Numerical and Analytical Methods with MATLAB for

Numerical and Analytical Methods with MATLAB® for Electrical Engineers

William Bober • Andrew Stevens

Bober • Stevens

ISBN: 978-1-4398-5429-7

9 781439 854297

9 0 0 0 0

K12515

Electrical Engineering

Combining academic and practical approaches to this important topic, Numerical and Analytical Methods with MATLAB® for Electrical Engineersis the ideal resource for electrical and computer engineering students. Based on a previous edition that was geared toward mechanical engineering students, this book expands many of the concepts presented in that book and replaces the original projects with new ones intended specifically for electrical engineering students.

This book includes:

• An introduction to the MATLAB® programming environment • Mathematical techniques for matrix algebra, root finding,

integration, and differential equations• More advanced topics, including transform methods, signal

processing, curve fitting, and optimization • An introduction to the MATLAB® graphical design environment,

Simulink®

Exploring the numerical methods that electrical engineers use for design analysis and testing, this book comprises standalone chapters outlining a course that also introduces students to computational methods and pro-gramming skills, using MATLAB® as the programming environment. Helping engineering students to develop a feel for structural programming—not just button-pushing with a software program—the illustrative examples and extensive assignments in this resource enable them to develop the necessary skills and then apply them to practical electrical engineering problems and cases.


Numerical and Analytical M

ethods with M

ATLAB® for Electrical Engineers

K12515_Cover_mech.indd 1 7/27/12 9:01 AM


CRC Series inCOMPUTATIONAL MECHANICS

and APPLIED ANALYSIS

Series Editor: J.N. ReddyTexas A&M University

Published Titles

ADVANCED THERMODYNAMICS ENGINEERING, Second EditionKalyan Annamalai, Ishwar K. Puri, and Miland Jog

APPLIED FUNCTIONAL ANALYSISJ. Tinsley Oden and Leszek F. Demkowicz

COMBUSTION SCIENCE AND ENGINEERINGKalyan Annamalai and Ishwar K. Puri

CONTINUUM MECHANICS FOR ENGINEERS, Third EditionThomas Mase, Ronald Smelser, and George E. Mase

DYNAMICS IN ENGINEERING PRACTICE, Tenth EditionDara W. Childs

EXACT SOLUTIONS FOR BUCKLING OF STRUCTURAL MEMBERSC.M. Wang, C.Y. Wang, and J.N. Reddy

THE FINITE ELEMENT METHOD IN HEAT TRANSFER AND FLUID DYNAMICS,Third Edition

J.N. Reddy and D.K. Gartling

MECHANICS OF LAMINATED COMPOSITE PLATES AND SHELLS:THEORY AND ANALYSIS, Second Edition

J.N. Reddy

MICROMECHANICAL ANALYSIS AND MULTI-SCALE MODELINGUSING THE VORONOI CELL FINITE ELEMENT METHOD

Somnath Ghosh

NUMERICAL AND ANALYTICAL METHODS WITH MATLAB®

William Bober, Chi-Tay Tsai, and Oren Masory

NUMERICAL AND ANALYTICAL METHODS WITH MATLAB®

FOR ELECTRICAL ENGINEERSWilliam Bober and Andrew Stevens

PRACTICAL ANALYSIS OF COMPOSITE LAMINATESJ.N. Reddy and Antonio Miravete

SOLVING ORDINARY AND PARTIAL BOUNDARY VALUE PROBLEMSIN SCIENCE and ENGINEERING

Karel Rektorys

STRESSES IN BEAMS, PLATES, AND SHELLS, Third EditionAnsel C. Ugural

CRC Press is an imprint of theTaylor & Francis Group, an informa business

Boca Raton London New York



MATLAB® and Simulink® are trademarks of The MathWorks, Inc. and are used with permission. The Math-Works does not warrant the accuracy of the text or exercises in this book. This book’s use or discussion of MATLAB® and Simulink® software or related products does not constitute endorsement or sponsorship by The MathWorks of a particular pedagogical approach or particular use of the MATLAB® and Simulink® software.

CRC PressTaylor & Francis Group6000 Broken Sound Parkway NW, Suite 300Boca Raton, FL 33487-2742

© 2013 by Taylor & Francis Group, LLCCRC Press is an imprint of Taylor & Francis Group, an Informa business

No claim to original U.S. Government worksVersion Date: 20120801

International Standard Book Number-13: 978-1-4665-7607-0 (eBook - PDF)

This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint.

Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmit-ted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers.

For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged.

Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe.

Visit the Taylor & Francis Web site athttp://www.taylorandfrancis.com

and the CRC Press Web site athttp://www.crcpress.com

v

Contents

Preface.............................................................................................................ixAcknowledgments...........................................................................................xiAbout.the.Authors........................................................................................ xiii

1 Numerical.Methods.for.Electrical.Engineers..........................................11.1 Introduction......................................................................................11.2 EngineeringGoals.............................................................................21.3 ProgrammingNumericalSolutions...................................................21.4 WhyMATLAB?................................................................................31.5 TheMATLABProgrammingLanguage............................................41.6 ConventionsinThisBook.................................................................51.7 ExamplePrograms.............................................................................5

2 MATLAB.Fundamentals.........................................................................72.1 Introduction......................................................................................72.2 TheMATLABWindows...................................................................82.3 ConstructingaPrograminMATLAB.............................................112.4 MATLABFundamentals................................................................122.5 MATLABInput/Output.................................................................212.6 MATLABProgramFlow.................................................................262.7 MATLABFunctionFiles................................................................322.8 AnonymousFunctions.....................................................................362.9 MATLABGraphics.........................................................................362.10 WorkingwithMatrices................................................................... 462.11 WorkingwithFunctionsofaVector................................................482.12 AdditionalExamplesUsingCharactersandStrings.........................492.13 InterpolationandMATLAB’sinterp1Function.........................532.14 MATLAB’stextscanFunction...................................................552.15 ExportingMATLABDatatoExcel.................................................57

vi ◾ Contents

2.16 DebuggingaProgram......................................................................582.17 TheParallelRLCCircuit.................................................................60Exercises....................................................................................................63Projects......................................................................................................63References..................................................................................................76

3 Matrices.................................................................................................773.1 Introduction................................................................................... 773.2 MatrixOperations.......................................................................... 773.3 SystemofLinearEquations.............................................................823.4 GaussElimination...........................................................................873.5 TheGauss-JordanMethod...............................................................923.6 NumberofSolutions.......................................................................943.7 InverseMatrix.................................................................................953.8 TheEigenvalueProblem................................................................100Exercises..................................................................................................104Projects....................................................................................................105Reference.................................................................................................108

4 Roots.of.Algebraic.and.Transcendental.Equations..............................1094.1 Introduction..................................................................................1094.2 TheSearchMethod.......................................................................1094.3 BisectionMethod.......................................................................... 1104.4 Newton-RaphsonMethod.............................................................1124.5 MATLAB’sfzeroandrootsFunctions...................................113

4.5.1 ThefzeroFunction........................................................ 1144.5.2 TherootsFunction....................................................... 118

Projects.................................................................................................... 119Reference.................................................................................................128

5 Numerical.Integration.........................................................................1295.1 Introduction..................................................................................1295.2 NumericalIntegrationandSimpson’sRule....................................1295.3 ImproperIntegrals.........................................................................1335.4 MATLAB’squadFunction..........................................................1355.5 TheElectricField...........................................................................1375.6 ThequiverPlot..........................................................................1415.7 MATLAB’sdblquadFunction...................................................143Exercises..................................................................................................146Projects....................................................................................................147

6 Numerical.Integration.of.Ordinary.Differential.Equations................1576.1 Introduction.................................................................................. 1576.2 TheInitialValueProblem..............................................................158

Contents ◾ vii

6.3 TheEulerAlgorithm......................................................................1586.4 ModifiedEulerMethodwithPredictor-CorrectorAlgorithm........1606.5 NumericalErrorforEulerAlgorithms...........................................1666.6 TheFourth-OrderRunge-KuttaMethod.......................................1676.7 SystemofTwoFirst-OrderDifferentialEquations.........................1696.8 ASingleSecond-OrderEquation...................................................1726.9 MATLAB’sODEFunction...........................................................1756.10 BoundaryValueProblems.............................................................1796.11 SolutionofaTri-DiagonalSystemofLinearEquations.................180

MethodSummaryformequations.................................................1816.12 DifferenceFormulas......................................................................1836.13 One-DimensionalPlateCapacitorProblem...................................186Projects....................................................................................................190

7 Laplace.Transforms.............................................................................2017.1 Introduction..................................................................................2017.2 LaplaceTransformandInverseTransform....................................201

7.2.1 LaplaceTransformoftheUnitStep..................................2027.2.2 Exponential......................................................................2027.2.3 Linearity...........................................................................2037.2.4 TimeDelay.......................................................................2037.2.5 ComplexExponential...................................................... 2047.2.6 Powersoft........................................................................2057.2.7 DeltaFunction................................................................ 206

7.3 TransformsofDerivatives..............................................................2097.4 OrdinaryDifferentialEquations,InitialValueProblem................2107.5 Convolution................................................................................. 2207.6 LaplaceTransformsAppliedtoCircuits.........................................2237.7 ImpulseResponse..........................................................................227Exercises................................................................................................. 228Projects....................................................................................................229References................................................................................................232

8 Fourier.Transforms.and.Signal.Processing..........................................2398.1 Introduction..................................................................................2398.2 MathematicalDescriptionofPeriodicSignals:FourierSeries........2418.3 ComplexExponentialFourierSeriesandFourierTransforms........2458.4 PropertiesofFourierTransforms...................................................2498.5 Filters.............................................................................................2518.6 Discrete-TimeRepresentationofContinuous-TimeSignals...........2538.7 FourierTransformsofDiscrete-TimeSignals.................................2558.8 ASimpleDiscrete-TimeFilter........................................................258Projects....................................................................................................269References................................................................................................273

viii ◾ Contents

9 Curve.Fitting.......................................................................................2759.1 Introduction..................................................................................2759.2 MethodofLeastSquares...............................................................275

9.2.1 Best-FitStraightLine........................................................2759.2.2 Best-Fitmth-DegreePolynomial.......................................277

9.3 CurveFittingwiththeExponentialFunction................................2799.4 MATLAB’spolyfitFunction....................................................2819.5 CubicSplines.................................................................................2859.6 TheFunctioninterp1forCubicSplineCurveFitting...............2879.7 CurveFittingwithFourierSeries..................................................289Projects....................................................................................................291

10 Optimization.......................................................................................29510.1 Introduction..................................................................................29510.2 UnconstrainedOptimizationProblems.........................................29610.3 MethodofSteepestDescent..........................................................29710.4 MATLAB’sfminuncFunction...................................................30110.5 OptimizationwithConstraints.....................................................30210.6 LagrangeMultipliers.................................................................... 30410.7 MATLAB’sfminconFunction...................................................307Exercises..................................................................................................316Projects....................................................................................................316Reference.................................................................................................322

11 Simulink..............................................................................................32311.1 Introduction..................................................................................32311.2 CreatingaModelinSimulink.......................................................32311.3 TypicalBuildingBlocksinConstructingaModel.........................32511.4 TipsforConstructingandRunningModels.................................32811.5 ConstructingaSubsystem.............................................................32911.6 UsingtheMuxandFcnBlocks......................................................33011.7 UsingtheTransferFcnBlock........................................................33011.8 UsingtheRelayandSwitchBlocks................................................33111.9 TrigonometricFunctionBlocks.....................................................334Exercises..................................................................................................337Projects....................................................................................................337Reference.................................................................................................339

Appendix.A:.RLC.Circuits...........................................................................341

Appendix.B:.Special.Characters.in.MATLAB®.Plots...................................353

ix

Preface

IhavebeenteachingtwocoursesincomputerapplicationsforengineersatFloridaAtlanticUniversity(FAU)formanyyears.Thefirstcourseisusuallytakeninthestudent’ssophomoreyear;thesecondcourseisusuallytakeninthestudent’sjuniororsenioryear.Bothcomputerclassesarerunaslecture-laboratorycourses,andtheMATLAB®softwareprogramisusedinbothcourses.Tofamiliarizestudentswithengineering-typeproblems,approximatelysixorsevenprojectsareassignedduringthesemester.Studentshave,dependingonthedifficultyoftheproject,eitheroneweekortwoweekstocompleteeachproject.Ibelievethatthebestsourceforstu-dentstocompletetheassignedprojectsineithercourseisthistextbook.

This book has its origin in a previous textbook, Numerical and Analytical Methods with MATLAB® by William Bober, Chi-Tay Tsai, and Oren Masory,alsopublishedbyCRCPress.Thepreviousbookwasprimarilyoriented towardmechanicalengineeringstudents.IandJonathanPlantofCRCPressenvisionedthatasimilartextwouldfillaneedinelectricalengineeringcurricula;asaresult,IenlistedDr.AndrewStevenstoreplacetheprojectsintheexistingtextbookwiththoseoriented towardelectrical engineering students.Thisnew textbook retainsthephilosophyofteachingthatexistsintheoriginaltextbook.

TheadvantageofusingtheMATLABsoftwareprogramoverotherpackagesisthatitcontainsbuilt-infunctionsthatnumericallysolvesystemsoflinearequa-tions,systemsofordinarydifferentialequations,rootsoftranscendentalequations,integrals,statisticalproblems,optimizationproblems,signal-processingproblems,andmanyothertypesofproblemsencounteredinengineering.AstudentversionoftheMATLABprogramisavailableatareasonablecost.However,tostudents,thesebuilt-in functionsareessentiallyblackboxes.Bycombininga textbookonMATLABwithbasicnumericalandanalyticalanalysis(althoughIamsurethatMATLAB uses more sophisticated numerical techniques than are described inthesetextbooks),themysteryofwhattheseblackboxesmightcontainissomewhatalleviated.ThetextcontainsmanysampleMATLABprogramsthatshouldprovideguidancetothestudentoncompletingtheassignedprojects.Manyoftheprojectsinthisbookarenon-trivialand,Ibelieve,willbegoodtrainingforagraduatingengineerenteringindustryorinanadvanceddegreeprogram.

x ◾ Preface

Furthermore,Ibelievethatthereisenoughmaterialinthistextbookfortwocourses,especiallyifthecoursesarerunaslecture-laboratorycourses.Theadvan-tageofrunningthesecourses(especiallythefirstcourse)asalecture-laboratorycourseisthattheinstructoris inthecomputerlaboratorytohelpthestudentsdebugtheirprograms.Thisincludesthesampleprogramsaswellastheprojects.

ThecommoncoreofthebookistheintroductiontotheMATLABprogram-mingenvironmentinChapter2.Then,dependingontheindividualcurriculum(buttypicallysophomoreyear),acoursemightproceedwithmathematicaltech-niques for matrix algebra, root finding, integration, and differential equations(Chapters3through6).Amoreadvancedcourse(perhapsinjuniororsenioryear)might include transform techniques (Chapters 7 and8) and advanced topics incurvefittingandoptimization(Chapters9and10).MATLAB’sgraphicaldesignenvironment,Simulink®,isintroducedinChapter11andcouldberelocatedelse-whereinthesyllabus.

We have tried to make each chapter stand alone so that each may be rear-rangedbasedonthepreferenceoftheinstructor.Inmanycases,wehaveusedtheresistor-inductor-capacitor(RLC)circuitasanexample,andwehaveputthebasicderivationofthiscircuitinAppendixAtominimizethechapterdependenciesandfacilitatereordering.Inallcases,wehaveattemptedtoprovideillustrativeexamplesusingmoderntopicsinelectricalengineering.

Allchapters(exceptforChapter1)containprojects,andsomealsocontainsev-eralexercisesthatarelessdifficultthantheprojectsandmightbeassignedpriortoaprojectassignment.Allprojectsrequirethestudenttowriteacomputerprogram,mostrequiringtheuseofMATLABbuilt-infunctionsandsolvers.

Additional materials, including down loadable copies of all examples in thetextbook,areavailablefromtheCRCpresswebsite:

http://www.crcpress.com/product/isbn/9781439854297MATLAB and Simulink are registered trademarks of The MathWorks,

Incorporated.Forproductinformation,pleasecontact

TheMathWorks,Inc.3AppleHillDriveNatick,MA01760-2098USATel:508-647-7000Fax:508-647-7001E-mail:[email protected]:www.mathworks.com

William.BoberFlorida Atlantic University

Department of Civil Engineering

xi

Acknowledgments

WewishtothankJonathanPlantofCRCPressforhisconfidenceandencourage-mentinwritingthistextbook.Inaddition,wewouldliketoexpressthankstoChrisLeGoffforhelpinpreparingtheartworkandtoPatrickFarrellandMichaelRohanfortheirtechnicalexpertise.

WealsowishtoexpressourdeepgratitudetoSelmaBober,TheresaStevens,andEmmaStevensfortoleratingthemanyhourswespentonthepreparationofthemanuscript,thetimethatotherwisewouldhavebeendevotedtoourfamilies.

xiii

About the Authors

William.Bober,.Ph.D., receivedhisB.S.degree in civil engineering from theCity College of New York (CCNY), his M.S. degree in engineering sciencefromPrattInstitute,andhisPh.D.degreeinengineeringscienceandaerospaceengineering from Purdue University. At Purdue University, he was on a FordFoundation Fellowship; he was assigned to teach one engineering course eachsemester. After receiving his Ph.D., he went to work as an associate engineer-ing physicist in the Applied Mechanics Department at Cornell AeronauticalLaboratoryinBuffalo,NewYork.AfterleavingCornellLabs,hewasemployedas anassociateprofessor in theDepartmentofMechanicalEngineeringat theRochester Institute of Technology (RIT) for the following twelve years. AfterleavingRIT,heobtained employment atFloridaAtlanticUniversity (FAU) intheDepartmentofMechanicalEngineering.Morerecently,hetransferredtotheDepartmentofCivilEngineering atFAU.While atRIT,hewas theprincipalauthor of a textbook, Fluid Mechanics, published by John Wiley & Sons. HehaswrittenseveralpapersforThe International Journal of Mechanical Engineering Education (IJMEE) and more recently coauthored a textbook, Numerical and Analytical Methods with MATLAB®.

Andrew.Stevens,.Ph.D.,.P.E.,receivedhisbachelor’sdegreefromMassachusettsInstituteofTechnology,hismaster’sdegreefromtheUniversityofPennsylvania,andhisdoctoratefromColumbiaUniversity,allinelectricalengineering.Hedidhis Ph.D. thesis work at IBM Research in the area of integrated circuit designforhigh-speedopticalnetworks.WhileatColumbia,helecturedacourseinthecoreundergraduatecurriculumandwontheIEEESolid-StateCircuitsFellowship.He has held R&D positions at AT&T Bell Laboratories in the development ofT-carriermultiplexersystemsandatArgonneNationalLaboratoryinthedesignofradiation-hardenedintegratedcircuitsforcollidingbeamdetectors.Since2001,hehasbeenpresidentofElectricalScience,anengineeringconsultingfirmspecializinginelectricalhardwareandsoftware.Hehaspublishedarticlesinseveralscientificjournalsandholdsthreepatentsintheareasofanalogcircuitdesignandcomputeruserinterfaces.

1

Chapter 1

Numerical Methods for Electrical Engineers

1.1 IntroductionAlldisciplinesofscienceandengineeringusenumericalmethodsfortheanaly-sisofcomplexproblems.However,electricalengineeringparticularlylendsitselfto computational solutions due to the highly mathematical nature of the fieldand its close relationship with computer science. It makes sense that engineerswho design high-speed computers would also use the computers themselves toaidinthedesign(aprocessknownasbootstrapping).Infact,theentirefieldofcomputer-aideddesign(CAD) isdedicated to thecreationand improvementofsoftwaretoolstoenabletheimplementationofhighlycomplexdesigns.

Thisbookdescribesvariousmethodsand techniques fornumerically solvingavarietyofcommonelectricalengineeringapplications,includingcircuitdesign,electromagneticfieldtheory,andsignalprocessing.Classicalengineeringcurriculateachavarietyofmethodsforsolvingtheseproblemsusingtechniquessuchaslinearalgebra,differentialequations,transforms,vectorcalculus,andthelike.However,inmanycases,thesearchforaclosed-formsolutionleadstoextremecomplexity,whichcancauseustolosephysicalinsightintotheproblem.Insolvingthesesameproblemsnumerically,wewilloftenreverttofundamentalphysicalrelations,suchasthedifferentialrelationshipbetweencapacitorcurrentandvoltageortheelectricfieldofapointcharge.Often,aproblemthatseemsintractablewhensolvedsym-bolicallycanbecometrivialwhensolvednumerically.Andsometimes,thesimplernumericalsolutioncanbeelusivebecausewearesousedtothinkingintermsofadvancedcalculus(aclassiccaseofnotbeingabletoseetheforestbutforthetrees).

2 ◾ Numerical and Analytical Methods with MATLAB

1.2 Engineering GoalsSomefundamentalgoalsinengineeringinclude

◾ Designnewproductsorimproveexistingones

◾ Improvemanufacturingefficiency

◾ Minimizecost,powerconsumption,andnonreturnableengineering(NRE)cost

◾ Maximizeyieldandreturnoninvestment(ROI)

◾ Minimizetimetomarket

Theengineerwill frequentlyusethe lawsofphysicsandmathematics toachievethesegoals.

Manyelectricalengineeringprocessesinvolveexpensivemanufacturingstepsthatarebothdelicateandtimeconsuming.Forexample,thefabricationofintegratedcir-cuitscaninvolvethousandsofmanufacturingsteps,includingwaferpreparation,maskcreation,photolithography,diffusionandimplantation,dicing,testing,packaging,andmore.Thesestepscantakeweeksormonthstoperformatsubstantialexpenseandincleanrooms.Anydesignmistakesrequirerepeatingtheprocess;thus,itisourjobasdesignerstomodelandsimulatedesignsasmuchaspossibleinadvanceofmanufacturetoeliminateflawsandminimizetheiterationsnecessarytoproducethefinalproduct.

Usingintegratedcircuitdesignasanexample,wemightusecomputersforthe

a.Designstage:Solvemathematicalmodelsofphysicalphenomena(e.g.,pre-dictingthebehaviorofPNjunctions)

b.Testingstage:Storeandanalyzeexperimentaldata(e.g.,comparingthelabo-ratory-measuredactualbehaviorofPNjunctionstotheprediction)

c.Manufacturingstage:Controllingmachineoperationstofabricateandtestsiliconwafersanddice

1.3 Programming Numerical SolutionsPhysicalphenomenaarealwaysdescribedbyasetofgoverningequations,andnumeri-calmethodscanbeusedtosolvethesetofgoverningequationsevenintheabsenceofaclosed-formsolution.Numericalmethodsinvariablyinvolvethecomputer,andthecomputerperformsarithmeticoperationsondiscretenumbersinadefinedsequenceofsteps.Thesequenceofstepsisdefinedintheprogram.Ausefulsolutionisobtainedif

a.Themathematicalmodelaccuratelyrepresentsthephysicalphenomena;thatis,themodelhasthecorrectgoverningequations.

b.Thenumericalmethodisaccurate. c.Thenumericalmethodisprogrammedcorrectly.

Thistextismainlyconcernedwithitems(b)and(c).

Numerical Methods for Electrical Engineers ◾ 3

Theadvantageofusingthecomputeristhatitcancarryoutmanycalculationsinafractionofasecond;atthetimeofthiswriting,computerspeedsaremeasuredinteraflops(trillionsoffloatingpointoperationspersecond).However,toleveragethispower,weneedtowriteasetofinstructions,thatis,aprogram.Fortheproblemsofinterestinthisbook,thedigitalcomputerisonlycapableofperformingarithmetic,logical, andgraphicaloperations.Therefore, arithmeticproceduresmustbedevel-opedforsolvingdifferentialequations,evaluatingintegrals,determiningrootsofanequation,solvingasystemoflinearequations,andsoon.Thearithmeticprocedureusuallyinvolvesasetofalgebraicequations.Acomputersolutionforsuchproblemsinvolvesdeveloping a computerprogram thatdefines a step-by-stepprocedure forobtainingananswertotheproblemofinterest.Themethodofsolutioniscalledanalgorithm.Dependingontheparticularproblem,wemightwriteourownalgorithm,orasweshallsee,wecanalsousethealgorithmsbuiltintoapackagelikeMATLAB®toperformwell-knownalgorithmssuchas theRunge-Kuttamethodfor solvingasetofordinarydifferentialequationsoruseSimpson’sruleforevaluatinganintegral.

1.4 Why MATLAB?MATLABwasoriginallywrittenbyDr.CleveMoleratUniversityofNewMexicointhe1970sandwascommercializedbyMathWorksinthe1980s.Itisageneral-purposenumericalpackagethatallowscomplexequationstobesolvedefficientlyand subsequently generate tabular or graphical output. While there are manynumerical packages available to electrical engineers, many are highly focusedtowardaparticularapplication(e.g.,SPICEformodelingelectroniccircuits).Also,MATLABisnottobeconfusedwithCADsoftwareforschematiccapture,layout,orphysicaldesign,althoughthissoftwareoftenintegrateswithanaccompanyingnumericalpackage.

Originally,MATLABwasacommand-lineprogramthatranonMS-DOSandUNIXhosts.Ascomputershaveevolved,sohasMATLAB,andmoderneditionsoftheprogramruninwindowedenvironments.Asofthetimeofthiswriting,MATLABR2011brunsnativelyonMicrosoftWindows,AppleMacOS,andLinux.Inthistext,we assume that you are running MATLAB on your local machine in a MicrosoftWindowsenvironment.Itshouldbestraightforwardfornon-Windowsuserstotranslatetheusagedescriptionstotheirpreferredenvironment.Inanycase,thesedifferencesarelargelylimitedtothecosmeticsandpresentationoftheprogramandnottheMATLABcommandsthemselves.AllversionsofMATLAB(onanyplatform)usethesamecom-mandset,andtheCommandWindowonallplatformsshouldbehaveidentically.

MATLABisofferedwithaccompanying“toolboxes”atadditionalcosttotheuser.Awidevarietyoftoolboxesareavailableinfieldssuchascontrolsystems,imagepro-cessing,radiofrequency(RF)design,signalprocessing,andmore.However,inthistext,welargelyfocusonfundamentalnumericalconceptsandlimitourselvestobasicMATLABfunctionalitywithoutrequiringthepurchaseofanyadditionaltoolboxes.


1.5 The MATLAB Programming LanguageTherearemanymethodologiesforcomputerprogramming,butthetasksathandboildowntothefollowing:

a.Studytheproblemtobeprogrammed. b.Listthealgebraicequationstobeusedintheprogrambasedontheknown

physicalphenomenaandgeometriesoftheproblem. c.Create a general design for the program flow and algorithms, perhaps by

creatingaflowchartorbywritinghigh-levelpseudocodetooutlinethemainprogrammodules.

d.Carryoutasamplecalculationbyhandtoprovethealgorithm. e.Writetheprogramusingthelistofalgebraicequationsandtheoutline. f.Debugtheprogrambyrunningitandfixinganysyntaxerrors. g.Testtheprogrambyrunningitusingparameterswithaknown(orintui-

tive)solution. h.Iterateoverthesestepstorefineandfurtherdebugthealgorithmandpro-

gramflow. i.Ifnecessary,revisetheprogramtoobtainfasterperformance.

Experienced programmers might omit some of these steps (or do them in theirhead),buttheoverallprocessresemblesanyengineeringproject:design,createaprototype,test,anditeratetheprocessuntilasatisfactoryproductisachieved.

MATLABmaybeconsideredaprogrammingorscriptinglanguageuntoitself,butlikeeveryprogramminglanguage,ithasthefollowingcorecomponents:

a.Datatypes,i.e.,formatsforstoringnumbersandtextintheprogram(e.g.,integers,double-precisionfloatingpoint,strings,vectors,matrices)

b.Operators(e.g.,commandsforaddition,multiplication,cosine,log) c.Controlflowdirectivesformakingdecisionsandperformingiterativeopera-

tions(e.g.,if,while,switch) d.Input/output(“I/O”)commandsforreceivinginputfromauserorfileand

forgeneratingoutputtoafileorthescreen(e.g.,fprintf,fscanf,plot,stem,surf)

MATLABborrowsmanyconstructs fromother languages.Forexample, thewhile,switch,andfprintfcommandsare fromtheCprogramming lan-guage(oritsdescendentsC++,Java,andPerl).However,therearesomefundamen-taldifferencesaswell.Forexample,MATLABstores functions (known inotherlanguagesas“subroutines”)inseparatefiles.Thefirstentryinavector(knowninmostotherlanguagesasanarray)isindexedbythenumber1andnot0.However,thebiggestdifferenceisthatallMATLABvariablesarevectors,thusprovidingtheabilitytomanipulatelargeamountsofdatawithatersesyntaxandallowingforthe

Numerical Methods for Electrical Engineers ◾ 5

solutionofcomplicatedproblemsinjustafewlinesofcode.Inaddition,becauseMATLABisnormallyruninteractively,itisalsorichinpresentationfunctionstodisplaysophisticatedplotsandgraphs.

1.6 Conventions in This BookWeusethefollowingtypographicalconventionsinthistext:

◾ Allinput/outputtoandfromMATLABareintypewriterfont.

◾ Incaseswhereyouaretypingdirectlyintothecomputer,thetypedtextisdisplayedinbold.

Weillustratethisinthefollowingexample,forwhichweuseMATLABtofindthevalue =

πsin4

x :

>>x = sin(pi/4) x= 0.7071

Inthiscase,>>representsMATLAB’sprompt,x=sin(pi/4)representstexttypedintotheMATLABcommandwindow,andx=0.7071representsMATLAB’sresponse.

1.7 Example ProgramsTheexampleprogramsinthisbookmaybedownloadedfromthepublisher’sWebsiteathttp://www.crcpress.com/product/isbn/9781439854297.Studentsmaythenruntheexampleprogramsontheirowncomputerandseetheresults.Italsomaybebeneficialforstudentstotypeinafewofthesampleprograms(alongwithsomeinevitablesyntaxandtypographicalerrors),therebygivingthemtheopportunitytoseehowMATLABrespondstoprogramerrorsandsubsequentlylearnwhattheyneedtodotofixtheproblem.

7

Chapter 2

MATLAB Fundamentals

2.1 IntroductionMATLAB• is a software program for numeric computation, data analysis, andgraphics.OneadvantagethatMATLABhasforengineersoverprogramminglan-guagessuchasCorC++ isthattheMATLABprogramincludesfunctionsthatnumericallysolvelargesystemsoflinearequations,systemsofordinarydifferentialequations,rootsof transcendentalequations, integrals, statisticalproblems,opti-mizationproblems,controlsystemsproblems,andmanyothertypesofproblemsencounteredinengineering.MATLABalsoofferstoolboxes(whichmustbepur-chasedseparately)thataredesignedtosolveproblemsinspecializedareas.

Inthischapter,thefollowingitemsarecovered:

◾ TheMATLABdesktopenvironment

◾ Constructingascript(alsocalledaprogram)inMATLAB

◾ MATLAB fundamentals and basic commands, including clear, clc,colon operator, arithmetic operators, trigonometric functions, logarithmicandexponential functions, andotheruseful functions suchasmax,min,andlength

◾ Input/outputinMATLAB,includingtheinputandfprintfstatements

◾ MATLAB program flow, including for loops, while loops, ifandelseifstatements,andtheswitchgroupstatement

◾ MATLABfunctionfilesandanonymousfunctions

◾ MATLABgraphics,includingtheplotandsubplotcommands


◾ Workingwithmatrices

◾ Workingwithfunctionsofavector

◾ Workingwithcharactersandstrings

◾ InterpolationandMATLAB’sinterp1function

◾ MATLAB’stextscanfunction

◾ ExportingMATLABdatatoothersoftware,suchasMicrosoftExcel

◾ Debuggingaprogram

Manyexamplescriptsare includedthroughoutthechapterto illustratethesevarioustopics.

2.2 The MATLAB WindowsUnderMicrosoftWindows,MATLABmaybestartedviatheStartmenuorclickingontheMATLABicononthedesktop.Onstartup,anewwindowwillopencontain-ingtheMATLAB“desktop”(nottobeconfusedwiththeWindowsdesktop),andoneormoreMATLABwindowswillopenwithinthedesktop(seeFigure 2.1forthe

Figure 2.1 MATLAB desktop windows. (From MATLAB, with permission.)

MATLAB Fundamentals ◾ 9

defaultconfiguration).ThemainwindowsaretheCommandwindow,CommandHistory,CurrentFolder,andWorkspace.YoucancustomizetheMATLABwindowsthatappearonstartupbyopeningtheDesktopmenuandchecking(orunchecking)thewindowsthatyouwishtoappearontheMATLABdesktop.Figure 2.1showstheCommandwindow(inthecenter),theCurrentFolder(ontheleft),theWorkspace(onthetopright),theCommandHistory(onthebottomright),andtheCurrentFolderbox(intheicontoolbar,secondfromthetop,justabovetheCommandwin-dow).Thesewindowsandthecurrentfolderboxaresummarizedasfollows:

◾ Command Window:IntheCommandwindowyoucanentercommandsanddata, make calculations, and print results. You can write a program in theCommand window and execute the program. However, writing a programdirectlyintotheCommandwindowisdiscouragedbecauseitwillnotbesaved,andifanerrorismade,theentireprogrammustberetyped.Byusingtheuparrow(↑)keyonyourkeyboard,thepreviouscommandcanberetrieved(andedited)forreexecution.

◾ Command History Window: ThiswindowlistsahistoryofthecommandsthatyouhaveexecutedintheCommandwindow.

◾ Current Folder Box: This box lists the active Current Folder (also calledtheCurrentDirectoryinolderversionsofMATLAB).To.run.a.MATLAB.script.(program),the.script.of.interest.needs.to.be.in.the.folder.listed.in.this.box..Byclickingonthedownarrowwithinthebox,adrop-downmenuwillappearthatcontainsnamesoffoldersthatyouhavepreviouslyused.Thiswillallowyoutoselectthefolderinwhichthescriptofinterestresides(seeFigure2.2).Ifthefoldercontainingthescriptofinterestisnotlistedinthedrop-downmenu,youcanclickontheadjacentlittleboxcontainingthreedots,whichallowsyoutobrowseforthefoldercontainingtheprogramofinterest(seeFigure2.3).

◾ Current Folder Window (on the left):Thiswindow lists all thefiles in theCurrentFolder.Bydoubleclickingonafileinthiswindow,thefilewillopenwithinMATLAB.

◾ Script Window (also called theEditorwindow inolderMATLABversions):Toopenthiswindow,usetheFilemenuatthetopoftheMATLABdesktopandchooseNewandthenScript(orinolderversionsofMATLAB,clickonNew M-File)(seeFigure 2.4).TheScriptWindowmaybeusedtocreate,edit,andexecuteMATLABscripts.Scripts are then savedasM-Files.Thesefileshavetheextension.m,suchascircuit.m.Toexecutetheprogram,youcanclicktheSave and Run icon (thegreenarrow) in theScriptwindowor return totheCommandwindowandtypeinthenameoftheprogram(withoutthe.mextension).


Figure 2.3 Drop-down menu for selecting folder containing program of interest. (From MATLAB, with permission.)

Figure 2.2 Drop-down menu in the Current Folder window. (From MATLAB, with permission.)


2.3 Constructing a Program in MATLABThislistsummarizesthestepsforwritingyourfirstMATLABprogram:

1.StarttheMATLABdesktopviatheWindowsStartmenuorbydoubleclick-ingontheMATLABicononthedesktop.

2.ClickonFile-New-Script.ThisbringsupanewScriptwindow. 3.TypeyourscriptintotheScriptwindow. 4.SavethescriptbyclickingontheSaveiconintheicontoolbarorclickingon

FileinthemenubarandselectingSaveinthedropdownmenu.Inthedialogboxthatappears,selectthefolderwherethescriptistoresideandtypeinafilenameofyourownchoosing.ItisbesttouseafolderthatcontainsonlyyourownMATLABscripts.

5.Beforeyoucanrunyourscript,youneedtogototheCurrentFolderboxatthe topof theMATLABdesktop, clickingon thedownarrowand in thedropdownmenu,selecting(orbrowsingto)thefolderthatcontainsyournewscript.

6.YoumayrunyourscriptfromtheScriptwindowbyclickingontheSave and Rungreenarrowintheicontoolbar(seeFigure2.5)oralternatively,fromthe

Figure 2.4 Opening up the Script window. (From MATLAB, with permission.)


commandwindowbytypingthescriptname(withoutthe.mextension)aftertheMATLABprompt (>>). For example, if theprogramhas been saved ascircuit.m, then typecircuit after theMATLABprompt (>>), as shownbelow:

>> circuit

Ifyouneedadditionalhelpgettingstarted,youcanclickonHelpinthemenubarintheMATLABwindowandthenselectProduct Helpfromthedrop-downmenu.ThiswillbringupthehelpwindowasshowninFigure 2.6.Byclickingonthelittle‘+’boxnexttotheMATLABlistingintheleftcolumn,youwillgetaddi-tionalhelptopicsasshowninFigure 2.7.Onceyouselectoneofthehelptopics,thehelpinformationwillbeintheright-handwindow.Youcanalsotypeinatopicinthesearchwindowtoobtaininformationonthattopic.

2.4 MATLAB Fundamentals

◾ Variablenames

− muststartwithaletter.

Save and Run button

Figure 2.5 Save and run button in the script window. (From MATLAB, with permission.)


− cancontainletters,digitsandtheunderscorecharacter.

− canbeofanylength,butmustbeuniquewithinthefirst19characters.

Note:Donotuseavariablenamethatisthesamenameasthenameofafile,aMATLABfunction,oraselfwrittenfunction.

◾ MATLABcommandnamesandvariablenamesarecasesensitive.Uselowercaselettersforcommands.

◾ Semicolons are usually placed after variable definitions and programstatementswhenyoudonotwantthecommandechoedtothescreen.Inthe absence of a semicolon, thedefined variable appears on the screen.For example, if you entered the following assignment in the commandwindow:

>>A = [3 4 7 6]

Figure 2.6 Product help window. (From MATLAB, with permission.)


Inthecommandwindow,youwouldsee

A= 3476 >>

Alternatively,ifyouaddthesemicolon,thenyourcommandisexecutedbut there is nothing printed to the screen, and the prompt immediatelyappearsforyoutoenteryournextcommand:

>>A = [3 4 7 6]; >>

◾ Percentsign(%)isusedforacommentline.

◾ AseparateGraphicswindowopenstodisplayplotsandgraphs.

◾ Thereareseveralcommandsforclearingwindows,clearingtheworkspaceandstoppingarunningprogram.

Figure 2.7 Getting started in product help window. (From MATLAB, with permission.)


clc clearstheCommandwindow clf clearstheGraphicswindow clear removesallvariablesanddatafromtheworkspace Ctl-C abortsaprogramthatmayberunninginaninfiniteloop

◾ thequitorexitcommandsterminateMATLAB.

◾ thesavecommandsavesvariablesordataintheworkspaceofthecurrentdirectory.Thefilenamecontainingthedatawillhave.matextension.

◾ User-definedfunctions(alsocalledself-writtenfunctions)arealsosavedasM-files.

◾ ScriptsandfunctionsaresavedasASCIItextfiles.Thus,theymaybewritteneitherinthebuilt-inScriptwindow,Notepad,oranywordprocessor(savedasatextfile).

◾ ThebasicdatastructureinMATLABisamatrix.

◾ Amatrix is surroundedbybrackets andmayhave anarbitrarynumberof

rowsandcolumns;forexample,thematrix =1 3

6 5A maybeenteredinto

MATLABas

>>A = [1 3<enter> 6 5];<enter>

or

>>A = [1 3 ; 6 5];<enter>

where the semicolon within the brackets indicates the start of a new rowwithinthematrix.

◾ Amatrixofonerowandonecolumnisascalar;forexample:

>>A = [3.5];

Alternatively,MATLABalsoacceptsA = 3.5(withoutbrackets)asascalar.

◾ Amatrixconsistingofonerowandseveralcolumnsoronecolumnandsev-eralrowsisconsideredavector;forexample:

>>A = [2 3 6 5] (rowvector)

>>A = [2 365](columnvector)


Amatrixcanbedefinedbyincludingasecondmatrixasoneoftheelements;example:

>> B = [1.5 3.1]; >> C = [4.0 B]; (thusC=[4.01.53.1])

◾ AspecificelementofmatrixCcanbeselectedbywriting

>>a = C(2);(thusa =1.5)

Ifyouwishtoselectthelastelementinavector,youcanwrite

>>a = c(end);.(thusa =3.1)

◾ Thecolonoperator(:)maybeusedtocreateanewmatrixfromanexistingmatrix;forexample:

if

5 7 10

2 5 2

1 3 1

A=

then gives521

x= A(:,1) x =

ThecolonintheexpressionA(:,1)impliesalltherowsinmatrixA,andthe1 impliescolumn1.

gives =

7 10

5 2

3 1

x= A(:,2:3) x

ThefirstcolonintheexpressionA(:,2:3)impliesalltherowsinA,andthe2:3impliescolumns2and3.

Wecanalsowrite

y=A(1,:),whichgivesy=[5710]

The1impliesthefirstrow,andthecolonimpliesallthecolumns.


◾ Acoloncanalsobeusedtogenerateaseriesofnumbers.Theformatis:

n=startingvalue:stepsize:finalvalue.Ifomitted,thedefaultstepsizeis1.Forexample:

gives 1 2 3 4 5 6 7 8 .n = 1:8 n [ ]=

Toincrementinstepsof2,use

gives 1 3 5 7n=1:2:7 n [ ]=

Thesetypesofexpressionsareoftenusedinaforloop,whichisdiscussedlaterinthischapter.

◾ Arithmeticoperators:

+ addition

- subtraction* multiplication/ division^ exponentiation

◾ Todisplayavariablevalue,justtypethevariablenamewithoutthesemico-lon,andthevariablewillappearonthescreen.

Examples(trytypingthesestatementsintotheCommandwindow):

clc; x = 5; y = 10; z = x + y w = x – y z = y/x z = x*y z = x^2

Notethatinthearithmeticstatementz=x+y,thevaluesforxandywereassignedinthetwopriorlines.Ingeneral,allvariablesontheright-handsideofanarithmeticstatementmustbeassignedavaluebeforetheyareused.

◾ Specialvalues:

pi π

i or j −1

inf ∞

ans the last computed unassigned result to an expression typed in the Command window


Examples(trytypingthesestatementsinthecommandwindow):

x = pi; z = x/0(givesinf )

◾ Trigonometricfunctions:

sin sinesinh hyperbolic sineasin inverse sineasinh inverse hyperbolic sinecos cosinecosh hyperbolic cosineacos inverse cosineacosh inverse hyperbolic cosinetan tangenttanh hyperbolic tangentatan inverse tangentatan2 four-quadrant inverse tangentatanh inverse hyperbolic tangentsec secant

sech hyperbolic secantasec inverse secantasech inverse hyperbolic secantcsc cosecantcsch hyperbolic cosecantacsc inverse cosecantacsch inverse hyperbolic cosecantcot cotangentcoth hyperbolic cotangentacot inverse cotangentacoth inverse hyperbolic cotangent

Theargumentsofthesetrigonometricfunctionsareinradians.However,theargumentscanbemadeindegreesifa“d” isplacedafterthefunctionname,suchassind(x).


clc; x = pi/2; y = sin(x)


z = atan(1.0) x = 30; w = sind(x) z = atand(1.0)

◾ Exponential,squareroot,anderrorfunctions:

exp exponential

log natural logarithm

log10 common (base 10) logarithm

sqrt square root

erf error function


clc; x = 2.5; y = exp(x) z = log(y) w = sqrt(x)

◾ Complexnumbers:

Complexnumbersmaybewrittenintwoforms:Cartesian,suchasz=x+yj;orpolar,suchasz=r*exp(j*theta).Notethatweusejfor −1throughoutthistext.However,MATLABallowstheuseofifor −1aswell.Note:iandjarealsolegalMATLABvariablenameswhichareoftenusedwithinloops.Toavoidconfusion,programswhichinvolvecom-plexnumbersshouldnotuseiorjasvariablenames.

abs absolute value (magnitude)

angle phase angle (in radians)

conj complex conjugate

imag complex imaginary part

real complex real part


clc; z1 = 1 + j; z2 = 2 * exp(j * pi/6) y = abs(z1)


w = real(z2) v = imag(z1) phi = angle(z1)

◾ Otherusefulfunctions:

length(X) Gives the number of elements in the vector X.

size(X) Gives the size (number of rows and the number of columns) of matrix X.

sum(X) For vectors, sum(X) gives the sum of the elements in X. For matrices, gives a row vector containing the sum of the elements in each column of the matrix.

max(X) For vectors, max(X) gives the maximum element in X. For matrices, max(X) gives a row vector containing the maximum in each column of the matrix. If X is a column vector, it gives the maximum value of X.

min(X) Same as max(X)except it gives the minimum element in X.

sort(X) For vectors, sort(X) sorts the elements of X in ascending order. For matrices, sort(X) sorts each column in the matrix in ascending order.

factorial(n) n ! = 1 × 2 × 3 × ⋯ × n

mod(x,y) modulo operator, gives the remainder resulting from the division of x by y. For example, mod(13,5)=3, that is, 13 ÷ 5 gives 2 plus remainder of 3 (the 2 is discarded). As another example,mod(n,2)gives zero if n is an even integer and one if n is an odd integer.


clc; A = [2 15 6 18]; length(A) y = max(A) z = sum(A) A = [2 15 6 18; 15 10 8 4; 10 6 12 3]; x = max(A) y = sum(A) size(A) mod(21,2) mod(20,2)

◾ Alistofthecompletesetofelementarymathfunctionscanbeobtainedbytypinghelp elfunintheCommandwindow.


◾ Sometimes,itisnecessarytopreallocateamatrixofagivensize.Thiscanbedonebydefiningamatrixofallzerosorones;forexample:

=

0 0 0

0 0 0

0 0 0

A = zeros(3)

0 0

0 0

0 0

B = zeros(3,2)=

1 1 1

1 1 1

1 1 1

C = ones(3)=

1 1 1

1 1 1D = ones(2,3)=

Thefunctiontogeneratetheidentitymatrix(maindiagonalofones;allotherelementsarezero)iseye;example:

1 0 0

0 1 0

0 0 1

I = eye(3)=

2.5 MATLAB Input/Output◾ Ifyouwishtohaveyourprogrampausetoacceptinputfromthekeyboard,

usetheinputfunction;forexample,toentera2by3matrix,use

Z=input('EntervaluesforZinbrackets\n');

thentypein:

[5.1 6.3 2.5; 3.1 4.2 1.3]


Thus,

=

5.1 6.3 2.5

3.1 4.2 1.3Z

Notethattheargumenttoinput()isacharacterstringenclosedbysinglequotationmarks,whichwillbeprintedtothescreen.The\n(newline)tellsMATLABtomove thecursor to thenext line.Alternatively,\t (tab) tellsMATLABtomovethecursorseveralspacesalongthesameline.

If youwish to enter textdata toinput, youneed to enclose the textwithsinglequotationmarks.However,youcanavoidthisrequirementbyentering a secondargumentof's' toinput as shown in the followingstatement:

response=input('Plotfunction?(y/n):\n','s');

Inthiscase,theusercanrespondwitheitherayorn(withoutsinglequo-tationmarks).


z = input ('Enter a 2x3 matrix of your choosing\n')

name=input('Enter name enclosed by single quote marks:')

response = input('Plot function? (y/n):\n', 's');

◾ Thedisp commandprints just the contentsof amatrixor alphanumericinformation;forexample(assumingthatmatrixXhasalreadybeenenteredintheCommandwindow):

>> x = [3.6 7.1]; disp(X); disp('volt');

Thefollowingwillbedisplayedonthescreen:

3.60007.1000 volt >>

◾ Thefprintf commandprints formatted text to the screenor toafile;forexample:

>>I = 2.2; >>fprintf ('The current is %f amps \n', I);


Thefollowingwillappearonthescreen:

Thecurrentis2.200000amps

The%freferstoaformattedfloatingpointnumber,andthedefaultissixdecimalplaces(althoughinearlierversionsofMATLAB,thedefaultfor%fformatwasfourdecimalplaces).

fprintfusesformatstringsbasedontheCprogramminglanguage[1].Thus,youcanspecifytheminimumnumberofspacesfortheprintedvariableaswellasthenumberofdecimalplacesusing

%8.2f

Thiswillalloweightspacesforthevariable,totwodecimalplaces.Youcanalsospecifyjustthenumberofdecimalplaces,forexample,saythreedecimalplaces,andthenletMATLABautomaticallydecidethenumberofspacesforthevariableusing

%.3f

However,tocreateneatlookingtables,itisbesttospecifythenumberofspacesintheformatstatementthatallowsforseveralspacesbetweenvariablesinadjacentcolumns.

Otherformatsareasfollows:

%i or %d used for integers

%e scientific notation (e.g., 6.02e23), with defaultsix decimal places

%g automatically use the briefest of %f or %e format

%s used for a string of characters

%c used for a single character

UnlikeC,theformatstringinMATLAB’sfprintfmustbeenclosedbysinglequotationmarks(andnotdoublequotes).

Sometimes,theformatpartofthefprintfcommandmaybetoolongtofitonasinglelineinyourscript.Whenthishappens,youcanusethreedots(...) totellMATLABtocontinuethestatementonasecondline,asshownbythefollowingexample:

fprintf('resistance=%fvolts=%fcurrent=%f\n',... R,V,I);


ThreedotsmaybeusedtocontinueanyMATLABstatementontoaddi-tionallinestoimprovethereadabilityofyourscript,asshownbythefollow-ingstatement:

a=(-4*pi^2*omega^2*r^2*(cos(arg1))^2/sqrt(arg2)+... 4*pi^2*omega^2*r^2*(sin(arg1))^2/sqrt(arg2))*... exp(j*pi*omega);

Analternativetousingthethreedotstoextendalengthyarithmeticstate-mentontomorethanonelineistobreakupthestatementintoseveralsmallerstatementsandthencombinethem.Asanexample,wecouldbreakupthepreviousstatementintothreeshorterstatementstocalculatea:

a1=-4*pi^2*omega^2*r^2*(cos(arg1))^2/sqrt(arg2); a2=4*pi^2*omega^2*r^2*(sin(arg1))^2/sqrt(arg2); a=(a1+a2)*exp(j*pi*omega);

◾ Printingtoafile:ItisoftenusefultoprinttheresultsofaMATLABpro-gramtoafile,possiblyforinclusioninareport.Inaddition,programoutputthat isprintedtoafilecansubsequentlybeeditedwithinthefile,suchasaligningoreditingcolumnheadingsinatable.Beforeyoucanprinttoafile,youneedtoopenafileforprintingwiththecommandfopen.Thesyntaxforfopenis

fo=fopen('filename','w')

Thus,foisapointertothefilenamedfilename,andthewindi-cateswriting.

Toprinttofilenameuse

fprintf(fo,'format',var1,var2,...);

wheretheformatstringcontainsthetextformatforvar1,var2,andsoon.Whenyouarefinishedwithallofyourprintstatements,youshouldclosethefilewiththefclosecommand,asshowninthefollowingexample.

Example(writethissimpleprogramintheScriptwindowandsaveitasio_example.m):

%io_example.m %Thisprogramisatestforprintingtoafile. clear;clc; V=120; I=2.2; fo=fopen('output.txt','w'); fprintf(fo,'V=%4ivolt,I=%5.2famp\n',V,I); fclose(fo);


The extension on the output file should be .txt (otherwise, MATLABwillstarttheimportwizard).TheresultingoutputfilecanbeopenedfromeithertheScriptwindowortheCommandwindow.Toaccesstheoutputfile,clickonFile-Open,whichbringsupthescreenshowninFigure 2.8.IntheboxlabeledFile name,typein*.txt. Thiswillbringupallthefileswiththeextension.txt.Toopenthefileofinterest,doubleclickonthenameoftheoutputfile(inthisexample,thefilenameisoutput.txt). InearlierversionsofMATLAB,youwouldnotbeabletoopentheoutputfilewithoutincludingthefclose(fo) statement in theprogram.None the less it is still goodpracticetoincludethefclosestatementafteralltheoutputstatementsintheprogramorattheendoftheprogramitself.

◾ Thefscanfcommandmaybeusedtoreadfromanexistingfileasfollows:

A=zeros(n,m); fi=fopen('filename.txt','r'); [A]=fscanf(fi,'%f',[n,m]);

wheren ×misthenumberofelementsinthedatafile.The‘r’inthefopenstatementindicatesthatthefileisforreadingdataintotheprogram.

Then ×mmatrixisfilledincolumnorder.Thus,rowsbecomecolumns,andcolumnsbecomerows.Tousethedatainitsoriginalorder,transposethe

Figure 2.8 Files with extension .txt. (From MATLAB, with permission.)


read-inmatrix.Totransposeamatrix AinMATLAB,simplytypeinA'.Thischangescolumnstorowsandrowstocolumns.

◾ Anexistingdatafilecanalsobeenteredintoaprogrambytheloadcom-mand.Theload command,unlikethefscanfcommand,leavesrowsasrowsandcolumnsascolumns.

Forexample: loadfilename.txt x=filename(:,1); y=filename(:,2);

Theinputfilemusthavethesamenumberofrowsineachcolumn.

2.6 MATLAB Program Flow◾ Thefor loopprovidesthemeanstocarryoutaseriesofstatementswithjust

afewlinesofcode.

Syntax:form=1:20

statement;⋮

statement; end

MATLABsetstheindexmto1,carriesoutthestatementsbetweentheforandendstatements,thenreturnstothetopoftheloop,changesmto2,andrepeatstheprocess.Afterthe20thiteration,theprogramexitstheloop.Anystatementthatisnottoberepeatedshouldnotbewithinthefor loop.Forexample,tableheadingsthatarenottoberepeatedshouldbeoutsidethefor loop.

Example(writethissimpleprogramintheScriptwindow,saveitasfor_loop_example.m,andrunthescript):

%for_loop_example.m clear;clc; %Tableheadings: fprintf('n y1 y2\n'); fprintf('-----------------------------------\n'); %Usingtheindexintheforloopinanarithmeticstatement forj=1:10 y1=j^2/10; y2=j^3/100; fprintf('%5i %10.3f%10.3f\n',j,y1,y2); end


Note:Iftheindexinaforloopisusedtoselectanelementofamatrix,thenthe indexmustbean integer.However, ifyouarenotusing theforindexasamatrixindex,thentheforindexneednotbeanintegerasshowninthefollowingexample(writethissimpleprogramintheScriptwindow,saveitasfor_loop_example2.m,andrunthescript):

%for_loop_example2.m %Usingarangeofnon-integerxvaluesinaforloop. %Therangeofxisfrom-0.9to+0.9instepsof0.1. clear;clc; %printthetableheaderoutsideofthe'for'loop: fprintf('x y1 y2\n'); fprintf('-----------------------------------\n'); forx=-0.9:0.1:0.9 y1=x/(1-x); y2=y1^2; fprintf('%5.2f %10.3f%10.3f\n',x,y1,y2); end

◾ Thewhile loop

Syntax: n=0; whilen<10 n=n+1; statement; ⋮ statement; end

Inthewhile loop,MATLABwillcarryout thestatementsbetweenthewhileandendstatementsaslongastheconditioninthewhilestatementissatisfied.Ifanindexintheprogramisrequired,theuseofthewhileloopstatement(unliketheforloop)requiresthattheprogramcreateitsownindex,asshownabove.Notethatthestatement“n=n+1”maynotmakesensealge-braicallybutdoesmakessenseintheMATLABlanguage.The“=”operatorinMATLAB(asinmanycomputerlanguages)istheassignmentoperator,whichtellsMATLABtofetchthecontentsofthememorycellcontainingthevariablen,putitsvalueintothearithmeticunitoftheCPU(centralprocessingunit),incrementby1,andputthenewvaluebackintothememorycelldesignatedforthevariablen.Thus,theoldvalueofnhasbeenreplacedbythenewvalueforn.

◾ ifstatement

Syntax:iflogicalexpression


statement; ⋮statement;

else statement;

⋮ statement;end

Ifthelogicalexpressionistrue,thenonlytheuppersetofstatementsisexecuted.Ifthelogicalexpressionisfalse,thenonlythebottomsetofstate-mentsisexecuted.

◾ Logicalexpressionsareoftheform

a==b; a<=b;

a<b; a>=b;

a>b; a~=b; (a not equal to b)

◾ Compoundlogicalexpressions

a>b && a~=c (a > b AND a ≠ c)

a>b || a<c (a > b OR a < c)

◾ Theif-elseifladder

Syntax:iflogicalexpression1

statement(s);elseiflogicalexpression2

statements(s);elseiflogicalexpression3

statement(s); else

statement(s);end

Theif-elseifladderworksfromtopdown.Ifthetoplogicalexpressionistrue,thestatementsrelatedtothatlogicalexpressionareexecuted,andthepro-gramwillleavetheladder.Ifthetoplogicalexpressionisnottrue,theprogrammovestothenextlogicalexpression.Ifthatlogicalexpressionistrue,theprogramwillexecutethegroupofstatementsassociatedwiththatlogicalexpressionandleavetheladder.Ifthatlogicalexpressionisnottrue,theprogrammovestothenextlogicalexpressionandcontinuestheprocess.Ifnoneofthelogicalexpres-sionsaretrue,theprogramwillexecutethestatementsassociatedwiththeelse


statement.Theelsestatementisnotrequired.Inthatcase,ifnoneofthelogicalexpressionsaretrue,nostatementswithintheladderwillbeexecuted.

◾ Thebreak commandmaybeusedwithina fororwhile looptoendtheloop;forexample:

form=1:20statement;

⋮statement;ifm>10

break;end

end

Inthisexample,whenmbecomesgreaterthan10,theprogramleavesthefor loopandmovesontothenextstatementintheprogram.

◾ Theswitchgroup

Syntax:switch(var)

casevar1statement(s);



otherwisestatement(s);

end

wherevartakesonthepossiblevaluesvar1,var2,var3,andsoon. Ifvar equalsvar1, those statementsassociatedwithvar1areexecuted,

andtheprogramleavestheswitchgroup.Ifvardoesnotequalvar1,thepro-gramtestsifvarequalsvar2,andifyes,theprogramexecutesthosestatementsassociatedwithvar2andleavestheswitchgroup.Ifvardoesnotequalanyofvar1,var2,andsoon,theprogramexecutesthestatementsassociatedwiththeotherwisestatement.Ifvar1,var2,andsoonarestrings,theyneedtobeenclosedbysinglequotationmarks.Itshouldbenotedthatvarcannotbealogi-calexpression,suchasvar1>=80.Forexample(writethissimpleprogramintheScriptwindow,saveitasswitch_example.m,andruntheprogram):

% switch_example.m %Thisprogramisatestoftheswitchstatement. clear;clc;


var='a'; x=5; switch(var) case'b' z=x^2; case'a' z=x^3; end fprintf('z=%6.1f\n',z);

Additionalexampleprogramsfollowthatdemonstrate theuseof thefor loop,nestedfor loops,thewhile loop,theif statement,theif-elseif statements, thebreak statement, theinput statement, and thefprintf statement.SinceanumberofthesesampleprogramsinvolvedeterminingexbyaTaylorseriesexpansion,westartwithadiscussionoftheseriesexpansionofex.

Example 2.1: Calculation of ex

The Taylor series expansion of ex to n terms is

= + + + + + +

n

e xx x x x

nx

n

Term index: 1 2 3 4

12! 3! 4! !

2 3 4

There are several possible programming algorithms to calculate this series. The simplest method is to start with sum=1 and then evaluate each term using exponentiation and MATLAB’s factorial function and add the obtained term to sum. An alternative approach would be to evaluate each term individually, save them in an array, and then use MATLAB’s sum function to add them all. However, for some series expressions it is best to compute each term in the series based on the value of the preceding term. For example, in the above series, we can see that the third term in the series can be obtained from second term by multiplying the second term by x and dividing by 3; that is, term3 = term2 ⋅ x/3. In general, term (n) = term(n–1) ⋅ x/n.

%Example_2_1.m%Thisprogramcalculatese^xbyseriesandbyMATLAB's%exp()function.The'for'loopendswhenthetermonly%affectstheseventhsignificantfigure.%Note:e^x=1+x+x^2/2!+x^3/3!+x^4/4!+...clear;clc;x=5.0;s=1.0;forn=1:100term=x^n/factorial(n);s=s+term;


if(abs(term)<=s*1.0e-7)break;endendex1=s;ex2=exp(x);fprintf('x=%3.1fex1=%8.5fex2=%8.5f\n',x,ex1,ex2);------------------------------------------------------------

Example 2.2: Nested Loops

%Example_2_2.m%Thisprogramillustratestheuseofnestedloops,i.e.%aninner'for'loopinsideanouter'for'loop.%Theprogramcalculatese^xbybothMATLAB's'exp'%command(variable'ex2'),andbyaTaylorseries%expansion(variable'ex1'),where-0.5<x<0.5.%Theouter'for'loopisusedtodeterminethex%values.TheinnerloopisusedtodeterminetheTaylor%seriesmethodforevaluatinge^x.Inthisexample,%term(n+1)isobtainedbymultiplyingterm(n)byx/n.%Thevariable'term'isestablishedasavectorsothat%MATLAB'sbuilt-in'sum'functioncanbeusedtosum%allthetermscalculatedintheTaylorseriesmethod.%Amaximumoffiftytermsisusedintheseries.%Programoutputissentbothtothescreenandtoa%file.Byprintingtheoutputtoafile,youcaneasily%edittheoutputfiletolineupcolumnheadings,%etc.(whichyoucan'tdowhenprintingtothescreen).%Note:e^x=1+x+x^2/2!+x^3/3!+x^4/4!+...clear;clc;xmin=-0.5;dx=0.1;fo=fopen('output.txt','w');%Tableheadingsfprintf('xex1ex2\n');fprintf('-------------------------------------\n');fprintf(fo,'xex1ex2\n');fprintf(fo,'-------------------------------------\n');fori=1:11x=xmin+(i-1)*dx;ex2=exp(x);term(1)=1.0;forn=1:49term(n+1)=term(n)*x/n;ifabs(term(n+1))<=1.0e-7break;endendex1=sum(term);fprintf('%5.2f%10.5f%10.5f\n',x,ex1,ex2);fprintf(fo,'%5.2f%10.5f%10.5f\n',x,ex1,ex2);endfclose(fo);------------------------------------------------------------


Note: In this example, the two statements

fori=1:11 x=xmin+(i-1)*dx;

could be replaced with the following single statement:

forx=-0.5:0.1:0.5

This approach is acceptable when the for loop index is not used to select an element of a matrix. In MATLAB, the variable selecting an element of a matrix must be an integer.

Example 2.3: While Loop

%Example_2_3.m%Calculationofe^xbybothMATLAB'sexpfunctionand%theTaylorseriesexpansionofe^x.Theinput()%functionisusedtoestablishtheexponentx.%A'while'loopisusedindeterminingtheseries%solution.A'break'statementisusedtoendtheloop%ifthenumberoftermsbecomesgreaterthan50.The%displaystatementisusedtodisplaythevaluesofx,%exandex2inthecommandwindow.%Note:e^x=1+x+x^2/2!+x^3/3!+x^4/4!+...clear;clc;x=input('Enteravaluefortheexponentx\n');n=1;ex=1.0;term=1.0;whileabs(term)>ex*1.0e-6term=x^n/factorial(n);ex=ex+term;n=n+1;ifn>50break;endendex2=exp(x);disp(x);disp(ex);------------------------------------------------------------

2.7 MATLAB Function FilesMATLABfunctionsareuseful ifyouhaveacomplicatedprogramandwish tobreakitdownintosmallerparts.Also,ifaseriesofstatementsistobeusedmanytimes,itisconvenienttoplacetheminafunction.Functionsareequivalenttosub-routinesinmostprogramminglanguages,butinMATLABtheyareusuallystoredinseparatefilesinsteadofinthemainprogram(althoughsmallfunctionscanbedefinedinthesamefileasyourmainscript.Thisisdescribedinthenextsection).


◾ Thefunctionfilenamemustbesavedasfunction_name.m.

◾ MATLABhasa template forwritinga function (seeFigure 2.9).Thefirstexecutablestatementinthefunctionfilemustbefunction.AscanbeseenfromFigure 2.9,thefunctiontemplateisoftheform:

function[outputarguments]= Untitled(inputarguments)

seeabovefigure

SomeexamplefunctiondefinitionsareshowninTable 2.1. Ifthefunctionhasmorethanoneoutputvalue,thentheoutputvariables

mustbeinbrackets.Ifthereisonlyoneoutputvalue,thennobracketsarenecessary.Iftherearenooutputvalues,useemptybrackets.

◾ AfunctiondiffersfromascriptinthatthereturnargumentsmaybepassedtoanotherfunctionortotheCommandwindow.Variablesdefinedandmanip-ulatedinsidethefunctionarelocaltothefunction.

Table 2.1 Example of Function Usage

Function Definition Line Function File Name

function[P,V]=power(i,v) power.m

functionex=exf(x) exf.m

function[]=output(x,y) output.m

Figure 2.9 Function template. (From MATLAB, with permission.)


◾ ManyofMATLAB’s“built-in”functionsareactuallyimplementedas.mfiles.For example,factorial is implemented in thefile factorial.m.Youcanfindfactorial.mbytypingwhich factorialintheCommandwindow.

Example 2.4: MATLAB Function Definition

%exf1.m(Example2.4)%Defineafunction,'exf1',thatevaluatese^x.%Thisfunctiontakes'x'asaninputargumentandis%calledeitherfromtheCommandWindoworfromanother%program.Theresultingoutput,'ex',isavailableto%beusedinanotherprogramorintheCommandWindow.%Inthisexample,term(n+1)isobtainedfromterm(n)%bymultiplyingterm(n)byx/n.%Note:e^x=1+x+x^2/2!+x^3/3!+x^4/4!+...functionex=exf1(x)term(1)=1.0;forn=1:99term(n+1)=term(n)*x/n;ifabs(term(n+1))<=1.0e-7break;endendex=sum(term);------------------------------------------------------------

To test out this function, run it from the Command window. Some examples follow:

>>exf1(1.0) ans=

2.7183>>y = exf1(5.0)y=

148.4132

Example 2.5: Using a MATLAB Function

%Example_2_5.m%Thisprogramusesthefunctionexf1(definedin%Example2.4)inseveralarithmeticstatements,i.eto%calculatew,yandz.Inthisprogram,theoutputis%senttoafile.clear;clc;fo=fopen('output.txt','w');fprintf(fo,'xywz\n');fprintf(fo,'----------------------------------\n');forx=0.1:0.1:2.0y=exf1(x);w=5*x*exf1(x);z=10*x/y;fprintf(fo,'%4.2f%8.3f%8.3f%8.3f\n',x,y,w,z);endfclose(fo);--------------------------------------------------------


Thefollowingexampledemonstratesthatthenamesoftheargumentsinthecall-ingprogramneednotbethesameasthoseinthefunction.Itisonlytheargumentlistinthecallingprogramthatneedstobeinthesameorderastheargumentlistdefinedinthefunction.Thisfeatureisusefulwhenafunctionistobeusedwithseveraldifferentscripts,eachscripthavingdifferentvariablenamesbutwitheachofthevariablenamescorrespondingtovariablesinthefunction.

Example 2.6: Defining and Using a MATLAB Function

%Example_2_6.m%Thisprogramusesthefunctionexf2tocalculate%q=e^w/z.wandzareinputvariablestofunction%exf2.Thefunctionexf2calculatesqandreturnsit%tothisprogram.Theinputvariablestofunctionexf2%neednotbethesamenamesasthoseusedinthe%callingprogram.Forone-to-onecorrespondence,the%argumentpositionsinthecallingprogramhavetobe%thesameastheargumentpositionsinthefunction.%Note:functionexf2(definedbelow)mustbecreated%beforethisprogramisexecutedbecausethefunction%exf2calculatesexyandreturnsittothecallingprogram.%qissetequaltoexy.clear;clc;z=5;forw=0.1:0.1:2.0q=exf2(w,z);fprintf('Valuesofw,zandqfrommainprogram:');fprintf('w=%.2fz=%.0fq=%.5f\n',w,z,q);end---------------------------------------------------------------------%exf2.m%ThisscriptisusedwithscriptExample_2_6.mandcreatesa%function'exf2'thatevaluatese^x/y.Inthis%function,xandyareinputvariablesthatneedtobe%definedinanotherprogram.Theoutputvariableinthe%functionexf2issentbacktothecallingprogram.%Thisexampledemonstratesthatthenamesofthe%argumentsinthecallingprogramneednotbethesame%asthoseinthefunction.Itistheorderlistinthe%callingprogramthatneedstobethesameastheorder%listinthefunction.%Inthisexample,term(n+1)isobtainedfromterm(n)by%multiplyingterm(n)byxanddividingbyindexn.The%programdoesitsownsummingofterms.%Note:e^x=1+x+x^2/2!+x^3/3!+x^4/4!+...functionexy=exf2(x,y)%Note:xandyaresetequaltowandzrespectively.s=1.0;term=1.0;forn=1:100term=term*x/n;s=s+term;ifabs(term)<=s*1.0e-5break;end


endexy=s/y;%Thesefprintfstatementsdemonstratetheequivalence%betweenfunctioninputvariables(x,y)andcalling%programvariables(w,z)intheprogramExample_2_6.m.%Comparetheprintoutofx,yandexywithw,zandq.fprintf('Valuesofx,yandexyfromexf2():');fprintf('x=%.2fy=%.0fexy=%.5f\n',x,y,exy);---------------------------------------------------------------------

2.8 Anonymous FunctionsSometimes, it ismoreconvenient todefinea function insideyour script ratherthaninaseparatefile.Forexample,ifafunctionisbrief(perhapsasingleline)andunlikelytobeusedinotherscripts,thentheanonymousformofafunctioncanbeused.Thiswillsaveyoufromhavingtocreateanother.mfile.Anexampleis

fh=@(x,y)(y*sin(x)+x*cos(y));

MATLABdefinesthe@signasafunction handle.Afunctionhandleisequiva-lenttoaccessingafunctionbycallingitsfunctionname,whichinthiscaseisfh.The (x,y) defines the input arguments to the function, and the (y*sin(x)+x*(cos(y))istheactualfunction.Anonymousfunctionsmaybeusedinascriptorinthecommandwindow.

