Mechanical Engineering - Control and Actuating Devices for Mechanical Systems

Completed notes of the course

Complete course

ControlTa r g e t s : OVER/24 : 15/09Displacementcontrol - >keep a quantity cust.- followtrajectoryJuserallytogether -> followlowof motionVilationcontrolFacecontrolFirstdecidethe reference - controltargetb ACTUATORS :· Hydraulic - highfeeces · Pneumatic -> safety · Electric-> high use they allhave limits whichhartobeconsideredduringtheproject .HowtodecidetheControl strategy :- feedfownd/feedback FeedtoNARD : identifythecontrol targets -> inputs69 z -tor s(syst . ) >seresponseMbmathematicalcontrolforce1modelcommandThere's Nocontrolontheresponce of the system ,whichmakesitfarkesbutnot really accurate &"¥ automaestaFBACK :us mathematicalmodel 69 - mIref -> o or - >Iresponce - &I - To s s o r Therespaceissloweraretotheanalysisofthesense.Itcan affect the stabilityofthesyst , whichtheFFcan'tsinceitdoesn'teffectthepositionof thesyst The3canstabiliseanrustablesyst .- havetobe careful "¥ STASILYmustbealways , granted e. feedbackcontrol. :· Ster I decidedbroder · wind S · road integrity mate The trajectory istheinputtothecontrol->drivermakedecisions speed aouspues & · positions · yourIt'spossibletomuixthetwo ] t pe ( ... ] I known fazufa · 7 daf so - st nmJaunknown Ta r g e t : -Mr "¥x Egofmotion : 3inputsyst ↑ Ux-> fdmixe + rx+kx = fu + fu + fdfc = ? &m- fa &Mi-> fie &wiFfC : -for sten A=perfectactuator ↑futfelfo can'tbesentof themathematicalmodelsinceit'sunknown - ~+x+ Ex = fc + fd in , I aIhavetobe found bbb fc = m+x+ Ex - fa 3 anhadhander aan fa wie+ zieking fa substitutezewithse of zofmi+Ex+kx = m +Eff EqofmotionofthezofPERFECT : Iforisnotaninputanymisefa ->om= mrk==theonlysolutionisse=xf "¥ withthisconditionofr can follow thereferencex(t) = xy(t) + xou(t) + xp(t) unknownb ~hornogenius-> "¥ feemotionofsteady-stateresponse set soit'sthetransientthesysonlyoreresponse1)x(t)-notemalface : mirex+her0 = = extdetconstanthavetofind ↓ : (md +ed + k) Ve n t alwaymjted +k = 0= 1- im x = Em = wohW . = naturalprequence 2h = ZmeWo dampingfactor ,howthesys , behaves Ju =- + We-J carbeeitherrealorcompletcouj , &xe(t) = Xax**+ Xae X- ztxl M ifh3,1 : d = +z =- X ~ stunderdampedauh--ifh : 11 =- +iW stW # Ka() = ect(Acoswt +Bsimw+)Eitherwaytheresponsetendertoo after awhichNote!:Itdoesn'tchangethetransientresponce , there'sNo influenceofthe fo, that'swhy Fordoesn'teffectthestabilitywer thetransientrepare of thesys.2) Koult) : Mi+2x+kx = fu ->kuknownrandom force "¥ sum of azmoiccomponentsif itdependsonthetime fan fui(cos(2 ;++ Pi) Ws former 6 is↓ bar i," I frj = j3mi+ex+kx = (fu(cos(r ++ 4) se -realpentoftherector Fu = Ifule"Ymi+ex+kx = k(fue +2 + ) = R((fu(c +(x + m)17/09Completeresponseofthesystem : x(t) = xult) +2pult) + Hpe(t)I doesn'teffectthe steady-stateresparen2) Su = Ifulcos (2 ++ 4) = Re(fue ht) =Harmoniccomponentbforce'sphase "¥ Fatifulent mi+Tx + kx = fu(t) = fuevnt ->thesolutionisalsoanharmoniefunction : irt xpu(t) = Yo u r e↓mr" +ier+ k)XoelirtI relationbetweentheimputforce andtheoutputforce : G(2) = T = IG(z)/e"4(b) knownasFRFprequencyrespons - mr+i2r + kfunctionbothcompe Koult) = be (anoe) = Re) irtre in => IG(r)llful cos(r++ y(r) + 4) "¥ theamplitudeand frequencydependboth fromtheinputface. G AYWa if isdrumping : & ·r= 0 = >IG)= · ~ &-·r= wo = >(G)-0 - 6 WoS ga ↑ ·r= /Gl=-mr quasiatic If weconsiderthe dumping therootswillbewe Wo rthfor2Woit'stheQUASI-STATICREGION , where onlythestifuenzules . Neartheresource , theresponce of the systemis amplified and90outofphase.forMssW o it'stheSESMICZoneandit's180out of zane.xpa-0 foreverydisturbancesowecanactivethe target force.It'sbetterto stay intheseismiczare , awayfrom theresonance , whichwouldcreateesbeamstothesystem.Thecapabilityof theFFCtorejectthedistantldependsonly onthepassivepartofthe system(m .2, k)whichisnotgood,3)mi+ex+kx = min + Exe + Exr = E with= n= H)thereference, knownfunctionof time.xr(t) = &(X+;/cos(2,++4;)Ur Mwecancounderjust - an Ill , areharmoniccomponentxr(t) = (X