Control Systems SecondEdition Dr.N.C.Jagan B.E.,ME., Ph.D.,MlSTE,FIE Retd. Professor in Electrical Engineering University College of Engineering Osmania University, Hyderabad. BSPBSPublications 4-4-309, Giriraj Lane,Sultan Bazar, Hyderabad - 500 095 - A.P. Phone: 040-23445605,23445688 Copyright 2008 by Publisher All rightsreserved Nopart of thisbook or parts thereof maybereproduced,storedinaretrievalsystemor transmittedinanylanguageorbyanymeans,electronic,mechanical,photocopying, recording or otherwisewithout theprior written permission of thepublishers. Published by : SSPBS Publications Printed at 4-4-309, Giriraj Lane,Sultan Bazar, Hyderabad - 500 095 - A.P. Phone: 040-23445605,23445688 e-mail: contactus@bspublications.net www.bspublications.net Adithya Art Printers Hyderabad ISBN:978-81-7800-177-2(HB) Contents 1Introduction1 1. 1Why AutomaticControl? ....................................................................1 1.2OpenLoopandClosedLoopControlSystems................................ 2 1.3OpenLoopVsClosedLoopControlSystems ................................... 3 1.4FeedbackControlSystems ................................................................ 4 1.5Classificationof ControlSystems...................................................... 4 1.5.1Continuous TimeFeedbackControlSystems ..................... 5 1.5.2DiscreteDataFeedbackControlSystems.......................... 5 1.5.3LinearControlSystems........................................................ 6 1.5.4Non-Linear ControlSystems................................................ 6 2Mathematical Modelling of Physical Systems8 2.1Introduction ............................................................., ........................... 8 2.2MathematicalModels of PhysicalSystems ........................................ 8 2.2.1ElectricalSystems ................................................................. 9 2.2.2DualNetworks ..................................................................... 14 2.2.3MechanicalSystems ........................................................... 15 2.2.4AnalogousSystems ............................................................ 18 2.2.5Gears ................................................................................... 24 2.2.6ThermalSystems ................................................................ 27 2.2.7FluidSystems ...................................................................... 29 Contents 2.3BlockDiagramsof ClosedLoopS y s t ~ m s........................................ 34 2.3.1BlockDiagramReductionTechniques ............................... 35 2.4SignalFlowGraphRepresentationof ControlSystems................. 42 2.4.1Constructionof a SignalFlowGraphfora System ........... 43 2.4.2Mason'sGainFormula....................................................... 44 2.5EffectsofFeedback......................................................................... 52 2.5.1Effect onsystemdynamics ................................................. 53 2.5.2Effectduetoparameter variations.................................... 54 2.5.3Effect onBandwidth ............................................................ 57 2.5.4Effect onNoiseSignals ....................................................... 58 2.6Modellingof Elementsof ControlSystems ...................................... 59 2.6.1DCServoMotor .................................................................. 59 2.6.2ACServoMotors ................................................................. 63 2.6.3Synchros............................................................................. 65 2.6.4ACTachoGenerator ........................................................... 68 2.6.5DCTachoGenerator .......................................................... 69 2.6.6Potentiometers .................................................................... 69 2.6.7StepperMotors................................................................... 71 Problems ........................................................................................~............ 73 3Time Response Analysis of Control Systems81 3.1Introduction ....................................................................................... 81 3.2StandardTestSignals ....................................................................... 81 3.2.1Impulse Signal ....................................................................... 81 3.2.2StepSignal ........................................................................... 82 3.2.3Rampsignal.......................................................................... 82 3.2.4Parabolicsignal.................................................................... 82 3.3Representationof Systems .............................................................. 83 3.4FirstOrderSystem........................................................................... 85 3.4.1Responsetoa UnitStepInput............................................. 85 3.4.2Response toa UnitRampInput or Unit VelocityInput........ 86 3.4.3ResponsetoaUnitParabolicor AccelerationInput ............ 87 3.5SecondOrder System...................................................................... 88 3.5.1Responsetoa Unit StepInput............................................. 88 3.5.2Responsetoa Unit RampInput ........................................... 92 3.5.3TimeDomainSpecificationsof a SecondOrder System ..... 93 Contents 3.6SteadyStateErrors.......................................................................... 99 3.6.1ErrorConstants.................................................................... 99 3.6.2Dependenceof SteadystateErroronTypeof theSystem 101 3.6.3GeneralizedErrorCoefficients- Error Series .................... 104 3.7DesignSpecificationsof a ControlSystem .................................... 113 3.7.1ProportionalDerivativeErrorControl(PDcontrol)...........114 3.7.2ProportionalIntegralController(PIControl) ...................... 115 3.7.3Proportional,IntegralandDerivativeController (PIDControl) ....................................................................... 116 3.7.4DerivativeOutputControl ................................................... 116 Problems ................................................................................................... 125 4Stability of Systems129 4.1Introduction ..................................................................................... 129 4.2StabilityPreliminaries ...................................................................... 130 4.3NecessaryConditionsforStability ................................................. 133 4.4Routh- HurwitzStability Criterion ................................................... 134 4.5SpecialCases ................................................................................. 137 4.6RelativeStability............................................................................. 147 Problems ................................................................................................... 149 5Root Locus Analysis151 5.1Introduction ..................................................................................... 151 5.2BasicIdea....................................................................................... 151 5.3Developmentof RootLocusTechnique........................................ 153 5.4Propertiesof RootLocus............................................................... 155 Problems ................................................................................................... 187 6Frequency Response of Control Systems , 190 6.1Introduction ..................................................................................... 190 6.2TimeDomain AnalysiS VsFrequencyDomain Analysis................ 190 6.3FrequencyResponseof a ControlSystem................................... 191 6.4CorrelationBetweenTimeResponseand FrequencyResponse ..................................................................... 194 Contents 6.5GraphicalRepresentationof TransferFunctions......................... 196 6.5.1Bodeplots .......................................................................... 196 6.5,2PolarPlots .......................................................................... 218 6.5.3LogMagnitudeVsPhasePlots ......................................... 222 6.6AllPassSystems............................................................................ 223 6.7Minimum Phase Systems ................................................................ 224 Problems ................................................................................................... 226 7Nyquist Stability Criterion and Closed Loop Frequency Response 229 7.1Introduction ..................................................................................... 229 7.2Principleof Argument..................................................................... 230 7.3Developmentof NyquistCriterion.................................................. 234 7.3.1NyquistContour ................................................................. 234 7.3.2NyquistStabilityCriterion ........................ " ......................... 234 7.3.3NyquistContour WhenOpenLoopPolesOccur on jro-axis ............................................................................ 236 7.3.4Nyquist Stability Criterionfor Systemswhichare OpenLoopStable .............................................................. 252 7.4RelativeStability............................................................................ 254 7.4.1Measuresof RelativeStability:GainMargin .................... 256 7.4.2PhaseMargin ..................................................................... 257 7.4.3Adjustment of GainK forDesiredValuesof Gainand PhaseMarginsUsingPolar Plots ...................................... 260 7.4.4GainMarginandPhaseMarginUsingBodePlot ............. 261 7.5ClosedLoopFrequencyResponse .............................................. 264 7.5.1ConstantM Circles............................................................ 267 7.5.2ConstantN circles .............................................................. 269 7.5.3NicholsCharts .................................................................... 272 Problems ................................................................................................... 277 8Design in Frequency Domain281 8.1Preliminariesof Design.................................................................. 281 8.2A Design Example ........................................................................... 282 8.3Realisationof CompensatorsandtheirCharacteristics ............... 285 Contents 8.3.1LeadCompensator ............................................................ 286 8.3.2LagCompensator .............................................................. 290 8.3.3Lag-LeadCompensator.................................................... 291 8.4CascadeCompensationinFrequencyDomainUsing BodePlots ...................................................................................... 293 8.4.1Designof LeadCompensator ........................................... 293 8.4.2Designof LagCompensator............................................. 304 8.4.3DesignofLag-LeadCompensator................................... 311 Problems ................................................................................................... 316 9State Space Analysis of Control Systems318 9.1Introduction ..................................................................................... 318 9.2StateVariables ................................................................................ 319 9.3StateEquationsforLinearSystems.............................................. 321 9.4CanonicalFormsof StateModelsof LinearSystems ................... 324 9.4.1PhaseVariableForm......................................................... 324 9.4.2DiagonalForm................................................................... 330 9.5TransferFunctionfromStateModel............................................. 334 9.6Diagonalisation .............................................................................. 337 9.6.1EigenvaluesandEigenvectors......................................... 