Solution 3.8.6.7 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 sr3867a.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, weadd0.048 seconds to the t vector so that t(69) becomes 0. Then wesety(69) equal to zero and t(69) equal to zero. Then wetruncate the t and y vectors to the entries in locations 72 through vehundred. The commands are: t=t+0.048;; t(69) = 0;; y(69) = 0;; t=t(69:500);; y=y(69:500);; plot(t,y);; print -deps sr3867b.eps;; 1 -0.5 0 0.5 1 1.5 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Figure 1: Rawstep response data The adjusted response is shown in Figure 2. The rationale for the adjust- ments is 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 four 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 = ;0:048 s. Wearenowinaposition to apply the identi cation techniques discussed in Chapter 3. As a rst step, wecreate the function y 1 = A;y(t) and then w(t)=ln(y(t));; as shown bytheMATLAB dialogue: EDU>max(y) ans = 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Figure 2: Adjusted step response 1.9100 EDU>y1 = 1.92-y;; EDU>w = log(y1);; EDU>plot(t,w) EDU>print -deps lny1.eps EDU> Note that 3.98 was increased to 1.92 to avoid ln(0). The result is shown in Figure 3. The slope is roughly ; 3:2 0:8 = ;4 So there is a pole around s = ;4. Our next goal is to nd B using the MATLAB dialogue: EDU>B = (y-1.92)./exp(-4*t);; EDU>plot(t(50:75),B(50:75)) EDU>mean(B(50:75)) 3 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 -5 -4 -3 -2 -1 0 1 Figure 3: Plot of ln(1;y(t) ans = -1.9662 EDU>print -deps B.eps EDU> Note that wehave used the MATLAB command: ./ to do an elementbyelement division using two matrices. The plotofB versus time 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);1:92+ 1:9662e ;4t : 4 0.18 0.2 0.22 0.24 0.26 0.28 0.3 -2.04 -2.02 -2 -1.98 -1.96 -1.94 -1.92 -1.9 -1.88 Figure 4: Comparison of adjusted response and y 1 (t)=1;e ;0:15t We then take the natural logarithm of y 2 and plot it versus time. as shown in Figure 5. We showthe range of slopes. Atoneendofthe range wehave ; 4:4;2:5 0:044 = ;43:2: In the middle of the range wehave ; 5:3;2:5 0:044 = ;63:63: At the other end of the range wehave ; 6:4;2:5 0:042 = ;92:8: Thus the second pole is somewhere between s = ;43 ands = ;96. Then C = ;(A + B)=;(1:91;1:9662) = 0:0562: Then our estimate of y(t)is ^y(t)=1:91;1:9662e ;4t +0:0562e ;43t : 5 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 -8 -7 -6 -5 -4 -3 -2 Figure 5: Plot of ln[y 2 (t)] A comparison of y(t) and ^y(t)isshown in Figure 6. As can be seen the t is very good. It should be clear that will get essentially the same t using anyofthe three values of p 2 that wecalculated. The last taskisto ndK.Wehave K p 1 p 2 =1:91;; whichinthe presentcasebecomes K =1:91443 = 328:5: Thus G(s)= 328:5 (s +4)(s+43) : 6 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Figure 6: Plot of ^y(t)versusy(t) 7