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 A01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_FM.indd 1 12/8/2014 11:15:06 AM 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. No part of this eBook may be used or reproduced in any manner whatsoever without the 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 NETWORK_ANALYSIS_AND_SYNTHESIS_Copyright_Page.indd 1 7/7/2015 9:58:24 AM To my mother Late Smti. Shaila Bala Bhattacharya —S. K. Bhattacharya To my mother Smti. Paramjeet Kaur —Manpreet Singh A01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_FM.indd 3 12/8/2014 11:15:06 AM Thispageisintentionallyleftblank 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 A01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_FM.indd 5 709 12/8/2014 11:15:06 AM Thispageisintentionallyleftblank 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 A01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_FM.indd 7 56 56 12/8/2014 11:15:06 AM viii 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 A01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_FM.indd 8 155 12/8/2014 11:15:06 AM Contents ix 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 A01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_FM.indd 9 12/8/2014 11:15:06 AM x 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 A01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_FM.indd 10 322 12/8/2014 11:15:06 AM Contents xi 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 A01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_FM.indd 11 384 409 12/8/2014 11:15:07 AM xii 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 A01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_FM.indd 12 12/8/2014 11:15:07 AM Contents xiii 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 A01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_FM.indd 13 12/8/2014 11:15:07 AM xiv 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 A01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_FM.indd 14 12/8/2014 11:15:07 AM Contents xv 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 A01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_FM.indd 15 693 709 12/8/2014 11:15:07 AM Thispageisintentionallyleftblank 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. A01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_FM.indd 17 12/8/2014 11:15:07 AM xviii 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. A01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_FM.indd 18 12/8/2014 11:15:07 AM 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 A01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_FM.indd 19 12/8/2014 11:15:07 AM Thispageisintentionallyleftblank 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. Thispageisintentionallyleftblank 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. M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 1 12/4/2014 11:33:52 AM 2 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. M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 2 12/4/2014 11:33:53 AM 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) M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 3 12/4/2014 11:33:53 AM 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 12/4/2014 11:33:55 AM 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 12/4/2014 11:33:56 AM 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 M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 6 12/4/2014 11:33:58 AM 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. M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 7 12/4/2014 11:34:00 AM 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 12/4/2014 11:34:01 AM 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. M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 9 12/4/2014 11:34:02 AM 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. M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 10 12/4/2014 11:34:04 AM 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 M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 11 12/4/2014 11:34:05 AM 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 M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 12 12/4/2014 11:34:07 AM 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. M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 13 12/4/2014 11:34:08 AM 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 M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 14 12/4/2014 11:34:09 AM 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. M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 15 12/4/2014 11:34:09 AM 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 M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 16 12/4/2014 11:34:11 AM 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. M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 17 12/4/2014 11:34:12 AM 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 M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 18 12/4/2014 11:34:12 AM 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. M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 19 12/4/2014 11:34:13 AM 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 M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 20 IL(SC) = IS 12/4/2014 11:34:14 AM 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. M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 21 12/4/2014 11:34:14 AM 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. M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 22 12/4/2014 11:34:16 AM 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? M01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH01.indd 23 12/4/2014 11:34:16 AM 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] 12/4/2014 11:34:18 AM 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. M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 26 11/25/2014 3:42:50 PM 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 M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 27 11/25/2014 3:42:52 PM 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 M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 28 11/25/2014 3:42:54 PM 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.) M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 29 11/25/2014 3:42:56 PM 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 11/25/2014 3:42:58 PM 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. M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 31 11/25/2014 3:42:59 PM 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. M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 32 11/25/2014 3:43:01 PM 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 11/25/2014 3:43:03 PM 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 M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 34 (2.10) 11/25/2014 3:43:03 PM 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. M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 35 11/25/2014 3:43:05 PM 36 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. M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 36 11/25/2014 3:43:07 PM 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. M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 37 11/25/2014 3:43:09 PM 38 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. M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 38 11/25/2014 3:43:11 PM 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. M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 39 11/25/2014 3:43:12 PM 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 M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 40 11/25/2014 3:43:14 PM 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. M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 41 11/25/2014 3:43:16 PM 42 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 M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 42 11/25/2014 3:43:17 PM 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. M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 43 11/25/2014 3:43:19 PM 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 M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 44 11/25/2014 3:43:23 PM 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) 11/25/2014 3:43:25 PM 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 M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 46 11/25/2014 3:43:26 PM 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 M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 47 11/25/2014 3:43:29 PM 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 M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 48 11/25/2014 3:43:30 PM 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 ) M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 49 11/25/2014 3:43:33 PM 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. M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 51 11/25/2014 3:43:37 PM 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) 11/25/2014 3:43:38 PM 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. M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 54 11/25/2014 3:43:42 PM 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 M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 55 11/25/2014 3:43:44 PM 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 11/25/2014 3:43:50 PM 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 M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 62 11/25/2014 3:44:00 PM 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 M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 70 11/25/2014 3:44:19 PM 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) 11/25/2014 3:44:21 PM 72 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 