337 9.7Solutionof StateEquation ............................................................. 345 9.7.1Propertiesof StateTransitionMatrix ................................. 346 9.7.2Methodsof ComputingState TransitionMatrix ................. 346 9.8Qualitative Analysisof ControlSystems ....................................... 358 9.8.1StateControllability ............................................................ 358 9.8.2Observability ...................................................................... 366 9.8.3DualityPrinciple ................................................................. 372 Problems ................................................................................................... 373 AnswerstoProblems ............................................................................. 377 MCQ'sfromCompetitiveExaminations............................................. 387 AnswerstoMCQ's .................................................................................. 481 -djJ-"This page is Intentionally Left Blank" 1Introduction 1.1Why AutomaticControl? Automaticcontrolof manydaytodaytasksrelievesthehumanbeingsfromperformingrepetitive manualoperations.Automaticcontrolallowsoptimalperformance ,of dynamicsystems,increases productivity enormously, removes drudgery of performing same task again and again. Imagine manual controlof asimpleroomheatingsystem.If theroomtemperatureistobemaintainedatadesired temperatureTOe,bycontrollingthecurrentinanelectricalheatingsystem,thecurrentmaybe adjusted by moving thevariable arm in a rheostat.The temperature of the room depends on a host of factors: number of persons in the room,the opening and closing of doorsdue topersonsmoving in andout,fluctuation~ nthesupplyvoltageetc.A humanoperatorhastocontinuouslymonitorthe temperature indicated by a thermometer and keep on adjusting the rheostat to maintain the temperature all the twenty four hours. The operator should be continuously alert and relentlessly perform a simple job of moving the arm of the rheostat. Any mistake on his part may result in great discomfiture to the personsintheroom. Now,imagine the same operation of measuring the temperature,estimating the error between the desired temperature and the actual temperature, moving the arm of the rheostat accurately by an automaticcontroller.Sinceerror. between the actual temperatureand the desired temperature iscontinuouslyobtainedandusedtoactivatethecontroller,anydisturbancescauseddueto movementsof personsoccupyingtheroom,supplyvariationsetc.willbeautomaticallytaken careof.Howmuchof arelief itis!Thisisonlyasimpletask,butmanycomplexindustrial processes,spacecraftsystems,missileguidancesystems,roboticsystems,numericalcontrol of machinetoolsemployautomaticcontrolsystems.Thereisnofieldinengineeringwhere automaticcontrolisnotemployed.Evenhumansystemisaverycomplexautomaticfeedback control system. The modem engineers and scientists must, therefore, have a thorough knowledge of theprinciplesof automaticcontrolsystems. 2ControlSystems Thefirstautomaticcontrolwasinventedby JamesWatt.Heemployedacentrifugalorflyball governor for the speed control of a steam engine in1770. But much of theadvances had towait for morethanahundredyears,untilMinorsky,HazenandNyquistcontributedsignificantlyinthe development of control system theory. Hazen coined the word "servo mechanisms" to describe feedback control systems in which the variable to be controlled is a mechanical position, velocity or acceleration of a given object. During 1940s, frequency response methods and root 'locus techniques were developed todesignlinear,stable,closedloopcontrolsystemswithgivenperformancemeasures.In later part of 1950s, much emphasis was given in designing systems, which not only satisfied given performance measures, but also provided optimum design in a given sense. As the systems became more and more complex with morenumber of inputsand outputs and with theadvent of digitalcomputers,modern controltheoryr ~ \ l e r t e dbacktomethodsbasedontimedomainanalysisandsynthesisusingstate variable representations. In the period between1960 and1980, to cope upwith the complexity and stringent requirements on accuracy, speed and cost adaptive control was developed. Both deterministic and stochastic systems were considered and controllers were designed which were optimal, adaptive and robust. The principles developed in automatic control theory were not only used in engineering applications, but also in non engineering systemslikeeconomic,socioeconomicsystems and biologicalsystems. 1.2Open Loopand Closed LoopControlSystems OpenLoopControl Systems:A systemin which theoutput hasnoeffecton thecontrol actionis knownasanopenloopcontrolsystem.For agiven input thesystemproducesacertainoutput.If thereare any disturbances,the out put changes and thereisno adjustment of the input tobring back the output to the original value. A perfect calibration is required to get good accuracy and the system should befreefromany externaldisturbances.Nomeasurements aremade at theoutput. A traffic control system isa good example of an open loop system.The signals change according toapresettimeandarenotaffectedbythedensityof trafficonanyroad.Awashingmachineis another example of an openloopcontrolsystem.Thequalityof wash isnotmeasured;every cycle likewash,rinse and dry' cyclegoesaccording toapreset timing. Closed Loop Control Systems:These are also known as feedback control systems. A system which maintains a prescribed relationship between the controlled variable and the reference input,and uses thedifferencebetweenthemasasignaltoactivatethecontrol,isknownasafeedbackcontrol system. The output or the controlled variable ismeasured and compared with the reference input and an error signal is generated. This isthe activating signal to the controller which, by its action, tries to reduce theerror.Thus the controlled variable iscontinuously fedbackand compared with theinput signal.If theerror isreduced tozero,theoutputisthedesiredoutput andisequal tothereference input signal. Introduction 3 1.3Open LoopVsClosed LoopControl Systems Theopenloopsystemsaresimpleandeasiertobuild.Stability,whichwillbediscussedinlater chapters,isnot aproblem.Openloopsystemsarecheaper andtheyshouldbepreferred whenever there is a fixed relationship between the input and the output and there are nodisturbances. Accuracy isnot criticalinsuchsystems. Closed loop systems are more complex, use more number of elements to build and are costly. The stability isa major concern forclosed loopsystems.Wehave toensure that the systemisstable and willnot cause undesirableoscillationsin theoutput.Themajoradvantageof closedloopsystemis thatitisinsensitivetoexternaldisturbancesandvariationsinparameters.Comparativelycheaper componentscanbeusedtobuildthesesystems,asaccuracyandtolerancedonotaffectthe performance. Maintenance of closed loopsystemsismoredifficult than open loopsystems.Overall gainof thesystem isalsoreduced. OpenLoopSystems Advantages 1.They aresimple and easy to build. 2.They arecheaper,astheyuselessnumber of components tobuild. 3.They are usually stable. 4.Maintenance iseasy. Disadvantages 1.Theyarelessaccurate. 2.If externaldisturbancesarepresent,output differssignificantly fromthe desired value. 3.If there arevariationsin theparametersof thesystem,theoutput changes. ClosedLoopSystems Advantages 1.Theyaremoreaccurate. 2.Theeffect of external disturbance signals can bemadeverysmall. 3.Thevariationsinparametersof thesystemdonot affecttheoutputof thesystemi.e.the output may be made less sensitive to variation is parameters. Hence forward path components canbeof lessprecision.Thisreducesthecostof thesystem. 4.Speedof theresponsecanbegreatlyincreased. Disadvantages 1.They are morecomplex and expensive 2.They requirehigherforwardpath gains. 3.The systemsareprone toinstability.Oscillationsintheoutput many occur. 4.Cost of maintenanceishigh. 4ControlSystems 1.4FeedbackControlSystems Fig.1.1representsafeedbackcontrolsystem. Controller Reference input r------------, I+e(t) - - - - - - ~ ~~ - - - - - - - - ~ orr(t)I-CommandI signal error or actuatingI signal_________ --1 Feedbackb(t) signal Fig. 1.1A feedback control system Manipulating signal Controlledoutput c(t) Afeedbackcontrolsystemisrepresentedasaninterconnectionof blockscharacterizedbyan inputoutputrelation.Thismethodof representingacontrolsystemisknownasablockdiagram representation.Whileothermethodsarealsousedtorepresentthecontrolsystem,thisismore popular.Theinputtotheentiresystemiscalledasareferenceinputoracommandinput,ret). An error detector senses the difference between the reference input and the feedback signal equal to or proportionaltothecontrolledoutput.Thefeedbackelementsmeasurethecontrolledoutputand convert or transfonn it toa suitable valuesothat itcan be compared with the referenceinput.If the feedbacksignal,bet),isequaltothecontrolledoutput,c(t),thefeedbacksystemiscalledasunity feedbacksystem. The difference between the reference input and the feedback signal is known as the error signal or actuating signal e(t), This signal isthe input to the control elements which produce a signal known as manipulated variable,u(t).This signal manipulates thesystem or plant dynamicssothat thedesired output isobtained.Thecontroller actsuntiltheerror between theoutput variable and thereference input iszero.If the feedback path isabsent, the system becomes an open loopcontrol system and is representedinFig.1.2. InputIIu(t)~ IPlantIOutpu.:. c(t) r(t)- - - - I . ~Controller........ Fig. 1.2Openloop control system 1.5ClassificationofControlSystems Dependingonthetypeof signalspresentatthevariouspartsof afeedbackcontrolsystem,the systemmaybeclassifiedasa(i)continuoustimefeedbackcontrolsystemora(ii)discretetime feedbackcontrolsystem. Introduction 5 1.5.1 Continuous Time Feedback Control Systems If the signals in all parts ofa control system are continuous functionsof time, the system isclassified ascontinuous time feedback control system. Typically allcontrol signals are of low frequency and if thesesignalsareunmodulated,thesystemisknownasad.c.controlsystem.Thesesystemsuse potentiometersaserrordetectors,d.camplifierstoamplifytheerrorsignal,d.c.servomotoras actuating device and d.ctachometers or potentiometers asfeedback elements. If thecontrol signal is modulated by an a.c carrier wave, the resulting system isusually referred toasan a.c control system. These systems frequently use synchros as error detectors and modulators of error signal, a.c amplifiers to amplify the error signal and a.c servo motors as actuators. These motors also serve as demodulators and produce an unmodulated output signal. 1.5.2 Discrete Data Feedback Control Systems Discretedatacontrolsystemsarethosesystemsinwhichatoneormorepansof thefeedback controlsystem,thesignalisintheformof pulses.Usually,theerrorinsuchsystemissampledat uniformrateandtheresultingpulsesarefedtothecontrolsystem.Inmostsampleddatacontrol systems, the signal is reconstructed as a continuous signal, using a device called 'hold device'. Holds of different orders are employed,but the most common hold deviceisa zero order hold.It holds the signalvalueconstant,atavalueequaltotheamplitudeof theinput timefunctionatthatsampling instant,untilthenextsamplinginstant.A typicaldiscretedatacontrolsystemisshowninFig.1.3 whichusesa sampler anda data hold. ret) c(t) Sampler Fig. 1.3Discrete data controlsystem Thesesystemsarealsoknownassampleddatacontrolsystems. Discreet data controlsystems,inwhich a digitalcomputer isused asone of theelements, are known as digital control systems. The input and output to the digital computer must be binary numbers and hence these systems require the use of digital to analog and analog to digital converters. A typical digitalcontrolsystem isshown inFig.1.4. c(t) Fig. 1.4Digital feedback controlsystem Digital devices may beemployed in thefeedback circuits asmeasuring elements. 6ControlSystems Afurtherclassification of control systems can bemadedepending on thenatureof thesystems, namely, 1.Linear controlsystems 2.Non-linear controlsystems 1.5.3 Linear Control Systems If a system obeys superposition principle, the system issaid to be a linear system. Let xl (t) and x2(t) be two inputs to a system and Yl(t) and Yit) be the corresponding outputs. For arbitrary real constants kland forinput klxl(t) + x2(t),if theoutput of thesystem is given by klYl(t)+ k2Y2(t), then thesystemissaid tobealinear system.Thereareseveralsimpletechniquesavailableforthe analysisanddesign of linear controlsystems. 