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An example is given on how to solve the automatic control task. Follow the steps and solve the automatic control system task.
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FacultyofMechanicsDepartmentofBiomechanicalEngineeringMEBIB17028AutomaticControl:HomeworkBookDr.JuliusGriÿskeviÿcius2022
Contents1Homework21.1Assignment......................................21.2Homeworksolutionexample.............................41
1Homework1.1AssignmentAcommonapplicationofcontrolsystemsisinregulatingthetemperatureofachemicalprocess.Theßowofachemicalreactanttoaprocessiscontrolledbyanactuatorandvalve.Thereactantcausesthetemperatureinthevattochange.Thistemperatureissensedandcomparedtoadesiredset-pointtemperatureinaclosedloop,wheretheßowofreactantisadjustedtoyieldthedesiredtemperature.Figure1showsthecontrolsystempriortotheadditionofthecontroller.Figure1:BlockdiagramofachemicalprocesscontrolsystemChooseparametersfromTable1.Performfollowingtasks:1.Obtainatransferfunctionofclosed-loopsystemandÞndcharacteristicequation.2.Evaluatetransientcharacteristicsoftheprocessalone(dampingratio,naturalfre-quency,settlingtime,maximumovershoot,risetime,timetoÞrstpeakifapplicable).3.DeterminetherangeofKvaluesforstabilityusingRouth-Hurwitzcriterion(assumeC(s)=1).4.SketchtheRootLocusoftheplant(seeRootLocusRules.pdfforguidelineshowto).5.SketchtheBodeplotoftheplant(seeBodeRules.pdfforguidelineshowto).Deter-mineGainandPhaseStabilityMargins.6.Supposeacontrolleroftheform:Gc(s)=Ks+zcs+pcistobeused.DeterminethemaximumsKforstabilityusingRouth-Hurwitzcrite-rion(choosezcandpcvaluesfromTable1).7.ChecksystemstabilityusingHurwitzandMikhaylovcriterions.2
Table1:Homeworkassignmentindividualparameters.Var.No.a1a2a3bc1c2zcpc11,181,960,240,430,0970,0622,3618,1921,711,340,480,400,1200,1151,6016,0031,401,220,270,400,0970,1251,7910,7840,681,370,490,370,1000,0721,5418,6050,771,330,340,370,1360,1052,5515,6161,771,770,300,420,1100,0511,8018,6571,171,650,160,380,0990,0861,5625,7581,931,490,160,350,1320,1424,7920,2291,891,040,350,390,0980,0701,6917,98101,911,740,330,420,0980,0812,3425,11110,611,930,220,420,0600,0783,5521,75120,911,880,170,420,0870,0792,9318,72131,751,150,400,360,1290,1401,5018,71140,911,530,240,380,0760,1092,0215,42151,671,740,320,430,1040,1202,1627,77160,761,800,420,400,0930,1013,7024,71170,921,240,400,370,0960,1314,7925,70181,741,490,100,420,0730,1081,3918,26190,751,710,080,410,0800,0891,8018,56201,191,270,400,360,0620,1362,8719,53210,181,880,390,380,0890,1353,8713,18221,821,790,260,390,1440,1374,3213,10231,751,630,130,380,1210,0532,9320,01240,921,180,370,400,0660,1424,0122,02251,021,870,440,420,1000,0542,9311,76261,541,040,120,420,0840,1194,1014,09270,521,980,080,400,1210,0611,9610,39280,471,790,300,400,0780,0662,0914,59291,311,630,450,360,0830,1173,3126,18301,681,290,350,350,1490,0971,8422,02310,951,820,370,380,0700,1034,2422,66320,251,280,470,410,0630,1141,5514,56330,151,140,430,370,0840,1241,3913,94341,881,360,140,420,0580,0834,9321,35350,271,540,280,450,0940,1341,8026,79360,591,200,020,430,1370,1481,8715,30371,121,450,460,410,1150,1304,8121,98380,791,150,340,430,0790,1411,9926,70390,861,340,080,440,1250,1272,3413,08400,71,70,250,40,10,1593
