Topic #3 16.31 Feedback Control Frequency response methods ? Analysis ? Synthesis ? Performance ? Stability Copy right 2001 by Jon at h an H ow. 1 Fall 2001 16.31 3–1 Introduction ? Rootlocusmethodshave: –Advantages: ?Goodindicatoriftransientresponse; ?Explicityshowslocationofallclosed-looppoles; ?Trade-o?sinthedesignarefairlyclear. –Disadvantages: ?Requiresatransferfunctionmodel(polesandzeros); ?Di?culttoinferallperformancemetrics; ?Hardtodetermineresponsetosteady-state(sinusoids) ? Frequency response methods are a good complement to the root locustechniques: –Caninferperformanceandstabilityfromthesameplot –Canusemeasureddataratherthanatransferfunctionmodel –Thedesignprocesscanbeindependentofthesystemorder –Timedelaysarehandledcorrectly –Graphicaltechniques(analysisandsynthesis)arequitesimple. Fall 2001 16.31 3–2 Frequency response Function ? Given a system with a transfer function G(s), we call the G(jω), ω ∈[0,∞)thefrequency response function (FRF) G(jω)=|G(jω)|argG(jω) –TheFRFcanbeusedto?ndthe steady-state responseofa systemtoasinusoidalinput. If e(t)→ G(s) → y(t) and e(t)=sin2t, |G(2j)| =0.3, argG(2j) = 80 ? , then the steady-stateoutputis y(t)=0.3sin(2t?80 ? ) ? The FRF clearly shows the magnitude (and phase) of the responseofasystemtosinusoidalinput ? Avarietyofwaystodisplaythis: 1.Polar(Nyquist)plot–Revs.ImofG(jω)incomplexplane. –Hardtovisualize,notusefulforsynthesis,butgivesde?nitive testsforstabilityandisthebasisoftherobustnessanalysis. 2.Nichols Plot – |G(jω)| vs. argG(jω), which is very handy for systemswithlightlydampedpoles. 3.Bode Plot–Log|G(jω)|andargG(jω)vs.Logfrequency. –Simplesttoolforvisualizationandsynthesis –Typicallyplot20log|G|whichisgiventhesymbol dB Fall 2001 16.31 3–3 ? Uselogarithmicsinceif log|G(s)| = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle (s+1)(s+2) (s+3)(s+4) vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle =log|s+1|+log|s+2|?log|s+3|?log|s+4| and each of these factors can be calculated separately and then addedtogetthetotalFRF. ? Canalsosplitthephaseplotsince arg (s+1)(s+2) (s+3)(s+4) = arg(s+1)+arg(s+2) ?arg(s+3)?arg(s+4) ? Thekeypointinthesketchingoftheplotsisthatgoodstraightline approximationsexistandcanbeusedtoobtainagoodprediction ofthesystemresponse. Fall 2001 16.31 3–4 Example ? DrawBodefor G(s)= s+1 s/10+1 |G(jω)|= |jω+1| |jω/10+1| log|G(jω)|=log[1+(ω/1) 2 ] 1/2 ?log[1+(ω/10) 2 ] 1/2 ? Approximation log[1+(ω/ω i ) 2 ] 1/2 ≈ ? ? ? ? ? 0 ω lessmuch ω i log[ω/ω i ] ω greatermuch ω i Twostraightlineapproximationsthatintersectat ω ≡ ω i ? Errorat ω i obvious,butnothugeandthestraightlineapproxima- tionsareveryeasytoworkwith. 10 ?2 10 ?1 10 0 10 1 10 2 10 0 10 1 10 2 Freq |G| Fall 2001 16.31 3–5 Toformthecompositesketch, –Arrange representation of transferfunction so that DC gain of eachelementisunity(exceptforpartsthathavepolesorzeros attheorigin)–absorbthegainintotheoverallplantgain. –Drawallcomponentsketches –Start at low frequency (DC) with the component that has the lowestfrequencypoleorzero(i.e. s=0) –Use this component to draw the sketch up to the frequency of thenextpole/zero. –Changetheslopeofthesketchatthispointtoaccountforthe newdynamics: -1forpole,+1forzero,-2fordoublepoles,... –ScalebyoverallDCgain 10 ?2 10 ?1 10 0 10 1 10 2 10 3 10 ?1 10 0 10 1 10 2 Freq |G| Figure1: G(s)=10(s+1)/(s+10)whichisa“lead”. Fall 2001 16.31 3–6 ? SinceargG(jω)=arg(1+jω)?arg(1+jω/10),wecanconstruct phaseplotforcompletesysteminasimilarfashion –Knowthatarg(1+jω/ω i )=tan ?1 (ω/ω i ) ? Canusestraightlineapproximations arg(1+jω/ω i )≈ ? ? ? ? ? ? ? ? ? ? ? ? ? 0 ω/ω i ≤0.1 90 ? ω/ω i ≥10 45 ? ω/ω i =1 ? Drawthecomponentsusingbreakpointsthatareatω i /10and10ω i 10 ?2 10 ?1 10 0 10 1 10 2 10 3 0 10 20 30 40 50 60 70 80 90 100 Freq Arg G Figure2:Phaseplotfor(s+1) Fall 2001 16.31 3–7 ? Thenaddthemupstartingfromzerofrequencyandchangingthe slopeas ω →∞ 10 ?2 10 ?1 10 0 10 1 10 2 10 3 ?80 ?60 ?40 ?20 0 20 40 60 80 Freq Arg G Figure3:Phaseplot G(s)=10(s+1)/(s+10)whichisa“lead”. Fall 2001 16.31 3–8 10 ?4 10 ?3 10 ?2 10 ?1 10 0 10 1 10 ?3 10 ?2 10 ?1 10 0 Freq (Hz) Magnitude Actual LF MF HF +1 0 ?2 ?2 +1 ?1 10 ?4 10 ?3 10 ?2 10 ?1 10 0 10 1 10 2 ?180 ?160 ?140 ?120 ?100 ?80 ?60 ?40 ?20 0 20 Freq (Hz) Phase (deg) Actual LF MF HF Bode for G(s)= 4.54s s 3 +0.1818s 2 ?31.1818s?4.4545 . The poles are at (-0.892, 0.886, -0.0227) Fall 2001 Non-minimum Phase Systems ? Bodeplotsareparticularlycomplicatedwhenwehavenon-minimum phasesystems –Asystemthathasapole/zerointheRHPiscallednon-minimum phase. –The reason is clearer once you have studied the Bode Gain- Phase relationship –Keypoint:Wecanconstructtwo(andmanymore)systems that have identical magnitude plots, but very di?erent phase diagrams. ? Consider G 1 (s)= s+1 s+2 and G 2 (s)= s?1 s+2 10 ?1 10 0 10 1 10 2 10 ?1 10 0 Freq |G| MP NMP 10 ?1 10 0 10 1 10 2 0 50 100 150 200 Freq Arg G MP NMP Figure4: Magnitudeplotsareidentical,butthephaseplotsaredramaticallydi?erent. NMPhasa180deg phaselossoverthisfrequencyrange.