Forexample,intheCommandwindow,typeinthefollowingtwolines:

>>fh = @(x,y)(y*sin(x)+x*cos(y));>>w = fh(pi,2*pi)w= 3.1416

Additional informationon anonymous functions canbeobtainedby typinghelp function _ handleintheCommandwindow.

2.9 MATLAB Graphics◾ Plotcommands:

plot(x,y) linear plot of y versus x

semilogx(x,y) semilog plot (log scale for x axis, linear scale for y axis)

semilogy(x,y) semilog plot (linear scale for x axis, log scale for y axis)

loglog(x,y) log-log plot (log scale for both x and y axes)


◾ Multipleplots:

plot(x,y,w,z) gives multiple plots on the same axes: in this case y versus x and z versus w.

Forexample,suppose

=

1 1 1 1

2 2 2 2A

t y z wt y z w

t y z wn n n n

LetT =A(:,1),Y =A(:,2),Z=A(:,3),andW=A(:,4),giving

= = = =

1

2

1

2

1

2

1

2T

tt

t

Y

yy

y

Z

zz

z

W

ww

wn n n n

Then,toplotYversusT,ZversusT,andWversusTallonthesamefig-ure,write

plot(T,Y,T,Z,T,W);

Ofcourse,wecouldhaveavoidedtheadditionalstepbywriting

plot(A(:,1),A(:,2),A(:,1),A(:,3),A(:,1),A(:,4))

◾ Customizingplots:

Youcanaddagridtotheplotwiththecommand,

grid

Tolabelthexandyaxesandtoaddatitletotheplot,addthefollowingcommandstoyourprogram:

xlabel('T'); ylabel('Y,Z,W'); title('Y,Z,WvsT');


Youcanalsoaddtexttotheplotwiththecommand

text(xposition,yposition,'textstatement');

Greeklettersandmathematicalsymbolscanbeusedinxlabel,ylabel,title,andtextbyspellingouttheGreekletterandprecedingitwitha‘\’(backslashcharacter).Thus,todisplayω,use\omega,andtodisplayγ,use\gamma.Forexample:

ylabel('\omega'),title('\omegavs.\gamma'),text(10,5,'\Phi');

For a complete list ofGreek symbols and additional special characters,seeAppendixB.Youmayalsooccasionallyneedtoprinta“'”characterinyourlabelortitle.Inthiscase,useadouble'to“escape”thesingle-quotecharacterinyourstring.Thus,togeneratetheplottitle“Signal'A'vs.Signal'B'”,youwouldtype

title('Signal''A''vs.Signal''B''')

Inmostcases,youcanletMATLABselectthescalingforthexandyaxesofatwo-dimensionalplot.However,ifyouareunsatisfiedwithMATLAB’sautomatic selection, thenyoucanuse theaxis command to set theplotlimitsexplicitly:

axis([xminxmaxyminymax])

wherexminandxmaxsetthescalingforthexaxis,andyminandymaxsetthescalingfortheyaxis.

◾ Interactivelyannotatingplots: As analternative to adding thexlabel,ylabel, andtitle com-

mands to yourprogram,you can create theplot, then clickon the InsertmenuintheplotwindowandchooseX Labelfromthedrop-downmenu.This will highlight a box in which you can type in the abscissa variablename.YoucanrepeatthisprocessfortheY Label andtheTitleoftheplot.OtheroptionsavailableintheInsertmenuareTextBox, Text Arrow, Arrow,andothers.Whenyouclickanyoneoftheseoptions,acrosshairwillappear,andyoucanthenmovetheitemtothelocationwhereyouwantittoappear,thenleftclickthemousetofixthelocation.Youcanthentypeinthedesiredtext.ToremovetheoutlinesofaTextBox,placethecursorintheTextBoxandrightclickthemouse.Thiswillbringupadrop-downmenu;selectLine Styleandthenleftclickonnone.

◾ Savingplots: Tosaveaplot,clickonFile intheplotwindowandselecttheSaveoption

fromthedrop-downmenu.Thisproducesawindowwhereyoucanenterafilename.Thedisadvantageofthismethodisthatifyoudecidetorerunthe


script,theitemsthatyoumanuallyinsertedwillnotbesaved.Ifyouwishtocopythefigureintoareport,youcanclickonEditintheplotwindow,thenselectCopy Figurefromthedrop-downmenu.Youcanthenpastethefigureintoyourreport. Ifyouneedamonochromeversionofyourplot (forbestreproductiononaphotocopier), youcanmakeallofyourcurvesblackbychoosingtheFilemenuandthenselectingExport Setupfromthedrop-downmenu.ThiswillopenawindowinwhichyouneedtoclickonRenderingandchangetheColorspacetoblack and white.

Therearemanymoreoptionsavailableintheplotwindow;however,weleaveituptothestudenttoexplorefurther.

Example 2.7: Creating Plots

%Example_2_7.m%Thisprogramcreatesasimpletableandasimpleplot.%First,thevectorsy1,y2andtarecalculated.%Nextatableofy1=n^2/10andy2=n^3/100iscreated.%Theny1andy2areplotted.clear;clc;forn=1:10t(n)=n;y1(n)=n^2/10;y2(n)=n^3/100;end%Bymakingt,y1andy2asvectors,theirvaluescanbe%printedoutoutsidethe'for'loopthatcreatedthem.%Printcolumnheadingsfprintf('ty1y2\n');fprintf('--------------------------------------\n');forn=1:10fprintf('%8.1f%10.2f%10.2f\n',...t(n),y1(n),y2(n));end%Createtheplot.y1isred,y2isingreen.plot(t,y1,'r',t,y2,'g');xlabel('t'),ylabel('y1,y2'),grid;title('y1andy2vs.t');%Plotidentificationisalsoestablishedbyaddingtext%totheplot.text(6.5,2.5,'y1');%Intheabovestatement,6.5istheabscissaposition%and2.5istheordinatepositionwherethe'y1'label%willbepositioned.text(4.2,2.4,'y2');%OtherMATLABplotoptions:%Availablecolortypes:%blue'b'%green'g'%red'r'%cyan'c'%yellow'y'%Curvesy1andy2canalsobedistinguishedbyusing


%differenttypesoflinesaslistedbelow.%Availablelinetypes:%solid(default)%dashed'--'%dashed-dot'-.'%dotted':'%Alternatively,youcancreateamarkerplotof%discretepoints(withoutaline)byusingoneofthese%markerstyles:%point'.'%plus'+'%star'*'%circle'o'%x-mark'x'%The'legend'commandmayalsobeusedinplaceofthe%textcommandtoidentifythecurves.Thelegendbox%maybemovedbyclickingontheboxanddraggingitto%thedesiredposition.----------------------------------------------------------------------------

The resulting plot is shown in Figure 2.10.

Example 2.8: The Subplot Command

Suppose you want to plot each of the curves in the previous example as a separate plot, but all on the same page. The subplot command provides

1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

8

9

10

t

y1,y

2y1 and y2 vs. t

y1y2

Figure 2.10 Example plot generated with the plot command.


the means to do so. The command subplot(m,n,p) breaks the page into an m by n matrix of small plots, and p selects the matrix position of the plot. The following example demonstrates the use of the subplot command:

%Example_2_8.m%Thisprogramisanexampleoftheuseofthesubplot%commandandtheelseifladder.Valuesofy1,y2,y3%andy4areconstructedasvectors.Separateplotsof%y1vs.t,y2vs.tandy3vs.tareplottedonthe%samepage.clc;clear;forn=1:11t(n)=n-1;y1(n)=t(n)^2/10;y2(n)=sin(pi*t(n)/10);y3(n)=exp(t(n)/2);y4(n)=sqrt(t(n));endforn=1:4subplot(2,2,n)ifn==1plot(t,y1),grid,title('y1vs.t');xlabel('t'),ylabel('y1');elseifn==2plot(t,y2),grid,title('y2vs.t');xlabel('t'),ylabel('y2');elseifn==3plot(t,y3),grid,title('y3vs.t');xlabel('t'),ylabel('y3');elseifn==4plot(t,y4),grid,title('y4vs.t');xlabel('t'),ylabel('y4');endend----------------------------------------------------------------------------

Example 2.9: The sprintf Function

Sometimes, we might wish to enter formatted data into the xlabel, yla-bel, or title of a graph. We can do this with the sprintf command, which allows data to be formatted using a similar mechanism as fprintf but sends the result to a function or variable instead of to the screen or a file. This is shown in the following example:

%Example_2_9.m%Thisprogramdemonstratestheuseofsprintfcommand%topassformatteddatatoylabelandtitleinagraph.%Thesubplotcommandisalsodemonstrated.clear;clc;%Definethedatatoplotx=-10:2:10;m=2:5;b=[1101001000];%Createtableandplotfori=1:4


fprintf('m=%i\n',m(i));%printtableheadingsfprintf('xyc\n');fprintf('----------------------\n');%(xtopowerm(i))dividedbyb(i)forj=1:length(x)yc(j)=x(j)^m(i)/b(i);fprintf('%5.0f%8.2f\n',x(j),yc(j));endfprintf('\n\n');%plotsubplot(2,2,i);plot(x,yc);xlabel('x');ylabel(sprintf('(xtopower%d)/%.0f',m(i),b(i)));grid,title(sprintf('m=%d,b=%d',m(i),b(i)));end-------------------------------------------------------------------------

See Figure 2.11 for the resulting plot.

–10 –5 0 5 100

20

40

60

80

100

x

(x to

pow

er 2

) / 1

m = 2, b = 1

–10 –5 0 5 10–100

–50

0

50

100

x

(x to

pow

er 3

) / 1

0

m = 3, b = 10

–10 –5 0 5 100

20

40

60

80

100

x

(x to

pow

er 4

) / 1

00

m = 4, b = 100

–10 –5 0 5 10–100

–50

0

50

100

x

(x to

pow

er 5

) / 1

000

m = 5, b = 1000

Figure 2.11 Example subplots and example use of the sprintf command in the plot titles and labels.


Example 2.10: The Stem Plot

In discrete-time signal processing, a stem plot is often used to display signal sequences. Given a signal x[n] that is defined for integer values of n, the com-mand stem(n,x) will plot x as vertical “lollipops” stemming from the horizon-tal axis. The following script generates a stem plot for the following data:

x[0] = 2.4, x[1] = 3.8, x[2] = −1.8, x[3] = 2.1, x[4] = 1.9

x[5] = 0.8, x[6] = 0.0, x[7] = −0.8, x[8] = 1.5

%Example_2_10.m%Thisscriptusesthe'stem'commandtoplotdiscrete%data.clc;clear;x=[2.43.8-1.82.11.90.80.0-0.81.5];n=[012345678];stem(n,x);xlabel('n'),ylabel('x[n]');axis([-19-35]);------------------------------------------------------------------------


–1 0 1 2 3 4 5 6 7 8 9–3

–2

–1

0

1

2

3

4

5

n

x[n]

Figure 2.12 A stem plot.


Example 2.11: Creating a Data File to be Used in an External Program

%Example_2_11.m%Thisprogramisusedtocreateadatafilethatcanbe%loadedinanotherprogramforuseinavarietyofways.clear;clc;forn=1:20t(n)=n;y1(n)=n^2/10;y2(n)=n^3/100;y3(n)=n^4/1000;endfo=fopen('ydata.txt','w');fori=1:20fprintf(fo,'%6.1f%10.5f%10.5f%10.5f\n',...t(i),y1(i),y2(i),y3(i));end%Runningtheprogramcreatesthedatafile'ydata.txt':fprintf('Dataissavedtofile''ydata.txt''\n');%Addingtheclearstatementremovesalldatafromthe%workspace:clear;-------------------------------------------------------------------------

Example 2.12: Loading Data Created in an External Program

%Example_2_12.m%Thisprogramdemonstratestheuseoftheloadcommand.%Adatafilenamedydata.txt(createdinExample_2_11.m)%isloadedintothisprogram.Then,variablest,y1,y2,y3%areextractedfromtheloadeddataandplottedallon%onegraph.clear;clc;fprintf('Readingdatafrom''ydata.txt''\n');loadydata.txtt=ydata(:,1);y1=ydata(:,2);y2=ydata(:,3);y3=ydata(:,4);plot(t,y1,t,y2,'--',t,y3,'-.');xlabel('t'),ylabel('y1,y2,y3'),title('y1,y2andy3vs.t'),grid;legend('y1','y2','y3');-----------------------------------------------------------------

An alternative to the load command is the dlmread command. This command will read an ASCII delimited file. All data in the file must be numeric. In Example 2.12, we could replace the lines starting with loadydata.txt and ending with y3=ydata(:,4) with

Y=dlmread('ydata.txt'); t=Y(:,1); y1=Y(:,2); y2=Y(:,3); y3=Y(:,4);


Example 2.13: Semilog Plots%Example_2_13.m%Thisprogramisademonstrationofasemilogplot%commandclc;clear;x=1:0.5:5;fori=1:length(x)y(i)=3.0*exp(x(i));endsemilogy(x,y),title('Semilogplotofyvs.x');xlabel('x'),ylabel('y'),grid;---------------------------------------------------------

Example 2.14: Plotting Trigonometric Functions%Example_2_14.m%Thisscriptcalculatesfunctionssin(2x/3)andcos(3x+pi)%for-pi<=x<=pi.Thexdomainissubdividedinto50%subdivisions.Thefunctionsateachdatapointis%calculatedandstoredinthevectors'fsin'and'fcos'.%Thenthetwovectorsareprintedasatabletoafile%named'output.txt'.Plotsofbothfunctionsonthesame%axisarecreated.Finally,themaximumvaluesof%'fsin'and'fcos'aredeterminedandprinted.clear;clc;xmin=-pi;dx=2*pi/50;fori=1:51x(i)=xmin+(i-1)*dx;arg1=2*x(i)/3;arg2=3*x(i)+pi;fsin(i)=sin(arg1);fcos(i)=cos(arg2);endfo=fopen('output.txt','w');%Tableheadingsfprintf(fo,'xfsinfcos\n');fprintf(fo,'--------------------------------------\n');fori=1:51fprintf(fo,'%10.5f%10.5f%10.5f\n',...x(i),fsin(i),fcos(i));endfprintf(fo,'--------------------------------------\n\n');%Printmaximumvalues:fsin_max=max(abs(fsin));fcos_max=max(abs(fcos));fprintf(fo,'fsin_max=%10.5f\n',fsin_max);fprintf(fo,'fcos_max=%10.5f\n',fcos_max);%Plotdata:plot(x,fsin,x,fcos,'--');xlabel('x'),ylabel('fsin,fcos'),grid,title('fsinandfcosvs.x');legend('fsin','fcos');-------------------------------------------------------------------------


2.10 Working with MatricesAdditionalexamplesfollowdemonstratingtheuseof

1.nested for loops forenteringa3×3matrixfromthekeyboard. 2.theinputcommand. 3. theloadcommand. 4. anexampleofaself-writtenfunction. 5.thefscanfcommand.

Example 2.15: Interactive Keyboard Input of a Matrix

%Example_2_15.m%Inputa3by3matrixfromthekeyboard.%Nestedloopsareusedtocreatea3by3matrix.%Anelementinthematrixisestablishedbytheindices%ofboththeinnerandouterloops.clear;clc;forn=1:3form=1:3fprintf('n=%im=%i',n,m);a(n,m)=input('Entera(n,m):');fprintf('\n');endend%Printtheresultingmatrix:a-----------------------------------------------------------

Example 2.16: Loading a Matrix from a File

%Example_2_16.m%MatricesAandBareloadedfromafilenamed%data216.txt,thenprintedtothescreen.MatricesAand%BarecombinedintomatrixC,whichissenttofunction%signal_out.ThisfunctionsubdividesmatrixCbackinto%matricesAandBandprintsthemouttothescreen.%Note:Beforerunningthisprogram,thedatafile%data216.txtandfunctionfilesignal_out.mneedtobe%created(seebelow).clear;clc;n=5;loaddata216.txtA=data216(:,1:n);B=data216(:,n+1);fprintf('MatrixAelements:\n');Afprintf('\n');fprintf('MatrixBelements:\n');BC=[AB];signal_out(C);-----------------------------------------------------------


%Datafile'data216.txt'(donotincludethisheaderline)8.772.405.661.551.0-32.044.931.214.481.101.0-20.073.531.462.921.211.0-8.535.054.042.512.011.0-6.303.541.043.471.021.0-12.04-----------------------------------------------------------%signal_out.m%ThisfunctionacceptsmatrixA&matrixBandprints%themtothescreenandisusedwithExample2.16.function[]=signal_out(C)n=5;A=C(:,1:n);fprintf('Thisoutputisfromsignal_out:\n\n');fprintf('MatrixAelements:\n')fori=1:nforj=1:nfprintf('%8.3f',A(i,j));endfprintf('\n');endfprintf('\n\nMatrixBelements:\n');B=C(:,n+1);fori=1:nfprintf('%8.3f\n',B(i));end----------------------------------------------------------

Example 2.17: Loading a Matrix with fscanf

%Example_2_17.m%ThisprogramusesthefscanfcommandtoenteranMbyN%matrixintoMATLAB.TheMbyNmatrixisenteredin%columnorder.Thus,rowsbecomecolumnsandcolumns%becomerows.ThescriptcreatesmatricesAandBand%thenprintsthemouttothescreen.%Note:Beforerunningthisprogram,thedatafile%data217.txtneedstobecreated(seedatabelow).clear;clc;n=5;A=zeros(n);B=zeros(5,1);fin=fopen('data217.txt','r');[A]=fscanf(fin,'%f',[n,n]);fprintf('MatrixAelements:\n');A[B]=fscanf(fin,'%f',[n]);fclose(fin);fprintf('\n');fprintf('MatrixBelements:\n');Bfprintf('\n');fprintf('MatrixCelements:\n');C=[A'B]%Comparingthedata217datafilewithmatricesAandB,


%weseethatthedata217datafilehasbeentransposed%(rowsbecomecolumnsandcolumnsbecomerows).----------------------------------------------------------%Datafile'data217'(donotincludethisheaderline)8.772.405.661.551.04.931.214.481.101.03.531.462.921.211.05.054.042.512.011.03.541.043.471.021.0-32.04-20.07-8.53-6.30-12.04----------------------------------------------------------

2.11 Working with Functions of a VectorMATLABallowsfunctionstohavevectorsasarguments.Thereturnvaluecanalsobeavector.Forexample(writethissimplescriptinthescriptwindow,saveitasvector_function.m,andrunthescript):

%vector_function.m clear;clc; %Definevectorx; x=0:30:360; %Lety1bethesineofavectorxwherexisindegrees. %Runningsindonavectorwillreturnavector: y1=sind(x); %Lety2(n)bethesineofthenthelementofx.Wewill %useaforlooptocalculateeachvaluey2(n)andthen %comparey1andy2. forn=1:length(x) y2(n)=sind(x(n)); end %Tableheadings fprintf('x y1 y2 \n'); fprintf('------------------------------------------------\n'); forn=1:length(x) fprintf('%5.1f %8.5f %8.5f\n',x(n),y1(n),y2(n)) end

Inthegeneratedoutput,doesy1=y2?Weseethatinsomescripts,wecouldreplacetheuseofaforloopbytaking

afunctionofavector,whichproducesavector,thusreducingthenumberoflinesinthescript.Thisconceptisdemonstratedaboveinthescriptvector_function.m.However,ifthescriptrequiresamathematicaloperationoftwofunctions(suchasaproductoftwovectorfunctions),thentheoperationwillrequireanelement-by-elementoperation.Element-by-elementoperationsarediscussedinChapter3.


2.12 Additional Examples Using Characters and StringsAdditionalexamplesfollowthatdemonstrate

◾ printingastringofcharacters

◾ theuseoftheswitchstatement

◾ theuseoftheif-elseifladder

◾ theuseofafor looptoselecttheproperintervalinamatrixforestablishingagradeorforinterpolation.

Example 2.18: Printing a Character String Matrix from a Loop

%Example_2_18.m%Sometimesyoumightwishtoprintoutastringof%charactersinaloop.Thiscanbedonebydeclaringa%2-Dcharacterstringmatrixasshowninthisexample.%Notethatallrowcharacterstringsmusthavethesame%numberofcolumnsandthatcharacterstringsmustbe%enclosedbysinglequotationmarks.clear;clc;parts=['Internalmodem''Graphicsadapter''CDdrive''DVDdrive''Floppydrive''Harddiskdrive'];fori=1:5fprintf('%16s\n',parts(i,1:16));end------------------------------------------------------------

Example 2.19: Printing Strings to the Screen or to a File Based on User Input

%Example_2_19.m%ThisexampleisamodificationofExample2.18.The%programaskstheuseriftheywanttohavetheoutput%gotothescreenortoafile.Thisexampleillustrates%theuseoftheswitchstatement.clear;clc;char=['Internalmodem''Externalmodem''Graphicscircuitboard''CDdrive''Harddiskdrive'];fprintf('Choosewhethertosendtheoutputtothe\n');fprintf('screenortoafilenamed''output.txt''.\n');fprintf('\n');fprintf('Enter''s''forscreenor‘'f''forfile\n');var=input('(withoutquotes):','s');switch(var)


case's'fori=1:5fprintf('%22s\n',char(i,1:22));endcase'f'fo=fopen('output.txt','w');fori=1:5fprintf(fo,'%22s\n',char(i,1:22));endfclose(fo);otherwisefprintf('Youdidnotenteran''s''oran');fprintf('''f'',tryagain.\n');end------------------------------------------------------------

Example 2.20: The if-elseif Ladder

%Example_2_20.m%Thisexampleusestheif-elseifladder.%Theprogramdeterminesalettergradedependingonthe%scoretheuserentersfromthekeyboard.clear;clc;gradearray=['A';'B';'C';'D';'F'];score=input('Enteryourtestscore:');fprintf('Scoreis:%i\n',score);ifscore>100fprintf('Error:scoreisoutofrange.\n');fprintf('Rerunprogram.\n');break;elseif(score>=90&&score<=100)grade=gradearray(1);elseif(score>=80&&score<90)grade=gradearray(2);elseif(score>=70&&score<80)grade=gradearray(3);elseif(score>=60&&score<70)grade=gradearray(4);elseif(score<60)grade=gradearray(5);endfprintf('Gradeis:%c\n',grade);------------------------------------------------------------

Example 2.21: Interactive User Input Establishing a Grade

%Example_2_21.m%Thisprogramdeterminesalettergradedependingonthe%scoretheuserentersfromthekeyboard.Thisversion%usesalooptodeterminethecorrectintervalof%interest.Foralargenumberofintervals,thismethod%ismoreefficient(fewerstatements)thanthemethodin%Example2.20clear;clc;gradearray=['A';'B';'C';'D';'F'];


sarray=[100908070600];score=input('Enteryourtestscore:');fprintf('Scoreis:%i\n',score);%Thefollowing2statementsareneededforthecase%whenscore=100.ifscore==100grade=gradearray(1);elsefori=1:5if(score>=sarray(i+1)&&score<sarray(i))grade=gradearray(i);endendendfprintf('Gradeis:%c\n',grade);------------------------------------------------------------

Example 2.22: Printing String Arrays

%Example_2_22.m%Thisprogramdeterminesthelettergradesofseveral%students.Student'snamesandtheirtestscoresare%enteredintheprogram.Thisexampleusesnested'for'%loopsandan'if'statementtodeterminethecorrect%lettergradeforeachstudent.clear;clc;gradearray=['A';'B';'C';'D';'F'];sarray=[100908070600];Lname=['Smith''Lambert''Kurtz''Jones''Hutchinson''Diaz'];Fname=['Joe''Jane''Howard''Martin''Peter''Carlos'];score=[84;72;93;64;81;75];%Thescore=100istreatedseparately.forj=1:6ifscore(j)==100grade(j)=gradearray(1);elsefori=1:5ifscore(j)>=sarray(i+1)&&score(j)<sarray(i)grade(j)=gradearray(i);endendendendfprintf('LastnameFirstnameGrade\n');fprintf('----------------------------------\n');forj=1:6


fprintf('%12s%10s%c\n',...Lname(j,1:12),Fname(j,1:10),grade(j));end------------------------------------------------------------

Example 2.23: A Self-Written Function and String Arrays

An important building block in program development is the self-written function. In the following, we construct the function func _ grade to determine the grade in the previous example. The input to func _ grade is the score, and the output is the corresponding grade. Remember that a user-defined function is saved in a separate .m file.%Example_2_23.m%Thisprogramusesfunc_grade.m(definedbelow)to%determinethegradeofastudent.Theinputtothe%functionisscoreandthefunctionreturnsthegrade%tothecallingprogram.clear;clc;Lname=['Smith''Lambert''Kurtz''Jones''Hutchinson''Diaz'];Fname=['Joe''Jane''Howard''Martin''Peter''Carlos'];%Thescoresofthestudentsnamedabovearegiveninthe%vectorscoreintheorderlistedinLname.score=[84;72;93;64;81;75];fprintf('LastnameFirstnameGrade\n');fprintf('--------------------------------\n');forj=1:6grade(j)=func_grade(score(j));fprintf('%12s%10s%c\n',...Lname(j,1:12),Fname(j,1:10),grade(j));end------------------------------------------------------------%func_grade.m%ThisfunctionworkswithExample_2_23.mfunctiongrade=func_grade(score)gradearray=['A';'B';'C';'D';'F'];sarray=[100908070600];ifscore==100grade=gradearray(1);elsefori=1:5if(score>=sarray(i+1)&&score<sarray(i))grade=gradearray(i);endendend------------------------------------------------------------


2.13 Interpolation and MATLAB’s interp1 FunctionFormanyphysicalphenomena,wefrequentlyfindthatapropertyofagivenmate-rialwillbecomputedasafunctionofsomeothervariable.Forexample,themobil-ityof electrons in silicon isdependenton temperatureand isoftenpresentedasatableoflaboratorymeasurementsforafewdistincttemperatures.Ifwewishtodeterminetheelectronmobilityatatemperaturethatisnotinthetable,weneedtointerpolate.Thegenerallinearinterpolationformulais

= +

− × −−

( ) ( )1

1 2 1

2 1

T TT T

whereT2andT1aretheclosestsurroundingmeasuredtemperaturestoT,andμ2andμ1aretheelectronmobilitiesattemperaturesT2andT1,respectively.Thenextexample illustrates thecomputationofthemobility foranarbitrarytemperaturebasedonatableofknownvaluesatknowntemperatures.

Example 2.24: Interpolation Using an Anonymous Function

%Example_2_24.m%Thisprogramusesananonymousfunctiontolinearly%interpolatetheelectronmobilityofsilicon.Measured%valuesofthemobility(cm^2/V-s)vstemperature(K)%aregiven[2].clear;clc;%Anonymousfunction(avoidscreatinganextra.mfile):Muf=@(T,T1,T2,Mu1,Mu2)(Mu1+(T-T1)*(Mu2-Mu1)/(T2-T1));%TablevaluesTt=[100150200250300350400450];Mut=[196507427372321801407972.0705.5531.8];fprintf('Thisprograminterpolatesfortheelectron\n');fprintf('mobility(Mu)ataspecifiedtemperatureT.\n');fprintf('Thevalidtemperaturerangeis100-450K.\n\n');T=input('EnterTatwhichMuistobedetermined:');%Determiningtheclosestsurroundingtemperaturesto%temperatureT.forn=1:length(Tt)-1ifT>=Tt(n)&&T<=Tt(n+1)T1=Tt(n);T2=Tt(n+1);Mu1=Mut(n);Mu2=Mut(n+1);endendMu=Muf(T,T1,T2,Mu1,Mu2);fprintf('T=%.1f(K)Mu=%.3f(cm^2/V-s)\n',T,Mu);------------------------------------------------------------

MATLABhasafunctionnamedinterp1thatperformsinterpolation.Thenextexampledemonstratestheuseofthisfunctionforinterpolatingforelectronmobility.Inaddition,MATLAB’sinterp1functioncandomultipleinterpola-tionsbyjustcallingthefunctiononce.


Thesyntaxforinterp1is:

Yi=interp1(X,Y,Xi)

whereXandYareasetofknownx,ydatapoints,andXiisthesetofxvaluesatwhichthesetofyvaluesYiaretobedeterminedbylinearinterpolation.ArraysX andYmustbeofthesamelength.Thefunctioninterp1canalsobeusedforinterpolationmethodsotherthanlinearinterpolation,andthisiscoveredinChapter9oncurvefitting.

Example 2.25: Interpolation Using interp1 and Saving to a .mat File

%Example_2_25.m%ThisprogramusesMATLAB'sfunctioninterp1to%interpolatefortheelectronmobilityofsilicon.%Measuredvaluesofthemobility(cm^2/V-s)vs%temperature(K)aregiven[2].Tablevaluesfor%temperaturearespecifiedinmatrixT_data.Table%valuesformobilityarespecifiedinMu_data.The%temperatureatwhichtheelectronmobilityistobe%determinedisenteredfromthekeyboard.clear;clc;T_data=[100150200250300350400450];Mu_data=[196507427372321801407972.0705.5531.8];fprintf('Thisprograminterpolatesfortheelectron\n');fprintf('mobility(Mu)ataspecifiedtemperatureT.\n');fprintf('Thevalidtemperaturerangeis100-450K.\n\n');T=input('EnterTatwhichMuistobedetermined:');Mu=interp1(T_data,Mu_data,T);fprintf('T=%.1f(K)Mu=%.3f(cm^2/V-s)\n\n',T,Mu);%ThedataT_dataandMu_datacanbesavedinadatafile%andloadedlaterinanotherprogram.Bytheuseofthe%'clear'commandthedatafileisremovedfromthe%workspace.Anotherscriptcanbringsthedatabackinto%theworkspacebytheloadcommand.Thescriptcanthen%usethedataforsomeotherpurpose.WewantT_dataand%Mu_datatobecolumnvectors,sosavethetranspose.T_Mu_data=[T_data'Mu_data'];saveT_Mu_data;%T_Mu_dataissavedina.matfileandwillbeused%inExample2.25.fprintf('Mobilityvs.temperaturedatasavedtofile');fprintf('T_Mu_data.mat\n');clear;%T_Mu_datahasbeenremovedfromtheworkspace%T_Mu_datawillbeusedinExample_2_26.------------------------------------------------------------

If we wish to interpolate for electron mobility at several different tem-peratures, we can modify Example 2.25 by replacing all lines after (and including) the lines


fprintf('Thisprograminterpolatesfortheelectron\n');fprintf('mobility(mu)ataspecifiedtempratureT.\n');

with the following lines:

T2=[432182343210385];fprintf('Thisprograminterpolatesfortheelectron\n')fprintf('mobility(Mu)ataspecifiedtempratureT.\n');fprintf('specifiedtemperatures,T2.\n');fprintf('Theallowabletemperaturerangeis100-450K.\n\n');Mu=interp1(T_data,Mu_data,T2);fprintf('T2(K)Mu(cm^2/V-s)\n');fprintf('---------------------------\n');fori=1:5 fprintf('%6.1f %9.3f\n',T2(i),Mu(i));end------------------------------------------------------------

Example 2.26: Interpolating Data Stored in a .mat File

%Example_2_26.m%Thisscriptisasampleofaplotprogramloadingdata%obtainedinanotherprogram.MakesureyourunExample%2.25beforerunningthisexample.clear;clc;fprintf('Loadingdatafromfile"T_Mu_data.mat"\n');loadT_Mu_dataT=T_Mu_data(:,1);Mu=T_Mu_data(:,2);plot(T,Mu);xlabel('T'),ylabel('Mu'),title('Muvs.T'),grid;------------------------------------------------------------

2.14 MATLAB’s textscan FunctionThere may be occasions when you wish to read a data file into a program thatcontainsbothnumericaldataandastringofcharacters.MATLAB’stextscanfunctionisbestsuitedforthisoperation.

Syntax: C=textscan(fid,format)

ThefunctionwillreaddatafromanopentextfileidentifiedbyfidintoacellarrayC.Theformatisoftheform%f,%d,%c,%s,andsoon.Thenumberoffor-

matspecifiersdeterminesthenumberofcellsinthecellarrayC.Eachcellwillcon-tainthenumberoflinescontainedinthedatafileandbeofthetypespecifiedbytheformatstatement.Stringspecifiersalsoinclude%q,whichisastringenclosedbydoublequotationmarks.Forasinglecharacter,use%c.

Note:Toreferencethecontentsofacell,enclosethecellnumberwith.Seethefollowingexamples.

IfyouwishtoreadinNlinesfromtheopendatafile,use

C=textscan(fid,format,N)


Example 2.27: Loading Mixed Text and Numerical Data from a File

%Example_2_27.m%Loadtheproductdatafrominv4.txtintothearrays%'sku','desc',and'cost',andprint.%sku,descandcoststandsforitem#,%itemdescriptionanditemcostrespectively.clear;clc;%Note:inventory227.txtisdefinedbelow.fid=fopen('inventory227.txt');C=textscan(fid,'%d%14c%f',5);%ContentsofcellblockCcontains1rowand3columnssku=C1;desc=C2;cost=C3;fclose(fid);count=length(sku)fori=1:countfprintf('%5i%14s%6.2f\n\n',...sku(i),desc(i,1:14),cost(i));end------------------------------------------------------------%inventory227.txtfile(donotincludethisheaderline)1'hammer'2.582'plier'1.203'screwdriver'1.564'solderingiron'3.705'wrench'2.60------------------------------------------------------------

Example 2.28: Another Example of Loading Mixed Text and Numerical Data from a File

%Example_2_28.m%Readproductinventoryintoamatrixusingtextscan.%Thefile'inventory228.txt'islistedbelow.clc;clear;fid=fopen('inventory228.txt');C=textscan(fid,'%d%q%f');%NeedtodesignatecontentsincellC2ascharacters.sku=C1;desc=char(C2);cost=C3;count=length(sku);fprintf('sku#descriptioncost\n');fprintf('-----------------------------------\n');fori=1:countfprintf('%4i%14s%6.2f\n\n',...sku(i),desc(i,1:14),cost(i));endfclose(fid);------------------------------------------------------------%inventory228.txtfile(Donotincludthisheaderline)1"hammer "2.582"plier "1.20


3"screwdriver "1.564"solderingiron"3.705"wrench "2.60------------------------------------------------------------

2.15 Exporting MATLAB Data to ExcelMATLAB can write its data directly to spreadsheet software such as MicrosoftExcel.Todothis,followthesesteps:

1.InyourMATLABprogram,createamatrixoftheoutputdata,sayM. 2.WritethedatatoanExcelfilewiththefollowingstatement: xlswrite('filename.xls',M);

ThiswillcreateanExcelfilenamedfilename.xlsofyourdata,M,whichyoucantheneditwithinExcel(seeFigure 2.13).

3.You can now use all of the formatting and data manipulation commandsavailable inExcel.For example, to add tableheadings,open theExcelfile,placethecursorinthetopleftcell(cellA1),andtypeintableheadingsineach

Figure 2.13 Example of sending output to Excel.


ofthecellcolumns.Toaddupthesecondcolumn,placethecursorincellB13andtype=SUM(B3:B12).Youcanalsoapplybold,italics,andcellborders.

2.16 Debugging a ProgramItiscommonwhenwritingaprogramtomaketypographicalerrors,suchasomittingaparenthesis,forgettingacommainatwo-dimensionalarray,andsoon.Thistypeofbugiscalledasyntax error.Whenthisoccurs,MATLABwillprovideanerrormessagepointingoutthelineinwhichtheerrorhasoccurred.However,therearecaseswhentherearenosyntaxerrors,buttheprogramstillfailstorunorgivesanobviousincorrectanswer.Whenthisoccurs,youcanutilizethedebugfeatureinMATLAB.Thedebugfeatureallowsyoutosetbreakpointsinyourprogram.Theprogramwillrunuptothelinecontainingthebreakpoint.Tosetabreakpoint,leftclickthemouseinthenarrowcolumnnexttothelinenumberthatyouwishtobeabreakpoint.AsmallredcirclewillappearnexttothebreakpointlineasshowninFigure 2.14.Thescriptwillnowrunupto,butnotincluding,thelinecontainingthebreakpoint,asshowninFigure 2.15(atthecommandtoprintmatrixA).TheK>>promptindicatesthatyouareinthedebugmode.YoucanthenclickonthedebuglistingintheMATLABtoolbarandselectfromseveraloptionsinthedrop-downmenu(seeFigure 2.16).Iftheprogramcontainsself-writtenfunctions,youcancontinueexecutiononelineatatimeinthefunction.Todothis,youfirstneedtoclearallthebreakpointsinthemainprogram,thenselecttheStep Inoptioninthedebugdrop-downmenu(seeFigure 2.16).TheStep Outoptionreturnsthecontrol

Debugging breakpoint

Figure 2.14 Setting a break point. (From MATLAB, with permission.)


Figure 2.15 Program execution is halted when the break point is reached in the program. (From MATLAB, with permission.)

Figure 2.16 The Debug drop-down menu. (From MATLAB, with permission.)


tothemainprogram.SelectingClear Breakpoints in All Fileswillremoveallbreakpointsfromtheprogram.

2.17 The Parallel RLC CircuitTheresistance,inductor,capacitor(RLC)circuitisafundamentalprobleminelec-trical engineering that we revisit several times in this text (see Figure 2.17 andAppendixA).Thissectionderivesthegoverningequationsdescribingthevoltagesandcurrentsforthecircuitcomponents.ThecircuitisdrivenbyacurrentsourceIo(t)viaaswitchthatisopenedattimet =0.Wedonot(forthemoment)concernourselveswith thedefinitionof Io(t) other than that it is used todetermine theinitialconditionsofthecircuitelements.Westarttheanalysiswiththeconstituentvoltage-currentrelationsforresistors,inductors,andcapacitors:

Resistor(Ohm’slaw):vR=RiR (2.1)

=Inductor: v L

d idtL

L (2.2)

=Capacitor: i C

dvdtCC (2.3)

whereR,L,andCarethecomponentvaluesfortheresistor(inohms), inductor(inhenries),andcapacitor(infarads),respectively,andthevoltagesvandcurrentsiareasdefinedintheFigure 2.17.Inaddition,weassumethatthestatefulcircuitelementsLandChaveknowninitialconditionsiL(0)andvC(0)atthemomentthattheswitchisopened.

Kirchhoff’s current law (KCL) states that the sumof currents at any circuitnodeiszero.ApplyingKCLatthetoprightnodeinthecircuitgives

+ + = 0i i iR L C (2.4)

CIo(t)

−

+vR

−

+vL

−

+vC

iL iCiR

LR

t = 0

Figure 2.17 Parallel RLC circuit.


Also,theparalleltopologyofthecircuitgives

= = ≡v v v vR L C (2.5)

SubstitutingEquation(2.5)intoEquation(2.2)gives

=

1didt L

vL (2.6)

andsubstitutingEquations(2.1)and(2.3)intoEquation(2.4)gives

= − −

1 1dvdt RC

vCiL (2.7)

Equations(2.6)and(2.7)areasystemoftwofirst-orderdifferentialequationswith two unknowns. We can combine these into one second-order differentialequationbydifferentiatingbothsidesofEquation(2.7):

= − −

1 12

2d vdt RC

dvdt C

didtL (2.8)

SubstitutingEquation(2.6)intoEquation(2.8)givesthesecond-orderdiffer-entialequation

+ + =

1 1 02

2d vdt RC

dvdt LC

v (2.9)

Tosolvethissecondorderhomogeneousdifferentialequation,weseekafunc-tionv(t)suchthatthederivatives ( )v t and ( )v t reproducetheformofv(t).Afunction that satisfies this condition is αAe t ,whereA andα are arbitrary con-stants.Ifweassumethat = αv Ae t ,then

= α = αα α( ) and ( ) 2v t Ae v t Aet t

SubstitutingthesetermsintothedifferentialEquation(2.9)gives

α + α + =α1 1 02

RC LCA e t (2.10)

Since ≠α 0e t ,solvingthisequationinvolvessolvingthequadraticpart

α + α+ =

1 1 02

RC LC (2.11)


Thesolutionsare

α = − ± −

12

12

12

RC RC LC (2.12)

Thus,therearetwosolutionsthatsatisfythedifferentialequationforthetwovaluesofαinEquation(2.12).ThegeneralsolutiontoEquation(2.9)isthesumofallknownsolutions:

exp 12

12

1 exp 12

12

1

or equivalently

exp 12

exp 12

1 exp 12

1

2 2

2 2

v ARC

tRC LC

t BRC

tRC LC

t

vRC

t ARC LC

t BRC LC

t

= − + − + − − −

= − − + − −

(2.13)whereexp(x)=ex,andAandBareconstantsthataredependentontheinitialcon-ditionsofthecircuit.

Notethatthissolutionhastworegionsofinterest:

a) Overdamped: If >1

212

RC LC,thenthesolutionsaredecayingexponen-

tialsovertime.

b) Underdamped: If <1

212

RC LC,thenthesolutionsaredecayingsinusoids

overtime.

Fortheunderdampedcase,wecanshowthesinusoidalbehaviorbyapplyingtheidentities = +cos sine x j xjx and = −− cos sine x j xjx toEquation(2.13),giving

= − − + −exp 12

cos 1 12

sin 1 12

2 2

vRC

t ALC RC

t BLC RC

t

(2.14)wherethecoefficientsA′andB′aretobedeterminedbyinitialconditions.

If =1

21 ,

2

RC LCthenthesquareroottermiszero,andthesystemissaid

tobecritically damped.Forthiscase[3],thesolutionis

= + −( )exp 1

2v A B t

RCt (2.15)


The equations developed in this section are used in projects starting withProject2.7.

ExercisesExercise2.1 ShowthattheTaylorseriesexpansionofcos xaboutx=0is

= − + − +cos( ) 1

2! 4! 6!. . .

2 4 6

xx x x

Exercise2.2 ShowthattheTaylorseriesexpansionofsin xaboutx=0is

= − + − +sin( )

3! 5! 7 !. . .

3 5 7

x xx x x

Exercise2.3 UsingtheTaylorseriesexpansionofexaboutx=0(seeExample2.1),showthat = +cos sine x j xjx andthat = −− cos sine x j xjx ,where = −1.j

ProjectsProject.2.1

UsingtheTaylorseriesexpansionofcos xaboutx =0(seeExercise2.1)developacomputerprogramin MATLABthatwillevaluatecosxfrom.−π ≤.x ≤.π.instepsof0.1π.Computetheseriesbydeterminingeachtermaccordingtotheequation

term(k+1)=term(k)×amultiplicationfactor

wherethemultiplicationfactorneedstobedetermined.Useasmanyas50terms;however,stopaddingtermswhenthelasttermonlyaffectsthesixthdecimalplaceintheanswer.Also,computethevalueofcos xusingMATLAB’sbuilt-incos()functionoverthesameintervalandstepsize.PrintatableofyourresultstoafileusingthetableformatinTableP2.1.Usesixdecimalplacesforcos x.

Also,createaplotofcos xfor–π ≤ x ≤ π.

Table P2.1 Table Format for Project 2.1

xcos x by cos()

cos x by series

Terms in the series

–1.0 π

–0.9 π

–0.8 π

⋮

0.9 π

1.0 π


Project.2.2

UsingtheTaylorseriesexpansionofsin xaboutx =0(seeExercise2.2),developacomputerprogramin MATLABthatwillevaluatesinxfrom.−π ≤.x ≤.π.instepsof0.1π.ComputetheseriesbyusingMATLAB’sfactorialfunctionanduseasmanyas50terms.However,stopaddingtermswhenthelasttermonlyaffectsthesixthdecimalplaceintheanswer.

Also, compute the valueof sin x usingMATLAB’sbuilt-insin() functionoverthesameintervalandstepsize.CreateatablesimilartothetableshowninProject2.1,exceptreplacecos xwithsin x.Createaplotofsin xfor–π ≤ x ≤ π.

Project.2.3

DevelopacomputerprograminMATLABthatwillevaluatethefollowingfunc-tionfor–0.9≤x≤0.9instepsof0.1by:

1. anarithmeticstatement.2. aseriesallowingforasmanyas50terms.However,stopaddingtermswhenthe

lasttermonlyaffectsthesixthdecimalplaceintheanswer.

Thefunctionanditsseriesexpansion,validforx2<1,is

= + = − +

⋅⋅

−⋅ ⋅⋅ ⋅

+⋅ ⋅ ⋅⋅ ⋅ ⋅

− +−( ) (1 ) 1 12

1 32 4

1 3 52 4 6

1 3 5 72 4 6 8

2 1/2 2 4 6 8f x x x x x x

Printoutatable(toafile)intheformatshowninTableP2.3;usesixdecimalplacesforf(x).

Project.2.4

DevelopacomputerprograminMATLABthatwillevaluatethefunction

= +( ) (1 )1/3f x x

for–0.9≤x≤0.9instepsof0.1by


x

f(x)(arithmetic statement)

f(x)(by series)

No. of Terms Used in the Series

–0.9

–0.8

–0.7

⋮

0.7

0.8

0.9


1. anarithmeticstatement.2. seriesallowingforasmanyas50terms.However,stopaddingtermswhenthe

lasttermonlyaffectsthesixthdecimalplaceintheanswer.

ByusingaTaylorseriesexpansion,itcanbeseenthattherelationshipbetweensuccessivetermsis

+ = × − − ×term( 1) term( ) 1

3( 1)k k k

xk

fork=1,2,3,....Printoutatable(toafile)intheformatshowninProject2.3;usesixsignificant

figuresforf(x).

Project.2.5

Thebinomialexpansionfor +(1 )x n ,wherenisaninteger,isasfollows:

(1 ) 1 ( 1)2!

( 1)( 2)3!

( 1)( 2)...( 2)( 1)!

2 3

1

x n x n n x n n n x

n n n n r xr

x

n

rn

+ = + +−

+− −

+ +

− − − +−

+ +−

ConstructaMATLABprogramthatwillevaluate +(1 )x n byboththepreced-ingseriesandanarithmeticstatementforn=10and1.0≤x ≤10.0instepsof0.5.PrintouttheresultsinatableasshowninTableP2.5.


x

(1 + x)n

(arithmetic statement)

(1 + x)n

(by series)

1.0

1.5

2.0

2.5

⋮

9.5

10.0


Project.2.6

Thisprojectexaminesa340Vpeak-to-peaksinusoidalwaveformat60HzthathasnoDCcomponent(i.e.,itiscenteredaround0V).

1. PlotthiswaveforminMATLABbyusingthesin()function.Plotovertheinterval0≤t ≤50msecinstepsof0.2msec.

2. Computeandprinttothescreentheroot-mean-square(RMS)voltageforthiswaveformbyperformingthefollowingoperationsinMATLAB:

a. Squaringit. b. Computingthemeanvalueofthesquaredwaveformbyaveragingit

overthe100-msectimeinterval. c. Takingthesquarerootoftheaverage.

3. In North America, residential electrical service is usually delivered as two340Vpeak-to-peaksinusoids,each180°outofphasewitheachother(FigureP2.6a),referredtoasone-phaseservice.Forhigh-powerappliances(e.g.,airconditionersandstoves),thecurrentisdrawnfromboth“legs”oftheservice.

–170

170

t

v Phase A

Phase B

1 sec60

(a)

–170

170

t

120°

v Phase A Phase B Phase C

120°

(b)

Figure P2.6 (a) One-phase power is delivered as two sinusoids that are out of phase by 180°. (b) Three-phase power is delivered as three sinusoids that are out of phase by 120°.


a. Plotthedifferencewaveformbetweenthetwolegsofone-phaseservice. b. Compute andprint to the screen theRMSvalueof thedifference

waveform(asinpart2).4. Inmanycommercialandindustrialbuildings,three-phaseelectricalserviceis

deliveredasthree340Vpeak-to-peaksinusoids,witheachleg120°outofphasewiththeothers(seeFigureP2.6b).

a. Plotthewaveformofthevoltagedifferencebetweenanytwoofthethreelegs

b. ComputeandprinttothescreentheRMSvalueofthiswaveform.Mosthigh-voltageappliancesareratedtoworkateitheroftheRMSvoltagescomputedinpart3orpart4.

Project.2.7

Given the followingcomponentvalues in theparallelRLCcircuitdescribed inSection2.17

R=100Ω,L =1mH,C =1μF

andthe following initialconditions,determine thecoefficientsA andB for thefollowingthreecases:

1. = =(0) 0.25A, (0) 6 Vi vL C

[Hint:convert (0)iL to (0)v withEquation(2.7).]

2. = − =(0) 0.25A, (0) 6 Vi vL C

3. = =(0) 0, (0) 6 Vi vL C

For each case, determine an expression for A and B in terms of R, L, C,(0) and (0)v iC L andwriteaMATLABprogramthatwill

a. Determinev(t)for0≤t ≤500μsecinstepsof1μsec. b. Printoutatableofvversustevery10μsec.

c. Plotvversust forallthreecasesonthesamegraph.LabeleachcurvewiththevalueofiL(0).

Project.2.8

Theenvelopeof thevoltagewaveformfor theparallelRLCcircuitdescribed inSection2.17isgivenby

= ± ( )−( ) 2v t Aet

RC (P2.8)

Giventhefollowingcomponentvalues

R=5000Ω,L =1μH,C =10pF

= =(0) 0 A, (0) 3.3Vi vL C

determinev(t)for0≤t ≤50nsecinstepsof1nsec.Plotvversustforboththeoscillatingfunctionanditsenvelopeonthesamegraph.


Project.2.9

InsteadofusingvasthestatevariableintheanalysisoftheparallelRLCcircuit,wecanuseiLbydifferentiatingbothsidesofEquation(2.6)andcombiningwithEquation(2.7)toobtainthedifferentialequation

+ + =

1 1 02

2d idt RC

didt LC

iL LL (P2.9a)

NotethatEquation(P2.9a)hastheidenticalformasEquation(2.12)andthushassolutionsoftheidenticalform.Fortheunderdampedcase,

= − α − +β −exp 12

cos 1 12

sin 1 12

2 2

iRC

tLC RC

tLC RC

tL .(P2.9b)

where αandβ are constants to be determined by initial conditions and haveamperesasunits.

ForthecomponentvaluesR=100Ω,L =1mH,C =1μF,andinitialcondi-tions =(0) 0iL and =(0) 6 VvC ,

1. Solveforthecoefficientsαandβ. 2. PlotiL(t)for0≤t ≤500μsecinstepsof0.5μsec

3. PlotiL(t)andvC(t)(solvedinProject2.7)onthesameaxes.NotethatwhenvCisatamaximum,iLiszeroandviceversa.Thisillustrateshowtheenergyinthecircuitoscillatesbackandforthbetweenthecapacitorandinductor.

Project.2.10

Theclassicalformforasecond-orderdifferentialequationis

+ ζω + ω =2 0

2

22d y

dtdydt

yn n (P2.10a)

whereζisthedamping factor,andωnisthenatural frequency.WecanmatchthetermsofEquation(P2.9a)withEquation(P2.10a)tofindthedampingfactorandnaturalfrequencyfortheparallelRLCcircuit.Thus,

ω = ζ =

ω=

1 and 12 2LC RC

LCRCn

n (P2.10b)

Notethatforζ <1thecircuitisunderdamped,forζ >1thecircuitisoverdamped,andforζ =1,itiscriticallydamped.

For the component values L = 1 mH, C = 1 μF, and initial conditions=(0) 0.25iL Aand =(0) 6vC V:

1. DeterminetheresistorvalueRcritthatmakesthecircuitcriticallydamped.2. Plot the inductor current iL versus time for the three values of R: R = Rcrit,

R = 5 Rcrit,and = 12R Rcrit .Assumetheinterval0≤t ≤500μsecinstepsof0.5

μsec.Plotallofthewaveformsononegraph.


Project.2.11

WewishnowtoconsidertheRLCcircuitdescribedinSection2.17tobesubjectedtoadrivingcurrentsuchthatinsteadofopeningtheswitchatt =0,weclosetheswitch.Inthiscase,thesumofthecurrentsatthetopleftnodeis

+ + − = 0i i i IR L C o (P2.11a)

NotethattheIotermisnegativebecauseIoisarbitrarilydefinedinFigure 2.17asflowing into thenodeinsteadofaway fromthenode(as istruefortheotherthreecurrents).

Thegoverningequationthenbecomes

+ + =

1 12

2d idt RC

didt LC

i IL LL o (P2.11b)

Thecompletesolutionofthisdifferentialequationisequaltothehomogeneoussolution(foundinProject2.9)plusaparticular solution.

Letusassumethatthedrivingcurrentisasinusoidoftheform = ω( ) sin .I t A to o o To obtain the particular solution ip, we seek a solution whose first and secondderivativescanbesummedtoequaltheformofthedrivingcurrent.Thus,assume

= ω + ωsin cosi A t B tp p o p o (P2.11c)

whereApandBpareconstants.Then,

=ω ω − ω( cos sin )i A t B tp o p o p o

and

=ω − ω − ω( sin cos )2i A t B tp o p o p o

SubstitutingtheseexpressionsintoEquation(P2.11b)gives

− ω − ω + ω + − ω + ω + ω = ω

1 1 sin 1 1 cos sin2 2ARC

BLC

A t BRC

ALC

B t A tp o o p p o p o o p p o o o

(P2.11d)

CollectingcoefficientsofthesineandcosinetermsontheleftsideofEquation(P2.11d)andequatingthemtothesineandcosinecoefficientsontherightsideofthatequationgives

−ω −

ω=

1 2

LCA

RCB Ao p

op o

ω+ −ω =

1 02

RCA

LCBo

p o p


SolvingthesetwoequationsforAp andBPgives

= ×−ω

ω+ −ω

1

1

2

22

2A A LC

RC LC

p o

o

oo

(P2.11e)

=− ×

ω

ω+ −ω

122

B A RC

RC LC

p o

o

oo

(P2.11f)

SubstitutingEquations(P2.11e)and(P2.11f)intoEquation(P2.11c)gives

= ×ω

+ −ω

−ω ω −ω

ω11

1 sin cos22

22i A

RC LCLC

tRC

tp oo

o

o oo

o (P2.11g)

WecanrewriteEquation(P2.11g)usingthetrigonometricidentity

αsinωt+βcosωt=γsin(ωt− ϕ)

where

γ = α +β2 2

and

φ=

βα

−tan 1

ApplyingtheserelationstoEquation(P2.11g)gives

= ×

−ω +ω

ω − φ1

1sin( )

22 2

i A

LC RC

tp o

oo

o (P2.11h)

UsingthedefinitionsfromProject2.10fornaturalfrequencyω =1LCn and

dampingfactor ζ=ω

12RC n

,wecanalgebraicallymanipulateEquation(P2.11h)into

=ω

−ωω

+ ζωω

ω −φ1

1 2

sin( )2 2

iA

tpo

no

n

o

n

o (P2.11i)


Theterm

ω−ωω

+ ζωω

1

1 22 2

Ao

no

n

o

n

istheamplitudeoftheoscillation.ForagivenωAo

n,

thelargertheterm−ωω

+ ζωω

1

1 22 2

o

n

o

n

,thelargeristheamplitude.

−ωω

+ ζωω

Let ampl = 1

1 22 2

o

n

o

n

1. ConstructaMATLABprogramtocreateaplotofamplversus ωω

o

nfor

valuesofζ=1.0,0.5,0.25,0.10,0.05,and <ωω

<0 2o

ninstepsof0.01.

2. Whathappensasω →ωo n?

Project.2.12

ModifyExample2.25asfollows:

1. Useaself-writtenfunctionsavedasan.mfiletodotheinterpolation.Thelinearinterpolationformulafory(x)=f(x)expressionis

= +

−−

× −( )11

2 12 1y y

x xx x

y y

wherepoints(x1,x2)arethenearestsurroundingpointstox and(y1,y2)aretheyvaluesat(x1,x2),respectively.UseaglobalstatementtobringT1,T2,μ1,andμ2intotheself-writtenfunction,whereT1,T2aretheclosesttemperaturetablevaluestothetempera-tureatwhichμistobedeterminedandμ1,μ2aretheμvaluesatT1,T2respectively.

2. Printoutthetabledatatothescreen,thenasktheuserifheorshewishestointer-polateforμatatemperatureotherthanthosetemperaturesgiveninthetable.Iftheanswerisyes,asktheusertoenterthetemperature,dotheinterpolation,andprintouttheresult.Usetheswitchgrouptodeterminewhethertodotheinterpolation.Iftheanswertoyourqueryisno,exittheprogram.

3. Iftheanswertothequeryinpart2isyes,thenasktheuserifheorshewishestoenteranothertemperature;ifyes,asktheusertoenterthetemperature,dotheinterpolation,andprintouttheresult.Theprocessistocontinueuntiltheanswertothequeryisno.Theuseristoenterthefollowingfivetemperatures(k),oneatatime:125,180,218,334,427.


Project.2.13

ThisprojectisamodificationofExample2.24inwhichthe“anonymousfunction”thatdoestheinterpolationistoconsistofargumentsof , , , ,1 2 1 2x x x y y .andusestheinterpolationformula

= +

−−

× −( )11

2 12 1y y

x xx x

y y

wherepoints(x1,x2)arethenearestsurroundingpointstox and(y1,y2)arethey values at (x1, x2) , respectively. In Example 2.6, it was shown that the inputvariablestoafunctionneednothavethesamenamesasthoseusedinthecallingprogram. For one-to-one correspondence, the argument positions in the callingprogrammustbethesameastheargumentpositionsinthefunction.TablevaluesforelectronmobilityμasafunctionoftemperatureTareasfollows:

T =[100150200250300350400450]

μ =[196507427372321801407972.0705.5531.8]

TheunitsforT are(K)andtheunitsforμare(cm2/V-s).

1. CreateaMATLABprogramthatcontainstheanonymousfunctionasdescribedandthatinterpolatesfortheelectronmobility μatthefollowingtemperaturesTT:

TT =[110165215365440]

2. PrintoutatableofvaluesofμattemperaturesTT.

Project.2.14

Indiscrete-timesignalprocessing,signalsarerepresentedasasequenceofdiscretenumericalvalues,asopposedtothe“smooth”waveformsincontinuous-timesig-nalprocessing.

FigureP2.14ashowsacontinuous-timewaveform(a1-kHzsinewave)anditsdiscrete-time counterpart,which is generatedby sampling the continuous-timesignalatarateof8000samples/sec(every125μsec).Thissamplingrateiscom-monlyusedforvoice-gradetelephonyequipment,andtheresultingdiscrete-timesequence might subsequently be transmitted by methods such as PAM (pulseamplitudemodulation)orPCM(pulsecodemodulation).

Now,supposewewanttousethissampledwaveforminamoviesoundtracktoberecordedonDVD.ThedigitalaudioonDVDstypicallyusesasamplingrateof48,000samples/sec(every20.8333μsec); thus,wewillneedtoupsampleour8000-sample/secwaveformtouseitonDVD.

Inupsampling,weaugmentagivendiscretewaveformbyaddingsamplestothe sequence to achieve the desired higher sampling rate. In this case, upsam-plingfrom8000to48,000samples/secwillrequireaddingfivenewsamplesforeveryoneintheoriginalwaveform.Therearemanymethodsfordoingthis,andonesimpleapproachistocomputevaluesfortheaddedsamplesbyinterpolatingbetweenadjacentvaluesoftheoriginalwaveform(FigureP2.14b).


DevelopaMATLABprogramthatwill

1. Computethesignalvaluesoftheoriginal8000-samples/secwaveformovertheperiodof0<t ≤2msec,foratotalof17samples.Assumesignalpeakvaluesof±1forthesinewave.

2. Createatableconsistingofsamplenumber, sampletime,andsamplesignalvalue.

3. UseMATLAB’sinterp1functiontocalculatetheupsampledwaveform(i.e.,atasamplerateof48,000samples/sec).Toavoidround-offerror,useΔt=20.833333μsec.Youshouldendupwithasamplesizeof97values.

t

v

1

–1

1 msec

125 µsec

(a)

t

v

1

–1

1 msec

20.833333 µsec

(b)

Figure P2.14 (a) A 1-kHz sine wave is sampled at 8000 samples/sec. (b) If we want to upsample the sine wave to 48,000 samples/sec, we need to interpolate five additional samples between points of the original waveform.


4. Plot theoriginaland theupsampledwaveformson the samegraph.For theoriginal sample waveform, use circles to display the data points. For theupsampledwaveform,use× stodisplaythedatapoints.

Project.2.15

This project involves determining the quality of the upsampled waveform ofProject2.14.This isaccomplishedbycomparingtheupsampledwaveformwithan ideal 1-kHz sinewave.A commonperformancemetric for audio systems isthetotal harmonic distortion(THD),whichisameasureofthenoiseatharmonicfrequenciesotherthanthefrequencyofinterest(1kHzinourcase).Becauseoursignalisperiodicandofknownfrequency,wecanassumethatalltheimperfec-tionswillcontributetotheTHD.DevelopaMATLABprogramthatwill

1. Compute the valuesof an ideal48,000-samples/secwaveformVideal over theperiodof0<t ≤2msec,foratotalof97samples.Assumewaveformpeakvaluesof±1forthesinewave.

2. Subtract this idealwaveformfromourupsampled48,000-samples/secwave-formtocalculate thewaveformerrorVerror for the interpolationmethodthatwasusedinProject2.14.

3. CalculatetheTHDoftheupsampledwaveformasfollows:

( )( )

=THD ,2

,2

V

Verror RMS

ideal RMS

(P2.15)

UsethemethoddescribedinProject2.6tocalculate( ),Verror RMS and( ),Videal RMS .

4. Printout ,Verror RMS , ,Videal RMS ,andTHD.

Project.2.16

Mathematician Joseph Fourier is credited with the theorem that any periodic waveformmaybeexpressedasasummationofpuresinesandcosines.Forexam-ple,thesquarewaveofFigureP2.16acanbewrittenasasumofsines:

∑

=π

π+

ππ

+π

π+

=π

π

=

∞

( ) 4 sin2 4

3sin

6 45

sin10

4 sin2

1odd

v tt

Tt

Tt

T

kk t

Tkk

FigureP2.16bshowsthefirstthreetermsoftheseriesandtheirsummation.

1. Write a MATLAB script that implements a self-written functionsqwave(n,T,i),whichtakesthefollowingarguments:

n the number of terms of the Fourier series

T the period of the square wave in seconds

i the number samples per period


Thefunctionshouldreturntwoarrays,tandV,eachcontainingielements,where

t=anarrayofitimepoints.V=anarrayoficomputedvaluesofthenth-degree approximatedsquarewave

2. Run yoursqwave(n,T,i) function and plot the results for the followingarguments:

T=1msec,i=1001,n=1,3,10,100

3. Forn=100,printoutatableof(t,V)atevery20timepoints.

Project.2.17

Bob’sHardwareStorewishestocreateanonlineprogramtosellinventoryitemsinitsstore.YouaretocreateaninteractiveMATLABprogramforthispurpose.Thepro-gramistocontainthefollowingitems:adatafile,amainscriptandabillingfunction.

t

v

1

–1

T = 1 ms(a)

t

v

2nd term

1st term

3rd term

Sum of 1st, 2ndand 3rd terms1

–1

(b)

Figure P2.16 (a) Square wave. (b) Three terms in a Fourier series of a square wave.


Datafile:Thedatafileistocontainadescriptionoftheinventoryitemsforsale,theircost,andthequantityavailableforsale.Thelistshouldcontainatleast10items.Thisdatafilewillbe read into themain script.Wheneveran item ispurchased, theinventoryavailableforsaleshouldbeupdatedinthedatafile.

Mainscript:1. Thescriptistoprinttothescreentheitemsforsale,theircost,andthequantity

availableforpurchase.2. Thescriptistoasktheuserifheorshewishestomakeapurchase.Ifyes,the

scriptistoasktheuserthelinenumberoftheitemheorshewishestopurchaseandthequantity.

3. Iftheuserdoesnotwishtomakeapurchase,exittheprogram.Iftheuserdoesnot respondcorrectly to thequery,printanerrormessage to the screenandinformtheusertorestarttheprogram.Then,exittheprogram.

4. Iftheuserdoesmakeapurchase,theprogramistocontinueaskingtheuserifheorshewishestomakeanotherpurchase.Ifyes,printtothescreenthelistofitems,theircost,andtheupdateditemsavailableforpurchase.Then,thescriptistoasktheuserthelinenumberoftheitemheorshewishestopurchaseandthequantity.Iftheanswerisno,exittheprogram.

Billingfunction:Whentheuserisfinishedmakingallofhisorherpurchases,themainscriptistocallthebillingfunctionthatprintsoutabillfortheitemspurchased.Thebillshouldcontainthenameandaddressofthestore,theuser’sfirstandlastnames,theuser’saddress,billheadings,alistofalltheitemspurchased,theitems’unitprices,thetotalpriceforeachitempurchased,andfinallythetotalpriceofalltheitemspurchased.

References 1. Kernighan,B.W.,andRitchie,D.M.,The C Programming Language,2nded.,Prentice-

Hall,EnglewoodCliffs,NJ,1988. 2. Sze,S.M.,Physics of Semiconductor Devices,2nded.,Wiley,NewYork,1981. 3. Kreyszig,E.Advanced Engineering Mathematics,8thed.,Wiley,NewYork,1999.

77

Chapter 3

Matrices

3.1 IntroductionInengineering,wearefrequentlyconfrontedwithmatricesoraprobleminvolvingasetoflinearequations.Inaddition,thebasicelementinMATLAB®isamatrix,anditisthereforeappropriatetohaveachapteronmatrices.First,wediscusssomebasicmatrixconceptsandoperationsandthetreatmentofmatricesinMATLAB,includingseveralsampleMATLABprogramsinvolvingmatrices.Thisisfollowedbyadiscussionofseveralmethodsforsolvingasystemoflinearequations,includ-ingtheuseofMATLAB’sinvfunction,Gausselimination,andtheGauss-Jordanmethods.Anexampleoftheuseofmatricestosolveaprobleminresistivecircuitsisincluded.Finally,thematrixeigenvalueproblemisdiscussed,includinganapplica-tiontoaresonantcircuitproblem.

3.2 Matrix Operations◾ Arectangulararrayofnumbersofthefollowingformiscalledamatrix.

=

11 12 1

21 22 2

1 2

A

a a aa a a

a a a

n

n

m m mn

Thenumbersaijinthearrayarecalledtheelementsofthem ×nmatrixA.


◾ Amatrixofmrowsandonecolumniscalledacolumnvector.

◾ Amatrixofonerowandncolumnsiscalledarowvector.

◾ Matricesobeycertainrulesofaddition,subtraction,andmultiplication.

◾ Additionandsubtraction:IfmatricesAandBhavethesamenumberofrowsandcolumns,then

= + =

+ + ++ + +

+ + +

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )

11 11 12 12 1 1

21 21 22 22 2 2

1 1 2 2

C A B

a b a b a ba b a b a b

a b a b a b

n n

n n

m m m m mn mn

= − =

− − −− − −

− − −

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )

11 11 12 12 1 1

21 21 22 22 2 2

1 1 2 2

C A B

a b a b a ba b a b a b

a b a b a b

n n

n n

m m m m mn mn

◾ AdditionandsubtractionofthematricesAandBareonlydefinedifAandBhavethesamenumberofrowsandcolumns.

◾ Multiplication:TheproductABisonlydefinedifthenumberofcolumnsinAequalsthenumberofrowsinB.IfC.=.AB,where

= =and11 12 13

21 22 23

11 12

21 22

31 32

A Ba a aa a a

b bb bb b

then

=

+ + + +

+ + + +C

( ) ( )( ) ( )

11 11 12 21 13 31 11 12 12 22 13 32

21 11 22 21 23 31 21 12 22 22 23 32

a b a b a b a b a b a ba b a b a b a b a b a b

◾ IfAhasmrowsandBhasKcolumns,thenC.=.ABwillhavemrowsandKcolumns.Ageneralexpressionfortheelementcijis

∑==1

c a bij ik kj

k

K

InMATLAB,themultiplicationofmatricesAandBisenteredasA * B.

Matrices ◾ 79

◾ Transpose:Thetransposeofamatrixistheoperationofexchangingtherowsandcolumns.If

=

2 5 17 3 84 5 2116 3 10

A

then

=2 7 4 165 3 5 31 8 21 10

AT

InMATLAB,thetransposeofmatrixAisenteredasA’.

◾ Summingtheelementsofavectororthecolumnsofamatrix:

= = + +If [ ], then sum( ) .1 2 3 1 2 3A Aa a a a a a

If

=11 12 13

21 22 23

31 32 33

Bb b bb b bb b b

then

= + + + + + +sum( ) [( ) ( ) ( )]11 21 31 12 22 32 13 23 33B b b b b b b b b b

InMATLAB,thesumofmatrixAisenteredassum(A).

◾ Dotproduct:Thedotproductoftwovectorsisdefinedas

∑⋅ =A B a bi i

Forexample,if = −[4 1 3]A and = −[ 2 5 2]B ,then

⋅ = − + − + = −(4)( 2) ( 1)(5) (3)(2) 7A B


Theresultofadotproductisalwaysascalar(i.e.,asinglenumber). InMATLAB,thedotproduct ⋅A B isenteredasdot(A,B).

◾ Identity:TheidentitymatrixIisamatrixwherethemaindiagonalelementsarealloneandallotherelementsarezero.