nolcos(r ++ 4) =I= he(Xzeirt) n-1 &Hs = G (t)- f = ( - mr + ir + h) Xroe int = fe int- far)The steady statesolutionisthesameasbefore.wi + rx+kx = ( , (2)eint =xpr(t) = Yo u e i t(mr +ir+ k)Xpert - Det Xan = False)G(r)=-me + i+if canextimate - maxier + kXeo-m -m, E =2, Y = kXa = XvoIRI al &&&thesteadystateresponce ~ isexactlythereference T -responcegaR = GWo-(2) = R(x)Xe Whenthe systemisintheresonamezome, theresponseof theactuatedis nearly tee , riceversethe highe thefrequencythe highn theactuatorresponce.WeTheffCisnotableto change thetransientresponseherabletorejectthedisturbanceresponse.FBC : There's no mathematicalmodelfaconsideringA-1goesbetweenfadfo its andS=1C = x2 - x s mi+rx+rx = f + fuConsideringfc = Kpe + kqi = Up(x-x) + ka(xr - x)↑↑PROPORT,ONALDERIVATIVEGAINGAINmi++x+ru = up(xx) + ka(xz - x) + fu mi+(r+ ka)x + (k +kp)x = kpxr + kaxe + fu FBCsysunu- - Otk+ x2(t)Zult)2inputsbothintimex(t) = xh(t) + xpr(t) + voult) ->insteadof2&kwe'rI,andhe1)xs(t)mi+ hi + kx =mIt * 2 =- ifup+(ka = 0)=Xc==- Wa t e r HKo =oWokeasmKat T --w, theresponseis going T&tobe fasterbuttheoscillationsare going tobeM larger , sincer, &weare movingawayfromeach int waother .A kd* ByincreasingId theoscillationsare going tobesmalles.ka*=heXc↑Haveto carefully chooseUpdasoIcanobtainthederivedresponse irt 2)xormi+ ex + kx = Kpxr(t) + Kaur(t)x =(t) = xneI- (Kp + irka) seirt xpz = xaeirtFutire + k)konSotirkalert Bors = Kp+irKol = LKor - mr + iyr+kek+ = k + kp If( = 1 = >xp = xr Wa na ne kpT ! ·r=0= = ·r-0 =L = 0gaif up -0Wa nUp=>W&Swastintheresonance ifK&KFO : WasgoesrightoftheKa = OhcbzorbutmuchbetteamplitudegetsbiggesbeforeziggerkameansbetteKat=heresponse by thesysmeasUp == theresourcezome.Nevertheleit'simpossibletomakeNon-M13)Kou(t)mi+ x + u = Re(fuer)Mou-Kpoeint ( - metire+K+ ) Kart-fut L = 2 T -> similartoG(2 , &4) - mr" + ir2++ k+Ital& aup= Wind · Ka =0 had - gaTa r g e t : Kpu-0=La = 0needtowork Wawithhigher r LAPLACETRANSFORM :24/09Itswitchesdifferentialequationsto algebraiceq . Itallowstomove fromthetimedomaintoLaplacedomain.t-S f(t)) = f(s) = 1 f( + )eat S = 5+ir↓complexfunction Iseir -- istformercers · ((r)-1fitte at - fromtimeto pequency domain6itonlyprovidesthesteadystateresponse , sinceisonly thecomplexpart. Inthe Laplace domainwecanhavebothtransientIn steadystaterespaces.Thecondition ofexistanceis : /PfH)e"1dt Inmechanicalsysit'simpossiblethat1. isn'ttrue2.lezeros(10 , helpolulcoS . INITIALLe : o f(t) = f(0) = 2f(s)6 . FINALvalue : o f(t)-est(s) - steadystatevalueIMPULSEFUNCTION :-n imp(6) · 1 -> t Oste Eto=> f(t)-imp(t) o+10 , teaf "¥ )(imp(t)) = ! impltest atE + j(t)dt-] jdt-Edt - Et)" - &(implt)) =mpHedt-mpt e-St=1 SEPFUNCTION :N1- · to st&(step(H) = step() at edt =- ** = mi+exe+kx = f withf(t) , x(t) f wewenttogototheLaplacedomi-x(t) 2(f( +)) = f(s)Let'sassumetheintialvalues : 2(n(+)) = X(s)x(0) = u(0) = i(0) = 0&(n(t)) = SX(s) - x(0)Resulting in :"¥ (ic(+)) = s X(s) - Sk(0) - x(0)ms-X + 2sX +kX = fNoticethatwemove from a differentialeq . toan algebraiceq , whichiseasiertohandle.G(s) =msrs +m transferfunction(1f)transientsteadystate↓x(t) = 2(X(s)) = 2"(G(s)f(s)) = x2(t) + xp(t)Whilethefefrepresentstheresponseofthesystem toaharmonicfunctiononlyandgivesthesteadystate.TheIfgivesthecompleteresponseofthesystemtoanyimputNoticethatthebots of thecharacteristeq . arethesameasthepolesofG(s) - 11,2= - Eresantationof f : IG)G(ir) = (G(r))e1y(r) i 1Glir) = mr + i2r +k for fefshal-2w- worl · Wo -T Bode diagram : IGlas&WoIG(r)lab = 20 log(G(r))logz so lo i theseare asymptoticdiagrams 2-D T40 dB/deaM=+= realdiagramsn9 - slogr - -T-- POLAR Or G(2) = 1G(r)/e:9/2NG(r)r = Ma2: 0-+0I ↓ not ess Increasing thevalueof r, Iobtainacontinuescurve , having a diagramholdingtheresponse of 3 . tem .& thesysIG) i . - &S - -T3*⑨T&Re · -2 j 2=0Z⑨rsWo- 3-TD NQUISTDIAGRAM : r(-00;+ a)Asthepolardiagrambutitgoesfrom - Ototo- Em&G(-r) = G*(1) + ]can plotthe diagramfee & -soto justbyplottingthe G simetricrespecttotherealexis.