Solution: 7.9.9.6 In Example 7.6.2 the plantisagain G p (s)= 1 (s + 1)(s +5) : The compensator, G p (s)= K c (s + 1)(s+ z 1 ) s(s +11) ;; Figure 1 shows howwe determine the location of the free zero bysolving the equation ; 1 ; 2 ;  3 = ;180  ;; or =  1 +  2 +  3 = ;180  = 135  +105:95  +60:255  = 121:2  : . Then XXX Re(s) Im(s) -5 -11 -7 7 -z 1 θθθ 123 α Figure 1: Determining location of zero z1 = 7+ 7 tan121:2  = 2:761 Then the gain to place the closed loop poles at s = ;7 j7is K c = jsjjs +5jjs +11j js +2:761j j s=;7+j7 = 71: Then, the step response, shown in Figure 2 is generated by the MATLAB dialogue: EDU>gcgp = zpk([-2.761],[0 -5 -11],71) Zero/pole/gain: 71 (s+2.761) -------------- s (s+5) (s+11) EDU>tc = feedback(gcgp,1) Zero/pole/gain: 71 (s+2.761) ------------------------- (s+2) (s^2 + 14s + 97.99) EDU>p = [1 14 97.99] p= 1.000000000000000e+00 1.400000000000000e+01 9.798999999999999e+01 EDU>roots(p) ans = -7.000000000000000e+00 + 6.999285677838846e+00i -7.000000000000000e+00 - 6.999285677838846e+00i EDU>step(tc,3) EDU> Wesee that the actual closed loop poles are very close s = ;7j7, and the damping ratio is almost exactly 0:707. The percentovershoot is under ten percent, so wearenot far o from the ideal response of T N2 (s). To examine the components of the response, we use partial fraction ex- pansion. Wehave C(s)= 1 s + B s +2 + M s +7; j 7 + M  s +7+j7 ;; where B = (s +2)C(s) j s=;2 Time (sec.) Amplitude Step Response 0 0.5 1 1.5 2 2.5 3 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 2: Unit step response = 71(s +2:761) s(s +7; j 7)(s +7+j 7) j s=;2 = ;0:3649 M = 0:48736 6 2:2805 rad: Thus, c(t)=[1; 0:3649e ;2t +0:9747e ;7t cos(7t +2:2805)]1(t): The residue associated with s = ;2isquite large, and it is negative. The net e ect is to cause the system to never overshoot.