CH03-3 stability and routh criterion

Ch.3TransientandSteady-StateResponseAnalysis,StabilityandRouthcriterion,Basicconceptofstability,stable,unstable,稳定性和劳斯判据,Definitionofstability,Allsystemwilldeviatefromtheoriginalequilibriumunderthedisturbance.Ifthesystemcanreturntotheequilibriumwhenalldisturbancevanish,thenitcanbecalledstablesystem.Stabilityisintrinsicforasystem.Foralinearsystem,Itreliesnotontheinitialandexternalconditions,butonthestructure,parameters.,Methodsinanalyzestability,Characteristicequationandcharacteristicroots(特征方程和特征根)Algebraiccriterion(代数判据)Rootlocus(根轨迹)Stabilitycriterioninfrequencydomain(频率稳定判据),Sincestabilityistheresearchofsystemafterremovingthedisturbances,andhasnothingtodowithinput,itcanberepresentedbythepulseresponsefunction.Ifthepulseresponseconvergences,thesystemisstable,andviceversa.,Krealcharacteristicroots,rpairsofconjugateroots,（1）ifitwillreturntooriginalequilibrium,theresoscillationbecauseofcomplexroots;（2）iftheresponseisexponentialattenuation;（3）ifor,when,itsunstable;（4）ifoneofequalsto0,when,thesystemcannotreturntotheoriginalequilibriumorshowsevenoscillation.Itsalsotreatedasunstable.,衰减振荡,指数衰减,Sufficientandnecessaryconditionsforstability,Allrootsofthecharacteristicequationhavenegativerealparts.Namelyallrootsmustlieonthelefthalfofthesplane.,特征方程的所有根都具有负的实部,充分必要条件,特征方程的所有根均在s平面的左半部分,所有闭环系统传递函数的极点都在s平面的虚轴的左半部分,一个复实根,原点上的单根,一个正实根,Marginallystablewillbetreatedasunstable,sincetherealwaysexistsapproximationinsystemmodeling.andtheparametersarealwayschanging.,一对共轭复根具有负实部,一对共轭虚根,一对共轭复根具有正实部,Ifabovemethodisadopted,allpolesmustbefiguredoutfortheclosed-looptransferfunction.Itisnearlyimpossibleforthesystemabove2-order.Itisanaturaldemandforasubstituteforthisdirectsolution.,判别系统稳定性的方法（1）劳斯判据和赫尔维兹判据（2）根轨迹法（3）奈奎斯特判据（4）李雅普诺夫直接法（5）利用计算机直接求根法,TheRouth-HurwitzCriterionTheRouthcriterioncanascertainthestabilitybydeterminingthesignoftherootswithoutsolvingthecharacteristicequation.ThenumberoftherootswithpositiverealpartsisequaltothenumberofchangesinsignofthefirstcolumnoftheRoutharray.Thisisnecessaryandsufficient!,正实数根的数量等于劳斯行列表中第一列的符号改变次数,Subsidiarystatements(1)Forastablesystem,thecoefficientsofthecharacteristicpolynomialmustbepositive(havethesamesign)andnonzero.Thisrequirementisnecessarybutnotsufficient.,(2)ItissufficienttoassurethestabilityofthecontrolsystemthatalltheentriesinthefirstcolumnoftheRoutharrayarepositive.,系统稳定的必要条件：特征方程式的所有系数必须都是正数。,(3)ItissufficienttodeterminethatthecontrolsystemisunstableiftherearepositiveandnegativeentriesinthefirstcolumnoftheRoutharray.(4)TheresnoinfluenceonRouthcriterionifallentriesofonerowintheRoutharraymultiplyonepositivenumber.,Firstorder:Ifa1,a0havethesamesign,thesystemisstable.,Secondorder:Ifa1,a2,a0havethesamesign,thesystemisstable.,Thirdorder:Ifa0,a1,a2,a3arepositive,anda1a2a3a0,thesystemisstable.,TheRoutharrayisscheduledas:,HowtoscheduletheRoutharrayconsiderthecharacteristicequationas:,Severaldistinctcases:Case1:Noentryinthefirstcolumniszero.Case2:Thereisazerointhefirstcolumn,andtherowhasatleastonenonzeroentry.Case3:Thereisazerointhefirstcolumn,andentriesinthecorrespondingrowareallzero.,充分必要条件：各项系数全为正，行列表的第一列都具有正号。第一列的系数中如果出现负号，则行列表的第一列的系数符号改变的次数就等于特征方程式的实部为正的根的数目。,Case1行列表的第一列系数均不为零,Example1,Thesystemisunstable,andhas2polesintherighthalfs-plane.,Example2(withparameter):thecharacteristicequationis,s3+s2+s+k=0;kisnot1or0,Thus:Thesystemisunstablewhenkislessthan0andwhenkisgreaterthan1;Else,thesystemisstable.,Adjustingtheparametermaychangethestability!,Case2某行第一列系数等于零，其他列不全等于零,用无穷小代替零，如果上面的系数符号与下面的符号相反，则表明这里有一个符号变化，则可判定系统不稳定。如果符号不变，则表示系统有共轭虚根，是临界稳定。,Case2,Example3Thecharacteristicequationis:,s4+s3+s2+s+k=0,Letapproachesto0,thereistheproductofkandminusinfiniteinthefirstcolumn.Thesystemisalwaysunstable.,Replacethezeroelementwithasmallpositive,Case3某行所有各项系数均为零,某行系数均为零，或者只有等于零的一项，说明存在对原点对称的实根、共轭虚根或共轭复数根。（1）将不为零的最后一行的各系数组成一个多项式，辅助多项式的次数为偶数。（2）求辅助多项式对s的导数，将其系数构成新行，代替全为零的那行。（3）继续进行计算。（4）对原点对称的根可有辅助多项式等于零求得。,Case3Zerorow,Example4Thecharacteristicequationis:,s4+5s3+7s2+5s+6=0,2.Buildtheauxiliarypolynomialusingtheaboverow.,1.Whentherearerootssymmetricallylocatedabouttheorigin,zerorowwillhappen.,s2+1=0,s1,2=j,Example5,Thecharacteristicequationis:,Determinetherangeofkanda,tomakethesystemstable.,RouthArray,Letk=40,thenweneeda0.639,Detectionoftheabundantdegreeofstability,Therouthcriterioncandecidetheabundantdegreeofstabilityaswell.Inthissituation,thecoordinatesystemcanbemovedandusetherouthcriterionagain.,Example6Thecharacteristicequationis:determinetherangeofktomakethesystemstable.Ifwerequirefurthertherootsareallintheleftsideofs=-1,howtoadjusttheparameterk?,Thecharacteristicequationcanberewrittenas:,ByRoutharray,wefindthattherangeofktomakethesystemstableis:0k13,fortherequirementthatalltherootsareintheleftsideofs=-1,substitutings=s1-1intothecharacteristicequation,wehave:,RoutharrayLettingalltheelementsinthefirstcolumntobepositiveyieldsmorestrictrange:,Meathodstocorrectthestructurallyunstablesystem,Nomatterhowtoadjustthestructuralparameter,somesystemsarestillunstable.Theyarecalledstructurallyunstablesystem.,Structurallyunstablesystem,1.Modifytheintegralelement,Employproportionalfeedbacktoembracetheintegralelement.,Inertialelement,Theclosedlooptransferfunction:,Characteristicequations:,Routharray:,Lettin