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HIGHER-ORDER CIRCUITS AND COMPLEX FREQUENCY
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Fig. 8-40
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A series RLC circuit contains R 1
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, L 2 H, and C 0:25 F. Simultaneously apply magnitude and frequency scaling, with Km 2000 and Kf 104 . What are the scaled element values Ans: 2000
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; 0:4 H; 12:5 mF At a certain frequency !1 , a voltage V1 25 08 V applied to a passive network results in a current I1 3:85 308 (A). The network elements are magnitude-scaled with Km 10. Obtain the current which results from a second voltage source, V2 10 458 V, replacing the rst, if the second source frequency is !2 103 !1 . Ans: 0:154 158 A In the circuit of Fig. 8-41 let R1 C1 R2 C2 10 3 . Find v2 for t > 0 if: (a) v1 cos 1000t u t , (b) v1 sin 1000t u t . Ans: a v2 sin 1000t ; b v2 1 cos 1000t In the circuit of Fig. 8-42 assume R 2 k
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, C 10 nF, and R2 R1 and v1 cos !t. following frequencies: (a) !0 5 104 rad/s, (b) !1 105 rad/s. Ans: a v2 2 sin !0 t; b v2 0:555 cos !1 t 146:38 Find v2 for the
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Fig. 8-41
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Noninverting integrators. In the circuits of Fig. 8-43(a) and 8-43(b) nd the relationship between v2 and v1 . Ans: a v1 RC=2 dv2 =dt; b v1 2RCdv2 =dt
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In the circuit of Fig. 8-44 nd the relationship between v2 and v1 . v2 R2 v1 = R1 R2 . Ans: R1 R2 C1 C2 dv2 dv R1 R2 v2 R1 R2 C1 1 R2 v1 dt dt
Show that for R1 C1 R2 C2 we obtain
CHAP. 8]
HIGHER-ORDER CIRCUITS AND COMPLEX FREQUENCY
Fig. 8-43 8.47 In the circuit of Fig. 8-44 let R1 9 k
9R2 , C2 100 pF 9C1 , and v1 104 t V. the switch is closed. Ans: i 1:0001 mA Find i at 1 ms after
Fig. 8-44
Lead network. The circuit of Fig. 8-45(a) is called a lead network. (a) Find the di erential equation relating v2 to v1 . (b) Find the unit-step response of the network with R1 10 k
, R2 1 k
, and C 1 mF. (c) Let v1 cos !t be the input and v2 A cos !t  be the output of the network of Part (b). Find A and  for ! at 1, 100, 331.6, 1100, and 105 , all in rad/s. At what ! is the phase at a maximum   dv2 R1 R2 dv 1 1 v ; b v2 1 10e 1100t u t Ans: a v2 1 dt R1 R2 C dt R1 C 1 11 (c) ! A  1 0.091 0.58 100 0.128 39:88 331.6 0.3015 56.48 1100 0.71 39.88 105 1 0.58
p Phase is maximum at ! 100 11 331:6 rad/s 8.49 Lag network. The circuit of Fig. 8-45(b) is called a lag network. (a) Find the di erential equation relating v2 to v1 . (b) Find the unit-step response of the network with R1 10 k
, R2 1 k
, and C 1 mF. (c) Let v1 cos !t be the input and v2 A cos !t  be the output of the network of Part (b). Find A and  for ! at 1, 90.9, 301.5, 1000, and 105 , all in rad/s. At what ! is the phase at a minimum   dv dv 10 90:91t e Ans: a v2 R1 R2 C 2 v1 R2 C 1 ; b v2 1 u t 11 dt dt
HIGHER-ORDER CIRCUITS AND COMPLEX FREQUENCY
[CHAP. 8
Fig. 8-45
! A 
1 1 0.58
90.9 0.71 39.88
301.5 0.3015 56.48
1000 0.128 39.88
105 0.091 0.58
p Phase is minimum at ! 1000= 11 301:5 rad/s 8.50 In the circuit of Fig. 8-46 nd the relationship between v2 and v1 for (a) k 103 , case nd its unit-step response; that is, v2 for v1 u t .   6 dv2 4 106 v2 4 107 v1 ; v2 10 1 e 4 10 t u t Ans: a dt   9 dv2 4 108 v2 4 109 v1 ; v2 10 1 e 4 10 t u t b dt (b) k 105 . In each
Fig. 8-46
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