1.5.4 Non-Linear Control Systems Any system which doesnotobey superposition principle issaid tobea non-linear system.Physical systemsareingeneralnon-lienarandanalysisof suchsystemsisverycomplicated.Hencethese systems areusuallylinearlised and well known linear techniques are used to analyse them. Thesesystemscan be furtherclassifieddepending on whether theparameters of thesystem are constants, or varying with respect to time. When the input to a system isdelayed by T seconds, if the output isalsodelayed by the same time T,thesystem issaid tobe a timeinvariant system.Thus x(t)__System1----.. y(t)x (t- T)System1----.. Y (t-' T) (a)(b) Fig. 1.5Time invariant system. On the other hand,if the output isdependent on the time of application of the input, the system is saidtobeatimevaryingsystem.Likenon-linearsystems,timevaryingsystemsalsoaremore complicated for analysis. In this text we will be dealing with linear time invariant continuous systems only. The layout of this book is as follows: The mathematical modelling of processes suitable for analysis and design of controllers isdiscussed inChapter 2.Typicalexamplesfromelectrical,mechanical,pneumaticandhydraulicsystemsare given. The transfer function of the overall system isobtained by block diagram and signal flow graph representationof thesystems.Theeffectsof feedbackontheperformanceof thesystemarealso discussed.. Inchapter3,timedomainspecificationsof thecontrolsystemaredefinedwithrespecttothe response of a typical second order system,for unit stepinput.Steady state errors are defined and the use of PID controllers isdiscussed to satisfy the design specifications of a control system.In chapter 4,the stability aspects of the system are discussed and algebraic criteria for obtaining the stability of the system aredeveloped. Introduction 7 The rootsof thecharacteristicequation of thesystemdeterminethebehaviour andstability of a controlsystem.Therootschangeastheparametersarechanged.Theconcept of thelocusof these roots as one of the parameters,usually the gain of theamplifier,ischanged from0to00isdiscussed in chapter 5.The design of a control system isrendered very easy by considering the response of the systemtosinusoidalsignals.Frequencydomainanalysisanddevelopmentof frequencydomain specificationsarediscussedin chapter 6. Relativestability aspects are considered in chapter 7.Nyquist stability criterion isdeveloped and measuresof relative stability, viz, gain margin on phase margin are dermed. In chapter 8,design of compensating RCnetworks to satisfy the design specifications of a control system in frequency domain isdiscussed.In chapter 9,state space representation of controlsystems isdeveloped,which enables modern techniquestobeusedin thedesign of controlsystems. - ~ -2Mathematical Modelling of PhysicalSystems 2.1Introduction Whenever a task istobeperformed,a set of physicalobjectsareconnected together anda suitable inputisgiventothem,toobtain thedesiredoutput.Thisgroupof objectsisusuallytermedasthe 'system'.Thesystemmayconsistof physicalobjectsanditmaycontaincomponents,biological economicalor managerialinnature.Inorder toanalyse,designorsynthesiseacomplexsystem,a physicalmodelhastobeobtained.Thisphysicalmodelmaybeasimplifiedversionof themore complexsystem.Certainassumptionsaremadetodescribethenatureof thesystem.Usuallyall physical systems in the world are nonlinear in nature. But under certain conditions these systems may be approximated by linear systems. Hence for certain purposes, a linear model may beadequate. But if stringentaccuracyconditionsaretobesatisfied,linear modelmay notbesuitable.Similarly,the parameters of the system may befunctionsof time.But if they are varying very slowly, they may be assumed tobe constant.In many engineering systems the elements areconsidered tobelumped and theirbehaviourisdescribedbyconsidering theeffect atitsendpointscalled terminals.Longlines transmitting electrical signals, may not be adequately represented bylumped elements. A distributed parameter representation maybecalledforinthiscase.Hencedependingon therequirementsof a given job, suitable assumptionshave tobemade anda'physical model'hasto befirst defined.The behaviour of thisphysical modelis then describedin termsof amodelsothat known techniquesof mathematical analysis can beapplied to the given system. 2.2MathematicalModelsofPhysicalSystems Thesystemmaybeconsideredtobeconsistingof aninterconnectionof smallercomponentsor elements,whosebehaviour canbedescribedbymeansof mathematicalequationsor relationships. Inthisbook,wewillbeconsideringsystemsmadeupof elementswhicharelinear,lumpedand timeinvariant.Anelementissaidtobelinearif itobeystheprincipleof superposition andhomogeneity.If theresponsesof theelementforinputsxj(t)andxit)areYj(t)andYit) respectively, theelement islinear if the response to theinput, kj xj(t) + Is xit) iskj Yj(t)+ Is Y2(t) asshown inFig.2.1. MathematicalModellingof PhysicalSystems 9 x,(t)I LInear element Fig. 2.1Definitionof linear element An element issaid to be 'lumped'if the physical dimensions or spacial distribution of the element does not alter the signal passing through it. The behaviour of such elements are adequately represented by the effects at the end points called terminals.The temperature of a body may be treated assame, at all points of the body under certain conditions.Similarly the mass of a body maybeconsidered as concentratedatapoint.Arotatingshaftmaybeconsideredasrigid.Anelectricalresistormaybe represented by a lumped element, since the current entering at one terminal leaves the other terminal withoutundergoinganychange.Thevoltagedistributioninthephysicalbodyof theresistorisnot considered.Only thevoltageacrossitsterminalsistaken foranalysis.These aresomeexamplesof lumped elements. If theparametersrepresenting theelementsarenot changing with respect totime,theelement is saidtobetimeinvariant.Thusif asystemiscomposedof linear,lumpedandtimeinvariant elements,itsbehaviourcanbemodelledbyeitherlinearalgebraicequationsorlineardifferential equations with constant coefficients. If theinput output relations arealgebraic, the system issaid to beastaticsystem.Ontheotherhand,if therelationsaredescribedbydifferentialequations,the systemissaidtobeadynamicsystem.Wearemostlyconcernedwithdynamicresponseof the systems and therefore, one ofthe ways by which a system is mathematically modelled is by differential equations.AnothermostusefulandcommonmathematicalmodelistheTransfer function'of the system.Itisdefinedastheratioof Laplacetransformof theoutput totheLaplacetransformof the input.Thisisillustrated inFig.2.2. R (s).LI __T_(_S)_---If---C-(.!) Fig.2.2Transfer function of a system InFig.2.2,thetransferfunctionis, C(s) T(s)=R(s) ..... (2.1) In defining the transfer function,it is assumed that allinitial conditions in the system arezero. Havingdefinedthetwocommonwaysof describinglinearsystems,letusobtainthe mathematical models of some commonly occuring Electrical, Mechanical, Thermal, Fluid, Hydraulic systemsetc. 2.2.1Electrical Systems Mostof theelectricalsystemscanbemodelledbythreebasicelements:Resistor,inductor,and capacitor.Circuits consisting of these threeelements areanalysed byusing Kirchhoff's Voltagelaw andCurrentlaw. 10 (a)Resistor:Thecircuit modelof resistorisshown inFig.2.3(a) i(t)R v(t)Fig. 2.3(a)Circuit model of resistor The mathematical model isgiven by the Ohm's law relationship, vet)=i(t)R; i(t)=vet) R (b)Inductor:Thecircuit representation isshowninFig.2.3(b) Fig. 2.3(b)Circuit model of inductor Theinput output relationsare given byFaraday'slaw, or ( ) - L diet) vt- --dt I i(t) =- Jvdt L whereJv dtisknownasthe fluxlinkages 'Y (t).Thus 'Y(t) i(t) =-L ControlSystems ..... (2.2) ..... (2.3) ..... (2.4) ..... (2.5) If onlyasinglecoilisconsidered,theinductanceisknownasself inductance.If avoltageis induced in a assecond coildue toa current in thefirstcoil,the voltageissaid tobeduetomutual inductance,asshowninFig.2.3(c). ++ Fig. 2.3(c)Mutual inductance MathematicalModellingof PhysicalSystems InFig.2.3(c), di v(t)=M_1 2dt (c)Capacitor:Thecircuit symbolof a capacitor isgiveninFig.2.3(d). ~ ~ L-V(t)---J Fig. 2.3(d)Circuit symbol of a capacitor vet)=~fidt C or dv i(t)=C-dt 11 ..... (2.6) ..... (2.7) ..... (2.8) In eqn.(2.7),Sidt isknown as thechargeon thecapacitor andisdenotedby'q'.Thus q =Sidt..... (2.9) q(t) andvet)=C..... (2.10) Another useful element, frequently used in electrical circuits, is the ideal transformer indicated inFig.2.4. + Fig. 2.4Model of a transformer Themathematicalmodelof a transformer isgiven by, V2=N2=.!.L VINI12 + .... (2.11) Electricalnetworksconsisting of theaboveelements areanalysed using Kirchhoff'slaws. Example 2.1 Consider the network in Fig.2.5.Obtain the relation between the applied voltage and the current in the formof (a) Differential equation(b)Transferfunction i v ~ ':}c Fig. 2.5AnR.L.C series circuit excited by a voltage source. 12ControlSystems Solution: (a)Writing downtheKirchhoff'svoltagelawequation for theloop,wehave diet)1t R i(t)+ L--+- fi (t)dt =v dtC-00 ..... (2.12) t Denotingfi(t)d(t)byq(t),wecanalsowriteeqn.(2.8)as -00 Ld2q(t) + R dq(t) + q(t)=v dt2 dtC ..... (2.13) Thisisa 2nd order linear differentialequation with constant coefficients. (b)Transfer function Taking Laplace transformof eqn.(2.13) with allinitial conditions assumed tobe zero,wehave 1 Ls2Q(s)+RsQ(s)+C Q(s)=Yes) Q(s)C Yes)Ls+Rs+lLCs2+RCs+l C ..... (2.14) .... (2.15) Thisisthetransferfunctionof thesystem,if q(t)isconsideredasoutput.Instead,if i(t)is considered astheoutput,taking Laplacetransformof eqn.(2.12),wehave, Example 2.2 I(s) RI(s)+LsI(s)+Cs=Yes) I(s)Cs =----,-=----;;;----yes)Ls + R + _1LCs2 + RCs + 1 Cs ..... (2.16) ..... (2.17) Consider the parallel RLC network excited by a current source (Fig.2.6).Find the (a) differential equation representation and(b)transferfunctionrepresentation of thesystem. vet) i(t) ic Fig. 2.6ParallelRLCcircuit excited by a current source MathematicalModellingof PhysicalSystems Solution: (a)Applying Kirchhoff'scurrent law at thenode, vet)Cdv(t)1. -+--+- Jvdt=:let) RdtL Replacing J v dtbyqJ(t),thefluxlinkages,wehave, Cd2 ~(t) + ~d\jf(t)+\jf(t)=:i(t) dtRdtL (b)Taking Laplacetransformof eqn.(2.19),wehave, Cs2 qJ(s)+.l s qJ(s)+~(t)=:I(s) RL qJ(s) I(s)2S1 Cs+-+-RL L LCs2 +LGs+l 13 ..... (2.18) ..... (2.19) ..... (2.20) If thevoltageistakenastheoutput,taking Laplacetransformof eqn.(2.18),weget 1 GV(s)+ CsV(s)+LsYes)=:I(s) Yes) = ----= -----Ls ..... (2.21 ) I(s)Cs +G + _1LCs2 +LGs+ 1 Ls Example 2.3 Obtain thetransferfunctionI(s)N(s)inthenetworkof Fig.2.7. Fig. 2.7Network for the Example 2.3 14 Solution: Writing the loop equations (3+ s)11(s) - (2+ s)11(s) Solving forI(s),wehave - (s+ 2)12(s) +(3+2s)12(s)- s.l(s) 3+s-(s + 2)yes) - (2 +s)(3 + 2s)0 0-s0 I(s)= , 3+s-(s + 2)0 - (2+s)(3+ 2s)-s 0-s (1+2S+ ;s) Yes)[s(s + 2)] s(s + 2) V(s).2s I(s)- ----'-:----'--:;-'--'-----;:----- -(s+3)(4s4 + 12s3+9s2 +4s+1) I(s)_2S2(S+2) Yes)-- (s+3)(4s4 +12s3t9s2 +4s+1) 2.2.2Dual Networks =Yes) =0 ControlSystems ..... (1) ..... (2) ..... (3) Consider the two networks shown in Fig.2.5and Fig.2.6 and eqns.(2.13) and (2.19).In eqn.(2.13) if the variables q and v and circuit constant RLC are replaced by their dual quantities as per Table 2.1 eqn.(2.19)results. Table2.1DualQuantities vet)Bi(t) i(t)Bvet) 1 RB-=G R CBL LBC q =fidtB'l' =fvdt MathematicalModellingof PhysicalSystems 15 The two networks, are entirely dissimilar topologically, but their equations areidentical. If the solution of oneequationisknown,thesolution of theother isalsoknown. ThusthetwonetworksinFig.2.5and2.6areknownasdualnetworks.Givenanyplanar network,a dualnetwork can alwaysbeobtained using Table2.1. 2.2.3Mechanical Systems Mechanicalsystemscan bedividedinto twobasicsystems. (a)Translationalsystemsand(b)Rotationalsystems Wewillconsiderthesetwosystemsseparatelyanddescribethesesystemsintermsof three fundamental linear elements. (a)Translational systems: 1.Mass:Thisrepresentsanelementwhichresiststhemotionduetoinertia.Accordingto Newton'ssecondlaw of motion, theinertia forceisequal tomasstimesacceleration. dvd2x f,=Ma=M.