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ELECTRIC CIRCUITS EIGHTH EDITION
W.Nilsson James Prcfessor Emeritus Iowa StateUnirelsitJ
Susan A. Riedet Marquette Unive$ity
Upper SaddleRiver,New Je$ey 07458
Dataot File Libtott of CongresCataloginginPublicatiafl
Vice President and EditoriatDilectot,Ecs: Marcia J Hotton Editorial Assistanr Carck Sntder Acquisitions Editor: Mt.rdel McDonald Executive Managing Editor. Vince O'Blien ManagingBditor: David A. Georye hoduction Editor Rore reldli/t Director of Creative ServicestPaul Belfanti Cte3tile Dilector: han Lopez Aft Djnector. Ionethan B oyldn Managing Editor, AV Management and hoduction: Patdctd au /ru AfiEditor Xidohong Zhu Photo Researcher:,tcon Drsarro Manu{acturing Manager:/4lerisHeydt Long ManuJacturing Buyer Zbd McDo'", Marketing Managei ?m Grrigan
O 2008,2m5,2001,2000.1996PearsonEducation,Inc. PeaftonPrenticeHall PeaNonEducation.Inc. Upper Saddle River, NJ 07458 All rights resemed.No parl of this book rnay be reproduced, in any form or by any means,without permission in witing from the publisher. Peanon Prentice Hall6 is a trademark of PearsonEducation,Inc The aurhor and publisher ofrhis book have used their best efforts in preparing this book.These efforts include the development, research,and testing of the theoiies and prograns to determine their effectiveness The author and publisher mate no warranty of any kind, exFessed or inplied. with regard to these programs or the documentation contained in this book. The aulhor and pubtisher shal not be liable in my event for incidental or consequential danages in connection with, or adsiDg out of, the fumishhg, performance,or use of these prcgams PSpi@ is a regislered trademark of Caden@ Design Systems. hinted in the United States of America 1 0 9 8 7 6 5 4 3 2 r
r s B N0  l , l  l , t s 1 ? 5  r Pedson Education Ltd., ro"dd Pedson Education Australia Pty. Ltd., S)dne] PearsonEducation Singapoie, fte. Ltd. PearsonEducation North Asia Ltd., Hong KonC PearsonEducation Canada lnc., ?r/o,to PearsonEducaci6n de Mexico, S.A. de C.V PearsonEducation Japan, k , PearsonEducation Malaysia, Pte. Ltd. Peanon Education, lnc., Upper Sadt e Rire,. NewJersey
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PhotoCredits coyer Image Courtesy of Coibis/RF, Royalty Free. Chapter 2 Practical Pe$pecaiy€r Courtesy of Corbis/NY, Ron Chapple. Figue 29: Couitesy of Corbis/NY, Thom Lang. Chapt€r 3 Practical Per6pective: Coufiesy of Getty Images/Creative Express. Chapler 4 Practical Perspective: Couitesy of Getty Images/Ciealive Express. Chapt€r 5 Practical PeNpectiver Courtesy o{ Getty lmag€s/PhotodiscRed, Akira Kaede. Chapter 6 Practical P€rspective: Courtesy of Getty Images,iPhotonica,Ron Rovtar. Chapter 7 Practicel Perspective: Courtesy ol Getty Images/PhotodiscGreen. Chapter 8 Pnctical P€Bpective: Courtesy of Getty Images/Image Source Royalty Free,Image Source Pink. Chspter 9 Prsctical P€rspectivs Courtesy of Getty Images/PhotodiscGreen, Steve Cole. Chapter 10 Practical Penp€crive: Courtesy of Alamy Images Royalty Free. Chapter 11 Practical Persp€ctive: Cou esy of Coibis/NY, Rolf Vennenbernd. Chapter 13 Practicsl Persp€ctive: Courtesy of Getty Images/PhotodiscBlue. Chapter 14 Praclical Persp€ctive: Courtesy of Corbis/RE Tom Grill. Chapter 15 Pncticrl P€rspective: Courtesy of Corbis^lY Dana Hoff.
Preface The eighth edition of Electlrc Cilclrr is a carefully planned revision to the most widely used introductory circuits text of the past 25 years.As this book has evolvedover the yearsto meet the changinglearningstylesof students,importantly,the underlying teachirg approachesand philoso phiesremainunchanged.Thegoalsare: . To build an understandingof conceptsand ideasexplicitlyin termsof previousleardng. . To emphasizethe relationshipbetween conceptualunderstanding and problemsolvingapproaches. . To provide studentswith a strong foundation of engineeringpractices
WHYTHISEDITION? When planr rg for the eighth editjon ievision of Electic Citcuits, carcfnl thought was given to how \ve should best update this classictext to improve upon the success o{ precedingeditionsandmakethe eightl edition ascom, pelliry as the first.Througha thoroughreviewprocessthat includedboth inshuctors and students who currendy use tlectli. C;rc&itsand those who use other texts, our revision plan was formed. wllat emerged from this exercisewas a clear pictue of what matters most to instructors and stu derts. With this feedback in mind, we made the following changes: . Problem solving is fundamental to the study of circuit analysis.The authorsput their pdmary eftbrt into updatingand addingnew endofchapter problems. The result is a ftesh text with approximately 80o/"new or revised problems comparedto the previous edition. Having a wealth ofnew problemsto assignand work is a key to suc cessin any circuits course. . The eighth edition iepresentsa major redesignto the text. Careful attentionwaspaid toward how to presentthe material tex1,figures, and aitworkin a clean,clear mannerthat would facilitatelearning and encouragereading.The seventhedition was the fi$t introductory circuits text to recognize the changing needs of today's students with a modern,fourcolordesign.The eightheditionrefinesthis color treatmentwith a morc pedagogjcallycoherentpresentation. ' Navigation was imprcved by the addition of page numbers to the chapterobjectives,less rcliance on icons whete nameswere more effective, ard the updated organization of endofchapter problems by sectionlevels Additionally,the layout was enhancedto limit the instarceswhereExamplesspill over onto multiple pages. . All artwork, pholos, and imageshave been modernized and enhanced to present a crisper illustration of the key elementsand application of circuit analysis.
. Recognizing ttrat moie classpreparation and studying is happening online wilh t}re use of additional resources, ttre development of online resources for tle eighth edition represents a sigdfica[t improvement from the seventh edition. From online, automatically graded homework, to study aids and an ebook, all of this and more is now available on an easy{onavigate website for students and College textbooks excel at presenting complicaled mateiial in a clear, straigltforward mannei, Authon and publishers spend countless hous developingthe bestpossibleleamingaid for studentsand teachhg aid for instructors. Prcntice Hall is committed to working with authors to create textbooksandsupporthg resourcesthat enablebetterteachingandbetter student leamhg. The eigh0l edition of tlectric Cir.cr.itr is one such example,It setthe standardfoi circuitseducation25 yean ago and il continues that trend today.
HALLMARK FEATURES Chapter Problems Userc of Electic C cuits have consistently rated the Chapter Problems as one ofthe book'smost attmctivefeatures.Inthe eighth edition,theie are ovei 1000problems with approximately 80% that are new or revised from the previous edition. Problems are organized at the end of each chapter by
PracticalPerspectives The eighth edition continues the use of Practical Perspective inftoduced with the chapter openers.They offer examplesof realworld circuits, taken from realworld devices.Most chapters begin wittr a bdef descdption of a practical application of the material that follows. Once the chapter mateda1 is presented,the chapter concludes witl a quantitative analysis of the application along with a Practical Perspective problem. This enables you to unde$tand how to apply the chapter contents to the solution of a realworld problem.
Assessment Problems Each chapter begins with a set of chapter objectives At key points in the chapter,you are askedto stop and assessyoul masteryof a particular problems.Ifyou are able to objectiveby solvingore or more assessment problems givel solve the assessment for a objective, you have mastered that objective.
Examples Every chapter includes many examples that illustrate the concepts presented in the text in the folm of a numeric example. There are over 130 examples in this text. The examples are intended to illustrate the applicationoI a particularconcept,and alsoto encouragegood problemsolving skilts.
Equations andConcepts Fundamental Throughout the text, you will seefundamental equatioN and concepts set apart from the main text.This is done to help you focus on some of the key piinciples in electric circuits and to help you navigate tbrough the important topics.
Tools Integrationof Computer Computer tools can assiststudents in the learning processby pioviding a visual representation oI a circuit's behavior, validating a calculated solu_ tion, rcducing the computational burden of morc complex circuits, and iterating toward a desired solution using parameter variation. This computational suppofi is often invaluable in the design process The eighth edition ircludes the support of PSpice, a popular computer tool Chapter problems suited for explomtion with Pspice are so marked.
Emphasis Design The eighth edition continues to support the emphasison the design of circuits in many ways First, several of the Practical Perspective discussions focus on the design aspects of the circuits. The accompanying Chapter Problems continue the discussion of the design issuesin these pnctical examples.Second,desiglodentedChapterPrcblemshave been labeled explicitly, enabling students and instructols to identify those problems with a design focus.Third, the identification of problems suited to explo_ ration wit}I Pspice suggestsdesign opporturdties using this soflware
Accuracy All text and problems in the eig.hthedition have undergone our strict hallmark triple accuracychecking process,to ensure the most errorftee book possible.
ANDINSTRUCTORS FORSTUDENTS RESOURCES ha[[.con/ni lsson www.pren The eighth edition of Elsctlic Circuits comeswith Prentice Hall's poweful new suite of student and instructol online resources
ForStudentsi Online homework and Dractice with immediate feedback and inte graredebookuslDgPH UradeAssl\l . Online Study Guide that highlights the key conceptsof electric circuits . Additional book and coursespecificresouces
,
ForInstructors: . Assignable, automatically graded online homework with PH cradeAssist . Digital version of all figur€s from the book . Interactive Learning classroomPowerPoint slides . Sample Chapter Tests . Additional book and coursespecificresources
ADDITIONAL OFFLINE RESOURCES ForStudents: . Student Study PackThis new resoutce teachesstudents techniques for solvingproblemspresentedin the text. Organizedby concepts, this is avaluableproblemsolvingresourcefor all levelsof students. . Introduction to Pspice ManualUpdated for the eighth edition, the manual comeswith the latest available releaseof the software on CD.
ForInstructors: . Instructor Solutions ManualFully worked out solutions to endofchapter problems. ' Instructor Problem Bank A tremendous new resoruce with many additional problems and conespo ding solutions to problems not found in the text.This is a great tool for qeating homework and exams.
0rdering0ptionsl Ek!:tric Citcuits wilh PH Grade Assist Online Homework Access: ISBN 0135142911 Elecftic Citcuits wft SfitdentStudt Pa&:ISBN 0,13 51,42903 Eled c Circuits \NithInrrcduction to Pspice Ma utl..ISBN 013,514292X
PREREQUISITES In writing tlte fint 12 chapters of the text, we have assumed tlat the reader has taken a course in elementary differential and integral calculus. We have also assumedthat the rcader has had an introductory physics coune, at either the high school or university level, that iritroduces the concepts ol energy,power, electric charge,electric current, electdc poten, tial, and electromagnetic fields. In writing the final six chapters,we have assumedthe studenthas had,or is enrolledin, an introductorycoune in differential equations.
c0uRsE 0m0Ns The lext hasbeen designedfor usein a onesemestei, twosemester, or a thieequartersequence. . SinglesemestetcourserAlter coveringChapters14 and (hapten 6 10 (omitting Sections?.7 and 8.5) the instructor can choosefrom Chaprer5 (operational amplifiels), Chapter 11 (tbreephasecircuirs).Chapters13
and 14 (hplac€ methods), and Chapter 18 (TkoPort Circuits) to develop the desiredemphasis . TtrosemestersequencerAssuming three lectuies per week, tle fiIst nine chapters can be covered during the first semester, leaving ChaptersI0 l8 for thesecond.emesler. ' Academicquartu schedrlerThe book can be subdividedinto three parts:Chapten 16, Chapters712,and Chapters1318. The introduction to operational amplifier circuits can be omitted vithout interfercnce by the rcader going to the subsequentchapten. For examplq if Chapter 5 is omitted, the instructor can simply skip Section7.7,Sectron85, Chapter 15, and those problerns and assessingobjective problems in the chaptercfolowing Chapter 5 that pertain to operational amplifiers. There are seveml appendixesat the end of the book to help readers make effective useof t]left matlematical background.Appendix A review$ Cramer's method of solving simultaneous lirear equations and simple matrix algebra;complex numbers are reviewed ir Appendix B;Appendix C contains additional mateiial on magnetically coupled coils and ideal transfomels; Appendi{ D contabs a biiel discussior of the decibeliAppendixE is dedicated to Bode diagEmsiAppendix F is devoted to an abbreviated table of trigonometdc identities that are useful in circuit analysis;and an abbreviated table of useful integrals is given in Appendix G Appendix H prcvides answersto selectedsuggestedprcblems
ACKNOWLEDGMENTS We continue to expressour appreciation for the contributions of Norman Wittels of Woicester Polytechnic Institute. His contributions to the Practical Perspectives geatly enhanced both this and the previous two editions. Jacob Chacko, a transmission and distribution engineer at the Ames Municipal Electric System, also contributed to t}le Practical Perspecrives. Special tlanks also to Robert Yahn (USAF), Stephen O'Conner(USAF), andwilliam Oliver (BostonUniversity)for their con tinued intercst in and suggestionsfor this book. There were many hardworking people behind the scenesat our publisher who deserve our thanks and gmtitude for their efforts on behalf of the eigl[h editiorl At Pretrtice Hall, we would like to thank Michael McDonald, Rose Kernan, Xiaohong Zhu, Lisa McDowell, Jonathan Boylan, David A. Georye, Tim Galigan, and Scott Disanno for their continued suppo and a ton of really hard work. The authon would also like to acknowledge the stafl at GEX Publishing Servicesfor their dedication and hard woik in typesetting this text. The many revisions of the text were guided by carelul and thorough reviews tom professors.Our heanJelt thanks to: . Paul PanayotatoER tgerc University . Evan Goldsrein, University ofwathington . Kalpathy B. Sundaram, Uniretsity of Central Flotidtl . Andrew K. Chan, feta, A&M Unitersity . A. S^taaiJazi,Viryinia PolytechnicInstitute and State University . Clifford H. Grigg, RoseHulman Institute of Technology
. Karl Bithdnger, Univetsit! ofwashington . Carl Wells, Iryr8fttnglonState Unirercity ' Aydir I.Karsilayan,kasA&M UnireNity . Ramakant Sdvastava,University of Florida . Michel M. Mala$iz, Univetsity of Mithigan, Ann Arbor . Christopher Hoople, Rocheskr Institute ofTechnolosy . SannasiRamanan, RochesterItlstitute ofTechnologj . Gary A. Hallock, U/.ivenitt) ol Texasat Austin The authom would also like to thanl Ramakant S vastava of the Udversityol !loflda.aDdlhe AccuraryRe!iewTeamal CEX Publisbing Senices, for their help in checking ttre text and all the problems in the eighthedition. We are deeply indebted to the many instructors and students who have offered positive feedback and suggestionsfor improvement. We use as many of these suggestionsas possible to continue to improve the content, the pedagogy, and the presentation. We are honorcd to have the opportunity to impacr the educational expeiience ol the many thousands of future engheers who will tum the pagesof this text. JaMEsW. NlssoN A, REDEL SUSAN
BriefContents Chapter 1 Chapter 2 Chapter 3 Chapter4 Chapter5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Appendix A AppendixB Appendix C Appendix D Appendix E Appendix F Appendix G Appendix H
List of Examplesxiii Preface xvii CircuitVariables 2 CircuitEtements 22 SimpteResistiveCircuits 56 Techniques of CircuitAnalysis 92 The0perational Amplifier 154 Inductance, Capacitance, andMutualInductance186 Response of FirstorderRLandRCCircuits 228 NaturalandStepResponses of RtCCircuits 284 Sinusoidal SteadyState Analysis 330 SinusoidalSteadyState PowerCalculations390 Balanced ThreePhase Circuits 432 Introductionto the LaplaceTransform 466 TheLaplaceTransform in CircuitAnalysis 506 Introductionto Frequency SelectiveCircuits 566 ActiveFitterCircuits 606 FourierSeries 656 TheFourierTransform 698 TwoPort Circuits 730 TheSotutionof LinearSimultaneous Equations759 Complex Numbers781 Moreon Magneticatly Coupted CoilsandldealTransformers 787 TheDecibel 797 BodeDiagrams799 An Abbreviated Tableof Trigonometric Identities 817 An Abbreviated Tableof Integrals 819 Answers to Selected Problems821 Index 839
Contents Listof Examptesxiii Prefacexvii Chapter1 CircuitVariables 2 1.1 1.2 1.3 1.4 1.5 1.6
Etectri.atEngineering: An 0verview 3 TheInterrationatSystemof Units 8 CircuitAnalysis! An overview t0 VoltageandCurrent 11 TheldeatEasicCircuitEtement 12 PowerandEnergy 14 Sunnoty 16 Prcblehs 17
Chapter 2 CircuitElements22 2.1 2,2 2,3 2.4 2.5
PrdcticqlPerspective: E[ectri.slSoIety 23 VoltageandCuffentSources24 Etectrical Resistan(e (0hm'sLaw) 28 Constrlction of a CircuitModet 32 Kirchhoff's Laws 36 Anatysis of a CircuitContaining Dependent Souraes42 ProcticolPetspective: ElectficolSolety 46 Sumndry 47 Prcblens 48
Chapter 3 SimpleResistive Circuits 56 ProcticolPerspedive,A neor Window DeIrcster57 nesistors in Series 58 Resistors in Paraltel 59 TheVoltageDivider andCurrentDivider LITCUIIS
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VottageDivisionandCurrentDivision 65 Measuring VoltageandCurrent 68 Measuring ResistanceThe Wheatstone Bridge 71 3 . 7 DeltatoWye (PitoTee) EquivatentCircuits 73 PructicalPe5pedive:A RearWindow Defrcster 76 Sumnory 79 Problems 80
Chapter 4 Techniques of CircuitAnalysis 92 Pnctical Pe6pedive:Circuitswith Reolistic Resistors 93 4.7 Terminology94 4.2 Introduction to the NodeVol.tage Method 97 4.3 TheNodeVottage Methodand0ependent Sourcest00 4.4 TheNodeVoltage Metbod:SomeSpecial Cases 101 4.5 Introduction to the MeshCurrent MethodJ05 4.6 TheMeshCurrent MethodandDependent SourcesJoZ 4.7 TheMeshCurrent Method:SorneSpecial Cases 109 the NodeVottage MethodVersus the MeshCurrentMethod 112 4.9 Source Transfoimations116 4.10 Th6v€nin andNortonEquivalents 119 4.11 Moreon Deriyinga ThAvenin Equivatent12J 4.12 Ma)dmum Powerfransfer126 4.13 Superposition 129 Pncticol Pe6pectivetCitcuitswith Realistic Resistors 133 Sunnary 137 Problens 138
Chapter 5 TheOperational Amplifier 154 5.1 5.2 5.3 5.4 5,5 5.6 5.7
PrdcEcqlPe6pective:StruinGqges 155 operational AmptifierTerninals 156 lerminalVottages andCurrents 156 TheInvertingAmptifier Circuit 161 TheSummingAmptifier Circuit 163 TheNoninvertingAmptifier Circuit 164 TheDifferenceA$plifier Cir.uit J65 A MoreReatistic Modelfor the 0perational Ampliliet 170 ProcticalPerspective: Sttuin 6ages 173 Sunnary 175 Problens 176
Chapter6 Inductance. Capacitance, andMutualInductance186 Pnctical Perspective: ProximitySwitches 187 TheInductor 188 TheCapacitor195 SeriesParaltel Cornbinations of Inductance andCapacitance200 6.4 MutuatInductance203 6.5 A CloserLookat MutualIndudance 207 PructicalPerspective: Prcxinity Sdtches 214 Sunmary 217 Prcblens 218
Chapter9 SinusoidaI SteadyState Analysis 330
6.1
9.1 9.2 9.3 9.4 9.5 9.6 9.7
Chapter7 Response of First0rderfl andflf, Circuits 228 7.1 7.2 7.3 7,4 7,5 7.6 7.7
PructicolPerspedive:A FloshingLight Citcuit 229 TheNaturalResDonse of an fl Circuit 230 TheNaturalResponse of an 8CCircuit 2j6 TheStepResponse ot fl. andRCCircuits 240 A Generat Solutionfor SteDandNatural Responses248 Sequentiat Switching 254 Unbounded Response258 TheIntegratingAmplifiet 260 ProdicdlPe6pective:A FlqshingLight Circuit 263 Sumnory 265 froDens
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Chapter8 NaturalandStepResponses of RLCCircuits 284 8.1 8,2 8.3 8.4 8.5
PtocticalPetspedive:Anlgnition Cir.uit 285 Introduction to the NaturalResDonse of a Palattet nIC Circuit 286 TheFormsof the NaturalResDonse of a Paratl.et RICCircuit 291 TheStepResponse of a Parallel RICCircuit 301 TheNaturatandStepResponse of a Series ilc Circuit 308 A Circuitwith TwoIntegratingAmplifiers 312 Ptodicol Percpective: An lgnition Circuit 317 Summam Szu
Probte;321
9.8 9.9 9.10 9.11 9.12
PracticdlPetspective: A Househotd Distribution Circuit 331 TheSinusoidal. Sowce 332 TheSinusoidal. ResDonse335 ThePhasor 337 ThePassive CircuitElements in the Freouencv Domain 342 Kirchhoff!Lawsin the Frequency Donain 346 Series.Parallel. andDeltatoWye SimDlifi(ations348 Source Transformations andTh6veninNorton Eouivalent Circuits 355 lhe NodeVottage Method 359 TheMeshCurrent Method 360 TheTransformer361 TheldealTransformer365 PhasorDiagramst72 Pnctical Percoective: A HouseholdDistribution Circait 375 Sunnary 375 rnoDtems J/b
Chapter10 Sinusoidal SteadyState PowerCalculations390 10.1 10.2 10.3 10.4 10.5 10.6
PructicolPerspedive:HeotingAppliancs 391 Instantaneous Povrer392 Average andReactive Power 394 ThermsValueandPowerCatculations399 Comptex Power 401 PowerCatcllations 403 Maximum Power Transfer 410 ProcticdlPerspedive:HeqtingApptisnces 417 Sumnoty 419 Prcblent 420
11 Balanced Chapter ThreePhase Circuits 432 11.1 11.2 11.3 11.4 11.5
ProcticolPe6pective:Trunsnission ondDlstibution oI Electic Powet 433 Balanced ThreePhase Voltages4j. ThreePhase VoltageSources435 Anatysis of the WyeWye Circuit 436 Anatysis of the WyeDelta Circuit 442 PowerCa[.ulations in Balanced ThreePhase Circsits 445
11.6 Measuring Average Powerin ThreePhase Llrcut$
452
Ptsdicol Perspedive:Transnission on.t DisttibutionoJElectic Powet 455 Su nary 456 ftootens
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Chapter 12 Introduction to the Laptace Transform 466 12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8 12.9
Definition of the LaplaceTransform 46l TheStep Function 468 TheImputseFunction 470 functionaITransfofids 474 0perationatTransfortf.s 475 Apptyingthe LaplaceTransform 481 InverseTransforms 482 Potesand Zerosof F(s) 494 Initial and FinatValue Theorems 495 Sumnaty 498 Ptoblens 499
Chapter 13 TheLaplace Transform in CircuitAnalysis 506 Pncticol Pe$pective:SurgeSuppressors507 1 3 . 1 CircuitElernents ifl the s Domain 508 13.2 f,ircuitAnatysis in the s Domain 5Jl 13.3 Applicatiohs512 13.4 TheTransfer function 526 TheTransfer Function in Partiatfraction Expansions528 13.6 TheTransfer Functionandthe Convotution Integrat 537 13,7 TheTransfer Fundionandthe Steadystate SinusoidalResponse 537 13.8 TheImpulsefunctionin Ciiclit Analysis 540 Procticd[Pe6pective:SurgeSuppressors548 Summary549 Probtems 550
Chapter 14 Introduction to Frequency SelectiveCircuits 566 PrccticalPe6pective:PushbuttonTelephone LtrcutEs 5b/
14.1 SomePretiminaries 568 14.2 LowPass Fillets 570 14.3 HighPass Filtets 577
14.4 Bandpass Fitters 582 14.5 Bandreject Fitters 593 ProcticqlPerspe.tive:PushbuttonTe[ephone Circuits 598 Sunnary 599 Problens 599
Chapter15 ActiveFilterCircuits 606 Prdcticdl PeBpedive: BqssVolume Confiol 607 15.1 Firstorder LowPass and HighPass 75.2 15.3 75,4 15.5
iitters 608 ScaLing612 0p AmpSandpass andBandreje€t Fitters 615 fligherorder0p AmpFitters 622 Narrowband Bandpass andBandreje€t ntters
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PructicolPerspective. EdssVolune Control 642 Sunndry 644 Prcblens 646
Chapter 16 Fourier Series 656 16.1 FourierSeriesAnalysis: An overview 658 16.2 ThefourierCoefficients659 16.3 TheEffectof Symmetry on the Fourier Coefficients662 16.4 An Alternative Trigonometric Formof the FourierSeries 668 16.5 AnApptiration670 16.6 AveragePower Calculations with Periodic funcDonso/5 16.7 ThermsValueof a PeriodicFunclion 678 16.8 TheExponentiat Formof the Fourier sede. 679 16.9 AmptitudeandPhaseSpe.tra 682 Sunnory 685 Prcblens 686
Chapter 17 TheFourier Transform698 17.1 TheDerivation of the FourierTransform699 17.2 TheConvergence of the FourierIntegral 70J 17.3 usingLaplace Transforms to FindFourier franstofifis 703 17.4 FourierTransforrns in the Limit 706 17.5 SomeMathematical Properties7r8 17.6 0perational Transfo(ms710
17.7 CircuitApplications714 17.8 Parseval! Theorem 717 Sunnary 724 Problens 725
Chapter18 TwoPort Circuits 730 gquations 731 18.1 TheTerminat 18.2 TheTwoPortPa'amelers 732 18.3 Anatysis of the Terminated Two,Port Cit.uit 741 18.4 Interconneded TwoPoft Ctr.uits 747 Sunnaty 751 Prcblens 752
AppendixA TheSolutionof Linear SimultaneousEquations 759 A.1 Pretiminary Steps /59 4.2 Crame/sMethod 760 4.3 TheCharacteristic Determinant760 4.4 TheNurnerator Determinant760 4.5 TheEvaluation of a Determinant761 4,6 Mati.es 764 A.7 MatrirAtgebra 765 A.8 Identity,Adjoint,andInverseMatri(es ZZ0 4.9 Partitioned Matrices 772 4.10 Apptications776
Appendix B Complex Numbers781 8.1 Notation 781 8.2 TheGraphirat Representation ot a Complex Nurnber 782 8.3 Arithmeti.operations /83 8.4 UsefulIdentities 785 8,5 TheIntegerPowerof a Complex Number 785 8.6 TheRootsof a Complex Number 286
AppendixC Moreon Magnetically Coupled CoilsandIdeal Transformers787 C,1 Equivatent Circuitsfor Magnetically Coupted Coils 787 C.2 TheNeedfor ldealTransformers in the Equivalent Circuits 792
Appendix D TheDecibet 797 Appendix E BodeDiagrams799 E.1 E.2 E,3 E.4 E.5 8.6 E.7 E.8
Reat,First0rder PolesandZeros 299 StraightlineAmptitudePlots 800 MoreAccurate AmplitudePtots 8ra StraightLine Phase AngtePlots 805 BodeDiagrarns: Complex PolesandZeros 807 AmotitudePtots 809 Correcting StraightLine AmplitudePtots 810 Phase AnglePtots 813
AppendixF An Abbreviated Table of Trigonometric Identities 817 AppendixG An Abbreviated Tableof Integrals 819 AppendixH Answers to Selected Problems821 Index 839
Listof Examptes 2 Chapter of IdeatSources26 2.1 lestingInterconne.tions of IdeatIndependent 2.2 Testinglnterconnedions Sources27 andDependent for a vottage, Current.andPower 2.3 Catcutating Circuit 31 SimpleResistive 2.4 Construding a CircuitModelof a Ftashtight33 a Circuit[{ode[Basedon Terminal 2.5 Constructing Measurements 35 CurrentLaw 39 2.6 UsingKirchhoff's 2.7 UsingKirchhofftVottageLaw 39 Laws 2.8 Applyingohm'sLawandKirchhoff's Current 40 to Findah Unknown 2.9 Constructing a Cir€uitl4odelBasedonTerninal Measurements 41 2.10 Applyingohm'sLawand(i(hhoff's Laws Vottage 44 to Findan Unknown Law 2.11 ApFlyingohm'sLawandKirchhoff's in an Arnplifiertircuit 45
3 Chapter 61 3.1 ApptyingSeriesParattetSirnptification Circuit 63 3.2 Anatyzing the vottageDivider Circuit 64 3.3 Analyzing a CurrentDivider 3.4 UsingVoltageDivisionandCurrentDivision to Solvea Circuit 67 Ammeter 69 3.5 Usinga d Arsonval Vottmeter Z0 3.6 Usinga dArsonvat Transform75 3.7 Applyinga DeltatoWye
4.6 4.7 4.8 4,9 4.10 4.11 4,12 4.13
Chapter5 an 0p AmpCircuit J60 5.1 AnatyzinE
6 Chapter 6.1 6.2 6.3 6.4 6.5 6.6
4 Chapter 4.1
4.4 4.5
ldentifying Node.Eranch,Mesh,and Loop in a Circuit 95 Method 99 Usingthe NodeVottage Methodwith Usingthe NodeVol.tage Dependent Sources 100 Meihod 106 Usingthe MeshCuffent Methodwith Usingthe MeshC{rrrent DependentSources 108
Method tlnderstanding the NodeVoltage Method 113 Versus MeshCurent andMeshCurrent [omparingthe NodeVottage Methods115 to Sotve Transformations UsingSource a Cit.uit 117 Transformation UsingSpecialSource 118 Techniques Equivalent of a Circuit Findingthe Th6venin Source 122 with a Dependent Equivatent Usinga Test Findingthe Thdvenin SoUwrce 124 for MadmumPower the Condition Catcutating Transfer 128 to Solvea Circuit 132 UsingSuperposition
Giventhe Current, Determining the Vottage, of an Indlctor 189 at the Terminats the Current,Giventhe Vottage. Determining at the Terminats of antndu.tor 191 Powet Vottage. Determining the Current, andEnergy for an Inductor 193 Power, and Detefmining Current,Vottage. 197 for a capacitor Energy Finding?).p, and,lr)Inducedby a Triangutar CurrentPutsefor a Capac.tlor198 for a Circuit Equations FindingMeshCurrent coits 206 with Magneticatty Coupled
7 Chapter of an the NaturalResponse Determining RLCircuit 234 of an the Natual Response 7.2 oetermining Inductors 235 iI Circuitwith Parallel of an the NaturalResponse 7,3 Determining nfCircuit 238
7.1
xiv
Lkt of Exanptes
7.4
Determining the NaturalResponse of an RCCircuitwith SeriesCapacitors239 7.5 Determining the StepResponse of an RLcir.uit 244 7.6 Determining the StepResponse of an RCCitcuit247 7.7 Usingthe General SolutionMethodto Findan RCCircuit'sStepResponse250 7.8 Usingthe Generat SotutionMethodwith Zero lnitiat Conditions251 7.9 Usingthe General SolutionMethodto Findan ,(LLrrcurl5 >Iepxesponsez5l 7.10 Determining the StepResponse of a Circuit with Magneticatl.y Coupted Coils 253 7.11 Anatyzing an Rl Circuitthat hasSequentiat Switching 255 7.72 Analyzing an RCCirclit that hasSequential Switching 257 7,13 Finding the Unbounded Response in an RCCircuit 259 7.14 Anatyzing an IntegratingAnptifier 261 7.15 Anatyzing an IntegratingAmptifierthat has Seqoentiat Switching 262
Chapter 8 8.1
Findingthe Rootsof the Characteristic Equation of a ParallelRLCCircuit 290 8.2 Findingthe overdamped NaturalResponse of a Paraltet RaCCircuit 293 8.3 Catrulating Bran(hCurrents in the Natural Response of a Panllel RLCCir.uit 294 8.4 Findingthe Underdamped NaturatResponse of a Parattet nlC Circuit 297 8.5 Findingthe CriticattyDamped Naturat ResDonse of a ParallelRICCircuit 300 8.6 Findingthe overdamped StepResponse of a Parattet f,lCCircuit 304 8.7 Findingthe Underdamped step Response of a Parattet RICCircuit 305 8.8 Findingthe CriticaltyDamped StepResponse of a Parattet faf Circuit 305 8.9 Comparing theThreeStep Response Forms 306 8.10 FindingStepResponse of a Paratl.et ilc Cir.uit with Initiat Storedlnergy 306 8.11 Finding the Underdamped NaturaI Response of a SeriesRaf,Circuit 310
8.12 Finding the Underdarnped StepResponse of a SeriesRtCCircuit ?11 8,13 Anatyzing TwoCascaded lntegrating Amptifiers 3l.t 8.14 Anatyzing TwoCascaded IntegratingAmplifiers with Feedback Resistors316
Chapter 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10 9.11 9.12 9.13 9.14 9.15 9.16
Findingthe Characteristi€s of a Sinusoidal Cufient 334 findingthe Characteristics of a Sinusoidal Vottage 334 Translating a SineExpression to a Cosine Expressiont.t4 Catcutating the rmsVatueof a Triangutar Waveforrn3J5 Adding Cosines UsingPhasors341 Impedances in Series 349 Combining Combining Impedances i'n Seriesandin Paraltet 351 Usinga DeltatoWye Transform in the F equency Domain 353 Perforrnihg lransformations Source ir the Frequency Domain 356 Findinga Th6venin Equivalent in the Frequency Domain 35l Usingthe NodeVottage Methodin the Frequenqr oomain 359 Usingthe MeshCurrent l4ethodin the Frequenqf Donain 360 Anatyzing a Linearlransformer in the Frequenqr Domain 364 Anatyzing an IdealTransformer Circuitin the trequency Domain 370 UsingPhasorDiagrams to Analyzea Citcuit 372 UsingPhasorDiagrams to AnalyzeCapacitive LoadihgEffects 373
Chapter 10 10.1 Cattulating Average andReadivePower 396 10.2 MakingPowerCalculations Involving HousehotdAppliances 398 10,3 oetermining Average PowerDetivered to a Resistor by a Sinusoidat Voltage 400
Listof Examphs xv
10.4 10.5 10.6 10,7
Power 402 Catcutating Complex Average Power 406 Calcutating andReactive Power in Pardllet Loads 40l Calculating Balancing PowerDelivered with Power Absorbed in an acCircuit 408 10,8 DeteminingMa$mumPower Transfer without LoadRestrictions41t 10.9 DeteminingMaximum Power Transfer with LoadImpedance Restri.llon 414 10.10FindingMaximum Power Transfer with I pedance AogleRestidions 414 10.11FindingMaximunPower Transfer in a Circuit with an IdealTransformer4J5
11 Chapter 11,1 Anatyzing a WyeWye Cit.uiI 440 11.2 Anatyzing a WyeDetta Circuit 444 11,3 Catcul.ating Powerin a ThreePhase WyeWye Circuit 449 11.4 Catculating Por,rer in a ThreePhase WyeDelta
14.4 14.5 14.6 14.7
Loading the Seriesna fiighPass Fitter 580 Designing a Bandpass Fitter 587 Designing a Parattet RaCBandpass Fitter 588 Determining Effed of a Nonideal Voltage Source on a fLC Bahdpass Fitter 589 14.8 Designing a SeriesRtCBandreject Fitter 596
15 Chapter 15.1 0esigning a LowPass 0p AmpFitter 609 15.2 Designing a HighPass 0p AmpFitter 611 15,3 Scal.ing a SeriesRLCCir.uit 613 15.4 Scating a Prototype LowPass 0p Arnp FiLlet 614 15.5 Designing a Broadband Bandpass 0p Anp Filtet 618 15.6 Designing Bandreject a Sroadband 0p Amp Filtet 621 15.7 Designing a Fourthorder LowPass 0p Amp fitrer
bz5
15.8 CaLcutating Butterworth Transfer Functions628 11.5 Catcllating ThreePhase Powerwith 15.9 Designing a Fourthorder LowPass an Unspecified Load 450 ButterworthFilter 631 11.6 Computing Wattmeter Readings in ThreePhase 15.10Determining the orderof a Eutterworth Citcuits 454 Filtet 634 15.11AnAlternate Approach to Determining 12 Chapter the orderof a Eutterworth fitter 634 a High4Bandpass Fitter 638 12.1 UsingSteptunctionsto Represent a Function 15.12Designing oT ilntre uuralon 4/u 15.13Designing a HighoEandreject Fitter 64, tltcull
45u
13 Chapter 13.1 Deriving the Transfer tundionof a Circuit 527 13.2 Anatyzing the Transfer Fundion of a Circuit 529 13.3 Usingthe Convolltior Integralto Find an oljtplt Signat 535 13.4 Usingthe Transfer Function to Find the SteadyState Sinusoidal Response539
Chaptor 14 14.1 Designing Filtet 574 a LowPass 14.2 Designing a SeriesfC LowPass titter 575 14.3 0esigning a SeriesRl HighPass tillet 579
Chapter 16 16.1 Findingthe FourierSeriesof a Triangutar Waveform with NoSymmetry660 16.2 Findingthe FourierSeriesof an odd Function with Symmetry667 16.3 Catcutating Formsof the Trigonometric Folrier Sedesfor Periodic Voltage 669 16.4 Calcutating Average Powerfor a Circlit with a Periodic VoltageSo[.e 677 16.5 Estirnating the rmsValueof a Periodic tun(tion 679 16.6 Findingthe Exponential Formof the Fourier Series 681
Slnusoidal Steadystate Response716 17.3 ApplyingParevaltIheoren 719 17.4 ApptyingParsevalt Theorem to an Ideal Eandpass tilter 720 17.5 AppMngParseval's fheoremto a Lowpass Fiatet 721
Measurements 235 18.3 Findingt parameters fromMeasurements ard Table18.1 738 18,4 Anatyzing a Terminated Twoport Circuit 246 18.5 Analyzing Cascaded Twoport Circuits Z5O
Variables Circuit 1.1 EtectricatEngineering: An overviewp. 3 1.2 TheInternationatSystemof Units p. 8 1.3 CircuitAnatysisrAn overviewp. 10 1.4 Voltageand (urrent p. 1? 1.5 TheIdeal BasicCircuitEtementp. 12 1.6 Powerand Energyp. 14
1 Unde6tand andbe abteto use5I unitsandthe standadprefi:esfor powersof 10. 2 Knowandbe abteto useihe dennitionsof 3 Knowafd be abt€to us€ihe defifitionsof 4 Beabteto usethe passive siqnconvention to caLcutaie the powerfor af idealbasiccjrcujt €tementgivenits vottageandcunent.
ElectricaI engineering is an excitingand challcngingprofcssion for anvone who has a genuine interest in, and aptitude lor, applied scienceand mathematics.Over the past century and a hall, electrical engineershave played a dominant rolc in thc developnrentof svstemsthat have cbangedthe v',aypeople live and work. Satellitecommu cation links,telephones,digital com puters, tcicvisions,diagnosticand surgical medical equipment, assembiylinerobots,and eleclricalpower tools are representative componentsof systemsthat define a modern technological socicty.Asan clcctricalengincer.you canparticipatein this ongoing technologicalrevolution by improvilg and refining these existingsystemsand by discovcdngand dcvclopingnew systems to meet the needsot our everchanging socjety. As ]'ou embark on the study of circuit analysis,you nccd to gain a feel for where tlris study fits into the hierarchyof topics that conlprisean introductionto electricalengineering.Hencewe begin by presentingan oveNiew of electricalelElineering,some ideasabout an engineeringpoint of view as jt relatesto circuit analysis.and a reviervof the internationalsystemof units. Wc then describegenerallywhat circuit analysisentails.Next, we intloducethe conceplsolvollage and current.Welbllow these conceptswith discussiolof an ideal basicelementand the necd for a polarity reference system.We conclude the chapter by describinghow current and voltagerclatc to powcr and cncrgy.
An0verview 1.1 EtectncatEngin€€nng:
An Overview Engineering: 1.1 Etectrical Electrical engineeringis the profession concetned with systemsthat oroduce.transmit, analmeasureelectric signals.Elect cal engineeing ;ombines the pbysicist'smodels of natural phenomena ith the mathematician'stools lor manipulatingthose modelsto producesystemsthat meet practicalneeds.Electiical systemspervadeour lives;theyare found ir homes, schools,workplaces, and transportation vehicles everywhere We beginby presentinga few examplesfrom eachofthe five major classificationsof electricalsystems: . commu catlonsyslems ' computel systems t control systems . power sysEms . signalprocessing systems Then we describehow electrical engineersanallze and designsuchsystems Conrmunication systems are electrical systems that generate' transmit. and distribute information. Wellknown examples include television equipment, such as cameras,transmitters, receivers,and VCRS; radio telescopes,used to explore the univeise; satellite systems,which return images of other planets and our own: radar systems,used to coordinateplane fl ights;ard telephonesystems. Figure 1.1depictsthe major componentsof a modern t€lephonesystem. Sta ing at the left of the figure,inside a telephone,a microphone turns sou d waves into electric signals.These signalsare caded to a s$ritchingcenter where they are combined with the sigmls from tens,hundreds, or thousands of othei telephones.The combined signals leave the switchingcenter;tleir form dependson the distancethey must travel.In our example,tley are sent througl wires in underground coaxial cablesto a microwavetransmissionstation.Here, the signalsare transformedinto microwavefrequenciesand broadcastfrom a tansmission antena through air and space,via a communicationssatellite,to a rcceivingantenna.The microwave receiving station translates the microwave signals into a folm suitablefor furtler transmission,perhapsas pulsesof light to be sent through fiberopticcable.On arival at the secondswitchingcenter,the combinealsignalsare separated,and each is routed to the approprrate telephone,where an earphoneacts as a speakerto convert the leceived electricsignalsback hto soundwaves.At eachstageof the process,electric ci.cuits operate on tlle signals. Imagine the challenge involved in designing, building, and operating each circuit in a way that guarantees that all of the hundredsof thousandsof simultaneouscallshave high quality Computer syslemsuse electric signalsto processinformation mrging from word processingto mathematicalcomputations.Systemsrange in size anal power from pocket calculato$ to personal computers to supercomputersthal perform suchcomplextasksas plocessingweather data and modelingchemicalinteractionsof complexorganicmolecules. Thesesystemsincludenetworksof microcircuits,or iniegratedcircuitssYstem. postagestampsized assembliesof hundreds,thousands,or millions of tigure1,1 A Atetephone
Circuit Vanabbs electricalcomponentsthat often operateatspeedsandpowerlevelsclose to fundamenlal physical limits, including th.3speedol light and the thermo dynamic laws, Control systemsuse electricsignalsto regulateprocesses, Examples include the control of tempemtures, pressures,and flow mtes in an oil refinery;the fuelairmixlure in a fuelinjectedautomobileengine;mechanismssuchasthe molors,doors,andlightsin elevators;andthe locksin the Panama Canal.The autopilot and autolanding systemsthat help to fly and land airyIanes aie also familiar control systems Powersyst€msgenerateand distributeelectricpower.Electricpower, which is the foundatior of our technologybased society,usuallyis generated in large quartities by nuclear, hydroelectric, and thermal (coal, oil, or gasfired)generato$.Poweris distributedby a grid of conductorsthat crissooss tlle country. A major challenge in designing and operating such a systemis to provide sufficient redurdancy and control so that failure of any piece of equipment does not leave a city, state, or region completely
figure1.2 a A tT scanofan adLrtt head.
Figure1.3 A Anajrplane
Signalprocessingsystens act on electric signals that represent information. They translorm the signalsand the informarion contained in tlem into a more suitable form. There are many different ways to process the signalsand their information. For example,imageprocessing systems gathermassivequantitiesof data from orbiting weathersatellites,reduce the amount of data to a manageablelevel,and transform the remaining data into a video imagefor the eveningnewsbroadcast.A computerized tomography(CT) scanis another€xampleofan imageprocessing system. It takessignalsgenemtedby a specialXray machineand transformsthem into an imagesuchas the one in Fig. 1.2.Although the original Xray signals are of little useto a physician,once they are processedinto a recognizableimagethe informationthey containcanbe usedin the diagnosisof djseaseand injury. Considerable interaction takes place among the engineering disci plines involved in designhg and operating these five classesol systems. Thus conmunications engineersuse digital computeis to control the flow of information,computers contain control systems!arrd control systems contail computers,Po\ver systemsrequire extensive communications sys temsto coordinatesafelyand reliably the operationof components, which may be spread acrossa continent. A signalprocessingsystemmay involve a conmunicationslink, a computer.and a control system. A good exampleof the interaction among systemsis a commercial airplane,such as the one showl in Fig. 1.3.A sophisticatedcommunicatiom systemenablesthe pilot and the air traffic controllerto monitor the plane'slocation.permittingthe air traffic controllerto designa safeflight path for a of the nearbyaircraftand enablingthe pilot to keep the plane on its designatedpath. On the newestcommercialairplanes,an onboard compulersystemis usedfor managingenginetunctions,implementingthe navigation and flight control systems,and gererating video information scrcensin the cockpit.A complexcontrol systemusescockpit commands to adjustthe positionand speedofthe airplane,producingthe appropriate signals to the engines and the control surfac€s (such as the wing flaps, ailerons, and rudder) to ensure the plane rcmains safely airbome and on the desiredflight path.Theplane must have its own power systemto stay aloft and to provide and distributethe electricpower neededto keep the
1.1
cabin lights on, make the coffee, and show the movie. Signalprocessing systems reduce the noise in air taffic communications and tlansform information about the plane's location into the more meaningful form of a video display in the cockpit. Engineering challengesabound in t}le design of each of these systemsand their integration into a coherent whole. For example,these systemsmust operate il widely varying and unpredictable environmental conditions. Perhaps the most importalt engineering challenge is to guarantee that sufficient redundanry is incorporated in the designsto ensurethat passenge$arive safelyand on time a! their desired destinations. Although electrical engineersmay be interestedprimarily in one area, they must also be knowledgeable il other areas that interact with this area of interest. This intemction is part of what makes electrical engi neering a challenging and exciting profession.The emphasisin engineer ing is on makingthingswork,soan engineelis free to acquireand useany technique,ftom anyfield, tlat helpsto get t}lejob done
CircuitTheory In a field asdiverseas electricalengineering,you might well askwhether all of its brancheshave anythingin common.The answeris yes rlectric circuits.An elect'riccircuit is a mathematicalmodel that approximates the behaviorof an actualelectricalsystem.Assuch,itprovidesan important foun{:lationfor leamingin your later coursesand as a practicing engineerthe details of how to design and opefate systemssuch as tnose iust described.The models,the mathematical techniques,and the language of circuit theory will form the intellectual framework for your future engi neering endeavors. Note that the term electtic circuit is commonly used to reter to an actual electrical system as well as to the model that represents it. In this text. when we talk about an electdc circuit, we always mean a model, unlessotherwisestated.Itis the modelingaspectof circuit theorythat has broad applications acrossengineering disciplhes. Circuit theory is a specialcaseof elestromagneticfield theory: the study of static and moving electdc charges.Altlough generalized Iield theorl might seemto be an approp ate starting point for investigating electric signals, its application is not only cumbersomebut also requires the use of advanced mathematics. Consequendy,a course in electromagnetic field theory is not a prerequisite to urderstanding the material in this book. We do, ho ever, assumethat you have had an introductory physics course ilr which electdcal and magneticphenomenawere discussed. Three basicassumptionspermit us to use circuit theory,rather than electromagnetic field theory to study a physical systemrepresented by an electric circuit. These assumptionsare as follows: 7. Electrical effects happen instantaneously throughout a system.We can make this assumption because we know that electric signals travel at or near the speed of light. Thus, if the system is physica y small, elecfic signals move through it so quickly that we can consider them to affect every point in the systemsimultaneously A system that is small enoughso that we call make this assumptionis system. calleda lumpedparamet€r
An0verview Engjneenng: EL€ciricat
2. The net charge on ewty component in the system is always zero. Thus no component can collect a net excess of charge, although some components,as you will learn latei, can hold equal but oppositeseparatedcharges. 3. Therc is no magnetic co pling betteeenthe componentsin a system. As we demonstrate later, magnetic coupling can occtrr within a componenL Thafs it; there are no other assumptions.Using circuit theory provides simple solutions (of sufficient accuracy) to problems that would become hopelesslycomplicatedif we were to use electromagneticfield theory These belefits are so great that engineeN sometimes specifically design electical systemsto ensurethat theseassumptioNare met. The importance of assumptions 2 and 3 becomes apparent after we inlroduce the basic circrlit elementsaltd the rules for analyzing interconnected elements. However,we needto take a closerlook at assumption1.The question r\ "How smalldoesa pb).icalsyslemhare ro be ro quatifyas a lumped parameter system?"We can get a quantitative handle on the question by noting that electdc signals propagate by wave phenomena. If the wavelength of the signal is large compared to the physical dimensions of the system, we have a lumpedparameter system.The wavelength I is the velocity dividedby the repetitionrate,or frequency,of the signal;thatis,,\ = c/f. The frequency/ is measuredin hertz (Hz). For example,power systems in the United States operate at 60 Hz. If we use t]le speed oI light 103m s) as lhe velocir) ot propagation. 1r  3 lhe walelenglhis 5 l0'm. If rbe po\rer(ystemot inrere\ric pbysically smatterlhan $is wavelength,we can representit as a lumpedparametersystemand use circuit theory to analyzeits behavior.How do we definemarel? A good N\e 1sthe rule of 1/10ri: if rhe dimensionoI the systemis 1/10th (or smaller.)of the dimension of the wavelength,you have a lumpedparameler system.fhus,aslong asthe physicaldimensionof the power systemis less than 5 x 10)m,we cantreat it as a lump€dparameter system. On the other hand, the propagation frequency of mdio signalsis or tlle order of 10vIIz. Thus the wavelength is 0.3 m. Using the rule of 1/10ttr, rtre relevant dimerNionsof a connunication syslemthat sendsor receivesmdio signalsmust be less tian 3 cm to qualify as a lumpedparameter system. menever any of the pertinent physical dimensionsof a systemurder study approachesthe wavelength of its signals,we must use electromagneticfield theory to aralyze that system. Thioughout this book we study circuits de ved ftom lumpedparameter systems.
ProblemSotving As a practicingengineer,you will not be askedto solve problemsthat have already been solved. Wlether you are trying to improve the perfomance of an existing systemor qeating a rew system,you will be working on unsolved problems.As a student, however, you witl devote much ol your attention to the discussionof problemsalready solved.By reading about and discussinghow theseproblemswere solvedin the past,and by solving related homework and eram problems on your own, you w.ill begin to develop the skills to successfully attack the unsolved problems you'll faceas a practicingengineei.
AnoveMew Engineering: 1.1 Etectrical Some general problemsolving prccedures are presented here Many perlainlo lhinkingaboutand organLing)our solulionslr0legy rhem ol proceedingwith calculations. 6elor€ 1.. Identify what'sSiven and what's to be found h problem solving,you need to know your destilation before you can selecta route tbr getting tlere. What is the problem asking you 10 solve or find? Sometimesthe goal of tle prcblem is obvious; other times you may need to paraphrase or make lists or tables of known and unknowr inJormation to seeyour objective. The problem statement may contain extraneous information tlat you need to weed out betore proceeding.On the other hand' it may offer incomplete information or more complexities thar can be handled given the solution methods at your disposal. In that case, you'll need to make assumptionsto fill in the missinginformation or simpliJy the problem context. Be prepared to circle back and reconsider supposedlyextraneousinformatior and/or your assumptionsif youl calculationsget boggeddown or produce an answerthat doesn't seemto make sense. 2. Sketcha cicuil diagram or other risual model. Translathg a verbal problem description into a visual model is often a useful step in the solution process.If a circuit diagam is aheady provided' you may need to add inlormation to it, such as labels, values' or leference directions.You may also want to redraw the circuit in a simpler, but equivalent, form. Later il this text you vrill Ieam the methods for developing such simplified equivalent clrcuits 3. Think of several solution methoh and decide on a w6Jrof choosing among theru'f\is colnse will help you build a collection of analytical tool$ several of which may work on a given pioblem. But one method may pioduce fewer equations to be solved than another, or it may require only algebra bstead of calculus to reach a solulion. Suchefficiencies,if you can anticipate them, can sfeamline your cal_ culations consideiably. Having an alternative method in mind also givesyou a path to pursue iJ your first solution attempt bogs down 4. Calculate a solution. Your planning up to tltis point should have helped you identify a good analytical method and the correct equationstor lhe probtem.Not\ comesrhe\olulionof lhoseequalions. Paperandpencil, calculator, and computer metlods are all available for performing t]le actual calculations of circuit analysis Efficiency and your insxuctor's prefeiences will dictate which tools you shoulduse. 5. UseWw creatirit!.ltyou suspectthat yolu answeris off baseor if the calculationsseemto go on and on without moving you toward a solu_ tion, you shotlld pause and consider altematives.You may need to revisit your assmlptions or selecta diffeient solution method. Or, you may need to take a lessconventionalproblem_solvingapproach'such asworking backwad from a solution.This text provides answen to all of the AssessmentPrcblems and many of the Chapter Problems so that you may wo* backward when you get stuck. ln the real world, you wonl be given answersin advance,but you may have a desired prcblem outcomein mind from which you can work backward.Other seative approachesinclude allowing yourself to see parallels witl
CircuiiVariabtes other t!?es of probleris you've successfullysolved, following your intuition or hunches about how to proceed, and simply setting the prcblem asidetemporadly and coming back to it later. 6. Test your sol tion. Ask yourself whether the solution you've obtained makes serlse.Does the magnitlde of the answer seemiea sonable?Is the solution physically realizable? You may want to go furtler and rework tle problem via an alternative method. Doing so will not only test the validity of your odghal answer,but will also help you develop your intuition about the most efficient solution methodsfor various kinds of problems.In the real \qorld, safetycdtical designsare always checked by several iDdependent means. Getting into the habit ol checking your answeis will benefit you as a studentand as a practicingengheer. Theseproblemsolving stepscannot be used asa recipe to solve every prob lem in this or any other cource.You may need to skip, changethe order of, oI elaborate on certail stepsto solve a particular problem, Use thesesleps as a guideline to develop a problemsolving style that works for you.
1.2 TheInternational Systemof Units Engineen comparc theoretical results to experimental results and compare competingengine€rhgdesignsusingquantitativemeasures. Modern engineering is a multidisciplinary profession in which teams of engineers work together on projects, and they can connnunicate their results in a meaningful way only if they all use the same units of measure. The Intemational System of Units (abbieviated SI) is used by aI the major engheeringsocietiesand most engifleersthroughoutthe world;hencewe useit in this book. Th€ SI units are basedon sevend€lired quantities: . length ' fime . elect c current . thermod]'namic tempemture . amount of substance . luminousintensity
:rI4*l:lrlillrri Qumtily
BasicUnit
Length Mass
kg
Tine Thermodynamic temperature AmouDt of substane
K
1.2
System ofUnits TheIntenationat
Thesequantities,along rvith the basicunit and s]'mbo1for each,are lisled in Tablc 1.1.Although not sldclly SI units,the familiar time unils of minute (60 s), hour (3600s), and so on arc often usedin engineeringcal culalions.In addirion.defined quanlitiesarc combinedto form deriv€d units.Some,suchas force,energl',power,and elcctriccharge.you already krow through previousphysicscou$es.Table 1.2lists the derived units usedin this book. Ir manycases, the SI unit is either ioo smallor too largeto useconveprefixcs niently. Standard correspondingto polve$ of 10, as ljsted in Table 1.3,are then appliedto the basicunit. All o{ lhescprefixesare correct.but engineenoftcn useonly the onesfor powersdivisibleb]' 3;thus centi,deci,deka,and hectoare usedrare\,.Also, engineersoflen selectthe prefix lhat places the base number in the range between I and 1000. Supposethat a time calculationyieldsa result of 10 ' s,that js,0.00001s. Most engineers would dcscribe this quantity as 10 /is. that is. j ps. 10 = 10 x 10j s,mlher than as0.01ms or 10,000,000
prefiresfor powersof 10 objective1lJnderstandand be abteto useSI units and the standard 1.1
1.2
How many dollarsper millisecondwould the federalgovemmenthaveto collectto retire a delicit of$100 billion in one year?
If a signalcantmvelin a cableat 80%of the speedof light,whatlengthof cable,ir inches, rcpreserts1 ns?
Answer:9.45'.
Answ€r: $3.17/ms. NOTE: Abo try ChaptetPrcblems1.1,1.3,and L6.
Pr€fixesto signify TABLE1.3 Standardized Prefu
Symbol
Pol€r
t0 r3 TABLE 1.2 DerivedUnits in SI Qufllity
UnitNrne (Slnbol)
pico
p
hetv (Hz)
106
newtoD(N) joulc (J)
N m
nilli
1 0 3 t0 l
J coulonb (c) Electricpotential
vo1l(V)
d
h C/V
Magnetic llux lrenry (H)
1o' t0l
kilo
sienens(s)
l0' l)
I/C
ohm (o) Electricconduclance
10 r' 10v
M
106
G
10e
T
l0''
10
1.3 CircuitAnalysis:An 0verview
modet forelechcatenqitiguret.4 A A conceptuaL needng design.
Before becoming involved in the details of circuit analysis, we need to take a broad look at engineeing design,specilicallythe designof electric circuits. The purpose of this oveNiew is to prcvide you with a perspective on where circuit analysis fits within the whole of circuit design. Even though this book focuseson circuil analysis,w€ try to provide opportudties for circuit design where appropriate. A11engineeringdesignsbegin with a need,as shownin Fig. 1.4.This need may come from the desire to improve on an existing desig , or it may be something brandnew A careful assessmentof the need results in design specifications,which are measumble characteristics of a proposed desigl. Once a designis proposed,the designsp€cificationsallow us to assesswhether or not the design actually meets the need. A concept for the design comesnext. fhe concept derives ftom a complete understandingof the designspeciJicationscoupled with ar insight into t]!e need,which comesfrom education and experience.The concept may be realized as a sketch, as a written description, or ir some other fom. Often t]le next step is to translate tle concept into a mathematical model. A commoily usedmathematicalmodel for electrical systemsis a circuit model. lhe elements fhat comprise the circuit model are called ideal circuit components. An ideal circuit component is a mathematical model of an actualelecldcalcomponent,like a battery or a light bulb. It is important for the ideal circuit component used in a circuit model to iepresent the behavior of the actual electrical component to an acceptable degree of accuracy.The tools of circuit snalysis, the focus of tlis book, are then applied to ttre circuit. Circuit analysisis basedon mathematical techniques and is used to predict the behavior of the circuit model and its ideal circuit components.A comparison between the desired behavior, from t}le design specifications,and the predicted behavior, from circuit analysis,may lead to rcfinements ir the c cuit model and its ideal circuit elemerts. Once the desiredandpredictedbehavioraie in agreement,a physicalprototypecan be constructed. The physical proroqpe is an actual electrical system,constnrcted ftom actualelectricalcomponentsMeasuiementtechniquesare usedto determine the actual,quantitativebehaviorof the physjcalsystem.This actual behavior is compared vrith the desired behavior from the design specifications and the predicted behavior from circuit analysis.The comparisons may result ir rcfhements to the physicalprototype,the circuit model,or botb Eventually, this iterative process,ir which models, components,and systems are coflthually rcfined, may produce a design that accurately matches the design specifications and ttrus meets the need. From this desciption, it is clear that circuit analysis plays a very importantrole in the designprocess.Becauseciicuit analysisis appliedto circuit models,practicingengineerstry to use mature circuit models so that the resultingdesignswill meet the designspecificationsin the filst iteration.In this book. we use modelsthat have been lestedfor between 20 and 100 years; you can assume that they are matuie. The ability to model actual electrical systemsvrith ideal circuit elements makes circuit theory extremely useful to engineers. Sayingthat the interconnectionof ideal circuit elementscan be used to quantitativety predict th€ behavior of a system implies that we can
1.4
Voltage andCLrent
describethe interconnectionwith mathematicalequation$For the mathematicalequationsto be useful,wemustwrite them in termsof measurable quantities.Inthe caseof circuits.thesequantitiesare voltageand current, whjch we discussin Section 1.4.The study of circuit analysisinvolves understandingthe behavior of eachideal circuit elementin terms of its voltage and current and understandingthe constraintsimposedon the voltageand currentas a resultof interconncctingthe ideal elcmcnts
1.4 VottageandCurrent The conceptof electricchargeis the basisfor describingall electricalpheof electriccharge. nomena.Let's review someimportant characteristics . The charge is bipolar, meaDingthat electrical effects are described in termsof positiveand negativecharges. . The electric chargeexistsin discretequanrities,which are integral multiplesof the electroniccharge,1.6022x 101eC. . Electical effectsare attributcdto both the sepafationof chargeand chargesin motion. ln circuit theory, the separation of charge creates an electric force (vol! age),andthe motion of chargecreatesan electricfluid (cunenr). The conceptsof voltageand cu ent are usetul from an enginee.ing point ol view becausethey can be expressedquantitatively.Whenever positiveand negativechargesare separated, energyis expended.Voltage is the energyper unit chargecreatedby the separation.We expressthis ratio in differential form as
(r.1)
4 Definitionof voltage
0 = the voltagein volts, 1, = the energyinjoules, 4 = the charge ir coulombs. The elect cal effectscausedby chargesin motion dependon the rate of charge flow The rate of charge flow is klown as the electric currcnl, which is expressedas
,=#,
(1.2) < Definitionof current
j : the currentin amperes, q = the chargein coulombs, I : the time in seconds. Equations1.1and1.2are definitionsfor the magnitudeofvoltage and cu ent,rcspectively.The bipolarnatureof electdcchargerequircsthat we assignpolarity referencesto thesevariables.we will do so in Section1.5.
11
t2 Although current is made up of discrete,moving elechons, w€ do not need to considerthem hdividualy becauseof the enormousnumber of them. Rather, we can think of electrons and thei corresponding charge as one smoothly flowing entity. Thus, i is heated as a continuous variable. One advantageof usingcircuit modelsis that we canmodel a compo, rent stricUy in terms of tie voltage and current at its terminals. Thus two physically different components could have the same relationship betweenthe teminal voltage and terminal current. lf they do, for purposesof circuit analysis,they are identical.Once we know how a component behaves at its terminals, we can anal'ze its behavior in a circuit. However,when developingcircuit models,we are interestedin a component's intemal behavior.We might want to know, for example,whether charge conduction is taking place because of free electrons moving through the dystal lattice structureof a metal or whetherit is becauseof electronsmovingwithin the covalentbondsof a semiconductormaterial. However, these concerns are beyond the rcalm of circuit theory In this book we usec cuit modelsthat have alreadvbeen develoDed: we do not di.cu\\ho$ compoDent modelsarede\eloped.
1.5 TheIdealBasicCircuitElement
Figure 1.5A Anid€albask circuit ehment.
An ideal basiccircuit el€menthasthree attributes:(1) it hasonly two terminals,which are points of connection to other circuit components;(2) it is describedmathematicallyin terms of current and/or voltage;and (3) it cannotbe subdividedinto other elements. We usethe word deal to imply that a basiccircuit elementdoesnot exist as a realizablephysicalcomponent. However,aswe discussed in Seclion1.3.ideal elementscan be connected in order to model actual devices and systems We use the word Dartcto imply that the circuit elementcannotbe further reducedor subdivided into other elements.Thus the basic circuit elemelts form the bui]ding blocks for constructing circuit models, but they themselvescannot be modeledwith any other type of element. Figur€ 1.5is a rcpresentation of an ideal basic circuit element.The box is blank becausewe are making IIo commitment at this time asto the type of circuitelementit is.In Fig.1.5,the voltageacrossthe teminals of the box is denoted by o, and t}le cunent in tlle circuit element is denoted by i. The polaiity reference for the voltage is indicated by the plus and minus signs, and the reference direction for the curent is shown by the arrow placed alongsidethe current. The interprctation of these rcIercnces given positive or negativenumerical valuesof o and i is sunmarized in Table 1.4.Note that
.u voltagedrop from terminal1 to terminal2
i
voltage rise from terminal 1 ro terminal 2
vollageise ftomterminal2to terminall
voltagedrop from terminal2 to terminal1
positivechargeflowing from terminal1 ro terminal2
positive charge tlowing from terminal 2 to terminal I
negativechargeilowing ftom terminal2 io tenninal1
negative charge flowi.g fiom terminal 1 to termilal2
1.5
TheldeatEasicCiftuitEtement
the notion ofpositivechargeflowingin one directionis equiv algebraically aleni lo the notion ofnegativechargeflowingin the oppositedircction Thc assignmcntsof the rcferencepolarily for voliage and the reftr encedircction for curreni irc entirely arbitrary.Howevel.olrccyou have assignedthe Iefercnces,you must wrilc all subsequentequations to agr;e \vith the choscnreferencesThe most widely Lrscdsign convention applied lo theseretcrencesis called the passivesign conv€niion,which we use throughoul thjs book The passivesign conventioncan bc staled
Whcneverthe referencedirectionIor the currentin an elementis in the directionof the re{ercncevoltagcdrop acrossthe elemcnt(asin Fig.1.5).usea positivesignjn any exprcssionthat relateslhc voltage ro Lhccuff 0.
Calculatethe total charge(in microcoulombs) enteringthe elementat its upper terninal
Answer: 4000/..C. NOTE: Also try Chapter Ptoblem 1.9
 ( . r . , le'" c.
Find lhe maximumvalueofthe curlenl entering the terminalifa = 0.03679sr.
Answer: l0 A.
14
1.6 PowerandEnergy Power and energy calculations also are important in circuit analysis.One reason is that although voltage and current are useful variables in the analysis and design of electrically based systems,the useful output of the systemoften is nonelectrical, and this output is conveniently expressedin terms of power or energy.Another reason is that all practical deviceshave Iimitations on the amount of power that they can handle. In the design process,therefore, voltage and current calculations by tlemselves are not sufficient. We now relate power and ene.gy to voltage and curent and at the sametime use the power calculation to illustrate ttre passivesEn convention. Recall from basicphysicsthat power is the time rate of expending or absorbing energy. (A water pump rated 75 kW can deliver more lite$ per secord than ore rated 7.5 kW.) Mathematically, energy per unit time is erDressedin the form of a derivative. or
,:'H,.,
Definitionof power>
(1.3)
the power rn watts, the energyin joules, the time in s€conds.
Thus1W is equivalent to 1 J/s. The powei associatedwith the flow of chargefollows directly from thedefinitionof voltageandcurrentin Eqs.1.1and 1.2,or
aw
/ aw\( aq\
dt
\ dq l\dt J'
Power equation>
(1.4)
p : t}le power in watts, 1)= the voltage in volts, i : the curent in amperes,
1.6
Equation 1.4 shows that the pow€r associatedwith a basic ckcuit element is simply the pmduct of the current in the element and the voltage adoss the element. Therefore, power is a quaDtity associatedwith a pair of tei_ minals, and we have to be able to tell from our calculation whether power is being delivered to the pair of teminals or extracted from it. This information comes ftom the corect application and interpretation of tle passive sign convention. If we usethe passivesign convention, Eq. 1.4 is correct if the rcference alirection for the current is in the direction of the rcference voltage drop acrossthe terminals. Otherwise, Eq. 1.4 must be written with a minus sign. In other words, if the current reference is il the direction of a refererce voltage rise acrossthe teminals, the expressionfor the power is
Power andEneqy
+ (b)p = ?)i
(1.5)
'The algebraic sign of power is based on charge movement tbiough voltage drcps and rises.As positive chargesmove ttrrough a drop in voltage,they lose energy,and as they move through a rise in voltage, they gain energy.Figure 1.6 summarizesthe relationship between the polarity rcferencesfor voltage and current and the expressiol for powerWe can now state the rule for interprcting the algebraic sign of power: If the power is positive (that is, if p > 0), power is being deiivered to the circuit iNide the box. If the power is negative (that is, iJ p < 0), power is being extracted from the circuit inside the box.
For example, suppose that we have selected the polarity refeiences shown h Fig. 1.6(b). Assume further that our calculations for the curent and voltage yield ttre following numeical results: j=4A
and D:
1 0V .
Then the power associatedwith the terminal pair 1,2 is
? = _(_10x4): 40w. Thus t}Ie circuit inside the box is absoibing 40 W. To take this analysis one step furthel, assumethat a colleague is solv_ ing the same problem but has chosen the reference polarities shown in Fig.l.6tcl.Theresuldngnumerical\aluesdre i:4A,
0:10V,
and
p:40w.
Note that interprcting these results in terms of this reference system gives the same conclusions that we proviously obtainednamely, that the circuit inside the box is absorbing 40 W In fact, any of the reference systems il Fig. 1.6 yields tlis sameresult.
,
I;:]
l   : l
*113
(c)p=d
41._stl}l .
L    I *f:] 1 (d)p = ,i
andth€expr€sion rcferences figure1.6a Potadty
signof power atgebraic { Interpreting
76
objective3Know andusethe definitionsofpowerandenergyt Objective 4Be abteto usethe passive sign 1.5
Assumethat a 20 V voltagedrop occursacross an clcmcnt from tcrminal2 to lcrninal l and that a curcnt of4 A entcrstcminal2 a) SpeciJythe valuesof d and t for the polarity ret 0
Answ€r: 1,140MW Celilo to Sylmar.
NOTE: Also try ChapterProblems1.12,1.17,1.24, and 1.26.
Srimnrary 'lhe
InternationalS]'stemof tjnits (SI) enablesengjneers to communicatc in a mcaningful way about quanlilalive rcsults.Tablc 1.1summarizcsthc bascSl unils;Table1.2 prcsentssomcuscill dcrivcdSI units,(Seepages8 and9.) Circuit analysisis basedon the variablesofvoltage and current.(Seepaser1.) Vollageis the erergy per unit chargecreatedbt' charge sepantion and has the SI unit of \olt Q) = dlNldll). (Seepage11.) Cunenl is the rate ofchargetlow and hasthc SI unit of ampere(, : d4ldr). (Sccpagc I l.) The id€albasiccircrit elemenlis a twoterminalcompoDert that cannot be subdivided;ii car be described malheDralically iD tenns ofits terminalvoltageard currenl. (Seepage12.)
The passivesign conventionusesa positivesign in the expressionthal relales lhe voltagc and cu cnt at thc le niials oI an elemenl when thc relcrcncc dircction Ior the cu enl lhroughthe elemenlisin thc dircclionof the relerence vollage drop across the element. (Scc page13.) Pofler is energy per unit of iime and is equal to the productofthe terminalvoltageand curren! it hasthe SI unit of$'att (t : dflrlll = ot). (Seepage1a) The algebraicsignof power is interpretedasfollows: . If p > 0, power is being deliveredto the circuit or cfcurt component, . If r, < 0, power is beingextracledtuomthe circuil or circuii component.(Seepage15.)
17
Problems Seclion 12 L1 There are apprcximately 250 million passenger vehicles rcgistered ir the United States. Assume that the batteiy in the average vehicle stores 440 watthours (Wh) of energy. Estimate (in gigawatthouis) tlle total energy stored in U.S.passengei vehicles, 1.2 The line desuibed in AssessmentProblem 1.7 is 845 mi in lengtl. The line contains lour conductors, eachweighing2526lb per 1000ft. How many kilo $ams of conductorare in the line? L3 The 4 gigablte (GB : 10ebytes) flash memory chip for an MP3 player is 32 rnm by 24 rnn by 2.1mm.This memory chip holds 1000theeminute songs a) How many seconds oI music fit into a cube whosesidesarc 1 mm? b) How many bytes of memory are stored in a cube whosesidesaie 100pm? 1,4 A handheld video player displays 320 x 240 picture elements(pixels)in eachframe of the video.Each pixel requires 2 bytes of memory. Videos are dis' played at a mte of 30 frames per second.How many minutes of video will fit in a 30 gigabyte memory? 1,5 Some species of bamboo can grow 250 nn/day. Assume individual cells in the plant are 10 pm long. a) How long, on average,does it take a bamboo stalkto $ow 1 cell length? b) How many cells are added in one week, on average? 1.6 One liter (L) oI paint covers approximately 10 m2 of wall. How thick is the layer before it dries? (/{tnr: 1L=1x106mm3.)
35 /rA curent, due to the flow of electrons.What is the averagenumber of elechonsper secondthat flow past a fixed reference qoss section that is perpendicular to the direction of flow? 1,9 The current enteriry tlle upper terminal of Fig. 1.5is i : 24 cos4000tA. Assume t}le charge at t}le upper terminal is zerc at the instantthe current is passingthrough its maximum value.Find the expressionfor q(t). L10 How much eneryy is extracted ftom an electron as it flows through a 6 V battery from tlle positive to tbe negarivererminal: F\pfess )ouf answef iD attojoules. Sections1.F1.6 1,11 Ore 9 V battery supplies 100 rA to a camping flashlight. How much energy does the battery supply in 5 h? 1,12 Two electiic circuits, repiesented by boxes A and B, are connectedas shown in Fig. P1.12.Therefer ence dhection foi ttre current i h the inteiconnection and the reference poladty for the voltage ?.r across the inter connectiotr are as shown in the figure.For eachof the following setsof nume cal values,calculate the power in ttre interconnection and state whether the power is flowing ftom A to B or a) i:5A, b) i: 8A, c) i:16A, d) t:
u: l20Y r:25OY t,: 1s0v
10A, 0:
480V
figuruP1.12 Section 1.4 A curent of 1200A exists in a copper wfue,wrth a circular ciosssection (iadius = 1.5 nrm). The current is due to free electrons moving through the wfue at aIt average velocity of o meten/second If the concentntion of free electrons is 1de elecrrons per cubic meter and iJ they are uniformly dispersed tbroughout the wire, then what is the averagevelocity of an electon? 1.8 In electronic circuits it is not unusual to encounter currents in the micioampere iange. Assume a
1.13 The references for the voltage and cu ent at the terminal oI a circuit element are as shown in Fig.1.6(d).ThenumedcalvaluesIor o andt are 40V and 10A a) Calculate the power at the teminals and state whether the power is being absorbedor deliveredby the elementin rhe box.
18
va'iabtes Circuit b) Given that the current is due to electronflow, statewhetherthe electronsare enteringor leaving terminal2. c) Do the electronsgain or loseenergyasthey pass thmugh the elementin the box?
1.14 RepeatProblem1.13with a voltageof
60V.
1.15 Wlen a car hasa dead battery it can often be started by connectingthe battery from another car acrossits terminals. The posilive terminals are connected together as are the negative terminals.The connection is illustrated in Fig. P1.l5. Assume the crment i in Fig.P1.15is measuedandfound to be 30A. a) Which car hasthe deadbattery? b) If this connectionis maintainedfor 1 min, how much energyis transferredto the deadbattery'?
a) Find tlre power absorbedby the element at 1=10ms. b) Find the total energyabsorbedby the element. 1,19 The voltage and current at the terminals of the circuit elementin Fig.1.5are shownin Fig.Pl.19. a) Sketchthe power versuslplot for 0 < t < 50 s. b) Calculalethe energydeliveredto the circuit ele ment at t : 4, 12, 36,ard 50 s" fiqureP1.19
1](v) t0
Figure P1.15 2 0 2
8 1
8 10
1,16 The manufacturer of a 9V diycell flashlight battery saystlat the battery will deliver 20 mA for 8u conrinuoushours.During thal time lhe vollageqill drop tromq v lo o V Assumethe drop in !ollageis linear witl time.How much energydoesthe battery deliver in tlis 80 h interval? 1,17 The voltage and current at th€ t€rminals of the cir cuit elementin Fig.1.5are zerolor , < 0. For , > 0 they are ?) = e
s00,
i : 30
?1500! v,
]0 Il.4 0
2024'2832364044
s2 s6 r(s)
1.0
mA 40e56'+ 10e1s00i
a) Find the power at r  1 ms. b) How much energyis deliveredto the circuit element between0 and 1 ms? c) Find the total energydeliveredto the element 1.18 The voltage and curent at the terminals of fte cir 6trG cuit elementin Fig.1.5aie zerofor I < 0.Fort > 0 they are
t' = 400eroo'sin 200t V, j = 5e1m'sin200tA.
1.20 lhe voltage and cu ent at the terminals of t}le cir 6pre cuit elementin Fig1.5are zerofor 1 < 0. For I > 0 r):75  75a1oauv, j = 50e rm/mA. a) Find the maximumvalue ofthe power delivered to the circuit. b) Find the total energydeliveredto the element.
Probt€ms19 1:1 The voltage and current at the teminals of the eleP*r€ ment inFig.1.5 are 36 sin 200rt V,
r':
1,23 The voltage and currcnt at t}Ie teminals of the ciitsi'c crit element in Fig. 1.5 are zero for t < 0. For t > 0 they are
i = 25 cos 2007t A. 1) (16,000r+ 20)eeo'V,
a) Find the maximum value of the power being delivered to the element. b) Find t]le maximum value of the power beirg erlracted frcm the element. c) Find the aveiage value of p iJl the interval 0 0.
a) Fird the time (in milliseconds) when the power deli\eredlo lhe ciJcuitelemeDlis maximum. b) Find the maximumvalue olp in milliwatts. c) Find the total energy delivered to the circuit element in millijoules.
20 '.26 The numerical values for the currents and voltages in the circuit in Fig.P1.26are givenin Table P1.26. Find the total power developed in the circuit P1.26 Figure
5.0 2.0 3.0 5.0
150 250 200 400
1.0
50 350 400
I
2.0
g
350
60
P1.26 TABI.E Etemenl
Currenr (A)
Yoltage (n\7)
150 150
0.6
100 250 300
0_8 0.8
rigureP1,28
2.0 7.2
300
f
1.28 The numerical values of the voltages and currents in the interconn€ction seenin Fig. P1 28 are grven in Table P1.28. Does tlle interconnection satisfy t]le Powercheck?
4 fl3
r27
Assumeyou are an engineerin chargeof a project and one of your subordinate engineen reports that the interconnection in Fig. P1.27 does noi pass the power check. The data for the interconnection are givenin TableP1.27. a) Is the subordinate corect? Explain your answer' b) If the subordinate is corect, can you find the error in the data? P1.27 Figure
,
+
Elem€nt
.
L
l
+
Volt.ge (\,)
36
250 250
2a
t
108 32
+ f g
48
h j
80 80
100 150 350 200  150 300
129 One method of checkirg calculationsinvolving interconnectedcircuit elementsis to see that the total power deiivered equals the total power absorbed (conservationofenergy principle). Witl this thought in mind, check the interconnection in Fig. P1.29and statewhether it satisfiesthis power check.The current and voltagevaluesfor eachelement arc given in Table P1.29.
1.30 a) In the circuit shown in Fig. P1.30, identity which elemenls are'absoibing power and which are detilefiDgpower. usi0g the pa$ive sign conventron. b) The numerical values of tle currents ard voltagesfor each element are given in Table P1.30. How much total power is absorbed and how much is delivered in this circuit?
Figure P1.29 figureP1.30
!
,t" "l
1.6 2.6 1,.2 1.8 f g j
80 60 50 2t)
1.8 36 3.2
30 40 30 20
2.4
30
L
+
Elemenl
volraee(mv)
b
300 100 200 2M 350
I g h
200 250 50
!
c
+
Curenl (irA)
25 10 15 35 25 10 35 10
Elements Circu'it Thereare five ideal basic circuit elements:voltascsourccs. 2.1 VoLtage and CunentSourcesp. 24 2.2 ELectric;L Resistance (Ohmt Law) p. 28 2.3 Con5trudionof a Circuitl4odetp. 32 2.4 Kir€hhoff'slaws p. .t6 2.5 Anatysisof a CircuitContainingDependent
Undertandth€ symbots for afd the behaviorof the fottowifgideatbasiccircujtel€ments; jndependent voltag€andcuir€ntsources, d€pefdertvotiageaid cuiientsourc€s, and Beabteto stat€0hmt taw,Kirchhoffscurrent ta& and (irchhoffsvottagetaw andbe abteto usetheselawsto anatyze simpt€cjrcujtr. Knowhowto caLculate th€ powerfor €ach elementjf a simpl€circujtandbe abteto detefmjiewhetheror not the powerbatarces
22
cunent sources,r'esistors, inductols.and capacito$.In this chaptcr wc discussthc chaaactcristicsof voltage sources,current Altlrough this ma,yseemlike a small num sources,and resistors. pracber of elementswith which to beginanalyzingcircuits,mal1y tical systemscan be modclcdwith iust sourcesand resistors.They are alsoa LlseftLl startingpoint becauseo[ their relativesimplicity; thc mathcmaticalrclationshipsbctwcen voltage and current ill sourcesand resistorsare algebraic.Thusyou will bc ablc to bcE:in leaning the basic techniquesoI cjrcuit analysiswith only algebraic manipulations. We will postponeintroducinginductols and capacitorsuntil Chapter6, becausetheir userequiresthal you solveintegral and However. the basicanalyticaltechniqucs differential ecluations. for solvingcircuitswith induclols and capacitorsare the sameas those introduced in this chapter.So, by the time you need tcr begirlmanipulatingmore difficult equations,you shouldbe vcry familiar with the methodsof writirg them.
PracticaI Perspective Etectrical. Safety
"DangerHigh VoLtage." Thiscommonty seenwarning is misteading. Attformsof energy, including etectricaL energy, can gut ifs not ontythe vottagethat harms. be hazardous. The staticetectriciry shockyou receive whenyou watkacross a carp€t andtoucha doorknob is annoying butdoesnotinjure. Yetthat sparkis caused by a vottagehundr€ds or thousands oftimeslargerthanthevoltages that cancause harm. Theetectricat energy that canactuatty cause injuryis due to etectricai currentandhowit ftowsthroughthe body.Why, then,doesthesignwarnof highvottage? Eecause ofthe way poweris produced eLectrical anddjstributed, it is easierto determjrevoLtages than cuffents.Atso, most etectdcaL produce sources constant, specified voltages. So the signs warnaboutwhatis easyto measure. Determining whether andunderwhatcondjtjons a sourcecansupptypotentiaLty dangerous currents is moredifficutt,asthisrequjres anunderstanding of etectrical engineering. Before wecanexamine thisaspect of eLectricaL safety, we haveto learnhowvoltages andcurrents areproduced andthe reLationship between them.Theetectrical behavior of objects,
suchas the humanbody,is quite comptexand often beyond comptetecomprehension. To atLowus to predjctand controt phenomena, eLectfical we usesimpLiryjng modelsjn whjchsimple mathematjcaL retationships betweenvottageand current jn reaLobjects.Suchmodapproximate the actuaL retationships elsandanalyticaL methods formthe coreofthe electrjcaL engineering technjques that wjLlattow usto understand attetectrjcaL oheromena. incldding oseretdLirgto electdcaI safety. At the end of this chapter,we witt usea sjmpteetectric circuitmodelto describehow andwhy peopleareinjuredby etectriccurrents,Eventhoughwe rnayneverdeveLop a compteteand accurateexptanation of the etectricaL behaviorof the humanbody,we can obtaina ctoseapproximation usjng simptecjrcuit modeLs to assessand improvethe safetyof etectrjcalsystemsand devices.DeveLoping modetsthat pro, videan understanding that h jmperfectbut adequate for solvinq practicalprobtems lies at the heartof engineering. I4uch of the at of etectricat engineering, whichyou wil[ Learnwith js in knowingwhen and how to sotvedifficutt experience, probtems by usingsimptiryingmodets.
23
24
2.1 VoltageandCurrentSources ideal voltageand curent sources,we need to consider Befoie aliscussing the general nature of electncal sources.An elechical sourc€ is a devjce that is capable oI converting nonelectric en€rgy to electric energy ard vice versa.A discharging battery converts chemical energy to electric energy, whereas a battery being charged converts electric energy to chemical energy.A dynamo is a machhe that conveits mechanical energy to electdc mode,it is energyandvice versa.If operalingin the mechanicaltoelectic energy, it is to mechanical ftom electric called a generator. If transforming about these to remember thing referred to as a motor. The impotant sourcesis tlat theY can eitler deliver or absorb elect c power, generally maintaining either voltage or current. This behavior is of particular interest for circuit analysisand led to the creation of the ideal voltage sourceand the ideal current sourceas basiccircuit elements.Thechallenge is to model practical sourcesin telms of the ideal basic circuii An ideal voltage source is a circuit element that maintains a prescibed voltage across its terminals regardless of tle curent flowing ilt those terminals. Similarly, an idesl cuEent source is a circuil element that maintains a presoibed current thmugh its terminals regardlessof the volf age aooss those teminals. These circuit elements do not exist as practical devicesthey are idealizedmodelsof actualvoltageand currentsources. Usingan idealmodelior currenlandvollagerourcesplacesan important restriction on how we may descdbe them mathematically. Becausean ideal voltagesourceprovidesa steadyvoltage,even if the current in the it is impossibleto speciry'the current in an ideal voltage elementchanges, source as a function of its voltage. Likewise, iJ t}le only inlormation you have about an ideal ctment sourceis the value of curent supplied,it is impossibleto determinethe voltageaooss that current sourceWe have sacrificed our ability to relate voltage and current in a practical source for the simplicity of using ideal sourcesin circuit analysis. Ideal voltageand curenl sourcescan be further describedas either independent sourcesor dependent sources An independent sowc€ estab lishes a voltage or current in a circuit without relying on voltages or cur rents elsewherein the circuil. The value of the voltage or current supplied is specifiedby the value of the independentsourcealore. In contrast'a avoltageor curent whosevaluedependson dep€ndenlsouce establishes tie value of a voltage or current elsewherein the circuit. You cannot spec,A (  ) ify the value of a dependent source unlessyou know the value of the volt ageor currenron whichil depends. The circuit symbolsfor the ideal independentsourcesale shown rn Fig.2.1.Note that a circle is usedto rcpresentan independentsourceTo completely specify an ideal independent voltage source in a circuit, you (b) G) must include the value of the supplied voltage and the reference polarity, ideal independforG)anideatinde as shownin Fig.2.1(a).Similarly,to completelyspecifyan Figure2.1^ Ihecircuiisymbob pendent vothge source and(b)anjdeatindependent ent cuirent source,youmust includeth€ valueofthe suppliedcurrentand its rcferencedirection,asshownfuFig.2.1(b).
Y L
(b Y
21 Vottag€ andCunent Sourc€s 25 The circuit symbolsIor the ideal dependentsourcesare shown in Fig. 2.2.A diamond is used to representa dependert source.Both the depeDdentcurrent sourceand the dependentvoltagesourcemay be controlled by either a voltage ot a current elsewherein the circuit, so there are a total of four variations,as indicated by the symbolsin Fig. 2.2. Dependent sourcesare sometimescalled controlled sources. To completely specify an ideal dependent voltagecontrolled voltage source!you must identify the controlling voltage,the equationthat permits you to compute the suppliedvoltage ftom the controlling voltage, and the referencepoladty for the suppliedvoltage.ln Fig.2.2(a),tlrecontrollirg voltage is named o,, the equation that determinesthe supplied
I, = Itrx.
? t G)
( c)
?
and the referencepolarity for ?r is as indicated.Note that ri is a multiplying constantthat is dimensionless. Similar requircments exist for completely specifying the other ideal (b) (d) dependentsources.InFig.2.2(b),the controllingcurrcnt is i' the equation Figure2.2 A Thecircujtsymbols for G)an ideat for the supplied voltage ," is Ds = p1\,
the relerencepolarity is as show4 and the multiplyingconstantp hasthe dimensionvolts per ampere.In Fig. 2.2(c),the conrroling volrageis o, the equationfor the suppliedcurrentrsis rs:
dl)r,
the referencedirectior is asshown,and themultiplying constantd hasthe dimensionamperesper volr. In Fig.2.2(d),the controling currentis j,, the equation for the supplied current i, is i, = Bi,, the referencedirection is as shown,and the multipllng constantp is dimensionless. Finally, in our discussionoI ideal souices,we note that they are examplesof activecircuit elements.Anactiveelementis one that models a device capableof generatingelectdc energy.Passiveetementsmodel physicaldevicesthat cannot generateelectric energy Resistors,inductorE and capacitors are examples of passivecircuit elements.Examples 2.1 and 2.2 illustrate how the characteristicsof ideal indepefldent and dependentsourceslimit the t'?es ofpermissibleinterconnectionsofthe
dependent voltagecontrothd vottas€ (b) an source, jd€atdependent currentcontrotted vothgesoufte,(c)an jdeatdependent vottageconhoLled currcnisource, and (d) anjdeatd€pendent cunent{ontrolLed cunentsource.
26
C cuitEtements
TestingInterconnections of ldealSources
Using t}le defidtions ofthe ideal independentvoltage and current sources,state which interconnections in Fig. 2.3 are pemissible and which violate tlle constraints imposed by t}le ideal sources.
10v
10v
Solution Conneclion(a) is valid. Each sourcesuppliesvolt age aooss the samepair of terminals,marked a,b. This requires that each sourcesupply the samevoltagewiti the samepolarity,which they do. Connection (b) is valid. Each source supplies current tbrough the same pair of terminals, marked a,b.This requiresthat eachsourcesupplythe same currentin the samedirection,which they do. Conrection (c) is not permissible. Each source supplies voltage across the same pair of terminals, marked a,b.This requires that each sourcesupply the samevoltage with the samepolarity, which they
(b)
+\
LOV
/+
5V
G)
Connection(d) is not permissible.Each source suppliescurrenr througl the samepair oI teminals, rnarked a,b.This requiresthat each sourcesupply the same current in the same direction, which they Connection(e) is valid.The voltagesourcesupplies voltage acrossthe pair of terminalsmarked a,b.The current sourcesupplies curent through the same pair of terminals.Becausean ideal voltage sourcesuppliesthe samevoltageregardlessof the current, and an ideal curent sourcesuppliesthe sane currentregardlessof the voltage,thisis a permissibleconnection.
fortuampte 2.1. Figur€2.3a Thec'rcuits
(d)
2.1
Voltage andCurentSources
TegtingInterconnections of IdeatIndependent andDependent Sources
Using the definitions of the ideal independent and depende0r sourcer. sralewhichiDterconneclrons in Fig. 2.4 are valid and which violate the consuamN imposedby the ideal sources.
T\\
),=3r,,
Solution b
Connection (a) is invalid. Both the independenr souce and t])e dependent souice supply voltage across the same pair of terminals, Iabeled a,b.This requircs that each source supply the same voltage with the samepolaity.The independent sourcesupplies 5 Y but the dependentsouce supplies 15V Connectiotr (b) is valid. The independent volragesource suppliesvoltage acioss the pair of terminals maiked a,b. The dependent current source suppliescurent through the sarnepair of terminals. Becausean ideal voltagesourcesuppliesthe same voltage regardless of cuirent, and an ideal currert source supplies the samecurent regardlessof voltage,this is an allowable contrection. Connection(c) is valid. The independenrcurrent source supplies crment tlrough the pait of termhals marked a,b.The dependentvoltage source supplies voltage aqoss the same pafu of terminals. Because an ideal cufient source supplies the same current regardless of voltage, and an ideal voltage source supplies the same voltage regaidless of currcnt, this is an allowable connection. Connection(d) is invalid. Both the irdependent sourceand the dependent sourcesupply cunent through ttre samepair of terminals, labeled a,b.This requkes that each source supply the same curent in the same reference dircction. The independent souce supplies 2 A, but the dependent source suppfies 6 A iII the opposite direction.
(.)
r ) /", ="3", , ( t +\
b (b)
t)
'.=4r.
G)
I
T i, b (d)
Figure2.4lr Thecircuitstor Examph 2.2.
28
Objective1understand ideal basiccircuit etements 2.7
2.2
For the circuit shown, a) what vdl e ofzr is rcquired in order for the interconnectionto bc valid? b' for lhi. valu
13.\2)
Note that the resistanceof the equivalent resistor is alwayssmaller than the rcsistance of the smallest rcsistor in the parallel comection. Sometimes, usirg conductancewhen dealing witl resistorsconnectedin parallel is more cotrvenient,In that case.Eo. 3.12b€comes
'Fl' jn palaLteL. Figure3.8, Tworcsistols connecied
+G2+
G " q => G i : G l
+Gk
(3.13)
Many times only two resistors are connected in pamllel Figure 3.8 illustrates this special case We calculate tle equivalent resistance ftom Eq.3.121
1 1 1 R*=&&:
n2+ R1 ' Rt&
R.R"
(3.14)
(3.15)
Thus for just two resistorc h parallel the equivalent resistarce equals the product of the resistaDcesdivided by the sum of the resistance$ Remember that you can only use this result in the special caseofjust two rcsistors in Darallel. Examole 3.1 illustrates the usefulnessof these results.
3.2
Relisto6inPanttet
61
ApplyingSeriesParall.et simpLifi cation Find i,. rr . aod i2 in lhe circuilshownin Fig.3.s.
40
t20v
+
ttl:18o ',{ 6 0
Solution We begin by noting that the 3 f,) resistor is in senes wirh rbe6 Q resistor.weLherelorereplacelhis series combination with a 9 O resistor, reducing the circuit to the one shown in Fig. 3.10(a).We now can replace the parallel combhation of the 9 O and 18 O resistors with a singleresistanceof (18 x 9)/(18 + 9), or 6 O. Egure 3.i0(b) shows t]Iis turtler reduction of the circuit.The nodesx and y marked on all diagmms facilitate tracing through t]le reduction of the circuit. From Fig. 3.10(b) you can veiify that iJ equals 120/10,or 12 A. Figure3.11showsthe iesult at this point in the analysis.We added the voltage ?)1to Using Ohm's helpcladfy the subsequentdiscussion. law we compute the value of ,1:
'72v. 1\ = (12)(6)
1)1 :72
9
v Figur€3.9 a Th€crrcujtfor E,dmph3.1.
4f)
r,l$rso r,,f9 f )
l20v y
G) 40
120V
x
+\ '"
60
(3.16)
But t 1is the voltage drop fmm node x to node y, so we can return to the circuit shown in Fig. 3.10(a) andagainuseOhm'slaw to calculatetr and ir. Thus,
 72 : 4 A ' 18 1 8
3()
v (b) jn Fjg.3.9. Figue3,10a A sinpLification ofthecjrcuitshown
(3.17)
4f)
+\ 12A 9
:8A.
120V
(3.18)
We have found t}le thrce specified currents by using seriesparallel reductions in combination with Ohm'slaw'
6()
v figuru3.r1 A lhecircuit of Fig.3.10(b) showins thenumedcaL
Belore leaving Example 3.1, we suggest that you take the tlme to show that the solution satisfies Kirchloffs curent law at every node and Kirchloff's voltage law around every closed path. (Note that there are thee closed paths that can be tested.) Showing t]Iat the power delivered by the voltage souce equalsthe total power dissipatedin the rcsistors also is informative. fsee Prcblems 3.3 and 3.4.)
62
SimphRsistiveCircuits
objective1Be abteto recognize resistors conneded in seriesandin parattel 3.1
Fortbecircuitshown,find (a) thevollageo, (b) thepowerdeliveredto thecircuilby the cufent source, and(c) thepowerdissipated in the 10() resistor.
7.211
101)
Answer: (a) 60Vr (b) 300w; (c) s7.6w NOTE: Alsotry Chapter Problems 3.1,3.2,3.5, and3.6.
3.3 The\foltageSivider
andeurrentSivider eireuits
(")
(b)
Figure3.12a (a)A votiasedivider circuitand(b)the vottagsdivids circuitwithcrrent ; indicaied.
Al timss especiallyin eleclronic circuils developingmore than one voltagelevel from a singlevollagesupplyis necessary. One way of doing this is by usinga voltsgedivid€rcircuil,suchasthe one in Fig.3.l2. We analyze this circuit by directly applying Ohm's la$' and Kirchhoff'slaws.Toaid the analysis, we introducethe curent i asshownin Fig.3.12(b).From Kilchhoff's current la\ Rr and R, carry the samecurrel1l.Applyilg Kilchhoff's voltagelaw aroundthe closedloop yields r.lRr+iRz,
Rr+R,
(3.20)
Nowwc canuseOhm'slawto calculale ul and,2: R r 01 =rr(1 : ?rnr + R2,
(3.21)
R, .0 2= u < 2 : x s R t + R,
13.22)
Equations3.21and 3.22show that ,r and o, are fractionsof o, Each ftaction is the ratio of the resistanceacrosswhich rhe divided voltageis definedto the sum ofthe two resistances. Becausethis ratio is alwayslcss than 1.0.the divided voltagesor and ?J2are alwaysiess than the source If you desirea particularvaluc of ?)2,and o, is specified,an infinile numberofcombinarionsof Rr and R, yicld thc propcr ratio.For example, supposethat !. equals15 V and 1J2 is to bc 5 V Thon r2/o, = i and,liom
3.3
TheVottage'Divider andCurrcntDjvider Circuiis
Eq.3.22,wefind that this mtio is satisfiedwhereverR, : ;R1.Other factors that may enter into the selection of R1, and hence R2, include the powei lossesthat occur in dividing the source voltage and the effects of connecting the voltagedivider circuit to other circuit components. Consider connecting a resistor Rz fu parallel witl R2, as shown in Fig. 3.13.The resistorR, acts as a load on the voltagedjvidercircuit.A load on any circuit consistsof one or more circuit elementsthat draw power from the circuit. With the load Rr connected,the expressionfor the output voltagebecomes '
'" Rr + R.q
\3.23) Figure3.13A A voLtage djvjder connected to a toadR..
RtRr R2+ RL
(3.21)
Substituting Eq.3.24into Eq.3.23yields R2
Rlt1 + (,R/R')l + R"' Note that Eq. 3.25 reduces to F'q.3./2 as Rz+c!, as it should. Equation 3.25showsthat, as long as R, >> R2,the voltageratio r,/?i" is essentially undisturb€d by t})e addition of the load on the divider. Atrother chaructedstic of the voltagedivider circuit of interest is the sensitivity of the divider to tie tolerances of the resistors.By roler.ar?ce we mean a range of possible values.The resistancesof commercially available resistorsalways vary within some percentageof theh stated value. Example3.2illustratesthe effectof resistortolerancesin a voltagedivider circuit.
Anatyzing the VoltageDivider Circuit The resistorsused in the voltagedividercircuit shownin Fig.3.14have a toleranceof t10%. Find the maximumand minimum valueofo,.
25kO
t00v
Sotution From Eq. 3.22,the maximum value of r)ooccurswhen R is l0oohighandRr is l0'o lo\\.and(beminimum value of t', occurswhen R2 is 10% low and Rr is 10% high.Thereforc
,lma\) = ,,(Illrn) =
100kI)
Figur€3,14 A Thecircujtfor Exampi€ 3.2.
(100v110) = 83.02 v. 1101(100x90) e0 +
: 76.60v.
Thus,in makingthe decisionto use10% resistorsin this voltage divider, we recognize that the noload orltput voltage will lie between 76.60and 83.02V
64
Simpte R€sistive Circuits
TheCurrentDivider Circuit
t
o , * 1,, '*
rigure3.15A Ihedrnentdivider circuit.
The curentdivider cfucuit showr in Fig 3.15consistsof two tesistors connectedin paralleladoss a current source.The current divider is designed to divide the current i. between R1 and R2. We find t]le relationship between ttre curent i, a.ndthe current in each resistor (that is,i1 and ir) by directly applying Ohm's law and Kirchhoffs current law. The voltage aoossthe parallelresistomis (3.26) Ffom Eq.3.26, ,"'. =
R, , R, + R7"'
13.27)
,"'. =
R' , R, + Rz''
(3.28)
Equations3.27and 3.28showthat the current dividesbetweentwo resis tors in parallel such that the curent in one resistor equals the curent entering the parallel pail multiplied by the other resistanceand divided by the sum of the rcsistors.Example 3.3 i ustratesthe use of the cuflentdivider equation.
Analyzing a CufientDivider Circuit Find the power dissipatedin the 6 f) resistor shown inFig.3.16.
and the power dissipated in the 6 O resistor is p=(3.2)'(6)=61.44w.
Solution
1.6r}
First,we mustfind the curent in the resistorby simplifying the circuit with seriesparallelrcductions ltus, the circuit shownir Fig. 3.16rcducesto the one shownin Fig. 3.17.We find the qrllent i, by usingthe formulafor currentdivision: 16
j" = _(10) tb+ 4
frro 3on
figure 3.16 A Theci,cuitforErampte l.l.
:8 A.
Note that i, is the cunert in the 1.6() iesistor in Fig. 3.16.We now car turther divide i, between the 6 O and 4 O resistols.The current in the 6 O resistoris
rd: _(8)
t)
: 3.2A,
10A
I )
160
4{)}
Figure3.17 A A sjmptjfication oftheciftuitshown in Fig.3.16.
3.4
Vottage Division andCrrentDivision
0bjective2*(now how to designsimplevottagedividerand cuff€ntdivid€rcircuits 3.2
^) Find the noloadvalueof?J,in the Find n, when nr is 150kO. c) How muchpower is dissipatedin thc 25 kO resistorifthe load terminalsare accidentally shoncircuited? d) What is the maximumpower dissipatedjn the ?5 kO resistor?
3,3
a) Find the valueof R that will cause4 A of currentto flow throughthe 80Q resistorin the circuit shown. b) How muchpower will lhe resistorR from part (a) nced to dissipale? c) How much power will the currentsource genemtelor the valueof R froln pafi (a)?
60() 100 20A
It 800
(a)150v; (b)133.33 V (c)1.6W (d)0.3w
Answ€r: (a) 30O; (b) 7680w; (c) 33,600 W.
NOTE: Also tty ChaptetPnblems 3.13,3.15,antl3.21.
3.4 tfqlltaqefiivisisn
a$deurrentllivision We canrow generalizethe resrlts from analyzingthc voltagedivider circuit in Fig.3.12and tlre currentdivider circuil in Fig.3.15.Thegeneralizations will yield two addiiional and very usctul circuit analysislechniques known asvoltagedivisionand currentdivision.Considerthe circuil shown inFig.3.18. The box on tlre left can cortain a singlevoltagesourceor any other combinationof basiccircuitelementsthat resultsin thc vollag€?)shownin the tigure.To the right of the box are r resistorsconncctcdin series.We division. are interestedin finding the voltagedrop ol acrossan arbitraryresistorRl figure3.18t Cjrcuitusd to ittustEievottage in terms of the voltager. We start by usingOhm's law !o calculatei, fte currentthrough all of the resistorsin serics,in lerms of the current o and
' n l + R 2 +  + R , , R.q
(3.2e)
SjmphResjstive Cjrcuitr
The equivalentresistance, Rcq,is the sum ol the r resistorvaluesbecause the resistorsare in series,as shownin Eq. 3.6.We apply Ohm's law a second time to calculatethe voltagedrop o/ acrossthe resistorRj. usingthe cureni t calculatedin Eq.3.29:
Vottaqedivision equation>
ur:
iRt :
RI =1)
(3.30)
Note that we used Eq.3.29 to obtain the righthand side oI Eq.3.30. Equation 3.30 is the voltage division equation.It saysthat the voltage drop ri acrossa singleresistorRi ftom a collectionof se esconnected resistorsis pioportionalto the total voltagedrop , acrossthe setofseriesconnected resislors.The coDstantof propor'tionality is the mtio of the single resistanceto the equivalert resistanceof the seriescomected set of resistors, or R'/Red. Now considerthe c cuit showr in Fig. 3.19.The box on the left can contain a single curent source or any other combination of basic circuit elemenls that results in the current i shown in the figure. To tlre right of the box are r resistorsconnectedin para el. We are inrerestedin finding the curent ii throughan arbitraryresistorRr in terms of the current i. We stafi by usingOhm'slaw to calculatez),thevoltagedrop acrosseachof the resistorsin parallel.in termsof the current; and the r resistors: 0 = (R1 R2 ... iln,) : rReq.
(3.31)
Theequivalent resistance ofn resistors in parallel,Rcq,canbe calculated usirgEq.3.12.We applyOhm'slawa second timeto calculate thecurrent l, throughtheresistorR;,usingthevoltageo calculated in Eq.3.31:
Currentdivision equationt
n
R,
Note that we used Eq.3.31 to obtain the righthand side of Eq 3.32. Equation 3.32is the current division equation.It saysthat the current i through a singleresistorRr {rom a collectior of parallelconnected resisto$ is proportionalto the total current i srpplied to the set of parallelconnectedresistors.The constantof proportionality is the ratio of the
tigure3.19A Circuii used to ittustrate cunentdivjsion.
3.4
Vottage Divisjon andCurrent Division
6l
equivalentrcsislanceof the parallelconnecled serofresistorsto the siDgle resrstance. or Rcq/&. Note that the constanlof proportionaliryin thc cul' rent divisionequationis rhe inverseof rhc constantof proportionalilyil1 the voltagedivisionequationl Examplc 3.4 usesvoltage divislon and current division to solve for voltagesand culrenlsin a circuit.
UsingVottageDivisionandCurrentDivisionto Solvea Circuit [Jsecurrenr divisionto find thc cullett i, and use voltagedivisionto find the voltager,, for the circuit ]n Fig.3.20.
Solution We san use Eq. 3.32if we can find the equivalcnl tcsislanceof the four parallel branchescontaining rcsislors.Symbolically,
Figur€3.20 ,",ThecircuitforExamph 3.4.
Req= (36 + 44) l0 (40 + 10 + 30) 24
= 8 01 0 8 0 2 4 
I * 1 * 1 lJ0 r o s 0
t
=60.
t
Applyhg Eq.3.32.
i, =
6
t(8A)
= 2A.
We can usc Ohm's law to find thc \,ottagedrop acrossthe 24 () resistor:
This is also tlle voltage drop acrossthe branch con tainingthe 40 Q, thc 10 O, and the 30 0 resistorsin . +Ucc.
Figure5.5 A Ie,minatcu,reni vadabe5
(5.1)
tigure5.6A Thevoliage transfer chancteristic ofan We seeftom Fig. 5.6 atrd Eq. 5.1 that the op amp has three distinct iegions of opemtiotr. When the magnitude of the input voltage difference ( o,  o, ) is small, t]Ie op amp behaves as a linear device, as the output voltage is a Linearfunction of tlle input voltages.Outside this linear region, t]Ie output of the op amp satumtes,and the op amp behavesas a nonlinear device, becausetlle output voltage is no longer a Lhear functiol of the input voltages.When it is opemting linearly, the op amp's output voltage is equal to the difference in its input voltages times the multiplying constant, or gaitr.,4. wllen we conline the op amp to its linear operaling region, a corstraint is imposed on the input voltages,o? and r'" . The constraint is based on tpical numerical values for ycc and ,4 in Eq. 5.1.For most op amps,t]le recommendeddc power supply voltages seldom exceed20 V and t]Ie gain, ,4, is mrely lessthan 10,000,or 104.We seefrom bottr Fig.5.6 and Eq.5.1 that in the linear region, the magnitude of the input voltage differerce ( l r ? , , I m u s l b el e s st b a o2 0 l 0 " . o r 2 m V
The0p€rabonat Amplifi er
Typically, node voltages in th€ circuits w€ study are much larger than 2 mV,so a voltagedifferenceoflessthan 2 mV meansthe two voltagesare essentiallyequal.Thus,when an op amp is constrainedto its linear operal ingregion andthe nodevoltagesarernuchlargerthan2 mV the constraint on the input voltagesof the op amp is
Inputvottageconstraint for idealop ampts
(5.2)
Note tlat Eq.5.2 characterizes the relationshipbetweenthe input voltages for an ideal op amp;thatiq an op ampwhosevalueof.4 is infinite. The input voltage constrajnt jn Eq.5.2 is called the yiltual short condition at the input of the op amp.It is natural to ask how the virtual short is maintained at the input of the op amp when the op amp is embeddedin a circuit, thus ensudnglinear operation.Theansweris that a signalis fed back from the output terminal to the inverting inpul ter minal. This configuration is known as negative feedback becausethe signal fed back ftom the output subtiacts from the input signal.The negative feedback causes the input voltage difference to declease. Becausethe oulput voltage is proportional to the input voltage difference,the outpul voltage is also decreased,and the op amp operatesin its linear region, If a circuit containing ar op amp does not provide a negative leedback path from lhe op amp output to the invertinginput, then the op amp will nonnally saturate.The difference in the input signals must be extremely smallto preventsaturationwith no negativefeedback.B t evenif the circuit providesa negativefeedbackpath for t}le op amp,linearoperationis not ensured. So how do we know whetler the op amp is operating in its linear region? The aNwer is,we don't!We deal with this djlemmaby assuminglin ear operation, peforming the circuit analysis,and then checking our results for coltradictions. For example,supposewe assumethat an op amp in a circuit is operating in its linear region, and we compute the output voltage of the op amp to be 10 V On examiningthe circuit, we discover that ycc is 6 Y resulting in a contradiction, becausethe op amp's output voltage can be no larger than Vcc, Thus oul assumption of linear operation was invalid, and the op amp output must be saturated at 6 V We have identified a constrainton the input voltagesthat is based on the voltage transfer characteristicof the op amp integrated circuit, the assumptionthat the op amp is restricted to its linear operating rcgion and to typical valuesfor ycc and.4. Equation 5.2 representsthe voltage constmint for an ideal op amp, that is, wittr a value of,4 that is infinite. We now tum oul attention to the constraint on the input currents. Analysis of the op amp integrated circuit reveals that the equivalent resistanceseenby the input termimls of tlre op amp is very large,tpically 1 M O
5.2 Te,minatVottages andcu'renis 159 or more. Ideally, the equivalent input resistanceis infinite, resulting in the
(5.3) < Inputcurrentconstraint for ideatop amp
Note ttrat the curent constraint is nor based on assuming the op amp is confined to its linear operating region as was t}te voltage constraht. Together, Eqs. 5.2 and 5.3 folm the constaints on terminal behavior that define our ideal op amp model. From Kirchhoffs current law we know that the sum of the currents entering the operational amplfier is zero, or
ip+ir
ia
i,
r. =u.
(5.4)
Substituting the constraint given by Eq.5.3 into Eq.5.4 gives
io:
Q , .+ i . ) .
The significance of Eq. 5.5 is that, even though the curent at the input terminals is negligible, there may still be appreciable curent at the output terminal. Beforc we starl analyzingcircuils containing op ampElet's further simplify tlle circuit symbol.w1len we know that the ampliEer is operaringwithin its linear rcgion, the dc voltages+Vcc do not enter into t}Ie circuit equations. In this case,we can remove the power supply teminals from tle s]'rnbol and the dc power suppliesfrom the circuit, as showt ir Fig. 5.7.A word of caution: Becausethe power supply teminals have been omitted, therc is a dangerof inJening from the symbol that i, + in + i, = 0. We have already noted that suchis not the case;that is,i, + i, + i, + i., + i. : 0. In other words! the ideal op amp model conshaint that i, : i, : 0 does no1 imply that i" : 0. tigure5.7 A The0pampsyrbotwiththe powersuppty Note that tle positive and negative power supply voltages do not have to be equal ir magnitude. In the linear operating region, z, must lie between the two supply voltages.For example, if V+=15V and y = 10V,then10V< D. < 15V.Be awarealsothatthevalueof.4 is not constant under all operating conditions. For now, however, we assumet]tat it is. A discussion of how and why the value of .4 can change must be delayed until after you have studied the electronic devices and componentsused to fab cate an amplifier. Example 5.1 ilustrates t}le judicious applicationof Eqs.5.2 ard 5.3. When we use these equations to predict the behavior of a circuit containing an op amp, in effect we are using an ideal model of the device.
160
lhe operationaL Amptifier
Anatyzing an 0p AmpCircuit Theop ampin thecircuitshownin Fig.5.8is ideal. a) Calculate 1,, if ?J! 1 V and ,b = 0 Vb) Repeat(a) for 0o : 1 V and ob : 2 V. c) Ifo, = 1.5V, specifythe rangeof rb that avoids amplifiersaturation.
t ! ! 1 0 0k O
The current constraint requires i, = 0. Substiluting the values for the three currerts into the nodevoltageequation,weobtain 7 25
1). 100
''
Hence, t, is 4V, Notc that because,, lies between j 10V, the op amp is in its linear region of opemtion. b) Using the sameprocessas in (a),we get ap:ah:un=)y,
'"
25
12 : 25
100
ff"'
Figure5.8 ..' Theci,cuitforEMmpte 5.1.
,25 :
1 25.^,
,1tl(]
Sotution a) Becausea legative feedback path exists from the op amp's output to its inverting irput througl the 100 kO resistor, let's assumethe op amp is confined to its linear operating region. We can write a rodevoltage equation at the inverting input terminal.The voltage at tlle irverting input terminal is 0, as r'r, : x'b = 0 from the connectedvoltage source, and uh:1)p from the voltage constraintEq. 5.2.The node'voltageequationat
Therefore, ?,o= 6 V. Again, ?, lies within + 10 V. c) As beforqo, = op = ob,andi25= itoo.Because ?]a:15V, 1 , 5 0 b
25
0o0t
100
Solving for ob as a function of o, gives I.'b=i(o+"i
From Ohm'slaw,
,,r = ru. ,,,)/25:
l

