6.31 ECTUR乇#2 DOMIN ANT POLES ROOT LOCUS BAsICS PERFORMANCE工sus DYNAMIC COMPENSA TION P.I.D SYNTHESIS REfERENCEs To FRANKLIN powELL CoPY r(GH千J.Hw,2ol
MG!∈R- ORDER5)sT6As ●RREL丹T(0sHP5 BETWEEN1MER6soN5毛 TAST∈P兵 ND THE POLE LOCAT(0NS心EK ALCULATE0 FoR A SECOND-ORDER SYSTEM GuES60工 NSIGHTS AcTuALLy Gooo APPRoXIMATIONS foR MAwy HIGHER- ORDER SYSTEMS BECAUSE THEIR TRANSIENT RESPONSE IS DOMINATED BY A PAIR oF COMPLEX-CONJUGATE POLES RELATIVE DOMINAACE oF CLOSED-LOOP ADLES DETERM4EDβ O RATIO oF REAL PARTS OF THE CLOSED-Loop PoLES R ④ RELATI∈ MAGNITU0 ES oF THE REs0UEs ENALUATED AT THE CLDSED-Loof PoLEs O SLOWER Po LES TEND To DoMINATE MoRE THAN FAST ONES(WRICH DECAY QUICKLY),ALL ELSE BEING EQUAL.( FACTOR oF 5) ② A ZERo川ER升 POLE WILL TE0T0 EDuCE ITs EfFECT oN THE SYSTEM RESPONSE (ANO RESULT IN A SM ALLER RESIOUE
EXAMPL Y G +s+5 U=s PARTIAL FRACTION EXPANSION (-54)( (s+)(5+5 S十 s+5 七 5七 1-125e 625e 25 LARGER THAN 5 5七 e SL0 WER THA小e ∴. EXPECT THIS TERM1 DOMINATE R∈ sPONSE 55(5+0 (s+)针+5) =2b=55(6+64)1+22+(2 (s+)s+5) ≤+l s↓5 5七 t)≤4+6.24e-125e MUCH SMA LLER CONTRIBUTION THIS TIME EXPECT e-s (FAST) TD DOMINATE THE 工NTAL咫 SPONSE→SHL0 THEN SEE THE CONTRI BUTION OF THE SLou MoDE
Slow dominant pole num=5 den=conv([11],[1 5])step(num, den, 6) Time(sec. Fast dominant pole num=5.5*[0.91];den=conv([111,[151);step(num,den,6) Step Response Time (sec
Slow dominant pole num=5;den=conv([1 1],[1 5]);step(num,den,6) Amplitude Time (sec.) Step Response 0 1 2 3 4 5 6 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Fast dominant pole num=5.5*[1 0.91];den=conv([1 1],[1 5]);step(num,den,6) Amplitude Time (sec.) Step Response 0 1 2 3 4 5 6 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Similar example, but with second order dynamics combined with a simple real pole Z.15; wn=;plist=lwn/2: 1: 10*wn nd=wn^2;dd=2* zwn wn^2l;t=[0:.25:20; tf(nd, dd; y]=step(sys, t); for p=plist; num=nd; den=conv(l/p ll, dd); sys=tf(num, den): templ=step(sys, t); stem end pot(ty(∷,1),d",t,y(:,2),+,t,y(:4),+',ty(:,8),v); labeld'step response), xlabel( time(sec)) legend(2, num2str(plist(I)), num2str(plist(3)), num2str(plist() ◇ 2nd 16 05 + 2.5 o28a Q 000 6◇ 0.2 (sec) For values of p=2.5 and 6.5, the response is very similar to the second order system. The response with p=0.5 is clearly no longer dominated by the second-order dynamics
Similar example, but with second order dynamics combined with a simple real pole. z=.15;wn=1;plist=[wn/2:1:10*wn]; nd=wn^2;dd=[1 2*z*wn wn^2];t=[0:.25:20]'; sys=tf(nd,dd);[y]=step(sys,t); for p=plist; num=nd;den=conv([1/p 1],dd); sys=tf(num,den);[ytemp]=step(sys,t); y=[y ytemp]; end plot(t,y(:,1),'d',t,y(:,2),'+',t,y(:,4),'+',t,y(:,8),'v'); ylabel('step response');xlabel('time (sec)') legend('2nd',num2str(plist(1)),num2str(plist(3)),num2str(plist(7))) 0 5 10 15 20 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 step response time (sec) 2nd 0.5 2.5 6.5 For values of p=2.5 and 6.5, the response is very similar to the second order system. The response with p=0.5 is clearly no longer dominated by the second-order dynamics
