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Network Analysis
and Synthesis
S. K. Bhattacharya
Director (Academics)
Shaheed Udham Singh Engineering College
Mohali, Punjab
India
Manpreet Singh
Assistant Professor
Department of Electrical Engineering
BBBS Engineering College
Fatehgarh Sahib, Punjab
India
Delhi • Chennai
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Copyright © 2015 Pearson India Education Services Pvt. Ltd
Published by Pearson India Education Services Pvt. Ltd, CIN: U72200TN2005PTC057128,
formerly known as TutorVista Global Pvt. Ltd, licensee of Pearson Education in South Asia.
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publisher’s prior written consent.
This eBook may or may not include all assets that were part of the print version. The publisher reserves the right to remove any material in this eBook at any time.
ISBN 978-93-325-4285-3
eISBN 978-93-325-4726-1
Head Office: A-8 (A), 7th Floor, Knowledge Boulevard, Sector 62, Noida 201 309, Uttar
Pradesh, India.
Registered Office: Module G4, Ground Floor, Elnet Software City, TS-140, Block 2 & 9,
Rajiv Gandhi Salai, Taramani, Chennai 600 113, Tamil Nadu, India.
Fax: 080-30461003, Phone: 080-30461060
www.pearson.co.in, Email: companysecretary.india@pearson.com
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To my mother Late Smti. Shaila Bala Bhattacharya
—S. K. Bhattacharya
To my mother Smti. Paramjeet Kaur
—Manpreet Singh
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Brief Contents
Preface
About the Authors
1. Basic Concepts
xvii
xxi
1
2. Kirchhoff’s Laws, Mesh and Nodal Analysis
26
3. Steady State Analysis of AC Circuits
90
4. R–L–C Circuits and Resonance
147
5. Network Theorems and Applications
172
6.Transient Response of Circuits Using
Differential Equations
226
7. Laplace Transform
271
8. Transient Response of Circuits Using
Laplace Transform
320
9. Three-Phase Systems and Circuits
358
10. Network Functions − s-Domain Analysis
of Circuits
408
11. Two-port Network Parameters
460
12. Network Synthesis and Realisability
499
13. Filters and Attenuators
585
Index
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Contents
Preface
xvii
About the Authors
xxi
1. Basic Concepts
1.1
1.2
1.3
1.4
1.5
1
Introduction 1
Voltage, Current and Resistance 2
Ohm’s Law 2
Electrical Power and Energy 3
Series and Parallel Connections of Resistors 3
1.5.1
1.5.2
1.5.3
1.5.4
Series Connection of Resistors 3
Parallel Connection of Resistors 4
Series–Parallel Circuits 5
Ladder Network 5
1.6 Basic Circuit Elements
9
1.6.1 Resistors 9
1.6.2Inductors—Self-Inductance and Mutual Inductance
1.6.3 Capacitors 14
10
1.7 Inductors and Capacitors in DC Circuits 17
1.8 DC Network Terminologies and Circuit Fundamentals 17
1.8.1 Network Terminologies 18
1.8.2 Voltage and Current Sources
1.8.3 Source Transformation 21
Review Questions
19
23
2.Kirchhoff’s Laws, Mesh and Nodal Analysis
2.1 Kirchhoff’s Laws
26
26
2.1.1 Kirchhoff’s Current Law 26
2.1.2 Kirchhoff’s Voltage Law 29
2.2
2.3
2.4
2.5
2.6
Mesh Analysis 30
Nodal Analysis 43
Super Nodal Analysis 53
Super Mesh Analysis 54
Methods of Solving Complex Network Problems
2.6.1 Numerical Problems Based on Kirchhoff’s Laws
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Contents
2.6.2Numerical Problems Based on Mesh
and Nodal Analysis 60
Review Questions 82
Multiple Choice Questions
Answers 89
86
3.Steady State Analysis of AC Circuits
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
90
AC Voltage Applied Across a Resistor 90
AC Voltage Applied Across an Inductor 92
AC Voltage Applied Across a Capacitor 95
R–L Series Circuit 99
Apparent Power, Real Power and Reactive Power 101
Power in R–L Series Circuit 101
Power Triangle of R–L Series Circuit 102
R–C Series Circuit 103
3.8.1 Power and Power Triangle of R–C Series Circuit
105
3.9 R–L–C Series Circuit 105
3.10 AC Parallel Circuits 108
3.10.1Phasor or Vector Method of Solving
Circuit Problems 108
3.10.2 Admittance Method of Solving Circuit Problems 110
3.10.3 Use of Phasor Algebra in Solving Circuit Problem 115
3.11 AC Series–Parallel Circuits 133
Review Questions 139
Multiple Choice Questions 143
Answers 146
4. R–L–C Circuits and Resonance
147
4.1
R–L–C Series Circuit with Variable Frequency
Input Voltage 147
4.2 Series Resonance 148
4.2.1Effect of Variation of Frequency on Current and Voltage
Drops 149
4.2.2Effect of Variation of Frequency on Impedance
and Power Factor 150
4.3Applications of R–L–C Circuits 151
4.3.1 Band-pass Filter 151
4.3.2 Band-stop Filter 154
4.4 Parallel Resonance 155
4.4.1 Ideal Tank Circuit
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4.4.2 Non-ideal Tank Circuit 156
4.4.3 Resonant Frequency 156
4.5 Parallel Resonant Filters
156
4.5.1 Band-pass Filter 157
4.5.2 Band-stop Filter 157
4.6 Applications of Resonant Circuits
4.6.1
4.6.2
4.6.3
4.6.4
157
Tuned Amplifier 157
Input to Receiver from an Antenna
Other Applications 158
Locus Diagram 167
Review Questions
158
169
5. Network Theorems and Applications
172
5.1 Introduction 172
5.2 Superposition Theorem 173
5.3 Thevenin’s Theorem 174
5.3.1 Procedure for Applying Thevenin’s Theorem
175
5.4 Norton’s Theorem 179
5.5 Millman’s Theorem 181
5.6 Maximum Power Transfer Theorem 182
5.7Maximum Power Transfer Theorem
for Complex Impedance Circuits 185
5.8 Reciprocity Theorem 186
5.9 Tellegen’s Theorem 188
5.10 Compensation Theorem 189
5.11 Star−Delta Transformation 190
5.11.1 Transforming Relations from Delta to Star
5.11.2 Transforming Relations from Star to Delta
190
191
5.12 Numericals on Network Theorems 195
Review Questions 223
6.Transient Response of Circuits
Using Differential Equations
226
6.1 Transient Condition in Networks 226
6.2Transient Response of R–L Series
Circuits Having DC Excitation 227
6.2.1Rise of Current Through R–L Series Circuit 227
6.2.2 Time Constant of R–L Series Circuit 230
6.2.3 Decay of Current Through R–L Series Circuit 230
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Contents
6.3Transient Response in R–C Series
Circuits Having DC Excitation 239
6.3.1 Case I: Capacitor is Getting Charged 239
6.3.2 Case II: Discharging of Capacitor 243
6.4Transient Response of R–L–C Series
Circuits Having DC Excitation 249
6.5 Sinusoidal Response of R–L Circuits 253
6.6 Sinusoidal Response of R–C Circuits 258
6.7 Sinusoidal Response of R–L–C Circuits 262
Review Questions 268
7.Laplace Transform
271
7.1 Concept of Laplace Transform 271
7.2 Laplace Transform of Standard Functions 272
7.3Laplace Transform Problems Based
on Standard Formula 280
7.4 Properties of Laplace Transform 286
7.4.1
7.4.2
7.4.3
7.4.4
Property 1: First Shifting Property 286
Property 2: Multiplication by t n 287
Property 3: Division by ‘t’ 288
Property 4 290
7.5 Summary of Useful Properties of Laplace Transform 291
7.6 Initial Value Theorem 292
7.7 Final Value Theorem 295
7.8 Inverse Laplace Transform 299
7.9 Convolution Theorem 310
Review Questions 315
Multiple Choice Questions 317
Answers 319
8.Transient Response of Circuits
Using Laplace Transform
8.1Steps to Find Transient Response
Using Laplace Transform 320
8.2 Circuit Elements in the s-Domain
320
321
8.2.1 Resistor in the s-Domain 321
8.2.2 Inductor in s-Domain 321
8.2.3 Capacitor in s-Domain 321
8.3 DC Response of R–C Series Circuit
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8.4
8.5
8.6
8.7
DC Response of R–L Series Circuit 324
DC Response of an R–L–C Series Circuit 326
Sinusoidal Response of R–L Series Circuit 329
Sinusoidal Response of R–C Series Circuit 333
Review Questions 355
9.Three-Phase Systems and Circuits
358
9.1 Introduction 358
9.2 Advantages of Three-Phase Systems 359
9.3 Generation of Three-Phase Voltages 360
9.3.1 Equation of Three-phase Voltages 360
9.3.2 Balanced Three-phase System 362
9.4 Terms Used in Three-Phase Systems and Circuits 363
9.5 Three-Phase Winding Connections 363
9.5.1 Star Connection 363
9.5.2 Delta Connection 364
9.5.3Relationship of Line and Phase Voltages
and Currents in a Star-connected System 365
9.5.4Relationship of Line and Phase Voltages
and Currents in a Delta-connected System 366
9.6 Active and Reactive Power 368
9.7Comparison Between Star Connection
and Delta Connection 369
9.8 Measurment of Power in Three-Phase Circuits 376
9.8.1One-wattmeter Method 376
9.8.2 Two-wattmeter Method 377
9.8.3 Three-wattmeter Method 379
9.8.4 Star to delta and Delta to Star Transformation
9.9More Numericals Basesd on Three-Phase
Balanced Load 387
9.10 Method of Solving Problems on Unbalanced Load
Review Questions 401
Multiple Choice Questions 405
Answers 407
10. Network Functions − s-Domain Analysis
of Circuits
10.1 Introduction
397
408
408
10.1.1 Terminals and Ports 408
10.1.2 Concept of Complex Frequency
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Contents
10.2 Transformed Impedances in s-Domain 411
10.2.1 Resistance 411
10.2.2 Inductance 412
10.2.3 Capacitance 412
10.3 One-Port Network 413
10.3.1 Driving Point Impedance and Admittance Functions
413
10.4 Two-Port Network 420
10.4.1 Network Functions of a Two-port Network
421
10.5 Transfer Function 422
10.6 Network Function in Generalised Form 423
10.7 Poles and Zeros of Network Functions 424
10.7.1 Poles of a Network Function 424
10.7.2 Zeros of a Network Function 424
10.8 Pole–Zero Diagram 424
10.9 Time-Domain Response from Pole–Zero Plot 427
10.10 More Examples on Network Function 437
10.11Poles and Zeros of Network Functions
and Their Significance 449
10.12 Stability Criterion for an Active Network 450
10.13 Examples Based on Pole–Zero Plot 452
Review Questions 458
11.Two-port Network Parameters
460
11.1 Introduction 460
11.2 Two-port Network Parameters 461
11.2.1Open-circuit Impedance-parameters or
Z-parameters 461
11.2.2 Short-circuit Admittance Parameters 462
11.2.3Relationship Between Impedance
and Admittance Matrix 464
11.2.4 Hybrid or h-parameters 466
11.2.5 Inverse Hybrid or g-parameters 467
11.2.6 Transmission Parameters 468
11.2.7 Inverse Transmission Parameters 469
11.3 Correlation of Two-Port Network Parameters 477
11.3.1 Conversion of Y-parameters to Z-parameters 477
11.3.2Conversion of A, B, C and D or t-parameters
to h-parameters 478
11.3.3 Conversion of h-parameters to Y-parameters 479
11.4 Two-Port Reciprocal and Symmetrical Networks 479
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11.4.1 Reciprocal Two-port Network 479
11.4.2 Symmetrical Two-port Networks 480
11.5Terminated Two-Port Network 480
11.6 Interconnected Two-Port Network 482
11.7
T-Circuit Representation of Two-Port Network 483
11.8
p -Circuit Representation of Two-Port Network 484
11.9 Image Impedance 484
11.10 More Solved Numericals 485
Review Questions 495
12. Network Synthesis and Realisability
499
12.1 Introduction 499
12.2 Hurwitz Conditions for Stability 500
12.3 Properties of Positive Real Functions 506
12.4Synthesis of Networks by Foster’s
and Cauer’s Methods 513
12.5 Foster and Cauer Forms 513
12.5.1 Synthesis of R–C Network 513
12.5.2Properties of the R–C Impedance
or R–L Admittance Function 513
12.5.3 Foster Form-I of R–C Network 513
12.5.4 Foster Form-II of R–C Network 514
12.5.5 Cauer Forms of R–C Network 515
12.5.6Synthesis of R–L Network 515
12.5.7Properties of R–L Impedance
Function/R–C Admittance Function 515
12.5.8 Foster Form-I of R–L Network 516
12.5.9 Foster Form-II of R–L Network 516
12.5.10 Cauer Form-I of R–L Network 517
12.5.11 Cauer Form-II R–L Network 517
12.5.12 Synthesis of L–C Networks 517
12.5.13 Properties of L–C Immittance 517
12.5.14 Foster Form-I of L–C Network 517
12.5.15 Foster Form-II of L–C Network 518
12.5.16 Cauer Form-I of L–C Network 518
12.5.17 Cauer Form-II of L–C Network 519
12.6More Numericals on Synthesis of L–C Network 548
Review Questions 576
Multiple Choice Questions 582
Answers 584
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Contents
13. Filters and Attenuators
13.1 Introduction
585
585
13.1.1 Measurement in Decibels
13.2 Types of Filters 587
13.3 Classification of Passive Filters
13.3.1
13.3.2
13.3.3
13.3.4
587
588
Low-Pass Filters 588
High-Pass Filters 588
Band-Pass Filters 588
Band-Stop or Band-Elimination Filter
589
13.4 Parameters of a Filter 589
13.4.1
13.4.2
13.4.3
13.4.4
Propagation Constant (g ) 590
Attenuation Constant 590
Phase Shift Constant (b ) 591
Characteristic Impedance (Z0 ) 591
13.5 Filter Networks 591
13.5.1 Formation of Symmetrical T-Network 591
13.5.2 Formation of Symmetrical p-Network 592
13.5.3 Ladder Network 592
13.6 Analysis of Filter Networks
593
13.6.1 Symmetrical T-Network 593
13.6.2 Analysis of p-Network 601
13.6.3 Summary of Parameters of Filter Network
13.7 Classification of Filters 604
13.8 Constant K-Type or Prototype Filters
13.8.1
13.8.2
13.8.3
13.8.4
13.8.5
13.8.6
13.8.7
604
605
Constant K-type Low-Pass Filters (Lpf) 605
Constant K-type High-Pass-Filters (HPF) 615
Comparison of Constant K-Type LPF and HPF 619
Constant K-type Band-Pass Filter 624
Constant K-type Band-Stop/Band-Elimination Filter 637
Comparison of Constant K-type Filters 645
Limitations of Constant K-type Filters 649
13.9 m-Derived Filters 649
13.9.1
13.9.2
13.9.3
13.9.4
13.9.5
13.9.6
13.9.7
13.9.8
13.9.9
m-Derived T-section 650
m-Derived p-section 651
m-Derived Low-Pass Filter 653
Summary of m-Derived Low-Pass Filter 657
m-Derived High-Pass Filter 659
Summary of m-Derived HPF 662
Comparison of m-Derived LPF and HPF 663
m-Derived Band-Pass Filter 665
m-Derived Band-Stop Filter 667
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13.10 Composite Filters
668
13.10.1 Composite Low-Pass Filter 669
13.10.2 Composite High-Pass Filter 670
13.11 Additional Solved Numericals on Filters
671
13.11.1 Problems on m-Derived Low-pass Filters 671
13.11.2 Problems on m-Derived High-pass Filters 674
13.11.3 Problems on Composite Filters 675
13.12 Attenuators
13.12.1
13.12.2
13.12.3
13.12.4
13.12.5
681
Introduction 681
T-type Attenuator 682
p-type Attenuator 685
Lattice Attenuator 688
Bridged T-type Attenuator
691
13.13More Solved Problems on Filters and Attenuators
Review Questions 702
Multiple Choice Questions 705
Answers 707
Index
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Preface
It is with great pleasure we present you with this comprehensive book, Network Analysis and
Synthesis, which has been developed over a period of time. The content of this book has been
decided on the basis of analysis of the syllabus prescribed by all the leading Indian universities
on the subject of ‘Network Analysis and Synthesis’ or ‘Circuits and Networks’. Although, a
range of books are available on this subject, this book is expected to satisfy the needs of students
of different learning abilities. Presentations of topics have been made so simple that even an
average student will be able to follow this book almost independently.
For each chapter, the objectives have been well-stated so as to guide students in selfevaluation after studying each chapter. After studying the entire book, it is expected that
students will be able to analyse, design, and synthesise electrical circuits and networks.
Although, the students might be familiar with some basic concepts and principles dealt with
in this book, a review of understanding of these basic concepts will facilitate understanding of
the advanced topics. We believe that basic knowledge of differential and integral calculus is
essential to understand this book.
Pedagogy
1. Defining the Learning Objectives Course Content provides simply an outline of the topics. It is passive and does not convey specifically what is to be learnt. Once the objectives
are stated, the students know what to learn and the teachers also know how to teach and
how to evaluate learning.
In this book, learning objectives have been stated to help both students and teachers.
2. Delivery of Concepts and Principles All the chapters have been defined by some key
features: simple language, examples to rule, known to unknown, simple to complex and
concrete to abstract. These presentations have been made in an interactive way.
3. Developing Higher Order Intellectual Abilities Ability to recall and reproduce is the
lowest level of intellectual attainment. Emphasis has been given to develop in students the
higher order intellectual abilities including abilities of application, analysis and synthesis
which together may be defined as problem solving ability.
4. Creating Ample Opportunities for Practice A large number of solved examples in each
chapter demonstrate the many ways to solve the problems. This is followed by review
questions which is composed of ample number of exercise problems with answers so that
students can practice and gain confidence. Exercises have been graded from simple to
complex to make learning motivating.
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Preface
5. Self Feedback in Achieving the Learning Objectives Self feedback provides students the
opportunity to evaluate their learning abilities. For this, a large number of short answer
type and multiple choice type review questions have been provided in each chapter.
Contents and Coverage
This book is divided into 13 chapters. The content and coverage of chapters are as follows:
Chapter 1 reviews the basic concepts related to electrical circuits, series parallel connections, functions of circuit elements, property of storing energy by inductors and capacitors,
voltage and current sources and source transformation.
Chapter 2 explains applications of Kirchhoff’s current and voltage laws in mesh and nodal
analysis of circuits.
Chapter 3 presents steady state analysis of R–L, R–C and R–L–C series and parallel circuits
with plenty of solved numericals.
Chapter 4 establishes the condition for series and parallel resonance. The chapter also deals
with the significance of bandwidth and quality factor. Also explains how series and parallel
resonant circuits are used in the field of electronics.
Chapter 5 defines and explains the network theorems like Superposition theorem, Thevenin’s
theorem, Millman’s theorem, Maximum Power transfer theorem, etc. The chapter solves a large
number of complex network problems using network theorems. Also shows methods of simplifying networks using star-delta transformation technique.
Chapter 6 explains the transient condition that may occur in electrical networks. The chapter
derives expressions for current and voltage under transient condition in R–L, R–C and R–L–C
series circuits. It also carries out transient analysis of R–L, R–C, and R–L–C series circuits with
sinusoidal inputs.
Chapter 7 provides explanation of Laplace transform and its application in solving circuit
problems.
Chapter 8 lists the steps to find transient response of electrical networks using Laplace transform. Determines transient response of electrical circuits by Laplace transform method.
Chapter 9 explains three-phase balanced and unbalanced systems. The chapter shows how to
calculate power and power factor of balanced three-phase loads. It solves numerical problems
related to balanced and unbalanced star and delta connected loads.
Chapter 10 explains the concept of complex frequency. The chapter shows how to convert
circuit parameters from time domain to s-domain. The chapter explains the method of finding transfer function of electrical networks. It also explains the concept of zeros and poles
of a transfer function. Shows how to write the characteristic equation and then apply Routh’s
­stability criterion.
Chapter 11 defines the two-port network parameters and shows how to represent them in the
form of matrix equation. Calculates Z-parameters and Y-parameters. Defines A, B, C, D transmission parameters. The chapter also shows how to calculate A, B, C, D parameters of a given
networks. Also calculates Z-parameters and Y-parameters of interconnected two-port networks.
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Preface
xix
Chapter 12 explains the concept of network Synthesis. The chapter shows how to synthesize
networks by Foster and Cauer methods.
Chapter 13 explains the basic function of a filter circuit. Draws and explains basic filter
networks in different sections. The chapter shows method of analysis of K-type and m-derived
filters. Composite filters using constant K-type and m-derived filters have also been discussed
in this chapter.
Acknowledgements
This work is the outcome of years of experience in teaching ‘Network Analysis and Synthesis’
and other related subjects to the students of electrical and electronics engineering. We would
like to thank all those who provided feedback in the form of their learning difficulties when we
were teaching the subject. Thanks are also due to the faculty of various engineering colleges/
universities, for reviewing the manuscript and giving valuable suggestions.
We appreciate the excellent editorial work done by the team members of Pearson Education.
Lastly, we would like to convey our special thanks to Mrs Sumita Bhattacharya and
Mrs Shivdeep Kaur for their patience and encouragement which helped us to complete this
big task.
S. K. Bhattacharya
Manpreet Singh
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About the Authors
S. K. Bhattacharya is director (academics) of Shaheed Udham
Singh Engineering College, Mohali. He was the principal of
Technical Teachers’ Training Institute, Chandigarh. He also served
as the director of National Institute Technical Teachers Training and
Research, Kolkata and director of Hindustan Institute of Technology,
Greater Noida.
Bhattacharya is a Ph.D. from Birla Institute of Technology and
Science, Pilani. He has undergone practical training in an electricity board and a transformer factory in the Netherlands. He was also
a Senior Fellow of Ministry of Human Resource Development,
Government of India besides being a Fellow of the Institution of
Engineers and Fellow of the Institution of Electronics and Telecommunication Engineers,
India. He has written a number of popular textbooks and published over a hundred technical
papers.
Manpreet Singh is presently working as assistant professor of electrical engineering in BBBS Engineering College, Fatehgarh Sahib,
Punjab. Earlier he worked as assistant professor in SUS College of
Engineering and Technology, Tangori, Mohali. He is a M.Tech. from
Guru Nanak Dev Engineering College, Ludhiana.
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Basic Concepts
1
Chapter objectives
After carefully studying this chapter, you should be able to do the following:
Explain the concept of voltage, curDetermine the factors on which resisrent, and resistance.
tance, inductance, and capacitance
depend.
State and explain Ohm’s law.
Explain how an inductor and a capaciDistinguish between electrical power
tor can work as energy-storing devices.
and energy.
Explain the role of magnetic core
Distinguish between series and parallel
material in an inductor.
connections of resistances.
Explain the role of dielectric material
Calculate branch currents and potential
in a capacitor.
drops in parallel and series–­
parallel
circuits.
Distinguish between series and parallel connections of resistors and
Reduce a ladder network into a single
capacitors.
equivalent resistance.
Be conversant with the network
Explain the basic function of circuit
terminologies.
elements like resistance, inductance,
and capacitance.
Distinguish between a voltage source
and a current source.
Distinguish between self-inductance
and mutual inductance and establish
Convert a voltage source into a current
their relationship.
source and vice-versa.
1.1 INTRODUCTION
In this chapter, we shall review the basic concepts related to electrical networks. The students
must have studied these concepts in their earlier classes. However, a brief review will help them
to understand the contents of chapters that would follow. We will review the basic concepts of
voltage, current, resistance, Ohm’s law, electrical power and electrical energy, series and parallel connections of resistors and division of currents and voltages, and circuit parameters such as
resistor, inductor and capacitor.
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Network Analysis and Synthesis
1.2 VOLTAGE, CURRENT AND RESISTANCE
When two oppositely charged bodies are connected by a wire, flow of electrons occurs from one
body to the other. This movement of electrons constitutes a flow of electric current. The wire
that allows the flow of electrons is called a conductor.
Electromotive force is an electric pressure that causes current to flow through a conducting wire. A conducting wire of large cross-sectional area allows more current to flow than
a conductor with small cross-sectional area. Therefore, a conductor with a large crosssectional area offers less resistance to flow of current than a conductor with a small crosssectional area. Therefore, we can say that resistance is the opposition to the flow of current
through a conductor.
Voltage or Electro motive force (EMF) or potential difference is an electric pressure that
causes the flow of current, that is, the movement of electrons through a conductor. The opposition offered by the conductor to the flow of current is called its resistance.
The direction of electron flow is from the negatively charged body to the positively charged
body. However, current flow is considered from the positive to the negative terminal.
The current that flows in one direction is called direct current (Dc). In alternating current
(ac), the flow of charge reverses directions alternately every fraction of a second (every halfcycle). The cycle repeats over and over again.
1.3 OHM’S LAW
In an electric circuit, current, voltage and resistance are related by an important law, called the
Ohm’s law. Ohm’s law is expressed as follows:
Voltage
Resistance
V
I = (1.1)
R
Current =
or
where I is in amperes, V is in volts and R is in ohms.
Ohm’s law states that the current flowing through a resistance is directly proportional to the
potential difference between its ends and inversely proportional to the value of the resistance,
provided the temperature remains constant.
By applying Ohm’s law, the current flowing through a resistor can be calculated by measuring the voltage drop across it, provided the value of the resistance is known.
Conductance (G) is the inverse of resistance (R). In terms of conductance, Ohm’s law can be
represented as follows:
I = VG
where
G=
1
R
The unit of G is siemens and that of R is ohms.
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Basic Concepts 3
1.4 ELECTRICAL POWER AND ENERGY
When electric current flows through a conducting material, work is done in moving the electrons. Further, heat is dissipated as a result of the work done.
Work is also done when electricity is converted into light, sound and heat as in an electric
lamp, a speaker and an electric heater, respectively.
Power (P) is defined as the time rate of doing work (W ). The power supplied to any electrical
equipment is as follows:
P = V I watts
By Ohm’s law,
V = IR
Therefore,
P = V I = IRI = I2 R watts
(1.2)
V V2
=
watts (1.3)
R
R
Energy consumed is expressed as watt-hour or kilowatt-hour (kWh). Energy consumed or work
done is as follows:
Energy = Power × Time
W=P×t
Wh or kWh;
The unit of work is joule. If unit of power is watt and unit of time is second,
1 watt-second = 1 joule;
or, joules/second = watt
(1.4)
Calorie is the unit of energy
1 calorie = 4.2 joules
We can convert electrical energy in kWh into calories as
P =V I =V
Further,
1 kWh = 1 × 1000 × 60 × 60 watt-second or joules
1 kWh =
36 × 105
= 860 × 103 calories
4.2
Thus, we have explained the relationship between work, power, and energy.
1.5 SERIES AND PARALLEL CONNECTIONS OF RESISTORS
A number of resistors need to be connected in series, parallel, or series-parallel in practical
circuits. It is often required to determine their equivalent resistance and also calculate current
and voltage distribution in these resistances. The steps are explained in the following sections.
1.5.1 Series Connection of Resistors
When a number of resistors are connected in series, there is only one path through which the current can flow. The magnitude of current is the same in all parts or components of a series circuit.
When a number of resistances are connected in series, their equivalent resistance is given as
follows:
Req = R1 + R2 + R3 + … + Rn(1.5)
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4
Network Analysis and Synthesis
I
V1
V
Series-connected resistors can be used as a voltage
divider as shown in Figure 1.1.
As in Figure 1.1,
R1
I
I
V2
R2
I=
Figure 1.1 Voltage Divider Circuit
V
R1 + R2
By applying Ohm’s law, we get the following:
V1 = IR1
V1 =
or,
V
× R1
R1 + R 2
and V 2 = IR 2 =
V × R2
R1 + R 2
Voltage Divider Rule in Series Circuits
From the above, the voltage divider rule is expressed as follows:
V1 = V
R1
(1.6)
R1 + R2
V2 = V
R2
(1.7)
R1 + R2
Similarly,
1.5.2 Parallel Connection of Resistors
Resistors are said to be connected in parallel when the same voltage is applied across each resistor. However, the current through each parallel branch will be different.
When a number of resistances are connected in parallel, their equivalent resistance is derived
as the following form:
R eq =
R1R 2 R 3 R n
1
(1.8)
=
1
1
1
1
R1 + R 2 + R 3 + + R n
+
+
++
R1 R 2 R 3
Rn
when two resistances, R1 and R2 are connected in series, Req = R1 + R2 and when connected in
parallel, Req =
RR
1
= 1 2 .
1
1
R1 + R2
+
R1 R2
Current Divider Rule in Parallel Circuit
Figure 1.2 shows two resistors, R1 and R2, which are connected in parallel across a voltage source V.
By applying Ohm’s law, we can write the following:
I1 =
M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 4
V
V
,I =
R1 2 R2
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Basic Concepts 5
Further, by assuming that current I is divided into currents
I1 and I2, we obtain the following:
I
V
1
R1 + R 2
1 
V
V
+
=V  +
I = I1 + I 2 =
 =V
R1R 2
R1 R 2
 R1 R 2 
or
and
I=
I1
I2
R1
R2
Figure 1.2 C
urrent Division in
a Parallel Circuit
V
V
=
R1R 2
R eq
R1 + R 2
I1 =
I R eq
R2
V
I R1R 2
=
=
=I
R1
R1
R1 R1 + R 2
R1 + R 2
I2 =
I R eq
R1
V
I R1R 2
=
=
=I
R1 + R 2
R2
R2
R 2 R1 + R 2
Therefore, the current divider rule in a parallel circuit as in Figure 1.2, is as follows:
Branch current,
I1 = I
R2
(1.9)
R1 + R2
I2 = I
R1
(1.10)
R1 + R2
Similarly,
Branch current,
1.5.3 Series–Parallel Circuits
A series–parallel circuit is shown in Figure 1.3. The current and voltage drops in the circuit are
calculated as
I1 = I2 + I3; V1 = I1R1; V2 = I2R2; V3 = I3R3
The voltage across terminals A and B is equated to the
voltage drops in the two parallel paths. Therefore,
I1
V2 = V3
V = V1 + V2 = V1 + V3
1.5.4 Ladder Network
Figure 1.4 shows a ladder network. This is a special case
of series–parallel network. By replacing series-connected
resistors into their equivalents, such circuits can be
reduced into a simple circuit.
To reduce the circuit into a simple circuit, we proceed in steps as shown in Figure 1.5. Across terminals
M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 5
R1
+
−
V1
A
V
I2
V2
I3
R2
R3
V3
B
I1
Figure 1.3 Series–Parallel Circuit
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6
Network Analysis and Synthesis
R1
R3
V
C and D, R5 and R6 are in series and connected in
parallel with R4 to form Rx. Then across terminals A
and B, R3 and Rx are in series and connected in parallel with R2 to form Ry. The equivalent resistance
is Ry.
R5
R2
R4
R6
Figure 1.4 A
Ladder Network of
Resistors
A
C
R1
A
R3
V
R1
R5
R2
R4
B
C
⇒
R6
A
R1
R3
V
R2
D
Rx
B
⇒
V
Ry
D
B
Rx = R4 || (R5 + R6) and Ry = R2 || (R3 + Rx)
Figure 1.5 Simplification of Ladder Network
Resistance of the whole network across the voltage V
is therefore, equals to R1 + Ry.
A
D
8Ω
Example 1.1 Calculate the current supplied by the
12 V battery in the network shown in Figure 1.6.
8Ω
12 V
Solution: Points A and C can be brought together.
Similarly, points B and D can be brought together. First,
B
8Ω
8Ω
C
Figure 1.6
8Ω
A, C
8Ω
8Ω
12 V
⇒
8Ω
4Ω
A, C
D
4Ω
12 V
B
B
A, C
D
⇒
So, the current supplied by
the 12 V battery is 6 A
6A
12 V
4Ω
4Ω
⇒
12 V
2Ω
B, D
Figure 1.7
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Basic Concepts 7
we will bring A and C together; then, we combine B and D
together as shown in Figure 1.7. The process is completed by
taking equivalents of parallel resistances.
Example 1.2 Calculate the resistance between the terminals P
and Q in the circuit shown in Figure 1.8.
A
Q
Q
8Ω
8Ω
B
C
8Ω
D
8Ω
8Ω
D
⇒
P
16 Ω
16 Ω
D
Q
C
8Ω
Figure 1.8
A
P
B
8Ω
Solution: The circuit is redrawn as shown in Figure 1.9. The
resistance between terminals P and Q is calculated by taking the
equivalent of series resistances.
P
8Ω
A
P
⇒
8Ω
Q
Figure 1.9
So, the equivalent resistence between terminals P and Q is 8 Ω.
Example 1.3 What will be the equivalent resistance between terminals P and Q of the ladder
network shown in Figure 1.10.
A
P
C
1Ω
5Ω
12 Ω
4Ω
16 Ω
3Ω
Q
E
16 Ω
3Ω
B
16 Ω
16 Ω
4Ω
D
F
Figure 1.10
Solution: By considering series and parallel connection of resistors, the equivalent resistance
across terminals P and Q is determined as shown in Figure 1.11.
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8
Network Analysis and Synthesis
C
A
P
1Ω
5Ω
12 Ω
4Ω
16 Ω
3Ω
Q
E
16 × 16 = 8 Ω
16 + 16
16 Ω
4Ω
3Ω
B
D
F
⇒
A
P
C
1Ω
5Ω
12 Ω
16 Ω
3Ω
Q
4 + 8 + 4 = 16 Ω
16 Ω
3Ω
B
D
⇒
A
P
C
1Ω
5Ω
12 Ω
3Ω
Q
16 × 16 = 8 Ω
16 + 16
16 Ω
3Ω
B
D
⇒
A
P
1Ω
12 Ω
16 Ω
5+8+3 ⇒
= 16 Ω
12 Ω
3Ω
3Ω
Q
P
P
1Ω
16 × 16 ⇒
16 + 16
=8Ω
1+8+3
= 12 Ω
Q
Q
B
12 Ω
Figure 1.11
So, the equivalent resistance, Req between terminals P
and Q is
R eq =
12 × 12
= 6Ω
12 + 12
Example 1.4 Four resistances of equal value, R are
connected as shown in Figure 1.12. Calculate the equivalent
resistance across terminals P and Q.
M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 8
R
P
A
R
Q
R
B
R
Figure 1.12
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Basic Concepts 9
R
P
AB
R
P
R×R = R
R+R
2
AB
⇒
R
Q
R
2
P
P
⇒
Q
R
Q
R×R = R
R+R
2
R
⇒
Q
R
2
Figure 1.13
Solution: We can redraw the circuit as shown in
Figure 1.13 and find the equivalent resistance as
shown using successive steps.
Thus, the equivalent resistance is R.
Example 1.5 In the circuit shown in Figure 1.14,
calculate values of branch currents I1 and I2.
Solution: In the circuit, total current I is divided into I1
and I2. We will first calculate the total current I and then
calculate the branch currents using current divided rule.
I1
R1
I
4Ω
I
I2
R2
3.6 Ω
6Ω
I
60 V
Figure 1.14
60
60
60
=
=
= 10 A
4×6
3.6 + 2.4
6
3.6 +
4+6
R2
6
=6A
I1 = I ×
= 10 ×
4+6
R1 + R 2
I=
v
=
R eq
I2 = I ×
R1
4
= 10 ×
=4A
R1 + R 2
4+6
1.6 BASIC CIRCUIT ELEMENTS
The basic circuit elements are resistors, inductors, and capacitors. These are described in brief
in the following sections.
1.6.1 Resistors
The resistors offer resistance to current flow. The common types of resistors are wire wound
type and carbon composition type. Resistors can be of fixed value or variable or adjustable
value. However, standard values of fixed resistors range from 2.7 Ω to 22 MΩ with tolerances
varying from ± 20% to ± 1%.
High-power resistors are generally wire wound and are mounted on ceramic tubes. Further,
carbon composition resistors are normally of low power rating that ranges from fractions of a
watt to a few watts.
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10 Network Analysis and Synthesis
i
Flux
v (ac)
e
Coil
Figure 1.15 E
MF is Induced
in a Coil Due
to Changing
Current Flowing
Through it
1.6.2 Inductors—Self-Inductance
and Mutual Inductance
When there is sudden growth (sudden increase) or decay (sudden decrease) in current flowing through a coil, a changing
flux will be produced. According to Faraday’s law, an EMF
is induced in a coil due to changing current flowing through
it. The amount of induced EMF in the coil depends upon the
self-inductance of the coil. The EMF will also be induced
in another neighbouring coil due to the changing current in
the coil which causes mutual inductance between the coils.
Figure 1.15 shows that EMF (e) is induced in a coil when a
changing magnetic flux, which is produced due to changing
current flow, links the coil. EMF is induced only when the flux
is changing. If the flux is constant, no EMF will be induced.
Self-inductance
A changing current produces an EMF of self-induction e in a coil that varies directly with the
time rate of change of current. The induced emf can be expressed as follows:
di
(1.11)
dt
where L is called the self inductance or simply inductance of the coil
e=L
e
di /dt
The inductance L of a coil is one henry, when an EMF of 1 V is induced by the changing current
of 1 A/s flowing through the coil.
If N is the number of turns, in terms of rate of change of flux linkage (Nf) with respect to i,
inductance L of the coil is expressed as the following:
Inductance
L=
L=
d (N f )
df
=N
(1.12)
di
di
N2A
N2A
NI
=m
, inductance, L = m0 mr
(1.13)
l
l
l
where, m = m0 mr is the permeability of the flux path; l is the length of the flux path; A is the area
through which flux passes; B is the flux density.
It is to be noted that inductance of a coil gets increased many times if the permeability of the
flux path is increased by placing a magnetic material as the core material for the coil.
with f = BA, B = m0 mr H, and H =
Mutual Inductance
When changing flux of one coil links another coil, that is, if the two coils are magnetically
coupled, an EMF is also induced in the second coil also. Two coils have a mutual inductance of
1 H when an EMF of 1 V is induced in one coil by the changing current at the other coil at the
rate of 1 A/second.
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Basic Concepts 11
The amount of flux linkage from one coil to the other
is indicated by a factor called coefficient of coupling.
Figure 1.16 shows two coupled coils having self-inductance of L1 and L2, respectively. According to equation
(1.13) stated earlier,
L1 = mN p2
A
and
l
L2 = m N s2
L1
A
l
Np
Ns
e1
e2
i1
M
L2
Figure 1.16 Mutual Inductance
between Two Coils
On multiplying L1 and L2, we obtain the following expression:
 A
L1 L2 = m 2 N p2 N s2  
l
2
 A
L1 L2 = m N p N s   (1.14)
l
The mutual inductance between the two coils is due to the EMF induced in the second coil for
the change in current in the first coil.
Let e2 be the EMF induced in the second coil, as shown in Figure 1.16.
e2 = M
di1
(1.15)
dt
where M is the mutual inductance and e2 is the EMF induced in the second coil; di1/dt is the rate
of change of current in the first coil.
According to Faraday’s law, the equation of induced EMF in the second coil is given as
follows:
e2 = N s
d ( Kf )
dt
where K is the coefficient of coupling between the two coils.
e2 = KN s
df
dt
(1.16)
From equations (1.15) and (1.16), we obtain the following expression:
df
KN s
e2
dt = KN df =
M=
s
di1
di1
di1
dt
dt
(1.17)
We know, flux f = BA, B is the flux density, and A is the area of the core of the coil.
Flux,
f = BA
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12 Network Analysis and Synthesis
= m HA
NI 
N p i1 A
 A 
=m
= m N p   i1  as B = m H and H = l 
l
l
 A
df = mN p   di1(1.18)
l
From equations (1.17) and (1.18), M can be expressed as the following:
 A
M = K mN p N s   (1.19)
l
Now, from equations (1.14) and (1.19), we derive the following expression:
L1 L2 =
M
K
or,
M = K L1 L2 (1.20)
Types of Inductors
The wave shape of power supplies and of a fluctuating dc can be
improved by using an inductor. An inductor opposes any change in the current level; hence, the
fluctuations are reduced due to the connection of an inductor in series.
Power Supply Inductors.
Two coupled coils can be used to make a variable inductor. The coils
may be connected either in series or in parallel. The total inductance value is changed by adjusting the position of one coil with respect to the other.
Laboratory Inductors.
Moulded Inductors. Small moulded inductors of range 1–10 mH are made with a maximum
current carrying capacity of 70 mA. Their values are colour-coded as is done for resistors.
These inductors are of minute size and are used in electronic circuits.
Energy Stored in an Inductor
Energy is stored in an inductor in the form of an electromagnetic field when a current (I amperes)
flows through the coil of inductance (L henries). The amount of energy stored (W ) is given by
the following expression:
Again,
Therefore,
W =
1 2
LI joules(1.21)
2
L=
mN 2 A
NI
; B = mH = m
l
l
I=
Bl
(1.22)
mN
W =
1 2 1 N 2 A B 2l 2
LI = m
2
2
l m 2N 2
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Basic Concepts 13
W =
B 2 Al
joules
2m
(1.23)
Example 1.6 Two coils having number of turns 1000 and 2000, respectively, are placed side
by side. The self-inductance of coil with 1000 turns is 800 mH. Only 80% of the flux produced
by coil with 1000 turns links the second coil. Calculate the emf induced in coil 2 when current
changes at the rate of 10 A/s in the coil with 1000 turns. In addition, calculate the mutual
inductance of the two coils.
Solution: The two coils having magnetic coupling is shown in Figure 1.17.
Flux linkage in coil 2
Change in current in coil 1
N 2 ( kf1 ) 2000 × 0.8 f1
=
=
I1
I1
Mutual inductance, M =
=
Coil 1
I1
Coil 2
f1
1600f1
f2 = Kf1
I1
Self-inductance, L1 =
or,
K = 0.8
f1
I1
N1 = 1000
N1 f1
Figure 1.17
I1
=
N2 = 2000
L1 800 × 10 -3
=
1000
N1
 f1  1600 × 800 × 10 -3
= 640 × 10 -3 H
M = 1600   =
1000
 I1 
= 1280 mH.
Emf induced in coil 2, e2 is given as
di
e2 = M 1 = 1280 × 10 -3 × 10 = 12.8 V.
dt
∴
Example 1.7 Two coils are placed side by side. A current of 4A amperes flowing through one
coil of 400 turns produces a flux of 2 mWbs. When the current in the coil is suddenly reversed
in 1 ms, an emf of 20 volts is induced in the second coil. The coefficient of coupling between
the two coils is 50%. Calculate the self and mutual inductance of the two coils.
Solution: Let induced emf in the second coil be e2.
e2 = M
di1
dt
Current in the first coil has been reversed from 4A to -4A. So, total change of current is 8A and
has taken place in 1 ms.
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14 Network Analysis and Synthesis
Substituting values in the above equation
20 = M
8
1 × 10 -3
20 × 1 × 10 -3
= 2.5 × 10 -3 H.
8
N f 400 × 2 × 10 -3
L1 = 1 1 =
I1
4
M =
or,
Self inductance,
= 0.2 H.
We know,
M = K L1L 2
Given
K = 0.5
So,
M 2 = K 2 L1 L 2
or,
L2 =
=
M2
K 2 L2
=
( 2.5 × 10 -3 ) 2
(0.5) 2 × 0.2
6.52 × 10 -6
0.25 × 0.2
= 125 × 10 -6 H.
1.6.3 Capacitors
A capacitor is made using two thin metal plates separated by a layer of insulating material.
The layer of insulating material between the capacitor plates is known as the dielectric. Typical
dielectric materials are mica, paper, rubber, air, etc. Figure 1.18 shows a capacitor that has the
ability to store electric charge.
A voltage V is applied across the capacitor by closing the switch S. The plates of the capacitor will aquire positive and negative charges as shown in Figure 1.18 (a). Free electrons from
the positive plate will be attracted by the positive terminal of the battery thereby leaving positive
Electron
flow
S
V
I
++++
+
−
Dielectric
−−−−
Electron
I
flow
(a)
S (Switch open)
++++
+
V
−
−−−−
Dielectric
(b)
Figure 1.18 E
nergy Stored in a Capacitor (a) During Charging from
a Voltage Source and (b) After Charging
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Basic Concepts 15
charge on it and electrons will be supplied from the negative terminal of the battery to the other
plate of the capacitor. These charges will be retained by the capacitor plates even after the
switch S is opened. This is because the dielectric material placed between the plates will oppose
any flow of charge through it. In this way, a capacitor stores charge.
It is interesting to note that both a voltage cell and a capacitor store charge. However, a
capacitor cannot be used to supply a load current. This is because when load current is drawn
from a voltage cell, the chemical action within the cell continuously creates free electrons to
supply the current. In a capacitor, such action does not take place, and therefore, a capacitor
cannot be considered as a voltage source.
Dielectric Strength
When a voltage is applied across the two plates of a capacitor, the dielectric material gets
subjected to electric pressure pressing for the flow of current through it. If the voltage applied
is too high, the stress becomes more and at a certain high voltage, the material may break
down. The dielectric strength of a given material is the voltage per unit thickness at which
the dielectric may breakdown and hence cannot prevent the current flow. Dielectric strength
of air is 3 kV/mm, whereas for mica, it is 200 kV/mm. So, mica is a better dielectric material
than air.
The capacitance of a capacitor is expressed as follows:
C=
eA
(1.24)
d
where A is the area of the plates, d is the distance between the plates and e is equal to e0er where,
e0 is the permittivity of free space and er is the relative permittivity and also called dielectric
constant. The value of e0 is 8.85 × 10−12 F/m approximately. The value of dielectric constant for
vacuum or air is 1 and for mica, it varies from 3 to 7. The value of dielectric constant for paper
varies from 4 to 6, and for ceramic, the value may be as high as 1000. If the space between the
two capacitor plates is filled by paper or polythene, or mica or other dielectric, the value of
capacitance gets increased.
Types of Capacitors
Air Capacitors. Air capacitors are made using two sets of metal plates; one set fixed and the
other set movable. Movable set of plates is adjusted by a rotatable shaft so that the area of the
plates opposite to each other is changed.
Paper Capacitors. Paper capacitors are made by placing a layer of paper between two layers
of metal foils. The metal foils and paper are rolled up and external connections are brought out
from the metal foils.
Plastic Film Capacitors. These are made the same way as paper capacitors except that the
paper is replaced by very thin film of plastic material like Polystyrene or Mylar.
Alternate layers of metal foil and mica are used to make mica capacitors.
Connections are taken out from metal foils. The whole of the assembly is encapsulated (placed
inside) in moulded plastic jacket.
Mica Capacitors.
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16 Network Analysis and Synthesis
Ceramic Capacitors. A ceramic disc is used as the dielectric. Thin films of metal are deposited on each side of the ceramic disc. Connections are taken out from the metallic sides. The
whole assembly is encapsulated in a container made of a plastic material.
Electrolytic capacitors are made the same way as paper capacitors.
Thin sheets of aluminium foils separated by an electrolyte is rolled up and enclosed in a small aluminium cylinder. Terminals are brought out from the two metallic foils. A direct voltage is applied
to the capacitor that causes a thin layer of aluminium oxide to form on the positive plate towards
the side of the electrolyte. The aluminium oxide formed becomes the dielectric. The electrolyte
and the foil on one side makes one plate while the positive sheet of foil on the other side forms
the second plate. Electrolytic capacitors have one terminal identified for positive connection.
An electrolytic capacitor should be connected with correct polarity; otherwise, the capacitor
may explode causing serious injuries to persons handling the device.
Electrolytic Capacitors.
Series and Parallel Connections of Capacitors
Series Connection. Series connection of capacitors results in increasing the total thickness of
the dielectric between the two outermost plates. For example, when three parallel-plate capacitors are connected in series, the total capacitance for three series-connected capacitors as shown
in Figure 1.19 is expressed as the following:
Cs =
A
d1
d2
d3
C1
C2
Cs
C3
or
A
Figure 1.19 Capacitors
Connected
in Series
d
A
C1
C2
A
C3
d + d2 + d3
d
d1
d
1
= 1
=
+ 2 + 3
Cs
e 0e r A
e 0e r A e 0e r A e 0e r A
1
1
1
1
or
=
+
+
Cs e 0e r A e 0e r A e 0e r A
d1
d2
d3
or
A
1
1
1
1
=
+
+ (1.25)
Cs C1 C2 C3
Parallel Connection. Parallel connection of capacitors is
equivalent to increasing the total plate area. When three capacitors are connected in parallel, as has been shown in Figure 1.20,
their total capacitance Cp is given as follows:
Cp =
e 0e r ( A1 + A2 + A3 )
d
or Cp =
e 0e r A1 e 0e r A2 e 0e r A
+
+
d
d
d
or Cp = C1 + C2 + C3 (1.26)
Cp
Figure 1.20 Capacitors
Connected
in Parallel
e 0e r A
d1 + d2 + d3
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Basic Concepts 17
Energy Stored in a Charged Capacitor
The capacitance of 1 farad is the capacitance of the capacitor that contains a charge of 1 coulomb when the potential difference between the plates is 1 volt.
From the definition of farad, charge Q can be expressed as follows:
Q = CV
(1.27)
where C is the capacitance in farad, V is the voltage between the plates in volts and Q
is the charge in coulombs. When a capacitor is charged, energy is supplied to it which
the capacitor stores as stored energy. Current is the rate of flow of charge, so that charge,
Q = I × t.
Energy supplied in charging the capacitor is,
1
W = V I t (1.28)
2
Again,
C=
Q It
=
V V
or
I=
CV
t
(1.29)
From equations (1.29) and (1.28), the energy stored, W can be calculated as follows:
1
W = V It
2
1 CV
= V
×t
2
t
or
W =
1
CV 2 joules(1.30)
2
1.7 INDUCTORS AND CAPACITORS IN DC CIRCUITS
When a circuit containing a resistance and an inductance in series is switched on to a dc voltage source, maximum level of counter EMF is induced in the inductor, thereby making the initial current to be zero. However, the counter EMF gradually falls to zero and hence the current
grows gradually to its maximum value.
When a circuit containing a resistance and a capacitance in series is switched on to dc voltage, the charging current is maximum initially and then the current falls to zero.
1.8 DC NETWORK TERMINOLOGIES AND CIRCUIT FUNDAMENTALS
Before discussing various laws and theorems, certain terminologies related to dc networks are
described first. These include the definition of terms used, voltage and current sources, series
and parallel circuits and voltage and current source transformations.
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18 Network Analysis and Synthesis
1.8.1 Network Terminologies
While discussing network theorems, laws, and electrical and electronic circuits, one often
comes across the following terms.
1. Circuit: A conducting path through which an electric current either flows or is intended to
flow is called a circuit.
2. Electric network: A combination of various circuit elements, connected in any manner, is
called an electric network.
3. Linear circuit: The circuit whose parameters are constant, that is, they do not change with
the application of voltage or current is called a linear circuit.
4. Non-linear circuit: The circuit whose parameters change with the application of voltage
or current is called a non-linear circuit.
5. Circuit parameter: The various elements of an electric circuit are called its parameters,
such as resistance, inductance, and capacitance.
6. Bilateral circuit: A bilateral circuit is one whose properties or characteristics are the same
in either direction, for example, transmission line.
7. Unilateral circuit: A unilateral circuit is one whose properties or characteristics change
with the direction of its operations, for example, diode rectifier.
8. Active network: An active network is one that contains one or more sources of EMF.
9. Passive network: A passive network is one that does not contain any source of EMF.
10. Node: A node is a junction in a circuit where two or more circuit elements are connected
together. For convenience, the nodes are labelled by letters.
11. Branch: The part of a network that lies between two junctions is called a branch.
12. Loop: A loop is a closed path in a network formed by a number of connected branches.
13. Mesh: Any path that contains no other paths within it is called a mesh. Thus, a loop
­contains meshes, but a mesh does not contain a loop.
For example, let us consider an electric circuit as shown in Figure 1.21.
A
i
R1
R2
B
+
E1
_
Source
Branch
C
Node
+
E3
Mesh I
Mesh II
_
E2
Source
D
Figure 1.21 Different Parts of an Electric Circuit
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Basic Concepts 19
1. Number of nodes (N) = 4 (that is, A, B, C, D)
2. Number of branches (B) = 5 (that is, AB, BC, BD, CD, AD)
3. Independent meshes (M ) = B − N + 1
= 5 − 4 + 1 = 2 (that is, ABDA, BCDB)
4. Number of loops = 3 (that is, ABDA, BCDB, and ABCDA). It is seen that a loop ABCDA
encloses two meshes, that is, mesh I and mesh II.
1.8.2 Voltage and Current Sources
A source is a device that converts mechanical, thermal, chemical, or some other form of
energy into electrical energy. There are two types of sources: voltage source and current
source.
Voltage Source
Voltage sources are further categorised as ideal voltage source and practical voltage source.
Examples of voltage sources are batteries, dynamos, alternators, etc. An ideal voltage source is
defined as the energy source that gives constant voltage across its terminals irrespective of current drawn through its terminals. The symbol of ideal voltage source is shown in Figure 1.22(a).
In an ideal voltage source, the terminal voltage is independent of the value of the load resistance
(RL) connected. Whatever is the voltage of the source, the same voltage is available across the
load terminals of RL, that is, VL = VS under loading condition as shown in Figure 1.22(b). There
is no drop of voltage in the source supplying current to the load. The internal resistance of the
source is therefore zero.
In a practical voltage source, voltage across the load will be less than the source voltage
due to voltage drop in the resistance of the source itself when a load is connected, as shown in
Figure 1.22(c).
V
IL
+
+
VS
_
_
+
VS
VS
Practical
source
VL < VS
VL Load
RL
_
(Load current)
(a)
Ideal source
VL = VS
(b)
IL
(c)
Figure 1.22 V
oltage Source and its Characteristics: (a) Symbol; (b) Circuit
and (c) Load Characteristics
Current Source
In certain applications, a constant current flow through the circuit is required. When the load
resistance is connected between the output terminals, a constant current (IL) will flow through
the load.
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20 Network Analysis and Synthesis
Some examples of current source are photoelectric cells, collector current in transistors, etc.
The symbol of current source is shown in Figure 1.23.
Practical Voltage and Current Sources
A practical voltage source like a battery has the drooping load characteristics due to some internal resistance. A voltage source has small internal resistance in series, whereas a current source
has some high internal resistance in parallel.
1. For ideal voltage source, Rse = 0.
2. For ideal voltage source, Rsh = ∞.
IL
IS
IS
Load
I
+ IS
IL = IS
Ideal source
Practical source
IL < IS
VL
_
VL
(a)
(c)
(b)
Figure 1.23 Current Source and its Characteristics: (a) Symbol; (b) Circuit
and (c) Characteristics
A practical voltage source is shown as an ideal voltage source in series with a resistance. This
resistance is called the internal resistance of the source, as shown in Figure 1.24(a). A practical current source is shown as an ideal current source in parallel with its internal resistance, as
shown in Figure 1.24(b).
IL
Rse
VS
VL
Ish
RL
IS
(a)
IL
Rsh
RL
(b)
Figure 1.24 R
epresentation of Practical Voltage and
Current Sources (a) Voltage Source and
(b) Current Source
From Figure 1.24(a), we can write the following expression:
VL (open circuit), that is, VL (OC) = VS
That is, when the load RL is removed, the circuit becomes an open circuit and the voltage
across the source becomes the same as the voltage across the load terminals.
When the load is short-circuited, the short-circuit current, IL(SC) = Vs/Rse
In the same way, from Figure 1.24(b), we can obtain the following expression:
VL(OC) = Ish Rsh
and
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IL(SC) = IS
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Basic Concepts 21
In source transformation as discussed in Section 1.8.3, we shall use the equivalence of opencircuit voltage and short-circuit current.
Independent and Dependent Sources
The magnitude of an independent source does not depend upon the current in the circuit or voltage across any other element in the circuit. The magnitude of a dependent source gets changed
due to some other current or voltage in the circuit. An independent source is represented by a
circle, while a dependent source is represented by a diamond-shaped symbol. Dependent voltage sources are also called controlled sources.
There are four kinds of dependent sources:
1.
2.
3.
4.
voltage-controlled voltage source (vcvs)
current-controlled current source (cccs)
voltage-controlled current source (vccs)
current-controlled voltage source (ccvs)
The dependent voltage sources find applications in electronic circuits and devices.
1.8.3 Source Transformation
A voltage source can be represented by an equivalent current source. Similarly, a current source
can be represented by an equivalent voltage source. This source transformation often provides
simplified solutions of circuit problems.
Transformation of Voltage Source into Current Source and Current
Source into Voltage Source
A voltage source is equivalent to a current source and vice-versa if they produce equal values
of IL and VL when connected to the load RL. They should also provide the same open-circuit
voltage and short-circuit current.
If voltage source is converted into current source as in Figure 1.25, we consider the shortcircuit current equivalence. The short-circuit currents in the two equivalent circuits are, respecV
tively, Vs /Rse and Is. Therefore, the equivalent current source is represented as I s = s .
R se
Rse
IL
Vs
RL
Voltage source
Is
Ish
IL
Rsh
RL
Current source
Figure 1.25 Equivalent Current Source
If the current source is converted into voltage source, as in Figure 1.26, we will consider the
open-circuit voltage equivalence. The open circuit voltage in the two circuits are, repectively, Ish
Rsh and Vs. Therefore, the equivalent voltage source is Vs = Ish Rsh.
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22 Network Analysis and Synthesis
Is
Ish
IL
Rsh
RL
IL
Rse
Vs
RL
Voltage source
Current source
Figure 1.26 Equivalent Voltage Source
The two circuits are equivalent when their open circuit voltages and short circuit currents are
the same.
A few examples will further clarify this concept.
Example 1.8 Convert a voltage source of 20 V with internal resistance of 5 Ω into an
equivalent current source.
Solution: Given Vs = 20 V, Rse = 5 Ω
V
20
The short-circuit current, I s = s =
= 4 A . The voltage source is represented by a curR
5
se
rent source of 4 A.
The internal resistance will be the same as Rse in both the cases. So, Rsh is shown as Rse.
The condition for equivalence is checked from the following Voc should be same and Isc
should also be same.
In Figure 1.27(a), Voc = 20 V; in Figure 1.27(b) Voc 4 A × 5 Ω = 20 V
5Ω
A
+
A
4A
Rse
20 V
−
Voc
4A
5Ω
Rsh
B
B
(a)
Voc
(b)
Figure 1.27 C
onversion of a Voltage Source into a Current Source (a) Voltage Source
and (b) Equivalent Current Source
Isc in Figure 1.27(a) is 4 A. This is because I sc = 20 5 = 4 A ; Isc in Figure 1.27(b) is 4 A. This
is because when terminals A and B are short-circuited, the whole of 4 A will flow through the
short-circuit path.
These two circuits are equivalent because the open-circuit voltage and short-circuit current
are same in both the circuits.
Example 1.9 Convert a current source of 100 A with internal resistance of 10 Ω into an
equivalent voltage source.
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Basic Concepts 23
Solution: Here, I = 100 A, Rsh = 10 Ω
For an equivalent voltage source, the following can be derived:
V = Ish × Rsh = 100 × 10 = 1000 V
Rsh = Rse = 10 Ω in series
The open-circuit voltage and short-circuit current are same in the two equivalent circuits as
shown in Figure 1.28(a) and 1.28(b), respectively.
A
Ish
+
I = 100 A
−
10 Ω
Rsh
10 Ω
Voc
1000 V
Rse
Voc
V
B
(a)
A
B
(b)
Figure 1.28 Conversion of a Current Source into an Equivalent Voltage Source
(a) Current Source and (b) Equivalent Voltage Source
R eview Q uestions
Short Answer Type
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
Define resistance, current and potential difference.
Distinguish between electrical power and electrical energy.
State Ohm’s law, and mention the conditions.
Distinguish between self-inductance and mutual inductance.
State the factor on which resistance and inductance of a coil depends.
Establish the relationship between self-inductance and mutual inductance.
Explain why the inductance of a coil increases when we place an iron bar inside it.
What do you mean by coefficient of coupling between two coils and what is its significance?
What are the various types of inductors and capacitors generally in use?
A resistor connected in series gives a higher value of equivalent resistance but a capacitor
connected in series gives a lower value of equivalent capacitance. Explain.
What do you mean by dielectric constant and what is its significance?
What is an electrolytic capacitor?
When capacitors are connected in parallel, the total capacitance increases. Explain how.
How do you calculate the energy stored in an inductor and in a capacitor?
Explain why a charged capacitor cannot be considered as a voltage source.
How do you convert a voltage source into an equivalent current source and vice–versa?
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24 Network Analysis and Synthesis
Numerical Questions
1. For the resistive circuit shown in Figure 1.29, calculate the current flowing through the 10 Ω
resistor and voltage drop across the 5 Ω resistor.
2Ω
5Ω
20 Ω
24 V
10 Ω
20 Ω
Figure 1.29
[Ans. 1 A, 10 V]
2. Calculate the resistance between the terminals A and B in the network shown in Figure 1.30.
What will be the resistance across the same terminals when terminals D and B are
short-circuited?
C
2Ω
3Ω
A
B
2Ω
3Ω
D
[Ans. 2.5 Ω, 1 Ω]
Figure 1.30
3. Find the value of V1 in the circuit shown in Figure 1.31.
2Ω
2Ω
1Ω
4Ω
12 V
4Ω
V1
3Ω
Figure 1.31
[Ans. 2.25 V]
4. Calculate the resistance across terminals AB of the network shown in Figure 1.32.
A
4Ω
B
10 Ω
10 Ω
2Ω
3Ω
10 Ω
6Ω
5Ω
Figure 1.32
M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 24
[Ans. 5 Ω]
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Basic Concepts 25
5. Calculate the value of resistance across terminals PQ in the network shown in Figure 1.33.
7Ω
P
7Ω
5Ω
Q
5Ω
Figure 1.33
[Ans. 6 Ω]
6. A voltage source of 20 V has an internal series resistance of 2 Ω. What will be its equivalent
current source?
[Ans. Is = 10 A, Rsh = 2 Ω]
7. Find the equivalent resistance between the terminals A and B of the network shown in
Figure 1.34.
A
1Ω 1Ω
2Ω
1Ω
2Ω
2Ω
B
1Ω
Figure 1.34
[Ans. 1 Ω]
8. Find the equivalent resistance between the terminals P and Q of the network shown in
Figure 1.35.
P
2Ω
8Ω
4Ω
4Ω
8Ω
6Ω
3Ω
2Ω
2Ω
4Ω
5Ω
Q
3Ω
Figure 1.35
M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 25
[Ans. 5.33 Ω]
12/4/2014 11:34:19 AM
Kirchhoff’s Laws,
Mesh and Nodal
Analysis
2
CHAPTER OBJECTIVES
After carefully studying this chapter, you should be able to do the following:
State and explain Kirchhoff’s current Solve network problems using the
and voltage laws.
method of nodal analysis.
Apply Kirchhoff’s current and voltage Carryout nodal analysis of AC circuits.
laws to solve network problems.
Calculate voltage at any point in a net Carryout mesh analysis by Correctly
work using node voltage analysis method.
framing the voltage equation’s for each Explain with the help of an example
loop in a circuit.
the concept of super nodal analysis.
Solve mesh equations using determi- Calculate current through any branch of a
nants.
network and also calculate voltage at any
point in the network using Kirchhoff’s
Explain the method of nodal analysis
with the help of a simple example.
laws or mesh or nodal analysis.
In this chapter, we will discuss the two important laws provided by Kirchhoff and their applications in solution of network problems with plenty of solved examples.
2.1 KIRCHHOFF’S LAWS
There are two Kirchhoff’s laws. They are Kirchhoff’s current law and Kirchhoff’s voltage law. By applying these two important laws, we are able to solve electrical network
problems.
2.1.1 Kirchhoff’s Current Law
This law is applicable to a node or junction in an electrical circuit.
Kirchhoff’s current law (KCL) states that the algebraic sum of all the currents meeting at a
junction or a node in a circuit is equal to zero.
In other words, the KCL can be stated as: The sum of currents entering into any node or junction is equal to the sum of currents leaving the node or junction.
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Kirchhoff’s Laws, Mesh and Nodal Analysis 27
A node is a junction point of two or more branches.
To understand the KCL further, let us consider the electric circuit shown in Figure 2.1.
At junction point B, current I1 is incoming and currents I2 and I3 are outgoing. The incoming current is
considered positive and the outgoing current is considered negative. Now, the algebraic sum of all the currents
is zero. Therefore, the following equation is obtained:
R1
V
+
I3 R3
B
I1
I2
R2
−
I6
R6
R4
I5
E
I7
R5
R7
I4
I1 + ( − I 2 ) + ( − I 3 ) = 0
I1 = I 2 + I 3
or
R8
That is, the incoming current at point B = the sum of
outgoing currents from point B.
Similarly, by applying KCL at point E in the circuit,
we get the following:
Figure 2.1 Kirchhoff’s Current
Law Applied in an
Electric Circuit
I 2 + I 5 + ( −I 6 ) + ( −I 7 ) = 0
I2 + I5 = I6 + I7
or
That is, the sum of incoming currents = the sum of outgoing currents.
Example 2.1 By applying Kirchhoff’s current law, calculate the current flowing through all
the branches in the circuit shown in Figure 2.2.
A
I
80 A
C
E
G
I1
I2
I3
I4
2Ω
3Ω
4Ω
5Ω
B
D
F
H
Figure 2.2
Solution: In the circuit shown in Figure 2.2, points A, C, E and G are at the same potential while
points B, D, F and H are at the same potential. Therefore, the potential difference between points A
and B, that is, VAB is the same as VCD, VEF and VGH. Therefore, VAB = VCD = VEF = VGH = V (say )
V
V
V
V
, I 2 = , I 3 = , and I 4 =
2
3
4
5
By applying KCL, we obtain the following equation:
Therefore, I1 =
I = I1 + I 2 + I 3 + I 4
On substituting the values in the above equation, we get the following:
V V V V
80 = + + +
2 3 4 5
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28
Network Analysis and Synthesis
 1 1 1 1
=V  + + + 
 2 3 4 5
= V (0.5 + 0.33 + 0.25 + 0.2)
80 = 1.283 V
or
V=
80
= 62.35 V
1.283
Therefore,
I1 =
V 62.35
=
= 31.17 A
2
2
I2 =
V 62.35
=
= 20.77 A
3
3
I3 =
V 62.35
V 62.35
=
= 15.58 A and I 4 = =
= 12.48 A
4
4
5
5
Thus, the sum of I1 + I 2 + I 3 + I 4 = 31.17 + 20.77 + 15.58 + 12.48 = 80 A = I .
Example 2.2 By applying Kirchhoff’s current law, determine the current flowing through all
the branches in the circuit shown in Figure 2.3.
A
I1
2Ω
I2
I
3Ω
I3
15 A
4Ω
I4
5Ω
10 A
Figure 2.3
Solution: By considering point or node A, and by applying KCL, we can say that
incoming currents = outgoing currents;
That is,
or
15A = I1 + I 2 + I 3 + I 4 + 10 A
I1 + I 2 + I 3 + I 4 = 5 A
By assuming the potential of point A as V, we get the following:
I1 =
V
V
V
V
, I = , I = , and I 4 =
2 2 3 3 4
5
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Kirchhoff’s Laws, Mesh and Nodal Analysis 29
Therefore,
V V V V
+ + + =5
2 3 4 5
or
or
or
So,
1 1 1 1
V + + + =5
 2 3 4 5
V [0.5 + 0.333 + 0.25 + 0.2] = 5
5
1.283V = 5 . Therefore, V =
= 3.9 V
1.283
I1 =
V 3.9
=
= 1.95 A
2
2
I2 =
V 3.9
=
= 1.3 A
3
3
I3 =
V 3.9
=
= 0.97 A
4
4
I4 =
V 3.9
=
= 0.78 A
5
5
Thus, I = I1 + I2 + I3+ I4 + 10 = 1.95 + 1.30 + 0.97 + 0.78 + 10 = 15 A
2.1.2 Kirchhoff’s Voltage Law
E2
R1
R2
Kirchhoff’s voltage law (KVL) is applicable to any closed path or
A
B
C
D
+
−
loop in an electric circuit.
I
KVL states that the algebraic sum of all branch voltages around E +
I
1
−
any closed path in a circuit is zero. In other words, the KVL is
stated as the algebraic sum of all the sources of EMFs in a closed
E
loop in an electric circuit is equal to the sum of all the voltage
Figure 2.4
drops in that closed loop.
This law is further explained with the help of an electric circuit
shown in Figure 2.4.
By applying KVL in the closed loop ABCDEA, all the branch voltages are equated to zero
as given in the following:
− IR1 − E2 − IR2 + E = 0
We have used certain sign conventions in the above voltage equation. These are as follows:
1. While moving in the direction of current flow, the voltage drops in the resistors have been
taken as negative. (If we move in opposite direction of current flow, these drops will be
taken as positive.)
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30
Network Analysis and Synthesis
2. While moving from positive terminal of the battery towards the negative terminal, the voltage is considered negative.
3. While moving from negative terminal of the battery towards the positive terminal, the voltage is considered positive.
These sign conventions have to be followed in formulating the voltage equations for the various
loops in an electric circuit.
We shall now discuss the analysis of circuits using the Kirchhoff’s laws. We will illustrate
separately the mesh analysis method and node analysis method.
2.2 MESH ANALYSIS
A mesh is a closed path in a circuit. However, there may be a number of meshes in a circuit.
Mesh analysis is used to find mesh currents. For this, KVL is applied in all the meshes and
mesh equations are formed. The number of mesh equations are always equal to the number of
meshes. From the mesh equations, mesh currents can be calculated. After finding the mesh currents, we can find the voltage drop and power dissipation in any element of the electric circuit.
Let us consider few examples of loop or mesh analysis.
Example 2.3 Use mesh analysis to find voltage across the 4 Ω resistance in the circuit shown
in Figure 2.5.
Solution: In the given electric circuit, there are three meshes. Let I1 , I 2 and I 3 be the currents
flowing in the three meshes in clockwise direction as shown in Figure 2.6.
20 V
2Ω
4Ω
1Ω
40 V
20 V
2Ω
10 V
1Ω
40 V
2Ω
4Ω
10 V
2Ω
I1
Mesh I
I2
Mesh II
I3
Mesh III
Figure 2.6
Figure 2.5
By applying KVL in mesh I, that is, by equating all the branch voltages to be zero, we get the
following:
−1 × I1 + 20 − 2( I1 − I 2 ) − 40 = 0
3I1 − 2 I 2 = −20
or
By applying KVL in mesh II, we obtain the following equation:
+ 40 − 2( I 2 − I1 ) − 10 − 2( I 2 − I 3 ) = 0
or
or
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 30
2( I 2 − I1 ) + 2( I 2 − I 3 ) = 40 + ( −10)
−2 I1 + 4 I 2 − 2 I 3 = 30
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Kirchhoff’s Laws, Mesh and Nodal Analysis 31
By applying KVL in mesh III, we arrive at the following forms:
−2( I 3 − I 2 ) + 10 − 4 I 3 = 0
or
2( I 3 − I 2 ) + 4 I 3 = 10
or
−2 I 2 + 6 I 3 = 10
or
− I 2 + 3I 3 = 5
Therefore, we have three mesh equations as follows:
3I1 − 2 I 2 = −20
−2 I1 + 4 I 2 − 2 I 3 = 30
− I 2 + 3I 3 = 5
(2.1)
(2.2)
(2.3)
To find the voltage across 4 Ω resistance, we need to find the current flowing in 4 Ω resistor.
From Figure 2.6, it is clear that current I3 is flowing through 4 Ω resistor.
Using Cramer’s rule, let us find the value of I3.
Therefore,
3 −2 0
∆ = −2 4 −2
0 −1 3
= 3(12 − 2) + 2( − 6 − 0) + 0
= 30 − 12
= 18
and
3 −2 −20
∆ 3 = −2 4 30
0 −1 5
= 3( 20 + 30) + 2( −10 − 0) − 20( 2 − 0)
= 150 − 20 − 40
= 90
When calculating ∆3, we have replaced the third column, that is, the coefficients of I3 by the
constants on the right-hand side of the three mesh equations.
Therefore, current I3 flowing through 4 Ω resistor is calculated as follows:
∆ 3 90
=
= 5 A.
∆ 18
The voltage across 4 Ω resistance = 4I 3 = 4 × 5 = 20 V
I3 =
Example 2.4 Use the loop method to find the current in 5 Ω resistance of the circuit of Figure 2.7.
Solution: The given circuit has there meshes. Let I1, I2 and I3 be the current flowing in the
three meshes in clockwise direction, as shown in Figure 2.8.
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32
Network Analysis and Synthesis
1Ω
1Ω
5Ω
20 V
20 V
4.5Ω
4Ω
4Ω
4.5Ω
5Ω
2Ω
1A
I1
Mesh I
Figure 2.7
2Ω
I2
Mesh II
1A
I3
Mesh III
Figure 2.8
By applying KVL in mesh I, we get the following equation,
−I1 − 5(I1 − I2) + 20 = 0
or
6I1 − 5I2 = 20
(2.4)
By Applying KVL in mesh II, we get,
−5(I2 − I1) − 4.5I2 − 2(I2 − I3) = 0
or
5(I2 − I1) + 4.5I2 + 2(I2 − I3) = 0
or
−5I1 + 11.5I2 − 2I3 = 0
(2.5)
From mesh III, it is apparent that I3 can be taken as follows:
I3 = −1 A.
(2.6)
We have to find the current through 5 Ω, which is, I1 − I2.
Therefore, let us find I1 and I2 from equations (2.4), (2.5) and (2.6).
6
−5 0
∆ = −5 11.5 −2
0
0
1
= 6(11.5 − 0) + 5(− 5 − 0) + 0
= 69 − 25
= 44
20 −5 0
∆1 = 0 11.5 −2
−1 0
1
= 20(11.5 − 0) + 5(0 − 2) + 0
= 230 − 10
= 220
6 20 0
∆ 2 = −5 0 −2
0 −1 1
= 6(0 − 2) − 20(−5 − 0) + 0
= −12 + 100
= 88.
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Kirchhoff’s Laws, Mesh and Nodal Analysis 33
Now, we can obtain the currents as,
∆1 220
=
= 5A
∆
44
∆
88
= 2A
I2 = 2 =
∆
44
I1 =
Therefore, the current through 5Ω resistor = I1 − I2
=5−2
= 3 A (downward)
Example 2.5 Using mesh analysis, find the current through 100 Ω resistance in Figure 2.9.
Solution: Let I1, I2 and I3 be the currents flowing in clockwise direction in the three meshes,
as shown in Figure 2.10.
From Figure 2.10, it can be observed that the current through 100 Ω resistance = I1 − I2
(in downward direction).
Therefore, let us determine I1 and I2
For this, we apply KVL in mesh I, that is, by taking voltage drop as equal to voltage rise as
20I1 + 100 (I1 − I2) = 50
or
120I1 − 100I2 = 50
or
12I1 − 10I2 = 5
(2.7)
By applying KVL in mesh II, we obtain the following;
− 100(I2 − I1) − 120(I2 − I3) = 0
100(I2 − I1) + 120(I2 − I3) = 0
−100I1 + 220I2 − 120I3 = 0
−5I1 + 11I2 − 6I3 = 0
or
or
or
(2.8)
By applying KVL in mesh III, we get the following;
− 120(I3 − I2) − 30I3 − 20 = 0
120(I3 − I2) + 30I3 = −20
−120I2 + 150I3 = −20
−12I2 + 15I3 = −2
or
or
or
20 Ω
+
−
50 V
20 Ω
30 Ω
100 Ω
120 Ω
(2.9)
50 V
20 V
+
−
I1
Mesh I
Figure 2.9
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 33
30 Ω
100 Ω
120 Ω
I2
I3
Mesh II
20 V
Mesh III
Figure 2.10
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34
Network Analysis and Synthesis
For solving mesh equations (2.7), (2.8) and (2.9) for I1 and I2 by Cramer’s rule, we proceed as
follows:
12 −10 0
∆ = −5 11 −6
0 −12 15
= 12(165 − 72) + 10( −75 − 0) + 0
= 1116 − 750
= 366
5 −10 0
∆1 (for I1 ) = 0
11 −6
−2 −12 15
= 5(165 − 72) + 10(0 − 12)
= 345
Therefore,
I1 =
∆1 345
=
= 0.942623 A.
∆ 366
12 5 0
∆ 2 (for I 2 ) = −5 0 −6
0 −2 15
= 12(0 − 12) − 5( −75 − 0) + 0
= 231
Therefore,
I2 =
∆ 2 231
=
= 0.6311 A
∆ 366
Therefore, the current through 100 Ω resistance = I1 − I2
= 0.942623 − 0.6311
= 0.3115 A.
Example 2.6 Find the current through branch BC, as shown in Figure 2.11, using mesh
analysis.
Solution: In the given circuit, as shown in Figure 2.12, there are three meshes. Let I1, I2 and
I3 be the mesh current.
Now, the current through branch BC = I2.
By applying KVL in mesh I, we get the following equation:
or
−2I1 − 1(I1 − I3) − 2I1 + 5 = 0
5I1 − I3 = 5
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(2.10)
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Kirchhoff’s Laws, Mesh and Nodal Analysis 35
C
D
2Ω
C
1Ω
D
2Ω
1Ω
5V
I2
1Ω
1Ω
I2
5V
Mesh II
A
B
E
2Ω
A
2Ω
2Ω
2Ω
1Ω
I1
5V
Mesh I
H
2Ω
G
2Ω
−
E
2Ω
1Ω
5V
B
5V
+
F
H
Mesh III
I1
G
2Ω
2Ω
I3
2Ω
5V
I3
F
Figure 2.12
Figure 2.11
By applying KVL in mesh II, the following equation is obtained:
or
−2I2 − 1I2 − 1I2 − 2(I2 − I3) + 5 = 0
6I2 − 2I3 = 5
(2.11)
By Applying KVL in mesh III, we obtain the following:
or
−1(I3 − I1) − 2(I3 − I2) − 2I3 − 2I3 −5 = 0
−I1 − 2I2 + 7I3 = −5
(2.12)
Let us find I2 from these mesh equations by Cramer’s rule.
5 0 −1
∆ = 0 6 −2
−1 −2 7
= 5( 42 − 4) + 0 − 1(0 + 6) = 184
5 5 −1
∆ 2 ( to find I 2 ) = 0 5 −2
−1 −5 7
= 5(35 − 10) − 5(0 − 2) − 1(0 + 5)
= 125 + 10 − 5 = 130
Therefore,
I2 =
∆ 2 130
=
= 0.706 A
∆ 184
Therefore, the current through branch BC = 0.706 A.
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Network Analysis and Synthesis
Example 2.7 Using mesh analysis, find the
current through branch DE in the circuit, shown
in Figure 2.13.
A
B
H
G
A
F
B
E
C
2Ω
1A
I1
2(I2 − I1) + 1I2 + 2(I2 − I3) = 0
−2I1 + 5I2 − 2I3 = 0
−
10 V
+
D
1Ω
− 2(I2 − I1) − 1I2 − 2(I2 − I3) = 0
or
or
2Ω
Figure 2.13
(2.13)
By applying KVL in mesh II, we write the following equation,
1Ω
2Ω
1A
I1 = 1 A
D
1Ω
Solution: Let I1, I2 and I3 be the currents flowing
in the three meshes in clockwise direction as
shown in Figure 2.14.
From the circuit, it is clear that current I1 is
the same as the current source.
So,
C
(2.14)
1Ω
2Ω
I2
−
10 V
+
I3
H Mesh I G Mesh II F Mesh III E
Figure 2.14
For mesh III, we get the equation,
or
or
− 2(I3 − I2) − 1I3 + 10 = 0
2 (I3 − I2) + 1I3 = 10
−2I2 + 3I3 = 10
(2.15)
To solve the three equations for I3, we use Crammer’s rule to calculate the value of ∆, ∆3
and I3:
1
0 0
∆ = −2 5 −2
0 −2 3
= 1(15 − 4) + 0 + 0 = 11
1
0 1
∆ 3 = −2 5 0
0 −2 10
= 1(50 − 0) + 0 + 1( 4 − 0)
= 50 + 4 = 54
∆
54
I3 = 3 =
= 4.9 A
∆ 11
Therefore, the current through branch DE = 4.9 A.
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Kirchhoff’s Laws, Mesh and Nodal Analysis 37
Example 2.8 Using mesh analysis, find the
power dissipation in 5 Ω resistance in the circuit
shown in Figure 2.15.
Solution: Let us consider I1 and I2 be the currents
flowing in the two meshes, as shown in Figure
2.16, in clockwise direction.
We have to find the power dissipation in 5 Ω
resistance. For this, firstly, we will have to find current in 5 Ω resistance in the circuit in Figure 2.16,
that is, I1.
By applying KVL in mesh I, we get
or
or
−5I1 − 2(I1 − I2) − 20 + 50 = 0
5I1 + 2(I1 − I2) = 50 + (−20)
7I1 − 2I2 = 30 (2.16)
5Ω
2Ω
3Ω
50 V
20 V
10 V
Figure 2.15
5Ω
50 V
2Ω
3Ω
20 V
I1
Mesh I
I2
10 V
Mesh II
Figure 2.16
By applying KVL in mesh II, we get
or
−2(I2 − I1) − 3I2 − 10 + 20 = 0
2(I2 − I1) + 3I2 = 20 + (−10)
−2I1 + 5I2 = 10
or
(2.17)
For solving these mesh equations for I1 using Cramer’s rule, we proceed as follows:
∆=
7 −2
−2 5
= 35 − ( 4) = 31
∆1 (for I1 ) =
30 −2
10 5
= 150 + 20 = 170
I1 =
∆1 170
=
= 5.48 A
∆
31
Therefore, power dissipation in 5 Ω resistance = I12 (5) = (5.48)2 × 5 = 150.25 W
Example 2.9 Find the current in the load resistance, RL in Figure 2.17 using mesh analysis.
Solution: Let the three mesh currents are I1, I2, and I3 as shown in Figure 2.18.
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Network Analysis and Synthesis
1Ω
1Ω
1Ω
10 V
10 V
5V
1Ω
5V
1Ω
1Ω
1Ω
RL = 2 Ω
1Ω I1
RL = 2 Ω
I2
2Ω
I3
2Ω
Mesh I
Figure 2.17
Mesh II
Mesh III
Figure 2.18
The three mesh equations are formulated as
−1I1 − 1(I1 − I2) −1I1 + 10 = 0
or
I1 + (I1 − I2) + I1 = 10
or
3I1 − I2 = 10
and
1I2 − 5 − 2(I2 − I3) − 1(I2 − I1) = 0
I2 + 2 (I2 − I3) + (I2 − I1) = −5
or
−I1 + 4I2 − 2I3 = −5
and
−2I3 − 2(I3 − I2) + 5 = 0
or
2I3 + 2(I3 − I2) = 5
or
−2I2 + 4I3 = 5
(2.18)
(2.19)
(2.20)
Thus, the three mesh equations are written as follows:
3I1 − I2 = 10
−I1 + 4I2 − 2I3 = −5
0 − 2I2 + 4I3 = 5
Now, let us solve the three mesh equations for I3 using Cramer’s rule.
3 −1 0
∆ = −1 4 −2
0 −2 4
= 3(16 − 4) + 1( −4 − 0) + 0
= 36 − 4 = 32
3 −1 10
∆ 3 (for I 3 ) = −1 4 −5
0 −2 5
Therefore,
= 3( 20 − 10) + 1( −5 − 0) + 10( 2 − 0)
= 30 − 5 + 20 = 45
I3 =
∆ 3 45
=
= 1.40625 A
∆ 32
Therefore, the current through load resistance = 1.40625 A.
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Kirchhoff’s Laws, Mesh and Nodal Analysis 39
Example 2.10 Find the current through
the 10 Ω resistance in the circuit, as shown in
Figure 2.19, using the loop method.
3Ω
+
30 V
Solution: Let I1 and I2 be the mesh currents
as shown in Figure 2.20.
By applying KVL in mesh I, we get
or
or
−3I1 − 10(I1 − I2) + 30 = 0
3I1 + 10(I1 − I2) = 30
13I1 − 10I2 = 30
−
3Ω
(2.21)
−
100 V
15 Ω
+
30 V
+
10 Ω
−
−
I1
(2.22)
100 V
I2
Mesh I
We now solve equations (2.21) and (2.22) for I1
and I2, by Cramer’s rule. We get the following:
∆=
+
10 Ω
Figure 2.19
By applying KVL in mesh II, we get.
−10(I2 − I1) − 15I2 −100 = 0
or
10(I2 − I1) + 15I2 = −100
or
−2I1 + 5I2 = −20
15 Ω
Mesh II
Figure 2.20
13 −10
−2 5
= 65 − 20 = 45
∆1 (for I1 ) =
30 −10
−20 5
= 150 − 200 = −50
I1 =
Therefore,
∆1 −50
=
= −1.11 A
∆
45
∆2 =
13 30
−2 −20
= −260 + 60 = −200
∆ 2 −200
=
= −4.44 A
45
∆
Current flowing through 10 Ω resistance is given as
I2 =
Therefore,
= I1 − I2
= −1.11 − (−4.44) = −1.11 + 4.44 = 3.33 A (in downward direction)
Example 2.11 In the circuit shown in Figure 2.21, given that current through 5V source is
zero, find the unknown voltage V by using mesh analysis.
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40
Network Analysis and Synthesis
1Ω
1Ω
1Ω
5V
1Ω
1Ω
1Ω
1 Ω Mesh II I2
1Ω
1Ω
Mesh III I3
1Ω
5 V Mesh I
1Ω
1Ω
I1
V
V
Figure 2.21
Figure 2.22
Solution: Let currents I1, I2 and I3 be flowing in three meshes in clockwise direction, as shown
in Figure 2.22.
By applying KVL in mesh I, we get the following equation
or
or
−1I1 − 1(I1 − I2) − 1(I1 − I3) − V + 5 = 0
1I1 + 1(I1 − I2) + 1(I1 − I3) = 5 − V
3I1 − I2 − I3 = 5 − V
(2.23)
By applying KVL in mesh II, the following equation is obtained.
or
or
−1(I2 − I1) − 1I2 − 1(I2 − I3) = 0
1(I2 − I1) + 1(I2 − I3) + 1I2 = 0
−I1 + 3I2 − I3 = 0
(2.24)
By applying KVL in mesh III, we obtain the equation,
or
or
−1(I3 − I1) − 1(I3 − I2) − 1I3 + V = 0
1(I3 − I1) + 1(I3 − I2) + 1I3 = V
−I1 − I2 + 3I3 = V
(2.25)
Now, let us determine I1 by Cramer’s rule:
3 −1 −1
∆ = −1 3 −1
−1 −1 3
Therefore,
= 3(9 − 1) + ( −3 − 1) − 1(1 + 3)
= 24 − 4 − 4 = 16
5 − V −1 −1
5 − V −1 −1
∆1 (for I1 ) = 0
3 −1
∆1 (for I1 ) = 0
3 −1
V
−1 3
−1 3
V
= (5 − V )(9 − 1) + 1(0 + V ) − 1(0 − 3V )
= (5 − V )(9 − 1) + 1(0 + V ) − 1(0 − 3V )
= 8(5 − V ) + V + 3V
= 8(5 − V ) + V + 3V
= 40 − 4V
= 40 − 4V
So
So
∆
40 − 4V 4(10 − V ) 10 − V
I1 = ∆1 = 40 − 4V = 4(10 − V ) = 10 − V
1
16
I1 = ∆ = 16 =
= 4
∆
16
16
4
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Kirchhoff’s Laws, Mesh and Nodal Analysis 41
Now, according to the question, current through 5 V source is zero.
That is, I1 = 0.
10 − V
=0
4
10 − V = 0
V = 10
So,
or
or
Therefore, unknown voltage is 10 V.
Example 2.12 In the network shown in Figure 2.23, find the current in the branch (2 + j3) Ω.
5Ω
2Ω
j3Ω
4Ω
j5Ω
30∠0° V
6Ω
35.36∠45°
Figure 2.23
Solution: The voltage sources are expressed as
30∠0° = 30 (cos 0° + j sin 0°) = 30 (1 + 0) = 30 V
and
35.36∠45° = 35.36 (cos 45° + j sin 45°)
= 35.36 cos 45° + j 35.36 sin45°
1
= 35.36 ×
2
+ j 35.36 ×
1
2
= ( 25 + j 25)V
The given circuit is redrawn as in Figure 2.24.
5Ω
2Ω
j3Ω
j5Ω
30 V
I1
4Ω
6Ω
I2
(25 + j 25)V
I3
Figure 2.24
We have to find the current through (2 + j3) Ω, that is, current I2.
For this, firstly, let us write the mesh equations.
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Network Analysis and Synthesis
By applying KVL in mesh I, we get the following:
or
or
−5I1 − j5(I1 − I2) + 30 = 0
5 I1 + j 5( I1 − I 2 ) = 30
I1 + j ( I1 − I 2 ) = 6
(1 + j )I1 − jI 2 = 6
or
(2.26)
By applying KVL in mesh II, the following set of equations are obtained.
−j5(I2 − I1) − 2I2 − j3I2 − 6(I2 − I3) = 0
or
j 5 ( I 2 − I1 ) + 2 I 2 + j 3I 2 + 6 ( I 2 − I 3 ) = 0
or
− j 5 I1 + (8 + j8) I 2 − 6 I 3 = 0
(2.27)
By applying KVL in mesh III, we get
−6(I3 − I2) − 4I3 − (25 + j25) = 0
or
6 ( I 3 − I 2 ) + 4 I 3 = −(1 + j )25
or
−6 I 2 + 10 I 3 = −(1 + j )25
(2.28)
By solving these three equations for I2 by Cramer’s rule, we can get the following:
−j
1+ j
0
∆ = − j 5 8 + j 8 −6
−6
0
10
= (1 + j ){10(8 + j 8) − 36} + j ( j 50 − 0) + 0
= (1 + j )((80 + j 80 − 36) + j 2 50
= (1 + j )( 44 + j 80) − 50
= 44 + j 80 + j 44 + j 2 80 − 50
= 44 + j1124 − 80 − 50
= −86 + j124
1+ j
∆ 2 (for I 2 ) = − j 5
0
6
0
0
−6
−(1 + j )25 10
= (1 + j ){0 − 150(1 + j )} − 6( − j 50 − 0) + 0
= (1 + j ){−150(1 + j )} + j 300
= −(1 + j ) 2 150 + j 300
= −(1 + j 2 + 2 j )150 + j 300
= −(1 − 1 + 2 j )150 + j 300
= − j 300 + j 300 = 0
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Kirchhoff’s Laws, Mesh and Nodal Analysis 43
Therefore, I 2 =
∆2
0
=
= 0. Current flowing through the (2 + j3) Ω branch is zero.
∆ −86 + j 24
2.3 NODAL ANALYSIS
A
R1
V1
B
R3
V2
C
R5
D
Nodal analysis is a technique used to find
I4 I
I1
I3
5
the voltage at various nodes of an electric
circuit. This can be done by using KCL at
R2
R4
2A
various nodes. The application of KCL at
I2
each node will give the node equation.
A node is a point in an electric circuit at
H
G
F
E
which current divides. For example, conFigure 2.25 Nodal Analysis of an Electric
sider the circuit shown in Figure 2.25.
Circuit
In this circuit, points A and D are not
nodes because current is not getting divided here. However, points B and C are nodes because
at these points current is dividing.
At node B, by applying KCL, we have, I1 = I 2 + I 3
At node C, by applying KCL, we have, I 3 = I 4 + I 5
In the nodal analysis, unknown voltages are assumed at the nodes in the circuit. Points E, F,
G and H in this circuit are considered as ground nodes and their potential is considered as zero.
Let the voltages at nodes B and C be V1 and V2 respectively.
Node voltage equations are formed by applying KCL, as given in the following.
At node B, current I1 is determined as,
I1 = I2 + I3
or
2=
V1 − 0 V1 − V2
+
R3
R2
(2.29)
(where I1 = 2 A; as current I2 is flowing from node B to node G, their corresponding potential
is V1 and 0. Further, current I3 is flowing from node B of potential V1 to node C of potential V2)
At node C, the current I3 is given as follows:
I3 = I4 + I5
or
V1 − V2 V2 − 0 V2 − 0
=
+
R3
R4
R5
(2.30)
By solving equation (2.29) and (2.30), we can determine V1 and V2 and hence I2, I3, I4, and I5.
Further, we can also calculate the dissipation of power in the circuit elements.
Example 2.13 Find the current in 50 Ω resistance in the network shown in Figure 2.26, using
the nodal analysis.
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44
Network Analysis and Synthesis
Solution: In the given circuit, there is one
node; that is, node A.
Let the voltage at node A be V1, as shown
in the circuit. Let I1, I2 and I3 be the currents
as shown in Figure 2.27.
At node A, current I1 is calculated as
follows:
20
40
+
120 V
−
60
25
50
+
60 V
−
+
40 V
−
Figure 2.26
I1 = I2 + I3
or
or
or
or
or
20
120 − V1 V1 − 60 V1 − 40
=
+
40 + 20
25
60 + 50
I1
40
120 − V1 V1 60 V1 − 40
=
− +
60
25 25
110
+
120 V
V
V 60 V
4
2− 1 = 1 − + 1 −
60 25 25 110 11
A(V1)
50
25
I2 +
60 V
−
0V
−
60
I3
+
40 V
−
Figure 2.27
1
1
1
60 4
+
+ ) = 2+
+
25 110 60
25 11
= 2 + 2.4 + 0.36
V1 (0.04 + 0.009 + 0.0167) = 4.76
V1 (
V1 (0.06584) = 4.76
or
V1 = 72.3 V
Current through the 50 Ω resistance, I3 =
V1 − 40 72.34 − 40
= 0.294 A .
=
110
110
Example 2.14 Using nodal analysis, determine the current in 2 Ω resistance in the circuit
shown in Figure 2.28.
8A
8Ω
100 V +
−
2Ω
4Ω
10 Ω
3Ω
5Ω
Figure 2.28
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Kirchhoff’s Laws, Mesh and Nodal Analysis 45
Solution: The circuit is redrawn with assumed current directions, as shown in Figure 2.29
I5
V2
V1
B
A
2Ω
I1
4Ω
D
10 Ω
I3
I2
V3
I6
C
8Ω
100 V +
−
8A
3Ω
I4
5Ω
I7
0V
Figure 2.29
Let us consider currents I1, I2, I3, I4, I5, I6 and I7 as shown in the Figure.
Clearly, at points B, C and D, current is dividing; therefore, these are nodes.
Let the voltages at nodes B, C and D be V1, V2 and V3, respectively.
By applying KCL at node B, we can obtain the following:
I1 = I 2 + I 3
or
or
or
or
or
100 − V1 V1 − 0 V1 − V2
=
+
4
2
8
100 − V1 V1 + 2(V1 − V2 )
=
4
8
100 − V1
= V1 + 2(V1 − V2 )
2
100 − V1 = 2V1 + 4 V1 − 4 V2
7V1 − 4V2 = 100
(2.31)
By applying KCL at node C, the following is obtained.
I3 = I 4 + I5 + I6
or
V −V
V1 − V2 V2 − 0
=
+ ( −8) + 2 3
2
3
10
or
V1 − V2 V2 − 24 V2 − V3
=
+
2
3
10
or
or
or
V1 − V2 10 V2 − 240 + 3V2 − 3V3
=
2
30
15 V1 − 15 V2 = 13 V2 − 3 V3 − 240
15 V1 − 28 V2 + 3 V3 = −240
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 45
(2.32)
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46
Network Analysis and Synthesis
By applying KCL at node D, we can obtain the following:
I5 + I6 = I7
−8 +
or
V2 − V3 V3 − 0
=
10
5
or
−80 + V2 − V3 V3
=
10
5
or
−80 + V2 − V3 = 2V3
or
V2 − 3V3 = +80
(2.33)
Now, according to the question, we have to find the current in 2 Ω resistance; that is, current I3
for which we need the values of V1 and V2.
Therefore, let us solve the node voltage equations (2.31), (2.32) and (2.33) for V1 and V2 by
Cramer’s rule:
7 −4 0
∆ = 15 −28 3
0
1 −3
= 7(84 − 3) + 4( −45 − 0) + 0
= 567 − 180 = 387
and
100
−4 0
∆1 = −240 −28 3
80
1 −3
= 100(84 − 3) + 4(720 − 240) + 0
= 810 + 1920 = 2730
V1 =
∆1 2730
=
= 7.05 V
∆
387
For V2, and ∆2 calculations are made follows:
7 100
0
∆ 2 = 15 −240 3
0
80
−3
= 7(720 − 240) − 100( −75 − 0) + 0
= 3360 + 7500 = 10860
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Kirchhoff’s Laws, Mesh and Nodal Analysis 47
Therefore, V2 can be determined as in the following:
V2 =
Therefore,
∆ 2 10860
=
= 28.06 V
∆
387
Current through 2 Ω resistance, I3 =
Example 2.15 Find the voltages of
nodes 1 and 2 in the network shown in
Figure 2.30.
Solution: Let voltages at nodes 1 and 2
be V1 and V2, as shown in Figure 2.31.
By applying KCL at node 1, we can
calculate I1 as follows:
V1 − V 2 7.05 − 28.06
=
= 10.5 A.
2
2
10 Ω
50 − V1 V1 − 0 V1 − V2
=
+
j5
2
10
−j 10 Ω
50∠0° V
j4Ω
Figure 2.30
V2
2
10 Ω
50 − V1 2V1 + j 5V1 − j 5V2
=
j10
10
or
j 50 − jV1 = 2V1 + j 5V1 − j 5V2
( 2 + j 6)V1 − j 5V2 = j 50
3Ω
V1
1
or
or
2
j5Ω
I1 = I2 + I3
or
2Ω
1
2Ω
I1
50∠0° V
j5Ω
3Ω
I3
I2
−j 10 Ω
I4
j4Ω
I5
Figure 2.31
(2.34)
By applying KCL at node 2, we obtain the following equation:
I3 = I4 + I5
or
V1 − V2 V2 − 0 V2 − 0
=
=
2
3 + j 4 − j10
or
V1 − V2 − j10V2 + (3 + j 4)V2
=
2
− j10(3 + j 4)
or
− j 5(3 + j 4)(V1 − V2 ) = − j10V2 + 3V2 + j 4V2
or
( − j15 + 20)(V1 − V2 ) = − j10V2 + 3V2 + j 4V2
− j15V1 + 115V2 + 20V1 − 20V2 = − j10V2 + 3V2 + j 4V2
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48
Network Analysis and Synthesis
( 20 − j15)V1 + ( −23 − j 21)V2 = 0
or
(2.35)
Now, we solve equations (2.34) and (2.35) for V1 and V2 using Cramer’s rule.
∆=
2 + j6
− j5
20 − j15 −23 − j 21
= ( 2 + j 6)( −23 − j 21) + j 5( 20 − j15)
= −46 − j 42 − j138 + 126 + j100 + 75
= 155 − j80
∆1 =
j 50
− j5
0 −23 − j 21
= j 50( −23 − j 21)
= 1050 − j1150
V1 =
∆1 1050 − j1150
=
V = 8.9∠− 20.3° V
∆
155 − j 80
∆2 =
2 + j6
20 − j15
j 50
0
= − j 50( 20 − j15)
= −750 − j1000
V2 =
Therefore,
∆ 2 −750 − j1000
=
V = 7.167∠154.3°V
∆
155 − j 80
Example 2.16 In the network of Figure 2.32, use the node voltage analysis to find Vx.
2Ω
2Ω
j2Ω
10∠0 V
2Ω
j2Ω
j 2 Ω Vx
Figure 2.32
Solution: In the given network, there are two nodes.
Let node voltages be V1 and V2 as shown in Figure 2.33.
I1 = I2 + I3
or
10 − V1 V1 − 0 V1 − V2
=
+
2
j2
2
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Kirchhoff’s Laws, Mesh and Nodal Analysis 49
V1
+ V1 − V2
j
or
10 − V1 =
or
10 − V1 = − jV1 + V1 − V2
or
( 2 − j )V1 − V2 = 10
2Ω
Node 1
2Ω
V1
I1
(2.36)
I3
j2Ω
10∠0 V
I2
By applying KCL at node 2, we get current
I3 as follows:
or
j2Ω
I4
j2Ω
I5
Figure 2.33
I3 = I4 + I5
or
Node 2
2Ω
V2
V1 − V2 V2 − 0 V2 − 0
=
+
2
j2
2 + j2
V2 V2
V1 − V2 =
+
j 1+ j
(1 + j )(V1 − V2 ) = − j (1 + j )V2 +V2
or
or
(1 + j )V1 − (1 + j )V2 = − jV2 + V2 + V2
(1 + j )V1 − (2 − j )V2 − (1 + j )V2 = 0
or
(1 + j )V1 + ( − 3)V2 = 0
or
(1 + j )V1 − 3V2 = 0
(2.37)
Therefore, the node voltage equations are given as in the following:
( 2 − j )V1 − V2 = 10
(1 + j )V1 − 3V2 = 0
Let us solve the two node voltage equations for V1 and V2 by Cramer’s rule.
∆=
2− j
1+ j
−1
−3
= −3( 2 − j ) + (1 + j )
= −5 + j 4
∆1 =
10 −1
0 −3
= −30
∆2 =
2 − j 10
1+ j 0
= 0 − 10(1 + j )
= −10(1 + j )
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50
Network Analysis and Synthesis
Therefore,
Now, I 5 =
=
V1 =
∆1
−30
=
and
∆ −5 + j 4
V2 =
∆ 2 −10(1 + j)
=
−5 + j 4
∆
V2
−10(1 + j )
−10(1 + j )
=
=
2 + j 2 ( −5 + j 4)( 2 + j 2) ( −5 + j 4)2(1 + j )
−5
−5
1
A=
=
−5 + j4
−5(1 − j 0.8) 1 − j 0.8
Therefore,
Vx = I5 ( j 2)
By substituting the value of I5 in the equation, we get the following:
Vx =
Vx =
1
× j2
1 − j 0.8
2∠90°
−1
1.28∠ − tan 0.8
=
2∠90°
= 1.56 V
1.28∠ − 39°
Example 2.17 Find the node voltages in the
network shown in Figure 2.34.
Solution: In the given network, there are two nodes
as shown in Figure 2.35.
Let I1, I2, I3, I4 and I5 be the currents flowing in the
circuit as shown in the Figure.
Let node voltages be V1 and V2.
By applying KCL at node 1, we can obtain current
I1 as follows;
5A
4Ω
6Ω
2Ω
10 A
Figure 2.34
I1 = I2 + I3
or
0 − V1 V1 − V2
=
+ ( −5)
2
4
or
−V2 V1 − V2 − 20
=
2
4
or
or
V1
Node 1
I2
2Ω
−2V1 = V1 − V2 − 20
3V1 − V2 = 20
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 50
I1
(2.38)
V2
Node 2
5A
I3
4Ω
I4
6Ω
I5
10 A
Figure 2.35
11/25/2014 3:43:36 PM
Kirchhoff’s Laws, Mesh and Nodal Analysis 51
By applying KCL at node 2, the following is obtained:
I2 + I3 = I4 + I5
V1 − V2
V −0
+ ( −5) = 2
+ ( −10)
4
6
V1 − V2 − 20 V2 − 60
=
or
4
6
or
6V1 − 6V2 − 120 = 4V2 − 240
or
6V1 − 10V2 = −120
Let us solve node equations (2.38) and (2.39) by Cramer’s rule.
or
∆=
(2.39)
3 −1
6 −10
= −30 + 6 = −24
∆1 =
20
−1
−120 −10
= −200 − 120 = −320
∆2 =
3 20
6 −120
= −360 − 120 = −480
V1 =
∆1 −320
=
= 13.33 V
∆
−24
V2 =
∆ 2 −480
=
= 20 V
∆
−24
Example 2.18 Find the current in the 15 Ω resistance in the circuit shown in Figure 2.36 using
nodal analysis.
30 Ω
10 Ω
20 Ω
5Ω
10 V
15 Ω
12 V
9V
Figure 2.36
Solution: In the given network, there are two nodes.
Let the node voltages be V1 and V2 as shown in Figure 2.37.
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52
Network Analysis and Synthesis
30 Ω
V1
10 Ω
I1
I3
5Ω
15 Ω
10 V
I2
20 Ω
V2
I4
I5
12 V
9V
Figure 2.37
Since, we have to find the current through the 15 Ω resistance; we need the value of V1.
Let us find V1 using the nodal analysis
By applying KCL, we can determine I1 at node 1 as follows:
I1 = I2 + I3
10 − V1 V1 − 0 V1 − V2
=
+
15
10
30
or
10 − V1 V1 V1 − V2
= +
6
3
2
or
10 − V1 2V1 + 3V1 − 3V2
=
6
6
or
10 − V1 = 5V1 − 3V2
or
6V1 − 3V2 = 10
(2.40)
By applying KCL at node 2, we can obtain the following:
I3 = I4 + I5
or
V1 − V2 V2 − 9 V2 − 12
=
+
10
5
20
or
V1 − V2 V2 − 9 V2 − 12
=
+
2
1
4
or
V1 − V2 4V2 − 36 + V2 − 12
=
2
4
or
2(V1 − V2 ) = 5V2 − 48
or
2V1 − 7V2 = −48
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 52
(2.41)
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Kirchhoff’s Laws, Mesh and Nodal Analysis 53
Let us solve equations (2.40) and (2.41) for V1 using the Cramer’s rule.
∆=
6 −3
2 −7
= −42 + 6 = −36
∆1 =
10 −3
−48 −7
= −70 − 144 = −214
V1 =
∆1 214
=
= 5.94 V.
∆
36
Current through the 15 Ω resistance,
I2 =
V1 − 0 5.94 − 0
=
= 0.396 A
15
15
2.4 SUPER NODAL ANALYSIS
If a voltage source comes in between the two nodes, then we remove the voltage source and
replace it by a short circuit and combine the two nodes; this is called a super node.
1Ω
4Ω
Let us consider an example.
A
B
C
D
Example 2.19 Determine the current
through 2 Ω resistance shown in Figure 2.38
using nodal analysis.
Solution: In the given network, there are
two nodes at B and C.
Let the voltages at nodes B and C be V1
and V2, respectively, as shown in Figure 2.39.
By applying KCL at node B, we can calculate current I1 as follows:
5V
2Ω
5Ω
H
G
I1 = I2 + I4 + I5
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 53
F
E
Figure 2.38
A
V1
B
1Ω
V2
C
2Ω
10 A
I3
D
I5
3Ω
I2
H
4Ω
5V
I1
I1 = I2 + I3
As there comes a voltage source of 5 V in
between nodes B and C, we will combine the
nodes and use super node analysis. Applying
KCL, we will have
3Ω
10 A
5Ω
I4
G
F
E
Figure 2.39
11/25/2014 3:43:40 PM
54
Network Analysis and Synthesis
or
10 =
V1 − 0 V2 − 0 V2 − 0
+
+
2
3
9
or
10 =
9V1 + 6V2 + 2V2
18
180 = 9V1 + 8V2
or
or
Further,
9V1 + 8V2 = 180
(2.42)
V1 − V2 = 5.
(2.43)
Now, let us solve the nodal equations (2.42) and (2.43) using Cramer’s rule.
∆=
9 8
1 −1
= −9 − 8
= −17
∆1 =
180 8
5 −1
= −180 − 40
= −220
∆2 =
9 180
1 5
= 45 − 180
= −135
Therefore,
Current in 2Ω resistance = I2
=
V1 =
∆1 220
=
= 12.94 V
∆
17
V2 =
∆ 2 135
=
= 7.94 V .
∆
17
V1 12.94
=
= 6.47 A
2
2
2.5 SUPER MESH ANALYSIS
If a current source comes between the two meshes, then we remove the current source and
replace it by a open circuit and combine the two meshes; this is called a supermesh.
Let us consider an example.
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Kirchhoff’s Laws, Mesh and Nodal Analysis 55
Example 2.20 Find the current through branch
BC in the network shown in Figure 2.40 using the
mesh analysis.
B
Figure 2.40
A
2(I1 − I2) = 5
2 I1 − 2 I 2 = 5
(2.44)
Now, as there is a current source of 5 A in
between mesh II and mesh III, we can remove
branch CF containing the current source and
consider the super mesh BCDEFGB as shown
in Figure 2.42.
By applying KVL on super mesh, we obtain
the following:
5V
1Ω
B
C
2Ω
+
−
G Mesh II
1Ω
I3
F Mesh III E
Figure 2.41
B
A
C
D
1Ω
2Ω
5V
H Mesh I
(2.45)
1Ω
I2
I3
I1
2(I2 − I1) + 1I2 + 1I3 = 0
−2I1 + 3I2 + I3 = 0
D
5A
I2
I1
H Mesh I
−2(I2 − I1) − 1I2 − 1I3 = 0
or
or
1Ω
5A
1Ω
−2(I1 − I2) + 5 = 0
or
2Ω
+
5V
−
Solution: Let I1, I2 and I3 be the currents flowing
in three meshes in clockwise direction as shown
in Figure 2.41.
By applying KVL in mesh I, we get the
following:
or
C
1Ω
G
F
E
Mesh II & Mesh III
Supermesh
Since there are three meshes in the given cirFigure 2.42
cuit, there must be three mesh equations.
Now, the third mesh equation can be
formed from branch CF (that was removed).
The current in branch CF is I2 − I3 (downward) or I3 − I2 (upward).
Further, given that the current in branch CF is 5 A (upward), we can state the following:
I3 − I2 = 5
−I2 + I3 = 5
or
(2.46)
Now, let us solve equations (2.44), (2.45) and (2.46) for I2 by Cramer’s rule.
2 −2 0
∆ = −2 3 1
0 −1 1
= 2(3 + 1) + 2( −2 − 0) + 0
= 8−4 = 4
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56
2 −2 0
∆ = −2 3 1
0 −1 1
Network Analysis and Synthesis
= 2(3 + 1) + 2( −2 − 0) + 0
= 8−4 = 4
2 5 0
∆ 2 (for I 2 ) = −2 0 1
0 5 1
= 2(0 − 5) − 5( −2 − 0) + 0
= −10 + 10 = 0
Therefore,
I2 =
∆2 0
= = 0.
∆ 4
Hence, current through branch BC = I2 = 0 A
2.6 METHODS OF SOLVING COMPLEX NETWORK PROBLEMS
In this section, we have taken up large number of complex network problems having either DC
or AC excitations (voltage or current).
2.6.1 Numerical Problems Based on Kirchhoff’s Laws
Example 2.21 Calculate the voltage Vs across the open switch
in the circuit of Figure 2.43.
A
Solution: By applying KVL, starting from point A in the
clockwise direction and considering the polarity of sources, we
get the following:
or
or
B
50 V
20 V
D
−50 + 30 + Vs + 10 − 20 = 0
− 30 + Vs = 0
Vs = 30 V
10 V
Solution: By applying KVL in loop ABCDEFA, we get the
following:
or
or
−16 × 3 − 4 × 2 + 40 − V1 = 0
−48 − 8 + 40 = V1
V1 = −16 V
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 56
−
Vs
+
C
Figure 2.43
A
Example 2.22 Find the unknown voltage V1 in the circuit
of Figure 2.44.
30 V
V1
−
B
C
R1
+
16 A
+
40 V
R2
F
3Ω
10 A
−
2
4A
E
D
Figure 2.44
11/25/2014 3:43:45 PM
Kirchhoff’s Laws, Mesh and Nodal Analysis 57
Example 2.23 Find the voltage between C
and E in the circuit shown in Figure 2.45.
Solution: The given circuit can be redrawn
as shown in Figure 2.46.
Let current i1 is flowing in mesh I and i2
in mesh II.
By applying KVL in mesh I, we obtain the
following:
6Ω
A
D
20i1 = −20
or
40 V
H
G
Figure 2.45
9Ω
D
i1 = −1A
Further, by applying KVL in mesh II, we get the
equation as follows:
−8i2 − 5i2 + 40 = 0
or
8i2 + 5i2 = 40
C
F
5Ω
10 V
5Ω
9Ω
8Ω
E
20 V
−9I1 − 5I1 − 6I1 − 20 = 0
or,
B
C
20 V i1
8Ω
E
5 Ω i2 40 V
5Ω
Mesh I
A
F
Mesh II
B
6Ω
10 V
H
G
Figure 2.46
40
or
i2 =
= 3.07 A
13
The direction of current i1 is opposite to the assumed direction.
Potential of E wrt. H
= 5i2 = 5 × 3.07
= 15.35V
Potential of C wrt. H = +10 − 5i1
= 10 − 5 × 1 = 5V
Therefore, voltage between C and E = 5 −15.35 = −10.35V. That is point E is at higher potential
than point C.
1Ω
Example 2.24 Determine the branch current in the
network, as shown in Figure 2.47.
1Ω
5V
Solution: We are assuming current directions in the
branches as shown in Figure 2.48.
1Ω
( I1 − I 3 ) + ( I 2 + I 3 ) = I1 + I 2 = I
The direction of currents are assumed, as shown in Figure 2.48.
By applying KVL in closed circuit CDGC, we obtain the
following:
5 − 1I1 − 1I 3 + 1I 2 = 0
or
I1 − I 2 + I 3 = 5
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 57
5V
1Ω
1Ω
1Ω
10 V
Figure 2.47
(2.47)
11/25/2014 3:43:48 PM
58
Network Analysis and Synthesis
D
I1 − I3
1Ω
5V
B C I1
I2
I
By applying KVL in closed circuit DEGD, we get the equation as follows:
1Ω
−1(I1 − I 3 ) + 5 + 1(I 2 + I 3 ) + 1I 3 = 0
5V
E
I3
1Ω
I2 + I3
1Ω
I
1Ω
or
I1 − I 3 − I 2 − I 3 − I 3 = 5
or
I1 − I 2 − 3I 3 = 5
By applying KVL in closed circuit BCGEFAB, the following can be obtained:
−1I 2 − 1(I 2 + I 3 ) + 10 − 1I = 0
G
A
1Ω
10 V
(2.48)
F
By substituting I = I1 + I2 in the equation, we get the
following:
Figure 2.48
I 2 + I 2 + I 3 + I1 + I 2 = 10
or
I1 + 3I 2 + I 3 = 10
(2.49)
By solving equations (2.47), (2.48) and (2.49) by Cramer’s rule, the branch currents can be
calculated as follows:
1 −1 1
∆ = 1 −1 −3
1 3 1
= 1( −1 + 9) + 1(1 + 3) + 1(3 + 1)
= 8 + 4 + 4 = 16
Therefore,
5 −15 1−1 1
∆1 = ∆
5 1 =−15 −−
31 −3
10 310 13 1
= 5( −1=+59()−+1 1+(59)++30
(15) + 110
) + 10)
1()5++130
(15
= 40 +=3540++2535=+100
25 = 100
So
So
∆
100
∆
100
I1 = I11 == 1 == 6.25=A6.25A
16
∆
∆
16
and
1 5 1
∆ 2 = 1 5 −3
1 10 1
= 1(5 + 30) − 5(1 + 3) + 1(10 − 5)
= 35 − 20 + 5 = 20
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 58
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Kirchhoff’s Laws, Mesh and Nodal Analysis 59
I2 =
∆ 2 20
=
= 1.25 A
∆ 16
1 −1 5
∆ 3 = 1 −1 5
1 3 10
= 1( −10 − 15) + 1(10 − 5) + 5(3 + 1)
= −25 + 5 + 20 = 0
I3 =
∆3
= 0.
∆
Current through branch CD = I1 = 6.25 A
Current through branch DE = I1 − I 3 = 6.25 − 0 = 6.25 A
Current through branch DG = I 3 = 0
Current through branch CG = I 2 = 1.25 A
Current through branch EG = I 2 + I 3 = 1.25 + 0 = 1.25 A
Current through branch EFAB = I = I1 + I 2 = 6.25 + 1.25 = 7.5 A
Example 2.25 Determine the value of I1 , I 2 and I in the
network using Kirchhoff’s Laws, as shown in Figure 2.49.
Solution: From the given circuit, and by applying KCL at
node K, we obtain the following:
I1 + I 2 = I
12 V
+ −
B
A
(2.50)
K
2Ω
I1
I
Now, by applying KVL in circuit BCEFB, the following
equations can be formed.
L
8V
+ −
F I2
C
1Ω
D
E
10 Ω
Figure 2.49
−12 + 2I1 − 1I 2 + 8 = 0
2I1 − I2 = 4
or
(2.51)
By applying KVL in closed circuit AKFEDA, we calculate the following:
−8 + 1I 2 + 10I = 0
By using equation (2.50), the equation can be rewritten as follows:
10( I1 + I 2 ) + I 2 = 8
or
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 59
10 I1 + 11I 2 = 8
(2.52)
11/25/2014 3:43:53 PM
60
Network Analysis and Synthesis
By solving equations (2.51) and (2.52) using Cramer’s rule, we can calculate the currents as follows:
2
10
4
∆1 =
8
∆=
∆2 =
I1 =
Therefore,
−1
= 22 − ( −10) = 32
11
−1
= 44 − ( −8) = 52
11
2 4
= 16 − 40 = −24
10 8
∆1 52
=
= 1.625 A
∆ 32
∆ 2 −24
=
= −0.75 A
∆
32
Therefore, the required currents are as follows:
I2 =
and
I1 = 1.625 A
I 2 = −0.75 A
I = I1 + I 2 = 1.625 − 0.75 = 0.875 A
2.6.2 Numerical Problems Based on Mesh and Nodal Analysis
Example 2.26 Using the mesh analysis, find the current in branch DG in the network shown
in Figure 2.50.
Solution: Let i1, i2 and i3 be the currents flowing in three meshes, as shown in Figure 2.51.
By applying KVL in mesh I, that is, by equating voltage drops to voltage rise, we get
1i1 + 1(i1 − i2 ) + 1(i1 − i3 ) = 5
or
D
5V
B
5V
C
1Ω
1i2 + 1(i2 − i3 ) + 1(i2 − i1 ) = 5
E
1Ω
or −i1 + 3i2 − i3 = 5
1Ω
1Ω
10 V
Figure 2.50
(2.54)
By applying KVL in mesh III, that is, by equating voltage drops to voltage rise, we get
G
A
(2.53)
By applying KVL in mesh II, that is, by equating voltage
drops to voltage rise, we get
1Ω
1Ω
3i1 − i2 − i3 = 5
F
1(i3 − i2 ) + 1i3 + 1(i3 − i3 ) = 10
or
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 60
−i1 − i2 + 3i3 = 10
(2.55)
11/25/2014 3:43:55 PM
Kirchhoff’s Laws, Mesh and Nodal Analysis 61
By solving equations (2.53), (2.54) and (2.55) using
Cramer’s rule, we calculate ∆, ∆1, ∆2 and ∆3 as follows:
3 −1 −1
∆ = −1 3 −1
−1 −1 3
D
1Ω
1Ω
5V
B
5V
C
i1
Mesh I
= 3(9 − 1) + 1( −3 − 1) − 1(1 + 3)
i2
1Ω
Mesh II
1Ω
= 24 − 4 − 4 = 16
1Ω
i3
A
1Ω
i3
Mesh III
G
5 −1 −1
∆1 = 5 3 −1
10 −1 3
E
F
10 V
Figure 2.51
= 5(9 − 1) + 1(15 + 10) − 1( −5 − 30)
= 40 + 25 + 35 = 100
3 −1 5
∆ 3 = −1 3 5
−1 −1 10
3 5 −1
∆ 2 = −1 5 −1
−1 10 3
= 3(15 + 10) − 5( −3 − 1) − 1( −10 + 5)
= 3(30 + 5) + 1( −10 + 5) + 5(1 + 3)
= 75 + 20 + 5 = 100
= 105 − 5 + 20 = 120
i1 =
∆
∆1 100
∆
100
120
=
A; i 2 = 2 =
A; i3 = 3 =
A
∆
∆
16
16
16
∆
Current in branch DG = i1 − i2 =
100 100
−
=0
16 16
Example 2.27 Using the mesh analysis, determine
the voltage across 5 Ω resistance in the network shown in
Figure 2.52.
9A
Solution: The circuit is redrawn as shown in Figure 2.53.
From mesh I, it is clear that i1 can be taken as follows:
i1 = 9 A
(2.56)
By applying KVL in mesh II, we obtain the following
equations:
or
−4(i2 − i1 ) − 5(i2 − i4 ) − 2(i2 − i3 ) = 0
+4i1 − 11i2 + 2i3 + 5i4 = 0
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 61
4Ω
2Ω
4Ω
+
12 V
−
5Ω
100 Ω
20 Ω
Figure 2.52
11/25/2014 3:43:57 PM
62
Network Analysis and Synthesis
9A
By substituting i1 = 9 A in the equation, we can get the
equation as follows:
i1
+36 − 11i2 + 2i3 + 5i4 = 0
Mesh I
4Ω
2Ω
+
12 V
i3
−
Mesh III
5Ω
100 Ω
Mesh IV
Figure 2.53
or
i4
(2.57)
By applying KVL is mesh III, the following equations
can be obtained:
Mesh II
4Ω
11i2 − 2i3 − 5i4 = 36
or
i2
−4i3 − 2(i3 − i 2 ) − 100(i3 − i 4 ) + 12 = 0
20 Ω
or
−4i3 − 2i3 + 2i 2 − 100i3 + 100i 4 + 12 = 0
or
2i 2 − 106i3 + 100i 4 + 12 = 0
−2i 2 + 106i3 − 100i 4 = 12
(2.58)
By applying KVL is mesh IV, the following equations can be calculated as follows:
−5(i4 − i2 ) − 20i4 − 100(i4 − i3 ) = 0
−5i2 − 100i3 + 125i4 = 0
or
(2.59)
Therefore, the three equations are as follows:
11i2 − 2i3 − 5i4 = 36
−2i2 + 106i3 − 100i4 = 12
−5i2 − 100i3 + 125i4 = 0
By solving equation (2.57), (2.58) and (2.59) using Cramer’s rule, the following set of equations
can be calculated as follows:
11
−2
−5
∆ = −2 106 −100
−5 −100 125
= 11(106 × 125 − 100 2 ) + 2( −250 − 500) − 5( 200 + 530)
= 30, 600
−5
36 −2
∆1 = 12 106 −100
0 −100 125
= 36(106 × 125 − 100 2 ) + 2(12 × 125 − 0) − 5( −1200 − 0)
= 126, 000
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Kirchhoff’s Laws, Mesh and Nodal Analysis 63
11
−2 36
∆ 2 = −2 106 12
−5 −100 0
= 11(0 + 1200) + 2(0 + 60) + 36( 200 + 530)
= 39, 600
i2 =
∆1 126000
∆
39600
= 1.29 A
=
= 4.11 A and i 4 = 2 =
∆
30600
∆ 30600
10 V
The voltage across 5 Ω resistance = 5 (i2 − i4) = 5
(4.1176 − 1.29) = 14.138 V.
+
−
− +
20 Ω
8V
3Ω
i1
Example 2.28 Determine the current in 5 Ω resistor
in the circuit shown in Figure 2.54 by mesh analysis.
2Ω
Solution: By applying KVL in mesh I, that is, by
equating voltage drops to voltage rise, we get
+ −
12 V
i2
2Ω
i3
5Ω
Figure 2.54
20(i1 − i2 ) + 2(i1 − i3 ) = 8
or
22i1 − 20i2 − 2i3 = 8
or
11i1 − 10i2 − i3 = 4
(2.60)
Similarly by applying KVL in mesh II, we get
20(i2 − i1 ) + 3i2 + 2(i2 − i3 ) = 10
or
−20i1 + 25i2 − 2i3 = 10
(2.61)
By applying KVL in mesh III, we obtain the equation as follows:
2(i3 − i1 ) + 2(i3 − i2 ) + 5i3 = 12
−2i1 − 2i2 + 9i3 = 12
or
(2.62)
Let us find i3 using Cramer’s rule as in the following:
11 −10 −1
∆ = −20 25 −2
−2 −2 9
= 11( 25 × 9 − 4) + 10( −180 − 4) − 1( 40 + 50)
= 501
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 63
11/25/2014 3:44:02 PM
64
Network Analysis and Synthesis
11 −10 4
∆ 3 = −20 25 10
−2 −2 12
= 11( 25 × 12 + 20) + 10( −240 + 20) + 4( 40 + 50)
= 1680
i3 =
∆ 3 1680
=
= 3.35 A
501
∆
Therefore, the current through 5 Ω resistor = i3 = 3.35 A.
Example 2.29 Find the current in branch BG
of network shown in Figure 2.55 using mesh
analysis.
A
1Ω
H
i1 = 1 A
C
2Ω
1A
Solution: Let i1, i2 and i3 be the mesh currents as
shown in Figure 2.56.
From mesh I, current i1 is taken as follows:
2Ω
B
D
1Ω
G
2A
F
E
Figure 2.55
From mesh III, the value current i3 is written as
the following:
A
1Ω
2Ω
B
C
D
i3 = −2 A
By applying KVL in mesh II, we get the following:
i1
2(i2 − i1 ) + 2i2 + 1(i2 − i3 ) = 0
or
2Ω
1A
−2i1 + 5i2 − i3 = 0
i2
2A
i3
H Mesh I G Mesh II F Mesh III E
Figure 2.56
Substituting the value of i3 in the above equation,
we obtain i2 as follows:
or
1Ω
−2(1) + 5i2 − ( −2) = 0
i2 = 0
Therefore, the current in branch BG = i1 − i2 =
1 − 0 = 1 A.
Example 2.30 Using the mesh analyses, find
the current through 1 Ω resistor in the network
shown in Figure 2.57.
Solution: Let i1, i2 and i3 be the mesh currents
as shown in Figure 2.58.
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 64
1A
4V
+
2Ω
3Ω
1Ω
−
3A
Figure 2.57
11/25/2014 3:44:05 PM
Kirchhoff’s Laws, Mesh and Nodal Analysis 65
From mesh I, it is clear that i1 = 1 A; and
from mesh III, i3 = −3 A.
By applying KVL in mesh II, that is, by
equating voltage drops to voltage rise, we get
Mesh I
i1
4V
2(i2 − i1 ) + 1(i2 − i3 ) = 4
or
1A
2Ω
+
3Ω
1Ω
i2
−
−2i1 +3i2 − i3 = 4
i3
Mesh II
3A
Mesh III
Figure 2.58
By substituting the value of i1 and i3 in the
equation, we get
−2(1) + 3i2 − ( −3) = 4
or
−2 + 3i2 + 3 = 4
or
3i 2 = 3, i.e., i 2 = 1A
Therefore, the current through 1 Ω resistor = i2 − i3 = 1 − ( −3) = 4 A
5Ω
Example 2.31 Find the current through 8 Ω resistor
in the network shown in Figure 2.59 using nodal
analysis.
4Ω
3Ω
8Ω
5V
2Ω
20 V
Solution: Let node voltages be V1 and V2 as shown
in Figure 2.60.
By applying KCL at node 1, we write
I1 = I2 + I3
so,
and
analysis ]
or
or
or
or
V1
I1 3 Ω
20 V
20 − V1 V1 − 5 V2 − 30 V2 − 5
=
+
+
5
3
8
2
5V
Figure 2.59
5Ω
I3 = I4 + I5
I1 = I 2 + I 4 + I 5 [Applying supernodal
30 V
5V
4Ω
I3
I2
8Ω
V2
5V
2 Ω I5
30 V
I4
5V
5V
Figure 2.60
20 − V1 8V1 − 40 + 3V2 − 90 + 12V2 − 60
=
5
24
480 − 24V1 = 40V1 − 200 + 15V2 − 450 + 60V2 − 300
64V1 + 75V2 = 1430
(2.63)
Further, from the Figure, it is clear that
V1 − V2 = −5
Let us find V1 and V2 using Cramer’s rule.
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 65
(2.64)
11/25/2014 3:44:08 PM
66
Network Analysis and Synthesis
∆=
64 75
1 −1
∆1 =
= −64 − 75 = −139
∆2 =
1430 75
−5 −1
= −1430 + 375 = −1055
64 1430
1
−5
= −320 − 1430 = −1750
Therefore, V1 and V2 can be calculated as follows:
V1 =
∆1 −1055
∆
−1750
=
= 7.58 V and V 2 = 2 =
= 12.58 V
∆
∆
−139
−139
Now, the current through 8 Ω resistor I 5 =
V2 − 30 12.58 − 30
=
= −2.17 A
8
8
Example 2.32 Find the voltage drop across 4 Ω
resistance in the network shown in Figure 2.61 using
mesh analysis.
2Ω
+
Solution: Let i1, i2 and i3 be the mesh currents as
shown in Figure 2.62.
By applying KVL in mesh I, we get the following:
−
2Ω
+
12Ω
12 V
1Ω
3Ω
+ −
24 V
4Ω
−
10 V
2i1 + 12(i1 − i2 ) + 1(i1 − i3 ) = 12
or
15i1 − 12i2 − i3 = 12
(2.65)
Figure 2.61
By applying KVL in mesh II, the following equation
is obtained:
12(i2 − i1) + 2i2 + 3(i2 − i3) = −10
or
−12i1 + 17i2 − 3i3 = −10
Further, by applying KVL in mesh III, the following
equation is formed
1(i3 − i1) + 3(i3 − i2) + 4i3 = 24
or
−i1 − 3i2 + 8i3 = 24
12 V
+
2Ω
Mesh I
i1
−
Mesh II
12Ω
(2.67)
+ −
24 V
i2
+
−
10 V
3Ω
1Ω
Mesh III
We now solve equations (2.65), (2.66) and (2.67) for
i3 using Cramer’s rule.
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 66
2Ω
(2.66)
i3
4Ω
Figure 2.62
11/25/2014 3:44:10 PM
Kirchhoff’s Laws, Mesh and Nodal Analysis 67
15 −12 −1
∆ = −12 17 −3
−1 −3 8
= 15(17 × 8 − 9) + 12( −96 − 3) − 1(36 + 17) = 664
15 −12 12
∆ 3 = −12 17 −10
−1 −3 24
= 15(17 × 24 − 30) + 12( −12 × 24 − 10) + 12(36 + 17) = 2730
i3 =
∆ 3 2730
=
= 4.11 A
664
∆
Voltage drop across 4 Ω resistor = 4i3 = 4 × 4.11 = 16.44 V
Example 2.33 Find currents i1, i2 and i3 in the
network shown in Figure 2.63 using loop analysis.
10 Ω
Solution: By applying KVL in mesh I, that is, by
equating voltage drops to voltage rise, we get
or
10 Ω
i1
30i1 − 10i2 − 10i3 = 100
3i1 − i2 − i3 = 10
10 Ω i 10 Ω
3
+
−100 V
10i1 + 10(i1 − i3 ) + 10(i1 − i2 ) = 100
or
10 Ω
(2.68)
50 V
i2
Figure 2.63
Similarly by applying KVL in mesh II, we get
10(i2 − i1 ) + 10(i2 − i3 ) = −50
or
−10i1 + 20i2 − 10i3 = −50
−i1 + 2i2 − i3 = −5
or
(2.69)
Similarly by applying KVL in mesh III, the following equation is obtained
10(i3 − i1 ) + 10i3 + 10(i3 − i2 ) = 0
or
i3 − i1 + i3 + i3 − i2 = 0
or
−i1 − i2 + 3i3 = 0
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 67
(2.70)
11/25/2014 3:44:12 PM
68
Network Analysis and Synthesis
We now apply Cramer’s rule to find i1, i2 and i3.
3 −1 −1
∆ = −1 2 −1
−1 −1 3
10 −1 −1
∆1 = −5 2 −1
0 −1 3
= 10(6 − 1) + 1( −15 − 0) − 1(5 − 0)
= 50 − 15 − 5 = 30
= 3(6 − 1) + 1( −3 − 1) − 1(1 + 2)
= 15 − 4 − 3 = 8
3 10 −1
∆ 2 = −1 −5 −1
−1 0 3
3 −1 10
∆ 3 = −1 2 −5
−1 −1 0
= 3( −15 − 0) − 10( −3 − 1) − 1(0 − 5)
= 3(0 − 5) + 1(0 − 5) + 10(1 + 2)
= −45 + 40 + 5 = 0
= −15 − 5 + 30 = 10
Therefore,
i1 =
∆
∆1 30
∆
0
10
=
= 3.75 A; i 2 = 2 = = 0; and i3 = 3 =
= 1.25 A
∆
∆
8
8
8
∆
Example 2.34 Find the current in 3 Ω resistance
for the network shown in Figure 2.64 using mesh
analysis.
Solution: Let i1 and i2 be the mesh currents as
shown in Figure 2.65. By applying KVL in mesh I,
that is, by equating voltage drops to voltage rise, we
get
3Ω
2Ω
2V
2Ω
2Ω
4V
4V
Figure 2.64
2i1 + 3i1 + 2(i1 − i2 ) = 4 + ( 2)
or
7i1 − 2i2 = 6
(2.71)
By applying KVL in mesh II, that is, by equating
voltage drops to voltage rise, we get
i2 − i1 + i2 = −2
or
−i1 + 2i2 = −2
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 68
2Ω
2V
2Ω
2Ω
4V
2(i2 − i1 ) + 2i2 = −4
or
3Ω
4V
i1
Mesh I
(2.72)
i2
Mesh II
Figure 2.65
11/25/2014 3:44:14 PM
Kirchhoff’s Laws, Mesh and Nodal Analysis 69
We solve equations (2.71) and (2.72) using Cramer’s rule, as,
∆=
7 −2
−1 2
∆1 =
= 14 − 2 = 12
i1 =
6 −2
−2 2
∆2 =
= 12 − 4 = 8
7 6
−1 −2
= −14 + 6 = −8
∆1 8
∆
−8
=
= 0.66 A and i 2 = 2 =
= −0.66 A
∆ 12
∆ 12
Current in 3 Ω resistor = i1 = 0.66 A.
3Ω
Example 2.35 Find the current through RL in
the network, as shown in Figure 2.66, using mesh
analysis.
+
6V
−
or
3i1 − 2i2 = 2
6Ω
3Ω
(2.73)
6V
By applying KVL in mesh II and III togather
(super mesh analysis), that is, by equating all
the voltage drops to voltage rise, we get
RL = 3 Ω
3A
Figure 2.66
6 = 3i1 + 6(i1 − i2 )
9i1 − 6i2 = 6
1Ω
I1
Solution: Let mesh currents be i1, i2 and i3 as
shown in Figure 2.67.
By applying KVL in mesh I, that is, by equating voltage rise to voltage drops, we get
or
2Ω
+
−
2Ω
6Ω
i1
Mesh I
1Ω
RL= 3 Ω
3A
i2
i3
Mesh II
Mesh III
Figure 2.67
6(i2 − i1 ) + 2i2 + i3 + 3i3 = 0
or
−6i1 + 8i2 + 4i3 = 0
or
−3i1 + 4i2 + 2i3 = 0
(2.74)
Further, it is clear from the Figure that the value of (i3 − i2) can be taken as,
i3 − i2 = 3 A
Let us use Cramer’s rule to find i3.
3 −2 0
∆ = −3 4 2
0 −1 1
= 3( 4 + 2) + 2( −3 − 0) + 0
= 18 − 6 = 12
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 69
and
(2.75)
3 −2 2
∆ 3 = −3 4 0
0 −1 3
= 3(12 − 0) + 2( −9 − 0) + 2(3 − 0)
= 36 − 18 + 6 = 24
11/25/2014 3:44:17 PM
70
Network Analysis and Synthesis
Therefore,
i3 =
∆ 3 24
=
= 2A
∆ 12
Current through RL = i3 = 2 A.
Example 2.36 Find the current through
5 Ω resistor between A and B as shown in
Figure 2.68 using mesh analysis.
Solution: Let i1, i2 and i3 be the mesh
currents as shown in Figure 2.69
From mesh III, current i3 can be taken as,
i3 = 1 A
3Ω
5Ω
A
+
20 V
−
B
4Ω
5Ω
1A
(2.76)
Figure 2.68
By applying KVL in mesh I, that is, by equating voltage rise to voltage drops, we get
3Ω
5Ω
A
B
20 = 3i1 + 4(i1 − i2 )
or
7i1 − 4i2 = 20
(2.77)
+
20 V
−
4Ω
i1
Similarly by applying KVL in mesh II, we
can write
Mesh I
i2
Mesh II
1A
i3
Mesh III
Figure 2.69
4(i2 − i1 ) + 5i2 + 5(i2 − i3 ) = 0
or
5Ω
−4i1 + 14i2 − 5i3 = 0
By substituting i3 = 1, we get the following:
−4i1 + 14i2 − 5 = 0
−4i1 + 14i2 = 5
or
(2.78)
Let us solve equations (2.77) and (2.78) for i2 using Cramer’s rule.
∆=
7 −4
−4 14
= 98 − 16 = 82
∆2 =
7 20
−4 5
= 35 + 80 = 115
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Kirchhoff’s Laws, Mesh and Nodal Analysis 71
i2 =
∆ 2 115
=
= 1.4024 A
∆
82
Now, current through 5 Ω (between A and B) = i2 = 1.4024 A.
Example 2.37 Find the current through the j3 Ω inductive reactance shown in Figure 2.70
using the mesh analysis.
j3Ω −j5Ω
+
50 ∠ 30° V
10 ∠ 60° V
j5Ω
−j2Ω
j3Ω
Figure 2.70
Solution: Let i1, i2 and i3 be the mesh currents as shown in Figure 2.71.
j3Ω −j5Ω
+
+
4.33 + j 2.5
i1
Mesh I
−j2Ω
j5Ω
5 + j 8.66
i2
i3
j3Ω
Mesh II
Mesh III
Figure 2.71
5∠30° = 5(cos 30° + j sin 30°)
= 4.33 + j 2.5
10 ∠60° = 10(cos 60° + j sin 60°) = 5 + j8.66
By applying KVL in mesh I, we get the following equation
4.33 + j 2.5 = − j 2(i1 − i2 )
or
− j 2i1 + j 2i2 = 4.33 + j 2.5
(2.79)
By applying KVL in mesh II, we write the equation as:
− j 2(i2 − i1 ) + ( j 3 − j 5)i2 + j 5(i2 − i3 ) = 0
or
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 71
j 2i1 + ji2 − j 5i3 = 0
(2.80)
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Network Analysis and Synthesis
By applying KVL in mesh III, we obtain the following:
j 5(i3 − i2 ) = 5 + j8.66
− j 5i2 + j 5i3 = 5 + j8.66
or
(2.81)
We now solve equations (2.79), (2.80) and (2.81) for i2 using Cramer’s rule.
0
− j2 j2
j
∆ = j2
− j5
0
− j5 j5
= − j 2( j 2 5 − j 2 25) − j 2( j 2 10 − 0) + 0
= − j 2( −5 + 25) − j 2( −10)
= − j 40 + j 20 = − j 20
− j 2 4.33 + j 2.5 0
∆2 = j2
0
− j5
0
5 + j8.66
j5
= − j 2(0 + j 5(5 + j8.66)) − ( 4.33 + j 2.5)( j 210 − 0) + 0
= − j 210(5 + j8.66) + 10( 4.33 + j 2.5)
= 10(5 + j8.66) +110( 4.33 + j 2.5)
= 50 + j86.6 + 43.3 + j 25
= 93.3 + j111.6
∆ 2 93.3 + j111.6 145.46 ∠50.10°
=
=
= 7.273∠ − 129.9°
∆
− j 20
20 ∠180°
= −4.66 − j 5.579
i2 =
Current through the j3Ω = i2 = −4.66 − j5.579
i2 = ( 4.66) 2 + (5.579) 2
j6Ω
= 7.19 A
Example 2.38 Find the current through 6 Ω
resistor shown in Figure 2.72 using the mesh
analysis.
Solution: Let i1 and i2 be the mesh currents as
shown in Figure 2.73.
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 72
6Ω
+
10∠60°
2∠0°
−j 8 Ω
Figure 2.72
11/25/2014 3:44:22 PM
Kirchhoff’s Laws, Mesh and Nodal Analysis 73
j6Ω
By combining mesh I and mesh II [taking up
super mesh], we write the equation as follows:
+
10∠60°
5 + j8.66 = j6i1 + 6i2 + (−j8)i2
or
6Ω
5 + j8.66 = j6i1 + (6 − j8)i2
(2.82)
5 +j 8.66
Further, the value of (i2 − i1) can be taken as
follows:
−j 8 Ω
2∠0°
i1
i2
2A
Mesh I
Mesh II
Figure 2.73
i2 − i1 = 2 A
(2.83)
We now solve equations (2.82) and (2.83) for i2 using Cramer’s rule as,
∆=
j 6 6 − j8
1
−1
∆2 =
= j 6 + (6 − j8)
= 6 − j2
i2 =
j 6 5 + j8.66
2
−1
= j12 + 5 + j8.66
= 5 + j 20.66
and
∆ 2 5 + j 20.66 21.25∠76.36°
=
=
= 3.36 ∠94.82°
∆
6 − j2
6.32∠ − 18.43°
Current through 6 Ω = i2 = 3.36 ∠94.82°A
5Ω
Example 2.39 Find the current
through the (4 − j4) Ω impedance shown
in Figure 2.74 using mesh analysis.
1Ω
4
j2Ω
−j 4 Ω
10 ∠ 90 ° +
−
Solution: Let i1 and i2 be the mesh
currents as shown in Figure 2.75.
Figure 2.74
5Ω
1Ω
10 ∠ 90 ° V +
−
= j 10
i1
j2Ω
Mesh I
4Ω
i2
−j 4 Ω
Mesh II
Figure 2.75
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Network Analysis and Synthesis
By applying KVL in mesh I, we can write the equation as follows:
5i1 + (1 + j 2)(i1 − i2 ) = j10
or
5i1 + i1 − i2 + j 2i1 − j 2i2 = j10
or
(6 + j 2)i1 − (1 + j 2)i2 = j10
(2.84)
By applying KVL in mesh II, we get the following equation
(1 + j 2)(i2 − i1 ) + ( 4 − j 4)i2 = 0
or
i2 − i1 + j 2i2 − j 2i1 + 4i2 − j 4i2 = 0
or
−(1 + j 2)i1 + (5 − j 2)i2 = 0
(2.85)
We now apply Cramer’s rule to find i2, as
∆=
−(1 + j 2)
6 + j2
−(1 + j 2) 5 − j 2
= (6 + j 2)(5 − j 2) − (1 + j 2) 2
= 30 −
= 30 −
= 34 −
= 37 −
j12 + j10 − j 2 4 − (1 + j 2 4 + j 4)
j 2 + 4 − (1 − 4 + j 4)
j2 + 3 − j4
j6
Further,
∆2 =
6 + j2
−(1 + j 2)
j10
0
= 0 + j10(1 + j 2)
= j10 + j 2 20
= −20 + j10
i2 =
−20 + j10 22.36 ∠153.43°
∆2
=
= 0.596 ∠162.64°
=
37 − j 6
37.48∠ − 9.21°
∆
Current through (4 − j4) Ω impedance = i2 = 0.596∠162.64° A.
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Kirchhoff’s Laws, Mesh and Nodal Analysis 75
Example 2.40 Find the current through 1 Ω resistance in the network shown in Figure 2.76
using nodal analysis.
j5Ω
+
j2Ω
1Ω
−j 4 Ω
5Ω
50∠0°
+
− 55.46∠33.69°
Figure 2.76
Solution: Let the V1 and V2 be the nodal voltages as shown in Figure 2.77.
j5Ω
V1
I1
+
−
50∠0°
j2Ω
1
I2
I3
5Ω
1Ω
V2
2
I4
I5
−j 4 Ω
+
− 55.46∠33.69°
= − 46.145 −j 30.76
Figure 2.77
By applying KCL at node 1, I1 is written as follows:
I1 = I 2 + I 3
50 − V1 V1 − 0 V1 − V2
=
+
j5
j2
5
50 − V1 j 2V1 + 5V1 − 5V2
=
j5
j10
j 500 − j10V1 = −10V1 + j 25V1 − j 25V2
( −10 + j 35)V1 − j 25V2 = j 500
(2.86)
By applying KCL at node 2, we get the following
I3 = I 4 + I5
or
(2.87)
V1 − V2 V2 − 0 V2 − ( −46.145 − j 30.76)
=
+
j2
− j4
1
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Network Analysis and Synthesis
or
V1 − V2 V 2 V2 + 46.145 + j 30.76
=
+
− j4
j2
1
or
V1 − V2 V2 − j 4V2 − j184.58 + 123.04
=
j2
− j4
or
V1 − V2 =
V2 − j 4V2 − j184.58 + 123.04
−2
−2V1 + 2V2 = V2 − j 4V2 − j184.58 + 123.04
or
−2V1 + (1 + j 4)V2 = 123.04 − j184.58
or
(2.88)
Let us apply Cramer’s rule to find V2.
∆=
−10 + j 35 − j 25
1+ j4
−2
= (1 + j 4)( −10 + j 35) − j 50
= −10 + j 35 − j 40 − 140 − j 50
= − j 55 − 150
= −150 − j 55
and
∆2 =
j 500
−10 + j 35
123.04 − j184.58
−2
= ( −10 + j 35)(123.04 − j184.58) + j1000
= −1230.4 + j1845.8 + j 4306.4 + 6460.3 + j1000
= 5229.9 + j 7152.2
.V2 =
∆ 2 5229.9 + j 7152.2
8860.35∠53.82°
=
= 55.46 ∠213.68 = −46.145 − j 30.76
=
∆
−150 − j 55
159.76∠ − 159.86°
Now, current through 1 Ω resistance = I5
V 2 − {−46.145 − j 30.76}
1
{−46.145 − j 30.76} − {−46.145 − j 30.76} 0
=
= = 0A
1
1
=
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Kirchhoff’s Laws, Mesh and Nodal Analysis 77
Example 2.41 Find the current through the (4 + j4) Ω impedance shown in Figure 2.78 using
nodal analysis.
5Ω
4Ω
j4Ω
2Ω
+
+
50∠ 0° V
j2Ω
j2Ω
26.26 ∠ − 156.8° V
Figure 2.78
Solution: Let the node voltage be V1 and V2 shown in Figure 2.79.
5Ω
V1 4 Ω
I1
+
50∠ 0° V
j 4 Ω V2
2Ω
I3
I5
j2Ω
j2Ω
+
26.26 ∠ − 156.8° V
I4
50 V
− 24.136 −j 10.35
I2
Figure 2.79
By applying KCL at node 1, we get the following equation
I1 = I 2 + I 3
or
or
or
or
or
or
50 − V1 V1 − 0 V1 − V2
=
+
j2
5
4 + j4
50 − V1 ( 4 + j 4)V1 + j 2(V1 − V2 )
=
5
j 2( 4 + j 4)
50 − V1 4V1 + j 4V1 + j 2V1 − j 2V2
=
j8 − 8
5
( j8 − 8)(50 − V1 ) = 20V1 + j 20V1 + j10V1 − j10V2
j 400 − j8V1 − 400 + 8V1 = 20V1 + j 30V1 − j10V2
(12 + j 38)V1 − j10V2 = −400 + j 400
(2.89)
By applying KCL at node 2, we get the following:
I3 = I4 + I5
or
or
or
V1 − V 2 V 2 − 0 V 2 − {−24.136 − j10.35}
=
+
j2
2
4 + j4
V1 − V 2 V 2 V 2 + 24.136 + j10.35
=
+
j
1
2 + j2
V1 − V 2 V 2 + jV 2 + j 24.136 − 10.35
=
2 + j2
j
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78
I = I +I
3 Synthesis
4
5
Network Analysis and
or
V1 − V 2 V 2 − 0 V 2 − {−24.136 − j10.35}
=
+
j2
2
4 + j4
V1 − V 2 V 2 V 2 + 24.136 + j10.35
+
=
j
1
2 + j2
V1 − V 2 V 2 + jV 2 + j 24.136 − 10.35
=
2 + j2
j
(V1 − V 2 ) j = 2V 2 + j 2V 2 + j 48.272 − 20.7 + j 2V 2 − 2V 2 − 48.272 − j 20.7
or
jV1 − jV 2 = j 31.572V 2 − 68.972
or
or
or
jV1 + j 32.572V2 = −68.972
or
(2.90)
We now apply Cramer’s rule to find V1 and V2.
∆=
12 + j 38
j
− j10
j 32.572
∆1 =
= j 32.572(12 + j 38) + j 210
= j 390.86 − 1237.736 − 10
= −1247.736 + j 390.86
and
−400 + j 400
−68.972
− j10
j 32.572
= ( −400 + j 400) j 32.572 − j 689.72
= − j13028.8 − 13028.8 − j 689.72
= −13028.8 − j13718.52
Further,
∆2 =
12 + j 38 −400 + j 400
j
−68.972
= −68.972(12 + j 38) − j ( −400 + j 400)
= −827.664 − j 2620.936 + j 400 + 400
= −427.664 − j 2220.936
∆1 −13028.8 − j13718.52 18919.49∠ − 133.52°
=
=
∆
−1247.736 + j 390.86
1307.523∠162.60°
= 14.4697∠ − 296.12° = 6.37 + j12.99
V1 =
V2 =
−427.664 − j 2220.936 2261.736 − j100.89
2263.98∠ − 2.55
∆2
=
=
=
1307
.
523
∠
162
.
60
°
1307
.
5
2
3
∠
162
.
60
°
1307
.523∠162.60°
∆
= 1.731∠ − 165.15° = −1.673 − j 0.44
Current through the (4 + j4) Ω impedance =
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 78
V1 − V2 8.043 + j12.55
=
A.
4 + j4
4 + j4
11/25/2014 3:44:31 PM
Kirchhoff’s Laws, Mesh and Nodal Analysis 79
Example 2.42 Find the current through
3 Ω resistance shown in Figure 2.80 using
the mesh analysis.
10 Ω
2Ω
4
Solution: Let i1, i2 and i3 be the mesh
currents as shown in Figure 2.81.
From mesh I, it is clear that
i1 = 1.732 + j A.
2∠ 30° A
3Ω
−j 4 Ω
j8Ω
(2.91)
Figure 2.80
By applying KVL in mesh II, we get the
following equation,
10 Ω
( 4 − j 4)(i2 − i1 ) + 10i2 + ( 2 + j8)(i2 − i3 ) = 0
or
−( 4 − j 4)i1 + (16 + j 4)i2 − ( 2 + j8)i3 = 0
By substituting the value of i1 from equation (2.91), we get the equation as follows:
2Ω
4
3Ω
2∠ 30° A
1.732 + j
i1
−j 4
Mesh I
i2
Mesh II
j 8 Ω i3
Mesh III
Figure 2.81
−( 4 − j 4)(1.732 + j ) + (16 + j 4)i2 − ( 2 + j8)i3 = 0
or
(16 + j 4)i2 − ( 2 + j8)i3 = ( 4 − j 4)(1.732 + j )
= 6.928 + j 4 − j 6.928 + 4
(16 + j 4)i2 − ( 2 + j8)i3 = 10.928 − j 2.928
(2.92)
By applying KVL in mesh III, we obtain the following:
( 2 + j8)(i3 − i2 ) + 3i3 = 0
−( 2 + j8)i2 + (5 + j8)i3 = 0
or
(2.93)
We now solve equations (2.92) and (2.93) for i3 using Cramer’s rule.
∆=
16 + j 4 −( 2 + j8)
−( 2 + j8)
5 + j8
= (16 + j 4)(5 + j8) − ( 2 + j8) 2
= 80 + j128 + j 20 − 32 − {4 − 64 + j 32}
= 48 + j148 + 64 − j 32
= 112 + j116
Now
∆2 =
16 + j 4 10.928 − j 2.928
−( 2 + j8)
0
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Network Analysis and Synthesis
= (2 + j8) (10.928 − j2.928)
Therefore, current through 3 Ω = i3 =
∆ 2 ( 2 + j8)(10.928 − j 2.928)
=
112 + j116
∆
93.19∠60.97°
161.24 ∠46°
= 0.577∠14.97°Α
=
10 Ω
Example 2.43 Find the current in 5 Ω
resistance as shown in Figure 2.82 using mesh
analysis.
Solution: Let i1 and i2 be the mesh currents as
shown in Figure 2.83.
By applying KVL in mesh I, we obtain the
following:
j2Ω
2
+
5Ω
50∠ 0° V
j4Ω
Figure 2.82
50 = 10i1 + ( 2 + j 4)(i1 − i2 )
or
(12 + j 4)i1 − ( 2 + j 4)i2 = 50
(2.94)
10 Ω
j2Ω
By applying KVL in mesh II, we get the equations as follows:
+
( 2 + j 4)(i2 − i1 ) + j 2i2 + 5i2 = 0
or
−( 2 + j 4)i1 + (7 + j 6)i2 = 0
(2.95)
We now solve equations (2.94) and (2.95) for i2
using Cramer’s rule.
∆=
2Ω
50 V
i1
Mesh II
Figure 2.83
∆2 =
2
i2 =
i2
j4Ω
Mesh I
12 + j 4 −( 2 + j 4)
−( 2 + j 4)
7 + j6
= (12 + j 4)(7 + j 6) − ( 2 + j 4)
= 84 + j 72 + j 28 − 24 − {4 − 16 + j16}
= 60 + j100 + 12 − j16
= 72 + j84
5Ω
50∠ 0° V
and
12 + j 4 50
−( 2 + j 4) 0
= 0 + 50( 2 + j 4)
= 100 + j 200
∆ 2 100 + j 200 223.6 ∠63.43°
=
=
= 2.02∠14.04°°A
∆
72 + j 84
110.63∠49.39°
Therefore, current through 5 Ω resistor = i2 = 2.02∠14.04°A.
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Kirchhoff’s Laws, Mesh and Nodal Analysis 81
Example 2.44 Find the current through
resistance (4 + j6) Ω impedance shown in
Figure 2.84 using mesh analysis.
Solution: Let i1 and i2 be the mesh
currents as shown in Figure 2.85.
By applying KVL in mesh I, we get the
following:
2Ω
−j 5 Ω
+
+
50∠ 0° V
100∠ 0° V
j6Ω
Figure 2.84
2Ω
(6 + j11)i1 − ( 4 + j 6)i2 = 100
(2.96)
By applying KVL in mesh II, the equations can be written as follows:
(4 + j6) (i2 − i1) + (3 − j5)i2 = −50
or
3Ω
4Ω
( 2 + j 5)i1 + ( 4 + j 6)(i1 − i2 ) = 100
or
j5Ω
j5Ω
−j 5 Ω
3Ω
4Ω
+
+
100∠ 0° V
100 V
−(4 + j6) i1 + (7 + j) i2 = −50
(2.97)
50 V
j6Ω
i1
Mesh I
i2
Mesh II
Figure 2.85
Let us solve equations (2.96) and (2.97) for i1 and i2 using Cramer’s rule.
∆=
6 + j11 −( 4 + j 6)
−( 4 + j 6 )
7+ j
= (6 + j11)(7 + j ) − ( 4 + j 6) 2
= 42 + j 6 + j 77 − 11 − {16 − 36 + j 48}
= 31 + j83 + 20 − j 48
= 51 + j 35
∆1 =
100 −( 4 + j 6)
−50
7+ j
= 100 + (7 + j ) − 50( 4 + j 6)
= 700 + j100 − 200 − j 300
= 500 − j 200
∆2 =
6 + j11 100
−( 4 + j 6) −50
= −50(6 + j11) + 100( 4 + j 6)
= −300 − j 550 + 400 + j 600
= 100 + j 50
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Network Analysis and Synthesis
i1 =
∆1 500 + j 200
=
A
∆
51 + j 35
i2 =
and
∆ 2 100 + j 50
=
A
∆
51 + j 35
Current through the (4 + j6) Ω impedance = (i1 − i2) A =
500 + j 200 100 + j 50
−
= 7A.
51 + j 35
51 + j 35
R E V IE W Q U E S T I O N S
Numerical Questions
1. Find the mesh currents of the circuit shown in Figure 2.86.
4Ω
1Ω
2Ω
28 V
I1
7V
I2
Figure 2.86
[Ans. 5 A, −1 A]
2. Find the mesh currents of the circuit shown in the Figure 2.87.
150 Ω
24 V
+
−
50 Ω
I1
I3
100 Ω
300 Ω
I2
250 Ω
Figure 2.87
[Ans. −93.793 mA, 77.241 mA, 136.092 mA]
3. Find the voltage across 8 Ω resistor shown in Figure 2.88 using mesh analysis.
1A
8Ω
15 Ω
84 Ω
4A
Figure 2.88
[Ans. −24 V]
4. Find the node voltages in the circuit shown in Figure 2.89.
Node 1 Node 2
+
60 V
−
6Ω
36 Ω
12 Ω
20 Ω
40 Ω
+
100 V
−
Figure 2.89
[Ans. 42 V, 60 V]
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Kirchhoff’s Laws, Mesh and Nodal Analysis 83
5. Find the current through 11 Ω resistance in Figure 2.90 using mesh analysis.
+
25 Ω
11 Ω
−
2Ω
10 Ω
56 Ω
50 V
Figure 2.90
[Ans. 2 A]
6. Find the voltage drop across 20 Ω in Figure 2.91 using nodal analysis.
− +
90 V
10 Ω
+
40 V
−
5Ω
20 Ω
100 Ω
2A
Figure 2.91
[Ans. 60 V]
7. Find the current in branch BC of the circuit shown in Figure 2.92.
B
A
+
14 Ω
C
2Ω
4Ω
24 V
E
−
8Ω
I
D
2Ω
5A
H
G
1Ω
F
Figure 2.92
[Ans. 1.76 A]
8. Find the voltage across resistance 8 Ω in the circuit shown in Figure 2.93 using mesh and
nodal analysis.
30 V
+ −
136 Ω
8Ω
+
50 V
−
19 Ω
1A
Figure 2.93
[Ans. 12 V]
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84
Network Analysis and Synthesis
9. Find the power dissipated in 8 Ω resistor in the current shown in Figure 2.94 using mesh
analysis.
+
10 V
15 V
10 Ω
−
+
10 Ω
+
− 20 V
10 Ω
5Ω
−
8Ω
Figure 2.94
[Ans. 7.4 W]
10. Find the current in 1.5 Ω resistance in the circuit shown in Figure 2.95 using nodal analyses.
+
−
2Ω
2Ω
2Ω
9V
3Ω
1.5 Ω
Figure 2.95
[Ans. 0.75 A]
11. Find the current through 5 Ω resistor in the circuit shown in Figure 2.96 using mesh analysis.
+
50 V
−
15 Ω
10 Ω
5Ω
10 Ω
15 Ω
Figure 2.96
[Ans. 0.59 A]
12. Find the voltage across the 16 Ω resistor in the circuit shown in Figure 2.97 using mesh
analysis.
5Ω
100 V
+
40 Ω
−
20Ω
20Ω
16 Ω
+ −
100 V
Figure 2.97
[Ans. 20 V]
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Kirchhoff’s Laws, Mesh and Nodal Analysis 85
13. Find the current flowing through the resistor 80 Ω in the circuit shown in Figure 2.98 using
nodal analysis.
20 Ω
100 Ω
2A
80 Ω
Figure 2.98
[Ans. 1 A]
14. Find the current flowing through the 15 Ω resistor in the circuit shown below using nodal
analysis.
5Ω
1Ω
30 A
5Ω
15 Ω
50 V
Figure 2.99
[Ans. 2.31 A]
15. Find the current flowing in branch BE in the circuit shown in Figure 2.100 using nodal analysis.
A
+
−
B
2Ω
6V
F
C
1Ω
2Ω
E
+
−
6V
D
Figure 2.100
[Ans. 2.25 A]
16. Find the current flowing through ZL in the circuit shown in Figure 2.101 using nodal analysis.
6Ω
10∠0 V
6Ω
j6Ω
3.6
−j 6 Ω
j 4.8
ZL
Figure 2.101
[Ans. 0.621∠−26.56°A]
17. Using mesh analysis, determine the current in 20 Ω resistor in the circuit shown in
Figure 2.102.
+ −
6Ω
100 V
36 V 18 Ω
36 Ω
20 Ω
Figure 2.102
[Ans. 2 A]
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Network Analysis and Synthesis
18. Calculate the power dissipation in 6 Ω resistor in the circuit shown in Figure 2.103 using a
mesh analysis.
A
3V
+
3Ω +
−
−
6Ω
6V
RL = 6 Ω
B
Figure 2.103
[Ans. −1.5 W]
M U LTI P L E C HO I C E Q U E S T I ON S
1. Current I3 in the following circuit is
I2 = 2 A
I1 = 7 A
I3
I4 = 3 A
Figure 2.104
(a) −8 A
(b) −2 A
(c) 2 A
(d) 8 A
2. The voltage V2 in the circuit is
V1 = 4 V
A
B
E = 20 V
V2
D
V3 = 6 V
C
Figure 2.105
(a) −10 V
(b) 10 V
(c) 12 V
(d) 18 V
3. Nodal analysis involves systematically applying Kirchhoff’s current law to the nodes within
a circuit.
(a) True
(b) False
4. Mesh analysis involves systematically applying Kirchhoff’s current law to the meshes
within a circuit.
(a) True
(b) False
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Kirchhoff’s Laws, Mesh and Nodal Analysis 87
5. How many meshes are present within the following circuit?
Figure 2.106
(a) 1
(b) 2
(c) 3
(d) 4
6. In KCL, incoming currents are equal to –––––––––
7. An electrical network with seven independent branches and independent nodes excluding
the reference node should preferably be solved by
(a) Mesh current analysis
(c) Node voltage analysis
(b) KCL
(d) KVL and KCL
8. Which one of the following laws of electrical network is used in the nodal analysis of
network?
(a) KVL
(b) KCL
(c) Faraday’s laws
(d) Ohm’s law
9. In the mesh current analysis, the number of mesh equations is equal to the number of
meshes.
(a) True
(b) False
10. With I1 = 2 A and I2 = 2 A directed into a node, current I3 coming out of the node must be
equal to 3 A.
(a) True
(b) False
11. In assigning the direction of branch currents,
(a) The directions are critical
(c) They must point into a node
(b) The directions are not critical
(d) They must point out of a node
12. The branch current method uses
(a)
(b)
(c)
(d)
Kirchhoff ’s voltage and current laws
Thevenin’s theorem and Ohm’s law
Kirchhoff ’s current law and Ohm’s law
The superposition theorem and Thevenin’s theorem
13. The voltage at node 1 in the Figure is
Node 1
49 Ω
12 V
80 Ω
24 Ω
6V
Figure 2.107
(a) 6 V
(b) 12 V
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 87
(c) 4.25 V
(d) 3 V
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Network Analysis and Synthesis
14. A super mesh is created if
(a)
(b)
(c)
(d)
a voltage source is common to two loops
a current source is common to two loops
a current source is common to two nodes
a voltage source is common to two nodes
15. A super node is created if
(a)
(b)
(c)
(d)
a voltage source is common to two loops
a current source is common to two loops
a current source is common to two nodes
a voltage source is common to two nodes
16. The number of effective loops in the circuit of the Figure 2.108 is
8Ω
4Ω
V
2Ω
6Ω
I
Figure 2.108
(a) 4
(b) 3
(c) 1
(d) 2
17. The number of nodes in the circuit of Figure 2.109 is
R1
V1
R2
R3
−+
V3
I1
R4
V2
Figure 2.109
(a) 3
(c) −1
(b) 2
(d) 4
18. The voltage at node 1 in the Figure 2.110 is
Node 1
5V +
−
2Ω
2A
Figure 2.110
(a) 2.5 V
(b) 5 V
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 88
(c) 3 V
(d) 2 V
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Kirchhoff’s Laws, Mesh and Nodal Analysis 89
19. Current I1 in the circuit shown in the Figure 2.111 is
2Ω
2Ω
2A
10 V
I1
Figure 2.111
(a) −5 A
(b) 5 A
(c) 2 A
(d) 1 A
20. Find V1 in the circuit shown in the Figure 2.112.
V1
2Ω
6V
4Ω
1A
8V
Figure 2.112
(a) −2.67 V
(b) 8 V
(c) 5.33 A
(d) 2.67 V
ANS W E RS
1. d
9. a
19. c
2. a
10. a
20. b
3. a
11. b
4. b
12. a
5. c
13. c
M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 89
6. outgoing currents
14. b
15. d
16. a
7. c
17. b
8. b
18. b
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Steady State Analysis
of AC Circuits
3
Chapter objectives
After carefully studying this chapter, you should be able to do the following:
Establish the phase relationship
Make steady state analysis of R–L–C
between voltage and current in a purely
series circuits and establish phasor relationresistive, purely inductive, and purely
ship between voltage drops and current.
capacitive circuits.
Use phasor algebra in solving AC cirExplain the concept of active and reaccuit problems.
tive powers.
Make steady state analysis of AC parMake steady state analysis of R–L
allel circuits.
series circuits and R–C series
Solve numerical problems on AC
circuits.
series–parallel circuits.
An electric circuit may consist of resistance, inductance and capacitance connected in series and
parallel combinations. The circuit may have any number of such components. Such circuits will
have a source of an AC supply. For example, we may have a resistance and an inductance connected in series across a source of supply voltage. Such a circuit will be called an R–L series circuit. When a resistance, an inductance and a capacitance are connected in series across a supply
source, the circuit is called an R–L–C series circuit. Similarly, we may have parallel combination
of connections of circuit elements. We may have to calculate the current in each branch of a circuit
and the total current, the power factor, the power consumed by each branch, etc. First, we will discuss the behaviour in terms of voltage, current, power and power factor of resistance, inductance
and capacitance in AC circuit independently, and then discuss R–L, R–C and R–L–C circuits.
3.1 AC voltage APPLIED ACROSS A RESISTOR
Figure 3.1 shows a resistor R connected across a source of AC supply. The Figure also shows the
wave forms of voltage, current and power consumed.
Let the supply voltage be v = Vm sin w t.
Current i =
v V m sin w t V m
sin w t
=
=
R
R
R
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steady state analysis of aC Circuits 91
i
v
R
p
p
v
i
v
i
V
t
(a)
(b)
I
(c)
Figure 3.1 Alternating Voltage Applied Across a Resistor (a) Circuit Diagram; (b) Wave
Shapes of v, i and p and (c) Phasor Diagram
i = Im sin w t
or
Vm
R
The rms value of current i is designated in capital letter as I
where
Im =
and
I=
Im
2
As shown in Figure 3.1(b), both voltage and current are in time-phase, that is, there is no phase
difference between the voltage and the current. In Figure 3.1(c), the rms values of voltage and
current have been shown and they are shown to be in phase. Power is the product of voltage and
current at every instant of time. The product of n and i have been calculated and has been shown.
The average value of the power in a purely resistive circuit can also be calculated as follows:
P=
1
2p
=
=
=
or
=
1
2p
2p
∫ vid (w t )
0
2p
∫ V m sin w t I m sin w t dw t
0
1
2p
2p
∫ V m sin q I m sin q dq
(expressing w t in radians, that is, substituting w t = q )
0
V mI m
4p
V mI m
4p
V mI m
4p
2p
∫ 2 sin q dq
2
0
2p
∫ (1 − cos 2q )dq
0
1 − cos 2q 