= =IA AI A

Forexample,forthecaseofathree-by-threematrixA,

=

=+ + + + + ++ + + + + ++ + + + + +

1 0 00 1 00 0 1

( 0 0) (0 0) (0 0 )( 0 0) (0 0) (0 0 )( 0 0) (0 0) (0 0 )

11 12 13

21 22 23

31 32 33

11 12 13

21 22 23

31 32 33

AIa a aa a aa a a

a a aa a aa a a

InMATLAB,youcangenerateann ×nidentitymatrixwitheye(n).

◾ Inverse:Theinverseofamatrix,denoted −A 1 ,isamatrixsuchthat

= = = ×− −A A AA I1 0 00 1 00 0 1

(for a 3 3 matrix)1 1

InMATLAB,thematrixinverse −A 1 isenteredasinv(A).

◾ Determinant:

Beforetheuseofcomputers,determinantsweredevelopedasamethodforobtainingasolutiontoasystemof linearequations.Computationally, it isonlypracticalforasysteminvolvingjustafewequations[1].However,deter-minantsdohaveanapplicationintheeigenvalueproblemthatisdiscussedattheendofthischapter.

Thedeterminantofa2×2matrixis

= = −

11 12

21 2211 22 12 21D

a a

a aa a a a

Notethatthedeterminantofamatrixiswrittenwithtwostraightlinesthatenclosethematrixelements.

Matrices ◾ 81

Thedeterminantofa3×3matrixis

= = − +

11 12 13

21 22 23

31 32 33

11

22 23

32 3312

21 23

31 3313

21 22

31 32

D

a a a

a a a

a a a

aa a

a aa

a a

a aa

a a

a a

InMATLAB,thedeterminantofthematrixAisenteredasdet(A).

◾ Dimensions:YoucanobtainthesizeofmatrixAbyrunningsize(A).Thiscommandisusefulwhenyourunascriptandyougetanerrormessagelike“Index exceeds matrix dimensions.” Adding thesize()command in thescriptwillhelpyoudeterminetheproblem.

◾ Element-by-elementoperations:Giventwomatricesofthesamedimensions,wecanperformelement-by-elementmultiplicationanddivisioninMATLABwiththe.*and./operators.

Given = 1 2 3A a a a and = 1 2 3B b b b ,then

= =. * 1 1 2 2 3 3C A B a b a b a b

= =. / 1

1

2

2

3

3D A B

ab

ab

ab

Also,notethedotproductcanbeexpressedasacombinationofanele-ment-by-elementmultiplicationandasum:

( )⋅ = = = + +sum( . * ) sum 1 1 2 2 3 3 1 1 2 2 3 3A B A B a b a b a b a b a b a b

◾ Productoffunctionsoftwovectors:

AsdescribedinChapter2,MATLABallowstakingafunctionofavectorandthattheresultisalsoavector.However,ifascriptrequiresamathemati-caloperationoftwofunctions(suchasaproductoftwovectorfunctions),thentheoperationwillrequireanelement-by-elementoperation.

In the followingexample,wecompute theproductof twovector func-tions,bothdirectlyandindirectly,byusingaforloopandmultiplyingtheelementsofeachfunction.Writethissimplescriptintoanewscriptwindow,saveitasmatrix_example.m,andrunthescript:


%matrix_example.m clear;clc; x=0:30:360; %y1istheproductoftwovectorfunctions y1=sind(x).*cosd(x); fprintf('x y1 y2\n'); fprintf('--------------------------------------\n'); forn=1:length(x); %y2(n)istheproductoftheelementsofthetwo %functions. y2(n)=sind(x(n))*cosd(x(n)); fprintf('%5.1f%8.5f%8.5f\n',x(n),y1(n),y2(n)); end -------------------------------------------------------

Dothetwodifferentmethodsforcomputingy1andy2givethesameanswer?

Example 3.1: Matrix Operations

The following example illustrates several matrix operations, including addi-tion, dot product, element-by-element multiplication and division, sum, and transpose.

%Example_3_1.m%ThisprogramdemonstratesmatrixalgebrainMATLABclear;clc;a=[159]b=[2612]c=a+bd=dot(a,b)e=a.*bf=a./bg=sum(a.*b)h=a*b'---------------------------------------------------------------------------

3.3 System of Linear EquationsInengineering,wearefrequentlyconfrontedwiththeproblemofsolvingasetoflinearequations.

Giventhesetofequations

+ + + + =

+ + + + =

+ + + + =

11 1 12 2 13 3 1 1

21 1 22 2 23 3 2 2

1 1 2 2 3 3

a x a x a x a x ca x a x a x a x c

a x a x a x a x c

n n

n n

n n n nn n n

(3.1)

Matrices ◾ 83

Thesetcontainsnequationsinnunknowns.Itisconvenienttoexpressthesetasamatrixequationoftheform

==AX C (3.2)

where

= = =, ,

11 12 1

21 22 2

1 2

1

2

1

2A X C

a a aa a a

a a a

xx

x

cc

c

n

n

n n nn n n

MatrixAhasnrowsandncolumns.MatrixXhasnrowsand1column.Note:ThenumberofcolumnsinAmustequalthenumberofrowsinX;other-

wise,matrixmultiplicationisnotdefined.MatrixChasnrowsand1columnInmatrixalgebra,Xcanbeobtainedbymultiplyingbothsidesby −A 1 ,thatis,

= = =–1 –1 –1A AI

X A C IX X A C

MATLABoffersthefollowingfunctionsforsolvingasetoflinearequations:

◾ Theinvfunction:

TosolveasystemoflinearequationsinMATLAB,youcanuse

X=inv(A)*C

Themethodofsolvingasystemoflinearequationsbyusingtheinvfunc-tionismorecomputationallycomplicatedthanamethodcalledGausselimi-nation,whichisdiscussednext.

◾ TheGausseliminationfunction:

MATLABallowsforthesolutionofthelinearsystem,AX=C,byGausseliminationbywriting

X=A\C

Note theuseofMATLAB’sbackslashoperator to solve forXbyGausselimination.


Example 3.2

The following example solves the three equation linear system shown and writes the results to a file:

+ − =

− + + =

− − = −

x x x

x x x

x x x

3 2 10

3 2 5

1

1 2 3

1 2 3

1 2 3

%Example_3_2.m%Thisprogramsolvesasimplelinearsystemofequations%byboththeuseofMATLAB'sinversematrixfunctionand%bytheGausseliminationmethod.clc;clear;A=[32-1;-132;1-1-1]C=[105-1]'%X1isthesolutionusingmatrixinversefunction:X1=inv(A)*C%X2isthesolutionusingGausseliminationmethod:X2=A\C%checksolution:A*X1%Usethesize()commandtodeterminethenumberofrows%andthenumberofcolumnsinmatrixA.[A_rowsA_cols]=size(A)%PrintmatrixA,matrixCandthesolutionofthelinear%systemofequationstothefileoutput.txt:fid=fopen('output.txt','w');fprintf(fid,'Amatrix:\n');fori=1:A_rowsforj=1:A_colsfprintf(fid,'%6.1f',A(i,j));endfprintf(fid,'\n');endfprintf(fid,'Cvector:\n');fori=1:length(C)fprintf(fid,'%6.1f',C(i));endfprintf(fid,'\n');fprintf(fid,'SolutionX1(usinginversematrix):\n');fori=1:length(X1)fprintf(fid,'%6.1f',X1(i));endfprintf(fid,'\n');fprintf(fid,'SolutionX2(usingGausselimination):\n');fori=1:length(X2)fprintf(fid,'%6.1f',X2(i));endfprintf(fid,'\n');fclose(fid);

Matrices ◾ 85

Example 3.3: An Example in Resistive Circuits

A typical example involving a system of linear equations can be found in the resistive circuit of Figure 3.1. The goal is to solve for the node voltages v1, v2 , and v3 as functions of the input voltages V1 and V2 and the input current I1.

Using Ohm’s law (v = iR) and Kirchhoff’s current law (KCL), we can write expressions for the sum of currents at each node of interest. KCL states that the total current entering any circuit node must sum to zero. In Figure 3.1, the resistor currents are written as in and are arbitrarily defined in the directions as drawn. Writing three KCL equations for the currents entering each node (and noting that currents drawn as leaving the node are written mathemati-cally as negative numbers) gives

At node ➀: + + =i i i 01 2 3 (3.3)

At node ➁: − + =I i i 01 2 4 (3.4)

At node ➂: − − + =i i i 03 4 5 (3.5)

We now rewrite these three equations in terms of voltages using Ohm’s law. Note that in these types of problems, it is often easier to write the equa-

+–

+–

R1

R2

R3

R4

R5i1

2

31

i2

i3

i4

i5

V1 V2I1

v1v2

v3

Figure 3.1 Resistive circuit.


tions in terms of conductances Gn instead of resistances (where =1/G Rn n), and Ohm’s law may be rewritten as i = vG. Thus,

=

−→ = −i

V vR

i V v G( )11 1

11 1 1 1 (3.6)

=

−→ = −i

v vR

i v v G( )22 1

22 2 1 2 (3.7)

=

−→ = −i

v vR

i v v G( )33 1

33 3 1 3 (3.8)

=

−→ = −i

v vR

i v v G( )43 2

44 3 2 4 (3.9)

=

−→ = −i

V vR

i V v G( )52 3

55 2 3 5 (3.10)

Substituting these expressions for i1 through i5 into Equations (3.3) to (3.5) gives

− + − + − =V v G v v G v v G( ) ( ) ( ) 01 1 1 2 1 2 3 1 3 (3.11)

− − + − =I v v G v v G( ) ( ) 01 2 1 2 3 2 4 (3.12)

( ) ( ) ( )− − − − + − =v v G v v G V v G 03 1 3 3 2 4 2 3 5 (3.13)

These expressions can be rearranged into a system of three equations with three unknowns:

+ + − −

− + −

− − + +

=

G G G G GG G G GG G G G G

vvv

VGI

V G

( )( )

( )

1 2 3 2 3

2 2 4 4

3 4 3 4 5

1

2

3

1 1

1

2 5

(3.14)

This equation has the form AX = C and may be solved by Guass elimination as shown previously in Example 3.2.

The following script solves for the node voltages for the following circuit values:

= = = = =

= = =

R R R R R

V V I

2200 , 10 k , 6900 , 9100 , 3300

12V, 3.3 V, 2 mA

1 2 3 4 5

1 2 1

Matrices ◾ 87

%Example_3_3.mResistiveCircuit%Thisprogramsolvesfortheinternalnodevoltagesfor%thecircuitshowninFigure3.1.%TheconductancesGareinunitsofsiemens%ThenodevoltageVareinunitsofvolts%ThecurrentsIareinunitsofampsclear;clc;g1=1/2200;g2=1/10000;g3=1/6900;g4=1/9100;g5=1/3300;V1=12;V2=3.3;I1=.002;%PuttheproblemintostandardformAX=C:A=[g1+g2+g3-g2-g3;-g2g2+g4-g4;-g3-g4g3+g4+g5];C=[V1*g1;I1;V2*g5];X=A\C;%printtheresultsfprintf('V1=%9.1fV\nV2=%9.1fV\nI1=%9.1eA\n',V1,V2,I1);fprintf('g1=%9.5fS\ng2=%9.5fS\ng3=%9.5fS\n',g1,g2,g3);fprintf('g4=%9.5fS\ng5=%9.5fS\n',g4,g5);fprintf('\n');fprintf('Node#v(volts)\n');fprintf('-------------------------\n');forn=1:length(C)fprintf('%3i%9.1f\n',n,X(n));end---------------------------------------------------

3.4 Gauss EliminationAspreviouslydiscussed,theGausseliminationmethodiscomputationallymoreefficient than the inverse matrix method or the method of determinants withCramer’srule.WewishtousetheGausseliminationmethodtosolvethesystemoflinearequationsdescribedbyEquation(3.1)andrepresentedinmatrixformbyEquation(3.2).IntheGausseliminationmethod,theoriginalsystemisreducedtoanequivalenttriangularsetthatcanreadilybesolvedbybacksubstitution.Thereducedequivalent.setwouldappearlikethefollowingsetofequations:

+ + + + =

+ + + =

+ + =

=

+ =

=

− − − − −

11 1 12 2 13 3 1 1

22 2 23 3 2 2

33 3 3 3

1, 1 1 1, 1

,

a x a x a x a x c

a x a x a x c

a x a x c

a x a x c

a x c

n n

n n

n n

n n n n n n n

n n n n (3.15)


wherethetilde(∼)variablesareanewsetofcoefficients (tobedetermined)andwherethenewcoefficientmatrix A isdiagonal(i.e.,allofthecoefficientsleftofthemaindiagonalarezero.Then,

=

= −−− −

− −1 ( )

etc.

,

11, 1

1 1,

x c a

xa

c a x

n n n n

nn n

n n n n

Todeterminethereducedequivalentset A and C ,itisconvenienttoaugmenttheoriginalcoefficientmatrixAwiththeCmatrixasshownbelow:

=AA

11 12 1

21 22 2

1 2

1

2

a a aa a a

a a a

cc

c

aug

n

n

n n nn n

(3.16)

Thefollowingprocedureisusedtoobtainthereducedequivalentset:

1.MultiplythefirstrowofEquations(3.1)bya21/a11andsubtractfromthesec-ondrow,giving

…= − × = − × = − ×, , , etc.21 21

21

1111 22 22

21

1112 23 23

21

1113a a

aa

a a aaa

a a aaa

a

and

= − ×2 2

21

111c c

aa

c

Thisgives = 021a

2.Forthethirdrow:multiplythefirstrowofEquations(3.1)by /31 11a a andsubtractfromrow3,giving

…= − × = − × = − ×, , , etc.31 31

31

1111 32 32

31

1112 33 33

31

1113a a

aa

a a aaa

a a aaa

a

Matrices ◾ 89

and

= − ×3 3

31

111c c

aa

c

Thisgives = 031a .

3.Thisprocessisrepeatedfortheremainingrows4,5,6,…,n.Theoriginalrow1iskeptinitsoriginalform.Allotherrowshavebeenmodifiedandthenewcoefficientsaredesignatedbya(′).Exceptforthefirstrow,theresultingsetwillnotcontainx1.

4.Fortheprecedingsteps,thefirstrowofEquations(3.1)wasusedasthepivot row,anda11wasthepivot element.Wenowusethenewrow2asthepivotrowandrepeatsteps1–3fortheremainingrowsbelowrow2.Thus,multiplythe

newrow2by 32

22

aa

andsubtractfromrow3,giving

…= − × = − × = − ×, , , etc.32 32

32

2222 33 33

32

2223 34 34

32

2224a a

aa

a a aaa

a a aaa

a

and

= − ×3 3

32

222c c

aa

c

Thisgives = 032a .

Similarly,multiplythenewrow2by 42

22

aa

andsubtractfromrow4,giving

…= − × = − × = − ×, , , etc.42 42

42

2222 43 43

42

2223 44 43

42

2224a a

aa

a a aaa

a a aaa

a

and

= − ×4 4

42

222c c

aa

c

Thisgives = 042a .

Thisprocessiscontinuedfortheremainingrows5,6,…,n.Exceptforthenewrow2,thesetdoesnotcontainx2.


5.Thenextrowisnowusedasapivotrow,andtheprocessiscontinueduntilthe(n – 1)throwisusedasthepivotrow.Whenthisiscomplete,thenewsystemistriangularandcanbesolvedbybacksubstitution.

Thegeneralexpressionforthenewcoefficientsis

… …= − × = + + = + +for 1, 2, , and 1, 2, ,a a

aa

a i k k n j k k nij iji k

kkkj

wherekisthepivotrow.

Theseoperationsonlyaffectthe aij and c j .Thus,weneedonlyoperateontheAandCmatrices.

Example 3.4: Given

− + = −

− + + = −

+ − =

x x x

x x x

x x x

3 4

3 4 3 10

2 3 2 18

1 2 3

1 2 3

1 2 3

The augmented coefficient matrix is

=−

−−

−−

=−a

a

augA1 3 13 4 32 3 2

41018

multiply row 1 by31and subtract from row 221

11

Row 2 becomes (−3 + 3), (4−9), (3 + 3), (−10−12) = (0, −5, 6, −22)The new matrix is

=−−

−

−−

=aa

aug EquivA1 3 10 5 62 3 2

42218


,

31

11

Row 3 becomes (2−2), (3 + 6), (−2−2), (18 + 8) = (0, 9, −4, 26)The new matrix is

=−−

−

−−aug EquivA

1 3 10 5 60 9 4

42226

,

Matrices ◾ 91

We now use row 2 as the pivot row.

=−−

−

−−

= −aa

aug EquivA1 3 10 5 60 9 4

42226


,

32

22

Row 3 becomes

− − × − + × − × = −(0 0), 9

95

5 , 495

6 , 2695

22 0, 0,345

,685

The new matrix is

=

−

−

−

−

−

aug EquivA

1 3 1

0 5 6

0 0345

4

22

685

,

The system is now triangular, and the system can be rewritten as

=Equiv EquivA X C

which gives

− + = −

− + = −

= −

x x x

x x

x

3 4

5 6 22

345

685

1 2 3

2 3

3

Solving:

= − = −

− + − = − =

− + − = − =

x x

x x

x x

345

685

2

5 6( 2) 22 2

3(2) ( 2) 4 4

3 3

2 2

1 1


There are two important considerations:

1. If akk is zero, where k is the pivot row, then the process cannot be car-ried out.

2. Greater accuracy in the solution is obtained if the pivot element is the absolute maximum available from the set. That is; if the pivot row is k, one compares the aik’s for i = k + 1, k + 2, … , n (see Figure 3.2). If ↑a ai k kk| | | |max , then the row containing the ai k| |max is interchanged with the kth row. This only affects the ordering of the equations and does not affect the solution.

If after row interchange is carried out and one of the akk’s remains zero, then the system is singular, and no solution can be obtained.

One last consideration: it can be shown that if the magnitude of the pivot element is much smaller than other elements in the matrix, the use of the small pivot element will cause a decrease in the accuracy of the solution. To check if this is the case, you can first scale the equations, that is, divide each equation by the absolute maximum coefficient in that equation. This makes the absolute maximum coefficient in that equation equal to 1.0. If akk << 1, then the solution is likely to be inaccurate.

3.5 The Gauss-Jordan MethodTheGauss-JordanmethodisamodificationoftheGausseliminationmethodandcan be used to solve a system of linear equations of the form AX. =. C. In thismethod,theobjectiveistoobtainanequivalentcoefficientmatrixsuchthat,exceptforthemaindiagonal,allelementsarezero.Themethodstartsout,asintheGausselimination method, by finding an equivalent matrix that is triangular. It thencontinues,assumingthatAisann ×nmatrix,usingthenthrowasthepivotrow,multiplyingthenthrowby −1, ,a an n n n andsubtractingtheresultfromrown–1,thusmakingthenew =− 01,an n .Theprocessisrepeatedforrowsn −2,n −3,…,1.

Comparefor

kth row

maxaik

Figure 3.2 Row interchange.

Matrices ◾ 93

Example 3.5

As an example, the Gauss-Jordan method is applied as follows: Starting with the triangular equivalent matrix of Example 3.4,

=

−

−

−

−

−

aug EquivA

1 3 1

0 5 6

0 0345

4

22

685

,

Multiply row 3 by = ×aa

6534

23

33 and subtract from row 2. Row 2 becomes

− × − − × − × − + × = − −03034

0 , 53034

0 , 63034

345

, 223034

685

(0, 5, 0, 10)

Next, multiply row 3 by =aa

534

13

33 and subtract from row 1. Row 1

becomes

− × − − × − × − + × = − −1

534

0 , 3534

0 , 1534

345

, 4534

685

(1, 3, 0, 2)

The new aug EquivA , becomes

=

−

−

−

−

−

aug EquivA

1 3 0

0 5 0

0 0345

2

10

685

,

Now, row 2 is used as the pivot row. Multiply row 2 by =aa

35

12

22 and

subtract from row 1. Row 1 then becomes

− × − + × − × − + × =135

0 , 335

5 , 035

0 , 235

10 (1, 0, 0, 4)


The new aug EquivA , becomes

= − −

−

aug EquivA

1 0 0

0 5 0

0 0345

4

10

685

,

Thus, the equivalent set of equations becomes

=

− = −

= −

x

x

x

4

5 10

345

685

1

2

3

And thus =x 41 , =x 22 , and = −x 23 , which is the same answer that was obtained previously by the Gauss elimination method.

3.6 Number of SolutionsSupposeaGauss eliminationprogram is carriedoutand the following resultsareobtained:

+ + + + =

+ + + =

+ + =

=

+ + =

=

=

=

+

+

0

0

0

11 1 12 2 13 3 1 1

22 2 23 3 2 2

33 3 3 3

1

2

a x a x a x a x c

a x a x a x c

a x a x c

a x a x c

c

c

c

n n

n n

n n

rr r rn n r

r

r

n

Matrices ◾ 95

wherer <nand …, , ,11 12a a arr arenotzero.Therearetwopossiblecases:

1.Nosolutionexistsifanyoneofthe +1cr through cn isnotzero. 2.Infinitelymanysolutionsexistif +1cr through cn areallzero.

Ifr =nand …, , ,11 12a a ann arenotzero,thenthesystemwouldappearasfollows:

+ + + + =

+ + + =

+ + =

=

=

11 1 12 2 13 3 1 1

22 2 23 3 2 2

33 3 3 3

a x a x a x a x c

a x a x a x c

a x a x c

a x c

n n

n n

n n

nn n n

Forthiscase,thereisonlyonesolution.

3.7 Inverse MatrixGivenamatrixsystemofequationsofthegeneralform

AX=C

Wecansolvebyfront-multiplyingbothsidesoftheequationbythematrixinverse −A 1 :

- −−= = =A AX IX X A C1 1

MATLAB’smethodofsolutionis

X=inv(A)*C (solvesby −1A )

or

X=A/C (solvesbyGausselimination)

Letusseewhatisinvolvedbydetermining −1A .


Let −B = A 1 .Then, =BA I.Wewilldemonstratefora3×3matrix:

=1 0 00 1 00 0 1

11 12 13

21 22 23

31 32 33

11 12 13

21 22 23

31 32 33

b b bb b bb b b

a a aa a aa a a

FirstrowofBA:

Element(1,1): + + = 111 11 12 21 13 31b a b a b a

Element(1,2): + + = 011 12 12 22 13 32b a b a b a

Element(1,3): + + = 011 13 12 23 13 33b a b a b a

Here, 11b , 12b , 13b aretheunknowns.Also,notethatthetransposematrix

=11 21 31

12 22 32

13 23 33

Aa a aa a aa a a

T

Let

=1

11

12

13

Bbbb

Then,

=100

1A BT

Solvefor 11b , 12b , 13b byGausselimination.SecondrowofBA:

Element(2,1): + + = 021 11 22 21 23 31b a b a b a

Element(2,2): + + = 121 12 22 22 23 32b a b a b a

Element(2,3): + + = 021 13 22 23 23 33b a b a b a

Matrices ◾ 97

Here, 21b , 22b , 23b aretheunknowns.

Let

=2

21

22

23

Bbbb

Then,

=010

2A BT

Solvefor 21b , 22b , 23b byGaussElimination.

ThirdrowofBA:

Element(3,1): + + = 031 11 32 21 33 31b a b a b a

Element(3,2): + + = 031 12 32 22 33 32b a b a b a

Element(3,3): + + = 131 13 32 23 33 33b a b a b a

Here, 31b , 32b , 33b aretheunknowns.

Let

=3

31

32

33

Bbbb

Then,

=001

3A BT

Solvefor 31b , 32b , 33b byGausselimination.

Analternative[1]tothemethoddescribedistoaugmentthecoefficientmatrixwith the identity matrix, then apply the Gauss-Jordan method, making the


coefficientmatrixtheidentitymatrix.Theoriginalidentitymatrixthenbecomes−1A .Thestartingaugmentedmatrixfora3×3coefficientmatrixisasfollows:

1 0 00 1 00 0 1

11 12 13

21 22 23

31 32 33

a a aa a aa a a

Example 3.6

This method is illustrated by the following example:

− + = −− + + = −

+ − =

x x xx x xx x x

3 43 4 3 102 3 2 18

1 2 3

1 2 3

1 2 3

Writing the augmented matrix:

=−

−−

augA1 3 13 4 32 3 2

1 0 00 1 00 0 1

Multiply row 1 by −31

and subtract from row 2, giving

(−3+3×1), (4−3×3), (3+3×1), (0+3×1), (1+3×0), (0+3×0) = (0,−5, 6, 3, 1, 0)

Multiply row 1 by =21

2 and subtract from row 3, giving

(2−2×1), (3−2×(−3)), (−2−2×1), (0−2×1), (0−2×0), (1−2×0) = (0, 9, −4, −2, 0, 1)

=−−

− −aug EquivA

1 3 10 5 60 9 4

1 0 03 1 02 0 1

,

Multiply row 2 by −95


+ × + × − − + × − + × + × + ×

=

095

0 , 995

( 5) , 495

6 , 295

3 , 095

1 , 195

0

0, 0,345

,175,95, 1

=

−−

aug EquivA

1 3 10 5 6

0 0345

1 0 03 1 0175

95

1,

Matrices ◾ 99

Multiply row 3 by 3034


− × − − × − × − × − ×

− × = − − −

03034

0 , 53034

0 , 63034

345

, 33034

175

, 13034

95

,

03034

1 0, 5, 0, 0,2034

,3034



− × − − × − × − × − ×

− × = − − −

1534

0 , 3534

0 , 1534

345

, 1534

175

, 0534

95

,

0534

1 (1, 3, 0,12,

934

,534

)

=

−

−

− −

− −aug EquivA

1 3 0

0 5 0

0 0345

12

934

534

02034

3034

175

95

1

,



− × − − × − − × − × − + ×

− + × =

135

0 , 335

( 5) , 035

0 ,12

35

0 ,934

35

2034

,

534

35

3034

1, 0, 0,12,334

,1334

= − − −aug EquivA

1 0 0

0 5 0

0 0345

12

334

1334

02034

3034

175

95

1

,

Divide row 2 by −5 and row 3 by 345

, giving

=aug EquivA

1 0 0

0 1 0

0 0 1

12

334

1334

0434

634

12

934

534

,


Thus,

=−A

12

334

1334

0434

634

12

934

534

1

It is left as a student exercise to show that =−AA I1 .

3.8 The Eigenvalue ProblemOneapplicationoftheeigenvalueproblemisinresonantcircuits.ConsidertheLCnetworkshowninFigure 3.3.Usingtheconstituentrelationsforvoltageacrossan

inductor =v Ldidt

andcurrentthroughacapacitor =i Cdvdt

,thegoverning

differentialequationsdescribingthetwonodevoltages 1v and 2v are

= − +

1 1212

1 11

1 12

d vdt LC

vLC

v (3.17)

= − +

1 1 1222

1 21

1 2 2 22

d vdt LC

vLC L C

v (3.18)

Wewishtodeterminethemodesofoscillationsuchthatbothvoltagesoscillateatthesamefrequency.Toobtainsuchasolution,set

= ω1 1v K e j t (3.19)

= ω2 2v K e j t (3.20)

L1

L2C1 C2

v1 v2

Figure 3.3 LC circuit.

Matrices ◾ 101

whereK1andK2areconstantstobedetermined.SubstitutingEquations(3.19)and(3.20)intoEquations(3.17)and(3.18)gives

−ω − =

1 1 01 1

21

1 12LC

KLC

K (3.21)

− + + −ω =

1 1 1 01 2

11 2 2 2

22LC

KLC L C

K (3.22)

Equations (3.21)and (3.22)are twohomogeneous linearalgebraic equationsintwounknowns.Linearalgebratellsusthattheonlywayfortwohomogeneouslinearalgebraicequationsintwounknownstohaveanontrivialsolutionisforthedeterminantofthecoefficientmatrixtobezero.Thus,

−ω −

− + −ω

=

1 1

1 1 101 1

2

1 1

1 2 1 2 2 2

2

LC LC

LC LC L C

(3.23)

Now,letω = λ2 andalsolet

=−

− +

1 1

1 1 11 1 1 1

1 2 1 2 2 2

ALC LC

LC LC L C

RewritingEquation(3.23)intermsofλandtheindividualelementsai j ofAgives

λ − + λ + − =( ) ( ) 0211 22 11 22 12 21a a a a a a (3.24)

ThesolutionofEquation(3.24)givestheeigenvalues λ1 and λ2 ofmatrixA,whichrepresentthesquareofthetwonaturalfrequenciesforthissystem.Theratiooftheamplitudesoftheoscillationofthetwovoltagescanbeobtainedbysubstitut-ingthevaluesofλintoEquation(3.21)or(3.22),whichgives

( )= − − λ = − λ12

111 1 12 1 1 1

KK

a a LC

forthefirstmode,and

= − − λ = − λ( ) 12

111 2 12 1 1 2

KK

a a LC

forthesecondmode.


Theeigenvector V1 associatedwith λ1 is

− −λ( )

1

1 11 1 12

KK a a

andtheeigenvector V2 associatedwith λ2 is

− −λ( )

1

1 11 2 12

KK a a

Since 1K isarbitrary,wecanselect = 11K ,andthus

=

− −λ1

( )111 1 12

V a a (3.25)

and

=

− −λ1

( )211 2 12

V a a (3.26)

IfVisaneigenvectorofamatrixA correspondingtoaneigenvalueλ,thensoisbV,wherebisanonzeroconstant[1].

MATLABhasabuilt-infunctionnamedeigthatfindstheeigenvaluesofasquarematrix.MATLAB’sdescriptionofthefunctionfollows:

E=EIG(X)isavectorcontainingtheeigenvaluesofasquarematrixX. [V,D]=EIG(X)producesadiagonalmatrixDofeigenvaluesanda

fullmatrixVwhosecolumnsarethecorrespondingeigenvectorssothatX*V=V*D.

Forthisproblem,thematrixArepresentsX.Thus,running

[V,D]=eig(A)

givestheeigenvectorsassociatedwith λ1 and λ2 .V(:,1)correspondsto λ1 ,andV(:,2)correspondsto λ .2

Matrices ◾ 103

Example 3.7

Suppose in Figure 3.3 the following parameters were given:

= = = =L L C C15mH, 25mH, 1 F, 33 F1 2 1 2

We wish to create a MATLAB script that will determine

1. the eigenvalues of the system by Equations (3.21) and (3.22) by MATLAB’s eig function.

2. the eigenfunctions by both Equations (3.21) and (3.22) and MATLAB’s [V,D] output.

The program follows.

%Example_3_7.m:Eigenvaluesandeigenvectors.%L=eig(A)isavectorcontainingtheeigenvaluesofa%squarematrixA.%[M,D]=eig(A)producesadiagonalmatrixDof%eigenvaluesandafullmatrixMwhosecolumnsarethe%correspondingeigenvectorssuchthatA*L=M*D.%UnitsareinSIunits.clear;clc;L1=15e-3;L2=25e-3;%henriesC1=1e-6;C2=33e-6;%faradsA(1,1)=1/(L1*C1);A(1,2)=-1/(L1*C1);A(2,1)=-1/(L1*C2);A(2,2)=1/(L1*C2)+1/(L2*C2);%Solveforeigenvaluesandeigenvectorsusingderivation%inEqn3.25:%lambda^2-(A(1,1)+A(2,2))%*lambda+(A(1,1)*A(2,2)-A(1,2)*A(2,1))=0%Solveforlambdausingquadraticformula:b=-(A(1,1)+A(2,2));c=A(1,1)*A(2,2)-A(1,2)*A(2,1);lambda1=(-b+sqrt(b^2-4*c))/2;lambda2=(-b-sqrt(b^2-4*c))/2;%FromEqns3.26and3.27:eigenvector1=[1;-(A(1,1)-lambda1)/A(1,2)];eigenvector2=[1;-(A(1,1)-lambda2)/A(1,2)];fprintf('EigenvaluessolvedviaEquation3.25:');fprintf('lambda1=%.0flambda2=%.0f\n',lambda1,lambda2);fprintf('EigenvectorssolvedviaEquation3.26');fprintf('and3.27:\n');eigenvector1eigenvector2%SolveforeigenvaluesandeigenvectorsusingMATLAB's%eigfunction.Callingeigandreturningresulttoa%scalarwillgivetheeigenvalues.L=eig(A);fprintf('EigenvaluesviaMATLABeigfunction:');fprintf('L(1)=%.0fL(2)=%.0f\n',L(1),L(2));%Callingeigandreturningresulttotwovariableswill%givetheeigenvectorsandadiagonalmatrixofthe%eigenvalues.[VD]=eig(A);


fprintf('EigenvectorsviaMATLABeigfunction:\n');V1=V(:,1)V2=V(:,2)D

The following results were generated by MATLAB:

EigenvaluessolvedviaEquation3.25:lambda1=68723140lambda2=1175850EigenvectorssolvedviaEquation3.26and3.27:eigenvector1=1.0000–0.0308eigenvector2=1.00000.9824EigenvaluesviaMATLABeigfunction:L(1)=68723140L(2)=1175850EigenvectorsviaMATLABeigfunction:V1=0.9995-0.0308V2=0.71340.7008D=1.0e+007*6.8723 0 0 0.1176

Examining the results, it can be seen that lambda1=D(1,1) and lambda2=D(2,2). Also, eigenvector1 is a scalar multiple of V1 and eigenvec-tor2 is a scalar multiple of V2.

Exercises

Exercise.3.1

UseGausseliminationtosolvethefollowingsystemsofequations:

a.+ − =− + = −+ + =

2 3 204 3 14

5 21

1 2 3

1 2 3

1 2 3

x x xx x xx x x

b.+ + =

− − + =+ + =

4 8 82 3 2 14

3 4 30

1 2 3

1 2 3

1 2 3

x x xx x xx x x

c.

+ + − =− + − =− + − =+ − + = −

2 3 15 2 5 4 5

2 3 3 33 8 1

1 2 3 4

1 2 3 4

1 2 3 4

1 2 3 4

x x x xx x x xx x x xx x x x

Matrices ◾ 105

ProjectsProject.3.1

ForthecircuitofFigure 3.1,solveforthefiveunknowncurrentsinsteadofthethreeunknownvoltages.SolvealgebraicallybyusingEquations(3.3),(3.4),and(3.5)andbyusingKirchhoff’svoltage law(KVL)towritetwoadditionalequations.KVLstatesthatthesumofthevoltagedropsaroundanyloopinthecircuitmustbezero.WriteKVLequationsforthesetwocircuitloops: → → → → →1 1 2 4 5 2V R R R R V and → →3 4 2R R R .Youshouldwindupwithasystemoffiveequationsandfiveunknowns.SolvewithMATLABusingtheinversematrixtechniqueandconfirmthatyouranswermatchestheresultsfoundinExample3.3.

Project.3.2

FigureP3.2showsaresistivecircuitknownasa“laddernetwork.”

1. UsingOhm’slawandKirchhoff’scurrentlaw,writeasystemofequationsforthesecond-orderandthird-ordernetworksofFiguresP3.2aandP3.2b.Inaddition,writetheequationsforafourth-orderladdernetwork.

2. Youshouldseeapatternemergingfromyoursolutionsinpart1.Usethispat-terntowritethematrixforaeighth-orderladdernetwork.SolveforallcircuitvoltagesusingMATLABforVref=5Vandthefollowingresistorvalues:

= =

= =

= =

= =

= =

= =

= =

= =

2200 , 2200

1200 , 6800

3900 , 2200

3300 , 5700

1800 , 5100

4700 , 3900

5700 , 6800

1200 , 2700

11 12

21 22

31 32

41 42

51 52

61 62

71 72

81 82

R R

R R

R R

R R

R R

R R

R R

R R

3. WriteaMATLABprogramtosolvethisproblemautomaticallyforarbitrarynandR0(asshowninFigureP3.2c)underthefollowingconditions:

= = = = = =

= = = = =

−

−2

0 11 21 31 ( 1) 1

0 12 22 32 ( 1)2

R R R R R R

R R R R R

n n

n

Findsolutionsforthefollowingtwocases:

n=16, R0=1000Ω, Vref=10

n=24, R0=100Ω, Vref=1


Project.3.3

One very important application of the eigenvalue problem is in the theory ofmechanicalvibrations.Considerthesystemoftwomasses(m1andm2)andthreesprings(withspringconstants 1k , 2k ,and 3k )showninFigureP3.3.

Thegoverningdifferentialequationsdescribingthemotionofthetwomassesare

= − −( )1 1 2 2 1 1 1m x k x x k x (P3.3a)

=− − −( )2 2 2 2 1 3 2m x k x x k x (P3.3b)

UsingMATLAB’seigfunction,determinethemodesofoscillationsuchthateachmassundergoesharmonicmotionatthesamefrequency.Usethefollowingvalues:

m1=m2=1500kg,k1=3250N/m,k2=3500N/m,k3=3000N/m

+–

R11

R12

R21

R22Vref

v1 v2

(a)

+–

R11

R12

R21

R22

R31

R32

v1 v2 v3

Vref

(b)

+–

R11

R12

R21

R22

R31

R32

Rn1

Rn2

vnv3v1 v2

Vref

(c)

Figure P3.2 (a) Second-order resistor ladder network. (b) Third-order resistor ladder network. (c) An nth-order resistor ladder network.

Matrices ◾ 107

Project.3.4

Suppose a manufacturer wishes to purchase a piece of equipment that costs$40,000.Heplanstoborrowthemoneyfromabankandpayofftheloanin10years in120 equalpayments.Theannual interest rate is 6%.Eachmonth, theinterestchargedwillbeontheunpaidbalanceoftheloan.Hewishestodeterminewhathismonthlypaymentwill be.Thisproblemcanbe solvedby a systemoflinearequations.

Letxjbetheamountinthejthpaymentthatgoestowardpayingofftheprin-cipal.Then,theequationdescribingthejthpaymentis

∑= = + −=

= −

th payment1

1

j M x P x Ij n

n

n j

(P3.4a)

where M=themonthlypayment P=theamountborrowed I=themonthlyinterestrate=one-twelfthoftheannualinterestrate

Thetotalnumberofunknownsis121(120xjvaluesandM).ApplyingEquation(P3.4a)toeachmonthgives120equations.The121stequation

is

∑==

=

1

120

P xnn

n

(P3.4b)

DevelopaMATLABprogramthatwill:

1. Ask the user to enter from the keyboard the amount of the loan P, theannualinterestrateI,andthetimeperiodY(inyears).

2. Setupthesystemoflinearequations,using ,An m asthecoefficientmatrixofthesystemoflinearequations.Thenrepresentstheequationnumber,andmrepresentsthecoefficientof xm inthatequation.Set =121x M .

m1 m2k1 k2 k3

x1 x2

m2m1k1x1 k3x2k2(x1 − x2)

Figure P3.3 Mechanical vibration problem with two degrees of freedom.


3. SolvethesystemoflinearequationsinMATLAB. 4. Printoutatableconsistingoffourcolumns.Thefirstcolumnshouldbethe

monthnumber,thesecondcolumnthemonthlypayment,thethirdcolumntheamountofthemonthlypaymentthatgoestowardpayingofftheprincipal,andthefourthcolumntheinterestpaymentforthatmonth.

Project.3.5

TheticandtoccommandsinMATLABallowyoutomeasuretheexecutiontimeforagivenseriesofcommands.Theticandtocbehaveasastopwatch;ticstartsthestopwatch,andtocstopsitandprintstheelapsedtime.Anexam-pleusageis

tic; a=factorial(25); toc;

Runningthisscriptwillcompute25!andthenprintthetotalexecutiontime:

Elapsedtimeis0.005509seconds.

Ofcourse,thetimethatyouseewilldependonthespeedofyourcomputer.

1. Write a program in MATLAB that creates a random n × n matrix Awhereeachelementisarandomintegerbetween − maxE and maxE .UseMATLAB’s rand and round functions to accomplish this. To learnhowtousethesefunctions,typehelp randandhelp round intothe MATLAB Command window. Initially, assume that n = 100 and

= 100maxE . 2. Also,createa1×ncolumnvectorCwithrandomelements,usingthesame

valuesfornand maxE above. 3. SolvetheequationAX.=.CforXintwoways:(a)byfinding −A 1 (withinv)

andmultiplyingitbyCand(b)byusingGausselimination(theMATLAB‘\’operator).Useticandtoctomeasuretheamountoftimeittakesforeachmethod.Dotheresultsmakesense?

4. Rerunyourprogram10timesandobtainanaverageexecutiontimeforthetwosolutionmethods.Alsofindaverageexecutiontimesforthefollowingval-uesofn:3,10,30,300,1000.Plottheexecutiontimesversusnonlog-logaxes.

Reference 1. Kreyszig,E.,Advanced Engineering Mathematics,8thed.,Wiley,NewYork,1999.

109

Chapter 4

Roots of Algebraic and Transcendental Equations

4.1 IntroductionIn the analysisof various engineeringproblems,weoftenare facedwithaneedtofindrootsofequationswhosesolutionisnoteasilyfoundanalytically.Typicalexamplesincludenthdegreepolynomialsandtranscendentalequationscontainingtrigonometricfunctions,exponentials,orlogarithms.Inthischapter,wereviewsev-eralmethodsforsolvingsuchequationsnumericallywithmathematicaltechniquesofvaryingcomplexityandaccuracy.Alsoincludedisasectiononthefzerofunc-tionofMATLAB®,whichmaybeusedtoobtaintherootsofequationsofthetypementionedpreviously.

4.2 The Search MethodTheequationwhoserootsaretobedeterminedshouldbeputintothefollowingstandardform: f(x)=0 (4.1)

Weproceedasfollows:First,asearchismadetoobtainintervalsinwhichrealrootslie.ThisisaccomplishedbysubdividingthexdomainintoNequalsubdivisions,giving

… = ++ +, , , and1 2 3 1 1x x x x x x xN i i


Then,locatewheref(x)changessign(seeFigure 4.1).Thisoccurswhenthesignsoftwoconsecutivevaluesoff(x)aredifferent,thatis,

<+( ) ( ) 01f x f xi i

Thesignchangeusuallyindicatesthatarealroothasbeenpassed.However,itmayalsoindicateadiscontinuityinthefunction.(Example:tanxisdiscontinuousatx=π/2.)

Oncetheintervalsinwhichtherootsliehavebeenestablished,wecanusesev-eraldifferentmethodsforobtainingtherealroots.

4.3 Bisection Method

Supposeithasbeenestablishedthatarootliesbetween xi and +1xi .Then,cuttheintervalinhalf(seeFigure 4.2),andthus

= ++ 2

12

x xx

i i

f (x)

xx1 x2

x3

f (x1)

f (x2)

f (x3)

xN+1xNxN–1x4

f (x4)

Figure 4.1 Selecting the interval in which a root lies. In this case, f(x2)f(x3) < 0, and thus the root lies between x2 and x3.

Roots of Algebraic and Transcendental Equations ◾ 111

Nowcompute +( ) ( )12

f x f xi i :

Case1:If <+( ) ( ) 012

f x f xi i ,thentherootliesbetween xi and + 12

xi .

Case2:If >+( ) ( ) 012

f x f xi i ,thentherootliesbetween + 12

xi and +1xi .

Case3:If =+( ) ( ) 012

f x f xi i ,then xi or + 12

xi isarealroot.

Forcases1and2,selecttheintervalcontainingtherootandrepeattheprocess.

Continue repeating theprocess, say r times, then =( )2

xx

f r ,whereΔx is theinitialsizeoftheintervalcontainingtherootbeforethebisectionprocessand( )x f isthesizeoftheintervalcontainingtherootafterr bisections.If ( )x f issufficientlysmall,thenaverygoodapproximationfortherootisanywherewithinthelastbisectedinterval,saythemidpointoftheinterval.

Example:For20bisections,

= ≈ × × −( )2

1.0 10206x

xxf

Programmethod:

=

=

= +

+

Set

12( )

1

x x

x x

x x x

A i

B i

C A B

f (x)

xxi+1xi xi+1/2

Figure 4.2 Selecting the interval containing the root in the bisection method.


If <( ) ( ) 0f x f xA C (indicatingthattherootliesbetweenxAandxC ),

=

= +

Then set

12( )

x x

x x x

B C

C A B

andrepeattheprocess.

Otherwise,if >( ) ( ) 0f x f xA C (indicatingthattherootliesbetween xB and xC),

=

= +

Then set

12( )

x x

x x x

A C

C A B

andrepeattheprocess.

If =( ) ( ) 0f x f xA C ,theneither xA or xC isaroot.

4.4 Newton-Raphson MethodTheNewton-Raphsonmethodusesthetangenttothecurvef(x)=0toestimatetheroot.

Weneedtoobtainanexpressionfor ( )f x andmakeaninitialguessfortheroot,say 1x (seeFigure 4.3). ( )1f x givestheslopeofthetangenttothecurveat 1x .

f (x1)

x1x2

f (x)

x

f (x2)

Slope = f '(x1)

Figure 4.3 Predicting root in the Newton-Raphson method.


Onthetangenttothecurve,

−−

=( ) ( ) ( )1 2

1 21

f x f xx x

f x (4.2)

Ournextguess(oriteration)fortherootisobtainedbysettingf(x2)=0andsolvingforx2,thatis,

= −

( )( )2 1

1

1x x

f xf x

(4.3)

Checkif < ε( )2f x ,whereεistheerrortolerance.Iftrue,thenquit. 2x istheroot,soprint 2x .Ifnottrue,thenset =1 2x x andrepeattheprocess.Continue repeating the process until < ε( )f xn , where n is the number of

iterations.Analternateconditionforconvergenceis

< ε

( )( )

1

1

f xf x

(4.4)

Usually,ε shouldbeavery smallnumber, suchas10–4,but it canbeeitherlargerorsmallerdependingonhowcloseyouwishtogettotherealroot.Youcanalwaystesttheaccuracyoftheobtainedrootbysubstitutingtheobtainedrootintothefunctionf(x)toseeifitissufficientlyclosetozero.

TheNewton-Raphsonmethodiswidelyusedforitsrapidconvergence.However,therearecaseswhereconvergencewillnotoccur.Thiscanhappenif

a. ( )f x changessignneartheroot. b.Theinitialguessfortherootistoofarfromthetrueroot.

If we combine the Newton-Raphson method with the search method forobtainingasmallintervalinwhichtherealrootlies,convergenceisgenerallynotaproblem.

4.5 MATLAB’s fzero and roots FunctionsMATLABhasbuilt-infunctionstodeterminetherealrootsofafunctionofonevariable, such as a transcendental equation or an nth-degree polynomial. Thefzerofunctionisusedfortranscendentalequations,andtherootsfunctionisusedforpolynomialequations.First,wediscussthefzerofunction.


4.5.1 The fzero FunctionTogetstarted,clickintheCommandwindowandtypein

>>help fzero

Thisgivesseveraldifferentdescription,thefirsttwoarepresentedhere.

FZEROSingle-variablenonlinearzerofinding. X=FZERO(FUN,X0)triestofindazeroofthefunctionFUN

nearX0,ifX0isascalar.ItfirstfindsanintervalcontainingX0wherethefunctionvaluesoftheintervalendpointsdifferinsign,thensearchesthatintervalforazero.FUNisafunctionhandle.FUNacceptsrealscalarinputXandreturnsarealscalarfunctionvalueF,evaluatedatX.ThevalueXreturnedbyFZEROisnearapointwhereFUNchangessign(ifFUNiscontinuous),orNaNifthesearchfails.

X=FZERO(FUN,X0),whereX0isavectoroflength2,assumesX0isafiniteintervalwherethesignofFUN(X0(1))differsfromthesignofFUN(X0(2)).Anerroroccursifthisisnottrue.CallingFZEROwithafiniteintervalguaranteesFZEROwillreturnavaluenearapointwhereFUNchangessign.

Asdescribed,fzerohas severaldifferentusagesdependingonwhetheryouhaveasingleinitialguessofX0(i.e.,X0isascalar)orifyouhaveaknownintervalforthezero(i.e.,X0isavectoroflength2).Thefirstusageisappropriateifyouhavesomeideaofwheretherootlies.Thesecondusageismoreappropriatewhenthereismorethanoneroot,andallrootsneedtobeobtained.Thesecondusageshouldbeusedincombinationwiththesearchmethoddescribedpreviously.

Thefirstusageoffzerofunctionis

X=FZERO(FUN,X0)

FUNisfunction handleforthefunctiontobesolvedandcanbethenameofafunc-tionfile(withthe.mextension)orcanbeananonymousfunction(asdescribedinSection2.8).X0istheinitialguessfortheroot,andXisthesolutioncalculatedbyMATLAB.Thus,supposethenameofthefunctionfilewhoserootistobeobtainedismyfun.m,andourguessfortherootis3.0.Then,wewouldwrite

X=fzero('myfun',3.0)


Analternativeandequivalentwaytorunthecommandis

X=fzero(@myfun,3.0)(nosinglequotationmarksneeded)

Ifnorootisfound,fzeroreturnsNaN(notanumber).Inthesecondusageoffzero,X0isavectoroflength2thatdefinesaninterval

overwhichFUNchangessign,thatis,thevalueofFUNatX0(1)isoppositeinsigntothevalueatX0(2).MATLABgivesanerrorifthisconditionisnotmet.

Insomeinstances,wewouldliketofindthezeroofafunctionoftwoargu-ments,sayXandP,wherePisaparameterandisfixed.Tosolvewithfzero,Pmustbedefinedinthecallingprogram.Forexample,supposemyfunisdefinedinanM-fileasafunctionoftwoarguments:

functionf=myfun(X,P) f=cos(P*X);

Thefzerostatementwouldneedtobeinvokedasfollows:

P=1000; root=fzero(@(X)myfun(X,P),X0)

whererootisthezerooffunctionmyfunwhenP=1000.NotethatPneedstobedefinedbeforethefzerofunctioniscalled.

An alternative to adding parameterP as an argument tomyfun is to useMATLAB’sglobalstatementtodefineparameterP infunctionmyfunasanexternallyassignedvariable.Theglobalstatementneedstobeusedinboththecallingprogramandthefunctionmyfun.

Example:Callingprogram:

globalP;P=1000;X0=10.0;root=fzero(@myfun,X0);

-----------------------------------------------------

Thefilemyfun.m:

functionf=myfun(x)globalP;f=cos(P*x);

-----------------------------------------------------

Examplesofbothmethodsfollow.


Example 4.1: Using the fzero function to Find the First Root of a Transcendental Function

In Section 2.17, the governing equation for the voltage v(t) of a parallel RLC

circuit was derived. For the case when <RC LC1

212

, the system is under-

damped, and the solution for v(t) has the form

= − − + −v tRC

t ALC RC

t BLC RC

t( ) exp1

2cos

1 12

sin1 1

2

2 2

(4.5)

Determine the the smallest time when v(t) is zero, that is, find the first zero crossing of v(t) on the t axis. Assume the following parameters:

= = = = = −− −R L C A B100 , 10 H, 10 F, 6.0V, 9.0V3 6

0 ≤ t ≤ 0.5 msec

%Example_4_1.m%Thisscriptdeterminestheinitialzerocrossingofthe%underdampedresponseofaparallelRLCcircuit.The%underdampedresponseis:%v(t)=exp(-t/(2RC))*(A*cos(sqrt(1/LC-(1/2RC)^2)t)%+B*sin(sqrt(1/LC-(1/2RC)^2)t))%Thecircuitcomponentvaluesandinitialconditions%are:%R=100ohms;L=1.0e-3henry;C=1.0e-6farads;%A=6.0volt,B=-9.0volt%Solveovertheinterval0<=t<=0.5e-3secclear;clc;R=100;L=1.0e-3;C=1.0e-6;A=6.0;B=-9;fprintf('Example4.1:Findfirstzerocrossingof');fprintf('underdampedRLCcircuit\n');tmin=0.0;tmax=0.5e-3;%splituptimespaninto100intervals:N=100;dt=(tmax-tmin)/N;%First,calculatetandv(t)ateachtimestepforn=1:N+1t(n)=tmin+(n-1)*dt;v(n)=func_RLC(t(n),A,B,R,L,C);fprintf('%10.4e%10.3f\n’,t(n),v(n));endplot(t,v),xlabel('t'),ylabel('v'),title('vvstforaRLCcircuit'),grid;%Next,usethesearchmethodtofindthefirsttimestep%wherethesignofv(t)changes.Whenfound,usefzero%tosolve.forn=1:N


sign=v(n)*v(n+1);ifsign<=0.0root_interval(1)=t(n);root_interval(2)=t(n+1);%Solveforthezeroofthemulti-argument%functionfunc_RLC.mbyinvokingfunc_RLCasan%anonymousfunctionofonevariable.root=fzero(@(t)func_RLC(t,A,B,R,L,C),...root_interval);break;endendfprintf('\n\n');ifn<Nfprintf('Firstzerocrossingisatt=%10.4es\n',...root);elsefprintf('Norootsliewithin%g<=t<=%gs\n',...tmin,tmax);end%Checksolution.Puttingrootintofunc_RLCshould%producezero.v=func_RLC(root,A,B,R,L,C);fprintf('v=%8.6e\n',v);----------------------------------------------------------------------------%func_RLC.m%ThisfunctionworkswithExample_4_1.mfunctionv=func_RLC(t,A,B,R,L,C)arg1=1/(2*R*C);arg3=sqrt(1/(L*C)-1/(2*R*C)^2);%Equation(4.5):v=exp(-arg1*t)*(A*cos(arg3*t)+B*sin(arg3*t));----------------------------------------------------------------------------

PROGRAM RESULTS

Firstzerocrossingisatt=1.8831e-005s

We can modify Example 4.1 and use the global statement to bring the parameters into the function. To do this,

1. Add the following global statement immediately after the clear;clc; statements in the calling program and immediately after the function statement in the function program:

globalRLCAB;

2. Replace the function statement:

functionv=func_RLC(t,A,B,R,L,C)


with

functionv=func_RLC(t)

3. Replace the following statement in the calling program:

v(n)=func_RLC(t(n),A,B,R,L,C);

with

v(n)=func_RLC(t(n));

4. Replace the following statement in the calling program:

root=fzero(@(t)func_RLC(t,A,B,R,L,C),root_inteval);

with

root=fzero('func_RLC',root_inteval);

Run each program and check to see if both programs give the same answer.

4.5.2 The roots FunctionMATLAB has a function to obtain the roots of a polynomial. The function isroots.Toobtaintheusageofthefunction,intheCommandWindowtype

>> help roots

Thisgives

ROOTSFindpolynomialroots. ROOTS(C)computestherootsofthepolynomialwhose

coefficientsaretheelementsofthevectorC.IfChasN+1components,thepolynomialisC(1)*X^N+...+C(N)*X+C(N+1).

Thus, to find the roots of the polynomial + + + + = 04 3 2Ax Bx Cx Dx E , runroots([ABCDE]).Therootsfunctionwillgivebothrealandimaginaryrootsofthepolynomial.

SomeadditionalusefulMATLABfunctionsare

poly(V) Finds the coefficients of the polynomial whose roots are V

real(X) Gives the real part of X

imag(X) Gives the imaginary part of X


Example 4.2: Using roots to Find the Zeros of a Polynomial

%Example_4_2.m%Thisprogramdeterminestherootsofapolynomialusing%thebuiltinfunction'roots'.%Thefirstpolynomialis:f=x^3-4.7*x^2-35.1*x+85.176.%Therootsofthispolynomialareallreal.%Thesecondpolynomialis:f=x^3-9*x^2+23*x-65.The%rootsofthispolynomialarebothrealandcomplex.%Complexrootsmustbecomplexconjugates.%ToobtainmoreinfooncomplexnumbersinMATLAB,run%"helpcomplex"intheCommandWindow.clear;clc;%Definecoefficientsoffirstpolynomial(realroots)C(1)=1.0;C(2)=-4.7;C(3)=-35.1;C(4)=85.176;fprintf('Thefirstpolynomialcoefficientsare:\n');Cfprintf('Therootsare:\n');V=roots(C)fprintf('Therecalculatedcoefficientsofthe');fprintf('polynomialwhoserootsareVare:\n');C_recalc=poly(V)fprintf('\n\n');%Definecoefficientsofsecondpolynomial(realand%complexroots)D(1)=1.0;D(2)=-9.0;D(3)=23.0;D(4)=-65.0;fprintf('Thesecondpolynomialcoefficientsare:\n');Dfprintf('Therootsare:\n');W=roots(D)fprintf('Therealandimaginarypartsofthe');fprintf('rootsare:\n');re=real(W)im=imag(W)fprintf('Therecalculatedcoefficientsofthe');fprintf('polynomialwhoserootsareWare:\n');W_recalc=poly(W)

Projects

Project.4.1

Thecurrent-voltagerelationshipofasemiconductorPNdiodecanbewrittenasfollows:

= −1i I eD S

qkT

vD (P4.1a)


whereiDandvDarethediodecurrentandvoltage,respectively,asdefinedinFigureP4.1;ISisaconstant(withunitsofamperes),whichisdeterminedbythesemicon-ductordopingconcentrationsandthedevicegeometry; = × −1.6 10 19q coulombistheunitelectriccharge; = × −1.38 10 23k joule/kelvinistheBoltzmannconstant;andTisabsolutetemperature(inkelvin).

Wewouldliketodeterminethecurrentandvoltagethroughthediodeforthiscircuit.UsingKirchhoff’svoltagelaw,weknowthatthesumofthevoltagedropsaroundthecircuitmustsumtozero:

− − = 0V v vin R D (P4.1b)

ApplyingOhm’slawforthecurrentthroughtheresistor =v i RR R andobserv-ing that the resistor current equals the diode current =i iR D , we can rewriteEquation(P4.1b)as

− − − =1 0V I e R vin S

qkT

vD

D (P4.1c)

Let

= − − −( ) 1f v V I e R vD in S

qkT

vD

D

1. CreateaMATLABfunctionforf(vD)andplotfortheinterval ʺ ʺ0 0.8vD Vfor10mVsteps(80subdivisionsonthevDdomain).

2. Use the searchmethod to obtain a small intervalwithinwhich the root ofEquation(P4.1c)lies.

3. UseMATLAB’sfzerofunctiontoobtainamoreaccuratevaluefortheroot.Usethefollowingparameters:

T =300K,IS=10–14A,Vin = 5V,R =1000Ω.

Project.4.2

WewishtodeterminetheDCtransfercharacteristicforthediodecircuitofFigureP4.1.WewillconsiderVinasaparameter,where 5V 12VVinʺ ʺ instepsof1V.WewishtofindthevalueofvDforallvaluesofVin.

−

+

vR

iR

R

vD

+

−

iD

Vin

+

−

Figure P4.1 Diode-resistor circuit.


WriteaMATLABprogramthatwillfindtherootsof ( )f vd =0,where

= − − −( ) 1f v V I e R vD in S

qkT

vD

D

Yourprogramshould

1. UseaglobalstatementstobringvaluesofVinintothefunctionf(vD). 2. Take ʺ ʺ0.2 0.8vD Vwith60subdivisionsonthevDdomain. 3. Usethesearchmethodtofindasmallintervalinwhichtherootoff(vD)=0lies. 4. UsetheMATLAB’sfzerofunctiontoobtainamoreaccuratevaluefortheroot. 5. ConstructatableconsistingofallvaluesofVinandthecorrespondingrootsof

f(vD)=0. 6. IfyoudidProject4.1,confirmthatyourprogramreturnsthesameresultasin

Project4.1when = 5Vin V.

Project.4.3

RepeatProject4.2,butthistimedonotuseglobalstatements.Instead,modifyyour ( )f vd functiontoincludetheargumentVin.Youwillalsoneedtomodifythefzerostatementappropriatelytouseafunctionwithmultiplearguments.

Project.4.4

Wewishtoconsiderthesettling time(describedinthefollowing)oftheinductorcurrentiLintheRLCcircuitofSection2.17andProject2.9.TheresistanceRcritthatmakesthecircuitcriticallydampedisgivenby

=

4R

LCcrit

For >R Rcrit ,thegoverningequationdescribingiLis

= − − + −exp 12

cos 1 12

sin 1 12

2 2

iRC

t ALC RC

t BLC RC

tL

ThecoefficientsAandBaredeterminedbytheinitialconditionsandthecom-ponentvalues.ThesolutionsforAandBare

=

=

−

× +

(0)

1

1 12

(0) (0)22

A i

B

LC RC

vL

iRC

L

L


Thesettlingtimeofawaveformisthetimetsettlethatittakesthewaveform(iL)toreachwithinapercentageofsomespecifiedvalueandstaywithinthatrange(seeFigureP4.4).Forthisproblem,wewilltakethesettlingtimeforiL(t)tobewithin±10%oftheinitialvalueiL(0)andstaywithinthatrange.WriteaMATLABprogramthatdeter-minesthesettlingtimeforthefollowingcircuitparameters:

= = = = = ≤ ≤4 , 1.0mH, 1.0 F, (0) 0.25A, (0) 6 V, 0 1000 sR R L C i v tcrit L

1. CreateaplotofiL(t)versust.Onthesameplot,createthelines0.1 (0)iL and−0.1 (0)iL .

2. Use the search method and MATLAB’sfzero function to determine thesettlingtimeofiL(t).Hint:Use | ( )|i tL andthe 0.1 (0)iL linetofindthemaxi-mumroot,thusdeterminingthesettlingtime.

Project.4.5

An n-channel metal-oxide-semiconductor field-effect transistor (MOSFET) isshowninFigureP4.5a.TheMOSFEThasthreeterminals:gate, source,anddrain.(Actually,MOSFETsalsohaveafourthbulkterminal,whichwedonotdiscusshere.)InMOSFETcircuits,weareusuallyconcernedwiththedraincurrentiDas

0.05

–0.05tsettle

t

iL

0.5

0

Figure P4.4 The settling time tsettle is defined as the time it takes for iL to reach and stay within some error band of its final value. This problem uses an error band of ±10%, but error bands of ±5% or ±2% are also common.


iD

vGS+

−

vDS

+

−

Drainterminal

Sourceterminal

Gateterminal

(a)

Typical Output Characteristics

30Drain to Source Voltage VDS (V)

4020 5010

Dra

in C

urre

nt I D

(A)

0

2

4

6

8

10

VGS = 10 V TC = 25°C

12

0

5

6

7

8

9

Pch = 100 W3

4

(b)

Figure P4.5 (a) An n-channel MOS transistor and its defined terminal param-eters. (b) Terminal characteristics of the 2SK1056 power MOSFET. (Courtesy of Renesas Electronics Corporation.)


afunctionoftheterminalvoltages.AcommonmathematicalmodelforthedraincurrentiDforann-channelMOSFET[1]isasfollows:

( )

( )

( )

=

≤

− − − λ > ≤ −

− − λ > > −

0 for

2 (1 ) for and ( )

1 for and ( )

2

2

i

v V

K v V v v v v V v v V

K v V v v V v v V

D

GS th

GS th DS DS DS GS th DS GS th

GS th DS GS th DS GS th

(P4.5a)

whereiDisdefinedinFigureP4.5a;vGSandvDS arethetransistorgate-to-sourceanddrain-to-sourcevoltages,respectively;Vthisthetransistor“thresholdvoltage”;andKandλareconstantsthataredependentonsilicondopinglevels,electronmobility,anddevicegeometry.FigureP4.5bshowsaplotoftypicaliDasafunctionofvDS(forvariousfixedvaluesofvGS)forthe2SK1056powerMOSFET.

1. CreateaMATLABfunctionid _ powerfetthattakestwoargumentsvds and vgs and computes the MOSFET current id as defined byEquation(P4.5a).Take

= = λ =0.2V, 0.2, 0.01V Kth

2. Plot iD as a function of vDS for ʺ ʺ0 50VvDS for values of=1,2, 3, 4, 5, 6, 7, and 8vGS V.Yourplotshouldlooksimilar(butnotidenti-

cal)totheoneinFigureP4.5b. 3. Becausethe2SK1056isratedforamaximumof100W,FigureP4.5bdoesnot

showvaluesofiDunderconditionswherethetotaldevicepower = ×P i vD DS exceeds 100 W. Modify your function definition for id _ powerfet tocheckthedevicepowerandreturnNaN(notanumber)ifthepowerexceeds100W.RerunyourplotandconfirmthatitnowresemblesFigureP4.5b.

4. UsethefzerofunctiontofindthevalueofvGSwhenvDS=5VandiD=2A.Hint:Forthisproblem,youwillneedtosolveid _ powerfetasafunctionofvGSwith andv iDS D asfixedparameters.

Project.4.6

The simplest digital logic gate is the inverter (Figure P4.6a), which converts aBoolean“true”into“false”andviceversa.Inelectronics,weoftenrepresentalogi-caltrueasahighvoltage(e.g.,+5V)andfalseasalowvoltage(e.g.,0V).Thus,fromanelectronicstandpoint,an inverter isacircuit thatacceptsa+5Vinputandoutputs0Vandacceptsa0Vinputandoutputs+5V.Asimplecircuitthatimplements these requirements is the resistor-MOSFETcircuitofFigureP4.6bandisaformofresistor-transistor logic(RTL).Inthiscase,ann-channelMOSFETisloadedwitha2-kΩresistor,VDDisthepowersupplyvoltage(typically5V),vinistheinput(andisequaltovGS),andvout istheoutput(andisequaltovDS).

UsingKirchhoff’svoltagelaw,wecanwriteanequationfortheoutputvoltageloop:

− − = 0V v vDD R DS (P4.6a)


ByOhm’slaw,

=v i RR D (P4.6b)

SubstitutingEquation(P4.6b)intoEquation(P4.6a)gives

− − = 0V i R vDD D DS (P4.6c)

whereiDisdeterminedbyEquation(P4.5a).CreateaMATLABprogramthatusesMATLAB’sfzerofunctiontodeter-

mine the roots of Equation (P4.6c). Use the following n-channel MOSFETparameters:VDD=5V,Vth=0.7V,K =0.0005,andλ =0.05.Take0≤vDS≤5Vinstepsof0.1Vand0≤vGS≤5Vinstepsof0.1V.Inactuality,arangeofvoltagesisconsideredavalid“low”oravalid“high”signal.FortheRTLinverterdescribedhere,assumethattherange0−1Vconstitutesavalidlowand4−5Vconstitutesavalidhigh.

1. Createatableofvoutversusvin(correspondingtovDSvs.vGS). 2. Plotvoutversusvin 3. Does thecircuitdescribedherebehave likean inverter; that is,doesahigh

inputproducealowoutputandviceversa? 4. Theinverternoise marginisdefinedastheamountofnoiseallowedonavalid

inputsuchthattheoutputremainsvalid.Forexample,ifvinraisedabovethelegalrangeforalow,sayto1.1V,doestheoutputremainavalidhigh?Findthehighestlowvinthatstillproducesavalidhighvout.Also,findthelowesthighvinthatstillproducesavalidlowvout.Printthesevaluestothescreen.

vin vout

(a)

R

iD

vGS+

−

vDS

+

−

vR+

−

vin

vout

VDD

(b)

Figure P4.6 (a) Logic inverter. (b) RTL inverter circuit using an n-channel MOS transistor.


Project.4.7

TheRTLlogicfamilyofProject4.6wastypicallyusedinthe1950swhentransis-torswereexpensive(andpackagedindividually)andresistorswerecheap.Whenintegratedcircuitswereinventedinthe1960s,theoppositebecametrue:transis-torsbecameverycheap, and resistorsbecamecomparatively expensive (becausetheyareinefficienttoimplementonachip).

FigureP4.7ashowsacomplementary MOSFET(CMOS)inverter.Inthiscase,theresistorisreplacedbyap-channelMOSFET.Wecanmathematicallymodel

vGS,n+

−

vDS,n

+

−

vin vout

VDD

vGS,p−

+vDS,p−

+iD,p

iD,n

(a)

vDS,p

−

+

vGS,p

−

+

iD,p

Sourceterminal

Gateterminal

Drainterminal

(b)

Figure P4.7 (a) CMOS inverter circuit. (b) A p-channel MOS transistor and its defined terminal parameters.


thep-channelMOSFETsimilarlytothen-channelMOSFETofProject4.5;thatis,thep-channeldraincurrentiD,Pisgivenby

=

≥

− − − − λ < ≥ −

− − − λ < < −

0 for

2( ) (1 ) for and ( )

( ) (1 ) for and ( )

,

, ,

, , , ,2

, , , , , ,

, ,2

, , , , , ,

i

v V


K v V v v V v v V

D p

GS p th p

p GS p th p DS p DS p p DS p GS p th p DS p GS p th p

p GS p th p p DS p GS p th p DS p GS p th p

(P4.7a)

where ,vGS p , ,vDS p ,and ,iD p areasdefinedinFigureP4.7b.Notethatthep-channelmodelmentionedissimilartothen-channelmodelofEquation(P4.5a)exceptforsomesignchanges.Inaddition,foragivenintegratedcircuittechnology,thevaluesofVth,K,andλwillbedifferentforp-versusn-channelMOSFETs(andthuswillbedenotedwithpandnsubscripts,respectively).Kirchhoff’svoltagelawisusedtorelatethevoltagesintheoutputsideofthecircuit:

=,v vDS n out (P4.7b)

= −,v v VDS p out DD (P4.7c)

Similarly,ontheinputside:

=,v vGS n in (P4.7d)

= −,v v VGS p in DD (P4.7e)

WecanalsoapplyKirchhoff’scurrentlawattheoutputnode:

+ = 0, ,i iD n D p (P4.7f)

wherethen-channelsourcecurrent ,iD n isdefinedas

( )=

≤

− − + λ > ≤ −

− + λ > > −

0 for

2( ) (1 ) for and ( )

( ) (1 ) for and ( )

,

, ,

, , , ,2

, , , , , ,

, ,2

, , , , ,

i

v V


K v V v v V v v V

D n

GS n th n

n GS n th n DS n DS n n DS n GS n th n DS n GS n th n

n GS n th n n DS n GS n th n DS n GS n th n

(P4.7g)

1. CreateaMATLABprogramusingthefzero function toobtaintherootsofEquation(P4.7f)byapplyingequations(P4.7a)and(P4.7g).TakeVDD=5V,Vth,n=0.7V,Vth,p=–0.8V,Kn=0.0005,Kp=0.00015,λn=0.05,andλp=0.1.Assumeʺ ʺ0 5vout Vinstepsof0.1Vand ʺ ʺ0 5vin Vinstepsof0.1V.