Ro LOcus BAsICS BASIC F∈∈06 CK SYSTEM GA PLANT TRANSFER FUNCTION CONTROLLER T.F. KNC 2.D, MoNIC. CONT ROL COMM ANOS PoLYNoMIALS工川s SYSTEM OUT PUT 「; REFERENCE INPUT(EN·) RESPONSE ERRoR WE WILL 0ISCVSS THE PER FORMANCE GOALS N№R0 ETAIL LATER,f0RNu可s7 CONCENTRATE ON LOCAT IDN oF SySTEM POLES THIS IS THE WITy FEEOBACK FORM WE COULD PUT THE CONTRoL G IN THE FeEOBACK ATH WITHOUT CHANGING THE POLE LOCATIONS DIST URBANCES AD0ED LATER
BASIC QUESTIONS: ANALYSIS GIVEN Nc 0c WHERE Do THE CLOSED-LooP POLES GO A5 A FUACTIO of THE GAIN K. I.E. WHAT K To AclE- YNTHESIS GⅣENNP,D HOW SHOULO WE PICk K,Ne,Dc To GeT THE CLDSED POLES WHERE WE WANT THEM 小DE:BoPE5M∈ THAT WE KNOW WHERE THE CLOSED-LooP POLES SHOULD BE LOCATED, THIS THEME WILL RUN TH ROUGHOUT THE ENTIRE COURSE uElL5PE川AL7 F TIME LooK|NGAT工T。 BLOCK DIAGRAM ANALYSIS EASy To SHow THAT y P L 十G CLOSED-LooP CL K Np Dp k Nc Np TRANSFER FUNCTIO THE DENOMINATOR IS CHLLED THE (4) CHARACTERISTIc EQvATIDN +THE RD0TS OF ARE CALLED THE CLOSED-LooP POLES. POLES ARE CLEARLY A FUNCTIDN of K (FOR GIVEN Ne, Np, De, Dp)+ A"LOCUS OF AooTp
OBSE RVATIONS LOCUS IS CONSISTENT WITH FIXING E COMPENSATOR DYN AM 0THE小 CHANGING THE GAIA K ROOT LOCUS ENABLES US TO DETERMINE KEy FEATU RES OF THE CLOSE0-L0oP /STEM TRANSIENT) RESPONSE (FROM THE POLE LOCATLONS GIVEN THE OPEN-LooP N FORMATION 0c.k WILL SEE THAT IT IS HARD To INFER SoME PERFoRMANCE PRo PERTIES FRoM THIS LOCUS sE800ETEc川uT00 PPROACH SUGGESTE D BY W.R. EVANS Evans, W.R. Graphical analysis of control systems, AIEE Transactions, vol 67,, 1948, pp 547-551; and Control system synthesis by root locus method, AIEE Transactions, vol. 69, 1950, pp. 88-89
2 ROOT LOCUS ANALYSIS IN GENERAL THE FULL ROOT LDCUS IS VERy COMPLEX AND REQUIRES MATLAB TOOLS LIKE LOCus(AM,E心) EVANS ORIGINAL PAfER USES A SoURCES +SINks RNAL⊙GY FULL PLOTTING RULES A FPE AAGE 260 → NEED T0 EVELof SoMEβsc0RWNG 5KLsD心 E CAN Do“ BACK oF THG ENVEL0作 0ESG小s BAsIc POINTS: ASSUME Nc, Dc KNOWN LET T帖EN8c-A+KLd K REAL, PosITIVE CAN USE THIS TO ANSWER BASIC QUESTo N 5s。0A1 HE ROoT LoCs2 As:0NL工F凵(s)。\30。360°·L L6,1,2, k PosITive 1飞0°LDcs NEGATIVE LOCUs
BASIC QUESTIONS O WHERE DoEs THE LOCUS START POLES ib WHERE DOEs THE LoCUs END? ZERo s ASYMPTOTES ③WNEN/wERE工 s THE LOCUS ON THE REAL LINE LOCUs POINTS ON THE REAL LIE ARE千0 E LEFT0fN ODD NUMBER OF REAL-AXIs PoL∈SAA0zER0s ④G叫小MAT5。工5 ON THE LOCUS HAT GAIN1sNEE0ED16T训S CLDSED-LDoP PoLE kls: k De