2

∵ sin q =
2
2p
V mI m
V mI m V m I m
sin 2q 

= VI
q − 2  = 4p × [( 2p − 0) − 0] = 2 =
2 2
0
The power factor is the cosine of the phase angle between the voltage and the current. Since the
voltage and current are in phase, the value of power factor angle f is equal to zero.
Therefore, the power factor in a purely resistive circuit is = cosf = cos 0° = 1.
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Network analysis and synthesis
3.2 AC voltAge APPlIeD ACRoSS AN INDUCtoR
In this case, let us assume a pure inductor. A pure inductor is one whose resistance is assumed
to be negligible. Let such an inductor be connected across a voltage source, v = Vm sin w t. As
a result of the application of the voltage, an alternating current i will flow through the inductor
coil. This alternating current will produce an alternating magnetic field around the inductor coil.
This alternating magnetic field will induce an emf in the coil which is given as follows:
di
dt
e=L
di
is the rate of change of the alternating current flowdt
ing through the inductor coil as shown in Figure 3.2.
where L is the inductance of the coil and
L
i
di
e=L
dt
p
v
i
p
v
i
V
p /2
0
p
90°
wt
v = Vm sin w t
(a)
(b)
I
(c)
Figure 3.2 Alternating Voltage Applied Across an Inductor (a) Circuit Diagram; (b) Wave
Shapes of Voltage, Current and Power and (c) Phasor Diagram
This induced emf, according to Lenz’s law, will oppose the voltage applied. Therefore, we can
write the equation as in the following:
di
v=e=L
dt
or
L di = v dt = Vm sinw t dt
Vm
sinw t dt
L
By integrating the equation, we get the following form:
or
di =
Vm
sin w t dt
L ∫
V
= m ( − cos w t )
wL
i=
Vm
p

sin  w t − 
wL 
2
or
i=
or
p

i = I m sin  w t − 

2
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steady state analysis of aC Circuits 93
where
Im =
Vm
wL
Thus, we observe that in a purely inductive circuit, v can be written as in the following:
v = V m sin w t
p

i = I m sin  w t − 

2
and
= I m sin(w t − 90°)
Further, current i is sinusoidal but lagging behind v by 90°. The voltage and current wave shapes
have been shown in Figure 3.2(b). The instantaneous power p is the product of v and i. The wave
shape of instantaneous power has also been shown in the Figure. The phasor diagram of rms
values of v and i has been shown in Figure 3.2(c). In a purely inductive circuit, current I lags the
p
voltage V by degrees, that is, 90°.
2
Power factor
cos f = cos 90° = 0
Average power
P=
=
=
=
=
=
=
1
2p
2p
p

∫ V m sin w tI m sin  w t − 2  d (w t )
0
V mI m
2p
V mI m
4p
V mI m
4p
V mI m
4p
V mI m
4p
2p
p

∫ sin w t sin  w t − 2  d (w t )
0
2p
p

∫ 2 sin w t sin  w t − 2  d (w t)
0
2p



p
p

p 
∫ cos  w t − w t + 2  − cos  w t + w t − 2   d (w t)
0
2p

p 
∫ cos 2 − cos  2w t − 2   d (w t )
0
2p
∫ (0 − sin 2w t ) d (w t )
0
V m I m cos 2w t
⋅
4p
2
2p
0
V I
 cos 4p − cos 0 
= m m ×

4p
2


=0
[∵ cos 4p = cos 0 = 1]
Average power in a purely inductive circuit P = 0.
Hence, the average power absorbed by a pure inductor is zero.
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Network analysis and synthesis
Im =
We had earlier taken
Vm
wL
The opposition to current is w L. This is called inductive reactance XL.
The value of XL = w L = 2p f L.
The opposition offered by an inductor to the flow of current is XL, that is, equal to w L.
This is called the inductive reactance and is expressed in ohms. Inductance L is expressed in
henry.
As mentioned earlier, the values of alternating quantities are expressed in terms of their
effective or rms values rather than their maximum values.
Therefore,
Im =
I=
or
Vm
wL
can be written as,
Im
2
=
Vm / 2
wL
V
, i.e., V = IX L
XL
If V is taken as the reference axis, we can represent V as a
phasor and can represent it as V∠0°.
90°
Since current I is lagging voltage V by 90°, we represent
the current as I∠−90° or −jI for a purely inductive circuit.
Again, if I is taken as the reference axis, then I and V can
I∠0°
I∠−90°
be represented as I∠0° and V∠+90° or +jV, respectively, as
or − jI
Figure 3.3 Phasor Diagram of shown in Figure 3.3.
Note that j is an operator that indicates the rotation of a
V and I in a Purely
phasor by 90° in the anti-clockwise direction from the referInductive Circuit
ence axis.
V∠0°
V∠+90°
or +jV
Concept of Reactive Power
Let us examine why the power absorbed by a pure inductive circuit is zero. Let us refer to the
inductive circuit shown in Figure 3.2. In Figure 3.2(b), it is observed that for one-half cycle,
the power is negative and for the next half cycle, the power is positive. The average value of
the power for a complete cycle, that is, the power consumed is zero. The positive power indicates
that the power is drawn by the circuit from the supply source. When the current rises in the
circuit, energy is utilised in establishing a magnetic field around the inductor coil. This energy
is supplied by the source and is stored in the magnetic field. As the current starts reducing, the
magnetic field collapses and the same stored energy is returned to the supply source. Thus, in one
half cycle, the power is drawn by the inductor and in the second half cycle, the power is returned
to the source. In this way the net power absorbed by the inductor over a complete cycle is zero.
The power, that is, being circulated from the source to the inductor and back to the source
is called the reactive power. The active and reactive power will also be discussed in a separate
section.
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steady state analysis of aC Circuits 95
3.3 AC voltAge APPlIeD ACRoSS A CAPACItoR
A sinusoidal voltage source has been shown connected across a pure capacitor in Figure 3.4(a).
When current starts flowing, the capacitor starts getting charged. Charge q of the capacitor in
terms of capacitance of the capacitor C and supply voltage v is expressed as in the following:
q = Cv
Current i is the rate of flow of charge. Therefore,
dq
dt
dv
=C
dt
d
= C Vm sin w t
dt
= w C Vm cos w t
i=
[By substituting q = Cv ]
or
p

i = w CV m sin  w t + 

2
or
p

i = I m sin  w t + 

2
I m = w CVm =
where
C
Vm
V
= m
1/w C X c
P
v
i
where X c =
1
wC
P
v
i
I
i
wt
90°
v = Vm sin w t
(a)
V
(b)
(c)
Figure 3.4 Alternating Voltage Applied Across an Capacitor (a) Circuit Diagram;
(b) Wave Shapes of Voltage, Current and Power and (c) Phasor Diagram
p

Hence, in a pure capacitive circuit, v = Vm sin w t and current i = I m sin  w t +  . Thus in a

2
purely capacitive circuit, the current leads the voltage by 90°. Since the current leads the voltage
by 90°, the power factor of the purely capacitive circuit is
P.f = cosq = cos 90° = 0
1
Xc =
is called the capacitive reactance of the capacitor.
wC
To express in terms of rms values,
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Network analysis and synthesis
we assume
Im
Therefore,
Im =
2
= I and
Vm
2
=V
Vm
Xc
V
Xc
The term Xc is the opposition offered by the capacitor to the flow of current and is called
capacitive reactance.
I=
or
Concept of Reactive Power
In a capacitor, similar to an inductor, the average power absorbed for a complete cycle is zero.
When voltage is applied, the capacitor starts getting charged, energy gets stored in the capacitor
in the form of electro-static field. When the applied voltage starts falling from its maximum
value, the energy starts getting returned to the supply. In this way, the power is absorbed from
and then returned to the supply source. The net power absorbed by a pure capacitor is zero. The
power, that is, being circulated from the source to the capacitor and back to the source is called
the reactive power.
The average or net power in a pure capacitor circuit can be calculated as given in the following:
Average power P =
=
=
=
=
=
1
2p
1
2p
2p
∫
P d (w t ) =
0
2p
∫
0
1
vi d (w t )
2p
2p
p

∫ V m sin (w t ) I m sin  w t + 2  d (w t )
0
V mI m
4p
V mI m
4p
V mI m
4p
V mI m
4p
2p

p
∫ 2 sin w t sin  w t + 2  d (w t )
0
2p



 p
p

p 
∫ cos  w t − w t − 2  − cos  w t + w t + 2   d (w t )
0
2p

p 
∫ cos  − 2  − cos  2w t + 2   d (w t )
0
2p
∫ siin 2w td(w t )
0
=0
Average power in a purely capacitive circuit P = 0.
Therefore, it is proved that the average power absorbed by a pure inductor and a pure capacitor is zero.
example 3.1 An inductor of 0.5 H is connected across a 230 V, 50 Hz supply. Write the
equations for instantaneous values of voltage and current.
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steady state analysis of aC Circuits 97
Solution:
V = 230 V, V m = 2 V = 1.414 × 230 = 324 V
X L = w L = 2p f L = 2 × 3.14 × 50 × 0.5 Ω = 157 Ω
I=
V
230 230
=
=
= 1.46 A
X L X L 157
I m = 2 I = 1.414 × 1.46 = 2.06 A
The equations are as follows:
n = V m sin w t = 324 sin w t = 324 sin 2p ft = 324 sin 314t
and
p
p


i = I m sin  w t −  = 2.06 sin  314t − 


2
2
example 3.2 A 230 V, 50 Hz sinusoidal supply is connected across the following:
(a) A resistance of 25 Ω; (b) An inductance of 0.5 H; (c) A capacitance of 100 µF.
Write the expressions for instantaneous current in each case.
Solution: Given
n = 230 V
V m = 2 V = 1.414 × 230 = 324.3 V
w = 2p f = 2. × 3.14 × 50 = 314 rad/s
Voltage equation is v = V m sinw t
or
v = 324.3 sin 314t
Inductive reactance
X L = w L = 314 × 0.5 = 157 Ω
Capacitive reactance
XC =
=
1
1
=
w C 314 × 100 × 10 −6
10 −6
= 32.2 Ω
314 × 100
When the voltage is applied across a 25-Ω resistor, the current will be calculated as follows:
i=
or
Vm
324.3
sin w t =
sin 314t
R
25
i = 12.97 sin 314t A
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Network analysis and synthesis
The current through the inductor is expressed as in the following:
i=
=
Vm
p

sin  w t − 

XL
2
324.3 
p
sin  314t − 

157
2
i = 2.06 sin(314t − 90°)A
or
The current through the capacitor is given as follows:
i=
Vm
p

sin  w t + 

Xc
2
324.3
sin(314t + 90°)
32.2
i = 10.07 sin(314t + 90°)A
=
example 3.3 An alternating voltage of rms value 100 V, and frequency 50 Hz is applied
separately across a resistor of 10 Ω, an inductor of 100 mH and a capacitor of 100 µF. Calculate
the current flow in each case. Draw the phasor diagrams for each case and explain the phasor
diagrams.
Solution:
R = 10 Ω
X L = w L = 2p f L = 2 × 3.14 × 50 × 100 × 10 −3 Ω
= 31.4 Ω
XC =
=
1
1
1
=
=
w C 2p fC 2 × 3.14 × 50 × 100 × 10 −6
106
= 31.8 Ω
314 × 100
Current through R =
100
= 10 A
10
Current through L =
100 100
=
= 3.18 A
X L 31.4
Current through C =
100 100
=
= 3.1 A
X C 31.8
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steady state analysis of aC Circuits 99
We know that in a resistive circuit, current is in phase with the applied voltage; in a purely
inductive circuit, current lags the voltage by 90°, while in a purely capacitive circuit, current
leads the voltage by 90°. The phasor diagrams have been shown in Figure 3.5.
I
V = IXL
V = IR
I
90°
90°
V = IXC
I
(a)
(b)
(c)
Figure 3.5 Phasor Diagrams of (a) Resistive Circuit; (b) Purely Inductive Circuit
and (c) Purely Capacitive Circuit
3.4 R–L SeRIeS CIRCUIt
Let us consider a resistance element and an inductor connected in series as shown in Figure 3.6.
A voltage of V and frequency f is applied across the whole circuit. The voltage drop across the
resistance is VR and across the inductor is VL. The current flowing through the circuit is I. The
same current is flowing through R and L. The voltage drops VR and VL are
C
I
L
Z
R
=I
I
VL
f
V, f
(a)
VL = IXL
V
VR
VR = IR B
A
I
(b)
Figure 3.6 R–L Series Circuit (a) Circuit Diagram and (b) Phasor Diagram
V R = IR , V L = IX L where X L = w L = 2p fL
We have to add VR and VL to get V. However, these are to be added vectorially as they are
not in phase; that is, these vectors are not along the same direction. To draw the current and
voltage phasor, we take current as the reference phasor as shown in Figure 3.6(b) as current I
is common to VR and VL; that is the same current is flowing through both the resistance and
the inductance. Therefore, we have chosen I as the reference phasor. The voltage drop across
the resistance and the current flowing through it are in phase. This is because, as we have
seen earlier that in a resistive circuit, voltage and current are in phase. The current flowing
through an inductor lags the voltage across it by 90°. This is to say, voltage drop across L,
that is, VL will lead the current by 90°. Again VL = IXL and XL = w L. The vector sum of VR
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Network analysis and synthesis
and VL is equal to V. The angle between V and I is called power factor angle f. The power
factor is cos f. Considering triangle ABC in Figure 3.6(b), we can express the equation as
follows:
V 22 =
=V
V R22 +
+V
V L22
V
R
L
or
V =
= V
V R22 +
+ V L22 =
= ((IR
IR ))22 +
+ ((IX
IX L ))22
V
R VL
L
or
V =
= II
V
or
I=
R 22 +
+X
X L22
R
L
V
2
R +X
2
L
=
V
or V = IZ
Z
Z = R 2 + X L2
where
Z is called the impedance of the total circuit. Triangle ABC in Figure 3.6 (b) is also called the
impedance triangle, which is redrawn as in Figure 3.7. From the impedance triangle, Z can be
expressed as follows:
C
C
or
Z = R 2 + X L2
Z
VL = IXL
XL
V
=I
Z
Z = R + jX L
f
Operator j indicates the rotation by 90° in anticlockwise direction.
f
A
B
VR = IR
(a)
A
R
Z
Z cos f = R
B
cos f =
R
(b)
Figure 3.7 Impedance Triangle for R–L
Circuit (a) In terms of Voltage
and (b) In terms of Resistance
and Reactance
Z sin f = XL
Figure 3.7(a) is the same as in Figure 3.7(b).
Since current I is common to all sides, it has
been taken out in Figure 3.7(b). Impedance Z can be represented as the vector sum of R and XL.
Since IXL is leading I by 90° and IR is in phase with I, we can write Z as follows:
R
Z
X
tan f = L
R
Z = R + jX L and cos f =
Power = VI cos f
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steady state analysis of aC Circuits 101
3.5 APPAReNt PoWeR, ReAl PoWeR AND ReACtIve PoWeR
Apparent power S = rms value of voltage × rms value of current
or
S = VI
The apparent power is expressed in volt-ampere, that is, VA or in kilo-volt ampere is kVA.
Real power or active power P or W = Apparent power × Power factor
or
P = VI cos f. Watts or kilo-Watts
Reactive power Q = VI sin f VAR or kilo-VAR.
These three types of powers are related as given in the following:
S 22 =
S =
S=
S=
KVA =
KVA =
or
or
P 22 + Q 22
P +Q
P 22 + Q 22
P +Q
( KW ) 22 + ( KVAR ) 22
( KW ) + ( KVAR )
3.6 PoWeR IN R–L SeRIeS CIRCUIt
Let us now develop a general expression for power in an AC circuit by considering the instantaneous values of voltage and current as shown in Figure 3.8.
p = vi
v
u or i
or p
+
−
90°
+
i
−
180°
270°
Pav = VI cosf
360°
f
T
Figure 3.8 Wave Forms of Voltage, Current and Power in an R–L Series Circuit
A sinusoidal voltage is expressed as follows:
v = V m sin w t
In a circuit, when the current is lagging the voltage by an angle f, current i is expressed as in
the following:
i = I m (w t − f )
The sinusoidal waveforms of voltage and current are shown in Figure 3.8. It is observed that the
current wave is lagging the voltage wave by an angle f, which is the power factor angle.
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Network analysis and synthesis
The expression for the voltage and current in series R–L circuit is as follows:
v = V m sin w t
i = I m sin (w t − f ), as I lags V
The power is the product of instantaneous values of the voltage and the current,
p = v ×i
= V m sin w t × I m sin (w t − f )
1
= V m I m [2 sin w t sin (w t − f )]
2
1
= V m I m [cos f − cos ( 2w t − f )]
2
1
1
= V m I m cos f − V m I m cos ( 2w t − f )
2
2
The average power over a complete cycle is calculated as follows:
1
2p
1
=
2p
Pav =
2p
∫0
2p
∫0
1
V I [cos f − cos ( 2w t − f )]d (w t )
2 m m
1 2p 1
1
V I cos( 2w t − f )d (w t )
V m I m cos f d (w t ) −
2p ∫0 2 m m
2
Now, the second term is a cosine term whose average value over a complete cycle is zero.
Hence, the average power consumed is calculated as follows:
1
2p
1
=
2p
1
=
2p
Pav =
2p
∫0
1
V m I m cos fd (w t )
2
1
⋅ V m I m cos f | w t |02p
2
1
⋅ V m I m cosf ⋅ 2p
2
V mI m
V
I
cos f = m × m cos f
2
2
2
Pav = V rms × I rms cos f = V I cos f W
Pav =
3.7 PoWeR tRIANgle oF R–L SeRIeS CIRCUIt
The voltage–current relationship is expressed in terms of power factor. Power factor is defined
as the cosine of the phase angle between the voltage and the current; cos f is known as the
power factor. The power factor can also be expressed as the ratio R/Z = resistance/impedance
= cos f.
In Figure 3.9, the power triangle diagram has been developed from the simple voltage–
current relationship in an R–L series circuit. First, we have shown that I is lagging V by the
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steady state analysis of aC Circuits 103
power factor angle f. The in-phase component of I is I cos f and quadrature component is I sin f,
as shown in Figure 3.9(a).
I cosf
f
A
V
I cosf
B
V
f
I sinf
I
I
I sinf
C
(a)
A
kVI cosf
f
A
B
kVI sinf
kVI
P = kW
f
S = kVA
C
B
Q = kVAR
C
(b)
Figure 3.9 Power Triangle Diagram Developed from Voltage–current Relationship
(a) Voltage–current Relationship and (b) Power Triangle
By multiplying all the sides of triangle ABC by kV (kilo-volt), we can draw the power triangle
as in Figure 3.9(b).
kVA cos f = kW
kVA sin f = kVAR
Active and Reactive Power
In the power triangle diagram, if f is taken as zero, that is, if the circuit is resistive, reactive
power Q becomes zero. If the circuit is having pure inductance or capacitance, f = 90, active
power P becomes zero. The reactive power will be present whenever there is inductance or
capacitance in the circuit. Inductors and capacitors are energy-storing and energy-releasing
devices. The energy is stored in an inductor and a capacitor in the form of magnetic and electric
fields, respectively, and are of importance in the field of electrical engineering.
3.8 R–C SeRIeS CIRCUIt
Consider a circuit consisting of a pure resistance R connected in
series with a pure capacitor C across an AC supply of frequency
f as shown in Figure 3.10.
When the circuit draws current I, then there are two voltage
drops.
1. Drop across pure resistance VR = I × R.
2. Drop across pure capacitance VC = I × XC.
Here, X C
1
=
and I , V R and V C are the rms values.
2p fC
M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 103
I
VR
I
I
I
90°
VC
R
C
VR
VC
u = Vm sinw t
Figure 3.10 R−C Series
Circuit
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Network analysis and synthesis
The phasor diagram for such a circuit can be drawn
by taking the current as a reference phasor repreIR
= 90° IX = V
sented by OA as shown in Figure 3.11. The voltage
C
C
VR
f
drop VR across the resistance is in phase with curV
rent and is represented by OB. The voltage drop
(a)
(b)
across the capacitor VC lags the current by 90° and
Figure 3.11 Phasor Diagrams of R–C is represented by BC. The phasor OC is the phasor
Circuit (a) Current, I as the sum of two voltages V and V . Hence, OC repR
C
Reference Phasor and (b)
resents the applied voltage. Thus, in a capacitive
Voltage, V as the Reference
circuit, current leads the voltage by an angle f. The
Phasor
same phasor diagram can be drawn by taking voltage V as the reference vector as shown in Figure 3.11(b).
In Figure 3.11(b), we have drawn V as the reference vector. Then, current I has been shown
leading V by an angle f. The voltage drop across the resistance VR = IR has been drawn in phase
with I. The voltage drop across the capacitance VC = IXC has been drawn lagging I by 90° (VC lagging I is the same as I leading VC). The length of VR and VC are such that they make an angle of 90°.
In an R–C series circuit, I leads V by an angle f or supply voltage V lags current I by an angle
f as shown in the phasor diagram in Figure 3.11(b).
O
VR = IR
f
I
V
=
IZ
B
VC =
IXC
C
A
I
tan f =
V C IX C X C
=
=
VR
IR
R
f = tan −1
XC
R
Applied voltage V = V R2 + V C2
= (IR ) 2 + (IX C ) 2
=I
R 2 + X C2
V = IZ
Z = R 2 + X C2 = impedance of the circuit
where
Voltage and current wave shapes of this circuit are shown in Figure 3.12, which indicates that
the current in a R−C circuit leads the voltage by an angle f. Angle f is called the power factor
angle.
P = VI cosf
u, i
f
u = Vm sinw t
i = Im sin(w t + f )
S = VI
Q = VI sinf
t
f
(a)
(b)
Figure 3.12 (a) Wave Forms of the Voltage and the Current and Their Phase Relationship
in an R–C Series Circuit and (b) Power Triangle Diagram
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steady state analysis of aC Circuits 105
3.8.1 Power and Power triangle of R–C Series Circuit
The expression for the voltage and the current is as follows:
v = V m sin w t
i = I m sin (w t + f ) as I leads V
The power is the product of the voltage and the current. The instantaneous power is given in the
following:
P = v ×i
= V m sin w t × I m sin(w t + f )
1
= V m I m [2 sin w t sin(w t + f )]
2
1
= V m I m [cos( −f ) − cos( 2w t + f )]
2
V I
1
= V m I m cos f − m m cos( 2w t + f )
2
2
The second term is a cosine term whose average value over a complete cycle is zero. Hence, the
average power is consumed by the circuit is given as follows:
V mI m
V I
cos f = m m cos f
2
2 2
Pav = V rms I rms cos f = V I cos f W
Pav =
The power triangle has been shown in Figure 3.12 (b).
Thus, various powers are expressed as in the following:
Apparent power S = VI volt amperes or VA;
Active power P = VI cos f W;
Reactive power Q = VI sin f VAR;
where cos f = power factor of the circuit.
Note that power factor cos f is lagging for a resistive-inductive circuit and is leading for a
resistive-capacitive circuit.
3.9 R–L–C SeRIeS CIRCUIt
Consider a circuit consisting of resistance R, inductance L and capacitance C connected in
series with each other across an AC supply. The circuit has been shown in Figure 3.13(a).
The circuit draws a current I. Due to the flow of current I, there are voltage drops across R,
L and C that are given by the following:
Voltage drop across resistance R is VR = IR; Voltage drop across inductance L is VL = IXL;
and Drop across capacitance C is VC = IXC
where I, VR, VL and VC are the rms values. The voltage–current relationships in the resistance, inductance, and capacitance have also been shown in Figure 3.13(a).
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Network analysis and synthesis
VL
I
I
VR
R
L
I
VR
VL = IXL
VC
I
VC = IXC
C
I
VL
V
VC
(VL − VC)
f
I
I
VR = IR
u = Vm sin wt
VC = IXC
(a)
(b)
Figure 3.13 (a) R–L–C Series Circuit and (b) Phasor Diagram
The phasor diagram of the R−L−C circuit depends on the magnitude of VL and VC which obviously depends upon XL and XC. Let us consider the different cases.
When XL > XC, that is, when inductive reactance is more than the capacitive reactance, the circuit will effectively be inductive in nature. When XL > XC, obviously IXL, VL is
greater than IXC, that is, VC. Therefore, the resultant of VL and VC will be VL − VC and V is the
phasor sum of VR and (VL − VC). The phasor sum of VR and (VL − VC) gives the resultant supply
voltage V. This is shown in Figure 3.13 (b) and again redrawn as in Figure 3.14.
CASE 1.
VL
VL − VC
f
O
VR
B
B
V
A
O
I
V
f
VR
VL − VC
A
VR = V cos f
I
VC
Figure 3.14 Phasor Diagram of Current and Voltage Drops in an R–L–C Circuit Where
XL > XC
OB = OA 2 + AB2
Applied voltage is,
V = V R2 + (V L − V C ) 2
= (IR ) 2 + (IX L − IX C ) 2
= I R 2 + ( X L − X C )2
V = IZ
or
where
Z = R2 + ( X L − X C )2
tan f =
VLVC I ( X L − X C ) ( X L − X C )
( X − XC )
=
=
, f = tan −1 L
R
VR
IR
R
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steady state analysis of aC Circuits 107
Note that when XL > XC, R–L–C series circuit will effectively be an inductive circuit where
current will lag the voltage V as shown in the phasor diagram of Figure 3.14.
When XL < XC, the circuit will effectively be capacitive in nature. When XL < XC,
obviously, IXL, VL is less than IXC (that is, VC). Therefore, the resultant of VC − VL will be
directed towards VC. Current I will lead (VC − VL). The phasor sum of VR and (VC − VL) gives
the resultant supply voltage V. This is shown in Figure 3.15.
CASE 2.
VL
VR
I
f
O
VC − VL
V
VR
A
VC − VL
f
V
I
VR = V cos f
B
VC
Figure 3.15 Phasor Diagram of an R–L–C Series Circuit When XL < XC
Applied voltage represented by OB = OA 2 + AB2
V = V R2 + (V C − V L ) 2
= (IR ) 2 + (IX C − IX L ) 2
= I R 2 + ( X C − X L )2
or
V = IZ
where
Z = R2 + ( X C − X L )2
tan f =
VC - VL I ( X C − X L ) X C − X L
 X − XL 
∴ f = tan −1  C
=
=


R
VR
IR
R
CASE 3. When XL = XC, then VL = VC. Therefore, VL and VC will cancel each other and their
resultant will be zero and hence VR = V. In such a case, the overall circuit will behave like a
purely resistive circuit. The phasor diagram is shown in Figure 3.16. The impedance of the circuit will be minimum, that is, equal to R.
VL
A
O
VC
VR = V
I
VR = V cosf
= V cos 0°
=V
f = 0, cos f = 1
Figure 3.16 Phasor Diagram of an R–L–C Series Circuit When XL = XC
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Network analysis and synthesis
3.10 AC PARAllel CIRCUItS
Parallel circuits are formed by two or more series circuits connected to a common source of supply.
The parallel branches may include a single element or a combination of elements in series.
The following three methods are available for solving AC parallel circuit problems:
1. Phasor or vector method
2. Admittance method
3. Using vector algebra (symbolic method or j-operator method)
These methods are explained with examples as follows.
3.10.1 Phasor or vector Method of Solving Circuit Problems
A parallel circuit consisting of three branches has been shown in Figure 3.17. Branch 1 consists
of R1, L1 and C1 in series. Branch 2 is resistive and capacitive and branch 3 is resistive and
inductive. Let the current be I1, I2 and I3 in branch 1, 2 and 3, respectively. The total current
drawn by the circuit is the phasor sum of I1, I2 and I3.
3
2
I3
R3
L3
I2
R2
C2
I1
R1
L1
Branch 1.
= R12 + ( X L1 − X C1 ) 2 = Z 1
Current I1 = V /Z 1
C1
The phase difference of this current with
respect to the applied voltage is given by
X − X C1
f 1 = tan −1 L1
.
R1
This current will lag the applied voltage by an
angle f 1, if X L1 > X C1 .
1
I
V, f
Figure 3.17
Impedance of branch 1
AC Parallel Circuit
In the case of X C1 > X L1, I1 will lead voltage V.
Branch 2.
It is a resistive–capacitive branch (I2 leads V ).
Impedance of branch 2
Z 2 = R2 2 + X C 2 2
Current
I 2 = V /Z 2
Branch current I2 leads applied voltage V by an angle f 2 , and it can be written as follows:
f 2 = tan −1
Branch 3.
XC2
R2
Resistive–inductive branch
Z3 = R32 + X L 32
Current
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steady state analysis of aC Circuits 109
This current will lag the applied voltage by an angle f 3, and it can be expressed as in the
following:
X
f 3 = tan −1 L3
R3
Choose a current scale and draw to the scale the current vectors with the voltage as the reference
axis. Let us add vectorially any two currents, say I1 and I2. The vector sum of I1 and I2 is OE,
as shown in Figure 3.18(a). Add vectorially OE with the other branch current, that is, with I3 to
get sum of the three currents as OF. Convert this length OF to amperes using the current scale
chosen earlier.
An alternate method is to show the three currents with the voltage as the horizontal reference
axis as shown in Figure 3.18(b). Calculate the sum of the horizontal components and vertical components of the currents and then determine the resultant. This has been explained in the following.
The branch currents with their phase angles with respect to V, as shown (not to the scale)
separately in Figure 3.18(b).
Resultant current I can be found out by resolving branch currents I1 , I 2 and I 3 into their X
and Y components as shown in Figure 3.18(b).
X component of I1 (OL) = I1 cos f 1
X component of I 2 (OM ) = I 2 cos f 2
X component of I 3 (ON ) = I 3 cos f 3
Sum of X components (active components) of branch currents = I1 cosf 1 + I 2 cosf 2 + I 3 cos f 3
Y component of I1 ( AL) = −I1 s i n f 1
Y component of I 2 ( BM ) = + I 2 sin f 2
Y component of I 3 ( DN ) = −I 3 sin f 3
I2
B
I2
I 1+ I 2
f2
O
f3
I3
I2 sinf 2
E
f
f1
I1
C
I
A
D
B
V
F
I1 + I2 + I3 = I
I 1+ I 2
(a)
O
f2 N
f3
L
M
f1
I3 sinf 3
I3
V
I1 sinf 1
I1 A
D
(b)
Figure 3.18 Phasor Diagrams of Parallel Circuit Shown In Figure 3.17 (A) Vector Addition
of the Three Currents to get the Total Current and (B) Addition of Active and
Reactive Components of the Currents to get the Total Current
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Network analysis and synthesis
Sum of Y component (reactive components) of the branch currents = − I1 sin f 1 + I 2 sin f 2 − I 3 sin f 3
n f 1 + I 2 sin f 2 − I 3 sin f 3
Active component of resultant current I = I cosf
Reactive component of resultant current I = I sinf
The active and reactive components of the resultant current must be equal to the sum of
active and reactive components of the branch currents.
∴
I cosf = I1 cos f 1+ I 2 cos f 2 + I 3 cos f 3
I sinf = −I1 sin f 1+ I 2 sin f 2 − I 3 siinf 3
Resultant current
I = ( I cos f ) 2 + ( I sin f ) 2
= ( I1 cos f 1+ I 2 cos f 2 + I 3 cos f 3) 2 + ( − I1 sin f 1+ I 2 sin f 2 − I 3 sin f 3) 2
tan f =
I sinf − I1 sin f 1+ I 2 sin f 2 − I 3 sin f 3
=
I cosf I1 cos f 1+ I 2 cos f 2 + I 3 cosf 3
f = tan −1
( − I1 sin f 1+ I 2 sin f 2 − I 3 sin f 3)
( I1 cos f 1+ I 2 cos f 2 + I 3 cos f 3)
The resultant current lags the applied voltage if f is negative, leads the voltage in case f is
positive.
The power factor of the circuit as a whole is expressed as follows:
cos f =
I cos f
I
cos f =
I1 cos f 1+ I 2 cos f 2 + I 3 cos f 3
I
=
sum of active components of branch currents
resultant currrent
3.10.2 Admittance Method of Solving Circuit Problems
Concept of Admittance and Admittance Method: Admittance is defined as the reciprocal of the
impedance. It is denoted by Y and is measured in unit mho or siemens.
Components of admittance are as follows:
1. If the circuit contains R and L, then Z = R + jXL;
2. If the circuit contains R and C, then Z = R − jXC.
3. On considering XL and XC as X, we can write Z = R ± jX.
Let us consider an impedance as given in the following:
Z = R ± jX
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steady state analysis of aC Circuits 111
A positive sign is for an inductive circuit and negative sign is for a capacitive circuit.
Y=
Admittance
1
1
=
Z R ± jX
Rationalising the expression, we get the following:
Y=
R ∓ jX
( R ± jX )( R ∓ jX )
=
R ∓ jX
R2 + X 2
=
R
X
∓j 2
2
R +X
R + X2
=
R
X
∓j 2
Z
Z2
2
Y = G ∓ jB
where G = conductance =
and B = susceptance =
R
Z2
mho
mho
Z2
The value of B is negative if the circuit is
inductive and the value of B is positive if
the circuit is capacitive. The impedance
triangle and admittance triangle for the
circuit have been shown in Figure 3.19.
Z
O
f
X
R
Impedance triangle
Figure 3.19
Consider
a parallel circuit consisting of two branches 1 and 2.
Branch 1 has R1 and L1 in series, while branch 2 has
R2 and C1 in series, respectively. The voltage applied
to the circuit is V volts as shown in Figure 3.20.
Total conductance is found by adding the
conductances of two branches. Similarly, the total
susceptance is found by algebraically adding the
individual susceptance of different branches.
B
Y
f
Admittance triangle
Impedance and Admittance
Triangles
Application of Admittance Method.
1
2
I
I1
R1
L1
R2
C1
I2
V
Figure 3.20
1
1
1
=
+
Z Z1 Z 2
I1 =
V
= VY1
Z1
Y = Y1 +Y 2
I2 =
V
= VY 2
Z2
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G
O
X
Parallel Circuit
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Network analysis and synthesis
Total conductance G = G1 + G2
Total susceptance B = (−B1) + B2 [remember, the value of B is negative if the circuit is
inductive]
Total current I = VY
Power factor cos f = G/Y
It is quite clear that this method requires calculations that are time consuming. To illustrate this
method, we will take one example.
example 3.4 Two impedances Z1 and Z2 are connected in parallel across a 230 V, 50 Hz
supply. Impedance Z1 consists of a resistance of 14 Ω and an inductance of 16 mH. Impedance
Z2 consists of a resistance of 18 Ω and an inductance of 32 mH. Calculate the branch currents,
line current and total power factor. Draw the phasor diagram indicating voltage and currents.
Solution: Let
Z1
R1 = 14 Ω, X L1 = w L1 = 2p f L1
1
Z 1 = R12 + X L12 = 14 2 + 52 = 14.9 Ω
R 2 = 18 Ω, X L 2 = w L 2 = 2p f L 2
= 2 × 3.14 × 50 × 32 × 10
−3
= 10 Ω
Z 2 = R 2 2 + X L 2 2 = 182 + 10 2 = 20.6 Ω
16 mH
14 Ω
= 2 × 3.14 × 50 × 16 × 10 −3 = 5 Ω
2
I
Z2
I1
18 Ω
32 mH
I2
230 V, 50 mH
Figure 3.21 Refers to Example 3.4
The phase angles of Z1 and Z2 are calculated from the impedance triangles shown in
Figure 3.22(a) as:
X
5
f 1 = tan −1 L1 = tan −1 = 19.6°
14
R1
X L2
10
= tan −1 = 29°
18
R2
Let the admittance of branch 1 is Y1 and the admittance of branch 2 is Y2 and they can be written
as follows:
1
1
=
= 0.067∠ − 19.6°
Y1 =
Z1 14.9∠19.6°
f 2 = tan −1
Y2 =
1
1
=
= 0.0485∠ − 29°
Z 2 20.6 ∠29°
On considering voltage V as the reference axis, we get the following set of equations:
I1 =
V
= V Y 1 = 230 ∠0 × 0.067∠ − 19.6°
Z1
= 15.41∠ − 19.6° A
I2 =
V
= V Y 2 = 230 ∠0° × 0.0485∠ − 29°
Z2
= 11.15∠ − 29° A
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I1 =
V
= V Y 1 = 230 ∠0 × 0.067∠ − 19.6°
Z1
steady state analysis of aC Circuits 113
= 15.41∠ − 19.6° A
I2 =
V
= V Y 2 = 230 ∠0° × 0.0485∠ − 29°
Z2
Z1
= 11.15∠ − 29° A
XL2
f2
f1
R2
R1
The phasor diagram indicating V, I1 and
I2 has been shown in Figure 3.22(b). The
sum of I1 and I2 gives total current I. The
cosine of angle between V and I gives
the value of total power factor.
By considering the cosine and sine components of the branch currents and the line
current are as follows:
(a)
f2
f1 f
V = 230 ∠0°
I1 = 15.41 A
I1
f 1 = 19.6°
I2 = 11.15 A
f 2 = 29°
A
I2
I
(b)
I cos f = I1 cos f 1+ I 2 cos f 2
Figure 3.22 (a) Impedance Triangle and
(b) Phasor Diagram
I sin f = I1 sin f 1+ I 2 sin f 2
and
Z2
XL1
By substituting values in the equation, we get the following:
I cos f = 15.41 × 0.942 + 11.15 × 0.335 = 18.24
I sin f = 15.41 × 0.325 + 11.15 × 0.485 = 10.4
tan f =
I sin f
10.4
=
= 0.57 and f = tan −1 0.57 = 30°
I cos f 18.24
Power factor = cos f = cos 30° = 0.866 lagging
I = ( I sin f ) 2 + ( I cos f ) 2 = (10.4) 2 + (18.24) 2 = 21 A
I = 21∠ − 30° A
Current I can also be calculated as follows:
I = VY
where
Y = Y1 + Y2
example 3.5 Figure 3.23 shows a parallel
circuit, an inductance L and a resistance R
connected across 200 V, 50 Hz AC supply.
Calculate the following:
1.
2.
3.
4.
The current drawn from the supply
Apparent power
Real power
Reactive power
M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 113
I
200 V
50 Hz
IR
IL
R = 40 Ω
L = 0.0637 H
Figure 3.23
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Network analysis and synthesis
Solution: Resistance of resistive branch R = 40 Ω
Inductive reactance of inductive branch can be given as follows:
X L = 2p f L
= 2p × 50 × 0.0637 = 20 Ω
V 200
=
= 5A
R 40
V
200
=
= 10 A
Current drawn by inductive branch I L =
X L 20
Current drawn by resistive branch I R =
V
f
IR = 5 A
I cos f = IR
I sin f = IL
1. Current drawn from the supply (see Figure 3.24)
I = I R2 + I L2
I
IL = 10 A
Figure 3.24 Phasor Diagram
= 52 + 10 2 = 11.18 A
2. Apparent powerS = V × I = 200 × 11.18 = 2.236 kVA
3. Real power P = VI cos f = VI R = 200 × 5 = 1.0 kW
4. Reactive power Q = VI sin f = V × I L = 200 × 10 = 2.0 kVAR
example 3.6 The parallel circuit shown in Figure
3.25 is connected across a single-phase 100 V, 50 Hz AC
supply. Calculate the following:
1.
2.
3.
4.
The branch currents
The total current
The power factor
The active and reactive power supplied by the
source.
1
2
I
I1
8Ω
j6Ω
6Ω
−j 8 Ω
I2
100 V, 50 Hz
Figure 3.25
Solution: It is assumed that the students are aware of the method of representation of a
complex number in the form of a + jb or a + ib. However, this has been explained after solving
this problem.
Z 1 = R + jX L = 8 + j 6 = 82 + 6 2 ∠ tan −1
6
= 10 ∠40°
8
Z 2 = R − jX C = 6 − j 8 = 6 2 + 82 ∠ − tan −1
I1 =
V
100 ∠0
=
= 10 ∠ − 40°A
Z 1 10 ∠40°
I2 =
V
100 ∠0
= 10 ∠48°A
=
Z 2 10 ∠ − 48°
8
= 10 ∠ − 48°
6
I = I1 + I 2 = 10 ∠ − 40° + 10 ∠48°
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steady state analysis of aC Circuits 115
I = 10 cos 40° − j10 sin 40° + 10 cos 48° + j 10 sin 48°
= (10 cos 40° + 10 cos 48°)) + j (10 sin 48° − 10 sin 40°)
= 10 × 0.766 + 10 × 0.669 + j (10 × 0.743 − 10 × 0.642)
= 7.66 + 6.69 + j (7.43 − 6.48)
= 14.35 + j 0.95
= (14.35) 2 + (0.95) 2
= 14.45 A
Power factor angle f = tan −1
0.95
= 4°
14.35
Power factor cos f = 0.99
Active power
= V I cos f
Reactive power
= 100 × 14.45 × 0.99
= 1330 W
= V I sin f
= 100 × 14.45 × 0.069
= 99.7 VAR
3.10.3 Use of Phasor Algebra In Solving Circuit Problem
Alternating quantities such as voltage
and current can be represented either
in the polar form or in the rectangular form on real and imaginary axis.
Figure 3.26 shows a voltage V represented in the complex plane. Voltage
V can be represented as V ∠ f. This is
called the polar form of representation.
Further, voltage V can be represented as
V = a + jb = V cos f + jV sin f . This is
called the rectangular form of representation using a j-operator.
Significance of Operator j
Y
Y-axis or imaginary axis
V
−X
b
f
a
→
V = V∠f
→
V = a + jb
X-axis
or real axis
−Y
Figure 3.26 Representation of a Phasor
Operator j used in the expression indicates a real operation. This operation when applied to phasor indicates the rotation of that phasor in the counter-clockwise direction through 90° without
changing its magnitude. As such it has been referred to as an operator. For example, let phasor
A drawn from O to A be in phase with X-axis, as shown in Figure 3.27(a). This phasor when
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116
Network analysis and synthesis
represented by jA shows that phasor A has been rotated in the anti-clockwise direction by an
angle 90° and as such its position now is along Y-axis. If operator j is again applied to phasor
jA, it turns in the counter-clockwise direction by another 90°, thus giving a phasor j2A, that is,
equal and opposite to phasor A, that is, equal to −A (see Figure 3.27(a)).
jA
Imaginary axis, ie
Y-axis
Im
B
5
j 2A = −A
O
f
Real axis, ie
X-axis
A
O
f = tan−1 3
4
= 37°
j3
A
4
Re
OB = 5 37°
= 4 + j3
j 3A = −jA
(a)
(b)
Figure 3.27 Use of Operator j to Represent a Phasor (a) Shows a Phasor Rotation by a
j Operator and (b) Representation of a Phasor Polar form as well as a+jb form
Thus, j2 can be seen as equal to −1. Therefore, the value of j becomes equal to
Hence,
j = + −1 ,
90° CCW rotation from OX-axis
j 2 = j × j = ( −1) 2 = −1,
3
3
4
4
2
180° CCW rotation from OX- axis
270° CCW rotation from OX- axis
j = ( −1) = − −1,
j = ( −1) = ( −1) = 1 ,
and
−1.
360 CCW rotation from OX- axis
From the above, it is concluded that j is an operator rather than a real number. However, it represents a phasor along Y-axis, whereas the real number is represented along X-axis.
As shown in Figure 3.27(b), phasor OB can be represented as 5∠37° in the polar form. In the
rectangular form, OB is represented as 4 + j 3.
VE
Imaginary axis
V1
OB =
jb1
V2
q1
O
q2
q
a1 A a 2 B
AB
OA
3
4
= 5∠37° = 5 cos 37° + j 5 sin 37°
= 5 × 0.8 + j 5 × 0.6 = 4 + j 3
= 4 2 + 32 ∠ tan −1
D
jb2
C
OA 2 + AB2 ∠ tan −1
Real axis
Figure 3.28 Addition of Phasor
Quantities
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The addition and subtraction of phasor quantities are
shown in Figure 3.28.
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steady state analysis of aC Circuits 117
Let
V1 = a + jb1 and V2 = a2 + jb2
V = V1 + V 2
By adding V1 and V 2 ,
= (a1 + jb1 ) + (a2 + jb2 )
= (a1 + a2 ) + j (b1 + b2 )
The magnitude of the resultant vector
V = ( a1 + a2 ) 2 + (b1 + b2 ) 2
Phase angle q = tan −1 =
(b1 + b2 )
( a1 + a2 )
By subtracting V1 from V2, we get the expression as follows:
V = V1 − V2
= ( a1 + jb1 ) − ( a2 − jb2 )
= ( a1 − a2 ) + j (b1 − b2 )
The magnitude of the resultant vector V = ( a1 − a2 ) 2 + (b1 − b2 ) 2
q = tan −1 =
Phase angle
(b1 − b2 )
( a1 − a2 )
The multiplication and division of phasor quantities can be given as follows:
Let
b 
V1 = a1 + jb1 = V1∠q 1; where q 1= tan −1  1 
 a1 
b 
V2 = a2 + jb2 = V2 ∠q 2; where q 2 = tan −1  2 
 a2 
The multiplication of V1 and V2 can be expressed as follows:
V1 = V1 × V 2 = V1∠q 1× V 2 ∠q
2
= V1 V 2 ∠(q 1+ q 2 ),
in the ploar form angles are added algebrically.
The division of V1 by V2 can be written as in the following:
V1 a1 + jb1 V1∠q 1 V1
=
=
∠q 1− q
=
V2 a2 + jb2 V2 ∠q 2 V2
2
angles are subtracted algebraically.
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Network analysis and synthesis
example 3.7 A coil having a resistance of 5 Ω and inductance of 30 mH in series are connected across a 230 V, 50 Hz supply. Calculate current, power factor and power consumed.
Solution: Given, R = 5Ω and L = 30 mH = 30 × 10−3 H
X L = w L = 2p f L
Inductive reactance
X L = 2 × 3.14 × 50 × 30 × 10 −3 Ω = 9.42 Ω
Z = R + jX L = 5 + j 9.42
Impedance
= 52 + (9.42) 2 ∠ tan −1
9.42
5
= 10.66 ∠ tan −1 1.884
= 10.66 ∠62° Ω
V
230 ∠0
=
= 21.57∠ − 62° A
Z 10.66 ∠62°
Current
I=
Magnitude of current,
I = 21.57 A
Current I is lagging voltage V by 62°. Power factor = cosf = cos 62° = 0.47 lagging.
Power consumed = VI cos f = 230 × 21.57 × 0.47 = 2331.7 W
The phasor diagram along with its circuit has been shown in Figure 3.29.
Z = R + jXL
R
L
V = 230° 0°
f = 62°
I
I = 21.57 A
230 V, 50 Hz
Figure 3.29
example 3.8 For the R–L–C series circuit shown
in Figure 3.30, calculate the current, power factor and
power consumed.
15 Ω
0.⊥5 H
100 µF
I
V = 230 V, 50 Hz
Solution: Inductive reactance
Figure 3.30
X L = 2p f L = 2 × 3.14 × 50 × 0.15
= 47.1 Ω
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steady state analysis of aC Circuits 119
Capacitive reactance
1
1
XC =
=
2p fC 2 × 3.14 × 50 × 100 × 10 −6
= 31.84 Ω
Impedance
Z = R + j(X L − X C )
= 15 + j ( 47.1 − 31.84)
= 15 + j15.26 Ω = 152 + (15.26) 2 ∠ tan −1
15.26
15
15.26
15
230 ∠0
230 ∠0°
V
I= =
=
Z 15 + j15.26 21.39∠ tan −1 15.26 /15
230 ∠0°
=
21.39∠ tan −1 1.01
230 ∠0°
=
21.39∠45.3°
= 10.75∠ − 45.3° A
= 21.39∠ tan −1
Current
This shows that the magnitude of current is 10.75 A and the current lags the voltage by 45.3°.
cos f = cos 45.3° = 0.703 lagging
Power factor
Power consumed
P = VI cos f
= 230 × 10.75 × 0.703 W
= 1738.16 W
example 3.9 The coils having impedance Z1 and Z2
are connected in series across a 230 V, 50 Hz power
supply as shown in Figure 3.31.
The voltage drop across Z1 is equal to 120 ∠30° V.
Calculate the value of Z2.
Solution:
We have
or
Z1 = 15 40°
Z2
V1
V2
V = 230 V, 50 Hz
Figure 3.31
V = V1 + V2
V 2 = V − V 1 = 230 ∠0 − 120 ∠30°
= 230(cos 0° + j sin 0°) − 120(cos 30° + j sin 30°)
= 230 − 120 × 0.866 − j120 × 0.5
60
= 126.1 − j 60 = 126.12 + 60 2 ∠ − tan −1
126.1
= 139.6 ∠ − 25.4°
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Network analysis and synthesis
Since this is a series circuit, the current flowing through the circuit components is the same.
The circuit current can be calculated by using any of the following relations:
Accordingly,
I=
V1
V
V
or I = 2 or I =
Z1 + Z 2
Z1
Z2
I=
V1 120 ∠30
=
= 8∠ − 10°A
Z1 15∠40
Since the same current will flow through Z2, it can be written as follows;
V2 139.6 ∠ − 25.4°
=
= 17.45∠ − 15.4°
8∠ − 10
I
= 17.45(cos 15.4° − j sin 15.4°)
= 17.45 × 0.964 − j17.45 × 0.2656
= 16.82 − j 4.6 Ω = R − jX c
Z2 =
example 3.10 An alternating voltage V = (160 + j170)V is connected across an L–R series
circuit. A current of I = (12 − j 5)A flows through the circuit. Calculate impedance, power
factor and power consumed. Draw the phasor diagram.
Solution:
V = 160 + j170 = (160) 2 + (170) 2 ∠ tan −1
170
= 233∠46.8°
160
233∠46.8°
V 160 + j170
=
=
I
12 − j 5
19.2∠ − 22.6°
= 12.13 ∠46.8° + 22.6°
= 12.13∠69.4° Ω
Z =
Impedance
Z = 12.13(cos 69.4° + j sin 69.4°)
= 12.13 × 0.35 + j 12.13 × 0.93
= 4.24 + j 11.28 = R + jX L
V = 233 46.8°
46.8°
22.6°
f=
69.
4°
ref axis
I = 19.2°∠−22.6°
Figure 3.32
The series circuit consists of a resistance of 4.24 Ω and an
inductive reactance of 11.28 Ω. The phasor diagram is drawn by
considering a reference axis. Let X-axis be the reference axis.
The voltage applied has a magnitude of 223 V and is making
46.8° with the reference axis in the positive direction, that is,
in the anti-clockwise direction. The current flowing is 19.2 A
lagging the reference axis by 22.6° as shown in Figure 3.32.
The angle between phasor V and phasor I is 69.4°. This is the
power factor angle.
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steady state analysis of aC Circuits 121
Power factor
cos f = cos 69.4°
= 0.35 lagging
P = VI cos f
Power consumed
P = 233 × 19.2 × 0.35
= 1565.76 W
If supply frequency is taken as 50 Hz, the value of L can be calculated from XL as follows:
X L = 11.28 Ω
X L = w L = 2p fL
or,
L=
XL
11.28
11.28
=
=
H
2p f 2 × 3.14 × 50
314
L=
11.28 × 103
mH = 35 mH
314
example 3.11 A sinusoidal voltage of v = 325 sin 314t when applied across an L–R series
circuit causes a current of i = 14.14 sin (314t − 60°) flowing through the circuit. Calculate the
value of L and R of the circuit. Further, calculate the power consumed.
Solution: Given
v = 325 sin 314t
comparing with
v = Vm sin w t
Vm = 325 V
RMS value
V=
Vm
2
w = 314
=
325
= 230 V
1.414
2p f = 314
314
f =
= 50 Hz
2p
or
given
i = 14.14 sin(314t − 60°)
comparing with
i = Im sin(w t − f )
Im = 14.14
rms value
I=
2
=
14.14
= 10 A
1.414
f = 60°
Power factor angle
Power factor
Im
cos f = cos 60° = 0.5 lagging
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Network analysis and synthesis
P = VI cos f = 230 × 10 × 0.5
Power
= 1150 W
V
230 ∠0
=
= 23∠60° Ω
I 10 ∠ − 60°
Impedance
Z=
In complex form
Z = 23(cos 60° + j sin 60°)
= 23 × 0.5 + j 23 × 0.866
= 11.5 + j 22.99
= R + jX L
Thus, resistance of the circuit R = 11.5 Ω
Inductive reactance
X L = 22.99 Ω
or
w L = 22.99
L=
22.99 22.99
22.99
=
=
w
2p f
2 × 3.14 × 50
22.99
22.99 × 103
H=
mH
314
314
= 73.21 mH
=
example 3.12 A variable resistance R and an inductance L of value 100 mH in series are
connected across a 50-Hz supply. Calculate at what value of R the voltage across the inductor
will be half the supply voltage.
Solution: Given
L = 100 mH
Z
X L = w L = 2p fL
R
L
VR
VL
I
= 2 × 3.14 × 50 × 100 × 10 −3
= 31.4 Ω
V, 50 Hz
Figure 3.33 Refers to Example 3.12
We have to find R for which
VL =
1
V
2
VL = IX L and V = IZ = I R 2 + X L2
Equating
VL =
1
V
2
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steady state analysis of aC Circuits 123
IX L =
or
1
I R 2 + X L2
2
or
R 2 + X L2 = 2 X L = 2 × 31.4 = 62.8 Ω
Equating
R 2 + X L2 = (62.8) 2
R 2 = (62.8) 2 − (31.4) 2
R 2 = 3943.8 − 985.9
R = 54.4 Ω
example 3.13 A voltage of v = 100 sin(314t + 0) is applied across a resistance and inductance
in series. A current of 10 sin (314t − p / 6) flows through the circuit. Calculate the value of R
and L of the circuit. Further, calculate the power factor.
Solution:
v = 100 sin 314t
= V m sin w t
V m = 100 V
V (RMS value) =
Vm
2
=
100
= 70.7 V
1.414
w = 314 or, 2p f = 314, f =
314
= 50 Hz
2p
i = 10 sin (w t − 30°)
= I m sin (w t − 30°)
Similarly
I m = 10, I =
Im
2
=
10
= 7.07 A
1.414
Current I is lagging V by 30°.
Power factor
cos f = cos 30°
= 0.866 lagging
Impedance
In rectangular form,
Z=
V
70.7∠0
=
= 10 ∠30°
I 7.07∠ − 30°
Z = 10(cos 30° + j sin 30°)
= 10 × 0.866 + j10 × 0.5
= 8.66 + j 5.0
= R + jX L
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Network analysis and synthesis
R = 8.66 Ω
XL = 5.0 Ω
Therefore,
and
X L = wL = 314 L
Again
XL
5
5000
=
H=
mH
314 314
314
= 15.92 mH
L=
example 3.14 The expression of applied voltage and current flowing through an AC series
L–R circuit are as follows:
p
p


v = 200 sin  314t +  and i = 20 sin  314t + 


3
6
Calculate the following for the circuit: (a) power factor; (b) average power; (c) impedance, and
(d) values of R and L.
Solution: Comparing voltage v and current i with the standard forms, i.e., n = Vm sin w t and
i = Im sin w t we get
Vm = 200, I m = 20, w = 314
RMS values,
V =
I=
Vm
2
Im
2
=
200
= 141.4 V
1.414
=
20
= 14.14 A
1.414
Figure 3.34 shows the voltage and current with respect to a common reference axis. The voltage V is leading the reference axis by
p /3°, that is, 60°, while current I is leading the reference axis by
30°. The phase angle between V and I is 30°. The current in the
circuit lags the voltage by 30°.
Power factor cos f = cos 30° = 0.866 lagging
Average power
P = VI cos f = 141.4 × 14.14 × 0.866
V
°=
60°
f
I
30
30°
ref. axis
Figure 3.34
= 1732 W
The impedance of the circuit
Z = R 2 + X L2
Again,
Z=
In the polar from,
Z=
V 141.4
=
= 10 Ω
I 14.14
V ∠60° 141.4 ∠60°
=
I ∠30° 14.14 ∠30°
= 10 ∠30° Ω
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steady state analysis of aC Circuits 125
As can be seen from the impedance triangle of Figure 3.35,
Z = Z cos f + jZ sin f
Z
= 10 cos 30° + jZ sin 30°
= 10 × 0.866 + j10 × 0.5
= 8.66 + j 5
= R + jX L
Z sinf = XL
f
Z cosf = R
Figure 3.35
Impedance
Triangle
R = 8.66 Ω
XL =wL
L=
XL
5000
5
H=
=
= 15.92 mH
w
314
314
example 3.15 In an L–R–C series circuit, the voltage drops across the resistor, inductor and
capacitor are 20V, 60 V and 30 V, respectively. Calculate the magnitude of the applied voltage
and the power factor of the circuit.
Solution: In Figure 3.36, the voltage drops across
the circuit components and their phase relationship
have been drawn. Since it is a series circuit, it is
always convenient to have current as the reference
axis. The voltage drop across the resistor and the
current are in phase. Voltage across the inductor will
lead the current, and the voltage across the capacitor
will lag the current. The circuit diagram and the
phasor diagram have been shown in Figure 3.36.
From triangle ABC, we get the following:
I
R
L
C
VR
VL
VC
V
VL
C
V
f
AC2 = AB2 + BC2
or
V 2 = (VR ) 2 + (VL − VC ) 2
= ( 20) 2 + (60 − 30) 2
= 1300
B
VC
I (reference axis)
VR = 20 V, VL = 60 V,
VC = 30 V
Figure 3.36 R–L–C Series Circuit and
Its Phasor Diagram
V = 36 V
or
Power factor
A
(VL − VC)
VR
= cosf =
Power factor angle
AB 20
=
= 0.55
AC 36
f = 56.6°
Current I is lagging V by an angle f = 56.6°. The
power factor is taken as a lagging power factor.
example 3.16 In the circuit shown in Figure 3.37,
calculate the value of R and C.
M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 125
I
R
L = 100 mH
C
VR
VL
VC
i = 14.14 sin (314t + p )
6
u = 325 sin 314 t
Figure 3.37
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126
Network analysis and synthesis
325 sin 314t , w = 314
Solution: v = 325 sin 314t , wv ==314
V
325
Vm V 325
= Vm =
= 230 V
230
=m = 325=, V
2 1.414
2 1.414
14.14 I = I m = 14.14 = 10 A
= 10 A 2 1.414
=
1.414
Vm = 325, V =
I=
Im
2
p
degrees.
6
By considering V as the reference voltage, we obtain the following:
V
230 ∠0
Z= =
= 23∠ − 30°
I 10 ∠ + 30°
In an L–R–C series circuit, if current I is leading voltage V, we have to
VL
consider the circuit as leading P.f circuit. This means the capacitive reactance is more than the inductive reactance (that is, the circuit is effectively
B VR
I
an R–C circuit). We will draw the phasor diagram by taking current on the
f
reference axis. Here, we see that V is lagging I by the power factor angle.
VC − VL
V
That is, I is leading V by an angle f. The phasor diagram, taking I as the
reference axis has been shown in Figure 3.38.
C
When we take V as the reference axis, the following can be obtained:
Current I is leading V by
A
VC
Figure 3.38
Z =
V ∠0
230 ∠0
=
= 23∠ − 30°
1∠ + 30° 10 ∠ + 30
Z = 23 cos 30° − j 23 sin 30°
= 23 × 0.866 − j 23 × 0.5
= 19.9 − j 11.5
Z = R − j (X C − X L )
R = 19.9 Ω
X C − X L = 11.5 Ω
X L = w L = 314 × 100 × 10 −3 = 31.4 Ω
X C − 31.4 = 11.5
X C = 42.9 Ω
X C = 42.9 =
C=
1
1
=
w C 314C
1
F = 72.23 µF
314 × 42.9
Power factor = cos f = cos 30° = 0.866 leading.
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steady state analysis of aC Circuits 127
I
In Figure 3.39, AB = IR = VR has
B
B
been drawn in the direction of current I.
R
I
V
−
V
C
L
R
=
X C − XL
I ( X C − X L ) is effectively a voltage drop,
VR
= I(XC − XL)
f = 30°
f = 30°
that is, capacitive in nature. I will lead
Z
C
A
C
V = IZ
I ( X C − X L ), or we can say that I ( X C − X L ) A
Impedance triangle
will lag I by 90°. BC has been shown lagFigure 3.39
ging AB by 90°. The sum of AB and BC is
AC for which the total voltage V = IZ. By taking out I from all the sides of the triangle ABC, the
impedance triangle has been drawn.
cosf =
Power factor,
IR R
=
IZ Z
example 3.17 A resistance of 15 Ω and an inductance of 100 mH are connected in parallel
across at 230 V, 50 Hz supply. Calculate the branch currents, line current and power factor.
Further, calculate the power consumed in the circuit.
Solution: The circuit diagram and the phasor diagram have been shown in Figure 3.40. We
note that in a parallel circuit, the voltage applied across the branches is the same. The current
in the resistive branch is in phase with the voltage, while current in the inductive branch lags
the voltage by 90°. The phasor sum of the branch currents gives us the total line current. Since
in a parallel circuit voltage, V is common to the parallel branches, we generally take V as the
reference axis while drawing the phasor diagram. Current through the resistive branch IR has
been drawn in phase with V. Current through the inductive branch IL is lagging V by 90°. The
sum of IR and IL gives I, as shown Figure 3.40.
cosf =
Power factor
IR
I
R = 15 Ω
IR
f
L = 100 mH
I
90°
IL
V = 230 V, 50 Hz
IL
V
IR
cos f =
IR
I
I
Figure 3.40
Now using the given values, the calculations are made as follows.
Inductive reactance
X L = w L = 2p fL
= 2 × 3.14 × 50 × 100 × 10 −3 Ω
= 31.4 Ω
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128
Network analysis and synthesis
V 230
=
= 15.33 A
R
15
V
230
IL =
=
= 7.32 A
X L 31.4
IR =
I = I R2 + I L2 = (15.33) 2 + (77.32) 2 = 288.89 = 17 A
= cos f =
Power factor
15.33
= 0.9 lagging
17
f = cos −1 0.9 = 25°
Power factor angle
Since the line current I is lagging the voltage V by 25°, the power factor is mentioned as lagging.
The students should note that while mentioning power factor, it is essential to indicate whether
the same is lagging or leading.
Power consumed
P = VI cos f = 230 × 17 × 0.9 = 3519 W
example 3.18 For the circuit shown in Figure
3.41, calculate the total current drawn from the
supply. Further, calculate the power and power
factor of the circuit.
Solution: For branch 1, impedance Z1 is calculated
as follows:
R1 = 5 Ω
1
2
I
L1 = 150 mH
I1
R2 = 50 Ω L2 = 15 mH
I2
230 V, 50 Hz
Figure 3.41
Z1 = R1 + jX L1
= 5 + jw L1 = 5 + j 2p × 50 × 150 × 10 −3
= 5 + j31.4
= 52 + 31.4 2 ∠ tan −1
31.4
5
= 31.8∠81° Ω
Similarly for branch 2, Z1 is expressed as follows:
Z 2 = R2 + jX L 2 = 50 + j 2p × 50 × 15 × 10 −3
= 50 + j 4.71
= (50) 2 + ( 4.71) 2 ∠ tan −1
4.71
50
= 50.22∠5.4° Ω
Current
I1 =
V
230 ∠0°
=
= 7.23∠ − 81°A
Z1 31.8∠81°
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steady state analysis of aC Circuits 129
Current
I2 =
230 ∠0
V
=
= 4.58∠ − 5.4°A
Z 2 50.22∠5.4°
I = I1 + I 2 = 7.23∠ − 81° + 4.58∠ − 5.4°
Total current
= 7.23 cos 81° − j 7.23 sin 81° + 4.58 cos 5.4° − j 4.58 sin 5.4°
= 7.23 × 0.156 − j 7.23 × 0.987 + 4.58 × 0.995 − j 4.58 × 0.09
= 1.127 − j 7.136 + 4.557 − j 0.414
= 5.68 − j 7.55 = (5.68) 2 + (7.55) 2 ∠ − tan −1
I = 9.44 ∠ − 53°A
Line current
The phasor diagram representing the branch currents
and the line current with respect to the supply voltage
has been shown in Figure 3.42. The line current lags the
applied voltage by an angle f = 53°.
Thus,
7.55
5.68
f 2 = 5.4°
f = 81°
V = 230 ∠0°
I2 = 4.58∠−5.4°
f = 53°
power factor = cos f = cos 53°
= 0.6 lagging
I1 = 7.23∠−81°
It may be noted that the branch 2 is more resistive and
less inductive than branch 1. This causes current I1 to be
more lagging than current I2.
I = 9.44∠−53°
Figure 3.42
P = VI cos f
Power
= 230 × 9.44 × cos 53°
= 230 × 9.44 × 0.6
= 1302.7 W
example 3.19 Two impedances Z1 = 10 + j12 and Z2 = 12 − j10 are connected in parallel
across 230 V, 50 Hz supply. Calculate the current, power factor and power consumed.
Solution: The two impedances are of the form Z1 = R1 + jX L and Z 2 = R2 − jX C
Z1 consists of a resistor and an inductor, while Z2 consists of a resistor and a capacitor.
Z 1 = 10 + j12
= (10) 2 + (12) 2 ∠ tan −1 12 /10
= 15.62∠ tan −1 1.2
I1
= 15.62∠50° Ω
Z 2 = 12 − j10
= (12) 2 + (10) 2 ∠ − tan −1 10 /12
= 15.62∠ − 40° Ω
M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 129
I
R1
L
R2
C
I2
230 V, 50 Hz
Figure 3.43 Refers to Example 3.19
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130
Network analysis and synthesis
V
V
and I 2 = , and then add I1 and I2 to get I. Alternately, we may find
Z1
Z2
V
the equivalent impedance of the circuit, Z and then find I =
Z
We may calculate I1 =
Z =
Z 1Z 2
15.62∠50° × 15.62∠ − 40°
=
Z 1 + Z 2 15.62∠50° + 15.62∠ − 40°
=
243.98∠10°
15.62 cos 50° + j15.62 sin 50° + 15.62 cos 40° − j15.62 sin 40°
=
243.98∠10°
15.62 × 0.64 + j15.62 × 0.76 + 15.62 × 0.76 − j15.62 − j15.62 × 0.64
243.98∠10°
243.98∠10°
=
21.86 − j1.88 21.94 ∠ − tan −1 1.88/ 21.86
243.98∠10
=
= 11.12∠15° Ω
21.94 ∠ − 5°
Z =
or
Total line current I =
V
230 ∠0°
=
= 20.68∠ − 15° A
Z 11.12∠15°
Current I lags voltage V by 15°.
Power factor = cos f = cos 15° = 0.96 lagging
Magnitude of current I = 20.68 A
Supply voltage V = 230 V
Power consumed
P = VI cos f
= 230 × 20.68 × 0.96
= 4566 W
= 4.566 kW
example 3.20 For the circuit shown in Figure 3.44,
calculate the total current, power and power factor of the
whole circuit. Further, calculate the reactive power and
apparent power of the circuit. Draw the phasor diagram.
Solution: XL = 2 × 3.14 × 50 × 50 ×
XC
10−3
= 15.57 Ω
1
1
106
=
=
=
= 63.7 Ω
w C 314 × 5 × 10 −6 314 × 50
R1 = 12 Ω L = 50 mH
I1
I
R2 = 50 Ω
C = 50 µF
I2
200∠30°
f = 50 Hz
Figure 3.44
Z 1 = R1 + jX L = 12 + j15.57 = 19.6 ∠52.2°
Z 2 = R 2 − jX C = 50 − j 63.7 = 80.9∠ − 52°
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steady state analysis of aC Circuits 131
I1 =
200 ∠30°
200 ∠30°
V
V
=
= 10.2∠ − 22.2°; I 2 =
=
= 2.47∠82°
Z1 19.6 ∠52.2°
Z 2 80.9∠ − 52°
I = I1 + I 2 = 10.2∠ − 22.2° + 2.47∠82°
= 10.2 cos 22.2° − j10.2 sin 22.2° + 2.47 cos 82° + j 2.47 sin 82°
= 10.2 × 0.92 − j10.2 × 0.37 + 2.47 × 0.14 + j 2.47 × 0.99
= 9.72 − j1.33
= 9.8∠ − 7.8°
I = 9.8 A
Total current
Voltage V is making an angle of +30° with the
reference axis as shown in Figure 3.45. Current
I2 is making 82° with the reference axis; current
I1 is making −22.2° with the reference axis. The
resultant of I1 and I2 is I. Current I is making an
angle of −7.8° with the reference axis. The phase
difference between V and I is 37.8°, as shown in
Figure 3.45.
Therefore, power factor cos f = cos 37.8° =
0.79 lagging
V = 200∠30°
I2 = 2.47∠82°
82°
f = 30° + 7.8° = 37.8°
30°
22.2°
ref. axis
7.8°
I = 9.8∠−7.8°
I1 = 10.2∠−22.2°
Figure 3.45
P = V I cos f
Active Power (kW)
= 200 × 9.8 × 0.79
= 1548.4 W = 1.5484 kW
= VI sinf
Reactive power (kVAR)
= 200 × 9.8 × sin 37.8°
= 200 × 9.8 × 0.61
= 1195.6 VARs = 1.1956 kVAR
R
= VI = 200 × 9.8
= 1960 VA
= 1.96 kVA
Apparent power (kVA)
We will now check the result as
( kVA ) 2 = ( kW ) 2 + ( kVAR ) 2
kVA = ( kW ) 2 + ( kVAR ) 2 = (1.5484) 2 + (1.1956) 2
= 1.96
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132
Network analysis and synthesis
example 3.21 Three impedances Z1, Z2 and Z3 are connected in parallel across a 230 V, 50 Hz
supply. The values are given as Z 1 = 12∠30°, Z 2 = 8∠ − 30° and Z 3 = 10 ∠60°. Calculate the
total admittance, equivalent impedance, total current, power factor and power consumed by the
whole circuit.
Z1 = 12∠30°
Solution: Admittance
Y1 =
1
1
=
Z 1 12∠30°
Y2 =
1
1
=
Z 2 8∠ − 30°
Y3 =
1
1
=
Z 3 10 ∠60°
Z2 = 8∠−30°
Z3 = 10∠60°
230 V, 50 Hz
Figure 3.46 Refers to Example 3.21
Total admittance Y = Y1 + Y2 + Y3
By substituting values we have,
1
1
1
Y =
+
+
12∠30° 8∠ − 30° 10 ∠60°
= 0.08∠ − 30° + 0.125∠30° + 0.1∠ − 60°
= 0.08(cos 30° − j sin 30°) + 0.125(cos 30° + j sin 30°) + 0.1(cos 60° − j sin 60°)
= (0.227 − j 0.064) mho
= 0.235∠ − 14° mho
1
1
Impedance
Z= =
= 4.25∠14°Ω
Y 0.235∠ − 14°
V
Total current
I = = V Y = 230 ∠0° × 0.235∠ − 14°
Z
= 54.05∠ − 14°A
cos f = cos 14° = 0.97 lagging
Power factor
P = V I cos f = 230 × 54.05 × 0.97 = 12058 W = 12.058 kW
example 3.22 For the circuit shown in Figure
3.47, calculate the current in each branch and
total current by the admittance method. Further,
calculate power and power factor of the total circuit.
Solution:
I
R1 = 12 Ω
XL = 12 Ω
I1
R2 = 8 Ω
XL = 16 Ω
I2
Y1 =
230 V, 50 Hz
1
1
1 1 1
1
=
= 0=.0589∠ − 45° = 0.0589∠ − 45° Figure 3.47
Y1 =
=
Z 1 12 + j12 16
1245
Z 1.96 ∠
+ °j12 16.96 ∠45°
Y2=
1
1
1 1 1
1
= 0=.0559∠ − 64° = 0.0559∠ − 64°
=
Y ==
=
4°j16 17.88∠64°
Z 2 8 + j16 2 17
Z.288∠86+
I1 = V Y 1 = 230 ×I01 .0589
− 45° = 13.54 ∠ − 45°A
= V Y∠
1 = 230 × 0.0589∠ − 45° = 13.54 ∠ − 45° A
I 2 = V Y 2 = 230 ×I 20.=0559
− 64° = 12.85∠ − 64°A
VY∠
2 = 230 × 0.0559∠ − 64° = 12.85∠ − 64°A
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steady state analysis of aC Circuits 133
I = I1 + I 2 = 13.54 ∠ − 45° + 12.85∠ − 64° = 13.54 cos 45° − j13.54 sin 45° + 12.85 cos 64°
I = I1 + I 2−j= 12.65
13.54 ∠
45° + 12.85∠ − 64°
sin− 64°
I = 15.2 − j 21 = 25.9∠ − 54°A
I = 15.2 − j 21 = 25.9∠ − 54°A
or,
= cos f = cos 54° = 0.58 lagging
= VI cos f = 230 × 25.9 × 0.58 = 3455 W = 3.455 kW
Power factor
Power
3.11
AC SeRIeS–PARAllel CIRCUItS
Consider the series–parallel circuit consisting of three branches A, B and C as shown in
Figure 3.48. We will explain as to how to
solve such a circuit problem.
Impedance of branch A
R2 B X2
A
IA
R1
IB
X1
IC R3
I
Z A = R1 + jX1
Impedance of branch B
C
X3
V
Figure 3.48
Z B = R2 + jX 2
Impedance of branch C
ZC = R3 + jX 3
Total impedance of the circuit Z is given as follows:
Z = ZA +
Total current
I=
Z B ZC
Z B + ZC
V
= IA
Z
Current
IB = I
ZC
(applying current divider rule )
Z B + ZC
Current
IC = I
ZB
Z B + ZC
By applying the admittance method, we can also solve the problem as given in the following:
YA =
I
I
I
; YB =
; YC =
ZA
ZB
ZC
Total admittance of the parallel branches B and C is as follows:
YBC = YB + YC
Impedance
Total impedance
M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 133
Z BC =
1
YBC
Z = Z A + Z BC
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134
Network analysis and synthesis
I=
Total current
V
V
=
Z Z A + Z BC
example 3.23 Determine the total
current drawn from the supply by the
series–parallel circuit shown in Figure 3.49.
Further, calculate the power factor of the
circuit.
10 Ω
0.0636 H
6Ω
Z1
8Ω
I
398 µF
0.0319 H
Z3
Z2
Solution:
230 V, 50 Hz
Figure 3.49
w = 2p f = 2 × 3.14 × 50 = 314
Z 1 = 10 + j 314 × 0.0638
= 10 + j 20 = 22.36 ∠64°
Z 2 = 8 − jX C
1
106
=
= 8Ω
w C 314 × 398
Z 2 = 8 − j 8 = 11.3∠ − 45°
Z 3 = 6 + j 314 × 0.0319 = 6 + j10
= 11.66 ∠59°
ZZ
Z = 1 2 + Z3
Z1 + Z 2
XC =
or
Equivalent impedance
22.36 ∠64° × 11.3∠ − 45°
+ 11.66 ∠59°
10 + j 20 + 8 − j8
252.7∠19°
=
+ 11.66 ∠59°
18 + j12
252.7∠19°
=
+ 11.66 ∠59°
21.63∠34°
= 11.68∠ − 15° + 11.66 ∠59°
= 11.21 − j 3 + 6 + j10
= 17.21 + j 7
= 18.58∠22° Ω
=
V
230 ∠0°
=
= 12.37∠ − 22°A
Z 18.58∠22°
Total current I = 12.37 A
Power factor cos f = cos 22° = 0.92 lagging
Current
I=
example 3.24 What should be the value of R for which a current of 25 A will flow through it
in the circuit shown in Figure 3.50.
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steady state analysis of aC Circuits 135
Solution:
5Ω
Z1
Z1 = 5Ω, w = 2p f = 2p × 50 = 314
Z2 10 Ω
Z 2 = 10 + j 314 × 50 × 10 −3
= 10 + j15.7
= 18.6 ∠57.5° Ω
or
50 mH
R
230 V, 50 Hz
Z3 = R Ω
Total impedance
Z3
Figure 3.50
Z
Z 1Z
Z2
Z
Z =
= Z 1+ Z2 +
+Z
Z 33
Z 11 + Z 22
55 ×
× 18
18..6
6∠
∠57
57..55°° R
=
= 5 + 10 + j15.7 +
+R
5 + 10 + j15.7
93
93..00 ∠
∠57
57..55°° R
=
= 15 + j15.7 +
+R
15 + j15.7
93
93..00∠
∠57
57..55°° R
=
= 21.7∠46.5° +
+R
21.7∠ 46.5°
Z
∠11
28∠
11°° +
Z =
= 44..28
+R
R
V 230
=
= 9.2
I
25
By equating the two expressions for Z, we get the following expression:
Again
or
Considering the real part
Z=
4.28∠11° + R = 9.2
R = 9.2 − 4.28∠11°
= 9.2 − 4.28(cos11° + j sin11°)
= 9.2 − 4.19 + j 0.8
= 5.01 + j 0.8
R = 5.01 Ω
example 3.25 In the series–parallel circuit
shown in Figure 3.51, the parallel branches
A and B are in series with branch C. The
impedances are ZA = (4 + j3)Ω, ZΒ = (10 − j7)Ω
and ZC = (6 + j 5)Ω . If the voltage applied
to the circuit is 200 V at 50 Hz, calculate the
following:
1. Current IA, IB and IC;
2. The power factor for the whole circuit.
4
6
IC
C
ZC
j5
IA
IC
10
IB
A
ZA
j3
−j 7
ZB
B
200 V, 50 Hz
Figure 3.51
Draw also the phasor diagram.
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Network analysis and synthesis
Solution:
Z A = ( 4 + j 3) = 5∠36.9° Ω
Z B = (10 − j 7) = 12.2∠ − 35° Ω
Z C = (6 + j 5) = 7.8∠39.8° Ω
Z A + Z B = 4 + j 3 + 10 − j 7
= 14 − j 4
= 14.56 ∠16°
Z AB =
Z AZ B
ZA + ZB
=
5∠36.9° × 12.2∠ − 35°
4 + j 3 + 10 − j 7
=
5∠36.9° × 12.2∠ − 35°
61∠1.9°
=
14 − j 4
14.56 ∠ − 16°
= 4.19∠17.9°
= 4.19(cos 17.9° + j sin 17.99°)
= 4 + j1.3
Z = Z C + Z AB = 6 + j 5 + 4 + j1.3
= 10 + j 6.3 = 11.8∠32.2°
V = 200 ∠0°
200 ∠0°
V
IC = =
= 16.35∠ − 32.2°A
Z 11.8∠32.2°
Let
Using the current divider rule, we obtain the currents as follows:
IA = IC
ZB
12.2∠ − 35°
= 16.35∠ − 32.2° ×
= 13.7∠ − 51.2 A
ZA + ZB
14.56 ∠ − 16°
IB = IC
ZA
5∠36.9°
= 16.35∠ − 32.2° ×
= 5.7∠20.7°A
ZA + ZB
14.56 ∠ − 16°
Phase angle between applied voltage V and line current I is −32.2°.
Hence, the power factor of the whole circuit = cos f = cos 32.2°
= 0.846 lagging
Voltage drop across series branch C, VC = IC ZC
= 16.3∠−32.2° × 7.8∠39.8°
= 127.53∠7.6°V
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steady state analysis of aC Circuits 137
Voltage drop across parallel branches,
VA = VB = IC ZAB
= 16.35∠−32.2° × 4.19∠17.9°
= 68.5∠−14.3° V
20.7°
51.2°
IB = 5.7∠20.7° V = 127.53∠7.6°
C
7.6°
V = 200∠0°
14.3°
32.2°
VB = 68.5∠−14.3°
Note that the voltage across the parallel
branches is also equal to I A Z A or I B Z B .
The complete phasor diagram is shown in
Figure 3.52.
example 3.26 In the circuit shown in
Figure 3.53, determine the voltage at 50 Hz to
be applied across terminals AB in order that a
current of 10 A flows in the capacitor.
5Ω
IC = 16.35∠−32.2°
IA = 13.7∠−51.2°
Figure 3.52 Phasor Diagram Representing
Voltages and Currents in the
Circuit of Figure 3.43
0.0191 H
Z1
I1
A
398 µF
7Ω
C
I
8Ω
0.0318 H
B
Z2
I2
Figure 3.53
Solution:
Z 1 = 5 + j 2p × 50 × 0.0191 = 5 + j 6 = 7.81∠50.2° Ω
1
Z2 = 7− j
= (7 − j 8) Ω = 10.63∠ − 48.8° Ω
2p × 50 × 398 × 10 −6
Z 3 = 8 + j 2p × 50 × 0.0318 = (8 + j10) Ω = 12.8∠51.34°Ω
Current in the capacitive branch, I 2 = 10 ∠0° = 10 + j 0 A
Voltage drop across the parallel branch, VAC = I2Z2
= 10 ∠0° × 10.63∠ − 48.8°
= 106.3∠ − 48.8° V
= (70.02 − j 79.98)V
Current in inductive branch is calculated as
I1 =
VAC 106.3∠ − 48.8°
=
= 13.6 ∠ − 99° A
7.81∠50.2°
Z1
= ( −2.13 − j13.44)A
Circuit current
I = I1 + I2 = 10 + j0 − 2.13 − j13.44 = 7.87 − j13.44
= 15.57∠−59.65° A
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Network analysis and synthesis
Voltage drop across series branch, VCB = IZ3
= 15.57∠ − 59.65° × 12.8∠51.34°
= 199.4 ∠ − 8.31° V
Voltage applied across terminals AB, VAB = VAC + VCB
= (267.33 − j1.8.8) V
= 288.62∠22.15° V
example 3.27 In a series–parallel circuit
shown in Figure 3.54, parallel branches A, B
and C are in series with branch D. Calculate
(a) The impedance of the overall circuit;
(b) Current taken by the circuit and (c) Power
consumed by each and the total power
consumed.
2Ω
IA
P
A
3Ω
IB
2Ω
I
=
Q
B
IC
Solution: The impedance of branch A is, ZA =
2 + j0 = 2 Ω
The impedance of branch B is, Z B = 3 + j 4
Z AB =
4Ω
2Ω
1Ω
1Ω
D
C
110 V, 50 Hz
Figure 3.54
ZA ZB
2(3 + j 4) 6 + j8
=
=
ZA + ZB 2 + 3 + j 4 5 + j 4
6 + j8 5 − j 4 62 − j16
×
=
= 1.51 + j 0.39
5 + j 4 5 − j 4 52 + 4 2
The impedance of branch C, ZC = (2 − j2) Ω
The equivalent impedance of the parallel circuit is given as follows:
Zp =
Z AB ZC
Z AB + ZC
(1.51 + j 0.39)( 2 − j 2)
(1.51 + j 0.39) + ( 2 − j 2)
= 1.136 − j0.118 = 1.142 Ω
=
The impedance of branch D,
ZD = 1 + j1
Total impedance of the overall circuit
Z = ZP + ZD = 1.136 − j0.118 + 1 + j1 = 2.136 + j0.882
V
Z
I
ID
= 2.311∠22.46° Ω
= 110 V
= 2.311 Ω
100
=
= 47.6 A
2.311
= 47.6 A
R = 1Ω
D
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steady state analysis of aC Circuits 139
V = 110 V
Z = 2.311 Ω
100
I=
= 47.6 A
2.311
I D = 47.6 A
RD = 1 Ω
Power consumed by branch
D = I D2 RD
= ( 47.6) 2 × 1
= 2265.8 W
Voltage drop across terminals PQ = IZP
Current in branch A,
IA =
IZ p
ZA
=
47.6 × 1.142
= 27.18 A
2
RA = 2 Ω
Power consumed by branch A = I A2 RA = ( 27.18) 2 × = 1477.5 W
Current in branch B,
IB =
IZ P 47.6 × 1.142
=
= 10.87A
ZB
5
RB = 3 Ω
Power consumed by branch B = I B2 R B = (10.87) 2 × 3 = 354.5 W
Current in branch C,
IC =
IZ P 47.6 × 1.142
=
= 19.21 A
ZC
2.83
RC = 2 Ω
Power consumed by branch C = I C2 RC = (19.21) 2 × 2 = 738 W
Total power consumed by circuit = 1477.5 + 354.5 + 738 + 2265.8
= 4835.8 W
R e v Ie W Q U e S t I o N S
Short Answer type
1. Explain frequency, time period, instantaneous value, maximum value and average value of
a sinusoidal voltage.
2. What do you understand by harmonic wave of a non-sinusoidal wave?
3. Why do we use rms value instead of average value for an alternating quantity?
4. Show that for a sinusoidal voltage rms value is 0.707 times its maximum value.
5. What is the value of Form Factor for a sine wave? What is the significance of the value of
Form Factor for alternating quantity?
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Network analysis and synthesis
6. The Form Factors for different kinds of voltage wave shapes have been calculated as 1.0,
1.11 and 1.15. Is it possible to predict the type of the voltage wave shapes?
7. What are inductive reactance, capacitive reactance and impedance of an L–R–C circuit?
8. What is meant by power factor of an AC circuit? What are its minimum value and its maximum value?
9. Prove that average power in an AC circuit is VI cosf, where V is the rms value of voltage,
I is the rms value of current and cosf is the power factor.
10. What is the significance of very low (poor) power factor of a circuit?
11. A resistance R, an inductance L and a capacitance C are connected in series across an
alternating voltage V current I flows through the circuit. Draw phasor diagram showing the
voltage drops across the circuit parameter with respect to V and I.
12. A resistance of 10 Ω and an inductive reactance of 10 Ω are connected in series. Calculate
the value of impedance and draw the impedance triangle.
13. Show that the current in a pure inductive circuit lags the voltage by 90°.
14. What is the power factor of a purely resistive circuit, purely inductive circuit and purely
capacitive circuit?
15. State the condition for maximum current in an L–R–C series circuit.
16. State the condition for series resonance in an L–R–C circuit.
17. State the condition for parallel resonance. How do we calculate the value of capacitor to be
shunted to create resonant condition?
18. What is the value of resonant frequency in case of series resonance and in case of parallel
resonance?
19. What is meant by Q-factor of a series resonant circuit? What does Q-factor signify?
20. What do you mean by Bandwidth in a series circuit?
21. A resonant circuit with high Q-factor is also called a tuned circuit, explain why.
22. Write the expression for resonant frequency, Q-factor and dynamic impedance for a parallel resonant circuit.
23. Explain what is meant by phase and phase difference of alternating quantities.
24. A sinusoidal current is expressed as i = 100 sin 314t. What are the maximum value, rms
value, frequency and time period of the alternating current?
25. A sinusoidal voltage v = 300 sin (314t + 30°) when connected across an AC series circuit
produces a current i = 20 sin (314t − 30°). What is the power factor of the circuit? Draw the
phasor diagram
26. Define apparent power, active power and reactive power of an AC circuit.
27. Define the terms: impedance, inductive reactance, capacitive reactance, admittance, active
power, reactive power and power factor for an AC circuit.
28. Two impedances Z1 = 10∠30° and Z2 = 20 ∠60° are connected in series. What is the value
of equivalent impedance?
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steady state analysis of aC Circuits 141
29. Two impedances Z1 = 10∠30° and Z2 = 20 ∠30° are connected in parallel. What is the
equivalent impedance?
30. A impedance of 10 + j10 is connected across a voltage of 230∠60° V. What are the magnitude of current and the value of power factor?
31. Two impedances Z1 = 10∠30° and Z2 = 10 ∠60° are connected in series. Calculate the
equivalent impedance.
32. What is resonant frequency? Why series resonance is called voltage resonance?
Numerical Problems
1. Calculate the rms value of an alternating current i = 20(1 + sinq ).
[Ans. 24.5 A]
2. Calculate the rms value of a half-wave rectified voltage of maximum value of 100 V.
[Ans. 50 V]
3. An alternating voltage is expressed as v = 141.1 sin 314t. What are the rms value, time
period and frequency?
[Ans. 100 V, 20 ms, 50 Hz]
4. An alternating current of frequency 50 Hz has its maximum value of 5 A and lagging the
voltage by 30°. Write the equation for the current.
[Ans. i = 5 sin (314t − 30°)]
5. An alternating voltage is expressed as v = 100 sin 314t. Determine the time taken for the
voltage to reach half its maximum value, and the time is counted from t = 0. At what time,
voltage will reach its maximum value?
[Ans. t = 1.66 ms; t = 5 ms]
6. Determine the average value of the voltage wave form shown in the Figure 3.55.
n
15 V
0
3
6
9
12
t
[Ans. 7.5 V]
Figure 3.55
7. An alternating voltage is defined as:
n = 100 sin q V
n=0V
0<q <p
p < q < 2p
What is the rms value of this voltage?
[Ans. 50 V]
8. Find the rms value of the sinusoidal voltage waveform shown in the Figure 3.56.
n
100V
0 p
4
p
Figure 3.56
M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 141
2p
T
[Ans. 47.6 V]
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Network analysis and synthesis
9. Find the rms value of the voltage wave shown in the Figure 3.57.
n
50 V
0
2
4
6
Figure 3.57
8
t
[Ans. 20.4 V]
10. A resistance of 50 Ω, an inductance of 0.1 H and a capacitance of 50 µF are connected in
series across a 230 V, 50 Hz supply. Calculate the following:
1. The value of impedance
2. Current flowing
3. Power factor
4. Power consumed
[Ans. Z = 59.5 Ω, I = 3.86 A, P.f. = 0.84 leading, P = 746.7 W]
11. In an R−L−C series circuit, the voltage across R is 160 V, across L is 240 V and power consumed is 1000 V when a voltage of 200 V at 50 Hz is applied across the circuit. Calculate
the value of the capacitor and the current flowing through the circuit.
[Ans. C = 165.8 µF, 6.25 A]
12. An impedance of Z = 50∠−60°Ω is connected across a 230 V, 50 Hz supply. Calculate the
values of circuit elements, current and power factor.
[Ans. R = 25, C = 73.54 µF, I = 4.6, P.f. = 0.5 leading]
13. A coil of resistance 5 Ω and inductance 20 mH is connected across a voltage of v = 230 sin
314t. Write an expression for the current flowing through the circuit.
[Ans. i = 28.75 sin (314t−51°)]
14. A resistance of 50 Ω and a capacitor of 100 µF are connected in series. The supply voltage
to the circuit is 200 V at 50 Hz. Calculate the voltage across the resistor and across the
capacitor. Further, calculate the current and power factor.
[Ans. VR = 168.7 V, VC = 107 V, I = 3.37 A, P.f. = 0.84 leading]
15. In an R−L−C series circuit, a maximum current of 0.5.A is obtained by varying the value
of inductance L. The supply voltage is fixed at 230 V, 50 Hz. When maximum current flows
through the circuit, the voltage measured across the capacitor is 350 V. What are the values
of the circuit parameters?
[Ans. R = 460 A, L = 2.229 H, C = 4.549 µF]
16. A 200 V, variable frequency supply is connected across an L−R−C series circuit with
R = 10 Ω, L = 10 mH and C = 1 µF. Calculate the frequency at which resonance will occur.
Further, calculate the Q-factor and bandwidth.
[Ans. f = 1591.5 Hz, Q-factor = 10, bandwidth = 159 Hz]
17. A coil of R = 10 Ω, L = 0.023 H is connected in parallel with another coil of R = 5 Ω, L =
0.035 H. The combination is connected across at 200 V, 50 Hz supply. Calculate the current
drawn from the supply and the power factor.
[Ans. I = 31.4 A, P.f. = 0.63 lagging]
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steady state analysis of aC Circuits 143
18. A coil of resistance 20 Ω and inductance of 300 mH is connected in parallel with a capacitance of 200 µF. The combination is connected across 200 V, variable frequency power supply. At what frequency, will the parallel circuit resonate and what would be the current at
resonance?
[Ans. 20.5 Hz, 2.66 A]
19. Calculate the value of R in the circuit shown in the Figure 3.58 such that the circuit will
resonate.
V, f
R
−j 2
10 Ω
j 10 Ω
[Ans. R = 6 Ω]
Figure 3.58
20. Calculate the half power frequencies of a series resonance circuit in which the resonant
frequency is 150 kHz and bandwidth 75 kHz.
[Ans. f1 = 117 kHz, f2 = 0.19 kHz]
M U ltI P l e C Ho I C e Q U e S t I oN S
1. The voltage and current in an AC circuit is represented by u = Vm sin (w t + 30°) and i = Im
sin (w t − 45°). The power factor angle of the circuit is
(a) 15°
(b) 75°
(c) 45°
(d) 30°
2. A current is represented by i = 100 sin (314t − 30°) A. The rms value of the current and the
frequency are, respectively
(a) 100 A and 314 Hz
(b) 100 A and 50 Hz
(c) 70.7 A and 314 Hz (d) 707 A and 50 Hz
3. A current of 10 A is flowing through a circuit. The power factor is 0.5 lagging. The instantaneous value of the current can be written as
(a) i = 10 sin 60°A
(c) i = 14.14 sin (w t − 60°)A
(b) i = 10 sin (w t − 30°)A
(d) i = 14.14 sin (w t + 60°)A
4. In a purely inductive circuit
(a) current lags the voltage by 90°
(c) voltage lags the current by 90°
(b) current leads the voltage by 90°
(d) current lags the voltage by 180°
5. Form factor of an AC wave indicates
(a)
(b)
(c)
(d)
Low sharp or steep the wave shape is
Low flat the wave shape is
Low symmetrical the wave shape is
The degree of its conformity to sinusoidal form
6. Power consumed by a pure inductor is
(a) infinite
(b) very high
(c) zero
(d) very small
7. If form factor of a sinusoidal wave is 1.11, then the form factor of a triangular wave will
(a) also be 1.11
(b) be less than 1.11
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(c) be more than 1.11
(d) be 1
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Network analysis and synthesis
8. A voltage of u = 100 sin (314t − 30°) is connected across a 10-Ω resistor. The power
dissipated in the circuit will be
(a) 100000 W
(b) 1000 W
(c) 500 W
(d) 250 W
9. The average value of a sinusoidal current is
(a)
2I m
p
(b)
Im
p
(c)
Im
2p
(d)
I m2
2p
10. Form factor of an alternating wave form is the ratio of
(a) rms value and average value
(c) Maximum value and average value
(b) Average value and rms value
(d) Maximum value and rms value
11. A capacitance of C farad is connected to a 230 V, 50 Hz supply. The value of capacitive
reactance is
(a) 314 C Ω
(b)
1
Ω
314C
(c) 628 C Ω
(d)
1
Ω
628C
12. The form factor of a square wave is
(a) 1.11
(b) 1.0
(c) 0
(d) 1.414
13. Two sinusoidal waves are represented as n1 = 100 sin (w t + 30°) and n2 = 200 sin (w t − 60°).
The phasor relationship between the voltages can be expressed as
(a) n1 lags n2 by 90°
(b) n2 lags n1 by 90° (c) n1 lags n2 by 30°
(d) n2 lags n1 by 60°
14. Inductive reactance of coil of 0.1 H at 50 Hz is
(a) 31.4 Ω
(b) 62.8 Ω
(c) 314 Ω
(d) 5 Ω
15. The power factor of a purely resistive circuit is
(a) 1.0
(b) 0
(c) 0.1
(d) 0.5
16. A sinusoidal voltage is represented as v = 141.4 sin (628t − p / 3) ; therefore, the rms value,
frequency and phase angle are, respectively,
(a) 141.4, 628, 60°
(b) 100, 100, −60°
(c) 141.4, 50, 60°
(d) 141.4, 100, 60°
17. In an R–L series circuit, the power factor of the circuit is increased if
(a) XL, the inductive reactance is increased
(c) R, resistance is decreased
(b) XL, the inductive reactance is decreased
(d) the supply frequency is increased
18. The power factor of an R–L circuit can be expressed as
(a) cosf =
R
Z
(b) cosf =
XL
Z
(c) cosf =
R
XL
(d) cosf =
XL
R
19. An R–L series circuit consists of R = 3 Ω and XL = 4 Ω. The impedance of the circuit is
(a) Z = 7 Ω
(b) Z = 1 Ω
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(c) Z = 5 Ω
(d) Z =
7Ω
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steady state analysis of aC Circuits 145
20. The impedance of a circle is 20∠30°. The value of resistance and inductive reactance of the
circuit, respectively, are
(a) 10 Ω and 17.32 Ω
(b) 17.32 Ω and 10 Ω
(c) 10 Ω and 10 Ω
(d) 10 Ω and 8.65 Ω
21. The impedance of an R–L series circuit is 628∠30°, when the supply frequency is 50 Hz.
The value of inductance L is
(a) 314 H
(b) 1 H
(c) 2 H
(d) 628 H
22. An impedance of Z = 6 + j8 Ω is connected across a 200 V, 50 Hz supply. The power factor
of the circuit is
(a) 0.6 lagging
(b) 0.6 leading
(c) 0.75 lagging
(d) 0.8 lagging
23. The current flowing through the circuit of a 200 V, 50 Hz supply is
(a) 10 A
(b) 20 A
(c) 14−28 A
(d) 2 A
24. An R–L series circuit has an impedance of 10 + j10 Ω. The power factor angle of the circuit is
(a) 30° lagging
(b) 30° leading
(c) 45° leading
(d) 45° lagging
25. An R–C series circuit has resistance of 10 Ω and capacitive reactance of 10 Ω. The phase
difference between the voltage and current in the circuit will be
(a) current will lead the voltage by 90°
(c) current will lead the voltage by 45°
(b) current will lag the voltage by 90°
(d) current will lag the voltage by 45°
26. The impedance of an R–L series circuit is (50 + j100) Ω at 50 Hz. When the supply frequency is 100 Hz, the value of impedance will be
(a) (50 + j1000) Ω
(b) (50 + j200) Ω
(c) (100 + j100) Ω
(d) (100 + j200) Ω
27. A voltage of u = 10 sin (314t + 15°) is applied across an R−L−C series circuit, where
R = 5 Ω, XL = 15 Ω, and XC = 10 Ω. The current flowing in the circuit will be
(a) 033 A
(b) 1 A
(c) 1.414 A
(d) 2 A
28. The resonant frequency in R−L−C series circuit is
(a) f 0 =
2p
LC
(b) f 0 =
LC
2p
(c) f 0 =
1
2p LC
(d) f 0 =
1
2p
L
C
29. A series RLC circuit has R = 50 Ω, L = 50 µH and C = 2 µF. The Q-factor of the circuit is
(a) 0.1
(b) 1
(c) 10
(d) 2
30. When a parallel circuit is in resonance, which of the following of the circuit is maximum?
(a) current
(b) impedance
(c) admittance
(d) power factor
31. In an R−L−C series circuit, if the frequency less than the resonance frequency,
(a) XL > XC
(b) XC > XL
(c) XC = XL
(d) X Ca
1
XL
32. In an R−L−C series circuit, if the frequency is made more than the resonant frequency, the
circuit will effectively be
(a) inductive
(b) capacitive
(c) resistive
(d) oscillatory
33. Two impedances Z1 = 4 + j4 Ω and Z2 = 4 − j4 Ω are connected in parallel. Their equivalent
impedance is
(a) 8 + j8 Ω
(b) 4 + j0 Ω
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(c) 8 − j8 Ω
(d) 8 + j0 Ω
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Network analysis and synthesis
34. When an inductance L and a resistance R are connected in parallel across an AC supply, the
current drawn by the two parallel branches will be out of phase by
(a) 0°
(b) 90°
(c) 180°
(d) 45°
35. When an inductance L and a capacitance C are connected in parallel across an AC supply,
the current drawn by the two parallel branches will be out of phase by
(a) 0°
(b) 90°
(c) 180°
(d) 45°
36. In an R−L circuit, XL = R. The power factor angle of the circuit is
(a) 30°
(b) 45°
(c) 60°
(d) 0°
37. In a series resonant circuit, a change in supply voltage will cause a change in
(a) the current drawn,
(c) the bandwidth of the circuit,
(b) the Q-factor of the circuit,
(d) the resonant frequency as well
38. Which of the following conditions is true for both series and parallel resonance?
(a) impedance is minimum
(c) power factor is zero
(b) power factor is unity
(d) power is low
39. The bandwidth of a series R−L−C circuit is
(a)
C
2p L
(b)
R
2p L
(c)
C
2p R
(d)
L
2p R
40. The product of voltage and current in an AC circuit is called
(a) active power
(b) apparent power
41. In a series resonance circuit
(a) L = C
(b) L = R
(c) average power
(d) reactive power
(c) XL = XC
(d) R = L = C
ANS W e RS
1.
11.
21.
31.
41.
b
b
b
b
c
2.
12.
22.
32.
d
b
a
a
3.
13.
23.
33.
c
b
b
b
4.
14.
24.
34.
a
a
d
b
5.
15.
25.
35.
M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 146
a
a
c
c
6.
16.
26.
36.
c
b
b
b
7.
17.
27.
37.
c
b
b
a
8.
18.
28.
38.
c
a
c
b
9.
19.
29.
39.
a
c
a
b
10.
20.
30.
40.
a
b
b
b
11/17/2014 4:48:23 PM
R–L–C Circuits
and Resonance
4
Chapter objectives
After carefully studying this chapter, you should be able to do the following:
State the condition of resonance in an
Explain how a series resonant cirR–L–C series circuit.
cuit can be used as a band-pass and
band-stop filter.
Explain the effect of variation of frequency of supply voltage on the curDistinguish between series resonance
rent, power factor, and voltage drops in
and parallel resonance.
an R–L–C series circuit.
Establish the condition for parallel resExplain how R–L–C series circuit at
onance in an ideal L–C circuit.
resonance can be used as filter element.
Explain the working of a non-ideal
Explain with the help of an example
tank circuit.
the application of R–L–C series resoDraw and explain parallel resonant circuit.
nant band-pass and band-stop
Explain the meaning of half-power
filters.
frequencies.
Explain the use of resonant circuit in a
radio receiver.
Define quality factor of a series resonant circuit.
Solve numerical problems on resonant
Draw and explain selectivity curve and
circuits.
bandwidth of a series resonant circuit.
Draw locus diagram of current in a
Establish relationship between bandR–L circuit with variable R.
width and quality factor.
4.1 R–L–C SERIES CIRCUIT with variable
frequency input voltage
In the previous chapter, we had discussed R–L–C circuits
with a constant frequency voltage source. We will now
consider a series R–L–C circuit with a variable frequency
voltage source, as shown in Figure 4.1.
M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 147
R
L
C
Vs
Figure 4.1 Series R–L–C
Circuit
11/17/2014 4:54:04 PM
148 Network Analysis and Synthesis
The inductive reactance (XL) causes the total current to lag the source voltage. The capacitive
reactance (XC) has the opposite effect. It causes the total current to lead, lag, or be phase with
the voltage source. Thus, XL and XC will tend to cancel each other’s effect. In case XL equals XC
at a particular frequency, they will cancel each other and the total impedance of the circuit will
be the resistance only. With the supply voltage of variable frequency, at a particular frequency,
XL will be equal to XC. Such a condition is called resonance. It is interesting and of practical
importance to study the effect of changing frequency of input voltage in circuits when the circuit elements are either connected in series or in parallel.
4.2 SERIES RESONANCE
When the R–L–C series circuit is connected across a voltage source of variable frequency, and
the frequency is increased, XL will increase but XC will decrease because
X L = 2p f L and X C =
1
2p f C
Series resonance will occur when XL = XC.
The frequency at which resonance occurs is
L
I
R
called the resonant frequency fr. At resonant
0V
VR
I
frequency fr, the inductive and capacitive
fr
reactances are equal in magnitude and they
VS
VC
effectively cancel each other, thus making
(a)
(b)
total impedance Zr = R, which is shown in
Figure
4.2. Since the current flowing through
Figure 4.2 Series Resonance: (a) Circuit
the
series
circuit is the same, voltage drop
­Diagram and (b) Phasor Diagram
across the capacitor will be equal to the voltage drop across the inductor. During any given cycle of the input voltage, polarities of the voltages across C and L are opposite. The equal and opposite voltages will cancel each other. As
shown in Figure 4.2, there is no voltage drop across points A and B but there is current flowing
through the circuit. Therefore, the total reactance of the circuit must be zero.
Since,
VAB = 0, I XTotal = 0
or
XTotal = 0
or
XL - XC = 0
Therefore, the condition for series resonance is XL = XC when f = fr.
The condition for resonance can be written as follows:
1
2p f r L =
2p f r C
VR
A +
C
VC
− −
VL
+ B
VL
or
or
M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 148
f r2 =
fr =
1
4p 2 LC
1
2p LC
11/17/2014 4:54:05 PM
R–L–C Circuits and Resonance 149
Example 4.1 A resistance of 10 Ω, a capacitor of 470 pF and inductor of 0.5 mH are connected
in series across a variable frequency voltage source. Calculate the value of frequency at which
the circuit will attain the resonance condition.
Solution: f r =
1
2p LC
=
1
2 × 3.14 0.5 × 10 -3 × 470 × 10 -12
= 328kHz
4.2.1 Effect of Variation of Frequency on Current
and Voltage Drops
In a series R–L–C circuit, at a frequency fr, XL = XC. Below this resonant frequency, XL is less
than XC as frequency is decreased from fr, XL decreases but XC increases (XL is directly proportional to frequency, while XC is inversely proportional to frequency).
Let us start from zero frequency, that is, at f = 0 Hz, (that is, when supply voltage in DC), the
capacitor shown in Figure 4.2 appears to be open, and hence, blocks the flow of current in the
R–L–C circuit. When no current flows, there is no voltage drop across R or L as in Figures 4.3(b)
and (d) respectively, and the supply voltage appears across the capacitor as shown in Figure 4.3(c).
The impedance of the circuit is infinite at zero frequency as XC is infinite and XL is zero.
As frequency increases, XC decreases but XL increases. The total reactance XC - XL decreases.
The impedance of the circuit is Z = R 2 + ( X C - X L ) 2 . As frequency increases, XC - XL decreases
and hence Z decreases. Since Z decreases, current I in the series circuit increases as has been
shown in Figure 4.3(a). As current increases, the voltage across the resistor increases [see fig
4.3(b)], and both VC and VL increases [see fig 4.3(c) and (d)]. The combined voltage across the
inductor and the capacitor decreases from its initial voltage VS as has been shown in Figure 4.3 (e).
I VS /R = Im
0
VR
f
fr
IXC
VC
VS
VS
0
f
fr
(a)
0
fr
(b)
IXL
VL
(c)
VCL
VS
0
f
VS
fr
(d)
f
0
fr
f
(e)
Figure 4.3 E
ffect of Variation of Frequency of Supply Voltage in an R–L–C Series Circuit:
(a) Current; (b) Voltage Across R; (c) Voltage Across C; (d) Voltage Across L
and (e) Voltage Across C and L
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150 Network Analysis and Synthesis
At resonant frequency fr, XL = XC, Z = R (minimum), and hence, the current is maximum.
The voltage across the capacitor and the inductor can be much larger than the supply voltage but
these two voltages oppose each other making their combined voltage as zero.
Resonance Condition
The variations of I, VR, VL, VC and VCL with variation of frequency in the series R–L–C circuit
have been shown in Figure 4.3. At resonance frequency fr, the impedance of the circuit is R only,
and hence, the maximum current flows in the circuit creating a resonance condition.
Under the resonance condition, the impedance of the circuit is minimum and is equal to the
resistance of the circuit; the current is the maximum in the circuit and the voltage across the
capacitance is equal and opposite to the voltage across the inductance, which is much more than
the source voltage. That is why a series resonant circuit is called a voltage resonant circuit. At
resonance, the voltage drop across the capacitor and the inductor will be many times more than
the source voltage.
As frequency is increased above resonant frequency, XL continues to increase and XC
­continues to decrease.
The total reactance XL - XC will increase, and hence, the current will decrease. As current
decreases, the voltage across the resistance decreases. Further, VL and VC will decrease but the
difference of VL and VC will be increasing. When frequency is increased to a very high value,
the circuit current will approach zero. Accordingly, VR and VC will approach zero, and VL will
approach the value of apply voltage VS. (See Figure 4.3 (b), (c) and (d)).
4.2.2 Effect of Variation of Frequency on Impedance
and Power Factor
The variation of XL, XC and Z with
frequency has been shown in Figure
P.f (Lagging)
4.4. It can be observed that at resonant
XL
frequency fr, the impedance of the circuit
is minimum, that is, equal to R.
Z
P.f.
At zero frequency, both XC and Z are
infinitely large and XL is zero. As the
XC
frequency increases, XL increases but
Z=R
XC decreases. Since XC is larger than
0
f(Hz)
fr
XL at frequencies lower than fr, Z also
decreases along with XC. At frequenFigure 4.4 Variations of XL, XC, Z and p.f as a
cies
higher than fr, XL increases and is
Function of Frequency in an R–L–C
higher
than XC and hence Z increases.
Series Circuit
At frequencies lower than resonant
frequency, XC is higher than XL and hence current I will lead the source voltage VS. At resonant
frequency, XL becomes equal to XC, and the impedance of the circuit is purely resistive and is
minimum. Therefore, the current will be maximum as both VS and I will be in phase, that is, the
power factor of the circuit will be unity. At frequencies higher than the resonant frequency, XL
is higher than XC, and hence, the current will lag the voltage.
XL, XC, Z
(Ω)
1.0
P.f
(Leading)
Unity P.f.
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R–L–C Circuits and Resonance 151
4.3 APPLICATIONS OF R–L–C
CIRCUITS
C
L
Vin (Input
signal)
R–L–C series circuits are often used as filters. The
common use of R–L–C circuits as band-pass and
band-stop filters are described in the following
sections.
R
Vout (Output
signal)
Figure 4.5 An R–L–C Series Resonant
Band-pass Filter
4.3.1 Band-pass Filter
A series R–L–C circuit with L–C part placed in between the input and the output has been
shown in Figure 4.5. The output is taken across the resistor as shown. At resonant frequency,
the capacitive reactance is equal to the inductive reactance and the two reactances cancel each
other. The circuit works as a band-pass filter. Signals at resonant frequency are allowed to pass
from input to output without any reduction in amplitude because the L–C part is not offering
any opposition. In fact, for a range of frequencies extending below and above the resonant
frequency, a significant strength of the input signal will pass to the output circuit. This band of
frequencies is called pass-band. The signals at frequencies lower or higher than the pass-band
appearing at the input are considered to be rejected by the filter.
The filtering of signals of frequencies higher or lower than the resonant frequency band is
due to the varying impedance characteristic of the filter circuit. The impedance is very low at
resonant frequency and very high at frequencies lower and higher than the resonant frequency.
Hence, the filter will block the current at these very low and very high frequencies.
Bandwidth of the Pass-band
The bandwidth (BW) of the pass-band filter is
the range of frequencies at which the ­current
in the circuit is equal to or greater than 70.7%
of its value at resonant frequency. The BW
is illustrated in Figure 4.6 where at resonant
frequency, the circuit current is the maximum
allowing 100% of the input signal appearing
at the output. The frequencies at which the
output of a filter is 70.7% of its maximum
value are called the cut-off frequencies f1 and
f2 as shown. Frequency f1 is called the lower
cut-off frequency and f2 is called the upper
cut-off frequency. These two frequencies f1
and f2 are also called band frequencies, -3 dB
frequencies or half-power frequencies.
Thus, BW = f2 - f1 Hz.
I
1.0
0.707
f1 fr
f2
f
BW
Figure 4.6 B
andwidth of a Series Resonant
Circuit
Example 4.2 A series resonant R–L–C circuit has a maximum current of 200 mA at the
resonant frequency. The lower and upper cut-off frequencies are 8 kHz and 10 kHz, respectively.
What is the value of current at cut-off frequencies? What is the BW?
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152 Network Analysis and Synthesis
Solution: Resonant current = 200 mA.
The current at cut-off frequencies is 70.7% of the resonant value. Hence, the current at cutoff frequencies of 8 kHz and 10 kHz is = 0.707 Im = 0.707 × 200 mA = 141.4 mA
f1 = 8 kHz and f2 = 10 kHz
Given,
BW = f2 - f1 = 10 kHz - 8 kHz = 2 kHz.
Half-power Frequencies
As mentioned earlier, the upper and the lower cut-off frequencies are called half-power frequencies. This is because the power from the source at these frequencies is one-half the power
2
R
delivered at resonant frequency. To establish this relation, at resonance, Pmax = I max
The power at f1 and f2 is as follows:
2
= (0.707 I max ) 2 R = 0.5 I max
R
= 0.5 Pmax
Therefore, the power at the cut-off frequencies is equal to half the power at resonant
frequency.
Quality Factor of a Resonant Circuit
Quality factor is the ratio of energy stored in the reactor (reactive power) to the true power in the
resistance of the circuit. It is the ratio of power in L to the power in R.
Q=
energy stored
reactive power
=
energy dissipated
True or acttive epower
I 2 XL
or
2
I R
=
XL
R
where XL is the reactance at resonant frequency fr
Bandwidth and Selectivity
I
Imax1
Greater
selectivity
Imax2
0.707 Imax1
Lower
selectivity
0.707 Imax2
BW1
BW2
Figure 4.7 S electivity Curves of a Band-pass
Filter
M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 152
f
Bandwidth (BW) is the range of frequencies for which the current is equal
or greater than 70.7% of its value at
resonant frequency.
Selectivity indicates how the resonant circuit discriminates certain frequencies against all other frequencies of
the signal. If the BW is narrow, selectivity is high. Figure 4.7 shows selectivity
curves of a band-pass filter circuit. It is
observed that if the curve is steeper, the
selectivity is higher.
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R–L–C Circuits and Resonance 153
Ideal and Actual Selectivity Curve
Ideally, at resonant frequency band ( f2 - f1), the filter circuit accepts signals and rejects or eliminates any signal outside these frequencies, as shown in Figure 4.8. Actually, the signals outside
the BW are not completely eliminated. If the selectivity curve is as shown in Figure 4.8(b),
there will be complete elimination of signals outside the BW, which is not the case in practice.
However, outside the pass-band, the signals are reduced to lower than 70.7% and are assumed
to be rejected.
I
Im
I
Im
0.707 Im
f
f1 fr f2
BW
f1
f
f2
Pass band
Pass band
(b)
(a)
Figure 4.8 Selectivity Curve of Pass-band Filter: (a) Actual and (b) Ideal
Relation between Quality Factor and Bandwidth
Higher quality factor indicates that the selectivity curve is sharper and hence narrower
is the BW. The BW, quality factor Q and resonant frequency fr are related as given in the
following:
BW =
fr
Q
Example 4.3 Calculate the BW of the filter circuit shown
in Figure 4.9.
1
Solution: Resonant frequency f r =
2p LC
r = 10 Ω
0.04 µF L = 10 mH
C
By substituting the values in the equation, we get the
following:
fr =
L
Vin
R V
o
50 Ω
Figure 4.9
1
2 × 3.14 10 × 10 -3 × 0.04 × 10 -6
= 79.6 × 103 H 3 = 79.6 kHz
At resonant frequency, the value of XL is calculated as follows:
XL = 2p frL = 6.28 × 79.6 × 103 × 10 × 10-13
= 5 kΩ
RTotal = r + R = 10 + 50 = 60 Ω
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154 Network Analysis and Synthesis
Q=
BW =
XL
5 × 1000
=
= 83.3
60
RTotal
f r 79.6 × 1000
=
= 955Hz.
Q
83.3
4.3.2 Band-stop Filter
When the output is taken from the L–C portion of an R–L–C circuit, the series resonant circuit
becomes a band-stop filter as shown in Figure 4.10.
Such a filter will reject the signals with frequencies between the lower and upper cut-off frequencies
R
and will pass those signals with frequencies lower
L
Vin
Vo
than lower cut-off frequency and higher than higher
C
cut-off frequency. This type of filter is also known as
band-reject or band-elimination filter. The response
curve for a band-stop or stop-band filter is shown in
Figure 4.10 A Series Resonant
Figure 4.11.
Band-stop Filter
It is seen that for the stop-band filter, the output
I
current and the output voltage are minimum at resoIm
nant frequency. At very low frequency, XC is very high
and
XL is very low, and hence, the combination of L–C
0.707 Im
network appears to be nearly open. Hence, almost the
whole of the input voltage will appear at the output.
At resonant frequency the impedance offered by the
f
L–C network is reduced to its lowest level, the whole
f1
fr
f2
of input signal is grounded and the output voltage
becomes negligible. As the frequency increases, the
Stop band
impedance of L–C network increases and the output
Figure 4.11 Response Curve for a
voltage increases.
Stop-band Filter
Example 4.4 An R–L–C circuit has the following values: R = 40 Ω, C = 0.01 µF, L = 100 mH
and the resistance of the coil is 10 Ω. An input voltage 120 mV of variable frequency is appearing
at the input. The output is taken out from the L–C combination. Calculate the output voltage at
resonance. Further, calculate the bandwidth.
Solution: The circuit is shown in Figure 4.12. At
resonance, XL = XC. Since XL and XC will cancel each
other, the circuit will be resistive only. The current in
the circuit will be Vin R + r . The output voltage will be
obtained as follows:
Vo = Vin
r
R+r
M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 154
R = 40Ω
120 mV
C = 0.01 µF
Vo
r = 10 Ω
L = 10 0 mH
Figure 4.12
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R–L–C Circuits and Resonance 155
By substituting the values in the equation, we get the following form:
10
Vo = 120 mV
= 24 mV
40 + 10
1
1
Resonant frequency f r =
=
2p LC 6.28 100 mH × 0.01µF
= 4825 Hz
XL = 2p frL
= 6.28 × 4825 × 100 × 10-3
= 3030 Ω.
Quality factor Q =
XL
3030
=
= 60.6
r + R 10 + 40
Bandwidth BW =
f r 4825
=
= 80 Hz
Q 60.6
4.4 PARALLEL RESONANCE
We will consider resonance in a parallel
L–C circuit, where the resistance of the
inductance coil is neglected. The parallel resonance occurs when XL = XC. The
frequency at which this occurs is called
the resonant frequency. An ideal parallel
resonant circuit is shown in Figure 4.13.
In the circuit shown at fr, XL = XC and
VS
IL =
fr =
1