- =M-M~~ 2 ..... (2.22) Wherea,vandxdenoteacceleration,velocityanddisplacementof thebodyrespectively. Symbolically,thiselement isrepresented by ablock asshown in Fig.2.8(a). (a)(b)(c) Fig. 2.8Passive linear elements of translationalmotion(a)Mass(b)Dashpot(c)Spring. 2.Dash pot:Thisisan element which opposes motion due to friction.If the frictionisviscous friction,thefrictionalforceisproportional tovelocity.Thisforceisalsoknown asdamp ling force.Thuswecanwrite dx f=Bv=B-Bdt ..... (2.23) WhereBisthedampingcoefficient.Thiselementiscalledasdashpotandissymbolically representedasinFig.2.8(b). 3.Spring:The third element which opposes motion is the spring. The restoring force of a spring isproportional tothedisplacement.Thus fK=Kx..... (2.24) WhereK isknownasthestiffnessof thespring or simply spring constant.Thesymbolused forthiselementisshowninFig.2.8(c). 16ControlSystems (b) Rotational systems:Corresponding to the three basic elements of translation systems, there are threebasicelementsrepresenting rotationalsystems. 1.Moment ofInertia:This element opposes the rotational motion due to Moment oflnertia. The opposing inertia torque is given by, drod2e T=Ja=J - =J-Idtdt2 ..... (2.25) Wherea,OJand()aretheangularacceleration,angularvelocityandangulardisplacement respectively.J isknownasthemoment of inertia of thebody. 2.Friction:Thedamping or frictionaltorque which opposes the rotational motionisgiven by, de T=Bro=B-Bdt Where B is the rotational frictional coefficient. ..... (2.26) 3.Spring:The restoring torqueof a spring is proportional to theangular displacement() and is given by, ..... (2.27) WhereKisthetorsimalstiffnessof thespring.The threeelements definedaboveareshown inFig.2.9. J Fig.2.9Rotationalelements Since the three elements ofrotational systems are similar in nature to those of translational systems noseparatesymbolsare necessary torepresent theseelements. Having defined the basic elements of mechanical systems, we must now be able to write differential equationsforthesystemwhenthesemechanicalsystemsaresubjectedtoexternalforces.Thisis donebyusingtheD' Alembert'sprinciplewhichissimilartotheKirchhoff'slawsinElectrical Networks.Also,thisprincipleisamodifiedversionof Newton'ssecondlawof motion.The D' Alembert's principlestatesthat, "For any body,the algebraic sum ofexternally applied forcesand the forcesopposing the motion inany given directioniszero". Toapplythisprincipletoanybody,areferencedirectionof motionisfirstchosen.Allforces acting in this direction are taken positive and those against this direction are taken asnegative. Let us apply this principle toa mechanical translation system shown in Fig.2.10. AmassMisfixedtoawallwithaspringKandthemassmovesonthefloorwithaviscous friction. An external forcef isapplied tothe mass.Let usobtain thedifferentialequation governing the motion of thebody. MathematicalModellingof PhysicalSystems 17 x f Fig.2.10A mechanical translationalsystem Let us take areference direction of motion of the body fromleft to right. Let the displacement of themassbe x.Weassume that themassisarigidbody,ie,every particleinthebody hasthesame displacement, x.Letusenumerate theforcesacting on the body. ( a)externalforce=f (b)resisting forces: (i)Inertia force,fM= _Md 2 ~ dt dx (ii)Damping force,fB= - Bdt (iii)Spring force,fK= - Kx Resisting forcesaretaken tobe negativebecause they actinadirectionoppositetothe chosen reference direction.Thus, using D' Alembertsprinciple we have, f-Md2x_Bdx-Kx dt2 dt =0 d2xdx M-+B-+Kx=f dt2 dt or..... (2.28) This isthe differentialequation governing themotion of themechanicaltranslation system.The transfer functioncan be easily obtained by taking Laplace transform of eqn (2.28).Thus, Xes) - - = - ~ - - -F(s)MS2+ Bs+ K If velocity ischosen as theoutput variable,we can write eqn.(2.28)as du M- + Bu + K Iu dt =f dt ..... (2.29) Similarly, the differential equation governing the motion of rotational system can also be obtained. For thesystemin Fig.2. I 1,we have d2ede J-2 + B-+ Ke=T..... (2.30) dtdt 18 The transfer functionof thissystemisgivenby 9(s) --=-----T(s)Js2+Bs+K ControlSystems J Fig. 2.11Mechanical rotatinal system Since eqn. (2.28) and eqn. (2.30)are similar, if the solution of one system isknown,the solution for the other system can be easily obtained.Such systems areknown asanalogoussystems.Further, theeqns.(2.15)and(2.19)of theelectricalsystemsshowninFig.2.5and2.6aresimilarto eqns.(2.28) and (2.30) of mechanical systems of Figures 2.10 and 2.11. Hence the electrical systems in Figures2.5and 2.6 arealsoanalogousto mechanicalsystemsin Figures2.1 0 and 2.11. 2.2.4Analogous Systems Analogoussystemshavethesametypeof equationseventhoughtheyhavedifferentphysical appearance.Mechanicalsystems,fluidsystems,temperaturesystemsetc.maybegovernedby the same types of equations as that of electrical circuits. In such cases we call these systems as analogous systems. A set of convenient symbols are already developed in electrical engineering which permits a complex system to be represented by a circuit diagram.The equations can be written down easily for these circuits and thebehaviour of these circuits obtained.Thusif an analogouselectrical circuit is visualised for a given mechanical system,it iseasy topredict thebehaviour of thesystem using the welldeveloped mathematical tools of electrical engineering.Designing and constructing a modelis easierinelectricalsystems.Thesystemcanbebuiltupwithcheapelements,thevaluesof the elements can be changed with ease and experimentation is easy with electrical circuits. Once a circuit isdesigned with the required characteristics, it can be readily translated into a mechanical system.It isnot only truefor mechanical systems but alsoseveral other systems likeacoustical,thermal,fluid andeveneconomicsystems. The analogous electrical and mechanical quantities are tabulated in Table 2.2. Table2.2Analogousquantitiesbasedonforcevoltageanalogy Electrical systemMechanical system TranslationalRotational VoltageVForcefTorqueT CurrentiVelocityuangular velocity (j) ChargeqDisplacementxangular displacement 9 InductanceLMassMMoment ofInertiaJ 11 CapacitanceCCompliance -Compliance -KK ResistanceRDamping coefficientBDamping coefficientB MathematicalModellingof PhysicalSystems 19 Comparing the eqn. (2.15) of the electrical system and eqn. (2.28) and eqn. (2.30) of mechanical systems it iseasy to see the correspondence of the variousquantities in Table 2.2.Since the external forceappliedtomechanicalsystem,fl'ismadeanalogoustothevoltageappliedtotheelectrical circuit, thisiscalled asForce- Voltageanalogy. If themechanicalsystemsarecomparedwiththeelectricalcircuitinFig.2.6,inwhichthe sourceisacurrentsource,forceappliedinmechanicalsystemisanalogoustothecurrent.The analogous quantitiesbased on force-current analogy are given in Table 2.3 The Table 2.3can be easily understood by comparing eqn.(2.19) of theelectrical system with eqns.(2.28)and(2.30)of mechanicalsystems. Table2.3AnalogousquantitiesbasedonForce - Current analogy Electrical systemMechanical system TranslationalRotational CurrentiForcefTorqueT VoltagevVelocityuangular velocity (0 Flux linkages'I'Displacement xangular displacemente CapacitanceCMassMMoment of InertiaJ ConductanceGDamping coefficientBRotational Damping coefficient B I1 InductanceLCompliance -Compliance -KK Nowletusconsidersomeexamplesof mechanicalsystemsandconstructtheir mathematical models. Example 2.4 Writetheequations describing the motion of themechanicalsystem shown in Fig.2.12. Alsofindthetransfer functionXj(s)lF(s). Fig. 2.12A Mechanical system for example 2.4 Solution: The first step is to identify the displacements of masses Mj and M2as Xjand x2 in the direction of the applied external force! Next we write the equilibrium equation for each of the masses by identifying theforcesactingon them.Let usfirstfindouttheforcesacting on massMj 20 Externalforce=f Restoringforces: d2x (i)Inertia forc.e,- Ml-2-1 dt ( .. ).fid2(XI-X2) 11Dampmgorce, - Bl2 dt (iii)Spring force,- Kl(xl- x2) (If xl> x2 thespring expandsandrestoringforceisupwards) ControlSystems (Note: Since top end of dash pot and spring are rigidly connected to mass M2, their displacement isx2 inthedownwarddirection.Similarly,thebottomendsof dashpotspringarerigidly connected themass Ml' they movedownwardby Xl'Therelativedisplacementof thelower end_upper end is(xl-x2)in the downward direction. Hence restoring forces are-Bl d (XI- x2) dt and - Kl(xl- x2) respectively due to thedash pot andspring). Hence the equation of motion is Now for mass M2 External force =Zero Restoringforces: d2 x (i)Inertia force,- M2-2_2 dt ..... (1) (li)Dampingforces, dX2 (a) - B2dt(since one end isfixed and the other isconnected toMlrigidly) (b) - B d(x2- XI)(Ifdx2 > dXI , the motion isin the downwarddirection and 1dtdtdt (iii)Spring forces: (a) - K2x2 thefrictionalforceisin the upwarddirection) (b) - Kl(Xz- Xl)(If Xz> Xl'thespring iscompressedandtherestoringforce isupward) MathematicalModellingof PhysicalSystems 21 Hence the equation of motion for M2is, d2 x2 dX2 d(x2- Xl) M2~ + B 2 ~ + B Idt+K2X2+KI(X2-XI)=0 .... (2) From eqns. (l) &(2), force voltage analogous electrical circuits can be drawn asshown in Fig.2.13 (a). v[t] Fig. 2.13 (a)Force - Voltage analogous circuit for mechanical system of Fig.2.12.Mechanical quantities areshowninparenthesis Force- currentanalogouscircuitcanalsobedevelopedforthegivenmechanicalsystem.Itis giveninFig.2.13(b).Notethatsincethemassisrepresentedbyacapacitanceandvoltageis analogous to velocity, one end of the capacitor must always be grounded so that its velocity is always referredwithrespect totheearth. XI(s) =--F(s) Transferfunction Taking Laplace transform of eqns.(1)and (2)and assuming zeroinitial conditions,we have MIs2XI(s)+BIs[XI(s)- X2 (s)]+KI[XI(s) - X2 (s)]=F(s) or(MI s2+Bls +KI)XI(s) - (BI S+KI)X2 (s) =F(s)..... (3) andM2s2X2 (s)+B2sX2(s) +BIs[Xis) - XI(s)]+K2X2 (s) +KI[ ( ~(s) - XI(s)]=0 or- (Bls +KI)XI(s) +[M2 s2+(B2+BI)s +(K2 +KI)Xis) =0..... (4) Fig. 2.13 (b)Force current analogous circuit for the mechanicalsystem of Fig.2.12 22ControlSystems Solving forXis) in eqn (4),we get Bjs+KjX() Xis) =M2S2+(Bj+ B2)s+(Kj + K2)js Substituting forXis) in eqn.(3) (M/+Bjs+Kj)Xj(s)- Xj(s)=F(s) M2s+(Bj +B2)s+Kj +K2 Xj(s)=M2s2+(Bj +B2)S+Kj +K2F(s) MjS2+ Bj s+KjM2 S2+(Bj + B2)s+(Kj + K2)-(Bj S+Kj)2 Thus Xj(s)M2s2+(Bj +B2}s+Kj +K2 T(s)=F(s)=(Mjs2 +BjS+Kj)[M2s2 +(Bj +B2)S+(Kj +K2)J-(Bjs+Kj)2 Which isthedesired transfer function. Example 2.5 Obtainthe f-vand f-ianalogouscircuitsforthemechanicalsystemshowninFig.2.14.Also write down the equilibrium equations. Fig. 2.14Mechanical system for Ex.2.5 Solution: Letfbe theforceacting on the spring instead of velocity u.The displacements are indicated in the figure. The equilibrium equations are: K(Xj=f B("2 - '(3)+ K (X2 - Xj)=0 B ("3 - '(2)+ M X3=0 Fromeqs(1)and(2),wehave B ("2 - '(3)=f Fromeqs(3)and(4),wehave MX3 =f ..... (1) ..... (2) ..... (3) ..... (4) ..... (5) MathematicalModellingof PhysicalSystems 23 Force Current Analogous Circuit Replacingtheelectricalquantitiesinequations(1),(4),and(5)bytheirforce-currentanalogous quantities using Table2.3,wehave or or or ~ f(v- v)dt=i L'2 G (q,2 - q,3) =i G(V2 - V3)=i .. C'f'3=i CdV3=i dt ..... (6) ..... (7) ..... (8) If iisproducedbyavoltagesourcev,wehavetheelectricalcircuitbasedon f-ianalogyin Fig.2.14(a). G[B] v[u] C[M] Fig. 2.14 (a)F-ianalogous circuit for mechanical systeminEx.2.5 ForceVoltage Analogous Circuit Usingforcevoltageanalogy,thequantitiesineqs(1),(4)and(5)arereplacedby themechanical quantities to get, or or 1 -(q-q )=v C'2 ~f (i, - i2)dt =v C Rh -qJ =v R (i2- i3)=v d2q3 L--=v dt2 ..... (9) ..... (10) 24ControlSystems or di3 L-=v dt ..... (11) If the voltage isdue to a current source i,we have the force voltage analogous circuit isshown in Fig.2.14(b) + v[f] R[B]L[M] Fig. 2.14 (b)Force voltage analogous circuit for the mechanical systemof Ex.2.5 2.2.5Gears Gearsaremechanicalcoupling devicesusedforspeed reduction or magnification of torque.These are analogous to transformers in Electrical systems. Consider two gears shown in Fig.2.15. The first gear, to which torque T 1 is applied,isknown as the primary gear and has N 1 teeth on it.The second gear,which iscoupled to this gear and is driving a load,is known asthe secondary gear and has N2 teeth on it. Fig. 2.15Geartrain The relationships between primary and secondary quantities are on the following principles. 