mA.
= (?Jd 0,)/100:0,/100 mA. 4o1r
Now, if the amplfier is to be within the linear regior of operation, 10 V < 0, < 10V. Substituting these limits on o, into the expression lor ?b,we seethat ob is limited to 0.8V 0 ; ? ,: 0 , t < 0 . d) Figure 6.4 shows the voltage waveform. e) No;the voltageis proportionalto dt/dt, not i. f) At 0.2 s,which coresponds to the moment when /t/d/ is pa$ing rhrougbzero and changingsign. g) YeE at t : 0. Note that the voltage can change instantaneouslyacross the teminals of an inductor,
Fisure5.2r Th€cjrcujt forb(ampl€ 6.1.
a) Sketch the curent waveform. b) At what instant of time is the current maximum? c) Er?ress the voltage adoss the lemrinals of the 100mH irductor as a function of time. d) Sketch tlle voltage waveform. e) Are the voltage and the clrlrsnt at a maximum at t])e same time? f) At what instant of time does the voltage change polarity? g) Is therc ever an instantaneouschange in voltage aqossthe inductor?If so,at what time?
r (A)
Flgure5.31Thecunentwaveform forExample 6.1.
?r(v)
Sotution a) Figue 6.3showsthe curent waveform. b) dildt = lo(stest + s 1= 10e 5'(1  sr) Als;dildt = 0 whent = s. (SeeFis.6.3.) l
Figure6.4
ThevoLtage waveform fo' E\ample 6.L
190
Inductance,Capacitance,and Mutuatlnductance
Currentin an Inductorin Termsof the Voltaae Acrossthe Inductor Equation6.1expresses the voltageaoossthe terminalsofan inductorasa function oI the curent in the inductor. Also desirable is the ability to expressthe curent as a function of the voltage.To find i as a function of o, we startby multiplyingboth sidesofEq.6l by a differentialtime dt:
=,(*),,
(6.2)
Multiplying the rate at which i vadeswith tby a differenrialchangein rime generatesa differentialchangein t, so we $,riteEq.6.2 as adt = Ldi.
{6.3)
We next integrateboth sidesof Eq. 6.3.For convenience, we interchange the two sidesofthe equationand write
"l*,'o'=1,"' ' o'
(6.4)
Nole that we useI and i as the variablesof irtearation. whereasI and I becomelimirsonlhe integral..Thcn. trom Lq.0.4.
Theinductori  z' equationb
1 r 4.t) : i I 1'!1r + j(ro),
(6.5)
where i(r) is the current coresponding to 1,and i(r0) is the value of the inductor curent when we initiate the integration,namely,t0. In many practicalapplications,to is zero and Eq 6.5becomes
iI'.*.
(0).
(6.6)
Equations6.1 and 6.5 both give the relationshipbetweenrhe volrage and current at the terminalsof an inductor. Equation 6_1exDresses the \olrageasa funcrionot currenr.u hereasLq.o.5i\pfesse\lhe curfcnlasa tunction of voltage.In bolh equationsthe referencedirectionfor the current is in the directionof the voltagedrop acrossthe teminals. Note that i(lo) cardesits own algebraicsign.If the initial cu enr is in the samedircc tion as the reference direction for i, it is a positive quantity. If the initial current is in the oppositedirection,it is a negativequantity.Example 6.2 illustratesthe applicationof Eq. 6.5.
6.1 TheInductor 191
Determining the Current,Giventhe Voltage. at the Terminals of an Inductor The voltage pulse applied to the 100 mI{ inductor shown in Fig. 6.5 is 0 for t < 0 and is given by the expfession
r=0, 100mH
1)= z\tetuV,t>0
a(t) = 20tu ntV Figure6.5
foit > 0.Also assumei : 0 for I < 0.
lhe circuitfor Exanpte 6.2.
a) Sketch the voltage as a function of time. b) Find the inductor currenaas a function of time. c) Sketch the current as a functiotr of time.
Solution a) The voltage as a lunction of time is shorm ln Fig.6.6. b) Thecunentin thehductor is 0 a = 0.Therefore, thecurrentfor t > 0is 1= t
2 I
I t0
10te10' s 10)A,
Flgure6.6 A ThevoLtage wavefo,n fortuample 6.2
r (A)
 2 r . x et c d t + o
t ",o' ll, = 2 0 0 11 m ( 1 0 + r 1 )l l . =2(1
r 0
c) Figure 6.7 showsthe cment asa function of time.
Figure5.7 Thecun€ntwavetorm forExampte 6.2.
Note ir Example 6.2 that i approaches a constant value of 2 A as t increases.We say more about this rcsult after discussingth€ energy stored in an hductor,
PowerandEnergyin the Inductor The power and energy relafionships for an inductor can be derived directly ftom the current al1d voltage relationships. ff the curent reference is in rhe dircction of t]le voltage drop across the terminals of the inductor.the Doweris
(6.7)
192
Inductance, CaDacitance, andlluiuatInductanc€
ReqgmbeIlhlt pqwg.r.istu wa!t!, voltageis in voltq atrd aurrent is in amperes.If we expressthe inductorvoltageas a function of tlte hductor currcnt,Eq. 6.7becomes
'i:lt:',;1.. :: ': :'i'i:Li:i!.. ':.
Power in aninductor >
..
\t dt .:i.
(6.8)
We can also expressthe curent it terms of the voltage:
o:,1!1,"',*. r,"t)
(6.e)
Equation 6.8 is useful in expressing the energy stored in the inductor. Power is the time mte of expending energy,so d.w
,.di
(6.10)
Multiplying both sidesof Eq.6.10 by a differential time gives the differential relationship dw : Li .1i.
(6.11)
Both sidesof Eq.6.11 are integiatedwith the understandingthat the refeietrce for zerc energy correspotrdsto zero curent h the inductor. Thus
I
Energy in anindudor>
dx=Llyd)J,
$u,iffi
(6.12)
As before,we use different s]'mbolsof integration to avoid confusion with the limits placedon the integrals.In Ed: 6.12,the energyis itr joules, inductanceis in henrys,and curent is in amperes. To illustrate the applicatiotrof Eqs.6.7and 6.12,we return to Examples6.1and6.2by meansof Example6.3.
6.1 lheInductor 193
Determining the Current, Voltage, Power. andEnergy for an Inductor a) Por Example6.1,plot i, o,p, and x, versustime. Line up the plots veftically to allow easy assess ment of each variable's behavior. b) In what time interval is eneryy being stoied in the inductor? c) In what time interval is energy being extmcted from the inductoi? d) What is the maximum energy stored in the hduclor? e) Evaluate tl]e integrals
f" J" oo'
and
e) Fiom Example6.1, i : 10t?5'A
and
? , : e 5 ' ( 1 5 4 v .
Therefore, p = 1)i = Tote 1or 5or2e 1orw.
/.l",oot, 0.6
and comment on thet significance. Repeat (a) (c) for Example 6.2.
0.8
1'(v)
In Example6.2,why is there a sustainedcurent ir the inductor asthe voltageapproaches zero?
Solution a) The plots of i, o,p, and ?.'follow direcdy from the er?ressions for i and t, obtained in Example 6.1 and are .hown^in I ig. 6.8.ln particulaf.p  ,t. andl/,  (;)Lt. b) An increasing energy cun'e indicates that energy is being stored.Thus erergy is being stored in the time intenal 0 to 0.2 $ Note that this correspondsto the irterval when p > 0. c) A decreasingenergy curve indicates that energy is being extracted.Thus energy is being extracted in rherimeinrerval0.2s lo oo.Nole thatrhiscorrespondsto t}le inte al when p < 0. d) From Eq.6.12 we seethat eneigy is at a naxlmum when current is at a maximum; glarcing at the gmphscontirms this.From Example 6.1,ma)omum curent : 0.736A. Therefore,unax : 2'7.07mJ.
u (nJ.)
figure 6.8 A Thevarjabtes t, r, p, andu venusr for Example 6.1.
194
Induchnce, Capacitance, aid l,lutuat Inductance
Thus
ta.2
f,tu lo.z rat : tul*tto'  tt l"
/
'{t#. *[#r,,
g) The application of tle voltage pulse stores ene4y in the inductor. Because the inductor is ideal,this energycannotdissipateafter the vottage subsidesto zero, Therefore, a sustained current circulates in the circuit. A losslessinductor obviously is an ideal circuit element. Practical inductors rcquire a rcsistor in the circuit model. (More about this laler.)
',]):'
= 0.2e) : 27.0'7tJ,l,
, (v) /
f "1at lm = IoLlmt to, t, Jate',t ]0.,
'{3#..e[#, *'
"]);
i (A)
: ^0.2e 2 = 21.M mJ.
Based on the definition of p, the area undei the plot of p ve^us r represents the energy erpeDded over the intenal of inregration. Hence tlte integmtion of t}le power between 0 and 0.2 s representsthe energystored in the inductor during tiis time interval. The inteFal of p ovei the interval 0.2 s co is the energy extracted. Note that in this time interval, all the energyodginallystoredis removed;thatis,after the cment peak haspassed,no energyis stoied in the inductor. f) The plots of o, 4p, and 1, follow dircctly from the er?ressions for ?,and i given in Example 6.2 and are shown in Fig. 6.9. Note that in this case the po$er is alwals posilire.and henceenergyis alwaysbeing stored during the volrage pulse.
? (w)
u (mr)
0.4 0.5 0.6 Figure5.9 l. Theva abtes 1',t, p, and?ove6usr for Exampte 6.2.
'
6.2
TlreCapacjtor 1 9 5
for vottage, current,power,andenergyin aninductor objective1Know andbeab[€to usethe equations 6.1
The curent sourcein the circuit showngeneratesthe curent pulse lr(t)=O,
t o.
Fnrd (a) u(0); (b) the iDstantof time,greater than zero,when the voltageo passesthrough zerc;(c) the expressionfor the power deiivered to the inductor:(d) the instantwhen the poNer deliveredto the inductoris maximum;(e) the naximum power;(t the instantof time when the storederergy is maximum;and(g) the maximum energystoredin the inductor.
An5wer (a) 28.8V (b) 1.s4ms; (c) 76.8e6ooi + 384e15oo'  307.2e2{otw. r > 0; (d) 411.0s i!s; (e) 32./2w; (f) i.54rns; (g) 28.s7mJ.
NOTE: Also try Chaptet Pnbkns 6.1 and 6.3.
6.2 Thefapacitor The circuit parameteroI capacitanceis representedby the leller C. is measuredin farads(F), and is synbolizedgraphicallyby tlvo short paral lel conductiveplates,as shown in Fig.6.10(a).Becausethe farad is an (a) extrernelylargequartity ofcapacitance.practicalcapacitorvaluesusually lic in the picofarad(pF) to microtarad(,.rF)range. C Thc graphic symbol for a capacitoris a reminder that capacitance occurswheneverelectricalcoDduclorsarc scparatedby a dielectric,or insulating,material. This condition implics that electric charge is nol lransportedthrough the capacitor.Although applying a voltage to the (b) terminalsofthe capacitorcannotmove a chargethroughthe dielectric.it a capacitor. can displacetI chargc within the dielectric.As lhe voltage varies with Figure6.10;]: (a)Ihecircuiisymbotfor (b)Assignjnq vottage and cu entto the rcfercnce time, the displacementot chargealso varies with lillle, causingwhat is thep$sivesignconvention. capacitor, totLowing knowr as the displecem€ntcunent. fiom a At the terminals.Lhedisplacementcurrentis indistinguishablc conductioncurcnt. The current is propolioDal 1()thc ralc at which the voltageacrossthc capacitorrarieswith time, or, malhematicallv,
t =r # ,
(6.13) n; capacitori
whcre i is measuredin amperes,C in larads.?rin volts,and tin seconds.
t, equation
196
Inductance, capacitance, andtlutuat Inductance Equation 6.13reflects the passivesign convention shown in Fig. 6.10(b); that is,the current referenceis in the direction of the voltaqe drop acrossthe capacilor.It tbe currenl referencei\ in fie direclion oirhe volraserise. Eq.o.lJ i\ wrirrenwitha mjDu.srgn. Two importantobsenationsfollow from Eq. 6.13.Iirst.voltagecarmot change instantaneouslyacrossthe terminals of a capacitor. Equation 6.13 indicares tbrl sucha changewouldproduceintinirecurrenr,a phrsical impossjbiiir).second. it the vollageacrosslhe lermtnal.is con.ranr,rhe capacrtorcurrent is zero.The reason is that a conduction current cannot be establishedin the dielectdc male al of the capacitor. Only a time_varying voltage can produce a displac€mentcurrent, Thus a capacitor behavesas an open circuit in the presenceof a constant voltage. Equation 6.13givestle capacitorcurent as a function of the capacitor voltage.Expressingthe voltageas a lunction of the cu ent is alsouseful. To do so,we multiply both sidesof Eq. 6.13by a differcnriattime dr and then integrate the resulting differentials:
id! = Cdo
l))'*=il,"''*
Carrying out the integralion of the lefthand side of the secondequa, tlon grves
Capacitor?) i equationb
,r,t=ll',0,*,
(6.28)
fquivatentcapacitance iniiial voltage>
(6.2e)
We leave tlle dedvation of the equivalent circuit for sedesconnected capacitors as an exercise.(SeePrcblem 6.30.) The equivalent capacitanceof capacitors connected in parallel is simply the sum of the capacitancesof the individual capacitors, as Fig. 6.18 and the follouring equation show:
Conbining capacitors in paraltet>
(6.30)
Capacito$ connected in parallel must carry the samevoltage.Therefore, if fiere is an initial voltage acrossthe origbal pamllel capacitors,this same idtial voltage appeaN acrossthe equivalent capacitance Ceo.The derivation of the equivalent circuit for pamllel capacitors is left ai an exercise. (Seehoblem 6.31.) We saymore about se esparallel equivalent circuits of inductors and capaciton in Chapter 7, where we interpret results based on their use.
;rr''T
+
cl
c1
q (ro)
IT
l + "^i"t^s (a)
.
 _),
;
1
_)" T
1
;1 l
(a)
_('.q L=!+ L+...*! cr c c"
cea,
1,(d = ?,(d + !(h) + " + 1'"(ro) (b)
C e 1 , =C r + C 2 +  + C n
Figure 5,17A Anequivatent cjrcuittor capacitors connected in (d)Ihesedes series. caDacirors. (b)Tteeqriva.ent crcur.
(b) figure6.18" Anequjvatent cjrcuitfor capacitors connected in palattet. (a) CapacjtoB (b)Ihe equivatent in paratteL. cncuii.
6.4
l4utuatlnductance203
objective3Be ableto combine inductors or capacitors in seriesandin paral(elto forma singLe equivaLent indudor 6.4
The initial valuesoflr and l2 in the circldt showl are + 3 A and 5 A, respective]y. The voltageat the terminalsofthe paralielinductorsfort > 0is 30e"mV. a) Il lhc paralleliDductorsare replaccdby a singicinductor',what is its irduclancc? b) Whal is Lheinitial currentand ils leference directior in the equivalcn1inductor? c) Use the equivalcntinduclor to find l(r). d ) I n d ; { , 1 a n Lr lt ) . \ e r i r } r r J . r r e . o l u r ' o n . fbr i1G),ir(,), and ,0) satisfyKirchhofl's
An5wer: (a) ,18mH; (b) 2 A,p; (c) 0.125e5, 2.125A,1 > 0; ( d )r r 0 ) = o l e s r + 2 . 9 4 . r > o , jy'r) = o.o25e5' 5.025A, r > o. 6.5
Thc currenl ai the terminalsof the two capaci lors showDis 240? "'l.A for I > 0. Thc inilial valuesofur andq are 10Vand 5V. respectivellCalcuiatethe total cncrgy trapped in the capacitoKasr + N. (H,irij Don'lcombine the capacitorsin series lind thc energy lrappedin each,and then add.)
240mH
Answer: 20AJ. NOTE: ALto try ChapterProbler8 6.21,6.22,6.26, urul6.27.
6.4 t4util;linducta*se The magneticijcld we consideredin orn studyof iiduolors in Secrion6.1 wasrestrictedto a singlecircuit.We saidthat induclancei! the parameter that relatcsa voilage 1()a time\,aryingcuflcnt in the samecircuit;thus, jnductanccis nore preciselyreferredto as scll inductance. We now considerthe situationin which lwo circuitsare linkcd by a  + v V  t , _ magnclic[e]d.In this case.the voltageinducedin rhe secondcircuil can bc rclaled lo the timevaryingcurrcnl in the first circuir bv a paramcter kno\n asnutual inductance. Thc circuil showr in Fig.6.i9 reprcscnlstlvo magnclicallycoupled coils.The scll nrductancesof the tr,o coils are labeled/,1 ard lr, and the mutual inductanccis labeledM. The doublc headedarrow adjacentto M indicatesthc pair oI coitswith this valuc of Fig!ie6.19 i. Twomagneticatty coupted coiLs. mutualinducQnce.Tlisrotation is necdcdparlicularlyin circ its conrrin ing morc than oDepair ofrnagneticallycouPledcoils. The easiestway to analvzecircuitsconlainhg murualinducranceis 1{) uscmeshcu ents.Theproblemis 1owdle thecircuitequationsthat describe lhe circuitin termsof thc coil currents.First,choosethc rclcrencedirection lbr eachcoil current.Figurc620lholvs arbitrarilysclcctcdrelcrencecurrerls.After choosingthc rcfcrcncedirectionsfor i and ;2.sum the voltages aroundeachclosedpath.Becauscol lhe mutualinductanceM. thcrc will be Figure6.20i Coitcurients tLandl2 usedto describe two vollagesacrosseachcoil.namcll a selfhducedvoltageand a nutually tlrecncuitshownin Fiq.6.19.
.,o ',1[,', i^'
204
Inductance,Capacitance,andllutualInduchnc€
r.d r. .l'rra 
I
*Qrl.,l ["? *" t . l Figurc5.21A Thecjrcuitof Fig.6.20withdotradded toth€ coitsindicating the potarity ofthe muiuatly
Dotconvention for mutuattycoupl€dcoits>
induced voltage.The self induced voltage is the product of the self, inductance of the coil and the 6rs1derivative of th€ cunent in tftat coil. The mutually induced voltage is the pioduct of tlrc mutual inductanceof the coils and the fust derivative of the curent in the other coil. Consider the coil on the left in Flg. 6.20whose selJitrductarcehasthe value la. The selJinduced voltage adoss this coil is ar(dildt) and the mutualy hduced voltage aooss this coit is M(di/dr). But what about the polariries of rheserwo voltages? Using the passivesign convention, the selfinduced voltage is a voltage drop in the direction of the curent producing the voltage. But the poladty of the mutually induced voltage dependson the way t]le coils are wound in relation to tte reference directior of coil clrllents. In general, showing ihe details of mutually coupled windings is very cumbersome.Instead,we keep track of the polarities by a method known as t}Ie ilot convention,in which a dot is placed on one lerminal of each whding, as shown in Fig.6.21.These dots carry the sign informatior and allow us to draw the coils schematically rather than showing how they wrap around a core structure. The rule for using the dot convention to determine the polarity of mutually induced voltage can be summaized as follows: When the reference direction for a curent enters the dotted teminal ol a coil, t]le reference polarity of the voltage that it induces in the othei coil is positive at its dotted terminal. Or, stated alternatively,
Dotconvention for mutuattycoupted coits (alternate)>
W}len tie reference dircction for a current leaves the dotled terminal of a coil, the reference pola ty of the voltage that it induces in the other coil is negative at its dotted terminal. For the most part, dot markings will be provided for you in the circuit diagrams in this text. The important skill is to be able to wdte the appropriate circuit equations given your underctanding of mutual inductance and the dot convention. Figu ng out where to place the polarity dots if they are not given may be possible by examining tie physical configura, tion of an actual circuil or by testing it in fhe laboratory. We will discuss these procedures afrer \i'e discussthe use of dot markings. In Fig. 6.21,the dot convention rule indicates that the reference polarity for the voltage induced in coil 1 by the current i2 is negative at tie do! ted terminal of coil 1.This loltaEe (Mdi2/dt) is a voltage dse with respecr to 4. The voltage i duced in coil 2 by the curent ilis M dh/.Lt, ^nd its rcferencepolarity is positive at the dotted terminal of coil 2. This voltage is a voltage ise in the direction of 4. Figue 6.22 sho*s t}le self and mulually inducedvoltagesaoosscoils 1 and 2 alongwith their polarity maiks.
M
+
Ll
tiguru6.22a Thes€lfandmutuattyinduced vottases appearjng across ihecoits shown in Fig.6.21.
6.4 Muiuatlnductance 205 Now let\ look at the sum of the voltagesaroundeachclosedloop.In Eqs.6.31and6.32,voltageises in the referencedircctioll ofa currentare negative:
u"  i,R, t t,d" " d t
t,*, * t,*
d
, o ' t.   o .
(6.31)
= o.
\6.3?)
,*
TheProcedure for Determining DotMarkings We shift now to two methods of determining dot markings. The first assumesthat we know the physicalarangement of the two coils and the mode of eachwindingin a magneticallycoupledcircuir.The following six steps,applied here to Fig.6.23,determinea set ofdot markirgs:
,l
a) Arbitradly select one terminal say,the D terminal of one coil and mark it with a dot. b) Assigna currentinto the dotted termiml and labelit rD. ?) c) Use the ighthand iule1 to determinethe direction of the magneric lsrel field establishedby iD iruide the coupl€dcoilsand label this field dD. rA.rbitrarily d) Arbitrarily pick one terminal of the secondcoilsay, terminal A and assigna curent into this terminal, showirg the current as iA. (Step1) e) Use the ghthandrule to detemine t}le direction of the flux established by ta ir ide the coupled coils and label this flux dA. Figure6.23A A setof coilsshowing a method for d€t€rmining a set ofdot nrarkings. f) Compare the direcrions of the two fluxes dD and dA. If the flrlxes have the samercference direction, place a dot on the terminal of the second coil wherethe test current (iA) enters.(In Fig.6.23,the fluxesdD and dA havethe samereferencedirection,and thereforea dot goeson terminal A.) If the fluxes have different reference directioni place a dot on the teminal of the secordcoil wheretie testcurrentleaves. The relativepolaritiesof magneticallycoupledcoilscan alsobe determhed experimentally. This capability is important becausein some situations,determininghow the coilsare wound on the core is impossible.One experimentalm€thodis to connecta dcvoltagesouice,a resistor,a switch, and a dc voltmetei to the pair oI coiiE as shownin Fig.6.24The shaded box coverjng the coils implies tlat physical inspection of the coils is nol possible.The resistorRLimitsthe magnitudeof tie currentsuppu€dby the dc voltagesource, Flgure 6.24A AnexperimentaL setup tordeteririning The coil terminal connected to the positive terminal of tle dc source via the switch and limiting resistor receives a polarity mark, as show, in Fig.6.24.Wlen the switchis closed,the voltmeterdeflectionis observed.If the momentarydeflectionis upscale,the coil teminal connectedto the positiveterminalof the voltmeterreceivesthe poladty mark.ffthe deflection is dowNcale, the coil terminal connected to the negative telmhal of ure voltmeter ieceives the polarity mark. Example6.6 showshow to usethe dot markingsto fomulate a set of circuitequationsin a circuit containingmagneticalycoupl€dcoils. ' s".ti*ion
otru.uauyttawonpage207.
206
Inductance,Capacitance,andMutualInductance
FindingMeshCurrent Equations for a Circuitwith Magenticatty Coupted Coits a) Write a set of meshcrment equations that descdbe the circuit in Eg. 6.25 in tems of the currents i1 and ,2. b) Verify that if there is no energy stored in tlle clrcuit at r  0 and if ic = 16  16e5'A, tie solutions lbr i1 and i2 are i 1= 4 + 6 4 a 5 t  6 8 e 4 t A ,
i1(0)=4+6468=0, i,(0)=1
i2= 7  52a5t+ 51e4tA..
t.) 8H
b) To check the validity of i1 and i2, we begin by testhg 1}leinitial and final values of i1 and i2.We know by h}?ot}lesis that t1(0) = ir(0) = 0. From the giver solutions we have
20f,l
reu f i'i aoo
5 2 + s 1= 0 .
Now we observe that as l approachesinlinity the source current (is) approachesa constatrt va.lue of 16 A, and ttrerefore the magnetically coupled coils behave as shon circuils.Hence al I  m lhe circuil feducesro rhat (howo in Fg. b.26. From Fig. 6.26 we see that at I : co the thrce rcsistors are in pamllel aqoss the 16 A source. The equivalent resistanceis 3.75 O and thus the voltageacrossthe 16 A curent sourceis 60V It tbllows that 4(co)
Figure6.25 Thecircuit forExample 6.6.
60 20
60 60
t(,:o): r @ = t o
Solution a) Summingtle voltagesaroundthe il meshyields 4d
 I3dtlis t2) .20(i i)
l5(;
r")=0.
Thesevaluesagreewith the final valuespredicted by t}Ie solutions for 4 and i2. Finally we check the solutions by seeing if they satisfy the differential equatiom dedved in (a). We will leave this fhal check to the readei via Problem6.37.
The i mesh equation is 20lir
) l,)  60i, + lb:{/, dt
i"r '
si, 8"1'  ". dt
Note that the voltage aqoss the 4 H coil due to the current (is  i), that is,gd(is  i)/dt,1s a voltage drop in the d ection of tt. The voltage induced in the 16 H coil by tle curent i1, that is, 8dtldl, is a voltage rise in tle direction of i2.
50
20a
Figure5.26A lhecircuiiof Lumpte 6.6when r = co.
6.5
A Ctoser Lookat l4utual Inductance 207
0bjective4Use the dot convention to writemeshcurrent equations for mutuallycoupted coils 6.6
a) Wite a set of meshcurtentequationslor the circuit in Example6.6if the dot on the 4 H inductor is at the righlhand telminal, the referelce djrectionof is is reversed,alrd the 60 O resistoris increasedro 780 (). b) Verify that ifthere is no energysroredin rhe circuitat r : 0, andif is  1.96 1.96e4tA, the solutionsto the differentialequations derived in (a) of this AssessmentProDremare
Ans$,er:(a) 4(dirldr) + 25ir + a(rlizlrh) 20i2 : sis s(disldt) 8(di/dt)  20ir + 16(rli2/ dt) + 800i2  16(di:r/dt); (b) vedfication.
ir = 0.,1 11.6e{'+ 12€5'A, /2 
0.Ul
0'1o,""
,' 5 A
NOTE: Also try ChaptetPrcblen 6.34.
6.5 A fl[oserLookat Mutual3*duetaalee In order to fully explainthe circuit paramercrmutual inductance,and to examinethc limilationsandassunptionsmadein the qualitativediscussion presentedin Section6.4,wc beginwith a more quanritativcdescdptionof selfinductance thanwasprcliousiy providcd
A Reviewof SetfInductance :lhe conceptofinductancecan be tracedto Michael Faraday. who did pio neeringwork in this area in t]re early 1800s.Faradaypostulatedthat a magnelicfield consistsof lines of force surroundingthe currentcarrying conductor.Visualizetheselines of lbrce as energystoring elastjcbands that closeon themselves.As the currentincreasesand decreases. the clastic bands(that is.thc liDesof force) spreadand collapseaboutrhe conductor.The voltagejnducedin the conductorisproportionalto rhe numberoI lines that collapseinto, or cut. thc conductor.This imageof inducedvolf ageis expresscdby what is callcdFaradays law;lnar N,
,:n'
(6.33)
where,\ is rcferredto asthc flux linkageand is neasuredin weber{urns. Howdowegetfrom Faraday'slaw to the definitionofinduclaDcepresentedin Scction61?Wecanbeginto draw thisconnectionusingFig.6.27 Figure6.27A Represeniation of a magnetjc fieldtink
208
Inductanc€, CaDaciiance, andlllutuat Inductance
The linestbreadingthe N tums and labeledd representthe magnetic lin€s of foice that mal(e up the magnetic field. The shength of th€ mag' netic field dependson the strengttrof the current,and the spatialorientation of ihe field depelds on the diection of the current. The righthand rule rclates the odentation of the field to the direction of the cment: w}len the fingers of tlle right hard are rrrapped around the coil so that the fingers point in the directior of tlle curent, tle thumb points in the direction of that potion of the magneticfieldinsidethe coil.Theflux linkageis the product of the magneticfield (0), measuredin webers(Wb), and the numberof turns linted bv the field fN\: 16.34) The magnitude of the flu\, d, is rclated to the magniiude of the coil currentby the relationship
O: sNt,
(6.35)
whereN is the numberof tums on the coil, and O is ttrepermeanceofthe space occupied by the flux. Permeance is a quantity that describes the magneticpropertiesof this space,and assuch,a detaileddiscussionofpermeancejs outsidethe scopeof this text.Here,we need only obsene that, when the spacecontainhg the flux is made up of magnetic materials (such as iron, nickel, and cobatt),the permeancevarieswith the flux, giving a nonlinearrelationshipbetween4 and t. But when the spacecontaining the flux is comprised of noimagnetic materialsr the permeance is constant, giving a linear relationship between d and i. Note from Eq. 6.35 that the flux is alsopropoitional to the numberof tums on tle coil. Here, we assumethat the core materialthe spacecontaining the flux is notunagnetic.Then, substituting Eqs.6.34 and 6.35into Eq. 6.33yields
d^
= N;:
d(No)
N;tENrl
= Nza!! _ t!!
(6.36)
which showsthat selfinductance is propoitional 10 tlle square of the number of tums on the coil.We make useof this observationlater. The polarity of the induced voltage in tie circuit in Fig. 6.27reflects the reaction of the field to the cwent creating the field. For example,when i is increasing, dt/dl is positive and o is positive. Thus energy is required to establishthe magnetic field. The product ?i givestlre rate at which energy is stored in the field. Wlen the field collapses,di/dt is negative,and again the polarity of the induced voltage is in opposition to the change.As the field collapsesabout the coil, energy is returned to the circuit. Keeping in mind this furtler insight into the concept of sefinductance, we now turn back to mutual inductance,
6.5 A ctoser Look at l{utuat Inductance209
TheConcept of MutualInductance Figure 6.28 shows two magnetically coupled coils.You should verily that the dot markings on the two coils agreewith the direction of the coil windings and curents shown.The number of turns on each coil are N, and N". respeclively.Coii I is energizedb) a trme+arling cunenr tour.. rlai establishesthe current ir itr the Nl turns. Coil 2 is not etrergized and is open. The coils arewound on a ronmagnetic core. The flux produced by the cment il can be divided into two components,labeled du and d21. The flu\ component d11is tlle flux produced by ir that links only the N1 Figure 6.28A Twonragneticatly coupted coib. turns.The component 421is t}le flrlx produced by i1 that links the N2 turns and the N1 tums. The tint digit in the subscript to the flux gives the coil number, and the second digit refers to the coil current. Thus dj r is a flux linking coil 1 and produced by a current in coil 1,whereasd2r is a flux lhking coil2 and producedby a currentin coil 1. The total flux linling coil 1 is dl, tle sum of d11and d\: .lt + 6*
et:
(6.37)
The flux dr and its components dx and d2r are related to the coil curent i1 asfolows:
6t = 94rir 0i
(6.38)
: 9l ,Ni1,
(6.3e) (6.40)
where 9r is the permeance of the spaceoccupied by the flux d1, 9rr is the permeanceof the spaceoccupied by the flux d1r, and 921is the permeance of t}tespaceoccupiedby the flux d21.SubsrirutingEqs.6.38,6.39, and 6.40 into Eq. 6.37 yields the telatioNhip between the permealce of the space occupied by the total llux dr and tle pemeances of rhe spacesoccupied by its componentsd11andd2r: 91=gr+@21.
(6.41)
We use Faraday's1awto derive erpressions for r)r and o2:
: N,*@u, ,' =# =o'ry;fu o,') : N]g.11+s^\*
= *p,*
= ",*
(6.42)
and d(Nzdzt) dt
= N,w,g,,\
=u.frg,,N,i,1 (6.43)
210
Induchnce,Capacitance,andl,4utuallnduciance
The coefficient of dt/dt i]) Eq. 6.42 is the selfinductance of coil 1. The coefficientof dtldl in Eq. 6.43is the mutual inductancebetweencoils 1 and 2.Thus \6.44)
The subscript on M specifiesan irductance that relates the voltage induced in coil2 to the curent in coil 1. The coefficient of mutual inductance sives D2= M2r
di. dt.
(6.45)
Note that tle dot convention is used to assignthe polarity rcference to ?)2 in Fig.6.28. For the coupled coils in Fig. 6.28, exciting coil 2 from a timevarying current source (i2) and Ieaving coil 1 open produces the circuit arrangement shown h Fig. 6.29. Again, the polarity reference assignedto or is basedon the dot convention. The total flux linking coil2 is Figure5.29A Themagnetkatly coupted coil50f Fig.6.28,withcoit2 exciied andcoill open.
6z:6zz+4o.
(6.46)
The flux d, and its components d22afil dD are related to the coil curent i2 as follows:
O2= g2N2i2,
\6.47)
4n : qrNzrz,
(6.48)
0D: grzN2i2.
\6.49)
The voltagesu2and ur afe
u=ff=uZo,ft=1,ff,
(6.50)
 Nfi2{to*. ,,:ff =fiw,ql
(6.51)
The coefficient of mutual inductancethat relates the voltage induced in coil 1 to the timevaryingcurrent in coil 2 is the coefficient of dizldt in Eq.6.51: Mp = N1N20p.
(6.52)
For nonmagnetic materials, the pemeances O12and @r are equal, and rherefore MD:
M2r = M.
(6.53)
Hence for linear circuits with just two magnetically coupled coils, attaching subscriptsto the coefficientofmutual inductanceis not necessary.
6.5 A Ctos€r Loolatl,iutuat Inductance211
MutualInductance in Termsof SelfInductance The value of mutual inductance is a furction of the seiJinductances.We derivetlis relationshipasfollows.From Eqs.6.42atrd6.50,
Lr = Nlgr,
(6.54)
L2 = N1!t2,
(6.55)
respectively. FromEqs.6.54 and6.55, LrL2 N?N1!trs2.
(6.56)
We now use Eq.6.41 and the corresponding expressionfor @2to write
L[2 = N1Ni(gr + @2)(922+ gn).
(6.57)
But for a linearsystem, O21: 9p, soEq.6.57becomes
Lr4= (N1N2s1)2(1. #)('.
#)
=*(,H)('.H)
(6.58)
Replacing the two terms involving permeances by a single consIant expresses Eq.658in a more meaningfulform:
i=("H)(.H)
(6.5e)
Substituting8q.6.59 into Eq.6.58yields
tf
= kzLrLz
(6.60) < Retating setfindudances andmutuar inductance usingcouptingcoeffiaient wherc the constantt is called t}le coefficientof coupling.Ac€ordingto Eq.6.59,1/fr'musrbe$eater than1,whichmeansthat ,t musrbetessthan1. In fact,the coefficientof couplingmustlie between0 and 1,or 0 0. b) Sketchtversuslloro< I < @.
figureP6.3 iL(.4
u,@
,1mH
prcbrems219 The curent in a 100/rH inductor is known to be iL = zotelt A Ior t>0. a) Find the voltage aqoss the inductor for t > 0. (Assume the passivesign convention.) b) Find the power (in microwarts) at the rerminals of the inductor when r : 100ms. c) Is the inductor absorbing or delivering powei at 100ms? d) Find the energy (in microjoules) srored in the inductor at 100ms. e) Find t}le maximum energy (in midojoules) slorediD rhe inducloraDdthe time (in micro seconds)when it occurs.
The current in and $e voltage acrossa 2.5 H inductor are known to be zero for t = 0. The voltage acrosslhe inductofis gjren bv the graph in Fig.po.5
lort > 0.
a) Derive the expression for the curent as a function of tim; in the intenals 0 = t < 2s, 2 s < r < 6 s , 6 s < r = 1 0 s ,1 0 s< , < 1 2 s , a n d 1 2 s= t < c o . b) For r > 0, what is the current in the ilductor when the voltage is zero? c) Sketchiversus,foro< t < c.o. figureP6.5
, (v)
6.7 a) Find the inductor curent in tle circuit in Fig. P6.7 if t) = 250 sin 10001V,l, = 50 mH, and (0) : 5 A. b) Sketch r', i, p, and tl) veNus r. In makhg rhese sketches,usethe fomat usedin Fig. 6.8.plot over one complete cycle of the voltage waveform. c) Describe tle subintervals in the lime inte al between 0 ard 27 ms when power is being abso$ed by the inductor.Repeat for the subintervals when power is being delivered by the inductor. Figure P6,7
fl 6.8 The current ir a 15 mH inductor is known to be t 0.
The voltage affoss the inductor (passive sign convention) is 60 V at I = 0. a) Find the e4)ression for the voltage across the inductor for r > 0. b) Fitrd the time, greater than zero, when the power at t]le terminals of the inductor is zero. 6.9 Assume in Problem 6.8 that the value of the voltage asoss the inductor aff = 0is 300 V insteadof 60V a) Find the nume cal expressions for j and o for
r>0.
b) Specify the time itrtervals when the inductor js storing eneryy and the time intenals when the inductor is delivedng eneryy. .) Show that the total energy extracted from the inductor is equal to t}le total energy stored.
6.10 The current in a 2 H inductor is 6,6 The curent in a 20 mH inductor is known ro oe 7 + (15 sin 140r 35 cos1zl0r)€20'mA for I > 0. Assume the passivesign convention. a) At what instant of time is tle voltage acrossthe ilductor maximum? b) What is the maximum voltage?
i =254,
t=0;
i = (81 cos 51 + 82 sir sl)e,A,
I > 0.
The voltage adoss the inductor (passive sign qjnvention) is 100V a : 0. Crlculate the power at the terminals of the hductor at t = 0.5 s. State whether the inductor is absorbingor deliverhg power.
220
Inductance,capacjtance,andMutuatlnductance
6.11 Initially there was no energy stored in the 20 H inductor in the circuit in Fig. P611when it was placed acioss the terminals of the voltmeter. At t = 0 the inductor was switched instantuneously to positionb whereitiemainedlor 1.2s beforereturning instantaneously10 position a. The d'Arsonval voltmeterhasa fullscalereadingof25V and a sensitivity of 1000 O/V. W}lal will the reading of the voltmeter be at the instant the switch returns to position a if the inertia of 1ie d'Arsonval movement is negligible?
Section6, 6.14 A 0.5pF capacitoris subjeciedto a voltagepulse having a duration of 2 s.The pulseis describedby the following equations: ( qot'v.
o=1 0.
Prcbtems 221 c) Is there an instantanoous change in the vottage aooss tle capacitor at I : 0? d) Is tlere an instantaneous change in the current in the capacitor at t = 0? e) How much energy (in mioojoules) is stored in the capacitorat, = co?
6,17 The curent shown in Fig. P6.17 is applied to a 6flft 0.25pF capacitot. The idtial voltage on the capacitor is zero. a) Fird the charge on the capacitor at t = 30 ps. b) Find the voltage on the capacitor a1r = 50 ps. c) How much energy is stored in the capacitor by this culrent? figureP6.17
6.19 The voltage across the terminals of a 0.4/r,F
t
fzsv,
r5m'+,42€ ]soev. I > 0 . [erre The initial qrllent ir the capacitor is 90 mA. Assume t}Ie passivesign convention. a) What is the initial eneigy srored ir the capacitor? b) Evaluatethe coefficients,4land.42. c) What is the expiessionfor the capacitor current? Section 6.3 6.20 Assume tlat the initial energy stored in the inductors of Fig. P6.20is zero. Find the equivalent inductance with respect to the terminals a,b. figureP6.20 21H
6.21 Assume that the initial energy stored in the inducton of Fig. P6.21is zero Find rhe equivalent iDductancewith respectto the terminalsa,b. 6.18 The initial voltage on the 0.2 rlF capacitor shown in 6tre Hg. P6.18(a)is 60.6 V. The capacitorcurrent has the wavefom shownin Fig.P6.18(b). a) How much energy, in miffojoules, is stored in the capacitorat t = 250ps? b) Repeat(a) for t = oo. Flgure P6.18
O.2pF 
60.6V
(a)
Flgure P5,21 12H
5H
14H
222
Inductance, Capacitance andl4utuat Inductance Figurc P6.23
622 The two parallel inductors in Fig. P6.22 are con
necledacrosslhe terminaL of a black bo),al /  0, The resulting voltage u for t > 0 is known to be 1800€ m' V. It is also known that ir(o) = 4 A and ir(0) = 16 4. a) Replacerhe originalinducror.with an equivalent inductor and find i(r) for I > 0. b) Find ir(t) for t > 0. c) Find i2(t) foi t > 0. d) How much energy is delivered to the black box in the time intenal0 < 1 < oo? e) How much energy was initially stored in the par' allel inductors? f) How much energy is tiapped in ttre ideal inductoft? g) Show that your solutions for tr ard t, a$ee with the answerobtainedin (0.
32H
624 For the circuit shown in Fig. P6.23,how many milliseconds after the switch is opened is the energy delivercdto the black box 80o/oof the total energy delivercd?
625 Find t}le equivalent capacitancewit}l respect to the terminals a,b for tlle circuit shown in Fig. P6.25. Figure P6.25 12p.F 8V 20 pF
FigureP6.22
rts?1PF  L 28 pF
24 p.F
5V
;,oi10H,?0)l 2Y 6.:26 Find the equivalent capacitancewith respect to the terminals a,b for the circuit shown in Fig. P6.26. figueP6.26 8nF
6.23 The thee inducton in tlle circuit in Fig. P6.23 are
30v +
conrected across the termimls of a black box at t = 0. The resulting voltage for I > 0 is knorn to be u. : 7250e25'Y If tr(0) = 10 A and i(0) : s A, ftrd
a),,(o); b) ',(r),r > 0;
18 DF
1 5 V+ 5.61F, //
40 Dr b.
+ 5V
6.n
l0v +
c) il(r),r > 0; d)ir(r),r>0;
The two series connected capacitors in Fig. P6.27 are connected to the terminals of a black box at I = 0. The resultingcurrent l(t) for t > 0 is known pA. to be 900e 15oot
e) the initial energy stored in the three inducto$; f) the total energydeliveredto the blackbox;and g) the energy tmpped in tlre ideal inductors
a) Replace'he origjnalcapacirors wirh an equi!dlent capacitor and find z"(r) for / > 0. b) Find 01(t)for t > 0.
ProbLems 223 c) Find r'r(r) for r > 0. d) How much energyis delivercdro the black box in the time interval0 < r < oo? e) How much energy was initially stored in the seriescapacitors? f) How much energy is trapped in the ideat capacitors? g) Showt]tat the solutionsfor or and o, agreewith the answerobtainedin (1). Figure P6.27
45V
15V +
6.?2 The four capacitorsin the circuit jn Fig.P6.28are connected acrossthe tenninals of a black box at 1 = 0. The resulting current i, for a > 0 is kllown to be
d) the percentageof the initial energystoredthat is deliveredto the black box and e) the time, in miliiseconds, it takes to deliver 5 /..J to the blackbox. 6.30 Derive the equivalert circuit Ior a seriesconnection of ideal capaciton.Assumethat eachcapacitorhas its o$,ninitial voltage.Denote theseinitial voltages as rr(r0), o2(r0),and so on. (Ht rr Sum rhe voltages acrossthe stdng of capacitors,recognjzingthat the seriesconnectionforcesthe current in cachcapacitor to be the same.) 6.31 Derive the equivalentcircuit for a parallel connection of ideal capacitors. Assumethat the initial volf age acrossthe paralleledcapacitorsis u(lo). (dir?rr Sum the currentsinto th€ string of capaciton,recognizing that the parallel connection forces the voltageacrosseachcapacitorto be the same.)
Sections6.1{,3 6.32 The cment ir the circuit in Fig. P6.32is knowl to be i, : 5043o0'(cos60001+ 2sin60001.1mA for / > 0+.Find ?,r(0+)and z,l0+). Figure P6.32
it = 50e^a'tt A. If o"(0) : is v, t.(0) = 45 v, and z'd(o): 40 V, tind the following for / > 0: (a) olr, (b) ?,,0),
320Q
(c)o.(1),(d)trlr, (e)n(0, and(f) ,,(,). Figure P6.28
6.33 At t = 0, a seriesconnectedcapacitotand inductoi are placedacrossthe terminalsof a black box, as shownin Fig.P6.33.For r > 0, ir is known that j, : e30'sin6ot A If0c(0) = tigureP6.33
6.29 For the circuitin Fig.P6.28,calculate a) the initial eneigy storedin the capacitors; b) the final energystoredin the capacitors; c) the total energydeliveredto the black box;
300V lind ", for I > 0.
224
Inductance. Calacitance. andlllutuat Inductance
Section 6.4 6.34 There is no energy srored in the circuit in Fig. P6.34 at the time the switchis opened. a) Derive the dilferential equation that govems the behaviorot 12 if Lr  l0 H. L,  40 H. M:5H,andRd:90O. b) show that when ie = 10;'  10 A, t > 0, tlle differential equation derived in (a) is satisfied when i2 = at  5e 225tA, t > lJ. c) Find the expressionfor the voltage ?1 acrossthe current source. d) what is the initial value of x'r? Does tlis make sensein terms of known circuit behavior?
Figure P6.34
l),
LI
6.35 Let o. represent the voltage acrossthe 16 H inductoi in the circuit in Fig. 6.25.Assume D, is positive at thedot.AsinExample6.6,i"= 16  t'n5' O. a) Can you find 2o without having to differentiate the expressionsfor the currents? Explain.
6.37 a) Showthat the differential equationsdedvedin (a) of Example6.6canberearangedasfollows: di.
4;
di"
+ 2sit  8i
di, 8dt
 2Oi2= si.  8
di"
dt:
di, di. 20r, I t6i + 80i, t6dt
dt
b) Show that the solutions for it, and 12given in (b) of Example6.6satisfythe differentialequations given in pa (a) of this problem. 6,38 The physical construction of four pairs of magneticaly coupled coils is shown in Fig. P6.38. (See page 225.) Assume that the magnetic flux is confined to the core material ir each structure. Show two possible locations for the dot markings on each pair of coils. 6.39 The polarity markings on two coils are to be determhed expeimentally. The experimental setup is shoM in Fig. P6.39.Assumethat tho terminal connected to the negative termhal of tle battery has been given a poladty mark as shown. When the switch is close4 the dc voltmeter kicks upscale. Where should the polarity mark be placed on the coil connected to the voltmeter? rigureP6.39
b) Derive tle expressionfor r',. c) Check your answer in (b) using the approp ate current derivatives and inductances,
6.36 Iet DsreFesentthevoltageacrossthe currentsource in the circuitin Fig.6.25.Thereferencefor ?,sis positive at theupperteminal of the currentsource. a) Findts as a Iunctionof time whents = 16b) wlat is the initial valueof os? c) Fitrd t])e expressionfor the powerdevelopedby the currentsource, d) How muchpoweris the currentsourcedeveloping whenI is infinite? e) Calculatethe power dissipatedin eachresistor wherlt is infinite.
6.40 a) Show that the two coupled coils in Fig. P6.40can be replaced by a single coil having an inductance of aab : a1 + L2 + 2M. (Hint: Express o^bas a tunction of tab.) b) Show that if the connectionsto the terminals of the coil labeled a2 are reveNed, Ldb= Lr + Ia 2M. Figure P6.40 M
a5:r._._ri'^_'b
Pmblems225 rigureP6.38
0)
G)
6.41 a) Show that the two magneticaly coupled coits in Fig.P6.41canbe feplaced by a singlecoithaving a.ninductance of
(d)
b) Show that if the magnetic polarity of coil 2 is reversed,then Lrlo  M'
M2 hlo Lr+ L2 2M
Lr+ L2+2M Fi$re P6.41
(Airi Let it and i2 be clockwise mesh curents in the left anddght "whdows" of Fig.P6.41,respectively. Sum t]te voltages around the two meshes. In mesh 1 let ?,abbe t})e unspecified applied voltage.Solve for d4/dt as a function of oab.)
226
Inductanc€, Capacitance, andl{otuat Inductance
S€.tion 6.5 6,42 TWo magnetically coupled coils are wound on a nonmagneticcore.The se[inductanceof coil 1 is 250mH,the mutualinductanceis 100mH, the coef ficient of coupling is 0.5, ard tlle physical structure ofthe coilsis suchthat 91r = 922. a) Find 12 and the turns ratio N/N2. b) If Nr = 1000,what is the valueofgr and 0r? 6.43 The selfinductances of two magnetically coupled coils are l,r = 400 riH and 12 : 900 &H. The coupling medium is nonmagnetic. If coil t has 250 tums and coil 2 has 500 tums, find 9x and 921 (in nanowebersper ampere) when lie coefficientof couplingis 0.75. 6.,14 Two magneticallycoupled coils have selfinductances of 52 mH and 13 mH, respectiveryThe mutual induc tance betweenthe coils is 19.5mlL a) What is the coefficient of coupling? b) For these two coils,what is the largest value that ,14can have? c) Assume that the physical structure of these coupled coils is such that Or : 92. What is the turns ratio N/N, if Nl is the number of tums on the 52 mH coil? 6.45 The selfinductances of two magtetically coupled coils are 288 mH and 162 InH, respectively. The 2B8 mH coil has 1000 tums, and the coefficient of coupling between the coils is Z. The coupling medium is nonmaglelic. When coil 1 is excited with coil2 open,thefluxlinking only coil 1 is 0.5aslarge as the flux linkhg coil 2. a) How manyturff doescoil2 have? b) What is the value of 02 in nanowebersper c) What is the value of 9n in nanowebersper ampere? d) What is the ratio (drxldrr?
6.47 The selfinductances of thl3 coils in Fig. 6.30 are L1 : 25 mH andL2 : 100mH. If the coefficientof couplingis 0.8, calculatethe energystored in the systemin millijouleswhen (a) t1 : 10 A, 4 : 15 A: (b) i1: 10A, i= 15A; (c) tr = 10A, '2 = 15 A; and (d) t1 = 10 A, i2 : 15 A. 6.48 The coefficient of coupling in Prcblem 6.47 is increasedto 1.0. a) If ,r equals10 A, what value of i resultsir zero storedenergy? b) Is there any physicallyrealizablevalue oft2 that canmake the storedenergynegative? Sections6.16.5 6.49 Rework the Practical Penpective example, except pll$lf?i! that this time, put the button on the hottom of tie divider circuit, as shown in Fig. P6.49.Calculate tlre output voltage o(t) when a finger is present. Figuru P6.49
t"(t)
6.50 Somelamps are made to tum on or off when the ,l$llJjhbaseis touched.Theseusea onetenDinalvariation of the capacitive s$itch circuit discussed in the PracticalPerspective. Figue P6.50showsa circuit model ol such a lamp. Calculate the change in the voltage t'(t) when a person touches the lamp. A(.umeall capacirorr are i0iliallyJischarged. Figure P6,50 10pF
6,46 a) Startingwith Eq.6.59,show Oratthe coefficient of couplingcanalsobe expressedas
' J(#X#) b) On the basis of the fracnons 62\/6r and.OdQ, explainwhy k is lessthan 1.0.
..{
=ti
Lamp Person10pF
Probtems 227 6.51 In the Practical Perspectiveexample,we calculated the output voltagewhen the elevarorbufton pPEIifiIHLE is the upper capacitor in a voltage divider. In Problem 6.49,we calculatedthe voltage when the button is the bottom capacitorin the divider, and we got the same tesult! You may wonder if this will be true for all suchvoltagedividers.Calculate the voltagedifference(fingerversustro finger) Ior the circuits in Figs. P6.51(a) and (b), which use two identicalvoltage sources.
FigucP6.51 Fixed
_p5 pF + )(t)
,"(r)
; \ ""(,)
f
No finger
?5pF ri^"a
capacitor
E5 pF + ?,(r) Finger
Response of First0rder Rl andRCCircuits 7.1 TheNaturatResponse of an RL[itcuirp.230 p.236 7.? TheNaturatResponse of anRCaircuit 7.3 TheSt€pResponse of f,l and RCCitcuits p. 24A 7.4 A GeneratSoLutionlor Stepand NaturaL p. 248 Responses 7.5 SequentiatSwitchingp 254 p. 258 7.6 UnboundedResponse 7.7 TheIntegnting Amplifier p. 26,
1 Beableto deiermireihe faturatfesponse of botlrRLaid RCcncuits. 2 Beabteto determine the steprcsponse of both 3 Knowhowto anatyze circuitswith sequentjat swiiching. 4 B€abt€to analyze op ampcncujtscontaining rcsistors anda sjngl€capacitof.
228
In Chapter6, we notedthatanimportanlalt buteo[ inductois and capacitors is their ability to storeenergy.We are now in a positionto determinethc currcntsandvoltagesthat alisewhen cnergyis either releasedor acquiredby an inductor or capacitor ln responseto all abrupt changein a dc voltaE:cor curent source, In this chaptcr.we will focus on circuits that consistonly of sources,resiston,and either (but nol both) inducton or capaci' tors. For brevity, such configurationsare called lla (.esistorinductor) andRC (resistorcapacitor) circuits. Our analysisof R, and RC circuitswill be dividedinto three phases.II1 the first phase,wc considcrthe curents and voltages that arisewhen storederergy in an inductor or capacitoris suddenly releasedto a resistivenetwork. This happcnswhcn thc inductor or capacitoris abruptlydisconnectedfronr its dc source. Thuswe canreducethe circrlit to one ofthe two equivalenttbrms shownin Fig.7.1on page230.Thecurents and voltagesthat arise in this configurationare leferred to asthe natuml tesponseof the circuit,to emphasizethat the natureof the circuit itsclf,not cxtcr nal sourcesof excitation.doterminesits behavior. Tn tbe secondphaseof our analysis,we considerthe cunents and voltagesthat arisewhenenergyis beingacquircdby an inductor or capacitordue to the suddenapplicationof a dc voltageor cunent source.This responseis referred to as the step rcsponse. The proccssfor finding both the natural and stepresponses is the same;thus.jnthe third phaseof our analysis, we developa general method that can be usedto find thc responseof RL and nC cjrcuitsto any abruptchangein a dc voltageor cuflent soulce. Figure7.2on page230showsthe lour possibilitiesfo. the gen eral configurationol RZ and RC circuits.Note that when there are no indcpendentsourcesin the circuit.the Th6veninvoltageor Norlon cunent is zero, and the circuit reducesto one of those shownin Fig.7.1;thatis,we havea naturalresponse problem. na and RC circuits are also known as firstorder circuits, becausetheir voltagesalrd culrents are describedby first ordcr ditferential cquations.No matter how complex a cjrcuit may
PracticaI Perspective A Fl.ashing LightCircuit Youcan probablythink of rnanydjfferentapplications that requirea flashingtight.A stitLcamerausedto takepicturesjn low light conditionsemptoysa brightftashof Lightto ittuminatethe sceneforjust tongenoughto recordthe inrageon fitm. GereralLy, the cameracannottakeanotherpjctureuntjl the circuitthat createsthe ftashof Lighthas'techarged." otherapplicatjons usefLashing tightsas warnjfgfor hazards,such as tat[ antennatowers,constructionsites, and secureareas.In designingcircuitsto producea flashof tight the engineermustkrow the requir€nr€nts of the apptication. Forexampte, the designengineerhas to knowwheth€rthe flashis controtLed manuatLy by operatinga switch(asin the caseof a camela)of if the flashis to repeatitsetfautomaticattyat a predetermined rate.Theengineeratsohasto knowif the flashinglight js a permanent fixture(ason ar antenna)or
a temporary instatlation(as at a construction sjte). Another qre\fionrhathas[o beanswFred'. power wherLrer a sorrcei. r€adiLy avaitabte. Manyofthe cjrcuitsthat areusedtodayto controlflashjng Lightsare basedon €tectroniccircujtsthat are beyondthe scopeof this text. Neverthetess we can get a feel for the jnvotvedin designing thoughtprocess a fLashing tight circuit by anatyzjng a circuitconsisting of a dc voLtage source, a reststor, a capacr'tor, and a lampthat is designed to discharge a jn flashoftight at a criticatvottage. Sucha cjrcuitisshown the figure.Weshatldjscuss this cjrcuitat the endofthe chapter.
229
230
Response of Fi6t0rder fi andRrCircuits
c.,r (a)
(b)
FiguE7.t A Thetu/oforms ofthecncuits fornatuGt (a)Rrcncujt. (b)nrcircuit. rcsponse.
appear,if it can be reduced to a Th6venir or Norton equivalent connected to the terminals of an equivalent inductor or capacitor, it is a firstorder circuit. (Note that if multiple inductors or capacitors exist in the original circuit, they must be intercoinected so that they can be replaced by a single equivalentelement.) Aftei introducing tlle techniques for analyzing the natural and step rcsponsesof fi$torder circuits, we discusssome special casesof interest. The first is that of sequentialswitching,involving circuits in which switching can take place at two or more hstants in time. Next is the unbounded rcsponse.Finally, we anallze a useful circuit called the integrating amplifier.
Rrh
I
l

