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RouthHurwitz Criterion Sp ecial Cases Zero only in the rst column When fo rming the Routh table replace the zero with small numb er and evaluate the st column fo ositive negative values of Problem De

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

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RouthHurwitz Criterion Sp ecial Cases Zero only in the rst column When fo rming the Routh table replace the zero with small numb er and evaluate the st column fo ositive negative values of Problem De






Presentation on theme: "RouthHurwitz Criterion Sp ecial Cases Zero only in the rst column When fo rming the Routh table replace the zero with small numb er and evaluate the st column fo ositive negative values of Problem De"— Presentation transcript:

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