
of a two-port network? Why are they so called?
The open-circuit impedance parameters represent the
relation between the voltages and the currents in the
two-port network.
The impedance parameter equations may be
written as
V1 = z 11I 1 + z 12 I 2
⎡V1 ⎤ ⎡ z 11 z 12 ⎤ ⎡ I 1 ⎤
⎢ ⎥=⎢
⎥ ⎢ ⎥ or,
V 2 = z 21I 1 + z 22 I 2
⎢⎣V 2 ⎥⎦ ⎢⎣ z 21 z 22 ⎥⎦ ⎢⎣ I 2 ⎥⎦
In this matrix equation, it is easily seen without even
expanding the individual equations, that
V
z 11 = 1
= Driving point impedance at Port-1
I 1 I =0
2
V
z 12 = 1
= Transfer impedance
I 2 I =0
1
V
z 21 = 2
= Transfer impedance
I 1 I =0
2
V
z 22 = 2
= Driving point impedance at Port-2
I 2 I =0
1
It can be seen that the dimensions of all the
parameters are impedance. All the impedances correspond to the driving point and transfer impedances at each port with the other port having zero
current (i.e., open circuit). For this reason, these
parameters are referred as the open-circuit impedance parameters.
399
Two-Port Network
2. Define y-parameters. Determine the relationship
between the z and y parameters.
The y-parameter equations may be written as
⎡ I 1 ⎤ ⎡ y 11
⎢ ⎥=⎢
⎢⎣ I 2 ⎥⎦ ⎢⎣ y 21
y 12 ⎤ ⎡V1 ⎤
⎥⎢ ⎥
y 22 ⎥⎦ ⎢⎣V 2 ⎥⎦
or,
I 1 = y 11V1 + y 12V 2
I 2 = y 21V1 + y 22V 2
The parameters y11, y12, y21, y22 can be defined in a similar manner, with either V1 or V2 on short circuit.
I
y 11 = 1
= Driving point admittance at Port-1
V1 V =0
2
I
y 12 = 1
= Transfer admittance
V 2 V 1= 0
I
y 21 = 2
= Transfer admittance
V1 V =0
2
I
y 22 = 2
= Driving point admittance at Port-2
V 2 V =0
1
It can be seen that the y-parameters correspond to the
driving point and transfer admittances at each port
with the other port having zero voltage (i.e., short circuit). For this reason, these parameters are also referred
as the short-circuit admittance parameters.
Relation between z and y parameters
The z-parameter equations are
V1 = z 11I 1 + z 12 I 2
V 2 = z 21I 1 + z 22 I 2
The y-parameter equations are
I 1 = y 11V1 + y 12V 2
Comparing Eqs. (1), (2) and (4), we get,
z 11 =
3. Why are the ABCD parameters termed ‘transmission parameters’?
The ABCD parameters represent the relation between
the input quantities and the output quantities in a
two-port network. They are thus voltage–current
pairs.
These parameters are known as transmission parameters as in a transmission line, the currents enter at one
end and leave at the other end, and we need to know a
relation between the sending-end quantities and the
receiving-end quantities.
4. What are transmission parameters? Where are they
most effectively used?
The transmission (or ABCD) parameters represent
the relation between the input quantities and the
output quantities in a two-port network. They are
thus voltage–current pairs.
However, as the quantities are defined as an input–
output relation, the output current is marked as going
out rather than as coming into the port.
I1
(1)
(2)
⎛ I
⎞
y
I 1 = y 11V1 + y 12 ⎜ 2 − 21 V1⎟
⎝ y 22 y 22 ⎠
or,
V1 =
y 22
y
I − 12 I 2
y 1
y
(3)
where,
y (y11y22 y12y21)
Substituting this value in the second equation of Eq. (2)
Port 1
V1
I2
Linear
passive
network
Port 2
V2
Fig. 6.126
I 2 = y 21V1 + y 22V 2
From Eq. (2), V = I 2 − y 21 V ; substituting this in the
2
y 22 y 22 1
first equation,
y 22
y
y
y
; z = − 12 ; z 21 = − 21 ; z 22 = 11
y 12
y
y
y
The transmission parameter equations may be written as
⎡V1 ⎤ ⎡ A
⎢ ⎥=⎢
⎢⎣ I 1 ⎥⎦ ⎣C
B ⎤ ⎡ V2 ⎤
⎥⎢ ⎥
D ⎦ ⎢⎣ − I 2 ⎥⎦
or,
V1 = AV 2 − B I 2
I 1 = CV 2 − DI 2
The parameters A, B, C, D can be defined in a similar
manner with either Port 2 on short circuit or Port 2 on
open circuit.
A=
V1
= Open-circuit reverse voltage gain
V 2 I =0
2
⎛y
⎞
y
I 2 = y 21 ⎜ 22 I 1 − 12 I 2 ⎟ + y 22V 2
y ⎠
⎝ y
V
= Short-circuit transfer impedance
B =− 1
I 2 V =0
y 21
y
I + 11 I 2
y 1
y
I
= Open-circuit transfer admittance
C= 1
V 2 I =0
or, V 2 = −
2
(4)
2
400
Network Analysis and Synthesis
I
= Short-circuit reverse current gain
D =− 1
I 2 V =0
I1
These parameters are most effectively used in
transmission lines. In a transmission line, the
currents enter at one end and leave at the other
end, and we need to know a relation between the
sending end-quantities and the receiving-end
quantities.
V1
I2
Za
Zb
2
5. How will you find the -equivalent of a given network when its y-parameters are known?
The configuration of a typical -network is shown in
Fig. 6.127.
I1
I2
Ya
V2
Yb
(V −V )Y +V Y = I ⇒ (Y +Y )V −Y V = I
V Y + (V −V )Y = I ⇒ −Y V + (Y +Y )V = I
2 b
2
1
c
2
a
c
c 1
1
c 2
b
c
2
2
)
y 11 = (Ya +Yb ; y 12 = y 21 = −Yc ; y 22 = (Yb +Yc
)
V1
I2 2
2
Za
⎛ Z − Za ⎞
I
I
∴ 1 Z a +V 2 = 1 Z b ⇒ V 2 = I 1 ⎜ b
⎟
2
2
⎝ 2 ⎠
1
Thus, the y-parameters are
(
V2
Assuming I2
0, the current I1 enters the bridge
at the point A and divides equally between the two
arms.
By KCL equations at the two nodes, we get
1
Zb
Za
To find the z-parameters, we redraw the network as
shown in Fig. 6.128 (b).
Fig. 6.127
1 a
I1
1
Here, Za are the series arms and Zb are the diagonal or
shunt arms.
2
c
Lattice network
Fig. 6.128 (b) Equivalent network
1
2
Fig. 6.128 (a)
1
2
YC
1
V1
Za
Zb
1
V1
Zb
(
)
)
Rearranging, Ya = y 11 + y 12 ; Yb = y 22 + y 12 ;
Yc = − y 12 = − y 21
From these equations of the admittances, we can find
out the -equivalent of a given network when its
y-parameters are known.
6. Write a technical note on derivation of short-circuit
admittance parameter y12 of a symmetrical and
reciprocal two-port lattice network.
A lattice network forms the basis of design of most
four-terminal networks like attenuators, filters, etc. It
consists of two identical impedances in series arm and
two identical impedances in shunt arm as shown in
Fig. 6.128 (a).
⎛ Z − Za ⎞
V
∴ z 21 = 2
= b
I 1 I = 0 ⎜⎝ 2 ⎟⎠
2
I
Also, V1 = 1 Z a + Z b
2
(
⎛ Z + Za ⎞
=⎜ b
⎟
⎝ 2 ⎠
1 I =0
V1
) ⇒ z =I
11
2
As the network is reciprocal and symmetrical,
⎛ Zb + Za ⎞
⎛ Z − Za ⎞
∴ z 21 = z 12 = ⎜ b
⎟ and z 11 = z 22 = ⎜ 2 ⎟
⎝
⎠
⎝ 2 ⎠
Hence we can write the y-parameter y12 for the lattice
network as follows.
⎛ Zb − Za ⎞
⎜⎝ 2 ⎟⎠
z 12
=−
y 12 = −
2
2
z
⎛ Zb + Za ⎞ ⎛ Zb − Za ⎞
⎜⎝ 2 ⎟⎠ − ⎜ 2 ⎟
⎝
⎠
1⎛ Z − Z b ⎞ 1⎛ 1
1⎞
= ⎜ a
=
−
2 ⎝ Z a Z b ⎟⎠ 2 ⎜⎝ Z b Z a ⎟⎠
401
Two-Port Network
7. What is a gyrator? Mention some properties of
an ideal gyrator. Show that a gyrator is a nonreciprocal device.
A gyrator is a two-port network that is designed to
transform a load impedance into an input impedance
where the input impedance is proportional to the
inverse of the load impedance. It is characterized by a
single resistance, R, known as the gyration resistance.
It can be shown that a gyrator is a non-reciprocal
device.
The symbol of a gyrator is shown in Fig. 6.129 (a) and
Fig. 6.129 (b). The arrow head indicates the direction
of gyration.
The v–i relationships for the gyrators of Fig. 6.122 (a)
and (b) are given below:
For Fig. 6.129 (a),
⎡v 1 ⎤ ⎡ 0
⎢ ⎥=⎢
⎢⎣v 2 ⎥⎦ ⎣ R
−R ⎤ ⎡ i 1 ⎤
⎥⎢ ⎥
0 ⎦ ⎢⎣ i 2 ⎥⎦
or, v1
Ri2 and v2
Ri1
R ⎤⎡ i1 ⎤
⎥⎢ ⎥
0 ⎦ ⎢⎣ i 2 ⎥⎦
or, v1
(a)
i1
v1
Ri2 and v2
4. If the output port of an ideal gyrator is terminated
with an inductor L, the input port behaves like a
capacitor.
5. If the output port of an ideal gyrator is terminated
with a voltage source, the input port behaves like a
current source.
6. If the output port of a gyrator is connected across
a current-controlled two-terminal resistor, i.e., v2
f( i2) then the input port becomes a voltage-controlled resistor. The resulting voltage-controlled
resistor is then the dual of the original current-controlled resistor.
A gyrator is a hypothetical device used for physical systems where the reciprocity condition does not
hold good.
8. What is negative impedance converter (NIC)? Show
that an NIC is a non-reciprocal device.
A negative impedance converter (NIC) is a twoport device that offers negative impedance, i.e., the
impedance seen at the input port is equal to the
negative of the load impedance with some conversion ratio.
For Fig. 6.129 (b),
⎡v 1 ⎤ ⎡ 0
⎢ ⎥=⎢
⎢⎣v 2 ⎥⎦ ⎣ − R
3. If the output port of an ideal gyrator is terminated with
a capacitor, the input port behaves like an inductor.
Ri1
(b)
R
i1
i2
R
v2 v1
i1
i2
v2
Fig. 6.129 Symbol of gyrator
i2
v1
NIC
v2
ZL
Fig. 6.130 Negative impedance
converter (NIC)
It is characterized by the v–i relationships:
From the v–i relationships, it is clear that for a gyrator,
z12 z21 and hence it is non-reciprocal.
Properties of a gyrator
1. A gyrator is a non-energic or passive element, i.e., at all
times the power delivered to the two-port is identically zero.
2. When a gyrator is terminated at the output port
with a linear resistance RL, the input port behaves as
⎛R2⎞
a linear resistor with resistance ⎜ ⎟ .
⎝ RL ⎠
v 1 = kv 2
i2 = k i1
or,
⎡v 1 ⎤ ⎡ 0
⎢ ⎥=⎢
⎢⎣ i 2 ⎥⎦ ⎣ k
k ⎤⎡ i1 ⎤
⎥⎢ ⎥
0 ⎦ ⎢⎣v 2 ⎥⎦
(1)
−k ⎤ ⎡ i 1 ⎤
⎥⎢ ⎥
0 ⎦ ⎢⎣v 2 ⎥⎦
(2)
or,
v 1 = − kv 2
i 2 = −k i 1
or,
⎡v 1 ⎤ ⎡ 0
⎢ ⎥=⎢
⎢⎣ i 2 ⎥⎦ ⎣ − k
where, k is the conversion ratio.
From equations (1) and (2), it is seen that h12 h21; hence
NIC is a non-reciprocal device.
402
Network Analysis and Synthesis
Exercises
1. Currents I1 and I2 entering at ports 1 and 2 respectively
of a two-port network are given by the following equations:
4. Find the open-circuit and short-circuit impedances of
the network shown in Fig. 6.133.
⎡ 31
−19 ⎤
44
44 ⎥ −1
;
[ y = ⎢⎢
⎥
−19
23
⎥
⎢⎣
44
44 ⎦
z-parameters do not exist]
( )
I 1 = 0.5V1 − 0.2V 2
I 2 = − 0.2V1 +V 2
where V1 and V2 are the voltages at ports 1 and 2
respectively. Find the y, z and ABCD parameters for the
network. Also find its equivalent -network.
[ y11 0.5 mho; y12
0.2 mho; y21
0.2 mho; y22
1 mho; z11 2.174 ; z12
0.435 ; z22 1.086 ;
A 5; B 5 ; C 2.3 mho; D 2.5; Y1 0.3 mho;
Y2 0.2 mho; Y3 0.8 mho]
2. Determine the z- and y-parameters of the networks
shown in Fig. 6.131.
(a)
j 80
j 40
4
3
1
V1
V2
2
Fig. 6.133
5. Find the z-parameters for the 2-port networks shown
in Fig. 6.134 containing a controlled source.
2
1
⎡ −2
[z =⎢
⎢1
⎣ 2
j 60
2
1
(b)
j 20
j 25
I1
2
1
1
3V1
−1 ⎤
⎥
3 ⎥
2⎦
( )]
I2
30
2
1
(c)
V1
1F
2H
1
2
V2
2
1
1
Fig. 6.134
2
1
Fig. 6.131
⎡ − j 120 − j 160 ⎤
[(a) z = ⎢
⎥
⎣ − j 160 − j 80 ⎦
(
⎡ 30 + j 40
(b) z = ⎢
⎢⎣ j 40
(
)
⎡ 1+ 2 s
⎢
(c) z = ⎢
⎢ 1
⎣
⎤
⎥
30 + j 80 ⎥⎦
j 40
(
)
)
⎤
⎥
⎛ 1⎞ ⎥
+
1
⎜⎝ s ⎟⎠ ⎥
⎦
1
( ),
6. A 2-port network made up of passive linear resistors
is fed at Port 1 by an ideal voltage source of V volt. It is
loaded at Port 2 by a resistor R.
( ),
(i) With V 10 volts and R 6 , currents at ports 1
and 2 were 1.44 A and 0.2 A respectively.
(ii) With V 15 volts and R 8 , current at Port 2 was
0.25 A.
( )]
3. Obtain the z-parameters for the circuit shown in Fig. 6.132
and hence draw the z-parameter equivalent circuit.
I1 2
2
I2
V1 1
2
V2
⎡14
5
[ z = ⎢⎢
2
⎣⎢ 5
2 ⎤
5⎥
6 ⎥
5 ⎥⎦
( )]
Determine the -equivalent circuit of the 2-port network.
[ YA
0.2 mho; YB
0.05 mho]
7. Calculate the T-parameters for the blocks A and B
separately and then using these results, calculate the
T-parameters of the whole circuit shown in Fig. 6.135.
Prove any formula used.
1
1
V1
2
Block A
Fig. 6.132
0.3 mho; YC
Fig. 6.135
1
1
2
Block B
V2
403
Two-Port Network
V1
11. Test results for a two-port network are
1
2
1
V2
2
2
(a) Port 2 open-circuited, I1
0.01 0 (A), V1
26.4 (V)
1.4 45 (V), V2 2.3
(b) Port 1 open-circuited, I2
0.01 0 (A), V1
53.1 (V)
1
90 (V), V2 1.5
⎡3
[For each block, T = ⎢⎢ 2
1
⎢⎣ 2
5 ⎤
2⎥ ;
3 ⎥
2 ⎥⎦
⎡7
or whole circuit, T = ⎢ 2
⎢3
⎢⎣ 2
15 ⎤
2⎥ ]
7 ⎥
2 ⎥⎦
The source frequency in both the test was 1000 Hz.
Find z-parameters.
⎡ 140 ∠45°
[⎢
⎣230 ∠ − 26.4°
⎡
⎤
[ ⎢10 3 ⎥
3
6
⎣
⎦
I1 4
( )
V1
2
2
2
13. For the network shown in Fig. 6.139, find the
y-parameters and also the equivalent T-network.
2
8
9. Find the z-parameters for the lattice network shown in
Fig. 6.137.
I2
Za
Zb
V1
I1
Zb
V2
V1
I2
Za
Zb
Zb
Fig. 6.139
⎤
⎡ ⎡ 62
⎤
−30
⎥
⎢ ⎢ 112
112 ⎥ , Z = 8
,
a
⎢ ⎢ −30
⎥
13 ⎥
38
⎥
⎢ ⎢⎣
112
112 ⎥⎦
⎥
⎢
⎥
⎢ Z b = 32
, Z c = 30
⎦
⎣
13
13
V2
⎛ Zb − Za ⎞
⎛ Z + Za ⎞
[ z 11 = z 22 = ⎜ b
⎟ , z 12 = z 21 = ⎜ 2 ⎟ and
⎝
⎠
⎝ 2 ⎠
11
2Z a + Z b
) ; z = 2Z Z
22
14. Find the h-parameters for the network shown in Fig. 6.140.
I1
Z b2 ]
, z 12 = z 21 =
2Z a + Z b
2Z a + Z b
a
I2
8
b
⎡
⎡ 0.5 −0.2 ⎤ −1
⎡2.174 0.435 ⎤
,
, z =⎢
⎢y = ⎢
⎥
⎥
1 ⎦
⎣ −0.2
⎣ 0.435 1.087 ⎦
⎢
⎢
⎡
5 ⎤
⎢T = ⎢ 5
⎥ ; Y = 0.3 , Yb = 0.8 , Yc = 0.2
−1
⎢⎣
.
2
3
2.5 ⎦ a
⎣
( )
⎤
⎥
⎥
⎥
−1 ⎥
⎥⎦
( )
16
16
V1
10. Currents I1 and I2 entering at Port-1 and Port-2 respectively of a two-port network are given by the follow0.2 V2, I2
0.2 V1 V2,
ing equations: I1 0.5 V1
where V1 and V2 are the voltages at Port-1 and Port-2
respectively. Find the y, z and ABCD parameters for the
network. Also find the equivalent -network.
( )
V2
2
Fig. 6.137
(
z =
4
1
V1
Za
Za
2 Za + Zb
V2
9
9
( )]
I2
6
Fig. 6.136
I1
( )]
12. Find the z-parameters for the network shown in Fig. 6.138.
8. Find out the z-parameters of the two-port network
shown in Fig. 6.136.
⎡6 2 ⎤
]
[z =⎢
⎥
⎣2 6 ⎦
2
100 ∠ − 90° ⎤
⎥
150 ∠ − 53.1° ⎦
V2
8
Fig. 6.140
[ h11 =
32
3
1
1
1
; h12 = ; h21 = − ; h22 = mho ]
3
3
12
15. The h-parameters of a two-port network are
h11
h22
35 ; h12 2.6
0.3 10 6 mho
10 4;
h21
0.98;
404
Network Analysis and Synthesis
The input terminals are connected to a 0.001-V sinusoidal source and a 104-ohm resistance is connected
across the output port. Find the output voltage.
[0.26 Volt]
16. Find the y and z-parameters for the network shown in
Fig. 6.141.
I1
1
V1
1H
1H
2
1F
1F
1
2
Fig. 6.144
⎡1+ 3s 2 + s 4
⎢
3
⎣ 2s + s
I2
1
1
1H
1
20. Determine the y parameters of the overall network,
considering two networks connected in parallel.
V2
12
3s + 4 s 3 + s 5 ⎤
⎥
1+ 3s 2 + s 4 ⎦
1
Fig. 6.141
⎡13
[⎢ 7
⎢2
⎢⎣ 7
2 ⎤
7⎥
3 ⎥
7 ⎥⎦
⎡− 3
( ) ; ⎢⎢ 2 5
⎢⎣ − 5
−2 ⎤
5 ⎥ mho ]
13 ⎥
5 ⎥⎦
)
(
17. Find the y-parameters for the network shown in Fig. 6.142.
0.2V2
I1 5
V1
[ y 11 = y 22 =
0.4I 2
I1
]
18. Find the transmission parameters of the network
shown in Fig. 6.143.
5
V1
0.3V1
V2 10 4
5
3
; y 12 = y 21 = −
4
3
]
21. Two identical sections of the circuit shown in fig. 6.146
are connected in series. Obtain the z-parameters of the
combination and verify by direct calculation.
V2
⎡ 0.2
− 0.24 ⎤
[⎢
⎥
⎣ − 0.333 0.4833 ⎦
V2
Fig. 6.145
Fig. 6.142
I1
1
I2
20
1
1
V1
V1
0.5 mho V2
0.2mho
Fig. 6.146
[z11
I2
I2
1mho
z22
6 ; z12
z21
4 ]
22. The z-parameters of a two-port network are
z11
V2
Fig. 6.143
⎡ 55
⎢
26
[⎢
⎢7
⎢⎣ 20
50 ; z22
30 ; z12
z21
20 ;
Calculate the y-parameters, ABCD parameters and the
image parameters of the network.
50 ⎤
⎥
13 ⎥
]
⎥
1⎥
⎦
19. Determine the T-parameters for the network shown in
Fig. 6.144 using the concept of interconnection of two
two-port networks.
⎤
⎡ y 11 = 0.0273 mho ; y 22 = 0.0454 mho ;
⎥
⎢
⎥
⎢ y 12 = y 21 = − 0.01818 mho ;
⎢ A = 2.5 ; B = 55 ; C = 0.05 mho ; D = 1.5 ⎥
⎥
⎢
⎥⎦
⎢⎣ Z i 1 = 42.82 ; Z i 2 = 25.69 ; = 1.28
23. For the symmetrical two-port network, calculate the
z-parameters and ABCD parameters. Hence or otherwise, find the characteristic impedance and propagation constant for this network.
405
Two-Port Network
I1
40
40
24. A two-port network has
I2
2
1
V1
20
1
V2
2
Fig. 6.147
⎡ z 11 = z 22 = 60 ; z 12 = z 21 = 20 ; A = D = 3 ;
⎤
⎢
⎥
Z
=
56
57
=
1
762
B
=
160
;
C
=
0
.
05
m
ho;
.
;
.
⎢⎣
⎥⎦
0
(i) at Port 1, driving point impedances of 60 and
50 with Port 2 open circuited and short circuited
respectively.
(ii) at Port 2, driving point impedances of 80 and
70 with Port 1 open circuited and short circuited
respectively.
Find the image parameters of the network. Derive the
expressions used.
[54.77 , 74.83 ]
Questions
1. (a) Consider a linear passive two-port network and
explain what are meant by i) open-circuit impedance parameters, and ii) short-circuit admittance
parameters.
4. What are transmission parameters? Where are they
most effectively used? Establish, for two-port networks,
the relationship between the transmission parameters
and the open-circuit impedance parameters.
(b) What are the open-circuit impedance parameters
of a two-port network? How can the transmission
parameters be obtained from open-circuit impedance parameters?
5. (a) Two two-port networks are connected in parallel.
Prove that the overall y-parameters are the sum of
corresponding individual y-parameters.
(c) Establish for two-port networks, the relationship
between the transmission parameters and the
open-circuit parameters.
(b) Two two-port networks are connected in cascade.
Prove that the overall transmission parameter
matrix is the product of individual transmission
parameter matrices.
(d) Define z and y parameters of a typical four-terminal network. Determine the relationship between
the z and y parameters.
(c) Two two-port networks are connected in series.
Prove that the overall z-parameters are the sum of
corresponding individual z-parameters.
(e) Express h-parameters in terms of z-parameters for
a two-port network.
6. (a) Define ‘transfer function’ and ‘driving point function’ of a two-port network.
(f ) Derive expressions for the y-parameters in terms of
ABCD parameters of a two-port network.
(b) Derive the expression of input impedance of a
two-port network terminated with a load-impedance ZL, in terms of its -parameters.
2. (a) What do you understand by a reciprocal network?
What is a symmetrical network?
(b) Write technical note on derivation of short-circuit
admittance parameter y12 of a symmetrical and
reciprocal two-port lattice network.
(c) How will you find the ␲-equivalent of a given network when its y-parameters are known?
3. (a) Explain what are meant by the transmission (ABCD)
parameters of a two-port network. Derive the conditions necessary to be satisfied for the network to
be i) reciprocal, and ii) symmetrical.
Or,
Prove that for a reciprocal two-port network, T
(AD BC ) 1
(b) Prove that for a symmetrical two-port network,
h (h11h22 h12h21) 1
(c) Derive the expression of output impedance of a
two-port network terminated with a load-impedance ZL, in terms of its transmission parameters.
7. What is a gyrator? Mention some properties of an
ideal gyrator. Show that a gyrator is a non-reciprocal
device.
8. What is negative impedance converter (NIC)? Show
that an NIC is a non-reciprocal device.
9. What are image parameters? Derive expression of
image parameters in terms of (i) ABCD parameters
(ii) open-circuit and short-circuit impedances.
10. What is a symmetrical network? Derive expressions for
characteristic impedance and propagation constant of
a symmetrical networks in terms of short-circuit and
open-circuit impedances.
406
Network Analysis and Synthesis
Multiple-Choice Questions
1. Which one of the following pairs is correctly
matched?
(i) Symmetrical two-port network: AD BC 1
(ii) Reciprocal two-port network: z11 z22
(iii) Inverse hybrid parameters: A, B, C, D
(iv) Hybrid parameters: (V1, I2) f (I1, V2)
2. What is the condition for reciprocity in terms of
h-parameters?
(i) h11 h22
ii) h12 h21 h11 h22
(iii) h12 h21 0
iv) h12 h21
3. For a reciprocal network, the two-port ABCD parameters are related as follows:
(i) AD BC 1
(ii) AD BC 0
(iii) AD BC
1
(iv) AC BD 1
4. For a symmetrical two-port network,
(i) z11 z22
(ii) z12 z21
(iii) z11z22 z122 0
(iv) z11 z22 and z12
z21
5. For a two-port network to be reciprocal, it is necessary
that
(i) z11 z22 and y12 y21
(ii) z11 z22 and AD BC 0
(iii) h21
h12 and AD BC 0
(iv) y12 y21 and h21
h12
6. A two-port network is symmetrical if
(i) z11 z22 z12 z21 1
(ii) AD BC 1
(iii) h11 h22 h12 h21 1
(iv) y11 y22 y12 y21
7. A two-port network is reciprocal if and only if
(i) z11 z22
(ii) BC AD
y21
(iv) h12 h21
(iii) y12
1
B
(ii) B
C
(iii) C
D
(iv) D
⎡ ( A1 A2 + C 1C 2
B 1B 2 ⎤
⎥ (iv) ⎢
D1D2 ⎥⎦
⎢⎣(C 1 A2 − D1C 2
) ( A A − B D ) ⎤⎥
) (C C + D D )⎥⎦
1 2
1
2
1 2
1
2
11. Consider the following statements:
For a bilateral network,
1. A D
2. z12
z21
Of these statements,
(i) 1, 2 and 3 are correct
(iii) 1 and 3 are correct
3. h12
h21
(ii) 1 and 2 are correct
(iv) 2 and 3 are correct.
12. In a two-port network containing linear bilateral passive circuit elements, which one of the following conditions for z parameters would hold?
(i) z11 z22
(ii) z12 z21 z11 z22
(iii) z11 z12 z22 z21
(iv) z12 z21
13. The relation AD BC 1, where A, B, C and D are the
elements of a transmission matrix of a network, is
valid for
(i) any type of network
(ii) passive but not reciprocal network
(iii) passive and reciprocal network
(iv) both active and passive network
14. When a number of two-port networks are connected
in cascade, the individual
(i) Zoc matrices are added
(ii) Ysc matrices are added
(iii) chain matrices are multiplied
(iv) H-matrices are multiplied
15. The h parameters h11 and h22 are related to z and y
parameters as
1
8. In terms of ABCD parameters, a two-port network is
symmetrical if and only if
(i) A
⎡ A1 A2
(iii) ⎢
⎢⎣C 1C 2
A
9. The condition for reciprocity of a two-port network
having different parameters are
i) h12
h21
ii) g12
g21
iii) A D
Choose the correct combination:
(i) 1 and 2 (ii) 1 and 3 (iii) 2 and 3 (iv) 1, 2 and 3.
10. Two two-port networks with transmission parameters
A1, B1, C1, D1 and A2, B2, C2, D2 respectively are cascaded.
The transmission parameter matrix of the cascaded
network will be
⎡ A1 B1 ⎤ ⎡ A 2 B 2 ⎤
⎡ A1 B1 ⎤ ⎡ A 2 B 2 ⎤
⎥⎢
⎥
⎥+ ⎢
⎥
(ii) ⎢
(i) ⎢
⎢⎣C 1 D1 ⎥⎦ ⎢⎣C 2 D2 ⎥⎦
⎢⎣C 1 D1 ⎥⎦ ⎢⎣C 2 D2 ⎥⎦
(i) h11
z11 and h22 =
1
z 22
(ii) h11
z11 and h22
y22
(iii) h11 =
z
1
and h22 =
z 22
z 22
1
and h22 y22
y 11
16. Two two-port networks and having A B C D parameters as
4 D
A
3 D
A
(iv) h11 =
B
5, C
3
and
B
4, C
2
are connected in cascade in the order of ␣, ␤. The
equivalent ‘A’ parameters of the combination is
(i) 17
(ii) 22
(iii) 24
(iv) 31.
407
Two-Port Network
17. With the usual notation, a two-port resistive network
3
4
satisfies the condition A = D = B = C
2
3
The z11 of the network is
(i) 5
(ii) 4
(iii) 2
(iv) 1
3
3
3
3
18. The reciprocal of a network function is
(i) an immittance function, if the original function is
an immittance function
(ii) a transfer function, if the original function is a
transfer function
(iii) never an immittance function
(iv) never a transfer function
19. A two-port network is defined by the relations I1
V2, I2 2V1 3V2 . Then z12 is
1
(iv)
(i) 2
(ii) 1
(iii)
2
2V1
1
4
20. Consider the following statements:
1. Transfer impedance is the reciprocal of transfer
admittance.
2. One can derive transfer impedance of a network
if its driving-point impedance and admittance are
known.
3. Driving-point impedance is the ratio of the Laplace
transform of voltage and current functions at the
input.
Of these statements
(i) 1, 2 and 3 are correct
(iii) 2 and 3 are correct
(ii) 1 and 2 are correct
(iv) 3 alone is correct
21. Consider the following statements:
1. The two-port network shown below does NOT
have an impedance matrix representation.
Z
1
1
22. If two two-port networks are connected in series, and
if the port current requirement is satisfied, which of
the following is true?
(i) The z-parameter matrices add
(ii) The y-parameter matrices add.
(iii) The ABCD-parameter matrices add.
(iv) None of these.
23. If two two-port networks are connected in parallel,
and if the port current requirement is satisfied, which
of the following is true?
(i) The z-parameter matrices add
(ii) The y-parameter matrices add.
(iii) The ABCD-parameter matrices add.
(iv) None of these.
24. If two two-port networks are connected in cascade,
and if the port current requirement is satisfied, which
of the following is true?
(i) The z-parameter matrices add.
(ii) The y-parameter matrices add.
(iii) The ABCD-parameter matrices add.
(iv) None of these.
25. The z11 and z22 parameters of the given network are
5
3
4
5
10
Fig. 6.150
(i) 8 , 7.75
(iii) 12 , 8.5
(ii) 13 , 9
(iv) none of the above
26. For the network shown, the parameters h11 and h21 are
I1
2
V1
2
I2
1
6
4
V2
Fig. 6.148
2. The two-port network shown below does NOT
have an admittance matrix representation.
1
2
Y
1
2
Fig. 6.149
3. A two-port network is said to be reciprocal if it satisfies z12 z21 or an equivalent relationship.
Of these statements
(i) 1 and 2 are correct
(iii) 1 and 3 are correct
(ii) 1 and 3 are correct
(iv) none is correct.
Fig. 6.151
(i) 5
(iii) 3.4
and − 2
3
(ii) 3.4
and − 2
5
and − 3
(iv) none of the above
5
27. The maximum value of the transmission parameter
A for a passive, reciprocal, linear two-port network is
(i) 1
(ii) 2
(iii) 3
(iv) none of the above
28. The unique feature of ABCD parameters as compared
to z, y and h parameters is
(i) none
(ii) short-circuit functions
408
Network Analysis and Synthesis
(iii) open-circuit functions
(iv) reverse transverse functions
29. The driving point impedance of the infinite ladder
network shown in Fig. 6.152 is:
R1
R1
R1
R2
R2
R2
R1
R1
R2
Infinity
2
⎤
⎥
⎥
rb + re ⎥⎦
⎡ rb + re
(ii) ⎢⎢
⎢ bc
⎣
⎤
⎥
1 ⎥
re + rd ⎥⎦
⎡ rb + re
(iii) ⎢
⎢
⎢ cb
⎣
⎤
bc
⎥
1 ⎥
re + rd ⎥⎦
⎡ bc
(iv) ⎢⎢
r +r
⎢b e
⎣
⎤
cb
⎥
1 ⎥
re + rd ⎥⎦
bc
and R2
1.5 )
(ii) 3.5
⎛
3 ⎞
(iii) 3
(iv) ln⎜ 1+ ⎟
3
.5 ⎠
⎝
3.5
30. A two-port network is described by the relations:
35. The y-parameter ‘y21’ of the network shown in Fig. 6.155
I1 6
4
V1 = 2V 2 + 0.5I 2
I 1 = 2V 2 + I 2
V1
What is the value of the h22 parameter of the network?
(i) 1 mho
(ii) 2
(iii) −2 mho
(iv) 4
31. What are the suitable values for Z1 and Z2, to make the
input impedance, Zin, of the network equal to R?
R
Z1
(
2 sin t −
(i) is 2 mho
(iii) is 3 mho
)
(iii) cos ␻t
1H
(ii) cos ␻t
(iv)
sin ␻t
(
2 sin t +
)
4
33. Which one of the following gives the h-parameter
matrix for the network shown in Fig. 6.154?
I1
rb
re
Fig. 6.154
V2
oc
⎛ 10 ⎞
(i) ⎜
⎟ ∠ − 45°
⎝ 2⎠
⎛ 10 ⎞
(ii) ⎜
⎟ ∠45°
⎝ 2⎠
(iii) 5∠45°
(iv) 5∠ − 45°
37. For the two-port network, the parameter y21 will be
I
rd
2
1
Y3
V1
I2
cb 1
2
Fig. 6.156
re
V1
(ii) is 6 mho
(iv) does not exist
i =10cos 2t
(ii) 2R and R
(iv) 4R and 4R
4
V2
R
32. The forward voltage transfer function of a two-port
s+
network is
. What will be the output voltage
s2 + 2
if the input voltage is ␦(t)?
(i)
14 I1
6
Fig. 6.155
Fig. 6.153
(i) R and R
(iii) 3R and 2R
I2
36. The phasor current through the inductance in the
circuit shown is
Z2
Z in
cb
34. In a two-port network, the output short-circuit current was measured while the source voltage at the
input was 1V; the value of the output current would
provide the parameter
(iii) h21
(iv) y21
(i) B
(ii) y12
Fig. 6.152
(given R1
(i) 3
⎡ 1
(i) ⎢⎢ re + rd
⎢
⎣ cb
Y1
gm
1
V2
Y2
V1
2
Fig. 6.157
(i) Y2
(iii) Y3
Y3
gm
(ii) gm
(iv) gm
Y3
Y2
Y3
409
Two-Port Network
38. For the given two-port network, z21 will be
2
2
Vi
2
1
V1
V2
2
1
1
Fig. 6.158
(i) 2
(ii) 3
(iii) 1
(iv) 4
5
5
5
5
39. The h-parameters for a two-port network are defined
⎡E ⎤ ⎡h
h ⎤⎡ I ⎤
by ⎢ 1 ⎥ = ⎢ 11 12 ⎥ ⎢ 1 ⎥ . For the two-port network
⎢⎣ I 2 ⎥⎦ ⎢⎣ h21 h22 ⎥⎦ ⎢⎣ E 2 ⎥⎦
shown in Fig. 6.159, the value of h12 is given by
4
2
2
I2
4
2
E1
(ii) 0.167
(iii) 0.625
(i) 1 V, , 10
(iii) 1 V, 0,
(ii) 1 V, 0, 10
(iv) 10 V, , 10
E2
(iv) 0.25
40. The z matrix of a two-port network is given by
⎡0.9 0.2 ⎤
⎢
⎥ . The element y22 of the corresponding y
⎣0.2 0.6 ⎦
matrix of the same network is given by
(i) 1.2
(ii) 0.4
(iii) 0.4
(iv) 1.8
v1
Z2
⎡Z
(iii) ⎢ 1
⎢⎣ Z 2
Z1 + Z2 ⎤
⎥
Z 2 ⎥⎦
⎤
⎥
Z 1 + Z 2 ⎥⎦
v2
⎡ Z1
(ii) ⎢
⎢⎣ Z 1 + Z 2
(iv) ⎡⎢ Z 1
⎣⎢ Z 1
Z1 ⎤
⎥
Z 2 ⎥⎦
⎤
⎥
Z 1 + Z 2 ⎥⎦
42. The parameters of the circuit shown in Fig. 6.161 are
Ri 1 M , R0 10 , A 106 V/V. If Vi 1 μV then
output voltage, input impedance and output impedance respectively are
Z2
V2
Fig. 6.162
⎡1 0⎤
(iv) z parameters, ⎢
⎥
⎣0 1⎦
44. The impedance parameters z11 and z12 of the two-port
network in Fig. 6.163 are
5
3
4
5
10
(i)
(ii)
(iii)
(iv)
Fig. 6.160
⎡ Z1
(i) ⎢
⎢⎣ Z 1 + Z 2
V1
Fig. 6.163
i2
Z1
I2
⎡0 0 ⎤
(iii) h parameters, ⎢
⎥
⎣0 0 ⎦
41. For the two-port network shown in Fig. 6.160, the
z-matrix is given by
i1
I1
43. The parameter type and
the matrix representation
of the relevant two-port
parameters that describe
the circuit shown are
(i) z parameters,
⎡0 0 ⎤
⎢
⎥
⎣0 0 ⎦
⎡1 0⎤
(ii) h parameters, ⎢
⎥
⎣0 1⎦
Fig. 6.159
(i) 0.125
AVi
Fig. 6.161
2
I1
R0
Ri
Z1
z11
z11
z11
z11
2.75 and z12 0.25
3 and z12 0.5
3 and z12 0.25
2.25 and z12 0.5
45. For the lattice circuit shown in Fig. 6.164, Za j 2
and Zb j 2 . The values of the open circuit imped⎡z
z 12 ⎤
ance parameters z = ⎢ 11
⎥ are
⎢⎣ z 21 z 22 ⎥⎦
1
Za
Za
2
Fig. 6.164
3
Zb
Zb
4
410
Network Analysis and Synthesis
I1
⎡1− j
(i) ⎢
⎣1+ j
1+ j ⎤
⎥
1+ j ⎦
⎡ 1− j
(ii) ⎢
⎣ −1+ j
1+ j ⎤
⎥
1− j ⎦
⎡1+ j
(iii) ⎢
⎣1− j
1+ j ⎤
⎥
1− j ⎦
⎡ 1− j
(iv) ⎢
⎣ −1− j
−1+ j ⎤
⎥
1− j ⎦
re
(i) re and ␤r0
(iii) 0 and ␤r0
r0
V2
V1
n :1
49. A two-port network is represented by ABCD parameters given by
(i)
A + BR L
C + DR L
(ii)
AR L + C
BR L + D
(iii)
DR L + A
BR L + C
(iv)
B + AR L
D + CR L
Fig. 6.165
1
(ii)
n
(iii) n
2
1
(iv) 2
n
47. The h-parameters of the circuit shown in Fig. 6.166 are
10
V1
(ii) 0 and ␤r0
(iv) re and ␤r0
⎡V1 ⎤ ⎡ A
⎢ ⎥=⎢
⎢⎣ I 1 ⎥⎦ ⎣C
B ⎤ ⎡ V2 ⎤
⎥⎢ ⎥
D ⎦ ⎢⎣ − I 2 ⎥⎦
If Port-2 is terminated by RL then the input impedance
seen at Port-1 is given by
I2
I1
I1
␤I1
Fig. 6.167
46. The ABCD parameters of an ideal n : 1 transformer
⎡n 0 ⎤
shown in Fig. 6.165 are ⎢
⎥ The value of X will be
⎣0 x ⎦
(i) n
I2
I2
20
Fig. 6.166
⎡ 0.1 0.1⎤
(i) ⎢
⎥
⎣ − 0.1 0.3 ⎦
50. An ideal gyrator is a
(i) passive reciprocal device
(ii) passive and non-reciprocal device
(iii) active and reciprocal device
(iv) active and non-reciprocal device
51. When two gyrators are connected, in cascade the
device acts as a/an
(i) negative impedance converter
(ii) ideal transformer
(iii) perfect transformer
(iv) none of the above
⎡30 20 ⎤
(iii) ⎢
⎥
⎣20 20 ⎦
52. If r1 and r2 are real numbers for a gyrator where r1
it is a
(i) positive impedance converter
(ii) positive impedance inverter
(iii) negative impedance converter
(iv) negative impedance inverter
⎡10
1 ⎤
(iv) ⎢
⎥
⎣ −1 0.05 ⎦
53. An active gyrator is one when
(i) r1 r2
(ii) r1
(iv) r1
(iii) r1 r2
⎡10 −1 ⎤
⎥
(ii) ⎢
⎣ 1 0.05 ⎦
48. In the two-port network shown in Fig. 6.167 below, z12
and z21 are, respectively,
r2,
r2
r2
54. An ideal impedance converter is a two-port network
which when terminated at one port by driving point
411
Two-Port Network
impedance ZL(s) offers at the other port an input
impedance that is
(i) directly proportional to ZL(s)
(ii) inversely proportional to ZL(s)
(iii) square root of ZL(s)
(iv) none of the above, at all frequencies
55. An ideal transformer cannot be described by
(i) h parameters
(ii) ABCD parameters
(iii) g parameters
(iv) z parameters
⎡ z 11
56. A network N with impedance matrix ⎢
⎢⎣ z 21
z 12 ⎤
⎥ is
z 22 ⎥⎦
followed by an ideal transformer with 1 : a ratio. The
overall impedance matrix is
⎡az 11
z 12 ⎤
⎥
(i) ⎢
2
⎢⎣ z 21 a z 22 ⎥⎦
⎡ z 11 az 12 ⎤
⎥
(ii) ⎢az
⎢⎣ 21 z 22 ⎥⎦
⎡z
az 12 ⎤
(iii) ⎢ 11
⎥
2
⎢⎣az 21 a z 22 ⎥⎦
⎡a 2 z 11 az 12 ⎤
(iv) ⎢
⎥
2
⎢⎣ az 21 a z 22 ⎥⎦
⎡ z 11
57. A network N with impedance matrix ⎢
⎢⎣ z 21
z 12 ⎤
⎥ is
z 22 ⎥⎦
preceded by an ideal transformer with 1 : a ratio. The
overall impedance matrix is
⎤
a ⎥
⎥
z 22 ⎥⎥
⎦
⎡az 11 z 12 ⎤
⎥
(i) ⎢ z
⎢⎣ 21 az 22 ⎥⎦
⎡ z 11
2
⎢
(ii) ⎢ a
z
⎢ 21
⎢⎣ a
z 12
⎡a 2 z 11 az 12 ⎤
⎥
(iii) ⎢
z 22 ⎥⎦
⎢⎣ az 21
⎡ z 11
(iv) ⎢ a
⎢
⎢⎣ az 21
⎤
az 12 ⎥
⎥
z 22 ⎥⎦
58. A network N with short-circuit admittance matrix
⎡ y 11 y 12 ⎤
⎢
⎥ is followed by an ideal transformer with
⎢⎣ y 21 y 22 ⎥⎦
1 : a ratio. The overall admittance matrix is
⎡ y 11
⎤
⎡ y 11 ay 12 ⎤
ay 12 ⎥
⎢
⎥
(ii) ⎢ a
(i) ⎢
⎥
⎢⎣ay 21 y 22 ⎥⎦
⎢⎣ ay 21 ay 22 ⎥⎦
⎡
⎢ y 11
(iii) ⎢
⎢ y 21
⎢⎣ a
⎤
a ⎥
⎥
y 22 ⎥
2⎥
a ⎦
y 12
⎡
⎢ y 11
(iv) ⎢
⎢ y 21
⎢⎣ a
⎤
a ⎥
⎥
ay 22 ⎥⎥
⎦
y 12
59. A network N with short-circuit admittance matrix
⎡ y 11 y 12 ⎤
⎢
⎥ is preceded by an ideal transformer with
⎢⎣ y 21 y 22 ⎥⎦
1 : a ratio. The overall admittance matrix is
⎡
⎢ y 11
(i) ⎢ y
⎢ 21
⎢⎣ a
⎤
a ⎥
⎥
a 2 y 22 ⎥⎥
⎦
y 12
⎡ y 11 ay 12 ⎤
⎥
(iii) ⎢
2
⎢⎣ay 21 a y 22 ⎥⎦
⎡a 2 y 11 ay 12 ⎤
⎥
(ii) ⎢
y 22 ⎥⎦
⎢⎣ ay 21
⎡ y 11
⎢
(iv) ⎢ a
⎢ ay
⎣⎢ 21
⎤
ay 12 ⎥
⎥
y 22 ⎥
a ⎥⎦
60. The h-parameters of a negative impedance converter
(NIC) with k as conversion factor are
⎡k
(i) ⎢
⎢0
⎣
0 ⎤
⎥
1 ⎥
k⎦
⎡ 0
(iii) ⎢
⎢1
⎣ k
k⎤
⎥
0⎥
⎦
⎡1
⎢ k
(ii) ⎢
⎣ 0
0⎤
⎥
k ⎥⎦
⎡0
(iv) ⎢
⎢0
⎣
1 ⎤
k⎥
k ⎥⎦
61. When a gyrator is connected in tandem with a passive
reciprocal network, the overall two-port network acts
as a兾an
(i) passive reciprocal network
(ii) active reciprocal network
(iii) passive non-reciprocal network
(iv) active non-reciprocal network
62. When a gyrator with gyration resistance r is terminated through a resistor R, the equivalent element at
the input terminals is
R
r2
(iv) 2
r
R
63. When a gyrator with gyration resistance r is terminated through a capacitor C, the equivalent element
at the input terminals is
(i) a capacitor with value r C
(ii) a capacitor with value r2C
(iii) an inductor with value r C
(iv) none of the above
(i) r 2 R
(ii) rR
(iii)
64. When a negative impedance converter (NIC) with
conversion ratio k is terminated through an impedance ZL, the equivalent element at the input terminals is
k2
(i) kZL
(ii) k 2 ZL
(iii) k ZL
(iv) −
ZL
412
Network Analysis and Synthesis
Answers
1. (iv)
2. (iii)
3. (i)
4. (i)
5. (iv)
6. (iii)
7. (ii)
8. (iv)
9. (i)
10. (ii)
11. (iv)
12. (iv)
13. (iii)
14. (iii)
15. (iii)
16. (ii)
17. (ii)
18. (i)
19. (iv)
20. (iv)
21. (ii)
22. (i)
23. (ii)
24. (iv)
25. (i)
26. (ii)
27. (iv)
28. (iv)
29. (i)
30. (iii)
31. (i)
32. (iv)
33. (iii)
34. (iv)
35. (iv)
36. (i)
37. (ii)
38. (i)
39. (iv)
40. (iv)
41. (iv)
42. (i)
43. (iii)
44. (i)
45. (iv)
46. (ii)
47. (iv)
48. (ii)
49. (iv)
50. (ii)
51. (ii)
52. (ii)
53. (ii)
54. (i)
55. (iv)
56. (iii)
57. (ii)
58. (iii)
59. (ii)
60. (iii)
61. (iii)
62. (iii)
63. (iv)
64. (iii)
7
Fourier Series
and Fourier Transform
PART I: FOURIER SERIES
Introduction
In 1807, the French mathematician Joseph Fourier (1768–1830) submitted a paper to the Academy of
Sciences in Paris. In it he presented a mathematical description of problems involving heat conduction.
Although the paper was at first rejected, it contained ideas that would develop into an important area
of mathematics named in his honour, Fourier analysis. One surprising ramification of Fourier’s work was
that many familiar functions can be expanded in infinite series and integrals involving trigonometric
functions. The idea today is important in modeling many phenomena in physics and engineering.
In this chapter, in the first part, we will discuss the basic concepts of Fourier series. Then we will apply
this concept to find the steady-state response of an electric circuit subject to a periodic excitation. A
function of time f(t) is said to be periodic if f(t) f (t nT ); where, n is a positive integer and T is the
period. Thus, a periodic function repeats itself every T second.
v(t )
V
2T
(a)
Fig. 7.1 Periodic function
T
0
T
2T
3T
4T
t
(b)
In the second part of this chapter, we will learn about another transform method, namely Fourier
transform, which is used to find the steady-state response of a network to aperiodic excitation.
414
Network Analysis and Synthesis
7.1
DEFINITION OF FOURIER SERIES
French mathematician J B J Fourier first studied the periodic function in 1822 and published his theorem
which states that
“Any arbitrary periodic function can be represented by an infinite series of sinusoids of harmonically
related frequencies.” This infinite series is known as Fourier series.
Thus, if f (t) is a periodic function then the Fourier series is
f (t ) = a0 + a1 cos t + a2 cos 2 t + ⋅⋅⋅+ an cos n t + ⋅⋅⋅+ b1 sin t + b2 sin 2 t + ⋅⋅⋅+ bn sin n t + ⋅⋅⋅
()
∞
(
∴ f t = a0 + ∑ an cos n t + bn sin n t
where,
7.2
n=1
)
(7.1)
2
T
th
n — the n harmonic of fundamental frequency
a0, an, bn—the Fourier coefficients
— the fundamental frequency
DIRICHLET’S CONDITIONS
The conditions under which a periodic function f (t) can be expanded in a convergent Fourier series, are
known as Dirichlet’s conditions.
