Solution 3.8.5.5 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 +0:01)(s+50)=s 2 +50:01s+0:5: Note that wehavetwo nonlinear equations in three unknowns, namely 1 R 2 C = 50:01 1 R 2 C 2 = 0:5: Squaring the rst equation and dividing by the second we obtain:  R R 2  2 = 50:01 2 0:5 =5002;; or R R 2 =70:7: If wechoose R 2 =16k ;; then R =70:716 k =1:1316 M : From the table wechoose R =1:1M : Wenowhavetohave 1 R 2 C =50:01;; or C = 1 50:01R 2 = 1 50:011610 3 = 1:2498  f: 1 Wecan come close to this value with the parallel combination of a 0:22  f, capacitors and a 1:0  fcapacitor, t0 obtain C =1:253010 ;6 f: We next checktheother coecientofthe denominator to see if it is correct. 1 R 2 C 2 = 1 16 2 10 6 1:2530 2 10 ;12 = 0:5264: Thus the denominator polynomial is s 2 +4888s+0:5264 = (s +49:87)(s+0:0106): and so the roots are very close to those wewant. Wenowturnour attention to the numerator and the gain. The gain is K =10= C 1 C ;; or C 1 =10C =101:25316 ;6 =12:53 mboxf: Wecan come very close with a parallel combination of a 10, two1,a0.22 and a 0.33  fcapacitors, to obtain C 1 =12:5510 ;6  f: Thus, assuming that C 1 =12:55 f;; wenext consider the coecients of the numerator of the transfer function whichis (s +0:1)(s+5)=s 2 +5:1s+0:5: Wethus have the twoequations (1=R 1 ;R 3 =RR 5 ) = 5:1C 1 1=RR 4 C 1 C = 0:5 From the second equation weseethat R 4 = 1 0:5RCC 1 = 1 0:51:110 6 11:25310 ;6 12:5510 ;6 = 115:62 k : 2 We can come very close with the series combination of a 91 k and a 15 k . Thus, welet R 4 =116k : Wenow turn our attention to the last coecientwhichisdetermined from the equation. (1=R 1 ;R 3 =RR 5 )=5:1C 1 ;; or 1 R 1 ; R 3 RR 5 =0:000064005: Wedoarough calculation on R 1 byletting R 1 = 1 5:1C 1 =15:624 k : If wechoose R 1 =15k ;; then R 3 RR 5 = 1 R 1 ;5:1C 1 =2:661710 ;6 : Then R 3 R 5 = R2:661710 ;6 =2:9278: All wehavetodonowischoose The proper ratio of R 3 and R 5 . Then, if we choose R 3 =47k and R 5 =16k ;; R 3 R 5 =2:9375: Then the numerator polynomial is s 2 +5:0993s+4:9984 = (s +0:0997)(s+4:996): Close enough, and wehavebeen able to set all the coecients of the numer- ator and denominator, as well as the gain. 3