-T*⑨T she 2=0 Noticethatthecurvehastobeclosed , and · there'sonly onewaytowalkit. e swe BLOCKDIAGRAMS : Agraphicalwaytodescribe a system . 29/09It'seasiertoseetheconnectionwithintheF(S)x(s)-> - systemandtheinput/outputs .RULES :· sunmingpoints : Sure of signals y - · sick-uppoints : duplicatesa sigual - BASICPROPERTIES : Y · SERIES: --stG Ez = G , x s y = G2 · G , x=G(s) = G . G=Giy = G2 . zThisisveryusedinthe feedforward control : f K"¥ SoIobtainthat : actuatorsys((s) = =A . GIn ordertogetthe target x = 12,the transferfunction(s)hastobetheclosestto1. In a perfectmodel : "=G andina perfectactuator : A =1. Withthesetwoassumption , ((S) = 1 . Eonlyinthequasi-static zone , · PARALLEL : - Y, = xG . & y = xGnY = x(G ,+ Gr) =G(s) = G ,+ G Wecannowsocea feedback-loop : gregativefeedbackY ~ o= +L == & ↑y = Ge = G(x - z) = G(x - Hy)If wesumzinstead ofsubtracting , wehave - -Back control : controlactuatorKKr- > e ↑ feel - I -L # sensor lyRAG CLOSEDLoop f = 1+RAGH= _ TRANSFER + FUNTION "¥weareabletoknowthe stabilityofthecontrolbystudying itsroots.GH = RAGH istheOpenLoopTr a n s f e rFUNCTION ."¥ I'mableto study the stabilityoftheFB-CS.SABILITY : Thecapabilityofthesystemtogobacktoits startingpointafter aperturbance . ⑨ If there'sno fiction : s & W- NOTASYMPTOTICSTABILITY=x - xe withfunction We s ASYMPTOTICSTABILITY If weconsiderthetopmore , wehaveathirdcase : " UNSTABLEstNotethatSTABILITYisNota propertyofthesystem , itdependsortheparticularconditionweare amflising . -UgriuM : 2 = 4 , x = i = 0Inthecase ofsteady star & o(carmoving , turbine).Ialsohavetoconsidethes ->se turba ... =x-2)aswellasthe forcefieldsf = f(x , ziis ,in-fofthesystem , since theychangethemass , dancingandspringcoefficients .Theseforces caneitherbeConservativeorNOTConservative . In case offeedbackcontrol , fc = Kp(x-x)+ Kalie-x) = f. (x , 2) , resulting :mi+ (2 + ka)x+ (kp +k)x = 0Wefirstanalyse thelinearsystemtherthenon-linear . FoltheWA P U N Otherom : ASYMP . UNSTABLENONASYMSTABILINY-SABILITYLINEARXXY To v u n e a r XX? 1.NanlinlarEDM.2. Defineconditionstoanalyse . 3.Linearisation(kg , 2).4. Analysethecenturbedfeenation,STABILITYlinearsystem S . Lyapunovthereu.1 dof : ↑ -mi + rx+kn = f(x , x) 2. Equibifium : KNo = f(x , 0) -nonlinearolgelianc equations.f(x) &"¥ therecanbemultipleeq points 3. Ta y l o r Series : f(x , i) f(x , 0 +-xa +mi+2x+kx = f(x , 0) - k-)) - Eji jemi + zi + ka + 0= fe , a) - k , - Tj mu + (n + 1) + (k +4+ ) = 0-zommi +r, i +k+2 = 0=ne ↓t4 = 5'm + 12 +k,= 0= 2 = Wo = I= -= wohe we Stability I dof : 9) m, + , lysoastmotoricalStable : ii NREALNCOMPLEXCONS ~ -Wus d ,=-X, h5 , 142 =- xb(4 , 5) = F , ( , a) + ( no) I= "¥ Ex . ) - By wecanconsiderthe perturbation =- . (i=& s + (Br+s(i + (1 ++ 1)* = 0 - >notethatthe global massmatrixstaysthesame. +Bi + Ex = 0 "¥wehaveto studytheForinordertounderstandhowIchangerthesys . resp. 9)= e*= (15 + 3) + 1)e+ n = 0 "¥ feenationresponse "¥ aut (15 + 1 + 3) = 0b9. In a + 9 0 nedealgera "¥ - STABILITY=SRe(d)0&di&if 1 dof >2polesif mdof=>Inpoles&Es + 1 = 1 Ka T &Est = (In Euentr)howcanIknowthe stabilityof thesystemwithouthaving tosolverthe&undereq . (case2 dop)? I : 1 . B , S,aresos(simmetric&positivedefinite) STABLE2 . B = 0 "¥ SDP : -simetric2.7conservative Kry-Kyn-150 =start - diagonaltermysob "¥ 10-STATICINSTABILITYKar& /Kyy OKay Kyr O1 . 2non-conservativeKay#Kyr - IMykynl(Kamika)2 =flusterinstabilistY ~ "¥ Wa s & 3 . B0 - >Sapthe damping candissipate the energyintroducedbythe "¥ ↓ stiffen . · if one ofEase ofZyywe'llhave1 dofDYNAMICINTABILITY .· ifLayTy p e=SFLUSTERINSTABILITY ."¥ inthecase of - We e anair failherethewindhasthesamefrequencyofthe wing -> flutterinstability Statesparepresentation : Zindofs D =+2 + 17=f() - b state rector si Wemoved from a2order differentialequationtoa15des.