Solution 3.8.6.3 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 sr3863a.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 1.1999 seconds to the t vector so that t(39) 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+1.1999;; t(39) = 0;; y(39) = 0;; t=t(39:500);; y=y(39:500);; plot(t,y);; print -deps sr3863b.eps;; 1 -5 0 5 10 15 20 25 30 35 40 45 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 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 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Figure 2: Adjusted step response 4.9000 EDU>y1 = 4.91-y;; EDU>w = log(y1);; EDU>plot(t,w) EDU>print -deps lny1.eps EDU> Note that 4.9 was increased to 4.91 to avoid ln(0). The result is shown in Figure 3. The slope is roughly ; 3:3 20 = ;0:165: So there is a pole around s = ;0:2. Our next goal is to nd B using the MATLAB dialogue: EDU>B = (y-4.9)./exp(-0.165*t);; EDU>plot(t(50:250),B(50:250)) EDU>mean(B(50:150)) 3 0 5 10 15 20 25 30 35 40 45 50 -5 -4 -3 -2 -1 0 1 2 Figure 3: Plot of ln(1;y(t) ans = -4.7483 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. This initial value of B is too small because we know jbj > jAj =4:9: So we try EDU>mean(B(150:200)) 4 0 5 10 15 20 25 -10 -9 -8 -7 -6 -5 -4 -3 Second estimate of B Figure 4: Comparison of adjusted response and y 1 (t)=1;e ;0:15t ans = -5.0380 This looks better. Having found B wecannowattempt to nd p 2 .Todo so weform the function y 2 (t)=y(t);(A + Be ;p 1 t = Ce ;p 2 t : In the presentcase y 2 (t)=y(t);4:9+5:038e ;0:165t : We then take the natural logarithm of y 2 and plot it versus time. as shown in Figure 5. There is not a lot of information but wecanmakearough estimate of the slope to be ; 1:5 0:28 = ;5:6: Thus the second pole is somewhere around s = ;5. 5 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 -6 -5.5 -5 -4.5 -4 -3.5 -3 -2.5 -2 -1.5 Figure 5: Plot of ln[y 2 (t)] Then C = ;(A + B)=;(4:9; 5:038) = 0:138: Then our estimate of y(t)is ^y(t)=4:9;5:0382e ;0:165t +0:138e ;5:6t : A comparison of y(t) and ^y(t)isshown in Figure 6. As can be seen the t is very good. The last taskisto ndK.Wehave K p 1 p 2 =4:9;; whichinthe presentcasebecomes K =4:9 0:165 5:6=4:53: Thus G(s)= 4:53 (s +0:165)(s+5:6) : 6 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Figure 6: Plot of ^y(t)versusy(t) 7