Solution 3.8.4.1 + R 2 C 2 R 1 C 1 Z 2 Z 1 V i V 0 V x Figure 1: Noninverting opamp For the noninverting opamp of Figure 1, the transfer function is V o (s) V i (s) = 1+ Z 2 (s) Z 1 (s) = Z 1 (s)+Z 2 (s) Z 2 (s) : It is worthing noting that this last experssion is the inverse of a voltage divider. Let Z 1 = R 1 + 1 C 1 s = R 1 C 1 s +1 C 1 s ;; and Z 2 = R 2 k C 2 + R 3 = R 2 R 2 C 2 s +1 + R 3 = R 2 R 3 C 2 +(R 2 + R 3 ) R 2 C 2 s +1 Then Z 2 + Z 1 = R 2 R 3 C 2 s +(R 2 + R 3 ) R 2 C 2 s +1 + R 1 C 1 s +1 C 1 s = (C 1 s)[R 2 R 3 C 2 s +(R 2 + R 3 )] + (R 1 C 1 s +1)(R 2 C 2 s +1) (C 1 s)(R 2 C 2 s +1) 1 Next, Z 1 (s)+Z 2 (s) Z 2 (s) = (C 1 s)[R 2 R 3 C 2 s +(R 2 + R 3 )]+(R 1 C 1 s +1)(R 2 C 2 s +1) (C 1 s)(R 2 C 2 s +1) R 1 C 1 s +1 C 1 s = (R 2 R 3 C 2 C 3 + R 1 R 2 C 1 C 2 )s 2 +[R 1 C 1 + R 2 C 1 +(R 2 + R 3 )C 1 ]s +[(R 2 + R 3 )C 1 +1] (R 1 C 1 s +1)(R 2 C 2 s +1) Wecannot achieveazero exactly at the origin with this con guratlon, but we can put a zero very close to the origin. Suppose we let the numerator be (s +5)(s +0:001) = s 2 +5:001+ 0:005: Then we essentially haveazero at s =0.Nextnote that (s + a)(s+ b)=s 2 +(a + b)s + ab;; and cs 2 + ds + e = c(s 2 +(d=c)s+(e=c): Then. given our expression for V 0 (s)=V i (s)wehave (R 2 + R 3 )C 1 +1 R 2 R 3 C 2 C 3 + R 1 R 2 C 1 C 2 ;; = ab [R 1 C 1 + R 2 C 1 +(R 2 + R 3 )C 1 ] R 2 R 3 C 2 C 3 + R 1 R 2 C 1 C 2 = a + b ab ;; (R 2 + R 3 )C 1 +1 = K: We see that with the noninverting con guration things are considerably more tedious. We next havetochoose some values. We rst let R 2 =10 6 ;; C 1 =10 ;6 f;; so that 1 R 2 C 2 = 1 10`10 ;6 = 1: Wenextnote that R 2 + R 3 R 2 R 3 C 2 =5;; 2 or 5R 2 R 3 C 2 = R 2 + R 3 ;; or R 3 (5R 2 C 2 ; 1) = R 2 ;; or nally R 3 = R 2 5R 2 C 2 ;1 : Since R 2 C 2 =1,wehave R 3 = R 2 4 =25010 3 : A nearbyvalue is 24010 3 . Then since the gain is R 3 =R 1 wemust have R 2 =24 10 3 : Finally,wemust have 1 R 1 C 1 =50;; or C 1 = 1 50R 1 = 1 502410 3 =0:83310 ;;6 : A nearbyvalue in the table is 0:82 f. Thus, the values weneedare R 1 =24k C 1 =0:82  f R 2 =1M C 2 =1 f R 1 =240k 3