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HIGHER-ORDER CIRCUITS AND COMPLEX FREQUENCY

[CHAP. 8

EXAMPLE 8.6 A series RL circuit, with R ¼ 10  and L ¼ 2 H, has an applied voltage v ¼ 10 e2t cos ð10t þ 308Þ. Obtain the current i by an s-domain analysis. v ¼ 10

308 est ¼ Ri þ L

di di ¼ 10i þ 2 dt dt

Since i ¼ Iest , 10 308 est ¼ 10Iest þ 2sIest

or

10 308 10 þ 2s

Substituting s ¼ 2 þ j10, I¼

10 308 10 308 ¼ ¼ 0:48 43:38 10 þ 2ð2 þ j10Þ 6 þ j20

Then, i ¼ Iest ¼ 0:48e2t cos ð10t  43:38Þ (A). EXAMPLE 8.7 A series RC circuit, with R ¼ 10  and C ¼ 0:2 F, has the same applied voltage as in Example 8.6. Obtain the current by an s-domain analysis. As in Example 8.6, ð ð 1 i dt ¼ 10i þ 5 i dt v ¼ 10 308 est ¼ Ri þ C Since i ¼ Iest , 10

308 est ¼ 10Iest þ

5 st Ie s

from which

10 308 ¼ 1:01 10 þ 5=s

32:88

Then, i ¼ 1:01e2t cos ð10t þ 32:88Þ (A).

Note that the s-domain impedance for the capacitance is 1=ðsCÞ. a series RLC circuit will be ZðsÞ ¼ R þ sL þ 1=ðsCÞ

8.7

Thus the s-domain impedance of

NETWORK FUNCTION AND POLE-ZERO PLOTS

A driving voltage of the form v ¼ Vest applied to a passive network will result in currents and voltages throughout the network, each having the same time function est ; for example, Ie j est . Therefore, only the magnitude I and phase angle need be determined. We are thus led to consider an s-domain where voltages and currents are expressed in polar form, for instance, V , I , and so on. Figure 8-12 suggests the correspondence between the time-domain network, where s ¼  þ j!, and the

Fig. 8-12

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An  
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