M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 73 11/25/2014 3:44:24 PM 74 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. M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 74 11/25/2014 3:44:26 PM 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 M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 75 11/25/2014 3:44:27 PM 76 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 = M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 76 11/25/2014 3:44:28 PM 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 M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 77 11/25/2014 3:44:30 PM 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 M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 79 11/25/2014 3:44:34 PM 80 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. M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 80 11/25/2014 3:44:37 PM 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 M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 81 11/25/2014 3:44:39 PM 82 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] M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 82 11/25/2014 3:44:41 PM 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] M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 83 11/25/2014 3:44:42 PM 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] M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 84 11/25/2014 3:44:42 PM 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] M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 85 11/25/2014 3:44:44 PM 86 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 M02_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH02.indd 86 11/25/2014 3:44:45 PM 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 11/25/2014 3:44:46 PM 88 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 11/25/2014 3:44:47 PM 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 11/25/2014 3:44:48 PM 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 90 11/17/2014 4:46:54 PM 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. M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 91 11/17/2014 4:46:56 PM 92 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 92 11/17/2014 4:46:57 PM 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. M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 93 11/17/2014 4:46:59 PM 94 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. M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 94 11/17/2014 4:47:00 PM 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, M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 95 11/17/2014 4:47:01 PM 96 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. M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 96 11/17/2014 4:47:02 PM 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 97 11/17/2014 4:47:04 PM 98 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 98 11/17/2014 4:47:05 PM 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 99 11/17/2014 4:47:08 PM 100 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 100 11/17/2014 4:47:09 PM 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. M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 101 11/17/2014 4:47:10 PM 102 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 102 11/17/2014 4:47:11 PM 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 11/17/2014 4:47:13 PM 104 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 104 11/17/2014 4:47:14 PM 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). M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 105 11/17/2014 4:47:15 PM 106 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 106 11/17/2014 4:47:16 PM 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 107 11/17/2014 4:47:17 PM 108 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 108 I 3 = V /Z 3 11/17/2014 4:47:19 PM 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 109 11/17/2014 4:47:21 PM 110 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 110 11/17/2014 4:47:22 PM 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 111 G O X Parallel Circuit 11/17/2014 4:47:24 PM 112 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 112 11/17/2014 4:47:25 PM 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 11/17/2014 4:47:28 PM 114 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° M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 114 11/17/2014 4:47:30 PM 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 115 11/17/2014 4:47:31 PM 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 116 The addition and subtraction of phasor quantities are shown in Figure 3.28. 11/17/2014 4:47:34 PM 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. M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 117 11/17/2014 4:47:36 PM 118 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 118 11/17/2014 4:47:38 PM 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° M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 119 11/17/2014 4:47:40 PM 120 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. M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 120 11/17/2014 4:47:42 PM 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 121 11/17/2014 4:47:44 PM 122 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 122 11/17/2014 4:47:46 PM 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 123 11/17/2014 4:47:47 PM 124 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 124 11/17/2014 4:47:49 PM 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 11/17/2014 4:47:52 PM 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. M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 126 11/17/2014 4:47:53 PM 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 127 11/17/2014 4:47:55 PM 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° M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 128 11/17/2014 4:47:57 PM 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 11/17/2014 4:47:59 PM 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° M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 130 11/17/2014 4:48:00 PM 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 131 11/17/2014 4:48:02 PM 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 132 11/17/2014 4:48:05 PM 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 11/17/2014 4:48:07 PM 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. M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 134 11/17/2014 4:48:09 PM 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. M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 135 11/17/2014 4:48:11 PM 136 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 136 11/17/2014 4:48:12 PM 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 137 11/17/2014 4:48:15 PM 138 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 138 11/17/2014 4:48:16 PM 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? M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 139 11/17/2014 4:48:17 PM 140 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? M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 140 11/17/2014 4:48:17 PM 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] 11/17/2014 4:48:19 PM 142 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] M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 142 11/17/2014 4:48:19 PM 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 143 (c) be more than 1.11 (d) be 1 11/17/2014 4:48:20 PM 144 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 144 (c) Z = 5 Ω (d) Z = 7Ω 11/17/2014 4:48:22 PM 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 M03_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH03.indd 145 (c) 8 − j8 Ω (d) 8 + j0 Ω 11/17/2014 4:48:22 PM 146 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 M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 149 11/17/2014 4:54:06 PM 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. M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 150 11/17/2014 4:54:06 PM 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? M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 151 11/17/2014 4:54:07 PM 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. 11/17/2014 4:54:08 PM 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 M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 153 11/17/2014 4:54:09 PM 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 11/17/2014 4:54:11 PM 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. M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 155 11/17/2014 4:54:12 PM 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. M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 156 11/17/2014 4:54:13 PM 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. M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 157 11/17/2014 4:54:14 PM 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? M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 158 11/17/2014 4:54:15 PM 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 11/17/2014 4:54:16 PM 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 M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 160 11/17/2014 4:54:17 PM 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 M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 161 11/17/2014 4:54:18 PM 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. M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 162 11/17/2014 4:54:20 PM 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? M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 163 11/17/2014 4:54:21 PM 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 M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 164 11/17/2014 4:54:22 PM 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 = M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 