Solution 7.9.3.13 Σ GR C + c G p Figure 1: Nonunity feedbackand cascade compensation For Figure 1 wehave G p (s)= 1 s(s +1) and G c (s)= K c (s +2) (s + p) : Weneedto nd K c and p so that the closed loop system has dominantpoles with  =0:8and! n =5rad/s. For  =0:8and! n =5,thedesired poles will be at s = ;! n j! n q 1; 2 = (0:8)(5)j(5)(0:6) = ;4 j3: Figure 2 shows the angles that must be calculated at s = ;4+j3to satisfy the angle condition at the desired location of the closed loop poles. That is wemust have ; 1 ; 2 ;  3 = ;180  ;; or  3 = +180  ;  1 ; 2 = [180  ; tan ;1 (3=2)]+ 180  ; [180  ; tan ;1 (3=4); [180  ; tan ;1 (3=3)] = 123:68  +180  ; 143:13  ;135  = 25:56  1 X X X -1 -p -2 -4 3 Re(s) Im(s) α θ 1 θ 2 θ 3 Figure 2: Satisfaction of angle condition at s = ;4+j4 Then p = 4+ 3 tan 3 = 4+ 3 tan(25:56  ) = 10:27: The gain is then K = jsjjs +1jjs +10:27j js +2j s=;4+j3 = 40:909: Then the MATLAB dialogue EDU>g = zpk([-2],[0 -1 -10.272727], 40.9091) Zero/pole/gain: 40.9091 (s+2) ----------------- s(s+1) (s+10.27) EDU>tc = feedback(g,1) 2 Zero/pole/gain: 40.9091 (s+2) ------------------------- (s+3.273) (s^2 + 8s + 25) EDU>step(tc) EDU>print -deps sr79313.eps produces the step response shown in Figure 3 Time (sec.) A mp li tu d e Step Response 0 0.5 1 1.5 2 2.5 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 3: Step response of compensated system 3