Solution 3.8.5.6 For the opamp circuit of Figure 3.27 wehave V o (s) V i (s) = ;(C 1 =C)[s 2 + [(1=C 1 )(1=R 1 ;R 3 =RR 5 )]s+1=RR 4 C 1 C s 2 +(1=R 2 C)s +1=R 2 C 2 Webegin with the denominator whichshould be (s +0:1)(s+10)=s 2 +10:1s +1: Note that wehavetwo nonlinear equations in three unknowns, namely 1 R 2 C = 10:1 1 R 2 C 2 = 1: Squaring the rst equation and dividing by the second we obtain:  R R 2  2 = 10:1 2 1:0 =102:01;; or R R 2 =10:1: If wechoose R 2 =10k ;; then R =10:110 k =101k : From the table wechoose R =100k : Wenowhavetohave 1 R 2 C =10:1;; or C = 1 10:1R 2 = 1 10:11010 3 = 9:9  f: 1 Wecancome close to this value bychoosing C =10f: We next checktheother coecientofthe denominator to see if it is correct. 1 R 2 C 2 = 1 100 2 10 6 10 2 10 ;12 = 1:0: Thus the denominator polynomial is s 2 +10s +100=(s +9:899)(s+0:1010): and so the roots are very close to those wewant. Wenowturnour attention to the numerator and the gain. The gain is K =0:1= C 1 C ;; or C 1 =0:1C =0:11010 ;6 =1mboxf: .Thus, C 1 =1f: We next consider the coecients of the numeratorof the transfer function whichis (s +1+j5)(s+1;j5) = s 2 +2ss +26: Wethus have the twoequations (1=R 1 ;R 3 =RR 5 ) = 2C 1 1=RR 4 C 1 C = 26 From the second equation weseethat R 4 = 1 26RCC 1 = 1 2610010 3 10 ;6 110 ;6 = 38:46 k : Wecancome very close by R 4 =39k : 2 Wenow turn our attention to the last coecientwhichisdetermined from the equation. (1=R 1 ;R 3 =RR 5 )=2C 1 ;; or 1 R 1 ; R 3 RR 5 =0:000002: Wedoarough calculation on R 1 byletting R 1 = 1 2C 1 =500k : If wechoose R 1 =470k ;; then R 3 RR 5 = 1 R 1 ;5:1C 1 =1;;276610 ;7 : Then R 3 R 5 = R2:661710 ;6 =0:0128: All wehavetodonowischoose The proper ratio of R 3 and R 5 . Then, if we choose R 3 =13k and R 5 =1:0M ;; R 3 R 5 =0:013: Then the numerator polynomial is s 2 +1:9977s+25:641 = (s +0:9988j4:9642)(s+0:9988;j4:9642): Close enough, and wehavebeen able to set all the coecients of the numer- ator and denominator, as well as the gain. 3