2π√LC
IC
IC
I=0
L
C
IL
VS
I=0
IL
Figure 4.13 Parallel Resonance Circuits
Vs
V
and I C = S
XL
XC
Therefore, IL = IC
As IL and IC are in opposite directions, total current I = 0.
At resonant frequency, XL and XC are equal and IL and IC cancel each other as these are equal
V
V
in magnitude and opposite in phase. Total current I is zero. The impedance Z = S = S = ∞
I
o
4.4.1 Ideal Tank Circuit
It is interesting to note that although current I is zero, there exists current in the inductor and
the capacitor. Such a parallel resonant circuit is often termed as tank circuit. Normally, a tank
stores water or some other liquid. Here, the circuit stores energy. At resonance, energy is stored
in the magnetic field of the current carrying inductive coil and in the electric field of the capacitor. This stored energy is transferred back and forth between the inductor and the capacitor on
alternate half cycles. On alternate half cycles, the inductor gets energised while the capacitor is
de-energised and vice-versa; and this process continues indefinitely.
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156 Network Analysis and Synthesis
4.4.2 Non-Ideal Tank Circuit
Now, we will consider the resonance in tank circuit by considering the resistance of the inductor
coil, as shown in Figure 4.14.
I=0
I
r
fr
C ⇒ fr
req
z=∞
Leq
C
L
Figure 4.14 Non-ideal Tank Circuit and its Equivalent Circuit
The equivalent inductance Leq and equivalent resistance req are given as follows:
 Q 2 + 1
Leq = L 

 Q2 
Q=
and req = r (Q 2 + 1)
XL
r
At parallel resonance, XL(eq) = XC.
The L–C branches act as an ideal tank. Since, current I = 0, the tank circuit will have infinite
impedance at resonance.
The total impedance of the circuit at resonance is equal to equivalent resistance req which
can be written as follows:
Zr = req (Q2 + 1)
4.4.3 Resonant Frequency
The resonant frequency of the circuit is derived as in the following:
XL(eq) = XC
 Q 2 + 1
1
2p f r L 
=
 Q 2  2p f r C
or
fr =
1
Q2
2p LC
Q2 + 1
For high values of Q, fr can be written as follows:
1
fr =
2p LC
Therefore, the resonant frequency for parallel resonant circuit is approximately the same as
series resonant frequency.
4.5 PARALLEL RESONANT FILTERS
Similar to the series resonant circuits, parallel resonant circuits are used as band-pass and band-stop
filters. Parallel band-pass filter and band-stop filters have been discussed in brief in this section.
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R–L–C Circuits and Resonance 157
4.5.1 Band-pass Filter
The basic circuit and frequency response curves of a parallel band-pass filter have been shown
in Figure 4.15.
Vo
R
Vin
Vo
C
L
Vmax
I
0.707Vmax
Imin
f
f1 fr f2
(a)
(b)
Figure 4.15 B
asic Parallel Resonant Band-pass Filter (a) Circuit
Diagram and (b) Frequency Response Curves
At resonant frequency, the impedance of the circuit reaches its maximum value and the output
voltage is also maximum. At low and high frequencies, the impedance of the tank circuit goes
low, and hence, the output voltage gets decreased.
4.5.2 Band-stop Filter
A basic parallel resonant band-stop filter has been shown in Figure 4.16 (a). Further, the bandstop filter response curve is shown in Figure 4.16 (b). The output is taken across the load
­resistance RL.
L
Vin
Vo
C
RL
Vo
0
(a)
fr
f
(b)
Figure 4.16 B
and-stop Filter (a) Circuit Diagram and (b) Response
Curve
4.6 APPLICATIONS OF RESONANT CIRCUITS
Some of the important applications of resonant circuit are mentioned in the following sections.
4.6.1 Tuned Amplifier
A tuned amplifier amplifies a signal within a specified band of frequencies. Such a tuned amplifier in its basic form is shown in Figure 4.17.
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158 Network Analysis and Synthesis
Vin
Vo
f
A
Input
signal
C
Output
signal
fr
f
Figure 4.17 A Basic Tuned Band-pass Filter and Its Output Response
The circuit comprises an amplifier and a parallel resonant circuit. The input signal with a wide
range of frequencies is accepted and amplified by the amplifier. The resonant circuit allows
(passes on) only a selected band of frequencies at the output. The variable capacitor allows the
tuning of the circuit for the desired frequency.
4.6.2 Input to Receiver from an Antenna
Radio signals in the form of electromagnetic waves are propagated through the atmosphere.
An antenna is used to receive the radio signals. A small voltage is induced in the antenna
when the electromagnetic waves cut across the antenna. The arrangement is shown in
Figure 4.18.
Antenna
Parallel
resonant circuit
Waves
L
C
To receiver
Coupling
transformer
Figure 4.18 Use of Resonant Circuit in a Radio Receiver
However, among a wide range of electromagnetic frequencies, only a particular range of
­frequency signal is received. Since such signals are weak, amplifiers are used to amplify the
signal.
4.6.3 Other Applications
Filter circuits are also used in a TV receiver, super heterodyne receiver, etc.
Example 4.5 Two impedances Z1 = 5 + j 10 and Z2 = 10 - j 20 are connected in parallel. The
parallel combination is connected in series with another impedance Z3 = 10 + jX. At what value
of X, the circuit will produce resonance?
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R–L–C Circuits and Resonance 159
Solution: Total impedance Z = Z3 +
Z1Z 2
Z1 + Z 2
By substituting the values in the equation, we get the following:
Z = 10 + jX +
or
(5 + j10)(10 - j 20)
5 + j10 + 10 - j 20
= 10 + jX +
50 - j100 + j100 + 200
15 - j10
= 10 + jX +
250
15 - j10
= 10 + jX +
250(15 + j10)
(15 - j10)(15 + j10)
= 10 + jX +
3750 - j 2500
225 + 100
= 10 + jX +
2500
3750
-j
325
325
Z = 21.54 + j (X - 7.69)
For resonance to occur, the imaginary part of Z must be zero. Hence, X - 7.69 = 0 or X = 7.69 Ω.
When the value of X will be adjusted to 7.69 Ω, the circuit will resonate.
Example 4.6 A resistor, an inductor and a capacitor
are connected in series across at a 100 V variable
frequency supply source, as shown in Figure 4.19. At
a frequency of 250 Hz, the circuit resonates and the
current is 0.8 A. At resonance, the voltage across the
capacitor is measured as 200 V. Determine the values
of r, L and C.
R = 20 Ω
r
L
C
200 V
I = 0.8 A
V = 100 V, f
Figure 4.19
Solution: Voltage drop across the capacitor VC = 200 when a current of 0.8 A flows.
VC = I X C = 0.8
or
M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 159
1
= 200 V.
wC
0.8
= 200
2p f r C
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160 Network Analysis and Synthesis
C=
0.8
F
0.2 × 3.14 × 100 × 200
0.8 × 106
µF
628 × 200
= 6.369 µF
=
or
At resonance, XL = XC and Z = R + r
Therefore,
V
= R+r
I
or
100
= 20 + r
0.8
r = 125 - 20 = 105 Ω
or
XC =
1
1
106
=
=
w C 2p × 150C 628 × 6.369
X C = X L = 2p f L =
L=
106
628 × 6.369
106
2p f × 628 × 6.369
106
629 × 628 × 6.369
100
=
0.28 × 6.28 × 6.369
=
= 0.396 H
= 398 mH
L = 0.1 H
Example 4.7 In the circuit shown in Figure 4.20, the
Q-factor of the coil is 10. What will be the value of the
capacitor and the coil resistance when resonant frequency is
1000 rad/s?
r
Solution: Given L = 0.1 H, Q-factor = 10 and wr = 1000 rad/s
Figure 4.20
We know that Q =
C
V
w r L 1000 × 0.1
=
= 10
r
r
or
100
= 10
r
or
r = 10 Ω
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R–L–C Circuits and Resonance 161
Again
wr =
or
w r2 =
or
C=
1
L × w r2
=
=
1
LC
1
LC
1
0.1 × (1000) 2
106
0.1 × (1000) 2
F
µF
=10 µF
Example 4.8 An R–L–C series circuit has R = 10 Ω, L = 0.1 H and C = 8 µF. Calculate the
resonant frequency and the Q-factor at resonance. Further, calculate half power frequencies
and BW.
Solution: At resonance, XL = XC
or
2p f o L =
or
fo =
1
2p f oC
1
2p LC
By substituting the values in the equation, we get the following:
1
fo =
or
2 × 3.14 0.1 × 8 × 10 -6
fo = 178 Hz
voltage across L I o X L 2p f o L
Q-factor =
=
=
supply voltage
R
Io R
=
2p L
1
1 L
=
R 2p LC R C
By substituting the values, we obtain the value as follows:
Q-factor =
1
0.1
= 11.2
10 8 × 10 -6
The half-power frequencies correspond to 0.707 of the resonant current. Let the frequencies be
f1 and f2, that is, the lower and upper frequencies forming the BW.
f1 = f o -
R
4p L
f 2 = fo +
R
4p L
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162 Network Analysis and Synthesis
By substituting the values, we can calculate the value of f1 and f2 as follows:
10
12.56 × 0.1
= 169 Hz
f 1 = 178 -
10
12.56 × 0.1
= 187 Hz
f 2 = 178 +
BW = f2 - f1 = 187 - 169 = 18 Hz
Example 4.9 An inductive coil has a resistance of 2.5 Ω and an reactance of 25 Ω. This
coil is connected in series with a variable capacitor. A voltage of 200 V at 50 Hz is applied
across this series circuit. Calculate the value of C at which the current in the circuit will be
maximum.
Solution: In an R–L–C series circuit, under resonance condition, XL = XC and Z = R.
1
= X L = 25
2p fC
XC =
C=
or
=
1
F
2 × 3.14 × 50 × 25
106
µF = 127.4µF
314 × 25
Example 4.10 Calculate the value of the inductance L for which the parallel circuit shown in
Figure 4.21 will be in resonance at a frequency of 2000 rad/s.
Solution: Admittance, Y = Y1 + Y2
=
1
1
+
Z1 Z 2
=
1
1
+
5 + jX L 10 - j12
=
5 - jX L
25 + X L2
+
V
10 + j12
10 Ω
L
12 Ω
Figure 4.21
10 2 + 122


5
10
=
+
+

2
2
2
 25 + X L 10 + 12 
5Ω
 12
XL 
-j
j

25 + X L2 
 244
For resonance, the imaginary part of Y will be zero.
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R–L–C Circuits and Resonance 163
Therefore,
XL
25 +
or
X L2
=
12
244
12 X L2 - 244 X L + 300 = 0
XL = 19 Ω or 1.32 Ω
or
XL = wL
L=
or
XL
19
=
H
w
2000
19 × 1000
mH
2000
= 9.5mH
=
L=
Further,
1.32 1.32 × 1000
=
mH = 0.66 mH
w
2000
Example 4.11 In the R–L–C series circuit shown in
Figure 4.22, resonance occurs when the value of C is
20 µF. The supply voltage is v = 20 sin 500 t. Find the
values of L and Q-factor.
R = 30 Ω
wL=
or
L=
or
L=
Q -factor =
C
υ = 100 sin 500 t
Solution: Supply voltage n = 20 sin 500 t is of the form
v = Vm sin w t. Therefore, w = 500 rad/s. At resonance,
the current in the circuit is maximum.
When XL = XC
or
L
Figure 4.22
1
wC
1
2
w C
=
1
2
(500) × 20 × 10
-6
=
1
25 × 10 × 20 × 10 -6
4
1
H = 0.2H
5
w L 500 × 0.2
=
= 3.33
R
30
Example 4.12 In a series R–L–C circuit, the supply voltage is 230 V at 50 Hz. The resonant
current is 2 A. Under the resonant condition, the voltage across the capacitor is measured to be
equal to 460 V. What are the values of R, L and C?
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164 Network Analysis and Synthesis
Solution: At resonance, XL = XC, and hence, the circuit impedance is equal to R only
V 230
=
= 115 Ω
I
2
The voltage across the capacitor at resonance is equal to 460 V.
Z =R=
Therefore,
VC = I XC
XC =
or
VC 460
=
= 230 W
I
2
1
= 230
wC
or
C=
or
1
1
=
w × 230 2p f × 230
1
F
2 × 3.14 × 50 × 230
106
µF
=
314 × 230
=
= 13.85 µF
XL = XC = 230
Again,
2p f L = 230
L=
230
230
=
= 0.732H
2 × 3.14 × 50 314
Example 4.13 A sinusoidal voltage V = 100 sin cot is applied across an R–L–C series circuit.
At resonant ­frequency, the voltage across the capacitor is measured as 400 V. The impedance
of resonance is 50 Ω, what is the value of resonant frequency? What are the values of L and C?
the BW is 500 rad/s.
Solution:
n = Vm sin w t
Vm = 100, V ( rms) =
Here,
Vm
2
= 70.7 V.
Voltage Vc across the capacitor is 400 V.
Z = R, w2 - w1 = 500 rad/s.
R = 50 Ω
At resonance,
Therefore,
Quality factor
Q=
Voltage across the capacitor 400
=
= 5.65
supply voltage
70.7
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R–L–C Circuits and Resonance 165
Resonant frequency
fr = quality factor × BW
or
fr = Q (f2 - f1)
or
wr = Q (w2 - w1)
fr =
or
BW, w 2 - w1 =
L=
or
Q(w 2 - w 1) 5.65 × 500
=
= 449.8 Hz
2p
6.28
R
L
R
50
=
= 0.1H
w 2 - w1 500
XL = XC
At resonance,
C=
or
or w L =
1
2
w L
=
1
wC
1
2
( 2p f r ) L
=
1
(6.28 × 44 + .8) 2 × 0.1
F
= 0.157 µF
Example 4.14 An inductive coil has resistance of 10 Ω and inductance of 100 mH. This coil
is connected in parallel with a capacitor of 20 µF. A variable frequency, 200 V is applied across
this parallel circuit. Calculate the frequency at which the circuit will resonate. Further, calculate
the Q-factor and resonant current.
Solution: For a parallel circuit, the resonant frequency fr is given as follows:
fr =
1
2p
1
R2
- 2
LC L
By substituting the given values, the equation can be expressed as follows:
fr =
1
2p
1
100 × 10
-3
× 20 × 10
-6
-
(10) 2
(100 × 10 -3 ) 2
= 113 Hz
Q-factor =
or
X L 2p f o L 2 × 3.14 × 113 × 100 × 10 -3
=
=
R
R
10
Q-factor = 7
Resonant current I r =
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V
Zo
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166 Network Analysis and Synthesis
100 × 10 -3
L
=
= 500 Ω
CR 20 × 10 -6 × 10
200
Ir =
= 0.4 A
500
Zo =
Example 4.15 A coil of resistance 5 Ω and inductive reactance of 10 Ω is connected across a
voltage source of 230 V at 50 Hz. Calculate the value of the capacitor which when connected in
parallel with the coil will bring down the magnitude of line current to a minimum.
Solution:
IL =
230
230
V
=
=
R + jX L 5 + j10 11.18∠64°
= 20.57∠ - 64°
cos f = cos 64 = 0.438
sin f = sin 64 = 0.895
When the capacitor is connected in parallel, it will draw a reactive current IC. If the value of IC
becomes equal and opposite to IL sin f, as shown in the phasor diagram in Figure 4.23, these
two currents will cancel each other and the resonant current will be IL cos f, that is, in phase
with the voltage. Line current I will be equal to IL cos f, which is the current at resonance.
C
IC = ILsinf
IC
I
R = 5Ω
XL = 10 Ω
IL
V = 230 V, 50 Hz
0
f
ILsinf
ILcosf
V
IL
Figure 4.23
I = IL cos f = 20.57 × cos 64 = 20.57 × 0.45
= 9A
IC = IL sin f = 20.57 × 0.895 = 18.4 A.
Again,
or
or
IC =
XC =
V
XC
V
230
=
= 12.5 Ω
I C 18.4
1
= 12.5
2p fC
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R–L–C Circuits and Resonance 167
C=
or
=
1
F
2p f × 12.5
106
µF
2 × 3.14 × 50 × 12.5
C = 255 µF
or
Example 4.16 A parallel circuit is shown in Figure 4.24.
Calculate the value of R in the circuit for which the circuit will
reasonate.
Solution: Equivalent impedance Z =
Z1Z 2
Z1 + Z 2
or
Z=
=
RΩ
6Ω
Z2
10 Ω
4Ω
V
Figure 4.24
( R + j 6)(10 - j 4)
=
( R + j 6) + (10 - j 4)
=
Z1
( R + j 6)(10 - j 4)
( R + 10) + j (6 - 4)
( R + j 6)(10 - j 4)[( R + 10) - j 2]
[( R + 10) + j 2][( R + 10) - j 2]
[(10 R + 24)( R + 10) + 2(60 - 4 R)] + j (60 - 4 R)( R + 10) - j 2(10 R + 24)
( R + 10) 2 + 4
At resonance, the imaginary part of Z will be zero.
Therefore,
or
j(60 - 4R)(R + 10) - j2(10R + 24) = 0
60R - 4R2 + 600 - 40R - 20R - 48 = 0
or
or
R2 = 138
R = 11.75 Ω
4.6.4 Locus Diagram
The magnitude and phase of current vector in an R–L–C circuit changes when the parameters R,
L and C are varied keeping frequency f and voltage V constant. The path traced by the tip of the
current vector is called the current locus. Locus diagrams are also drawn when the frequency
is variable.
Let us consider a simple R-L series circuit. The applied voltage and the reactance XL are
constant. We will draw the locus of I, when R is varied from 0 to ∞.
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168 Network Analysis and Synthesis
V
. This is the maximum current that will flow. Since the circuit will be purely
XL
inductive, the current will lag the voltage by 90 that has been shown as OC in Figure 4.25. When
R is gradually increased, the magnitude of current I becomes less than its maximum value and
f becomes less than 90 . When R becomes infinitely large, I will be zero and f will be zero.
It is observed that the tip of the current vector will lie on a semi-circle.
From Figures 4.25 and 4.26, we get the following expression:
If R = 0, I =
Z = R 2 + X L2
I=
V
XL
I
V
X
R
V
, sin f = L , cos f =
Z
Z
Z
IY
I
Z
R
f
Ix
O
I
C
V XL
f
XL
R
Figure 4.26 I mpedance
Triangle of an
R-L Circuit
Figure 4.25 L ocus of Current in a R-L Circuit with
Variable R: (a) R-L Circuit and (b) Locus of
Current is a Circle
The horizontal component of I is Ix and its vertical component is IY
VX
V X
I X = I sin f = × L = 2L
Z Z
Z
(4.1)
V R VR
I Y = I cos f = × = 2
(4.2)
Z Z Z
From (4.1) and (4.2), we can write the expression as follows:
I X2 + I Y2 =
=
=
or
I X2 + I Y2 =
V 2 X L2
+
V 2 R2
Z4
Z4
V 2 × Z2
=
V 2 ( R 2 + X L2 )
Z4
Z2 × Z2
V2
Z2
V2
R 2 + X L2
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R–L–C Circuits and Resonance 169
IX = I sin f
=
IX =
Z2 =
or
I X2 + I y2 =
I X2 + I Y2 -
or
V XL
Z Z
VX L
Z2
VX L
(4.4)
IX
V2
Z2
=
V2
V
=
IX
VX L/I X
XL
V
IX = 0
XL
2
 V 
Adding 
on both sides, we get the following form:
 2 X L 
2
2

 V 
V 
2
I
I
+
=
y
 X 2 X 
 2 X 
L
L
This equation is of the form:
(x - a)2 + y2 = r2
This is the equation of a circle.
 V

V
In this case, the radius of the circle is
and the centre of the circle is at 
, O .
2XL
 2XL

This proves that the locus of the current in an R-L series circuit, where R is variable is a
circle. This locus is also called the circle diagram. For an R-C circuit with variable R, the locus
of current will also be a circle.
Similarly, one can draw the locus diagram for a constant resistance and variable inductance
series circuit. The circle diagrams are useful in analysing the equivalent circuits of induction
machines, transmission lines, etc.
R e v ie w Questi o n s
Short Answer Type
1. Show the variations of XL and XC versus frequency for an R–L–C series circuit and identify
the resonant frequency.
2. Define resonance frequency for a series R–L–C circuit.
3. Draw graphs of impedance, current and phase angle versus frequency for a series R–L–C
circuit.
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170 Network Analysis and Synthesis
4. Draw the phasor diagram for a series R–L–C circuit at resonance. Further, draw the waveforms of VL, VC and VR at resonance.
5. The Q-factor is a measure of quality of a resonance circuit. Explain.
6. Establish the relationship between Q-factor and BW of a resonant circuit.
7. Deduce the formula for Q-factor of an R–L–C resonant circuit.
8. Show that resonant circuit frequencies f1 and f2 are those frequencies at which the power
delivered to the circuit is half the power delivered at resonance.
9. Show that a parallel L–C circuit has a maximum impedance at the resonance frequency.
10. Draw the graph of impedance versus frequency and current versus frequency for a parallel
resonance circuit.
11. Show that the resonance frequency for a series and parallel resonance circuit is the same
when the Q-factor is high.
12. Prove that for a series resonance circuit, the resonance frequency and Q-factor are
1
1 L
fr =
and Q =
R C
2p LC
13. Define Q-factor for a series resonance circuit.
14. Show how an R–L–C series circuit can be tuned to resonate at a range or frequencies.
15. Derive an equation for the Q-factor of a series resonant circuit in terms of R, L and C.
16. With the help of suitable graphs, define half-power frequencies and BW as applied to a
resonant circuit.
17. Derive an equation for the BW of a resonance circuit in terms of the resonance frequency
and the Q-factor of the circuit.
18. Draw and explain the phasor diagram for the component currents in a parallel L–C circuit
at resonance.
19. A parallel resonant circuit is called a rejector circuit. Explain.
20. An R–L–C series resonant circuit is called an acceptor circuit. Explain.
21. Show that frequency for a parallel resonance circuit is approximately the same as that of a
series resonance circuit.
22. Give three examples of applications of resonance circuits.
Numerical Questions
1. Determine the resonance frequency for a series R–L–C circuit with R = 100 Ω, L = 0.085 mH
and C = 298 pF connected across a 10 V variable frequency supply source. Further, calculate the circuit currents at 0.25 fr, 0.5 fr, 0.75 fr, fr, 1.5 fr and 2 fr and plot the current versus
frequency graph.
[Ans. 1 × 106 Hz]
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R–L–C Circuits and Resonance 171
2. Determine the Q-factor for a series R–L–C circuit with R = 25 Ω, L = 0.1 mH and C = 1000 pF
connected across an AC supply source. What will be the value of capacitance required for
resonance at 500 kHz when the inductance is doubled and what will be the new Q-factor?
[Ans. 12.6, 500 pF, 25]
3. An R–L–C series circuit has R = 10 Ω, L = 30 mH and C = 1 µF and is connected across
at a 10 V variable frequency supply source. Calculate the frequency for which the voltage
developed across the capacitor is maximum and what will be its value?
[Ans. 920 Hz, 173 V]
4. An R–L–C series circuit is connected across a 300 mV variable frequency supply source.
The maximum current of 5 mA in the circuit is obtained at a frequency of 6 kHz. The
Q-factor at this frequency is 105. Calculate the values of R, L and C. Further, calculate the
voltage across the capacitor.
[Ans. 60 Ω, 0.167 H, 4220 pF, 31.5 V]
5. An inductive coil of L = 0.12 H and r = 12 Ω is connected in parallel with a capacitor of
60 µF across a 100 V variable frequency supply. Calculate the frequency at which the circuit will resonate. Draw the phasor diagram.
[Ans. 57.2 Hz]
6. Determine the quality factor of a coil for a series resonance circuit having R = 10 Ω,
L = 0.1 H and C = 10 µF.
[Ans. 10]
7. The Q-factor of an R–L–C series circuit is 5 at 50 rad/s. The supply voltage is 100 V and the
resonance current is 20 A.
[Ans. R = 5 Ω, L = 0.5 H, C = 800 µF]
8. A circuit of R = 4 Ω, L = 0.5 H and a variable capacitance C in series is connected across a
100 V, 50 Hz supply. Calculate the following:
(i) the value of C for which resonance will occur;
(ii) the voltage across the capacitor at resonance and the Q-factor.
[Ans. 20.3 µF, 2925 V, 39.25]
9. In an R–L–C series circuit, a maximum current of 0.5 A is obtained by varying the value
of inductance L. The supply voltage is fixed at 230 V, 50 Hz. When maximum current flows
through the circuit, the voltage measured across the capacitor is 350 V. What are the values
of the circuit parameters?
[Ans. R = 460 Ω, L = 2.23 H, C = 4.55 µF]
10. Calculate the half frequencies of a series resonance circuit in which the resonance f­ requency
is 150 kHz and BW is 75 kHz.
[Ans. f 1 = 117 kHz, f 2 = 19 kHz]
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Network Theorems
and Applications
5
Chapter objectives
After carefully studying this chapter, you should be able to do the following:
State and explain Superposition theoEstablish the condition for maximum
rem with an example.
power transfer in a complex impedance
circuit.
Solve
circuit
problems
using
Superposition theorem.
State and explain, with the help of an
example, the Reciprocity theorem.
State and explain Thevenin’s theorem.
State Tellegen’s theorem and apply the
Apply Thevenin’s theorem to calculate
theorem to solve network problem.
current flowing through any branch of
an active electric network.
State and explain Compensation
theorem.
Explain with the help of an example
the procedure for applying Thevenin’s
Simplify electrical circuits using star–
theorem to an electric network.
delta transformations.
State and explain Millman’s theorem.
Establish transforming relations of
star–delta transformation of resistance.
Solve network problems using
Millman’s theorem.
Solve complex network problem by
using network theorem suitably.
State and explain maximum power
transfer theorem.
5.1 INTRODUCTION
In a previous chapter, we had described the methods of solving network problems using mesh
current analysis and nodal voltage analysis. The procedure involves solving a number of equations. For a complex network, the number of equations becomes large. However, many networks
often require only restricted analysis, for example, finding the current through a particular
branch or a particular circuit element.
A number of theorems for circuit analysis have been developed to solve circuit problems
with ease. When all the theorems and techniques will be known, we will be in a position to
apply a particular theorem or a technique that reduces the time for solving a given problem. In
this chapter, the following theorems and techniques have been discussed.
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173
Network Theorems and Applications
1.
2.
3.
4.
5.
6.
7.
8.
9.
Superposition theorem
Thevenin’s theorem
Norton’s theorem
Millman’s theorem
Maximum power transfer theorem
Reciprocity theorem
Tellegen’s theorem
Compensation theorem
Star−delta transformation
5.2 SUPERPOSITION THEOREM
An electrical network may contain more than one source of supply. The sources may be voltage
sources or current sources or a combination of both. In solving circuit problems having multiple sources of supply, the effect of each source is considered separately with all other sources
replaced by their internal resistances every time. The combined effect of all the sources is then
taken into consideration to calculate the current in any branch.
The superposition theorem can be stated as follows: In a linear network containing more
than one source, the current in any branch or the potential difference across any two points can
be found by considering each source separately and then by adding their individual effects.
While considering each source, the other sources are replaced by their internal resistances.
If the value of internal resistances of the sources are not given, the voltage sources are shortcircuited and the current sources are open-circuited.
The procedure is illustrated through a few examples.
Example 5.1 Using superposition theorem, calculate the currents in the network shown in Figure 5.1.
Solution: Let us consider separately the effect of
each voltage source by short-circuiting the other
source as shown in Figure 5.2.
4Ω
4Ω
4Ω
24 V
12 V
Figure 5.1
4Ω
4Ω
4Ω
4Ω
24 V
4Ω
24 V
4Ω
4Ω
I1
2Ω
= 4A
I1 = 424
+2
12 V
= 2A
I2 = 412
+2
4Ω
2Ω
12 V
I2
Figure 5.2
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174 Network Analysis and Synthesis
Now, we will combine the effect of the two voltage
sources.
Current supplied by the 24 V source = 4 A
Current through branch AB = 4 A
Current supplied by the 12 V source = 2 A
Current through branch CB = 2 A
Current through branch BD = 4 + 2 = 6 A.
4Ω
A
I1
4A
24 V
I1
I1 =
2A
I2
4Ω
12 V
I1 + I2
I2
D
4Ω
6Ω
Current through 6 Ω resistor across BD = I1 − I2 = 2.4 − 0.8
= 1.6 A
2A
D
Figure 5.4
24
= 2.4 A
4+6
6
I3 = 2 A ×
= 1.2 A
4+6
2Ω
B
24 V
Now, we will consider the effect of current source by shortcircuiting the voltage source, as shown in Figure 5.6.
By applying current divider rule, we get the following
form:
4
I2 = 2 A ×
= 0.8 A
4+6
Further,
C
Figure 5.3
Example 5.2 Calculate the current through the 6 Ω
resistor shown in Figure 5.4 using superposition
theorem.
Solution: We will first consider the effect of voltage
source by open-circuiting the current source, as shown in
Figure 5.5.
4Ω
B
4Ω
2Ω
B
I1
24 V
6Ω
D
Figure 5.5
4Ω
2Ω
B
I3
6Ω
I3
I2
2A
2A
D
Figure 5.6
5.3 THEVENIN’S THEOREM
It may sometimes be required to calculate the current through any circuit element in an active
electrical network when the value of the circuit element is changed to different values, keeping
all other circuit elements unchanged. We can use Thevenin’s theorem to determine the current
flowing through any circuit element in an active network.
According to Thevenin’s theorem, current I through a resistor R connected across any two
points A and B of an active network containing one or more sources of emf is as follows:
Voc
VTh
=
R + r R + RTh
where Voc is the potential difference across the terminals A and B with R disconnected; and r
is the resistance of the network between A and B with R disconnected and the sources of emfs
replaced by their internal resistances (if the value of internal resistance of the voltage source is
not provided, the source terminals have to be shown short-circuited.)
I=
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Network Theorems and Applications
175
Thevenin’s theorem can be stated as follows: Any two terminals A and B of an active network
can be replaced by a constant voltage source having an emf E and an internal resistance r. The
value of E is equal to the open-circuit potential difference between terminals A and B, that is, Voc
and r is the resistance of the entire network measured or calculated between A and B with the resistance R between A and B disconnected and sources of emfs replaced by their internal resistances.
5.3.1 Procedure for Applying Thevenin’s Theorem
The procedure is illustrated through an example.
Consider a network as shown in Figure 5.7.
Let us assume that we are required to calculate the
current through the 3 Ω resistor. Let this resistance be
called the load resistance. We will remove the load
resistance and place the rest of the network in a dotted
box as shown is Figure 5.8.
A
4Ω
2Ω
3Ω
12 V
6V
B
Figure 5.7
A
A
P
4Ω
2Ω
12 V
Load
3Ω
6V
4Ω
2Ω
I
12 V
Q
B
(a)
VAB = ?
6V
B
(b)
Figure 5.8
Now, remove the load resistance of 3 Ω temporarily and find the open-circuit voltage across the
terminals A and B.
From Figure 5.8(b), by applying KVL in the loop, we obtain the following:
12 − 4I − 2I − 6 = 0
6I = 6
I = 1A
or
VAB = VPQ = 12 − 4I = 12 − 4 × I = 8 V
VTh = VAB = 8 V
Further,
RTh =
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4×2 8
=
4+2 6
= 1.33 Ω
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176 Network Analysis and Synthesis
Now, short-circuit the voltage sources and calculate the resistance
of the network across terminals A and B. The Thevenin’s equivalent
circuit consists of a voltage source VTh, which is 8 V, in series and a
resistance RTh, which is 1.33 Ω. Thevenin’s equivalent circuit has been
shown in Figure 5.9.
The current through the load resistance is calculated by connecting
the load resistance across terminals A and B of the Thevenin’s equivalent circuit. The load current is calculated as
A
RTh
VTh
1.33 Ω
Load
3Ω
8V
B
Figure 5.9
IL =
VTh
8
8×3
=
=
= 1.846 A
RTh + RL 4
13
+3
3
Example 5.3 Applying Thevenin’s theorem, calcu­
late the current flowing through the 10 Ω resistor in
the circuit shown in Figure 5.10.
Solution: We will convert the current source into
an equivalent voltage source and remove temporarily
the resistance of 10 Ω from terminals AB as shown
in Figure 5.11. Then, we will calculate open-circuit
voltage VTh across AB. Then, we will calculate the
resistance of the network, that is, RTh across A and B
by short-circuiting the voltage sources.
Applying Kirchhoff’s voltage law in the loop, the
value of I is calculated as follows:
−2I + 12 − 4I −2I − 4 = 0
12 V
B
Figure 5.10
P
12 V 4 Ω
2Ω
2Ω
3Ω
A
VTh =?
I
4V
B
Q
Figure 5.11
I = 1A
or
10 Ω
2Ω
2A
A
2Ω
8I = 8
or
3Ω
4Ω
Note that since the terminals AB are open-circuited no current can flow through the 3 Ω resistor.
The voltage across AB is the same as the voltage across terminals PQ. VTh = VAB = VPQ = 4 + 2I
= 4 + 2 × 1 = 6 V.
RTh is calculated as shown in Figure 5.12.
RTh = (6 || 2) + 3
6×2
+3
6+2
= 1.5 + 3
= 4.5 Ω
=
3Ω
4Ω
2Ω
A
2Ω
RTh
B
Figure 5.12
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Network Theorems and Applications
Now, we can draw the Thevenin’s equivalent circuit and
connect the load resistance of 10 Ω across terminals AB
to determine the load current as shown in Figure 5.13.
IL =
=
177
A
4.5 Ω
RTh
6V
VTh
VTh
RTh + RL
Load
10 Ω = RL
IL
B
Figure 5.13
6
= 0.414 A
4.5 + 10
Example 5.4 Applying Thevenin’s theorem, calculate the current through the load resistance
RL = 10 Ω in the circuit shown in Figure 5.14.
4Ω
24 V
A
4Ω
8Ω
2A
RL = 10 Ω
B
12 V
Figure 5.14
Solution: We temporarily remove the resistance RL from the terminals AB. To calculate VTh,
that is, VAB, we will apply superposition theorem and calculate the total the current flowing
through the 8 Ω resistor and then calculate the voltage drop across it. The voltage drop across
the 8 Ω resistor is equal to the open-circuit voltage VAB, which is the VTh.
We will then calculate RTh across terminals AB by open-circuiting the current source and
short-circuiting the voltage sources. First, we consider the 24 V source, then the 12 V source,
and then the 2 A source. Later, we calculate the current through the 8 Ω resistor. Keeping in
view the current direction in each case, we calculate the total current through the 8 Ω resistor,
as shown in Figure 5.15.
1. Considering the 24 V source, we open-circuit 2 A current source and short-circuit the 12 V
voltage source.
A
4Ω
24 V
I1 (from A to B) =
24
= 1.5A
4+ 4+8
A
4Ω
8Ω
I1
B
B
2.Considering the 12 V source, we open-circuit the 2 A current source and short-circuit the
24 V voltage source.
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178 Network Analysis and Synthesis
A
4Ω
A
4Ω
8Ω
I2
B
B
12 V
I 2 (from B to A ) =
12
= 0.75
8+ 4+ 4
3. Considering the 2 A current source, and short-circuiting the 12 V and 24 V voltage sources.
A
4Ω
A
4Ω
2A
8Ω
I3
I 3 (from A to B) = 2 ×
B
B
=
Figure 5.15
4
4+8+ 4
8
= 0.5A
16
Following the principle of the superposition, we consider the combined effect of all the sources
at which the current flows through the 8 Ω resistor.
I = I1 + I3 − I2 = 1.5 + 0.5 − 0.75 = 1.25.
VTh = VAB = 1.25 × 8 = 10 V.
RTh is calculated by calculating the equivalent resistance of the circuit across terminals A and B
and by shorting the voltage sources and keeping open the current source as shown in Figure 5.16.
4Ω
A
4Ω
8Ω
RTh = ?
B
Figure 5.16
8×8
8+8
= 4Ω
RTh =
Thus, the Thevenin’s equivalent circuit and current through the load resistance is shown in
Figure 5.17.
A
4Ω
10 V
RTh
VTh
IL
RL = 10 Ω
B
Figure 5.17
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IL =
VTh
10
=
RTh + RL 4 + 10
= 0.714 A
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Network Theorems and Applications
179
5.4 NORTON’S THEOREM
In Thevenin’s theorem, we have seen that a two-terminal active network is converted into a voltage source and an equivalent series resistance which is connected across the load through which
the current is to be calculated.
Another method of analysing a network is provided by Norton’s theorem. In applying
Norton’s theorem, a two-terminal network with current and voltage sources is converted into a
constant current source and a parallel resistance and is connected across the load through which
the current is to be calculated.
Norton’s theorem is stated as follows: Any two-terminal linear network consisting of voltage and/or current sources can be converted into a constant current source and a parallel resistance. The value of the current source is the current that will flow if the two terminals are
short-circuited. The value of the parallel resistance is the equivalent resistance of the whole
network viewed from the open-circuited terminals after all the sources are replaced by their
internal resistance. (Note that if the internal resistances are not given, the voltage sources are
short-circuited and current sources are open-circuited.)
Example 5.5 Using Norton’s theorem, calculate the
current through the 10 Ω resistor in the network, as shown
in Figure 5.18.
A
4Ω
Figure 5.18
A
I1
I2 6 Ω
I2
ISC
6V
12 V
B
A
4Ω
10 Ω
6V
Solution: We short-circuit the terminals AB as in Figure
5.19(a) and calculate the short-circuit current. Then,
we calculate the resistance across the open-circuited
terminals A and B by short-circuiting the voltage sources
as shown in Figure 5.19(b).
I1
6Ω
4Ω
12 V
6Ω
R=?
B
B
(a)
(b)
Figure 5.19
6 12
I SC = I1 + I 2 = + = 1.5 + 2 = 3.5 A
4 6
Resistance R across the open-circuited branch across terminal A and B is given as in the following:
R=
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4×6
= 2.4 Ω
4+6
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180 Network Analysis and Synthesis
Norton’s equivalent current source with parallel resistance along with the 10 Ω resistance across
terminals A and B is shown in Figure 5.20. Using current divider rule, current through the load
is calculated as follows:
A
ISC = 3.5 A
R=
2.4 Ω
IL
2.4
10 + 2.4
2.4
= 3.5 ×
12.4
= 0.677 A.
I L = 3.5 ×
10 Ω
B
Figure 5.20
A
Example 5.6 Using Norton’s theorem, calculate
the current through the 10 Ω resistor in the network
shown in Figure 5.21.
Solution: We short-circuit terminals AB and
calculate the short-circuit current. Then, we calculate
the resistance across the open-circuited terminals
A and B.
Since terminals AB are shorted, no current
will flow through the branch containing 6 Ω and I = 10 A
4 Ω resistors. See Figure 5.22.
Using current divider rule, the current is calculated as follows:
I SC = I ×
6Ω
2Ω
8 Ω IL
10 A
10 Ω
4Ω
B
Figure 5.21
A
10 A
8Ω
6Ω
2Ω
ISC
4Ω
B
Figure 5.22
8
8
= 10 × = 8 A
8+ 2
10
The open-circuit resistance across terminals A and B is calculated as shown in Figure 5.23.
A
A
6Ω
2Ω
8Ω
R=?
4Ω
B
10 Ω
10 Ω
RAB =
10 × 10 100
=
= 5Ω
10 + 10 20
B
Figure 5.23
Now, we draw the Norton’s equivalent circuit as a current source with a parallel resistance and
connect the 10 Ω resistor across its terminals as shown in Figure 5.24.
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Network Theorems and Applications
I L = I SC ×
A
5
5 + 10
5
15
= 2.66 A
ISC
8A
=8×
5Ω
R
10 Ω
IL
B
Figure 5.24
5.5 MILLMAN’S THEOREM
When a number of voltage sources form parallel branches, the common voltage across their
terminals can be calculated by applying Millman’s theorem.
This theorem is illustrated through an example. Let us assume that three voltages V1, V2 and
V3 are connected in parallel across terminals A and B. R1, R2 and R3 are, respectively, their internal resistances as shown in Figure 5.25(a). We convert the voltage sources into equivalent current
sources as shown in Figure 5.25(b). By combining the current sources and the resistances, the
equivalent circuit will be as shown in Figure 5.26.
A
R1
R2
V1
V2
A
R3
V3
B
R1
V1
I1 =
R1
V2
I2 =
R2
(a)
R2
R3
V3
I3 =
R3
B
(b)
Figure 5.25
I = I1 + I 2 + I 3 =
V1 V2 V3
+
+ = V1G1 + V2G2 + V3G3
R1 R2 R3
Resultant voltage across A and B, that is,
1
VAB = I Req = (V1G1 + V2G2 + V3G3 ) ×
(G1 + G2 + G3 )
Millman’s theorem states that in any network,
if there are a number of voltage sources V1, V2, V3,
… Vn in parallel with their internal resistances R1,
R2, R3, … Rn, respectively, then, these sources can
be replaced by a single voltage source with its internal resistance as V ′ and R′, respectively, where
V′ =
V1G1 + V2G2 + + Vn Gn
G1 + G2 + + Gn
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and
A
I
1
1 1 1
= + +
Req R1 R2 R3
I = I 1 +I 2 + I 3
= G1 + G2+ G3
B
Figure 5.26
R′=
1
G1 + G2 + + Gn
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182 Network Analysis and Synthesis
Example 5.7 Three voltage sources of 12 V,
13 V and 14 V having internal resistances of
1 Ω, 2 Ω and 3 Ω, respectively, are connected in
parallel as shown in Figure 5.27. What will be the
voltage available across their terminals?
A
1Ω
2Ω
12 V
3Ω
13 V
Solution:
14 V
B
V V V 
1
VAB =  1 + 2 + 3  ×
 R1 R2 R3  1 + 1 + 1
R1 R2 R3
Substituting the values, we calculate the voltage as follows:
Figure 5.27
1
 12 13 14 
VAB =  + +  +
 1 2 3   1 1 1
 + + 
1 2 3
1
= (12 + 6.5 + 4.66) ×
1 + 0.5 + 0.33
1
= 23.16 ×
1.83
= 12.7 V
5.6 MAXIMUM POWER TRANSFER THEOREM
The power is supplied from the source to the load. Let the internal resistance of the source be
Ri and the load resistance be RL.
Applying maximum power transfer theorem, we find out at what value
of
load (load resistance or impedance) maximum power will be transI
Ri
ferred from the source to the load. Let us assume that a source having an
RL
emf E and internal resistance Ri is connected to a load of resistance RL as
E
shown in Figure 5.28.
The current flowing in the circuit is given as follows:
Source
Load
E
Figure 5.28
I=
Ri + RL
Power delivered is equal to power consumed assuming no line loss. Power delivered P is
expressed as in the following:
2
 E 
E 2 RL
2
P = I RL = 
R
=
L
 R + R 
( R + R )2
i
L
i
L
To determine the value of RL at which P will be maximum, we differentiate P with respect to
RL and equate to zero.
dP
d  E 2 RL 
=

=0
dRL dRL  ( Ri + RL ) 2 
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183
Network Theorems and Applications
or
or
d
E 2 [ RL ( Ri + RL ) −2 ] = 0
dRL
E 2 [1⋅ ( Ri + RL ) −2 + RL ( −2)( Ri − RL ) −3 ] = 0
or

2 RL 
1
E2 
−
=0
2
( Ri + RL )3 
 ( Ri + RL )
( RL + RL ) − 2 RL = 0
or
Ri − RL = 0
or
RL = Ri
or
This shows that the maximum power will be transferred from the source to the load when the
value of load resistance becomes equal to the internal resistance of the source.
The maximum power transfer theorem is stated as follows: Maximum power is transferred to the load, when the load resistance is equal to the source resistance.
When a complex network is analysed for maximum power transfer, the circuit is first converted into a voltage source with an internal resistance by applying Thevenin’s theorem.
The value of maximum power is calculated as follows:
P(max) = I 2 RL
=
E 2 RL
( RL + RL ) 2
=
E2
E2
=
when Ri = RL
4 RL 4 Ri
In power systems, maximum power transfer from the generator to the load is not tried because
of poor efficiency of transmission and poor voltage regulation.
In the field of electronics, maximum power transfer from the source to the load is achieved
through impedance matching.
In electronics, we often deal with small power. For example, maximum power transfer is
desirable from the output amplifier to the speaker
of an audio sound system. A TV antenna receives
2Ω
1Ω
power from radio waves. The power collected by
A
the antenna is very small. The TV receiver circuit
RL
is designed to make maximum use of the power
8V
4Ω
Source
Load
delivered by the antenna.
Example 5.8 A 8 V battery is supplying power
through a network to a load, RL, as shown in Figure
5.29. Calculate the value of RL at which the power
transfer will be maximum.
Solution: The circuit is converted into a Thevenin’s
equivalent circuit across terminals AB as shown in stages
in Figures 5.30, 5.31 and 5.32.
8
= 1A
2+4+2
VTh = VAB = VPQ = 4 × 1 = 4 V
I=
M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 183
2Ω
B
1Ω
Figure 5.29
2Ω
8V
A
4Ω
I
2Ω
1Ω
P
Q
1Ω
B
Figure 5.30
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184 Network Analysis and Synthesis
Note that when terminals AB is open-circuited, no current will flow through the 1 Ω resistors
and hence there will be no voltage drop in these resistors. The equivalent resistance RTh is given
as follows:
A
A
2Ω
1Ω
1Ω
4Ω
RTh = ?
1Ω
2Ω
RTh = ?
2Ω
B
1Ω
RTh = 4 Ω
B
Figure 5.31
A
RTh = 4 Ω
RL
Voc = 4 V
B
Thus, the Thevenin’s equivalent circuit is represented as shown in
Figure 5.32.
For maximum power transfer, resistance can be calculated as in the
following:
RL = RTh = 4 Ω
The value of maximum power transfer is as follows:
Figure 5.32
Pmax =
V 2
E2
42
= oc =
= 1 Watt
4 RL 4 RL 4 × 4
Example 5.9 In the circuit shown in Figure
5.33, calculate the value of load resistance RL at
which the maximum power will flow through the
load. Further, calculate the transmission efficiency
when maximum power transfer occurs.
6Ω
3Ω
24 V
RL
6A
6Ω
Figure 5.33
Solution: We redraw the circuit as shown in
Figure 5.34, and convert the current source into its equivalent voltage source.
6Ω
24 V
3Ω
RL
6A
6Ω
A
6Ω
RL
24 V
6Ω
3Ω
18 V
B
Figure 5.34
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Network Theorems and Applications
VTh and RTh are calculated as shown in the following. As
in Figure 5.35, we remove the load resistance RL and then
proceed to calculate VAB, that is, VTh.
Applying KVL, we calculate the following;
A
6Ω
24 V
3Ω
I
18 V
6Ω
24 − 6I − 3I − 18 − 6I = 0
B
Figure 5.35
6 = 15 I
or
185
2
A = 0.4 A
5
VTh = VAB = 18 + 3I = 18 + 3(0.4) = 19.2 V.
I=
The Thevenin equivalent resistance between terminals A and B is calculated by short-circuiting
the voltage sources.
(6 + 6) × 3 36 12
=
=
RTh =
= 2.4 Ω
(6 + 6) + 3 15 5
Thevenin’s equivalent circuit is shown in Figure 5.36. In Figure 5.36,
A
I
for maximum power transfer,
R
E
19.2
RL = Ri = 2.4 Ω, I =
=
=4A
Ri + RL 2.4 + 2.4
Maximum power = I 2 RL = 42 × 2.4 = 38.4 W.
Power supplied to the load
Power transfer efficiency, h =
Power supplied by the source
=
=
i
E
2.4 Ω
19.2 V
RL
B
Figure 5.36
I 2 RL
I 2 Ri + I 2 RL
I 2 RL
2 I 2 RL
( ∵ Ri = R L )
1
2
100
=
= 50%
2
=
5.7 MAXIMUM POWER TRANSFER THEOREM FOR COMPLEX
IMPEDANCE CIRCUITS
Maximum power transfer theorem can be applied to complex impedance circuits.
For a complex source impedance, maximum power transfer occurs when the load impedance
is the complex conjugate of the source impedance. Let us consider a circuit with a source supplying power to a load as shown in Figure 5.37. ZS is the source impedance and ZL is the load
impedance.
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186 Network Analysis and Synthesis
RS
I
Source
jXS
ZS
RL
VS
ZL
jXL load
Figure 5.37
Current I =
=
Magnitude of I =
VS
RS + jX S + RL + jX L
VS
( RS + RL ) + j ( X S + X L )
VS
2
( RS + RL ) + ( X S + X L ) 2
Power delivered to the load, P = I 2RL
P=
VS2 RL
( RS + RL ) 2 + ( X S + X L ) 2
Power is maximum with constant value of RL when XL = −XS. Then, the maximum power is
P=
VS2 RL
( RS + RL ) 2
However, when RL is variable, maximum power occurs when RL = RS.
Thus, for maximum power transfer, the following conditions are obtained:
RL = RS and XL = − XS.
This shows that maximum power transfer occurs when Z L = RL + jX L = RS − jX S = ZS∗
That is, load impedance is equal to the complex conjugate of the source impedance. That is,
ZL = ZS∗
5.8 RECIPROCITY THEOREM
The reciprocity theorem is stated as follows: In a linear bilateral network, if a voltage source
V in a branch A produces a current I in any branch B, then if the same voltage source is removed
and inserted in branch B, it will produce a current I in branch A. In other words, we can say that
voltage V and I are interchangeable between branch A and B.
This theorem is illustrated through an example.
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Network Theorems and Applications
Example 5.10 Verify reciprocity theorem in the
circuit shown in Figure 5.38.
A
24 V
Solution: The 24 V voltage source in branch AB
is causing a current I flowing through branch CD.
We have to show that same current I will flow
in branch AB when 24 V source is removed and
inserted in branch CD.
We calculate the total current supplied by the
24 V battery as shown in Figure 5.39.
24
IT =
= 3A
2 + 3+ 3
C
3Ω
6Ω
2Ω
2Ω
I
B
D
Figure 5.38
3Ω
24 V
4Ω
6Ω
6Ω
2Ω
I
24 V
3Ω
3Ω
2Ω
Figure 5.39
The current through branch CD is calculated by using current divider rule as in Figure 5.40(a).
I = 3×
6
= 1.5A.
12
The current flowing in branch CD has been calculated as 1.5 A.
Now, we place the 24 V source in branch CD, as
shown in Figure 5.40(b).
The total current supplied by the 24 V battery
is as follows:
IT =
=
A
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3Ω
1.5 A
C
1.5 A
4Ω
1.5 A
6Ω
2Ω
2Ω
I
B
D
Figure 5.40(a)
IT = 11/4 A
A
I
6
11 6
I = IT ×
= × = 1.5 A
(3 + 2) + 6 4 11
The same amount of current, that is, 1.5 A is
flowing through branch AB.
Thus, the reciprocity theorem is established.
3A
24 V
24
24 × 11
24
=
=
30
5×6
96
2+4+
2+4+
5+6
11
11
A
4
3A
2Ω
11/4 A
C
4Ω
3Ω
6Ω
B
24 V
2Ω
D
Figure 5.40(b)
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188 Network Analysis and Synthesis
5.9 TELLEGEN’S THEOREM
Tellegen’s theorem states that the algebraic sum of powers in all branches in a network at any
instant is zero. This theorem is valid for any network that may be linear or non-linear, active or
passive and time varying or time invariant, and all branch currents and voltages in the network
must satisfy Kirchhoff’s laws.
According to Tellegen’s theorem, the rate of supply of energy by the active elements of a
network equals the rate of energy dissipated or stored by the passive elements of the network.
Tellegen’s theorem is stated mathematically as follows:
b
∑ Vk I k = 0
K =1
Vk and Ik should satisfy KVL and KCL, respectively. In the above expession b indicates the
number of branches.
Example 5.11 In the network shown in
Figure 5.41, the branch voltages and currents
have the following values. Verify Tellegen’s
theorem for the network shown in the Figure.
V0 = 20 V, V1 = 16 V, V2 = 2 V, V3 = 4 V,
V4 = 14 V, V5 = 6 V
I4
+
+
A
−
−
B
V1
+
I0
I1
+
V3
−
V0
I 0 = −16 A, I1 = 12 A, I 2 = 2 A, I 3 = 14 A,
I 4 = 4 A, I 5 = 2 A
−
V4
I3
I2
D
+ V −
2
C
I5
+
V5
−
E
Figure 5.41
Solution: We will first verify if the data provided satisfy Kirchhoff’s laws.
1. By applying KVL for loop ABCA, + V0 − V1 − V3 = 0
or + 20 − 16 − 4 = 0
2. For loop BDECB, + V2 −V5 + V3 = 0
or + 2 − 6 + 4 = 0
3. For loop ABDECA, −V1 + V2 − V5 + V0 = 0
or − 16 + 2 − 6 + 20 = 0
4. For loop ABDA, − V1 + V2 + V4 = 0
or −16 + 2 + 14 = 0
To verify the applicability of KCL, the values at nodes A, B and C are calculated as follows: at
node A, −I0 = I1 + I4
or + 16 = 12 + 4
At node B, I1 + I2 = I3
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Network Theorems and Applications
189
or 12 + 2 = 14
At node C, I3 + I5 = −Io
or 14 + 2 = 16
Now, we apply Tellegen’s theorem to show that the sum of instantaneous power of all the
branches is zero.
5
∑ Vk I k = V0 I 0 + V1I1 + V2 I 2 + V3 I 3 + V4 I 4 + V5 I 5
K =0
= 20 × ( −16) + 16 × 12 + 2 × 2 + 4 × 14 + 14 × 4 + 6 × 2
=0
5.10 COMPENSATION THEOREM
This theorem is useful in determining the changes in current or voltage when the value of resistance gets changed in the circuit. If a small change in resistance in a network takes place from
R to R + ∆ R, this will cause a change in current in all branches.
According to Compensation theorem, the change in current in all the branches is equal to
the current produced by a voltage source of magnitude (I ∆ R) placed in series with the resistance whose value has changed.
Example 5.12 In the circuit shown in Figure 5.42,
the resistance of the branch having 5 Ω resistance has
changed to 6 Ω due to the connection of an ammeter
having an internal resistance of 1 Ω. Determine the
value of compensation source voltage and verify
results.
P
+
20 V
I
3Ω
5Ω
−
Solution: Total current delivered by the source can be
given as follows:
IT =
2.5 Ω
2Ω
Q
Figure 5.42
20
20
=
= 4A
5×5
5
2.5 +
5+5
Current through the 5 Ω resistor is I = 2 A.
When resistance has changed to 6 Ω, current I T′ through the branch is calculated as in the
following:
20
20
=
= 3.825 A
6 × 5 2.5 + 2.73
2.5 +
6+5
5
I ′ = 3.325 ×
= 1.74 A
5+6
dI = I ′ − I = 1.74 − 2 = −0.26 A
I T′ =
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190 Network Analysis and Synthesis
2.5 Ω
1Ω
3Ω
5Ω
VC
2Ω
2V
Figure 5.43
Compensation voltage VC = I ∆R = 2 × 1 = 2 V.
The circuit is drawn as in Figure 5.43 with the
compensation voltage source. The compensation
source is inserted in series with R + ∆ R.
Now, the current through the 5 Ω branch is calculated with the compensation voltage source in
place and replacing the voltage source by its internal
resistance.
2
2
I CV =
=
= 0.26 A
5 × 2.5 6 + 1.66
6+
5 + 2.5
Thus, the result is verified.
5.11 STAR−DELTA TRANSFORMATION
In any network, resistance elements may be seen as connected in series, in parallel, in star
formation or in delta formation. To simplify such circuits, star connected resistances can be
converted into equivalent delta connected resistances and vice-versa. In such transformation,
the circuit conditions are not changed.
5.11.1 Transforming Relations from Delta to Star
Let us consider three resistances RAB,
RBC and RCA connected in delta formation
between the terminals AB, BC and CA,
RA
RCA
RAB
respectively, as shown in Figure 5.44(a).
These three resistances can be converted into
RB
equivalent star forming resistances RA, RB
RC
and RC as shown in Figure 5.44(b).
RBC
C
B
C
B
For the purpose of equivalence, we will
(a)
(b)
equate the resistances of the two networks
Figure 5.44
across terminals AB, BC and CA as shown in
equations (5.1), (5.2) and (5.3), respectively.
RAB ( RBC + RCA )
RA + RB =
RAB + RBC + RCA
(5.1)
RBC ( RAB + RCA )
RB + RC =
RAB + RBC + RCA
(5.2)
RCA ( RBC + RAB )
RC + RA =
RAB + RBC + RCA
(5.3)
Subtracting equation (5.2) from equation (5.1), we obtain the following:
R R − RBC RCA
RA – RC = AB CA
RAB + RBC + RCA
(5.4)
A
A
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191
Network Theorems and Applications
Adding equation (5.4) and equation (5.3), we get the following form:
RAB RCA
RA =
RAB + RBC + RCA
Similarly, the values of RB and RC can be calculated.
When delta connected resistances are changed to star connected resistances, their values are
given as follows:
RA =
RAB RCA
(5.5)
RAB + RBC + RCA
RB =
RAB RBC
(5.6)
RAB + RBC + RCA
RC =
RBC RCA
(5.7)
RAB + RBC + RCA
Example 5.13 Three resistance of values 1 Ω, 2 Ω and 3 Ω are connected in delta formation
between terminals AB, BC and CA, respec­tively, as shown in Figure 5.45. Calculate the equivalent
star connected resistances.
A
Solution:
RA =
RAB RAC
1× 3
3 1
=
= = Ω
RAB + RBC + RCA 1 + 2 + 3 6 2
A
3Ω
RA
1Ω
RC
RAB RBC
1× 2
2 1
RB =
=
= = Ω
RAB + RBC + RCA 1 + 2 + 3 6 3
2Ω
C
RBC RCA
2×3
6
RC =
=
= = 1Ω
RAB + RBC + RCA 1 + 2 + 3 6
B
RB
C
B
(a)
(b)
Figure 5.45
5.11.2 Transforming Relations from Star to Delta
Now, let us consider three resistances RA,
RB and RC connected in star. The equivalent
delta connected resistances are RAB, RBC
and RCA as shown in Figure 5.46(a) and (b).
The basic equations guiding the transformation will be the same, that is, equations (5.1), (5.2) and (5.3).
From the basic equations, we derived
equations (5.5), (5.6) and (5.7). We will use
equation (5.5), (5.6) and (5.7) to determine
RAB, RBC and RCA in terms of RA, RB and RC.
M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 191
A
A
RA
RCA
RAB
RC
RB
C
B
C
(a)
RBC
B
(b)
Figure 5.46
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192 Network Analysis and Synthesis
Multiplying equation (5.5) by equation (5.6), we get the following:
RA RB =
2
RAB
RBC RCA
(5.8)
( RAB + RBC + RCA ) 2
Multiplying equation (5.5) by equation (5.7), the following form is obtained:
RA RC =
2
RCA
RAB RBC
( RAB + RBC + RCA ) 2
(5.9)
Multiplying equation (5.6) by equation (5.7), we write the equation as follows:
RB RC =
2
RBC
RAB RCA
( RAB + RBC + RCA ) 2
(5.10)
Adding equation (5.8), (5.9) and (5.10), we obtain the equation as follows:
RA RB + RB RC + RC RA =
or
RA RB + RB RC + RC RA =
RAB RBC RCA ( RAB + RBC + RCA )
( RAB + RBC + RCA ) 2
RAB RBC RCA
( RAB + RBC + RCA )
Earlier, we had equation (5.7) as in the following:
RC =
RBC RCA
RAB + RBC + RCA
By substituting the equation, we get the following form:
RA RB + RB RC + RC RA = RAB RC
Dividing both sides by RC, we write RAB as follows:
RAB = RA + RB +
RA RB
RC
Similarly, RBC and RCA can be calculated. When star connected resistances are changed to delta
connected resistances, their values are given as follows:
RAB = RA + RB +
RA RB
(5.11)
RC
RBC = RB + RC +
RB RC
(5.12)
RA
RCA = RC + RA +
RC RA
(5.13)
RB
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Network Theorems and Applications
Example 5.14 Three resistances each of
3 Ω value are connected in star formation.
Calculate their equivalent delta.
Solution: The three star connected
resistances and their equivalent delta
forming resistances between the terminals
A, B and C are shown in Figure 5.47.
Using the transforming equations, the
following can be obtained:
A
RA
A
3Ω
RC
RCA
RAB
3Ω
3 Ω RB
C
B
RBC
C
B
Figure 5.47
RAB = RA + RB +
RA RB
3×3
= 3+ 3+
= 9Ω
RC
3
RBC = RB + RC +
RB RC
3×3
= 3+ 3+
= 9Ω
RA
3
RCA = RC + RA +
RC RA
3×3
= 3+ 3+
= 9Ω
RB
3
Example 5.15 Calculate the equivalent resistance
of the network shown in Figure 5.48 across
terminals A and B using star−delta transformation
where necessary.
D
A
3Ω
C
3Ω
4Ω
2Ω E
F
4Ω
B
Solution: From Figure 5.48, it is seen that point D
and E in the network is joined together and hence
2Ω
considered as the same point. Between E and F,
Figure 5.48
the two 4 Ω resistors are in parallel. Therefore,
the network is redrawn and shown in Figure 5.49(a). The same is redrawn by arranging the
resistance elements as in Figure 5.49(b).
A
3Ω
C
3Ω
F
2Ω
D, E
2Ω
B
A
3Ω
C
3Ω
2Ω
2Ω
2Ω
D, E
2Ω
B
(a)
(b)
Figure 5.49
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194 Network Analysis and Synthesis
Now converting the delta forming resistances of 2 Ω values across terminals CBD into equivalent star the network is redrawn as shown in Figure 5.50(a). Through series−parallel conversion, the equivalent resistance between terminals AB is calculated as shown in Figure 5.50(b),
(c) and (d).
A
C
3Ω
3 + 2 = 11 Ω
3
3
D, E
3Ω
RC
= 2/3 Ω
3 + 2 = 11 Ω
3
3
RD = 2/3 Ω
N
RB = 2/3 Ω
A
2
Ω
3
B
N
B
(a)
(b)
2×2
=
2+2+2
2×2
=
RD =
2+2+2
2×2
=
RB =
2+2+2
RC =
2
Ω
3
2
Ω
3
2
Ω
3
11/3 Ω
A
N
11/3 Ω
A
11/6 Ω
N
A
2/3 Ω
2/3 Ω
B
B
(c)
B
2.5 Ω
(d)
Figure 5.50
A
Example 5.16 Six resistances, each of value R are connected
as shown in Figure 5.51. Calculate the equivalent resistance of the
network across terminals BC.
Solution: There are many ways of solving this problem. However,
we will convert the star forming resistances across terminals ABC
with star point at N and then make series−parallel conversions.
M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 194
R
N
R
R
B
R
R
R
C
Figure 5.51
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195
Network Theorems and Applications
A
A
×
3R
3R =
R
+
R
R
RAB RAC
3 R
R = 4
×3
R + 3R
R
3
4 R
R
RBC
B
R × 3R = 3 R
R + 3R 4
B
C
R
C
R×R
= 3R
R
R×R
= R+ R+
= 3R
R
R×R
= R+ R+
= 3R
R
RAB = R + R +
RBC
RCA
Figure 5.52
As shown in Figure 5.52, between terminals AB, resistance R and RAB are in parallel. Similarly,
between terminals AC, resistances R and RAC are in parallel. Again, between terminals BC,
resistances R and RBC are in parallel. The equivalent resistance between terminals BC is calculated as shown in Figure 5.53.
A
3
R
4
3
R
4
B
3
R
4
C
3
B
3
R
2
R
4
C
3R × 3R 9R2
8
4
RBC = 2
=
= RΩ
2
3R + 3R
9R
2
4
4
Figure 5.53
Note: The time taken to solve this problem will be less if the delta forming resistances are converted to equivalent star.
5.12 NUMERICALS ON NETWORK THEOREMS
Having studied all the network theorems, we will now solve some more network problems.
Example 5.17 Using Thevenin’s theorem,
calculate the current flowing through the 8 Ω
resistor connected across the terminals A and
B in Figure 5.54.
Solution: We will first convert the current
source of 5 A with the parallel resistance of 1 Ω
M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 195
C
A
4Ω
12 V
2Ω
1Ω
8Ω
5A
D
B
Figure 5.54
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196 Network Analysis and Synthesis
C
A
4Ω
2Ω
1Ω
8Ω
12 V
5V
D
B
Figure 5.55
P
4Ω
12 V
A
2Ω
1Ω
into an equivalent voltage source and then redraw
the circuit as shown in Figure 5.55.
To apply Thevenin’s theorem, we will temporarily remove the 8 Ω resistor through which the
current is to be calculated. Then we will calculate open-circuit voltage VAB, which is called VTh.
We will then calculate RTh, that is, the resistance
of the entire network across terminals AB. Then,
we will draw the Thevenin’s equivalent circuit and
calculate current through the load resistance of
8 Ω. We draw the circuit with 8 Ω resistor removed,
as shown in Figure 5.56.
By applying KVL in the loop, as shown in the
Figure, we get the loop equation as follows:
I
12 − 4I − 1I − 5 = 0
5V
Q
B
5I = 7
or
Figure 5.56
7
I = = 1.4 A
5
Since, no current will flow through the 2 Ω resistor, there will be no voltage drop across it.
Voltage across PQ will be equal to VAB. To calculate VPQ, we move from P to Q. The battery voltage is taken as positive and voltage drop in the 1 Ω resistor is also taken as
positive. Thus,
VPQ = VAB = 5 + 1 I = 5 + 1 × 1.4 = 6.4 V
Point P is at higher potential than point Q. The equivalent resistance of the whole network across
terminal AB with the voltage sources short-circuited is calculated as:
RAB = RTh = 2 +
1× 4
= 2.8 Ω
1+ 4
The Thevenin’s equivalent circuit is drawn and the load maintenance of 8 Ω is placed across AB,
as shown in Figure 5.57.
RTh
2.8 Ω
VTh
6.4 V
A
I
RL
8Ω
I=
VTh
RTh + RL
6.4
2.8 + 8
= 0.6 A
=
B
Figure 5.57
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Network Theorems and Applications
197
Example 5.18 Find the current in the 3 Ω resistor in the circuit of Figure 5.58 using
Thevenin’s theorem.
j6Ω
j5Ω
3Ω
10∠0°
j4Ω
−j 6 Ω
−j 4 Ω
Figure 5.58
Solution: Step 1: Remove the branch through which the current is to be calculated. Therefore,
we remove the 3 Ω resistor and redraw the circuit as shown in Figure 5.59.
j6Ω
A
10 ∠ 0°
j5Ω
B
−j 6 Ω
j4Ω
i1
−j 4 Ω
i2
Figure 5.59
Step 2: Let us find the voltage across open-circuited terminals, that is, A and B. From Figure 5.59,
it is clear that i1 =10 A and i2 = 0.
VA = −j6i1 = −j6(10) = −j60 V
Therefore,
VB = 0
VTh = VA − VB = −j60 − 0 = − j60 V.
Therefore,
Step 3: Let us find (ZTh), that is, the equivalent impedance of the network removing all the
sources, as seen from open-circuited terminals A and B. The current source has been kept open
as shown in Figure 5.60.
Between terminal BB′, we can get the following form:
Z BB′ = j 4 (j 5 − j 4)
j6Ω
A
j5Ω
B
= j4 j
j4 ⋅ j
−4
=
j4 + j
j5
4
= j
5
−j 6 Ω
=
A′
j4Ω
−j 4 Ω
B′
Figure 5.60
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198 Network Analysis and Synthesis
ZTh = Z AB = − j 6 + j
4
5
= − j 5.2
Step 4: Therefore, the Thevenin’s equivalent circuit will be as shown in Figure 5.61.
Now, the current through 3 Ω resistance is calculated as follows:
i=
A
ZTh = − j 5.2 Ω
+
−
VTh = − j 60 V
3Ω
I
B
VTh
ZTh + 3
Figure 5.61
− j 60
− j 5.2 + 3
+60 ∠ − 90°
=
6 − 60°
= 10 ∠ − 30°A
=
5Ω
A
100∠0° V
Example 5.19 Find the current through (5 + j4) Ω
impedance shown in Figure 5.62 using Thevenin’s
theorem.
5Ω
+
Ω
j1
10
j4Ω
6Ω
B
−j
8Ω
Figure 5.62
Solution: Step 1: Remove the branch (5 + j4) Ω
impedance, as shown in Figure 5.63(a).
The same circuit is redrawn as shown in Figure 5.63(b).
I
Ω
I1
5Ω
j1
10
+
+
A
100∠0° V
6Ω
B
8
−j
I2
10
100∠0° V
100 V A
VTh
6
Ω
(a)
j 15
B
100 V
−j 8
(b)
Figure 5.63
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Network Theorems and Applications
199
Now, equivalent impedance of the network is as follows:
Zef = (10 + 6) ( j15 − j8)
= 16 j 7
=
112∠90°
(16)( j 7)
j 112
=
=
16 + j 7 16 + j 7 17.46 ∠23.63°
= 6.415∠66.37°
100
100
=
= 15.58∠ − 66.37°A
I=
Zef 6.4150 ∠66.37°
= 6.25 − j14.27A
Now, by using current divider rule, we calculate the current as follows:
( j15 − j8) 15.58∠ − 66.37°( j 7)
=
16 + j 7
17.46 ∠23.63°
= (0.89∠ − 90°)(7∠90°
= 6.23A
I1 = I
and
I2 = I − I1
= 15.58∠−66.37 − 6.23
= 6.25 − j14.27 − 6.23
= 0.02 − j14.27
Now
VA = 100 − Voltage drop across 10 Ω
= 100 − 10I1
By substituting the value of I1 in the equation,
VA = 100 − 10 (6.23)
= 100 − 62.3
= 37.7 V.
Thus,
VB = 100 − Voltage drop across j15 Ω resistor.
= 100 − j15 I2
= 100 − j15 (0.02 − j14.27)
= 100 − j0.3 − 214.05
= −114.05 − j0.3
10 Ω
VTh = VA − VB
= 37.7 − (−114.05 − j0.3)
6Ω
= 151.75 + j0.3
= 151.75 ∠ 0.11
Step 2: Let us find ZTh across terminals A and B by shortcircuiting the voltage source, as shown in Figure 5.64.
M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 199
A
B
j 15 Ω
−j 8 Ω
Figure 5.64
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200 Network Analysis and Synthesis
ZTh = (10 6) + ( j15 − j8)
=
10.6 ( j15 − j8)
+
10 + 6
j15 − j8
60 120
+
16
j7
= 3.75 − j17.14 Ω
=
ZTh = 3.75 −j 17.14
−
VTh = 151.75 ∠ 0.11°
Step 3: The Thevenin’s equivalent circuit is shown
in Figure 5.65.
151.75∠0.11°
3.75 − j17.14 + 5 + j 4
151.75∠0.11°
=
8.75 − j13.14
151.75∠0.11°
=
15.786 ∠ − 56.2°
= 9.61∠56.31°
=
+
5Ω
j4Ω
100∠0° V
6Ω
−j
8Ω
Figure 5.66
10
+
100∠0° V
6
Solution: Note that this problem has already been
solved using Thevenin’s theorem. Here, we will use
Norton’s theorem and verify the result.
Step 1: Short-circuit the branch in which response is to be
determined as shown in Figure 5.67.
This diagram can be redrawn as shown in Figure 5.68.
Applying KVL in closed circuit CDEHC, the following
form is obtained:
100 = 10I1 + 6 (I1 − IN)
or
16I1 − 6IN = 100
or
8I1 − 3IN = 50
(5.14)
Applying KVL in closed circuit CFGHC, the equation can be
written as follows:
100 = j 15I2 − j8 (I2 + IN)
M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 200
Ω
5
Example 5.20 Find the current through (5 + j4) Ω
impedance in the network shown in Figure 5.66 using
Norton’s theorem.
10
j1
Current through the (5 + j4 Ω) impedance is
= I = 9.61∠56.45° A.
Figure 5.65
5Ω
VTh
ZTh + 5 + j 4
j4
I
j1
I=
5
+
−j
8
Figure 5.67
C
I
D
+
F
I2
I1
j 15
10
10 0 V A
IN
I1 −IN
6
E
B
I2 +IN
−j 8
G
H
Figure 5.68
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Network Theorems and Applications
j7I2 − j8IN = 100
or
201
(5.15)
Applying KVL in closed circuit DFBAD, we get the equation as in the following:
10I1 − j15I2 = 0
(5.16)
Step2: Let us solve equations 1, 2 and 3 for IN using Cramer’s rule
8
0
−3
∆= 0
j 7 − j8
10 − j15 0
= 8(0 − j 2 120) + 0 − 3(0 − j 70)
= 960 + j 210
∆N
8
= 0
0
0
j 7 100
10 − j15 0
= 8(0 + j1500) + 0 + 50(0 − j 70)
= j12000 − j 3500
= j 8500
IN =
∆N
j8500
j850
850°∠90°
=
=
=
= 8.65∠77.47°
∆
960 + j 210 96 + j 21 98.27∠12.33°
Step 3: Let us find ZN (Norton equivalent impedance) as
shown in Figure 5.69.
ZTh = (10 6) + ( j15 − j8)
10 Ω
10.6 ( j15)( − j8)
=
+
10 + 6
j15 − j8
60 120
=
+
16
j7
= 3.75 − j17.14
= 17.54 ∠ − 77.65° Ω
A
B
6Ω
j 15 Ω
−j 8 Ω
Figure 5.69
Step 4: let us draw Norton’s equivalent circuit as shown in Figure 5.70.
I
IN = 8.65 ∠ 77.47°
ZN = 17.54 ∠ − 77.65°
5
j4
Figure 5.70
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202 Network Analysis and Synthesis
Now, by current divider rule, the current through the impedance (5 + j4) is given as follows:
I = IN
ZN
Z N + 5 + j14
{8.65∠77.47°}{17.54 ∠ − 77.65°}
3.75 − j17.14 + 5 + j 4
151.721∠ − 0.18°
=
8.75 − j °13.14
151.721∠ − 0.18°
=
15.786 ∠ − 56.34°
= 9.61∠56.26 A
=
9A
Example 5.21 Find the current in 5 Ω resistance
in the network shown in Figure 5.71 using
superposition theorem and then verify the result
using Thevenin’s theorem.
Solution: In the given circuit, there are two
sources. We will consider one source at a time.
Let us first see the effect of 9 A current source.
The circuit is redrawn with 9 A current source
and the voltage source short-circuited, as shown in
Figure 5.72.
Let us find current in 5 Ω resistor due to 9 A current source, acting alone using the mesh analysis.
Current through 5 Ω resistor = i3 − i2
(from left to right)
Now, let us find i2 and i3.
From mesh I,
I1 = 9 A
4Ω
2Ω
4Ω
9A
I1
4Ω
Substituting i1 = 9, we get the equation as follows:
M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 202
(5.18)
Mesh I
4Ω
2Ω
−4i1 + 11i2 − 5i3 − 2i4 = 0
20 Ω
Figure 5.71
(5.17)
4(i2 − i1) + 5(i2 − i3) + 2(i2 − i4) = 0
11i2 − 5i3 − 2i4 = 36 100 Ω
+
12 V
−
Applying KVL in mesh II, we get the following
form:
or
5Ω
Mesh II
I2
5Ω
100 Ω
I4
Mesh IV
20 Ω
I3
Mesh III
Figure 5.72
12/3/2014 7:58:23 PM
Network Theorems and Applications
203
Applying KVL in mesh III, we write the equation as in the following:
5 (i3 − i2) + 20i3 + 100 (i3 − i4) = 0
− 5i2 + 125i3 − 100i4 = 0
or
(5.19)
Applying KVL in mesh IV, the equation can be calculated as follows:
2(i4 − i2) + 100(i4 − i3) + 4i4 = 0
−2i2 −100i3 + 106i4 = 0
or
(5.20)
Let us solve equations (5.18), (5.19) and (5.20) using Cramer’s rule for i2 and i3
11
−5
−2
∆ = −5 125 −100
−2 −100 106
= 11(125 × 106 − 100 2 ) + 5( −5 × 106 − 200) − 2(500 + 250)
= 35750 − 3650 − 1500
= 30, 600
11 36 −2
∆ 3 (for i3 ) = −5 0 −100
−2 0 106
= 11(0 − 0) − 36( −5 × 106 − 200) − 2(0 − 0)
= 26280
36 −5
−2
∆ 2 (for i2 ) = 0 125 −100
0 −100 106
= 36(125 × 106 − 100 2 ) + 5(0 − 0) − 2(0 − 0)
= 117000
Therefore
∆ 2 117000
=
= 3.82 A
30600
∆
∆
26280
= 0.8588 A
i3 = 3 =
∆ 30600
i2 =
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204 Network Analysis and Synthesis
Therefore, the current due to the 9 A current source alone through the 5 Ω resistor is equal to
(i2 − i3), that is, (3.82 − 0.8588) A. That is, a current of 2.9612 A will flow from right to left.
Now, we will have to find current through 5 Ω resistor due to 12 V source acting alone.
We keep the 12 V source in the circuit and open-circuit the current source, as shown in
Figure 5.73(a).
4Ω
4Ω
2Ω
I
5Ω
4Ω
4Ω
100 Ω
Mesh I
2Ω
20 Ω
+
12 V
−
i1
100 Ω
III
+
12 V
−
5Ω
i3
i2
Mesh II
(a)
20 Ω
II
Mesh III
(b)
Figure 5.73
Applying KVL in mesh I, we get the following form:
4i1 + 5(i1 − i2) + 2(i1 − i3) = 0
11i1 − 5i2 − 2i3 = 0
or
(5.21)
Applying KVL in mesh II, we can write the equation as follows:
5(i2 − i1) + 20i2 + 100(i2 − i3) = 0
or
−5i1 + 125i2 − 100i3 = 0
or
−i1 + 25i2 − 20i3 = 0
(5.22)
Applying KVL in mesh III, the equation can be calculated as follows:
2(i3 − i1) + 100(i3 − i2) + 4i3 = 12
or
−2i1 − 100i2 + 106i3 = 12
or
−i1 − 50i2 + 53i3 = 6
(5.23)
Let us solve equations (5.21), (5.22) and (5.23) for i1 and i2.
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Network Theorems and Applications
205
11 −5 −2
∆ = −1 25 −20
−1 −50 53
= 11( 25 × 53 − 1000) + 5( −53 − 20) − 2(50 + 25)
= 3060
0 −5 −2
∆1 = 0 25 −20
6 −50 53
= 0 + 5(0 + 120) − 2(0 − 150)
= 600 + 300
= 900
11 0 −2
∆ 2 = −1 0 −20
−1 6 53
= 11(0 + 120) + 0 − 2( −6 − 0)
= 1320 + 12
= 1332
∆1
900
=
= 0.294 A
∆ 3060
∆
1332
i2 = 2 =
= 0.435 A
∆ 3060
i1 =
Current through 5 Ω resistance due to 12 V source = i2 − i1 (from right to left) = 0.435 − 0.294
= 0.141 A.
By superposition theorem, we can say the following:
Current through 5 Ω resistances = Due to 9 A source + Due to 12 V source
= 2.9612 + (0.141)
= 3.1022 A.
Using Thevenin’s theorem, the following steps are to be followed:
Step 1: Open-circuit the branch containing 5 Ω resistance, as shown in Figure 5.74.
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206 Network Analysis and Synthesis
Step 2: Let us find VTh (voltage across AB)
The circuit can be redrawn as in Figure 5.75.
From mesh I, the current can be given as follows:
9A
i1 = 9 A
Applying KVL in mesh II, we get the following
form:
4Ω
2Ω
4(i2 − i1) + 20i2 + 100(i2 − i3) + 2(i2 − i3) = 0
A B
4Ω
20 Ω
100 Ω
+
12 V
−
or
−4i1 + 126i2 − 102i3 = 0
or
−2i1 + 63i2 − 51i3 = 0
Substituting i1 = 9, we can write the equation as
follows:
Figure 5.74
63i2 − 51i3 = 18
9A
Applying KVL in mesh III, the equation can
be calculated as in the following form:
Mesh I
i1
4Ω
2Ω
A
12 V
+
−
i2 B
i3
Mesh III
4i3 + 2(i3 − i2) + 100(i3 − i2) = 12
or
−102i2 + 106i3 = 12
or
−51i2 + 53i3 = 6
20 Ω
100 Ω
∆=
Figure 5.75
(5.25)
Solving equation (5.24) and (5.25), we get the
equation as follows:
Mesh II
4Ω
(5.24)
63 −51
= 63 × 53 − 512 = 738
−51 53
∆1 =
18 −51
= 1260
6 53
∆2 =
63 18
= 1296
−51 6
Therefore, the value of i2 and i3 can be calculated as in the following:
∆1 1260
=
= 1.707 A
738
∆
∆
1296
i3 = 2 =
= 1.756 A
738
∆
i2 =
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207
Network Theorems and Applications
VA = 100 (i3 − i2) = 100(1.756 − 1.707)
Now,
= 4.9 V
VB = 20i2 = 20(1.707) = 34.14 V
VTh = VA − VB = 4.9 − 34.14
Therefore,
= −29.24 V
Step 3: Let us find RTh using Figure 5.76(a) and (b).
4Ω
4Ω
2Ω
4Ω
A
2Ω
B
100 Ω
20 Ω
A
4Ω
B
20 Ω
100 Ω
(a)
(b)
Figure 5.76
RTh = {2 4 100} + {20 4}
8
 80
=  100 +
6
 24
= {1.33 100} + 3.33
= 1.312543 + 3.33
= 4.64 Ω
Now, let us draw Thevenin’s equivalent circuit as shown in
Figure 5.77.
Current through 5 Ω resistance
29.24
4.6425 + 5
29.24
=
9.6425
= 3.03 A
=
M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 207
RTh = 4.64 Ω
+
−
29.24 V
A
5Ω
i
B
Figure 5.77
12/3/2014 7:58:28 PM
208 Network Analysis and Synthesis
Example 5.22 Find the current through
the load resistance RL in the circuit
shown in Figure 5.78 using Thevenin’s
theorem and Norton’s theorem.
j3Ω
5Ω
4Ω
A
5Ω
Solution: Using Thevenin’s theorem,
the following steps are to be followed:
25 ∠ 0 ° A
B
Figure 5.78
5Ω
j3Ω
4Ω
A
5Ω
−j 3 Ω
25 A
− j 75
75∠90°
=
i2 =
9 + j 5 10.29∠29.05°
RL = 1 Ω
j5Ω
Step 1: Open-circuit the terminals AB,
as shown in Figure 5.79.
Step 2: Let us find voltage (VTh) across
open-circuited terminals
From mesh I, we get i1 = 25 A
From mesh II, we obtain −j3 (i2 − ii) + j3i2 +
4i2 + (5 + j5) i2 = 0
or
+ j3i1 + (9 + j5) i2 = 0
By substituting i1 = 25, we obtain the
following:
(9 + j5) i2 = − j75
−j 3 Ω
i1
VTh
i2
Mesh I
j5Ω
Mesh I
B
Figure 5.79
= 7.288∠ − 119.05 A
VTh = (5 + j 5)i2 = (7.07∠45°)i2
= (7.07∠45°)(7.288∠ − 119.05)
= 51.53∠ − 74.05°
j3Ω
4Ω
A
5Ω
5Ω
−j 3 Ω
Step 3: Now, let us find ZTh from circuits shown
in Figure 5.80.
j5Ω
B
ZTh = ( 4 + j 3 − j 3) (5 + j 5)
= 4 5 + j5
4(5 + j 5)
=
4 + 5 + j5
20 + j 20
=
9 + j5
28.28∠ 45°
=
10.29∠29.05°
= 2.75∠15.95°
M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 208
4Ω
j3Ω
A
5Ω
−j 3 Ω
j5Ω
B
Figure 5.80
12/3/2014 7:58:30 PM
Network Theorems and Applications
Let us draw Thevenin’s equivalent circuit, as shown in
Figure 5.81.
A
ZTh
2.75 ∠ 15.95°
+
VTh = 51.53 ∠ −74.05°
IL = current through RL
209
1Ω
IL
=
VTh
ZTh + RL
=
51.53∠ − 74.05°
2.75∠15.95° + 1
=
51.53∠ − 74.05°
2.644 + j 0.7556 + 1
=
51.53∠ − 74.05°
3.644 + j 0.7556
=
51.53∠ − 74.05°
3.72∠11.71°
B
Figure 5.81
= 13.85∠ − 85.76°A
By using Norton’s theorem, the following steps are performed.
Step 1: Short-circuit the terminals AB as shown in Figure 5.82 and draw an equivalent circuit.
Step 2: Let us use mesh analysis to find IN
From mesh I, i1 = 25 A
and from mesh II, we get the following:
j3Ω
4Ω
− j 3(i2 − i1 ) + j 3i2 + 4i2 = 0
5Ω
j 3i1 + 4i2 = 0
25 A
or
or
4i2 = − j 3i1
put
i1 = 25
A
5Ω
−j 3 Ω
j5Ω
B
equivalent to
4i2 = − j 75
j3Ω
or
i2 = − j
75
4
= − j18.75
i2 = 18.75∠ − 90ο
I N = i2 = 18.75∠ − 90ο A
M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 209
4Ω
A
5Ω
25 A
−j 3 Ω
i1
i2
Mesh I
Mesh II
IN
B
Figure 5.82
12/3/2014 7:58:32 PM
210 Network Analysis and Synthesis
Step 3: To find ZN refer to Figure 5.83
Z N = ( 4 + j 3 − j 3) (5 + j 5)
j3Ω
= 4 5 + j5
4(5 + j 5)
4 + 5 + j5
20 + j 20
=
9 + j5
28.28∠ 45°°
=
10.29∠29.05°
= 2.75∠15.95°
j5Ω
B
j3Ω
51.56 ∠ − 74.05°
2.644 + j 0.755 + 1
=
51.56 ∠ − 74.05°
3.644 + j 0.755
=
51.56 ∠ − 74.05°
3.72∠11.771°
= 13.85∠ − 85.76°A.
5Ω
−j 3 Ω
j5Ω
B
Figure 5.83
A
IN = 18.75 ∠ − 90°
Step 1: We remove the load impedance
ZL and redraw the circuit as shown in
Figure 5.86.
M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 210
ZN = 2.75∠ 15.95°
1Ω
(RL)
IL
B
Figure 5.84
2Ω
−j 10 Ω
3Ω
10∠ 0°
ZL
J5Ω
Example 5.23 Find the current thro­
ugh ZL in the circuit shown in Figure 5.85
using Thevenin’s theorem and Norton’s
theorem
Solution: Using Thevenin’s theorem,
the following steps are to be followed:
4Ω
A
{By current divider formula}
=
5Ω
−j 3 Ω
The Norton’s equivalent circuit is shown in
Figure 5.84.
Current through RL is IL and it can be
expressed as follows:
I
IL = IN N
Z N +1
(18.75∠ − 90°)( 2.75∠15.95°)
2.75∠15.95° + 1
A
5Ω
=
=
4Ω
Figure 5.85
A
i1
Mesh I
VTh
−j 10 Ω
3Ω
10∠ 0°
i2
B
Mesh II
Figure 5.86
12/3/2014 7:58:35 PM
Network Theorems and Applications
211
Step 2: Let us find VTh using mesh analysis
From mesh I, i1=10 A
From mesh II and by applying KVL, we get the following equations:
3(i2 − i 1) + (−j10) i 2 = 0
− 3i 1 + (3 − j10) i 2 = 0
or
Substituting i1 = 10, the current and voltage are calculated as follows:
(3 − j10) i2 = 30
30
30
=
3 − j10 10.44 ∠ − 73.3°
= 2.87∠73.3°
i2 =
VTh = − j10i2
= − j10( 2.87∠73.3°)
= {10 ∠ − 90°}{2.87∠73.3°}
= 28.7∠ − 16.7°
ZTh = 3 − j10
−j 10 Ω
3Ω
We find ZTh by open-circuiting the current source as
shown in Figure 5.87.
A
B
Figure 5.87
3( − j10)
3 − j10
30 ∠ − 90°
− j 30
=
=
3 − j10
3 − j10
30 ∠ − 90°
=
10.44 ∠ − 73.3°
= 2.87∠ − 16.7°
=
The Thevenin’s equivalent circuit is shown in
Figure 5.88.
IL =
VTh
ZTh + Z L
28.7∠ − 16.7°
2.87∠ − 16.7° + 2 + j 5
28.7∠ − 16.7°
=
2.75 − j 0.825 + 2 + j 5
28.7∠ − 16.7°
=
4.75 + j 4.175
=
M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 211
28.7∠ −16.7°
ZTh
+
VTh = 28.7∠ −16.7°
IL
ZL = 2 + j 5
Figure 5.88
12/3/2014 7:58:37 PM
IL =
VTh
ZTh + Z L
212 Network Analysis and Synthesis
28.7∠ − 16.7°
2.87∠ − 16.7° + 2 + j 5
28.7∠ − 16.7°
=
2.75 − j 0.825 + 2 + j 5
28.7∠ − 16.7°
=
4.75 + j 4.175
28.7∠ − 16.7°
=
6.32∠41.31°
I L = 4.54 ∠ − 58.01°A
=
Using Norton’s theorem, the following steps are to be performed.
Step 1: We short-circuit the load terminals A and B
as shown in Figure 5.89. The whole of current 10 A will
flow through the short-circuited path, so that IN = 10 A.
Step 2: To find ZN across terminals A and B, we
open-circuit the current source as shown in Figure
5.90. Across terminals A and B, 3 Ω resistor and
A
IN
−j 10 Ω
3Ω
10 A
B
Figure 5.89
Z N = 3 − j10
3( − j10)
3 − j10
− j 30
=
3 − j10
30 ∠ − 90°
=
10.44 ∠ − 73.3°
= 2.87∠ − 16.7°
=
A
−j 10 Ω
3Ω
B
Figure 5.90
Norton’s equivalent circuit is shown in Figure 5.91.
Using current divider rule, current can be calculated as follows:
A
IN = 10 A
ZN = 2.85∠ −16.7°
IL
ZN = 2 + j 5
B
Figure 5.91
M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 212
IL =
I N ⋅ ZN
Z N + ZL
10( 2.87∠ − 16.7°)
2.87∠ − 16.7° + 2 + j 5
28.7∠ − 16.7°
=
2.75 − j 0.825 + 2 + j 5
28.7∠ − 16.7°
=
4.75 + j 4.175
28.7∠ − 16.7°
=
6.32∠41.31°
= 4.54 ∠ − 58.01°A
=
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213
Network Theorems and Applications
Example 5.24 Determine the cur­
rent through (4 − j8) Ω branch in the
circuit shown in Figure 5.92 using
Thevenin’s theorem and Norton’s
theorem.
3Ω
+
j4Ω
+
4Ω
100∠ 0°
50∠ 90° V
−j 8 Ω
Figure 5.92
Solution: Using Thevenin’s theorem,
the steps to be followed are given as follows:
Step 1: Open-circuit the branch through
which current is to be determined as shown in
Figure 5.93.
3Ω
+
100 V
Step 2: To find VTh, we find the current through
the circuit as follows:
i=
=
j4Ω
A
B
VTh
+
i
J 50 V
Figure 5.93
100 − j 50
3 + j4
111.80 ∠ − 26.56°
5∠53.13°
= 22.36 ∠79.67°
≈ 4 − j 21.998
≈ 4 − j 22 A
VTh = 100 − 3i = 100 − 3 (4 − j22)
= 100 − 12 + j66
3Ω
= 88 + j66
= 110∠63.86 V
A
B
To find ZTh, we short-circuit the voltage sources as shown
in Figure 5.94.
ZTh
3( j 4)
j12
j12
= 3 j4 =
=
=
3 + j 4 3 + j 4 5∠53.13°
12∠90°
=
5∠53.13°
= 2.4 ∠36.87°
Thevenin’s equivalent circuit will be as shown in
Figure 5.95.
M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 213
j4Ω
Figure 5.94
ZTh = 2.4∠ 36.87°
+
A
4
VTh = 110∠63.86°
−j 8
I
B
Figure 5.95
12/3/2014 7:58:42 PM
214 Network Analysis and Synthesis
110 ∠63.86°
2.4 ∠36.87° + 4 − j8
110 ∠63.86°
=
1.9199 + j1.44 + 4 − j8
110∠63.86°
=
5.9199 + j 6.56
110∠63.86°
=
8.836 ∠ − 47.93
= 12.44 ∠84.79°A
Current through ( 4 − j8) Ω resistor =
3Ω
Using Norton’s theorem, the following steps are to
be followed:
j4Ω
IN
+
+
A
100 V
i1
Mesh I
j 50 V
B
i2
Mesh II
Step 1: Short-circuit the branch through which
current is to be determined, as shown in Figure 5.96.
Let us find IN. From mesh I, 100 − 3i1 = 0
i1 =
or
Figure 5.96
100
= 33.33A
3
From mesh II, − j 4i 2 − j 50 = 0
i2 =
or
− j 50
= −12.5 A
j4
IN = i1 − i2
= 33.33 − (− 12.5)
= 45.83 A
ZN = 3 j4 =
=
j12
3 + j4
j12
12∠90°
=
= 2.4 ∠36.87° Ω
5∠53.13° 5∠53.13°
Norton’s equivalent circuit is shown in Figure 5.97.
By current divider formula, we can calculate IL as in the following:
IL =
4
IN = 45.83 A
ZN = 2.4∠ 36.87°
IL
−j 8
Figure 5.97
M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 214
ZL
I N ⋅ ZN
Z N + ZL
( 45.83)( 2.4 ∠36.87°)
2.4 ∠36.87° + 4 − j8
109.992∠366.86°
=
1.9199 + j1.44 + 4 − j8
=
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215
Network Theorems and Applications
109.992∠36.89°
5.9199 − j 6.56
109.992∠36.86°
=
8.836 ∠ − 47.93
= 12.44 ∠84.79°A
=
Example 5.25 Verify the reciprocity theorem for
the network shown in Figure 5.98.
A
C
I
+
Solution: Let us find current in branch CD, that is,
I due to the voltage source in branch AB as in
Figure 5.99.
Applying KVL in mesh I, we get the following:
2Ω
−j 10 Ω
3Ω
j5Ω
B
D
Figure 5.98
A
3 (i1 − i2) = 10
or
10∠ 0°V
C
i
+
10
i1 – i2 =
3
i1 − i2 = 3.33
i1
(5.26)
B
Mesh I
Applying KVL in mesh II, the following form can
be obtained:
2Ω
−j 10 Ω
3Ω
10 V
i2
i3
Mesh II
Mesh III D
j5Ω
Figure 5.99
3(i2 − i1) + (−j10) (i2 − i3) = 0
or
− 3i1 + (3 − j10) i 2 + j10 i 3 = 0
(5.27)
Applying KVL in mesh III, the equation can be written as follows:
− J10 (i3 − i2) + (2 + j5) i3 = 0
j10i2 + (2 − j5) i3 = 0
or
(5.28)
Let us solve equations (5.26), (5.27) and (5.28) using Cramer’s rule for i3.
1
−1
0
∆ = −3 3 − j10
j10
0
j10
2 − j5
= 1{(3 − j10)( 2 − j 5) − j 2 100} + 1{−3( 2 − j 5) − 0} + 0
= {6 − j15 − j 20 − 50 + 100} + {−6 + j15}
= 50 − j 20
and
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216 Network Analysis and Synthesis
1
−1
3.33
∆ 3 = −3 3 − j10
0
0
0
j10
= 1{0 − 0} + 1{0 − 0} + 3.33{− j 30 − 0}
= − j 99.9
i3 =
∆3
− j 99.9
99.9∠ − 90°
=
=
∆ 50 − j 20 53.85∠ − 21.80°
= 1.855∠ − 68.2° = I
V
10
=
= 5.39∠68.2°(5.29)
I 1.855∠ − 68.2°
Now, let us insert the voltage source in branch CD
A
C
and calculate the current through branch AB, as
shown in Figure 5.100.
2Ω
Applying KVL in mesh I, we get the following:
−j 10 Ω
3Ω
I
j5Ω
(2 − j5) i1 + j10i2 = 10
(5.30)
+
Applying
KVL
in
mesh
II,
the
following
form
is
i2
i1
i3
10 V
obtained:
B Mesh III
Mesh II
Mesh I D
j10i1 + (3 − j10)i2 − 3i3 = 0
(5.31)
Figure 5.100
Applying KVL in mesh III, the equation can be
written as follows:
−3i2 + 3i3 = 0 or −i2 + i3 = 0
(5.32)
Now, let us solve equations (5.30), (5.31) and (5.32) for i3.
j10
2 − j5
0
∆ = j10 3 − j10 −3
−1
0
1
= ( 2 − j 5){3 − j10 − 3} − j10{ j10 − 0} + 0
= ( 2 − j 5)( − j10) − j 2 100
= − j 20 − 50 + 100
= 50 − j 20
and
∆3 =
2 − j5
j10
10
j10
0
3 − j10
−1
0
0
= ( 2 − j 5){0 − 0} − j10{0 − 0} + 10{− j10 − 0}
= − j100
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Network Theorems and Applications
217
Therefore,
∆3
− j100
100 ∠ − 90°
=
=
∆ 50 − j 20
50 − j 20°
100 ∠ − 90°
=
53.85∠ − 21.80°
i3 =
= 1.855∠ − 68.2° = I (5.33)
V
10
=
= 5.39∠68.2°(5.34)
I 1.855∠ − 68.2°
V
A
j4Ω
j5Ω
From equations (5.29) and (5.34), it is clear that
ratio in
I
both the cases is same, and hence the reciprocity theorem is
100∠ 0°
verified.
Example 5.26 Determine the maximum power delivered
to the load in the circuit shown in Figure 5.101.
B
Figure 5.101
Solution: Firstly, let us find the Thevenin’s equivalent
voltage source across terminals A and B.
Open-circuit the terminals AB as in Figure 5.102.
j4Ω
A
VTh
j3Ω
10 0 V
i
B
VTh = j3i = j3 (−j14.28)
Therefore,
j5Ω
+
100
i=
= − j14.28 A
j7
Now
ZL
j3Ω
Figure 5.102
= − j2 42.84
j4
j5
A
= 42.84 V
j3
To find ZTh, we consider the circuit with short-circuiting the
voltage source as in Figure 5.103.
ZTh
j 2 12
= ( j 4 j 3) + j 5 =
+ j5
j 4 + j3
−12
=
+ j 5 = j 7.714 + j 5
j7
= j 6.714 Ω
B
Figure 5.103
ZTh = j 6.714
+
ZL
VTh = 42.84∠ 0°
IL
Therefore, the given circuit can be redrawn as shown in
Figure 5.104.
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Figure 5.104
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218 Network Analysis and Synthesis
Now, to get the maximum power delivered to the load impedance, the load impedance must
be the complex conjugate of the source impedance.
ZL = − j6.714
VTh
VTh
IL =
=
=0
ZTh + Z L
j 6.714 − j 6.714
That is,
Therefore,
Therefore, maximum power transferred to the load is = I L2 Z L = 0
Example 5.27 Determine the maximum power
delivered to the load in the circuit, as shown in
Figure 5.105.
Solution: Firstly, let us find the Thevenin’s
equivalent circuit across terminals AB. We will
calculate VTh and RTh. Figure 5.106 shows the
diagrammatic representation of the circuit with
load impedance removed.
Applying KVL in mesh I, we get the following
form:
4Ω
+
2Ω
+
−
j6Ω
10∠ 0° V
+
2Ω
+
i
−
j6Ω
5∠ 90° V
A
−
VTh
−j 4 Ω
B
Voltage across the capacitor = −j4 × i
= −j4 (1.25 + j1.25)
= −j5 + 5
= 5 − j5
VTh = − 10∠0 + 5∠90 − (5 − j5)
= − 10 + j5 − 5 + j5
= −15 + j10
= 18.027∠146.30
2Ω
ZL
−j 4 Ω
4Ω
j5
j5
5∠90°
=
=
2 + j 2 2.828∠45° 2.828∠45°
= 1.768∠45° = 1.25 + j1.25
A
5∠ 90° V
B
i=
4Ω
A
−
Figure 5.105
−j4i + (2 + j6) i = 5∠90
(2 + j2) i = j5
or
10∠ 0° V
Figure 5.106
To find ZTh, we short-circuit the voltage sources and find equivalent impedance of the circuit across the terminals AB as in
Figure 5.107.
−j 4 Ω
ZTh = {( 2 + j6) ( − j 4)} + 4
j6Ω
B
Figure 5.107
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=
− j 4( 2 + j 6)
+4
2 + j6 − j 4
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Network Theorems and Applications
219
− j8 + 24
+4
2 + j2
12 − j 4
=
+4
1+ j
12 − j 4 + 4 + j 4
=
1+ j
ZTh =
16
1+ j
16 ∠0°
=
1.414 ∠45°
= 11.315∠ − 45°
= 8 − j8 Ω
=
The Thevenin’s equivalent circuit is shown in Figure 5.108.
Now, to get the maximum power delivered to load
impedance, the load must be equal to complex conjugate
of the source impedance.
Therefore, for maximum power to be delivered to load,
ZL is calculated as follows:
ZTh = (8−j 8) Ω
+
VTh = 18.027∠ 146.30°
ZL
Figure 5.108
ZL = 8 + j8
IL =
VTh
ZTh + Z L
18.027∠146.30°
(8 − j8) + (8 + j8)
18.027∠146.30°
=
16
= 1.126 ∠146.30°A
=
Therefore, maximum power transferred to load PL = I L2 RL
= (1.126)2 (8)
= 10.14 Watts.
Example 5.28 Determine the value of load impedance,
ZL, for which maximum power will be delivered to this load
from the source in the circuit shown in Figure 5.109.
Solution: Open-circuit voltage across terminals A and B after
removing the load impedance is calculated using Figure 5.110.
M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 219
j 10 Ω
+
A
5Ω
ZL
100∠ 0°
−j 10 Ω
B
Figure 5.109
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220 Network Analysis and Synthesis
j 10 Ω
+
100∠ 0° V
i
Applying KVL in the loop in the circuit of Figure 5.110,
we get the following form:
A
5Ω
100 ∠0° − j10i − 5i − ( − j10)i = 0.
VTh
−j 10 Ω
or
B
i=
100
100
=
= 20A
5
j10 + 5 − j10
Figure 5.110
j 10 Ω
VTh = (5 − j10) i
= (5 − j10) 20
= 100 − j200 V
A
To find ZTh, we consider the circle across open-circuited
terminals A and B by short-circuiting the voltage source as
shown in Figure 5.111.
ZTh = j10 (5 − j10)
5Ω
−j 10 Ω
B
+
Figure 5.111
=
(20 + j 10) Ω
ZTh
=
VTh = (100 − j 200) V
A
ZL
j10(5 − j10)
j10 + 5 − j10
j 50 − j 2 100
5
100 + j 50
=
5
= 20 + j10
B
Therefore, Thevenin’s equivalent circuit is shown in
Figure 5.112.
Now, according to the maximum power transfer theorem,
10 Ω
ZL = Complex conjugate of ZTh
A
= (20 − j10) Ω
4Ω
2Ω
Figure 5.112
Example 5.29 Determine the maximum power
delivered to the load in the circuit shown in
Figure 5.113.
Solution: Let us draw Thevenin’s equivalent circuit
of the given circuit. Firstly, we calculate the VTh after
removing the load impedance as shown in Figure 5.114.
From mesh I,
I1 = 2∠30° = 2 cos 38 + j 2 sin 30° = (1.732 + j1) A
Applying KVL in mesh II, we get the following:
(4 − j4) (i2 − i1) + 10 i2 + (2 + j8) i2 = 0
or
−(4 − j4) i1 + (16 + j4) i2 = 0
M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 220
ZL
2∠ 30°A
−j 4 Ω
j8Ω
B
Figure 5.113
10 Ω
A
4Ω
2Ω
VTh
2∠ 30°
i1
Mesh I
−j 4 Ω
i2
Mesh II
j8Ω
B
Figure 5.114
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Network Theorems and Applications
221
We substitute the value of i1 in the equation, we obtain the equation as follows:
− (4 − j4) (1.732 + j) + (16 + j4) i2 = 0
or
(16 + j4) i2 = (4 − j4) (1.732 + j)
( 4 − j 4) (1.732 + j )
or
i2 =
(16 + j 4)
(5.65∠ − 45°)( 2∠30°)
16.5∠14.036°
= 0.6848∠ − 29.036°
VTh = (2 + j8) i2
= (2 + j8) (0.6848∠−29.036 )
= (8.246∠75.96) (0.6848∠−29.036 )
= 5.6468∠46.924
=
To find ZTh, we open-circuit the current source as in Figure 5.115.
ZTh = (10 + 4 − j 4) ( 2 + j8)
10 Ω
= (14 − j 4) ( 2 + j8)
=
(14 − j 4)( 2 + j8)
14 − j 4 + 2 + j8
(14.56 ∠ − 15.94°)(8.24 ∠75.96°)
=
16 + j 4
119.97∠60.02°
=
16.49∠14.036°
= 7.27∠45.984°
= 5.05 + j 5.228
A
4Ω
2Ω
−j 4 Ω
j8Ω
B
Figure 5.115
Thevenin’s equivalent circuit of the given circuit can be drawn as shown in Figure 5.116.
According to the maximum power transfer theorem, ZL = complex conjugate of ZTh
= 5.05 − j5.228
VTh
IL =
ZTh + Z L
5.6468
=
(5.05 + j 5.228) + (5.05 − j 5.228)
5.6468
=
10.1
= 0.559 A
Maximum power transferred to load, PL = IL2 RL
= (0.559)2 × 5.05
= 1.5785 Watts
M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 221
5.05 + j 5.228
ZTh
+
VTh = 5.6468 ∠ 46.924°
IL
ZL
(5.05 − j 5.228)
Figure 5.116
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222 Network Analysis and Synthesis
Example 5.30 Calculate the current flowing through the load impedance of the circuit shown
in Figure 5.117 applying Thevenin’s theorem.
Solution: We open-circuit the load
terminals by removing the load impedance
as shown in Figure 5.118 and calculate the
VAB, that is, VTh.
We apply KVL in the loop and calculate
i as follows:
or
or
3Ω
XL =
−
2Ω
5∠90°
IL
4Ω
ZL
3Ω
B
Figure 5.117
j5
2 + j2
VTh = VAB = voltage drop across PQ + voltage rise across QP + voltage drop across
SA.
Note that no voltage will drop across the
3 Ω resistor as no current is flowing.
A
−
XC = 4 Ω
5∠90° − ( 2 + j 6)i − ( − j 4)i = 0
( 2 + j 6)i − j 4i = j 5
i ( 2 + j 2) = j 5
i=
+
+
6Ω
12∠0°
3Ω
R
2Ω
i
−
Q
12∠0°
+
+
j6Ω
S
−
A
5∠90°
−j 4 Ω
B
P
Figure 5.118
VTh = −( − j 4)i + 5∠90° − 12∠0°
=−
( − j 4) j 5
+ j 5 − 12
2 + j2
=−
20
+ j 5 − 12
2 + j2
=−
20 ∠0
+ j 5 − 12
2.828∠45°
= −7.07∠ − 45° + j 5 − 12
= −7.07 cos 45° + j 7.07 sin 45° + j 5 − 12
= −7.07 × 0.707 + j 7.07 × 0.707 + j 5 − 12
= −5 + j 5 + j 5 − 12
= −17 + j10
= 19.78∠150°
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223
Network Theorems and Applications
To calculate ZTh, we short-circuit the voltage
sources and calculate the equivalent impedance
from the circuit shown in Figure 5.119.
( − j 4)( 2 + j 6)
ZTh = 3 +
− j 4 + 2 + j6
24 − j8
= 3+
2 + j2
3Ω
A
j6
−j 4
2Ω
B
Figure 5.119
25.3∠ − 18ο
2.83∠45°
= 3 + 8.33∠ − 63°
= 3 + 8.33 cos 63° − j8.33 sin 63°
= 3 + 4.82 − j 7.33
= (7.82 − j77.33) Ω
= 3+
The Thevenin’s equivalent circuit is
shown in Figure 5.120.
VTh
IL =
ZTh + Z L
19.78∠150°
7.82 − j 7.33 + 4 + j 3
19.78∠150°
=
11.82 − j 4.333
19.78∠150°
=
12.29∠ − 20°
= 1.609∠150° + 20 = 1.609∠170°A.
A
IL
ZTh = 7.82 −j 7.33
VTh = 19.78∠150°
ZL
4
j3
=
B
Figure 5.120
R E V IE W Q U E S T I O N S
Short Answer Type
1. State and explain Thevenin’s theorem
2.With a simple example, show how by applying Thevenin’s theorem, current flowing through
a branch of an electrical network can be calculated.
3. Write the steps of application of Thevenin’s theorem.
4. State and explain Norton’s theorem.
5. Distinguish between Thevenin’s theorem and Norton’s theorem.
6. What is maximum power transfer theorem? Prove the theorem.
7. Explain reciprocity theorem with the help of a suitable example.
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224 Network Analysis and Synthesis
8.
9.
10.
11.
12.
Explain Tellegen’s theorem with an example.
State and explain Millman’s theorem.
Explain superposition theorem with an example.
Write the conversion formula for delta to star conversion of three resistors.
Write the relationship of star−delta transformation of the resistors.
Numerical Problems
1. Calculate the current flowing through the 5 Ω resistor as shown in Figure 5.121
3Ω
2Ω
12 V
4 Ω
1Ω
5Ω
[Ans. 0.663 A]
Figure 5.121
2. Calculate the current flowing through the 2 Ω resistor connected across terminals A and B
in the network shown in Figure 5.122 by applying the following:
(i) Kirchhoff’s laws
(ii) Thevenin’s theorem
(iii) Nodal voltage analysis
Compare the time taken in each case.
A
2Ω
2V
1Ω
3Ω
12 Ω
2Ω
4V
[Ans. I = 0.817 A; applying Kirchhoff’s
laws takes maximum time.]
B
Figure 5.122
3. Apply Norton’s theorem to calculate the current through the 5 Ω resistor in the circuit shown
in Figure 5.123. Further, verify by applying Thevenin’s theorem.
4.5 Ω
10Ω
15 Ω
24V
A
I
5Ω
B
Figure 5.123
M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 224
[Ans. I = 1 A]
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Network Theorems and Applications
225
4. Calculate the value of RL for which maximum power will be transferred from the source
to the load in the network shown in Figure 5.124. Further, calculate the value of maximum
power transferred.
4Ω
12 V
3Ω
RL
2Ω
[Ans. RL = 7.33 Ω; Pmax = 0.545 W]
Figure 5.124
5. By using superposition theorem, calculate the current flowing through the 10 Ω resistor in
the network shown in Figure 5.125.
10 A
2.5 Ω
5Ω
10 Ω
10 V
25Ω
40 V
[Ans. 0.054 A]
6. Apply Thevenin’s theorem to calculate the current flowing through the 30 Ω resistor connected across terminals A and B in the network shown in Figure 5.126.
Figure 5.125
A
15 Ω
150 V
B
60 Ω
13 A
30Ω
40 Ω
60 V
[Ans. IAB = 1.25 A]
Figure 5.126
7. For the circuit shown in Figure 5.127, determine the load impedence which will dissipate
maximum power. Also, claculate the maximum power.
6Ω
j8 Ω
(12 + j0)V
−j6 Ω
ZL
Load
Figure 5.127
[Ans. ZL = (5.4 + j7.8) Ω
Pmax = 6.0 W]
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Transient Response
of Circuits Using
Differential Equations
6
CHAPTER OBJECTIVES
After studying, this chapter, you should be able to do the following:
sudden application of a DC
Explain transient condition that may
voltage.
occur in a network.
Explain transient condition in a R–L Distinguish between steady state condition and transient condition with respect
series circuit.
to voltage and current in a R–L and R–C
Derive expressions for rise and decay
series circuit with DC excitation.
of current in a R–L series circuit under
transient condition.
Derive expressions for current and
voltage under transient condition in a
Explain the meaning of time-constant
R–L–C series circuit on sudden appliof R–L series circuit.
cation of a DC excitation.
Solve numerical problems of R–L
Solve numerical on transient R–L–C
series circuit under transient condition.
series circuits.
Explain transient response of R–C
Carryout sinusoidal response i.e.,
series circuit.
response with a sinusoidally varying
Derive expression for charging current
input in a R–L series circuit and R–C
of a capacitor and a resistor in series on
series circuit.
application of a DC voltage.
Analyse transient condition in a R–L–C
Make a complete analysis of transient
series circuit with sinusoidal input.
condition of R–C series circuit on
6.1 TRANSIENT CONDITION IN NETWORKS
Electrical networks contain resistors, inductors, and capacitors. Inductors and capacitors are
energy storing devices. Energy stored in these circuit elements cannot change instantaneously.
The period of adjustment during which the stored energy in these elements changes from their
initial level to the final level is called the settling time, which is generally a fraction of a second.
The time taken by a circuit to change from one steady state condition to another steady state
condition is called transient time.
A transient condition in networks occurs due to switching operations. During the transient
period, the current and voltages change from their initial values to the new values.
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Transient Response of Circuits Using Differential Equations 227
When a circuit is switched on, there exist two sources of energy in the circuit. One source is
the initial stored energy in inductances and capacitances at the time of switching. The second
energy source is the energy, that is, applied externally in the form of voltage or current sources.
The complete response of a circuit representing a system can be represented in two parts,
namely, its forced response or steady state response and transient response. The transient
response or solution shows the way the circuit responds when a forcing function (a voltage) as
input changes in energy state. Transient response depends upon the circuit parameters and their
initial charge condition.
Transient response of electrical circuits can be determined either by using differential equations or by using Laplace transform.
In this chapter, we will determine transient response of R–L, R–C, and R–L–C circuits using
differential equations.
6.2 TRANSIENT RESPONSE OF R–L SERIES CIRCUITS HAVING
DC EXCITATION
We will consider the transient response of R–L circuit, R–C circuit and R–L–C circuit one by
one due to the switching on DC voltage or sinusoidal voltage.
6.2.1 Rise of Current Through R–L Series Circuit
Let us consider that DC voltage V is switched on to the
series R–L circuit at t = 0, as shown in Figure 6.1.
Let i(t) be the current flowing through the circuit
after closing the switch.
Let us find an expression for current i(t) after closing switch k at t = 0.
Applying KVL to the series R–L circuit, we get the
following:
R
L
Switch
k
V
i(t )
Figure 6.1 Transient Response of
R–L Circuit
di
V = Ri + L
dt
Dividing both sides by L, we obtain the equation as follows:
di R V
+ i=
dt L
L
(6.1)
Equation (6.1) is a first-order linear equation of the following form:
di
+ Pi = Q
dt
whose solution is given as follows:
i(I.F) = ∫ Q(I.F)dt + C
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228
Network Analysis and Synthesis
where I.F is integrating factor, and it can be written in the following form:
Pdt
I.F = e ∫
Now, on considering equation (6.1), we obtain the equation as follows:
di  R 
V 
+ i =  
L
dt  L 
where P =
R
V
and Q =
L
L
R
R
t
Pdt
∫ dt
Therefore, I.F = e ∫
= e L = eL
Further, its solution will be written as follows:
i(I.F) = ∫ Q(I.F)dt + C
or
i
(e ) = ∫ VL ⋅ e
R
t
L
R
t
L ⋅ dt
+C
R
=
t
V
e L dt + C
∫
L
=
or
i
R
t
eL
V
⋅
+C
L R
L
(e ) = VR ⋅ e
or
R
t
L
i=
R
t
L
+C
V
C
+ R
R
t
eL
R
i=
− t
V
+ Ce L
R
(6.2)
Now, in order to find the value of C, we will apply the initial condition.
Since, the inductor does not allow sudden change of current through it, at t = 0, i = 0
t = 0
Applying the initial condition, at
 in equation (6.2), we get the equation as follows:
i = 0
V
+ Ce 0
R
V
0 = +C
R
V
C=−
R
0=
or
or
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Transient Response of Circuits Using Differential Equations 229
Now, substituting this value of C in equation (6.2), we can write the equation as in the following:
R
V V − Lt
− e
R R
R
− t
V
i = 1 − e L 
(6.3)
R

This gives an expression for the rise in current i through an inductor when DC voltage is applied
R
V
V − t
to it. This expression has two parts, the steady state part , and the transient part e L .
R
R
The graphical representation of changing current in R–L circuit is shown in Figure 6.2.
Steady-state current or iss is the final current that remains in the circuit when the transient
dies out. The time taken for the current to reach its steady-state value from its initial value is
called transient time.
i=
iss = Lt i(t )
and steady state current,
t →∞
V
1 − e
t →∞ R 

iss = Lt
or
−R 
t
L 

V
V
V
= 1 − e −∞  = (1 − 0) =
R
R
R
V.
Therefore, the steady-state current in R–L series circuit, iss =
R
The voltage across the resister, VR = Ri = R ×
(
VR = V 1 − e
or
(
R
− t
V
1− e L
R
R
− t
L
)
)
R
The voltage across the inductor, VL = L
di
V R − t
=L × e L
dt
R L
or
R
− t
L
VL = Ve
iss =
i
Current
O
V
, is the steady state value
R
i(t ) =
R t
−
V
L
1−e
R
Time(t )
Figure 6.2 Graphical Representation of Transient Current in R–L Circuit
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Network Analysis and Synthesis
6.2.2 Time Constant of R–L Series Circuit
The ratio of L and R in the R–L circuit is called time constant.
Let t is represented by t where
L
t =
R
L
Now, substituting t = t = = time constant in equation (6.3), we get the following:
R
R L
− ⋅ 
V
i = 1 − e L R 
R


=
or
V
1 − e −1 
R
i=
V  1
1−
R  e 
i=
V
[1 − .368]
R
= 0.632
V
V
= 63.2% of
R
R
= 63.2% of final current, that is, the steady-state current.
Therefore, the time constant of R–L series circuit may be defined as the time at which the current through the R–L series circuit rises to 63.2% of the steady-state value. The transient part of
the current reaches 36.8% of its initial value.
6.2.3 Decay of Current Through R–L Series Circuit
Let us consider that after closing the switch, the current in the circuit has reached the steadystate value. Now, suddenly the voltage is disconnected by opening switch K and connecting it to
V
K1 as shown in Figure 6.3. At t = 0, Current i has been equal to .
R
K
+
V
−
R
L
K1
i (t )
Figure 6.3 Decay of Current in an R–L Circuit
Now, let us find the expression for current i(t) as in the following.
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Transient Response of Circuits Using Differential Equations 231
By applying KVL, we obtain the equation as follows:
di
Ri + L = 0
dt
di R
+ i=0
or
dt L
By comparing this equation with the form,
di
+ Pi = Q
dt
R
P = , and Q = 0
We have
L
R
R
t
Pdt
∫ L dt
∫
L
Therefore, I.F = e
=e
= e and its solution will be given as follows:
i(I.F) = ∫ Q(I.F)dt + C
i
or
(e ) = O + C
R
− t
L
i=
or
C
R
− t
e L
i(t ) = Ce
R
− t
L
(6.4)
Now applying the initial condition
that is,
at t = 0, i =
Substituting the value in equation (6.4), we get
V
R
V
= Ce 0
R
V
. By substituting this value in equation (6.4), we can obtain the following form:
R
R
V − t
i (t ) = e L
(6.5)
R
This is the expression for decay in current through the R–L series circuit when it is suddenly
disconnected from supply. This shows that the current decays exponentially.
L
Now, when t = t = ( time constant), then
R
R L
V − ⋅
i (t ) = e L R
R
V −1
= e
R
V
= (.368)
R
V
i(t ) = 36.8% = 36.8% of initial current.
R
or C =
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Network Analysis and Synthesis
V
R
The time constant of R–L series circuit may also be
defined as the time at which the current in the R–L series
circuit decays to 36.8% of the initial value.
The graphical representation of decay of current
through R–L series circuit shown in Figure 6.4.
Steady-state current after the decay of current in R–L
series circuit is calculated as follows:
Initial value
i
Current
Time(t )
iss = Lt i(t )
Figure 6.4 Decay of Current in an
R–L Series Circuit
t →∞
R
V − Lt
e
t →∞ R
= Lt
V −∞
e =0
R
This shows that under steady state condition, the current in the circuit becomes zero.
=
Example 6.1 The values of R and L in a series R–L circuit are 10 Ω and 40 H, respectively. At
the instant of closing the switch, the current rises at the rate of 5 A/s. Calculate the following:
1. the value of applied voltage
2. rate of growth of current when 6A flows in the circuit
3. find the energy stored in the inductor.
Solution: The
Figure 6.5.
Here,
given
circuit
is
shown
in
R = 10 Ω
L = 40 H
K
R = 10 Ω
L = 40 H
V
i (t )
Further, at the instant of closing the switch, that is,
at t = 0, the following changes take place.
Current rises at the rate of 5A/s, that is,
di
Figure 6.5
= 5A/s
dt
Further, we know that in the case of R–L series circuit, the value of current at t = 0, i = 0.
Now, by applying KVL in the circuit, we get the following form:
di
V = Ri + L
(6.6)
dt
Now at t = 0, i = 0

. By substituting these values in equation (6.6), we can calculate V as follows:
di
and
= 5 (given) 
dt

V = 0 + 40 (5)
V = 200 V ⇒ applied voltage
 di 
The rate of growth of current   when i = 6A flows in the circuit.
 dt 
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Transient Response of Circuits Using Differential Equations 233
By applying KVL in the circuit, we can get the following form as follows:
di
dt
By substituting V = 200 V, R = 10 Ω, L = 40 H and i = 6 A in the equation, we get the following
form:
di
200 = 10(6) + 40
dt
di
200 = 60 + 40
dt
di
or
40 = 140
dt
di 140
or
=
= 3.5 A/s
dt
40
di
= 3.5 A/s. This value is taken as the required rate of growth of the current.
dt
1
Energy stored in the inductor is the final energy stored in the induction and is given by LI 2
2
where,
V = Ri + L
V 200
=
= 20 A
R 10
1
Energy stored = ( 40) ( 20) 2
2
= 20( 20) 2 = 8000 joules
I
(Final value)
=
Example 6.2 A DC voltage of 20 V is applied in an R–L circuit where R = 5 Ω and L = 10 H.
Calculate (a) the current i; (b) voltage across resistor and voltage across the inductor; (c) the
time constant and (d) the maximum value of stored energy.
Solution: The given circuit is shown in Figure 6.6.
(
R
(
)
5
)
− t
− t
V
20
−0.5 t
)A
i = 1− e L =
1 − e 10 = 4 (1 − e
R
5
VR = iR = 20 (1 − e
VL = L
0.5 t
)V
d
di
= 10  4 (1 − e −0.5 t ) = 20e −0.5t V
dt
dt
Time constant =
L 10
=
= 2s.
R 5
M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 233
R = 5Ω
L = 10 H
K
V = 20 V
Figure 6.6
12/3/2014 8:01:32 PM
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Network Analysis and Synthesis
iss = I =
Maximum stored energy =
V 20
=
= 4Α
R 5
1 2 1
LI = 10 × 4 2 = 80 joules.
2
2
Example 6.3 A 100 V DC source is applied across the coil of R = 10 Ω and L = 10 H. Find
the energy supplied to the coil in the first 5 s. If the coil is disconnected and immediately shortcircuited, then find the energy dissipated.
Solution: The given circuit is shown in Figure 6.7.
V = 100 V
R
− 
V
Lt
i = 1 − e 
R

− t
100 
1 − e 10 
10 

10
=
= 10 (1 − e
−t
K
R = 10 Ω
L = 10 H
100V = V
)
i (t )
The energy supplied in first 5 s:
Figure 6.7
5
= ∫ V i dt
0
5
= ∫ 100 × 10(1 − e − t ) dt
0
5
= 1000∫ (1 − e − t ) dt
0
5
 e −t 
= 1000 t −

−1 

0
= 1000 (5 − 0) + (e −5 − e 0 ) 
= 1000 [5 + e −5 − 1] joules = 4006.8 jooules
When the coil is disconnected and immediately short-circuited, then i = 10 (1 − e − t )
Now, the coil is short-circuited at t = 5 s.
i = 10 (1 − e −5 )
= 10(1 − 0.0068)
= 9.93 A
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Transient Response of Circuits Using Differential Equations 235
1 2
Li
2
1
= (10)i 2
2
1
= × 10 × (9.93) 2 = 493 joules.
2
Energy dissipated =
Examples 6.4 A coil has resistance of 1 Ω and an inductance of 1 H. It is suddenly connected
to 6 V DC voltage source. Calculate the following:
1. Initial and final values of current
2. Time constant
3. Rate of change of current at t = 0 and t = ∞
4. Voltage across inductance at t = 0 and t = ∞
5. Voltage across resistance at t = 0 and t = ∞ and
6. Current at t = 1 s
Solution: Given R = 1 Ω, L = 1 H and V = 6 V
The circuit is shown in Figure 6.8.
Initial and final values of the current: We know
current flowing through R–L series circuit at a
time, when it is connected to a DC source of supply is given by
i (t ) =
1Ω
6V
1H
At
t=0
switch
is closed
R
1

− t
− t
V
1 − e L  = 6 1 − e 1 
R
1 

Figure 6.8
= 6 (1 − e − t ) A
Now, the initial current is calculated as follows:
Lt i(t ) = Lt 6(1 − e − t )
t →0
t →0
= 6(1 − e 0 )
= 6(1 − 1) = 0 A
Further, the final current is calculated as in the following equation:
= Lt i(t )
t →∞
= Lt 6(1 − e − t )
t →∞
= 6(1 − e −∞ ) = 6(1 − 0) = 6 A
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Network Analysis and Synthesis
= Lt i(t )
t →∞
= Lt 6(1 − e − t )
t →∞
= 6(1 − e −∞ ) = 6(1 − 0) = 6 A
Time constant =
L 1
= = 1s
R 1
di
dt
We have the following forms:
Rate of change of current =
i = 6 (1 − e − t )
di
= 6 0 − e − t ( − 1) 
dt
= 6e − t A/s
di
= 6e 0 = 6 A/s
dt t = 0
and
di
= 6e −∞ = 6(0) = 0
dt t =∞
The voltage across the inductor can be written as follows:
VL = L
di
d
= 1⋅ 6 (1 − e − t )
dt
dt
= 1.6e − t = 6e − t V
Now, the voltage is obtained as in the following:
VL
VL
t =0
t =∞
= 6e 0 = 6V
= 6e − ∞ = 0
Further, the voltage across the resistor can be calculated as follows:
VR = Ri = 1× 0.6(1 − e − t ) = 6(1 − e − t )V
Therefore,
Current at
VR
t =0
= 6(1 − e 0 ) = 6(1 − 1) = 0
VR
t =∞
= 6(1 − e −∞ ) = 6(1 − 0) = 6 V
t = 1 s, i = 6(1 − e − t ) = 6(1 − 0.368)
or
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= 6(1 − 0.368)
i = 3.792 A
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Transient Response of Circuits Using Differential Equations 237
Example 6.5 A coil having a resistance of 100 Ω and an inductance of 20 H is connected
to a 200 V DC source. Suddenly, the coil is disconnected from the battery and short-circuited.
Calculate the following:
1.
2.
3.
4.
The current in the coil at t = 0
Rate of change of current at t = 0
Time constant and
Time taken for the current to decrease to 0.25A
Solution: When R–L series circuit is disconnected from the battery, then the decay of current
through the series circuit is given by the following:
R
i=
− t
V
−e L
R
100
=
200 − 20 t
−e
100
= 2e −5t A
Given
R = 100 Ω; L = 20 H and V = 200 V
The current in the coil at t = 0 will be calculated as follows:
i t = 0 = 2e 0 = 2 A
The rate of change of current, that is,
di
is calculated as in the following:
dt
di d ( −5t )
=
2e
= 2( − 5)e −5t
dt dt
= −10e −5t A/s
Therefore,
di
dt
= −10e 0 = −10 A/s.
t =0
Time constant can be calculated as follows:
L
R
20
=
100
= 0.2 s
t =
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Network Analysis and Synthesis
L
R
20
=
100
= 0.2 s
t =
The time taken for the current to decrease to 0.25A is obtained in the following equations:
We have i = 2e −5 t (from equation (6.6))
Substitute i = 0.25, we get the following:
0.25 = 2e −5t
0.25
e −5t =
2
e −5t = 0.125
or
−5t = loge 0.125
or
1
t = − loge 0.125
5
= 0.416 s.
or
Example 6.6 The field winding of a DC motor has 80 Ω resistance and 12 H inductance. It
is connected to a 240 V DC source. At t = 0, the supply is disconnected and a field discharge
resistance of 80 Ω is connected across the field winding. Find the rate of change of current in
the winding at t = 0.
Solution: Given that at t = 0, supply is disconnected.
Before disconnection of the supply, the maximum current flowing through the circuit
would be
I=
V 240
=
= 3A
R 80
Now, at t = 0, the supply is disconnected by changing the position of switch from position S to
position S′ as shown in Figure 6.9.
The voltage equation will be given as follows:
S
S′
Ri + L
80 Ω
80 Ω
240 V
12 H
S
(Field
discharge
resistor)
S′
Figure 6.9
M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 238
di
=0
dt
Substituting the values in the equations, we will
have the following form:
160i + 12
di
=0
dt
Now, at t = 0 and i = 3 A
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Transient Response of Circuits Using Differential Equations 239
Therefore, the above equation can be rewritten as 160(3) + 12
or
480 + 12
or
12
di
=0
dt
di
=0
dt
di
= −480
dt
di
480
=−
= −40Α/s
dt
12
or
6.3 TRANSIENT RESPONSE IN R–C SERIES CIRCUITS
HAVING DC EXCITATION
Consider an R–C series circuit, as shown in Figure 6.10.
Let us consider two cases, that is, when the capacitor is getting charged and when the capacitor is getting discharged.
6.3.1 Case I: Capacitor is Getting Charged
When switch K is closed, the supply voltage V gets connected to the R–C series circuit at
t = 0. Let us assume that initially the capacitor is fully discharged. Therefore, at t = 0, there is no
charge stored in the capacitor. The capacitor can be assumed as short circuit initially, and hence
V
the current flowing in the series R–C circuit is .
R
Thus, in a series R–C circuit, when initially the circuit is switched on to DC voltage source,
V
the current flowing in the circuit is .
R
V
That is, at t = 0, i =
R
Now, by applying KVL in the circuit of Figure 6.10,
we get the following form:
K
1
V = Ri + ∫ idt
C
Differentiating both sides with respect to time, we
obtain the equation as follows:
0=R
di 1
+ i
dt C
M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 239
R
+
V
−
C
Figure 6.10 Transient Response of
R–C Series Circuit
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Network Analysis and Synthesis
1
di
+
i=0
dt RC
or
(6.7)
Comparing it with the Leibnitz equation, we can write the equation as in the following:
di
+ Pi = Q
dt
We get the equation as follows:
1
and Q = 0
RC
P=
Therefore, integrating factor (I.F) = e ∫
1
Pdt
=e
1
∫ RC dt
= e RC
t
Now, the solution of equation (6.7) will be given as follows:
i(I.F) = ∫ Q(I.F)dt + K1
( )
t
or
i e RC = 0 + K1 ,
where K1 is the constant of integration
or
i=
K1
t
e RC
i = K1e
−
t
RC
(6.8)
Now, in order to find the value of K1, apply the initial condition in the equation:
t=0
V

That is, substituting at V  in equation (6.8), we get = K1e 0
R
i= 
R
or
K1 =
V
R
Now, substituting the value of K1 in equation (6.8), we get the following form:
t
V −
i = e RC
R
(6.9)
This is the expression for the current flowing in the R–C circuit at any time t, after the switch
is closed at t = 0.
The graphical representation of response of the circuit flow in the R–C circuit is shown in
Figure 6.11.
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Transient Response of Circuits Using Differential Equations 241
Time Constant of R–C Series Circuit
t = RC is the time constant of R–C circuit
Now, when t = t = RC , then we substitute t in Current
equation (6.9) and we get the following:
(i )
V
=i
R
RC
V −
i = e RC
R
V −1
= e
R
t=0
Time
t
Figure 6.11 Charging Current in an R–C
Series Circuit
V 1
⋅
R e
V
= .368 = 36.8% of initial current
R
=
The time constant of R–C series circuit is that time at which the current flowing in the circuit
reaches 36.8% of initial value.
Voltage Across the Capacitor
1
idt
C∫
By substituting the value of i from equation (6.9), we can write the equation as follows:
VC =
t
VC =
1 V − RC
e
dt
C∫R
t
−
V
e RC dt
=
∫
RC
−
t
V e RC
=
⋅
+ K2
RC −1
RC
where K2 is the constant of integration.
VC = −Ve
Now, at
−
t
RC
+ K2
(6.10)
t = 0
 we assumed the capacitor to be fully discharged initially.
VC = 0
Therefore, we get the following from equation (6.10):
0 = −Ve 0 + K 2
or
K2 = V
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Network Analysis and Synthesis
By substituting the value of K2 in equation (6.10), we can write the equation as follows:
VC = −Ve
−
t
RC
+V
t 

−
VC = V 1 − e RC 


(6.11)
This is the expression for the capacitor voltage.
Equation (6.11) is indicating that the capacitor gets charged exponentially when a DC voltage is applied across it.
The graphical representation of charging of
the
capacitor when a DC voltage is switched
V
on is shown in Figure 6.12.
VC
It is concluded that when a series R–C
(Capacitor
t
−
circuit
is supplied with a DC voltage, then
voltage)
VC = V 1 − e RC
capacitor starts charging and capacitor current starts decaying as depicted graphically in
Figure 6.13. From the graphs in Figure 6.13,
Time (t )
it is observed that initially VC = 0 (minimum)
Figure 6.12 Gradual Building Up of Voltage
V
Across the Capacitor
and i = (maximum).
R
However, with the passage of time, i decays
V
exponentially and VC rises exponentially.
Finally, when capacitor charges to maxiV
mum value V, current i through the circuit
R
Capacitor voltage becomes zero.
The instantaneous values of current and
Capacitor current
capacitor voltage are as follows:
o
t
V − RC
e
R
t 

−
VC = V 1 − e RC 


i=
Time (t )
Figure 6.13 Graphical Representation
of Capacitor Voltage and
Capacitor Current
Steady-State or Final Values
The steady-state value of current is given as follows:
iss = Lt i(t )
t →∞
t
V − RC
e
t →∞ R
V
= e−∞
R
=0
= Lt
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iss = Lt i(t )
t →∞
Transient Response of Circuits
Using Differential Equations 243
t
V − RC
e
t →∞ R
V
= e−∞
R
=0
= Lt
The steady-state value of capacitor voltage can be written as in the following:
Vss = Lt VC (t )
t →∞
t 