&Stateinourvector ↓ MATRIX2MX1b "¥ indusvectorSTAEMATRIX1 = 12 + 34isthestatespaceoftheSYSTEMImxIn - representationofthesystem. stability -> fermotion - >15 = Amste * o1 = 402 ~ eig(t) = det (x5-1) = 0 +> poesofthesysI+ 1 - 5Caseoffeedbackcontrol : - astaterector-f = 3c(uy - z)= [-i = Au + 30 = 14 + 3(4 - x) u Ac - thestabilityof thesystemIII iseffectedby thepresence - > eig(A) = oofthecontroll "¥ Stable Re(eig(t !)Othecontrolactionchangesthepositionof the poles . E : Stu Hmz+ xz + kz = f is = 12 + B - &= -Fuz + Emm x = 4 z = zn = feig(t) = det165-t) = 1 m )= j + m + E - 0 = - f = kp(ty - z)+ ka(zy - z) = (krp]((() f ="¥ (f - x)Ac = 1-3-eig() = det (+) = ClosedLoop : 8/10GH = RGH ->openlooptransferfunction A · "¥ GH-closedloopI - -Ta s 6 L = - themoreisneartotheebetterthesysisbetter&tracking therefereverSTABILITY :->directmethods ->evaluatethepolesoftheIfL "¥ L(rootsofthech . eg .= eig(tc) I ↓ "¥ ifallpoleshavecobepartsthesysisstableRootLowsfeedback "¥ gives a representationofthepolesmovingbythe changingundirectmethods : - ofthecontrolsystem. "¥ GH & NyquisttheylookoftheopenloopIftostudythestabilityoftheBodeclosedloopsystem&QUISTCRITIRION :Feedback1)mapping : consideringavariableS = 5+ik ->it increasingfrom SointheclockingSm(s)wisedirection .XSXSoscanassumevaluesonly intheCsX & Re(s)patter XXSCs&in thiscase1zero&3polesinsideCS-planeSm(f)f(s)-> analyticalfunctionof sindescrete&-(So)numberofpoints.·asRe(f) d N = 3 - 1 = 2 a-TC=mapofCF-plane2)Cauchysincipleofargument : the#ofincirclementoftheoriginoftheFplaceis :ppocesoffec(v)N = z - Pwithzzeros offECs(3) ifNSOtheincirclementsareclockwise. if NCOtheincirclementsareanti-clockwise.It's define the function+ : f(s) = dn(s) = 1 += RH-Mantdat &GH "¥ terosof-= PolesofL(f(s) =o(=>d. (s) = 0) selectingC,withR=tooweare &includingallmustablepoles&zeos. PolesofF=PolesofGHIsistheSur &NyquistpathIf Cisinthepositiverealpartthesys,isunstable .= Cf- D ↓ "¥ themapC,istheNyquistdiagramoffontheSunaxis : Sihac( -0;+ a) E "¥ StabilityofL:unstableZerosN = z - Punstablepolesoff&z =D D ofFbN == P10b "¥ instablepoles unstablepolesofGHnopolesofLcan layof LCs- = 1+GH-GH = f1=wecarlook&the NyquistdiagramofGH.5)m(f)Sm(GH)Sonowwehavetocounttheincirclements E Re(f)= G Re(GH)ofGH-peacearoundthepoint1-10) - bF-planeGH-place -1Thesystemisstable if theanticlockwise3cases : incirclementsare equal tothecust. · N = 0 -> GHdoesn'thaveanymust . P.ofGH. · NSO ->thesystem is cust. · NCO->the systemisstable( = >N =- P. &X : Zesth = xp( - x) - 1 -I =mstrstKp = RGH = RG = RPr ,n= -aFit-s ifthesys . isstable&ms2+25+kunderdamped. "¥ Nyquisdiagram(afterBode'sIGHdB& ↑ more Sms(GH)& -T · Men a I - z = 0 , p = 0 , N = 0=Stableifupincreases , nothingchangeinthestabilityofthesystem.5m(S)GH =G - SantSaut e umS(sm " bu +---+b , )Sur(s) D 50 : GH == Jere iseiaM -stodigne "¥ =Set with act) - =Sm(GH)80 : GH-Dev8 ec L 4GH:- Whatif wehavezpoles&theorigin?wehavetodrawSoGH = = Be oas manyhalf circlesSoGH- 120 =IGH1 =aasmanyis2. "¥ GHEX : upGH = s(s + 25 + 2) Denti => SpassivesystemisstabeIsP = 0IGHlas - zaUp &↑ -60 aclosingtheNyquist & - W diagram-&· - Do a - 3πUp if weincreaseuptoomuch , we'llhavethatN = 2#P,andsothesyswillbecustable. Nyquistcriteriorcanalso highlight the stabilityofthe13/10system . Sur "¥ INDICATORSORRELATIVESTABILITYRe 6↓ f PHASEMARGINPurGAINMARGIN -- angulardistance ofmagnituddistance of the diagramfrom 5-1,03the diagramfrom 5-1,03Gar from CrossoverFrequency : IGHlaB IGH(rg)) = 0aB & g "¥m=+ 4GH(rg) - 9 Sur Imisrelatedtothe - feg , wherethelldsoutRetheIlivePm>O · HigIGHldBGmCOGmS0 Su · timel ↑ mGH o with&GH(rp) =- T2p- if Emisrelatedtothe fea /Gm >Othatcutthephasee-T Gulas = 200g (rol) = 20gIGH(2pl "¥ & IGH)/1s IGH(mplar 30 = Gulag OInordertohaveRobustSTABIL194 : GulasGaB "¥ NeverthelesstheMinimumPhaseSYS . S Pm>30 - 60findsGm&Ommuchmoreimportant - MumPHASeSYSTEMS : GHhasnocustablepolesezeros . & & rustStabilitytheFBCsysisstableif wehavepositivemargins. ifthere'snoincirclements E aroundthepoint1-1 , 03 "¥ ifnegative theFBCsysthesysisobsstableisunstable&CRITERION : for theopenloopIFGHitonlyappliesonly underCertainHp : BothIm&Guhavetobesotohave1.GHhasnounstablesoles aStablesys . &sufficientconditions forunstability : 2.IGHlascrosses only onesO&Baxis. & IGH(mp)1c1IGH(mp)las >OdB&fBCsysisUNSTABLE .