165 V Zo 11/17/2014 4:54:23 PM 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 M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 166 11/17/2014 4:54:24 PM 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 ∞. M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 167 11/17/2014 4:54:25 PM 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 M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 168 (4.3) 11/17/2014 4:54:26 PM 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. M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 169 11/17/2014 4:54:27 PM 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] M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 170 11/17/2014 4:54:27 PM 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] M04_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH04.indd 171 11/17/2014 4:54:28 PM 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. M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 172 12/3/2014 7:57:19 PM 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 M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 173 12/3/2014 7:57:19 PM 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= M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 174 12/3/2014 7:57:23 PM 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 = M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 175 4×2 8 = 4+2 6 = 1.33 Ω 12/3/2014 7:57:24 PM 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 M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 176 12/3/2014 7:57:28 PM 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. M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 177 12/3/2014 7:57:30 PM 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 M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 178 IL = VTh 10 = RTh + RL 4 + 10 = 0.714 A 12/3/2014 7:57:33 PM 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= M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 179 4×6 = 2.4 Ω 4+6 12/3/2014 7:57:35 PM 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. M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 180 12/3/2014 7:57:39 PM 181 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 M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 181 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 12/3/2014 7:57:41 PM 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 M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 182 12/3/2014 7:57:43 PM 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 12/3/2014 7:57:45 PM 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 M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 184 12/3/2014 7:57:47 PM 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. M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 185 12/3/2014 7:57:50 PM 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. M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 186 12/3/2014 7:57:51 PM 187 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 M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 187 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) 12/3/2014 7:57:53 PM 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 M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 188 12/3/2014 7:57:54 PM 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′ = M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 189 12/3/2014 7:57:56 PM 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 M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 190 12/3/2014 7:57:59 PM 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 12/3/2014 7:58:00 PM 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 M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 192 12/3/2014 7:58:02 PM 193 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 M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 193 12/3/2014 7:58:04 PM 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 12/3/2014 7:58:05 PM 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 12/3/2014 7:58:07 PM 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 M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 196 12/3/2014 7:58:10 PM 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 M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 197 12/3/2014 7:58:12 PM 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 M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 198 12/3/2014 7:58:15 PM 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 12/3/2014 7:58:16 PM 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 12/3/2014 7:58:19 PM 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 M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 201 12/3/2014 7:58:21 PM 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 = M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 203 12/3/2014 7:58:24 PM 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. M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 204 12/3/2014 7:58:24 PM 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. M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 205 12/3/2014 7:58:25 PM 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 = M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 206 12/3/2014 7:58:27 PM 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 = 12/3/2014 7:58:39 PM 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 = 12/3/2014 7:58:45 PM 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 M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 215 12/3/2014 7:58:46 PM 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 M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 216 12/3/2014 7:58:48 PM 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. M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 217 Figure 5.104 12/3/2014 7:58:51 PM 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 M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 218 = − j 4( 2 + j 6) +4 2 + j6 − j 4 12/3/2014 7:58:53 PM 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 12/3/2014 7:58:55 PM 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 12/3/2014 7:58:58 PM 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 12/3/2014 7:59:00 PM 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° M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 222 12/3/2014 7:59:02 PM 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. M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 223 12/3/2014 7:59:05 PM 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] 12/3/2014 7:59:06 PM 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] M05_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH05.indd 225 12/3/2014 7:59:08 PM 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. M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 226 12/3/2014 8:01:15 PM 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 M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 227 12/3/2014 8:01:16 PM 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 M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 228 12/3/2014 8:01:19 PM 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 M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 229 12/3/2014 8:01:22 PM 230 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. M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 230 12/3/2014 8:01:24 PM 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 = M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 231 12/3/2014 8:01:27 PM 232 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 M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 232 12/3/2014 8:01:30 PM 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 234 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 M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 234 12/3/2014 8:01:34 PM 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 M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 235 12/3/2014 8:01:36 PM 236 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 M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 236 = 6(1 − 0.368) i = 3.792 A 12/3/2014 8:01:38 PM 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 = M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 237 12/3/2014 8:01:40 PM 238 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 12/3/2014 8:01:42 PM 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 12/3/2014 8:01:44 PM 240 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. M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 240 12/3/2014 8:01:46 PM 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 M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 241 12/3/2014 8:01:49 PM 242 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 M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 242 12/3/2014 8:01:51 PM 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 M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 243 12/3/2014 8:01:53 PM 244 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. M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 244 12/3/2014 8:01:54 PM 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 M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 245 12/3/2014 8:01:56 PM 246 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 M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 246 12/3/2014 8:01:59 PM 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 M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 247 12/3/2014 8:02:00 PM 248 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 M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 248 12/3/2014 8:02:02 PM 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 M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 249 (6.17) 12/3/2014 8:02:04 PM 250 