they oth indicate the system is unstable with oles in the rhp brPage 3br Another less computationally exp ensive metho to use when zero ccurs in the 57519rst column is to create the Routh table using the olyno mial that has the recip ro cal ro ots o ID: 27110
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Routh-HurwitzCriterion:SpecialCasesZeroonlyinthe¯rstcolumnWhenformingtheRouthtable,replacethezerowithasmallnumber"andevaluatethe¯stcolumnforpositiveornegativevaluesof".Problem:Determinethestabilityoftheclosed-looptransferfunction:T(s)=10s5+2s4+3s3+6s2+5s+3(1)s5135s4263s3¡13262=60!"¡15232=72¡10202=0s2¡26"72"=6"¡7"¡23"0"=3¡20"0"=0s1¡"726"¡7"36"¡7"=42"¡49¡6"212"¡14¡"06"¡7"06"¡7"=0¡"06"¡7"06"¡7"=0s03001 LabelFirstColumn"=+"=¡s51++s42++s3"+¡s26"¡7"¡+s142"¡49¡6"212"¡14++s03++Thechangesinsignoccurinbothinstancessoitdoesn'tmatterwhichwechoose.theybothindicatethesystemisunstablewithtwopolesintherhp.2 Another,lesscomputationallyexpensivemethodtousewhenazerooccursinthe¯rstcolumnistocreatetheRouthtableusingthepolyno-mialthathasthereciprocalrootsoftheorigi-nalpolynomial.i.e.T(s)=10s5+2s4+3s3+6s2+5s+3(2)byreplacingswith1=s.D(s)=3s5+5s4+6s3+3s2+2s+1(3)s5362s4531s34:21:40s21:3310s1¡1:7500s0100sincetherearetwosignchanges,thesystemisunstableandhastworighthalfplane(rhp)poles.Thisisthesameastheresultobtainedbyusing"fortheoriginalpolynomial.3 EntirerowiszeroProblem:Determinethenumberofrighthandplanepolesintheclosed-looptransferfunc-tion:T(s)=10s5+7s4+6s3+42s2+8s+56(4)s5168s467!1642!6656!8s30!0!0s2s1s0Nowwearefacedwiththeproblemofzerosinthethirdrow.4 1.Formanewpolynomialusingtheentriesintherowabovezeros.Thepolynomialwillstartwithpowerofsinthatrow,andcontinuebyskippingeveryotherpowerofs,i.e.P(s)=s4+6s2+8(5)2.Nextwedi®erentiatethepolynomialwithrespecttosandobtaindP(s)ds=4s3+12s+0(6)3.FinallytherowwithallzerosintheRouthtableisreplacedwiththecoe±cientsinEq.(6),andcontinuethetable.s5168s467!1642!6656!8s360!64!160!612!360!0s2380s11300s0800Weseenosignchangeshencenorhppoles.5 ²Whydoesanentirerowofzerosoccur?Whenapurelyoddorevenpolynomialisafactoroftheoriginalpolynomial.(s4+6s2+8isanevenpolynomialasitonlyhasevenpowerofs.)sjwjwwjsquadrantal and symmetricalimaginarysymmetrical and symmetrical and reals²Somepolynomialonlyhaverootssymmetri-calabouttheorigin.²Routhtablefromtheevenpolynomial(s4!s0)isatestoftheevenpolynomial.²Therowsofzerosindicatesthepossilbilityofj!roots.6 Lookatalargerexample:T(s)=20s8+s7+12s6+22s5+39s4+59s3+48s2+38s+20(7)s8112394820s712259380s6¡610!¡1¡620!¡2610!1620!20s5620!1660!3640!200s413200s360!64!260!66!360!000s2322000s1130000s0400001.Rowsofzerosofs3,Formanewpolynomialusingtheentriesintherowabovezeros,i.e.s4.P(s)=s4+3s2+2(8)7 2.Nextwedi®erentiatethepolynomialwithrespecttosandobtaindP(s)ds=4s3+6s+0(9)3.FinallytherowwithallzerosintheRouthtableisreplacedwiththecoe±cientsinEq.(9),andcontinuethetable.²Astheentriesfroms4tos0arelookingattheevenpolynomialsandtherearenosignchanges.Soallfourpolesmustexitsonthej!axis.²Therearetwosignchangesfroms8tos4,sothereare2righthandplanepolesand2lefthandplanepoles.²Finalresults:2rhp,2lhp,4j!poles.8