chapter plex roots and the characteristic eqn章節(jié)根特征方程

H62SPCChapter5SingleLoopFeedbackSystemandRootsoftheCharacteristicEquationTransferFunctionConvenientrepresentationofalinear,dynamicmodel.Atransferfunction(TF)relatesoneinputandoneoutput:Thefollowingterminologyisused:xinputforcingfunction“cause”youtputresponse“effect”InverseLaplaceTransform

Thefollowingtransformallowsustomovebackintothetimedomaintisaconstantforthepurposeofintegration‘s’isacomplexvariableandthereforemakestheintegrationprocesscomplicated-ajobformathematicians!Fortunately,itispossibletousetablesofinverseLaplacetransformstoconvertbackintothetimedomainwithouthavingtocomputethecomplexintegrals.Thisisthemethodthatweuse.LaplaceTransformtablesPartialFractions:Revision

Findthetransformfunctionh(t)/H(s)Steps:KeepallresistancesRthesameReplaceallvoltagesv(t)

byV(s)Replaceallcurrentsi(t)byI(s)ReplaceallinductancesLbysL

ReplaceallcapacitorsCby1/sC

ApplyKirchhoff’sLawstodetermine fromwhichthefrequencyresponse,amplitudeand phaseresponsescanbeobtained.Kirchoff’sCurrentLawsConsidertheKCLintimedomain:ApplytheLaplacetransform:Kirchoff’sVoltageLawsConsidertheKVLintimedomain:ApplytheLaplacetransform:BasicSingleLoopFeedbackSystemG(s)–Forwardpathtransferfunction；

H(s)–FeedbackpathtransferfunctionC(s)–ControlledoutputsignalR(s)–ReferenceinputsignalE(s)–ActuatingerrorsignalC(s)R(s)G(s)+-H(s)E(s)ClosedLoopTransferFunctionC(s)R(s)G(s)+-H(s)E(s)

ClosedloopTransferFunction(CLTF)Feedback:PositiveandNegativePositiveFeedbackSystemsMorebegetsmoreLessbegetsless.Examples?NegativeFeedbackSystemsMorebegetslessLessbegetsmore.Examples?TransferFunctionC(s)R(s)G(s)+-H(s)

Frombefore:Thebelowequationisastheclosed-looptransferfunctionofthissystem(CLTF):UnityFeedbackControlSystemC(s)R(s)G(s)+-E(s)G(s)–Forwardpathtransferfunction； Open-looptransferfunction C(s)–ControlledoutputsignalR(s)–ReferenceinputsignalE(s)–ActuatingerrorsignalWhatisitsCLTF???????CharacteristicEquationofatransferfunctionC(s)R(s)G(s)+-H(s)

Frombefore:Thecharacteristicequationofalinearsystemisobtainedbyequatingthedenominatorpolynomialofthetransferfunctiontozero,thusthecharacteristicequationofthissystemis:

Therootsofthisequationdeterminetheresponseofthesystemtochangesinthereference,R(s)TheserootsareZEROS!1stand2ndsystem:TransferfunctionandcharacteristicEquationWriteoutthetransferfunctionandcharacteristicequationsforbothcircuits!ResponsetochangesininputtimeVoltage10VOnetypeofresponse(2ndOrder)Anothertypeofresponse(1stOrder)Referencevalue:Responsetoastepchangeininput,R(s):TheshapeofthisresponseisdeterminedbyfindingtherootsoftheCharacteristicEquationExampleIC(s)R(s)

Oursystem:OurInput:Astepinputof10unitstime10

Provethisyourself!

OurOutput:InputTransferFunction

Solvingtheseweget:A=10,B=-10,C=-20 [Proveyourself!]

NowweneedtomanipulatethisabitsothatwecanuseInverseLaplaceTransformTables

ThisoneiseasyThisoneneedsabitofworkExampleII

ExampleIII

1

FromInverseLaplaceTransformTablesComparingthesolutionwiththeCharacteristicEquation

Ourcharacteristicequationwas:

C(s)R(s)

Notethatc(t)hastheexponentialdecayrateequaltotherealpartoftheroots(-1),andthattheoscillatorypartc(t)hasafrequencyequaltothecomplexpartoftheroots(2π/2=π)ThisiswhythedenominatorofthetransferfunctioniscalledtheCharacteristicEquation

-Therootsofitdeterminetheshapeofthesystemresponse.RealandComplexRootsofthecharacteristicequationRealRootsoftheC.E

Thisgivesrisetoanexponentialdecaycomponentinthetimedomain.i.e.K=Constant

=DecayrateKf(t)timeComplexRoots

Thisgivesrisetoatransientterminthetimedomainofthetype:

Hence,forapairofcomplexroots,definedbys2+as+b=0,thedecayrateandthedampedfrequencycanbefoundby:

BOTH

anddeterminetheresponseofthesystemResponseExamplesttttSame,differentSame,differentComplexRoots:Relationships

Ifwecomparethesetwoforms:

Comparingthecoefficients:

ComplexRoots:RelationshipsII

ProveYourself!Example:Second-orderCircuit

InitialConditions

Zero-inputResponsetocharacteristicequation

Rootsofthesecond-ordersystem

PracticalCases:

CaseA:tworeal,unequalroots

CaseB:tworeal,equalrootsCaseA:IF(????)2?4????=0,tworeal,equalroots(??1=??2=???)R=4Kohm??1=??2=?2000Correspondingly,thetime-domainsolutionshouldbe:??????=??1???2000??+??2?????2000??;??≥0Takingtheinitialconditionsintotheconsideration:????0=15=??1??????(0)????=??0??