1 Solution 9.10.4.3 Σ Σc (s)G G p (s) R(s) C(s) + _ + + D(s) Figure 1: UnityFeedbackwith Disturbance at the Output For the system of Figure 1, G p (s)= 4 (s +4) 2 and wewant T c (s)= ! 2 n s 2 +2! n s + ! 2 n ;; To ndG c wesetD =0andnotethat T c = G c G p 1+G c G p : Solving rst for G c G p ,wehave G c G p = T c 1;T c : Then G c = 1 G p T c 1;T c : The closed loop transfer function between D and C with R =0,canbe found as follows. With R =0, C d = D;C d G;; whichcanberearranged as C d = D 1+G ;; 2 or T d = C d D = 1 1+G = 1 1+G c G p = 1 1+ T c 1;T c = 1;T c : Wethus havetwoexpressions for T d : T d = 1 1+G c G p =1;T c : Since T c (s)= ! 2 n s 2 +2! n s + ! 2 n : Then T c 1;T c = ! 2 n s 2 +2! n s + ! 2 n 1; ! 2 n s 2 +2! n s + ! 2 n = ! 2 n s 2 +2! n s : Then G c (s) = T c 1;T c 1 G p = ! 2 n s 2 +2! n s (s +4) 2 4 = (! 2 n =4)(s+4) 2 s(s +2! n ) : Thus, we see that the compensator cancels the twopoles of the plant and replaces them with poles at s =0ands = ;2! n . 3 Wenow nd T d (s) = 1;T c (s) = 1; ! 2 n s 2 +2! n s + ! 2 n = s(s +2! n ) s 2 +2! n s + ! 2 n : If wechoose  =0:8and! n =5rad/s, then T d (s) = s(s +2! n ) s 2 +2! n s + ! 2 n = s(s +2(0:8)(5) s 2 +2(0:8)(5)s+5 2 = s[s +8] s 2 +8s +5 2 = s(s +8) (s +4;j3)(s+4+j3) G c (s) = (5 2 =4)(s+4) 2 s[s +2(0:8)(5)] = (25=4)(s+4) 2 s(s +8) : b. The frequency response of T d (j!)isshown in Figure 2 The disturbance rejection at 10 r./s. is ver poor, about 0 dB. Wecanimprove the situation by increasing ! n In fact, we can get anynoiseattentuation wewantat10 rad/s bymerely increasing ! n .The price wewillpayisincreased gain in our compensator, since the gain of the compensator is ! n =2. As long as gain is not a problem, noise suppression is not a problem. If, however, there is a limitation on gain, then there will be a limitation on noise suppression. For instance, suppose we let  =0:8 and ! n =1000rad/s. Then T d (s) = s(s +2! n ) s 2 +2! n s + ! 2 n = s(s +2(0:8)(1000) s 2 +2(0:8)(1000)s+1000 2 = s(s +1600) s 2 +1600ss+10 6 4 10 -1 10 0 10 1 10 2 10 3 -60 -50 -40 -30 -20 -10 0 10 20 Figure 2: Frequency Response of T d (j!) . 5 10 -1 10 0 10 1 10 2 10 3 -60 -50 -40 -30 -20 -10 0 10 20 Figure 3: Frequency Response of T d (j!) . = s(s +8) (s +800;j600)(s+800+j600) G c (s) = (1000 2 =4)(s +4) 2 s[s +2(0:8)(1000)] = 250;;000(s+4) 2 s(s +1600) : The Bode magnitude plot of T d is shown in Figure 3. Wehave plentyof noise suppression around 10 rad/s, but weneedagain of 250,000 in our compensator. If we use a more realistic value of ! n =100rad/s, then wesee from Figure 4 that wecanachieveabout 15 dB of noise suppression. This is not great, but it is de nitely an improvement. The gain requirementisa more realistic 2500. 6 10 -1 10 0 10 1 10 2 10 3 -60 -50 -40 -30 -20 -10 0 10 20 Figure 4: Frequency Response of T d (j!) .