2. Createatableofvoutversusvin. 3. Plotvoutversusvin. 4. Doesthecircuitdescribedherebehavelikeaninverter?Thatis,doesahigh

inputproducealowoutputandviceversa?

Reference 1. A.S. Sedra and K.C. Smith, Microelectronic Circuits, 3rd ed., Saunders College,

Philadelphia,1989.

129

Chapter 5

Numerical Integration

5.1 IntroductionInthischapter,wederiveSimpson’sruleforapproximatingthesolutionofdefiniteintegrals.Wethendemonstrateusageofthebuilt-inMATLAB®functionsquadanddblquadforevaluatingdefiniteintegrals.Finally,weapplythesefunctionstosolvesomecommonproblemsinelectromagneticfieldtheory.

5.2 Numerical Integration and Simpson’s RuleWewanttoevaluatetheintegralIwhere

∫= ( )I f x dxA

B

(5.1)

ThestepsforcalculatingInumericallyareasfollows:

◾ Subdividethexaxisfromx =Atox =BintoNsubdivisions,whereNisaneveninteger.

◾ Simpson’sruleconsistsofconnectinggroupsofthreepointsonthecurvef(x)bysecond-degreepolynomials(parabolas)andsummingtheareasundertheparabolastoobtaintheapproximateareaunderthecurve(seeFigure 5.1).


Weproceedbyfirstexpandingf(x)inaTaylorseriesaboutxiusingthreeterms,thatis,

= − + − +( ) ( ) ( )2f x a x x b x x ci i (5.2)

wherea,b,andcareconstantstobedetermined.Then,theareaundertwoadjacentstripsA2stripsiscomputedbyintegratingf(x)

betweenxi–1andxi+1:

∫ ∫= = − + − +

−

+

−

+

( ) [ ( ) ( ) ]2 strips2

1

1

A f x dx a x x b x x c dxx

x

i i

x x

x x

i

i

i

i

(5.3)

Let ξ = −x xiThen,

dξ =dx

When = − ,x x xi ξ =−Δx,andwhen = + ,x x xi ξ =Δx.MakingthesesubstitutionsintoEquation(5.3)gives

( )

3 2

3( )

3( )

2( )

2( ) ( ) ( )

2 strips

2

3 2

3 3 2 2

1

1

A f x dx

a b c d

a bc

ax

ax

bx

bx c x c x

x

x

x

x

x

x

i

i

∫

∫

=

= ξ + ξ + ξ

=ξ

+ξ

+ ξ

= − − + − − + − −

−

−

−

+

x1

f1

xi

fi

xi−1

fi−1

xi+1

fi+1

xN+1

fN+1

∆ x ∆ x

x

. . . . . .

. . .

. . .. . .

. . .

A2strips

f (x)

Figure 5.1 Arbitrary three points on the curve f(x).

Numerical Integration ◾ 131

Thus,

= +23( ) 22 strips

3Aa

x c x

Now,define = ( )f f xi i .SolvingEquation(5.2)forfi, +1fi ,and −1fi gives

= =

= = + +

= = − + − +

+ +

− −

( )

( ) ( )

( ) ( ) ( )

1 12

1 12

f x f c

f x f a x b x c

f x f a x b x c

i i

i i

i i

Addingthelasttwoequationsgives + = ++ − 2 21 12f f a x ci i .

Solvingforagives

= + −+ −

12

[ 2 ]2 1 1ax

f f fi i i

Then,

= × + − + = + − ++ − + −23

12

[ 2 ]( ) 23[ 2 6 ]2 strips 2 1 1

31 1A

xf f f x f x

xf f f fi i i i i i i i

or

= + +− +3

[ 4 ]2 strips 1 1Ax

f f fi i i (5.4)

ToobtainanapproximationfortheintegralI,weneedtosumallthetwo-stripareasunderthecurvefromx =Atox =B(seeFigure 5.2),thatis,

= + +

= + +

= + +

= + +− +

3[ 4 ]

3[ 4 ]

3[ 4 ]

3[ 4 ]

1 1 2 3

2 3 4 5

3 5 6 7

/2 1 1

Ax

f f f

Ax

f f f

Ax

f f f

Ax

f f fN N N N


Thus,

∫= = + + + + + + +=

=

+

+

( )3[ 4 2 4 2 4 ]1 2 3 4 5 1

1

1

I f x dxx

f f f f f f fx A

x B

N N

N

(5.5)

ThisisSimpson’sruleforintegration.

Example 5.1

Solve by Simpson’s rule:

∫= + − +I x x x dx( 3.2 3.4 20.2)3 2

0

10

%Example_5_1.m%ThisprogramcalculatesanintegralbySimpson'sRule%Theintegrandis:x^3+3.2*x^2-3.4*x+20.2%Thelimitsofintegrationarefrom0-10.clear;clc;A=0;B=10;N=100;dx=(B-A)/N;%Computevaluesofxandfateachpoint:x=A:dx:B;f=x.^3+3.2*x.^2-3.4*x+20.2;%Usetwoseparateloopstosumuptheevenandoddterms%ofSimpson'sRule.Also,excludeendpointsintheloop.sum_even=0.0;fori=2:2:Nsum_even=sum_even+f(i);end

f3f2f1

fN

x1 x2 x3 xN

A1 A2A3

f (x)

x. . .

. . .

f6f5f4 f7. . .

x7x6x5x4 xN+1xN−1

fN−1

fN+1

AN/2

Figure 5.2 Integration areas for Simpson’s rule.


sum_odd=0.0;fori=3:2:N-1sum_odd=sum_odd+f(i);end%CalculateintegralasperEquation5.5I=dx/3*(f(1)+4*sum_even+2*sum_odd+f(N+1));%Displayresultsfprintf('Integrand:x^3+3.2*x^2-3.4*x+20.2\n');fprintf('Integrationlimits:%.1fto%.1f\n',A,B);fprintf('I=%10.4f\n',I);

--------------------------------------------------------------------------

PROGRAM RESULTSIntegrand:x^3+3.2*x^2-3.4*x+20.2Integrationlimits:0.0to10.0I=3598.6667

5.3 Improper IntegralsAnintegralisimproperiftheintegrandapproachesinfinityateitheroftheend-points.Inmanycases,theintegrationwillstillresultinafinitesolution.

Example 5.2

∫=

+I

xx

dxlog(1 )

0

1

(5.6)

This integral is improper since both the numerator and denominator are zero at the lower limit (x = 0). The exact value of I can be obtained by residue

theory in complex variables and in this case evaluates to =π

=I12

0.822467.2

Let

∫=

+= +I

xx

dx I Ilog(1 )

0

1

1 2

where

∫=

+

ε

Ix

xdx

log(1 )1

1

and

∫=

+ε

Ix

xdx

log(1 )2

0


where ε << 1. To evaluate I2, expand log(1 + x) in a Taylor series about x = 0, giving

+ = − + − + − +x x x x x xlog(1 )

12

13

14

15

2 3 4 5

then

+= − + − + − +

xx

x x x xlog(1 )

112

13

14

15

2 3 4

∫= − + − + − + − +

ε

I x x x x x x dx112

13

14

15

16

17

22 3 4 5 6

0

= ε − ε + ε − ε + ε − ε + ε − +I

14

19

116

125

136

149

22 3 4 5 6 7 (5.7)

Evaluate I1 by Simpson’s rule and evaluate I2 by Equation (5.7). The fol-lowing program illustrates the method:

%Example_5_2.m%Thisprogramevaluatestheimproperintegral:%log(1+x)/xThelimitsofintegrationarefrom0to1.%Theintegrandisundefined(0/0)atx=0.Thus,the%integralisbrokenupinto2parts:I1andI2.%I1isevaluatedfromepsilonto1.%I2isexpandedinaTaylorSeriesandevaluatedfrom0%toepsilon.clear;clc;epsilon=0.0001;A=0;B=1;N=100;dx=(B-epsilon)/N;%EvaluateI1.First,evaluatelog(1+x)/xforeachvalue%ofxovertheinterval[epsilon,1].x=epsilon:dx:B;f=log(1+x)./x;%Next,calculatetheevenandoddtermsforSimpson's%Rule.sumeven=0.0;fori=2:2:Nsumeven=sumeven+f(i);endsumodd=0.0;fori=3:2:N-1sumodd=sumodd+f(i);end%AsperEquation(5.5),I1istheweightedsumofthe%evenandoddterms,plustheendterms.


I1=dx/3*(f(1)+4*sumeven+2*sumodd+f(N+1));fprintf('Integrand=log(1+x)/x\n');fprintf('I1limitsofintegrationare');fprintf('from%.4fto%.4f\n',epsilon,B);fprintf('I2limitsofintegrationare');fprintf('from%.4fto%.4f\n',A,epsilon);fprintf('I1=%15.6f\n',I1);%CalculateI2viafirst4termsofTaylorseries%expansion.I2=epsilon-1/4*epsilon^2+1/9*epsilon^3...-1/16*epsilon^4;fprintf('I2=%15.6f\n',I2);I=I1+I2;fprintf('I=I1+I2=%10.6f\n',I);I_exact=0.822467;fprintf('I_exact=%10.6f\n',I_exact);

--------------------------------------------------------------------------

PROGRAM RESULTSIntegrand=log(1+x)/xI1limitsofintegrationarefrom0.0001to1.0000I2limitsofintegrationarefrom0.0000to0.0001I1= 0.822367I2= 0.000100I=I1+I2=0.822467I_exact= 0.822467

As we see, the method works quite well.

5.4 MATLAB’s quad FunctionThe MATLAB function for evaluating an integral is quad. A description ofthefunctioncanbeobtainedbytypinghelp quadintheCommandwindow(excerptedinthefollowing):

Q=QUAD(FUN,A,B)triestoapproximatetheintegralofscalar-valuedfunctionFUNfromAtoBtowithinanerrorof1.e-6usingrecursiveadaptiveSimpsonquadrature.FUNisafunctionhandle.ThefunctionY=FUN(X)shouldacceptavectorargumentXandreturnavectorresultY,theintegrandevaluatedateachelementofX.

Q=QUAD(FUN,A,B,TOL)usesanabsoluteerrortoleranceofTOLinsteadofthedefault,whichis1.e-6.LargervaluesofTOLresultinfewerfunctionevaluationsandfastercomputation,butlessaccurateresults.TheQUADfunctioninMATLAB5.3usedalessreliablealgorithmandadefaulttoleranceof1.e-3.


Thus, the quad function takes as arguments a function handleFUN thatdefines the integrand in a .m file and the limits of integration. Alternatively. iftheintegrandisnotverylargeandcanbeexpressedinasingleline,thenyoucandefinetheintegrandwithinyourscriptwithananonymousoraninlinefunction(seeExamples5.4and5.5).Iftheintegrandinvolvesverysmallnumbersorverylargenumbers,youmightwishtochangethedefaulttolerancebyaddingathirdargumenttoquad(asshowninthesecondusagedescription).Thequadfunctionisabletoevaluatecertainimproperintegrals(seeExercises5.2d,5.2e,and5.2f).

Example 5.3

We now repeat Example 5.1, but this time we use MATLAB’s quad function to do the integration. The integral I in Example 5.1 is

∫= + − +I x x x dx( 3.2 3.4 20.2)3 2

0

10

The program follows:

%Example_5_3.m%Thisprogramevaluatestheintegralofthefunction%'f1'betweenAandBbyMATLAB'squadfunction.Since%thefunction'f1'isjustasingleline,wecanusethe%anonymousformofthefunction.clear;clc;f1=@(x)(x.^3+3.2*x.^2-3.4*x+20.2);A=0.0;B=10.0;I=quad(f1,A,B);fprintf('Integrationoff1over[%.0f,%.0f]',A,B);fprintf('byMATLAB''squadfunction:\n');fprintf('f1=x^3+3.2*x^2-3.4*x+20.2\n');fprintf('integral=%10.4f\n',I);

--------------------------------------------------------------------------

PROGRAM RESULTSIntegrationoff1over[0,10]byMATLAB'squadfunction:f1=x^3+3.2*x^2-3.4*x+20.2integral=3598.6667

We see that the results are the same as those obtained in Example 5.1.

Example 5.4

∫ + +

tt t

dtEvaluate:13

0

1


%Example_5_4.m%Thisprogramevaluatestheintegraloffunction"f2"%betweenAandBbyMATLAB'squadfunction.The%integrandcanbeexpressedinasinglelineandthuswe%canusetheanonymousformofthefunction.clear;clc;A=0.0;B=1.0;f2=@(t)t./(t.^3+t+1.0);I2=quad(f2,A,B);fprintf('Integrationoff2over[%.0f,%.0f]',A,B);fprintf('byMATLAB''squadfunction:\n');fprintf('f2=t/(t^3+t+1)\n');fprintf('integral=%f\n',I2);

--------------------------------------------------------------------------

PROGRAM RESULTSIntegrationoff2over[0,1]byMATLAB'squadfunction:f2=t/(t^3+t+1)integral=0.260068

Let us repeat Example 5.2, in which we evaluated the improper integral described by Equation (5.6), but this time we evaluate the integral using MATLAB’s quad function with limits from 0 to 1. Recall that the function log(1 + x)/x is undefined at x = 0. In addition, we use the inline method for specifying the function.

Example 5.5

%Example_5_5.m%Thisprogramevaluatestheimproperintegrallog(1+x)/x%withlimitsfrom0to1usingquad.clear;clc;A=0;B=1;fprintf('Thisprogramusesthequadfunctionto\n');fprintf('evaluatetheintegraloflog(1+x)/xfrom');fprintf('%.0fto%.0f.\n',A,B);I=quad(inline('log(1+x)./x'),A,B);fprintf('I=%10.6f\n',I);--------------------------------------------------------------------------

This program produces the following result:

Thisprogramusesthequadfunctiontoevaluatetheintegraloflog(1+x)/xfrom0to1.I=0.822467

This answer is the same as shown in Example 5.2.

5.5 The Electric FieldElectricfieldproblemsofteninvolveintegration,andbeforeweproceed,wereviewsomebasicconceptsaboutelectromagneticfields.

Webeginby assuming that apoint chargeofQ coulombs is located at thecoordinates ( , , )x y zp p p infreespace.WedenotethelocationofQbythevector rp


(seeFigure 5.3).TheelectricfieldE attheobservationpoint ro (correspondingto

thecoordinates ( , , )x y zo o o )duetothechargeat rp isdefinedas

=

πε4ˆ2E

re

Q

or (5.8)

where εo isthepermittivityoffreespace( × −8.85 10 12 farad/m), r isthevector

fromthepointchargetothepointofmeasurement, er istheunitvectorinthe r

direction,andtheresultingE isinvolts/meter.FromFigure 5.3,wecanobserve

thevectorsum = +r r ro p ,andthus

= −r r ro p .Then,Equation(5.8)becomes

=

πε −

−

−42E

r r

r rr r

Q

o o p

o p

o p (5.9)

wherewehaveappliedthedefinitionoftheunitvector=errr .

Wenowbreakdown ro and rp intotheirx,y,andzcomponentssothatwecanexpress them vectorially in Cartesian coordinates as

= + +ˆ ˆ ˆr e e ex y zo o x o y o z and

= + +ˆ ˆ ˆr e e ex y zp p x p y p z ,where e x , e y,and ez areunitvectorsinthex,y,andzdirections,respectively.Then,

− = − + − + −( ) ( ) ( )2 2 2r r x x y y z zo p o p o p o p ,andEquation(5.8)becomes

( )=

πε

− + − + −

− + − + −4( ) ( ) ( )

( ) ( ) ( )2 2 232

Ee e eQ x x y y z z

x x y y z zo

o p x o p y o p z

o p o p o p

(5.10)

Thus,Equation (5.10)gives theelectricfieldat the location ( , , )x y zo o o due toapointchargeofmagnitudeQlocatedat ( , , )x y zp p p .

y

x

z

+

Point charge Qlocated at rp

Measurementlocation ro

r

ro

E

rp

Figure 5.3 The electric field E at observation point

ro due to a point charge

located at rp .


Wenowwishtocalculatetheelectricfieldduetoalineofpointchargesextend-ingalongthez-axisfrom−∞to∞(seeFigure 5.4).Weshallassumethatthechargesareevenlyspacedalongthelinewithalinear charge density representedbythesym-bolλwithunitscoulomb/meter.WecanrewriteEquation(5.10)indifferentialformtoobtain an expression for

Ed in termsof thedifferentialquantity = λdQ dz p,

where dz p representsaninfinitesimalsegmentofthelinechargeinthezdirection:

( )=λ

πε

− + − + −

− + − + −4( ) ( ) ( )

( ) ( ) ( )2 2 232

Ee e e

ddz x x y y z z

x x y y z z

p

o

o p x o p y o p z

o p o p o p

(5.11)

SeparatingtheindividualdirectionalcomponentsofEquation(5.11)andinte-gratingfrom−∞to+∞ gives

∫ ( )

=λ

πε

−

− + − + −−∞

∞

4 ( ) ( ) ( )2 2 232

Edz x x

x x y y z zx

p

o

o p

o p o p o p

(5.12)

∫ ( )

=λ

πε

−

− + − + −−∞

∞

4 ( ) ( ) ( )2 2 232

Edz y y

x x y y z zy

p

o

o p

o p o p o p

(5.13)

∫ ( )

=λ

πε

−

− + − + −−∞

∞

4 ( ) ( ) ( )2 2 232

Edz z z

x x y y z zz

p

o

o p

o p o p o p

(5.14)

MeasurementlocationLine charge of density

λ coul/m along ±z axis

Differential linecharge element λdz

x

z

r

ro

rp

dE

Figure 5.4 The differential electric field

dEr at observation point ro due to a

differential line charge element λdz located along the z axis.


Example 5.6

Given a line charge located along the z axis with density λ = × −2 10 coul/m,9 calculate the electric field at the location =x y zo o o( , , ) (0.005, 0, 0).

%Example_5_6.m%Thisprogramusesthequadfunctiontoevaluatethe%integralsofequations(5.12),(5.13)and(5.14)to%findEx,EyandEz.Thefollowingparametersareused%intheintegration:lambda=2e-9coul/m,xo=5e-3,yo=0%andzo=0.%Assumethat10isagoodapproximationforinfinity.clc;clear;globalxoyozolambdaepsilon0;lambda=2e-9;epsilon0=8.85e-12;xo=5e-3;yo=0;zo=0;Ex=quad(@linecharge_Ex,-10,10);Ey=quad(@linecharge_Ey,-10,10);Ez=quad(@linecharge_Ez,-10,10);fprintf('E-fieldduetoalinecharge');fprintf('at[%5.3f,%5.3f,%5.3f]:\n',xo,yo,zo);fprintf('[Ex,Ey,Ez]=[%10.4e,%5.2f,%5.2f]V/m\n',...Ex,Ey,Ez);

--------------------------------------------------------------------

%linecharge_Ex.m:FunctiontocalculateExat(xo,yo,zo)%foralinechargeonthezaxisfunctionEx=linecharge_Ex(zp)globalxoyozolambdaepsilon0;Ex=lambda/(4*pi*epsilon0)*xo..../(xo.^2+yo.^2+(zo-zp).^2).^1.5;

------------------------------------------------------------------------

%linecharge_Ey.m:FunctiontocalculateEyat(xo,yo,zo)%foralinechargeonthezaxisfunctionEy=linecharge_Ey(zp)globalxoyozolambdaepsilon0;Ey=lambda/(4*pi*epsilon0)*yo..../(xo.^2+yo.^2+(zo-zp).^2).^1.5;

------------------------------------------------------------------------

%linecharge_Ez.m:FunctiontocalculateEzat(xo,yo,zo)%foralinechargeonthezaxisfunctionEz=linecharge_Ez(zp)globalxoyozolambdaepsilon0;Ez=lambda/(4*pi*epsilon0)*(zo-zp)..../(xo.^2+yo.^2+(zo-zp).^2).^1.5;

------------------------------------------------------------------------

PROGRAM RESULTSE-fieldduetoalinechargeat[0.005,0.000,0.000]:[Ex,Ey,Ez]=[7.1934e+003,0.00,0.00]V/m


5.6 The quiver PlotAquiverplotisatwo-dimensionalplotofavectorfunctionthatusessmallarrowstorepresentthelocalvectormagnitudeanddirection.Forexample,aquiverplotcanbeusedtoshowanelectricormagneticfield(i.e.,theoutputfromthevectorfunc-tion)asafunctionofposition(i.e.,theinputtothevectorfunction).TheMATLABcommandisquiver(X,Y,U,V).Toseethesyntax,runhelp quiver intheCommandwindow:

QUIVER(X,Y,U,V)plotsvelocityvectorsasarrowswithcomponents(u,v)atthepoints(x,y).ThematricesX,Y,U,Vmustallbethesamesizeandcontaincorrespondingpositionandvelocitycomponents(XandYcanalsobevectorstospecifyauniformgrid).QUIVERautomaticallyscalesthearrowstofitwithinthegrid.

Beforewecanusethequiverfunction,weneedtoestablishan(X,Y)gridofobservationpoints,andthesimplestwaytoaccomplishthisistouseMATLAB’smeshgridfunction.Themeshgridcommandisinvokedasfollows:

[X,Y]=meshgrid(x,y)

wherexandyarevectorsofvaluesoverwhichtocreatethegrid.Thefollowingexamplecreatesagridoverthearea−2≤x≤2and−2≤y≤2withastepsizeof1byusingthemeshgridcommand.Thegridthatiscreatedwillconsistoftwoarraysof25x-y coordinates (onearrayofx coordinatesanda secondarrayofycoordinates).

Example 5.7

%Example_5_7.m:exampleusageofmeshgrid()fprintf('Thisprintoutisfrommeshgrid():\n');x=-2:1:2;y=-2:1:2;[XY]=meshgrid(x,y)------------------------------------------------------------------------

PROGRAM RESULTSThisprintoutisfrommeshgrid():X= -2 -1 0 1 2 -2 -1 0 1 2 -2 -1 0 1 2 -2 -1 0 1 2 -2 -1 0 1 2


Y= -2 -2 -2 -2 -2 -1 -1 -1 -1 -1 0 0 0 0 0 1 1 1 1 1 2 2 2 2 2

Thus, meshgridis a very compact way of generating the coordinate sets.As can be seen, the number of elements in the grid will equal the number of elements in xtimes the number of elements in y. For each grid point, there is an X element and a Y element.

Example 5.8

To demonstrate the quiver plot, we will calculate the electric field on the x-z plane resulting from the line charge along the z axis of Example 5.6.

%Example_5_8.m%ThisprogramplotstheelectricfieldontheX-Zplane%foralinechargelocatedontheZaxisandextending%toinfinityinbothZdirections.Thecalculationis%madeovertheintervalxo=[-.01,.01]mandzo=%[-.01,.01]mwithastepsizeof0.002m.Thisexample%reusesthefunctionsalreadydefinedinExample5.6.clc;clear;globalxoyozolambdaepsilon0;lambda=2e-9;epsilon0=8.85e-12;%formeasurementsontheX-Zplane,yowillalwaysbe0:yo=0;%FirstcreateagridofX-Zlocationsforwhich%calculatetheelectricfield[X,Z]=meshgrid(-10e-3:2e-3:10e-3,-10e-3:2e-3:10e-3);%CalculatetheE-fieldateachX-Zlocation,andstore%theresultinthematricesExandEzfori=1:length(X(:,1))forj=1:length(Z(1,:))xo=X(i,j);zo=Z(i,j);%NOTE:don'tcalculatethefieldwhensituated%ONthelinechargebecauseitwillbeinfinity.%Instead,usethevalue"NaN"(whichwillcause%nopointtobeprintedonthequiverplot).ifxo==0Ex(i,j)=NaN;Ez(i,j)=NaN;elseEx(i,j)=quad(@linecharge_Ex,-10,10);Ez(i,j)=quad(@linecharge_Ez,-10,10);endendend%Plottheelectricfield


quiver(X,Z,Ex,Ez);title('E-fieldduetoalinechargealongtheZaxis');xlabel('xaxis');ylabel('zaxis');

------------------------------------------------------------------------

Figure 5.5 shows the resulting quiver plot.

5.7 MATLAB’s dblquad FunctionTheMATLABfunctionfornumericallyevaluatingadoubleintegralisdblquad.Adescriptionofthefunctioncanbeobtainedbytypinghelp dblquadintheCommandwindow:

Q=DBLQUAD(FUN,XMIN,XMAX,YMIN,YMAX)evaluatesthedoubleintegralofFUN(X,Y)overtherectangleXMIN<=X<=XMAX,YMIN<=Y<=YMAX.FUNisafunctionhandle.ThefunctionZ=FUN(X,Y)shouldacceptavectorXandascalarYandreturnavectorZofvaluesoftheintegrand.

-0.015 -0.01 -0.005 0 0.005 0.01 0.015-0.015

-0.01

-0.005

0

0.005

0.01

0.015Electric field due to a line charge along the Z axis

x axis

z ax

is

Figure 5.5 Vector plot of the electric field in the x-z plane due to a line charge along the z axis.


Non-squareregionscanbehandledbysettingtheintegrandtozerooutsideoftheregionofinterest.Forexample,thevolumeofahemisphereofradiusRcanbedeterminedbythedblquadfunctionbysetting-R<=x<=Rand0<=y<=Randsettingz=0forpoints(x,y)thatlieoutsidethecircleofradiusRaroundtheorigin.

Theusageofdblquadissimilartoquad,except

◾ FUNmustbeafunctionoftwovariables(insteadofone):avectorofvaluesinthexdirectionandasinglevalueintheydirection.

◾ Theremustbetwosetsofintegrationlimits(oneforthexdirection,andonefortheydirection).

Example 5.9

Calculate the surface area and volume of a hemisphere with radius of 1 m.To find the surface area, we define a differential area element = φ θ φdA R d dsin2 (as shown in Figure 5.6) and then double integrate over

the intervals φ = π[0, 2] and θ = [0,2π] using dblquad. To find the volume,

θ

φ

R

R sinφdθ

dφ

x

y

z

dA = (R sinφ dθ)(Rdφ)

R

Figure 5.6 To find the surface area of a hemisphere, we define a differential surface element dA and then do a double integration over dϕ and dθ.


we define a differential volume element = − −dV R x y dxdy2 2 2 (as shown in Figure 5.7) and double integrate over the intervals x = [−R,R] and y = [−R,R].

%Example_5_9.m%Thisprogramcalculatesthesurfaceareaandvolumeof%ahemisphere(withradius=1)usingdblquad.Theexact%valuesfromformulaearealsocalculated.clear;clc;globalR;R=1;%calculatesurfaceareaSA=dblquad('hemisphere_dA',0,pi/2,0,2*pi);SA_exact=2*pi*R^2;%calculatevolumeV=dblquad('hemisphere_dV',-R,R,-R,R);V_exact=2/3*pi*R^3;%printresultsfprintf('SurfaceareaSAofahemisphereof');fprintf('radius%.4fm\n',R);fprintf('SAbyDBLQUAD=%.4fm^2\n',SA);fprintf('SAexact=%.4fm^2\n',SA_exact);fprintf('VolumeVofahemisphereofradius%.4fm\n',R);

R

differential volume element

dxdy

x

y

z

R2 – x2 – y2

R2 – x2 – y2dV = dxdy

Figure 5.7 To find the volume of a hemisphere, we define a differential volume element dV and then do a double integration over dx and dy.


fprintf('VbyDBLQUAD=%.4fm^3\n',V);fprintf(‘Vexact=%.4fm^3\n',V_exact);

-----------------------------------------------------------------------

%hemisphere_dA.m:Thisfunctiondefinestheintegrand%fordeterminingthesurfaceareaofahemisphereandis%usedinMATLAB'sdblquadfunction.function[dA]=hemisphere_dA(phi,theta)globalR;dA=R^2*sin(phi);

-----------------------------------------------------------------------

%hemisphere_dV.m:Thisfunctiondefinestheintegrand%z(x,y)fordeterminingthevolumeofahemisphereand%isusedinMATLABísdblquadfunction.Fromanalytical%geometry,theequationofasphereisX^2+Y^2+Z^2=%R^2.Thus,dV=Z*dX*dY,whereZ=sqrt(R^2-X^2-Y^2).Note%thatthehemisphereisonlydefinedfor(X,Y)points%whichliewithinacircleofradiusRaroundthe%origin,theregionthatistheprojectionofthe%hemisphereontotheX-Yplane.An'if'statementis%usedtosetZ=0for(X,Y)pointsoutsidethecircle%ofradiusR.Note:XisavectorandYisascalar.functionZ=hemisphere_dV(X,Y)globalR;fori=1:length(X)ifX(i)^2+Y^2<=R^2Z(i)=sqrt(R^2-X(i)^2-Y^2);elseZ(i)=0;endend-----------------------------------------------------------------------

PROGRAM RESULTSSurfaceareaSAofahemisphereofradius1.0000mSAbyDBLQUAD=6.2832m^2SAexact =6.2832m^2VolumeVofahemisphereofradius1.0000mVbyDBLQUAD=2.0944m^3Vexact =2.0944m^3

Exercises

Exercise.5.1

UseaTaylorseriesexpansionoflog(1+x)aboutx=0toshowthat

+ = − + − + − +log(1 ) 1

213

14

15

2 3 4 5x x x x x x


Exercise.5.2

UseMATLAB’squadfunctiontoevaluatethefollowingintegrals;notethatinte-gralsd,e,andfareimproperintegrals:

∫= + −a.5 23 3

0

3

Idx

e ex x

∫=

−−π

π

b.sin1 4sin2

/2

/2

Ix dx

x

∫= −

π

c. (sinh cos )0

I x x dx

∫= −

d.31 2

0

1

Ie dx

x

x

∫= −

e.log(1 )

0

1

Ix dxx

∫= −

f.log(1 )2

0

1

Ix dxx

Projects

Project.5.1

ForthelinechargeofExample5.8,calculateandplot(usingquiver)theelectricfieldinthex-yplane(insteadofthex-zplane)fortheinterval−50≤x ≤50mmand−50≤y ≤50mmwithastepsizeof10mm.Remembernottocalculatethefieldatthelocationofthelinecharge.

Project.5.2

WenowmodifythelinechargeofExample5.8tobeofafinitelengthℓ =0.02malongthezaxis,startingatz =−0.01mandendingatz =+0.01m.Assumeλ = × −2 10 9 coul/m.

1. Calculateandplot(usingquiver)themagnitudeoftheelectricfieldfortheinterval−0.05≤x ≤0.05mand−0.05≤z ≤0.05mwithastepsizeof0.01m.


2. Printtothescreeneveryfourthvalueofx0,z0,andthecorrespondingvaluesofEx,Ez ,and themagnitudeof

E .

3. Now,assumethatthetotalchargeofthelinesegment = λ = × −0.04 10 9Q coulisconcentratedasapointchargeattheorigin.CalculatethemagnitudeoftheelectricfieldforthispointchargeoverthesameintervalaspreviouslyusingEquation(5.8).NotethatbecauseQisapointcharge,youcanuseEquation(5.8)directly,andnointegrationisrequired.

4. CalculatethepercentageerrorinusingthepointchargemethodinthevaluesofExandEzateachpointascalculatedinparts1and3.Usethefollowingformula forcalculating thepercentageerror inEx anda similar formula forcalculatingthepercentageerrorinEz.

=

−×% error in 100

,1 ,2

,1E

E EEx

x x

x

where

Ex,1=Excalculatedinpart1.

Ex,2=Ex calculatedinpart3.

Printtothescreeneveryfourthvalueofxo,zo,andthecorrespondingvaluesofthepercentageerrorinthevaluesofExandEz.

5. Atwhatdistance from theorigin is thepercentage error inusing thepointchargemethodlessthan5%?Inmanycases,ifwearefarenoughawayfromthecharge,wecanavoidintegratingoveritsgeometryandinsteadjusttreatitasapointcharge,therebymakingourcalculationsmucheasier.Thisiscalledthefar-field approximation.

Project.5.3

Apositivesurface charge densityofmagnitude σ = × −4 10 coul/m12 2 extendsinfi-nitelyinthex-yplaneasshowninFigureP5.3a.UsingamethodsimilartowhatwasdescribedinSection5.4(showyourwork),defineadifferentialchargeelementdQ =σdydxandthenfindtheresultingelectricfieldbyintegratingoverthesheetofchargefor = −∞ ∞[ , ]x p , = −∞ ∞[ , ]y p ,and = 0z p :

∫∫ ( )

=σ

πε

−

− + − + −−∞

∞

−∞

∞

4 ( ) ( ) ( )2 2 232

Edy dx x x

x x y y z zx

p p

o

p

o p o p o p

(P5.3a)

∫∫ ( )

=σ

πε

−

− + − + −−∞

∞

−∞

∞

4 ( ) ( ) ( )2 2 232

Edy dx y y

x x y y z zy

p p

o

p

o p o p o p

(P5.3b)

∫∫ ( )

=σ

πε

−

− + − + −−∞

∞

−∞

∞

4 ( ) ( ) ( )2 2 232

Edy dx z z

x x y y z zz

p p

o

p

o p o p o p

(P5.3c)


1. WriteaMATLABprogramusingthedblquadfunctiontocalculateEx,Ey,andEz atthefollowingpoints:

xo, yo, zo=[0,0,2× 10–3] (abovetheplate)

= × − × ×− − −, , [ 30 10 , 5 10 , 12 10 ]3 3 3x y zo o o (belowtheplate)

Assumethat10isagoodenoughapproximationforinfinity.Notethatbysymmetry,thexandycomponentsofthefieldshouldintegratetozero(orclosetoit).Confirmthisresult.

2. Now, assume that there are two sheet charges in the planes parallel to thex-yplane: thefirst sheet locatedat = +1z p mmwith surfacechargedensityσ = × −4 10 coul/m12 2 andthesecondsheetlocatedat = −1z p mmandoppo-sitelychargedwithsurfacechargedensity σ = − × −4 10 coul/m12 2 (seeFigureP5.3b).Findtheelectricfieldviasuperpositionbycalculatingtheelectricfield

y

x

zMeasurement point

Sheet charge with surfacecharge density σ coul/m2

Differential chargeelement σdxdy

r ro

rp

dE

(a)

y

x

z

Sheet charge with surfacecharge density σ coul/m2

located at zp = 1 mm

Sheet charge with surfacecharge density –σ coul/m2

located at zp = –1 mm

(b)

Figure P5.3 (a) The differential electric field Ed at observation point

ro due to a

differential sheet charge element σdxdy located on the x-y plane. (b) Two parallel and oppositely charged sheet charges located on the planes zp = +1mm and zp = –1mm.


separatelyforeachsheetandthenaddingthemtogether.FindEx,Ey,andEz forthethesethreepoints:

xo,yo,zo=[0,0,2× 10–3] (abovetheplates)

xo,yo,zo=[10× 10–3,0,–0.5× 10–3] (betweentheplates)

= × − × ×− − −, , [ 30 10 , 5 10 , 12 10 ] (below the plates)3 3 3x y zo o o

Printouttheresults.Dotheseresultsmakesense?

Project.5.4

TheBiot-Savartlawrelateselectricalcurrenttomagneticfield(seeFigureP5.4a)andisdefinedforcurrentflowthroughwiresas

4 2d

Ido rBe

r

=π

× (P5.4)

whereBd isthedifferentialmagneticfield(avector), o isthepermeabilityof

free space ( π × −4 10 7 henry/m), I is the current (in amperes),

d is adifferen-tialwirelengthwithdirectioncorrespondingtothecurrentflow,

r isthevector

fromthecurrentelementtothemeasurementpoint, er istheunitvectorinther direction,and×representsthevectorcrossproduct.InFigureP5.4a,wealsodefine

rp asthelocationofthecurrentelementand

ro asthepointofmeasure-

mentsuchthat r r ro p= − .

WewishtocalculatethemagneticfieldforaninfinitewireextendingalongthezaxisandcarryingacurrentofI= 1mAinthe+zdirection.Notethat,inreality,aninfinite-

I

y

x

z

Wire

Measurementlocation

r

ro

rp

dB

Differential currentelement Id

Figure P5.4a The differential magnetic field element Bd at observation point

ro

due to a differential wire element carrying current

Id .


lengthcurrent-carryingwireisimpossiblebecauseitviolatesconservationofcharge.Fornow,itissufficienttopresumethatthewireisconnectedtoacurrentsourceat±∞.

1. AsshowninFigureP5.4b,forawireinthezdirection, ed dz p zˆ= .Duetothe

crossproductinEquationP5.4,if

d isz directed,thentheresultingBd will

onlyhavecomponentsinthexandydirections.Writetwoseparateequationsfor dBx and dBy bywriting er intermsofits e x, e y,and ez components,computing thecrossproductsbetween

d and er and then separating the

xandycomponentsofBd intoseparateequations.Hint:Derive e x and e y

from er usingthemethodemployedinEquation(5.10). 2. Write two MATLAB functions for dBx and dBy that compute the x- and

y-directedmagneticfieldsintermsofasingleargumentzp.Useglobalvariablesforthemeasurementpositionparametersxo,yo ,andzo .Usethequadfunctiontointegrateovertheinterval = −∞ ∞[ , ]z p andcalculatethemagneticfieldatthesepoints:

= × ×− −, , [1 10 , 1 10 , 0]4 4x y z (firstquadrant)

= − × ×− −, , [ 1 10 , 1 10 , 0]4 4x y z (secondquadrant)

x,y,z=[–1×10–4,–1× 10–4,0] (thirdquadrant)

x,y,z=[1×10–4,–1× 10–4,0] (fourthquadrant)

Assume that 10 is a sufficient approximation for infinity. Print out the results.Commentonanysymmetryyoufindinthesesolutions.

3. Use the meshgrid function to create a set of points forro where

− ≤ ≤200 200xo mm, − ≤ ≤200 200yo mm, and = 0zo . Use a step size

I

x

z

Wire along z axiswith current I

Differential currentelement Idzpe z

r

ro

Magnetic field dB(directed into page) atmeasurement location rorp

Figure P5.4b The magnetic field due to a wire along the z axis will have only x and y components.


of50mmforbothxoandyo.ComputethemagneticfieldBxandBy ateachvalueof

ro andplotwithquiver.Wheniteratingoveryourvaluesfor

ro ,

besuretoskipthepointattheorigin.

Project.5.5

FigureP5.5ashowsacircularwirewithradiusRcenteredattheoriginandcarry-ingacurrentI.UsingtheBiot-SavartlawfromEquation(P5.4),wecanwritethedifferentialmagneticfieldintermsofadifferentialcurrentelement eId IRd pˆ

= θ θ ,

where θRd p isthedifferentiallength(incylindricalcoordinates)ofthecurrentele-mentatlocation

rp ,and eθ istheunitvectorinthe θ p direction.Inthistypeof

problemwithcylindricalsymmetry,thesimplestapproachistoconvertthecylin-dricalunitvectorsinto(constant)cartesianunitvectorsbutstillperformtheinte-grationusingthecylindricalvariables.FigureP5.5bshowshow

rp and eθ canbe

expressedintermsoftheirxandycomponentsusingtheunitvectors e x and e y :

ˆ cos ˆ sinr e eR Rp x p y p= θ + θ (P5.5a)

ˆ ˆ sin ˆ cose e ex p y p= − θ + θθ (P5.5b)

WecannowrewriteEquation(P5.4)as

4 2 2 2 3

2

( ˆ sin ˆ cos ) (ˆ ( cos ) ˆ ( sin ) ˆ ( ))

(( cos ) ( sin ) ( ) )B

e e e e ed

IRd

x R y R z z

x R y R z zo

px p y p x o p y o p z o p

o p o p o p

=π

θ− θ + θ × − θ + − θ + −

− θ + − θ + −

(P5.5c)

where o isthepermeabilityoffreespace( π × −4 10 7 H/m).

y

x

z

Wire loopwith radius R

Iθp

Measurementlocation

Differential currentelement IRdθpe θ

rro

rp

dB

Figure P5.5a The differential magnetic field element

dB at observation point ro

due to a wire loop located in the x-y plane.


1. Calculate the vector cross product in Equation (P5.5c) and split the resultintothreeseparateequationsforthex,y,andzcomponentsofthedifferentialmagneticfield.Use the resultingexpressions fordBx,dBy, anddBz towritethreeMATLABfunctionsthattakeoneargument(θp)andcanbeusedwithMATLAB’squadfunctiontodetermineBx,By,andBz .

2. UseMATLAB’smeshgridfunctiontocreateasetofpointsinthex-zplanefor–8≤x0≤8mm(withastepsizeof1mm)and–2≤z0≤2mm(withastepsizeof0.5mm).DetermineBxandBz ateachpointintheset(besuretousethecorrectintegrationlimits)andplotwithquiver.AssumeR =2.5mm,I =1mA.

Project.5.6

Asolenoidisacoilofwireusedtocreateamagneticfieldtoactivateanactuator.Typicalusesincludeelectricalrelays,electricallycontrolledwatervalves,andauto-motivestartergears.FigureP5.6ashowsasolenoidwithradiusR,lengthD,turncountN,andcurrentI.

WemodelthesolenoidinMATLABasacylindricalsheetcurrentwithdensity

ˆK eNID

= θ A/mwith thecurrentdensitydirected in theθdirection.TheBiot-

Savartlawforasurfacecurrentis

=π

×4 2B

K e

rd

dAo r (P5.6a)

wheredAisatwo-dimensionaldifferentialsurfaceelement,r isthevectorfrom

thedifferentialsurfacetothemeasurementpoint, er istheunitvectorinther

direction,×representsthevectorcrossproduct,and o isthepermeabilityoffreespace( π × −4 10 7 H/m).FromFigureP5.6b,weseethatincylindricalcoordinates,

y

x

ey cosθp

exR cosθp

ˆ−ex sinθp

eθ

eyR sinθp

Wire loopwith radius R

θp

I

rp

Figure P5.5b rp and ˆ

peθ may be decomposed into their x and y components to solve for the magnetic field with MATLAB’s quadfunction.


I

N turnsD

2R

Figure P5.6a A solenoid consisting of N turns, radius R, length D, and current I.

y

z

x

D

2R

θpMeasurement location

Differential surface elementdA= Rdθpdzp

Cylindrical sheet currentwith current densityK amp/m in the θ direction

r

ro

rp

dBK

Figure P5.6b The solenoid can be modeled as a sheet current of magnitude K A/m.


= θdA Rd dzp p ,whererp isthelocationofthedifferentialsurface(withcylindri-

calcomponents rp , θ p ,and z p),and eθ istheunitvectorinthe θ p direction.Wealsodefinetheobservationpoint

ˆ ˆ ˆr e e ex y zo o x o y o z= + + .Fromthegeometryoftheproblem,wecanwrite

KdA and

r intermsofCartesianunitvectors:

ˆ ( ˆ sin ˆ cos )K e e edA

NID

Rd dzNID

Rd dzp p p p x p y p= θ = θ − θ + θθ (P5.6b)

ˆ ( cos ) ˆ ( sin ) ˆ ( )r r r e e ex R y R z zo p x o p y o p z o p= − = − θ + − θ + − (P5.6c)

SubstitutingbackintoEquation(P5.6a)gives

4( ˆ sin ˆ cos ) ˆ ( cos ) ˆ ( sin ) ˆ ( )

( cos ) ( sin ) ( )2 2 232

Be e e e e

dNIRD

d dzx R y R z z

x R y R z z

op p

x p y p x o p y o p z o p

o p o p o p( )( )

=π

θ− θ + θ × − θ + − θ + −

− θ + − θ + −

(P5.6d)

1. Calculate the vector cross product in Equation (P5.6d) and split the resultintothreeseparateequationsforthex,y,andzcomponentsofthedifferentialmagnetic field. Use the resulting expressions for dBx, dBy, and dBz to writethreeMATLABfunctionsthattaketwoarguments(θpandzp)andcomputethedifferentialBfieldinthex,y,andzdirections,respectively.

2. AssumethatthesolenoidiscenteredattheoriginwithradiusR=2.5mmandD=15mm.Usemeshgridtocreateathree-dimensionalpointset[X,Y,Z]ofmeasurementpointsfor

ro .Notethatmeshgridtakesanoptionalthirdargu-

menttocreatethree-dimensionalpointsets(seehelpmeshgridforusagedetails).Usetheselimitsfor ( , , )x y zo o o : − ≤ ≤3 3xo mm(withastepsizeof1.5mm), ʺ ʺ0 3yo (withastepsizeof1.5mm),and 10 10zo− ≤ ≤ mm(withastepsizeof5mm).

3. UsedblquadtocalculateBx,By,andBz ateachpointintheset(besuretousethecorrectintegrationlimits).Assume:N = 500,I =1mA.

4. Create a three-dimensional quiver plot of your results using MATLAB’squiver3function(runhelpquiver3forusagedetails).

157

Chapter 6

Numerical Integration of Ordinary Differential Equations

6.1 IntroductionInthischapter,weexamineseveralmethodsforsolvingordinarydifferentialequa-tions(ODEs).ODEscanbebrokenupintotwocategories:

a.Initial value problemsinwhichweknowthenecessaryinitialconditions,forexample,thevalueofacircuitnodevoltage(oritsslope)att =0;

b.Boundary value problemsinwhichweknowthevalueofthedependentvariableatspecificcoordinatesintheproblemgeometry,forexample,theelectricpoten-tialatbothendsofaconductor.

For initial value problems, we examine several numerical integration meth-ods, including theEulermethod, themodifiedEulermethod, theRunge-Kuttamethod,andthebuilt-inode45functionofMATLAB.Forboundaryvalueprob-lems,weemployimplicitfinite-differencemethodsthatreduceODEstoasystemoflinearalgebraicequations,whichturnouttobeatri-diagonalsystemthatcanbesolvedalgebraically.


6.2 The Initial Value ProblemInaninitialvalueproblem,thevaluesofthedependentvariableandthenec-essaryderivativesareknownatthepointatwhichtheintegrationbegins.Webeginwithafirst-orderdifferentialequationofthegeneralformsuchthatthederivativeisaknownfunctionofxandy,andtheinitialcondition,y(0)=YI,isknown:

=

=

( , )

(0)

dydx

f x y

y YI

(6.1)

Thereareseveraltechniquesforsolvingthistypeofproblem,includingEuler’smethod,whichissimplebutnotusedverymuch;themodifiedEulermethod;theRunge-Kuttamethod;andothers.Eachtechniquehasprosandconswithrespecttosimplicity,accuracy,andcomputationalefficiency.

6.3 The Euler AlgorithmThegeneralapproachtosolvingdifferentialequationsnumericallyistosubdividethexdomainintoNsubdivisionsgiving +, , ,... ,1 2 3 1x x x xN andthen“march”inthex directionovertheintervalwhilecalculating +, , , ...,2 3 3 1y y y yN .Notey1 is

xi xi+1

yi yi+1

y

xx1

y1 yN+1

xN+1

Figure 6.1 To solve a first-order differential equation y′ = f(x, y), we divide the interval of interest into N subdivisions and then estimate the next value of y by using an approximation based on a Taylor series around the known point.

Numerical Integration of Ordinary Differential Equations ◾ 159

specifiedand isequal to the initialconditionYI (seeFigure 6.1).ATaylor seriesexpansionaboutanarbitrarypointxigives

( ) ( )= + − + − +( ) ( ) ( )

1!( )2!

2y x y xy x

x xy x

x xii

ii

i (6.2)

Ifweuseonlythefirsttwotermsoftheseries,wecanapproximatethevaluefory(xi+1)(whichwedenoteasyi+1)as

≈ + −

≈ +

+ +( ) ( )1

( )1 1y y xy x

x x

y y h

i ii

i i

i i (6.3)

where = −+1h x xi i,whichisdefinedasthestep size, and yi istheslopeofthecurveofy(x)atxi.SubstitutingEquation(6.1)into(6.3),weobtain

= ++ ( , )1y y h f x yi i i i (6.4)

AscanbeseenfortheconfigurationshowninFigure 6.2,thepredictionofyi+1byEuler’smethodovershootsthetruevalueofyi+1.Thus,startingwithi =1,weobtain

= + = +( , ) ( , )2 1 1 1 1y y h f x y Y h f x YI I

Similarly,y3canbedeterminedbysubstitutingi=2intoEquation(6.4):

= + ( , )3 2 2 2y y h f x y

xi xi+1

yi

y

x

yi+1,Euler

yi+1,exact

Figure 6.2 In the Euler method, we use a first-order Taylor series approximation (i.e., a straight line) to approximate the next value of y.


The process is continued, obtaining , , ,...,4 5 6 1y y y yN + . The Euler method isanexampleofanexplicitmethodbecauseeachnewvalueforyi+1dependssolelyonpreviousvaluesofxandy.

6.4 Modified Euler Method with Predictor-Corrector Algorithm

ThemodifiedEulermethod is amore accurateway for calculating the value ofyi+1thanthesimplelinearapproximationoftheEulermethodintheprevioussec-tion. Again, we consider a differential equation of standard form = ( , )y f x y withinitialcondition =(0)y YI ,andagainwesubdividethexdomainintoNsub-divisions and march in the x direction. We saw that in the Euler method, tak-ing = + = ++ ( , )1y y hy y hf x yi i i i i i overshoots the true value of yi+1 (as shown inFigure 6.2).Now,supposewewereabletodetermine +( )1y xi ,whichwouldbetheslopetothecurveat + .1xi Ifweweretopredictyi+1byusing +( )1y xi inEquation6.3,thatis,

( , )1 1 1 1y y h y y h f x yi i i i i i= + = ++ + + + (6.5)

thenwewouldundershootthetruevalueofyi+1.ThisisshowninFigure 6.3.Here,wehaveconstructedatangenttothecurveatyi+1anddrawnaparallellinepassing

yi

xi xi+1

yi+1

y

x

Slope = yi

Slope = yi+1

yi + hyi+1

yi + hyi

h

Figure 6.3 In the modified Euler method, we use the average of two straight-line approximations to predict yi+1.


through point ( , )x yi i to obtain the predicted value for yi+1. Since using yi inEquation(6.3)overshootsthetruevalueofyi+1andusing +1yi inEquation(6.3)undershoots thetruevalueofyi+1,weseethatabetterestimate foryi+1wouldbeobtainedbyusinganaverageofthetwoderivativesinEquation(6.3),thatis,

= +

++

+

211y y h

y yi i

i i (6.6)

Unfortunately,Equation(6.6)isnolongerexplicitbecausewedonotknowthevalueof +1yi .TheuseofEquation(6.6)insolvingthedifferentialEquation(6.1)isanexampleofanimplicitmethod.However,wecanapproximateavaluefor +1yi byusingthepredictor-correctormethod.Toapplythismethod,werewriteEquation(6.6)asfollows:

( )= +

++

+

211y y h

y yiC

ii i

P

(6.7)

wherethePsuperscriptindicatesthepredictedvalue,andtheCsuperscriptindicatesthecorrectedvalue.Equation(6.7)iscalledthecorrector equationandcanbeusedtoiterativelycalculatethevalueforyi+1.SubstitutingEquation(6.1)into(6.7)gives

( )= + ++ + +2

( , ) ( , )1 1 1y yh

f x y f x yiC

i i i i iP (6.8)

Thepredictor-correctortechniqueproceedsasfollows:

1.UsetheEulermethodtodetermineafirstpredictedvalueforyi+1,called ,11yiP+

thatis,

= ++ ( , )11y y h f x yiP

i i i

2.Calculatethefirstcorrectedvalue +11yiC byusing 1

1yiP+ inEquation(6.8).

3.Use +11yiC asthenewpredictedvalue 1

2yiP+ inEquation(6.8).

4.Calculateanewcorrectedvalue +12yiC .

5.Repeatsteps3and4until − < ε+ ++

1 11y yi

CiCn n ,whereεisanerrortolerancethat

dependsonthedesiredaccuracyandistypicallyafractionofapercentofthelastcorrectedvalue,forexample, ε = × +0.01% 1yiCn .


Example 6.1

In this example, we find the exact solution of a first-order differential equa-tion and then compare with the corresponding solutions found by the Euler and modified Euler methods.

A simple RC circuit is “driven” by a voltage source VD as shown in Figure 6.4. Writing Kirchhoff’s voltage law around the loop gives

− − =V v vD R C 0 (6.9)

Applying Ohm’s law =v iRR( ) and the constituent relation for capacitor

current =i CdvdtC , we can rewrite Equation (6.9) as

− − =V RC

dvdt

vDC

C 0

Rearranging gives

+τ

=τ

dvdt

vVC

CD1

(6.10)

where τ = RC is the time constant for the RC circuit. Equation (6.10) is a first-order differential equation of vC with respect to time. We assume that at t = 0, the initial capacitor voltage is zero, and that the driving voltage is a ramp function, that is =V t V tD o( ) . We first solve Equation (6.10) analytically so that we can have a basis for comparison with numerical solutions. We begin by guessing a particular solution vC,P having the same form as the driving function:

= α + βv tC P, (6.11)

where α and β are constants to be determined. Substituting vC,P into Equation (6.10), we obtain

( ) ( )α + β +

τα + β =

τddt

t tV to1

+–

VD vC

+

−

C

R

+ vR −

Figure 6.4 RC circuit.


which gives

ατ+ α +

βτ

=τ

t V to (6.12)

Matching terms between the left and right sides of Equation (6.12), we

see that α = Vo and β = − τVo . Thus,

= − τv V t VC P o o,

Next, we find the complementary solution, vC,H, to the homogeneous equation, which is Equation (6.10) when the driving function is zero.

+τ

=dvdt

vC HC H

10,

, (6.13)

Equation (6.13) is separable. Thus,

= −

τ→ = −

τ+ γ → = γ

− τdvv

dtv

tv eC H

C HC H C H

tln ln,

,, , (6.14)

where γ is a constant to be determined. We now take the complete solution as the sum of the particular and homogeneous solutions:

= +

= − τ + γ−τ

v v v

V t V e

C C P C H

o ot

, ,

(6.15)

Applying the initial condition =vC (0) 0 to Equation (6.15), we find that γ = τVo . Thus, the complete solution for the ramp response of an RC circuit is

( )= − τ −

−τv t V t eC ot

( ) 1 (6.16)

We now solve Equation (6.10) numerically for the following parameter values: R = 1 kΩ, C = 1 μF, and =Vo 10 V . First, we rearrange the governing differential equation (6.10) into the form of Equation (6.1):

( ) ( )=

τ− =

τ− =

dvdt

V v V t v f t vCD C o C C

1 1( , ) (6.17)

Next, we choose a time interval and time step. For the RC circuit, we know that the interesting transient behavior will occur within a few mul-tiples of the time constant, and thus we arbitrarily choose an interval of t =

[ 0, 10τ ]. We also divide the interval into 20 steps, and thus =τ

h1020

.


A MATLAB script that carries out the Euler method and the modified Euler method follows:

%Example_6_1.m%CalculaterampresponseofRCcircuitvia3methods:%1.exactsolution%2.byEulermethod%3.bymodifiedEulermethodclc;clear;%Circuitparameters:R=1e3;%ohmsC=1e-6;%faradstau=R*C;%secondsVo=10;%voltsf=@(tm,Vc)(1/tau)*(Vo*tm-Vc);%Equation6.17%ParametersforEuleralgorithminterval=10*tau;steps=20;h=interval/steps;%Createsetoftimepointstocalculatet=0:h:interval;%InitialconditionsVc_exact(1)=0;Vc_euler(1)=0;Vc_mod_euler(1)=0;%ExactandEulermethod:marchintimefori=1:length(t)-1%ApplyEquation6.16Vc_exact(i+1)=Vo*(t(i+1)-tau+tau*exp(-t(i+1)/tau));%ApplyEquation6.4Vc_euler(i+1)=Vc_euler(i)+h*f(t(i),Vc_euler(i));end%ModifiedEuler:applycorrectorequationtolerance=1e-4;%.01percentfori=1:length(t)-1%usetheEulermethodtocalculateaninitialguess%forthepredictedvalueVc_predicted=Vc_mod_euler(i)...+h*f(t(i),Vc_mod_euler(i));%Repeatedlyapplythecorrectorequationuntilthe%predictorandcorrectorareequal(orwithin%tolerance)forj=1:50Vc_corrected=Vc_mod_euler(i)...+(h/2)*(f(t(i),Vc_mod_euler(i))...+f(t(i+1),Vc_predicted));ifabs(Vc_corrected-Vc_predicted)/Vc_predicted...<tolerancebreak;elseVc_predicted=Vc_corrected;endend%ifwereachedtheiterationlimitof50,thenprint%warning:ifj==50


fprintf('Warningatstep');fprintf('%d:reachediterationlimit\n',i);endVc_mod_euler(i+1)=Vc_corrected;end%Plotresultssubplot(1,2,1),plot(t,Vc_euler,'x',t,Vc_exact),legend('Euler','Exact','Location','NorthWest');title('Eulersolutioncomparedtoexact');xlabel('seconds');ylabel('Vc(volts)');text(0.006,0.005,sprintf('%dsteps',steps));subplot(1,2,2),plot(t,Vc_mod_euler,'x',t,Vc_exact),legend('ModifiedEuler','Exact','Location','NorthWest');title('ModifiedEulersolutioncomparedtoexact');xlabel('seconds'),ylabel('Vc(volts)');text(0.006,0.005,sprintf('%dsteps',steps));

The resulting solutions for Example 6.1 are graphed in Figure 6.5. Note that for N = 20 subdivisions in the time domain, both the Euler and modified Euler solutions are almost indistinguishable from the exact solution.

Figure 6.6 shows the same example where we have increased the time step by reducing N to 6 (accomplished by changing steps=20 to steps=6in the program). Note that in this case, the Euler solution deviates substantially from the exact solution and also oscillates above and below the exact solu-tion. On the other hand, the solution for the modified Euler method remains accurate and stable even for small N. Ideally, we would like to reduce N to as small as possible (to reduce computation time), but at the same time we want to avoid numerical instability. Thus, the modified Euler method is a compro-mise that has some additional computational operations (due to the iterative predictor-corrector algorithm) but yet stays stable for large time steps.

0 0.005 0.010

0.02

0.04

0.06

0.08

0.1

seconds

Vc

(vol

ts)

Euler solution compared to exact

20 steps

0 0.005 0.010

0.02

0.04

0.06

0.08

0.1

seconds

Vc

(vol

ts)

Modified Euler solution compared to exact

20 steps

EulerExact

Modified EulerExact

Figure 6.5 Plot of VC versus t for RC circuit for 20 steps over the interval 0 ≤ t ≤ 10τ: (a) Euler and exact solutions; (b) modified Euler and exact solutions.


6.5 Numerical Error for Euler AlgorithmsSolutionstodifferentialequationsusingtheEulerandmodifiedEuleralgorithmsarebydefinitionanapproximationtotheexactsolutionbecausethederivationsofthesealgorithmshavetheirbasisinatruncatedTaylorseries.Ifwewishtodeter-minetheaccuracyofaparticularnumericalscheme,wecantestitonadifferentialequationforwhichanexactsolutionisavailable.Unfortunately,inmostcases,wedonotknow(orcannotsolvefor)theexactsolutionofthedifferentialequation;thisisusuallywhyweresortedtoanumericalmethodinthefirstplace.However,inExample6.1,wedidderivetheexactsolution;thus,wecancalculatetheglobal percentage error ,globalEi foragivennumericalalgorithmattimestepias

=

−,global

,exact ,numerical

,exactE

y yyi

i i

i (6.18)

Analternativemeasureofaccuracyisthelocal truncation error(LTE),inwhichwecalculatethehigher-ordertermsoftheTaylorseries(oratleastthefirstdiscardedterm)anduseittodecidewhetherwehavechosentoolargeatimestep.Forexample,

0 0.005 0.010

0.02

0.04

0.06

0.08

0.1

seconds

Vc

(vol

ts)

Euler solution compared to exact

6 steps

0 0.005 0.010

0.02

0.04

0.06

0.08

0.1

seconds

Vc

(vol

ts)

Modified Euler solution compared to exact

6 steps

EulerExact

Modified EulerExact

Figure 6.6 Plot of VC versus t for RC circuit for six steps over the interval 0 ≤ t ≤ 10τ. (a) Euler and exact solutions. Note that the Euler solution oscillates. (b) Modified Euler and exact solutions. Note that the modified Euler method is stable even for a large step size.


inEquation (6.3),wediscarded the second-orderandhigher terms in theTaylorseriestoderivetheEulermethod;thus,theLTEis

= +

2higher-order termsEuler

2LTEyhi

Ifweneglectthehigher-orderterms,thenthetruncationerrorisontheorderofh2,whichiscommonlywrittenasO(h2).Qualitatively,thismeansthatifwereducethestepsizebyafactoroftwo,thentheLTEisreducedbyapproximatelyafactoroffour,resultinginamoreaccuratesolution.

Inpractice,thesimplestmethodforchoosingthestepsizeistostartwithsomefractionofthefastestknownparameterinthesystem.Forexample,whensolving

anRCcircuit, startwitha stepsizeof1/100of the timeconstant,or =100

hRC .

Then,doublethestepsizeandreruntheprogram.Ifthetwosolutionsareidenti-cal,thenthestepsizeissufficientlysmall,andiftheprogramrunsslowly,thenyoumightconsiderredoublingyourstepsizeandcomparingagaintoobtainafasterexecutiontimewithoutlosingaccuracy.

6.6 The Fourth-Order Runge-Kutta MethodThefourth-orderRunge-Kuttamethodusesaweightedaverageofderivativeesti-mateswithin the interval of interest to calculate a value for yi+1.Weagain startwiththefirst-orderdifferentialequationofEquation(6.1)withknowninitialcon-

ditions. In the modified Euler method, we used = + ++ +2( )1 1y y

hy yi i i i . In the

Runge-Kuttamethod,insteadweuse

( )= + + + ++ 6

2 21 1 2 3 4y yh

k k k ki i (6.19)

where

=

= + + +

= + + +

= + + +

( , ) (value of at )

2,

2estimate of at

2

2,

2a second estimate of at

2( , ) (estimate of at )

1

2 1

3 2

4 3 1

k f x y y x

k f xh

yhk y x

h

k f xh

yhk y x

h

k f x h y hk y x

i i i

i i i

i i i

i i i

Runge-KuttaisanexplicitalgorithmandthusissimpletocomputewithMATLAB.


Example 6.2

Solve the RC circuit of Example 6.1 using the Runge-Kutta method for N = 20 and N = 6 steps. Also, find the worst-case global error for each case. The MATLAB program follows:

%Example_6_2.m%CalculaterampresponseofRCcircuitviatwomethods:%1.exactsolution%2.Runge-Kuttamethodclc;clear;%circuitparametersR=1e3;%ohmsC=1e-6;%faradstau=R*C;%secondsVo=10;%voltsf=@(tm,Vc)1/tau*(Vo*tm-Vc);%Equation6.16%parametersforEuleralgorithminterval=10*tau;steps=[206];forj=1:length(steps)h=interval/steps(j);%createsetoftimepointstocalculatet=0:h:interval;Vc_exact=zeros(length(t),1);Vc_RK=zeros(length(t),1);%initialconditionsVc_exact(1)=0;Vc_RK(1)=0;fori=1:length(t)-1%CalculateexactsolutionVc_exact(i+1)=Vo...*(t(i+1)-tau+tau*exp(-t(i+1)/tau));%CalculateRunge-Kuttasolutionk1=f(t(i),Vc_RK(i));k2=f(t(i)+h/2,Vc_RK(i)+h/2*k1);k3=f(t(i)+h/2,Vc_RK(i)+h/2*k2);k4=f(t(i+1),Vc_RK(i)+h*k3);Vc_RK(i+1)=Vc_RK(i)+h/6*(k1+2*k2+2*k3+k4);end%plotresultssubplot(1,length(steps),j)plot(t,Vc_RK,'x',t,Vc_exact),legend('Runge-Kutta','Exact','Location','NorthWest');title('RampResponseforRCcircuit');xlabel('seconds'),ylabel('Vc(volts)');text(0.005,0.005,sprintf('%dsteps',steps(j)));%calculateglobalerrorglobal_error_RK=abs((Vc_exact-Vc_RK)./Vc_exact);fprintf('Worstcaseglobalpercenterrorfor');fprintf('Runge-Kuttafor%2dsteps:%4.2f%%\n',...steps(j),100*max(global_error_RK));end--------------------------------------------------------------------------


PROGRAM RESULTS

WorstcaseglobalpercenterrorforRunge-Kuttafor20steps:0.23%

WorstcaseglobalpercenterrorforRunge-Kuttafor6steps:9.73%

Note that the fewer the steps, the larger the corresponding global error. Results are plotted in Figure 6.7.

6.7 System of Two First-Order Differential EquationsConsiderthefollowingtwofirst-orderODEs.

= =

= =

( , , ); (0)

( , , ); (0)

0

0

dudt

f t u v u u

dvdt

g t u v v v (6.20)

Tosolveasystemoftwofirst-orderdifferentialequationsbytheRunge-Kuttamethod,take

= + + + +

= + + + +

+

+

6( 2 2 )

6( 2 2 )

1 1 2 3 4

1 1 2 3 4

u uh

k k k k

v vh

l l l l

i i

i i

(6.21)

0 0.005 0.010

0.02

0.04

0.06

0.08

0.1

Seconds

Vc

(vol

ts)

Ramp Response for RC Circuit

20 steps

Runge-KuttaExact

0 0.005 0.010

0.02

0.04

0.06

0.08

0.1

Seconds

Vc

(vol

ts)

Ramp Response for RC Circuit

6 steps

Runge-KuttaExact

Figure 6.7 Plot of VC versus t for RC circuit via Runge-Kutta method: (a) 20 steps; (b) 6 steps.


where

( , , ) (estimate of at )


2,

2,

2first estimate of at

2

2,

2,


2

2,

2,

2second estimate of at

2

2,

2,


2



1

1

2 1 1

2 1 1

3 2 2

3 2 2

4 3 3

4 3 3

k f t u v u t

l g t u v v t

k f t h u h k v h l u t h

l g t h u h k v h l v t h

k f t h u h k v h l u t h

l g t h u h k v h l v t h

k f t h u hk v hl u t h

l g x h u hk v hl v t h

i i i i

i i i i

i i i i

i i i i

i i i i

i i i i

i i i i

i i i i

=

=

= + + + +

= + + + +

= + + + +

= + + + +

= + + + +

= + + + +

(6.22)

and = = −+1h t t ti i .

Example 6.3

Solve the following system of first-order differential equations by the Runge-Kutta method over the interval 0 ≤ t ≤ 10:

= − =

= − + =

−drdt

re ry f t r y

dydt

y ry g r y

t2 2 ( , , )

( , )

0.1

(6.23)

Initial conditions: r(0) = 1.0, y(0) = 3.0.

%Example_6_3.m%Thisprogramsolvesasystemof2firstorderordinary%differentialequationsbytheRunge-KuttaMethod.%First,definetwofunctionstocomputedr/dtanddy/dt:rprime_func=@(t,r,y)2.0*r*exp(-0.1*t)-2.0*r*y;yprime_func=@(r,y)-y+r*y;%Openafiletowriteresults:fo=fopen('Example_6_3_output.txt','w');fprintf(fo,'try\n');fprintf(fo,'--------------------------\n');%Definetimeintervalandstepsizetmax=10;steps=1000;h=tmax/steps;%Initialconditions:r(1)=1.0;y(1)=3.0;t(1)=0.0;


%Estimateofderivativesandmarchingintime.fori=1:stepst(i+1)=i*h;K(1)=rprime_func(t(i),r(i),y(i));L(1)=yprime_func(r(i),y(i));K(2)=rprime_func(t(i)+h/2,r(i)+h/2*K(1),...y(i)+h/2*L(1));L(2)=yprime_func(r(i)+h/2*K(1),y(i)+h/2*L(1));K(3)=rprime_func(t(i)+h/2,r(i)+h/2*K(2),...y(i)+h/2*L(2));L(3)=yprime_func(r(i)+h/2*K(2),y(i)+h/2*L(2));K(4)=rprime_func(t(i)+h,r(i)+h*K(3),y(i)+h*L(3));L(4)=yprime_func(r(i)+h*K(3),y(i)+h*L(3));r(i+1)=r(i)+h/6*(K(1)+2*K(2)+2*K(3)+K(4));y(i+1)=y(i)+h/6*(L(1)+2*L(2)+2*L(3)+L(4));ifmod(i-1,10)==0fprintf(fo,'%.2f%10.4f%10.4f\n',t(i),r(i),y(i));endendfclose(fo);plot(t,r,t,y,'--'),xlabel('t'),ylabel('r,y');grid,legend('r','y');title('Runge-KuttaSolutionFor2nd-OrderSystem');

PROGRAM OUTPUT

The resulting MATLAB plot is shown in Figure 6.8.

0 2 4 6 8 100

0.5

1

1.5

2

2.5

3

3.5

t

r, y

Runge-Kutta Solution For 2nd-Order System

ry

Figure 6.8 Solution of two first-order differential equations by the Runge-Kutta method.


