Solution: 7.9.9.5 In Example 7.3.2 we adjusted the gain slightly for the near cancellation of the pole and zero. In realitywe probably would not do that. Thus, in this problem we merely wanttoseethe e ect of leaving the compensator gain at 1.8 and the overall gain at 18. The step response, shown in Figure 1 is determined from the MATLAB dialogue: EDU>gcgp = zpk([-1.1],[0 -1 -6],18) Zero/pole/gain: 18 (s+1.1) ------------- s (s+1) (s+6) EDU>tc = feedback(gcgp,1) Zero/pole/gain: 18 (s+1.1) ------------------------------- (s+1.145) (s^2 + 5.855s + 17.3) EDU>p = [1 5.855 17.3] p= 1.000000000000000e+00 5.855000000000000e+00 1.730000000000000e+01 EDU>roots(p) ans = -2.927500000000000e+00 + 2.954613976478146e+00i -2.927500000000000e+00 - 2.954613976478146e+00i EDU>step(tc,3) EDU>print -deps sr7995.eps EDU> Wesee that the actual closed loop poles are very close s = ;3j3, 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). 1 Time (sec.) A mp li tu d e 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 1: Unit step response 2