1.The number of teeth on the gear isproportional to the radius of thegear r\N\ -=-2.Thelinear distance traversed along thesurface of each gear issame. If the angular displacements of the two gearsare()1and82 respectively,then r191=r292 r2 9\ ..... (2.31) ..... (2.32) MathematicalModellingof PhysicalSystems 3.Thework donebyone gear issameasthat of the other. If T 1 and T 2arethetorqueson thetwogears, TI 01=T202 Usingeqs(2.31),(2.32)and(2.33),wehave, 1T2N2 -=-=-Recall that,foranidealtransformer of primary tums N 1andsecondary turns N2, VI12NI -=-= V2 IIN2 25 ..... (2.33) ..... (2.34) ..... (2.35) Eqn.(2.35)issimilar to eqn.(2.34) and hence a gear train in mechanical system isanalogous to an ideal transformer in theelectricalsystem. Writing the equilibrium equations for themechanicalsystem of Fig.(2.14),we have ..... (2.36) ..... (2.37) WhereT Misthemotor torque driving theshaft of the gear,T 1isthetorqueavailablefor the primary gear, T2 is the torque at the secondary gear, TL is the load torque and J2 is the moment of inertia of secondary gear and the load. Using eqn.(2.34) in eqn.(2.37),wehave ..... (2.38) Substituting forTIin eqn.(2.36),using eqn.(2.38), But, and JI '91 + BI91 + KI1 +~ .(J2'0'2+ B292 + K2O2 + TL)=TM..... (2.39) N2 = ~ O 2NI 2 26ControlSystems Using theserelationsin eqn.(2.39),we get ..... (2.40) Replacing and Wehave, JI eqe\+ BI9\+ Kleq el + TLeq =TM..... (2.41) ThustheoriginalsysteminFig.2.15canbereplacedbyanequivalentsystemreferredto primary sideasshown in Fig.2.16. Fig. 2.16Equivalent systemreferedtoprimary side . The moment of inertia J2,frictionalcoefficient B2and torsional spring constant K2of secondary side arerepresented by their equivalents referred to primary side by, J12J, r MathematicalModellingof PhysicalSystems The load torquereferred to primary sideisgiven by, _Nl TLeq-TL-N2 27 These equationscan be easily seen to be similar to the corresponding electrical quantitiesin the ideal transformer. All the quantities can also be referred to the secondary side of the gear. The relevant equations are .lzeq+ B2eq 92 + K2eq 92 + TL =TMeq ..... (2.42) where (::r(::J and 2.2.6Thermal Systems Thermal systems are those systems in which heat transfer takes place from one substance to another. They can be characterised by thermalresistanceandcapacitance,analogoustoelectricalresistance andcapacitance.Thermalsystemisusuallyanonlinearsystemandsincethetemperatureof a substance is not uniform throughout the body, it is a distributed system. But for simplicity of analysis, the system isassumed tobelinear and is represented by lumped parameters. (a)Thermal resistance There are two typesof heat flowthrough conductors: Conduction or convection and radiation. Fig.2.17Thermalresistance For conduction of heat flowthrough aspecificconductor,according toFourierlaw, ...... (2.42) 28 where, q=Heat flow,Joules/Sec K=Thermalconductivity,J/sec/m/degk A=Area normal toheat flow,m2 =Thicknessof conductor,m 8=Temperature in oK Forconvectionheattransfer, q=HA (81 - 82) whereH=Convectioncoefficient, k The thermal resistanceisdefined by, d8R=-=-dqKA (Conduction) HA (Convection) The unit of Risdeg seclJ ControlSystems ..... (2.43) ..... (2.44) ..... (2.45) Forradiationheattransfer,theheatflowisgovernedbyStefan-Boltzmannlawforasurface receiving heat radiation from a black body: q=KAE (841 - 842) =A cr(8\ - 842)..... (2.46) where, cris aconstant,5.6697x10-8 J/sec/m2/K4 Kisa constant E is emissivity Aissurface in m2 The radiation resistanceisgiven by d81 R=-d =3degsec/J..... (2.4 7) q4Acr8a where 8a is the average temperature of radiator and receiver. Since eqn. (2.46) is highly nonlinear, itcan be used only forsmall rangeof temperatures. (b)Thermal Capacitance Thermalcapacitance istheabilitytostorethermalenergy.If heat issupplied toabody,itsinternal energyraises.For thesystemshowninFig.2.18, Fig. 2.18ThermalCapaCitance MathematicalModellingof PhysicalSystems where where de C- =q dt Cisthe thennalcapacitance C=WCp W~weight of block in kg Cp ~specificheat at constant pressureinJ/deg/kg 2.2.7'FluidSystems 29 ..... (2.48) ..... (2.49) Fluidsystemsarethosesystemsinwhichliquidorgasfilledtanksareconnectedthroughpipes, tubes,orifices,valvesandother flowrestricting devices.Compressibility of afluidisan important property which influences the perfonnance of fluidsystems.If the velocity of sound in fluidsis very high,compared to thefluidvelocity,thecompressibility can be disregarded.Hencecompressibility effects are neglected in liquid systems. However compressibility plays an important role in gas systems. The type offluid flow,laminar or turbulent, is another important parameter in fluid systems. If the Reynolds number is greater than 4000, the flowissaid tobe turbulent and if the Reynolds number is less than 2000, it issaid tobelaminar flow. For turbulent flowthrough pipes,orifices,valvesand other flowrestricting devices,theflowis foundfromBernoulli's law and isgiven by q =KA~ 2 g ( h l-h2) ..... (2.50) whereq~liquid flow rate,m3/sec K~aflowconstant A~area of restriction,m2 gisacceleration due togravity,mlsec2 hhead of liquid, m Theturbulentresistanceisfoundfrom ..... (2.51) It can be seen that theflowresistance depends on hand q and therefore itisnon linear.It has to be linearised around theoperating point and used over a small range around this point. Thelaminar flow resistanceis found fromthe Poisseuille - Hagen law" 128j..lL hi- h2=nyD4q ..... (2.52) whereh~head,m L~length of the tube,m D~inside diameter of the pipe,m 30 qliquid flow rate,m3/sec mabsoluteviscosity,kg-sec/m2 yfluiddensity kg/m3 ControlSystems Since the flowislaminar,headisdirectly proportional to theflowrateand hence,laminar flow resistanceisgiven by R=128sec/m2 1tYD4 Liquid storage tanks arecharacterised by thecapacitance and isdefined by, dv C=- m2 dh ..... (2.53) ..... (2.54) where vvolume of the liquid tank in m3.Hence thecapacitance ofa tank isgiven by itsarea of cross section at agiven liquid surface. Gassystemsconsistingof pressurevessels,connectingpipes,valvesetc.maybeanalysedby using thefundamentallaw of flowof compressible gases.Again,we have toconsider twotypes of flow: turbulent and laminar flow.For turbulent flow through pipes,orifices,valvesetc.,wehave where (pl-P2)Y 0)flowrate,kg/sec Kflowconstant Aarea of restriction,m2 ygasdensity,kg/m3 ppressurein kg/m2 Turbulent gasflowresistanceis therefore given by R=sec/m2 dw ..... (2.55) ..... (2.56) This is not easy to determine since the rational expansion factor depends on pressure. Usually the resistanceisdetermined fromaplot of pressure against flowrate foragiven device. The laminar gas flow resistanceisobtained using eqn.(2.50). ______________ resistanceR Fig. 2.19Gasresistance andcapacitance The capacitance parameter forpressure vesselsisdefined as dv C=--dp CapacitanceC ..... (2.57) MathematicalModelling0/ PhysicalSystems wherev-+weight of gasinvessel,kg p-+pressurekg/m2 The flowrateand the gaspressure arerelated by thecontinuitylaw: dp C- =0) dt WhereOJistheflowrateinkg/sec. 31 ..... (2.58) Let usnow consider somethennalandfluidsystemsand obtain their transfer functions. Example 2.6 Find thetransferfunctionC(s)of thethennalsystemshowninFig.2.20.Heatissuppliedby yes) convection toacopper rodof diameter O. Fig. 2.20Thermal System Solution: u(temperature) -.. Convection Copperrod C(temperature) 4+--TheThennal resistanceof thecopper rod,fromeqn.(2.45),is; I R=-HA Here Ais thesurface area of therod. HenceA = 7tOL where L isthe length of the rod. I . .R =7t OLdegsec/J The thennalcapacitance of therod,fromeqn.(2.49),isgivenby: C=WC p 7t02L =-4-pCp where,Cp isthe specificheat of copper andpisthe densityof copper. Fromeqn.(2.43),wehave,q=HA(u - c)and fromeqn.(2.48), dc C- =q dt Combining these twoequations, dcI C-=- (u-c) dtR 32 CdcdtRR dc RC- +c=u dt whereRC=Tis the timeconstant of thesystem. dc T-+c=u dt 11t02L T=RC=--x--pCp 1tDLH4 But =DpCp 4H Thusthetransfer functionof thesystemis, C(s) --=-- where U(s)Ts+l DpCp T=--4H ControlSystems Example 2.7 Obtain the transfer functionC(s)for the system shown in Fig. 2.21.c is the displacement of the U(s) piston with massM. u,pressure B,damping f-M,mass Gas

Fig. 2.21A fluid system Solution:Thesystemisacombinationof mechanicalsystem with mass M,damping B and agas systemsubjectedtopressure. The equilibrium equation is Me+Bc=A[U-Pg]..... (2.1) Where Pg is the upward pressure exerted by the compressed gas. For a small change in displacement of mass,thepressure exerted isequal to, P P=- Ac gV where, Pis thepressureexerted by the gaswith avolumeof gasunder the piston tobeV ButPV= WRT where,R isthegasconstant. Mathematical Modellingof Physical Systems andTisthetemperatureof the gas. WRT p =---v-WRT P=--Ac gy2 Substituting eqn.(2)ineqn.(1)wehave, WRT Me+Bc+--2- A 2c=Au Y Thetransfer functionis, C(s)A U(s)MS2+Bs+K WRT2 whereK = --P:. y2 Example 2.8 Obtain the transfer functionfor the liquid level system shown in Fig.2.22. c Fig. 2.22Liquid level system Solution:Thecapacitance of thevessel isC=A where Aisthearea of crosssection of thevessel. The outflow q is equal to, c q=-R where Ris the laminar flow resistanceof the valve. c A c =(u - q)=u - -R c Ac+ - =u R The transfer functionis C(s)R --=---u(s)RAs+l R 33 ...... (2) 34ControlSystems 2.3Block Diagramsof Closed LoopSystems Thetransfer functionof anysystemcan berepresentedbyablock diagramasshownin Fig.2.23. u(s)I G(s) Fig.2.23Block diagram of a system C(s) output A complex system can be modelled as interconnection of number of such blocks. Inputs to certain blocksmaybederivedbysumming or substracting twoor moresignals.Thesumordifferenceof signalsisindicated in the diagram by asymbol called summer,asshown in Fig.2.24(a). + (a)(b) Fig. 2.24 (a)Symbol of a summer (b)Pick off point On the other hand,asignal may be taken fromthe output of a block and given to another block. This point fromwhich thesignal is tapped isknown aspick off point and isshown in Fig.2.24 (b). Asimple feedback system with theassociated nomenclatureof signalsisshown in Fig.2.25. R(s) + Summer E(s) pick off point B(s) Fig.2.25Simple feedback system C(s) G(s)=E(s); Forwardpath transfer function,or plant transferfunction B(s) R(s) =E(s); Transfer functionof the feedbackelements where,R(s);Referenceinput or desired output C(s);Output or controlled variable B(s);Feedback signal E(s);Error signal G(s)R(s);Loop transferfunction This closed loop system can be replaced by asingle block by finding the transfer function Mathematical Modellingof Physical Systems But C(s)=G(s)E(s) E(s)=R(s) B(s) =R(s) R(s)C(s) C(s)=G(s)[R(s) R(s)C(s)] C(s)[1=+= G(s)R(s)]=G(s)R(s) C(s)G(s) --= --'--'--R(s)1 + G(s) R(s) Thistransfer functioncan berepresentedbythesingleblock showninFig.2.26. R(s)--------1.. ao1I1 Ir------...-C(s) Fig. 2.26Transfer functionof the closedloop system 35 ..... (2.59) ..... (2.60) ..... (2.61) ..... (2.62) ..... (2.63) (Note:If thefeedbacksignalisaddedtothereferenceinputthefeedbackissaidtobepositive feedback. On the other hand if the feedback signal is subtracted from the reference input, the feedback issaid tobenegativefeedback). For themostcommoncaseof negativefeedback, C(s) R(s) G(s) 1 + G(s) R(s) ..... (2.64) Theblock diagramof acomplexsystemcanbeconstructedbyfindingthetransfer functionsof simple subsystems.The overall transfer functioncan befoundby reducing theblock diagram,using block diagramreduction techniques,discussedin thenextsection. 2.3.1Block Diagram Reduction Techniques A block diagram with several summers and pick off points can be reduced to a single block, by using block diagramalgebra,consisting of the following rules. Rule1 : TwoblocksG1 andG2 in cascadecan bereplacedbya single block asshown inFig.2.27. Fig. 2.27Cascade of twoblocks Rere z - =G xI' r=G z2 zyy -.- =- =GG xzXI2 y 36ControlSystems Rule 2 : A summing point can be moved from right side of the block to left side of the block as shown in Fig.2.28.

