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in series with the 50 V. That is, i 1 50 2:5 A 20 vC 1 2:5 A 10
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HIGHER-ORDER CIRCUITS AND COMPLEX FREQUENCY
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Fig. 8-19 Knowing the end conditions on vC , we can write vC vC 0 vC 1 e t= vC 1 25 1 e t=10 wherein t is measured in ms. The current in the capacitor is given by iC C and the current in the parallel 10-
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resistor is i10
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Hence, vC 2:5 1 e t=10 10
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A A dvC 5e t=10 dt A V
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2:5 1 e t=10
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The problem might also have been solved by assigning mesh currents and solving simultaneous di erential equations.
For the time functions listed in the rst column of Table 8-2, write the corresponding amplitude and phase angle (cosine-based) and the complex frequency s.
See columns 2 and 3 of the table. Table 8-2 Time Function i t 86:6 A 3 i t 15:0e 2 10 t A v t 25:0 cos 250t 458 V v t 0:50 sin 250t 308 V i t 5:0e 100t sin 50t 908 A i t 3 cos 50t 4 sin 50t A A 8 86:6 08 A 15:0 08 A 25:0 458 V 0:50 608 V 5:0 08 A 5 53:138 A s 0 2 103 Np/s j250 rad/s j250 rad/s 100 j50 s 1 j50 rad/s
For each amplitude and phase angle in the rst column and complex frequency s in the second column in Table 8-3, write the corresponding time function.
See column 3 of the table. Table 8-3 A 8 10 08 2 458 5 908 15 08 100 308 s j120 j120 2 j50 5000 j1000 0 Time Function 10 cos 120t 2 cos 120t 458 5e 2t cos 50t 908 15e 5000t cos 1000t 86.6
CHAP. 8]
HIGHER-ORDER CIRCUITS AND COMPLEX FREQUENCY
p An amplitude and phase angle of 10 2 458 V has an associated complex frequency s 50 j 100 s 1 . Find the voltage at t 10 ms.
p v t 10 2e 50t cos 100t 458 At t 10
s, 100t 1 rad 57:38, and so p v 10 2e 0:5 cos 102:38 1:83 V
A passive network contains resistors, a 70-mH inductor, and a 25-mF capacitor. Obtain the respective s-domain impedances for a driving voltage (a) v 100 sin 300t 458 V , (b) v 100e 100t cos 300t V .
(a) Resistance is independent of frequency. At s j300 rad/s, the impedance of the inductor is sL j300 70 10 3 j21 and that of the capacitor is 1 j133:3 sC (b) At s 100 j300 s 1 , sL 100 j300 70 10 3 7 j21 1 1 40 j120 sC 100 j300 25 10 6
For the circuit shown in Fig. 8-20, obtain v at t 0:1 s for source current a i 10 cos 2t (A), (b) i 10e t cos 2t (A).
Zin s 2 (a) At s j2 rad/s, Zin j2 3:22 7:138
. 2 s 2 s 3 4 s 4 s 4
Then, or v 32:2 cos 2t 7:138 V
V IZin 10 08 3:22 7:138 32:2 7:138 V and v 0:1 32:2 cos 18:598 30:5 V. (b) At s 1 j2 s 1 , Zin 1 j2 3:14 11:318
. V IZin 31:4 11:318 V and v 0:1 31:4e 0:1 cos 22:778 26:2 V. or
Then v 31:4e t cos 2t 11:318 V
Fig. 8-20
Fig. 8-21
Obtain the impedance Zin s for the circuit shown in Fig. 8-21 at (a) s 0, (c) jsj 1.
(b) s j4 rad/s,
HIGHER-ORDER CIRCUITS AND COMPLEX FREQUENCY   4 s2 3s 4 s Zin s 2 2 2 4 s s 2 2 s 1 s 2 s 1 (a) Zin 0 4
, the impedance o ered to a constant (dc) source in the steady state. b Zin j4 2 j4 2 3 j4 4 2:33 29:058
j4 2 j4 2
[CHAP. 8
This is the impedance o ered to a source sin 4t or cos 4t. (c) Zin 1 2
. At very high frequencies the capacitance acts like a short circuit across the RL branch.
Express the impedance Z s of the parallel combination of L 4 H and C 1 F. frequencies s is this impedance zero or in nite
Z s 4s 1=s s 4s 1=s s2 0:25
At what
By inspection, Z 0 0 and Z 1 0, which agrees with our earlier understanding of parallel LC circuits at frequencies of zero (dc) and in nity. For jZ s j 1, s2 0:25 0 or s j0:5 rad=s
A sinusoidal driving source, of frequency 0.5 rad/s, results in parallel resonance and an in nite impedance.
The circuit shown in Fig. 8-22 has a voltage source connected at terminals ab. The response to the excitation is the input current. Obtain the appropriate network function H s .
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