- erwecau saita aif thegainmarginis10&up , thesys , isSurkpTunstable F ocus : RepresentationofpolesofL> roset YRe J G = RG Y - * brauch = G = 6d(s) = 0 - >poles of MGH+ &GH = 01+GH = 0 "¥ Man+GH = 0- &GH↓↑G+(s) = kpm(s) - kpm + = 0↓conte Kp : 0 -S+inedetofindthepoles1)kp = 0=a,= da ++kpm * + = da +–"¥ polesofL = polesof GH->branchesof therootLocusstart from here2)kp++du = dan + kp * M Œ &possibleonlywhen G = 0m = #zeros of GH-mbranchesendingtothezero ofGH "¥ q = m-mebraucher tending toa 3)simmetrywithRe-axis . 4)fa = * "¥ withk = 1 , 3 , 5, ..., 99 · WaEPi-1z :"¥ poles&zerosofGH5)realaxissegmentstothe leftof anoddnumber of realpoles/7belong tothezootcocus.SuXPa#110X111111111/IIX :SuGH = 2L = 1+GHP =- 1G · Pez =- S m(9 + 2P. = 0Ps =- 2 &x = = ka = + ev = =fora large upthetworeswillgrowfromeachotherintherealaxis ."¥ Dynamicinstability.We found theR-Locusinthecase ofnegativefeedback , incaseofpositivefeedback : first theclearethesame 4)On=*2k * withU = 1 , 2,3,% , ...,9thecatestaysthesame5)realaxissegmentstothe rightof anoddnumber of realpoles/7belong tothezootcocus. CONTROLS : CONTROL : · -8 m , z , kso=>Hp : stable&undamped.1dof : *mi+ex+kx = ko) withseeP,2=-x+iThe target is : 1Nof1.Timedomain :‘ mi+ex+ 1 Kpx = Kpk thestiffnesofthesyste is oaimputfunctionof Up. "¥ wecanstabilizeanunstablesystemwiththep-controAPCcanstabiliseonly a STATICINSTABILITY &if and onlyif les0 -staticinstability ? X 1unstablepole . Let'sconsidernowthatthesystemisstabe , insposingthe fee motionsolution : ↓tx = x2=> (m) +2)+k+ kp) = 0 ->maracteristeg . ofthesys. "¥ --* J with =Im h = ↓ ↓WaJ2 = - Wif h,91therootsarereal&the systemis orrdoped , ifhethesys , isunderdamped.kpt =>Wo&theresponseisfaster. - h ,b if westartfrom c . con;rootstheresponsewillhavelargeroscitationwithhigherfrequency.ofwestartfromrealrotstheresponsewillstarthavingoscilations.The serformanceof thep-controlisgivenby hourcloseIistokef xrefx = xn + up = 11 , t - Habit non t A Ht 6. xn - up=(kp + k)xp = kpkz = kpun = e*(ycos(w ,+ y)) n Let'snowseesomeindicators : 1)STEADYSTATEERROR : 20 = 1 - Nohowcloseisthesys - responsetothe rof &thesteadystate2)RiseTIMEtrtimeneededtothesys . toenterthe90%zone of the zef , how fastthesys-response.K 3)Po = krax - No percentageovershot.x 9)SECTELINGtimets : timeneededtothesystoreachthe steadytstate . Theseindexesare usefulonlyif theenorc = zop-k issmall enough .Rp&=>no = 1debutI=>wat = s+zb II = >h= b=00a If wecontradastabensys - westillobtain a ne stabensys. 2 . Laplacedomain : Lma+2x+kx = f - (ms +25+ k)X = F-coutz . F = Kp(Xzy - X)r(s) = E = kptransferfunctionoftheregulature. Els=* =is Yeste If ofthepassivesysBlockdiagram : * Fut &-> Hp : sensor&actuatorideal - testn kpOpenLoop[F : GH = RG = ms2+es+1ClosedCoop[F : L = kp -> performanceofthesys : mSz+25+k+kp(E1 = )z= xufR = kpIRldB => ke24 -2->there'sno influenceof thecontroletothephaseforopenLoop Bode criteriousPolesofGH = poles : P.,2=- X Fiw T - Wehaveno infos · - Sys . isminphaseInszeos/cust - Poles). Pi ,z -SincetheIdoen'tcross-# : Gm D"¥ SS - thesys - isstable -T PuIf therespectionalgarkpincreases: kp4-Pmb(10) =>thesysisstable-kp guitario. · lStable&kpBlocus : \kp : 0-0=>therealpartstaysnegative ↑ andthereforethesys , isstable*kp .-X · RePi ,z= -CFiWpolesof GH.Y... =q = -m=2branchestending toosincethere'sasynthots & nozacos J & = withk = 1....,q= 0x= FG = 3 =27i= & =- x9Assaidbeforetherealsartstaysthesame , whiletheSupartincreasesasthekpincreases. "¥ damping isreducedh,b- = h , Wa Performan : ILlaB ( ... 15/10)"I, es we SimplifieddiagramferLstartingfromGH : - - 39 ~To , GHlarge , L = K2+8,GH0 , L= 1 G-G2 aB If Kp&,Cob(+0GH Mo &B - >bothIGH&Illhavethesamestope - todiaL2w > kp()4sincetheirpolesarethesame "¥ kpa= Wo =34 Ei finalvalue : No=Cu()= Xof=(witStep). "¥= >X = (Xm Vo =((s)X(y = ((0) = 6Go = 1 - No 27/18PD-CONTROL :1 def : seHp : Stable , underdamped , Im m, 2,0 - s - - f Imi+2x+kx = f P ,2=x+iw "¥x e I = Kok) o Ta r g e t :x Kofmixe+2xi + kx = kp(xnf - x) + ky(22f - x)min+ (e + Ks)2 + freshet - she controlactisis = effectingboththe globaldamping & stiffen1.Timedomain : Im , 2,,k=50=Stablesince2dkso&thegains arepositivetoo. Ingeneral : SK8=Static - >wecanstabiliseR>0 instabilitythissystemwithpcontroller(1custablepole) . 