-0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 Figure 1: Rawstep response data Solution 8.8.3mtr4cy The rst step is to get the data into MATLAB. The MATLAB statements load mtr4stepc t=mtr4stepc(1:500,1);; y=mtr4stepc(1:500,2);; plot(t,y) print -deps sr883mtr4cya.eps seperates the data into a column vector of times and a column vector of responses, and then plots and saves the step response. shown in Figure`1 As can be seen the time response, whichwas taken o the oscilloscope, does not begin exactly at t =0,andnot quite at y =0.The MATLAB program t=t+0.0599;; y=y(45:500);; t=t(45:500);; y(1) = 0;; y(2) = 0.05;; 1 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 Figure 2: Adjusted step response y(3) = 0.1;; y(4) = 0.15;; y(5) = 0.2;; t(1) = 0;; plot(t,y) print -deps sr883mtr4cyb.eps produces the adjusted reponse shonwinFigure 2. We are nowinaposition to apply the identi cation techniques discussed in Chapter 3. As a rst step, the MATLAB statements maxy = max(y) y1 = (max(y) + 0.0001) -y;; w=log(y1);; plot(t(1:300),w(1:300));; print -deps 883mtr4cylny1.eps produces the response shown in Figure 3. Note that the maximum resposne was increased to0.001 to avoid ln(0). 2 0 0.5 1 1.5 2 2.5 3 -5 -4 -3 -2 -1 0 1 2 Figure 3: Plot of ln(1;y(t) 3 0 0.5 1 1.5 -14 -12 -10 -8 -6 -4 -2 Figure 4: Comparison of adjusted response and y 1 (t) The slope is roughly ;4:55=2=2:275: So there is a pole around s = ;2:3. Our next goal is to nd B using the MATLAB statements: s=-4.55/2 Bv = (y - max(y))./exp(s*t);; plot(t(1:150),Bv(1:150)) print -deps 883mtr4cyB.eps B=mean(Bv(70:150)) A=mean(y(250:455)) Note that wehave used the MATLAB command: ./ to do an elementbyelement division using two matrices. The plotofB versus time is shown in Figure 4. The values obtained are A =4:0696 and B = ;6:8974: 4 0 0.5 1 1.5 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 Figure 5: Plot of ln[y 2 (t)] These values havetheright relationship, at least, because weknow jBj > jAj Having found B we can nowattempt to nd p 2 .Todosoweform the function y 2 (t)=y(t);(A + Be ;p 1 t = Ce ;p 2 t : In the presentcase y 2 (t)=y(t);(4:0696; 6:8974e ;2:275t : We then take the natural logarithm of y 2 and plot it versus time. as shown in Figure 5. This plot has a slope of about -2.5. We conclude that the second pole is either very far to the left, or very close to the rst pole. We will haveto\ sh" a bit here. So wegoaheadand nd C = ;(A + B)=;(4:0696;6:8974) = 2:8278: and place the second pole at s = ;8. Weknow the pole is farther out, so we start with this guess. Then our estimate of y(t)is ^y(t)=4:0696;6:8974e ;2:275t +10:6972e ;8t : 5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 Figure 6: Plot of ^y(t)versusy(t) A comparison of y(t) and ^y(t)isshown in Figure 6. As can be seen the estimated responses goes south initially.However, the overall t is not that bad. The culprits are probably are estimates of B and C.Rather than try to re ne our estimates of B and C,whichdon't really interest us anyway, weuse some of the conveientMATLAB commands to nish up the analysis. If we use the MATLAB statments K=A*abs(s)*abs(s1) g=zpk([],[s s1],K) T=linspace(0,0.4,449);; [yhat1 T] = step(g,T);; plot(t,y,'k-',T,yhat1,'k--') print -deps sr883mtr4cyd.eps weget the responses shown in Figure 7, and the t is very prettygood. The last taskisto ndK.Wehave K p 1 p 2 =4:0696;; 6 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 Figure 7: Plot of ^y(t)versusy(t) whichinthe presentcasebecomes K =4:06962:275 8=74:0661: However, Figure 8 shows the armature voltage for the recorded step re- sponse. Thus, we need to divide K by36toachieve the correct gain. Finally G(s)= 2:0574 (s +2:275)(s+8) : 7 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 0 5 10 15 20 25 30 35 40 Figure 8: Recorded armature voltage load mtr4stepc t=mtr4stepc(1:500,1);; y=mtr4stepc(1:500,2);; plot(t,y) print -deps sr883mtr4cya.eps t=t+0.0599;; y=y(45:500);; t=t(45:500);; y(1) = 0;; y(2) = 0.05;; y(3) = 0.1;; y(4) = 0.15;; y(5) = 0.2;; t(1) = 0;; plot(t,y) print -deps sr883mtr4cyb.eps maxy = max(y);; y1 = (max(y) + 0.0001) -y;; 8 w=log(y1);; plot(t(1:300),w(1:300));; print -deps 883mtr4cylny1.eps s=-4.55/2 Bv = (y - max(y))./exp(s*t);; plot(t(1:150),Bv(1:150)) print -deps 883mtr4cyB.eps B=mean(Bv(70:150)) A=mean(y(250:455)) y2 = y-A +B*exp(s*t);; w2 = log(y2);; plot(t(1:150),w2(1:150)) print -deps 883mtr4cylny2.eps C=-(A + B) s1 = -8 yhat = A + B*exp(s*t) + C*exp(s1*t);; plot(t,y,'k-',t,yhat,'k--') print -deps sr883mtr4cyc.eps K=A*abs(s)*abs(s1) g=zpk([],[s s1],K) T=linspace(0,4.5,455);; [yhat1 T] = step(g,T);; plot(t,y,'k-',T,yhat1,'k--') print -deps sr883mtr4cyd.eps load mtr4avc tav = mtr4avc(1:500,1);; av = mtr4avc(1:500,2);; plot(tav,av) print -deps 883mtr4cyav.eps 9