fl"* Y
7.1 TheNaturat Response of an RLCircuit
.j, I
G) vrh
t
L
(b)
Yrh Rft
The natural responseof an Ra circuit can best be desc bed in terms of the circuit shown in Fig. 7.3.We assumethat the independent current source generatesa constantcurent of 1"A, and that t]re switch has been in a closed position for a long time. We define the phrase a long time morc accumtely later in this section.For now it meansttrat all currents and voltageshave rcached a constant value.Thus oDly constant, or dc, currents can exist in the circuit just prior to t]le switch's being opened, and therefore the inductor appears as a shot c cuit (adi/dl : 0) prior to the release of the stored energy. Becausethe inductor appears as a short circuit, the voltage acrossthe inductive branch is zero, and there €an be no current in either Ro or R. Therefore, all the source current 1, appearsin the inductive branch. Fhding the natuial response requires finding the voltage and curent at the terminals of the resistor after tlle switch has been opened, that is, after the source has been disconrected and the inductor begins releasing energy.If we let t = 0 denotethe instantwhen the switchis opened,the problem becomesone offinding ?(r) and t(t) for I > 0. For t > 0, the cir, cuit shownin Fig.7.3reducesto tlle one showr in fig.7.4 on the next page.
Derivingthe Expression for the Current (d)
tigure7,2 A Fourpossible fi'storder circuits. (a)Aninductor connected t0 a Th6venin equjvateni. (b)Aninductor connected to a Norionequjvatent. connected to a Th6venin equjvatent. k) A capacitor (d)A capacitor connect€d to a Norton equivaLent.
Figure 7.3l' AnRtcircuit.
To find i(t), we use Kirchhoff's voltage Iaw to obtain an expressioninvolving i,R, and a. Summingthe voltagesaroundthe closedloop gives
t4n^r=o,
(7.1)
wherewe uselhepassivesigncorvention.Equation7.1is known asa firs! oider ordinaly differential equation, becauseit contains terms involvhg the ordinary derivative of the unknown, that is, dy'dt. The highest order derivativeappeaing in the equationis 1;hencethe telm fitstotdet. We can go one step fu her in descdbing this equation. The coeffi cients in the equation, R and a, are constants; that is, ttrey are not ftrnctionsof either the dependentvariableior the independentvariablet.Thus ttre equation can also be described as an ordinary differcntial equation witl constant coefficients.
7.1
To solve Eq. 7.1,we divide by a, transposerhe rerm involving i ro the righthand side, and then multiply both sidesby a differenrial time dr. The resuttrs di,
4tot
TheNatumt RespoNe ofanil Circuit 231
,=.'8,
1 7 . 2 ) F i g u r7e, 4A t t p , , . 1 ' ts o w i  t g . / . ? .f o .,  0 .
Next,we recognizethe lelrhard sideofEq.7.2 as a differertial changein the current i, that is, di. We now divide through by j, gettitg di
i
R, = 70,
(7.3)
We obtair an explicit expressionfor i as a function of r by integrating both sidesofEq.7.3. Usingr and I asvariablesof integiationyields
:1,"'*, li,,'+=
17.4)
in which i(lo) is the cu[ent corespondingto time r0,and j(1) is the current correspondingto time t. Here, r0 = 0. Therefore,carrying out the indicaledintegrationgives
' ( 0 (0)
R L"
(7.5)
Basedon the definitionofthe naturallogarithm, i\t) = i\o)e tR Ltt
(7.6)
Recall from Chapter 6 that an instantaneous change of current cannot occur in an inductor. Therefore, in the first instant after the switch has beenopened,the curent in the inductor remainsunchanged.Ifwe use0 to denotethe timejustprior to switching,and 0+for the time immediately Figure 7.5A Thecurrent response forthecncuit shown Iollowing switching, then jn Fjg.7.4. ,(0):(0+)=10,
d Initial inductorcurrent
where,asin Fig.7.1,10denotesthe initial curent in the inductor.The initial current ir the inductor is oriented in the same directiorl as t]re reference directionof i. HenceEq.7.6 becomes itJ\:
r^" lR/L)t
(7.7) { Naturalresponse of annt circuit
which showsthat the currentstartsfrom an initial value10and decreases exponentiallytowardzerc as r increases. Figure7.5showsthis response. We derivethe vollageacrossthe rcsistorin Fig.7.4ftom a direct appli cationof Ohm'slaw: r:
iR=
IoRe\RtL)t, t >0+.
(7.8)
232
Response of Fnst0rd€r RtandRrCjrcuits Note that in contrust to the expressiotrfor the curent shown in Eq. 7.7, the voltage is defined only for I > 0, not at t = 0. The reason is that a step change occurs in the voltage at zero. Note that for r < 0, the derivative of the current is zerq so the voltage is also zero. (This result follows hom 1)=Ldi/dt\.Jmus
(7.e)
"(0):0, t (0) = 1oR,
(7.10)
wherer{0+) is obtainedfrom Eq.7.8 withr = 0+.1Withthis stepchangeat an hstant in time, the value of the voltage at 1 = 0 is unknowr. Thus we use I > 0 in defining the rcgion of validity for these solutions. We dedve the power dissipated in the resistor ftom any of the follow
p = izR,
(7.11)
Whicheverform is used,theresultingexpressioncan be reducedto p
lzRe2(RtL)t,
t > 0+,
17.12)
The energy delivered to the rcsistor during any intenal of time after the swilchhasbeenopenedis
=  ,,": fo'na",,'',,a,
 "u'*'", :4fu13*1t = L*tZrt  e2tR/Lr),t >0.
(7.13)
Note ftom Eq. 7.13 tlat as t becomesiniinite, the energy dissipated in the iesistor approachesthe initial energy stored in t]te inductor.
TheSignificance of the TimeConstant The e\pressions lof j(1)(tq 7.7iand d tEq. 7.8)includea termot rhe "'. " plorm The coefLicienrot lnamel]. n L determinesrbe rate al which the cwent or voltage approacheszero. The reciprocal of this ratio is the time constsntof the c cuit. denoted
Timeconstant for Bl circuit>
\7.11)
t we can de6ne the er?ressio.s 0 and 0+ more fomaly. The exprcsion r(0 refeB to the ) linit or the variable x as t +0 fron tne left. oi lron negalive tine. The expresion r(0+) relers to the limil of ihe rariable x as r + 0 fron the right, or fton posilive tine
7.1 TheNatumt Response ofanfit Circuii 233 Using the timeconstantconcept,we write the expressionsfor curtent, voltage, power, and energy as ilt)= Ioea',
t>0,
1 ) ( t )= I o R e  ' / 1 , p = I1Re2th,
*=
t

rut^r,
t>0+, t>0+,
eo,,t. t>0.
(7.15) (7.16) \7.17) (7.18)
The time constant is an important parametet fot firsForder circuits, so mentioning several of its charactedstics is worthwhile. First, it is convenient to fhink of the time elapsedafter switching in terms of integral multiples of r. thus one time constant after the inductor has begun to release its storedeneigy to the resistor,the curent has been reducedto tl, or apprcximately 0.37of its initial value. Table 7.1 gives the value of ?t" for integal multiples of r from 1 to 10. Note that when the elapsed time exceedsfive time constants,the current is less than 1ol" of its initial value. Thus we sometimes say that five time constants after switching has occuned, the currents and voltages have, for most practical purposes,rcached tlle final values. For single timeconstant circuits (firstorder circuits) with 1% accuacy, the phnse a /org dme implies that tive or more time constants have elapsed.Thus the existenceof current in the RL circuit shown in Fig.7.1(a) is a momentary event and is refered to as the tramient r€sponse of the circuit. The response that exists a long time after the switching has taken place is called the steadystateresponse.The phrase d long ,tne then also means the time it takes the circuit to reach its steadystatevalue. Any firstoder circuit is characterized,in part, by the value of its time constant.If we have no method for calculating the time constant of such a circuit (perhaps becausewe don't know the values of its components), we candetermineits valuefmm a plot of the circuit'snaturalresponse. That's becauseanother important characteristic of the time constant is that it givesthe time required for the current to reach its final value if the curent continues to change at its initial rate. To illustmLe, we evalrratedi/ dt at 0* and assumethat the cunent continuesto changeat this mte:
firot=lr,: lt
2r 3r 5r
3.6788x 1.3534x 4.9'787x x 1.8316
10_1 10 1
10n 10, 6.7379\ 10 3
2.4i88x 10 3 9.1188x 10+ 8r 3.3546x 1or 9r 1.2341x 1or 10t 4.5400x 10 5 6r '7r
(7.19')
Now, if i starts as 10 and decreasesat a constant rate of 1o/r amperesper second.the exDressionfor i becomes Io
11.zoJ
Equation 7.20 indicates that i would reach its fhal value of zero in r seconds.Figue 7.6 showshow this graphic interpretation is useful in esti. mating tle time conslant of a circuit from a plot of its natural rcsponse. Such a plot could be generated on an oscilloscopemeasuring output currcnt. Drawing the tangent to the nalural responseplot at t : 0 and readhg Figure7,6 A A g€phicjntenretatjon of thetimecor the value at which the tangent intersects the time axis givesthe value of r. stantoftheRt cjrcuitshown in Fig.7.1.
234
Response of Fnstoder Rlandfc Cncuits Calculating the natural responseof an RL circuit can be summarized
Calculating the naturalresponse of Rl circuitts
1. Find the initial current,10,throughthe inductor. 2. Find the time constant o{ the circuit, r : a/F 3. Use Eq. 7.15,Ioe '/', to genemtet(l) from 10and i A other calculations of interest follow ftom knowing i(1). Examples7.1 and 7.2 illustratethe numericalcalculatioNassociated with the natuml responseofan Rt circuit.
Determining the NaturalResponse of anil Circuit The switch in th€ circuit shown in Fig. 7.7 has been closedfor a long time before it is opened at t : 0. Find a) iro) for t > 0, b) i,(t) for t > 0*, c) ?,o(,)for r > 0+, d) the percentage of the total energy slored tn the jn the 10 O resistor. 2 H inductorthat is dissipated 2Q
0 . 1 o, i , l 2 H
)'"
b) We find the current in the 40 O resistor most easilyby usingcurrent division;lhatis, '"
'"10
10 + 40
Note that this er?ression is valid for I > 0* . The inductorbehavesas becausejo = 0att:0 a shor circuit prior to the s$ritch behg opened, producing ar instantaleous changein the cufent i". Then, t,(r)= 4d5rA,
100
40f)
Figure7.7 4 ThecircuiiforExampte 7.1.
r>0*.
c) We find the voltage r, by direct applicationof Ohm's law: D.(t) = 40i": 160e 5'v,
r > 0'.
d) The power dissipatedin the 10 O resistorls
Solution a) The switchhasbeenclosedfor a long time p or to r :0, so we know the voltage across the inductor must be zero at r : 0 . Therefore the initial cuirent in the inductor is 20 A at r = 0 . Hence,iz(O) also is 20 A. becausean instantaneou.changein rbe cuffenrcannotoccufiD an inductor. We replace the resisrive circuit connectedto the terminalsof the inductor wu a singleresistorof 10 (): R.q=2+(40 10)=10O. The time constantof the c cuit is I/Rcq, or 0.2s, giving the expressionfor the inductor current as ,r(r)  20e5,A, r>0.
,? D N ' u r = ?  2 5 b 0 lf0 W . t(l
/'0'
The total energydissipatedin the 10 O resistoris u )^ " ( t \ :
/I 2 5 o 0 P r u ' d: l  2 5 b 1 . .lo
The initial energystoredin the 2 H inductorrs l _ 1 ru(O)= Zi'(0)::(2)(4n0)
= 40uJ.
Thereforethe percentageoI energydissipatedin the 10 O resistoris
=o.to.. ffirroot
7.1 TheNatumt ResDons€ ofanNlCircuit 235
Determining the NaturalResponse of anBl.Ci]cuitwith ParallelInducto]s In the circuit shown in Fig. 7.8,the initial currents in inducto$ 1,1 and Z2 have been established by sourcesnot shown.Theswitchis openedat I = 0.
:
7.6
t > 0,
= 5 . 7 o zP A . / > 0 + .
i,: 
a) Find i1,i2,and ir for I > 0. b) Calculate the inilial erergy stored in the paiallel
2.4e2' A,
Nole that the expressionsfor the inductor cunents it and ,2 are valid lor r > 0, whereasthe expression for the resistorcurent i3 is valid for t > 0*.
c) Determine how nuch energy is stored in tl]e inductorsast+co. d) Sbowthar$e roralenergydelivefedro lhe re.istive network equals the dilference between the resultsobtainedin (b) and (c).
,o1
8()
Solution
jn Fig.7.8. tisure7.9 A simptification ofthecircuit shown
a) The key to finding curents ir, i2, and i3 lies in knowing the voltage o(t). We can easily find o(r) if we reduce the circuit shown in Fig. 7.8 to the equivalent folm shown in Fig. 7.9. The parallel inductors simplify to an equivalent inductance of 4 H. carryingan i0ilialcurrentof l2 A.The resi* tive network reduces to a single rcsistance of 8 O. Hence the initial value of i0) is 12 A and rhelimeconslanr is 4/8.or n.5s. Iherelore
b) The initialenerg)sroredin rheinduclor(h 1 1 d: (5X6a) + (20)(16):320J.
i(t)=12e)1A,
c)As t+co, j1'1.6A and i21.6A. Thereforc, a long time after the switch has been opened, the energy stored in the two inductors is t _ t d J2J. ^(5Xl.o) :(20X l.ol
t>0.
d) we obtainlbe toralenergydeliveredlo the resisrive network by integrating the expression for powerfromTerolo infinir): lhe ioslcnraneous
Now z'(l) is simply tlle prcduct 81,so .,(t) = 96e2'Y,
t>o+.
u::
The circuit showsthat o(l) : 0 at t = 0 , so the expressionfor D(r) is valid for r > 0*. After obtaining o(,), we can calculatei1,i, and 4:
"4t )"
ln
i,' =: I 96e2'dr  8 :Jo =1.69.6e2A, t>0,
i,:j l'sa"^a',t 4{}
(5H)
,ln
= 1152:,l +
1 I l
L1
I pdt:
lo I r = 0 r2QoH) 1
Figurc 7.8A Thecncuit forExamph 7.2.
40{))
,rrn
"l
/.m
I 1rs2e4,dt
.la
: 288J.
This result is the dillerence between the initially stored eneryy (320 J) and the energy trapped in the paftllel induclors (32 J). The equivalent inductor for the parallel inductorc (which predicts the terminal behavior of the parallel combination) has an initial energy of 288 J; that is, the energy stored in the equivalent inductor representsthe amount of energythat will be delivered to the rcsistive netwoik at the terminals of the original inductors. 10r)
236
Response ofFirstorder 8l andRrCircuits
objective1Be ableto determine the naturaL response of both [l andRCcircuits 7.1
The swirchin thc circuil shorvnhasbeenclosed for a lo g time and is openedat t  0. a) Calcular.3 the initial valueof;. b) Calcularethe inirial cncrgv sloredin the
7,2
c) w}lal is the time corstant of the circuit lbr t>0'l d) Wlat is thc nunericat expressionfor ;0) for e) What percentageof the initial energystored hasbeendissipatedin the 2 r) resislor5 ms afler the switchhasbeenopcned?
120V
Answer: (a) 12.5 Ar (b) 62snJ; (c) a nis: (d) 12.5a1sorA.r > 0r (e) e1.8o/.. At r : 0, the s*itch in the circuit shownnoves instantaneously from posiliona to posilionb. a) Calculate,. for t > 0'. b) What percentageof the initial cncrgyslored in the inclucroris eventuallydissipaledin the ,1O resistor'?
8c "'V. t > 0: (b) 80%. (a)
NOTE: Also try ChapterPrabkn$7.1 7.3.
ResBrnne 7 . 2 lheN*tsrraI *'i'e* R{ eirer"rit
F i g r r e7 . 1 0 , 1 A nf C c i r c u i i .
rigurc7,11;! Thecir.uitshownin
As mcntionedin Section7.1,the naturalresponscot an RC circuil is aral, ogousto thal ofan Rl, circuit.Consequentl],. wc don't treal lhe RC circuit in thc sane detail aswe did the nl circuit. The Daturalresponseof an RC circuil is dcvclopedfroD the circuit shownin Fig.7.10.Webeginb]' assumingthal lhc ss,ilchhasbeenin position a lbr a long time,allowilrgthe loop nade up of thc dc vollagesource ys.the resistorn1. aDdthe capacitorC to reacha steadystalc condilion. Recall fionr Chapler6 that a capacitorbehavesas an opcn circuil il1 the presenccof a constantvoltage.'lhusdre voltage sourcccaniot suslaiDa currcnt,and so lhe sourcevoltageappearsacrossthe capacitortenninals. In Scction7.3,we will discusshow thc capacitorvollageactuallybuildsto the sleadystate value ofthc dc voltagesource,but for now the important pojnl is that when the switch is moved from posiiioDa to position b (at i = 0), the voltageon the capaciloris y. Becausethere can be no instantancouschangein the voltageat the tcrminalsoI a oapacitor.the problem rcduces1lrsolvingthe circujtshownin Fig.7.l1.
7.2 TheNaturaL Response ofannCCircuit 237
Derivingthe Expression for the Voltage We can easily find tlle voltage ?(t) by thinldng in tems of node voltages. Using the lower junction between R and C as the reference node and summing tlle curents away from the upper junction between R and C gives
c4 * {:0.
(7.21)
Comparing Eq. 7.21 with Eq.7.1 shows that tlte same mathematical techniques can be used to obtair the solutior for z(t). We leave it to you to showthat D(i) = u(o\e1tRc, t0.
\7.22)
As we have already roted, t]le initial voltage on tle capacitor equals the voltage sourcevoltage %, or
(7.23) < Initial capacitor vottage where yo denotesthe ioitial voltage on the capacitor.The time constant for the nC circuit equals the product of the resistarce and capacitance, namely,
i:rli,;#!l
(7.24) < Tine constant for BCcircuit
SubstitutingEq$ 7.23arrd'7.24n\b Eq.1.22yields (7.25) < NalulalresDonse of annc circuii which indicates that tle natural rcsponse of an RC circuit is an exponential decay of the initial voltage. The time constant RC governs tle mte of decay.Figure 7.12 shows the plot of Eq. 7.25 and the $aphic interpretation of the time constant. After determiningD(l), we can easilydedve the expressionsfor i,p, and 1,:
44 = !!
=\t1,
,  ()*,
vo
1,(t,= ve r' a(t) :vo
17.26J Flgure 7.12 ThenaiuraLrcsponse ofanif circuit.
pDi
= 2 R e  2 ' , , t. > O + .
:?,a,,
17.27)
17.28)
238
Response of Fi6t0rder Rrandfr Cjrcujts Calculating the natural responseol an RC circuit car be summarized as followsi
Catcutating the naturalresponse of an fC circuit>
l . Find the initial voltage, y0, acrossthe capacitor. 2. Find the time constant of the circuit, r = RC. 3. Use Eq.7.25,?(t) : U& 't', to Eerleifre1)(1)from yo and r. Al1 other calculations of interest follow from knowing t)0). Examples7.3 and 7.4 illustratethe numericalcalculationsassociated rvith the naturalresponseofan RCcircuit.
Determining the NaturalResponse of an fC Circuit The switchin the circuit shownin Fig.7.13hasbeen in position r for a long time. At I : 0, the swltch moves instantaneously to position y. Find a) oc(t) for l > 0, b) 0,(t) for r > 0*, c) to(4 for r > 0*, and d) the total energydissipatedin the 60 kO resistor.
b) The easiestway to find o,(r) is to note that the resistive circuit forms a voltage divider across the teminals of the capacitor. Thus
, r')
48 _  b0. z'v. s;rcfl)
r> u .
This expression for ,"(t) is valid for / > 0+ becausetro(0)is zero.Thuswe havean instantaneouschang€in the voltageacrossthe 240kO
c) We find the current i,(t) ftom Ohm's law: Figure7.13A Thecircujt forExampte 7.3.
i"(t\ :
r,"(t) : e 2 5 r mA, 60 x 103
t>0+.
Solution a) Becausethe switch has been in positionr for a long time, the 0.5/rF capacitor will charge to 100V and be positive at the upper terminal. We can replace ttre resistive network conlected to the capacitorat a : 0' with an equivalentresist, ance of 80 kO. Hence the time constantof rhe circuiris (0.5 x 101(80 x 1d) or 40 ms.Then, "c(t) = 100e' 5'V, / > 0.
d) Thepowerdissipatedin the 60kf,l tesistoris p60ko0): t3(r)(60x 103) 60e50,mw, r > 0. The total energydissipatedis
7.2 ThelaructRespolse o'ar nr('(uit
239
Determining the NaturalResponse of an 8CCircuitwith SeriesCapacitors The initial voltages on capacitors C1 and C2 io the circuit shown in Fig. 7.14 have been established by sourcesnot shown.The switch is closed at t : 0. a) Find o(t, or(t), and o(t) for / > 0 and t(t) for r>0. b) Calculare rbe inirialenergysroredin lhe capacitors C1and Cr. c) Determine how much energy is stored in tlle capaqlonast+oo, d) Show tlat the total energy delivered to the 250kO resistor is ttre difference between the resultsobtainedir (b) and (c).
r(, Cl(5pF) "(r)
+
1'(r_)250ko
+ 24V
c2Q0 t/F) nzl)
Figure7.14A Thecjrcuit forE\ampte 7.4.
Solution
l(4
a) Once we know o(t), we can obtain the cunent t(t) from Ohm's law. After determining i(l), we can calculateq(1) and ?rr(l) becausethe voltage aooss a capacitor is a function of the capacitor cunent. To find ?0), we roplace the seriesconnected capacitoN with an equrvalent capacitor. It has a capacitanceof 4 pF and is chargedto a voltage of 20 V Therefore, the circuit shoM ir Fig. 7.14 reduces to the one shown in Fig. 7.15, which revealsthat the initial value of o(t is 20V and that the time constantof the circuit is (4)(250) x 10 r, or 1 s.Thus the exprcssionfor o(r) is r(t) = 20e'V,
250ko Figure7.15 A A simphfication of theckcuitshown in Fjg.7.14.
b) Theinitial energystoredir C1is 1 t,j  :(5 \ 10n)(16) = a0 sJ. The initial energy stored in C, is
t>0. w2:
The current i(t) is ,/,1 i ( r l= : * : 8 o e i I)U,WU
!A.
/>o+.
 (16e' i.'6
, r ( r )=  =
20) V,
4
t > 0,
tt
I 80 . to6e'dr + U
&'o : 40 + 5760 : 5800pJ.
rr+
20V
and u2+ +20V.
Ttere[orelhe energysloredin rbe two capaci fols is
.\t.ta
= ( 4 e . + 2 0 ) V ,r > 0 .
x 10 )(s76) = 5760pJ.
The loralenergysroredin rherwocapacilors is
Knowing i(t), we calculate the expressionsfor 0(t) and or0): r0b /' r,(t) =  80 . l06e "dx 5 ./o
;Q0
rr& =
1 , + 20) x 10"(400): 5000pJ. ;(5
240
Response of Firstorder RLandRCCircujts
d) The totalenergydelivcredto the250kO resistoris f 4onu 2' r l t R t . ' r u J lpd! l^^^ / ./ r Z)lr.(rlru Comparingthe results oblained in (b) and (c) 800riJ : (5800
Thc enelgy storedin the equivalcntcapacitorin Fig.7.l5 is +(,1x 10 6)(400),or 800/..J.Because this capacilorpredictsthe terminal bchavior of the original seriesconnectedcapacilorsjthe encrgyslored in the equivalentcapacitolis the encrgydeliveredto the 250kO rcsislor.
5000)/1J.
objective1Be abteto determin€ the naturalresponse of bothnt andnCcircuits 7.3
The switchin the circuit shownhasbeen closed for a long rime and is openedat r : 0. Find a) the initial valueof?r(/), b) the time constantlol r > 0c) the numericalexptessionfor ?r(l)after the switchbasbeenopened. d) thc inilial energystoredirl the capacitor,and e) the lengthoI timc requiredto dissipate757o of ihe initially storedenergy
7,4
The switchin the circuit shownhasbee! croseo for a long time beforebejngopenedat r = 0. a) Find 1r,,(l)for r > 0. b) Whal percentageofthe initjal encrgystored in thc circuit hasbeendissipatedafter lhe svilch hasbeen open for 60 rns?
50kJ)
rsv )r, Answen (a) 200Y (b) 20 ns; (c) 200€ 5o'V, r > 0j (d) 8 mJ;
''uloo*,
Answer: (a) 8e '5' + 4e Iu v. L > 0; (b) 81.05o/o.
(c) 13.86ms. NOTE: ALto try ChapterPrcbleft 7.21und7.21.
7.3 TheStenll{esponse*f ffA
ai'dfff e!rcu'its We arc now readyto discussthc problemof firding the currentsand vol! agcsgeneratedin firstordcrna or RC circuitswhen either dc vollageor currenl soulcesare suddcnlyapplied.Theresponseof a circuit to the sudden applicationofaconstantvollageorcurrent sourccis referredto asthe
7.3
lhe SteDResDome offt andRCCncuits 241
steprcsponseof ttre circuit.In presentingthe stepresponse, we showhow the circuit rcspotrdswhenenergyis beingstorediI) the inductoror capacitor. Webeginwith the stepresponseof an RL circuit.
TheStepResponse of an nl Circuit To begin, we modiry ttre firsrorder circuit shown in Fig. 7.2(a) by adding a switch.We usethe resultingcircuit, shownin Fig.7.16,in developingthe stepiesponseof an Ra circuit.Energy storcdin the inductor at the time the switch is closed is given in terms of a nonzero initial current i(0). The task is 10find t]le eeressions for the current in the c cuit and for the voltage acrossthe inductor after the s$ritch has been closed.The procedure is the sameas that usedin Section7.1;we usecircuit analysisto derive the differential equation that descdbes the cftcuit in terms of the variable of intercst,and then we useelementarycalculusto solvethe equation. AJter the switchin Fig.7.16has been closed,Kirchhoffs voltagelaw rcquires that
v":Ri+L+,
11.2e)
which can be solved for the current by separating the variables t and r ard then integrating.The first stepin this approachis to solveEq.7.29for the deivative di/ dtl
Ri+u" dt
, * )
L
(7.30)
Next, we multiply both sides of Eq. 7.30by a differential time dl. This step reduc€s the leffhand side ot the equation to a differential change in the ctrllent, Thus
a \
.h*
R/.
,
dr
L\i
 ; )v"\ 0,,
v.\ R)/lt
We now separatethe variablesin Eq.7.31to get di
R,
ttu/R)
L
"''
Figure 7.15A A circujt used to ittustrate ihestep tesponse ofantstorder ru cncuit.
\7.31)
242
Response of Fiuiorder Rl andRrCircuiis and then integate both sidesof Eq. 7.32.Using .r and y asvariables for th€ intesration. we obtain [i(t)
dx
R
f .
1 , " ,  , v 1 4 =7 1 , ' t '
(7,33)
where 10 is the crment at I = O and j(t) is the curent at any t > 0. Performing tlle integration called for in Eq.7.33 generatesthe exprcssion . n
i(t)  (4/R)
hlvr$
=
R
Lr'
\7.34)
from which
i(t) (ulR) I o  ( WR )
Stepresponse of & circuit>
(7.35)
When the initial energy in the inductor is zero, 10 is zeio. Thus Eq. 7.35
i0 =f; \^"t*ro'
(7.36)
Equation 7.36 indicates that after tlle switch has been closed,the currcnt hcreases exponentially from zero to a final value of y,/R. The time constalt of the circuit. Z/R. detemines the mte of hclease. One time constant after the switch has been closed. the cunent will have reached approximately 6370of its final value, or
u,':\\^"'=o.o:zr*.
17.37)
If ihe currentwereto continueto increaseat its initial rate.it wouldreach its final valueat t = 7: that is.because di clt

4( '\.,,,
u ,,.
R \ , , /
L
(7.38)
7.3
TheStepResponse ofRt andffCCjrcuits
the initial rate at which i(1) inoeases is
_u L' firot
\7.39)
If the current were to continue to increaseat this rate, the erpression for i would be
'=l''
(7.40)
from which, at t = 7,
. uL
u
I, R
R.
\1.41)
Equations7.36and ?.40are plottedin Fig.7.1?.The valuesgivenby Eqs.7.37 and7.41arealsoshownin thisfigure. anitrductoris ady'dt,sofromEq.?.35,forI > 0, Thevoltageacross
TX.E), r"r,r, R)'
= (v"  ry!.\RtL,.
Q.4z)
The voltage across the inductor is zero before the switch is closed. Equation 7.42 indicates fiat the inductor voltage jumps to I{  1oRat the instant the switch is closed and then decaysexponentially to zeio. Doesthe valueof o at t = 0* make sense?Becausethe initial current is 10 and the inductor prevents an instantaneous change i[ curent, the Figure7.17.. Thest€presponse 0fthenLcircujt current is 10 in the instant after the switch has been closed.The voltage shownin Fiq.7.16when10 = 0. drop aooss the resistor is 1oR,and t]le voltage impressedacrossthe inductor is the souce voltage minus t]le voltage drop, that is, ys 10R. When the initial inductorcurent is zero,Eq.7.42sjmplifiesto r : Vse IRILY
17.13)
If the initial cuirent is zerq the voltage acrossthe inductor jumps to 14.We also expect the inductor voltage to approachzero as1inoeases,becausethe curent in the circuit is approachhg the constantvalue of VJR.Figure 7.18 showsthe plot of Eq.7.43 and the relationship between the time constant and the initial rate at which the hductor voltage is deueasing. If there is an initial current in the inductor, Eq. 7.35 gives the solution for it. The algebraic sign of 10is positive if the initial current is in ttre same 0 ? 2 t 3 r 4 r directionasi;otherwise,lo carriesa negativesign.Example7.5 illustrates ve66 tjme. the application of Eq.7.35 to a specfic circuit. figure7.18A Inductorvottag€
244 Response of FjrstOrderRL andfr Circujts
Determining the StepResponse of an Rl Circuit The switchin the c cuit shownin Fig.7.19hasbeen in position a for a long time. At I = 0, the swltcn movesfrom positiona to positionb.The switchis a makebeforebreak type; that is, the connectionat position b is establishedbefore the conneclionat position a is broken,so there is no interruption of currentthroughthe inductor.
b) The voltageacrossthe induclor is
a) Find the expressionfor t0) for, > 0. b) What is the initial volrageacrossthe irducrorjust afterthe switchhasbeenmovedto positionb? c) How many milliseconds after the switch has been moveddoesthe inductorvoltageequat24V? d) Does this initial voltagemake sensein terms or circuitbehavior? e) Plot both t(r) and ,l(r) versusr.
The initial inductorvoltageis
dt 7 ' = L ,
: 0.2(200e10,) :40e lo'V, t>0+. 2(0.) = 40 V. c) Yes; in the instant after the switch has beefl movedto positionb, the inductor sustainsa current of 8 A counterclockwisearound the newly formed closedpath. This current causesa 16 V drop acrossthe 2 O resistor.This voltage drop rddc ro lhe dfop acros.lhe.ource.producinga 40 V drop acrossthe inductor. d) We find the time at which the inductor voltage €quals24V by solvirg the expression 24 : 4De tol tor t:
':
1 , 4 0
1ot
= 51.08x 10 l  51.08ms. figure 7.19 A ThecircuitforExamph 7.5.
Solution a) The switchhasbeenin positiona for a long time, so the 200 mH inductor is a short circuit across the 8 A current source.Therefore,the irductoi carriesan initial curent of 8 A. This currenr is orientedoppositeto the referencedirectionfor i; Jhus10is 8 A. w}len the switchis in posiuonD, the final valueof i will be 24l2,or 12A.T\e time constaDtof the circuit is 200/2, or 100 ms. Substitutingthesevaluesinto Eq.7.35gives
r = 12 + (8 =12
e) Figure7.20showsthe graphsoIt0) and r,(r) versus/. Note that the instantof time when the cur rert equalszero correspondsto the instant of time when the inductorvoltageequalsthesource voltageof24V aspredicredby Kirchhoff,svolrage law 1J(v)t(A) 40 32 21 1(r 8
, (nt
12)€4'rr
2 0 et n A , t > 0 .
figure7.20.d Thecurrent andvottaqe wavetorns for
7.3
TheSiepRespoNe of{L andRfCircujts 245
of bothnl' andnCcircuits obj€ctive2Be abteto determine ihe stepresponse 75
\O7 f:
Assumethal the switchin the circuit showDro Fig.?.19hasbeenin positionb for a long timeand at r = 0 ir moves10positiona.Find (a);(o*); (tr) z(o); (c) r, t > 0: (d) i(r), r > 0; and (e) ,l0), t > 0+. Al,o rt Chaplrt Prohlrn'
Answer: (a) l2A: (b) 200v: (c) 20ms; (d) 8 + 20?' A, I > (e) 200eso'V,I> 0 .
..1.1.13.
We can also describethc voltage?J(r)acrossthe inductor in Fig.7.16 directly,Dotjusl in lemrsol lhe circuit current.Webeginby noling thal thc voltage acrossthe resistoris the ditferencebetweenthe sourcevollage and the inductorvollage.Wc$.rite 11.44) $ h c r cv i . a c o n \ t d n r . D i r t e f e n l i rbtoj nt h! r i d c \ u i r hc \ t e c lr o r i ' n e ) i e l d \
dt
Rdt
(7.45)
Theq iI we mulliply eachside of Eq. 7.15by the induclalce a, wc gct an expressionfor thc voltageacrossthe inductor on the lefl hand sidc,or (7.46)
PuttingEq.7.46into standardIoIm yiclds d x R
(7.47)
You shouidvedly (in Problen17.40)that the solutionto Eq.7.47i! idcnti cal io that givenin Eq.7.42. At this poinl, a generalobservationabout the step responseoI an Rl, circuil is pertincnt.(This observationwill prove helplul latcr.) When we derivedlhe dificrcntial equatior for the inductorcurrenl,we obtained Eq.7.29.Wenow rcwritc Eq7.29as (li A
R. t
L
V" L
\1.48)
Obscrvc that Eqs.7.47 and 7.48 have Lhesame form. Specifically,each equatcsthc sum ofthe firsi derivativeoI lhe variablcand a constanttimes the variableto a constantvalue.In Eq.7.47,thcconstanton the righthaDd sidehappensto be zero;hencethisequationtakcson the sameform asthe
246
Response of Firstorder Rl andfftCi(uits
natuml rcsponseequationsin Section7.1.Ir bolh Eq.7.47 a$d,Es.7.48. rbe consla0lmultiplyinglhe dependent !ariableis the rec;procal of tbe time constant, that is, R/a = 1/r. We encounter a similar situation iII the de vationsfor the stepresponseof an RC circuit.In Section7.4.we will use these observations to develop a general approach to finding the llatural and step responsesof Ra and RC circuits.
TheStepResponse of an nCCircuit We can find the step responseof a firsforder RC c cuit by analyziry the circuit shownin Fig.7.21.For mathematicalconvenience, we choosethe Norton equivalent of the retwork comected to the equivalett capacitor. SummingrhecuJrenFa\rayfiom tbe lop nodein Fig.r.:t generiresrhe differcntial equation Figure 7.21AA circuit used to iltustnterhe step response of a firstoder Rrcjrcuit.
,!; ,3 =,"
(7.49)
Division of Eq.7.49by C gives
dt