These are as follows:
(i) f (t) is a single-valued function.
(ii) f (t) has a finite number of discontinuities in each period, T.
(iii) f (t) has a finite number of maxima & minima in each period, T.
T
T
0
0
(iv) The integral, ∫ f (t ) dt exists and is finite or in other way, ∫ ⎡⎣ f (t ) ⎤⎦ dt < ∞.
T
Note
2
2
If f (t) is current or voltage, ∫ ⎡⎣ f ( t ) ⎤⎦ dt represents energy which would be supplied by the source in one cycle.
0
That means the energy in the waveform for each cycle must be finite. All physical waveforms would, of course,
satisfy this criterion.
Therefore, in practical engineering problems, it is not necessary to check whether a function satisfies the
Dirichlet condition.
7.3
CONVERGENCE OF FOURIER SERIES
There are three factors involved in the convergence of Fourier series, viz.,
• Can we find the co-efficients, an and bn?
• Can we sum the resulting series for f (t)?
• Can we approximate f ((t) with a small number of terms of the series?
415
Fourier Series and Fourier Transform
Weak dirichlet conditions These are the conditions for being able to find an and bn. These conditions
do not restrict f (t) to be finite. In particular, they allow impulses to be present which have infinite value but
finite areas under them.
Whenever f (t) is infinite, Fourier series will not converge at that point. Therefore, if f (t) satisfies weak
Dirichlet condition it is possible to find an and bn, but it may not be possible to sum the series
Strong Dirichlet Conditions These are conditions for the convergence
of f (t) everywhere. For these, f (t) must be finite. If a function satisfies these
conditions, it is possible to find the series and to find its sum.
f (t )
T/2
For example, consider f (t) as the square wave.
f (t ) =
f −1 (t ) =
f ′(t ) =
⎞
4⎛
cos 3t cos 5t cos 7t
⎜⎝ cos t − 3 + 5 − 7 + ⋅⋅⋅⎟⎠
1
f (t )1
)
0
1
(
t
1
T
− sin t + sin 3t − sin 5t + sin 7t − ⋅⋅⋅
Functions f (t) and f 1(t) satisfies strong D. conditions; and the series for
them are uniformly convergent. But, f (t) satisfies only week D. conditions
and the series for it is not convergent at point t T/2, 3T/2, 5T/2, …
7.4
T
f (t)
1
⎞
4⎛
sin 3t sin 5t sin 7t
⎜⎝ sin t − 32 + 52 − 72 + ⋅⋅⋅⎟⎠
4
t
0
T/2
2T
t
Fig. 7.2 Illustration of
strong dirichlet’s condition
FOURIER ANALYSIS
This involves two operations:
1. The evaluation of the coefficient a0, an and bn.
2. Truncation of the infinite series after a finite number of terms so that f (t) is represented within allowable error.
7.4.1 Evaluation of Fourier Coefficients
∞
(
f (t ) = a0 + ∑ an cos n t + bn sin n t
n=1
)
(7.2)
From (7.2),
T
T
∞ T
0
0
n=1 0
)
(
∫ f (t )dt = a0 ∫ dt + ∑ ∫ an cos n t + bn sin n t dt = a0T
⎧⎪
⎨
⎩⎪
t0 + T
t0 + T
t0
t0
∫ sin m tdt = 0 for all m; and
∴ a0 =
⎫⎪
∫ cos n tdt = 0 for all n;⎬
1 T
f (t )dt
T ∫0
⎭⎪
(7.3)
416
Network Analysis and Synthesis
This shows that a0 is the average value of f (t) over a period; therefore, it is called the dc value of the signal.
Now from Eq. (7.2),
T
∞ T
T
∫ f (t )cos k tdt = ∫ a cos k tdt + ∑ ∫ ( a cos k t cos n t + b cos k t sin n t )dt = 0 + a 2 + 0
0
0
n=1 0
0
⎧
⎪
⎪
⎨
⎪
⎪
⎩
T
n
n
k
t0 + T
t0 + T
⎫
t0
t0
⎬
⎪
T
=
for n = m ⎪
⎭
2
∫ sin n t sin m tdt = 0 for m ≠ n and ∫ cos n t cos m tdt = 0 for n ≠ m ⎪⎪
T
=
for n = m
2
T
∴ ak =
T
T
0
0
2
f (t )cos k tdt
T ∫0
∞ T
(7.4)
)
(
Again from Eq. (7.2), ∫ f (t )sin k tdt = ∫ a0 sin k tdt + ∑ ∫ an sin k t cos n t + bn sin k t sin n t dt = 0 + 0 + bk
n=1 0
T
2
T
∴ bk =
Example 7.1
2
f (t )sin k tdt
T ∫0
(7.5)
For the periodic waveform shown in Fig. 7.3, find the Fourier series expansion.
Solution Here, v(t)
V, for 0 t T/2
0, for T/2 t T
T
v (t )
T
V
1
1 2
V
a0 = ∫ v (t )dt = ∫ Vdt =
2
T0
T 0
0
Fig. 7.3 Periodic function
of Example 7.1
T
T
⎛ 2 ⎞
2
2 2
an = ∫ v (t )cos n tdt = ∫ V cos ⎜ n ⎟ ddt 0
T0
T 0
⎝ T ⎠
T
T
⎛ 2 ⎞
2
2 2
V
bn = ∫ v (t )sin n tdt = ∫ V sin ⎜ n ⎟ dt =
1 − cos n
T0
T 0
n
⎝ T ⎠
(
and,
T/2 T 3T/2
); n = ±1, ± 2, ± 3, ⋅⋅⋅
= 0; for even n
V
; for odd n
=
n
⎡1 2
⎤
2
2
So, the Fourier series of the square wave is given as v (t ) = V ⎢ + sin t +
sin 3 t +
sin 5 t + ⋅⋅⋅⎥
3
5
⎣2
⎦
7.4.2 EXPONENTIAL FORM OF FOURIER SERIES
∞
(
We have the trigonometric Fourier series, f (t ) = a0 + ∑ an cos n t + bn sin n t
n=1
)
t
417
Fourier Series and Fourier Transform
We know that, sin n t =
e jn t − e − jn t
e jn t + e − jn t
and cosn t =
2j
2
Thus,
) (
(
)
⎡ e jn t + e − jn t
e jn t − e − jn t ⎤
⎢
⎥
f (t ) = a0 + ∑ an
+ bn
2
2j
⎥
n=1 ⎢
⎣
⎦
∞ ⎡⎛
a − jbn ⎞ jn t ⎛ an + jbn ⎞ − jn t ⎤
= a0 + ∑ ⎢⎜ n
⎥
⎟ e +⎜ 2 ⎟ e
⎠
⎝
n=1 ⎢
⎥⎦
⎣⎝ 2 ⎠
∞
⎤
⎛
b ⎞
b ⎞
1 ⎡⎛
= a0 + ∑ ⎢⎜ an + n ⎟ e jn t + ⎜ an − n ⎟ e − jn t ⎥
j⎠
j⎠
⎝
n=1 2 ⎢
⎥⎦
⎣⎝
∞
⎛ a − jbn ⎞
⎛ a + jbn ⎞
C0 = a0 , Cn = ⎜ n
and Cn* ( or C− n ) = ⎜ n
⎟
⎟
⎝ 2 ⎠
⎝ 2 ⎠
Let,
∞
f (t ) = C0 + ∑ ⎡⎣Cn e jn t + C− n e − jn t ⎤⎦
Thus the series becomes,
n=1
∞
f (t ) = C0 + ∑ Cn e jn t
or,
(7.6)
n=− ∞
This is the exponential form of the Fourier series.
Now,
C n=
T
⎤ 1T
an − jbn 1 ⎡ 2 T
2
= ⎢ ∫ f (t )cos n tdt − j ∫ f (t )sin n tdt ⎥ = ∫ f (t ) cos n t − j sin n t dt
2
2 ⎢⎣ T 0
T0
⎥⎦ T 0
Cn =
1
f (t )e − jn t dt
T ∫0
)
(
T
Thus,
(7.7)
This equation is valid for both positive, negative and zero values of n.
Example 7.2 For the square wave shown in Example 7.1, find the exponential Fourier series.
Solution f (t)
v(t) V, for 0 t
0, for T/2 t T
T/2
T
So,
Cn =
1
1 T
f (t )e − jn t dt = ∫ 2 Ve − jn t dt
∫
T0
T 0
n
V
1 2
0, C0 = ∫ Vdt =
T 0
2
T
For
T
For
n
0
1 2
V 1 ⎡ − jn T 2 ⎤ jV
⎡ e − jn − 1⎤
Cn = ∫ Ve − jn t dt =
e
− 1⎥ =
⎦
T 0
T − jn ⎢⎣
⎦ 2 n⎣
(since T
2 )
418
Network Analysis and Synthesis
Cn = 0 for even n
Or,
=−
jV
for odd n
n
Thus, the exponential Fourier series becomes,
jV − j 5 t jV − j 3 t jV − j t V jV j t jV j 3 t jV j 5 t
e −
e
e
v (t ) = ⋅⋅⋅+
e
+
e
+
e
+ −
−
− ⋅⋅⋅
5
3
2
3
5
7.4.3 Amplitude and Phase Spectrum
From the trigonometric Fourier series,
∞
)
(
∞
(
f (t ) = a0 + ∑ an cos n t + bn sin n t = A0 + ∑ An cos n t −
n=1
where,
A0 = a0 , An = an 2 + bn 2 ;
n
n=1
n
)
⎛b ⎞
= tan −1 ⎜ n ⎟
⎝ an ⎠
Also, for exponential form, Cn is complex and we may write it as,
Cn = Cn e n and Cn =
j
The quantities An and
n
A
1
an 2 + bn 2 = n and
2
2
n
⎛ b ⎞
= tan −1 ⎜ − n ⎟
⎝ an ⎠
are called the amplitude and the phase of the nth harmonic, respectively.
• Variation of An with n (or n ) is known as the amplitude spectrum or frequency spectrum.
• Variation of n with n (or n ) is known as the phase-spectrum of the signal.
As both An and n occurs at discrete values of the frequency, i.e., n 1, 2, 3, etc., these spectra we called
line spectra.
A
Since Cn = n ; there is a scale factor of ½ for the amplitude spectrum for exponential form of the Fourier
2
series compared to the trigonometric form for all lines except the one for n 0. Also, in the case of exponential form spectral lines one drawn for both for positive and negative values of n.
Example 7.3 For the square wave shown in Example 7.1, draw the amplitude and phase spectra.
Solution From the results of Example 7.1, we have,
⎡1 2
⎤
2
2
v (t ) = V ⎢ + sin t +
sin 3 t +
sin 5 t + ⋅⋅⋅ ⎥
3
5
⎣2
⎦
Amplitude
V/2
0
(a)
1 2 3
4 n (or n )
Amplitude spectrum
Phase
2V
2V
V
∠90 ; V2 = 0; V3 =
∠90
Magnitudes V0 = ∠0 ; V1 =
2
3
[since the cosine components are all zero, the phase angle will be
⎛b ⎞
tan ⎜ n ⎟ = tan −1 ∞ = 90° ]
⎝ 0⎠
−1
( )
So, the line spectra are shown in Fig. 7.4.
/2
0
1 2
3
4
5 n (or n )
(b) Phase spectrum
Fig. 7.4 Amplitude and phase
spectra of Example 7.3
419
Fourier Series and Fourier Transform
Significance for Line Spectra The amplitude-spectrum renders valuable information as to where to
truncate the infinite series and yet maintain a good approximation to the original waveform.
7.4.4 Effective Value of a Periodic Function
The effective (or rms ) value of a periodic function f (t) is defined as
Feff ( Frms ) =
T
T
∞
2
1
1 ⎡
⎡
⎤
f
(
t
)
dt
=
A
+
∑ A cos n t −
⎢
⎦
T ∫0 ⎣
T ∫0 ⎣ 0 n=1 n
⎤
(
∞ ⎛
A ⎞
Feff ( Frms ) = A0 2 + ∑ ⎜ n ⎟
n=1 ⎝
2⎠
n
2
⎡
∞
⎤
)⎥ dt = T1 ⎢ A T + ∑ A T2 ⎥
⎦
⎣
2
0
n=1
2
n
⎦
2
(7.8)
This shows that the effective value of a periodic function is the square root of the effective values of the
harmonic components and the square of the dc value.
7.5
WAVEFORM SYMMETRY
There are few methods by which the evaluation of Fourier coefficients is simplified by symmetry consideration.
These methods reduce the amount of labour involved in finding out the coefficients.
T
⎡ 0
⎤
T
2
1
1⎢
⎥
+
Now,
a0 = ∫ f (t )dt =
f
(
t
)
dt
f
(
t
)
dt
∫0
⎥
T0
T ⎢ −T∫
⎣ 2
⎦
Putting t
x in the first integrand and t x in the second integrand, we get
⎡T
⎤
1⎢ 2
⎥
⎡
⎤
a0 =
f
(
x
)
f
(
x
)
dx
+
−
⎦ ⎥
T ⎢ ∫0 ⎣
⎣
⎦
⎡T
⎤
0
T
2
2⎢ 2
2
Now,
an = ∫ f (t )cos n tdt =
f (t )cos n tdt + ∫ f (t ) cosn tdt ⎥ = ⎡⎣ I1 + I 2 ⎤⎦
∫
⎢
⎥
T0
T 0
T
−T
⎣
⎦
2
Since the variable ‘t’ in I1 and I2 integrals is dummy variable, let x t in I1 and x
t in I2.
T
⎡T
⎤
2
2⎢ 2
⎥
∴ an =
f
(
x
)cos
n
xdx
−
f
(
−
x
)cos
n
x
(
−
dx
)
∫0
⎥
T ⎢ ∫0
⎣
⎦
T
Thus,
2 2
an = ∫ ⎡⎣ f ( x ) + f ( − x ) ⎤⎦ cos n xdx
T 0
Similarly,
2 2
bn = ∫ ⎡⎣ f ( x ) − f ( − x ) ⎤⎦ sin n xdx
T 0
T
The following symmetries are considered
1. Odd or rotation symmetry,
2. Even or mirror symmetry,
420
Network Analysis and Synthesis
3. Half- wave, or alternation symmetry, and
4. Quarter-wave symmetry.
V (t )
V
Odd symmetry A function f (x) is said to be odd if,
f (x)
f ( x)
Hence, for odd functions a0
0
and
an
T/2
0 and
T
0 T/4
V
T/2
T
T
t
4 2
f ( x )sin n x
Fig. 7.5 Odd function
T ∫0
Thus, the Fourier series expansion of an odd function contains only the sine terms, the constant and the
cosine terms being zero.
bn =
Even symmetry A function f (x) is said to be even, if
f (x)
f ( x)
T
∴ a0 =
f ( t)
2 2
f ( x )dx
T ∫0
V
T
T/2 0
4 2
an = ∫ f ( x )cos n xdx
T 0
t
T/2
V
Fig. 7.6 Even function
and bn 0
Thus, the Fourier series expansion of an even periodic function contains only the cosine terms plus a
constant, all sine terms being zero.
Half-wave or alternation symmetry A periodic function f (t) is said to have half-wave symmetry if it
satisfies the condition
f (t)
f (t
T
),
2
where T is the time period of the function
T
⎡ 0
⎤
2
1⎢
⎥ = 1 ⎡I + I ⎤
(
)
(
)
∴a0 =
f
t
dt
+
f
t
dt
∫0
⎥ T ⎣ 1 2⎦
T ⎢ −T∫
⎣ 2
⎦
For I1, let x
(t
T/2); so, f (t)
f (x
T/2)
f (x)
T
0
2
and
dt
x
T/2
0
t
0
T/2
dx
T
2
∴ I1 = ∫ f (t )dt = ∫ − f ( x )dx = − ∫ f ( x )dx
−T
2
0
0
T
T
⎡ T
⎤
⎡T
⎤
2
2
1⎢ 2
1⎢ 2
⎥
⎥=0
(
∴ a0 = − ∫ f ( x )dx + ∫ f (t )dt = =
f
x
)
dx
−
f
(
x
)
dx
∫0
⎥ T ⎢ ∫0
⎥
T⎢ 0
0
⎣
⎦
⎣
⎦
T
⎡T
⎤
⎡ 0
⎤
2
2⎢ 2
2
⎥= ⎢
⎥ = 2 ⎡I + I ⎤
∴ an =
f
(
t
)cos
n
tdt
f
(
t
)cos
n
tdt
+
f
(
t
)cos
n
tdt
∫0
⎥ T ⎢ T∫
⎥ T ⎣ 1 2⎦
T ⎢ −T∫
⎣ 2
⎦
⎣− 2
⎦
421
Fourier Series and Fourier Transform
Again putting x
(t
T/2) and following the same procedure,
T
0
T
2
I1 = ∫ f (t )cos n tdt = ∫ − f ( x )cos n
−T
T
0
2
T
2
( x − T 2 )dx = ∫ − f ( x)cos(n x − n )dx
2
0
2
= ∫ − f ( x )cos n cos n xdx = ∫ − f (t )cos n cos n tdt
0
0
T
an =
(
2
1 − cos n
T
) ∫ f (t )cos n tdt
2
0
0; for even n, and
T
4 2
= ∫ f (t )cos n tdt , for odd n.
T 0
Similarly, bn
0, for even n; and
T
4 2
= ∫ f (t )sin n tdt , for odd n.
T 0
Thus, the Fourier series expansion of a periodic function having half-wave symmetry contains only odd harmonics, the constant term being zero.
Quarter–wave symmetry The symmetry may
be regarded as a combination of the first three kinds
of symmetry provided that the origin is properly
chosen.
For Fig. 7.7 (a), the wave has alternation and odd
symmetry; thus the Fourier series consists of odd sine
terms only.
t
t
Fig. 7.7 (a) sin t:
Fig. 7.7 (b) cos t:
combination of half-wave combination of half-wave
and odd symmetry
and even symmetry
T
a0
0; an
0; and
8 4
bn = ∫ f (t )sin n tdt , n being odd only.
T 0
For Fig. 7.7 (b), the origin, having chosen one quarter cycle away, as in Fig. 7.7 (a), the wave has alternation
and even symmetry; thus the Fourier series consists of odd cosine terms only.
T
a0
Note
0; bn
0; and
8 4
an = ∫ f (t )cos n tdt , n being odd only.
T 0
(i) The sum or product of two or more even functions is an even function, and with the addition of a constant,
the even nature of the function is still preserved.
(ii) The sum of two or more odd functions is an odd function, but the addition of a constant removes the odd
nature of the function. The product of two odd functions is an even function.
422
Network Analysis and Synthesis
7.6
TRUNCATING FOURIER SERIES
When a periodic function is represented by a Fourier series, the series is truncated after a finite number of terms.
So, the periodic function is approximated by a trigonometric series of (2N 1) terms as,
N
(
S N (t ) = a0 + ∑ an cos n t + bn sin n t
n=1
)
(7.9)
such that the coefficients a0, an and bn are chosen to give the least mean square error.
The truncation error is
() ()
()
eN t = f t − S N t
(7.10)
So, the mean square error or figure of merit or the cost criterion for optimal minimal error is
T
EN = eN2 (t ) =
2
1
⎡⎣ eN (t ) ⎤⎦ dt
∫
T0
(7.11)
where, EN is a function of a0 , an and bn, but not of ‘t’.
Example 7.4 Show that the mean square error is a minimum if the coefficients in the approximated trigonometric series SN (t) are the Fourier coefficients.
Solution In order to make ‘EN’ minimum, the necessary conditions are
and
∂EN
= 0 for n = 0, 1, 2
∂an
(7.12a)
∂EN
= 0 for n = 1, 2
∂bn
(7.12b)
These two equations give (2N 1) equations from which (N
number of bn for n 1, 2, …, N can be determined.
From equations (7.11) and (7.12a)
1) number of an for n
0, 1, 2, …, N and ‘N’
T
∂EN 2 T
∂e (t )
2
= ∫ eN (t ) N dt = ∫ ⎡⎣ f (t ) − S N (t )⎤⎤⎦ cos n tdt = 0
∂an T 0
∂an
T0
T
T
T
N
⎡
⎤
T
f
(
t
)cos
n
tdt
=
S
(
t
)cos
n
tdt
=
a
+
a
cos
n
t
+
b
si
n
n
t
cos
n
tdt
an cos 2 n tdt = an
=
⎥
n
n
∫0
∫0 N
∫0 ⎢⎣ 0 ∑
∫
2
n=1
⎦
0
T
or,
)
(
T
or,
∴ an =
2
f (t )cos n tdt
T ∫0
(n
0, 1, 2, …, N)
T
Similarly, from Eq. (7.12b), we get
∴ bn =
2
f (t )sin n tdt
T ∫0
(n
1, 2, …, N)
Therefore, it is proved that a Fourier series with a finite number of terms represents the best approximation
for a given periodic function by any trigonometric series with the same number of terms.
However, there is no analytical method for the evaluation of estimation of error due to truncation of
infinite series; i.e., we cannot predict the number of minimum terms to be retained in the series within a
423
Fourier Series and Fourier Transform
prescribed accuracy. The minimization of error is done by trial and error method, using more terms until
specifications are met.
It is observed that when a periodic waveform is truncated by a Fourier series with a finite number of terms,
there is a considerable amount of error near the points of discntinuity of the wave. The amount of error is
decreased with the increase of number of terms included in the truncated Fourier series. This phenomenon is
known as Gibb’s phenomenon.
For example, we consider a square wave as shown in Fig. 7.8. A general approximation of the wave can be
obtained by taking more and more number of terms of the Fourier series expansion.
Figure 7.8 shows the wave-shapes taking the first term, first 5 terms, first 11 terms and first 49 terms,
respectively. The rate of oscillation of ripples increases near the points of discontinuity as the contribution
of more harmonics is taken into consideration. The wave-shape tends to perfectly match the given waveform
when a large number of harmonics is considered.
If we consider a point where the waveform f (t) is discontinuous, with different limits to the right and left
of as f ( ) and f ( ), respectively then the value of the function at will be,
1
K=1
0
1
t
1
K=5
0
t
1
1
1
K = 11
0
t
K = 49
0
t
1
1
Fig. 7.8 Fourier series approximation of square wave; number of terms in Fourier sum is
indicated as K in each plot
( )=
f
( + ) + f ( − ) or, f
( ) ( ) ( +)− f ( )
−f − =f
2
The truncated Fourier series must pass through these three points, f ( ), f (
tion of the wave.
f
) and f (
8
sin ␻t , i.e., the first
␲2
term in the Fourier series, find the mean square error.
Example 7.5
f (t)
V
If f (t) is approximated by
Solution Truncation error, eN = f (t ) −
Mean square error,
T
8
2
) for correct representa-
T/2
T
0 T/4
V
T/2
sin t
Fig. 7.9 Waveform of Example 7.5
2
T
2
⎤
⎤
1
4 4⎡
8
4 4 ⎡ 4t 8
EN = eN = ∫ eN 2 (t )dt = ∫ ⎢ f (t ) − 2 sin t ⎥ dt = ∫ ⎢ − 2 sin t ⎥ dt = 0.0047
T0
T 0⎣
T
T
⎦
⎦
0 ⎣
T
2
T
t
424
Network Analysis and Synthesis
7.7
STEADY-STATE RESPONSE OF NETWORK TO PERIODIC SIGNALS
∞
The voltage (periodic) is
(
v (t ) = A0 + ∑ An cos n t −
n=1
n
)
We want to find out the steady state current, i(t). Phasors corresponding to terms in right-hand side are,
V0 = A0 e j 0 and Vn = An e
−j n
Let, Z( j ) Impedance phasor of the network at any frequency .
So, the current phasors are,
V
A e j0
I0 = 0 = 0
= I0 e j 0
Z ( j 0) Z ( j 0)
−j
Vn
Ae n
−j
In =
= 0
= In e n
Z( j ) Z( j )
i(t ) = I 0 + I1 + I 2 + ⋅⋅⋅
By superposition principle, the net current phasor is
∞
So, transforming from frequency domain to time domain,
(
i(t ) = I 0 + ∑ I n cos n t −
n=1
n
)
7.7.1 Average Power Calculation
∞
n=1
Here,
(
v (t ) = V0 + ∑Vn cos n t −
Let,
∞
n
) and i(t ) = I + ∑ I cos( n t − )
0
n=1
n
n
V0
Vn
dc voltage component
the amplitude of the nth harmonic voltage,
the phase angle of the nth harmonic voltage,
n
I0 dc current component,
In the amplitude of the nth harmonic current,
the phase angle of the nth harmonic current
n
Instantaneous power, P(t) v(t) i(t)
Average power,
Pav =
T
T
∞
1
1 ⎡⎛
v
(
t
)
i
(
t
)
=
V
+
⎢
∑V cos n t −
T ∫0
T ∫0 ⎣⎜⎝ 0 n=1 n
∞ T
(
Pav = V0 I 0 + ∑ ∫Vn I n cos n t −
or,
n=1 0
n
∞
VI
Pav = V0 I 0 + ∑ n n cos
n=1 2
or,
7.8
⎞⎛
(
n
⎞⎤
∞
)⎟⎠ ⎜⎝ I + ∑ I cos( n t − )⎟⎠ ⎥ dt
0
n=1
n
n
⎦
)cos( n t − )dt
n
(
n
−
n
)
(7.13)
STEPS FOR APPLICATION OF FOURIER SERIES TO CIRCUIT ANALYSIS
1. Fourier series of the given periodic excitation function is obtained.
2. The circuit elements are transformed from time domain to frequency domain (i.e., R → R, L → j nL,
1
C→
for nth harmonic).
j nC
425
Fourier Series and Fourier Transform
3. The Fourier series of the dc and ac components of the response are calculated.
4. Using superposition, the Fourier series of the response is obtained by summing up the individual dc
and ac response components.
7.9
POWER SPECTRUM
It is the distribution of the average power over the different frequency components.
Let Pn the average power for the nth harmonic component.
Note
Pn is always positive so that only a magnitude spectrum is possible.
Another form of a line spectrum for power is also possible [Fig. 7.10 (b)]; obtained by assuming half of Pn to the
positive frequency n and half to the negative frequency.
Pav
Pav
P1
P0
0
P2
P3
P0
P4
v 2v 3v 4v v
3v 2v v 0
v 2v 3v v
(a) Power
spectrum for
positive
(b) Power spectrum
for both positive and
netgative
Fig. 7.10 Power spectra
PART II: FOURIER TRANSFORM
Introduction
The Fourier series representation of a period function describes the function in the frequency domain in
terms of amplitude and phase spectra. The Fourier transform extends this frequency domain description
to functions that are not periodic.
Fourier transform is a powerful tool in the study of power spectra, correlation functions, noise and
other advanced problems
7.10
DEFINITION OF FOURIER TRANSFORM
The Fourier transform or the Fourier integral of a function f (t) is denoted by F ( j ) and is defined by
∞
( ) = F ⎡⎣ f (t )⎤⎦ = ∫ f (t )e
F j
−j t
dt
(7.14)
−∞
and the inverse Fourier transform is defined by
∞
∞
1
j t
j2 f
=
f t = F −1 ⎡⎣ F j ⎤⎦ =
F
(
j
)
e
d
∫ F j 2 f e df
2 −∫∞
−∞
()
( )
Equations (7.14) & (7.15) form the Fourier transform pair.
(
)
(7.15)
426
Network Analysis and Synthesis
Explanation
∞
f (t ) = ∑ Cn e jn t
Consider the exponential Fourier series,
(7.6)
−∞
T
where,
1 2
Cn = ∫ f (t )e − jn t dt
T −T
(7.7)
2
If the period T becomes infinite, the function does not repeat itself and becomes aperiodic or non-periodic.
2
= n +1 − n = =
So, the interval between adjacent harmonic frequencies is
T
1
or,
(7.16)
=
=
T 2
2
(
As T → ,
)
→ d and the frequency goes from a discrete variable over to a continuous variable.
1 d
→
and n →
(7.17)
T
2
∞
CnT → ∫ f (t )e − j t dt
From (7.7) and (7.17),
−∞
∞
F ( j ) = F ⎡⎣ f (t ) ⎤⎦ = ∫ f (t )e − j t dt
This is the Fourier transform of f (t) i.e., F ( j ).
−∞
So, from Eq. (7.6),
∞
⎛ 1⎞
f (t ) = ∑ CnT e jn t ⎜ ⎟
⎝T⎠
−∞
As T → , CnT → F ( j ), n →
(
)
and
1
d
→
T
2
(7.18)
and ∑ → ∫ (summation approaches integration). Thus,
from (7.18),
∞
f (t ) =
Spectra Let, F ( j )
1
F ( j )e j t d
2 −∫∞
F (j ) ej␾ (␻)
The variation of F (j ) with ‘ ’ is referred to as the amplitude spectrum.
The variation of ␾ ( ) with ‘ ’ is returned to as the phase-spectrum.
Since F (j ) is a continuous function, the corresponding amplitude and phase spectra are continuous spectra.
7.11
CONVERGENCE OF FOURIER TRANSFORM
When f (t) is a singlevalued function and is different from zero over an infinite interval of time, the behavior of
f (t) as t →
determines the convergence of the Fourier transform.
∞
The Fourier transform will exist if ∫ f (t ) dt < ∞
−∞
427
Fourier Series and Fourier Transform
7.12
FOURIER TRANSFORM OF SOME FUNCTIONS
f (t) ⴝ Aeⴚat u(t), a > 0 Fourier transform will exist if a
∞
∴ F ( j ) = F ⎡⎣ f (t ) ⎤⎦ = ∫ f (t )e
−j t
−∞
∞
0
dt = A ∫ e e
− at
e( )
dt = A
− a+ j
− a+ j
−j t
0
(
Phase,
⎛ ⎞
( j ) = − tan −1 ⎜ ⎟
⎝ a⎠
a2 +
0
F( jv)
⎤ = Ke − a t e − j t dt = Ke ( a− j )t dt + Ke −( a+ j )t dt
∫
∫0
⎦ −∫∞
−∞
0
2 Ka
K
K
+
= 2
a− j
a+ j
a + 2
Thus the Fourier transform of the double exponential function has zero phase
for all values of ␻ and the magnitude spectrum is shown in Fig. 7.11
time,t
(a) Double exponential
function
∞
0
=
Note
K
2
f (t) ⴝ Keⴚa t , for all Values of t (Double Exponential Function)
−a t
0
A
a+ j
f (t )
F( j ) =
F ( j ) = F ⎡ Ke
⎣
)
=
A
Amplitude,
∞
∞
t
2K
a
v
(b) Fourier transform
Fig. 7.11 Double
exponential function and its
Fourier transform
There are some important functions which do not have Fourier transforms in a strict sense; because they do
∞
not satisfy the Dirichlet’s condition, i.e., ∫ f (t ) dt is infinite (such as, the step function and sinusoidal function).
−∞
However, the Fourier transform of these function are evaluated by approximating these functions in time
domain as the limiting value of another function which possesses Fourier transform.
Fourier transform of some constant, K; for all values of t Here, we can approximate the constant as
f (t ) = Lt ⎡ Ke
a →0 ⎣
−a t
∴ F ⎡⎣ K ⎤⎦ = 0; for
∞
⎤ ∴ F ⎡ K ⎤ = Lt Ke − a t e − j t dt = Lt 2 Ka
⎣ ⎦ a →0 ∫
⎦
a →0 a 2 + 2
−∞
≠0
= ∞ ; for
=0
[by L Hospital’s rule, i.e., differentiating both numerator and denominator with respect to ‘a’]
Thus, F [K ] is an impulse function at
0. The strength (amplitude) of the impulse function is
obtained as
∞
∞
2 Ka
∫ F ⎡⎣ K ⎤⎦ d = ∫ a +
2
−∞
2
d =2 K
−∞
∴ F ⎡⎣ K ⎤⎦ = 2 K
( )
428
Network Analysis and Synthesis
Thus, F [K ] is an impulse function at ␻ 0. The strength
(amplitude) of the impuse function is obtained as,
∞
f (t)
2pKd( v)
K
∞
2 Ka
d =2 K
2
a
+ 2
−∞
∫ F [ K ]d = ∫
−∞
F ( jv)
0
time,t
0
v
(a) Constant K
∴ F [ K ]d = 2 K ( )
Fig. 7.12
Hence, Fourier transform of a constant K is an
impulse of magnitude 2␲K as shown in Fi.g 7.12
(b) Magnitude spectrum
of constant K
Constant K and its magnitude
Unit impulse function or dirac delta function, ␦(t) Some problems involve the concept of an impulse,
which may be intuitively thought of as a force of very large magnitude impacting just for an instant.
∞
∴ F ⎡⎣ (t ) ⎤⎦ = ∫ (t )e − j t dt
−∞
We use shifting property of impulse function as explained below.
The product of any arbitrary function f (t) with unit impulse function ␦(t) provides the function ␦(t) to exist only at t 0. Mathematically,
F (j v)
d(t)
∞
∫ f (t ) (t ) = f (t )
0
t =0
Time, t
(a) Impulse function
−∞
This shifting property can also be applied at any instant of time, say t
that we can write,
F ( jv)
t0 , so
1
∞
∫ f (t ) (t − t )dt = f (t ) t =t = f (t0 )
0
−∞
Using this property, we have the Fourier transform of unit impulse function as,
∞
∴ f [ (t )] = ∫ (t )e − j t dt = e 0 = 1
−∞
0
v
(b) Fourier transform of
impulse function
Fig. 7.13 Impulse
function and its Fourier
transform
Thus, Fourier transform of an impuse function is unity as shown in Fig. 7.13
F (t )
Fourier transform of signum function, sgn( t)
Sgn(t)
A signum function is defined as
1
0
for t 0
for t 0
1 for t 0
1
0
Time, t
1
Transform is not
∞
∴ ∫ SSgn(t)dt is infinite, direct evaluation of Fourier transform is not possible.
−∞
Fig. 7.14 (a)
Sgn(t)
Therefore, the given function has to be expressed as a limiting case of some other function and then the
Fourier transform is computed. Let the Sgn(t) be multiplied by eⴚa t and a → 0.
∞
F ⎡⎣Sgn (t ) ⎤⎦ = Lim ∫ e
a →0
−∞
−a t
∞
⎡ 0 a− j t
⎤
⎡ −1
1 ⎤
− a+ j t
Sgn(t )e − j t dt = Lim ⎢ − ∫ e ( ) dt + ∫ e ( ) dt ⎥ = Lim ⎢
+
a →0
a →0 a − j
a + j ⎥⎦
⎣
⎢⎣ − ∞
⎥⎦
0
429
Fourier Series and Fourier Transform
F ( jv)
2
j
Fig. 7.14 shows the magnitude and phase spectrum of Signum function.
F ⎡⎣ Sgn(t ) ⎤⎦ =
or,
2
jv
0
Fourier Transform of Unit Step Function, u(t)
u(t)
1 for t
0 for t
time,t
(b) Magnitude spectrum of
Sgn(t)
0
0
F( jv)
∞
Since ∫ u(t )dt is infinite, direct evaluation of Fourier transform is imposible.
p
2
−∞
p
2
1 1
Let, u(t ) = + Sgn(t)
2 2
()
( ) + j1
F ⎡⎣ u t ⎤⎦ =
F ( j␻)
Thus, the amplitude of unit step function u(t) in Frequency domain will be a
combination of rectangular hyperbola and impulse function (of strength ␲ at
␻ 0) as shown in Fig. 7.15.
7.13
␣F{f (t)
␤g(t)}
␤F{g(t)}
provided the Fourier transform of f (t) and g(t) exist.
Scaling If, F{f (t)}
F (␻) and c ⑀ R, then
{
Time shifting
If F{f (t)} F(␻) F(␻) and t0 ⑀ R, then
F{f (t t0)} e jvt0 F(␻)
Proof
∞
{ (
F f t − t0
)} = ∫ f (t − t )e
0
−j t
dt = e
∞
− j t0
−∞
−∞
Frequency shifting
If F{f (t)}
F
F (␻) and
( − ) = F {e
Proof
0
{
F e
j 0
j 0
␻ ⑀ R, then
}
f (t )
}
∞
f (t ) = ∫ e
−∞
j 0t
f (t )dt =F
∫ f ( u )e
( − )
0
−j u
⎛
} 1c F ⎜⎝ c ⎞⎟⎠
F cf (t ) =
du
␣F (␻)
1
j␻
␲
0
time,␻
Fig. 7.15 Magnitude
spectrum of unit step function
PROPERTIES OF FOURIER TRANSFORMS
Linearity If ␣, ␤ ⑀ C then F{␣f (t)
v
(c) Phase spectrum of
Sgn(t)
Fig. 7.14 Signum function
and its magnitude spectrum
⎡1⎤
⎡1
⎤
1 2
1
∴ F ⎡⎣ u(t ) ⎤⎦ = F ⎢ ⎥ + F ⎢ Sgn (t ) ⎥ = 2 × ( ) + ×
2 j
2
2
2
⎣ ⎦
⎣
⎦
or,
0
␤G(␻)
430
Network Analysis and Synthesis
Symmetry
If F{f (t)}
Proof
F (␻), then F {F (t)}
2␲f ( ␻)
Use the formula for the inverse Fourier transform
()
{ ( )}
f t = F −1 F
( )
2 f −
Then
∞
1
F
2 −∫∞
=
∞
( )
∞
e j td =
∞
()
()
1
F x e jxt dx
2 −∫∞
{ ( )}
()
= ∫ F x e − jxt d = ∫ F t e − j t dt =F F t
−∞
−∞
Modulation
If F {f (t)}
F (␻) and ␻0 ⑀ R, then
{ ( ) ( t )} = 12 ⎡⎣ F ( + ) + F ( − )⎤⎦
1
F { f ( t )sin ( t )} = ⎡⎣ F ( + ) − F ( − ) ⎤⎦
2
F f t cos
Proof
0
0
0
0
0
0
Use the frequency-shifting theorem to get
{ ( ) ( t )} = 12 ⎡⎣ F {e
F f t cos
0
j 0t
( )}
{
f t +F e
− j 0t
( )}
1
f t ⎤ = ⎡⎣ F
⎦ 2
( + ) + F ( − )⎤⎦
0
Differentiation in time
0
()
Let n ⑀ N, and suppose that f (n) is piecewise continuous. Assume that Lim f ( ) t = 0 , then
k
{
( )} = ( j ) F ( )
F f( ) t
n
{ ( )} = j F ( )
F { f ′′ ( t )} = − F ( )
F f′ t
In particular
2
and
Proof
n
Assume n
1. The general case can be proved by induction.
∞
()
()
−j t
−j t
∫ f ′ t e dt = f t e
−∞
∞
−∞
∞
( )(
−∫ f t −j
−∞
Frequency differentiation
{
In particular
)e
−j t
dt = j F
}
F t n f (t ) = j n F ( )
Let n ⑀ N and suppose that f is piecewise continuous. Then
and
t →∞
{ ( )} = jF ′( )
F {t f ( t )} = − F ′′ ( )
F tf t
2
n
( )
( )
431
Fourier Series and Fourier Transform
Proof
We will prove the theorem for n
∞
F′
1. The argument for larger n is a repetition of this.
( ) = ∫ ⎡⎣ f (t )e
−j t
−∞
∞
{ ( )}
()
⎤dt = − j ∫ ⎡tf t e − j t ⎤dt = − jF tf t
⎦
⎣
⎦
−∞
These properties can be tabulated as follows.
Table 7.1 Properties of Fourier Transforms
∞
∞
Sl No.
Time domain f ( t ) =
1
F ( j )e j t dt
2 −∫∞
Frequency domain F ( j ) = ∫ f ( t )e − j t dt
−∞
1
2
3
4
f (t) real
f (t) even, f (t)
f (t) odd, f (t)
y(t) tn f (t)
F (j )
F (j )
F(j )
5
y(t)
f (at)
6
y(t)
f (t
t0)
Y( j )=e
7
y (t ) =
d n f (t )
dt n
Y( j )= j
8
y (t ) = ∫ f (t )dt
Y( j )=
y (t ) = f (t )e
Y ( j ) = F ⎡⎣ j
f ( t)
f ( t)
Y ( j ) = ( j )n
dnF( j )
d n
1 ⎛j ⎞
Y( j )= F⎜ ⎟, a>0
a ⎝ a ⎠
− j t0
F( j )
( ) F( j )
∞
−∞
9
F* ( j )
F ( j ), F ( j ) is real
F ( j ), F ( j )is imaginary,
j 0t
n
F( j )
j
( − )⎤⎦
0
Example 7.6 Show that when f (t) is an even function of t, its Fourier transform F (j␻) is a function of and is
real; while when f (t) is an odd function of t, its Fourier transform F (j␻) is an odd function of ␻ and is imaginary.
Solution From the definition,
∞
∞
−∞
−∞
)
(
∞
∞
−∞
−∞
F ( j ) = ∫ f (t )e − j t dt = ∫ f (t ) cos t − j sin t dt = ∫ f (t )cos tdt − j ∫ f (t )sin tdt = P( ) + jQ( )
∞
where,
P( ) = ∫ f (t )cos tdt = Even f unction of
i.e., P( ) = P( − )
−∞
∞
and
Q( ) = ∫ f (t )sin tdt = Odd f unction of
−∞
Now,
j
F( j ) = F( j ) e ( )
i.e., Q( ) = −Q( − )
432
Network Analysis and Synthesis
F ( j ) = P1 ( ) + Q 2 ( ) = Even f unction of
( ) ⎤⎥ = Odd f unction of
⎢⎣ ( ) ⎥⎦
⎡Q
( ) = tan ⎢ P
−1
and
When f (t) is an even function
– f (t) cos t is an even function
– f (t) sin t is odd function.
∞
∴P
( ) = 2 ∫ f (t )cos tdt
0
Q
( )=0
( ) = P ( ) = Even and Real
So,
F j
• When f (t) is an odd function
– f (t) cos ␻t is an odd function
– f (t) sin ␻t is an even function
P(␻)
∴Q
and
( )
0
∞
= −2 ∫ f (t )sin tdt
0
( ) = jQ( ) = Odd and Imaginary (Proved)
F j
So,
7.14
ENERGY DENSITY AND PARSEVAL’S THEOREM
This theorem states that the energy content (W) of a waveform (periodic or non-periodic) over the whole
frequency band is
∞
W = ∫ f 2 (t )dt =
−∞
Proof
We have,
∞
∞
∞
⎡ 1 ∞
⎤
j t
W = ∫ f 2 (t )dt = ∫ f (t ) ⋅⎡⎣ f (t )dt ⎤⎦ = ∫ f (t ) ⎢
F
j
)
e
d
(
⎥ dt
∫
⎢⎣ 2 − ∞
⎥⎦
−∞
−∞
−∞
∞
∞
∞
∞
⎡
⎤
2
1
1
1
j t
(
)
(
)d
F
(
j
)
f
(
t
)
e
dt
d
F
j
F
j
d
=
F( j ) d
=
⋅
−
=
⎢∫
⎥
∫
∫
2
2
2 −∫∞
⎢⎣ − ∞
⎥⎦
−∞
−∞
∞
or,
∞
2
1
F( j ) d
∫
2 −∞
W = ∫ f 2 (t ) dt ==
−∞
∞
2
1
F( j ) d
2 −∫∞
(Proved)
433
Fourier Series and Fourier Transform
Note
(i) Since F ( j ) is an even function of ,
∞
W = ∫ f 2 (t )dt ==
1
−∞
(ii) Since
∞
2
∫ F( j ) d
0
2 f, where f is the frequency,
∞
∞
−∞
−∞
∞
2
2
W = ∫ f 2 (t )dt == ∫ F ( j 2 f ) df = 2 ∫ F ( j 2 f ) df
0
The quantity F ( j2 f) 2 df is the energy in an infinitesimal band of frequency df. It represents the energy density
in the frequency domain and has a unit of Joule/Hertz.
Total energy content within the frequency band f1 and f2 is
f2
(
Wb = 2 ∫ F j 2 f
) df
2
f1
For the integration range
to , the total energy is,
− f2
(
Wb = ∫ F j 2 f
− f1
(iii) If f (t) is the voltage across a 11- energy.
f2
) df + ∫ F ( j 2 f ) df
2
2
f1
resistance or current through the same resistance, then Wb is known as
Example 7.7 The current in a 10- resistor is i (t ) =10 e −2 t u (t ) ( A ) . What is the energy associated with the
frequency band 0
2 rad/s?
f (t ) = i(t ) = 10e −2 t u(t )
Solution Here,
∴ F( j ) =
10
2+ j
So, the energy associated with the given frequency band is
W=
7.15
10
2
2
10 100d
103 ⎡ 1 −1 ⎛ ⎞ ⎤ 103 ⎡ ⎤
F
(
j
)
d
=
=
⎢ tan ⎜ ⎟ ⎥ =
⎢ 8 ⎥ = 125 Joule
∫0
∫0 4 + 2
⎝ 2 ⎠ ⎦0
⎣ ⎦
⎣2
2
2
COMPARISON BETWEEN FOURIER TRANSFORM AND LAPLACE TRANSFORM
The defining equations are
∞
∞
0
−∞
F ( s ) = ∫ f (t )e − st dt and F ( j ) = ∫ f (t )e − j t dt
The Followings are some differences and similarities
1. Laplace transform is one-sided in the interval 0 t ∞ and Fourier transform is double-sided in the
interval ∞ t ∞. Thus, Laplace transform is applicable for positive time function, f (t), t 0;
while Fourier Transform is applicable for functions defined for all times.
434
Network Analysis and Synthesis
2. Laplace transform includes the initial conditions and is applicable for transient analysis; while Fourier
transform is only applicable for steady-state analysis.
∞
3. For functions f (t)
0 for t
0 and ∫ f (t ) dt < ∞, the two transforms are related as F ( j ) = F ( s ) s = j
0
Thus, Laplace transform is associated with the entire s-plane, while, Fourier transform is restricted to
the imaginary (j ) axis.
4. Laplace transform is applicable to a wider range of functions than the Fourier transform. On the other hand,
Fourier transforms exist for signals that are not physically realizable and have no Laplace transform.
7.16
STEPS FOR APPLICATION OF FOURIER TRANSFORM TO CIRCUIT ANALYSIS
By Fourier transform, we can find the response of a circuit due to non-periodic functions. The general procedure is described below.
1. Fourier transform of the given excitation function is obtained.
1
2. Fourier transform of the circuit elements is obtained (i.e., R → R, L → j L, C →
).
j C
Y( j )
3. The transfer function in Fourier transform domain is defined as, H ( j ) =
or, Y ( j ) =
X(j )
H ( j ) ⋅ X ( j ) ; where, Y(j ) is the response transform and X(j ) is the excitation transform.
4. Taking the inverse Fourier Transform of the product H ( j ) ⋅ X ( j ) , we get the response y(t).
Solved Problems
PART I FOURIER SERIES
Problem 7.1 Determine the Fourier series for the square waveform shown below and plot the magnitude and the phase spectra.
Solution The waveform,
()
f t =V ; 0 < t < T
= −V ; T
4
4
< t < 3T
4
= V ; 3T
<t <T
4
Obviously, the given function is an even function.
bn
T
Now,
a0 =
T
()
0
f (t )
V
T
2 2
2 4
2 2
f t dt = ∫ Vdt = − ∫ Vdt = 0
∫
T 0
T 0
TT
T
T/ 2 0
T/ 2
T
t
4
T
⎡T
⎤
2
4
4⎢ 4
⎥
an = ∫ f t cos n tdt =
V
cos
n
tdt
−
V
cos
n
tdt
∫
⎥
T 0
T ⎢ ∫0
T
⎣
⎦
4
T
2
()
=
V
Fig. 7.16
⎛n ⎞⎤
⎛n T⎞
⎛ n T ⎞ ⎤ 4V ⎡
4V ⎡ ⎛ n T ⎞
− sin ⎜
+ sin ⎜
⎥=
⎢ 2 sin ⎜ ⎟ ⎥
⎢sin
n T ⎣ ⎜⎝ 4 ⎟⎠
⎝ 2 ⎠⎦
⎝ 2 ⎟⎠
⎝ 4 ⎟⎠ ⎦ n2 ⎣
[ ␻T
2␲]
435
Fourier Series and Fourier Transform
4V
n
4V
sin
=
; for n 1, 5, 9, . . .
n
2 n
4V
; for n 3, 7, 11, . . .
=−
n
=
0;
for n
2, 4, 6, . . .
⎞
4V ⎛
1
1
1
1
f t =
⎜⎝ cos t − 3 cos 3 t + 5 cos 5 t − 7 cos 7 t + 9 cos9 t ⋅⋅⋅⎟⎠
()
So,
Magnitude spectra
C1
F
C3
C5
C7
0 1␻ 2␻ 3␻ 4␻ 5␻ 6␻ 7␻
␻
Fig. 7.17
Phase spectra
Phase, ␾
␲
0 1␻
3␻
5␻
␻
7␻
Fig. 7.18
Problem 7.2 Find the Fourier series of the function whose
periodic waveform is shown in Fig. 7.19 and plot its frequency
spectra.
Solution
The function is even
T
()
bn
V
t
0
T
T
2 2
2 4
2V T V
∴ a0 = ∫ f t dt = ∫ Vdt =
× =
T 0
T 0
T 4 2
T
f (t )
T
()
T/ 2
0
T/ 2
T
n
1, 5, 9 …
Fig. 7.19
()
4 2
4V 4
∴ an = ∫ f t cos n tdt =
f t cos n tdt
T 0
T ∫0
T
⎡
⎤
4V ⎢⎛ sin n t ⎞ 4 ⎥
⎡
=
T ⎢⎜⎝ n ⎟⎠ 0 ⎥ ⎣
⎣
⎦
=
T = 2 ⎤⎦
4V ⎡ ⎛ n T ⎞ ⎤ = 4V ⋅ sin n = 2V
;
⎥
⎢sin
2 n
n T ⎣ ⎜⎝ 4 ⎟⎠ ⎦ 2 n
=−
2V
;
n
n = 3, 7, 11, …
436
Network Analysis and Synthesis
⎛
( ) V2 + 2V ⎜⎝ cos t − 13 cos 3 t + 15 cos 5 t − 17 cos 7 t + ⋅⋅⋅⎞⎟⎠
∴f t =
Line spectra
C1
F
Phase,
C3
C0=V/2
C5
␲
C7
␻
0 1␻ 2␻ 3␻ 4␻ 5␻ 6␻ 7␻
Fig. 7.20
0 1␻
Here, v(t)
V; for 0 < t < T
()
␻
7␻
f (t )
V
2
T
0; for < t < T
2
T
5␻
Fig. 7.21
Problem 7.3 Find the Fourier series for the train of pulses shown in
Fig. 7.22 and draw the amplitude and the phase spectra.