20 = > Static/dynamic - >wecanrecovertheinstabilitystability withtwopols12must.pol x I - - 2+ ze] si = is= Lutfa, statoeRAPRESENTATION eigst)-(und ↓ -withW da2--wat if k (Wongala thesysgetsfasterTb regulatoif kh "¥ thesyshaslenoscilations. Performance : KWo-tztA ① Koea Iwho Rob 400) , I +kpsys - reachesthesteady stateinthesametime , K ⑨ kathatPodak4(kps)kpaWas ↑ Co ++tsb An stfor a stepimport : Ref = 1= King =+ &+ = 0 "¥ 2 . Lapadomain : mi+rx + kx = f = ) (ms +2+ k)X = == = =) = ms- Ye s + h f = Kol P1 ,2=- Xtiwe � 2with2(e) = Ef = kpz + kySE = (kp + kps)(Xy - X) = Re-RregulaturTimeconstant[d : Kn = Ta k p =>R = kp(1 + Ta s ) "¥ 3 ariq[ I cou conte a e IRlaBduetothezero D Increasing kpdoesnot change thephase,but onlytheamplitude.Z = - So I comchaya Akp S 4Etheshapedoesn'te92 changeeitherEX das) this + R - OpenloopTF : GH = RGH = Kp(z+ Td s )P. (GH) =- x+iwz = T ms"+25+kRCloseLoop[F : L = 1+RGH J kp(1+ Td S )ms" + (2+kp)S + (x + kp)BodeCriterion:& IGHlaspolsofGH = polesofG*diagrammadiGH, * fossestatoLsaribbe↑= = Ea 90) StableKP52 - -YPm >0 04+ Pmb/ , got -T"¥it'smeistheone &ofIGHlarwhencrosses"¥ iflettheHasdiagramcrossesthes theOlive Liveafter , butthephaseneverreaches -I Nyquist : SeIf kprthe diagon neveroverlapthepoint3-1 , 03 · N =- P Stable -botcocus-NSmTo fixed Kp : 0+ a2x-wm= 2m= 1 -X-sh9 = 1 E nX-w 2 = + = πStableofUp Performance : ab& + "¥ ot th 27 &P-,2 L->kpz = - Puz(w)ak+= p- IGH/dB&what if Pe ,2 letifz< P1 ,2 & Gu =+-48-2 >2 Im 30P1.22 Lsmaller34bandwidtha4- - SoI'mreducingstability aswellas-T↑Pm >0 performance . kn = Ta k psmallslowerUpalower "¥=- Edeffect onthesys.PHASELEAD : distentionof thephaseinsidethe system ."¥ responseis fast ,butthenthetimetoreach steadystateis longerI everwithverycouroscilations. 3/4PI-CONTROL : 1 def : seHp : Stable , underdamped ,m, 2,0 # = f mi+2x+kx = f P ,2=x+iw 38 # "¥ Ta r g e t :x Kofz non allowustonou // 1.Timedomain : eviz I mi + 2x+(k + kp)x +k , (xdt = kpxy +k , /xdt6 . Y-Jadt = y = x , y = x , ij = 2 mij+zij + (k + kp)y +k , y = kpYo + k , Yz =Iorderdiff . eq. "¥ p += Wa s h w a "¥ thesys , responseisfasterwithhigheroscillations. Lifet tomoretoa y = i SPACE We havetoadd2eqs.statesecte : z = / = (4) a) - I I ** + 34 , +I1000statematrixofthefeedbackcontrolsys eig(Ac) = det(15 - A) = 0 &13 + 5 + more "¥ thesys , isstable2 . Laplacedomain : thisis onlythesystem1mi+ex+kx = f =(ms +2s+ k)X = 7 = G = = ms+es+kIloS . c . f = mp(ky - x)+k , ((ky - x)dt = + = kp2 + = = (kp+) =I2(Jeat)=(xy - X)= = REThe integratinggain isgivenby a "¥ M = up+ 4 = ko(v+) = wo)) bintrocingIpole&theorigin andazero in-Fi Iflas& kp4↑So increasingthekedoesnot charge-2 thephase."¥ introducing thepoleinthe wigi S2weareabletoreducetheZ steadystateeverto zero .b4 2intheIScontrollerwewouldhaveS - tohavealp- a u = Tis(ms" +2+ k) + m(1 +4 , 5) R "¥ Ensu ↑ - - no(T , S + 1)OpenLoopIf : GH = RGH = no)v + s)montes +u = < , s(ms +es+ r)CloseLop[f : L = G = 40 (5 , S+ )↑(ms +es+ (k + kp)Sk , )Bode :- as Kp4P1 =0 -2↑P2 ,3=- Xtre -+S z = - i>2P,z02 ,3 =p . (1)(w) 4 >2 Thesys ,is &minimumphase : · I Gm =+00=> Stable+kp -T- T& m>0 ↑ Pm>O ifkn= 0mb(0)Nyquist : P = 0 m ostate BootLocus : kp : 0 ->+o TifixedPolesGH = polesof2Imm= 3 ↑ AX-W P. =Gz = - - E -X D-XsheP2 .3=- Xtiwm= 1 & --w q = m - m = 2 2 = n = 1 , 3XStable+kp2x = + -Mini Zufomance : 2 = 1 - 1020 = CultSX(s) = e(s) =* I step -> Xy = > Ms ↓ for ae regulator ea =0+K .a1lasGHkp4 kowswa ↑ WcI>2ha = zmwb[00 P,2L92 ,3-> kp4 >2 -->4s -T- n If we placethezer&veryhighrequency , wewould havelarger bro-X&StthanX , making thesysmustable. & alGHlas -WAlsowithBodewecanseethat -60thesys , isstabeonlyforCourke. - 4032 If thezew is &thesoles ,one of P,+,&themiscancelledout.b4 >2 I↓ sys - unstable-wehavetoplacethezeroonthe left - of the poles . EFFECTofDiSTURBANCES ameasurementsNose : 12/11&disturbanceNot + Ex-> te Aswecanseethesyscontrolles t nowhas3imputs : - measurementY 1tMnoise # 04 f , d , m .tSensor y = (x+ n)H x = (f + d) = ( y) + d)G = (r(xy - (x + m(H) + a) =3 I &= Gd+ RxG - RxHG - RnH6 Gd + GRRGHx - 1+ RGH1+ RH2mf - 1+ RGHm . withGH = GopenloopIf---Ifdist . IfimputIfraiseoutputoutput . b &orton X allItshavethesamedenomin,x = (d + (xy + hmm same sysandsame poles ,thereforethesame stabilityproperties . 