RC=V
(7.50)
CompadngEq. 7.50vrith Eq.7.48rcvealsthat the form of the solutionfor oc is the same as that for the curent in the inductive circuit. namelv. Fq. 7.15.Tberelore. b) simpl)substitlrling rhe appropriare variabtes anl coefficients, we can wdte the solutiotr for ?c directly. The translation requircs that 1, replace 14,C replace a, 1/R reptace R, and y0 replace 10. We get
Stepresponse of anfC circuit>
(7.51)
A similar derivation for the curent in the capacitor yields the differential equation d dt
i
l RC'
^ ''
\7.52)
Equation ?.52has the sameform as Eq. ?47,hencethe solution for j is obtained by usjng the same translations used for the solution of Eq.7.50.Thus
i(r.\\t,'^",,6.,
\7.53)
where y0 is the initial value of oc, the voltage acrossthe capacitor. We obtainedEqs.7.51and 7.53by usinga mathematicalamlogy to the solution for the step rcsponse of the inductive circuit. Let's see whether these solutions for the RC circuit make sensein tems of known circuit behavior.From Eq.7.51, note tlat the initial voltage affoss the capacitoris yo,the final voltageacrossthe capacitoris IrR, and t]te time
ofil andRCCncuits 247 7.3 TheStepR€sponse
constantof the circuit is RC.Also note that the solutionfor oc is valid for t > 0. fhese obse ationsare consistentwith the behaviorof a capacitor in parallelwith a resistorwhendrivenby a constantcurrentsource. Equation7.53predictsthat the curent in the capacitorat t  0 is 1"  yo/R.Thispredictionmakessense because thecapacitor voltagecannot changehstantaneously, andthereforethe initial curent in the rcsistor is yo/R,The capacitorbranchcurent changeshstantaneouslyftom zero att = 0 tols  yo/Rat I : 0+.Thecapacitorcuirentis zeroat I : l)o. Also rote that the final valueof 1, = 1"R. Example?.6illustrates howto useEqs.7.51and7.53to fhd the step resDonse of a firstorderRC circuit.
Determining the StepResponse of annCCircuit The switch in t}Ie circuit shown in Fig.7.22 has been in position 1 for a long time. At t = 0, t})e switch moves to position 2. Find a) 0.(t) for / > 0and b) i"(r) for t > 0*.
The value of the Norton current sourceis the ratio of the opencircuit voltage to the Th6venin resistance,or 60/(40 x 10) : 1.5 mA. The resullrng Norronequivalenlcifcuit is sboqn in Fig. 7.23. From Fig.'7.23, I,R = 60V and RC:10ms. We have already noted tiat o,(0) : 30 V, so the solutionfor z'. is .t). = =
Figule 7.22A lhecircuiitor Exampte 7.6.
vo,
io :
160x 103
(7s) = 60 v (40+ 160)< 103
( 60)]rlm,
60 + 90e1mrv, ,>0.
b) We write the solution for i, directly from and Eq. 7.53 by noting that 1, = 15mA : (30/40) x 0.7s mA: 10r, or U/R
Solution a) The switch has been in position 1 for a lorg time, so the initial value of ?, is 40(60/80),or 30 V To take advantageof Eqs.7.51and 7.53,wefind the Norton equivalent with respect to the terminals of the capacitorfoi I > 0.To do this,we beginby computing the open'circuit voltage, which is given by the 75 V source divided across the 40kO and 160kO resistors:
60+ [30
2.25ermt mA, t > o+.
We check the consistencyof the solutions foi o, and L by noting that t,,
C?
(0.25
t11
=
l0 6)(o000er'"',
2.25elau mA.
Becausedp.,(D)/dt:0, the expressionfor i, clearly is valid only for t > 0.
Ner1, we calculate the fh6venin iesistance, as seento the right of the capacitor, by shoning t}le 75V source ard making sedes ard pamllel combinatioN of tlle resistors: RTl' : 8000 + 40,000lL160,000= 40 kO
Figlre 7.23 a Theeqoivatent circuiifort > 0 forthecircuit s h o wjnn F j g . 7 . 2 2 .
248
Response of Fnsi,order Rl andffCircuih
objective28e abteto determine the stepresponse of bothnl andfC circuits 7.6
a) Find the expressionfor the voltagcacross the 160kO resistorin the circuirshownm Fig.7.22.Let this volragebe denoleuui, anu assumethat the reference polarity for the voltageis posjtivea1the upper termillalof rhe 160kf) resistor.
b) Specifythe interval of time for which the expressionobtainedin (a) is valid. Answer: (a) 60 + 72e 1t)rtyl ( b )I > 0 ' .
NOTE: Also try ChaptetPtublems7.50und 7.51.
7.4 A SeneralSolutiog'l for $tep ftndNsturelResponses arr ,  l )
V* .J /G\
G)
The generalapproachLo finding cither the natural responseor the step responseof the first order Rl, and RC circuitssbownin Fig.7.24is based on their differential equationshaving the sanrclorm (compareEq.7.48 and Eq.7.50).Togeneralizethe solutionofthesc lour possiblecircuits,we let r(r) representthe unknown quanriry,givingr(,) four possiblevatues.Ir canrepresentthe currentor voltageat the lcrminalsof an inductoror rhe current or voltage at the terminalsof a capacitor.From Eqs.7.47.7.4tj. 7.50.aDd7.52.wc knolv that thc differentialcquariondescribingany one of rhe four circuilsin Fig.7.24rakesthe fornr
i/p L
\7.51)
dt; (b)
wherethe value ofthe constant1( canbe zero.Becauscthe sourcesin rhe circuit are conslantvoltagcsaDd/orcurrcnls,the final value of .r will be conslantlthal js,Lhefinal valuemust satisfyEq. 7.54,and.rvhenx reaches its firralvaluc.lhe derivativedr/dl must be zero.Hencc r, = Kr. where.r/ represeDts the final valueofthe variable. We soivcEq.7.54by separatingthevariabtes,beginnil1gby solvingtor thc first derivative:
frr
(x
..
Figure7.24a Fourpo$jbhfirt,ordercircuits. connected to a Th6venin equivatent. {a)Anjnductor (b)Aninductorconnected to a Nodonequivateni. connected to a Th6venjn equivateni. G)A capacjtor (d)A capacitor connected to a No,tonequjvahnt.
Kr)
1x
r1) (1.56)
In w ling Eq.7.56,we usedEq. 7.55to subsriruter, for Kr. We now mul tiply both sidesofEq.7.56 by dr and divideby.r .r/ to olrtain
o'
:1at.
17.51)
7.4
A GenentSotuiion forSteDandNaturat ResDon5es249
Next,we integrateEq. 7.57.To obtair asgenerala solutionaspossible,we usetime lo asthe lorqerlimit and I asthe upperlimit. Time t0 corresponds to the time of the switchingor other change.Previouslywe assumedthat r0 : 0, but t}Iis changeallows the switchingto take place af any time. Using& and o asslanbolsof integration,we get
f't') a"
t J'1';ut7
1' f
r du. 1J"
(7.58)
Canying out the integiation called for in Eq.7.58 gives
(7.59) < ceneralsolutionfor naturalandsteo resDonses of f[ andfC cirauits The importanc€of Eq.7.59becomesappaientif we write it out in words: the unknown the final variable as a  value of the variable function of time tlre idtial thefinaLl value of the  vaiueof rhe x I variable variablel
€
rlne$tulrd)
(7.60)
In many cases,the time of switching that is, t0 is zero, W}len computing the step and natural rcsponses of circuits, it may help to follow thesesteps: 1. Identify the va able of interest for t}le circuit.For RCcircuits,it is most convenient to choose the capacitive voltage;for Ra circuits, it is best to chooset}le inductive cuffent 2. Detemine tlle initial value of the variable, which is its value at 10. Note that if you choosecapacitive voltage or inductive current as your variable of interest, it is not recessary to distinguish berween t : l; and I : td.2This is becausethey both are continuous variables.If you choose another variable, you need to remember that its initial value is defined at , = 16. 3. Calculatetle fhal value of the variable,which is its value asI  m. 4. Calculate the time constant for the circuit. With these quantitieE you can use Eq. 7.60 to produce an equation describing the variable of interest as a function of time. You can then find equations lor other circuit variables using the circuit analysis techniques introduced in Chapters 3 alld 4 or by rcpeating the preceding stepsfor the other vadables. Examples 7.77.9 illustrate how to use Eq. 7.60 to find the step respoff e of an RCor Rl, circuit. 'Tne expressions r; andrf areanalogous to (' and0'. ttus r(rt) is the linit of r(r) asr + ,0 lron the left, andrG is the linit of r(t) ast  rofron the rieh!.
< Calcutating the naturalor steprcsponse of Rl or nCcircuits
250
Response of Fi6t order8r andfC (i(uits
Usingthe General SolutionMethodto Findannf CircuifsStepResponse The switchin the circuit showr in Fig.7.25hasbeen in position a for a long time. At r = 0 the switch is moved to position b. a) What is the initial valueof 0c? b) What is the final valueof 0c? c) What is the time constantofthe circuit when the switchis in positionb? d) Wlat is the expressionfor z,c(r) when I > 0? e) Wrat is the expressionfor i(r) when r > 0+? t) How long after the switch is in position b does the capacitorvoltageequalzero? g) Plot "c(1) and i(r) venus r.
Solution a) The switch has been in position a for a long time, so the capacitor looks lite an open circuit. Thercfore the voltage acioss the capacitor is the voltage acmsstle 60 (} resistor.From tie voltage, divider rule, the voltageacrossthe 60 O resistor is a0 x [60/(60 + 20)], or 30 v As rhe refer ence for oc is positiveat the upper lerminal of the capacitor,wehavez,c(o): 30 V. b) After the switchhasbeenin positionb for a long time,the capacitorwill look like an open circuit in terms of the 90 V source.Thus the final value ofthe capacitoivoltageis + 90V. c) The time constantis T:RC
= (400x 103x0.5 x 10{) = 0.2s. d) Substitutingthe appropriatevaluesfor z/, t'(0), and r irto Eq.7.60yields 0c(t)=90+( 30 90)e{' = 90 120ts'V, I > 0. e) Here the value for r doesn't chaflge.Thus we need to find only the initial and final values for the curent in the capacitor.When obtainingthe initial value, we must get t}le value of t(0+), becausethe currett in the capacitorcan change instantaneously. This current is equal to the current in the iesistor, which from Ohm,s law is [90 ( 30)]/(400 x 103)= 300/,A. Nore that whenapplyingOhm'slawwe recognizedrharthe
Figure7.25A Thecircuit forExample 7.7. capacitorvoltagecannotchangeinstantaneously. The fiMl value of i(t)  0, so i0)0+(3000)e5. = 300esrpA, r>0+. We could have obtahed this solution by diffbrentiating tlle solution in (d) and mulriplying by the capacitance.You may want to do so for yourseli Note that this altemative approachto finding r(r)alsopfedrclc lhedisconlinu] dl I = 0. f) To find how long the switch musr be in position b before the capacitorvoltage becomeszero, we solve the equation derived in (d) for rhe rime when 0c(t) : 0: 120t5r = 90 or

,.
720 90'
, =1r,/\43\ / = 57.54ms. Note that when ,c = 0, i = 225pA and the voltagedropacross the400kO resistoris 90V g) Figure7.26showsthe graphsof oco) and i0) , (i.A) t)cry) 300 250 200 150 100 50
Figure7.26a Thedrr€ntandvottage waveforms tor Exanrpte 7.7.
7.4 A CeneratSotutjon torStepandNaturalResponses 251
Usingthe General SolutionMethodwith ZeroInitiaLConditions The switch in t]le circuit shown in Fig.7.27 has been open for a long time. The initial charge on the capacitor is zero. At 1 = 0, the swirch is closed.Find the expressior for
Substitutingtlese valuesirto Eq.7.60genemtes
(, =o+(3 o\ett5tn: a) i(r) for t > 0 and
: 3 e 2 o o ' m Ai,> 0 + . b) ,0) when t > 0.
30ko
b) To find ?,(1),we note ftom the circuit that it equalsthe sum of the voltage acrossthe capacitor and the voltageacrossthe 30 kO resistor.To find the capacilor volrage (which is a drop in the direcrion of the current), we note that its inirial value is zero and its final value is (7.5)(20),or 150VThe time constantis the same as before, or 5 ms. Therefore we use Eq. 7.60 to wnte
Figure7.27A Thecircujttor ExampLe 7.8.
z'c(l) 150+ (0
SoLution
150)e'mr
= (150 150e,oo,)v,
a) Becausethe initial voltage on the capacitoris zero,at the instantwhen the switchis closedthe currentin the 30 kO branchwill be
(7.5X20) (0) =
I > 0.
Hencethe expression for thevoltagez'(,)is z(t) = 153 lsOeioo'+ (30x3)eioo'
50
: 050
6oa'rir) v,
I > 0*.
= 3 mA.
The final value of the capacitor current will be zero because the capacitor eventually will appear as an open circuit in terms of dc cwent. Thus if : 0. The time constant of the circuit will equal the product of the Th6venin resistance(as seen ftom the capacitor) and the capacitance. Thercforer = (20 + 30)10'(0.1)x 10 " : 5ms.
As one checkon this expression, note that rt predictsthe hitial valueof the voltageacross the 20O resistoras 150 60, or 90 V The instant t}re switch is closed,the curent in the 20ko resisroris (7.5)(30/50), or 4.5 nA. This curent producesa 90V drop acrossthe 20kl) resistor,confirming the value predictedby the solutiofl,
252
andRCCircuits Response 0f First0rderft
SolutionMethodto Findan RLCircuifsStepResponse Usingthe General The switchin the circuit shownin Fig.7.28hasbeen open for a long time.At I : 0 the switchis closed. Find the expressionfor
constantis 80/1,or 80 ms.we useEq. 7.60to foi 0(t): wite theexpression oO=0+(15
o)e'l3orrr'
a) "0) whenI > 0 and : 15t12stV,
I > 0+.
b) i(t) when r > 0. b) We have alreadynoted that tlle initial value of the inductor cuirent is 5 A. After the switchhas been closedfor a loDgtime, the inductor current reaches20/1,or 20A. The circuittime constantis 80 mq so the expressionlor i(t) is
i(t) : 20 + (s  20)e r25t
= (20  15?125,) A,
r > 0.
7.9. Figure7.28 A ThecjrcujtforExampLe
We determine t}lal the solutions for t'(l) and i(l) agree by noting that
Solution a) The switch has been open for a long time, so the initial current i]l th.3 inductor is 5 A, onented lrom top to bottom. Immediately after the swrtch closes,the current still is 5 A, and tierefore the initial voltage across the inductor bemmes 20 5(1),or 15V The fhal valueofthe inductor voltageis 0 V WitI the switch closed,the time
NOTE:
1' 5i] :80 x 10 3[15(12.5)e : 15e l'r'v,
r > ot.
Assessyour ande$anding of the seneral solution method b) trrng Chapter Problems 7 53 a d 754
Example7.10showsthat Eq.7.60 can even be used lo find th€ step responseof somecircuitscontainingmagneticallycoupledcoils
7.4
A GenentSotution forStepandNaturaL Responses253
Determining StepResponse of a Circuitwith Magnetical.ty Coupted Coils There is no energystoredin the circuit in Fig.7.29 at the time the switchis closed. a) Find the solutions for i., u., ir and i2. b) Show that the solutions obtained in (a) make sensein termsoI known circuit behavior,
'.3 H
120V
Figure7.29a Thecircuit forExampte 7.10.
Solution a) For the circuit in Fig.7.29,the magneticallycoupled coils can be replacedby a singleinductor having an inductance of LrL2  Ml Lr+ L2 2M (SeeProblem6.41.)It follows thal the circuit in Fig.7.29canbe simplifiedasshownin Fig.7.30. By hypothesisOleinitial valueof i, is zero. From Fig. 7.30we seethat the final value of i. will be 120/7.5or 16A. The time constantof the circuit is 1.5/7.5or 0.2 s.It follows directly from Eq.7.60that io=16
1 6 e  s t A ,t > 0 .
fhe voltage r,, follows from Kirchloff's voltage law.Thus, D.: 120
7 5i"
 12(le5tV.t>O+. To find i1 and i2 we first note from Fig.7.29that ^air
.diz
. dir
dt
at
dt
di
i :
.1t
IT
j j
"diz /11
d;
n,ttn Therefore
Figure7.30,d ThecjrcuitinFjg.7.29wjththe magneiicaLty coupLed coitsreptaced byanequivatent cojt.
Because i2(0) is zero we have
a=
.ltj

because
ale " dx
8+8€s'A,
r>0.
Using Kircbioffs current law we get iI:24 2 4 es 'A , t > 0. b) First we observethat i,(0), ir(0), and ir(0) are all zero, which is consistent with the statement that no energyis stored in the circuit at the instant the switch is closed. Next we obsene r',(0) = 120V, which is consistentwith the fact that i,(0) = 0. Now we observethe solutions for ir and 4 are consistentwith the solution for o, by observing dt dt '
.lt
follows from Fig. ?.29 tlat + iz,
1.5H
t1l
At
= 36oe st  24oe sl : 720esty, t>O*, di.
u:6:+15::
di"
: '720e 5t  600e 5l : 120e5tv,
t > 0'
254
Response of Fnstorder Rl andnr Cncuitr The final values of 4 and i2 can be checked using flux linlages. The flux linking the 3 H coil (11) must be equal to the flux linldng the 15 H coil (12),because
The tinal valueoflr is
n(co) andthetioalvalueofi is
', : dn'
ir(oo) = The consistencybetween these final values for i1 and i2 and the final value of the flux Linkage can be seenfrom the expressions: 3i1 + 6i2 Wbturns
^r(co) = 3i,(@) + 6i(,.o)
= 3(A) + 6( 8) : z wb{urm,
6i1 + 15i, Wbtums. Regardless of which expression we use, we obtain
= 6Qa) + E(
,\1 = i2 : 24  24ts' Wbtums. Note the solution for It or 12 is consistent with the solution for rro. The final value of the flux linking either coil 1 or coil 2 is 24 Wbtums, that iE I(co) = r,(oo) : 24 wbtums.
^,(o.) : 6i1(oo)+ 15ir(,tro) 8) : 24 wbtums.
It is worth noling that the final values of 11 ard i2 can only be checked via flu\ linlage becauseat t : oo the two coils are ideal short circuits. The division of curent between ideal short circuits cannot be found from Ohm's law.
NOTE: Assessyour understanding of this material bt using the genersl solution method to sotve Chaptet froDlems /.ol and /.o /.
7.5 Sequentiat Switching Whenever switching occuN more that once in a circuit, we have sequential switching.For example,a single,twoposition switch may be switched back and fo ]\ or multiple switchesmay be opened or closed in sequerce.The time reference for all switchings cannot be I : 0. We detemine tlle voltages and cu ents generated by a switching sequenceby using tle techniques described prcviously in this chapter. We derive the er:pressionsfor o(t) and i(, for a given position of the switch or switches and rhen use these solutions to determine the initial conditions for the next position of the switah or switches. With sequential switching problems,a premium is placed on obtaining the inilial value r0o). Recall that an)'thing but inductive curents and capacitive voltages can change instantaneously at the time of switching. Thus solving fi$t for itductive currents and capacitive voltages is even more pertinent in seque ial switching problems. Drawing the circuit that pefiains to each time irterval in such a problem is often helpful in the Effmples 7.11 and 7.12 illustrate the analysis techniques for circuits with sequential svritching.The fircl is a natual responseproblem with two switching times,and the secondis a step responseproblem.
SequentjatSwjtchjng 255
7.5
Analyzing an ff Circuitthat hasSequential. Switching The two switches in the circuit shown in Fig. 7.31 havebeenclosedfor a long time.At r  0, switch1 is opened.Then,35ms later,switch2 is opened. a) Find ir(l) for 0 < , < 35 ms. b) Find iz for t > 35 ms. c) Wlat percentage of the initial energy stored in the 150 mH inductor is dissipatedin the 18 O r€sistor? d) Repeat(c) for the 3 O resistor. e) Repeat(c) lor the 6 O resistor.
F i g u r € 7 . 3 2 l^t e L i ' L i . s h o w l i n  i g . 7 . l l . r o '0r .'
150mH
,:35ms
40 \ )60v
1i
18()
Figure 7.33A Thecncujtshown in Fig.7.31, for0 < r =35 ms.
3r} 2
: 1 , 2 A 1 6ll
tt. 150inH
18()
b) When t:35ms,
the value of the inductor
ir = 6e rL = 1.48A Figure7.31 A Thecircujtfor Exampte 7.11.
Thus, when switch 2 is opened, the circuit rcducesto the one shown in Fig. 7.34,and the time constant changesto (150/9) X 103, or 16.67ms The expressionfor i. becomes
Solution
oo3s) iL : 1.48e4411 A,
a) For t < 0 both switches are closed, causing the 150nH induclor ro shorrflrcurlthe l8 O resistoi. The equivalent circuit is shown in Eg. 7.32.We determine the initial current in the inductor by solving for ir(o) ir the circuit showr in Fig.7.32. AJter making several soutc€ transformations,we find 4_(0) to be 6 A. For 0 < t < 35 ms, switch 1 is open (switch 2 is closed),which disconnectsthe 60 V voltage souice and the 4 O and 12 O resis' torc from the circuit. The inductor is no longer behaving as a short circuit (becausethe dc source is no longer in the circuit), so t}Ie 18 O rcsistor is no lorger shortcircuited. The equivalent circuit is shown in Fig.7.33.Note tlat t}le equivalent rcsistance across the teminals of the inductor is the parallel combhation of 9 O and 18 O, or 6 O. The time constantof the circuit is (150/6) x 10 3, or 25 ms,Theieforc the expressionfor ir is
r > 35 ms.
Note that the exponentialfunction is shiftedin time by 35 ms.
150mH l, (0.035)= 1.48A
tigure7.34A Thecjrcuitshown in Fig.7.31, forr > 35ms. c) The 18 O resistor is in the circuit only during the first 35 ms of the switching sequence.During tlfs intelval, tle voltage auoss the rcsistor is ) u, = o.ts:(6Pro') tll
iL = 6e Nt A,
0 < , < 35 ms.
=36eNtY,
0 15 ms. b) Figure 7.36 showsthe plot of o versus r. c) The plot in Fig. 7.36 rcveals that the capacitor voltage will equal 200 V at two different times: once in the inte al between 0 and 15 ms and once after 15 ms.We find the fiISl time by solving
200=400400tlmr',
400)ermr
400erm) v,
0 < , < 15 ms.
Note thal. becauserhe swirch remainsin position b for only 15 ms,this expressionis valid or y for t]le time interval from 0 to 15 ms.Aftcr ure switch has beer in this position for 15 ms, the voltage on the capacitor will be ?r(15mt = 400  400ers: 310.75V. Therefore,when the switch is moved to position c, the initial voltage on the capacitor is 310.75V Wit}l the switch in position c, the final value of the capacitor volaageis zero, and the time constantis 5 m$Again, we useEq.7.59to write the er?ression for the capacitor voltage:
which yields 11: 6.93ms. We find the second time by solving the expression oors) 200 : 31O;75etno(t In this case,t2: 17.20ms.
r,(v) 300 200
/:310.75.
ao(/ od:J
100
?, = 0 + (310.75 o)€ 20r(!ools)  310.75?200(r0015) V, 15ms < r.
Flgure 7.35^ Thecapacitor votiage fortxampte 7.12.
258
Response of FifstOrder nl andRCCjrcujis
objective3Know howto anatyze circuitswith sequential switching 7.7
In the circuit sho$'n.switchI hasbeenclosed r n d . $ i r c h) h , \ . . c n o p c l o r : rl o ' r gl i r n eA .l t  0. s$,itch1 ls opcncd.Ther l0 ms later, switch2 is closcd.Find
S$ir(hJ iI rhecicuir.h^qn hl. bccnop 1s.
7,8
a) r.(r) for 0 < r < 0.01s. b) ,tJ.(/)for r = 0.01s. c) dre total cncrgydissipatedin the 25 kQ d) rhe lolal energydissipatedin thc 100ko
20 AFFrI
b
 1
0.8o
9ll
l f ok ( l
\
i3o
r:(l 2H
ii 3o
60
Answer: (a) 80.rrrrlv; oorrVl (b) 53.63e5o1r Answer: (a) (3 3c t'5')A,0 < , < 1s; (b) (4.8 + 5.98r r'25(/r) A. r > I s.
(c) 2.e1nJ; (d) 0.2enrJ. NOTE: Aha iry ChapterProbl ts7.72and7.76.
7.6 i,inbcu*ti{:d RrsFs*se A circuil responsemay grox'.rather lhan decay.exponentiallywith timc. This ttpe of responsc. calledan unbomded response.is possiblcif rhe cir cuit containsdcpcndcn! soulces.Ir that case,the Th6vcnin cquivalenl resistance with rcspccl10the lerminalsofeither an inductoror acapacilor nav be negaiivc.This negadveresistancegeneratcsa ncgativctine con stant.and thc rcsullingcurerts and voltagesincreascwithoul limi1.Tn aD actual circuit. thc rcsponseeventuallyreachesa limiting valuc whcn a componentbreaksdo\\n or. goesinto a saturationstate,prohibiling rur ther increasesin voltagcor cult'enl. wlren we considcrunboundedresponses, the conceptof a tinal laluc is confusirg.Hcncc,ralher lhan usingthe stepresponsosolutiongi!en in Eq.7.59.wedcrivc lhe differentialequationthat describcsthc circuil con, taining thc negalile resistanceand then solvc it using the separariorof variablcs lcchnique.Example 7.13 prcscnts an exponenlialll,growing responscin termsoI lhe vollageacrossa capacitor
Response 259 Unbounded
7.6
Response in an Rccircuit Findingthe Unbounded a) When the switch is closed in the circuit shown m Fig.7.37.lhe \ollageon the capaciloris l0 V Find the expressionfor ?),for t > 0. b) Assume that the capacitor shortcilcuits when its terminal voltage reaches 150 V How many millisecondselapsebefore the capacitorshorG circuits?
metfod r e d t of r d R n F i g u r €7 . 3 8^ T f er e s r  l o i ( e
5 ko 7.13. Figure7.37 l Th€crrcujtfor Exampt€
jn ofthecjrcujt shown Fig!rc7.39l A simptjfication fig.7.31.
Sotution a) To find the Th6venin equivalent resistance with respectto the capacitor teminals, we use the tes! souce method describedin Chapter 4Figure 7 38 shows the {esulting circuit, where ?,r is the test voltage ald ir is the test current. For th expressed in volts,we obtail .
rr:
UT
10
 /l )rL2 0 t +
1)T
,
For , > 0, the ditferential equation describing the circuit shownin Fig.7.39is
_3 15 1s_oyy_:g 111
,a
r0_r: o.
Dividing by the coefficient of the first derivative yields
2mA.
4
Sol\ingfof tbe rario ,r"/ir yieldsthe fie\enin
Rrr, :
5
= 5 kO
d
t
,'r,u^=o. "
We now usethe separationof variablestechnique to find 0"(r): q(t):10e4utY,
with this Th6veniil resistance,we car simplify lo lheone\hownin thecircuirsho\rnin Fig.7.37 Fig.7.39.
t > 0.
b) o, : 150V when?au' 15.Therefore,4ot= ln 15, and/ = 67.70ms.
NOTE: AssessyoLr understandingof thk material by trying Chapter Probkms 7 86 and 7.87.
The fact that hterconnected circuit elements may lead to everhcreasing currents and voltages is impofant to engineels. If such inter: connections are unhtended, the resulting ckcuit may experienc€ unexpected,and potentially dangerous,component failures'
260
Response of Fi6t0rder fi andRrCircuiis
7.7 TheIntegratingAmplifier
tigurc7.40e Animeg,aLing anplifier.
Recall from the introduction to Chapter 5 tlat one reason for our interest in t})e operational ampMier is its use as an integrating amplifier. We are now ready to analyze an integratingampiifier circuit, which is shown in Fig. 7.40. The purpose of such a circuit is to generate an output voltage proportionalto the integal of the input voltage.InFig.7.40,we addedthe branch curents it and i,, along with the node voltages o, and o?, to aid our analysis. We assumetlat the operational amplifier is ideal Thus we take advantageof the constraints \7.61) \7.62)
Becauseu? : 0. (7.63)
dD^
(7.64)
Hence,from Eqs.7.61,7.63, and'7.64,
du.
n:
1
R{r,,"
\7.65)
Multipling both sidesof Eq.7.65 by a differentialtime dl and thetrintegrating from lo to I generatesthe equation 1 R.
t"'
,,r d) + ?,o(10).
(7.66)
In Eq. 7.66,t0representstlle instantin time whenii'ebeginthe inte$ation. Thus ?r,(to)is the value of the output voltage at that time. Also, because bn: tp: 0, t"00) is identical to the idtial volrage on tlre feedback caDacitorC,, Equation 7.66 states that the ouQut voltage of an inregrating amplifier equals the initial value of the voltage on the capacitor plus an inverted (minus sign), scaled (1/&Cl) replica of tle inregral of the input voltage. ff no energyis storedin the capacitorwhenintegation commences, Eq.7.66
ol0 
1 l nq J,"o'ot
\7.67)
7.7
TheIntegBiingAmptifier 261
If 0Jis a step changeifl a dc voltage level, the output voltage will vary linearly with time. For example, assumetiat the input voltage is the rectangular voltagepulseshownin Fig.7.41.Assume alsothat the initial valueof ,"(1) is zero at the hstant or steps from 0 to y,. A direct application of Eo.7.66vields
."=
l,v^t
+ 0, 0 0. h) Find z'r(0 ). i) Find r'r(0*). j) Fird ?,.(co). k) Wite tlle expressionfor !'r(t) for t > 0'. l) Wdte the expressionfor i"(l) for I > 0*.
rent source, o represent tlre fracrion of initial energy srored ir the inductor that is dissipated in to seconds,ard a represent the inductance. a) Show that
L1LI10  o)l 2t. b) Test the expression derived in (a) by using it to find the value of R in Problem 7.5.
Figure P7.3 50(l
7.6 In the €ircuit in Fig. P7.5,Iet 1s represent tlle dc cur
100()
2000
7,4 ln the circuit in Fig. P7.4, the voltage and current expressionsare o = 100t30!v,
r >0+;
i = 4e4'A,,
t>0.
Find a) R. b) r (in miltiseconds). c) L. d) tle idtial energy stoied in the inductor. e) the time (in milliseconds)it takes to dissipate 80% of the initial stored energy. Figure P7.4
7,7 The switch ir ttre circuit in Fig. P7.7 has been open 6fl( for a long time. At , = 0 t]le switch is closed. a) Determinei,(0") and i,(m). b) Determinei,(t) for t > 0*. c) How many millisecondsafter the switch has been closedwill t]le curent in the switch equal 3.8A? figureP7.7
12o"
160
80v
'(r=0
+\
20nH 40

8()
7,8 The switch in the circuit in Fig. P7.8hasbeen closeda 6dc long time At I : 0 it is opened.Find "o(r) for l > 0. FigueP7.8 15O \ )80v
50O r
i s oo
lf} 0.2H o.a:60r)
20o"
2A The switch in tlle circuit seen in Fig. P7.5 has been in posirion 1 for a long time. At r : 0, the svritch moves instantaneously to position 2. Find the value of R so tlar 50ol. of the initial energy stored in the 20 In}I inductoris dissipatedin R in 10lrs.
7.9 Assume that the switch in the circuit in Fie. P7.8has been open for one time constant. At this instant, what percentage of t}le total energy stored in the 0.2 H inductor has been dissipated in ttre 20 O
Prcbtems ?67
7.10 In the circuit shown in Fig. P7.10,the switch has 6fl( beenin positiona for a longtime.AtI :0,itmoves instantaneously from a to b. a) Find r"(t) for I > 0. b) W}lat is the total energydeliveredto the 1 kO rcsistor? c) How manytime constantsdoesit taketo deliver 95%of the energyfound in (b)?
7.14 The switch in the circuit in Fig. P7.14 has been 6tr4 closed for a long time before openirg at t = 0. Find ?o(t) for r > 0'. tigur€P7,14
50nH
tigureP7.10
7.11 The switch in the circuit in Frg. P7.11 has been in 6nc position 1 for a lory time. At t = 0, the switch moves instantareously to position 2. Find ?o(t) for t > 0*. figlre P7.11 7A 1
7.15 The switchin Fig.P7.15hasbeenclosedfor a long time beforeopeningat t : 0. Find a)iL(t),t>o. b)rL$),t>o. c)iA(r),r>0. Figure P7.15 3r)
50
96mH
')s.rv
2 15()
5(}
50 ia
4.5f,)
T 0;
b
7.19 The two switches shom in the circuit in Fig. P7.19 6tG operate simultaneously,Prior to t : 0 each switch has been in its indicated position for a long time. A1 I : 0 the two switches move instantaneously to their new positions. Find a) 0"0), t > 0. b) ,o(r),I> 0. FigueP7.r9
j  5e1000! mA,
I > 0+
Find a) R.
b) c. c) ' (in minisecondt. d) the initial €nergystoredin the capacitor. e) how many microsecondsit takes to dissipate 80o/, of the initial eneryy stored in the capacitor. Figj'Je ?7.22
80 mH "o
i^
7.23 The switch in the circuit in Fig P7.23 is closed at ED.r I = 0 aft€r beingopen for a long time. 7.m For the circuit seenin Fig. P7.19,find a) the total energydissipatedin the2.5 kO resistor. b) the energy trapped in t]le ideal inductors
a) Find tr(r) and i(0 ).
u) nna 4(o+)ana;z(o+). c) Explahwhyil(o) = t1(0*).
Probtems 269
d) Explain why i(0 ) + i(0+). e) Findil(t) for t > 0. l) End t (t) for t > 0*. tigure P7.23
726 In tllle circuit sho*n in Fig. W.26, both switches operate together; tlat is, they either open or close at the same time. The switches are closed a Iong time before opening a1I  0. a) How many microjoules of energy have been dissipatedin the 12 kO resistor 2 ms after the switchesopeD? b) How long does it take to dissipate95% of rhe initialy stored erergy? Figur€ P7,26
7.24 The switch in the circuit in Fig. P7.24 has been in position a for a long time. At r : 0, the swifch is thown to positionb. a) Find t,(t for t > 0+. b) What percentage of the initial energy stored in the capacitoris dissipatedin the 4 kO resistor 250 /,csafter the switch has been tfuown?
7.25 Both switches in the ckcuil in Fig. P7.25have been sflc closed for a long time.At t = 0, both switchesopen simultaneously. a) Find roc) lor t > 0+. b) Find ,"(1) for , > 0. c) Calculate tle energy (in microjoules) trapped in the circuit.
7.27 The switch in the circuit in Fig. P7.27has beenin En@posirion 1 for a long time before moving to position 2 a  0. Findl,(t) foi t > 0t. figtr'e P7.21
7.28 The switch in the circuit seen in Fig. P7.28has been in position x for a long time. At r : 0, the switch moves instantaneously to posirion y. a) Find d so that the time constantfor I > 0is1ms. b) For the a found in (a), find t,a. Flgure P7.28
20ko Figure P7.25
10ko
270
Response of Fnst0rder Rtandrr Cinuits
7.29 a) In Problem 7.28, how many microjoules of energy !!re generated by the dependent curent source during the time the capacitor discharyes to 0V? b) Show that for t > 0 the total energy stored and genemted in the capacitive circuit equals t]Ie total energy dissipated.
730 After the circuit in Fig. P7.30has beer in operation tsd.r for a long time! a screwdriver is inadvertently connected aqoss the terminals a,b. Assume the resis! ance of ttle screwdriver is negligible. a) Find the current in the screwdriver at I : 0+ and b) Derive the expression foi the curent ir the screwdriver for t > 0+.
7.32 At the time the switch is closed in the circuit in Fig. P7.32,the voltage acrosstlle paralleled capacitors is 30 V and the voltage on the 200 nF capacitor is 10V a) Wlat percentage of the initial energy stored in the three capacitors is dissipated in the 25 kO resistor? b) Repeat(a) for the 625O and 15 kO resistois. c) What porcentage of the initial energy is tapped in the capacitors? tigureP7,32
l0 nF
tigureP7.30 7.33 The switch in the circuit seen in Fig. P7.33has beer arft closed for a long time. The switch opens at 1 = 0. Fhd the numerical expressions tor i,(t) and oo(t) when t > 0*. Figure P7.33
7,31 At the time the switch is closed in the circuit shown in Fig. P7.31,the capacitors are charged as shown. a) Find ?J,(t)for t > 0'.
60ko
,J4
b) Wlat percentage of the total ene4y initially stored ir the three capacitors is dissipated in the
25kO rcsistor? 0 Find 2,1(tfor t > 0. d ) Find ?r20)for t > 0. e) End the energy (in milijoules) trapped itr the idealcapacitois. Figure P7,31
+ + 0.6pF "l
" 25kO
Section 7.3 7,34 The switch ir the circuit sho$n in Fig. P7.34 has Eflft been closed foi a long time before opening at I : 0. a) Find tlle numerical expressions for i.(t) and ,"(l)fort > 0. b) Find tlre numedcal values of 0r(0+) and o,(0+). FlgureP7.34
Probtems 271 7.35 The switch in tle ckcuit shown in Fig. P7.35 has Btrc been in position a for a long time. At r:0, the switch moves instantaneously to position b. a) Find the numerical e4ression for io(t) when
Figur€ P7.38
b) Find the numerical expression for o,(r) for l>0*. FiqueP7,35
7.36 After the switch in the circuit of Fig. P7.36has been open for a long time, it is closed at I : 0. Calculate (a) the initial value of i; (b) the final value of i; (c) the time constant for t > 0; and (d) the nunedcal expressionfor i(t) when I > 0. Flgur€ P7.36 5kO
4kO
75mH
7.39 The switch in the circuit iII Fig. P7.39 has been closed for a long time. A student abruptly opens the switch and repo s to her instructor that when the switch opened, an electdc arc with noticeable per, sistence was established across the switch, and at the same time the voltmeter placed aqoss the coil was damaged. On the basis of your aralysis of the circuit in koblem 7.38,can you explainto the student why this happened? Pigure P7.39
20kO
737 The currert and voltage at the teminals of tlle inductorin the circuit in Fig.7.16are
(D : (10 10r5m') A, r>0; u(t\ : 20oesm'v, l>0*. a) Specify the numerical values of %, R, 1,, and I. b) How many miliseconds after the switch has beencloseddoesthe energystoredin tlle inducior reach 25% of its final value? ?.38 The svritch in the circuit shown in Fig. P7.38 has been closed for a long time. The switch opens at l:0.Foit>0*i a) Find I'o0) as a function of 1s, Rr, R2,and a. b) Explain what happers to ?,(t) as R2 gets larger and larger. c) Find osw as a function of 1s, Rt, R2,ard L. d) Explain what happens to osw as R2 gets larger and larger.
7.4O a) Derive Eq.'7.4'7byf st conveting the Th6venin equivalent in Fig. 7.16 to a Norton equivalent and then summing the curents away from the upper [ode, using the inductor voltage o as the variable of interest. b) Use the separation of variables technique to find the solution to Eq. 7.47.Verily that your solution agees with the solution given in Eq 7.42. 7.41. The swilch in tle circuit in Fig. P7.41has been open 6ntr a long time before closing ar r = 0. Find ,"G) for t>0. Figure P7.41 87.2mH
20()
10O
09?]i
40O
272
Rtandlr Circujts Response of Fitstod€r
7.42 The swilch in the circuit in Fig. P7.42has been open 6xa a long time before closhg at I = 0. Filld o,(1) for r>0. P7.42 Figure 100
),,o
150 5 0 " i2mH
:
l'*(
,n ot^(
7.43 There is no energy stored in the inductors a1 and az at the time the switch is opened in the circuit shown in Fig. P7.43. a) Derive t]le exp.essionsfor the curents i1(t) and i20)forl > 0. b) Use the expressionsderived in (a) to find i1(x') and i2(oo).
7,46 The makebeforebreak switch in the circuit of 6r,t! Fig.P7.46hasbeen in positiol a for a long time At t = 0, tle switch moves instantaneously to position b. Find a)?,,(,),r>0*. b)t10),t>0. c)4(t),t>0. Figure ?7.46
15o r,+
?.47 Theswitchir thecircuitin Fig P747hasbeenopen 6ire a long time before closingat t = 0. Find o,(t) for
FiglreP7.43
u
iz,.l)l
figureP7.47
t=0 40 mH
5rI 10A
7.44 The switch in the circuit in Fig. P7.44 has been in 6xd position 1 for a long time. At I = 0 it moves instantaleously to position 2. How many milliseconds after the switch opentes does ?t"equal 80 V?
60 nH
ligve P7.44 7.48 The switch iII the circuit in Fig. P7.48 has been in tsdG position i for a long time. The initial charge on the 15 nF capacitor is zero At t : 0, the switch moves iNfantaneouslyto positiony. a) Find 0.0) for t > 0". b) Fhd 1)(, for t > 0. tigur€P7.48 ?,45 For the circuit in Fig. P7.44,find (in joules): a) the total energy dissipated in [he 80 O resisfor; b) the energy trapped in the inductors; c) tle idtial energy stored in tlle inductors.
7.49 For the circuit in Fig. P7.48,find (ia microjoules) a) the energydeliveredto the 200kO resistor; b) the energytmpped in the capacito$; c) the initial energy stored in the capacitors. 7.50 The switch in tle circuit shown in Fig. P7.50 has antr been closed a long time before opening at r = 0. For, > 0+,find a) ro\t). b) i,(r).
c) 4(r). d) i2O. e),r(0). figueP7,50
19
4ko
Figure P7.53
1g.ko" 120V
4;
6125tO ,:0 150ko 40nF
7.54 The switch in the circuit of Fig. P7.54 has been in position a for a Iong time. At t  0 the switch is movedto positionb. Calculate(a) the initial voltage on tlre capacito! (b) the fhal voltage on the capacitor; (c) tlre time constant (in micmseconds) for t > 0; and (d) tle length of time (in microseconds) required for the capacitor voltage to reach zero after the switch is moved to position b. Figure P7.54
20ko 7.51 Thecircuit in Fig.P7.51hasbeenin operationfor a 6trc Iong time.At t = 0, the voltagesouce dropsfrom 100V to 25V andthe currentsourcereversesd ection.Find r,,(r) for , > 0. Figrre P7.51 5ko
15kO 75V 10k()
2ko
100v
7.52 The curent and voltage at le terminals of the capacitor in the ckcuit in Fig. 7.21 are i(r) : 50e'sm'mA' v. ,{r}  (80  80P25m')
t > 0t; r > r.
a) Specify t}le numerical values of 1", U, R, C, and r, b) How many midoseconds after the switch has been closed does the energy stored in the capacitor reach64% of its final value? 7.53 Assume that the switch in the circuit of Fig. P7.53 has been in position a for a Iong time and that at 1 = 0 it is moved to position b. Find (a) oc(o+); (b) ?,c(oo);(c) r for I > 0; (d) (0*); (e) oc, r > 0; and (f) ,, t > 0+.
755 The switch in the circuit shownin Fig. P7.55has 6rc beencloseda long time beforeopeningat I : 0. a) Whatis the initial valueof i,(l)? b) Whatis thefinalvalueof i,(tx c) Whatis tlle time constantof the circuitfor t > 0? d) What is the numericalexpressionfor i.(t) when r>0? e) What is the numericalexpressionforu,,(t) when
figureP7.55 6.8kO 75V
274
Response of Firstord€r nt andfr Circujts
7.56 The switch in the circuit seenin Eg. P7.56hasbeen in Edc position a for a long time. At t : 0, the swrtchmoves instantaneouslyto position h For t > 0*, find
b),,(r). c) r.lt).
7.59 The switch in the circuit shown in Fig. P7.59 has EdG been in the oFF position for a long time. At 1 = 0, the svdtch moves instantaneously to the oN position. Find o,(l) fo. t > 0. Figure P7.59 30 x ldia
d) ,g(0).
7,60 Assume that the switch in the circuit of Fig. P7.59 6na has been in ttre oN position for a long time before switching instantaneously to the oFF position at t = 0. Find ?r,(t) for 1 > 0.
7.57 The switch in the circuit seenin Fig. P7.5?has been 6tra in position a lor a long time. At t = 0, the switch moves instantaneously to position b. Find oo(r) and i,(t)forr>0'. Figure P7.57
.i,r,r! ,,,,"
7.61 a) Derive Eq.7.52 by tust converting the Norton equivalent circuil shown in Eg. 7.21to a Th6venin equivalent and then summingthe voltagesaround the closedlooA usilg the capacitorcurrent i asthe relevant vadable, b) Use the separalion of variables technique to find the solution to Eq. 7.52.Verify that your solution agees with that of Eq.7.53. 7.62 There is no energy stored in tlle capacitors C1 and C2at the time the switch is closed in the circuit seen in Fig.P7.62. a) Dedve the expressions for o(t and zJr(t) for t>0. b) Use t})e expressionsderived in (a) to find o(oo) and 02(').). Figure P7.62
,r(t) 758 The switch in the circuit shown in Fig. P7.58opensat mPi.r , : 0 alter behg closed for a long time. How many milliseconds after the switch opens is the energy stored in the capacitor 90% of its final value? rig'lle P7.58
?,n, 7.63 The switch in the circuit of Fig. P7.63 has been in sP,m position a for a long time. At t : 0, it moves instantaneously to position b. For t > 0*, find a) polt).
b) c) d) e)
i,(r). ?,(0. ,10. theenergyfappedin thecapacitonast + oo
Figure P7.63
7.67 There is no energy stored in the circuit in Fig. P7.67 EPj.Eat the time the switch is closed. a) Find i,0) for t > 0. b) Ftnd u"0) for r > 0*. c) Fird ir(,) for r > 0. d) Find i(D for I > 0. e) Do your answen make sensein terms of knowlr circuit behavior?
100v
I
Figure P7.57 +i"
7.64 The switch in the circuit in Fig. P7.64 has been in 6Ee position a for a Iong time. At I = 0, the switch moves instantaneously to position b. At the instant the switch makes contact $rith terminal b, switch 2 opens.Find o,(r) for t > 0.
t+
.
10v
7.68 There is no energy stored in the circuit in Fig. P7.68 Edc at the time the switch is closed. a) Find (i) for I > 0. b) Fhd r,(t for t > 0*. c) Find r,r(r) for 1 > 0. d) Do your answe$ make sensein terms of known circuit behavior? Figure P7.68
Section 7,4 7.65 Therc is no energy stored in t}le circuit of Fig. P7.65 at the time the switch is closed. a) Fird io(t) for I > 0. b) Find 0o(t)for t > 0*. c) Find t1(t for I > 0. d) Find irQ) for t > 0. e) Do your answersmake sensein terms of known circuit behavior? FiWleP7.55 ,/10riH\.
200v
8H
d'
7.69 Repeat Problem 7.68 if the dot on the 8 H coil is at 6nc the top of the coil. Section 7.5 7.70 In the circdt in Fig. P7.70,switch A has been open E ic and switch B has been closed for a long time. At t = 0, switch A closes One second after switch A closes,switchB opens.Find tr(t) for t > 0. Figure P7.70
7,66 Repeat (a) and (b) in Example 7.10if the mutual inductance is reduced to zero,
l0 nH
276
Response of First0rderfl andRCCncujts
7.71 There is no energy stored in the capacitor in the cir 6trr! cuit ir Fig. P7.71when switch 1 closesat r :0. Three microsecondslater, switch 2 closes.Find od(t) fort > 0.
ligurc P7.74
tigureP7,71 r:0 + 50/rs
+
7.72 The action of the two switchesin the circdt seenin *rI( Fig. It.72 is asfollows. For t < 0, switch 1 is in position a and switch 2 is open.This state has existed for a long time. At t = 0, switch 1 moves instantaneously frcm position a to position b, while switch 2 remains open. Tivo hundred fifty microseconds after switcb I operales.switch 2 closes.remain( closed for 400 ps, and then opens.Find o,(t) 1 ms after switch 1 moves to position b.
7.75 The switch in the circuit shom in Fig. P7.75has EPicbeenin positiona for a long time.At I :0, the switchis movedto positionb, whereit remainsfor 100ps.The switchis then movedto positionc, whereit remainsindefinitely.Find a) (0). b) t(25ps). c) i(200ps). d) 0(100 i.is). e) r(100+ps). FigureP7.75
IrgurcP7.72
500v
7.73 For the circuit in Fig.P7.72,how manymilliseconds after switch 1 movesto position b is the energy sroredin the inductor4% of its initial value?
+ J
96()
480f)
20nH
276 The suritch in the cftcuit in Fig. P7.76 has been in Edc position a for a long time. At I = 0, it moves instantaneouslyto positionb, whereit remainsfor 250ms before moving instantaneouslyto position c. Find I,otbrr > 0. tigureP7.76
7.74 The capacitor in the circuir seen in Fig. P7.74 has Efld been charged to 494.6mV Aft = 0, switch 1 closes, causingthe capacitor to discharye hto the resistive network. Switch 2 closes50 &s after switch 1 closes. Find t])e magnitude and direction of the current in the second switch 100 ps after switch 1 closes.
120()
4{}
50A
7.77 In the ckcuit itr Fig. P7.77,switch t has been in posisPI.r tion a and switch 2 has been closed for a long time. At t  0, switch 1 moves instantareously to position b. Four hundred microseconds later, switcb 2 opens,rcmahs open for 1ms, and then recloses. Find ,, 1.6 ms after switch 1 makes contact with teminal b.
0+40Ops
35kO
Figure P7.80
I J sko
7.81 The voltage waveform shown in Fig. P7.81(a) is Erla applied to the circuit of Fig. P7.81(b). The initial voltage on the capacitor is zero. a) Calculate0o(.). b) Make a sk€tchof r',(t) ve$us t Figure P7.81
7.78 For the circuit in Fig. P7.77,what percentage of the initial energy stored in the 50 nF capacitor is dissipatedin the 10kO resistor?
7.79 The voltage wavefonn shown in Fig. P7.79(a) is am applied to the circuil of Fig. P7.79(b). The initial curent in the inductor is zero. a) Calculate0"(l). b) Make a sketch of ?,,(t) venus L c) Find io at I : 4 &s. FlgueP7.79 ?,"(v)
20
..1
1t)
6 (a)
t(nd (b)
7.82 The voltage signalsourceh dre cncuit in Fig.P7.82(a) is geneiatingthe signalsholvl in Fig.P7.82(b).Thereis no stored energyat t = 0. a) Derive tlre expressionsfor o,(t) that apply in the i n r e r l a lrs. 0 : 0 = t < l 0 m s ; 1 0 m s < r < 20ms;and20ms < t 20 /.s,what would you expect the ouFut voltage to be? Er?lain. Figure P7.95
10mV e0) 15
25nF
7.94 The voltage pulse shown in Fig. P?.94(a) is applied Edc to the ideal integrating amplifier shown in Fig.P7.94(b).Derive the nume cal expressions for o,(t) when t,(0) : 0 for the time intenals a)t "'.
0(t) = looa'misin 97980r v,
t > 0.
d) Figure8.9 showsthe plot oI t (t) ve$us I for the fkst 11 ms after the stored eneryy is released.It clearly indicatesthe damped oscillatorynature of the underdampedresponse.The voltage 2{t) approachesits final value,alternating between values that are greater than and less than the fiml value.Fu hemore, theseswingsabout the final value deqease exponentialy wittr time.
Therefore, the responseis underdamped. Now,
a6: ^,/S ]:
r,,iFJxld:
roor,66
: 979.80radls, st:
a + jod:
s2 = a  jad =
200 + j97980rad/s, 200
j979.80rad/s.
For t}le underdamped case,we do not ordinadly solvefor J1 and 12becausewe do not usethem
, (v) 80 60 40 20 0 20 40
7 8 9
tesponse for Exampte 8.4. Figure8.9 A ThevoLtage
,(*)
298
Naturatand StepResponses ofntaCircuih
Characteristics of the Underdamped Response The underdampedresponsehasseveralimporranl characterisrics. Firsr.as the dissipatiyelossesin the circuil decrease. rhe pcrsistenceof thc oscillations incrcaser and the frcquencv of the oscillalionsapproachcso0. I'r other words.asR+,6. thcdissipationin thecircuir in Fig.8.ltapproaches zero bccausep : i)7R. As R co. a+ 0. which tells us lhat o,j + rr0. When a : 0. the maximum amplirudeof rhc voltage rentainsconslant: thusthe oscillationat .r0is sustained.In Example8.4,if /t wcrc increascd to infinily, the solutionfor o(r) \\'ouldbccorne 1Jt) : 98 sin 1000rV.
r > 0.
'lhus,
iD this casethe oscillarior is sustaincd,the maximum amplitudeof thc voltageis 98 V and the frequencyof oscillationis 1000rad/s. Wc Draynow describequalitarivelythc differencebetwccnan undcrdiDped and an ovcrdampedresponsc.In an underdampcdsystem.thc lcspoNe oscillates,or "bounces,"aboul ils fiDal value.This oscilLalionis also rcferred to as rinqing. In an ovcrdamped system,the responsc approachesits final value without ringing or in what is sometjmcs dcscibed asa "sluggish"manner.Whcn specifyingthe dcsircdresponscof a sccondordersystcmryou mav want to reachthe final value in the sho(csl tiDe possiblc.and you may not be concernedwith smaLloscillations about that final value.If so,you would designthe systcmcomponentsto achievean underdampedresponse.On the other hand.you nlay bc concornedthat thc responsenot exc€edits final value.pcrhapsto ensurcthat oomponentsarc not damaged.Insucha case,youwould dcsignthe systcm componentsto achievean overdampedresponse,and you would havc to :rc(€pri relalivcl)\lu\ ri.e ro rhclinJlvalu(.
obj€ctivelBe able to detemine the naturaland the st€p responseof paraltelilc circuits 8.4
A 10 trI}I inductor,a 1pF capacitor,and a variableresistorare connectedin parallelin the circuit shown.The resistoris adjustedso thal the roots ofthe characteristicequationarc 8000 ,r i6000 rad/s.The initial volrageon the capacitoris 10Y and the initial cuffent in thc inductor is 80 mA. Find a) Ri b) dr(o+)ldt; c) Br and 82 in the solutionfor ?);and
a);r(r). NOTE: Aho tty ChdpterProblems8.3 antl8.20.
Answ€r: (a) 62.5O; (b) 240,000v/s; (c) Br: 10v, B, = 80/3 V; (d) rr.(r)= 10e3mo,[8cos6000t + (82/3)sin600011 mA whenI > 0.
Response of a PanLtel RrcCncLrit 299 8.2 TheForms ofthe NaturaL
TheCriticattyDamped VoltageResponse The secondordercircuit in Fig.8.8is critically dampedwhen ofr : o', or o0 : a. When a circuit is critically damped,the responseis on the verge of oscillating.In addition, the two roots of the characteristicequation are real and equalt that is,
1 2RC
(8.32)
When this occun, the solution for the voltage no longer takesthe form of Eq. 8.18.This equation breaksdown becauseif rr = 12 = a, it predictsthat D = \ A 1+ 4 ) e d :
AGd,
(8.33)
where ,40 is an arbitrary constant. Equation 8.33 cannot satisfy two inde' pendent initial conditions (y0, 10) with only one arbitrary constant, A0. Recallthat the circuitparametenR and Cfix d. We can trace this dilemmaback to the assumptionthat the solution takesthe form of Eq.8.18.wllen the roots of the characteristicequation are equal, the solution for the differential equation takes a different folm, namely
/8 14) < Vottage naturalresponsecriticall, parattelf,lCcircuit damped Thus in the caseof a repeated root, the solution involves a simple exponenlial telm plus the product of a linear and an exponential term. The justification of Eq. 8.34 is left for an intrcductory course in differential equations.Finding the solution involves obtaining ,1 and D2 by follovring the same pattem set in tle overdamped and underdamped cases:We use the initial values of ttre voltage and the derivative of the voltage with respect to time to write two equations containing Dr andlot D2. From Eq. 8.34,the two simultaneousequationsneededto detemine D, ard D, are
a(.r)=uo:D2, do(o) _ ic(o*)_ Dr
(8.35)
aD2.
(8.36)
As we can see,in the case of a critically damped response,both the equation for o(l) and the simultaneous equations for the constants,r and D2 diJfer ftom tlos€ lor over and underdamped responses,but the general approach is the same.You will rarcly encounter critically damped systems in practicg largely becauseoo must equal a €xactly.Both of these quartities depend on circuit parameten, and in a real circuit it is very difficult to choosecomponent values that satisfy an exact equality relationship. Example8.5illustratesthe approachfor finding the criticallydamped resoonseof a oarallelRIC circuit,
300 NatuDtandStepResponses ofiltOrcurts
Findingthe CriticattyDamped NaturaIResponse of a Paratlel ntc Circuit a) For thc circuit in Example8.4 (Fig.8.8),find the valucoIn thairesultsin a criticallydampedvolt
Eqs.8.35and 8.36,r, : 0 and ,i = 98,000V/s. Substjtutinglhesevaluesfor a. ,1, and ,2 inlo Eq.8.34gives
b) Calculate!(r) for r > 0. c) Irlot 1r(l)vcrsusrfor 0 < t = 7 ms.
? ) ( i )= 9 8 , 0 0 0 t i ? r o m ' rv>, 0 . c) Figure 8.10 showsa plot of 2,0) versus1in thc intervall)= I < 7ms.
Sotution  r f f o m I \ J n r n , c8 . 4 .$ < k n o w r h " r u ; Thereforefor crilical damping.
l0'.
1 )RC.
"(v) 12 21
106 : 4000(). (2000)(0.12s)
8 0
b) Fron the solutionof Example8.4.we know that u(0*)  0 and dr(0+\ldt = 98,000v/s. From
1
, (mt
Figure8.10,i llre vottage rcsponse for Exanrph 8.5.
obiedive 186 abteto determin€the naijraland the st€p r€sponse of paralletf,lCcircuits 8,5
The resistorin ihe circuit in Assessment PfoblemR.4i. idtu.led tof crilicaldamping The inductanccand capacitance valuesare 0.4H and 10 /..F,respectively. Ihe inirial energy storedin thc circuit is 25 mJ ard is distributed equallybetweenthe ind ctor and capacitor. Find (a) R:(b) yor(c) 10;(d) Dr and ,, in the soiution Ior t';and (e) in. I > 0*.
Answer: (a) 100O:
(b)s0v; (c) 2s0mA; (d) 50.000v/s,50 V; (e) tn(r) = (500re500.+0.50rs00)A, t=t)'
NOTE: AIso *j ChapterProblems8.4dnrl8.21.
A Summary of the Results We concludeour discussionofthe parallcl R/,Ccircuir\ naturalresponse with a bricf sunmary of the results.The first slep iD finding the natural responscis to calculatethe roots of the characteristicequation.You thcn knowimmediatelywhetherthe rcsponseis overdamped.underdampcd,or criticallydamped. If the roots are real and dislincl (o3 < a2), the responseis over dampcdand the voltageis
n(.t): A'd."'+ Aze''.
8.1 TheStepRespo6e ofa Pa,altetRtC Circuit
\/7', 1 2RC'
" 
l
"u
Lc
The values of 41and ,42 are determined by solving the following simultaneousequatlons: D(0*)= A1 +A2, dr'(0+r l.{0+l d
t
L
If the ioots are complex ..rA> d' t}le response is underdamped ard the voltageis D(t)  B1e "'cosadt + B2e"'sinadt,
,
ra
"
The values of Br and B, are found by solving t}le following simultaneous 1)(0):uo:Br, dDt\+\  ' i.tq+  ( =aB u"B dr li the roots of the characteristicequation are real and equal (dA : a'), the volfage responseis a(t) = D1rca'+ Dp d', where a is as in the other solution forms, To determine values for the constants,1 and D2,solvethe folowing simultaneousequations: a(O):Uo=D2,
/u(01
ic(u+)
8.3 TheStepResponse of a Paraltel
RICCircuit Finding the step responseof a parallel RaC circuit involvesfinding rhe voltage acrossthe parallel branchesor the current in the individual branchesas a result of the sudden application of a dc current source.Therc may or may not be energy stored in the circuit when the current source is applied.The task is representedby the circuit shown in Fig. 8.11.To Figlre8.111A circuitLrsed to de(fibethestep of a paGlletfr circuit. develop a general approach to finding the step responseof a secondorder rcsponse
301
102
Natunt andSiepResponses of,?rrCircujts circuit, we focus on finding the current in the inductive branch (ir). This current is of particular interest becauseit does not approachzero as I increases.Rather, after the switch has been open for a long time, the inductoi current equals Uredc sourcecurrent I Becausewe want to focus on tlle lechdque for finding rhe step response,we assumethat tie initial energy stored in tlle circuit is zero.This assumptionsimplifies t]le calculations and doesn't alter the basic processinvolved. In Example 8.10 we will see how the presenceof idtially stored energy enters into the general procedure. To find the inductor current ir, we must solve a secondorder differ, ential equation equated to the forcing lunction 1, which we derive as fol lows.From Kirchhoff's curent law. we have iL+iR+ic:1,
i,r!*c!:r.
(8.37)
Because  cllL
(8.38)
we get (Lu dt
 r12i, dt2'
(8.3e)
SubstitutingEqs.8.38and 8.39into Eq.8.37gives L (li, iL+=:1
i LC.:L dr
(8.40)
For convenienc€,we divide througl by LC and rearrange terms: d2i, dt!
I
dit
RC d!
i,
t
LC
LC
(8.41)
Compadng Eq. 8.41 with Eq. 8.3 reveals that the presence of a nonzero term on the dghthand side of the equation alters the task. Before showing how to solve Eq. 8.41 directly, we obtain the solution indirectly. When we know the solution of Eq.8.41, explai ng the direct approach will be easier.
TheIndirectApproach We can solve foi ir indhectly by first finding the voltage o. We do this with the techniquesintroducedin Section8.2,because the differentialequation that ?Jmust satisfyis identicalto Eq. 8.3.To seetlis, we simply return to Eq. 8.37and expressi, as a functioDof r; thus
l[',0,*i*rolt='
18.42)
8.3 lheSteD ResDonse ofa Panttet flrCircuit 303 Differentiating Eq. 8.42 once with respecl to t reduces the righfhand side to zero because1 is a constant.Thus o L
d2u
ldu t< dt
^dz\ ar
1 cla RC dt
^
u LC
(8.43)
As discussed in Section8.2,the solutionfor t, dependson the roots ofthe characteristic equation. Thus the three Dossiblesolutions are b=Aretn+A2est,
(8.44)
b = Bpn'
(8.45)
a=
Dle"t
cos tJdt + B2e"'sin adt, + Dze"t.
\8.46)
A word of caution: Because there is a souce in the circuit for l > 0, you must tate into account the value of the sourcecurrent at t = 0* when you evaluate the coefficients h Eqs.8.4ll8.46. To fhd the three possiblesolutions for ir, we substitute Eqs.8.448.46 into Eq. 8.37.You should be able to vedry, when this has been done, that the three solutions for i, vrill be iL:I+Aid't+Aie"1t, iL:
I + B'Ie'cosaat + Bie "Isind,1t.
iL=I+Dita"t+Die"t,
\8.47) (8.18) (8.4e)
where Ai, Ai, Bi, Bi, Di, and Di, arc arbitrary constants. Ir each case,the primed constantscan be found indiiectly in tems of the arbitrary constantsassociatedwith the voltage solution. However, this approach is cumbe$ome.
TheDired Approach It is much easier to find tle primed constants directly in tems of the initial valuesof the responsefunction.Foi the circuit being discussed,we would find the pimed constants from iz(o) and dir(0)/dt. The solution for a secondorder differential equation witl a constant forcing function equals the foiced responseplus a responsefunction identical in form to the natural response.Thuswe can alwayswrite the solution for the steo resoonsein the form I lunctionof lhe samefolm I I = r/ + asthenat"*i *"p.*.J' 
(8.50)
304 Naturat andStepResponses of Rtf Circujts
ol thesameformI ' = . vr + lfunctron I astheDot"r*"p.'*J'
(8.51)
where 1t and yi represent tbe final value of the response function. The fillal value may be zero, as was,for example,the casewith the voltage x' in tlle c cuit in Fig.8.8. Examples 8.G8.10 illustrate ttre technique of finding the step responseofa parallelRaCcircuit usingthe direct approach.
Findingthe overdamped StepResponse of a ParalteI flc Circuit The initial energystoredin the circuit in Fig.8.12is zero. At t : 0, a dc current source of 24 mA is applied to the circuit. The value oI the resistor is 400 (). a) Wlat is the initial value of ir? b) What is the initial value of diLldt? c) wllat are the roots of the characteristicequation? d) W}lat is t}le numerical expression for iz(r) when t>0?
c) From tlle circuirelements.weobtain
.a== o
r h. r 0 " ,
a' =25r103. Becauseofr < l, fie roots oI the chamcteristic equationarercal anddistinct.Thus rad/s, ,1 : 5 x 104+3 x 104= 20,000 .r2:
Figure8.12.AThecircuiitorExampte 8.6.
^:=
\z))\z)) 1 loe ')(4uu,(25, 'RC t
5x10*
3 x 10'= 80,000 rad/s.
d) Because the rootsof the characteristic equation arerealanddistinct,theinductorcurent response will be overdamped. Thusir(l) takest}le form of Eq.8.47,namely, iL= If + A\es)t+ Aietxt
Solution a) No erergy is stored in the circuit prior to the applicationofthe dc curent source,so the initial current in the inductor is zero. The inductor prohibits ar instantaneouschangein inductor current; therefore il(o) = 0 immediately after the switchhasbeenopened. b) The initial voltage on the capacitor is zero before the switchhas been opened;ther€foreil will be zero immediately after. Now, because u: LdiL/dt,
!:tot : o.
> Indudorcurrentin overdamped parattet RICcircuitstepresponse Hence,ftom thissolution,the two simultaneous equations thatdetermine A\ andAi are i,(0)=./,+A'.+A\=0. ;(0)
= sL,ai+ 'rAi:0
Solvingfor ,4i and Ai gives , 4 j: 3 2 m A a n d A i = 8 m A . Thenumerical solutionfor ilo) is iLo : Qa , 32e20,m& + 8es0,1\J0' ) mA, I > 0.
8.1 lhesteoResoonse ola Para.teL flt LtrLut
305
Findingthe Underdamped StepResponse of a Parattet RICCircuit The resistorin the circuit in Example8.6 (Fig.8.12) is increasedto 625O. Find ,z(1)for I > 0.
Here, d is 32,000 rad/s, rrr is 24,000 rad/s, and is 24 mA. 4 As in Example8.6,Bi andBi aredetermined from the initial conditioN Thusthe two simultare
Solution rr(0)=1f+Bi=0,
BecauseI, and C remain fixed, oA has the same value as in Example 8.6; that is, o; = 16 x 103. Increasing n to 625O decreases a to 3.2 x 10amd/s. With ofr > a', the roots of the charactedsticequationare complex.Hence
*Q)
3.2 x 10"+ j2.4 x 10"rad/s,
= dBt  "B't: o.
B't : 21m4
3.2 x 10 j2.4 x 10"rad/s. The cu{ent responseis now underdamped and givenby 8q.8.,18:
Bi:
32 mA.
Thenume cal solutionfor ir(t) is itul = It + Btc"'cosd.t. Be't:inuut.
paratler > Inductorcurrentin underdamped ruCcircuitstepr€sponse
iLO = (24
24e 32ac4t cos24,0}0t
 32e'2ooo'sin 24.ooor) mA. r > o.
Findingthe CriticatlyDamped StepResponse of a ParallelRlc circuit The resistorin the circuitin Example8.6 (Fig.8.12) is setat 500O. Find ir for I > 0.
Again, Di and Di are computed from initial i L ( O )= I r + D ! 2 = 0 ,
Sotution We know tlat oAremainsat 16 x l0s.WithRsetat 500O, d becomes4 x 104s 1,which corresponds to c tical damping. Therefore tlre solution for i_(1) takesthe form of Eq.8.49:
!!,or=o',oo,,:0. d
'
Thus ,i =
itltl:
l
mA/s and Di: 960,000
24m1..
It + D\te "' + Diad.
Thenumericalexpression for iz(t) is > Inductorcunentin criticalLydampedpanlleL nlc circuit step response
24t10m0') mA, r>0. iLO : Q1  960,000rerq0m'
306
NatuEtand SiepResponses 0fRlfCncuits
Comparing the ThreeStep Response Forms a) Plot on a single graph, over a range lrom 0 to 220!,s, the overdamped, underdamped, and cdtically damped responses derived in Examples8.68.8. b) Use the plots of (a) to find the time requiredfor tr to reach90o/.ofits final value. c) On the basisolthe rcsultsobtainedin (b), which responsewould you specityh a designt}lat puts a premiumon reaching90% ofthe final valueof tlte output in the shortest time? d) W}lich responsewould you specify in a design that must ensure that the final value of the current is neverexceeded?
Sotution a) SeeFig.8.13. b) The final value of ir. is 24 mA, so we can read the timesoffthe plots correspondiq to ir = 21.6mA. Thusrd,l= 130/.rs,r.r = 91 ps,andt"d : 74 ps. c) The underdampedresponseieaches90o/oof the final value in the fastesttime,so it is the desired responsetype when speedis the most importart design specification.
(/l = 625O) Underdanped
26 22 18 l4 10
1n=+oooy fi licalltlo'!'a'p.a = (1l daDpcd 500O)
tC
2 i)
140
180
I(it
Figure8.13.rr Thecun€niptotsfor Example 8.9.
d)From the plot, you can see that the underdamped responseove$hoots the final value of current, whereasneither the critically damped nor the overdampedrespoNe producescurents in excessof 24 mA. Although specifying either of the latter two responseswould meet the design specificatior, it is best to use the overdamped fesponse. Ir $ould be impraclicalto requirca design to achievethe exact component values tlat ensurea criticallydampedresponse.
FindingStepResponse of a ParallelRLCCircuitwith Initial StoredEnergy Energy is stored in the circuit in Example 8.8 (Fig.8.12,withR = 500 O) at the instantthe dc cur, rent source is applied.The initial current in the inductoris 29 mA, and the initial voltageacrossthe capacitor is s0 v Find (a) iL(o); (b) diL(o)/dt: (c) tr(r) for I > 0; (d) u(D for t > 0.
b) The capacitor holds the initial voltage acrossthe inductor to 50V Therefore
r;(o)
: 50,
lo1l:fl
" 103:2oooA/s.
So[ution a) Therecannotbe an instartaneouschangeof cu. rent in an inductor, so the initial value of ir in the first instant after the dc curent source has been appliedmust be 29 mA.
") From the solution of Example 8.8,we kno\q that the currentresponseis citically damped.Thus iL(t) = Ir + D\te "' + Die ""
8.3 TheStepResponse ofa Panttetilrcncujt 307 Thus the nume cal expressionfor ir(l) is d
^:^
4 0 . 0 0r0a d / s a n d / ,  ' 4 m A .
Notice that the effect of the nonzero initial stoied energy is on the calculations for the constants Di and ,i, which we obtain tom the nntial conditions.First we use the initial value of the inductorcurent:
iLC) = (.24+ 2.2 x l06reroooor + seaoooo) mA,
t > 0.
d.)We can get the expressionfor ?J(t),t > 0 by using the relationshipbetweenthe voltage and cment in an inductor:
tr(0) = 1r + Di  29mA., T(, : L;
from which we get
Di:29
24
' 1o r) 2.2
Thesolutionfor ,i is
106)(4o.ooo)reroooo'
+ 2,2 X I06e4aIUt
=(0)
+ (5)( 40,000)?ro!m/l x 10 3 :
+ 50? {,000,V, I > 0_ 2.2 x 106tua1,auJt
D'r=2000+aDtz : 2000+ (40,000)(5 x 10 ) : 2200Als : 2.2 x 106nA/s.
To check this result, let's verify that the initial voltage acrossthe inductor is 50 V:
D()t)
2.2x 106(ox1) + 500):50v.
StepResponses of RtCCircuits 308 NatuGtand
8.4 TheNaturalandStepResponse of a SeriesRICCircuit +
Figure 8.14A A circuiiused to itLustrate thenaturat response ofa series Rlrcircuit.
The procedures for finding the natural or step respoNes of a seriesRZC circuit are the same asthose used to find the natural or step responsesof a parallel RIC circuit, becauseboth circuits are describedby differertial equationsthat have t}le sameform. We begin by sumnine the voltages aroundthe closedpath in the circuit shownin Fig.8.14.Thus ,ii
Ri
i
L",'
fl
^lidttvaj.
dt
(8.52)
cJo
We now differentiateEq.8.52oncewitl respectto I to get _ d' i  i  o .tr c
_di ar
(8.53)
which we can reanange as dzi
Rdi
i
,t,'iAi
"'
18.54J
ComparingEq. 8.54with Eq. 8.3 revealsthat tiey have the sameform. Therefore,lofind the solutionof Eq.8.54,we follow the sameprocessthat led us to the solutionofEq.8.3. From Eo.8.54. the characteristic eouation for the sedesRIC circuit is
Characteristic equationseries f,aCcircuit >
(8.55)
R 2t.
/RY \i)
(8.56)
(8.57)
The neperfrequency(d) for the sedesRIC circuit is Neperfrequenq'series Rlf circuit>
. = , ,"0u.
(8.58)
andtheexpression for theresonant radianfrequency is Resonant radianfrequencyseries nlc circuit>
(8.5e)
Note tlat tle neper frequency of the seriesRIC circuit differs ftom that of the parallelRaC circuit,but the resonantmdian frequenciesare the same.
8.4 TheNatuaLand StepResponse ofa5e esffrCjrcujt 309 The currentresponsewill be overdamped,underdamped,or critically damped accordjng to whether rofr < c', rofr > c', or tafr = a', respectively. Thusthe threepossiblesolutionsfor the curent are asfollows: i(t\ = nt""'
+ "\rer,t (overdamped), i(t, : Bf"' cos.Dat+ B2e "'sinadt (underdamped), i(t) = 1;rt" "r + Dr€"! (criticallydamped).
(8.60)
f8.ol) < CurrentnaturaL response formsin series , _ l(ta orcults
When you have obtainedthe natural current response,you can find the natural voltage responseacrossany circuit element. To verifl that the procedure for finding the step responseof a series RaCcircuit isthe sameasthatfor a parallelRIC circuit,we showthat ihe differentialequationthat descibes the capacitorvoltagein Fig. 8.15has the sameform as the differential equation l]rat describesthe inductor current in Fig.8.11.For convenience, we assumethaazero energyis storedin the circuii at the instantthe switchis closed. Applying Kirchhoff'svoltagelawto the circuitshowr in Fig.8.15gives Figure8.15A A circuitusedto ittushate thesiep rcsponse of a se.ies fircncuit.
V  R i + L ! + D . '.
(e.63)
/11
Th€ current(i) is relatedto the capacitorvoltage(oc) by the expression (8.64)
from which (8.65)
Substitute Eqs.8.64 and 8.65 into Eq.8.63 and write the resulting
d?t4
R dua
*  i a 
:!_ LC
V I.C
(8.66)
Equation8.66hasthe sameform as Eq. 8.41;thereforethe procedurefor finding ,c parallelsthat for finding ir. The three possiblesolutionsfor z'c are asfollows:
D .  V + A\e"lt+ A?\r (oveidamped),
(8.67)
r c = V f + Bie "' cosodt + Bie"' sln @,1 (undeidamped), (8.68) { Capacitor vottagestepresponse formsin series nlC circuits "t D c : u f + Dtte + Diad (criticallydamped), (s.6e) whereyl is the final valueof rc. Hence,from the circuil shownin Fig.8.15, the final value of z'c is t}le dc source voltage y. Example8.11and 8.12illustratethe mechanicsof finding the natural and steDresDonses ofa seriesRIC circuit.
310
NaturaLand StepResponses ofnlf Circuits
Findingthe Underdamped NaturalResponse of a SeriesRLCCircuit The 0.1pF capacitorin the circuit shown rn Fig.8.16is chargedto 100V At l : 0 the capacitor is dischargedthrougha seriescombinationof a 100mH inductoranda 560O resistor. a) Findj(r) for r > 0. b) Findoc(r)for r > 0.
the initial conditions.The inductor current is zero before the switch has been closed, and henceit is zeroimmediatelyafter.Therefore (0)=0=81. Io find 82.we evalualeJir0 l Jr. Fromlhe cil cuit, we note that,becausei(0) : 0 immediately alter the switch has been closed,there vall be no voltagedrop acrcssthe resistor.Thus the initial voltageon the capacitorappearsacrossthe ter minals of the inductor, which leads to the
di(0]' Figur€8.16 A lhe circuiifor E)Gnpte 8.11.
di(o)
Solution a) The filst step to finding i0) is to calculatethe root\of rhecharacleri(lic equadon, For rbegi\en
@ 6 =_
(ld)(r01 (100x0.1)
= 1000A/s. Because B1 : 0, co,so00r Tsinab0o/j. :  a00?,cuu'(24
Ll1
Thus d(0)
= 960082,
R " 2 L 8, :
= !!L ^ n' 2(100)
: 2800rad/s. Next,wecornparearfrto a2andnotethat oA > 02, l=7.84x106 = 0.0784x 103. At thispoinl.$e know lbal lbe re\poDse is underdamped andthat the solutionfor i(t) is of the folm i(t) : B( "t cosa,tt+ B2a"t sinridt, where o = 2800rad/s and od: 9600md/s. Thenumericalvaluesof B1 and ,, comefrom
=T=rr*'t"
t000 s 0.1M2 A. 9600
The solution foi i(t) is t(/)
go00/A. r j 0. 0.104)e2m'\in
b) To find oc(t), we can use either of the followitrg relationships:
"= tl''o'*'ooo' ,1,
. = t R + L 1 . llt
Wichever expressionis used (the secondis reconmended), the result is oc(r) = (100cos9600r+ 29.17sin 9600t)e'm'V,
r > 0.
8.4 TheNatunL andStepResponse of a sedesSLrCjrcuii 311
Findingthe Underdamped Step Response of a Seriesffc Circuit No energyis storedin thc 100mII inductor or the 0.4r.F capacitor when the switch in the circuit shownin Fig.B.l7 is closed.Find z,c(t)for t = 0.
The roots are complex.so the voltagc respoise is underdamped.'nrus 0c0):
'18 + Bid lroo'cos4800r + rle rao(r'sin4800r,r > 0.
No energyis stored in thc circuil initially. so both ,c(0) and d?rc(o)/lt arc zcro.Ther, 0c(0):0=,18+Bi, F i g u r e8 . 1 7h T h e. i r . u i tf o rE a n r p l8e. i 2 .
dt).(0'\ __:t)
Soh'ingfor ai and Bi yields
5olution
ai : 18v,
The roots ofthe characteisticequationare
:
=4'rn0Ri 1400R
280
//zso\. 
+ 1 / l