RC

= Lt V 1 − e
t →∞ 


= V [1 − e − ∞ ]
= V (1 − 0)
=V
6.3.2 Case II: Discharging of Capacitor
Now, let us consider a discharging case, that is, when the fully charged capacitor is disconnected
from DC supply. Let the switch is moved to position K1, as shown in Figure 6.14.
Now, the direction of current is opposite to that
K
of the previous case.
Let us apply KVL, we get the following:
1
Ri + ∫ idt = 0
C
Differentiating the equation with respect to time,
we can form the equation as follows:
V
R
K1
i
+
C
−
Figure 6.14 Discharging of the
Capacitor when the
Switch is Moved from
Position K to K1
di 1
R + i=0
dt C
di
1
+
i=0
dt RC
or
(6.12)
By comparing equation (6.12) with the Leibnitz equation, we write it in the following
form:
di
+ Pi = Q .
dt
Therefore, we get
P=
1
,Q=0
RC
1
Further,
1
t
∫ dt
Pdt
I.F = e ∫
= e RC = e RC
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Network Analysis and Synthesis
and solution of equation (6.12) will be
i(I.F) = ∫ Q(I.F) dt + K 3
where K3 is the constant of integration
t
i ⋅ e RC = 0 + K 3
or
−
t
or
i = K3e RC .
Now, in discharging case,
at t = 0,
−V
i=
(negative sign is due to the opposite direction of current)
R
By applying the initial condition in equation (6.13), we get the following:
(6.13)
V
= K3e 0
R
V
K3 = −
R
Substituting this value in equation (6.13), we get
the following form:
t
V −
i = − e RC
(6.14)
R
t
This is the expression for transient discharging
current.
The graphical representation of discharging current has been shown in Figure 6.15. The direction
of discharging current has been shown as negative.
−
or
i
O
i= −
−V
t
V −RC
e
R
R
Voltage Across Capacitor During
Discharging
The transient voltage across the capacitor during
discharging is given as follows:
Figure 6.15 Discharging Transient
Current in R–C Circuit
VC =
1
idt
C∫
t
=
1
V − .dt
− e RC
∫
C
R
−
t
V e RC
=−
⋅
+ K4
1
RC
−
RC
where K4 is the constant of integration.
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Transient Response of Circuits Using Differential Equations 245
VC = Ve
Therefore,
−
t
RC
+ K4
(6.15)
Now, applying the initial condition, that is, at t = 0 and VC = V (maximum) in equation (6.15),
we have the following.
V = Ve 0 + K 4
K 4 = 0.
By substituting the value of K4 in equation (6.15), we get the form as follows:
VC = Ve
−
t
RC
The equation indicates that the capacitor voltage
decays exponentially and the graphical representation of capacitor voltage during discharging has
been shown in Figure 6.16.
V
−
VC (t ) = Ve
Example 6.7 A circuit has resistance of 1000 Ω
and a series capacitance of 0.1 µF. At t = 0, it is
connected to a 12 V battery. Find the following:
1.
2.
3.
4.
5.
t
RC
VC
o
The current at t = 0
Rate of change of current at t = 0
Rate of change of capacitor voltage at t = 0
Current at t = 0.1 ms
Voltage across capacitor at 0.1 ms
Time
Figure 6.16 Capacitor Voltage During
the Period of Discharging
1000 Ω
t =0
Solution: Given
0.1 × 10−6 F
12 V
R = 1000 Ω
C = 0.1 µF = 0.1 × 10−6 F
V = 12 V
Figure 6.17
The circuit is shown in Figure 6.17.
Current at t = 0.
V
12
it = 0 = =
= 12 × 10 −3 = 12 mΑ. The rate of change of current at t = 0, we have the
R 1000
following:
t
i=
V − RC
e
R
t
Therefore,
t
di V  1  − RC
V −
e
= −
= − 2 e RC

dt R  RC 
R C
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Network Analysis and Synthesis
or
V
V
12
 di 
= − 2 e0 = − 2 = −
 
2
dt t = 0
R C
R C
(1000) × 0.1 × 10 −6
=−
12
= −120 Α/s
0.1
The rate of change of capacitor voltage at t = 0 can be given as follows:
t 

−
VC = V 1 − e RC 


Therefore,
  1  − t 
dVC
RC 
= V 0 −  −
e
dt
  RC 

t
=
V − RC
e
RC
and
12
V 0
V
 dVC 
=
=
e =


dt  t = 0 RC
RC 1000 × 0.1 × 10 −6
12
12
=
= −4
−3
0.1 × 10
10
= 12 × 10 4
= 120000 V/s
The current at t = 0.1 × 10 −3 s.
Therefore, we have the following form:
t
i (t ) =
V − RC
e
R
0.1×10 −3
(i )t = 0.1×10 −3
12 −1000 × 0.1×10 −6
=
e
1000
−10 −4
12 10 −4
=
e
1000
12 −1
12
× 0.368
=
e =
1000
1000
= 0.012 × 368
= 4.4 × 10 −3 A
The voltage across the capacitor at t = 0.1 × 10 − 3 s
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Transient Response of Circuits Using Differential Equations 247
Therefore, we have
t 

−
VC = V 1 − e RC 


0.1 × 10


−
1000 × 0.1 × 10 −6 

= 12 1 − e




−3
VC
t = 0.1 × 10 −3
= 12[1 − e −1 ]
= 12[1 − 0.368]
= 12[0.632]
= 7.584 V
Example 6.8 A 5 µF condenser is connected through a 1000 kΩ resistance to a DC source
of 10 V. After being charged for half a minute, the condenser is disconnected and discharged
through a resistor R. Determine the energy dissipated in R.
Solution: Energy dissipated in R = energy stored in the capacitor during the charging time.
=
1
CV 2
2 C
Voltage build up across the capacitor after half minute is calculated as follows:
t 

−
VC = V 1 − e RC 


Now, at t = 30 s, the voltage can be given as in the following:
30


− 6
−6
VC = 10 1 − e 10 × 5×10 


30

− 
= 10 1 − e 5 


= 10 1 − e −5 
= 10(0.99326)
= 9.3 V
Energy dissipated in the resistor =
1
1
CV 2 = (5 × 10 −6 )(9.3) 2
2
2
= 246.64 × 10 −6 joule
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Network Analysis and Synthesis
Example 6.9 The constant voltage of 100 V is applied at t = 0 to a series R–C circuit
having R = 5 MΩ, C = 20 µF . By assuming no initial charge to the capacitor, find current i and
the voltage across R and C.
Solution:
V = 100 V
R = 5 MΩ = 5 × 106 Ω
C = 20 µF = 20 × 10 −6 F
Now, current can be expressed as follows:
t
t
i=
=
Voltage across R
100 − 5×106 × 20 ×10 −6
V − RC
e
=
e
R
5 × 106
20
106
e
−
t
100
= 20 × 10 −6 e
= R×i
−
t
100 Α
= 5 × 106 × 20 × 10 −6 e
= 100e
Voltage across capacitor
−
−
t
100
t
100 V
= V – Voltage across resistor
= 100 − 100e
−
t
100
t 

−
= 100 1 − e 100  V


Example 6.10 A capacitor of 12 pF is to be charged and is designed such that the time
taken by the capacitor voltage to reach 90% of its final value should not exceed 3 ns. Find the
maximum value of resistance that may be used.
Solution: Given
C = 12 pF = 12 × 10 −12 F
t = 3 ns = 3 × 10 −9 s
VC = 90% of V = 0.9 V
We have
R=?
t 

−
VC = V 1 − e RC 


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Transient Response of Circuits Using Differential Equations 249
Substituting all the given values in the equation, we get,
0.9V = V 1 − e −3 ×10
or
0.9
−9
/R ×12 ×10 −12

−3 × 10 −9
−12
R
×
= 1 − e 12 × 10
−1
or
−3
e R× 4×10 = 1 − 0.9
or
e
or
−
−
250
R
= 0.1
250
= loge 0.1
R
R=−
250
= 108.7 Ω
loge 0.1
6.4 TRANSIENT RESPONSE OF R–L–C SERIES CIRCUITS
HAVING DC EXCITATION
Consider an R–L–C series circuit as shown Figure
6.18. Let us assume that the capacitor and inductor
are initially uncharged, that is, at t = 0, there is no
charge on L or C.
Let us find the expression for current in the circuit when switch S is closed at t = 0. By applying
di
1
KVL, we get V = Ri + L + ∫ idt
dt
C
Differentiating the above equation with respect to
t, we get the following:
di
d 2i 1
0=R +L 2 + i
dt
C
dt
Rearranging the equation, we can rewrite it as follows:
R
K
L
V
i
C
Figure 6.18 Transient Response of
R–L–C Series Cicruit
d 2i R di
1
+
+
i=0
2
L
dt
LC
dt
(6.16)
The above is a second order linear differential equation with only complementary function.
Substituting,
d
d2
for D and
for D 2
2
dt
dt
We get the symbolic form of equation (6.16) as
1 
 2 R
 D + D +
i=0
L
LC 
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(6.17)
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Network Analysis and Synthesis
Now, the characteristics equation of (6.17) is given by the following:
R
1
D2 + D +
=0
L
LC
The roots of characteristic equation are as follows:
2
−
D=
R
 1 
 R
±   − 4(1) 
 LC 
 L
L
2(1)
2
−
=
4
R
 R
±   −


L
L
LC
2
2
4
 R
  −
−R
L
LC
=
+
4
2L
2
=−
R
1
 R
±   −
 2L 
LC
2L
Let us consider the following two cases.
2
1
 R
, then roots are real and unequal and are given as follows:
Case I: when   >
 2L 
LC
2
a=−
1
R
 R
+   −
 2L 
2L
LC
b=−
R
1
 R
−   −
 2L 
2L
LC
2
Then solution will be written as in the following:
i = C1e at + C2e bt
i
Over-damped
response
o
t
Figure 6.19 Overdamped Response
of Current in R–L–C Series
Circuit
M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 250
(6.18)
In this case, the overdamped response is obtained.
The overdamped response is shown in Figure 6.19.
2
1 ,
 R
then roots are comCase II: when   ≤
 2L 
LC
plex conjugate and provide the underdamped
response as shown in Figure 6.20.
Now,
2
D=−
R
1
 R 
±   −
 2L 
2L
LC
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Transient Response of Circuits Using Differential Equations 251
2
1
 R
, then
If   ≤
 2L 
LC
i
Underdamped
response
2
1
 R
will be a negative quantity
  −
2L
LC
o
t
=−
=−
 1   R  2 
R
± − 
 −   

2L
2 L 
 LC
R
 1   R
±j 
−
 LC   2 L 
2L
Figure 6.20 Underdamped
Response of Current in
R–L–C Series Circuit
2
= a ± jb
i
Then, solution is given by the following:
i = ea t (C1 cos b t + C2 sin b t )
2
1 ,
 R
Case III: when   =
then roots are equal and
 2L 
LC
they provide a critical damped response as shown in
Figure 6.21.
D=−
t
Figure 6.21 Critical Damped
Response of R–L–C
Series Circuit
R
R
,−
2L
2L
Therefore, the response is given by the following:
i = C1e
−
R
t
2L
+ C2 e
−
R
t
2L
Example 6.11 For the circuit shown in Figure 6.22, find the transient current when switch S
is closed.
Solution:
Applying KVL in the circuit, we write the following form:
V = Ri + L
di 1
+
idt
dt C ∫
M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 251
S
5Ω
1H
1
24 V
2
F
Figure 6.22
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Network Analysis and Synthesis
Substituting values we get the following form:
24 = 5i +
24 = 5i +
or
di
1
+ 1 ∫ idt
1
dt
2
di
+ 2∫ idt
dt
Differentiating both sides with respect to t, we get the form as follows:
0=
or
5di d 2i
+
+ 2i
dt dt 2
d 2i 5di
+
+ 2i = 0
dt 2 dt
Symbolic form of the equation is given as in the following:
( D 2 + 5 D + 2)i = 0
or
D=
=
−5 ± 25 − 4(1)( 2) −5 ± 25 − 8
=
2(1)
2
−5 ± 17
2
−5 ± 4 −5 + 4 −5 − 4
,
≈
2
2
2
≈ −0.5, −4.5
D≈
Therefore, the solution will be given as follows:
i = C1e −0.5t + C2e −4.5t
Now, applying the initial condition, that is, at
By applying the condition,
Voltage across inductor
that is,
or
at
(6.19)
t = 0
 we get the form as follows:
i = 0
0 = C1 + C2
t=0
= 24 V
di
= 24
dt
di
1. = 24
dt
(6.20)
L
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{∴ L = 1H}
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Transient Response of Circuits Using Differential Equations 253
Now, differentiating equation (6.19), we can write the equation as in the following:
di
= −0.5C1e −0.5t − 4.5C2e −4.5t
dt
Now applying the condition, the following can be obtained:
At t = 0

 we get
di
= 24 
dt

By substituting the values in the equation, we get the form as follows:
24 = −0.5C1 , −4.5C2
(6.21)
By solving equation (6.20) and (6.21), we can get the value as in the following:
C1 + C2 = 0
(6.22)
−0.5C1 − 4.5C2 = 24
(6.23)
From equation (6.22), C2 = −C1; by substituting this value in equation (6.23), we get the value
as follows:
−0.5C1 + 4.5C1 = 24
or
4C1 = 24
or
C1 = 6
From equation (6.20), the value of C2 is calculated as follows:
C 2 = −C1
= −6
Substituting the value of C1 and C2 in equation (6.19), we get
i = 6e −0.5t − 6e −4.5t
or
i = 6 e −0.5t − e −4.5t  A
6.5 SINUSOIDAL RESPONSE OF R–L CIRCUITS
We will now study the transient response of circuits of R, L, and C when a sinusoidally varying
voltage is applied across such circuits.
Let us consider a series R–L circuit shown in Figure 6.23.
When the switch S is closed, a sinusoidal voltage Vm sin(w t + q ) is applied to series R–L
circuit. Here, q is the phase angle and Vm is the amplitude of line voltage wave.
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Network Analysis and Synthesis
By applying KVL, we get the form as follows:
K
Vm sin(w t + q ) = Ri + L
R
i
L
V = Vm sin(w t + q )
di
dt
V
di R
+ i = m sin(w t + q )
dt L
L
or
(6.24)
Its symbolic term is given as follows:
Figure 6.23 Transient Response of
R–L Circuit to Sinusoidal
Input Voltage
Vm
R

sin(w t + q )
 D +  i =
L
L
d.
dt
The complementary function of equation (6.24) can be obtained from characteristic equation. The characteristic equation of equation (6.24) is given as in the following:
Here, D =
D+
or
R
=0
L
D=−
R
L
Therefore, the complementary function of the solution i is calculated as follows:
iC = K1e
R
− t
L
(6.25)
Particular solution: let us obtain particular solution by using the method of undetermined
co-efficient.
Let
iP = A cos (w t + q ) + B sin(w t + q )
(6.26)
diP
= − Aw sin(w t + q ) + Bw cos(w t + q )
dt
(6.27)
Now
Substituting the value of equation (6.26) and equation (6.27) in equation (6.24), we get the
following form:
R
[ − Aw sin(w t + q ) + Bw cos(w t + q )] + L [ A cos(w t + q ) + B sin(w t + q )] =
Vm
sin(w t + q )
L
V
R 
R 


or  − Aw + B sin(w t + q ) +  Bw + A cos(w t + q ) = m sin(w t + q )


L 
L 
L
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Transient Response of Circuits Using Differential Equations 255
Equating the coefficients of sin(w t + q ) and cos(w t + q ) on both sides, we get the equation as
follows:
V
R
− Aw + B = m
(6.28)
L
L
R
A= 0
L
(6.29)
V 
V
1 R
R
B− m =
B− m

wL
wL
L  wL
(6.30)
Bw +
From equation (6.25), A is written as follows:
A=
Substituting this value of A in equation (6.29), we get the following:
Bw +
V 
R  R
⋅
B− m =0
L  wL
wL 
Bw +
or
R2
2
wL
B−
Vm ⋅ R
wL2
=0
or

R2  V ⋅ R
B w + 2  = m 2
wL  wL

or
w 2 L2 + R 2  V ⋅ R
= m
B
2
wL
wL2
B=
or
Vm ⋅ R
2
R + w 2 L2
(6.31)
Substituting the value of B from equation (6.31) in equation (6.30), we get the following
form:
A=

V R
V
V 
R
R2
⋅ 2 m 2 2− m = m 2
− 1
2
2
w L R + w L w L wL  R + w L

=
Vm  R 2 − R 2 − w 2 L2 


wL  R 2 + w 2 L2 
=
Vm  −w 2 L2 


wL  R 2 + w 2 L2 
A=
−wLVm
(6.32)
R 2 + w 2 L2
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Network Analysis and Synthesis
Substituting the value of A and B from equation (6.31) and (6.32) in equation (6.26), we obtain
the following:
iP =
=
=
=
−w LVm
2
2 2
R +w L
Vm ⋅ R
cos(w t + q ) +
Vm
2
R + w 2 L2
2
R + w 2 L2
sin(w t + q )
× [ −w L cos(w t + q ) + R sin(w t + q ) ]


−wL
R
×
cos(w t + q ) +
sin(w t + q ) 

R 2 + w 2 L2  R 2 + w 2 L2
R 2 + w 2 L2


Vm
R
wL
×
sin(w t + q ) − cos(w t + q ) ⋅

R 2 + w 2 L2  R 2 + w 2 L2
R 2 + w 2 L2 
Vm
Substituting
R 2 + w 2L 2
R
2
R + w 2 L2
(6.33)
= cos f
and
wL
wL
2
R + w 2 L2
j
= sin f , we get
tan f =
R
wL
R
f = tan −1
or
wL
R
Substituting the above values in equation (6.33), we get the equation as follows:
Vm
iP =
2
R + w 2 L2
Vm
=
R 2 + w 2 L2
× [sin(w t + q ) ⋅ cos f − cos(w t + q ) ⋅ sin f ]
× [sinn(w t + q − f )]
 
−1 w L  
sin  w t + q − tan R  

R +w L 
Vm
iP =
2
2 2
(6.34)
Therefore, the complete solution is as follows:
i = iC + iP
R
t
i = K1e L +
wL 

sin  w t + q − tan −


R
R +w L
Vm
2
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(6.35)
2 2
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Transient Response of Circuits Using Differential Equations 257
Using equation (6.25) and equation (6.34) and applying the initial condition,
that is, at
t = 0
 (6.32), we get
i = 0
By substituting the values in the equation, we get the form as follows:
wL 

sin q + 0 tan −1
R 

R +w L
Vm
wL 

sin q − tan −1
K1 = −

2
2 2

R
R +w L
0 = K1e 0 +
or
Vm
2
2 2
Substituting this value in equation (6.34), we get the following equation:
R
−Vm
Vm

w L  − Lt
w L


⋅e
+
i=
sin q − tan −1
sin  w t + q − tan −1


 (6.36)
2
2 2


R 
R
R 2 + w 2 L2
 R +w L
This is the final expression for current.
Example 6.12 A coil a resistance of
50 Ω and an inductance of 0.2 H. Find the
expression for the current if the coil is
excited by a voltage 150 sin 500 t.
K
50 Ω
i
Solution:
With the following given values, the circuit is drawn as shown in Figure 6.24.
0.2 H
V =150 sin 500t
Figure 6.24
V m = 150 V,
w = 500 rad/s,
R = 50 Ω,
L = 0.2 H, is drawn as in Figure 6.24.
Now, we have the following form:
i=
−Vm
R
Vm
w L
w L  − Lt


sin q − tan −1
+
sin  w t + q − tan −1
 ⋅ e

2
2 2
2
2 2


R 
R
R +w L
R +w L
Vm = 150 

w = 500

q = 0  equation, we get
R = 50 

L = 0.2 
M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 257
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Network Analysis and Synthesis
Substituting the values in the equation, we write the equation as in the following:
500(.2)  − .2t

sin  0 − tan −1
e

50 
50 2 + (500) 2 (.2) 2
+
=
500(0.2) 

sin  500t + 0° − tan −1


50 
50 + (500) (.2)
150
2
−150
i=
or
50
−150
i=
12500
2
2
sin  − tan −1 2 e
(
−
50
.2 t
+
150
12500
(
sin 500t + 0 − tan −1 2
)
)
+150
150
sin tan −1 2 ⋅ e −250t +
sin (500t − 63.4°)
111.80
111.8
= 1.34 sin (63.4°) e −250t + 1.34 sin(500t − 63.4°)
i = 1.2e −250t + 1.34 sin(500t − 63.4°)A
or
6.6 SINUSOIDAL RESPONSE OF R-C CIRCUITS
K
R
i
C
Vm sin(w t + q )
Let us consider a series RC circuit where the
input voltage is a sinusoidal one, as shown in
Figure 6.25.
Let us consider at t = 0, switch S is closed
and sinusoidal voltage Vm sin(w t + q ) is
applied to the circuit.
Let us find the expression for the current.
By applying KVL, we get the form as follows:
Figure 6.25 Transient Response of R–C
Circuit with Sinusoidal Voltage
Vm sin(w t + q ) = Ri +
1
idt
C∫
Differentiating the equation with respect to t, we rewrite the form as in the following:
Vmw cos(w t + q ) = R
or
di 1
+ i
dt C
V w
di
1
+
i = m cos(w t + q )
dt RC
R
Its characteristic equation is as follows:
D+
or
M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 258
1
=0
RC
D=−
1
RC
12/3/2014 8:02:27 PM
Transient Response of Circuits Using Differential Equations 259
Therefore, the complementary function will be given as follows:
iC = K1e
−
1
t
RC
(6.37)
Now, the particular solution can be obtained by the method of undetermined coefficient.
Let
iP = A cos(w t + q ) + B sin (w t + q )
(6.38)
Now,
diP
= − Aw sin(w t + q ) + Bw cos(w t + q )
dt
(6.39)
Substituting the values of equation (6.38) and equation (6.39) in equation (6.34), we get the
following form:
1
[ − Aw sin(w t + q ) + Bw cos(w t + q )] + RC [ A cos(w t + q ) + B sin(w t + q )] =
or
Vmw
cos(w t + q )
R
Vmw
B 
A 


cos(w t + q )
 − Aw +
 sin(w t + q ) +  Bw +
 cos(w t + q ) =
RC
RC
R
Equating the coefficients of sin (w t + q ) and cos(w t + q )on both sides, we obtain the equation
as follows:
B
B
B
− Aw +
= 0 or, Aw =
or, A =
(6.40)
RC
RC
w RC
Bw +
and
A Vmw
=
RC
R
(6.41)
Substituting the value of A from equation (6.40) in equation (6.41), we write the equation as
follows:
Bw +
V w
B
1
⋅
= m
wRC RC
R
1  Vmw
+
= R
wR 2 C 2 
or

B w

or
 w 2 R 2C 2 + 1 Vmw
B
=
2 2 
R
 wR C 
or
B=
=
Vmw 2 R 2C 2
(
R 1 + w 2 R 2C 2
)
=
Vmw 2 RC 2
1 + w 2 R 2C 2
Vm ⋅ R
Vm ⋅ R
=
2 2 2
1
1+ w R C
R2 + 2 2
2 2
w C
w C
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Network Analysis and Synthesis
B=
or
Vm ⋅ R
R + (1/w C )
2
2
(6.42)
Now, using equation (6.40), A can be given as in the following:
A=
Vm R
1
B
=
⋅
wRC wRC R 2 + (1/w C )2
Vm
A=
(6.43)
2
w C  R + (1/w C ) 


Substituting the value of A and B from equation (6.42) and (6.43) in equation (6.38), we obtain
the following form:
2
Vm
Vm ⋅ R
⋅ sin(w t + q )
cos(w t + q ) +
2
2
2
2


+
(
1
/
w
)
R
C
w C R + (1/w C )




Vm

1 
= 2
× sin(w t + q ) ⋅ R + cos(w t + q ) ⋅
2
w C 
R + (1/w C )

iP =
=

1 
× sin(w t + q ) ⋅ R + cos(w t + q ) ⋅
w C 

R + (1 w C ) ⋅ R + (1 w C )
=

1


Vm
R
wC
× sin(w t + q ).
+ cos(w t + q ) ⋅
2
R 2 + (1/w C ) 2 
R 2 + (1/w C ) 2
2  1 
+
R

 w C 

Vm
2
2
2
2




 (6.44)


Now, the sine, cosine and tangent values are as in the following:
R
= cos f
2
R + (1/w C ) 2
1/w C
and
2
R + (1/w C ) 2
= sin f
tan f =
then
1
w CR
1
w CR
By substituting the above values in equation (4.44), the equation can be written as follows:
or f = tan −1
iP =
=
Vm
R 2 + (1/w C ) 2
Vm
2
R + (1/w C ) 2
× [sin(w t + q ).cos f + cos(w t + f ). sin f ]
× sin(w t + q + f )
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Transient
V Response of Circuits Using Differential Equations 261
m
iP =
2
R + (1/w C ) 2
Vm
=
2
R + (1/w C ) 2
Vm
=
 1 
R2 + 
 w C 
2
× [sin(w t + q ).cos f + cos(w t + f ). sin f ]
× sin(w t + q + f )

1 
sin  w t + q + tan −1
w CR 

Now, the complete solution will be in the following form:
i = iC + iP
i = K1e
−
t
RC
Vm
+
2

1 
sin  w t + q + tan −1
w CR 

(6.45)
 1 
R2 + 
 w C 
In order to find K1,we apply the initial condition. That is, at t = 0, we obtain i as given in the
following:
Vm
sinq , By substituting l in the equation, we obtain the from as follows
R
Vm
Vm

1 
sin q = K1e 0 +
sin q + tan −1
2
R

w CR 
R2 + 1 / w C
i=
(
or
K1 =
)
Vm
Vm

1 
sin q + tan −1
si n q −

2
2

w
CR
R
R + (1/w C )
Substituting this value of K1 in equation (10), we form the following equation:
 − t
V
Vm

−1 1  
m

i=
sin q −
sin q + tan
e RC
2
2

w CR  
 R
+
(
1
/
w
)
R
C


Vm

1 
+
sin  w t + q + tan −1

2
2

w
CR
R + (1/w C )
Example 6.13 In the circuit shown in
Figure 6.26, obtain the expression for current.
Solution:
Given
R
K
2
R = 10Ω a + b
(6.46)
10 Ω
2
C = 10 µf = 10 × 10 −6 F
Vm = 10 V
10 sint
i
10 µF
C
w =1
q = 00
M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 261
Figure 6.26
12/3/2014 8:02:35 PM
262
Network Analysis and Synthesis
Now, from equation (6.46), we have the following form:
V
Vm

1 

 m sin q −
sin q + tan −1
2
2

w CR  
 R
+
(
1
/
w
)
R
C


Vm

1 
+
sin  w t + q + tan −1

2
2

w
CR
R + (1/w C )
i=e
−
t
RC
Substituting all the given values, we obtain the equation as follows:






10
1
− t /10 ×10 ×10 −6 10
−1
i=e
sin  0° + tan
10 sin 0° −

2
1 × 10 × 10 −6 × 10  

1


2 
10 + 


 1 × 10 × 10 −6 


10
1


+
sin  t + 0° + tan −1

2

1 × 10 × 10 −6 × 10 
1

2
10 
 1 × 10 × 10 −6 

4 
10
10
= e −10 t  0 −
sin(tan −1 10 4 ) +
sin(t + tan −1 10 4 )A
2
25
2
25


10 + 10
10 + 10
6.7 SINUSOIDAL RESPONSE OF R–L–C CIRCUITS
Let us consider a circuit R–L–C series circuit as shown in Figure 6.27.
At t = 0, the switch is closed, and sinusoidal
R
voltage Vm sin(w t + q ) is applied to the
K
R–L–C series circuit.
Applying KVL, we get the following form:
L
i
Vmsin(w t + q )
C
Ri + L
Differentiating both sides with respect to
time, we obtain the equation as follows:
Figure 6.27
R
or
di 1
+
idt = Vm sin(w t + q )
dt C ∫
d 2i
dt
2
+
di Ld 2 i 1
+
+ i = Vmw cos(w t + q )
dt dt 2 C
V w
R di 1
+ i = m cos(w t + q )
L dt C
L
M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 262
(6.47)
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Transient Response of Circuits Using Differential Equations 263
The corresponding characteristic equation is,
D2 +
1
R
D+ =0
L
C
2
−R
1
 R
±   −4 ×1×


L
L
C
D=
2
or
2
D=−
or
1
R
 R
±   −
 2L 
L
2L
Let us consider the following three cases.
2
Case I: when  R  > 1 , then roots will real and unequal.
 2L 
LC
2
D=
1
−R
 R 
±   −
 2L 
2L
LC
D=
1
−R
 R 
+   −
=a
 2L 
2L
LC
or
D=
−R
1
 R
−   −
=b


LC
2L
2L
Therefore,
ic = C1e at + C2 e bt
Let
2
2
(6.48)
2
Case II: when  R  < 1 , roots will be complex conjugate.
 2L 
LC
D=−
 1  R 2
R
± −
−  
2L
 LC  2 L  
1  R
R
=−
±j
− 
2L
LC  2 L 
2
= a ± jb
Therefore,
ic = e at [C1 cos bt + C2 sin bt ]
M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 263
(6.49)
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264
Network Analysis and Synthesis
2
Case III: when  R  = 1 and D = − R , − R (real and equal), then the following form
 
2L
2L
2L
LC
can be obtained:
ic = C1e
−
R
t
2L
+ tC2 e
−
R
t
2L
(6.50)
Particular solution:
iP = A cos(w t + q ) + B sin(w t + q )
Let
and
diP
= − Aw sin(w t + q ) + Bw cos(w t + q )
dt
d 2 iP
2
(6.51)
(6.52)
= − Aw 2 cos(w t + q ) − Bw 2sin(w t + q )
dt
Substituting these values in equation (6.47), we can write the equation as follows:
 − Aw 2 cos(w t + q ) − Bw 2 sin(w t + q )  + R [ − Aw sin(w t + q ) + Bw cos(w t + q ) ]

 L
V w
1
+
[ A cos(w t + q ) + B sin(w t + q )] = mL cos(w t + q )
LC
wR
A

2
+
 − Aw + B
 cos(w t + q )
L
LC 
or
or
V w
wR B

+  − Bw 2 − A
+  sin(w t + q ) = m cos(w t + q )


L
L
Lc
 wR
  1
wR 
 1
2
2 
 A  LC − w  + L B  cos(w t + q ) +  − L A +  LC − w  B  sin(w t + q )




Vmw
=
cos(w t + q ).
L
By comparing the coefficient of cos(w t + q ) and sin(w t + q ) on both sides, we get
−
V w
 1
 wR
A
−w 2 +
B= m
 LC
 L
L
(6.53)
wR
 1

A+ 
−w 2 B = 0
 LC

L
(6.54)
From equation (6.54), we can obtain the following form:
wR

 1
A =
−w 2 B

 LC
L
or
M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 264
A=
L  1

−w 2 B


wR LC
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Transient Response of Circuits Using Differential Equations 265
or
wL
 1
−
A =
 B
 wRC
R 
or
 1 − w 2 LC 
A =
B
 wRC 
(6.55)
Substituting the value of A in equation (6.53), B can be written as follows:
 1 − w 2 LC   1
Vmw
2  wR
 wRC  B  LC − w  + L B = L


or
 1 − w 2 LC   1 − w 2 LC  wR  Vmw
B 
=

+
L
 wRC   LC  L 
 (1 − w 2 LC )(1 − w 2 LC ) wR  Vmw
+
B
=
L 
L
wRLC 2

or
 (1 − w 2 LC )(1 − w 2 LC ) + w 2 R 2C 2  Vmw
B
=
L
wRLC 2


B=
B=
=
=
Vmw 2 RC 2
(1 − w 2 LC )(1 − w 2 LC ) + w 2 R 2C 2
Vm ⋅ w 2 RC 2
(1 − w 2 LC ) + w 2 R 2C 2
Vmw 2 RC 2
2
 1

L2C 2 
− w 2  + w 2 R 2C 2
 LC

Vmw 2 RC 2
2
 1
w 2 R2 

L2C 2 
−w 2 + 2 

L 
 LC
w 2R
L2
B=
2
w 2 R2
 1

−w 2 + 2


LC
L
Vm ⋅
(6.56)
Now, from equation (6.55), we write A as given in the following:
A=
(1 − w
M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 265
2
LC
wRC
)⋅B
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Network Analysis and Synthesis
Here, substituting the value of B, we obtain the equation as follows:
w 2R
 1
w 2R
2
⋅
−
w
LC
V

 m 2
(1 − w 2 LC )
LC
L
L2
A=
=
⋅
2
2
2
2
wLC
 1
w R
 1

w 2 R2 

−w 2 + 2
wRC . 
−w 2 + 2 



LC
L
L 
 LC
Vm
w 2 RC  1

.
−w 2

L  LC
2
 1
w 2 R2 

−w 2 + 2 
w RC 

L 
 LC
Vm
or

 1
Vmw 
−w 2

 LC
(6.57)
2
2 2
 1
w
R

−w 2 + 2 
L 

L 
 LC
Substituting the values of A and B from equation (6.56) and (6.57) in equation (6.51), we get
the following form:
A=
w 2R
L2
sin(w t + q )
iP =
cos(w t + q ) +
2
2
 1
w 2 R2
 1
w 2 R2 
2
2
−w  + 2
−w  + 2 
L 



LC
L
L 
 LC

 1
Vmw 
−w 2

 LC
Vm

 1
w 2R
−w 2
Vmw 
Vm 2

 LC
L
sin(w t + q )
=
cos(w t + q ) +
2
2
2 
2 





w
w  1
1

− wL  + R 2 
L ⋅ 2 
− w L + R 2 


L2  w C
L  w C



 1
−w 2
Vm L 
Vm R
LC


=
cos w t + q ) +
sin(w t + q )
2
2

 1
 1


2
2
− wL  + R 
w 
 w L − w L + R

 w C

 1

Vmw L 
−w 
LC
w
Vm R

 cos(w t + q ) +
=
sin(w t + q )
2
2

 1


1

2
− w L + R 2 
w 
 w C − w L + R

 w C


 1
Vm L 
−w 
LC
w
Vm R
 cos(w t + q ) +

ip =
sin(w t + q )
2
2
 1

 1

2
2
 w C − w L + R
 w C − w L + R
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
 1
Vmw L 
−w 
LC
w
Vm R
 cos(w t + q ) +

=
sin(w t + q )
2
2
Response
of Circuits
Using Differential
Equations 267

 1 Transient


1

2
− w L + R 2 
w 
 w C − w L + R


 w C

 1
Vm L 
−w 
LC
w
Vm R
 cos(w t + q ) +

ip =
sin(w t + q )
2
2
 1

 1

2
2
 w C − w L + R
 w C − w L + R
or

 1
− w L
Vm 
Vm R
 cos(w t + q ) +
w C
=
sin(w t + q )
2
2
 1



1
2
2
 w C − w L + R
 w C − w L + R
=
=
or
iP =
Vm
2
 1

2
 w C − w L + R
 1


− w L cos(w t + q ) + R sin(w t + q ) 


 w C

Vm
2
2
 1


2  1
2
 w C − w L  + R  w C − w L  + R


 1

− w L  cos(w t + q ) 
 R sin(w t + q ) + 
wC



Vm
 1

 w C − w L 
2

 1


−w L


wC

R
× sin(w t + q ) ⋅
+ cos(w t + q ) ⋅
2
2


 1


1

 w C − w L 
 w C − w L 

Now, the values of sine, cosine and tangent can be written as follows:
R
 1

 w C − w L 
and
2
 1

 w C − w L 


1
 w C − w L 
2
M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 267







(6.58)
= cos f
 1

 w C − w L 
= sin f and tan f =
R
12/3/2014 8:02:51 PM
268
Network Analysis and Synthesis
By substituting these values in the equation, we obtain the following form:
The complete solution is given as,
i = iC + iP
R E V IE W Q U E S T I O N S
Numerical Problems
1. Find the current flowing in the circuit.
20 Ω
[Ans. e−t A]
20 V
0.05 F
i (t )
2. Find the voltage drop across the resistor in the following circuit.
5Ω
[Ans. 80 (1 − e−2t)A]
80 V
+
−
2.5 H
i (t )
3. Find the current flowing in the following circuit.
R
S
20 Ω
0.05 H
[Ans. 2.04e−200t sin 979.8 t]
100 V
20 µF
i
4. Find the current flowing in the following circuit and calculate the voltage drop across the
resistor.
2Ω
S
[Ans. i(t) = 5e−5000t A, VR = 10e−5000t V]
100 µF
10 V
i (t )
5. Find the current flowing in the following circuit and also calculate the voltage drop across
the resistor.
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Transient Response of Circuits Using Differential Equations 269
K
10 sin10t
1Ω
[Ans. −0.99 e−t + 0.99 cos 10 t + 0.99 sin 10 t]
1F
i
6. In the circuit shown V = 100 V and C1 = 1 µF. Determine the value of R1 if C1 is to get
charged to 50 V in 20 ms. Also calculate the value of P2 that will limit the maximum discharge current to 1 mA.
R1
[ Ans. R1 = 29.9 k Ω, R2 = 100 k Ω]
R2 C1
V
7. Find the current flowing through circuit, i(0) = 0 A.
K
2R
V 

 Ans. 2 R A 


L
V
i (t )
8. If i(0) =
V
A, find the current flowing through the following circuit.
2R
2R
3R

V − Lt
e
 Ans.

2R


L
V
i (t )
9. Given i(0) = 2 A and
di
= 12A/s. Find the value of R in the following circuit.
dt
20 V
10 H
[Ans. 120 Ω]
R
40 H
M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 269
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270
Network Analysis and Synthesis
10. Find the expression for the current in the following circuit.
K
1Ω
0.1H
10e−100t
10  −10t


− e −100t  
 Ans. 9 e


11. A resistance of 2.2 kΩ and a capacitor of 10 µF are connected in series. A voltage of 50 V
is switched on across the circuit. Calculate the instantaneous value of capacitor voltage at
intervals of 15 ms up to 105 ms time and then plot the graph of voltage across capacitor
versus time.
[Ans. 24.7 V, 37.2 V, 43.5 V, 46.7 V, 48.3 V, 49.2 V, 49.6 V]
12. An R–C series circuit has R = 10 Ω and C = 0.1 F. A DC voltage of 20 V is switched on to the
circuit at t = 0. Determine (a) the expression for the current; (b) voltages across the resistor
and the capacitor.
[Ans. i = Ce−t, VR = 20e−t V and VC = 20(1 − e−t) V ]
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Laplace Transform
7
Chapter objectives
After carefully studying this chapter, you should be able to do the following:
Provide a explanation of the concept of
Use the properties of Laplace transLaplace transform.
form to solve problems on determining
Laplace transform of different functions.
Distinguish between the functions of
Laplace transform and inverse Laplace
State and explain initial value theorem.
transform.
State and explain final value theorem.
Determine the Laplace transform of
Solve problems using initial value and
standard functions.
final value theorems.
Determine Laplace transform of cerUse standard formula to carryout inverse
tain functions using standard transforLaplace transform of different functions.
mation formula.
State and explain convolution theorem.
7.1 CONCEPT OF LAPLACE TRANSFORM
In this chapter, we introduce the Laplace transform which is used for providing the solution of
network problems.
We have known that Fourier transform is used in finding solutions of large variety of engineering problems. Fourier transform enables understanding of system behaviour in frequency
domain. This is done by expressing a signal f(t) as a continuous sum of complex exponentials.
The Fourier transform is defined as follows:
F ( jw ) =
∞
∫
f (t ) e − jw t dt
−∞
There are some special functions where Fourier transform is not possible. The introduction
of a convergence factor in the form of e−s t, where s is a real number, makes the integral
convergent.
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272 Network Analysis and Synthesis
The introduction of a convergence factor into the Fourier transform creates a new transformation, which is called Laplace transform. Thus, the Laplace transform is defined as in the
following:
∞
∞
0
0
F ( s) = ∫ f (t ) e −s t e − jw t dt = ∫ f (t )e − (s + jw )t
It is to be noted that the lower limit of the integral has been taken as 0 instead of −∞. This is because
with s > 0, the convergence factor e−s t will diverge rather than converge when t tends to −∞.
If we substitute s = s + jw, then
∞
F (s ) = ∫ f (t ) e − st dt
0
F(s) is called the Laplace transform of f (t).
Laplace transform is a mathematical tool to convert the function from time domain to frequency
domain. Let f(t) be any time domain function, then its Laplace transform is defined as follows:
∞
L{ f (t )} = F ( s) = ∫ e − st f (t )dt
0
where s is a complex frequency (s = −s + jw).
The integral 0 to ∞ does not take care of any information contained in f (t), when t tends to −∞.
Thus, the Laplace transform changes the time domain function f (t) to frequency domain
function F(s).
Inverse Laplace transformation converts a frequency domain function into time domain
function.
The Laplace transformation method of solving differential equations has number of
advantages.
The solution follows a systematic procedure where the initial conditions are taken care automatically in the transform operation, and a complete solution is obtained in one operation.
7.2 LAPLACE TRANSFORM OF STANDARD FUNCTIONS
We will now find Laplace transform of some standard functions.
Example 7.1 Determine the Laplace transform of unity, that is, f (t) = 1.
Solution: By definition, we can write the following equation:
∞
L[ f (t )] = ∫ e − st f (t ) dt
0
Here,
f(t) = 1
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Laplace Transform
273
Therefore, the following is obtained:
∞
L [1] = ∫ e − st 1dt
0
∞
= ∫ e − st dt
0
∞
 e − st 
=

 −s  0
e −∞ − e 0
−s
0 −1 1
=
=
−s
s
=
1
L [1] = .
s
Therefore,
Example 7.2 Find the Laplace transform of exponential function f (t ) = e at.
Solution: By definition, we can get the following:
∞
L[ f (t )] = ∫ e − st f (t ) dt
0
Substituting f (t) =
e at,
we get the following:
∞
L e at  = ∫ e − st ⋅ e at dt
0
∞
= ∫ e − st + at dt
0
∞
= ∫ e − ( s − a )t dt
0
 e − ( s − a )t 
=

 −( s − a) 
−∞
∞
0
0
e −e
−( s − a)
0 −1
=
−( s − a)
=
=
M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 273
1
s−a
12/4/2014 2:06:23 PM
0
 e − ( s − a )t 
=

 −( s − a) 
0
0
e −e
−( s − a)
0 −1
=
−( s − a)
1
=
s−a
=
274 Network Analysis and Synthesis
Therefore,
−∞
∞
L e at  =
1
s −a
Example 7.3 Calculate the Laplace transform of t.
Solution: By definition, the equation can be written as follows:
∞
L[ f (t )] = ∫ e − st ⋅ f (t ) dt
0
Substituting f (t) = t, we get the following:
∞
L[t ] = ∫ e − st ⋅ t dt
0
∞
= ∫ t ⋅ e − st dt
0
Integrating by parts, the equation can be written as follows:
∞
d


L [t ] = t ∫ e − st dt − ∫ (t ) ⋅ ∫ e − st dt 
dt

0
∞
∞
 t ⋅ e − st 
e − st
dt
1
=
⋅
−

∫
 −s  0 0 −s
∞
 e − st 
= 0−
2
 ( −s )  0
 e −∞ − e 0 
= −

2
 s

 0 − 1 1
= − 2  = 2
 s  s
Therefore,
L [t ] =
M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 274
1
s2
12/4/2014 2:06:24 PM
Laplace Transform
275
Example 7.4 Calculate the Laplace transform of t n.
Solution: By definition, the following form is obtained:
∞
L [t n ] = ∫ e − st ⋅ t n dt
0
∞
= ∫ t n ⋅ e − st dt
0 ↓ ↓
I II
Integrating by parts, we get the equation as follows:
t n ⋅ e − st
=
−s
= 0+
∞
0
∞
e − st
dt
−s
− ∫ nt n −1
0
∞
n n −1 − st
t e dt
s ∫0
↓ ↓
I II
Again integrating by parts, we get the equation as follows:
n  t n −1e − st
= 
s  −s

∞
0
∞
− ∫ ( n − 1)t n − 2
0
=
n
( n − 1) n − 2 − st 
t e dt 
0 −
−s ∫0
s 

=
∞

n  ( n − 1)
⋅
t n − 2e − st dt 

∫
s  s

0
=
e − st 
dt
−s 

∞
n ( n − 1)
s2
∞
∫t
n − 2 − st
e
dt
0
=
=
n ( n − 1)( n − 2)1
sn
n!
s
n
M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 275
∞
⋅ ∫ t 0 ⋅ e − st dt
0
∞
⋅ ∫ e − st dt
0
12/4/2014 2:06:24 PM
276 Network Analysis and Synthesis
n ! e − st
⋅
s n −s
=
n! e − e0 
⋅

−s 
sn
=
=
=
L [t n ] =
Therefore,
∞
=
0
−∞
n !  0 − 1


s n  −s 
n!
sn ⋅s
n!
s n +1
n!
s n +1
Example 7.5 Calculate the Laplace transform of sin at.
Solution: By definition,
∞
L[sin at ] = ∫ e − st ⋅ sin at dt
0
Now using the formula, we get the form as follows:
∫e
ax
sin bx dx =
e ax
a2 + b2
[a sin bx − b cos bx ],
Then, we have the following form:
∞
L[sin at ] = ∫ e
0
Hence,
∞
− st

 e − st
sin at dt = 
× [ − s ⋅ sin at − a cos at ]
2
2
 ( − s) + a
0
∞
1
= 2
× e − st ( − s ⋅ sin at − a cos at ) 
2
0
s +a
1
= 2
× e −∞ ( − s ⋅ sin ∞ − a cos ∞) − e 0 ( − s ⋅ sin 0 − a cos 0) 
s + a2 
1
= 2
× [0 − 1(0 − a)]
s + a2
a
= 2
s + a2
L [sin at ] =
a
2
s + a2
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Laplace Transform
277
Example 7.6 Calculate the Laplace transform of cos at.
Solution: By definition, we get the following:
∞
L[cos at ] = ∫ e − st ⋅ cos at dt
0
∞
 e − st

=
× ( − s ⋅ cos at + a ⋅ sin at ) 
2
2
 ( − s) + a
0


e ax
ax
× (a cos bx + b sin bx ) 
∵ ∫ e cos bxdx = 2
2
a +b


Therefore,
L[cos at ] =
=
=
=
=
1
2
s +a
∞
2
1
2
s + a2
1
2
s + a2
1
2
s + a2
× e − st ( − s ⋅ cos at + a ⋅ sin at )  0
0 − e 0 ( − s ⋅ cos 0 − a ⋅ sin 0) 
× [0 − 1⋅( − s ⋅1 − 0)]
× [ s]
s
s2 + a2
Hence,
L [cos at ] =
s
s 2 + a2
Example 7.7 Calculate the Laplace transform of sinh at.
Solution: By definition, we get the following:
∞
L[sinh at ] = ∫ e − st ⋅ sinh at dt
0
∞
 e at − e − at 
= ∫ e − st ⋅ 
 dt
2


0
=

eq − e −q 
∵ sinh q =

2


∞
1
{e − ( s − a )t − e − ( s+ a )t }dt
2 ∫0
1  e − ( s − a )t e − ( s+ a )t 
=
−


M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd
2 −( s − a) 277
−( s + a)
∞
12/4/2014 2:06:27 PM
∞
L[sinh at ] = ∫ e − st ⋅ sinh at dt
0
278 Network Analysis and Synthesis
∞
 e at − e − at 
= ∫ e − st ⋅ 
 dt
2


0
=

eq − e −q 
sinh
=
∵
q


2


∞
1
{e − ( s − a )t − e − ( s+ a )t }dt
2 ∫0
1  e − ( s − a )t e − ( s+ a )t 
= 
−

2  −( s − a) −( s + a) 
∞
0
=
1 e − e
e −e 
−


2  −( s − a) −( s + a ) 
=
1  0 −1
0 −1 
−

2  −( s − a) −( s + a) 
=
1 1
1 
−

2  s − a s + a 
=
1  s + a − ( s − a) 
2  ( s − a)( s + a) 
=
1 a+a 
2  s 2 − a 2 
−∞
0
−∞
0
1  2a 
2  s 2 − a 2 
a
= 2
s − a2
=
L [sinh at ] =
Hence,
a
2
s − a2
Example 7.8 Calculate the Laplace transform of cosh at.
Solution: By definition,
∞
L[cosh at ] = ∫ e − st ⋅ cosh at dt
0
∞
 e at + e − at 
= ∫ e − st 
 dt
2


0

eq + e −q 
∴ cosh q =

2


∞
=
1
{e − ( s − a )t + e − ( s + a )t }dt
2 ∫0
=
1  e − ( s − a )t e − ( s + a )t 
+


2  −( s − a) −( s + a) 
M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 278
∞
0
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Laplace Transform
=
1  e −∞ − e 0 e −∞ − e 0 
+


2  −(s − a) −(s + a) 
=
1  0 −1
0 −1 
+
2  −(s − a) −(s + a) 
=
1 1
1 
+

2  s − a s + a 
=
1  s +a+s −a 
2  (s − a)(s + a) 
=
=
1  2s 
2  s 2 − a2 
s
s 2 − a2
L [cosh at ] =
Hence,
279
s
2
s − a2
The summary of formulae for Laplace transform is given in a tabular form in Table 1.
Table 7.1 Laplace Transform of Certain Functions
f(t)
L[f(t)]
1. L[1]
1
s
2. L[t]
1
3. L[t n]
s2
n!
s n +1
4. L[e at]
1
s −a
5. L[e -at]
1
s+a
6. L[sin at]
7. L[cos at]
a
2
s + a2
s
s 2 + a2
(Continued )
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280 Network Analysis and Synthesis
Table 7.1 (Continued)
f(t)
L[f(t)]
a
8. L[sinh at]
2
s − a2
s
9. L[cosh at]
s 2 − a2
7.3 LAPLACE TRANSFORM PROBLEMS BASED ON STANDARD
FORMULA
Some more examples based on Laplace transform are provided as follows:
Example 7.9 Evaluate L [3e −5t + 8 cos 3t + 2 sinh 2t − 5t 3 ]
Solution:
L [3e −5t + 8 cos 3t + 2 sinh 2t − 5t 3 ]
= L [3e −5t ] + L [8 cos 3t ] + L [2 sinh 2t ] − L [5t 3 ]
= 3L [e −5t ] + 8L [cos 3t ] + 2L [sinh 2t ] − 5L [t 3 ]
1
3!
s
2
= 3⋅
+ 8⋅ 2
+ 2⋅ 2
− 5 ⋅ 3+1
2
2
s +5
s +3
s −2
s
3
8s
4
30
=
+
+
−
s + 5 s 2 + 32 s 2 − 22 s 4
Example 7.10 Find the Laplace transform of sin2 3t.
Solution:
1 − cos 2(3t ) 
L [sin 2 3t ] = L 

2


∵ We have,
1 − cos 2q
sin 2 q =
2
+
cos
1
2q
cos 2 q =
2
1 − cos 6t 
=L

2


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281
Laplace Transform
=
=
=
=
=
1
L[1 − cos 6t ]
2
1
[ L[1] − L[cos 6t ]]
2
1 1
s 
− 2

2  s s + 6 2 
1  s 2 + 36 − s 2 


2  s( s 2 + 36) 

1
36

 2
2  s( s + 36) 
Therefore,
L [sin 2 3t ] =
18
2
s + 36
Example 7.11 Evaluate L[cos2 5t]
Solution:
1 + cos 2(5t ) 
L[cos 2 5t ] = L 

2


1 + cos10t 
= L

2


1
= L[1 + cos10t ]
2
1
= [ L[1] + L[cos10t ]]
2
L [cos 2 5t] =
Therefore,
1 1
s

+ 2


2
2  s s + 10 
Example 7.12 Find the Laplace transform of sinh2 3t.
Solution:
 e 3t − e −3t  2 
L [sinh 3t ] = L 
 

2



2
 e 6t + e −6t − 2e 3t ⋅ e −3t 
=L 

4


1  6t
= L e + e −6t − 2
4
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281

eq − e −q 
∵ sinh q =



2
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 e 3t − e −3t  2 
L [sinh 3t ] = L 
 

2



 e 6t + e −6t − 2e 3t ⋅ e −3t 
=L 

4


1
= L e 6t + e −6t − 2
4
282 Network Analysis and Synthesis
2
Therefore,
L [sinh 2 3t ] =

eq − e −q 
∵ sinh q =



2
1 1
1
2
+
−
4  s − 6 s + 6 s 
Example 7.13 Find the Laplace transform of cosh2 4t.
Solution:
 e 4t + e −4t  2 
L [cosh 2 4t ] = L 
 

2



−
8
t
8
t
 e + e + 2e 4t e −4t 
=L

4


1
= L [e 8t + e −8t + 2]
4
Therefore,
L [cosh 2 4t ] =

eq + e −q 
∵ coshq =


2

1 1
1
2
+
+
4  s − 8 s + 8 s 
Example 7.14 Evaluate L[sin3 2t]
Solution:
 3 sin 2t − sin 3( 2t ) 
L[sin 3 2t ] = L 

4


 3 sin 2t − sin 6t 
= L

4


1
= L[3 sin 2t − sin 6t ]
4
1
= [3L[sin 2t ] − L[siin 6t ]]
4
1
2
6 
= 3 ⋅ 2
− 2
2
4 s +2
s + 6 2 
6 1
1 
=  2
−
4  s + 4 s 2 + 36 
3  s 2 + 36 − s 2 − 4 
= × 2

2  ( s282+ 4)( s 2 + 36) 
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3 sinq − sin 3q 

3
∴ sin q =

4


12/4/2014 2:06:32 PM
 3 sin 2t − sin 6t 
= L

4


1
= L[3 sin 2t − sin 6t ]
4
1
= [3L[sin 2t ] − L[siin 6t ]]
4
Laplace Transform 283
1
2
6 
= 3 ⋅ 2
−
4  s + 22 s 2 + 6 2 
6 1
1 
=  2
− 2
4  s + 4 s + 36 
3  s 2 + 36 − s 2 − 4 
= × 2

2  ( s + 4)( s 2 + 36) 

48
3 
32
= × 2
=
2
2  ( s + 4)( s + 36)  ( s 2 + 4)( s 2 + 36)
Example 7.15 Evaluate L[cos3 4t]
Solution:
 3 cos 4t + cos 12t 
L[cos3 4t ] = L 

4


=
1
L[3 cos 4t + cos 12t ]
4
=
1
[ L[3 cos 4t ] + L[cos12t ]]
4
=
1
[3L[coss 4t ] + L[cos12t ]]
4
=
1
s
s

3⋅
+
4  s 2 + 4 2 s 2 + 122 
=
s 3
1 
+ 2

2
4  s + 16 s + 144 
=
s  s 2 + 144 + s 2 + 16 


4  ( s 2 + 16)( s 2 + 144) 
=

s 
2 s 2 + 160
× 2

4  ( s + 16)( s 2 + 144) 
=
s
2( s 2 + 80)
× 2
4 ( s + 16)( s 2 + 144)
=
3 cos q + cos 3q 

3
∴ cos q =

4


s( s 2 + 80)
2( s 2 + 16)( s 2 + 144)
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284 Network Analysis and Synthesis
Example 7.16 Evaluate L[cosh3 2t].
Solution:
L[cosh 3 2t ] = L[(cosh 2t )3 ]
 e 2t + e −2t  3 
= L 
 

2



 (e 2t )3 + (e −2t )3 + 3 ⋅ e 2t ⋅ e −2t (e 2t + e −2t ) 
= L

8


[∵ ( a + b)3 = a3 + b3 + 3ab( a + b)]
 e6t + e −6t + 3(e 2t + e −2t ) 
= L

8


1
=  L[ee6t ] + L[e −6t ] + 3L[e 2t ] + 3L[e −2t ]
8
1
1
1 
1 1
= 
+
+ 3⋅
+ 3⋅
8 s −6 s +6
s−2
s + 2 
=
1  s + 6 + s − 6 3( s + 2 + s − 2) 
+

8  s 2 − 36
s2 − 4

=
1  2s
6s 
+ 2

2
8  s − 36 s − 4 
Example 7.17 Evaluate L[sin 2t cos 3t]
Solution:
L [sin 2t cos3t ]
1
= L [2 sin 2t cos 3t ]
2
1
= L [sin( 2t + 3t ) + sin( 2t − 3t )]
2
1
= L [sin 5t + sin( −t )]
2
1
= L [sin 5t − sin t ]
2
1
= [L {sin 5t } − L {sin t }]
2
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[ ∵ 2 sin A cos B = sin( A + B ) + sin( A − B )]
12/4/2014 2:06:33 PM
Laplace Transform
=
1 5
1 
−
2  s 2 + 52 s 2 + 12 
=
1  5(s 2 + 1) − (s 2 + 25) 


2  (s 2 + 25)(s 2 + 1) 
=
1  5s 2 + 5 − s 2 − 25 


2  (s 2 + 25)(s 2 + 1) 
=

1
4s 2 − 20

 2
2
2  (s + 25)(s + 1) 
=
285
2s 2 − 10
2
(s + 25)(s 2 + 1)
Example 7.18 Evaluate L[cos 3t cos 2t]
Solution:
L [cos 3t cos 2t ]
1
L [2 cos 3t cos 2t ]
[∵ 2 cos A cos B = cos( A + B ) + cos( A − B )]
2
1
= L [cos(3t + 2t ) + cos(3t − 2t )]
2
1
= L [cos 5t + cos t ]
2
1
= [L {cos 5t } + L {cos t }]
2
=
=
s 
1 s
+
2  s 2 + 52 s 2 + 12 
=
1  s (s 2 + 1) + s (s 2 + 25) 


2  (s 2 + 25)(s 2 + 1) 
=
s  s 2 + 1 + s 2 + 25 


2  (s 2 + 25)(s 2 + 1) 
=

2s 2 + 26
s
 2

2
2  (s + 25)(s + 1) 
=
s (s 2 + 13)
2
(s + 25)(s 2 + 1)
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286 Network Analysis and Synthesis
7.4 PROPERTIES OF LAPLACE TRANSFORM
In this section, we will state the various properties of Laplace transform and solve problems
using these properties.
7.4.1 Property 1: First Shifting Property
L [L
f ([tf )](t=)] f=(sf )(s )
If
L [e at f at(t )] =
L [e f (t )] f=(sf (−sa−) a)
then
Example 7.19 Evaluate L[e−3t cos 2t].
Solution: We have
L [cos 2t ] =
s
2
s + 22
Now, by the first shifting property, we get the following:
L [e −3t cos 2t ] =
s+3
[Replace s by s + 3]
(s + 3) 2 + 22
Example 7.20 Evaluate L[e2tt3]
Solution: Now,
L [t 3 ] =
=
L [e 2t t 3 ] =
Therefore, by the shifting property,
3!
s 3+1
3!
s4
3!
( s − 2) 4
Example 7.21 Evaluate L[sinh 2t cos 3t]
Solution: Now,
 e 2t − e −2t 

L[sinh 2t ⋅ cos 3t ] = L 
cos 3t 

2



1  2t
L (e − e −2t ) ⋅ cos 3t 
2 
1
=  L{e 2t cos 3t} − L{e −2t ⋅ cos 3t}
2
1  ( s − 2)
( s + 2) 
= 
−

2  ( s − 2) 2 + 32 ( s + 2) 2 + 32 
286
=
M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd
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 e 2t − e −2t 

L[sinh 2t ⋅ cos 3t ] = L 
cos 3t 

2



Laplace Transform 287
1  2t
−2t

= L (e − e ) ⋅ cos 3t 
2
1
=  L{e 2t cos 3t} − L{e −2t ⋅ cos 3t}
2
1  ( s − 2)
( s + 2) 
= 
−

2
2
2  ( s − 2) + 3 ( s + 2) 2 + 32 
[Using first shifting property]
7.4.2 Property 2: Multiplication By t n
If
then
L[ f (t )] = f ( s)
L[ f (t )] = f ( s) n
d
L[t n f (t )] = ( −1) n d nn [ f ( s)]
n
n ds
L[t f (t )] = ( −1)
[ f ( s)]
ds n
Example 7.22 Evaluate L[t cos 3t]
Solution: We have
L [cos 3t ] =
s
2
s + 32
Therefore,
L[t cos 3t ] = ( −1)1
=−
d1  s 


ds1  s 2 + 32 
d  s 
ds  s 2 + 9 
1⋅ ( s 2 + 9) − s ⋅ ( 2 s + 0) 
= −

( s 2 + 9) 2


 s 2 + 9 − 2s 2 
= − 2
2 
 ( s + 9) 
 9 − s2 
= 2
2
 ( s + 9) 
=
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s2 − 9
( s 2 + 9) 2
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288 Network Analysis and Synthesis
Example 7.23 Evaluate L [t2e−3t sin2t]
Solution: First, we will find
L [sin 2t ] =
2
2
s + 22
Then, we will use multiplication by tn property.
L [t 2 sin 2t ] = ( −1) 2
d2  2 


ds 2  s 2 + 4 
d2  1 


ds 2  s 2 + 4 
d  0 ⋅ (s 2 + 4) − 1⋅ ( 2s + 0) 
=2 

ds 
(s 2 + 4 ) 2



−2s
d
=2  2

ds  (s + 4) 2 
=2
= −4

d 
s

 2
2
ds  (s + 4) 
1⋅ (s 2 + 4) 2 − s ⋅ 2(s 2 + 4) ⋅ 2s 
= −4 

(s 2 + 4 ) 4


=
=
−4(s 2 + 4)[s 2 + 4 − 4s 2 ]
(s 2 + 4 ) 4
−4( 4 − 3s 2 )
(s 2 + 4 )3
At last, we will use the first shifting property and we get the equation as follows:
L[e −3t t 2 sin 2t ] =
−4[4 − 3( s + 3) 2 ]
[Replace s by s + 3]
[( s + 3) 2 + 4]3
7.4.3 Property 3: Division By ‘t’
Property 3 states the following:
If
L[ f (t )] = f ( s)
∞
then
∞
f (t ) 
L 
= ∫ f ( s) ds

 t  s
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Laplace Transform
289
 e −t − e −2t 
Example 7.24 Evaluate L 

t


Solution: Firstly, we will find
L [e −t − e −2t ] = L {e −t } − L {e −2t }
1
1
=
−
s +1 s + 2
Now, we will use division by ‘t’ property, and the equation can be written as follows:
 e − t e −2t  ∞  1
1 
L
−
 = ∫
 ds

 t  s s + 1 s + 2
= [ log | s + 1 | − log | s + 2 |]s
∞
∞

s +1 
= log
s + 2  s

 1
s 1 + 
 s
= log
 2
s 1 + 

s
∞
s

1
1+

s
= log
2

1+

s
∞




 s
1
s = 0 − log s + 1 = log s + 2
= log 1 − log
2
s+2
s +1
1+
s
1+
sin 2t 

.
Example 7.25 Evaluate L e −t
t 

Solution: Firstly, we will find
L [sin 2t ] =
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2
s 2 + 22
.
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290 Network Analysis and Synthesis
Now, we will use the division by ‘t’ property to get the following:
∞
2
 sin 2t 
L
ds
=∫ 2

 t  s s + 22
∞
= 2∫
s
= 2⋅
1
2
s + 22
ds
∞
1
s
tan −1
2
2s
= tan −1
∞
s
2s
= tan −1 ∞ − tan −1
s
p
− tan −1
2
2
s
= cot −1
2
s
2
=
At last, we will use the first shifting property and the following is obtained:
sin 2t 

−1 (s + 1)
L e −t
 = cot
2
t


7.4.4 Property 4
If
then
L
L[[ ff ((tt )]
)] =
=

 tt
L
L  ∫ ff ((tt ))dt
dt  =
=
 00

ff (( ss))
ff (( ss))
ss

t
Example 7.26 Evaluate L  ∫ cos 2t dt 
 0

Solution: Now,
L [cos 2t ] =
Therefore
s
2
s + 22
t
 1
s
L  ∫ cos 2t dt  = ⋅ 2
s
s + 22
 0

1
= 2
s + 22
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291
Laplace Transform
 t sin t 
dt 
Example 7.27 Evaluate L  ∫
 0 t

Solution: We have
L [sin t ] =
1
2
s + 12
∞
1
 sin t 
L
 = ∫ 2 2 ds
t

 s s +1
Now,
∞
=
1
s
tan −1
1
1s
= tan −1 s
∞
s
−1
= tan ∞ − tan −1 s
p
= − tan −1 s = cot −1 s
2
 t sin t  1
L ∫
dt  = ⋅ cot −1 s
 0 t
 s
Therefore,
7.5 SUMMARY OF USEFUL PROPERTIES OF LAPLACE TRANSFORM
L[ f (t )] = f ( s), then
If
Property 1: First shifting property
L [e at f (t )] = f (s − a)
Property 2: Multiplication by tn
L[t n f (t )] = ( −1) n
dn
[ f ( s)]
ds n
Property 3: Division by t
∞
 f (t ) 
L
 = ∫ f ( s) ds
 t  s
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292 Network Analysis and Synthesis
Property 4:
 1
t
L  ∫ f (t ) dt  = ⋅ f (s )
 0
 s
7.6 INITIAL VALUE THEOREM
It states that Lt f (t ) or f (0), that is, the initial value of a function f (t) in any time domain is
t →0
equal to Lt sF ( s) . That is, it can be expressed as follows:
s →∞
Lt f (t ) = Lt sF ( s)
t →0
s→∞
F (s ) = L [ f (t )]
Where
Example 7.28 Verify the initial value theorem for the function e4t cos 2t.
Solution: Given f(t) = e4t cos 2t
Now,
L[ f (t)] = L[e4t cos 2t]
( s − 4)
( s − 4) 2 + 2 2
s−4
=
( s − 4) 2 + 4
= F ( s)
=
[Using first shifting property]]
Now, the L.H.S of the initial value theorem can be given as follows:
Lt f (t ) = Lt e 4t ⋅ cos 2t
t →0
t →0
= e 0 ⋅ cos 0
=1
(7.1)
The R.H.S of the initial value theorem is as follows:
s−4
( s − 4) 2 + 4
4
1−
s
= Lt s 2 ⋅
2
s→∞

4
 4
s 2 1 −  + 2 
s
s 

292
Lt sF ( s) = Lt s ⋅
s→∞
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s→∞
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s−4
s→∞
( s − 4) 2 + 4
4
1−
2
s
= Lt s ⋅
2
s→∞

4
 4
s 2 1 −  + 2 
s
s 

4
1−
s
= Lt
2
s→∞ 
4
4
1 −  + 2
s
s
1− 0
=
=1
(1 − 0) 2 + 0
Laplace Transform
293
Lt sF ( s) = Lt s ⋅
s→∞
(7.2)
From equations (7.1) and (7.2), it is clear that Lt f (t ) = Lt sF ( s) . Hence, the initial value
t →0
t →0
theorem is verified.
Example 7.29 Find the initial value of 2e−3t cos t using the initial value theorem.
Solution: Given f(t) = 2e−3tcos t
Firstly, let us find L[ f (t)] for which
s
L [cos t ] =
2
s + 12
Now, using the first shifting property, we get the following form:
L [e −3t cos t ] =
(s + 3)
(s + 3) 2 + 1
and
L[2e −3t cos t ] =
2( s + 3)
= L[ f (t )] = F ( s)
( s + 3) 2 + 1
Now, by the initial value theorem,
Initial value of f (t ) = Lt s F ( s)
s →∞
Substituting the value of F ( s)
= Lt s
s →∞
= Lt
s →∞
M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 293
2( s + 3)
( s + 3) 2 + 12
 3
s 2 2 1 + 
 s
 3  2 1 
s 2 1 +  + 2 
s 
 s 
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Initial value of f (t ) = Lt s F ( s)
s →∞
Substituting the value of F ( s)
294 Network Analysis and Synthesis
= Lt s
s →∞
= Lt
2( s + 3)
( s + 3) 2 + 12
 3
s 2 2 1 + 
 s
 3  2 1 
s 2 1 +  + 2 
s 
 s 
 3
2 1 + 
 s
2(1 + 0)
=
=2
= Lt
2
2
s →∞
1 (1 + 0) + 0
 3
1 +  + 2
s
s
s →∞
Example 7.30 Find the initial value of the function, where the Laplace transform is given as
follows:
(s + 1)(s + 2)
s (s + 3)(s + 4)
Solution: Given
L[ f (t )] = F ( s) =
( s + 1)( s + 2)
s( s + 3)( s + 4)
Now, by the initial value theorem, we get the following:
s f ( s)
Initial value of f (t) = sLt
→∞
Substituting the given value of F (s), the following form is obtained:
= Lt s
s →∞
( s + 1)( s + 2)
s( s + 3)( s + 4)
= Lt s
s →∞
( s + 1)( s + 2)
s( s + 3)( s + 4)
 1  2 
1 +  1 + 
s
s
= Lt
s →∞ 
3  4 
1 +  1 + 
s
s
=
(1 + 0)(1 + 0)
=1
(1 + 0)(1 + 0)
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Laplace Transform
295
7.7 FINAL VALUE THEOREM
Final value theorem states that Lt f (t ) or the final value of a function f (t) is expressed as in
t →∞
the following:
f (∞) = Lt s ⋅ F ( s)
s→0
where F (s) = L[ f (t)] of any time domain function f (t).
The numericals based on the final value theorem are given as follows.
Example 7.31 Verify the final value theorem for a function
f (t) = 4 + e−t sin t
Solution: Given f (t) = 4 + e−t sin t
and
L [ f (t)] = 4 L [1] + L [e−t sin t]

1 
1
= 4⋅ + 

2
2
s  (s + 1) + 1 
Using the first shifting property, the form is obtained as follows:
=
4
1
+
s (s + 1) 2 + 12
Now, the L.H.S of the final value theorem can be written as follows:
Lt f (t ) = Lt [4 + e − t sin t ]
t →∞
t →∞
= 4 + e −∞ sin ∞
= 4 + 0.sin ∞
=4
(7.3)
Further, the R.H.S of the final value theorem can be given as in the following:

4
1
Lt sf ( s) = Lt s ⋅  +

2
s→ 0
s→ 0
 s ( s + 1) + 1


s
= Lt  4 +

2
2
s→0
( s + 1) + 1 

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296 Network Analysis and Synthesis
= 4+
= 4+
0
(0 + 1) 2 + 1
0
2
=4
(7.4)
From equations (7.3) and (7.4), it is clear that Lt f (t ) = Lt sf ( s) . Hence, the final value theot →∞
s→ 0
rem is verified.
Example 7.32 Find the final value of f (t) = 8 (2 − e−4t) using the final value theorem.
Solution: Given
f (t) = 8 [2 − e−4t]
Now,
L[ f (t )] = 8  L{2} − L{e −4t }
1 
2
= 8 −
 = f ( s)
 s s + 4
Now, by the final value theorem, we obtain the final value of f (t) = Lt s f ( s)
s →0
1 
2
= Lt s ⋅ 8  −

s→0
 s s + 4
s 

= Lt 8  2 −
s→0 
s + 4 
0 

= 8 2 −
0 + 4 

= 8[2 − 0]
= 16
Example 7.33 Find the value of a function whose Laplace transform is given as follows:
(s + 1)(s + 2)
s (s + 3)(s + 4)
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Laplace Transform
297
Solution: Given
L[ f (t )] = f ( s) =
( s + 1)( s + 2)
s( s + 3)( s + 4)
Now, by the final value theorem, we get the value of f (t) = Lt s f ( s)
s →0
( s + 1)( s + 2)
= Lt s s( s + 3)( s + 4)
s→ 0
= Lt
s→0
( s + 1)( s + 2)
( s + 3)( s + 4)
=
(0 + 1)(0 + 2)
(0 + 3)(0 + 4)
=
1
6
Example 7.34 Find the Laplace transform of unit step function.
Solution: The unit step function is defined as follows:
0; t < 0
u (t ) = 
1; t ≥ 0
Now,
∞
L [u (t )] = ∫ e − st ⋅ u (t )dt
0
∞
= ∫ e − st ⋅1dt
0
∞
= ∫ e − st dt
0
e − st
=
−s
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∞
0
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298 Network Analysis and Synthesis
Therefore,
=
e −∞ − e 0
−s
=
0 −1 1
=
−s
s
L [u (t )] =
1
s
Example 7.35 Find the Laplace transform of unit ramp function.
Solution: The unit ramp function is defined as follows:
0 ; t < 0
r (t ) = 
t ; t ≥ 0
Therefore,
∞
∞
∞
0
0
0
t [r (t )] = ∫ e − st ⋅ r (t )dt = ∫ e − st ⋅ t dt = ∫ t ⋅ e − st dt
↓↓
I II
Integrating by parts, we get the following:
t .e − st
=
−s
=0−
Therefore,
Lt [r (t )] =
∞
0
∞ − st
e
dt
−s
0
−∫
e − st
( − s) 2
∞
0
1
s2
Example 7.36 Find the Laplace transform of impulse function.
Solution: We know the following:
Unit impulse function = time derivative of unit step function
or
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s (t ) =
du (t )
dt
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Laplace Transform
Therefore,
299
 du (t ) 
L [s (t )] = L 

 dt 
= s L [u (t )] − u (0)


 dy 
∵ L  dt  = sL[ y ] − y(0) 
 


= s L [u (t )] − 0
Now, u (0) = 0, and therefore
= s L [u (t )]
1
= s⋅
s
=1
L[impulse function ] = 1
Therefore,
∵ L [u (t )] =
1
s
7.8 INVERSE LAPLACE TRANSFORM
Some useful formulas of inverse transform are provided in a tabular form.
f (s)
L−1[ f (s)]
1
s
1
1
T
s2
1
s
n
t n −1
( n − 1)!
1
s+a
e−at
1
s+a
e at
f (s)
a
2
s + a2
s
s 2 + a2
a
2
s − a2
s
2
s − a2
L−1[ f (s)]
sin at
cosh at
sinh t
cosh at
 2s 
Example 7.37 Evaluate L−1  2

 4s − 16 
Solution:
 2 s  1 −1  s  1 −1  s 
 2s 
−1
L 2
= L  2
= L  2
=L  2

s − 4 2
 4 s − 16 
 s − 22 
 4( s − 4 )  2
1
= cosh 2t
2
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300 Network Analysis and Synthesis

 −1  s 
= cosh at 
∵ L  2
2
s − a 


 1 
Example 7.38 Evaluate L−1 
3
 (s − 1) 
Solution:
 1 
1
L−−11  1 3  = e tt L−−11  13 
L  (s − 1)3  = e L  s 3 
s 
 (s − 1) 
∵ By
at −1
second
shifting
property,
L−−11[ f (s − a)]
∵ By using second shifting property,
L−1usin
[ f (sgg−second
a)] = e shifting
L [ f (property,
s )]
∵ By
usin
L
[ f (s − a)]


2
t
= et t 2
= e t 2!
2!
 1 
1
= e t L−1  3 
L−1 
3
s 
 (s − 1) 
= et
t2
2!
 −1  1 
t n −1 
∵ L  n  =

 s  ( n − 1)! 

1


Example 7.39 Evaluate the inverse Laplace transform of  2
.
 s + 2 s + 3 
1


L−1  2
=?
 s + 2s + 3 
That is,
Solution:
1
1



−1 
L−1  2
=L  2

2
2
 s + 2s + 3 
 s + 2s + 1 − 1 + 3 
(Making perfect square, by adding and subtracting the squarre of coefficient of s.)




1
1
1
−1 
−1 
= L−1 
=L 
=L 
2
2
2
2
2
 ( s + 1) − 1 + 3 
 ( s + 1) + 2 
 ( s + 1) + ( 2 ) 
Using the second shifting property, we get the following:


 e − t −1 
1
2
L  2
= e − t L−1  2
=
2
2
2
 s + ( 2) 
 s + ( 2) 
=
e −t


 a 
= sin at 
sin 2t ∵ L−1  2
2
2
s + a 


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Laplace Transform
301
s


Example 7.40 Evaluate L−1  2

 s + 2s + 5 
Solution:

s
s
s



−1 
−1 
L−1  2

=L  2
=L 
2
2
2
 s + 2s + 5 
 s + 2s + 1 − 1 + 5 
 ( s + 1) − 1 + 5 


s
= L−1 

2
 ( s + 1) + 4 


s
= L−1 
2
2
 ( s + 1) + 2 


s +1
1
= L−1 
−
2
2
2
2
(ss + 1) + 2 
 ( s + 1) + 2

 −1 

s +1
1
−L 
= L−1 
2
2
2
2
 ( s + 1) + 2 
 ( s + 1) + 2 
Using the second shifting property, the following can be obtained:
 s  −t −1  1 
= e −t L−1  2
 −e L  2

 s + 22 
 s + 22 
−t
 s  e
 2 
= e −t L−1  2
−
L−1  2 2 

2
2
s + 2 
s + s 
= e −t cos 2t −
e −t
sin 2t
2
 s+3 
Example 7.41 Evaluate L−1  2

 s + 4s + 4 
Solution:
s+3
 s+3 

−1 
L−1  2
=L  2

 s + 4s + 4 
 s + 2s + 2s + 4 


s +3
= L−1 

 (s + 2)(s + 2) 
 s+3 
= L−1 
2
 ( s + 2) 
 s + 2 + 1
= L−1 
2
 ( s + 2) 
 s+2
1 
+
= L−1 

2
( s + 2) 2 
 (s + 2)
M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 301
 1
1 
+
= L−1 
2
 ( s + 2) ( s + 2) 
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 s + 4s + 4 
302 Network Analysis and Synthesis
 s + 2s + 2s + 4 


s +3
= L−1 

(
+
2
)(
+
2
)
s
s


 s+3 
= L−1 
2
 ( s + 2) 
 s + 2 + 1
= L−1 
2
 ( s + 2) 
 s+2
1 
= L−1 
+

2
(s + 2) 2 
 (s + 2)
 1
1 
+
= L−1 
2
 (s + 2) (s + 2) 
 1 
 1 
= L−1 
+ L−1 

2
s + 2
 ( s + 2) 
1
= e −2t + e −2t L−1  2 
s 
= e −2t + e −2t ⋅ t
 s+3 
−2t
L−1  2
 = e (1 + t )
 s + 4s + 4 
 4s + 3 
Example 7.42 Evaluate L−1  2

 s + 5s + 6 
Solution:
4s + 3
 4s + 3 

−1 
L−1  2

=L  3
 s + 5s + 6 
 s + 3s + 2s + 6 
(7.5)
4s + 3 
−1 
=L 

 (s + 2)(ss + 3) 
Let
f (s ) =
4s + 3
(s + 2)(s + 3)
Let us make partial fractions of the following form:
4s + 3
A
B
=
+
(7.6)
(s + 2)(s + 3) s + 2 s + 3
To find the values of A and B, we write the equation as follows:
4s + 3
A (s + 3) + B (s + 2)
=
(s + 2)(s + 3)
(s + 2)(s + 3)
or
4 s + 3 = A (s + 3) + B (s + 2) (7.7)
Substitute s = −2 in equation (7.7), and we obtain the value of A.
−8 + 3 = A (−2 + 3) ⇒ –5 = A
In order to find B, we substitute s = –3 in equation (7.7), we get the value as follows:
–12 + 3 = 0 + B (–3 + 2)
⇒ –9 = –B or B = 9
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303
Substituting the values of A and B in equation (7.6), the following form is obtained:
4s + 3
−5
9
=
+
.
(s + 2)(s + 3) s + 2 s + 3
By substituting this value in equation (7.5), we get the equation as follows:
9 
 4s + 3 
−1  −5
L−1  2
 = L s + 2 + s + 3
 s + 5s + 6 


 −5 
 9 
= L−1 
+ L−1 


s + 2
 s + 3
 1 
−1  1 
= −5L−1 
 + 9L  s + 3 
s + 2


By taking the inverse Laplace transform, the equation can be written as follows:
 4s + 3 
− 2t
− 3t
L−1  2
 = –5e + 9e
 s + 5s + 6 


s
Example 7.43 Evaluate L−1 
2
2
 (s − 1) (s + 2) 
Solution: Now firstly, we will make partial fractions equation as in the following:
s
2
(s − 1) (s + 2)
=
A
B
C
+
+
(7.8)
2
(s − 1) (s − 1)
s+2
To find the values A, B and C
s
(s − 1) 2
or
=
A (s − 1)(s + 2) + B (s + 2) + C (s − 1) 2
(s − 1) 2 (s + 2)
S = A (s − 1) (s + 2) + B (s + 2) + C (s − 1)
(7.9)
Substituting s = 1 in equation (7.9), we get the following:
1 = 0 + B (1 + 2) + 0 ⇒ 3B = 1 ⇒ B =
1
3
Substituting s = –2 in equation (7.9), we obtain the value as follows:
−2 = 0 + 0 + C ( −2 − 1) 2 ⇒ −2 = 9C ⇒ C =
−2
9
To find the value of A, let us expand equation (7.9) as in the following:
s = A (s 2 + s − 2) + Bs + 2B + C (s 2 + 1 − 2s )
or
s = A s 2 + A s − 2A + Bs + 2B + Cs 2 + C − 2Cs
or
s = ( A + C )s 2 + ( A + B − 2C )s + ( −2A + 2B + C )
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304 Network Analysis and Synthesis
Equating the co-efficient of s2, s and constant terms on both sides, we get the following set of
equations:
A + C = 0
A + B − 2C = 1
−2A + 2B + C = 0
(7.10)
(7.11)
(7.12)
The value of A can be found from any of the above equations.
From equation (7.10), we get the following value:
A = −C
 −2  2
A = −  =
 9 9
That is,
Substituting the values of A, B and C in equation (7.5), the following form is obtained:
1
2
−2
3
9
=
+
+ 9
(s − 1) 2 (s + 2) s − 1 (s − 1) 2 s + 2
s
Taking the inverse Laplace transform of the equation, we get the following:
 2 −1  1  1 −1  1  2 −1  1 

s
− L 
L−1 
= L 
+ L 

2
2
 s − 1 3
s + 2
 (s − 1)  9
 (s − 1) (s + 2)  9
2
1
1 2
= e t + e t L−1  2  − e −2t
9
3
s  9
2
1
2
= e t + e t ⋅ t − e −2t
9
9
3


s
Example 7.44 Evaluate L−1  2

2
 (s + 1)(s + 4) 
Solution: Firstly, let us make partial fractions as in the following:
s
(s 2 + 1)(s 2 + 4)
=
As + B
s2 +1
+
Cs + D
s2 + 4
(7.13)
To find the values of A, B, C and D
s
(s 2 + 1)(s 2 + 4)
or
=
( A s + B )(s 2 + 4) + (s + D )(s 2 + 1)
(s 2 + 1)(s 2 + 4)
s = ( A s + B )(s 2 + 4) + (s + D )(s 2 + 1)(7.14)
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305
As there is no simple fraction in this case, all the fractions are quadratic. Hence, we cannot find
any value directly. We will have to expand equation (7.14) as follows:
s = A s 3 + 4 A s + Bs 2 + 4 B + Cs 3 + Cs + Ds 2 + D
s = (A + C )s 3 + ( B + D )s 2 + ( 4 A + C )s + ( 4 B + D)
Equalising the co-efficient on both sides, we get the following form:
A + C = 0
B + D = 0
4A + C = 1
4B + D = 0
(7.15)
(7.16)
(7.17)
(7.18)
From equations (7.16) and (7.18), we get the values of B and D as follows:
B + D = 0
⇒ B= D=0
4 B + D = 0
Now, solving equations (7.15) and (7.17), the values of A and C are obtained as follows:
A+C = 0
4A+ C = 1
(7.15)
1
⇒ A=
3
(7.17) 
C=
−1
3
Now substituting the values of A, B, C and D in equation (7.13), we get the equation as follows:
1
−1
s+0
s+0
3
3
=
+
(s 2 + 1)(s 2 + 4) s 2 + 1
s2 + 4
s
1
s
1
s
= ⋅ 2 2− ⋅ 2
3 s + 1 3 s + 22
Now, taking the inverse Laplace transform on both sides, the following form can be obtained:
 1 −1  S  1 −1  s 

s
L−1  2
= L  2 2− L  2

2
S + 1  3
 S + 22 
 (s + 1)(s + 4)  3
1
1
= cos t − cos 2t
3
3
1
= (cos t − cos 2t )
3
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

1
Example 7.45 Evaluate L−1 

2
 (s − 1)(s + 1) 
Solution: Let us make partial fractions of
A
Bs + C
1
=
+ 2
2
( s − 1)( s + 1) s − 1 s + 1
↓
Simple
factor
(7.19)
↓
Quadratic
factor
Let us find the values of A, B and C.
1
(s − 1)(s 2 + 1)
or
=
A (s 2 + 1) + ( Bs + c )(s − 1)
(s − 1)(s 2 + 1)
1 = A (s 2 + 1) + ( Bs + c )(s − 1)(7.20)
In this case, there is one simple factor (s − 1), and therefore, one value can be obtained by substituting s = 1 in equation (7.20)
Substituting s = 1 in equation (7.20), we get the equation as follows:
1 = A (12 + 1) + 0
or
2A = 1
or
A=
1
2
As there is no other simple factor, and therefore, other values will be obtained by expanding
equation (7.20).
1 = A s 2 + A + Bs 2 − Bs + Cs − C
L = ( A + B )s 2 + ( − B + C )s + ( A − C )
Equating the co-efficient on both sides, we get the form as follows:
A + B = 0
−B + C = 0
(7.21)
(7.22)
A–C=1
1
−1
Substituting the value of A = in equation (7.21), we get B =
2
2
Further, substituting the value of B in equation (7.22), we get C =
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(7.23)
−1
2
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307
Now, substituting the values of A, B and C in equation (7.20), we get the form as follows:
−1  −1
1
s+ 
 2
1
2 + 2
=
2
2
1
s
−
(s − 1)(s + 1)
s +1
1
1 s +1
= 2 − 2
s −1 2 s +1
1 1
s +1 
−
= 
2  s − 1 s 2 + 1
=
1 1
1 
s
− 2
− 2 

2  s − 1 s + 1 s + 1
Now, taking the inverse Laplace transform on both sides, the equation can be written as in the
following:
 1  −1  1 

1
−1  s 
−1  1  
L−1 
 − L  2  − L  2 
 = L 
2
 s + 1 
 s − 1
 s + 1
 (s − 1)(s + 1)  2 
=
1 t
e − cos t − sin t 
2


1
Example 7.46 Evaluate L−1  2

2
 (s + 2s + 2)(s + 2s + 5) 
Solution: Firstly, let us make partial fractions
1
2
2
(s + 2s + 2)(s + 2s + 5)
=
As + B
2
s + 2s + 2
+
Cs + D
2
s + 2s + 5
(7.24)
Let us find the values of A, B, C and D.
1
(s 2 + 2s + 2)(s 2 + 2s + 5)
or
=
( A s + B )(s 2 + 2s + 5) + (Cs + D )(s 2 + 2s + 2)
(s 2 + 2s + 2)(s 2 + 2s + 5)
1 = ( A s + B )(s 2 + 2s + 5) + (Cs + D )(s 2 + 2s + 2) (7.25)
Since in this case, there is no simple factor. Both the factors are quadratic. Therefore, all the
values will be obtained by expanding equation (7.25)
Expanding equation (7.25), we get the form as follows:
1 = As3 + 2 As 2 + 5 As + Bs 2 + 2 Bs + 5 B + Cs3 + 2Cs 2 2Cs + Ds 2 + 2 Ds + 2 D
or
1 = ( A + C )s 3 + ( 2A + B + 2C + D )s 2 + (5A + 2B + 2C + 2D )s + (5B + 2D )
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308 Network Analysis and Synthesis
Equating the coefficients on both sides, the following form is obtained:
and
A + C = 0
2A + B + 2C + D = 0 ⇒ 2( A + C ) + B + D = 0
⇒ 0+B +D = 0
[ using equation (7.26)]
(7.26)
⇒ B + D + 0(7.27)
and
and
5A + 2B + 2C + 2D = 0 ⇒ 5A + 2C + 2( B + D ) = 0 using equation ( 4)
⇒ 5A + 2C + 0 = 0
⇒ 5A + 2C = 0(7.28)
5B + 2D = 1
(7.29)
From equation (7.26) and (7.28), we get A = C = 0.
1
−1
By solving equations (7.27) and (7.29), we get B = and D =
3
3
Substituting the values of A, B, C and D in equation (7.24), we get the following:
1
1
1
1
0+
0+ 0− 3 0−
1
3
1
3
3
=
+
+
=
( s 2 + 2 s(+s 22+)(2s 2s +
+ 22)(
s +s 25+) 2 ss+2 5+)2 s +s 22+ 2ss2++22s +s 25+ 2s + 5
1
1
1
1
1
= . 2 = 1.
−1 . 2 − 1 .
3 s + 23s +s 22+ 23s +s 2 + 23s +s 25+ 2 s + 5
1
1
1
1
= . 2 = 1. 2 2 1 − . 2 − 1. 2 2 1
3 s + 23s +s 21+ −21s ++122 − 132 +s 2 + 23s +s 21+ −21s ++125 − 12 + 5
1
1
1
1
= .
− .
2
2
2
3 (s + 1) − 1 + 2 3 (s + 1) − 12 + 5
1
1
1
1
= .
−
3 (s + 1) 2 + 12 3 (s + 1) 2 + 22
=

1
1
1
−


2
2
2
2
3  (s + 1) + 1 (s + 1) + 2 
Taking the inverse Laplace transform, the equation is written as follows:
 1  −1 



1
1
1

−1 
L−1  2
−L 

 = L 
2
2
2
2
2
 (s + 1) + 1 
 (s + 2s + 2)(s + 2s + 5)  3 
 (s + 1) + 2 
Using the second shifting property, we get the following form:
1
 1 
 1 
= e − t L−1  2 2  − e − t L−1  2

3
s +1 
 s + 22 

e −t
1
sin 2t 
= e − t sin t −
3
2

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309
1
1
Example 7.47 Evaluate L−1  ⋅ sin 
s
s
Solution:
7
3
5
 

 1
 1
 1
 













1
1  1 

1
s
s
s
+ … 
L−1  ⋅ sin  = L−1    −
+
−
s  s 

3
!
5
!
7
!
s
s

 


 
 

x 3 x 3 x 7 …
+ 
+
−
∵ sin x = x −
3! 5! 7!


 1 1
1
1
1

= L−1   − 3
+ 5
− 7
+ …
s
s
s ⋅ 3! s ⋅ 5 ! s ⋅ 7 !

 
1
1
1
1
1

+
−
+…
= L−1  2 −
+
3!s 4 5!s 6 5!s 6 7 !s 8
s
1 1
1 1
1 1
1
= L−1  2  − L−1  4  + L−1   − L−1  8  + …
3
5
56
!
!
7
!
s 
s 
 
s 
=t −
=t −
1 t3 1 t5 1 t7 …
⋅ + . − . +
3! 3! 5 ! 5 ! 7 ! 7 !
t3
(3!) 2
+
t5
(5!) 2
−
t7
(7 !) 2
+…
1
1
Example 7.48 Evaluate L−1  cos 
s
s
Solution:
   1 2  1 4

   




 s
1
1   s

1
L−1  cos  = L−1  1 −
+
+  
s

2!
4!
s
s

 
 
 


x2 x4
+
+ 
∵ cos x = 1 −
2! 4 !


1 
1
1

= L−1  1 − 2 + 4 + 

 s  s .2 ! s .4 !
2
1
1
= L−1  −
+
+
3
s
2!s
4 !s 5

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310 Network Analysis and Synthesis
1 1
1 1
1
= L−1   − L−1  3  + L−1  5  + 2
!
4
!
s
 
s 
s 
= 1−
= 1−
1 t2 1 t4
. + . +
2! 2! 4 ! 4 !
t2
( 2 !) 2
+
t4
( 4 !) 2
+
7.9 CONVOLUTION THEOREM
The theorem states the following:
If
L−1[ f ( s)] = f (t )
and
L−1[ g ( s)] = g (t )
then
L−1[ f (s ) ⋅ g (s )] = f (t ) * g (t )
That is, convolution of f (t) and g(t) is as follows:
t
∫ f (u) g (t − u) du
0


1
Example 7.49 Evaluate L−1  2
using the convolution theorem.
2
 (s + 1) 
Solution:
 1


1
1 
= L−1  2
⋅ 2
L−1  2

2
 (s + 1) (s + 1) 
 (s + 1) 
⇓
⇓
Let
f (s )
f (s ) =
g (s ) =
Then
and
g (s )
1
s2 +1
1
s2 +1
 1 
L−1 f (s )  = L−1  2  = sin t = f (t )
 s + 1
 1 
L−1[g (s )] = L−1  2  = sin t = g (t )
 s + 1
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Laplace Transform
Now, by the convolution theorem, we get the following form:
L−1 f (s ) ⋅ g (s )  = f (t ) * g (t )
t
Therefore,
or
1 
 1
L−1  21 ⋅ 21 
−1  s + 1 s + 1 
L  2
⋅ 2 
 s + 1 s + 1
= ∫ f (u ) ⋅ g (t − u ) du
0
t
t
= ∫ sin u ⋅ sin (t − u ) du
= ∫0 sin u ⋅ sin (t − u ) du
0
t


1
L−1  2 1 2  =
−1 ( s + 1)
L  2
=
2
 ( s + 1) 
1t
12 ∫ 2 sin u sin(t − u ) du
0 2 sin u sin(t − u ) du
2 ∫0
t
1
= ∫ {cos(u − t + u ) − cos(u + t − u )}du
20
[∵ 2 sin A sin B = cos( A − B ) − cos( A + B )]
t
=
1
{cos( 2u − t ) − cos t }du
2 ∫0
=
1  sin( 2u − t )

2 
2

− cos t ⋅ | u |u = 0 t 

u =0
 1  2 sin t
1  sin t − sin( −t )

− t cos t 
= 
− cos t (t − 0) = 
2
2

 2 2
=
1
[sin t − t cos t ]
2
t


1
Example 7.50 Evaluate L−1 
 using the convolution theorem.
 (s + 1)(s + 2) 
Solution: Let
1
1
f ( s) =
and g ( s) =
s +1
s+2
Then
 1 
−t
L−1[f (s )] = L−1 
 = e = f (t )
 s + 1
 1 
−2t
L−1[g (s )] = L−1 
 = e = g (t )
s + 2
Now, by the convolution theorem, the following equations are obtained:
and
t
L−1  f ( s) ⋅ g ( s)  = f (t ) * g (t ) = ∫ f (u ) − g (t − u ) du
0
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t
or
1 
 1
− u −2( t − u )
L−1 
du
⋅
 = ∫ e ⋅e
 s +1 s + 2 0
t
or


1
−u − 2t + 2u
L−1 
du
 = ∫e
 (s + 1)(s + 2)  0
t
= ∫ e −2t + u du
0
= e −2t + u
t
0
= e −2t +t − e −2t + 0
= e − t − e −2t
Example 7.51 If f 1 (t ) = e −a t and f 2 (t ) = e − b t , then evaluate the convolution of two functions.
Solution: Given
f 1 (t ) = e −a t
f 2 (t ) = e − b t
and
Convolution of f1(t) and f2(t) = f1(t) * f2(t)
t
= ∫ f1 (u ) f 2 (t − u ) du
0
t
= ∫ e −a u ⋅ e − b ( t −u ) du
0
t
= ∫ e −a u − b t + b u ⋅ du
0
t
= ∫ e − b t + ( b −a )u du
0
t
 e − b t + ( b −a )u 
=

 ( b − a ) u =0
=
e − b t + ( b −a ).t − e − b t + 0
b −a
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Laplace Transform
=
313
e −a t − e − b t
b −a


1
Example 7.52 Evaluate L−1  2
2
 (s + 2s + 5) 
Solution:


1
1
1


= L−1  2
⋅ 2
L−1  2

2
 s + 2s + 5 s + 2s + 5 
 (s + 2s + 5) 
⇓
⇓
f (s )
Let
f (s ) =
g (s ) =
g (s )
1
2
s + 2s + 5
1
s 2 + 2s + 5
1
1



−1 
L−1 f (s )  = L−1  2
=L  2

2
2
 s + 2s + 5 
 s + 2s + 1 − 1 + 5 


1
= L−1 

2
2
 (s + 1) − 1 + 5 


1
= L−1 

2
 (s + 1) + 4 
Using the second shifting property, we get the following form:
Now,
 1 
= e −t L−1  2

 s + 22 
=
e −t −1  2 
L  2

2
 s + 22 
=
e −t
sin 2t = f (t )
2
Further,
g (t ) =
e −t
sin 2t
2
Now, by the convolution theorem, the equation can be written as follows:
t
L−1  f ( s) ⋅ g ( s)  = ∫ f (u ) g (t − u ) du
0
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and
t
e−u
e − ( t −u )sin(t − u )
1
1


L−1  2
sin
u
du
. 2
2
=
⋅
 ∫
2
 s + 2s + 5 s + 2s + 5  0 2
or
 1 t −u

1
= ∫ e sin 2u ⋅ e − ( t −u )sin( 2t − 2u ) du
L−1  2
2
(
s
s
)
+
+
2
5
 40

t
=
1 − u −t + u
e
⋅ sin 2u ⋅ sin( 2t − 2u )du
4 ∫0
=
1 −t
e ⋅ sin 2u ⋅ sin ( 2t − 2u ) du
4 ∫0
=
e −t
sin 2u ⋅ sin( 2t − 2u ) du
4 ∫0
=
e −t
2 sin 2u ⋅ sin ( 2t − 2u ) du
8 ∫0
=
e −t
{cos( 2u − ( 2t − 2u )) − cos( 2u + ( 2t − 2u ))]} du
8 ∫0
=
e −t
{cos( 4u − 2t ) − cos 2t} du
8 ∫0
=

e − t  sin( 4u − 2t ) t
t
− cos 2t u u = 0 

u=0
8 
4

=

e − t   sin( 4t − 2t ) − sin(0 − 2t ) 

 − cos 2t ⋅ {t − 0}
8  
4


=
e − t  sin 2t − sin( −2t )

− t cos 2t 
4
8 

=
e −t
8
=
e − t  2 sin 2t

− t cos 2t 

8  4

=
e − t  sin 2t

− t cos 2t 
8  2

t
t
t
t
t
M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 314
 sin 2t + sin 2t

− t cos 2t 

4


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Laplace Transform
315
R evie w Q uestions
Short Answer Type
1. What is convolution in time domain? What is the Laplace transform of convolution of two
time domain functions?
2. State the advantage of using Laplace transform in networks. Given ‘s’ domain representations for resistance, inductance and capacitance.
3. State and prove convolution theorem.
Numerical Questions
1. Define Laplace transform of a function f (t). Find the Laplace transform for the function


w
 Ans. =
2
2
(s + a) + w 

f (t ) = e − at sin w t ⋅ u(t ) 2. Determine the Laplace transform of the function


a
a
−
 Ans. 2
2
2
2
(s + a ) + a 
s +a

f (t ) = (1 − e −a t ) sin a t 3. Using convolution theorem, find the inverse laplace transform of
1
(s + 2)(s + 3)
[Ans. et −e−2t]
4. Define Laplace transform of a time function. Determine Laplace transforms for
(i) The impulse function
(ii) The unit step function
n!
1


 Ans. 1, s ,
n +1 
s
a
(
−
)


(iii) tn eat
5. Find the inverse Laplace transform for
3s 2 + 4
s (s 2 + 4 )
[Ans. (1 + 2cos 2t)]
6. State the initial value theorems in the Laplace transform. What is the value of the function
at t = 0, if
F (s ) =
4(s + 25)
[Ans. Initial value = 4]
s (s + 0 )
7. Find the Laplace transform of the function cosh w t.
M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 315
s


 Ans. 2

s −w 2 

12/4/2014 2:07:21 PM
316 Network Analysis and Synthesis
8. Find the Laplace transform of the function
n! 

 Ans. − n +1 
s 

F (t) = tn
9. Using the partial fraction method, obtain the inverse Laplace transform of
I (s ) =
10000
s (s + 250)
[Ans. − 40(1 − e −250t )]
10. State the final value theorem and using this theorem find the final value of the function
where Laplace transform is
s+6
s (s + 3)
[Ans. Final value = 2]
11. Find the Laplace transform of the function

s 2 − 16 
 Ans. − 2

(s + 16) 2 

F(t) = t cos 4t
12. Find the convolution of f1 (t) and f2 (t) when f1 (t) = e−at and f2 (t) = t


e − at
Ans.
−
[at e at − e at + 1]

2
a


13. Find the Laplace transform of the functions

4s 

 Ans. − 2
(s + 4 ) 2 

F (t) = t sin 2t 14. Find the inverse Laplace transform of
s 2 + 7s + 14
s 2 + 3s + 2
[Ans. 1 − 4e−2t + 8e−t]
15. Define the unit step, ramp and impulse function. Determine the Laplace transform for three
functions.
16. Find the inverse Laplace transform of
F (s ) =
7s + 2
3
s + 3s 2 + 2s
[Ans. 1 + 5e−t −6e−2t]
17. Find the Laplace transform of
f (t ) = e −q t cos w t 

s +q
 Ans. −

(s + q ) 2 + w 2 

18. Find the final value of the function whose Laplace transform is
s+9
[Ans. Final value = 1.8]
s (s + 5)
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317
Laplace Transform
M ultiple C hoice Q uestions
1. Laplace transform of a unit impulse function is
(a) S
(b) 0
(c) e-s
2. Laplace transforms of a damped sine wave e
(a)
1
s
(b)
(s + a ) 2 + f 2
(s + a ) 2 + f 2
3. The final value of f (t) for a given F (s ) =
(a) Zero
(b) 1/15
4. Laplace transforms of the function e-2t is
1
(a)
(b) (s + 2)
2s
5. The Laplace transform of a function is
(a) E sinw t
(b) Eeat
6. If f (t ) = r (t − a ), f ( s) =
(a)
e −a s
(d) 1
sin (q t ) ⋅ u(t ) is
(c)
f
(d)
(s + a ) 2 + f 2
f2
(s + a ) 2 + f 2
s
(s + 4)(s + 2)
(c) 1/8
(c)
(d) 1/6
1
s+2
(d) 2s
1
E e − as . The function is
s
(c) Eu(t − a)
a
s +a
(b)
2
−a t
(c)
(d) E cos w t
1
s +a
(d)
e −as
s
s
7. The integral of a step function is
(a) A ramp function (b) An impulse function
(c) Modified ramp function
(d) A sinusoidal function
8. Laplace transform of the function f (t ) = (1 − e −a t ) sin a t , where a is a constant is
(a)
(c)
s
2
s +a
2
a
s2 + a 2
−
−
s +a
2
(s + a ) + a
a
1
2
s +a
2
−
a
(s + a ) 2 + a 2
(d) NOT
(s + a ) 2 + a 2
9. Laplace inverse equation
(a) e −t − e −2t
(b)
2
1
is
(s + 2)(s + 3)
(b) e t − e −2t
(c) e t − e 2t
(d) NOT
10. Laplace transform equation tn eat is
(a)
n!
s
n +1
(b)
n!
s
n −1
M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 317
(c)
n!
(s − a)
n −1
(d)
n!
(s + a) n +1
12/4/2014 2:07:29 PM
318 Network Analysis and Synthesis
2s + 3
11. Inverse Laplace transform for
(a) 1 − e −3t
is
s 2 + 3s
(b) 1 + e −3t
3s 2 + 4
12. Inverse Laplace transform for
(a) 1 + cos 2t
s (s 2 + 4 )
(b) 1 + 3cos 2t
(c) 1 + e 3t
(d) 1 − e 3t
(c) 1 + 2cos 2t
(d) 1 + 3cos t
is
6s
is
(s + 2)(s + 1)(s + 4)
13. Inverse Laplace transform for
(a) −2e −t + 6e −2t + 4e −4t
(b) −2e −t + 6e −2t − 4e −4t
(c) −2e −t + 6e 2t + 4e −4t
(d) −2e −t + 6e −2t + 4e 4t
14. The value of function f (s ) =
(a) 10
4(s + 25)
at t = 0 is
s (s + 10)
(b) 4
(d) ∞
(c) 0
15. Laplace transforms of tn u (t) is
(a)
n!
(b)
sn
n!
(c)
s n −1
( n − 1)!
s n −1
(d)
n!
s n +1
16. Laplace transform of cosh w t u(t) is
(a)
s
2
s +w
2
(b)
w
2
s +w
(c)
2
17. Obtain the inverse Laplace transforms of I (s ) =
(a) 40(1 + e 250t )
(b) 40(1 − e 250t )
18. The final value of the function I (s ) =
(a) 1
(b) 2
w
2
s −w
2
(d)
s
2
s −w 2
10 4
s (s + 250)
(c) 40(1 − e −250t )
(d) −40(1 + e 250t )
s+6
is
s (s + 3)
(c) 3
(d) 0
19. The Laplace transform of t cos 4t is
(a)
s 2 − 16
(s 2 + 16) 2
(b)
s 2 + 16
(s 2 + 16) 2
M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 318
(c)
s2
(s 2 + 16) 2
(d) NOT
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319
Laplace Transform
20. If f1 (t ) = e − at and f 2 (t ) = t , then convolution of f1 (t) and f2 (t0) is
(a)
e at 
at e at + e at + 1
2
a
(c)
e − at 
at e at − e at + 1
a2 
e − at  − at
at e − e at + 1
2 
a
e − at  at
(d) 2 at e − e at − 1
a
(b)
21. Laplace transform of cos2 t is
(a)
2s + 4
2s (s 2 + 4)
(b)
s2 + 4
2s (s 2 + 4)
(c)
2s 2 + 4
s (s 2 + 4 )
(d)
2s 2 + 4
2s (s 2 + 4)
22. Laplace transform of t sin 2t is
(a)
4s
2
(s + 4 )
23. If V (s ) =
2
s 2 + 7s + 14
s 2 + 3s + 2
(a) 1 + 4e −2t + 8e −t
24. If I (s ) =
(b)
s
2
(s + 4 )
2
(c)
4s
(d)
2
s +4
4s
2
(s + 4 )3
, then V (t ) = ?
(b) 1 − 4e −2t + 8e −t
(c) 1 − 4e −2t − 8e −t
(d) 1 − 4e −2t + 8e −t
(c) −e −t + 4e −4t
(d) −e −t − 4e −4t
3s
, then i (t) is
(s + 1)(s + 4)
(a) e −t + 4e −4t
(b) e −t + 4e 4t
25. The inverse Laplace transforms for f (s ) =
(a) 1 + e −t − 6e −2t
(b) 1 + 5e −t − 6e −2t
7s + 2
3
s + 3s 2 + 2s
is
(c) 1 − 5e −t − 6e −2t
(d) 1 + 5e t + 6e −2t
ANS W E RS
1. d 2. c 3. a 4. c 5. c 6. a 7. a 8. c 9. b
11. b 12. c 13. b 14. b 15. d 16. d 17. c 18. b 19. a
21. d 22. a 23. b 24. c 25. b
M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 319
10. c
20. c
12/4/2014 2:07:40 PM
Transient Response
of Circuits Using
Laplace Transform
8
Chapter objectives
After carefully studying this chapter, you should be able to do the following:
List the steps to find transient response
DC excitation by Laplace transform
method.
of electrical networks using Laplace
transform.
Solve numerical problems on tranWrite differential equations of circuit
sient analysis of R–C, R–L, and R–L–C
variables in time domain and convert
series circuit using Laplace transform
them into Laplace transform form.
method.
Determine transient response of R - C
Make analysis of R–C and R–L series
circuit using Laplace transform and
circuits with sinusoidal input by Laplace
appreciate the method.
transform method.
Determine transient response of R - L
Solve numerical on transient analycircuit using Laplace transform with
sis of circuits with sinusoidal input
DC excitation.
voltage by Laplace transform method
and appreciate the method over timeCarryout transient response analydomain analysis method.
ses of R–L–C series circuit with
8.1 steps to find transient response using laplace
transform
Laplace transform is a very useful mathematical tool used in determining the transient response
of electrical networks.
The basic procedure in the form of steps to find the transient response using Laplace transform are mentioned in the following:
Step 1:
Step 2:
Step 3:
Step 4:
write the differential equation of the circuit in time domain, using KVL.
take Laplace transform of the differential equation.
find I (s) or V(s).
take the inverse Laplace transform to find i(t) or v(t).
We will first express the circuit elements in s-domain and then proceed to write the differential
equation of circuits and take their Laplace transform.
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Transient Response of Circuits Using Laplace Transform
8.2 CIRCUIT ELEMENTS IN THE s-DOMAIN
i(t )
+
8.2.1 Resistor in the s-Domain
321
R
ν(t )
Figure 8.1 R
esistor in Time
Domain
Consider a resistor R in a circuit, as shown in Figure 8.1, where
the current and voltage are shown in time domain as i(t) and
v(t), respectively.
By ohm’s Law, we get the following form:
V = iR
Taking Laplace transform on both sides, we get the following equation:
L[v] = RL[i]
or
V(s) = RI(s)
(8.1)
This equation shows that the resistance element does not change
while going from t-domain to s-domain.
The s-domain equivalent circuit of Figure 8.1 is shown in
Figure 8.2.
I(s)
R
+
ν(s)
−
Figure 8.2 Resistor in
s-Domain
8.2.2 Inductor in s-Domain
Let us consider an inductor with initial current I0 as shown in Figure 8.3. When a changing
­current is flowing through the inductor.
di
We can write,
v=L
dt
Taking the Laplace transform of the equation, we get the following equation:
V(s) = L[sI(s) − i(0)]
i
a +
V(s) = L[sI(s)−I0]
V(s) = sLI(s) − LI0(8.2)
Equation (8.2) can be implemented using the circuit shown in
Figure 8.4.
If I0 is taken as 0, the equivalent of inductor in s-domain will
be as shown in Figure 8.5.
+
Now, in t-domain, i = C
dv
dt
M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 321
−
b
SL
LI0
−+
I(s)
V(s)
−
Figure 8.4 Inductor in
s-Domain
8.2.3 Capacitor in s-Domain
Consider a capacitor in a circuit where voltage and current
are expressed in t-domain, as shown in Figure 8.6. Let the
initial voltage on the capacitor be v0 volts
ν
Figure 8.3 Inductor in Time
Domain
Now, we have assumed i(0) = I0
Therefore,
I0
L
V(s)
SL
+
I (s)
−
Figure 8.5 Inductor in s-Domain
Under Zero Initial
Condition
11/17/2014 5:19:28 PM
322 Network Analysis and Synthesis
+
+
V0
C
−
−
I(t ) V(t )
Figure 8.6 C
apacitor Placed in
t-Domain Circuit
I(s)
V0
s
+−
1
Cs
V
v(0) = V0
I(s) = C[sV(s) − V0]
V
I(s)
I(s) = C[sV(s) − v(0)]
Substitute
I(s) = Cs V(s) − C V0
Figure 8.7 Capacitor in
s-Domain
+
Taking the Laplace transform on both sides, we get the
following:
 dv 
L [i(t )] = CL  
 dt 
V ( s) =
or
−
I ( s) V0
+ (8.3)
Cs
s
This equation can be represented by the circuit shown in
Figure 8.7.
If
V0 = 0,
1
Cs
Figure 8.8 Capacitor in s-Domain
when Initial Voltage is
Zero
1
I ( s) (8.4)
Cs
If initial capacitor voltage is zero, that is, V0 = 0, then the equivalent circuit in s-domain will be
as shown in Figure 8.8.
Table 8.1 provides a summary of representation of circuit elements in t-domain and s-domain.
V ( s) =
Table 8.1 Circuit Elements
Element
t-Domain Circuit
R
Resistor
i(t )
+
i(t )
R
−
V(t )
L
Inductor
R
+
−
I(s) V(s)
I0
+ V(t ) −
CV (t )
Capacitor
s-Domain Equivalent
Circuit (When Initial
Condition is Zero)
s-Domain Circuit
+ −
i(t ) V
0
+
+
−
I(s) V(s)
sL
LI0
−+
I(s)
V(s)
I(s)
V
+−
1
Cs
V0
s
sL
−
+
i(s )
−
V
+
I(s) 1
Cs
−
8.3 DC RESPONSE OF R–C SERIES CIRCUIT
Let us consider a series R–C circuit as shown in Figure 8.9. When switch K is closed, the initial
transient response will be there. The transient response will die down and steady-state values of
current and voltage will exist.
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323
Transient Response of Circuits Using Laplace Transform
Let switch K is closed at t = 0.
Let the initial capacitor voltage is zero. The time varying current is
expressed as i(t).
Applying KVL, the following can be obtained:
1
i(t )dt
C∫
Taking the Laplace transform on both sides, we get the following:
V = Ri(t )
K
R
+
V
−
i(t )
C
Figure 8.9 R–C Series
Circuit
V
1 I ( s)
= RI ( s) +
s
C s
or
or
V 
1
= R +  I ( s)
s 
Cs 
V  RCs + 1
=
I ( s)
s  Cs 
Cs V
RCs +1 s
I ( s) =
CV
RCs + 1
=
CV
I ( s) =
1 

RC  S +


RC 
Taking the inverse Laplace transform, we get the following form:


V -1  1 
i (t ) = L 

R
s+ 1 
RC 

-t
i (t ) =
V RC
(8.5)
e
R
This is the required transient response of DC series R–C circuit in time domain.
Example 8.1 Find the time domain current response i(t) of the circuit shown in Figure 8.10.
s
Solution: Apply KVL, we get the equation as follows:
1
20 = 10i(t ) +
idt
0.1 ∫
M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 323
10 Ω
0.1 F
20 V
i(t )
Figure 8.10
11/17/2014 5:19:34 PM
324 Network Analysis and Synthesis
Taking the Laplace transform on both the sides, we get the equation as follows:
20
I ( s)
= 10 I ( s) + 10
s
s
or
20 
10 
= 10 +  I ( s)
s 
s
or
20 (10 s + 10)
I ( s)
=
s
s
or
20 = 10 (s + 1) I(s)
or
I (s) =
20
10( s + 1)
I ( s) =
2
s +1
Taking the inverse Laplace transform on both the sides, the following form is obtained:
 1 
L-1[ I ( s)] = 2 L-1 

 s + 1
or
i(t) = 2e−t A
8.4 DC RESPONSE OF R–L SERIES CIRCUIT
Let us consider a series R–L circuit as shown in Figure 8.11. Let at t = 0, switch s is closed and
DC voltage V can be applied.
Let at t = 0, i = 0
R
L
Applying KVL, we get the equation as follows:
s
di(t )
dt
V
i(t )
Taking the Laplace transform of the equation, we get the following form:
Figure 8.11 R–L Series
V
Circuit
= RI ( s) + L[ sI ( s) - i(0)]
s
Substitute i(0) = 0, the equation is written as follows:
V = Ri(t ) + L
V
= RI ( s) + L[ sI ( s) - 0]
s
= (R + Ls) I (s)
or
I ( s) =
V
s( Ls + R)
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Transient Response of Circuits Using Laplace Transform
=
I ( s) =
325
V
R

Ls  s + 

L
1
V
R
L 
ss + 

L


V  L /R L /R 
= 
R
L s
s+ 

L
(using the partial fractions)


1 
V L 1
=
 R
L R s
s+ 

L
Taking the inverse Laplace transform, we get the following form:


1 
V -1  1
i (t ) = L  
R
s s + R 

L
or
i (t ) =
(
R
- t
V
1- e L
R
)
(8.6)
Example 8.2 A series R–C circuit with R = 30 Ω and L = 15 H has a constant voltage V = 60
V applied at t = 0 as shown in Figure 8.12. Determine current i.
Solution: Applying KVL, we get the equation as follows:
60 = 30i(t ) + 15
di(t )
dt
Taking the Laplace transform on both sides, we obtain the equation
as follows:
30 Ω
s
60
= 30 I ( s) + 15sI ( s)
s
60
= (30 + 15s) I ( s)
s
M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 325
60 V
15 H
i (t )
Figure 8.12
11/17/2014 5:19:37 PM
326 Network Analysis and Synthesis
I ( s) =
or
60
s(15s + 30)
=
60
15s( s + 2)
=
4
s ( s + 2)
B 
A
I ( s) =  +
 (8.7)
 s s + 2
4
4
=
= 2 (using the partial fractions)
s ( s + 2) s = 0 s + 2 s = 0
Now,
A = s⋅
and
B = ( s + 2) ⋅
4
4
4
=
=
= -2
s( s + 2) s=-2 s s=-2 -2
Substituting the values of A = 2 and B = −2 in equation (8.7), we get the following:
I ( s) =
2
2
s s+2
Taking the inverse Laplace transform, the following form is obtained:
1
 1 
i(t ) = 2 L-1   - 2 L-1 

s
 s + 2
= 2 × 1 − 2 × e−2t
= 2 − 2 × e−2t
i(t) = 2(1 − e−2t ) A
8.5 DC RESPONSE OF AN R–L–C SERIES CIRCUIT
Applying KVL in the circuit shown in Figure 8.13, we can write the equation as follows:
V = Ri(t ) + L
R
L
C
s
V
i (t )
Figure 8.13 R–L–C Series
Circuit
di 1
+ idt (8.8)
dt c ∫
Taking the Laplace transform on both sides, we get (assuming
initial conditions to be zero) the following form:
V
1 I ( s)
= RI ( s) + LS I ( s) +
s
C s
V 
1
or
= R + sL +  I ( s)
s 
Cs 
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327
V  RCs + s 2 LC + 1
=
 I ( s)
s 
Cs

( s 2 LC + RCs + 1)
I ( s)
V=
C
or
I ( s) =
or
=
VC
2
s LC + RCs + 1
VC
1 
 2 R
LC  s + s +


L
LC 
V
=
L( s + a )( s + b )
=
V
1
⋅
L ( s + a )( s + b )
=
K 
V  K1
+ 2
L  s + a s + b 
Let
s2 +
R
1
= ( s + a )( s + b )
s+
L
Ls
(using the partial fractions)
Taking the inverse Laplace transform, the equation can be written as follows:
i (t ) =
 1 
V
-1  1 
+ K 2 L-1 
 K1 L 


L
s +a 
 s + b 
i (t ) =
V
K e -a t + K 2 e - b t 
L 1
(8.9)
Example 8.3 The circuit shown in Figure 8.14 consists of resistance, inductance and
capacitance in series with a 100 V constant source. When the switch is closed at t = 0, find the
transient current.
Solution: Applying KVL, the following form can be obtained:
100 = Ri + L
di 1
+
idt
dt C ∫
= 20i + 0.05
di
1
+
idt
dt 20 × 10 -6 ∫
R
s
6
di 10
idt
= 20i + 0.05 +
dt 20 ∫
di 1000000
= 20i + 0.05 +
idt
dt
20 ∫
M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 327
20 Ω
100 V
i (t )
L
0.05 H
C
20 µF
Figure 8.14
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328 Network Analysis and Synthesis
di
+ 50, 000∫ idt
dt
Taking the Laplace transform on both sides, the equation can be written as follows:
100 = 20i + 0.05
100
I ( s)
= 20 I ( s) + 0.05sI ( s) + 50, 000
s
s
50, 000 

= I ( s)  20 + 0.05s +
s 

100  20 s + 0.05s 2 + 50, 000 
=
 I ( s)
s
s


or
or
I ( s)[0.05s 2 + 20 s + 50, 000] = 100
I ( s) =
=
=
=
=
=
=
=
100
2
0.05s + 20 s + 50, 000
100
20
50, 000 

0.05  s 2 +
s+
.
0
05
0.05 

100
2
0.05( s + 400 s + 106 )
2000
2
s + 400 s + 106
2000
s 2 + 400 s + ( 200) 2 - ( 200) 2 + 10, 00000
2000
2
( s + 200) - 40000 + 10, 00000
2000
( s + 200) 2 + 9, 60, 000
2000
( s + 200) 2 + (979.8) 2
Taking the inverse Laplace transform on both sides, the following form can be obtained:


2000
i(t ) = L-1 
2
2
 ( s + 200) + (979.8) 
Using the second shifting property, the equation can be written as follows:


2000
= e -200t L-1  2
2
 s + (979.8) 
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Transient Response of Circuits Using Laplace Transform


1
= 2000e -200t L-1  2
2
 s + (979.8) 
1
 1  1
= 2000e -200t
× sin 979.8t ∵ L-1  2
 = sin at
979.8
 s + a2  a
=
2000 -200t
e
sin 979.8t
979.8
i(t) = 2.04 e−200t sin 979.8t
8.6 SINuSOIDAL RESPONSE OF R–L SERIES CIRCUIT
We will now find the transient response of circuits to sinusoidal inputs.
Consider the R–L circuit shown in Figure 8.15.
Applying KVL, the following equation is obtained:
Vm sinwt = Ri(t ) + L
R
di
dt
s
Vmsin w t
Taking the Laplace transform on both sides, we get the
equation as follows:
Vm L [sin w t ] = RI ( s) + LsI ( s)
or
Vm
w
2
s +w 2
=
or I ( s) =
wVm
L
Figure 8.15 Transient Response
of R–L Circuit to
Sinusoidal Input
= ( R + sL) I ( s)
I ( s) =
or
L
i(t )
wVm
( s 2 + w 2 )( R + sL)
wVm
( s + jw )( s - jw )( R + sL)
1
R

( s + jw )( s - jw )  s + 

L


wVm 
A
B
C 
=
+
+


R 
L  ( s + jw ) ( s - jw ) 
s
+



L  
(8.10)
(using the partial fractions)
Now, using the partial fractions, we get the equation as follows:
1
R

( s + jw )( s - jw )  s + 

L
M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 329
=
A
B
C
+
+
R
( s + jw ) ( s - jw ) 
 s + 
L
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330 Network Analysis and Synthesis
Therefore, the value of A = ( s + jw ) ⋅
=
=
=
=
A=
=
=
B=
R

( s + jw )( s - jw )  s + 

L
1
R

( s - jw )  s + 

L
s = - jw
s = - jw
1
R

( - jw - jw )  - jw + 

L
1
R

- j 2w  - jw + 

L
=
1
 jw L + R 
- j 2w 


L
L
L
=
- j 2w ( - jw L + R ) - j 2w ( R - jw L )
L
- j 2w ( R - jw L )
The value of B = ( s - jw ) ⋅
=
1
1
R

( s + jw )( s - jw )  s + 

L
1
R

( s + jw )  s + 

L
s = jw
s = jw
1
R

( jw + jw )  jw + 

L
1
 jw L + R 
j 2w 


L
L
j 2w ( R + jw L )
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Transient Response of Circuits Using Laplace Transform
R

The value of C =  s +  ⋅

L
=
1
R

( s + jw )( s - jw )  s + 

L
s= -
R
L
1
( s + jw )( s - jw ) s = - R
L
=
=
1
s2 + w 2
s= -
1
2
 R
2
  + w
L
R
L
=
L2
2
2 2
R +w L
or C =
L2
2
R + w 2 L2
Substituting the values of A, B and C in equation (8.10), we get the equation as follows:


L
L
L2

wVm - j 2w ( R - jwL) j 2w ( R + jwL) R 2 + w 2 L2 


I ( s) =
+
+
R 
s - jw
s + jw
L 
s
+

L 

1
1


wL
 - j 2( R - jwL) j 2( R + jwL)
2
2 2 
= Vm 
+
+ R +w L 
R 
s + jw
s - jw

s+

L 
Vm
Vm
VmwL
I ( s) =
+
+
R
- j 2( s + jw )( R - jwL) j 2( s - jw )( R + jwL)

( R 2 + w 2 L2 )  s + 

L
Taking the inverse Laplace transform of the equation on both sides, the following forms are
obtained:
R
i (t ) =
- t
Vm
Vm
V wL
⋅ e - jw t +
⋅ e jw t + 2 m 2 2 ⋅ e L
- j 2( R - jw L)
j 2( R + jwL)
(R + w L )
R
=
- t
Vm
Vm ( R - jw L)
V wL
( R + jw L)
⋅
e - jw t +
e jw t + 2 m 2 2 e L
- j 2 ( R - jw L)( R + jw L)
j 2( R + jw L)( R - jw L)
(R + w L )
= j
Vm ( R + jw L)
2( R 2 + w 2 L2 )
e - jw t - j
Vm ( R - jw L)
2( R 2 + w 2 L2 )
e jw t +
Vmw L
( R 2 + w 2 L2 )
e
R
- t
L
R
- t
 j ( R + jw L) - jw t
V (w L)
( R - jw L) jw t 
=
-j
e  + 2m 2 2 e L
e

 ( R + w L )
2 R 2 + w 2 L2  R 2 + w 2 L2
R 2 + w 2 L2
Vm
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332 Network Analysis and Synthesis
Substitue
R + jw L
2
2 2
R +w L
= e jq = cos q + i sin q , we get the following equation as follows:
cosq =
R
R 2 + w 2 L2
wL
sin q =
R 2 + w 2 L2
wL
q = tan -1
R
or
Further,
R - jw L
2
=
=
Vm
2 R 2 + w 2 L2
Vm
2 R 2 + w 2 L2
Vm
2
2 R + w 2 L2
+
=
=
=
wL
R
= e - jq
2 2
R +w L
=
and tan q =
Vm sin q
R 2 + w 2 L2
×  jeiq e - jw t - je - jq e jw t  + Vm ⋅
Vm
R 2 + w 2 L2
substituting q = tan -1
R 2 + w 2 L2
e
e
R
- t
L
R
- t
L
× j[cos(q - w t ) + j sin(q - w t ) - {cos(q - w t ) - j sin(q - w t )}]
e
2 ( R 2 + w 2 L2 )
R 2 + w 2 L2
R 2 + (w L) 2 R 2 + (w L) 2
Vm sin q
× j[e j (q -w t ) - e - j (q -w t ) ] +
Vm
Vm
wL
R
- t
L
× j[2 j sin(q - w t )] +
× [ - sin (q - w t )] +
sin (w t - q ) +
Vm sin q
R 2 + w 2 L2
Vm sin q
R 2 + w 2 L2
Vm sin q
( R 2 + w 2 L2 )
e
e
e
R
- t
L
R
- t
L
R
- t
L
wL
, we get the following equation:
R
R
Vm
w L
w L - Lt


i (t ) =
sin  w t - tan -1
sin  tan -1
e
 +
2
2 2
2
2
2


R
R 
R +w L
R +w L
This is the expression for the transient response.
Vm
M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 332
(8.11)
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Transient Response of Circuits Using Laplace Transform
8.7 SINUSOIDAL RESPONSE OF R–C SERIES CIRCUIT
Consider the R−C circuit with sinusoidal input as shown in Figure 8.16.
Applying KVL, we get the following form:
Vm sin w t = Ri(t ) +
1
idt
C∫
R
s
Taking the Laplace transform on both sides, the equation
can be written as follows:
Vm ⋅
w
s2 + w 2
= RI ( s) +
1 I ( s)
⋅
C s
1

=  R +  I ( s)
Cs


Vm ⋅
or
Vmsin w t
C
i(t )
Figure 8.16 R
–C Series Circuit
with Sinusoidal
Input
 RCs + 1
=
 I ( s)
 Cs 
V ⋅ w Cs
I ( s) = 2 m2
( s + w )( RCs + 1)
w
s2 + w 2
=
Vm ⋅ w Cs
1 

RC ( s 2 + w 2 )  s +


RC 
=
Vmw
⋅
R
=
Vmw
⋅
R
I ( s) =
S
1 

( s2 + w 2 )  s +


RC 
S
1 

( s + jw )( s - jw )  s +


RC 
Vmw  A
B
C 
⋅
+
+
1 
R
s + jw s - jw
s+


RC 

(8.12)
Firstly, let us find the values of A, B and C
The value of A = ( s + jw ) ⋅
=
M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 333
S
1 

( s + jw )( s - jw )  s +


RC 
S
1 

( s - jw )  s +


RC 
s = - jw
s = - jw
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334 Network Analysis and Synthesis
- jw
1 

( - jw - jw )  - jw +


RC 
- jw
=
1 

- j 2w  - jw +


RC 
=
1
(1 - jw RC )
2
RC
RC
A=
2(1 - jw RC )
=
or
The value of B = ( s - jw ) ⋅
=
S
1 

( s + jw )( s - jw )  s +


RC 
S
1 

( s + jw )  s +


RC 
s = jw
s = jw
jw
=
1 

( jw + jw )  jw +


RC 
jw
=
jw RC 

j 2w 1 +


RC 
B=
RC
2(1 + jw RC )
1 

The value of C =  s +
⋅

RC 
=
=
S
1 

(s + jw )(s - jw )  s +


RC 
S
2
s +w 2
-1
RC
1
w2
R 2C 2
M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 334
s=-
=
s=-
1
RC
1
RC
-1
RC
- RC
1 + w 2 R 2C 2
=
R 2C 2
1 + w 2 R 2C 2
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Transient Response of Circuits Using Laplace Transform
335
Now, substituting the values of A, B and C in equation (8.12), the following form can be
obtained:
- RC


RC
RC
V m ⋅ w  2(1 - jw RC ) 2(1 + jw RC ) 1 + w 2 R 2C 2 

⋅
+
+
I (s ) =
1
s - jw
R
 s + jw

s
+


R
C


V mw C
V mw C
V mw c
1
1
=
⋅
+
⋅
⋅
2(1 - jw RC ) s + jw 2(1 + jw RC ) s - jw 1 + w 2 R 2C 2
1
s+
1
RC
Taking the inverse Laplace transform on both sides, the following equation can be obtained:
1
i (t ) =
Vmw C
Vmw C
Vmw C
t
e - jw t +
e jw t e RC
2
2
2
2(1 - jw RC )
2(1 + jw RC )
1+ w R C
t
=
=
Vmw C (1 + jw RC )
Vmw C
Vmw C
(1 - jw RC ) jw t
e - jw t +
⋅
e
e RC
2
2
2
2(1 - jw RC )(1 + jw RC )
2(1 + jw RC ) 1 - jw RC
1+ w R C
Vmw C (1 + tw RC )
2(1 + w 2 R 2C 2 )
e - jw t +
Vmw C (1 - jw RC )
2(1 + w 2 R 2C 2 )
e jw t -
Vmw C
1 + w 2 R 2C 2
e
-
t
RC
or
i (t ) =
=
=
=
=
t 
 (1 + jw RC )e - jw t (1 - jw RC )e jw t
RC 

×
+
e
2
2
1 + w 2 R 2C 2 

Vmw C
t 
 (1 + jw RC )e - jw t (1 - jw RC )
Vmw C
×
+
e jw t - e RC 
1 
2
2


w 2C 2  R 2 + 2 2  
w C 

t 
 (1 + jw RC )e - jw t (1 - jw RC )
×
+
e jw t - e RC 
1 
2
2


w C  R 2 + 2 2  
w C 

Vm
Vm
R2 +
1
w 2C 2
t
 (1 + jw RC )
1 - RC 
(1 - jw RC ) jw t

e - jw t +
e
e
×
2w C
wc
 2w C

t
 1  1


 1

1 - RC 

e
×  
+ jR e - jw t + 
- jR e jw t  1   2  w C
wC

wC

 2


 R + 2 2  
w C
Vm
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336 Network Analysis and Synthesis
 

1

 

 - jw t
R
wC
 

+j
+
e
 

1
1 
2
2
R + 2 2
  R + 2 2



Vm
w C
w C
1

× 
=


2
1
1
1



R2 + 2 2  
t
 jw t

R
wC
wC
w C
 
-j
e RC  
e
 

1
1
1

  R 2 + 2 2
R 2 + 2 2 
R2 + 2 2
 
w C
w C 
w C
 

1



 e - jw t + e jw t 
 e - jw t - e jw t  
R
wC

+ j2




2
2j
1 
1 


 2
2
R + 2 2
 R + 2 2

C
C
w
w
Vm

=
×


1
1
R2 + 2 2 

t
wC
w C

e RC


1
2
R + 2 2


w C




1
1

t 
Vm
R
wC
wC
⋅ cos w t = j
=
×
e RC 
sin w t 

1
1
1
1

R2 + 2 2
R2 + 2 2
R 2 2 2  R 2 +  2 2 
w C 
w C
w c
w C


1
wC
Substitute
= sin q
1
2
R + 2 2
w C
Further,
R
= cos q
1
2
R + 2 2
w C
1
tan q =
w CR
or
1
q = tan -1
w CR
We can write the following equation as follows:
t 

Vm
i (t ) =
× sin q ⋅ cos w t + cos q sin w t - sin q e RC 
1


R2 + 2 2 
w C
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Transient Response of Circuits Using Laplace Transform
=
Vm
R2 +
Substituteq = tan -1
or
i (t ) =
1
w C2
2
337
t 

× sin(w t + q ) - sin q e RC 


1
, we get the equation as follows:
w CR
t
 
 -1 1  - RC 
1 

sin
tan
e
× sin  w t + tan -1

2
w CR 
w CR 




 1 
R2 + 

wC
Vm
(8.13)
The equation is the transient response of R−C series circuit to sinusoidal input.
Example 8.4 Find the current through the resistor and the
capacitor using Laplace transform for the circuit shown in
Figure 8.17. The switch is closed at t = 0 and initial change in
the capacitor is zero.
Solution: Applying KVL to the given circuit, the following
equation can be obtained:
10 = 2i +
1
100 × 10 -6
s
2Ω
100 µF
10 V
Figure 8.17
∫ idt
Taking the Laplace transform on both sides, we get the equation as follows:
or
10
1 I ( s)
= 2 I ( s) + -4
s
s
10
10
I ( s)
= 2 I ( s) + 10000
s
s
10
10000


= I ( s)  2 +
s
s 

or
10
 2 s + 10000 
= I ( s) 

s
s


or
10 = I(s)[2s + 10000]
10 = 2(s + 5000)I(s)
10
I ( s) =
2( s + 5000)
5
or
I ( s) =
s + 5000
Taking the inverse Laplace transform, the equation can be written as follows:
or
I(t) = 5e−5000t A
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338 Network Analysis and Synthesis
Example 8.5 For the given circuit (as shown in Figure 8.18), find the complete solution
for current i(t) using Laplace transformation. Assume zero charge across the capacitor before
switching.
Solution: By applying KVL, the equation is written as follows:
10 sin(10t ) = 1⋅ i(t ) +
2
1
1
idt
1∫
Taking the Laplace transform on both sides, the following form can
be obtained:
1Ω
K
10 ⋅
1F
10 sin(10t )
= I ( s) +
I ( s)
s
 1
= I ( s) 1 + 
s + 10
 s
 s + 1
= I ( s) 

 s 
2
2
100 s
I ( s) =
or
s + 10
2
100
or
Figure 8.18
10
2
( s + 1)( s 2 + 10 2 )
Using the partial fraction of the equation, the following form is obtained:
=
A
Bs + C
(8.14)
+ 2
s + 1 s + 100
Let us find the values of A, B and C.
100 s
2
( s + 1)( s + 100)
or
=
A
Bs + C
+
s + 1 s 2 + 100
100s = A (s2 + 100) + (Bs + C) (s + 1)
= As2 + 100 A + Bs2 + Bs + Cs + C
100s = (A + B) s2 + (B + C) s + (100A + C)
By comparing both sides, we get the following equation:
A + B = 0
(8.15)
B + C = 100
(8.16)
100 A + C = 0
(8.17)
Substituting B = −A (from equation (8.15)) in equation (8.16), the equation can be written as
follows:
B − 100 A = 100
or
−100 A + B = 100
M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 338
(8.18)
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339
Transient Response of Circuits Using Laplace Transform
By solving equations (8.15) and (8.18), we get the form as follows:
A+B=0
−100 A + B = 100
101A = −100
-100
= -0.99
101
or
A=
or
A = − 0.99
From equation (8.17), we get the value of C.
C = −100A
= −100 (−0.99)
C = 99
Further, from equation (8.15), the value of B is given as follows:
B = − A; B = 0.99
Substituting the values of A, B and C in equation (8.14), the equation is written as follows:
I ( s) =
-0 ⋅ 99 0 ⋅ 99 + 99
+ 2
s +1
s + 100
99
-0 ⋅ 99 0 ⋅ 99 s
+
+
s + 1 s 2 + 10 2 s 2 + 10 2
Taking the inverse Laplace transform, we get the following form:
I ( s) =
s
 1 

 1 
-1 
i(t ) = -0.99 L-1 
+ 99 L-1  2

 + 0.99 L  2
2
s
+
1
 s + 10 2 


 s + 10 
99
= -0.99e - t + 0.99 cos 10t + sin 10t
10
= − 0.99e−t + 0.99 cos 10t + 0.99 sin 10t
Example 8.6 In the circuit shown in Figure 8.19, switch is closed at t = 0. Find the current in
the circuit at any time t using Laplace transform.
Solution: Applying KVL to the circuit, we get the following
form:
V = Ri + L
di
dt
M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 339
20 V
+
s
3Ω
0.5
−
Figure 8.19
11/17/2014 5:19:55 PM
340 Network Analysis and Synthesis
di
dt
Taking the Laplace transform on both sides, the equation can be written as follows:
20
= 3I ( s) + 0.5sI ( s)
s
20
or
= (3 + 0.5s) I ( s)
s
20
I ( s) =
or
s(3 + 0.5s)
20
=
3 

0.5s  s +


0.5 
40
=
s( s + 6 )
Using the partial fractions of the equation, we get the following form:
A
B
I ( s) = +
s s+6
40
40
A=
s=0=
6
s+6
Now,
20
=
3
40
40
s = -6 =
B=
s
-6
-20
=
3
By substituting the values, we obtain 20 = 3i + 0.5
Substituting the values of A and B in the equation of I(s), we get the equation as follows:
20 -20
I ( s) = 3 + 3
s s+6
20  1
1 
=
3  s s + 6 
Taking the inverse Laplace transform, the equation can be written as follows:
i (t ) =
20
(1 - e -6t )A
3
Example 8.7 In the circuit shown in Figure 8.20, switch s is
closed at t = 0. Find the voltage across the inductor as a function of
time using the Laplace transform.
M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 340
s
1Ω
1V
1H
i(t )
Figure 8.20
11/17/2014 5:19:56 PM
Transient Response of Circuits Using Laplace Transform
341
Solution: Applying KVL to the circuit, V can be calculated as follows:
V = Ri + L
di
dt
di
dt
Taking the Laplace transform on both sides, we get the following form:
Substituting values 1 = Li + 1
1
= I ( s) + sI ( s)
s
1
or
= (1 + s) I ( s)
s
1
or
I ( s) =
s( s + 1)
Using the partial fractions of the equations, the following equation is obtained:
1
1
I ( s) = s s +1
Taking the inverse Laplace transform, we get the equation as follows:
1
 1 
i(t ) = L-1   - L-1 

s
 s + 1
Therefore,
= 1 - e -t
di
dt
d
= 1⋅ (1 - e - t )
dt
= (0 − e−t (−1))
Voltage across inductor = L
= e−t V.
R
Example 8.8 In the circuit shown in Figure 8.21, the inductor is
initially relaxed. Find the current flowing in the circuit.
Solution: Let the current flowing through the circuit be i(t).
Applying KVL, we get the following form:
d (t ) = Ri(t ) + L
d (t )
L
Figure 8.21
di
dt
Taking the Laplace transform on both sides, the equation can be written as follows:
 d (t ) 
L[d (t )] = RL[i(t )] + L  L

 dt 
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342 Network Analysis and Synthesis
1 = RI(s) + Ls I(s)
[∴ Laplace transform of impulse function is 1 so L[d (t)] = 1.]
1
sL + R
1
I ( s) =
R

Ls + 

L
Taking the inverse Laplace transform, the following form is obtained:
I ( s) =
or
R
i (t ) =
1 - Lt
e A
L
Example 8.9 For the circuit as shown in Figure 8.22, let at t = 0,
i = 2 A.
Find an expression for i(t) for t > 0 using the Laplace transform
method.
Solution: Applying KVL to the circuit, we get the following form:
0 = 50 × 10 -3
or
0 = 0.05
50 mH
i(t )
200 Ω
Figure 8.22
di
+ 200i
dt
di
+ 200i
dt
Taking the Laplace transform on both sides, the equation can be written as follows:
0 = 0.05[sI(s) − i(0)] + 200 I(s)
Given
i(0) = 2 A
0 = 0.05[sI(s) − 2] + 200I(s)
0 = 0.05sI(s) − 0.1 + 200 I(s)
or
(0.05s + 200)I(s) = 0.1
or
0.05 (s + 4000) = 0.1
or
( s + 4000) I ( s) =
0.1
0.05
2
s + 4000
Taking the inverse Laplace transform, we get the equation as follows:
or
I (s) =
2


i(t ) = L-1 

 s + 4000 
i(t) = 2e−4000t A.
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343
Transient Response of Circuits Using Laplace Transform
Example 8.10 In the circuit shown in Figure 8.23, switch
s is closed at t = 0. Find the voltage across the inductance at
t > 0, using s-domain analysis.
s
+ 10 V
−
4F
4Ω
4H
4Ω
Solution: The circuit is shown in Figure 8.24.
Figure 8.23
The s-domain equivalent circuit of the circuit will be as
shown in Figure 8.25.
Now, the driving point impedance of the given networks is as follows:
 1