I----.-y= w--u}l W-------I Fig.2.28Moving a summingpoint to left of the block For the twosystems tobe equivalent the input and output of the system should be the same.For the left handsidesystem,wehave, y=G[x]+ W For thesystemon rightsidealso, y=G[x + [W]]=G[x]+ w Rule3:Asumming pointcan bemovedfromleft sideof theblock torightsideof theblock as showninFig.2.29. + x---.t + y=G(x+w) =G[x]+G[w] y Fig. 2.29Moving a summer to the right side of theblock x Y = G[x]+ G[w] Rule 4:APick off point can be moved fromthe right sideof the block toleft side of theblock as shown in Fig.2.30. x

w y= G[x] w=y y= G[x] w= G[x]= y Fig. 2.30Moving a pick off point to left side of the block Rule5:APick off point can be moved fromleft side of theblock to the right side of the block as shown inFig.2.3 1. w ........I------1- ...----1QJtiJ w ........I-------' y= G[x]y=G[x] w=x w=i[Y]=x Fig. 2.31Moving a pick off point to the right side of the block Mathematical Modellingof Physical Systems 37 Rule6:A feedbackloop can be replaced by a singleblockasshown in Fig 2.32. x ~ x-- Y l+GH Fig. 2.32Feedback loopisreplacedby a singleblock Usingthesesixrules.whicharesummerisedinTable2.4,anycomplexblockdiagramcanbe simplified toasingleblockwithaninputandoutput.Thesummingpointsandpickoff pointsare moved totheleft or right of theblocksso thatwehaveeither twoblocksincascade,which canbe combined,or asimplefeedbackloopresults,whichcanbereplaced by a single block. This is illustrated in the following examples. Table 2.4Rulesof Block diagramalgebra Ruleno.GivensystemEquivalentsystem 1.Blocks in cascade x-1 G1 H G2 ~ Y X.. 1G1 G21 .. Y 2.Moving summing point toleft x-1 I + w ~ G.yy GY w X ~ ' 3.Moving summing point to right x ::f 1 G ~ Y wG x-1 G I ~y :iJ4rl G ~ y ' w:J 4.Moving pick off point toleft w G x w;J I G ~ Y X ~ ' 5.Moving pick off point toright w J.. G 6.Absorbing a loop xL: [.: x-1 I+:H~ Y 38ControlSystems Example 2.9 Find theoveralltransfer functionof thesystemshowninFig.2.33usingblockdiagramreduction technique. + R(s)--.....-t C(s) Fig. 2.33Block diagram of a system Solution: StepI:Using rule6,thefeedback loop with G2 and H2isreplaced by a single block asshown inFig.2.33(a). R(s)B(s) Fig. 2.33 (a) G2 -----=--l+GH 22 Step 2:The twoblocksin theforwardpath arein cascade and canbereplaced using ruleIas shown inFig.2.33(b). +

Fig. 2.33 (b) Step 3:Finally,the loopin Fig.2.33(b)isreplaced by a singleblock using rule6. G1G2 R(s) .. 1+ G2 H2 .. G1 G2H1 1+ 1---.----C(s) 1+ G2H2 Fig. 2.33 (c) The transfer functioncan besimplified as, Mathematical Modellingof Physical Systems 39 Example: 2.10 Obtain theoverall transfer functionof thesystemshowninFig.2.34(a). + R(s)----I~ - C ( s ) Fig. 2.34 (a)Block diagram of a system for Example 2.10 Solution: Step1 :Moving the pick off point (2) to the right of block G3 and combining blocks G2 and G3 in cascade,wehave, + R(s)----{,1--------'---.--. C(s) Fig. 2.34 (b) Pick off point (2) moved to right of block G3 Step 2:Moving the pick off point (1) to the right of block G2 G3 R(s)---i 1------'--.-----. C(s) ~ - - - - - - - ~ ~ ~ - - - - - - - - ~ G2G3 Fig. 2.34 (c)Moving pick off point (1) to right of G2 G3 40ControlSystems Step3.'AbsorbingtheloopwithGz G3 andH3,andcombiningitwithblockG1 incascade, wehave, R(s) +IG1 Gz G3 C(s) - l+GGH z33 HIH2 --+-GG G3 23 Fig. 2.34 (d) Absorbing the loop and combiningit withG1 HIH2 The feedbackpathsGG. and-G havethesameinputs and both aresubtracted fromR(s)at 233 the summer 1.Hence they can be added and represented by a single block as shown in Fig.2.34 (d). Step 4.'Finally theclosed loopin Fig.2.34(e) isabsorbed and the simplified block isgiven by GIG2G3 1+G2G3H3 R(s)----+I ____---=c.-::c.-::___ 1---_.---C(s) 1 +GIG2G3 (HI+ H2G2) 1 + G2G3H3G2G3 Fig. 2.34 (e)Finalstepintheblock diagram reduction Thetransfer functionof thesystemis, C(s)GI G2 G3 --= -------'----"'---'''-----R(s)1+G2 G3 H3+GI HI+GI G2 H2 Example 2.11 C(s) Reduce theblock diagram and obtainR(s)in Fig.2.35(a). R(s)--....., I---r-- C(s) Fig. 2.35 (a)Block diagram of a system for Ex.2.11 Mathematical Modellingof Physical Systems 41 Solution: Step1 :Move pick off point (2) toleft of G2 and combine G2 G3 in cascade.Further G2 G3 and G4 havesameinputsandtheoutputsareaddedatsummerIII.Hencetheycanbeaddedand representedby asingleblock asshown inFig.2.35(b). R(s) C(s) Fig.2.35 (b)Block diagram after performing step1 Step 2: Moving thepick off point (1)toright of block (G2 G3 +G4),we have R(s)AI-------r--.--C(S) GG+G 234 Fig. 2.35 (c)Block diagram after step 2 Step 3: Absorbing loop with (G2 G3 + G4)and H2 R(s) +G1 (G2 G3 +G4) ... -1+H2(G2 G3 +G4) C(s) GH 21 '""'-GG+G 234 Fig. 2.35 (d)Block diagram after step 3 The transfer functionof thesystemis Fig. 2.35 (e)Simplifiedblock of the sytem 42ControlSystems C(s)=___G_I_C_G_2_G_3_+_G_4_) __ RCs)1+H2 CG2 G3 +G4)+HI GI G2 2.4Signal Flow Graph Representation of Control Systems Anotherusefulwayof representatingasystemisbyasignalflowgraph.Althoughblockdiagram representationof asystemisasimplewayof describingasystem,itisrathercumbusometouse block diagramalgebra andobtain itsoverall transfer function. A signalflowgraph describeshow a signal gets modified as it travels from input to output and the overall transfer function can be obtained veryeasilybyusingMason'sgainformula.Letusnowseehowasystem can berepresentedbya signal flow graph. Before wedescribe a system using a signal flow graph,let us define certain terms. 1.Signal flow graph:It isa graphical representation of the relationships between the variables of asystem. 2.Node:Every variable in a system is represented by a node. The value of the variable isequal to the sum of the signals coming towards the node.Its value is unaffected by the signals which aregoingaway fromthenode. 3.Branch:Asignaltravelsalongabranchfromonenodetoanothernodeinthedirection indicated on the branch. Every branch is associated with a gain constant or transmittance. The signal gets multiplied by this gain asit travelsfrom one node to another. In the example shown in Fig. 2.36, there are four nodes representing variables xi' x2'x3and x4. Thenodesarehereafterreferredbytherespectivevariablesforconvenience.Forexample, nodes1 and 2 are referred to asnodes Xland x2 respectively. The transmittance or gain of the branches area12,a23 anda42In thisexample, x2 =al2 Xl+ a42 x4 Fig.2.36Exampleshowsnodes,branchesandgainsof tHebranches The value x2 isunaffected by signal going away fromnode Xltonode x3Similarly x3=~ 3x2 4.Input node:It isa node at which only outgoing branches are present.In Fig.2.37 node Xlis an input node.It isalsocalled assource node. Xlal2 o-__~__~__~__~__~ ~ ~ ~__~__~ X 5 Fig. 2.37Anexample of a signal flow graph Mathematical Modellingof Physical Systems 43 5.Output node:Itisa nodeatwhich onlyincoming signalsarepresent. Node Xsisan output node.In some signal flow graphs,this condition may not besatisfied by any of the nodes.We canmakeanyvariable,representedbyanode,asanoutputvariable.Todothis,wecan introduceabranchwithunitgain,goingawayfromthisnode.Thenodeattheendof this branch satisfies the requirement of an output node.In the example of Fig.2.37,if the variable x4 is to be made an output variable, a branch isdrawn at node x4 as shown by the dotted line in Fig.(2.37) to create a node x6.Now node x6has only incoming branch and itsgain isa46 =1. Therefore x6isan output variable and since x4 =x6'x4 is alsoan output variable. 6.Path:It isthe traversalfromone node to another node through thebranches in thedirection of thebranchessuch thatnonodeistraversedtwice. 7.Forward path:It isa path frominput node tooutput node. In the example of Fig. 2.37, Xl- x2 - x3 - x4 - Xsis a forward path.Similarly xl- x2 - Xsis also aforwardpath 8.Loop:It isa path starting and ending on the same node.For example,in Fig.2.37, x3- x4 - x3 isa loop.Similarly x2 - x3- x4 - x2 is also a loop. 9.Non touching loops:Loops which have nocommon node, are said to benon touching loops. 10.Forward path gain: Thegain product of the branchesin theforwardpathiscalledforward path gain. 11.Loop gain:Theproduct of gains of branchesintheloopiscalled asloopgain. 2.4.1Construction of aSignal Flow Graph foraSystem A signal flowgraph for a given system can be constructed by writing down the equations governing the variables.Consider thenetworkshown inFig.2.38. + R2Vo(s)output Fig.2.38A network for constructinga signalflowgraph Identifying the currentsandvoltagesinthebranchesasshowninFig.2.38;wehave Vis)=[Il(s)- I2(s)]sL I2(s)=[V2(s) - Vo(s)]Cs Vo(s)=12(s)R2 44ControlSystems Thevariables11(s),12(s),V1(s),V2(s)andVo(s)arerepresentedbynodesandthesenodesare interconnectedwithbranchestosatisfytherelationshipsbetweenthem.Theresultingsignalflow graphisshowninFig.2.39. Fig. 2.39Signal flow graph for the network inFig.2.38 Now,consider theblock diagramgiveninFig.2.40. .......--(C(s) Fig. 2.40Block diagram of a system The relationship between the various variables indicated on the diagrams are: X1(s)=R(s)- H1(s) - His) Xis) X2(s)=G1(s)X1(s)- H3(S)C(s) X3(s)=Gis) X2(s) C(s)=G3(s)X2(s) Signal flowgraph can beeasily constructed forthisblock diagram,asshown in Fig.2.41. -H3 R(s)C(s) Fig. 2.41Signal flow graph for the systeminFig.2.40 2.4.2Mason's Gain Formula The transfer function (gain) of the given signalflowgraph can be easily obtained by using Mason's gain formula.Signal flowgraphs wereoriginated byS.l. Mason and hehasdeveloped aformula to obtain the ratioof output to input,called as gain,for the given signal flowgraph. Mathematical Modellingof Physical Systems 45 Mason's gain fonnula is given by, output variable1:Mk~ k T=input variable=k~..... (2.65) whereMkisthe J(hforwardpath gain,~isthe detenninant of the graph,given by, ~=1 -1: Pml+ 1:Pm2.