DefinitionofstabilityTheRouth-HurwitzStabilityCriterionDesignExamplesTheproblemofstabilityofthecontrolsystemiscentraltocontrolsystemdesignAnunstableclosed-loopsystemisgenerallyofnopracticalvalueThebounded-inputandbounded-outputstabilityThischaptersolvestwoproblemsDefinitionofstabilityHowtodeterminethestabilityofasystemAstablesystemisdefinedasasystemwithabounded(limited)systemresponsetoaboundedinput.Linearsystemisstableifandonlyiftheabsolutevalueofitsimpulseresponse,g(t),integratedoveraninfiniterange,isfinite.Thelocationinthes-planeofthepolesofasystemindicatestheresultingtransientresponse.Thepolesintheleft-handportionofthes-planeresultinadecreasingresponse-stablesystemThepolesofthejω-axisandintheright-handplaneresultinaneutralandanincreasingresponse-unstablesystemIntermsofthelinearsystems,weknowthatthestabilityrequirementmaybedefinedintermsofthelocationofthepolesoftheclosed-looptransferfunctionTheclosed-loopsystemtransferfunctioniswrittenaswhereq(s)=Δ(s)=0isthecharacteristicequationwhoserootsarethepolesoftheclosed-loopsystemTheoutputresponseforanimpulsefunctioninput(wheren=0)is:Toobtainaboundedresponse,thepolesoftheclosed-loopsystemmustbeintheleft-handportionofthes-plane.Anecessaryandsufficientconditionforafeedbacksystemtobestableisthatallthepolesofthesystemtransferfunctionhavenegativerealparts.Thesystemisnotstableifnotallpolesareintheleft-hands-plane.Ifthecharacteristicequationhassimplerootsontheimaginaryaxis(jω-axis)andallotherrootsintheleft-hands-plane,thesteady-stateoutputwillbesustainedoscillationsforaboundedinputToascertainthestabilityofafeedbackcontrolsystem,onecoulddeterminetherootsofthecharacteristicpolynomialq(s)TherearethreeapproachestothequestionofstabilityThes-planeapproachThefrequencyplane(jω)approach3.Thetime–domainapproachTheRouth-HurwitzstabilitymethodprovidesananswertothequestionofstabilitybyconsideringthecharacteristicequationofthesystemThecharacteristicequationintheLaplacevariableiswrittenastoascertainthestabilityofthesystem,itisnecessarytodeterminewhetheranyoneoftherootsofq(s)liesintheright-halfofthes-plane.IfEq.(6.3)iswritteninfactoredform,wehaveWewillfindthatWenotethatallthecoefficientsofthepolynomialmusthavethesamesignifalltherootsareintheleft-handplaneAllthecoefficientsarenonzeroTheserequirementsarenecessarybutnotsufficient.q(s)=(s+2)(s2-s+4)=(s3+s2+2s+8),unstable.TheRouth-Hurwitzcriterionisbasedonorderingthecoefficientsofthecharacteristicequation

intoanarrayorscheduleasfollowsFurtherrowsoftheschedulearethencompletedasfollowsTheRouth-Hurwitzcriterionisanecessaryandsufficientcriterionforthestabilityoflinearsystems.TheRouth-Hurwitzcriterionstatesthatthenumberofrootsofq(s)withpositiverealpartsisequaltothenumberofchangesinsignofthefirstcolumnoftheRoutharrayForthestablysystem,theremustbenochangesinsigninthefirstcolumnTherearefourdistinctcasesreconsiderednoelementinthefirstcolumniszerothereisazero,butsomeotherinthesamerowarenonzerothereisazero,buttheotherinthesamerowarealsozeroasin3withrepeatedrootsonthejω-axissecond-ordersystemCase1:noelementinthefirstcolumniszeroTherefore,therequirementforastablesecond-ordersystemissimplythatallthecoefficientsbepositiveorallthecoefficientsbenegative.Third-ordersystemForthethird-ordersystemtobestable,itisnecessaryandsufficientthatthecoefficientsbepositiveanda2a1>a0a3.Theconditionwhena2a1=a0a3