Solution
3␻
0
T/2
3T/2 2T
T
Fig. 7.22
T
1
1 2
V
∴a0 = ∫V t dt = ∫ Vdt =
2
T0
T 0
an =
and
and
bn =
T
T /2
2
2
2V ⎡ ⎛ n T ⎞ ⎤
V
(
t
)cos
n
t
d
t
=
V cos n tdt =
⎥ = 0 [ ␻T
⎢sin
∫
∫
T0
T 0
n T ⎣ ⎜⎝ 2 ⎟⎠ ⎦
T
T /2
⎛n T⎞⎤ V
2
2V
2V ⎡
1 − cos n
V
(
t
)sin
n
tdt
=
sin n tdt =
⎢1 − cos ⎜
⎥=
∫
∫
T0
n T⎣
T 0
⎝ 2 ⎟⎠ ⎦ n
(
bn
) [ ␻T
0, for n even
2V , for n odd
=
n
⎡1 2
⎤
2
2
∴V (t ) = V ⎢ + sin t +
sin 3 t +
sin 5 t + ⋅⋅⋅⎥
3
5
⎣2
⎦
Amplitude spectra
C1
F
C3
C0=V/2
C5
C7
0 1␻ 2␻ 3␻ 4␻ 5␻ 6␻ 7␻
␻
Fig. 7.23
Phase spectra
Phase, ␾
␲2
0 1␻
Fig. 7.24
3␻
5␻
7␻
␻
2␲]
2␲]
t
437
Fourier Series and Fourier Transform
Problem 7.4 For the periodic function shown in Fig.7.25, determine the
exponential form of Fourier series and show the line spectra. Also, find its
trigonometric form.
f (t)
Solution The function is defined as,
f (t)
0
V
V
vt
V,
0 t
[T
V,
t
2 ]
p
2p
3p
Fig. 7.25
2
Since the function is odd, the coefficients Cn will be purely imaginary.
∴ Cn =
2
2
⎤
1
1 ⎡
− jn t
− jn t
f
(
t
)
e
dt
=
Ve
dt
−
V
e − jn t dt ⎥ ; for
⎢
∫
∫
∫
2 0
2 ⎢⎣ 0
⎥⎦
2
⎤ V ⎡ 1 − jn t ⎤
V
1 ⎡ V
e − jn t ⎥ −
=
⎥ = j2 n
⎢ − jn e
2 ⎢⎣ − jn
2
⎦0
⎣
⎦
=
(
)
(
V
V
1 − e − jn +
e − jn 2 − e jn
j2 n
j2 n
Now,
(1− e
( )
( )
n
2V ⎡
1 − −1 ⎤ ; n
⎢
⎥⎦
j 2n ⎣
Cn =
n
− jn 2
− e jn
)[ T
n
and
e − j 2 n = cos 2 n − jsin2 n = 1
0
2V
; for n odd
jn
0;
for n even
2V
jn
C− n = −
For
) + j 2V n ( e
0
)
e − jn = cos n − j sin n = −1
∴ Cn =
− jn
n
0, C0 =
2
2
⎤
1
1 ⎡
f
t
dt
=
Vdt
−
Vdt ⎥ = 0
⎢∫
∫
∫
2 0
2 ⎢⎣ 0
⎥⎦
()
exponential form of Fourier series is,
∞
( ) 2jV ∑ 1n e
f t =
jn t
; n odd only
n=1
=
⎤
2V ⎡ j t 1 3 j t 1 j 5 t 1 j 7 t
e + e + e
+ e
+ ⋅⋅⋅⎥
j ⎢⎣
3
5
7
⎦
To find trigonometric form,
a0
(
) jn2V − jn2V = 0
0, a0 = 0, an = Cn + C− n =
⎡ 2V 2V ⎤ 4V
bn = j Cn − C− n = j ⎢
+
=
jn ⎥⎦ n
⎣ jn
(
∴ f (t ) =
)
⎤
4V ⎡
1
1
⎢sin t + 3 sin 3 t + 5 sin 5 t + ⋅⋅⋅⎥
⎣
⎦
for n odd.
2 ,
1]
438
Network Analysis and Synthesis
Amplitude spectra
F
C1
C3
C5
C7
0 1 2 3 4 5 6 7
Fig. 7.26
Phase spectra
Phase,
2
0 1
3
5
7
Fig. 7.27
Problem 7.5 The waveform shown in Fig. 7.28 is used
as ‘sweep’ in radar and television circuits. Find the Fourier
series and plot the line spectra.
V
t ; 0<t <T
T
T
1 V
∴ Cn = ∫ te − jn t dt ; n ≠ 0
T 0T
v(t )
V
Solution The function, V (t ) =
0
T
2T
3T
4T
T
⎡
⎤
T
− j 2n
−1 ⎤
V ⎡ te jn t
e − jn t ⎤
V ⎢ Te − jn T e − jn T ⎥ V ⎡ T 2 e − j 2 n e
⎢
⎥
= 2⎢
+∫
=
+
−
⎥ = 2⎢
⎥ T 2 ⎢ − j 2n
2
jn ⎦0 T
− jn
T ⎣ − jn
n2 2 ⎥
jn
⎢
⎥
⎣
⎦
0 ⎦
⎣
[
T
2 ]
2T
Fig. 7.28
(
( )
=
Since,
(
T
)
V
V
jV − j 2 n
e
+ 2 2 e− j 2n − 2 2
2n
4n
4n
e − j 2 n = cos 2 n − j sin 2 n
) =1
∴ Cn =
jV
; for n ≠ 0
2n
T
For
n
0, C0 =
V
V
tdt =
2 ∫
2
T 0
exponential form,
v (t ) = ⋅⋅⋅−
jV − j 3 t jV − j 2 t jV − j t V jV j t jV j 2 t jV j 3 t
e
−
e
−
e
+ +
e +
e
+
e
+ ⋅⋅⋅
6
4
2
2 2
4
6
• To convert into trigonometric form;
jV
jV
, C− n = −
Here, Cn =
2n
2n
t
439
Fourier Series and Fourier Transform
)
(
)
(
V
V
, an = Cn + C− n = 0 and bn = j Cn − C− n = −
n
2
⎤
1
V V⎡
1
∴V t = − ⎢sin t + sin 2 t + sin 3 t + ⋅⋅⋅ ⎥
2
2
3
⎣
⎦
∴ a0 = C0 =
()
Line spectra
Phase, f
F
p2
V 2
4
3
2
1
0
4 3 2 10
n
1 2 3 4
1
2
3
4
v
p2
Fig. 7.29
Problem 7.6 Find the trigonometric Fourier series for the waveform shown in Fig. 7.30 and sketch the spectra.
V
Solution Here, f t =
t ; for 0 < t <
and
()
0 ; for
V
⇒ a0 =
4
1 V
⇒ an = ∫
< t<2
t cos n td
V
( t)
vt
0
0
2V ; for n odd
n2 2
0; for n even
=−
⇒ bn =
1 V
∫
v (t)
p
2p
3 p 4p
Fig. 7.30
t sin n td ( t )
0
V
; for n even
n
V
; for n odd
n
−
∴ f (t ) =
⎤ V⎡
⎤
V 2V ⎡
1
1
1
1
− 2 ⎢cos t + cos 3 t + cos 5 t + ⋅⋅⋅⎥ + ⎢sin t − sin 2 t + sin 3 t − ⋅⋅⋅⎥
4
9
25
2
3
⎣
⎦
⎣
⎦
Line spectra
The even harmonic amplitudes are given directly by bn coefficients, since there are no even cosine terms.
But, the odd harmonic amplitudes are given by computation:
C
F
Cn = an 2 + bn 2
2
⎛ 2V ⎞ ⎛ V ⎞
∴ C1 = ⎜ 2 ⎟ + ⎜ ⎟
⎝ ⎠ ⎝ ⎠
C3 (0.109)V,
1
V/4
2
C3
C5
(0.377) V
C5 (0.064)V
␻
0 1␻ 2␻ 3␻ 4␻ 5␻
Fig. 7.31
440
Network Analysis and Synthesis
Problem 7.7 Find the Fourier series expansion of the rectified sine waveforms shown in Fig.7.32.
()
Solution Here, f t = Asin t ;
0< t <
for
= − Asin t ;
f (t )
1
< t<2
for
vt
() ( )
Since, f t = f −t ⇒ The function is even.
bn
0
0
2p
3p
4p
Fig. 7.32
T
()
4 2
f t cos n td
T ∫0
∴ an =
p
( t)
=
4
A
A
Asin t cos n td ( t ) = ∫ 2 sin t cos n td ( t ) = ∫ ⎡⎣sin( n + 1) t − sin( n − 1) t ⎤⎦ d ( t )
∫
2 0
0
0
=
A ⎡ − cos( n + 1) t cos( n − 1) t ⎤
+
⎢
⎥ ; for n ≠ 1
n +1
n −1
⎣
⎦0
A ⎡⎛ 1
1 ⎞ ⎛ 1
1 ⎞⎤
+⎜
−
+
⎢⎜ −
⎟ ⎥ ; n ≠1 = 0
⎟
⎣ ⎝ n + 1 n + 1⎠ ⎝ n − 1 n − 1⎠ ⎦
For odd n;
an =
For even n;
an =
For
4 2
a1 = ∫ f t cos td ( t )
T 0
A ⎡ 2n − 2 − 2n − 2 ⎤
A ⎡⎛ 2 ⎞ ⎛ −2 ⎞ ⎤
=
⎢
⎥=−
+
⎥
⎢⎜
⎟ ⎜
⎟
⎢⎣ n + 1 n + 1 ⎥⎦
⎣ ⎝ n + 1⎠ ⎝ n − 1⎠ ⎦
(
T
n = 1,
=
)(
)
4A
( n − 1)
2
()
4
A
A
A
Asin t cos td ( t ) = ∫ sin 2 td ( t ) = −
⎡⎣cos 2 t ⎤⎦ = −
⎡⎣cos 2 − 1⎤⎦ = 0
∫
0
2 0
2
2
0
T
2 2
2
A
2A
a0 = ∫ f (t )dt =
Asin td ( t ) = − ⎡⎣cos t ⎤⎦ =
∫
0
T 0
2 0
Also,
So, the Fourier series is
⎛
⎞
n t 2A 4A 1
1
1
=
−
( ) 2 A − 4 A ∑ cos
⎜⎝ 3 cos 2 t + 15 cos 4 t + 35 cos 6 t + ⋅⋅⋅⎟⎠
n −1
f t =
n = 2 , 4 ,6
(
2
)
Spectra
F
C2
2 A/␲
C4
C6
␻
0 1␻ 2␻ 3␻ 4␻ 5␻ 6␻
Fig. 7.33
441
Fourier Series and Fourier Transform
Problem 7.8 Determine the Fourier series of voltage response obtained at the output of a half-wave
rectifier shown in Fig. 7.34. Plot the discrete spectrum of the waveform.
Solution
Here, time period T
V (t )
0.4 second;
1
2
2
= 2.5Hz;
=
= 5 rrad/s
T
T 0.4
The function v(t) Vm cos 5 t ; 0 ≤ t ≤ 0.1
Vm
∴f=
0.4
; 0.1 ≤ t ≤ 0.3
0
0.2 0.1
0
0.1 0.2
0.4
t
Fig. 7.34
Vm cos 5 t ; 0.3 ≤ t ≤ 0.4
If the period extending from t
integrals.
0.1 to t
0.3 is taken, it will result in fewer equations and hence, fewer
()
∴ v t = Vm cos5 t ; − 0.1 ≤ t ≤ 0.1
0.1 ≤ t ≤ 0.3
0;
0.1
0.3
⎤ V
1
1 ⎡
v
t
dt
=
V
dt
+
cos
5
0 dt ⎥ = m
⎢∫ m
∫
∫
0.4 −0.1
0.4 ⎢⎣ −0.1
⎥⎦
0.1
0.3
a0 =
()
()
0.3
∴ an =
2
V cos 5 ntdt ;n ≠1
0.4 −∫0.1 m
0.1
0.1
1
= 5Vm ∫ cos 5 t cos 5 ntdt = 5Vm ∫ ⎡⎣cos 5
2
−0.1
−0.1
0.1
For, n
1,
a1 = 5Vm ∫ cos 2 5 tdt =
−0.1
Similarly, bn
(1 + n)t + cos 5 (
)
1 − n t ⎤⎦ dt =
2Vm cos
( n 2 ) ; n ≠1
1 − n2
Vm
2
0 for any value of n, and the Fourier series thus contains no sine terms.
V V
2V
2V
2V
∴ v t = m + m cos 5 t + m cos10 t − m cos 20 t + m cos 30 t − ⋅⋅⋅
35
2
3
15
()
Spectra
F
0.5Vm
0.4Vm
0.3Vm
0.2Vm
0.1Vm
0 2.5 5
10
15
20
25
f (H z )
Fig. 7.35
Problem 7.9 Find the trigonometric Fourier series for the half-wave rectified sine-wave shown in Fig.
7.36 and sketch the spectrum.
Solution
Here, the wave is, f (t ) = V sin t ; 0
0;
t
t
2
442
Network Analysis and Synthesis
∴ a0 =
1
V
V sin t =
∫
2 0
∴ an =
1
f (t)
V
∫V sin t cos n td ( t ) n ≠1
0
(
V
⎡sin 1 + n
2 ∫0 ⎣
=
V ⎡ − cos 1 + n
⎢
2 ⎢
1+ n
⎣
=
=
For n
1, a1 =
(
V
(
1 − n2
)
(1− n )
2
1
2
1− n
1
0;
for n odd
; for n even.
1, b1 =
1
∫V sin t cos td ( t ) = 0
V/p
∫V sin t sin n td ( t ) = 0; n 1
0
0
1 2
4
6
8
10
n
Fig. 7.37
( t ) = V2
2
∫V sin td
V/2
F
0
For n
t
4
⎥⎦0
0
Similarly, bn =
3
Fig. 7.36
) t − cos(1 − n) t ⎤⎥
(1 + cos n ) ; n ≠1
2V
0
) t + sin (1 − n) t ⎤⎦ d t
=
⎛
( ) V + V2 sin t − V ⎜⎝ 23 cos 2 t + 152 cos 4 t + 352 cos66 t + ⋅⋅⋅⎞⎟⎠
So the series is, f t =
Problem 7.10 A square wave has a value 10 from − ␲ to ␲ , zero from ␲ to 3 ␲ , 10 from 3 ␲ to 5 ␲
2
2
2
2
2
2
and so on. Find the Fourier series expansion of the wave.
Solution For the square wave given, time period is 2 . The Fourier coefficients are evaluated as
a0 =
1 2
f
2 −∫
( )d = 21 ∫ 10d = 5
2
an =
−
−
2
∫ f ( )cos nd = ∫ 10cos n d = n sin n
2
1
2
2
1
−
2
10
20
; a2 = 0; a3 = −
bn =
1
∫ f ( )sin nd = ∫ 10sin n d = − n cos n
−
2
2
20 ⎛ n ⎞
sin ⎜ ⎟
n
⎝ 2 ⎠
2
∴a1 =
2
=
2
−
20
; a4 = 0
3
2
1
−
2
10
=0
2
−
2
443
Fourier Series and Fourier Transform
Therefore, the Fourier series is given as
v = 5+
20
cos −
20
20
cos 3 +
cos 5 − ⋅⋅⋅
3
5
Problem 7.11 State and prove Parseval’s theorem useful in computing the effective value of a given
periodic function, f(t).
Or,
A periodic function f( ) with period 2␲ is expressed in Fourier series as follows:
∞
a
f ( ) = 0 + ∑ (a n cos n + bn sin n )
2 n =1
1
⎡f
2 −∫ ⎣
Prove that,
f
Solution
()
2
2
⎛a ⎞ 1 ∞
⎤ d = ⎜ 0 ⎟ + ∑ a n 2 + bn 2
⎦
⎝ 2 ⎠ 2 n =1
(
)
∞
( ) = 2 + ∑ ( a cos n + b sin n )
a0
n=1
n
n
Since, ∫ cos n sin n d = ∫ cos n d = ∫ sin n d = 0
−
−
1
⎡f
∴
2 −∫ ⎣
()
−
2
⎡
∞
∞
1 ⎢ ⎛ a0 ⎞
2
2
⎤ d =
+
+
a
cos
n
bn 2 sin 2
⎟
⎜
∑
∑
⎦
2 −∫ ⎢⎝ 2 ⎠ n=1 n
n=1
⎢⎣
2
⎤
⎥d
⎥
⎥⎦
2
an 2
bn 2
1 ⎛ a0 ⎞
1
1
2
+
2
+
2 siin 2 n d
=
d
cos
n
d
⎜ ⎟
∑
∑
2 −∫ ⎝ 2 ⎠
2 n=1 2 −∫
2 n=1 2 −∫
2
=
an 2 1
bn 2 1
1 ⎛ a0 ⎞
⋅
+
+
2
.
1
cos
2
+
⋅ ⋅ ∑ 1 − cos 2 n d
n
d
∑
⎜ 2⎟
2 2 n=1 −∫
2 2 n=1 −∫
⎝ ⎠
)
(
(
2
)
2
⎛a ⎞
⎡ 2 ⎧ sin 2 n ⎫
1
sin 2 n ⎫ ⎤ ⎛ a0 ⎞
1
2⎧
⎡ an 2 2
=⎜ 0⎟ +
⎬ + bn ⎨ −
⎬ ⎥ =⎜ ⎟ +
⎢ an ⎨ +
∑
∑
⎣
2n ⎭ −
2
2
4
n
−
⎝ 2 ⎠ 4 n=1 ⎣ ⎩
⎝
⎠
n=1
⎭ ⎦
⎩
2
1
⎡f
∴
2 −∫ ⎣
Note
()
⎛a ⎞ 1 ∞
⎤ d = ⎜ 0 ⎟ + ∑ an 2 + bn 2
⎦
⎝ 2 ⎠ 2 n=1
2
(
( ) + b ( 2 )⎤⎦
2
n
) (Proved)
For statement and proof of this theorem consult the text earlier.
Problem 7.12 If v ( t ) =10 + 6 cos( t + 45 ° ) +1.8 cos(2 t −10 ° ) volt and i( t ) = 3 +1.4 cos( t + 20 ° ) + 0.5 cos 2 t
mA, calculate the average power in watt. Determine also the effective voltage and effective current.
Solution
Average power
V I
VM 1 I M 1
V I
cos 1 + M 2 M 2 cos 2 + M 3 M 3 cos 3
2
2
2
444
Network Analysis and Synthesis
= 10 × 3 +
6 × 1⋅ 4
1⋅ 8 × 0 ⋅ 5
cos 45° − 20° +
cos10°
2
2
Effective voltage
10 +
Effective current
32 +
2
)
(
34.25W
( ) = 12.58 V
6 2 + 1⋅ 8
2
2
)
(
1
1⋅ 4 2 + 0 ⋅ 52 = 3⋅178 A
2
Problem 7.13 Determine the effective voltage, effective current, and average power supplied to a
passive network if the supplied voltage is, v ( t ) =100 + 50 cos(10 t + 30 ° ) + 25 cos(30 t + 60 ° ) V and the resulting
current is, i( t ) = 2 cos(10 t + 75 ° ) + 3 cos(30 t + 78 ° ) A .
v(t )
V
Solution Same as Prob. 7.12. Ans: 107.53V; 2.55A; 71.02 W
Problem 7.14 (a) Find the trigonometric Fourier series of the triangular waveform shown in Fig. 7.38
8V
(b) If this voltage is approximated by 2 sin t , find the mean-square
error.
T/2
T/2
0 T/4
T
T
t
V
Fig. 7.38
(c) If this voltage waveform is applied to the network in Fig. 7.39, then find the current i(t) and draw the
magnitude and phase spectra of i(t). Take ␻0 1 radian/second for the waveform.
Solution a) The wave is an odd function and has half-wave symmetry.
∴ an = 0
1
a0 = 0
and
4V
t ; 0<t <T
4
T
3T
4V
= − t + 2V ; T < t <
4
4
T
V (t)
V (t ) =
Now,
T
i (t )
Fig. 7.39
()
8 4
∴ bn = ∫ f t sin n tdt ; n is odd only.
T 0
T
T
T
⎡
⎤
8 4 4V
32V ⎡ −t cos n t
n
sin n t 4 ⎥
cos n t ⎤ 4 16V ⎢ T
t sin n tdt = 2 ⎢
dt ⎥ =
− cos
+
= ∫
+∫
⎥
T 0 T
n
n T⎢ 4
2
n
T ⎣ n
⎦0
0
⎣
⎦
=
T
n ⎤ 8V
n
16 ⎡ T
− ×0+
sin
= 2 2 sin
⎢
⎥
n T⎣ 4
2n
2 ⎦ n
2
∴ bn =
8V
,n
n2 2
=−
8V
,n
n2 2
1, 5, 9, …
3, 7, 11, …
{
T
2 }
1F
445
Fourier Series and Fourier Transform
Hence,
V (t ) =
(b) The error is
⎞
8V ⎛
1
1
1
sin t − 2 sin 3 t + 2 sin 5 t − 2 sin 7 t + ⋅⋅⋅⎟
2 ⎜
⎠
⎝
3
5
7
(t ) = v (t ) − 8V sin t
2
T
EN =
The main square error is
()
1 2
t dt
T ∫0
T
()
Since the wave has half-wave symmetry,
4 4
∴ EN = ∫ 2 t dt
T 0
Now,
v t =
( ) 4TV t; for 0 < t < T 4
T
2
⎤
4 4 ⎡ 4V 8V
∴ EN = ∫ ⎢ t − 2 sin t ⎥ dt = 0.0047 V2
T 0⎣T
⎦
(c) Here,
V ( n ) V ( n ) nV n
( ) Z n = j = ( ) ∠ tan ( 1 n )
( ) 1−
1+ n
i n =
i
i
−1
i
2
n
( )
n
∴i n =
1 + n2
( )
8V
sin ⎡ nt + tan −1 1 ⎤ ; for n
n ⎥⎦
n2 2 ⎢⎣
2
( )
2
( )
sin ⎡ nt + + tan −1 1 ⎤ ; for n
n ⎦⎥
⎣⎢
n 1+ n
8V
2
2
∴ i1 =
∴ i3 =
∴ i5 =
∴ i (t ) =
8V
2
2
3 10
8V
2
8V
2
5 26
)
(
sin t + 45° = 0 ⋅ 707
8V
2
(
1, 5, 9, …
sin ⎡ nt + tan −1 1 ⎤
n ⎦⎥
⎣⎢
n 1+ n
8V
=
and i(n ) =
×
8V
(
2
3, 7, 11, ...
sin t + 45°
)
)
8V
siin 3t + 198 ⋅ 44°
2 2
3
(
sin 3t + 180° + 18 ⋅ 44° = 0 ⋅ 949
)
sin 5t + 11⋅ 31° = 0 ⋅ 98
(
8V
sin 5t + 11⋅ 31°
2 2
5
(
)
)
⎡⎣0.707 sin(t + 45° ) + 0.105sin( 3t + 198.44° ) + 0.039 sin(5t + 11.31° ) + ⋅⋅⋅⎤⎦
Problem 7.15 A series RL circuit with R 10 ohms and L 5 H contains a current
i( t ) =10 sin1000 t + 5 sin 3000 t + 3 sin 5000 t A
Find the effective voltage and the average power.
446
Network Analysis and Synthesis
Solution Here,
1000 rad/s and it contains three harmonics:
For fundamental harmonic R1 = 10 , X L1 = L = 1000 × 5 = 5000
)
(
∴ Z1 = R1 + jX L1 = 10 + j 5000 = 5000∠89.88°
For third harmonic
R3 = 10 , X L 3 = 3 L = 15000
(
)
∴ Z 3 = 10 + j15000 = 15000 ⋅ 003∠89.96°
For fifth harmonic R5 = 10 , X L 5 = 5 L = 25000
∴ Z 5 = (10 + j 25000) = 25000 ⋅ 001∠89.977°
∴ v (t ) = 10 Z1 sin(1000t − 89.88° ) + 5 Z 3 sin( 3000t − 89.96° ) + 3 Z 5 sin(5000t − 89.977° )
= 5000 ⋅ 01sin(1000t − 89 ⋅88° ) + 75000 ⋅ 015sin( 3000t − 89 ⋅ 96° ) + 75000 ⋅ 003sin(5000t − 89 ⋅ 977° )
∴ effective voltage, V =
Average power
=
Pav =
1
2
1
⎡(5000 ⋅ 01)2 + (75000 ⋅ 015)2 + (75000 ⋅ 003)2 ⎤ 2 = 8 ⋅ 291 × 104 V = 82 ⋅ 91 kV
⎣
⎦
Vm1 I ns
V I
V I
cos 1 + mL m 2 cos 2 + m 3 m 3 cos 3
2
2
2
5000 ⋅ 01 × 10
75000 ⋅ 015 × 5
75000 ⋅ 003 × 3
cos89 ⋅88° +
cos89 ⋅ 96° +
cos89 ⋅ 977° = 691⋅ 6595 W
2
2
2
Problem 7.16 A periodic current source, i(t) 10 6cos (100t 45 ) 3cos (200t 10 ) 2.1cos (300t 35 )
is the input to a parallel RC circuit with R 0.5 ohm and C 0.02 F. Calculate the steady-state response
v(t) of the circuit.
Solution {same as Prob. 7.15}
Z1 = 0 ⋅ 35∠ − 45° ; Z 2 = 0 ⋅ 22 ∠ − 63.43° ; Z 3 = 0 ⋅158∠ − 71.56°
∴ v (t ) = 5 + 2 ⋅121cos100t + 0 ⋅ 671cos( 200t − 73⋅ 43° ) + 0 ⋅ 332 cos( 300t − 36 ⋅ 56° )
Problem 7.17 The square wave source, v(t) shown in Fig. 7.40 excites
a series RL circuit with R 2 ohm and L 2 H. Determine the current
␲
volt.
response i(t), taking ␻ 1 radian/second and V
4
Solution [same as Prob. 7.14]
Here, from Prob.7.1
⎞
4V ⎛
1
1
1
v (t ) =
⎜⎝ cos t − 3 cos 3 t + 5 cos 5 t − 7 cos 7 t + ⋅⋅⋅⎟⎠
Here,
V=
4
V
T
v( t )
T/ 2
T/ 2
0
V
Fig. 7.40
V
1
1
∴ v (t ) = cos t − cos 3 t + cos 5 t − ⋅⋅⋅
3
5
1
Y ( jn) =
⇒ Y1 = 0 ⋅ 353∠ − 45° ; V1 = 1∠0°
2 + j 2n
T
t
447
Fourier Series and Fourier Transform
1
Y3 = 0 ⋅158∠ − 71.565° ; V3 = ∠ − 180°
3
1
Y5 = 0 ⋅ 098∠ − 78.69° ; V5 = ∠0°
5
∴ I1 = V1Y1 = 0 ⋅ 353∠ − 45°
∴ I 3 = V3Y3 = 0 ⋅ 0527∠108.435°
I 5 = V5Y5 = 0 ⋅ 0196 ∠ − 78.69°
and
∴ i(t ) = 0 ⋅ 353cos(t − 45° ) + 0 ⋅ 0527 cos( 3t − 251⋅ 6° ) + 0 ⋅ 0196 cos(5t − 78 ⋅ 69° ) + ⋅⋅⋅
Problem 7.18 Determine the Fourier series of repetitive waveform of
Fig. 7.41 up to 5th harmonic, when time of repetition, T 20 ms. Calculate
the fundamental frequency current in the circuit of Fig. 7.42, where
R ⴝ 10 ohms and L 0.0318 H with voltage transform of the waveform.
Solution The wave has half-wave symmetry.
an bn 0 ; for n even ; and
For n odd, an =
T
2
v(t )
100V
0
100V
T/2
T
t
Fig. 7.41
T
2
4
4
f (t )sin n tdt and bn = ∫ f (t )sin n tdt
∫
T0
T0
and
a0
Now,
v(t)
0
200
T
t; 0 ≤ t ≤
T
2
T /2
R
L
V (t)
Fig. 7.42
4 200
∴ an = ∫
t cos n tdt
T 0 T
T
⎡
⎤
800 ⎢ T sin n
cos n t 2 ⎥
×
+
n
T2 ⎢2
n2 2 0 ⎥
⎣
⎦
800
400
( −2 ) − 2 2
n2 4 2
n
sin n t ⎤
800 ⎡ t sin nwt
−∫
dt ⎥
⎢
n
T2 ⎣ n
⎦
800
⎡cos n − 1⎤⎦
n 2T 2 ⎣
2
T
4 2 200
200
bn = ∫
t sin n tdt =
T0 T
n
∴ v (t ) = −
400
2
(cos t +
1
1
1
200
1
cos 3 t + 2 cos 5 t + ⋅⋅⋅) +
(sin t + sin 3 t + sin 5 t + ⋅⋅⋅)
2
5
3
5
3
The fundamental frequency voltage is
Impedance,
⎛ 200
⎞ 400
400
Vf = ⎜
sin t − 2 cos t ⎟ = 2 ( )2 + 1
2
⎝
⎠
Z = ( R + j L ) = 10 + j (0 ⋅ 0318)
Current due to fundamental frequency,
If =
Vf
Z
=
2
⎛
⎞
400
sin t − cos t ⎟
⎜
⎠
(10 + j 0 ⋅ 0318 ) ⎝ 2
448
Network Analysis and Synthesis
If =
or,
Here,
=
2
2
=
= 100
T 20 × 10−3
400
2
1
0.0318
( )2 + 1 ×
∠ tan −1
2
2
2
10
(10) + (0 ⋅ 0318 )
rad/s
∴ I f = 5 ⋅ 33∠ − 44.9°
Putting this value,
∴ I f ( rms ) =
5 ⋅ 33
2
A = 3⋅ 76 A
Problem 7.19 An RLC series circuit with R = 25 ohms, L = 1 H, and C = 10 microfarads is energized with a
voltage source, V ( t ) =15 sin100 t +10 sin 200 t + 5 sin 300 t (V)
Find the expression for the current i(t). Determine the effective value of the current, and the average power
consumed by the circuit.
Solution
[Same as Prob. 7.16]
⎛
1 ⎞
= 900 ⋅ 3∠ + 88 ⋅ 4°
Z1 = R + j ⎜ L −
C ⎟⎠
⎝
⎛
1 ⎞
= 41⋅ 62 ∠53⋅1°
Z2 = R + j ⎜ 3 L −
3 C ⎟⎠
⎝
1 ⎞
⎛
Z 3 = R + j ⎜ 3ω L −
= 41 ⋅ 62∠53 ⋅ 1°
⎝
3ωC ⎟⎠
∴ i (t ) =
15
10
5
sin100t + sin 200t + sin 300t
Z1
Z2
Z3
= 0 ⋅ 0167 sin(100t + 88 ⋅ 4° ) + 0 ⋅ 0332 sin( 200t + 85 ⋅ 2° ) + 0 ⋅12 sin( 300t + 53⋅1° ) + ⋅⋅⋅
Irms =
1
1
1
⎡ I12 + I 2 2 + I 32 ⎤ 2 = 1 ⎡(0 ⋅ 0167)2 + (0 ⋅ 0332 )2 + (0 ⋅12 )2 ⎤ 2 = 0 ⋅ 088 A = 88 mA
⎣
⎦
⎣
⎦
2
2
∴ Pav =
15 × 0 ⋅ 0167
10 × 0 ⋅ 0332
5 × 0 ⋅12
cos88 ⋅ 4° +
cos85 ⋅ 2° +
cos 53⋅1° = 0.197 W
2
2
2
Problem 7.20 Determine the expression for current in an impedance of R 10 ohms, L 0.0318 H with
applied emf, e( t ) = 200 sin 314 t + 40 sin(942 t + 30 ° ) +10 V
Also, calculate the rms value of voltage and current as well as the power factor of the circuit.
Solution
[Same as Prob. 7.19]
i1 =
200 sin 314t
= 14 ⋅14 sin 314t ∠ − 44.95°
10 + j 314 × 0 ⋅ 0318
i2 =
40 sin( 942t + 30° )
= 1⋅ 28 sin( 942t + 30° )∠ − 71⋅ 54°
10 + j 942 × 0 ⋅ 0318
449
Fourier Series and Fourier Transform
10
=1
10
i(t ) = 14 ⋅14 sin( 314t − 44 ⋅ 95° ) + 1⋅ 28 sin( 942t − 41.54° )
i0 =
200 + 40
V 2 + V2 2
= 102 +
= 144 ⋅ 568 volts
∴Vrms = V0 + 1
2
2
2
2
2
I 2 + I22
14 ⋅14 2 + 1⋅ 282
∴ I rms = I 0 2 + 1
= 12 +
= 10.089 A
2
2
∴ power factor =
Averag Power
Apparent Power
VI
VI
200 × 14 ⋅14
40 × 1⋅ 28
V0 I 0 + 1 1 cos 1 + 2 2 cos 2 10 × 1 +
cos 44 ⋅ 95° +
cos 71⋅ 54°
2
2
2
2
=
=
= 0.69
Vrms × I rms
144 ⋅ 568 × 10 ⋅ 089
Problem 7.21 In a two-element series network, voltage v(t) is applied, which is given by
v ( t ) = 50 + 50 sin 5000 t + 30 sin10000 t + 20 sin 20000 t (V)
The resulting current is given as
i( t ) =11.2 sin(5000 t + 63.4 ° ) +10.6 sin(10000 t + 45 ° ) + 8.97 sin(20000 t + 26.6 ° ) (A)
Determine the network elements and the power dissipated in the circuit.
Solution
Power dissipated,
50 × 11⋅ 2
30 × 10 ⋅ 6
8 ⋅ 97 × 20
Pav = 50 × 0 +
cos 63⋅ 4° +
cos 45° +
cos 26 ⋅ 6° = 318 W
2
2
2
In the expression of current i(t), the dc term is missing though it is present in the applied voltage, v(t). Hence,
in the series network, there must be a capacitor which blocks dc components. Again from the expression of i(t),
we see that the current is leading by an angle less than 90 . Hence, the conclusion is the presence of a resistive
element in series with the capacitor (RC).
I eff =
Now,
11⋅2 +10 ⋅ 6 2 + 8 ⋅ 972
= 12 ⋅ 6 A
2
∴ Pav = I eff 2 R ⇒ R =
Again, at
318
=2
(12 ⋅ 6 )2
⎛ 1 ⎞
= 45° = tan −1 ⎜
⎝ CR ⎟⎠
10,000 rad/s,
⇒ C=
1
1
=
= 50 F
R 20, 000
Problem 7.22 In a linear circuit consisting of R 9 ⍀ and L 8 mH, a current, i = 5 + 100 sin(1000t + 45° ) +
100 sin(3000t 60 )A is flowing. Find the equation of applied voltage.
Solution
Here, R
9
and L
(
)
(
8 mH, i = 5 + 100 sin 1000t + 45° + 100 sin 3000t + 60°
)A
450
Network Analysis and Synthesis
For dc component
Current, I0 5A, Z0
R
9
∴V0 = I 0 × R = 5 × 9 = 45 V
For first-harmonic component
Current, I1 = 100∠45° A
Impedance,
Z1 = R + jω L = 9 + j 2π × 1000 × 8 × 10 −3 = ( 9 + j8 ) = 12.04 ∠41.663° ( Ω )
( )
∴ V1 = I1 Z1 = 100∠45° × 12.04 ∠41.63° = 1204 ∠86.63° V
For third-harmonic component
Current, I 3 = 100∠60° A
Impedance,
)
(
Z 31 = R + j 3 L = 9 + j 2 × 3 × 1000 × 8 × 10−3 = 9 + j 24 = 25.63∠69.44°
( )
( )
∴V3 = I 3 Z 3 = 100∠60° × 25.63∠69.44° = 2563∠129.44° V
applied voltage is given as
v = 45 + 1204 sin 1000t + 86.63° + 2563sin 3000t + 129.44°
)
(
(
)(V)
Problem 7.23 Calculate the impedance consisting of R and L and the power factor of a circuit whose
expression for voltage and current are
v (t ) = 250 sin 314t + 50 sin(942t + 30° ) (V)
i (t ) = 17.7 sin(314t − 45° ) + 1.583 sin(942t − 41.6° ) (A)
Solution
The fundamental frequency current, I1 =
The third harmonic current, I 3 =
Equating the magnitudes of (i),
250 sin 314t
= 17 ⋅ 7 sin( 314t − 45° )
R+ j L
50 sin( 942t + 30° )
= 1⋅ 583sin( 942t − 41⋅ 6° )
R + j3 L
250
R +
2
2
2
= 17.7 ⇒ R 2 +
2
(i)
(ii)
L2 = 199.495
(iii)
L
Equating the angles of (i)
314t − tan −1
Putting in (iii),
L
L
L
= 314t − 45° ⇒ tan −1
= 45° ⇒
=1 ⇒
R
R
R
⇒ ( L )2 = 99 ⋅ 747 ⇒
L=R
L = 9 ⋅ 987 = R
9 ⋅ 987
= 0 ⋅ 0318
314
∴ R = 9 ⋅ 987 L = 0 ⋅ 0318H
∴L=
Power factor
250 × 17 ⋅ 7
50 × 1⋅ 583
VI
V1 I1
cos 45° +
cos 71⋅ 6°
cos 1 + 3 3 cos 3
Average power
2
2
2
2
=
= 0.69
=
Apparent power
V12 V32
I12 I 3 2
2502 + 50°
17 ⋅ 72 + 1⋅ 5832
×
+
×
+
2
2
2
2
2
2
451
Fourier Series and Fourier Transform
PART II FOURIER TRANSFORM
Problem 7.24 Determine the Fourier transform of one cycle of sine wave, f(t) = Asin␻0t.
∞
Solution
T
∴ F ( j ) = ∫ f (t )e − j t dt = A ∫ sin
−∞
0
te − j t dt = I (say)
f (t)
0
A
T
⎡
⎤
T
⎛ cos 0 t ⎞
cos 0 t
−j t
⎥
= A⎢e− j t ⎜ −
−
j
e
d
t
⎟
∫
⎢
⎥
⎝
⎠0 0
0
0
⎣
⎦
)
(
⎡ 1 −j T
= A⎢−
e
cos
⎢⎣ 0
T
j ⎧⎪
T −1 −
⎨ ∫ cos
0
0 ⎪
⎩0
)
(
t
0
Fig. 7.43
⎪⎫ ⎤
te − j t dt ⎬ ⎥
0
⎪⎭ ⎥⎦
T
⎡
⎡
⎤⎤
T
⎛ sin 0 t ⎞
1 −j T
j ⎢ ⎧⎪ − j t ⎛ sin 0 t ⎞ ⎫⎪
−j t
⎢
⎥⎥
−
−
=A
(e
+ 1) − 2 ⎨e ⎜
j
e
dt
⎟
⎟⎬
⎜
∫
⎢
⎢
⎥⎥
⎠ ⎭⎪
⎠
⎝
0⎝
0
0
0 ⎢⎪
0
⎥⎦ ⎥⎦
⎢⎣ 0
⎣⎩
T
⎤
A −j T
A ⎡
j
( cos 0T
=
+1 + j
sin 0 te − j t dt ⎥
e
⎢0 +
∫
⎢
⎥⎦
0
0 ⎣
0 0
)
(
)
(
=
A
(e
−j T
)
+1 + I
0
2
2
0
⎡
or, I ⎢1 −
⎢⎣
2
2
0
⎤ A −j T
+ 1) ⇒ I =
⎥ = (e
⎥⎦
0
A 0
−
0
2
2
(e
−j T
Solution
Here, f(t)
Fig. 7.44
a
−a
dt + ∫ − Ae
0
−j t
a
⎡ je − j t 0
je − j t ⎤ jA ⎡
⎢
⎥ = ⎣1 − e + j a − e − j a + 1⎤⎦
dt = A
−
⎢
⎥
−a
0⎦
⎣
( ) = j 2 A (1 − cos a )
⇒ F j
( )
Amplitude is F j
A
A
−∞
= ∫ Ae
f (t )
a
∞
−j t
)
a
∴ F ( j ) = ∫ f (t )e − j t dt
0
⎛ a⎞
sin 2 ⎜ ⎟
⎝ 2 ⎠
2A
2A 2 ⎛ a⎞ A a
=
1 − cos a =
sin ⎜ ⎟ =
2 ⎛ a⎞ 2
⎝ 2 ⎠
⎜⎝ 2 ⎟⎠
)
(
(
)
The amplitude is zero when 1 − cos a = 0 ⇒
1)
+1
Problem 7.25 Find the Fourier transform of the single pulse shown in Fig. 7.44 draw the
continuous magnitude and phase spectra.
A ; − a ≤ t ≤ 0;
A ; 0≤t ≤a
0 ; for all other values of t
cos
a = 2n
⇒
=
2n
a
t
452
Network Analysis and Synthesis
( )
Phase is ∠F j = + 90° when
> 0 = − 90°
The spectra are shown in the figures below.
F( j )
<0
when
A a
2
F( j )
90
90
0
4
a
Fig. 7.45
2
a
0
2
a
4
a
Amplitude spectra
Phase spectra
Problem 7.26 Find the Fourier transform of the single triangular pulse shown in Fig. 7.46 and draw the
continuous spectra.
Solution
i.e.,
The wave is, f(t)
⎡ 2 ⎤
V0 ⎢1 − t ⎥
⎣ a ⎦
⎡ 2 ⎤
f (t ) = V0 ⎢1 − t ⎥ ; for t > 0
⎣ a ⎦
∞
∴ F ( j ) = ∫ f (t )e
−∞
−j t
f (t )
V0
⎡ 2 ⎤
f (t ) = V0 ⎢1 + t ⎥ ; for t < 0
⎣ a ⎦
and
a/2
0
Fig. 7.46
∞
⎡ 2 ⎤
dt = ∫ V0 ⎢1 − t ⎥ e − j t dt
⎣ a ⎦
−∞
a
a
⎡
⎫⎤
⎧0
a
2
2
V
2
V
2
V
⎢
⎪
⎪⎥
−j t
−j t
−j t 2
−j t
j t
0
0
0
= V0 ∫ e dt −
t e dt =
−
e
⎨ ∫ −te dt + ∫ te dt ⎬ ⎥
⎢
a
∫
−
a a
−j ⎢
a ⎪ a
a
0
2
⎪⎥
−
−
⎭⎦
2
2
⎩− 2
⎣
a
2
a
V ⎛ − j a + j a ⎞ 2V 0
2V 2
= 0 ⎜ e 2 − e 2 ⎟ + 0 ∫ te − j t dt − 0 ∫ te − j t dt
a 0
−j ⎝
⎠ a a
−
2
a
a
a
⎡
⎤
⎡
⎤
⎛ j a −j a ⎞
0
2V0 ⎜ e 2 − e 2 ⎟ 2V0 ⎢ te − j t 2
e − j t ⎥ 2V0 ⎢ te − j t 2 2 e − j t ⎥
=
+
−
dt ⎥ −
−
dt ⎥
⎜
⎟ a ⎢ − j a ∫a − j
2j
a ⎢ − j 0 ∫0 − j
−
⎢
⎥
⎢
⎥
−
⎝
⎠
2
⎣
⎦
⎣
⎦
2
=
2V0
=
2V0
a
⎡
⎤
0 ⎤
⎡
⎛ a ⎞ 2V0 ⎢ ⎪⎧ a e + j al2 ⎪⎫ e = j t ⎥ 2V0 ⎢ ⎪⎧ a e − j al2
⎪⎫ e = j t 2 ⎥
sin ⎜ ⎟ +
− 0⎬ + 2 ⎥
−
⎬+ 2
⎨
⎨0 +
a ⎢ ⎩⎪ 2 − j
⎝ 2 ⎠ 2 ⎢ ⎩⎪ 2 − j ⎭⎪
a⎥
⎭⎪
0 ⎥
− ⎥
⎢
⎢⎣
2⎦
⎣
⎦
a
⎞ V
+j
2V
⎛ a ⎞ V + j a 2V ⎛
sin ⎜ ⎟ − 0 e 2 + 02 ⎜ 1 − e 2 ⎟ + 0 e − j a / 2 − 02 e − j a / 2 − 1
2
j
j
⎝ ⎠
a ⎝
a
⎠
(
)
a/ 2
t
453
Fourier Series and Fourier Transform
=
2V0
⎛ a ⎞ 2V ⎛ e − j a / 2 − e + j a / 2 ⎞ 2V0
+ j a/2
− e − j a/2 + 1
sin ⎜ ⎟ + 0 ⎜
⎟ + a 2 1− e
2j
⎝ 2 ⎠
⎝
⎠
=
2V0
⎛ a ⎞ 2V
⎛ a ⎞ 2V
sin ⎜ ⎟ − 0 sin ⎜ ⎟ + 02 2 − e − j a / 2 − e j a / 2
⎝ 2 ⎠ a
⎝ 2 ⎠
=
)
(
(
)
⎛ e + jω a / 2 − e − jω a / 2 ⎞ ⎤ 4V0 ⎡
4V0 ⎡
⎛ a ⎞ ⎤ 4V
2 ⎛ a⎞
1− 2⎜
⎥ = 2 ⎢1 − cos⎜ ⎟ ⎥ = 02 × 2 sin ⎜ ⎟
2 ⎢
⎟
2
2
aω ⎣
⎝ ⎠⎦ a
⎝
⎠⎦ a ⎣
⎝ 4 ⎠
∴ F( j ) =
8V0
a
2
⎛ a⎞
sin 2 ⎜ ⎟
⎝ 4 ⎠
⎪F (j )⎮
V0a
2
⎛ a⎞
sin
V0 a ⎜⎝ 4 ⎟⎠
Bringing it into standard form, F ( j ) =
2 ⎛ a⎞ 2
⎜⎝ 4 ⎟⎠
Its continuous amplitude spectrum is shown. The first
zero occurs when a =
i.e., a = .
4
4
0
8
a
4 0 4 8
a
a a
Fig. 7.47
Problem 7.27 Find the Fourier transform of the existing voltage
v(t) V0e t, t 0
0, t 0
and sketch approximately its amplitude and phase spectrum.
Solution
∞
∞
∞
−∞
−∞
−∞
F ( j ) = ∫ f (t )e − j t dt = ∫ V0 e − t e − j t dt = V0 ∫ e − (1+ j )t dt =
The amplitude and phase are F ( j ) =
V0
1+
2
and
v
( j ) = − tan −1 ( )
Phase,
F
∞
V0
V
⎡ e − (1+ j )t ⎤ = 0
⎣
⎦
− ∞ 1+
(1 + j )
+j
1.5
0.5
0
0.5
0
10 8
Fig. 7.48
1
6
4
2 0
2 4
6 8 10
1.5
10 8 6 4 2 0 2 4 6 8 10
454
Network Analysis and Synthesis
Problem 7.28 In the figure, Vi(t) 10sgn(t) volt. Using the Fourier transform method, find VC (t ) and
sketch VC (t ) versus time, t. Given: R 5 ohms, C 1F.
Solution
vi (t ) =10 sgn(t )
R
∴Vi ( j ) = 10 ×
2 20
=
j
j
Vc ( j ) =
Vi ( j )
× Xc
z( j )
V i( t )
C
V C ( t)
Transfer function of the circuit
V (j )
1/ j C
1
Fig. 7.49
H( j )= c
=
=
Vi ( j ) R + 1 / j C 1 + j RC
where, Vc( j ) is the Fourier transform of Vc(t)
V(j )
20
20
∴Vc ( j ) = H ( j ) × Vi ( j ) = i
× XC =
=
Z( j )
j (1 + j RC ) j (1 + j 5)
20
100
2
1
=
−
= 10( ) − 20
1 + ( j 5)
j
j
(1 / 5) + j
Taking inverse Laplace transform,
vC (t ) = 10 sgn(t ) − 20e
−t
5
u(t ) volt
To plot this curve, we follow the following steps:
• From
t
0, vi (t )
10 V,
10 V;
vC (t)
• At t
0, vi (t ) jumps from
10 V to 10 V
and thus, vC (t) approaches its final value of
10 V exponentially with time-constant of 5
seconds.
10 V
Voltage
5V
0V
Time
5V
3.0s 2.0s 1.0s 0s 1.0s 2.0s 3.0s 4.0s 5.0s 6.0s 7.0s
Fig. 7.50
Problem 7.29 Find the response voltage in the network
shown in Fig.7.51 Use Fourier transform method.
Solution By KCL,
Given:
v (t )
dv
i1 (t ) = 2 + 0.5 2
1
dt
1
I1(t) = 2e tu(t) (A)
0.5F V2
Fig. 7.51
−t
i1 (t ) = 2 e u(t )
Taking Fourier transform,
⎡ 1 ⎤
2
= V2 ( j ) ⎢1 + j ⎥
1+ j
⎣ 2 ⎦
⎡ 1
4
1 ⎤
V2 ( j ) =
= 4⎢
−
(1 + j )( 2 + j )
2 + j ⎥⎦
⎣1+ j
I1 ( j ) = V2 ( j ) +
Taking inverse Fourier transform
1
j V2 ( j )
2
or,
v2 (t ) = ( 4 e − t − 4 e −2 t )u(t )
V( t)
A
Problem 7.30 Find the Fourier transform of the sine pulse shown in Fig. 7.52
and sketch the amplitude and phase spectra. This voltage is applied to a series
RL circuit with R 1 ohm and L 1.0 H. Determine the amplitude and phase
spectra for the resulting current, i(t).
t (second)
0
Fig. 7.52
455
Fourier Series and Fourier Transform
Solution
⎡ 1 + e − j ⎤ A(1 + cos
=
[from Prob. 7.25] ⇒ V ( j ) = A ⎢
2 ⎥
1− 2
⎦
⎣ 1−
∴V( j ) = A
(1 + cos
)2 + sin 2
1−
)
Asin
1− 2
2(1 + cos
1− 2
=A
2
−j
)
= 2A
cos(
2
1−
)
2
⎞
− ⎞ −
−1 ⎛
⎟⎠ = tan ⎜⎝ tan 2 ⎟⎠ = 2
⎛ − sin
∴ ( j ) ⎡⎣ Angle of V ( j ) ⎤⎦ = tan −1 ⎜
⎝ 1 + cos
The amplitude and phase spectra are shown.