6.8 A Single Second-Order EquationForasinglesecond-orderODE,themethodofsolutionistoreducetheequationtoasystemoftwofirst-orderequations.Giventhefollowingsecond-orderdifferentialequationwithinitialconditions

= =

= =

( , , )

(0) and (0)

2

2

0 0

ud udt

f t u u

u u u u (6.24)

Let =u v,then = = = ( , , )udvdt

v f t u v .Also, = =( , , )u g t u v v ,giving

=

= ( , , )

u v

v f t u v (6.25)

Thus,wehaveconvertedEquations(6.24)intotwofirst-orderdifferentialequa-tionsofthesameformasEquations(6.20);thus,thesamesolutiontechniquescanbeapplied,thatis,

=

= =

= + + + +

= + = + +

= + + + +

= + = + +

= + + + +

= + = + +

( , , ) ( at )

( ) ( at )

2,

2,


2

2 2first estimate of at

2

2,

2,


2

2 2second estimate of at

2


( ) (estimate of at )

1

1

2 1 1

2 1

3 2 2

3 2

4 3 3

4 3

k f t u v v t

l u t v u t

k f th

uhl v

hk v t

h

l u th

vhk u t

h

k f th

uhl v

hk v t

h

l u th

vhk u t

h

k f t h u h l v hk v t h

l u t h v hk u t h

i i i i

i i i

i i i i

i i i

i i i i

i i i

i i i i

i i i


Thevaluesofu andv atthenexttimesteparegivenby

( )

( )

= + + + +

= + + + +

+

+

62 2

62 2

1 1 2 3 4

1 1 2 3 4

u uh

l l l l

v vh

k k k k

i i

i i

Example 6.4

We illustrate this method by applying it to the series RLC circuit of Appendix A. Equation (A.6) gives the capacitor voltage vC (with no driving voltage) as

+ + =

d vdt

RLdvdt LC

vC CC

10

2

2

We wish to solve for vC for the component values R = 10 Ω, L = 1 mH, C =

2 μF, and initial conditions =vC (0) 10 and =dvdt LCC (0)

6.

We first reduce this single second-order differential equation into two first-order equations.

Let =dvdt

vC2 . Then, the two coupled equations are

=

= = − −

dvdt

v

dvdt

d vdt LC

vRLv

C

CC

1

2

22

2 2

Assuming an underdamped system, the exact solution has the form

= − α − +β −vRLt

LCRL

tLC

RL

tC exp2

cos1

2sin

12

2 2

For the given initial conditions, α = 0 and β =

−

dvdt

LCRL

C (0)

12

2.


The program follows:

%Example_6_4.m%Thisprogramsolvesasinglesecondorderordinary%differentialequationbytheRunge-KuttaMethod.The%differentialequationdescribesthecapacitorvoltage%inaseriesRLCcircuitclear;clc;%Componentvalues:R=10;L=1.0e-3;C=2.0e-6;%Initialconditions:vc(1)=0;v2(1)=6*sqrt(1/(L*C));%Definetimeintervalandstepsize.Sincethenatural%frequencysqrt(1/LC)is3559Hz,chooseatimeinterval%tocaptureseveraloscillations.3559Hzhasperiodof%281usec,sochoose1000usecforinterval.tmax=1000e-6;steps=100;h=tmax/steps;t=0:h:tmax;%Definev2'=f(t,x,y):v2_prime=@(t,x,y)-(1/(L*C))*x-(R/L)*y;%DoRunge-Kuttaalgorithm:fori=1:stepsK1=v2_prime(t(i),vc(i),v2(i));L1=v2(i);K2=v2_prime(t(i),vc(i)+h/2*L1,v2(i)+h/2*K1);L2=v2(i)+h/2*K1;K3=v2_prime(t(i),vc(i)+h/2*L2,v2(i)+h/2*K2);L3=v2(i)+h/2*K2;K4=v2_prime(t(i),vc(i)+h*L3,v2(i)+h*K3);L4=v2(i)+h*K3;vc(i+1)=vc(i)+h/6*(L1+2*L2+2*L3+L4);v2(i+1)=v2(i)+h/6*(K1+2*K2+2*K3+K4);end%Calculateexactsolutionbeta=v2(1)/sqrt(1/(L*C)-(R/(2*L))^2);vc_exact=exp(-R/(2*L)*t)*beta....*sin(sqrt(1/(L*C)-(R/(2*L))^2)*t);%Printresultstoscreen.fprintf('Runge-Kuttasolutionforasingle');fprintf('secondorderODE:\n\n');fprintf('tR-KSol''nExactSol''n\n');fprintf('-------------------------------------\n');fori=1:10:steps+1%justprinteverytenthstepfprintf('%10.3e%8.4f%8.4f\n',...t(i),vc(i),vc_exact(i));end%Plotresults:plot(t,vc_exact,t,vc,'x');xlabel('t(seconds)'),ylabel('volts');title('CapacitorvoltageofseriesRLCcircuit');legend('Exact','Runge-Kutta');

--------------------------------------------------------------------------

A comparison of the Runge-Kutta solution with the exact solution is shown in Figure 6.9.


6.9 MATLAB’s ODE FunctionMATLAB has several built-in ODE functions that solve a system of first-orderODEs, includingode23 andode45. In this chapter,wedemonstrateode45,which is based on fourth- and fifth-order Runge-Kutta methods. A descriptionoftheode45functionfollows(thisdescriptioncanbeobtainedbytypinghelp ode45intheCommandwindow):

ODE45Solvenon-stiffdifferentialequations,mediumordermethod. [TOUT,YOUT]=ODE45(ODEFUN,TSPAN,Y0)withTSPAN=[T0

TFINAL]integratesthesystemofdifferentialequationsy'=f(t,y)fromtimeT0toTFINALwithinitialconditionsY0.ODEFUNisafunctionhandle.ForascalarTandavectorY,ODEFUN(T,Y)mustreturnacolumnvectorcorrespondingtof(t,y).EachrowinthesolutionarrayYOUTcorrespondstoatimereturnedinthecolumnvectorTOUT.ToobtainsolutionsatspecifictimesT0,T1,...,TFINAL(allincreasingoralldecreasing),useTSPAN=[T0T1...TFINAL].

0 0.2 0.4 0.6 0.8 1

× 10–3

–3

–2

–1

0

1

2

3

4

5

t (seconds)

Vol

tsCapacitor Voltage of Series RLC Circuit

ExactRunge-Kutta

Figure 6.9 Solution of a single second-order differential equation by the Runge-Kutta method.


Thus,ode45takesasargumentsahandletoafunctiondescribingthedifferentialequations(ODEFUN),avectordescribingatimeinterval(TSPAN),andavectordescrib-ingtheinitialconditions(Y0).ThefunctionODEFUNmusttaketwoarguments:atimetandavectorofvalues , ,...,1 2y y yn ,tobepassedtothedifferentialequations.Thesystemofndifferentialequationsmustbeinstandardform,thatis,nequationsoftheform = ( , , ,..., )1 2y f t y y yn n .ODEFUNmustreturnavectorofderivativesofthe form , ,...,1 2y y yn .The time intervalTSPAN is typically a two-element vectorcontainingastartandendtime;ode45willautomaticallychooseanappropriatetimestep(andmightevenvarythetimestepwithintheinterval).ode45willreturntwovectors:alistoftimepointsTOUTandthesolutionYOUTateachtimepoint.Ifyouwanttoforceode45tosolvethesystematspecifictimepoints,thenyoucanexplicitlyspecifythetimepointsinTSPAN(insteadofjustthestartandendtimepoints).

Example 6.5

Solve the following system of three first-order differential equations using MATLAB’s ode45 function:

=

= −

= −

y y y t

y y y

y y y0.51

1 2 3

2 1 3

3 1 2

Initial conditions: =y (0) 01 , =y (0) 1.02 , and =y t( ) 1.03 .

%Example_6_5.m%Thisprogramsolvesasystemof3ordinarydifferential%equationsbyusingMATLAB'sode45function.%y1'=y2*y3*t,y2'=-y1*y3,y3'=-0.51*y1*y2%y1(0)=0,y2(0)=1.0,y3(0)=1.0clear;clc;initial=[0.01.01.0];tspan=0.0:0.1:10.0;[t,Y]=ode45(@dydt3,tspan,initial);y1=Y(:,1);y2=Y(:,2);y3=Y(:,3);fid=fopen('output.txt','w');fprintf(fid,'ty1y2y3\n');fprintf(fid,'-------------------------------------\n');fori=1:2:101fprintf(fid,'%6.2f%8.4f%8.4f%8.4f\n',...t(i),y1(i),y2(i),y3(i));endfclose(fid);plot(t,y1,t,y2,'-.',t,y3,'--');xlabel('t'),ylabel('y1,y2,y3');title('(y1,y2,y3)vs.t'),grid;


text(5.2,-0.8,'y1'),text(7.7,-0.25,'y2');text(4.2,0.85,'y3');

--------------------------------------------------------------------------

%dydt3.m%ThisfunctionworkswithExample_6_5.m%y1'=y2*y3*t,y2'=-y1*y3,y3'=-0.51*y1*y2%y1=Y(1),y2=Y(2),y3=Y(3).functionYprime=dydt3(t,Y)Yprime=zeros(3,1);Yprime(1)=Y(2)*Y(3)*t;Yprime(2)=-Y(1)*Y(3);Yprime(3)=-0.51*Y(1)*Y(2);

PROGRAM RESULTS

The calculated solutions for y1, y2 , and y3 are shown in Figure 6.10.

Example 6.6

Solve the parallel RLC circuit of Section 2.17 using MATLAB’s ode45 func-tion. Assume = = = = =R L C v iL50 , 1 H, 10 nF, (0) 3.3V, (0) 0 A . Plot the results for 0 ≤ t ≤ 4 μsec.

0 2 4 6 8 10–2

–1.5

–1

–0.5

0

0.5

1

1.5

2

t

y1, y

2, y

3(y1, y2, y3) vs. t

y1

y2

y3

Figure 6.10 Solution of three differential equations by MATLAB’s ode45 function.


%Example_6_6.m%Solveasystemoftwofirst-orderordinarydifferential%equations(Equations2.6and2.7).%Assume:y(1)=vandy(2)=iL%Assume:R=50,L=1e-6,C=10e-9,v(t=0)=3.3,i(t=0)=0.%Equation2.6:yprime(2)=(1/L)*y(1)%Equation2.7:yprime(1)=-1/(R*C)*y(1)-(1/C)*y(2)clear;clc;initial=[3.30];tspan=0:.01e-6:4e-6;[t,y]=ode45('dydt_rlc',tspan,initial);v=y(:,1);iL=y(:,2);fprintf('tviL\n');fprintf('---------------------------------------\n');%printeverytenthvalueofvandiLfori=1:10:length(t)fprintf('%1.10f%10.4f%10.4f\n',...t(i),v(i),iL(i));end%PlotvandiLtogether.Useplotyy()tocreatea%second'y'axisontherightside.plotyy()returns%graphicshandlestotheaxesandlinessothattheir%stylecanbeadjusted.[axis,line1,line2]=plotyy(t,v,t,iL);xlabel('t'),ylabel('v(volts)');title('vandiLvs.t'),grid;%setlabelforright'y'axisset(get(axis(2),'Ylabel'),'String','i_L(amps)');%plotiLwithadashedlineset(line2,'LineStyle','--');legend('v','i_L');

--------------------------------------------------------------------------

%dydt_rlc.m%ThisfunctionworkswithExample_6_6.m%Assume:v=y(1)andiL=y(2)functionyprime=dydt_rlc(t,y)yprime=zeros(2,1);R=50;L=1e-6;C=10e-9;yprime(1)=-1/(R*C)*y(1)-1/C*y(2);yprime(2)=(1/L)*y(1);

PROGRAM RESULTS

The results are plotted in Figure 6.11. Note the complementary nature of v(t) and iL(t) such that when either is zero, the other is always near a maximum. This behavior demonstrates how the energy in the RLC circuit oscillates back and forth between the capacitor and inductor.


6.10 Boundary Value ProblemsWhen an ODE involves boundary conditions instead of initial conditions, thenweuseadifferentapproachtosolvetheproblem.Inaboundaryvalueproblem,weessentiallyneedto“fit”asolutionintotheknownboundaryconditionsasopposedtosimplyintegratingfromtheinitialconditions.Anexampleofthistypeofproblemistodeterminetheelectricfieldbetweentheplatesofacapacitorwithaknownchargedensitybetweentheplatesandafixedvoltageacrosstheplates.Thefixedvoltageistheboundarycondition,andthesolutionmaybefoundbynumericallysolvingasecond-order,nonhomogeneousODEusingfinite difference formulas.Withcertaintypesofboundaryconditions,thenumericalmethodwillreducetosolvingasetoflinearequationsthatfallintothecategoryofatri-diagonalmatrix.Thesolutionofasetoflinearalgebraicequationsthatareclassifiedasatri-diagonalsysteminvolvesfewercalculationsthanthesolutionbytheGausseliminationmethoddescribedinChapter3,whichisimportantifthesystemofequationsislarge.Thesolutionofatri-diagonalsystemisdiscussednext.

0 0.5 1 1.5 2 2.5 3 3.5 4× 10–6

–4

–2

0

2

4

t

v (v

olts

)

v and iL vs. t

–0.4

–0.2

0

0.2

0.4

i L (

amps

)

viL

Figure 6.11 Solution of RLC circuit with the ode45 function.


6.11 Solution of a Tri-Diagonal System of Linear Equations

Atri-diagonalsystemofequationshasthefollowingform:

−

− −

− −

− −

−

=

1 0 0 01 0 0

0 1 00 0 10 0 0 1

1

2 2

3 3

4 4

5

1

2

3

4

5

1

2

3

4

5

ab a

b ab a

b

xxxxx

ccccc

(6.26)

whereai,bi,andciareconstantterms,andxiaretheunknowns.Thetermtri-diag-onalreferstothefactthattheelementsofthecoefficientmatrixontheleftsideofEquation(6.26)isnonzeroforthemaindiagonalandthediagonalsimmediatelyaboveandbelowthemaindiagonal.Asidefromthethreediagonals,therestoftheelementsinthecoefficientmatrixareallzero.BymultiplyingoutthematricesinEquation(6.26),thesetofequationsforthisfifth-ordersystembecomes

− =1 1 2 1x a x c (6.27)

− + − =2 1 2 2 3 2b x x a x c (6.28)

− + − =3 2 3 3 4 3b x x a x c (6.29)

− + − =4 3 4 4 5 4b x x a x c (6.30)

− + =5 4 5 5b x x c (6.31)

Thegeneralconceptforsolvingatri-diagonalsystemisasfollows:

We can solve Equation (6.27) for x1 and substitute the result into Equation(6.28),givinganequationinvolvingonlyx2andx3,whichwedesignateasEquation(6.28′).

WecanthensolveEquation(6.28′)forthex2intermsofx3andsubstitutetheresultintoEquation(6.29).Thisgivesanequationonlyinvolvingx3andx4,whichwedesignateasEquation(6.29′).

Thisprocess iscontinueduntil the lastequation.Whenx4 is substituted intoEquation(6.31),anequationonlyinvolvingx5isobtained,thusallowingustosolvealgebraicallyforx5.

Then,bybacksubstitution,wecanobtainalltheotherxivalues.


Method Summary for m equationsArrangethesetofequationsintothegeneralform

= + ++ −1 1x a x b x ci i i i i i (6.32)

Note:forthefirstandlastequations(correspondingtotheupper-leftandlower-rightcoefficientsinthetri-diagonalmatrix), = 01b and = 0am .

Bythesubstitutionprocedureoutlined,wecanobtainasetofequationsoftheform

= + +1x d e xi i i i (6.33)

wherediandeiarethecoefficientsforthe“prime”equations.Notethatem=0.Then,

=

= +

= +

− − −1 1 1

1 1 1 2

x d

x d e x

x d e x

m m

m m m m

(6.34)

Ifgeneralexpressionsfordiandeicanbeobtained,thenwecansolvethesystemforallxi.

WestartbyrewritingEquation(6.33)toputthe(i –1)thequationintotheform

= +− − −1 1 1x d e xi i i i (6.35)

SubstitutingEquation(6.35)in(6.32)gives

= + + ++ − −( )1 1 1x a x b d e x ci i i i i i i i

Solvingforxigives

=

+−

+−

−

−

+

−

( )(1 ) (1 )

1

1

1

1x

c b db e

a xb ei

i i i

i i

i i

i i (6.36)

MatchingtermsbetweenEquations(6.36)and(6.33),weobtain

=+−

=−

−

−

−

1

1

1

1

1

dc b d

b e

eab e

ii i i

i i

ii

i i (6.37)

whichisvalidfori =2,3,…,m.


Theveryfirstequationinthesystemisalreadyintheform = + −1x d e xi i i i ;thus,matchingtermsbetweenEquations(6.27)and(6.33),weobtain

=

=

1 1

1 1

d c

e a (6.38)

We can now successively apply Equation (6.37) to find …, , ,2 3d d dm and…, , ,2 3e e em .

Then, =x dm m ,andbybacksubstitution

= +

= +

= +

− − −

− − − −

1 1 1

2 2 2 1

1 1 1 2

x d e x

x d e x

x d e x

m m m m

m m m m

Example 6.7

Solve the following system of equations for all xi:

+ =+ + = −

+ − =− + + =

+ − = −+ =

x xx x x

x x xx x x

x x xx x

2 7 (6.39)4 3 14 9 (6.40)

11 6 3 (6.41)3 2 31 1 (6.42)

4 8 19 (6.43)20 11 (6.44)

1 2

1 2 3

2 3 4

3 4 5

4 5 6

5 6

First, we need to put the equations in the form of Equation (6.32), which is

= + ++ −x a x b x ci i i i i i1 1

This gives

=− +x x2 71 2

=− − −x x x

143

43

93

2 3 1

= − +x x x6 11 33 4 2

=− + +x x x

312

32

12

4 5 3

= − −x x x8 4 195 6 4

=− +x x20 116 5


We see that

= − − −

= − − − −

= − −

a

b

c

2 14 / 3 6 31/ 2 8 0

0 4 / 3 11 3 / 2 4 20

7 3 3 1/ 2 19 11

A MATLAB program to solve Equations (6.39)–(6.44) for all xi follows:

%Example_6_7.m%Exampleofsolvingatri-diagonalsystemofequationsa=[-2-14/36-31/280];b=[0-4/3-113/2-4-20];c=[7-331/2-1911];m=length(a);%Computedandecoefficients:d(1)=c(1);e(1)=a(1);fori=2:md(i)=(c(i)+b(i)*d(i-1))/(1-b(i)*e(i-1));e(i)=a(i)/(1-b(i)*e(i-1));end%Computex:x(m)=d(m);fori=(m-1):-1:1x(i)=d(i)+e(i)*x(i+1);end%Displaythesolution:x

PROGRAM RESULTSx= 37.3362-15.1681-8.0600-29.65151.1653-12.3051

Note that the example contains only arithmetic operations and no expensive matrix operations (e.g., solving by Gauss elimination or finding an inverse matrix). Thus, the algorithm will be extremely fast, even for systems with many equations (e.g., order of m = 1000 or more).

6.12 Difference FormulasTo solve a boundary value problem involving an ordinary, linear differentialequationnumerically,weneed thedifference formulasobtainedbyTaylor seriesexpansion.Thiswillenableustoreducedifferentialequationstoasetofalgebraic


equations.AswesawinSection6.3,wecanexpandy =g(x)inaTaylorseriesexpan-sionaboutpointxi,thatis,

( ) ( )= + − + − + − +( ) ( ) ( )( ) ( )

2!( )3!

2 3y x y x y x x xy x

x xy x

x xi i ii

ii

i

Wedefinethestepsize = −h x xi ,whichgives

+ = + + + +( ) ( ) ( ) ( )

2!( )3!

2 3y x h y x y x hy x

hy x

hi i ii i

Let + = +( ) 1y x h yi i and =( )y x yi i , =( )y x yi i ,andsoon.Then,theTaylorseriesexpansionequationcanbewrittenas

= + + + ++ 2! 3!1

2 3

y y y hy h y h

i i ii i (6.45)

WenowrewriteEquation(6.45)forthepoint = −x x hi ,thatis,theequidis-tantpointontheothersideofxi.Thisgives

− = = + − + − + − +−( ) ( )

2!( )

3!( )1

2 3y x h y y y hy

hy

hi i i ii i

or

= − + − + −− 2! 3!1

2 3

y y y hy h y h

i i ii i (6.46)

WeuseEquations(6.45)and(6.46)toderiveseveraldifferenceformulas.WestartbyusingthefirsttwotermsofEquation(6.45):

= ++1y y y hi i i

Solvingfor yi gives

=

−+1yy y

hii i (6.47)

Thisistheforwarddifferenceformulafor yi witherroroforderh. Similarly,fromEquation(6.46),usingthefirsttwotermsintheexpansiongives

=

− −1yy y

hii i (6.48)

Thisisthebackwarddifferenceformulafor yi witherroroforderh.


Now,letussubtractEquation(6.46)fromEquation(6.45)andkeepthefirstthreetermsineachequation.Thisgives

− =+ − 21 1y y y hi i i

Solvingfor yi gives

=

−+ −

21 1y

y yhi

i i (6.49)

Thisisthecentraldifferenceformulafor yi withanerroroforder 2h (theerrorisorder2h becausethreetermsintheTaylorserieswereusedtoobtaintheformula).Ifweadd

Equations(6.45)and(6.46)andkeepthefirstthreetermsineachequation,weobtain

+ = ++ − 21 12y y y y hi i i i

Solvingfor yi gives

=

+ −+ − 21 12y

y y yhi

i i i (6.50)

Thisisthecentraldifferenceformulafor yi witherroroforder 2h .The four difference formulas derived are summarized in Table 6.1. We now

showhowtheyareappliedtomodelaboundaryvalueproblemthatleadstoatri-diagonalsystemoflinearequations.

Table 6.1 Summary of Finite Difference Formulas for Boundary Value Problems

=−+y

y yh

ii i1

First-order forward difference formula. Used for y’ boundary condition at beginning of the domain.

=− −y

y yh

ii i 1

First-order backward difference formula. Used for y’ boundary condition at end of the domain.

=−+ −y

y yh2

ii i1 1

First-order central difference formula. Used for first-order differential equation at interior points.

=+ −+ −y

y y yh

2i

i i i1 12

Second-order central difference formula. Used for second-order differential equation at interior points.


6.13 One-Dimensional Plate Capacitor ProblemFigure 6.12showsaparallelplatecapacitorwithconstantappliedvoltageVoandafixedchargedensityρbetweentheplates.Forcaseswithplanarsymmetrysuchastheparallelplatecapacitorforwhichthechargedensityonlychangesinthexdirection(i.e.,thereisnoyorzdependency),thenPoisson’sequationdescribingtheelectricpotentialΦreducestoanODE:

Φ=−ρε

( ) ( )2

2d xdx

x (6.51)

whereΦ(x)istheelectricpotential(involts),ρ(x)isthex-dependentchargedensity(inC/m3),andεisthedielectricconstantforthematerialbetweentheplates.

Example 6.8

Solve for Φ(x) between the plates of the capacitor of Figure 6.12 with a plate separation of D meters, ρ = ρ −x x Do( ) ( )2 , and boundary conditions Φ(0) = 0 and Φ =D Vo( ) .

Substituting the expression for ρ(x) into Equation (6.51), we obtain

Φ=−ρε

− =−ρε

− +ddx

x D x Dx Do o( ) ( 2 )2

22 2 2 (6.52)

Conducting plates

Material betweenplates with dielectric εand charge density ρ(x)

Vo

+

−Dx

Figure 6.12 Parallel plate capacitor with plate separation of D meters and fixed charged density ρ(x) C/m3 between the plates.


As we did for the RC circuit in this chapter, we solve Equation (6.52) using both classical and numerical techniques and then compare the solu-tions. We can solve this simple differential equation by integrating twice on both sides, giving

Φ =

−ρε

− + + α + βx xDx

Dx xo( )

112 3 2

4 32

2 (6.53)

where α and β are constants of integration that are determined by applying the boundary conditions. Applying the first boundary condition Φ(0) = 0 gives

=−ρε

− + + α + βD Do0

112

030

20 04 3

22

Thus, β = 0.

For the second boundary condition Φ =D Vo( ) , we get

=−ρε

− + + αV DD

DD

d Doo 1

12 3 24

33

22

Solving for α gives

α =

−ερ

−V

DDo

o 4

3

Thus, the exact solution to Equation (6.52) is

Φ =

−ρε

− + −ερ

+x xDx

Dx

VD

Dxo o

o( )

112 3 2 4

4 32

23

(6.54)

To solve Equation (6.52) numerically, we subdivide the region between the plates into N intervals and apply the finite difference formulas of Table 6.1. Dividing the region 0 ≤ x ≤ D into N intervals will result in N + 1 values for x and thus N + 1 values for Φ. Our goal is to determine the coef-ficients in an (N + 1) × (N + 1) tri-diagonal matrix.

We begin by applying the second-order central difference formula from Table 6.1 to Equation (6.52), giving

( )Φ =

Φ + Φ − Φ=−ρε

− ++ −

hx Dx Di

i i i oi i

221 1

22 2

Solving for Φi gives

( )Φ = Φ + Φ +

ρε

− ++ −h

x Dx Di i io

i i12

12 2

21 1

22 2 (6.55)


This equation is valid for i = 2, 3, 4, … , N. Note that Equation (6.55) is in the same general form as Equation (6.32). We can identify the elements of a tri-diagonal matrix as

=ai12,

=bi12,

( )=ρε

− +ch

x Dx Dio

i i2

22

2 2

We now use the boundary conditions to determine the values for the upper-left and lower-right corners of the coefficient matrix. First, we rewrite the boundary conditions as

Φ = → Φ =(0) 0 01 (6.56)

Φ = → Φ =+D V Vo N o( ) 1 (6.57)

Equation (6.56) gives

= = =a b c 01 1 1

Equation (6.57) gives

= = =+ + +a b c VN N N o0, 0,1 1 1

A MATLAB program to solve Equation (6.51) numerically follows:

%Example_6_8.m%Findtheelectricpotentialbetweenaparallelplate%capacitorwithfixedchargedensitybetweentheplates%andknownboundaryconditions.clc;clear;N=40;%stepcountD=.0004;%plateseparation(meters)h=D/N;%stepsizerho_o=1e4;%coulomb/m^3epsilon=1.04e-12;%dielectricbetweentheplatesVo=5;%voltageacrossplates%First,calculateallvaluesofx.Alsocalculate%theexactsolution.x=0:h:D;%Calculateexactsolution(Equation6.54)phi_exact=(-rho_o/epsilon)*(x.^4/12-(D/3)*x.^3...+(D^2/2)*x.^2-((epsilon*Vo/(D*rho_o))+(D^3/4))*x);%Next,solvenumericallyusingatri-diagonalmatrix.%Define'a','b',and'c'vectors:a=.5*ones(1,N+1);b=.5*ones(1,N+1);c=(rho_o*h^2/(2*epsilon))*(x.^2-2*D*x+D^2);%Setboundaryconditionsfor'a','b',and'c'vectorsa(1)=0;b(1)=0;c(1)=0;a(N+1)=0;b(N+1)=0;c(N+1)=Vo;%Solvetri-diagonalmatrixbycalculating"d"and%"e"vectorsd(1)=c(1);e(1)=a(1);fori=2:N+1d(i)=(c(i)+b(i)*d(i-1))/(1-b(i)*e(i-1));


e(i)=a(i)/(1-b(i)*e(i-1));end%calculatephibasedondande,alsocalculate%globalerrorphi(N+1)=d(N+1);fori=N:-1:1phi(i)=d(i)+e(i)*phi(i+1);end%Plotnumericalandexactsolutionsforphi:plot(x,phi,'x',x,phi_exact);xlabel('\itx(meters)'),ylabel('\Phi(volts)');grid;legend('numericalsolution','exactsolution',...'Location','SouthEast');%Findworst-caseglobalerror:error=abs((phi-phi_exact)./phi_exact);fprintf('Maxglobalerrorfor%dsteps:%5.3f%%\n',...N,100*max(error));

--------------------------------------------------------------------------

PROGRAM RESULTSMaximumglobalerrorfor40steps:0.029%

The MATLAB plot of Φ as a function of x is shown in Figure 6.13.

0 1 2 3 4

x 10-4

0

2

4

6

8

10

12

x (meters)

Φ (

volts

)

numerical solutionexact solution

Figure 6.13 Plot of the electric potential Φ(x) between the plates, including exact solution and finite difference solution for N = 40.


ProjectsProject.6.1

SolvetheRCcircuitofFigure 6.4forvC(t)wherethedrivingvoltagehasthefol-lowingform:

=

ω ≥

<( )

sin 00 0

V tV t t

tD

o

Assumetheinitialcondition =(0) 0vC andparametervaluesR =1kΩ,C =1μF,ω =2000rad/sec,andVo=10V.

1. FindtheexactsolutionforvC(t)byassumingaparticularsolutionoftheform= α ω +β ω( ) sin cos,v t t tC P and using Equation (6.14) for the homogenous

equationsolution,whichis = γ − τ,

/v eC Ht .Solvefortheconstantsbyα,β,and

γbysubstitutingback intothegoverningdifferentialequationandapplyingtheinitialcondition.

2. UsetheEuleralgorithmtosolvenumericallyforvC(t).Solveovertheintervalt =[0,5]mswithatimestepof0.1ms.

3. UsetheModifiedEuleralgorithmtosolvenumericallyforvC(t)forthesameintervalaspart2.

4. OnonegraphplotvC(t)versustobtainedbyboththeEulermethodandtheexactsolution.Onaseparategraph,plotvC(t)versustobtainedbothbythemodifiedEulermethodandtheexactsolution.

5. Increase the step size in your numerical solutions until you start to seeinstability.Whatisareasonableruleofthumbtoensurenumericalstabilityinthisfirst-ordersystemwithsinusoidalinput?

Project.6.2

FigureP6.2ashowsathird-orderRLCcircuit.Torunatime-domaintransientanal-ysis,wetransformthecircuitintodifferentialequationsusingthefollowingmethod:

RL

iL iC2iC1

1

C2C1

vS

+

−vL

+

−

vR

+

−

iR

2

VC1+ −VC2− + 21

Figure P6.2a Third-order RLCC circuit containing two capacitors and an inductor.


First,writeKirchhoff’scurrentlawforeachnodeinthecircuit:

atnode1: + − = 01 2i i iC C L (P6.2a)

atnode2: + = 02i iR C (P6.2b)

Second,writeKirchhoff’svoltagelawforthetwoloopsinthecircuit:

forloop1: − − = 01v v vS C L (P6.2c)

forloop2: + − = 02v v vL C R (P6.2d)

Third,writetheconstituentrelationsforeachcircuitelement:

for 1C : =1 1 1i C vC C (P6.2e)

for 2C : =2 2 2i C vC C (P6.2f)

forL: =v LiL L (P6.2g)

forR: =v i RR R (P6.2h)

Wechoosethethreevoltagesandcurrentswithderivativeterms(vC1,vC2,andiL )asthestate variablesforthisproblem.

1. Rearrangeequations(P6.2a)–(P6.2h)intothefollowingstandardform:

= ( , , , ( ))1

1 2dvdt

f v v i v tCC C L S (P6.2i)

= ( , , , ( ))2

1 2dvdt

g v v i v tCC C L S (P6.2j)

= ( , , , ( ))1 2

didt

h v v i v tLC C L S (P6.2k)

wherethefunctionsf,gandhinvolveparametersC1,C2,L,andR.

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

vS(t)

5 V

t (μsec)

Figure P6.2b Input pulse for the RLCC circuit.


2. SolveEquations(P6.2i)–(P6.2k)withtheRunge-Kuttaalgorithmforthefol-lowingcircuitparameters:C1=1μF,C2=0.001μF,R=100kΩ,L=0.01mH.Useatimeintervalof0≤t ≤10μsandastepsizeof0.01μs.Assumevs(t)isa5-Vpulsestartingattimet=0withrisetimeof0.1μs,an“on”timeof0.8μs,andfalltimeof0.1μs(asshowninFigureP6.2b).Initialconditions:

=(0) 01vC , =(0) 02vC ,and =(0) 0iL . 3. Plotonseparategraphs: ( ), ( ), ( ) and ( )1 2v t v t i t v tC C L S versustime.

Project.6.3

TheSallen-Keycircuit(FigureP6.3)iscommonlyusedtoimplementsecond-order(orhigher)filters.Althoughthereareseveralwaystomodelthiscircuit,weanalyzeitheredirectlyinthetimedomainusingdifferentialequations.Ifweassumethattheopampisideal,thenv2=voutandi5=0.ApplyingKirchhoff’scurrentlawatthenodeslabeledv1andv2andusingtheconstituentrelationsforresistors(vR=

iRR)andcapacitors =i CdvdtCC ,weget

Atnode 1v : + + =−

+−

+−

=0 ( ) 01 2 31

1

2 1

21

1i i iv vR

v vR

Cd v v

dtin out

Atnode 2v : + =−

+ =0 02 42 1

22

2i iv vR

Cdvdt

Rearrangingintoasystemoftwofirst-orderODEsandsetting =2v vout ,weget

=

−+

1 12 2 2 2

1dvdt R C

vR C

voutout (P6.3a)

= − + − − +

1 1 1 1 1 11

2 1 2 2 2 2 1 1 2 11

1 1

dvdt R C R C

vR C RC R C

vRC

vout in

(P6.3b)

+–i4

i3

i2i1C2

C1

R1 R2vin vouti5

v2v1

Figure P6.3 Sallen-Key circuit.


1. Solve for vout and v1 using MATLAB’s ode45 function. Assume that theinputtothecircuitvinisastepvoltagethatchangesfrom0Vto1Vattime= +0t .Assume the followingvalues for the circuit elements: = 5000 ,1R = 50002R , = 22001C pF, = 11002C pF. Use a time interval of t =

[0,100μs]andassume = =(0) (0) 01v vout . 2. FindtheimpulseresponseofthecircuitbyfirstcreatingaMATLABfunction

pulse(t)thatreturnsthefollowingvalues:

pulse(t)=< < −10 for 0 10

0 otherwise

6 6t

Then, solve for vout and v1 using MATLAB’s ode45 function where vi =pulse(t).Use thesamecomponentvalues, time interval,and initialcondi-tionsasinpart1.

3. Plotthestepresponse(frompart1)andtheimpulseresponse(frompart2)onthesamesetofaxes.Whatrelationshipcanyouseebetweenthetwo?

4. WhatisoneadvantageoftheSallen-Keycircuitversustheordinarysecond-orderRLCcircuitofAppendixA?

Problem.6.4

BycascadingtheSallen-Keycircuit,wecancreatehigher-orderfilters.FigureP6.4showsanexampleofafourth-orderfilter.

UsingEquations(P6.3a)and(P6.3b),writeasystemoffourfirst-orderdiffer-entialequationstodescribethefourth-orderSallen-Keycircuit.Rememberthatvoutofthefirststagewillbeequaltovin ofthesecondstage.

1. SolveforvoutusingMATLAB’sode45function.Assumethattheinputtothecircuitvinisastepvoltagethatchangesfrom0Vto1Vattime = +0t .Assumethe following values for the circuit elements: = = = = 50001 2 3 4R R R R ,

= 180001C pF, = 150002C pF, = 4.73C μF,and = 104C pF.Usea timeintervaloft =[0,0.005]secandassume = = = =(0) (0) (0) (0) 01 2 3v v v vo .

2. Plotvoutandvinwithrespecttotonthesamesetofaxes.

+–

C2

C1

R1 R2vin v1 v3

v2 v4vout

+–

C4

C3

R3 R4

Figure P6.4 Fourth-order filter using cascaded Sallen-Key circuits.


Project.6.5

Acommonboundaryvalueprobleminelectricalengineeringissolvingfortheelec-tricpotentialinthedepletionregionofaPNjunction.PNjunctionsareformedbytwoadjacentregionsofoppositelydopedsemiconductor(typicallysilicon)andareusedtocreateelectronicdevicessuchasbipolartransistors,MOSFETs,(metal-oxide-semiconductorfield-effecttransistors)andJFETs(junctionfield-effecttran-sistor).Herewewillanalyzethesimplestdevice,thePNdiode.

FigureP6.5ashowsthecircuitrepresentationofadiodewithappliedvoltageVAandresultingcurrentiD.FigureP6.5bshowsaphysicalrepresentationofthediodeas

iD

VA

+

−

Figure P6.5a Diode circuit.

p n

iD

++

+

+

+

++

++

++

– –

–

––

–

–

–

–

–

–

n-type silicon withelectron density ND

p-type silicon withelectron density NA

depleted n-type siliconwith net positive charge

depleted p-type siliconwith net negative charge

x

Figure P6.5b Close-up of the depletion region of a PN junction.


adjacentregionsofp-typesemiconductor(wherecurrentisprimarilycarriedbymobileholeswithdensityNAcarriers/cm3)andn-typesemiconductor(wherecurrentispri-marilycarriedbymobileelectronswithdensityND carriers/cm3).Intheareaimmedi-atelysurroundingthejunction(showninthefiguredetail),themobilechargecarriersofthe(normallyneutral)p-typeandn-typeregionswillrecombine,leavinganetfixedchargedensitylayeraroundthejunctioncalledthedepletion region.Thechargeden-sityinthedepletionregionisdependantonthedoping profile,andanexampledopingprofileisshowninFigureP6.5c.Inthisfigure,wehavemadetwoapproximations:(a)thatthesilicondopinglevelschangeabruptlyfromNA toNDatthejunction(i.e.atx=0),and(b)thatthedepletionregionendsabruptlyat–xP ontheleftsideandatxN ontherightside.Thus,thechargedensitiescanbeapproximatedasstraightlines:

ρ = –qNP A for − < < 0x xP (P6.5a)

qNN Dρ = for < <0 x xN (P6.5b)

where ρP and ρN arethechargedensitiesinthep-andn-typeregionsrespectively,andqistheunitelectriccharge( × −1.6 10 19 coulomb).(NotethatthechargedensityonthePsideofthedepletionregionisnegativebecauseitisdepletedofpositivecar-riers.Similarly,theNsidewillhaveapositivechargedensity.)

The governing equation for the voltage potential Φ in the p-type depletionregionis

Φ=

ε

2

2ddx

qN A (P6.5c)

and thegoverningequation for thevoltagepotentialΦ in then-typedepletionregionis

Φ=−

ε

2

2ddx

qND (P6.5d)

Wesetx =0attheexactpointoftheP-to-NtransitionandarbitrarilysetthevoltagepotentialΦ =0Vatx =0;thatis,

Φ(0)=0 (P6.5e)

( )P xρ

( )N xρ

Nxx

ρ

P-side of junction N-side of junction

++

+

+

––

––

DqN

AqN−

Px−

Figure P6.5c Charge densities ρN(x) and ρP(x) at the PN junction.


Thesecondboundarycondition is that theelectricfieldmustbezero intheneutralregionoutsidethedepletionregionbecausethereisnonetchargeinthat

region.Because = −Φ

Eddx

,thisconditionimpliesthat

Φ = 0 at = −x xP (P6.5f)

Φ = 0 at =x xN (P6.5g)

AssumethatwecanmodelthePNjunctionasaparallelplatecapacitorwiththeplates locatedat =x xN and = −x xP .Use the techniquedemonstrated inExample6.8tocalculatetheelectricpotentialinthedepletionregion.Assumethefollowingparameters:

q= × −1.6 10 19 coulomb, = 1016ND carriers/cm3, = ×0.3 1016NA ,

= 0.15 mxN , = 0.5 mxP ,Δx=0.005μm, ε = × −1.04 10 12 F/cm.

1. SolveforΦinthep-typeregionovertheinterval − ≤ ≤ 0x xP withthebound-aryconditionsΦ − =( ) 0xP andΦ(0)=0andusingthefinitedifferenceequa-tionstodefineatri-diagonalmatrixandthensolvingthematrixproblem.

2. SolveforΦinthen-typeregionovertheinterval ʺ ʺ0 x xN withtheboundaryconditionsΦ(0)=0and Φ =( ) 0xN usingthesamemethodasdescribedinpart1.

3. Plot your results for both thep- andn-type regions onone graphover theinterval − ≤ ≤x x xP N .

Problem.6.6

ForthePNjunctioninProblem6.5,wearbitrarilysetthegroundpointΦ =0tobeattheexactpointoftheP-to-Nboundary.SupposeinsteadwesetthegroundpointtobeattheP-to-intrinsicboundary,thatis, Φ − =( ) 0xP .Theproblem(forthep-typeregion)becomes

Φ=

ε

2

2ddx

qN A (P6.6a)

Φ − =( ) 0x p (P6.6b)

Φ − =( ) 0xP (P6.6c)

Forthen-typeregion,

Φ=−

ε

2

2ddx

qND (P6.6d)

Attheintersectionofthetworegions,wherex=0,

Φ =Φ− +(0 ) (0 ) (P6.6e)


and

Φ=

Φ− +(0 ) (0 )ddx

ddx

(P6.6f)

1. FindanexactsolutionforEquations(P6.6a)–(P6.6f)bydirectintegration. 2. Find a numerical solution for Equations (P6.6a)–(P6.6f) using MATLAB’s

ode45function.Assume: 0.3 1016NA = × carriers/cm3, 1016ND = carriers/cm3,ε = × −1.04 10 12 F/cm,−xP ≤x≤xN,where = 0.5 mxP , = 0.15 mxN ,andΔx=0.005μm.

3. Plotyourexactandode45solutionsonthesameaxes. 4. Findandprinttheglobalerrorintheode45solution.

Problem.6.7

A heat sink is often attached to electronic components to dissipate excess heatgeneratedbythedevicetopreventitfromoverheating(FigureP6.7a).AsketchofoneheatsinkfinisshowninFigureP6.7b,wherethetrapezoidalfinhasabottomwidthW1,topwidthW2,heightL1,anddepthD,andthecoordinatesxandyareasdrawninthediagram.Thetemperaturedistributioninthefinisgovernedbyaone-dimensionalanalysisoftheheatequationinasolid.Theanalysisincludesanempiricalrelationthatinvolvestheconvectiveheattransfercoefficienth.Thisempirical relation is introducedtoseparate theheatconductionprobleminthesolidfromtheheattransferprobleminthesurroundingfluid.Theconvectiveheattransfercoefficienthisdeterminedbyeitherexperimentoranalytical/numericalmethods.Thegoverningequationforthefinis

= − ∞( )d

dxAdTdx

hPk

T T (P6.7a)

Figure P6.7a Typical fin heat sink used to cool an electronic component such as a microprocessor.


or

+ = − ∞( )

2

2dAdx

dTdx

Ad Tdx

hPk

T T (P6.7b)

whereTisthetemperatureinthefinatpositionx,T∞isthesurroundingambientairtemperature,A isthefincross-sectionalarea,histheconvectiveheattransfercoefficient,kisthethermalconductivityofthefinmaterial,andPisitsperimeter.Webeginbywritingequationsfortheareaandperimeterintermsofthemeasure-mentsofthefin:

A=yD

P = 2y + 2D

Bythegeometryofsimilartriangles,wecandeterminethat =−

1

yW

L xL

,andthustheareaandperimeterasfunctionsofxare

= −

= − +

( ) ( )

( ) 2 ( ) 2

1

1

A xW DL

L x

P xW DL

L x D

Thegoverningdifferentialequationbecomes

− − = − + − ∞( ) 2 ( ) 2 ( )1

2

21 1W D

LL x

d Tdx

W DL

dTdx

hk

WL

L x D T T (P6.7c)

Cross section:

Device surface

L

L1

y

xD

W1

W2

Side view:

Figure P6.7b Side and cross-sectional views of a single fin of the heat sink with trapezoidal height L1, bottom width W1 , top width W2 , and depth D. The trap-ezoid is formed from a truncated triangle of height L.


Thefirstboundaryconditionisthatatthesurfaceofthedevice,thetempera-tureT W isknown:

=(0)T TW (P6.7d)

Toobtainthesecondboundarycondition,weemploytheconceptofenergyconservation,thatis

RatethatHeatLeavestheFinat(x=L1)perUnitSurfaceAreabyConduction =RatethatHeatIsCarriedAwaybyConvectionperUnitSurfaceArea

Thisstatementcanbewrittenmathematicallyas

− = − ∞( ) [ ( ) ]1 1k

dTdx

L h T L T (P6.7e)

Wewishtosolvethisproblemnumericallyusingthefinitedifferencemethod.First,subdividethexaxisintoNsubdivisions,giving … +, , ,1 2 3 1x x x xN .Take

thetemperatureatxitobeTi.Thefinitedifferenceformulasfor2

2d Tdx

anddTdx

(seeTable 6.1)are

=

+ −+ −( ) 22

21 1

2d Tdx

xT T T

xii i i

=

−+( ) 1dTdx

xT T

xii i

ThefinitedifferentialformofEquation(P6.7c)is

−

+ −−

−= − + −+ − +

∞( ) 2 2 ( ) 2 ( )1 1 12

1 1 1W DL

L xT T T

xw DL

T Tx

hk

WL

L x D T Tii i i i i

i i

SolvingforTigives

( ) ( )

( ) ( ) ( )

= − − + − +

× − − + − + − +

−

+ − ∞

2 2 2

2 2

1 12

11

1 11

11

21

TW DL

L xW DL

xh xk

WL

L x D

W DL

L xW DL

x TW DL

L x Th xk

WL

L x D T

i i i

i i i i i

(P6.7f)

Equation(P6.7f)isvalidfori =2,3,…,N.ThefinitedifferenceformforEquation(P6.7e)is

[ ]−

−= −+

+ ∞1

1kT T

xh T TN N

N


Solvingfor +1TN gives

=

+++ ∞

1

11T h x

k

Th xk

TN N

Also, =1T TW .Alltheheattransferfromthefintothesurroundingairpassesthroughthebase

ofthefin.Thus,therateofheatlossQthroughthefinisgivenby

=− −( )1

2 1QkAxT T

whereA1isthecross-sectionalareaatthebaseofthefin,thatis,atx=0+.UsingthemethoddescribedinSection6.11foratri-diagonalsystemofequa-

tions,writeaMATLABprogramthatwillsolveforthetemperaturedistributioninthefinandtherateofheatlossthroughthefin.Usethefollowingvalues:

200 C, 40 C, 204 W/m-K, 60 W/m -K,

50, 5 cm, 20 cm, 5 cm, 0.2 cm

2

1 1

T T k h

N W L L D

W = ° = ° = =

= = = = =

∞

Theoutputofyourprogramshouldincludethevaluesof °, , , ,h k T T Qw ;atableofTversusxatevery0.1cm;andaplotofTversusx.

201

Chapter 7

Laplace Transforms

7.1 IntroductionTransformtechniquesinvolveapplyingamathematicaloperationtotheequationsofaproblem,solvinginthe(presumablyeasier)transformdomain,andthentakingtheinversetransformtoobtainananswerinthedomainofinterest.Laplace trans-formsarefrequentlyusedtosolveordinaryandpartialdifferentialequations.Themethodreducesanordinarydifferentialequationtoanalgebraicequation,whichcanbemanipulatedtoaformsuchthattheinversetransformiseasilyobtained.In circuit theory, any circuit containing capacitors or inductors will contain atleastonedifferentialrelationship,andalargecircuitmaycontainmanymore.InChapter6,wesolvedcircuitsdirectlyinthetimedomainbyintegratingthedif-ferentialequationsinMATLAB.WenowdescribetheLaplacetransformapproach[1,2];weconvertthetime-domainequationstotheLaplacedomain,solvealgebra-ically,andthentaketheinversetransformbacktothetimedomaintoobtainthefinalanswer.Sometimes,wedonotevenbotherwith thefinal stepbecause theresultintheLaplacedomainissufficienttofullysolvetheproblemofinterest.

7.2 Laplace Transform and Inverse TransformLetf(t)beacausalfunctionsuchthatithassomedefinedvalueforallt≥0andiszerootherwise,thatis,f(t)=0fort <0.TheunilateralLaplacetransformisdefinedas

L ∫= = −

∞

( ( )) ( ) ( )0

f t F s e f t dtst (7.1)


F(s) iscalled theLaplace transformof f (t),andthe symbol L isused to indi-catethetransformoperation.Byconvention,weuselowercaselettersfortimedomain functions and uppercase for their Laplace domain (or “s-domain”)counterparts.

Theinverse Laplace transformofF(s)isdefinedtobethefunctionf(t)suchthat

L =− ( ( )) ( )1 F s f t (7.2)

Tables have been created that contain both f(t) and the corresponding F(s)(see Table 7.1attheendofthischapter).Often,weneedtorearrangethetransformtoaformorseveralformsthatareavailableinthetableofLaplacetransforms.WenowderivetheLaplacetransformforsomecommonfunctions.

7.2.1 Laplace Transform of the Unit StepWedefinetheunitstepfunctionu(t)mathematicallyas

( )1 for 00 for 0

u ttt

=≥

<

Then

L ∫=

= −

=

−

∞

− ∞

( ( )) (1)

1

0

0

u t e dt

es

s

st

st

(7.3)

7.2.2 ExponentialLet

( )f t e at=

Laplace Transforms ◾ 203

whereaisaconstant.Then,

L ∫

∫

=

=

= −−

=−

−

∞

− −

∞

− − ∞

( )

( )

1

0

( )

0

( )

0

e e e dt

e dt

es a

s a

at at st

s a t

s a t

(7.4)

7.2.3 LinearityLet

f(t)=Ag(t)+Bh(t)

whereg(t)andh(t)arefunctions,andAandBareconstants.Then,

L

L L

∫

∫∫

+ = +

= +

= +

−

∞

− −

∞∞

( ( ) ( )) ( ( ) ( ))

( ) ( )

( ( )) ( ( ))

0

00

A g t B h t Ag t Bh t e dt

A g t e dt B h t e dt

A g t B h t

st

st st

(7.5)

Thus,theLaplacetransformofasumisequaltothesumofthetransformsoftheaddends,andtheLaplacetransformofascaledfunctionisequaltothetrans-formoftheunscaledfunctionmultipliedbythescalingfactor.

7.2.4 Time DelayLet

f(t)=g(t −T )


wheref(t)issimplyatime-delayedversionofg(t)withdelaytimeTseconds(withTpresumedalwayspositive).Then,

L ∫− = − −

∞

( ( )) ( )g t T g t T e dtst

T

wherethebottomintegrationlimithasbeensettoTbecauseg(t −T )ispresumedtobezerofort <T.Then,letτ =t −T(andthusdτ =dt).Then,

L ∫

∫

− = τ τ

= τ τ

=

− τ+

∞

− − τ

∞

−

( ( )) ( )

( )

( )

( )

0

0

g t T g e d

e g e d

e G s

s T

sT s

sT (7.6)

Thus,theLaplacetransformofadelayedfunctionisequaltothetransformoftheundelayedfunctionmultipliedbythedelayfactor e sT− .

7.2.5 Complex ExponentialLet

( )f t e j t= ω

Then,applyingEquation(7.4),weget

L =− ω

=− ω

×+ ω+ ω

=+ ω+ω

=+ω

+ω+ω

ω( ) 1

1

2 2

2 2 2 2

es j

s js js j

s js

ss

js

j t

(7.7)


Byapplyingtheidentity cos sine x j xjx = + ,wealsoknow

L L

L L

= ω + ω

= ω + ω

ω( ) (cos sin )

(cos ) (sin )

e t j t

t j t

j t

(7.8)

Equating the real and imaginary components of Equation (7.7) andEquation (7.8)givestheLaplacetransformsforsineandcosine:

L ω =

+ω( ) 2 2cos t

ss

(7.9)

L ω =

ω+ω

( ) 2 2sin ts

(7.10)

Generalizingtothecasewhentheexponenthasbothrealandimaginaryparts,

L

L L

=− + ω

=− − ω

=−

− +ω+

ω− +ω

= ω + ω

+ ω( ) 1( )1

( )

( ) ( )( cos ) ( sin )

( )

2 2 2 2

es a j

s a js a

s ajs a

e t j e t

a j t

at at

Wemayobservefromthisthatmultiplicationofthefunction ( )f t e j t= ω bye atinthetimedomainproducesansshiftintheLaplacedomain.Thismaybegeneri-callystatedas

L = −( ( )) ( )e f t F s aat (7.11)

7.2.6 Powers of tLet

( ) 1f t t n= +

Then,

L ∫=+ +

∞

−( )1 1

0

t t e dtn n st


Tosolve,integratebyparts.Let 1u t n= + anddv e dtst= − .Then, ( 1)du n t dtn= +

and = −−

ves

st

.Applying udv uv vdu∫ = − ∫ ,weget

L∫ ∫= − +

+=

++ −

∞

+−

=

∞

−

∞1 1 ( )1

0

1

0 0

t e dt tes

ns

t e dtns

tn st nst

t

n st n (7.12)

FromEquation(7.12),wecanseethat

L L

L L

L L L

=

=−

= = =

−

− −

( ) ( )

( ) 1 ( )

( ) 1 ( ) 1 (1) 1

1

1 2

1 02

tns

t

tns

t

ts

ts s

n n

n n

(7.13)

7.2.7 Delta FunctionThedelta functionwasformalizedbyphysicistPaulDiracandisdefinedasfollows:

( ) for 0

0 otherwiset t

δ =∞ =

and

( ) 10

t dt∫ δ =

∞

Althoughwe cannotphysically realizeδ(τ) in an actual circuit (though it isoften approximated), it has many useful applications in circuit theory, and itsLaplacetransformis

L ∫δ = δ = =−

∞

− ∞( ( )) ( ) 1

00

t t e es t s t

Example 7.1

Find the Laplace transform of the pulse p(t) as shown in Figure 7.1, where A is the amplitude, T1 is the turn-on time, and T2 is the turn-off time.


We can express the pulse mathematically as the sum of two delayed unit steps with amplitude A:

p t Au t T Au t T( ) ( ) ( )1 2= − − −

We know from Equation (7.3) that the Laplace transform of the unit step is

s1 . Also, from Equation (7.5), the transform of a sum is simply the

sum of the transforms. In addition, from Equation (7.6), the transform of a time-delayed function is simply e sT− times the transform of the undelayed function. Thus, the transform of the pulse is

P s A es

A es

Ase e

sT sT

sT sT

( ) ( )( )1

( )( )1

( )

1 2

1 2

= −

= −

− −

− − (7.14)

Example 7.2

Find the inverse Laplace transform of the following function:

F s

es

T eT s

sT sT

( )1 2

42

2

2 2 2=−

+π+ π

− −

(7.15)

We start by decomposing the first term of Equation (7.15):

es s

es

sTsT1 1 1

2 2 2

−= −

−− (7.16)

We know from Equation (7.13) that L =ts

( )12 and from Equation (7.6)

that multiplication by e sT− in the Laplace domain corresponds to delay

T1 T2t

A

p(t)

Figure 7.1 Pulse with amplitude A, turn-on time T1, and turn-off time T2.


in the time domain. Thus, the inverse transform of Equation (7.16) is the subtraction of a ramp f t t( ( ) )= and a time-delayed ramp and is shown graphically in Figure 7.2a.

For the second term of Equation (2.15), we rearrange as follows:

T eTs

TeT

T s

Te T

sT

Tes

sTsT

sT

sT

2( ) 4

2(2 )

2

2

2

2 2 2 2 2

22

2 2

π+ π

=π+ π

=

π

+π

=ω+ ω

−−

−

− (7.17)

By inspection, Equation (7.17) corresponds to a time-delayed sine wave of

amplitude T and frequency ω =π

T2

and is illustrated in Figure 7.2b. The

complete inverse transform is shown in Figure 7.2c.

( )f t t=

t Tt

T

tT

− =

( ) ( )f t g t−( )g t t T= −for Tt ≥

(a)

Tt

T

−T T

t

T

2T

(b) (c)

Figure 7.2 Example of inverse Laplace transform. (a) The inverse transform of

− −

s2e sT

s1 12 is the subtraction of a ramp and a time-delayed ramp. (b) The inverse

transform of ω+ ω

−

s2 2Te sT is a time-delayed sine wave. (c) The complete inverse

transform is the sum of the graphs in (a) and (b).


7.3 Transforms of DerivativesLet

( ) ( )

f tdf tdt

=

Then,

L ∫ ∫= =−

∞

−

∞

( )0 0

fdfdt

e dt e dfst st

Integratingbyparts, letdv =dfand u e st= − .Then,v = fand du se dtst= − − .Applying udv uv vdu∫ = − ∫ ,weget

L∫ ∫= + = − +−

∞

−

=

∞ −

∞

(0) ( )0

00

e df fe s fe dt f s fst stt

st (7.18)

Thus,theLaplacetransformofaderivativeisthetransformoftheoriginalfunc-tionmultipliedbysandminusthevalueofthefunctionattimet = 0(i.e.,lesstheinitialcondition).

Similarly,forthesecondderivative,

L ∫= −

∞

( ) ( )0

f f t e dtst

L ∫ ∫( ) = =−

∞

−

∞

0 0

fdfdt

e dt e dfst st

Let dv df= and u e st= − . Then, v f= and du se dtst= − − . Integrating byparts,weget

L

L

∫ ∫= +

= − +

= − − +

−

∞

−

=

∞ −

∞

(0) ( )

(0) (0) ( )

00

0

2

f e dt f e s f e dt

f s f

f s f s f

st s tt

st

(7.19)

ByEquations(7.18)and(7.19),wecanseethepatternfortheLaplacetransformofthenthderivative,thatis:

L L= − − − −− −( ) ( ) (0) (0) (0)( ) 1 2 ( )f s f s f s f fn n n n n (7.20)


7.4 Ordinary Differential Equations, Initial Value Problem

ConsiderthedifferentialequationarisingfromtheparallelRLCcircuitofAppendixAinEquation(A.7).Thegoverningdifferentialequationfortheinductorcurrentiis

1 1 ( )i

RCi

LCi

I tLCo+ + = (7.21)

whereListheinductance(inhenries),Ristheresistance(inohms),Cisthecapaci-tance(infarads),andIoisthedrivingcurrent.

Wealsoassumetheinitialconditionstobe

i(0)=α and (0)i =β

First,weconvertthisequationintothefollowinggeneralform:Lety(t)=i(t),

= = =1 , 1 and ( ) ( )

pRC

qLC

f tI tLCo .Then,Equation(7.20)becomes

( )y py qy f t+ + = (7.22)

TheLaplacetransformofeachofthetermsinEquation(7.22)isasfollows:

L

L

L

L

( )

= − α −β

= −α

=

=

( ) ( )

( ) ( )

( ) ( )

( ) ( )

2y s Y s s

py p sY s

qy qY s

f F s

Thus,Equation(7.22)becomes

( ) ( ( ) ) ( ) ( )

( ) ( ) ( )

2

2

s Y s s p sY s qY s F s

Y ss p F s

s ps q

− α −β + − α + =

→ =+ α +β +

+ +

Let

( ) 1

2H ss ps q

=+ +

Then,

( ) [( ) ] ( ) ( ) ( )Y s s p H s F s H s= + α +β + (7.23)


BydoingapartialfractionexpansionofEquation(7.23)andusingLaplacetransformtables,wecanobtainy(t)byfindinginversetransformofY(s),thatis, ( ( )) ( )1 Y s y tL =− .

Example 7.3

Given the following differential equation (no damping and an exponentially decaying driving function), determine the solution for y(t):

y y e

y y

t5

(0) 2, (0) 0

+ =

= =

−

(7.24)

This problem fits the general form of Equation (7.22), with p = 0, q = 1, and f e t5= − . Thus,

Y(s) = (2s)H(s)+F(s)H(s)

where

H s

s( )

112=

+ and

L= =

+−F s e

st( ) (5 )

51

Thus,

Y s

ss s s

( )2

15

( 1)( 1)2 2=+

++ +

(7.25)

For the second term of Equation (7.25), we need to decompose by the method of partial fractions:

s sAs

Bs Cs

A s Bs C ss s

5( 1)( 1) 1 1

( 1) ( )( 1)( 1)( 1)

2 2

2

2

+ +=

++

++

=+ + + +

+ +

Matching terms in the numerator,

A B C B A C

A B C

0, 0, 5

52,

52,

52

+ = + = + =

→ = = − =

Thus,

s s sss

5( 1)( 1)

52( 1)

52( 1)( 1)2 2+ −

=+

−−+

and substituting back into Equation (7.25), we get

Y s

ss s

ss s

( )2

( 1)5

2( 1)5

2( 1)5

2( 1)2 2 2=+

++

−+

++


From Table 7.1, we observe

ss

t1

cos12L+

=−

st

11

sin12L+

=−

se t1

11L

+=− −

Therefore, by applying inverse transforms, the solution to Equation (7.24) is

y t t t e t( )

12cos

52sin

52

= − + + −

A plot of y versus t is shown in Figure 7.3.

Example 7.4

Determine the solution of the following differential equation:

y y y t

y y

3 2 5sin2

(0) 1, (0) 4

+ + =

= = − (7.26)

0 5 10 15 20−3

−2

−1

0

1

2

3

4

t(s)

y(m

)

y vs. t

Figure 7.3 Plot of y versus t for Example 7.3.


Using the same form as Equation (7.22),

p

q

f t tt

3

2

( ) 5sin2 10sin22

=

=

= =

In the Laplace domain,

H s

s s s s( )

13 2

1( 2)( 1)2=

+ +=

+ +

L= =

+F s

ts

( ) 10sin22

1042

Thus,

Y s s H s F s H s

ss s s s s s s

( ) ( 3)(1) 4 ( ) ( ) ( )

( 2)( 1)1

( 2)( 1)10

( 4)( 2)( 1)2

[ ]= + − +

=+ +

−+ +

++ + +

(7.27)

We now make use of the MATLAB function residue function, which can be used to solve partial fraction expansions. The residue function takes two arguments that are arrays containing the coefficients of the numerator and denominator polynomials to be expanded. It then returns the numerator coefficients (the “residues”), the denominator roots (the “poles”), and any nonfractional term (which will always be zero in the usual case where the denominator is of higher order than the numerator).

To solve with residue, we rewrite the first term of Equation (7.27) in polynomial form:

ss s

ss s( 2)( 1) 3 22+ +

=+ +

(7.28)

By the use of partial fractions, we can determine that

ss s s s( 2)( 1)

22

11+ +

=+

−+

Now, let us use MATLAB’s residue to reduce the right-hand side of Equation (7.28) to the sum of two fractions. In the numerator of the right-hand side of Equation (7.28), we see that the first polynomial term (for s1) has a coefficient of 1, and the second term (for s0) has a coefficient of 0. This is expressed in MATLAB as [10]. Similarly, the denominator polynomial has


an s2 coefficient of 1, an s1 coefficient of 3, and an s0 coefficient of 2, which is expressed as [132]. Then, in MATLAB we run

>> [r, p, k] = residue([1 0],[1 3 2])

r= 2 -1p= -2 -1k= []

This result shows that the residues are 2 and −1, and their respective poles are −2 and −1, giving

ss s s s3 2

2( 2)

1( 1)2 + +

=+

−+

(7.29)

which is the same that was obtained by the use of partial fractions.For the second term in Equation (7.27), we run residue([-1],[132])

to obtain

s s s s

s s

1( 2)( 1)

13 2

1( 2)

1( 1)

2

−+ +

=−

+ +

=+

−+

(7.30)

For the third term in Equation (7.27), we first rewrite as a polynomial:

s s s s s s s10

( 4)( 2)( 1)10

3 6 12 82 4 2 2+ + +=

+ + + +

Then, in MATLAB we obtain

>> [r, p, k] = residue([10],[1 3 6 12 8])r= -0.3750+0.1250i -0.3750-0.1250i -1.2500 2.0000p= 0.0000+2.0000i 0.0000-2.0000i -2.0000 -1.0000k= []


Thus, the residues are at −0.375 ± 0.125j, −1.25, and 2. The correspond-ing poles are at ±2j, −2, and −1. Thus,

+ + +=

+ + + +

=− +

−+− −

+

−+

++

s s s s s s s

js j

js j

s s

10( 4)( 2)( 1)

103 6 12 8

0.375 0.125( 2 )

0.375 0.125( 2 )

1.25( 2)

2( 1)

2 4 2 2

(7.31)

The final answer is the sum of the inverse transforms of Equations (7.29), (7.30), and (7.31). From Table 7.1,

L

L

L

L

+=

+=

−=

+=

− −

− −

−

− −

se

se

s je

s je

t

t

jt

j t

12

11

12

12

1 2

1

1 2

1 2

Collecting all the terms in Y(s) and applying these inverse transforms gives the complete solution for Equation (7.26):

L ( )= = − − + −− − − −y t Y s e e e je jet j t j t j t j t( ) ( ) 1.75 0.375 0.375 0.125 0.1251 2 2 2 2 2

Applying the identities e e tj t j t 2cos+ = ωω − ω and e e j tj t j t 2 sin− = ωω − ω , we obtain

y t e t tt( ) 1.75 0.75cos 2 0.25sin22= − −− (7.32)

It is illustrative to compare the Laplace-obtained solution with a direct solution of Equation (7.26) obtained via MATLAB’s ode45 function. The program follows, and the output is shown in Figure 7.4:

%Example_7_4.m:ThisprogramsolvesEquation7.26by%convertingitintoasystemoftwodifferential%equationsandsolvingwithode45.%Letv=y',Y=[yv],andY'=[y'v'].%Y(1)=y,Y(2)=v,Y'(1)=Y(2),Y'(2)=5*sin2t-3Y(2)-2Y(1)%Initialconditions:y(0)=1.0,y'(0)=v(0)=-4.0clear;clc;%Defineinitialconditions:Y0=[1.0-4.0];%Definetimeintervaltosolve:dt=0.1;Tstop=10;Tspan=0.0:dt:Tstop;%solvetheODEdirectlywithode45:[t,Yode45]=ode45('dydt',Tspan,Y0);%computethesolutionfoundviaLaplaceTransforms


%(Equation7.32):Ylaplace=1.75*exp(-2*Tspan)...-0.75*cos(2*Tspan)-0.25*sin(2*Tspan);plot(t,Yode45(:,1),'x',Tspan,Ylaplace),axis([0Tstop-11]);xlabel('t'),ylabel('Yode45,Ylaplace');title('ODEsolutionsusingode45andLaplaceTransform');legend('ode45','Laplace','Location','SouthWest');-------------------------------------------------------------------

%dydt:thisfunctionworkswithExample7.4.functionYprime=dydt(t,Y)Yprime=zeros(2,1);Yprime(1)=Y(2);Yprime(2)=5*sin(2*t)-3.0*Y(2)-2.0*Y(1);

Example 7.5

Determine the solution of the following differential equation:

+ + =<

≥

= =

y y yt t

t

y y

3 25 , for 2

0, for 2

(0) 1, (0) 0

(7.33)

0 2 4 6 8 10−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

t

Yod

e45,

Yla

plac

eODE solutions using ode45 and Laplace Transform

ode45Laplace

Figure 7.4 Comparison between solution obtained by Laplace transforms and MATLAB’s ode45 for Example 7.4.


Applying the standard form of Equation (7.22), let p = 3, q = 2, and

f t

t t

t( )

5 , for 2

0, for 2=

<

≥

We begin by writing f(t) into a form that is recognizably transformable, that is, in terms of the unit step function:

r t t u t u t

t u t t u t

( ) 5 ( ) ( 2)

5 ( ) 5 ( 2)

( )= − −

= − − (7.34)

The first term in Equation (7.34) is easily transformable L =t u ts

(5 ( ))52 ,

but the second is less recognizably so. However, for the second term, let 5t = 5(t − 2 + 2). Then,

f t t u t t u t

t u t t u t u t

( ) 5 ( ) 5( 2 2) ( 2)

5 ( ) 5( 2) ( 2) 5(2) ( 2)

= − − + −

= − − − − −

Then,

L ( )= − − − − −

= − −− −

F s tu t t u t u t

se

se

ss s

( ) 5 ( ) 5( 2) ( 2) 10 ( 2)

5 5 102

22

2

Thus,

Y s s H s

se

s sH ss( ) [( 3)(1)] ( )

5 5 10( )2

22= + + − +−

H s

s s s s( )

13 2

1( 2)( 1)2=

+ +=

+ +

=+ +

++ +

++ +

−+ +

−+ +

− −

Y ss

s s s s s s s

es s s

es s s

s s

( )( 2)( 1)

3( 2)( 1)

5( 2)( 1)

5( 2)( 1)

10( 2)( 1)

2

2

2

2

Let Y s Y Y Y Y Y( ) 1 2 3 4 5= + + + + , where

=+ +

=+ +

=+ +

= −+ +

= −+ +

−

−

Ys

s s

Ys s

Ys s s

Ye

s s s

Ye

s s s

s

s

( 2)( 1)

3( 2)( 1)

5( 2)( 1)

5( 2)( 1)

10( 2)( 1)

1

2

3 2

4

2

2

5

2


By partial fraction expansion via MATLAB’s residue function, we can determine that

L

L

L

L

L

= −

= − +

= − − +

= − − − − + −

= − + − −

− − −

− − −

− − −

− − −

− − −

Y e e

Y e e

Y t e e

Y t t e e e e u t

Y e e e e u t

t t

t t

t t

t t

t t

( ) 2

( ) 3 3

( )52

154

54

5

( )52

154

54

5 ( 2)

( ) 1012

12

( 2)

11

2

12

2

13

2

14

4 2 2

15

4 2 2

Summing these five terms gives

y t

t e e t

e e e e t

t t

t t

( )

52

154

94

7 , for 2

94

154

(7 5 ) , for 2

2

4 2 2

=

− − + <

− + + + ≥

− −

− −

(7.35)

This problem can also be solved numerically by the use of the ode45 function in MATLAB. The MATLAB program used to solve this problem follows:

%Example_7_5.m:ThisprogramsolvesEquation7.33by%convertingitintoasystemoftwodifferential%equationsandsolvingwithode45.%Letv=y',Y=[yv],andY'=[y'v'].%Y(1)=y,Y(2)=v,Y'(1)=Y(2).%Y'(2)=5*t-3Y(2)-2Y(1)fort<2%Y'(2)=-3Y(2)-2Y(1)fort>=2%Initialconditions:y(0)=1,y'(0)=0clear;clc;%Defineinitialconditions:Y0=[1.00.0];%Definetimeintervaltosolve:dt=0.1;Tstop=4;Tspan=0:dt:Tstop;%SolvetheODEdirectlywithode45:[t,Yode45]=ode45('dydt2',Tspan,Y0);%ComputethesolutionfoundviaLaplaceTransforms%(Equation7.35)fori=1:length(Tspan)ifTspan(i)<2Ylaplace(i)=2.5*Tspan(i)-15/4-...9/4*exp(-2*Tspan(i))+7*exp(-Tspan(i));else


Ylaplace(i)=-(9/4+15/4*exp(4))...*exp(-2*Tspan(i))...+(7+5*exp(2))*exp(-Tspan(i));endendplot(t,Yode45(:,1),'x',Tspan,Ylaplace),title('ODEsolutionsusingode45andLaplaceTransform');legend('ode45','Laplace');-------------------------------------------------------------------

%dydt2:ThisfunctionworkswithExample7.5.%Y(1)=y,Y(2)=v%Y1'=Y(2),Y2'=5*t-3Y2-2Y1,ift>2,Y2'=-3Y2-2Y1,ift>=2functionYprime=dydt2(t,Y)Yprime=zeros(2,1);Yprime(1)=Y(2);ift<2Yprime(2)=5*t-3.0*Y(2)-2.0*Y(1);elseYprime(2)=-3.0*Y(2)-2.0*Y(1);end

A comparison of the ode45 and the Laplace transform solution is shown in Figure 7.5.