0.2 v\0.2/
Bi:
106
tn.r)(rl4)
 14V.
Therefore,the solutionfor ,(.(r) is
: ( 1400+ i1800)rad/s.
Dc(t): (48  48eraoo'cos4s00r
r, = (1100  j1800)rad/s.
 14r r{orsin 4800r)V.
objective2Be ableto determine the naturalresponse andthe stepresponse of s€riesftf circuits 8.7
The switchin the circuitshownhasbeenin positiona tor a long 1ime.At t = 0 it movesto positionb. Find (a) l(0+);(b) "c(0+); (c) dl(0')/dr; (d) .rl, rr:and (e) t(D for I > 0.
Answer: (a) 0;
(b)s0Y (c) 10,000 A/s; (d) ( 8000+ 16000)radls, ( 8000 j6000)rad/s; (e) (1.67€Mo'sin60001) A for r > 0. NOTE: ALto try Chteter Problems8.158.17
Findoc(l) for I > 0 for thecircnitin Assessment Problem8.7. Answer: [100 e 3mi(50cos 6000t + 66.67sin60m01V for r > 0. 8,8
r > 0.
312
Natunt andStepR€sponses offtr Circuits
8.5 A Circuitwith TwoIntegrating Amplifiers A chcuit containirg two inregraring amplifiers coDnectedin cascadel is also a secondorder circuit; that is, the output voltage of the second integrator is related to t}le input voltage of the first by a secondorder ditrerential equation.We begin our analysisof a circuit containing two cascaded amplifierswith tle circuit shownin Fig.8.18. We assumethat the op amps are ideal. The task is to derive tlle differential equation that eslablishes the relationship between ?, and Dp.We beeh tle derivation by summirg the currents at the inverting input terminal of the first inte$ator. Becausethe op amp is ideal,
0u" n =K tj + c , d; tt o
u . , ,:)0 .
(8.70)
FromEq.8.70, dDot
dt
I
(8.71)
&Cl"g'
Now we sum the cullents away ftom the inverting input teminal of t}le second integrating amplifierl
0z), R, dbo
rJt Differcntiating Eq.8.73gives Euo dt'
c,frtoo"t: o. 1
l.8.t2)
(8.73)
&C)"" 7 doo1 R{2 dL
(8.74)
We find the differetrtial equation that govems the relationship betweetr o, and ?rsby substitutingEq.8.71into Eq.8.741
d2u" : 1 1 d,' &(\R C:*
(8.75)
CI
+
Figure8.18A Twointegratjng ampLifieu connected in Gscade. ' In a cascademnnection, the output siglal oI the ii6r mplifrd signal tor the secondanplifier.
(ror in Fig. 8.18) is the inpul
8.5 A CiftuitwiihTwoInt€grating AmpLjfiels 313
Example 8.13 illustrates tlle step responseof a circuit containing two cascaded integrating amplifierc.
Analyzing TwoCascaded IntegratingAmplifiers No energyis storedin the circuit shownin Fig.8.19 when the input voltage ?s jumps instantareously from 0 to 25 mV
0 . 1F F
1/,F
9V
a) Derive the expressionfor r'"0) for 0 < t< r""t. b) How long is it before the circuit satuiates?
Sotution
Figure 8.r9A Thecjrcuit forExampLe 8.13.
a) Figure 8.19irdicates that the amplifier scaling
1
1000
n1c1 (2s0x0.1) 1 RrC,
1000 (500)0)

No\ becauset's : 25mV for I > 0, Eq. 8.75

= (40X2)(25. 10 )) : 2.
To solvefor z',,we let da^ dt' then, dp(r\ ::2,
a n d d R 1 )= 2 d t .
becausethe energy stored in the circuit ini tially is zero, and the op amps are ideal. (See Problem 8.53.)Then, dr^ i=2t,
and Do: ( + 1).(0).
But ,"(0)  0, so the experssionfor rn becomes p.:t2, o 0 is 'noor + 20P 2oolro'V ,(l, 5e
8.7 The resistance in Problem 8.6 is increased to m'j.r 2.5kO. Find the expressionfor ?,(/)for t > 0. 8.8 The resistance in Pioblem 8.6 is increased to EFlc 12500/3 O. Find the expressionfor o(t) for r > 0.
a) Determine the numerical values of R, a, a, b) CalculatetRo), ir(t), and ic(t) for t > 0*. 8.3 The natural voltage response of the circuit in Fig.8.1is o(t) : 125ar0mr(cos 30001 2sin3000Dv, r>0, when the capacitor is 50nF. Fird (a) t; (b) n; (c) Yo;(d) 10;and (e) ir0). 8.4 The voltage responsefor the circuit in Fig. 8.1 is known to be D(t) = Dfe4aoot+ D2eaatu, t > o. The initial current in the inductor (It is 5 mA, and the initial voltage on the capacitor (yd is 25 V The inductor has an hductance of 5 H. a) Fitrd the value of R, C, D\, and.D2. b) Find ic(t) for / > 0+. 8.5 The idtial value of the voltage ?Jin the circuil in Fig. 8.1 is zero, and the initial value of the capacilor cunent, ic(O),is 15 mA. The expressionfot the capacitor currert is known to be i"(t) = AF16t+
A2eat' t>o+,
when R is 200 O. Find a) the valueofa, (,o,a, C,Aband A)
( ,,,,.0''!o)=
.liL(0)
r?ip(o) u(0) d t :
1 rc(o)\
L  R
c
)
b) the expressionfor r(t), t > 0, c) the expressionfor in(t) > 0, d) the expressionfor ir(l) > 0. 8.6 The ciicuit elementsin the circuit in Fig.8.1 aie 3 ' c R : 2 k O , C = 1 0 n F ,a n d a : 2 5 0 m H . T h e i n i tial inductor curert is 30mA, and the initial capacitorvoltageis 90V a) Calculatethe initial current in each branch of the circuit. b) Find 0(t) for t > 0. c) Find iz(r) for r > 0.
8.9 The natuml responsefor the circuitshownin Fig.8.1 is known to be o'.'300)v. u ( / )  1 2 ( r />. If C : 18 pF, find i1(0') in milliamperes. 8,10 In the circuil shown in Fig. 8.1, a 5 H inductor is 6tr4 shunted by a 8 nF capacitor, the resistor R is adjusled for c tical damping, Vo25V, and 1o = 1 mA. a) Calculate tle numedcal value of R. b) Calculate r'(t) Ior t > 0. c) Find 0(t) when tc(t) = 0. d) What percertage of the initially stored energy remains stored in the circuit at the instant ic(r) is 0? 8.11 In the circuit in Fig. 8.1, R = 2r}, a = 0.4H, Bdo c:0.25F,U.J:0V,and10 = 3A. a) Find r,(t) for t > 0. b) Find the filst three values of I for which do/dt is zero. Let these values of I be denoted t1, t2, and 13. c) Showthat 13 11= Zt. d) Show thaft2  11: Zd/2. e) Carcuhte?,(11), ?,(tr),and 0(t3). f) Sketch0(l) versustforO < t < t2. 8,12 a) Find 2(l) for I > 0 in the circuit in koblem 8.11 if the 2 O resistor is removed from the circuit. b) Calculate the ftequency of t (t) in hertz. c) Calculatethe ma,\imum amplitude of r,(t) in volts 8.13 Assume tlle underdamped voltage response of the circuitin Fig.8.1is *ritten as r(t) = (Ar + A)a"'cos a,i + j(At  Az\etsinadt The initial value of the inductor current is 10, and t}le initial value of the capacitor voltage is y0. Show tlat ,4, is the conjugate of,41. (lllrt Use the same processas oudinedin the text to find,41 and ,42.)
Prob.€ms 323 8.14 Show that the results obtained from Problem 8.13rharis.rhe expression\ tor,4r atrd.4rare consistent with Eqs.8.30and 8.31in the text
820 The resistor in the circuit of Fig. P8.19is inffeased tstrc from 1.6 kO to 2 kO and t]le inductor is deffeased from 1 H to 640 mH. Find zJ.(t) for t > 0.
8.15 The resistor in the circuit in Example 8.4 is changed strr! to 4000/\2 o.
8.21 The resistorir the c cuit of Fig.P8.19is deqeased 6dc from 1.6kO to 800O, andtle inductoris deqeased from 1 H to 160 mH. Find oo(t) for I > 0.
a) Find the numerical expressionfor o(1) when t>0. b) Plot rr(1) ve$us t for the time intenal 0 < t < 7 ms. Compare this response with the one in Example 8.4 (R=20kO) and Example8.5 (R  4 kO). In particular,compare peak values oI o(t) and the times when these peak values occur. 8,16 The switch in the circuit of Fig. P8.16 has beetr in 6plft position a for a long time. At t = 0 the switch moves instantaneously to position b. Find ?r,(t) for t>0. Figffe P8.16
t=0
Section 8.3 8.22 For the circuit in Example 8.6, find, for 1> 0, 6*E (a) ,0); (b) iR(t); and (c) tc(t). 8.23 For the circuit in Example 8.7, find, for , > 0, Em (a) ,,0) and (b) ic(1). 8.24 For the circuitin Example8.8,find o(t) for t > 0. 8.25 Assume that at fhe instant the 15 mA dc current Er!4 souce is applied to the circuit in Fig. P8.25,the initial curent in t}le 20 H inductor is 30 mA, and the initial voltage otr the capacitor is 60 V (positive at the upper terminal). Find the expressionfor ir(r) for t > 0 ifR equals800O. Figure P8.25
8.17 The capacitor in the circuit of Fig. P8.16 is decreasedto 1 rF and the itrductoris mueasedto 10H. Find o,(t) for 1 > 0. E.18 The capacitor in the circuit of Fig. P8.16 is decreasedto 800pF and t]le inductor is increasedto 12.5H. Find t"(t) for r > 0. 8.19 The two switchesin the circuit seen in Fig. P8.19 6flc operate synchronously.When switch 1 is in position a, switch 2 is in position d. When switch 1 moves to position b, switch 2 moves to position c. Svritch t has beenin positiona lor a long rime.Ar r 0. rhe switchesmove to their altenate positions.Find !,aG)forr>0.
l) '.,ullr" *,t 826 The rcsistancein the circuit in Fig. P8.25is changed Atrc ro 1250O. Fird ir(, for r > 0. 827 The resistancein the circuit in Fig. P8.25is changed tsdi! to 1000O Find ir(r) Ior, > 0. 8.28 The switch in the circuit in Fig. P8.28has been open E .r for a long time before closing at t = 0. Find o"(t) for r > 0. figureP8.28
Figure P8.19
1ko 60v 62.5nF
1.6kO
6.25pF
324
Natuntand SieoResDonses of{r Cncuits
8.29 a) For the circuit in Fig.P8.28,find i, for I > 0. """ b) Showthat your solutionfor i, is consistentwith the solutionfor r,. in Pioblem8.28.
Figure P8.33
1kO
830 Thereis no eneryystoredin the circuit in Fig.P8.30 6pr.rwhenthe switchis closedat r = 0. Find r"(t) for t>0.
,r 10/,r'
1.6H
FigoreP8.30
r:0'

834 Use the circuit in Fig. P8.33 'strcr a) Find the total energy deliveied to the incluctoi. b) Find tlle total energy deiivered to the equivalent
+
6.25pF,1c) Find the total energy delivered to the capacitor. d) Find the total energydelivered by the equivalenl cturenl source, e) Check the results of parts (a) through (d) against the conseraation of energy pdnciple.
8.31 a) For tle circuit in Fig. P8.30,find i, for I > 0. """ b) Show that your solution for i. is consistent with the solution for 0, in Problem 8.30. 8.32 The switch in the circuit in Fig. P8.32 has been 6trc open a long time before closing at t = 0. At the time the switch closes,the capacitorhas no stored energy.Find tto for t > 0.
8.35 The switch in the circuir in Fig. P8.35 has been EPI.[ open a long time before closingat t  0. Find iL(l) fort > 0. tigureP8.35
Figure P8.32
3000tr+ 31.25 rnH=500 72V
1 25rlF
8.33 The switch in the circuit in Fig. P8.33has been open ^"" a long time before closing at t : 0. Fhd a) ?o(t)for, > 0+, b) iro) foi t > 0.
8.36 Switches 1 and 2 in the circuit in Fig. P8.36are syn6trt! chronized. Wlen switch 1 is opened, switch 2 closes and vice versa. Switch t has been open a long tirne beforeclosirg at. : 0. Find ir0) for r > 0.
rigureP8.35 5kO
\ )240v
J w r c o1r v Switch
1
\,=0 Switch 2
5 sF
trl 8 0 H
1ko
(
60mA
PlobLems325
Section 8,4 8.37 The initial energy stoied in the 50 nF capacitor in the circuit in Fig. P8.37 is 90 /iJ. The initial energy stored in the inductoi is zero. fhe roots of the characteristic equation that describesthe natuml behavior of the current i are 1000 ai and 4000 s1. a) Find the numerical values of R and a. b) Find the numerical values of i(0) and di(0)/d1 lrnmediately after the switch has been closed. c) Find t(t) for t > 0. d) How many miqoseconds after the switch closes does the curent reach its maximum value? e) What is the maximum value of i h milliamperes? f) Fbd 0r(t) for r > 0. FlgureP8,37
FigueP8.40
{rvJ 72Ov
g.41 In the circuit in Fig. P8.41,the resistoris adjusted Mc for critical damping. The initial capacitor voltage is 90 V and the initial inductor current is 24 mA. a) Find the numerical value of R. b) Find the numerical values of i and dt/d/ immediately after the switch is closed. c) Find 0c0) for t > 0.
Flgure P8,4r
8,38 The current in the circnit in Fig. 8.3 is knom to be = Bre3m!cos600t + B2?{e sin 600t, I > 0. The capacitor has a value of 500 /rF; ttre initial value of the curent is zero; and the initial voltage on the capacitor is 12 V Fhd the values of R, a, 81, ard ,2.
8,39 Fhd the voltage aqoss the 500 ,.F capacitor for the circuit describedin Prcblem 8.38.Assume tle refererce polariry for the capacitor voltage is positive at t]le upper termiftl.
8.42 fhe switch in the circuit in Fig. P8.42has been In position a for a long time. At I = 0, the swrtch moves iNtartaneously to position b. a) What is the initial valueof?)d? b) what is the initial valu.eof dDJ dt? c) What is the nume cal expressionfor 2,,(t)for t>0?
FlgoreP8.42 8.40 The switch in the circuit shown in Fig. P8.40 has 6rft been closed for a long time. The switch opells at I : 0. Find a) t"(l) for t > 0, b) 0"(r) for t > 0.
326
Natu6tand StepResponses ofRlCCircuiis
8.43 The makebeforebreak switch in the circuit shown 6PIcr in Fig. P8.43has been in position a for a long time. At , = 0, the switch is moved instantaneouslyto position b. Find i(t) for t > 0. FigureP8.43
100mH
8.48 The switchin the circuit shownin Fig.P8.48has beenclosedfor a long time beforeit is openedat t  0. Assumethat the circuit parametersare such that the rcsponseis underdamped. a) Derive tle expressionfor o,(t) as a function of us, ct,od, C, andR fot t > O. b) Derive the expression for the valueof I when the magnitudeof r', is ma\imum. FlglreP8.48
8.44 The switch in the circuit shown in Fig. P8.44has B{c been closed for a long time. The switch opens at t : 0. Find ?a(t) for t > 0. tigun P8.44
30rI 10()
8()
10f}
100v
8.45 The initial energy stored in tle circuit in Fig. P8.45 6n(r
i. ?er^
Fin,l
r /,\ +^r,
>
n
8.49 fhe cftcuitparametenin the circuitof Fig.P8.48 a r e R = 1 2 0 O , a : 5 m H , C : 5 0 0 n F F ,a n d 0s = 600 V a) Expressr'"(t) numericallyfor t > 0. b) How manymiqosecondsafter the switchopens is t}le inductorvoltagemaximum? c) Wlat is the maximumvalue of the inductor voltage? d) Repeat(a)(c)withR reducedto 12O. 8.50 The circuit showtr in Fig. P8.50 has been in opera6'IG tion for a long time. At t = 0, the sourcevoltage suddenly drops to 100V Find o,(r) for r > 0.
tigureP8.45 Figure P8.50 4()
40 mH
8.,16the capacitor in the circuit shown in Fig. P8.45 is changed to 100 nF. The idtial energy stored is still zero Find od(t) for r > 0.
8.47 The capacitor in the circuit shown in Fig. P8.45 is changedto 156.25nF. The initial energy stored is still zerc. Find 0o(l) for t > 0.
8.s1 The swilch in the circuil of Fig. P8.51 has been in position a for a long time. At I = 0 the switch moves instantaneously to position b. Find
a) r'10') b) du.(o')ldt c) ?,'o(t)for t > 0.
tigureP8.51
the switch has been closed.At the time when the switch is closed, tlere is no energy stored in either the capacitoror the inductor. Find the rume cal values of Rand L. (H,nL Work Problem 8.53firsr.) 12kO
,.,,, mv
28v **+
8.52 The two switchesin ttre circuit seen in Fig. P8.52 sflft operate s]'nchronously.men switch 1 is in position a, switch 2 is closed.wlten srvitch1 is in position b, switch 2 is open. Svritch t has been in position a for a Iong time. At t : 0, it movesinstantaneouslyto position b. Find ?.G) for t > 0.
8.55 Show that, if no energy is stored in the circuit showr in Fig. 8.19 at the instant rJsjumps in value, then db./dt eqlralszeroatt =0. 8.56 a) find the equarion foruo(r) for 0 < r < r.urin the circuit sho$n in Fig.8.19if o,1(0) : 5 V and ?,,(0) = 8 V. b) How long does the circuit take to reach satumtion? 8.5? a) Rework Example 8.14with feedbackresistors R1 and R2removeil. b) Rework Example 8.14with r',(0) : 2v ^nd
0,(0)= 4v.
Figule P8.52
200r} 400O 100mH b\

+
160v1600o
8.53 Assume that te capacitor voltage in tle circuit of Fig. 8.15 is underdamped.Also assumethat no energyis stored in the circuit elementswhen the switch is closed. a) Showthat doc/dr = (SolaiVet sn(l.at. b) Show that doc/dt :0 when t = nr/r,,r, where n : O,7,2,.... c) Let t"= nnld,j, and show that !'c(t,)
:v  v (1)"e*"1"
d) Showthat "
1. oc1iV Td' uc(t)  V
whereZd = t3  tr,
854 The voltage acrossa 200 nF capacitor in the circuit of Fig. 8.15 is described as follows: After the switch has been closedfor severalseconds,the voltage is constant at 50 V The fbst time the voltage exceeds 50 Y it reaches a peak of 63.505V This occurs ?r/12 ms after the switch has been closed.The second time the voltage exceeds50 V it reachesa peak of 50.985V This secondpeak occursd4ms aftei
Section 8,5 E.5E The voltage signal of Fig. P8.58(a) is applied to tsr.r the cascaded integrating amplifiers shown in Fig. P8.58(b). There is no energy stored in the capacitorsat the instant the signalis applied. a) Derive the numerical expressionsfor to(t) and r',l(f) for the time intervals 0 < t < 0.2s and 0.2s
Figrre9,26Ir ThefrcqLrenrydomain version of a Norton
SinusoidaL St€ady.State Anatysis
PerfornlngSource Tnnsformations in the Frequency oonain Use the concept ofsource transformation to find tle phasor voltage Vo in the circuit shown in Fi& 9.27.
j3a
lo
o2o io6o
O2o i0.6o
Figure9.28 a Thefilstsiepjn reducing th€ circujtshov/n in Fig.9.27.
tigure9,27,t Theci(uitforExanple 9.9,
1.8O j2.4O
O.2A j0.64
Solution Wecanreplacethe sedescombination of the voltagesource(404) andtheimpedance of I  i3 O with the parallel combinationof a current source andthe 1 + j3 O impedance.The sourcecurrentis zt{l
40 = r ! _____::r + J r ;(l lu
_ /3) = a _it2A.
Thuswe canmodily the circuit shownin Fi9.9.27to the one shownitr Fig. 9.28.Note that the polarity referenceof the ,10V sourcedeterminesthe referencedirectionfor I. Next, we combirc the two parallel branches into a sitrgleimpedalce,
( 1+ i 3 x e  j 3 ) = 10
Also note that we have reduced the circuit to a simple series circuit. We calculate the current Io by dividitrg the voltage of the source by the total series
1.8+ j2.4 A,
which is in parallel with the cu[ent source of 4 i12 A. Another source tmnsformation converts this parallel combination to a se es combination consisting of a voltage source in serieswith the impedanceof 1.8 + t2.4 O.The voltageofthe voltage source is
v = (4
Figure9,29 ^ Thesecond st€pin reducing thecircuitshown in Fig.9.27.
36 jr2 12 jr6
. 
=
r2(3 jD 4(3  j4)
'1 0 + i ) 7 : = 1.56+ ,1.08A. It
j12X1.8+ j2.4) 36  j12V.
Using this sourcetransfomation, we rcdraw the circuitasFig.9.29.Notethe polarityof the voltage source.We addedthe current Io to the circuit to expeditethe solutionfor Yo.
We now obtainthe valueof V0by multiplying lo by theimpedance 10  i19: V0= (1.56+ i1.08)(10 j19) :36.12  jr8.84v.
9.i
Source Tmnsfomations andThaveninNorton Equivatent Circuiis 357
Findinga Th6venin Equivatent in the Frequency Donain Find the Th6veninequivalentcircuitwith respectto terminalsa.b for the circuit shownin Fig.93w.
j44{r
j40 (l
risure9.31A A sjmptified veBion ofthecircuit shown jn Fjg.9.30.
We relateth€ controllingvoltageY to the currentI by noting ftom Fig.9.31thal
tigure 9.30 A lhe circuitfo, ENample 9.10.
v,=10010L Then,
Solution We first determinethe Thdvenin equivalenl volf age.This voltageis the opencircuitvoltageappearingat terminaisa,b.We choosethe referencefor the Th6veninvoltageas positiveat terminal a.We can make two source transformations relative to the 120V 12 O, and 60 O circuit elementsto simplify this pofiion of the circuit.At the sametime, tiese transformationsmust prese e the identity of the controlling voltage \ becauseof the dependert voltagesource. We determinethe two sourcetransformations by first replacing t}le se es combination of the 120 V sourceand 12 O resistorwith a 10A currenl sourceill parallel with 12 O. Next, we replacethe parallel combinatior of the 12 and 60 O resistors $rith a single 10 O resistor. Fina y, we replace the 10 A souce in parallel with 10 O with a 100 V souce in serieswith 10 O. Figure 9.31 showsthe resultingcircuit. We addedthe currert I to Fig.9.31to aidturther discussion, Note that once we know the current I, we cancomputethe Th6veninvoltage.Wefind I by summingthe voltagesaroundthe closedpath in the circuitshownin Fig.9.31.Hence 100 : 10I
j40l + 120I + 10v":
(130  j10)I + 10v..
I :
30
900 : 18/ 126.8'7'A. J40
we now calculate va: Y,:
100
180/126.87' :20A + jt44V.
Finally,we note from Fig.9.31that Yrh=10V,+120I = 2080 + 11440 + 120(18)/  784 j)qQ, 815).
126.87"
)0.t7'V.
To obtainthe Th6veninimpedance,wemay use any of the techniquespreviouslyused to find the Thdvenin resistance.We ilustrate the testsource method in this example.Recall that in using this mcthod, we deactivate all independent sources Irom the circuit and then apply either a test voltage sourceoI a test current sourceto the terminaisof interest.The ratio of the voltage to the current at the sourceis the Th6veninimpedance.Figure 9.32 showsthe result of applyjngthis techniqueto the cncuit shownin Fig.9.30.Note that wc chosea lest voltage sourceVr. Also note that we deactivated the hdependentvoltagesourcewithan appropriate \horrcircuir andpresenedlhe idenrir)of V.
358
Steady StateAnatysis Sinusoidal
_ v.(e + i4)
i4o o
120(1 j4)
Ir=I.+It
9+i4\ 1 2 )
v? (. 10 140\ v?(3 12(10 YZrh:7:
i4) j40)'
0 12
/384O
equivateni forcatcutating thelh6venin Figure9.32a Acjrcuit Figure9.33depictstle Theveninequivalentcircuii. The branch curents I, and Ib have been added 10 the circuil to simplify the calculation of l? By straightforwardapplicationsof Kirchhoff's circuit laws,you should be able to verify the following relationships: T
=
10
138.4O
v  10L, cncuitshown equjvahntforthe Figure9.33 a TheTh€venin Fig. in 9.30.
lr' =
to sotvea circuitin the frequeicydomain techniques objective3Know howto useorcuit anatysis in expre.sionfor L'o(r) 9.10  ind lhe steadyslale lhe circuitshowoby u.inglhe lechniqueol the .inusoidalvollage sourcerranslormalions. solucesare
wilh respecr lo 9.11 FindrheThe!eni0equivalenl
. t6rmiuls a,bin thd circuit shown.
z,1= z0cos(4ooor+ 5:li3")v, zb : 96 si! 40001V. 15mH

,!.rl
"''" *: + 36.87')v . Answer: 48 cos(4000t ^'*:' ) 'IOTL: Al'o try Chapw ProblPnta 40.att. ando 1
Vir,= v"h: r0/4s' v;
9.8
TheNodeVoltage Method 359
9.8 TheNodeVoltage Method In Sectio s 4.24.4,we intioduced ttre basicconceptsof the nodevoltage method of circuit analysis.The same concepts apply when we use the nodevoltage method to analyze frequencydomain circuits. Example 9.11 illustrates the solution of such a cicuit by the nodevoltage technique. AssessmentProblem 9.12 and many of the Chapter Problems give you an opportunityto usethe nodevoltagemethodto solvefor steadystate sinu soidalresponses,
Usingthe NodeVottage Methodin the Frequency Domain Use the nodevoltage method to lind the branch currents Ia, Ib, and t in the circuit shown in Fig.9.34.
1() 10.6/E
iza
t l$ { 1r0ooo
v1(1.1+ 10.2) y2:10.6 + j21.2. Summingthe curents awaylrom node 2 gives
t ,
i)
5f,}
Multiplying by 1 +/2 and collecting the coefficientsof Y1and V2generatesthe expression
l t b
r.,+i5r) 20t,
v,
v2.v2
!

r+ tl
20I, ^
=
f
)
/)
The controlling current I* is Figure9.34 A Thecjrcujtfor Erampt€ 9.11.
I,:
Solution We can describe the circuit in terrns of two node voltagesbecauseit containstbree essentialnodes. Four branchesterminateat the essentialnode that stretchesaqoss the bottom of Fig.9.34,so we useit as the relerencenode.The remainingtwo essential nodesare labeledl and 2, and the appropriatenode voltages are designatedV and V2. Figuie 9.35 reflectsthe choiceof referencenode and the terminal labels.
1
j2A 2
J ; + , , . to

s0
20r,
Summingthe curents awayfrom node 1 yields
v
vv,
10.6++r_^:=0 ru r+ tz
1+j2'
Substituting tlis expression for I, into the node 2 equation, muitipling by 1 + j2, and collecting coefficientsof V1and V2producesthe equatron
sY+(4.8+J0.6)Yr:0. Thesolutionsfor V andV2are v = 68.40 j16.80V, v, = 68 j26 V. Hencetie bmnchcurrentsare I. :
t0
= 6.84 y'.68A,
v v
I, = ,I a Ib =
v,
v
Figur€9.35L lhecircuit shown in Pig.9.34,withthenode vottages defined.
YV
I,.: =
lz
20r=
......
J]
: 3.76+ lr.bSA, 1 44
j1192 A,
= 5 . 2+ 1 1 3 . o A .
To checkour work, we note that Ia + I. = = Ir = =
6.84  j1.68 + 3.76 + j1.68 10.6A, Ib + !c: 1.44 j11.92+ 5.2 + j13.6 3;76+ j1.68A.
360
Sieady StateAnatysis Sinusojdal
objective3Know howto lse circuitanatysis techniqlesto sotvea circuitin the frequency domaih 9.12 Use lhe nodc voltagemethodto lind the sleady fbr o0) in the circuitshown.The slateexpression sirusoidalsourcesare i = l0cos(,tArDd L,  100si d/ V. $ lrer 0, the step occurs to the ght of the origin, and if a < 0, the step occun to the left of the origin.Figure 12.4illustratesEq. 12.5.Note that the step function is 0 when the argumentI  /I is negative,and it is K when the argumentis positive. A step function equal to Kfor t < d is written as Ku(d  r). Thus Ku(a
t):K.
t €, the derivative is aa'(' (). Figure 12.9 shows these observations graphically. As € approaches zero,the valueof /'(t) betweenf€ approaches infhity. At the same time, the dumtion of this large value is apprcaching zero. Furthermore,the areaunder/'(t) between+€ remainsconstantas € +0. In this example, the area is unity. As € approacheszero, we say that the funcrion betweenre approachesa unit impulsefunclion. denored5(). Thus the derivative of/(/) at the origin approachesa unit impulse function as € approacheszerc, o{ /'(0) 6G)
:r + 0.5
Figure12,84 A magnified vjewofthe disco'rtjnujty in € Fis.12.1(b), assuftring a lineartransjtjon between
f'(! )
as€ +0.
If the area under the impulse function curve is other thar unity, the impulse tunction is denoted K6(t), where ( is the area.1 0 is sin dl; hen€ethe Iaplace transform is
e{sind1}:
l"
ut.un"' o,
= [("'';"'^)'.
Fiquer2.r5 A A sinEoidat tunction fo' I > 0.
I t l
=  t 
dt
2j
t

r \ " + lr)
112.20) Table 12.1 gives an abbreviatedlist of Laplace transform pairs. It includes the functions of most interest in an introductory coune on circuit aDDlications,
12.5 0peBtionat Tnnsfoms 475
Tlre
,f(, c> o) 6(r)
(imp'ilse)
u(t)
rG) 1
1 1 ;'
(ramp)
I
(exponenlial)
(cosine)
1
(danped ranp) (damped sine)
G+;Fia
12.5 0perationalTransforms Operational transforms indicate how mathematical operations oerformed oDeirher/(ri or F(,) areconvenedinto lheopposiridomajn.iteoperations of pdmary interest are (1) multiplication by a constantt (2) addirion (subhaction); (3) differentiatio{ (4) integmrion; (5) rranslarion in the time domain; (6) translation in the hequency domain;and (?) scalechanging.
Muttiptication by a Constant Fiom the defining integral, if
e{f(t)J = F(s),
s tKf (t)j : KF(r.
(12.21)
476
Introduction tothe Laplace T6nsfom
Thus,multiplicationof /0) by a constantcorespondsto multiplying F(s) by thesameconstant,
Addition(Subtraction) Addition (subtraction)in the time domair rranslatesinto addition (subtraction)in the ftequencydomain.Thusif
s{l(r} : r(r), s{/r(r)} = rr(r), s{/3(r)}= r3(t, then g{ft(t)
+ f2O  /:(4}
= r(")
+ Fls)  F3(s),
(12.2?)
which is dedved by simply substituting the algebraic sum of timedomain functions into the defining integral.
Differentiation Differentiation in the lime domain conesponds to multiplying F(r) by r and ttren subtracting t}le initial value of f(tthat is, /(0 )fton this prcduct:
' { + } = r F ( s )  r { o) .
112.23)
which is obtahed directly from the definition of the Laplace transfom, or
'{+\fl+)"",,
lrz.24)
We evaluate the integral h. Eq. 12.24 by integrating by parts. Letting u = e ", ajnddt = ld f (t)ldtldt yletds
.\+\ : n'',rl*I*',u,*''0,
\12.25)
Because we areassuming that /(t) is Laplacetransformable, the evalua tion of eY0) at I : oois zero.Thereforethe righthandsideof Eq. 12.25
f((D + sr
rr(r) /(0). ftr)e'.crt:
12.5 operationat Transtorms477 This observation completes the derivation of Eq. 12.23.It is an imporant result becauseit states that differentiation in the time domain teduces to an algehaic operation in the s domain. We determine the Laplace tmnsform of higherorder dedvatives by using Eq. 12.23 as the starting point. For example, to find rhe Laplace transtbm of the second dedvative of /(t), we first let
. .
df(t) 112.26)
dt
NowweuseEq.12.23to write G(r) = rF(s)
/(0 ).
\12.27)
But because ds(t) _ d2f(t\ ' dt dtz
"{ot:.")
 "{t'!!)
l d t )
l d f
)
scrr)  B(0).
(1?.28)
Combhing Eqs.12.26,12.27,and12.28gi'tes I atrr,tl .t O I ri=rj'i  "'rr', ,trot "'),'.
1tz.zs:
We find the Laplace ffansfom of the rth derivative by successively applying the preceding process,which leads to the general result
* l o ' t ' ! \ = s , F r , ) r , , / { o1  , ^ z l l t o t I d," I
d!
J,3af9) ,
d" lfg )
':,
Lrz,ru)
Integration Integration in the time domain correspondsto dividing by r ir the s domain. As before,we establishthe relationship by the defining integral:
'\lro^ =Ill[to,,,1,"""
(12.31)
478
Intrcductjon t0 theLaplace Transform We evaluatethe integralon the righthandsideofEq.12.31 by integrating by partq first letting
Then
: f(t)dt,
The htegratiorbypadsfomulayields
;f'r,r,,l;. .{[:o,*1= f 1,,r0,,,,, The first telm on the right'hand side of Eq. 12.32is zero at bot}l the upper and lower limits. The evaluation at t]le lower limit obviously is zero, whereasthe evaluation at the upper limit is zero becausewe are assuming that /(l) has a Laplace tiansfolm. The secondierm on the dghthand side of Eq. 12.32is F(s)/s; t}leiefore
'\lrta*j:?,
(12.33)
which rcveals that the operation of inte$ation in the time domain is transfomed to the algebraic operation of multiplyiDg by 1/r in the r domain. Equation 12.33and Eq. 12.30form the basis of the earlier statement that ttre Laplace transform translates a set of integrodiffeiential equations into a set of algebraic equations.
Translation in the TimeDomain IJ we start witll any function /(t),l(t), we can represetrt the samefunction, tra0slated in time by rhecon n, and an improp€r rational fir€tion if m < ,?. Only a proper iational function can be exparded as a sum of partial ftactions. This restriction posesro problem, as we show at the end ol this section.
PartiaIFractionExpansion: ProperRationa[Functions A prcper rational function is expanded hto a sum of partial fraclions by writing a telm or a sedes of tems for each rco1 of D(s). Thus D(s) must be in factored folm before we can make a partial fraction er?ansion. For each disainctroot of ,(r), a shgle term appearsin the sum of partial fraction$ For each multiple root of D(r) of multiplicity r, the expansion contains r tems. For examole.in t}re rational function
s + 6 r(r+3)(s+1)' ' the denominatorhasfour rcots.Tho of theseroots are distinctnamely, atr = 0ands  3. A multiplercot of multiplicity2 occursat s = 1. Thusthe pa ial fraction expansionof this functiotrtakesthe form Kz _ = Kr tl ,r . . l
s(r+3)(s+1),
Kr
K4 r ' l s r t l  '
1 1 2 1 3 )
The key to the partial ftaction technique for finding inveBe transforms lies in recognizing the /(t) coresponding to each telm in the sum of partial fractions. Frcm Thble 12.1you should be able to ve fy that
s+6 .^  + 3Xr + 1)' [s(s 3)(r = 1Y, + Kre3'+ K{e  + K4et)u(t).
(12.44)
All that remains is to establish a technique for detemhing tlle coefficients ((1, 1 0.
Use tlle results ftom Problem 12.32 and the circuit shownin Fig P12.32to a) Find 4(1) and i,(/). b) Find t1(6) and tr(co). c) Do the solutions for i1 and i make sense? Explain.
5(r' +8s+5) F(r) :
s' +4s+5
0
s' +25r+150 F(,) = s+20
12.43 Find /(1) for each of the following functions.
100(r+ 1) r,(s, + 2s + 5)'
a)
12.40 Find/(t) for eachof the followingtunctions: a)
18s' +66r+54 (r+1)(s+2)(r+3) 8r3+8912+311r+3oo
b)
r("): ( r20c +1f
0
s(s+2)(r' +8r+1s) c.)
o,,
17s2+172t+700
(r+2xr' +12r+100) 56s'+112r+5000 s(s' +r4r+625)
12.41 Find /(r) for each of the following functions.
4
s(r' ss+s0) rG)=
d)
e.)
sG+42 s4(r + 1)
12.,14 Derive ttre transfom pair given by Eq. 12.64.
12.45 a) Derive the transformpair givenby Eq.12.83. b) Derive the translormpair givenby Eq.12.84.
s'(s+ 10)
10(3s' +4s+4) r(r + 2)' ")
.7)
40G+ 2) r(s + 1)3
Sections12.80.9 12.46 a) Use the initialvalue theorem to find the initial valueof ?,in Problem12.26.
13 6s' + 15s+ 50 r' (f+4s+5)
b) Can tlre finalvalue theorem be used to find th€ steadystate valueof r'?W}ly?
f+os+s (s + 2)l t6s3 + 12tz + 216s
(s' +2s+5)'
728
Apply t}le hitial and finalvalue theorems to each transform pair in Problem 12.40.
.
12.8 Apply the inifial and finalvalue theoremsto each transfom pair in Problem12.41.
12.49 Apply the initial and finalvalue theorems to each tmnsformpai in Problem12.42.
probLems 505
12.51 Use the initial and finalvaluetheoremsto check the initial and final values of the current and vol! agein Problem12.28.
!2.52 Use the initial and finalvalue theoiems to check the initial and final values of the current in Problem12.30.
12.50 Use the initial and fhalvalue theoremsto check the initial and final values of the curent and vol! agein ?roblem 12.27.
Apply t}le initial ard finalvalue theorems to each transformpair in Problem12.43.
Transform TheLaplace in C'ircuit Analysis 13.1Circuitttenentsin thes Donainp. 508 in the 5 Domainp. 511 13.2CircuitAnatysis p. 512 13.3AppLications p. 526 13.4TheTransfer Function Funciion in PartiatFnction 13.5TheTransfer p. 528 Expansions Funchon andthe Convotution 13.6TheTransfer Integratp. 5J1 tunctionandthe 13.7lhe Transfer p. 5i7 SinusoidalResponse SteadyState in Circuit 13.8TheImputseFuncbon
1 Beabteto tGnsforma circuitinto the 5 donain usingLaptace t€nsforms;be sureyou howto rcpresent the initiat undeGtand in the conditjons on energystoEge €tements 2 Knowhowto anatyze a circujtin the s'domain ard be abLeto tlanfom an s'domainsotution backto the tine domain. the definitionandsiqnifica.ceof 3 Unde6tand ih€ transferfunctionandbe abt€to catcutate the hansferfurctior for a circujtusjng technjques. sdomain 4 Knowhowt0 usea cncuifstansferfunctionto jis calculate th€ cjrcuj{surjt inpulseresponse, andits steadystate unit steprcsponse, input. rcsponse to a sjnusoidal
506
The Laplace transform has two characteristicsthat make it an attractivetool in circuit analysis.Fi$t,it transformsa setof linear constantcoefficient differential equations into a set of linear polynomialequations,which are easierto manipulate.Second,it automaticallyintroducesinto the polynomialequationsthe initial valuesof the current and voltage variables.Thus,initial conditions are an inherent part of the transform process.(This contrasts with the classicalapproachto the solution of diflerential equations,in which initial conditions are consideredwhen the unknowncoefficientsare evaluated.) We beginthis chapterby showinghow we car skip the stepof w ting timedomainintegrodifferentialequationsand transfbrming them into the r domain. In Sectio[ 13.1,we'll develop the rdomaincircuit modelsfor resistors,inductors,and capacitorsso that we can write .rdomainequationsfor all circuits directly. Section13.2reviewsOhm's and Kirchhoff'slawsin the contextof the,t domain.After establishilgthesefundamentals,weapply the Laplace transform method to a variety of circuit problems in Section13.3. Analytical and simplificatior techniquesfirst introducedwith resistivecircuitssuch as meshcurlentand nodevoltagemeth ods and sourcetransformationscan be usedin the s domain as well. After solving for the circuit responsein the s domain,we inversetransformback to the time domain,usingpartial fraction expansion(asdemonstratedin the precedingchapter).Asbefore, checkingthe final timedomainequationsin terms of the initial conditionsand final valuesis an impo ant step in the solution process. The sdomaindescriptionsof circuit input and output lead us, in Section13.4,to the colcept of the transferfunction.Thetransfer function lor a particular circuit is the ratio of the Laplace transformof its output to the Laplacetransform of its input. In Chapters14 and 15,we'll examinethe designusesof the transfet function, bu1 here we focus on its use as an analyticaltool. We continuethis chapter with a look at the role of partial fraction
PracticaI Perspective Surge Suppressors personal Withthe adventof homebased computers, modems, fax machines, andothersensitiveeLectfonic equipment, it is necessary to provideprotectjonfromvoLtage surgesthat can occurjn a househotd circujtdueto switchinq. A comnefcialtyavaiLabte surgesuppressor is showf jn the accompanyjng figure.
Howcanfljppinga switchto turn on a ijght or turn off a hair dryercausea voltaqesurge?At the end of thjs chapter, we witl answerthat questjonusjng LapLace transformtech niquesto analyzea circuit.WewjlLitlustratehow a voltage surgecanbe createdby swjtchingoff a resistjveloadin a circuit operatingin the sinusojdaI st€adystate.
507
508
lhe tapa.eTEnsfo,m in CkcLrit Anatysis
er?ansion(Section13.5) and the convolution integral (Section13.6)in employingthe transfer function in circuit analysis.We concludewith a look at the impulsefuflctionin circuit analysis.
in the s Domain 13.1 CircuitElements The procedurelor developingan.rdomainequivalert circuit foi eachcir' cuit element is simple.First, we write the timedomain equation that relatesthe teminal voltage to the terminal culrent. Next, w.3lake the Laplace transform of the timedomainequation.This step generatesan algebraicrelationship between the sdomain current and voltage.Note that the dimension of a tmnsfomed voltage is voltseconds,and the dimension of a transformed curent is ampereseconds.A voltagetocurenl ratio in the r domain caries the dimensionof volts per ampere.An impedance in the s domair is measuredin ohms.and an admittanceis measuredin siemens.Finally,we constructa circuit model that satisfiesthe relation ship betweenthe sdomaincurrent and voltage.We use the passivesign convention in all tie deivations.
A Resistor in the s Domain We begin$riththe resistanceelement.From Obm's law, 1) :
Ri.
(13.1)
Because R is a constant, ttre Laplace transfom of Eq. 13.1 is
V=RI.
?
?
v:e.{D}
(13.2)
and I=gIi}.
"j n , ' t ! n j t I
Equation 13.2statesthat the r domain equivalentcircuit of a resistoris simply a resistanceof R obms thal caries a current of I ampereseconds and hasa terminal voltageof yvolt seconds. (a) (b) Figure 13.1 shows the time and frequencydomaincircuits of the resistor. Note that going froD the time domain to the frequencydomaiD Fig(re13.1Alhercsjstance ehnrent. G)rimedomain. doesnot changethe resistanceelemcnt.
t
i
. Ii
t tlllo ri
J
figure13.2A AnjnductoroflhenrXs canying an initialcunent of 10ampercs.
An Inductorin thes Domain Figure 13.2showsan inductor carrying an initial current of /o amperes. The timedomainequationthat relatesthe terminal voltageto the termi
"=,#.
(r3.3)
13.1 Circuit Etenents in ther Domain 509 The LaplacetransformofEq.13.3 gives
Y = rlsl
t(u )l = sLI
Lto.
(13.4)
TWo different circuit configurations satisfy Eq. 13.4.The first consists of an impedance of .rl, ohms in serieswith an independent voltage source ofalo vollseconds,asshownin Fig. 13.3.Note that the polarity marks on the loltage sourcealo agreewith the minus sign in Eq. 13.4.Note also that 110 carriesits own algebraicsign;that is, if the initial value of j is opposite to the reference direction for i, then 10has a negative value. The secondrdomainequivalert circuit that satisfiesEq.13.4consists Figure 13.3A Iheseries €quivatent cjrcujttor an oI an impedanceof rL ohms in parallel with an independentcurent inductor ofL henrys canying aniniijatcunent of 10 sourceof 1o/r ampereseconds, as shownin Fig. 13.4.We can dedve the alternativeequivalentcircuit shownin Fig. 13.4in severalways Ore way is simplyto solveEq.13.4for the currentl aIld then constructthe circuitto satisfythe resultingequation.Thus
', :
v+LIo
"z
= JV ' ; I o
!!
Two otier waysare:(1) {ind the Norton equivalentof the circuit shownin Fig.13.3and (2) startwith the inductor currert as a function ofrhe inductor voltage and then fiird the Laplace tmnsform of the resu]ting integral b equation.Weleavethesetwo approachesto Problems13.1and13.2. figure13.4 paraltelequivalent circuittor an Ifthe initial energystoredin the inductor is zero,that is,if10  0, rhe jnductorofA Th€ a henrys canying aninjtiatcurcntof10 sdomain€quivalentcircuit ofthe inductorreducesto an inductorwith an impedanceofsa ohms.Figure 13.5showsthis circuil.
*1 )
A Capacitor in thes Domain An initially charged capacitor also has two rdomain equivalent circuits. Figure 13.6showsa capacitorinitially chargedto y0 volts.The terminal current is
vlsLtl
I tigure13.5A TheJdomain circuitforaninductor rvhen theinitiatcurentisz€ro.
,
'Iiallsforming Eq. 13.6yields
? ,
't"
1 : C[sY  ?,0 )]
I Figure13.5A A capaciior ofC farads injtiattycharged to ro votts.
I=sCVCVa,
(13.7)
which indicatesthat the rdomain current1is the sum of two bmnch currents (Jnebranchconsistsof an admittanceofrC siemens. and the second branchconsistsof an independentcurrentsourceofCyo arDpereseconds. Figure13.7showsthis parallelequivalentcircuit.
510
IheLaptace Iransfom in circuitAnatysis
we derive the seriesequivalentckcuit for the charyedcapacitorby solvingEq.13.7for y:
(t)'.7
thc
b
rigure13.7A ThepaEltetequivaL€nt cjrcujt fora capacitor iniiialty charu€d to vovotts.
.r.
ri
hc
"T'
A Y
(13.8)
Figure 13.8showsthe circuit that sarisfiesEq. 13.8. In the equivalent circuits shown in Figs. 13.7 and 13.8, yo carries its own algebraic sign. In other wordi if the polarity of % is opposite to tlre reference polarity for o, yo is a negative quantity. If the initial voltage on the capacitor is zero, both equivalent c cuits reduce to an impedance of 1/rC ohms,asshowr in Fig.13.9. In this chapter, an important first problemsolving step will be to choose between the parallel or series equivalents when inductors and capacitors arc present. With a little forethought and some experience,the correctchoicewill often be quite evident.Theequivalentcircuitsare summarizedin Table13.1.
va/'
figure13.8a Theseries equivateni circuiifora jnjtiatly capaciior chaqed io v0votts.
*1, , *, t,,
FREQUENCYDOMAIN
TIME DOMAIN
.1" rl
,:R
.1" tl r5R
t,
,ln
V=RI
?=Ri
I
I b
Figure 13.9a Theidomain cjrcujttor a capacitor js zerc. when theinitiatvothge
.1" il ,lzlro
rl
I,
._r
i =!f" ,a' * 4
V=sLILIt'
* * T" ,, . +cyo
.l
" =lIo ia' + vo
lsc thc
A
I
Y
'
r
1
C
_. vr ZI as
V
r
"
13.2 Cncuit Anatysis in thesDomain511
13.2 CircuitAnalysis in the s Domain Befbre illustratinghow to usethe rdomainequivalentcircuitsin analysis, we needto lay somegroundwork, F st, we know that if no energy is stored in the inductor or capacitor, the relationshipbetweenthe teminal voltageandcurrentfor eachpassive element takes the form:
V=ZI,
(13.9) < ohm'stawin theJdomain
whereZ refersto the.rdomainimpedanceof the element.Thusa resistor hasan impedanceof R ohms,an inductor has an impedanceofra ohms, and a capacitorhas an impedanceof 1/sC ohms.The relationshipcon tained in Eq. 13.9 is also contained in Figs. 13.1(b), 13.5, and 13.9. Equation 13.9is sometimesreferredto as Ohm'slaw for thes domain. The reciprocal of the impedanceis admittance.Therefore, the s domain admittance of a resistor is 1/R siemens,an inductor has an admittance of l/sa siemens, and a capacitorhasan admittanceol sC siemens The rulesfor combiningimpedancesand admittancesin the.r domain are the sameas thosefor frequencydomaircircuits.Thus seriesparallel simplificationsand AtoY convetsionsalso are applicableto sdomain In addition,Kirchhoff'slawsapplyto sdomaincurrentsand voltages. Their applicability stems from the operational transfom stating that the Laplacetransformof a sum of timedomainfunctionsis the sum oI the transformsof the individualfunctions(seeTabte12.2).Becausethe algebraic sum of the curr€ntsat a node is zero in the time domain.the alsehrdic.um or rbe rran.formedcuffentcis dlso,/ero.A simrtarsrateminr holds for the algebraic sum of the transfomed voltages around a closed palh.Thesdomain verrionor Kirchhoifsla$ri.
alg>1 = 0,
(13.10)
alg>Y : 0.
(13.11)
Becausethe voltageand currentat the terminalsofapassiveelement are related by an algebmicequation and becauseK chhoff's laws still hold, all the techniquesof circuit analysisdcvelopedfor pure resistive networks may be used in r domah analysis.Thus rode voltages,mesh curents! sourcetransformations,and Th6venin,Nortonequivalentsare all valid techniques,even when erergy is storedinitially in the inductors andcapacitors. Initially storedeneryyrequiresthat we rnodifyEq.13.9by simply adding jndependentsourceseither in seriesor parallel with the element impedances.The addition ol these sources is governed by Kirchhoff'slaws.
512
jn CjrcujtAnatysjs Th€Laptace Tnnsform
transforms obiective1Be abteto transforma circlit into the 5 domainusinqLaDtace 13.1 A 500 O resistor.a 16 mH inductor,and a 25 nF capacitorare connectedin parallela) Expressthe adniltance oI lhis parallelcom' biMtion ofelementsas a ratioral function b) Computethe n mericalvaluesof tlle zeros and poles. Answer: (a) 25 x 10 e(s2+ 80,000r+ 25 x 103)/sr (b) z1 : 10,000 j30,000; : 40,000 + j30.000rpr = Lr. z2
13.2 The parallelcircuit in Asses$nenlProblem13.1 is placedin serieswith a 2000f) rcsistoI. a) Expressthc impedanceofthis seriescombinalion as a lational function ofj. b) Computethe Dumedcalvalueso[ thc zcros and poles. Answer: (a.)2000(.r+ 50,000)'/(." + 80.000s + 25 x 103); (b) .zr = ?, = 50,000; 40'000  j30,000, Pr = pr= 40,000+ /30,000.
NOTE: Also try ChapterProblems13.1and 13.5.
13.3 App!ie*t'i*iis we Dolvilluslralehow to usethc Laplacctransformto determineihe transient behavior oI severallinear lumped paramctcr circuits.We start b]' analyzingfamiiiar circuilsliom Chaptcrs7 and 13bccanscthey representa simple slarling piaceand bccauscthcy show that the Laplacetransform the easeof manipuapproachyieldslhe sameresulls.In all thc cxan1ples. lating algebraicequalions inslead of difibrcntial equations should be apParent.
TheNaturalResponse of an RCCircuit
Figurc13.10.,tThecapacitor djscharge circuit.
vo
we first revisit the natural responseof an RC circuit (Fig. 13.10)via Laplace transform techn'ques.(You may want to review the classical analysjsofthis samecircuit in Section7.2). Thc capacitoris initially chargedto vovolts.and we are interestediD the lime domain expressions ibr i and r.wc start by finding L ln transferring the circuitin Fig.13.10to thc r domain,wchavea choiceofrwo equivaienl circuilslor the chargedcapacitor.Bccrusewe are interestedin ihe culletrt.thesedesequivalenlcircuiIis more attractivc:it rcsultsin a singlenesh circuil in thc ircqucnc!,domain.Thuswe constructthe sdomaincircuil shownin Fig.13.11. Summingthc voltagcsaroundthe meshgeneratesthe expression
+:+,
+ RI.
fiqure13.11.ir AnJ domajn equivatent circuitfortlre circuitshownin Fig.13.10.
( 1 3r .2 )
Sol\ingEq.13.12for / yields I =
CVI
uulR
R C s + 1 .' + (fRC)
(13.13)
1 3 . 3 A p p l i c a t i o n s5 1 3
Note that the expressionfor 1is a proper rarionatfunctionoI s and canbe inversetransformed by inspcctioni i =
'iR':uO, ;e
(13.14)
which is equivalentto thc expressionfor the currentderivedby the classi cal mcthodsdiscussed in Chapter7.ln rhal chapter,rhc currentis givenby Eq.7.26,where7 is uscdin placeofRC. Aller we havc fouid i, the easiestwav to dctcrmine t) is simpllr to apply Ohm'slawrthat is,Irom the circuil, Rt = V,f 'R'ult).
u:
(13.15)
We Dowillustratca way to find o from rhecircuitwirhoutfirstfindingi. In this alternativeapproach,we retum to the original circuit of Fig. 13.10 cYo and lransferjt to the s domainusingthe parallelequjvalentcircuit for the chargedcapacitor.Using the parallel equivalentcircuit is atrractivenow becausewecandesc be the resultingcircuitin rcrmsof a singlenode volf Figure11.12s Anrdomain equivalent circuitforihe age.Figure13.12showsthe new.r domain equivalcntcircuit. circuitshownin Fiq.13.10. The nodevoltageequationthat describesthe ncw circuit is
t
lo, "c, = cu".
$ n
(13.16)
SolvingEq. 13.16tor Y gives
,,
uo s + (l/nc)
113.71)
InversetransformiDg Eq.13.17lcadsto the samecxpressionfor 1)givenby Eq.13.15, namciy. o = Ve 'tRc = Vae'tu(,t). (13.18) Our purposein deriving by direct use of the transformmethodis to showthat thc choiceof which.! domain equivalentcircuit io useis influerced by rvhichresponsesignalis ofinterest.
objective2l(now howto analyze a aircuitin ther domainandbeableto transform anJ domainsotutionto the trmedomain (b) t = 20? l'5oruG)mA, o1 = 80e r'5o'a(r)V, u2 = 20e 125o'u(t) Y.
13.3 The swilchin thc circuitshownhasbeenin positior a for a long lime.At I : 0, the smtch is thrown to positionb. a) Find 1,I4, and y, as rationalfunctionsof r. b) Findlh 0. e) Find iL for I > 0.
Pmbt€ms551 Figure P13.10 10nF
13.13 The switch in the circuit in Fig. P13.13 has been strd closed for a long time before opening at t = 0. a) Conshuct the rdomain equivalent circuit for
r>0. b) Find%. /1b.r=0
c) Find ?], for r > 0.
tigureP13.13 13.11 The makebeforebreak swilch ir the circuit it EIIG Fig.P13.11hasbeenin positiona for a long time.At t : 0, it moves instantaneously to position b. Find ?)ofort> 0
50f)
100
Figure P13,11
5H 15kO
l3.1rl The nakebeforebreak switch in the circuit seen in tstrG Fig. P13.12 has been in position a for a long time before moving instantaneouslyto position b a = 0. a) Construct the sdomain equivalent circuit foi t>0. b) Find 1, and t,. c) Find % and z',.
13.14 The switch in the circuit seen ir Fig. P13.14has 6Ec beenin positiona for a long time.At , : 0,itmoves instantaneously to position b. a) Find %. b) Fhd o..
FiguIe P13.12 tigureP13.14
!,, 5ILF
Iransfom in circujtAnatysjs 552 ThetapLace 13.15 Thereis no energystoredin the circuitin Fig.P13.15 Eoe at thetimetheswitchis closed.
FigureP13.18
a) Find lJ, for t > 0. b) Does your solution make sense in terms oI known circuit behavior? Explain. figurePl3.15
13.19 The switch in the circuit in Fig. P13.19has been EDft closedfor a long time. At r : 0, the switch is opened. a) Find ?).(t) for t > 0. b) Find j,(t) for t > 0. Fig(reP13.19
13.16 The switchin the circuit in Fig. P13.16has been in 6trft positiona for a long time.At,  0, it movesinstartaneously from a to b. a) Constructthesdomaincircuit for I > 0.
+
b) Find 1"(r). c) Find i,(t) for t > 0.
1)o
Figurc Pl3.16
i
0.5 H
13.20 Find % and o, in the circuil shownin Fig. P13.20if EPla the initial energy is zero and the switch is closed
Figure P13.20 L2A
5/"F
l20v
20mH 13.17 Find x'" in the circuit shown in Fig. P13.17 if "PIc ts  154(t)A.There is no energystoredin fhe circuitatr:0. tigureP13.17
200mH
13.21 Repeat Problem 13.20if the initial voltageon the 6trc capacitor is 20 V positive at the lower terminal. 13.22 a) Find the sdomain expression for % in the circuir in Fig.P13.22. b) Use the rdomain expressionderived in (a) to predict the initial and finalvalues of r,,. c) Find the timedomainer?ressionfor ?,,. Figure P13.22
7ko 13.18The switchin the circuitin Fig. P13.18has been 6r.r closedfor a longtimebeforeopeningat I : 0. Find ?),fort> 0,
.6u(t)mA.T3.2nF
13.23 Find the timedomain expiession for the curent in 64c Urecapacitorin Fig. P13.22.Assumethe reference direction for ic is down.
FlgufeP13.26
30tu(r)v 1324 There is no energy storcd in the capacitors in the E ic circuit in Fig. P13.24ar rhe time the switch is closed. a) Construct the sdomain circuit for t > 0_ b) Find 1r,I{, ard Yr. c) Find ir,01, and t2. d) Do your answersfor i1, 01,and 1,2make sensein terms of known circuit behavior? Explain.
13.2? There is no eneryy stored in the ctucuitin Fig. P13.27 6trG at the time the sourcesare energized. a) Find 11(s)atrd1,(s). b) Use the initial and 6na1valuetheoiems to check t}le initial and fhalvalues of llc) and tr(t. c) Find 4(1) and ir() for r > 0.
figureP13,24
Figur€ P13.27 4OO
*lr
2.5rF 1OH
5qTF
1s4(r)A
60,(r)V
13.25 There is no eneryy stoied in rhe circuir in Fig. p13.25 6rlc at the time the voltage sourceis energized. a) Find % and 1". b) Find ?)oand i, for, > 0. 1328 There is no eneryy stored in the circuit in Fig. P13.23 EPr.r at the time the voltage souce is tumed on, and 'os= 75u(t) Y a) Find % and 1,. b) Find 0" and to.
Figure P13.25 400()
50,000r"3or(r)
+
5H r,
c) Do tle solutions ior 1). a'nd,i. make sense in tems of knowl circuit behavior? Explain. figllreP13.28 4mF
13.26 There is no energy stored in the circuir in Fig. p13.26 a) Use the node voltage method to find 0o. b) Findrhetimedomaine\pression lor 1,. c) Do your answeNin (a) and (b) make sensein tems of known circuit behavior? Explain.
554
jn Circuit Analvsis Transtom TheLaptace
1319 The initial energy in the circuit in Fig. P1329 is xnc zero.Theideal.voltagesourceis 600&(t)V
tiglre P13.31
a) Flnd %(r). b) Use the initial and finalvalue theorems 10 find t'o(o') and 0,(oo). c) Do the values obtained in (b) agree with kno*r circuit behavior? Er?larn. d) Find 0,(4. Figurc Pt3.29 at 4
2oH
1oo __L_

lT"*
'
140 f,l
d) Find i,(1)
13.30 Thereis no eneryystoledin t}Iecircuit ir Fig P13.30 Edc
att:0
13.32 There is no eneryy stored in t}!e circuit in Fig. P13 32 Btrc at the time the current source turns on. Given that js = 100'(, A: a) Find 1,(s). b) Use the initial and final_value theorems to find i(0") and,.(co) c) Detemine if the results obtained ill (b) agree vrith known circuit behavior.
.
a) Find %. b) Fhd t',. c) Does your solution for 1'omake sensein tems of kDown circuit behavior? Er?lain
P13.32 Figure 20 i4
;
25H
zs()
P13.30 Figure
13,33 Beqinninawith Fq. l3.b5,.bowlhal lhe capacitor
3&(r)A
15a(,)v
cu;e01 ; the cjr;uil in Fig. 13 L9 is posidte lor 0 < I < 200ps and negahvefor I > 200ps Also show that at 20Ops, the current is zero and that t\is correspondslo when due/dr is teto
13.34 The switch ir the circuit shown in Fig. P13 34 has 13.31 There is no energy stored in the circuit in Fig. P13 31 Ed( at the time the current souce is energized. a) Find 1aand Ib. b) Find i. and tb. c) Find %, Vb,and %. d) Find t)., tb, and ttc. e) Assume a capacitor will break down whenever its terminal voltage is 1000V How long after the current soluce tulns on will one of t]re capacltors break down?
been open for a long time. The voltage of the sinusoidal sourceis o, = y..io1tt + d) The switch closesat t = 0. Note that tlte angle d in the voltage exprcssion detemines the value of the voltage at thi moment when the switch closes, tllat is, 0s(0)  Y sin C. a) Use the Laplace tmnsform method to find ifort > 0. b) Using the expiession derived in (a), write the expression for the current after the switch has beenclosedfor a long time.
Probtems 555 c) Using the e4ression derived in (a), write tle expressiotrfor the transient componelt of j. d) Find the steadystate expression for i using the phasor merhod.Verify that 'our eypressionis equivalentto that obtainedir (b). e) Specify the value of d so tlat the circuit passes directly hto steadystate operation when the switch is closed.
13,37 a) Find the cunent in t]le 4 kO resistor in the circuit in Fig. P13.36.The reference direction for the current is dowll through the resistor. b) Repeat part (a) if the dot on the 12.5H coil is revefsed,
13.38 The magneticaly coupled coils in the circuit seen trPI" in Fig. P13.38 carry initial currents of 3 and 2 A,
Figue Pt3.34
r:o
a) Find the initial energy stored ir the circuit. b) Find 11and 12. c) Find il and iz. d) Find the total energy dissipared in the 600 and 1350O resistors. e) Repeat (a) (d), witl the dot on the 90 mH inductor at the lower teminal.
rJ
13.35 The two switchesin the circuit shown in Fig. P13.35 E?'.r operate simultaneously.There is no energy stored ir the circuit at the instant the switches close.Find ilt) fot t > 0' by first finding the sdomain Th6venin equivale[t of the circuit to the left oI the terminalsa,b.
tigureP13.38 30nH
f
/\ I r / r' ) 4 0 m H )
6ooot
I
Figurc P13.35
.]r\l. 1
.]l  /\  ( 9 o m r t rr , ) l

