Solution 3.8.5.4 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+50)=s 2 +50:1s +5: Note that wehavetwo nonlinear equations in three unknowns, namely 1 R 2 C = 50:1 1 R 2 C 2 = 5: Squaring the rst equation and dividing by the second we obtain:  R R 2  2 = 50:1 2 5 =502:002;; or R R 2 =22:4: If wechoose R 2 =12k ;; then R =22:405412 k =268:9k : From the table wechoose R =270k : Wenowhavetohave 1 R 2 C =50:1;; or C = 1 50:1R 2 = 1 50:11210 3 = 1:66  f: 1 Wecan come close to this value with the parallel combination of three 0:22  fcapacitors and a 1:0  fcapacitor. Wenextchecktheother coe- cientofthe denominator to see if it is correct. 1 R 2 C 2 = 1 270 2 10 1 01:666 2 10 ;12 = 4:958: Thus the denominator polynomial is s 2 +50:1s +4:958 = (s +50)(s+0:0992): 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 : Wecancome very close with a parallel combination of a 10, 4.7, and two1  fcapacitors. Thus, assuming that C 1 =16:7 f;; wenext consider the coecients of the numerator of the transfer function whichis (s + 1)(s+5)=s 2 +6s +5: Wethus have the twoequations (1=R 1 ;R 3 =RR 5 ) = 6C 1 1=RR 4 C 1 C = 5 From the second equation weseethat R 4 = 1 5RCC 1 = 1 527010 3 1:6610 ;6 16:710 ;6 = 26:67 k : This is quite close to the value 27 k from the table. Then 1 RR 4 CC 1 =4:9383 2 Wenow turn our attention to the last coecientwhichisdetermined from the equation. (1=R 1 ;R 3 =RR 5 )=6C 1 ;; or 1 R 1 ; R 3 RR 5 =0:0001002: If wechoose R 1 =10k ;; then all wehavetodoismake R 3 RR 5 verysmall: Suppose wechoose R 3 =10k ;; and R 5 =180k ;; Then R 3 RR 5 = 10 4 1:810 5 2:710 5 = 2:0510 ;7 : Then the numerator polynomial is s 2 +4:99s +4:9383 = (s +0:9906)(s+4:99): Close enough, and wehavebeen able to set all the coecients of the numer- ator and denominator, as well as the gain. 3