Solution 3.8.4.3 Let Z 1 = R 1 k C 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 k C 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 2 =2 1 R 1 + R 2 R 1 R 2 (C 1 +C 2 ) = 5 1 R 2 C 2 = 0:1 From the rst equation we see that C 1 = C 2 : Our choices are fairly limited. If wechoose C 2 =10 f;; and R 2 =1:0M wegenerate the desired pole. Then with C 1 =100 f;; wegetagainof 2 1 =2 whichiswhatwewant. If wenow solvethe second equation for R 1 we obtain R 1 = R 2 5R 2 (C 1 + C 2 );1 = 10 6 510 6 1:110 ;4 ;1 = 10:1k : a convenientvalue is 10 k : Then G(s)= 2(s+5:05) s+0:1 : 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