14a &WaWmax , refsi Ta r g e t x- xu↑ 30s lyByL = 1, La = Lu = 0 To - Hp : H = 1(idealsensor) . -d1 - G = ms+2 + 1HP : Prz =- &tiw � S - x P-control : f = kp(kn - y) ~ Gr= ↑mBN4=>104d3 Ta r g e t L=1 :Akp↑&rw=G31=LF1 - 2xw-G(1 = )L=G -48L&G(11kp ko L&-GH = ms+2 + kwo wa "¥& kp "¥- ms2+2s + (k + up) final value Zi up -> therem -T– up 20 = 1- wo ⑨ : up Wa t "¥ he zomb oe12 "¥ e. b(e . +0 & ⑨ : Katheto - · Sin "¥ e o== u START92 ① : kit =>C. = 0Was worr(B4=(s12 . L = -= = 0= 3 M Ta r g e t( = 0(4) = 1(m)& k(m = T+ 4) "¥ When thishappensthesameconsiderations- = - we'vedonewithIholdforInE- xu = (1 - 1) with= f Sd = O S97?? x = (xm + (um = (xn - LM 2W = =(5 - 1=E +m= >themoristhenoise , affecting theMIlmeasurementsreventingtofollowthe -L &reference.1(l = 121, KLm =- #rxW == )(n = Lf - G= s ((m) = (G), k(u = ++4Gd =0= >x = (xy + tum = Lxz - LM&xx -x= (1 - 2)xof +2n refereuertracking =L= (ZnI conflicting requireaNoiserejection(50= (K HultradeoffchoiceisR>Bu =>thespeof the amplitudediagramof Lmustbethe ↓ highestpossibleneeded.ascontrol - >reducestheslopeehighfrequencies.NOTE!:garrallymeasurementsnoiseis&highfrequencies .Thechoiseof wemustbemadebetweenthecapabilityofcontrollinghighfrequencin andthecapabilityofthe sysofreducing measuremnoise3. == 199 , 95,Lab Ta r g e tLa +0 & 2Wo=G= La = T & Kal = t 4(a =- 4R~W==G221=L+(a) = IGnothingtodo& 4Ld = KG -ThepassivesysGsiltenmoise . dB A up&Wo&UpR↑ &G , >2wo-2-kp ‘-T-10kpnino I L D : Kpt = W ak "¥ BN(teb) "¥ Kalr=0)1 0 x ⑤ : Up == = htPob "¥ nab(004)sin , but wa w ka↑ ① ki = 0= ((d(r = 0)) = 0 SENSORDELAY:x(t) - ay(t)Y &J #-tE "¥ time alay FI mi+rxi+kx = fTa r g e t : x -xn aD Ap : P ,2= -XFic x Pcontrol -> f = kp(x - y) 1 E = 7 = ms + 1s+kY = HX H =e- S 2(f(t - c)) = f(s)e s t f(t) = x(t - 2) -y(s) = HX(s)H(s) = H(0) + S = 1 - ze% 5 = 1 - Es=H=1 - Es "¥ incarisationdiagrou · GH = RH = ko(r-Es)·ar test ms? + 25+k i &openloopz = &sesenso tiw Karl 24 . c. polsup · L = GHms+(2 - 24p)s + 2 + upclosedloop.~uBode : Mzk+2cc . pols *GH allo &II&2 z = EI smallD- -AD-2 - -- >h-Zπ Nyquist :· kolarge : N = -2-E Su "¥ unstable · SmallN = P = 0bstable . Howtodesign acontrollerfor a2 defsystem.17/17 S 6theactuatesisnotplaced wenowhave2 2DDEuse + 2, 12 - 2,(x ,- x2) + ka22 - kz(z - x) = fu = - 13)f = 14) =thisintroducesthecontroller. (mo 12 + B2 + 47 = fSince1 , &&Iare symmetrical and500 , thesys . isstableI soit'sthepassivesys.looking&the freeundampedsys.:S - + 17 = 0=> (Mx + 4)ze" - o= = zebt at)15 + 4) = 0 = 35m ,+ k + kmI=>characteristicequation .- kzdmy+ kz +42 = m = 23 = UsLet'ssupposethatm .= My = M , 2 = 2 = 1,1 = 42 = K↓ 2 m+2x - kI - kJm +2kI = jm = + 4kmj2 + 342 = 0 & Wa n "¥ wa = Moderofvibration : 15) (, Themanesmove togheterwiththesame amplitude , sotheseeing "¥ G(s)isthematrixofIs .A'(s) I ms + 22S + Mr S+M IG (S) = def (A(S))zus+1ms" + ZuS+Un "¥ raracteristiceq - ,tiw , wa P2,=- Xtiw , Wa Passivesys.wewanttocontrol : ↑Gener - a>x = G , f ,+ GizFz E Er # ->K ,= Grafz + Guf ,"¥ Designofthecontrolles :· COLOCATED : F ,= kp(kng - 2)UnfEz = 0 - - Sincewedon'tcareon controlling42 , theupperpartof theblockdiagram canbe ignored . da paginaS1 G = = ms+2es+2 - + 4, det(A)det(A)ms2+22s+2kOpenloop : GH = RG , H = UpPiz->Wa = m det(A)Pa, Wo r closedLoop : -G 22 IGHIdB- -4x-18PizZi&342S4 &=2 - inso -T · NonCOLOCATED : · - Gr =- Openloop : GH = RGH = Ko Piz->Wa = mPa, Wo r closedLoop : -G z, = - IGHldBX e --41-80-60PizPat72S14=2-T- ->unstable 2 - π- Im -2π- DCMOTORS : 19/11ragneticfield canbereducedby awirerunedby acurrent.