Solution: 7.9.1.4 For the system of Figure 1 wehave R + C G c G p Σ - Figure 1: Cascade Compensation with UnityFeedback G p = 1 (s+1)(s+5) and G c = K(s+0:1) s+b Weare assuming that the root locus passes through the point s = ;3+j3 when in fact it only passes close to this point. However, as pointed oout in the text it is a very good approximation and so we make it. The following MATLAB program computes the gain and the steady state error for a range of pole positions, based on this assumption. xstart = 0.09 xend = 0.0001 dx = 0.001 x(1) = xstart itop = round( (xstart - xend)/0.001 ) i=1 s=-3+j*3 for i = 1:itop if i > 1 x(i) = x(i-1)-dx;; end k(i) = ( abs(s + x(i) )*abs(s + 1)*abs(s + 5) )/ (abs(s + 0.1) );; kp(i) = k(i)*0.1 / (1*5*x(i));; ess(i) = 1 / (1 + kp(i) );; end plot(x,ess) print -deps lag4a.eps Figure 2 is a plot of steady state error versus the absolute value of the pole position. Wesee that for jbj < 0:01 weachieve the desired steady state 1 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 Figure 2: e ss versus absolute value of pole position 2 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 13.02 13.04 13.06 13.08 13.1 13.12 13.14 13.16 13.18 13.2 13.22 Figure 3: K versus absolute value of pole position error. Figure 3 is a plot of the gain K versus the absolute value of the pole position. As wesee the gain is essentially constant. Figure 4 is shows the step response for G c G p 9s)= 13(s+0:1) (s+0:01)(s+1)(s+5) 3 Time (sec.) Amplitude Step Response 0 5 10 15 20 25 30 35 40 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Figure 4: 4