20
V(j )
Phase,
15
10
2A
5
0
5
10
15
0
20
10
0
8
6
4
2
0
2
4
6
8 10
Fig. 7.53
The current in the RL series circuit,
I( j )=
V( j ) ∠ ( j )
V( j )
=
R+ j L
= 2A
∴ I( j ) = 2A
cos
1−
cos(
1−
R2 +
/2
2
2
2
2
1
1+
)
L2 ∠ tan −1
2
1
1+
2
∠−
2
and
L
R
=
V( j ) ∠ ( j )
1+
2
∠ tan −1
− tan −1 = I ( j ) ∠ ( j )
( j )=−
2
− tan −1
Problem 7.31 The current in a 10-ohm resistor is i(t) 10 e 2tu(t) A. Calculate the total energy W dissipated in the resistor during the time interval t 0 to . What is the energy W1 associated with the frequency
band 0
2 rad/s
Solution The instantaneous power,
Total energy dissipated
p(t ) = i 2 (t ) ⋅ R = 10 × 100e −4 t ; t > 0
∞
∞
∞
⎡ e −4 t ⎤
1000
⎡0 − 1⎤⎦ = 250 J
W = ∫ p(t )dt = ∫ 1000e −4 t dt = 1000 ⎢
⎥ =−
−
4
4 ⎣
⎣
⎦0
−∞
0
456
Network Analysis and Synthesis
The Fourier transform of i(t)is
10
2+ j
I( j )=
The energy associated,
W1 =
2
10
∫ I( j ) d
10
2
∞
2
1
⎪⎧
⎨ 1 Energy is, W1 = ∫ F ( j ) d
⎪⎩
−∞
2
0
=
⎪⎫
⎬
⎪⎭
2
1000 ⎡ 1 −1 ⎤ 500
⎡ tan −1 (1) − tan −1 (0) ⎤ = 500 × = 125J
=
⎢ 2 tan 2 ⎥ =
⎣
⎦
4
⎣
⎦0
100
∫0 4 + 2 d
Problem 7.32 A voltage, v(t) 100e 25t u(t) volt is applied to the input of an ideal low-pass filter having
a cut-off frequency of 25 rad/s. Calculate the percentage of the total energy transmitted through the filter.
Solution Fourier transform of v (t )
V( j )=
100
25 + j
2
∴V( j ) =
Total 1-
energy available at the filter input is
The 1-
104 d
104
d
Wi1 = ∫
=
∫
2
0 625 +
0 625 +
energy available at the filter output is
∞
1
W01 =
1
∫ V( j ) d
2
=
104
0
2
∞
∞
25
104
625 +
2
25
d
∫0 625 +
d
⎤ 104 1
104 ⎡ 1
−1
=
× × = 200 J
⎢ 25 tan 25 ⎥ =
25 2
⎣
⎦0
25
⎤
104 ⎡ 1
−1
104 1
=
× × = 100 J
⎢ 25 × tan 25 ⎥ =
2
25 4
⎣
⎦0
percentage of the input energy appearing at the output,
W01
100
× 100 =
× 100% = 50%
Wi1
200
Problem 7.33 A voltage, v(t) 4e 3t u(t) volt is applied to the input of an ideal band-pass filter having a
pass-band defined by 1 < f < 2 Hz. Calculate the total 1- energy available at the output of the filter.
Solution Let the output voltage is v0 (t ). The energy in v0(t) will be equal to the energy of that part of v(t),
having frequency components in the intervals, 1 f 2 and 2 f
1.
∞
Fourier transform of input,
∞
V ( j ) = 4 ∫ e −3t u(t )e − j t dt = 4 ∫ e (
−∞
or,
energy in the input signal is,
Wi1 =
16
∞
d
∫0 9 +
)t
u(t )dt =
−∞
∞
So, the total 1-
− 3+ j
=
2
4
3+ j
∞
8
W1 = ∫ v 2 (t )dt = 16 ∫ e −6 t dt = J
3
0
−∞
16
∞
∞
d
∫0 9 +
=
2
16 ⎡ 1 −1 ⎤ 16 1
8
⎢ 3 tan 3 ⎥ = × 3 × 2 = 3 J
⎣
⎦0
457
Fourier Series and Fourier Transform
−2
−2
1
16 d
16
d
=
W0 =
∫
∫
2
2 −4 9 +
2 −4 9 +
Total energy in the output is
=
−2
⎤
16 ⎡ 1
= ⎢ tan −1 ⎥
2
3 ⎦− 4
⎣3
⎛2 ⎞⎤
16 1 ⎡ −1 ⎛ 4 ⎞
− tan −1 ⎜
× × ⎢ tan ⎜
⎥ = 0.358 J
⎟
3 ⎣
⎝ 3 ⎟⎠ ⎦
⎝ 3 ⎠
Problem 7.34 The voltage, Vi (t) 5e 5t u(t) volt is applied to the input of the RC circuit shown in Fig. 7.54.
Determine the percentage of the 1-⍀ energy that is transmitted to the output.
10k
1
1
=
= 4
= 110 rad/s
c
RC 10 × 10 × 10−6
Solution Here, the cut-off frequency,
Vi( t )
Fourier transform of vi(t)
Vi ( j ) =
2
∴ Vi ( j ) =
Total 1-
V 0( t)
Fig. 7.54
25
25 + 2
energy available at the filter input is
1
Wi1 =
The 1-
5
5+ j
10uF
∞
25d
∫0 25 + 2 d
=
25
∞
∞
d
∫0 25 +
d
2
=
25 ⎡ 1 −1 ⎤
25 1
⎢ 5 tan 5 ⎥ = × 5 × 2 = 2.5 J
⎣
⎦0
energy available at the filter output is
W01 =
1
10
∫V (j ) d
2
i
=
25
0
10
10
d
∫0 25 +
=
2
⎤
25 ⎡ 1
25 1
−1
⎢ 5 × tan 5 ⎥ = × 5 × 1.107 = 1.762 J
⎣
⎦0
percentage of the input energy appearing at the output,
W01
1.762
× 100 =
× 100% = 70.48%
Wi1
2.5
Problem 7.35 (a) Find the Fourier transform of the function,
)
f (t = Ae a for t ≥ 0 = 0 for t < 0
0 for t 0
(b) Use the above transform to find the output voltage V0 in the
Fig. 7.55.
−t
3
t
i (t)=e u(t )( A)
Fig. 7.55
Solution (a) Fourier transform of the function is
∞
∞
( ) = ∫ f (t )e
I j
−∞
−j t
∞
dt = ∫ Ae
0
−t
∞
a
⎛1
⎞
−⎜ + j ⎟ t
⎠
e − j t dt = A ∫ e ⎝ a
−∞
dt = A
⎛1
⎞
−⎜ + j ⎟ t
⎝a
⎠
e
Aa
=
1+ j a
⎛1
⎞
−⎜ + j ⎟
⎠0
⎝a
1F V ( t )
0
458
Network Analysis and Synthesis
V0 ( j ) V0 ( j )
⎡1+ 3 j ⎤
+
= V0 ( j ) ⎢
⎥
1
3
⎣ 3 ⎦
j
b) By KCL,
I( j )=
Here,
( ) = 1 +1j
I j
(from result of (a) with A
1 and a
1)
⎡1+ 3 j ⎤
1
= V0 ( j ) ⎢
⎥
1+ j
⎣ 3 ⎦
or,
⎡ 3
3
3
2
⎢
∴V0 ( j ) =
=
− 2
(1 + j )(1 + j 3 ) ⎢ 1 + j
1+ j
⎢⎣ 3
⎤
⎥
⎥
⎥⎦
3 −t
3
V0 (t ) = e 3 − e − t
2
2
Taking inverse Fourier transform
Problem 7.36 (a) For the pulse shown in Fig. 7.56, prove that.
sin
F ( j ) =V
f (t )
V
2
t
2
(b) Draw the frequency spectra of this waveform and explain how you would use this
result to estimate the bandwidth required for the transmission of such a signal.
(c) Calculate the percentage of energy associated with this pulse that lies in the
dominant portion of the amplitude spectrum.
Solution
a) The pulse is
f (t ) = V , −
2
<t <
ⴚ Ⲑ2 0
Ⲑ2
Fig. 7.56
2
So, the Fourier transform,
∞
∞
−∞
−∞
F ( j ) = ∫ f (t )e − j t dt = ∫ Ve − j t dt =V
( )
∴F j
e
j
2
−e
j
−j
2
= 2V
⎛ ⎞
sin ⎜ ⎟
⎝ 2⎠
⎛ ⎞
sin ⎜ ⎟
⎝ 2⎠
= 2V
×
2
⎛ ⎞
⎜⎝ 2 ⎟⎠
F(j )
⎛ ⎞
sin ⎜ ⎟
⎝ 2⎠
=V
⎛ ⎞
⎜⎝ 2 ⎟⎠
V
The plot of sin x versus x (here, x =
) is shown in Fig.7.57.
2
x
b) The function goes through zero when x =
is an integral multiple of .
2
The function is unity at x 0. This form is called sampling function or interpolating function or filtering function, and it occurs frequently in modern
communication theory.
6
Fig. 7.57
4
2 0 2
4 6
459
Fourier Series and Fourier Transform
From the figure, we see that the major portion of the amplitude spectrum of the rectangular pulse spreads over
2
2
to
. If the pulse is carried through a transmission system, the bandwidth
the frequency range from −
(BW) of the system must accommodate the major portion of the amplitude spectrum for reasonable fidelity in
2
transmission; i.e., the cut-off frequency of the system must be at least, C =
.
⎡
2 ⎤
Thus, C × = 2 ⎢ BW =
⎥
⎣
⎦
product of the bandwidth and pulse width is a constant.
(c)
We know that the dominant portion of the amplitude spectrum lies in the frequency range 0 ≤ ≤
2
.
The Fourier transform of the rectangular voltage pulse is
The portion of the total 1spectrum is
⎛ ⎞
sin ⎜ ⎟
⎝ 2 ⎠
V ( j ) =V
⎛ ⎞
⎜⎝ 2 ⎟⎠
energy associated with v(t) that lies in the dominant portion of the amplitude
2
W1′Ω =
1
∫V
0
=
=
=
⎛ ⎞
sin 2 ⎜ ⎟
⎝ 2⎠
2 2
⎛ ⎞
⎜⎝ 2 ⎟⎠
2
d =
2V 2
sin 2 x
∫ x
0
2
⎫
⎧
dx ⎨let , x =
,∴dx = d ⎬
2
2
⎭
⎩
sin 2 x ⎤
2V 2 ⎡ 2 ⎧ 1 ⎫
sin 2 x ⎤ 2V 2 ⎡
⎢sin x ⎨− ⎬ + ∫
dx ⎥ =
dx ⎥
⎢0 + ∫
⎢⎣
⎥⎦
⎢⎣ 0 x
⎥⎦
⎩ x ⎭0 0 x
4V 2
2V 2
sin 2 x
2V 2
d
x
=
∫0 2 x
∫
sin
0
d
1
[ Let , = 2 x,∴dx = d ]
2
× 1.418
[The value of the integral as found from the table of sine integrals is 1.418]
2V 2
∴W1′ =
× 1.418
Total 1-
energy for v(t) is
W1 = ∫V 2 d = V 2
0
Hence the percentage of total energy contained in the dominant portion of the amplitude spectrum is
W1′
2 × 1.418
× 100 =
× 100 = 90.2%
W1
460
Network Analysis and Synthesis
Summary
1. A function of time f (t ) is said to be periodic if it repeats
itself every T seconds i.e. f (t ) f (t nT ) where n is a
positive integer and ‘T ’ is the period. Thus, a periodic
function repeats itself every T seconds.
2. The conditions under which a periodic function f (t )
can be expanded in a convergent Fourier series, are
known as Dirichlet’s conditions
3. Any non-sinusoidal periodic function can be represented by Fourier series expansion as
)
f (t ) = a 0 + (a1 cos t + a 2 cos 2 t + ⋅⋅⋅
(
(
= a 0 + ∑ a n cos n t + bn sin n t
n =1
∞ ⎛
A ⎞
Feff ⎡⎣or Fr ms ⎤⎦ = A0 2 + ∑ ⎜ n ⎟
⎝
n =1
2⎠
)
)
∞
( ) = F ⎡⎣f (t )⎤⎦ = ∫ f (t )e
T
and
bn =
()
∞
n =− ∞
)
(
= ∫ F ( j 2 f e j 2 f df
−∞
10. If f (t ) is a single-valued function and is different from
zero over an infinite interval of time then Fourier trans-
∫ f (t ) dt < ∞.
−∞
T
where,
( )
form will exist if
4. For an odd function, a0 0 and an 0, for an even
function, bn 0 and for a function with half-wave symmetry, a0 an bn 0 for even values of n.
5. The exponential form of Fourier series expansion is
∞
and
∞
2
f (t )sin n tdt
T ∫0
f (t ) = C 0 + ∑ C n e j n t
dt
inverse Fourier transform is defined as
∞
1
f t = F −1 ⎡⎣ F j ⎤⎦ =
F ( j )e j t d
∫
2 −∞
T
2
a n = ∫ f (t )cos n tdt
T 0
−j t
−∞
where, the Fourier coefficients are given as
1 T
∴ a 0 = ∫ f (t )dt
T 0
2
8. Steady-state response of a circuit with non-sinusoidal
periodic excitation can be found using Fourier series
representation.
9. Fourier transform is used for aperiodic functions. It
is defined as F j
+ ⋅⋅⋅ + b1 sin t + b2 sin2 t + ⋅⋅⋅ + ⋅⋅⋅
∞
7. Effective or rms value of a periodic function is given as,
Cn =
1
f (t )e − jn t dt .
T ∫0
6. If a periodic function is written as
∞
f (t ) = A0 + ∑ An cos ( n t −
n =1
n
)
then variation of An with n (or n ) is known as the amplitude spectrum or frequency-spectrum and variation of
n with n (or n ) is known as the phase-spectrum of
the signal.
11. Fourier transform of a constant is an impulse, i.e.,
∴ F ⎡⎣ K ⎤⎦ = 2 K
. Fourier transform of an impulse
( )
∴ F ⎡⎣ K (t ) ⎤⎦ = K .
12. Fourier transform of a unit step function u(t ) is a combination of rectangular hyperbola and impulse func1 .
tion (of strength at
0), i.e. F ⎡u t ⎤ =
+
⎣
⎦
j
13. Some important properties of Fourier transform are
listed in Table 7.1.
14. The relation between a function f (t ) and its Fourier
transform is given by Parseval’s theorem, given as
is a constant, i.e.
()
∞
W = ∫ f 2 (t )dt =
−∞
( )
∞
2
1
F(j ) d .
2 −∫∞
Short-Answer Questions
1. What are the conditions which a periodic function
must satisfy to have its Fourier series expansion?
The conditions under which a periodic function f(t) can
be expanded in a convergent Fourier series are known
as Dirichlet’s conditions.
These are as follows:
(i) f (t ) is a single-valued function.
(ii) f (t ) has a finite number of discontinuities in each
period, T.
(iii) f (t ) has a finite number of maxima and minima in
each period, T.
T
∫ f (t ) dt exists and is finite or in
(iv) The integral,
0
T
another way, ⎡f (t ) ⎤2dt < ∞ .
∫⎣ ⎦
0
461
Fourier Series and Fourier Transform
2. Derive an expression for the effective value of a
non-sinusoidal periodic waveform.
T
2 2
a n = ∫ ⎡⎣f ( x ) + f ( − x ) ⎤⎦ cos n xdx
T 0
Thus,
Or,
T
bn =
2 2
⎡f ( x ) − f ( − x ) ⎤⎦ sin n xdx
T ∫0 ⎣
Discuss the method of computing the effective
value of a non-sinusoidal periodic waveform.
Similarly,
The effective (or rms) value of a periodic function f (t ) is
defined as
For an odd function f (x ),
f (x )
f ( x)
Hence, for odd functions a0 0 and
Feff ( Fr ms )
an
0
T
=
T
T
∞
2
1
1 ⎡
⎡⎣f (t ) ⎤⎦ dt =
⎢ A0 + ∑ An cos n t −
∫
∫
T 0
T 0⎣
n =1
(
2
⎤
dt
n ⎥
⎦
)
∞
1⎡ 2
2T ⎤
=
⎢ A T + ∑ An ⎥
T ⎣ 0
2⎦
n =1
⎛A ⎞
( Fr ms ) = A0 + ∑ ⎜ n ⎟
n =1 ⎝ 2 ⎠
∞
Feff
bn =
and
Thus, the Fourier series expansion of an odd function
contains only the sine terms, the constant and the
cosine terms being zero.
4. Show that the Fourier-series expansion of a periodic function with even (mirror) symmetry contains only the cosine terms plus a constant.
2
2
This shows that the effective value of a periodic function is the square root of the effective values of the harmonic components and the square of the dc value.
3. Show that the Fourier series expansion of a periodic function with odd (rotation) symmetry contains only the sine terms.
The Fourier coefficients are obtained as follows.
T
⎤
⎡ 0
T
2
1
1⎢
a 0 = ∫ f (t )dt =
f (t )dt + ∫ f (t )dt ⎥
∫
⎥
T 0
T ⎢ −T
0
⎥⎦
⎣⎢ 2
Putting t
x in the first integrand and t
second integrand, we get
The Fourier coefficients are obtained as follows.
a0 =
T
⎡ 0
⎤
T
2
1
1⎢
⎥
f
(
t
)
dt
f
(
t
)
dt
f
(
t
)
dt
=
+
∫0
⎥
T ∫0
T ⎢ −T∫
⎥⎦
⎣⎢ 2
Putting t
x in the first integrand and t
second integrand, we get
⎡T
⎤
1 2
a 0 = ⎢ ∫ ⎡⎣f ( x ) + f ( − x ) ⎤⎦dx ⎥
⎥
T ⎢0
⎢⎣
⎥⎦
x in the
Now,
T
an =
2
f (t )cos n tdt
T ∫0
⎡T
⎤
0
2 2
2
= ⎢ ∫ f (t )cos n tdt + ∫ f (t ) cos n tdt ⎥ = ⎡⎣ I 1 + I 2 ⎤⎦
⎥ T
T ⎢0
T
−
⎢⎣
⎥⎦
2
Since the variable ‘t ’ in I1 and I2 integrals is a dummy
t in I2.
variable, let x t in I1 and x
∴ an =
4 2
f ( x )sin n x
T ∫0
T
⎡T
⎤
2
2⎢ 2
f
(
x
)cos
n
xdx
−
f ( − x )cos n x ( −dx ) ⎥
∫
∫
⎥
T ⎢0
0
⎢⎣
⎥⎦
a0 =
x in the
⎡T
⎤
1⎢ 2
⎡⎣f ( x ) + f ( − x ) ⎤⎦dx ⎥
∫
⎢
⎥
T 0
⎢⎣
⎥⎦
Now,
T
an =
2
f (t )cos n tdt
T ∫0
⎡T
⎤
0
2⎢ 2
2
=
f (t )cos n tdt + ∫ f (t ) cos n tdt ⎥ = ⎡⎣ I 1 + I 2 ⎤⎦
∫
⎥
T
T ⎢0
−T
⎢⎣
⎥⎦
2
Since the variable ‘t ’ in I1 and I2 integrals is a dummy
t in I2.
variable, let x t in I1 and x
T
⎡T
⎤
2
2⎢ 2
∴ an =
f ( x )cos n xdx − ∫ f ( − x )cos n x ( −dx ) ⎥
∫
⎥
T ⎢0
0
⎢⎣
⎥⎦
T
Thus,
an =
2 2
⎡f ( x ) + f ( − x ) ⎤⎦ cos n xdx
T ∫0 ⎣
T
Similarly,
bn =
2 2
⎡f ( x ) − f ( − x ) ⎤⎦ sin n xdx
T ∫0 ⎣
462
Network Analysis and Synthesis
For an even function f (x ),
f (x )
T
0
2 2
f ( x )dx
T ∫0
T
an =
2
4
f ( x )cos n xdx and bn 0
T ∫0
Thus, the Fourier series expansion of an even periodic
function contains only the cosine terms plus a constant, all sine terms being zero.
an =
5. Show that the Fourier series expansion of a periodic function with half-wave symmetry contains
only the odd harmonics.
X T/2
0
A periodic function f (t ) is said to
t
0
T
T
2
t/2
time period of the
T
⎡ 0
⎤
2
1⎢
1
∴ a0 =
f (t )dt + ∫ f (t )dt ⎥ = ⎡⎣ I 1 + I 2 ⎤⎦
∫
⎥
T ⎢ −T
T
0
⎢⎣ 2
⎥⎦
For I1, let x (t T 2); so, f (t ) f (x T 2)
and dt dx
0
0
2
2
) ∫ f (t )cos n tdt
(
2
1− cos n
T
0; for even n, and
0
T
=
4 2
f (t )cos n tdt , for odd n.
T ∫0
Similarly, bn
0, for even n; and
T
4 2
= ∫ f (t )sin n tdt , for odd n.
T 0
Thus, the Fourier-series expansion of a periodic function having half-wave symmetry contains only odd
harmonics, the constant term being zero.
6. What is Gibb’s Phenomena? Explain.
In mathematics, the Gibb’s phenomenon (also known as
ringing artifacts), named after the American physicist
J Willard Gibbs, is the peculiar manner in which the Fourier series of a piecewise continuously differentiable
periodic function f (t ) behaves at a jump discontinuity.
f(x)
1
K=1
2
∴ I 1 = ∫ f (t )dt = ∫ −f ( x )dx = − ∫ f ( x )dx
−T
0
T
have half-wave symmetry if it
satisfies the condition
f (t )
f (t T 2), where T
function
2
= ∫ −f ( x )cos n cos n xdx = ∫ −f (t )cos n cos n tdt
f ( x)
T
∴ a0 =
T
2
0
t
0
1
T
⎡ T2
⎤
2
1
∴ a 0 = ⎢ − ∫ f ( x )dx + ∫ f (t )dt ⎥
⎥
T ⎢ 0
0
⎢⎣
⎥⎦
T
T
⎡
⎤
2
1 2
= ⎢ ∫ f ( x )dx − ∫ f ( x )dx ⎥ = 0
⎥
T ⎢0
0
⎢⎣
⎥⎦
1
K=5
0
t
1
⎡
⎤
2⎢ 2
f (t )cos n tdt ⎥
∫
⎥
T ⎢ −T
⎢⎣ 2
⎥⎦
T
⎡ 0
⎤
2
2
2⎢
=
f (t )cos n tdt + ∫ f (t )cos n tdt ⎥ = ⎡⎣ I 1 + I 2 ⎤⎦
∫
⎥
T ⎢ −T
T
0
⎥⎦
⎣⎢ 2
T
∴ an =
Again putting x
procedure,
(t
T 2) and following the same
T
0
2
I 1 = ∫ f (t )cos n tdt = ∫ −f ( x )cos n
−T
T
2
0
2
(
= ∫ −f ( x )cos n x − n
0
)dx
(
x −T
2
)dx
1
K = 11
0
t
1
1
K = 49
0
1
Fig. 7.58 Fourier series approximation
of square wave; number of terms in Fourier
sum is indicated as K in each plot
t
463
Fourier Series and Fourier Transform
It is observed that when a periodic waveform is
truncated by a Fourier series with a finite number of
terms, there is a considerable amount of error near
the points of discontinuity of the wave. The amount of
error is decreased with the increase of number of terms
included in the truncated Fourier series. This phenomenon is known as Gibb’s phenomenon.
For example, we consider a square wave as shown in
Fig. 7.58. A general approximation of the wave can be
obtained by taking more and more number of terms of
the Fourier series expansion.
The figures show the wave-shapes taking the first
term, first 5 terms, first 11 terms and first 49 terms,
respectively. The rate of oscillation of ripples increases
near the points of discontinuity as the contribution of
more harmonics is taken into consideration. The waveshape tends to perfectly match the given waveform
when a large number of harmonics is considered.
If we consider a point where the waveform f (t ) is
discontinuous, with different limits to the right and left
of as f (
) and f (
), respectively then the value of
the function at will be,
T
Cn =
where,
1 2
f (t )e − j n t dt
T −T∫
If the period T becomes infinite, the function does not
repeat itself and becomes aperiodic or non-periodic.
So, the interval between adjacent harmonic frequencies is
2
= n +1 − n = =
T
1
or,
(iii)
=
=
T 2
2
As T → or
→ d , and the frequency goes from a
discrete variable over to a continuous variable.
)
(
1 d
and n →
→
T
2
From (2) and (4),
This is the Fourier transform of
f (t ) i.e., F( j ).
−∞
∞
F ( j ) = F ⎡⎣f (t ) ⎤⎦ = ∫ f (t )e − j t dt
So, from Eq. (1),
The truncated Fourier series must pass through these
three points, f ( ), f ( ) and f ( ) for correct representation of the wave.
As T →
−∞
∞
⎛ 1⎞
f (t ) = ∑ (C nT e jn t ⎜ ⎟
⎝T ⎠
−∞
)
7. When do we use Fourier transform?
Discuss that Fourier integral is the limit of
Fourier series, as time period T of a repetitive
wave approaches infinity as the limit.
Or,
and ∑
→
from (5),
(v)
1 d
→
T
2
∫ (summation approaches integration). Thus,
, CnT → F( j ), n
→
, and
∞
f (t ) =
1
f ( j )e j t d
2 −∫∞
This is the inverse Fourier transform.
How would you obtain Fourier integral from Fourier
series?
Fourier transform is an integral method for studying the
steady-state behaviour of linear circuits. This transform
is used for analyzing non-periodic functions. Periodic
functions are analyzed by Fourier series expansion. But
a Fourier series becomes a Fourier transform when the
time period of the function becomes very very large,
i.e., T → or → . Under this condition, the discrete
line spectra become continuous spectra.
Fourier transform as a limit of Fourier series
Consider the exponential Fourier series,
∞
−∞
(iv)
∞
C nT → ∫ f (t )e − j t dt
f ( +)+ f ( −)
)=
2
or, f ( ) − f ( − ) = f ( + ) − f ( )
f(
f (t ) = ∑C n e j n t
(ii)
2
(i)
8. What is the difference between a Fourier series and
Fourier integral?
a) Fourier series is applicable for periodic function
whereas Fourier integral (transform) is applicable
for non-periodic functions.
b) Amplitude spectrum in case of Fourier series is a
line spectrum whereas, in case of Fourier transform,
the amplitude spectrum is a continuous spectrum.
9. How does Fourier transform differ from Laplace
transform?
The defining equations are,
∞
∞
0
−∞
F ( s ) = ∫ f (t )e − st dt and F ( j ) = ∫ f (t )e − j t dt
464
Network Analysis and Synthesis
The following are some differences and similarities:
(a) Laplace transform is one-sided in the interval
0 t
and Fourier transform is double-sided in
the interval
t
. Thus, Laplace transform
is applicable for positive time function, f (t ), t 0;
while Fourier transform is applicable for functions
defined for all times.
(b) Laplace transform includes the initial conditions
and is applicable for transient analysis; while Fourier
transform is only applicable for steady-state analysis.
(c) For functions f (t )
0 for t
and ∴ P
10. Show that when f (t) is an even function of t, its Fourier transform F( j␻) is an even function of ␻ and
is real; while when f (t) is an odd function of t, its
Fourier transform F( j␻) is an odd function of ␻ and
is imaginary.
From the definition of Fourier transform,
−∞
)
F ( j ) = ∫ f (t )e − j t dt = ∫ f (t )( cos t − j sin t dt
∞
∞
−∞
−∞
= ∫ f (t )cos tdt − j ∫ f (t )sin tdt = P ( ) + jQ ( )
where,
∞
P ( ) = ∫ f (t )cos tdt = Even function of
−∞
i.e. , P ( ) = P ( − )
∞
and
Q ( ) = ∫ f (t )sin tdt = Odd function of
−∞
i.e., Q ( ) = −Q ( − )
Now,
j
F(j )= F(j )e ( )
F ( j ) = P 1 ( ) + Q 2 ( ) = Even function of
and
0
0
entire s-plane, while, Fourier transform is restricted
to the imaginary ( j ) axis.
(d) Laplace transform is applicable to a wider range of
functions than the Fourier transform. On the other
hand, Fourier transforms exist for signals that are not
physically realizable and have no Laplace transform.
−∞
( ) = 2 ∫ f (t )cos tdt
∞
0 and ∫ f (t ) dt < ∞,
Thus, Laplace transform is associated with the
∞
∞
∴P
Q( ) 0
So, F( j ) jQ( ) even and real
• When f (t ) is an odd function
- f (t ) cos t is an odd function
- f (t ) sin t is an even function
P( ) 0
the two transforms are related as F ( j ) = F ( s ) s = j .
∞
- f (t ) sin t is odd function
( ) ⎤⎥ = Odd function of
⎢⎣ ( ) ⎥⎦
∞
( ) = 2 ∫ f (t )sin tdt
0
So,
F( j )
jQ( )
11. Prove that the Fourier transform of the convolution
of two time-varying functions is equal to the product of the Fourier transform of each function.
According to convolution integral, if h(t ) is the impulse
response of a linear network, then the response of the
same network y(t ) subject to any arbitrary input w(t ) is
given by the convolution integral as,
∞
∞
−∞
−∞
y (t ) = ∫ h ( )w (t − )d = ∫ w ( )h (t − )d
Y(
) = F ⎡⎣ h (t ) * w (t )⎤⎦ = H ( )W ( )
(ii)
i.e., convolution in time-domain corresponds to multiplication in frequency-domain.
Proof
we get,
Taking Fourier transform of both sides of (i),
Y(
∞
) = ∫ ⎡⎣ h ( )w (t − )d ⎤⎦e
−j t
dt
−∞
Exchanging the order of integration and factoring h( )
which is independent of t, we get,
Y(
∞
∞
−∞
−∞
) = ∫ h ( ) ∫ ⎡⎣w (t − )e
For the integral within bracket, let,
t (
), and dt d
∴Y
∞
∞
−∞
−∞
( ) = ∫ h ( ) ∫ ⎡⎣w ( )e
⎡Q
• When f (t ) is an even function
- f (t ) cos t is an even function
(i)
If W( ), H( ) and Y( ) are the Fourier transforms of w(t ),
h(t ) and y(t ), respectively, then
∞
( ) = tan ⎢ P
−1
Odd and Imaginary
= ∫h
−∞
Y(
( )e
−j
−j
∞
d
) = H ( )W ( )
−j t
dt ⎤⎦d
(t
(+)
d ⎤d
⎦
∫ w ( )e
−∞
), so that,
−j
d
465
Fourier Series and Fourier Transform
12. When a complex voltage wave is applied to a pure
capacitor, the current wave has more harmonics
than the applied voltage wave. Explain why.
We consider a voltage as given below be applied to a
pure capacitor C.
v V1m sin t V2m sin 2 t V3m sin 3 t
The capacitance reactances for different harmonics are
as given.
1
; for fundamental
C
1
=
; for second harmonic
2 C
1
=
; for third harmonic, and so on.
3 C
XC =
I r ms =
=
1
2
1
2
1
2
V1m 2 +V 2 m 2 +V 3 m 2 + ⋅⋅⋅
(V
C
1m
) + (2V
2
2m
C
) + (3V
2
3m
C
) + ⋅⋅⋅
2
( V + 4V + 9V + ⋅⋅⋅⋅ ) C
2
1m
2
2m
)
(
)
(
)
(
From v and i, it is seen that the percentage harmonics in the current wave is less than that in the voltage
wave. For nth harmonic, the percentage harmonic in the
current wave is 1/n -times than in the voltage wave.
Respective rms values of the voltage and current are
given as,
Hence, the current waveform is obtained by the principle of superposition considering the different harmonic components.
i V1m ( C )sin( t 90 ) V2m (2 C )sin(2 t 90 )
V3m (3 C )sin(3 t 90 )
From v and i, it is seen that the percentage harmonics
in the current wave is more than that in the voltage
wave. For nth harmonic, the percentage harmonic in the
current wave is n times than in the voltage wave.
Respective rms values of the voltage and current are
given as,
V r ms =
Hence, the current waveform is obtained by the principle of superposition considering the different harmonic components.
V
V
∴ i = 1m sin t − 90° + 2 m sin 2 t − 90° +
L
2 L
V3m
sin 3 t − 90° + ⋅ ⋅ ⋅
3 L
2
3m
From the above discussion, we conclude that when a complex voltage wave is applied to a pure capacitor, the current
wave has more harmonics than the applied voltage wave.
13. When a complex voltage wave is applied to a pure
inductor, the current wave has lesser harmonics
than the applied voltage wave. Explain why.
We consider a voltage as given below be applied to a
pure inductor L.
v V1m sin t V2m sin 2 t V3m sin 3 t
The inductance reactances for different harmonics are
as given.
L; for fundamental
XC
2 L; for second harmonic
3 L; for third harmonic, and so on.
V RMS =
1
I RMS =
1
=
2
V1m 2 +V 2 m 2 +V 3 m 2 + ⋅⋅⋅
2
2
2
⎛ V1m ⎞ ⎛ V 2 m ⎞ ⎛ V 3 m ⎞
⎜
⎟ +⎜
⎟ +⎜
⎟ + ⋅⋅⋅
2 ⎝ L ⎠ ⎝ 2 L⎠ ⎝ 3 L⎠
2
⎞ 1
1 ⎛
V 2
2 V
⎜ V1m + 2 m 4 + 3 m 9 + ⋅⋅⋅⎟
2⎝
⎠ L
From the above discussion, we conclude that when a
complex voltage wave is applied to a pure inductor,
the current wave has lesser harmonics than the applied
voltage wave.
14. If a voltage wave containing a dc component is
applied to a series RC circuit, the current wave
does not contain the corresponding dc component.
Explain why.
We consider a voltage wave as given by
v V0 V1m sin t V3m sin 3 t V5m sin 5 t
be applied to a series RC circuit. Here, V0 is the dc component of the voltage wave.
The impedance of the circuit at any frequency is
⎛
1 ⎞
Z (n = ⎜ R +
jn C ⎟⎠
⎝
)
so that the current for different harmonics will be
I=
V0
Z (0
+
V1
+
V3
+
V5
) Z (1) Z (3) Z (5)
+ ⋅⋅⋅
Now, the impedance corresponding to the dc component is Z(0)
Hence the dc component of the current is
I0 =
V0
Z (0
)
=
V0
=0
∞
466
Network Analysis and Synthesis
Hence, we see that if a voltage wave containing a dc
component is applied to a series RC circuit, the current wave does not contain the corresponding dc
component.
The plot of
sin x
versus x (here, x =
) is shown in
2
x
Fig. 7.60.
F(j )
15. Find the amplitude-frequency distribution of a
single non-repetitive voltage pulse of duration one
microsecond and explain how its frequency-bandwidth is estimated.
V
Or,
Consider a periodic voltage pulse waveform of
period T (second) and width T0 (second). Find an
expression for the frequency-spectra of this waveform and explain how you would use this result to
estimate the bandwidth required for the transmission of such a signal.
Or,
f( t)
(a) For the pulse shown in
Fig. 7.59, prove that
V
␻␦
2
F ( j ␻ ) =V ␦
␻␦
2
sin
ⴚ␦Ⲑ2 0
2
<t <
∞
∞
−∞
−∞
F ( j ) = ∫ f (t )e − j t dt = ∫ Ve − j t dt =V
( )
∴F j
⎛ ⎞
sin⎜ ⎟
⎝ 2 ⎠
=V
⎛ ⎞
⎜⎝ 2 ⎟⎠
2
4
6
(b) The function goes through zero when x =
2
So, the Fourier transform,
= 2V
0
Fig. 7.59
f (t ) = V , −
⎛ ⎞
sin⎜ ⎟
⎝ 2 ⎠
4
Fig. 7.60
␦Ⲑ2
(b) Draw the frequency spectra of this waveform
and explain how you would use this result
to estimate the bandwidth required for the
transmission of such a signal.
(a) The pulse is,
6
0
t
e
j
⎛ ⎞
sin⎜ ⎟
⎝ 2 ⎠
= 2V
×
2
⎛ ⎞
⎜⎝ 2 ⎟⎠
2
−e
j
−j
2
is
2
. The function is unity
an integral multiple of
at x 0. This form is called sampling function or
interpolating function or filtering function, and
it occurs frequently in modern communication
theory.
From Fig. 7.60, we see that the major portion
of the amplitude spectrum of the rectangular
pulse spreads over the frequency range from
2
2
−
to
. If the pulse is carried through a
transmission system, the bandwidth (BW) of the
system must accommodate the major portion of
the amplitude spectrum for reasonable fidelity
in transmission; i.e. the cut-off frequency of the
system must be at least,
Thus,
C
× =2
C
=
2
.
⎡
2 ⎤
⎢ BW =
⎥
⎣
⎦
product of the bandwidth and pulse width is a
constant.
467
Fourier Series and Fourier Transform
Exercises
Fourier Series
1F
1. Find the Fourier series expansion for the following
functions and sketch the frequency spectrum.
(a)
f (t )
v (t)
10
0
A
2H
v(t )
t
20
3
2
Fig. 7.63 (a)
Fig. 7.63 (b)
t
0
T
(b)
T
4. Find the Fourier series expansion for the waveforms
shown in Fig. 7.64.
2T
f (t )
⎡
⎤
1
1
1
[(a) v = −2 ⎢ sin x + sin2 x + sin3 x + sin 4 x + ⋅⋅⋅⎥
2
3
4
⎣
⎦
T/2
T T/2 0 T/2 T
2T
4V
V 4V
4V
cos 3 x +
cos 5 x + ⋅⋅⋅ ]
(b) v = 2 + 2 cos x +
2
2
5
3
t
( )
f (t)
(c)
(a)
1
v
2
0
2
2
0
( )
4
x
3
t
Fig. 7.56
(b) f (x )
A ∞ A
[Ans: (a) f (t = + ∑
sin n t
2 n =1 n
)
V
( ) T4 − 2T ⎡⎢cos t + 31 cos 3 t + ⋅⋅⋅⎤⎥
(b) f t =
2
2
⎣
∞
( ) 1 + 21 sin t − 2 ∑ 4 n1 − 1cos2n t ]
(c) f t =
n =1
0
⎦
2
2. A periodic waveform as shown in Fig. 7.62 feeds an RL
1
load with R 10 ohm and L
H. Calculate the power
2
at the fundamental frequency supplied to the load.
3
x
Fig. 7.64
5. A triangular wave increases linearly from 0 to Vm during
the interval 0 to . The wave has zero value during the
interval to 2 and this cycle is repeated. Find the
Fourier series representation of the wave.
[v =
V m 2V m ⎛
⎞
1
− 2 ⎜ cos x + cos 5 x + ⋅⋅⋅⎟ +
4
25
⎝
⎠
⎤
Vm ⎛
⎞
1
1
1
sin x − sin2 x + sin3 x − sin 4 x + ⋅⋅⋅⎟ ]
2
3
4
⎝⎜
⎠⎦
f (t )
A
t
0
2
T
2T
Fig. 7.62
3. A waveform of the shape shown in Fig. 7.63 (a) is
applied to the network shown in Fig. 7.63 (b). Calculate the power dissipated in a 20- resistor. Take
1 rad/s.
[1.17 W]
6. A wave has a constant value Im during the interval −
2
3
. This cycle
2
2
2
is repeated in the next intervals. Find the Fourier series
for the wave.
to
and
Im during the interval
to
⎡ 4I m ⎛
⎞⎤
1
1
1
⎢i =
⎜⎝ cos − 3 cos 3 + 5 cos 5 − 7 cos 7 + ⋅⋅⋅⎠⎟ ⎥
⎣
⎦
468
Network Analysis and Synthesis
7. (a) Find the trigonometric Fourier series for the voltage
wave shown in Fig. 7.65.
12. The voltage source in Fig. 7.68 is an exponentially
decaying pulse,
v (t )
v(t )
for t
for t
t
0
0
Find the output voltage V0.
1.0
0
0
e
⎛ 1 ⎞ − t RC
RC − t
−
e
e
[V0 = ⎜
1− RC
⎝ 1− RC ⎟⎠
for t
0.]
0.5 1.0 1.5 2.0 t(second )
C
Fig. 7.65
(b) If this voltage is applied to a capacitor of 1F, find the
current.
8. A series RLC circuit with R
has an applied voltage
v(t)
150 sin 1000t
5
,L
100 sin 2000t
5 mH, C
75 sin 3000t (V)
Determine the effective current and average power.
[16.58 A; 1374 W]
Fourier Transform
9. Find the Fourier transform of the following functions:
(i) f (t ) e at u(t ), a 0
(ii) f (t ) e a t , for all values of t
(iii) f (t ) 1
(iv) Unit impulse function, (t )
(v) Signum function, sgn(t )
(vi) Unit step function, u(t )
R
v (t)
50 F
Fig. 7.68
13. A cosine pulse v
Vm cos t is zero for all time except
−
≤ t ≤ . Find the Fourier transform of the pulse
2
2
and sketch the continuous amplitude spectrum and
⎡ 2V m
⎛
⎞⎤
cos ⎜
⎢
2
⎟⎠ ⎥
2
⎝
−
1
⎣
⎦
phase spectrum.
14. Find the Fourier transform of the triangular pulse
shown in Fig. 7.69.
10. Determine the output voltage response across the
capacitor to a current source excitation i(t ) e tu(t ), as
shown in Fig. 7.66.
[v (t ) e t e 2 t (V )]
f (t )
1.0
⎡
⎛ t⎞ ⎤
⎥
⎢ sin2 ⎜
⎝ 4 ⎟⎠ ⎥
⎢ t
⎢2
2 ⎥
⎛
t⎞ ⎥
⎢
⎜
⎟
⎢
⎝ 4 ⎠ ⎥⎦
⎣
t
0
i (t)
0.5
t
1 F v(t)
Fig. 7.69
Fig. 7.66
11. The current source in Fig. 7.67 is i(t ) 4e t for t 0.
Find the voltage V0 using Fourier transform method.
[v (t ) 8e t 8e 2 t (V )]
V0
15. Determine the response of the network shown in
Fig. 7.70 (b) when a voltage having the waveform
shown in Fig. 7.70 (a) is applied to it, by using Fourier
transform method.
v (t)
1
i (t )
1
0.5 Fv (t
1
v (t )
0
0
Fig. 7.67
Fig. 7.70 (a)
t
Fig. 7.70 (b)
1F
469
Fourier Series and Fourier Transform
Questions
1. (a) What are the conditions which a periodic function
must satisfy to have its Fourier series expansion?
(b) Write the trigonometric form of the Fourier series
for a function f(t) and explain, by deriving necessary relations, how the values of various coefficients are obtained.
Or,
What do you understand by Fourier series? Outline
the general procedure of determining Fourier series
of periodic waveform.
(c) Give the exponential form of Fourier series for a
periodic function.
2. Derive an expression for the effective value of a nonsinusoidal periodic waveform
Or,
Discuss the method of computing the effective value
of a non-sinusoidal periodic waveform.
3. (a) Explain clearly the significance of the following
terms used in determining Fourier series of a given
waveform:
i. Odd symmetry or rotation symmetry
ii. Even symmetry or mirror symmetry
iii. Half-wave symmetry or alternation symmetry
iv. Quarter-wave symmetry
(b) Show that the Fourier series expansion of a periodic function with odd (rotation) symmetry contains only the sine terms.
(c) Show that the Fourier series expansion of a periodic function with even (mirror) symmetry contains only the cosine terms plus a constant.
(d) Show that the Fourier series expansion of a periodic function with half-wave symmetry contains
only the odd harmonics.
ii. When a complex voltage wave is applied to a
pure inductor, the current wave has lesser harmonics than the applied voltage wave.
iii. If a voltage wave containing a dc component is
applied to a series RC circuit, the current wave does
not contain the corresponding dc component.
6. (a) Give the definitions of a Fourier transform pair and
illustrate its use in network analysis with one example.
(b) Explain clearly the difference between Fourier
transform and Laplace transform and discuss briefly
their importance in analyzing electrical network.
Or,
Define Fourier’s transform. How does Fourier transform differ from i) Fourier integral, and ii) Laplace
transform?
(c) Write a brief note on the use of Fourier transform and
Fourier integrals in the analysis of circuits excited by
ideal sources of non-sinusoidal waveforms.
(d) Discuss the important properties of Fourier transforms.
7. Explain briefly the inter-relation between Fourier
series, Fourier transforms and Laplace transforms.
8. When do we use Fourier transform?
Discuss that Fourier integral is the limit of Fourier
series, as time period T of a repetitive wave approaches
infinity as the limit.
Or,
How would you obtain Fourier integral from Fourier
series?
9. Find the amplitude-frequency distribution of a single
non-repetitive voltage pulse of one-microsecond
duration and explain how its frequency-bandwidth is
estimated.
Or,
4. Discuss in brief the following:
i. Fourier series and its applications to network
analysis
ii. Method of analyzing the complex waveform by
Fourier series
iii. Frequency and phase spectra of periodic waveform
iv. Truncating Fourier series
v. Gibb’s phenomenon
10. State and prove Parseval’s theorem for a periodic function.
5. Explain why:
i. When a complex voltage wave is applied to a
pure capacitor, the current wave has more harmonics than the applied voltage wave.
11. Show that when f (t ) is an even function of t, its Fourier transform f ( j ) is an even function of and is real;
while when f(t) is an odd function of t, its Fourier transform f ( j ) is an odd function of and is imaginary.
Consider a periodic voltage pulse waveform of period
T (second) and width T0 (second). Find an expression
for the frequency-spectra of this waveform and explain
how you would use this result to estimate the bandwidth required for the transmission of such a signal.
470
Network Analysis and Synthesis
Multiple-Choice Questions
1. A current consists of a fundamental component of
amplitude I1, and a third harmonic of amplitude I3. The
rms value of current will be
(I + I )
(i)
1
(iii)
(I + I )
(ii)
3
1
2
I 12 + I 32
(iv)
3
2 2
(I + I )
2
1
2
3
2
2. The Fourier series expansion of a periodic function
with half-wave symmetry contains only
(i) sine terms
(iii) odd harmonics
(ii) cosine terms
(iv) even harmonics
3. A periodic function f (t ) is said to have a quarter wave
symmetry, if it possesses
(i) even symmetry at an interval of quarter of a wave
(ii) even symmetry and half-wave symmetry only
(iii) even or odd symmetry without the half-wave
symmetry
(iv) even or odd symmetry with the half-wave symmetry
4. If f (t ) is a periodic waveform with even symmetry then
its Fourier series expansion does not contain
(i) sine terms
(iii) odd harmonics
(ii) cosine terms
(iv) even harmonics
5. Periodic signal that obeys Dirichlet’s condition can be
represented by
(i) Fourier series
(ii) Fourier transform
(iii) Inverse Fourier transform (iv) none of these
6. Which of the following conditions is true for an even
function?
(
(i) f (t ) = −f t ±T
(iii) f (t )
2
)
f( t)
(ii) f (t )
f (t )
(iv) f (t )
f (T )
7. Which of the following conditions is true for an odd
function?
(
(i) f (t ) = −f t ±T
(iii) f (t )
2
)
f( t)
(ii) f (t )
(iv) f (t )
f( t)
f (T )
8. A periodic function f (t ) having a time period T repeats
itself after half-time period T/2. The Fourier series of
f (t ) would contain
(i) cosine terms only
(ii) sine terms only
(iii) odd harmonic terms only
(iv) even harmonic terms only
9. Which of the following statements is true for a delayed
step function u(t T )?
(i) It has an infinite Fourier series.
(ii) It has no Fourier series.
(iii) It has a finite Fourier series.
(iv) Its Laplace transform is 1s.
10. Which one of the following is the correct Fourier transform of the unit step signal u(t )?
1
(i)
( )
(ii)
j
(iii)
1
+
j
( )
(iv)
1
+2
j
( )
11. If f (t )
f ( t ) and f (t ) satisfy the Dirichlet’s conditions then f (t ) can be expanded in a Fourier series containing
(i) only sine terms
(ii) only cosine terms
(iii) cosine terms and a constant term
(iv) sine terms and a constant term
12. The Fourier transform F( j ) of an arbitrary signal has
the property:
(i) F ( j )
(iii) F ( j )
F( j )
F*( j )
(ii) F ( j )
(iv) F ( j )
F( j )
F*( j )
13. The Fourier series expansion of an odd periodic function contains
(i) cosine terms
(ii) constant terms only
(iii) sine terms.
14. For the expansion of f ( t ) in the Fourier series
a0 a1 cos t
an cos n t
b1 sin t
bn
sin n t, if f ( t ) f (
t ) then
(i) an
(iii) a0
0
0
(ii) bn
(iv) an
0 for all n
0 for all n except n
0.
