Solution 8.8.4 Re(s) Im(s) X X X -94.3 -28.3 s + 28.3 s s + 94.3 Figure 1: Calculating gain along negativerealaxis Figure 1 shows howtocalculate the gain at a point along the negativereal axis. That it, K = jsjjs +28:3jjs +94:3j: The MATLAB program p1 =0 p2 = 28.3 p3 = 94.3 x=linspace(-15,-10,500);; K=abs(x + p1).*abs(x+p2).*abs(x+p3) plot(x,K) print -deps 884b.eps plots and saves the graph shown in Figure 2. As we mighthavesuspected, the break-out occurs very near the midpointbetween the poles at s =0and the pole at s = ;28:3. K(s = ;13) = 16;;171: The MATLAB program p1 =0 p2 = 28.3 p3 = 94.3 x=linspace(-15,-10,500);; K=abs(x + p1).*abs(x+p2).*abs(x+p3) plot(x,K) 1 Re(s) Im(s) X X X -94.3 -28.3 s + 28.3 s s + 94.3 Figure 2: Finding break-out point print -deps 884b.eps s=-13 K=abs(s + p1)*abs(s+p2)*abs(s+p3) g=zpk([],[0 -28.3 -94.3],16171) tc = feedback(g,1) T=linspace(0,1,200);; u=3.15*ones(1,200);; [Y,T] = lsim(tc,u,T);; plot(T,Y,'k-') print -deps sr884.eps plots and saves the step response shown in Figure 3. By comparing this response to Figure 8.22, weconlcude that wearepretty close to the break- out point. The gain of the plantis289,sothe gain required from the FAC and the power ampli er is 16;;181 289 =56: Since the rails of the power ampli er are 30 V, wecannot quite get to the break-out point. However, we knowthat the nonlinear e ects of the motor speed up the response, and that is why the tworesults are so close. 2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.5 1 1.5 2 2.5 3 3.5 Figure 3: Theoretical critically damped response 3