Z (s ) =  + 40
 4s

4
4s
i(t ) s
 1 + 16s  4.4s
=
 4s  4 + 4s
+ 10 V
−
 1 + 16 s  16 s
=
 4 s  4 + 4 s
=
=
=
=
i2(t )
i3(t )
4Ω
4H
I2(s)
I3(s)
4Ω
Figure 8.24
 1 + 16 s   16 s 

 ⋅

4s   4 + 4s 
=
16 s
1 + 16 s
+
4s
4 + 4s
=
i1(t )
4F
(1 + 16 s)(16 s)
(1 + 16 s)( 4 + 4 s) + 64 s 2
I(s)
+ 10
− s
I1(s)
1
4s
4
4
4s
Figure 8.25
16 s + 256 s 2
4 + 45s + 64 s + 64 s 2 + 64 s 2
16 s(16 s + 1)
128s 2 + 68s + 4
16 s(16 s + 1)
4(32 s 2 + 175 + 1)
45(16 s + 1)
32 s 2 + 17 s + 1
Now, the current can be calculated as follows:
10
I ( s) = s
Z ( s)
10
s
=
4 s(16 s + 1)
32 s 2 + 17 s + 1
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344 Network Analysis and Synthesis
=
=
=
10 32 s 2 + 17 s + 1
×
s
4 s(16 s + 1)
10(32 s 2 + 17 s + 1)
4 s 2 (16 s + 1)
5(32 s 2 + 17 s + 1)
2 s 2 (165 + 1)
(8.19)
By the current divider formula, we get the following form:
I 3 ( s) =

1
I ( s) ⋅   4 + 
4s 

1

 4 + 
4s

4

4 + 4s
  16 s + 1 
  4 s  4 

I ( s) 
  16 s + 1

  4 s  + 4 


=
(16 s + 1)
⋅4
4s
+ 4s
(16 s + 1)
+4
4s
 4(16 s + 1) 
I ( s) 
16 s + 1 + 16 s 
=
4(16 s + 1)
+ 4s
16 s + 1 + 16 s
I ( s)[4(16 s + 1)]
=
4(16 s + 1) + 4 s(16 s + 1 + 16 s)
=
=
=
=
I ( s)[4(16 s + 1)]
64 s + 4 + 4 s(325 + 1)
I ( s)[4(16 s + 1)]
64 s + 4 + 128s 2 + 4 s
4(16 s + 1) I ( s)
128s 2 + 68s + 4
4(16 s + 1) I ( s)
4(32 s 2 + 17 s + 1)
(16 s + 1)
=
I ( s)
(32 s 2 + 175 + 1)
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345
Transient Response of Circuits Using Laplace Transform
Substituting the value of I(s) from equation (8.19), the following can be obtained:
5(32 s 2 + 17 s + 1)
(16 s + 1)
I 3 ( s) =
⋅
2
(32 s + 17 s + 1)
2 s 2 (16 s + 1)
I 3 ( s) =
5
2s2
Taking the inverse Laplace transform, the equation is written as follows:
5
 1
i3 (t ) = L-1  2 
s 
2
5
or
i3 (t ) = t
2
di3
Therefore, voltage across the inductor = L
dt
d 5 
= 4⋅  t
dt  2 
5
=4
2
= 10 V
Example 8.11 For the network shown in Figure 8.26, the
initial position of switch (s) is ‘1’. After steady state, if the
position of switch is changed to ‘2’, the find current i(t) for
t ≥ 0 using Laplace transform technique.
Solution:
Case I: when the position of switch‘s’ is 1
Applying KVL, we get the equation as follows:
1
s
2R
2
V
L
i
R
Figure 8.26
di
dt
Taking the Laplace transform on both sides, the equation is written as follows:
V = 2 Ri + L
or
or
V
= 2 RI ( s) + LsI ( s)
s
V
= ( 2 R + Ls) I ( s)
s
V
s( Ls + 2 R)
V
=
2R 

Ls  s +


L
I ( s) =
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346 Network Analysis and Synthesis
V
1
⋅
2R 
L 
ss +


L
Using the partial fraction of the equation, we get the equation as in the following:


V A
B 
= ⋅ +
2 R 
L s
s+
L 

or
I ( s) =
and
A=
B=
1
2R
s+
L
1
s
s=
s=0
-2 R
L
-L
2R
1
L
=
=
2R 2R
L
Substituting the values of A and B, the following form is obtained:
L 
 L
V  2R
R 
2
= ⋅
2 R 
L  s
s+

L 


1 
V 1
=
 2 R 
2R  s
s+

L 
Taking the inverse Laplace transform, we get the following form:
2R
- t
V 
L
1
e
i (t ) =


2R 

=
2R
- t
V 
L
1
e


t -∞
t -∞ 2 R


V
V
-∞
(1 - e ) =
=
2R
2R
Now, this steady-state current will act as the initial current for case II.
Case II: when position of switch is changed to position 2.
V
At t = 0, i =
2R
Steady state current = Lt i(t ) = Lt
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347
Transient Response of Circuits Using Laplace Transform
Now, applying KVL to the circuit (as shown in Figure 8.27),
we get the following form:
di
3Ri(t ) + L = 0
dt
Taking the Laplace transform on both sides, the equation is
written as follows:
or
or
2R
L
2
V
R
i(t )
Figure 8.27
3RI ( s) + L[ sI ( s) - i(0)] = 0
Substitute i(0) =
S
1
V
, we get the following form:
2R
V 

3RI ( s) + L  sI ( s) =0
2
R 

VL
3RI ( s) + LsI ( s) =0
2R
VL
(3R + Ls) I ( s) =
2R
VL
2 R( Ls + 3R)
VL
=
3R 

2 RL  s + 

L
V
I ( s) =
or
3R 

2R  s + 

L
Taking the inverse Laplace transform, the following form is obtained:


V -1  1 
L 
i (t ) =

2R
 s + 3R 
3R

L 
V -Lt
is the final expression for i(t)
i (t ) =
e
2R
I ( s) =
or
Example 8.12 In the circuit shown in Figure 8.28, at t = 0+, the voltage across the coil is 120 V.
Find the value of resistance R using the Laplace transform.
Solution:
Given, at
20 Ω
1
t = 0 +,
di
di
L = 120 ⇒ 10 = 120
dt
dt
⇒
di
A
= 12
dt
s
M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 347
2
120 V
K
10 H
R
40 Ω
Figure 8.28
11/17/2014 5:20:08 PM
348 Network Analysis and Synthesis
Case I: initially, the switch is at position 1, and by applying KVL, we get the following form:
di
60i(t ) + 10 = 120
dt
Applying the Laplace transform, the equation is written as follows:
120
60 I ( s) + 10 sI ( s) =
s
120
or
(60 + 10 s) I ( s) =
s
120
or
I ( s) =
s(10 s + 60)
120
=
10 s( s + 6)
12
s( s + 6 )
A
B
= +
s s+6
2
2
= s s+6
Taking the inverse Laplace transform, the equation is given as follows:
=
i(t ) = 2 - 2e -6t
= 2(1 − e−6t) A
Steady state current = Lt i(t )
t →∞
= 2(1 - e -∞ )
= 2A
This current will act as the initial current for case II.
Case II: when the switch is at position 2, t = 0+. By applying KVL, we get the following form:
(60 + R)i(t ) + 10 ⋅
di
=0
dt
di
= 12
dt
Given
(60 + R) i(t) + 120 = 0
Taking the Laplace transform, the equation is written as follows:
(60 + R) I ( s) +
or
120
=0
s
I ( s) =
M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 348
-120
s(60 + R)
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349
Transient Response of Circuits Using Laplace Transform
Taking the inverse Laplace transform, we get the following form:
-120 -1  1 
⋅L  
i (t ) =
60 + R
s
-120
i (t ) =
60 + R
Now, at
t = 0, i(t) = 2 A.
Substituting the values, R can be calculated as follows:
-120
60 + R
120 + 2R = −120
2=
or
Therefore, R = −120 Ω (taking magnitude), R = 120 Ω.
Example 8.13 A series R–L circuit shown in Figure 8.29
experiences an exponential voltage v = 10e−100t after closing
the switch at t = 0. Assume that R = 1 Ω and L = 0.1 H. Find the
expression for current using Laplace transform.
Solution: Applying KVL, we get the equation as follows:
s
1Ω
+ 10e−100t
−
0.1 H
i
Figure 8.29
di
dt
Taking the Laplace transform on both sides, we get the following form:
10e -100t = 1i + 0.1
or
10
= I ( s) + 0.1 sI ( s)
s + 100
10
(1 + 0.1s) I ( s) =
s + 100
10
I ( s) =
(1 + 0.1s)( s + 100)
10
=
0.1( s + 10)( s + 100)
100
=
( s + 10)( s + 100)
Using the partial fractions of the equation, the equation is obtained as follows:
A
B
+
s + 10 s + 100
100/90 -100/90
=
+
s + 10
s + 100
100  1
1 
=

90  s + 10 s + 1000 
I ( s) =
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350 Network Analysis and Synthesis
Taking the inverse Laplace transform, we get the following form:
i (t ) =
10 -10t
[e
- e -100t ] A
9
Example 8.14 A step voltage of 10 V is applied at t = 0 in
a series R–C circuit (as shown in Figure 8.30) when R = 1 Ω
and C = 2 F. The initial charge of the capacitor is zero. Find i(t)
using the Laplace transform.
s
10 V
Solution: Applying KVL, we get the following form:
10 = 1i +
1Ω
i(t )
2F
Figure 8.30
1
idt
2∫
Taking the Laplace transform on both sides, the following equation is obtained:
10
1 I ( s)
= I ( s) +
s
2 s
or
10 =
or
I ( s) =
=
( 2 s + 1)
I ( s)
2
20
2s + 1
20
1

2 s + 

2
10
1
s+
2
Taking the inverse Laplace transform, the equation can be written as follows:
=
t
i(t ) = 10e -1/ 2 A
Example 8.15 For a given circuit shown in Figure 8.31, find the current flowing in the circuit
using both differential equation method and Laplace transform method.
Solution:
(a) By differential equation method and applying KVL to
the given circuit, we get the following form:
5i + 10
di
= 100
dt
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s
1 00 V
5Ω
+
−
i
10 H
Figure 8.31
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Transient Response of Circuits Using Laplace Transform
351
di
+ 0.5i = 10
dt
or
Comparing it with the Leibnitz equation
di
+ Pi = Q , the following form is obtained:
dt
P = 0.5
Q = 10
Pdt
0.5 dt
I.F = e ∫
= e∫
= e 0.5t
Therefore,
Hence, the required solution will be as follows:
i(I.F) = ∫ Q(I.F)dt + c
i(e 0.5t ) = ∫ 10e 0.5t dt + c
= 10
e 0.5t
+c
0.5
i(e 0.5t ) = 20e 0.5t + c
or
i = 20 + ce -0.5t
Applying the initial condition, that is, at t = 0 and i = 0, we get the following form:
0 = 20 + ce0 ⇒ c = −20
Substituting c = −20, i is calculated as follows:
i = 20 − 20 e− 0.5t
i = 20 (1 − e− 0.5t) A
(b) Using the Laplace transform method and by applying KVL, we get the equation as in the
following:
5i + 10
di
= 100
dt
Taking the Laplace transform, we get the following form:
5I(s) + 10[sI(s) − i(0)] = 100
Substituting i(0) = 0, the equation can be written as follows:
100
s
100
(5 + 10 s) I ( s) =
s
100
I ( s) =
s(100 + 5)
100
=
10 s( s + 0.5)
5 I ( s) + 10 sI ( s) =
or
or
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352 Network Analysis and Synthesis
I ( s) =
10
s( s + 0.5)
2 
2
= 10  
 5 s + 0.5 
1 
1
I ( s) = 20  0.5 
s
s
+

Taking the inverse Laplace transform, the equation is as follows:
i(t) = 20 (1 − e− 0.5t) A
Example 8.16 For a given circuit shown in Figure 8.32,
find the current flowing through the circuit using differential
equation method and Laplace transform method.
Solution: By differential equation method and by applying
KVL to the given circuit, we get the following form:
80 = 4i +
1
1 × 10 -6
s
4Ω
+
80 V
−
i
1µF
Figure 8.32
∫ itd
Differentiating both sides with respect to t, the equation is as follows:
0=4
or
4
di
+ 1000000i = 0
dt
di
+ 500000i = 0
dt
P = 500000
Q=0
or
Here,
Now,
di
+ 106 i
dt
i(I.F) = ∫ Q(I.F)dt + C
i(e500000t) = 0 + C
I = ce− 500000t
or
Now, at
t = 0, i =
V
80
==
= 20
R
4
Therefore, we get 20 = ce0 or c = 20 and i = 20 e− 500000t A
By the Laplace transform method and by applying KVL, we get the following form:
4i +
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1
10 -6
∫ idt = 80
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Transient Response of Circuits Using Laplace Transform
353
Taking the Laplace transform, the equation is written as follows:
4 I ( s) +
1 I ( s) 80
=
s
10 -6 s
or

106 
80
 4 + s  I ( s) = s


or
80
 4 s + 1000000 

 I ( s) =
s
s
or
4( s + 500000) I ( s) = 80
or
80
4( s + 500000)
20
I ( s) =
s + 500000
I ( s) =
Taking the inverse Laplace transform, we get the following form:
1


i(t ) = 20 L-1 

 s + 500000 
i(t ) = 20e -500000t A
Example 8.17 Determine the current in the circuit given
below at t ≥ 0. The initial current i(0) = 1.
Solution: Given at t = 0 and i(0) = 1
Applying KVL, we get the following equation for the given
circuit.
V = R × i (t ) + L
10 V
+
−
4Ω
i
2H
Figure 8.33
di
(t ) - i ( 0 )
dt
Taking Laplace transform, we obtain the equation as follows:
V
= R × I ( s) + L[ sI ( s) - i(0)]
s
Substituting the values, we get the following equation:
10
= 4 I ( s) + 2 sI ( s) - 2
s
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354 Network Analysis and Synthesis
10 = 4sI(s) + 2s2I(s) − 2s
or
or
I ( s) =
10 + 2 s
2(5 + s)
5+ s
=
=
2 s 2 + 4 s 2( s 2 + 2 s) s( s + 2)
Using partial fraction of the equation, we get the following:
I ( s) =
A
B
+
s s+2
A= s
5+ s
5
=
s ( s + 2) s = 0 2
B = ( s + 2) ×
3
5+ s
=2
s( s + 2) s = - 2
Substituting,
I ( s) =
5 2 -3 2
+
s
s+2
Taking inverse Laplace transform, we get the following:
5 3

i(t) =  - e -2t  A
2 2

Example 8.18 In the parallel circuit shown, calculate the
branch currents.
Solution: The equivalent impedance of the parallel
R × Ls 4 × 1s
4s
branches, Zeq(s) =
=
=
R + Ls 4 + 1s s + 4
A
i1
i = 4A
4Ω
i2
1H
B
Figure 8.34
Let the voltage across the parallel branches be VAB.
VAB ( s) = I ( s) Zeq ( s) =
Current
I 2 ( s) =
4
4s
16
×
=
s s+4 s+4
VAB ( s)
16
16
=
=
sL
( s + 4 ) s × 1 s( s + 4 )
Using partial fraction, we get the following equation:
I 2 ( s) =
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16
A
B
= +
s( s + 4 ) s s + 4
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355
Transient Response of Circuits Using Laplace Transform
A = s×
16
So, A = 4
s( s + 4 ) s = 0
B = ( s + 4) ×
Substituting,
I 2 ( s) =
16
So, B = -4
s( s + 4 ) s = - 4
4 -4
+
s s+4
Taking inverse Laplace transform, we get the following:
i(t) = (4 − 4e−4t) A
I1 ( s ) = I ( s ) - I 2 ( s ) =
=
4
16
s s( s + 4 )
4
s+4
Taking inverse Laplace transform, we get the following:
i1(t) = (4 − 4e−4t) A
REV IE W Q UESTIONS
Numerical Questions
1. Using the Laplace transform method, find the current flowing in the following circuit
(i(0) = 0)
10 Ω
40 H
K
200 V
i
[Ans. 20(1 − e− 0.25t)A]
2. A coil has resistance of 1 Ω and an inductance of 1 H. When it is connected to 6 V DC voltage source, calculate initial and final values of current using Laplace transform method.
[Ans. i(0) = 0 A, i(∞) = 6 A]
3. The field winding of a DC motor has 80 Ω resistance and 12 H inductance. It is connected
to a 240 v DC source. At t = 0, the supply is disconnected and a field discharge resistance of
80 Ω is connected across the field winding. Find the rate of change of current in the winding at t = 0 using the Laplace transform method.
[Ans. −40 A/s]
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356 Network Analysis and Synthesis
4. A circuit has resistance of 1000 Ω and capacitance of 0.1 mF. At t = 0, it is connected to a
12 V battery. Find the current at t = 0, using the Laplace transform method.
[Ans. −12 mA]
5. For the following circuit, find the current flowing through circuit, voltage across resistor
and capacitor using the Laplace transform method. Switch is closed at t = 0.
5 MΩ
s
20 µF
100 V
[Ans. 20 × 10-6 e−t/100A VR = 100 e−t/100V
Vc = 100 (1 − e−t/100)V.]
6. Find the transient current for the following circuit when switch s is closed, using the Laplace
transform method.
5Ω
1H
s
1
F
2
24V
[Ans. 6[e− 0.5t − e− 4.5t]]
7. Find the expression for current if the switch is closed at t = 0, using Laplace transform
method.
s
50 Ω
150 sin 500t
0.2 H
i
[Ans. 1.2−250t + 1.34 sin (500t − 63.4°) A]
8. Determine the current i for t ≥ 0 if Vc (0) = 4 V for the circuit shown below.
4Ω
20 V
1F
8
[Ans. i = 4e-2t]
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Transient Response of Circuits Using Laplace Transform
357
9. Determine the voltage across the resistor, v0 in the circuit shown. Assume zero initial
condition.
+
νο
4Ω
+
−
−
2H
1F
2
1
4
4
[Ans. v0 = te - t + e - t - e -4t]
3
9
9
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Three-phase Systems
and Circuits
9
Chapter objectives
After carefully studying this chapter, you should be able to do the following:
State the advantages of three-phase
Distinguish between active power and
system over single-phase system.
reactive power.
Explain how three-phase voltages are
Calculate current, power, and power
generated.
factor of balanced three-phase loads.
Distinguish between a balanced supply
Make connection diagram and meaand unbalanced supply system.
sure power and power factor in a threephase balanced and unbalanced load.
Distinguish between a balanced load
and an unbalanced load.
Convert a star-connected load into an
equivalent delta connected load.
Write the relationship between phase
and line quantities in a star- and deltaSolve numerical problems of unbalconnected system.
anced star- and delta- connected loads.
9.1 INTRODUCTION
Generation, transmission and distribution of electricity is done by three-phase electrical networks consisting of generators, transformers and transmission and distribution lines forming
the power system. In a three-phase system, we have three independent voltages induced in the
three windings of the generator. To understand the difference between a single-phase voltage
and a three-phase voltage system, let us consider how these voltages are generated in AC generators. We have known that EMF is induced in a coil if it cuts lines of force. In Figure 9.1, we
have placed one coil in slots of a hollow cylindrical stator core. A two-pole magnet is rotated
at a particular speed by some means. The flux lines will cut the conductors and EMF will be
induced in the coil. Since flux of North and South poles will cut the conductors alternately, an
alternating single-phase voltage will be induced in the coil.
Here, we have seen that the coil is stationary and the magnetic field is being rotated.
However, we could have had the field stationary and the coil rotating by placing the field magnets on the stator and the coil on a cylindrical rotor. What is required is to produce a relative
motion between the magnetic field and the conductor. The magnitude of the induced EMF will
depend upon the number of coils, the strength of the magnetic field and the speed of rotation of
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Three-phase Systems and Circuits 359
A
Slot
N
Rotor
tor
duc
Con
e
A
Coil
t
S
A′
Stator
core and frame
A′
r
ucto
d
Con
Figure 9.1 Generation of Single-phase Voltage
the magnet. The frequency of the induced EMF will depend upon the number of magnetic poles
confronted by the coils per revolution. Normally, the number of magnetic poles and the speed
of rotation of the magnetic poles by a drive, usually a turbine, are kept constant. The number
of coils and number of turns used in each coil are kept as per design and are constant once
the machine is constructed. That is the reason why the magnitude of the induced EMF and its
frequency are constant. Thus, we get a single-phase voltage from the single-phase windings
that can be used to supply an electric circuit comprising resistance, inductance and capacitance
elements.
In a three-phase system, we will have three-phase voltages induced in the three-phase windings of the generator. In all generating stations, three-phase generators are installed.
9.2 ADVANTAGES OF THREE-PHASE SYSTEMS
A three-phase system has a number of advantages over a single-phase system. Some of the
advantages are mentioned below.
1. The output of a three-phase machine generating electricity is more than the output of a
single-phase machine of the same size.
2. The most commonly used three-phase induction motors are self starting. For single-phase
motors, a separate starting winding is required.
3. Electrical power transmission from the generating station to the places of use is done by
transmission lines. It has been seen that three-phase power transmission is more economical than single-phase power transmission.
4. The power factor of three-phase systems is better than that of the single-phase systems.
5. Single-phase supply can also be obtained from a three-phase supply.
6. The instantaneous power in a single-phase system is fluctuating with time giving rise to
noisy performance of single-phase motors. The power output of a symmetrical three-phase
system is steady.
7. For rectification of AC into DC, the DC output voltage becomes less fluctuating if the
number of phases is increased.
Thus, we see that from generation, transmission, distribution and utilisation points of view,
three-phase systems are preferred over single-phase systems due to a number of reasons mentioned above.
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360 Network Analysis and Synthesis
9.3 GENERATION OF THREE-PHASE VOLTAGES
Due to a number of practical considerations, generators are built to generate poly-phase voltages.
Commercial generators are built to generate three-phase voltages.
In three-phase generators, three separate windings are made. Windings are made of coils.
These windings are placed in stator slots at an angle of 120° apart, as shown in Figure 9.2. RR′ is
one phase winding.
R
R
Stator
Y′
R
N
R′
B′
B′
R′
Y′
Y′
Y
B
S
Y
B
R′
Y
B′
B
Figure 9.2 Generation of Three-phase Voltages in the Three-phase Windings
YY′ is the second phase winding and BB′ is the third phase winding. The three-phase windings are placed at an angular distance of 120°. For simplicity, only one coil per phase has been
shown. In practice, a number of coils connected in series make one phase winding.
When the magnetic poles are rotated by a prime mover (say a turbine), the magnetic flux of
North and South poles will cut the windings in sequence. For clockwise rotation, flux will be
cut by coil RR′ first, then by coil YY′ and then by a coil BB′. Therefore, EMF will be induced in
these coils in sequence. There will be a time phase difference between the EMFs induced in these
coils (windings). The time phase difference will be 120°. In terms of time, the phase difference
will be the time taken by the magnetic poles to rotate by 120°, that is, one-third of a revolution.
Thus, across the three-phase windings, we will get three voltages that are equal in magnitude
and frequency but having a time phase difference of 120° between them, as shown in Figure 9.3.
9.3.1 Equation of Three-phase Voltages
ER
The equations of voltages are as follows:
eR = Em sin w t(9.1)
240°
120°
eY = Em sin (w t − 120°)(9.2)
eB = Em sin (w t − 240°)(9.3) E
EY
B
Resultant EMF = Em sin w t + Em sin (w t − 120°) + Em sin
(w t − 240°) = 0
Since the three voltages are equal in magnitude but displaced in
EY + EB
time phase by 120°, their phasor sum is zero as shown in Figure 9.3.
Three-phase supply is required for large-capacity electrical Figure 9.3 Three-phase
Voltages
loads. These loads could be three-phase motors used in indusDisplaced in Time
trial, commercial, agricultural and other sectors. For example,
Phase by 120°
the water pump used for irrigation purpose is invariably a
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Three-phase Systems and Circuits 361
three-phase motor-driven pump requiring a three-phase supply. Therefore, similar to the threephase supply, we will have three-phase loads. Three-phase supply will be supplying electrical
power to three-phase loads. A number of terms are used in connection with three-phase supply
and three-phase loads. These are described as follows. Further, the three-phase windings can
be connected together in the form of star or delta. The voltage between the two-phase windings
and current flowing through the phase windings, and the supply line will be different in star and
delta connections. These will be studied in detail.
By now, we must have realised that by phase we mean a winding. A phase difference between two
windings is the physical angular displacement between them. In a three-phase winding, the phase
difference between the windings is 120°. Phase sequence is the order in which maximum voltage is
induced in the windings. For example, if the magnetic fields cut the conductors of the phase RR′ first
and then cut the conductors of phase YY ′, and lastly cut the conductors of phase BB′, then EMF will
be induced in all the phases of equal magnitude but their maximum value will appear in a sequence
RYB. Then, we call the phase sequence of EMF as RYB. If the magnetic system of Figure 9.2 rotates
in the anticlockwise direction, the phase sequence of EMF induced in the three phases will be RBY.
Elementary Three-phase Generator
Let us consider an elementary three-phase generator as shown in Figure 9.4(a). Three-phase
windings are placed in slots in the stator. For simplicity, only one coil per phase has been used.
R–R′ is one coil making R-phase windings. Y–Y ′ is another coil forming Y-phase winding. B–B′ is
one coil forming B-phase winding. These three-phase windings are placed in the stator slots at an
angle of 120° in space. For simplicity, only one coil per phase has been shown. In actual practice,
a number of coils are connected together to form each phase winding. The rotor carries the magnetic poles that produce the magnetic field. Direct current is supplied to the field windings so that
the field magnets are excited. When the rotor is rotated by a turbine, the magnetic flux are cut by
the stator coils RR′, YY ′ and BB′ in sequence. The EMFs induced in these coils are sinusoidal in
nature because of the nature of flux distribution. The voltage induced in the three-phase windings
will be identical in nature but they will be displaced in time-phase by 120° as has been shown.
The order in which the phase voltages attain their maximum value or peak value, Vm is called the
Vm VR
Field winding
VB
VY
R
N
Y′
B′
ν
If
B
Rotor
0
S
120°
240°
360°
Y
R′
wt
Stator
(a)
(b)
Figure 9.4 G
eneration of Three-phase Voltages (a) Three-phase Two-pole Generator
and (b) Three­-phase Voltages Induced in the Windings
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362 Network Analysis and Synthesis
phase sequence. If the rotor rotates in the clockwise
direction, voltages in the phases will be induced in
the sequence: VR, VY and VB. If the rotor poles rotate
in the opposite direction, the phase sequence of the
induced voltage will change from RYB to RBY.
The phase voltages should supply power to electrical loads for which the two-end terminals of each
phase are to be connected to loads as shown in
Figure 9.5. Six wires are required to be taken out from
the generator to the load. Instead of taking out six wires
from the generator and connecting them to the loads
separately as shown in Figure 9.5, the three-phase
windings are connected either in star or in delta so that
only three wires are to be taken out from the generator to the load. The loads are also connected either in
star or in delta. In the case of the star connection, a
fourth wire may be taken out from the neutral point.
Load
R
VR
R′
Load
Y
VY
Y′
Load
B
VB
B′
Figure 9.5 T he Three-phase Windings
are Connected to the
Load Independently
Through Six Wires
9.3.2 Balanced Three-phase System
In a balanced three-phase supply system all the
phase ­voltage are equal in magnitude but displaced
by 120°. A balanced star-connected supply system is
shown in Figure 9.6. The phase voltage and the line voltages have been represented through
phases in the Figure.
VBR
VBN = 230∠−240°
−VYN
VRY
R
30°
240°
VRN = 230∠0°
120°
N
VYN = 230∠−120°
Y
VYB
B
(a)
(b)
Figure 9.6 R
epresentation of a Balanced Star-connected Supply
System
Here, let
VRN = 230 ∠0°
Then
VYN = 230 ∠ − 120°
VBN = 230 ∠ − 240°
The corresponding line voltages are equal to 3 times the phase voltage and leading the corresponding phase voltage by 30°. We draw the line voltage by assuming a phase sequence of RYB.
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Three-phase Systems and Circuits 363
VRY = VRN − VYN = 3 × 230 ∠ + 30° = 400 ∠30°
VYB = VYN − VBN = 3 × 230 ∠ − 90° = 400 ∠ − 90°
VBR = VBN − VRN = 3 × 230 ∠ − 210° = 400 ∠ − 210°
Like phase voltages, the line voltages have also a phase difference of 120°.
9.4 TERMS USED IN THREE-PHASE SYSTEMS AND CIRCUITS
The following are some of the terms used while describing a three-phase system. These are as
follows:
1. Balanced supply: A set of three sinusoidal voltages (or currents) that are equal in magnitude but
has a phase difference of 120° constitute a balanced three-phase voltage (or current) system.
2. Unbalanced supply: A three-phase system is said to be unbalanced when either of the
three-phase voltages are unequal in magnitude or the phase angle between the three phases
is not equal to 120°.
3. Balanced load: If the load impedances of the three phases are identical in magnitude as
well as phase angle, then the load is said to be balanced. It implies that the load has the
same value of resistance R and reactance XL and/or XC in each phase.
4. Unbalanced load: If the load impedances of the three phases are neither identical in magnitude nor in phase angle, then the load is said to be unbalanced.
5. Single phasing: When one phase of the three-phase supply is not available then the condition is called single phasing.
6. Phase sequence: The order in which the maximum value of voltages of each phase appear
is called the phase sequence. It can be RYB or RBY.
7. Coil: A coil is made of conducting wire, say copper, having an insulation cover. A coil can
be of a single turn or many number of turns. Normally, a coil will have a number of turns.
A single turn of a coil will have two conductors on its two sides called coil sides.
8. Winding: A number of coils are used to make one winding. Normally, the winding coils are
connected in series. One winding forms one phase.
9. Symmetrical system: In a symmetrical three-phase system, the magnitude of three-phase
voltages is the same but there is a time phase difference of 120° between the voltages.
9.5 THREE-PHASE WINDING CONNECTIONS
A three-phase generator will have three-phase windings. These phase windings can be connected in two ways:
1. Star connection
2. Delta connection
9.5.1 Star Connection
The star connection is formed by connecting the starting or finishing ends of all the three windings together. A fourth conductor, that is, taken out of the star point is called the neutral point.
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364 Network Analysis and Synthesis
The remaining three ends are brought out for connection to load. These ends are generally
referred to as R–Y–B, to which the load is to be connected. The star connection is shown in
Figure 9.7(a). This is a three-phase, four-wire star-connected system. If no neutral conductor is
taken out from the system it gives rise to a three-phase, three-wire star-connected system.
IL = IPh
R
R1
VPh
VL
R2
Y1
B1
To
load
N
Y2
B2
N
IL
R2
R1
Y2
Y1
B2
B1
R
Y
Y
IL
B
(a)
To
load
B
(b)
Figure 9.7 Star Connection of Phase Windings (a) Three-phase Four-wire
System and (b) Three-phase Three-wire System
The current flowing through each line conductor is called line current IL. In the star connection, the
line current is also the phase current. Similarly, the voltage across each phase is called phase voltage VPh. Voltage across any two line conductors is called line voltage VL. When a balanced threephase load is connected across the supply, terminals R, Y, B currents will flow through the circuit.
The sum of these currents, that is, IR, IY and IB will be zero. The neutral wire connected between
the supply neutral point and the load neutral point will carry no current for a balanced system.
9.5.2 Delta Connection
The delta connection is formed by connecting the end of one winding to the starting end of
the other and connections are continued to form a closed loop. In this case, the current flowing
through each line conductor is called line current IL and the current flowing through each phase
winding is called phase current IPh. However, we find that the phase voltage is the same as the line
voltage in a delta connection. The delta connection of windings has been shown in Figure 9.8.
R1
R
R2
IL
IPh
VPh = VL
IPh
Delta
B1
Y2
R
R1
B2
IPh
R2
Y1
IL
IL
(a)
To
load
Y
B
Y
Y1
Y2
To
load
B
B1
B2
(b)
Figure 9.8 Delta Connection of Three-phase Windings (a) Three-phase Three-wire Delta
Connection and (b) Connection Scheme of Windings Forming a Delta
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Three-phase Systems and Circuits 365
9.5.3 Relationship of Line and Phase Voltages and Currents
in a Star-connected System
Consider the balanced star-connected system as shown in Figure 9.9.
Suppose, the load is inductive and, therefore, current will lag the applied voltage by an angle f.
Consider a balanced system so that the magnitude of current and voltage of each phase will be
the same.
That is, phase voltages, VR = VY = VB = VPh
Line current IR = IY = IB = IL
Line voltage VL = VRY = VYB = VBR
Phase current IPh = IR = IY = IB
IL = IPh for star connection as the same phase current passes through the lines to the load.
VRY
IR
VR
VB
VPh VRY = VL
N
VRB
IY
VYB
IB
f
IB
N
VY
IR
R
VB
(a)
N
VYB
60° 30°
f
IY
Y
B
f
−VB
VY
−VR
VBR
(b)
Figure 9.9 B
alanced Star-connected System (a) Three-phase Sypply System and
(b) Phasor Diagram
To derive the relation between VL and VPh, consider line voltage VRY.
VRY = VRN + VNY
VRY = VRN + (−VYN)
Similarly,
VYB = VYN + (−VBN)
and
VBR = VBN + (−VRN)
The procedure for drawing the phasor diagram of Figure 9.9(b) is as follows.
Draw three phasors VR, VY and VB representing the phase voltages. These voltages are of
equal magnitude but displaced by 120°. The line voltage phasors VRY , VYB and VBR are drawn
by vectorially adding the phase voltages. For example, to draw line voltage VRY , we have to add
the phase voltages as follows:
VRY = VRN + VNY = VRN + (−VYN)
The phasor VYN is obtained by reversing VNY. VRY is obtained by vectorially adding VRN and
VYN, as shown in Figure 9.9(b). Similarly, the other line voltages have been drawn. The phase
currents IR, IB and IY have been shown lagging the phase voltages by the power factor angle f.
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366 Network Analysis and Synthesis
From the phasor diagram shown in Figure 9.9(b), the phase angle between phasors VR and
(− VY) is 60°.
\ V RY = V R 2 + V Y 2 + 2V RV Y cos 60°
V RY = V L = V Ph 2 + V Ph 2 + 2V PhV Ph ×
1
2
V L = 3V Ph
V L = 3V Ph
(9.4)
Thus, for the star-connected system, the following can be stated:
Line voltage = 3 × phase voltage
Line Current = phase current
Power: Power output per phase = VPh I Ph cos f
Total power output = 3VPh I Ph cos f
= 3×
VL
3
× I L cos f
P = 3VL I L cosf
(9.5)
Power = 3 × line voltage × line current × power factor
9.5.4 Relationship of Line and Phase Voltages and Currents
in a Delta-connected System
Consider the balanced delta-connected system as shown in Figure 9.10.
In a delta-connected system, the voltage across the winding, that is, the phases is the same
as that across the line terminals. However, the current through the phases is not the same as
through the supply lines.
Therefore, in the case of the delta-connected circuit, the phase voltage is equal to the line
voltage, but the line current is not equal to the phase current.
Line voltage
VL = VRY = VYB = VBR
Line current
IL = IR = IY = IB
Phase voltage
VPh = VRY = VYB = VBR
Phase current
IPh = IRY = IYB = IBR
\ VPh = VL for delta-connected load.
(9.6)
Figure 9.10(a) shows a three-phase delta-connected supply system connected to a three-phase
delta-connected load. The line currents are IR, IY and IB. The phase currents are IRY, IYB and
IBR. The phasor diagram in Figure 9.10(b) has been developed by first showing the three-phase
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Three-phase Systems and Circuits 367
VBR
IBR
−IRY
IY
IB
f
IYB
R
IR
VRY
IBR
Z Ph
VYB
Z
B
Y
B
IB
IBR
ZPh
f
VYB
IRY
Ph
VBR
−IYB
f
R
IYB
IY
−IBR
IRY
VRY
IRY
Y
IR
(a)
(b)
Figure 9.10 Delta-connected System (a) A Three-phase Delta-connected Load Supplied
From a Delta-connected Supply Source and (b) Phasor Diagram of Voltages
and Currents
voltages VYB, VBR and VRY of equal magnitude but displaced by 120° from each other. Then,
the phase currents IYB, IBR and IRY have been shown lagging respective phase voltages by power
factor angle f. The line currents are drawn by applying KCL at the nodes R, Y, B and adding
the phasors, as shown.
To derive the relation between IL and IPh, apply KCL at node R, as shown in Figure 9.10(b)
IR + IBR = IRY
\ IR = IRY − IBR
Similarly, at node Y and B, we can write the following:
IY = IYB − IRY
IB = IBR − IYB
Since the phase angle between phase currents IRY and −IBR is 60°, the following can be obtained
as follows:
\ I R = I RY 2 + I BR 2 + 2I RY I BR cos 60
I R = I L = I Ph 2 + I Ph 2 + 2I Ph I Ph ×
1
2
I L = 3 × I Ph
Thus, for a three-phase delta-connected system,
Line current = 3 × phase current
Line voltage = phase voltage
Power: power output per phase = VPh IPh cosf
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368 Network Analysis and Synthesis
Total power output = 3 VPh IPh cosf
= 3 × VL ×
IL
3
cos f
= 3VL I L cos f
Power = 3 × line voltage × line current × power factor
For both star-connected and delta-connected systems, the total power P is
P = 3V L I L cosf
(9.7)
If per phase power is Ph and total power is PT , then
PT = 3Ph
(9.8)
9.6 ACTIVE AND REACTIVE POWER
The expression for average power in an AC circuit is
P = V (I cos f).
The quantity in brackets is the projection of current phasor I into voltage phasor V . Accordingly
in-phase component of IPh along V has been shown in Figure 9.11 as IPh cos f and the perpendicular component as IPh sin f. If we multiply all the sides of the triangle ABC by VPh, the
triangle becomes a power triangle where AB = VPh IPh cos f is called the active power, BC =
VPh IPh sin f is called the reactive power and VPh IPh is called the apparent power.
A
IPh cosf
B
V
f
IPh
A
IPh sinf
VPh × IPh cosf
f
V
kVA cosf = kW
B
VPh × IPh sinf
f
B
kVA sinf
= kV AR
kVA
C
VPh × IPh
(a)
(b)
(c)
Figure 9.11 P
ower Triangle Showing the Relationship Between (a) Active Power,
(b) Reactive power, and (c) Apparent Power
Apparent power = kVA
(9.9)
Apparent power × cos f = Active power = kW
(9.10)
Apparent power × cos f = Reactive power = kVAR
(9.11)
103,
Multiplying all the sides of the power triangle by
that is, expressing the power in terms
of ‘kilo’, the power triangle is redrawn, as shown in Figure 9.11(c) (kVA is kilo Volt Ampere).
kVA cos f = kW (kilo-Watt)
kVA sin f = kVAR (kilo-VAR)
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(9.12)
(9.13)
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Three-phase Systems and Circuits 369
9.7 COMPARISON BETWEEN STAR CONNECTION
AND DELTA CONNECTION
As mentioned earlier, the three windings of a generator can be connected either in star or in
delta. Same is the case with the transformers. Three-phase electrical loads and the windings of
three-phase motors can also be connected in star or in delta.
The relationship between voltages, current and their phase relationship along with some
other related factors for star and delta connections have been compared and are presented in
Table 9.1.
Table 9.1 Comparison Between Star and Delta Connections
Star Connection
Delta Connection
1. Line current is the same as phase cur- Line current is 3 × the phase current,
rent, that is, IL = IPh
that is, I L = 3I Ph
2. Line voltage is 3 the phase voltage, Line voltage is the same as phase voltage, that is, VL = VPh
that is, VL = 3VPh
3. Total power = 3V L I L cosf
Total power = 3V L I L cosf
4. Per phase power = VPh IPh cos f
Per phase power = VPh IPh cos f
5. Three-phase three-wire and three-phase Three-phase three-wire system is
four-wire systems are possible
possible
6. Line voltages lead the respective phase Line currents lag the respective phase
voltages by 30°
currents by 30°
Example 9.1 A 400 V, three-phase, 50 Hz power supply is applied across the three terminals
of a delta-­connected three-phase load. The resistance and reactance of each phase is 6 Ω and
8 Ω, respectively. Calculate the line current, phase current, active power, reactive power and
apparent power of the circuit.
Solution: The load is delta connected. Hence, the following can be calculated as follows:
VPh = VL = 400 V
Z Ph = R + jX = 6 + j8 = 6 2 + 82 ∠ tan −1
I Ph =
Power factor
8
= 10 ∠53° Ω
6
VPh 400 ∠0°
=
= 40 ∠ − 53°A
Z Ph 10 ∠53°
cos f = cos 53° = 0.6 lagging
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370 Network Analysis and Synthesis
I L = 3 I Ph = 1.732 × 40 = 69.28 A
R 6
=
= 0.6 lagging
Z 10
sin f = 0.8
cos f =
Power factor
= 3VL I L cos f = 1.732 × 400 × 69.28 × 0.6
Active power
= 28798 W = 28.798 kW
28.8 kW
= 3VL I L sin f = 1.732 × 400 × 69.28 × 0.8
Reactive power
= 38397 VAR = 38.397 kVAR
38.4 kVAR
= 3V Ph I Ph = 3V L I L
Apparent power
Active power
= 28.8 kW
O
= 1.732 × 400 × 69.28
= 47997 VA = 47.997 kVA
48 kVA
A
The power triangle is shown in Figure 9.12.
ra
pe
Ap
f = 53°
nt
Reactive
power
= 28.4 kVAR
r=
we
po
Apparent Power in kVA =
( Active power ) 2 +
( Reactive power ) 2
48
.8
= ( 28..8) 2 + (38.4) 2
A
kV
= 829.44 + 1474.56
B
= 2304 = 48
Figure 9.12
Example 9.2 A balanced star-connected load of (8 + j6) Ω per phase is connected to a balanced
three-phase, 400 V supply. Find the line current, power factor, power and total volt-amperes.
Solution:
Phase voltage
VP =
Impedance per phase
Z P = R 2 + X L2 = 82 + 6 2 = 10 Ω
Phase current
IP =
Line current
Power factor
LineVoltage, V L
3
=
400
3
= 231V
V P 231
=
= 23.1 A
ZP
10
IL = IP = 23.1 A
cos f =
8
R
=
= 0.8 (lagging)
Z 10
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Three-phase Systems and Circuits 371
P = 3V L I L cos f
Total power
= 3 × 400 × 23.1 × 0.8
= 12, 800 W
Total volt amperes
= 3V L I L
= 3 × 400 × 23.1 = 16, 000 VA
Example 9.3 A three-phase four-wire supply system has a line voltage of 400 V. Three noninductive loads of 16 kW, 8 kW and 12 kW are connected between R, Y and B phases and the
neutral, respectively. Calculate the current flowing through the neutral wire.
Solution: Loads connected between the different phases
and the neutral are of 16 kW, 8 kW and 12 kW, respectively
as shown in Figure 9.13.
The current through the neutral wire line is the phasor
sum of all the line currents. We will first calculate the line
currents and then add them vectorially.
For star connection VPh
VL
400
=
=
= 231 V
3
3
IR
R
VPh VL = 400 V
16 kW
N
B
The three-phase voltages are equal but have a phase
difference of 120° between them.
Iy
Y
IB
8 kW
12 kW
Figure 9.13
VR = 231∠0°, VY = 231∠ − 120°, and VB = 231∠ − 240°
IR =
16 × 1000 16 × 1000
=
= 69.3∠0°
231∠0
VR
IY =
8 × 1000
8 × 1000
= 34.6 K ∠120°
=
231∠ − 240°
VY
IB =
12 × 1000
12 × 1000
=
= 52∠240°
VB
231∠ − 240°
Current through the neutral wire, IN is
IN = IR + IY + IB
= 69.3∠0° + 34.6 ∠120° + 52∠240°
= 69.3 (cos 0° + j sin 0°) + 34.6 (cos120° + j sin 120°) + 52(cos 240° + j sin 240°)
= 69.3 (1 + j0) + 34.6 (−0.5 + j 0.866) + 52 (−0.5 + j 0.866)
IN = 69.3 − 17.3 + j 30 − 26 + j 45
= 26 + j 75
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372 Network Analysis and Synthesis
I N = 26 2 + 752 = 6301
= 79.4 A
Example 9.4 A three-phase load has a resistance and reactance of 6 Ω each for all the three
phases. The load is connected in star. A 400 V, 50 Hz, three-phase supply is connected across the
load. Calculate phase voltage, phase current, power factor, power consumed per phase and the
total power consumed by the load.
Solution: The circuit is shown is Figure 9.14.
Z Ph
VL
400
= 231 V
3
3
= 6 + j 6 W = 8.48∠45°Ω
VPh =
I Ph =
IPh
=
R
IL
VPh
6Ω
VL = 400 V
6Ω
N
VPh
231∠0°
=
= 27.2∠ − 45° A
Z Ph 8.48∠45°
6Ω
6Ω
I Ph = I L = 27.2 A
B
Y
Angle between VPh and IPh is 45°.
6Ω
6Ω
Figure 9.14
Power factor = cos 45° = 0.7 lagging
Power absorbed by each phase of the load = VPh IPh cos f
= 231 × 27.2 × 0.7
= 4398 W
Total power consumed = 3 × 4398 = 13194 W
Example 9.5 A 400 V, 50 Hz, three-phase supply is provided to a three-phase star-connected
load. Each phase of the load absorbs a power of 200 W. The load power factor is 0.8 lagging.
Calculate the total power supplied to the load and the line current.
Solution:
Power consumed by each phase = 2000 W
Power consumed by all three phases = 3 × 2000 W
= 6000 W
Total power supplied = total power consumed.
VL = 400 V,VPh =
VL
3
=
400
3
= 231 V
Power consumed per phase = VPh IPh cos f
= VPh IPh cos f = 2000
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Three-phase Systems and Circuits 373
I L = I Ph =
2000
= 10.82 A
231 × 0.8
Example 9.6 A 400 V, 50 Hz, three-phase supply is provided to a three-phase delta-connected
load. The resistance and inductance of each phase of the load is 8 Ω and 0.04 H, respectively.
Calculate the phase current and the line current drawn by the load. Further, calculate the total
power consumed.
Solution: The impedance of load per phase is given as follows:
ZPh = R + jw L = 8 + j2w × 50 × 0.04
= 8 + j12.56 Ω
Since the load is delta connected, VPh = VL = 400 V
I Ph =
VPh
400
400∠0
=
=
= 26.86∠ − 58°
Z Ph 8 + j 12.56 14.89∠ tan −1 (12.56 /8)
I L = 3 I Ph = 1.73 × 26.86 = 36 A
Total power consumed = 3 VPh IPh cos f
= 3 × 400 × 26.86 cos f W
= 3 × 400 × 26.86 cos 58°
= 3 × 400 × 26.86 × 0.52 = 1736 W
Example 9.7 A three-phase star-connected load consumes a total of 12 kW at a power factor
of 0.8 lagging when connected to a 400 V, three-phase, 50 Hz power supply. Calculate the
resistance and inductance of load per phase.
Solution: Total power consumed = 12 kW
Per phase power consumed = 4 kW
Therefore
VPh IPh cos f = 4000 W
I Ph =
or
M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 373
I Ph =
4000
V Ph cosf
4000
( 400 / 3 ) × 0.8
= 21.6 A
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374 Network Analysis and Synthesis
Z
I Ph =
V Ph
Z Ph
Z Ph =
V Ph
231
=
= 10.7 Ω. Power factor cos f = 0.8, sin f = 0.6
I Ph 21.6
X
f
R
Figure 9.15
As shown in Figure 9.15,
R = Z Ph cos f = 10.7 × 0.8 = 8.56 Ω
X = Z Ph sin f = 10.7 × 0.6 = 6.42 Ω
X = 6.42
2p f L = 6.42
Now,
6.42
= 20.4 × 10 −3 H
2 × 3.14 × 50
= 20.4 mH
L=
Example 9.8 A balanced three-phase star-connected load of 8 + j6 Ω per phase is supplied
by a 400 V, 50 Hz supply. Calculate the line current, power factor, active and reactive
power.
Solution:
VL = 400 V, for star connection, VPh =
Z Ph
I Ph
VL
3
6
= 8 + j 6 = 82 + 6 2 ∠ tan −1 = 10 ∠37°
8
VPh
231
=
=
= 23.1∠ − 37°
Z Ph 10 ∠37°
=
400
3
For star connection, the following form is obtained:
IPh = IL = 23.1 A
Angle of lag of IPh with VPh is 37°
Power factor
= cos f = cos 37° = 0.8 lagging
Active power
= 3 V L I L cosf
= 1.732 × 400 × 23.1 × 0.8
= 12802 W = 12.802 kW
Reactive power
= 3 V L I L sinf
=1.732 × 400 × 23.1 × 0.6
= 9602 VAR = 9.602 kVAR
Example 9.9 A delta-connected three-phase motor load is supplied from a 400 V, three-phase,
50 Hz supply system. The line current drawn is 21 A. The input power 11 kW. What will be the
line current and power factor when the motor windings are delta connected?
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Three-phase Systems and Circuits 375
Solution: Line voltage VL = 400 V, VPh = VL for delta connection
Line current I L = 3 I Ph = 21 A
Impedance of each winding, Z Ph =
VPh
400
400 × 3
=
=
= 33 Ω
21
I Ph I L / 3
P = 3V L I L cos f = 3 × 400 × 21 × cos f
or
cos f =
P
3 × 400 × 21
=
11 × 1000
3 × 400 × 21
= 0.756
When the motor windings are star connected, the equation can be written as follows:
VL
400
=
= 231 V
3 1.732
ZPh will remain the same as the same windings are connected in star.
VPh =
I Ph =
V Ph 231
=
= 7A
Z Ph
33
In star connection, the line current is the same as phase current. Hence, IL = IPh = 7 A.
Power factor depends on the circuit parameters
R
Z
Since both R and Z remain unchanged, the power factor will remain the same at 0.756. Students may
note that the line current in a star connection is one-third of the line current in a delta connection.
cosf =
Example 9.10 A balanced star-connected load of 4 + j6 Ω per phase is connected across
a 400 V, three-phase, 50 Hz supply. Calculate line current, phase current, line voltage, phase
voltage, power factor, total power and reactive power.
Solution:
Z /Ph = 4 + j 6 = 7.21∠56°
For star connection,
V Ph =
I Ph =
Power factor
VL
3
=
400
3
= 231 V
V Ph
231∠0
= 32∠ − 56°
=
Z Ph 7.21∠56°
cos f = cos 56° = 0.56 lagging
IPh = 32 A
IPh = IL = 32 A
VL = 400 V
VPh = 231 V
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376 Network Analysis and Synthesis
Total power = 3 VL I L cosf
= 3 VPh IPh cos f
= 3 × 231 × 32 × 0.56
= 12418 W
= 12.418 kW
Total reactive power = 3 V L I L sinf
= 3 VPh IPh sin f
= 3 × 231 × 32 × 0.83
= 18406 VAR
= 18.406 kVAR
9.8 MEASURMENT OF POWER IN THREE-PHASE CIRCUITS
We have known that in DC circuits, power is measured as the product of voltage and current, that
is, power P = VI. DC power can be measured using a voltmeter and an ammeter. In AC circuits,
power P = VI cos f. In three-phase AC circuits, total power is three times the power per phase.
Wattmeter is an instrument used for the measurement of power in AC circuits. Wattmeters are
available as single-phase wattmeters and three-phase wattmeters. Single-phase wattmeters can
be used to measure the three-phase power. In the case of star-connected balanced load with
neutral connection, only one single-phase wattmeter can be used to measure the three-phase
power. The three-phase power is three times the single-phase power. For unbalanced three-phase
loads, that is, currents in the three phases are not the same, and hence two wattmeters are to be
used to measure the three-phase power. These methods are described in the following sections.
9.8.1 One-wattmeter Method
In this method, only one single-phase wattmeter can be used to measure the total three-phase
power. In this method, the current coil (CC) of the wattmeter is connected in series with any
phase and the pressure coil (PC) is connected between that phase and the neutral, as shown
in Figure 9.16. One-wattmeter
Single-phase
method has a de-merit that even
Wattmeter
a slight degree of unbalance in
CC
the load produces a large error
R
in the measurement. In this
Three-phase
PC
method, one wattmeter will meabalanced
Load
sure only the power of one phase.
Three-phase
N
Hence, the total power is taken
balanced supply
as three times the wattmeter
reading.
Y
B
Figure 9.16 O
ne-wattmeter Method of Measuring
Power of a Star-connected Balanced
Three-phase Load
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\ Total power = 3 × VPh IPh cos f
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Three-phase Systems and Circuits 377
9.8.2 Two-wattmeter Method
This method requires only two wattmeters to measure three-phase load for balanced as well as
unbalanced loads. In this method, two wattmeters are connected in two phases and their pressure
coils are connected to the remaining third phase, as shown in Figure 9.17.
R
IR
W1
CC
R
i
PC
IR
W1
CC
PC
i
i
N
IB
B
IY
Y
i
i
PC
Y
W2
CC
B
IB
i
IY
PC
W2
CC
Figure 9.17 T wo-wattmeter Method of Measuring Power for Star- and
Delta-connected Load
This method of measurement is useful for balanced and unbalanced loads.
Let us consider the measurement of three-phase power of a star-connected load using two
single-phase wattmeters as shown in Figure 9.18(a). We will calculate the power measured by
the two wattmeters separately. Let W1 and W2, respectively, be the two-wattmeter readings. The
current flowing through the current coil of wattmeter W1 is IR. The voltage appearing across its
pressure coil is VRB. The wattmeter reading will be equal to W1 = VRB IR cos of angle between
VRB and IR. Similarly, the wattmeter reading W2 will be equal to W2 = VYB IB cos of angle
between VYB and IB. We will now draw the phasor diagram, and calculate W1 and W2.
From the phasor diagram, as shown in Figure 9.18(b), we get the equation as follows:
W1 − VRB I L cos(30 − f ) = 3 VPh cos(30 − f ) = VL I L cos(30 − f ) (9.14)
Further,
W2 = VYB I Y cos(30 + f ) = 3VPh cos(30 + f ) = VL I L cos(30 + f ) (9.15)
We know that the total power in a three-phase circuit is 3VPh IPh cos f or equal to 3 V L I L cosf .
R
W1
R
VRB = VR − VB
IR
VRB
VR
IR
B
Y
IY
VYB
B
W2
(a)
30°
f
Y
f
VB
30 − f −VB
30°30 + f
VYB
IY VY
(b)
Figure 9.18 M
easurement of Three-phase Power Using Two Single-phase Wattmeters
(a) Circuit Diagram and (b) Phasor Diagram
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378 Network Analysis and Synthesis
Let us add the two wattmeter readings, that is, W1 and W2.
= W1 + W2 = 3 VPh I Ph cos (30 − f ) + 3 VPh I Ph cos (30 + f )
= 3VPh I Ph [cos (30 − f ) + cos (30 + f )]
= 3 VPh I Ph 2 cos f cos 30°
= 3 VPh I Ph 2 cos f
3
2
= 3VPh I Ph cos f
or
W1 + W 2 = 3V L I L cosf (9.16)
Thus, it is proved that the sum of the wattmeter readings is equal to the three-phase power.
Now, when the two wattmeter readings are subtracted from each other, we obtain the following form:
W1 − W2 = 3 VPh I Ph [cos (30° − f ) − cos(30° + f )
= 3 VPh I Ph 2 sin f sin 30°
or
3 (W1 − W2 ) = 3VPh I Ph sin f
or
3 (W1 − W 2 ) = 3V L I L sin f
(9.17)
Dividing equation (9.17) by equation (9.16)
3 (W1 − W 2 )
=
W1 + W 2
or
Power factor
3 V L I L sin f
3 V L I L cos f
f = tan −1
= tan f
3 (W1 − W2 )
W1 + W2
cos f = cos tan −1
3(W1 − W2 )
(9.18)
W1 + W2
Thus, from the two wattmeter readings, we can calculate the total active and reactive powers and
the power factor of the circuit.
Effect of Change in Power Factor on Wattmeter Readings
We will now study the effect of change in load power factor on the wattmeter readings. Let us
rewrite the wattmeter readings as follows:
W1 = 3 VPh I Ph cos (30 − f )
W2 = 3 VPh I Ph cos (30 + f )
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We will consider a power factor of unity, 0.5, less than 0.5 and 0 and study the effect on the
wattmeter readings.
1. At unity power factor, when cos f = 1, that is, f = 0
W1 = 3 VPh I Ph cos 30°
W2 = 3 VPh I Ph cos 30°
Thus, at power factor = 1, both the wattmeter readings will be positive and of equal value.
2. At 0.5 power factor, cos f = 0.5, that is, f = 60°
W1 = 3VPh I Ph cos( −30°)
= 3 VPh I Ph cos 30°
W2 = 3 VPh I Ph cos(30° + 60°) = 0
Thus, at power factor equal to 0.5, one of the wattmeters will give zero reading.
3. When the power factor is less than 0.5, that is, f > 60. Let us observe the wattmeter readings.
W1 = 3 VPh I Ph cos (30 − f )
W2 = 3 VPh I Ph cos (30 + f )
When f > 60, W1 will give positive readings but W2 will give a negative reading. Thus, for
power factor less than 0.5, that is, for f > 60°, one of the wattmeters will give a negative
reading.
4. When the load is purely inductive or capacitive, the power factor will be zero, that is, f = 90°
W1 = V L I L cos (30° − 90°) = V L I L cos 60°
W 2 = V L I L cos (30° + 90°) = −V L I L sin 30°
Both the wattmeters show equal but opposite readings. Hence, the total power consumed
will be zero.
9.8.3 Three-wattmeter Method
In this method, three wattmeters are used to measure three-phase power. Three wattmeters
are connected in each phase, as shown in Figure 9.19, and their pressure coils are connected
between each phase and the neutral. This method is valid for the measurement of three-phase
power for balanced and unbalanced loads. The main drawback of this method is the requirement
of three wattmeters.
Example 9.11 In the two-wattmeter method of power measurement for a three-phase
load, the readings of the wattmeter are 1000 W and 550 W. What is the power factor of the
load?
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380 Network Analysis and Synthesis
Solution:
W1 = 1000 W, W2 = 550 W
Power factor of load
cos f = cos tan
= cos tan −1 3
or
N
W −W2
3 1
W1 + W 2
−1
W1
R
i1
W2
B
1000 − 550
1000 + 550
cos f = cos tan −1 0.5 = 0.9
Y
i3
i2
W3
Figure 9.19 M
easurement of Three-phase
Balanced or Unbalanced
Example 9.12 In the measurement
Power Using Three Single-phase
of three-phase power by the twoWattmeters
wattmeter method, for a certain load,
one of the wattmeters reads 20 kW and the other 5 kW after the current coil connection of
one of the wattmeters has been reversed. Calculate the power and power factor of the
load.
Solution:
W1 = 20 kW
W2 = – 5 kW
P = W1 + W2 = 20 – 5 = 15 kW
Power factor of the load
= cos tan −1
W1 − W 2
3
W1 + W 2
= cos tan −1
20 − ( −5)
3
20 + ( −5)
= 0.3273 lagging
Example 9.13 Draw the connection diagram for the measurement of power in a three-phase
star-connected load using the two-wattmeter method. In one such a measurement, the load
connected was 30 kW at 0.7 pf lagging. Find the reading of each wattmeter.
Solution: The connection diagram for the measurement of power in a three-phase Y-connected
load using the two-wattmeter method has been shown in Figure 9.17. We know that the reading
of the two wattmeters will be 3VPh I Ph cos (30 − f ) and 3 VPh I Ph cos (30 + f ), respectively.
For star connection,
3 V Ph = V L and I Ph = I L
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Three-phase Systems and Circuits 381
The total load
P = 30 kW
Power factor cos f = 0.7 lagging
Phase angle
f = cos−1 (0.7) = 45.57° lagging
V LI L =
Therefore,
=
P in kW × 1000
3 cos f
30 × 1000
3 × 0.7
= 24743.6 VA
W1 = VL IL cos (30 – f )
Reading of wattmeter
= 24743.6 cos (30 – 45.57°)
= 23.835 kW
W2 = VL IL cos (30 + f )
Reading of wattmeter
= 24743.6 cos (30 + 45.57°)
= 6.165 kW
Thus, the total power is calculated as P = W1 + W2 = 23.835 + 6.165
= 30 kW
Example 9.14 A three-phase balanced load connected across a three-phase, 400 V AC supply
draws a line current of 10 A. Two wattmeters are used to measure the input power. The ratio of
two wattmeter readings is 2:1. Find the readings of the two wattmeters.
W
Solution: Let the ratio of wattmeter readings be X, that is, 2 = X
W1
W −W2 
tanf = 3  1
 W1 + W 2 
and
Now, we will divide both numerator and denominator by W1. Then, tan f will be given as
follows:
 1 − W2 /W1 
1− X 
= 3
= 3

 1 − X 
 1 + W2 /W1 
and power factor
cos f =
=
1
1
=
sec f
1 + tan 2 f
1
1 + 3([1 − X ) /[1 − X )]2
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382 Network Analysis and Synthesis
W2 1
= = 0.5
W1 2
Substituting
cos f =
1
1 + 3 (1 − 0.5 1 + 0.5)
= 0.866
f = cos–1 (0.866)
= 30°
Wattmeter reading
W1 = VL IL cos (30° – 30°) = 400 × 10 × cos 0° = 4000 W
Wattmeter reading
W2 = VL IL cos (30° + 30°) = 400 × 10 × cos 60° = 1000 W
Example 9.15 Three equal impedances, each consisting of R and L in series are connected in
star and are supplied from a 400 V, 50 Hz, three-phase, three-wire balanced supply system. The
power input to the load is measured by the two-wattmeter method and the two wattmeters read
3 kW and 1 kW,respectively. Determine the values of R and L connected in eachphase.
Solution:
Reading of wattmeter 1
W1 = 3 kW
Reading of wattmeter 2
W2 = 1 kW
P = W1 + W2 = 3 + 1 = 4 kW
Total power
Power factor of the circuit,
cos f = cos tan −1
W1 − W 2
3
W1 − W 2
3 −1
3
3 +1
= cos 40.89
= 0.7559 lagging
= cos tan −1
Line current
P
IL =
3 V L cosf
4 × 1000
= 7.64 A = Phase current I P
3 × 400 × 0.7559
=
Impedance of the circuit per phase Z =
V P 400 3
=
IP
7.64
= 30.237 W
R = Z cos f = 30.237 × 0.7559 = 22.856 W
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Three-phase Systems and Circuits 383
Reactance per phase
X L = Z 2 − R2
= (30.237) 2 − ( 22.856) 2
= 19.796 Ω
Inductance per phase
L=
XL
2p f
19.796
2p × 50
= 0.063 H
= 63 mH
=
Example 9.16 The power input to a three-phase motor is measured by two single-phase
wattmeters. The total input power has been measured as equal to 15 kW and the power factor
calculated as 0.5. What have been the readings of the two wattmeters?
Solution: Total power = W1 + W2 = 15 kW
We have to calculate W1 and W2
When we measured three-phase power by the two-wattmeter method, the readings of the two
wattmeters are as follows:
W1 = VL IL cos (30 – f )
and
W2 = VL IL cos (30 + f )
cos f = 0.5, f = 60°
3 V L I L cos f = 15 kW
V LI L =
15
3 × 0.5
= 17.3 kVA
W1 = V L I L cos (30 − f )
= 17.3 cos (30 − 60°)
= 17.3 × 0.866 = 15 kW
W 2 = V L I L cos (30 + f )
= 17.3 cos 90°
= 17.3 × 0
=0
Total power
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384 Network Analysis and Synthesis
Thus, at a load power factor of 0.5, one of the wattmeters has given zero reading. This has been
explained earlier under the effect of change in power factor on wattmeter readings.
9.8.4 Star to Delta and Delta to Star Transformation
Load impedances may often be connected in either star or in delta. It may be necessary to convert
star connected impedances to equivalent delta or delta connected impedances into equivalent
star for the purpose of circuit analysis. The formulae for such conversion are given as follows.
Conversion of Star Load into Delta Load
R
R
ZR
ZB
ZBR
ZRY
ZY
B
B
Y
Y
ZYB
Figure 9.20 Star to Delta Transformation
The equivalent delta impedances in terms of star impedances across the three terminals as
shown in Figure 9.20, are as follows:
Z RY = Z R + Z Y +
ZR ZY
ZB
Z YB = Z Y + Z B +
ZY ZB
ZR
Z BR = Z B + Z R +
ZB ZR
ZY
Conversion of Delta Load into Star Load
Load impedances may also be connected in delta. For the conversion of delta connected impedances into equivalent star connected impedances, as shown in Figure 9.21, the following relations will hold good.
R
R
ZBR
ZR
ZRY
ZB
B
ZYB
Y
B
ZY
Y
Figure 9.21 Delta to Star Transformation
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Three-phase Systems and Circuits 385
ZR =
Z RY Z BR
Z RY + Z YB + Z BR
ZY =
Z RY Z YB
Z RY + Z YB + Z BR
ZB =
Z YB Z BR
Z BR + Z YB + Z BR
Example 9.17 A three-phase 400 V, 50 Hz
supply is connected across a three-phase load as
shown in Figure 9.22. Calculate the equivalent
delta load.
Solution: The equivalent delta load is shown
by assuming clockwise phase sequence, that is,
phase sequence of RYB as shown in Figure 9.23
are calculated as
R
ZR = 2 + j 3
400V
400V
ZY = 1 − j 2
ZB = 3 + j 4
Y
B
400V
Figure 9.22
R
R
ZR = 2 + j 3
ZBR
ZRY
ZY = 1 − j 2
ZB = 3 + j 4
B
B
Y
ZYB
Y
Figure 9.23
Z RY = Z R + Z Y +
ZR Zr
ZB
= 2 + j3 + 1 − j3 +
= 3+
( 2 + j 3) (1 − j 3)
3 + j4
22 + 32 ∠ tan
−1
3
2×
12 + 32 ∠ − tan −1
4
3
3.61 ∠56.3° × 3.162 ∠ − 66°
= 3+
5 ∠53.13°
= 3 + 2.23∠ − 62.83°
= 3 + 2.23[cos 62.83° − j sin 62.83]
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1
32 + 4 2 ∠ tan −1
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Z RY = Z R + Z Y +
ZR Zr
ZB
= 2 + j3 + 1 − j3 +
( 2 + j 3) (1 − j 3)
3 + j4
22 + 32 ∠ tan
−1
386 Network Analysis and Synthesis
= 3+
3
2×
12 + 32 ∠ − tan −1
4
3
3.61 ∠56.3° × 3.162 ∠ − 66°
= 3+
5 ∠53.13°
= 3 + 2.23∠ − 62.83°
= 3 + 2.23[cos 62.83° − j sin 62.83]
= 4.02 − j 1.998 = 4.48∠ − 26°
3
1
32 + 4 2 ∠ tan −1
Z YB = Z Y + Z B +
Z YZ B
ZR
(1 − j 3) (3 + j 4)
2 + j3
1
4
12 + 32 ∠ − tann −1 × 32 + 4 2 ∠ tan −1
3
3
= 4 + j1 +
3
= 1 − j3 + 3 + j 4 +
22 + 32 ∠ tan −1 2
3.162 ∠ − 66° × 5 ∠53.13°
= 4 + j1 +
3.6 ∠58°
= 4 + j1 + 4.39∠ − 66° + 53.13° − 58°
= 4 + j1 + 4.93 ∠ − 71.87°
= 4 + j1 + 4.93(cos 71.87° − j 71 − 87°).
= 5.53 − j 353.3 = 353.4∠ − 89°
Z BR = Z B + Z R +
ZB ZR
ZY
= 3 + j 4 + 2 + j3 +
(3 + j 4)( 2 + j 3)
1 − j3
= 5 + j7 +
5∠53.13° × 3.61∠56.3°
3.162∠ − 66°
= 5 + j7 +
18.05 ∠109.43
3.162∠ − 66°
= 5 + j 7 + 5.72 ∠175.43°
= 5 + j 7 + 5.73 (cos175.43° + j sin175.43°)
= −0.707 + j 7.46 = 7.49∠95.4°
Example 9.18 A 400 V, 50 Hz, three-phase supply is connected across a delta-connected load
having identical impedance of 4 + j3 for each phase. What should be the values of equivalent
star impedances so as to keep the line currents unaltered?
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Three-phase Systems and Circuits 387
Solution: The equivalent star impedances as shown in Figure 9.24 are calculated as
R
R
4 + j3
ZR
4 + j3
ZB
B
ZY
Y
4 + j3
B
Y
Figure 9.24
ZR =
=
( 4 + j 3) ∗ ( 4 + j 3)
( 4 + j 3)( 4 + j 3) 4 + j3
=
=
( 4 + j 3) + ( 4 + j 3) + ( 4 + j 3)
3( 4 + j 3)
3
4 2 + 32 ∠ tan
3
−1
3
4
= 1.67 ∠40° h Ω
The values of ZY and ZB will be the same as ZR.
9.9 MORE NUMERICALS BASED ON THREE-PHASE
BALANCED LOAD
Example 9.19 A star-connected three-phase load has a
resistance of 4 Ω and a capacitive reactance of 10 Ω in each
phase. If it is fed from a 400 V three-phase balanced supply, find
the line current, total volt amperes, active and reactive power.
Solution: The circuit is drawn as shown in Figure 9.25.
V
Given,
VL = 400 V; VP = L
3
400
VP =
= 230.94 V
∴
3
and
Z = 4 − j10
= 10.77 ∠−75.77°
IL = IP =
a
Ia
4Ω
−j 10 Ω
−j 10 Ω
b
c
Ib
−j 10 Ω
4Ω
4Ω
Ic
Figure 9.25
VP 230.94
=
= 21.44 A
| Z | 10.77
Volt ampere = 3V L I L = 3 ( 400 × 21.44)
= 14855.64 VA
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388 Network Analysis and Synthesis
P (active power) = VA cos f = 14855.64 cos (−75.77°)
= 5518.58 W
Q (reactive power) = VA sin f = 14855.64 sin (−75.77°)
= − 13792.5 VAR
or
13792.5 VAR (capacitive)
Example 9.20 A delta-connected three-phase load
has a resistance of 4 Ω and a capacitive reactance of
5 Ω in each phase. It is fed from a 400 V three-phase
balanced supply as shown in Figure 9.26. Find the
line current, total volt ampere, active and reactive
power.
a
Ia
4Ω
−j 5 Ω
three-phase
supply
−j 5 Ω
b
4Ω
Ib
4Ω
Solution: Given
c
−j 5 Ω
Ic
VL = 400 V
Figure 9.26
VP = VL = 400 V (∵ of Delta connection)
Z = (4 - j5) Ω
= 6.403∠− 57.04°
IP =
VP
400
=
= 62.47 A
| Z | 6.403
Line current I L = 3 I P = 3 (62.47) = 108.19 A
Volt ampere (S ) = 3 V L I L = 3 (108.19) ( 400)
= 74960.46 VA
Active power ( P ) = 3 V L I L cos f
= 74960.49 cos ( −57.04°)
= 46831.7W
Reactive power (Q ) = 3 V L I L sin f
or
= 74960.46 sin( −57.04)
= −58530.85 VARs
58530.85 VARs capacitve.
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Three-phase Systems and Circuits 389
Example 9.21 Three identical impedances of 5∠30° each are connected in star and another
set of three identical impedances 9∠60° are connected in delta. If both these sets of impedances
are connected across a balanced three-phase 400 V supply, find the line current, total volt
amperes, active power and reactive power.
Solution: The three delta impedances can be converted equivalent star impedances, the value
of each impedance is equal to the following:
Zy =
Z ∆ 9∠60°
=
= 3∠60° Ω
3
3
Now, the load across each phase consists of two impedances 5∠30° and 3∠60° in parallel.
Thus, equivalent load impedance across each phase is given as follows:
Z ef =
(5∠30°) (3∠60°)
5∠30° + 3∠60°
=
15∠90°
4.45 + j 2.26 + 1.76 + j 2.43
=
15∠90°
6.21 + j 4.69
=
15∠90°
7.78∠41.18°
= 1.93∠48.83° Ω
VP =
VL
(∵ of star connection )
3
400
=
3
= 230.94 V
Line current I L = I P =
230.94
= 119.65 A
1.93
Total volt ampere (S ) = 3 V L I L
= 3 ( 400) (119.65)
= 82899.08 VA
Active power ( P ) = 3 V L I L cos f
= 82899.08 cos ( 48.83°)
= 59685.85 W
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390 Network Analysis and Synthesis
Reactive power (Q ) = 3 V L I L sin f
= 82899.08 sin ( 48.83°)
= 57531.35 VARs.
Example 9.22 Three impedances each 4 – j5 Ω are connected
in star across 400 V three-phase supply as shown in Figure 9.27.
1. Find line currents Ia, Ib and Ic
a
Ia
4Ω
400 V,
n
3−f
−j 5 Ω
supply
2. Find total volt amperes
Solution: Given
−j 5 Ω
−j 5 Ω
4Ω
4Ω
VL = 400 V
400
VP =
3
= 230.94 V
Van = 230.94 ∠0
Vbn = 230.94 ∠ −120°
b
c
Ib
Ic
Figure 9.27
Vcn = 230.94 ∠ +120°
V
230.94
230.94
I a = an =
=
Z
4 − j 5 6.403∠ − 57.04°
= 36.067∠57.04°
Ib =
V bn 230.94 ∠ − 120°
=
= 36.067∠ − 62.96°A
Z
6.403∠ − 57.04°
Ic =
V cn
230.94 ∠120°
= 36.067∠177.04°A
=
Z
6.403∠ − 57.04°
Total volt ampere (S ) = 3 V L I L
= 3 ( 400) (36.067)
= 24987.2 VA
Active power (P ) = 3 V L I L cos f
= 24987.2 cos( −57.04°)
= 15610.8 W
Reactive power (Q) = 3 V L I L sin f
= 3 V L I L sin( −57.04°)
= 24987.2 sin( −57.04°)
= −19510.58 VARs
= 19510.58 (capacitive).
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Example 9.23 A balanced 440 V three-phase system
has delta-connected load with ZAB = 15∠90°, ZBC =
10∠30° and ZCA = 10∠0° as shown in Figure 9.28.
Find phase and line currents.
A
Solution: Let
B
VAB = 440 ∠0 V
IA
IAB
10∠0°
C
VBC = 440 ∠−120°V
15∠90°
ICA
IC
10∠30°
IBC
IB
Figure 9.28
VCA = 440 ∠120°V
440 ∠0°
= 29.3∠ − 90° A
15∠90°
440 ∠ − 120°
=
= 44 ∠ − 150° A
10 ∠30°
440 ∠120°
=
10 ∠0°
= 44 ∠120°
I AB =
I BC
I CA
Therefore, the line currents are given as follows:
IA = IAB − ICA = − j 29.3 − 44∠120°
= − j29.3 − (13.59 + j41.85)
= −13.59 + j71.15
= 72.44 ∠−112.01 A
IB = IBC − IAB
= 44∠−150° − (− j29.3)
= 156.32 − j81.83 + j29.3
= 156.32 − j52.53
= 164.91 ∠− 20.63 A
and
IC = ICA − IBC
= 44∠120° − 44∠−150°
= 13.59 + j41.85 − (156.32 − j81.83)
= 142.73 + j123.68
= 188.86 ∠154.54 A
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392 Network Analysis and Synthesis
Example 9.24 A three-phase 415 V symmetrical supply is feeding a balanced threephase load. The line current is 80 A. The resistance between any two pair of terminals
is 5 Ω. Find the resistance, reactance and power factor per phase if the load is star
connected.
Solution: Let the impedance of each phase be R1 + jX1
Now, as seen from Figure 9.29, resistance between any two
terminals is 2R1, and therefore, 2R1 = 5
or
R1 = 2.5 Ω
Given
IL = 80 A
Therefore,
Further,
IP = IL = 80 A
a
R1
j X1
j X1
j X1
R1
R1
b
VL
415
VP =
=
3
3
c
Figure 9.29
Further, we know the following:
VP = IP ZP
= IP (R1 + jX1)
R1 + jX 1 =
or
Vp
IP
R1 + jX 1 =
415/ 3
80
R1 + jX 1 = 2.995
R12 X 12 = 2.995
R12 + X 12 = 8.97
or
X 12 = 8.97 − R12
= 8.97 − 2.52
= 8.97 − 6.25
= 2.72
or
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X 1 = 1.649 Ω
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Three-phase Systems and Circuits 393
R
R1
2.5
Power = 1 =
=
.
j1.649
2
5
Z
R
+
jX
+
factor
1
1
2.5
=
2.99∠37.12
2.5
=
= 0.836 lagging
2.999
Example 9.25 Three similar coils, arranged symmet­rically in
space are fed from a 400 V, three-phase 50 Hz supply. The coils
are connected in star and each coil has a resistance of 50 Ω and
inductive reactance of 120 Ω. The mutual inductance wM between
each pair of coils is 60 Ω. Find the current taken by each coil.
Solution: The circuit is shown in Figure 9.30.
Given
VL = 400 V
Therefore,
Vp =
400
3
a
Ia
50 Ω
j 120 Ω
N
j 120 Ω
50 Ω
b
c
= 230.94 V
j 120 Ω
50 Ω
Ib
Ic
Figure 9.30
Further, the voltage across phase a is given as follows:
Ia (50 + j120) + Ib ( jwM) + Ic ( jwM) = 230.94
or
Ia (50 + j120) + Ib ( j60) + Ic ( j60) = 230.94
or
Ia (50 + j120) + j60(Ib + Ic) = 230.94
(9.19)
Now, for balanced system, the equation can be written as follows:
Ia + Ib + Ic = 0
Ib + Ic = − Ia
or
Substituting this value in equation (9.19), we get the following:
Ia (50 + j120) − Ia ( j60) = 230.94
or
Ia (50 + j60) = 230.94
Ia =
or
230.94
50 + j 60
230.94
78.10 ∠55.77
= 2.96 ∠ − 55.77A
=
Power factor = cos (−55.77°)
= 0.640 lagging
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394 Network Analysis and Synthesis
Example 9.26 Three identical coils connected in star take 8 kW at 0.8 power factor when
connected to 440 V, three-phase three-wire supply. One of the coil is short-circuited. Find the
line currents.
Solution: Given
3 V L I L cosf = 8000
IL =
8000
3VL cos f
8000
3( 440)(0.8)
= 13.12A
= IP
=
Z=
=
VP
IP
440 / 3
13.12
= 19.36 ∠ cos −1 0.8
= 19.36 ∠40.96
= (15.48 + j11.61)Ω
After one coil is short-circuited as shown in Figure 9.31, let us assume the following: Vab = 440
∠0°V
a
Vbc = 440 ∠−120° V
Ia
Z
Vca = 440 ∠120° V
V ab
Z
440 ∠0°
=
15.48 + j11.61
440 ∠0°
=
19.36 ∠40.96°
= 22.72∠ − 40.96°A
= 18.177 − j13.63A
V
I c = cb
Z
−V bc
=
Z
Ia =
Z
b
c
Ib
Ic
Figure 9.31
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Three-phase Systems and Circuits 395
−440 ∠ − 120
19.36 ∠40.96
= −22.72∠ − 160.96 A
= −( −18.58 − j13.07) A
= 18.58 + j13.07 A
=
Ib = − (Ia + Ic)
= −[18.177 − j13.63 + 18.58 + j13.07]
= −[36.757 + j0.56]
= 36.76 ∠199.03°A
Example 9.27 In a balanced delta-connected load fed from three-phase 440 V system, the
phase current is 30 A and power factor angle is 50° lagging. Find the following:
1. Line current
2. Active power
3. Reactive power
Solution: Given IP = 30 A, VL = 440 V, f = 50°
In the of delta connection, the following can be calculated as follows:
1. I L = 3I p = 3 (30) = 51.96 A
2. Active power
P = 3 V L I L cos f
= 3 ( 440)(51.96) cos 50°
= 27999.78 W
3. Reactive power
Q = 3 V L I L sin f
= 3 ( 440)(51.96) sin50°
= 27999.78 VARs
Example 9.28 A balanced three-phase star-connected load is fed from 400 V, three-phase
50 Hz supply. The line current is 50 A and total power is 25 kW. If power factor is leading, find
the load impedance and its components.
Solution: Given
IL = IP = 50 A; VL = 400 V and P = 25 kW
We know
P = 3 V L I L cosf
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396 Network Analysis and Synthesis
P
cosf =
3V L I L
25 × 103
=
Therefore,
Z=
3 ( 400)(50)
= 0.722 leading
VP 400 / 3
=
50
IP
= 4.618 Ω
R = Z cos f = 4.618(0.722) = 3.33 Ω
X C = Z sinf = 4.618 1 − cos 2 f
= 4.618 1 − 0.7222 = 3.195 Ω
1
wC
1
1
C=
=
w X c 2p fX c
Xc =
or
=
1
= 9.96 × 10 −4 F
2p (50)(3.195)
Example 9.29 A three-phase alternator is supplying 150 kW at 1100 V and at 0.8 power factor.
The alternator is star connected. Find active and reactive components of the line current. If the
power factor is raised to 0.9 keeping the voltage and current constant, find the new power output.
Solution: Given cosf = 0.8,
sinf = 1 − cos 2 f
= 1 − (0.8) 2
= 0.6
P = 150 kW; VL = 1100 V
Now,
or
P =P 3= V L3I LVcos
L I Lfcos f
P P
I L =I L =
V Lfcos f
3V L 3cos
3
=
150
150 ×
103× 10
=
98A.41 A
= 98=.41
3 (1100
)(0.8)() 0.8)
3 (1100
Active component of current = ILcos f
= 98.41 (0.8) = 78.73 A
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Three-phase Systems and Circuits 397
Reactive component of IL = IL sin f
= 98.41 (0.6) = 59.046 A
When the power factor is raised to 0.9, the value of P is calculated as follows:
P = 3V L I L cos f
= 3 (1100) (98.41) (0.9)
= 168.74 kW
9.10 Method of Solving PROBLEMS ON UNBALANCED LOAD
In reality, it is difficult to maintain a perfect balanced three-phase load on a power system. Unbalance
load condition occurs due to un even loading of the three phases, causing either unequal magnitude
of the three-phase impedances or the impedance angles. Unbalance load condition will occur due
to unbalanced voltage supply also. In this section, we will take up examples of unbalanced load.
Example 9.30 A 400 V three-phase four-wire system has loads of 4 − j7, 3 + j4, 4 + j5 Ω
connected in star.
Find the line currents, neutral current and total power.
Solution:
Let Van =
400
3
= 230.95∠ 0 V; Vbn = 230.95∠ − 120 V and Vcn = 230.95∠ + 120 V
Ib =
Ia =
Vbn
Zb
230.95∠ − 120°
3 + j4
230.95∠ − 120°
=
5∠59.03
= 46.191∠ − 179.03 A
Van 230.95∠0
=
Za
4 − j7
=
230.95∠ 0
8.062∠ − 66.95
= 28.65∠66.95 A
= (14.21 + j 24.87) A
=
= −43.7 − j14.94 A
Ic =
Vcn
Zc
230.95∠120°
4 + j5
230.95∠120°
=
6.403∠57.04°
= 36.06 ∠62.96°A
= (19.82 + j 30.12)A
=
Current in neutral wire = − (Ia + Ib + Ic)
= − [14.21 + j24.87 − j14.94 + 19.82 + j30.12]
= 9.67 − j40.05 A
= 41.2 ∠−84.9°A
Total power = (230.95) (28.65) cos 66.95°+ (230.95)(46.191) cos
179.03 + (230.95)(36.06) cos 62.96°
= 3282.82 + (−10094.28) + 4576.6
= 2234.79 W.
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398 Network Analysis and Synthesis
Example 9.31 For the unbalanced deltaconnected load shown in Figure 9.32, find the phase
currents, line current and the total power consumed
by the load when phase sequence is abc.
Solution: Let
Vab = 100 ∠0°V
Iab
3Ω
5Ω
100 V
c
Vbc = 100 ∠−120°V = (− 50 − j86.6) V
a
a
j4Ω
−j 3 Ω
Ica
c
4Ω
j3Ω
Ibc
b
b
Vca = 100∠120°V = (− 50 + j86.6) V
Figure 9.32
\Phase currents can be calculated as follows:
I ab =
V ab
100
100
=
=
= 20 ∠ − 59.03°A = (12 − j15.99)A
Z ab 3 + j 4 5∠59.03°
I bc =
V bc 100 ∠ − 120° 100 ∠ − 120°
= 20 ∠ − 160.96°A = ( −16.35 − j11.51)A
=
=
5∠40.96°
Z bc
4 + j3
I ca =
Vca 100 ∠120°
100 ∠120°
=
=
3
5.83∠ − 34.4
Zca
5− j
= 17.15∠154.4° A
= ( − 12.93 + j11.26)A
\Line currents can be given as follows:
Ia′a = Iab− Ica = 12 − j15.99 − (− 12.93 + j11.26)
= 24.93 − j27.25
= 36.93∠− 52.82 A
Ib′b = Ibc − Iab= − 16.35 − j11.51 − (12 − j15.99)
= − 28.35 + j4.48
= 28.7 ∠190° A
Ic′c = Ica −Ibc= − 12.93 + j11.26 − (−16.35 − j11.51)
= 3.42 + j22.77
= 23.02 ∠90.5° A
\ Power consumed can be given as in the following
P = Iab2Rab + Ibc2Rbc + Ica2Rca
= 202 (3) + 202 (4) + (17.15)2 (5)
= 4.27 kW.
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Three-phase Systems and Circuits 399
Example 9.32 A three-phase, four-wire system having 250 V line-to-neutral has the following
loads connected between the respective lines and the neutral:
ZR = 20 ∠0° Ω; ZY = 20∠35° Ω, ZB = 20∠−55° Ω
Calculate the current in the neutral wire and the power taken by each phase when the phase
sequence is RYB.
Solution: Let
VRN = 250∠0° V
VYN = 250∠−120° V
VBN = 250∠+120° V
I R = I RN =
VRN
250
=
= 12.5A
ZR
20 ∠0°
I Y = I YN =
VYN 250 ∠ − 120°
= 12.5∠ − 155°A
=
ZY
20°∠35°
= ( −9.5 − j8.2)A
V
250 ∠ − 120°
I B = I BN = BN =
= 12.5∠175°A
ZB
20°∠ − 55°
= ( −11.54 + j 4.78)A
Neutral current = −(IR + IY + IB)
= −[12.5 − 9.5 − j8.2 − 11.54 + j4.78]
= (8.54 + j3.42)A = 9.2 ∠ 24.24°A.
Resistances of three phases are as follows:
RR = 20 Ω; RY = 20 cos 35°= 17.05 Ω and RB = 20 cos 55°= 12.98 Ω
\ Power taken by each phase is given as follows:
PR = 12.52 (20) = 3125 W
PY = 12.52 (17.05) = 2664.06 W
PB = 12.52 (12.98) = 2028.125 W
Example 9.33 A 400 V, three-phase supply feeds an unbalanced three-wire, star-connected
load. The branch impedances of load are ZR = (2 + j4) Ω, ZY = (1 + j2) Ω and ZB = (8 + j10) Ω.
Find the line currents.
Solution: The given star-connected load and its equivalent delta have been shown in Figure 9.33.
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400 Network Analysis and Synthesis
R
I1
R
IR
ZR = 2 + j 4
ZB = 8 + j 10
ZBR
ZY = 1 + j 2
IB
B
IR
IB
I2 B
Y
ZRY
ZYB
IY
Y
I3
IY
Figure 9.33
Z RY =
ZR ZY + ZY ZB + ZB ZR
ZB
( 2 + j 4)(1 + j 2) + (1 + j 2)(8 + j10) + (8 + j10)( 2 + j 4)
8 + j10
−6 + j8 − 12 + j 26 − 24 + j 52 −42 + j86
=
=
8 + j10
8 + j10
95.7∠128.9°
=
12.8∠57.04°
= 7.48∠ 71.86°Ω
=
Z RY
Similarly,
Z YB =
95.7∠128.9° 95.7∠128.9° 95.7∠128.9°
=
=
ZR
2 + j4
4.47∠70.48°
= 21.4 ∠58.42°Ω
Z BR =
95.7∠128.9°
ZY
95.7∠128.9°
1+ j2
95.7∠128.9°
=
2.24 ∠ 70.48°
= 42.72∠58.42°Ω
=
Now, taking VRY as reference, VRY = 400∠0
VYB = 400 ∠−120°
VBR = 400 ∠−240°
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Three-phase Systems and Circuits 401
IR =
V RY
400 ∠ 0°
=
= 53.47∠ − 71.86°Α
Z RY 7.48∠ 71.86°
IY =
V YB 400 ∠ − 120°
= 18.69∠ − 178.42°Α
=
Z YB 21.4 ∠58.42°
IB =
V BR
400∠ − 240°
=
= 9.36 ∠ − 298.42°Α
Z BR 42.72 ∠58.42°
Line current in delta load are written as follows:
I1 = IR − IB = 53.47∠−71.86°− 9.36 ∠−298.42°
= 22.87 − j48.33 + 0.232 − j9.357
= 23.102 − j57.68
= 62.13∠−75.74°A
Similarly, we can find other currents as in the following:
I2 = IY − IR
I3 = IB− IY
R E V I E W QU E S T I ON S
Short Answer Type
1. What is the difference between a single-phase winding and a three-phase winding?
2. Draw wave shapes of a three-supply.
3. What is the difference between a balanced load and an unbalanced load?
4. Show that the phasor sum of the three-phase balanced voltages is zero.
5. What do you mean by phase sequence of three-phase voltages?
6. Distinguish between the star connection and the delta connection of three-phase windings.
7. Derive the relationship between line current, line voltage, phase current and phase voltage
in case of star and delta connection of three-phase windings.
8. Prove that the power in a three-phase circuit is equal to 3 V L I L cos f.
9. Distinguish between the active power and the reactive power in a three-phase system.
10. What is the significance of low power factor of any load on the system?
11. Draw the circuit diagram for the measurement of three-phase power with two single-phase
wattmeters.
12. At what value of load power factor, the reading of one of the wattmeters in the two-wattmeters method of measurement of three-phase power will be zero?
13. What are the advantages of the three-phase system over the single-phase system?
14. Write the relationship between the phase voltage and the current in a delta-connected load.
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402 Network Analysis and Synthesis
15. Draw the connection diagram for three-phase resistive–inductive balanced load connected
across a three-phase supply. Further, draw the phasor diagram showing voltages and currents.
16. Discuss the concept of three-phase voltages. What is phase sequence and what is its
significance?
17. Are three-phase loads always balanced? Explain
18. Discuss the phenomenon of neutral shift. In which type of system, does it occur and why?
Derive an expression for calculating the neutral shift voltage.
19. If a load is unbalanced, how are the line currents calculated in a three-phase three-wire
system.
20. What are the advantages of three-phase system?
21. Derive the relationship between the phase and the line voltages and currents in a threephase delta-connected circuit. Draw phasor diagrams also.
22. Derive the relationship between the phase and line voltages and currents in a three-­phase
star-connected circuit. Draw phasor diagrams also.
23. Show that the total power in a three-phase balanced load is P = 3V L I L cosf
24. Compare three-phase star- and delta-connected systems.
25. Why is an unbalanced load not normally used on a three-phase, three-wire system?
Numerical Questions
1. Three coils having same resistance and inductance are connected in star. A three-phase
400 V supply is connected across the three coils. The power consumed by each coil is 800 W
and the load power factor is 0.8 lagging. What is the total power consumed by the coils.
If now the coils are connected in delta across the supply, what would be the total power
consumed? Further, calculate the line current of a delta connected load.
[Ans. 2400 W, 7200 W, 13 A]
2. A balanced three-phase star-connected load supplied from a 400 V, 50 Hz, three-phase supply system. The current drawn by each phase is 20 ∠ − 60° A. Calculate the line current,
phase voltage and total power consumed.
[Ans. 20 A, 230.94 V, 6928 W]
3. A delta-connected load has a resistance of 15 Ω and inductance of 0.03 H per phase. The
supply voltage is 400 V, 50 Hz. Calculate line current, phase current, phase voltage and
total power consumed.
[Ans. 39.1 A, 22.5 A, 400 V, 22.94 kW]
4. Three identical coils of resistance 20 Ω and inductance 500 mH are connected first in star
and then in delta across a 400 V, 50 Hz power supply. Calculate phase current, line current,
phase voltage and power consumed per phase.
[Ans. 1.46 A, 1.46 A, 230.94 V, 42.6 W; 2.53 A, 4.38 A, 400 V, 127.8 W]
5. Calculate the phase current and line current of a delta-connected load drawing 75 kW at
0.8 power factor from a 440 V, three-phase supply.
[Ans. 71 A, 122.97 A]
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Three-phase Systems and Circuits 403
6. The power consumed by a three-phase balanced load has been measured by two singlephase wattmeters. The readings of the two wattmeters are 8.2 kW and 7.54 kW. Calculate
the total power consumed and the load power factor.
[Ans. 15.7 kW, 0.997 lagging]
7. In the measurement of three-phase power by two single-phase wattmeters, it has been
observed that the ratio of the two wattmeters readings is 3:1. What is the power factor of
the load?
[Ans. 0.75]
8. Three identical coils are connected in star across a three-phase 415 V, 50 Hz supply. The
total power drawn is 3 kW at a power factor of 0.3. Calculate the resistance and inductance
of each coil.
[Ans. 5.16 Ω, 52.3 mH]
9. Two single-phase wattmeters are used to measure three-power. The readings of the
two wattmeters are 2000 W and 400 W, respectively. Calculate the power factor of
the circuit. What would be the power factor if the reading of the second wattmeters is
negative?
[Ans. 0.65, 0.36]
10. Three identical coils each having a resistance of 10 Ω and inductive reactance of 10 Ω are
first connected in star and then connected in delta across a 400 V, 50 Hz power supply. Calculate the line current in each case and the readings of two wattmeters connected for the
measurement of power.
[Ans. 16.33 A, 49 A, 6309 W and 1690 W, 18931 W, 5072 W]
11. A three-phase balanced star-connected load is connected to a 240 V three-phase supply. The
total volt amperes are 6000 VA and total active power is 4800 W. Find the resistance and
reactance of load in each phase.
[Ans. 7.68 Ω, 5.76 Ω]
12. Three impedances 5 ∠ 30°, 10 ∠ 45° and 10 ∠ 60° Ω are connected in star across a threephase three-wire 440 V system. Find the line currents.
[Ans. 35.7 A, 32.8 A, 27.7 A]
13. A three-phase four-wire 400 V system supplies three star-connected loads ZR = 10∠0°,
ZY = 15 ∠ 30° and ZB = 10 ∠ 30° Ω. If the phase sequence is RYB, find the neutral current.
[Ans. 10.94 A]
14. A balanced three-phase star-connected load has impedance 6 + j8 in each phase. Find the
total power if the voltage is 220 V.
[Ans. 2903.2 W]
15. Three loads (80 + j60) Ω, (31 + j59) Ω and (30 − j40) Ω are connected in delta to a threephase, 200 V supply. Find the phase currents and line currents.
[Ans. 3∠− 62.28° A, 4 ∠−66.8° A, 2 ∠ 83.1°;
4.784 ∠− 76° A, 5.82 ∠ 103.2°, 1.037∠−80.2°]
16. An unbalanced star-connected load is supplied from a symmetrical three-phase, 440-V,
three-wire system. The branch impedances of the load are
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404 Network Analysis and Synthesis
Z1 = ∠30°
Z2 = 10∠ 45°
Z3 = 10∠ 60°
Determine the line currents.
[Ans. 35.75∠−71.3° A, 32.7∠156.1° A, 27.67∠48° A]
17. Three identical coils of (8 + j6) Ω impedance are connected in delta across 400 V mains.
Calculate the power consumed.
[Ans. 38.4 kW]
18. The power consumed in a three-phase balanced star-connected load is 2 kW at a power
factor of 0.8 lagging. The supply voltage is 400 V, 50 Hz. Calculate the resistance and reactance of each phase.
[Ans. RPh = 51.3 Ω
XPh = 38.5 Ω]
19. A symmetrical three-phase 100 V three-wire supply feeds an unbalanced star-connected
load, with impedance of the load as ZR = 5∠0° Ω, ZY = 2∠90° Ω and ZB = 4∠−90° Ω. Find
the line currents.
[Ans. 27.06 ∠− 8.671°A, 19.7∠238.85°, 26.7∠128.33°]
20. A three-phase three-wire unbalanced load is star-connected. The phase voltages of two of
the arms are
VR = 100∠−10°; VY = 150∠100°.
Calculate voltage between star point of the load and the supply neutral.
[Ans. 31.37∠−68.63°]
21. A three-phase three-wire unbalanced load is star-connected. The phase voltages are
VR = 10 ∠ 0° V
VY = 8 ∠ 30° V
VB = 5 ∠ 45° V.
Assume the line voltage as 415 V, find line currents.
[Ans. 26.8 ∠ 132°A,
− 38.225∠10.26°A,
−32.766 ∠−125.9°A]
22. A three-phase four-wire, 380 V supply is connected to an unbalanced load having phase
impedances of ZR = 4 + j3, ZY = 4 − j3 and ZB = 2 Ω. The impedance of the neutral wire is
Zn = (1 + j2) Ω. Find the phase currents.
[Ans. 48.18 ∠−40.76°A,
51.26 ∠ 283.51°A,
97.55 ∠ 117.10°A]
23. Three impedance 20∠0° Ω, 16∠20° Ω and 25∠90° Ω are connected in delta across a balanced supply of 250∠0° V. Find the phase currents, active powers and reactive powers in
each phase.
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Three-phase Systems and Circuits 405
[Ans. Phase currents: 10∠−90°, 15.6 ∠−140°, 12.5 ∠ 120°;
Active powers: 0 W, 3662 W, 3125 W;
Reactive powers: 2500 VAR, 1333 VAR, 0 VAR]
24. Given a balanced three-phase, three-wire system with y-connected load for which line voltage is 230 V and impedance of each phase is (6 + j8) Ω. Find the line current and power
absorbed by each phase.
[Ans. IL = 13.3 A,
Power = 1061 W]
25. A three-phase 400-V, 50 Hz, AC supply is feeding a three-phase delta-connected load with
each phase having a resistance of 25 Ω , an inductance of 0.15 H and a capacitor of 120 µF
in series. Determine the line current, apparent power, active and reactive power.
[Ans. Line current = 21.4 A;
P = 11.45 W;
S = 14.83 kVA;
Q = 9.43 kVAR lagging]
M ultiple C hoice Q uestions
1. Three-phase system is used
(a)
(b)
(c)
(d)
for transmission of electrical power.
for generation of electrical power
for distribution of electrical power
for generation, transmission and distribution of electrical power.
2. In a three-phase system, the phase sequence indicates
(a)
(b)
(c)
(d)
the amplitude of voltages
the order in which the voltages obtain their maximum values.
the phase difference between the three voltages.
the frequency in which the phase voltages are changing.
3. In a star-connected system, the relationship between the phase and line quantities are
(a) VPh = VL
(b) VPh = 3VL
(c) 3IPh = IL
(d) IPh = IL
4. In a delta-connected system, the relationship between the phase and the line quantities are
(a)
3VPh = VL
(b) VPh = 3VL
(c) I Ph = I L
(d) I L = 3I Ph
5. Line currents drawn by a three-phase star-connected balanced load is 12 A when connected
to a balanced three-phase four-wire system. The neutral current will be
(a) 36 A
(b) 4 A
(c) 0 A
(d) 3 A
6. Power in a balanced three-phase system circuit is
(a)
3V LI L cosf
(b)
3VL I L
(c) 3VLIL cos f
(d) 3VP IP cos f
(c) 3VP IP
(d) 3VP IP cos f
7. Reactive power of a three-phase circuit is
(a)
3V LI L
(b)
3VL I L sinf
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406 Network Analysis and Synthesis
8. Power in single-phase AC circuit can be expressed as
(a) V I sin f
(c) V I cos f
(b) V I
(d) 3V I
9. One single-phase wattmeter can be used to measure power in a three-phase circuit when
(a)
(b)
(c)
(d)
the load is balanced
the load is delta-connected and is balanced
the load is balanced, star-connected and the neutral wire is available.
the load is balanced and star-connected.
10. A balanced three-phase sinusoidal power supply means
(a) three sinusoidal voltages of the same frequency and maximum value displaced in 120 time
phase
(b) three sinusoidal voltages of any frequency but having the same maximum value
(c) three sinusoidal voltages of any frequency and maximum value with no time phase
displacement between them.
(d) three sinusoidal voltages of any value but having a time phase displacement of 120 between
them.
11. An unbalanced three-phase supply system will have
(a)
(b)
(c)
(d)
three unequal voltages
three voltages having unequal time phase displacement between them
three voltages of unequal magnitude and angular displacement among them
all of the above.
12. In the two-wattmeter method of measuring three-phase power, the reading of the two wattmeters will be equal when the power factor of the circuit is
(a) 0
(b) 1
(c) 0.5
(d) 0.866
13. In the two-wattmeter method of measuring three-phase power, the reading of one of the
wattmeters can be negative when the power factor angle is
(a) more than 60
(b) less than 60
(c) more than 30
(d) less than 30
14. Four equal resistance of 100 Ω, each connected in delta is supplied from 400 V three-phase
star-connected supply, the line current drawn will be
(a) 12 A
(b) 4 A
(c) 6.928 A
(d) 13.856 A
15. A three-phase load is balanced when
(a)
(b)
(c)
(d)
magnitudes of three impedances are equal
the magnitudes of currents drawn by the three loads are equal
all the three loads are equal in magnitude and phase angle.
the three impedances are pure resistances.
16. In a three-phase balanced star-connected load, the neutral current is equal to
(a) line current
(b) phase current
(c) zero
(d) none of these.
17. The relationship between the line and the phase voltage of delta-connected load is given by
(a) VL = 3VP
(b) VL =
VP
3
(c) VL = VP
(d) VL =
2VP
p
18. The power measurement in balanced three-phase circuit can be done by
(a) one-wattmeter method only
(c) three-wattmeter method only
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(b) two-wattmeter method only
(d) any one of the above.
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Three-phase Systems and Circuits 407
19. In a three-phase system, the EMFs are
(a) 30° apart
(b) 45° apart
(c) 60° apart
(d) 120° apart
20. Power absorbed by a three-phase load is given by
(a)
3VP I P sinf
(b)
3VL I L sinf
(c)
3VL I L cosf
(d)
3VP I P cosf
21. The phenomenon of neutral shift occurs in
(a)
(b)
(c)
(d)
three-phase three-wire system feeding balanced star-connected load
three-phase three-wire system feeding an unbalanced star-connected load.
three-phase four-wire system
none of these
22. Neutral shift voltage is given by
23.
24.
25.
26.
27.
28.
(a) V on =
V an + V bn + V cn
Za + Zb + Zc
(b) V on =
V an + V bn + V cn
Ya + Yb + Yc
(c) V on =
V anY an + V bnY b + V cnY c
Za + Zb + Zc
(d) V on =
V anY a + V bnY b + V cnY c
Ya + Yb + Yc
In a balanced three-phase system, the neutral current is zero (T/F).
The two wattmeter method can be used only for balanced loads (T/F).
Phase sequence is immaterial if load is __________
Apparent power in three-phase system = __________
Three-phase three-wire system does not have __________ conductor.
Phase reversal of a four-wire unbalanced load supplied from a balanced three-phase supply
changes
(a)
(b)
(c)
(d)
magnitudes of phase currents
the power consumed
only the magnitude of neutral current.
magnitudes as well as phase angle of neutral current.
29. If positive phase sequence of a three-phase load is RYB, the negative sequence will be
(a) YRB
(b) BYR
(c) RBY
(d) all of (a), (b) & (c)
ANS W E RS
1. d 2. b 3. d 4. d 5. e 6. a 7. b 8. c 9. c
11. d 12. b 13. a 14. c 15. c 16. a 17. c 18. d 19. d
21. b 22. d
27. neutral
23. True
28. d 29. d
24. False
M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 407
25. Passive
26.
10. a
20. c
3V L I L
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Network Functions −
s-Domain Analysis of
Circuits
10
Chapter OBJECTIVES
After carefully studying this chapter, you should be able to do the following:
¾¾ Explain the concept of complex ¾¾ Explain the concept of zeros and poles
frequency.
of a network function (transfer function) and plot them in s-plane.
¾¾ Convert circuit parameters from timedomain to s-domain.
¾¾ Formulate, from the pole-zero diagram, the network function.
¾¾ Calculate the driving point impedance or admittance of one-port ¾¾ Find the time-domain response of a
network.
network output from the given network
function.
¾¾ Calculate the network functions of a
two-port network.
¾¾ State and explain the Routh-Hurwitz
theorem defining the stability criterion
¾¾ Find the transfer functions of R-C and
of an active network.
R-L networks.
¾¾ Express a transfer function of networks ¾¾ Write the characteristic equation of a
network function and apply Routh's
in generalised form.
stability criterion.
10.1 INTRODUCTION
Transfer functions are commonly used to characterise the input–output relationship of networks
and systems. The transfer function of a linear time invariant system is defined as the ratio of Laplace
transform of the output to the Laplace transform of the input, with all initial conditions being zero.
This chapter aims at introducing the concept of transfer function which is mathematically
identical to impedance and admittance function (also called network functions). We will determine the transfer function of circuits in terms of poles and zeros and then use Routh–Hurwitz
stability criterion to analyse the circuit behaviour.
10.1.1 Terminals and Ports
Each end of a circuit element is called a terminal. At least two terminals are required for connecting a network to a source or to a load. A pair of terminals is called a port. We will determine
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Network Functions − s-Domain Analysis of Circuits 409
the impedance and admittance function for a one-port and two-port networks. Because of the
similarity of impedance and admittance, these two quantities are given a common name called
‘immittance’. Therefore, an immittance is an impedance or an admittance.
Network parameters are converted in s-domain before determining the impedance or admittance functions. We will first understand the concept of s which is called the complex frequency.
10.1.2 Concept of Complex Frequency
The solution of differential equations for networks is expressed in the following form:
i(t) = I0 e s nt
where sn is a complex number, that is, the root of the characteristic equation and is expressed
as follows:
sn = s n+ jwn
This complex number has two parts, namely the real part and the imaginary part. The complex
quantity sn = sn + jwn is defined as the complex frequency. The complex frequency appears in
exponential form. The exponential function increases exponentially for sn > 0 and decreases
exponentially for sn < 0. The concept of complex frequency is explained further as follows.
The exponential function e ± jw t can be considered in terms of a rotating vector of unit length
as shown in Figure 10.1. The positive sign stands for counter clockwise rotation, while the negative sign stands for clockwise rotation.
For the counter clockwise rotation, the real part
Im
of e ± jw nt , that is, the projection on the real axis is
wn
equal to cos wnt and the imaginary part, that is, the
wt
projection on the imaginary axis is equal to sin wnt.
Re
s nt jw nt
The product e e
is visualised as a rotating
vector whose magnitude is not constant at unity but
changes continuously. The projection of such a notating vector for sn < 0 will show a damped or gradually
reducing sinusoid. For sn > 0, the projection will show Figure 10.1 Concept of Complex
Frequency Illustrated
increasing oscillations, as shown in Figure 10.2.
Output, c (t )
Output, c (t )
t
Output, c (t )
t
Here sn = 0 (no change in magnitude) s n < 0 (magnitude decreasing)
w n = 7 (cycles/sec)
wn = 7
t
s n > 0 (magnitude increasing)
wn = 7
Figure 10.2 Effect of Change in sn on Output in Time Domain
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410 Network Analysis and Synthesis
Thus, we can say that the imaginary part of the complex frequency, that is, ± jwn shows the
sinusoidal oscillations whereas the real part of the complex frequency, that is, ± sn shows the
exponential rise or decay of the oscillations.
We will again return to the concept of complex frequency s and explain its significance.
Complex frequency (s) is given by the following form:
s = s + jw
It consists of two parts, that is, real and imaginary parts.
Real part of s, that is, s is called Neper frequency and imaginary part of s, that is, w is called
radian frequency.
Radian frequency (w) represents the number of oscillations per second
and
Neper frequency (s ) represents the magnitude of the oscillations.
Let us consider
i(t) = I0 es t
Substituting the value of s = s + jw, we get
i(t) = I0 e(s t+ jw) = I0 es t e jw t = I0 es t (cos w t + j sin w t)
i(t )
Case I: When s = 0, then i(t) = I0 (cos w t +
i(t )
wt
wt
sin w t
cos w t
Figure 10.3 V
ariation of Current, i(t) with
s=0
j sin w t); that is, there will be no attenuation
or variation in the magnitude of oscillation.
Figure 10.3 shows the variation of real output
with s = 0.
Case II: When s < 0, then i(t) = e-s t (cosw +
j sin w t).
Then, oscillations will be damped as shown
in Figure 10.4(a).
Case III: When s > 0,
i(t) = I0 es t (cos w t + j sin w t)
Then, growing oscillations will be there as shown in Figure 10.4(b).
e−s tcosw t
e−s tsinw t
0
wt
Variation of real part
with s < 0
0
wt
Variation of imaginary part
with s < 0
Figure 10.4(a) Variation of Current, i(t) with s < 0
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Network Functions − s-Domain Analysis of Circuits 411
es tcosw t
es tsinw t
0
wt
0
wt
Variation of real and image part
with s > 0
Figure 10.4(b) Variation of Current i(t) with s > 0
Therefore, we can summarise the concept of the complex frequency s as shown.
s = s + jw
Neper frequency
Radian frequency
Signifies only
the amplitude
of oscillations
Signifies the number of
oscillations per second
w = No. of oscillations
Second
When
Output is:
s =0
Undamped oscillations
s <0
Damped oscillations
s >0
Gradual growing (underdamped)
oscillations
We will now proceed to explain network functions in s-domain.
10.2 TRANSFORMED IMPEDANCES IN s-DOMAIN
In this section, we will write the transformed equation in s-domain of impedance and admittance representations of the circuit elements, that is, resistance, inductance and capacitance.
10.2.1 Resistance
In time-domain current and voltage are, respectively, represented as i(t) and v(t), and in
s-domain, they are represented as I(s) and V(s).
In time domain, voltage and current are related, according to ohm’s law, as v(t) = R i(t).
The corresponding equation in s-domain is written as
V(s) = R I(s)
The transformed impedance, that is, the impedance in the s-domain is given as
Z (s ) =
V (s )
=R
I (s )
The resistance does not change when it is transformed from time domain to s-domain.
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412 Network Analysis and Synthesis
10.2.2 Inductance
For an inductance, voltage–current relation in time domain is given as follows:
v (t ) = L
di(t )
dt
The transform equivalent equation is as follows:
v( s) = L[ sI ( s) - i(0)] = sL I ( s) - LI 0
v( s) = L[ sI ( s)] with zero initial condition
v ( s)
= Z ( s) = sL
I ( s)
or
10.2.3 Capacitance
The time-domain equation for capacitance is
i (t ) = C
dv (t )
dt
The equivalent transform equation in s-domain is
I (s ) = C [sV (s ) - V (0)]
I(s) = CsV(s) with zero initial condition
or
The transform impedance for the capacitor is
V (s )
1
= Z (s ) =
.
I (s )
Cs
The circuit elements in time domain and their representation in s-domain are shown in
Table 10.1.
Table 10.1 Circuit Elements in Time Domain and Their Values in s-domain
Circuit Elements in Time Domain
Circuit Element in s-domain
Resistance R
R
Inductance L
sL
Capacitance C
1
Cs
An inductance in the s-domain with initial conditions is represented by an impedance of
sL ohms in series with an independent voltage source of LI0 volt-seconds as shown in
Figure 10.5(a).
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Network Functions − s-Domain Analysis of Circuits 413
A capacitor in the s-domain is represented as shown in Figure 10.5(b) with initial charge V0.
1
+
SL
−
LIo
Cs
+
−
i
+
Vo
s
−
i
(a)
(b)
Figure 10.5 R
epresentation in s-domain with Initial Stored Energy (a) inductor;
(b) Capacitor
10.3 ONE-PORT NETWORK
It is a network that has only one port, that is, one pair of
terminals. It is shown in Figure 10.6.
The block diagram representation of one-port network as shown has only one port 1-1′. v(t) is the voltage at the input port and i(t) is the input current.
For the one-port network, we can determine either
the impedance or the admittance. These are defined as
follows.
1
v (t )
i(t )
One-port
network
1′
Figure 10.6 One-port Network
Having a Pair of
Terminals
10.3.1 Driving Point Impedance and Admittance Functions
It is defined as the ratio of Laplace transform of voltage to the Laplace transform of current at
input port. It is denoted by Z(s).
Z (s ) =
Mathematically,
V (s )
I (s )
The driving point admittance function is the reciprocal of the driving point impedance
function.
That is,
Y (s ) =
1
I (s )
=
Z (s ) V (s )
2F
Example 10.1 Find the driving point impedance Z(s) of the network
shown in Figure 10.7.
v (t )
Solution: Before finding driving point impedance, let us convert
the given circuit parameters into s-domain. For this, we will have to
remember, the conversion table shown in Table 10.2.
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1
H
4
2F
Figure 10.7
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414 Network Analysis and Synthesis
Table 10.2 Circuit Parametics Conversion Table
Parameter in t-domain
R
L
C
v(t)
i(t)
Parameter in s-domain
R
sL
1
Cs
V(s)
I(s)
Therefore, s-domain equivalent of the given circuit is shown in
Figure 10.8.
From the diagram, the driving point impedance Z(s) can be
obtained as:
1 s
1
in parallel with 
+
Z ( s) =
2 s  4
2s 
s 1
×
1 4 2s
=
+
2s s 1
+
4 2s
1
1
=
+ 8
2s s 2 + 2
4s
1 1
4s
=
+ × 2
2s 8 (s + 2)
=
=
∴ Z(s) =
v (s) sL =
s
4
1
1
=
Cs 2s
Figure 10.8
1
s
+
2s 2(s 2 + 2)
s2 + 2 + s2
2s (s 2 + 2)
2s 2 + 2
2s (s 2 + 2)
=
2(s 2 + 1)
2s (s 2 + 2)
=
s2 +1
s (s 2 + 2)
Example 10.2 Find the driving point impedance Z(s) of the
given network as shown in Figure 10.9.
Solution: Referring to the table of conversion from time
domain to s-domain, we redraw the given circuit with s-domain
parameters as shown in Figure 10.10.
As observed from the circuit, the three impedances are in parallel. Their equivalent impedance Z(s) is calculated as follows:
Z (s ) =
1
1
=
Cs 2s
1 4s  4s 16 
 +

s 9  15 15s 
M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 414
1F
4
H
9
4
H
15
15
F
16
Figure 10.9
1
s
4s
9
4s
15
16
15s
Figure 10.10
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Network Functions − s-Domain Analysis of Circuits 415
1 4s
⋅
 4 s 2 + 16 
= s 9 
1 4 s  15s 
+
s 9
4s
4 s 2 + 16
= 9s 2
15s
9 + 4s
9s
=
=
=
=
=
=
∴ Z(s) =
=
4 s 2 + 16
4 s 2 + 9 15s
2
 4s   4s + 16 
 2
 ⋅ 

4s + 9  15s 
4s
4s
4s 2 + 16
+
2
15s
4s + 9
4s ( 4s 2 + 16)
15s ( 4s 2 + 9)
60s 2 + ( 4s 2 + 9)( 4s 2 + 16)
15s ( 4s 2 + 9)
4s ( 4s 2 + 16)
60s 2 + ( 4s 2 + 9)( 4s 2 + 16)
4s ⋅ 4(s 2 + 4)
60s 2 + 16s 4 + 64s 2 + 36s 2 + 144
16s (s 2 + 4)
16s 4 + 160s 2 + 144
16s (s 2 + 4)
16[s 4 + 10s 2 + 9]
s (s 2 + 4 )
s 4 + 10s 2 + 9
=
s (s 2 + 4 )
s 4 + 9s 2 + s 2 + 9
=
s (s 2 + 4 )
(s 2 + 1)(s 2 + 9)
1H
2
Example 10.3 Find the driving point admittance of
the one-port network shown in Figure 10.11.
Solution: We draw the s-domain equivalent network
of the given network as shown in Figure 10.12.
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1.0 F
3H
2
4F
3
1H
3
Figure 10.11
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416 Network Analysis and Synthesis
1
s
2
C
1
s
Y(s)
D
Now, let us find the driving point impedance of the
3
s
2
A
network. It is observed that between terminals A and
3
3s
are in series and their comB, impedances and
3
s
2
s bination is in parallel with the impedance 3 . Thus,
4s
ZAB is calculated as follows:
3
4s
B
 3   3s 3 
Z AB =    +  .
 4s   2 s 
Figure 10.12
s
1
Now, ZAB and are in series and the combination is in parallel with impedance . Therefore,
2
s
the driving point impedance Z(s) is given as in the following:
Z ( s) =
1
s
 s  3  3s 3   
 +
 +  
 2  4s  2 s  
=
1
s
 s  3 3s 2 + 6  
 +

 2  4 s 2 s  

 3  3s 2 + 6   

 ⋅
 
 s  4 s  2s   
 2 +  3 3s 2 + 6  



+
 4s
2s  




2

3  3s + 6  
⋅


 s 4 s  2s  
 2 + 3 + 6 s 2 + 12 


4s


=
1
s
=
1
s
=
1 s
3(3s 2 + 6) 
 +

s  2 2 s(6 s 2 + 15) 
1 s
9( s 2 + 2) 
 +

s  2 6 s( 2s 2 + 5) 
1 s
3( s 2 + 2) 
=  +

s  2 2 s( 2 s 2 + 5) 
=
1  s ⋅ s( 2 s 2 + 5) + 3( s 2 + 2) 


s 
2 s( 2 s 2 + 5)

3
2
2
1  2 s + 5 s + 3s + 6 
= 

s 
2 s( 2 s 2 + 5)

=
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Network Functions − s-Domain Analysis of Circuits 417
=
=
=
=
=
Z (s ) =
1  2s 3 + 8s 2 + 6 


s  2s ( 2s 2 + 5) 
1 ( 2s 3 + 8s 2 + 6)
⋅
s 2s ( 2s 2 + 5)
1 2s 3 + 8s 2 + 6
+
s 2s ( 2s 2 + 5)
2s 3 + 8s 2 + 6
2s 2 ( 2s 2 + 5)
2( 2s 2 + 5) + 2s 3 + 8s 2 + 6
2s ( 2s 2 + 5)
2s 3 + 8s 2 + 6
2s 2 ( 2s 2 + 5)
×
2s ( 2s 2 + 5)
2( 2s 2 + 5) + 2s 3 + 8s 2 + 6
2s 3 + 8s 2 + 6
1
× 2
s
4s + 10 + 2s 3 + 8s 2 + 6
2(s 3 + 4s 2 + 3)
2s 4 + 12s 3 + 16s
or Y ( s) =
2 s 4 + 12 s3 + 16 s
2( s3 + 4 s 2 + 3)
Example 10.4 Find the driving point admittance of the one-port network shown in Figure 10.13.
3
Ω
2
1
Ω
6
1Ω
Y(s) = ?
2
F
3
2F
Figure 10.13
Solution: s-domain equivalent of the given network is shown in Figure 10.14 we first find
3
3
1
1
Z(s). In this circuit, and
are in parallel. Similarly, and
are in parallel. These parallel
2
2s
6
2s
equivalents and 1 Ω resistor are all in series across the input terminals 1-1′. Therefore, Z(s) is
calculated as follows:
3 3  1 1 
Z ( s) = 
+
 +1
 2 2s   6 2s 
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418 Network Analysis and Synthesis
3
2
1
6
1
1
Y(s) = ?
1
2s
3
2s
1′
Figure 10.14
=
=
=
=
=
=
=
=
3 3
1 1
⋅
⋅
2 2s + 6 2s + 1
3 3 1 1
+
+
2 2s 6 2s
1
9
4s + 12s + 1
3s + 3 s + 3
6s
2s
9
2s
1 6s
+
⋅
+1
⋅
4s 3s + 3 12s s + 3
9
1
+
+1
2(3s + 3) 2(s + 3)
9
1
+
+1
6(s + 1) 2(s + 3)
1
3
+
+1
2(ss + 1) 2(s + 3)
3 + (s + 1) + 2(s + 1)(s + 3)
2(s + 1)(s + 3)
3s + 9 + s + 1 + 2(ss 2 + 4s + 3)
2(s + 1)(s + 3)
Z(s) =
2s 2 + 12s + 16 2(s 2 + 6s + 8) (s + 2)(s + 4)
=
=
2(s + 1)(s + 3) 2(s + 1)(s + 3) (s + 1)(s + 3)
Y(s) =
( s + 1)( s + 3)
( s + 2)( s + 4)
Example 10.5 Find the driving point impedance of the one-port
network shown in Figure 10.15.
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1Ω
4Ω
1
F
2
1
F
6
Figure 10.15
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Network Functions − s-Domain Analysis of Circuits 419
Solution: Now, s-domain equivalent circuit of the given circuit is
shown in Figure 10.16.
By considering series-parallel combinations of the circuit elements, we calculate the driving point impedance Z(s). Between
6
terminals A and B, impedance 4 and are in series and the series
s
2
combination is in parallel with . With this parallel combination,
s
we add the 1 Ω resistor to get Z (s).
or,
Therefore,
1
A
4
Z (s) = ? 2
s
6
s
B
Figure 10.16
2 4s + 6
⋅




s
+
2
6
2
4
6


s
s
Z(s) = 1 +   4 +   = 1 + 
=
1
+

2
4
s
+6


s
s
s
s




+
s
s
2( 4 s + 6)
2( 4 s + 6)
s
2( 4 s + 6)
4 s( 2 s + 3)
s2
s2
Z(s) = 1 +
= 1+
= 1+
⋅
= 1+ 2
2
2 + 4s + 6
4s + 8
4s + 8
s
4 s ( s + 2)
s
s
2s + 3
Z(s) = 1 +
s (s + 2)
=
s (s + 2) + 2s + 3 s 2 + 2s + 2s + 3
=
s (s + 2)
s ( s + 2)
=
s 2 + 4s + 3 (s + 1)(s + 3)
=
s (s + 2)
s (ss + 2)
Example 10.6 Find the driving point admittance of the one-port network shown in
Figure 10.17.
1Ω
2
1Ω
4
5Ω
4
1H
8
5
H
24
Figure 10.17
Solution: The s-domain equivalent of given network is shown in Figure 10.18.
1
2
1
4
5
4
s
8
5s
24
Figure 10.18
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420 Network Analysis and Synthesis
By considering the series–parallel combinations of the circuit elements, the driving point
impedance Z(s) is calculated as follows:
1  1 s   5 5s 
+
+
2  4 8   4 24 
1 s
5 5s
⋅
⋅
1 4 8
= +
+ 4 24
2 1 s 5 5s
+
+
4 8 4 24
s
25s
1
= + 32 + 96
2 2 + s 30 + 5s
24
8
s
1
25s
= +
+
2 4( s + 2) 4(5s + 30)
Z ( s) =
or,
=
1
s
25s
+
+
2 4(s + 2) 20(s + 6)
=
1
s
5s
+
+
2 4(s + 2) 4(s + 6)
=
2(s + 2)(s + 6) + s (s + 6) + 5s (s + 2)
4(s + 2)(s + 6)
=
2s 2 + 16s + 24 + s 2 + 65 + 5s 2 + 10s
4(s + 2)(s + 6)
Z ( s) =
Z ( s) =
8s 2 + 32 s + 24 8( s 2 + 4 s + 3) 2( s 2 + 4 s + 3)
=
=
4( s + 2)( s + 6) 4( s + 2)( s + 6) ( s + 2)( s + 6)
1
2( s + 1)( s + 3)
( s + 2)( s + 6)
∴ Y ( s) =
=
( s + 2)( s + 6)
Z ( s) 2( s + 1)( s + 3)
10.4 TWO-PORT NETWORK
It is a network having two ports, that is, two pairs of terminals. Representation of a two-port
network is shown in Figure 10.19 (b). For the sake of comparison, one-port network has been
shown in Figure 10.19(a).
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Network Functions − s-Domain Analysis of Circuits 421
i1(t )
1
v1(t )
1′
1
v1(t )
One-port
network
1′
i2(t )
Two-port
network
2
v2(t )
2′
(b)
(a)
Figure 10.19 Representation of Networks (a) One-port Network, (b) Two-port Network
The pair of terminals 1−1′ that are connected to the supply source or energy source. It is the
driving force for the network and is called the driving point of the network.
The network shown in Figure 10.19 (b) has two ports 1-1′ and 2-2′. The port 1-1′ is assumed
to be connected to the energy source and port 2−2′ to a load. Following are the network functions for this two-port network.
10.4.1 Network Functions of a Two-port Network
1. Driving point impedance function
Z 11 (s ) =
V 1 (s )
(driving point input impedance function)
I 1 (s )
Z 22 (s ) =
V 2 (s )
(driving point output impedance function)
I 2 (s )
2. Driving point admittance function
Y 11 (s ) =
I 1 (s )
(driving point input admittance function)
V 1 (s )
Y 22 (s ) =
I 2 (s )
(driving point output admittance function)
V 2 (s )
3. Transfer impedance functions
Z 12 (s ) =
V 1 (s )
I 2 (s )
Z 21 (s ) =
V 2 (s )
I 1 (s )
Y 12 (s ) =
I 1 (s )
V 2 (s )
Y 21 (s ) =
I 2 (s )
V 1 (s )
4. Transfer admittance functions
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422 Network Analysis and Synthesis
5. Voltage transfer function
G 21 (s ) =
V 2 (s )
V 1 (s )
G12 (s ) =
V 1 (s )
V 2 (s )
a 21 (s ) =
I 2 (s )
I 1 (s )
a 12 (s ) =
I 1 (s )
I 2 (s )
6. Inverse voltage transfer function
7. Current transfer function
8. Inverse current transfer function
10.5 TRANSFER FUNCTION
In simple words, transfer function of any network is defined as the ratio of Laplace transform of
output to the Laplace transform of input with all initial conditions to be zero. That is,
Transfer function =
Laplace transform of output
Laplace transform of input
In case of two-port network, the output quantities are V2(s) and I2(s) and the input quantities are
V1(s) and I1(s). There are four basic transfer functions of a two-port network.
1.
2.
3.
4.
Transfer impedance function Z12(s) and Z21(s)
Transfer admittance function Y12(s) and Y21(s)
Voltage transfer function G21(s) and G12(s)
Current transfer function a21(s) a12(s)
We will now determine the transfer function of some given networks.
Example 10.7 Find the transfer function of the network shown
in Figure 10.20.
Solution: The s-domain equivalent network of the given
network is shown in Figure 10.21.
By applying KVL in the circuit, we get the following equation:
1
V i (s ) = R I (s ) +
I (s )
Cs
1

or,
V i (s ) =  R +  I (s ) (10.1)
Cs


Further,
1
Vo (s) = Voltage across capacitor =
I ( s)(10.2)
Cs
M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 422
R
vi (t )
C
vo(t )
i (t )
Figure 10.20
R
1
Cs
Vi (s )
Vo (s )
I (s )
Figure 10.21
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Network Functions − s-Domain Analysis of Circuits 423
Dividing equation (10.2) by equation (10.1), we get the equation as follows:
1
 Cs  I ( s)
Vo ( s)
=  
1
Vi ( s) 
 R + Cs  I ( s)
1
1
1
= Cs = Cs =
RCs + 1 RCs + 1
1
R+
Cs
Cs
Vo ( s)
1
=
is the required transfer function.
Vi ( s) RCs + 1
Example 10.8 Determine the transfer function of the network
shown in Figure 10.22.
Solution: The s-domain equivalent network of the given
network is shown in Figure 10.23.
Applying KVL to the given network, we get the following:
Vi (s) = RI (s) + sL I (s)
= (R + sL) I (s)(10.3)
Moreover,
Vo (s) = sL I (s)(10.4)
Dividing equation (10.4) by equation (10.3), the equation can be
written as follows:
Vo ( s)
sL I ( s)
=
Vi ( s) ( R + sL) I ( s)
Vo ( s)
sL
=
is the required transfer function of the network.
Vi ( s) R + sL
R
vi (t )
L
vo (t )
i (t )
Figure 10.22
R
Vi (s)
sL
Vo (s)
I (s)
Figure 10.23
10.6 NETWORK FUNCTION IN GENERALISED FORM
A network function, also called transfer function, in generalised form can be represented as follows:
F ( s) =
=
N ( s) am s m + am -1s m -1 + am - 2 s m - 2 + … + a1s + a0
=
D ( s)
bn s n + bn -1s n -1 + bn - 2 s n - 2 + … + b1s + b0
K ( s - Z1 )( s - Z 2 )… ( s - Z m -1 )( s - Z m )
( s - P1 )( s - P2 )… ( s - Pn -1 )( s - Pn )
where K = am/bn and the roots Z1, Z2… and P1, P2… are the zeros and poles, respectively. The
location of the roots in the s-plane provides an insight into the nature of the network function
and this information is used in network analysis and synthesis.
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424 Network Analysis and Synthesis
10.7 POLES AND ZEROS OF NETWORK FUNCTIONS
From the denominator of the transfer function, we get the poles, and from the numerator of the
transfer function, we get the zeros. These poles and zeros are then plotted in the s-plane. The
positions of the poles and zeros provide us information about the stability condition of the network. In this section, we study poles and zeros of a network function.
10.7.1 Poles of a Network Function
Poles are those complex frequencies, that is, values of ‘s’ for which the network function
becomes infinite.
Consider a network function
Z (s ) =
(s + 1)
s (s + 2)(s + 3)
Now, for s = 0, −2 and −3, the function will become infinite. Therefore, s = 0, −2 and −3 are the
poles of the network function.
In simple way, we can say that poles of the transfer function can be determined by equating
its denominator to zero. The denominator when equated to zero is called the characteristic equation of the transfer function.
10.7.2 Zeros of a Network Function
Zeros are those complex frequencies, that is, values of ‘s’ for which the network function
becomes zero.
Consider a same network function
Z (s ) =
(s + 1)
s (s + 2)(s + 3)
Now, for s = −1, the function will become zero. Therefore, s = −1 is the zero of the network
function. In simple way, we can say that zeros of the network function can be determined by
equating its numerator to zero.
10.8 POLE–ZERO DIAGRAM
The poles and zeros of a network function can be plotted in s-plane.
For example, let us plot the poles and zeros of the network function represented as
F (s ) =
s (s + 3)(s + 7)
(s + 1)(s + 5)
For determining the poles, substitute denominator of the function to zero. That is, (s + 1) (s + 5) = 0.
Therefore, poles are at s = -1 and -5.
For determining the zeros, equate the numerator to zero. So, s (s +3) (s + 7) = 0. Therefore,
zeros are at s = 0, −3 and −7.
Pole-zero diagram of a network function has been drawn in s-plane, that is, complex frequency plane as shown in Figure 10.24.
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Network Functions − s-Domain Analysis of Circuits 425
A pole is represented by a cross, whereas a
zero is represented by a dot.
As s = s + jw, the abscissa of s-plane is real
axis or s-axis and ordinate of s-plane is imaginary axis or jw-axis.
All the poles and zeros of the network function are shown in s-plane in Figure.
As shown in Figure 10.24, P1 and P2 are the
two poles and Z1, Z2 and Z3 are the three zeros of
the network function.
j w axis or imaginary axis
s-plane
Z3 P2 Z2
P1
−7 −5 −3
−1
Z1
Real axis
or
s -axis
Example 10.9 Draw the pole–zero diagram
for the network function Z(s).
Z (s ) =
Figure 10.24 s-Plane Representation
of Poles and Zeros of a
Network Function
(s + 1)(s + 3)(s + 5)(s + 7)
(s + 2)(s + 4)(s + 6)(s + 8)
Solution: For determining the poles, equate the denominator of the above function to zero.
(s + 2) (s + 4) (s + 6) (s + 8) = 0
This is also called the characteristic equation.
The poles are at s = −2, −4, −6 and −8
For zeros, equate the numerator to zero.
(s + 1) (s + 3) (s + 5) (s + 7) = 0
The zeros are at s = −1, −3, −5 and −7
Therefore, required pole–zero diagram is shown in the s-plane as in Figure 10.25.
jw
Poles
s-plane
s=
s=
s=
s=
− 8 − 7 − 6 − 5 − 4 − 3 − 2 −1
0
s
Zeros
Figure 10.25 Pole–zero Diagram of a Network Function
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426 Network Analysis and Synthesis
Example 10.10 Draw the pole–zero diagram of the impedance transformed function
Z (s ) =
s (s 2 + 3)(s 2 + 7)
(s 2 + 1)(s 2 + 5)
Solution: Given the network function is as follows:
Z (s ) =
s (s 2 + 3)(s 2 + 7)
(s 2 + 1)(s 2 + 5)
For poles, substitute denominator = 0
(s2 + 1) (s2 + 5) = 0
That is,
s2 = −1
s = ± j or
s=±j
Therefore, the poles are at
For zeros, substitute numerator = 0
∵ j = -1 or j 2 = -1


s2 = −5
or
s=±j 5
± j 2.24
and
s(s 2 + 3) (s 2 + 7) = 0
Therefore, the zeros are at
s = 0, s 2 = −3
s 2 = −7
or
s = 0, s = ± j 3 , s = ± j 7
and
Therefore, zeros are at
s = 0; ± j 1.732
and
± j 2.65
Now, let us draw pole–zero diagram, as shown in Figure 10.26. By examining the positions of
poles and zeros, we should be able to predict the nature of output of the network.
jw
s-plane
s = j 2.65
s = j2.24
s = j1.732
s = j1.0
s
s=0
−s
s = − j1.0
s = − j 1.732
s = − j 2.24
s = − j 2.65
Figure 10.26 Pole–zero Diagram
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Network Functions − s-Domain Analysis of Circuits 427
Example 10.11 Draw the pole–zero diagram for the network function.
Z (s ) =
5s + 4
(s - 1)(s 2 + 2s + 4)
Solution: Given network function is
Z (s ) =
5s + 4
(s - 1)(s 2 + 2s + 5)
For poles, substitute (s – 1) (s 2 + 2s + 5) = 0
For s − 1 = 0 or s = 1 and for s 2 + 2 s + 5 = 0,
s=
-2 ± 22 - 4(1)(5)
2(1)
-2 ± 4 - 20
2
-2 ± -16
=
2
-2 ± j 4
=
2
= -1 ± j 2
=
jw
s-plane
s = −1 + j 2
s = j2
Therefore, the given network function has the following three poles:
s= j
s = 1; −1 + j2 and −1 − j2
For zeros of the given network function, substitute
5s + 4 = 0
or
or
s = −0.8
s = −1
-4
= -0.8
5
The pole–zero diagram is shown in Figure 10.27.
s=1
s
s = −j
5s = −4
s=
0
s = −1− j 2
s = −j2
Figure 10.27 Pole–zero Diagram
10.9 TIME-DOMAIN RESPONSE FROM POLE–ZERO PLOT
The time-domain response can be obtained from the pole–zero plot of a network function.
Let us have V(s) or I(s), and we want to find the time-domain response, that is, v(t) or i(t)
from poles and zeros.
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428 Network Analysis and Synthesis
Let V(s) or I(s) has m number of poles and n number of zeros. That is, the poles and zeros
are, respectively, as follows:
P1, P2, P3, …, Pm and Z1, Z 2, Z 3, …, Zn
V (s ) or I (s ) =
Let
(s - Z 1 )(s - Z 2 )… (s - Z n )
(s - P1 )(s - P2 )… (s - Pm )
The procedure for determining v(t) and i(t) is given in the following:
Step 1: Using the partial fractions of the given network functions, we can write the equation
as follows:
V ( s) or I ( s) =
K3
K1
K2
Km
+
+
++
s - Pm
s - P1 s - P2 s - P3
Step 2: Find the values of K1, K2, K3, …, Km. They can be calculated as follows:
K1 means residue at s = P1 =
( P1 - Z1 )( P1 - Z 2 )( P1 - Z3 )… ( P1 - Z n )
( P1 - P2 )( P1 - P3 )( P1 - P4 )… ( P1 - Pm )
and
K 2 means residue at s = P2 =
( P2 - Z1 )( P2 - Z 2 )( P2 - Z3 )… ( P2 - Z n )
, etc
( P2 - P1 )( P2 - P3 )( P2 - P4 )… ( P2 - Pm )
Step 3: Take the inverse Laplace transform to obtain V(t) or i(t).
The procedure will be easy to understand when we take up few examples.
Example 10.12 Draw the pole–zero diagram of given network function and hence obtain
time-domain response.
V (s ) =
2s 2 + 80s + 1000
s (s + 10)(s + 30)
Solution: Given network function is
V (s ) =
2s 2 + 80s + 1000 2(s 2 + 40s + 500)
=
s (s + 10)(s + 30)
s (s + 10)(s + 30)
For poles, substitute denominator = 0.
That is, s(s +10) (s + 30) = 0.
s = 0, −10 and −30 are the poles of the given network function.
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Network Functions − s-Domain Analysis of Circuits 429
For zeros, equate the numerator to zero.
That is,
or
2(s 2 + 40 s + 500) = 0
s 2 + 40s + 500 = 0
or
s=
-40 ± ( 40) 2 - 4(1)(500) -40 ± 1600 - 2000
=
2(1)
2
or,
s=
-40 ± -400 -40 ± j 400 -40 ± j 20
=
=
= -20 ± j10
2
2
2
s = -20 ± j10 are the zeros of the given network function.
The pole–zero diagram is shown in Figure 10.28.
jw
s = j 30
s = j 20
Z1
s = −20 + j 10
P3
s-plane
s = j 10
P2
P1
s = −30 s = −20 s = −10
s = 0 s = 10 s = 20
s
s = −j 10
s = −20 − j 10
Z2
s = −j 20
s = −j 30
Figure 10.28
Now, let us find time-domain response.
Given network function has poles and P1, P2, P3 and zeros Z1, Z2
Here, P1 = 0;
Given, V (s ) =
P2 = −10;
2
P3 = −30
and
Z1 = − 20 + j10;
Z2 = −20 − j10
2
2s + 80s + 1000 2(s + 40s + 500)
=
s (s + 10)(s + 30)
s (s + 10)(s + 30)
Step 1: Using the partial fractions, V(s) can be calculated as follows:
V (s ) =
M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 429
K3
K1
K2
+
+
(10.5)
s s + 10 s + 30
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430 Network Analysis and Synthesis
Step 2: Find the values of K1, K2 and K3 from poles and zeros as,
K1 ( residue at s = 0 that is, P1 ) =
H ( P1 - Z 1 )( P1 - Z 2 )
( P1 - P2 )( P1 - P3 )
=
2{0 - ( -20 + j10)}{0 - ( -20 - j10)}
{0 - ( -10)}{0 - ( -30)}
=
2( 20 - j10)( 20 + j10)
(10)(30)
=
2( 20 2 + 10 2 ) 2(500) 10
=
=
300
300
3
Value of K 2 ( residue at s = -10 that is, P2 ) =
H ( P2 - Z1 )( P2 - Z 2 )
( P2 - P1 )( P2 - P3 )
=
2( -10 + 20 - j10)( -10 + 20 + j10)
( -10 - 0)( -10 + 30)
=
2(110 - j10)(10 + j10)
-10( 20)
=
2(10 2 + 10 2 ) 2( 200)
=
= -2
-200
-200
Value of K3 ( residue at s = -30 that is, P3 ) =
H ( P3 - Z1 )( P3 - Z 2 )
( P3 - P1 )( P3 - P2 )
=
2( -30 + 20 - j10)( -30 + 20 + j10)
( -30 - 0)( -30 + 10)
=
2( -10 - j10)( -10 + j10)
-30( -20)
=
2(10 2 + 10 2 ) 400 2
=
=
600
600 3
Step 3: Substituting the values of K1, K2 and K3 in equation (10.5), we get the following
form:
10
2
-2
3
V ( s) =
+
+ 3 (10.6)
s s + 10 s + 30
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Network Functions − s-Domain Analysis of Circuits 431
Step 4: Taking the inverse Laplace transform of equation (10.6), the following equation can
be obtained:
L-1[V ( s)] =
v(t ) =
or
10 -1  1 
 1  2 -1  1 
+ L 
L   - 2 L-1 
3
 s + 30 
 s + 10  3
s
10
2
(1) - 2e -10t + e -30t
3
3
The required time-domain response is as follows:
v (t ) =
10
2
- 2e -10t + e -30t
3
3
2(s + 4)
draw the pole–zero diagram of the network
(s + 3)(s + 8)
function, and hence, find time-domain response, that is, i(t).
Example 10.13 Given I (s ) =
Solution: The network function is
I (s ) =
2(s + 4)
(s + 3)(s + 8)
For poles, substitute (s + 3) (s + 8) = 0
Therefore, poles are at s = −3 and −8
For zeros of the given network function, consider s + 4 = 0. So, zero is at s = -4
Pole–zero diagram of the given network function is shown in Figure 10.29.
jw
s-plane
P2
s = −8
Z1
−7 −6
P1
s = −3
s
−5 s = −4 s = −3 s = −2 s = 0
s = −1
Figure 10.29
Let us find i(t).
Given
I (s ) =
2(s + 4)
(s + 3)(s + 8)
It has two poles at s = -3 and -8 and one zero at s = − 4
Therefore,
P1 : s = − 3;
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P2 : s = − 8;
Z1 : s = − 4
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432 Network Analysis and Synthesis
Step 1: Using the partial fractions of the given network function
I (s ) =
K1
K
+ 2 (10.7)
s +3 s +8
Step 2: Let us find values of K1 and K2
Value of K1 ( that is, residue at s = -3, P1 ) =
H ( P1 - Z 1 )
( P1 - P2 )
2{-3 - ( -4)}
{-3 - ( -8)}
2{-3 + 4} 2
=
=
( -3 + 8)
5
=
Now value of K 2 ( that is, residue at s = -8, P2 ) =
H ( P2 - Z 1 )
( P2 - P1 )
2( -8 - ( -4))
( -8 - ( -3))
2( -4) 8
=
=
-5
5
=
Step 3: Substituting the values of K and K2 in equation (10.7), we get the following form:
2
8
5
I ( s) =
+ 5 (10.8)
s+3 s+8
Step 4: Taking the inverse Laplace transform of equation (10.8), the following can be obtained:
2 -1  1  8 -1  1 
L 
+ L 
5
 s + 3  5
 s + 8 
2
8
i (t ) = e -3t + e -8t A
5
5
i (t ) =
or
Example 10.14 Draw the pole–zero diagram of the given network function
V ( s) =
5s
( s + 1)( s 2 + 4 s + 8)
and hence obtain v(t ).
Solution: Given network function is as follows:
V (s ) =
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5s
(s + 1)(s 2 + 4s + 8)
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Network Functions − s-Domain Analysis of Circuits 433
For poles of the given network function, substitute (s + 1) (s2 + 4s + 8) = 0
That is,
s = -1
or
s=
=
or s2 + 4s + 8 = 0
-4 ± 4 2 - 4(1)(8)
2(1)
-4 ± 16 - 32 -4 ± -16
=
= -2 ± j 2
2
2
Therefore, given network function has three poles that are as follows:
s = -1 : P1;
s = -2 + j2 : P2;
s = -2 - j2 : P3
For zeros of the given network function, substitute numerator = 0
5s = 0
That is,
or
s=0
Therefore, given network function has one zero at s = 0
Therefore, pole–zero diagram of a given network function is shown in Figure 10.30.
jw
P2
s = −2 + j 2
s-plane
s =j 2
s=j
P1
s = −2
Z1
s=0
s = −1
s
s = −j
s = −j 2
s = −2 − j 2
P3
Figure 10.30
Let us find time-domain response, that is, v(t)
Given
5s
V (s ) =
(s + 1)(s 2 + 4s + 8)
5s
=
(s + 1){s - ( -2 + j 2)}{s - ( -2 - j 2)}
or
=
5s
(s + 1)(s + 2 - j 2)(s + 2 + j 2)
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434 Network Analysis and Synthesis
Its poles are given as follows:
P1: s = -1;
P2: s = -2 + j2;
P3: s = -2 -j2
And zeros are Z1: s = 0
To find v(t)
Step 1: Using the partial fractions of V(s), the equation can be written as follows:
V (s ) =
5s
(s + 1)(s + 2 - j 2)(s + 2 + j 2)
V (s ) =
K3
K1
K2
(10.9)
+
+
s +1 s + 2 - j2 s + 2 + j2
Step 2: Let us find the values of K1, K2 and K3
Value of K1 (That is, residue at s = -1, P1 ) =
H ( P1 - Z1 )
( P1 - P2 )( P1 - P3 )
5( -1 - 0)
{-1 - ( -2 + j 2)}{-1 - ( -2 - j 2)}
-5
=
( -1 + 2 - j 2)( -1 + 2 + j 2)
=
=
=
-5
( -1 - j 2)(1 + j 2)
-5
( -1) 2 + 22
[ ∴ ( a + b)( a - jb) = a 2 + b 2 ]
-5
= -1
1+ 4
H
-Z
H (( P
P22 Z11 ))
Value
That is,
is, residue
residue at
at ss =
=-22 +
+ jj 2,
2, P
=
Value of
of K
K 22 ((That
P22 )) =
P22 -P
P33 ))
(( P
P22 -P
P11 ))(( P
55(( -22 +
+ jj 22 - 00))
=
=
-22 +
+ jj 22 - (( -11)}{
)}{-22 +
+ jj 22 - (( -22 - jj 22)}
)}
{{-22 +
+ jj 22))
55(( =
=
(( -2
-2
2
+
j
2
+
1
)(
2
+ jj 22 +
+ 22 +
+ jj 22))
2 + j 2 + 1)( -2 +
5( -2 + j 2)
= 5( -2 + j 2)
=
(( -11 +
+ jj 22)(
)( jj 44))
=
=
=
M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 434
5( -2 + j 2)
- j4 + j2 8
Substitute j 2 = -1