15. Two complex waves will have the same waveform if:
(i) they contain the same harmonics
(ii) the harmonics are similarly spaced with respect
to the fundamental
(iii) the ratio of corresponding harmonics to their
respective fundamentals is the same
(iv) all of the above
16. The complex wave is symmetrical when
(i) it contains only even harmonics
(ii) it contains only odd harmonics
471
Fourier Series and Fourier Transform
(iii) it contains both odd and even harmonics
(iv) the phase difference between even harmonics and
3
fundamental is either
or
2
2
17. An even waveform when expressed in exponential
Fourier series will contain
(i) only imaginary coefficients
(ii) only real coefficients
(iii) both (i) and (ii)
(iv) none of these
()
∞
(iv) 270 W
()
) ( )d
) (
25. An
f (t = 1 for −
t
Ⲑ2 0
Ⲑ2
Fig. 7.72
(i)
(iii)
( )
⎛ ⎞
sin⎜ ⎟
⎝ 2⎠
⎛ ⎞
⎜⎝ 2 ⎟⎠
(ii)
(
)
()
(
)
(
(
sin 2
input
voltage
=2
(2 k ) ;
t)
(
(
)
2
2
)
)
)
)
v (t = 10 2 cos (t + 10° +
of resistance R 1 and an inductance L 1 H. The
resulting steady-state current i(t ) in amperes is
(
)
(
(
)
3
cos 2t + 55°
2
(
)
(
(
)
3
cos 2t − 35°
2
(i) 10 cos t + 55° + 10 cos 2t + 10° + tan−1 2
(ii) 10 cos t + 55° + 10
The value of F( ) is
sin
1
2
10 5 cos 2t + 10° V is applied to a series combination
f ( t)
1
≤t ≤
2
2
= 0 otherwise
()
x(
(iv) x t = x t −T = x t −T
)d
20. Fourier transform of the
gate function as shown in
Fig. 7.72 is
x(t T )
x(T t )
(iii) x t = x T − t = − x t −T
−∞
1
(iv) f t =
exp − j t F − j
2 −∫∞
(where is the width of the
gate function)
(iv)
(ii) has Fourier transform but not Laplace transform
(i) x(t )
(ii) x(t )
) ( )d
∞
)
(iii) 1
satisfies the equation
( ) 21 ∫ exp( − j t )F ( + j )d
(
)
T
k = 1, 2, … Also, no sine terms are present. Then x(t )
(iii) f t =
∞
2
(ii) (2
2
sion contains no terms of frequency
(ii) f t = 1 exp + j t F j
2 −∫∞
()
1
(2 + )
24. x(t ) is a real-valued function of a real variable with
period T. Its trigonometric Fourier series expan-
−∞
(
(f )
(iv) none of these
) = ∫ exp( − j t )f (t )dt is
∞
(iv) j f
(iii) has both Laplace and Fourier transforms
(iii) 135 W
∞
(
(f )
1
+
j f
(i) has Laplace transform but not Fourier transform
9 t (Second)
6
(i) f t = ∫ exp + j t F j
−∞
1
j f
(ii)
23. A ramp function
19. The inverse Fourier transform of
F(j
(iii)
(i)
form in a pure resistor of 10 is shown
in Fig. 7.71. The
0 3
power dissipated in
Fig. 7.71
the resistor is
(ii) 52.4 W
(i) j f
22. The Fourier transform of the unit impulse function (t )
would be
9
18. The current wavei(A)
(i) 7.29 W
21. The Fourier transform of a signum function is given by
)
2
⎛ ⎞
sin⎜ ⎟
⎝ 2⎠
(iv)
2 ⎛ ⎞
⎜⎝ 2 ⎟⎠
(
)
(iii) 10 cos t − 35° + 10 cos 2t + 10° − tan−1 2
(iv) 10 cos t − 35° + 10
26. Choose the function f (t ),
rier series cannot be defined.
(
t
(i) 3 sin(25t)
(ii) 4 cos(20t
(iii) exp( t )sin(25t) (iv) 1
)
)
)
, for which a Fou3)
2sin(710t)
472
Network Analysis and Synthesis
Answers
1. (iv)
2. (iii)
3. (iv)
4. (i)
5. (i)
6. (iii)
7. (ii)
8. (iii)
9. (iii)
10. (iii)
11. (i)
12. (iii)
13. (iii)
14. (ii)
15. (iv)
16. (ii)
17. (ii)
18. (iv)
19. (ii)
20. (iii)
21. (ii)
22. (iii)
23. (i)
24. (iv)
25. (iii)
26. (iii)
8
Sinusoidal Steady
State Analysis
Introduction
While dc circuit analysis is carried out by solving algebraic equations, the analysis of ac circuits composed
of capacitors, inductors as well as resistors will require solving differential equations. The solution of a
differential equation represents the response of the circuit to both the external input and the initial state,
and is composed of two parts:
• homogeneous solution representing the transient/natural response caused by the initial
condition, and
• particular solutions representing the steady-state/forced response caused by the external input.
A sinusoidal excitation function provides both the transient and steady-state responses. When the
transient response dies out, the circuit is said to be in sinusoidally steady state.
In this chapter, we will discuss the basics of alternating quantities and the analysis of different electrical
circuits under sinusoidally steady state.
8.1
ADVANTAGES OF USING ALTERNATING CURRENTS IN ELECTRICAL
ENGINEERING
Alternating current has a number of advantages over dc. Some of the advantages are given below.
1. Alternators (generators designed for ac operation) do not require the slip-rings and commutators (brushes)
upon which their dc cousins depend.
2. An even greater advantage of ac is that its voltage can be stepped up to higher levels with a transformer,
sent great distances through high-tension wires, and stepped down at its destination.
3. Alternators at power stations produce three-phase electricity; they have three coils equally spaced around
their primary coil, each of which is induced to produce a 50-Hz alternating current for three circuits.
Three-phase electricity can supply as much current through three thin wires as it would normally take
two thick wires to carry. The advantage in using a thinner wire is to minimize the electrical resistance
that a thick wire would produce.
4. Also, line losses are lower for ac than dc for a given wattage delivery and wire diameter.
474
Network Analysis and Synthesis
8.2
BASICS OF SINUSOIDS
A sinusoid is a signal that has the form of sine or cosine function.
We consider a sinusoidal voltage, v (t ) Vm sin t
where, Vm is the amplitude,
t is the argument of the sinusoid,
is the angular frequency of the sinusoid in rad/s 2 f
T is the time period of the sinusoid.
As the sinusoid is periodic, it repeats itself; such that
⎛ 2 ⎞
v t = v t + T = Vm sin ⎜ t +
⎟⎠ = Vm sin
⎝
() (
)
2
T
( t + 2 ) = V sin
m
A shifted sinusoid can be written as v (t ) Vm sin ( t
where,
is the phase of the sinusoid.
Thus, we see that
sin t sin ( t 180 )
cos t cos ( t 180 )
cos t sin ( t 90 )
sin t cos ( t 90 )
)
Fig. 8.1
Sinusoid
8.2.1 Advantages of Sinusoidal Waveforms
1. Sinusoidal waveforms produce minimum disturbance in electrical circuits during operation.
2. Sinusoidal waveforms produce electromagnetic torque which is free of noise and oscillations.
3. Sinusoidal waveforms cause less interference to nearby communication lines (telephones, etc.)
4. The iron and copper losses with sinusoidal waveforms are low in transformers and rotating ac machines.
Therefore, the machines operate with higher efficiency with sinusoidal waveforms.
5. The possibility of resonance is much reduced with the use of sinusoidal waveforms compared to other
non-sinusoidal waveforms containing harmonic frequencies.
8.3
TERMINOLOGIES
We consider the following terminologies for alternating quantities.
Waveform and Waveshape Alternating quantities
may be represented graphically. The shape of the
curve obtained by plotting the values of the function
at different instants is known as the waveform or
waveshape.
In electrical engineering, any alternating voltage or
current may have any waveshape. However, any waveform can be represented by the various combinations of
sinusidal waves. Thus, sinusoid is the basis of all alternating quantities.
i
Im
0
Im
Fig. 8.2
b
c
f
d
a
e
1 cycle
1 cycle
g
h
Periodic waveform
Period and frequency The time taken by an alternating quantity to complete one cycle is known as the
time period. It is measured in seconds and denoted by T.
475
Sinusoidal Steady State Analysis
The number of cycles completed per second by an alternating quantity is known as the frequency. It is measured in Hertz or cycles per second and is denoted by f.
The relation between T and f is f 1冫T .
Phase Phase is a frequently used term for alternating quantities. The word comes from a Greek word which
originally referred to the eternally regular changing appearance of the moon through each month, and then
was applied to the periodic changes of some quantity, such as the voltage in an ac circuit. Electrical phase is
measured in degrees, with 360° corresponding to a complete cycle. A sinusoidal voltage is proportional to the
cosine or sine of the phase.
The phase of an oscillation or wave is the fraction of a complete cycle corresponding to an offset in the displacement from a specified reference point at time t 0.
The concept of phase can be readily understood in terms of simple harmonic motion. Simple harmonic
motion is a displacement that varies cyclically, as depicted below and
Displacement
Period
described by the formula x(t ) A sin(2 f t
)
where A is the amplitude of oscillation, and f is the frequency. A motion
Amplitude
1
Time
with frequency f has period T = where t is the elapsed time,
f
and is the phase of the oscillation. It determines or is deterFig. 8.3 Periodic function
mined by the initial displacement at time t 0.
Phase Shift Here, is sometimes referred as a phase shift, because it represents a shift from zero phase. But
a change in is also referred as a phase shift.
A
B
B
Phase shift = 90 degrees
A is ahead of B
(A ‘leads’ B)
A
Phase shift = 90 degrees
B is ahead of A
(B ‘leads’ A)
B
Phase shift = 180 degrees
A and B waveforms are
mirror images of each other
A B
Phase shift = 0 degrees
A and B waveforms are
in perfect step with each other
A
Fig. 8.3
Examples of phase shifts
476
Network Analysis and Synthesis
For infinitely long sinusoids, a change in is the same as a shift in time, such as a time delay. If x(t) is delayed
1
(time-shifted) by of its cycle, it becomes:
4
⎛
⎛ T⎞ ⎞
⎛ T⎞
⎛
⎞
x ⎜ t − ⎟ = Asin ⎜ 2 f ⎜ t − ⎟ + ⎟ = Asin ⎜ 2 ft − + ⎟ whose ‘phase’ is now − . It has been shifted by − .
2
4
2
2
⎠
⎝
⎠
⎝ 4⎠
⎝
⎝
⎠
Phase Difference When two alternating quantities of the same frequency are considered simultaneously,
they may not pass through a particular point at the same instant. One may pass through its maximum value at
an instant while the other may pass through its value other than the maximum. These two quantities are said to
have a phase difference.
A B
A B
Phase difference is measured by the angular distance between the
A B
points where the two alternating waves cross the reference line in the
A B
same direction.
A B
A B
The quantity ahead in phase is said to lead the other quantity
Fig. 8.5 Two alternating waves with
while the second quantity is said to lag behind the first quantity. If
phase difference
two quanties have zero phase difference, they are said to be in phase
with each other. In Fig. 8.5, the wave B is lagging behind the wave A or the wave A is leading the wave B.
Example 8.1 Find the amplitude, phase, period and frequency of the sinusoid given as f( t )
100 cos(50t 45 ).
Solution The amplitude is Fm
The phase is,
45
The angular frequency is,
The period is,
T=
The frequency is,
100
50 rad/s
2
=
2
= 0.1257 s
50
f=
1
1
=
= 7.958 Hz
T 0.1257
Example 8.2 Calculate the phase angle between the two currents:
i1
4sin(377t
25 )
and i2
5cos(377t
40 )
Does i1 lead or lag i2?
Solution i1
4 sin(377t 25 )
i2 5 cos(377t 40 )
phase angle between i1 and i2 is,
8.4
4 cos(377t
115
25
90 )
4 cos(377t
115 )
( 40 )
155
Here, i2 lags behind i1.
SOME VALUES OF ALTERNATING QUANTITIES
The magnitude of an alternating quantity changes with time. Four different types of values are specified for an
alternating quantity:
1. Instantaneous value,
2. Peak or maximum or crest value,
477
Sinusoidal Steady State Analysis
3. Average or mean value, and
4. Effective or RMS value.
Instantaneous Value The value of an alternating quantity at any instant of time is known as the instantaneous value.
It is denoted by small (lower case) letters. For example, instantaneous value of an alternating current is
denoted by i.
Peak or Maximum or Crest Value The maximum value of an alternating quantity, attained in each cycle
is known as the peak or maximum or crest value.
For example, for the alternating voltage given by v(t ) Vm sin t; the peak value is Vm.
Average or Mean Value The average value of an alternating quantity over a given time interval is the
summation of all instantaneous values divided by the number of values taken over that interval.
In other words, the average value of a waveform is the area under the curve divided by the length of the base
of the curve. Mathematically,
T
Vav =
1
vdt , where T is the time period of the quantity.
T ∫0
In electrical sense, the average value of an alternating current is the equal dc current that transfers across a circuit the same amount of charge as that transferred by an ac current during a given interval. Note that the average
value of a purely sinusoidal waveform is always zero.
Effective or rms Value In mathematics, the root mean square (abbreviated rms or rms), also known as
the quadratic mean, is a statistical measure of the magnitude of a varying quantity. It is especially useful when
variates are positive and negative, e.g., sinusoids.
It can be calculated for a series of discrete values or for a continuously varying function. The name comes
from the fact that it is the square root of the mean of the squares of the values.
The rms of a collection of n values {x1, x2, x3, . . ., xn} is
xrms =
x 2 + x2 2 + x32 + ⋅⋅⋅+ xn 2
1 n 2
xi = 1
∑
n
n i =1
The corresponding formula for a continuous function f (t ) defined over the interval T1
t
i (t )
i2
T
f rms =
()
2
1 2
⎡ f t ⎤ dt
∫
⎣
⎦
T2 − T1 T
1
T
or,
()
2
1
⎡ f t ⎤ dt
f rms =
⎦
T ∫0 ⎣
The rms of a periodic function is equal to the rms of one period of the function. The rms value of a continuous function or signal can be approximated by
taking the rms of a series of equally spaced samples as follows.
T2
i1
in
T
Fig. 8.6
rms value
i12 + i2 2 + i32 + ⋅⋅⋅+ in 2
n
In electrical sense, the rms or effective value of an alternating current or voltage is that constant current or
voltage which when applied to a resistance will produce the same average power dissipation as that produced
by the alternating current or voltage in the same resistance.
For this current waveform, the rms value is obtained as
I rms =
478
Network Analysis and Synthesis
8.4.1 Form Factor
It is defined as the ratio of the rms value to the average value for an alternating wave.
rms value
∴ formfactor K f =
average value
( )
For a sinusoidal wave its value is 1.11.
Form factor is used to determine the effective or rms value of an alternating quantity whose average value
over half of a period is known.
8.4.2 Peak Factor
It is defined as the ratio of the peak value to the rms value for an alternating wave.
value maximum value
=
( ) peak
rms value
rms value
∴ peak factor K p =
For a sinusoidal wave, its value is 1.414.
Peak factor is used to find the value of dielectric strength of an insulating material since the dielectric stress
developed is proportional to the peak value of the applied voltage.
These two factors indicate the shape of an alternating wave. For a more pointed wave near the peak, the values
of these factors will be more. For a rectangular wave, both the factors are equal to unity, i.e., K f
Kp 1.
Example 8.3 Calculate the rms value, average value, form factor, and peak factor of a periodic current
having following values for equal time intervals changing suddenly from one value to next as 0, 2, 4, 6, 8, 10,
8, 6, 4, 2, 0, 2, 4, 6, 8, 10, 8, . . .
Solution
The average value of the current is given as
0 + 2 + 4 + 6 + 8 + 10 + 8 + 6 + 4 + 2
=5A
10
The rms value of the current is given as
Average Value =
rms value =
8.5
02 + 2 2 + 4 2 + 6 2 + 82 + 102 + 82 + 6 2 + 4 2 + 2 2
= 5.83 A
10
Form factor =
5.83
RMS value
=
= 1.17
5
average value
Peak factor =
peak value 10
=
= 1.71
rmsvalue 5.83
COMPLEX NUMBER SYSTEMS
Complex numbers allow mathematical operations with phasor quantities and are useful in analysis of ac circuits.
With the complex number system, you can add, subtract, multiply, and divide quantities that have both magnitude and angle, such as sine waves and other ac circuit quantities that will be studied alter.
A complex number can be represented in two different formats in either the Euclidean and polar coordinate
system.
479
Sinusoidal Steady State Analysis
j
z
y
z
Euclidean Representation z x jy where x and y are the real (horizontal) and
imaginary (vertical) part of complex variable z, respectively.
Polar Representation z
冷z冷e j z where 冷z冷 and z are the magnitude and phase
angle, respectively.
The two representations can be converted from one form to the other:
Complex Number Conversion
Rectangular to polar forms
z
x
Fig. 8.7
Complex
number
representation
⎧ z = x 2 + y 2 magnitude
⎪
⎨
⎞
−1 ⎛ y
⎪∠z = tan ⎜⎝
⎟⎠ phase angle
x
⎩
Polar to rectangular forms z 冷z冷e j z 冷z冷(cos z jsin z)
due to Euler identity, i.e., x 冷z冷cos z real part
y 冷z冷sin z imaginary part
x
jy
Mathematical Operation of Complex Numbers
Complex numbers can be added, subtracted, multiplied and divided. The arithmetic operations of two complex
numbers z x j y 冷z冷e j z and w u j v 冷w冷e j are listed below.
Addition
Complex Numbers Must be in the Rectangular Form in Order to Add Them
• Add the real part of each complex number to get the real part of the sum. Then add the j parts of each
complex number to get the j part of the sum.
Subtraction
Complex Numbers Must be in the Rectangular Form in Order to Subtract Them
• Subtract the real part of each complex number to get the real part of the difference. Then subtract the j
parts of each complex number to get the j part of the difference.
z w (x u) j(y v),
z w (x u) j(y v)
Multiplication
Multiplication of Two Complex Numbers is Easier When Both Numbers are in Polar Form
• Multiply the magnitudes, and add the angles algebraically.
zw (x j y)(u jv ) (xu yv ) j(xv
冷z冷冷w冷e j ( z
yu)
w)
Division
Division of Two Complex Numbers is Easier When Both Numbers are in Polar Form
• Divide the magnitude of the numerator by the magnitude of the denominator to get the magnitude of the
quotient. Then subtract the denominator angle from the numerator angle to get the angle of the quotient.
j ∠z −∠w )
ze (
xu + yv + j yu − xv
z x + jy x + jy u − jv
=
=
=
=
w u + jv
u + jv u − jv
u2 + v 2
w
(
(
)(
)(
) (
)
) (
)
480
Network Analysis and Synthesis
Rotation A complex number (vector) z
j
by an angle of . In particular, as e
complex number multiplied by j or
冷z冷e j z multiplied by e j will become ze j z
冷z冷e j( z ), i.e., rotated
−j
= j and e 2 = − j , they can be considered as 90 rotation factors. Any
j will be rotated counter clockwise or clockwise by 90 degrees.
2
Complex Conjugate The complex conjugate of z x j y 冷z冷e j z is z* x j y 冷z冷e j z. In general,
z* can be obtained by negating every j in the expression of z (replacing j by j ). The magnitude of a complex
number z x j y can be found by
zz * =
( x + jy )( x − jy ) = x + y
2
2
Reciprocal
z −1 =
8.6
⎛ 1⎞
1
1
, ∠ z −1 = ∠ ⎜ ⎟ = 0 − ∠z = −∠z
=
2
2
⎝ z⎠
z
x +y
( )
PHASOR REPRESENTATION
8.6.1 Introduction to Phasors
For analysis of alternating circuits, a sinusoidal quantity (voltage or current) is represented by a line of definite
length rotating in anti-clockwise direction with the same angular velocity as that of the sinusoidal quantity. This
rotating line is called the ‘phasor’.
Sinusoidal quantities are scalar quantities varying periodically with time. According to the definition of
a vector, these are not vectors. Voltage is the work done per unit charge and current is the flow of electrons
through a wire and these are not vectors. However, as a sinusoid is specified by its amplitude and phase angle,
they are termed as ‘phasor’, keeping some similarity with the term ‘vector’, where the amplitude is considered
as the magnitude and phase angle as the direction of the vector.
8.6.2 Transformation of Sinusoid into Phasor
To represent a dc voltage or current, only its amplitude I or V is needed. However, to represent a sinusoidal volt) or current i(t ) Im cos( t
), three values are needed:
age v(t ) Vm cos( t
• Amplitude, the peak value Vm or Im
• Frequency
2 f
• phase
To simplify the computation of a sinusoidal variable, it is often represented by a complex variable (vector in
complex plane) which can be more conveniently dealt with, as various mathematical operations (addition/
subtraction, multiplication/division, etc.) on exponential functions can be much more easily carried out than
sinusoidal functions.
We consider a function, f (t ) re j t (r cos t jr sin t)
If ‘ ’ is constant, this function will rotate in counter-clockwise direction at constant angular velocity, . The
variation is shown in Fig. 8.8.
The projection of this rotating line segment on both the real and imaginary axis will be the cosine and sine
components, i.e., Re[ f (t )] r cos t
and
Im[ f (t )] r sin t
481
Sinusoidal Steady State Analysis
Similarly, in electric-circuit theory, the voltages and currents can be represented by a rotating function characterized by a magnitude (radius, r) and a phase with respect to a reference angle. Such a rotating function is
termed ‘phasor’.
Specifically, a sinusoidal voltage can be represented as
()
( t + ) = Re ⎡⎣V (t )⎤⎦ = Re ⎡⎣V e e ⎤⎦ = Re ⎡⎣V e
()
( t + ) = Im ⎡⎣V (t )⎤⎦ = Im ⎡⎣V e e ⎤⎦ = Im ⎡⎣V e
v t = Vm cos
v t = Vm sin
where
()
V t = Vm e (
j
j
j t
m
j
j t
m
t+
)
= 2Vrms e (
j
t+
j
rms
rms
j
2 e j t ⎤ = Re ⎡V 2 e j t ⎤
⎦
⎣
⎦
2 e j t ⎤ = Im ⎡V 2 e j t ⎤
⎦
⎣
⎦
)
is the complex variV
able, Vm is the peak magnitude of the voltage, Vrms = m
is
2
the effective value (rms), and the phasor representing the voltage is defined as
sine function
v
r
V = Vrms e j = Vrms ∠
0
0
2
t
Vm
cosine function
The frequency
2 f is not explicitly represented by the
phasor, as all currents and voltages in the circuit considered here
have the same frequency—same as that of the energy source or
input of the circuit.
For example, a 120-V, 50-Hz ac voltage given as
)
A
0 v
• the magnitude, the rms (effective) value V, and
• the phase .
(
Vm
t
in terms of
()
Vme j t
j
(
2
Fig. 8.8
t
Phasor representation of sinusoid
)
(
v t = 120 2 cos 2 ft + 60° = 120 2 cos 2 × 50 × t + 60° = 170 cos 314t + 60°
)
is expressed as V Vrms
120 60 with rms value Vrms 120 and
60 , and the implied frequency
f 50 Hz. All sinusoidal signals, currents as well as voltages can be represented by phasors.
Note that physically voltages and currents are not complex quantities, rather they are sine or cosine functions
of time. They are converted into complex quantities for simplification of the solution of electric problems. The
physical solution is obtained from the complex solution by taking the real or imaginary component.
8.6.3 Difference Between Time Domain and Phasor Domain
1. v (t ) is the instantaneous or time-domain representation and V苳 is the frequency or phasor-domain representation.
2. v (t ) is time-dependent and V苳 is not.
3. v (t ) is always real with no complex term, but V苳 is generally complex.
8.6.4 Transformation from Time Domain to Phasor Domain
If an instantaneous voltage is described by a sinusoidal function of time such as
v (t ) = Vm cos
where, Vm
amplitude of the voltage; V
( t + ) = 2V cos( t + )
effective value of the voltage,
482
Network Analysis and Synthesis
The phasor transform of the sinusoid is given by
{ ( )} = P {V cos( t + )} = P { 2V cos( t + )} = Ve = V ∠
V =P v t
j
m
8.6.5 Transformation from Phasor Domain to Time Domain
The inverse phasor transform of a phasor is given by
()
{ }
{
v t = P −1 V = P −1 Vm e j
} = Re{V e e } = Re{V e ( ) } = V cos( t + )
j
m
j
j t
m
t+
m
Note that phasor analysis is applicable only when the frequency is constant. For circuits with multiple sources
of different frequencies, phasor analysis is inapplicable.
Note
By convention or custom, we write the time variables as lower-case letters to remind us that they are functions
of time. We express the phasor quantity by upper-case and bold letters. That is, we will write the voltage as v to
indicate v(t ) and we will write the current as i to indicate i(t ). Similarly, the phasor voltage is written as V and
the phasor current as I.
When we say the voltage varies sinusoidally with time, we immediately think of writing it as a sine
function, v v (t ) V0 sin t
But, when we decide to set t 0, this same variation could be written as a cosine function, v v (t )
V0 cos t.
There is no real difference in these two forms. Either sine or cosine can be taken as the reference for a particular application and there will be no difference in the results.
8.6.6 Advantages of Using Phasor
A sinusoidal waveform has two attributes, magnitude and
Phasor
Waveform
phase, and thus sinusoids are natural candidates for reprec
Imaginary
sentation by phasors. One reason for using such a represen0
axis
0
C
tation is that it simplifies the description since a complete
spatial or temporal waveform is reduced to just a single point
C
Real axis
Time
represented by the tip of a phasor’s arrow. Thus changes in
0
the waveform are easily documented by the trajectory of the
Fig. 8.9 Phasor representation of sinusoids
point in the complex plane.
The second reason is that it helps us to visualize
Geometric
Algebraic
how an arbitrary sinusoid may be decomposed into
Temporal waveform:
Imaginary
the sum of a pure sine and pure cosine waveform. To
v(t) C cos(t
)
axis
perform the decomposition using trigonometry is a
Acos(t ) Bsin(t )
Real
B C sin( ) C
axis Phasor representation:
tedious business. However, if the sinusoid is repreC Ce i
sented by a phasor then the same method used for
A C cos( )
C Ccos( ) iC sin( ) A iB
decomposing vectors into orthogonal components
Fig. 8.10 Phasor decomposition of Sinusoids
may be used for decomposing the given sinusoid
into its orthogonal sine and cosine components. This method is illustrated in Fig. 8.10.
The phasor C can be represented algebraically in either of two forms. In the polar form, C is the product
of the amplitude C of the modulation with a complex exponential ei which represents the phase of the waveform. In the Cartesian form, C is the sum of a ‘real’ quantity (the amplitude of the cosine component) and an
483
Sinusoidal Steady State Analysis
‘imaginary’ quantity (the amplitude of the sine component). The advantage of these representations is that the
ordinary rules of algebra for adding and multiplying may be used to add and scale sinusoids without resorting
to tedious trigonometry.
Example 8.4 Transform the sinusoids into phasors:
(a) i 5 cos( t 75 )
(b) v 10 sin( t 12 )
Solution
(a) In phasor form, I 5
75
(b) v 10 sin( t 12 ) 10 cos( t
In phasor form, V 10
102
12
90 )
10 cos( t
102 )
Example 8.5 Express the phasors into sinusoids:
(a) I
3
j4
(b) V
j(8
j6)
Solution
( a) I (3 j4) 5 126.87
i 5 cos( t 126.87 )
(b) V j(8 j6) 6 j8 10 36.87
v 10 cos( t 36.87 )
Example 8.6 Find the resultant of the three voltages e1, e2 and e3 given by
⎛
⎛
⎞
⎞
e1 = 20 sin t , e2 = 30 sin ⎜ t − ⎟ , e 3 = 40 cos ⎜ t + ⎟
4⎠
6⎠
⎝
⎝
Solution
e1 = 20 sin t
⎛
⎞
e2 = 30 sin ⎜ t − ⎟ = 30 sin( t − 45° ))
4⎠
⎝
⎛
⎛
⎛
⎞
⎞
2 ⎞
= 40 sin( t + 120°))
e3 = 40 cos ⎜ t + ⎟ = 40 sin ⎜ t + + ⎟ = 40 sin ⎜ t +
3 ⎟⎠
6⎠
2 6⎠
⎝
⎝
⎝
In phasor form, the voltages are written as
e1 = 20∠0° = 20
(
)
e2 = 30∠ − 45° = 21.213 − j 21.213
(
)
e3 = 40∠120° = −20 + j 34.641
Therefore, the resultant of the three voltage is
(
)
(
) (
)
( )
e = e1 + e2 + e3 = 20 + 21.213 − j 21.213 + −20 + j 34.641 = 21.213 + j13.428 = 25.106 ∠ 32.33° V
In time form, the resultant voltage can be written as e
25.106 sin( t
32.33 ) (V)
484
Network Analysis and Synthesis
8.7
THE J OPERATOR
The j operator comes from complex numbers and states that.
In terms of circuit theory, the symbol j is an operator that rotates a phasor by 90 in the anti-clockwise
direction without changing its magnitude.
90°
Thus jVm sin t Vm cos t and jVm cos t
Vm sin t.
jV
Also, multiple operations of j rotate the phasor as given below:
2
j V
V
1, rotates the phasor by 180 in the anti-clockwise direction,
j2
0°
180°
3
j, rotates the phasor by 90 in the clockwise direction,
j
j 3V jV
j2 1, rotates the phasor by 180 in the clockwise direction,
270° ( 90°)
j, rotates the phasor by 90 in the anti-clockwise direction and so on.
j3
Fig. 8.11 Signifi-
8.8
PHASOR DIAGRAMS
cance of j notation
The graphical representation of the phasors of sinusoidal quantities taken all at the same frequency and with
proper phase relationships with respect to each other is called a phasor diagram.
In electrical engineering, alternating voltage and current phasors are represented in phasor diagrams.
Phasor diagrams can be drawn in terms of either the maximum or rms values. However, as the rms values are
of much more practical importance in electrical engineering, phasor diagrams are generally drawn in terms
of rms values.
8.8.1 Conventions for Drawing Phasor Diagrams
1. Rotation of phasor in the counter-clockwise direction is taken as, a positive direction of rotation, i.e., a
phasor rotated in the counter-clockwise direction is said to lead a given phasor while a phasor rotated
in the clockwise direction is said to lag the given phasor.
2. For a series circuit, where the current is same in all parts of the circuit, the current phasor is generally
taken as the reference phasor.
3. For a parallel circuit, where the voltage is same in all parts of the circuit, the voltage phasor is generally
taken as the reference phasor.
4. This is not necessary to draw the voltage and current phasors to the same scale. But, if several voltage
phasors or several current phasors are to be drawn in the same phasor diagram, they must be drawn to
the same scale.
8.9
CIRCUIT RESPONSE TO SINUSOIDS
We will see how the phasor transform can simplify the voltage–current relationship for inductors and capacitors, eliminating the need for derivatives and integrals. In fact, the voltage–current relationship for resistors,
inductors and capacitors in the phasor domain looks just like Ohm’s law, where voltage equals current times
a scaling constant. We call the scaling constant impedance. It serves the same role as resistance, but in the
phasor domain. It is a constant like resistance, but turns out to be complex-valued and varies with the frequency of the signal involved.
8.9.1 Necessity of Phasor Transform
Sinusoids are special signals. Note that the integral and derivative of a sinusoid is a sinusoid. Thus, the voltage–
current relationships for inductors and capacitors, which are characterized by integrals and derivatives, tell us
485
Sinusoidal Steady State Analysis
that a sinusoidal current produces a sinusoidal voltage. The only difference between the sinusoidal voltage
across and current through these devices is possibly the amplitude and phase. The frequency of the current
will be the same as the frequency of the voltage. Thus, if we only consider sinusoidal signals, all we need to
keep track of is magnitude and phase of the voltages and currents. This is where the phasor transform comes
in.
8.10
KIRCHHOFF’S LAWS IN PHASOR DOMAIN
We will now consider Kirchhoff’s voltage and current laws in the phasor (frequency) domain.
8.10.1
Kirchhoff’s Voltage Law
For KVL, let v1, v2, …, vn be the voltages around a closed path.
v1 v2 … vn
In sinusoidal steady state, these voltages can be written as,
Vm1 cos
or,
or,
or,
0
( t + ) +V cos( t + ) + ⋅⋅⋅+V cos( t + ) = 0
1
m2
2
mn
⎤ + ⋅⋅⋅ + Re ⎡V e
⎦
⎣ mn
j
j
j
Re ⎡ Vm1e 1 + Vm 2 e 2 + ⋅⋅⋅ + Vmn e n e j t ⎤ = 0
⎣
⎦
j t
Re ⎡⎣ V1 + V2 + ⋅⋅⋅+ Vn e ⎤⎦ = 0
Re ⎡⎣Vm1e e
j 1
j t
(
⎤ + Re ⎡Vm e
⎦
⎣ 2
j 2
e
j t
e
j t
)
)
(
n
j n
⎤=0
⎦
V1 + V2 + ⋅ ⋅ ⋅ + Vn = 0
e j t ≠ 0; ⇒
This shows that KVL holds good for phasors.
8.10.2
Kirchhoff’s Current Law
Applying KCL to n sinusoidal currents having, in general, different magnitudes and phases (but the same frequency) we get
i1 i2 … in 0
In sinusoidal steady state, these currents can be written as
I m1 cos
or,
or,
or,
( t + ) + I cos( t + ) + ⋅⋅⋅+ I cos( t + ) = 0
1
m2
2
mn
⎤ + ⋅⋅⋅ + Re ⎡ I e
⎦
⎣ mn
j
j
j
j
t
Re ⎡ I m1e 1 + I m 2 e 2 + ⋅⋅⋅ + I mn e n e ⎤ = 0
⎣
⎦
j t
⎡
⎤
Re ⎣ I1 + I 2 + ⋅⋅⋅+ I n e ⎦ = 0
Re ⎡⎣ I m1e e
j 1
(
j t
⎤ + Re ⎡ I m e
⎦
⎣ 2
)
(
e j t ≠ 0; ⇒
j 2
e
j t
)
n
j n
e
j t
⎤=0
⎦
I1 + I 2 + ⋅ ⋅ ⋅ + I n = 0
This shows that KCL holds good for phasors.
8.11
VOLTAGE AND CURRENT PHASORS IN SINGLE-ELEMENT CIRCUITS
In this part, we always assume that ac current or voltage is sinusoidal. When sinusoidal signals are applied
to ideal R, L or C elements, or to any series or parallel combination of these elements, the response is also
486
Network Analysis and Synthesis
sinusoidal. The response of circuit elements to sinusoidal voltages or currents can be obtained by considering
the defining element equations.
We consider the three elements:
1. resistor,
2. inductor, and
3. capacitor.
8.11.1
Resistor
( t + ) + jV sin ( t + )
and assume the complex current response i ( t ) = I e ( ) = I cos( t + ) + jI sin ( t + )
We apply a complex voltage
()
v t = Vm e (
j
)
t+
= Vm cos
j
m
t+
m
m
m
so that in time domain, by Ohm’s law,
()
() ⇒ Ve(
j
v t = Ri t
m
⇒ Vm e = RI m e
j
j
t+
)
= RI m e (
j
t+
)
; dropping t he e j t term
iR
⇒ Vm ∠ = I m ∠
⇒ V = RI
Fig. 8.12
Thus, v–i relationship in phasor form for a resistor has the same form as in the time
domain.
The quantity of R is called the ac resistance
Time form
Phasor form
and is measured in ohms ( ).
vR
vR
VR
Vm
I
iR
Conclusion The voltage across a resistance
I
I
m
is in-phase with the current through it. When
VR
R
the instantaneous value for current is zero, the
0
2 p vt
p
instantaneous voltage across the resistor is also
zero. Likewise, at the moment in time where the
Fig. 8.13 Wave diagram and phasor diagram for a resistor
current through the resistor is at its positive peak,
the voltage across the resistor is also at its positive
peak, and so on. At any given point in time along the waves, Ohm’s law holds true for the instantaneous values
of voltage and current.
Example If a voltage v(t)
10 cos (50t
45 ) (Volt) is applied to a resistor R
10
then the current will be
V 10∠ − 45°
=
= 1∠ − 45°
R
10
∴ i t = cos(50t − 45 ) A
I=
()
8.11.2
( )
Inductor
Inductors do not behave the same way as resistors. Whereas resistors simply oppose the
flow of electrons through them (by dropping a voltage directly proportional to the current),
inductors oppose changes in current through them by dropping a voltage directly proportional to the rate of change of current. In accordance with Lenz’s law, this induced voltage is
always of such a polarity as to try to maintain current at its present value.
iL
L
vL = L diL/dt
Fig. 8.14
487
Sinusoidal Steady State Analysis
Expressed mathematically, the relationship between the voltage dropped across the inductor and rate of current change through the inductor is as such:
()
v t =L
Assuming again a complex voltage, v(t)
we get
Vm e j ( t + ) = L
V me
j( t
)
()
di t
dt
and a complex current response, i(t)
Ime j ( t
)
,
d
⎡ I e j ( t + ) ⎤ = j LI m e j ( t + ) ⇒ Vm e j = j LI m e j
⎦
dt ⎣ m
⇒
V = j LI
Thus, the time domain differential equation becomes an algebric equation in phasor domain.
v V ∠90°
LI ∠90°
The opposition to current flow is Z L = =
=
= L ∠90° = j L = jX L
i
I ∠0°
I ∠0°
The magnitude of the impedance is called the inductive reactance XL
L and is measured in ohms ( ).
Time Form
Phasor Form
The voltage phasor for L is 90 ahead of the current phasor.
vL leads iL by 90°
vL
Vm
iL
VL
I
Im
冦
3/2
2
90° 0
2
2
Fig. 8.15 (a) Wave diagram for an inductor
t
VL
L
I
90°
Fig. 8.15 (b) Phasor diagram
for an inductor
Conclusion The voltage across an inductive reactance leads the current through it by 90 . The voltage
dropped across an inductor is a reaction against the change in current through it. Therefore, the instantaneous
voltage is zero whenever the instantaneous current is at a peak (zero change, or level slope, on the current sine
wave), and the instantaneous voltage is at a peak wherever the instantaneous current is at maximum change (the
points of steepest slope on the current wave, where it crosses the zero line). This results in a voltage wave that is
90 out of phase with the current wave.
8.11.3
Capacitor
Capacitors do not behave the same as resistors. Whereas resistors allow a flow of electrons
through them directly proportional to the voltage drop, capacitors oppose changes in voltage
by drawing or supplying current as they charge or discharge to the new voltage level. The
flow of electrons through a capacitor is directly proportional to the rate of change of voltage
across the capacitor. This opposition to voltage change is another form of reactance, but one
that is precisely opposite to the kind exhibited by inductors.
iC
C
Fig. 8.16
vC
488
Network Analysis and Synthesis
Expressed mathematically, the relationship between the current through the capacitor and rate of voltage
change across the capacitor is as such:
dv (t )
i (t ) = C
dt
Assuming again a complex voltage, v(t)
Ime j( t+ ) = C
Vm e j( t
)
and a complex current response, i(t)
d
⎡V e j ( t + ) ⎤ = j CVm e j ( t + )
⎦
dt ⎣ m
⇒
I = j CV or V =
Im e j ( t
)
, we get,
I
j C
The opposition to current flow is
ZC =
⎡ 1 ⎤
1
v V ∠0°
V ∠0°
=
∠ − 90° = − j ⎢
=
=
⎥ = − jX C
iC I C ∠90°
CV ∠90°
C
⎣ C⎦
The magnitude of the impedance in this case is called the capacitive reactance, XC and is measured in ohms ( ).
Time Form
Phasor form
The current phasor for C is 90 ahead of the voltage phasor.
C : ic leads vc by 90°
I
Vm
vc
Im
冦
2
ic
90° 0
2
I
3/2
2
t
VC
90°
C
VC
Fig. 8.17 (a) Wave diagram for an
capacitor
Fig. 8.17 (b) Phasor wave
diagram for a capacitor
Conclusion The current through a capacitive reactance leads the voltage across it by 90 . The current
through a capacitor is a reaction against the change in voltage across it. Therefore, the instantaneous current
is zero whenever the instantaneous voltage is at a peak (zero change, or level slope, on the voltage sine wave),
and the instantaneous current is at a peak wherever the instantaneous voltage is at maximum change (the
points of steepest slope on the voltage wave, where it crosses the zero line). This results in a voltage wave
that is 90 out of phase with the current wave.
8.12
PHASOR ANALYSIS OF R-L SERIES CIRCUIT
When a sinusoidal voltage is applied to any type of RL circuit, each resulting voltage drop and the current in
the circuit are also sinusoidal and have the same frequency as the applied voltage.
The inductance causes a phase shift between the voltage and the current that depends on the relative values
of the resistance and the inductive reactance.
We consider an RL series circuit with a sinusoidal voltage source, as shown in Fig.8.18.
489
Sinusoidal Steady State Analysis
R
V
0
L
冦
v (t)
VR
I lags VL by 90°.
VR and I are in phase.
Amplitudes are arbitrary.
VL
I
90°
i(t)
Fig. 8.18 RL
series circuit
Fig. 8.19 Voltage and current wave diagram
for an RL series circuit
Current
By KVL,
VR + VL = V ⇒ RI + j LI = V ⇒ I =
V
R+ j L
⇒
I=
Vm
R +
2
2
L2
∠ − tan −1
( L R)
• The current is the same through both the resistor and the inductor.
Current Phasor Diagram
(R2)
V
⎛ WL ⎞
tan ⎜
⎝ R ⎟⎠
IL IR
−1
Fig. 8.20 Current
phasor diagram for an
RL series circuit
Voltage
• The resistor voltage VR is in phase with the current.
• The inductor voltage, VL leads the current by 90 .
• There is a phase difference of 90o between the resistor voltage, VR, and the inductor voltage, VL .
⎛V ⎞
• The source voltage, Vs, can be expressed as VS = VR + jVL = VR2 + VL2 ∠ tan −1 ⎜ L ⎟
⎝ VR ⎠
Voltage Phasor Diagram
VL
VL
Vs
VL
90
VR
VR
Fig. 8.21
Impedance It is given as Z
90
Voltage phasor diagrams for an RL series circuit
ZR
ZL
R
j L
R
jXL
• The magnitude of the is impedance is Magnitude Z = R 2 + X L2
⎛ XL ⎞
⎟
⎝ R⎠
• The phase angle is expressed as = tan −1 ⎜
I
VR
490
Network Analysis and Synthesis
• The impedance triangle is shown below.
XL
Z = 兹R 2 X 2L
XL
)
/R
X L
Z
1(
90
=
R
ta
u
R
(a)
(b)
Fig. 8.22
XL
n
R
(c)
Voltage traingle for an RL series circuit
Example 8.7 Find the current response of an RL series circuit with R 2 , L 1 H if an alternating
voltage given as v(t) 10sin 3t (V) is applied to it. Also, find the rms value of the current. What is the power
factor of the circuit?
Solution Here, the impedance of the circuit is Z = R + j L == 2 + j 3 × 1 = 2 + j 3 = 13∠56.3°
The supply voltage is, v(t)
current, I =
10sin 3t
90 )
V
10
90
10
V 10∠ − 90°
=
∠ − 146.3°
=
Z
13∠56.3°
13
()
instantaneous current, i t =
10
13
rms value of the current, I rms =
power factor of the circuit
8.13
10cos (3t
( )
(
cos 3t − 146.3
10
13
×
1
2
cos (56.3 )
)( A )
= 1.96 A
0.555 (lagging)
PHASOR ANALYSIS OF RC SERIES CIRCUIT
When a sinusoidal voltage is applied to any type of RC circuit, each
resulting voltage drop and the current in the circuit are also sinusoidal and have the same frequency as the applied voltage.
The capacitance causes a phase shift between the voltage and
the current that depends on the relative values of the resistance and
the capacitance reactance.
Current
By KVL,
⇒
VR + VC = V ⇒ RI +
I=
Vm
1
R2 + 2 2
C
1
V
I =V ⇒ I =
j C
R+ 1
VR
VC
I
0
I leads VC by 90 .
VR and I are in phase.
Amplitudes are
arbitrary.
90
V
Fig. 8.23 Voltage and current wave
diagram for an RC series circuit
j C
( CR)
∠ tan −1 1
• The current is the same through both the resistor and the capacitor.
491
Sinusoidal Steady State Analysis
Current Phasor Diagram
IC
IR
⎛ 1 ⎞
tan−1 ⎜
⎝WCR⎟⎠
V
Fig. 8.24 Current
phasor diagrams for
an RC series circuit
Voltage
• The resistor voltage VR is in phase with the current.
• The capacitor voltage VC lags the current by 90 .
• There is a phase difference of 90 between the resistor voltage VR and the capacitor voltage VC.
⎛V ⎞
• The source voltage Vs can be expressed as VS = VR − jVC = VR2 + VC2 ∠ − tan −1 ⎜ C ⎟
⎝ VR ⎠
Voltage Phasor Diagram
VR
VR
I
u
90
90
VC
VC
Fig. 8.25
VS
VR
VC
Voltage phasor diagrams for an RC series circuit
Impedance
It is given as
⎛
⎛ 1 ⎞⎞
⎛ 1 ⎞
ZT = Z R + ZC = R + j 0 + ⎜ 0 − j ⎜
⎟ = R − j ⎜⎝ C ⎟⎠ = R − jX C
⎟
C
⎠
⎝
⎝
⎠
)
(
• The magnitude of the impedance is
Magnitude ZT = R 2 + X C2
⎛ XC ⎞
⎟
⎝ R ⎠
• The phase angle is expressed as = − tan −1 ⎜
• The impedance triangle is shown below.
R
R
R
= tan 1(XC /R )
90
Z
XC
XC
(a)
Fig. 8.26
Z = 兹R 2 X 2C
(b)
Impedance triangle for an RC series circuit
(c)
XC
492
Network Analysis and Synthesis
Example 8.8 In an RC series circuit with R 1 , the voltage across the resistor is 1.59 cos(2t 125 ) (V)
when a supply voltage 7.68 cos(2t 47 ) (V) is applied to it. Determine the value of the capacitor C.
Solution Here, R
1
,
vR
1.59 cos(2t 125 ) (V)
v
current through the resistance, i = R = 1.59 cos 2t + 125° A
1
In phasor form, I 1.59 125
Supply voltage, V 7.68 47
V 7.68∠47°
impedance of the circuit, Z = =
= 4.83∠ − 78°
I 1.59∠125°
)( )
(
⎛
⎛ 1 ⎞
j ⎞ ⎛
j ⎞
1
Z =⎜ R−
= 1−
= 1 + 2 ∠ − tan −1 ⎜
C ⎟⎠ ⎜⎝ 2C ⎟⎠
⎝
⎝ 2C ⎟⎠
4C
Here,
2
⎛
1 ⎞
∴ ⎜ 1 + 2 ⎟ = 4.83 ⇒ C = 0.106 F
⎝ 4C ⎠
(
8.14
)
PHASOR ANALYSIS OF RLC SERIES CIRCUIT
A series RLC contains both inductance and capacitance. Since inductive reactance
and capacitive reactance have opposite effects on the circuit phase angle, the total
reactance is less than either individual reactance.
A series RLC circuit is shown in Fig. 8.27.
The total impedance for the series RLC circuit is stated as
Z ZL R Zc R j(XL XC)
In terms of magnitude and phase,
(
)
Reactance
8.14.1
)
V
L
C
i(t )
Fig. 8.27 RLC
series circuit
⎛ X − XC ⎞
2
Z = R + j X L − X C = R 2 + X L − X C ∠ tan −1 ⎜ L
R ⎟⎠
⎝
(
R
Analysis of a Series RLC Circuit
Figure 8.28 shows that for a typical series RLC circuit, the total
reactance behaves as follows:
Capacitive: Inductive:
XC
XL
XL XC
XC
XL
• Very low frequency, XC is high and XL is low. The circuit is
predominantly capacitive.
• As the frequency increases, XC decreases and XL increases.
• Until a value is reached where XC XL and the two reactances cancel, making the circuit is purely resistive. This
condition is called series resonance.
• As the frequency increases further, XL becomes greater
than XC, and the circuit is predominantly inductive.
XL
XC
0
f
Series resonance
Fig. 8.28 Variation of reactance with
frequency in an RLC series circuit
493
Sinusoidal Steady State Analysis
Impedance The impedance Z is minimum at resonance (i.e., Zr
increases in value above and below the resonant point.
R
) and
Z( )
• At frequencies below fr , XC XL : the circuit is capacitive.
• At resonant frequency fr , XC XL : the circuit is purely resistive.
• At frequencies above fr , XC XL : the circuit is inductive.
XC
Z
Z
R
0
f
fr
Current and Voltages in a Series RLC Circuit
• At the series resonant frequency, the current is maximum (i.e., Imax Vs /R).
• Above and below resonance, the current decreases because the impedance
increases.
VR
I
VS /R
f
(a) Current
Fig. 8.30
VR
VS
fr
Fig. 8.29 Variation
of Impedance, and
reactances with freqnency
VC
IXC
IXC
VS
VS
f
fr
XL
fr
f
f
fr
(b) Resistor voltage (c) Capacitor voltage (d) Inductor voltage
Variation of current and voltages with frequency in an RLC series circuit
Phase Angle of a Series RLC circuit
XL → the current leads the source voltage. The phase angle
decreases as the frequency approaches the resonant value and is 00 at resonance.
• At frequencies above resonance, XC XL → the current lags the source voltage. The phase angle increases
as the frequency approaches 90 .
• At frequencies below resonance, XC
90 (l lags VS)
I
VS
VS
I
u
I
0
fr
90 (l leads VS)
冦
冦
0
VS
XC XL
Capacitive:
I leads VS
f
XL XC
Inducitive:
I lags VS
(c) Above fr, I lags Vs. (d) Phase angle versus frequency.
(a) Below fr, I leads Vs. (b) At fr, I is in
phase with Vs.
Fig. 8.31 Variation of phase angles with frequency in an RLC series circuit
Example 8.9 A series circuit consisting of a 25- resistor, 64-mH inductor and an 80- F capacitor is connected to a 110-V, 50-Hz single-phase supply. Calculate the current and voltage across each element and the
overall power factor of the circuit.
Solution Here, R
25
, L
64 mH, C
80 F, V
110 V, f
50 Hz
494
Network Analysis and Synthesis
Impedance of the circuit is
⎛
j ⎞
Z =⎜ R+ j L−
C ⎟⎠
⎝
j
2 × 50 × 80 × 10−6
= 25 + j 20.1 − j 39.78 = 25 − j19.68
= 25 + j 2 × 50 × 64 × 10−3 −
)( )
(
current in the circuit, I =
( )
110
V
=
= 3.46 ∠38.2° A
Z 25 − j19.68
( )
Voltage across resistance VR = I × R = 3.46 ∠38.2° × 25 = 86.45∠38.2° V
( )
Voltage across inductance VL = I × j L = 3.46 ∠38.2° × j 20.1 = 69.5∠128.2° V
⎛
⎛ j ⎞
j ⎞
= 134.1∠ − 51.9° V
= 3.46 ∠38.2° × ⎜ −
Voltage across capacitance VC = I × ⎜ −
⎟
⎝ 39.78 ⎟⎠
⎝ C⎠
( )
Power factor of the circuit
8.15
cos(38.2 )
0.786 (leading)
STEPS FOR SINUSOIDAL STEADY-STATE ANALYSIS (PHASOR APPROACH
TO CIRCUIT ANALYSIS)
For circuits containing passive elements R, L, and C and sinusoidal sources, the phasor approach to circuit
analysis is as follows:
1. Convert voltages v and currents i to phasors V and I, respectively.
2. Convert R’s, L’s and C’s to impedances.
3. Use the rules of circuit analysis to manipulate the circuit in the phasor domain.
4. Return to the time domain for voltages and currents, etc., by using the inverse-phasor-transformation
forms: i = Re ⎡⎣ Ie j t ⎤⎦ and v = Re ⎡⎣Ve j t ⎤⎦ .
