Solution 3.8.5.2 For the opamp circuit of Figure 3.26 wehave V o (s) V i (s) = ;(1=R 1 C 1 )s s 2 +(1=R 2 C 1 +1=R 2 C 2 )s +1=R 1 C 1 R 2 C 2 Wewishtomake the denominator (s +1;j2)(s+1+j2) = s 2 +2s +5 Letting R 2 = aR 1 ;; C 2 = bC 1 ;; and =1=R 1 C 1 ;; weobtain V o (s) V i (s) = ; s s 2 + (1=a +1=ab)s+ 2 =ab ;; with =1. Then wehavethetwoequations [(b+1)=ab] = 2 2 =ab = 5 Dividing the rst equation bythesecond yields b+1 =0:4;; or b =0:4 ;1 Since =5then b =1;; and a = 2 b1000 = 25 15 = 5: Then wemust have 1 R 1 C 1 = =5;; 1 or R 1 C 1 =0:2 Wechoose R 1 =200k and C 1 =1 f: Then R 2 =5R 1 =1:0M and C 2 =1 f 2