Solution 9.10.5.5 Σ Σ c (s)G G p (s) R(s) C(s) + _ + + D(s) H(s) Figure 1: Distubance and Input to Plant For the system of Fig. 2 wehave G p (s)= 1 s(s +1) ;; and the two compensators G c1 (s)= K c (s +3) s + b) G c2 (s)= K c (s +1:2) s + b : (a) The angle contribution to place the poles at the desired location has the same form both compensators, namely = +180  ; 1 ; 2 ;; as shown in Fig. 2 For G c1 wehave = tan ;1 (3=2:8)+ 180  ;[180  ; tan ;1 (3=4);[180  ; tan ;1 (3=3)] = 133  +180  ; 143:1  ;135  = 35  : The pole is then at b = 4+ 3 tan(35  ) =8:3: XXXO θ 1 θ 2 α β Re(s) Im(s) Figure 2: Angle Calculation to Place Pole of Compensator The gain to place the poles is K c = jsjjs +1jjs +8:3j js +1:2j s=;4+j3 = p 25 p 18 p 27:49 p 16:84 = 27:1 Thus, G c1 (s)= 27:1(s+1:2) s +8:3 : For G c2 wehave = [180  ; tan ;1 (3=1)]+ 180  ; [180  ; tan ;1 (3=4); [180  ; tan ;1 (3=3)] = 108:4  + 180  ; 143:1  ; 135  = 10:33  : The pole is then at b = 4+ 3 tan(10:33  ) =20:46: The gain to place the poles is K c = jsjjs+1jjs+20:46j js +3j s=;4+j3 = p 25 p 18 p 279:93 p 10 = 112:2 Thus, G c2 (s)= 112(s +3) s +20:46 : (b) To nd T d (s)= C d (s) D(s) ;; Set R =0.Then C d =(D;G c C d )G p = DG p ;G c G p C d ;; or T d (s)= G c (s) 1+G c (s)G p (s) : For T d (j!)tobesmall wemust either have G p (j!)very large or G c (j!) very small. Wecan"tdo anything about G p (j!), but wedohavecontrol of G c (j!). For G c1 wehave T d1 = 27:1(s+1:2) s(s +1)(s+8:3)+ 27:1(s+1:2) = 27:1(s+1:2) s 3 +9:3s 2 +35:4s +32:52 = 27:1(s+1:2) (s +1:3)(s+4;j3)(s+4+j3) For G c2 wehave T d1 = 112(s +3) s(s +1)(s+20:46)+ 112(s +3) = 112(s+3) s 3 +21:46s 2 +132:46s+336 = 112(s+3) (s +13:5)(s+4;j3)(s+4+j3)