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 Download Pdf
Associate Professor. Mehran University of Engineering & Technology Jamshoro, Pakistan. email: . imtiaz.hussain@faculty.muet.edu.pk. URL :. http://imtiazhussainkalwar.weebly.com/. Lecture-17. Routh-Herwitz Stability Criterion.
Routh . Herwitz. Stability Criterion. Routh-Hurwitz Stability Criterion. It . is a method for determining continuous system . stability.. It can tell whether there are roots in the RHP of the characteristic polynomial of a control system without solving it..
Module Leader: Dr Muhammad . Arif. Email: . muhammadarif. 13. @hotmail.com. Batch: . 10. BM . Year: . 3. rd. Term: . 2. nd. . Credit Hours (Theory): . 4. Lecture Timings: Monday (. 12:00-2:00. ) and Wednesday (.
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Data. Jae-Gil Lee. 2*. . Gopi. Attaluri. 3. Ronald Barber. 1. . Naresh. Chainani. 3. Oliver Draese. 3. Frederick Ho. 5. . Stratos. Idreos. 4*. Min-Soo Kim. 6*. Sam Lightstone. 3. Guy Lohman.
Data. Jae-Gil Lee. 2*. . Gopi. Attaluri. 3. Ronald Barber. 1. . Naresh. Chainani. 3. Oliver Draese. 3. Frederick Ho. 5. . Stratos. Idreos. 4*. Min-Soo Kim. 6*. Sam Lightstone. 3. Guy Lohman.
N Queens Problem by Max Groover, Chris Wehunt , Forest Gates Introduction The N - Queens problem is finding the placement of N queens on an N×N chessboard so that no two queens attack each other.
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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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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
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