Solution 3.8.3.2 Let Z 1 = R 1 C 1 s R 1 + 1 C 1 s = R 1 R 1 C 1 s+1 ;; and Z 2 = R 2 + 1 C 2 s = R 2 C 2 s+1 C 2 s ;; Then Z 2 Z 1 = R 2 C 2 s+1 C 2 s R 1 R 1 C 1 s+1   (C 2 s)(R 1 C 1 s+1) (C 2 s)(R 1 C 1 s+1)  = (R 2 C 2 s+1)(R 1 C 1 s+1) R 1 C 2 s = (R 2 =C 1 )(s+(1=R 1 C 1 )(s+(1=R 2 C 2 ) s : We next havetochoose some values. We knowthat R 2 C 1 = 10 1 R 1 C 1 = 0:1 1 R 2 C 2 = 1: If wechoose R 1 =10 5 ;; C 1 =10 ;4 f;; ;; R 2 =10 5 and C 2 =10 ;5 f: Then G(s)= ;10(s+0:1)(s+1) s ;; and wehaveachieved a PID compensator. 1