Solution 8.8.3mtr3cy The rst step is to get the data into MATLAB. The MATLAB statements load mtr3stepc t=mtr3stepc(1:500,1);; y=mtr3stepc(1:500,2);; plot(t,y) print -deps sr883mtr3cya.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 -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 Figure 1: Rawstep response data not begin exactly at t =0,andnot quite at y =0.The MATLAB program t=t+0.0434;; y(1) = 0;; y(2) = 0.05;; y(2) = 0.1;; y(3) = 0.15;; 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 Figure 2: Adjusted step response y(4) = 0.2;; t(1) = 0;; plot(t,y) print -deps sr883mtr3cyb.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 883mtr3cylny1.eps produces the response shown in Figure 3. Note that the maximum resposne was increased to0.001 to avoid ln(0). The slope is roughly ;3:6=1:5=2:4: So there is a pole around s = ;2:4. 2 0 0.5 1 1.5 2 2.5 -6 -5 -4 -3 -2 -1 0 1 2 Figure 3: Plot of ln(1;y(t) Our next goal is to nd B using the MATLAB statements: s=-3.6/1.5 Bv = (y - max(y))./exp(s*t);; plot(t(1:150),Bv(1:150)) print -deps 883mtr3cyB.eps B=mean(Bv(60:100)) A=mean(y(250:454)) 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 =3:7789 and B = ;6:0539: These values havetheright relationship, at least, because weknow jBj > jAj 3 0 0.5 1 1.5 -14 -12 -10 -8 -6 -4 -2 Figure 4: Comparison of adjusted response and y 1 (t)=1;e ;24:9t 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); (3:7789;6:0539e ;2:4t : 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)=;(3;;7789; 6:0539) = 2:2749: and place the second pole at s = ;10. Weknow the pole is farther out, so we start with this guess. Then our estimate of y(t)is ^y(t)=3:7789;6:0539e ;2:4t +2:2749e ;10t : A comparison of y(t) and ^y(t)isshown in Figure 6. As can be seen the 4 0 0.5 1 1.5 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 Figure 5: Plot of ln[y 2 (t)] 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 Figure 6: Plot of ^y(t)versusy(t) 5 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 Figure 7: Plot of ^y(t)versusy(t) our estimate goes negativeinitially.However, the overall t is not that bad. The problem is probably our choices of B and C.Ratherthan try to adjust B andC,wenowuse some of the MATLAB commands to nish up the analysis. If weusethe 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 sr883mtr3cyd.eps we get the responses shown in Figure 7, and the t is prettygood, which means our estimates of B and C are proably the problem. The last taskisto ndK.Wehave K p 1 p 2 =3:7789;; whichinthe presentcasebecomes K =3:77892:4 0=90:6943: 6 -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 Figure 8: Recorded armature voltage However, Figure 8 shows the armature voltage for the recorded step re- sponse. Thus, we need to divide K by32toachieve the correct gain. Finally G(s)= 2:8342 (s +2:4)(s++10) : 7 The entire MATLAB program that does this analysis is s=-3.6/1.5 Bv = (y - max(y))./exp(s*t);; plot(t(1:150),Bv(1:150)) print -deps 883mtr3cyB.eps B=mean(Bv(60:100)) A=mean(y(250:454)) 8