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 12/3/2014 8:02:07 PM 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 12/3/2014 8:02:10 PM 252 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 M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 252 {∴ L = 1H} 12/3/2014 8:02:12 PM 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. M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 253 12/3/2014 8:02:15 PM 254 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 M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 254 12/3/2014 8:02:17 PM 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 wL 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 M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 255 12/3/2014 8:02:20 PM 256 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 M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 256 (6.35) 2 2 12/3/2014 8:02:22 PM 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 12/3/2014 8:02:24 PM 258 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 M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 259 12/3/2014 8:02:30 PM 260 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 ) M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 260 12/3/2014 8:02:33 PM 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) 12/3/2014 8:02:38 PM 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) 12/3/2014 8:02:40 PM 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 12/3/2014 8:02:44 PM 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 12/3/2014 8:02:47 PM 266 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 M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 266 12/3/2014 8:02:49 PM 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. M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 268 12/3/2014 8:02:53 PM 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 12/3/2014 8:02:58 PM 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 ] M06_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH06.indd 270 12/3/2014 8:02:58 PM 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. M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 271 12/4/2014 2:06:21 PM 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 M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 272 12/4/2014 2:06:22 PM 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 M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 276 12/4/2014 2:06:25 PM 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 12/4/2014 2:06:28 PM 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 ) M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 279 12/4/2014 2:06:29 PM 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 M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 280 12/4/2014 2:06:30 PM 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 M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 281 eq − e −q ∵ sinh q = 2 12/4/2014 2:06:31 PM 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) M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 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) M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 283 12/4/2014 2:06:33 PM 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 M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 284 [ ∵ 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) M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 285 12/4/2014 2:06:34 PM 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 12/4/2014 2:06:35 PM 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) = M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 287 s2 − 9 ( s 2 + 9) 2 12/4/2014 2:06:35 PM 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 M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 288 12/4/2014 2:06:36 PM 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 ] = M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 289 2 s 2 + 22 . 12/4/2014 2:06:37 PM 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 M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 290 12/4/2014 2:06:38 PM 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 M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 291 12/4/2014 2:06:40 PM 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→∞ M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd s→∞ 12/4/2014 2:06:41 PM 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 12/4/2014 2:06:42 PM 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) M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 294 12/4/2014 2:06:44 PM 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 M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 295 12/4/2014 2:06:45 PM 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) M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 296 12/4/2014 2:06:47 PM 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 M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 297 ∞ 0 12/4/2014 2:06:48 PM 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 M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 298 s (t ) = du (t ) dt 12/4/2014 2:06:50 PM 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 M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 299 12/4/2014 2:06:53 PM 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 M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 300 12/4/2014 2:06:56 PM 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) 12/4/2014 2:06:57 PM 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 M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 302 12/4/2014 2:06:59 PM Laplace Transform 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 ) M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 303 12/4/2014 2:07:01 PM 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) M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 304 12/4/2014 2:07:02 PM Laplace Transform 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 M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 305 12/4/2014 2:07:04 PM 306 Network Analysis and Synthesis 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 = M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 306 (7.23) −1 2 12/4/2014 2:07:06 PM Laplace Transform 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 ) M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 307 12/4/2014 2:07:07 PM 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 M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 308 12/4/2014 2:07:10 PM Laplace Transform 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 M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 309 12/4/2014 2:07:11 PM 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 M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 310 12/4/2014 2:07:13 PM 311 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 M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 311 12/4/2014 2:07:15 PM 312 Network Analysis and Synthesis 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 M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 312 12/4/2014 2:07:16 PM 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 M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 313 12/4/2014 2:07:18 PM 314 Network Analysis and Synthesis 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 12/4/2014 2:07:19 PM 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) M07_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH07.indd 316 12/4/2014 2:07:24 PM 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 12/4/2014 2:07:35 PM 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. M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 320 11/17/2014 5:19:26 PM 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. M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 322 11/17/2014 5:19:32 PM 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) M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 324 11/17/2014 5:19:35 PM Transient Response of Circuits Using Laplace Transform = I ( s) = 325 V R Ls s + L 1 V R L ss + 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 M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 326 11/17/2014 5:19:38 PM Transient Response of Circuits Using Laplace Transform 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 11/17/2014 5:19:40 PM 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) M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 328 11/17/2014 5:19:42 PM 329 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 11/17/2014 5:19:43 PM 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 ) M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 330 11/17/2014 5:19:45 PM 331 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 M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 331 11/17/2014 5:19:46 PM 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) 11/17/2014 5:19:47 PM 333 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 11/17/2014 5:19:49 PM 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 11/17/2014 5:19:49 PM 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 M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 335 11/17/2014 5:19:50 PM 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 M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 336 11/17/2014 5:19:51 PM 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 M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 337 11/17/2014 5:19:53 PM 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) 11/17/2014 5:19:54 PM 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 M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 341 11/17/2014 5:19:58 PM 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 Ls + 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. M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 342 