8.16 CONCEPT OF REACTANCE, IMPEDANCE, SUSCEPTANCE AND ADMITTANCE
AS PHASORS
Impedance Impedance (Z) of any two-terminal network is the ratio of the phasor voltage (V ) to the
phasor current (I).
V
Z=
I
Since it is the ratio of voltage to current, its unit is ohm ( ). Practically, it represents the obstruction that the
device exhibits to the flow of sinusoidal current
As a complex variable, the complex impedance Z can be written as: Z =
The magnitude and phase angle of Z are Z = R 2 + X 2 ∠Z = tan −1
X
R
(
)
V
= R + jX = Z e j∠Z = Z ∠Z
I
495
Sinusoidal Steady State Analysis
The real part of impedance Re[Z] R is called resistance.
The imaginary part of impedance Im[Z]X is called reactance.
Impedance, resistance and reactance are all measured by the same unit, ohm ( ).
In particular, the impedanes of the three types of elements R, L and C are
Z = R;
for resistor
= j L; for inductor
1
=
; for capacitor
j C
Admittance The reciprocal of the impedance Z is called admittance. So, it is the ratio of phasor current to
phasor voltage.
Y=
1
1
R − jX
= G + jB
=
=
Z R + jX R 2 + X 2
which contains real and imaginary parts:
R
R +X2
X
The imaginary part of admittance is called susceptance: B = Im ⎡⎣Y ⎤⎦ = − 2
R +X2
Unlike R and X, G and B do not correspond to any particular circuit elements. The magnitude and phase of
complex admittance are,
The real part of admittance is called conductance:
Y = G 2 + B2 =
1
R +X
2
2
G = Re ⎡⎣Y ⎤⎦ =
; ∠Y = tan −1
2
B
−X
= tan −1
= −∠Z
G
R
Admittance, conductance and susceptance are all measured by the same unit, siemen (S).
Impedance Z and admittance Y = 1
are both complex variables. The real parts, R and G, are, always posiZ
tive, while the imaginary parts, X and B, can be either positive or negative. Therefore, Z and Y can only be in the
1st or the 4th quadrants of the complex plane.
In particular, the admittances of the three types of elements R, L and C are
1
;
for resistor
R
1
=
; for inductor
j L
= j C ; for capacitor
Y=
Ohm’s law can also be expressed in terms of phasor admittance or impedance as I = VY or V = Z I
Note
(i) G ≠
1
; but if X
R
0, then G =
1
.
R
(ii) The term ‘immittance’, a combination of ‘impedance’ and ‘admittance’, is used as a general term for both
impedance and admittance.
496
Network Analysis and Synthesis
8.17
AC POWER ANALYSIS
We introduce the following four types of power associated with alternating voltages and currents:
1. Instantaneous power
2. Average power
3. Apparent power
4. Complex power
8.17.1
Instantaneous Power
It is the power at any instant of time. It is the product of the instantaneous voltage v(t) and the instantaneous
current i(t), i.e.,
p(t) v(t) i (t) (in watt)
We consider the sinusoidal voltage and current in a two-terminal device as
v(t) Vm cos(␻t ␪v) and i(t) Im cos(␻t ␪i)
Thus, the instantaneous power is p(t) Vm Im cos(␻t ␪v)cos(␻t ␪i)
()
1
p t = Vm I m cos
2
or,
( − ) + 12 V I cos( 2 t + + )
v
i
m m
v
i
This shows that the instantaneous power has two components, one constant and the other varying with time at a frequency double of the supply frequency. The instantaneous power changes with time and it is difficult to measure.
8.17.2
Average or Real or True or Active Power
It is the average of the instantaneous power over a time interval. It is the power consumed by the resistive
loads in an electrical circuit.
T
∴P=
()
1 2
p t dt
T2 − T1 T∫
1
For a periodic function, f (t)
f (t
nT ), and thus, the average power may be computed as,
t +T
P=
()
11
p t dt T is time period
T ∫t
1
T
=
()
1
p t dt
T ∫0
In sinusoidally steady state, the average power is
1 ⎡1
V I cos
T ∫0 ⎢⎣ 2 m m
T
P=
1
= Vm I m cos
2
( − ) + 12 V I cos( 2 t + + )⎤⎥ dt
v
i
m m
T
v
i
⎦
T
( − ) T1 ∫ dt + 12 V I T1 ∫ cos( 2 t + + )dt = 12 V I cos( − ) + 0
v
i
m m
0
v
i
m m
0
1
⇒ P = Vm I m cos
2
( − ) ( in watt )
v
i
v
i
{ ␻T
2 }
497
Sinusoidal Steady State Analysis
Now, V
Vm ␪v
and I
Im ␪i
or,
I*
Im
␪i
1
Thus, the average power or real power is P = Vm I m cos
2
Note
(i) When
v
(ii) When ( v
8.17.3
i
( − ) = V I cos( − ) = 12 Re ⎡⎣VI *⎤⎦
v
i
rms rms
v
i
in (watt)
1
1 2
VmIm = I R .
2
2
90 , i.e for a purely reactive circuit, P 0.
, i.e. for a resistive circuit, P =
i
)
Apparent Power or Total Power
We assume that a sinusoidal voltage, v (t ) = Vm cos( t + v ) is applied to a network and the resultant current
is i(t ) = I m cos( t + i ).
)
(
(
)
(
)
1
The average power delivered to the network is P = Vm I m cos v − i = Vrms I rms cos v − i = S cos v − i
2
where, S VrmsIrms, is the apparent power.
So, apparent power is the product of the rms (effective) values of voltage and current (in VA).
For dc circuits, the average power will be simply the product of voltage and current. If we apply this concept in the sinusoidal steady state then the product ‘VrmsIrms’ is also a power that is not really absorbed in the
device, but is ‘apparent’. For this reason, it is so called. The real power is VrmsIrms cos(␪v ␪i).
Note that the total power delivered by an alternating source is the apparent power. Part of this apparent
power, called true power, is dissipated by the circuit resistance in the form of heat. The rest of the apparent
power, called reactive power, is returned to the source by the circuit inductance and capacitance.
Power Factor It is the ratio of the real or average power to the apparent power.
∴ PF =
average power P
= = cos
apparent power S
( − )
v
i
Power factor is a number always between 0 and 1. The angle (␪v ␪i) is called the power-factor angle.
8.17.4
Complex Power and Reactive Power
For an ac load with voltage phasor V
by the load is given by
Vm ␪v and current phasor I
1
S = VI * = Vrms I * = Vrms I rms ∠
rms
2
∴ S = Vrms I rms cos
Im ␪i, the complex power S absorbed
( − )
v
i
( − ) + jV I sin ( − ) = ( P + jQ )
v
i
rms rms
v
i
where, P Vrms Irms cos(␪v ␪i) is the real or average or active power (in watt)
and Q Vrms Irms sin(␪v ␪i) is the reactive power (in VAR).
Reactive power is the power consumed in an ac circuit because of the expansion and collapse of magnetic (inductive) and electrostatic (capacitive) fields. Unlike true power, reactive power is not useful power
because it is stored in the circuit itself. This power is stored by inductors, because they expand and collapse
their magnetic fields in an attempt to keep the current constant, and by capacitors, because they charge and
discharge in an attempt to keep the voltage constant. Circuit inductance and capacitance consume and give
498
Network Analysis and Synthesis
back reactive power. Reactive power is a function of a system’s amperage. The power delivered to the inductance is stored in the magnetic field when the field is expanding and returned to the source when the field
collapses. The power delivered to the capacitance is stored in the electrostatic field when the capacitor is
charging and returned to the source when the capacitor discharges. None of the power delivered to the circuit
by the source is consumed; all is returned to the source. The true power, which is the power consumed, is
thus zero. We know that alternating current constantly changes; thus, the cycle of expansion and collapse of
the magnetic and electrostatic fields constantly occurs.
Therefore, we conclude that reactive power is the rate of energy flow between the source and the reactive
components of the load (i.e., inductances and capacitances). It represents a lossless interchange between the
load and the source.
Note
(i) Q 0, for resistive load (power factor is unity)
(ii) Q 0, for capacitive load (leading power factor)
(iii) Q 0, for inductive load (lagging power factor)
8.17.5
Power Triangle
The relationship between real power, reactive power and apparent power
Apparent Power (VA)
Reactive
S
can be expressed by representing the quantities as vectors. Real power is
Power
VAR
represented as a horizontal vector and reactive power is represented as a
Q
vertical vector. The apparent power vector is the hypotenuse of a right
Real Power (W)
triangle formed by connecting the real and reactive power vectors. This
P
representation is often called the power triangle, as shown in Fig. 8.32.
Fig. 8.32 Power triangle
Using the Pythagorean theorem, the relationship among real, reactive
and apparent power is
(Apparent power)2 (Real power)2 (Reactive power)2
or,
S2
8.18
P2
Q2 or,
(VA2
(watt)2
(VAR)2
POWER CALCULATIONS IN DIFFERENT ELECTRICAL ELEMENTS
We consider the power calculations in the following elements or circuits:
1. in a purely resistive circuit,
2. in a purely inductive circuit,
3. in a purely capacitive circuit,
4. in R-L series circuit,
5. in R-C series circuit, and
6. in RLC series circuit.
8.18.1
Power in a Purely Resistive Circuit
Let the instantaneous voltage, v
and the instantaneous current, i
Vmax sin ␻t
Imax sin ␻t
instantaneous power, p = vi = Vmax I max sin 2 t =
Vmax I max
2
(1 − cos 2 t )
499
Sinusoidal Steady State Analysis
average power, P =
T
T
Vmax I max T
Vmax I max
1
1 Vmax I max
pdt
=
−
cos
t
dt
=
1 − cos 2 t dt =
= Vrms I rms
1
2
∫
∫
∫
T0
T0
2T 0
2
2
)
(
)
(
The phasor diagram and power curves for a purely resistive circuit are shown in Fig. 8.33.
p = V max I max
V = V max sin ␻t
i = I max sin ␻t
pv i
Sin2 ␻t
Pmax = VmaxImax
P = Pmax /2
0
I = V/R
␲/2
V
␲
Time
3␲/2
2␲
(a) Phasor
(b) Power curves
diagram
Fig. 8.33 (a) Phasor diagram (b) Power curves
From the plots it is observed that
(i) the power has a frequency twice that of the voltage or current
(ii) the power is always positive and varies between zero and a maximum VmaxImax
(iii) the average value of the power is a constant, VmaxImax
8.18.2
Power in a Purely Inductive Circuit
Let the instantaneous voltage, v Vmax sin ␻t
t
Vmax
Vmax
Vmax t
1
sin tdt = −
cos t =
sin
current, i = ∫ vdt =
∫
L
L 0
L0
L
where,
I max =
( t − 2 ) = I sin( t − 2 )
max
Vmax
L
instantaneous power, p = vi = Vmax I max sin t sin
average power,
P=
( t − 2 ) = − V 2I 2sin t cos t = − 12V I sin 2 t
T
T
⎤
1
1 ⎡ Vmax I max
pdt
=
sin 2 t ⎥ dt = 0
⎢−
∫
∫
T0
T 0 ⎢⎣
2
⎥⎦
max max
max max
{ ␻T
2␲}
The phasor diagram and power curves for a purely inductive circuit are shown in Fig. 8.34.
From the plots, it is observed that
(i) the power has a frequency twice that of the voltage and current.
(ii) when v and i are both increasing or decreasing, the power is positive and energy is delivered from the
source to the inductance
(iii) when either v is increasing and i is decreasing or v is decreasing and i is increasing, the power is negative and energy is returning from the inductance to the source
(iv) the average value of the power is zero; the reason is explained here
500
Network Analysis and Synthesis
V = V max Sin␻t
␲⎞
2 ⎟⎠
⎛
⎝
i = I max Sin ⎜␻t
pv i
Vmax Imax
2
0
␲
3␲
2
␲
2
2␲
Time
V
90
P = V max I max Sin2␻t
2
I = V /vL
(a) Phasor diagram (b) Power curves
Fig. 8.34 (a) Phasor diagram (b) Power curves in purely inductive circuit
During the second quarter of a cycle, the current and the magnetic flux of the inductor increases and the
inductor draws power from the supply source to build up the magnetic field. The power drawn is positive.
1
The energy stored in the magnetic field during the building up is LI max 2 .
2
In the next quarter, the current decreases. However, the emf of the inductor tends to oppose this decrease.
The inductor acts as a generator and returns energy to the supply. The power is negative.
This event repeats and a proportion of power is continually exchanged between the field and the inductive
circuit and the average power consumed by the purely inductive circuit becomes zero.
8.18.3
Power in a Purely Capacitive Circuit
Here let the instantaneous voltage be v
current, i = C
where,
I max =
Vmax sin ␻t
Vmax
dv
= CVmax cos t =
sin
1
dt
C
( t + 2 ) = I sin( t + 2 )
max
Vmax
1
C
instantaneous power,
average power, P =
p = vi = Vmax I max sin t sin
( t + 2 ) = V 2I 2sin t cos t = 12V I sin 2 t
T
T
⎤
1
1 ⎡Vmax I max
pdt
=
sin 2 t ⎥ dt = 0
⎢
∫
∫
T0
T 0 ⎢⎣ 2
⎥⎦
max max
max max
{
T =2
}
The phasor diagram and power curves for a purely capacitive circuit are shown in Fig. 8.35.
From the plots, it is observed that
(i) the power has a frequency twice that of the voltage and current
(ii) when v and i are both increasing or decreasing, the power is negative and returning from the capacitance to the source
(iii) when either v is increasing and i is decreasing or v is decreasing and i is increasing, the power is positive and energy is delivered from the source to the capacitance
(iv) The average value of the power is zero; The reason is explained here
501
Sinusoidal Steady State Analysis
P = V max Imax
2 sin2␻t
v = vmax sin␻t
␲⎞
⎛
i = I max Sin ⎜␻t
pv i
⎝
2 ⎟⎠
Vmax Imax
2
0
I = V␻CV
90
V
(a)
Phasor diagram
(b)
2␲
3␲
2
␲
␲
2
Time
Power curves
Fig. 8.35 (a) Phasor diagram (b) Power curves for a purely capacitive circuit
In the first quarter cycle, the energy is taken from the supply and stored in the capacitor. The energy stored
1
is CVmax 2 . In the next quarter cycle, this energy is returned to the supply. This process is repeated and thus
2
the average power is zero.
8.18.4
Power in RL series circuit
Here, let the instantaneous voltage be v
⎛ L⎞
= tan −1 ⎜
where,
⎝ R ⎟⎠
Vmax sin ␻t and the current, i
p = vi = Vmax I max sin t sin
instantaneous power,
( t − ) = 12 V I
max max
T
average power,
P=
␾)
(
⎡cos − cos 2 t −
⎣
{
1
1
pdt = Vmax I max cos = Vrms I rms cos
2
T ∫0
Imax sin(␻t
T =2
}
The phasor diagram and power curves for an RL series circuit are shown in Fig. 8.36.
pv i
p = vi
V L = IXL
v
i
V = IZ
0
90
␾
(a)
Phasor diagram
Fig. 8.36
8.18.5
VR IR
␲ ⫺
3␲ 2␲
2
Time
(b)
Power curves
(a) Phasor diagram (b) Power curves for an RL series circuit
Power in RC Series Circuit
Here, let the instantaneous voltage be v
where,
⫺ ␲
2
␾
Average
power
P = VI Cos ␾
⎛ 1 ⎞
= tan −1 ⎜
⎝ RC ⎟⎠
Vmax sin ␻t and the current, i
Imax sin(␻t
␾)
)⎤⎦
502
Network Analysis and Synthesis
instantaneous power,
p = vi = Vmax I max sin t sin
( t + ) = 12 V I
max max
(
⎡cos − cos 2 t +
⎣
)⎤⎦
T
1
1
pdt = Vmax I max cos = Vrms I rms cos
{ ␻T 2␲}
2
T ∫0
The phasor diagram and power curves for an RC series circuit are shown in Fig. 8.37.
average power,
P=
p = vi
p iv
v
V R = IR
f
90
V C = IXC
0
i
␲ 3␲
2
␲
2
␾
V = IZ
2␲
Time
(a) Phasor diagram
(b) Power curves
Fig. 8.37 (a) Phasor diagram (b) Power curves for an RC series circuit
8.18.6
Power in RLC Series Circuit
Here, let the instantaneous voltage be v
(
)
Vmax sin ␻t and the current, i
Imax sin(␻T
␾)
⎡ L− 1
⎤
⎢
⎥
C
where,
= tan ⎢
⎥
R
⎢
⎥
⎣
⎦
Three cases may appear:
−1
a) When
b) When
c) When
1
, the phase angle ␾ will be negative and the circuit behaves as an RL series circuit.
C
1
L=
, the phase angle ␾ will be zero and the circuit behaves as a purely resistive circuit.
C
1
L<
, the phase angle ␾ will be positive and the circuit behaves as an RC series circuit.
C
L>
⎛
1 ⎞
The applied voltage is V = VR 2 + VL 2 + VC 2 = I R 2 + ⎜ L −
C ⎟⎠
⎝
⎛
1 ⎞
The impedance of the circuit is Z = R + ⎜ L −
C ⎟⎠
⎝
2
2
2
The power factor of the circuit is cos =
R
=
Z
VL
IXL
VL VC
V
IZ
VR
IR
R
⎛
1 ⎞
R +⎜ L−
C ⎟⎠
⎝
2
␾
2
Power consumed in the circuit is P I2R VI cos ␾
The phasor diagram is shown in Fig. 8.38.
VC
IXC
Fig. 8.38 Phasor diagram for
an RLC series circuit
I
503
Sinusoidal Steady State Analysis
Example 8.10 Two coils of impedance 25.23 37 and 18.65 68 ohms are connected in series across
a 230-V, 50-Hz supply. Find the total impedance, current, power factor, apparent power, active power and
reactive power.
(
)
Z = 18.65∠68° = ( 6.98 + j17.29 )
Z = Z + Z = ( 20.15 + j15.18 ) + ( 6.98 + j17.29 ) = 27.13 + j 23.47 = 42.32 ∠50.12° ( )
Solution Impedances are Z1 = 25.23∠37° = 20.15 + j15.18
1
total impedance,
1
2
V
230∠0°
=
= 5.43∠ − 50.11° A
Z 42.32 ∠50.12°
power factor cos( 50.11 ) 0.64 (lagging)
apparent power VI 230 5.43 1250 VA
active power VI sin ␾ 1250 sin(50.11 ) 959.2 VAR
reactive power VI cos ␾ 1250 0.64 901.5 W
current,
8.19
( )
I=
SINUSOIDAL STEADY-STATE RESPONSE OF PARALLEL AC CIRCUITS
Parallel ac circuits are very frequently used in transmission and distribution systems. Analysis of parallel ac
circuits is based upon the concept that for parallel circuits, the voltage across the parallel branches is the same
and the total current is the summation of all the currents flowing through the parallel branches.
The following general procedures may be followed to determine the sinusoidal steady-state response of
parallel circuits:
1. As the voltage across each element is the same for a parallel network, the voltage phasor is taken as the
reference for drawing the phasor diagrams.
2. The current in every branch is the ratio of the voltage to impedance.
(i) If the branch is purely resistive, the current will be in phase with the voltage.
(ii) If the branch is purely inductive, the current will lag the voltage by 90 .
(iii) If the branch is purely capacitive, the current will lead the voltage by 90 .
(iv) If the branch is inductive with some resistance, the current will lag the voltage by some angle
greater than 0 but less than 90 .
(v) If the branch is capacitive with some resistance, the current will lead the voltage by some angle
greater than 0 but less than 90 .
3. The total current is the phasor summation of all the branch currents.
We consider the sinusoidal steady-state analysis of the following parallel circuits:
1. Parallel RL circuit,
2. Parallel RC circuit, and
3. Parallel RLC circuit.
8.19.1
i
iR
Parallel RL Circuit
For the parallel circuit shown in Fig. 8.39 (a), let the supply voltage be, v(t) =
Vm sin␻t
v t 1
v t dt
+
By KCL, i t = iR t + iL t =
R L∫
() () ()
()
()
iL
v(t )
R
Fig. 8.39 (a)
RL circuit
Parallel
L
504
Network Analysis and Synthesis
In sinusoidal steady state,
IR
V sin t 1
V sin t Vm
cos t
i t = m
+ ∫Vm sin tdt = m
−
R
L
R
L
()
2
2
⎛ 1⎞ ⎛ 1 ⎞
= ⎜ ⎟ +⎜
V sin
⎝ R ⎠ ⎝ L ⎟⎠ m
V
␾
IL
l
( t− )
Fig. 8.39 (b) Voltage and current
phasor diagram for RL parallel circuit
⎛ R⎞
= tan −1 ⎜
⎝ L ⎟⎠
where,
The phasor diagrams for the currents and voltage in an RL parallel circuit are shown in Fig. 8.39 (b). Two
cases may appear:
(i) If R
L then ␾ → 90
In this condition, the current drawn by the resistive branch is negligibly low and the total current is
V
almost equal to the inductor current, i.e., i t = iL t = m sin t − 90°
L
(ii) If R
L then ␾ → 0
In this condition, the current drawn by the inductive branch is negligibly low and the total current is
almost equal to the resistance current, i.e.,
() ()
8.19.2
(
)
Parallel RC Circuit
For the parallel circuit shown in Fig. 8.40, let the supply voltage be v(t)
Vm sin ␻t
i
( ) + C dv (t )
i (t ) = i (t ) + i (t ) =
dt
R
v t
By KCL,
R
iR
v(t )
R
In sinusoidal steady state,
()
i t =
Vm sin t
d
+ C Vm sin t
R
dt
(
)
C
Fig. 8.40 (a) Parallel
2
=
iC
C
Vm sin t
⎛ 1⎞
+ CVm cos t = ⎜ ⎟ +
R
⎝ R⎠
( C ) V sin ( t + )
2
m
I
IC
f
where, ␾ tan 1 (␻RC)
V
The phasor diagrams for the currents and voltage in an RC parallel
IR
circuit are shown in Fig. 8.40 (b). Two cases may appear:
Fig. 8.40 (b) Voltage and current
1
phasor diagram for RC parallel circuit
(i) If R >>
then ␾ → 90
C
In this condition, the current drawn by the resistive branch is negligibly low and the total current is almost
equal to the capacitor current, i.e., i(t) = ic(t) = cvm sin(␻t+ 90 )
(ii) If R <<
1
, then ␾ → 0
C
In this condition, the current drawn by the capacitive branch is negligibly low and the total current is
V
almost equal to the resistance current, i.e., i t = iR t = m sin t
R
() ()
505
Sinusoidal Steady State Analysis
8.19.3
Parallel RLC Circuit
i
For the parallel circuit shown in Fig. 8.41, let the supply voltage be v(t)
() () () ()
By KCL, i t = iR t + iL t + iC t =
( ) + 1 v t dt + C dv (t )
()
R L∫
dt
v(t )
v t
In sinusoidal steady state,
V sin t 1
d
i t = m
+ ∫Vm sin tdt + C Vm sin t
dt
R
L
()
(
iR
Vm sin ␻t
2
2
V sin t 1
⎛ 1⎞ ⎛
1 ⎞
= m
−
Vm sin t + CVm cos t = ⎜ ⎟ + ⎜ C −
V sin
R
L
L ⎟⎠ m
⎝ R⎠ ⎝
where,
R
Fig. 8.41
circuit
)
iL
iC
L
C
Parallel RLC
( t+ )
⎡ ⎛
1 ⎞⎤
= tan −1 ⎢ R ⎜ C −
⎥
L ⎟⎠ ⎦
⎣ ⎝
Three cases may appear:
1
, the circuit behaves as a parallel capacitive circuit. The current phase angle is positive, i.e., the
(i) If C >
L
current leads the voltage.
1
(ii) If C <
, the circuit behaves as a parallel inductive circuit. The current phase angle is negative, i.e.,
L
the current lags the voltage.
1
(iii) If C =
, the circuit behaves as a parallel resistive circuit with inductor and capacitor currents cancelL
ing each other. The circuit under this condition is said to be a parallel resonant circuit.
Example 8.11 The following circuit of Fig. 8.42 shows a parallel RL arrangement connected across 200-V, 50-Hz ac supply.
Calculate
(i) the current drawn from the supply,
(ii) apparent power,
(iii) real power, and
(iv) reactive power.
200 V, 50 Hz
40
Fig. 8.42 Circuit of Example 8.11
Solution Here, R 40 , L 0.0637 H, V
XL 2␲fL 10␲ 0.0637 20
(i)
current drawn from the supply,
2
200 V, f
2
50 Hz
2
2
⎛ 1⎞ ⎛ 1 ⎞
⎛ 1⎞ ⎛ 1⎞
I = IR + IL = ⎜ ⎟ + ⎜
V = ⎜ ⎟ + ⎜ ⎟ × 200 = 11.18 A
⎟
⎝ R⎠ ⎝ L⎠
⎝ 40 ⎠ ⎝ 20 ⎠
2
(ii)
apparent power, S
(iii)
real power,
(iv)
reactive power,
2
VI 200 11.18 2236 VA
V 2 2002
P = VI R =
=
= 1000 W = 1 kW
R
40
Q = VI L =
R L
2.236 kVA
V 2 2002
=
= 2000 VAR = 2 kVAR
20
XL
0.0637 H
506
Network Analysis and Synthesis
Example 8.12 In the parallel circuit shown in Fig. 8.43, the resistance R dissipates a power of 10 W. If the magnitude of the supply current is |I| 1A, find the
current drawn by the capacitor and its value.
2
I
200 V,
50 Hz
2
V
200
Solution The value of the resistance is R =
=
=4k
P
10
V
200
current through the resistance, I R = =
= 0.05 A
R 4 × 103
Since,
)
(
C
R
Fig. 8.43 Circuit of
Example 8.12
2
I = I R 2 + I C 2 ⇒ I C = I 2 − I R 2 = 12 − 0.05 = 0.9987 A
C=
value of the capacitor,
IC
0.9987
=
= 15.9 F
2 f V 100 × 200
Example 8.13 The current in the resistive branch of a parallel RLC circuit is given by iR 100 cos(500t
45 ) (A) . What is the current in the inductive and capacitive branches? Take R 10 , L 10 mH, C 10 ␮F.
Solution Here, R
10
,L
voltage across the circuit,
10 mH, C
10 F, ir
(
100 cos(500t
45 ) (A)
)
(
v = vR = iR × R = 10 × 10 cos 500t − 45 = 1000 cos s 500t − 45
)(V ) = v = v
L
C
Inductive current will lag this voltage by 90 .
)
(
(
1000
v
=
cos 500t − 45 − 90 = 200 cos 500t − 135
L 500 × 10 × 10−3
∴ iL =
)( A )
Capacitive current will lead the voltage by 90 .
v
∴ iC =
= 1000 × 500 × 10 × 10− 6 cos 500t − 45 + 90 = 5cos 500t + 45
1
C
(
)
(
)( A )
8.20 SINUSOIDAL STEADY-STATE RESPONSE OF SERIES–PARALLEL AC CIRCUITS
A series–parallel circuit consists of several combinations of circuit components connected in series and/or parallel.
The method of analysis of such a circuit is based upon the knowledge of the analysis of series and parallel circuits.
Example 8.14 For the circuit shown in Fig. 8.44, determine the total current
and the power factor of the circuit.
Solution Impedance of the RL branch, ZRL (6 j8)
Impedance of the RC branch, ZRC (4 j3)
Equivalent impedance of the circuit,
6 + j8 × 4 − j 3 48 + j14
Z Z
Z = RL RC =
=
= 4.47∠ − 10.3°
Z RL + Z RC
10 + j 5
6 + j8 + 4 − j 3
(
(
total current in the circuit,
power factor of the circuit
) (
) (
)
)
8
( )
V
230∠0°
= 51.43∠10.3° A
=
Z 4.47∠ − 10.3°
cos(10.3 ) 0.98 (leading)
I=
6
230 V,
50 Hz
( )
Fig. 8.44 Circuit of
Example 8.14
4
3
507
Sinusoidal Steady State Analysis
Solved Problems
Problem 8.1 Determine the average and rms values of a sinusoidal current having Im as the maximum value. Hence, find the values of the peak
factor and form factor.
Solution The average value of a sine wave over a complete cycle is zero. So,
we consider the half-cycle average value.
i(t) Im sin ␻t
∴ I av =
1
Im
∫ I sin td ( t ) =
m
⎡⎣ − cos t ⎤⎦ =
0
2 Im
Im
2p
0
p
t
Fig. 8.45
= 0.637 I m
0
I av =
⇒
2
I m = 0.637 I m
The rms value of the sine wave is obtained as
∴ I rms =
1
2
2
∫ I m sin td
0
0
Im ⎡
I ⎡⎛
⎤
⎞ ⎛
⎞⎤ I
1
1
1
t − sin 2 t ⎥ = m ⎢⎜ − siin 4 ⎟ − ⎜ 0 − sin 0⎟ ⎥ = m = 0.707 I m
⎢
2 ⎣
2
2 ⎣⎝
2
2
⎠ ⎝
⎠⎦
2
⎦0
2
=
Im2
( t ) = 2 ∫ (1 − cos 2 t )d ( t )
Im
= 0.707 I m
2
I
peak value
= m = 2 = 1.414
peak factor =
rms value I m
2
⇒
I rms =
2
Im
rms value
2=
=
= 1.11
form factor =
average value 2 I m
2
Problem 8.2 Calculate the average and root mean square values, the form factor, and peak factor of a
periodic current wave having the following values for equal time intervals over half-cycle, changing suddenly from one value to the next:
0, 40, 60, 80, 100, 80, 60, 40, 0
8
∑ ii 0 + 40 + 60 + 80 + 100 + 80 + 60 + 40
I av = i =1 =
= 57.5 A
8
8
Solution Average value,
8
The rms value,
I rms =
∑i
i =1
8
i
2
=
0 + 402 + 602 + 802 + 1002 + 802 + 602 + 4002
= 64.42 A
8
508
Network Analysis and Synthesis
Here, peak value
100 A
I rms
form factor =
I av
I max
peak factor =
I rms
=
64.42
= 1.12
57.5
=
100
= 1.554
64.42
Problem 8.3 Determine the rms value of a triangular wave in which
the average rises uniformly from 0 to V volts and completes the cycle
by falling instantaneously to zero.
v (t )
V
Or,
0
Determine the effective value of a sawtooth wave.
T
2T
3T
4T
t
2␲
3␲
Fig. 8.46
Solution The rms or effective value is given as
2
T
T
T
2
1
1 ⎡V ⎤
V2 T3 V
V2 2
⎡
⎤
=
=
t
dt
Vrms =
v
t
dt
t
dt
=
=
⎢
⎥
⎦
T ∫0 ⎣
T ∫0 ⎣ T ⎦
T3 3
T 3 ∫0
3
()
Problem 4 Calculate the average and rms value for a half-wave rectified sinusoidal quantity. Also, find the peak factor and form factor.
v(t)
vm
Solution Average value,
2
t
() ( )
1
Vav =
v t d
2 ∫0
1
t =
V sin td
2 ∫0 m
V
V
t = m ⎡⎣ − cos t ⎤⎦0 = m
2
( )
0
␲
Fig. 8.47
rms value,
2
Vrms =
=
( ) ( t ) = 21 ∫V sin
2
1
⎡v t ⎤ d
∫
⎣
⎦
2 0
2
m
td
( t)
0
) ( t ) = 4 ⎛⎜⎝ t − 12 siin 2 t ⎞⎟⎠ = 4
Vm 2
1 − cos 2 t d
2 × 2 ∫0
(
2
Vm 2
Vm 2
=
0
Vm
2
Vnm
Vrms
2 = = 1.57
form factor =
=
2
Vav Vm
peak factor =
Vmax
V
= m =2
Vrms Vm
2
f(␪)
Fm
Problem 8.5 The half-cycle of an alternating signal is as follows
It increases uniformly from zero at zero to Fm at , remains constant
from to(180 − ), decreases uniformly from Fm at (180 − ␣) to zero
at 180 . Calculate the average and effective values of the signal.
0
Fig. 8.48
␣
␪
(180
␣)
509
Sinusoidal Steady State Analysis
Solution
The function is given as
f
( )=
Fm
0< <
;
= Fm ;
=
Fm
< <
( − )
( − ); ( − ) < <
The average value is
−
⎤ 1⎡ F
⎤
F
1⎡
d ⎥ = ⎢ ∫ m d + ∫ Fm d + ∫ m − d ⎥
⎢∫ f
⎢⎣ 0
⎢⎣ 0
⎥⎦
⎥⎦
−
⎡
2
−
F ⎧⎪
1 F 2
− −
+
= ⎢ m
+ Fm − 2 + m ⎨ × −
⎢
2
2
2
⎪⎩
⎣
2
2
2
2
⎤ F
F ⎡
= m ⎢ + −2 +
−
+ −
+
+ − ⎥= m −
2
2
2
⎣2
⎦
()
Fav =
)
(
)
(
)
(
(
) ⎫⎪⎤⎥
(
) = F ⎛⎜⎝ 1 − ⎞⎟⎠
2
⎬⎥
⎪⎭ ⎦
m
The effective value is
Frms 2 =
1
∫ ⎡⎣ f
0
⎡ Fm 2
( )⎤⎦ d = 1 ⎢ ∫
2
2
⎢⎣ 0
⎡ 2
1 ⎢ Fm 3
= ⎢ 2
+ Fm 2
3
⎢
⎣
−
2
d + ∫ Fm 2 d + ∫
Fm 2
2
−
( −2 )
⎧
Fm 2 ⎪
+ 2 ⎨
⎪
⎩
( − )
3
3
⎤
( − )d⎥
2
⎥⎦
⎫⎤
2
⎤
⎪ ⎥ Fm ⎡
=
⎬⎥
⎢ 3 + −2 + 3⎥
⎣
⎦
⎪⎥
− ⎭⎦
⎛ 4 ⎞
⎛ 4 ⎞
= Fm 2 ⎜ 1 − ⎟ ∴ Frms = Fm ⎜ 1 − ⎟
⎝ 3 ⎠
⎝ 3 ⎠
Problem 8.6 Find the average and rms values of the periodic function shown in Fig. 8.49.
= 60° =
Solution From Problem 8. 5, with
5
and Vrms =
V
3 m
2
, Vav = Vm
3
3
v(u)
Vm
u
0
p/3
Fig. 8.49
Problem 8.7 Transform the following sinusoids into phasors:
(a) i(t)
4 sin (10t
10 ),
(b) v(t)
7 cos (2t
40 )
Solution Taking sine as reference,
(a) i(t ) = 4 sin(10t + 10 ) ∴ I = 4 ∠10°
()
(
)
(
)
(
(b) v t = −7 cos 2t + 40 = 7 sin 2t + 40 − 90 = 7 sin 2t − 50
) ∴V = 7∠ − 50°
2p/3
p
4p/3 5p/3
2p
510
Network Analysis and Synthesis
Problem 8.8 Find the sinusoids represented by the following phasors:
(a) I
3
j4
j8e j20
(b) V
()
( t − 53.13 )
= 8∠ − 20° ∴ v ( t ) = 8 sin ( t − 20 )
Solution (a) I = −3 + j 4 = 5∠ − 53.13° ∴ i t = 5sin
V = j8e − j 20
(b)
Problem 8.9 Find the sum of the five emf’s:
e1 = 20 sin t e2 = 10 sin
( t + 3 ) e = 15cos t e = 10sin( t − 6 ) e = 25cos( t + 2 3 )
3
4
5
Solution
e1 = 20 sin t = 20∠0° = 20
e2 = 10 sin
( t + 3 ) = 10∠ 3 = (5 + j8.66)
e3 = 15cos t = 15sin
( t + 90 ) = 15∠90= j15
( t − 6 ) = 10∠ − 6 = (8.66 − j5)
e = 25cos( t + 2
= 25sin ( t + 2
+
= 25∠210° = −21.65 − j12.5
3)
3
2)
e4 = 10 sin
5
The sum of the five emf is
(
e = e1 + e2 + e3 + e4 + e5
)
(
)
(
∴ e = 13.49 sin ( t + 27.17 )
) (
)
= 20 + 5 + j8.66 + j15 + 8.66 − j 5 + −21.65 − j12.5 = 12 + j 6.16 = 13.49∠27.17°
Problem 8.10 The voltage across an ideal element is v = 3 cos 3 t (V) and the associated current through
the element is i = −2 sin(3 t +10 ° ) (A) . Determine the phase relationship between the average voltage and
current.
Solution Given: v 3 cos 3t (v) and
I leads v by an angle 100 .
(
)
)
(
Problem 8.11 Find the response ‘i’ of an RL series circuit if R
2
,L
2
,L
1 H, and input voltage, v(t) 10 sin 3t (V)
∴ X L = j L = j × 3 ×1= j3
∴I =
)(A)
1 H and the input voltage, v(t)
10 sin 3t.
Solution Given: R
(
i = −2 sin 3t + 10 = 2 cos 3t + 10 + 90 = 2 cos 3t + 100
V 10∠0°
10
=
=
∠33.69° = 2.77∠33.69°
Z 2 + j3
13
511
Sinusoidal Steady State Analysis
()
( )
( )
i t = 2.77 sin( 3t + 33.69° ) A = 2.77 cos( 3t − 146.31° ) A
So, the current is
Problem 8.12 Find the steady-state voltage v for the RC circuit shown in Fig. 8.50
100 rad/s.
when i 10cos t (A), R 1 , C 10 mF, and
Solution
Given: R
1
,C
10 mF, ␻
100 rad/s
i
and
R
C
( ) ⇒ I = 10∠90°
i = 10 cos t = 10 sin( t + 90 ) A
Fig. 8.50
R
1
10∠90° 10
=
∠45°
∴V = I × Z = I ×
= 10∠90° ×
=
−3
1 + j RC
1+ j
1 + j100 × 1 × 10 × 10
2
( ) 10 sin( t + 45 ) = 10 cos( t − 45 ) ( A )
∴v t =
2
2
Problem 8.13 Two circuits having the same numerical ohmic impedance are joined in parallel. The
power factors of the circuits are 0.8 and 0.6. What is the power factor of the circuit?
Solution Let the resistances and reactances of the two circuits be R1, R2 and X1 and X2.
For the 1st circuit
Power factor 0.8
∴ R1 = Z cos = 0.8 Z and X 1 = Z sin = 0.6 Z
For the 2nd circuit
Power factor 0.6
∴ R1 = Z cos = 0.6 Z and X 1 = Z sin = 0.8 Z
(
Impedance of the circuit ZT = R1 + jX 1
(0.8Z + j 0.6 Z × (0.6 Z + j 0.8Z
) ( R + jX ) = (0.8Z + j 0.6 Z )) + (0.6 Z + j 0.8Z ))
2
2
⎛
j1 ⎞
=⎜
Z = 0.505Z ∠45°
+
1
.
4
j1.4 ⎟⎠
⎝
power factor of the circuit is cos 45
0.707
Problem 8.14 Obtain the expression for the time-domain
currents i1(t) and i2(t) in the circuit shown in Fig. 8.51.
Solution
Here,
3
i1
= 103 rad/s;
10cos103t (V)
−3
∴ X L = j L = j10 × 4 × 10 = j 4
3
and
XC =
−j
−j
= − j2
=
C 103 × 500 × 10−6
)
and
2
1
2
1
i2
2i1
Fig. 8.51
( 3 + j 4 ) I − j 4 I = 10
− j 2 I + 2 I + j 4( I − I ) = 0 ⇒ ( 2 − j 4 ) I + j 2 I = 0
(
By KVL in phasor domain, 3 I1 + j 4 I1 − I 2 = 10∠00 ⇒
4 mH
500 ␮F
1
2
1
2
(i)
(ii)
512
Network Analysis and Synthesis
Solving for I1 and I2,
I1 =
and,
2
10 − j 4
0
j2
=
− j 20
− j 20
=
= 1.24 ∠29.745°
j 6 − 8 + j8 + 16 8 + j14
(3+ j4) − j4
(2 − j4) j2
( 3 + j 4 ) 10
( 2 − j 4 ) 0 = −20 + j 40 = 2.77∠56.31°
I =
( 3 + j 4 ) − j 4 j 6 − 8 + j8 + 16
(2 − j4) j2
∴ i ( t ) = 1.24 cos(10 t + 29.7 ) ( A ) and i ( t ) = 2.77 cos(10 t + 56.31 ) ( A )
3
3
1
2
Problem 8.15 For the circuit, find the node voltages vA and
vB using node voltage method. The source current is given as
1000 rad/s.
is(t) 10cos t (A),
Solution
VA
5
is(t )
= 1000 rad/s;
Here,
VB
10
C = 100 ␮F
∴ X L = j L = j1000 × 5 × 10−3 = j 5
Fig. 8.52
−j
−j
XC =
=
= − j10
and
C 1000 × 100 × 10−6
By KCL in phasor domain
V
V −V
At the node (A),
−10∠0° + A + A B = 0 ⇒ VA 1 + j − VB = 100
− j10
10
(
and at the node (B),
Solving for VA and VB,
(
VA =
(1 + j )
(
VB =
−1
3− j2
−1
) = 300 − j 200 = 100( 3 − j 2 ) = 87.446∠ − 47.73°
1 + j1 100
−1
0
(1 + j )
−1
+ 2 −1
3 + j1+
4 + j1
=
−100
−100
= 24.253∠ − 165.964°
=
3 + j1 + 2 − 1 4 + j1
(3− j2)
∴ v ( t ) = 87.45cos(1000t − 47.7 ) ( A ) ⎞
⎟
v ( t ) = 24.25cos(1000t − 165.9 ) ( A )⎟⎠
−1
A
and
)
(3− j2)
−1
and,
)
VB VB VB − VA
+ +
= 0 ⇒ − VA + VB 3 − j 2 = 0
5 j5
10
100
0
B
L = 5 mH
(i)
(ii)
513
Sinusoidal Steady State Analysis
Problem 8.16 Find the voltage vx if, v1(t)
20cos 1000t (V) and v2(t)
10 mH vx 0.1mF
20sin 1000t (V).
Solution
= 103 rad/s;
Here,
v1
25
v2
−3
∴ X L = j L = j10 × 10 × 10 = j10
3
XC =
and
Also, V1 = 20∠0°
−j
−j
= 3
= − j10
C 10 × 0.1 × 10−3
Fig. 8.53
and V2 = 20∠ − 90°
By KCL in phasor domain
Vx − 20∠0° Vx − 20∠ − 90° Vx
+
+ =0
j10
− j10
25
At the node (x),
⎡ 1
1 ⎤ 20
20
1
⇒ Vx ⎢
−
+ ⎥=
∠0° −
∠ − 90°
j10
⎣ j10 j10 25 ⎦ j10
V
⇒ x = 2 − j 2 ⇒ Vx = 50 − j 50 = 70.71∠ − 45° V
25
)
(
()
( )
(
∴ v x t = 70.71cos 1000t − 45
)(V)
Problem 8.17 (a) A current I 10 30 flows through an impedance, Z 20
22 . Find the average
power delivered to the impedance.
(b) Calculate the average power absorbed by an impedance, Z (30 j70) when a voltage, V 120 0
is applied across it.
Solution (a) Given: I
10 30 , Z
average power delivered,
20
V
IZ
200 8 (V)
)
(
1
1
Pav = Vm I m ∠ v − i = 200 × 10 cos 8 − 30 = 927.18 W
2
2
)
(
22
(b) Here, Z = 30 − j 70 = 76.16 ∠ − 66.8
∴I =
;V
120 0 V
V
120∠0°
= 1.576 ∠66.8 A
=
Z 76.16 ∠ − 66.8°
( )
The average power is given as
1
∴ Pav = Vm I m cos
2
( − ) = 12 × 120 × 1.576 × cos(0° − 66.8° ) = 37.24 W
v
i
Problem 8.18 Calculate the average power absorbed by the resistor
and inductor. Find the average power supplied by the voltage source.
Solution
Here, V
(
)
8 45 (V), Z = 3 + j1 = 3.16 ∠18.43°
V
8∠45°
∴I = =
= 2.53∠26.565 A
Z 3.16 ∠18.43°
( )
3
8 45 (V)
Fig. 8.54
j1
514
Network Analysis and Synthesis
1
The power supplied by the source is ∴ Pav = Vm I m cos
2
( − ) = 12 × 8 × 2.53 × cos( 45 − 26.565 ) = 9.6 W
v
i
1
1
1
Average power absorbed by the resistor, PR = VR I R = × 3 I R I R = × 3 × 2.53 × 2.53 = 9.6 W
2
2
2
This is equal to the power supplied by the source such that the average power absorbed by the inductor is
zero.
Problem 8.19 Calculate the average power absorbed by
each of the five elements in the circuit.
Solution
)
(
By KVL, 8 − j 2 I1 + j 2 I 2 = 40
j4
8
I1
I2
j2
40 0 (V)
20 90 (V)
j 2 I1 + j 2 I 2 = − j 20
and
Fig. 8.55
Solving for the currents,
I1 =
I2 =
40
− j 20
j2
j2
j2
j2
(8 − j 2 )
40
j2
− j 20
j2
j2
(8 − j 2 ) j 2
=
(8 − j 2 ) j 2
j80 − 40 −5 + j10
= 5∠53.13° A
=
j16 + 4 + 4 1 + j 2
( )
=
− j160 − 40 − j80
5 + j 30
=−
= −13.775∠17.1° A
j16 + 4 + 4
1+ j 2
( )
)
(
average power supplied by the 40-V source is,
1
P40 V = × 40 × 5cos 0 − 53.13 = 60 W
2
average power supplied by the 20-V source is,
1
P20 V = × 20 × 13.75cos 90 − 17.1 = 40 W
2
(
)
1
1
P8 = × 8 I1 × I1 = × 8 × 5 × 5 = 100 W
2
2
average power absorbed by the resistor
Problem 8.20 Obtain the power factor and the apparent power of a load whose impedance is
Z (60 j40) when the applied voltage is v(t) 150cos(377t 10 )(V).
Solution Here, V
(
∴I =
power factor = cos
)
150 10 (V), Z = 60 + j 40 = 72.11∠33.69°
V
150∠0°
=
= 2.08∠ − 23.69° A
Z 72.11∠33.69°
( − ) = cos(10 − 23.69 ) = 0.832 ( lag )
v
i
1
1
apparent power = Vm I m = × 150 × 2.08 = 156 VA
2
2
( )
515
Sinusoidal Steady State Analysis
Problem 8.21 Given a circuit with an impedance Z (3 j4) and an applied voltage, V 100 30
(volt), determine the apparent, real, and reactive power. What will be the power factor of the circuit?
Solution
)
(
100 30 (V), Z = 3 + j 4 = 5∠53.13°
Here, V
∴I =
V 100∠30°
=
= 20∠ − 23.13° A
Z 5∠53.13°
( )
1
1
apparent power = Vm I m = × 100 × 20 = 1000 VA
2
2
1
real power = Vm I m cos
2
( − ) = 1000 × cos( 30 + 23.13 ) = 600 W
v
1
reactive power = Vm I m sin
2
power factor = cos
i
( − ) = 1000 × cos( 30 + 23.13 ) = 800 W
v
i
( − ) = cos( 30 + 23.13 ) = 0.6
v
i
Problem 8.22 A resistance and an inductance are connected in series across a voltage, v(t) 283sin 314t.
. Find the value of the inductance and the power factor.
The current expression is given by 40 sin 314 t −
4
What is the power drawn by the circuit?
)
(
Solution Here, V
283 0 (V), I
40 −45 (A),
∴ Z = R2 +
Also, ␻
314 rad/s or f
2
L2 =
283
= 7.075
40
50 Hz
⎛ L⎞
∴ tan −1 ⎜
=
⇒
⎝ R ⎟⎠ 4
From the equation (i),
(i)
L=R
2 R = 7.075 ⇒ R = 5
∴L=
R
5
=
= 0.0159 H
2 f 100
power factor = cos 45° = 0.707
1
1
power drawn = Vm I m cos 45° = × 283 × 40 × 0.707 = 4000 W
2
2
Problem 8.23 In an RL series circuit, a voltage of 100 V at 25 Hz produces 1A while the same voltage at
75 Hz produces ½ A. Draw the diagram and insert the values of R and L.
Solution
Since the voltage is constant
∴
I1 Z 2
=
I 2 Z1
516
Network Analysis and Synthesis
)
)
(
(
∴
⇒ 4 R 2 + 4 × 2500
Now,
)
)
(
(
2
2
R2 + 2 f2 L
R 2 + 2 × 75 L
1
=
=
2
2
1
R 2 + 2 f1 L
R 2 + 2 × 25 L
2
2
L2 = R 2 + 22500
2
L2 ⇒ 3R 2 = 12500
2
L2
(i)
100
I1 = 1 =
(
R + 50 L
2
)
2
⇒ R 2 + 2500
2
L2 = 10000
(ii)
Solving (i) and (ii),
R
79.05
and
L
0.3898 H
Problem 8.24 A resistance of 20 , inductance of 0. 2 H and capacitance of 150 F are connected in
series and are fed by a 230-V, 50-Hz supply. Find XL, XC, Z, Y, pf, active power and reactive power.
