Solution 3.8.4.6 Let Z 1 = R 1 kC 1 = R 1 =(C 1 s) R 1 +(1=C 1 s) = R 1 R 1 C 1 s +1 ;; and Z 2 = R 2 kC 2 = R 2 =(C 2 s) R 2 +(1=C 2 s) = R 2 R 2 C 2 s +1 Then Z 1 +Z 2 = R 1 R 1 C 1 s+1 + R 2 R 2 C 2 s +1 = (R 1 R 2 C 2 s+ R 1 )+(R 1 R 2 C 1 s +R 2 )) (R 1 C 1 s+1)(R 2 C 2 s +1) = R 1 R 2 (C 1 +C 2 )s +(R 1 +R 2 ) (R 1 C 1 s+1)(R 2 C 2 s +1) Then Z 1 + Z 2 Z 1 = R 1 R 2 (C 1 +C 2 )s+(R 1 +R 2 ) (R 1 C 1 s+1)(R 2 C 2 s+1) R 1 R 1 C 1 s+1 = R 1 R 2 (C 1 + C 2 )s+(R 1 + R 2 ) R 1 (R 2 C 2 s +1) =  (C 1 + C 2 ) C 2  s+ R 1 +R 2 R 1 R 2 (C 1 +C 2 ) s + 1 R 2 C 2 Wenexthavetochoose some values. Wenotethatwehave C 1 +C 2 C 1 =10 1 R 1 + R 2 R 1 R 2 (C 1 + C 2 ) = 0:1 1 R 2 C 2 = 0:01 From the rst equation we see that C 1 =9C 2 : Our choices are fairly limited. If wechoose C 1 =100 f and C 2 =10 f wegetagainof11,whichisclose to what wewant. Then if wechoose R 2 =10M ;; the pole will be at ; 1 10 7 )10 ;5 = ;0:01;; Wecaneasilymakea10M resistor from the values in Table 3.1, and modern JFET opamps can tolerate a resistor this big, since the o sets have become very small, particularly if wearewilling to payaboutabuckfora quad opamp. If wenow solvethe second equation for R 1 we obtain R 1 = R 2 0:1R 2 (C 1 +C 2 );1 = 10 7 0:110 7 1:110 ;4 ;1 = 10 7 109 = 91:7k ! 91 k : Then G(s)= 11(s+0:1008) s+0:01 : Given that wenormally use 5% components, this compensator is very close to the desired one. The gain is 10% high but easily adjusted at the power ampli er that will of needs be present. 2