11/17/2014 5:19:59 PM 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 M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 343 11/17/2014 5:20:01 PM 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) M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 344 11/17/2014 5:20:02 PM 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) = M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 345 11/17/2014 5:20:05 PM 346 Network Analysis and Synthesis V 1 ⋅ 2R L ss + 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 M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 346 11/17/2014 5:20:06 PM 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) 11/17/2014 5:20:09 PM 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) = M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 349 11/17/2014 5:20:11 PM 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 M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 350 s 1 00 V 5Ω + − i 10 H Figure 8.31 11/17/2014 5:20:13 PM 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 M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 351 11/17/2014 5:20:14 PM 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 + M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 352 1 10 -6 ∫ idt = 80 11/17/2014 5:20:16 PM 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 M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 353 11/17/2014 5:20:17 PM 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) = M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 354 16 A B = + s( s + 4 ) s s + 4 11/17/2014 5:20:19 PM 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] M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 355 11/17/2014 5:20:19 PM 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] M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 356 11/17/2014 5:20:20 PM 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 M08_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH08.indd 357 11/17/2014 5:20:21 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 358 11/17/2014 5:33:24 PM 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. M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 359 11/17/2014 5:33:25 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 360 11/17/2014 5:33:25 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 361 11/17/2014 5:33:26 PM 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. M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 362 11/17/2014 5:33:28 PM 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. M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 363 11/17/2014 5:33:28 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 364 11/17/2014 5:33:28 PM 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. M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 365 11/17/2014 5:33:29 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 366 11/17/2014 5:33:29 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 367 11/17/2014 5:33:30 PM 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) M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 368 (9.12) (9.13) 11/17/2014 5:33:31 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 369 11/17/2014 5:33:32 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 370 11/17/2014 5:33:33 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 371 11/17/2014 5:33:34 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 372 11/17/2014 5:33:35 PM 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 11/17/2014 5:33:36 PM 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? M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 374 11/17/2014 5:33:37 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 375 11/17/2014 5:33:38 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 376 \ Total power = 3 × VPh IPh cos f 11/17/2014 5:33:39 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 377 11/17/2014 5:33:40 PM 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 ) M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 378 11/17/2014 5:33:41 PM Three-phase Systems and Circuits 379 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? M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 379 11/17/2014 5:33:42 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 380 11/17/2014 5:33:43 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 381 11/17/2014 5:33:44 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 382 11/17/2014 5:33:45 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 383 = W1 + W2 = 15 + 0 = 15 kW 11/17/2014 5:33:45 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 384 11/17/2014 5:33:46 PM 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] M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 385 3 1 32 + 4 2 ∠ tan −1 11/17/2014 5:33:47 PM 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? M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 386 11/17/2014 5:33:47 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 387 11/17/2014 5:33:48 PM 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. M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 388 11/17/2014 5:33:49 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 389 11/17/2014 5:33:50 PM 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). M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 390 11/17/2014 5:33:51 PM Three-phase Systems and Circuits 391 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 391 11/17/2014 5:33:52 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 392 X 1 = 1.649 Ω 11/17/2014 5:33:52 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 393 11/17/2014 5:33:53 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 394 11/17/2014 5:33:54 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 395 11/17/2014 5:33:55 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 396 11/17/2014 5:33:56 PM 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. M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 397 11/17/2014 5:33:56 PM 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. M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 398 11/17/2014 5:33:57 PM 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. M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 399 11/17/2014 5:33:57 PM 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° M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 400 11/17/2014 5:33:58 PM 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. M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 401 11/17/2014 5:33:58 PM 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] M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 402 11/17/2014 5:33:58 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 403 11/17/2014 5:33:59 PM 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. M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 404 11/17/2014 5:33:59 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 405 11/17/2014 5:34:00 PM 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 M09_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH09.indd 406 (b) two-wattmeter method only (d) any one of the above. 11/17/2014 5:34:00 PM 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 11/17/2014 5:34:01 PM 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 M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 408 11/17/2014 5:36:27 PM 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 M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 409 11/17/2014 5:36:29 PM 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 M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 410 11/17/2014 5:36:31 PM 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. M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 411 11/17/2014 5:36:31 PM 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). M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 412 11/17/2014 5:36:32 PM 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. M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 413 1 H 4 2F Figure 10.7 11/17/2014 5:36:33 PM 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 11/17/2014 5:36:35 PM 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. M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 415 1.0 F 3H 2 4F 3 1H 3 Figure 10.11 11/17/2014 5:36:36 PM 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) = M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 416 11/17/2014 5:36:37 PM 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 M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 417 11/17/2014 5:36:39 PM 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. M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 418 1Ω 4Ω 1 F 2 1 F 6 Figure 10.15 11/17/2014 5:36:40 PM 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 M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 419 11/17/2014 5:36:42 PM 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). M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 420 11/17/2014 5:36:42 PM 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 M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 421 11/17/2014 5:36:43 PM 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 11/17/2014 5:36:45 PM 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. M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 423 11/17/2014 5:36:47 PM 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. M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 424 11/17/2014 5:36:47 PM 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 M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 425 11/17/2014 5:36:48 PM 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 M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 426 11/17/2014 5:36:49 PM 