自动控制原理（中英文对照李道根）习题3.题解.pdf

自动控制原理（中英文对照李道根）习题3.题解.pdf

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Solutions13SolutionsP3.1Theunitstepresponseofacertainsystemisgivenbytteetc21）（,0t（a）Determinetheimpulseresponseofthesystem.（b）Determinethetransferfunction）（）（sRsCofthesystem.Solution:

Theimpulseresponseisthedifferentialofcorrespondingstepresponse,i.e.tteetttctk22）（d）（d）（AsweknowthatthetransferfunctionistheLaplacetransformofcorrespondingimpulseresponse,i.e.2324221112）（）（）（222sssssseetLsRsCttP3.2ConsiderthesystemdescribedbytheblockdiagramshowninFig.P3.2（a）.DeterminethepolaritiesoftwofeedbacksforeachofthefollowingstepresponsesshowninFig.P3.2（b）,where“0”indicatesthatthefeedbackisopen.Solution:

Ingeneralwehave）（sR）（sCsk2（a）Blockdiagramsk1000.10t）（tc0.10t）（tc0.10t）（tc0.10t）（tcAsymptoticline0.10t）（tcParabolic（b）Unit-stepresponses

（1）

（2）（3）（4）（5）FigureP3.2Solutions1421020221）（）（kkskskksRsCNotethatthecharacteristicpolynomialis210202）（kksksswherethesignofsk2isdependedontheouterfeedbackandthesignof21kkisdependedontheinterfeedback.Case

（1）.Theresponsepresentsasinusoidal.Itmeansthatthesystemhasapairofpureimaginaryroots,i.e.thecharacteristicpolynomialisintheformof212）（kkss.Obviously,theoutletfeedbackis“”andtheinnerfeedbackis“0”.Case

（2）.Theresponsepresentsadivergedoscillation.Thesystemhasapairofcomplexconjugaterootswithpositiverealparts,i.e.thecharacteristicpolynomialisintheformof2122）（kkskss.Obviously,theoutletfeedbackis“+”andtheinnerfeedbackis“”.Case（3）.Theresponsepresentsaconvergedoscillation.Itmeansthatthesystemhasapairofcomplexconjugaterootswithnegativerealparts,i.e.thecharacteristicpolynomialisintheformof2122）（kkskss.Obviously,boththeoutletandinnerfeedbacksare“”.Case（4）.Infactthisisarampresponseofafirst-ordersystem.Hence,theoutletfeedbackis“0”toproducearampsignalandtheinnerfeedbackis“”.Case（5）.Consideringthataparabolicfunctionistheintegralofarampfunction,boththeoutletandinnerfeedbacksare“0”.P3.3Considereachofthefollowingclosed-looptransferfunction.Byconsideringthelocationofthepolesonthecomplexplane,sketchtheunitstepresponse,explainingtheresultsobtained.（a）201220）（2sss,（b）61166）（23ssss（c）224）（2sss,（d）5）（52（5.12）（2ssssSolution:

（a）10）（2（20201220）（2sssssByinspection,thecharacteristicrootsare2,10.Thisisanoverdampedsecond-ordersystem.Therefore,consideringthattheclosed-loopgainis1k,itsunitstepresponsecanbesketchedasshown.（b）3）

（2）（1（661166）（23sssssssByinspection,thecharacteristicrootsare1,2,3.Obviously,allthreetransientcomponentsaredecayedexponentialterms.Therefore,itsunitstepresponse,withaclosed-loopgain1k,issketchedasshown.0.10t）（tc0.10t）（tcSolutions15（c）1）1（4224）（22ssssThisisanunderdampedsecond-ordersystem,becauseitscharacteristicrootsarej1.Hence,transientcomponentisadecayedsinusoid.Notingthattheclosed-loopgainis2k,theunitstepresponsecanbesketchedasshown.（d）5（21（5.12）5）（52（5.12）（222ssssss）Byinspection,thecharacteristicrootsare21j,5.Since51.0,thereisapairofdominantpoles,21j,forthissystem.Theunitstepresponse,withaclosed-loopgain5.0k,issketchedasshown.P3.4Theopen-looptransferfunctionofaunitynegativefeedbacksystemis）1