5( -2 + j 2) 10( -1 + j ) 5 ( -1 + j )
=
=
- j4 - 8
4( -2 - j ) 2 ( -2 - j )
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Network Functions − s-Domain Analysis of Circuits 435
Rationalising, we get the following:
5 ( -1 + j ) -2 + j
×
2 ( -2 - j ) -2 + j
5 ( -1 + j )( -2 + j )
=
2 ( -2) 2 + 12
=
=
5  2 - j - j2 + j2 

2 
4 +1

5 ( 2 - j 3 - 1)
2
5
1 - j3
K2 =
2
1 - j3
, converting into polaar form
=
2
re jf
=
2
=
r = ( Real part ) 2 + (Imaginary part ) 2
where
and
f = tan -1
Image part
Real part
Here,
r = 12 + 32 = 10 = 3.16
 -3 
f = tan -1   = -71.56°
 1
K2 =
Similarly,
3.16e - j 71.56°
= K2 = 1.58e-j71.56°
2
K3 =1.58e j71.56°
[∴ K3 is the residue at s = -2 -j, that is, P3 and P3 is complex conjugate of P2]
Therefore, the value of K3 will also be the complex conjugate of K2. Therefore, there will be
just change in sign of f .
Step 3: Substituting the values of K1, K2 and K3 in equation (10.9), we get the following
form:
V (s ) =
M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 435
-1 1.58e - j 71.56° 1.58e + j 71.56°
+
+
s +1
s + 2 - j2
s + 2 + j2
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436 Network Analysis and Synthesis
Taking the inverse Laplace transform of the equation, the equation can be written as follows:
1
1

 1 



+ 1.58e - j 71.56° L-1 
V (t ) = - L-1 
+ 1.58e j 71.56° L-1 



 s + 1
 s + ( 2 - j 2) 
 s + 2 + j2 
V (t ) = -e - t + 1.58e - j 71.56 e - ( 2 - j 2)t + 1.58e j 71.56 e - j ( 2 + j 2)t A
Example 10.15 The pole–zero diagram of a network function is shown in Figure 10.31.
Using the pole–zero diagram, find the time-domain response by graphical method.
jw
P2
j3
P1
−4
Z2
−3
−2
Z1
−1
−j 3
P3
Figure 10.31
Solution: From the pole–zero diagram provided, we notice that the zeros are at the origin and
at s = − 1. The poles are at s = −4, s = −3 + j3 and s = −3 − j3.
Thus, there are two zeros and three plots. The network transfer function, but V(s) is written
as follows:
V (s ) =
2Ks (s + 1)
(s + 4)(s + 3 - j 3)(s + 3 + j 3)
Here, K = 2
Using the partial fraction of V(s), we can write the equation as follows:
V (s ) =
K3
K1
K2
+
+
s + 4 s + 3 - j3 s + 3 + j3
The values of K1, K2 and K3 are calculated graphically as for the pole at P1 at s = − 4.
K1 = K
Product of all the directed lines from all zeros to P1
Product of all the directed sines from the other poles to P1
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Network Functions − s-Domain Analysis of Circuits 437
( P1 - Z1 )( P1 - Z 2 )
( P1 - P2 )( P1 - P3 )
K1 = 2
( -4 - 0)[4 - ( -1)]
( -4 - 3 - j 3)(-4 - 3 + j 3)
=2
K1 = 2
-4 × -3
2 × 12
2×2
=
=
= 2.4
2
2
( -1 + j 3)( -1 - j 3) ( -1) + ( j 3)
1+ 9
For the pole P2 at s = −3 + j3
K2 = K
=2
( P2 - Z1 )( P2 - Z 2 )
( P2 - P1 )( P2 - P3 )
( -3 + j 3 - 0)( -3 - j 3 + 1)
( -3 + j 3 + 4)( -3 - j 3 + 3 + j 3)
1
= ( -1 + j8)
5
For the pole P3 at s = − 3 − j3
1
K3 = Conjugate of K 2 = ( -1 - j8)
5
Thus, substituting the values of K1, K2 and K3,
V (s ) =
K3
K1
K2
+
+
s - P1 s - P2 s - P3
Taking the inverse Laplace transform, we get the following form:
 2.4
0.2( -1 + j 8) 0.2( -1 - j 8) 
V (t ) = L-1V (s ) = L-1 
+
+
s + 3 + j 3 
s + 3 - j3
 (s - 4)
V (t ) = 2.4e -4t + 0.2( -1 + j 8)e ( -3+ j 3)t + 0.2( -1 - j 8)e ( -3- j 3)t
or
(
)
(
)
8

 1
V (t ) = 2.4e -4t + e -3t  - e j 3t + e - j 3t + j e j 3t - e - j 35 
5
 5

10.10 MORE EXAMPLES ON NETWORK FUNCTION
Example 10.16 Find the driving point impedance for the network shown in Figure 10.32.
Solution: s-Domain equivalent of the network can be drawn as shown in Figure 10.33.
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438 Network Analysis and Synthesis
 6
Z (s ) = 6s  + 2s
 s
6
(6s )
s + 2s
=
6
6s +
s
36
36s
6s
= 2
+ 2s = 2
+ 2s = 2
+ 2s
s +1
6s + 6
6s + 6
=
=
6H
2H
1
H
6
Zi
Figure 10.32
6s
2s
6s + 2s 3 + 2s
s2 +1
Z(s )
2s 3 + 8s
s2 +1
1 =6
1 s s
6
Figure 10.33
Example 10.17 Find the driving point admittance function for
the network shown in Figure 10.34.
Solution: s-domain equivalent of the given network can be drawn
as shown in Figure 10.35.
Now
8H
3
F
32
Figure 10.34
32 
8
Z (s ) = 8s  s + 
3
3s 
 8 s82s 2+ +3232

= =8s8s 

  3s3s  
8
H
3
Z(s ) = ?
8s
 8 s82s 2+ +3232

8s8s 

Figure 10.35
  3s3s  
=
=
2
8s82s + +3232
8s8s+ +
3s3s
2

 8 s8s 2+ +3232
8s8s 

2
2
3
s
  3s   8s8(s8(s82s + +3232
) ) 6464
s (ss(2s + +4)4)
=
== 2 2 2 2
==
=
2
2
s 2s + +3232 3232
1)1)
(s(2s + +
s s + +8s8s + +3232 3232
2424
3
s
3s
8
s
3
1 = 32
3
3s
s
32
3
2s 2 + 4)
ss(s
2s2(s
+ 4) 2s23s + +8s8s
== 2 2
== 2 2
s s ++
11
s s ++
11
or
Y ( s) =
s2 + 1
2 s3 + 8s
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Network Functions − s-Domain Analysis of Circuits 439
Example 10.18 Find the driving point impedance function for the network shown in Figure 10.36.
Zi
5
H
144
3
H
16
16
F
5
16
F
3
Figure 10.36
Solution: The s-domain network of the given network can be drawn as shown in Figure 10.37.
Z(s )
5s
144
3
s
16
5
16s
3
16s
Figure 10.37
Now
 5 s 5   3s 3 
Z ( s) = 
+

144 16 s  16 16 s 
 5 s   5   3s   3 


  

144   16 s   16   16 s 
=
+
5ss
3s
3
5
+
+
16 16 s
144 16 s
25
9
(144)(16) (16)(16)
+
=
5s 2 + 45
3s 2 + 3
16 s
144 s
25
9
(144)(16)
=
+ 256
(5s 2 + 45) 352 + 3
16 s
144 s
25s
9s
=
+
2
16(5s + 45) 16(3s 2 + 3)
=
25s(3s 2 + 3) + 9 s(5s 2 + 45)
16(5s 2 + 45)(3s 2 + 3)
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440 Network Analysis and Synthesis
=
=
s[25(3s 2 + 3) + 9(5s 2 + 45)]
16.5.3( s 2 + 9)( s 2 + 1)
s(120 s 2 + 480)
240( s 2 + 9)( s 2 + 1)
=
=
s[75s 2 + 75 + 45s 2 + 405]
240( s 2 + 9)( s 2 + 1)
120 s( s 2 + 4)
240( s 2 + 9)( s 2 + 1)
=
s( s 2 + 4 )
2( s 2 + 9)( s 2 + 1)
Example 10.19 Find the driving point impedance
function for the network shown in Figure 10.38.
Zi
Solution: s-domain equivalent of the given network can
be drawn as shown in Figure 10.39.
Z (s ) =
=
2s  2s
8  1
 +

9 15 15s  2s
2s 2s 2 + 8 1
9 15s 2s
 2s + 8   1 

 
2s  15s   2s 
=
9
2s 2 + 8 1
+
15s
2s
2s
9
=
2s
9
=
2s 2s 2 + 8
9 4s 3 + 31s
=
=
Z(s )
2s
15
2s
9
8
15s
1
2s
Figure 10.39
2s 2 + 8
30s 2
2
4s + 16 + 15
30s
2s 2 + 8
30s 2
4s 2 + 31
30s
=
=
2F
15
F
8
Figure 10.38
2
=
2
H
15
2
H
9
2
 2s   2s + 8 
   3
9  4s + 31s 
2s
2s 2 + 8
+ 3
9 4s + 31s
2 s( 2 s 2 + 8)
9( 4 s3 + 31s)
2 s( 4 s3 + 31s) + 9( 2 s 2 + 8)
9( 4 s 2 + 31s)
2 s( 2 s 2 + 8)
8s 4 + 62 s 2 + 18s 2 + 72
4 s( s 2 + 4 )
8s 4 + 80 s 2 + 72
=
4 s( s 2 + 4 )
8( s 4 + 10 s 2 + 9)
M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 440
=
s( s 2 + 4 )
2( s 2 + 9)( s 2 + 1)
11/17/2014 5:37:02 PM
=
=
=
2 s( 2 s 2 + 8)
9( 4 s3 + 31s)
Network Functions − s-Domain Analysis of Circuits 441
2 s( 4 s3 + 31s) + 9( 2 s 2 + 8)
9( 4 s 2 + 31s)
2 s( 2 s 2 + 8)
4
8s + 62 s 2 + 18s 2 + 72
4 s( s 2 + 4 )
8s 4 + 80 s 2 + 72
=
4 s( s 2 + 4 )
8( s 4 + 10 s 2 + 9)
=
s( s 2 + 4 )
2( s 2 + 9)( s 2 + 1)
and
Y ( s) =
1
2( s 2 + 9)( s 2 + 1)
=
Z ( s)
s( s 2 + 4 )
Example 10.20 Find the driving point impedance function of the network shown in
Figure 10.40.
Solution:
Z ( s) =
=
1 11s  121

+
+ 55s 


2s  2
2s
1 11s
+
2 s  2
121
2s
 121 + 110 s 2  


2s

 
Z(s )
1
2s
 11s(121 + 110 s 2 ) 

1 
4s
=
+
2
2 s  11s 121 + 110 s 
+
 2

2s
 121s(11 + 10 s 2 ) 

1 
4s
=
+ 2
2
2 s  11s + 121 + 110 s 


4s
=
1 121s(11 + 10 s 2 )
+
2s
121s 2 + 121
=
1 121s(11 + 10 s 2 )
+
2s
121( s 2 + 1)
=
1 s(11 + 10 s 2 )
+
2s
s2 + 1
=
=
11s
2
55s
Figure 10.40
s 2 + 1 + 2 s 2 (11 + 10 s 2 )
2 s( s 2 + 1)
s 2 + 1 + 22 s 2 + 20 s 4
2 s( s 2 + 1)
M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 441
=
20 s 4 + 23s 2 + 1
2 s( s 2 + 1)
=
20 s 4 + 23s 4 + 1
2 s( s 2 + 1)
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442 Network Analysis and Synthesis
Example 10.21 Find the driving point impedance function of the network shown in
Figure 10.41.
1
2
Z(s )
1
4
5
4
s
8
5s
24
Figure 10.41
Solution:
1  1 s   5 5s 
+
+

2  4 8   4 24 
 1   s   5   5s 
1  4   8   4   24 
+
= +
1 s
5 5s
2
+
+
4 8
4 24
s
25s
1
= + 32 + 96
2 2 + s 30 + 5s
8
24
Z ( s) =
=
1
s
25s
+
+
2 4( 2 + s ) 4(30 + 5s )
=
1
s
5s
+
+
2 4(s + 2) 4(s + 6)
=
2(s + 2)(s + 6) + s (ss + 6) + 5s (s + 2)
4(s + 2)(s + 6)
=
2s 2 + 16s + 24 + s 2 + 6s + 5s 2 + 10s
4(s + 2)(s + 6)
=
8s 2 + 32 s + 24 8( s 2 + 4 s + 3)
=
4( s + 2)( s + 6) 4( s + 2)( s + 6)
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Network Functions − s-Domain Analysis of Circuits 443
Example 10.22 Find the driving point impedance
function of the network shown in Figure 10.42.
1Ω
2
Solution: s-domain equivalent of the given network is
shown in Figure 10.43.
Z ( s) =
1
H
3
1  s  8  5s 5 
+  +

2  3  7  49 14 


1 s
= +
2 3


1  s
= +
2 3


  5s   5    
 8   49    

14 

 +  5s 5   
7



+

 49 14 
8Ω
7
5 Ω
14
5 H
49
Figure 10.42
1
2
Z(s )

 25s 
 8  ( 49)(14) 

 +
 7  70 s + 245 
 ( 49)(14) 

8
7
s
3
5
14
5s
49
Figure 10.43
1 s  8
25s  
= 1 +  s  8 + 25s  
= 2 +  3  7 + 70s + 245  
2  3  7 70s + 245  
1 s  8
5s  
= 1 +  s  8 + 5s  
= 2 +  3  7 + 14s + 49  
2  3  7 14s + 49  
1 s  8
5s  
= 1 +  s  8 +
5s  
= 2 +  3  7 + 7( 2s + 7)  
2  3  7 7( 2s + 7)  
1  s  8( 2s + 7) + 5s  
= 1 +  s  8( 2s + 7) + 5s  
= 2 +  3  7( 2s + 7)  
2  3  7( 2s + 7)  
1  s 211s + 56 
= 1 +  s 211s + 56 
= 2 +  3 7( 2s + 7) 
2  3 7( 2s + 7) 
1  s 3s + 8 
= 1 +  s 3s + 8 
= 2 +  3 2s + 7 
2  3 2s + 7 
 s  3s + 8  
s (3s + 8)


 s  3s + 8  
s (3s + 8)


1  3  2s + 7   1 
3( 2s + 7)



= 1 + 3  2s + 7  = 1 + 
3( 2s + 7)
= 2 +  s 3s + 8  = 2 +  s ( 2s + 7) + 3(3s + 8) 
2  s + 3s + 8  2  s ( 2s + 7) + 3(3s + 8) 
3( 2s + 7)
7


 33 + 22ss +
+ 7 
3( 2s + 7)



1
s (3s + 8)
= 1+
s (3s + 8)
= 2 + s ( 2s + 7) + 3(3s + 8)
2 s ( 2s + 7) + 3(3s + 8)
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444 Network Analysis and Synthesis
=
=
=
s (3s + 8)
1
+ 2
2 2s + 16s + 24
s 2 + 8s + 12 + s (3s + 8)
2s 2 + 16s + 24
4s 2 + 16s + 12
2
2(s + 8s + 12)
=
2(s + 1)(s + 3)
(s + 2)(s + 6)
1
4
Example 10.23 Find the driving point admittance
function for the network shown in Figure 10.44.
Solution: First, let us find the driving point
impedance Z(s).
Z ( s) =
1  s  4  5s 5   
+  +

4  6  7  98 28   


1
s
= +
4 6



  5s   5    
 4   98      
28 

 +  5s 5   
7

 

+

 98 28   


1 s
= +
4 6

25s 

 4 (98)( 28) 
 + 10 s + 3s 
7

196 

=

1  s 4
25s

  +
4  6  7 14(10 s + 3s)  
=
1  s 8(10 s + 35) + 25s 
+
4  6 14(10 s + 35) 
=
1  s 80 s + 280 + 25s 
+
4  6
70( 2 s + 7) 
=
1  s  105s + 280  
+

4  6  70( 2 s + 7)  
=
1  s 5( 21s + 56) 
+
4  6 70( 2 s + 7) 
=
1  s 21s + 56 
+
4  6 14( 2 s + 7) 
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Z(s )
4
7
s
6
5
28
5s
98
Figure 10.44
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Network Functions − s-Domain Analysis of Circuits 445
=
1  s 3s + 8 
+
4  6 2( 2 s + 7) 
s  3s + 8 
s(3s + 8)
1 6  2( 2 s + 7)  1
12( 2 s + 7)
= +
= +
3s + 8
4 s
4 s(22 s + 7) + 3(3s + 8)
+
6( 2 s + 7)
6 2( 2 s + 7)
=
1
s(3s + 8)
1
3s 2 + 8 s
+
= +
4 2{s( 2 s + 7) + 3(3s + 8)} 4 2( 2 s 2 + 16 s + 24)
=
1
3s 2 + 8s
s 2 + 8s + 12 + 3s 2 + 8s 4 s 2 + 16 s + 12
+
=
=
4 4( s 2 + 8s + 12)
4( s 2 + 8s + 12)
4( s 2 + 8s + 12)
=
So,
Y (s) =
s2 + 4s + 3
s 2 + 8s + 12
s 2 + 8s + 12
s2 + 4s + 3
Example 10.24 Find the driving point admittance function for the network shown in
Figure 10.45.
1
Ω
3
2
Ω
3
1
s
Z(s )
2
3s
10
Ω
3
5
3s
10
s
Figure 10.45
Solution: First, through the series–parallel equivalent calculations, we calculate Z(s) and then
find Y(s).
Z ( s) =
1
5
 1  2  2  5 10 10 
 +   +   + 
s 
 3  3s  3  3s  3
=
1
s
 1  2  2  5  10 s + 30  
 +  + 
 
 3  3s  3  3s  3s  




1 1  2
=  +
s  3  3s


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
  5   10 s + 30  
 2   3s   3s  



 +  5 10 s + 30 

3 
+
3s
 3s


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446 Network Analysis and Synthesis



1  1  2
=  +
s  3  3s



 5(10s + 30) 
 2 

9s 2
 +

 3  5 + 10s + 30 


3s

=
1
s
 1  2  2  5(10s + 30) 

 +  +
3  3s  3  3s (10s + 35) 


=
1
s
 1  2  2 10s + 30 
 +  +

 3  3s  3 3s ( 2s + 7) 
=
1  1  2 2s ( 2s + 7) + 10s + 30 
 +

s  3  3s
3s ( 2s + 7)

=
1
s
 1  2 4s 2 + 24s + 30 
 +

3s ( 2s + 7) 
 3  3s
2

 2   4s + 24s + 30  

 

1  1  3s   3s ( 2s + 7)  
=  +

s 3
2 4s 2 + 24s + 30 
+


3s
3s ( 2s + 7)




2( 4s 2 + 24s + 30)


2
1 1
9s ( 2s + 7)

=  +

s  3 2( 2s + 7) + 4s 2 + 24s + 30 


3s ( 2s + 7)


 1

2( 4 s 2 + 24 s + 30)
 +

2
 3 3s{2( 2 s + 7) + 4 s + 24 s + 30}
1  1 2( 4 s 2 + 24 s + 30) 
=  +

s  3 3s( 4 s 2 + 28s + 44) 
=
1
s
=
1 s( 4 s 2 + 28s + 44) + 2( 4 s 2 + 24 s + 30)
s
3s( 4 s 2 + 28s + 44)
=
1 4 s3 + 36 s 2 + 92 s + 60
s 3s( 4 s 2 + 28s + 44)
=
3
2
 1   4 s + 36 s + 92 s + 60 
  
2
s  3s( 4 s + 28s + 44) 
1 4 s3 + 36 s 2 + 92s + 60
+
s
3s( 4 s 2 + 28s + 44)
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Network Functions − s-Domain Analysis of Circuits 447
=
=
=
=
Y (s ) =
and
4s 3 + 36s 2 + 92s + 60
3s 2 ( 4s 2 + 28s + 44)
3( 4s 2 + 28s + 44) + 4s 3 + 36s 2 + 92s + 60
3s ( 4s 2 + 28s + 44)
4s 3 + 36s 2 + 92s + 60
2
s[3( 4s + 28s + 44) + 4s 3 + 36s 2 + 92s + 60]
4 s3 + 36 s 2 + 92 s + 60
s[4 s3 + 48s 2 + 176 s + 192 ]
4[ s3 + 9 s 2 + 23s + 15]
4[ s 4 + 12 s3 + 44 s 2 + 48s]
=
s3 + 9 s 2 + 23s + 15
s 4 + 12 s3 + 44 s 2 48s
s 4 + 12s 3 + 144s 2 + 48s
s 3 + 9s 2 + 23s + 15
Example 10.25 Find the driving point impedance function
of the network shown in Figure 10.46.
Solution: s-domain equivalent of the given network is drawn
as shown in Figure 10.47.
Consider the series–parallel combination of circuit elements, the equation can be written as follows:
1Ω
16
1Ω
4
F
15
Figure 10.46
15  1 1 
+1
Z (s ) =
+
4s  16 4s 
 1 1
15  16   4s 
=
+
+1
1
1
4s
+
16 4s
1
15 64s
+
+1
=
4s s + 4
16s
15
1
=
+
+1
4s 4(s + 4)
15(s + 4) + s + 4s (s + 4)
=
4s ( s + 4 )
=
4F
1
16
1
15
4s
1
4s
Figure 10.47
4 s 2 + 32 s + 60 s 2 + 8s + 15 ( s + 3)( s + 5)
=
=
4 s( s + 4 )
s( s + 4 )
s( s + 4 )
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448 Network Analysis and Synthesis
Example 10.26 Calculate the current in the circuit
shown in Figure 10.48. Consider all initial conditions
to be zero.
R = 3Ω
L = 1H
V = 25 e−3t
Solution: First, we draw the transformed circuit in
s-domain as shown in Figure 10.49.
C = 0.5 F
2
Z (s ) = 3 + 1.0s +
s
Figure 10.48
3s + s 2 + 2
s
25
V (s ) =
s+3
R = 3Ω
Z (s ) =
I (s ) =
V(s) =
25
SL = 1.0s
s+3
25s
V (s )
=
Z (s ) (s + 3)(s 2 + 3s + 2)
25s
=
2
(s + 3)(s + 2s + s + 2)
25s
I (s ) =
(s + 3)(s + 2)(s + 1)
1
Cs
=
1
0.5s
=
2
s
Figure 10.49
Using the partial fractions, we get the following equation:
I (s ) =
K3
K1
K2
+
+
(s + 1) (s + 2) (s + 3)
Coefficients K1, K2 and K3 are calculated as follows:
K1 = I ( s)( s + 1) s = -1
or
K1 =
25s
= -12.5
( s + 2)( s + 3) s = -1
K 2 = I ( s)( s + 2) s = - 2
or
K2 =
25s
= 50
( s + 3)( s + 1) s = - 2
K3 = I ( s)( s + 3) s = - 3
=
25s
= -37.5
( s + 1)( s + 2) s = - 3
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Network Functions − s-Domain Analysis of Circuits 449
Substituting the values of K1, K2 and K3, the equation is written as follows:
I (s ) =
-12.5
50
-37.5
+
+
(s + 1) (s + 2) (s + 3)
Taking the inverse Laplace transform, the following form is obtained:
i (t ) = -12.5e -t + 50e -2t - 37.5e -3t A
10.11 POLES AND ZEROS OF NETWORK FUNCTIONS
AND THEIR SIGNIFICANCE
We have calculated network functions in the form of driving point impedance or admittance
functions. As mentioned earlier, a network function, also called transfer function, is represented
as follows:
N ( s) =
a s n + a1s n -1 + … + an -1s + an
P ( s)
= om
Q( s) bo s + b1s m -1 + … + bm -1s + bm
where a0, a1, …, an and b0, b1, …, bn are the coefficients of the polynomials P(s) and Q(s),
respectively.
By factorising the numerator and the denominator, the network function is written as follows:
N ( s) =
a ( s - z1 )( s - z2 )… ( s - zn )
P ( s)
= 0
Q( s) b0 ( s - p1 )( s - p2 )… ( s - pm )
Z1, Z2, …, Zn are called the zeros and P1, P2, …, Pm are called the poles. If the poles or zeros are
repeated, then the function has multiple poles and multiple zeros.
Poles and zeros provide useful information about the network function including its stability
condition. For example, the driving point impedance is represented as in the following:
Z (s ) =
V (s )
I (s )
The dominator equated to zero provides the poles. A pole of Z(s) implies zero current when
voltage is finite. This is nothing but an open-circuit condition. A zero of Z(s) implies no voltage
for a finite current. This means a short-circuit condition.
Properties of Driving Point Functions
1. If the poles and zeros are not repeated, they are called simple poles and simple zeros, otherwise they are called multiple poles and multiple zeros.
2. All poles and zeros, if complex, must occur in conjugate pairs.
3. The real parts of all poles must be negative and any pole on the jw-axis must be simple.
Now, we will consider the stability criterion of an active network on the basis of positions of
poles and zeros.
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450 Network Analysis and Synthesis
10.12 STABILITY CRITERION FOR AN ACTIVE NETWORK
A network is said to be stable only when all the poles lie on the left half of the s-plane. The
location of the poles provides an idea about the stability of the network. Let us consider the
denominator polynomial of the network function. The denominator when equated to zero gives
the characteristic equation,
Q(s) = b0 sm + b1 sm-1 + … + bm
According to Routh Stability Criterion, for a stable system, all the roots of the characteristic
equation must have negative real parts. There should not be any positive pole.
It should, however, be noted that simply by having all the coefficient of the polynomial as
positive and real may not ensure that the roots are lying on the negative half in the s-plane. For
example, the polynomial can be given as follows:
Q(s) = s3 + 4s2 + 15s + 100
Here, all the coefficients are positive and real but may not satisfy the Routh’s stability criterion.
Routh's Stability Analysis
The Routh stability criterion is illustrated first by forming the Routh’s array. Let the polynomial
under consideration be as in the following:
Q(s) = b sm + b s m-1 + … + bm
0
1
We prepare the Routh array by writing the first row coefficients b0, b2, b4, etc., and second row
coefficients b1, b3, b5, etc., as
Routh array
First Column
Second Column
Third Column
First Row
b0
b2
b4
Second Row
b1
b3
b5
Third Row
c1
c2
Fourth Row
d1
d2
The quantities of third row c1, c2, etc, fourth row d1, d2, etc., are calculated as
c1 =
b1b2 - b0 b3
bb -b b
; c2 = 1 4 0 5
b1
b1
d1 =
c1b3 - b1c2
c b -0
; d2 = 1 5
, etcc
c1
c1
According to the Routh-Hurwitz stability criteria, for the system to be stable, there should not
be any change in sign of the digits in the first column of the Routh array. If there is change in sign
of the digits, that would indicate that some root of the characteristic equation lies on the right
side of the jw axis in the s-plane. The number of changes in the sign of digits in the first column
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Network Functions − s-Domain Analysis of Circuits 451
of the Routh’s array moving from top to bottom is equal to the number of roots of the polynomial
with positive real part. For a stable system, all the roots will have negative real parts only.
The condition for stability is that all the coefficients of the denominator polynomial should
have same sign; and there is no missing term between the highest and the lowest power of s,
that is, none of the coefficients should be absent. These two conditions can be determined by the
inspection of the polynomial. However, these conditions, although necessary but not sufficient.
For evaluating the necessary and sufficient conditions, we have to do Routh’s tabulation.
Routh’s stability criterion states that for a system to be stable, there should not be any change
in the sign of the digit of the first column of the Routh array. This is the necessary and sufficient
condition for the stability of a system.
An active network is said to be unstable if its output (voltage or current) is oscillatory and
increases in magnitude indefinitely. The output oscillations of a stable network die out quickly
and the output reaches its steady state value.
Example 10.27 The denominator of the network function is expressed as Q(s) = s3 + s2 + 3s + 7.
Apply Routh’s stability criterion and comment on the stability.
Solution: Routh’s array can be written as follows:
s3
1
3
s
2
1
7
s
1
s
0
-4 0
7
0
By examining the Routh’s array, it is seen that there are two sign changes in the first column,
that is, from +1 to -4 and from -4 to +7. Thus, there are two roots with positive real part. For
the system to be stable, the condition is that all roots should have negative real part only. For
which there should not be any sign change in the first column of Routh array. In this case, the
system is not stable.
Example 10.28 The denominator of a network function is expressed as a polynomial
Q(s) = s3 + 2s2 + 3s + K
For what value of K, the system will be stable?
Solution: Routh Array is written as follows:
s3
1
3
2
2
6-K
2
K
K
s
s1
s0
0
0
If the value of K is less than 6, all the terms of the first column will be positive and hence there
will be no change in sign. Therefore, for K < 6, the network representing a system will be stable.
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452 Network Analysis and Synthesis
10.13 EXAMPLES BASED ON POLE–ZERO PLOT
Some more solved examples on network function are provided in the following.
Example 10.29 A network function is given in the following:
p (s ) =
Obtain the pole–zero diagram.
2s
(s + 2)(s 2 + 2s + 2)
Solution: For poles: Substitute the denominator to zero.
That is,
(s + 2) (s2 + 2s + 2) = 0
⇒ Either s + 2 = 0
or s2 +2s + 2 = 0
s = -2 or
-2 ± j 2
2
s = -1 ± j
-2 ± 4 - 8
s=
2
−1 + j
=
1
1+j
−1
−2
1
−1 − j
Therefore, poles one situated at s = -2, -1 + j, -1 -j
and zeros are situated at s = 0
The pole–zero diagram is shown in Figure 10.50.
jw
s
2
−1
Figure 10.50
3s
(s + 1)(s + 4)
Draw the pole–zero plot in the s-plane and obtain the time-domain response.
Example 10.30 A transformed voltage is given by V (s ) =
Solution: Given that
3s
(s + 1)(s + 4)
From this function, it is clear that the function has
poles at s = -1 and s = -4 and zero ats = 0. Accordingly,
the pole–zero plot is shown in Figure 10.51.
Let us find time-domain response
V (s ) =
K
K
3s
V ( s) =
= 1 + 2
( s + 1)( s + 4) s + 1 s + 4
Taking the inverse Laplace transform, we get the
following:
V(t) = K1e-t + K2e-4t(10.10)
jw
−2
P2
−4
P1
−3
−2
−1
−1
Z1
−3
1
2
3
s
−4
Figure 10.51
K1 = V ( s)( s + 1) s = -1
=
3s
s + 4 s = -1
= -1
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Network Functions − s-Domain Analysis of Circuits 453
and
K 2 = V ( s)( s + 4) s = -4
3s
=4
s + 1 s = -4
Substituting the values of K1 and K2 in equation (10.10), we get the equation as follows:
=
v(t) = -e-t + 4e-4t V.
Example 10.31 Calculate the driving point impedance Z(s) of the network shown in Figure
10.52. Plot the poles and zeros of the driving point impedance function on the s-plane.
Solution: s-Domain equivalent of the given network can be drawn as Figure 10.52 shown in
Figure 10.53.
By considering the series-parallel connections of the circuit elements, we calculate Z(s) as
follows:
s
Z (s ) = 4 + 
2
s
= 4+
2
s + 2

s 
 s   s + 2
  

2
s 
= 4+
s s+2
+
s
2
s+2
= 4+ 2 2
s + 2(s + 2)
2s
= 4+
Z ( s) =
s ( s + 2)
s2 + 2s + 4
=
0.5 F
4Ω
 2 
1 +  
s 
0.5 H
Zi
1Ω
Figure 10.52
1 = 2
0.55 s
4
0.5 s = 1 s
2
Z(s )
1Ω
4 s 2 + 8s + 16 + s 2 + 2 s
s2 + 2s + 4
Figure 10.53
5s 2 + 10 s + 16
s2 + 2s + 4
For zeros, substitute 5s2 + 10s + 16 = 0
-10 ± 100 - 320
= -1 ± j 22 = -1 ± j1.48
10
-s2 + 2s + 4 = 0
s=
or
and for poles,
s=
-2 ± 4 - 16 -2 ± j 2 3
=
= -1 ± j 3 = -1 ± j1.732
2
2
The required pole–zero plot is shown in Figure 10.54.
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454 Network Analysis and Synthesis
jw
j2
s = −1 + j 1.732
s = −1 + j 1.48
−2
s-plane
j
−1
o
s = −1 − j 1.48
s = −1 − j 1.732
1
2
s
−j
−j 2
Figure 10.54
Example 10.32 Express the impedance Z(s) for the network
shown in Figure 10.55 in the form:
N (s )
. Plot its poles and zeros and also infer
D (s )
about the stability of the system.
1H
Z (s ) = K
 6
=s+
 5
 12s 2 + 18  


5s  

  6   12s 2 + 18  
  

5s  
 s  
=s+

2
 6 + 12s + 18 
 s

5s


1
F
6
Z(s )
Solution: s-domain equivalent of the given network is shown
in Figure 10.56:
 6  12s 18  
Z (s ) = s +  
+ 
 5  5 5s  
12 H
5
5
F
18
Figure 10.55
12s
5
s
Z(s)
6
s
18
5s
Figure 10.56
 6(12s 2 + 18) 




5s 2
=s+

2
 30 + 12s + 18 


5s
 6(12 s 2 + 18) 
= s+

2
 s(30 + 12 s + 18) 
= s+
= s+
= s+
36( 2 s 2 + 3)
s(12s 2 + 48)
36( 2 s 2 + 3)
12 s( s 2 + 4)
3( 2 s 2 + 3)
s( s 2 + 4 )
s4 + 4s2 + 6s2 + 9
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2
 6(12 s + 18) 
= s+

2
 s(30 + 12 s + 18) 
= s+
36( 2 s 2 + 3)
s(12s 2 + 48)
Network Functions − s-Domain Analysis of Circuits 455
= s+
= s+
=
z ( s) =
36( 2 s 2 + 3)
12 s( s 2 + 4)
3( 2 s 2 + 3)
s( s 2 + 4 )
s4 + 4s2 + 6s2 + 9
s( s 2 + 4 )
s 4 10 s 2 + 9
s( s 2 + 4 )
=
( s 2 + 9)( s 2 + 1)
s( s 2 + 4 )
The equation is in the form of the following:
Z (s ) = K
Here,
N (s )
D (s )
N(s) = (s2 + 9) (s 2 + 1)
D(s) = s(s2 + 4) and K = 1
Now, poles are found by equating the denominator to zero: ie
s(s2 + 4) = 0
therefore, poles are at
s = 0, ± j2
and
The zeros are calculated by equating the numerator os z(s) to zero.
s(s 2 + 9) (s 2 + 1) = 0
Therefore, zeroes are at
s = ± j, ± j3
The pole–zero diagram for the given network is shown in Figure 10.57.
jw
j3
j2
j
−j
s
−j 2
−j 3
Figure 10.57
Stability: From pole–zero plot, it is clear that the given function Z(s) has three poles at s = 0,
± j2 and zeros at s = ± j and ± j3. All the poles and zeros are on jw-axis. For a system to be stable,
all the poles and zeros should lie on the left-hand side of s-plane. Therefore, this system is oscillatory, that is, unstable.
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456 Network Analysis and Synthesis
Moreover, for the function Z(s), the degree of numerator is 4 which is exceeding the degree
of the denominator (that is, 3). Hence, the system is unstable. (For a stable system, the degree,
that is, the power of s, of the numerator should be less than the degree of the denominator.)
Example 10.33 Calculate the current flowing in
the circuit of Figure 10.58 when the initial voltage
across the capacitor is 4 V.
R = 3Ω
V = 12 V
Solution: The s-domain transformed circuit of the
Figure is shown in Figure 10.59.
Applying KVL, the voltage equation is written as
follows:
12
4
4
- 3I - I - = 0
s
s
s
or
+
−
−
3Ω
+
−
−
+
+
I
−
+
−
or
8
 3s + 4 
=I
 s 
s
or
I=
C= 1F
4
Figure 10.58
12
s
4
 12 4 

 -  - I  3 +  = 0
s s
s
+
1 = 4
s
Cs
V(0) 4
s = s
Figure 10.59
8
8
2.67
=
=
4
3s + 4

 (s + 1.33)
3 s + 

3
Taking the inverse Laplace transform, we get the following equation:
i(t) = 2.67e−1.33t A.
Example 10.34 The output y(t) of a network function is given as y(t) = te−2t. Determine input
or excitation required. Given that the impulse response of the network is e−t + e−2t.
Solution: Let x(t) be the input or excitation and x(s) be its laplace transform. We know the
Laplace transform of
1
1
+
s +1 s + 2
Y (s )
1
1
s + 2 + s +1
G (s ) =
=
+
=
X (s ) s + 1 s + 2 (s + 1)(s + 2)
e -t + 2 -2t =
Transfer function
=
2s + 3
(10.11)
(s + 1)(s + 2)
The output should be y(t) = t e− 2t
That is,
Y (s ) =
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1
( s + 2) 2
(10.12)
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Network Functions − s-Domain Analysis of Circuits 457
From (10.11) and (10.12), we get the following equation:
(s + 1)(s + 2)
Y (s )
1
=
×
2
2s + 3
G ( s ) ( s + 2)
s +1
=
(s + 2)( 2s + 3)
1
1
=
s + 2 2s + 3
X (s ) =
X (s ) =
or
1
1
s + 2 2(s + 1.5)
Taking the inverse Laplace transform, we calculate the input function as follows:
x (t ) = L-1X (s ) = e -2t - 0.5e -1.5t
Example 10.35 Apply Routh’s criterion to find the stability of the system whose characteristic
equation is given as s4 + 2s3 + 8s2 + 10s + 15 = 0.
Solution: Routh array:
s4
s
3
s2
s1
1
8
15
2
2 × 8 - 10 × 1
=3
2
3 × 10 - 15 × 2
=0
3
10
2 × 15 - 1 × 0
= 15
2
0
Auxiliary equation
0
In the fourth row, all the terms are zero. In such a case, we write the auxiliary equation as
A(s) = 3s2 + 15 = 0
Derivative of A(s) with respect to s is as follows:
dA (s )
= 6s
ds
6s + 0 = 0
Therefore,
We will use this equation to replace the zeros of s1 row and continue to complete the Routh
array as:
s1
s
0
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0
15
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458 Network Analysis and Synthesis
By looking at the complete array, we observe that there is no sign change in the elements of first
column and hence there is no root of the characteristic equation with positive real part.
Solving the auxiliary equation, the roots are evaluated as follows:
3s2 + 15 = 0
s2 = -5 or s = ± j 5
or
Hence two roots lie on the imaginary axis of the s-plane. Hence the system is limitedly stable.
R eview Q uestions
Short Answer Type
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
Explain the concept of complex frequency.
Explain various network functions for one-port network.
What is a two-port network? Explain its various network functions?
What do you mean by poles and zeros of a network function?
Explain, how can we determine the time-domain response from pole–zero plot.
What are the properties of driving point functions?
What is the significance of poles and zeros of network functions.
Explain the stability criterion for an active network.
Explain Routh–Hurwitz stability criterion with the help of an example.
What do you mean by driving point impedance function?
What do you mean by a stable system and an unstable system?
Numerical Questions
1. An inductance of 1 H and a capacitance of 0.25 F are connected in parallel. The combination
is connected in series with a resistor of 3 Ω. Calculate the total impedance of the circuit.

3s 2 + 4s + 12 
 Ans. Z =

s2 + 4 

2. Calculate the current flowing through the circuit when a voltage of v = 40 e−4t is applied
across an RLC, circuit R = 3 Ω, L = 1 H and C = 0.5 F with initial conditions to be zero.
40 -t 80 -4t 

-2t
 Ans. i = 40e - 3 e - 3 e 


3. Calculate the current in the circuit in an RC series with R = 4 Ω and C = 0.125 F when a
voltage of 20 V is switch on. The initial voltage across the capacitor is 4 V.
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[Ans. i = 4 e−2t]
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Network Functions − s-Domain Analysis of Circuits 459
4. Calculate the current in the R–L series circuit with R = 4 Ω and L = 2 H. A voltage of 10 V
is applied across the R−L circuit with i (0) = 1 A.

 5 3 -2t  
 Ans. i =  2 - 2 e  


5. The transform current I(s) of a network is given by I (s ) =
3s (s + 2)
.
(s + 1)(s + 4)
Plot the poles and zeros in the s-plane and obtain the time response.
[Ans. i = 3 − e−t −8 e−4t A]
6. Find the range of values of k for which the system represented by the following characteristic equation is stable.
s3 + 30 s2 + 600 s + 600 K = 0
[Ans. 0 < K < 30]
7. Using the Routh–Hurwitz stability criterion, ascertain the stability of the system whose
characteristic equation is given as
s6 + s5- 2s4 - 3s3- 7s2 - 4s - 4 = 0
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Two-port Network
Parameters
11
chapter Objectives
After carefully studying this chapter, you should be able to do the following:
Define
the
two-port
network
Calculate for a given network the A, B,
parameters.
C, D parameters.
Write in terms of matrix equation the
Establish correlation between the nettwo-port network parameters.
work parameters. Convert A, B, C, D
parameters into h-parameters.
Establish relationship between impedance and admittance matrix of a twoExplain two-port reciprocal network
port network.
and symmetrical network.
Calculate Z-parameters and Y-parameCalculate the Z-parameters and
ters of two-port networks.
Y-parameters of interconnected twoport networks.
Define and represent hybrid or
h-parameters of two-port networks.
Represent two-port networks as
T-circuit and p -circuit.
Define and represent inverse hybrid or
g-parameters.
Explain the concept of image impedance of a network.
Define transmission parameters or A,
B, C, D parameters.
11.1 INTRODUCTION
An electrical network may be represented by a rectangular block having pairs of terminals.
Each pair of terminals is called a port. A pair of terminals at which signal enters is called a port.
Similarly, a pair of terminals through which signal leaves is called another port. In a two-port
network, signal enters the input port and after getting processed come out at the other port. For
example, consider a Public Address System (PA system). The system comprises a microphone,
an amplifier and a loud speaker. The audio-amplifier receives through the microphone signals
and sends the amplified signals through the loud speaker. The audio-amplifier is, therefore,
a two-port network.
A port is defined as any pair of terminals into which signal is received or from which signal
is taken out.
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Two-port Network Parameters 461
I1
I2
Figure 11.1 shows a two-port network having two
2
1
ports, namely 1−1′ and 2−2′. The block may repreTwo-port
V1
V2
sent a communication system, an electronic system,
network
2′
1′
an electrical transmission or distribution system, etc.
Terminals 1−1′ is a pair that constitutes a port. Figure 11.1 Block Diagram
Similarly, the pair of terminals 2−2′ constitutes
Representation of
another port. The voltage and current at the input tera Two-port Network
minals are V1 and I1 and at the output port are V2 and
I2, respectively. It has been assumed that both I1 and I2 are entering the network through terminals
1 and 2, respectively. Thus, there are four variables, that is, V1, I1, V2 and I2. Out of these four
variables, two are dependent variables and the other two are independent variables. In a two-port
network, a variable pair is related to the other variable pair by a pair of linear equations formed
with the help of network parameters. The different network parameters that can be used for the
formulation of simultaneous equations are presented in the following sections.
11.2 TWO-PORT NETWORK PARAMETERS
In a two-port network, there are four variables. Arbitrarily, any two of these variables can be
taken as independent variables. Open-circuit impedance parameters or Z-parameters are calculated by considering V1 and V2 as dependent variables and I1 and I2 as independent variables.
Similarly, I1 and I2 can be taken as dependent variables and V1 and V2 are taken as independent
variables to determine the short-circuit admittance or Y-parameters. In this section, we will
determine the two-port network parameters of different circuit configurations.
11.2.1 Open-circuit Impedance-parameters or Z-parameters
A two-port network is redrawn as in Figure 11.2.
Here, we consider V1 and V2 as dependent variables and
I1 and I2 as independent variables. Let Z11, Z12, Z21 and Z22
are the Z-parameters. The voltage V1 and V2 in terms of I1
and I2 are expressed as follows:
1
I1
V1
1′
I2
Two-port
network
2
V2
2′
Figure 11.2 Two-Port network
V1 = Z11I1 + Z12 I 2 (11.1)
V2 = Z 21I1 + Z 22 I 2 (11.2)
In terms of matrix equations, the following can be obtained:
V1   Z11
V  =  Z
 2   21
Z
where matrix  11
 Z 21
Z12   I1 
 
Z 22   I 2 
Z12 
is called the impedance matrix.
Z 22 
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462 Network Analysis and Synthesis
The individual Z-parameters can be found as indicated in the following. When the output
port is open-circuited, that is, when I2 = 0, the equation (11.1) will be as follows:
V1 = Z11 I1 + 0 that is Z11 =
V1
I1
From equation (11.2), we will have the following form:
V2 = Z 21 I1 + 0 that is Z 21 =
V2
I1
When the input port is open-circuited, that is, when I1 = 0, then equation (11.1) will be given
as follows:
V1 = 0 + Z12 I 2
that is Z12 =
V1
I2
and from equation (11.2), we will have the following equation:
V2 = 0 + Z 22 I 2
that is
Z 22 =
V2
I2
Therefore, the Z-parameters of a two-port network are as follows:
Z11 =
V1
V
, Z 21 = 2
I1 I 2 = 0
I1 I 2 = 0
Z12 =
V1
V
, Z 22 = 2
I 2 I1 = 0
I 2 I1 = 0
All these parameters are also called open-circuit impedance parameters.
Z11 is known as the driving-point impedance at the input port, when output port is opencircuited.
Z21 is known as the transfer impedance at the input port, when output port is open-circuited.
Z22 is known as the driving-point impedance at the output port, when input port is opencircuited.
Z12 is known as the transfer impedance at the output port, when input port is open-circuited.
11.2.2 Short-circuit Admittance Parameters
The admittance parameters are also called Y-parameters. To determine Y-parameters, V1 and
V2 are taken as independent variables and I1 and I2 as dependent variables. Port currents I1
and I2 are expressed in terms of the voltages V1 and V2. The network equations are written as
follows:
I1 = Y11 V1 +Y12 V2 (11.3)
I2 = Y12 V1 +Y22 V2 (11.4)
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Two-port Network Parameters 463
The equations can be represented in matrix form as follows:
 I1  Y11 Y12  V1 
 
 =
 I 2  Y21 Y22  V2 
where Y11, Y12, Y21 and Y22 are admittance parameters and these can be determined as in the
following.
When output port is short-circuited, that is, when V2 = 0, equation (11.3) will give the
following form:
I1 = Y11V1 + 0
or
Y11 =
I1
V1
and equation (11.4) will be calculated as follows:
I 2 = Y21V1 + 0
or Y21 =
I2
V1
When the input port is short-circuited, that is, when V1 = 0, equation (11.3) can be written as in
the following:
I1 = 0 + Y12V2
or Y12 =
I1
V2
Y22 =
I2
V2
and equation (11.4) will be given as follows:
I 2 = 0 + Y22V2
or
These Y-parameters are also called short-circuit admittance parameters, as these are found by
short-circuiting the input port or the output port.
Summary of short-circuited admittance parameters is provided below.
I1
is also known as the driving-point admittance at input port with output port
V1 V2 = 0
short-circuited.
Y11 =
I2
is also known as the transfer admittance at input port with output port
V1 V2 = 0
short-circuited.
Y21 =
I1
is also known as the transfer admittance at output port with input port
V 2 V1 = 0
short-circuited.
Y 12 =
I2
is also known as the driving-point admittance at output port with input port
V2 V1 = 0
short-circuited.
Y22 =
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464 Network Analysis and Synthesis
11.2.3 Relationship between Impedance
and Admittance Matrix
[Z ] = [Y ]−1
[Y ] = [Z ]−1
or
That is,
 Z11 Z12  Y11 Y12 

=

 Z 21 Z 22  Y21 Y22 
=
−1
Y11
adjoint of 
Y21
Y11
Y12
Y21
Y22
Y12 

Y22 
Y22 − Y12   Y22 −Y12 
 

Y 
−Y
Y11   Y

=
=  21
 −Y21 Y11 
Y


Y 
 Y
where | Y| is the determinant of Y-matrix.
Similarly,
Y11

Y21
Y12   Z11
=
Y22   Z 21
Z12 

Z 22 
−1
 Z11 Z12 
adjoint of 

 Z 21 Z 22 
=
 Z11 Z12 


 Z 21 Z 22 
T
 Z 22 − Z 21 


−Z
Z11 
=  12
Z
Z 
 Z 22 − Z12   Z 22
− 12 
  Z

Z
Z
Z
−
11 

=  21
=
 − Z 21
Z11 
Z


Z 
 Z
Summarising,
 Z11

 Z 21
 Y22 −Y12 
Y 
Z12   Y

=

Z 22   −Y21 Y11 


Y 
 Y
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Two-port Network Parameters 465
and
 y 11

 y 21
where Y =
and Z =
Y11
Y12
Y21
Y22
Z11
Z12
Z 21
Z 22
 z 22
y 12   z
=
y 22   − z 21

 z
− z 12 
z 

z 22 

z 
, that is, determinate of Y-matrix.
, that is, determinant of Z-matrix.
Example 11.1 For the circuit shown in Figure 11.3, find
Z-parameters and Y-parameters.
Solution: Firstly, let us find Z-parameters. When I2 = 0,
the circuit of Figure 11.3 can be represented as shown in
Figure 11.4.
Z11 is the driving-point impedance at input port with output port open-circuited
Z11 =
V1
= Z1 + Z3
I1 I 2 = 0
Z 21 =
V2
= Z3
I1 I 2 = 0
Z 12
I1
+
Z1
Z2
I2
Z3
V1
+
V2
−
−
Figure 11.3
+
I1
V1
Z1
I2 = 0
Z3
−
Figure 11.4
V
= 1
= Z3
I 2 I1 = 0
Z 22 =
V2
= Z2 + Z3
I 2 I1 = 0
Z-parameters are as follows:
and

 Z1 + Z3 Z3
[Z ] = 

+
Z
Z
Z
2
3
 3
Z1 + Z3 Z3
= ( Z1 + Z3 )( Z 2 + Z3 ) − Z32
Z =
Z 2 + Z3
Z3
= Z1Z 2 + Z1Z3 + Z 2 Z3 + Z32 − Z33
= Z1Z 2 + Z 2 Z3 + Z3 Z1
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466 Network Analysis and Synthesis
Now, let us find Y-parameters as given in the following:
Y11 =
Z 2 + Z3
Z 22
=
Z
Z1Z 2 + Z 2 Z3 + Z3 Z1
Y12 =
− Z3
− Z12
=
Z
Z1Z 2 + Z 2 Z3 + Z3 Z 1
Y21 =
− Z3
− Z 21
=
Z
Z1Z 2 + Z 2 Z3 + Z3 Z1
Y22 =
Z1 + Z3
Z11
=
Z
Z1Z 2 + Z 2 Z3 + Z31
11.2.4 Hybrid or h-Parameters
Hybrid parameters: V1 and I2 are taken as dependent variables.
In this case, network equations can be written as follows:
V1 = h11 I1 + h12 V2 I2 = h12 I1 + h22 V2 (11.5)
(11.6)
Or the matrix form is given as in the following:
V1   h11 h12   I1 
 =
 
 I 2   h21 h22  V2 
h-parameters are found as in the following: It is clear from equations (11.5) and (11.6),
h-­parameters can be found by substituting I1 = 0 or V2 = 0, that is, input port open-circuited or
output port short-circuited.
When output is short-circuited, that is, V2 = 0.
Equation (11.5) gives V1 = h11 I1 + 0. Hence, h11 =
V1
I1
Equation (11.6) gives I 2 = h21 I1 + 0. Here, h21 =
I2
I1
When input is open-circuited , that is, I1 = 0
Equation (11.5) gives V1 = 0 + h12V2 . Hence, h12 =
Equation (11.6) gives I 2 = 0 + h22V2 . Here, h22 =
V1
V2
I2
V2
Therefore, the h-parameters are as follows:
h11 =
V1
is also known as short-circuited input impedance parameter, in Ω.
I1 V2 = 0
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Two-port Network Parameters 467
h12 =
V1
is also known as open-circuited reverse voltage gain, dimensionless.
V2 I1 = 0
h21 =
I2
is also known as short-circuited current gain, dimensionless.
I1 V2 = 0
I2
is also known as open-circuited output admittance in mho, .
V2 I1 = 0
Since all the four parameters are of different types, these are known as hybrid or h-parameters.
h22 =
11.2.5 Inverse Hybrid or g-Parameters
When I1 and V2 are taken as dependent variables. The network equations can be written as in
the following:
I1 = g11V1 + g12 I2 (11.7)
V2 = g21V1 + g22 I2 (11.8)
Or the equations can be given in the matrix form as follows:
 I1   g11 g12  V1 
 =
  
V2   g21 g22   I 2 
The inverse h-parameters can be defined/determined by the following:
Substituting
V1 = 0 or I2 = 0
That is, input port short-circuited or output port open-circuited.
When output port is open-circuited, that is, when I2 = 0
Equation (11.7) gives
I1 = g11V1 + 0
I1
V1
or
g11 =
Equation (11.8) gives
V 2 = g 21V1 + 0
or
g21 =
V2
V1
When input port is short-circuited, that is, when V1 = 0
Equation (11.7) gives
I1 = 0 + g12 I 2
I1
I2
or
g12 =
Equation (11.8) gives
V 2 = 0 + g 22 I 2
or
g22 =
M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 467
V2
I2
11/19/2014 4:16:28 PM
468 Network Analysis and Synthesis
Therefore, g-parameters are shown as in the following:
I
g11 = 1
V1 I 2 = 0
g12 =
I1
I 2 V1 = 0
g21 =
V2
V1 I 2 = 0
g22 =
V2
I 2 V1 = 0
The h and g matrices are the inverse of each other.
[h] = [ g ]−1
That is,
−1
 h11 h12   g11 g12 

=

 h21 h22   g21 g22 
 g22 − g12 
 g
g 

=
 − g21 g11 


g 
 g
Where g is the determinant, for example, matrix [g]
or
11.2.6 Transmission Parameters
When V1 and I1 are taken as dependent variables, the network equations can be written as
follows:
V1 = AV2 − BI2(11.9)
I1 = CV2 −DI2 (11.10)
Or they can be represented in matrix form as follows:
V1   A B  V2 
 =


 I1  C D   − I 2 
The A, B, C and D parameters or transmission parameters (t-parameters) are defined as in the
following:
From equation (11.9) and equation (11.10), we get the following form:
1 V2
, this is also known as open-circuit voltage gain.
=
A V1 I 2 = 0
1 −I2
=
, this is called as short-circuit transfer admittance.
B V1 V2 = 0
1 V2
=
, this is known as open-circuit transfer impedance.
C I1 I 2 = 0
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Two-port Network Parameters 469
1 −I2
=
, this is called as short-circuit current gain.
D
I1 V2 = 0
11.2.7 Inverse Transmission Parameters
When V2 and I2 are taken as dependent variables: The two-port network equations can be
written as in the following:
V2   A1 B1  V1 


 = 1
1 
 I 2  C D   − I1 
where A1, B1, C1 and D1 are inverse transmission parameters.
From the matrix form, inverse transmission parameters can be defined as follows:
1
A1
1
B
1
1
C
1
1
D
1
=
V1
V 2 I1 = 0
=
−I1
V 2 V1 = 0
=
V1
I 2 I1 = 0
=
−I1
I 2 V1 = 0
Example 11.2 For the network shown in Figure 11.5, find the A, B, C and D parameters.
Solution: Applying KVL, we get the following form:
or
V1 − 4sI1 − 2(I1 + I2) = 0
V1 = (4s + 2) I1 + 2I2(11.11)
1
V2 − I 2 − 2( I1 + I 2 ) = 0
s
1

V2 = 2I1 +  2 +  I 2 
s
and
or
(11.12)
The equation (11.12), can be written as follows:
I1 =
1
1 

V2 −  2 +  I 2  (11.13)


s 
2
+
M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 469
1
s
4s
I2
+
I1 + I2
Substituting this value in equation (11.11), we get the following form:

1 

V 2 − I 2  2 + s  
 ( 4s + 2) + 2I 2
V1 = 

2

I1
V1
2Ω
−
V2
−
Figure 11.5
11/19/2014 4:16:31 PM
470 Network Analysis and Synthesis
or

1 

V1 = V2 − I 2  2 +   {2 s + 1} + 2 I 2

s 

or


1
V1 = ( 2 s + 1)V2 − I 2  2 +  ( 2 s + 1) − 2
s


 ( 2s + 1) 2

= ( 2s + 1)V2 − I 2 
− 2
s


2
 4 s + 1 + 4 s − 2 s 
= ( 2s + 1)V2 − I 2 

s


 4 s 2 + 2 s + 1
V1 = ( 2 s + 1)V2 − 
 I 2(11.14)
s


From equations (11.13) and (11.14), the following equation can be obtained:
( 4 s 2 + 2 s + 1) I 2
s
1
 2 s + 1
I1 = V2 − 
I
 2 s  2
2
V1 = ( 2 s + 1)V2 −
Comparing these equations with the general equations,
V1 = AV2 −BI2
I1 = CV2 − DI2.
we get,
4 s2 + 2s + 1
s
1
2s + 1
C= , D=
2
2s
A = ( 2 s + 1), B =
Example 11.3 Find transmission parameters of the network shown in Figure 11.6. Further,
prove that AD − BC = 1.
Solution: Applying KVL, the equation can be written as
follows:
V1 = 4I1 + 3I2 V2 = 3I1 + 5I2 M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 470
V1
(11.15)
V2 = 2I2 + 3(I1 + I2)
or
1Ω
2Ω
I2
+
I1 + I2
V1 = 1I1 + 3(I1 + I1)
or
I1
+
3Ω
−
(11.16)
V2
−
Figure 11.6
11/19/2014 4:16:33 PM
Two-port Network Parameters 471
From equation (11.16), we can write the following form:
3I1 = V2 − 5I2
1
5
or
I1 = V2 − I 2(11.17)
3
3
Substituting this value of I1 in equation (11.15), the equation can be given as follows:
 V − 5I 2 
V1 = 4  2
+ 3I 2

3 
4
20
I + 3I 2
= V2 −
3
3 2
4
11
V1 = V2 − I 2(11.18)
3
3
From equations (11.18) and (11.17), the following forms are obtained:
4
11 
V1 = V2 − I 2 
3
3 

1
5
I1 = V2 − I 2 

3
3
Comparing these equations with general form of equations,
V1 = AV1 − BI1
I1 = CV2 − DI2
we get
4
11
, B=+
3
3
5
1
C= , D=+
3
3
A=
Now,
 4   5   11  1 
AD − BC =    +  −  +   
 3   3  3   3
+20  11
−+ 
 9
9
+20 11
=
−
9
9
9
= =1
9
=
Hence proved, AD − BC = 1.
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472 Network Analysis and Synthesis
Example 11.4 Find the Z-parameters of the network
shown in Figure 11.7.
+
1
s
1
s
I2(s)
+
I1 + I2
Solution: Applying KVL, the following form can be
obtained:
V1(s)
1
V1 = ⋅ I1 + 2 s( I1 + I 2 )
s
or
I1(s)
V2(s)
2s
−
−
Figure 11.7
1

V1 =  + 2 s I1 + 2 sI 2(11.19)
s

Applying KVL in the output loop, we get the following equation:
1
V2 = ⋅ I 2 + 2 s( I1 + I 2 )
s
1

or
V2 = 2 sI1 +  + 2 s I 2 ( s) (11.20)
s

Comparing equations (11.19) and (11.20) with the equations in general form, the equation can
be written as follows:
V1 = Z11 I1 + Z12 I2
V2 = Z21 I1 + Z22 I2
where
1
Z11 = 2 s + ,
s
Z12 = 2 s,
Z 21 = 2 s,
Z 22 = 2 s +
1
s
+
We can write the equations in matrix form as follows:
1


2s 
 2s + s
[Z ] = 

1
 2s
2s +

s 
Example 11.5 Calculate the transmission param­
eters of the network shown in Figure 11.8.
Solution The network with current directions is
shown in ­Figure 11.9.
Case I: When output is open-circuited, that is, I2 = 0
M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 472
I1 1 Ω
2Ω
I2
1
2
3Ω
V1
4Ω
+
V2
−
2′
1′
−
Figure 11.8
I1 1 Ω
+
I3 2 Ω
3Ω
V1
I1
I2
4Ω
+
V2
I3
−
−
Figure 11.9
11/19/2014 4:16:40 PM
Two-port Network Parameters 473
Applying KVL for the first loop, we can write the equation as follows:
V1 = 1I1 +3(I1 − I3) = 4I1 − 3I3(11.21)
Applying KVL for the second loop, we get the following form:
0 = 3 (I3 − I1) + 2 I3 + 4 I3
9I3 − 3I1 = 0
I1 = 3I3
I
I3 = 1
3
or
or
or
and substituting this value in equation (11.21), the equation can be written as in the following:
I 
V1 = 4 I1 − 3  1 
 3
= 4 I1 − I1
V1 = 3I1
Also,
V2 = 4 I 3
1  4
= 4  I1  = I1
3  3
Therefore,
C=
I1
3
= Ω
V2 I 2 = 0 4
( By definition )
A=
3I
V1
9
= 1 =
4
V2 I 2 = 0 4
I1
3
Case II: When output is short-circuited , that is, V2 = 0.
Applying KVL for the first loop, we get the equation as follows:
V1 = I1 + 3 (I1 − I3) = 4I1 − 3I3(11.22)
+
Applying KVL for the second loop, the equation can
be written as in the following:
0 = 3 (I3 − I1) + 2I3
I1
1Ω
2 Ω I2
1
2
3Ω
V1
I1
I3
−
2′
1′
or
5I3 = 3I1
M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 473
(11.23)
+
−
Figure 11.10
11/19/2014 4:16:41 PM
474 Network Analysis and Synthesis
Further,
I2 = −I3(11.24)
Substituting I3 = −I2 in equation (11.23), we get the following form:
−5I2 = 3I1
I1
5
= =D
− I 2 V2 = 0 3
or
Therefore, we have the equations as follows:
A=
5
3
9
; D = ;C =
4
3
4
Now, next we have to find B.
Substituting I1 =
5
I from equation (11.23) in equation (11.22), we get the value of V1:
3 3
 5
V1 = 4   I 3 − 3I 3
 3
=
20
I 3 − 3I 3
3
=
11
I3
3
From equation (11.24), I3 = −I2.
V1 = −
V1
11
= =B
− I 2 V2 = 0 3
Therefore,
Hence B =
11
I
3 2
11
.
3
Example 11.6 Determine A, B, C and D parameters
for the network shown in Figure 11.11.
M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 474
1
1Ω
+
V1
11
I1
0.5 Ω
−
2
I2
+
0.5 Ω
V2
−
21
Figure 11.11
11/19/2014 4:16:42 PM
Two-port Network Parameters 475
Solution: We know the following:
A=
V1
V2 I 2 = 0
B=
−V1
I 2 V2 = 0
C=
I1
V2 I1 = 0
D=
− I1
I 2 V2 = 0
Case I: When output is short-circuited, that is, V2 = 0
In the first loop by applying KVL, the following equation
can be obtained:
V1 = 0.5 (I1 − I3)
= 0.5 I1 − 0.5 I3
and
I3 = −I2
I1
+
I3
I2
1Ω
V1 0.5 Ω
−
V1 = 0.5I1 + 0.5I2(11.25)
Applying KVL in the second loop, we get the following
equation:
0.5 (I3 − I1) + 1I3 = 0
I1
I3
Figure 11.12
or
1.5I3 = 0.5I1(11.26)
Substituting the value of 0.5 I1 in equation (11.25), the equation can be written as follows:
V1 = 1.5I3 + 0.5I2
Substituting I3 = −I2, we get V1 as follows:
V1 = −1.5I2 + 0.5I2
= −1I2
or
−V1
= 1Ω = B
I 2 V2 = 0
Hence, B = 1 Ω and from equation (11.26), we have 1.5I3 = 0.5I1
Substituting I3 = −I2, we get −1.5I2 = 0.5I1.
or
−I1
I2
=
1.5
=3= D
0.5
V2 = 0
Hence, D = 3.
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476 Network Analysis and Synthesis
+
1Ω
I1
V1
0.5 Ω
I1
I2 = 0
0.5 Ω
Case II: To find A and C, we will have to open circuit
the output port.
Applying KVL in the first loop, we get the following
form:
0.5(I1 − I3) = V1
+
V2
I3
−
−
or 2V1 = I1 − I3 (11.27)
Applying KVL in the second loop, the following equation can be written as follows:
Figure 11.13
0.5(I3 − I1) + 1I3 + 0.5I3 = 0
2 I 3 − 0.5 I1 = 0
0.5
or
I3 =
I (11.28)
2 1
Substituting this value in equation (11.27), the equation can be given as in the following:
0.5
2V1 = I1 −
I
2 1
or
2V1 =
Further,
1.5
1.5
I1 or V1 =
I (11.29)
2
4 1
V2 = 0.5I3
I3 =
Substituting
0.5
I from equation (11.28)
2 1
V2 = 0.5 ×
0.5
I (11.30)
2 1
Now, from equation (11.29) and (11.30), we can obtain the following:
1.5
V1
1.5
2
4
=
=
×
4 0.5 × 0.5
V2 I = 0 0.5 × 0.5
2
2
3× 2
=
=3= A
4 × 0.5
Hence, A = 3 and from equation (11.30)
I1
2
=
V2 I 2 = 0 0.5 × 0.5
=8
=C
Hence, A = 3, B = 1, C = 8 and D = 3.
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Two-port Network Parameters 477
Example 11.7 Find the Z-parameters of the given network
shown in Figure 11.14.
Z1 = j40
+
Solution: Applying KVL in the input loop, we get the
equation for V1 as follows:
I1
Z2 = j80
I1 + I2
−
V1 = − j120I1 − j160I2 +
Z3 = −j160 V2
V1
V1 = j40I1 + (−j160) (I1 + I2)
I2
−
Figure 11.14
(11.31)
Applying KVL in the output loop, the equation can be written as in the following:
V2 = j80I2 −j160 (I1 + I2)
V2 = −j160I1 − j80I2 (11.32)
Comparing equations (11.31) and (11.32) with the general equations, we can calculate the
following:
V1 = Z11 I1 + Z12 I2
V2 = Z21 I1 + Z22 I2,
we get
Z11 = −j120 Ω
Z12 = − j160 Ω
Z21 = − j160 Ω
Z22 = − j80 Ω
11.3 CORRELATION OF TWO-PORT NETWORK PARAMETERS
11.3.1 Conversion of Y-parameters to Z-parameters
We can find the Z-parameters from Y-parameters as follows:
[I ] = [Y ][V ]
and
[V ] = [ Z ][ I ] = [Y ]−1[ I ]
So
[Z ] = [Y ]−1
Hence
 Z11

 Z 21
Z12  Y11
=
Z 22  Y21
where
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Y =
Y12 

Y22 
Y11
Y12
Y21
Y22
−1
=
1
Y
Y22 − Y12 


 −Y21 Y11 
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478 Network Analysis and Synthesis
11.3.2 Conversion of A, B, C and D or t-Parameters
to h-Parameters
Let us find t-parameters in terms of h-parameters.
We have the following form:
V1 = h11I1 + h12V2(11.33)
I2 = h21I1 + h22V2(11.34)
From equation (11.34), we get I1 as in the following:
I1 = −
h 22
I
V2 + 2 (11.35)
h 21
h 21
Substituting equation (11.35) in equation (11.33), V1 can be calculated as follows:
 −h
I 
V1 = h 11 22 V2 + 2  + h 12V2
h 21 
 h 21
h
− h 11h 22
V2 + 11 I 2 + h 12V2
=
h 21
h 21
 −h h

h
=  11 22 + h 12  V2 + 11 I 2
h 21
 h 21

or
 −h h + h h 
h
V1 =  11 22 12 21  V2 + 11 I 2(11.36)
h 21
h 21


Comparing equation (11.35) and (11.36) with the following general equations of A, B, C and D
parameters, we get the following:
V1 = AV2 − BI2
where
I1 = CV2 − DI2
− h 11h 22 − h 12 h 21 − h 
A=
=

h 21
h 21 
 from equation (111.36)
− h 11

B=

h 21
− h 22 
C=
h 21 
 from equation (11.35)
1 
D=−
h 21 
h =
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h 11
h 12
h 21
h 22
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Two-port Network Parameters 479
11.3.3 Conversion of h-Parameters to Y-Parameters
We have the equation as follows:
I1 = Y11V1 + Y12V2 (11.37)
I2 = Y21V1 + Y22V2 (11.38)
From equation (11.37), we can calculate V1 as follows:
Y
1
I1 − 12 V2 (11.39)
V1 =
Y11
Y11
Substituting this value of V1 from equation (11.39) in equation (11.38), we get the following
forms:
 1

Y
I 2 = Y21 
I1 − 12 V2  + Y22V2
Y11 
 Y11
Y
Y Y
= 21 I1 − 21 12 V2 + Y22V2
(11.40)
Y11
Y11
I2 =
Y21
Y Y −Y Y
I 2 + 11 22 21 12 V2
Y11
Y11
Comparing equations (11.39) and (11.40) with the general equations of h-parameters, we have
the following:
V1 = h11I1 + h12V2
I2 = h21I1 + h22V2
where
1 
Y 11 
 from equation (11.39)
−Y 12 
h 12 =
Y 11 
Y

h21 = 21

Y 11

 from equation (11.40)
Y 
Y 11Y 22 − Y 21Y 12
h22 =
=
Y 11
Y 11 
h 11=
11.4 TWO-PORT RECIPROCAL AND SYMMETRICAL NETWORKS
11.4.1 Reciprocal Two-port Network
A two-port network is reciprocal if the ratio of excitation to response remains unchanged when
the position of source and response are interchanged.
Network containing constant passive elements (that is, R, L, C ) are always reciprocal. However,
networks containing active electronic elements and dependent sources are not reciprocal.
The principle of reciprocity is depicted in Figure 11.15.
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480 Network Analysis and Synthesis
I1
+
I2
I1
Two-port
network
Vs
−
I 2′
I2
+
V
− s
Two-port
network
I 1′
Figure 11.15 Reciprocal of Two-port Networks I 2′ = I1′
The two networks will be reciprocal if I 2′ = I1′ .
11.4.2 Symmetrical Two-port Networks
A two-port network is symmetrical if there is no change in electrical behaviour external to the network when the ports are interchanged. Figure 11.16 depicts the condition for symmetrical network.
I2 = 0
I1
+
Vs
−
Two-port
network
I1 = 0
I2
+
Vs
−
Two-port
network
V2 V1
Figure 11.16 Symmetrical Two-port Networks
Vs
V
= s
I 1 I 2 = 0 I 2 I1 = 0
The relationship of parameters for reciprocal and symmetrical two-port network are given in
the following Table 11.1.
The network will be symmetrical if
Table 11.1 Relationship of Parameters for Reciprocal and Symmetrical
Two-port Network
Parameter
Condition for Reciprocity
Condition for Electrical Symmetry
Z
Z12
Z11 = Z2
Y
Y12 = Y21
Y11 = Y22
h
h12 = −h21
h11 h22 − h12 h21 = 1
g
g12 = −g21
g11 g21 = 1
t
AD − BC = 1
A=D
t′
A′ D ′ − B ′ C ′ = 1
A′ = D ′
11.5 TERMINATED TWO-PORT
NETWORK
Figure 11.17 shows a two-port network with
an ideal generator at the input port and load
impedance ZL connected at the output port.
The input impedance of the network in terms
of the parameters of the network is c­ alculated
as follows.
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I1
V1
+
−
I2
1 Two-port 2
network
1′
V2
ZL
2′
Figure 11.17 Driving-point Impedance
at the Input Port of a Load
Terminated Network
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Two-port Network Parameters 481
The network equations are as follows:
V2 = − ZL I2
V1 = Z11 I1 + Z12 I2
(11.41)
(11.42)
V2 = Z21 I1 + Z22 I2
(11.43)
Substituting (11.41) in (11.43), we get the following:
−ZL I2 = Z21 I1 + Z22 I2
− I1Z 21
or
(11.44)
I2 =
Z L + Z 22
Substituting (11.44) in (11.42), we obtain the equation as follows:
 − I1Z 21 
V1 = Z11I1 + Z12 
 Z L + Z 22 

Z Z 
V1 = I1  Z11 − 12 21 
Z

L + Z 22 
The driving-point impendence at 1−1′ is the ratio of V1 and I1. Hence,
V
Z Z
Z11 = 1 = Z11 − 12 21 (11.45)
I1
Z L + Z 22
From equation (11.45), we can determine the driving-point impendence when the output port is
open- and short-circuited.
or
1. When output port is open, ZL = ∞, then we get the following:
V
Z Z
Zin = 1 = Z11 − 12 21 = Z11
I1
∞ + Z 22
2. When output port is shorted, ZL = 0, then we obtain the equation as follows:
V
Z Z
Zin = 1 = Z11 − 12 21
1 I1
I1
0 + Z 22
Zs
Two-port
V
Z Z − Z12 Z 21
1
Vs +
V1
−
Zin = 1 = 11 22
=
network
Z 22
Y11
I1
I2
2
V2
Now, let us calculate the driving-point impedance at
1′
2′
the output port with source impendence at the input
Figure 11.18
port, as shown in Figure 11.18.
Let I1 is the current due to source voltage Vs at port 1−1′. The equations are as follows:
V1 = Vs − I1 Zs
(11.46)
V2 = I2 Z22 + I1 Z21
(11.47)
Further,
V1 = I1Z11 + I2Z12
From (11.46) and (11.48), we get the following form:
I1 Z11 + I2 Z12 = Vs − I1 Zs
(11.48)
I1 (Z11 + Zs) = Vs − I2 Z12
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482 Network Analysis and Synthesis
I1 =
V s − I 2 Z 12
Z 11 + Z s
Substituting the value of I1 in equation (11.47), V2 can be calculated as follows:
V2 =
(V s − I 2 Z 12 )Z 21
+ I 2 Z 22
Z 11 + Z s
When the source Vs is short-circuited at port 1−1′, Vs = 0.
Then,
− Z 21Z 12
I + Z 22 I 2
V2 =
Z s + Z 11 2
Driving-point impendence at port 2−2′ is
V2
I2
V 2 Z 22 Z s + Z 22 Z 11 − Z 21Z 12
=
I2
Z s + Z 11
Hence,
If the input port is open-circuited, Zs → ∞.
V2
= Z 22
I2
Then,
11.6 INTERCONNECTED TWO-PORT NETWORK
Two-port networks may be connected in cascade (series) and in parallel. The parameters of such
interconnected network can be expressed in terms of Z-parameters or Y-parameters. It can be
shown that each Z-parameter of the interconnected series network is the sum of the corresponding parameters of the individual networks. If X and Y are the two series-connected networks,
then the parameters would be given as follows:
Z 11 = Z 11X + Z 11Y ; Z 12 = Z 12 X + Z 12 Y
Z 21 = Z 21X + Z 21Y ; Z 22 = Z 22 X + Z 22 Y
For parallel networks, we calculate the Y-parameters of individual networks. It can be shown
that each Y-parameter of the parallel network is the access of the corresponding parameters of
the individual networks.
Example 11.8 Two networks shown in Figure 11.19 are connected in series. Determine the
Z-parameters of the cascaded network.
1
4Ω
4Ω
2
3
8Ω
1′
8Ω
8Ω
16 Ω
2′ 3′
(a)
4
4′
(b)
Figure 11.19 (a) X-network and (b) Y-network
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Two-port Network Parameters 483
Solution: For the X-network, the Z-parameters are Z11X = 4 + 8 = 12 Ω, Z12X = Z21X = 8 Ω and
Z22X = 4 + 8 = 12 Ω.
The Z-parameters of the Y-network can be written as follows:
Z11Y = 8 + 16 = 24 Ω, Z12Y = Z21Y = 16 Ω and Z22Y = 8 + 16 = 24 Ω.
The Z-parameters of the series-connected network will be the sum of the parameters of the
individual network.
Z11 = Z11X + Z11Y = 12 + 24 = 36 Ω
Z12 = Z12X + Z12Y = 8 + 16 = 24 Ω
Z21 = Z21X + Z21Y = 8 + 16 = 24 Ω
Z22 = Z22X + Z22Y = 12 + 24 = 36 Ω
Now, we will connect the two networks in series as shown
I1
I2
in Figure 11.20.
1
2
Z11 =
Z 22 =
4Ω
V1
= 36 Ω
I1 I 2 = 0
4Ω
V1
V2
= 36 Ω
I 2 I1= 0
V2
16 Ω
8Ω
1
Z12 = Z21 = 24 Ω
8Ω
2
Figure 11.20
11.7 T-CIRCUIT REPRESENTATION
OF TWO-PORT NETWORK
1
A T-connected two-port network is shown in
Figure 11.21
Applying KVL, we get the following form:
V1 = I1 (Z1 + Z2) + I2 Z2 (11.49)
V2 = I1 Z2 + I2 (Z2 + Z3)
8Ω
(11.50)
I1
V1
11
Z1
Z3
Z2
I2 +
2
V2
−
21
Figure 11.21 T-circuit
Representation of a
Two-port Network
Comparing equations (11.49) and (11.50) with the general
equations of Z-parameters
V1 = Z11 I1 +
Z12 I2
and
V2 = Z21 I1 + Z22 I2
We get the following,
or
and
or
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Z11 = Z1 + Z2
Z12 = Z2= Z21
Z22 = Z2 + Z3
Z2 = Z12 = Z21
Z1 = Z11 − Z12
Z3 = Z22 − Z12
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484 Network Analysis and Synthesis
11.8 o-CIRCUIT REPRESENTATION
OF TWO-PORT NETWORK
I1
+
V1
A two-port network can also be represented by an
equivalent p circuit as shown in Figure 11.22.
Applying KCL, the nodal equation can be written as in the following:
−
I2
Y2
Y1
Y3
+
V2
−
Figure 11.22 p
-Representation of a
Two-port network
I1 = V1 (Y1 + Y2 ) − V2Y2 

I 2 = −V1Y2 + V2 (Y2 + Y3 ) 
Comparing the equations with the general equations of Y-parameters, we get the following forms:
Y11 = Y1 + Y2
Y12 = Y1 = −Y2
Y22 = Y2 + Y3
Therefore,
Y2 = −Y12 = −Y21
Y1 = Y11 + Y12
Y3 = Y22 + Y12
11.9 IMAGE IMPEDANCE
Image impedance is a concept used in electronic
Z
network design and analysis and most often in filter
design. Zi1 and Zi2 are the image impedances of a twoZi1
Zi 2
Y
port network. These values are such that if port 1 − 1′
of the network is terminated in Zi1, the input impedance of port 2 − 2 ′ is Zi2. Similarly, if port 2 − 2 ′ is terminated in Zi2, the input impedance at port 1 − 1′ is Zi1. Figure 11.23 An Illustration of the
Concept of Image
Figure 11.23 shows the simple ‘L’ network with
Impedance
series impedance Z, shunt admittance Y and image
impedances Zi1 and Zi2.
Now, by definition, Z11 can be calculated as follows:
Z 11 = Z + Z i 2
=Z+
1
Y +
1
Z + Z i1
and by solving Z11, we get the following form:
Z i12 = Z 2 +
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Z
Y
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Two-port Network Parameters 485
Similarly,
Y i22 = Y 2 +
Y
Z
Hence, the two-image impedances are related to each other by the following:
Z i1 Z
=
Y i2 Y
11.10 MORE SOLVED NUMERICALS
Example 11.9 Determine the Z-parameters of the network
shown in Figure 11.24.
I1
I2
Z1
1
V1
Z2
2
V2
Solution: Applying KVL, we get the following form:
V1 = (I1 + I2) Z2 = I1Z2 + I2Z2
or
and
V1 = Z2I1 + Z2I2
11
21
Figure 11.24
(11.51)
V2 = I2Zi + (I1 + I2) Z2
= I1Z2 + I2 (Z1 + Z2)
(11.52)
We have the Z-parameter equations as follows:
V1 = Z11 I1 + Z12 I2
(11.53)
V2 = Z21 I1 + Z22 I2
(11.54)
Comparing equations (11.51) with equation (11.53) and equation (11.52) with equation (11.54),
we get the following:
Z11 = Z 2
Z12
Z 21
Z 22


= Z2


= Z2

= Z1 + Z 2 
Example 11.10 Given A, B, C and D parameter of a two-port network, determine its
Z-parameters.
Solution: We know that A, B, C and D parameters of a two-port network are represented by
the following:
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486 Network Analysis and Synthesis
V1 = AV2 − BI2(11.55)
and
I1 = CV2 − DI2 Equation (11.56) can be written as follows:
V2 =
(11.56)
D
1
I1 + I 2(11.57)
C
C
Now, substituting the value of V2 from equation (11.57) in equation (11.55), we get the following:
D 
1
V1 = A  I1 + I 2  − BI 2
C 
C
A
AD − BC
= I1 +
I 2 (11.58)
C
C
Further, Z-parameters of a two-port network are represented by the equation as follows:
(11.59)
V1 = Z11 I1 + Z12 I2
V2 = Z21 I1 + Z22 I2
Comparing equations (11.58) and (11.59), we obtain the following form:
Z11 =
A
C
and
Z12 =
(11.60)
AD − BC
C
Comparing equations (11.57) and equations (11.60), we have the following form:
Z 12 =
1
C
and Z 22 =
D
C
Example 11.11 Find the Y-parameters for the network shown in Figure 11.25.
1
V1
1Ω
I1
2Ω
2Ω
4Ω
I2
6Ω
2
V2
21
11
Figure 11.25
Solution: With port 2−2′ open-circuited, the circuit may be redrawn as shown in Figure 11.26.
Therefore, the output voltage
V2 = I3 (6)
(11.61)
From mesh 3 in Figure 11.26, applying KVL, we get the following:
(I3 − I2) 4 + (6 + 2) I3 = 0
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Two-port Network Parameters 487
1Ω
1
V1
11
2Ω
I1
Mesh 1
2Ω
I2
4Ω
Mesh 2
I2 = 0
6Ω
I3
Mesh 3
2
V2
21
Figure 11.26
or
4 I3 − 4 I2 + 8I3 = 0
or
12I3 = 4I2
For mesh 2, applying KVL, the equation can be written as follows:
2 (I2 − I1) + 1 I2 + (I2 − I3)4 = 0
or
2I2 − 2I1 + I2 + 4I2 − 4I3 = 0
or
7I2 − 2I1 − 4I3 = 0
(11.62)
(11.63)
and
V1 = ( I1 − I 2 )2
or
From equation (11.62), we have 12I3 = 4I2
V1 = 2 I1 − 2 I 2 (11.64)
4
I
12 2
1
= I 2 (11.65)
3
By substituting value of I3 in equation (11.63), we get the following form:
I3 =
or
1 
7 I 2 − 2 I1 − 4  I 2  = 0
3 
I2 

∵ I 3 = 3 


4
I = 2 I1
3 2
 17 
or
  I 2 = 2 I1
3
6
or
I 2 = I1(11.66)
17
From equation (11.64), the following form can be obtained:
2I1 − 2I2 = V1
6 

6 
2 I1 − 2  I1  = V1 ∵ I 2 = I1 
 17 
17 

7I 2 −
or
or
 34 − 12 

 I =V
17  1 1
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488 Network Analysis and Synthesis
I1
17
=
= 0.7727
V1 V2 = 0 22
Y11 =
or
12
From equation (11.61), we have V2 = I3 (6) and from equations (11.62), I 2 = I 3 = 3I 3
4
From equation (11.63), we have the following form:
7 I 2 − 2 I1 − 4 I 3 = 0
7(3I 3 ) − 2 I1 − 4 I 3 = 0
21I 3 − 4 I 3 = 2 I1
or
2
I
17 1
By substituting the value of I3 in equation (11.61), the equation can be written as follows:
I3 =
or
V2 = I 3 (6)
2
I (6)
17 1
12
= I1
17
=
Y 21 =
or
I1
17
=
= 1.4166
V2 V2 = 0 12
Now, with port 1−1′ open-circuited, the circuit may be redrawn as shown in Figure 11.27.
+
1
V1
−
1Ω
I1 = 0
2Ω
Mesh 1
I1
Mesh 2
I2
2Ω
4Ω
6Ω
I3
Mesh 3
2
I2
V2
Mesh 4
11
+
21
−
Figure 11.27
V2 = (I2 − I3)6
or
6I2 − 6I3 = V2(11.67)
For mesh 3 in Figure 11.27, by applying KVL, we get the equation as follows:
( I1 − I 3 )4 + (1 + 2) I1 = 0
or
4 I1 − 4 I 3 + 3I1 = 0
or
7 I1 − 4 I 3 = 0
or
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I1 =
4
I 7 3
(11.68)
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Two-port Network Parameters 489
For mesh 2 in Figure 11.27, by applying KVL, the following can be obtained:
4(I3 − I1) + 2I3 + 6(I3 − I2) = 0
or
−4I1 − 6I2 − 12I3 = 0
or
4 
−4  I 3  − 6 I 2 + 12 I 3 = 0
7 
(11.69)
4 

∵ I1 = 7 I 3 


16 

12 −  I 3 = 6 I 2
7
or
I3 =
or
42
I (11.70)
68 2
From equation (11.67), we get the following form:
6I2 − 6I3 = V2
or
 42 
6 I 2 − 6  I 2  = V2
 68 
or
 408 − 252 

 I 2 = V2
68
or
 156 

 I = V2
68  2
Y22 =
or
42 

∵ I 3 = 68 I 2 


I2
68
=
= 0.436
V2 V1= 0 156
From Figure 11.27, I1 = 0 and therefore, we get the following form:
V1 = I1 × 2 (11.71)
Now, from equation (11.68), the following equation is obtained:
7 I1 = 4I3
7
or
I 3 = I1
4
From equation (11.69), we get the equation as follows:
12I3 − 6I2 − 4I1 = 0
7 
12  I1  − 4 I1 = 6 I 2
4 
or
or
7 

∵ I 3 = 4 I1 


21I1 − 4I1 = 6I2
17I1= 6I2 or I1 =
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6
I
17 2
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490 Network Analysis and Synthesis
Now, substituting the value of I1 in equation (11.71), the equation can be written as in the
following:
V1 = I1 (2)
6
I ( 2)
17 2
12
= I2
17
I
17
Y 21 = 2
=
V1 V 1 = 0 12
=
or
Example 11.12 For the symmetrical two-port
network shown in Figure 11.28, find the Z-parameters
and A, B, C and D parameters.
I2
I1
30 Ω
V1
30 Ω
40 Ω
V2
Solution: Using KVL for loops, we get the equation
as follows:
Figure 11.28
V1 = 30 I1 + 40( I1 + I 2 )
V1 = 70 I1 + 40 I 2
or
and
(11.72)
V2 = 30 I 2 + 40( I1 + I 2 )
= 40 I1 + 70 I 2
(11.73)
Let us find the Z-parameters
Case I: When I2 = 0 from equations (11.72) and (11.73), we get the following:
V1 = 70 I1 and V2 = 40 I1
or
V1
70
V1
V
= 1 = 70 Ω
∴ Z11 =
I1 I 2 = 0  V1 
 70 
I1 =
Z 21 =
40 I1
V2
=
= 40 Ω
I1 I 2 = 0
I1
Case II: When I1 = 0, from equations (11.72) and (11.73), we obtain the following:
V1 = 40 I2
∴ Z 12 =
and
V2 = 70 I2
40I 2
V1
=
= 40 Ω
I 2 I1 = 0
I2
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Two-port Network Parameters 491
Z 22 =
V2
V
I1 = 0 = 2 = 70 Ω
V2
I2
70
Let us find A, B, C and D parameters.
A=
Z11 70
=
= 1.75
Z 21 40
B=
Z 22 Z11 − Z12 Z 21 70(70) − 40( 40) 3300
=
=
= 82.5
40
40
Z 21
C=
1
1
=
= 0.025 Ω
Z 21 40
D=
Z 22 70
=
= 1.75
Z 21 40
Example 11.13 Determine the Y-parameters for the network shown in Figure 11.29.
Solution: Let I3 be the current in the middle loop. By applying KVL, we get from the Figure
as follows:
V1 = 2(I1 − I3)
and
(11.74)
2(I3 − I1) + 1I3 + 3V1 + 1.5 (I3 + I2) = 0
4.5 I3 = 2I1 − 1.5 I2 − 3V1
or
(11.75)
V2 = 1.5 (I2 + I3)
and
(11.76)
From equations (11.74), (11.75) and (11.76), we have the following forms:
 2 I 1.5 I 2 3V1 
V1 = 2 I1 − 2  1 −
−
4.5 4.5 
 4.5
= 2 I1 − 2[0.44 I1 − 0.33I 2 − 0.666V1 ]
= 2 I1 − 1.112 I1 − 0.666 I 2 − 1.333V1
or
V1 = 1.112 I1 − 0.666 I2 − 1.333V1
2.333V1 = 1.112 I1 − 0.666 I 2
or
V1 = 0.476 I1 − 0.285 I 2 (11.77)
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+
V1
I1
1Ω
+ 3V −
1
I2
+
1.5 Ω V2
2Ω
−
−
Figure 11.29
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492 Network Analysis and Synthesis
 2 I 1.5 I 2 3V1 
V2 = 1.5 I 2 + 1.5  1 −
−
4.5 4.5 
 4.5
Further,
= 1.5 I 2 + 0.666 I1 − 0.499 I − V1
= 0.666 I1 − I 2 − [0.476 I1 − 0.285I 2 ]
V2 = 0.19 I1 + 1.285 I 2(11.78)
[∵V1 = 0.476 I1 − 0.285 I 2 from equation (11.77)]
Therefore, from equations (11.77) and (11.78), we have the following:
0.476 −0.285
Z=

 0.19 1.285 
Now, Y-parameters can be found as follows:
[Y ] = [ Z ]−1
0.476
=
 0.19
 −1.285
=
 0.19
−0.285
1.285 
−1
−0.285
−0.476 
Example 11.14 Find the Z-parameters of the network shown in Figure 11.30.
1Ω
+
1
2
1Ω
1H
1H
V1
Zin
1Ω
+
2F
1Ω
4
11
2F
V2
21
Figure 11.30
Solution: s-domain equipment circuit of the given circuit can be drawn as shown in
Figure 11.31.
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Two-port Network Parameters 493
1
+
Z1(s)
1
2
1
+
Z2(s)
s
s
1
2s
1
V1
1
4
Zin
V2
1
2s
11
21
Figure 11.31
Now,
 s × 1
 s + s + 1 × 1
s2 + 2s
= 2
Z1 ( s) =
 s × 1
s + 3s + 1
 s + s + 1 + 1
and
Z 2 ( s) =
 1 1 

 4 × 2s  1 

 +  ×1
 1 + 1  2s 
 4 2 s 


 1 1 
 4 × 2s  1 

 +  +1
 1 + 1  2 
 4 2 s 

2s + 2s + 4
=
2 s + 2 s + 4 + 4 s 2 + 8s
s +1
4s + 4
= 2
= 2
4 s + 12 s + 4 s + 3s + 1
The circuit is drawn as shown in Figure 11.32.
The Z-parameters are found as shown in the
following:
Z11 ( s) = Z1 ( s) + Z 2 ( s)
=
2
s + 2s
2
s + 3s + 1
Z12 ( s) = Z 2 ( s) =
Z 21 = Z 2 ( s) =
+
+
s +1
2
s + 3s + 1
=1
s +1
s 2 + 3s + 1
s +1
1
V1
−
Z1(s)
s2 + 1
3s + 1
s2 +
Z2(s)
1′
2
+
s+ 1
V2
s 2 + 3s + 1
2′
−
Figure 11.32
s 2 + 3s + 1
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494 Network Analysis and Synthesis
Z12 ( s) = Z 2 ( s) =
Z 21 = Z 2 ( s) =
Z 22 ( s) =
Example 11.15 The network of
Figure 11.33 contains a current-controlled
current source. For the network, find
Z-parameters.
s +1
2
s + 3s + 1
s +1
s 2 + 3s + 1
s +1
2
s + 3s + 1
1
2Ω
I1
V1
I2
1Ω
Solution: First, we convert the current
source in equivalent voltage source as 1′
shown in ­Figure 11.34.
The loop equations for Figure 11.34 are
as follows:
V1 = 1 (I1 − I3)(11.79)
6I1 = 1 (I3 − I1) + 2I3 + 2(I3 + I2)
or
6I1 − I3 + I1 − 2I3 − 2I3 - 2I2 = 0
or
7I1 − 2I2 − 5I3 = 0
(11.80)
Further,
+2
2Ω
V2
3I1
−2′
Figure 11.33
2Ω
I1
V1
I2
2Ω
1Ω
V2
I3
+
−
6I1
Figure 11.34
V2 = 2(I2 + I3) − 6I1
= 2I2 + 2I3 − 6I1
V2 = −6I1 + 2I2 + 2I3(11.81)
From equation (11.80), we have the following:
7I1 − 2I2 = 5I3
7I − 2I 2
or
= 1.4 I1 − 0.4 I 2
I3 = 1
5
I3 = 1.4I1 − 0.4I2(11.82)
By substituting the value of I3 from equation (11.82) in equation (11.79), we get the following:
V1 = I1 − (1.4I1 − 0.4I2)
= I1 −1.4I1 + 0.4I2
V1 = −0.4I1 + 0.4I2 (11.83)
By substituting the value of I3 from equation (11.82) in equation (11.81), the following equation
can be obtained:
V2 = −6I1 + 2I2 + 2 (1.4I1 − 0.4I2)
= −6I1 + 2I2 + 2.8I1 − 0.8I2
V2 = −3.2I1 + 1.2I2 M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 494
(11.84)
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Two-port Network Parameters 495
From equations (11.83) and (11.84), the equations are written as follows:
Z11 =
V1
= −0.4 Ω
I1 I 2 = 0
Z 21 =
V2
= −3.2 Ω
I1 I 2 = 0
Z12 =
V1
= 0.4 Ω
I 2 I1 = 0
Z 22 =
V2
= 1.2 Ω
I 2 I1 = 0
R E V IE W Q U E S T I O N S
Short Answer Type
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
What are Z-parameters?
Establish the relationship between impedance and admittance parameters.
What are h-parameters? Why these are called so?
Derive inverse hybrid parameters for a two-port network.
Define A, B, C and D parameters of a two-port network.
How h-parameters can be converted into Y-parameters?
What do you mean by symmetrical two-port networks?
What do you mean by a terminated two-port network?
Explain the concept of image impedance.
Explain the relationship between various network parameters of a two-port network.
Numerical Questions
1. Find Z-parameters
1
100 Ω
j 200 Ω
1
−j 300 Ω
11
400 Ω
500 Ω
21
Figure 11.35
[Ans. Z11 = 500 − j300 Ω, Z22 = 500 − j100 Ω, Z12 = Z21 = − j300 Ω]
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496 Network Analysis and Synthesis
2. Find Y- and Z-parameters.
2 Ω 0.5 F
1
2
2H
0.5 F
1Ω
11
21
Figure 11.36

 Ans. Y -parameters:


3s 2 + 3s + 1
s ( s + 2)


=
,
=
Y
Y
11


2( s + 1) 22
2 s( s + 1)




−s
Y12 = Y21 =


2( s + 1)




Z -parameters:




2
2
s
(
s
+
)
2
3
3
1
2
s
s
(
)
+
+


Z
Z11 =
,
=
22

3s 2 + 5s + 2 
s(3s 2 + 5s + 2)


2s


Z12 = Z 21 = 2


3s + 5s + 2


3. Find admittance parameters
1Ω
1
2
1F
1F
1H
11
0.5 Ω
21
Figure 11.37
s ( s + 2)


Y 11 =


s +1


−s 

 Ans. Y 12 = Y 21 = s + 1 



3s 2 + 3s + 1
Y 22 =


s (s + 1) 

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Two-port Network Parameters 497
4. Find Y- and Z-parameters
3Ω
3Ω
4Ω
Figure 11.38
 Ans. Z -parameters:




Y -parameters:




Z11 = 7 = Z 22 

Z12 = Z 21 = 4 

7
Y11 = Y22 = 
33 

−1 
Y12 = Y21 =
11 
5. Find Z-parameters
1
(10 + j 15)Ω
(8 − j 20)Ω
Za
Zb
3
Zc (12 + j 6)Ω
2
4
Figure 11.39
[Ans. Z11 = 22 + j21, Z12 = Z21 = 12 + j6, Z22 = 20 − j14]
6. Find Z-parameters
2F
3F
3H
Figure 11.40


6s 2 + 1
= Z 22 
Z 11 =

2
6s
 Ans.



Z 12 = Z 21 = 3s


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498 Network Analysis and Synthesis
7. For the given circuit, find A, B, C and D parameters
2F
2F
3H
Figure 11.41


 Ans.



8. A two-port network has parameters
6s 2 + 1
6s 2
1
3s
A = D = 0.98 and B = j60 Ω. Find the ratio
V2 if I 2 = 0
V1
12s 2 + 1

12s 2 
6s 2 + 1 

6s 2

[Ans. - 0.98]
9. Find Z-parameters
(10 − j 4) Ω
(16 + j 8) Ω
(16 + j 8) Ω
(10 − j 4) Ω
Figure 11.42
10. Find Z-parameters for the circuit shown in Figure 11.43
L1
 Ans. 13 + j 2 3 + j 6 

3 + j 6 13 + j 2

L2
C
Figure 11.43
1

sL +

Cs
 Ans.
1

Cs

M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 498
1
Cs 

1 
sL2 +
Cs 
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Network Synthesis
and Realisability
12
Chapter objectives
After carefully studying this chapter, you should be able to do the following:
Explain the concept of network synthesis.
it is a positive real function and hence
realizable.
Explain the Hurwitz conditions for
stability.
Synthesize networks by Foster method.
Test polynomials for stability applying
Synthesize networks by Cauer method.
Hurwitz stability criteria.
From the driving point impedance or
State the properties of a positive real
admittance function realize the netfunction.
work in Foster and Cauer form I and II,
respectively.
Test a network function to find whether
12.1 INTRODUCTION
Network analysis deals with finding out the output response, using various techniques, when the
excitation signal (input signal) and the network are known.
Network synthesis deals with the realisation of the network from the given excitation and output
response.
The network synthesis provides alternate solutions. The network synthesis technique is used
in filter design where computations, innovations and judgement are required. In the network
synthesis, we first determine the driving point function in impedance and admittance form. The
driving point admittance/immittance is given by the following form:
F (s ) =
R (s )
E (s )
where R(s) is the output response and E(s) is the excitation response.
There are two basic considerations to be taken care of. These are: (i) realisability and
(ii) stability.
It is determined first whether the function is realisable or not; that is, to find out whether it
is possible to obtain a passive network. This is termed as realisability. For stability, there are
certain conditions that must be satisfied by the network function.
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500 Network Analysis and Synthesis
A network function is tested first for its realisability. It is then tested for its stability. After
that, we proceed to realize the network using Foster and Cauer methods.
In this chapter, we will examine the network function for their stability condition. We will
also explain the process of realisation of the network, when the driving point network function
is given.
12.2 HURWITZ CONDITIONS FOR STABILITY
The driving point impedance or admittance is expressed as the ratio of two polynomials, P(s)
and Q(s). These polynomials are tested for Hurwitz stability conditions. If they satisfy the conditions, then we say that they are Hurwitz.
Consider a polynomial:
P ( s) = an s n + an −1s n −1 + an − 2 s n − 2 + …+ a1s + a0
Then, for the polynomial to be Hurwitz, it has to satisfy the following Hurwitz’s conditions for
stability:
Condition 1: All the coefficients a0, a1, a2 … an must be positive.
Condition 2: No powers of ‘s’ should be absent between the highest and lowest degree term
of ‘s’ unless all the even or all the odd terms are missing. For example, P(s) =
s5 + 3s3 + 2s2 + 3s + 1 is not Hurwitz as the term s4 is missing. The polynomial
P(s) = s5 + 2s3 + 3s is Hurwitz as all coefficients are positive and all the even
terms are missing.
Condition 3: The continued fraction expansion of the ratio of the odd to even parts or even to
odd parts of Hurwitz polynomial gives all positive quotient terms.
Now, consider a polynomial
P(s) = s4 + s3 + 2s2 + 3s + 2.
The even parts of the polynomial P(s), that is, e(s) = s4 + 2s2 + 2,
while the odd parts of the polynomial P(s), that is, e(o) = s3 + 3s.
By continued fraction expansion, if it is seen that all the quotient terms are positive, then the
given polynomial P(s) is Hurwitz.
If a polynomial satisfies the conditions of Hurwitz for stability, we say that the polynomial
must be Hurwitz. The testing of a polynomial for satisfying the conditions of Hurwitz is further
explained through examples.
Example 12.1 Check whether the given polynomial is Hurwitz or not
P(s) = s4 + s3 + 2s2 + 3s + 2
Solution: Given
P(s) = s4 + s3 + 2s2 + 3s + 2.
All the coefficients (that is, s4, s3, s2 and s are positive). There are no terms missing between
the highest and lowest degree terms. Both the conditions, that is, condition 1 and condition 2
are satisfied.
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Network Synthesis and Realisability 501
For condition 3 we will perform the continued fraction expansion.
Now,
Even part of
Odd part of
P(s) = s4 + s3 + 2s2 + 3s + 2
P(s) = s4 + 2s2 + 2 = e(s)
P(s) = s3 + 3s = o(s)
The continued fraction expansion of e( s) is given by the following:
o( s)
)
(
s 3 + 3s s 4 + 2s 2 + 2 s
4
2
s + 3s
− s2 + 2


− s 2 + 2  s 3 + 3s  −s


s 3 − 2s
 s

5s  − s 2 + 2  −

 5
−s 2
  5s
2  5s 
  2
5s
x
Since all the quotient terms are not positive, the given polynomial is not Hurwitz.
Example 12.2 Test if the polynomial s3 + 6s2 + 12s + 8 is Hurwitz.
Solution: Given
P(s) = s3 + 6s2 + 12s + 8
We have to test for stability with respect to all the three conditions namely
1. All the coefficients must be positive.
2. No powers of s should be absent and
3. Continued fraction expansion should give all positive quotient terms
The given polynomial satisfies the first two conditions. We will test for the third condition.
Now, even part of P(s) = 6s2 + 8 = e(s) and odd part of P(s) = s3 + 12s = o(s)
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502 Network Analysis and Synthesis
We will use the continued fraction expansion as follows:
o (s ) s 3 + 12s
=
e ( s ) 6s 2 + 8
)
s
6s 2 + 8 s 3 + 12s 
6
4
3
s + s
3
32  2
 9
s  6s + 8  s
 16
3 
6s 2
 32  4s
8 s 
 3  3
32
s
3
x
o (s )
Since all the quotient terms of
are positive, the given polynomial is Hurwitz.
e (s )
Example 12.3 Test if the polynomial s4 + 8s2 + 32 is Hurwitz.
Solution: In this case, only even part is given in the following:
e(s) = s4 + 8s2 + 32
To get odd part, we will differentiate even part e(s) as follows:
o(s) = e′(s) = 4s3 + 16s. We find continued fraction expansion of even to odd parts as
)
s
4s 3 + 16s s 4 + 8s 2 + 32 
4
s 4 + 4s 2
4s 2 + 32 ) 4s 3 + 16s (s
4s 3 + 32s
 s

−16s  4s 2 + 32  −

 4
4s 2
 s

32  − 16s  −

 2
−16s
x
Since all powers of s are not present and all the quotient terms are not positive, and the polynomial is not Hurwitz.
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Network Synthesis and Realisability 503
Example 12.4 Test whether the polynomial s5 + 5s3 + 4s is Hurwitz.
Solution: In this case, only odd part is given.
o(s) = s5 + 5s3 + 4s
That is,
To get the even part, we will differentiate the odd part.
e(s) = o′(s) = 5s4 + 15s2 + 4
That is,
The continued fraction expansion is given by the following:
)
s
5s 4 + 15s 2 + 4 s 5 + 5s 3 + 4s 
5
s 5 + 3s 3
5

2s 3 + 4s  5s 4 + 15s 2 + 4  s

2
5s 4 + 10s 2

2
5s 2 + 4  2s 3 + 4s  s

5
8
2s 3 + s
5
12  2
 25
s  5s + 4  s
 12
5 
5s 2
 12  12
4 s  s
 5  20
12
s
5
x
Since all the quotient terms are positive, the polynomial is Hurwitz.
Example 12.5 Find the limits of K so that the polynomial s3 + 14s2 + 56s + K may be Hurwitz.
Solution: By inspection, we see that the first two conditions are satisfied.
Odd part of the given polynomial, that is, o(s) = s3 + 56s
Even part of the given polynomial, that is, e(s) = 14s2 + K
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504 Network Analysis and Synthesis
The continued fraction expansion is given by the following:
)
14s 2 + K s 3 + 56s  s

Ks  14
s3 +

14
K 

2
 56 −  s 14s + K  14s
14
 14s 2

K

 56 −
14


K


K
K 

 56 −  s   56 − 14  s
14

K
K 

 56 −  s
14
x
Now, for the polynomial to be Hurwitz, quotient terms should be positive.
14
>0
K
56 −
14
K
56 −
14 > 0
K
K
56 −
14 > ∞
14
That is,
and
or
or
56 −
This gives no value for K.
K
>0
14
K
14
or
56 >
or
K < 56 × 14
or
K < 56 × 14
or
K < 784
Therefore, the limit of K is 0 < K < 784.
Example 12.6 Test whether the following polynomial is Hurwitz or not.
F ( s) =
2 s 4 + 6 s3 + 11s 2 + 10 s + 5
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Solution: Given
F (s ) =
2s 4 + 6s 3 + 11s 2 + 10s + 5
s 4 + 5s 3 + 8s 2 + 9s + 2
=
P (s )
Q (s )
Let us first check for P(s):
P(s) = 2s4 + 6s3 + 11s2 + 10s + 5
P(s) = 2s4 + 11s2 + 5
Even part of
Odd part of
P(s) = 6s3 + 10s
The continued fraction expansion is given by the following:
)
s
6s 3 + 10s 2s 4 + 11s 2 + 5 
3
10
2s 4 + s 2
3
23 2  3
 18s
s + 5 6s + 10s 

 23
3
90
6s 3 + s
23
140  23 2
23
529
 23
s
s s + 5 
s × s2 =
 140
420
23  3
3
23 2
s
3
 140  1 140s 28s
5
s ×
=
 23  5 23
23
140
s
23
x
Since all the quotient terms of P(s) are positive, P(s) is Hurwitz.
Now, let us check for Q(s):
Q(s) = s4 + 5s3 + 8s2 + 9s + 6
Even part of Q(s) = s4 + 8s2 + 6 and Odd part of Q(s) = 5s3 + 9s
The continued fraction expansion is carried out as follows:
5s 3 + 9s s 4 + 8s 2 + 6  1 s
5
9
s4 + s2
5
31 2
25

 5
s + 6 5s 3 + 9s 
× 5s 3 = s

 31s 2
5
31
150
5s 3 +
s
31
129  31 2
 31 31 2 961s
s s + 6 
× s =

 129s 5
31
5
645
31 2
s
5
 129  1 129s 129s
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s ×
=
6
)
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)
5s 3 + 9s s 4 + 8s 2 + 6  1 s
5
9 2
4
s + s
5
31 2
25

 5
s + 6 5s 3 + 9s 
× 5s 3 = s
2


5 Synthesis
31
506 Network Analysis and
31s
3 150
5s +
s
31
129  31 2
 31 31 2 961s
s s + 6 
× s =
 129s 5
31  5
645
31 2
s
5
 129  1 129s 129s
s ×
=
6
 31  6
31
186
129
s
31
x
All the quotient terms of Q(s) are also positive, and therefore, Q(s) is also Hurwitz.
P ( s)
Since P(s) and Q(s) both are Hurwitz, F ( s) =
is also Hurwitz.
Q ( s)
12.3 PROPERTIES OF POSITIVE REAL FUNCTIONS
Positive real functions (PRF) represent physically realisable networks.
The driving point impedance function Z(s) and driving point admittance function Y(s) of a
network can be expressed as the ratio of two polynomials.
a s n + a1s n −1 + a2 s n − 2 + … + an
P ( s)
If F ( s) =
is an immittance function, it is a PRF if
= 0m
Q( s) b0 s + b1s m −1 + b2 s m − 2 + … + bm
the following conditions are satisfied:
1. All the coefficients of P(s) and Q(s) are real and positive.
2. Q(s) is a Hurwitz polynomial.
3. All the terms in the numerator and denominator polynomial are present. There is no missing term.
4. All the poles and zeros of F(s) lie in the left half of the s-plane.
Testing of a function to be positive real or not is illustrated through a few examples.
The necessary and sufficient conditions for a rational network function F(s) to be positive
real (that is, physically realisable) are stated as follows:
Condition 1: F(s) must not have any pole in the right-hand side of s-plane; that is, both P(s)
P(s )
and Q(s) of F (s ) =
are Hurwitz.
Q (s )
This condition can be checked by continued fraction expansion of the odd to even parts or
even to odd parts of P(s) and Q(s).
Condition 2: real part of F(s), that is,
Re[F( jw)] ≥ 0, for all w
or
A(w 2) = M1 ( jw ) × M2 ( jw ) − N1 ( jw ) × N2 ( jw ) ≥ 0 for all w .
Where M1(s) and N1(s) are even and odd parts of P(s) respectively. M2(s) and N2(s) are the even
part and odd part of Q(s) respectively.
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Condition 3: This condition is tested when the poles of F(s) lie on the jw -axis. This is done
by using the partial fraction expansion of F(s) to check whether the residues of the poles on the
jw -axis are positive and real.
s2 + 6s + 5
Example 12.7 Determine whether the function F ( s) = 2
is a positive real functions
s + 9 s + 14
(PRF) and hence realisable.
Solution: Given
s 2 + 6s + 5
F (s ) =
2
s + 9s + 14
=
P(s )
Q (s )
The following necessary conditions are first applied to the given function as
1. All the coefficients of P(s) and Q(s) are real and positive.
2. All the terms in the numerator and denominator polynomial are present. There is no missing term.
Hence, the function F(s) may be a PRF.
To check whether it is truly PRF or not, we will check for the sufficient conditions.
Step 1: firstly, we will check whether F(s) is Hurwitz or not
Given
F (s ) =
s 2 + 6s + 5
2
s + 9s + 14
=
P(s )
Q (s )
Let us consider P(s) as follows:
P(s) = s2 + 6s + 5
Even part of
P(s) = s2 + 5 and odd part = 6s
The continued fraction expansion is carried out for P(s) and Q(s) as follows:
)
s
6s s2 + 5 
6
s2
 6
5 6s  s
 5
6s
x
Since all the quotient terms of P(s) are positive, P(s) is Hurwitz.
Now, let us consider Q(s) as in the following:
Q(s) = s2 + 9s + 14
Even part of Q(s) = s2 + 14 and the odd part of Q(s) = 9s
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The continued fraction expansion is as follows:
)
s
9s s 2 + 14 
9
2
s
  9
14  9s  s
  14
9s
x
All the quotient terms of Q(s) are positive. Therefore, Q(s) is also Hurwitz.
P(s )
is also Hurwitz.
Q (s )
Step 2: to check whether Re|F( jw)| ≥ 0, for all w, that is,
A(w 2) = M1 ( jw) × M2 ( jw) − N1 ( jw) × N2 ( jw) ≥ 0 for all w.
Since P(s) and Q(s) are both Hurwitz, F (s ) =
P ( s ) s 2 + 6s + 5
=
Q (s ) s 2 + 9s + 14
F (s ) =
Now,
P(s) = s2 + 6s + 5
The even part is M1(s) = s2 + 5 and its odd part is N1(s) = 6s
Q(s) = s2 + 9s + 14
The even part is M2(s) = s2 + 14 and its odd part is N2(s) = 9s
Consider
or
or
A (w 2 ) = M 1 (s ) × M 2 (s ) − N 1 (s ) × N 2 (s )
2
2
2
2
2
2
A (w ) = (s + 5)(s + 14) − (6s )(9s )
A (w ) = (s + 5)(s + 14) − 54s
4
2
4
2
4
2
2
= s + 14s + 5s + 70 − 54s 2
= s + 19s + 70 − 54s
= s − 35s + 70
s = jw
s = jw
2
2
s = jw
s = jw
s = jw
s = jw
4
= ( jw ) − 35( jw ) 2 + 70
= j 4w 4 − 35 j 2w 2 + 70
= 1⋅ w 4 − 35( −1)w 2 + 70 [∵ j 2 = −1, j 4 = 1]
= w 4 + 35w 2 + 70
Since A(w 2) ≥ 0 for all w, the given function is a positive real functions (PRF) and the function
is physically realisable network.
3
3
s2 + s +
4 is a positive real functions (PRF) and
Example 12.8 Test whether F (s ) = 2 4
s +s+4
represents a physically realisable network.
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Solution:
Step 1: check whether F(s) is Hurwitz or not
Now,
3
3
s2 + s +
P(s )
4
4
F (s ) = 2
=
Q (s )
s +s+4
Now, consider P(s) as in the following:
3
3
P(s ) = s 2 + s +
4
4
3
3
and its odd part N 1 (s ) = s
4
4
Continued fraction expansion is as follows:
The even part M 1 (s ) = s 2 +
)
3
34
s s2 +  s
4
43
s2
3 3 
 ss
4 4 
3
s
4
x
Since all the quotient terms are positive, P(s) is Hurwitz.
Now, take
Q(s) = s2 + s + 4
The even part is M2(s) = s2 + 4 and its odd part is N2(s) = s
Continued fraction expansion is given as follows:
)

s s2 + 4  s

s2
 s
4 s 
 4
s
x
Quotient terms are positive ad Q(s) is also Hurwitz.
Since both P(s) and Q(s) are Hurwitz, and therefore, F(s) is also Hurwitz.
Step 2: consider A (w 2 ) = M 1 (s )M 2 (s ) − N 1 (s )N 2 (s )
3

3 
=  s 2 +  (s 2 + 4 ) −  s  (s )

4 
4
3
3
= s 4 + 4s 2 + s 2 + 3 − s 2
4
4
= s 4 + 4s 2 + 3
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s = jw
s = jw
s = jw
s = jw
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consider A (w 2 ) = M 1 (s )M 2 (s ) − N 1 (s )N 2 (s )
510 Network Analysis and Synthesis
3

3 
=  s 2 +  (s 2 + 4 ) −  s  (s )

4 
4
3
3
= s 4 + 4s 2 + s 2 + 3 − s 2
4
4
= s 4 + 4s 2 + 3
s = jw
s = jw
s = jw
s = jw
A (w 2 ) = ( jw ) 4 + 4( jw ) 2 + 3
= j 4w 4 + 4 j 2w 2 + 3
A (w 2 ) = w 4 − 4w 2 + 3
or,
(i)
Since all terms in A(w 2) are not positive, we have to check whether A(w 2) ≥ 0 for all values of w .
Let us apply Sturm test to check A(w 2) ≥ 0. (It is obtained by considering the derivative of P0(x).)
For this, we substitute w 2 = x, in (i) as
Then,
P0(x) = x2 - 4x + 3
Derivative of P0(x), that is, P1(x) = 2x − 4
Now, dividing P0(x) by P1(x), we get the following:
)
x
2x − 4 x 2 − 4 x + 3  + 1
2
2
x − 2x
− 2x + 3
2x − 4
− 1 = − P2 ( x )
P0(x) = x2 - 4x + 3
P1(x) = 2x - 4 and P2(x) = 1
Now, we find the sign changes of P0(x), P1(x), P2(x) as x changes from 0 to ∞
Limits
Sign of P0(x)
Sign of P1(x)
Sign of P2(x)
Sign Changes
x=0
positive
negative
positive
2
x=∞
positive
positive
positive
No sign change
Since there are two sign changes, A(w 2) is always positive for all values of w . The given function is not PRF.
(overall sign changes = 2 - 0 = 2)
Example 12.9 Test whether the function A(w 2) = 3w 4 - 12w 2 + 4 is positive real.
Solution: Given A(w 2) = 3w 4 - 12w 2 + 4. To check whether A(w 2) ≥ 0 for all values of w, let
us apply Sturm test.
Substitute
Then,
and
w 2 = x.
P0(x) = 3x2 - 12x + 4
P1(x) = 6x - 12 (derivative of P0(x))
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Now, divide P0(x) by P1(x)
)
x
6 x − 12 3x 2 − 12x + 4  − 1
2
3x 2 − 6 x
− 6x + 4
6 x + 12
− 8 ⇒ − P2
P0(x) = 3x2 - 12x + 4
Now,
P1(x) = 6x - 12
P2(x) = +86
We find sign changes of P0(x), P1(x), P2(x) for x = 0 and x = ∞.
P0(x)
P1(x)
P2(x)
Sign Changes
x=0
positive
negative
positive
2
x=∞
positive
positive
positive
No sign change
Since there are two sign changes, the function is not positive real.
(Overall sign changes = 2 - 0 = 2)
Example 12.10 Test whether Y(s ) =
2s 3 + 2s 2 + 3s + 2
is a positive real function.
s2 +1
Solution:
Step 1: let us check whether Y(s) is Hurwitz or not.
Now,
2s 3 + 2s 2 + 3s + 2
Y( s ) =
2
s +1
=
P(s )
Q (s )
Consider P(s) as in the following:
Its odd part N1(s) =
Now,
2s3
P(s) = 2s3 + 2s2 + 3s + 2
+ 3s and even part M1(s) = 2s2 + 2
(
)
2s 2 + 2 2s 3 + 3s s
2s 3 + 2s
(
)
s 2s 2 + 2 2s
2s
2
 s
2 s 
 2
s
x
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Since all the quotients are positive, P(s) is Hurwitz.
Now, consider Q(s) = s2 + 1
Its even part M2(s) = s2 + 1 and there is no odd part; that is, N2(s) = 0
Therefore, odd part (derivative of M2(s)) = 2s
)
s
2s s 2 + 1 
2
Now,
s2
) (
1 2s 2s
2s
x
Since all the quotients are positive, Q(s) is also Hurwitz.
Now, both P(s) and Q(s) are Hurwitz, and therefore, Y(s) is also Hurwitz.
Step 2: let us find A(w 2 ) = M1 ( s) M 2 ( s) − N1 ( s) N 2 ( s)
s = jw
= ( 2 s 2 + 2)( s 2 + 1) − ( 2 s3 + 3s)(0)
4
2
4
2
2
= 2s + 2s + 2s + 2 − 0 − 0
= 2s + 4 s + 2
Now,
s = jw
s = jw
s = jw
A(w 2) = 2( jw )4 + 4( jw )2 + 2
= 2j4w 4 + 4j2w 2 + 2
= 2w 4 - 4w 2 + 2
Let us apply Sturm test to check whether A(w 2) ≥ 0 or not
Substitute
Then,
and
w2 = x
P0(x) = 2x2 - 4x + 2
P1(x) = 4x - 4
Now, divide P0(x) by P1(x), we get the following:
)
x 1
4 x − 4 2x 2 − 4 x + 2  −
2 2
2x 2 − 2x
− 2x + 2
−2x + 2
0
= −P2 ( x )
Therefore,
P0(x) = 2x2 - 4x + 2
P1(x) = 4x - 4
P2(x) = -0
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Limits
Sign of P0(x)
Sign of P1(x)
Sign of P2(x)
Sign Changes
x=0
x=∞
positive
positive
negative
positive
negative
negative
1
1
Therefore, overall sign changes = 1 - 1 = 0.
Since there is no overall sign change, the given function is a PRF, that is, positive real functions.
12.4 SYNTHESIS OF NETWORKS BY FOSTER’S
AND CAUER’S METHODS
The Foster’s method of network synthesis uses the partial fraction expansion of the driving point
immittance function. Foster is of two types: Foster form-I and Foster form-II. When the driving
point function is an impedance, it is referred to as Foster form-I. Foster form-II is used for driving point admittance function.
The Cauer’s methods employ continued expansion approach to synthesise a given immittance function. In Cauer form-I, the terms of both numerator and denominator are arranged in
descending degree of s. In Cauer form-II, the terms in the numerator and denominator polynomials of the driving point immittance function are arranged in ascending order.
Foster and Cauer methods of network synthesis are presented in details.
12.5 FOSTER AND CAUER FORMS
12.5.1 Synthesis of R–C Network
The impedance Z(s) of R–C network is of the type as in the following:
Z ( s) =
H ( s + s 1 )( s + s 3 )
( s + s 2 )( s + s 4 )
Further, for R–C impedance function, Z(0) ≥ Z(∞).
12.5.2 Properties of the R–C Impedance or R–L Admittance
Function
1. The poles and zeros of the ZRC(s) or YRL(s) lie on the negative real axis of the complex
s-plane.
2. The poles and zeros are interlacing, that is, they alternate along the negative real axis.
3. The poles and zeros are simple. There are no multiple poles and zeros.
4. ZRC(0) > ZRC(∞)
12.5.3 Foster Form-I of R–C Network
It is obtained by using the partial fraction expansion of ZRC(s).
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The partial fraction expansion of ZRC(s) is given as follows:
Z RC ( s) =
P0
P2
P4
+
+
+ + P∞
s s +s2 s +s4
⇓
⇓
represents represents
capacitor parallel
comb. of
R and C
C0 =
1
P0
R2 =
P2
s2
C2 =
1
P2
⇓
parall el
comb.oof
R and C
R4 =
P2
s4
C4 =
⇓
represents
resistor
R∞ = P∞
1
P4
Accordingly, the Foster form-I of R–C network is shown in Figure 12.1.
R2
R4
R∞
C0
C2
Z(s)
C4
Figure 12.1 Foster Form-I of R–C Network
12.5.4 Foster Form-II of R–C Network
It is obtained from the following:
Y RC (s ) =
1
Z RC (s )
Ps
Y RC (s ) = P0 + 2 + + P∞ s
s +s2
⇓
⇓
⇓
represents represents repres ents
resistor
series
capacitance
comb. of
⇓
R and C
⇓
1
1
Ro =
R2 =
C ∞ = P∞
P0
P2
P
C2 = 2
s2
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Foster from-II of R–C network is shown in Figure 12.2.
Y(s)
R0
R2
R4
C2
C4
C∞
Figure 12.2 Foster Form-II of R–C network
12.5.5 Cauer Forms of R–C Network
The R–C network for Cauer form-I and Cauer form-II is shown in Figure 12.3 (a) and (b),
respectively. Note that in Cauer form-I resistors are in series and capacitors are in parallel. In
Cauer form-II capacitors are in series while the resistors are in parallel. Method of calculation
of their values will be illustrated through an example.
R1
Z(s)
R2
C2
C1
C1
Z(s)
R3
C2
R1
Rn
C3
C3
R2
Cn
Cn
R3
Rn
Figure 12.3 ( a) Cauer Form-I of R–C Network and
(b) Cauer Form-II of R–C Network
12.5.6 Synthesis of R–L Network
General driving point impedance of R–L network is given by the following:
Z RL =
H ( s + s 1 )( s + s 3 )( s + s 5 )....
( s + s 2 )( s + s 4 )( s + s 6 ).......
In the case of R–L network, ZRL(∞) > ZRL(0).
12.5.7 Properties of R–L Impedance
Function/R–C Admittance Function
1. The poles and zeros are simple. There are no multiple poles or zeros.
2. The poles and zero interlace (that is, alternate) each other along the negative real axis.
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3. The poles and zeros lie on the negative real axis of the complex s-plane.
4. Z(∞) > Z(0)
12.5.8 Foster Form-I of R–L Network
The partial fraction expansion of ZRL(s) is given as follows:
P2 s
Ps
+ 4 + + P∞ s
s +s2 s +s4
⇓
⇓
⇓
⇓
R 0 = P0 R 2 = P2 R 4 = P4
L ∞ = P∞
P
P
L2 = 2 L4 = 4
s2
s4
Z (s ) = Po +
Figure 12.4 shows the Foster form-I of R–L network.
R2
R4
RO
L∞
L2
Z(s)
L4
Figure 12.4 Foster Form-I of R-L Network
12.5.9 Foster Form-II of R–L Network
Consider the foster form-II of R–L network shown in Figure 12.5.
The network is designed by using the
partial fraction method. Here, we consider
the partial fraction expansion of Y(s).
Y(S )
R2
R4
L2
L4
L0
R∞
Figure 12.5 Foster Form-II of R–L Network
Y (s ) =
P0
P2
P4
+
+
+ + P∞
s s +s2 s +s4
⇓
⇓
⇓
L0 =
s
1
R2 = 2
Po
P2
L2 =
1
P2
L4 =
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R4 =
⇓
s4
P4
R ∞ = P∞
1
P4
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12.5.10 Cauer Form-I of R–L Network
Figure 12.6(a) shows the representation of cauer form-I for R–L network.
L1
L2
L3
R1
R2
R3
Figure 12.6(a) Schematic Diagram Shows Cauer Form-I of R–L Network
12.5.11 Cauer Form-II R–L Network
Figure 12.6(b) shows the representation of cauer form-II for R–L network.
R1
R2
R3
L1
R4
L2
L3
L4
Figure 12.6(b) Schematic Diagram Shows Cauer Form-II of R–L Network
12.5.12 Synthesis of L–C Networks
LC immittance is given by
Z (s ) =
H (s 2 + w12 )(s 2 + w 32 )…
s (s 2 + w 22 )(s 2 + w 42 )…
12.5.13 Properties of L–C Immittance
1. The immittance function (that is, Z(s) or Y(s)) is a ratio of odd to even polynomial or even
to odd polynomial.
2. The highest order terms in the numerator and denominator differ by one.
3. There is a pole or zero at origin
4. There is a pole or zero at infinity
5. The poles and zeros occur in complex conjugate pairs.
12.5.14 Foster Form-I of L–C Network
The partial fractions of ZLC(s) are given by
Z (s ) =
P0
2P s
2P s
+ 2 2 2 + 2 4 2 + + P∞ s
s s + w2 s + w4
⇓
⇓
⇓
C0 =
1
P0
C2 =
1
2P2
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 517
L2 =
2P2
L4 =
w 22
C4 =
⇓
2P4
w 42
L ∞ = P∞
1
2P4
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Z (s ) =
P0
2P s
2P s
+ 2 2 2 + 2 4 2 + + P∞ s
s s + w2 s + w4
⇓
⇓
518 Network Analysis and Synthesis
1
C0 =
P0
C2 =
⇓
L2 =
1
2P2
2P2
L4 =
w 22
C4 =
⇓
2P4
w 42
L ∞ = P∞
1
2P4
The Foster form-I of the L–C network is shown in Figure 12.7.
L2
C0 =
L4
P∞
1
P0
C2
Z(s)
C4
Figure 12.7 Foster Form-I of L–C Network
12.5.15 Foster Form-II of L–C Network
Y (s ) =
P0
2P2 s
2P4 s
+
+
+ ....... + P∞ s
s s 2 + w 22 s 2 + w 4 2
⇓
L0 =
C2 =
⇓
⇓
1
P0
2P2
w2
1
2P2
L2 =
2
C4 =
⇓
L4 =
1
2P4
C ∞ = P∞
2P4
w 42
The Foster form-II for the L–C network is shown in Figure 12.8.
L2
Y(s)
L4
L0
C∞
C2
C4
Figure 12.8 Foster Form-II of L–C Network
12.5.16 Cauer Form-I of L–C Network
The Cauer form-I of L–C network is shown in Figure 12.9(a) below.
L1
L2
C1
L3
C2
C3
Figure 12.9(a) Cauer Form-I of L–C Network
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Network Synthesis and Realisability 519
12.5.17 Cauer Form-II of L–C Network
Cauer form-II of the L–C network is shown in Figure 12.9(b).
C2
C1
C3
L1
L2
L3
Figure 12.9(b) Cauer Form-II of L–C Network
Example 12.11 Show whether the following driving point functions represent a L–C or a
R–C network. Give reasons for your answer.
1. Z ( s) =
2. Z ( s) =
s 2 + 8s + 12
s2 + 6s + 5
s3 + 9 s
4
s + 5s 2 + 4
Solution:
For function 1:
Z ( s) =
=
s 2 + 8s + 12
s2 + 6s + 5
=
s 2 + 6 s + 2 s + 12
s 2 + 5s + s + 5
( s + 6)( s + 2)
( s + 1)( s + 5)
In the above, all the poles and zeros are real, and therefore, it may be R–C or R–L impedance
function. We have to check whether it is R–C or R–L. For this, we have to find Z (∞).
 6  2
s 1 +  s 1 + 
 s   5
( s + 6)( s + 2)
Z ( s) =
=
( s + 1)( s + 5)
 1  5
s 1 +  s 1 + 
 s  s
 6  2
1 +  1 + 
s
s
=
1
5



1 +  1 + 
s
s
Substituting s = ∞, we get the following:
Z(∞) =
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(1 + 0)(1 + 0)
=1
(1 + 0)(1 + 0)
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520 Network Analysis and Synthesis
That is, Z(∞) = 1
Z(0) =
Now,
(0 + 6)(0 + 2) 12
=
(0 + 1)(0 + 5)
5
Thus, Z(0) > Z(∞). Therefore, the given impedance function is of R–C type. The property of
R–C impedance as stated in section 12.5.2, is that ZRC(0) > ZRC(∞).
Z ( s) =
For function 2:
=
=
=
s3 + 9 s
s 4 + 5s 2 + 4
s ( s 2 + 9)
s4 + 4s2 + s2 + 4
s( s 2 + 9)
s 2 ( s 2 + 4) + 1( s 2 + 4)
s( s 2 + 9 )
( s 2 + 1)( s 2 + 4)
⇓
Since Z(s) has quadratic factors and there is a difference of one between degree of numerator
and denominator, it can be written as follows:
Z ( s) =
s3 + 9 s
s 4 + 5s 2 + 4
is of L–C type. (See properties of L–C impedance section in 12.5.13)
Example 12.12 Synthesise all the four forms of the R–C driving point function represented as
Z RC ( s) =
2( s + 2)( s + 4)
( s + 1)( s + 3)
Solution: To synthesis the Foster form-I circuit, let us consider the given driving point
function as
Z ( s) =
2( s + 2)( s + 4)
( s + 1)( s + 3)
The partial fraction expansion of Z(s) will be as follows:
A
B
+
(12.1)
s +1 s + 3
[∵ degree of numerator and denominator is same]
Let us find the values of A and B
= 2+
2( s + 2)( s + 4)
A
B
= 2+
+
( s + 1)( s + 3)
s +1 s + 3
2( s + 2)( s + 4) 2( s + 1)( s + 3) + A( s + 3) + B( s + 1)
=
( s + 1)( s + 3)
( s + 1)( s + 3)
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Network Synthesis and Realisability 521
or
2(s + 2) (s + 4) = 2(s + 1) (s + 3) + A(s + 3) + B(s + 1)
To find A, substitute s = -1 in equation (12.2), we get
or
2(-1 + 2) (-1 + 4) = 0 + A (-1 + 3) + 0
2(1) (3) = A (2)
or
A=3
(12.2)
To find the value of B, substitute s = -3 in equation (12.2)
2(-3 + 2) (-3 + 4) = 0 + 0 + B(-3 + 1)
or
2(-1) (1) = B(-2)
or
-2 = -2B
or
B =1
Now, substituting the values of A and B in equation (12.1), we get the following form:
1
3
+
s +1 s + 3
P2
P4
Z ( s) =
+
+ P∞
s +s 2 s +s 4
Z ( s) = 2 +
P
In this case, term 0 is missing, and therefore, C0 is absent. (For reference, see Figure 12.1)
s
3
Term
represents the parallel combination of R2 and C2.
s +1
P
3
where
R2 = 2 = = 3Ω
s2 1
C2 =
1 1
= F
P2 3
General form of Foster form-I is represented as follows. The corresponding network is shown
in Figure 12.10.
Z ( s) =
P0
P2
+
+ P∞
s
s +s2
↓
↓
↓
represents represents represents
cappacitor R and C in resistor
1
C0 =
parallel
R∞ = P∞
P0
R2 =
P2
s2
C2 =
1
P2
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 521
R2
R∞
C0
Z(s)
C2
Figure 12.10
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522 Network Analysis and Synthesis
1
represents the parallel combination of R4 and C4.
s+3
P
1
where
R4 = 4 = Ω
s4 3
Term
C4 =
1 1
= = 1F
P4 1
Required Foster form-I is shown in Figure 12.11. Note that in the network, C0 is absent.
1Ω
3
3Ω
2Ω
Z(s)
1F
1 F
3
Figure 12.11
Foster form-II circuit is developed from Y(s) of the given impedance function as
1
( s + 1)( s + 3)
=
Z ( s) 2( s + 2)( s + 4)
1 ( s + 1)( s + 3)
(12.3)
= ⋅
2 ( s + 2)( s + 4)
Since the degree of numerator and denominator is same, the partial fraction will be as follows:
Y ( s) =
Y ( s) =
1
A
B
+
+
(12.4)
2 s+2 s+4
Let us find the values of A and B.
1 ( s + 1)( s + 3) ( s + 2)( s + 4) + A ⋅ 2( s + 4) + B ⋅ 2( s + 2)
Now,
⋅
=
2 ( s + 2)( s + 4)
2( s + 2)( s + 4)
or
(s + 1) (s + 3) = (s + 2) (s + 4) + A ⋅ 2(s + 4) + B ⋅ 2(s + 2)
(12.5)
To find A, substitute s = -2 in equation (12.5) [∵ in the denominator of A factor is s + 2]
(-2 + 1) (-2 + 3) = 0 + A ⋅ 2(-2 + 4) + 0
(-1) (+1) = A ⋅ 2(2)
(-1) = 4A
−1
or
A=
( negative)
4
Here, none of the coefficient can be negative. In such cases, we will use the partial fractions as
in the following:
Y ( s)
( s + 1)( s + 3)
=
s
2 s( s + 2)( s + 4)
or
or
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Network Synthesis and Realisability 523
Now, degree of denominator > degree of numerator, and therefore, partial fraction will be used
as follows:
A
B
C
Y ( s) = +
+
(12.6)
s s+2 s+4
(s + 1)(s + 3)
(s + 1)(s + 3)
(0 + 1)(0 + 3)
=
=
Value of A = s ⋅
2s (s + 2)(s + 4) s = 0 2(s + 2)(s + 4) s = 0 2(0 + 2)(0 + 4)
1× 3
2×2×4
3
=
16
B = Y (s )(s + 2) s = − 2
=
=
(s + 1)(s + 3)
2s (s + 4) s = − 2
( −2 + 1)( −2 + 3)
2 × ( −2)( −2 + 4)
1
=
8
=
C = (s + 4 ) ⋅
(s + 1)(s + 3)
(s + 1)(s + 3)
=
2s (s + 2)(s + 4) s = − 4
2s (s + 2) s = − 4
( −4 + 1)( −4 + 3)
2( −4)( −4 + 2)
( −3)( −1)
3
=
=
2( −4)( −2) 16
3
C=
16
Now, substituting the values of A, B and C in equation (12.6), we get the following:
3
1
3
Y (s ) 16
8
16
=
+
+
s
s s+2 s+4
1
3
s
s
3
or,
Y (s ) =
+ 8 + 16 (12.7)
16 s + 2 s + 4
3
Here,
P0 =
16
1
P2 = , s 2 = 2
8
3
P4 = , s 4 = 4
16
=
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524 Network Analysis and Synthesis
General partial fractions of YRC (s) are
YRC ( s) = P0 +
P2 s
Ps
+ 4 + + P∞ (12.8)
s +s2 s +s4
Therefore, comparing equations (12.7) and (12.8), we get the following:
1 16
3
= Ω
term
will represent resistor of value R0 =
3
P0
16
1
s
represents series combination of R2 and C2
Term 8
s +s2
1
R2 =
= 8Ω
where
P2
P
1
C2 = 2 = F
s 2 16
3
s
Term 16 represents series combination of R4 and C4.
s+4
Y(s)
R0
16 Ω
3
R2
8Ω
R4
C2
1
F
16
C4
where R4 =
1 16
= Ω
3
P4
C4 =
P4
3
F
=
s 4 64
16 Ω
3
3
F
64
Therefore, required Foster form-II, is shown
in Figure 12.12.
In Cauer form-I, we arrange numerator
and denominator of Z(s) in descending power of s and then carry out continued fraction expansion to identify the circuit elements from the quotient terms.
Given
2( s + 2)( s + 4)
Z ( s) =
( s + 1)( s + 3)
Figure 12.12
=
2( s 2 + 4 s + 2 s + 8)
=
2 s 2 + 12 s + 16
s 2 + s + 3s + 3
s2 + 4s + 3
s 2 + 4 s + 3 2 s 2 + 12 s + 16 2 ⇒ R1 (Constant term represents resistor ) → Z1
)
(
2 s 2 + 8s + 6
s

