Solution 3.8.3.1 Let Z 1 = R 1 + 1 C 1 s = R 1 C 1 s+1 C 1 s ;; and Z 2 = R 2 C 2 s R 2 + 1 C 2 s = R 2 R 2 C 2 s+1 : Then Z 2 Z 1 = R 2 R 2 C 2 s+1 R 1 C 1 s+1 C 1 s   (C 1 s)(R 2 C 2 s+1) (C 1 s)(R 2 C 2 s+1)  = R 2 C 1 s (R 1 C 1 s +1)(R 2 C 2 s+1) = (1=R 1 C 2 )s (s+1=R 1 C 1 )(s+1=R 2 C 2 ) : We next havetochoose some values. Welet R 1 =10 6 ;; C 1 =10 ;5 f;; ;; R 2 =5110 3 and C 2 =10 ;6 f: Then G(s)= s (s+10)(s+19:6) : Given that wenormally use 5% components, this compensator is very close to the desired one. 1