（1）（sssGDeterminetherisetime,peaktime,percentovershootandsettingtime（usinga5%settingcriterion）.Solution:

Writingheclosed-looptransferfunction2222211）（nnnsssssweget1n,5.0.Sincethisisanunderdampedsecond-ordersystemwith5.0,thesystemperformancecanbeestimatedasfollows.Risingtime.sec42.25.0115.0arccos1arccos22nrtPeaktime.sec62.35.011122nptPercentovershoot%3.16%100%100225.015.01eepSettingtime.sec615.033nst（usinga5%settingcriterion）P3.5Asecond-ordersystemgivesaunitstepresponseshowninFig.P3.5.Findtheopen-looptransferfunctionifthesystemisaunitnegative-feedbacksystem.Solution:

Byinspectionwehave%30%100113.1pSolvingtheformulaforcalculatingtheovershoot,（s）0.13.101.0t）（tcFigureP3.55.00t）（tc0.20t）（tcSolutions163.021ep,wehave362.0lnln22ppSince.sec1pt,solvingtheformulaforcalculatingthepeaktime,21npt,wegetsec/7.33radnHence,theopen-looptransferfunctionis）4.24（7.1135）2（）（2sssssGnnP3.6AfeedbacksystemisshowninFig.P3.6（a）,anditsunitstepresponsecurveisshowninFig.P3.6（b）.Determinethevaluesof1k,2k,anda.Solution:

Thetransferfunctionbetweentheinputandoutputisgivenby2221）（）（kasskksRsCThesystemisstableandwehave,fromtheresponsecurve,21lim）（lim122210kskasskkstcstByinspectionwehave%9%10000.211.218.2pSolvingtheformulaforcalculatingtheovershoot,09.021ep,wehave608.0lnln22ppSince.sec8.0pt,solvingtheformulaforcalculatingthepeaktime,21npt,wegetsec/95.4radnThen,comparingthecharacteristicpolynomialofthesystemwithitsstandardform,wehave00.218.208.0t）（tc）（sR）（sC）（2assk1k（a）（b）FigureP3.6Solutions1722222nnsskass5.2495.4222nk02.695.4608.022naP3.7Aunitynegativefeedbacksystemhastheopen-looptransferfunction）2（）（kssksG（a）Determinethepercentovershoot.（b）Forwhatrangeofkthesettingtimelessthan0.75s（usinga5%settingcriterion）.Solution:

（a）Fortheclosed-looptransferfunctionwehave222222）（nnnsskskskshence,byinspection,wegetsec/radkn,22Thepercentovershootis%32.4%10021ep（b）Since9.022,letting.sec75.025.033ktns（usinga5%settingcriterion）resultsin2275.06k,i.e.32kP3.8FortheservomechanismsystemshowninFig.P3.8,determinethevaluesofkandathatsatisfythefollowingclosed-loopsystemdesignrequirements.（a）Maximumof40%overshoot.（b）Peaktimeof4s.Solution:

Fortheclosed-looptransferfunctionwehave22222）（nnnsskskskshence,byinspection,wegetkn2,kn2,andnnk22Takingconsiderationof%40%10021epresultsin280.0.Inthiscase,tosatisfytherequirementofpeaktime,412npt,wehave）（sR）（sC2skas1FigureP3.8Solutions18.sec/818.0radnHence,thevaluesofkandaaredeterminedas67.02nk,68.02nP3.9Theopen-looptransferfunctionofaunityfeedbacksystemis）2（）（ssksGAstepresponseisspecifiedas:

peaktimes1.1pt,andpercentovershoot%5p.（a）Determinewhetherbothspecificationscanbemetsimultaneously.（b）Ifthespecificationscannotbemetsimultaneously,determineacompromisevalueforksothatthepeaktimeandpercentovershootarerelaxedthesamepercentage.Solution:

Writingtheclosed-looptransferfunction222222）（nnnssksskswegetknandk1.（a）Assumingthatthepeaktimeissatisfiedsec1.1112ktnpweget16.9k.Then,wehave33.0and%5%33%10021epObviously,thesetwospecificationscannotbemetsimultaneously.（b）Inordertoreducepthegainmustbereduced.Choosingsec2.221ppttresultsin04.31k,57.01,%102%3.111ppRechoosingsec31.21.22ppttresultsin85.21k,59.01,%10.51.2%0.101ppLettingsec255.205.23ppttresultsin941.23k,583.03,%10.2505.2%5.103ppInthisway,acompromisevalueisobtainedas941.2kP3.10Acontrolsystemisrepresentedbythetransferfunction）13.04.0）（56.2（33.0）（）（2ssssRsCEstimatethepeaktime,percentovershoot,andsettingtime（%5）,usingthedominantpolemethod,ifitispossible.Solution:

RewritingthetransferfunctionasSolutions193.0）2.0）（56.2（33.0）（）（22sssRsCwegetthepolesofthesystem:

3.02.021js，,56.23s.Then,21，scanbeconsideredasapairofdominantpoles,because）Re（）Re（321ss，.Method1.Afterreducingtoasecond-ordersystem,thetransferfunctionbecomes13.04.013.0）（）（2sssRsC（Note:

1）（）（lim0sRsCks）whichresultsinsec/36.0radnand55.0.Thespecificationscanbedeterminedassec0.42112npt,%6.12%10021epsec67.2011ln12nstMethod2.Takingconsiderationoftheeffectofnon-dominantpoleonthetransientcomponentscausebythedominantpoles,wehavesec0.8411）（231npsst%6.13%10021313essspsec6.232ln1313ssstnsP3.11Bymeansofthealgebraiccriteria,determinethestabilityofsystemsthathavethefollowingcharacteristicequations.（a）02092023sss（b）025103234ssss（c）0210922345sssssSolution:

（a）02092023sss.Allcoefficientsofthecharacteristicequationarepositive.UsingL-Ccriterion,01609120202DThissystemisstable.（b）025103234ssss.Allcoefficientsofthecharacteristicequationarepositive.UsingL-Ccriterion,0153110025301103DThissystemisunstable.Solutions20（c）0210922345sssss.（ItsbettertouseRouthcriterionforahigher-ordersystem.）Allcoefficientsofthecharacteristicequationarepositive.EstablishtheRoutharrayasshown.Therearetwochangesofsigninthefirstcolumn,thissystemisunstable.P3.12Thecharacteristicequationsforcertainsystemsaregivenbelow.Ineachcase,determinethenumberofcharacteristicrootsintheright-halfs-planeandthenumberofpureimaginaryroots.（a）0233ss（b）0160161023sss（c）04832241232345sssss（d）0846322345sssssSolution:

（a）0233ss.TheRoutharrayshowsthattherearetwochangesofsigninthefirstcolumn.Sothattherearetwocharacteristicrootsintheright-halfs-plane.（b）0160161023sssThe1s-rowisanall-zerooneandanauxiliaryequationismadebasedon2s-row0162sTakingderivativewithrespecttosyields02sThecoefficientofthisnewequationisinsertedinthe1srow,andtheRoutharrayisthencompleted.Byinspection,therearenochangesofsigninthefirstcolumn,andthesystemhasnocharacteristicrootsintheright-halfs-plane.Thesolutionoftheauxiliaryare4js,thesystemhasapairofpureimaginaryroots.（c）04832241232345sssss.TheRoutharrayisestablishedasfollows.The1s-rowisanall-zerooneandanauxiliaryequationbasedon2s-rowis042sTakingderivativewithrespecttosyields02sThecoefficientofthisnewequationisinsertedinthe1srow,andtheRoutharrayisthencompleted.Byinspection,therearenochangesofsigninthefirstcolumn,andthesystemhasnocharacteristicrootsintheright-half5s1914s21023s402s1021s-0.800s23s1-32s0021s2300s23s1162s101160161s0200s165s112324s3124848613s411642s411641s0200s4Solutions21s-plane.Thesolutionoftheauxiliaryare2js,thesystemhasapairofpureimaginaryroots.（d）0846322345sssss.TheRout