Solution 3.8.6.5 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 sr3865a.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 0.0454 seconds to the t and set t(39) = 0, and y(39 = 0. Then we truncate the t and y vectors to the entries in locations 40 through vehundred. The commands are: t=t+0.0454;; t(39) = 0;; y(39) = 0;; t=t(39:500);; y=y(39:500);; plot(t,y);; print -deps sr3865b.eps;; 1 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 -0.2 0 0.2 0.4 0.6 0.8 1 1.2 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>max(y) ans = 0.9720 EDU>y1 = 0.973-y;; EDU>w = log(y1);; EDU>plot(t,w) 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Figure 2: Adjusted step response EDU> The result is shown in Figure 3. The slope is roughly ; 3:8;1 0:95;0:25 = ;4:05: So there is a pole around s = ;4. Wenowattempt to nd B.Wegenerate B,andplot of B,with the MATLAB statements: B=(y-0.972)./exp(-4*t);; plot(t(100:150),B(100:150) print -deps sr3865d.eps Figure 4 shows a plot of B versus t for 0:4 <t<0:6. Figures 5 through 8 show the calculation of B for p 1 =4:1;;4:15;;4:2and 4.3 respectively. We can use the MATLAB commnand: mean(B(100:200)) std(B(100:200)) 3 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 -7 -6 -5 -4 -3 -2 -1 0 Figure 3: Adjusted step response 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 -1.3 -1.25 -1.2 -1.15 -1.1 -1.05 -1 -0.95 -0.9 -0.85 Figure 4: B for 0:4 <t<0:6secondsp 1 =4:0 4 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 -1.4 -1.35 -1.3 -1.25 -1.2 -1.15 -1.1 -1.05 -1 -0.95 -0.9 Figure 5: B for 0:4 <t<0:6secondsp 1 =4:1 p 1 4 4.05 4.1 4.15 4.2 4.3 Mean -1.014 -1.045 -1.08 -1.11 -1.144 -1.215 STD 0.065 0.07 0.075 0.082 0.089 0.105 Table 1: Means and Standard deviations for various values of p 1 5 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 -1.5 -1.4 -1.3 -1.2 -1.1 -1 -0.9 Figure 6: B for 0:4 <t<0:6 secondsp 1 =4:15 6 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 -1.5 -1.4 -1.3 -1.2 -1.1 -1 -0.9 -0.8 Figure 7: B for 0:4 <t<0:6secondsp 1 =4:2 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 -1.7 -1.6 -1.5 -1.4 -1.3 -1.2 -1.1 -1 -0.9 Figure 8: B for 0:4 <t<0:6secondsp 1 =4:3 7 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 -0.2 0 0.2 0.4 0.6 0.8 1 1.2 Figure 9: Measured reponse versus y 1 (t) to nd the mean and standard deviation of B. Tablel 1 summarizes this data. If wechoose p 1 =4:05, then B = ;1:045: Then C = ;(0:972;1:045) = 0:073: We next nd y 1 (t) = A+ Be ;4t = 0:972;1:045e ;4t ;; and compare it to the measured response, as shown in Figure 9. as shown in Figure 9. The t is excellent. We next form y 2 (t) = y(t);y 1 (t) = Ce ;p 2 t ;; 8 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 -8 -7 -6 -5 -4 -3 -2 Figure 10: Measured reponse versus y 1 (t) and then plot the natural logarithm of this function verus time as shown in Figure 10 This places the second pole somewhere between s = ;40 and s = ;24. If wechose p 2 =24,then K p 1 p 2 = K 4:0524 =0:972: Thus K =0:9724:0524 = 94:5 If wechose p 2 =40,then K p 1 p 2 = K 4:0540 =0:972: Thus K =0:9724:0540 = 148: Figures 11 and 12 compare the step response of these twotransfer functions to the measured response. As can be seen the t is prettygoodinboth cases. 9 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Figure 11: Measured reponse versus y 24 (t)=0:972;1:045e ;4:05t +0:073e ;24t By wayofreference the actual transfer function implemented was G(s)= 150 (s+5)(s+30) : Weare not far o , and either transfer function would be very serviceable. the second. 10 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Figure 12: Measured reponse versus y 24 (t)=0:972;1:045e ;4:05t +0:073e ;40t 11