Solution: 7.9.9.4 G pG c + - R C Referring to Figure 1 wecanwrite XX X Re(s) Im(s) -5 3 -1.732 Figure 1: Location of desird closed loop poles Im(s) = 3tan60  = 5:196: The closed loop poles will not be exactly at s = ;3j5:196;; but as discussed in Example 7.4.1, they will be close enoughttothis location that wecan goahead and assume that they are. A little later wewill compute the exact locations of the closed loop poles to verify that our assumption is correct. Then the gain required to place the poles at the desired locaations is K c = jsjjs +1jjs +5j js +0:1j j s=;3+j5:196 1 = 65:5685:568 5:951 = 31:257 The step response, shown in Figure 2 is determined from the MATLAB dialogue: EDU>gcgp = zpk([-0.1],[0 -1 -5],31.257) Zero/pole/gain: 31.257 (s+0.1) -------------- s (s+1) (s+5) EDU>tc = feedback(gcgp,1) Zero/pole/gain: 31.257 (s+0.1) ---------------------------------- (s+0.08746) (s^2 + 5.913s + 35.74) EDU>p =[1 5.913 35.74] p= 1.000000000000000e+00 5.913000000000000e+00 3.574000000000000e+01 EDU>roots(p) ans = -2.956500000000000e+00 + 5.196066565201027e+00i -2.956500000000000e+00 - 5.196066565201027e+00i EDU>step(tc,30) EDU>print -deps sr7994.eps EDU> We see that the actual closed loop poles are very close to the poles we used to compute the gain. The step response shows that bylowering the damping ratio, we can mitigate to some extenttheslowcreep towards steady state. 2 Time (sec.) A mp li tu d e Step Response 0 5 10 15 20 25 30 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 2: Unit step response 3