Solution 3.8.3.3 Let Z 1 = R 1 k C 1 + R3 = R 1 C 1 s R 1 + 1 C 1 s + R 3 = R 1 R 3 C 1 s +(R 1 + R 3 ) R 1 C 1 s +1 ;; 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 R 3 C 1 s+(R 1 +R 3 ) R 1 C 1 s+1   (R 1 C 1 s +1)(R 2 C 2 s +1) (R 1 C 1 s +1)(R 2 C 2 s +1)  = R 2 (R 1 C 1 s +1) [R 1 R 3 C 1 s +(R 1 + R 3 )](R 2 C 2 s +1) = (1=R 3 C 2 )[s +(1=R 1 C 1 )] [s +(R 2 + R 3 )=(R 1 R 3 C 1 )][s+(1=R 2 C 2 )] : We next havetochoose some values. We knowthat 1 R 3 C 2 = 10 R 2 + R 3 R 1 R 3 C 1 = 1 1 R 1 C 1 = 0:1 1 R 2 C 2 = 10 Let R 2 =10 4 and C 2 =10 ;5 f: 1 Then R 3 = 1 10C 2 =10 4 : Looking at the second equation wenowhave R 2 + R 3 R 1 R 3 C 1 =1;; or R 2 = R 1 C 1 R 3 ;R c =10R c ;R c =9R c Thus R 2 =90k : A convenientvalue from the Table is 91 k . The last step is to nd C 2 . Since R 2 C 2 =0:1;; C 2 = 0:1 R 2 =1;;1  f: If wechoose 1  f, then the pole will be at s = ; 1000 91 = ;11;; whichisclose enough. 2