Solution 8.8.1 The rst step is to get the data into MATLAB. The MATLAB program load mtr6step size(mtr6step) t=mtr6step(1:500,1);; y=mtr6step(1:500,2);; plot(t,y) print -deps sr881a.eps produces the reponse ans = 500 2 and seperates the data into a column vector of times and a column vector of responses, and then prints the step response EDU>t = tfid(1:500,1);; EDU>y = tfid(1:500,2);; EDU>plot(t,y) EDU>print -deps sr881a.eps EDU> shown in Figure`1 As can be seen the time response, whichwas taken o the oscilloscope, does not begin exactly at t =0,andnot quite at y =0.The MATLAB program y(34) y(35) t(34) t=t-t(34);; y(34) = 0;; t=t(34:500);; y=y(34:500);; plot(t,y) print -deps sr881b.eps produces the response 1 -0.05 0 0.05 0.1 0.15 0.2 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 Figure 1: Rawstep response data y34 = -0.00437500000000 y35 = 0.00203123000000 t34 = 0.00143619000000 and produces the adjusted reponse shonwinFigure 2. We are nowinaposition to apply the identi cation techniques discussed in Chapter 3. As a rst step, the MATLAB program max(y) 2 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0 0.5 1 1.5 2 2.5 3 3.5 Figure 2: Adjusted step response y1 = 3.2566 -y;; plot(t,log(y1));; print -deps 881lny1.eps produces the response shown in Figure 3. Note that 3.25656 was increased to 3.2566 to avoid ln(0). The slope is roughly ; [1:3;(;0:52)] 0:073 = ;24:93: So there is a pole around s = ;25. Our next goal is to nd B using the MATLAB program: load mtr6step size(mtr6step) t=mtr6step(1:500,1);; y=mtr6step(1:500,2);; plot(t,y) print -deps sr881a.eps 3 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 -12 -10 -8 -6 -4 -2 0 2 Figure 3: Plot of ln(1;y(t) y34 =y(34) y35 =y(35) t34 =t(34) t=t-t(34);; y(34) = 0;; t=t(34:500);; y=y(34:500);; plot(t,y) print -deps sr881b.eps s=-(1.3 -(-0.52))/0.073 max(y) y1 = 3.2566 -y;; plot(t,log(y1));; print -deps 881lny1.eps B=(y - max(y))./exp(s*t) plot(t(100:150),B(100:150)) print -dpes 881B.eps mean(B(100:150) 4 0.035 0.04 0.045 0.05 0.055 0.06 -4.1 -4.05 -4 -3.95 -3.9 -3.85 -3.8 -3.75 -3.7 -3.65 -3.6 Figure 4: Comparison of adjusted response and y 1 (t)=1;e ;24:9t 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 looks to be about right B because weknow jBj > jAj 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); (3:25656; 3:8992e ;24:9t : 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 5 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 Figure 5: Plot of ln[y 2 (t)] estimate of the slope to be ; 2:5 0:0225 = ;111: Thus the second pole is somewhere around s = ;100. Then C = ;(A + B)=;(325656; 3:8992) = 0:64264: Then our estimate of y(t)is ^y(t)=3:25656;38992e ;24:9t +0:64264e ;111t : 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 =3:25656 24:9111 = 9021:2: 6 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0 0.5 1 1.5 2 2.5 3 3.5 Figure 6: Plot of ^y(t)versusy(t) However, Figure 7 shows the armature voltage for the recorded step re- sponse. Thus, we need to divide K by27toachieve the correct gain. Finally G(s)= 334 (s +24:9)(s++111) : This is quite close to the result wegotinthis chapter using the identi cation algorithm presented inthis chapter. 7 -0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0 5 10 15 20 25 30 Figure 7: Recorded armature voltage 8