Solution 3.8.4.4 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 + 1 C 2 s = R 2 C 2 s +1 R 2 C 2 s Then Z 1 + Z 2 = R 1 R 1 C 1 s +1 + R 2 C 2 s +1 C 2 s = (R 1 C 2 s)+(R 1 C 1 s + 1)(R 2 C 2 s +1) (R 1 C 1 s +1)(C 2 s) = R 1 R 2 C 1 C 2 s 2 +(R 1 C 1 + R 2 C 2 + R 1 C 2 )s +1 (R 1 C 1 s +1)(C 2 s) Then Z 1 + Z 2 Z 1 = R 1 R 2 C 1 C 2 s 2 +(R 1 C 1 + R 2 C 2 + R 1 C 2 )s +1 R 1 C 2 s Wenexthavetochoose some values. Before doing so, we note that the form of the compensator is G c (s) = ( s + 1)( s+1) s = ( s 2 +( + )s +1) s Then wehavethe equations R 1 C 1 R 2 C 2 = (R 1 C 1 + R 2 C 2 + R 1 C 2 = + R 1 C 2 = 1 Wenotethat the dc gain is 1= .The desired transfer function is G(s) = 10(s +1= )(s +1= ) s = (10= )( s + 1)( s+1) s : Wenow write R 1 R 2 C 1 C 2 s 2 +(R 1 C 1 + R 2 C 2 + R 1 C 2 )s +1 R 1 C 2 s = (10= )( s+ 1)( s+1) s : In this problem =1 ;; =0:2;; and 10 =50= 1 R 1 C 2 ;; or R 1 C 2 =0:02 Then wecan write the equations R 1 C 1 R 2 C 2 = =0:2 R 1 C 1 + R 2 C 2 +0:02 = 1:2;; or R 1 C 1 R 2 C 2 = =0:2 (R 1 C 1 + R 2 C 2 = 1:18;; Letting R 1 C 1 = a and R 2 C 2 = b;; wecannow write ab = =0:2 a + b = 1:18;; and wenowhavetwo nonlinear equations in two unknowns. solving the second equation for b and substituting in the rst yields a(1:18; a)=0:2;; 2 or a 2 ; 1:18a+0:2=0;; with solutions a =0:2052 and a =0:9748: Wehavetwochoices. WenowhavetouseTable 3.1. to nd the appropriate values for the two resistors and capacitors. Since wehavemoreresistors than capactors we start bychoosing R 1 and C 2 to satisfy R 1 C 2 =0:02 Wechoose R 1 =200k and C 2 =0:1  f;; then the short MATLAB program a=0.2045 b=0.2/a C1 = a/R1 R2 = b/C2 a=0.9748 b=0.2/a C1 = a/R1 generates the output a= 0.2045 b= 0.9780 C1 = 1.0225e-06 3 R2 = 9.7800e+06 a= 0.9748 b= 0.2052 C1 = 4.8740e-06 R2 = 2.0517e+06 Choosing a =0:9748 is the better choice because we can choose R 2 =2M and C 1 =4:7  f: Then the transfer function is G c (s)= 10(s +1:064)(s+5) s ;; whichisvery close to that speci ed. 4