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. M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 427 11/17/2014 5:36:50 PM 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. M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 428 11/17/2014 5:36:50 PM 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 11/17/2014 5:36:51 PM 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 M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 430 11/17/2014 5:36:52 PM 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; M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 431 P2 : s = − 8; Z1 : s = − 4 11/17/2014 5:36:53 PM 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 ) = M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 432 5s (s + 1)(s 2 + 4s + 8) 11/17/2014 5:36:53 PM 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) M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 433 11/17/2014 5:36:54 PM 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 ) 11/17/2014 5:36:55 PM 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 11/17/2014 5:36:56 PM 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 M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 436 11/17/2014 5:36:57 PM 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. M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 437 11/17/2014 5:36:58 PM 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 8s8s 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 M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 438 11/17/2014 5:37:00 PM 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) M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 439 11/17/2014 5:37:00 PM 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) 11/17/2014 5:37:03 PM 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) M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 442 11/17/2014 5:37:03 PM 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) M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 443 11/17/2014 5:37:04 PM 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) M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 444 Z(s ) 4 7 s 6 5 28 5s 98 Figure 10.44 11/17/2014 5:37:05 PM 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 M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 445 5 10 s + 30 2 3s 3s + 5 10 s + 30 3 + 3s 3s 11/17/2014 5:37:06 PM 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) M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 446 11/17/2014 5:37:06 PM 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 ) M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 447 11/17/2014 5:37:09 PM 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 M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 448 11/17/2014 5:37:11 PM 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. M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 449 11/17/2014 5:37:11 PM 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 M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 450 11/17/2014 5:37:12 PM 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. M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 451 11/17/2014 5:37:12 PM 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 M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 452 11/17/2014 5:37:14 PM 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. M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 453 11/17/2014 5:37:16 PM 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 M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 454 11/17/2014 5:37:18 PM 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. M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 455 11/17/2014 5:37:18 PM 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 ) = M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 456 1 ( s + 2) 2 (10.12) 11/17/2014 5:37:20 PM 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 M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 457 6 0 15 11/17/2014 5:37:21 PM 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. M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 458 [Ans. i = 4 e−2t] 11/17/2014 5:37:21 PM 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 M10_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH10.indd 459 [Ans. unstable] 11/17/2014 5:37:22 PM 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. M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 460 11/19/2014 4:16:21 PM 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 M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 461 11/19/2014 4:16:23 PM 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) M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 462 11/19/2014 4:16:23 PM 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 = M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 463 11/19/2014 4:16:24 PM 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 M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 464 11/19/2014 4:16:25 PM 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 M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 465 11/19/2014 4:16:26 PM 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 M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 466 11/19/2014 4:16:27 PM 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 M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 468 11/19/2014 4:16:29 PM 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. M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 471 11/19/2014 4:16:33 PM 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. M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 475 11/19/2014 4:16:43 PM 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. M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 476 11/19/2014 4:16:44 PM 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 M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 477 Y = Y12 Y22 Y11 Y12 Y21 Y22 −1 = 1 Y Y22 − Y12 −Y21 Y11 11/19/2014 4:16:46 PM 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 = M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 478 h 11 h 12 h 21 h 22 11/19/2014 4:16:47 PM 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. M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 479 11/19/2014 4:16:47 PM 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. M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 480 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 11/19/2014 4:16:49 PM 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 M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 481 11/19/2014 4:16:50 PM 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 M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 482 11/19/2014 4:16:51 PM 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 M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 483 Z11 = Z1 + Z2 Z12 = Z2= Z21 Z22 = Z2 + Z3 Z2 = Z12 = Z21 Z1 = Z11 − Z12 Z3 = Z22 − Z12 11/19/2014 4:16:52 PM 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 + M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 484 Z Y 11/19/2014 4:16:54 PM 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: M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 485 11/19/2014 4:16:55 PM 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 M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 486 11/19/2014 4:16:56 PM 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 M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 487 11/19/2014 4:16:59 PM 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 M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 488 I1 = 4 I 7 3 (11.68) 11/19/2014 4:17:00 PM 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 = M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 489 6 I 17 2 11/19/2014 4:17:02 PM 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 M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 490 11/19/2014 4:17:04 PM 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) M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 491 + V1 I1 1Ω + 3V − 1 I2 + 1.5 Ω V2 2Ω − − Figure 11.29 11/19/2014 4:17:05 PM 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. M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 492 11/19/2014 4:17:06 PM 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 M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 493 11/19/2014 4:17:07 PM 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) 11/19/2014 4:17:09 PM 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] M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 495 11/19/2014 4:17:09 PM 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) M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 496 11/19/2014 4:17:10 PM 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 M11_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH11.indd 497 11/19/2014 4:17:11 PM 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 11/19/2014 4:17:12 PM 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. M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 499 11/17/2014 6:13:48 PM 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. M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 500 11/17/2014 6:13:48 PM 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) M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 501 11/17/2014 6:13:48 PM 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. M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 502 11/17/2014 6:13:49 PM 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 503 11/17/2014 6:13:49 PM 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 504 s 4 + 5s3 + 8s 2 + 9 s + 6 11/17/2014 6:13:50 PM Network Synthesis and Realisability 505 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 505 s × = 6 ) 11/17/2014 6:13:51 PM ) 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. M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 506 11/17/2014 6:13:51 PM Network Synthesis and Realisability 507 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 507 11/17/2014 6:13:52 PM 508 Network Analysis and Synthesis 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. M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 508 11/17/2014 6:13:53 PM Network Synthesis and Realisability 509 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 34 s s2 + s 4 43 s2 3 3 ss 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 509 s = jw s = jw s = jw s = jw 11/17/2014 6:13:53 PM 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)) M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 510 11/17/2014 6:13:54 PM Network Synthesis and Realisability 511 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 511 11/17/2014 6:13:54 PM 512 Network Analysis and Synthesis 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 512 11/17/2014 6:13:55 PM Network Synthesis and Realisability 513 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). M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 513 11/17/2014 6:13:55 PM 514 Network Analysis and Synthesis 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 514 11/17/2014 6:13:56 PM Network Synthesis and Realisability 515 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. M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 515 11/17/2014 6:13:57 PM 516 Network Analysis and Synthesis 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 = M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 516 R4 = ⇓ s4 P4 R ∞ = P∞ 1 P4 11/17/2014 6:13:58 PM Network Synthesis and Realisability 517 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 11/17/2014 6:13:59 PM 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 518 11/17/2014 6:13:59 PM 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(∞) = M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 519 (1 + 0)(1 + 0) =1 (1 + 0)(1 + 0) 11/17/2014 6:14:01 PM 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) M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 520 11/17/2014 6:14:02 PM 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 11/17/2014 6:14:03 PM 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 522 11/17/2014 6:14:05 PM 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 = M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 523 11/17/2014 6:14:06 PM 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 11/17/2014 6:14:08 PM 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 525 11/17/2014 6:14:09 PM 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 526 11/17/2014 6:14:10 PM 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) M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 527 11/17/2014 6:14:11 PM 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 528 11/17/2014 6:14:12 PM 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) M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 529 11/17/2014 6:14:12 PM 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). M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 530 11/17/2014 6:14:14 PM 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’. M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 531 11/17/2014 6:14:14 PM 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(∞) M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 532 11/17/2014 6:14:15 PM 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 = M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 533 ( s + 1)( s + 3) 1 = s 2 s = −2 11/17/2014 6:14:16 PM 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) M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 534 11/17/2014 6:14:18 PM 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 535 11/17/2014 6:14:19 PM 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 536 1 2 s 5 11/17/2014 6:14:20 PM 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. M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 537 11/17/2014 6:14:21 PM 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 538 11/17/2014 6:14:22 PM 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 539 11/17/2014 6:14:22 PM 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) = M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 540 5/12 1/ 4 1/ 3 + + s s+4 s+6 11/17/2014 6:14:23 PM 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) M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 542 11/17/2014 6:14:25 PM 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 543 11/17/2014 6:14:26 PM 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 11/17/2014 6:14:27 PM 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 11/17/2014 6:14:29 PM 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 547 11/17/2014 6:14:30 PM 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 ) M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 550 11/17/2014 6:14:34 PM 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 551 11/17/2014 6:14:34 PM 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 11/17/2014 6:14:40 PM 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 11/17/2014 6:14:42 PM 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 11/17/2014 6:14:43 PM 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 11/17/2014 6:14:45 PM 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 11/17/2014 6:14:50 PM 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 564 11/17/2014 6:14:52 PM 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 11/17/2014 6:14:53 PM 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. M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 568 11/17/2014 6:14:56 PM 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 11/17/2014 6:14:57 PM 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 11/17/2014 6:15:05 PM 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 ) M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 577 2 11/17/2014 6:15:07 PM 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] M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 578 11/17/2014 6:15:08 PM 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] M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 579 11/17/2014 6:15:09 PM 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 580 11/17/2014 6:15:10 PM 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 581 11/17/2014 6:15:12 PM 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 582 11/17/2014 6:15:13 PM 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. M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 583 11/17/2014 6:15:13 PM 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 M12_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH12.indd 584 4. c 10. a 16. d 5. d 11. d 17. d 6. b 12. b 18. a 11/17/2014 6:15:14 PM 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. M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 585 12/3/2014 8:20:07 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 586 12/3/2014 8:20:08 PM Filters and Attenuators 587 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. M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 587 12/3/2014 8:20:09 PM 588 Network Analysis and Synthesis 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 ∞. M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 588 12/3/2014 8:20:09 PM Filters and Attenuators 589 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 589 12/3/2014 8:20:10 PM 590 Network Analysis and Synthesis 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 590 12/3/2014 8:20:11 PM Filters and Attenuators 591 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 591 Network is Made from 12/3/2014 8:20:11 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 592 12/3/2014 8:20:12 PM Filters and Attenuators 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 593 12/3/2014 8:20:12 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 594 12/3/2014 8:20:14 PM Filters and Attenuators 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) M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 595 12/3/2014 8:20:15 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 596 2 12/3/2014 8:20:17 PM Filters and Attenuators 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 597 g = 2 q q e + e = cosh q (13.13) ∵ 2 ∵ coshq = 1 + 2 sinh 2 q 2 Z1 4Z2 12/3/2014 8:20:20 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 598 sin b = 2 Z1 4Z2 b = 2 Z1 4Z2 j sin 12/3/2014 8:20:22 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 599 cos Z b = 1+ 1 2 2Z2 12/3/2014 8:20:24 PM 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. M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 600 12/3/2014 8:20:25 PM Filters and Attenuators 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 = M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 601 4 Z 12 Z 22 Z 12 + 4 Z 1Z 2 12/3/2014 8:20:27 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 602 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 12/3/2014 8:20:30 PM Filters and Attenuators 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. M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 603 12/3/2014 8:20:31 PM 604 Network Analysis and Synthesis 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. M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 604 12/3/2014 8:20:33 PM Filters and Attenuators 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 605 12/3/2014 8:20:34 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 606 jw L 4 jw C 12/3/2014 8:20:37 PM 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] M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 607 12/3/2014 8:20:39 PM 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 = M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 