(-ll 1:Pmr. (2.66) whereP mris theproduct of thegains of mth possible combination of r non touching loops. or~=1 - (sum of gains of individual loops) + sum of gain products of possiblecombinations of 2non touching loops) - (sum of gain productsof allpossible combinationsof 3 nontouchingloops)+ .......... (2.67) and ~ kis the value of ~for that part of the graph which is non touching with kthforward path. To explain the use of this fonnula, let us consider the signal flow graph shown in Fig. 2.42, and findthetransferfunctionC(s)/R(s). R(s)gl2 Fig.2.42Signalflowgraph toillustrate Mason'sgainformula Let usenumerate theforwardpaths andloopsof this flowgraph. 1.Forward paths 2.Loops MI =gl2 g23g34g4SgS6g67 M2 =gl2g23g36g67 M3 =gl2g23g34g46 LI= g23h32 ~=g23g34h42 L3=~ shS4 L4=gS6h6s Ls=g46h6shS4 L6=g23g36h6shS4h42 &7C(s) 46ControlSystems 3.Twonon touching loops Out of thesix loops there are 3 combinations of two non touching loops.Their gain products are: P12= LIL3= g23g4Sh32hS4 P22= LIL4= g23gS6h32h6s P32= LILs= g23g46h32h6shS4 4.Threenon touching loops Thereisnocombinationsof 3 non touching loopsor 4nontouchingloopsetc. ..Pm3=Pm4=O Let usnowcalculate thedeterminant of theflowgraph. Ll=1 - (LI + L2+ L3+ L4+ Ls+ L6) + PI2 + P22 + P32 =1 - (g23h32+ g23g34h42+ g4shS4+ gS6h6S + g46h6shS4 + g23g36h6shS4h42) + (g23g4Sh32hS4+ g23gS6h32h6s + g23g46h32h6shs4) andLlI= valueof Llwhich isnon touching with MI' Eliminate all terms in Llwhich have any node in common with forward path MI' All loops LI to L6 haveatleast one nodein common with MIandhence, LlI=1 Similarly Ll2= valueof Llwhich is non touching with M2. The loop L3isnot having any common node withMzandhenceeliminating allloopsexcept L3 wehave, Ll2=1 - L3=1 - g4ShS4 Similarly,Ll3=1 Applying Mason's gain formula (eqn.2.65) wehave C(s)M[ Ll[+ M2Ll2+ M3Ll3 T(s)=R(s)=Ll Example 2.12 Vo(s) Find thetransferfunctionV;(s)forthenetwork shownin Fig.2.38. Solution:The network and its signal flow graph are reproduced in Fig. 2.43(a) and (b) for convenience. Step1 :Forward paths: Thereisonly oneforwardpath. 1 PI=- . sL.Cs.Rz R[ LCs2R2 R[ Mathematical Modellingof Physical Systems (a)The network Step 2:(i) Loop gains Fig. 2.43 L=_Ls IRI ~=Cs(- sL)=- s2LC L3=- ~Cs (ii) Product of gainsof two non touching loops P12= LI.L3 sL =- - (- R2cs) Rl R22 =R. LCs 1 Step 3 :Thedeterminant of the graph (b)Itssignalflowgraph LsR2 Ll=1 +_+S2 LC+R2 Cs+- LCs2 RIRI andLlI=1 Step4:Thetransfer functionis LCs2 R2 Vo(s)_RI VI (s)- 1 +Ls+ s2LC + RCs +R2LCs2 R2R II 47 48ControlSystems Example 2.13 Find thetransfer functionfortheblock diagramof Fig.2.40. Solution: Theblock diagramanditssignalflowgrapharereproducedinFig.2.44(a)and (b). R(s)-.......--t1--'---' C(s) Fig.2.44 (a)Block diagram of a system R(s) 0-___---0-.......-0-___---0-.......---+--0 C(s) Fig.2.44 (b)Signal flow graphof block diagram inFig.2.44 (a) Step 1 :Forward path gains PI=GIG2G3 Step2:Loop gain LI=-GIHI L2=-GI G2H2 L3=-G2 G3 H3 Twoor morenon touching loopsarenotpresent hence P mr=0forr=2,3,..... Step 3:Thedeterminant of the graph Ll=1 + GI HI+ GI G2 H2+ G2 G3 H3 Lli=1 Step'Transfer function Mathematical Modellingof Physical Systems 49 Example 2.14 DrawthesignalflowgraphfortheblockdiagramgiveninFig.2.45(a)andobtainthetransfer functionC(s)/R(s). R(s) Fig. 2.45 (a)Block diagramof a system R(s)o---t_-O-__-0:It is denoted by f(t)=Au (t).If A =1,itiscalled aunit step function. 3.2.3Ramp signal Arampsignal isshown in Fig.3.3. A f(t) 1.0~ t Fig. 3.3A Ramp Signal. Itiszerofort 2,it isdifficult tofindthe rootsanalytically.Numerical methods forrootdetermination forhigher order polynomialsisquitecumbersome.Hence,algebraiccriteria are developed to find out if any roots lie in the right half of s-plane. The actual location of the roots is unimportant. 4.3NecessaryConditionsforStability In section 4.2 we have seen that the system will be stable if the roots of the characteristic equation lie in theleft half of s-plane.Thefactorsof the characteristic polynomial D(s) can have termslike (s+ 0), (s+O"k)2+rof where0"1arepositiveand real.Thus D(s) =ao sn+al sn - 1 +.....+an ..... (4.4) =ao IT(s + 0"1)IT{(s+ O"k)2+ rof}..... (4.5) Since0"1andO"kareallpositiveandrealtheproductineqn.(4.5)resultsin allpositiveandreal coefficientsinthepolynomialof s.Thusif thesystemisstableallthecoefficients,ai'mustbe positive and real. Further,since thereare nonegative termsinvolved in theproduct of eqn.(4.5),nocancellations can occur and hence nocoefficient can be zero.Thus none of thepowers of s in between the highest andlowest powers of s must bemissing.But,if a root ispresent at theorigin aniszero. 134ControlSystems There isone exception to thiscondition,namely,all theodd powers of s,or all the even powers of s,may bemissing.Thisspecialcaseoccursif thecharacteristicequation containsrootsonlyon thejaraxis. If there isnoroot at theorigin,thecharacteristic polynomial isgiven by O(s) =IT(s2+ ro?) which will yield a polynomial with even powersof s only.On the other hands,if it hasa root at the origin,in addition torootsonjaraxis only,O(s) isgiven by, O(s)=sIT(s2+ ro?) andO(s)willhaveonlyoddpowersof s.Sincesimplerootson jaraxisarepermitted,wemay conclude thatif any power of s ismissing in O(s),alleven powersor alloddpowersof s maybe missing forstablesystems.In conclusion,it can bestated that the necessary conditions forstability of asystem are: 1.The characteristic polynomial O(s) =0 must haveall coefficients real and positive. 2.None of thecoefficients of thepolynomial should be zero except the constant an. 3.Noneof thepowersof sbetweenthehighestandlowestpowersof sshouldbemissing. However allodd powersof s,or alleven powers of s may be missing. It is to be emphasised that these are only necessary conditions but not sufficient. All stable systems must satisfy theseconditions but systemssatisfying all these conditions need not be stable. For example, O(s) =s3+ s2+ 3s + 24 =0 =(s- 1 + j2.J2)(s - I - j2.J2)(s + 3)=0 Eventhough,O(s)hasallitscoefficientspositiveandnopowerof sismissing,the complexrootshaverealpartswhicharepositive.Hencethesystemwiththeabovecharacteristic equation is not stable. The necessary conditions helpustoeliminate polynomials with negativecoefficientsor missing powers of s by visual inspection only. If the characteristic equation satisfies all the necessaryconditions, itisapossiblecandidateforexaminingfurther,forstability.A.HurwitzandE.J.Routhhave independentlyestablishedtheconditionsforstabilityof asystemwithoutactuallyfindingout the roots.Thecriteria isknown asRouth-Hurwitz criterion forstability. 4.4Routh- HurwitzStabilityCriterion This criteria determines how many roots of the characteristic equation lie in the right half of s-plane. This test also determines all the rootson thejaraxis, so that their multiplicity can be found out. The polynomial coefficients are arranged in an array called Routh array.Let the characteristic equation be given by, Stabilityof Systems 135 The Routh array is constructed asfollows.Each row is identified in the descending order of powers of s.Thefirstrow containscoefficientsof alternatepowersof sstarting with thehighest power sn. The secondrowcontains coefficientsof alternatepowersof sstartingwiththesecondhighest power sn - I.Theother rowsare constructed in a systematic way asindicated in the procedure given below. sn _.a4 an sn-I al_ ... as sn-2 - yb3 sn- 3 ccc2'"c3 sn-4 dldz Sl11 sOa n The coefficients in sn - 2 row are obtained asfollows: b= a,a2 -aOa3 I a, b= a,a4 -aoas 2 a, a, Thecoefficientsof sn - 3rowareobtainedinasimilarway,considering thecoefficientsof the previous tworowsasfollows: b,a3 b2 c= I b, blas- and soon.c= 2 bl Similarly,the coefficients of any particular row can be obtained by considering the previous two rowsandforming theproducts asbefore.If any element in arow ismissing,it isregarded aszero. There willbe only 2 entries in the s2row and one element each in s I and sOrows. The Routh array is constructed untilthe sOrow iscomputed.The Routh Hurwitz criterion isstated asfollows: For a system to be stable, it is necessary and sufficient that each entry in the first column of Routh array,constructedfromthecharacteristicequation,bepositive.If anyentryinthefirstcolumnis negative,thesystemisunstableandthenumber of rootsof thecharacteristicequationlyingin the righthalf of s-plane isgivenby,the number of changesin thesign of entriesin thefirstcolumn of Routharray. 