4 s + 10  s 2 + 4 s + 3  ⇒ C1 → Y1

4
s2 +
5
s
2
3

8
s + 3 4 s + 10  ⇒ R2 → Z 2

3
2
4s + 8
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 524
3
3
2  s + 3  s ⇒ C2 → Y2
2
2
3
s
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2 s 2 + 8s + 6
s

4 s + 10  s 2 + 4 s + 3  ⇒ C1 → Y1

4
s2 +
5
s
2
Network Synthesis and Realisability 525
3

8
s + 3 4 s + 10  ⇒ R2 → Z 2

3
2
4s + 8
3
3
2  s + 3  s ⇒ C2 → Y2
2
2
3
s
2
 2
3  2  ⇒ R3 → Z3
 3
2
x
Note that in Cauer form-I realization of the circuit element
from Z(s), quotient terms with s in the numerator represents
the capacitor. The first quotient element will be Z1 since we
started with Z(s).
We know, in Cauer form-I of R–C, resistors are in series
and capacitors are in shunt, that is, parallel.
Therefore, required Cauer form-I is shown in Figure 12.13.
For Cauer from-II , we arrange the numerator and denominator in the ascending power of ‘s’ so that
Z ( s) =
Now,
2Ω
8Ω
3
1F
Z(s) 4
2Ω
3
3F
4
Figure 12.13
16 + 12 s + 2 s 2
3 + 4s + s2
)
 16
3 + 4 s + s 2 16 + 12 s + 2 s 2 
 3
16 +
−
64
16
s + s2
3
3
28 10 2 
 9
s + s  3 + 4s + s2  −

 285
3
3
↓
30
3+ s
28
here the quotient is negative
Since we got a negative quotient as above, we will have to find Cauer form-II circuit by using
the function Y(s)
Y (s) =
1
3 + 4s + s2
=
Z ( s) 16 + 12 s + 2 s 2
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526 Network Analysis and Synthesis
Let us find continued expansion
1
1

 3
16 + 12 s + 2 s 2  3 + 4 s + s 2  ⇒ =

 16
R1 Y1
3+
9
3
s + s2
4
8
1
1
7
5 
 64
s + s 2  16 + 12 s + 2 s 2  ⇒
=
 7s
C1 Z1
8
8 
16 +
40
s
7
5  49
1
1
44
7
s + 2s2  s + s2 
⇒
⇒
4
7
8  176
R2
Y2
7
49 2
s+
s
4
88
1
1
6 2  44
 44 × 88
s 
s + 2s2 
⇒
=
 42 s
C2 Z 2
88  7
44
s
7
1
1
 6 2 3
=
2s2  s  ⇒
 88  88
R3 Y3
6 2
s
88
x
1
. In Cauer form-II, capacitors are in series and resistors are in parallel,
Y1
as shown in Figure 12.14.
First element will be
42
F
44 × 88
C1
7
16
3
Ω
R1
64
C2
F
R2
176
49
Ω
R3
88
3
Ω
Figure 12.14 Cauer Form-II
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Network Synthesis and Realisability 527
Example 12.13 Synthesise all the four forms of the driving point function represented as
Z (s ) =
(s + 1)(s + 5)
(s + 3)(s + 7)
Solution: To find Z(∞), we write the following:
 1  5
s 1 +  s 1 + 
 s  s
Z ( s) =
 3  7 
s 1 +  s 1 + 
 s  s
 1  5
1 +  1 + 
s
s
=
 3  7 
1 +  1 + 
s
s
1
5

1 +  1 + 
∞
∞
Z(∞) =
3
7

1 +  1 + 
∞
∞
Therefore,
=1
Further, to find Z(0), substitute s = 0 in Z(s)
Z ( 0) =
(0 + 1)(0 + 5)
5
=
(0 + 3)(0 + 7) 21
Now, Z(∞) > Z(0)
Therefore, given impedance is of R–L type.
For Foster from-I, we start write Z(s) as
Z (s ) =
(s + 1)(s + 5)
(s + 3)(s + 7)
Since degree of numerator is equal to the degree of the denominator, the partial fractions of Z(s)
will be given as follows:
Z ( s) =
A
B
+
+ 1 (12.9)
s+3 s+7
Let us find values of A and B
( s + 3)( s + 5)
A
B
=
+
+1
( s + 3)( s + 7) s + 3 s + 7
( s + 1)( s + 5) A( s + 7)) + B( s + 3) + ( s + 3)( s + 7)
=
( s + 3)( s + 7)
( s + 3)( s + 7)
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528 Network Analysis and Synthesis
or
(s + 1) (s + 5) = A(s + 7) + B(s + 3) + (s + 3)(s + 7)
(12.10)
To find A, substitute s = −3 in equation (12.10), we get the following:
or
or
(−3 + 1) (−3 + 5) = A(−3 +7) + 0+ 0
(−2)(2) = A(4)
−4 = 4A
A = −1 (negative value)
Since none of the coefficient should be negative, and we will use the partial fractions of
Z (s )
as
s
Z ( s)
( s + 1)( s + 5)
=
s
s( s + 3)( s + 7)
Now, the degree of denominator > degree of numerator, and the partial fractions will be used
as follows:
Z ( s) A
B
C
(12.11)
= +
+
s
s s+3 s+7
A=
( s + 1)( s + 5)
s( s + 3)( s + 7) s = 0
=
( s + 1)( s + 5)
( s + 3)( s + 7) s = 0
=
5
21
B = ( s + 3)
( s + 1)( s + 5)
( s + 1)( s + 5)
=
s( s + 3)( s + 7) s = − 3
s( s + 7) s = − 3
( −3 + 1)( −3 + 5)
−3( −3 + 7)
( −2)( 2) 1
=
=
−3( 4)
3
=
C=
(s + 7)(s + 1)(s + 5)
s (s + 3)(s + 7) s = − 7
=
(s + 1)(s + 5)
s (s + 3) s = − 7
=
3
7
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Network Synthesis and Realisability 529
Substituting the values of A, B and C in equation (12.11), we get the following form:
3
1
5
Z
(
s
)
7
3
21
1 = 3+
5
+
s3
s7 s + 3 s + 7
Z ( s) 21
=
+
+
1 parameter as,
3
s values
s ofsFoster
+ 3 sfrom-I
+ 7 circuit
Writing Z(s) as, we find the
s
s
5
3
7
1
3
+
+
Z ( s) = s
s +s4
s + s7 2s
5
3 21
Z ( s) =
+
+
s + s↓
21
s + s 2↓
4
↓
↓
5 ↓
1 ↓
3
R4 = P4 = Ω
Ω R2 = P2 = Ω
7
21
3
5
1
3
R0 = Ω R2 = P2 = Ω
R4 = P4 = Ω
21
3L = P2 = 1 7L = P4 = 3
2
4
P
1 s 2 3P×4 3 3 s 4 7 × 7
L2 = 2 =
L4 =
=
s2 3×3 1
s4 7×7 3
H
= H
=
9
49
1
3
H
= H
=
9
49
The required Foster form-I circuit is shown in Figure 12.15.
R0 =
1
5
21
3
Ω
3
7
Ω
Ω
1
Z (s)
9
H
3
49
H
Figure 12.15
Foster form-II circuit is realised using the function Y(s)
Y ( s) =
1
( s + 3)( s + 7)
=
Z ( s) ( s + 1)( s + 5)
Here, degree of denominator = degree of numerator. Therefore, partial fractions will be as
follows:
A
B
Y (s ) = 1 +
+
(12.12)
s +1 s + 5
Let us find values of A and B
( s + 3)( s + 7)
A
B
= 1+
+
( s + 1)( s + 5)
s +1 s + 5
( s + 3)( s + 7) ( s + 1)( s + 5) + A( s + 5) + B( s + 1)
=
( s + 1)( s + 5)
( s + 1)( s + 5)
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530 Network Analysis and Synthesis
or
(s + 3) (s + 7) = (s + 1) (s + 5) + A(s + 5) + B(s + 1)
To find A, substitute s = −1 in equation (12.13)
(12.13)
(−1 + 3) (−1 + 7) = 0 + A(−1 + 5) + 0
A=
12
=3
4
To find the value of B, substitute s = −5 in equation (12.13).
(−5 + 3) (−5 + 7) = 0 + 0 + B (−5 + 1)
(−2) (2) = −4B or B = 1
Now, substituting the values of A and B in equation (12.12), we get the following form:
or
Y (s ) = 1 +
Y (s ) =
3
s +s2
P2 = 3,
+
P4 = 1,
↓
1
Y(s)
3
1
3
3
1
+
s +1 s + 5
1
+ 1
s +s4
↓
P∞ = 1
R ∞ = P∞ = 1 Ω
↓
R2 =
s2 1
= Ω
P2 3
R4 =
s4 5
= = 5Ω
P4 1
L2 =
1 1
= H
P2 3
L4 =
1
= 1H
P4
Ω
5Ω
H
1H
1Ω
The required Foster form-II is shown in Figure
12.16.
For Cauer form-I, we start with,
Z (s ) =
Figure 12.16
=
(s + 1)(s + 5)
(s + 3)(s + 7)
s2 + 6s + 5
(Arranging the polynomials
s 2 + 10 s + 21
in ascending power of ‘s’)
Let us find continued fractions as,
)
(
s 2 + 10s + 21 s 2 + 6s + 5 1
2
s + 10s + 21
)
(
− 4s − 16 s 2 + 10s + 21 − s / 4 ⇒ Quotient term is negative,
s 2 + 4s
Therefore, we will visualized the Cauer form-I circuit using Y(s).
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Network Synthesis and Realisability 531
Y (s ) =
=
Y (s ) =
1
Z (s )
s 2 + 10s + 21
s 2 + 6s + 5
s 2 + 10s + 21
s 2 + 6s + 5
Now we divide as,

1

s 2 + 6s + 5  s 2 + 10s + 21 1 ⇒
⇒ first Quotient will be Y as we have visualized it from Y

Y1

s 2 + 6s + 5
s

4s + 16  s 2 + 6s + 5  ⇒ Z 1

4
s 2 + 4s
1


2s + 5  4s + 16  2 ⇒


Y2
4s + 10
s

6  2s + 5  ⇒ Z 3

3
2s
1
 6
5 6  ⇒
  5 Y3
6
x
The Cauer form-I circuit is shown in Figure 12.17
1
H
4
Y(s)
1Ω
1
H
3
1
Ω
2
5 Ω
6
Figure 12.17 Cauer form-I
For Cauer form-II, we arrange the polynomials in descending power of ‘s’.
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532 Network Analysis and Synthesis
Z ( s) =
5 + 6s + s2
( s + 1)( s + 5)
s2 + 6s + 5
= 2
=
( s + 3)( s + 7) s + 10 s + 21 21 + 10 s + s 2
Now we divide as,
)
 5
21 + 10 s + s 2 5 + 6 s + s 2  ⇒ Z1 , because we are visualizing the circuuit using Z ( s)
 21
5+
50
s
21
 21 × 21 441 1
76

s + s 2  21 + 10 s + s 2 
=
=

21
76 s Y1
 76 s
21 +
441
s
76s
76 s 5776
319
 76
 76
s + s2 
s + s2
×
=
= Z2

 21
319 s 21 6699
76
76
5776 2
s+
s
21
6699
 6699 319 s 2136981 1
923 2  319
s 
s + s2 
×
=
=
47348s Y2
6699  76
76
 623s 2
319
s
76
 923 2  923
s
= Z2
s2 
 6699  6699
923 2
s
66999
x
The Cauer form-II circuit is shown in Figure 12.18.
Example 12.14 Find the Foster form-I and Foster
form-II of the given expression for Z(s).
Z ( s) =
( s + 1)( s + 3)
s( s + 2)
5
21
Ω
5776
6699
76
441
H
Ω
923
6699
Ω
47348
2136981
Figure 12.18 Cauer Form-II
Solution: First, let us check whether the given
impedance is of R–C type or R–L type. Remember, for R–C, Z(0) > Z(∞) and for R–L,
Z(∞) > Z(0).
For this, let us find Z(∞)
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Network Synthesis and Realisability 533
Z(s) can be written as follows:
 1  3
s 1 +  s 1 + 
 s  s
Z ( s) =
 2
s × s 1 + 

s
 1  3
1 +  1 + 
s
s
=
 2
1 + 
s
Substituting s = ∞, we get the following equation:
Z(∞) =
(1 + 0)(1 + 0)
=1
(1 + 0)
Further, to find Z(0),
substitute s = 0 in
Z (s ) =
Now,
Z(0) = ∞
(s + 1)(s + 3)
s (s + 2)
(0 + 1)(0 + 3)
Z(0) =
=∞
0( 0 + 2)
Z(∞) = 1
Clearly, Z(0) > Z(∞), and therefore, given impedance is of R–C type.
For realizing the Foster Form-I network,
(s + 1)(s + 3)
Z (s ) =
s (s + 2)
we write,
Here, degree of denominator = degree of numerator, and therefore, we will find partial fractions
as follows:
B
A
Z (s ) = 1 + +
(12.14)
s s+2
Let us find the values of A and B
A = Z (s) s s = 0
=
( s + 1)( s + 3) s
s( s + 2)
=
s=0
3
2
B = Z (s) ( s + 2) s = − 2
=
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( s + 1)( s + 3)
1
=
s
2
s = −2
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534 Network Analysis and Synthesis
Now, substitute the values of A and B in equation (12.14),
3
1
2
2
Z ( s) =
+
+ 1
s
s+2
↓
↓
↓
3
1
, P2 = ,
P∞ =1
2
2
P
1
, R2 = 2 , R∞ = P∞ = 1 Ω
C0 =
s2
P0
P0 =
=
R2
C0
2
F
3
1 Ω
4
2F
C2
Z(s )
Figure 12.19
R∝
1Ω
2
F
3
1
2×2
1
= Ω
4
1
C2 =
= 2F
P2
=
The Foster form-I circuit is shown in Figure 12.19.
For Foster form-II, we write
1
s ( s + 2)
Y (s ) =
=
Z (s ) (s + 1)(s + 3)
Let us use the partial fractions as follows:
A
B (12.15)
s( s + 2)
Y ( s) =
=
+
( s + 1)( s + 3) s + 1 s + 3
Let us find values of A and B
s ( s + 2)
A( s + 3) + B( s + 1)
=
( s + 1)( s + 3)
( s + 1)( s + 3)
or
s(s + 2) = A (s + 3) + B(s + 1)
(12.16)
To find A, substitute s = −1 in equation (12.16),
−1(−1 + 2) = A (−1 + 3) + 0
−1(1) = 2A
1
A= −
or
2
Since none of the coefficient of Z(s) should be negative, we will obtain the partial fractions of
Y (s ) .
s
Y ( s)
s( s + 2)
=
s
s( s + 1)( s + 3)
Y (s )
(s + 2) (12.17)
=
s
(s + 1)(s + 3)
(Here, degree of denominator > degree of numerator)
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Network Synthesis and Realisability 535
Therefore, partial fractions will be given as follows:
Y ( s)
A
B (12.18)
=
+
s
s +1 s + 3
Let us find values of A and B
A = ( s + 1)
( s + 2)
( −1 + 2) 1
=
=
( s + 1)( s + 3) s = −1 ( −1 + 3) 2
B = ( s + 3)
( s + 2)
s+2
−3 + 2 −1 1
=
=
=
=
( s + 1)( s + 3) s = − 3 s + 1 s = − 3 −3 + 1 −2 2
A=
1
1
;B =
2
2
Substituting the values of A and B in equation (12.18), we get the following:
Y (s) 1/ 2 1/ 2
=
+
s
s +1 s + 3
Y ( s) =
s
s
2
2
+
(s + s 2 )
(s + s 4 )
1
↓
2
1
R2 =
= 2Ω
P2
P2 =
C2 =
P2
1
=
s 2 2 ×1
↓
R4 =
1
= 2Ω
P4
C4 =
P4
1
=
s4 2×3
1
1
= F
F
2
6
Therefore, required Foster form-II circuit is drawn as shown in Figure 12.20.
=
Example 12.15 Obtain the Cauer-I and Cauer-II form of
(s + 1)(s + 3)
network for Z (s ) =
s (s + 2)
Solution: From the given Z(s), Z(∞) = 1, Z(0) = ∞; Clearly, Z(0) >
Z(∞), and therefore, the given impedance function is of R–C type.
Cauer-I: To obtain Cauer-I,
We arrange the numerator and denomination in descending
power ‘s’ and carry out the division work. The circuit elements are
identified from the division work as,
Z ( s) =
Y(s)
R2
2 Ω R4
C2
1 C4
F
2
2Ω
1
F
6
Figure 12.20
( s + 1)( s + 3) s 2 + 4 s + 3
= 2
s( s + 2)
s + 2s
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536 Network Analysis and Synthesis
)
(
s 2 + 2 s s 2 + 4 s + 3 11 ⇒ Z ⇒ R1 = 1 Ω
s2 + 2s
1
1

2 s + 3  s 2 + 2 s  s ⇒ Y ⇒ C1 = F

2
2
3
s2 + s
2
1 

s 2 s + 3  4 ⇒ Z ⇒ R2 = 4 Ω

2 
2s
R1 = 1 Ω
Z(s)
C1
4Ω
R2
1
F
2
C2
1
F
6
1
 1 1
3  s  s ⇒ Y ⇒ C2 = s
 2 6
6
1
s
2
x
Cauer form-I of representation of the circuit has been shown
in Figure 12.21. In the circuit the resistors are in series and
capacitors are in parallel.
For Cauer Form-II representation of the circuit for the
given Z(s), we arrange the numerator and denominator in
ascending order of s.
Figure 12.21
( s + 1)( s + 3) s 2 + 4 s + 3 3 + 4 s + s 2
=
= 2
s( s + 2)
2s + s2
s + 2s
We carry out the division work to identify the circuit elements from the quotient terms.
Z (s) =
)
 3
1
2
2s + s 2 3 + 4 s + s2  ⇒
⇒ C1 = F
Z1
3
 2s
3+
3
s
2
4
5
1
5

s + s 2  2 s + s 2  ⇒ ⇒ R1 = Ω

2
Y1
4
5
2s +
4 2
s
5
1
1
1 2 5
 25
s  s + s2  ⇒
⇒ C2 =
F


25
5
2
2s
Z2
5
s
2
1
 1 1
s2  s2  ⇒
⇒ R2 = 5 Ω
 5 5
Y2
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1 2
s
5
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2
s + s  2s + s

2s +
 5 ⇒ Y ⇒ R1 = 4 Ω
1
4 2
s
5
1 2 5
1
1
 25
s  s + s2  ⇒
⇒ C2 =
F
 2s
5 2
Z2
25
5
s
2
Network Synthesis and Realisability 537
1
 1 1
s2  s2  ⇒
⇒ R2 = 5 Ω
 5 5
Y2
1 2
s
5
x
Required Cauer form-II circuit for Z(s) is shown in Figure 12.22 where the capacitors are in
series and the resistors are in parallel.
Example 12.16 Synthesise the impedance function Z(s) in
Foster form-II where Z(s) is given as
Z ( s) =
( s + 5)( s + 7)
( s + 1)( s + 6)( s + 8)
C1 =
2
F
3
R1
Z(s)
Solution: We find Z(∞) and Z(0) to check the nature of the
circuit being represented by Z(s)
2
F
25
C2
5
Ω
4
5Ω
R2
Figure 12.22
 5  7
s 1 +  s 1 + 
 s  s
( s + 5)( s + 7)
=
Z ( s) =
( s + 1)( s + 6)( s + 8)
 1  6   8
s 1 +  s 1 +  s 1 + 
 s  s   s
 5  7
1 +  1 + 
s
s
=
1
6

  8

s 1 +  1 +  1 + 
 s  s   s
5 
7

1 +  1 + 
(11 + 0)(1 + 0) 1
∞
∞
=
= =0
Z(∞) =
1
6
8
∞(1)(1)(1)
∞

∞ 1 +  1 +  1 + 
 ∞  ∞  ∞
35
(0 + 5)(0 + 7)
=
. Since Z(0) > Z(∞), the given impedance function repre(0 + 1)(0 + 6)(0 + 8) 48
sents an R–C network.
and Z (0) =
Foster form-II circuit is designed using Y(s) as in the following:
Y ( s) =
1
( s + 1)( s + 6)( s + 8)
=
Z ( s)
( s + 5)( s + 7)
Here, degree of denominator < degree of numerator.
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538 Network Analysis and Synthesis
For the partial fractions, degree of denominator ≥ degree of numerator.
Therefore, to make degree of denominator = degree of numerator, we divide both sides of
r(s) by s as follows:
Y ( s) ( s + 1)( s + 6)( s + 8)
=
s
s( s + 5)( s + 7)
Y ( s)
A
B
C
= 1+ +
+
(12.19)
s
s s+5 s+7
Now,
or
( s + 1)( s + 6)( s + 8) s( s + 5)( s + 7) + A( s + 5)( s + 7) + B ⋅ s( s + 7)) + C ⋅ s( s + 5)
=
s( s + 5)( s + 7)
s( s + 5)( s + 7)
(s + 1)(s + 6)(s + 8) = s(s + 5)(s + 7)+A(s + 5)(s + 7)+B⋅s(s + 7) + C⋅s(s + 5) (12.20)
Substitute s = 0 in equation (12.20) to get A as
or
(0 + 1)(0 + 6)(0 + 8) = 0 + A (0 + 5) (0 + 7) + 0
48 = 35A
or
A=
48
35
To find B, substitute s = −5 in equation (12.20)
(−5 + 1) (−5 + 6) (−5+8) = 0 + 0 + B (−5) (−5 + 7)
or
(−4) (1) (3) = B (−5) (2)
−12 = −10B
or
12 6
=
10 5
Substitute s = −7 in equation (12.20) to get C as
B=
or
or
or
(−7 + 1)(−7 +6)(−7 + 8) = 0 + 0 + 0+C(−7)(−7 + 5)
(−6) (−1)(1) = C(−7)(−2)
6 = 14C
6 3
=
14 7
Now, substitute the values of A, B and C in equation (12.19), we get the following form:
C=
or
Y ( s)
3/ 7
48/ 35 6 / 5
= 1+
+
+
s
s
s+5 s+7
48 6 / 5s 3/ 7 s
Y ( s) = s +
+
+
35 s + 5 s + 7
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Network Synthesis and Realisability 539
6
3
s
s
48
5
7
Y ( s) =
+
+
+ 1s
35
s+ 5
s+ 7
48
6
3
← P0 =
, ← P2 = , ← P4 = , P∞ = 1;s 2 = 5, s 4 = 7
35
5
7
The circuit elements are cal culated as
1
1 5
1 7
= Ω
= Ω C∞ = P∞ = 1F
R0 =
R2 =
R4 =
P0
P2 6
P4 3
=
35
Ω
48
C2 =
=
P2
6
=
s2 5×5
6
F
25
C4 =
=
P4
s4
3
3
=
F
7 × 7 49
Therefore, the required Foster form-II is shown in Figure 12.23. The students many compare the
s circuit with the standard Foster from-II of R–C network shown in Figure 12.2.
Y(s)
35 Ω
48
R0
R2
5 Ω R4
6
7 Ω
3
C2
6 F C4
25
3 F
49
C∝
1F
Figure 12.23
Example 12.17 A function F(s) has poles and zeros as follows:
Poles at s = 0, −4, −6
Zeros at z = −2, −5
Synthesise F(s) (a) as an impedance in Foster form; (b) as an admittance in Cauer form
Solution: Since the numerator equated to zero given the zeros and the denominator equated to
zero given the poles, the impedance function F(s) is written as,
F ( s) =
( s + Z1 )( s + Z 2 )
( s + P1 )( s + P2 )( s + P3 )
F ( s) =
( s + 2)( s + 5)
s( s + 4)( s + 6)
Substituting the given values,
(a) We calculate Z(∞) and Z(0) to identify the nature of the network in Z(s) as synthesis of F(s)
as impedance in Foster form. We first
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540 Network Analysis and Synthesis
 2  5
s 1 +  s 1 + 

( s + 2)( s + 5)
s  s
Z ( s) =
=
s( s + 4)( s + 6)
 4  6
s 1 +  s 1 + 
 5  s 
(1 + 0)(1 + 0)
∞(1 + 0)(1 + 0)
=0
(0 + 2)(0 + 5)
Z ( 0) =
=∞
0(0 + 4)(0 + 6)
Z (∞) =
and
Z(0) > Z(∞), and therefore, Z(s) represents an R–C network.
Using Z(s), we can now proceed to design Foster form-I circuit.
We consider the partial fraction of Z(s) as in the following:
Z (s ) =
Now,
B
C
A
+
+
(12.21)
s s+4 s+6
A = s⋅
( s + 2)( s + 5)
( s + 2)( s + 5)
( 2)(5) 5
=
=
=
s( s + 4)( s + 6) s = 0 ( s + 4)( s + 6) s = 0 ( 4)(6) 12
B = ( s + 4) ⋅
( s + 2)( s + 5)
( s + 2)( s + 5)
=
s( s + 4)( s + 6) s = − 4
s( s + 6 ) s = − 4
( −4 + 2)( −4 + 5)
−4( −4 + 6)
( −2)(1) 1
=
=
−4( 2) 4
=
and
C = ( s + 6) ⋅
( s + 2)( s + 5)
( s + 4)( s + 6) s = − 6
=
( −6 + 2)( −6 + 5)
−6( −6 + 4)
=
( −4)( −1) 1
=
−6( −2)
3
Substituting the values of A, B and C in equation (12.21), we get the following:
Z ( s) =
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5/12 1/ 4
1/ 3
+
+
s
s+4 s+6
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Network Synthesis and Realisability 541
Z ( s) =
5/12 = P0 1/ 4 = P2 1/ 3 = P
+
+
s
s+4
s+6
s2 = 4 s4 = 6
From the above we find the circuit elements as,
C0 =
P
1 12
1
1
1
= 4 F;
= F; R2 = 2 =
= Ω; C2 =
5
s 2 4 × 4 16
P2
P0
R4 =
P4
1
1
1
=
= Ω and C4 =
= 3F
P4
s 4 3 × 6 18
The required Foster form-I is shown in Figure 12.24.
1
Ω
16
1
Ω
18
R2
R4
C2
4F
C4
3F
C0
12
F
5
Z(s)
Figure 12.24
(b) We now write F(s) as an admittance and visualize the
Cauer form-I circuit.
Y ( s) =
1
s( s + 4)( s + 6) s3 + 10 s + 24 s
=
= 2
Z ( s) ( s + 2)( s + 5)
s + 7 s + 10 s
We now carry out the division work and determine the circuit elements from the quotient as,
)
(
s 2 + 7 s + 10 s3+ 10 s 2 + 24 s s ⇒ Y ⇒ First element will be Y as we are deesigning from Y ( s)
s3 + 7 s 2 + 10 s

1
3s 2 + 14 s s 2 + 7 s + 10  ⇒ Z

3
s2 +
14
s
3
7

9
s + 10 3s 2 + 14 s  s ⇒ Y

7
3
90
s
3ss 2 +
7
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 541
8 7
 49
s s + 10 
⇒Z
 24
7 3
7
s
3
11/17/2014 6:14:24 PM
3s + 14 s s + 7 s + 10  ⇒ Z

3
s2 +
14
s
3
7

9
s + 10 3s 2 + 14 s  s ⇒ Y

7
3
542 Network Analysis and Synthesis
90
3ss 2 +
s
7
8 7
 49
s s + 10 
⇒Z
 24
7 3
7
s
3
8  8
10 s  s ⇒ Y
 7  70
8
s
7
x
In Cauer from-I of R–C network Z stands for R and Y stands for C.
Therefore, required Cauer form-I circuit is shown in Figure 12.25. Note that in the circuit the
first element starts with C1 and not R1 as we started with Y(s) and not Z(s).
Y(s )
C1
1
Ω
3
49
Ω
24
R2
R3
1F
C2
9
F
7
C3
8
F
70
Figure 12.25
Example 12.18 An impedance function is given as,
Z ( s) =
s( s + 2)( s + 5)
( s + 1)( s + 4)
Find Foster form-I, Foster form-II, Cauer form-I and Cauer form-II of circuits, which this
impedance function represents.
Solution: Given,
s( s + 2)( s + 5)
( s + 1)( s + 4)
Firstly, we will check whether the given impedance is of R–L or R–C type. Z(∞) can be calculated as follows:
Z (s) =
Now,
 2   5
 2   5
s ⋅ s 1 +  s 1 +  s 1 +  1 + 


s  s
s  s
Z ( s) =
=
 1  4 
 1  4 
s 1 +  s 1 + 
1 +  1 + 
 s 
s
s
s
∞(1)(1)
=∞
Z(∞) =
(1)(1)
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Network Synthesis and Realisability 543
0(0 + 2)(0 + 5)
=0
(0 + 1)(0 + 4)
Clearly, Z(∞) > Z(0), and therefore, the given impedance is of R–L type
Foster form-I circuit is designed using Z(s) as follows:
and
Z(0) =
Z ( s) =
s( s + 2)( s + 5)
( s + 1)( s + 4)
Here, the degree of s in the denominator < degree of s in the numerator. For the partial fractions,
the degree of denominator ≥ numerator.
Therefore, to make degree of denominator = degree of numerator, we will divide both sides
of Z(s) by s. Accordingly.
Z ( s) s( s + 2)( s + 5) ( s + 2)( s + 5)
=
=
s
s( s + 1)( s + 4) ( s + 1)( s + 4)
Its partial fractions will be as follows:
Z ( s)
A
B
= 1+
+
(12.22)
s
s +1 s + 4
Let us find values of A and B
( s + 2)( s + 5)
A
B
= 1+
+
( s + 1)( s + 4)
s +1 s + 4
( s + 2)( s + 5) ( s + 1)( s + 4) + A( s + 4) + B( s + 1)
=
(12.23)
( s + 1)( s + 4)
( s + 1)( s + 4)
To find A, substitute s = −1 in equation (12.23)
(−1 + 2)(−1 + 5) = 0 + A(−1 + 4) + 0
(1)(4) = 3A
4
or
A=
3
To find B, substitute s = −4 in equation (12.23)
(−4 + 2) (−4 + 5) = 0 + 0 + B(−4 + 1)
(−2)(1) = B(−3)
2
or
B=
3
Now, putting the values of A and B in equation (12.22), we get the following:
Z ( s)
4 /3 2/3
= 1+
+
s
s +1 s + 4
4
2
s
s
3
Z ( s) = s +
+ 3 . (i)
s +1 s + 4
4
2
P2 = , P4 = , P∞ = 1
3
3
s 2 = 1, s 4 = 4
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2
4
s
s
Z ( s) = s + 3 + 3 .
s +1 s + 4
4
2
Here,
P2 = , P4 = , P∞ = 1
3
3
s 2 = 1, s 4 = 4
Z(s) can be written in the following form:
544 Network Analysis and Synthesis
P2 s
P4 s
(ii)
Z (s ) =
+
+ P∞ s
s + sP22s
s + sP44s
+
+ P∞ s
Z (s ) =
s4 + s 2
s + s24
From (i) and (ii), then
circuit
parameters
are
found
as
R 2 = P2 = Ω
R 4 = P4 = Ω L ∞ = P∞ = 1H
34
32
R 2 = P2 = Ω
R 4 = P4 = Ω L ∞ = P∞ = 1H
P2
P4
43
23
=
=
L2 =
L4 =
P
P
4
3 ×1
3 × 42
s
s
L2 = 2 2 =
L4 = 4 4 =
4 s 2 3 ×1
1s 4 3 × 4
= H
= H
34
61
= H
= H
3
6
The Foster form-I circuit is shown in Figure 12.26.
Z(s)
4
Ω
3
2
Ω
3
R2
R4
1H
L2
L4
L∞
4
H
3
1
H
6
Figure 12.26
Foster form-II circuit is designed using Y(s).
1
( s + 1)( s + 4)
Now,
Y (s) =
=
Z ( s) s( s + 2)( s + 5)
Let us use the partial fractions.
Since degree of denominator > degree of numerator, the partial fractions will be given as
follows:
A
B
C
Y ( s) = +
+
(12.24)
s s+2 s+5
Now,
A= s
( s + 1)( s + 4)
( s + 1)( s + 4)
(1)( 4) 2
=
=
=
2(5)
5
s( s + 2)( s + 5) s = 0 ( s + 2)( s + 5) s = 0
B = ( s + 2)
( s + 1)( s + 4)
( −2 + 1)( −2 + 4) ( −1)( 2) 1
=
=
=
( −2)( −2 + 5)
( −2)(3) 3
s( s + 2)( s + 5) s = − 2
C = ( s + 5)
( s + 1)( s + 4)
( −5 + 1)( −5 + 4) ( −4)( −1)
=
=
s( s + 2)( s + 5) s = − 5
( −5)( −3)
−5( −5 + 2)
C=
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 544
4
15
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A= s
( s + 1)( s + 4)
( s + 1)( s + 4)
(1)( 4) 2
=
=
=
2(5)
5
s( s + 2)( s + 5) s = 0 ( s + 2)( s + 5) s = 0
B = ( s + 2)
( s + 1)( s + 4)
( −2 + 1)( −2 + 4) ( −1)( 2) 1
=
=
=
( −2)( −2 + 5)
( −2)(3) 3
s( s + 2)( s + 5) s = − 2
C = ( s + 5)
( s + 1)( s + 4)
( −5 + 1)( −5 + 4) ( −4)( −1)
=
=
s( s + 2)( s + 5) s = − 5
( −5)( −3)
−5( −5 + 2)
Network Synthesis and Realisability 545
4
15
Substituting the values of A, B and C in equation (12.24)
4 /5
2 / 5 1/ 3
Y ( s) =
+
+
s
s+2 s+5
Y(s) is written in the standard from as
P
P22
P44
Y ( s) = 00 +
+
s
s + s 22
s + s 44
C=
( s),circuit
Comparing Comparing
the above two
oftwo
Y(s),from
we find
elements
as elements as
thefrom
above
of Ythe
we find
the circuit
s4
s
1
L00 =
R22 = 22
R44 = 4
P44
P00
P22
=
5
H
2
=
2
1/ 3
5
4 /15
75
=
Ω
4
1
C44 =
P44
=
=6Ω
L22 =
1
P22
= 3H
=
15
F
4
With these values, we draw the Foster form-II
circuit as shown in Figure 12.27.
Now, we will find the Cauer form-I circuit.
We have,
Z ( s) =
3
L0
Y(s)
2
s( s + 2)( s + 5) s + 7 s + 10 s
= 2
( s + 1)( s + 4)
s + 5s + 4
R2
6Ω
R4
75 Ω
4
L2
3H
L4
15
H
4
Foster Form –II
Figure 12.27
Now we will carry out the division work and
determine the circuit elements from the quotient as
)
5H
2
(
s 2 + 5s + 4 s3 + 7s 2 +10 s s⇒ Z ⇒ L1 = 1H ⇒ First element will be Z as we are designing from Z ( s)
s3 + 5s 2 + 4 s
1

1
2s 2 + 6 s s 2 + 5s + 4  ⇒ ⇒ R1 = 2 Ω

2 Y
s 2 + 3s
)
(
2 s + 4 2 s 2 + 6 s s ⇒ Z ⇒ L2 = 1H
2
2s + 4 s
1


2s 2 s + 4 1 ⇒ ⇒ R2 = 1 Ω


Y
2s
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 545

1
1
11/17/2014 6:14:28 PM
s 3 + 5s 2 + 4 s
1

1
2s 2 + 6 s s 2 + 5s + 4  ⇒ ⇒ R1 = 2 Ω

2 Y
s 2 + 3s
)
(
546 Network Analysis and2 sSynthesis
+ 4 2 s 2 + 6 s s ⇒ Z ⇒ L2 = 1H
2
2s + 4 s
1


2s 2s + 4 1 ⇒ ⇒ R2 = 1 Ω


Y
2s
1
 1
4 2 s  s ⇒ Z ⇒ L3 = H
 2
2
2s
x
Z(s)
L1
L2
L3
1H
1H
1
H
2
R1
2Ω
R2
1Ω
Cauer form-I
Accordingly, the Cauer form-I circuit is drawn as in
Figure 12.28.
Now, we will find the Cauer form-II circuit.
Given,
Z (s) =
Figure 12.28
s( s + 2)( s + 5) s3 + 7 s 2 + 10 s
= 2
( s + 1)( s + 4)
s + 5s + 4
For Cauer form-II, we arrange the polynomials in ascending powers of ‘s’ as in Z(s) as
Z (s ) =
10s + 7s 2 + s 3
4 + 5s + s 2
We now carry out the division work and determine the circuit elements from the quotient as
 10
s ⇒ In Cauer from-II, the quotient terms shouuld have ’s’in th
4 + 5s + s 2 10 s + 7 s 2 + s3  4


’s’in the denominator, however, here, s is in the numerator.
)
10s +
50 2 10 3
s + s
4
4
22
6 

− s 2 − s3 4 + 5s + s 2


4
4


⇓
The terms are negative, and hence, we cannot design Cauer form-II circuit using Z(s).
Let us find the circuit using Y(s).
Y (s) =
4 + 5s + s 2
10 s + 7 s 2 + s3
The division work is carried out and the circuit elements are found from the quotients as
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 546
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Network Synthesis and Realisability 547
1
10

 4
10s + 7s 2 + s 3 4 + 5s + s 2 
⇒ ⇒ L1 = H ⇒ First element will be Y

 10s
Y
4
as we are designing from Y (s )
4+
14
2
s + s2
5
5
11
3 
50
 50
Ω
s + s 2  10s + 7s 2 + s 3  ⇒ Z ⇒ R 2 =


5
5
11
11
33 2
s
11
10s +
47 2 3  11
3  121
1
235
s + s  s + s2 
⇒ ⇒ L2 =
H
 5
11
5  235s
121
Y
11
121 2
s+
s
5
235
 235 47

×
 20 11



47 2 3  47 × 47
20s 2 + s 3 
s +s =
 4 × 11
 11


 = 2209 ⇒ Z ⇒ R = 2209 Ω

3


44
44
47 2
s
11
1
235
 20 2  20
s 3
s
⇒ ⇒ L3 =
H
 235  235s
20
Y
20 2
s
235
x
In Cauer form-II of R–L circuit, Z stands for R and Y stands for L. The Cauer form-II circuit is
shown in Figure 12.29
Y(s)
L1
50
Ω
11
2209
Ω
44
R2
10
H L2
4
R3
235
H L3
121
235
H
20
Cauer form-II
Figure 12.29
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548 Network Analysis and Synthesis
12.6 MORE NUMERICALS ON SYNTHESIS OF L–C NETWORK
Example 12.19 Synthesise the function Z(s) using Foster forms
Z ( s) =
s( s 2 + 10)
( s 2 + 4)( s 2 + 16)
Solution: Foster form-I
Z (s ) =
s (s 2 + 10)
2
(s + 4)(s 2 + 16)
Partial fraction expansion of generalised L–C impedance or admittance from results in
Z ( s) =
Z ( s) =
For the given function, P0
2 P2 s
2 P4 s
+
+ + R∞ s.
+
2
2
2
s
s + w2
s + w 42
(
2 P2 s
2
s +4
2 P2 s = ( s 2 + 4)
Now,
2P2 =
or
or
2P4 =
+
2 P4 s
s + 16
(12.25)
2
s + 16
( s 2 + 4)( s 2 + 16) s2 = − 4
s 2 + 10
2
)
s( s 2 + 10)
=
s2 = − 4
2 P4 s = ( s 2 + 16)
Further,
) (
−4 + 10 6 1
=
=
−4 + 16 12 2
s( s 2 + 10)
2
( s + 4)( s 2 + 16) s2 = −16
s 2 + 10 −16 + 10 −6 1
=
=
=
−16 + 4 −12 2
s2 + 4
From equation (12.25), we get the following form:
Z ( s) =
2 P2 s
2
s +4
+
L2 =
=
w
2
2
s 2 + 16
w 24 = 16
w 22 = 4,
⇓
2 P2
2 P4 s
⇓
=
1
2×4
1
H
8
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 548
L4 =
C4 =
2 P4
w
2
4
=
1
1
=
H
2 × 16 32
1
1
=
= 2F
2 P4 1/ 2
11/17/2014 6:14:31 PM
Network Synthesis and Realisability 549
and
C2 =
1
1
=
= 2F
2 P2 1/ 2
Required Foster form-I is shown in Figure 12.30.
Foster form-II is obtained from Y(s).
Now,
2
2
1
(s + 4)(s + 16)
=
Z (s )
s (s 2 + 10)
Y (s ) =
1
H
8
1
H
32
L2
L4
(s 2 + 4)(s 2 + 16)
Y (s ) =
C2
2
s (s + 10)
2F
Z (s)
Now, degree of denominator < degree of
numerator
Y ( s) =
=
=
s( s3 + 10 s) + (10 s 2 + 4)
s( s3 + 10 s)
s( s 2 + 10)
s( s3 + 10 s)
3
s + 10 s
= s+
+
+
10 s 2 + 4
Now,
=
( s 2 + 4)( s 2 + 16)
s( s 2 + 10)
s 4 + 20 s 2 + 64
s3 + 10 s
Divide Numeratorr by Denominator
s( s 2 + 4 )
10 s 2 + 4
s3 + 10 s) s 4 + 20 s 2 + 64( s
s( s 2 + 4 )
s 4 + 10 s 2
10 s 2 + 4
10s 2 + 64
s( s 2 + 4 )
Y ( s) =
Y ( s) =
2F
Figure 12.30
Y ( s) =
s( s 2 + 10)
C4
10 s 2 + 4
s( s 2 + 4 )
+ s (12.26)
10 s 2 + 64
s( s 2 + 10)
+s
⇓
Now, degree of denominator > numerattor.
=
Now,
P0
2P s
+ 2 2 + s(12.27)
s s + 10
P0 = s
(10 s 2 + 64)
s( s 2 + 10)
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 549
=
s=0
10 s 2 + 64
s 2 + 10
=
s=0
0 + 64 32
=
0 + 10
5
11/17/2014 6:14:32 PM
550 Network Analysis and Synthesis
2 P2 s = ( s 2 + 10)
and
or
2P2 s =
2 P2 =
or
(10 s 2 + 64)
s( s 2 + 10)
10 s 2 + 64
s
s 2 = −10
10 s 2 + 64
s2
s 2 = −10
10( −10) + 64
−10
=
s 2 = −10
−100 + 64 −36 18
=
=
−10
−10 5
32
Therefore,
P0 =
5
18
2 P2 =
5
Now, considering equation (12.27), we get the following form:
=
Y ( s) =
P0
2 P2 s
+
+ s ⇒ C∞ = 1 F
s s 2 + 10
↓
w 22
⇓
1
L0 =
P0
=
5
H
32
⇓
L2 =
1
5
= H
2 P2 18
C2 =
2 P2
w 22
=
18
9
=
F
5 × 10 25
Required Foster form-II is shown in Figure 12.31.
Example 12.20 Synthesise the impedance
function as follows:
Z ( s) =
2
2
( s + 2)( s + 4)
2
2
s( s + 3)( s + 5)
Solution:
Y (s)
5
H
32
L2
L0
5
H
18
C2
9
F
25
1F
C∞
Figure 12.31
Z (s ) =
2
2
(s + 2)(s + 4)
s (s 2 + 3)(s 2 + 5)
Here the degree of denominator > numerator
We express, Z ( s) =
P0
2 P6 s
2P s
2P s
+ 2 2 2 + 2 4 2
+ + P∞ s
2
5 (s + w 2 ) (s + w 4 ) (s + w 62 )
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Network Synthesis and Realisability 551
Here, Z ( s) =
C0 =
2 P2 s
+
2
s + 32
+
w2
⇓
⇓
1
P0
L2 =
15
F
8
1
= H
18
=
P0
s
2 P4 s
Now
2
s + 52
w4
⇓
2 P2
w2
L4 =
2
1
6.3
3
=
H
50
and
1
C2 =
2 P2
=
=
=
2 P4
w 42
P0 =
3
10.5
=
( s 2 + 2)( s 2 + 4)
s( s 2 + 3)( s 2 + 5) s = 0
(0 + 2)(0 + 4)
(0 + 3)(0 + 5)
8
15
and 2 P2 s = ( s 2 + 3) ⋅
( s 2 + 2)( s 2 + 4)
s( s 2 + 3)( s2 + 5) s2 = − 3
or
and
1
C4 =
2 P4
=6F
P0 = s
2P2 =
10
F
3
=
( s 2 + 2)( s 2 + 4)
s 2 ( s 2 + 5)
s2 = −3
( −3 + 2)( −3 + 4) −1
=
−3( −3 + 5)
−6
1
6
and
2 P2 =
2 P4 s = ( s 2 + 5)
( s 2 + 2)( s 2 + 4)
s( s 2 + 3)( s 2 + 5) s2 = − 5
or
2 P4 =
=
( s 2 + 2)( s 2 + 4)
s 2 ( s 2 + 3)
s2 = −5
( −5 + 2)( −5 + 4) 3
=
10
−5( −5 + 3)
Required Foster form-I is shown in Figure 12.32.
1
H
18
3
H
50
L2
L4
C0
15
F
8
Z (s)
6F
C2
10
F
3
C4
Figure 12.32
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552 Network Analysis and Synthesis
Example 12.21 The driving point impedance of a one-port LC network is given by the
following:
Z ( s) = 3
( s 2 + 1)( s 2 + 16)
s( s 2 + 9)
Obtain the first and second Foster forms.
Solution: Foster form-I:
Z (s ) =
3(s 2 + 1)(s 2 + 16)
s ( s 2 + 9)
Since degree of denominator < numerator, we write the equation as follows:
Z (s ) =
=
=
3s (s 3 + 9s ) + ( 24s 2 + 48)
s (s + 9)
3s (s 3 + 9s )
2
s (s + 9)
3s (s 3 + 9s )
(s 3 + 9s )
24s 2 + 48
2
s (s + 9)
+
+
)
24s 2 + 48
s ( s + 9)
24s 2 + 48
s (s 2 + 9)
+
2 P2 s
2
s + 92
1
P0
L2 =
⇓
2 P2
w 22
L∞ = 3H
56
=
3× 9
56
H
27
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 552
(
2
( 24s 2 + 48)
s (s 2 + 9)
s=0
0 + 48
=
0+9
16
P0 =
3
and
Now,
+ 3s
w2
⇓
3
=
F
16
=
3s + 27s
( −) ( −)
Now, P0 = s
+ 3s
⇓
C0 =
4
24s 2 + 48
s (s 2 + 9)
Using the partial fractions, we get the following:
P0
s
s 3 + 9s
s 3 + 9s 3s 4 + 51s 2 + 48 3s
2
⇓
Now, degree of denominator > numerator
⇓
=
3s 4 + 51s 2 + 48
24s 2 + 48
= 3s +
or Z (s ) =
Z (s ) =
2
Value of 2p2
2 P2 s = ( s 2 + 9)
24 s 2 + 48
s( s 2 + 9)
or
2 P2 =
24 s 2 + 48
s2
s2 = − 9
24( −9) + 48
−9
168
=
9
56
2 P2 =
3
=
11/17/2014 6:14:36 PM
Network Synthesis and Realisability 553
and
1
3
=
F
2 P2 56
C2 =
The Foster form-I circuit, using the calculated data, has been shown in Figure 12.33.
For Foster form-II,
Value of 2P2
2 P2 s = ( s 2 + 1) ⋅
s ( s 2 + 9)
2
3( s + 1)( s + 16) s2 = −1
or
2P2 =
L2
3
F
16
2
56
H
27
3H
C0
s2 + 9
3( s 2 + 16) s2 = −1
L∞
C
3 2
F
56
Z(s)
−1 + 9
8
=
3( −1 + 16) 45
8
or 2P2 =
45
Foster form-I
=
Figure 12.33
s ( s 2 + 9)
Y (s ) =
3(s 2 + 1)(s 2 + 16)
⇓
Now, degree of denominator > numerator.
2P s
2P s
= 2 2 + 2 4
s + 12 s + 162
w2
w4
⇓
1
L2 =
2 P2
45
=
H
8
2P
C2 = 22
w2
8
=
F
45
Value of 2p4
⇓
1
2 P4
45
=
H
7
L4 =
and
C4 =
2 P4
2 P4 s = ( s 2 + 16)µ
or 2P4 =
s( s 2 + 9)
3( s 2 + 1)( s 2 + 16) s2 = −16
s2 + 9
3( s 2 + 1) s2 = −16
−16 + 9
−7
=
3( −16 + 1) −45
7
2 P4 =
45
=
w 42
7
7
=
=
45.16 720
The required Foster form-II circuit is shown in Figure 12.34
Example 12.22 Find Foster forms of the following:
Z ( s) = 2
s( s 2 + 4 )
( s 2 + 1)( s 2 + 25)
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 553
Y(s)
45 H
8
45 H
7
8 F
45
7 F
720
Figure 12.34
11/17/2014 6:14:37 PM
554 Network Analysis and Synthesis
Solution: This is a question for practice for the students. Please check your result with the
solution provided in Figure 12.35.
1
H
4
7
H
100
4F
4
F
7
Z(s)
Y(s)
8
H
25
8
H
63
63
F
32
1F
2
Figure 12.35
Example 12.23 Find Cauer form-I and Cauer form-II of the following:
Z ( s) =
s( s 2 + 3)( s 2 + 5)
( s 2 + 2)( s 2 + 4)
Solution: Cauer form-I can be given as follows:
Z ( s) =
=
)
(
s( s 2 + 3)( s 2 + 5)
( s 2 + 2)( s 2 + 4)
s5 + 8s3 + 15s
s4 + 6s2 + 8
s 2 + 6 s 2 + 8 s5 + 8s3 + 15s s ⇒ Z1
s5 + 6 s3 + 8s

s
2 s3 + 7 s s 4 + 6 s 2 + 8  ⇒ Y1

2
s4 +
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 554
7 2
s
2
5 2  2
4
s + 8 2 s + 7 s  s ⇒ Z 2

5
2
32
2s2 + s
5
3 5 2
 25
s s + 8  s ⇒ Y2
 6
2 2
5 2
s
2
11/17/2014 6:14:38 PM

s
2 s3 + 7 s s 4 + 6 s 2 + 8  ⇒ Y1

2
s4 +
7 2
s
2
5 2  2
4
s + 8 2 s + 7 s  s ⇒ Z 2

5
2
Network Synthesis and Realisability 555
32
2
2s + s
5
3 5 2
 25
s s + 8  s ⇒ Y2
 6
2 2
5 2
s
2
3  3
8 s  s ⇒ Z3
 5  40
3
s
5
x
The required Cauer form-I is shown in Figure 12.36.
For Cauer Form-II,
Z ( s) =
s5 + 8s3 + 15s
4
2
s + 6s + 8
For Cauer form-II, we arrange the terms of ‘s’ in ascending order.
=
1H
Z(s)
4
H
5
1
F
2
3 H
40
25 F
6
Figure 12.36
15s + 8s3 + s5
8 + 6s2 + s4

 15
8 + 6s 2 + s 4  15s + 8s 3 + s 5  s

8
45 3 15 5
s + s
8
4
13 3 7 5
− s − s
4
8
⇓
15s +
The coefficients should not be negative, and hence, we cannot design Cauer form-II circuit
using Z(s). Let us design the circuit using Y(s).
Y ( s) =
8 + 6s2 + s4
15s + 8s3 + s5
1

 8
15 + 8s 2 + s 2 8 + 6 s 2 + s 4 
⇒

 15s
Y
8
64
8 + s2 + s4
15
15
26 2 7 4 
225
1
 15
s + s  15s + 8s3 + s5 
× 15s =
⇒
2
 26 s
15
15 
26 s
Z
105 3
15s +
s
26
103 3 5  26 2 7 4  26 26 s 2
676
1
s +s 
s + s 
×
=
⇒
3
 15
26
15
15
1545
s
Y
 10 s
26 2 676 4
s +
s
15
1545
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 555
11/17/2014 6:14:39 PM
15 + 8s + s  8 + 6 s + s 
⇒

 15s
Y
8
64
8 + s2 + s4
15
15
26 2 7 4 
225
1
 15
s + s  15s + 8s3 + s5 
× 15s =
⇒
2
 26 s
15
15 
26 s
Z
556 Network Analysis and Synthesis
105 3
15s +
s
26
103 3 5  26 2 7 4  26 26 s 2
676
1
⇒
s +s 
s + s 
×
=
3
 15
s
Y
26
15
15
1545
 10 s
26 2 676 4
s +
s
15
1545
75 4  103 3 5  103 103s3 10609 1
s 
s +s  4 ×
=
=
23
78s
Z
1545  26
 3s
or
3
103 3
s+
s
103
26
1
 3 4 3
⇒
s5
s
 103  103s
Y
3 4
s
103
x
In Cauer form-II, Y corresponds to L and Z corresponds to C.
Accordingly, the circuit is drawn as shown in
Figure 12.37.
26
F
225
15
H
8
Example 12.24 Find two faster realisations of the
given function
Z ( s) =
2 s3
78
F
10609
1545
H
676
103
H
3
Figure 12.37
s2 + 1
Solution: Foster form-I circuit is designed using Z(s)
Z ( s) =
2 s( s 2 + 4 )
; clearly, the given impedance is of LC type.
s2 + 1
Using the partial fractions, we get the following equation:
Z ( s) =
=
2 s( s 2 + 4 )
s2 + 1
2
( s + 1)2 s + 6 s
2
= 2s +
=
5 +1
6s
s2 + 1
6s
2
s +1
2s
s 2 + 1 2 s3 + 8s
)
2 s3 + 2 s
6s
+ 2s
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 556
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Network Synthesis and Realisability 557
Comparing the equation with the general form, we can write it as Z ( s) =
We get the following:
P0
2P s
+ 2 2 2 + … + P∞ s
s s +w 2
2P2 = 6
w22 = 1
P∞ = 2
L2 =
Foster form-I circuit is shown in Figure 12.38.
Foster form-II circuit is designed using function Y(s) as
follows:
Y ( s) =
=
1
s2 + 1
= 3
Z ( s) 2 s + 8s
2H
Z(s )
s2 + 1
2 s( s 2 + 4 )
P0 = s ⋅ =
s2 + 1
2
2 s( s + 4 ) s
2 P2 s = ( s 2 + 4) ⋅ =
or
2 P2 =
s2 + 1
2s
2
C2 = 1 = 1 F
2P2 6
Figure 12.38
P
2P s
= 0+ 2 2 2
s s +w 2
Now,
2P2 6
= = 6H
w 22 1
=
=0
s2 + 1
2
2( s + 4) s
s2 + 1
2 s( s 2 + 4 ) s = − 4
=
s2 = − 4
=
=
=0
s2 + 1
2s
0 +1
= 1/8
2(0 + 4)
s2 = − 4
−4 + 1 −3
=
= 3/8
2( −4) −8
Therefore, Foster form-II will be as shown in Figure 12.39.
Y(s )
L0
1
= 8H
P0
L2
8
1
=
H
2P2
3
C2
2P2 3/8
3
=
=
F
4
32
w 22
Figure 12.39
Example 12.25 Find the two Foster realisations of the following function:
Z (s ) =
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 557
s 3 + 4s
2s 4 + 20s 2 + 18
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558 Network Analysis and Synthesis
Solution: Foster form-I can be obtained as follows:
Z ( s) =
=
s3 + 4 s
2( s 4 + 10 s 2 + 9)
2 P2 s
2
s + 9
↓
w
Now,
+
=
or
2 P2 =
or
2 P4 =
(12.28)
2
4
s3 + 4 s
2( s 2 + 9)( s 2 + 1)
s2 = − 9
s3 + 4 s
2( s 2 + 1)
s( s 2 + 4 )
s2 = − 9
2( s 2 + 1)
s2 = − 9
s2 + 4
s2 = − 9
2
2 P4 s = ( s 2 + 1) ⋅
=
2( s 2 + 9) + ( s 2 + 1)
s2 + 1
↓
w
2( s + 1)
5
=
16
and
s3 + 4 s
2 P4 s
2
2
2 P2 s = ( s 2 + 9) ⋅
=
=
=
−9 + 4
−5
=
2( −9 + 1) 2( −8)
( s3 + 4 s)
2( s 2 + 9)( s 2 + 1)
( s3 + 4 s)
2( s 2 + 9) s2 = −1
s2 + 4
2
2( s + 9) s2 = −1
=
=
s 2 = −1
s( s 2 + 4 )
2( s 2 + 9) s2 = −1
−1 + 4
3
=
2( −1 + 9) 16
The required Foster form-I circuit is shown in Figure 12.40.
Z(s )
2P2 5/16
5
=
=
H
9
144
w 22
2P4 3/16
3
=
=
H
1
16
w 42
1
16
=
F
2P2
5
16
1
=
F
2P4
3
Figure 12.40
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 558
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Network Synthesis and Realisability 559
Foster form-II circuit is designed using Y(s).
Y ( s) =
=
2( s 2 + 9)( s 2 + 1)
s3 + 4 s
2s
s3 + 4 s 2 s 4 + 20 s 2 + 18
)
2 s( s3 + 4 s) + (12 s 2 + 18)
= 2s +
( s3 + 4 s)
( −) 2 s 4 + 8s 2
12 s 2 + 18
12 s 2 + 18
s3 + 4 s
Y (s ) =
or
Y ( s) =
12s 2 + 18
s (s 2 + 4 )
+ 2s
P0
2P s
+ 2 2 + 2s
s s + 4
↓
↓
w 22 H
↓
representing
C∞ = 2 F
P0 = s ⋅
Now,
12s 2 + 18
2
s (s + 4 )
2P2 s = (s 2 + 4) ⋅
=
2P2 s =
or
12s 2 + 18
2
s +4
s=0
=
s=0
18 9
=
4 2
(12s 2 + 18)
s (s 2 + 4 )
s2 = −4
(12s 2 + 18)
s
s2 = −4
(12s 2 + 18)
2
s (s + 4 )
=
s2 = −4
12( −4) + 18 −48 + 18
=
−4
−4
−30 15
=
−4
2
Therefore, the required Foster form-II circuit is shown in Figure 12.41.
=
Y(s)
1
2
=
H
P0 9
L2
L0
C2
2
1
=
H
2P2 15
2P2
15
=
F
8
w 22
2F
C∞
Figure 12.41
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 559
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560 Network Analysis and Synthesis
Example 12.26 Synthesise first and second Foster form of L–C network for the impedance
Z ( s) =
( s 2 + 1)( s 2 + 16)
s( s 2 + 4 )
Solution: Foster form-I is calculated as follows:
Z ( s) =
=
( s 2 + 1)( s 2 + 16)
s( s 2 + 4 )
s( s3 + 4 s) + (13s 2 + 16)
s( s 2 + 4 )
= s+
=
13s 2 + 16
s( s 2 + 4 )
13s 2 + 16
s( s 2 + 4 )
+s
s
s3 + 4 s s 4 + 17 s 2 + 16
)
↓
represents
L∞ = 1H
=
13s 2 + 16
2
s( s + 4 )
2 P2 s = ( s 2 + 4) ⋅
1
1
=
F
P0 4
C0
Z(s)
C2
1
1
=
F
2P2
9
=
s =0
13s 2 + 16
2
s +4
=
16
=4
4
=
−52 + 16 −36
=
=9
−4
−4
s =0
2
13s + 16
13s 2 + 16
s
s( s 2 + 4 )
s2 = − 4
s2 = − 4
or
2P2
9
=
H
4
w 22
L2
13s 2 + 16
P0
2 P2 s
+
s s 2 + w 22
P0 = s ⋅
=
( −) s 4 + 4 s 2
2 P2 =
1H
L∞
13s 2 + 16
s
s2 = − 4
Foster form-I circuit will be as shown in Figure 12.42.
Foster form-II circuit is designed using Y(s).
Figure 12.42
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 560
Y ( s) =
s( s 2 + 4 )
1
= 2
Z ( s) ( s + 1)( s 2 + 16)
11/17/2014 6:14:46 PM
Network Synthesis and Realisability 561
=
2 P2 s
s2 + 1
2 P4 s
+
s 2 + 16
2 P2 s = ( s 2 + 1) ⋅
Now,
=
or
2 P2 =
s( s 2 + 4 )
( s 2 + 1)( s 2 + 16) s2 = −1
s( s 2 + 4 )
( s 2 + 16)
s 2 = −1
s2 + 4
=
2
s + 16
s 2 = −1
2 P4 s = ( s 2 + 16) ⋅
and
2 P4 =
or
s( s 2 + 4 )
( s 2 + 1)( s 2 + 16) s2 = −16
s2 + 4
2
s +1
−1 + 4
3
=
−1 + 16 15
=
s 2 = −16
−16 + 4 −12 4
=
=
−16 + 1 −15 5
Therefore, required Foster Form-II circuit will be as shown in Figure 12.43
1
15
=
H L4
2P2
3
L2
Y(s)
2P2 3/15
=
1 C
w 22
4
3
=
F
15
C2
5
1
=
H
2P4
4
2P4 4/5
1
=
F
=
20
16
w 24
Figure 12.43
Z ( s) =
Example 12.27 Realise
s4 + 7s2 + 9
s( s 2 + 4 )
in the form of Cauer L−C network
Solution: Cauer form-I can be calculated as follows:
Z (s ) =
s 4 + 7s 2 + 9
s 3 + 4s
(
)
s3 + 4 s) s 4 + 7 s 2 + 9 s ⇒ L1 = 1H
s4 + 4s2
1

s
3s 2 + 9  s3 + 4 s  → C1 = F

3
3
s 3 + 3s
)
(
s 3s 2 + 9 3s → L2 = 3H
3s
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 561
2
1
 s
9  s  → C2 = F
 9
9
11/17/2014 6:14:48 PM
(
)
s3 + 4 s) s 4 + 7 s 2 + 9 s ⇒ L1 = 1H
s4 + 4s2


1
s
→ C1 = F
3
3
562 Network Analysis and Synthesis
3s 2 + 9  s3 + 4 s 
s 3 + 3s
)
(
s 3s 2 + 9 3s → L2 = 3H
3s
2
1
 s
9  s  → C2 = F
 9
9
s
x
Therefore, the required Cauer form-I circuit is shown in
Figure 12.44.
For Cauer form-II
Z ( s) =
1H
3H
L1
L2
C1
Z(s)
9 + 7s2 + s4
4 s + s3
1 C
F 2
3
1
F
9
Figure 12.44
)
4
 9
4 s + s3 9 + 7 s 2 + s 4  ⇒ C1 = F
 4s
9
9+
9 2
s
4
19 2 4 
s + s  4 s + s3

4
4s +
19
 4
× 4 s ⇒ L1 = H

2
16
19 s
16 3
s
19
3 3  19 2 4  19 19 s 2 381
12
F
s  s +s  3 ×
=
⇒ C2 =
1
19  4
4
12
s
38
 35
19 2
s
4
19
 3  3
s 4  s3 
⇒ L2 = H
 19  19 s
3
3 3
s
19
x
4
F
9
Therefore, Cauer form-II circuit will be as shown in
Figure 12.45.
Example 12.28 Find two Cauer realisations of the following
function:
Z ( s) =
C1
Z(s)
L1
12
F
381
C2
19
H L2
16
19
H
3
10 s 4 + 12 s 2 + 1
2 s( s 2 + 1)
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 562
Figure 12.45
11/17/2014 6:14:49 PM
Network Synthesis and Realisability 563
Solution: Cauer form-I can be obtained as follows:
(
)
2 s3 + 2 s 10 s 4 + 12 s 2 + 1 5s ⇒ L1 = 54
10 s 4 + 10 s 2
)
(
2 s 2 + 1 2 s3 + 2 s s ⇒ C1 = 1 F
2 s3 + s
)
(
s 2 s 2 + 1 2 s ⇒ L2 = 2H
2s
2
)(
1 s s ⇒ C2 = 1 F
s
x
Therefore, the required Cauer form-I circuit will be as shown in
Figure 12.46.
Z(s)
Cauer form-II can be obtained as follows:
Z ( s) =
1 + 12 s 2 + 10 s 4
5H
2H
L1
L2
C1
1F C2
1F
Figure 12.46
2 s + 2 s3
)
 1
2 s + 2 s3 1 + 12 s 2 + 10 s 4  ⇒ C1 = 2F
 2s
1 + s2
2
11

 2s
11s 2 + 10 s 4  2 s + 2 s3  2 =
⇒ L1 = H

 11s
11s
2
2s +
20 s3
11
2 3
121
2
 11
s  11s 2 + 10 s 4  3 × 11s 2 =
⇒ C2 =
F
 2s
11 
2s
121
11s 2
3
2 s3
1
 2s  1
10 s 4 
×
=
⇒ L2 = 55H

 11  10 s 4 11 55s
2 3
s
11
x
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 563
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564 Network Analysis and Synthesis
Therefore, required Cauer form-II is shown in Figure 12.47.
2
F
121
2F
Example 12.29 Find the Foster form-I and the Cauer form-II
of the function represented as
C1
Z(s )
2(s + 1)(s + 3)
Z (s ) =
(s + 2)(s + 6)
L1
C2
11
H L2
2
55 H
Figure 12.47
Solution:
Given Z (s) =
Z(0) =
2( s + 1)( s + 3)
( s + 2)( s + 6)
6 1
=
12 2
Z(0) = 2; now, Z(∞) > Z(0), and therefore, the given impedance is of R–L type.
Foster form-I can be calculated as follows:
Z ( s) =
2( s + 1)( s + 3)
( s + 2)( s + 6)
= 2+
A
B
+
(Partial fractions)(12.29)
s+2 s+6
Let us find values of A and B.
2(s + 1)(s + 3) 2(s + 2)(s + 6) + A (s + 6) + B (s + 2)
(12.30)
=
(s + 2)(s + 6)
(s + 2)(s + 6)
Substituting s = −2 in equation (12.30), we get the following:
2(−1)(1) = 0 + A(4) + 0
1
A = − . Since the coefficient should not be negative, we need to use the partial fractions of
2
Z ( s) 2( s + 1)( s + 3)
=
s
s( s + 2)( s + 6)
=
and
P0
P
P
+ 2 + 4 (12.31)
s s+2 s+6
P0 = s
2( s + 1)( s + 3)
s( s + 2)( s + 6) s
=
=0
2( s + 1)( s + 3)
( s + 2)( s + 6) s
=0
2(1)(3) 1
=
=
( 2)(6) 2
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Network Synthesis and Realisability 565
P2 = ( s + 2) ⋅
2( s + 1)( s + 3)
s( s + 2)( s + 6) s = − 2
=
2( s + 1)( s + 3)
s( s + 6 ) s = − 2
=
2( −1)(1) 1
=
−2( 4)
4
P4 = ( s + 6) ⋅
2( s + 1)( s + 3)
s( s + 2)( s + 6) s = − 6
=
2( s + 1)( s + 3)
s( s + 2) s = − 6
=
2( −5)( −3) 5
=
−6( −4)
4
Substituting the values of P0, P2 and P4 in equation (12.31), we get the following equation:
1
1
s
s
Z ( s) 1 4
= +
+ 4
s
2 s+2 s+6
5
1
s
s
1 4
Z ( s) = +
+ 4
2 s+2 s+6
Ps
Ps
= P0 + 1 + 4
s +s 2 s +s 4
or
Therefore, the required Foster form-I circuit is shown in Figure 12.48.
P2 =
P0 =
1
Ω
2
P4 =
5
Ω
4
R2
R4
L2
L4
P2 1/4 1
σ2 = 2 = 8 H
P4 5/4 5
σ4 = 6 = 24 H
R0
Z (s)
1
Ω
4
Figure 12.48
For Cauer form-II, we calculate Z(s) can be calculated as follows:
Z ( s) =
2 s 2 + 8s + 6
s 2 + 8s + 12
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 565
=
6 + 8s + 2 s 2
12 + 8s + s 2
11/17/2014 6:14:53 PM
566 Network Analysis and Synthesis
)
1
12 + 8s + s 2 6 + 8s + 2 s 2 
2
6 + 4s +
4s +
s2
2
3s 2 
3
12 + 8s + s 2 
s
2 
12 +
95
2
7s 2 
3s 2  2
8
+ s  4s +
× 4s =


11
2  7s
7
4s +
8s 2
7
5 2  7 s 2  14 7 s 49
s  +s  2 ×
=
 5s
14  2
2 5s
7s
2
 5  5
s 2 s 2 
 14  14
5 2
s
14
x
1
Ω
2
Therefore, the required Cauer form-II circuit is shown
in Figure 12.49.
Example 12.30 Find the circuit in second Cauer form
of the following function
Z ( s) =
R1
Z (s)
s2 + 4s + 3
5/14 Ω
8/7Ω
L1
1
H
3
R2
L2
R3
5
H
49
Figure 12.49
s 2 + 8s + 12
Solution: Given network function is as follows:
Z ( s) =
Now,
Z(0) =
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 566
s2 + 4s + 3
s 2 + 8s + 12
3 1
0+0+3
=
=
0 + 0 + 12 12 4
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Network Synthesis and Realisability 567
Further, Z(∞) can be calculated as follows:
 4 3
4 3
s 2 1 + + 2  1 + + 2
 s s 
s 5
=
Z (s ) =
8 12
8 12 
2
s 1 + + 2  1 + + 2
 s s 
s s
1+ 0 + 0
=1
1+ 0 + 0
Z(∞) > Z(0)
Therefore, the given impedance is of R–L type.
Now, Cauer form-II can be obtained as in the following:
3 + 4s + s2
Z ( s) =
12 + 8s + s 2
1
12 + 8s + s3 3 + 4 s + s 2 
4
s2
3 + 2s +
4
3 2
6
2 s + s  12 + 8s + s 2 
s
4 
9
12 + s
2
7s 2 
3  2
4
+ s  2s + s2  × 2s =

2
4  7s
7
4s2
2s +
7
5s 2  7 s 2  28 7 s 98
+s  2 ×
=
 5s
28  2
2 5s
7s
2
 5 2 5
s 2
s
 28  28
5 2
s
28
x
Z(∞) =
)
The Cauer form-II circuit is shown in Figure 12.50.
1
Ω
4
4
Ω
7
Example 12.31 For a given function
R1
R2
1
L2
H
6
Z(s)
Z ( s) =
( s + 1)( s + 3)( s + 5)
s( s + 2)( s + 4)( s + 6)
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 567
L1
5
Ω
28
R3
5
H
98
Figure 12.50
11/17/2014 6:14:55 PM
568 Network Analysis and Synthesis
Find the circuit in the following:
(a) Foster form-I and II and
(b) Cauer form-I
Solution: Firstly, let us check whether the given impedance function is of R–C or of R–L type.
Z(0) =
(1)(3)(5)
=∞
0( 2)( 4)(6)
Z(∞) = 0
In this case, Z(0) > Z(∞)
Therefore, the given impedance is of R–C type.
Foster form-I can be obtained as in the following:
( s + 1)( s + 3)( s + 5)
s( s + 2)( s + 4)( s + 6)
P
P
P
P
= 0 + 2 + 4 + 6 (12.32)
s s+ 2 s+ 4 s+ 6
↓
↓
↓
s2
s4
s6
Z ( s) =
Now,
P0 = s ⋅
(s + 1)(s + 3)(s + 5)
(1)(3)(5)
5
=
=
s (s + 2)(s + 4)(s + 6) s = 0 ( 2)( 4)(6) 16
P2 = ( s + 2) ⋅
=
( s + 1)( s + 3)( s + 5)
( s + 1)( s + 3)( s + 5)
=
s( s + 2)( s + 4)( s + 6) s = − 2
s( s + 4)( s + 6) s = 2
( −1)(1)(3) 3
=
−2( 2)( 4) 16
P4 = (s + 4) ⋅
(s + 1)(s + 3)(s + 5)
( −3)( −1)(1) 3
=
=
s (s + 2)(s + 4)(s + 6) s = − 4 −4( −2)( 2) 16
P6 = ( s + 6) ⋅
( s + 1)( s + 3)( s + 5)
( s + 1)( s + 3)( s + 5)
=
s( s + 2)( s + 4) s = − 6
s( s + 2)( s + 4)( s + 6) s = − 6
=
( −5)( −3)( −1) 5
=
−6( −4)( −2) 16
Therefore, the required Foster form-I circuit is drawn as in Figure 12.51.
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Network Synthesis and Realisability 569
1
P0
=
16
F
5
C0
Z(s)
P2
3
Ω
=
σ2
32
P4
3
Ω
=
σ4
64
P6
5
Ω
=
σ6
96
R2
R4
R6
C2
C4
1
16
=
F
P2
3
C6
16
1
=
F
P4
3
16
1
=
F
P6
5
Figure 12.51
Foster form-II can be calculated as follows:
Y ( s) =
or
s( s + 2)( s + 4)( s + 6)
s
( s + 1)( s + 3)( s + 5)
Y ( s) ( s + 2)( s + 4)( s + 6)
=
s
( s + 1)( s + 3)( s + 5)
= 1+
P
P2
P
+ 4 + 6 (12.33)
s +1 s + 3 s + 5
Let us find values of P2, P4 and P6
( s + 2)( s + 4)( s + 6) ( s + 1)( s + 3)( s + 5) + p2 ( s + 3)( s + 5) + p4 ( s + 1)( s + 5) + p6 ( s + 1)( s + 3)
=
( s + 1)( s + 3)( s + 5)
( s + 1)( s + 3)( s + 5)
or
(s + 2) (s + 4) (s + 6) = (s + 1) (s + 3) (s + 5) + P2(s + 3) (s + 5)
+ P4(s + 1) (s + 5) + P6(s + 1) (s + 3)
(12.34)
Substituting s = −1 in equation (12.34), we get the following:
(1)(3)(5) = 0 + P2 ( 2)( 4) + 0 + 0
15 = 8 P2 or
P2 =
15
8
Substituting s = −3 in equation (12.34), P4 can be calculated as follows:
( −1)(1)(3) = 0 + 0 + P4 ( −2)( 2) + 0
3
4
Substituting s = −5 in equation (12.34), we get the following:
or
P4 =
( −3)( −1)(1) = 0 + 0 + 0 + P6 ( −4)( −2)
or
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 569
P6 =
3
8
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570 Network Analysis and Synthesis
Substituting the values of P2, P4 and P6 in equation P, the following equation can be obtained:
3
15
3
8
8
4
Z ( s) = 1 +
+
+
s +1 s + 3 s + 5
3 P6
3 P4
s
s
Z ( s) = 8 + 4
+8
+ 1P∞
s+ 1 s+ 3 s+ 5
P2 15
or
s
s2
s4
s6
The required Foster form-II circuit is shown in Figure 12.52.
R2
Y(s)
1 = 8 ΩR4
P2 15
1 = 4 ΩR6
3
P4
1 = 8 Ω
3
P6
P2 15
σ2 = 8 FC
4
P4 1
σ4 = 4 FC
6
P6 3
σ6 = 40 F
C2
C∞
P∞ = 1F
Figure 12.52
Cauer form-I can be calculated as follows:
Z ( s) =
=
( s 2 + 4 s + 3)( s + 5)
s( s 2 + 6 s + 8)( s + 6)
=
s3 + 4 s 2 + 3s + 5s 2 + 20 s + 15
s( s3 + 6 s 2 + 8s + 6 s 2 + 36 s + 48
s3 + 9 s 2 + 23s + 15
s 4 + 12 s3 + 44 s 2 + 48s
)
1
s 4 + 12 s3 + 44 s 2 + 48s s3 + 9 s 2 + 23s + 15 
s
s3 + 12 s 2 + 44 s + 48
− 3s 2 − 21s − 33
⇓
Since none of the coefficents should be negative,
we will consider Y(s) as follows:
)
s 3 + 9s 2 + 23s + 15)s s 4 + 12s 3 + 44s 2 + 48s (s
4
s + 9s 3 + 23s 2 + 15s

1
3s 3 + 21s 2 + 33 s  s 3 + 9s 2 + 23s + 15 
3

s 3 + 7s 2 + 11s

3
2s 2 + 12s + 15 3s 3 + 21s 2 + 33s  s

2
21s  3s
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 570
2
3
2
45
11/17/2014 6:14:57 PM
)
s 3 + 9s 2 + 23s + 15)s s 4 + 12s 3 + 44s 2 + 48s (s
s 4 + 9s 3 + 23s 2 + 15s
Network Synthesis and Realisability 571

1
3s 3 + 21s 2 + 33 s  s 3 + 9s 2 + 23s + 15 
3

s 3 + 7s 2 + 11s
)

3
2s 2 + 12s + 15 3s 3 + 21s 2 + 33s  s

2
5s + 15 3s 2 +
21s  3s

2 5
3s 3 + 18s 2 +
3s 2 +
3s 2 + 9s
45
s
2
2
21s 
2
 2s + 12s + 15 
3
2 
2s 2 + 7s
5s + 15
Continued to LHS.
3 
 10
s  5s + 15 
 3
2 
5s
 3  5
15 s 
 2  10
3
s
2
x
Therefore, Cauer form-I circuit is shown in Figure
12.53.
1
Ω
3
Example 12.32 Find second Cauer form of the
network whose function is
Z ( s) =
Y(s)
( s + 3)( s + 6)
( s + 1)( s + 5)
1F
3
F
2
10
Ω
3
3
F
5
1
F
10
Figure 12.53
Solution: Given
Z (s ) =
2
Ω
3
(s + 3)(s + 6)
(s + 1)(s + 6)
Z(0) =
Now,
(3)(6)
=3
(1)(6)
Z(∞) = 1
and
Since Z(0) > Z(∞), the given function is of R–C type.
For finding the circuit in Cauer form-II, the process is as follows
Z ( s) =
s 2 + 9 s + 18
2
s + 6s + 5
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 571
=
18 + 9 s + s 2
5 + 6s + s2
11/17/2014 6:14:59 PM
572 Network Analysis and Synthesis
)
5 + 6 s + s 2 18 + 9 s + s 2
 18

5
108s 18s
+
5
5
−63s
5
18 +
Since, the coefficient is negative, we will now consider Y(s) as
Y ( s) =
5 + 6s + s2
18 + 9 s + s 2
and proceed
)
5
18 + 9 s + s 2 3 + 4 s + s 2 
 18
5
5 2
5+ s+ s
2
18
7
13 2 
36
 7
s+
s  18 + 9 s + s 2  × 18 =
 75

2
18
7s
26
18 +
s
7
7
7 s 49
37
13  7
×
=
s + s2  s + s2 
2
2
18  37 s 2 74
7s
49 2
s+
s
2
74
40 2  37
 24642
s 
s + s2 
 280
 7
666
37 s
7
 40 2  40
s 2
s
 666  666
40 2
s
666
x
Therefore, required Cauer form-II circuit is
shown in Figure 12.54.
7
F
36
Example 12.33 Find two Foster realisation
of the network function
Y(s)
Z ( s) =
( s + 3)( s + 5)
s( s + 4 )
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 572
18
Ω
5
24642
F
280
74
Ω
49
40
Ω
666
Figure 12.54
11/17/2014 6:15:00 PM
Network Synthesis and Realisability 573
Solution:
Now,
Z ( 0) = ∞
Z (∞) = 0,
and therefore, Z(0) > Z(∞)
The given impedance is the R–C type.
Foster form-I can be calculated as follows:
( s + 3)( s + 5)
s( s + 4) (12.35)
P
P
= 1+ 0 + 2
s s+4
Z ( s) =
Let us find P0 and P2
( s + 3)( s + 5) s( s + 4) + P0 ( s + 4) + sP2
=
s( s + 4 )
s( s + 4 )
(s + 3) (s + 5) = s(s + 4) + P0(s + 4) + sP2
or
(12.36)
To find P0, substitute s = 0 in equation (12.36)
(3) (5) = 0 + P0 (4) + 0
P0 = 15/ 4
or
To find P2, substitute s = −4 in equation (12.36)
(−1)(1) = 0 + 0 − 4P2
or
P0 = 1/ 4
Substituting the values of P0 and P2 in equation (12.35), we get the following:
15/ 4 1/ 4
15/ 4 1/ 4Z (s ) = 1 +
+
+
s
s+4
s
s+4
P0
15/ 4
1/ 4 P2
15/ 4 P0 1/ 4 P2
=
+
+ 1H
=
+
+ 1H s
s+ 4
s
s+ 4
s2
Z (s ) = 1 +
s2
P2
1
Ω
=
σ2
16
Therefore, required Foster form-I circuit is shown in
Figure 12.55
Foster form-II can be calculated as follows:
Y ( s) =
=
s( s + 4 )
( s + 3)( s + 5)
P2
P
+ 4 (12.37)
s+3 s+5
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 573
4
1
F
=
P0 15
Z(s)
H
1Ω
1
= 4F
P2
Figure 12.55
11/17/2014 6:15:01 PM
574 Network Analysis and Synthesis
Let us find P2 and P4.
P ( s + 5) + P4 ( s + 3)
s( s + 4 )
= 2
( s + 3)( s + 5)
( s + 3)( s + 5)
s(s + 4) = P2(s + 5) + P4(s + 3)
or
(12.38)
Substituting s = -3 in equation (12.38), we get the following:
−3(1) = P2 (2) + 0
−3
or
P2 =
2
Since, the coefficient cannot be negative, and therefore, we will do the partial fraction of the
following:
Y ( s)
s+4
=
s
( s + 3)( s + 5)
=
P2
P
+ 4 (12.39)
s+3 s+5
P2 = ( s + 3) ⋅
Now,
=
s+4
s+5
=
s = −3
P4 = ( s + 5) ⋅
and
=
( s + 4)
( s + 3)( s + 5) s = − 3
1
2
( s + 4)
( s + 3)( s + 5) s = − 5
( s + 4)
−1 1
=
=
( s + 3) s = − 5 −2 2
Now, substituting the values of P2 and P4 in equation (12.39), we get the following form:
Y ( s) 1/ 2
1/ 2
=
+
s
s+3 s+5
or
Y (s ) =
1/ 2 P2 s 1/ 2 P4 s
+
s+ 3
s+ 5
s2
s4
Therefore, Foster form-II circuit is shown in Figure
12.56.
Example 12.34 Find Cauer form-I and Cauer
form-II for the network function
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 574
Y(s)
1
= 2Ω
P2
1
= 2Ω
P4
P2
1
F
=
σ2
6
P4
1
F
=
σ4
10
Figure 12.56
11/17/2014 6:15:03 PM
Network Synthesis and Realisability 575
Z ( s) =
s 2 + 5s + 4
s2 + 2s
Solution: Given network function is as follows:
Z ( s) =
s 2 + 5s + 4
s2 + 2s
Z (0) = ∞; Z (∞) = 1
Here,
That is, Z(0) > Z(∞), and therefore, given network function is of R–C type
For Cauer form-I, the procedure is followed as follows:
Z ( s) =
s 2 + 5s + 4
s2 + 2s
(
)
s 2 + 2 s s 2 + 5s + 4 1 ⇒ R1 = 1Ω
s2 + 2s

1
s
3s + 4 s 2 + 2 s  ⇒ C1 = F
3

3
s2 +
4
s
3
2 
9
9
s 3s + 4  ⇒ R2 = Ω
2
3 
2
3s
2 s
1
4 s  ⇒ C 2 = F
 3 6
6
2
s
3
x
Therefore, required Cauer form-I circuit is shown in
Figure 12.57.
For Cauer form-II, the calculation are as follows:
1Ω
Z(s)
Z ( s) =
9Ω
2
1
F
3
1
F
6
4 + 5s + s 2
2s + s2
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 575
Figure 12.57
11/17/2014 6:15:04 PM
576 Network Analysis and Synthesis
)
1
2
2 s + s 2 4 + 5s + s 2  ⇒ C1 = F
s
2
4 + 2s

3
2
3s + s 2  2 s + s 2  ⇒ R1 = Ω


3
2
2s +
2 2
s
3
1
1 2
9
s  3s + s 2  ⇒ C 2 = F


9
s
3
3s
1
F
2
Z(s)
 1 1
s 2  s 2  ⇒ R2 = 3Ω
 3 3
1
F
9
3
Ω
2
Figure 12.58
1 2
s
3
x
3Ω
Therefore, required Cauer form-II circuit is shown in Figure 12.58.
R E V IE W Q U E S T I O N S
Short Answer Type
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
Differentiate network analysis and network synthesis.
What are Hurwitz conditions for stability?
What are the properties of a positive real Function?
Explain the process of synthesis of R–C network using Foster method.
Explain Cauer’s method to synthesising R–C network.
What are the properties of R–C admittance function?
What are the properties of R–L impedance function?
What are the properties of L–C immittance?
Explain how one-port R–L network can be synthesised using Foster method.
Explain how one-port R–L network can be synthesised using Cauer method.
Draw Foster forms of L–C network.
12. Draw Cauer forms of L–C network
13. Explain the synthesis of L–C network using Foster method.
14. What do you mean by a Hurwitz polynomial?
M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 576
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Network Synthesis and Realisability 577
Numerical Questions
s 2 + a s + a0
1. Determine the condition for which the function F ( s) = 2 1
is positive real. It is
s + b1s + b0
given that a0, b0, a1 and b1 are real and positive.
2. Express the impedance Z(s) for the network shown in Figure 12.59
12
H
5
1H
1
F
6
Z(s)
5
F
18
Figure 12.59
N ( s)
Z ( s) = K
. Plot its poles and zeros. From the pole-zero plot, what can you infer about
D ( s)
the stability of the system?

( s 2 + 9)( s 2 + 1) 

 Ans. Z ( s) =
s( s 2 + 4 ) 

Z ( s) is unstable.
3. Find voltage-transfer function for the network shown in Figure 12.60
2F
2F
I1
V1
2Ω
1F
1Ω
V2
Figure 12.60


V2 ( s)
8s 2
=
 Ans.

V1 ( s) 12 s 2 + 12 s + 1

4. Test whether the polynomial F ( s) = s 4 + s3 + 2 s 2 + 3s + 2 is Hurwitz
[Ans. Not Hurwitz]
5. Test whether the function
Ks
F ( s) =
is positive real, where a and k are positive constants
2
s +a
[Ans. Positive real]
6. Given the network function F ( s) =
s +1
, plot the zero and poles on s-plane
s + 2s + 5
N (s )
7. Find the driving point impedance Z (s ) = K ⋅
, for the network shown in Figure 12.61
D (s )
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578 Network Analysis and Synthesis
Z(s)
1
F
2
1
H
3
1Ω
Figure 12.61
Verify that Z(s) is positive real and that the polynomial D(s) + K.N(s) is Hurwitz
[Ans. Z ( s) =
s2 + 2s + 6
s 2 + 3s
, Z(s) is PRF, D(s) + K.N(s) is Hurwitz]
3( s + 2)( s + 4)
s( s + 3)
[Ans. as in Figure 12.62]
8. Design a one-port R–L network to realise the driving point function F ( s) =
1
H
9
1
Ω
3
Y(s)
1
H
72
1
Ω
27
Figure 12.62
9. Draw the pole-zero plot in s-plane for network function F ( s) =
4s
2
s + 2s + 2
P ( s)
10. Express the driving point admittance Y(s) in the form Y ( s) = K ⋅
for the network
Q
( s)
shown in Figure 12.63 and also verify that Y(s) is PRF.
2Ω
Y(s)
1Ω
3
3
F
2
Figure 12.63
[Ans. Y ( s) =
7s + 2
is a PRF]
2s + 4
2(s + 1)(s + 3)
. Determine the corresponding
(s + 2)(s + 6)
(b) R–L network;
11. Consider the system function Z (s ) =
(a) R–C network;
[Ans. as in Figure 12.64]
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Network Synthesis and Realisability 579
1Ω
2
Z(s)
1Ω
4
5Ω
4
1
F
8
5
F
24
8
Ω
7
1
Ω
2
5
Ω
14
1
H
3
Z(s)
(a)
5
H
49
(b)
Figure 12.64
12. Determine if the function F ( s) =
s3 + 5s 2 + 9 s + 3
s3 + 4 s 2 + 7 s + 9
is positive real
[Ans. F(s) is a PRF]
13. Realise the impedance Z ( s) =
2( s 2 + 1)( s 2 + 9)
s( s 2 + 4 )
15
H
8
[Ans. as in Figure 12.65]
2
F
9
2H
Z(s)
in three different ways
16 H
3
Y(s)
2
F
15
3 F
16
(a)
16 H
5
5
F
144
(b)
24
H
5
2H
1
F
12
Z(s)
5
F
36
(c)
Figure 12.65
14. List out the properties of LC immittance function and then realise the network having the
2 s5 + 12 s3 + 16 s
driving point impedance function Z ( s) =
by continued fraction method.
s4 + 4s2 + 3
[Ans. as in Figure 12.66]
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580 Network Analysis and Synthesis
8
H
3
2H
2
H
3
3
F
4
1
F
4
Figure 12.66
2( s + 1)( s + 3)
15. For the network function Y ( s) =
, synthesise in one Foster and one Cauer
( s + 2)( s + 4)
form
[Ans. as in Figure 12.67]
1
H
4
3Ω
4
3
H
16
1Ω
2
1
Ω
2
1
Ω
2
3Ω
4
Z(s)
1
Ω
6
1
H
4
1H
(b)
(a)
Figure 12.67
16. Consider the function F ( s) =
s 2 + 1.03
s 2 + 1.23
. Plot its poles and zeroes
17. Determine whether the function F ( s) =
s3 + 2 s 2 + 3s + 1
s3 + s 2 + 2 s + 1
is positive real or not.
[Ans. F(s) is a PRF]
18. Synthesise a one-port LC network in Cauer form-II whose driving point impedance is
Z ( s) =
6 s3 + 2 s
12 s 4 + 8s 2 + 1
[Ans. as in Figure 12.68]
5
F
2
2H
10 F
6
H
25
Figure 12.68
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Network Synthesis and Realisability 581
19. Given I ( s) =
4 s ( s + 2)
, draw its pole-zero plot. Further, find i(t) from it.
( s + 1)( s + 3)
[Ans. i(t) = −2e-t−6e-3t A]
20. Find the voltage transfer function for the network shown in Figure 12.69.
I1
V1
I2
3H
1
H
2
1
F
4
2F
V2
Figure 12.69
V2 ( s)
s2
= 4
]
V1 ( s) 6 s + 57 s 2 + 8
21. Determine the range of constant ‘K’ for the polynomial to be Hurwitz
P(s) = s3 + 3s2 + 2s + k
[Ans. 0 < k < 6]
22. Synthesise an R–C ladder and an R–L ladder network to realise the function
s2 + 4s + 3
F ( s) = 2
as an impedance or an admittance.
s + 8s + 12
[Ans. as in Figure 12.70]
[Ans.
1Ω
4
5Ω
8
1Ω
8
1
Ω
4
5
F
48
1
F
16
4
Ω
7
1
H
6
(a)
5
Ω
28
10
H
196
(b)
Figure 12.70
23. Calculate the driving point admittance of the network shown in Figure 12.71
5F
Z
4H
1
F
8
12 H
Figure 12.71

5s ( 2 s 2 + 1) 
 Ans. Y ( s) =

30 s 4 + 22 s 2 + 1

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582 Network Analysis and Synthesis
24. A transform voltage is given by
3s
. Make the pole-zero plot in the s-plane and obtain the time domain
( s + 1)( s + 4)
response.
[Ans. i(t) = −e−t + 4e−4t]
V ( s) =
25. Check if the given driving point impedance Z(s) represents a passive one-port network
Z ( s) =
s4 + s2 + 1
s3 + 2 s 2 − 2 s + 10
[Ans. No]
M U LTI P L E C HO I C E Q U E S T I ON S
1. A network function is said to have simple pole or simple zero if
(a)
(b)
(c)
(d)
The poles and zeros are on the real axis
The poles and zeros are repetitive
The poles and zeros are complex conjugate to each other
The poles and zeros are not repeated.
2. A function H(s ) =
(a)
(b)
(c)
(d)
2s
2
s +8
will have a zero at
s = ± j4
Anywhere on the s-plane
On the imaginary axis
On the origin
3. The network shown in Figure 12.72 has zeros at
R
Z(s)
C
L
Figure 12.72
(a) s = 0 and s = ∞; (b) s = 0 and s = −
4. Which of the following is a PRF
(a)
(b)
1
R
R
; (c) s = ∞ and s = − ; (d) s = ∞ and s = −
CR
L
L
( s + 1)( s + 2)
( s 2 + 1) 2
( s − 1)( s + 2)
s2 + 1
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Network Synthesis and Realisability 583
(c)
(d)
s4 + s2 + 1
( s + 1)( s + 2)( s + 3)
s −1
s2 − 1
5. A network function can completely be specified by:
(a)
(b)
(c)
(d)
Real parts of zeros
Poles and zeros
Real parts of poles
Poles, zeros and a scale factor
6. In the complex frequency s = σ + jw, while w has the unit of rad/s and σ has the unit of
(a) Hz
(b) Neper/s
(c) Rad/s
(d) Rad
7. Which of the following property relates to L–C impedance or admittance functions:
(a) The poles and zeros are simple and lie on the jw -axis
(b) There must be either a zero or a pole at origin and infinity.
(c) The highest (or lowest) powers of numerator or denominator differ by unity.
(d) All of the above.
8. If a network function has zeros only in the left-half of the s-plane, then it is said to be
(a)
(b)
(c)
(d)
A stable function
A non-minimum phase function
A minimum phase function
An all-pass function.
9. Zeroes in the right half of the s-plane are possible only for
(a)
(b)
(c)
(d)
d.p impedance function.
d.p admittance functions.
d.p impedance as well as admittance functions.
transfer functions.
10. An L–C impedance or admittance function:
(a)
(b)
(c)
(d)
has simple poles and zeros in the left half of the s-plane.
has no zero or pole at the origin or infinity.
is an odd rational function.
has all poles on the negative real axis of the s-plane.
11. The Laplace-transformed equivalent of a given network will have
(a)
5
8s
(b)
5s
8
(c)
8s
5
5
F capacitor replaced by
8
8
(d)
5
12. If a network function contains only poles whose real-parts are zero or negative, the network is
(a)
(b)
(c)
(d)
always stable.
stable, if the jw -axis poles are simple.
stable, if the jw -axis poles are at most of multiplicity 2.
always unstable.
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584 Network Analysis and Synthesis
13. Which of the following kind of network has the same admittance and impedance properties?
(a) L–C type
(b) R–L type
(c) R–C type
(d) R–L–C type
14. Both odd and even parts of a Hurwitz polynomial P(s) have roots
(a)
(b)
(c)
(d)
in the right-half of s-plane
in the left-half of s-plane
on the σ-axis only
on the jw -axis only
15. The impedance of a network is given as Z ( s) =
(a)
(b)
(c)
(d)
not a PRF
R–L network
R–C network
L–C network
s ( s + 2)
. The function is
( s + 3)( s + 4)
16. If F1(s) and F2(s) are positive real, then which of the following are positive real?
(a)
1
1
and
(b) F1 ( s) + F2 ( s)
F1 ( s)
F2 ( s)
(c)
F1 ( s) F2 ( s)
F1 ( s) + F2 ( s)
(d) All of these.
17. Which of these is not a positive real Function?
(a) F(s) = Ls
(b) F(s) = R
(c) F ( s) =
(d) F ( s) =
k
s
s +1
s2 + 2
18. A stable system must have
(a)
(b)
(c)
(d)
zero or negative real part for poles and zeros.
at least one pole or zero lying in the right-half s-plane
positive real part for any pole or zero
negative real part for all poles and zeros.
ANS W E RS
1. d
7. d
13. a
2. d
8. c
14. d
3. c
9. d
15. a
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4. c
10. a
16. d
5. d
11. d
17. d
6. b
12. b
18. a
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Filters and
Attenuators
13
Chapter objectives
After carefully studying this chapter, you should be able to do the following:
Explain the basic function of a filter
Solve problems on constant K-type
circuit.
filters.
Distinguish between a passive filter
Make comparison of parameters of
and an active filter.
constant K-type low-pass and highpass filters.
Classify passive filters and explain
function of each type of filter.
State the limitations of constant K-type
filters.
Explain the parameters of a filter.
Modify constant K-type filters to obtain
Draw and explain basic filter networks
m-derived filters.
in T and p sections.
Analyse all types of m-derived filters.
Carryout analysis of filter networks in
both T-section and p -section.
Develop composite filters using constant K-type and m-derived filters.
Make analysis of constant K-type or
proto-type filters.
13.1 INTRODUCTION
A filter blocks unwanted signals or noise signals and passes wanted or desired signals. A filter is
basically frequency selective network that allows signals of a particular band of frequencies and
rejects or attenuates signals of other frequencies. For example, a low-pass filter (LPF) passes
low frequency signals and attenuates all frequencies above a selected cut-off frequency.
A high-pass filter (HPF) network will pass only those input signals whose frequencies are
above the selected cut-off frequency.
A band-pass filter (BPF) passes signals of a selected frequency band, while a band-stop filter
(BSF) blocks frequencies within its band. The band of frequencies passed by a filter is called
pass band and the band of frequency that separates pass band and stop band of a filter is called
cut-off frequency. Figure 13.1 shows various types of filters, their input and output wave forms
and also their frequency response.
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586 Network Analysis and Synthesis
An LPF passes low frequency signals and blocks or attenuates signals that have frequencies above a given cut-off frequency ( fc). The input wave form for an LPF, as shown in Figure
13.1(a), is composed of a low frequency signal and high frequency unwanted signal. The filter
will allow the low frequency signal. This low frequency signal will appear at the output of the
filter. However, the high frequency unwanted signal (noise signal) will be stopped or drastically
reduced at the output.
Gain
Input signal
Output signal
Low-pass
filter
Vi
Vo
Passband
Vi
Vo
f
fc
Input signal
Output signal
Low-pass filter network
Frequency response
∝
Attenuation
Pass-band
fc
f
Figure 13.1(a) Low-Pass Filter
V 
The gain  o  versus frequency ( f ) response graph for the filter and the attenuation versus
 Vi 
frequency response have been shown in the Figure.
An HPF allows high frequency input signals and stops or attenuates the low frequency signals as shown in Figure 13.1(b).
Vi
High-pass
filter
Input signal
Vo
Vo
Vi
Output signal
fc
f
Frequency response
Figure 13.1(b) High-Pass Filter
Wave forms of BPF and BSF have been shown in Figure 13.1(c) and (d), respectively.
Band-pass
filter
Vo
Vi
f1
f2
f
Figure 13.1(c) Band-Pass Filter
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Vo
Vi
Band-stop
filter
f1
f2
f
Figure 13.1(d) Band-Stop Filter
13.1.1 Measurement in Decibels
The change in the output from a filter is measured in decibels (dB). The decibel is one-tenth of
a bel. Since bel is large, the unit of decibel is used. Let the output voltage of a filter unit changes
from v1 to v2 as the frequency changes. The change in power is expressed as the log of the ratio
of power that changes from p1 to p2. The change in power (DP) is expressed as follows:
P 
∆P = log  2  Bel
 P1 
P 
∆P = 10 log  2  dB (13.1)
 P1 
If power is assumed to be dissipated in the load resistance RL, then DP is given as in the following:
 v 2 ÷ RL 
∆P = 10 log  22
 dB
 v1 ÷ RL 
2
or
v 
∆P = 10 log  2  dB
v 
1
or
∆P = 20 log
V2
dB (13.2)
V1
Assuming power as i2 RL, DP can be expressed as follows:
i2
dB (13.3)
i1
The change in output power, voltage and current can be measured using the above three
­expressions respectively.
∆P = 20 log
13.2 TYPES OF FILTERS
Basically filters are of two types: Active filters and Passive filters.
Active filters are the filters having active elements like OP-AMP and transistor, in addition
to resistor and capacitor. An active filter not only passes or stops a particular band of frequency
but also amplifies the signal that passes through it.
Passive filters are made up of only passive components like inductor and capacitor. Such
­filters cannot amplify the signal that passes through them.
In this chapter, we will describe passive filters only.
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13.3 CLASSIFICATION OF PASSIVE FILTERS
On the basis of functions they perform, passive filters are classified as follows:
1. Low-Pass Filters (LPF)
2. High-Pass Filters (HPF)
3. Band-Pass Filters (BPF)
4. Band-Stop Filters (BSF)
13.3.1 Low-Pass Filters
These are the filters that pass all the frequencies lower than the selected cut-off frequency fc
and attenuate/stop/suppress the signals whose
frequency is greater than fc. Attenuation characteristic for an ideal LPF is shown in Figure
13.2. From the attenuation characteristic of the
LPF, it is clear that attenuation is zero in pass
band and attenuation of signal is maximum in
stop band.
It is to be noted that the characteristic shown
is for an ideal LPF. For the low-pass filter the
pass-band is from 0 to fc and stop-band is from
fc to ∞.
Attenuation
Pass band
0
Stop band
fc
Cut off frequency
Frequency
Figure 13.2 Attenuation Characteristic
of an LPF
13.3.2 High-Pass Filters
It is a filter that passes all the signals whose frequency is higher than the cut-off frequency and
stops the signals whose frequency is less than fc.
The attenuation characteristics of HPF is shown
in Figure 13.3.
Therefore, for a high-pass filter (HPF) the pass
band is from fc to ∞ and the stop-band is from 0 to fc.
Attenuation
Stop band
0
Pass band
fc
Frequency
Figure 13.3 Attenuation
Characteristic of an HPF
13.3.3 Band-Pass Filters
It is a filter that passes a particular band of frequencies and stops all other frequencies.
It has two cut-off frequencies: lower cut-off frequency ( f1) and higher cut-off frequency ( f2).
This filter passes all those signals whose frequency lies inside the band f1 to f2 and stops all other
frequency signals. The attenuation characteristics for BPF is shown in Figure 13.4. Therefore,
for a BPF, the following are given:
Pass band: f1 to f2
Stop band: 0 to f1 and f2 to ∞.
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Attenuation
Stop band
0
Pass band
f1
Stop band
f2
Frequency
Figure 13.4 Attenuation Characteristic of a BPF
13.3.4 Band-Stop or Band-Elimination Filter
It is a filter that stops particular band of frequencies and passes all other frequencies. It is just
opposite to that of a band-pass filter (BPF). The attenuation characteristics for a band-stop filter
is shown in Figure 13.5.
Attenuation
Pass band
0
Stop band
f1
Pass band
f2
Frequency
Figure 13.5 Attenuation Characteristic of a Band-Stop Filter
Therefore, for a BPF, the following are given:
Pass band: 0 to f1 and f2 to ∞
Stop band: f1 to f2.
13.4 PARAMETERS OF A FILTER
There are four important parameters that are necessary to analyse the performance of a filter
network. They are as follows:
1. Propagation constant g
2. Attenuation constant a
3. Phase shift constant b
4. Characteristic impedance Z0
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13.4.1 Propagation Constant (f )
For any two-port network terminated by characteristic
impedance Z0, as shown in Figure 13.6, we can write the
following:
I1 V1
=
= eg
I2 V2
g = loge
I1
V
= loge 1
I2
V2
I1
V1
I2
Two-port
network
V2
Z0
Figure 13.6 A Two-port
Network
Terminated by
Characteristics
Impedance Z0
where g is known as propagation constant. Propagation
constant determines the propagation performance of any two-port network.
Moreover,
g = a + j b (13.4)
where a is real part of g and is known as attenuation constant of the filter, and b is imaginary
part of g and is known as phase constant.
13.4.2 Attenuation Constant
Whenever a signal passes through a passive network/filter, it gets attenuated, because passive
components like capacitors and inductors consume some of the signal energy. The attenuation
constant determines the attenuation of the signal when it passes through the filter.
Units of Attenuation
Attenuation can be expressed in decibels or nepers.
Neper: It is defined as the natural log of the ratio of input current or voltage or power to the
output current or voltage or power.
Neper( N ) = loge
I1
V
P
1
= loge 1 = loge 1
I2
V2 2
P2
Decibel: It is defined as the ten times the common log of the ratio of input current/voltage/power
and output current/voltage/power.
Decibel (D) can be written as follows:
D = 20 log10
I1
V
P
= 20 log10 1 = 20 log10 1 (13.5)
I2
V2
P2
Relation Between Nepers and Decibel
Attenuation in Nepers =
Attenuation in decibels
(13.6)
8.686
= 0.115 × Attenuation in decibels
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13.4.3 Phase Shift Constant (a )
When the signal passes through the filter, it gets some shift in phase. Phase shift constant signifies the phase shift in the signal when it passes through the filter.
Units of Phase Shift
The unit of phase shift is radians or degrees. The relation between radians and degrees can be
written as follows:
p radians = 180°
13.4.4 Characteristic Impedance (Z0)
Characteristic impedance is the image impedance of a two-port network. For symmetric networks, the image impedance at port 1−1′ is equal to the image impedance at port 2−2′. They are
equal to the characteristic impedance Z0.
13.5 FILTER NETWORKS
A filter is constructed from reactive elements such as inductors and capacitors. Filter produces
no attenuation to pass band or transmission band and provides complete attenuation to all
other frequencies called attenuation band or stop band. Filters are used in the communication
systems.
Filters are made of symmetrical T- or p-Sections. T- and p-Sections are the combination of
asymmetrical L networks, as shown in Figures 13.7 and 13.8, respectively.
13.5.1 Formation of Symmetrical T-Network
Z1
2
Z1
2
+
2Z2
Asymmetrical L
network for T-section
2Z2
Asymmetrical L
network for T-section
Z1
2
Z1
2
2Z2
2Z2 =
2Z2 × 2Z2
2Z2 + 2Z2
=
4Z22
4Z2
= Z2
Symmetrical T-section
Figure 13.7 S
ymmetrical and T-section
Asymmetrical L-networks
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is
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592 Network Analysis and Synthesis
13.5.2 Formation of Symmetrical o-Network
Z1
2
Z1
2
+
2Z2
Asymmetrical L-network
2Z2
Asymmetrical L-network
Z1 Z1
= Z1
+
2
2
2Z2
2Z2
Asymmetrical p-Network
Figure 13.8 Symmetrical p-Network Made from Asymmetrical L-networks
13.5.3 Ladder Network
A ladder network is a cascade or series connection of many T- and p-Sections. There are two
types of ladder networks.
1. T-section ladder network.
2. p-Section ladder network.
These are shown in Figures 13.9 and 13.10, respectively.
T-section Ladder Network
It is formed by connecting many T-sections in series, as shown in Figure 13.9.
Z1
2
Z2
Z1
2
Z1
2
T-section
Z2
Z1
2
Z1
2
T-section
Z1
2
Z1 Z1
= Z1
+
2
2
Z2
Z2
Z1
2
T-section
Z1
2
Z1
Z2
Z2
Figure 13.9 T-section Ladder Network
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593
p-Section Ladder Network
It is formed by connecting several sections of p-networks in cascade, as shown in Figure 13.10.
Z1
Z1
2Z2
2Z2
2Z2
2Z2
2Z2
Z1
Z1
2Z2
Z1
2Z2
2Z2 = Z2
2Z2
Z1
2Z2
Z2
Figure 13.10 p -Section Ladder Network
13.6 ANALYSIS OF FILTER NETWORKS
In this section, we will explain how the basic parameters of T- and p-networks are determined.
13.6.1 Symmetrical T-network
We know that the basic parameters of a filter network are its
characteristic impedance, propagation constant, attenuation
constant and phase shift constant. These are calculated as
follows.
Characteristic Impedance (Z0)
2
1
Z1
2
Z1
Z2 2
1′
Z0
2′
If a two-port network is symmetrical, the image impedance
Figure 13.11 A T-network
Zi1 at port 1-1′ is equal to the image impedance Zi2 at port
2-2′ and that image impedance is called the characteristic
impedance Z0.
When the network is terminated with Z0 (characteristic impedance), then input impedance
Zin = Z0(13.7)
Consider the T-network shown in Figure 13.11.
The input impedance of the T-network shown in figure is written as follows:
Z 

Z 2  Z0 + 1 



Z
Z 
Z
2

Zin = 1 +  Z 2  Z0 + 1   = 1 +
Z

2 
2  2
Z 2 + Z0 + 1
2
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594 Network Analysis and Synthesis
Z1Z 2
Z1 Z 2 Z0 + 2
=
+
Z
2
Z0 + Z 2 + 1
2
Z1
ZZ
Z 2 Z0 + 1 2
2
= 2 +
Z
1
Z0 + Z 2 + 1
2
Z1 
Z1 
ZZ
Z0 + Z 2 +  + Z 2 Z0 + 1 2



2
2
2
=
Z1
Z0 + Z 2 +
2
2
Z1Z0 Z1Z 2 Z1
ZZ
+
+
+ Z 2 Z0 + 1 2
2
2
4
2
Zin =
Z1
Z0 + Z 2 +
2
Now, substitute Zin = Z0 (using equation (13.7))
Z1Z0 Z1Z 2 Z12
ZZ
+
+
+ Z 2 Z0 + 1 2
2
2
4
2
Characteristics impedance Z0 =
Z1
Z0 + Z 2 +
2
2
Z Z
ZZ
ZZ
Z
ZZ
Z02 + Z0 Z 2 + 0 1 = 1 0 + 1 2 + 1 + Z 2 Z0 + 1 2
2
2
2
4
2
Z1Z 2 Z12 Z1Z 2
+
+
2
4
2
2
Z
Z02 = 1 + Z1Z 2
4
Z02 =
Z1 2
+ Z1Z 2 (13.8)
4
The expression is for characteristic impedance of a T-network.
Z0 (characteristic impedance) can also be expressed
in terms of ZOC (open-circuit impedance) and ZSC (short-­
circuit impedance).
Let us find open-circuit impedance of T-network. From
Figure 13.12, it is observed that the open-circuit ­impedance
is written as follows:
Z
ZOC = 1 + Z 2 (13.9)
2
Z0 =
ZOC
Z1
2
Z1
Z2 2 I = 0
Open
circuit
Figure 13.12 Open-circuit
Impedance of a
T-Network
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From Figure 13.13, the short-circuit impedance of the
T-network is calculated as follows:
ZSC =
ZSC
Z1  Z1

+
Z
2  2 2 
Z1
2
Z1
Z2 2
595
Short
circuit
Figure13.13 Short-circuit
Impedance of a
T-network
 Z1 
⋅Z
Z1  2  2
=
+
Z1
2
+ Z2
2
Z1
Z1Z 2
2
= 2 +
1 Z1
+ Z2
2
Z12 Z1Z 2 Z1Z 2
+
+
2
2
= 4
Z1
+ Z2
2
or
ZSC
Z12
+ Z1Z 2
(13.10)
= 4
Z1
+ Z2
2
Multiplying equations (13.9) and (13.10), we get the following:
ZOC ZSC

 Z12
 4 + Z1Z 2 

Z
 
=  1 + Z2  ⋅
Z1
 2

+ Z2
2
ZOC ZSC =
or
ZOC ZSC =
Z12
+ Z1Z 2
4
Z12
+ Z1Z 2
4
= ZOT
(using equation (13.8))
Therefore, the characteristic impedance for a T-network, ZOT is given as follows:
ZOT = ZOC ZSC (13.11)
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596 Network Analysis and Synthesis
1
V1
I
Propagation Constant (g )
Z1
2
Z1
2
Z2
I1
1′ Mesh I
II
As shown in Figure 13.14, I1, I2, V1 and V2 are input
current, output current, input voltage and output
voltage, respectively.
Applying KVL in mesh II, as shown in Figure
13.14(a), we get the following:
2
Z0
V2
I2
Mesh II 2′
Figure13.14(a) A Symmetrical
T-network
Connected to a
Load Impedance Z0
or
or
Z1
I +I Z =0
2 2 2 0
Z
− I1Z 2 + I 2 Z 2 + 1 I 2 + I 2 Z0 = 0
2
Z 2 ( I 2 − I1 ) +
or
Z


I1Z 2 = I 2  Z 2 + 1 + Z0 


2
Z1
I1 Z 2 + 2 + Z0
(13.12)
=
I2
Z2
Now, by definition, we get the following form:
I1
= eg
I2
Substituting this value in equation (13.12), the following form can be obtained:
eg =
or
Z2 +
Z2 +
Z1
+ Z0 = eg Z 2
2
Z1
+ Z0
2
Z2
Z1
2
Z
Z0 = Z 2 (eg − 1) − 1
2
Z0 = eg Z 2 − Z 2 −
or
Z12
+ Z1Z 2 (for T-network)
4
Substituting this value in the equation, we get the following form:
Z0 =
We have
Z12
Z
+ Z1Z 2 = Z 2 (eg − 1) − 1
4
2
Squaring both sides, the equation can be written as follows:
Z 12
Z 

+ Z 1Z 2 =  Z 2 (eg − 1) − 1 
4
2

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or
or
or
597
Z12
Z2
Z Z2
+ Z1Z 2 = Z 22 (eg − 1) 2 + 1 − 2 Z 2 (eg − 1) ⋅ 1 + 1
4
4
2
4
Z1Z 2 = Z 22 (eg − 1) 2 − Z1Z 2 (eg − 1)
Z1Z 2 + Z1Z 2 (eg − 1) = Z 22 (eg − 1) 2
Z1Z 2 [1 + eg − 1] = Z 22 (eg − 1) 2
Z1Z 2 eg = Z 22 (eg − 1) 2
or
(eg − 1) 2 =
or
(eg − 1) 2 =
or
or
e 2g + 1 − 2eg =
e −g (e 2g
Z1Z 2 e l
Z 22
Z1 g
Z1
e =
Z2
Z 2 e −g
Z1
Z 2 e −g
Z
+ 1 − 2eg ) = 1
Z2
or
eg + e −g − 2 =
or
eg + e −g =
Z1
Z2
Z1
+2
Z2
Dividing both sides by 2, we get the following:
Z
eg + e −g
= 1 +1
2
2Z2
cosh g = 1 +
or
Z1
2Z2
Z
g
= 1+ 1
2
2Z2
or
1 + 2 sinh 2
or
2 sinh 2
Z
g
= 1
2 2Z 2
or
sinh 2
Z
g
= 1
2 4Z2
or
sinh
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g
=
2
q
q
 e + e

= cosh q  (13.13)
∵
2


∵ coshq = 1 + 2 sinh 2
q
2
Z1
4Z2
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598 Network Analysis and Synthesis
or
Z1
g
= sinh −1
2
4Z2
or
g = 2 sinh −1
Z1
(13.14)
4Z 2
This is the required expression for propagation constant of symmetrical T-network.
Attenuation Constant (a ) and Phase Shift Constant (b )
From equation (13.14), propagation constant is a complex function represented as g = a + jb.
The real part a is a measure of change in the magnitude of the current or the voltage in the
network and the imaginary part b is a measure of difference in the phase between the input and
output currents or voltages.
sinh
g
=
2
Z1
4Z2
Substituting g = a + jb, in the equation, we get the following:
 a + jb 
=
sinh 
 2 
Z1
4Z2
b
a
sinh  + j  =
2
2
Z1
4Z2
a
a
 b
 b
cosh  j  + cosh sinh  j  =
 2
 2
2
2
Z1
4Z2
or
or
sinh
[∵ sin( A + B) = sin A cos B + cos A sin B]
sinh
a
b
a
b
cos + jcosh sin =
2
2
2
2
Z1
(i)
4Z 2
∵ cosh( jq ) = cosq 
 sinh( jq ) = jsinq 


Case I: When a = 0, equation (13.15) can be written as follows:
0 + j cos 0° ⋅
or
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sin b
=
2
Z1
4Z2
b
=
2
Z1
4Z2
j sin
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Filters and Attenuators
599
Considering the magnitude,
or
sin
Therefore,
b
=
2
Z1
4Z2
b
= sin −1
2
or,
b = 2 sin −1
Z1
4Z2
Z1
(13.15a)
4Z2
Case II: To get expression for a, substitute b = p in equation (i)
0 + j cosh
a
×1 =
2
Z1
4Z2
a
=
2
Z1
4Z2
a
=
2
Z1
4Z2
or
j cosh
or
cosh
a
= cosh −1
2
or
That is,
a = 2 cosh −1
Z1
4Z2
Z1
(13.15b)
4Z2
Cut-off Frequency (fc )
We have
coshg = 1 +
Z1
2Z2
(from equation (13.12))
For pass band, a = 0
That is,
g = a + jb = 0 + jb = jb
Substituting g = jb in the equation, we get the following form:
Z
 jb 
= 1+ 1
cosh 
 2 
2Z2
or
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cos
Z
b
= 1+ 1
2
2Z2
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600 Network Analysis and Synthesis
Now, the limits of cos q are ±1
−1 ≤ cos
That is,
b
≤1
2
Therefore, in pass band, the equation can be written as follows:
−1 < 1 +
Z1
<1
2Z2
or
−2 <
Z1
<0
2Z2
or
−1 <
Z1
< 0 ⇒ condition for pass band
4Z2
Cut-off frequency can be obtained by substituting the following:
Z1
=0
4Z2
a
Stop band
Pass band
−1
Z
= −1 or
4Z2
Stop band
or
0
Z1
4Z2
Figure 13.14(b)
or
Z1 + 4Z2 = 0
Z1 = 0
Z1 = −4Z2
(ii)
The graphical representation of the equations is shown
in the following Figure 13.14(b).
Summary of Equations of T-network
1. ZOT =
Z12
+ Z1Z 2 : characteristic impedance
4
2. g = 2 sinh −1
3. a = 2 cosh −1
4. b = 2 sin −1
Z1
: propagation constant
4Z2
Z1
: attenuation constant
4Z2
Z1
4Z2
: phase shift
5. Z1 + 4 Z 2 = 0 : equation to obtain cut-off frequency.
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13.6.2 Analysis of p-Network
Let us find the characteristic impedance for p-network
shown in Figure 13.15.
From the figure, the input impedance of the network is
given as follows:
Z in = 2Z 2 [Z 1 + [2Z 2 Z 0 ]]
601
2
1
Z1
2Z2
Zin
2Z2
1′

2Z 2 ⋅ Z0 
= 2 Z 2  Z1 +

Z 2 + Z0 
2

Zo
2′
Figure 13.15 A p -network
 2 Z Z + Z1Z0 + 2 Z 2 Z0 
= 2Z2  1 2

2Z 2 + Z0


 2 Z Z + Z1Z0 + 2 Z 2 Z0 
2Z2  1 2

2Z 2 + Z0

=
2 Z Z + Z1Z0 + 2 Z 2 Z0
2Z2 + 1 2
2Z 2 + Z0
=
=
Zin =
 2 Z Z + Z1Z0 + 2 Z 2 Z0 
2Z2  1 2

2Z 2 + Z0

4 Z 2 2 + 2 Z 2 Z0 + 2 Z1Z 2 + Z1Z0 + 2 Z 2 Z0
2Z 2 + Z0
2 Z 2 ( 2 Z1Z 2 + Z1Z0 + 2 Z 2 Z0 )
4 Z 2 2 + 2 Z1Z 2 + Z1Z0 + 4 Z 2 Z0
4 Z1Z 22 + 2 Z1Z 2 Z0 + 4 Z 22 Z0
4 Z 2 2 + 2 Z1Z 2 + Z1Z0 + 4 Z 2 Z0
By definition, Zin = Z0, substituting this value in the equation, we get the following:
4 Z1Z 22 + 2 Z1Z 2 Z0 + 4 Z 22 Z0
Z0 =
4 Z 2 2 + 2 Z1Z 2 + Z1Z0 + 2 Z 2 Z0
4 Z 22 Z 0 + 2Z 1Z 2 Z 0 + Z 1Z 0 2 + 2Z 2 Z 0 2 = 4 Z 1Z 22 + 2Z 1Z 2 Z 0 + 4 Z 2 2 Z 0
Z 1Z 0 2 + 2 Z 2 Z 0 2 = 4 Z 1Z 2 2
or
Z 0 2 ( Z 1 + 2 Z 2 ) = 4 Z 1Z 2 2
Z 02 =
or
4 Z 1Z 2 2
Z 1 + 2Z 2
Multiplying and dividing by Z1, we get the following:
Z 02 =
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602 Network Analysis and Synthesis
Z02 =
4 Z12 Z 22

Z2
4  1 + Z1Z 2 
 4

Z 0 = Z 0p
=
Z12 Z 22
Z12
+ Z1Z 2
4
Z1Z 2
(13.16)
Z12
+ Z1Z 2
4
Now, we will prove the following:
Zop = ZOC ZSC
where ZOC and ZSC are, respectively, the open-circuit and short-circuit impedance of the
p-Network.
Open-circuit Impedance (Zoc) of p-Network
From Figure 13.16,
Z1
ZOC
2Z2
2Z2
ZOC = 2 Z 2 ( Z1 + 2 Z 2 )
Open
circuit
Figure 13.16 Determination
of Open-circuit
Impedance of a
p -Network
=
2 Z 2 ( Z1 + 2 Z 2 )
2 Z 2 + Z1 + 2 Z 2
=
2 Z 2 ( Z1 + 2 Z 2 )
(13.17)
Z1 + 4 Z 2
Let us find short-circuit impedance of p-network:
From Figure 13.17 and its equivalent circuit shown in Figure 13.18, we get,
ZSC = Z1 2 Z 2
Z1
ZSC
2Z2
2Z2
Short
circuit
Figure 13.17 Determination
of Short-circuit
Impedance of a
p -Network
=
ZSC =
2 Z1Z 2
(13.18)
Z1 + 2 Z 2
Multiplying equation (13.17) and (13.18), we get the
following:
Z1
Z OC Z SC =
ZSC
Z1 ⋅ 2 Z 2
Z1 + 2 Z 2
2 Z2
Figure 13.18 Equivalent
Circuit of
Figure 13.17
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2Z 2 ( Z 1 + 2Z 2 )
2 Z 1Z 2
×
Z 1 + 2Z 2
Z 1 + 4Z 2
=
2 Z 2 ( 2 Z 1Z 2 )
Z 1 + 4Z 2
=
4 Z 1Z 22
Z 1 + 4Z 2
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603
Multiplying and dividing by Z1, we write the equation as follows:
ZOC ZSC =
=
ZOC × ZSC =
or
ZOC ZSC =
Z12 + 4 Z1Z 2
4 Z12 Z 22
Z2

4  1 + Z1Z 2 
 4

Z12 Z 22
Z12
+ Z1Z 2
4
Z1Z 2
Z12
4
Therefore,
4 Z12 Z 22
= ZO = ZOπ
+ Z1Z 2
ZOp = ZOC ZSC
The relation between the characteristic impedances of T-network ZOT and p-network ZOp are
given as in the following:
We have
ZOT =
and
ZOπ =
Z12
+ Z1Z 2 (13.19)
4
Z1Z 2
Z12
4
Substituting the value of
the following form:
+ Z1Z 2
Z12
+ Z1Z 2 = ZOT from equation (13.19) in equation (13.20), we get
4
Z Oπ =
or
(13.20)
Z 1Z 2
Z OT
Z OT Z Oπ = Z 1Z 2
Hence, the product of characteristic impedance of T-network and p-network is equal to the
product of series and shunt impedances.
Here, Z1 is equal to the total series impedance of the filter network and Z2 is equal to the total
shunt impedance of the filter network.
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13.6.3 Summary of Parameters of Filter Network
Parameters of filter network are given in Table 13.1.
Table 13.1
Parameters of T-network and p -network
Parameter
Characteristic impedance
p -network
T-network
ZOT =
Z12
+ Z1Z 2
4
ZOT = ZOC ZSC
ZOπ =
Z1Z 2
Z12
+ Z1Z 2
4
ZOπ = ZOC ZSC
Propagation constant
g = 2 sinh −1
Z1
4Z2
g = 2 sinh −1
Z1
4Z2
Attenuation constant a
a = 2 cosh −1
Z1
4Z2
a = 2 cosh −1
Z1
4Z2
Phase shift b
b = 2 sinh −1
Z1
4Z2
b = 2 sin −1
Equation to obtain cut-off
frequency
Z1 + 4 Z 2 = 0
Z1
4Z2
Z1 + 4 Z 2 = 0
13.7 CLASSIFICATION OF FILTERS
So far, we have seen four types of filters, that is, LPF, HPF, BPF and BSF. When a number of
signals are transmitted along a line, filters need to be used to separate them. For example, a
speech channel using a carrier frequency of 50 kHz requires a bandwidth of say, 46 to 54 kHz.
A band-pass filter (BPF) should pass freely signal of any frequency within this band of frequencies, that is, 46 kHz to 54 kHz and reject all other signals outside this range.
Thus, the output of a filter varies with the frequency. This is called the frequency response
of a filter. If the range of frequency or the range of variation of the signal amplitude is large,
logarithmic scale is used to plot the frequency response.
We have also known that filters are of two basic types: passive filters and active filters. In
both cases, filters are designed to select or reject a band of frequencies. This is achieved by
using series or parallel combination of R, L and C. In active filters, operational amplifiers or
transistors and R, L and C are used.
Filters may be classified to be of constant K-type or m-Derived type.
Depending upon the relationship between Z1 and Z2, filters are classified as follows:
1. Constant K-type or prototype filters.
2. m-Derived filters
These are described in this section.
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605
13.8 CONSTANT K-TYPE OR PROTOTYPE FILTERS
A constant K-type filter is a filter that satisfies the following relationship:
Z1Z2 = K2(13.21)
where Z1 is the series arm impedance.
Z2 is the shunt arm impedance.
K is the design impedance or nominal impedance or zero characteristic impedance.
Constant K-type filters can be of T-type or p -type. These filters are also called prototype filters
because other complex type of filters can be derived from constant K-type filters. Constant
K-type filters may be of low-pass type, high-pass type, band-pass type or band-stop type. These
are discussed in the following sections.
13.8.1 Constant K-type Low-Pass Filters (LPF)
In LPF, the series element is inductor and shunt element is capacitor. T- and p-Section for constant K-type LPF are shown in Figure 13.19.
L
2
L
2
Z1
2
Z1
2
C
(Z1)
L
C
2
(2Z2)
(Z2)
C
2
(2Z2)
Figure 13.19 Constant K-type LPF
Expressions for Different Parameters of Constant K-type LPF
Design Impedance (K). In constant K-type LPF,
Total series impedance, Z1 = jwL and
1
jw C
Further, for constant K-type filters, we have, Z1Z 2 = K 2
Substituting the values of Z1 and Z2 in the equation, we get the following:
Total shunt impedence, Z2 =
 1 
= K2
( jw L) 
 jw C 
L
= K2
C
or
or
K=
L
This is the expression that will be used in design of constant K-type LPF(13.22)
C
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606 Network Analysis and Synthesis
Cut-off Frequency (fc).
We have limits of pass band as given in the following:
−1 <
Z1
=0
4Z2
or
Z1
= −1
4Z2
Z1 = 0
or
Z1 = −4 Z 2
jw L = 0
or
Z1 + 4Z2 = 0
j 2p f c L = 0
or
jw L +
This means, either
or
fc = 0 or
or
Now, we have fc = 0 and the condition
Substituting j2 = -1, we have −
4
=0
jw C
j 2w 2 LC + 4
=0
jw C
j 2w 2 LC + 4
= 0.
jw C
w 2 LC + 4
=0
jw C
or,
−w 2 LC + 4 = 0;
or,
2p f c =
Therefore, f c = 0 and f c =
Z1
<0
4Z2
2
;
LC
or, w =
or,
fc =
2
LC
1
p LC
1
(13.23)
p LC
Attenuation Constant (a ). To find attenuation constant, we have the following relation:
a = 2 cosh −1
Substituting Z1 = jw L and Z 2 =
Z1
4Z2
1
in the equation, we get the following:
jwC
a = 2 cosh −1
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jw L
4
jw C
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Filters and Attenuators
= 2 cosh −1
j 2w 2 LC
4
= 2 cosh −1
−w 2 LC
4
= 2 cosh −1
607
\w 2 = -1
w 2 LC
4
w = 2p f
Substitute
( 2p f ) 2 LC
4
a = 2 cosh −1
4p 2 f 2 LC
= 2 cosh −1 p 2 f 2 LC (13.24)
4
= 2 cosh −1
Now, from equation (13.23), we have the following form:
1
; or f 2 = 1
fc =
c
p LC
p 2 LC
1
or
p 2 LC = 2 (13.25)
fc
Substituting this value in equation (13.24), we get the following forms:
a = 2 cosh −1
f 2p 2 LC
= 2 cosh −1
 1 
f 2 2
 fc 
= 2 cosh −1
f2
f c2
 f 
a = 2 cosh −1  
 fc 
and attenuation in pass band a = 0.
Phase Constant b. In stop band, b = p and in pass band,
b = 2 sin −1
Z1
4Z2
 f 
= 2 sin −1  
 fc 
[derivation is same as that for attenuation constant]
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608 Network Analysis and Synthesis
The performance characteristic of a constant K-type LPF has been shown in Figure 13.20.
p
∞
a
(Attenuation)
b
Pass
band
O
Pass
band
Stop
band
Stop
band
Frequency
fc
O
fc
∞
Frequency
Variation of
phase shift
Variation of attenuation
with frequency
Figure 13.20 Performance Characteristic of a Constant K-type LPF
Characteristic Impedance (Z0). We have
ZOT =
Substituting Z1 = jw L and Z 2 =
1
,we get the following:
jw C
ZOT =
=
or
ZOT =
ZOT =
 1 
( jw L) 2
+ ( jw L) 
4
 jw C 
−w 2 L2 L
+
4
C
L w 2 L2
−
C
4
w 2 LC
( 2p f 2 ) LC
L  w 2 LC 
L
L
1−
=
1−
=
1−


C
C
C
4 
4
4
L
=K
C
Now,
or,
Z12
+ Z1Z 2
4
ZOT = K 1 −
4p 2 f 2 LC
= K 1 − f 2p 2 LC
4
ZOT = K 1 −
f2
fc2
Further, for LPF
fc =
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1
(13.26)
p LC
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Filters and Attenuators
or
609
f c 2 = p 2 LC
1
p 2 LC = 2
fc
or
2
 f 
ZOT = K 1 −   (13.27)
 fc 
and
Z1Z 2
ZOπ =
Z12
+ Z1Z 2
4
ZZ
(13.28)
= 1 2
ZOT
Substituting the value of Z1 = jwL
1
xc
jw L
 1 
( jw L) 
 jw C 
=
ZOT
Z2 =
ZOπ
ZOπ
L
K2
= C =
ZOT ZOT
L
L
= K, = K 2
C
C
Since,
Using equation (13.26), we get the following form:
ZOπ =
K2
 f 
K 1−  
 fc 
2
=
K
 f 
1−  
 fc 
2
Design Parameters. We have
L
L
= K , that is, K 2 = (i)
C
C
1
1
or f c 2 = 2
(ii)
p LC
p LC
and
fc =
From equation (i),
L = K2C
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(iii)
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610 Network Analysis and Synthesis
L=
and from equation (ii), we get
1
(iv)
p f c2C
2
Equating the two equations, that is, equation (iii) and (iv), we get the following form:
K 2C =
1
1
or, C 2 = 2 2 2
2
p K fc
p fc C
2
That is, C =
1
(13.29)
p Kf c
Substituting this value in equation (iii), the equation can be written as follows:
1
L = K2 C = K 2
p Kf c
K
L=
Therefore,
(13.30)
p fc
Summary of Constant K-type LPF
L
C
Design parameters: L = K
p fc
1
C=
K p fc
1
Cut-off frequency: fC =
p LC
Attenuation:
 f 
a = 2 cosh −1   in stop band
 fc 
= 0 in pass band
Phase constant:
1. Design impedance:
2.
3.
4.
5.
K=
 f 
b = 2 sin −1   in pass band
 fc 
= p in stop band
6. Characteristic impedance:
Z OT
f 
= K 1−  
 fc 
Z Oπ =
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 610
2
K
f 
1−  
 fc 
2
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611
Filters and Attenuators
Solved Numerical on Constant K-type LPF
Example 13.1 Design a low-pass T-section filter having a cut-off
frequency of 1.5 kHz to operate with a design impedance of 600 W.
Solution: Given
L
2
(63.66 mH)
fc = 1.5 kHz = 1500 Hz
K = 600 W
Now, for constant K-type LPF we write the equation as follows:
C=
(63.66 mH)
C(0.353 µF)
T-section constant K-LPF
Figure 13.21(a)
K
600
=
= 0.12732H = 127.32 mH
L=
p f c p (1500)
and
L
2
1
1
=
= 3.53 × 10 −7 = 0.353 × 10 −6 F
K p f c 600 ⋅ p ⋅1500
0.02 H
= 0.353 µF
0.02 H
0.06 µF
The required design is shown in Figure 13.21a.
Example 13.2 Figure 13.21(b) shows a passive filter section. Find
its cut-off frequency and characteristic impedance at f = 0.
Solution: Given filter is T-section LPF of Constant K-type. By
comparing Figure 13.21(b) with Figure 13.21(c), we get,
and
L
= 0.02 H ⇒ L = 0.04 H
2
C = 0.06 µF ⇒ C = 0.06 × 10 −6 F
L
2
L
2
C
Figure 13.21(c)
Now, for LPF, fc can be written as follows:
fc =
Figure 13.21(b)
1
1
=
= 6497.47 Hz
p LC p 0.04 × 0.06 × 10 −6
and
ZOT
Now,
 f 
= K 1−  
 fc 
K=
L
=
C
2
 0
f c = 816.49 1 −  
 fc 
0.04
0.06 × 10 −6
2
f c = 816.49 Ω
= 816.49 Ω
Example 13.3 Design a constant K-type LPF having a cut-off frequency of 2000 Hz
and a zero-frequency characteristic impedance of 200 W. Draw T- and p-Section of the
filter.
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612 Network Analysis and Synthesis
Solution: Given
fc = 2000 Hz
K = 200 W
Now, for an LPF, L and C can be calculated as follows:
L=
K
200
=
= 0.03183 H = 31.83 mH
p f c p ( 2000)
C=
1
1
=
= 7.957 × 10 −7 = 0.7957 × 10 −6 F = 0.7957 µF
K p f c 200(p × 2000)
Required design is given in Figure 13.22.
L
L
2
2
= 15.915 mH = 15.915 mH
L = 31.83 mH
C
2
= 0.398 µF
C = 0.7957 µF
C
2
= 0.398 µF
p -Section
T-section
Figure 13.22
Example 13.4 A constant K-type LPF composed of T-section has 63.6 mH inductance in
each series arm and 0.088 μF in the shunt arm. Find (1) cut-off frequency and (2) attenuation
in b at 5000 Hz.
Solution: The circuit is given in Figure 13.23. For comparison, the general design has also
been shown.
63.6 mH
L
2
63.6 mH
0.088 µF
Compare it with
general design as
L
2
C
Figure 13.23
We get
L
= 63.6 mH, that is, L = 127.2 mH = 0.1272 H
2
C = 0.088 µF = 0.088 × 10 −6 F
and
1. Cut-off frequency f c =
1
p LC
=
1
p 0.1272 × 0.088 × 10 −6
= 3008.6 Hz
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613
Filters and Attenuators
2. Attenuation at 5000 Hz, that is, f = 5000 Hz and we have
 f 
 5000 
a = 2 cosh −1   = 2 cosh −1 
 3008.6 
 fc 
= 2 × 1.09503
= 2.19006 Nepers
= 8.686 × 2.19006 dB
= 19.02 dB.
Example 13.5 Each arm of a symmetrical T-section LPF consists of 6 mH inductor, while the
shunt arm is a 0.03 μF capacitor. Find the design impedance and cut-off frequency.
Solution: Given circuit is shown in Figure 13.24
That is,
L
= 6 mH ⇒ L = 12 mH
2
Now for LPF, the design impedance
L
=
C
12 × 10 −3
0.03 × 10 −6
(6 mH)
(6 mH)
T-section LPF
Figure 13.24
= 632.455 Ω
and
Cut-off frequency
L
2
C = 0.03 µF
C = 0.03 µF
K=
L
2
fc =
=
1
p LC
1
× 0.03 × 10 −6
1
=
=
= 16776.40 Hz
−5
5.96 × 10 −5
p × 1.8973 × 10
p 12 × 10
1
−3
Example 13.6 Design the T- and p-Section of a constant K-type LPF having a cut-off
frequency of 10 kHz and design impedance of 450 W. Further, find its characteristic impedance
and phase constant at 5 kHz as well as determine the attenuation at 12 kHz.
Solution: Given
K = 450 W
fc = 10 kHz = 10,000 Hz
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614 Network Analysis and Synthesis
Now for LPF, L and C can be calculated as follows:
L=
K
450
=
= 0.01432 H = 14.32mH
p f c p (10, 000)
C=
1
1
=
= 7.0735 × 10 −8 F
K p f c 450(p )10, 000
= 0.0707 × 10 −6 F = 0.0707 µF.
The constant K-type LPF for both T- and p-Sections are shown in Figure 13.25.
L
2
= 7.16 mH
L
2
= 7.16 mH
L = 14.32 mH
C =
2
0.03535 µF
C = 0.0707 µF
C =
2
0.03535 µF
p-Type LPF
T-type LPF
Figure 13.25
Characteristic impedance at f = 5 kHz = 1500 Hz
We have
 f 
ZOT = K 1 −  
 fc 
2
 5000 
= 450 1 − 
 10, 000 
2
= 450 1 − (0.5) 2 = 450(0.8666) = 389.71 Ω
and
ZOp =
We now calculate Phase shift at
K
 f 
1−   2
 fc 
=
450
= 519.615 Ω
0.8666
f = 5 kHz = 5000 Hz as
 5000 
 f 
b = 2 sin −1   = 2 sin −1 
 10, 000 
 fC 
= 2 sin −1 (0.5) = 60°
= 60 ×
p
= 1.047 radians
180
Attenuation a at f = 12 kHz, is calculated as below
For constant K-type LPF, we have the value of a as,
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Filters and Attenuators
615
 f 
a = 2 cosh −1  
 fC 
 12000 
= 2 cosh −1 
= 2 cosh −1 (1.2) = 2 × 0.622
 10000 
= 1.244 N
= 1.244 × 8.686 = 10.81 dB.
13.8.2 Constant K-type High-Pass-Filters (HPF)
In an HPF, the series element is a capacitor and the shunt arm element is an inductor, that is, in
case of HPF.
1
and Z 2 = jw L
Z1 =
jw C
The circuit configuration of constant K-type HPF, both T-type and p-type, have been shown in
Figure 13.26.
2C
C(Z1)
2C
Z1
2
L
Z1
2
2Z2
(2L)
(Z2)
2Z2
(2L)
p -Section
T-section
Figure 13.26 Circuit Configurations of HPF
Expression of parameters for HPF are given as follows.
Design Impedance (K)
We know that for constant K-type HPF, the following can be written as follows:
Z1 Z2 = K2
Substituting the values of Z1 and Z2, we get the following:
 1 
2
 jw C  ( jw L) = K
or
L
= K2
C
K=
L
C
Cut-off Frequency (fc )
In pass band, we get the following:
−1 <
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Z1
<0
4Z2
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616 Network Analysis and Synthesis
That is,
Z1
= 0 or Z1 = 0
4Z2
∴
or
1
1
= 0 or = 0
jw C
w
1
1
= 0 or
= 0 or f c = ∞
2p f c
fc
or
1 = −4( −1)w 2 LC
Z1
= −1
4Z 2
or
Z 1 = −4 Z 2
or
 1 
 jw C  = −4 jw L
or
or
1 = −4 j 2w 2 LC
or
w2 =
1
1
or, w =
4 LC
2 LC
∴
2p f c =
or
fc =
1
2 LC
1
4 LC
Attenuation Constant (a )
In pass band, a = 0 and in stop band
Z1
4Z 2
a = 2 cosh −1
Z1 =
Substituting
1
and Z 2 = jw L
jw L
We get
a = 2 cosh −1
= 2 cosh −1
1
= 2 cosh −1
jw C
4 jw L
1
2
−4w LC
1
2
4 j w LC
; = 2 cosh −1
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 616
2
; = 2 cosh −1
1
2
4( 2p f ) LC
1
−4w 2 LC
; = 2 cosh −1
∵ j 2 = −1
1
2 2
4.4p f LC
12/3/2014 8:20:55 PM
Filters and Attenuators
or a = 2 cosh −1
1
2
16p f LC
= 2 cosh −1
617
1
f 16 p 2 LC (13.31)
2
2
Now for HPF, we have the following:
1
1
1
or, f c2 =
or, 16p 2 LC = 2 (13.32)
16p 2 LC
fC
4p LC
Substituting this value in equation (13.31), we get:
fc =
a = 2 cosh −1
1
= 2 cosh −1


1
f 2 2
 fc 
f c2
f2
 f 
= 2 cosh −1  c 
 f 
2
 f 
∴ a = 2 cosh −1  c 
 f 
Phase Constant (b )
In pass band, b = p
In stop band,
∞
Z1
b = 2 sin −1
∞
a
(Attenuation)
4 Z 2 [derivation is same as that of
 f 
attenuation]
b = 2 sin −1  C 
 f 
Stop
band
0
O
Performance of HPF
The variation of a with frequency and variation of b
with frequency have been shown in Figure 13.27 and
Figure 13.28, respectively.
We have the following:
Z1 =
Substituting
Z 12
+ Z 1Z 2
4
b
(Phase
shift)
1
and Z 2 = jw L
jw C
−p
2
ZOT =
1
2
2 2
j 4w c
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 617
+
L
=
C
∞
Frequency
∞
fc
O
 1 
 jw c 
 1 
+
( jw L) =
4
 jw c 
fc
Frequency
Figure 13.27 Variation of
Attenuation
with Frequency
Characteristic Impedance (Z0 )
Z OT =
Pass
band
−1
4w 2 c 2
Stop
band
Pass
band
−p
Figure 13.28 Variation
Phase
Shift with
L
Frequency
+
C
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618 Network Analysis and Synthesis
L
1
L
1 
L
1
−
=
1−
1 −
 =
2
2
2
C 4w c
C
C
4w LC
4w 2 LC
=
Substituting
L
=K
C
ZOT = K 1 −
= K 1−
1
4w 2 LC
= K 1−
1
16p 2 f 2 LC
1
4( 2p f ) 2 LC
= K 1−
1
4( 4p 2 f 2 ) LC
1
= K 1−
f 2 16p 2 LC
(From equation (13.32) we had 16p 2 LC = 1/fC2
1
Therefore, ZOT = K 1 −
1
f 2.
fc
 f 
ZOT = K 1 −  c 
 f 
or,
ZOπ =
and
Z1Z 2
=
ZOT
= K 1−
fc
f2
2
K2
 f 
K 1−  c 
 f 
=
2
K
 f 
1−  c 
 f 
2
Design Parameters
We have derived the following for HPF:
K=
and
fc =
1
4p LC
L
L
⇒ K 2 = ⇒ L = K 2C (13.33)
C
C
⇒ f c2 =
1
2
16p LC
and L =
1
2
16p f c2C
(13.34)
From equations (13.33) and (13.34), it is clear that C can be written as follows:
K 2C =
or
C2 =
or
C=
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 618
1
2
16p f c2C
1
2
K 16p 2 f c2
1
K 4p f c
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Filters and Attenuators
619
Substituting this value in equation (13.33), we get the value of L as follows:
L = K2 ⋅
1
K 4p f c
K
4p f c
L=
Summary of Constant K-type HPF
Design impedance:
K=
Cut- off frequency:
fc =
L
C
1
4p LC
Design parameters:
L=
K
4p f c
C=
1
K 4p f c
Attenuation:
 f 
a = 2 cosh −1  C  in stop band and in pass band a = 0.
 f 
Phase constant:
b = p in stop band
 f 
= 2 sin −1  C  in pass band
 f 
Characteristic impedance:
 f 
ZOT = K 1 −  C 
 f 
ZOP =
2
K
 f 
1−  C 
 f 
2
13.8.3 Comparison of Constant K-type LPF and HPF
We have, so far, discussed the constant K-type low-pass and high-pass filters in both p-section and
T-section. The design parameters of these filters are shown in a consolidated manner in Table 13.2.
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620 Network Analysis and Synthesis
Table 13.2
and HPF
Comparative Table of Design Parameters of Constant K-type LPF
Parameters
LPF
Circuit
configuration
L
C
2
C
2
Design
impedance
K=
L
C
Design
parameters
L=
K
p fC
Cut-off
frequency
C=
1
K p fC
Phase short
fC =
L
2
C
2L
2C
L
p -Section
T-section
T-section
L
C
K
L=
4p fC
K=
C=
1
a = 0 in pass band
Characteristic b = p in stop band
impedance
 f 
= 2 cosh −1  C  in stop band
 f 
b = p in stop band
 f 
= 2 sin −1   in pass band
 fC 
 f 
ZOT = K 1 −  
 fC 
1
4p LC
a = 0 in pass band
 f 
a = 2 cosh −1   in stop band
 fC 
2
K
 f 
1−  
 fC 
1
K 4p fC
fC =
p LC
Z 0p =
2C
2L
C
p -Section
Attenuation
HPF
L
2
2
 f 
= 2 sin −1  C  in pass band
 f 
 f 
ZOT = K 1 −  C 
 f 
Z 0p =
2
K
 f 
1−  C 
 f 
2
Solved Numericals on Constant K-type HPF
Example 13.7 Design a constant K-type HPF having a cut-off frequency of 5500 Hz and a
design impedance of 750 W. Draw T-section filter and p-Section filter.
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Filters and Attenuators
621
Solution: Given
fc = 5500 Hz
K = 750 W
Now for an HPF, L and C can be calculated as follows:
and
L=
K
750
=
= 0.01085 H = 10.85 mH
4p f C 4p (5500)
C=
1
1
=
= 1.929 × 10 −8 F
K ⋅ 4p f C 750( 4p )(5500)
= 0.01929 µF
Therefore, the T-section and p -section filters are as shown in Figure 13.29.
2C =
2C =
0.03858 µF 0.03858 µF
C = 0.01929 µF
2L =
21.7 mH
L = 10.85 mH
2L =
21.7 mH
p -Section filter
T-section filter is
Figure 13.29
Example 13.8 A T-section HPF has a cut-off frequency of 3000 Hz and infinite frequency
characteristic impedance of 500 W. Find characteristic impedance at 5000 Hz.
Solution: Given
fC = 3000 Hz
K = 500 W
We have to find
ZOT at f = 5000 Hz
Now for HPF, we get the following:
2
 f 
 3000 
ZOT = K 1 −  c  = 500 1 − 
 5000 
 f 
2
2
 3
ZOT = 500 1 −   = 500 0.64 = 400 Ω
 5
Example 13.9 Figure 13.30 shows an HPF section. Find the cutoff frequency and characteristic impedance at f = ∞.
Solution: Given circuit is a T-type HPF and we know the general
circuit configuration for T-type HPF is shown in Figure 13.31.
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 621
0.03 µF
0.03 µF
0.04 H
Figure 13.30
12/3/2014 8:21:05 PM
622 Network Analysis and Synthesis
2C
Therefore, given
2C = 0.03 mF or C = 0.015 mF
and
L = 0.044 H
2C
L
L
0.04
=
= 1632.993Hz
C
0.05 × 106
K=
Therefore,
and
Figure 13.31
fc =
Cut-off frequency is 1
4p LC
1
=
4p 0.04 × 0.015 × 10 −6
= 3248.73 Hz
Characteristic impedance at f = ∞
2
2
 f 
 f 
ZOT = K 1 −  c  = K 1 −  c  = K 1 = 0
 ∞
 f 
K = 1632.993 Hz
Example 13.10 Figure 13.32 shows a high-pass filter section.
Find cut-off frequency and characteristic impedance at f = ∞.
Solution: The general circuit configuration of p-Section high-pass
filter is as shown in Figure 13.33.
By comparing the general configuration with the given p-Section
we get
0.05 µF
0.1 H
Figure 13.32
C = 0.05 mF and L = 0.05 H
Cut-off frequency
( fc ) =
1
4p LC
and
K=
K
 f 
1−  c 
 f 
2
=
C
1
=
4p 0.05 × 0.05 × 10 −6
= 1591.54 Hz
and
Characteristic impedance at f equal to É is calculated as
ZOp =
0.1 H
K
 591.54 
1− 
 ∞ 
2
=
K
1− 0
2L
2L
Figure 13.33
=K
L
0.05
= 1000 Ω
=
C
.05 × 10 −6
Example 13.11 Design the T- and p-Section of a constant K-type high-pass filter having cutoff frequency of 20 kHz and design impedance of 450 W. Also, find its characteristic impedance
and phase constant at 25 kHz as well as determine the attenuation at 4 kHz.
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Filters and Attenuators
623
Solution: Given
fc = 20 kHz = 20,000 Hz;
Now for an HPF, we get the following:
K = 450 W
L=
450
K
=
= 1.79 × 10 −3 H = 1.79 mH
4p f c 4p ( 20, 000)
C=
1
1
= 8.841 × 10 −9 F = 0.008841 µF
=
K 4p f c 450( 4p )( 20, 000)
Therefore, the designs are shown in Figure 13.34.
2C =
2C =
0.017682 µF 0.017682 µF
C = 0.008841 µF
2L =
3.58 mH
2L =
3.58 mH
L = 1.79 mH
p -Section
T-section
Figure 13.34
The characteristic impedance at f = 25,000 Hz
2
 f 
 20, 000 
ZOT = K 1 −  C  = 450 1 − 
 25, 000 
 f 
2
2
 20 
= 450 1 −   = 450 × 0.6 = 270 Ω
 25 
ZOp =
K
 f 
1−  C 
 f 
2
=
=
450
 20, 000 
1− 
 25, 000 
450
 20 
1−  
 25 
2
=
2
450
= 750 Ω
0.6
The phase constant at 25 kHz, that is, at f = 25,000 Hz.
We have
 f 
 20, 000 
 20 
b = 2 sin −1  c  = 2 sin −1 
= 2 sin −1  
 25 
 25, 000 
 f 
= 2 × 53.13 = 106.26° = 106.26 ×
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 623
p
= 1.854 radians
180
12/3/2014 8:21:09 PM
624 Network Analysis and Synthesis
The attenuation at f = 4 kHz, that is, at 400 Hz
For constant K-type HPF, we have the following:
 f 
a = 2 cosh −1  c 
 f 
 20, 000 
= 2 cosh −1 
 4, 000 
= 2 cosh −1 (5) = 2 × 2.2924 = 4.5848 Nepers = 4.58486 × 8.686 dB
a = 39.824 dB.
13.8.4 Constant K-type Band-Pass Filter
+
Vin
LPF
+
HPF
Vout
−
−
Figure 13.35 B
lock Diagram of
a Constant K-type
Band-pass Filter
A band-pass filter can be obtained by connecting a LPF
and a HPF in cascade as shown in Figure 13.35.
The circuit configuration of constant K-type BPF has
been shown in Figure 13.36.
T-section
After connecting the T-sections in cascade (series arm is
made up of series resonant circuit and short arm is made
up of parallel resonant circuit), we get a band-pass filter as
shown in Figure 13.37.
L1
2
L1
2
2C1
2C1
L2
C2
T-section of constant-K LPF
T-section of constant-K HPF
Figure 13.36 A
Band-pass Filter Obtained by Connecting in Series an
LPF and an HPF
L1
2
2C1
C2
2C1
L1
2
L2
T-section of BPF
Figure 13.37 A
Band-pass Filter in T-section Obtained by Connecting in Series
an LPF and an HPF
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Filters and Attenuators
625
p-Section
After connecting p -sections in cascade (series), we get band-pass filter in p-section as shown
in Figure 13.38.
L1
C2
2
C1
C1
L1
C2
2
2L2
2L2
C2
2
p-Section of
constant-K HPF
p-Section of
constant-K LPF
2L2 C2
2
2L2
p-Section of BPF
Figure 13.38 A Band-pass Filter in p-Section Obtained by Connecting in Series a
p-Section of Constant K-type LPF and a Constant K-type HPF
Analysis of Constant K-type BPF
1
1
L2C2
L1C1
In band-pass filter, the resonant frequency of series arm is equal to the resonant frequency of
shunt arm.
1
1
=
or L1C1 = L2C2 (13.35)
That is,
L1C1
L2C2
For series arm, resonant frequency =
and for shunt arm, resonant frequency =
L2 L1
=
(13.36)
C1 C2
or,
Design Impedance (K)
Z1 (total series arm impedance) = jw L1 +
1
(13.37)
jw C1
and Z2 (total shunt arm impedance)
 1 
( jw L2 ) 
 jw C2 
1
= ( jw L2 ) ||
=
1
jw C2
jw L2 +
jw C2
jw L
jw L2
= 2 2 2
=
jw C2
j w L2C2 + 1
j 2w 2 L2 c2 + 1
jw C2
∴
Z2 =
jw L2
1 − w 2 L2C2
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 625
(13.38)
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626 Network Analysis and Synthesis
Now, for constant K-type filters, Z1 Z2 = K2
Substituting the value of Z1 and Z2 in the equation, we get the following:

jw L2 
1 
2
 jw L1 + jw C   1 − w 2 L C  = K


1
2 2
or,
 j 2w 2 L1C1 + 1 
jw L2 
2


=K

2
j
w
C
  1 − w L2C2 

1
or,
jw L2 
(1 − w 2 L1C1 ) 
2
×
=K
2
jw C1
 1 − w L2C2 
Using equation (13.35), that is, L1C1 = L2C2, we get the following:
(1 − w 2 L2C2 )
jw L2
×
= K2
jw C1
1 − w 2 L2C2
jw L2
= K2
jw C1
or
or,
K=
or, K =
L2
C1
L2 L1
=
,
C1 C2
Using equation (13.36), that is, using
We get,
L2
= K2
C1
L2
L1
=
C1
C2
Cut-off Frequencies
In BPF, there are two cut-off frequencies: low cut-off frequency (f1) and higher cut-off frequency (f2). Let us derive the expressions for f1 and f2 for pass band,
−1 <
Z1
<0
4Z2
Cut-off frequency can be obtained by the following equations:
Z1
= −1
4Z 2
Z 1 = −4 Z 2
or
Multiplying both sides by Z1, we get
Now, for constant K-type filters,
Therefore,
Z12 = -4Z1 Z2
Z1 Z2 = K2
Z12 = -4K2
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 626
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Filters and Attenuators
627
Z1 = ± −4 K 2
or
Z1 = ±j2K
Case I: Take
Z1 = j2K.
For constant K-type BPF,
Z1 = jw L1 +
1
jw C1
Therefore,
jw L1 +
1
= j 2K
jw C1
j 2w 2 L1C 1 + 1
or
jw C1
= j 2K
or
1 − w 2 L1C1
= j 2K
jw C1
or
1 − w 2 L1C1 = j 2 2w C1 K
j2 = -1
Substitute
or
1 - w 2L1 C1 = -2w C1K
or
L1C1 w 2 - 2C1 Kw - 1 = 0
w =−
=
( −2C1 K ) ± ( −2C1 K ) 2 − 4( L1C1 )( −1)
2C1 K ± 4C12 K 2 + 4 L1C1
2 L1C1
2C1 K ±
=
=
2( L1C1 )
4C12
 2 L1 
 K + C 
1
2 L1C1

L 
2C1 K ± 4C12  K 2 + 1 
C1 

∴ For quardratic equal equation of type


2
 ax + bx + c = 0



2
 x = −b ± b − 4aac



2a
2 L1C1
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628 Network Analysis and Synthesis
2C1K ± 2C1 K 2 +
=
L1
C1
2L1C1

L 
2C1  K ± K 2 + 1 
C1 

=
2L1C1
K + K2 +
w=
L1
C1
L1
K + K2 +
2p f1 =
(Taking positive value only)
L1
C1
L1
K + K2 +
f1 =
or
L1
C1
2p L1
This is the one cut-off frequency of BPF.
Case II: When Z1 = - j2K
Then

1 
 jw L1 + jw C  = − j 2 K
1
or
j 2w 2 L1C1 + 1
= − j 2K
jw C1
or
j 2w 2 L1C1 + 1 = − j 2 2w C1 K
1 − w 2 L1C1 = 2w C1 K
or
w 2 L1C1 + 2C1K w −1 = 0
or
or
[∴ j 2 = −1]
w=
−2C1 K ± ( 2C1 K ) 2 − 4( L1C1 )( −1)
2( L1C1 )
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 628
=
−2C1 K ± 4C12 K 2 + 4 L1C1
2 L1C1
12/3/2014 8:21:17 PM
629
Filters and Attenuators
=

L 
−2C1 K ± 4C12  K 2 + 1 
C1 

2 L1C1
−2C1 K ± 2C1 K 2 +
=
L1
C1
2 L1C1

L 
L
2C1  − K ± K 2 + 1  − K ± K 2 + 1
C
C

1
=
1
=
L1
2 L1C1
Taking the negative value, we get the following:
−K − K 2 +
w=
−K − K 2 +
2p f 2 =
or
L1
C1
L1
L1
C1
L1
, that is,

L1
2
K + K +
C
1
f2 = −

2p L1




.