Solution Given: R
20
,L
0.2 H, C
150 F, V
230 V, f
50 Hz
X L = 2 × 50 × 0.2 = 62.8
XC =
1
= 21.22
2 × 50 × 150 × 10−6
(
Z = R2 + X L − X C
Power factor,
) = 20 + (62.8 − 21.22 ) = 46.168
2
Y=
1
1
=
= 0.0217 S
Z 46.168
cos =
R
20
=
= 0.4332
Z 46.168
The current in the circuit =
2
2
( lag ) ⎡⎣ X > X ⎤⎦
L
C
V
230
=
= 4.982 A
Z 46.168
active power = VI cos = 230 × 4.982 × 0.4332 = 496.37 W
reactive power = VI sin = 230 × 4.982 × 0.901 = 1032.72 W
Problem 8.25
Fig. 8.56.
i1(t )
1
Find the currents i1(t) and i2(t) in the circuit shown in
Solution Given: L
1
1 H, C = F
9
∴ X L = j L = j3
i2(t )
3
and ␻
and X C =
3 rad/s
−j
= − j3
C
5cos3t
1/9 F
1H
Fig. 8.56
517
Sinusoidal Steady State Analysis
By KCL at the node (x), we get,
I1(t )
1
5∠0° − Vx
V
V
⎡ 1 j j⎤
= x + x ⇒ Vx ⎢1 + − + ⎥ = 5∠0°
1
3+ j3 − j3
⎣ 6 6 3⎦
I2(t )
x
3
5 0 (V)
⎡7 j ⎤
30∠0°
⇒ Vx ⎢ + ⎥ = 5∠0° ⇒ Vx =
= 3 2 ∠ − 8.13°
7+ j
⎣6 6 ⎦
j3
j3
Fig. 8.57
5∠0 − Vx
∴ I1 =
= 5∠0 − 3 2 ∠ − 8.13 = 5 − 4.158 − j 0.598 = 1∠38.9 A
1
V
3 2 ∠ − 8.13°
I2 = x =
= 2 ∠81.87 A
− j3
3∠ − 90°
( )
and
( )
() (
) ( A ) ⎫⎪
⎬
i ( t ) = 2 cos( 3t + 81.87 ) ( A ) ⎪⎭
Thus, the currents are given as i1 t = cos 3t + 38.9
2
Problem 8.26 Find the sum of the three currents:
()
()
i1 t = 20 sin t , i2 t = 10 sin
( t + 6 ) , i (t ) = 25cos ( t + 2 3 )
3
Solution The sum of the three currents is given as,
)
(
I = I1 + I 2 + I 3 = 20∠0° + 10∠30° + 25∠210°
= 20 + 8.66 + j 5 − 21.65 − j12.5 = 7 − j 7.5 = 10.259∠ − 46.97°
()
∴ i t = 10.259 sin
( t − 46.97 ) ( A )
Problem 8.27 Two sources, e1 = 200 sin t (V ) and e2 = 200 sin ( t + 30° ) (V ) are in series supplying power
to a circuit of impedance (8 j6) . Calculate the total source voltage, current and power supplied.
Solution
The sources are given as
)
(
E1 = 200∠0°
and E2 = 200∠30°
( )
total source voltage, E = E1 + E2 = 200 + 200∠30° = 200 + 173.2 + j100 = 386.37∠15° V
∴ e = 386.37 sin
The current in the circuit, I =
( t + 15° ) ( V )
E 386.37∠15° 386.37∠15°
=
=
= 38.6637∠ − 21.87° A
Z
8 + j6
10∠36.87°
( )
∴ i = 38.637 sin
( t − 21.87° ) ( A )
power supplied, P = EI * = 386.37∠15° × 38.637∠21.87° = 1.194 kW
Problem 8.28 An alternating voltage (80 + j60)V is applied to a circuit and the current flowing is
( −4 + j10) A . Find the impedance of the circuit, the power consumed and the phase angle.
Solution
(
)
Given: V = 80 + j 60 = 100∠36.87°
and
(
)
I = − 4 + j10 = 10.77∠ − 68.199° A
518
Network Analysis and Synthesis
Z=
impedance of the circuit,
)
(
1
P = Vrms I rms cos = × 100 × 10.77 × cos 105.069 = 140 Watt
2
power consumed,
Phase angle, ␾
V
100∠36.87°
=
= 9.285∠105.069°
I 10.77∠ − 68.199°
105.069
Problem 8.29 A R = 5 resistance and a L = 30 mH inductance are connected in parallel across a voltage,
v =100 sin (1000 t + 50 ° ) volt. Obtain the total current, i.
Solution
Given: R
5
,L
30 mH, ␻
1000 rad/s, V
100 50
100∠50°
The current through the resistor I R =
= 20∠50° = 20 × 0.6428 + j 0.766
5
(
The current through the inductor
)
(
I L ==
)
100∠50°
= 1.33∠ − 40° = 1.33 × 0.766 − j 0.6428
j 30
(
)
(
(
6428
total current is I = I R + I L = 20 × 0.6428 + j 0.766 + 1.33 × 0.766 − j 0.6
)
)
= 13.875 + j14.465 = 20.04 ∠46.19°
(
∴ i = 20.04 sin 1000t + 46.19°
)(A)
Problem 8.30 Find the expression for the current and calculate the power, when a voltage of
283 sin 314 t is applied to a coil of R 50 and L 0.159 H.
v
Solution Given: R
50
I=
the current in the circuit,
Power,
0.159 H, ␻
314 rad/s, V
283 0
Z = 50 + j 314 × 0.159 = 50 + j 50 = 50 2 ∠45°
Impedance of the circuit,
Current in the circuit,
,L
V
283∠0°
=
= 4 ∠ − 45° A
Z 50 2 ∠45°
( )
()
(
i t = 4 sin 314t − 45
)(A)
1
P = Vrms I rms cos = × 283 × 4 × cos 45 = 400 W
2
Problem 8.31 Two impedances (14 + j5) and (18 + j10) are connected in parallel across a 200-V, 50c/s supply. Determine the capacitance which when connected in parallel with the original circuit will make
the resultant power factor unity.
)
(
Solution Given: Z1 = 14 + j 5 = 14.87∠19.65°
V 200V, f 50 Hz
(
1
= 0.0634 − j 0.0226
14.87∠19.65°
1
= 0.0424 − j 0.0236
Y2 =
20.59∠29.05°
Admittances are given as Y1 =
)
and Z 2 = 18 + j10 = 20.59∠29.05°
519
Sinusoidal Steady State Analysis
) (
(
) (
) (
)
the circuit admittance is, Y = Y1 + Y2 = 0.0634 − j 0.0226 + 0.0424 − j 0.0236 = 0.1058 − j 0.0462 S
)( )
(
total current in the circuit, = 200Y = 21.16 − j 9.24 A
Now, the capacitance current must be equal to the imaginary part of the total current so that the resultant
power factor of the circuit becomes unity.
∴ I C = 9.24 ⇒
× C × V = 9.24 ⇒ 2 f × C × V = 9.24 ⇒ C =
9.24
9.24
=
= 147 F
2 f × V 2 × 50 × 200
Problem 8.32 A series ac circuit has a resistance of 15 and an inductive reactance of 10 . Calculate the value of a capacitor which is connected across this series combination so that the system has unity
power factor. The frequency of ac supply is 50 Hz.
Solution
Here, R
15
, XL
10
G=
conductance of the series branch,
15
15
R
= 2
=
2
2
R + X L 15 + 10 325
2
XL
10
10
=
2
R + X L 15 + 10 325
The power factor of the system will be unity if the susceptance of the capacitor to be connected in parallel is
equal to the susceptance of the series branch, i.e.,
B=
susceptance of the series branch,
2
2
C=B ⇒ C=
=
2
B
10
=
= 98 F
2 f 2 × 50 × 325
Problem 8.33 An ac voltage of 200 V is applied to a series circuit consisting of a resistor, an inductor, and
a capacitor. The respective voltages across these components are 170 V, 150 V, and 100 V and the current is
4 A. Find the power factor of the inductor and also of the circuit. Draw the phasor diagram.
Solution Given: Vab
170 V, Vbc
150 V, Vcd
100 V,
and
I
4A
Vbc 150
100
and X C =
=
= 37.5
= 25
I
4
4
If the inductive coil has a resistance RL and a reactance x then
⇒
(
(i)
)
(
170
Z=
+ RL + j x − 25 = 42.5 + RL + j x − 25
4
)
(
)
200
= 42.5 + RL + j x − 25 ⇒ 42.5 + RL + j x − 25 = 50 ⇒
4
By (ii) − (i),
( 42.5 + R ) + ( x − 25) − R + x = 2500 − 1406.25
2
2
L
2
Fig. 8.58
)
( 42.5 + R ) + ( x − 25) = 2500
2
2
L
(ii)
2
L
(
)
⇒ 85 RL + 1806.25 − 50 x + 625 = 1093.75 ⇒ 85 RL − 50 x = −1337.5 ⇒ RL = 0.588 x − 15.73
From (iii) and (i),
d
200 V
RL 2 + x 2 = 37.52 = 1406.25
(
c
4A
∴ XL =
Again, impedance of the circuit,
b
a
(0.588 x − 15.73) + x = 1406.25 ⇒ x = 28.856 ( taking possitive root )
2
2
(iii)
520
Network Analysis and Synthesis
From (iii),
)
(
RL = 0.588 x − 15.73 = 1.238
cos L =
power factor of the inductor,
Total resistance,
RL 1.238
=
= 0.033
X L 37.5
R = R1 + R2 = 42.5 + 1.238 = 43.738
cos =
the power factor of the circuit,
R 43.738
=
= 0.875
Z 37.550
Problem 8.34 A 230-V, 50-c/s voltage is applied to a coil of L 5 H and R
C. What value must C have so that the pd across the coil should be 250 V?
Solution Given: V
230 V,
total impedance of the coil,
current,
I=
f
50 Hz, L 5 H, R 2 ,
∴ = 2 f = 2 × 50 = 100
∴ X L = L = 500 = 1570
(
) (
VC
2
in series with a capacitor
250 V
)
Z L = R + j L = 2 + j1570 = 1570.8∠89.93°
250∠0°
V
=
= 0.159∠ − 89.93° A
Z 1570.8∠89.93°
( )
Z=
Now, impedance of the circuit,
230
= 1446.54
0.159
)
(
Hence, the impedance of the capacitance is ZC = 1570 − 1446.54 = 12425
∴
1
1
= 124.25 ⇒ C =
= 25.62 F
C
2 × 50 × 124.25
Problem 8.35 A 159.23- F capacitor in parallel with a resistance R draws a current of 25 A from 300-V,
50-Hz mains. Using phasor diagrams, find the frequency f at which this combination draws the same current
from a 360-V mains.
Solution When f
50 Hz,
capacitive current, I C =
resistive current,
V
300 V,
C
159.23 F,
supply current, I
V
300
=
= 15 A
X C 2 × 50 × 159.23 × 10− 6
I R = I 2 − I C 2 = 252 − 152 = 20 A
value of the resistance,
R=
V 300
=
= 15
I R 20
For the new frequency f, the supply current will remain the same, i.e., I
Resistive current,
I R′ =
V ′ 360
=
= 24 A
R 15
new capacitive current,
I C′ = I ′ 2 − I R′ 2 = 252 − 24 2 = 7 A
25 A.
25 A.
521
Sinusoidal Steady State Analysis
X C′ =
new capacitive reactance,
f=
new supply frequency,
V ′ 360
=
= 51.43
7
I C′
1
1
=
= 19.4 Hz
2 CX C 2 × 159.23 × 10−6 × 51.43
For phasor diagram, see Section 8.19.
Problem 8.36 A voltage v(t) 400 sin 314t(V) is applied to the circuit shown in
Fig. 8.59. Find the currents and their phase angles wrt the voltages for the three
branches. C 50 F; R1 100 ; R2 50 ; L 0.1 H.
Solution Here, C 50 F; R1
V 400 90 (V), ␻ 314 rad/s
100
IR =
current in the resistive branch,
1
current in the capacitive branch,
; R2
50
;
L
C
R1
R2
0.1 H,
Fig. 8.59
V 400∠0°
=
= 4 ∠0° A
100
R1
IC =
V
1
L
= 400∠0° × 314 × 50 × 10− 6 ∠90° = 6.28∠90° A
j C
current in the RL branch,
IC =
V
=
Z
400∠0°
R2 +
2
⎛ L⎞
L ∠ tan ⎜ ⎟
⎝ R2 ⎠
( )
2
−1
=
400∠0°
⎛ 314 × 0.1⎞
50 + 314 × 0.1 ∠ tan ⎜
⎝ 50 ⎟⎠
2
(
)
2
= 6.77∠ − 32.13° A
−1
Problem 8.37 An inductive circuit in parallel with a resistive circuit of 20 is connected across 50-Hz
supply. The inductive current is 4.3 A and the resistive current is 2.7 A. The total current is 5.8 A. Find (a)
power absorbed by the inductive branch, (b) inductance, and (c) power factor of the combined circuit. Also,
draw the phasor diagram.
Solution Here, f 50 Hz, R 20 , IL 4.3 A; IR
The circuit is shown in Fig. 8.60.
Supply voltage is V = I R × R = 2.7 × 20 = 54 V
The phasor diagram is shown in Fig. 8.61 below
2.7 A; I
V
54
Impedance of the RL branch, Z =
=
= 12.56
I RL 4.3
Now, from the phasor diagram, the phase angle of the impedance of the
RL branch is obtained as AD 2 = AB 2 + BD 2 + 2 AB × BD cos
AD 2 − AB 2 − BD 2 5.82 − 2.72 − 4.32
=
⇒ cos =
= 0.3385
2 AB × BD
22.7 × 4.3
resistance of the RL branch,
reactance of the RL branch,
R = Z cos = 12.56 × 0.3385 = 4.25
X = Z 2 − R 2 = 12.56 2 − 4.252 = 11.82
5.8 A
5.8 A
4.3 A
2.7 A
v (t)
R
20
L
Fig. 8.60
IR = 2.7 A B
A
C
V
f
IRL = 4.3 A
I = 5.8 A
D
Fig. 8.61 Phasor diagram
522
Network Analysis and Synthesis
(a) Power absorbed by the inductive branch
(b) Inductance,
L=
X
2 f
=
I RL 2 × R = 4.32 × 4.25 = 78.6 W
11.82
= 37.6 mH
2 × 50
(c) Power of the combined circuit is obtained as
pf =
AC AB + BC 2.7 + BD cos
2.7 + 4.3 × 0.3385
= 0.716 (Lagging)
=
=
=
5.8
AD
AD
5.8
Problem 8.38 An inductive coil of 15- resistance and 42- inductive reactance is connected in parallel
with a capacitive reactance of 47.6 . The combination is energized from a 200-V, 33.5-Hz ac supply. Find
the total current drawn by the circuit and its power factor.
Solution The circuit is shown in Fig. 8.62.
Current drawn by the inductive branch,
200 V,
33.5 Hz
V 200∠0
I RL = =
= 4.48∠ − 70.346° A = 1.508 − j 4.223 A
Z 15 + j 42
( ) (
)( )
total current drawn by the circuit is
(
) (
47.6
42
Fig. 8.62
Current drawn by the capacitive branch,
IC =
15
V
200
=
= 4.202 ∠90° = j 4.202 A
X C − j 47.6
)
(
)
( )
I = I RL + I C = 1.508 − j 4.223 + j 4.202 = 1.508 − j 0.02 = 1.5∠ − 0.81° ≈ 1.5∠0° A
power factor of the circuit
cos = cos(0 ) = 1.0
Problem 8.39 Two currents in each branch of a two-branched parallel circuit is given as
⎛
⎛
⎞
⎞
i a = 8.07 sin ⎜ 314t − ⎟ ; i b = 21.2 sin ⎜ 314t − ⎟
4⎠
3⎠
⎝
⎝
and supply voltage is 354 sin 314 t. Calculate (i) total current in the same form, and (ii) ohmic values of components in each branch.
()
Solution Here, supply voltage v t = 354 sin 314t
⎛
⎛
⎞
⎞
Currents, ia = 8.07 sin ⎜ 314t − ⎟ ; ib = 21.2 sin ⎜ 314t − ⎟ .
4⎠
3⎠
⎝
⎝
It is seen that the current ia lags behind the voltage by an angle
. Therefore, the branch a consists of a
4
resistance and a pure inductance in series.
Similarly, it is seen that the current ib lags behind the voltage by an angle
. Therefore, the branch b also
3
consists of a resistance and a pure inductance in series.
(
)( )
I = 21.2 ∠ − 60° = (10.6 − j18.36 )( A )
The currents in sin phasor form are given as I a = 8.07∠ − 45° = 5.706 − j 5.706 A
b
523
Sinusoidal Steady State Analysis
(i)
) (
(
) (
total current is given as I = I a + I b = 5.706 − j 5.706 + 10.6 − j18.36
( )
)
= 16.306 − 24 = 29.07∠ − 55.88° A
(
i = 29.07 sin 314t − 55.88
Thus, the supply current is,
Za =
(ii) Impedance of the branch a,
)( A )
354
= 43.866
8.07
)
(
resistance of the branch a,
Ra = Z a cos a = 43.866 × cos 45° = 31
reactance of the branch a,
X a = Z a sin
Zb =
Impedance of the branch b,
a
(
)
= 43.866 × sin 45° = 31
354
= 16.698
21.2
( )
= 16.698 × sin ( 60 ) = 14.461
resistance of the branch b,
Rb = Z b cos b = 16.698 × cos 60 = 8.349
reactance of the branch b,
X b = Z b sin b
Problem 8.40 A resistance of 12 and an inductance of 0.025 H are connected in series across a 50-Hz
supply. What values of resistance and inductance when connected in parallel will have the same resultant
impedance and pf? Find the current in each case when the supply voltage is 230 V.
Solution For series circuit,
f = 50 Hz, Rs = 12 ; L = 0.025 ⇒ X s = 2 × 50 × 0.025 = 7.854
Let Rp be the resistance and Xp the reactance in parallel circuit.
The resultant impedance of the series and parallel circuits will be same if,
conductance of series circuit conductance of parallel circuit
and
susceptance of series circuit susceptance of parallel circuit
∴
Rs
Rs 2 + X s
=
2
R 2 + X s 2 12 2 + 7.854 2
1
⇒ Rp = s
=
= 17.14
12
Rs
Rp
and
Xs
Rs 2 + X s
=
2
R 2 + X s 2 12 2 + 7.854 2
1
⇒ Xp = s
=
= 26.188
Xp
Xs
7.854
the value of the inductor for the parallel circuit is given as Lp =
Xp
2 f
Thus, the value of resistance and inductance for parallel circuit are, Rp
( )
V
230
= 16.037∠33.2° A
current in each case is given as I = =
Z 12 + j 7.854
=
26.188
= 83.361 mH
2 × 50
17.14
Lp = 83.36 mH
524
Network Analysis and Synthesis
Problem 8.41 Two circuits the impedances of which are given by Z1 (10 j15) and Z2 (6 j8) , are
connected in parallel. If the total current supplied is 15 A, what is the power taken by each branch?
Solution Current through impedance Z1 is
I1 = I ×
Z2
⎛ 6 − j8 ⎞
⎛
⎞
6 − j8
= 8.589∠ − 76.76° A
= 15 × ⎜
= 15 × ⎜
Z1 + Z 2
⎝ 16 + j 7 ⎟⎠
⎝ 10 + j15 + 6 − j8 ⎟⎠
( )
Current through impedance Z2 is
I2 = I ×
Z1
⎛ 10 + j15 ⎞
⎛
⎞
10 + j15
= 15.484 ∠ − 732.68° A
= 15 × ⎜
= 15 × ⎜
⎟
Z1 + Z 2
⎝ 16 + j 7 ⎟⎠
⎝ 10 + j15 + 6 − j8 ⎠
( )
power taken by the branch Z1 is P1 = I12 × R1 = 8.5892 × 10 = 738 W
power taken by branch Z2 is P2 = I 2 2 × R2 = 15.484 2 × 6 = 1438 W
Problem 8.42 Figure 8.63 shows a series–parallel circuit. Find
(i) Admittance of each parallel branch
(ii) Total circuit impedance
(iii) Supply current and power factor
(iv) Total power supplied by the source
1.6
L
N
Solution
1
(i) Admittance of RL branch, YRL =
= 0.2 ∠ − 36.87 mho
4 + j3
Admittance of RC branch, YRC =
4
j3
6
j8
j 7.2
100-V,
50-Hz
supply
Fig. 8.63
1
= − 0.1∠53.13 mho
6 − j8
(ii) Impedance of the parallel branches is,
( 4 + j 3) × (6 − j8) = 48 − j14 = 4..472135955∠10.3° = 4.4 + j 0.8
(
)
( 4 + j 3) + (6 − j8) 10 − j 5
total circuit impedance is Z = (1.6 + j 7.2 ) + ( 4.4 + j 0.8 ) = ( 6 + j8 ) = 10∠53.13° ( )
Zp =
(iii) Supply current, I =
(
( )
V
100
=
= 10∠ − 53.13° A
Z 6 + j8
)
Power factor = cos −53.13 = 0.6 (lagging)
(iv) Total power supplied by the source,
P = I 2 × R = 102 × 6 = 600 W
sin2t
Problem 8.43 For the circuit shown in Fig. 8.64, find the voltage v1.
Solution Here,
1
1
= 2 rad/s; L1 = H; L2 = 1 H; C = F
2
2
∴ X L1 = j1 ; X L 2 = j 2 ; X C = − j1
1
4cos2t
Fig. 8.64
v1
1 /2 F
1 /2H
2v1
1H
525
Sinusoidal Steady State Analysis
By KCL at the Node (1),
V1
V −V
− 4 ∠0° − 1∠ − 90° + 1 2 = 0
−j
j1
⎛ 4− j⎞
⇒ V2 = ⎜
= 1 + 4 j = 4.123∠75.96° V
⎝ − j ⎟⎠
)
(
( )
1
90
21
1
By KCL at the Node (2),
V2 − V1
V
− 2V1 + 2 + 1∠ − 90° = 0
− j1
1+ j 2
4 0
⎛1
2⎞
⇒ V2 j1 − V1 j1 − 2V1 + ⎜ − j ⎟ V2 − j1 = 0
5⎠
⎝5
( ) ( )
V1
j1
j1
2V1
j2
Fig. 8.65
⎛1
⎛1
⎞
⎞
2
2
⇒ V1 2 + j1 = V2 ⎜ − j + j1⎟ − j1 = 1 + 4 j ⎜ − j + j1⎟ − j1
5
5
⎠
⎠
⎝5
⎝5
(
)
(
)
)
(
11 2
+j
5
5
11 2
− +j
5 = −11 + j 2 2 − j1
⇒ V1 = 5
5×5
2 + j1
⇒ V1 2 + j1 = −
)(
(
)
⎛ 20
15 ⎞
4
3
= ⎜ − + j ⎟ = − + j = 1∠ − 36.87°
25 ⎠
5
5
⎝ 25
(
)
(
Thus, the voltage is given as v1 = cos 2t − 36.87 = cos 2t + 143.13
)(V)
Problem 8.44 Given that the voltages VAB and VBC in the circuit
are 100 V each, find R, L and C in the circuit and the power consumed. The line current is 5 A.
Solution Since VAB
VBC
will be as shown in Fig. 8.67.
VAC
100 V, the phasor diagram
∴ I R = 5cos 30° = 4.33 A and I C = 5sin 30° = 2.5 A
R
A
5A
B
L
C
5A
C
100 V, 50 Hz
Fig. 8.66
C
To find R
R=
100 100
=
= 23
I R 4.33
60
To find C
Here,
Now,
5A
IC
= 2 f = 2 × 50 = 100
XC =
A
IR
Fig. 8.67
100 100
=
= 40
IC
2.5
∴
1
1
= 40 ⇒ C =
= 79.6 F
C
100 × 40
B
526
Network Analysis and Synthesis
To find L
100
= 20
5
XL =
∴ L = 20 ⇒ L =
20
= 63.66 mH
100
Problem 8.45 A load as shown in Fig. 8.68 has an impedance of ZL (100 j100) .
Find the parallel capacitance required to correct the power factor to unity. Assume
377 rad/s.
that the source is operating at
=
100 X C2
(
+j
2
100 + 100 + X C
2
(
)
C
0.707
(
)
1002 X C + 100 + X C 100 X C
(
100 + 100 + X C2
2
)
(
X 100 + 100 + X C 100 100 + 100 + X C
=
=
100 X C
R
XC
2
∴
)
)
(
(
100 + j100 × jX C
Z L jX C
=
Z L + jX C 100 + j100 + jX C
ZL
Fig. 8.68
Solution Here, ZL (100 j100)
Hence, the original load has a lagging power factor of cos(45 )
Parallel combination of ZL and capacitance reactance XC is
Z=
vs
)
)
Here, the corrected power factor angle must be zero ( unity power factor).
∴
(
100 + 100 + X C
) = tan (0 ) = 0 ⇒ X = −200
C
XC
−1
−1
=
= 13.3 F
X C 377 × −200
⇒ C=
)
(
Problem 8.46 An industrial load takes 4 kW at a lagging pf of 0.8 when connected to a 200-V, 50-Hz
supply. Find the value of the parallel capacitance necessary to improve the pf to unity.
Solution
Here, load current, I L =
Power factor
cos
4 × 103
= 25 A
200 × 0.8
0.8 (lagging) ⇒
36.87
IL 25
36.87 A (20 j15) A
When C is connected in parallel, the current should be in phase with the voltage. Total current,
I
For unity power factor, IC
(IL
IC)
(20
j15
IC)
j15
∴ j CV = j15 ⇒ C
15
= 238.73 F
2 × 50 × 200
527
Sinusoidal Steady State Analysis
Summary
1. For transmission and distribution, alternating current
has a number of advantages over direct current.
2. A sinusoid is a signal that has the form of a sine or
cosine function and in general can be written as
)
v (t ) = V sin( t + ) where, V is the amplitude,
v (t = V m sin t . A shifted sinusoid can be written as
m
m
2
, T is the time period
T
of the sinusoid and is the phase of the sinusoid.
Use of sinusoids has several advantages like minimum
disturbance in electrical circuits, less interference to
nearby communication lines and less iron and copper
losses.
The value of an alternating quantity at any instant of
time is known as the instantaneous value.
The maximum value of an alternating quantity attained
in each cycle is known as the peak or maximum or crest
value.
The average value of an alternating quantity over a
given time interval is the summation of all instantaneous values divided by the number of values taken
T
1
over that interval. Mathematically, V av = ∫vdt , where
T 0
T is the time period of the quantity.
the angular frequency
3.
4.
5.
6.
is
2 f
7. The rms or effective value of a continuous periodic
t
T2 is
function f(t) defined over the interval T1
T
f rms =
T
2
2
1 2
⎡f t ⎤ dt or, f = 1 ⎡f t ⎤ dt
∫
⎣
⎦
rms
∫0 ⎣ ⎦
T 2 − T1 T
T
1
()
()
8. Form factor is the ratio of the rms value to the average
value for an alternating wave.
rms Value
( ) average
value
∴ form factor K f =
For a sinusoidal wave, its value is 1.11.
9. Peak factor is the ratio of the peak value to the rms
value for an alternating wave.
value maximum value
=
( ) peak
rms value
rms value
∴ peak factor K p =
For a sinusoidal wave its value is 1.414.
10. A phasor is a complex quantity that represents both
the magnitude and phase angle of a sinusoid. For a
sinusoid given as v t = V m cos t + , the corre-
()
(
)
sponding phasor is written as, v (t ) = V cos ( t + ) .
m
11. The graphical representation of the phasors of sinusoidal quantities taken all at the same frequency and with
proper phase relationships with respect to each other
is called a phasor diagram.
12. Both KVL and KCL hold good in phasor domain, i.e.,
V1 +V 2 +V 3 + ⋅⋅⋅ +V n = 0 and I 1 + I 2 + I 3 + ⋅⋅⋅ + I n = 0 .
13. The voltage and current in different circuit elements
have definite phase relations. For a resistor, the voltage
and current are always in phase, i.e., the phase angle is
zero. In a pure inductor, the current lags behind the
voltage by 900 and in a pure capacitor, the current
leads the voltage by 900.
14. Impedance (Z) of any two-terminal network is the
ratio of the phasor voltage (V) to the phasor current (I )
V
j ∠Z
= Z ∠Z . The real part
i.e. Z = = R + jX = Z e
I
of impedance Re[Z]
R is called the resistance. The
(
)
imaginary part of impedance Im[Z]
X is called the
reactance. Impedance, resistance and reactance are all
measured by the same unit, ohm ( ).
Z = R ; for aresistor
= j L ; for aninductor
1
=
; for a capacitor
j C
15. The reciprocal of the impedance Z is called admittance. So, it is the ratio of the phasor current to the
phasor voltage, i.e. . The real part of admittance
R
. The
is called conductance, G = Re ⎡⎣Y ⎤⎦ = 2
R +X 2
imaginary part of admittance is called susceptance,
X
B = Im ⎡⎣Y ⎤⎦ = 2
. Admittance, conductance and
R +X 2
susceptance are all measured by the same unit, siemen (S).
16. Instantaneous power absorbed by an element is the
product of the instantaneous voltage v(t) and the
instantaneous current i(t), i.e., p(t) v(t) i(t) (in watts).
17. Average or real or active power (in watts) is the
average of the instantaneous power over a time
T
1
interval, i.e., P = ∫ p t dt . For the sinusoidal voltT 0
age and current given as v t = V m cos t + v and
()
()
)
)
(
)
i (t = I m cos ( t + i , the average power is given as
1
P = V m I m cos ( v − i = V rms I rms cos ( v − i .
2
)
)
528
Network Analysis and Synthesis
18. The product of rms voltage and current is known as
apparent power (in VA), i.e., S Vrms Irms.
19. The ratio of average power to apparent power is known as
average power P
power factor, i.e., PF =
= = cos v − i .
apparent power S
20. Reactive power (in VAR) is the product of the applied
voltage and reactive component of the current. For
an ac load with voltage phasor, V = V m ∠ v and cur-
(
)
rent phasor, I = I m ∠ i , the reactive power is written as
)
Q = V rms I rms sin( v − i .
21. Complex power is the product of the rms voltage
phasor and the complex conjugate of the rms phasor
current. For an ac load with voltage phasor, V = V m ∠ v
and current phasor, I = I m ∠ i , the complex power S
absorbed by the load is given by,
)
1
S = VI * = V rms I * = V rms I rms ∠( v − i
rms
2
= V rms I rms cos ( v − i + jV rms I rms sin( v − i
(
= P + jQ
)
)
)
where P is the real or average or active power and Q is
the reactive power.
22. Although the inductor and capacitor take instantaneous power, the average power consumed in these
reactive elements is always zero.
23. For drawing a phasor diagram in a series circuit, generally the current is taken as the reference; while drawing a phasor diagram in a parallel ac circuit, voltage is
taken as the reference.
Short-Answer Questions
1. What are the advantages of generating electrical
energy as ac?
Alternating current has a number of advantages over
dc. Some of the advantages are given below.
1. Alternators (generators designed for ac operation) do not require the slip-rings and commutators
(brushes) upon which their dc cousins depend.
2. An even greater advantage to ac is that its voltage
can be stepped up to higher levels with a transformer,
sent great distances through high-tension wires,
and stepped down at its destination.
3. Alternators at power stations produce three-phase
electricity; they have three coils equally spaced
around their primary coil, each of which is induced
to produce a 50-Hz alternating current for three circuits. Three-phase electricity can supply as much
current through three thin wires as it would normally take two thick wires to carry. The advantage
in using a thinner wire is to minimize the electrical
resistance that a thick wire would produce.
4. Also, line losses are lower for ac than dc for a given
wattage delivery and wire diameter.
2. Explain the concept of phasor and vector. Is impedance a phasor quantity? If not, then how is it
expressed in phasor or complex form?
Or,
Explain the difference between impedance and a
phasor. What role does impedance play in phasor
diagrams?
• Concept of phasor and vector Vector is a multidimensional quantity; it has both magnitude and
direction. Phasor is a two-dimensional vector and is
used in electrical technology that relates sinusoidal
voltage and current.
For analysis of alternating circuits, a sinusoidal
quantity (voltage or current) is represented by a
line of definite length rotating in anti-clockwise
direction with the same angular velocity as that of
the sinusoidal quantity. This rotating line is called
the ‘phasor’. In other words, a phasor is a complex
representation of the magnitude and phase of a
sinusoid.
For a sinusoid,
)
v (t = V m cos ( t +
) = Re ⎡⎣V e (
j
m
t+
)⎤
⎦
= Re ⎡⎣ Ve j t ⎤⎦
where, V Vme j Vm
is the phasor representation of the sinusoid.
Sinusoidal quantities are scalar quantities
varying periodically with time. According to the
definition of a vector, these are not vectors. Voltage is the work done per unit charge and current
is the flow of electrons through a wire and these
are not vectors. However, as a sinusoid is specified by its amplitude and phase angle, it is termed
‘phasor’, keeping some similarity with the term
‘vector’, where the amplitude is considered as the
magnitude and phase angle as the direction of
the vector.
529
Sinusoidal Steady State Analysis
• Impedance vs phasor In phasor domain, impedance is defined as the ratio of the voltage phasor (V)
to the current phasor (I), i. e,
Z
V
I
Here, although Z is a frequency-dependent quantity and ratio of two phasors, it is not a phasor,
because it does not correspond to a sinusoidally
varying quantity.
In circuit theory, impedance in phasor domain is
expressed by generalized Ohm’s law, as given by
V ZI
where,
Z = R , for aresistor
= j L , for aninductor
1
=
, for a capacitor
j C
Impedance Z is a complex quantity and thus, can be
written as
Z (R jX) 冷Z冷
where, R 冷Z冷 cos is the resistance and X 冷Z冷 sin
is the reactance.
As impedance is not a phasor, it is not shown in a
phasor diagram. In a phasor diagram, the role of impedance is to change the magnitudes and phase angles of
different voltage and current phasors in the circuit.
4. Explain the method of representing alternating quantities as phasor quantities. What are the
advantages of phasor representation?
For analysis of alternating circuits, a sinusoidal
quantity (voltage or current) is represented by a line
of definite length rotating in anti-clockwise direction with the same angular velocity as that of the
sinusoidal quantity. This rotating line is called the
‘phasor’.
• Transformation of sinusoid into phasor
To represent a dc voltage or current, only its amplitude I or V is needed. However, to represent a sinu) or current i(t) Im
soidal voltage v(t) Vm cos( t
cos( t
), three values are needed:
• Amplitude, the peak value Vm or Im
• Frequency
2 f
• phase
To simplify the computation of a sinusoidal variable, it
is often represented by a complex variable (vector in
complex plane) which can be more conveniently dealt
with, as various mathematical operations (addition/
subtraction, multiplication/division, etc.) on exponential functions can be much more easily carried out than
sinusoidal functions.
(r cos t
We consider a function, f(t)
re j t
jr sin t)
If ‘ ’ is constant, this function will rotate in counterclockwise direction at constant angular velocity, . The
variation is shown in Fig. 8.69.
3. Why is a sinusoidal wave shape insisted for voltages and currents while generating, transmitting
and utilizing ac electric power?
A sinusoidal wave shape is insisted for voltages and currents while generating, transmitting and utilizing ac electric power because it has the following advantages:
1. Sinusoidal waveforms produce minimum disturbance in electrical circuits during operation.
2. Sinusoidal waveforms produce electromagnetic
torque which is free of noise and oscillations.
3. Sinusoidal waveforms cause less interference to
nearby communication lines (telephones, etc.)
4. The iron and copper losses with sinusoidal waveforms are low in transformers and rotating ac
machines. Therefore, machines operate with higher
efficiency with sinusoidal waveforms.
5. The possibility of resonance is much reduced with the
use of sinusoidal waveforms compared to other nonsinusoidal waveforms containing harmonic frequencies.
Vme j t
j
Vm
t
sine function
r
0
A
0
2
t
0
Vm
cosine function
2
t
Fig. 8.69 Phasor representation of a sinusoid
The projection of this rotating line segment on
both the real and imaginary axis will be the cosine and
sine components, i.e.,
Re[f(t)] r cos t and Im[f(t)] r sin t
530
Network Analysis and Synthesis
Similarly, in electric-circuit theory, the voltages and
currents can be represented by a rotating function
characterized by a magnitude (radius, r) and a phase
with respect to a reference angle. Such a rotating function is termed ‘phasor’.
Specifically, a sinusoidal voltage can be represented as
()
(
) = Re ⎡⎣V (t )⎤⎦ = Re ⎡⎣V e e ⎤⎦ =
Re ⎡Vrms e j
⎣
2e j t ⎤ = Re ⎡V 2e j t ⎤
⎦
⎣
⎦
v t = Vm cos t +
j
()
(
Im ⎡Vrms e j
⎣
2e j t ⎤ = Im ⎡V 2e j t ⎤
⎦
⎣
⎦
v t = Vm sin t +
()
where V t = Vm e (
j
t+
j t
m
) = Im ⎡⎣V (t )⎤⎦ = Im ⎡⎣V e e ⎤⎦ =
j
j t
m
)
= 2Vrms e (
j
t+
)
V = Vrms e j = Vrms ∠
in terms of
• the magnitude, the rms (effective) value V, and
• the phase
The frequency
2 f is not explicitly represented by
the phasor, as all currents and voltages in the circuit
considered here have the same frequency—same as
that of the energy source or input of the circuit.
• Advantages of using a phasor
A sinusoidal waveform has two attributes, magnitude
and phase, and thus sinusoids are natural candidates for
representation by phasors. One reason for using such a
representation is that it simplifies the description since a
complete spatial or temporal waveform is reduced to just
a single-point represented by the tip of a phasor’s arrow.
Thus changes in the waveform are easily documented by
the trajectory of the point in the complex plane.
Phasor
Waveform
Imaginary
axis
c
0
Time
5. Why is impedance represented by a complex
number? How is complex impedance dependent
on frequency?
In phasor domain, impedance is defined as the ratio of
the voltage phasor (V) to the curXC
X
rent phasor (I), i. e.,
XL
V
Z
I
is the complex variable, Vm is the peak magnitude
is the effective value (rms),
of the voltage, Vrms = Vm
2
and the phasor representing the voltage is defined as
c
0
To perform the decomposition using trigonometry
is a tedious business. However, if the sinusoid is represented by a phasor then the same method used for
decomposing vectors into orthogonal components
may be used for decomposing the given sinusoid into
its orthogonal sine and cosine components.
c
0
Real axis
Fig. 8.70 Phasor representation of sinusoids
The second reason is that it helps us to visualize
how an arbitrary sinusoid may be decomposed into
the sum of a pure sine and pure cosine waveform.
R
0
f
Here, although Z is a frequencydependent quantity and a ratio Fig. 8.71
of two phasors, it is not a phasor, Frequency
because it does not correspond to variation
of complex
a sinusoidally varying quantity.
In circuit theory, impedance in impedance
phasor domain is expressed by the
generalized Ohm’s law (V ZI). Impedance Z is a complex quantity and is given by,
(
)
Z = R ± jX = Z ∠ ±
where, R 冷Z冷cos is called the resistance and R 冷Z冷sin
is the reactance. The positive sign is taken for inductive
reactance and negative sign for capacitive reactance.
Z = R,
for aresistor
= j L, for aninductor
1
=
for a capacitor
j C
From the above explanation, we see that complex impedance has two components; the real component is the
resistance which is frequency independent. But, the imaginary part of the impedance is frequency dependent.
L2 fL
The inductive reactance is given as XL
1
2 fC
where f is the supply frequency. Therefore, the
inductive reactance is directly proportional to the frequency and the capacitive reactance is inversely proportional to the frequency.
The capacitive reactance is given as X C =
6. While drawing a phasor diagram in a parallel ac circuit, which quantity should be taken as reference
and why?
531
Sinusoidal Steady State Analysis
While drawing a phasor diagram in a parallel ac circuit,
voltage should be taken as the reference.
As the voltage across each element is the same for
a parallel network, the voltage phasor is taken as the
reference for drawing phasor diagrams.
7. In an ac parallel circuit, is it possible that the magnitude of a branch current is larger than the current drawn from the supply? Explain.
In an ac parallel circuit, the current in every branch is the
ratio of the voltage to impedance. If the branch is purely
resistive, the current will be in phase with the voltage. If the
branch is purely inductive, the current will lag the voltage
by 90 . If the branch is purely capacitive, the current will
lead the voltage by 90 . If the branch is inductive with some
resistance, the current will lag the voltage by some angle
greater than 00 but less than 90 . If the branch is capacitive
with some resistance, the current will lead the voltage by
some angle greater than 0 but less than 90 . The total current is the phasor summation of all the branch currents.
Now, if two parallel branches contain an inductor
and a capacitor respectively and if their reactances
are of equal magnitude then the phenomena of resonance occurs and the current in each of these parallel
branches may be larger than the current drawn from
the supply. However these two currents will be 180 out
of phase and will cancel each other.
8. Prove that the active power consumed in any
purely reactive circuit is zero.
Let the instantaneous voltage, v Vmax sin t
∴current in case of a purely inductive circuit,
t
i=
=
L
(
sin t −
2
max
and, current in case of a purely capacitive circuit,
i =C
2
Vmax Imax
1
=
V I sin2 t
2 sin t cos t =
2 max max
2
∴average power,
P=
(
Vmax
dv
sin t +
= CVmax cos t =
2
1
dt
C
(
= Imax s in t +
2
)
)
V
V
where, Imax = max or , max
1
L
C
In general for a purely reactive circuit, the current can
be written as
(
i = Imax sin t ±
2
)
T
T
⎤
1
1 ⎡ Vmax Imax
pdt
=
sin2 t ⎥ dt = 0 {
⎢
∫
∫
T0
T 0⎣
2
⎦
T
2 }
Hence, we conclude that the active power consumed
in any purely reactive circuit is zero.
9. Does an inductance draw instantaneous power as
well as average power?
For a pure inductance,
let the instantaneous voltage, v Vmax sin t
t
V t
V
1
∴current, i = ∫ vdt = max ∫ sin tdt = − max cos t =
L0
L 0
L
=
Vmaxx
L
(
sin t −
2
) = I sin( t − 2 )
max
Vmax
where, Imax = L
∴instantaneous power,
(
p = vi = Vmax Imax sin t sin t −
2
)=
V I
1
− max max 2 sin t cos t = − Vmax Imax sin2 t
2
2
∴average power,
P=
) = I sin( t − 2 )
)
(
p = vi = Vmax Imax sin t sin t ±
t
Vmax
Vmax
1
vdt =
sin tdt = −
cos t
L ∫0
L ∫0
L
Vmax
∴instantaneous power,
T
T
⎤
1
1 ⎡ V I
pdt = ∫ ⎢ − max max sin2 t ⎥ dt = 0
∫
T0
T 0⎣
2
⎦
{
T
2 }
Thus, we can see that an inductance draws
some instantaneous power which may be positive
or negative; but the average power drawn by an
inductance is always zero. The reason is explained
below.
When v and i are both increasing or decreasing, the power is positive and energy is delivered
from the source to the inductance. When either v is
increasing and i is decreasing or v is decreasing and
i is increasing, the power is negative and energy is
returning from the inductance to the source.
During the second quarter of a cycle, the current
and the magnetic flux of the inductor increases and the
inductor draws power from the supply source to build
up the magnetic field. The power drawn is positive. The
532
Network Analysis and Synthesis
energy stored in the magnetic field during the building
1
up is LImax 2 .
2
In the next quarter, the current decreases. However,
the emf of the inductor tends to oppose this decrease.
The inductor acts as a generator and returns energy to
the supply. The power is negative.
This event repeats and a proportion of power is
continually exchanged between the field and the
inductive circuit and the average power consumed by
the purely inductive circuit becomes zero.
10. Explain why the phasor of voltage across the
inductor leads the current phasor by 90 and the
phasor of voltage across the capacitor lags its
current by 90 .
For inductor The relationship between the voltage
dropped across the inductor and the current flowing
through it can be written as
()
v t =L
()
di t
dt
()
Assuming again a complex voltage, v t = Vm e (
j
t+
)
and a complex current response
j t+
i t = I e ( ) , we get
()
m
d ⎡ j( t + ) ⎤
j t+
I e
= j LIm e ( )
⎦
dt ⎣ m
⇒ Vm e j = j LIm e j
Vm e (
j
t+
)
=L
V = j LI
Or,
V = LI∠90°
Thus, the voltage across an inductive reactance
leads the current through it by 90 . The voltage
dropped across an inductor is a reaction against the
change in current through it. Therefore, the instantaneous voltage is zero whenever the instantaneous
current is at a peak (zero change, or level slope, on the
current sine wave), and the instantaneous voltage is
at a peak wherever the instantaneous current is at
maximum change (the points of steepest slope on
the current wave, where it crosses the zero line). This
results in a voltage wave that is 900 out of phase with
the current wave.
For capacitor The relationship between the current
through the capacitor and the voltage across it can be
written as
()
i t =C
()
dv t
dt
()
Assuming again a complex voltage, v t = Vm e
and a complex current response,
j t+
i t = I e ( ) , we get
()
( t+ )
m
Im e (
j
⇒
Or,
j
t+
)
=C
d⎡
j t+
j t+
V e ( ) ⎤ = j CVm e ( )
⎦
dt ⎣ m
I = j CV or V =
V=
I
j C
I
∠ − 90°
C
Thus, the current through a capacitive reactance
leads the voltage across it by 90 . The current through
a capacitor is a reaction against the change in voltage
across it. Therefore, the instantaneous current is zero
whenever the instantaneous voltage is at a peak (zero
change, or level slope, on the voltage sine wave), and
the instantaneous current is at a peak wherever the
instantaneous voltage is at maximum change (the
points of steepest slope on the voltage wave, where it
crosses the zero line). This results in a voltage wave that
is −90 out of phase with the current wave.
11. Define resistance, reactance, impedance and
admittance.
• Resistance It is that property of an object that
opposes the flow of electric current through it. It is
expressed in ohms ( ).
• Reactance It is the property of an object by which
it can store energy in either electrostatic or magnetic
forms. It is also expressed in ohms ( ). Reactance is
of two types—inductive and capacitive.
Inductive reactance can store energy when current flows
through it. It opposes the instantaneous change in current;
when the current changes, an emf is induced in it. Inductive
reactance is highly resistive to ac but does not oppose dc.
Capacitive reactance can store energy when a voltage is applied across it. It opposes the instantaneous
change in voltage. Capacitive reactance is highly resistive to dc but does not oppose ac.
• Impedance Impedance (Z) of any two-terminal
network is the ratio of the phasor voltage (V) to the
phasor current (I).
Z=
V
I
Since it is the ratio of voltage to current, its unit is ohm
( ). Practically, it represents the obstruction that the
device exhibits to the flow of sinusoidal current.
533
Sinusoidal Steady State Analysis
As a complex variable, the complex impedance Z can
be written as
Z=
(
)
V
= R + jX = Z e j∠Z = Z ∠Z
I
The magnitude and phase angle of Z are
Z = R 2 + X 2 ∠ Z = tan−1
X
R
The real part of impedance Re[Z] R is called resistance. The imaginary part of impedance Im[Z]=X is
called reactance. In particular, the impedances of the
three types of elements R, L and C are
Z = R;
for resistor
= j L; for inductor
1
=
; for capacitor
j C
• Admittance The reciprocal of the impedance Z is
called admittance. So, it is the ratio of phasor current
to phasor voltage.
Y=
1
1
R − jX
= G + jB
=
=
Z R + jX R 2 + X 2
In particular, the admittances of the three types of elements R, L and C are
1
Y = ; for resistor
R
1
; for inductor
=
j L
= j C ; for capacitor
12. Define conductance and susceptance.
The real part of admittance is called conductance (B) and
the imaginary part of admittance is called susceptance (B).
Y=
1
1
R X
=
=
=G
Z R ± jX R 2 + X 2
∴ G = Re ⎡⎣Y ⎤⎦ =
R
R2 + X 2
∴ B = Im ⎡⎣Y ⎤⎦ =
X
R2 + X 2
jB
The susceptance is said to be inductive if its sign is negative and is said to be capacitive if its sign is positive.
13. Explain active and reactive power. What is the
physical significance of reactive power?
• Active power It is the average of the instantaneous
power over a time interval. It is the power consumed
by the resistive loads in an electrical circuit.
In sinusoidally steady state, the active power is given as
1
1
P = Vm Im cos( v − i ) = Vrms Irms cos( v − i ) = Re[VI*](in watts )
2
2
• Reactive power It is defined as the product of the
applied voltage and reactive component of the current. It expressed as volt–ampere reactive (VAR).
For an ac load with voltage phasor V Vm v and current phasor I Im i, the reactive power is written as
( − )
Q = Vrms Irms sin
v
i
• Physical significance of reactive power Reactive poweris the power consumed in an ac circuit
because of the expansion and collapse of magnetic
(inductive) and electrostatic (capacitive) fields.
Unlike true power, reactive power is not a useful
power because it is stored in the circuit itself. This
power is stored by inductors, because they expand
and collapse their magnetic fields in an attempt
to keep the current constant, and by capacitors,
because they charge and discharge in an attempt
to keep the voltage constant. Circuit inductance and
capacitance consume and giveback reactive power.
Reactive power is a function of a system’s amperage.
The power delivered to the inductance is stored in
the magnetic field when the field is expanding
and returned to the source when the field collapses. The power delivered to the capacitance in
the electrostatic field when the capacitor is charging and returned to the source when the capacitor discharges. None of the power delivered to the
circuit by the source is consumed;all is returned to
the source. The true power, which is the power consumed, is thus zero. We know that alternating current constantly changes; thus, the cycle of expansion and collapse of the magnetic and electrostatic
fields constantly occurs.
Therefore, we conclude that reactive power is the rate
of energy flow between the source and the reactive components of the load (i.e., inductances and capacitances).
It represents a lossless interchange between the load
and the source.
14. Explain the difference between apparent power
and real power.
• Apparent power It is the product of the rms (effective) values of voltage and current (in VA).
For a sinusoidal voltage, v(t) Vm cos( t
v) applied
to a network resulting in current, i(t) Im cos( t
i),
the apparent power is given as
S VrmsIrms.
534
Network Analysis and Synthesis
• Real power (active power) It is the average of the
instantaneous power over a time interval. It is the
power consumed by the resistive loads in an electrical circuit.