->& = MHpermeability ( b-magneticfieldmagneticfield DENSITYINTENSITY (9)(Hm] "¥ canbemeasured by theApencircuitallaw : i � xdl = MiExample : ↑iBisconstant I 32nd = Mari · theliveisacircleLid – s MariB = 2nd=> righthandruletofindthedirectionofB Ti Lorentzforce :XXXXi= = = il&/ilBsin8 = -↓XXXXB =coust.seK9) with(8directionisgivenby Lthe righthand-BB aF Twowinesrunbythesameiattracteach other.r+&i Faradayaw : Magneticflax B > i "¥ p = &Ai(Nb) -> - "¥ scalarproduct ->9 & -->B e-erf(fau) (1) ATAvariableacreatesan electromagneticforce.Va r i a t i o n s :- B(i)ofa - A=e changes - 8Elementary electricmachine : justtheoreticalB = constX ex 7XX ! = 2xAm = 3A = Bexe- x2)x aiF I J->ee == = BeXXXX&Thusacurrentisinduced I different ↓N& S ofpotential) ."¥ · weare producingelectricity==ilns/f = ilB ->inordertohaveIwehavetoimposethesame face , elseitwon'twork S need a mechanical "¥ if wedon'timposeaPowerforcesf , itwon'twork Pre-EX = ilso they are equal &Pr = ci = ilbv If we supplythesys - withamentwehaveanelectricmoter. Motor :B = coust .&XX7XXf = ilb exi fe-lBv - >wehaveabackcurrent udue bythegenerationof I JXX ↓ electromotiveface .2wehavetocounterit. ↓&N Pre-fu = ilBVSelectricpowerbecomes Pes-ei = ilbrmechanic- = K,ie = k,0 Up-be =>hastobethesaveRealelectricmotor :↑& - F ALA& · & i &F - SConque : Im iSee=> =Zilb sing = ilsesim(rt) "¥ itstarts rotating r = Pee-eiarthetorquechangerdirectioneveryhalf period. CAP&S , an "¥ Howalway thethecurrenthasto e sturedirection. Wecanresolvetheproblemwithacoil: - &Tu rCOMMUTATORis · & srtai % brushestouchthewicesand – givethecurrent, everywolf rotationthewilchanges. SThetequestillisn'tconstant , adding loopsmalthetorqueSmoother . ↑ m= kilbr = kdi->coust . (b)sivament*up = kamagnet smotor . e = kar "¥ S & ↓ variable separat a "¥ dependsonthe#ofloops. P = BA "¥ Bcoust .Permanentmagnet XCmotees:statoz Mater (rustorshaft 7-· In loops ⑨S↑· g "¥ sc voltagegenerator -26/11Tn = kpia -d additionalvoltageisneededtokeepc = k p w [in theroterestatingiaNaElectricbalance for the circuit-Kirchoff's eq : sa prin i I Va = Raa + La +e Kirchoff Ie = kpwFaraday's Law- it tokeepthemotorDWTaTm - KpiaLorentzforcesrunningahastobedeltwithPa = Va i a = Raia + La 2 &i that tfea -PourePowerstoredbytheinductancethemechanicalbss lissipated / back unaeia = kpiqw = Tu W = Precsteady state : I w= w . (coust . ) =>w = 0di-aVa = Wa s ( c o n s t . ) =>powerstorediszerosinceat"Oia = ias(coust . )ia=-RW = [m = Kota CharacteristiccurveRa J ofthemotor.↑ mA VaBaRa 2 kaTa&inthisregionthetwqueapplied / opposite , sothemoteractsW&brakeWo Ta r g e t : ODuf Genui (Sm + mL) + En8 = Tw = Kpia M&5i+TmW = Tm = kpia horizontalplace La Raa=k eMBut· To Let'sgototheLaplacedomain:Jr+ 2mr = Tm = Kiawith : r = 2(w) LI (Las + Ra(Sa =a- kpr =Sr = SDfirsttwoeg . sarecoupled .G = Em = 1Js+Im5a= - Mon a ELECTRO - MECHANICALSYSTEM : Na +Va - kpz-> it III - ↑ - d Ifofallthesys : kbG-m= Ga (LaS+Ra)(Ss + em) + 4)kb=>theactuatorintroducesG+ = - = s()(aS + Ra)(Ss + em) + 4) anadditionalpole(inthe origin) Sincethere'smo ! stiffen ofthesys , is fee torotate . · regulate : Nakako - J - & · inte ⑤ Go isboth sysd& actuator- > feed-backcontrol -M I Of + L- sys . #TeWecannotacidetheaffectoftheactuator from thesys. OpenLoop : kpkpGH = RG-H = s()LaS + Ra)(Ss + em) + 4) ClosedLoop : = GH S Fortheopenloop :pocesarereal1 = -X ,, Pa =- XBootLocus : q = m - m = 3~ 5027, I3 · 2a = + k = 1 , 3 instability & 8a = + ,T Increasing thekpwillintroduceIthe dynamicsofAintroducejust If thedym.of themotorisfastenthanthe dynamicsof thesys-theeffectofthemotzedynamics onthesys . issmalles ,soIcan neglect Ef Jtheactuator.N "¥ Sincethesys-hasapoleintheorigin , there'sus steady stateeno, m therefore a11controlsis uselen . If thepolesarereal , theresponsewillhavenooscillations.stAhas dynamicinstability Wecanalsocontrolthespeed : 5sr+ 2mr = Tm = Kia · I(Las + Ra(Sa =a- kprkbbaying on Gor = (LaStRal(Ss + em) + 4 a Pregulator : Va = kp(ag - r) = R= Fent ↑ - openLoop : GH = MGoka +em) + 4ClosedLoop : G + + GHFortheopenloop :