608 1 (13.26) p LC 12/3/2014 8:20:41 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 609 (iii) 12/3/2014 8:20:43 PM 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 12/3/2014 8:20:45 PM 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. M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 611 12/3/2014 8:20:47 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 612 12/3/2014 8:20:48 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 613 12/3/2014 8:20:50 PM 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, M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 614 12/3/2014 8:20:51 PM 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 < M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 615 Z1 <0 4Z2 12/3/2014 8:20:53 PM 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 12/3/2014 8:20:57 PM 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 12/3/2014 8:20:59 PM 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. M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 619 12/3/2014 8:21:01 PM 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. M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 620 12/3/2014 8:21:04 PM 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. M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 622 12/3/2014 8:21:07 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 624 12/3/2014 8:21:10 PM 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) 12/3/2014 8:21:12 PM 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 12/3/2014 8:21:13 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 627 12/3/2014 8:21:15 PM 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 12/3/2014 8:21:19 PM 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] M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 630 12/3/2014 8:21:20 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 631 12/3/2014 8:21:23 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 632 12/3/2014 8:21:25 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 642 12/3/2014 8:21:46 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 644 12/3/2014 8:21:50 PM 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. M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 645 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 2 648 Network Analysis and Synthesis 12/3/2014 8:22:00 PM Filters and Attenuators 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 649 12/3/2014 8:22:01 PM 650 Network Analysis and Synthesis 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 650 m 2 Z12 + mZ1Z 21 4 12/3/2014 8:22:03 PM Filters and Attenuators 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. M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 651 12/3/2014 8:22:06 PM 652 Network Analysis and Synthesis 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 652 12/3/2014 8:22:08 PM Filters and Attenuators 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 653 12/3/2014 8:22:11 PM 654 Network Analysis and Synthesis 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 654 −1 Z 4Z2 fc < f < f∞ Z1 4Z2 f > f∞ f C2 f ∞2 Z1 = jw L 1 jw C [from T -section ] Z2 = 12/3/2014 8:22:14 PM Filters and Attenuators 655 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) 12/3/2014 8:22:17 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 656 12/3/2014 8:22:19 PM Filters and Attenuators 657 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 657 12/3/2014 8:22:20 PM 658 Network Analysis and Synthesis 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 12/3/2014 8:22:22 PM Filters and Attenuators 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 659 2 12/3/2014 8:22:25 PM 660 Network Analysis and Synthesis 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 12/3/2014 8:22:28 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 661 12/3/2014 8:22:30 PM 662 Network Analysis and Synthesis 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 662 12/3/2014 8:22:32 PM Filters and Attenuators 663 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. M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 663 12/3/2014 8:22:34 PM 664 Network Analysis and Synthesis 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 ) M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 664 12/3/2014 8:22:37 PM 665 Filters and Attenuators 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. M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 665 12/3/2014 8:22:40 PM 666 Network Analysis and Synthesis 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− M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 666 ( f 2 − f1 ) 2 ( f ∞ 2 − f ∞1 ) 2 12/3/2014 8:22:42 PM 667 Filters and Attenuators 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 667 12/3/2014 8:22:43 PM 668 Network Analysis and Synthesis 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. M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 668 12/3/2014 8:22:45 PM Filters and Attenuators 669 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 669 12/3/2014 8:22:47 PM 670 Network Analysis and Synthesis 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. M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 670 12/3/2014 8:22:48 PM 671 Filters and Attenuators 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 671 12/3/2014 8:22:50 PM 672 Network Analysis and Synthesis 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 672 12/3/2014 8:22:52 PM Filters and Attenuators 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. M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 673 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 12/3/2014 8:22:54 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 674 12/3/2014 8:22:56 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 675 12/3/2014 8:22:58 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 676 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 12/3/2014 8:23:02 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 677 12/3/2014 8:23:04 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 678 12/3/2014 8:23:07 PM Filters and Attenuators 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 12/3/2014 8:23:10 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 680 12/3/2014 8:23:12 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 681 12/3/2014 8:23:15 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 682 ∵ from equation (13.52) R + R + R 2 0 1 =N R2 12/3/2014 8:23:16 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 683 12/3/2014 8:23:18 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 684 12/3/2014 8:23:20 PM 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) M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 685 R1 R0 R2 R2 R0 Figure 13.105 12/3/2014 8:23:23 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 686 12/3/2014 8:23:26 PM Filters and Attenuators 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 687 12/3/2014 8:23:29 PM 688 Network Analysis and Synthesis 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 688 Figure 13.110 12/3/2014 8:23:30 PM Filters and Attenuators 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 689 12/3/2014 8:23:33 PM 690 Network Analysis and Synthesis 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 690 12/3/2014 8:23:36 PM Filters and Attenuators 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 ] M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 691 R0 R1 / 2 R0 2R2 Short Circuit Figure 13.115 12/3/2014 8:23:39 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 692 12/3/2014 8:23:41 PM 693 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. M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 693 C =1µF 2 C =1µF 2 Figure 13.119 12/3/2014 8:23:44 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 694 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 695 Z1 = 800 − 400 = 400 Ω 2 12/3/2014 8:23:52 PM 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) M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 696 12/3/2014 8:23:54 PM Filters and Attenuators 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 = M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 697 12/3/2014 8:23:57 PM 698 Network Analysis and Synthesis 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. M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 698 12/3/2014 8:24:00 PM Filters and Attenuators 699 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: M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 699 12/3/2014 8:24:02 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 700 12/3/2014 8:24:05 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 701 12/3/2014 8:24:08 PM 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] 12/3/2014 8:24:09 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 703 12/3/2014 8:24:11 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 704 12/3/2014 8:24:12 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 705 (c) R–C network (d) L–C network. 12/3/2014 8:24:12 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 706 (c) ∞ (d) (b) Below the cut-off frequency (d) Infinitely high 12/3/2014 8:24:13 PM 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 M13_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_CH13.indd 707 20. b 21. d 22. a 23. c 12/3/2014 8:24:14 PM 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 12/4/2014 11:32:33 AM 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 12/4/2014 11:32:33 AM 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 Z01_NETWORK_ANALYSIS_AND_SYNTHESIS_XXXX_INDEX.indd 711 12/4/2014 11:32:33 AM 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 12/4/2014 11:32:33 AM 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 12/4/2014 11:32:33 AM 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 12/4/2014 11:32:33 AM