Solution 3.8.3.6 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 3 = R 2 R 2 C 2 s +1 + R 3 = R 2 + R 2 R 3 C 2 s + R 3 R 2 C 2 s +1 = R 2 R 3 C 2 s +(R 2 + R 3 ) R 2 C 2 s +1 : Then Z 2 Z 1 = R 2 R 3 C 2 s +(R 2 + R 3 ) R 2 C 2 s +1 R 1 C 1 s +1 C 1 s   (C 1 s)(R 2 R 3 C 2 s +(R 2 + R 3 )) (R 1 C 1 s +1)(R 2 C 2 s +1)  = (R 3 =R 1 )s(s + R 2 +R 3 R 2 R 3 C 2 (s + 1 R 1 C 1 )(s + 1 R 2 C 2 ) : We next havetochoose some values. We rst let R 2 =10 6 ;; C 1 =10 ;6 f;; so that 1 R 2 C 2 = 1 10`10 ;6 = 1: Wenextnote that R 2 + R 3 R 2 R 3 C 2 =5;; or 5R 2 R 3 C 2 = R 2 + R 3 ;; 1 or R 3 (5R 2 C 2 ;1) = R 2 ;; or nally R 3 = R 2 5R 2 C 2 ;1 : Since R 2 C 2 =1,wehave R 3 = R 2 4 =25010 3 : A nearbyvalue is 24010 3 . Then since the gain is R 3 =R 1 wemust have R 2 =2410 3 : Finally,wemust have 1 R 1 C 1 =50;; or C 1 = 1 50R 1 = 1 502410 3 =0:83310 ;;6 : A nearbyvalue in the table is 0:82 f. Thus, the values weneedare R 1 =24k C 1 =0:82  f R 2 =1M C 2 =1 f R 1 =240k 2