Solution: 7.9.3.11 G pG c + - R C For the system shown above, wehave G p (s)= 1 s 2 and G c (s)= K c (s + a) s + b The compensator is G c = K(s +0:1) s +6 The root locus is shown in Fig 1. The gain calculation is made on the assumption that the locus will pass very close to s = ;3+j3. Indeed, we assume the locus passes through s = ;3+j3, and that the zero at s = ;0:1 cancels one of the poles at s =0.Then K = p 3 2 +3 2  p 3 2 +3 2 = 18: Thus the nal compensator is G c = 18(s +0:1) s +6 : The following MATLAB dialogue generates the step response of the closed loop system EDU>g = zpk([-0.1],[0 0 -6],18) Zero/pole/gain: 18 (s+0.1) ---------- s^2 (s+6) EDU>h=1 1 Re(s) Im(s) 2 -6 3 Figure 1: Root Locus of Compensated System 2 Time (sec.) Amplitude Step Response 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 2: Step Response of Compensated System h= 1 EDU>tc = feedback(g,h) Zero/pole/gain: 18 (s+0.1) --------------------------------- (s+0.1035) (s^2 + 5.896s + 17.39) EDU>step(tc,2) EDU>print -deps 79310sr.eps EDU> The step response is shown in Figure 2. 3