0 0.5 1 1.5 2 2.5 3 3.5 4

0.8

1

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6ODE solutions using ode45 and Laplace Transform

ode45Laplace

Figure 7.5 Comparison between solution obtained by Laplace transforms and MATLAB’s ode45 for Example 7.5.


7.5 ConvolutionGiven two functions f(t) andg(t)withknown transformsF(s) andG(s), respec-tively,thentheinversetransformoftheproductF(s)G(s)istheconvolutionintegral

L ∫ ∫( ) = τ − τ τ = τ − τ τ− ( ) ( ) ( ) ( ) ( ) ( )1

0 0

F s G s f g t d g f t dt t

(7.36)

Proof:

( ) ( ) ( ) ( )

( ) ( )

0 0

0

( )

0

F s G s f p e dp g e d

g f p e dp d

ps s

p s

∫ ∫

∫ ∫

= τ τ

= τ τ

−

∞

−τ

∞

∞

− +τ

∞

(7.37)

Lett =p +τ.Then,whenp =0,t =τandwhenp =∞,thent =∞.Also,dp =dt.Then,

( ) ( ) ( ) ( )

( ) ( )

0

F s G s g f t e dt d

g f t e dt d

t s

R

t s

∫∫

∫∫

= τ − τ τ

= τ − τ τ

−

τ

∞∞

− (7.38)

whereR istheregionbelowthelinet =τ asshowninFigure 7.6.TheintegrationorderofthemultipleintegralaswritteninEquation(7.38)is

tointegratewithrespecttotfirst,thenwithrespecttoτ.Inthiscase,theorderof

Region R

t

τt = τ

Figure 7.6 Integration region for convolution in the (τ, t) plane.


integrationdoesnotmatter,sowecanintegratewithrespecttoτ first,thenwithrespecttot.Thisgives

L

∫∫

∫∫

∫

= τ − τ τ

= τ − τ τ

= τ − τ τ

−

−

∞

( ) ( ) ( ) ( )

( ) ( )

( ) ( )

00

0

F s G s g f t e dt d

e g f t d dt

g f t d

R

s t

s t

t

t

Thus,

L ∫= τ − τ τ− ( ( ) ( )) ( ) ( )1

0

F s G s g f t dt

(7.39)

Alternatively,wecanreversetherolesoff andg,giving

L ∫= − τ τ τ− ( ( ) ( )) ( ) ( )1

0

F s G s g t f dt

Convolutionisdenotedbythe“*”operator;thus,

L ( ) = ∗ = ∗− ( ) ( ) ( ) ( ) ( ) ( )1 F s G s f t g t g t f t

Example 7.6

We illustrate the Laplace transform of the convolution function by studying the waveforms of Figure 7.7a where we have two pulses f(t) and g(t):

f tt T

g tT t T

( )for 0

0 otherwise

( )for

0 otherwise

1

2 3

=α ≤ <

=β ≤ <

Figure 7.7b shows the convolution operation where we take the mir-ror image f(τ − t) of f(t) and slide it along the t axis (by varying τ) while


T1t

T3T2t

α

β

f(t)

g(t)

(a)

T3T2t

g(t)

f (t − τ)for varying τ

(b)

T3T2t

T1+T2 T1+T3

f (t)*g(t)αβ

(c)

T1

T2 T1 + T2

T1 + T3

e−sT1

s2e−s(T1 + T3)

s2

+ + +

++ +–e−sT2

s2 s2–e−s(T1 + T2)–1 –1 –1 –1

(d)

Figure 7.7 Graphical representation of convolution. (a) The pulse waveforms f(t) and g(t). (b) We perform the convolution graphically by taking the mirror image of f(t), sliding it on the t axis, and integrating with g(t). (c) The result of graphi-cally convolving f(t)*g(t) is a trapezoidal waveform. (d) The identical solution is obtained by taking the inverse transform of F(s)G(s).


integrating the two functions. The result of the convolution f(t)*g(t) is the trapezoidal waveform shown in Figure 7.7c:

f t g t

T Tt T T t T

T t T T

T Tt T T T T t T T

( ) ( )

( )( ) for

for ( )

( )( ) for ( ) ( )

0 otherwise

3 22 2 3

3 1 2

3 21 3 1 2 1 3

∗ =

αβ−

− ≤ <

αβ ≤ < +

−αβ−

− − + ≤ < +

(7.40)

We now calculate the Laplace transforms of f(t) and g(t), multiply them in the s domain, and then inverse transform to obtain the identical answer. We previously calculated the transform of a pulse in Equation (7.14). Thus,

F ss

e

G sse e

sT

sT sT

( ) (1 )

( ) ( )

1

2 3

=α

−

=β

−

−

− −

Multiplying,

F s G s

se e e esT sT s T T s T T( ) ( ) 2

( ) ( )2 3 1 2 1 3( )=αβ

− − +− − − + − + (7.41)

By inspection, the inverse transform of Equation (7.41) consists of the sum of four time-delayed ramp functions and is shown graphically in Figure 7.7d. The sum of the four ramps equals the same trapezoid as in Figure 7.7c.

7.6 Laplace Transforms Applied to CircuitsIncircuittheory,capacitorsandinductorshaveconstituentrelationsthatparticu-larlylendthemselvestoanalysisintheLaplacedomain.Startingwiththecapacitor,

i C

dvdtCC=

whereiCisthecapacitorcurrent,vCisthecapacitorvoltage,andCisthecapacitorvalue (in farads). Applying Equation (7.18), we can rewrite this equation in theLaplacedomainas

( ) ( ) (0)I s C sV s vC C C( )= −

Ifweneglectforthemomenttheinitialcondition,thisbecomes

( ) ( ) 1V s I s

CsC C= (7.42)


NotethatEquation(7.42)resemblesOhm’slaw ( )v iR= ,andwethusdefinethecomplex impedance ZCofacapacitortobe

( ) 1Z s

CsC = (7.43)

Bysimilaranalysiswiththeconstituentrelationfor inductors =v LdidtLL , the

compleximpedanceZLofaninductorL(neglectinginitialconditions)isshownto be

( )Z s LsL = (7.44)

Finally,forresistors,theLaplacetransformofOhm’slawimplies

( )Z s RR = (7.45)

BecausewehavepreviouslyproventhatLaplacetransformsarelinear,wecannowsolvecomplicatedlinearcircuitsdirectlyintheLaplacedomainbyusingthesamealge-braictechniquesusedtosolvetheinput/outputrelationshipsforresistor-onlycircuits.Forexample,theparallelRLCcircuitasredrawninFigure 7.8withinputcurrentIo(s),outputvoltageVc(s),andcompleximpedancescanbesolvedwithoneequation:

( ) ( ) ( ) || ( ) || ( )V s I s Z s Z s Z sC o R C L( )=

where theparallel operator ( ) indicates the calculation for resistors (or imped-ances)inparallel.Then,

( )( )

11 1 1

11 1

11 12

V sI s

Z Z Z

RCs

Ls

Cs

sRC

sLC

C

o

R C L

=+ +

=+ +

= ×+ +

(7.46)

Io(s) ZR ZL CZ VC(s)+

−

Figure 7.8 RLC circuit using impedances.


Wealsodefine

( ) ( )

( )H s V s

I sC

o= (7.47)

asthetransfer functionofthecircuit,thatis,theratiooftheoutputvariable(inthiscaseVC )totheinputdrivingfunction(inthiscaseIo).

Letusnowassumeasinusoidalformforio(t)suchthat

( )i t C eoj t= ω

where ( cos sin )e t j tj t = ω + ωω isasumofsinesandcosinesoffrequencyω.Forconvenience,wehavealsoaddedaconstantfactorofCtosimplifythemathasweprogress.Then,

( ) 1

I s Cs jo =− ω

and

( ) ( ) ( )

( )( )( )1 2

V s H s I s

ss s s s s j

C o=

=+ + − ω

(7.48)

where s1and s2arethezerosofthedenominatorofEquation(7.46)andarecal-culatedfromR,L,andC.Wealsoassumeforthisanalysisthats1ands2arereal.RearrangingEquation(7.48)bypartialfractions,weget

=

+ − ω+

++ − ω

++

ωω + ω +

− ω( ) ( )( ) ( )( ) ( )( )

1

1 2 1

1

2

2 1 2

2

1 2V s

ss s s j

s s

ss s s j

s s

jj s j s

s jC

Byinversetransform,weget

( )

( )( ) ( )( )( )1

1 2 1

2

2 1 2

1 2v ts

s s s je

ss s s j

e H j eCs t s t j t=

+ − ω+

+ − ω+ ω− − ω (7.49)

AlthoughEquation(7.49)lookscomplicated,notethatthefirsttwotermscontainnegativeexponentialsandthusinthesteady state(t →∞),thisequationreducesto

lim ( ) ( )v t H j et

cj t= ω

→∞

ω (7.50)


ThemeaningofEquation(7.50)isthatforanysinusoidalinput(offrequencyω)tothecircuit,wecancalculatethevalueofthesteady-stateoutput(alsooffre-quencyω)bycalculatingH( jω).Thisderivationcanbegeneralizedfordetermin-ingthefrequency responseofanylinearcircuitbyfindingthetransferfunctionH(s)analyticallyandthenlettings =jωtofinditsfrequency-dependentbehavior.

Frequency response is traditionally plotted as two graphs showing the mag-nitude | ( )|H jω andphase∠H( jω),bothversus frequency.Themagnitudeplotistypicallyonlog-logaxes,andthephaseplotistypicallyonsemilogaxes(linearphasevs.logfrequency).Together,theplotsarereferredtoasaBode plot.

Example 7.7

Let

H s

s( )

10001000

=+

Plot the magnitude and phase of H(jω) over the interval [1 10 , 1 10 ]1 6ω = × ×− rad/sec. The solution is shown in Figure 7.9.

mag

H(jw

)

Bode Plot

100 101 102 103 104 105 106

100

100

10−2

101 102 103 104 105 106

−100

−80

−60

−40

−20

0

Frequency (rad/sec)

phas

e H

(jw)

Figure 7.9 Bode plot for ωω ==ωω ++

( )1000

1000H j

j.


%Example_7_7.m:Plotmagnitudeandphaseof%H(jw)=1000/(s+1000)forw=[1e-1,1e6].%First,createalogarithmicpointsetforthefrequency%rangefrom1to1e6rad/secwithtenpointsineach%decade.fork=1:60w(k)=10^(k/10);end%ComputethemagnitudeforH(jw)H_mag=abs(1000./(j*w+1000));%ComputethephaseforH(jw),andconverttodegreesH_phase=(180/pi)*angle(1000./(j*w+1000));%Plotmagnitudeonlog-logaxessubplot(2,1,1);loglog(w,H_mag);axis([10^010^610^-32*10^0]);ylabel('magH(jw)');title('BodePlot');%Plotphaseonlog-linearaxessubplot(2,1,2);semilogx(w,H_phase);axis([10^010^6-10010]);xlabel('Frequency(rad/sec)');ylabel('phaseH(jw)');

7.7 Impulse ResponseInSection7.2.7,wediscussedthedeltafunctionanditsproperties.WhathappensifweuseadeltafunctionastheinputtoacircuitwithknowntransferfunctionH(s)?InEquation(7.47),wedefinedH(s)astheratioofthecircuitoutputtothecircuitinput.Ifweassumethattheinputisadeltafunction,then

( ) ( ) ( )

( ) 1

V s H s I s

H s

C o

( )

=

= (7.51)

where we have assumed ( ) ( )i t to = δ with corresponding Laplace transform ofunity.IfweinversetransformEquation(7.51),weget

( ) ( )v t h tC =

Inotherwords,ifthecircuitinputisadeltafunction,thenthecircuitoutputis simply the inverse transform of the transfer function h(t), also known as theimpulse response.Conversely,ifwecanaccuratelycalculate(ormeasure)theimpulseresponseh(t)ofacircuit,thenwecanobtainthetransferfunctionH(s)andthuspredictthecircuitbehaviorforanyarbitraryinputwaveform.Thisconceptisillus-tratedinExample7.8.


Example 7.8

A circuit has an impulse response of

h t e tt( ) 10 cos102= −

What is the unit step response r(t) of the circuit?Transforming h(t) into H(s), we obtain

H s

ss

( ) 102

( 2) 1002=+

+ +

where we have applied Equations (7.9) and (7.11) to determine the Laplace transform. To obtain the step response, we first multiply H(s) by the unit step

transform s1

to obtain R(s):

R s

ss s

( ) 102

( 2) 1001

2=+

+ +

We now calculate the inverse transform as follows:

L

L

L

=

=+

+ +

=− −

+ −+− −

+ ++

×

= − + −

− +

= − − +

−

−

−

− + − + − −

− −

− − −

r t R s

ss s s

js j

js j s

e je e

je u t

e e t e t u t

t j t t j t t j t

t j t

t t t

( ) ( )

102

4 104

100.0096 0.0481

(2 10 )0.0096 0.0481

(2 10 )0.0192

(calculated using residue)

10( 0.0096 0.0481 0.0096

0.0481 0.0192 ( ))

1.92 ( cos10 5 sin10 ( ))

1

13 2

1

2 10 2 10 2 10

2 10

2 2 2

Exercises

Exercise7.1 Determine L ω(sin )2 t .

Exercise7.2 Determine L( cos )2e tt .

Exercise7.3 Determine L −( sin2 )3e tt

Exercise7.4 DeterminetheLaplacetransformofthefunctionshowngraphicallyinFigureE7.4.


Exercise7.5 Determine L+

+ −− 2 1

( 9)( 2)1

2s

s s.

Exercise7.6 Determine L+ +

− 1( 5 6)

.12s s

Exercise7.7 Determine L+ + +

− 1( 1)( 3 2)

12s s s

by convolution in the timedomain.

Exercise7.8 SolvethefollowingdifferentialequationbyLaplacetransforms:

3 2 3 , (0) 1, (0) 2y y y e y yt+ − = = =−

Projects

Project.7.1

FigureP7.1showsaseriesRLCcircuitdrivenbyaninputvoltagevi(t)andthreedefinedoutputvoltagesvR(t),vL(t),andvC(t).

t

1

−11

3

f (t)

Figure E7.4 f(t) consists of one cycle of a sine wave over the interval 1 ≤ t ≤ 3 sec.

C

−

+vR

−

+vL

−

+vC

L

R

+–

vi

Figure P7.1 Series RLC circuit.


1. Solvethecircuitusingcompleximpedancestofindthethreetransferfunc-

tions =( ) ( )( )

H sV sV sRR

i, =( ) ( )

( )H s

V sV sLL

i,and =( ) ( )

( )H s

V sV sCC

i.Putthetransfer

functionsinthefollowingstandardform:

( ) 1

22 3

24 5

H sK s K s Ks K s K

=+ ++ +

wheretheconstantsK1throughK5arereal.Notethatthefiveconstantsarescaledsuchthatthehighest-orderterminthedenominatorshouldhavenoconstant.

2. CreatetheBodeplotsfor ( )H jR ω , ( )H jL ω ,and ( )H jC ω .Usethesecom-ponentvalues:R=1kΩ,L=0.1mH,andC=0.1µF.Plotovertherange

[1, 1 10 ]10ω = × rad/sec.3. Thethreetransferfunctionsrepresentthreetypesoffilters:high-pass,low-pass,

andbandpass.Whichtransferfunctioncorrespondstowhichfiltertype?

Project.7.2

Acommon emitter amplifier is showninFigureP7.2awith inputvoltagevinandoutputvoltagevout.Inthesetypesofsingle-transistorcircuits,thevoltagesofinter-estrideatopDCcomponentsVBIASandVOFFSET,whichareusedtobiasthetransis-torintoitslinear(i.e.,amplification)mode.FigureP7.2bshowsthesmall-signalhybrid-pimodelofthebipolartransistor[3].Inthesmall-signalmodel,theDCcomponents (VCC, VBIAS, and VOFFSET) are ignored, and we can use Kirchhoff’s

+–+–

vin

VBIAS

LR

VCC

vout + VOFFSET

IBIAS

Figure P7.2a Common emitter amplifier. VBIAS is the nonvarying (“DC”) compo-nent of the input, which is adjusted to obtain the desired operating point (IBIAS ) of the transistor. VOFFSET is the DC component of the output. The voltages of interest are the varying components of the input vin and the output vout.


current law to solve for the transfer function in termsof complex impedances.WritingKCLequationsatnodesvπandvoutgives

−− +

−=π π

ππ

π( )1

( )1 0v v

rv

rc s

v v

c s

in

b

out (P7.2a)

−

+ + =ππ

( )1 0v v

c s

g vv

r Rout

mout

o L (P7.2b)

wherevπisthesmall-signalbase-emittervoltage;rπisthesmall-signalbase-emitterresistance;rb is thebaseresistance;cπandcμarethebase-emitterandcollector-emittercapacitances,respectively;roisthesmall-signalcollector-emitterresistance;gmisthesmall-signaltransconductance;andRListheloadresistance.Solvingtheseequationsgivesthetransferfunctionofthesystem:

( )

1

112

H svv

rc s g

c c sc

r Rc

r R r rg

sr r r R

out

in

bm

o Lo L b

m

b o L

( )

( )( )

= =−

+ + +ππ

ππ

(P7.2c)

Ifweassumethatthetransistor ismodel2N3904,wecanobtainnumericalvaluesforthecircuitparametersfromthemanufacturerdatasheetforthiscom-ponent [4]. These are summarized in Table P7.2. We also assume that VBIAS isadjustedtoobtainatransistorcollectorcurrent(IBIAS)of1mA,andthatRL=5 kΩ.Becausethe2N3904isadiscretedevice(andnotpartofanintegratedcircuit),we

alsoassumeasmallbaseresistancerb=25Ω.Also,assumethat =gIVm

o

T,where

= = 0.026VkTqT mV.

+–vin vout

+

−

RLrogmvπ

cµ

cπrπvπ

+

−

rb

Figure P7.2b Small-signal model of the common emitter amplifier. rb is the series base resistance, rπ is the small-signal base-emitter resistance, gm is the transcon-ductance (the transistor “gain”), ro is the small-signal collector-emitter resistance, and cπ and cμ are the base-emitter and collector-emitter capacitances, respec-tively. RL is the load resistance of the amplifier.


1. Create the Bode plot for the transfer function of Equation (P7.2a)using the parameters in Table P7.2. Plot over the frequency interval

[1 10 , 1 10 ]5 11ω = × × rad/sec.2. AcommontrickforhandanalysisofthiscircuitistousetheMiller approxima-

tion;weassumealternativevaluesforcμandcπasfollows:

0

( ) 1

c

c c g r R cm o L( )

=

= + +π π

SubstitutetheseapproximatedvaluesforcμandcπintoEquation(P7.2a)andfindtheapproximatetransferfunctionHMiller(s).

3. PlotHMiller(s)onthesamesetofaxes thatyoualreadyusedtoplotH(s).UnderwhatcircumstancesistheMillerapproximationreasonabletouse?

References 1. Churchill,R.V.,Operational Mathematics,2nded.,McGraw-Hill,NewYork,1958. 2. Kreyszig,E.,Advanced Engineering Mathematics,8thed.,Wiley,NewYork,1999. 3. Sedra, A.S., and Smith, K.C., Microelectronic Circuits, 3rd ed., Saunders College,

Philadelphia,1991. 4. 2N3904/MMBT3904/PZT3904 NPN general purpose amplifier, Fairchild

Semiconductordatasheet,http://www.fairchildsemi.com,2011.

Table P7.2 Parameters for 2N3904 NPN transistor

Transistor Parameter Typical Value

hFE (= β) 125

Cobo (= cμ) 3 pF

Cibo (= cπ) 4 pF

hie (= rπ) 4 kΩ

hoe (= 1/ro) 10 µmho


Table 7.1 Common Laplace Transforms

f(t) F(s)

1 Unit step: 1s1

2 Ramp: ts12

3 Power of t: tn( 1)!

n 1

−

−

sn

1( 1,2,...)n =

4 Derivative: df tdt( )

sF(s)−f(0)

5 Time delay: f(t−T) e−sT F(s)

6 Exponential: eat f(t) F(s−a)

7 Convolution: f(t)∗g(t) F(s)G(s)

8 Delta function: δ(t) 1

9t

1≠ s

1

10t

2≠ s

32

−

11t

n2

1 3 5 (2 1)

n n 1/ 2

…× × × × − π

−

s n( 1,2,...)n

12 =

− +

12 eat

s a1−

13 teat

s a1

( )2−

14n

t e1

( 1)!n at1

−−

s an

1( )

( 1,2,...)n−=

15 t ek at1− ks a

k( )

( )( 0)k

Γ−

>

(continued)


Table 7.1 Common Laplace Transforms (Continued)

f(t) F(s)

16a b

e e1

( )( )at bt

−−

s a s ba b

1( )( )

( )− −

≠

17a b

ae be1

( )( )at bt

−−

ss a s b

a b( )( )

( )− −

≠

18b a e c a e a b e

a b b c c a( ) ( ) ( )

( )( )( )

at bt ct

−− + − + −

− − − s a s b s c1

( )( )( )− − −

19a

at1sin

s a1

2 2+

20 cos a ts

s a2 2+

21a

at1sinh

s a1

2 2−

22 cosh a ts

s a2 2−

23a

at1

(1 cos )2 −s s a

1( )2 2+

24a

at at1

( sin )3 −s s a

1( )2 2 2+

25a

at at at12

(sin cos )3 −s a

1( )2 2 2+

26ta

at2

sins

s a( )2 2 2+

27a

at at at12

(sin cos )+s

s a( )

2

2 2 2+

28 tcos ats as a( )

2 2

2 2 2

−+



f(t) F(s)

29at btb a

cos cos2 2

−−

ss a s b

a b( )( )

( )2 2 2 22 2

+ +≠

30be bt

1sinat

s a b1

( )2 2− +

31 e btcosat s as a b( )2 2

−− +

32 e eat at

cos3

23 sin

32

atat2− −− a

s a3 2

3 3+

33 sin at cosh at − cos at sinh ata

s a44

3

4 4+

34a

at at12

sin sinh2

ss a44 4+

35a

at at12

(sinh sin )3 −s a

14 4−

36a

at at12

(cosh cos )2 −s

s a4 4−

37 a t at at at(1 )sin cos2 2+ −a s

s a8

( )

3 2

2 2 3+

38 L ten

ddt

t e( )!

( )n

t n

nn t= −

sss

1 1 n−

39te at

1(1 2 )at

π+

s

s a( )32−

40t

e e1

2( )bt at

3π− s a s b− − −

(continued)



f(t) F(s)

41t

ae a t1

erfc( )a t2

π−

s a1+

42t

ae a t1

erfc( )a t2

π+ s

s a2−

43ae a t

1erf( )a t2

s s a1

( )2−

44 e a terfc( )a t2

s s a1

( )+

45b a

e b a t1

erf ( )at

−−−

s a s b1

( )+ +

46 eba

a terf ( ) 1a t2 −b a

s s a s b( )( )

2 2

2

−− +

47 J at( )0s a

12 2+

48 J kt(2 )0se

1 k s/−

49t

kt1

cos 2≠ s

e1 k s/−

50 tkt

1cosh2

≠ se

1 k s/

51k

kt1

sin2≠ s

e1 k s3/ 2

/−

52k

kt1

sinh2≠ s

e1 k s3/ 2

/

53k

t

kt2

exp43

2

π− e k( 0)k s >−



f(t) F(s)

54kt

erfc2 s

e k1

( 0)k s ≥−

55t

kt

1exp

4

2

π−

se k

1( 0)k s ≥−

56t k

tk

kt

2 exp4

erfc2

2

π− − s e k( 0 )k s3/ 2 ≥− −

57 e e a tkt

kt

erfc2

erfc2

ak a t2− + + aes a s

k( )

( 0)k s

+≥

−

58 e e a tkt

erfc2

ak a t2 + es a s

k( )

( 0)k s

+≥

−

59te e

1( )bt at−

s as b

log−−

60t

at2

1 cos( )−s as

log2 2

2

+

61t

at2

1 cosh( )−s as

log2 2

2

−

62t

kt1sin

ks

tan 1−

239

Chapter 8

Fourier Transforms and Signal Processing

8.1 IntroductionInelectroniccircuits,weareoftenpresentedwithvoltageandcurrentsthatrepre-sentsignalsthatcarrymeaningfulinformation.Examplesinclude

◾ audio,includinghi-fiandvoice

◾ video,includingtelevisionandsurveillance

◾ telemetryfromtemperaturesensors,straingauges,medicalinstrumentation,powermeters,andthelike

◾ radar,loran,andGPSsignalsfordetectionandguidancesystemsonaircraftandships

◾ modulatedsignalsasAM,FM,VHF,UHF,CDMA(codedivisionmultipleaccess),GSM(Groupespécialemobile),andWi-Fiforbroadcastandpoint-to-pointtransmissionofaudio,video,voice,anddata

Therearecountlessreasonsforprocessingsignals.Foraudioandvideo,wemightwant tofilteroutnoiseor reducebandwidth.For telemetry signals,wemaywanttoexaminethedatafrommultiplesensorstoachieveagoalsuchaspredictingthe


weather,reducinganautomobile’sfuelconsumption,orinterpretingsensordatafromanelectrocardiogram.Inthecaseofcommunicationsystems,wecanusemodulationtomultiplexaudioorvideosignals(fromradioorTVstations)ontohigh-frequencycarriers,therebyallowingmultiplesignalstobecarriedoveracommonmedium(e.g.,acoaxialcableoropticalfiber)or throughtheairwaves. Inaddition,we typicallywanttoaccountforthelinklossandintersymbolinterference(ISI)introducedbythechannel.

TheprimarymathematicaltoolusedinsignalprocessingistheFourier trans-form. This transform provides a method for describing a signal in terms of itsfrequency components (i.e., in the “frequency domain”) instead of in the morefamiliar and tangible “timedomain.”Table 8.1 lists several typesof signals andtheircorrespondingfrequencyranges(i.e.,therangeoffrequencycomponentsthatmightbepresentinthewaveforms).Fortherestofthischapter,weuseaudiosignalsinourexamplesbecausetheyarelikelytobefamiliartomostreaders.However,all of the techniques introduced in this chapter are applicable to signals of anyfrequencyrange.

Figure 8.1showsanexamplewaveformforsomeofthesignalsofTable 8.1.Ingeneral,the“sharpness”oftheshapeofeachsignalcorrespondstoitshigh-frequencycomponents,andthemorejaggedthesignal,thehigherthefrequencyrange.Thus,overa25-msecinterval,thehigh-fidelityaudiosignalshowninFigure 8.1bismorejagged than the speech signal in Figure 8.1a. Periodic signals (e.g., the musicalnoteshowninFigure 8.1c)consistprimarilyofasinglefundamentalfrequencyplussmallerharmoniccomponents thatcause thewaveformtodeviate fromaprecisesinusoidalshape.

Table 8.1 Example Signals and Their Corresponding Frequency Ranges

Signal Type Frequency Range

Speech 200 to 3200 Hz

High-fidelity audio (range of human ear) 20 to 20,000 Hz

AM radio 535 to 1705 kHz (North America) or 526.5 to 1606.5 kHz (Europe)

FM radio 87.5 to 108.0 MHz

802.11b Wi-Fi networking 2.401 to 2.473 GHz

White noise 0 to ∞ Hz

A above middle C on the piano 440 Hz (fundamental)

880 Hz, 1320 Hz, 1760 Hz, … (harmonics)

Fourier Transforms and Signal Processing ◾ 241

8.2 Mathematical Description of Periodic Signals: Fourier Series

Webeginwitha signalx(t),whichmightrepresentaudible soundthathasbeenconvertedtoavaryingvoltageusingatransducer(e.g.,amicrophone).Ifx(t)isperi-odicasshowninFigure 8.1casasustainedAnoteonapiano,thenFouriertheorytellsusthatthewaveformmaybeexpressedasaweightedsumofpuresinusoids:

… …∑ ∑= + π ⋅ ⋅ + π ⋅ ⋅= =

( ) cos(2 ) sin(2 )0

1,2,3, 1,2,3,

x t A A nf t B nf tn o

n

n o

n

(8.1)

t

x(t)

0.025

(a)

t

y(t)

0.025

(b)

t

z(t)

0.025

(c)

Figure 8.1 Example of band-limited waveforms: (a) speech; (b) high-fidelity audio; (c) A above middle C on the piano.


Thefrequenciesnfointhesecondandthirdtermsrepresentallpossiblesineandcosinecomponentsofx(t).Then =1componentisthefundamentalfrequencyfo,andcomponentswheren >1arecalledtheharmonics.ThefactorsAnandBnarecalledtheFourier coefficients,andA0isthe“DC”componentofx(t).Wecandeter-mineA0,An,andBnmathematicallybymultiplyingx(t)byeachpossiblefrequencycomponentandthenintegratingoveroneperiodofx(t).Forexample,tofindtheFouriercoefficientforthemthcosineterm,wemultiplybothsidesofEquation(8.1)

bycos(2πmfot)andintegrateovertheinterval − ≤ ≤2 2T

tT

,where =1

Tfo

isthe

timeperiodforonecycleoftheperiodicwaveform:

∫

∑∑∫

π

= + π + π π

−

−

( )cos(2 )

cos(2 ) sin(2 ) cos(2 )

/2

/2

0

/2

/2

x t mf t dt

A A nf t B nf t mf t dt

o

T

T

n o n o o

T

T

(8.2)

Becausesineandcosineareorthogonalfunctions,allofthetermsontherightsideofEquation(8.2)willintegrateouttozeroexceptforthecosine-cosinetermswhenm =n,thatis,

∫

∫

∫

π π = =

π π = ≠

π π =

−

−

−

cos(2 )cos(2 )2,

cos(2 )cos(2 ) 0,

cos(2 )sin(2 ) 0, all ,

/2

/2

/2

/2

/2

/2

mf t nf t dtT

m n

mf t nf t dt m n

mf t nf t dt m n

o

T

T

o

o

T

T

o

o

T

T

o

(8.3)

Therefore,theintegrationoperationwill“filter”outallcomponentsofx(t)exceptforthefrequencywherem =n.ApplyingEquation(8.3)toEquation(8.2)gives

∫

∫

( )

( )

= π

= π

−

−

2 ( )cos 2

2 ( )sin 2

/2

/2

/2

/2

AT

x t nf t dt

BT

x t nf t dt

n o

T

T

n

T

T

o

(8.4)


TocalculatetheDCcoefficientA0,weintegrateEquation(8.3)form =0toobtain

∫=

−

1 ( )0

/2

/2

AT

x t dtT

T

(8.5)

Example 8.1

Find the Fourier series for the periodic pulse train shown in Figure 8.2 with

period T, frequency =fT

o1

, “on” time of τ, and “off” time of T − τ.

By centering x(t) on the y axis, we ensure that x(t) is an even function. Thus, by symmetry, all of the odd terms in Equation (8.1) (i.e., the sine terms) will be zero. We begin by finding the DC term (for n = 0) by applying Equation (8.5) and integrating x(t) over one cycle:

∫= ⋅

=τ

−τ

τ

AT

dt

T

110

/2

/ 2

For the higher-order terms:

∫ ( )

( )

( )

= ⋅ π

=π

π

=π

π τ

−τ

τ

−τ

τ

AT

nf t dt

Tnf tnf

nnf

n o

o

o

o

21 cos 2

2 sin 22

2sin

/2

/ 2

/ 2

/ 2

1

t (millisec)−τ/2 τ/2−T T

Figure 8.2 A pulse train with “on” time of τ and period T.


Thus, the solution for the first few terms of the Fourier series for the pulse train is

( ) ( )

( )

=τ+π

πτπ +

ππτ

π

+π

πτπ +

x tT T

f tT

f t

Tf t

o o

o

( )2sin sin 2

1sin

2sin 4

23

sin3

sin 6

The following MATLAB program calculates the Fourier coefficients for the pulse train where T = 1 ms and τ = 0.25 ms and then reconstructs the pulse train based on the Fourier coefficients.

%Example_8_1.m%ComputetheFourierseriesforapulsetrain.T=1e-3;%pulseperiodfo=1/T;%frequencyofpulsetraininHztau=0.25e-3;%"on"timeofasinglepulsen=1:10;%wewillcomputethefirst10coefficients%computetheDCcoefficient:A0=tau/T;%computecoefficients1thrunfori=1:length(n)An(i)=(2/(pi*n(i)))*sin(pi*n(i)*fo*tau);end%Plotthecoefficients:subplot(2,1,1);stem([0fo*n],[A0An]);xlabel('Hz');ylabel('Fouriercoefficients');axis([-20010200-.2.6]);%UsetheFouriercoefficientstoreconstitutethepulse%trainovertheinterval-1.5<t<1.5millisecwithn%coefficientst=-1.5e-3:1e-6:1.5e-3;waveform=zeros(1,length(t));%AddintheDCcoefficientwaveform=waveform+A0;%Addinsignalcomponentsfor1thrunthcoefficientsfori=1:length(n)waveform=waveform+An(i)*cos(2*pi*n(i)*fo*t);endsubplot(2,1,2);plot(t*1000,waveform);xlabel('t(millisec)');ylabel('reconstitutedx(t)');

-------------------------------------------------------------------

The first 10 calculated Fourier coefficients are shown in Figure 8.3a, and the reconstituted pulse using the 10 coefficients is shown in Figure 8.3b.


8.3 Complex Exponential Fourier Series and Fourier Transforms

Generally, aFourier seriesmaybe comprisedof theweighted sumofany set oforthogonal functions. Although sines and cosines are a common choice due totheirfamiliarity,thepreferredchoicefromamathematicalperspectiveisthesetofcomplexexponentialfunctions:

∑= π

=−∞

∞

( ) 2x t C en j n f t

n

o (8.6)

0 2000 4000 6000 8000 10000−0.2

0

0.2

0.4

0.6

Hz

Four

ier

coef

ficie

nts

(a)

−1.5 −1 −0.5 0 0.5 1 1.5−0.5

0

0.5

1

1.5

t (millisec)

reco

nstit

uted

x(t

)

(b)

Figure 8.3 (a) Fourier coefficients for the pulse train of Figure 8.2 for T = 1 msec and τ = 0.25 msec. (b) Pulse train reconstituted using the first 10 Fourier coef-ficients. Note that the missing higher-order terms cause the reconstituted wave-form to be inaccurate, especially at the sharp transitions.


Knowingtheidentities =+ −

cos2

xe ejx jx

and =− −

sin2

xe e

j

jx jx

,itcanbeproven

that Equation (8.6) is equivalent to Equation (8.1) where =−2

,CA jB

nn n

=+

− 2C

A jBn

n n ,and =0 0C A .Multiplyinganexponentialbyitscomplexconju-

gateandthenintegratingoveroneperiodimpliesequivalentorthogonalityrelations:

∫ =

=

≠π − π

−

,0,

2 2

/2

/2

e e dtT m n

m nj mf t j nf t

T

T

o o (8.7)

IntegratingbothsidesofEquation(8.6)overoneperiodandapplyingEquation(8.7)givesanexpressionforthecomplexexponentialFouriercoefficientsforanyperiodicx(t):

∫= − π

−

1 ( ) 2

/2

/2

CT

x t e dtnj nf t

T

T

o (8.8)

Wenowmake twomodifications toEquation (8.8).First,wenormalize the

Fouriercoefficientsbymovingthe 1T

factorfromtherightsidetotheleft.Then,we define X(f ) as a function to represent the normalized Fourier coefficients.Equation(8.8)becomes

∫= = − π

−

( ) ( ) 2

/2

/2

X nf TC x t e dto nj nf t

T

T

o (8.9)

Next,wereplacenfowiththevariablef(forfrequency)andletT →∞,therebyimplying that x(t) becomes an aperiodic signal. Then, the integration limits inEquation(8.9)becomeinfinity,andX( f )becomesacontinuousfunction:

∫= − π

−∞

∞

( ) ( ) 2X f x t e dtj ft (8.10)

Equation(8.10)isthedefinitionoftheFourier transformofthetime-domainsignalx(t).

Example 8.2

Plot the normalized complex exponential Fourier coefficients for the pulse train shown in Figure 8.2. Assume τ = 0.25 msec and generate three plots for T = 1, 2, and 10 msec.


As for Example 8.1, the DC component is =τ

CT

0 . For the n ≠ 0 terms,

∫= ⋅

=− π

=−

− π

=π τπ

− π

−τ

τ

− π

−τ

τ

− π τ π τ

CT

e dt

Tej nf

e ej n

nfn

nj nf t

j nf t

o

j nf j nf

o

o

o

o o

11

12

2

sin

2

/ 2

/ 2

2

/ 2

/ 2

A program that computes and plots the coefficients follows:

%Example_8_2.m%ComputenormalizedexponentialFourierseries%coefficientsforapulsetraintau=0.25e-3;%"on"timeofasinglepulseT=[1e-32e-310e-3];%pulsetrainperiodsfo=1./T;%frequencyofpulsetraininHzn=-150:150;fori=1:length(T)Cn(i,:)=(1./(pi*n)).*sin(pi*n*fo(i)*tau);%Findthe"zero"entryinnandcomputeit%separately(becauseithasadifferentformulathan%therest):zeroeth_element=find(n==0);Cn(i,zeroeth_element)=tau/T(i);%Normalizethecoefficients:Cn_normalized(i,:)=Cn(i,:).*T(i);end%findreasonablelimitsforaxessothatall3plots%havethesamemin/maxx_axis_limits=[-3.8/tau3.8/tau];y_axis_limits=[1.5*min(Cn_normalized(1,:))...1.25*max(Cn_normalized(1,:))];%PlotseparatelyforeachvalueofT:subplot(3,1,1);stem(n*fo(1),Cn_normalized(1,:));ylabel('Fouriercoefficients';'forT=1ms');axis([x_axis_limitsy_axis_limits]);title('ExponentialFourierSeriesforPulse');subplot(3,1,2);stem(n*fo(2),Cn_normalized(2,:));axis([x_axis_limitsy_axis_limits]);ylabel('Fouriercoefficients';'forT=2ms');subplot(3,1,3);plot(n*fo(3),Cn_normalized(3,:));axis([x_axis_limitsy_axis_limits]);ylabel('Fouriercoefficients';'forT=10ms');xlabel('f(Hz)');

-------------------------------------------------------------------


Figure 8.4 shows the three sets of Fourier coefficients computed in Example 8.2 for pulse periods of T = 1, 2, and 10 msec. Note that by keeping the pulse width constant and increasing the pulse period, the Fourier coefficients get closer and closer together. When T = 10 msec, they are so close together that we have plotted them as a continuous function. In the limit T → ∞, Figure 8.4c corresponds to the Fourier transform of the pulse.

–1.5 –1 –0.5 0 0.5 1 1.5

×104

0

1

2

3×10–4

Fou

rier

Coe

ffici

ents

for

T =

1 m

s

Exponential Fourier Series for Pulse

0

1

2

3×10–4

0 5000 10000–10000 –5000 15000–15000

Fou

rier

Coe

ffici

ents

for

T =

2 m

s

–1.5 –1 –0.5 0 0.5 1 1.5

×104

0

1

2

3×10–4

Fou

rier

Coe

ffici

ents

for

T =

10

ms

f (Hz)

Figure 8.4 Plot of the Fourier series coefficients for a pulse train with 0.25 msec on time and (a) period of 1 msec, (b) period of 2 msec, and (c) period of 10 msec. For the last case, the coefficients are so close together that they are plotted as a continuous function.


8.4 Properties of Fourier TransformsQualitatively, theFourier transformisadescriptionof the frequencymakeupofx(t). Figure 8.5 shows the three signals of Figure 8.1 transformed into the fre-quencydomain.Notethatwhendescribingtheirproperties,theactualshapesoftheFouriertransformsareingeneralunimportant(andinthiscasewehavedrawnthem as rounded rectangles). For the moment, we are more interested in theirbeginningandendpoints.Notethatthetransformsarealwayssymmetricaroundtheyaxisbecausewealwaysassumethatx(t)originatesfromanactualwaveformthatisrealizableandreal(andthushasnoimaginarycomponent).Nevertheless,

3200–3200f

200–200

X( f )

(a)

f–20 20–20000 20000

Y( f )

(b)

f

880

1320

1760

2200

2640

3080440

–880

–132

0–1

760

–220

0–2

640

–308

0

–440

Z( f )

(c)

Figure 8.5 Fourier transforms for common audio signal types: (a) speech; (b) high-fidelity audio; (c) A above middle C on the piano.


Fouriertransformsdohaveanimaginarycomponent,andinFigure 8.5,wehaveonlyplottedthemagnitude(andnotthephase).Theideaofasignalhavingcom-plexor“negativefrequency”mayappearconfusing,butitissimplyanartifactofourusingthecomplexexponentialas theorthogonal function.Whennecessary,wecanalwayschangeacomplexfrequencybacktoarealfrequencybyapplyingthe identity = +cos sine x j xjx ; inthemeantime,wecantakeadvantageof thesimplifiedmaththatcomes from“bundling”thesinesandcosines together intooneexponentialterm.

Mathematically,theFouriertransformmaybeconsideredasubsetofthebilat-eral Laplace transformwheres =j2πf[1].ItdoesnotprovideascompleteananalysisastheLaplacetransform(e.g.,itcannotbeusedtopredictsystemstability),butitssimplicityallowsconvenient insight into signalbehavior.Aswithany transformtechnique,wecantransformfromonedomaintotheotheratwilltotakeadvan-tageoftheconveniencesofeach.

Wecanderivetheinverse Fourier transformbysubstitutingthedenormalized

Fouriertransform ( )X fT

backintoEquation(8.6)toobtain

∑

∑

=

=

π

=−∞

∞

π

=−∞

∞

( ) 1 ( )

( )

2

2

x tT

X nf e

X nf enfn

oj nf t

n

oj nf t o

n

o

o (8.11)

Letting →nf fo and →fn

df ,inthelimitthesumbecomesanintegral:

∫= π

−∞

∞

( ) ( ) 2x t X f e dfj f t (8.12)

Equation(8.12)isthedefinitionoftheinverseFouriertransformofX( f ).Some importantFourier transformpairsare summarized inTable 8.2.Note

thatthesimilarityoftheforwardandinverseFouriertransforms(Equations(8.10)and(8.12)onlydifferbythesignoftheexponential)leadstothedualitypropertyinwhich each entry inTable 8.2 canbe considered a reversiblepair (hence thebidirectional arrow for each entry in the table).For example, given that apulse

(ofwidth2τ)inthetimedomaintransformsto π τπ

sin2 ff

inthefrequencydomain,

wecanalsoinferthata“pulse”inthefrequencydomain(ofwidth2Fo)transforms

to ≠≠

sin2 F tt

o inthetimedomain.


8.5 FiltersFiltering is the application of a frequency-dependent function on a waveform. ThevarioustypesoffiltersareclassifiedandcharacterizedbytheshapeoftheirfrequencyresponseintheFourierdomainandaresummarizedinTable 8.3.Forexample,alow-passfilterhasaFouriertransformthathasavalueofoneatlowfrequenciesandavalueof

Table 8.2 Common Fourier Transform Pairs

Property Time Domain x(t) ↔ Frequency Domain X(f)

Linearity x(t) = Ay(t) + Bz(t) ↔ X(f) = AY(f) + BZ(f)

Duality X(t) ↔ x(−f)

Impulse x(t) = δ(t) ↔ X(f) = 1

Cosine = πx t f t( ) cos 2 o ↔ =δ − + δ +

X ff f f f

( )( ) ( )

2o o

Sine = πx t f t( ) sin2 o ↔ =δ − − δ +

X ff f f f

j( )

( ) ( )2

o o

Pulse =≤ τ

x tt

( )1

0 otherwise↔ =

π τπ

X ff

f( )

sin2

Ideal low-pass filter

=ππ

x tf t

t( )

sin2 o ↔ =≤

X ff f

( )1

0 otherwise

o

Convolution/multiplication

y(t) = x(t)∗h(t) ↔ Y(f) = X(f)H(f)

Causal decaying exponential

= −αy t u t e( ) ( ) t ↔ =π + α

Y fj f

( )1

2

Causal decaying cosine

= π−y t u t e f t( ) ( ) cos 2ato ↔ ( )

=π +

π + + πY f

j f a

j f a f( )

( 2 )

2 (2 )o2 2

Causal decaying sine

= π−y t u t e f t( ) ( ) sin2ato ↔ ( )

=π

π + + πY f

f

j f a f( )

2

2 (2 )o

o2 2


zero(ornearlyzero)athighfrequencies.Inthiscase,thelow-frequencyregioniscalledthepassbandbecausesignalsarepassedthroughthefilterunchanged,andthehigh-frequencyregioniscalledthestopbandbecausesignalsinthisregionarestoppedbythefilter.Conversely,ahigh-passfilterhasthestopbandatlowfrequenciesandthepassbandathighfrequencies.Inabandpassfilter,thepassbandhasbothalowandahighlimit.Aband reject(oppositeofbandpass)hasalowandhighlimitonthestopband.Anequal-izerhasmultiplepassbandsofvaryingmagnitudesandistypicallyusedtocompensatefornonlinearfrequencybehaviorinanamplifierorcommunicationchannel.

Figure 8.6ashowsanexampleofanideallow-passfilterwithcutofffrequencyfo and corresponding frequency response HI( f ) (where the I subscript denotes“ideal”).IfweapplythisfiltertoawaveformwithFouriertransformX( f )contain-ingfrequencieswithintherange < <1 2f f f ,thenthefilteroutputistheproduct

( ) ( )IH f X f ,andallfrequenciesgreaterthanfoareremoved.Unfortunately,thereisafundamentalproblemwiththisideal“brickwall”filterin

thefrequencydomain.WeknowfromTable 8.2thatmultiplicationinthefrequencydomaincorrespondstoconvolutioninthetimedomain.IfweexamineHI(f )inthe

time domain, we see that the corresponding function =ππ

( ) sin2h t

f ttI

o extends

infinitelyforall±t.Thus,toconvolvex(t)with ( )h tI ,weneedtoknowallvaluesofx(t),

Low-pass:

High-pass:

Bandpass:

Band reject:

Equalizer:

HEQ( f )

HLP( f )

HBP( f )

HHP( f )

HBR( f )

f

f

f

f

f

Table 8.3 Frequency Reponse for Common Filter Types


includingthosefort <0.Afilterthatdependsonvaluesfort <0iscalledanoncausalfilterandisimpossibletoimplement,bothwithnumericalalgorithmsandrealcircuitcomponents.

Given that we cannot implement an ideal filter, there are various alternativeapproachesthatcanbeusedinstead.Oneapproachistoapplyawindow functionto

( )h tI tomakeitcausal,andthesimplestwindowfunctionistomultiply ≠≠

sin2 f tt

o

byarectangularwindowtotruncatetheinfinitetailsof ( )h tI ,thusforcingittobecausal.AnotherapproachistouseaknownrealizablefiltercharacteristicsuchastheButterworthorChebyshev responses.Ineithercase,thenetresultistoremovethedis-continuityinthefrequencyresponseat f o .Thecorresponding ( )H fR (wheretheRmeans“realizable”)isasmoothfilterthatcanactuallybeimplemented.ThisisshowninFigure 8.6b,where ( )H fR nowhascurvededges,andtheresultingfilteredoutputalsohassimilaredges.

8.6 Discrete-Time Representation of Continuous-Time Signals

Indigital signal processing(andinMATLAB),werepresentacontinuous-timesignalas a sequenceofnumbersby theprocessof sampling.Figure 8.7a showsa typicalcontinuous-timesignalx(t)(withthetimeargumentdenotedbyparentheses)anditsdiscrete-timecounterpartx[n](wherethediscreteargumentisdenotedwithsquarebrackets).Wederivex[n]fromx(t)byextractingthevaluesofx(t)atanevenlyspaced

f1 f2−f1−f2 fo−fo −fo fof1−f1

X( f ) HI ( f ) X( f )HI ( f )

(a)

X( f ) X( f )

f1 f2−f1−f2 fo−fo −fo fof1−f1

HR( f ) HR( f )

(b)

Figure 8.6 (a) An ideal “brick wall” low-pass filter with cutoff frequency fo. (b) A realizable low-pass filter necessarily has a smooth transition band.


intervalofsizeT,calledthesampling time(orequivalentlyatasampling rate =1 ).fTS

Mathematically,weconvertfromcontinuoustodiscretetimewiththeformula x[n]=x(nT ) (8.13)wherenisaninteger.Notethatadiscretesignalx[n]doesnotinherentlycarryanytimeinformationbutratherisjustaseriesofvaluesforxindexedbythedimension-less integern.Thus,oncewesampleacontinuous-timesignal, it is importanttorememberthesampletimeTifweeverwishtoconvertthediscreteversionbacktocontinuoustime.

Wemustalsotakecaretochooseanappropriatesamplerate.Figure 8.7bshowswhatcanhappenifwechooseTtoolarge.Thecorrespondingdiscretesignaly[n]

ismissingthehigh-frequencysinecomponent(withfrequencyofexactly 1 )T

that“rides” atop y(t). Thus, sampling a continuous-time signal does not guaranteeuniquenessinthediscretedomain,asevidencedinFigure 8.7bythefactthatx[n]andy[n]areidentical.The sampling theoremstatesthattofullydescribeacontin-uous-timesignaldiscretely,wemustchooseasamplingratethatisat leasttwice

–1

1

–1

1

x[n]x(t)

t n20 40 60 80 100 2 4 6

8 10T

(a)

20 40 60 80

–1

1

2 4 68 10

–1

1

y(t) y[n]

t n100

(b)

Figure 8.7 (a) A typical continuous-time signal x(t) and its corresponding dis-crete-time signal x[n], which is derived by sampling x(t) every T seconds. (b) If y(t) has an additional high-frequency component with a period of an exact fraction of T, then that component is invisible in the corresponding y[n].


asfastasthehighest-frequencycomponentoftheoriginalsignal.Inpractice,weusuallychooseasamplingratethatisslightlyhigherthantheminimumrequiredto account for circuit imperfections and component variations. For example, inaudioCDs,whichcanreproducefrequenciesupto20,000Hz,thesamplingrateis44,100samples/sec(insteadoftheminimumvalueof40,000samples/sec).

Therearemanymotivationsforconvertingcontinuous-timesignalstodiscretetime,butourimmediatereasonissowecanuseadigitalcomputertocarryoutthesignalprocessingwithaprogramlikeMATLAB.

8.7 Fourier Transforms of Discrete-Time SignalsThecontinuous-timeFouriertransform(whichwenowrefertoastheCFT)intro-ducedinSection8.3canbeextendedtodiscrete-timesystemsasthediscrete-time Fourier transform(DTFT)X(ϕ)andisdefinedas

∑φ = − π φ

=−∞

∞

( ) [ ] 2X x n e j n

n

(8.14)

ThecorrespondinginverseDTFTiscalculatedby

∫= φ φ−

π φ[ ] ( )1/2

1/2

2x n X e dj n (8.15)

NotethattheDTFTinEquation(8.14)isdefinedasasummation(insteadofanintegral)duetothediscretenatureofx[n].Inaddition,becausex[n]isdimensionless,thecorrespondingX(ϕ)isalsodimensionless.Equation(8.14)isperiodicinϕduetothe − π φ2e j n term;thus, it issufficienttoexamineX(ϕ)overtherange − ≤ φ ≤1

212

insteadofover all frequencies, as for theCFT. Ifx[n]wasderivedby sampling acontinuous-timesignalx(t)atasamplingratefS,thenϕcorrespondstothefrequency

range − ≤ ≤2 2f

ffS S ,which corresponds to the rangeof useful frequencieswith

respecttothesamplingtheorem.ItisimportanttorealizethattheDTFTisacontinuousfunctionandnota

discreteone.Asanexample,supposea1000-Hzsinewaveissampledat44100samples/sec.Thisimpliesthatthecorrespondingdiscreteversionofthe1000-Hzsinewavewillrepeatevery44.1samples,andthecorrespondingX(ϕ)willhave

nonzerocomponentsat φ = ± 100044100

,which isclearlynotawholenumber.If

x[n]containsavarietyoffrequencycomponents,thentheywillallberepresentedinX(ϕ)attheircorrespondingratiotofSandwillgenerateacontinuousfunction.

BecauseEquation (8.14) isperiodicwith respect toϕ, thevaluesofX(ϕ) forφ > 1

2 are merely repeated copies of the values in the region − ≤ φ ≤12

12 . The


implicationofthisisthatwecantreatx[n]asaperiodicfunctionthatcanberep-resentedbyaFourierserieswhosecoefficientscanbefoundbysamplingX(ϕ).ThisisareverseapplicationofthesametechniqueweusedinSection8.2toderivetheCFTasthe“smear”oftheFouriercoefficientsofaperiodiccontinuous-timewave-form.Mathematically,the“unsmear”oftheDTFTintoitsFouriercoefficientsiscalledthediscrete Fourier transform(DFT)andisdefinedas

∑= − π

=

−

[ ] [ ] 2 /

0

1

X k x n e j nk N

n

N

(8.16)

∑= π

=

−

[ ] 1 [ ] 2 /

0

1

x nN

X k e j nk N

k

N

(8.17)

whereNisthenumberofpointsweusedtosampleX(ϕ).WehaveusedtheletterktorepresenttheindexoftheDFTinthedimensionless“discretefrequency”domain

(correspondingtothephysicalfrequencies0throughfSinstepsof −1

1NfS ).

Note thatwhereas theDTFTuses the range − ≤ φ ≤12

12

and is symmetric

aroundtheyaxis,byconventiontheDFTusestherange0≤k ≤N −1andthusisdefinedfork ≥0,thatis,intherighthalfplane.ThereasonforthisisthatadiscretewaveformcanonlybesymmetricaroundtheyaxisifithasanoddvalueofN,andtheDFTisoftencalculatedusinganevenvalueforN(usuallyapoweroftwo).Theimplication(asweshallseeintheexample)isthattheresultingDFTwillbesym-

metricabout =−12

kN insteadofk =0.

Theprecedingderivationsmayappearcomplicatedandabstract,butthereisamajoradvantagetousingtheDFTinsteadoftheDTFT:TheDFTcanbecom-putedusing the fast Fourier transform (FFT)algorithm[2].TheFFTprovides acomputationally efficientway to calculateEquation (8.16)byusingonly integeradditionsandmultiplications,thusenablingfrequency-domaincomputationstobeeasilyimplementedwithalow-costdigital signal processor(DSP).WhilethetermFFTisoftenusedinterchangeablywithDFT,intruththeFFTissimplyoneofseveralknownmethodsthatmaybeemployedforcomputingtheDFT.

MATLABprovidesthefftfunctionforcalculatingFFTsofdiscretesignals.ExecutingX=fft(x)willreturnthen-pointFFTofthevectorx,wherenisthenumberofelementsinx.Alternatively,youmayexplicitlyspecifythenumberofpointsintheFFTbyaddingasecondargumenttofft.Thus,runningX=fft(x,64)willcalculatea64-pointFFTonthevectorx;ifxhasgreaterthan64elements,thentheextraelementswillbediscardedintheFFTcalculation,andifxhasfewerthan64elements,thenxwillbeautomaticallyextendedto64elements(bypaddingwithzeros)priortotheFFTcalculation.MATLABalsohasacorre-spondingifftfunctionforcomputingtheinverseDFT.


Example 8.3

Let x(t) be the triangular waveform in Figure 8.8 with width 1 msec, height 1 V, and total waveform length of 4 msec. Assuming a sampling rate of 10 kilosamples/sec:

1. Calculate φX( ) (the DTFT magnitude) directly from its definition in Equation (8.14).

2. Calculate X k[ ] (the DFT magnitude) directly from its definition in Equation (8.16).

3. Calculate X k[ ] using MATLAB’s fft() function.

%Example_8_3.m%CalculatetheDTFTandDFTforatriangularwaveform.%First,definex(t)triangle_width=1e-3;triangle_height=1;T=1e-4;triangle_slope=triangle_height/(triangle_width/2);fs=1/T;t=0:T:4e-3;n=t/T;N=length(n);%definediscretetrianglewaveformx[n]x=zeros(1,N);x(1:6)=0:triangle_slope*T:triangle_height;x(6:11)=1:-triangle_slope*T:0;%definephiforDTFTwitharbitrarygranularityof200%stepsphi=-.5:.005:.5;%definekforDFTassamelengthasx[n]k=0:N-1;%initializeDTFTandDFTvarsX_DTFT=zeros(1,length(phi));X_DFT=zeros(1,length(k));%calculateDTFTandDFTbasedondefinitions(sumof%complexexponentials)fori=1:length(n)X_DTFT=X_DTFT+x(i)*exp(-j*2*pi*n(i)*phi);X_DFT=X_DFT+x(i)*exp(-j*2*pi*n(i)*k/N);end%Calculate41-pointFFTwithMATLAB'sfft()functionX_FFT=fft(x);

1 2 3 4

1

t (ms)

x(t)

Figure 8.8 Triangular continuous-time waveform x(t).


%plotresultssubplot(4,1,1);stem(n,x);xlabel('n');ylabel('x[n]');subplot(4,1,2);plot(phi,abs(X_DTFT));xlabel('\phi');ylabel('X(\phi)(DTFT)');subplot(4,1,3);stem(k,abs(X_DFT));xlabel('k');ylabel('X[k](DFT)');subplot(4,1,4);%plotFFThorizaxisasproportionofsamplingratestem(k/(N-1)*fs,abs(X_FFT));xlabel('f=fs*(k/(N-1))(Hz)');ylabel('X[k](FFT)');

-------------------------------------------------------------------

Results are shown in Figure 8.9. Some observations are as follows:

◾ The discrete-time version of the triangle function is denoted x[n] and is shown in Figure 8.9a.

◾ The DTFT of x[n] is shown in Figure 8.9b and has the shape of x

xsin2

2 .

This makes sense because a triangle waveform can be constructed by convolving two rectangular pulses, and thus in the frequency domain

looks like the product of two xx

sin functions.

◾ The results for the DTFT (Figure 8.9b) and DFT (Figure 8.9c) are iden-tically shaped, but the DFT plots the “negative” frequencies on the right half of the plot instead of the left. However, since the DFT is always symmetrical, we can simply discard the right half of the plot (or at least ignore it in our heads).

◾ The results for the sum-of-complex-exponentials DFT (Figure 8.9c) and the MATLAB-calculated FFT (Figure 8.9d) are identical, as expected. However, in Figure 8.9d, we chose to label the x axis in

terms of −k

NfS

1 (in Hertz) instead of the abstract dimensionless k parameter, thus providing more context for the plot.

8.8 A Simple Discrete-Time FilterAsdescribedinSection8.3,afilterisasysteminwhichtheinput/outputrelation-shipissomefrequency-dependentfunction.Incontinuous-timefilterdesign,ananalog circuit containing capacitors, inductors, and amplifiers is typically usedto implement the desired response. The filter is characterized by its frequencyresponseH( f )andcorrespondingimpulseresponseh(t),providinganinput/out-put relationshipdeterminedbyY( f )=H( f )X( f ) in the frequencydomainory(t)=h(t)∗x(t)inthetimedomain(wherethe“*”indicatesconvolution).Ifx(t)istheDiracdeltafunctionδ(t),theny(t)=h(t)becauseanyfunctionconvolvedwithδ(t)isitself.


0 5 10 15 20 25 30 35 400

0.5

1

n

x [n

]

(a)

–0.5 0 0.50

2

4

6

φ

X (

φ) (

DT

FT

)

(b)

0 5 10 15 20 25 30 35 400

2

4

6

k

X [k

] (D

FT

)

(c)

f = fs*(k/(N – 1)) (Hz)

0 2000 4000 6000 8000 100000

2

4

6

X [k

] (F

FT

)

(d)

Figure 8.9 (a) Discrete-time version of the triangle waveform in Figure 8.8. (b) The DTFT magnitude |X(𝛟)|. (c) The DFT magnitude |X[k]|. (d) The DFT as com-puted by MATLAB’s fft() function. Note that in the last case, the frequency axis has been rescaled from 0 to fS instead of zero to N.


Incontinuous-timefilters,h(t)andδ(t)arelargelyamathematicalconveniencebecauseit is impossibletocreateadeltafunctionexactlyinthelab(thoughit isoftenapproximatedbyashortpulse).Ontheotherhand,indiscrete-timesystems,itispossibletorepresentthediscretedeltafunctionδ[n]accurately(alsocalledtheunit samplefunctionorKronecker deltafunction).Thediscretedeltafunctionδ[n]issimplyasequencewithavalueofoneatn =0andzerootherwise(seeFigure 8.10)andthusistriviallyimplementable.Inaddition,anydiscretesequencex[n]canbeexpressedasasumofdelayeddeltas.Forexample,thex[n]showninFigure 8.11acanbewrittenmathematicallyas

x[n]=0.5δ[n]+1.5δ[n −1]+δ[n −2]−0.5δ[n −3]

Therefore, ifwecandesignalinearsystemwithknownbehaviorforδ[n]withacorrespondingimpulseresponseh[n],thenwecanusethatsystemtoprocessanyx[n].Inthiscase,weknowtheoutputpreciselybecausex[n]isjustaweightedsumofdeltafunctions.

ThisconceptisillustratedbythesysteminFigure 8.11b,whichimplementstheimpulseresponseofFigure 8.11a.Conceptually,thissystemisbuiltfromamplifi-ers,anadder,anddelay elements(denotedby“D”inthediagram)whosepurposeistostorethevalueofasignalforexactlyone“tick”ofn.Inpractice,weusecomputerregistersforthedelayelementsandcomputerarithmeticforthemultiplicationandaddition.Thesystemworksasfollows:

1.Initially,assumethatallofthedelayelementscontainthevaluezero;thus,theoutputiszero

2.At“time”n =0,theinputdeltafunctionisone;thus,thesummedoutputofthesystemis0.5

3.Atn =1,theinputreturnstozero.However,theonefromn =0isstoredinthefirstdelayelement;thus,thesummedoutputis1.5

4.Atn =2,theinputremainszero,andtheoneisshiftedtothenextdelayele-ment,causingthesummedoutputtobe1.0.

–1 1 2 3 4 5

1

δ[n]

n

Figure 8.10 The discrete-time version of the delta function consists of a single sample of unity height at n = 0 and zeros elsewhere.


5.Atn =3, theonegetsshiftedtothefinaldelayelement,andthesummedoutputis−0.5.

6.Finally,forn >3,allofthedelayelementsagaincontainzeroandtheoutputreturnstozero.

Thus,theoutputsequenceofthissystemwithrespecttonduetoadeltafunctioninputwillbe0.5,1.5,1.0,−0.5,whichcorrespondstothesamesequenceash[n]inFigure 8.11a.

ThistypeoffilterisgeneralizedinFigure 8.12forNdelayelementsandiscalledanNth-order(orequivalentlyanN +1“tap”)finite impulse response(FIR)filter.Mathematically,thissystemcalculatestheconvolutiony[n]=h[n]∗x[n].

ThereareseveralmethodsfordesigningFIRfilters[3],andthesimplestmethodwasdescribed inSection8.5:Beginwithan idealbrickwall frequency response

–1 1 23

4 5

1

2

h[n]

n

–1

(a)

D

D

D

+

0.5

1.5

1.0

–0.5

δ[n]

0.5 for n = 01.5 for n = 11.0 for n = 2

−0.5 for n = 30 otherwise

y[n]=

Figure 8.11 (a) Typical impulse response h[n] for a discrete-time filter. (b) A circuit to implement a filter with impulse response h[n] consists of three delay elements, four multipliers, and an adder.


( )H fI , convert ( )H fI to the time domain to obtain a noncausal impulseresponse ( )h tI ,thenapplyawindowfunctiontotruncatethetailsof ( )h tI ,thuscreatingarealizableimpulseresponse ( )h tR .Wealsoshift ( )h tR intimesothatitbeginsatt =0.Thus,foranideallow-passfilter(showninFigure 8.13a)withcutoff

frequency f o ,theidealimpulseresponseis =ππ

( ) sin2h t

f ttI

o ,andthecorrespond-ingrealizableimpulseresponseis

=

π −

π −≤

( )

sin22

2

for

0 otherwise

h t

f tt

tt

t tR

ow

ww

(8.18)

wheretwisthewidthoftherectangulartruncationwindow(seeFigure 8.13b).Toimplementasadiscrete-timefilterwithsamplingratefS(andcorrespondingsam-

plingtime =1

TfS

),weconvert ( )h tR to [ ]h nR byapplyingt =nT;thus,

( )( )=πφ −

π −≤

[ ]sin2

for

0 otherwise

2

2h nn

nn W

R

oW

W

where φ = f fo o S isthediscrete-timefiltercutofffrequency,and =W t Tw isthewindowwidth(seeFigure 8.13c).NotethatthenumberoftapsintheFIRfilterisW +1,correspondingtoWtapsforn =1,2,3,…,Wplusonemoreatn =0.

D

D

+

x[n]

y[n]

a0

a1

aN

Figure 8.12 General circuit to implement an N + 1 tap FIR filter.


t

hI(t)

2 fo

1fo

(a)

t

hR(t)

tw

(b)

n

hR[n]

W

(c)

Figure 8.13 (a) Ideal impulse response hI(t) for an ideal low-pass filter. (b) By shifting and truncating hI(t), we obtain a realizable impulse response hR(t). (c) By sampling hR(t), we obtain hR[n], which can be implemented as a discrete-time FIR filter.


Example 8.4

Design a low-pass, discrete-time, rectangular-window FIR filter with cutoff frequency of =fo 1000 Hz. Assume a sampling rate of =fS 40000 samples/sec and use 401 taps.

A 401-tap filter implies that the impulse response should have a length of W = 400T. Thus, for a rectangular window, we create a realizable version of this filter by shifting h tI ( ) to the right by 200T and then truncating the tails for t < 0 and t > 400T to obtain hR (t). The resulting realizable 401-point hR[n] is

( )

( )=

π −

π −≤ ≤h n

ff

n

nnR

o

S[ ]sin2 200

200for 0 400

0 otherwise

Next, we create an input signal x(t) to send into our filter. We want an input signal that contains components in both the passband and stopband; thus, we arbitrarily choose an x(t) containing components at 500, 1500, and 3000 Hz. Thus,

x(t) = sin 2π 500t + sin 2π 1500t + sin 2π 3000t

In discrete time, the corresponding x[n] is

= π + π + πx n n n n[ ] sin 2

50044100

sin 2150044100

sin 2300044100

A MATLAB program to implement this FIR filter follows:

%Example_8_4.m%ThisexampleimplementsanFIRlow-passfilterusinga%truncatedsincfunction(equivalenttoanidealfilter%witharectangularwindow)clc;clear;%definesamplingrate:fs=40000;T=1/fs;%definecutofffrequencyoflow-passfilter:fo=1000;%definefrequenciesofinputsignalx(t):input_freqs=[50015003000];%definefilterorder(N):taps=401;N=taps-1;%Note:fromthispointforward,wewilloperateinthe%dimensionlessdiscrete-time"n"domain.First,we%redefinethefiltercutofffrequencyintermsofthe%dimensionlessPhiparameteroftheDTFT:phi_o=fo/fs;%Definehr[n]asashiftedsincfunctionover401tapsfori=1:tapsn=i-1;shifted_n=(n-N/2);hr(i)=sin(2*pi*phi_o*shifted_n).../(pi*shifted_n);%Note:sincesin(x)/xissingularforx=0,catch


%thiscaseandinserttheappropriatelimitvalue.%Otherwise,ourhr[n]won'tworkbecauseitwill%haveasingle"NaN"valueatthepeak.ifshifted_n==0hr(i)=2*phi_o;endend%Runthefilterforalongtimesothatwecanbesure%toreachthesteadystate.Arbitrarilychoose20000%samples.max_n=fs/2;n=0:max_n;%Definex[n].First,rescaletheinputfrequenciestobe%fractionsoffs:input_phis=input_freqs/fs;x=zeros(1,length(n));fori=1:length(input_phis)x=x+sin(2*pi*input_phis(i)*n);end%Shiftx[n]to"turnon"atatimegreaterthanthe%numberoftapsofthefilter.Thisisequivalentto%settingtheinitialconditionsofthefiltertozero.x_shifted=zeros(1,length(n));x_shifted(taps:length(n))=x(1:length(n)-N);%Computey[n]bysimulatingthedelaycircuitofFigure%8.12.Thisisequivalenttoconvolvingx[n]withhr[n].y=zeros(1,length(n));fori=taps:length(n)forj=1:taps%calculatethevalueforeachtapandaddtothe%runningtotaly(i)=y(i)+hr(j)*x_shifted(i-j+1);endend%Convertthediscretesignalshr[n],x[n],andy[n]to%thetimedomainbyrescalingthehorizontalaxisand%connectingthedots.interesting_interval=1400*T;subplot(3,1,1);plot(T*n(1:taps),hr);%rescaletheaxestoseetheinterestingparts:axis([0interesting_interval-.02.06]);ylabel('h_R(t)');title('FIRfilterw/rectangularwindow,timedomain');subplot(3,1,2);plot(T*n,x_shifted);axis([0interesting_interval-33]);ylabel('x(t)');subplot(3,1,3);plot(T*n,y);ylabel('y(t)');xlabel('t(sec)');axis([0interesting_interval-1.51.5]);figure;%Seehowgoodourfilterisbycomputinga4096-point%DFTofx[n]andy[n]andplottingonsemilogaxes.%Rescalethehorizontalaxistobeinunitsoffrequency


%insteadofthedimensionless"k"parameteroftheDFT.fft_points=4096;scaled_freqs=0:fs/(fft_points-1):fs;interesting_freq=6000;subplot(3,1,1);HR=fft(hr,fft_points);semilogy(scaled_freqs,abs(HR));ylabel('|H_R(f)|');%rescaletheaxestoseetheinterestingparts:axis([0interesting_freq1e-41e1]);title('FIRfilterw/rectangularwindow,freqdomain');subplot(3,1,2);%Calculatethefftontheendpartofx[n]andy[n]so%thatwearesurethatweareinthesteadystate%portion.x_end=x(end-fft_points+1:end);X=fft(x_end);semilogy(scaled_freqs,abs(X));axis([0interesting_freq1e-11e4]);ylabel('|X(f)|');subplot(3,1,3);y_end=y(end-fft_points+1:end);Y=fft(y_end);semilogy(scaled_freqs,abs(Y));axis([0interesting_freq1e-11e4]);ylabel('|Y(f)|');xlabel('f(Hz)');

-------------------------------------------------------------------

The time-domain signals hR(t), x(t), and y(t) for this filter are shown in Figure 8.14. Note that these plots transform the discrete-time signals back to continuous time by rescaling the horizontal axis by a factor of nT. In addition, although we calculated y(t) for 0.5 sec, we only plotted the “inter-esting” part of the plot (for t < 35 ms). As expected, the multifrequency signal x(t) is reduced by the filter to its low-frequency component at 500 Hz, as can be seen in the plot of y(t).

