Solution 3.8.6.1 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 sr3861a.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 wehavesomechoices 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 0.319989 seconds to the t vector so that t(38) becomes zero. Then wesety(38) equal to zero and t(38) equal to zero. Then wetruncate the t and y vectors to the entries in locations 40 through ve hundred. The commands are: t=t+0.319989;; t(38) = 0;; y(38) = 0;; t=t(38:500);; y=y(38:500);; plot(t,y);; print -deps sr3861b.eps;; 1 0 2 4 6 8 10 12 14 16 18 20 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 A;y(t)asshown bytheMATLAB dialogue: EDU>max(y) ans = 1.01200000000000 EDU>y1 = 1.013-y;; EDU>w = log(y1);; EDU>plot(t,w) 2 0 2 4 6 8 10 12 14 16 18 20 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 2: Adjusted step response EDU>print -deps lny1.eps EDU> The slope of this line is ; 2:9 3:6 = ;0:8: Thus, there is a pole in the vicintyofs = ;1. Our next goal is to nd B using the MATLAB dialogue: EDU>B= (y-1.012)./exp(-0.8*t);; EDU>plot(t(1:250),B(1:250)) EDU>print -deps Ba.eps EDU> A plot of B versus time for o<t<10 s is shown in Figure 3. Note that we have used the MATLAB command: ./ to do an elementbyelementdivision using twomatrices. We see from Figure 3 that B is nearly constant out to 4s, and reasonably constant for 4 <t<6s.Weusethe MATLAB dialogue: 3 0 1 2 3 4 5 6 7 8 9 10 -70 -60 -50 -40 -30 -20 -10 0 Figure 3: B for 0 <t<10s EDU>mean(B(100:150)) ans = -1.89616247864868 EDU>mean(B(1:100)) ans = -1.08425536791520 EDU>print -deps Bb.eps EDU>mean(B(50:200)) ans = -2.85276785406747 4 EDU>mean(B(50:100)) ans = -1.13470594960238 EDU>mean(B(100:125)) ans = -1.55002798284992 EDU>mean(B(75:125)) ans = -1.37774061572837 EDU>mean(B(75:100)) ans = -1.21094912760319 EDU>mean(B(50:75)) ans = -1.05752427643543 EDU> There is a slowbysteady progression in the mean value of B.Wewould like to be sure the e ects of the second pole havedecayed out. On that basis we would conclude that 1:05 <B<1:2: Wechoose B =1:1. Wenow compare the measured response to ~y(t)=1:012;1:1e ;0:8t : 5 0 2 4 6 8 10 12 14 16 18 20 -0.2 0 0.2 0.4 0.6 0.8 1 1.2 Figure 4: Comparison of adjusted response and ~y(t)=1:012;e ;0:8t The comparison is shown in Figure 4. 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);1:012+ 1:1e ;0:8t : We then takethe natural logarithm of y 2 and plot it versus t for 0 <t<2s, as shown in Figure 5. As can be seen, there are two distinct slopes. Figure 6 shows the plot of ln[y(t)] for 0 <t<1s. Then wecan nd of range of values for p 2 from the three slopes shown. Wehave ; ;2:24;(;3:2) 0:17 = ;4:7 ; ;2:23;(;3:2) 0:25 = ;3:88 ; ;2:23;(;3:52) 0:46 = ;2:8 6 0 0.5 1 1.5 2 2.5 3 -7 -6.5 -6 -5.5 -5 -4.5 -4 -3.5 -3 -2.5 -2 Figure 5: Plot of ln[y 2 (t)] for 0 <t;;3s 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -3.8 -3.6 -3.4 -3.2 -3 -2.8 -2.6 -2.4 -2.2 Figure 6: Plot of ln[y 2 (t)] for 0 <t;;1s 7 0 2 4 6 8 10 12 14 16 18 20 -0.2 0 0.2 0.4 0.6 0.8 1 1.2 Figure 7: Plot of ^y(t)versusy(t) Thus 2:8 <p 2 < 4:7: Then C = ;(A + B)=;(1:012;1:1) = 0:088: Then our estimate of y(t)is ^y(t)=1:012;1:1e ;0:8t +0:088e ;4:7t : A comparison of y(t) and ^y(t)isshown in Figure 7. As can be seen the t is very good. The last taskisto ndK.Wehave K p 1 p 2 =1:012;; whichinthe presentcasebecomes K =1:0120:84:7=3:81: Thus G(s)= 3:81 (s +0:8)(s+4:7) : 8