,'"1 \,L
/4r:soo
I
13.39 In t})e circuit in Fig. P13.39,switch 1 closesat I : 0, o"a and the matebeforebreak switch moves instantaneously lrom position a to position b. a) Construct the sdomain equivalent circuit for l>0. 1336 Therc is no ene4y stored in the circuit in Fig. P13.36 BtrG at t]le time the switch is closed. a) Find 11. b) Use the initial and finalvalue theorems to fhd ,1(0*)and 4(co). c) Find i1.
b) Find 1r. c) Use the initial and finalvalue theorems to checklhe initialandtinai!aluesoi i2. d) Find i2 foi t > 0*. Figure P13.39
Figure P13.36 1.5H 0.5H
1ko
1H
556
TheLaptace TEnsforn in Cncuit Anatysk
13.40 The makebeforebreak svritch in t}le circuit seenin 5m Fig.P13.40hasbeenin positiona for a long time.At t = 0, it movesinstantaneouslyto position b. Find t,forl > 0.
13.44 Find rJ,(t) in t})e circuit shown in Fig. P13.44if the Erc ideal op amp operates within its linear range and .,s = 400u(r)mV.
figureP13.44
Flgure P13.40
4nF
13.41 The switch in the circuit seen in Fig. P13.41has 6flft beenclosedfor a long time beforeopeningat, = 0. Use the Laplace transform method of analysisto find i,. 13.45 The op amp in t}le ciicuit showr in Fig. P13.45 is aac ideal. There is no eneigy stored h the circuit at the time it is energized.If 1," = Z6,06gtr1t,t, t.a (a) %, (b) ?,.,(c) how long it takesto satuiatethe operational amplifier, and (d) how small the rate of inqeasein 0s must be to preventsatumtion.
FigureP13.41
72Y
2H
50f,)
Figure P13.45 10nF
13,42 There is no energy stoied in the circuit seen in 6PIa Fig. P13 42 at the time the two so[ces are energized. a) Use the pinciple oI superpositionto find %. b) Find t), foi t > 0.
FigrreP13,42 2ko 60a(r)V
50 
1H
soonF ( I )l2(rlnA 4 k o
13.92for /2 13.43V€rifythatthesolutionofEqs 13.91and yieldsthesameexpression asthatgivenby Eq.13.90.
13.46 The op amp in the circuit shown in Fig. P13.46 is E rc icleal.Thereis no energystoredin the capacitorsat the instantthe clrcuit is energized. a) Find 2,,if Isr = 16!10)V and z)s,: 8u(r) V. b) How mary millisecondsafter the two voltage saturate? sourcesare turned ondoesthe op aIl1lp
figureP13.46
Seciions 13.+13.5 10 nF
13.49 Find the numerical expressionfor the transfer func, tion (y"/y, oI each circuit in Fig. P13.49 and give the numerical value of the poles and zercs of each traNferfunction. figureP13.49 25kO
800
rl
15kO
13,47 The op amps in the circuit show]) in Fig. P13.47are sPIt! ideal. There is no energy stored in t}le capacito$ at t : 0.If0s = 180r(t) mV, how manymilliseconds elapsebefore an op amp saturates?
5nH
Figure P13.47
800ko 400k()
i 13.48 The op amp in the ctucuit seen in Fig. P13.48 is tsdc ideal.There is no energy stored in the capacitors at the time the circuit is energized. Determine (a) %, (b) r,,, and (c) how long it takes to saturatethe operational amplifier. Figure P13.48 100nF
13.S0The operational amplifier in the circuit in Fig. P13.50 isideal. a) Find tlre numerical expression for the transfer tncrion H(s) : U"/Vs. b) Give rhe numericalvalue of eachzeio and pole of 11(s). Figure P13.50
100nF
250nI
l0 ko s09,nF 400ko
6u(r)V
400ko
1 2 . 5V
i
t
;
Transform in CircujtAnatysjs 558 Th€Laptace
13,51 The opemtion amplifier in the cncuit in Fig. P13.51 is ideal. a) Derive the nume cal er?ressionof the transfer funcrion H(r:U/vs for the circuit in Fig.P13.51. b) Give the numedcal value of each pole and zero of H(s). tigureP13,51
Figure P13.53
400ko + 62.5nF 400ko 800kf)
1354 In the circuir of Fig. P13.54 i, is ttre output signal
8nF
and ?rsis the input sEnal. Find the poles and zeros of tlre traNfer tunction.
62.5kO
Flgure P13.54
;
13.52 The operational amplfier in the circuit in Fig. P13.52 is ideal. a) Find fie numerical expression for the ttansfer lnncnon H(s)  WVg b) Give the numedcal value of each zero and pole of 11(r). rigureP13.52 Cr=,lnF
Rr:25kO
13.55 There is no energy storcd in the circuit in Fig. P13.55 Epr.r at uhetime the switch is opened.The sinusoidalcurient sourceis generatingtho signal60 cos4000t mA. The responsesignalis the current L. a) Find the tiansfer function 1,/1s. b) Find 1,(s). c) Descdbethe naturc ol the transientcomponent of i,(t) without solving for to0). d) Desoibe the mture of the steadystatecomponent of io(r) without solvhg for t"0). e) Verify tle observations made in (c) and (d) by finding i,(r). Figure P13.55
13.s6a) Find the transfer function /o//s as a function of 13.53 a) End the numerical exprcssionfor the tiansfer tunction H(s) = U.lU for the circuit tn Fig.P13.53. b) Give the numericalvalueof eachpole and zero of 11(s).
o
/r.for the circuit seen in Fig. P13.56. Find the largest value of p that will prcduce a bounded output signalfor a bounded input signal. Find i" for rl = 0.5, 0, 1, 1.5, and 2 if is : r0&(,) A.
Pobtems559 FigreP13.56 0.5k()
t ) '" Srko
10H
c) Repeat(a) when ft(l) changesto the rectangular pulsesho*n in Fig. P13.60(c). d) Sketch ](r) venus I for (a)(c) on a single graph. e) Do lhe skercbe. iD(d) makesen5e? E\plain. Figure P13.60
o(') I
Section 13.6 13.57 A rcctangular voltage pulse ?'i  [!r0)  u0  1)] V is applied to the circuit in Fig. P13.57.Use rhe convolution integal to find 0o.
orl
F1
4 0 r
4 0 t (a)
Figure P13.57 1H
13.58 Interchange
the inductor and resistor in Problem 13.57 and again use the convolution inte$al to find I'o.
13.59 a) Find ,(t) * r(r) when ,(r) and r(t are rhe rectangularpulsesshownin Fig.P13.59(a). b) Repeat (a) when r(t) chatrgesto the rectangular pulseshov,nin Fig.P13.59(b). c) Repeat(a) when n(t) changesto tle iectangular pulseshom in Fig.P13.59(c). Figur€ P13.50
13,61 The voltage impulse iesponseof a circuit is showr in Fig. P13.61(a).The input signal to the circuit is the rectangulai voltage pulse sho$,nilt Fig. P13.61(b). a) Dedve the equations for the output voltage. Note the range of time for which each equation is applicable. b) Sketch ?J,for 0 < t < 27 s. figurcP13.61
/'(t)(\t (a)
v,(^)(v) 13.60a) Giveny0) : nG) * "0), fi"d y(r) whent(r) and r0) are the rectangularpulses shown in Fig.P13.60(a). b) Repear(a) whenft(t) changesto the rectangular pulsesho$,nin Pig.P13.60(b).
560
lhe taptace T€nsfom in Circuii AnaLysis
13.62 Assume the voltage impulse response oI a circuit can be modeled by the tdangular wavefom sho\ln in Fig. P13.62.The voltage input signal ro fiis circuit is the step function 4r0) V. a) Use the convolution integral to derive the expiessionsfor the output voltage. b) Sketch tle output voltage over the intenal 0to25s. c) Repeat parts (a) and (b) if the area under the voltage mpulse responsestaystlle samebut the width of the impulse responsenarrows to 5 s. d) Which output waveform is closer to replicating ttre input waveform: (b) or (c)? Explah. Flgure P13.52
13.63 a) Assume the voltage impulse responseof a circuit is h(t):
{l;,'
tigureP13.54
I (pt
13.65a) Repeat Problem 13.64,given that the resistor in the circuit in Fig. P13.49(d)is increasedto 400 O. b) Does inqeasing the resistor ircrease or deqease the memory of the circuit? c) mich circuit comes closest to transmitting a replica of the inpur volrage?
l:t.66 a) Use the convolution integml to find i" in tlle cilcuit in Fig. P13.66(a)iI is is the pulseshown ln Fig.P13.66(b). b) Use t}le convolution integral to find z,o. c) Showrbaryoursolulions for r, andi, areconsistent by calculating o. a'r.d i" at 1 ms, 1 ms, 4 ms,and 4 ms. Figure P13.55
t 0+. b) Does your solulion make sense in tenns of known circuit behavior? Explain.
Figure P13.88 40nF
Figure Pt3.85
Seciioff 13.113,8
t3a7 fhere is no energy stored itr the circuit itr Fig. P13.87 at t]le time the impulse voltage is applied. a) Find ir for t > 0*. b) Find i for t > 0+. c) Find 0, for t > 0.
1:1.89Assumethe linetoneutralvoltageV, in the 60 Hz p:llflljl}lcircuit of Eg. 13.59is 120l! v(rms). Load R, is absorbhg1200W; load Rr is absoibing1800W and load Y, is absorbing350 magnetiziq VAR. The inductivereactanceof the line (X1) is 1 O. Assume Vsdoesnot changeafter the switchopens. a) Calculatethe inirial valueof t,0) and i,(t). cirb) FindV,, o,(t),and?,(0*)usingthesdomain cuitof Fig.13.60. c) Testthe steadystate componentof z', usingphasor domainanalysis, d) Using a computerprogram of your choiceplot 1lovs.tfor0 = t = 20ms.
d) Do 'our solurioD.for ir. r). and ,o makesensein termsofkDoqDcircuilbehariorltyplain.
FlgureP13.87
mD(r) v
13.90Assumethe switch in the circuit in Fig. 13.59 opensat the instant the sinusoidalsteadystate J#;lI#jE voltage oo is zero and going positive, i.e., uo : 120\/t sin 120ntV . a) Findto0) for t > 0. b) Using a computerprogiam of your choiceplot r,"(t)vs.tfor 0 < I < 20ms. c) Conpare the disturbancein the voltage in part (a) with tlat obtained in part (c) of Problem13.89.
Problems 565 13.91 The purpose of this problem is to show that the pi$fi*iiElinetoreutral voltage in the circuit in Fig. 13.59 can go directly into steady state it the load R6 is disconnectedfrom the circuit at precisely the right time. Let ?, = V^cos(I2jnt d') V, where y = 120\4. Assume ?s doesnot changeafter R, is disconnected.
a ) Find the value of 0 (in degrees) so tlat o, goes
directly into steadystate operation when the load Rbis discomected. b) For the value of d fourd in part (a), fird ?,,(1)Ior t>0. c) Using a computer program of your choice plot, on a single graph, for 10ms < l < 10ms, o,(t) befoie and after load R1,is disconnected.
Introduction to Frequency )elecuveLrrcurts 7
14.1 SomePretiminari€sp. 568 14.2 LowPasslilters p. 57A 14.3 tlighPassFilters p. 577 14.4 Bandpass fiLters p. 582 145 Bandrdectfitters p. 59t
1 Kiowthe ffr and8f circujtconfiguratjons that act as tow'pass fitteruandbe abteto desigi Rl andRtcncujt component vatuesto meeta specifi ed cutofffr€quency. 2 (nowthe {L andRf cncuitconfiguntjoisthat act as hjgh'passfittersard be abteto desjgn Rl andff cjrcuitcomponent vatuesto meeta specifi ed cutofffrequency. 3 Xnowthe Rrf cncuitconfig!rations ihat act as bandpass fitters,understand the definjtlonof andr€tationship amofqthe certerfiequency, cutofffrcqLrencj€s, bardwidth,andquatity ractorof a bafdpassfitter,andbe abteto desiqnRifcncuit compofentvatuesto meet designspecificatjons. 4 Knowthe RrCcncuitconfigurations that act as bandreject fitters,understand ihe definitionof andrctationshjp amongthe certerfrequency, cutofffieqlencj€s,bandwjdth, andquaLity ractorof a bafdrejectfitter,and be abteto designRtrcircuit componeitvaluesto me€t desiqnspecifications.

I
.
t^.
Up to this point in our analysisofcircuitswith sinusoidalsources, the sourcehequencvwas held constant.In this chapter,we analyze the effectof varyingsourcefrequencyon circuit voltagesand currents.Thc rcsult of this analysisis the frequencyresponseof a clrcurl. We've seen in previous chaptcrs that a circuit's response dependson the typesof elementsin the cjrcuit, the way the elements are connected. and the impedance of the elements, Although varying thc frcquencyof a sinusoidalsourccdoesnot changethe elementtypes or their connections,it does aLtelthe impedanceofcapacito$ and inductors,becausethe impedanceof theseelemcntsis a function offrequency.Aswe will see,thecareful choiceof circuit elenel1ts,their values,and their connections to other elementsenablesus to constructcircuitsthat passto the output only those input signalsthat residein a desiredrange of frequencies. Such circuits are called frequency.selective circuits. Many devicesthat communicatcvia clcct c signals,suchas tclcphones, radios, televisiolls,and satellites,employ fiequencyselectivecircuits. Frequencyselective circuits ale also calledfilters becauseof their ability to liller oLltcertair input signalson the basisof frequellcy.Figure 1,1.1on page 568 representsthis ability in a sim plistic way.To be more accurate,we shouldnote that [o practical liequency selectivecircuit cafl pefcctly or completelyfilter out selectedftequencies.Rather,filtersattenuatethat is,weakenor lessenthe effect ofany input signalswith tiequenciesoutside frequenciesoutside a particular frequency band. Your homc stereosystcmmay havea graphicequalizer,which is an excellent exampleof a collectionof lilter circuils.Each band ill the graphic equalizeris a lilter that amplifiessounds(audiblc ftcqucncics)in thc flequencyrangeof lhe band and attenuateslrequenciesoutside of that band. Thus the graphic cqualizcr cnablcs you to changcthe soundvolume in eachfiequencyband.
PracticaI Perspective Pushbutton Telephone Circuits In this chaptet we examinecircuitsin whichthe sourcefrequencyvaries.The behavjorof these cjrcuitsvariesas the sourcefrequency varj€s,because the impedance of the reactive components is a functjonof the sourcefrequency. These frequency dependent circujtsare caLLed filteB ano are useo ir manycommonetectricaldevices. In radios,fitte15areused to setectoneradiostation'ssignalwhilerejectjfgthe signats from otherstransmjttingat differentfrequencjes. In stereo systems, fittersareusedto adjusttheretativestfengthsofthe Lowand highfrequeicy components of the audjosignaL. Fr'tters areatsousedthroughouttetephone systems. prcducestonesthat you hear A pushbuttonteLephone whenyou pressa button.Youmayhavewonder€d aboutthese tones.Howareth€y usedto tetl the tetephone systernwhjch button was pushed? Whyare tonesusedat att?Whydo the tonessoundmusr'cal? Howdoesthe phonesystemtel[ the dif ferencebetweenbuttontonesandthe normatsoundsof peo pletatkingor sjnging?
The tetephonesystemwas designedto handleaudio sjgnatsthosewith frequencjes between300 Hz ard 3 kHz. Thus,alt signalsfrom the systemto the user have to be audibte incLuding the diattoneandthe busysignal.Similarly, att signabfrom the userto the systemhaveto be audjbLe, incLuding the sjgratthat the userhaspressed a button.It js importantto djstjngujsh buttonsignalsfromthe normaL audjo (DT[4F) 5ignat,so a duattonemuttiptefrequency d€signis employed. Whena numberbuttonis pressed, a uniquepairof sinusoidaL toneswjth ve[/ preckefrequencies is sent by the phoneto the telephone systern. TheDTI4F frequency andtiming specifrcations makeit unLikety that a humanvoicecouLdpro ducethe exacttone pairs,evef jf the personweretryjng.In the centralteLephone facitjty, electriccjrcuitsmonitorthe audiosignat,ljsteningfor the tonepairsthat signaL a number. In the Practical Perspective exampte at the endofthe chapter, we witLexamine the designof the DTMF filtersusedto determinewhjchbuttonhasbeenpush€d.
t
568
Introduction to Frequency S€lective Cjrcuits
We begin this chapter by anallzing circuits fiom each of the four major categoriesoffiltels: low pass,high pass,bandpass,and band reject. The tmnsfei function of a circuit is the starting point for the frequency figure14.1A Theactionofa fitieronaninputsiqnat responseanalysis Pay close attention to the similarities among ttre transtesuLrs in anoutputsignat. fer functions of circuits that pedorm the same filtedng function. We will employ these similarities when designing filter circuits in Chapter 15. Inout sgnrr
tl
ourout srgnaL
14.1 SomePretiminaries Recall from Se€tion 13.7that the txansferfunction of a cicuit provides an easyway to compute the steadystateresponseto a sinusoidalinput. Therc, we consideredonly fixedftequency souices.To study the ftequency response of a circuit, we replace a fixedfrequency sinusoidal source with a varyingfrequencysinusoidalsourc€.The hansfer function is still an imnensely useful tool becausethe magnitude and phaseof the output signal depend only on the magnitude and phaseof the transfer function fl(j.r). Note that tlte approach just ouUined assumesthat we can vary tlle frequency of a sinusoidal souce vrithout changing its magnitude or phase angle.Therefore, the amplitude and phase of the output will vary only if those of the transfer function vary as the frequency of tle sinusoidal sourceis chalged. To turther simplily this first look at frequencyselective circuits, we will also restrict our attention to caseswhere both the input and output signals are sinusoidal voltages,as illustrated in Fig 14.2.Thus, the transfer function of interest to us will be the ratio of the Laplace transform of the ouQut voltage 1o the Laplace tramform of the input voltage, or  %(s)/I4(s).We shouldkeepin mind,however,thatfor a particular Figure 14.2 A circujt withvottage inputandoutput. H(s) application, a current may be either the input signal or output signal of intercst. The signals passedfrom the irput to the output fall withit a band of frequencies called the passbatrd.Input voltages outside this band have their magnitudesattenuatedby the circuit and are thus effectivelyprevented ftom reaching the output terminals of the circuit. Frequencies not in a circuit's passbandare in its stopband.Frequencyselectivecircuits arc categodzed by tlle Iocation of the passband. One way of identifying the t)?e oI frequencyselective circuit is to examhe a ftequency responseplot. A frcquency responseplot showshow a ckcuit's transfer function (both amplitude and phase) changes as the sourceftequency changesA frequency responseplot has two parts. One is a graph of l1t(io)l versus frequency d. This part of the plot is called the magnitude plot.The other part is a graph of du(o) ve$us frcquency .). This part is called the phas€ angle plot. The ideal ftequenry responseplots for tle four major categoriesof filtels are shown in Fig. 14.3.Pa s (a) and (b) ilustrate the ideal plots for a lowpass and a highpass filter, rcspectively. Both filten have one passband and one stopband, which are defined by the cutofi ftequency that separatesthem. The names low pass and high pasr are derived from the magnitude plots: a low.pass filler passessignals at frequencies lower than the cutoff frequency from the input to the output, and a highpass filter
14.1 SomePretiminanes569
H(i.)l I
elia) o(i.") 0"
I
e0",)
o(i.)
o(i0\) 0'
0'
e(ia")
Figure plotsofthefourq,pes 14.3& Ideatfrequency r€sponse offiLter circujts. (a)Anideatlow+ass fitter.(b)Anjdeathighpass fitter.(c)AnideaLbandpass fittel (d)Anideatbandrej€ct fitter. passessignalsat frequencieshigher than the cutoff frequency.Thus the termslon and ft8, asusedhere do not refer to any absolutevaluesof frequency,but rather to relative values with respect to the cutoff frequency. Note from the graphs for botl these filters (as well as those for the bardpassand bandrejectfilters) that the phaseaflgleplot for an ideat fil tervades lin€arlyin the passband.Itis of no inlerestoutsidethe passband becausetherethe magnitudeis zero.Linear phaseva ation is necessary to avoidphasedistortion(seeChapter16,pp.770773.) The two remainingcategoriesof filters eachhavetwo cutoff frequencies.Figure 14.3(c) illustrates the ideal ftequency responseplot of a bandpassfiller, which passesa source voltage to the output only when the sourcelrequency is within the band defined by the two cutoff frequencies. Figure 14.3(d)showsthe ideal plot of a bandrej€ctfflter, which passesa sourcevoltageto the output only when the sourcefrequencyis outsidethe band dclined by the two cutoff frequencies The bandreject filter tius rejects,or slops,the sourcevoltagefrom reachingthe output when its frequency is within the bard defined by the cutoff frequencies. In specifying a realizable filter using any of the circuits from this chapter,itis importartto note that the magrlitudeandphaseanglecharacterislics are not irdependent.In other words,the charactedsticsof a circuit that result in a particular magnitude plot will also dictate the form of the phaseangle plot and vice versa.For example,once we selecta desired fbrm for the magnitude responseof a circuit, the phase angle iesponse is also determined.Alternatively,if we selecta desiredlbrm lor the phase
570 Intrcduchonto Frcquency Setective Cncuits angle response,the magnitude responseis also determined.Although there are some ftequencyselective c cuits for which the magnitude and phaseanglebehavior can be independentlyspecified,these circuits are not presentedhere, The next sections present examples of circuits from each of the four Iilter categories.They are a few of the many circuits that act as filters. You should focus your attention on tying to identify what properties of a circuit determine its behavior as a filter. Look closely at tle form of the transfer function for circuits that perform the same filtedtg functions Identifying tie folm of a filter's transfer tunction will ultimately help you in designing filtering circuits for particular applications. All of the filteis we will consider in this chapter are passivefiilters, so calledbecausetheir filtering capabilitiesdependonly on the passiveelements:rcsistors,capacitors,and inductoN.The largestoutput amplitude such filters can achieve is usualy 1, and placing an impedance in se es with the sourceor in parallel with the load will decreasethis amplitude. Because many pmctical lilter applications require inoeasing the amplitude of the output, passivefilters have som€ significant disadvantages.The only passivefilter desoibed in this chapter that can ampMy i1s output is the seriesRLC resonant filter. A much greater selection of ampliJying filters is found among the active filter circuits, the subjecl of Chapter 15.
14.2 LowPass Filters Here, we examinetwo circuitsthat behaveas lowpassfilters,the series RL circuit and ttre series RC circuit, and discover what characteistics of lhesecifcujlsdelefminerhecurofffrequenc).
L
R
(a) L
R
(b)
Figure 14.4a (a)A series Rl lowpass fitter.(b)rhe equjvatent cncuit ato : 0. andG)Th€€quivatent
TheSeriesRl.CircuitQualitativeAnalysis A series RI, circuit is shown in Fig. 14.4(a).The c cuit's input is a sinusoidal voltage source with varying frequency. The circuit's output is defined as the voltage across the resistor. Suppose the frequency of the sourcestartsvery low and increasesgradually.We krow that the behavjor of the ideal rcsistorwill not change,becauseits impedanceis independent of frequency.But consider how the behavior of the inductot changes. Recall that the impedance of atr inductor is joa. At low ftequencies, the inductor's impedanceis very small compared with the rcsistor\ impedance, and the inductor effectively functiors as a short citcuit. The term low frequencies thus refen ro any frequencies for which ol, > n. The equiv, alent circuir for (d = oo is shown in Fig. 14.4(c),where the ourpur volrage magnitude is zero. The phase angle of the output voltage is 90' more negative than that of the input voltage. Based on t]le behavior of the output voltage magnitude, this seiies Ra a0,) circuit selectively passeslowfrequency inputs to the output, and it blocks highfrequency inputs from reaching the output. This circuit's responseto 1.0 varying input frequency thus has the shape shom in Fig. 14.5.These two plots comprise t}le frequency response plots of the se es Rl, circuit in Fig. 14.4(a).The upper plot shows how ld(j@) vades with frequency.The lower plot showshow 9(j(o) vaiies as a function of hequency.We present a 0 more formal method for constiuctiry these plots in Apperdix E. e(i.) We have also supedmposed the ideal magnitude plot for a lowpass 0" filter from Fig. 14.3(a)on the magnitudeplot of the Ra filter in Fig. 14.5. There is obviously a difference between the magnitude plots of an ideal filter and the frequency rcsponse of an actual Ra filter. The ideal filter exhibits a discontinuity in magnitude at the culoff frequency, (,., which 90' createsan abrupttansition hto and out of the passband. While thisis,ideally, how we would like our filte$ to perfom, it is not possible to use real pLotforthe response componentsto constnrci a circuit that has this abrupt tansition in magni Figurc14.5A Thefrequency senes ft circuitin Fig. 14.4(a). tude. Circuits acting as lowpass filters have a magnitude response that changesgraduallyfrom the passbandto the stopband.Hence the magnitude plot of a real circuit requires us to define what we mean by the cutolf ftequency, o..
Definingthe CutoffFrequency We need to define the cutoft frequency, da, for realistic filter circuits when the magnitude plot does not allow us to identify a single frequency that divides tie passbandand the stopband.The definition for cutoff frequencywidely usedby electricalengineersis the frequencyfor which the transfer Iunction magnitude is decreased by the factor 1/\,, from its
(14.1) < Cutofffrequenc!' definition where }1ax is the maximum magnitude of the transfer function. It follows from Eq. 14.1 that t}le passband of a realizable filter is defined as tle range of lrequencies in which tle amplitude of the output voltage is at least ?0.7% of the ma\imum DossibleamDlitude. The constant1/\4 usedin defining ttri cutomtequency may seemlike an arbitrary choice.Examining another consequenceof the cutoff liequency will make this choice seem morc reasonable.Recall from Section 10.5that the averagepower delivered by any circuit to a load is proportional to Vl, where yr is t}te amplitude of the voltage drop aqoss the load: 1V? 2 R
(11.2)
572
rntroduction to Frequ€ncy Setectiv€ Cjrcujts ffthe circuit hasa sinusoidalvoltagesource,V(jo), then the load voltage is also a sinusoid, ard its amplitude is a function of the frequency (r. Define P"" as the value of tlre averagepower deliveredto a load when t})e magnitude of the load voltage is maximum:
1 v 1.^u* 2
R
(14.3)
If we vary the frcquencyof the sinusoidalvoltagesource,I40o),the load voltage is a maximum when the magnitude of the circuit's transfer furction is also a maximum: (14.4)
Now considei what happens to the average power when the frequency of the voltagesouce is o.. Using Eq. 14.1,we determinethe magnitudeof the load voltaseat o^ to be
vLtj@)l = lH(tu)lUl
=Lu ,, \,5'^*' 114.5) SubstitutingEq. 14.5into Eq.14.2,
P(*,)=:v+d I
\iw*)
2
R
1 v" J2 2 P* 2
(r1.6)
Equation 14.6 shows that at the cutotr frequency o., the average power delivered by the circuit is one half the maximum averagepower.Thus,r.r.is also called the halfTow€r ft€quency.Thereforc, in the passband,tle average po\ter deliveredto a loadis at leasr50o. ot rhe maRimumaveragepower. sL
Y,(r)
TheSeriesnl CircuitQuantitativeAnalvsis R
%(,
Figure14.6A Therdomajn equivatentfor thecircuit in Fig.14.4{a).
Now that we have defined the cutoff lrequency for real filter circuits, we can analyzethe seriesna circuit to discoverthe relationshipbetweenthe component values and the cutofl frequency for this lowpass filter. We beginby constructingth€ rdomainequivalentof the c cuit in Fig.14.4(a), assuminginitiat conditions of zero. The resulting equivalent circuit is shownin Fig.14.6.
14.2 LowPass Fitters 573 Tbe volraeetransferfunctronlor thiscircuil,,
H(,) =
R/L
s+R/L
\14.7)
To study the frequencyresponse,we make the substitutions  jd in Eq.14.'7. R/L jo + R/L
(14.8)
We can now separateEq. 14.8into two equations.The fi^t definesthe transfer functior magnitude as a function of frequency; the seconddefines the hansfer function phase angle as a function of frequency:
n(i,) =
RlL
G+GB'
d(ia,) =  tan
\n/
(14.e)
\r4.\a)
Close examination of Eq. 14.9 provides the quantitative support for the magnitudeplot shownin Fig.14.5.W}lenl, = 0, tlre denominatorand the numeratorare equal and lH(i0)l = 1. This meansthat at o = 0, the input voltage is passed to the output terminals without a change in the voltagemagnitude. As the ftequencyincieases,the numentor of Eq. 14.9is unchanged, but the denominator gets laiger. Thus l}I(ia)l decreasesas the frequency increases, as shown in the plot in Fig. 14.5.Likewise,as the frequency increases, the phaseanglechangesfrom its dc valueof0', becomingmore n€gative,as seenfrom Eq.14.10. When a,: c., t}le denominator of Eq. 14.9 is infinite and lH(ico) : 0, as seenin Fig. 14.5.At o : oo, the phaseanglereachesa limil of 90', as seenfromEq.14.10 and the phaseangleplot in Fig.14.5. Using Eq. 14.9,we can computethe cutoff ftequency,od. Remember that .r. is defined as the frequency at which lH(Ja ") : (I/\n)H ^ ". Fol the lowpassfilter, l1max= lH(jo) , asseenin Fig.14.5.Thus,foi the circuit in Fig.i4.4(a),
: jx1u11.,1
R/L
\,/A;snj
(14.11)
Solvirg Eq.14.11for o., we get
(14.12) { Cutofffrequency for ru fitters Equation 14.12provides ar impo ant result. The cutoff ftequency, o", can be set to any desiied value by appropdately selectingvalues for R and L. We can therefore design a lowpassfilter wilh whatever cutoff frequency is needed.Examplel4.l demonshatesthe designpotentialofEq.14.12.
574 Intoductionto Frcquency Setective Cncuitr
Designing a LowPass Filter Electiocardiology is the study of the electric signals produced by the hea{. These signals maintain the heart's rhltlmic beat, and they ate measuredby an instrument called an electrocardiograph.This instru, ment must be capable of detecting pedodic signals whose frequency js about 1 Hz (the normal hearr rale is 72 beats per minute). The instnrment must operate in the presenceof sinusoidalnoise consisting ot signalstrom rhe suffoundingelecrical environ ment! whose fundamental fiequency is 60 Hz the frequency at which electdc power is supplied. Choosevalues for R and I, in the circuit of Fig. 14.4(a)suchthat the resultingcircuit could be used ir an electrocardiographto filter out any noise above 10 Hz and pass the electric signals from the hea at or neat 1 Hz. Then compute the magnitudeof % at 1 Hz, 10 Hz, and 60 Hz to see how well the filter pefolms.
Solution The problem is to select values for R ard L that yield a lowpass filter with a cutoff frequency of 10 Hz. From Eq. 14.12,we seethat R and a cannot be specified independently to generate a value for o.. Therefore, let's choose a conmonly availabl€ value of 1,, 100 mH. Before we use Eq. 14.12to compute the value of R needed to obtain the desired cutoff frequency, we need to convert the cuto{f frequencyfrom hertz to radiansper second: o, = 2r(L0) = mn r^dls.
We can compute the magnitude of % using the equarion% : H(ja) .lul
u"(.) :
R/L
\/"j +1.R8 20t
\,ll + cw.;
lu tut.
Table 14.1 summarizesthe computed magnitude valuesfor the frequencies1 H4 10 Hz, and 60 Hz. As expected,the input and outputvoltageshavethe samemagnitudesat the low hequency,becausethe circuit is a lowpass filter. At the cutoff frequency, the output voltagemagnitudehasbeen reducedby U\D. ftorr. the unity passband magnitude.At 60 H4 the output voltage magnitude has been reduced by a factor of about 6, achieving the desiredattenuationof the noisethat could corrupt the signal the electrocardiographis d€signedto
NoW solve for the value of R which, together with a : 100mH, will yield a lowpassfilter with a cuf off frequencyof 10 Hz: R:a,L
: (20?r)(100 x 103) : 6.28().
I tL) 60
1.0 t,0 1.0
0.995 0.707 0.164
A SeriesRCCircuit
figuE14.7A A sedes fr towpass fitter.
The series RC circuit shown in Eg. 14.? also behavesas a lowpass filter. We can verify this via the same qualitative analysis we used previously. In fact, such a qualitative examination is an important problemsolving step that you should get in the habit of performing when analyzing filters. Doing so wlll enable you 10 prcdict the filtering charactedstics (low pass,high pass,etc.) and thus also prcdict the general form of the lransfer function. If
the calculated transfer function matches the qualitatively predicted folm, you have an important accuracycheck. Note that tle circuit's output is defined as the output acrossthe capacitor. As we did in the previous qualitative anal)ris, we use three frequency regions to develop the behavior of the setiesRC circuit in Fig. 14.7: L Zero ftequency (a = 0): The impedance of the capacitor is infinite, and the capacitor acts as an open circuit. The input and output voltagesare thus the same, 2. Frequenciesincrcasingfrom zeroi Tf\e impedance of the capacitor decreases relative to the impedanceofthe resistor,and the soutce voltage divides betweenthe resistiveimpedanceand the capacitive impedance.Theoutput voltageis thus smallerthan the source voltage. 3. Infinite frequency (.,, = oo): The impedance of the capacitor is zero,and the capacitoiactsas a short circuit.The output voltageis tlus zero, Based on this anal)sis of how tle output voltage changesas a function of trcquency,t}le seriesRC circuit functions as a lowpassfilter. Example 14.2 exolorcs this circuit ouantitativelv
Designing a SeriesfC LoryPass Fitter For the seriesRC circuit in Fig. 14.7: a) Find the transfei function between the soutce voltage and the output voltage. b) Determine an equation {or the cutoff frequency in t]le sedesRC circuit. c) Choose values for R and C that will ield a low' passfilter with a cutoff fiequency of 3 kHz.
Sotution a) To de ve an expressionfor the transfer function, we first constructtherdomain equivalentofthe cicuit in Fig. 14.7,as showr in Fig. 14.8. Using .rdomain voltage division on the equivalent circuit, we find I
H(r) :
RC I RC
Now' substitute r : jo atrd compute the magnitude of the resulthg complex expression: I RC =
laU,)1
(".t)'
x(r)
Y,(J)
Figure14.8^ Th€s'domain equivatentforthe circuit jn Fig.14.7.
b)At the cutoff frequency (,c, lg(jro)l is equat to (1,1\/2)HM. For a lowpass filter, H"  H(jo), and foi t}le circuit in Fig 14.8, H(J0)  1. We can then descdbethe relationship amongthe quantitiesR, C, and o.:
lH(ia,\=+0\
1 RC / 1 \ 2
l__: I \ nc,i
Soi,vingthisequationfor (,", we get
576
Inttoduction to Frequenq Setective Cjrcujts
c) From the resultsin (b), we seethat the cutoff frequency is determined by the values of R and C. BecauseRand C cannotbe computedindependently,let'schooseC : 1 pF. Given a choice,we will usualiy specify a value for C first, rather than for R or a, because the number of available capacitor values is much smaller than the number of resistor or inductor values, Remember
that we have to conve the specified cutoff ftequencyftom 3 kHz to (2r)(3) krad/s:
'^
1 a"C
(2,X3 x 103)0x 10) : 53.05 {).
Figure 14.9summarizesthe two lowpassfilter citcuits we have examined. Look carcfully at the tansfer functions. Notice how similar in form they arethey differ only in the terms that specifJ the cutoff ftequency. In fact, we can state a genenl form for the tmnsfer functions of these two lowpassfilten:
Transfer functionfor a lowDass fitter >
'L
+ vi
R
R/L ,trt=,+NL
v"
+ , r ( r ' : r +Ir/RC fcv" o,= rlRC Figure 14.9A Twotowpass fiLters. thes€ries nLand thesed€s RCtogetherwith thenhansfer tunctions and
114.t3)
Any circuit with the voltagemtio in Eq.14.13would behaveas a lowpass filtei wiih a cutoff ftequency of (,c.The problems at the end of the chaptei give you other examples of circuits with this voltage ratio.
Retatingthe Frequency Domainto the TimeDomain Finally, you miglt have roticed one orher important relationship. Remember our discussion of the natural responsesof the fustorder Rt and RC circuits in Chapter 6.An important pammeter for these circuits is the time constant, 7, which characterizes the shape of the time response. For the Ra circuit,the time constanthasthe value a/R (Eq.7.14);for rhe RCcircuit,the time constantisnCGq.7.24). Comparethe time consranrs to the cutoff fiequencies for these circuits and notice that 114.14) This iesult is a direct consequence of the relationship between the time rcsponse of a circuit and its frequency response,as revealed by the Laplace transform The discussion of memory and weighting as representedin the convolutjonhtegral of Section13.6showsthat as (,.+,]., rhe filter hasno memory, and the output approachesa scaledrcplica of the input; that is,no filtedng hasoccu ed.As o. + 0, the filter hasincreased memory and the output voltage is a distortion of the input, becausefilter
1 4 . 3 HiglrPds Fitters 577
objective1Know the 8l andRCcircuitconfigurations that actas Lowpass filters 14.1 A seriesllclow,pass filter iequircs a culolf frequcncyof 8 kHz. Use R  10 kO and computc the valueof C required.
14.2 A seriesRZ lowpassfiher with a cutofl frequcncyof2 kHz is needed.tlsing n = 5 kO, compure(a) I;(b) H0o) ar 50 kH4 and (c) d(id) at 50 kHz. Answer: (a) 0.40Hr (b) 0.{)4; (c) 87.71'.
NOTE: Also try ChaptetPtubkns11.1and 11.2.
14.3 lirglhFass Filt*rg We rext examinc two circuils that function as highpassfiliers. Oncc again,they are thc scriesRa circuit and the scriesRC circuir.We will see that the samcscricscilcuitcanact aseithcr a low passor a highpassfilrcr, dependingon Nhcre lhe output voltageis dctined.Wewill alsodeterminc the relationshipbellveenthe componentvaluesaDdthe cutoff frequcncy of thesefilters.
TheSeries RCCircuitQualitative Anatysis A scdesRC circuit is shorn in FiS.f4.f0(a).In contrast10 ils low pass counlerpartin Fig.147,th o outpul voltagehere is detincdacrossthe resislor,not the capacitor.BecauscoI this.the effectofthe changingcapacitive impedaDce is clifferenrthan it wasit the lowpassconfiguration. A1 d = 0, the capacitorbehaveslike an open circuil. so there is no currentflowingin thc rcsislor.Tlis is illustratedin the equivalentcircuitin Fig.14.10(b).In this circuit.thereis no voltagcacro,isthe resistor,and thc circuit filters out fie low lrequencvsourccvoltagebefore it reachesthc
G) C
As the frcqucnc) of the voltage sourccincreases, the impedanccoi the capacitordecrcascsrclalive to tire impedanccoI the resistof.and rhe C sourcevoltageis now divided betlveenthe capacilorand the resistor.fhe oulput voltagemagnitudcthusbegiDsto increasc. R When the frequcncyoI lhe sourceis infinitc ((, = c.), the capaciror behavesasa short circuil.and thusthereis no vollageacrossrhe capacitor. This is iliustratcdin the equivaientcircuir in Fig. 14.10(c).In this circuii, the input voltageand output voltagearc the sane. Thc phaseaDgledifferencebctx'cen the soulce and output voltages Figure 14,10t (a)A se esrr highpass fitter(b)the alsovaricsas lhe Irequencyof the sourcechaDges. For o = 6, thc outpu! equivatent ciruii ato  0: and(c)theequivatent voltagcis thc sameas the input voltagc,so the phaseanglediffcrcncc is
578
Introduction to Frequency Setective cncuits zero. As the fiequency of the source deoeases and the impedance of the capacitor hdeases, a phase shift is infoduced between the voltage and the current in the capacitor. This ueates a phase difference between the source and oulput voltages.The phase angle of the output voltage leads that of t])e source voltage. When a = 0, this phase angle difference reachesits maximumof +90'. Based on our qualitative analysis, we see that when the output is defhed as fie voltage acrossthe resistor, the seriesRC circuit behavesas a highpass filter. The components and €onnections are identical to the Iowpass series RC circuit, but the choice of output is differcnl. Thus, we have confimed the earlier obse ation that the filtedng charactedstics of a cfucuit depend on the definition of the output as well as on circuit comDonents.values.and comections. Figure 14.11shows the ftequency responseplot for the seriesRC highpass filter. For reference, the dashed lines indicate the magnitude plot for an ideal highpassfilter. We now tum to a quantitative analysisof this samecircuit.
H0.)l 1.0
TheSeriesfiCCircuitQuantitativeAnalysis 0
To begin, we conshuct the rdomain equivalent of the c cuit in Fig. 14.10(a).This equivalentis shown in Fig. 14.12.Applying sdomain voltage division to the circuit, we *Tite the transfet function:
o(i.) +90.
s + I/RC Making the substitution s = io results in Flgue14.U l. Thefrequency ptotforthe response series fr cncuit in Fig.14.10(a).
L JC
Yi(s)
+ Y,(r)
fltt,)t=
(11.15)
Next,we separate Eq. 14.15into two equationsThefirst is the equation describingt]le magdtudeof the tmnsferfulctioq tlle secondis the equation describingthe phaseangleof the transferfurction:
nA,) Figure14.12a Theidomajn equjvatent ofthe circuii jn Fig.11.10(a).
jd+ IlRC
' \E +nlRcf
o(i.) :90"
tan loRC.
(14.16)
114.17)
A closelook at Eqs. 14.16and 14.17confirms the shapeof the frcquencyresponseplot in Fig. 14.11.Using Eq. 14.16,we can calculatethe cutoff frequency for the sedesRC highpassfilter. Recall that at the cutoff frequency, the magnitude of t}le tra$fer funcrion is (l/\4)Itmax. For a highpassfiltei, H* = lZl(t.).=  fl(ico)i, as seenfrom Eg. 14.11. We can construct an equation for a)" by setting tlle lefthand side of Eq. 14.16 to (y\'A)lH(j'x'\, noting that for this series RC circuir, Ifi(jco) = 1:
t2
\Gll!RC),
(14.18)
1 4 . 3 N i q hP a 5tsi t e 6
579
SolvingEq.14.18for (,",we get 1
(74.7e)
Equation 14.19presents a familiar result.The cutoff ftequencyfor the seriesRC circuit has the value 1/RC, whether tle circuit is configured as a lowpassfilter in Fig. 14.?or as a highpassfilter in Fig. 14.10(a).This is perhapsnot a surprisingrcsult,as we have alreadydiscovereda connection between the cutoff ftequency, (,., and the time constant,r, of a circuit. Example 14.3 analyzes a sedes Ra circuit, this time configured as a highpassfilter. Example14.4examinesthe effectofadding a load resistor in panllel with the inductor.
Designing a SeriesRl HighPass Fitter Show that the seriesRa circuit in Fig. 14.13also acts like a highpassfiltefl a) Derive an expression for the circuit's transfer function, b) Use t}le result frorn (a) to determine an equation Ior the cutoff frequency in tlre seriesRL circuit. c) Choosevaluesfor R and I that will yield a highpassfilter wirh a cutoff frequency of 15 kHz. R
R Y(s)
v,(t
Flgur€14.14 A Therdomain equivatent ofthecjrcuitin Fig.14.13.
b) To find an equation for the cutoff frequency,first compute the magnitude of H(/.0): L
Figure14.13a Thecircuittor Example 14.3.
Solution a) Begin by constructing tle .rdomain equivalent of the seriesri, circuit, as shown in Fig. 14.14. Thenuserdomainvolragedir i.ionon rheequiralenl circuit lo conslrucllhe lransler[unction:
s+R/L Making the substjtution r : jd, se get
nUot :
sL
ja + RIL'
Notice that this equation has the same form as Eq.14.15for the seriesRChighpassfilter.
E\i.):
\/"1+ gnj
Then, as before, we set the leffhand side of this equation to (11\5\H^*, basedon the defmtion of the cutoff frequency .,,.. Remember that 1/mx = lH(J@)l for a highpass filter, and for the seriesRL circuit,lH(joo)l = 1. we solvethe resulting equation for the cutoff frequencyi \/, This is tle same cutoff frequency we computed for the sedesRa lowpass filter. c) Using t}le equation for (,. computed in (b), we recognize that it is not possible to specify values for R and l, independently. Therefore, let's arbitrarily selecta value of 500 O for R. Remember to convert the cutoff frequency to radians per second:
,
n
5
0 0 "'" (2,,)(15,000)
580 Intrcductjon to Frequency S€lective Cjrcujll
Loading the Seriesfl HighPass Fitter Examine the effect of placing a load resistor in parallel with t}le inductor ir the Rl, highpass lilrer showninFig.14.15: a) Detemine the transfei furction for the circulr m Fig.14.15.
magnitude is (1)(112)= 112,andthe curofffrequencyis (15,000X1/2)= 7.5kHz. A sketchot the magnitudeplots of the loadedand unloaded circuitsis shownin Fig.14.17.
b) Sketch the magnitude plot for the loaded Rl, highpass filter, usitrg the values for R and a from the circuit in Example 14.3(c)and letting R/  R. On lhe ramegraph..kercbfie mag0itude plot fo{ the unloaded RL highpassfilter of Example14.3(c).
R L'
RL
Figure14.15A Thecircuit forExanrpte 14.4.
Solution a) Begin by transfoming the cbcuit in Fig. 14.15ro the sdomain,as shownin Fig.14.16.Use voltage division across the parallel combination of irductor and load resistorto computethe transfer function:
RLSL
ll(r) =
Rz+sa
^*^3? R+Rr
/ R , \ \R+R/./ R.
'  \ /R + R , /\ R t
, a.:
Kt r+@.'
KR/L.
Note that (). is the cutoff frequency of the loadedfilter. b) For the unloaded Ra highpassfilrer from Example 14.3(c),the passbandmagnitudeis 1, and the cutoff hequencyis 15 kHz. For rhe loadedRa highpassfilter, R = Rr = 500O, so 1( = 1/2.Thus,for the loadedfilter, tlle passband
R
+
v,(s)
SLi
RL Y"(,)
Figure14.16a Th€rdomain equivat€nt of thea,cuitin Fig.14.15.
r
a 1 2'r2
0
30 40 f",ro f" 2D Frequency (kHz)
50
flgure14.17A Thenagnjtude ptotstortheunloaded Rl high,pass fiLter of Fig14.13 andthetoaded Rthiqhpass filter of Fig.14.15.
14.1 HjshPass Fjtte6
581
Compadng the transfer functions of t}le unloaded filter in Example 14.3 and t})e loaded filter in Example 14.4is uselul at this point. Both transfer functions are in the form: '''"
t(J
r
_!_
A ( i RL )
JC
with /( : 1 for the unloaded filter and ( : nz/(R + Rz) for t})e Ioaded filter. Note that the value of ( for the loaded circuit reduces to the value of 1
\14.?5)
Next, calculatethe cu1offftequencieq(,.t and oa2.Remember that at the cutoff frequenciesjthe magdtude of the rransfer tunction is (1/\,t )Anax. Becauseil,, = lHQo") , we can calculareflnax by subsrituting Eq. 14.25 ir\to Eq.1.4.22. Htu
= H(ja,)
oo@/L)
\IOtrcl ;,*+AAn \,(!lraFtL) ' /   . '
V ltl,lLC) tr/LC)1,  +\/t L
:1.
LCxR/L\
14.4 Bandpass Fitters 585
Now set the lefthand side of Eq. 14.22to (U^VDH^ 1/V2) aDdprepareto solvefor (,":
(which equals
a,(R/L) \/,
vrctn ,. ;O"nUj tItzn;
on,nc;l + t
\14.26)
We can equate the denominato$ of the two sidesof 8q.14.26 to get
L +. l, . . ^
1
114.27)
ReanangingEq. 14.27resultsin the folowing quadmticequation:
'lt+,n yc =o.
(14.28)
Thesolutionof Eq. 14.28yieldsfour valuesfor the cutoff frequercy.Only they identwo of thesevaluesare positiveand havephysicalsignificance; of this filter tifv the Dassband
(14.29)
< cutofffrequencies, seriesruCfittec (14.30)
We cair use Eqs. 14.29and 14.30to confirm that the center frequercy, &),, is the seomeldcmeanol the iwo culoll freouencies:
< Retationship between centerfrequenay andcutofffrcquencies
/l R
r r n r\rrtcr,t/ tl lnz r' t r n v r r r t * Vfzr V\:rl V\:r/ \icll (14.31)
586
(iftuiis IntrodLrction to Frcquency Setective
Recall that the bandwidth of a bandpassIilter is defined as the difference between the two cutoff ftequencies, Becaused.2 > ().1 we can compute the bandwidthby subtractingEq.14.29from Eq.14.30: Relationshipbetweenbandwidthand cutoff freouencies>
fl ^" + l
/ n Y
\n)
R
(+)[ + .\n)  (r"t) /nY
i'
\14.32)
The quality factor, the last ol the five characteristic parameters,is defined as the ratio of centerfrequencyto bandwidth.Using Eqs.14.25and 14.32:
ouatityfactor> (1/LC) (R/ L) L
(14.33)
cP
We now have five parametenthat characterizethe seriesRaC bandpassfilter: two cutoff frequencies,a,.1and d.2, which delimit the passband; the center fiequency, do, at which the magnitude of the transfer function is maximum;thebandwidth,B, a measureofthe width of the passband;and the quality factor,O, a secondmeasureof passbandwidth.As previously noted, only two of these patameters can be specfied independently in a design. We have already observed that the quality factor is specified in terms of tie center frequency and the bandwidth. We can also rewrite the equations for the cutoff frcquercies in terms of the center frequency and the bandwidth:
_ p 2
: L + 2
(!)'
\14.34)
(BY
\t)
114.35)
Alteinative foms for these equations express the cutoff ftequercies in terms of t]le quality factor and the center frequency: ..t:
o.'
(+) l
I  . tro Ir
/ r \rl ( V t r a lI
114.36)
I
Also seeProblem14.17at the end of the chaptei.
\r4.37)
I4.4
Bandpas5 Fjlte6
587
The examples that follow illustrate the design of bandpass filters, introduce another RLC ciicuit that behaves as a bandpass filter, and examine the eflects of souce iesistanc€ on the characteristic parameters ofa seriesRaCbandpassfilter.
Designing a Eandpass Filter A graphic equalizer is an audio amplifier thar allows you to select different levels oI amplification within differe0t frequenc) fegionsUsing the series RaC circuit in Fig.14.19(a),choosevaluesfor R, L, and C that yield a bandpasscircuit able to select inputs witiin the 110 kllz frequency band. Such a circuitmight be usedin a graphicequalizerto select this ftequency band from the larger audio band (generany 020 kHz) prior to amplification.
Sotution We need to compute valuesfor R, a, and C that producea bandpas\liler qirh curoff frequencie( of 1 kHz and 10 kHz There are many possible appioachesto a solutior. For instancq we could useEqs 14.29and 14.30,which specifyd.1 and (d"2 ir terms of R, a, and C. Becauseof the form of theseequations,the algebraicmanipulationsmight get compiicaled.Illstead,we will use the fact that ttre center frequency is the geometric mean of the cutoff fiequencies to compute .rd, and we will then use Eq. 14.31 to compute a and C from a,,. Next we will use the definition of quality factor to compute O, and last we will use Eq. 14.33to compute .R.Even though this approach involves more individual computational sfeps, each calculatron rs fairly simple. Any approach we choose will provide only two equations hsufficient to solve for the three unknowns because of the dependencies among the bandpassfilter charactedstics.Thus, we need to select a value for eithei R, a, or C and use the two equalionswe ve cho fundionfor f,LCbandreject Transfer
(14.55)
Equation 14.55is useful in filter design,becauseany circuit with a transler function in this form can be used as a bandreject filter'
14.5 Bandr€ject Fitters 597
vo
L
sc s, + t/Lc s'+(R/L)s+1/LC @.=.v/1/LC
a.= yI/LC
B=R/L
p= 1/Rc
rigure14.31 Twoilr bandreject filt€rs,together wiihequations forthetnnstur tunction, cent€r frequency, andbandwidth ofeach.
598
Introduction to Frequency Setective Cncujts
PracticaIPerspective Pushbutton Tetephone Circuits In the PracticatPerspectjve at the stat of this chapter,we described the (DTMF) duattonemuttipLefrequency systemusedto signalthat a button hasbeenpushedon a pushbuttontelephone.A keyetem€ntof the DTMF systemjs the DTMF receivera circuitthat decod€s the ton€sproduced by pushinga buttonanddetermines whichbuttonwaspushed. In orderto designa DTI4F reciever, we needa betterunderstanding of the DTlvlF systefir.As you canseefrornFig.14.32,the buttonson the tetephoneareorganized into rowsandcolunns.Thepair of tonesgenerated by pushinga buttondepends on the button'srowandcotumn.Thebutton'srow determjnes its lowfrequency tone, andthe button'scotumndetermines its "6" pressing highfrequency tone.1Forexampte, the button produces sinusoidattoneswith the frequencies'770Il2 and14'77HL At the teLephone switchjngfacitity, bandpassfilters in the DTI4F receiverfirst detectwhethertonesfrom both the Lowfrcqu€ncy and highgroupsare simuttaneousLy present.Thistest rejectsmanyextrafrequency neousaudiosignaLs that are not DTI4F. If tonesare presentjn both bands, otherfiLtelsare usedto setectamongthe possibl€tonesin eachbandso jnto a uniquebuttonsignat.AdditionaI that the frcquencies canbe decoded testsareperformed to preventfatsebuttondetection.Forexarnpte, ontyone tone per frequencybandis attowed;the high and lowbandfrequencies muststartandstopwithin a few firitliseconds of oneanotherto be consideredvatid; and the hjgh and towbandsignalamplitudesmust be suffi' cienttyctoseto eachother. You may wonderwhy bandpassfiLtersare used instead of a high passfjtter for the highfrequency group of DTI4F tones and a Lowpass gfoup of DTI4F fitter for the towfrequency tones.The reasonis that the tetephonesystemusesfrequenciesoutsideof the 3003 kHz band for purposes, othersjgnating suchas dngingthe phonekbetL.Bandpass fitters preventthe DTlvlF receiverfrom enoneousLy detectingthese other signals.
v *v 7.
t'l
Figue14.324 Tones generated bythercwsand pushbultons. columns oftehphone
NjfE: Assessyaur undestanding of this PracticdLPespective by hying Chapter Problen s 14.431 4.45.
l Afourth hiqh frequency toneis Eservedat 1633Nz.Ihi5toneis usedinfiequenttvand is not produced bya standad12buttontelephone.
Probtems599
5ummary . A frequency s€lective circuit, or fflt€r, enablessignalsat certain ftequercies to reach the output, and it attenuates signals at other frequencies to prevent them from r€aching the output. The passband contains the fre, quencies of those signals that are passed;t}le stopband contahs the frequencies of those signals that are atten, uated.(Seepage568.)
. Bandpassfilters and bandreject lilten eachhave two cut off foequencies,(d.1and rrr. Thesefilters aie further characterizedby their center ft€qu€ncy (.,,,), bandwidth (B), and quality factor (O). Thesequantities are defined as o"= ta1n6 p = o.2
0+1,
o: alp. The cutoff ftequency, o. , identifies t]le location on t}le trequency axis that separates the stopband from the passband. At the cutoff frequency,the magnitudeof the transferfunctior equals(1/1U )H"* (Seepage571.)
A lowpass ffller passesvoltages at frequencies below o. and attenuatesftequenciesabove .()..Any circuit with the lransfer function fl("):
(see pagesJ65 )6b.) . A bandpassfilter passesvoltages at frequencies within the passband,which is betweer @crand (,.2. It attenuates frequenciesoutside of the passband.Any circuit with the transfer function
Ps
3+Bs+4 functionsas a bandpassfilter. (Seepage591.) . A bandreject fflter attenuates voltages at frequencies within the stopband,which is betweeno"r and o.2. It passesfrequenciesoutsideof the stopband.Anycircuit with the transfer function
functions as a lowpass filter. (Seepage 576.)
A highpassfill€r passesvoltages at frequencies above o. and attenuates voltages at frequenciesbelow o.. Any circuit with t}le transfer funclion
rl(s) =  L Iunctionsas a high passfilter. (Seepage58i.)
sr+j st+8"+1, functioN as a bandrejectfilter. (Seepage596.) . Adding a load to the output of a passivefilter changes its filtering properties by altering the location and magnitude of the passband.Replacing an ideal voltage sourcewith one whosesourceresistanceis nonzeroalso changesthe filtering propertiesof the iest of the circuir, again by alterhg the locatjon and magnitude of the passband.(Seepage589.)
Problems Section14.2 14.1 a) Find the cutoff frequency in hertz for the Rl, filter shownin Fig.P14.1.
250mH
CalculateH(id) at o.,0.3o., and 3o". c) If o, : 50 cos ol V, D.rite the steady,stateexpres
sion for ?o when .r:
figurcP14,1
(r", o  0.3(d., and
600
Inhoduction to Freq0ency Sehctive Cjrcuitr
l2 Use a 25 mH inductorto designa lowpass, Ra,passive lilter with a cutoff frequency of 2.5 kHz. !1!B!!!1 a) Specify the value of the iesistoi. b) A loadhavinga resistanceof ?50O is connected acrossthe output teminals of tle filter. What is the corner, or cutofl lrequency of the loaded filter in hertz?
14,5 A resistoi denoted as Rr is connected in paiallel with the capacitor in the circuit in Fig. 14.7. The loaded low passfilter circuit is shown in Fig. P14.5. a) Derive the expression for ttre voltage transfer function V,/I4. b) At what frequencywill the magnitudeof H(Jo) be maximum? c) What is the maximum value of the magnitude ot H(jo)? d) At whar ftequency wi]l the magnitude of g0o) equal its maximum value divided by \2? e) Assumea rcsistanceof 300ko is addedin parallel with the 4 nF capacitor in the circuit in Fig. P14.4.Find a., nQI), nUQ, aQ0.za,). and 1i(j8a").
143 A resistor, denoted as Rr , is added in serieswith the inductor in the circuit in Fig. 14.4(a).The rew lowpassfilter circuit is shown in Fig. P14.3. a) Derive tle expression for d(r) wher€
H(s): WV.
b) At what hequencywill the magnitudeof fl(Jd) be maximum? c) Wlat is the ma\imumvalue of the magnitude oflt(io)? d) At what frequency will tlle magnitude of H(jo) equal its maximum value divided by \U? e) Assume a resistance of 75 O is added ir series witl the 250 mH inductor in the circuit in Fig. P14.1.Find a,, H(j0), H(ja.), H(j0.3a,\. and 1t(j3.d.).
Figure P14.5 R
t4s li!rM!
Figure P14.3
&
L
1,4 a) Find t}le cutoff frequency (in heitz) of the lowpassfilter sho\lll in Fig. P14.4. b) CalculaleH(io) at .r,,0.2o., ar.d8a". €) If ?,i:48ocosotmv, write the steadystate expressionfor ?,,when o = d.,0.2o., and 8o.. Figure P14.4
20ko
Use a 25 nF capaciror ro design a lowpass passrve filter with a cutoff frequency of 160krad/s. a) Specifythe cutoff tequency in hertz. b) Specify tle value of the filter resistor. c) Assume tlle cutoff frequency cannor increase by more than 8o/".What is the smallestvalue of load resistancethat can be connectedauoss the output teminals of the filter? d) If the resistor found in (c) is connected across the output terminals, what is the magnitude of H(/l,)wheno=0?
14,7 Design a passiveRC low passfilter (seeFig. 14.7) with a cutoff fiequency of 500 Hz using a 50 nF capacitor. a) What is the cutoff frequency in rad/s? b) What is the valueofthe resistor? c) Draw your circuit, labelingthe componentvalues and output voltage. d) What is the transfer function of the filter ir palt (c)? e) ff the filter ir part (c) is loaded witl a resisror whose value is the same as the resisror part (b), what is the transfer tunction of this loaded filter?
Problens601 f) What is the cutoft fiequency of the loaded filrer ftom part (e)? g) Wlat is the gain in the passband of rhe loaded filter from part (e)?
14.8 Study the circuit shown in Fig. P14.8 (wirhout the load resistor). a) As .,  0, the inductor behaveslil(e what circuit component?what valuewill the output voltage 0ohave? b) As (,+ co, the inductor behaveslike whar circuit component?Whaf value will the output voltageoohave? c) Basedon parts (a) and (b), what rypeof filrering doesthis circuit exhibit? d) What is the transfer function of the unioaded filter? e) If R : 1 kO and I, = 20 mH, what is rhe curoff tequency of the filter in rad/s?
S€ction 14.3 14.10 a) Find the cutoff frequency (in hertz) for rhe highpassfilter shownin Fig.P14.10. b) Find H(j(d) at o., 0.1.d.,and 10o.. c) If o, = Soocos(()tmv, write the steadystate expressionfbr oo when o : .d., (1)= 0.1o., and tD = 10ac. FigureP14.10 2.5 DF
:

.
14.11 A resistor,denoted as R., is connectedin series with the capacitorin the circuirin Fig.14.10(a).The new high,passfilter circuitis shownin Fig.P14.11. a) Derive the expression for l1(s) where
H(") = vJU.
b) At what fiequencywill the magnitudeof H(jo) be maximum? c) What is the maximum value of the magnitude ot H(ja)? d) At what frequencywill the magnitudeof t110rr) equalits maximumvaluedividedby \4? e) Assume a resistanceof 10kO is connectedin serieswith the 2.5 nF capacitorin the circuit in Fis. P14.10.Calculate (,., H(ja,), H(io.l.,.,), and H(j10o.).
Figure P14.8 I I R
: I
_l
RL
l1(,):#
Flg(reP14.11
14.9 Supposewe wish to add a load resistorin parallel with the resistorin the circuit shownin Fig.P14.8. a) What is the transfer lunction of the loaded filter? b) Comparethe transferfunction of the unloaded filter (part (d) oI Problem 14.8)afld the trans fer function of the loaded filrei (part (a) of Problem14.9).Arethe cutoff frequenciesdifferent? Are the passband gains different? c) What is the smallestvalueof load resistalcethat canbe usedwitl the filter from Problem14.8(e) such that the cutoff frequency of the resulting filter is no more than 10% different from me unloadedfilter?
t4.x2 Using a 20 nF capacitor, design a highpasspassive l!0!!!11
filter with a cutoff frequency of 800 Hz. a) Specifythe valueofR in kilohms. b) A 68 kO resistoris connectedadoss the output terminals of the filter. What is rhe curoff flequercy,in hertz,ofthe loadedfilter?
602
Circuits to Fr€qu€ncy Selective Introduction
14.13 Using a 25 mH lductor, design a highpass,RL, p?iTfll{passivefilter with a cutoff frequency of 160krad/s. _;tr4 a) Specify the value of the resistance. b) Assume theJilter is cornected to a purc resistive load. The cutoff frequency is not to drcp below 150krad/s. What is the smallest load resistor lhat can be contecled acrossthe outpul lerminalsof the filter?
14.14 Design a passiveRC higl passfilter (seeFig. 14.10[a]) with a cutoff ftequency of 300 Hz using a 100 nF capacitor. a) What is the cutoff ftequency in rad/s? b) wltat is the value of the resistor? c) Draw your circuit, labeling the comporent values and output voltage. d) Whar is fte transfer function of the filter in Part (c)? e) If the filter iII part (c) is loaded witl a resistor whose value is the same as the resistor h (b), what is tle transfer futrction of tlis loaded filter? I) What is the cutoff frequency of the loaded filter ftom pafi (e)? g) What is t])e gain in the pass band of the loaded filter from part (e)?
14.15 Consider the circuit shown in Fig. P14.15. a) With the input and output voltagesshown in the figurg this circuir behavesiike what t,?e of filte b) what is the transfer function, H(s) : %(s)/I4(r), of t}Iis filter? c) What is t})e cutoff frequency of this filter? d) wlat is the magnitude of the fillei's transfer tunctionats: rar.?
14.16 Supposea 1kO loadresistoris attachedto the filter itrFig.P14.15. a) Whatis the transferfunction,l/(s) = %(r)/I4(s), of this filter? b) What is the cutoff frequencyof this filter? c) How doesthe cutoff lrequencyof the loadedfilter comparewith the cutoff jlequency ot the unloadedfilter in Fig.P14.15? d) What elseis differentfor tlese two filters?
Section 14.4 14.f
Show that the altemative forms for tle cutoff frequencies of a bandpass filter, given in Eqs 14.36 and 14.37,canbe derivedfrom Eqs.14.34and 14.35.
14,18 Calculate the center frcquency, fie baldwidth, and the quality factor of a bandpass filter that has an upper cutoff frequency of 200 krad/s and a lower cutoff frequency of 180 krad/s. 14,19 A bandpassfilter hasa center,or rcsonant,tequency of 80 krad/s and a quality factor of 8. Find the bandwidth, the upper cutoff ilequency, and the lower cut ofI frequency.Expressall answersit kilohertz 1420 Use a 20 nF capacitor to design a seriesRIC bandsu cenliker.asshownal rhetoDofliq. l4.27.The 'lli'Ii Dass ier tequencyot I he riiref ic 20 kltz'and rhe qualir' factor is 5. a) Specify the values of R and a. b) what is the lower cutoff frequency in kilohertz? c) What is the upper cutoff frequency in kilohertz? d) What is tlle bandvddth of t}le filter in kilohertz? 1421 For the bandpass filter shown in Fig. P14.21,find
",
G) .,", (b) t, G) o , @)a"b G) f ,b $) aa, G) f a,
and (h) B.
Pl4.t5 Figure 1ko
tiglre P14.21 300 {)
50mH
Probt€ms 603 14.22 Using a 25 nF capacitor in the bandpass circuit shown in Fig. 14.22,design a ftlter with a qualiry fac*lsJPM _tsini tor of 10 and a center fiequency of 50 krad/s. a) Specifythe numericalvaluesofR and a. b) Calculate the upper and lower cutoff frequenciesin kilohertz. c) Calculatethe bandwidthin hertz. 14.23 For the bandpassfilter shownin Fig.P14.23,calcuEtrc late the following: (a) /,; (b) O; (c) /.r; (d) /.,; and (e) B. tigureP14.23
14,24 The input voltage in the circuit in Fig. P14.23is 800cos(,t mV. Calculat€ t}le output voltage when G) (,, : (,.; (b) o  ro"1;and(c) o = "2. 14.25 Design a seriesRIC bandpassfilter (seeFrg.14.19[a]) with a quality ol 5 ard a center ftequency of 20 krad/s, using a 0.05 I,F capacitoi. a) Draw your circuit, labelingthe componentvaluesand output voltage. b) For the filter in part (a),calculatethe bandwidth and lhe !aluesof the tqo cutolifrequeDcie\ 1416 The input to the sedes RLC bandpass tilter designedin Problem 14.25is 200coso1mV. Find rhe voltage drop adoss the resistor when (a)o  o,; (b)., = ac1:k) a = o,,;(d) o : 0.1o.; (e) o = 10a.. 1417 The input to the series RaC bandpass filter designedin Problem 14.25is 200cosot mV Find the voltage drop acrossthe seriescombinationof the inductor and capacitor when (a) ., = o,; (b) o: o.1; G) o = o,,; (d) 4, = 0.1o.; (e) o : 10o". 1428 A block diagram of a system consisting of a sinusoidal voltagesouice,an RIC sedesbandpassfiltei, and a load is shownin Fig.P14.28.Theinremal impedanceof tle sinusoidalsourceis 36 + J0 O, and the impedanceofthe load is 320 + i0 O.
The RIC series bandpassfilter has a 5 nF capacitor,a center frequencyof 250krad/s, and a quality factor of 10. a) Draw a circuit diagram of the system. b) Specify the rumerical values of a and R for rhe filter section of the system. c) Wlat is t}le quality factor of the interconnected system? d) What is the bandwidth (in hertz) of the inter, connectedsystem? Fig!reP14,28
1429 The purposeofthis problem is to investigatehow a resistiveload connectedacrossthe output terminals of the bandpasslilter shown in Fig. 14.19 affectsthe quality factor and hencethe bandwidth of the filterirg syst€m.The loaded filrer circuit is shownin Fig.P14.29. a) Calculate the transfer function y,/I{ for the cir cuit sbownin Fig.P14.29. b) What is the er?ressionfor the bandwidth of the system? c) What is the expressionfor the loaded bandwidth (pr) as a function of the unloadedbandwidtl (Bu)? d) What is tie expression for the quality factor of tbe system? e) What is the expressionfor the loaded quality factor (Cr) as a function of t}le unloaded quality tactor (qu)? f) What are the expressionsfor the cutoff frequencieso.t and o.2? figurcP14.29 R
604
Intrcductjon to Frequencr Setective Cncuits
14.30 Consider the circuit shown in Fig. P14.30. """ a) Find r,.,.. b) Find P. c) Find O. d) Find the steadystate expiession foi t', when 0, = 750cosoJ mV. e) Show tlar if RL is expressedin megohms the O of the circuit in Fig.P14.30is ^
2
14.33 For the bandrejectfilter in Fig. P14.33,calculate Em (a) o.;(b) fo;(c) O; (a) oa;(e) /a; (0 zoa;G)fa; and (h) B in kilohertz. tigureP14,33
5
u= 1+n5/R, f) Plot O venus RL for 1 MO < RL < 40 MO. Figure P14.30 1.25MO
14,31 The parameten in the c cuit in Fig. P1,1.30are R = 100kO, C=4pF, and a=400pH. The quality factor of tle circuit is not to drop below 9. What is t}le smallest permissible value of tle load resistorRL?
14,34 Use a 100nF capacitorto designa bandrcjectfilter, p?iii6lras shown in Fig. P14.34.The filter has a center frea;i;i quency of 50 kHz and a quality factor of 8. a) Specify the rumerical values of R and L. b) Calculatethe upper and lower corner,or cutofl tequenciesin kilohertz. c) Calculate the filter bandwidth in kilohertz. FiguleP14,34
100nF
Section14.5 14.32 a) Show (via a qualitativeanalysis)that the c cuit inFig.?14.32is a bandrejectfilter. Supporrlhe qualilari!eanal].isoi(a) b) I nd ng the voltage transfer function of the filter. c) Derive the expression foi the center fiequency of the filter. Derive the expressions for the cutoff frequene) wlat is th€ expressionfor the bandwidthof the filter? f) What is the expression for the quality faclor oI t}le circuit?
14.35 Assume the bandrejectfilter in Problem 14.34is 6dc loadedwith a 932O resistor. a) What is tlle quality factor of the loaded circuit? b) wlat is the bandwidth (in kilohertz) oI the loadedcircuit? c) w}lat is the upper cutoff lrequency in kilohefiz? d) Wlat is the lower cutoff frcquency in kilohertz?
tigureP14,12 14.36 Designan RaC bandrejectfiltei (seeFig. 14.28[a]) with a quality of 2/3 and a center hequency of 4 krad/s,usinga 80 nF capacitor. a) Diaw your circuit,labelingthe componentvaluesand output voltage, b) For the filter in part (a), calculate the bandvidth a0drhevalue.o( lhe l\ o culoii frequencies.
14,37 The input to tlle RIC bandrejectfilter designedin Problem 14.36is 125cosotmV. Find the vottage drop acrossthe seriescombinationof the inductor and capacitor when (a) ,) = aa; $) a = b.1; (c) a, = ar.,;(d) o : 0.1a.t(e) @ : 10!,".
14.38 The input to the RaC bardrejecl lilter designedin Pioblem 14.36is 125cos.rrmV. Find t}Ie voltage drop across the resistor when (a) o = ro,; (b) o  o"1; (c)(d = @d,;(d)o : 0.1a.;(e)o = 1u,t..
14.41 The load in the bandreject filter circuit shown in Fig.P14.39is 36 kO.The centerfrequencyofthefilter is 1 Mrad/s, and the capacitor is 400 pF. At very low and very high frequencies,the amplitude of the sinusoidaloutput voltageshouldbe at least96010 of the amplitudeofthe sinusoidalinput voltage. a) Specify the numerical values of R and a. b) What is the quality factor of the circuit?
Sections14.114.5 1439 The purpose of this problem is to investigate how a resistiveload connectedacrcssthe outputterminals of the bandreject filter shown in Fig. 14.28(a)affecrs the behaviorof the filter. The loadedfilter circuitis shownin Fig.P14.39. a) Find the voltage transfer function yo/14. b) Wlat is the expressionfor the centerfrequency? c) W}lat is the expiessionfor the bandwidth? d) What is the er?ression foi the quality factor? e) Evatuatefl(jo.). f) Evaluatea(jo). g) EvaluateH(/co). h) Wlal are the er?ressions for the corner frequencies1,.l and o.r? Figurc P14.39
14.40 The paramerers in rhe circuir in Fig. p14.39 s P r Ga r c R : 5 k O , t = 400mH, c:250pF, and Rr = 20ko a) Find,r,, B (in kilohertz),and O. b) Find fl(iO) and l1(jco). c) Find f., and /.r. d) Showtiat iJ RL is expressedin kilohmsrhe O of the circuitis
o:8[1 + (s/RL)]. e) Plot C versusRL for 2 kO < RL < 50 kO.
14,42 Giventhe following voltagetransferfunction:
v H@_i 4x106 r' +500r+4x106 a) At what frequencies(in radiansper second)is the ratio of Yo/I4equal to unity? b) At what ftequency is the ratio maximum? c) What is the maximumvalue of the ratio? 14.,fi} Designa seriesRaC bandpassfilrer (seeFie. 14.27) p?$l,jii, for detecting t.helowfrequency rone generated by iE.;a; pushinga telephonebutton as shownin Fig.14.32. a) Calculatethe valuesof l, and C that place the cutoff frequencies at ihe edges of the DTMF lowfrequencyband.Note that the resistancein standardtelephone circuils is alwaysR = 600 O. b) What is the output amplitude of this circuir at eachof the lowbard frequencies, relativeto the peak amplitude of the bandpassfilter? c) W}Ial is the output amplitude of this circuit at the lowestofthe highbandftequencies? 14.44 Design a DTMF highbandbandpassfilter simitar liglffitr to the lowband filter design in Problem 14.43.Be lone. .0.,.n sure lo iDcluderhe lourlh hiphtrequenc) r{tJJH,/.rDyourdesign. whar is lhe response amplitude of your filter to t}le highest of the lowfrequencyDTMFtones? 14.45 The 20 Hz signalthat drgs a telephone'sbell hasto t:l;Ill#r have a very larye amplitude to produce a loud enoughbell .igoal.Ho$ muchlargercan rhe ring .!r.. '' iDg sienalamplituLle be. relari\e lo lhe loqbank DTMF signal,so that the rcsponseof the filrer in Problem 14.43is no more than half as large as the largestof the DTMF tones?
ActiveFilterCircuits 15.1 fnstOrdefLowPass and HighPass 15.2 S.aling p. 612 15.3 0p Anp Sandpass and Bandreject Hlters p. 615 15.4 Higherord€r0p Amp FiLtersp. 622 15.5 NarowbandBandpass and Bandreje.t fiterc p. 636
I Knowthe op ampcjrcuitsthat behaveasfi6tandhishpass fittersandb€ abL€ orderLowpass componentvatues for thesecircuits to catcuLate of cutofffrcquency and t0 m€etspecifications B€abl€to desigffiLtercjrcuitssia'tjngwith a prototypecircuitandusescatingto achjeve desnedfrequency Esponsecharacterktics and howto us€cascaded fi6t and Undelsland secondoder Butt€rwothfitten to imptement afd bandreject tow pass,high pass,bandpass, to cakuLate 8e abteto usethe designequations for prototypenarcwband, component vaLues nLtetsio meetdesjred bandpasr, ard bandreject
lJp to this point, we haveconsideledonly passivefiltcr circuits, that is,filter circuitsconsistingof resistors,inductors,and capaci1ols.ll1ere are areasof application,however,where active circuits,those lhat employ op amps.have cefiain advantagesovcr passivefilters.Fo1instancc,activecircuitscan producebandpass and bandrejestfilters without using inductors.This is desirable becauscinductoft are usuallylarge,heavy,costly,and they may introduce electromagneticfield effccts that compromise the desiredhequencyresponsechalacteristics. Examinc the transtertunctionsof all the lilter circuits from Chapter 14 and you will noticc that the maximum magnilude docs not exceed 1. Even though passiveresonant filters can achievevoltage and curent amplification at the resonant liequency,passiveliheN in genelal are incapableof amplification, bccausethe output magnitudedoesllot exceedthe input magni tude.This is not a surpdsingobscrvation,as many of the transfer tunctions in Chapter 14 were derived using voltage or curcnt division.Active filtcrc provide a control over anpiification noi availablein passivelilter circuits. Finally, rccall that both the cutotTfrequency and the pass band magnitudeof passivefilters wcre altered with the additio of a resistiveload at the output oI the filter. This is not thc case with activefilters,due to thc propertiesof op amps.Thus,we use activecircuitsto implemeDtlilter designswben gain,load variation, and physicalsizc are impofiant parameteLsin the design specifications. ln this chapter,we examinea few of the many filtcr circuits that cmploy op amps.As you will see,theseop anlp circuitsovcr come the disadvantagesof passivcfilter circuits.Aiso, we will show how the basic op amp filler circuils can be combincd to achievespecificftequencyrcsponsesand to attain a more nearlv ideal tilfer response.Note that throughout this chaptcr l{e assumcthat cvcry op amp behavesas an ideal op amp.
606
PracticaI Perspective Bass Votume Control In this chapter,we continueto examinecircuitsthat arefre, quencys€tedjve.As described in Chapter14,this meansthat the behavjorof the circuit dependson the frequencyof its sinusoidal input. Mostof the circuitspresented herefatLjnto oneofthe four categorjes identifiedin Chapter14tow,pass fitters,highpass fitters,bandpass fitters,and bandreject fitters.But whereas the circuitsin Chapter14 wereconstructed usingsources,resjstors,capacitors,and inductors,the circuitsin thjs chapterempLoy op amps.Weshatlsoonseewhat advantages areconferred to a fitter cifcujt constructed using op amps. Audjoelectronjcsystems suchasradios,tapeptayers, and ptayers CD often providesepafatevotumecontrotslabeled "trebte"and"bass." TheseconkoLspermitthe userto select
the votumeof high frequencyaudiosjgraLs("treble")independent of the volume of tow frequencyaudio signats ("bass").Theability to independentty adjustthe amountof amptification(boost)or attenuatjon(cut) in thesetwo frequencybandsaLtows a tistenerto customjze the soundwith moreprecisionthan wouldbe providedwith a singtevoLume conkol. Hencethe boost and cut control circuit is aLso referred to as a tone controlcircuit, The PracticalPerspective exanrpteat the end of this chapterpresents a cifcuit that impLements bassvotumecontrot usinga singLeop amptogetherwith resistors andcapacitors. An adjustabLe resistorsuppfiesthe necessary controt overthe ampLification in the bassfrequency range.
607
15.1 FirstOrder LowPass Fitters andllighPass
fittel Figurer5.r A A filst orderLowpass
Consider ttre circuit in Fig. 15.1. Qualitatively, when the ftequency of the sourceis varied,only the impedarceof the capacitoris afrected.At very low frequencies,the capacitor acts like an open circuit, and the op amp circuit acts like an amplifier with a gain of R /Rr.At very high frequencies, the capacitor acts like a sho circuit, thereby connecting the output of the op amp circuit to gound. The op amp circuit in Fig. 15.1thus functions as a lowpassfilter wittr a passbandgain of Rr/Rr. To confirm this qualitative assessment,we can compute the hansfer function H(r) = %(r)/I1(s). Note that the circuit in Fig.15.1has the general form of the circuit shown il Fig. 15.2, where the impedance in the input path (Z) is the resistorR1,and the impedancein the feedbackpath (Zf) is the parallel combination of the rcsistor R2 ard tlle capacitor c. The circuit in Fig. 15.2 is analogous to the inverting amplifier circuit from Chapter 5, so its trarsfer functjon is Zt/Zi. Therefore, the transfer tunction for the circuit in Fig. 15.1is
zl Z,
" (;"1) R1
Figure15.2A A genemLop ampcjrcuit.
(15.1)
R.
(15.2)
and
1 Note that Eq. 15.1has the sameIorm as the generalequation for low passfilters given in Chapter 14, with an irDportantexception:Thegain in the passband,i(, is set by the ratio RzlR1.The op amp lowpassfilter thus permits the passbandgain and the cutoff ftequencyto be specified independentry
Plots A NoteAboutFrequency Response Frequencyresponseplots,introduced in Chapter 14,provide valuableinsight into the way a fiIter circuit functions Thus we make extensive use of frequency responseplots in this chapter, too. The frequency responseplots il Chapter 14 comprised two separateplots a plot of the transfer function mag!.itude versusfrequercy, and a plot of the transfer functior phaseanglg in degreeqversus frequency.w}len we use both plots' they are normally stackedon top of one another so that they can sharethe sameilequency a,\is.
lc.l
F i J  0 d q l o w  P dar n dP : g h  P af ist r e $
609
In this chaptor, we uso a special type of frequency response plots called Bode plots. Bode plots are discussedin detail in Appendix E, which includes detailed information about how to construct these plots by hand. You will probably use a computer to consrructBode plots,so here we summarize the special features of these plots. Bode plols differ from the frequency rcsponseplots in Chapter 14 in two impoitanl ways. First, instead of using a linear axis for the fiequency values, a Bode plot uses a logarithmic axis.This permits us to plot a wider turge of frequencies of hterest. Normally_we plot three or four decadesof frequencies,say from 1o'rad/s to 10"iad/s, or 1 kHz to 1MH4 choosingthe frequency mnge where the tmnsfer function charactedstics are changing. If we plot botl tle magnitude and phase angle plots, they agair share the frequency axis, Second, instead of plotting ttre absolute magnitude of the transfer function ve$us fuequency,the Bode magnitude is plotted in decibels (dB) versus the log of the frcquency. The decibel is discussedin Appendix D. Bdefl$ if t})e magnitude of the transfer function is lH(/'o) , its value in dB is given by ,4dB: 20 loglo H(jir). It is important to remember that while lli(jo) is an unsigned quantity, ,4dBis a signed quantity. When 2{dB: 0, the transfer function magnitude = 0. When 2{dB< 0, ttre traNfer function magniis 1, since201og1o(1) tude is between 0 and 1, and when ,4dB > 0, the transfer function magnitudeis greaterthan 1.Finally,note that : 20log1oL1/\41
3dB.
Recal tlat we deline tlle cutoff frequency of filters by determining t}Ie ftequency at which the maximum magnitude of the tmnsfer function has been reduced by 1/rZ. If we translate this definition to magnitude in dB, we define the cutolf frequency of a filter by determidng the frequency at which the maxjmum magnitude of the transfei function ill dB has been reduced by 3 dB. For example, if the magnitude of a lowpass filter in its passband is 26 dB, the magnitude used to fi[d the cutoff frequency is 26 3:23d8. Example 15.1 illustmtes the desigr of a fi$torder low pass filter to meet desircd specifications of passband gain ard cutoff frequency, and also illustrates a Bode magnitude plot of the filter's transfer function.
Designing a LowPass 0p AmpFilter
Solution Using the circuit shown in Fig. 15.1,calculate values for C and R2that,togetherwith n1 : 1 O, produce a lowpass filter having a gain of 1 in the passband and a cutoff fiequency of l rad/s. Construct the transfer function foi this filter and use it to slclc! a Bode magnitude plot of the filter\ frequency response.
Equation 15.2 gives tle passband gain in terms of Rl and R2,so it allowsus to calculatethe required valueof R2: R2: iiR1
: (1X1) =1O.
610
ActiveFitterCncuits
Equation15.3then permitsus to calculateC to meetthe specifiedcutoff frequency: ^'
1 &a. I
(1X1) = 1F. The tansfer function for the low passfilter is givenby Eq.15.1: fl(s) = _
r1 s + 1
TheBodeplot of l}I(jd)l is shownin Fig.15.3.This is the socalledprototne lowpassop amp filter, becau.e it u.esa resisror \alueof I O anda capaci tor valueof 1 R and it providesa cutoff frequency of 1 rad/s.As we shallseein the nextsection, prototypefilteN providea usefulsta ing pointfoi the designof filters by usingmore realisticcompotrent valuesto achievea desiredfrcquencyrcsponse,
vi
f*
1
0.5
1.0 @GadA)
5.0
IO
ptotofthetowpasr Flgue15.3 TheBod€ masnitud€ filterfrom ExampLe 15.1.
You may have recognized the circuit in Fig. 15.1 as the integrating amplifiei circuit introduced in Chapter 7. They are indeed the same circuit, so integralion in t]Ie time domain correspondsto lowpass filtering in the fiequency domain. This relationship berween integralion and lowpass filterilg is further contumed by the operational Iaplace transform for integrationderivedir Chapter12. The circuit in Fig. 15.4is a fi$torder highpassfilter. This circuit also hasthe generalform of the circuit in Fig. 15.2,only now tle impedancein the input path is the seriescombination of R1and C, and the impedance in ttre feedback path is the resistor R2.The transfer function lor the circuit in Fig 15.4is thus nltt: _
7. _ zi R
2
^
Flgure 15.4 Afirslorder hjghpass fitter.
l 115.4)
,,
Rz Rr'
(15.5)
and a . : *
I
(15.6)
15.1 Firstorde. LowPass andHigh,Pass Fitters 511 Again,the form of the transferfunctiongivenin Eq.15.4is the sameasthal givenin Eq.14.20,the equationfor passivehighpassfilrers.And again,the active filter permits t}le design of a passbandgain greater than 1. Example 15.2 considers the design of ar active highpassfilrer which must meet ftequencyresponsespeciJications from a Bode plot.
Designing a HighPass 0p AmpFitter Figure 15.5 shows the Bode magnirudeplot of a highpassfilter. Using the active high passfilter c1rcuit in Fig15.4,calculatevaluesof R1 and R2 that produce the desired magnitude response.Use a 0.1pF capaciror.Ifa 10 kO loadrcsistoris addedto thisfilter, how wil the magnituderesponsechangc?
30 20 10
Solution Begin by writing a transter function that has the magnitudeplot shownin Fig. 15.5.To do this,note that the gain in the passbandis 20 dB; therefore, ( : 10.Also note that the 3 dB point is 500rad/s. Equatiofl 15.4is the tmnsfer function for a highpassfilter, so tle traNfer function that has the magnitude responseshownin Fig.15.5is givenby
10r s+500 We cancomputethe valuesof R1and R2neededto yield this transfer function by equating the transfer functionwith Eq. 15.4:
'
(Rzln1)r l0r s + 5 0 0 s + (1/R1C)
3
tll
20 30
10i
5 10
50 100 o (rad/t
s001000 500010,000
Figure15,5,L TheBodemagnitude ptotofthe highpass fiLterfor
Equating the numerators and denominatorsand then simplifying,we get two equations: R,
10:;.
l
s 0 0 =_ .
Using the specifiedvalueof C (0.1pF),we find Rr = 20kO, R2 : 200kO. The circuit is shownin Fig.15.6. Becausewe have made the assumptionthat the op amp in this highpassfilte{ circuir is ideal, the addition of any load resistor,regardlessof its resistance, has no effect on the behavior of the op amp.Thus,the magnituderesponseof a highpass filter with a load resistor is the same as that of a highpassIiltei with no load resistor, which is depictedin Fig.15.5.
200ko
Figure15.6 r. Thehjghpass filterfor Exampte 15.2.
612
ActiveFitterCiLcuits
andhighpass fi(tersaldbeableto asfirst orderlowpass objectivelKnow the op ampcircuitsthat behave values catcutate theircornDonent 15,1 Computethe valuesfor R?and Cthat ]'ield a highpassfiller with a passbandgain of 1 and a cutoff frequency of 1 .ad/s if Rl is 1 O . (Notdr This is the prololype highpassfilter.)
15.2 Compuiethe resisiorvaluesneededfor the lowpassfilter circuitin Fig.15.1to producethe lmnsler function 20.000
.'1G)= .,+ sooo
Usea5l[Fcapacilor. Answer: R1: 10O, n, : 40O.
A n s w e rR: 2 : 1 O , C = l F . NOTE: Alro bt ChaptetPrcblens15.1and 15.7.
15.2 Scal.5ng Il1thc dcsignand analysisof both passiveand activefilter circuits,working lvilh elcmcnt valuessuch as 10, 1 H, and 1 F is convenient.Although Lhesevalucs are unrealistic for specifyingpractical components,they greatlysinrplili computations. After nal> o.t. Beginwith the Iowpassstage.From Eq.15.12,
RtCt
! 7958O. Finally,we need the gain stage.From Eq. 15.15,we see there are two unknowts, so one of the rcsistors can be selectedarbitrarily. Let's select a 1 kO resistor for Rj. Then,from Eq.15.15, Rf = 2(1000)
: 2,(10000),
:2000f}2ko. 1
[2i00000)](0.2x 106) ! 80O.
The resulting circuit is shown in Fig. 15.11.We leave to you to verily that t]le magnitude of this circuit's transfertunction is rcducedby 1/\4 at both cutoff frequencies, veiifying tle validity of the assumplrono.2 >> (r.1.
O.2 ttF
tigure15.1r Thecascaded opanpbandpass fitrerdesjsned in Exampte 15.5.
We can use a component approach to the design of op amp bandreject filters too, as illustratedin Fig. 15.12.Lile rlre bandpassfilrer, rhe bandreject filter consists of three sepamte components.There are important differences,however: 1. The ur tygain lowpassfilter has a cutoff frequency of .r"1,which is t]le smaller of the two cutoff frequercies. 2. The unitygain highpass filter has a cutoff frequency of o.2, which is lhe largerol the lwo culofffrequencies. 3. The gain component provides the desired level of gain in the passbands.
15.3 0p AmpBandpass andBandreject Filrels 619
30 20 10
/1 )ll
e
t
\
20
tl
Hi g h
30 40
5 10
l{l
s0 100 5001000 500010,000 o (rad/9
Figure ptotofabandrejedfitter. 15.12A Constructing iheBode magnitude
The most importantditterenceis that thesetluee comporentscannot be cascaded in serieqbecause they donot combineadditivelyonthe Bode plot Instead,we use a parallel connectionand a summingamplifier,as shownboth in block diagramform and asa circuitin Fig.15.13.Again,it js assumedthat the two cutoff fiequenciesare widely separated, so that the resultingdesignis a broadbandbandrejectfilter, and o.2 >> d.1. Then eachcomponentof the paralleldesigr car be createdindependently, and tlle cutofl tequency specifications will be satisfied. The transler function ofthe resultingcircuitis the sumof the lowpassand highpassfilter trans fer functions.From Fig.15.13(b),
n , . ' ' I ff i)l l' L r + ;,,(
, 't
'
l
(,"(r +.,.r) + r(r + o.r)\
G +..,X" + ..')
/ (15.16)
;,( (r + o.r(r + d.,) / Using the samerationaleas for the cascadedbandpassfilter, the two cutoff ftequenciesfor the transferfunction in Eq. 15.16are .r.r ard d., only if .r.2 >> {r" . Then the cutoff ftequencies are given by the equations I
R t Ct '
1 RsCs
(15.17) (15.18)
620
rigure15.13r. A pa€iLet opampbandreject fitter.(a)IhebLock djasEm. (b)rhecncuit. In the two passbands(as s + 0 and r + oo), the gain of the transfei function is R/ /R,. Therefore, (15.19)
As vrith the design of the cascadedbandpassfilter, we have six unknowns and three equationsTlpically we choosea cornrnerciallyavailable capacitor valuefor Cr and Cr. Then Eq$ 15.17and 15.18permit us to calculate]Rt and Rr to meet tle specified cutofl ftequencies.Finally, we choose a value for eitler Rf or & and tlen useEq.15.19to computethe otherrcsistance. Note the magnitudeof the transferlunction ir Eq. 15.16at the center rrequency,oo  "\/oc1,ocz:
l R ,/ + 2a"1(ja.) + r1(r,"): n' (j@.)z l;( \ 0.)" + ( o d + a d l j o . ) Rt
o,pe
)l
+ o"1@a
2r.,
Rr 2a.
(15.20)
Ifo.2 >> o.1,rhen H(jr,)l o.t. Begin with the prototype lowpassfilter and usescaling to meet the specifications for cutoff frequency and capacitor value. The ti€quency scalefactor,tf is 100,which shifts the cutofTfrequenc)from I rad s lo 100rdd/.. tte magnitude scalefactor f,, is 20,000,which pemits the use of a 0.5pF capacitor.Using these scale factors resultsir the following scaledcomponenlvalues: R' : 20 kO' C1 = 0.5pF. The resulting cutoff frequency of the low passfilter I RLCt 1 (20 x 10)(0.s x 10 6) : 100radls. We use the same approach to design thc high passfilter, startingwiti the prototpe highpassop amp filter. Here, the frequency scale factor is ,t/  2000, and the magnitude scale factor is ,ta : 1000,resultingin the following scaledcom ponentvaluesi R/r : 1 kO, Cg : 0.5s.F. Finally, because the cutoff frequencies are widely separated,we can use the ratio Rl/Rr to
0
l
".)
a  s
15 20 25 3 t1
50 100
5001000
500010,000
ptotforth€ Figure15.14"{ TheBode magnitud€ circujtto bed€signed jn Exampte 15.6. establisht}Ie desiredpassbandgain of 3. Let\ choose & = 1kO, as we are aheadyusingthat resistance R. R,tor RH. Then R/ JkQ. and ( : parallel op amp band3000/1000 3 The resulting rejectfilter circuitis sho$nin 8g.15.15. Now let's check our assumplion that a).2>> @.1by calculatingthe actual gain at the specifiedcutoff frequenciesWe do tlis by making the substitutionss : i2,(100) and r : J2T(2000) into the tmnsferfunctionfor the parallelbandreject filter,Eq.15.16and calculatingthe resultingmagnitude. We leave it to the reader to verify that the magnitude at the specitied cutoff ftequetcies is 2.024, which is less than the magnitude of 3l\a = 212 that we expect.Therefote, our rejecting band is somewhatwider than specifiedin the problem staiement
622 0.5r.F
in Exampte 156 bandrcject fittercjrcuild€sisned rigure15.15A Theresuttins your undelstandingof thismaterialbr trying ChapterPrcblems15.30and 15.31. NOTE: Assess
15.4 Higher0rder0p AmpFil.ters You have probably noticed that all ol the filler circuits we have examined so far, both passiveand active, are ronideal. Rem€mber tiom Chaptei 14 that an ideal filter has a discontinuityat the point of cutofl which sharplydividesthe passbandand the stopband.Allhough we cannot hope to constructa circuit with a discontinuousfrequencyresponse, we canconstructcircuitswith a sharper,yetstill continuous,transitionat the cutoff frequency.
ldenticalFilters Cascading How can we obtain a sharpei transition between the passband and the stopbandl One approachis suggestedby ttre Bode magnitudeplots in Fig. 15.16.This figure showsthe Bode magnitudeplots of a cascadeof idenaicalprototwe low passfiltersand includesplots ofjust onefilter,lwo It is obviousthat asmore threein cascade. and {our in cascade. in cascade. transition ftom the passbandto the filters are addedto the cascade,the lor constructing Bode plots (fiom stopbandbecomessharper.The rules filter, the [ansition occun with an Appendjx E) tell us that with one (dB/dec). pei Because circuitsin decade asymptoticslopeof 20 decibels plot, two filters a cascade with cascadeare additiveon a Bode magnitude = 40 dB/dec; for + ol 20 20 has a transition with an asymptoticslope four filters, it is is dB/dec, and for tbree filters. the asymptotic slope 60 80 dB/dec, asseenin Fig.15.16.
15.4 Higher0rder 0p Anp Fitt€rs
20 10 3 \
lLl
I
I {irst
i\
20
\
.g 30
I
\
40 ird o
50
\
r..l"nl.ol)
60 70 80 0.1
0.5
1
5
l0
tigure15.15A TheEodemagniiude ptotof a cascad€ ofidentjcat prototypendorder fi tt€rs.
In general,an nelement cascadeof identical lowpassfilten will tmnsitionfrom the passbandto the stopbardwith a slopeof 20l, dB/dec. Both the block diagram and the circuit diagram for such a cascadearc shown in Fig. 15.17.It is easy to compute the transfer function for a
C
risure 15.17 A Acascade t"*p.:l)nft',.{"1Ihe blockdiag€m.(b) Ihe cjrcujt. oria"",i*r*i vq,rrl
623
cascadeof ,, prototwe lowpassfiltenwe just multiply the individual transfer functions:
'"'
/  t \ /  t
\
/
\.l*1/\r+1/
r \
\s+1/
= (r(ry
(r5.21)
+ 1)'
The order of a filter is determined by tlle number of poles in its tmnsfer function.From Eq. 15.21,we seethat a cascadeof firstorderlowpass filters yields a higher order filter. In fact, a cascadeof n firstorder filters produces an ,thorder filter, having ,1poles in its transfer function and a final slopeof 20n dB/dec in the transitionband. There is an important issue yet to be resolved, as you will see if you look closelyat Fig.15.16.Asthe order of the lowpassfilter is indeasedby adding protot]?e lowpassfilters to t}le cascade,the cutoff frequency also changes. For example,in a cascadeof two firstorderlowpassfilters,the magnitudeofthe resultingsecondorderfilter at.r" is 6 dB,so the cutoff ftequenry oI the secondorder filter is not .d". In fact, the cutoff ftequency As long as we are able to calculate the cutoff ftequency of the highei order filten formed in tbe cascadeof firstorder filten, we can use frequency scaling to calculate component values that move the cutoff frequency to its specified location. If we start with a cascadeof n prototype lowpass filteis, we can compute the cutoff ftequency for t]te resulting nthorder low passfilter. We do so by solving for the value of o"" that results
in H(ja) : Y\nl
HG)=(,*rr€,
= : t,(*",)t la.: t), +, \r2'
.7"+ 
\\a)
'
95 = ,,1^+ r, (t5.2?) To demonshatethe use of Eq. 15.22,let's compute the cutoff f.e' quency of a fourthorder unitygain lowpassfilter constructed from a cascadeof four prototypelowpassfiltels:
V{4
= o.+:s,ua7..
(15.23)
ThuE we can design a fourth order lowpassfilter with any arbitrary cutoff frequency by starting with a fourthorder cascadecoflsisting of protot!?e
15.1 Nisher order 0pAmpFitters 625 lowpassfilters and then scaling the componentsby k, =.,16435 to placethe cutoff frequencyat any valueofo. desired. Note that we can build a higherorder lowpassfilter with a nonunity gain by adding an inverting amplifier circuit to the cascade.Example 15.7 illustratesthe designof a fourth order lowpassfilter with nonunitygain.
Designing a Foufthorder LowPass 0p ArnpFitter Desigr a fourthorder lowpass filter vrith a cutoff frequemy of 500Hz and apassbandgain of 10.Use 1 t[F capacitors. Sketch the Bode magnitude plot for this filter.
Solution
Finally, add an irverting amplifier stagewith a gain of Rf/Ri = 10.As usual,we can arbitrarily selectone of the two rcsistor values. Becausewe a.re akeady using138.46O resistorsslet R; = 138.46O; then, R, : 10tRi= 1384.6O.
We begin our designwith a cascadeof four prototype lowpass filters. We have already used Eq. 15.23 to calculate t}le c$toff frequency for the resultingfourthorderlowpassfilter as0.435md/s. A frequency scale factor of /./ : 7222.39will scale the component values to give a 500 Hz cutoff frequency.A magnitudescalelactor of fr", = 138.46 permitsthe useof 1 f.F capacitors. The scaledcomponentvaluesare thus
R : 1 3 B . 4 6 O ;C = 1 p F . 1pF
The circuit for this cascadedthe fourth,order lowpassfilter is shown in Fig. 15.18.It has the traNfer function
n(s):
t r0l
rrrr ro ::::=:
lr l
The Bode magniludeplot for this transferfunc, tion is sketchedin Fig.15.19. 1pF
tigure15.18A Thecascade circujtforihe fourthorderlowpass fiLterdesjgned in Lyampte 15.7.
30 t0
50 100
5001000 I (Hz)
500010,000
figure15.19A TheBodemagnjtude ptotforthefouth,order tow,pass fitt€rdesigned in txampte 15.7.
By cascadingidentical lowpassfiltels, we can increasethe asymptotic slope in the transition and control the location of the cutoff frequency,but our approach has a serious shortcoming: Th€ gain of the filter is not constant between zerc and the cutoff ftequency (d". Remember that in an ideal lowpassfilter, tie passbandmagnitude is 1 for all fiequencies below t]le cutoff frequency.But in Fig. 15.16,we seethat the magnitudeis less tlan 1 (0 dB) for frequencies much lessthan rhe curoff ftequency. This nonidealpassbardbehavioris bestundeistoodby looking at the magnitudeofthe transferfunctionfor a unitygaifllowpass,thorder cascade.Because
H(t :
(t + o.")''
the magnitudeis givenby
aQa)l I t/o' + a'  1 /
\
n'
115.21)
(V(d/o.,f + 1J As we can see liom Eq. 15.2, when o > (r"r), so that rhe lowpass and highpassfilter circuitscan be designedindependently ofone anotler. (Seepage620.) Highcr order activefillers have multiple polesin their transfer furctions, resulting in a sharper transition from the passband to the stopband and thus a more nearly ideal frequencyresponse. (Seepage622) The tiansfer function of an nthorder Butterworth lowpass filter with a cutoff frequencyof 1 rad/s can be determined from the equation
H(s)H(r) 
1 + (1)'sh
by ' tinding the roots of the denominatorpolynomial . assigningthe lefthalf plane roors ro fl(r) . writirg the denominatorof H(s) asa productof firsrand secondorderfactors (Seepage627628.) Tlle fundamental prcblem in the design of a Butterworthfilter is to determinethe order ofthe Iilter. The filter specification usually delines the sharpnessof the transitionband in termsof the quantities,4",o_,A,, and (,". From thesequantiti€s,we calculatethe'smlles't integer large. than the solution to either Eq. 15.42or Eq.15.46. (Seepase633.) A cascadeof secondorderlow pass op amp Iilters (Fig.15.21)with 1 O resistorsand capacitorvaluescho, sen to prcduce each factor in the Butterwortl polynomialwillFoduce an evenorderButterworthlowpass filter. Adding a prototype lowpassop amp filter will prc duce an oddo.der Butterwo h lowpass filter. (See page629.) A cascadeof secondorderhjgh pass op amp filters (Fig.15.25)with f F capacitorsand resisrorvaluescho, sen to produce each factor jn the Butterworth poly, nomial vill produce an even order Butterwo h highpassfiltei. Adding a prorotype highpassop amp filter will producean oddorderButterworth high pass filter. (Seepage635.) For both bigh and lowpassButterworth filters, frequency and magnitude scaling can be used to shift the cutoff frequency from 1 rad/s and to include realistic componentvaluesin the design.Cascadingar inverting amplifier will produce a nonunity passbandgain. (See page630.) Butterworth low,passand high,passfilters can be cas cadedto produce Butterworth bandpassfilte$ o{ any order . Buttelworth lowpassand highpassfilters can be combined in parallel with a summing amplifier to producea Butterworth bandrejectfiller of any order r. (Seepage635.) If a highQ,or nar(owband,bardpass,or bandrejectfilter is needed,the cascadeor parallel combinationwill not work. lnstead,the circuitssho$n in Figs 15.26and 15.29are used with the appropdatedesignequation$ Typically, capacitor values are chosen from those com mercially available,and the designequationsale used to specilythe resistorvalues.(Seepage637.)
646
ActiveFilterCircuits
Probtems Section 15.1
tigurcP15.3
15,1 Find the transfer function %/q for the circuit shown in Fig. P15.1if Zt is the equivalentimpedance of tle feedback circuit, Zi is the equivalent impedanceof the input circuit,and the opemtional amplifieris ideal.
Figure P15.1
r5,4 a) Using the circuit in Fig. 15.1,designa lowpass filter with a passbandgain of 15 dB and a cutoff frequenc)ot In kH,,.A..ume a 5 nf capaciror is available. Diaw the circuit diagram and label all components.
152 a) Use the resultsof Problem15.1to find the tmnsfer lunclionol lhe crrcuilqhownin fig. P15.2. b) What is the gain of the circuit as o  0? c) What is the gain of the circuit as o + N ? d) Do your ans$erslo (b) dnd (c) makesensein terms of known circuit behavior?
15.5 Designan op ampbasedlow passfilter with a cutofl frequency of 500 Hz and a passbandgain oI 10 using a 50 nF capacitor. a) DIaw your circuit, labelingthe componentvalues and output voltage. b) II the value of the feedback resistor in the filter is changed but the value of the resistor m the forward path is unchanged, what charactenshc of the filter is changed?
FigureP15,2
15.6 The input to the low pass filter designed in Problem15.5is 200cosolmV a) Supposethe power suppliesare +%.. What is the smallestvalue of %. that will still causethe op amp to operatein i1sLinearregion? b) Find the output voltage when (, : (,.. c) Find the output voltagewhen o = 0.1.10. d) Find the output voltagewhen o : 10o0.
RepeatProblem15.2,ushg the circuit shown in Fig.P15.3.
I5.7 a) Use the circuit in Fig. 15.4to designa highpass filter with a cutoff lrequenry of 40 kHz and a passbandgain of 12 dB. Use a 680 pF capacitor in the design. Draw the circuit diagram of the filter and label all the components.
Ploblenis647
15.8 Desgn an op ampbasedhigh passfilter with a cutoff frequency of 300 Hz and a passband gain of 5 using a 100 nF capacitor. a) Draw your circuit, labeling the component values and the ou4)ut voltage. b) If the value of tle feedback resistor in the filter is changed but the value of the resistor it the forward path is unchanged, what chaiacteristic of the filter is changed?
15.9 The input to the high pass filter designed in Problem15.8is 150cos/dtmv a) Suppose ttre power supplies are fy.d. Wiat is t]le smallest value of y.. that will still causethe op amp to opemte in its linear region? b) Find tle output voltage when a, : oc. c) End the output voltage when o  0.1a)0. d) Flnd the output voltage when (, : 10(,\).
15.11 The voltage tlallsfer function for either highpass prololypefiltef sho$Titr Fig.Pl5.l I is rItrl = :_Show that if either circuit is scaled in botlt magnitude and frequency,the scaled ffansfer function is (s/ kr\ H ' ( s ): . i  . ls/ Kl  + l
Flgsre P15.11 C=1F
!t
R=1O
Section152 15.10 The voltage tiansfer function of either lowpass prototypefilter showntu Fig.P15.10is I/(s) _
I
_.
Show that if either circuit is scaled in both magnitudo and frequency,the scaledtansfer funclion is l H'(s) : t s / K l t+  Figule P15.10 R=1O
15.12 The voltage tamfer function of the prototype bandpassfilter shown in Fig. P15.12is
a(r):
G)"
u.(i)".'
Show that if the circuit is scaled in both magnitude and ftequencv. t}le scaledtansfer function is
(r(;)
("')'. (iXi).' L=lH
Flgu'€P75.72 1Ft=1H
648
AciiveFiltercircuits
15.13 a) SpeciJyt}Ie component values for t}Ie prototype passivebandpassfiltei describedh Pmblem 15.12 iflhe qualiryfactoroftbe filleri.25. b) Specifythe componentvaluesfor the bandpass filrer describedin Problem 15.12if the quality factor is 25; t}le center,or resonart, frequency is 100krad/s; and t}le impedanceat resonance i s 3 . 6k O c) Draw a circuit diagram of the scaledfilter and label all the components. 15,14 An altemative to the prototype bandpass filter ilustrated h Fig. P15.12is to make o, = l rad/s, R = 1O, and I = I henrys. a) Wlat is the value of C in the prototype filter circuit? b) w}lat is the transfei funcfion of the prototype filter? c) Use the altemative prototlpe circuit just described to designa passivebandpassfrlter that has a quality factor of 20, a center frequency of 50 krad/s, atld an impedanceof 5 kO at resonance. d) Draw a diagramof the scaledfilter and label all the components. e) Use the iesults obtained in Problem 15.12to wdte the transfer function of the scaled circuit,
15.15The passivebandpassfilter illustratedin Fig. 14.22 has two prototype circuits.In the first prototype c i r c u i t ,o o : l r a d / s , C : 1 F , l,1H, and ROohms. In the second prototype circuit, od=1rad/s, R:1O, f a r a d s ,a n d C=O L: (1lQ)henrys. a) Use one of these prototype circuits (your choice)to designa passivebandpassfilter thal has a quality factor of 16 and a center frequency of80 krad/s.The resistorR is 80 kO. b) Draw a circuit diagram of the scaledfilter and label all components.
Show that tle obse ation made in Problem15.16 with respect to the transfor function for the circuit in Fig. 14.28(a)also appliesto the bandrejectfilter circuit (lower one) in Fig.14.31.
15.f
15.18 The passive bandrcject filter illustrated in Fig. 14.28(a) has the two prototlpe circuits shom in Fig.P15.18. a) Showthar for both cncuitE the transfer function is 12+1
u .( i ) ' . ' b) Wdte the transfer function for a bandreject filter that has a center trequency of 50 kmd/s and a quality factor of 5. Fig!reP15.18
't o! "" "
15.19The two prototype venions of the passive band reject filter shown in Fig. 14.31(lower circuit) are shownin Fis.P15.19(a)ard (b). Show that the transfer function for either s2+1 (;),.' FigureP15.19
'a'
,11
15.16 The tiansfer function for tie bandreject filter shownin Fig.14.28(a)is
(+)
(f)".(#) Showthat if the circuit is scaledin both magnitude and ftequency,the transfer function of the scaled circuit is equal to the transfer function of the unscaled circuitwith r replaced by (s/,tl),where,tl is the frequencyscalefactor.
E.m
Scalethe bandpassfilter in Problem 14.21so that the cenferfoequencyis 250kHz and the qualityfactor is 7.5,using a 10 nF capacjtor.Determine the valuesofthe resistor,theinductor,ard the two cuf off frequenci€sof the scaledfilter.
probtems649 15.21 Scalethe bandrejectfilter in Problem14.33to get a center tequency of 500krad/s, using a 50pH inductor.Determine the valuesof the resistor,the capacitor, and the bandwidttr of the scaled flltei. 15.22 The circuit in Fig. P13.26is scaled so that the 4 kO resistoris replacedby a 20 kO resistorand the 5 nF capacitor is replaced by a 100pF capacitor. a) What is tle scaledvalueofZ? b) W}Iat is the er?ression for io in the scaledcircuit? 15.23 Scalethe circuit in Problem 13.29so that the 10 O resistor is increased to 1 k0 and the frequ€ncy of the voltage responseis increasedby a factor of 1000.Find t,0). 15.24 a) Show that iJ tlle lowpassfilter cilcuit illustrated in Fig. 15.1is scaledin both magdtude and frequency,the transfer function of the scaledcircuit is the sameas Eq 15.1with s replacedby s/k/, where q is the frequency scalefactor. b) In the prototrye version of the lowpass filter circuit in Fig. 15.1, rr" = l rad/s, C = 1F, Rz = 1O, and Rr = 1/tri ohms. Wlat is the transfer function of the prototype circuit? c) Using the result obtained in (a), derive the tmnsferfunctionofthe scaledfiltei. 15.25 a) Show that if the highpass filter illustrated in Fig. 15.4 is scaledin both magnitude and frequency, the tmNfer function is the same as Eq. 15.4wirh r replac€dby r/t , where &l is the Irequency scalefactor. b) In the prototwe version of the highpassfilter circuit in Fig. 15.4, (,,. : l rad/s, Rl  1 O, C = 1 F, and n2 : tri ohms.Wlat is the transfer function of tlle prototype circuit? c) Using the result in (a), derivethe transferfunction of the scaled filter. Section 15.3 15J6 a) Using 20 nF capacitors,design an active broadband firstorder bandpassfilter that has a lower ,'fj:i" cutoff frequency of 2000 IIz, an upper cutoff fre;;iai quency of 8000 Hz, and a passband gain of 10 dB. Use prototype versionsof the lowpass and highpass filten in the design process (see Problems15.24and 15.25). b) Write the transfer function for the scaledfilter. c) Use the transfer function derived in part (b) to find li(/o.), where.d, is the centerftequencyof the filter.
d) What is ttre passbandgain (in decibels) of the file) Using a computerprogramof youi choice,make a Bode magnitudeplot of the filtei.
E.n
a) Using 5 nF capacitors, design an active broadband firstorder bandrejectlilter with a lower cutoff frequency of 1000IIz, an upper cutoff frequency of 5000 Hz, and a passband gain of 10 dB. Use t]le prototpe filter circuits inttoducedin hoblems 15.24and 15.25in the design process, b) Draw tle circuit diagmm of the filter and label all the components. c) What is the transfei function of the scaledfilter? d) Evaluatethe transferfunction derived in (c) at the center frequency of the filter. e) What is the gain (in decibels) at the center fiequency? f) Using a computerprogramof your choice,make a Bode magnitude plot of the lilter transfer lunction.
1528 Show that the circuit in Fig. P15.28behavesas a bandpass filter. (Hr?t find the transfer tunction for this circuitand showthat it hasthe sameform as ttre transfer function for a bandpassfilter. Use the resultfrom Problem15.1.) a) FiDd the center ftequency, bandwidth and gain for this bandpassfilter. b) Find the cutoff frequencies and the quality for this bandpassfilter. figuruP15.28 10 pF
650
ActiveFiLterCircuits
8.29 For circuits consisting of resistoG capacitoq inductorE and op amps,IH(i,) ' involves only even powers of (d.To illustmte this, compute lH(Jo) l' for the tlree circuits in Fig. P15.29when V H(') : :.
Pl5,29 Figure R
Sectiotr 15.4 15.32 The puryose oI this problem is to illustrate the advantageof an,?thorder lowpassButterworttr filter over the cascadeof r identical lowpass sections by calculating the slope (in decibels per decade) of each magnitude plot at the comer frequency o.. To facilitate the calculation, let y represent the magnitude of 1ie plot (in decibels), and let "1 = logld,. Then calculate d/dr at (o"for each plot. a) Show that at the coner ftequency ((," : 1 rad/s) of an ,?thorder lowpassprototype Butterworth filter, dy