In Figure 8.15, we show the frequency-domain results for the filter by running the fft function on hR(n), xn, and y[n]. In this case, we run a 4096-point FFT on all three signals. Then, these discrete-frequency DFT results are rescaled to the frequency domain and plotted on semilog axes for the interesting portion (f < 6000 Hz). Note the following:

◾ When viewed in the frequency domain, our truncated ideal impulse response causes ripples in both the passband and stopband and causes a nonzero stopband.

◾ The magnitude of the input signal X(f) contains three identical com-ponents at 500, 1500, and 3000 Hz.

◾ The magnitude of the output signal Y(f) contains the passband com-ponent at 500 Hz plus reduced (but nonzero) components at 1500 and 3000 Hz.

Dependingontheapplication,thisFIRfiltermayormaynotbesufficienttoprovidethedesiredresults.Forexample,inaudioapplications,thepassbandripple


maycauseaudibleinaccuraciesintheresultingfilteredoutput.Ontheotherhand,inotherapplications,aminimaltransitionregionbetweenpassbandandstopband(called the transition band) may be the most desirable, regardless of the ripple.Onemethodforalteringthefilterfrequencyresponseisbyapplyingdifferentwin-dowfunctionstotheidealfilterresponse.AcomparisonofwindowsisshowninFigure 8.16,whichshowstheidealimpulseresponsehI(n),threedifferentwindowsw[n],thecorrespondingrealizableimpulseresponsehR(n),andtheresultingFouriertransform.The three examples shownare the rectangularwindow (also calleda“boxcar”),whichwehavealreadydiscussed;aBartlettwindow(atriangle);andaHammingwindow,whichhasacosineshapewithanaddedDCcomponent.Thethreewindowsareexpressedmathematicallyas

=

= − −

= −π

rectangular: [ ] 1

Bartlett: [ ] 22 2

Hamming: [ ] 0.56 0.46cos 2H

w n

w nN

Nn

N

w nn

N

R

B

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035–0.02

0

0.02

0.04

0.06

h R(t

)FIR Filter Using Rectangular Window, Time Domain Results

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035

–2

0

2

x(t)

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035

–1

0

1

y(t)

t (sec)

Figure 8.14 MATLAB-generated plot of an FIR filter and its input and output signals. By applying hR(t) to x(t), we obtain the resulting y(t), which has the com-ponents above 1000 Hz filtered out.


wheren =0,1,2,…,N.Theresultingfrequencyresponsesarealllowpass,butwithvaryingshapes.Therectangularwindowgivesthemostrippleinthepassbandandstopbandbutalsohasthesharpesttransitionband.TheBartlettwindowcauseslessripplethantherectangularwindowbuthasawidertransitionband.TheHammingwindowhasthelargestdifferentialbetweenthepassbandandstopband;thatis,ithasthelargestdropbetweenthepassbandandthefirstlobeofthestopband.

0 1000 2000 3000 4000 5000 600010–4

10–2

100

|HR(

f)|

FIR Filter Using Rectangular Window, Frequency Domain Results

0 1000 2000 3000 4000 5000 6000

100

102

104

|X(

f)|

0 1000 2000 3000 4000 5000 6000

100

102

104

|Y(

f)|

f (Hz)

Figure 8.15 Fourier transforms of the signals of Figure 8.14 plotted on semilog axes. The top plot shows that the frequency response of the shifted and trun-cated ideal impulse response has ripple in the passband and stopband. The two rightmost peaks visible in |X(f)| (middle plot) are substantially reduced in |Y(f)| (bottom plot).


ProjectsProject.8.1

In telephony,DTMF (dual-tone multifrequency) is used to encode and transmitkeypressesfromalocalhandsettoaremotereceiveroverthetelephonenetwork.FigureP8.1showsa simplifiedrepresentationofa telephonekeypadwithsevenswitchesandsevensinusoidalvoltagesources.Theswitchesarearrangedsuchthatonanykeypress,twoswitchesforthecorrespondingrowandcolumnwillclose,

Rectangular (boxcar) window:

Bartlett (triangular) window:

Hamming (raised cosine + DC) window:

idealimpulse

responsewindow

realizableimpulse

response

frequencyresponse

HR(φ)hI[n] w[n]

wH[n]

wB[n]

wR[n]

hR [n] = hI [n]w [n]

Figure 8.16 An example of some of the window functions w[n] that may be applied to an ideal low-pass impulse response hI[n] to create a realizable impulse response hR[n] for use in an FIR filter. Note that to make the plots more readable, the discrete-time waveforms are drawn as continuous (by “connecting the dots”). The corresponding realizable DTFT frequency responses HR(𝛟) are also shown.


therebyconnectingtwoofthevoltagesourcestotheadderonthelowerright.TheresultingoutputvoltageVDTMFwillcontainthesumoftwodistinctfrequenciesthataresenttotheremotereceiver.Forexample,thefactthat“2”isinthetoprowandcentercolumnofthebuttonarraywillcausesignalsof697Hzand1336Hztobesummedandsenttotheremotereceiver.Onreceivingthetransmittedsignal,thereceivercandeterminewhichbuttonwaspressedbydeterminingthetwouniquefrequenciesinthereceivedwaveform.

1. Write a MATLAB function named dtmf.m that generates the appropriateDTMFtonesforthenumerals0through9(donotworryaboutthe*and#buttonsfornow).Yourfunctionshouldtakeoneargument(anumberfrom0to9)andgenerateavectoroftheappropriatelengthcontainingthewaveform.Assumethefollowingparameters:

Samplingrate:8000samples/sec Toneduration:0.25sec Maximumsignalamplitude:1

ABC DEF

JKLGHI MNO

TUVPRS WXY

697 Hz

770 Hz

852 Hz

941 Hz

1209 Hz 1336 Hz 1477 Hz

VDTMF

Figure P8.1 A telephone keypad consists of 12 buttons, 7 switches, 7 sine wave generators, and an adder.


2. Use your DTMF function to create a single waveform that dials your owntelephonenumber.Youwillneedtorunyourfunctionmultipletimes(foreachdigitinyourphonenumber)andconcatenatetheresultsintoasinglevector.

3. Save thewaveform toafilewithMATLAB’swavwrite() command.Thewavwrite()commandcanberunseveralways(runhelp wavwriteformoreinformation),butinthesimplestcasetakestwoarguments:avector(presumedtobe8000samples/sec)andafilename.ConfirmthatyoucanplaytheresultingWAVfilethroughyourcomputerspeakers.

4. ThewaveformyougenerateddoesnotgenerateaconformingDTMFsequence[4] because it does not contain the appropriate amount of silence betweentheindividualtones.Modifyyourdtmf.mfunctiontoadd0.25secofsilenceat theendofeachgenerated tone.Re-createyourWAVfilewith theaddedsilencebetweenthetones.Tryplayingtheresultingaudiointoyourowntele-phone(byholdingthehandsetuptoyourcomputerspeaker)andseeifyoucangetthegeneratedtonestodialyourphone.(Notethatthiswillonlyworkonlandlineswithadialtone,notonamobilephone.)

Project.8.2

DesignasystemtodetecttransmittedDTMFkeypresses.AssumethatfS=8000samples/sec.Proceedasfollows:

1. Create10testsignalsx0(t)throughx9(t)correspondingtothe10numericbut-tonsontheDTMFkeypadofFigureP8.1.Eachsignalshouldbe0.25seclongandshouldcontainthesumoftwosinewaves(eachwithanamplitudeofone)correspondingtotherowandcolumnfrequenciesforeachindividualbutton.Concatenateallofyourtestsignalstogethertoobtainonelongsignalof2.5secthatcontainsall10simulatedkeypressesinsuccession.(Alternatively,ifyoudidProject8.1,usethesignalthatyougeneratedinthatproblem.)

2. WriteaMATLABprogramthatdoesa1024-pointFFToneachsuccessivesetof1024elementsinyourinputsignalanddetectsthepresenceofanyofthesevenDTMFfrequencies.PrintyourdetectionresultstothescreenforeveryFFToperation.

3. WhathappenswhentheFFToverlapstwokeypresses?Add0.25secofsilenceinbetweeneachkeypresstoavoidthisproblemandrerunyourdetectionpro-gramtoseeifanyerroneousentriesareremoved.

4. ReducethesizeofyourFFTwindowto512,256,128,64,and32.WhatisthesmallestwindowthatyoucanuseandstillreliablydetectallsevenDTMFfrequencies?

Project.8.3

Designahigh-passFIRfilterwiththefollowingspecifications:

fS=8000samples/secfo=200HzTaps=201

Window:Bartlett(triangle)


Notethatahigh-passfiltercanbedecomposedintothedifferenceofanall-passfilterandalow-passfilter(seeFigureP8.3).Anall-passfilterhasavalueofoneforallfrequencies.

1. Find the impulse response of an all-pass filter by explicitly calculating theinverseDTFTaccordingtoEquation(8.15) forvaluesofn =−2,−1,0,1,2.Doesyoursolutionmakesense?

2. CalculatetheidealimpulseresponsehI[n]forthehigh-passfilterbysubtract-ing the corresponding low-pass impulse response from the all-pass impulseresponse.

3. Createarealizablehigh-passimpulseresponsehR[n]bymultiplyingtheidealresponsebyatriangularwindowwhosewidthisthenumberoftapsandwhoseheightisone.

4. Createatestinputsignalx(t)containingsinewavesat60,300,and1000Hz.Applyx(t) toyourhigh-pass impulseresponseandobtaintheoutputy(t)bysimulatingtheFIRfilterofFigure 8.12.

5. Performa1024-pointFFTonhR(t),x(t),andy(t)andplotthemagnitudesofthe transformson semilogaxes (usea linearhorizontal axis and logverticalaxis).Whatfrequencycomponentsmakeitthroughthehigh-passfilter?

Project.8.4

DesignabandpassFIRfilterthatdetectskeysfromthesecondrowoftheDTMFkeypadshowninFigureP8.1.AssumethatfS=8000samples/sec,andafilterwith401taps.Followthisprocedure:

1. First,createanidealbandpassresponseinthefrequencydomainsuchthatthepassband iscenteredon770Hzandhasa lowcutofffrequency fLandhighcutofffrequencyfH.ThefrequenciesfL andfH aredeterminedbythenecessitytorejecttheadjacentDTMFfrequenciesof697and852Hz.

2. BecauseFouriertransformsarelinear,abandpassresponsecanbecreatedbysubtractingtwolow-passresponses.Decomposeyourbandpassresponseintoalow-passresponsewithcutofffrequencyfHandasecondlow-passresponsewithcutofffrequencyfL.Createanidealimpulseresponsebysubtractingtheidealimpulseresponsesofthetwolow-passfilters.

3. CreatearealizableimpulseresponsebymultiplyingyouridealresponsebyoneofthewindowsinFigure 8.16.Explainwhichwindowyouchoseandwhyitisoptimalforthisapplication.

fo−fo fo−fo

1 1 1

HAP( f ) HLP( f ) HHP( f )

Figure P8.3 A high-pass frequency response HHP(f) can be constructed by sub-tracting a low-pass response HLP(f) from an all-pass response HAP(f) .


4. Create two test signals for yourfilter corresponding to the2 and5buttonsonthekeypadandrunthemthroughyourfilterbysimulatingtheFIRfiltercircuitofFigure 8.12.Plotthetime-domainresultsandconfirminyourfilteroutputthatyourfilterdetectsthe5andnotthe2.

Project.8.5

Use the technique employed inExample8.2andfind the complex exponentialFouriercoefficientsforthetriangularpulseofFigureP8.8.

Project.8.6

RerunExample8.4forthelow-passFIRfilter.Makethefollowingchangestothefilterparameters(oneatatime)andcommentqualitativelyontheeffectontheoutputplots.

1. ChangethesamplingratefSfrom40,000to44,100samples/sec. 2. Changethenumberoffiltertapsto101. 3. Changethenumberoffiltertapsto1601.

4. Changethethreeinputfrequenciesto200,500,and39,200Hz.

References 1. Siebert,W.M.,Circuits, Signals, and Systems,MITPress,Cambridge,MA,1986. 2. Cooley,J.W.,andTukey,J.W.,Analgorithmforthemachinecalculationofcomplex

Fourierseries,Mathematics of Computation,19:297–301,1965. 3. Oppenheim, A.V., and Schafer, R.W., Digital Signal Processing, Prentice-Hall,

EnglewoodCliffs,NJ,1975. 4. Multifrequencypush-buttonsignalreception,ITU-TRecommendationQ.24,http://

www.itu.int.

275

Chapter 9

Curve Fitting

9.1 IntroductionTherearemanyoccasionsinengineeringthatrequireanexperimenttodeterminethebehaviorofaparticularphenomenon.Theexperimentmayproduceasetofdatapointsthatrepresentsarelationshipbetweenthevariablesinvolvedinthephenom-enon.Wemaythenwishtoexpressthisrelationshipanalytically.Amathematicalexpressionthatdescribesthedataiscalledanapproximating function.Therearetwoapproachestodetermininganapproximatingfunction:

1.Theapproximatingfunctiongraphsasasmoothcurve.Inthiscase,theplottedcurvewillgenerallynotpassthroughallthedatapoints,butweseektominimizetheresultingerrortogetthebestfit.Aplotofthedataonlinear,semilog,orlog-logcoordinatescanoftensuggestanappropriateformfortheapproximatingfunction.

2.Theapproximatingfunctionpassesthroughalldatapoints(asdescribedinSection9.3).However,ifthereissomescatterinthedatapoints,thisapproxi-matingfunctionmaynotbesatisfactory.

9.2 Method of Least Squares9.2.1 Best-Fit Straight LineInthemethodofleastsquares,weseektofindthebest-fitstraightlinegivenasetofndatapoints( , ), ( , ), ,( , )1 1 2 2x y x y x yn n… .


Wewishtorepresenttheapproximatingcurveycasastraightlineoftheform

1 2y c c xc = + (9.1)

wherec1andc2areunknownconstantstobedetermined.LetDbethesumofthesquareoftheerrorsbetweentheapproximatinglineandtheactualpoints.Then,

( ) 2

1

D y y xi c i

i

n

∑[ ]= −=

(9.2)

( )1 22

1

y c c xi i

i

n

∑[ ]= − +=

(9.3)

[ ( )] [ ( )] [ ( )]1 1 2 12

2 1 2 22

1 22y c c x y c c x y c c xn n= − + + − + + + − + (9.4)

To obtain the best-fit straight-line approximating function, minimize D by

taking ∂∂

= 01

Dc

and ∂∂

= 02

Dc

.TakingthepartialderivativeofEquation(9.3)with

respecttoc1gives

0 2[ ( )][ 1]1

1

1 2Dc

y c c xi

n

i i∑∂∂

= = − + −=

01

2

1

1y c x nci

i

n

i

i

n

∑ ∑= − −= =

or

∑ ∑+ == =

1

1

2

1

nc x c yi

i

n

i

i

n

(9.5)

TakingthepartialderivativeofEquation(9.3)withrespecttoc2gives

0 2[ ( )][ ]2

1 2

1

Dc

y c c x xi i i

i

n

∑∂∂

= = − + −=

0 1 1

1

22

11

x y c x c xi i

i

n

i

i

n

i

n

∑ ∑∑= − −= ==

Curve Fitting ◾ 277

or

∑ ∑ ∑+ == = =1

12

1

2

1

x c x c x yi

i

n

i

i

n

i i

i

n

(9.6)

Equations(9.5)and(9.6)describeasystemoftwoalgebraicequationsintwounknownsthatcanbesolvedbythemethodofdeterminants(Cramer’srule):

1

2

2

2

2c

y x

x y x

n x

x x

y x x x y

n x x x

i i

i i i

i

i i

i i i i i

i i i

∑ ∑∑ ∑

∑∑ ∑

∑ ∑ ∑ ∑∑ ∑ ∑

( )( ) ( )( )( )( )

= =−

− (9.7)

2

2

2c

n y

x x y

n x

x x

n x y x y

n x x x

i

i i i

i

i i

i i i i

i i i

∑∑ ∑

∑∑ ∑

∑ ∑∑∑ ∑∑( )( )( )( )

= =−

− (9.8)

9.2.2 Best-Fit mth-Degree PolynomialWecangeneralizethisapproachforanmth-degreepolynomialfit.Inthiscase,taketheapproximatingcurveyctobe

y c c x c x c x c xc 1 2 32

43

m 1m= + + + + + + (9.9)

wherem ≤n −1,andnisthenumberofdatapoints.Themeasuredvaluesare( , )x yi i fori =1,2,…,n.


Let ( ),y y xc i c i= betheapproximatedvalueofyiatthepoint ( , )x yi i .Then,

∑ ∑ ( )[ ]= − = − + + + +

=

+

=

,2

1

1 2 32

12

1

D y y y c c x c x c xi c i

i

n

i i i m im

i

n

TominimizeD,take

…

∂∂

=∂∂

=∂∂

=+

0, 0, , 01 2 1

Dc

Dc

Dcm

Then,

∑

∑

∑

∑

( )

( )

( )

( )

[ ]

[ ]

∂∂

= = − + + + −

∂∂

= = − + + + −

∂∂

= = − + + + −

∂∂

= = − + + + −

=

+

=

+

=

+

+ =

+

0 2 1

0 2

0 2

0 2

1 1

1 2 1

2 1

1 2 1

3 1

1 2 12

1 1

1 2 1

Dc

y c c x c x

Dc

y c c x c x x

Dc

y c c x c x x

Dc

y c c x c x x

i

n

i i m im

i

n

i i m im

i

i

n

i i m im

i

m i

n

i i m im

im

Thissetofequationsreducesto

∑ ∑ ∑ ∑

∑ ∑ ∑ ∑ ∑

∑ ∑ ∑ ∑ ∑

( ) ( ) ( )( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

+ + + + =

+ + + + =

+ + + + =

+

++

+ ++

1 22

3 1

12

23

31

1

11

22

32

1

nc x c x c x c y

x c x c x c x c x y

x c x c x c x c x y

i i im

m i

i i i im

m i i

im

im

im

im

m im

i

(9.10)

Equation(9.10)canbesolvedbyGausselimination(asdescribedinChapter3).


MATLABusesmeansquareerror (MSE)asameasureof theprecisionwithwhichthefittedpolynomialdescribesthedata.MSEisdefinedby

MSE= 1 ( ), 2

1n

y yi c i

i

n

∑ −=

(9.11)

whereyc,i=yc(yi).Toobtainthebestpolynomialdegree,runyourprogramforseveraldifferentm

valuesandusetheonewiththelowestMSE.

9.3 Curve Fitting with the Exponential FunctionManysystems(e.g.,semiconductordevices)canbemodeledwithexponentialfunc-tions.Ifyourexperimentaldataappearstofallintothiscategory,itcanbefittedwithafunctionoftheform

12y ecx=α −α (9.12)

where 1α and 2α arerealconstants.Letusassumethatasetofnmeasureddatapoints( , ), ( , ), ,( , )1 1 2 2x y x y x yn n…

exists.Then,let lnz yi i= and ln ln 1 2z y xc c= = α −α andlet ln1 1c = α and .2 2c = −α Then,Equation(9.12)becomeslinearinzc ,thatis,

1 2z c c xc = + (9.13)

Forthedatapoints( , ), ( , ), ,( , )1 1 2 2x y x y x yn n… , thenewsetofdatapointsbecomes( , ), ( , ), ,( , )1 1 2 2x z x z x zn n… .

Aswederivedintheprevioussection,thebest-fitapproximatingstraight-linecurvebythemethodofleastsquaresgives

1

2

22c

z x x x z

n x x

i i i i i

i i

∑ ∑ ∑ ∑∑∑

( )( ) ( )( )( )

=−

− (9.14)

and

22

2cn x z x z

n x x

i i i i

i i

∑ ∑∑∑∑

( )( )( )

=−

− (9.15)


Then,α = exp( )1 1c and 2 2cα = − .TheMSEcanbedeterminedbyEquation9.11.Thisanalysiscanbeusedtodeterminethedampingconstantζ inaparallel

RLCcircuit(asanalyzedinAppendixA).Thiscanbeaccomplishedbyexaminingtheoscilloscopeoutputofthestepresponse(seeFigure 9.1).Thegoverningequationoftheenvelopeoftheresponseisoftheform

0y y e tn= −ζω (9.16)

where

capacitor voltage

initial capacitor voltage (dependent on initial conditions)

damping factor

natural frequency = 1

y

y

LC

o

n

=

=

ζ =

ω =

ComparingEquation(9.15)withEquation(9.10),weseethat

1 0

2

y

n

α =

α = ζω

(t1, y1)(t2, y2)

(t3, y3)

Figure 9.1 Oscilloscope screen shot of the step response of the capacitor voltage of a parallel RLC circuit. (With permission from Tektronix, Inc.)


withtreplacingx.Therefore

22 LC

nζ=

αω

= α (9.17)

Bymeasuringndatapointsontheenvelope,i.e.,(t1,y1),(t2,y2)… (tn,yn)wecandeterminethebestfitvalueforα2,givingourbestestimateforthedampingfactor,ζ.

9.4 MATLAB’s polyfit FunctionMATLABreferstocurvefittingwithapolynomialas“polynomialregression.”Thepolyfitfunctionreturnsavectorofm + 1coefficientsthatrepresentthebest-fitpolynomialofdegreemforthe( , )x yi i setofdatapoints.Thecoefficientordercor-respondstodecreasingpowersofx,thatis,

1 21

32

1y a x a x a x a x acm m m

m m…= + + + +− −+ (9.18)

The vector ( , , , , )1 2 1A a a a am m…= + containing the coefficients is obtained byrunningA=polyfit(X,Y,m),where ( , , , )1 2X x x xn…= and ( , , , )1 2Y y y yn…= .

MATLAB provides thepolyval(a,x) function to calculate the approxi-matedvalues ,yc i at( , , , )1 2x x xn… forthefittedcoefficients … +( , , , )1 2 1a a am :

, 1 21

31

1y a x a x a x a x ac i im

im

im

m i m…= + + + +− −+

ForcalculatingtheMSE,polyvalisusefulasseeninExample9.1.

Example 9.1

Given the following set of data points, find the best fit to polynomials of sec-ond, third, fourth, and fifth degree. Also, calculate the mean square error.

x y( , ) ( 10, 980), ( 8, 620), ( 6, 70), ( 4,80),

( 2,100), (0,90), (2,0), (4, 80), (6, 90), (8,10), (10,220)

= − − − − − − −

− − −

%Example_9_1.m%Thisprogramdeterminesthebestfitpolynomialcurve%passingthroughagivensetofdatapointsfor%polynomialdegrees2thru5.clear;clc;


%ProvideddatasetX=[-10-8-6-4-20246810];Y=[-980-620-7080100900-80-9010220];%Fittopolynomialsofdegreem:m=2:5;%Createasecondsetofxpointsoverthesameinterval%inordertoplotourfittedcurveX2=-10:0.5:10;%VectortokeeptrackofMSE(toprintatend)MSE=zeros(length(m));fori=1:length(m)fprintf('m=%i\n',m(i));%anmth-degreepolynomialhasm+1coefficients:A=zeros(m(i)+1);%dothefitA=polyfit(X,Y,m(i));%computetheMSEYc=polyval(A,X);MSE(i)=(1/length(X))*sum((Y-Yc).^2);%printtablefprintf('xyyc\n');fprintf('-----------------------------\n');forj=1:length(X)fprintf('%5.0f%5.0f%8.2f\n',...X(j),Y(j),Yc(j));endfprintf('\n\n');%plotsubplot(2,2,i);plot(X2,polyval(A,X2),X,Y,'o');xlabel('x');ylabel('y');grid;axis([-1010-1500500]);title(sprintf('Degree%dpolynomialfit',m(i)));end%printcomputedMSEfprintf('mMSE\n')fprintf('----------------------\n');fori=1:length(m)fprintf('%g%10.2f\n',m(i),MSE(i))end

-------------------------------------------------------------------

PROGRAM.OUTPUT

m MSE------------------- 2 32842.37 3 2659.97 4 2342.11 5 1502.95

As would be expected, the MSE decreases as the order of the fitted polyno-mial increases. The polynomial fits are plotted in Figure 9.2.


Example 9.2

Figure 9.3 shows an RC circuit, and Table 9.1 shows a set of steady-state voltage amplitudes and phases that were measured in the lab for VIN and VOUT at the given frequencies.

We wish to fit the data to the best-possible first-order frequency response.The RC circuit can be considered a first-order low-pass filter with a fre-

quency response of the form

H f

nj f d

( )2

0

0

=π + (9.19)

–10 0 10–1500

–1000

–500

0

500

x

yDegree 2 Polynomial Fit

–10 0 10–1500

–1000

–500

0

500

x

y

Degree 3 Polynomial Fit

–10 0 10–1500

–1000

–500

0

500

x

y


–10 0 10–1500

–1000

–500

0

500

x

y


Figure 9.2 Polynomial fits using MATLAB’s polyfit function for second-, third-, fourth-, and fifth-order polynomials.

+

−

C

R

VIN VOUT

Figure 9.3 RC circuit with input sinusoidal voltage VIN and output VOUT.


where H = VOUT / VIN is the frequency-dependent output-to-input ratio, and n0 and d0 are the numerator and denominator coefficients, respectively, to be determined. Because Equation (9.19) has the polynomial in the denomi-nator, we start by rearranging it into a polynomial of the form of Equation (9.18) so that we can apply polyfit. This gives

= π +

H nj f

dn

1 12

0

0

0

(9.20)

Matching terms between Equations (9.20) and (9.18) shows that a j n21 0= π and a d n2 0 0= ; thus, the inverse of the frequency response H1 can be fitted by using A=polyfit(f,1./H,1), where f is the vector of measured frequencies, and H is the vector containing measured values of V VOUT IN . The MATLAB program follows:

%Example_9_2.m%Fitlab-measuredfreqresponseto1storderresponseclear;clc;%Lab-measureddatafromTable9.1:f=[02e34e36e38e310e3];V_in=[555555];V_out=[52.11.3.8.6.5];Phi_degrees=[0-60-75-80-83-85];%RatioofV_outtoV_inisthemagnitudeofthe%frequencyresponse:H_mag=V_out./V_in;%Thephaseisthemeasuredphasedifference(converted%toradians)H_phase=Phi_degrees*pi/180;%ExpressthecomplexfreqresponseHas:%mag*exp(j*phase)H=H_mag.*exp(j*H_phase);%Dopolynomialfitto1/HA=polyfit(f,1./H,1);%Computecoefficientsoffittedfreqresponsebased%polyfitresults

Table 9.1 Laboratory-Measured Frequency Response of RC Circuit

Frequency (Hz)

VIN (volts peak-to-peak)

VOUT (volts peak-to-peak)

Phase Difference between VOUT and VIN

0 5 5 0

2,000 5 2.2 –60

4,000 5 1.2 –75

6,000 5 0.8 –80

8,000 5 0.6 –83

10,000 5 0.5 –85


n0=2*pi*j/A(1);d0=2*pi*j*A(2)/A(1);%Printfittedfrequencyresponse:fprintf('Fittedfrequencyresponse:\n');fprintf('H(f)=%.0f/(j*2*pi*f+%.0f)\n',n0,d0);fprintf('Polelocation:%.0frad/sec(%.0fHz)\n',...d0,d0/(2*pi));%Forcomparisonpurposes,calculatethefittedcurve%withfineprecision.Use100log-spacedpointsfrom1Hz%to1MHzf_fine=logspace(0,6,100);H_fit=n0./(j*2*pi*f_fine+d0);H_fit_mag=abs(H_fit);H_fit_phase=angle(H_fit);%CreateBodeplot:subplot(2,1,1);loglog(f,H_mag,'o',f_fine,H_fit_mag);axis([1e01e61e-32]);legend('measured','fittedfunction');ylabel('|H(f)|');title('Fitofmeasureddatato1st-orderfreqresponse');subplot(2,1,2);semilogx(f,(180/pi)*H_phase,'o',...f_fine,(180/pi)*H_fit_phase);axis([1e01e6-10010]);legend('measured','fittedfunction');ylabel('\angleH(f)(degrees)');xlabel('frequency(Hz)');

-------------------------------------------------------------------

PROGRAM.RESULTS

Fittedfrequencyresponse: H(f)=6204/(j*2*pi*f+6842)Polelocation:6842rad/sec(1089Hz)


9.5 Cubic SplinesSupposeforagivensetofdatapoints,allattempteddegreepolynomialapproxi-mating curves produced points that were not allowed. For example, suppose itis known that a particular property representedby thedata (such as voltage orabsolutetemperature)mustbepositive,andalltheattemptedpolynomialapproxi-mating curves produced some negative values. For this case, the polynomialapproximating functionwouldnot be satisfactory.Themethodof cubic splineseliminatesthisproblem.

Givenasetofn +1datapoints( , )x yi i fori =1,2,…,n +1,themethodofcubicsplinesdevelopsasetofncubicfunctionssuchthaty(x)isrepresentedbyadifferentcubicfunctionineachofthenintervals,andthesetofcubicspassesthroughthen + 1datapoints.


This isaccomplishedby forcing the slopesandcurvatures tobe identical foreachpairofcubicfunctionsthatjoinatagivendatapoint.

Note: Curvature

1

2

2

232

K

d ydx

dydx

=±

+

(9.21)

Weseethatifwesatisfythefollowingsetofequations,Equations(9.22),wewillautomaticallysatisfyboththeslopesandthecurvatureattheconnectingpoints.Thus,set

( ) ( )

( ) ( )

( ) ( )

int 1 int

int 1 int

int 1 int

y x y x

y x y x

y x y x

i i i i

i i i i

i i i i

[ ] [ ]

[ ] [ ]

[ ] [ ]

=

=

=

−

−

−

(9.22)

Atthepoint xi ,thevalues yi , yi ,and yi mustbethesameforboththeleftinterval(inti −1),andtherightinterval(inti).

100

10–2

|H(f

)|Fit of Measured Data to 1st-Order Frequency Response

MeasuredFitted function

–100

–80

–60

–40

–20

0

Frequency (Hz)

H(f

) (d

egre

es) Measured

Fitted function

100 101 102 103 104 105 106

100 101 102 103 104 105 106

Figure 9.4 Fitted frequency response for a first-order RC circuit.


Inthe(i−1)interval, 1x x xi i≤ ≤− (seeFigure 9.5),and

( ) ( ) ( ) ( )1 1 1 1 12

1 13y x A B x x C x x D x xi i i i i i i= + − + − + −− − − − − − − (9.23)

where 1Ai− , 1Bi− , 1Ci− ,and 1Di− arethecoefficientsofthefittedcubicpolynomialbetween 1xi− andxi .

Inthe(i)interval, 1x x xi i≤ ≤ + ,and

( ) ( ) ( ) ( )2 3y x A B x x C x x D x xi i i i i i i= + − + − + − (9.24)

whereAi,Bi,Ci,andDiarethecoefficientsofthefittedcubicpolynomialbetweenxiand 1xi+ .

Thisgivesn −1equationsandn +1unknowns.

Thus,valuesfor2

2d ydx

at 1x and 1xn+ mustbeassumedtoobtainthefinaltwoequations.Severalalternativesexistforchoosingtheselasttwoequations:

a.Assume ( ) ( ) 01 1y x y xn= =+ .This iswidelyused and forces the splines toapproachstraightlinesattheendpoints.

b.Assume ( ) ( )1 2y x y x= and ( ) ( )1y x y xn n=+ . This forces the splines toapproachparabolasattheendpoints.

9.6 The Function interp1 for Cubic Spline Curve FittingMATLAB’sbuilt-in functioninterp1 interpolatesbetweendatapointsbythecubicsplinemethod.

Usage:Yi=interp1(X,Y,Xi,'spline')

whereX,Yarethesetofdatapoints,andXiisthesetofXvaluesatwhichthesetofvaluesYiistobereturned.

yi–1

xi–1 xi+1xi

yi+1yi

y

x

inti

inti – 1

Figure 9.5 Two adjacent intervals in a cubic spline interpolation.


Example 9.3

%Example_9_3.m%Thisprogramusesinterpolationbycubicsplinesto%upsampleanaudiosignal.clear;clc;%Defineoriginaldatapointsfory(t)(timeinmicrosec)orig_t=[048121620242832364044485256];orig_y=[.7.9.9.7.30-.3-.7-.7-.30.3.7.7.3];%Defineupsampledtimepointsupsampled_t=0:60;%Calculateinterpolateddatapointsusingcubicsplinesinterp_y=interp1(orig_t,orig_y,upsampled_t,'spline');%Printoutputtoscreenfprintf('UPSAMPLINGVIACUBICSPLINEFIT\n');fprintf('time(microsec)origsampleupsample\n');fprintf('--------------------------------------\n');fori=1:length(upsampled_t)%Trytofindthecorrespondingtimepointinthe%originaldatasetforthisupsampledtimepointwith%MATLAB's"find"function.j=find(orig_t==upsampled_t(i));ifj>0%ifwefoundoriginalvalue,printbothorigand%interpolated.fprintf('%8.2f%10.1f%10.3f\n',...upsampled_t(i),orig_y(j),interp_y(i));else%otherwise,justprintinterpolatedvaluefprintf('%8.2f%10.3f\n',...upsampled_t(i),interp_y(i));endendplot(orig_t,orig_y,'o',upsampled_t,interp_y);xlabel('t(microsec)');ylabel('y(t)');grid;title('UpsamplingwithCubicSplineInterpolation');legend('original','upsampled','Location','SouthWest');

-------------------------------------------------------------------

PROGRAM.OUTPUT.(EXCERPT)

UPSAMPLINGVIACUBICSPLINEFIT

time(microsec) originalsample upsample 0.00 0.7 0.700 1.00 0.770 2.00 0.827 3.00 0.870 4.00 0.9 0.900 5.00 0.918 6.00 0.923 7.00 0.917 8.00 0.9 0.900 9.00 0.872


10.00 0.830 11.00 0.774 12.00 0.7 0.700 13.00 0.608 14.00 0.506 15.00 0.400 16.00 0.3 0.300 . . . .

9.7 Curve Fitting with Fourier SeriesSupposeanexperimentaldatasetu(t)producesaplotasshowninFigure 9.6,anditisdesiredtoobtainananalyticalexpressionthatcomesclosetofittingthedata.Letusassumethatthedatarangesoverthesymmetricinterval[−L,L].Ifnot,wecanmakeitsobyshiftingtheorigin.Inthiscase,theoriginalabscissadatawerefor0≤t≤10.5sec,andthedatawereshiftedbylettingx =t −5.25.Then,thexdomainissubdividedinto70equalintervalswithΔx =10.5/70=0.15sec.

Thus, 1x xi i−+ isuniformovertheentiredomain.Anattempttofitapolyno-mial-approximatingcurvetothesedatadidnotproduceanapproximatingcurvethatresembledthedataandwasthusnotsuccessful.However,theuseofaFourierseriesgivesareasonableanalyticalexpressionapproximatingthedata.If uc istheapproximatingcurve,thenbyaFourierseries,

∑= +

π+

π

=

∞

( ) cos sin0

1

u x a am xL

bm xLc m m

m

(9.25)

where

∫=−

12

( )0a Lu x dx

L

L

(9.26)

∫=

π

−

1 ( )cosaL

u xm xL

dxmL

L

(9.27)

∫=

π

−

1 ( )sinbL

u xm xL

dxmL

L

(9.28)

Using30termsintheseriesandSimpson’sruleofintegration,anapproximatingcurveasshowninFigure 9.7isobtained.


6

5

4

3

2

1

0

–1–6 –4 –2 0

x2 4 6

u c(x)

×104

Figure 9.7 Fourier series fit uc versus x for the data in Figure 9.6.

6

5

4

3

2

1

0

–1–6 –4 –2 0

x2 4 6

u(x)

×104

Figure 9.6 Experimental data of u versus x.


Theπ

+π

cos sinam xL

bm xLm m termscanbesimplifiedbyapplyingthe

trigonometricidentity cos sin sin( ),2 2a b a bβ + β = + β+ giving

2 2c a bm m m( )= +

Aplotofcmversus≠mL

(neglectingthephasefactorφ)isshowninFigure 9.8.

ProjectsProject.9.1

FigureP9.1showsaresistor-diodecircuitusingatype1N914silicondiode(D1)anda10kΩresistor(R1).TableP9.1showsalistoflaboratorymeasurementsofV2forvariousappliedvoltagelevelsofV1atroomtemperature(300K).

7000

6000

5000

4000

3000

Four

ier C

oeffi

cien

t Val

ue C

m

2000

1000

00 2 4 6 8 10

n*pi/L12 14 16 18

Figure 9.8 Fourier series coefficient amplitudes versus ≠nL

.

+

− D11N914

R110 K

+

−V2

V1

Figure P9.1 Diode-resistor circuit for laboratory measurement of diode I-V curve.


1. UseKirchhoff’svoltagelawtocalculatethediodecurrentiDintermsofV1andV2. 2. UsethetechniquedescribedinSection9.3tofindtheleast-squarefitofiDandvD(=V2)

totheformula

expi I v

VD S

D

T= (P9.1)

Table P9.1 Laboratory Measurements of Resistor-Diode Circuit

V1 (volts) V2 (volts)

0.189 0.189

0.333 0.317

0.393 0.356

0.819 0.464

1.067 0.487

1.289 0.501

1.656 0.518

1.808 0.522

2.442 0.541

3.949 0.566

4.971 0.579

6.005 0.588

6.933 0.595

7.934 0.602

9.014 0.607

10.040 0.613

11.009 0.619

15.045 0.634

19.865 0.647

24.64 0.657

29.79 0.666


Inthiscase,youareusingtherawdatatofindbest-fitvaluesforISandVT.Plotboththelabdataandyourfittedcurveonthesameaxes.

3. V kT qT = isknownasthethermal voltage(wherekistheBoltzmannconstantandqistheunitelectriccharge).Foryourbest-fitvalueforVT,whatisthecorrespondingtemperaturevalueT(inkelvin)?Doesthisvalueseemreasonable?

4. Amoreaccurateequationtomodelthediodeis

= expi I

vnVD S

D

T

wherenistheideality factor.Findthebest-fitvaluefornwhenT =300K.

5. ThecompleteI-Vmodelforthediodeis

= −exp 1i I

vnVD S

D

T

Isitreasonablethatweneglectedthe−1terminourcurvefit?Why?

Project.9.2

Themeasuredopen-loopfrequencyresponseofthe741opampisshowninFigureP9.2.

1. Find themagnitude (indecibels) andphase (indegrees)of the frequency responsebymeasuringdirectlyfromtheplot.Notethatthephasereadings(ontherightsideoftheplot)aredenotedas“PhaseLag,”whichmeansthatthereadingsshouldbeinterpretedasnegativedegrees.EnterthereadingsastwovectorsinMATLAB.

2. AmplifiergainAisusuallyexpressedindecibels(dB),where

20 logdBA A=

Convertyourdecibelreadingstoabsolutegainsbydividingby20andthenraisingto the power of 10. Also, convert your degree readings to radians. Then, express thefrequencyresponseincomplexpolarform.Rememberthat = φ| |z z e j ,where | |z isthemagnitudeandϕisthephaseangle.

3. UseMATLAB’spolyfitfunctiontofindafittoafrequencyresponseoftheform

( )

( 2 ) 20

21 0

H f dj f n j f n

=π + π +

Hint:getyourfitworkingforf≤1000handthenaddthepointsforthehigherfre-quenciesoneatatime.Becausethepolynomialisinthedenominator,trytofit1 ( )H f insteadofH( f ).


4. Find the locations of the poles for your fitted H( f ) by solving for the roots of thedenominator.

5. Plotthemeasuredresultsandthefittedresultsonthesamesetofaxes.Doesthefitseemreasonable?Atwhatfrequenciesdoes,thefitfail,andwhy?

Project.9.3

Reusethedata fororig _ tandorig _ y fromExample9.3toperformacurvefitusingFourierseries.Beginbyshiftingtsothatthedataaresymmetricaroundtheoriginsuchthat−L ≤t ≤L(whereL =28μsec).Then,proceedasfollows:

1. WriteaMATLABprogramtocalculatethefirstmFouriercoefficients.UseSimpson’srule(asdescribedChapter5)tocalculatethecoefficients 0a , , ,...,1 2a a am,and , ,...,1 2b b bm asinEquations(9.26)–(9.28).

2. GeneratethreesetsofFouriercoefficientsform =2,m =4,andm =6. 3. Generatethreefittedcurves(form =2,4,6)bysubstitutingyourcoefficientsbackinto

Equation(9.25).Plotonthesamesetofaxes.Whenplottingthefittedcurves,use150points on the horizontal axis over the range −L ≤ t ≤ L (instead of the 15 points inorig _ t).Alsoplotorig _ yonthesameaxes,usingthe×symboltoindicatethepoints.

4. Explainqualitativelywhathappensforthecasewhenm =6.

120

100

80

60

40

20

0

Ope

n-lo

op G

ain

(dB)

0

45

90

135

180

Phas

e La

g (D

egre

es)

0.1 1 10 100 1k 10k 100k 1M 10MFrequency (Hz)

GAIN

PHASE

VS = ±15 VRL = 2 kΩTA = 25°CCL = 100 pF

Figure P9.2 Frequency response of the 741 op amp. (With permission from Analog Devices, Inc.)

295

Chapter 10

Optimization

10.1 IntroductionTheobjectiveofoptimizationistomaximizeorminimizesomefunctionf(calledtheobject function).Youareprobablyfamiliarwithdeterminingthemaximaandmin-imaoffunctionsusingtechniquesfromdifferentialcalculus.However,theproblembecomesmorecomplicatedwhenweplaceconstraintsontheallowablesolutionstof.Asanexample,supposethereisanelectronicscompanythatmanufacturesseveraldifferenttypesofcircuitboards.Eachcircuitboardmustpassthroughseveraldif-ferentdepartments(suchasdrilling,pick-and-place,testing,etc.)beforeshipping.Thetimerequiredforeachcircuitboardtopassthroughthevariousdepartmentsisalsoknown.Thereisaminimumproductionquantitypermonththatthecompanymustproduce.However,thecompanyiscapableofproducingmorethanthemini-mumproductionrequirementforeachtypeofcircuitboardeachmonth.Theprofitthecompanywillmakeoneachcircuitboarditproducesisknown.Theproblemistodeterminetheproductionamountofeachtypeofcircuitboardpermonththatwillresultinthemaximumprofit.Asimilartypeofproblemmaybeoneinwhichtheobjectistominimizethecostofproducingaparticularproduct.Thesetypesofoptimizationproblemsarediscussedingreaterdetailinthischapter.

Inmostoptimizationproblems,theobjectfunction fwillusuallydependonseveralvariables, , , , ,1 2 3x x x xn… .Thesearecalledthecontrol variablesbecausetheirvaluescanbeselected.Optimizationtheorydevelopsmethodsforselectingoptimalvaluesforthecontrolvariables , , , ,1 2 3x x x xn… thateithermaximizeorminimizetheobjectivefunctionf.Inmanycases,thechoiceofvaluesfor , , , ,1 2 3x x x xn… isnotentirelyfreebutissubjecttosomeconstraints.


10.2 Unconstrained Optimization ProblemsIncalculus,itisshownthatanecessary(butnotsufficient)conditionforftohaveamaximumorminimumatpointPisthateachofthefirstpartialderivativesatPbezero,thatis,

( ) ( ) ( ) 0

1

fx

P fx

P fx

P2 n

∂∂

=∂∂

= =∂∂

= (10.1)

where the notation∂∂

( )fx

Pi

indicates the partial derivative with respect to xi

evaluatedatpointP,thatis,∂∂ =

fxi x P

.Ifn =1andtheobjectfunctionisy =f(x),

thenanecessaryconditionforanextremum(maximumorminimum)at 0x isfor( ) 00y x = .

Fory tohavealocalminimumat 0x , ( ) 00y x = and ( ) 00y x > .Fory tohavealocalmaximumat 0x , ( ) 00y x = and ( ) 00y x < .Forfinvolvingseveralvariables,theconditionforftohavearelativeminimumis

morecomplicated.First,Equation(10.1)mustbesatisfied.Second,thequadraticform

∑∑= ∂

∂ ∂− −

==

( ) ( ( ))(( ( ))2

11

Qf

x xP x x P x x P

i jj

n

i i j j

i

n

(10.2)

must be positive for all choices of xi and x j in the in the vicinity of point P,andQ=0onlywhen ( )x x Pi i= fori =1,2,…,n.ThisconditioncomesfromaTaylor series expansionof ( , , , )1 2f x x xn… aboutpointPusingonly termsup to∂

∂ ∂( ).

2 fx x

Pi j

Thisgives

… ∑ ∑∑ [ ]= +∂∂

− +∂∂ ∂

− −= ==

( , , , ) ( ) ( )[ ( )] ( ) ( ( ))( ( ))1 2

1

2

11

f x x x f Pfx

P x x Pf

x xP x x P x x Pn

ii i

i

n

i ji i j j

j

n

i

n

(10.3)

If ( , , , )1 2f x x xn… hasarelativeminimumatpointP,then∂∂

=( ) 0fx

Pi

for

i =1,2,…,nand … − >( , , , ) ( ) 01 2f x x x f Pn forall( , , , )1 2x x xn… inthevicinityofpointP.But, … − =( , , , ) ( )1 2f x x x f P Qn .Thus,for f x x xn( , , , )1 2 … tohavearelativeminimumatpointP,Qmustbepositiveforallchoicesof xi and x j inthevicinityofpointP.

Sincethisanalysisisquitecomplicatedwhenf isafunctionofseveralvariables,aniterativeschemeisfrequentlyusedasamethodofsolution.Onesuchmethodisthemethodofsteepestdescent.Inthismethod,wefirstneedtoguessforapointwhereanextremumexists.Usingagridtoevaluatethefunctionatdifferentvaluesofthecontrolvariablescanbehelpfulinestablishingagoodstartingpointfortheiterationprocess.

Optimization ◾ 297

10.3 Method of Steepest DescentConsiderafunctionfofthreevariablesx, y,and z.Fromcalculus,weknowthatthegradientoff,writtenas∇f,isgivenby

=∂∂

+∂∂

+∂∂

ˆ ˆ ê e effx

fy

fzx y z

where ˆ , ˆ , and ê e ex y z areunitvectorsinthex,y,andzdirections,respectively.At ( , , )0 0 0x y z ,wealsoknowthat ( , , )0 0 0f x y z pointsinthedirectionofthe

maximumrateofchangeoffwithrespecttodistance.Aunitvectorêgthatpointsinthisdirectionis

êg

=| |

ff

where =∂∂

+∂∂

+∂∂

| |2 2 2

ffx

fy

fz

.

Tofindarelativeminimumviathemethodofsteepestdescent,westartatsomeinitialpointandmoveinsmallstepsinthedirectionofsteepestdescent,whichis–êg.Let x y zn n n( , , )1 1 1− − − bethenewpositiononthenthiterationand x y zn n n( , , )theoldposition.Then,

( , , )

| ( , , ) |

( , , )

| ( , , ) |

( , , )

| ( , , ) |

1

1

1

x x

fx

x y z

f x y zs

y y

fy

x y z

f x y zs

z z

fz

x y z

f x y zs

n n

n n n

n n n

n n

n n n

n n n

n n

n n n

n n n

= −

∂∂

= −

∂∂

= −

∂∂

+

+

+

(10.4)

whereΔsissomesmalllength.


Example 10.1

Given

f x x x x x x x x x x x( , ) 4 4.5 4 2 2 21 2 1 2 12

22

1 2 14

12

2= + − + + − + −

Determine the minimum of f by the method of steepest descent, starting at point x x( , ) (6,10)1 2 = . Use a Δs = 0.1 and a maximum of 100 iterations.

%Example_10_1.m:%Thisprogramdeterminesarelativeminimumbythe%methodofsteepestdecent.Thefunctionis:%f(x1,x2)=4+4.5*x1-4*x2+x1^2+2*x2^2%-2*x1*x2+x1^4-2*x1^2*x2%Note:thisfunctionhasknownminimaat%(x1,x2)=(1.941,3.854)and(-1.053,1.028).The%functionalvaluesattheminimapointsare0.9855and%-0.5134respectively.clear;clc;%Definefunctionanditspartialderivativesfx_func=@(x1,x2)4+4.5*x1-4*x2+x1^2+2*x2^2...-2*x1*x2+x1^4-2*x1^2*x2;dfx1_func=@(x1,x2)4.5+2*x1-2*x2+4*x1^3-4*x1*x2;dfx2_func=@(x1,x2)-4+4*x2-2*x1-2*x1^2;%Definesteppingparametersds=0.1;max_iterations=100;%Firstguessx1=6;x2=10;fx=fx_func(x1,x2);%Printheadingsandinitialguesstoscreenfprintf('x1x2fx\n');fprintf('0:%7.4f%7.4f%10.4f\n',x1,x2,fx);forn=1:max_iterations%computepartialderivativesdfx1=dfx1_func(x1,x2);dfx2=dfx2_func(x1,x2);%computemagnitudeofgradientgradf_mag=sqrt(dfx1^2+dfx2^2);%computenextvaluesofx1andx2asperEqn10.4x1n=x1-dfx1/gradf_mag*ds;x2n=x2-dfx2/gradf_mag*ds;fxn=fx_func(x1n,x2n);fprintf('%2d:%7.4f%7.4f%10.4f\n',...n,x1n,x2n,fxn);%ifnewvalueislargerthanprevious,thenminimum%hasbeenfoundif(fxn>fx)fprintf('Aminimumhasbeenfoundafter');fprintf('%diterations\n\n',n);break;%otherwise,storenewvaluesovercurrentones%andcontinueelse


x1=x1n;x2=x2n;fx=fxn;endendifi>=max_iterationsfprintf('Error:nosolutionfoundafter');fprintf('%diterations\n',i);elsefmin=fx_func(x1,x2);fprintf('Therelativeminimumoccursat');fprintf('x1=%.4fx2=%.4f\n',x1,x2);fprintf('Theminimumvalueforf=%.4f\n',fmin);end

-------------------------------------------------------------------

PROGRAM OUTPUT (EXCERPT)

x1 x2 fx0: 6.0000 10.0000 683.00001: 5.9003 10.0077 622.71922: 5.8006 10.0155 566.25283: 5.7009 10.0233 513.4609 . . .89: 1.9936 4.0234 1.000190: 1.9638 3.9279 0.988391: 1.9495 3.8289 0.9901

Aminimumhasbeenfoundafter91iterationsTherelativeminimumoccursatx1=1.9638x2=3.9279Theminimumvalueforf=0.9883.

To obtain a more accurate result, we could rerun the program with revised starting values (from iteration 91) and use a smaller Δs. Alternatively, instead of starting at some arbitrary point as specified in the example, we can first use a grid program to establish a good starting point. In addition, a grid program may indicate that there is more than one relative minimum point.

Example 10.2

%Example_10_2.m:Thisprogramcalculatesthevaluesof%aspecifiedfunctionf(x1,x2)of2variablesfor%determiningagoodstartingpointforthemethodof%steepestdecent.Therangeofinterestisfrom%-10.0<=x1<=10.0and-10.0<=x2<=10.0.clear;clc;%Definefunctionofinterest:fx_func=@(x1,x2)4+4.5*x1-4*x2+x1^2+2*x2^2...-2*x1*x2+x1^4-2*x1^2*x2;%Definegridendpointsandstepsizex1min=-10.0;x1max=10.0;dx1=2.0;


x2min=-10.0;x2max=10.0;dx2=2.0;%Definegridandcalculatef(x1,x2)ateachpointx1=x1min:dx1:x1max;x2=x2min:dx2:x2max;fori=1:length(x1)forj=1:length(x2)f(i,j)=fx_func(x1(i),x2(j));endend%Printheadingfprintf('-----------------------------------');fprintf('-----------------------------------');fprintf('------------------------------------\n');fprintf('x2|');fprintf('x1\n');fprintf('-----------------------------------');fprintf('-----------------------------------');fprintf('------------------------------------\n');fprintf('|');fori=1:length(x1)fprintf('%7.1f',x1(i));endfprintf('\n');fprintf('-----------------------------------');fprintf('-----------------------------------');fprintf('------------------------------------\n');%Printvaluesoff(x1,x2)forj=1:length(x2)fprintf('%6.1f|',x2(j));fori=1:length(x1)fprintf('%8.1f',f(i,j));endfprintf('\n');end

-------------------------------------------------------------------

PROGRAM RESULTS

————————————————————————————————————————————————————— x2| x1————————————————————————————————————————————————————— |-10.0-8.0-6.0-4.0-2.00.02.04.06.08.010.0—————————————————————————————————————————————————————-10.0|12099.05488.02149.0738.0295.0244.0393.0934.02443.05880.012589.0-8.0|11659.05184.01949.0610.0207.0164.0289.0774.02195.05512.012069.0-6.0|11235.04896.01765.0498.0135.0100.0201.0630.01963.05160.011565.0-4.0|10827.04624.01597.0402.079.0 52.0129.0502.01747.04824.011077.0-2.0|10435.04368.01445.0322.039.0 20.073.0390.01547.04504.010605.00.0|10059.04128.01309.0258.015.0 4.033.0294.01363.04200.010149.02.0|9699.03904.01189.0210.07.0 4.09.0214.01195.03912.09709.04.0|9355.03696.01085.0178.015.0 20.01.0150.01043.03640.09285.06.0|9027.03504.0997.0162.039.0 52.09.0102.0907.03384.08877.08.0|8715.03328.0925.0162.079.0100.033.070.0787.03144.08485.010.0|8419.03168.0869.0178.0135.0164.073.054.0683.02920.08109.0

Since the functional value at (x1,x2)=(0,2) is lower than the functional values at surrounding points, we suspect that there is a relative minimum in the


vicinity of (x1,x2)=(0,2). A second relative minimum appears to be in the vicinity of (x1,x2)=(2,4).

10.4 MATLAB’s fminunc FunctionMATLABprovidesthefunctionfminunc(FUN,X0)tosolveunconstrainedopti-mizationproblems.Thetwoargumentstofminuncarethefunctiontooptimize(FUN)andaninitialguess(X0).FUNcanbeafunctiondefinedinaseparate .mfileormaybedefinedanonymouslywithinthescript.Theoutputoffminuncistheminimumthatisclosesttotheinitialguess.NotethatMATLABdoesnothaveaseparatefunctiontofindmaxima.Tofindamaximum,redefineyourfunctiontoreturnthenegativevalueandthenusefminunctofindtheminimum.

Example 10.3

Given

y x x x x( ) 40 200 123 2= + − +

Determine the minima and maxima.

%Example_10_3.m:Findtheminimaandmaximaof%y=x^3+40x^2-200x+12clc;clear;%First,plotthefunctionsothatwecanmakesome%usefulinitialguesses:x=-40:40;fori=1:length(x)y(i)=x(i)^3+40*x(i)^2-200*x(i)+12;endplot(x,y);xlabel('x'),ylabel('y'),grid,title('yvsx');%Next,findtheminimumandmaximumusingMATLAB's%anonymousfunctionmethod(findfirstguessesby%eyeballingplot).[xmin,minvalue]=fminunc(@(x)x^3+40*x^2-200*x+12,2);%Note:Tofindamaximum,insteadfindtheminimumof%thenegativeofthefunction.[xmax,maxvalue]=fminunc(@(x)-(x^3+40*x^2-200*x+12),...-30);%Printresultsfprintf('xmin=%.1fminvalue=%.1f,',xmin,minvalue);fprintf('xmax=%.1fmaxvalue=%.1f\n',xmax,-maxvalue);

-------------------------------------------------------------------

PROGRAM OUTPUT

xmin=2.3minvalue=-224.2,xmax=-29.0maxvalue=15063.0Determinemaximumandminimumvaluesofyandcomparewiththeresultsfromthefminuncfunction.


10.5 Optimization with ConstraintsInmanyoptimizationproblems,thevariablesinthefunctiontobemaximizedorminimizedarenotall independentbutarerelatedbyoneormoreconditionsorconstraints.Asimpleexamplethatillustratestheconstraintconceptfollows.

Example 10.4

Suppose we wish to determine the maximum and minimum values of the objective function

f(x,y) = 2x + 3y (10.5)

with the following constraints:

Lower bounds (LB): x ≥ 0, y ≥ 0Upper bounds (UB): x ≤ 3, y ≤ 3Constraint 1 (L1): x + y ≤ 4Constraint 2 (L2): 6x + 2y ≥ 8Constraint 3 (L3): x + 5y ≥ 8

The bounds are graphed in Figure 10.1. The MATLAB code that produces Figure 10.1 follows. Letters at the points of intersection were obtained by using the insert option on the MATLAB figure.

0 0.5 1 1.5 2 2.5 3–5

–4

–3

–2

–1

0

1

2

3

4

L1

L2

L3

UB

LB

x

y

AB

CD

Figure 10.1 Allowable region for control variables in determining maximum and minimum points of the object function.


%Example_10_4.m%Todrawlines,specifytwopointsontheline%upperboundonyx(1)=0;y(1)=3;x(2)=3;y(2)=3;%upperboundonxx1(1)=3;y1(1)=0;x1(2)=3;y1(2)=3;%lowerboundonyx2(1)=0;y2(1)=0;x2(2)=3;y2(2)=0;%lowerboundonxx3(1)=0;y3(1)=0;x3(2)=0;y3(2)=3;%lineL1x4(1)=0;y4(1)=4;x4(2)=3;y4(2)=1;%lineL2x5(1)=0;y5(1)=4;x5(2)=3;y5(2)=-5;%lineL3x6(1)=0;y6(1)=8/5;x6(2)=3;y6(2)=1;%plottheallowableregionplot(x,y,x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x6,y6)xlabel('x'),ylabel('y'),title('yvsx');

For each constraint, we replace the inequality with an equal sign to draw lines for LB, UB, L1, L2, and L3 and then add an arrow to indicate the region corresponding to the inequality. This specifies the allow-able region in which to consider the maximum and minimum values for function f.

We see in the graph that the allowable region for (x,y) in determining the maximum and minimum values for the object function f is the region enclosed by the polygon ABCD. For the specified object function, the minimum value of f is either on line L3 (between points D and C) or on line L2 (between points A and D). We can see this if we place a point P on line L3 and a point Q directly above it; then, the x values at points P and Q will be the same, but the y value at point Q will be larger than the y value of point P. Thus, f will be greater at point Q than f at point P. Thus, the minimum value of f is either on line L3 or on line L2.

The equation describing line L3 is

y

x85 5

= − (line L3)

Substituting this expression for y into Equation (10.5) gives the value of f on line L3, which is

f

x245

75

= + (on L3)

The minimum value for f occurs at the lowest allowable x value on L3, which is at point D. Point D results from the intersection of line L3 and line L2. The equation describing line L2 is

y = 4 − 3 x

(line L2)


The x coordinate at point D can be obtained by equating the y expressions

for line L1 and line L2 giving the x value at point D, or = =xD1214

0.8571. At point D, the value of f is 6.00.

Substituting the expression for y on line L2 into Equation (10.5) gives the equation for f on line L2, which is

f = 12 − 7x

(on line L2)

The minimum value for f on line AD occurs at the maximum allow-able value for x on AD, which is point D. Thus, in the allowable region, f 6.000.min =

From Figure 10.1, we see that the maximum value of f will be either at point B or somewhere else on line L1. The equation describing line L1 is

y = 4 − x

(on line L1)

Substituting this expression for y into Equation (10.5) gives

f = 12 − x

(on line L1)

The maximum f occurs where x is a minimum on line L1, which is at point B. At point B, x = 1. Thus, fmax = 11.0.

10.6 Lagrange MultipliersA more general and mathematical discussion of the optimization problem withconstraintsfollows.Supposewearegiventheobjectfunction f x x x xn( , , , , )1 2 3 … inwhichthevariablesx x x xn, , , ,1 2 3 … aresubject toNconstraints,say

( , , , , ) 0

( , , , , ) 0

( , , , , ) 0

1 1 2 3

2 1 2 3

1 2 3

x x x x

x x x x

x x x x

n

n

N n

…

…

…

Φ =

Φ =

Φ =

(10.6)

Theoretically,theNx’scanbesolvedintermsoftheremainingx’s.Then,theseN variablescanbeeliminatedfromtheobjectivefunctionf bysubstitution,andtheextremeproblemcanbesolvedasiftherewerenoconstraints.Thismethodisreferredtoastheimplicit methodandinmostcasesisimpractical.


ThemethodofLagrange multipliersprovidesthemeansforsolvinganextremaproblemwithconstraintsanalytically.Suppose f x x x xn( , , , , )1 2 3 … istobemaxi-mizedsubjecttoconstraints ( , , ),1 1 2x x xn…Φ ( , , ), ,2 1 2x x xn… …Φ ( , , )1 2x x xn n…ΦasinEquation(10.6).WedefinetheLagrange functionFas

… … …

… …

= +λ Φ

+ λ Φ + + λ Φ

( , , , , ) ( , , , , ) ( , , , , )

( , , , , ) ( , , , , )

1 2 3 1 2 3 1 1 1 2 3

2 2 1 2 3 1 2 3

F x x x x f x x x x x x x x

x x x x x x x x

n n n

n N N n

where iλ aretheunknownLagrangemultiplierstobedetermined.Wenowset

…

…

∂∂

=∂∂

=∂

=∂∂

=

Φ = Φ = Φ = Φ =

0, 0, 0, , 0

0, 0, 0, , 0

1 3

2 3

Fx

Fx

Fdx

Fx2 n

1 N

(10.7)

(Note:∂∂λ

= 0Fj

implies j 0Φ = .)

Thissetofn + Nequationsgivesallpossibleextremaoff[1].

Example 10.5

A silo is to consist of a right circular cylinder of radius R and length L, with a hemispherical roof (see Figure 10.2). Assume that the silo is to have a specified volume V = 8400 m3. Find the dimensions R and L that make its surface area S a minimum. Assume that the silo has a floor of the same

material. Note: = πV R43

sphere3 and S R4sphere

2= π .

R

L

Figure 10.2 Silo consisting of a right circular cylinder topped by a hemisphere.


Solution:

= π +πV R R L23

3 2 (10.8)

R R L V

23

03 2Φ= π +π − = (10.9)

S RL R R

RL R

2 2

2 3

2 2

2

= π +π + π

= π + π (10.10)

In this case, the surface area S is the function to be minimized, and the vol-ume V is the constraint. The Lagrange function is

F S

RL R R R L V2 323

2 3 2

= + λΦ

= π + π +λ π +π − (10.11)

The variables are R, L, and λ. Taking partial derivatives with respect to R and L gives

FR

L R R RL2 6 (2 2 ) 02∂∂

= π + π + λ π + π = (10.12)

FL

R R RR

2 0 2 or22∂

∂= π + λπ = → λ = − λ = − (10.13)

Substituting the value of λ from Equation (10.13) into (10.12) gives

L R

RR RL2 6

2(2 2 ) 02π + π − π + π = (10.14)

Equation (10.14) reduces to R − L = 0 or R = L.Substituting this result into Equation (10.8) gives

V R R

R

23

53

3 3

3

= π + π

= π


For V 8400 m3= ,

R

8400 35

11.7065m1/3

=×π

=

Substituting the values for R and L into Equation (10.10) gives S = 2152.6 m2.

10.7 MATLAB’s fmincon FunctionMATLAB’sfunctionforsolvingoptimizationproblemswithconstraintsisfmincon.Thefminconfunctiontakesasargumentsauser-definedfunctionFUN,astartingpointX0,plusadditionalargumentsthatdependonthetypeofconstraintsdefinedintheproblem.

ThefunctionFUN definestheobject functiontobeminimized.ThevectorX0isaninitialguessofthecontrolvariablesthatminimizestheobjectfunction.MATLAB’sfminconhandlesfourtypesofconstraints:

1.Inequality constraints:The constraints involveoneormore inequalities oftheform ʺ*A X B .Supposetheproblemspecifiesthree linear inequalityconstraintsoftheform

11 1 12 2 1 1

21 1 22 2 2 2

31 1 32 2 3 3

a x a x a x b

a x a x a x b

a x a x a x b

n n

n n

n n

+ + + ≤

+ + + ≤

+ + + ≤

TheseconstraintswouldbespecifiedinMATLABforusewithfminconas

A=[a11a12…a1n ;a21a22…a2n ;a31a32…a3n] B=[b1;b2;b3]

2.Equalityconstraints:Theconstraints involveoneormoreequalitiesof theform =*A X Beq eq.Supposetheproblemspecifiestwolinearequalitycon-straintsoftheform

a x a x a x b

a x a x a x b

n n

n n

11 1 12 2 1 1

21 1 22 2 2 2

+ + + =

+ + + =


where aij and bi are theelementsof A eq and Beq, respectively.Thesecon-straintswouldbespecifiedinMATLABforusewithfminconas

Aeq=[ 11 12 1a a a n; 21 22 2a a a n]

Beq=[ 1b ;

2b ]

3.Bounds:Theconstraintsinvolvelowerorupperlimitsoftheform ≥X LBand ʺX UB.Then,youwouldspecifytheboundsinMATLABforusewithfminconas

LB=[ 1 2l l ln] UB=[ 1 2u u un ]

wherex l x l x ln n, , ,1 1 2 2 …≥ ≥ ≥ and x u x u x un n, , ,1 1 2 2 …ʺ ʺ ʺ .

4.Nonlinearconstraints:Theconstraintsaredefinedbythefunction[C,Ceq] =NONLCON(X),whereNONLCON is a (user-defined)MATLABfunction thatspecifies the nonlinear inequality constraints C and the nonlinear equalityconstraintsCeq.Supposetheproblemspecifiestwononlinearinequalitycon-straints andonenonlinear equality constraint.Thenonlinear inequality andequalityconstraintfunctionsf1,f2,andf3needtobesetupsuchthat

≤ ≤ =( , ,..., ) 0, ( ,..., ) 0 and ( ,..., ) 01 1 2 2 1, 2 3 1, 2f x x x f x x x f x x xn n n

Then, in the function NONLCON, set C(1) f x x xn( , ,..., )1 1 2= , C(2)f x x xn( , ,..., )2 1 2= ,andCeq(1) f x x xn( , , ..., ) .3 1 2=

Adescriptionofthemanyinvocationsoffminconcanbeobtainedbytypinghelp fminconintheCommandwindow,someofwhicharedescribednext.

Usage 1X=FMINCON(FUN,X0,A,B)startsatX0andfindsaminimumXtothefunctionFUN,subjecttothelinearinequalitiesA*X<=B.FUNacceptsinputXandreturnsascalarfunctionvalueFevaluatedatX.X0maybeascalar,vector,ormatrix.

Thisfirstversionisusedonlyifalltheconstraintsarelinearinequalities;thatis,A*X<=B.Inthiscase,thescriptstartsatX0andfindscontrolvariablesXthatminimizetheobjectfunctioncontainedinFUNandreturnstheXvaluestothecallingprogram.TheinitialguessX0isacolumnvector,andA,Bdescribeasystemofequationsdefiningtheinequalityconstraints.


Usage 2X=FMINCON(FUN,X0,A,B,Aeq,Beq)minimizesFUNsubjecttothelinearequalitiesAeq*X=BeqaswellasA*X<=B.(SetA=[]andB=[]ifnoinequalitiesexist.)

Thesecondversionisusediftherearealsolinearequalityconstraints;thatis,Aeq*X=Beq,whereAeqandBeqdescribeasystemofequationsdescribingtheequalityconstraints.Iftherearenoinequalityconstraints,thenuse[]forAandB.

Usage 3X=FMINCON(FUN,X0,A,B,Aeq,Beq,LB,UB)definesasetoflowerandupperboundsonthedesignvariables,X,sothatasolutionisfoundintherangeLB<=X<=UB.UseemptymatricesforLBandUBifnoboundsexist.SetLB(i)=-InfifX(i)isunboundedbelow;setUB(i)=InfifX(i)isunboundedabove.