This is the second cut-off frequency of BPF.
Attenuation (a )
a = 2 cosh −1
We know
Z1
4Z2
Substituting the values of Z1 and Z2 from equations (13.37) and (13.38)
a = 2 cosh −1
4.
∴
Now,
So,
1
jw C1
jw L2
jw L1 +
= 2 cosh −1
1 − w 2 L2C2
a = 2 cosh −1
j 2w 2 L1C1 + 1
= 2 cosh −1
jw C1
4. jw L2
1 − w 2 L1C1
jw C1
4. jw L2
1 − w 2 L2C2
1 − w 2 L2C2
(1 − w 2 L1C1 ) (1 − w 2 L2C2 )
×
jw C1
4 jw L2
L1C1 = L2C2
a = 2 cosh −1
(1 − w 2 L1C1 )(1 − w 2 L1C1 )
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 629
j 2 4w 2 L2C1
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630 Network Analysis and Synthesis
(1 − w 2 L1C1 ) 2
a = 2 cosh −1
4w 2 L2C1
Further,
resonant frequency w 0 =
That is,
L1C1 =
a = 2 cosh −1

w2 
1
−

2
 w0 
1
L1C1
1
w0
2
1
=
L2C2
and
L 2C 2 =
1
w 02
2
4w 2 L2C1
where w is signal frequency and
w0 is resonant frequency
Phase Constant (b )
b = 2 sin −1
 w2
1 − 2 
 w0 
Z1
= 2 sin −1
4Z2
2
4w 2 L2C1
Resonant Frequency (fo )
We know cut-off frequencies occur when Z1 + 4Z2 = 0
Multiplying both sides by Z1, we get the following:
or,
Z1 = -4Z2
Z12 = - 4Z1 Z2
= - 4K2
Z1 = ± j2K
Let
Z1 = j2K
Z2 = -j2K
[\Z1 Z2 = K2]
at frequency f1
at frequency f2
That is, Z1 at f1 = Z1 at f2 [\Magnitude of Z1 is same at f1 and f1, the difference is only of
equation]
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Filters and Attenuators
jw1L1 +
631
1
1
= jw 2 L1 +
jw1L1
jw 2C1
j 2w12 L1C1 + 1 j 2w 22 L1C1 + 1
=
jw1C1
jw 2C1
1 − w12 L1C1 1 − w 22 L1C1
=
w1
w2
w 
(1 − w12 L1C1 ) =  1  (1 − w 22 L1C1 )
 w2 
Now, L1C1 =
1
w 02
when w0 is resonant free
 w12   w1   w 22 
1 − 2  =   1 − 2 
 w0   w2   w0 
(w 02 − w12 )
or
w 02
 w  w 2 −w 2 
=  1  . 0 2 2 
 w2   w0

w 
w 02 − w12 =  1  w 02 − w 22
 w2 
(
or
(
)
)
or
w 02w 2 − w12w 2 = w1 w 02 − w 22
or
w 02w 2 − w1w 02 = w12w 2 − w1w 22
w 02 (w 2 − w1 ) = w1w 2 (w1 − w 2 )
or
w 02 = w1w 2
Therefore,
or w 0 = w1w 2
or f 0 = f 1 f 2
Characteristic Impedance (Z0)
For a band-pass filter, the following equations can be written as,
Z 1 = jw L1 +
and
and
Putting,
Z2 =
Z OT =
j 2w 2 L1C1 + 1 1 − w 2 L1C1
1
=
=
jw C1
jw C1
jw C1
jw L 2
1 − w 2 L 2C 2
Z 12
+ Z 1Z 2 =
4
(1 − w 2 L1C1 ) 2
4 ⋅ ( jw C1 ) 2
 1 − w 2 L1C1  
jw L 2 
+



2
 jw C1   1 − w L 2C 2 
L 2C 2 = L1C1
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632 Network Analysis and Synthesis
Z OT =
=
=
(1 − w 2 L1C1 ) 2
−4w
C12
(1 − w 2 L1C1 ) 2
(1 − w 2 L1C1 )( jw L 2 )
+
−4w 2C12
 w2
1 − 2 
 w0 
2
jw C1 (1 − w 2 L1C1 )
L 2 (1 − w 2 L1C1 ) 2
−
C1
4w 2C12
L2
=
C1
2
 w2
1 − 2 
1 
 w0  
2
∵ L, G = 2 
= K −
2 2 
w0 
4w C1 
L2
−
C1
4w 2C12
= K2 −
ZOπ =
+
2
 w2
1 − 2 
 w0 
2
 w2
1 − 2 
 w0 
2
 w2
1 − 2 
 w0 
2
L2
= K2 − K2 ⋅
⋅
= K 1−
C1 4w 2 L2C1
4w 2 L2C1
4w 2 L2C1
K2
Z1Z 2
=
ZOT
K 1−
 w2
1 − 2 
 w0 
2
K
=
1−
4w 2 L2C1
 w2
1 − 2 
 w0 
2
4w 2 L2C1
Design Parameters
Now, at f1, that is, higher cut-off frequency
= j2K
that is,
Z1
2
1 − w1 L1C1
or
1- w 12 L1C1 = j2 2w1C1K
or
1- w 12 L1C1 = -2w1C1 K
jw1C1
= j 2K
L1C1 =
Now,
1
w 02
Therefore, we get
1−
or
f 12
f 02
w12
w 02
= −2w1C1K
or
w12
w 02
− 1 = 2w1C1K
− 1 = 2w1C1K
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Filters and Attenuators
f1 f 2
or
f 02 = f1 f 2
Therefore,
f12
− 1 = 2( 2p f1 )C1 K
f1 f 2
or
f1
− 1 = 4p f1C1 K
f2
or
f1 − f 2
= 4p f 1C1K
f2
f0 =
Now,
or ∴ C1 =
633
(f 1 − f 2 )
K ⋅ 4p f 1 f 2
where f1 is upper cut-off frequency and f2 is lower cut-off frequency.
Further, we have to calculate the value of K.
or
K=
L2
C1
or
K2 =
L2
C1
or
L2 = K 2C1
Substituting the value of C1 in the equation, we get the following form:
= K2 ⋅
L2 =
( f1 − f 2 )
K 4p f1 f 2
K ( f1 − f 2 )
4 f1 f 2
Further, we know
1
w0 =
or
L1C1 =
or
L1C1 =
L1C1
1
w 02
1
∵ w 0 = w1w 2
w1w 2
1
( 2p f1 )( 2p f 2 )
=
1
L1C1 =
2
4p f1 f 2
1
1
C
4p f1 f 2 1
L1 =
or
.
2
Substituting the value of C1 in the equations, we get the value of L1.
L1 =
L1 =
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 633
1
2
4p f1 f 2
.
1
( f1 − f 2 )
K 4p f1 f 2
K
p ( f1 − f 2 )
12/3/2014 8:21:27 PM
634 Network Analysis and Synthesis
and we have
K=
C2 =
or
L1
C2
L1
K
2
=
or
L1
= K2
C2
K
2
K ⋅ p ( f1 − f 2 )
1
C2 =
K ⋅ p ( f1 − f 2 )
Summary of Band-Pass-Filter:
Circuit Configuration.
2C1
2C1
L1
L1
2
C1
L1
2
L2
C2
C2
2
2L2
2L2
C2
2
p-Section
T-section
Figure 13.39 Band-pass Filter in T-section and in p -section
Design Impedance.
L1
=
C2
K=
L2
C1
Cut-off Frequencies.
K + K2 +
f 1 ( upper cut-off frequency) =
L1
C1
2p L1
K + K2 +
=−
and f 2 (lower cut-off frequency)
L1
C1
2p L1
Resonant Frequency.
f0 =
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 634
f1 f 2
12/3/2014 8:21:28 PM
635
Filters and Attenuators
a = 2 cosh −1
Attenuation.
 w2
1 − 2 
 w0 
4w 2 L 2C1
and
Phase Constant (b ).
b = 2 sin −1
 w2
1 − 2 
 w0 
2
4w 2 L2C1
Characteristic Impedances.
ZOT = K 1 −
 w2
1 − 2 
 w0 
2
2
4w L2C1
K
ZOp =
1−
 w2
1 − 2 
 w0 
2
4w 2 L2C1
Design Parameters.
C1 =
f1 − f 2
K ( f1 − f 2 )
L2 =
K ⋅ 4p f 1 f 2
4p f 1 f 2
L1 =
1
K
C =
p ( f1 − f 2 ) 2 K ⋅p ( f1 − f 2 )
Performance of Constant K-type BPF.
Variation of attenuation with frequency
Stop
band
a
O
Pass
band
Variation phase shift with frequency
Stop
band
f1
f2 f0
(Lower (Upper)
cut-off
frequency)
p
∞
f
Figure 13.40 a Versus f Characteristic
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 635
Stop
band
b
−p
Stop
band
f2
f
f1
Figure 13.41 b Versus f Characteristic
12/3/2014 8:21:30 PM
636 Network Analysis and Synthesis
Numericals on BPF
Example 13.12 Design the T-section and p-section of a constant K-type BPF that has a pass band
from 1500 to 5500 Hz and characteristic resistance of 200 Ω. Further, find resonant frequency of series
and shunt arms.
Solution: Given
5500 Hz
f1 (lower cut-off frequency) = 1500 Hz; f2 (upper cut-off frequency) =
and K = 200 Ω
Now, for a constant K-type BPF,
L1 =
C2 =
200
K
=
= 0.01915 H = 15.915 mH
p ( f 2 - f1 ) p (5500 − 1500)
1
= 3.9788 × 10 −7
Kp ( f 2 − f1 )
= 0.39788 × 10 −6 F = 0.39788 µF
L2 =
K ( f 2 − f1 ) 200(5500 − 1500)
=
= 7.716 mH
4p f1 f 2
4p (5500)(1500)
C1 =
f 2 − f1
= 1.9291 × 10 −7 F = 0.1929 F
K 4p f1 f 2
Required design is given as shown in Figure 13.42.
(7.9575 mH) (0.3858 µF) (7.9575 mH)
L1
L1
2C1
2C1
2
2
(15.915 mH) (0.1929 µF)
C1
L1
(0.3858 µF)
(0.19894 µF)
(0.39788 µF)
C2
(7.716 mH)
L2
2L2
(15.432 mH)
C2
2
2L2
(0.19894 µF)
C2
2
(15.432 mH)
p -Section
T-section
Figure 13.42
Now, resonant frequency of series arm
=
1
2p L1C1
=
1
2p 15.915 × 10
Resonant frequency of shunt arm=
1
2p L2C2
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 636
−5
=
× 0.1929 × 10 −6
= 2872.43 Hz
1
−3
2p 7.716 × 10 × 0.39788 × 10 −6
= 2872.42 Hz
12/3/2014 8:21:32 PM
Filters and Attenuators
637
13.8.5 Constant K-type Band-Stop/Band-Elimination Filter
By interchanging the series and shunt arms of the band-pass-filter, we can obtain the band-stopfilter as has been shown in Figure 13.43(a) and (b).
Circuit Configuration
L1
2
2C1
2C1
L2
L1
2
L1
2
After
C2
L1
2
2C1
interchanging
series and
shunt arms
Band-pass filter
L2
2C1
C2
Band-stop filter
(a)
L1
L1
C1
After
2L2
C2
2
2L2
C2
interchanging
2
series and
shunt arms
Band-pass filter
2L2
C1
C2
2
2L2
C2
2
Band-stop filter
(b)
Figure 13.43 Band-stop Filter Developed from Band-pass Filter (a)
T-section; (b) p -section
That is, series elements have been connected in parallel and shunt elements have been placed in series.
Further, in this filter, the components, that is, L1, C1, L2 and C2 are so selected that their
resonant frequencies are same. This frequency is known as the resonant frequency or centre
frequency or rms frequency of the filter. So, we will have
1
1
1
1
=
= w 0 ; or w 02 =
=
L
C
L
L1C1
L 2C 2
2C 2
1 1
1
or,
L1C1 = L2C2 = 2 (13.39)
w0
For a band-stop filter, Z1 and Z2 can be calculated as:
jw L1
1
( jw L1 )
jw C1
jw C
1
Z1 = ( jw L1 )
=
= 2 2 1
1
( jw C1 )
j w L1C1 + 1
jw L1 +
jw C1
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 637
12/3/2014 8:21:34 PM
638 Network Analysis and Synthesis
\
and
Z1 =
Z 2 jw L 2 +
jw L1
1 − w 2 L1C1
(13.40)
j 2w 2 L 2C 2 + 1
1
=
jw C 2
jw C 2
1 − w 2 L2C2
(13.41)
jw C2
The various parameters of a band-stop filter are calculated as,
Design impedance (K ): Know that for constant K-type filters, Z1 Z2 = K2
Substituting the value of Z1 and Z2 from equations (13.40) and (13.41), we get the following
form:
Z2 =
∴

jwL1   1 − w 2 L2C2 
2

  jw C
=K
2
 1 − w L1C1  

2
Substituting, L1C1 = L2C2 from equation (13.39), in the above, we get
( jw L1 )(1 − w 2 L1C1 )
( jw C 2 )(1 − w 2 L1C1 )
= K2
or
L1
= K2
C2
From equation (13.39), L1C1 = L2C2
or
or K =
L1
C2
L1 L 2
=
C 2 C1
Substituting this value in the equation, we get the value of K.
K=
L1
=
C2
L2
C1
Cut-off frequency: Cut­-off frequencies can be obtained from the equation
Z 1 + 4 Z 2 = 0 or Z 1 = −4 Z 2
Multiplying both sides by Z1, we get Z12 = −4 Z1Z 2
Z1Z 2 = K 2
Now,
Case I: When Z1 = j2K (let us assume that it is at f1)
Then, using equation (13.40), we have the following form:
Z1 =
or
or
jw L1
2
1 − w L1C1
= j 2K
wL1 = 2K − 2KL1C1w 2
w=
or
w L1
1 − w 2 L1C1
or 2KL1C1w 2 + L1w − 2K = 0
− L1 ± L12 − 4( 2KL1C1 )( −2K )
2( 2KL1C1 )
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 638
= 2K
w=
− L1 ± L12 + 16K 2 L1C1
4KL1C1
12/3/2014 8:21:36 PM
Filters and Attenuators
− L1 ± L1 1 + 16 K 2
=
C1
L1
4 KL1C1
4 KC1
−1 + 1 + 16 K 2
2p f 1 =
or
C1
L1
−1 ± 1 + 16 K 2
=
639
C1
L1
4 KC1
(Taking positive value only)
−1 + 1 + 16 K 2
or Expression for lower cut-off frequency of BPF, f 1 =
C1
L1
8p kC1
Case II: When Z1 = - j2K (let us assume that it is at f2), then the following equation can be
obtained
jw L1
2
1 − w L1C1
or
= − j 2K
or
w L1
= −2K
1 − w 2 L1C1
w L1 = −2K + w 2 2L1C1K , that is, 2L1C1K w 2 − L1w − 2K = 0
From the above equation,
1 + 16 K 2
∴w = 1 ±
C1
L1
4C1K
w=
L1 ± L12 + 16 K 2 L1C1
4 L1C1K
1 ± 1 + 16 k 2
or 2p f 2 =
C1
L1
C1
L1
4C1K
1 + 1 + 16 K 2
f2 =
Therefore,
= L1 ± L1 1 + 16 K 2
C1
L1
8p C1 K
Considering the positive value only, we get f2 as follows:
1 + 1 + 16 K 2
Expression for higher cut-off frequency,
f2 =
C1
L1
8p C1 K
Attenuation (a ): We have
a = 2 cosh −1
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 639
Z1
4Z2
12/3/2014 8:21:39 PM
640 Network Analysis and Synthesis
Using equations (13.40) and (13.41), we get
a = 2 cosh −1
a = 2 cosh −1
Putting,
 jw L1 


2
 1 − w L1C1 
2
4(1 − w L 2C 2 )
jw C 2
( jw L1 )( jw C 2 )
4(1 − w 2 L1C1 )(1 − w 2 L 2C 2 )
L1C1 =
1
w 02
wC 2 )
( jw L1 )( jw
= 2 cosh −1
4( −w 2 L1C1 )(1 − w 2 L 2C 2 )
w 2 L1C 2
= 2 cosh −1
4(1 − w 2 L1C1 ) 2
since L1C1 = L 2C 2
from equation (13.39), we get
w 2 L1C2
a = 2 cosh −1
 w2
4 1 − 2 
 w0 
2
Phase shift (b ):
b = 2 sin −1
Resonant frequency ( f0):
We have,
Substituting values,
We have,
or
or
w 2 L1C2
Z1
= 2 sin −1
4Z2
 w2
4 1 − 2 
 w0 
2
Z1 at f1 = - Z1 at f2
jw1 L1
1 − w12 L1C1
w1
1 − w12 L1C1
=
w1
w2
1 − 12
w0
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 640
=
− jw 2 L1
1 − w 22 L1C1
−w 2
1 − w 22 L1C1
=
−w 2
w2
1 − 22
w0
∵ L1C1 =
1
w 02
12/3/2014 8:21:41 PM
Filters and Attenuators
 w 22  -w 2
1 − 2  =
w1
 w0 
or
w 02 − w 22
w 02
=
w 02 − w 22 =
or
 w12 
1 − 2 
 w0 
−w 2 (w 02 − w12 )
w1w 02
−w 2 2
(w 0 − w12 )
w1
or
w 02w1 − w 22w1 = −w 2w 02 + w 2w12
or
w 02 (w1 + w 2 ) = w1w 2 (w1 + w 2 )
w 02 = w1w 2
Therefore,
641
or w 0 = w1w 2
or f 0 = f 1 f 2
Characteristic impedance (Z0): We have
Z12
+ Z1Z 2
4
ZOT =
Using equations (13.40) and (13.41), we get the following
Z OT =
Zop =
4(1 − w 2 L1C1 ) 2
ZOT = K 2 −
Therefore
and
( jw L1 ) 2
Z1Z 2
K2
=
=
ZOT
ZOT
+ K2 =
−w 2 L12

w 
4 1 − 22 
 w0 
2
+ K2
w 2 L12
 w2
4 1 −

 w0 
2
K2
K2 −
w 2 L12
 w2
4 1 − 2 
 w0 
2
Design parameters:
We had,
-Z1 = j2K at f1
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 641
12/3/2014 8:21:43 PM
642 Network Analysis and Synthesis
jw1 L1
Therefore,
1 − w12 L1C1
w1 L1
or
1 − w12 L1C1
= 2K
w1 L1 = 2 K (1 − w12 L1C1 )
or
Putting
= j 2K
L1C1 =
1
w 02
we get
 w2
w1L1 = 2K 1 − 12 
 w0 
or
w 2 −w 2 
or w1L1 = 2K  0 2 1 
 w0 
 f 2 − f 2
w1 L1 = 2 K  0 2 1  (i)
 f0 
f02 = f1 f2
We had,
 f f − f 2  2Kf 1 (f 2 − f 1 )
w1L1 = 2K  1 2 1  =
f 1f 2
 f 1f 2 
Therefore,
2 K ( f 2 − f1 )
f2
or
w1 L1 =
or
2p f1 L1 =
or
L1 =
Further, we have
L1
=K
C2
or
So,
C2 =
Therefore
C2 =
we have
L1C1 =
2 K ( f 2 − f1 )
f2
K ( f 2 − f1 )
p f1 f 2
L1
= K2
C2
L1
1 K ( f 2 − f1 )
=
⋅
p f1 f 2
K2 K 2
f 2 − f1
Kp f1 f 2
1
w 02
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Filters and Attenuators
C1 =
or
1
w 02 L1
=
w 02
C1 =
C1 = ( 2p f )( 2p
1
( 2p f1 )( 2p
1
C1 =
1f −
4
p
K
(
C1 =
2
4p K ( f 2 −
or
So,
L2
=K
C1
and finally
L2 = K 2
So,
1
1 K (f 2 − f )
=
⋅
K (f 2 − f 1 ) w1w 2 p f 2 f 2
⋅
p f1 f 2
1
1
1 K( f − f ) =
1K ( f − f )
f 2 ) ⋅ K ( f 2 − f1 ) = 4p 2 f1 f 2 ⋅ K ( f 2 − f1 )
4p 2 f1 f 2 ⋅ p 2f1 f 2 1
f 2 ) ⋅ p 2f1 f 2 1
p f1 f 2
p f1 f 2
f1 )
f1 )
L2
= K2
C1
or
643
or
L2 = K 2C1
1
K
L2 =
4p K ( f 2 − f1 )
4p ( f 2 − f1 )
Summary of Band-Stop Filter
Circuit configuration:
L1
2
L1
2
2C1
L2
L1
2C1
2L2
C1
2L2
C2
2
C2
C2
2
p-Section
T-section
Figure 13.44 Band-stop Filter in T and p-sections
Design impedance:
K=
L2
L1
=
C1
C2
Cut-off frequencies:
−1 + 1 + 16 K 2
f1 (lower cut-off frequency) =
8p KC1
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 643
C1
L1
: and
12/3/2014 8:21:48 PM
644 Network Analysis and Synthesis
1 + 1 + 16 K 2
f2 (upper cut-off frequency) =
C1
L1
8p C1 K
Attenuation (a ):
w 2 L1C2
a = 2 cosh −1
 w2
4 1 − 2 
 w0 
2
Phase shift (b ):
b = 2 sin −1
w 2 L1C2
 w2
4 1 − 2 
 w0 
2
Resonant frequency ( f0): f 0 =
f1 ⋅ f 2
Characteristic impedance (Z0):
Z OT = K 2 −
w 2 L12
 w2
4 1 − 2 
 w0 
2
K2
; Z 0p =
K2 −
w 2 L12
 w2
4 1 − 2 
 w0 
2
Design parameters:
L1 =
K (f 2 − f 1 )
1
; C1 =
p f 1f 2
4p K (f 2 − f 1 )
L2 =
f −f
K
; C2 = 2 1
4p (f 2 − f 1 )
K p f 1f 2
Example 13.13 Design a passive constant K-type BSF having a design impedance of 200 Ω
and cut-off frequency 2000 Hz and 6000 Hz.
Solution: Given
K= 200 Ω;
f1 = 2000 Hz; f2 = 6000 Hz
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Filters and Attenuators
645
Now, for a constant K-type BSF, we have the following form:
L1 =
K ( f 2 − f 1 ) 200(6000 − 2000)
=
p f1f 2
p (6000)( 2000)
= 0.02122 H = 21.22 mH
C2 =
f2 − f2
= 5.30516 × 10 −7 F
Kp f1 f 2
= 0.53051 µF
L2 =
K
200
200
=
=
4p ( f 2 − f1 ) 4p (6000 − 2000) 4p ( 4000)
= 3.97 mH
C1 =
1
= 9.947 × 10 −8 F
K 4p ( f 2 − f 1 )
= 0.09947 µF
Required design is shown as in Figures 13.45 and 13.46.
L1
2
= 10.61 mH
L1
2
= 10.61 mH
2C1
= 0.1989 µF
2C1
= 0.1989 µF
L2 = 3.97 mH
C2 = 0.53051 µF
T-section
Figure 13.45
L1 = 21.22 mH
C1 = 0.09947 µF
2L2 = 7.94 mH
2L2 = 7.94 mH
C2
2
= 0.2652 µF
C2
2
= 0.2652 µF
p -Section
Figure 13.46
13.8.6 Comparison of Constant K-type Filters
After discussing all the constant K-type filters, we now present their circuit parameters in a
consolidated way in Table 13.3.
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12/3/2014 8:21:51 PM
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 646
2. Design
impendence (K)
C
L/2
K=
L
C
p -Section
C
2
L
T-section
L/2
LPF
C
2
2L
L
2C
K=
L
C
p -Section
C
T-section
2C
HPF
2L
Parameters of Constant K-type Filters
1. Circuit
Configuration
Parameters
Table 13.3
2L2
C2
L1
=
C2
p -Section
C2
2L2
2
L1
L2
C1
C1
L2
2C1
T-section
2C1
K=
L1
2
BPF
L1
2
C2
2
C2
2
2L2
L1
2
L2
L1
=
C2
p -Section
C1
L1
C2
2
2L2
L2
C1
C2 2C1
T-section
K=
2C1
L1
2
BSF
646 Network Analysis and Synthesis
12/3/2014 8:21:54 PM
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 647
4. Design
parameters
3. Cut-off
frequency (fc)
L=
C=
1
K p fc
C=
fC =
K
p fc
1
p LC
L=
fC =
L1
C1
2p L1
K2 +
2p L1
L2 =
C1 =
K ( f1 − f 2 )
4p f1 f 2
f1 − f 2
K ⋅ 4p f1 f 2
L2 =
C1 =
C2 =
1
K p ( f1 − f 2 )
C2 =
1
K 4p f c
L1 =
f2 (upper) =
f1 (lower) =
L1 =
K
p ( f1 − f 2 )
f2(lower) = − K +
f1 (upper) =
L1
C1
K
4p f c
4p LC
1
K + K2 +
C1
L1
C1
L1
(Continued )
1
4p K ( f 2 − f1 )
K
4p ( f 2 − f1 )
f 2 − f1
K p f1 f 2
K ( f 2 − f1 )
p f1 f 2
8p C1 K
1 + 1 + 16 K 2
8p KC1
−1 + 1 + 16 K 2
Filters and Attenuators
647
12/3/2014 8:21:57 PM
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 648
Z 0p =
f 
1−  
 fc 
K
2
f 
Z 0T = K 1 −  
 fc 
 f 
b = sin −1  
 fc 
6. Phase shift (b )
7. Characterstic
impedance (Z0)
 f 
a = 2 cosh −1  
 fc 
(Continued)
5. Attenuation (a )
Table 13.3
2
Z 0p =
f 
1−  
 fc 
K
2
f 
Z 0T = K 1 −  c 
f 
 f 
b = 2 sin −1  c 
 f 
 f 
a = 2 cosh −1  c 
 f 
2
Z 0p =
K2
Z 0T
Z 0T = K 1 −
b = 2 sin
−1
2
2
w 2 
 2
 w0 
2
4w 2 L2C1
 w2
1 − 2 
 w0 
4w 2 L2C1
4w 2 L 2C1
a = 2 cosh −1

w2 
1 − 2 
 w0 
Z 0p =
K2
Z 0T
Z 0T = K 2 −
b = 2 sin −1
a = cosh −1

w2 
4 1 − 2 
 w0 
w 2 L12

w2 
4 1 − 2 
 w0 
w 2 L1C2

w2 
4 1 − 2 
 w0 
w 2 L1C2
2
2
2
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649
13.8.7 Limitations of Constant K-type Filters
The constant K-type filters have mainly the following two limitations:
1. The characteristic impedance of the filter circuit does not remain constant over the pass
band and it is a function of frequency, that is, it varies with frequency.
We know filter gives ideal performance only if it is terminated by a resistance equal to
the characteristic impedance.
Since in constant K-type filters, characteristic impedance is different at different frequencies, a mismatch occurs.
2.Its attenuation does not rise abruptly beyond the cut-off frequency as shown in Figure 13.47
and Figure 13.48 for a low-pass filter.
a
(Attenuation)
∞
Pass
band
Stop
band
∞
Frequency
fc
O
a = 2 cosh−1 f
fc
It is a fx n of
frequency of signal
Figure 13.47 a Versus f Characteristic (Actual Case)
a
(Attenuation)
∞
Pass
band
Stop band
fc
Ideally
a = ∞ in
stop band
Frequency
Figure 13.48 a Versus f Characteristic (Ideal Case)
To overcome these limitations of constant K-type filters, m-derived filters are used.
13.9 m-DERIVED FILTERS
In m-derived filters, prototype or constant K-type filter is used.
Following modifications are done to obtain m-derived filters from constant K-type filters.
1. The series impedance is multiplied by a factor m
2. The shunt impedance is divided by a factor m
3. An additional impedance of the opposite sign is added either in series or in parallel
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The resonant frequency of the shunt arm is so chosen such that it is slightly higher than that of
the constant K-type filter.
f∞
> fc
( resonant frequency )
That is,
At the frequency f∞, the impedance of shunt branch is zero, that is, the attenuation is infinite.
This can be seen from the expression for attenuation as
Z1
4Z2
a = 2 cosh −1
At f = f∞, Z2 = 0 and hence a = ∞. This produces very sharp cut-off or attenuation.
Here, m is a constant and 0 < m < 1.
13.9.1 m-Derived T-section
Z1
2
As mentioned earlier, m-derived T-section as shown in Figure 13.50
can be obtained from constant K-type filter shown in Figure 13.49
by making the following modifications:
Z1
Z2 2
Figure 13.49 Constant
K-type
mZ1
2
mZ1
2
Z1
by m.
2
Z1
mZ1
That is, replace
by
.
2
2
2. Impedance Z2 is replaced by Z21, such that the value of Z0
(characteristic impedance) should be same for both the cases.
1. Multiply series impedance, that is,
Now, for constant K-type, T-section, we have the following:
Z 21
ZOT =
Figure 13.50 m
-Derived
T-Section
Z 1OT =
=
Z12
+ Z1Z 2 (13.42)
4
For m-derived T-section, we have the following:
( mZ 1 ) 2
+ ( mZ 1 )Z 12
4
[Multiply series impedance by m]
m 2 Z12
+ mZ1Z 21(13.43)
4
Now, equating the characteristic impedances,
1
ZOT = ZOT
That is,
Z12
+ Z1Z 2 =
4
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m 2 Z12
+ mZ1Z 21
4
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651
Z12
m 2 Z12
+ Z1Z 2 =
+ mZ1Z 21
4
4
or
Z12 m 2 Z12
−
+ Z1Z 2 = mZ1Z 21
4
4
or
mZ1Z 21 = (1 − m 2 )
or
Z12
+ Z1Z 2
4
Z2
(1 − m 2 ) Z12 Z1Z 2  1 − m 2 
+
=
 Z1 + m
mZ1
mZ1 4
m
4


Z  1 − m2 
Z 21 = 2 + 
Z1
Therefore,
m  4 m 
or
Z 21 =
The m-derived T-section is as shown in Figure 13.51. Also transformation from constant K-type
to m-derived type has been illustrated in Figure 13.52.
mZ1
2
mZ1
2
mZ1
2
mZ1
2
Z2
m
Z2′
1− m 2
Z1
4m
Figure 13.51 Final m-Derived T-Section
Multiple senies impedence by m
Z1
2
Z1
2
Z2
mZ1
2
Divide shunt impedance by m
Constant-K
T-section
mZ1
2
Z2
m
1 − m2
Z1
4m
Additional
impedance
added in
parallel
m-Derived T-section
Figure 13.52 Transformation from Constant K-type to m-Derived Type Filter
13.9.2 m-Derived p-Section
The constant K-type filter and the corresponding m-derived filter in p-sections are shown in
Figure 13.53.
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Z1′
Z1
2Z2
2Z2
m
2Z2
Constant K-type p -Section
2Z2
m
Equivalent m-derived p -Section
Figure 13.53 Constant K-type Filter and its m-Derived in p -section
Now, we have, for constant K-type filters, Z Op =
Z 1Op
and for m-derived filters,
Considering, Z Op =
or
or
or
or
Z 1Op
we get,
Z1Z 2
Z12
+ Z1Z 2
4
( Z1Z 2 ) 2
Z12
+ Z1Z 2
4
( Z1Z 2 ) 2
Z12
+ Z1Z 2
4
=
=
=
=
Z 1Z 2
(13.44)
Z 12
+ Z 1Z 2
4
Z 11
Z2
m
(13.45)
2
Z 11
Z
+ Z 11 2
4
m
Z11
Z2
m
Z12
Z
+ Z11 2
4
m
 1 Z2 
 Z1

m
2
2
Z11
Z
+ Z11 2
m
4
 1 Z2 
 Z1

m
2
1
Z1 
Z11Z 2  + 1 
 m 4Z2 
( Z1Z 2 )
( Z11Z 2 )
=
Z

Z11 
2 1
1+ 1
+
m
 m 4Z 
4Z2

2
Z1
Z11
=
Z
Z 1m 2
1+ 1
m+ 1
4Z2
4Z2
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mZ1 +
or
or
mZ1 = Z11 +
Therefore, Z11 =
or,
Z11 =
653
Z1Z11m 2
Z Z1
= Z11 + 1 1
4Z2
4Z2


Z Z1
Z
Z1Z11 Z1Z11m 2
= Z11 + 1 1 (1 − m 2 ) = Z11 1 + 1 (1 − m 2 ) 
−
4Z2
4Z2
4Z2
 4Z2

mZ1
1
1
=
=
Z
Z1
Z (1 − m 2 )
1
1+
(1 − m 2 ) 1 + 1 (1 − m 2 )
+ 1
4Z2
4Z2
mZ1
4 Z 2 mZ1
mZ1
1
 1 − m2  1
1
+
⋅
mZ1  4 m  Z 2
=
mZ1
1
1
1
+
mZ1  4 m 

 Z2
1 − m2 
 4m 
So, Z11 is the parallel combination of mZ1 and 
 Z 2.
1 − m 2 
Therefore, final m-derived p-Section is redrawn as shown in
Figure 13.54.
13.9.3 m-Derived Low-Pass Filter
2Z2
m
4m
1 − m2
Z2
2Z2
m
Figure 13.54 m-Derived
p -section
Filter
In a Low-pass filter, series element is inductor, that is, Z1 = jwL and shunt element is the capaci1
tor, that is, Z 2 =
jw C
1
Representing Z1 as L and Z2 as the circuit configurations are drawn as shown in Figures 13.55
C
and 13.56.
Circuit Configurations
mZ1
2
mZ1
2
mL
2
Z2
m
mL
2
mC
1 − m2
L
4m
1 − m 2 Z1
4m
T-section
Figure 13.55 m-Derived Low-pass Filter in T-section
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mZ1
2Z2
m
mL
4m
1 − m2
Z2
1 − m2
C
4m
mC
2
2Z2
m
General p -section m-derived
mC
2
p -Section for m-derived
p -Section
Figure 13.56 p-Section for m-Derived Low-Pass Filter
Analysis of m-derived LPF is given in the following:
Frequency of infinite attenuation ( f∞):
1
1
1
=
=
f∞ =
 1 − m2 
1− m2 
(1 − m 2 )
LC
p
2
⋅
p
mL
C
2
L
2p mC ⋅ 
 4m 

4


 4m 
=
We know
1
p LC
1
 2 1 − m2 

 p LC


2
=
1
p LC 1 − m 2
= f c (cut-off frequency for LPF)
f∞ =
fc
, that is, f ∞ > f c (13.46)
1 − m2
Equating both sides, equation (13.46) we get
Thus,
f ∞2 =
or
f c2
1 − m2
f2
f2
1 − m 2 = c2 or, m 2 = 1 − c2
f∞
f∞
Expression to find the value of m for m-derived LPF, is m = 1 −
Attenuation (a ):
For m-derived LPF,
a = 2 cosh −1
= 2 sinh
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−1
Z
4Z2
fc < f < f∞
Z1
4Z2
f > f∞
f C2
f ∞2
Z1 = jw L
1
jw C
[from T -section ]
Z2 =
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Now, for m-derived LPF,
Z1
=
4Z2
or,
mwL

 1
1− m
wL 
4
−
4m
 mw C

2
Z1
=
4Z2
=
=
=
=
mwL

1
1 − m2 
4wL 
−

4m 
 mw C (wL)
m

 Now for LPF,


1


fC =

p LC 


or f C2 = 1 

p 2 LC 


1
or
= p 2 f C2 
LC


 1
1 − m2 
4
−

2
4m 
 mw LC
m
4
−
mw 2 LC
m
1 − m2
m
4p 2 fC2 1 − m 2
−
mw
m
m
4p
2
fC2
m( 2p f ) 2
Therefore, Z1 =
4Z2
−
1− m
m
2
=
m
4p
2
f C2
2 2
m ⋅ 4p f
−
1− m
m
2
=
m
f C2
2
mf
−
1 − m2
m
m



2 
1 − (1 − m ) 
mf 
f c2 


f 2 

f c2
2
 f 
 f 
For LPF, m = 1 −  c  or, m 2 = 1 −  c 
 f∞ 
 f∞ 
2
2
 f 
or, 1 − m 2 =  c 
 f∞ 
2
Substituting the value of (1-m2),
Z1
=
4Z 2
m2
1−
 f 
 f 
c
2
1− m2
 fc 
 
f
2
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 655
m2
=
 f 
 f 
c
2
2
f 
f 
1−  c  ÷  c 
f 
 f∞ 
2
=
 f 
m2  
 fc 
2
 f 
1−  
 f∞ 
2
(13.47)
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656 Network Analysis and Synthesis
m
Z1
=
4Z2
 f 
 f 
c
 f 
1−  
 f∞ 
Alteration Constant, a = 2 cosh −1
(13.48)
2
 f 
m 
 fc 
Z1
= 2 cosh −1
4Z2
 f 
1−  
 f∞ 
2
where f C < f < f ∞
Phase shift (b ) is calculated as,
b = 2 sin −1
Z1
4Z2
Z1
from equation (13.48) in the equation, we get
4Z2
Now, substituting the value of
m
b = 2 sin −1
f
fc
 f 
1−  
 f∞ 
2
Variation of attenuation (a ) and phase shift (b ) with frequency are shown in Figure 13.57.
m-Derived
p
a
Attenuation
0
Constant-K
fc
b
0
f∞
f
fc
f∞
f
Figure 13.57 Shows the Variation of Attenuation and Phase Shift of m-Derived Filter
Therefore, m-derived LPF produces very sharp attenuation but the attenuation falls off beyond
f∞. This is the disadvantage of m-derived filters.
Characteristic impedance: For T-section, the following equation can be obtained:
ZOT =

Z12
Z 
+ Z1Z 2 = Z1Z 2 1 + 1 
4
 4Z2 
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Using equation (13.47), we have the following form:
ZOT

f2 
m2 2 

fC 
= Z1Z 2 1 +
2
 1−  f  
 f  


∞
ZOT

f2 
m2 2 

fC 
= K 2 1 +
2
 1−  f  
 f  


∞
m2
ZOT = K 1 +
or,
Putting Z1Z 2 = K 2
f2
f C2
 f 
1−  
 f∞ 
2
For p-Section, we get the following form:
ZOp =
K2
Z1Z 2
=
ZOT
m2 ⋅
K 1+
f
f C2
 f 
1−  
 f∞ 
K
=
2
m2
1+
2
f2
f C2
 f 
1−  
 f∞ 
2
13.9.4 Summary of m-Derived Low-Pass Filter
Circuit configurations:
1−m2
C
4m
mL
2
mL
2
mL
mC
mC
2
mC
2
1−m2
L
4m
T-section
p -Section
Figure 13.58 m-Derived Low-pass Filters in T and p -Sections
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Expression for m:
2
 f 
m = 1−  C  ;
 f∞ 
where fC → cut-off frequency
0 < m < 1 f∞ → frequency of infinite attenuation
Cut-off frequency:
fC =
1
= 1 − m2 f∞
p LC
Design parameters:
L=
K
p fC
C=
1
Kp fC
Attenuation (a ):
m
a = 2 cosh −1
f
fC
 f 
1−   2
 f∞ 
Phase shift ( b ):
m
b = 2 sin −1
f
fC
 f 
1−   2
 f∞ 
Characteristic impedance:
m2
ZOT = K 1 +
 f 
1−  
 f∞ 
2
K
ZOp =
m2
1+
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 658
f2
f C2
f2
f C2
 f 
1−  
 f∞ 
2
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659
13.9.5 m-Derived High-Pass Filter
Circuit configurations:
4m
1−m2
2C
m
2C
m
L
m
L
C
m
2L
m
2L
m
4m
C
1−m2
p -Section
T-section
Figure 13.59 m-Derived High-pass Filters in T and p -sections
Expression for f∞: At f∞, the transmission through filter is zero and attenuation is infinite.
Now,
f∞ =
1
 L   4m 
2p   
C
 m   1 − m 2 
1
=
2p
4 LC
1 − m2
=
(1 − m 2 )
2p 2 LC
for T -section
=
1 − m2
4p LC
f∞ = 1 − m2 fc
Now, for constant K-type HPF, we get the following form:
fc =
1
4p LC
That is, f∞ < fC Expression for m: Now, we have
f ∞ = 1 − m 2 f c or, f ∞2 = (1 − m 2 ) f c2
2
or
 f∞ 
 f∞ 
2
2
 f  = 1 − m or, m = 1 −  f 
C
c
2
 f 
m = 1−  ∞ 
 fC 
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Attenuation (a ):
Firstly, let us find from T-section:
Z1 =
Therefore,
w L 1 − m2
m
; Z2 =
−
wC
m 4 mw C
m
Z1
wC
=
4Z2
 wL 1 − m 2 
−
4

 m 4 mw C 
=
=
m
wC
4  w 2 LC 1 − m 2 
−
4 m 
w C  m
m
 w LC 1 − m 2 
−
4
4 m 
 m
2
Z1
m
=
4 Z 2 4w 2 LC 1 − m 2
−
m
m
or
=
=
=
=
=
m2
4w 2 LC − (1 − m 2 )
m2
4( 2p f ) 216p 2 f C2 − (1 − m 2 )
m2
4.4p f 2
(1 − m 2 )
16p 2 f C2
2
m2
2
 f 
2
 f  (1 − m )
C
m2
2

 Now for HPF


1


fC =

4p LC 


1
or f C2 =


p 216 LC 


1
or LC =


16p 2 f C2 
 f   f∞ 
 f   f 
C
C
2
=
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 660

 Further, for HPF,


2

 f∞  
m
1
=
−

 f  


c

2 
 f  

m2 = 1 −  ∞  
or
 fc  



2
or 1 − m 2 =  f ∞ 

 f 


c


m2
 
2
 f   
 f  1 −
C
 
 

2
f∞  

f C  
2
f  
f C  
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Filters and Attenuators
Z1
=
4Z2
Z1
=
4Z2
or
Therefore, attenuation a = 2 cosh −1
 f 
m2  C 
 f 
2
 f 
1−  ∞ 
 f 
f
m C
f
2

1− 

(13.49)
f∞ 
f 
(13.50)
2
Z1
4Z2
= 2 cosh
661
m
−1
fC
f
 f 
1−  ∞ 
 f2 
2
Phase shift ( b ): We have
Z1
4Z2
b = 2 sin −1
Using equation (13.50), we get the value of b.
m
b = 2 sin −1
fC
f
 f 
1−  ∞ 
 f 
2
The variations of a and b with frequency have been shown in Figure 13.60.
Stop band
m-Derived
0
Constant-K
a
0
f∞
b
-p
fc
f
f∞
fc
f
Figure 13.60 Variation of Attenuation and Phase-shift with Frequency
Characteristic impedance (Z0): We have the following form:
ZOT =

Z12
Z 
+ Z1Z 2 = Z1Z 2 1 + 1 
4
 4Z2 
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Using equation (13.49), we have
ZOT
2 

 f 

m2  C  
 f  

= Z1Z 2 1 +
2
 f∞  

1−   

 f  

Now, Z1Z2 = K2
ZOT
2
2 

 f 
 f 

m2  C 
m2  c  
 f 
 f  

= K 1+
and
= K 2 1 +

2
2

 f∞ 
 f∞  
1−  
1−   

 f 
 f  

K2
Z1Z 2
=
Z 0T
ZOp =
 f 
m  C
 f 
2
2
k 1+
 f 
1−  ∞ 
 f 
2
K
=
2
f 
m  c
 f 
1+
2
 f 
1−  ∞ 
 f 
2
13.9.6 Summary of m-Derived HPF
Circuit configurations:
4m
2C
m
1−m2
2C
m
L
m
2L
m
4m
1−m2
L
C
m
2L
m
C
T-section
p -Section
Figure 13.61 m-Derived High-pass Filters in T and p -sections
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Expression for m:
 f 
m = 1−  ∞ 
 fc 
2
Cut-off frequencies:
fc =
f∞
1− m
2
=
1
4p LC
Component values:
L=
1
K
;C =
4p f c
K 4p f c
Attenuation (a ):
a = 2 cosh −1
m
 f 
1−  ∞ 
 f 
Phase shift (b ):
b = 2 sin −1
fC
f
m
2
fC
f
 f 
1−  ∞ 
 f 
2
Characteristic impedance:
m2 ⋅
ZOT = K 1 +
f C2
f2
 f 
1−  ∞ 
 f 
2
K
; ZOp =
1+
 f 
m2 ⋅  C 
 f 
2
 f 
1−  ∞ 
 f 
2
13.9.7 Comparison of m-Derived LPF and HPF
After discussing m-Derived LPF and HPF, we are now in a position to present these filters with
their design parameters in a consolidated way as in Table 13.4.
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Table 13.4
Design Parameters of m-derived Low-pass and High-pass Filters
Parameter
m-Derived LPF
Circuit configuration
m-Derived HPF
mL
2
mL
2
2C
m
2C
m
L
m
mC
1−m2
L
4m
4m
1− m 2
T-section
T-section
1−m2
C
4m
4m
1− m 2
mL
mC
2
mC
2
2
 f 
m = 1−  c  ;
 f∞ 
Cut-off frequency
fc
f∞ =
2
;
1− m
that is, f ∞ > f c
fc =
1
p LC
2
= 1 − m f∞
Component values
2L
m
p -Section
2
 f 
m = 1−  ∞  ;
 fc 
0 < m <1
Expression for resonant
frequency f∞
L
C
m
2L
m
p -Section
Expression for m
C
0 < m <1
f ∞ = 1 − m 2 f c ; that is,
f∞ < fc
fc =
=
1
4p LC
f∞
1 − m2
L=
K
p fc
L=
K
4p f c
C=
1
Kp fc
C=
1
K 4p f c
(Continued )
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Table 13.4
(Continued)
Parameter
Attenuation (a )
Phase shift ( b )
m-Derived LPF
m-Derived HPF
f


 m. f

C

a = 2 cosh −1 
2

 f  
 1−

 f  

∞

fC


 m. f

−1

a = 2 cosh 
2 

f
 
 1−  ∞  
 f  


f
 m.
fC

b = 2 sin −1 
 1−  f 
 f 

∞
Characteristic impedance
Z OT = K
f 
1+ m  
 fC 
 f 
1−  
 f∞ 






2
Z OT = K 1 +
2
K
Z op =
f 
m  
 fC 
2
 f 
1−  
 f∞ 
2
2
1+


 m ⋅ fC 


f
b = 2 sin −1 

2

 f∞  
 1 −  f  


f 
m2  C 
f 
2
f 
1−  ∞ 
 f 
2
K
Z op =
1+
f 
m2  C 
f 
2
f 
1−  ∞ 
 f 
2
13.9.8 m-Derived Band-Pass Filter
Circuit configuration:
mL1
2
mL1
2
2C1
m
2C1
m
mC2
L2
m
1− m 2
L1
4m
4m C
1
1− m 2
T-section of m-derived LPF
T-section of m-derived HPF
Figure 13.62 m
-Derived Band-pass Filters Obtained by Cascading m-Derived LPF
and HPF in T-sections
After connecting in cascade T-section of m-derived low-pass filter and T-section of m-derived
high-pass filter, we will obtain the m-derived band-pass filter as has been shown in Figure 13.63.
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2C1
m
mL1
2
2C1
m
mL1
2
1− m 2 L
1
4m
4m C
1
1− m 2
L2
m
mC2
Figure 13.63 T-Section of m-Derived BPF
After cascading T-section of m-derived low-pass filter with p-section of m-derived low-pass filter as
shown in Figure 13.64(a), we get a p-section band-pass filter as has been shown in Figure 13.64(b).
1− m 2 C
2
4m
4m L
2
1− m 2
mL1
mC2
2
C1
m
2L2
m
mC2
2
2L2
m
p-Section of
m-derived LPF
T-section of
m-derived LPF
(a)
mL1
mC2
2
2L2
m
1− m 2 C
2
C1 4m
m
4m L
2
1− m 2
mC2
2
2L2
m
p -Section of BPF
(b)
Figure 13.64 A
Band-pass Filter in p-section as in Figure (b) is Formed by Cascading
T-section and p-section of m-Derived Low-pass Filters as Shown in Figure (a).
Various useful results for m-derived BPF are as follows:
Expression for m:
m = 1−
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( f 2 − f1 )
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( f ∞ 2 − f ∞1 ) 2
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where f2 (upper cut-off frequency) and f1 (lower cut-off frequency)
f∞ 2 
 are the frequency of infinite attenuation.
f ∞1 
Component values:
L1 =
f −f
K ( f 2 − f1 )
1
K
;C =
;L =
; C1 = 2 1
p ( f 2 − f1 ) 2 K p ( f 2 − f1 ) 2
4p f1 f 2
K 4p f1 f 2
13.9.9 m-Derived Band-Stop Filter
m-derived band-stop filters can be made from m-derived band-pass filters both in T-section and
p-section after changing their series and shunt arms as shown in Figure 13.65 and Figure 13.66,
respectively.
mL1
2
2C1
m
2C1
m
mL1
2
mL1
2
1− m 2 L
1
4m
4m C
1
1− m 2
L2
m
mC2
mL1
2
2C1 1− m 2
L1
4m
m
After changing
series
and
shunt
arms
2C1
m
4m C
1
1− m 2
L2
m
mC2
T-section of BPF
T-section of BSF
T-section
Figure 13.65 m-Derived Band-stop Filter in T-section Obtained from T-section Band-pass Filter
mL1
mC2
2
2L2
m
C1
m
1− m 2 C
2
4m
4m L
2
1− m 2
mC2
2
p -Section of m-derived BPF
mL1
2L2
m
2L2
m
mC2
2
p -Section
C1
m
1− m 2 C
2
4m
4m L
2
1− m 2
2L2
m
mC2
2
p -Section of m-derived BSF
Figure 13.66 m
-Derived Band-stop Filter in p-section Obtained from p -section of
m-Derived Band-pass Filter
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Some useful results for m-derived BSF are as follows:
 f − f ∞1 
m = 1 −  ∞2
 f 2 − f1 
2
Component values:
L1 =
f −f
1
K
K ( f 2 − f1 )
;C =
;C = 2 1
; L2 =
4p ( f 2 − f1 ) 1 4p K ( f 2 − f1 ) 2 K p f1 f 2
p f1 f 2
13.10 COMPOSITE FILTERs
We have just studied constant K-type and m-derived filters. It has been observed that a constant
K-type filter does not give sharp cut-off/attenuation. We have also seen that an m-derived filter
has a sharp-cut-off or attenuation, but its attenuation decreases for frequencies beyond f∞. The
decrease in attenuation beyond f∞ is a limitation of m-derived filters that can be overcome by
cascading the m-derived filter with that of constant K-type filter having a rising attenuation
beyond cut-off.
Attenuation characteristics of constant K-type filter cascaded with m-derived filter are shown
in Figure 13.67.
Comps
ite filter
a
(Attenuation)
Constant-K
a=0
Pass
band
O
m-Derived
fc
f∞
Frequency
(For LPF)
Figure 13.67 Attenuation Characteristics (a versus f) of Constant K-type Filters
Therefore, the composite filter is a filter that consists of a number of m-derived and prototype sections with two terminating half sections. Terminating half sections are used for proper
impedance matching. Terminating half sections are used at both source end and load end with
m = 0.6.
A composite filter consists of the following:
1. One or more constant K-type sections to produce cut-off between the pass band and the
stop band at a specified frequency fc
2. One or more m-derived sections to given infinite attenuation at a frequency f∞ near to fc
3.Two terminating m-derived half section with m = 0.6 to provide matching with the source
and the load.
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Block diagram representation of a composite filter is shown in Figure 13.68.
Input
Terminating
half
section
Prototype
section
Terminating
half
Output
section
m-Derived
section
with
m = 0.6
with
m = 0.6
Figure 13.68 A Composite Filter Shown in Block Diagram Form
13.10.1 Composite Low-Pass Filter
Figure 13.69 shows a composite low-pass filter T-section which consists of two terminating half
sections at both source and load, one constant K-type section, and one m-derived section. Figure
13.70 shows a similar composition of a composite low-pass filter in p-section.
mL
2
L
2
L
2
mC
2
Constant-K
LPF
mL
2
m′L
2
C
1− m 2 L
2m
Terminating
half section
with m = 0.6
m′L
2
m′C
mC
2
1− m′2 L
4m′
1− m 2 L
2m
Terminating
half section
m = 0.6
m-Derived LPF
f
m′ = 1 − c
f∞
T-section
2
Figure 13.69 Composite LPF in T-section
1− m′2 C
4m′
1− m 2 C
2m
1− m2 C
2m
L
m′L
mL
2
mC
2
C
2
Terminating
half section
with m = 0.6
m′C
2
C
2
Constant-K
LPF
m-Derived section
m′ = 1−
p -Section
mL
2
m′C
2
Fc
F∞
2
mC
2
Terminating
half section
with m = 0.6
Figure 13.70 Composite LPF in p-section
Procedure to Design Composite LPF
Step 1: Design constant K-type section.
For this, find L =
K
1
and C =
p fc
K p fC
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and design T- or p-section using following configurations as shown in Figure 13.71.
L
2
L
2
L
C
2
C
C
2
p-Section
T-section
Figure 13.71 Constant K-section of the Composite Filter in T and p-sections
Step 2: Design m-derived sections.
For this, firstly find m′ = 1 − ( f c /f ∞ ) 2 , and then, find the component values and hence design
T- or p-Section as shown in Figure 13.72.
m′L
2
1− m′2
C
4m′
m′L
2
m′L
m′
m′C
2
1− m′2
L
4m′
m′C
2
p -Section
T-section
Figure 13.72 m-Derived Sections of the Composite Low-pass Filter in T and p-sections
Step 3: Design terminating half section using m = 0.6 as shown in Figure 13.73.
1− m 2
C
2m
mL
2
mC
2
mL
2
mC
2
1− m 2
L
2m
Terminatly half
section for T-type LPF
Terminatly half
section for p-type LPF
Figure 13.73 Terminating Half sections of Composite Low-pass Filter in T and p-sections
Step 4: Assemble all the sections of the designed composite filter.
13.10.2 Composite High-Pass Filter
On the same line of designing composite LPF, a high-pass composite filter can be developed in
both T and p-sections as shown in Figures 13.74 and 13.75, respectively.
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2C
m
2C
2C
2L
m
2C
m′
2C
m′
2C
m
L
m′
L
2L
m
4m′
C
1−m′2
2m C
1− m 2
Terminating
Constant-K T-type m-Derived T-type
half section
HPF
HPF
for T-type
L= K
F 2
4pfc
m′ = 1− ∞
HPF (m = 0.6)
Fc
C= 1
K4pfc
T-section
2m C
1− m 2
Terminating
half section
for T-type
HPF (m = 0.6)
Figure 13.74 A Composite High-pass Filter in p-section
4m′
L
1−m′2
2m
L
1− m 2
2L
m
C
2C
m
C
m′
2L
2L
m′
2L
p-Type HPF
(constant-k)
Terminating half
section for p-type
HPF (m = 0.6)
2m
L
1−m 2
p-Type HPF
(m-Derived)
p -Section
2C
m
2L
m′
2L
m
Terminating half
section for p-type
HPF (m = 0.6)
Figure 13.75 A Composite High-pass Filter in p-section
13.11 ADDITIONAL SOLVED NUMERICALS ON FILTERS
13.11.1 Problems on m-Derived Low-pass Filters
Example 13.14 An m-derived LPF has a cut-off frequency of 2000 Hz. If m = 0.3, find the
frequency of infinite attenuation.
Solution: Given fc = 2000 Hz
m = 0.3
Now, for m-derived LPF, we get the following equation:
f∞ =
fc
f∞ =
fc
1 − m2
2000
2000
2000
2000 =
f2000
=
= 2096.56 Hz
∞ = =
2096
56
f
=
=
=
.
Hz
Substituting values, ∞
0.91 0.9539
1 −0(.9539
0.3) 2
2
.
0
91
1 − (0.3)
1 − m2
2000
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Example 13.15 An m-derived LPF has fc = 1500 Hz and f∞ = 2000 Hz. Find the parameter.
Solution: Given
fc = 1500 Hz; f∞ = 2000 Hz
Now, for m-derived LPF, we have the following equation:
 f 
m = 1−  c 
 f∞ 
2
2
 1500 
Substituting values, m = 1 − 
= 1 − (0.75) 2 = 0.4375 = 0.6614
 2000 
Example 13.16 Design the T-section and p-section of m-derived LPF having a design
impedance of 500 W, cut-off frequency 2200 Hz and frequency of infinite attenuation of 2500 Hz.
Solution: Given
K = 500 W
fc = 2200 Hz
f∞ = 2500 Hz
Now, for m-derived LPF, we can write the equation as follows:
2
2
 f 
 2000 
m = 1−  c  = 1− 
= 1 − (0.88) 2 = 0.2256 = 0.4749
 2500 
 f∞ 
Now, for LPF, we can calculate L and C as follows:
L=
K
500
=
= 0.07234 H = 72.34 mH
p f c p ( 2200)
C=
1
= 2.8937 × 10 −7 F
Kp fc
= 0.28937 × 10 −6 F
= 0.28937 µF.
Now, the design of T-type m-derived LPF and component values are as follows:
mL 0.4749(72.34)
=
= 17.17 mH
mL = 17.17 mH mL = 17.17 mH
2
2
2
2
mC = 0.4749(0.28937)
= 0.1374 µF
mC = 0.1374 µF
 1 − m2 
 1 − 47492 
 4 m  L =  4(0.4749)  × 72.34 = 29.49 mH




1− m 2
L = 29.49 mH
4m
Figure 13.76
The required T-section is shown in Figure 13.76
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The design of p-type m-derived LPF and the component
values are given as follows:
mL = 2 × 17.17 = 34.34 mH
673
1− m2
C = 0.1179 µF
4m
mC 0.1374
=
= 0.0687 µF
2
2
 1 − m2 
 4 m  C = 0.1179 µF


mL = 34.34 mH
mC
2
(0.0687 µF)
The required p-Section is shown in Figure 13.77.
mC
2
(0.0687 µF)
Figure 13.77
Example 13.17 Design an m-derived LPF with
T-section having cut-off frequency of 7.2 kHz and infinite attenuation at 7.5 kHz and design
impedance of 500 W.
Solution: Given
fc = 7.2 kHz = 7200 Hz
f∞ = 7.5 kHz = 7500 Hz
K = 500 W
Now, for LPF, we calculate the value of L and C as follows:
L=
K
500
=
= 0.02210 H = 22.10 mH
p f c p (7200)
C=
1
= 8.84 × 10 −8 = 0.0884 × 10 −16 F = 0.0884 µF
Kp fc
and
2
2
 f 
 7200 
m = 1−  c  = 1− 
= 0.28
 7500 
 f∞ 
Now, let us find the component values of T-section m-derived LPF as in the following:
mL 0.28( 22.10)
=
= 3.094 mH
2
2
mC = (0.28)(0.0884)
= 0.024752 µF
 1 − m2 
 4 m  L = 18.18 mH


The required design of m-derived low-pass filter in
T-section is shown in Figure 13.78.
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mL = 3.094 mH mL = 3.094 mH
2
2
mC = 0.024752 µF
1− m 2
L = 18.18 mH
4m
Figure 13.78
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674 Network Analysis and Synthesis
13.11.2 Problems on m-Derived High-pass Filters
Examples 13.18 An m-derived HPF has fc = 1400 Hz and f∞ = 1300 Hz. Find the value of m.
Solution: Given
fc = 1400 Hz
f∞ = 1300 Hz
Now, for m-derived HPF, m can be calculated as follows:
We know,
f 
m = 1−  ∞ 
f 
2
C
2
 1300 
Substituting values, m = 1 − 
= 1 − (0.9285) 2 = 1 − 0.8622 = 0.1377 = 0.3711
 1400 
Example 13.19 An m-derived HPF has fc = 1700 Hz and m = 0.6. Find the frequency of
infinite attenuation.
Solution: Given
fc = 1700 Hz; m = 0.6
Now, for HPF, f∞ can be calculated as follows:
f ∞ = 1 − m 2 f c = 1 − 0.6 2 (1700) = 0.8(1700) = 1360 Hz
Example 13.20 Design m-derived HPF having a design impedance of 300 W, cut-off
frequency of 2000 Hz and frequency of infinite attenuation of 1700 Hz.
Solution: Given
K = 300 W; fc = 2000 Hz; f∞ = 1700 Hz
Now, for m-derived HPF, the following can be obtained:
2
2
f 
 1700 
m = 1−  ∞  = 1− 
= 0.5227
 2000 
 fc 
K
300
L=
=
= 0.047746 H = 47.75 mH
p f c p ( 2000)
1
C=
= 5.305 × 10 −7 F = 0.5305 µF
Kp f c
T-section:
Component values for T-section are as follows:
2c 2(0.5305)
=
= 2.01 µF
m
0.527
L 47.75
=
= 90.60 mH
m 0.527
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Filters and Attenuators
2C = 2.01 µF
m
 4m 
 4(0.527) 
C=

 × 47.75 = 2.9185 × 0.5305
2

1− m
 1 − 5272 
= 1.548µF
Therefore, the required design is shown in Figure 13.79.
p-Section:
Firstly, let us find component values as in the following:
675
2C = 2.01 µF
m
L
m = 90.60 mH
4m
C = 1.548 µF
1− m 2
Figure 13.79
2L
= 2(90.60) = 181.2 mH
m
C 2.01
=
= 1.005 µF
m
2
 4m 
 4(0.527) 
L=

 × 47.75 = 139.36 mH
2

1− m
1 − 0.5272 
4m L = 139.36 mH
1− m 2
The required design is shown in Figure 13.80.
13.11.3 Problems on Composite Filters
Example 13.21 Design a composite LPF to work
into 150 W with cut-off frequency of 1.5 kHz and very
high attenuation at 2 kHz.
2L
m
(181.2 mH)
C
m = 1.005 µF
2L
m
(181.2 mH)
Figure 13.80
Solution: Given
K = 150 W; fc = 1.5 kHz = 1500 Hz; f∞ = 2000 Hz
Step 1: Let us design constant K-type filter
Now, for constant K-type LPF, component values are as follows:
L=
K
150
=
= 0.03183 H = 31.83 mH
p f c p (1500)
C=
1
1
=
= 3.55367 × 10 −6 F = 0.3537 µF
K p f c 150(p )(1500)
Therefore, designs are as shown in Figure 13.81:
L
2
L
2
(15.915 mH)
(15.915 mH)
L (31.83 mH)
C
(0.3537 µF)
C
2
(0.17685 µF)
T-section
C
2
(0.17685 µF)
p -Section
Figure 13.81
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676 Network Analysis and Synthesis
m′L
2
= 10.5 mH
m′L
2
= 10.5 mH
Step 2: Let us design m-derived LPF and for m-derived
LPF.
2
1− m′2
L = 6.804 mH
4m′
T-section: Component values are as follows.
m ′L 0.66(31.83)
=
= 10.50 mH
2
2
Figure 13.82
1− m′2
4 m′
m′C = 0.66 (0.3537) = 0.2334 mH
 1 − m′2 
1 − 0.66 2 
L
=

 × 31.83
 4m ′ 


 4(0.66) 
C = 0.0756 µF
m′L = 21 mH
m′C
2
(0.1167 µF)
2
 f 
 1500 
m′ = 1 −  c  = 1 − 
= 0.66
 2000 
 f∞ 
m′C = 0.2334 µF
m′C
2
(0.1167 µF)
= 0.21378 × 31.83 = 6.804 mH
The circuit designed is shown in Figure 13.82.
p-Section:
Component values of the p-section as in Figure 13.83 are
as follows:
m1L = 2(10.50) = 21 mH
m 1C 0.2334
=
= 0.1167 µF
2
2
Figure 13.83
 1 − m 12 

 C = 0.21378 × 0.353
 4m1 
mL = 9.5495 mH
2
1− m 2
L = 16.976 mH
2m
mC = 0.1061 µF
2
= 0.0756 µF
Step 3: Let us design terminating half sections.
Terminating half section for T-type as shown in Figure
13.84: For terminating half sections, m = 0.6
mL 0.6(31.83)
=
= 9.549 mH
2
2
Figure 13.84
1− m2 
1 − 0.6 2 
L
=

 × 31.83 = 16.976 mH
 2m 


 2(0.6) 
mC 0.6(0.3537)
=
= 0.10611 µF
2
2
mL = 9.549 mH
2
1− m 2
2m
C = (0.18864 µF)
mC = (0.10611 µF)
2
Figure 13.85
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Terminating half section for p-type as shown in
Figure 13.85: Now,
1 − 0.6 2 
1− m2 
=
C

 × 0.3537
 2m 


 2(0.6) 
= 0.18864 µF
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Filters and Attenuators
677
Step 4: Now assemble the composite low-pass filter in T-section as shown in Figure 13.86 and
in p-section as shown in Figure 13.87.
15.91
mH
9.549 mH
15.91
mH
10.5 mH 10.5 mH
9.549 mH
0.2334 µF
16.976 mH
0.3537 µF
6.804 mH
0.1061 µF
Terminating half
section of
m-derived T-section
with m = 0.6
T-section of
constant-K
LPF
16.976 mH
0.1061 µF
Terminating half
section of
m-derived T-section
with m = 0.6
T-section of
m-derived LPF
m′ = 0.66
T-section
Figure 13.86
0.0756 µF
0.18864 µF
0.18864 µF
31.83 mH
9.549 mH
0.10611
µF
Terminating
half section of
m-derived p-Section
with m = 0.6
0.17685
µF
0.17685
µF
p-Section of
constant-K
LPF
21 mH
0.1167
µF
9.549 mH
0.1167
µF
p-Section of
m-derived
LPF m′ = 0.6
p -Section
0.10611
µF
Terminating
half section of
m-derived p-Section
with m = 0.6
Figure 13.87
Example 13.22 Design and draw the T-section of a composite LPF having the following
specifications:
1. Required cut-off frequencies: 2 kHz; 2. Design impedances: 500 W
Assume suitable value of m for m-derived full section and half section.
Solution: Given
fc = 2000 Hz
K = 500 W
Step 1: Let us design constant K-type LPF.
For constant K-type LPF, the value of L and C can be calculated as follows:
L=
K
500
=
= 0.07957 H = 79.57 mH
p f c p ( 2000)
C=
1
= 3.183 × 10 −7 f = 0.3183µF
Kp fc
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678 Network Analysis and Synthesis
L
2
Therefore, required T-section of constant K-type LPF is shown
in Figure 13.88
L
2
(39.785)
mH
(39.785) mH
Step 2: Design m-derived LPF (T-section)
C
(0.3183 µF)
For m-derived full section, if value of m′ is not given to us, then
it is assumed as 0.4.
Therefore, let m′ = 0.4
Now,
Figure 13.88
m′L
2
m ′L 0.4(79.57 mH)
=
= 15.914 mH
2
2
m ′C = 0.4(0.3183 µF) = 0.12732 µF
m′L
2
(15.914 mH)
(15.914 mH)
m′C (0.12732 µF)
1− m′2
4m′
 1 − m 12 
1 − 0.4 2 
L
=

 × 79.57 = 41.77 mH


 4m1 
 4(0.4) 
L = 41.77mH
Therefore, required T-section of m-derived LPF filter is
shown in Figure 13.89.
Figure 13.89
Step 3: Design the terminating half section of T-section
m-derived LPF with m = 0.6 as shown in Figure 13.90.
Component values are as follows:
mL (23.871mH)
2
1− m2
(42.43 mH)
2m
mL 0.6(79.57)
=
= 23.871 mH
2
2
mC 0.6(.3183)
=
= 0.09549 µF
2
2
1− m2 
1 − 0.6 2 
 2m  L =  2(0.6)  × 79.57




= 42.43mH
mC (0.09549 µF)
2
Figure 13.90
Step 4: Assemble the designed composite filter as shown in Figure 13.91.
23.871mH
42.43 mH
39.785
mH
39.785
mH
15.914
mH
0.12732
µF
0.318 µF
41.77 mH
0.09549
µF
Terminating
half section
with m = 0.6
Constant-K
LPF
15.914
mH
m-Derived section
with m′ = 0.66
23.871
mH
42.43 mH
0.09549
µF
Terminating
half section
with m = 0.6
Figure 13.91
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679
Example 13.23 Design a composite HPF to work into 500 W resistances with cut-off
frequency of 1500 Hz and with high attenuation of 1400 Hz.
Solution: Given
K = 500 W; fc = 1500 Hz; f∞1 = 1400 Hz; f∞2 = 1200 Hz
Now, for m-derived FPF, the following can be calculated:
2
2
f 1 
 1400 
m1 = 1 −  ∞  = 1 − 
= 0.359

 1500 
 fc 
2
2
f 2
 1200 
m2 = 1 −  ∞  = − 1 − 
= 0.6
 1500 
 fc 
Therefore, clearly, for full m-derived section, we will use m = 0.359 and for terminating half
section of m-derived filter, we will use m = 0.6
Step 1: Design constant K-type HPF.
For HPF, L and C can be calculated as follows:
2C
500
K
= 0.02653 H = 26.53 mH
=
4p f c 4p (1500)
1
= 1.061 × 10 −7 F = 0.1061µF
C=
K 4p f c
L=
2C
(0.2122 µF) (0.2122 µF)
L
(26.53 mH)
2C = 2(0.1061) = 0.2122 mF
Therefore, the required design of constant K-type HPF in T-section
and in p -section are shown, respectively, in Figures 13.92 and
13.93.
Figure 13.92
C (0.1061 µF)
Step 2: Design m-derived HPF.
For this, we will use m = 0.359.
T-section: Component values are as follows.
2c 0.2122
=
= 0.591µF
m
0.359
L 26.53
=
= 73.89 mH
m 0.359
 4(0.359) 
 4m 
C=

 × (0.1061) = 0.1749 µF
2
1− m 
1 − 0.3592 
The design is shown in Figure 13.94.
p-Section: Component values are as follows.
C 0.1061
=
= 0.2955 µF
m 0.359
2 L 2( 26.53)
=
= 147.79 mH
0.359
m
 4m 
 4(0.359) 
L=

 × 26.35 = 43.73mH
2
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd

1− m
1 − 0.3592  679
2L
(53.06 mH)
2L
(53.06 mH)
Figure 13.93
2C
m
2C
m
(0.591 µF)
(0.591 µF)
L
m (73.89 mH)
4m
C = 0.1749 µF
1−m 2
Figure 13.94
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680 Network Analysis and Synthesis
C 0.1061
=
= 0.2955 µF
m 0.359
2 L 2( 26.53)
=
= 147.79 mH
m
0.359
 4m 
 4(0.359) 
L=

 × 26.35 = 43.73mH
2

1− m
1 − 0.3592 
4m L = 43.73 mH
1− m 2
2L
m
(147.79 mH)
C
m = 0.2955 µF
The design is shown in Figure 13.95.
2L
m
(147.79 mH)
Step 3: Design of terminating half section with
m = 0.6.
For T-section: Component values can be calculated
as follows.
2C 2(0.1061)
=
= 0.3536 µF
m
0.6
2 L 2( 26.53)
=
= 88.43 mH
m
0.6
 2m 
 2(0.6) 
C=

 × 0.1061 = 0.1989 µF
2

1− m
1 − 0.6 2 
Figure 13.95
2C = 0.3536 µF
m
2L = 88.43 mH
m
2m
C = 0.1989 µF
1− m 2
Figure 13.96
2C = 0.3536µF
m
2m
L = 49.74µF
1− m 2
2L (88.43 mH)
m
The design is shown in Figure 13.96.
For p-section: The calculation of component values are
given as follows.
2C
= 0.3536 µF
m
2L
= 88.43mH
m
 2m 
 2(0.6) 
L=

 × 26.53 = 49.74 mH
2
1− m 
1 − 0.6 2 
The design is shown in Figure 13.97.
Design of Composite HPF
Figure 13.97
Now we are able to make the composite high-pass filter with two terminating half sections, the
constant-K section and a m-derived section in both T-section and p-section with the calculated
values as has been shown in Figure 13.98 and Figure 13.99, respectively.
0.3536 µF
0.2122
µF
0.2122
µF
0.5914
µF
88.43 mH
0.5914
µF
0.3536
µF
73.89mH
88.43 mH
0.1749 µF
0.1989
µF
26.53 mH
0.1989
µF
Terminating
half section
Constant-K
section
m-Derived
section
Terminating
half section
T-section
Figure 13.98 Composite High-pass Filter Network in T-section
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681
Filters and Attenuators
49.74 mH
0.3536 µF
88.43 mH
Terminating
half section
43.73 mH
0.1061
µF
53.06
mH
Constant-k
section
53.0
6 mH
p -Section
0.2955 µF
147.79 mH
49.74 mH
0.3536 µF
147.
79 mH
m-Derived
section
88.43 mH
Terminating
half section
Figure 13.99 Composite High-pass Filter Network in p-section
13.12 ATTENUATORS
Attenuators are used to reduce the signal level required in various applications in the field of
electronics. For example, attenuators may be used as volume controllers in a radio station or to
obtain a small voltage required in a testing of laboratory, etc.
13.12.1 Introduction
I1
I2
Attenuator is a network, normally a two-port resistive network that reduces the signal level to a desired value. It is
Attenuator
V1 Source
Load V2
inserted between a source and a load to reduce current,
voltage and power.
(signal of
reduced
Attenuators may be symmetrical or asymmetrical. In
level)
this section, we will discuss only symmetrical resistance
Figure 13.100
attenuators. A resistance attenuator is a network consisting
of resistors and is designed to reduce the voltage, current or
power by known amount. The block diagram of an attenuator has been shown in Figure 13.100.
Units of Attenuation Produced by Attenuator
Attenuation produced by an attenuator may be expressed in decibels (dB) or in nepers (N)
V1
I
P
= 20 log10 1 = 10 log10 1
V2
I2
P2
I
V
Attenuation in nepers = 1 = 1
I 2 V2
Attenuation in dB = 20 log10
Relation between decibels and nepers can be given as follows:
dB = 20 log10N
dB
or
log10 N =
20
 dB 
or
N = anti log  
 20 
dB
neper =
= 0.115dB
8.686
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682 Network Analysis and Synthesis
Basically, there are four types of attenuators. They are as follows:
1. T-type Attenuator;
3. Lattice Attenuator;
2. p-Type Attenuator
4. Bridged T-type Attenuator
These four types of attenuators are described in the following sections.
R1
13.12.2 T-type Attenuator
Symmetrical T-type attenuator is shown in Figure 13.101.
Here, R0 is the design impedance.
Let us find some parameters of T-type attenuator.
R0
Attenuation in Nepers (N)
By definition,
R2
I1
Mesh I
N=
R1
R0
I2
Mesh II
Figure 13.101 T-type Attenuator
I1
I2
(13.51)
Now, applying KVL in mesh II, we get the following:
-R2 (I2 - I1) - R1 I2 - R0 I2 = 0
or
R2 I1 -R2 I2 = R1 I2 + R0 I2
or
R2 I1 = R1 I2 +R2 I2 + R0 I2
or
R2 I1 = I2 (R1 + R2 + R0)
I1 R1 + R 2 + R 0
=
= N (13.52)
I2
R2
or
Characteristic Impedance (R0)
Now,
Rin = R1 + R2|| (R1 + R0)
(input resistance of T-type network from Figure)
R2 ( R1 + R0 )
R2 + R1 + R0
R1 + R0
= R1 +
R1 + R2 + R0
R2
= R1 +
R + R0
∴ Rin = R1 + 1
N
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∵ from equation (13.52) 
R + R + R

2
0
 1

=N
R2


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Filters and Attenuators
By definition,
683
Rin = R0
R1 + R0
= R0
N
R1 N + R1 + R0
= R0
N
R1N + R1 + R0 = NR0
(N + 1) R1 = NR0 - R0
R1 +
or
(N + 1) R1 = (N -1) R0
R0 =
or
N +1
R
N −1 1
(13.53)
Design Parameters
From equation (13.53), we have the following:
R1 =
N −1
R (13.54)
N +1 0
Now, to find the value of R2, let us consider equation (13.52),
R1 + R2 + R0
=N
R2
R1 + R2 + R0 = NR2
or
R1 + R0 = NR2 -R2
or
(N - 1 ) R2 = R1 + R0
or
Now, substituting the value of R1 from equation (13.53) in the equation, we get the following:
 N − 1
( N − 1)R 2 = 
R + R0
 N + 1 0
( N − 1)R 0 + ( N + 1)R 0
=
N +1
(N − 1)R 2 =
or
=
or
R2 =
NR 0 − R 0 + NR 0 + R 0
N +1
2NR 0
N +1
2NR 0
2NR
, that is, R 2 = 2 0
( N − 1)( N + 1)
N −1
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684 Network Analysis and Synthesis
Summary of T-type Attenuator
1. N =
R1 + R2 + R0
R2
( N + 1) R1
N −1
3. Design parameters are as follows:
2. R0 =
R1 =
R2 =
( N − 1) R0
N +1
2 NR0
N 2 −1
Example 13.24 Find the characteristic impedance for the
T-network shown in Figure 13.102.
Solution: Given a T-attenuator, let us compare it with the general
network of T-type attenuator as shown in Figure 13.103.
We get
R1 = 100 W
R2 = 400 W
Now, for T-type attenuator, we have the following.
By definition,
Rin = R0
100 Ω
100 Ω
400 Ω
R0
Figure 13.102
R1
R1
R2
R0
Figure 13.103
100 + 400 || (100 + R0) = R0
100 +
or
50000 + 100 R 0 + 40000 + 400 R 0
= R0
500 + R 0
90000 + 500 R0 = 500 R0 + R02
or
or
400(100 + R 0 )
= R0
400 + 100 + R 0
90000 = R02
R 0 = 90000 = 300 W
Example 13.25 Design a T-type actuator to given attenuation of 20 dB and to work in a line
of 200 W impedance.
Solution: Given
Now, we have
D = 20 dB;
R0 = 200 W
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Filters and Attenuators
685
 D
N = anti log  
 20 
 20 
= anti log  
 20 
= anti log (1) = 10
Further, for T-type attenuator, we get the following forms:
( N − 1)
R0
N +1
(10 − 1)
=
200
10 + 1
9( 200) 1800
=
=
= 163.63 Ω
11
11
R1 =
and
R2 =
=
=
2 NR0
R1 = 163.63 Ω
N 2 −1
R1 = 163.63 Ω
2(10)( 200)
10 2 − 1
4000
4000
=
= 40.4
100 − 1
99
R2 = 40.4 Ω
R0
R0 = 200 Ω
(T-type attenuator)
Therefore, the required attenuator is shown in Figure
13.104.
Figure 13.104
13.12.3 o-Type Attenuator
Circuit Configuration
Now, we should know that for an attenuator shown in Figure 13.105, the following can be written:
R OC
( N + 1)
(13.55)
= R0
(open circuited impedance)
N −1
RSC
( N − 1)
(13.56)
= R0
(short circuited impedance)
( N + 1)
Ro
(design impedance)
= R OC ⋅ R SC (13.57)
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R1
R0
R2
R2
R0
Figure 13.105
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686 Network Analysis and Synthesis
To find the values of R1 and R2, firstly, let us bisect the p-network as shown in Figure 13.106.
R1
R2
R1/2
R2
R2
Figure 13.106
Now,
ROC = R2
Open-circuit impedance of half section
R1R 2
R
R2 ⋅ 1
R SC
R1
2 =
2
= R 2 ||
=
R1 2R 2 + R1
( half selection )
2
R2 +
2
2
R1R 2
=
R1 + 2R 2
Now, we have general equations (13.55) and (13.56)
( N + 1)
(13.58)
N −1
R ( N − 1)
R1R 2
=
= 0
(13.59)
R1 + 2R 2
N +1
That is,
R OC = R 2 = R 0
and
R SC
Substituting the value of R2 from equation (13.58) in equation (13.59), we get the following:
R ( N + 1)
R1 ⋅ 0
R ( N − 1)
N −1
= 0
R 0 ( N + 1)
N +1
R1 + 2 ⋅
N −1
or
R1R 0 ( N + 1)
( N − 1)
N −1
= R0
R1 ( N − 1) + 2R 0 ( N + 1)
N +1
N −1
or
R1R 0 ( N + 1)
( N − 1)
= R0
R1 ( N − 1) + 2R 0 ( N + 1)
N +1
or
R1 ( N + 1)
N −1
=
R1 ( N − 1) + 2R 0 ( N + 1) N + 1
or
R1 ( N − 1) + 2R 0 ( N + 1) N + 1
=
R1 ( N + 1)
N −1
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or
687
N − 1 2R 0 N + 1
+
=
N + 1 R1
N −1
2R 0 N + 1 N − 1
=
−
R1
N −1 N +1
or
=
=
( N + 1) 2 − ( N − 1) 2
( N − 1)( N + 1)
N 2 + 1 + 2N − ( N 2 + 1 − 2N )
N 2 −1
2R 0
4N
= 2
R1
N −1
R1
N 2 −1
=
4N
2R 0
or
2( N 2 − 1) R0 ( N 2 − 1) R0
=
(13.60)
4N
2N
From equation (13.58), we have the following:
or
R1 =
R2 = R0
( N + 1)
(13.61)
N −1
Summary
Therefore, the design equations for p-type attenuators are as follows:
R1 =
R ( N + 1)
( N 2 − 1)
R0 ; R2 = 0
2N
N −1
Example 13.26 Design a p -type attenuator to give an attenuation of 3 N and to work in a line
impedance of 375 W.
Solution: Given
Now, for p -type attenuator, we get
R0 = 375 W;
R1 =
N=3
( N 2 − 1)
R0
2N
Substituting the values of N and R0, we get
R1 =
We have
(32 − 1)
8
(375) = × 375 = 500 Ω
2×3
6
R2 = R0
( N + 1)
N −1
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Substituting N = 3 and R0 = 375 W, the following are calculated.
R2 = 375
(3 + 1)
( 4)
= 375
= 2(375) = 750 Ω
3 −1
2
Therefore, the required p-type attenuator is shown in Figure 13.107.
R1= 500 Ω
R2 = 750 Ω
R0 = 375 Ω
R2 = 750 Ω
R0
Figure 13.107
13.12.4 Lattice Attenuator
Circuit configuration for symmetric lattice attenuator is shown in Figure 13.108.
R1
A
B
I
R2
R2
I′
2
C
R1
R0
D
2′
Figure 13.108
Now, the network shown in Figure 13.108 can also be drawn as shown in Figure 13.109.
I1
A
1
R2
V1
R1
R0
D
2
I1 − I ′
I′
I2
R2
R1
B
2′
I ′ + I2
R1
1′
R2
R1
R2
C
Figure 13.109
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Figure 13.110
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689
We calculate can the open-circuit impedance ROC from Figure 13.110.
(open-circuit impedance ROC
) = ( R11 + R22 ) || ( R11 + R22 )
OC
( R + R22 ) ⋅ ( R11 + R22 )
= 11
( R11 + R22 ) + ( R11 + R22 )
R2
R1
Short circuit
R1
( R11 + R22 ) 22 R11 + R22
=
2( R11 + R22 )
2
R1 + R 2
(13.62)
2
To find short-circuit impedance, we refer to Figure 13.111
RSC = (R1 || R2) + (R1 || R2)
R OC =
That is,
R2
Figure 13.111
=
=
R SC =
R1 R2
RR
+ 1 2
R1 + R2 R1 + R2
2R1R 2
(13.63)
R1 + R 2
Further, from Figure 13.109,
V1 = I1R0 = (I1 - I′) R1 + I2 R0 + (I′ + I2) R1
or
I1R0 = I1R1 - I′R1 + I2R0 + I′R1 + I2R1
or
I1R0 - I1R1 = I2R0 + I2R1
or
I1(R0 - R1) = I2 (R0 + R1)
or
N =
I1 R 0 + R1
=
(13.64)
I 2 R 0 − R1
Now, we have
R 0 = R OC ⋅ R SC
Substituting the values of ROC and RSC from equations (13.62) and (13.63) in the equation, we
get the following:
 R + R2  2 R1 R2
⋅
=  1
 2  R1 + R2
= R1 R2
or
R20 = R1R2(13.65)
Now, from equation (13.64), we get the following:
N R0 + R1
=
1 R0 − R1
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Applying components and dividend, we get
N − 1 ( R 0 + R1 ) − ( R 0 − R1 )
=
N + 1 ( R 0 + R1 ) + ( R 0 − R1 )
N − 1 2R1
=
N + 1 2R 0
or
 N − 1
R1 = 
R
 N + 1 0
or
Now, from equation (13.65), R2 can be calculated as follows:
R2 =
R 02
R1
Here, substituting the values of R1, we get the following equation:
=
R0 2
 N − 1

R
N + 1 0
 N + 1
R2 = 
R
 N − 1 0
Summary
Therefore, for lattice attenuator, design equations are as follows:
 N − 1
R
R1 = 
 N + 1 0
 N + 1
R
R2 = 
 N − 1 0
Example 13.27 Design a symmetrical lattice attenuation to give an attenuation of 20 dB and
to work in a line of 200 W impedance.
Solution:
Given
D = 20 dB; R0 = 200 W
Therefore,
D
N = antilog  
 20 
 20 
= anti log  
 20 
= antilog (1)
= 10
Now, for lattice attenuation, we get the following equation:
 N − 1
R1 = 
R
 N + 1 0
Substituting the value of N and R0 in the equation, we have the following equation:
9
1800
 10 − 1
R1 = 
= 163.63 W
× 200 = × 200 =
 10 + 1
11
11
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691
 N + 1
R2 = 
R
 N − 1 0
and
11
2200
 10 + 1
=
200 = ( 200) =
= 244.4 Ω

 10 − 1
9
9
Therefore, required lattice attenuator is shown in Figure 13.112.
R1 = 163.63 Ω
R0
R2 = 244.4 Ω
R2 = 244.4 Ω
R0 = 200 Ω
R1
R1 = 163.63 Ω
R0
Figure 13.112
R0
13.12.5 Bridged T-type Attenuator
Circuit configuration: The circuit configuration of a
bridged T-type attenuator has been shown in Figure
13.113.
Bisected half section of the bridged T-network is
shown in Figure 13.114.
Now, from Figure 13.114,
( N + 1)
ROC = R0 + 2 R2 = R0
(13.66)
N −1
and
R SC =
R2
Figure 13.113 Bridged T-network
R1/2
R0
R1
|| R 0
2
2R2
R1
2
=
R1
R0 +
2
R0 ⋅
or,
RSC = R0
( N − 1)
N +1
Figure 13.114 Bisected Half Section
of Bridged T-network
(13.67)
Circuit to find RSC is shown in Figure 13.115.
Now, from equation (13.67) and the circuit shown in Figure
13.115, we can write
R
R0 ⋅ 1
( N − 1)
2
[Since RSC is parallel equivalent
R SC = R 0
=
R1
N +1
of R1/2 and R0 as has been shown
R0 +
2
in Figure 13.116 ]
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R0
R1 / 2
R0
2R2
Short
Circuit
Figure 13.115
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692 Network Analysis and Synthesis
R1 / 2
R0
R1
( N − 1)
2
=
or
2R 0 + R1
N +1
2
RSC
Figure 13.116
R1
N −1
=
N + 1 R1 + 2R 0
or(N-1) (R1 + 2R0) = (N + 1)R1
or
(N-1) R1 + 2(N - 1)R0 = (N + 1) R1
or
[(N - 1) - (N + 1) ]R1 = -2(N - 1) R0
or
[-1 - 1] R1 = -2(N - 1) R0
or
or
-2R1 = - 2(N - 1) R0
R1 = (N - 1) R0
Further, from equation (13.66), we get the following equation:
R0 + 2 R2 = R0
( N + 1)
N −1
Therefore, we get the following set of equations:
R0 ( N + 1)
− R0
N −1
R ( N + 1) − ( N − 1) R0
= 0
N −1
R0 [ N + 1 − N + 1]
=
N −1
R
2 R0
2 R2 =
or R2 = 0
N −1
N −1
2 R2 =
Therefore, for bridged T-type attenuator, the equation is written as follows:
R1 = (N -1) R0
R
R2 = 0
N −1
Example 13.28 Design a bridged T-type attenuator for attenuation of 10 dB and load
resistance of 200 W.
Solution: Given
a = 10 dB;
Therefore,
R0 = 200 W
 D
 10 
N = antilog   = anti log   = 3.16
 20 
 20 
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Filters and Attenuators
R1 = 432 Ω
Now, a bridged T-type attenuator we can write as
R1 = (N - 1) R0
= (3.16 - 1) 200 = 2.16 (200) = 432 W
and
R0 = 200 Ω
R0
N −1
200
200
=
= 92.59 Ω
=
3.16 − 1 2.16
R2 =
R0
R0 = 200 Ω
R2 = 92.59 Ω
R0 = 200 Ω
Figure 13.117
Therefore, the required bridged T-type attenuator is shown in Figure 13.117.
13.13 MORE SOLVED PROBLEMs ON FILTERS AND ATTENUATORS
Example 13.28 Given two capacitors of 1mF each and coil L of 10 mH, compute the following:
1. Cut-off frequency and characteristic impedance at infinite frequency for an HPF.
2. Cut-off frequency and characteristic impedance at zero frequency for an LPF.
Draw the constructed sections of filters using these elements.
Solution: Given HPF is shown in Figure 13.118.
ZOT at infinite frequency = R0
Given
2C = 1 µF ⇒ C =
2C = 1µF
2C = 1µF
L = 10 mH
1
µF
2
L = 10 mH
Figure 13.118
L
=
C
=
Cut-off frequency
10 × 10 −3
= 20 × 103 = 1.414 × 10 2 Ω
1
−6
× 10
2
fc =
=
1
4p LC
L = 10 mH
1
4p 10 × 10
−3
1
× 10 −6
2
= 1.12 KHz
LPF from the given capacitors and inductor can be drawn as
shown in Figure 13.119.
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C =1µF
2
C =1µF
2
Figure 13.119
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694 Network Analysis and Synthesis
L = 10 mH = 10 × 10-3 H
C
= 1 µF C = 2 µF = 2 × 10 −6 F
2
Zop at infinite frequency = R0
L
10 × 10 −3
=
= 707 Ω
C
2 × 10 −6
=
fc =
Cut-off frequency
=
1
p LC
1
p 10 × 10 −3 × 2 × 10 −6
=
1
p 2 × 10 −8
=
1
p 2 × 10 −4
= 2250 Hz
Example 13.29 For the network shown in Figure 13.120, find the
characteristic impedance
j 100 Ω
−j 400 Ω
Solution: To find Z0:
Z OC (Open-circuit impedance) = j100 − j 400 = − j 300 Ω
and
j 100 Ω
Figure 13.120
Z SC (Short-circuit impedance ) = j100 + ( j100 − j 400)
= j100 +
( j100)( − j 400)
j100 − j 400
= j100 +
− j 240000
− j 300
= j100 + j133.33 = j 233.33 Ω
Now, we know the following:
Zo = ZOC × ZSC = ( − j 300)( j 233.33) = 264.57 Ω
Example 13.30 A symmetrical T-section has the following data:
ZOC = 800 Ω; ZSC = 600 Ω
1
Determine the T-section parameters and represent the two-port
network.
Solution: We have to find Z1 and Z2 of the network shown
Figure 13.121
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Z1
2
Z1
Z2 2
1′
2
2′
Figure 13.121
12/3/2014 8:23:47 PM
Filters and Attenuators
695
Z OC =
Z1
+ Z 2 (13.68)
2
Z SC =
Z1  Z1

+
Z2


2
2
Z 1Z 2
Z1
2
=
+
2 Z1
+ Z2
2
From the expression of ZSC and ZOC as above, we get
Z SC
Subtituting,
Z 1Z 2
Z1
=
+ 2
2
Z OC
Z1
= Z0 C − Z 2 in the expression for ZSC, we get
2

Z 
ZSC = ( Z0 C − Z 2 ) 1 + 2 
 ZOC 
or
or
or
or
22
Z
Z 222
Z
Z
Z
+
Z
−
Z
−
2
SC =
0
C
2
2
Z
Z
Z
=
+
Z
−
−
Z C + Z 22 − Z 22 − Z
Z SC
SC = 00 C
Z
Z 000 CCC
2
2
Z2
Z 22
Z SC =
=Z
Z 0C −
− Z
Z
=
−
C
Z SC
Z
0
Z
SC
0C
Z
Z 000 CCC
2
22
Z
Z
Z 222 =
=Z
−Z
SC
= Z 0C −
− Z SC
Z
Z 000 CC Z 00 CC Z SC
Z
C
2
Z
Z
SC )))
Z 22222 =
Z 000 CCC (Z
=Z
(Z
Z 00 CCC −
−Z
Z SC
Z
=
Z
(Z
Z
−
Z
SC
0
Z
Z
Z
))
2 =
0 C (( Z
0C −
Z
=
Z
Z
−
Z SC
Z 2 = Z 0C (Z 0C − Z
SC )
2
0C
0C
SC
Substituting the given values, we can calculate the value of Z2. as
Z 2 = 800(800 − 600) = 800( 2) = 1600 = 400 Ω
From equation (13.67), we obtain the following form:
Z1
= Z 0C − Z 2
2
Substituting the values of Z0C and Z2 in the equation, we get the following set of equations:
or
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Z1
= 800 − 400 = 400 Ω
2
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696 Network Analysis and Synthesis
Therefore, required T-section parameters are as follows:
Z1
= 400 Ω and Z 2 = 400 Ω
2
Example 13.31 Design a T-type symmetrical attenuation network that offers 40 dB
attenuation with a load of 400 Ω.
Solution: Given attenuation is 40 dB
 D
N = antilog  
 20 
(attenuation in nepers)
 40 
= antilog   = antilog (2) =100
 20 
R0 = 400 Ω
Series arm resistance can be given as follows:
( N − 1)
N +1
(100 − 1)
= 400
= 392.07Ω
(100 + 1)
R1 = R0
R0 = 392.07 Ω
R0 = 392.07 Ω
Shunt arm resistance can be written as follows:
R2 = 8.008 Ω
R2 = R0 .
Figure 13.122
2N
N 2 −1
[2 × 100]
= 400 ×
100 2
= 400 × 0.02 = 8.0008 Ω
The designed network in shown in Figure 13.122.
Example 13.32 Design a constant K-type BPF section having a cut-off frequency of 2 kHz
and 5 kHz and a nominal impedance of 600 Ω. Draw the configuration of the filter.
Solution: Now, for constant K-type BPF, we get the value of L1, L2, C1 and C2 as follows:
L1 =
R0
600
=
= 63.68 mH
p ( f 2 − f1 ) p (5000 − 2000)
C1 =
f2 − f2
5000 − 2000
=
4p R0 f1 f 2 4p (600)(5000)( 2000)
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L2 =
R0 ( f 2 − f1 ) 600(5000 − 2000)
=
= 14.33 mH
4p f1 f 2
4p (5000)( 2000)
C2 =
1
1
=
= 0.1769 µF
p R0 ( f 2 − f1 ) p × 600(5000 − 2000)
697
Therefore, the required configuration is shown in Figure 13.123.
L1
2
2C1
2C1
L1
2
(31.84 mF)
(0.0762 µF)
(0.0762 µF)
(31.84 mF)
C2
(0.1769 µF)
L2
(14.33 mF)
Figure 13.123
Example 13.33 Design the symmetrical bridge T-type attenuator with an attenuation of
40 dB and an impedance of 600 Ω.
Solution: Given
R0 = 600 Ω; D = 40 dB
Therefore,
D
 40 
N = antilog   = antilog   = 100
 20 
 20 
We have the following form;
R 2 R 3 = R12 = R 02
Therefore,
R1 = R 0 = 600 Ω
R1 = 600 Ω
That is,
and
R0
600
=
= 6.06 Ω
N − 1 100 − 1
R 3 = R 0 ( N − 1) = 600(100 − 1) = 59.4 Ω
R2 =
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Therefore, the required symmetrical bridge T-type attenuator is shown in Figure 13.124.
R3 = 59.4 Ω
R1 = 600 Ω
R1 = 600 Ω
R0 = 600 Ω
R2 = 6.06 Ω
R0
Figure 13.124
Example 13.34 Design an m-derived T-section (high-pass) filter with a cut-off frequency
fc = 20 kHz, f∞ = 16 kHz and a design impedance R0 = 600 Ω.
Solution: We know for m-derived filter, the following equation can be obtained:
2
2
 f 
 16000 
m = 1−  É  = 1− 
= 0.6
 20000 
 fc 
For HPF, L and C can calculated as follows:
L=
R0
600
=
= 2.39 mH
4p f c 4p ( 20000)
C=
1
1
=
= 0.007 µF
4p R0 f c 4p ( 20000)(600)
The components of T-section (m-derived HPF) are given as in the following:
0.024 µF
1 2.39
=
= 3.98 mH
m
0.6
0.024 µF
2C 2 × 0.007
=
= 0.024 µF
m
0.6
3.98 mH
0.026 µF
 4m 
 4 × 0.6 
C=
× 0.007 = 0.026 µF

2
 1 − 0.6 2 
1− m 
Figure 13.125
Therefore, required filter is shown in Figure 13.125.
Example 13.35 Design a symmetrical T-section having parameters of ZOC = 1000 Ω & ZSC = 600 Ω.
ZOC = 1000 Ω & ZSC = 600 Ω.
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Solution: Given
Z OC = 1000 Ω
Z SC = 600 Ω
Now, for a symmetrical T-section, we have the following:
Z 2 = ZOC ( ZOC − ZSC )
= 1000(1000 − 600)
= 1000( 400) = 400000 = 632.45 Ω
and
Z1
= Z OC − Z 2
2
or
Z1 = 2( ZOC − Z 2 )
or
Z1
Z
= 367.54 Ω 1 = 367.54 Ω
2
2
1
2
Z2 = 632.46 Ω
1′
= 2(1000 − 632.45) = 735.08 Ω
2′
Figure 13.126
Z1
= 367.54 Ω
2
Therefore, required symmetrical T-section is shown in Figure 13.126.
Example 13.36 Design an m-derived LPF (T- and p -Section) having a design impedance of
500 Ω and cut-off frequency 1500 Hz and an infinite attenuation frequency of 2000 Hz.
Solution: Given
R0 = 500 Ω; f c = 1500 Hz; f ∞ = 2000 Hz
We know, for an LPF, m can be calculated as follows:
f 
m = 1−  c 
 fÉ 
2
2
 1500 
= 1− 
= 0.661
 2000 
and
L=
R0
500
=
= 106.103 mH
p f c p (1500)
C=
1
1
=
= 0.424 µF
p R 0f c p (500)(1500)
Now, components of T-section (m-derived LPF) can be given as follows:
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700 Network Analysis and Synthesis
mL 0.0661 × 106.103
=
= 35.067 mH
2
2
mc = 0.424 × 0.661 = 0.280 µF
2
1 − 0.6612 
1− m 
 × 106.103 = 22.596 mH

 L = 
4m
 4 × 0.661 
The values of elements of p-Section (m-derived LPF) can be written as follows:
mc 0.661 × 0.424
=
= 0.140 µF
2
2
mL = 106.103 × 0.661 = 70.134 mH
1− m2 
1 − 0.6612 
c
=

 × 0.424 = 0.09 µF
 4m 


 4 × 0.661 
Therefore, configurations in T-section and in p-section are shown in Figure 13.127.
0.09 µF
35.067 mH
35.067 mH
70.134 mH
0.280 µF
0.140 µF
0.140 µF
22.569 mH
T-section
p -Section
Figure 13.127
Example 13.37 Design a symmetrical bridged T-type attenuator to provide attenuation of
60 dB and to work into a line of characteristic impedance of 600 Ω.
Solution: Given
R0 = 600 Ω
and
Now,
D
 60 
N = antilog   = antilog   = 1000
 20 
 20 
R1 = R0 = 600 Ω
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Filters and Attenuators
R2 =
701
R0
600
=
= 0.601 Ω
N − 1 1000 − 1
R 3 = R 0 ( N − 1) = 600 × 999 = 599400 Ω
Therefore, required attenuator is shown in Figure 13.128.
R3 = 599400 Ω
Example 13.38 In a symmetrical T-network,
if the ratio of input and output power is 6.76.
Calculate the attenuation in neper and dB.
Further, design this attenuator operating between
source and load resistance of 100 Ω.
R1 = 600 Ω
R1 = 600 Ω
R2 = 0.601Ω
R0
Solution: Given
Pin
= 6.76 and R0 = 1000 Ω
Pout
Figure 13.128
Pin
= 6.76 = 2.6
Pout
Now,
N=
and
D = 20 log10 N
D = 20 log10 N
= 20 log10 2.6 = 8.299 dB
= 20 log10 2.6 = 8.299 dB
Attenuation in dB
R0 = 600 Ω
Series arm resistance is given as follows:
( N − 1)
N +1
( 2.6 − 1)
= 1000
= 444 Ω
( 2.6 + 1)
R1 = R0
Shunt arm resistance is calculated as in the following:
R2 = R0
2N
N 2 −1
( 2 × 2.6)
= 1000
= 902.77 Ω
( 2.6) 2 − 1
Therefore, the required symmetrical T-type network is shown in
Figure 13.129.
R1 = 444 Ω
R1 = 444 Ω
R2 = 902.77Ω
Figure 13.129
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702 Network Analysis and Synthesis
R E V IE W Q U E S T I O N S
Short Answer Type
1. Design an m-derived LPF (T- and p-Section) having a design resistance of R0 = 500 Ω and
the cut-off frequency ( fc) of 1500 Hz and an infinite attenuation frequency of 2000 Hz.
[Ans. As shown in Figure 13.130]
0.09 µF
35.1 mH
35.1 mH
70.14 mH
0.28 µF
0.140 µF
22.6 mH
0.140 µF
m-Derived p -Section
m-Derived T-section
Figure 13.130
2. Design a symmetrical bridged T-type attenuator to provide attenuation of 60 dB and to
work into a line of characteristic impedance 600 Ω.
[Ans. As shown in Figure 13.131]
R3 = 599.4k Ω
R1 = 600 Ω R1 = 600 Ω
R2 = 0.60 Ω
R0
R0 = 600Ω
Figure 13.131
3. Design a symmetrical bridge T-type attenuator with an attenuation of 40 dB and an impedance of 600 W
[Ans. As shown in Figure 13.132]
R3 = 59.4 Ω
R1 = 600 Ω R1 = 600 Ω
R2 = 0.60 Ω
600 Ω
Figure 13.132
4. Design a m-derived T-section (HPF) with a cut-off frequency fc = 20 kHz, f∞ = 16 kHz and
a design impedance 600 W
M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 702
[Ans. As shown in Figure 13.133]
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Filters and Attenuators
0.024 µF
703
0.024 µF
0.026 µF
3.98 mH
Figure 13.133
5. Design a symmetrical T-section having parameters of ZOC = 1000 W and ZSC = 600 W
[Ans. As shown in Figure 13.134]
Z1
2
Z1
Z2 2
Z1
= 367.54 Ω
2
Z2 = 632.46 Ω
Figure 13.134
Z
1 = 367.54 Ω
2
Z2 = 632.4 W
6. Design a T-type symmetrical attenuator that offers 40 dB attenuation with a load of 400 W
[Ans. As shown in Figure 13.135]
R1 = 392.07 Ω R1 = 392.07 Ω
R2 = 8.0008 Ω
Figure 13.135
7. Design a constant K-type BPF section having cut-off frequencies of 2 kHz and 5 kHz and a
nominal impedance of 600 W. Draw the configuration of the filter.
[Ans. As shown in Figure 13.136]
31.84 mF
0.0762 µF 0.0762 µF
0.1769 µF
31.84 mF
14.33 mH
Figure 13.136
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704 Network Analysis and Synthesis
8. Given the network for HPF
1µF
1µF
10 mH
Figure 13.137
Find the cut-off frequency and characteristic impedance at infinity frequency for an HPF.
[Ans. 1.12 kHz, 141.4 W]
9. Given p -Section LPF
10 mH
1µF
1µF
Figure 13.138
Compute the cut-off frequency and characteristic impedance at zero frequency for an LPF.
[Ans. 4.5 kHz, 707 W]
10. Design a constant K-type BPF T-section having cut-off frequencies 2 kHz and 5 kHz and a
nominal impedance of 600 W. Draw the configuration of the filter.
[Ans. As shown in Figure 13.139]
0.0762 µF 31.84 mH
0.1769 µF
31.84 mH
0.0762 µF
14.33 mH
Figure 13.139
11. Design a prototype LPF, assuming cut-off frequency fc
12.
13.
14.
15.
State advantages of m-derived filters.
What is an attenuator? Classify and state its applications.
Derive the relationships between neper and decibel units
Derive the expression for characteristic impedance of a symmetrical bridged T-type
network
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Filters and Attenuators
705
16. What are the disadvantages of the prototype filters? How are they overcome in composite
filters?
17. Discuss the characteristics of a filter.
18. Differentiate between the attenuator and the amplifier. List the practical applications of
attenuators.
19. With the help of frequency response curves, give the classification of filters.
20.
21.
22.
23.
Derive an expression for design impedance of a symmetrical T-type attenuator.
Explain how the reactance and impedance of an HPF varies with frequency.
What is an attenuator? Define the terms decibel and neper. Derive the relation between the two.
Write short notes on the following
1. LPF and its design
2. T- and p-Section attenuators.
24. What are the requirements of an ideal filter? Distinguish between an LPF and an HPF.
25. What is the propagation constant of a symmetrical T-section and symmetrical p-Section?
What is its significance?
26. How do you choose the resonant frequency of an m-derived LPF and m-derived HPF.
27. What is a band-elimination filter?
M ultiple C hoice Q uestions
1. A 26 dB output in watts is equal to
(a) 2.4 w
(b) 0.26 w
(c) 0.156 w
(d) 0.4 w
2. The characteristic impedance of an LPF in attenuation band is
(a) Purely imaginary
(b) Zero
(c) Complex frequency (d) Real value
3. The real part of the propagation constant shows:
(a)
(b)
(c)
(d)
Variation of voltage and current on basic unit
Variation of phase shift/position of voltage
Reduction in voltage, current value of signal amplitude.
Reduction of only voltage amplitude.
4. The purpose of an attenuator is to
(a) Increase signal strength
(c) Decrease value of signal strength
(b) Decrease reflections
(d) Provide impedance matching
5 In a transmission line terminated by characteristic impedance, Z0
(a)
(b)
(c)
(d)
There is no reflection of the incident wave
The reflection is maximum due to termination.
There are a large number of maximum and minimum on the line.
The incident current is zero for any applied signal.
6. An attenuator is a
(a) Resistive network
(b) R–L network
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(c) R–C network
(d) L–C network.
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706 Network Analysis and Synthesis
7. If ‘a’ attenuation in nepers, then
a
(a) Attenuation in dB =
0.8686
(c) Attenuation in dB = 0.1a
(b) Attenuation in dB = 8.686a
(d) Attenuation is dB = 0.01a
8. For a constant K-type high-pass p -Section filter, characteristic impedance Z0 for f < fc is
(a) Resistance
(c) Capacitive
(b) Inductive
(d) Inductive or capacitive
9. An ideal filter should have
(a) Zero attenuation in the pass band
(c) Infinite attenuation in the pass band
(b) Zero attenuation in the attenuation band
(d) Infinite attenuation in the attenuation band
10. For an m-derived HPF, the cut-off frequency is 4 kHz and the filter has an infinite attenuation at 3.6 kHz, the value of m is
(a) 0.436
(b) 4.36
(c) 0.34
11. If ZOC = 120 Ω and Zsc = 30 Ω, the characteristic impedance is
(a) 30 Ω
(b) 60 Ω
(c) 120 Ω
12. If Zoc = 100 Ω and Zsc = 64 Ω, the characteristic impedance is
(a) 400 Ω
(b) 60 Ω
(c) 80 Ω
(d) 0.6
(d) 150 Ω
(d) 170 Ω
13. The frequency of infinite attenuation ( f∞) of a low-pass m-derived section is
(a) Equal to fc (c) Close to but greater than fc
(b) f∞ = ∞
(d) Close to but less than fc of the filter
14. A constant K-type BPF has pass band from 1000 Hz to 4000 Hz. The resonance frequency
of shunt and series arm is a
(a) 2500 Hz
(b) 500 Hz
(c) 2000 Hz
(d) 3000 Hz
15. A constant K-type low-pass T-section filter has Z0 = 600 Ω at zero frequency. At f = fc, the
characteristic impedance is
(a) 600 Ω
(b) 0 W
(c) ∞ W
(d) more than 600 Ω
16. In m-derived terminating half section, m is equal to
(a) 0.1
(b) 0.3
(c) 0.6
(d) 0.9
17. In a symmetrical T-type attenuator with attenuation N and characteristic impedance R0, the
resistance of each series arm is equal to
2N
N
(a) R0
(b) (N - 1) R0
(c)
R0
(d)
R0
2
2
N −1
N −1
18. For a transmission line, open-circuit and short-circuit impedance are 20 Ω and 5 Ω. The
characteristic impedance of the line is
(a) 10 Ω
(b) 20 Ω
(c) 25 Ω
(d) 100 Ω
19. For a prototype LPF, the phase constant b in the attenuation band is
p
2
20. In the m-derived HPF, the resonant frequency is to be chosen so that it is
(a) p
(b) 0
(a) Above the cut-off frequency
(c) Equal to the cut-off frequency
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(c) ∞
(d)
(b) Below the cut-off frequency
(d) Infinitely high
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Filters and Attenuators
707
21. In a symmetrical p -section attenuator with attenuation N and characteristic impedance R0,
the resistance of each shunt arm is equal to
(b) (N - 1) R0
(a) R0
(c)
N −1
R
N +1 0
(d)
N +1
R
N −1 0
22. Bridged T-type network can be used as
(a) Attenuator
(b) LPF
(c) HPF
(d) BPF
(b) 8.686 dB
(c) 0.115 dB
(d) 86.86 dB
23. One neper is equal to
(a) 0.8686 dB
24. Terminating half sections used in composite filters are built with the following value of m
(a) 0.3
(b) 0.8
(c) 0.1
(d) 0.6
25. A transmission line works as
(a) Attenuator
(c) HPF
(b) LPF
(d) None of (a), (b) and (c).
ANS W E RS
1. d 2. a 3. c 4. c 5. a 6. a 7. b 8. d 9. a
10. a
Because for m-derived HPF
2
f 
 3.6 
m = 1−  ∞  = 1− 
= 0.436
 4 
 fc 
11. b
Because
Z o = ZOC × ZSC = 120 × 30 = 3600 = 60 Ω
12. c 13. c
14. c
Because f1 = 1000
f2 = 4000
and f 0 = f 1f 2 = 2000 Hz
15. b 16. c 17. c 18. a 19. a
24. d 25. b
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20. b
21. d
22. a
23. c
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Thispageisintentionallyleftblank
Index
A
C
AC parallel circuits 108
AC series–parallel circuits 133
Active and reactive power 103, 368
Active network 18
Active power 368
Admittance function 408, 413
Admittance method 110
Admittance triangle 111
Air capacitors 15
Alternating current (AC) 2
Antenna 158
Apparent power 368
Attenuation constant (a ) 598
Attenuators 681
Capacitance 412
Capacitive reactance 96
Capacitor in s-Domain 321
Capacitors 14
Air 15
Ceramic 15
Electrolytic 15
Mica 15
Paper 15
Plastic film 15
Types of 15
Cauer form of L–C network 518–519
Cauer form of R–L network 517
Cauer forms of R–C network 515
Ceramic capacitors 15
Characteristic impedance (Z0) 591
Circle diagram 169
Circuit 18
B
Balanced load 363
Balanced supply 363
Balanced three-phase system 362
Band-elimination filter 589
Band-pass filter (BPF) 151, 157, 585, 588
Band-stop filter 154, 157, 589
Bandwidth (BW) 152
Bandwidth and selectivity 152
Bandwidth of the pass-band 151
Bilateral circuit 18
Branch 18
Branch currents 114
Bridged T-type attenuator 691
Z01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_INDEX.indd 709
Circuit elements in the s-Domain 321
Circuit parameter 18
Coefficient of coupling 11
Coil 363
Comparison between star and delta
connections 369
Comparison of constant K-type filters 645
Compensation theorem 189
Complex network problems 56
Composite filters 668
Composite high-pass filter 670
Composite low-pass filter 669
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710 Index
Conductor 2
Connection scheme of windings forming
a delta 364
Constant K-type filter 604, 605
Conversion of delta load into star load 384
Conversion of h-parameters to y-parameters 479
Conversion of star load into delta load 384
Conversion of t-parameters to h-parameters 478
Conversion of Y-parameters to Z-parameters 477
Convolution theorem 310
Correlation of two-port network parameters 477
Critical damped response of R–L–C series
circuit 251
Current coil (CC) 376
Current IL 364
Current locus 167
Current source 19
Cut-off frequencies 151
D
DC network 17
DC response of an R–L–C series circuit 326
DC response of R–C series circuit 322
DC response of R–L series circuit 324
Decay of current through R–L series circuit 230
Delta connection 364
Dielectric 14
Dielectric constant 15
Dielectric strength 15
Direct current (DC) 2
Discharging of capacitor 243
Driving point impedance 413
Driving point of the network 421
E
Electric circuit 90
Electric network 18
Electro motive force (EMF) 2
Z01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_INDEX.indd 710
Electrolytic capacitors 15
Elementary three-phase generator 361
F
Faraday’s law 10
Filter networks 591, 593, 604
Filters and attenuators 585–701
Final value of capacitor 242
Final value theorem 295
Foster and Cauer forms 513
Foster Form of L–C Network 517–518
Foster Form of R–C Network 513–515
Foster Form of R–L Network 516
G
Generation of three-phase voltage 360
Graphical representation of capacitor
Voltage and capacitor current 242
H
Half-power frequencies 152
High-pass filter (HPF) 585, 588, 615
Hurwitz conditions for stability 500
Hybrid or h-parameters 466
I
Ideal and actual selectivity curve 153
Ideal tank circuit 155
Image impedance 484
Immittance 408
Impedance and admittance function 408
Impedance triangle 111
Inductance 412
Inductor in s-Domain 321
Inductors 10, 92
Laboratory 12
Moulded 12
Power supply 12
Types of 12
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711
Index
Initial value theorem 292
Interconnected two-port network 482
Inverse hybrid or g-parameters 467
Inverse Laplace transform 299
Inverse transmission parameters 469
K
Kirchhoff ’s current law (KCL) 26
Kirchhoff ’s laws 26
Kirchhoff ’s voltage law (KVL) 29
Measurment of power in three-phase
circuits 376
Mesh 18, 30
analysis 30
Mica capacitors 15
Millman’s theorem 181
Moulded inductors 12
Mutual inductance 10
N
Laboratory inductors 12
Ladder network 5, 592
p −section 593
T-section 592
Laplace transform 271–314
Concept of 271–272
problems based on standard formula 280–285
Properties of 286–291
standard functions 272–280
Steps to find transient response using 320
Summary of useful properties of 291
Transient response of circuits using 320–355
Lattice attenuator 688
Limitations of constant K-type filters 649
Linear circuit 18
Locus diagram 167
Loop 18
Low-pass filter (LPF) 585, 588
Network analysis 499
Network functions 408–458
in generalised form 423
Poles and zeros of 424, 449
Poles of 424
of a two-port network 421
zeros of 424
Network synthesis 499
Network synthesis and realisability 499–576
Network terminologies 18
Network theorem 172
Nodal analysis 43
Node 18, 43
Non-ideal tank circuit 156
Non-linear circuit 18
Norton’s theorem 179
Numerical problems based on
Kirchhoff’s laws 56–60
Numerical problems based on mesh
and nodal analysis 60–82
Numericals on network theorems 195
M
O
Maximum power transfer theorem 182, 185
m-derived band-pass filter 665
m-derived band-stop filter 667
m-derived filters 604, 649
m-derived high-pass filter 659
Ohm’s law 2
One-port network 413
One-wattmeter method 376
Open-circuit impedance-parameters
or Z-parameters 461
L
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712 Index
P
Q
Parallel resonance 155
Paper capacitors 15
Parallel band-pass filter 157
Parallel circuits 108, 111
Parallel connection of resistors 4
Parallel resonance circuits 155
Parallel resonant filters 156
Parameters of a filter 589
Pass-band 151
bandwidth of 151
Passive network 18
p-circuit representation of two-port
network 484
Phase sequence 363
Phase shift constant (b ) 598
Phase voltage vph 364
Phase windings 363
Plastic film capacitors 15
p −network 601
Poles and zeros of network
functions 424, 449
Poles of a network function 424
Pole–zero diagram 424
Port 408
Positive real functions (PRF) 506
Potential difference 2
Power factor 102, 114
Power factor angle 104
Power in R–L series circuit 101
Power supply inductors 12
Power triangle of R–L series circuit 102
Propagation constant (g ) 590, 596
Properties of L–C immittance 517
Properties of the R–C impedance 513
Prototype filters 604, 605
p −section ladder network 593
p −Type attenuator 685
Quality factor of a resonant circuit 152
Z01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_INDEX.indd 712
R
R–C series circuit 103
Power triangle of 105
Reactive power 94, 96, 368
Reciprocal two-port network 479
Reciprocity theorem 186
Relationship of line and phase voltages
and currents in a delta-connected
system 366
Relationship of line and phase voltages and
currents in a star-connected system 365
Resistance 2, 411
Resistor in the s-Domain 321
Resistors 9
Resonance condition 150
Resonant circuits 157, 158
applications of 157
Resonant frequency 150, 156
Rise of current through R–L series circuit 227
R–L admittance function 513
R–L series circuit 99
R–L–C series circuit 90, 105
Applications of 151
R–L–C series circuit with variable frequency
input voltage 147
Routh’s stability analysis 450
S
Selectivity curve of pass-band filter 153
Self-inductance 10
Series circuit 111
Series connection of resistors 3
Series resonance 148
Series–parallel circuit 5, 133
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713
Index
Short-circuit admittance parameters 462
Single phasing 363
Single-phase system 359
Sinusoidal response of R–C circuits 258
Sinusoidal response of R–C series circuit 333
Sinusoidal response of R–L circuits 253
Sinusoidal response of R–L series circuit 329
Sinusoidal response of R–L–C circuit 262
Source 19
Electrical 19
transformation 21
Voltage 19
Stability criterion for an active network 450
Star connection 363
Star to delta and delta to star transformation 384
Star to delta transformation 384
Star–delta transformation 190
Steady-state value of capacitor 242
Steps to find transient response using
Laplace transform 320
Supermesh 54
analysis 54
Superposition theorem 173
Symmetrical p-network 592
Symmetrical system 363
Symmetrical T-network 591, 593
Symmetrical two-port networks 480
Synthesis of L–C networks 517, 548
Synthesis of networks 513
Synthesis of R–C network 513
Synthesis of R–L network 515
T
Tank circuit 155
T-circuit representation of two-port network 483
Tellegen’s theorem 188
Terminal 408
Terminated two-port network 480
Z01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_INDEX.indd 713
The Cauer’s methods 513
The Foster’s method 513
Thevenin’s theorem 173
Procedure for applying 175
Three-phase system 359
Advantages of 359
Balanced 362
Three-phase systems and circuits 358–401
Three-phase three-wire delta connection 364
Three-phase two-pole generator 361
Three-phase voltage 360, 361
Equation of 360
Three-phase winding connections 363
Three-phase windings 360, 361
Three-wattmeter method 379
Time constant of R–C series circuit 241
Time constant of R–L series circuit 230
Total current 114
Transfer function 422
Transformed impedances in s-Domain 411
Transforming relation from star to delta 191
Transforming relations from delta to star 190
Transient condition in networks 226
Transient response in R–C series circuits 239
Transient response of circuits using
differential equations 226–268
Transient response of circuits using
Laplace transform 320–355
Transient response of R–L circuit
to sinusoidal input voltage 254
Transient response of R–L series circuits 227
Transient response of R–L–C series
circuits 249
Transient time 226
Transmission parameters 468
T-section ladder network 592
T-type attenuator 682
Tuned amplifier 157
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714 Index
Two-port network 420, 421
Two-port network parameters 460–495
Two-port reciprocal and symmetrical
networks 479
Two-wattmeter method 377
U
Unbalanced load 363
Unbalanced supply 363
Unilateral circuit 18
Upper cut-off frequency 151
Z01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_INDEX.indd 714
V
Voltage 2
Voltage across capacitor during discharging 244
Voltage source 19
W
Winding 363
Z
Zeros of a network function 424
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