Solution 3.8.6.8 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 sr3868a.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 0.0399872 seconds to t and set t(77) = 0, and y(77) = 0. Then we truncate the t and y vectors to the entries in locations 40 through vehundred. The commands are: t=t+0.04399872;; t(77) = 0;; y(77) = 0;; t=t(77:500);; y=y(77:500);; plot(t,y);; print -deps sr3868b.eps;; 1 -1 0 1 2 3 4 5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 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 = 4.0600000000000000 EDU>y1 = 4.06-y;; EDU>w = log(y1);; EDU>plot(t,w) 2 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 2: Adjusted step response EDU> The result is shown in Figure 3. The slope is roughly ; 1:4+2:125 1:5 = ;2:35: Wenowattempt to nd a good estimate of B.We generate B over a range of values of t,andplot B,withthe MATLAB statements: B=(y-4.06)./exp(-2.35*t);; plot(t(50:250),B(50:250) print -deps sr3868d.eps Figure 4 shows a plot of B versus t for 0:5 <t<2:5seconds. Note that the curveis at for 0:5 <t<1:25, and \on average " at for 1:25 <t<1:75. We will use these tworegions to try to nd an estimate of B.Thereason that we use both regions is simply that wemust have jBj > jAj =4:06: We nd our estimates of B using the MATLAB dialogue: 3 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 Figure 3: Adjusted step response EDU>mean(B(50:75)) ans = -3.64249102381882 EDU>mean(B(150:175)) ans = -4.78636034776858 EDU>mean(B(125:175)) ans = -4.41616801718234 EDU>plot(t(125:175),B(125:175)) 4 EDU>print -deps Bb.eps EDU>mean(B(115:150)) ans = -3.87770352545026 EDU>mean(B(120:150)) ans = -3.93368201732749 EDU>mean(B(125:140)) ans = -4.08642988473146 EDU>mean(B(130:150)) ans = -4.06638038526051 EDU>mean(B(135:150)) ans = -4.13770851393740 EDU>plot(t(135:150),B(135:150)) EDU>mean(B(135:160)) ans = -4.39212730667767 EDU>plot(t(135:160),B(135:160)) EDU>mean(B(135:170)) 5 0 0.5 1 1.5 2 2.5 -30 -25 -20 -15 -10 -5 0 Figure 4: B for 0:5 <t<2:5seconds ans = -4.55674832530445 EDU>plot(t(135:170),B(135:170)) EDU>print -deps Bd.eps EDU>plot(t(135:160),B(135:160)) EDU>print -deps Bc.eps EDU> This dialogue represents the searchforagoodestimate of B.Notethat our initial estimates are too low(less than 4.06). In the process of looking for B we also plot B versus time. What wearelooking for is a mean slope of zero. Figure 4 is the global picture. Wesee that the plot of B B is very at out to t =1:5s,andrelatively at again for 1:25 <t<1:75. Figures 5 is a plot of B for 1:25 <t<1:75 s. Here the average slope is slightly downward. Figure 6shows B for 1:35 <t<1:60 s, and Figure 7 B for 1:35 <t<170. We see that the average slope is still relatively at in FIgure 6, but tilted 6 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 -9 -8 -7 -6 -5 -4 -3 -2 -1 Figure 5: B for 0:125 <t<1:75 s downward somewhat in Figure 7. Therefore, a reasonable guess is B =4:4: It maybealittle smaller, say4.1 or 4.2, but it probably isn't muchbigger. This is, of course, educated guesswork, but educated nonetheless. We next nd y 1 (t) = A+Be ;4t = 4:06;4:4e ;2:35t ;; and compare it to the measured response, as shown in Figure 8. The t is excellent. The t is prettygood.If weletp 1 =2:5wegettheplotinFigure 9 The t is muchbetter here. If weletP ;1=3weget the comparison shown in Figure 10 Thus, p 1 =2:5looks likeaslightly better estimate. Wenow de nitely knowthat 2:35 <p 1 < 3: We next form y 2 (t) = y(t);y 1 (t) = Ce ;p 2 t ;; 7 1.3 1.35 1.4 1.45 1.5 1.55 1.6 1.65 -6.5 -6 -5.5 -5 -4.5 -4 -3.5 -3 -2.5 Figure 6: B for 1:35 <t<1:6s 1.3 1.35 1.4 1.45 1.5 1.55 1.6 1.65 1.7 1.75 -9 -8 -7 -6 -5 -4 -3 -2 -1 Figure 7: B for 1:35 <t<1:7s 8 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 Figure 8: Measured reponse versus y 1 (t), p 1 =2:35 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 Figure 9: Measured reponse versus y 1 (t), p 1 =2:5 9 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 Figure 10: Measured reponse versus y 1 (t), p 1 =3 and then plot the natural logarithm of this function verus time as shown in Figures 11 and 12. From Figure 12 we see that the P 2 must be quite large, because the slope very quickly becomes quite shallow, indicating that the e ect of the pole at s = ;p 2 is over very quickly.From Figure 11, wecantrytogetareasonable estimate of p 2 .Wecompute three slopes: ;0:4;(;1:81) 0:07 = 20:4 ;0:4; (;1:8) 0:08 = 17:5 ;0:4;(;1:72) 0:106 = 12:45 Wecannowsaywith some con dence that 12 <P 2 < 20: 10 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 Figure 11: Plot of ln[y 2 (t)], 0 <t;;0:75 s 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 -2.6 -2.4 -2.2 -2 -1.8 -1.6 -1.4 -1.2 -1 -0.8 Figure 12: Plot of ln[y 2 (t)], 0 <t;;1:25 s 11 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 Figure 13: Measured versus estimated step reponse We know that C = ;(A+ B)=;(4:06;4:4) = 0:34: If wechose p 2 = ;17:5, then our estimate of y(t)is ^y(t)=4:06; 4:4e ;2:5t +0:34e ;17:5t : This estimate is compared to the measured response in Figure 13. The last task is to estimate K p 1 p 2 =4:06;; or K =4:062:517:5=177:6 The transfer function is then G(s)= 177:6 (s+2:5)(s+17:5) : 12