ThethirdversionisusedifyouwishtosetlowerandupperlimitsLBandUBtothecontrolvariables,whereLBandUBarevectorscontainingtheboundsonX.

Usage 4X=FMINCON(FUN,X0,A,B,Aeq,Beq,LB,UB,NONLCON)subjectstheminimizationtotheconstraintsdefinedinfunctionNONLCON.ThefunctionNONLCONacceptsXandreturnsthevectorsCandCeq,representingthenonlinearinequalitiesandequalitiesrespectively.FMINCONminimizesFUNsuchthatC(X)<=0andCeq(X)=0.(SetLB=[]and/orUB=[]ifnoboundsexist,dothesameforA,B,Aeq,Beqiftherearenolinearinequalityorlinearequalityconstraints.)

ThefourthversionisusediftherearenonlinearconstraintsasdefinedinthefunctionNONLCONofXasdemonstratedinExamples10.7and10.8.

Notethatinallusagesprovided,wehaveinvokedfminconasreturningasinglevector,thatis,theformX=fmincon(FUN,X0,...).However,forconvenience,we can also use the form[X,FVAL] = fmincon(FUN,X0,...), which alsoreturnsFVAL,whichistheminimumvalueoftheobjectfunction.

Example 10.6

In this example, we use fmincon to determine the maximum and minimum values of the object function of Example 10.4 subject to the constraints specified in that example.


%Example_10_6.m:Linearprogrammingexampleusing%MATLAB'sfminconfunctionforoptimization%Objectfunction:2*x(1)+3*x(2)%Lowerbounds%x(1)>=0x(2)>=0%Upperbounds%x(1)<=3x(2)<=3%Inequalityconstraints%x(1)+x(2)<=4%6*x(1)+2*x(2)>=8%x(1)+5*x(2)>=8clear;clc;lb=[0;0];%Lowerboundub=[3;3];%Upperboundx0=[0;0];%InitialguessAeq=[];%NolinearequalityconstraintsBeq=[];%NolinearequalityconstraintsA=[11;-6-2;-1-5];%LinearinequalityconstraintsB=[4;-8;-8];%Linearinequalityconstraints[xmin,fvalmin]=fmincon(@object_fmin,...x0,A,B,Aeq,Beq,lb,ub);[xmax,fvalmax]=fmincon(@object_fmax,...x0,A,B,Aeq,Beq,lb,ub);%Printminandmaxvalues:fprintf('MIN\n');fprintf('-------------------------------\n');fprintf('xmin(1)=%.4fxmin(2)=%.4f\n',xmin(1),xmin(2));fprintf('fmin=%.4f\n\n',fvalmin);fprintf('MAX\n');fprintf('-------------------------------\n');fprintf('xmax(1)=%.4fxmax(2)=%.4f\n',xmax(1),xmax(2));fprintf('fmax=%.4f\n',-fvalmax);%object_fmin.m%ThisfunctionworkswithExample_10_6.mfunctionf=object_fmin(x)f=2*x(1)+3*x(2);%object_fmax.m%ThisfunctionworkswithExample_10_6.mfunctionf=object_fmax(x)f=-(2*x(1)+3*x(2));

-----------------------------------------------------------------------

PROGRAM RESULTS

MIN-------------------------------xmin(1)=0.8571xmin(2)=1.4286fmin=6.0000MAX-------------------------------xmax(1)=1.0000xmax(2)=3.0000fmax=11.0000


Example 10.7

Solve for the minimum surface area of the silo in Example 10.4 using MATLAB’s fmincon function.

%Example_10_7.m:Thisprogramminimizesthematerial%surfaceareaofasilo.Thesiloconsistsofaright%cylindertoppedbyahemisphere.Solvewherethesilo%volumeis8400m^3.%DefinethevectorX=[R,L]whereRisradiusandLis%length(asdrawninFigure10.1)%Thefunctionobjfun_silocalculatesthesilosurface%areaandistheobjectivefunctiontobeminimized.%Thefunctionconfun_silodefinestheconstraintthat%thesilovolumebefixedat8400.clear;clc;globalV;V=8400;%Definebounds:radiusandlengthmustbepositive.LB=[0,0];UB=[];%InitialguessesforRandL:Xo=[10.020.0];fprintf('Thisprogramminimizesthesurfacearea\n');fprintf('ofasiloconsistingofarightcylinder\n');fprintf('toppedbyahemisphere.Thesilovolume\n');fprintf('isfixedat%.0fm^3.\n\n',V);%Runtheoptimization.Wehavenolinearconstraints,%sopass[]forthosearguments:[X,fval]=fmincon(@objfun_silo,Xo,...[],[],[],[],LB,UB,@confun_silo);%Printresults:fprintf('Silovolume:%9.3fm^3\n',V);fprintf('Minimumsurfacearea:%9.3fm^2\n',fval);fprintf('Optimumradius:%9.3fm\n',X(1));fprintf('Optimumlength:%9.3fm\n',X(2));----------------------------------------------------------------------%objfun_silo.m(objectfunctionforExample10.7)functionSA=objfun_silo(X)%Calculatesurfaceareaofsilo.Thevariablesare:%radiusR=X(1),lengthofcylinderL=X(2)SA=2.0*pi*X(1)*X(2)+3.0*pi*X(1)^2;-----------------------------------------------------------------------%confun_silo.m(constraintfunctionforExample10.7)function[c,ceq]=confun_silo(X)%Variablesare:%radiusR=X(1),lengthofcylinderL=X(2).globalV;%Nonlinearequalityconstraints:ceq=pi*X(1)^2*X(2)+2.0/3.0*pi*X(1)^3-V;%Nononlinearinequalityconstraints:c=[];

-----------------------------------------------------------------------


PROGRAM RESULTS

OptimizationProblem:Thisprogramminimizesthesurfaceareaofasiloconsistingofarightcylindertoppedbyahemisphere.Thesilovolumeissetat8400 m^3Silovolume: 8400.000m^3Minimumsurfacearea: 2152.651m^2Optimumradius: 11.707 mOptimumlength: 11.706 m

Example 10.8

Suppose we now reverse Example 10.7 and determine the maximum volume of the silo described in Example 10.7 subjected to the constraint that the surface area of the silo is to be less than a specified amount, say 2152.6 m2. The solution to the problem using MATLAB’s fmincon follows:%Example_10_8.m:Thisprogrammaximizesthevolumeofa%silo.Thesiloconsistsofarightcylindertoppedbya%hemisphere.Thevariablestooptimizeareradius'R'%andcylinderheight'L'.%LetX(1)=RandX(2)=L.%Eqnforsurfacearea:S=3*pi*X(1)^2+2*pi*X(1)*X(2);%Constraint:3*pi*X(1)^2+2*pi*X(1)*X(2)<=2152.6m^2.%Weneedtosetuptheconstraintintheform:%3*pi*X(1)^2+2*pi*X(1)*X(2)-2152.6<=0%Wewishtomaximizethevolumeundertheconstraint.%Theequationforthevolume,V,ofthesiloisgivenby%V=2.0/3.0*pi*X(1)^3+pi*X(1)^2*X(2).%SincewewishtomaximizeV,wewillneedtoplacea%minussigninfrontoftheexpressionforVandthen%minimize.clear;clc;globalSAmax;SAmax=2152.6;%Definebounds:radiusandlengthmustbepositive.LB=[0,0];UB=[];%TakeaninitialguessatthesolutionXo=[10.020.0];fprintf('Thisprogrammaximizesthevolumeofasilo\n');fprintf('consistingofarightcylindertoppedbya\n');fprintf('hemisphere.Themaximumsurfaceareais\n');fprintf('fixedat%.1fm^3.\n\n',SAmax);%Runtheoptimization.Wehavenolinearconstraints,%sopass[]forthosearguments:[X,fval]=fmincon(@objfun_silo2,...Xo,[],[],[],[],LB,UB,@confun_silo2);SA=3*pi*X(1)^2+2*pi*X(1)*X(2);%Printresults:fprintf('Maximumsurfacearea:%9.3fm^2\n',SA);fprintf('Maximumvolume:%9.3fm^3\n',-fval);


fprintf('Optimumradius:%9.3fm\n',X(1));fprintf('Optimumlength:%9.3fm\n',X(2));

-----------------------------------------------------------------------

%objfun_silo2.m(objectfunctionforExample10.8)functionV=objfun_silo2(X)%Computevolumeofsilo.Thevariablesare:%radiusR=X(1),lengthofcylinderL=X(2)V=-(2.0/3.0*pi*X(1)^3+pi*X(1)^2*X(2));

-----------------------------------------------------------------------

%confun_silo2.m(constraintfunctionforExample10.8)function[c,ceq]=confun_silo2(X)globalSAmax;%Variablesare:%radiusR=X(1),lengthofcylinderL=X(2)%Nonlinearinequalityconstraint:c=3*pi*X(1)^2+2*pi*X(1)*X(2)-SAmax;%Nononlinearequalityconstraints:ceq=[];

-----------------------------------------------------------------------

PROGRAM RESULTS

OptimizationProblem:Thisprogrammaximizesthevolumeofasiloconsistingofarightcylindertoppedbyahemisphere.Themaximumsurfaceareaissetat2152.6m^3Maximumsurfacearea:2152.600m^2Maximumvolume: 8399.701m^3Optimumradius: 11.706mOptimumlength: 11.706m

Example 10.9

Two machine shops, Machine Shop A and Machine Shop B, are to manu-facture two types of motor shafts, shaft S1 and shaft S2. Each machine shop has two turning machines: Turning Machine T1 and Turning Machine T2. The following table lists the production time for each shaft type on each machine and at each location:

Time to Manufacture (minutes)

Machine Shop A Machine Shop B

Shaft S1 Shaft S2 Shaft S1 Shaft S2

Turning Machine T1 4 9 5 8

Turning Machine T2 2 6 3 5


Shaft S1 sells for $35, and shaft S2 sells for $85. Determine the number of S1 and S2 shafts that should be produced at each machine shop and on each machine that will maximize the revenue for 1 hour of shop time.

%Example_10_9.m:ShaftProductionProblem%Thisprogrammaximizestherevenue/hrforthe%productionoftwotypesofshafts,typeS1andtypeS2.%Therearetwomachineshopsproducingtheseshafts,%shopAandshopB.Eachshophastwotypesofturning%machines,T1andT2,capableofproducingtheseshafts.%ShopA:%MachineT1takes4minutestoproducetypeS1shafts%and9minutestoproducetypeS2shafts.%MachineT2takes2minutestoproducetypeS1shafts%and6minutestoproducetypeS2shafts.%ShopB:%MachineT1takes5minutestoproducetypeS1shafts%and8minutestoproducetypeS2shafts.%MachineT2takes3minutestoproducetypeS1shaft%and5minutestoproducetypeS2shafts.%ShaftS1sellsfor$35andshaftS2sellsfor$85.%WewishtodeterminethenumberofS1&S2shaftsthat%shouldbeproducedateachshopandbyeachmachine%thatwillmaximizetherevenue/hrforshaftproduction.%Let:%X(1)=#ofS1shaftsproduced/hrbymachineT1atshopA%X(2)=#ofS2shaftsproduced/hrbymachineT1atshopA%X(3)=#ofS1shaftsproduced/hrbymachineT2atshopA%X(4)=#ofS2shaftsproduced/hrbymachineT2atshopA%X(5)=#ofS1shaftsproduced/hrbymachineT1atshopB%X(6)=#ofS2shaftsproduced/hrbymachineT1atshopB%X(7)=#ofS1shaftsproduced/hrbymachineT2atshopB%X(8)=#ofS2shaftsproduced/hrbymachineT2atshopB%Letr=totalrevenue/hrforproducingtheseshafts:%r=35*(X(1)+X(3)+X(5)+X(7))+85*(X(2)+X(4)+X(6)+X(8))clear;clc;%Objectivefunction:totalrevenueperhourfor%manufacturedshafts.Havethefunctionreturna%negativenumberbecausewearemaximizinginsteadof%minimizing.revenue=@(x)-(35*(x(1)+x(3)+x(5)+x(7))+85*(x(2)+x(4)…+x(6)+x(8)));%TakeaguessatthesolutionXo=[00000000];%Lowerandupperbounds:LB=[00000000];UB=[];%Wehavelinearinequalityconstraints.Werequirethat%eachmachinemakeanintegralnumberofshaftsper60%minutes,so:%4*X(1)+9*X(2)<=60


%2*X(3)+6*X(4)<=60%5*X(5)+8*X(6)<=60%3*X(7)+5*X(8)<=60A=[49000000;00260000;00005800;00000035];B=[60606060]';%Wehavenolinearequalityconstraints,sopass[]for%thosearguments.[X,rmax]=fmincon(revenue,Xo,A,B,[],[],LB,UB);fprintf('OptimizationResults:\n\n');fprintf('No.ofS1shaftsproducedatshopA');fprintf('bymachineT1:%2.0f\n',X(1));fprintf('No.ofS2shaftsproducedatshopA');fprintf('bymachineT1:%2.0f\n',X(2));fprintf('No.ofS1shaftsproducedatshopA');fprintf('bymachineT2:%2.0f\n',X(3));fprintf('No.ofS2shaftsproducedatshopA');fprintf('bymachineT2:%2.0f\n',X(4));fprintf('No.ofS1shaftsproducedatshopB');fprintf('bymachineT1:%2.0f\n',X(5));fprintf('No.ofS2shaftsproducedatshopB');fprintf('bymachineT1:%2.0f\n',X(6));fprintf('No.ofS1shaftsproducedatshopB');fprintf('bymachineT2:%2.0f\n',X(7));fprintf('No.ofS2shaftsproducedatshopB');fprintf('bymachineT2:%2.0f\n\n',X(8));fprintf('Themaxrevenueperhour:$%.0f/hour\n',-rmax);

---------------------------------------------------------

PROGRAM OUTPUT

OptimizationResults:No.ofS1shaftsproducedatshopA,T1:0No.ofS2shaftsproducedatshopA,T1:7No.ofS1shaftsproducedatshopA,T2:30No.ofS2shaftsproducedatshopA,T2:0No.ofS1shaftsproducedatshopB,T1:0No.ofS2shaftsproducedatshopB,T1:8No.ofS1shaftsproducedatshopB,T2:0No.ofS2shaftsproducedatshopB,T2:12Themaxrevenueperhour:$3274/hour

Note: When the program is run, MATLAB gives diagnostic warnings to the screen that can be ignored. If a satisfactory solution is obtained, MATLAB will inform you by saying

Localminimumfoundthatsatisfiestheconstraints.


ExercisesExercise.10.1

UseMATLAB’sfminunc() function tofind themaximaandminimaof thefollowingfunctions:

1. ( ) 10 20 154 3f x x x x= + − −

2. =−

− +( ) sin ( 0.75)

1.5 0.5625

2

2f xx

x x 3. = −( ) sin2 (for ≥ 0)0.5f x e x xx

Exercise.10.2

Usethemethodofsteepest-descenttoobtaintheapproximatepositionthatmakesf arelativeminimumwhere

( , ) 8 20 17 32 401 2 12

1 2 22

1 2f x x x x x x x x= − + − +

Use( , ) (6,4)1 2x x = asthestartingpoint,ds =0.05,and200steps.

Exercise.10.3

UseLagrangemultiplierstofindthevolumeofthelargestboxthatcanbeplacedinsidetheellipsoid

+ + = 1

2

2

2

2

2

2xa

yb

zc

sothattheedgeswillbeparalleltothecoordinateaxis.

ProjectsProject.10.1

AsiloconsistsofarightcircularcylindertoppedbyarightcircularconeasshowninFigure P10.1.Theradiusofboth thecylinderand thebaseof thecone isR.ThelengthofthecylinderisL,andtheheightoftheconeisH.Thecylinder,thecone,andthesilofloorareallmadeofthesamematerial.WriteaprogramusingMATLAB’sfminconfunctiontodeterminethevaluesofR,H,andLthatwillresultintheminimumsurfacesiloareaforaninternalsilovolumeof7000m3.

Note:Forarightcircularcone,

3

2

2 2

VR H

S R R H

=π

= π +


Project.10.2

Aretailstoresellscomputerstothepublic.Thereareeightdifferentcomputertypesthatthestoremaycarry.TableP10.2liststhetypeofcomputer,thesellingprice,andthecosttothestore.Thestoreplanstospend$20,000permonthpurchasingthecomputers.

Thestoreplanstospendnomorethan30%ofitscostsoncomputertypesC1andC2,nomorethan30%oncomputertypesC3andC4,nomorethan10%oncomputertypesC5andC6,andnomorethan30%oncomputertypesC7andC8.Thestoreestimatesthatitcansell30%moreoftypeC1thanC2,20%moreoftypeC3thantypeC4,20%moreoftypeC5thanC6,and60%moreoftype

L

H

R

Figure P10.1 Silo consisting of a right circular cylinder topped by a right cir-cular cone.

Table P10.2 Retail Price and Manufacturing Cost for Computer Types C1 through C8

Computer Type Selling Price ($) Cost ($)

C1 675 637

C2 805 780

C3 900 874

C4 1025 990

C5 1300 1250

C6 1500 1435

C7 350 340

C8 1000 1030


C7thanC8.UsethefminconfunctioninMATLABtodeterminethenumberofeachtypeofcomputerthatwillprovidethestorewiththemostprofit.Printoutthenumberofeachtypeofcomputerthestoreshouldpurchasepermonth,thetotalprofitpermonthandthetotalcostpermonthtothestore.

Project.10.3

The Jones Electronics Corporation has a contract to manufacture four differentcomputercircuitboards.Themanufacturingprocessrequireseachoftheboardstopassthroughthefollowingfourdepartmentsbeforeshipping:etchingandlamina-tion(etcheswiringintoboard);drilling(drillsholestosecurecomponents);assem-bly(installstransistors,microprocessors,etc.);andtesting.ThetimerequirementinminutesforeachunitproducedanditscorrespondingprofitvaluearesummarizedinTableP10.3a.

Eachdepartmentislimitedto3daysperweektoworkonthiscontract.TheminimumweeklyproductionrequirementtofulfillacontractisshowninTableP10.3b.

WriteaMATLABprogramthatwill

1. Determinethenumberofeachtypeofcircuitboardforthecomingweekthatwillprovidethemaximumprofit.Assumethatthereare8hoursperday,5daysperweekavailableforfactoryoperations.Note:Notalldepartmentsworkonthesameday.

2. Determinethetotalprofitfortheweek.

Table P10.3b Minimum Weekly Production Requirement for Various Circuit Board Types

Circuit Board Minimum Production Count

Board A 10

Board B 10

Board C 10

Board D 10

Table P10.3a Manufacturing Time for Various Circuit Board Types in Each Department

Circuit Board

Etching and Lamination (minutes)

Drilling (minutes)

Assembly (minutes)

Testing (minutes)

Unit Profit ($)

Board A 15 10 8 15 12

Board B 12 8 10 12 10

Board C 18 12 12 17 15

Board D 13 9 4 13 10


3. Determinethetotalnumberofminutesittakestoproducealltheboards. 4. Determinethetotalnumberofminutesspentineachofthefourdepartments. 5. Printtherequestedinformationtoafile.

Project.10.4

TheXYZoilcompanyoperatesthreeoilwells(OW1,OW2,OW3)andsuppliescrudeoiltofourrefineries(refineryA,refineryB,refineryC,refineryD).Thecostofshippingthecrudeoilfromeachoilwelltoeachoftherefineries,thecapacityofeachofthethreeoilwells,andthedemand(equalityconstraint)forgasolineateachrefineryaretabulatedinTableP10.4a.Thecrudeoilateachrefineryisdistilledintosixbasicproducts:gasoline,lubricatingoil,kerosene,jetfuel,heatingoil,and

Table P10.4a Cost of Shipping per 100 Liters ($)

Oil WellRefinery

ARefinery

BRefinery

CRefinery

DOil Well

Capacity (liters)

OW 1 9 7 10 11 7000

OW 2 7 10 8 10 6100

OW 3 10 11 6 7 6500

Demand (liters of gasoline)

2000 1800 2100 1900

Table P10.4b Cost of Distillation per 100 Liters ($)

Oil Well Refinery A Refinery B Refinery C Refinery D

OW 1 15 16 12 14

OW 2 17 12 14 10

OW 3 12 15 16 17

Table P10.4c Distillation Products per Liter of Crude Oil

Product Percentage per Liter from Distillation

Oil Well Gasoline

Lubricating Oil Kerosene Jet Fuel

Heating Oil Plastics

OW 1 43 10 9 15 13 10

OW 2 38 12 5 14 16 15

OW 3 46 8 8 12 12 14


plastics.ThecostofdistillationperliterateachrefineryfromeachoftheoilwellsisgiveninTableP10.4b.ThepercentageofeachdistilledproductperliteristabulatedinTableP10.4c.TheprofitfromeachproductistabulatedinTableP10.4d.

UsingthefminconfunctioninMATLAB,determinethelitersofoiltobepro-ducedateachoilwellandshippedtoeachofthefourrefineriesthatwillsatisfythegasolinedemandandwillproducethemaximumprofit.Printoutthefollowingitems:

1. Thelitersproducedateachoilwell 2. Thelitersofgasolinereceivedateachrefinery 3. Thetotalcostofshippinganddistillationofallproducts 4. Thetotalrevenuefromthesaleofalloftheproducts 5. Thetotalprofitfromalloftheproducts

Project.10.5

Asecond-ordercontrolsystemhasafrequencyresponseofthefollowinggeneralform:

( )

2

2

2 2H ss s

n

n n=

ω+ ζω +ω

(P10.5a)

where nω isthenaturalfrequency(inradians/second),andζisthedampingfactor.Thestep responseofasystemisdefinedasitstime-domainresponsetoaunitstep

input.Inthes domain,thiscorrespondstotheproductH(s)U(s),where =( ) 1U s

s

istheLaplacetransformoftheunitstepfunctionu(t).Tofindthetime-domainstepresponsey(t),wefindtheinverseLaplacetransformofH(s)U(s):

L

L

( )=

=ω

+ ζω +ω×

−

−

( ) ( ) ( )

21

1

12

2 2

y t H s U s

s s sn

n n

(P10.5b)

Byperformingapartial-fraction expansiononEquation (P10.5b) andusinginverseLaplacetransformtables,weobtain

( )( )=

− ζ − ω +ζ

ζ −ζ − ω ζ >

− −ω ζ =

− −ζ ω +ζ

− ζ− ζ ω ζ <

−ω

−ω

−ω

( )

1 cosh 11sinh 1 1

1 1 1

1 cos 11

sin 1 1

22

2

22

2

y t

e t t

e t

e t t

tn n

tn

tn n

n

n

n

(P10.5c)

Table P10.4d Product Revenue per Liter ($)

Gasoline Lubricating Oil Kerosene Jet Fuel Heating Oil Plastics

0.40 0.20 0.20 0.50 0.25 0.15


Thesettling time ts isdefinedasthetimeittakesfory(t)tosettletoitsfinalvaluewithinsomearbitraryrange,typically5%or10%.FigureP10.5showsthestepresponseforthreevaluesofζforω = 1n .

Inthisproblem,wewishtofindtheoptimumvalueofζtominimizetheset-tlingtime.

1. WriteaMATLABfunctionnamedstep _ response(zeta,wn,t)thatcalculatestheunitstepresponseofH(s)asperEquation(P10.5c)asafunc-tionofζand nω overthegivenintervalt.Confirmthatyourfunctioniscorrectbyplottingthestepresponseforζ =0.2,ζ =1,andζ =2.Assumeω = 1n andusethetimeinterval0≤t ≤20secinstepsof0.1sec.YourresultsshouldmatchFigureP10.5.

2. Write another MATLAB function named find _ settling _ time(zeta,wn,t,Etol)thatfindsthesettlingtimetSofawaveformgener-atedbystep _ response()tothetolerancespecifiedbytheEtolargument.Thewaveformy(t)maybeconsideredsettledwhenthemagnitudeofitsdeviationfromthefinalvalueislessthanthedesiredtolerance,thatis,

( ) ( )( )

y t yy

Etol− ∞∞

< (P10.5d)

5 10 15 20

1

ζ = 0.2

y(t)

t (sec)

ζ = 1

ζ = 2

Figure P10.5 Settling time of an RLC circuit for ζ = 0.2, ζ = 1, and ζ = 2.


wherey(∞)isthefinalsettledvalueofs(t),andEtol isthedesirederrortolerance(e.g.,5%or10%).Thus,forexample,tofindthetimetSittakesfory(t)tosettleto5%ofitsfinalvalue,thefunctionshouldusethetechniquesdescribedinChapter4tofindtherootoftheequation

( ) 11

0.05 0y t −− = (P10.5e)

Notethatintheunderdampedcase,y(t)mayoscillateinandoutoftolerancebeforefinallysettling.Thus,youshouldsearchfortherootofEquation(P10.5e)startingfromtherightsideofthewaveform(i.e.,startingatt =∞)ratherthanstartingatt =0.Then,youaresuretofindthefinalsettledvalueaftertheoscilla-tionshavediedaway.

3. Plottsfor0.2≤ζ ≤2instepsof0.01forasettlingtoleranceof5%.Assumethatω = 1.n Youmightseesomediscontinuitiesinyourplotforζ ≤1.Whatisthecauseofthediscontinuities?

4. UseMATLAB’sfminunc()functiontofindthevalueofζthatminimizesthesettlingtime.Comparethisvaluewithyourplotfrompart3above.Doesyouranswerlookcorrect?

5. Usethecurve-fittingtechniquesofChapter8tofityourdatafortsversusζtoafourth-orderpolynomialwithMATLAB’spolyfit()function.Rerunyourminimizationonthefittedpolynomial.Doestheanswerseemmoreaccurate?

Reference 1. Wylie,C.R.,Advanced Engineering Mathematics,McGrawHill,NewYork,p.596.

323

Chapter 11

Simulink

11.1 IntroductionSimulink®isusedwithMATLAB®tomodel,simulate,andanalyzedynamicsys-tems.Commonusesareforsolvingdifferentialequations,modelingfeedbacksys-tems,andsignalprocessing.

WithSimulink,modelscanbebuiltfromscratchoradditionscanbemadeto existing models. Simulations can be made interactive, so a change in para-meterscanbemadewhilerunningthesimulation.Simulinksupportslinearandnonlinearsystems,modeledincontinuoustime,sampletime,orcombinationsofthetwo.

Simulinkprovidesagraphicaluserinterface(GUI)forbuildingmodelsasblockdiagrams,usingclick-and-dragmouseoperations.Byusingscopesandotherdis-play blocks, simulation results can be seen interactively while the simulation isrunning.Theprogramincludesacomprehensivelibraryofcomponents(“blocks”)from which to construct models. Additional application-specific “toolboxes” arealsoavailable.

11.2 Creating a Model in Simulink 1.ClickontheSimulinkicononthemenubarintheMATLABwindowor

typesimulink in theMATLABCommandwindow.ThisbringsuptheSimulinkLibraryBrowserwindow(seeFigure 11.1).


2.ClickonFileintheSimulinkLibraryBrowserwindow.SelectNew–Model(foranewmodel)orOpen foranexistingmodel.Thiswillbringupanuntitledmodelwindow(forthecaseofanewmodel)oranexistingmodelwindow.

3.Tocreateanewmodel,youneedtocopyblocksfromtheLibraryBrowserwindow into thenewmodelwindow.This canbedonebyhighlighting aparticularblockanddraggingitintothemodelwindow.Tosimplifythecon-nectionsofblocks,youmayneedtorotateablock90°or180°.Todothis,highlighttheblock,clickonFormatinthemenubar,andthenselectRotate Block(for90°)orFlip Block(for180°).

Simulinkhasmanycategoriesfordisplayingthelibraryblocks;thoseofinterestforthischapterareCommonlyUsedBlocks,Continuous,Discontinuities,MathOperations,Ports&Subsystems,SignalRouting,Sinks,Sources,andUser-DefinedFunctions.

CommonblocksthatwillbeusedrepeatedlyinthischapterareConstant,Clock (from the Sources library); Product, Gain, Sum (from the MathOperations library); Integrator (fromtheContinuous library);Scope,Display,

Figure 11.1 Simulink Library Browser and untitled model window. (From MATLAB, with permission.)

Simulink ◾ 325

ToWorkspace(fromtheSinklibrary);Relay(fromtheDiscontinuitieslibrary);Switch,Mux(fromtheSignalRoutinglibrary),andFcn(fromtheUser-DefinedFunctionslibrary).

11.3 Typical Building Blocks in Constructing a Model 1.Additionoftwoconstantswithdisplayedoutput(seeFigure 11.2).Tosetthe

valueforaconstant,doubleclickontheblockandedittheconstantvalue.Toruntheprogram,clickonSimulation inthemenubarandselectStart;alternatively,clickthePlaybutton()inthemenubar.

2.Subtraction(seeFigure 11.3).TomaketheSumblockperformadifference,doubleclickontheblockandeditthelistofsigns.

3.Productoftwoblocks(seeFigure 11.4). 4.Divisionoftwoblocks(SeeFigure 11.5).

Display

50

Constant1

20

Constant

30

Figure 11.2 Constants, Sum, and Display blocks for addition.

Display

10

Constant1

20

Constant

30

Figure 11.3 Constants, Sum, and Display blocks for subtraction.


5.Integrateasinewave(seeFigure 11.6).Inthesinewaveblockparameters,setthefrequencyto2rad/s.Toviewtheoutputwaveform,firstclickthePlaybutton()andthendoubleclickonthescopeblocktoopentheplot.Then,rightclickontheplotandselectautoscale.Theresultingwaveformistheintegralofsin 2x:

∫ = − = − − = −sin2 dx 1

2cos2 1

2(cos2 1) 1

2(1 cos2 )

0 0x x t t

t t

(11.1)

Notethatbecausetheintegratorhasa(default)initialconditionofzero,theresultingwaveformwillvaryfromzerotoone.

Product Display

600

Constant1

20

Constant

30

Figure 11.4 Constants, Product, and Display blocks for multiplication.

Product Display

1.5

Constant1

20

Constant

30

Figure 11.5 Constants, Product, and Display blocks for division.

Sine Wave ScopeIntegrator

1s

Figure 11.6 Sine wave, Integration, and Scope blocks.

Simulink ◾ 327

6.Solutionofasimpleordinaryfirst-orderdifferentialequation.ThemethodofsolutionisillustratedinExample11.1,whichconsidersthestepresponseofanRCcircuit.

Example 11.1

For the series RC circuit of Figure 11.7, we examine the capacitor voltage vC with respect to time in response to a unit step input voltage. The governing equation (as derived in Example 6.1) is

dvdt RC

v vCi C

1( )= − (11.2)

where

R = 10 kΩ C = 4.7 μF

=

≤

>v

t

ti

0 V for 0

1V for 0

The block diagram for Equation (11.2) is shown in Figure 11.8, and the resulting scope output is shown in Figure 11.9.

+vi vC

+

−

CR

i

–

Figure 11.7 First-order RC circuit.

vc

vidvc/dt

1/RCStep

ScopeProduct Integrator

1s

Constant

21.28

Figure 11.8 Block diagram for solving Equation (11.2).


11.4 Tips for Constructing and Running Models 1.Toconnectlinesfromtheoutputofablocktotheinputofasecondblock,

placethecursorontheoutputofthefirstblock,rightclickonthemouse,anddragthelinetotheinputofthesecondblock.

2.Toconnectapointonalinetotheinputofablock,placethecursorontheline,rightclickonthemouse,anddragthelinetotheinputoftheblock.

3.Toaddalphanumericinformationabovealine,usetheleftbuttonanddou-bleclickabovethelineandatextboxwillappear.Typeinthedesiredlabelandclickelsewhereto complete.

4.Toviewtheresultsonascope,doubleclickonthescopetomakeagraphappear.Toselectthegraphaxis,rightclickonthegraphandselectaxis prop-ertiesorautoscale.Inmostcases,selectingautoscaleissufficient.Youmayalsoclickonthebinocularsicontoautoscalethegraph.

5.Tosetinitialconditionsforanintegrator,doubleclickontheblockandedittheinitialconditionline.

6.By default, Simulink runs over a time interval of 0 to 10 sec. Thesetimes are inappropriate for many models (e.g., high-frequency circuits).To edit the start and stop times, click on Simulation in the menu bar,select Configuration Parameters, and edit the start and stop time boxes.Alternatively,youcanadjustthestoptimeinthemenubar(thestarttimedefaultstozero).

Figure 11.9 Scope output of RC circuit.

Simulink ◾ 329

7.Torunthesimulation,clickonSimulationinthemenubarandchooseStart.Alternatively,clickthePlaybutton()inthemenubar.

11.5 Constructing a SubsystemSupposewebuildalargesystemconsistingofmanyblocks,andwewishtoreducethenumberofblocksappearingintheoverallblockdiagram.Thiscanbedonebycreatingasubsystem.Thesubsystemwillappearasasingleblock.Tocreateasub-system,placethecursorinthevicinityoftheregionthatistobecomeasubsystemandrightclickthemouse.Thisproducesasmalldashedboxthatcanbeenlargedbydraggingthemouseovertheregionenclosingthenumberofblockstobeincludedinthesubsystem(Figure 11.10).Whenthemousebuttonisreleased,adrop-downmenuappears.SelectCreate Subsystemtocausethemultipleblockstobereplacedwithasinglesubsystemblock.

In Figure 11.11, the constant, product, and integrator blocks have beencombined into the subsystem. This particular subsystem has one input andone output, but in general, a subsystem may have multiple inputs and out-puts.Bydoubleclickingonthesubsystem,youmayviewitscomponents(seeFigure 11.12).

vc

vidvc/dt

1/RCStep

ScopeProduct Integrator

1s

Constant

21.28

Figure 11.10 Constructing a subsystem.

vc

vi

Subsystem

In1 Out1Step

Scope

Figure 11.11 Block diagram with subsystem for solving Equation (11.2).


11.6 Using the Mux and Fcn BlocksAnalternativetotheblockdiagramforExample11.1isshowninFigure 11.13.Inthissolution,Simulink’sMuxandFcnblocksareusedtosolvetheproblem.TheMuxblock allows you to select amongmultiple inputs (to adjust thenumber ofinputs,doubleclickontheblockandeditthenumberofinputs).Theuppermostinputisdesignatedu[1],theonebelowisdesignatedu[2],andsoon.Inthisexample,theoutputfromtheMuxblockshouldgototheinputoftheFcnblock.

ThepurposeoftheFcnblockistoallowarbitrarymathematicalandMATLABfunctionstobedefinedwithinthemodel.Inthiscase,themathexpressionintheFcnblockisusedtoimplementEquation(11.2)intermsoftheu[]’s.

11.7 Using the Transfer Fcn BlockAcommonmethodfor solvingcircuitproblems is to substitutecomplex imped-ancesforthecapacitorsandinductorsandthensolvelikearesistivecircuit.FortheRCcircuit,theimpedanceoftheresistorissimplyZ RR = ,andtheimpedanceof

1/RC

vi

vc

dvc/dt

Step

ScopeIntegrator

1s

Fcn

u(2)*(u(1) – u(3))

Constant

21.28

Figure 11.13 Block diagram for solving Equation (11.2) using the Mux and Fcn blocks.

dvc/dt

1/RCOut1

1

Product Integrator

1s

Constant

21.28

In1

1

Figure 11.12 Subsystem model.

Simulink ◾ 331

thecapacitoris 1Z CsC = (wheres =jω).Then,thecapacitorvoltageissimplytheoutputofaresistivedivider:

1

1

1

1

( )

v ZZ Z

v

CsR Cs

v

RCs RC

v

H s v

CC

R Ci

i

i

i

=+

=+

=+

=

where =+

( ) 1/1/

H sRC

s RC is commonly referred to as the transfer function. In

Simulink, theTransferFcnblockallowsdirect entryof a transfer function intoyourmodelasshowninFigure 11.14.

11.8 Using the Relay and Switch BlocksRelaysandswitchesareusedindesignstoenablealow-powerdevice(e.g.,anelec-troniccontroller)tocontrolahigh-powersystem(e.g.,aboilerorfurnace).SimulinkhasRelayandSwitchblocksthatcanbeusedtosimulatethesetypesofsystems.

Example 11.2

In a home heating system, a temperature sensor is used to switch the boiler on and off to heat the house to a comfortable temperature. However, because most boilers do not turn on and off instantaneously (i.e., they take a few minutes to heat up after turning on and take time to cool after turning off), the control of the room temperature requires some hysteresis to avoid cycling the boiler on and off too often (which causes excessive wear on the boiler). This concept can be represented by a simple differential equation in

Transfer Fcn

21.28

s + 21.28

Step Scope

Figure 11.14 Block diagram using the Transfer Fcn block.


which the temperature T is set to fluctuate at a constant rate between 20°C and 22°C :

= =≤

− ≥dTdt

c cT

Twhere

30 if 20

30 if 22

The block diagram for this system consists of an Integrator, a Constant, a Relay, a Product, and a Scope (Figure 11.15). The Relay is used to invert the sign on c depending on whether the relay is on or off. The Relay parameters may be edited by double clicking on the Relay block and setting the param-eters to the following values:

Switch point on = 22Switch point off = 20Output when on = –1Output when off = +1

We assume that the initial room temperature is T = 18°C, and this value is entered into the parameters for the Integrator block. At the start of the simu-lation, T ≤ 20; thus, the Relay switch is off, and the output of the Relay is +1, causing T to increase. The Relay output will remain +1 until T reaches 22, at which point the Relay will turn on and its output will be −1, causing T to decrease. The Relay output will remain −1 until T reaches 20, at which point the Relay will turn off and its output becomes +1 again. This process will continue until the simulation end time is reached. The output of the Scope is shown in Figure 11.16.

Example 11.3

Some problems may involve a function that varies in time for t t0 1ʺ ʺ and is constant for t t t1 2< ≤ . This type of function can best be modeled with the Switch block that implements a DPST (double-pole, single-throw) switch with an additional terminal to control the opening and closing of the switch.

Scope

Relay

Product Integrator

1s

Constant

30

Figure 11.15 Block diagram using the Relay block from Example 11.2.

Simulink ◾ 333

Suppose

yt t

t

5 for 0 10

50 for 10 20=

≤ ≤

< ≤

The Simulink model for this problem is shown in Figure 11.17. The parameters for the Switch are as follows (note that terminal u1 is the top switch input, u2 is the middle [control] input, and u3 is the bottom switch input):

Figure 11.16 Scope output of Example 11.2.

Switch Scope

Gain

5

Constant

50

Clock

Figure 11.17 Block diagram using the Switch block from Example 11.3.


Criteria for passing first input: u2 ≥ ThresholdThreshold: 10

The resulting output is shown in Figure 11.18. Note that we use the Clock block to obtain a value for y that is proportional to the elapsed time.

11.9 Trigonometric Function BlocksFunctionssuchassine,cosine,andtangentcanbeobtainedviatheTrigonometricFunctionblock located in Simulink’sMathOperations library.The input tothisblockistheargumenttothedesiredtrigfunction.Iftheargumentinvolvesthe independentvariable t (as in sin ωt), thenwecanuse theClockblocktoobtainthevalueoft.ThisisshowninFigure 11.19,wherewecomputethevalueofcos2t.

Example 11.4: Simulation of an RLC Circuit

In Section 2.17, we studied the parallel RLC circuit. If we use a sinusoidal input current to the circuit in Figure 2.14 and assume that the switch closes at t = 0, the governing equation is

d idt RC

didt LC

iLC

I tL LL o

1 1 1sin

2

2 + + = ω (11.3)


Simulink ◾ 335

where the circuit is “driven” by a sinusoidal current of frequency ω and magnitude I0. The Simulink model in Figure 11.20 solves this second-order differential equation with the following circuit values and initial conditions:

C = 1 μFL = 10 mHR = 2000 ΩIo 5mA=ω = 2000 radian/seciL(0) 2mA=iL(0) 0=

The output of the simulation is shown in Figure 11.21. Simulation and workspace parameters for this example are as follows:

Simulation time:Start time: 0.0Stop time: 0.015

Solver options:Type: Variable stepSolver: ode45 (Dormand-Prince)Maximum step size: autoMinimum step size: autoInitial step size: autoRelative tolerance: 1e-3Absolute tolerance: auto

Workspace parameters:Variable name: iLimit data points to last: infDecimation: 1Sample time (-1 for inherited): -1

Save format: array

Note that you can export Simulink output values into MATLAB variables for further manipulation or for printing. You can do this with the To Workspace

cos 2t2t

Trigonometricfunction

cos

ScopeProduct

Constant

2

Clock

Figure 11.19 Block diagram using the Clock, Product, and Trigonometric Function blocks.


d2i/d

t2di

/dt

i

t

1/LC

1/R

C

Io/L

C

i(0)

= 2e

– 3

w

Io *

sin

wt/L

C

Trig

onom

etric

func

tion

sin

To

wor

kspa

ce

i

Sco

pe1

Sco

pe

Pro

duct

3

Pro

duct

2

Pro

duct

1

Pro

duct

Inte

grat

or1

Inte

grat

or

1 s1 s

Con

stan

t3

2000C

onst

ant2

5e –

3/(

10e

– 3*

1e –

6)

Con

stan

t1

1/(2

000*

1e –

6)

Con

stan

t

1/(1

0e –

3*1

e –

6)

Clo

ck

Figu

re 1

1.20

B

lock

dia

gram

for

solv

ing

the

seco

nd-o

rder

dif

fere

ntia

l equ

atio

n of

Equ

atio

n (1

1.3)

.

Simulink ◾ 337

block as shown in Figure 11.20. After the simulation is run, those vari-ables become available for use in any MATLAB program. When using To Workspace, be sure that the Save Format property is set to Array.

ExercisesExercise.11.1

ImplementthefollowingfunctioninSimulink:

= + +5 4 22y t t

1. UsingoneClockblock,threeConstants,twoMultipliers,andaSum. 2. UsingoneClockblock,threeConstants,aMuxblock,anFcnblock,aScope

block,andtwoToWorkspaceblocks.Thescopeblockshouldgiveaplotofyversust.Also,createaMATLABprogramthatconstructsatableofyversustintheCommandwindow.

3. UsingoneConstantandtwoIntegrators.

ProjectsProject.11.1

Adecision circuitisusedinacommunicationsystemtodecidewhetherareceiveddigital signal is a logicaloneor zero. It thenoutputs itsdecisionusingonboardlogiclevels.Forexample,intheRS-232serialcommunicationprotocol[1], thenominallogiclevelstorepresentalogicaloneorzeroare±12V.However,typicallogiclevelsonacomputermotherboardare+5Vand0V.Wecanusearelayblock



inSimulinktoexaminethebehaviorofadecisioncircuitthatimplementstheserequirementsinthepresenceofnoisethatisinducedintheRS-232cableduetofluorescentlights,radiowaves,andsoon.

1. UseSimulink’sSignalGeneratorblock(fromtheSourceslibrary)togenerateabinarysignalofalternatingonesandzerosusingRS-232logiclevelsatarateof1200bits/sec.(Note:1200bits/secisnotthesameas1200Hz.)

2. Add white noise to your RS-232 signal with a Band-limited White Noiseblock (from the Sources library) and a Sum block. Set the parameters onthewhitenoisetobe2 10 4× − forthenoisepowerand1 10 5× − forthesampletime.ViewtheresultingnoisybinarysignalwithaScopeblockandconfirmthatitstilllookslikeabinarysignal(albeitwithnoise).

3. UseaRelayblocktoconvertyourRS-232binarysignalfrom+12V(foraone)and–12V(forazero)into+5V(foraone)and0V(forazero).Use0Vforboththeswitch-onandswitch-offpointsfortherelay.Runthesimulationfor0.01secandviewtheresultingoutputfromtherelaywithascope.Youshouldseesomeerrorsontherelayoutputduetothenoiseontheinput.

4. Wenowquantifythebiterrorsontherelayoutputasfollows: a. First,weneed togeneratea“perfect” (i.e.,noiseless)versionof the

relay output for comparison purposes. Thus, make a copy of yourexistingblocks(includingSignalGenerator,Relay,andScope)withinthecurrentmodelbutomittheNoiseGeneratorinthecopy.Then,runthesimulationagainfor0.01secandconfirmthatyouhavebothperfectandnoisyversionsattheoutputsofyourtworelays.

b. Next,createanerrorsignalbysubtractingthetworelayoutputswithaSumblock.

c. Next,generatethemagnitudeoftheerrorsignalwithanAbs(abso-lutevalue)block(fromtheMathOperationslibrary).Thepurposeofthisistomakesurethattheerrorsignalisalwayspositivesothatitmaybeeasilycounted.

d. Finally, send the output of your Abs block to an Integrator block(withaninitialconditionofzero)anduseaDisplayblocktomoni-tortheoutputofyourintegrator.Thepurposeoftheintegratoristoprovidearunningtotalofallofthedetectederrors.

5. Runthesimulationfor0.01seconyourcompletemodelcontainingdualrelayswithcumulativeerrordetection.Attheendofthesimulation,youshouldwindupwithapositivevalueattheoutputoftheintegrator.Also,tryrunningtheintegrationfor1secandseethatyouhaveevenmoreerrors.

6. Onemethodofimprovingtheerrorperformanceofthisdecisioncircuitistoinserthysteresisintotherelaytoincreasethenoisemargins.Toobservethis,modifytherelayswitch-onandswitch-offvaluesto+0.5and–0.5andrerunthesimulationfora1-secintervalandconfirmthatyouseefewererrors.Trythe various sets of switch-on and switch-off values listed in Table 11.1 andcompletethetable.Whathappensifyouusetoomuchhysteresis?

Project.11.2

RepeatProject6.3foraSallen-Keycircuitwithastepandimpulseinput,butthistimeuseSimulinktoconstructasimulationofthesystem.Scopeoutputshouldbeforvoutandv1versust.Settheendtimetot = 0.0001secandprintouttheblockdiagram,impulseresponse,andstepresponse.

Simulink ◾ 339

Project.11.3

Aswingingkitchendoorinarestaurantweighsm =20kgandhasaspringtocauseittocloseautomatically(withspringconstantk =300N/m)andapistondampertopreventitfromslamming(withdampingcoefficientc =150N-sec).WewishtosimulatethisdoorasanRLCcircuitinSimulink.Assumethatthedoorpositionxvariesovertherange–1≤x ≤1meter,where–1and1correspondtothedoorbeingfullyopenineitherdirection,andzerocorrespondstotheclosedposition.

1. SetupamodelinSimulinktodescribethesecond-orderdifferentialequationforaseriesRLCcircuit(refertoAppendixAfortheproperequation).AssumethatthedoorpositionxisanalogoustothecapacitorvoltagevC.ForR,L,andC,usetheanalogousvaluesc,m,and1/k,respectively.Usetheinitialconditionx =1(i.e.,thedoorisfullyopenatt =0)andnoforcingfunction(analogoustozeroinputvoltage).Runthesimulationfor5sec.Isthisdoorunderdampedoroverdamped?

2. Now,assumethatthepistondamperbreaksoffthedoor,causingazerovalueforthedampingcoefficientc.Whathappenswhenyourerunthesimulation?Isthiswhatwouldhappeninreallife?Whyorwhynot?

3. Afterreplacingthedamagedpistondamper,therepairpersonmistakenlysetsthedampingcoefficientto50N-sec.Whathappenswhenyourerunthesimu-lation?Doesthismatchwhatwouldhappeninreallife?

4. Interpret these results back to the RLC circuit. What happens in an RLCcircuitforL =20,C=1/150,andaninitialcapacitorvoltageof1VwhenRchangesvaluefrom150Ωto0Ωto50Ω?

Reference 1. McNamara, J., Technical Aspects of Data Communication, 3rd ed., Butterworth-

Heinemann,Oxford,UK,1988.

Table 11.1 Cumulative Errors for the Decision Circuit of Problem 11.1 for Various Amounts of Hysteresis

Switch-on and Switch-off Values

Cumulative Error Detected at Integrator Output (per second)

0, 0

+0.5, −0.5

+2, −2

+5, −5

+8, −8

341

Appendix A: RLC Circuits

Theresistor-inductor-capacitor(RLC)circuitisusedrepeatedlyasanexampleinthistextanditsbehaviorisderivedindetailinthisappendix.Inelectricalengi-neering,thissimplesecond-ordersystemisuseddirectlyincircuittheory,signalprocessing,andcontroltheory.Inaddition,manynon-circuitproblemslendthem-selvestobeingmodeledwithRLC,includingphase-lockloops,transmissionlines,and non-electrical problems such as the spring-dashpot problem in mechanicalengineering,theresonanceofsoundwavesinacoustics,andeventheflowofauto-mobilesonahighwayduringatrafficjam.

TherearetwobasictopologiesofRLCcircuits:seriesandparallel(seeFigureA.1).Allotherconfigurationscangenerallybereducedtoeitherofthesetwobytransform-ingwithThéveninorNortonequivalentcircuits[1](FigureA.2),soweonlyanalyzethebasictwohere.Intheseriescase,theinputvariableisavoltageandtheoutputisthecurrent;intheparallelcase,theoppositeistrue(currentin,voltageout).

A.1 Direct Analysis with Differential EquationsStartingwiththeseriescircuit,wemodeltheresistor,inductor,andcapacitorvolt-age-currentrelationsusingtheirrespectiveconstituentrelations:

=v i RR R (A.1)

=v L

didtLL (A.2)

=i C

dvdtCC (A.3)

342 ◾ Appendix A

C

−

+vR

−

+vL

−

+vC

L

R

+–vin

C

iL iCiR

LRiin

(a) (b)

Figure A.1 (a) Series RLC circuit, (b) Parallel RLC circuit.

C

L

R

+–

C

LRiin iin × R

(a)

CLR

CL

R

+–vin

vinR

(b)

Figure A.2 Any RLC network can usually be converted to either the series or the parallel configuration by applying (a) Thévenin’s and (b) Norton’s theorems to con-vert a series voltage source and resistor into a parallel current source and resistor.

Appendix A ◾ 343

where vR, vL and vc are the voltages across the resistor, inductor, and capacitorrespectively, iR, iL and iC are their respective currents, andR,L andC are theirrespectivecomponentvaluesinohms,farads,andhenries.

ApplyingKirchhoff’svoltagelawaroundtheRLCloopgives

− − − = 0V v v vin R L C (A.4)

Intheseriescase,allofthecomponentshavetheidenticalcurrentandthuswedefine = = =I i i iout R L C . SubstitutingEquations (A.1) and (A.2) into (A.4) andrearranginggives

+ + =I R L

dIdt

v Voutout

C in (A.5)

NowsubstitutingEquation(A.3)into(A.5)gives

+ + =

2

2RCdvdt

LCd vdt

v VC CC in

+ + =

12

2d vdt

RLdvdt LC

vVLC

C CC

in (A.6)

Equation (A.6) is a second-order differential equation for vC which can besolvedviastandardtechniques.The“drivingfunction”Vin(t)andthetwoinitialconditionsvC(0)andiL(0)arenecessarytoobtainacompletesolution.TosolveforvC(t),wechooseaparticularsolutionwhichsatisfiesthedifferentialequationfornonzeroVin(t),andthenwechooseahomogenoussolution(bysettingVin(t)=0)whichallowsustosatisfytheinitialconditions.ThisapproachisdemonstratedinExample6.1.Finally,oncewehavesolvedEquation(A.6)forvC,wecanobtainIoutbyapplyingEquation(A.3).

By a similar analysis, we can solve the parallel RLC circuit by applyingKirchhoff’scurrentlawtoobtain

+ + =

1 12

2d idt RC

didt LC

iILC

L LL

in (A.7)

NotethatEquations(A.6)and(A.7)havesimilarformandthusbysolvinga single differential Equation, we can solve both the series and parallel RLCcircuits.

344 ◾ Appendix A

A.2 Analysis via Circuit TheoryAcommonmethodforsolvingRLCcircuitsistoapplysimplecircuittheorywherewemodeltheresistor,inductor,andcapacitorwiththeircompleximpedancesZR,ZL,andZC :

=

=

=1

Z R

Z Ls

ZCs

R

L

C

wheres =jωisthecomplexfrequency.ApplyingOhm’slawtotheseriesRLCcir-cuitofFigureA.1agives

=+ +

=+ +

1

IV

Z Z Z

V

R LsCs

outin

R L C

in

=+ +

1

12

IV

Ls

sRLs

LC

out

in (A.8)

Similarly,applyingOhm’slawintheparallelcasegives

( )=

=+ +

1 1

V I Z Z Z

I

R LsCs

out in R L C

in

=+ +

1

1 12

VI

Cs

sRC

sLC

out

in (A.9)

Notethattheinput-outputrelationsinEquations(A.8)and(A.9)havesimilarformandareconsideredtobe“second-order”systemsbecausethehighesttermin

Appendix A ◾ 345

theirdenominatorsiss2.Wecanrewritetheseequationsintothegeneralformforsecond-ordersystems:

=

+ ζω +ω( )

22 2H sKs

s sn n (A.10)

whereH(s)isthetransfer function, ωn isthenatural frequency,ζ isthedamping factor,andKisaconstantscalingfactor.MatchingtermsbetweenEquations(A.3),(A.2),and(A.1),wesee:

ω = ζ = =

ω = ζ = =

Series RLC circuit: 1 ,2

, 1

Parallel RLC circuit: 1 , 12

, 1

LCR C

LK

L

LC RLC

KC

n

n

Thephysicalmeaningsfornaturalfrequencyanddampingfactorbecomeappar-entwhenweobservethefrequencybehaviorforH(s).TheBodeplotforEquation(A.10) is shown inFigureA.3,whereweassumethat ω = 1000n andζhas the

–140

–120

–100

–80

–60

–40

Mag

nitu

de (d

B)

100 101 102 103 104 105 106–90

–45

0

45

90

Phas

e (de

g)

Bode Diagram

Frequency (rad/sec)

ζ = 0.1

ζ = 0.1

ζ = 1

ζ = 1

ζ = 10

ζ = 10

Figure A.3 Bode plot of an RLC circuit for 𝛚n = 1000 rad/s and ζ values of 0.1, 1, and 10.

346 ◾ Appendix A

values0.1,1,and10.Notethattheplotsarecenteredaroundωnandfalloffsym-metricallyasω→0andω→∞.Thus,H(s)isabandpass filterwithapassbandcen-teredaroundωn .ThisisconfirmedwhenweexamineEquation(A.10)atthelowandhighfrequencylimitsandat = ωs j n :

ω =+ ζω +ω

=

ω =∞

∞ + ζω ∞ +ω=

ω ω =ω

ω + ζω ω +ω=

ζω

ω→

ω→∞

at low frequencies: lim ( ) (0)0 2 (0)

0

at high frequencies: lim ( ) ( )2 ( )

0

at : ( )( ) 2 ( ) 2

0 2 2

2 2

2 2

H jK

H jK

H jKj

j jK

j n n

j n n

n nn

n n n n n

Thus,H(s)approacheszeroatlowandhighfrequenciesandhasapeakvalueatthenaturalfrequencywhichisdependentonthedampingfactor.Traditionally,thevaluesforζareclassifiedintothreetypes:underdamped(ζ<1),overdamped(ζ>1),andcriticallydamped(ζ =1).FortheunderdampedcaseinFigureA.3(ζ =0.1),H(s)hasatallnarrowpeakthatindicatesresonanceatthenaturalfre-quency,i.e.,onlyinputsignalsat(orverynearto)thenaturalfrequencywillbepassedbythefilter.Fortheoverdampedcase(ζ =10),thereisnopeakatallbutratherthere issimplyawideflatpassband.Thecriticallydampedcase(ζ =1)hasthenarrowestpassbandwithoutresonance,the implicationsofwhichwillbecomeclearmomentarilywhenweexaminethetransientresponse.

AnalternativerepresentationofthestandardformofEquation(A.10)is

=+ω

+ω( )

2 2H s

Ks

sQ

snn (A.11)

whereQisthequality factorandisanindicationoftheheightandwidthofthereso-nantpeakintheBodeplot.NotethatbycomparingEquations(A.11)and(A.10),

weseethat =ζ12

Q andthusthequalityfactoranddampingfactoraresimplytwo

differentwaysofexpressingthesameidea.ThechoicebetweenwhethertouseQorζisdependentontheapplication,withQusuallybeingusedinfilterdesignandζincontroltheory.Ahigh-Qfilter( > 1/2Q )isonewhichhashighresonanceandistypicallyusedinoscillatordesignortunablecircuits,andalow-Qfilter( <1/2)Q hasnoresonanceatall.Anintermediate Q factor( = 1/2Q )correspondstoacriti-callydampedresponse.

Thedampingfactorisfurtherillustratedbyexaminingthestepresponses(t).Fromsystemtheory,weknowthat

s(t)=h(t)*u(t)

Appendix A ◾ 347

whereh(t)isthesystemimpulseresponse(i.e.,theinverseLaplacetransformofH(s)),u(t)istheunitstepfunction,andthe“*”operatorindicatesconvolution.Inordertoavoiddoingtheconvolution,wesolveintheLaplacedomainwheretheconvolutionoperatortransformstomultiplication.Thus,thes-domainstepresponseS(s)is

= ×( ) ( ) 1

S s H ss

wherewehaveusedthetransformrelation L( ) =( ) 1u t

s.Byfactoringthedenomi-

natorofH(s),weobtain

( )( )( ) ( )

=+ ω ζ + ζ − +ω ζ − ζ −

×( )1 1

12 2

S sKs

s s sn n

(A.12)

Performingapartial-fractionexpansionforthecasewhereζ≠1gives

( ) ( )

=ω ζ −

−

+ω ζ + ζ −+

+ω ζ − ζ −( )

2 11

1

1

12 2 2S s

K

s sn n n

(A.13)

KnowingtheLaplacetransformrelation L = −( ) 1e s aat ,wecaninversetrans-

formEquation(A.13)toobtain

=

ω ζ −− +

( ) ( )− ζ+ ζ − ω − ζ− ζ − ω( )

2 12

1 12 2

s tK

e en

t tn n (A.14)

In the casewhereζ >1,Equation (A.14) canbe algebraically rearranged toobtainasinhfunction,andinthecasewhereζ <1,Equation(A.14)canberear-rangedintoasinefunction.

Inthespecialcasewhereζ =1,Equation(A.12)becomes:

=

+ω×( )

( )1

2S sKs

s sn (A.15)

This equation can be inverse transformed with the Laplace transform relation

L =−

( ) 1( )2

tes a

at.

348 ◾ Appendix A

Summarizing,thestepresponseoftheRLCcircuitforallvaluesofζandzeroinitialconditionsis

=

ω −ζ− ζ ω ζ <

ζ =

ω ζ −ζ − ω ζ >

−ζω

−ζω

−ζω

( )2 1

sin 1 for 1

for 1

2 1sinh 1 for 1

22

22

s t

Ket

KteKe

t

t

nn

t

t

nn

n

n

n

(A.16)

ThestepresponseisplottedinFigureA.4whereweassumethatω = 1000n andζhasthevalues0.1,1,and10.Notethatfortheunderdampedcase(ζ =0.1),thetran-sientresponsehasanoscillatorycomponent(duetothesineterm)whichdiesoutovertime(duetotheexponentialterm).Intheoverdampedcase(ζ =10),theresponsehasnooscillatorybehaviorandisasumofdecayingexponentials.Thecriticallydampedcase(ζ =1)canbeseentohavethefastestdecaywithoutanyoscillatorycomponent.

NotethatinEquation(A.16),thesinetermintheunderdampedcasehasan“effective”frequencythatisdependentonthedampingfactorandthenaturalfre-quency.Wemaydefinethis“dampedfrequency”ωd as

ω = −ζ ω1 2d n (A.17)

0 0.01 0.02 0.03 0.04 0.050

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8Step Response of 2nd-Order System for ζ =0.1,1,10

Time (seconds)

s(t)

ζ = 0.1

ζ = 1

ζ = 10

Figure A.4 Step response of an RLC circuit for 𝛚 = 1000 rad/s and ζ values of 0.1, 1, and 10.

Appendix A ◾ 349

andthusfortheunderdampedcase,thestepresponsemayberewrittenas

=

ωω ζ <

−ζω

( )2

sin for 1s tKe

tt

dd

n

(A.18)

Becausethedampingfactorcanbehardtomeasuredirectly,thedampedfre-quencycanbeusefulformeasuringζbecauseitmaybeeasilyreadfromtheoscil-latory step response on an oscilloscope screen. If both the damped and naturalfrequenciesareknown,thenfromEquation(A.17),

ζ = −

ωω

1 d

n

A.3 Application: Signal ProcessingIntheprevioussection,wederivedthebandpassresponsefor thestandard-formtransferfunctionofEquation(A.10).However,itisalsopossibletoobtainalow-passandhighpassresponsefromanRLCcircuit.FigureA.5showstheseriesRLCcircuitwherethesystemoutputsaretakenasthevoltagesvR,vL ,andvC acrossthethreecircuitelements.Applyingtherelation foravoltagedivider for impedancecircuits,wecanobtain

=+ +

=+ +

1 12

vV

R

R LsCs

RLs

sRLs

LC

R

in (A.19)

=+ +

=+ +

1 1

2

2

vV

Ls

R LsCs

s

sRLs

LC

L

in (A.20)

=+ +

=+ +

1

1

1

12

vV

Cs

R LsCs

LC

sRLs

LC

C

in (A.21)

NotethatEquations(A.19),(A.20),and(A.21)allhavethesamedenominator(andthusthesamenaturalfrequencyanddampingfactor)butdifferentnumera-tors.TheresultingBodemagnitudeplots(FigureA.5)showthattheresistorvolt-agehasabandpass response, the inductorvoltagehasahigh-pass response, andthecapacitorvoltagehasalow-passresponse.Fortheunderdampedcase,allthreeresponseswillhavearesonantpeakatthenaturalfrequency.

350 ◾ Appendix A

100 101 102 103 104 105 106

100 101 102 103 104 105 106

100 101 102 103 104 105 106

10–2

10–1

100

Series RLC Circuit for ζ = 0.1, 1, 10

10–6

10–4

10–2

100

102

10–6

10–4

10–2

100

102

Frequency (rad/sec)

ζ = 0.1

ζ = 0.1

ζ = 0.1

ζ = 10

ζ = 10

ζ = 10

ζ = 1

ζ = 1

ζ = 1

|vR|

|vL|

|vC|

Figure A.5 Bode plot (magnitude portion only) showing the bandpass, high-pass, and low-pass responses for 𝛚 = 1000 rad/s and ζ values of 0.1, 1, and 10.

Appendix A ◾ 351

A.4 Application: Spring-Dashpot SystemAproblem frommechanical engineeringwhichmaybemodeledwith anRLCcircuitisthe“spring-dashpot”systemconsistingofamass,aspring,andadamper.Acommonhouseholdexampleof a spring-dashpot system is apneumaticdoorclosingmechanism(FigureA.6)wherethedooristhemass,aspringclosesthedoor, and a pneumatic piston is used to adjust the closing speed. If thepistonis incorrectly adjusted, then the door either closes too slowly (overdamped) ortoo rapidlywitha fewbounces (underdamped). If adjusted just right (criticallydamped),thenthedoorclosesasquicklyaspossiblewithoutslammingandwith-outanyoscillation.

Fromphysics,theforceappliedtoamassduetoanattachedspringisdefinedas

= −springF kx (A.22)

x

Spring with constant k

Piston with dampingcoefficient c

Door withmass m

Door position

Figure A.6 Example of a spring-dashpot system: automatic door-closing mechanism.

352 ◾ Appendix A

wherekisthespringconstant(inunitsofnewton/meter)andxisthepositionofthedoorattachedtothespring(inmeters).

The damping force in the pneumatic piston is proportional to the velocity

=xdxdt

ofthedoor:

= −damperF cx (A.23)

wherecisthedampingcoefficientofthepiston(inunitsofnewton-seconds).Thesumoftheforcesonthedoor = +total spring damperF F F mustobeyNewton’s

secondlaw =total

2

2F md xdt

,andthus

= − −mx kx cx

+ + = 0xcmx

kmx (A.24)

Equation(A.24)hasthesameformasEquation(A.6)whereR =c,L =m,and= 1C k .Thus,wecanmodelthedoor/spring/dampersystemasaseriesRLCcir-

cuit(whichwehavealreadysolvedmathematically)wherethepistoncorrespondstotheresistor, themassof thedoorcorrespondstothe inductor,andthespringcorrespondstothecapacitor.

Reference 1. Johnson,D.E.,Hilburn,J.L.,andJohnson,J.R.,Basic Electric Circuit Analysis, 4th ed.,

PrenticeHall,1990.

353

Appendix B: Special Characters in MATLAB® Plots

MATLAB®allowstheuseofGreekandspecialcharactersinitsplotheadingsandlabels.ThemethodfordoingthisbasedontheTeXformattinglanguage[1]andissummarizedinthisappendix.

MATLABprovidesthefunctionstitle,xlabel,ylabel,andtextforadd-inglabelstoplots.TheselabelscanincludeGreekandspecialcharactersbyapplyingthecharactersequencesasshowninTableB.1.Thesesequencesallbeginwiththebackslashcharacter(\)andcanbeembeddedinanytextstringargumenttotitle,xlabel,ylabel,andtext.Subscriptsandsuperscriptsmayalsobeappliedbyusingthe_andôperators.Forexample,thesequenceVoiswrittenasV _ o,and106iswrittenas10 6.Ifthesubscriptorsuperscriptismultiplecharacters,thenusecurlybracestodelimitthestringtobesubscripted,forexample,VoutisgeneratedwithV _ out.

Example B.1

The following MATLAB script shows how to use special characters in a plot.

%Example_B_1.m%Thisscriptshowsexampleusageofspecialcharactersin%MATLABplots.%Plota1MHzsinewaveovertheinterval0<t<2microsect=0:2e-8:2e-6;fo=1e6;xout=sin(2*pi*fo*t);plot(t*1e6,xout);title('Plotofsin(2\pif_ot)forf_o=10^6Hz');xlabel('time(\museconds)');ylabel('x_out(t)');text(1.5,0.3,'\omega=2\pi\timesf_o');---------------------------------------------------------------------------

TheresultingplotisshowninFigureB.1.

354 ◾ Appendix B

Table B.1 Special Symbols for Use in MATLAB Plots

Character Sequence Symbol


CharacterSequence Symbol

\alpha α \upsilon υ \sim ∼

\beta β \phi ϕ \leq ≠

\gamma γ \chi χ \infty ∞

\delta δ \psi ψ \clubsuit ♣

\epsilon ε \omega ω \diamondsuit ♦

\zeta ζ \Gamma Γ \heartsuit ♥

\eta η \Delta Δ \spadesuit ♠

\theta θ \Theta Θ \leftrightarrow ↔

\vartheta ϑ \Lambda Λ \leftarrow ←

\iota ι \Xi Ξ \uparrow ↑

\kappa κ \Pi Π \rightarrow →

\lambda λ \Sigma Σ \downarrow ↓

\mu μ \Upsilon Υ \circ °

\nu ν \Phi Φ \pm ±

\xi ξ \Psi Ψ \geq ≥

\pi π \Omega Ω \propto ∝

\rho ρ \forall ∀ \partial ∂

\sigma σ \exists ∃ \bullet •

\varsigma ς \ni ∍ \div ÷

\tau τ \cong ≅ \neq ≠

\equiv ≡ \approx ≈ \aleph ℵ

\Im ℑ \Re ℜ \wp ℘

\otimes ⊗ \oplus ⊕ \oslash ∅

\cap ∩ \cup ∪ \supseteq ⊇

\supset ⊃ \subseteq ⊆ \subset ⊂

\int ∫ \in ∈ \o ∘

Appendix B ◾ 355

Reference 1. Knuth,D.E.,The TeXbook,AddisonWesley,Boston,1984.

0 0.5 1 1.5 2–1

–0.8

–0.6

–0.4

–0.2

0

0.2

0.4

0.6

0.8

1Plot of sin(2π fot) for fo = 106 Hz

Time (µseconds)

x out

(t)

ω = 2π × fo

Figure B.1 Example usage of Greek letters, special characters, subscripts, and superscripts in a MATLAB plot.

Table B.1 Special Symbols for Use in MATLAB Plots (Continued)



CharacterSequence Symbol

\rfloor ⌋ \lceil ⌈ \nabla ∇

\lfloor ⌊ \cdot ⋅ \ldots ⋯

\perp ⊥ \neg ¬ \prime ′

\wedge ∧ \times × \0 ∅

\rceil ⌉ \surd √ \mid

\vee ∨ \varpi ϖ \copyright ©

\langle \rangle



Bober • Stevens

ISBN: 978-1-4398-5429-7

9 781439 854297

9 0 0 0 0

K12515

Electrical Engineering

Combining academic and practical approaches to this important topic, Numerical and Analytical Methods with MATLAB® for Electrical Engineersis the ideal resource for electrical and computer engineering students. Based on a previous edition that was geared toward mechanical engineering students, this book expands many of the concepts presented in that book and replaces the original projects with new ones intended specifically for electrical engineering students.

This book includes:

• An introduction to the MATLAB® programming environment • Mathematical techniques for matrix algebra, root finding,

integration, and differential equations• More advanced topics, including transform methods, signal

processing, curve fitting, and optimization • An introduction to the MATLAB® graphical design environment,

Simulink®

Exploring the numerical methods that electrical engineers use for design analysis and testing, this book comprises standalone chapters outlining a course that also introduces students to computational methods and pro-gramming skills, using MATLAB® as the programming environment. Helping engineering students to develop a feel for structural programming—not just button-pushing with a software program—the illustrative examples and extensive assignments in this resource enable them to develop the necessary skills and then apply them to practical electrical engineering problems and cases.


Numerical and Analytical M

ethods with M

ATLAB® for Electrical Engineers

K12515_Cover_mech.indd 1 7/27/12 9:01 AM

Documents

Numerical and Analytical Methods with MATLAB for