Solution 3.8.6.2 The rst step is to get the data into MATLAB. The followng commands do that. EDU>load tfid EDU>size(tfid) ans = 500 2 EDU> the rst column is the time, and the second column is the step response. Weusethe following commands to put the data in a form wecan plot the step response versus time. EDU>t = tfid(1:500,1);; EDU>y = tfid(1:500,2);; EDU>plot(t,y) EDU>print -deps sr3862a.eps EDU> The time response is shown in Figure`1 As can be seen the time response, whichwas taken o the oscilloscope shows negativetimebeforethe step is applied and also the system does not start exactly at zero. The downward trend just before t =0iswherethe switchwas thrown in the analog circuit to initiate the step. So wehave some choices to make. We need to clean the data up a bit to apply the identi cation technique described in this chapter. First of all, we add 0.24 seconds to the t vector so that t(46) becomes t(40). Then wesety(40) equal to zero and t(40) equal to zero. Then wetruncate the t and y vectors to the entries in locations 40 through vehundred. The commands are: t=t+0.8999;; t(42) = 0;; y(42) = 0;; t=t(42:500);; y=y(42:500);; plot(t,y);; print -deps sr3862b.eps;; 1 -5 0 5 10 15 20 25 30 35 40 45 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 1: Rawstep response data The adjusted response is shown in Figure 2. The rationale for the adjust- ments are as follows. The response was generated bycreating a second order comepensator on the Feedback and Control box(FAC). The system response at steady state is almost exactly one in reponse to a 1 V input. Thus, all we wanttodoisget zero to be the pointatwhich the switchonthe FACwas thrown to initiate the step and the response goes down close to zero. This turns out to be approximately t = ;0:24 s. Wearenowinaposition to apply the identi cation techniques discussed in Chapter 3. As a rst step, wecreate the function 1;y(t)asshown bytheMATLAB dialogue: EDU>size(t) ans = 459 1 EDU>z(1:461)=1.005;; EDU>z=z';; EDU>y1=1-y;; 2 0 5 10 15 20 25 30 35 40 45 50 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 2: Adjusted step response EDU>w = log(y1);; plot(t,w);; The result is shown in Figure 3. The slope is roughly ; 2:25 15 = ;0:15 So there is probably a pole around s = ;0:15. Figure 4 shown the adjusted step response obtained experimentally compated to the function y 1 (t)=1;e ;0:15t : Figure 5 compares the adjusted response to y 2 (t)=1;e ;0:18t As can be seen the t is excellent. The other pole must haveatime constant much smaller than 5.56s, and hence a very good estimate of the plantis G(s)= 0:18 s +0:18 : 3 0 5 10 15 20 25 30 35 40 45 50 -7 -6 -5 -4 -3 -2 -1 0 1 Figure 3: Plot of ln(1;y(t) Indeed the actual transfer function implemented was G(s)= 1 (s +0:2)(s+5) : When the ratio of the second pole to the rst pole is this large, wecannot see anye ect from the second pole. 4 0 5 10 15 20 25 30 35 40 45 50 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 4: Comparison of adjusted response and y 1 (t)=1;e ;0:15t 5 0 5 10 15 20 25 30 35 40 45 50 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 5: Comparison of adjusted response and y 1 (t)=1;e ;0:18t 6