Solution 3.8.5.3 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)(s +100) 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 : Then wehavethetwo equations [(b+1)=ab] = 101 2 =ab = 100 Dividing the rst equation bythesecond yields b +1 =1:01;; or b =1:01 ;1 If =10then b =9:1;; and a = 2 b100 = 100 9:1100 = 0:1099 Then wemust have 1 R 1 C 1 = =10;; or R 1 C 1 =0:1 1 Wechoose R 1 =100k and C 1 =1 f: Then R 2 = 0:1099R 1 = 0:1099100 k = 10:99 k ! 1110:99 k : Also C 2 = 9:1C 1 = 9:11  f = 9:1  f: The closest value in Table 3.1 is 10  f. If wechoose this value then we obtain the denominator (s +0:9175)(s+99:1): If this is not close enough then wecancombine some of the capacitors in Table 3.1 to obtain a value closer to 9:1  f, for instance, 10  finseries with 100  ffor an equivlentcapacitance of 9:09  f. 2