1.2HomeworksolutionexamplePlanttransferfunctionG(s)=b1s+b0s(a2s2+a1s+a0)Chooseparametersfromtable:Table2:Homeworkassignmentindividualparameters.Var.No.a2a1a0b1b0ap4124342621.Drawablockdiagramoftheplantunderproportionalcontrol.Figure2:Blockdiagranoftheplantunderproportionalcontrol,whereKp=12.Obtainatransferfunctionofclosed-loopsystemandÞndcharacteristicequation.T(s)=KpG(s)1+KpG(s)Characteristicequation:1+KpG(s)=0s(2s2+4s+3)+Kp(4s+2)=02s3+4s2+(3+Kp)s+Kp2=03.Evaluatetransientcharacteristicsfromastepresponsewhentheplantisintheopen-andclosed-loopconÞguration(assumeKp=1).Whentheplantisintheopen-loopconÞguration,thestepresponseis:Y(s)=G(s)?R(s)=4s+2s(2s2+4s+3)1sThepolesofthetransferfunctionare:s1,2=0;s3,4=1±j0.707Y(s)=K1s+1j0.707+K2s+1+j0.707+A1s+A2s2K1,2=0.222±j0.7857;A1=0.444;A2=0.6667y(t)=49+23t49et cos p22t!+2.5p2sin p22t!!4
Judgingfromtheaboveexpression,itcanbenotedthatstepresponseisincreasingin-Þnitely,so,therearenotransientcharacteristicstobecalculatedfromstepresponse.TransientcharacteristicscouldbeestimatedfromimpulseresponseusingMATLAB.Whentheplantisinclosed-loopconÞguration,thestepresponseis:Y(s)=G(s)?R(s)=4s+2(2s3+4s2+7s+2)1sThepolesofthetransferfunctionare:s1=0.348;s2,3=0.8296±j1.4988;s4=0Y(s)=K1s+0.348+K2s+0.8296j1.4988+K3s+0.8296+j1.4988+K4sK1=0.376;K2,3=0.312±j0.2154;K4=1y(t)=10.376e0.348t(0.312±j0.2154)e(0.8296±j1.4988)tJudgingfromthepolesofthe3rdordersystem,therealpoleandcomplexpoleseparationisabout2.2times,thereforetheplantcannotbeapproximatedas2ndordersystemandtransientcharacteristicscanbecalculatedusingonlyMATLAB:Tr=1.5492sec;Ts=8.6927sec.Figure3:Stepresponseoftheplant(fromMATLAB)4.SketchtheRootLocusoftheplant(seeRootLocusRules.pdfforguidelineshowto).(a)Characteristicequation1+KpN(s)D(s)=0,whereN(s)=4s+2;D(s)=s(2s2+4s+3)Orderofcharacteristicequationn=3(numberofbranchesofthelocus),numeratororderm=1,thereforenumberqofinÞnitezeroswhens!1q=mn=2(b)Determinepolesandzeros:Zeros:N(s)=0;4s+2=0;4s=2;sz=0.5Poles:D(s)=0;s(2s2+4s+3)=0;sp1=0;sp2,3=1±j0.7075
(c)PointonrealaxiswhereasymptotesintersectrealaxisA=nXi=1pimXi=1ziq=(1+j0.7071j0.707)+0.52=0.75Asymptotesradiateoutwithangles?(0)=12180=90?(1)=2+12180=270=90Rootlocusexistsontheleftofoddnumbersofpolesandzeros.Onebranchofthelocusisontherealaxisbetweensp1=0andsz=0.5.Figure4:Beginsketchingtherootlocus(d)Break-away/-inpointsonrealaxisexistswhereN(s)D0(s)N0(s)D(s)=0(4s+2)(2s3+4s2+3s)0(4s+2)0(2s3+4s2+3s)=0(4s+2)(6s2+8s+3)4(2s3+4s2+3s)=016s3+28s2+16s+6=0WeneedtoÞndroots(youcanuseMATLABcommandroots([1628166]):s1=1.169;s2,3=0.291±j0.486Onlyonerootisreal(s1),butthisparticularrootisnotonrootlocus,thereforerootlocuswillnotcrosstherealaxis.Onesegmentoflocusisbetweenpoleatoriginandzeroat-0.5.TherestofthebranchesoflocusstartfromcomplexpolesandgotoinÞnityapproachingasymptotes(at90degrees).6