londB/dec.
Show that for a cascadeof r identical lowpass protot)?e sections,the slopeat o. is 'l"
')(tr()t n
1\
dB/dec. 2r/" c) Compure dy/dx for each type of filter for n:1,2,3,4,and.n. d) Discussfhe significanceof the resultsobtained tII paft (c). 15.33 a) Derermine the order of a lowpass Butterworth filter that has a cutoff frequency of 1000 Hz and a gain of at least40 dB at 4000H2. b) What is the actualgain,in decibels,ar 4000It? 15.34 The circuit in Fig. 15.21 has the tansfer function given by Eq. 15.34.Show that if ttre c cuit in Fig. 15.21 is scaled in both magnitude and frc quency,the transfer function of tle scaledciicuit is PC1C,
H'(") =
/ s \ 2 , 2 / r \ , 1530 Design a unitygain bandpassfilter, using a cascade connection, to give a center ftequercy of 50 krad/s and a bandwidtl of 300 krad/s. Use 150 nF capacitors. Specify /.1, /.r, Rz, ard Rl'.
15.31 Design a parallel bandreject filter with a center frcquencyof 5 kHz, a bandwidthof 30 kHz, and a passband gain of 4. Use 250 nF capacitors,and speciJy all resistor values.
\krl
1
Rc,\4/ Rtc'r;
15,35 a) Write the hansfer function for the prototype lo$'pass Butterworth filter obtained in Problem15.33(a). b) Write t}le traNfer function for the scaled filter in (a) (seekoblem 15.34). c) Check the expression derived in pafi (b) by using it to calculate tle gain (in decibels) at 4000 Hz. Compare your result with that found in Problem15.33(b).
probLems 651 15.36 a) Using 2 kO resisto^ and ideal op amps,design a circuit that vdll implement the lowpass Butterworth filter speciJiedin Problem 15.33. The gain in the passbandis one. b) Construct the circuit diagram and label all componentvalues, 15,37 a) Using 25 nF capa€itors and ideal op amps, design a highpass unity,gain Butter,vorth filrer ,ffiTi$ with a cutoff frequency of 5 kHz and a gain of at least 25 dB at 1kHz. b) Draw a c cuit diagram of t]Ie filter and label all component value$
t5.42 a) Use 300 pF capacitors in the circuit in Fig. 15.26 to design a bandpassfilter with a quatty factor of 20, a center frequency of 8 kHz, and a passband gain of 40 dB. b) Draw the c cuit diagram of the filter and label all the components.
15.43 Show that if o. = 1 rad/s and C : 1F in the circuir in Fig. 15.26,t.heprototype values of Rl, R2, and R3are
^
o
a
15.38 Veri!, the entries ir Thble 15.1 for rl = 5 and
2Q2 15.39 The circuit in Fig. 15.25 has the transfer function given by Eq. 15.47.Show tlat if the circuil is scaled in both magnitude and ftequency, the transfer function of the scaled circuit is ("')' d'(,)
=
2 /r\ 1  n,c \4/ R,Ra' \4/
/s\'
Hence the hansfer function of a scaled circuit is obtained from ttre transfer function of an unscaled circuit by simply replacing s in the urNcaled transfer function by r/&r, where ,t, is the frequency scaling factor.
15.40 a) Using3 kn fesistors andidealop amps.designa lowpassunitygain Butterwonh filter that has a cutoff frequency of 20 kHz and is down at least 25 dB at 100kHz. b) Draw a circuit diagram of the tiltei and label aU tlte componenfs,
15.41The highpass filter designed in Problem 15.37 is cascaded with the lowpass filter designed in Problem15.40. a) Describe the twe of filter fomed by this interconnectiotr. b) Specify tlre cutoff fiequencieq tle midfrequency, and the quality factor of the filter. c) Use tlle resultsof Problems15.33and 15.38to dedve the scaledtmnsfer futction of the filter. d) Check the deiivation of (c) by usingit to calculare It0oo), wheie .r" is the midhequencyof the Iilter.
K'
Rz = 2Q'
15.44 a) Design a broadband Butterwonh bandpassfilter with a lower cutoff frequency of 1000TIz and an upper cutolf frequency of 8000 Hz. The pass, band gain of the filter is 10 dB. The gain should be down at least 20 dB at 400 Hz and m kHz. Use 50 nF capacitors in the highpasscircuit and 5 kO resistors in the lowpass circuit. b) DIaw a circuir diagram of the filter and label all the components.
15.45 a) Derive the exptession for the scaled fiansfer tunction for the filtei desigred in Problem 15.44. bJ U.ing rheexpressioD deri!edin fa L find thegain (in decibels)at 500H2 and 5000Hz. c) Do the values obtained in part (b) satisi, the fil" tedng specifications given in Problem 15.44?
15.46 Derive the prototype transfer function for a fifthorder highpass Butterworth filter by firct writing the transfer function for a fifthorder prototype lowpass Butterworth filter and then replacing s by 1/r in the lowpassexpression.
15.47 The fifthorder Butterworh filter in Problem 15.46 is used in a system where the cutoff frequency is 10krad/s. a) Wlat is ttre scaledtransfer function for the filter? b) Testyour er?ressionby fhding the gain (in decibels) at the cutoff frequency.
15.48 The purpose of this problem is to guide you plfJfltr through the analysis necessaryto establish a design procedure lor derermining t}le circuit components in a filter circuit. The circuit to be analyzedis shown in Fig.P15.48.
Fiqure P15.48
a) Analyze the circuit qualitatively and convince yourselfthat the circuit is a lowpassfilter with a passbandgain of RzlR1. b) Support your qualitative analysisby deriving the transfer function U.IU. @int: In deriving the transfer function, represent the resistors witll theirequivalentconductances,tlatiqGr: 1/Rr, ard so forth.) To make tlle transfer function useIul in terms of the entriesh Table 15.1,put it in the form
Kb.
c) Now obse e that we have five ciicuit components Rt, iR2,R3,C1,and C2 and three tlansfer function constraints(, 61, and b,. At first glance, it appears we have two free chorc€s among the five component$ However, when we hvestigale ttre relationships between the circuit componentsand the transfer function constraints, we seethat if C2is chosen,there is ar upper limit on C1 in order for R2(Gr) to be realizable.With tlis in mind, show that iJ C, = 1F, the three corductancesare given by the expressioff Gr:
part of a thirdorder lowpass Buttenvorth filter havinga passbandgainof B. a) If C2 = 1F in the prototype secondordersection, what is the upper limit on C1? b) If tlre limiting value of Ct is chosen,what are the prctotype valuesof Rr, Rr, and R3? c) If the comer frequencyof the filter is 50 kHz and C2 is chosen to be 250 pE calculatethe scaledvaluesofC1, R1,rR2,and Rr. d) SpeciJythe scaledvaluesof the resistonand the capacitor in the firslorder section of the filter. e) Construct a circuit diagam of the filter and label allthe componentvalueson the diagam.
KG2l
c, = (g\c,, tt, + t/ttt,
t41t + r1q
2(r+ r() For G2 to be rcalizable,
'
15.49 Assume the circuit analyzedin Problem 15.48is
h? 4b.(1+ K)'
d) Basedon fie resultsobtainedin (c), outline the design procedure for selecting the circuit componentsoncer, b., ard b1are known.
15.50 Interchange the Rs and Cs in the circuit in p?::TixFig.P15.48;thatis, replaceRl with Cr, R2 with C2, R3with C3,Cr with R1,and C2with R2. a) Describe the type of filter implemented as a resultof the interchange. Confirm the filter type descibed in (a) by deriving the transfertunction %/ q. Write tle transler lunction jn a form that makesit compatible with Table15.1. c ) Set C2 : C3 = 1 F and derive the expressions for C1,Rl, and R2in tems ofl., bl, and b,. (See Problem15.48for the definitionof br and bo.) d) Assumethe filter describedin (a) is usedin the sametype oftlirdorder Butterworth filter that has a passbandgain of 8. WitI C, = C3 = 1F, calculatethe prototypevaluesof Cl, R1,and R2 in the secondordersectionof the filter.
ftobhns 653
15.51a) Use the circuitsanalyz€din Problems15.48and 15.50to implementa broadbandbandrejectfilter having a passbandgajn of 20 dB, a lower cornei hequency of 800 IIz, an upper corner frequencyof 7200Hz, and an attenuationof at least 20 dB at both 1500Hz and 13.5kHz. Use 50 nF capacitors whenever possible. b) Draw a circuit diagram of the filter and label a the components, a) Derive the transfer function Ior the bandreject filter describedin Problem15.51. b) Use the transfei function derived in part (a) to Iind the attenuation(in decibels)at the center Irequency of the filter.
15.53 The puryose of this problem is to develop the designequarions ior rhecircuiliD Fig.Pl5.)1.(see Pioblem 15.48for suggestions on the development of designequations.) a) Basedon a qualitativeanalysis, describ€the type of filter implementedby the circuir. b) Vedfy the conclusion rcached in (a) by deriving the transfer tulrction V"IU. Write the tralrsfer function in a form that makes it compatible with the entdesin Table15.1. c) How many ftee choicesare there in the selec tion of tlle circuit components? d) Derive the expressions for the conductances G1 : 1/n1 and Gz = llRz in terms of C1, Cr, and the coefficientsb, and b1. (See Problem 15.48for the definition of b, and d.) e) Are thereany restictions on Cl or C2? f) Assume the circuit in Fig. P15.53is used to design a fourth order lowpass unity gain Butterworth filtei. Specify the prototype values of Rl ard R2 in each secondorder section if 1 F capacitorsare usedin the prctotypecircuit. figurcPr5.53
15.54 The fourthorder lowpass unitygain Butterworth ,?nlfll{ filter in hoblem 15.53is used in a systemwhere the cutoff frequency is 25 kHz. The filtei has 750 pF caPacitors. a) Specify the numedcal values of Rr and R2 in each section of the filter. b) Draw a circuit diagramof the filter and label all the components.
15.55 Interchange the Rs and Cs i]) the circuit in Fig.P15.53,that is, replaceR1 with C1,R2 with C2, plisJ:Jl{ anclvrcevema. a) Analyze the circuit qualitatively and predict the type offilter implementedby the circuit. b.) Verify the conclusion reached in (a) by deiving the transfer function U.IU. Writ€ the transfer functior in a form that makes it compatible with t}le enrriesin Table15.1. c) How many free choicesare there in the selection of the circuit components? d) Find R1and R2asfulctions of 6., 4, C1,and C2. e) Are there ary restrictions on C1and C2? f) Assume the circuit is used in a thirdorder ButteNortl lilter of the t}pe found in (a). Speciry the prctotype values of Rt and n2 in the secondorder sectionof the filter it Ct = gt : 1 , .
15.56 a) The c cuit in Problem 15.55is used in a thirdorder highpassButterworthfilter that hasa cul ,liTiJ" off lrcquency of 40 klIz. Specify the values of Rl and R2if360 pF capacitorsare availableto con struct the filter. b) Specifytlre valuesof resistanceand capacitance in the finforder section of the filter . c) Draw th€ circuit diagram and label all the componenlS, d ) Give the numerical expressionfor the scaled lransfer function of the filter. e) Use the scaledtransferfunction de ved in (d) to find the gain in dB at the cutolf frequency.
654
ActiveFitterCircuit5 rigureP15.62
S€ction 155 1557 a) Show that tle transfer function for a prorotype bandrejectfilter is
1lsCI
12+1
s' +([lQ\s+1
(1 o)Rz aRz
(ld)R,
Y
dR,
b) Use tlle result found in (a) to find the transfer tunction of the filter designedin Example 15.13 R4+ 2R3
1s.s8a) Using the circuit shown in Fig. 15.29, design a na[owband bandreject filter having a center frequency of 4 kHz and a quality factor of 15. Basethe designon C : 150nF.
0
L rq
q
L rC1
Rt [r ] t".rrlri
Draw the circuit dia$am of the filter and label all component values on the diagram. What is the scaledtnnsfer function of the filtei?
nr
S€€{iotrs15.115.5 1559 Using tle circuit in Fig. 15.32(a) design a volume pll$l*ii, control c cuit to give a maximum gain of 20 dB and dfii a eaiDol 17 dB al a ffeoue0cvof 40 Hz. Use an "*"" ttt t o resisror and a lOdko potenriometer. Iest your design by calculating the maximum gair at a, = 0 and t]le gain at rr: 1/R2C1 using the selectedvalues of n1, R2,and C1, 15.60 Use the circuit in Fig. 15.32(a)to design a bassvolft;Ijii€ ume control circuit that has a mardmum gain of ,e.o*, 13.q8dB that drops otf 3 dB ar 50lt. 15.61 Plot the maximum gain in decibels versus d when = 0 for the circuitdesigredin Problern1s.59.Iet tifflHlEo a vary from 0 to 1 in incrementsof 0.1. 15.62 a) Show that the circuitsin Fig. P15.62(a)and (b) are eoui!alent. b) Show that the points labeled x and y in
Fig.P15.62(b)are alwaysat the samepot€ntial. c) Using the information in (a) and (b), showthat lhecircuitin lig.I5.33cdnbedrdwna(shownin Fig.P15.62(c). Showtlat the circuit in Fig. P15.62(c)is ir the form of the circuit in Fig.15.2,where Rr+(1  d)R2 + n1R2C1s 1 + Rrclr
zf
Rl+dR2+RrlR2Cls 1 + R2C1s
15.63 An engineeing project manager has received a ;ffmi, proposalfrom a subordjnale\rho claimsrhe circuil shoqnin Fig.P15.63 couldbe u.edas a rreble!olume control circuit if Ra >> Rt + R3 + 2R2. The subordinatefu her claimsthat the voltagetranster function for the circuit is
vItG) : ; {(2R3+ n4) + (1
B)& + n.l(BRa+ ntc,s}
{ ( 2 R r + R 4 ) + ( 1 B ) & + R 3 l ( B R+a R ) C r r } where R, = R1 + R3 + 2R2. Fortunately the projecf engineer has an electdcal engineering undergraduate student as an intern arld therefore asks t]le student to check the subordinate's claim. The student is askedto check tle behavior of tlle fansfer function as (,+0; as o+oo; a]Id the behavior when (.) = oo and B variesbetween0 and 1. Based on your testing of the hansfer function do you tlink the circuit could be used as a heble volume control? Explain.
Prcbtems 655 figuruP15.63 Rr R2
'0 C,
c) Is Rasignificantly greater rhan R,? d) When B = 1, what is the boostin decibelswhen a : L/&c2'l e) When B  0, whar is the cut in decibels when a : 1/&C2? 0 Based on the results obtained in (d) and (e), what is the significance of rhe frequency 1/R3C, when R4 >> Ro?
(1B)n4I pR4 R3
15.64 In the circuit of Fig. P15.63the componenrvalues plglfiiir are n1 = Ru = 20 kO, R: = 5.9kO, Ra : 500kO, and C2 = 2.7 nF . a) Calculatethe maximumboostin decibels. b) Calculate the maximum cut in decibels.
15.65 Using the component values given in pr#flI#hProblem 15.64,plot the maximum gain in decibels versus B when a = 0. LeL P vary from 0 to 1 in incrementsof0.1.
Fourier Series 16.1 Fo!rier SeriesAnatysis: An oveffie$ p. 658 16.2 The FourierCoefficientsp. 659 16.3 TheEffectof Symmetryon the Fourier Coetfi.ients p. 662 16.4 An AlternativeTrigonom€tric form of the FourierS€riesp. 668 16.6
AnAppti.ationp. 6/0 AveragePower [at(utations with Periodic
In the preceding chapters,we devoteda considerableamount oI discussionto steadystatesinusoidalanalysis.One reasonfol this interestin the sinusoidalexcitationfunction is that it allows us to find the steadystateresponseto nonsinusoidal,but peri odic, excitations.A periodic funct'ionis a lunction that repeats itself every I secondsFor example,the triangular wave illustrated in Fig. 16.1on page 657 is a nonsinusoidal,but periodic, break waveform. A periodiclunction is one that satisfiesthe relationship f(t) = ftt x nr1,
16.7 Therms VaLueoI a Periodic fun.iion p 678 16.8 TheExponentialFormof the Fourier 16.9 Amplitudeand PhaseSpecvap. 682
(16.1)
\rhere '? is an integer (1.2,3, . . .) and I is the period.The function shownin Fig.16.1is pe odic because
f(rd = ,f(ro T\ : f(ta + T) = f(to + 2r) : 1 Beabteto catcutate the trigonometric formof the Fouriercoefficients for a periodicwaveform lsjng the definitionofthe coeffjcieitsandthe sjmptificationspossibte if the wavefomexhibits oneor morctypesof symmeht 2 Krowhowto analyze a circLrjys response to a periodicwaveform usingFouriercoefficjents powerdeLivered 3 BeabLeto estimatethe average numbe'of Fourier to a r€sistorusinsa smaLL fom of the 4 Beabteto catcutate the exponentiaL Fouriercoefficjents for a penodicwaveform and magnitude usethemto generate andphase ptotsfor that waveform. spectrum
for any arbitrarily chosenvalue of a0.Note that Zis the smallest time interval that a periodic tunction may be shifted (in either direction)to producea function that is identical1()itsell Why the il1terestin periodic functions?One reason is that periodicuave man)clcctricalsourceiol practical\alLregenerale forms.For example,nonfilteredelectronicrectifieN driven from a sinusoidalsourceproducesrectified sinewavesthat are nonsinusoidal,but periodic.Figures16.2(a)and (b) on page657showthe waveformsof the fullwave and halfwavesinusoidalrectifiers, respectivcly. The sweepgeneratorusedto control the electronbeam ol a producesa periodictriangularwavelike cathoderayoscilloscope the one shownir Fig.16.3. Electronicoscillators,which are usefulin laboratorytestingof equipmenl, are designed to produce nonsinusoidal periodic waveforms.Function generators,which are capableof producing wavefbrms, squarewave, t angularwave, and rectangularpulse are found in most testinglaborato es.Figure 16.4illustratesq.pical waveforms.
Another practical problem that stimulates intercst in periodic functions is that power generators,althoughdesignedto producea sinusoidal waveform,cannotin practicebe made to producea pure sine wave.The distortedsinusoidalwave,however,is periodic.Engineersnatuially a.re interest€din ascertainingthe consequences of exciting power systems with a slightly distorted sinusoidal vollage. T h + 2 7 Interest in pe odic functions also stemsfrom the geneml obsenation that any nonlinea ty ir an otherwiselinear circuit createsa nonsinusoidal pedodic function. The rectifier circuit alluded to eartier is one example of Figule16,1,6 A periodkwavdo this phenomenon.Magnetic saturation,which occus in both machines and transfomen, is another example of a ronlinearity that generates a nonsinusoidalperiodicfunction.An electronicclippingcircuit,which uses transistorsaturation,is yet anotherexalnple. Moreover, nonsinusoidal pedodic functions are important in the analysisof nonelectricalsystems.Problemsinvolving mechanicalvibration, fluid flow, and heat flow all make use of Deriodic functions. In fact. the study and analysis of heat flow in a metal iod led the French mathe maticianJeanBaptisteJosephFourier (1768 1830)to the trigonometric seriesrepresentationof a periodicfunction.Thisseriesbearshis nameand of a nonfiltercd sinuis the startirg point for findhg the steadystatercsponse to periodic exci Figure16.2A 0utputwaveforms soidatrectjfier. Futlwave rcctjficabon. (b) Natfwave G) lationsof electriccircuits.
tigule 16.3A ThetdanguLar waveform of a cathode'my generatoL oscitloscope sweep
vn
produced Figurc16.4a Wavetorms byfunctjon used in taboratory testing.(a)Square wave s€nelator (b)Irjanqutar wave.G) Rectansutar pulse.
16.1 FourierSeriesAnatysis: An Overview What Fourier discovered in investigating heatflow problems is that a periodic tunction can be rcpresented by an infinite sum oI sine or cosine functionsthat are harmonicallyrelated.In other words,the period of any trigonometric term in the infinite series is an integral multiple, or harmonic, of the Iundamental period Z of the pedodic function. Thus for peiodic/(r), Fouriershowedthat/(t) canbe expressedas
Fourier seriesrepresentation of a periodic fundion >
(16.2)
where,?is the integersequence1,2,3, . , , . 1n8q.76.2, a., a", and b" are kno*n as the Fou er co€fficients and are calculated ftom /(t). The telm (.)o(which equals 2?r/7) represents the fundamentelfrequ€ncyof the periodic function f(l). The integral multiplesof d0 that is,2d0,3o0,4.r0,and so on aie knorpnas the harmonic frequenciesof/(t). Thus 2o0 is the secondharmonic,3.d0 is the third harmonis and nrro is the nth harmonic of f(t). We discuss the detemination of the Fourier coefficients in Section16.2.Beforepursuingthe detailsof usinga Fouder seriesin circuit analysis, we first need to look at tlre process in general terms. From an applicarions point of vie\ we can er?ress all the periodic functions of interest in terms of a Fourier series.Mathematically, the conditions on a periodic function /(1) that ensure expressing/(,) as a convergent Fourier series rknown as Dirichl€t's conditions) are that 1. /(t) be singlevalued, 2. f(t) have a finite number of discontinuities in t})e periodic inteival,
3. /(r) havea finite numberof maximaand minimain the periodic interval, 4. theintegral I
f(t)ldt
Anypedodic functiongenemredby a physicallyrealizablesourcesatisfies Dirichlet'scondjtions.Theseare sufficientconditions,not necessaryconditions.Thus if f0) neets these requircments,we know that we can expressit asa Fouder series.However,if/(t) doesnot meettheserequ ements,we still may be able to expressit as a Fouder series.The necessary conditionson /(t) are not known. After we have determined /(t) and calculated the Fouder coefficients (d,, a,, and 6,), we resolve the periodic source into a dc source (ao) plus a sum of sinusoidal sources(r, and 6"). Becausethe periodic souce is driv ing a lhear circuit, we may use the principle of superposition to find the steadystateresponse,Inparticulai,we fust calculatethe rcsponseto each sourcegeneratedby the Fouriersedesrcprcsentationof/(t) and then add the individual rcsponsesto obtain the total response.The steadystate
16.2 lhe Fourier Coefficients 659
response owing to a specific sinusoidal source is most easily found with the phasormelhodofaraltsit The procedue is straightforward and involves no new techniques of circuit analysis. It produces the Fouder series repiesentation of lhe steadystateresponse;coirsequently, the actual shape of the response is unknom. Futlemrore, the responsewaveform can be estimated only by adding a sufricient trumber of tems together. Even though.the Fouier seriesapproach to finding the steady:state.respolsedoes have some drawbacks,it introduces a way of thinking about a problem that is as important as getting quantitative results. In fact, the conceptual picture is even more important in some rcspectsthan the quantitative one.
16.2 TheFourierCoefficients AJter defining a periodic function over its fundamental period, we determine the Fourier coefficients from the rcIationships
(16.3)
(16.4) < Fouriercoefficient5 (i6.5)
In Eqs. 16.4 and 16.5, the subscdpt k indicates the kfi coefficient in the integersequence1,2,3, . . . . Note that a, is the averagevalueof /(t), dr is twice the averagevalue of /(1) cos,/rol, and 6k is twice the avemge value of /(r) sin toot. We easily derive Eq$ 16.316.5fiom Eq. 16.2by recalling the following hteglal relationships,which hold when m and n are intoge$:
J,,,
s]nme)otdt = 0.
Ior all fi,
(16.6)
'o"^'$ o' =
0,
for all m,
\16.7)
0,
for all m and n,
(16.8)
'"o"^ '"'on'ro' : f','" f,,^"
"io^,0, "n no,n, a,
fotallm+n,
T 2'
(16.e)
660
' ,o' ^,,,,,or r.o, a!
f'
o.  t ,T
to(a)tm  n. . I o fm = n .
(16.10)
We leaveyou to vedfy Eqs 16.616.10inProblem16.5. To derive Eq. 16.3,we simply integrateboth sidesoI Eq. 16.2over one pedod: f,. r
J,.
,,,,0, 
for,
J"
\""
6
lio+?
= I
: \ )a,cosn6r b.sinno!ldt
a,&+>l
/h+?
\ a , c o sn d a t b a s i n n u d ) d r
= a;f + O.
(16.11)
lqualronl6.J followsdirecllytrom tq. 16.IL To dedve t}le expiession for the ft1h value of a,, we first multiply Eq. 16.2by coskol and then irrtegiate botl sidesover one period of /(t): l'"rr I
f(t)coskaotdt:
J
J
>/
lh+r axcoska& dt I
h
 at cosnutt!coskuol + bnsinnuutcoskdutldt
= 0 + a // (Tt\r  0 .
r i o z. )
SolvingEq.16.12for ak yieldsthe er?ressioninEq.16.4. We obtail the expression for the frth value of b, by filst multiplying borh sidesof Eq. 16.2by sin /..rot and then integrating each side over one period of /(t). Exampte16.1showshow to useEqs.16.316.5to find t}le Fouder coefficients for a specific periodic function.
Findingthe FourierSeriesof a Triangular Waveform with NoSymmetry Find the Fouder series for the periodic voltage shownin Fig.16.5.
Figure16.5 A Thepeiodicvottage for Example 16.1.
Solution When usingEq$ 16.!16.5 to find a,, dr, and b*, we may choose the value of t0. For the periodic voltage of Fig. 16.5,the best choice for to is zem. Any other choice makes the rcquired integntions moie cumbe^ome.fhe expiessionfor o(t) between0 and fis
*r:(?)
16.2 ]lreFoudef Coeffici€nts661 The equationfor a, is
Thc equationfor the kth valueof D, is
,.:il'(?)'"=in"
b k= :
I
l+)L\i\k,tdl
2v^l
1 = __I_:_cln/.dur
Thisis cle"rlylhA,,cos(naot 0"),
(16.38)
where,4,,andd,,aredefinedby thecomplexquanlity
o^ jt,, = {d, + 41_!" = t,t a".
(16.39)
We dcriveEqs.16.38and 16.39usingthe phasormelhodto add the cosine and sinc terms in Eq. 16.2.We begin b]' expressilgthe sine functionsas cosinefunctions;thatis.we rewdte Eq.I6.2 as f(.i):
a,+ )a,cosn,,'t + b,,cos("oo1 90").
(1640)
Addlng the termsunder lhe summationsign by usingphasorsgives E{a,cos naot} : a, p
0{b, co.1rer q0)} : 0,,l lq =
(16.41)
1n,,.
116.42)
Then 2l{a,,cos(naxt+ ,, cos(roi1
90')} : a,
jb"
(16.13)
Senes 669 Tisonometric Form 0ftheFouder 16.4 AnAtternahve When we invene+ransfom Eq.16.43,weget
 tt',:= + b,cos(nont ancosn.,at ^,)2rr'^l r,r. nr.*, SubstirutingEq. 16.44into Eq. 16.40yields Eq. 16.38.Equation 16.43 cofesponds to Eq. 16.39.If the periodic function is either even or odd, ,4,,reducesto either a, (even) or 6i (odd), and 0, is either 0" (even) or 90'(odd). The denvation of the altemative form of the Fouder seriesfor a given periodic function is illustrated in Example 163.
voltage fourierSeriesfor Periodic calculatingFormsof the Trigonometric a) Dedve the expressionsfor a&and ,l for t}le peri odic functionshowninFig.16.11. b) wrire lhe first lour rerm. ol lhe Fourierseries representation of ?J(t) using the format of Eq. 16.38.
and )
bk:
Irl4
;.lr
vsinka.atdt coska6t r/a\
: r2V\ /
k , " " )
=( rF\ ( '  .t 9/ ) b) The averagevalueof t(l) is
T r3 f 4
2
1
T sT 3f '7f 2r 1 2 4
for ExampLe 16.3. fundoon Figure15.11A lhe pedodic
u^g/4)
jbklor k:1,2,and3arc
Thevaluesof ar
..
vr,
',Evv.45',   t .v
a,
tD
t
ib 0
Solution a) The voltage z,(r) is neither even nor odd, nor does it have halfwave symmetry Therefore we useLqs.lo.4and 16.5lo frnd,, andb . Choo.ing toaszero,we obtain
a ,  i..b 
a t9O
t''
v^ ^
.v^ \a2v
i:
^
/
t3s".
Thus lhe lirsl four term. in lhe Fourier.eries representationof ,0) are
a,] a,* l',,{o)"o"r,.,,, *: ll"'''r^"'"0.", rr  "' ,.,1r1
4
t
' "/iw 2c651ud n
+ lijilqrs(3@ot
^5')
:: co5(2@0/ q0')
135') + ...
670
FoderSenes
0bjediv€1Be abteto catculate the trigonometric formof the Fouriercoefficients for a periodicwaveform 16.4 a) Conputc /1 ,4sand dL 0\ for the pcriodic functionshownif Y,, = 9f; V. b) Using the format of Eq. 16.38,wdte the Fourier seriesfor o(r) up 1()and including the fiftir harmonicassuningZ : 125.66ms. Answer: (a) 10.4,5.2,0,2.6.2.1V and 120", 60', not defincd, 120', 60'; (b) 0(, = 21.99+ lo.4cos(5or 5.2cos(100t 60")+ 2.6 cos(200t 120') + 2.1cos(250t 60")V.
120') +
NOTE: Ako tt)' ChaptetPrcblem16.18.
16.5 An A*ulieatioll Now wc illuslrale how lo usea Fouricr seriesreDreseDtution of a neriodic c \ . ' , r l r o nt u n ( r i u lno l i n dr h c c r c a o\ y( a l er e \ p o ; . c u at i n e a r c . r i u iIr}.e RC circuil shownin Fig. 16.12(a)wjll provide our example.'$c circuil is cncrgizedwith the periodic squarcwave voltageshown in Fig. 16.12(b). Thc vollageacrossthe capacjioris the desiredresponse. or output,signal. The fint stepin finding thc steadyslareresponseis to represcntthe pcriodic excitationsourcewith ils For.der series.After noting that the sourcehas odd,halfwave.and quarter \vavesymmetrywe know !ha! Lhe Fouder coefficientsreducoto ,r, \rilh k restrictedto odd integervalues:
s y'ia bk: ; I V,,sinkart dt t,ln AV
= + iDl Figure16.12?, AnRf circuitexcited bya penodic (a)Ihe8r series v0ttaqe. cncuit.(b)Ihe squaewave
(r is odd).
(16.45)
Then the Fouricr seriesrepresentationof % is
,"
''in,.,.
AV
_* >
\16.46)
Wriling the seriesin expandcd[orm. lve have 4V .
4V^. ^ it
4V* + sin
4l_ 5dl + _ sinTdnr+
11,6.47)
The voltagc sourceexpressedby Eq. 16.:17is the equivaleDtof infidtely many sericsconnectedsinusoidalsourccs,cach sotrce having its orvn amptitudeand ficquercy.To fird the contributionof cach sourceto lhe output voltage,wcusethe principleof superpositiorl.
16.5 AnApplicatjon 671
For any one of the sinusoidal souces, the phasordomain expression for the output voltage is "
(16.48)
1+ i@RC
Al1 the voltage sources are expressedas sine functions, so we interpret a phasor in terms of the sine instead of the cosine.In other wordq wher we go from the phasor domain back to the time domain, we simply write the timedomain expressionsas sin(o, + d) instead of cos(ol + d). The phasor output voltage owing to the fundamental ftequency of the sirusoidalsourceis
v , =
(4u,JrJ /O'
(16.49)
1+ i@oRC
Writing Vl in polar folm gives
v,r :
$u^)/  9r
(16.50)
nY1 + u6R'C'
91 :
(16.51)
tan'ooRc.
From Eq. 16.50,the timedomain expressionfor the fundamentalfrequency component of ?J.is
trv 1 + @6R'C'
in(@d
Br).
\16.52)
We derive the tlirdhamonic component of the output voltage ir a simiphasorvolrageis lar manner. The rhirdbarmonic (4v/3n\,t0" "' I + 73oxRC
4V.
(16.53)
/Ft,
3,\h + s"?rfrC 13oonc. 83 : tan
(16.54)
Thetimedomainexpressionfor the t]Iirdharmonicoutput voltageis sin(Jdnr P r' ,  !h . 3:'V  + guiR'C'
(1655r
Hence the expressionfor the kthharmoniccomponentof the output voltageis
4V^
sin(koot k ^,4TF&Fe
Br:
tan 'koonc
 Bk)
(r is odd).
(k is odd), (16.56)
(16.57)
We now write down the Fou er se es represertationof the output voltage: "
. .
4Y : sin(n@o/ B,) t ,=(ls...nt/t + (rdoRc)2
(16.58)
The derivation oI Eq. 16.58was not difficult. But, althoughwe have an analyticexpressionfor the steadystateoutput,what?r,G)looks lile is not immediately apparcnt from Eq. 16.58.As we mentioned ea.rlier,this shortcomingis a problem with the Fourier sedesapproach.Equation 16.58is not useless, however,becauseit givessomefeel for the steadystate waveform of x'o(t), if we focus on the frcquency response of the circuit. For example,if C is large, 1/r?ooc is small for the higher order harmonics.Thus the capacitorshort circuitsthe highfrequencycomponentsof tle input wavelbrm,and the higher order harmonicsin Eq. 16.58are negligible comparedto the lower order harmorlics.Equatior 16.58reflecasthis condition in that.for larue C.
"#"e+"in('oo1 eo")
"#. F.;""'"'
(16.59)
Equatior 16.59showsthat the amplitudeof the harmonicin the output is decreasingby 1/n', comparedwith 1/n for the input harmonics.IfC is so large that only the fundamentalcomponentis significant,then to a first
Do(r, n
_cosoiy',
(16.60)
and Fouiier analysistells us that the squarewaveinput is defomed into a sinusoidaloutput. Nowlet's seewhathappensasC 0.The circuit showsthat ooard ?s are the same when C = 0, b€cause the capacitive branch looks like ar open c cuit at all frequencies.Equation 16.58predictsthe same result because, as C + 0, x^:
>
srnndnr.
(16.61)
But Eq.16.61isidenticalto Eq.16.46,and theieforeod rs asC +0. ThusEq. 16.58hasprovenusefulbecauseit enabledus to predictthat the output will be a highly distorted replica of the input waveform if C is large,and a reasonablereplica if C is small.In Chapter 13,we looked at the distortionbetweenthe input and output in tems of how much memory the systemweightingfunction had.In the frequencydomain,we look at the distortion betweenth€ steadystate inpul and output in terms of how the amplitude and phaseof the harmonicsare altered as they are transmitted through the circuit. w}len the network significantly alters the amplitude and phaserelationships among the harmonics at tlle output relative to that at the input, the output is a distortedversion of the input. Thus, in the frequencydomain, we speak of amplitude distortion and phasedistortion.
16.5 AnApptication 673
For the circuit here,amplitude distortion is prcsent becausethe amplitudesof the input harmonicsdecreaseas 1/n, whereasthe amplitudesof the output harmonics decreaseas
" \/1 + (rr"Raj' This circuit also exhibits phase distortion becauset}le phase angle of each input hamonic is zero, whereasthat of the flth harmonic in the output sig nal is  tan ' ,?ooRC.
An Applicationof the DirectApproach to the SteadyState Response For the simple RC circuit shown in Fig. 16.12(a),we can derive the expression for the steadystate responsewithout resortingto the Fourier series representation of the excitation function. Doing this extra analysis here adds to our understanding of the Fourier seriesapproach. To find the steadystate er?ression for o" by straightforward circuit analysis,we reason as follows. The squarewave excitation function alternates betweer charging the capacitor toward +ya and y. After the circuit reaches steadystateoperation, this altemate charging becomes pedodic.We know from the analysisof the singletime'constantRC circuit (Chapter7) that the rcsponseto abnpt changesin the driving voltageis exponential.Thus the steadystatewaveform ol the voltage aooss the capacitorin the circuitshownin Fig.16.12(a)is asshownin Fig.16.13. The analyticexpressionsfor o.(t) in the iime htelvals 0 < t < Z/2 andT /2 =  = T arc
Toward+I4,,
Figure16.13/\ Thesteadystate wavefom ofr, forthe
q = v + (vt  V)e1txc ,
( 1 6 . 6 2 )c j r c u i t i F n i q1. 6 . 1 2 ( a ) .
, t ) .=
(16.63)
O=t=T/2; V  + ( V 2 + U ^ ) e V ( t 2 ) ) / R CT, l 2 < t < 7 .
We dedveEqs.16.62and 16.63by usingthe methodsofChapterT, assummarized by Eq. 7.60.We obtain the valuesof y1 and /2 by noting ftom Eq.16.62that
v2:v^+
(V  v;ert2Rc
(16.61)
andfromEq.16.63 that rt2Rc U : v + (v2+ v.\e SolvingEqs.16.64and16.65for Vrandy2yields v1t er 2Rc 1 vt: V=
(16.65)
(16.66)
Substituting Eq.16.66 into Eqs.16.62 and16.63gives
u": v^
t)/ r=tr J\
x coslnoat
ei)dt.
(16.76)
Now' usingthe trigonometricidentily co\aco(P
t 2coi{o' P)
l cos(o B}. 2
we simplify8q.16.76to P  vd.td. :>
,'
 0., + cos(2r?a,ol
lcosro., n...1
/
qi')lrlt.
(16.1t)
The secondterm under the integralsig! integratesto zero,so
+ i?*,(pP : v,r.Id.
e,,1.
(16.78)
Equalion 16.78is particularly important becauseit statesthat in the case of an interactior betlveena periodic voltage and the correspondingperiodic current! the total averagepower is the st1lltof the averagepowers obtained from the interaction of currents and voltagesof tlle samefrequency.Currents and voltages of different frequencies do not interact to produce avetage power. Therefore, in averagepowercalculations involving periodic furclions, the total average power is the superposition of the average powe$ associatedwith each harmonic voltage ard curent Example 16.4illustrates the computation of avcragepower involving a periodic voltage.
Functions677 16.6 AveraqePower Catculations withPenodic
Average Power for a Circuitwith a Periodic VoltageSource Catculating Assume that the periodic squarewave voltage in Example 16.3is applied aqoss the terminalsoI a 15 O resistor. The value of y is 60 Y and that of f is 5 ms. a) wrile lhe {irstfivenonTerolermso[ thefourier cefle\ represenrarion of r(r). Lse rhe rrigonometric form given in Eq. 16.38. b) Calculate the average power associatedwith eachterm in (a). c) Calculate the total avemge power delivered to the 15 O resistor. d) Wlat percentageof the total power is delivered by the first five terms of the Fourier series?
Thus,using the first five nonzeroterms of the Fourierseries, ?r(r): 15 + 27.01cos(400,r 45") + l9.l0cos(800,r 90') + 9.00cos(1200rt 135') + 5.40cos(20002r 45') +.
V.
rsatpliedrolhelefminal. oI a resi' b) tte \ olrage tor, so we can lind tlre power associatedwith eachtermasfollows: 152
P,,.===lsW. l5
",=ff="o.t"*.
Solution
119.10'= 12.16 W, 2 t 5
a) The dc component of ?,(t) is
(60)(z/4) d,=
T
:15V. P3:
From Example16.3we have
A2: 60lr = 19.10v,
;1s
= 2.70W.
'p' = : i : a : 2 1 5
A1  .,A6olt = n.01v, dr : 45"
l q 2
n 0 7w
c) To oblair the total average power delivered to the 15 O resistor, we fi$t calculate the Ims value of z'(t):
0z: 90''
ltrr)'Q/4) : v900: 30v.
4=20\f2ln:9.00v, The total averagepower deliveredto the 15 O
d3 = 135" At=0,
3n2
P' = :
or= o',
t5
d) The total power delivered by the fint five nonzerotermsis
ds:45"' 21 do: t:
b0w.
2r(1000)= a U Urt a d / s . 5
P:
Pa"+ Pr+ P2+ P3+ Ps : 55.1W 5 .
This is (55.15/60X100), or 91.9270of the total.
16.7 ThermsValueof a Periodic Function Tbe rms value of a periodic function can be exprcssed in tems of the Fourier coefficients; by definition,
; I
(16.7sJ
fl)'dt
Representing f(t) by its Fouder sedesyields
n ",
lt t"l
Vrl,
N
2
(16.80) >A.costndae r "t )dt.
f"
The integral of the squared time function simplifiesbecausethe ody terms to survive integration ovei a pedod are the product of the dc term and the harmonic products of the samefrequency.All other products integrateto zero.ThereforeEq.16.80reducesto  ?
' /
\
;\dr + >t,All t:4 sl:1l
I
?'\\,2.)
'
(16.81)
Equation 16.81states that the rms value of a periodic function is tlle squareroot of the sum obtainedby addingthe squareofthe rms value of each harmonic to the squa.reoI tle dc value.For example,let's assumethat a periodicvollageis fepresenled by lhe linileseries r'  10 + 3Ocos(rrd 01)+ 2'cos(2aot  a2) + 5 cos(32,,0, 93) + 2cos(5dor 05). The rms valueof this voltageis
1G + Qol\//)' + (2V\r42 + (5/\"r2)2+ (2/^"2\2
\,564s: n.65v.
16.8 TheExponentiat Fornrofthe FourierSeries 679 Usually,infinitely many terms are requiredto representa pedodic function by a Fourier series,and thereforeEq. 16.81yieldsan estimateof the true rms value.We i ustratethis resultir Example16.5.
Estimating the rmsVatueof a PeriodicFunction Therefore,
UseEq.16.8110estimatethe lms valueof the voltage h Example16.4.
Solution
,v, ' n [ ^ , 1 3 1 1 1 ' u r , 1 '  r i o o ] '  , r . , o l ' V \rr/ \'r/\'z/'\':/
FromExample16.4, Ydc: 15V, U = 21.01/\,4v,
= 28.76v.
the Ims valueof the fundamenlal From Example 16.4,the true rms value is30VW.3 a p p r o a crhh i s v a l ube\ i n c l u d i n g m o r e am nd oreharmon l n ( r m s \ a l u e o l h e l h r f dh a r m o n r c i ; . i n F q . r b . 8 r . F oeri a m p l e . i f w e i n c l u d e r h e h a r m o n i c i throughk : 9, the equationyieldsa valueof 29.32V the rms valueof the fifth harmonic.
V2= 19.101^'ay, the rms valueof the secondharmonic, v quu v2v V. = 5.401",4V,
NOTE: AssessJow understandins ofthb materisl by trying Chaptet Prcblems 16.36and 16.37.
Form 16.8 TheExponential of the FourierSeries The exponential form of the Fourier series is of interest becauseit allows exDonentialform ofthe seriesis us to exDressthe seriesconciselv.The
f(t) = > c"et*"t,
c,;
(16.82)
1 t4'r fo)enbr'dt I
(16.83)
To deriveEqs.16.82and 16.83,we return to Eq. 16.2and replace thl] cosineandsinefunctionswith their exponentialequivalents: (16.84)
2
(16.85)
2j SubstirutingEqs.16.84and 16.85into Eq.16.2gives
f(t) = '" + ;?G,^' , e, '1, \1",'"
= ".
n, 1
* ('" +rjb.)",^, oa.wt 2(";'o^)r.^,
Now we define C" as  tt^
.
"4
A i^\ 
t"n,
) "n
'/
ab
n _ t.2.3.
(16.8i)
) 4
Fromthe definition of c", tf)
c.
t'ot
+l+ z L t JIn :;
1 fh+r
I , Jr, 1
=+ I r
tttr+T
Jtr
r
r"
I  i', It h ftrtsin r,rdrI tutco,nunrdr tJt I
f1)@osn..iat isinna{)d.t
ft)e i""'rdt,
(16.88)
which completes the deivation of Eq. 16.83.To complete the derivation of Eq.16.82,wefiist observeftomEq.16.88 that co=;
1 ftr+T ,,i(t)dt = a.. I I
(16.8e)
Jtt
Next we note that C.'
I t " ' I  ib,t. Jtttc"d d! C)  :ta. r I
(lo.eo,
J,,
yields 16.89, and16.90inroEq.16.86 Substituting Eqs.16.87, f (r) = Ca + >C"d^r
+ Cie j"'"t)
= >Creitu] + >C,iitu.
(16.e1)
Note that the secondsummationon the dghthandside of Eq. 1691 is C,e/t'dfrom 1tocc'netnad = >cnenb&.
(16.e2)
Becausethe summationfiom 1 to co is the sameas the summation ftom oCFt^c + >chetnad n=0 = >c"er*4, which completes the derivation of Eq. 16.82.
(16.e3)
'6.8
rormoI lhe loJderSe:e. T\e kponeoridl
681
We may also expressthe Ims value of a pedodic function in terms of the complexFourier coefficients. From Eqs.16.81,16.87,and 16.89,
(16.e4)
\/a;4 Ld :
d;'
(16.95)
(16.96)
Substituting Eqs. 16.95 and 16.96 into Eq. 16.94 yields the desired
(16.97)
Example 16.6 lllustrates the process of finding the exponential FourierseriesreDresentation ofa Deriodicfunction,
findingthe Exponential Formof the FourierSeries Find the exponential Fourier sedesfor the pedodic voltageshownh Fig.16.15.
= J:!\c
t"r, 2  et"u, 2)
= 2U. no'1t nPsln
rl2
0r/2
T ./2 T T+r/2
Figure16.15Alhe pedodicvottage forExample 16.6.
Here,because o(t) hasevensymmery,bn = 0 for all n, and hencewe expectC, to be real. Moreover,tbe amplitudeof C, foljowsa (sin.v)/.r indicated distribution,as whenwe rewrite 
Lr:
Solution Using ,2 as the starting point for the integiatior\ we have,from Eq.16.83, a
=11
vh T
v ainvnl
sln (naor/2) Vr t
1
nanrtL
We saymore about this subjectin Section16.9. represenralron The exponenl'alseries oi rC) is
44\'in ("0"/2)/",," ,,,,, ' = S / T t na\rtt2 ,,="\ = 1u_,) S \ r 1,1
sjn(nott/2r,^u ' nor/2
682
objective4Be abteto catcuhtethe exponential formof the Fouriercoefficients for a periodicwaveform 16,8 Derive the cxlressionfor the Fburier coclli cienisCi fbr the pedodicfunction shown. llirr:Takc advanlageof symmetryby usjngthe fact that C, = (a"  jb")i2.
16,9 a) Calculatethc rms valLreof the periodiccnrreni in AsscssmentProblem168. b) Using CrCn, estimatethe rms valuc. c) What is the perccntageo[ error in the value obtainedin (b),basedon thc true value found in (a)? d) For this periodicfunction,couldfewer terms be usedto estimatethe rms valueand siill insurcthc error is lessthan l%?
I (ns)
Answer: (a) \'/34 A; (b) s;7'7'7 Al 0.93 (c) %! ( d ) y e . : i r( I ( e a r eu \ c d . l h e f f o fi . 0.9870.
Answer: C, = j!(1 +3cos?).nodd NOTE: Ako trt ChapterProblens 16.11and16.45.
16.9 Asnglituete andFiras*Spe*trs Aperiodicrime functionis detiiedbyits Fouriercoefficientsand ils period. ID other words.$,hcnwc know d,, d,, 0,,,and Z. we can construcl/(r), at leasttteoretically.When we know d,,and b,,,we alsoknow the amplitude (A,) anclphaseanglc( d,,)of eachharmonic.Again,wc cannol,in genelal. visualizewhat thc pedodicfunctiontooks like in thc lime domair from a description()1lhe coefficienrsand phaseangles;ncvcrlheless, we recognize that thesequantiliescharacterizeihe periodic funclion oompletelnThus. rlith sufficicntcomputingtime, we can synthesizcthc lime domain wavelorm from thc amplitudeand phaseangledata.Also,whena periodicdriving tunction is exciiirg a circuit that is higtrly hequeDcyseleciive.the Foudel seriesof tlle stead\,statc rcsponseis dorninatedbyjust a fe$Jtcrms. Thus the descriptionof the rcsponsein tems of anplitude and phascmay providean uDderstanding ofthc outpul $aveform. We can presentgraphicallythe descriptionof a periodic function in tcrnrsor the anpiitude and phascanglcol eachterm in the Fourier series of ft).'Ihe plot of the amplitudc of cach term versusthe frequencyis callcdthc lmplitrde speclnlmof/C), and lhe plot oI the phaseangleversusthc frequercyis calledthe phasespectrumol f(r). Becausethe amplitudc and phaseangledata occur al disclelevaluesof the frequency(that is,at rr0.2rd0.3o0, . . .),lhcsc plols alsoare Ieferled to asline spectra.
An lltustrationof AmplitudeandPhaseSpectra Amplitude ard phasespectraplol\ are basedon either Eq.16.3E(,4,,and 0,) or Eq.16.82(C,,).we focuson Eq.16.82aDdleavethe plots basedon Eq. 16.38to Problemi6.4ll.To illuslrale the amplitudeand phasespcctra,
16.9 AmpLitude andPhase Spectra 683
which are based on the exponential form of the Fourier sefes, we use the pe odic voltage of Example 16.6.To aid the discussion,we assumethat V = 5Vandr = f/5. Frcm Example16.6, (^ r :
sin(naaal2)
vhr ; = I noar/z
(16.98)
which for the assumed values of y and r reduces to
(r6.9e) Figure16.16illustrates theplot of themagnitude ofC, fromEq.16.99lor valuesof r rangingfrom 10 to +10.The figue clearlyshowsthat the ampliludespectrumis boundedby the envelopeof the (sinr)/.rl function.Weusedthe orderof the harmonicasthe frequencyscalebecausethe numedcalvalueof Z is not specifled.lvhen we know I, we alsoknow o0 andthe trequency correspo0ding lo eachharmonic. Figure 16.17providesthe plot of ( sinr)/rl versusr, wherc r is in mdians It showsthat the function goesthrough zero wheneverr is an Flglre16.16A TheptotofC, versus u whenr = I/5, irtegral multiple of n'. FromEq. 16.98, for thepenodjc vottaqe for Exampte 16.6.
".o
(16.100)
From Eq. 16.100,we deduce that the amplitude spectrum goes ttrrough zero whenever n4r is an integer. For example,in the plot, "/Z is 1/5, and thereforethe envelopegoesthrough zero at n : 5, 10,15, 10, 15,and so on. In othor words, the fifth, tenth, fifteentl,... harmonics are all zero. As 2n l.sn n 0.5n0 the reciprccal of /f becomesan increasingly larger integer, the number of harmonics between every ,7 radians inueases If nr/f is not an inreger, (sinr/, versus 16.17 lhepLotof '. the amplitude spectrumstill follows the l(sinr)/xl envelope.However, Fiqure the envelope is not zerc at an integral multiple of oo. Because C, is rcal for all ,, the phase angle associatedwith Cn is either zeroor 180",dependhgon the algebmicsignot (sin .''l5)/(n1tl5). For example.lhe phaseangleis zero for /?  0.  l. +2. 3. and 14.Il i< not defined at ,r = +5, becauseC+sis zero.The phaseangle is 180' at n = t6, t7, t8, and +9, and itis not definedat +10.Thispattem repeaK itself as,r takes on largei integei values.Figure 16.18showsthe phase angleofC, givenby Eq.16.98.
180' 90' 1 5 1 3 1 1 9 7 5 3 1 74 72 1,0 A 6 4 2 Flgure16.18A Thephase angteof C,.
3 5 7 911 13 15 2 4 6 8 t O 7 2 1 4t 6
684 Now' what happcns 1o lhe amplitude and phase speclm if t(t) is shiftedalong the time axis?To Iind out, we shift the periodic vollagein E r a m p l (1 6 . 6 , un n l \ t o r h . r i g h l B ) h ) p o r h e . i \ .
,,()
: s
(16.101)
Therefore
,(r
to):
>
C,p)no\( ti =
t
C,,e t"qt'Ej"qt,
(16.102)
which ildicatesthat shiftingthe odgin hasno effecton the ampliludespec
(16.103)
However. referenceto Eq. 16.87revealsthat the phase spectrumhas changcdto (An+ no0t0)rads.For example,lel's shift the periodicvoltage in Example16.1r/2 units to thc right.As before,we assumethat r = 7/5: thcn the new phaseangle81,is
ai=
( 0 h +n r / s ) .
(16.104)
We havepio$cd Eq. 16.104in Fig.16.19fou rangingfron 8 ro +8. Note that no phascangleis associated with a zeroamplitudecoefficient. You may lvonder why we have devoted so much attenrion ro the amplitudcspeclrumof the periodic pulsc in Example 16.6.The reasonis that this partic! ar periodicwaveformprovidesan excellentway io illus irate thc transitionfrom the Fouricr seriesrepresentationof a periodic lunction to the Fouder transfom representarionof a nonpcdodicluncFigure16.19r! Theptoiof0;,versus n for Eq.16.104'. tion.We discussthe Fouricr translonnin Chaprer17. "
objettive4Be ableto calc!latethe exponential forrnof the Fouriercoefficients Jora periodicwaveform 16.10 The fuDctionin Assessment Problem16.8is shifl 0.
nn'""', rur,1  \ 4. , /ll\ i
17.2 The Fouder transfom of/(t) is givenby
r(,,)  0, r@) : 4, F(a\ : L. F(a)  4, tq(l,) = 0,
30. r>0,
c) f(4  te'',
r>0, r 0 and 1for t < 0.The signumfunctionis denotedsgn(1)and canbe expressedin termsofunit'stcp functions,or sgn(r)
sgn(t)= (t) ,(r.
\r7.?4)
Figure17.6showsthe functjon graphicaily. To find the Fourier lransformoI lhe signumfunction.we first crcatea Iunctionthat approachesthe signumfunctionin the limit:
sgn(i)= liqk ''u(t) FiguE17.6A Thesignum functjon.
I\4
171.25)
The function insidethe brackets.plotted in Fig. 17.7,has a Fourier transibrm becauscthe Fouder integralconverges. Becausc/(r) is an odd function. we useEq.17.23lo find its Fourier transform:
s{/(,)}:i;1" ,*1"=,, I
Fisure17,7A A function thatapproaches sgn(l) as
e"'u( r)1, € > 0.
1
2j. \11.26)
17.4 Folrier Transfoms in theLimit 707
As €  0,/(1)  sen(t),ands\f(t)I + 2lia.'I}Jercrorc,
sfte'(r)]: 3.
\17.27)
TheFourierTransform of a Unit StepFunction To find the Fourier tmnslorm of a ud1 step function, we use Eqs.17.18and 1'7.21.We do so by recognizing tlBt the unitstep function can be
I
I
'l{r)=t+tssn(r).
117.2s)
Thus,
="{1} . *{f"*,,,} *r,,,,r = .a@), !.
l.17.29)
TheFourierTransform of a CosineFunction To fiDd t]re Fourier transform of cosdot, we retum to the hvene{ransform integal ofEq.17.9 and obse ethatif F(a):2r6(aad,
(17.30)
then
rc =lv,ao 2a/(az+e]),a>o
positive' and negativeline eponential
2n6(a  @o) ,[6(@+ed+6(0ab)] j,[6(@ +.qr)  6(0  aro)]
17.5 SomeMathematical Properties The first mathematical property we call to your attention js that F(o) is a complei'quantity and can be. e4)ressed in eithet rectangular or polar form. Thus from the defining integral, f(t e i't dt
F(') :
= ./
/(tXcos,t  i sin'",,t)dt
i : l)ot.*,,., l
fosinatdt.
(17.36)
17.5 Some l,4aihemahcat Prcpedies
e1,y=
"o",t at l 71t1
n*,1:
ly6si^,tat.
(11.31)
(17.38)
Thuqusingthe definitionsgivenby Eqs.17.37and 17.38in Eq.17.36,weget F(o) = A(a) + jB(o\ 
F(,o)leia\o.
\17.39t
The following observations about F(.d) are pertinent: . The rcal part of F(rr)that iq,4(o)is an even function of d;in other words,A(o) = ,4(/d). . The imaginary pa of F(@)that is, B(o)is an odd turction oI o; in otler wordq B(o) : A( (,). . The magnitude of F(.d) that is, V,4'(o) + B2(.d)is an even func. The phase angle of F(/,,)that is, 0(t:,) : tan lB(o)lA(a);s an odd function of .",. . Replacing .,, by o generatesthe conjugate of F(o); in other words,
F(4 = r'@).
Hence, if /(r) is an even function, F(o) is real, and if /(t) is an odd function,F((o) is imaginary.If/(1) is even,from Eqs.17.37atrd17.38,
4.) =zl
t)cosatat
\17,40)
and ,(@) : o.
117.41)
A(o) = 0
\17.4?J
ff f(t) is an odd tunction,
and
oa,t: z
I
ttn"in.ra,.
117.43)
derivationsof Eqs. 17.40 17.43for you as Problems17.10
and17.11.
709
7X0
IheFounerTransforn If /(r) is an even function, its Fouder tnnsform is an even function, and if f(l) is an odd function,its Fou er transformis an odd function. Moreover,if/(t) is an evenfuflctior, from the inverseFourier hregral, r
tV)
r

1
1
tt
Jr.
:
*f',',,*"'
l/
6
l
;a)P"'d
t_6
+ is".dt\d'n
r 1 . y c o s , r a+, o
: =
Figure 17.8A A reciangutar frequenq spectrum.
f
tvr''d^
^ l zn
I
Ala) cost d.
117.11)
Now compareEq. 17.44with Eq. 17.40.Note that, exceptfor a factor of 1/22, these two equations have the same folm. fhus, the waveforms of A(d,) a'rd f(4 become interchangeable if f0) is an even function. For example,we have alreadyobservedthat a rectangularpulse in the time domain produces a frequency spectrum of the form (sin o)/zo. SpecificanX Eq. 17.11e4ressesthe Fourier transformof the voltagepulse shownin Fig.17.1.Hencea rectangulaipulsein the frequencydomainmustbe generated by a time domain function of the form (sin r)/i. we can illustrate tllis requirement by finding tie timedomain function f(/) corresponding to the frequencyspectrumshownin Fig.17.8.From Eq. 17.44,

tU):^
2 tt
I znJr
zMt\inutJ"
M c o s u t d u = l  I tn\

 a
:+1.*y) =*(^\#)
\11.45)
We say more about the ftequency spectnm of a rectangular pulse in the time domain versusthe rectangularftequency spectrumof (sint/t after we introduce Parseval'stheorem.
17.6 0perationaI Transforrns Fourier transforms,like Laplace tmnsforms, can be classified as functional and operational. So far, we have concentrated on the functional tnnsform$ We now discusssomeof the importantoperationaltmnsforms.With regardto the Laplacetransform,theseoperationaltransformsare similar to thos€ discussedin Chapter 12. Hence we leave tieir proofs to you as Problems17.1217.19.
17.6 operationat Tnnsfoms 711
Multiplicationby a Constant From the defidng integral, if
s{f o} = F(a\,
slxf(t\j = rr@)
'46t 117
Thus,multiplication of /(t) by a constant conesponds to multiplying F(o) by that sameconstant.
Addition(Subtraction) Addition (subtraction)in the time domaintanslates into addition (subtmction)in the frequencydomain.Thusif s{f t(t)} = Fla), tt{f2(t)} : F2@), s{f 3G)}: F3@),
s{f{t)
f2(t) + /3(r} = r(.d)  Fr(a) + F3G)),
117.47)
which is derivedby substitutingthe algebraicsum of timedomainfunctronsinto the definhg integal.
Differentiation The Fou er transfoft of the firclderivative of /(t) is
cl!+l
I d . J
 i.r@t.
tt7.1l)
The/ltl derivativeof/(t) is !j\
I a' "',;' trti I d t
I )
ti@YF(@).
Equations17.48and 17.49are valid if/(l) is zero at +co.
Integration
sa\= l';ata,,
(17.1e)
tlten
=+. s{s(tr\
(17.50)
EquatioD17.50is valid if
f(x) drc: 0.
ScateChange Dimensionally, time and frequency arc rcciprocals. Therefore, when time is shetched out, frequency is compressed(and vice versa), as reflected in the functional transform
e t l { d / rl l r ( 1 ) . a \a /
,' u.
(17.51)
Note that when 0 < d < 1.0,time is shetched out, whereaswhen a > 1.0, time is compressed.
Translation in the TimeDomain The effect of translating a function in the time domaitr is to alter the phase spectrum and leave the amplitude spectrum untouched. Thus  a)} : er" e@).
s{f(t
117.52)
ffl' is positivein Eq. 17.52,the time function is delayed,and if d is negative, the time furction is advanced.
Tnnstationin the Freguency Domain Tfanslation in the frequency domai[ coresponds to multiplication by the complex exponential in rhe time domain: ei{ej.&fG)} : F(a,  at).
(17.53)
Modulation Amplitude modulation is the process of varying the amplitude of a sinusoidalcarier. If the modulatingsignalis denoted/(t), the modulatedcarder becomes/(1) cos.dot.The amplitude spectrum of this carier is onehalf the amplitude spectrum of /(t) centered at tl')o, that is. .}{/fr) co( dor} 
I 2Ftur\
I 2ltd
uot.
Ot.r)
Convo[ution in the TimeDomain Convolution in the time domain coresponds to multiplication in tie hequency domain. Ir other words, )(t)

/I r(^)h(t ^)d^
J 
17.6 0perationatlransfoms 713
e{)(t)} = Y(o) : x(a\H(a).
117.55)
Equation 17.55 is important in applications of the Fou er transform, becauseit states that the transform of the response function y(.r) is the product of tle input ffansform X(l,)) and the system function l1(@). We saymore aboul this relatiotrship in Section 17.7.
Convotution in the Frequency Domain Convolution in the ftequency domain coresponds to finding the Fouder transform of the product of two time functions.Thus if
f(tt : f{t)f2(t),
' 6 1= j f ' ' , 1 , y r 7 .  a o ,
(17.56)
Table 17.2 summarizesthese ten opemtional transforms and another operational transform that we introduce in Prcblem 17.18.
^n
n(@)
Kf(t)
KF(o)
/n,  ,f,t) + R(r)
F1@)
.t^f(t)/dt'
0o)"F(o)
flat)
1p{q1.,'o
Fr(@\ + Fx@)
t(at 1 ;F(a
/,t^rrt,
^ir^
1 ad +;F(a
+ ao)
x(@)H(@)
flt)ho
i J *FI@F{'
{f(t\
u)'i;
')d"
714
TheFourier Transfom
obiective1Be abteto calcutate the Fourier transform of a function 17.4 Supposcf(r) is definedasfollows:
11= t 14 t + e .
l=
t = tt,
0 rl
 /,il
l= qio.,r,*l',iul
lzssl
l:osl
The transpose of the sum of two matices is equal to the sum of the rransposes.lhal is. (A + B)r = Ar + B7.
(A.i5)
fhe transpose of the product of two matrices is equal to the product of the transposestaker in reverse order. In other words, IABII:
BrAr.
(A.16)
llatrixALgebG 769
770
TheSolutjon of Lin€ar Sinruttaneous Equations
Equation A.46 can be extended to a product of any number of matri
IABCD]I = DrCrB/Ar.
(4.47)
If A : Ar, tne marrix is saidto be symm€tric.Only squarematrices canbesynmetric.
A.8 ldentity,Adjoint,andInverse Matrices An idenlity rnatrix is a square matrix where at : 0 for i + j, aJJdaii : I for i = j. In other words, all the elements in an identity matrix are zero exceptthosealorg the main diagonal,wheretley are equalto 1.Thus
0 1 0 0
0 0 1 0
t ; i lt i : l l "  [ i are all identity matrices.Note that identity matricesare alwayssquare.We will usethe slmbol U for an identity matrix. The adioint of a matrix A of orde X ,?is defined as
adjA: tArrl"x,,
(A.48)
where Aij is the cofactor of a,;. (See Section A.5 for the definition of a cofactor.) It follows from Eq. A.48 that one caII t]rinl of firding the adjoint of a square matdx as a twostep pmcess.First construct a matrix made up of the cofactors of A, and tlen transposethe matrix of cofacton. As an examplewe will find the adioint ofthe 3 x 3 matrix
[ 1 2 3 l A=l 3 2 11. L 1 1 5l The cofacto$ of the elements in A are
A 1 r 1 ( 1 0 1 ) : 9 , A r , : 1 ( 1 5+ 1 ) : 1 6 , a13:1(3+2):5, Azr=1(103):7, Arr:1(5+3):8, Ar: 1 ( 1+ 2 ) = 3 , : 431 1(2 6): 4, A3' = 1(1 9) = 8, 433=1(26)=4.
A.8
The matrix of cofacton is
[e R=l7
16 sl 8  31 . 8
L4
4l
It follows that the adjoint of A is
l s  t  + l
a d j A = B r =l  1 6 8 8 1 . l s  3  4 1 One can check the arithmetic of finding the ad.ioint of a matdx by using the tleorem adjA.A:
detA.U.
{A.4e)
Equation A.49 tells us that the adjoint of A times A equals the determinant of A times the identity matrix, or for oul example, detA : 1(e) + 3(7)  1(4) = 8. If we let C = adj A . A and use the technique illuslrated in Sectiotr A.7, we 6nd the elements oI C to be
q r = 9 2 1+ 4 =  8 , q2=18744=0, cB=27'720:O, c2r16+248:0, c22=32+16+8=8, cx=48+8+40=0, c a =l 5  9 + 4 : 0 , caz:10640, ca3:15320= 13.
Therefore
o ol
o o'l
[8 [t c = l 0  8 o l :  8 1o 1 o l L 0 o  8 1 L 0 o 1 l = detA'U. A squarematdx A has an inve$e, denoted as Ar, if
A1A=AAr=u.
(4.50)
Identjty,Adjoint,andInve6ellaticet
77!
772
lhe Solutjon of Linear Simulianeous Equations
Equation A.50 tells us that a matrix either piemulripLied or posrmultiplied by its inverse generates the idertity matrix U. For the inve^e matrix to exist,it is necessary that the determinantofA not equalzero.Only square matriceshaveinverses,and the inverseis alsosquare. A foimula for findina the inverse of a mat x is ,
adi A
(4.51)
The formula in Eq. A.51 becomes very cumbersome if A is of an order laiger than 3 by 3., Todaythe digital computereliminaresthe drudgeryoI havingto find the inverseof a matrix in numericalapplicationsof matdx algebra. It foilowsfrom Eq.A.51 that the inverseofthe matrix A in the prev! ousexampleis
: "'['i_il 7 8 3
: l
o.s;s I r.rzs ) 1  o.orio.rri
:rl
You shouldveriJythatA tA= AA r:U.
A.9 Partitioned Matrices It is often corvenient in matrix manipulations to partition a given matrl\ into submatrices.The o ginal algebraicopemtionsare then carriedout in terms of the submatdces.In partitioning a matdx, the placement of the partitionsis completelyarbitrary,with the oire restrictionthat a partition must dissectthe entire matrix.In selectingthe pa itions,it is also necessary to make sure the submatdcesare conformableto the mathematical operationsin which they are involved. For example, consider using submatrices to find the product C : AB.where
ti L:
2 3 4 3 0 2 1  1 2 1
;:l
? You can learn altemarive ncthods ior inding ihe inveEe in dy introdudory text on natrix theort. See,lor eranple,Fnnz E. Hohn Elcmetuat! Mari,,lkbla (Ncw york:
I'latrices 773 A.9 Patitioned and
" [ : ]
Assume that we decide 10 partition B into two submatdceq 811 and Bzt;thus
":fn'l
"
[B",.]'
Now sinceB hasbeen paftitioned into a two'row column matrix, A must be partitioned irto at least a twocolumn matrix; otheiwise the multipiication ca rot be pedomed. The location of the vertical pa itions of the A matdx will dependon the definitions of 811and Bn. For example,if
T r_ l 8 ", , : l


f1l

ttJ
0 l a n d 8 ', , = 1 , .
L lt
then Att must coltain tlree columns, and A12 must contain two colurms, Thus the partitioning shown in Eq. A.52 would be acceptable for executins the Droduct A3:
l 5t r4 c=l_10
3
2
3
i:l I
2
L:;i
.
(A.52)
3 0
Il on the other hatrd. we DartitioD the B matrix so that
"". ""=[i] "":[3] then A11must contain two columns, and AD must contain thrce columns In this casethe partitloning shown in Eq. A.53 would be acceptablein executins the Droduct C : AB:
iil
[11 131
c=l1 0
2 0 I 3 0
(4.53)
774
TheSolution ofLinear Simulhneous Eqoations For puiposes of discussion,we will focus on the partitionhg given in Eq.A.52 and leaveyou to verify that the partitionirg in Eq.A.53leads to the sameresult. From Eq. A.52 we can write
 tB"l A118.. 4'12821.
g
1l ' e'rl
la.,'J
(A.54)
It followsfrom Eqs.A.52 andA.54 that
"^:li 1i]nl"i] ":[.i:]'li] and
"[j] The A matri{ could also be pa itioned horizontally once the vertical partitioning is made consistent with tie multipiication operation. In this simple problem, the horizontal paltitions can be made at the disdetion of the analyst. Therefore C could also be evaluated using the partitionhg shownin Eq.A.55:
1 5
c=
2 4
3 3
4 2
5 1
2 0 I

1 O 2 l  3 1 0 1  1 1 0 1 0 2 1 1 2 0
3
.
(A.55)
A,9 Fdftitioned l,latrices t75
crrArlBi+A,rB1. c2ta2tilllArB2.
You shouldverif] thal
't
tr 2 1
c , ' = l;; ; l l o l 'f l4; s,lll lll ; l Lal
_ f  ' l + f u l _I 1 1 l L 7l
['
L 6l
o
L13l'
2l[ 2l
r'l
[i ll c , r  ol   , l l o l l 9 ' l ; ] L o 2 1 l L  1 1L 2 o . l ' trl t t
t t sl t t r:'l t l
= l 1 l + l0 l : l 1 1 . L11 L61 L 7l and
'
[".]
I 13'l c = l  1 31 .
L;l
776
Ihesotution of Ljnear Simuttaneous Equations
A.10 Apptications The following examplesdemonstrate some applications of matdx algebra in circuit analysis.
Use the matdx method to solvefor the node volt ages?]1and 02in Eqs.4.5and 4.6.
Solution
It follows frcm Eq. A.62 that the solutions for x'r and 1,2are obtain€d by solving for the mat x productArI. To find the invene of A, we first find the cofactorsof A.Thus
^,r( 1)10.6):0.6, ^1r=( 1f( o.s):0.5,
The first stepis to rewrite Eqs.4.5and 4.6in matix notation. Collectingthe coefficientsof r,1 and t2 and at the same time shifting tlle constant terms to t}le righthandsideofthe equationsgivesus 1.7rr
0.522: 10, (A.57)
d,1 : ( 1)3( 0.5) : 0.5,
Ln= ( r)\1.1):1,.i. The matrix of cofacton is
. 13i ?tl
0.5ttt+0.622:2. It followsthat in matrix notation,Eq. A.57 becomes
I r.7
o . s tl u , l t l o l
L u . s o . oLl o l L z l ' AV=1,
(A.5e)
lnh rrrl a d' j A  B '  l  . ,:. LU.) 1./l
A:
v:
L U.J
U.OJ
n sl l r 1 . 7 , { 0 . o r , 0 . 2 5 r 0 .  ? . v ul
(A.60)
From Eqs"A.65 and A.66, we can write the nverse of the coefficient malrix, that is,
t ,' .,t .
^ ,
tb2)
trol t = l: .
1 fo.o o.sl olz os tz
To fiDdrhe elemenrs ot lhe v malfi\. we premultiply both sidesof Eq. A.59 by the inverseof A; thus (A.60)
EquationA.60 reducesto
o.sllrol o ',' : z1oo[o.o o . sr . z z l 1oo[7] 77 8.4I
 e.oel ' I i0.91
(A.68)
It follows direcdythat (4.61)
It,l
l,')
v = ArL
(467)
Now the productA1I is found:
t z
Uv=A1I,
I t7  ".
.._1,
A'Av=Art.
(4.65)
The determinantof A is de'A
0.51
(A.64)
and the adjointofA is (A.58)
 1;7
(A63)
(A.62)
I e.oel
110.911' or ?)1: 9.09V and z)2= i0.91v.
(A.6e)
A.10 Applicatjons 777
1)'(45  80): 12s,
Usethe matdxmethodto find the threemeshcur rentsin thecircuitin Fig.4.24.
Ar1:(
Sotution
^,r = (1)5(1oo 2s) : 12s,
t}le me.hcuffenr equationrlhat de\cribelhe cifcuit in Fig.4.24are givenin Eq.434Theconstraint equationimposedby the currentcontrolledvoltage sourceis givenin Eq.4.35.Wlen Eq.4.35 is substituted into Eq. 4.34, the following set of equations
^i = (1f(20 + 200\:220,
L.n : ( 1)4Q2s 1oo): 125,
ar:
( 1)'( 1oo1oo)=200,
r\I
( 1 1 6 ( 2 5205 )  2 2 5 .
Thecofactormatrix is 25ii
5i2
20i1= 50,
6s 7ol
 ,o = B  125 125 r2s L220 200 22s)
5ii+10ir4t3=0, 5i1 4i2+9l:Q
( A7 0 )
from which we can write the adjoint of A:
In matrix notation,Eqs.A.70 reduceto AI = V,
r,
fra (A.71)
rrs
d d i A B /   " t
ltnl
, r i , o o
ps ns)
ln
5 2ol
(A.14)
,o,n
The determinant of A is 2U 25 5 5 : 10 4 detA
I A=l s 10 4, e) L s 4
s
ti'l
r= ; .
4
9l
: 25(90 16)+ s(45  80) s(20+ 200)= 125.
;,
It followsfrom Eq. A.73 that
and
12s 2zol
. f74 A r = ;l 6s 12s 2001.  '70 125 225)
['l vl 0l L 0l It follows from Eq. A.71 that the solution lbr I is
t=A'v.
(4.72)
We find the inverseofA by usingthe relationship , A'=
adi A :.. det A
(A.73)
To find the adjoint of A, we first calculate tie cofac toIS of A. fhus All:(
1 F ( 9 0 1 6 t ) = ' 7 4 ,
Arz: ( 1)3(4s  20):6s, dr3:(1)4(20+50):70,
(A.76)
The solutionfor I is
,
+ r:s )ol so  [2,].oo'l
I  r r { l b s 1 2 5 , , n 0 '"1o
0l=
r2s )251L ol
2 6 . 0 0] . ( A . 7 / )
L28.oo.l
The meshcunents follow directly from Eq. A.77.Thus
I i;l Izo.ol ,, ,r
126.0  128.01
{A.78)
or i1 = 29.6 A.,i2: 26 A, and13: 28 A. ExampleA.3 illustrates the application of the matrix method when the elements of the matrix are complex
778
Ihe sotution of Linear sinrultaneo6 Equations
Usethe matrix methodto fhd the phasormeshcur, rents11and 12in the circuit in Fig.9.37.
The cofactormatrix B is B
Solution
26 [( t 02
i13) I 27  lrb)l j16) (13 lia) ]
(A85)
Summingthe voltagesaroundmesh1 generatesthe
The adjoint of A is
(1 + j2)rl + 02  j16)(rr r) = 1501!:. (A.7e)
 B / f l : : t : l : l l :  l : : ' ( A 8 o ) adiA L( )7 jr6) (r3 jr4r)
Summingthe voltagesaroundmesh2 producesthe equation
The delerminantofA is
(12 y'6xr, 11)+ (1 + j3)r' + 39r,:0.(A.s0) The current controlling the dependent voltage
t, : (rr
Iu).
fl3  illlf2h  ,lJr
(A.81)
AJter substitutingEq. A.81 hto Eq. A.80, the equationsare put into a matix format by lirst co ecting,in eachequation,t}le coefficientsof 11and Ir; thus fll li4)l r 2 7 i l b r l
.","=lflit',ill,[i.';,2; : 60  j4s.
AI=V,
(4.83)
l12 jt6rl "^ _ f D / 1 4 127+ tt6 (26+ jrll l' fr , l  = l;'1. andv: Lr2.l
L
u
It followsfrom Eq.A.83 that (A.s4)
The inverse of the coefficient matdx A is fourd usingEq.A.73.In this case,thecofactorsofA are ^u=( 1 ) , (  2 6 j r 3 ) : 26 j13, Liu= e\3(.2'7 + j16) = 2'7  j16, ^I,l : (1)i(12 + j16) = 12  j16.
Ar, = (1f03
j14)= t3
(60  i45)
14.
(4.88)
EquationA.88 canbe simplifiedto
6 0+ i 4 s f t  2 6 l 1 J ) 0 2 l l b ) l 'a r 5 b 2 sL l 2 7 j l o ) ( r 3 j 1 4 ) l I  65  ,r3o 96
t28l
:'sl oo lr+s o+, rr;]
(A8e)
SubstitutjngEq.A.89 into A.84 givesus
I
I = ArV.
(A.87)
[(26 113) 02  i16)l t(27 j16) (13 j14))

l 'r' s. n1/.r r l
jtb)
Theinve$e of the coefficientmatrix is
rll l l o r t /= l 5 n 0 " . , ^  . 1 6o / i n. t 2 b j l r l )
No\using matrix notation.Eq.A.82 is written
I l2  jtn)Q
t l , l  I f ( 0 5 1 1 3 0 ,( q o 1 2 8 )l l s o 0 r.:l :S foO llasl (qa lt7) U ] l( 26  js2\l (A.e0) l(24 jsst)' It followstrom Eq.A.90 that j 5 2 \ 5 8 . 1 4 , l l o . 5 7 ' A . lt  ( )6 ,r o , ( 2a  /58)  r.2.'7, 122.48'A. ' '' l
In the fiist three exampleEthe matrix elementshavebeennumbenreal numbersin ExamplesA.1 and A.2, and complexnurnbersin Example A.3. It is also possiblefor t}le elementsto be fimctions Example A.4 illustrates the use of matrix algebrain a circuit problem where the elementsin the coefficient matrix are functions
A.10 Apptkations779
Use the matrix method to derive expressions for the node voltagesy1and y, in the circuitin Fig.A.1.

Sotution Summingthe curents away ftom nodes 1 and 2 gereratesthe followingset of equations:
u v "  j31 K
+ llsc + (V  y,)sc :0.
u + (v2 v)sc + (/, ;
/'4's2)
Letting C = 1/R and collecting the coefficients of Y1and Y2gives us
(G+2sc)vrscvz:Gvs, sCU+ (G + 2sc)v2= scv{
(A.e3)
Writing Eq.A.93 in matrix notationyields
AV:t.
fG + 2sc
A=L
rc
A.4. tiguruA.1 d Ihe circLrit for E/"ampte
Ys)sc:0
(A.e1)
Thedeteminantof A is rC I lC+2sC derA  ^ ^ ^ ^ l  G ' + t : C C + 3!C2. t s L (A.98)
The inverseof the coefficientmatrix is
[c + 2rc
sc I
G+zlcl L rC (G2+4scc+3s2c2)
sc I
c+2{c]'
(A.ee)
It foLlowsfrom Eq. A.95 that
n =Lh) f l l . " ' a r =fL.r( cyr ll i l
lG+ 2sc
It followsfrom Eq.A.94 that
v=ArI.
ttl:
L sc
(4.e5)
As before, we find the inverse of the coefficient matrix by first finding the adjoint of A and the determinant of A. The cofactors of A are All=( lflc + 2sCl : 6 1 2tg. ^r,=(1)3(sc)=rc, lzr=(r)3(sc):sc, 4,, : (1)4[6 + 2scl = G + 2sC. .a I ^ "+ ' , ^lsLl . Cr
(G' +4rcc+Ji2c2)
(A.100)
Carryiflgout the matdx mu]tiplicationcalledfor in Eq.A.100 gives
1 tvll = l(e +2scc + Cc'\v"1 q'cc + tCCll 1z"cc+ z?c'\v, )' lv,] fc, + (A.101)
Now the expressions for l{ and y, can be written d ectlyftom Eq. A.101;thus
The cofactoi matdx is l t: + ) " a B= ^"' L .rc
sc llcysl
G+2sc)lsc\)
lG2+z\cc+?c' lvs (A.102) ' (.G2 + 4sCG+ 3s2cz)
\A.96J
and therefore the adjoint of the coefficient matrix is
and
2(sCG+ ?c2)vs (G'+4tcc+3"tC\
(A.103)
780
TheSolutjon of Lin€ar Simultaneous Eq0ations
In o r filral example,we illustralehow matrix algebracan be usedto analvzethe cascadeconnectionof two twooc,ncrrcuns.
Show by meansof matrix algebrahow the input variablesyr and 1r can be describedas functiom of the output vadablesy2 and 12 in the cascadeconnectionshowr in Fig.18.10.
Theseconstraintrelationshipsare substitutedillto Eq.A.104.Thus
lrlt:;: Illl l s:n a\zllvrl latl at I r\ )'
Solution We begin by expressing, ir matri\ notation, the relationshipbetweenthe irput and outputvariables of eachtwopo circuit.Thus
(A.107)
The relationshipbetweenthe hput variables(14,1r) and the output variables (y2, 1, is obrained by substitutingEq.A.1O5into Eq.A.107.Theresultis
f"'l : fai, ,i,lf,i, ,i,1f",l { A ' 0 8 )
trl
_ l^i., ,:,,11v51 (4.104) L.t d.ll ril
trt .at
at) lai
, , ' , 1 / . , 
After multiplying the coefficient malrices, we have
tlll
lAli]
(A'05)
Now the cascadeconnection imposesthe constraints (4.106)
ttl
(a\1ai,+ a:labJlv,] + l(a\pL o\d!,) + abqh t(41a'i1 ) U,) @iph + ai2ah)
(A.10e)
Note that Eq. A.109 correspondsto writing Eqs.18.72 and 18.73in matrix form.
Appendix
Complex Numbers
Complex numben were hvented to pemit the enraction of the squarcroots of negative numben Complex numbers simplify the solution of prcblems tlat would otherwise be very difficult. The equation I + 8x + 41 : 0, for example, has no solution in a number system that excludes complex numbe$. Thesenumbeis, and the ability to manipulate them algebraically, are extremely useful in circuit allalysis.
8.1 Notation There are two ways to designatea complex nurnbei: wit]l the cartesian,or rectangulai,form or with the polar, or trigonometriq folm.In the r€ctarg ar Iornr, a complexnumber is written in tems of its real and imaginary components;hence n:a+
jb,
(8.1)
whered is the real component,b is the imaginarycomponent,andj is by definitionVl.1 In tlle polar form, a complex number is written in terms of its magdtude (or modulus)and angle(oi argument);hence (8.2) where c is the magnitude, d is ttre argle, e is the base of t]le natural logaithm, and, as befoie,j = V1. In the literature,the symboll:q is frequentlyusedin placeof erd;thatis,the polar form is written
,t = c1!.
(8.3)
Although Eq. B.3 is more convenient il printing text material, Eq. B.2 is of pfmary importance in mathematical opemtions because the rul€s for manipulating an exponential quantity are well known. For example,because * O)" = l', then (et1' = pto; lgsausey = 1/r,, $et: eig  1/ei0; and so foith. Becausethere are two ways of expressingthe samecomplex number, we need to relate one form to the other. The transition from the polar to the iectangularform makesuseofEuler's identity: (8.4)
r You may be nore fanilid wirh the notatior ; = \/= In electrical engineering, i is used as the s)dbol lof curent. and tence in eleclrical engineering liierature,l is Ned to derote
781
742 A complex number in polar form can be put in rectangular form by writing cel:
c ( c o s d+ j s i n d ) = ccosd + icsind)
(8.5)
The tmnsition from rcctangular to polar form makes use of the geometry of the dght tdangle, namely, / _\ a+tD=lva'+t1'lep \ /
(8.6)
tan9 : b/a.
(B.7)
It is not obvious ftom Eq.8.7 in which quadranfthe angle p lies.The ambiguity can be resolvedby a graphical represertation of the complex number.
8.2 TheGraphicaI Representation of a Comptex Number A complex number is represented graphically on a complex number plane, which usesthe horizontal axis for plotting the real component and tigure8,14 ThegEphical rcpresenrarion of' + i, lhe \erricala"\istor plollinglhe imdginar)componeDt.The angleot rhe when , andb arcbothpositive. complex number is measuredcounterclockwise ftom the Dosifive real axis. Ile graphicalplot of lhe complexnumber,  a + 1i t 7o",;f *e assume rhata andb arebolh positive. is sho\D i0 Fig.B.t. This plot makesvery clear tlle relationship betweenthe rectangular and polar foms. Any point in the complexnumberplane is uniquely defhed by giving eithei its distance from each axis (that iE a and b) or its iadial dis tance from the origin (c) and the angle of the mdial measurementp. It followsfrom Fig.B.l that d is in the first quadrantwhen d and 6 are 4+j3 = 5tr0!?" 4+j3:5/143.13. borhpo