resultsinamarginalstabilitycase,andonepairofrootsliesontheimaginaryaxisinthes-plane.Case2:azero,butsomeotherinthesamerowarenonzeroTherearetwosignchangesduetothelargenegativenumberinthefirstcolumn,c1=-12/?.Therefore,thesystemisunstable,andtworootslieintherighthalfofthes-plane.unstablesystemdeterminethegainKthatresultsinmarginalstability.Therefore,foranyvalueofKgreaterthanzero,thesystemisunstable.Also,becausethelastterminthefirstcolumnisequaltoK,anegativevalueofKwillresultinanunstablesystem.Consequently,thesystemisunstableforallvaluesofgainK.Case3:Zerosinthefirstcolumn,andotherelementsoftherowcontainingthezeroarealsozeroThisconditionoccurswhenthepolynomialcontainssingularitiesthataresymmetricallylocatedabouttheoriginofthes-planeCase3occurswhenfactorssuchas(s+σ)(s-σ)or(s+jω)(s-jω)occursUsingtheauxiliarypolynomialU(s),whichimmediatelyprecedesthezeroentryintheRoutharrayInthiscase,theorderoftheauxiliarypolynomialisalwaysevenandindicatesthenumberofsymmetricalrootspairs.ForastablesystemWhenk=8,wehavetworootsonthejω-axisandamarginalstabilitycaseWhenk=8,weobtainarowofzerosTheauxiliarypolynomialU(s)istheequationoftherowprecedingtherowofzeros.Whenk=8,thefactorsofthecharacteristicpolynomialareCase4:Repeatedrootsofthecharacteristicequationonthejω-axisIfthejω-axisrootsofthecharacteristicequationaresimple,thesystemisneitherstablenorunstable;itisinsteadcalledmarginallystableIfthejω-axisrootsarerepeated,thesystemresponsewillbeunstable,withaformtsin(ωt+Φ)TheRouth-HurwitzcriteriawillnotrevealthisformofinstabilityConsidersystemwhere,notetheabsencesignchangesTheauxiliarypolynomialatthelineisandtheauxiliarypolynomialatlineisFifth-ordersystemwithrootsonthejω-axisTheauxiliarypolynomialisForequationThetwochangessigninthefirstcolumnindicatethepresenceoftworootsintheright-handplane.Example6.4WeldingrobotcontrolTheRouth-HurwitzcriterionascertainstheabsolutestabilityofasystemItisdesirabletodeterminetherelativestabilityTherelativestabilityofasystemisindicatedbythelocationoftherootsofthecharacteristicequationInordertousetheRouth-Hurwitzcriterion,weshiftthes-planeaxis,Ashiftoftheverticalaxisinthes-planeto-σIfthesystemisabsolutestable,itisdesirabletodeterminetherelativestability.TherelativestabilityofasystemcanbedefinedasthepropertythatismeasuredbytherelativerealpartofeachrootorpairofrootsTherootr2isrelatively morestablethanthe root,Thecorrectmagnitudetoshifttheverticalaxisσmustbeobtainedonatrial-and-errorExample6.6AxisshiftCharacteristicequationFirsttry,letsn=s+2andnotethatweobtainaRouth-HurwitzwithoutazerooccurringinthefirstcolumnHowever,uponsettingtheshiftedvariablesnequaltos+1WeobtainCondition:adjustthetwoparametersKandaGoal:thesystemisstableandsteady-stateerrorforarampinputislessorequalto24%ThecharacteristicequationoffeedbacksystemisWehaveEquation:Routharray whereFortheelementoffirstcolumntobepositive,werequirethatKa,b3andc3bepositive.Thesteady-stateerrortorampinputr(t)=At,t>0isess=A/KvwhereThereforewehaveWhenessisequalto23.8%ofA,werequirethatK

a=42Thiscanbesatisfiedbytheselectionk=70anda=0.6Anotherselectionisk=50anda=0.84Routh-HurwitzStabilityConsiderthecharacteristicequationRoutharrayThesystemisnotstableIfthecharacteristicequationcontainstheunknownparameter,theRouth-HurwitzmethodcanbeutilizedtodeterminetherangeofvaluesthatthesystemisstableForexample,ThecharacteristicequationisExample6.11TrackedvehiclecontrolThedesignobjectistofindaandKsuchthatthesystemisstableandsteady-stateerrorforarampinputislessthanorequal24%ofthecommandTheclosed-loopcharacteristicequationisUsingtheRoutharrayisstableisstableIfr(t)=At,t>0thenthesteady-stateerrorisGiventhesteady-stateess<0.24A,thenForexample,K=70anda=0.6willsatisfyallthedesignrequirementsK=70anda=0.6Example6.11StabilityregionforanunstableplantAssumethez>0andp>0,thesystemisopen-loopunstable(withoutfeedback)Thecharacteristicequationoftheclosed-loopsystemisThegoalistodeterminetheregionofstabilityforK,pandz.TheRoutharrayisWhereFromtheRouth-HurwitzcriterionwefindthatwerequireKz>0,b2>0andp>1Settingb2=0,wehaveThereforewerequirethatConsiderthreecases:z≥p-1:thereisno0<K<∞thatleadstostabilityz<p-1:any0<k<∞satisfyingstabilityconditionforagivenpandzwillresultin