136ControlSystems Observe that the Routh Hurwitz criterion tells uswhether thesystem isstable or not.It does not giveany indication of the exact location of the roots. Example 4.1 Consider the characteristicequation, D(s) =S4+ 2s3 + 8s2 + 4s + 3 =0 Comment on its stability. Solution: Let usconstruct theRoutharray. S4 1 s3 2 s2 2x8-1x4 2 sl 6x4-2x3 6 sO 3 =6 =3 8 4 2x3-1xO 2 =3 3 o(.:Since thereis no entry in the3rd column it istaken aszero) (The entry in this row will always be an) Consider theentries in thefirst column,1,2,6,3,3. All arepositive and therefore,the system is stable. Example 4.2 Examine the characteristic equation D( s)=s4+ 2s3 + s2+ 4s+ 2 =0 for stability. Solution: Constructing theRoutharray,wehave S4 112 s3 24 S2 2-4 --=-1 2 2 SI 8 sO 2 The firstcolunm entriesare1,2,-1, 8 and2.Oneof thecoefficientsisnegativeand hencethe systemisunstable.Also,therearetwosignchanges,2to - 1 and-1to+8.Hencetherearetwo rootsof the characteristicequation in the right half of s-plane . Stabilityof Systems 137 4.4SpecialCases There are twospecial caseswhich occur in theconstruction of Routharray.Whenever they occur it isnotpossibletocompletethetable,inaroutineway.Letusseehowthesespecialcasescanbe tackled in completing the table. (i)First case: Sometimes, the first entry in a particular row may tum out tobe a zero.Since the calculation of next row entries involves division by this zero, the process of constructing the Routh array stops there.Toovercome this difficulty, thefollowingproceduresmay beadapted. (a)First method: Replace the zero in that row byE.Proceed with the construction of the table. Consider the entries of the firstcolumn of thearray by lettingE~0 fromthe positiveside. Example 4.3 Consider the characteristic equation D(s) =s5+ S4+ 3s3 + 3s2 + 6s+ 4 Comment on the stability. Solution: Construct theRoutharray. s5 1x3-1x3 1 =0 3 3 6-4 -1- =2 6 4 Now thereisa zeroin the s3row in the first column.Replace this byE andproceed. S5 36 S4 34 s3 E2 s2 3E-2 E 4 6E-4 ---4E 6E-4- 4E2 sl E 3E-23E-2 --E sO 4 138ControlSystems Letting E~0for the entriesin thefirstcolumn we have, It E =0 E ~ O It 3E-2 E ~ O -- =-00 E lt 6E-4-4E2 E ~ O =2 3E-2 sO 4 Hencetheelementsof pt columnof Routharrayare1,1,0,- 00,2and4.Therearetwosign changes and hence there are two rootsin theright haIf of s-plane.Thesystem thereforeisunstable. (b)Second method : (i)Replace s by.!..in D(s) and apply the Routh criterion for the resulting equation in z. z Considering Ex.4.3again and replacing s by.!..,we have, z 11336 -+-+-+-+- + 4 =0 Z5Z4Z3Z2Z Rearranging theterms, 4z5 + 60 + 3z3 + 3z2 + z + 1 = 0 Routharray. z54 o6 18-12 --=1 6 3-2 --=1 1 1 --1 _3- =-2 1 3 3 3 1 Therearetwosign changesin thefirstcolumnof theRoutharrayforthismodified characteristicequatipn.Hencetherearetworootsin therighthalf of z-plane.This implies that there are also two rootsin the right half of s-plane for theoriginal system. Stabilityof Systems 139 (ii)Secondcase: Forsomesystems,aparticularrowmaycontainallzeroentries.Thishappenswhenthe characteristic equation contains roots which are symmetrically located about real and imaginary axes, namely: (i)oneor morepairs of rootson the j m-axis (ii)oneor morepairs of realroots with oppositesigns,and (iii)oneormorepairsof complexrootswiththeirmirrorimagesaboutthe jm-axis,together forming quadrates in the s-plane. The polynomial whose coefficients are the entries in the row above the row of zeros is called an auxiliary polynomial and the roots of thispolynomial give these symmetrically located roots. Since a row contains all zero entries, the Routh table cannot be constructed further.To overcome this, the row of zeros is replaced with the coefficients of the differential of the auxiliary equation. Thisauxiliaryequationisalwaysapolynomialwithevenpowersof s,sincetherootsof this polynomial occur alwaysin pairs.The procedure isillustrated with an example. Example 4.4 Consider O(s) =S6+ S5+ 6s4 + 5s3 + 10s2+ 5s + 5 Obtain the number of rootsin the RHSof s-plane. Solution: Routh Table s6 6105 s5 55 S4 6-5 -=1 55 1 x3 00 TheRouthtableconstructionprocedurebreaksdownhere,sincethes3rowhasallzeros.The auxiliary polynomial coefficients are given by the s4 row.Therefore the auxiliary polynomial is, A(s)=S4+ 5s2 + 5 dA(s) -- =4s3 + lOs ds 140ControlSystems Replacing the s3rowin theRouth table with the coefficientsof~ ~ s ), wehave S6 6105 .s5 55 S4 55 s3 410 s2 20-10 4 =2.55 sl 25-20 2.5 =2 sO 5 Examining the first column of this table we see that there are no sign changes. But since there are symmetrically located roots we have tofind these rootsfor concluding about the stability.Factoring the auxiliary polynomial, we have the root as, j 1.1756 andj 1.902 These are roots on thejli}-axis and are simple.Therefore the system has no roots in the right half plane and the roots onjli}-axisare simple. Hence the system is limitedly stable. Example 4.5 Comment on the stability of thesystem with the following characteristic equation. D(s) = s6+ s5+ 7s4 + 6s3 + 31s2 + 25s+ 25 Solution: Since D(s) satisfies allnecessary conditionslet usconstruct theRouth table. 1 o 7 6 6 o 31 25 25 25 SincetheRouthtableterminatesprematurelyandthereisarowof zeros,letusconstructthe auxiliary polynomial. A(s) = s4+ 6s2 + 25 d ~ ~ S )=4s3+12s Stabilityof Systems Continuing the Routh table s67 S5 6 s4 6 s3 412 s2 325 Sl -64 -3 sO 25 31 25 25 25 141 There are two sign changes in the first column of the Routh table and hence the system is unstable. Let usfindthesymmetrical rootspresent,byfactoring theauxiliary polynomial A(s). A(s) =S4+ 6s2 + 25 =(S2+ 5)2+ 6s2 - 10s2 =(s2+ 5)2 - 4s2 =(s2+ 2s + 5)(s2 - 2s + 5) =(s + 1 + j2) (s + 1 - j2) (s- 1 + j2) (s- 1 - j2) Hence wehave symmetrically placed rootsout of which two arein the right half of s-plane. The location of therootsareshown in Fig.4.3. Fig. 4.3Roots formirga quadrate jro j2s-plane -111+1(j :Ie:Ie j2 If after completing the Routh table, there are no sign changes, the auxiliary polynomial will have roots only on the j lV-axis. They can be found out by factoring the auxiliary polynomial. The procedure todeterminestabilityof asystemandfindthenumberof rootsintherighthalf of s-plane,using RouthHurwitzcriterion, issummarised asfollows. Step 1:Thecharacteristic equation isexamined fornecessary conditions viz. (i)All the coefficients must be positive (ii)Nocoefficientof D(s)iszerobetweenthehighestandlowestpowersof s.If these conditions aresatisfied,gotostep2. 142ControlSystems Step 2:ConstructtheRouthtable.Examinethefirstcolumn .of thetable.If thereare'm'sign changesin thecolumn,thereare mrootsin theright half of s-plane and hence thesystem isunstable. If in any row,thefirstentry isazeroandatleast oneother element isnot a zero,it isnot possibletoproceedfurtherincompletingthetable.Onceazeroor anegativeentryis present in the first column, it can beconcluded th.at the system isunstable. If it isrequired tofmd the number of roots in the RHS of s-plane go to step 3. If all the entries in a row are zero,gotostep4. Step 3:Replace thezeroby asmallpositivenumberE.Completethe table.LetE~0in allthe elements of first column entries involvingE. Find the signs of these elements. The number of sign changes gives thenumber of rootsin theRHSof s-plane. Step 4:If all the elements of a row are zeros, it indicates that there are roots which are symmetrically situatedin thes-plane.Thelocation of theserootscan bedeterminedbyconsidering the row above the row of zeros. An auxiliary polynomiaIA(s) is constructed with the coefficient as the entries in this row. It is invariably an even polynomial in s. The roots ofthis polynomial givesthesymmetricallysituatedroots.Replacetherowof zeroswiththecoefficientsof the differential of the polynomial A(s). Complete the table now. Step5:If there are m sign changesin the first column of the table, there are m rootsin the RHS of s-planeandthesystemisunstable.Thereisnoneedtofindthesymmetricallysituated roots.If therearenosignchangesgotostep6. Step6:Factorisethepolynomial A(s).Sincetherearenorootsinthe RHSof s-plane.A(s)will contain rootson the jw-axisonly.Find these roots.Ifthese rootsaresimple, thesystem is limitedly stable. If any of these roots is repeated, the system isunstable. This concludes the procedure. Inmanysituations,theopenloopsystemisknownandevenif theopenloopsystemisstable, once the feedback loop is closed there is a chance for losing the stability. The forward loop invariably includes an amplifier whose gain can becontrolled. It istherefore desirable to know the range of the valuesof thisgain Kformaintainingthesysteminstablecondition.An exampleisconsidered to explain the procedure. Example 4.6 Findtherangeof valuesof KfortheclosedloopsysteminFig.4.4toremainstable.Findthe frequency of sustained oscillations under limiting conditions. R(s) +K C(s) S(S2+ S + l)(s + 3)(s + 4) Fig. 4.4A closed loop system for Ex.4.6. Stabilityof Systems Solution: The closedloop transfer functionisgiven by 0 17.625 (238.875+K)(96-K)-17.625 K> 0 141 K>O 12 K 143 144 ControlSystems If condition (iii) is satisfied condition (i)is automatically satisfied. Let us find out for what values of K, condition (ii) will be satisfied. - K2- 142.875K + 22932 - 2485.125K> 0 orK2+ 2628K- 22932 < 0 (K - 8.697)(K +2636.7) < 0 SinceK>0,theabovecondition issatisfiedforK< 8.697.Thustherangeof valuesof Kfor stability is0 < K < 8.697. IfK> 8.697 the entry in slrow will be negative and hence these will be two roots in the RHSof s-plane.Thesystemwillbeunstable.If K 0 (ii) (iii) or 100K-10>0orK>O.l 20K [lOOK -10 - 600K2) 10K-1>0 600 K2- 100 K + 10 < 0 60 K2- 10K + 1 < 0 Thisisnotsatisfied forany realvalue of Kas can be seen by factoring the expression. Hence for novalueof K,the system isstable. 145 Let us consider an open loop system which is unstable and find the values of the amplifier gain to obtain closed loop stability. Example4.8 Comment on the stability of theclosed loop system asthe gain K ischanged in Fig.4.5. R(s) + K(s + 4)1 C(S r--(S2+ 5s + 6)(s -1)- s+2 Fig. 4.5Closedloop system for Ex.4.8 Solution: The open loop system is clearly unstableasit has a pole s =1 in the RHSof s-plane. Let usexamine whether it can be stabilised under closed loop operation by a suitable choice of theamplifier gain K. Thecharacteristicequation isgiven by, D(s)= 1 + G(s)H(s)= 0 K(s + 4) 1+=0 (s + 2)(S2+ 5s + 8)(s -1) S4+ 76s3 + 11s2+ s(K - 2) + 4(K - 4) =0 146 Routhtable: S4 111 S3 6(K-2) S2 68-K 6 4(K-4) ( 68K6-10}K - 2) - 24(K - 4) 68-K 6 4(K- 4) For stability (i)68 - K > 0i.e.,K < 6