In sinusoidally steady state, the active power is given as
1
P = Vm Im cos( v − i ) = Vrms Irms cos( v − i )
2
= S cos( v − i ) (in watts )
15. What is a power triangle? Draw and explain.
The relationship between real power, reactive power
and apparent power can be expressed by representing
the quantities as vectors. Real power is represented as a
horizontal vector and reactive power is represented as a
vertical vector. The apparent power vector is the hypotenuse of a right triangle formed by connecting the real
and reactive power vectors. This representation is often
called the power triangle, as shown in Fig. 8.72.
Apparent power (V A)
Apparent
power
VAR
S
Q
Real power (W)
P
Fig. 8.72 Power triangle
Using the Pythagorean Theorem, the relationship
among real, reactive and apparent power is
(Apparent Power)2 (Real Power)2 (Reactive Power)2
S 2 = P 2 + Q 2 or
( VA ) = ( watt ) + ( VAR )
2
2
2
Questions
1. What are the advantages of generating electrical
energy as ac?
2. Define the following terms pertaining to ac wave:
(a) Amplitude (b) Frequency (c) Time period (d) Phase
(e) Phase difference (f) Phase-shift
3. Briefly discuss what you understand by average value
of a periodic function. Determine the effective value of
a sinusoidally varying function of time.
7. Explain the method of representing alternating quantities as phasor quantity. What are the advantages of
phasor representation?
8. What do you understand by ‘phase lag’ and ‘phase
lead’? Explain with the help of examples.
9. Explain the terms impedance, reactance, susceptance
and admittance. Draw the reactance and susceptance
curve for inductance and capacitance.
4. (a) Explain the terms ‘rms value’ and ‘average value’ of
an ac sinusoidal current.
10. Explain the concept of phasor and vector. Is impedance a phasor quantity. If not, then how is it expressed
in phasor or complex form?
(b) Distinguish between average value and rms value
of an alternating waveform.
Or,
(c) Derive the rms value and average value of ac sinusoidal current having Im as maximum value.
(d) Calculate the same for a half-wave rectified sinusoidal quantity.
(e) Do waves other than sine waves have effective value?
5. Why sinusoidal wave shape is insisted for voltages and
currents while generating, transmitting and utilizing ac
electric power?
6. (a) Define form factor and peak factor. Differentiate
between form factor and peak factor
(b) Derive the values of form factor and peak factor of
a sinusoidally varying quantity.
Explain the difference between impedance and a
phasor. What role does the impedance play in phasor
diagrams?
11. Why is impedance represented by a complex number?
How is complex impedance dependent on frequency?
12. Prove that the power consumed in a i) purely inductive circuit, and ii) purely capacitive circuit is zero when
an alternating voltage is applied. Draw the phasor diagrams for V and I.
13. Derive the relationship between the voltage and current for i) a purely inductive circuit, and ii) a purely
capacitive circuit. Also, show that the average power
consumed by the circuit is zero.
535
Sinusoidal Steady State Analysis
14. Prove that the active power consumed in any purely
reactive circuit is zero.
15. Does an inductance draw instantaneous power as well
as average power?
16. Develop an expression for the mean power consumed
over a cycle of a single-phase sinusoidal supply delivering power to a load comprising of i) a resistance R in
series with an inductance L, ii) a resistance R in series
with a capacitance C, and iii) RLC series circuit. Also,
draw the phasor diagrams.
17. Explain why the phasor of voltage across the inductor
leads the current phasor by 90 and the phasor of voltage across capacitor lags its current by 90 .
18. Draw the wave shapes for instantaneous voltage,
current and power in a series RL circuit. Why is the
power positive during some intervals and negative in
others? What is the effect of these positive and negative power regions on the total power consumed by
the circuit?
19. Draw the wave shapes for instantaneous voltage, current
and power in a series RC circuit. Why is the power positive during some intervals and negative in others? What
is the physical significance of the negative power?
20. Explain the following terms:
a) Apparent power
b) True power or average power
c) Complex power
d) Active power
e) Reactive power
f ) Power factor
21. What are active and reactive powers? Draw the power
triangle.
22. While drawing a phasor diagram in a parallel ac circuit,
which quantity should be taken as reference and why?
23. In an ac parallel circuit, is it possible that the magnitude of a branch current is larger than the current
drawn from the supply? Explain.
Exercises
1. Calculate the average and root mean square values
and the form factor of a periodic current wave having
ampere values for equal time intervals, changing suddenly from one value to the next: 0, 30, 45, 70, 90, 70,
45, 30, 0, 30, 45, 70, etc. What should be the average and the root mean square values of a sine wave
having the same peak value? [47.5, 54.5, 57.3, 63.6]
2. A current has the following steady state values in
amperes for equal intervals of time changing instantaneously from one value to the next: 0, 10, 20, 30, 20, 10,
0, 10, 20, 30, 20, 10, 0, etc. Calculate the rms
value of the current and its form factor. [17.8 A, 1.19]
3. a) Given i1(t) 4 cos( t 30 )
120 ), find their sum.
b) If v1(t)
10 sin( t
45 ), find v1 v2.
and i2(t)
30 ) and v2(t)
5 sin( t
5. Find the steady-state current in an RLC series circuit with
R 9 , L 10 mH and C 1 mF when a voltage,v(t) =
100cos(100t)(V ) is applied to it. Use phasors.
[i(t)
7.86 cos(100t
45 ) (A)]
6. The input to a series RL circuit with R 3 and L 0.54 H
is the voltage source, vs(t)
7.28 cos(4t
77 ) (V).
Determine the steady-state output voltage v0(t) across
the inductor.
[v0(t) 4.25 cos(4t 311 ) (V)]
7. Find the resultant emf of the following four emf’s:
( 6)
e = 40 sin( t +
e = 50 sin( t + 3
4)
3)
e1 = 50 sin t
2
e3 = 20 sin t −
4
[e
20 cos( t
80.45 sin( t
34.75 )]
8. Find the sum of the following voltages:
[a) 3.218 cos( t 56.97 ) (A)
b) 10.66 cos( t 36.95 ) (V)]
4. In a particular RL series circuit, a voltage, of 10 V at 50
Hz produces a current of 700 mA while the same voltage at 75 Hz produces a current of 500 mA. What are
the values of R and L in the circuit?
[6.88 , 0.04H]
( 3)
e = 20 sin( t +
e = 50 cos ( t + 2
3)
6)
e1 = 40 sin t
2
e4 = 20 sin t −
5
e3 = 30 cos t
[e
24.2 sin( t
0.096 )]
536
Network Analysis and Synthesis
9. Two coils are connected in parallel across a 200-V,
50-c/s supply. At the supply frequency their impedances are 8 and 10 , respectively, and their resistances are 6
and 4 , respectively. Find (a) the
current in each coil, (b) the total current, and (c) the
total power factor
[25 A, 20 A, 43.9 A, 0.609]
10. A sinusoidal 50-c/s voltage of 200-V rms, supplies the
following three circuits which are in parallel: (a) a coil of
0.03-H inductance and 3- , resistance (b) a capacitor
of 400 μF in series with a resistance of 100 , (c) a coil
of 0.02-H inductance and 7- resistance in series with
a 300-μF capacitor. Find the total current supplied and
draw a complete phasor diagram.
[29.4 A]
11. For the circuit shown in Fig. 8.73, find the magnitudes
of V1 and V2 and the current. Also, calculate the power
factor of the circuit and draw a complete phasor
diagram.
[149 V, 115 V, 4.53 A, 0.679]
10
0 .1 H
b) v(t) 80 cos(10t 20 ) (V) and i(t) 15 sin(10t
60 ) (A)
[a) 344.2 60cos(754t 35 ) (W), 344.2 W;
b) 385.7 600cos(20t 10 ) (W), 385.7 W]
15. For the circuit, find the average power supplied by
the source and the average power absorbed by the
resistor.
[2.5 W; 2.5 W]
4
Fig. 8.76
16. Determine the power generated by each source and
the average power absorbed by each passive element.
[P60V
207.8 W; P4A
20
0.05 H 40 F
20
V2
4 0 (A)
200 V, 50Hz
supply
Fig. 8.77
V1
Fig. 8.73
12. For the circuit shown in Fig. 8.74, find the node voltage
‘v’ in its sinusoidal steady-state form.
( ) 10 cos(10t + 63.4° ) ( V ) ]
[v t =
i
Vs(t) 10cos 10t (V)
5
10
v
10i
10 mF
10
5
Fig. 8.74
3i1
i1
30 mH
5 mF
Fig. 8.75
14. Calculate the instantaneous power and average power, if
a) v(t) 120cos(377t
(377t 10 ) (A)
45 ) (V) and
i(t)
160 W; PL
PC
0]
j5
j10
60 30 (V)
⎛ ⎞
17. Given the time-domain voltage v t = 4 cos ⎜ t ⎟ V ,
⎝6 ⎠
find both the average power and an expression for the
instantaneous power that result when the corresponding phasor voltage V 4 0 (V) is applied across an
impedance Z 2 60 .
()
( )
⎛
( ) ; 2 + 4 cos ⎜⎝ 3 t − 60 ⎞⎟⎠ ( W )]
13. For the circuit shown in Fig. 8.75, find i1(t) for
100 rad/s.
[ii(t) 1.05 cos(100t 71.6 ) (A)]
vs(t) 10兹2 cos ( t 45 )(V)
367.8 W; PR
⎛
⎞
[2 W; 2 cos ⎜ t − 60 ⎟ A
⎝6
⎠
0.5 H
3
j2
5 30 (V)
10cos
18. A resistor R in series with a capacitor C is connected
to a 50-Hz, 240-V supply. Find the value of C so that R
absorbs 330-W at 100 V. Find also the maximum charge
and maximum stored energy.
[43.77 μF;-9.55
10 3 C; 1.0417 J]
19. A coil of R
10
and L
0.1 H is connected in
series with a capacitor of 150-μF across a 200-V, 50-Hz
supply. Find XL, XC, Z, pf, current and voltage across the
capacitor.
[31.4 ; 21.2 ; 14.284 ;
0.7 (lagging); 14 A; 296.8 V]
20. Determine the rms value of the current in each branch
and the total current of the circuit shown in Fig. 8.78.
Draw the phasor diagram.
[9.76
46.32 (A); 5.64 57.86 (A); 10
13.2 (A)
537
Sinusoidal Steady State Analysis
Vrms
15
20
0.05 H
100 F
212 V
23. The parallel circuit shown in Fig. 8.79 is connected
across a single-phase 100-V, 50-Hz ac supply. Calculate (i) the branch currents, (ii) the total current, (iii) the
supply power factor and (iv) the active and reactive
powers supplied by the supply.
[10
Fig. 8.78
36.87 (A); 10 53.13 (A); 14.42 8.13 (A);
0.99 (leading); 1400 W; 200 VAR]
21. Two currents in each branch of a two-branched parallel circuit is given as
⎛
⎛
⎞
⎞
i a = 7.07 sin⎜ 314t − ⎟ ; i b = 21.2 sin⎜ 314t + ⎟
4
3
⎠
⎝
⎠
⎝
100 V, 50 Hz
and the supply voltage is 354sin314t. Derive a similar
expression for the supply current and calculate the
ohmic values of components assuming two pure components in each branch. State whether the reactive
components are inductive or capacitive.
Fig. 8.79
i
RC
20.54 sin (314t 40.58 ); RL
8.35 ; XC 14.46 ]
35.36 ; XL
35.36 ;
6
8
j6
j8
24. Find the impedance, current, power and power factor
of the following series circuits and draw the corresponding phasor diagrams: (i) R and L; (ii) R and C;
(iii) R, L and C. In each case, the applied voltage is
200 V, the frequency is 50 Hz; R 10 , L 50 mH and
C 100 μF.
22. The impedances of two circuits are given by Z1 (10
j15) and Z2 (6 j8) are connected in parallel. If
the total current supplied is 20 A, what is the power
taken by each branch?
[1312 W; 2556 W]
[(i) 18.62 , 10.74 A, 1.153 kW, 0.537 (lag)
(ii) 33.365 , 5.994 A, 359 W, 0.2997 (lead)
(iii) 18.97 , 10.54 A, 1.111 kW, 0.527 (lead)]
Multiple-Choice Question
1. The polar form of v (t) = 100 cos ( t 90 ) is
(i) V 100 90
(ii) V 100
90
(iii) V 100
45
iv) V 100 45
2. In an RLC circuit supplied from an ac source, the reactive power is proportional to
(i) average energy stored in the electric field
(ii) average energy stored in the magnetic field
(iii) sum of the average energy stored in the electric
field and that stored in the magnetic field
(iv) difference between the average energy stored in the
electric field and that stored in the magnetic field
3. The real part of admittance is … and the imaginary
part is …
(i) impedance, resistance
(ii) resistance, impedance
(iii) susceptance, inductance
(iv) conductance, susceptance
⎛I ⎞
4. The value of ⎜ rms ⎟ for the wave form shown is
⎝ I max ⎠
i
2
(i)
(ii) 1.11
(iii) 1
(iv)
5.
1
t
1
2
1
The phasor diagram
for an ideal induct- Fig. 8.80
ance having current I
through it and voltage V across it is
(i)
I
(iii)
V
V
I
Fig. 8.81
(ii)
(iv)
I
V
I
V
538
Network Analysis and Synthesis
6. The average power absorbed by a passive network
(i) is always zero
(ii) is always positive
(iii) is always negative
(iv) may be positive or zero but never negative
7. For the circuit shown in Fig. 8.82, the current i(t) will be
(i) 7.5 sin (1000t) A
(ii)
7.5 sin (1000t) A
(iii) 7.5 cos (1000t) A
(iv)
7.5 cos (1000t) A
i(t)
0.02 H
150 sin1000t
17. The form factor is the ratio of
(i) average value to rms value
(ii) rms value to average value
(iii) peak value to average value
(iv) peak value to rms value
18. The peak factor is the ratio of
(i) average value to rms value
(ii) rms value to average value
(iii) peak value to average value
(iv) peak value to rms value
19. The form factor for dc supply voltage is always
(i) zero
(ii) unity
(iii) infinity
(iv) any value between 0 and 1
V
Fig. 8.82
8. The unit of admittance is
(i) weber
(iii) ohm
16. In an RL series circuit, the power factor is
(i) leading
(ii) lagging
(iii) zero
(iv) unity
(ii) mho
(iv) ampere
9. The impedance of a 1-henry inductor at 50 Hz is
(i) 1
(ii) 31.4
(iii) 50
(iv) 314
10. In an RL series circuit, the phase angle difference
between voltage and current is
(i) 30
(ii) 90
(iii) 180
(iv) greater than zero but less than 90
11. What is the phase angle between inductor current
and applied voltage in a parallel RL circuit?
(i) 0
(ii) 45
(iii) 90
(iv) 30 .
12. The active power dissipated in an ac circuit is
(i) VI
(ii) VI*
(iii) VI cos
(iv) VI sin
13. The power factor of a practical inductor is
(i) unity
(ii) zero
(iii) lagging
(iv) leading
14. A circuit of zero lagging power factor behaves as
(i) an inductive circuit
(ii) a capacitive circuit
(iii) an RC circuit
(iv) an RL circuit
15. Power loss in an electrical circuit can take place in
(i) inductance only
(ii) capacitance only
(iii) inductance and resistance
(iv) resistance only
20. A voltage V is applied to an ac circuit resulting in the
delivery of a current I . Which of the following expressions yield the true power delivered by the source?
1. Real part of V I *
2. Real part of VI
3. I2 times the real part of
V
I
Select the correct answer using the codes given
below:
(i) 1 alone
(ii) 1 and 3
(iii) 2 and 3
(iv) 3 alone
21. The mean value of the current i
20 sin t from t = 0 to
t=
2 is
(i) 40π
(ii) 40/π
(iii) 1/40
(iv) π/40
22. A constant current of 2.8 A exists in a resistor. The rms
value of current is
(i) 2.8 A
(ii) about 2 A
(iii) 1.4 A
(iv) undefined
23. An alternating voltage e 200 sin 314t is applied to
a device which offers an ohmic resistance of 20 to
the flow of current in one direction while entirely preventing the flow in the opposite direction. The average value of current will be
(i) 5 A
(ii) 3.18 A
(iii) 1.57 A
(iv) 1.10 A
24. A 50-Hz ac voltage is measured with a moving iron
voltmeter connected in parallel. If the meter readings
539
Sinusoidal Steady State Analysis
are V1 and V2 respectively and the meters are free from
calibration errors then the form factor of the ac voltage may be estimated as
(i)
V1
V2
(iii) 2
(ii) 1.11
V1
V2
(iv)
V1
V2
V1
2 V2
25. A boiler at home is switched on to the ac mains supplying power at 230 V, 50 Hz. The frequency of instantaneous power consumed is
(i) 0 Hz
(ii) 50 Hz
(iii) 100 Hz
(iv) 150 Hz
26. A circuit component that opposes the change in circuit voltage is
(i) resistance
(ii) capacitance
(iii) inductance
(iv) all of the above
27. An instantaneous change in voltage is not possible in
(i) a resistor
(ii) an inductor
(iii) a capacitor
(iv) a current source
28. A circuit component that opposes the change in circuit current is
(i) resistance
(ii) capacitance
(iii) inductance
(iv) conductance
29. The power factor of an ordinary electric bulb is
(i) zero
(ii) unity
(iii) slightly more than unity
(iv) slightly less than unity
30. The power factor of an ac circuit is equal to
(i) cosine of the phase angle
(ii) sine of the phase angle
(iii) unity for a resistive circuit
(iv) unity for a reactive circuit
31. A series circuit containing passive elements has the
following current and applied voltage:
V 200 sin(2,000t 50 ), i = 4 cos (2,000t 13.2 )
The circuit elements
(i) must be resistance and capacitance
(ii) must be resistance and inductance
(iii) must be inductance, capacitance and resistance
(iv) could be either resistance and capacitance or
resistance, inductance and capacitance
32. In an ac circuit, if voltage V = (a + jb) and current I = (c +
jd) then the power is given by
(i) ac + ad
(ii) ac + bd
(iii) bc ad
(iv) bc + ad
33. In a parallel R-L circuit if IR is the current in the resistor
and IL is the current in the inductor then
(i) IR lags IL by 90
(ii) IR leads IL by 270
(iii) IL leads IR by 270
(iv) IL lags IR by 90
34. In an R-L-C parallel circuit, admittance is defined as the
reciprocal of
(i) resistance
(ii) reactance
(iii) impedance
(iv) susceptance
35. The unit of susceptance is
(i) farad
(iii) henry
(ii) ohm
(iv) mho
36. In an ac circuit having R, L and C in series and operating on lagging pf increase in frequency will
(i) reduce the current
(ii) increase the current
(iii) both (i) and (ii) are possible
(iv) have no effect on current drawn
37. In a network the sum of currents entering a node is
5 60 . The sum of currents leaving the node is
(i) 5 60
(ii) 5
60
(iii) 5 240
(iv) 15 A
38. In the circuit shown in
Fig. 8.83, if the power consumed by the 5-ohm resistor is 10 W then the power
factor of the circuit is
(i) 0.8
(ii) 0.6
(iii) 0.5
(iv) zero
L 10
5
V = 50cos vt
Fig. 8.83
39. Which of the following is true of the circuit in
Fig. 8.84?
1. V R = 100 2V
2. I = 2 A
3. L = 0.25 H
100
VR
250 冑 2 sin 300t
L
I
Fig. 8.84
150V
540
Network Analysis and Synthesis
Select the correct answer using the codes given below:
(i) 2 and 3
(ii) 1 and 2
(iii) 1 and 3
(iv) 1, 2 and 3
40. A series RLC circuit, consisting of R = 10 ohms, XL = 20
ohms, XC = 20 ohms is connected across an ac supply
of 100 V (rms) . The magnitude and phase angle (with
reference to supply voltage) of the voltage across the
inductive coil are respectively
(i) 110 V; 90
(ii) 100 V; 90
(iii) 200 V; 90
(iv) 200 V; 90
41. In a two-element series circuit, the applied voltage
and the resulting current are respectively
v(t) 50 50sin(5 103t)V and i(t)
103t 63.4 ) A
The nature of the elements would be
(i) R-L
(ii) R-C
(iii) L-C
(iv) neither R, nor L, nor C
5
15
冑3
1/3
I
V =3 0
Fig. 8.87
(i) 1 90
(iii) 5 90
(ii) 3 90
2
(iv)
45
45. For the given circuit, if v(t) 160 sin(␻t 10 ) and i(t)
5 sin(␻t 20 ) then the reactive power absorbed by
the black box N is given by
i
N
Fig. 8.88
(i) 50 VAR
(iii) 400 VAR
(ii) 100 VAR
(iv) 200 VAR
46. An ac sinusoidal voltage source is connected across
a series circuit consisting of a resistor and a capacitor.
The rms value of the voltage across the resistor and
capacitor are 100 V and 200 V respectively. The rms
value of the voltage of the source is
(i) 300 V
(ii) 100 5 V
10
Fig. 8.85
1. I = 2 A
2. the total impedance of the circuit is 5
3. cos␾ 0.866
Which of these statements are correct?
(i) 1 and 3
(ii) 2 and 3
(iii) 1 and 2
43 What is the power consumed by the 1the circuit shown in Fig. 8.86?
(iii) 100 3 V
(iv) 100 V
47. For the circuit shown in Fig. 8.89, the total impedance is
3.18 mH
(ii) 50 W
(iv) 130 W
j4
3
j4
Fig. 8.89
(i) (7
(iii) (0
Fig. 8.86
3
17 6
resistor in
1
(i) 30 W
(iii) 100 W
I C =4 90
v
V=10冑 6 V
10 冑 2 sin 314t
1
11.2 sin(5
42 Consider the following statements regarding the circuit shown Fig.8.85. If the power consumed by the
5- resistor is 10 W then
i
44. For the given circuit, ␻ = 3 rad/s. If V is taken as reference, the phasor of I is given by
j0)
j8)
(ii) (5
(iv) (7
j0)
j10)
48. Energy stored in an inductance and in a capacitance
over a complete cycle when excited by a purely sinusoidal ac source is
(i) zero and maximum respectively
(ii) zero and zero respectively
(iii) half of that due to a dc source of equal magnitude
(iv) maximum and maximum respectively
541
Sinusoidal Steady State Analysis
49. The rms value of the periodic waveform given in
Fig. 8.90 is
(i)
5
u
8 RMS
(ii)
2
u
3 RMS
(iii)
8
u
5 RMS
6A
t
T/2 T
6A
Fig. 8.90
3
u
2 RMS
54. For the triangular waveform shown in Fig. 8.93, the
rms value of the voltage is equal to
(iv)
(i) 2 6 A
(ii) 6 2 A
(iii)
4
3
(iv) 1.5A
A
V(t)
1
50. In Fig. 8.91, the admittance values of the elements in
siemens are YR = 0.5 + j0, YL = 0 j1.5, YC = 0 + j0.3
respectively. The value of I as a phasor when the voltage E across the elements is 10 0 V is
YR
I
YL
YC
52. The rms value of the voltage u(t)
17 V
(iii) 7 V
1
6
(ii)
1
3
(iv)
51. The rms value of the resultant current in a wire which
carries a dc current of 10 A and a sinusoidal alternating current of peak value 20 A is
(i) 14.1 A
(ii) 17.3 A
(iii) 22.4 A
(iv) 30.0 A
(i)
(i)
(iii)
1.5 + j0.5
5 j18
0.5 + j1.8
5 j12
3
4 cos(3t) is
(ii) 5 V
(iv)
3T/2
2
3
()
L
Fig. 8.92
t
i (t)
55. The circuit shown in Fig. 8.94, with
1
1
,L
H, C 3 F has
R
3
4
input voltage v(t)
sin 2t. The
resulting current i(t) is
(i) 5 sin(2t 53.1 )
(ii) 5 sin(2t 53.1 )
(iii) 25 sin(2t 53.1 )
(iv) 25 sin(2t 53.1 )
age is v t = 2 sin10 3 t .
R
2T
1
3
56. For the circuit shown in
Fig. 8.95, the time constant
RC = 1 ms. The input volt-
(3 + 2 2 ) V
53. The RL circuit of Fig. 8.92 is fed from
a constant magnitude, variable frequency sinusoidal voltage source
vIN. At 100 Hz, the R and L elements
each have a voltage drop uRMS. If the
frequency of the source is changed
to 50 Hz then the new voltage drop
across R is
T
Fig. 8.93
E=10 0 V
Fig. 8.91
(i)
(ii)
(iii)
(iv)
T/2
The output voltage v0(t) is
equal to
(i) sin(103t 45 )
(ii) sin(103t 45 )
(iii) sin(103t 53 )
(iv) sin(103t 53 )
R
v (t)
L
C
Fig. 8.94
R
v1(t )
Fig. 8.95
C v0(t )
542
Network Analysis and Synthesis
Answers
1
2
3
4
5
6
7
8
9
10
11
12
(ii)
(iv)
(iv)
(iii)
(iii)
(iv)
(iv)
(ii)
(iv)
(iv)
(iii)
(iii)
13
14
15
16
17
18
19
20
21
22
23
24
(iii)
(i)
(iv)
(ii)
(ii)
(iii)
(ii)
(ii)
(ii)
(i)
(ii)
(ii)
25
26
27
28
29
30
31
32
33
34
35
36
(iii)
(ii)
(iii)
(iii)
(ii)
(i)
(iv)
(ii)
(iv)
(iii)
(iv)
(i)
37
38
39
40
41
42
43
44
45
46
47
48
(iii)
(ii)
(i)
(iv)
(ii)
(i)
(iii)
(i)
(iii)
(ii)
(i)
(ii)
49
50
51
52
53
54
55
56
(i)
(iv)
(ii)
(i)
(iii)
(ii)
(i)
(i)
9
Magnetically Coupled
Circuits
Introduction
The circuits we have considered so far may be termed as conductively coupled in the sense that one
coil affects the adjacent coils by current conduction. But when two or more coils are very close to each
other, then the current in one coil will affect the emf induced in other coils and these coils are said to be
mutually coupled or magnetically coupled coils.
In this chapter, we will first discuss the concepts of magnetic coupling and dot conventions required
to write KVL equations with correct polarities. Then we will learn the theoretical aspects of transformers
and tuned circuits.
9.1
SELF INDUCTANCE
Consider a coil consisting of N turns and carrying a current I in the
counterclockwise direction, as shown in Fig. 9.1. If the current is
steady then the magnetic flux through the loop will remain constant.
However, suppose the current I changes with time. Then according to
Faraday’s law, an induced emf will arise to oppose the change. The
dI
> 0 , and counterclockwise if
dt
dI
< 0 . The property of a loop in which its own magnetic field opposes
dt
any change in current is called ‘self-inductance’, and the emf generated
is called the self-induced emf or back emf, which we denote as L. All
current-carrying loops exhibit this property. In particular, an inductor
is a circuit element which has a large self-inductance.
induced current will flow clockwise if
Mathematically, the self-induced emf can be written as
= −N
L
I
Fig. 9.1 Magnetic flux through the
current loop
d B
d → →
= − N ∫∫ B⋅ d A
dt
dt
544
Network Analysis and Synthesis
dI
dt
N
The two expressions can be combined to yield L =
and is related to the self-inductance L by
L
= −L
B
I
Physically, the inductance L is a measure of an inductor’s ‘resistance’ to the change of current; the larger the
value of L, the lower the rate of change of current.
Example 9.1 Self-inductance of a solenoid
Compute the self-inductance of a solenoid with turns N, length l, and radius R with a current I
flowing through each turn, as shown in Fig. 9.2.
Z
R
I
Solution Ignoring edge effects and applying Ampere’s law, the magnetic field inside a
N turns
I
→
NI ៣
I
៣
0
k = 0 nIk
solenoid is given by B =
l
N
where n =
is the number of turns per unit length. The magnetic flux through each
Fig. 9.2 Solenoid
l
2
2
turn is ␾ BA ␮0nI (␲R ) ␮0nI␲R
N
= 0 n2 R 2 l
Thus, the self-inductance is L =
I
We see that L depends only on the geometrical factors (n, R and l) and is independent of the current I.
9.2
COUPLED INDUCTOR
When the magnetic flux produced by an inductor links another inductor, these inductors are said to be coupled. Coupling is often undesired but in many cases, this coupling is intentional and is the basis of the transformer. When inductors are coupled, there exists a mutual inductance that relates the current in one inductor
to the flux linkage in the other inductor. Thus, there are three inductances defined for coupled inductors:
L11—the self inductance of the inductor 1
L22—the self inductance of the inductor 2
L12 L21—the mutual inductance associated with both inductors
When either side of the transformer is a tuned circuit, the amount of mutual inductance between the two
windings determines the shape of the frequency response curve. Although no boundaries are defined, this is
often referred as loose-, critical-, and over-coupling. When two tuned circuits are loosely coupled through
mutual inductance, the bandwidth will be narrow. As the amount of mutual inductance increases, the bandwidth continues to grow. When the mutual inductance is increased beyond a critical point, the peak in the
response curve begins to drop, and the centre frequency will be attenuated more strongly than its direct sidebands. This is known as over-coupling.
9.3
MUTUAL INDUCTANCE
Mutual inductance is the ability of one inductor to induce an emf across another inductor placed very close to it.
Suppose two coils are placed near each other, as shown in Fig. 9.3.
→
The first coil has N1 turns and carries a current I1 which gives rise to a magnetic field B1 . Since the two
coils are close to each other, some of the magnetic field lines through the coil 1 will also pass through the
545
Magnetically Coupled Circuits
coil 2. Let ␾21 denote the magnetic flux through one turn of the coil 2
due to I1. Now, by varying I1 with time, there will be an induced emf
associated with the changing magnetic flux in the second coil:
21
= −N2
→
→
d 21
d
= − ∫∫ B1 ⋅ d A2
dt
dt coil2
The time rate of change of magnetic flux ␾21 in the coil 2 is proportional to the time rate of change of the current in the coil 1:
d
dI
N 2 21 = M 21 1
dt
dt
where the proportionality constant M21 is called the mutual inducN
tance. It can also be written as M 21 = 2 21
I1
COil 2
N2
Coil 1
N1
21
I1
B1
Fig. 9.3 Changing current in the coil 1
produces changing magnetic flux in
the coil 2
The SI unit for inductance is henry (H).
1 henry 1 H 1 T-m2/A
The mutual inductance M21 depends only on the geometrical properties of the two coils such as the number
of turns and the radii of the two coils.
In a similar manner, suppose instead there is a current I2 in the
B2
second coil and it is varying with time (Fig. 9.4). Then the induced
emf in the coil 1 becomes
= − N1
12
→
→
d 12
d
= − ∫∫ B2 ⋅ d A1
dt
dt coil1
and a current is induced in the coil 1.
This changing flux in the coil 1 is proportional to the changing
current in the coil 2,
d
dI
N1 12 = M12 2
dt
dt
where the proportionality constant M12 is another mutual inductance
N
and can be written as M12 = 1 12
I2
COil 2
N2
I2
Coil 1
N1
12
Fig. 9.4 Changing current in the coil 2
produces changing magnetic flux in the
coil 1
Using the reciprocity theorem which combines Ampere’s law and the Biot–Savart law, one may show that
the constants are equal:
M12 ⬅ M21 ⬅ M
(9.1)
9.4
MUTUAL INDUCTANCE BETWEEN TWO COUPLED INDUCTORS
Let
L1, L2—two inductors placed very close to each other
v2(t)—open circuit voltage induced in L2 by a current i1(t) in L1
v1(t)—open circuit voltage induced in L1 by a current i2(t) in L2
546
Network Analysis and Synthesis
So, when only i1(t) is flowing, the magnetic flux emerging from L1 is given as
␾1 ␾11 (linkage with L1) ␾12 (linking with L2)
d
d di
di
∴ v1 = N1 1 = N1 1 1 = L1 1
dt
di1 dt
dt
where,
L1 = N1
d 1
di1
v2 = N 2
and
d 12
d
di
di
= N 2 12 1 = M 21 1
dt
di1 dt
dt
d 12
= mutual inductance of the coil L2 with respect to the coil L1
di1
Now, when only i2(t) is flowing, the magnetic flux emerging from L2 is given as ␾2 ␾21 (linkage with L1)
␾22 (linking with L2)
d
d di
di
∴ v2 = N 2 2 = N 2 2 2 = L2 2
dt
di2 dt
dt
d
where,
L2 = N 2 2
di2
d
d
di
di
and
v1 = N1 21 = N1 21 2 = M12 2
dt
di2 dt
dt
d
where, M12 = N1 21 = mututal inductance of the coil L1 with respect to the coil L2
di2
where,
9.5
M 21 = N 2
DOT CONVENTION
Mutual inductance is a positive quantity; but the sign of emf induced by it depends on the direction of winding
of the coils.
In circuit analysis, the dot convention is a convention used to denote the voltage polarity of the mutual
inductance of two components. Two good ways to think about this convention:
1. If a current enters the dotted terminal of one coil then the polarity of the emf induced in the second coil
will be positive at the dotted terminal of the second coil.
2. If a current leaves the dotted terminal of one coil then the polarity of the emf induced in the second coil
will be negative at the dotted terminal of the second coil.
Following these conventions, we find the four possible combinations:
Combination (1)
M
v2(t) = M di1(t )
dt
I1
Fig. 9.5
Combination (2)
M
I1
Fig. 9.6
v2(t) =
M di1(t )
dt
547
Magnetically Coupled Circuits
Combination (3)
M
I1
(t )
M di1
dt
v2(t ) =
Fig. 9.7
Combination (4)
M
I1
v2(t) = M
di1(t )
dt
Fig. 9.8
If we assume the current flowing in both the coils then we have the following combinations:
Combination (1)
M
I2
I1
v1(t)
()
v1 t = L1
v2(t)
( ) + M di (t )
di1 t
2
dt
di2 t
dt
di1 t
dt
dt
()
v2 t = L2
( )+M ( )
Fig. 9.9
Combination (2)
M
v1(t)
I1
I2
()
v1 t = L1
v2(t)
()
v2 t = L2
Fig. 9.10
Combination (3)
M
v1(t)
I1
I2
()
v1 t = L1
v2(t)
()
v2 t = L2
( ) − M di (t )
di1 t
2
dt
di2 t
dt
di1 t
dt
dt
( )−M ( )
( ) − M di (t )
di1 t
2
dt
di2 t
dt
di1 t
dt
dt
( )−M ( )
Fig. 9.11
Combination (4)
M
v1(t)
I1
I2
()
v1 t = L1
v2(t)
()
v2 t = L2
Fig. 9.12
( ) + M di (t )
di1 t
2
dt
di2 t
dt
di1 t
dt
dt
( )+M ( )
Also, for series connection of inductors, as shown:
M
M
i
i
L1
Fig. 9.13
L2
i
i
L1
Fig. 9.14
L2
548
Network Analysis and Synthesis
9.6
DETERMINATION OF COEFFICIENT OF COUPLING FROM ENERGY
CALCULATIONS IN COUPLED CIRCUITS
To find the energy stored in the coupled circuit, we consider two cases:
Case (1) We assume i2
M
0 and let i1 increase from 0 to I1.
v1(t )
( ) ( ) ( ) dt i
and power in L , p ( t ) = 0 ( i = 0 )
∴ power in L1, p1 t = v1 t i1 t = L1
2
2
I2
v2(t )
di1
1
Fig. 9.15 Coupled
circuit
2
t
I1
()
∴energy stored in the circuit, w1 = ∫ p1 t dt = ∫ L1i1
0
0
Case (2) We assume i1
I1
di1 1
= L1 I12
dt 2
0 and let i2 increase from 0 to I2.
()
() ()
()
() ()
di2
i
dt 2
di
and power in L1, p1 t = v1 t i1 t = M12 2 I1
dt
∴ power in L2, p2 t = v2 t i2 t = L2
t
)
(
I
)
(
2
2
1
∴ energy stored in the circuit, w2 = ∫ p1 + p2 dt = ∫ L2 i2 di2 + M12 I1di2 = L2 I 2 2 + M12 I1 I 2
2
t
0
1
From Case (1) and Case (2), the total energy stored in the coupled circuit when both i1 and i2 have reached
constant values of I1 and I2 is
)
(
1
1
W = w1 + w2 = L1 I12 + L2 I 2 2 + M12 I1 I 2
2
2
(9.2)
Now, if we reverse the order in which the currents reach their final values (i.e., first i2 increases from 0 to
I2 with i1 0 and then i1 reaches from 0 to I1 with i2 I2) then the total energy will be
1
1
W = L1 I12 + L2 I 2 2 + M 21 I1 I 2
2
2
From (9.2) and (9.3), we get,
M12 = M 21 = M
1
1
∴ total energy, W = L1 I12 + L2 I 2 2 + MI1 I 2
2
2
and for any instantaneous values,
()
()
()
() ()
1
1
w t = L1i12 t + L2 i2 2 t + Mi1 t i2 t
2
2
If the dotted terminals are in opposite sides then
1
1
W = L1 I12 + L2 I 2 2 − MI1 I 2
2
2
(9.3)
549
Magnetically Coupled Circuits
In general,
1
1
W = L1 I12 + L2 I 2 2 ± MI1 I 2
2
2
(9.4)
To find the limiting value of M Energy stored cannot be negative.
)
(
1
1
1
∴ L1 I12 + L2 I 2 2 − MI1 I 2 ≥ 0 ⇒
L I 2 + L2 I 2 2 − 2 L1 L2 I1 I 2 + L1 L2 I1 I 2 − MI1 I 2 ≥ 0
2
2
2 1 1
2
2
1
1
L1 I1 − L2 I 2 + L1 L2 − M I1 I 2 ≥ 0
⇒
L1 I1 − L2 I 2 + L1 L2 I1 I 2 − MI1 I 2 ≥ 0 ⇒
2
2
) (
(
)
) (
(
)
The squared term is never negative.
∴ L1 L2 − M ≥ 0
⇒
M ≤ L1 L2
(9.5)
Therefore, the maximum possible value of the mutual inductance is the geometric mean of the self-inductances
of the two coils.
Coefficient of coupling The degree to which the mutual inductance approaches its maximum value is
given by the coefficient of coupling (k), defined as
k=
M
(9.6)
L1 L2
So, 0 ≤ k ≤ 1 or, 0 ≤ M ≤ L1 L2
Note
9.7
(i) For k 1, the coils are called perfectly coupled coils.
(ii) For k ≤ 0.5, the coils are called loosely coupled coils.
(iii) For k ≥ 0.5, the coils are called tightly coupled coils.
INDUCTIVE COUPLING
When two coils are connected in series or parallel, mutual inductance exists between them. Depending upon
the type of connection, the voltage equation will be different.
9.7.1 Series Coupling
When two coils of self-inductances L1 and L2 are connected in series, two types of connections are possibles.
Series-aiding connection
In this connection, the two coils are connected in series in such a way that their induced emf’s are of same polarities.
M
M
i
i
L1
L2
Fig. 9.16 Series-aiding connections
i
i
L1
L2
550
Network Analysis and Synthesis
Here, total inductance
(L1
L2
M
2M)
i
Derivation By KVL,
di
di
di
di
v t = L1 + L2 + 2 M = L1 + L2 + 2 M
dt
dt
dt
dt
()
)
(
(
∴ Leq = L1 + L2 + 2 M
i
L1
)
L2
v(t)
Fig. 9.17 Series-aiding connection
Series-opposing connection In this connection, the two coils are connected in series in such a way that
their induced emf ’s are of opposite polarities.
M
M
i
i
i
L1
i
L2
L1
L2
Fig. 9.18 Series-opposing connections
Here, total inductance
(L1
L2 − 2M)
M
9.7.2 Parallel Coupling
When two coils of self-inductances L1 and L2 are connected in parallel,
two types of connections are possible.
L1
Parallel-aiding connection In this connection, the two coils are
connected in parallel in such a way that their induced emf ’s are of
same polarities.
Here, total inductance =
L2
M
L1
L2
Fig. 9.19 Parallel-aiding connections
L1 L2 − M 2
L1 + L2 − 2 M
M
Derivation By KVL,
L1
( ) and M dt + L dt = v (t )
di1
di
+M 2 =v t
dt
dt
di1
di2
i1
i2
In sinusoidally steady state,
j L1 I1 + Mj I 2 = V and j MI1 + j L2 I 2 = V
Solving for I1 and I2, we get
I1 =
V
V
j M
j L2
j L1
j M
j M
j L2
=
( L − M )V
j
2
2
(M − L L )
2
1 2
j L1 V
and
I2 =
j M V
j L1
j M
j M
j L2
=
j
2
(
( L − M )V
1
M 2 − L1 L2
)
L2
L1
2
Fig. 9.20
551
Magnetically Coupled Circuits
)
(
∴ total current, I = I1 + I 2 =
( L + L − 2 M )V
j
1
M
2
(M − L L )
( M − L L ) = j ⎡⎢ L L − M ⎤⎥
2
1 2
∴ input Impedance, Z =
V
=
I j
2
M
2
2
L2
L1
2
1 2
1 2
( L + L − 2M )
1
L2
L1
⎢⎣ L1 + L2 − 2M ⎥⎦
2
Fig. 9.21 Parallel-opposing
connections
L1 L2 − M
L1 + L2 − 2 M
2
Thus, the equivalent inductance is, Leq =
Parallel-opposing connection In this connection, the two coils are connected in parallel in such a way
that their induced emf ’s are of opposite polarities.
L1 L2 − M 2
L1 + L2 + 2 M
It can be derived in the same way as done for parallel-opposing connection.
Here, total inductance =
9.8
LINEAR TRANSFORMER
A transformer is a four-terminal device comprising of two (or more) magnetically coupled coils. It is composed
of two coils:
R
R
M
1
• a primary coil of resistance R1 and self-inductance L1
• a secondary coil of resistance R2 and self-inductance L2
L1
V1
A transformer is said to be linear if the coils are wound on a magnetically linear material for which the magnetic permeability is a constant.
Some linear materials are air, plastic, Bakelite and wood.
Circuit representation of a linear transformer is shown in Fig. 9.22.
I1
2
L2
ZL
I2
Fig. 9.22 Circuit representation of
a linear transformer
Calculation of input and reflected impedances By KVL for the two meshes,
)
0 = − j MI + ( R + j L + Z ) I
(
V1 = R1 + j L1 I1 − j MI 2
1
From (9.8), I 2 =
2
2
L
(9.7)
2
(9.8)
j MI1
R2 + j L2 + Z L
Putting this value in (9.7),
)
(
V1 = R1 + j L1 I1 −
2
j M × j MI1
M 2 I1
= R1 + j L1 I1 +
R2 + j L2 + Z L
R2 + j L2 + Z L
(
)
2
V
M2
∴ input impedance, Z in = 1 = R1 + j L1 +
I1
R2 + j L2 + Z L
(
(
)
)
Here, R1 + j L1 = Impedance of Primary Winding
(9.9)
552
Network Analysis and Synthesis
2
ZR =
and,
M2
R2 + j L2 + Z L
(9.10)
where,
Z R = Impedance due to coupling between primary and secondary, knwon as reflected impedance.
Note
The input impedance and reflected impedance value do not change with the position of dots on the winding,
as the same result is obtained by replacing M by −M.
9.9
DETERMINATION OF EQUIVALENT T AND CIRCUIT OF LINEAR
TRANSFORMER (CONDUCTIVELY EQUIVALENT CIRCUIT OF A
MAGNETICALLY COUPLED CIRCUIT)
A linear transformer can be replaced by an equivalent T or ␲ network.
A linear transformer with a source in the primary and a load in the secondary
is shown in Fig. 9.22. If we separate the resistances from the transformer, there
remains only a pair of mutually coupled inductors, as shown in Fig. 9.23.
By KVL for the two meshes,
di
di
di
di
v1 = L1 1 + M 2 and v2 = M 1 + L2 2
dt
dt
dt
dt
)
(
(
V1 = jω L1 I1 + jω MI 2 = jω L1 − M I1 + jω M I1 + I 2
or,
)
(
)
M
L1
V1
L2
I1
V2
I2
Fig. 9.23 Circuit representation
of a linear transformer without
resistances
)
(
and V2 = jω MI1 + jω L2 I 2 = jω M I 2 + I1 + jω L2 − M I 2
Equivalent T Circuit The above two equations can be written as
(L − M )I + j M (I + I )
V = j MI + j L I = j M ( I + I ) + j ( L − M ) I
V1 = j L1 I1 + j MI 2 = j
and
2
1
1
2 2
1
2
1
1
I1 L M
1
2
2
2
Therefore, the equivalent T network for the linear transformer is shown in Fig. 9.24.
Note that if the dots of any one of the windings are placed in the opposite end of the
coil, the mutual term becomes negative and the equivalent circuit can be obtained by
replacing M by −M. In that case, the three inductances are L1 M, − M, and L2 M.
Equivalent
Circuit Using the concept of T–␲ conversion or, star–delta
conversion, we get the equivalent ␲ circuit of a linear transformer as follows.
The three inductances of the equivalent
LA =
2
1
2
Similarly,
L L −M
LB = 1 2
M
Fig. 9.24 Equivalent
T network of a linear
transformer
I1
I2
LB
L L −M
and LC = 1 2
L1 − M
2
v1 LA
LC
v2
1 2
2
2
2
M
circuit are
( L − M ) M + M ( L − M ) + ( L − M )( L − M ) = L L − M
L −M
(L − M )
1
L2 M I 2
2
Fig. 9.25 Equivalent
network of a linear
transformer
553
Magnetically Coupled Circuits
Here also, if any one dot changes its location on the winding, the sign of M will change and in that case,
the three inductances will be
LA =
9.10
L1 L2 − M 2
,
L2 + M
L L −M2
L L −M2
LB = − 1 2
and LC = 1 2
M
L1 + M
IDEAL TRANSFORMER
A transformer is said to be ideal if it has the following properties:
1. Primary and secondary coils are lossless (i.e., R1 R2 0).
2. Primary and secondary coils have very large reactances compared to any connected impedance
(i.e., L1, L2, M → ∞)
3. Coupling between primary and secondary coils is perfect, i.e., k 1 or the leakage flux is zero.
An ideal transformer is a useful approximation of a very tightly coupled transformer (k ≈1) in which both
the primary and secondary inductive reactances are extremely large compared to the load impedance.
9.10.1 Calculation of Input Impedance for Ideal Transformer
I1
The circuit symbol of an ideal transformer is shown in Fig. 9.26.
By KVL,
V1 L1
V1 = j L1 I1 − j MI 2
0 = − j MI1 + j L2 + Z L I 2
From (9.12), I 2 =
(9.12)
j M
I
j L2 + Z L 1
⎛−
=⎜
⎝
2
L1 L2 + j L1 Z L +
j L2 + Z L
2
⎛−
j M
I1 = ⎜
j L2 + Z L
⎝
2
L1 L2 + j L1 Z L +
j L2 + Z L
L1 L2 ⎞
⎡
⎤
⎟ I1 ⎣ k = 1, ∴ M = L1 L2 ⎦
⎠
⎛ j L1 Z L ⎞
=⎜
⎟ I1
⎝ j L2 + Z L ⎠
∴ input impedance
V
j L1 Z L
j L1 Z L
⎡ L >> Z L ; for idealtransformer ⎤⎦
Z in = 1 =
≈
I1 j L2 + Z L
j L2 ⎣ 2
⎛L ⎞
⎛N ⎞
= ZL ⎜ 1 ⎟ = ZL ⎜ 1 ⎟
⎝ L2 ⎠
⎝ N2 ⎠
2
⎡
⎣
L ∝ N 2 ⎤⎦
L2 V
2
Fig. 9.26 Circuit
symbol of an ideal
transformer
Putting this in (9.11), we get
V1 = j L1 I1 − j MI 2 = j L1 I1 − j M
I2
(9.11)
)
(
M
2
M2⎞
⎟ I1
⎠
554
Network Analysis and Synthesis
2
⇒
⎛N ⎞
Z
Z in = Z L ⎜ 1 ⎟ = 2L
n
⎝ N2 ⎠
(9.13)
N2
is the turns ratio. Thus, the load impedance is approximately transferred as the square of
N1
turns ratio. This input impedance is also known as the reflected impedance as the load impedance is reflected
to the primary side.
This property of an ideal transformer to transform a given impedance into another impedance is used in
impedance matching, which is very useful in different applications involving maximum power transfer.
where,
n=
9.10.2 Calculation of Voltage and Current Transformation Ratio for Ideal Transformer
I1 =
From (9.12),
j L2 + Z L
I2
j M
Putting this in (9.11), we get,
⎛−
⎛ j L2 + Z L ⎞
V1 = j L1 I1 − j MI 2 = j L1 ⎜
I 2 − j MI 2 = ⎜
⎟
⎝ j M ⎠
⎝
⎛−
=⎜
⎝
2
L1 L2 + j L1 Z L +
j M
2
L1 L2 ⎞
⎟ I2
⎠
2
L1 L2 + j L1 Z L +
j M
2
M2⎞
⎟ I2
⎠
⎡ k = 1, ∴ M = L L ⎤
1 2 ⎦
⎣
⎛ L ⎞
⎛L ⎞
1
= ZL ⎜ 1 ⎟ I2 = ZL ⎜
⎟I
⎜⎝ L L ⎟⎠ 2
⎝M⎠
1 2
V1 = I 2 Z L
L
L1
= V2 1
L2
L2
voltage transformation ratio,
V2
L
N
= 2 = 2 =n
V1
L1 N1
(9.14)
where, n is the turns ratio. Depending upon the value of the turns ratio, three types of transformers are obtained.
Case (I): n > 1 In this case, the secondary voltage is greater than the primary voltage and the transformer
is termed step-up transformer.
Case (II): n < 1 In this case, the secondary voltage is less than the primary voltage and the transformer is
termed step-down transformer.
Case (III): n 1 In this case, the secondary voltage is equal to the primary voltage and the transformer is
termed isolation transformer.
Also,
I1 =
=
j L2 + Z L
j L2
I2 ≈
I ⎡ L >> Z L ; for idealtrransformer ⎤⎦
j M
j M 2 ⎣ 2
L2
L1 L2
I2 =
L2
I
L1 2
555
Magnetically Coupled Circuits
I2
L
N 1
= 1 = 1=
I1
L2 N 2 n
∴ current transformation ratio,
(9.15)
where n is the turns ratio. Thus, the ratio of the primary current to the secondary current is the turns ratio. It
must be noted that if any one dot changes its location on the winding, the current ratio wi