(e)Angleofdeparturefromcomplexpole:?z=6(?depz1)=6((1+j0.707)(0.5))=6(0.5+j0.707)==tan1?0.7070.5?=18054=126?p1=6(?depp1)=6((1+j0.707)0)=6(1+j0.707)==tan1?0.7071?=18035=144?p3=6(?depp3)=6((1j0.707)(1j0.707))=6(j1.414)==tan1?1.4140?=90?dep=180+X?ziX?pi?dep=180+126(14490)=72Sincetherearenocomplexzeros,sothereisnoangleofarrivalatcomplexzero.Figure5:Sketchoftherootlocus(f)Doesthelocuscrossestheimaginaryaxis?WeneedtoÞndcriticalvalueofKpusingRouthcriterion.1+Kp=0s(2s2+4s+3)+Kp(4s+2)=07
2s3+4s2+(3+Kp)s+Kp2=0Table3:Routhtables32(3+Kp)0s24Kp20s13Kp+300s0Kp20Fromthesystemofinequalities:?3Kp+3>0Kp2>0WeseethatKpmustbemorethan0and1forsystemstability.CriticalvalueofKp=0.Complexbranchesdonotcrossimaginaryaxis,onlypoleatorigin.Figure6:RangeofKpvaluesforstabilityFigure7:ExactrootlocusviaMATLABforreferenceApproximatesketchofRootLocusisinFigure5.5.SketchtheBodeplotoftheplant(seeBodeRules.pdfforguidelineshowto).DetermineGainandPhaseStabilityMargins.8
(a)TransferfunctionisG(s)=4s+2s(2s2+4s+3).RewritetransferfunctionintoBodeform:G(s)=4s+2s(2s2+4s+3)=2s0.5+1s23s2+43s+1=23s0.5+1s23s2+43s+1(b)Componentsofthetransferfunction:Constant23,realzerosz1=0.5,realpolesp1=0,complexconjugatepoles(2s2+4s+3)=s2+2?!0s+!20where!20=32;!0=q32=1.22;2?!0=2;?=1p!0=0.8165.s!j!:G(j!)=23j!0.5+1j!23!2+43j!+1=23j!0.5+1j!123!2+43j!Calculatingmagnitudeandphaseoftransferfunction.Magnitude:20log10|G(j!)|=20log10?23?+20log10?j!0.5+1?20log10(j!)20log10?123!2+43j!?|G(j!)|=pRe2+Im2,6G(j!)=tan1ImReThevariableinBodeplotisfrequency!anditispossibletocalculatemagnitudeandphaseoffrequencytransferfunctionatparticularvaluesofthefrequency,like0,10Ð2,1,10etc.20log10|G(j!=0)|=3.5218+20log100@s12+?00.5?21A20log10?p02+02?20log100@s?123!2?2+?43j!?21A=3.521820log10|G(j!=102)|=3.5218+20log100@s12+?1020.5?21A20log10?p(102)2+02?20log100@s?123(102)2?2+?43102?21A=36.54Phase:6G(j!)=6?23?+6?j!0.5+1?6(j!)6?123!2+43j!?6G(j!=102)=tan1023+tan1 1020.51!tan1?1020?tan1?43102123(102)2?=89.69
(c)DrawasymptotesofeachcomponentaccordingtotheBoderulestable.TermMagnitudePhaseConstant:K20logK>0Realzero:s!0+1=s0.5+1lowfreq.asymptoteat0dBhighfreq.asymptoteat+20dB/decConnectasymptoticlinesat!0=0.5Lowfreq.asymptoteat0Highfreq.asymptoteat+90Connectwithlinefrom0.1!0to10!0,0.05to5Poleatorigin:1s-20dB/dec;from0dBat!=1-90forall!Underdampedpoles:1?s!0?2+2??s!0?+11(s1.22)2+1.634(s1.22)+1Lowfreq.asymptoteat0dBHighfreq.asymptoteat-40dB/decConnectasymptoticlinesat!0Drawpeakatfreq!0H(j!)=20log10(2?)Lowfreq.asymptoteat0Highfreq.asymptoteat180Connectwithstraightlinefrom!=!010?to!010?!1=1.22?100.8165=0.1861!1=1.22?100.8165=7.9958Figure8:SketchoftheBodeplot10
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