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nonzero, the function is a damped cosine. Only negative values of are considered. If and ! are zero, the result is a constant. And nally, with ! 0 and nonzero, the result is an exponential decay function. In Table 8-1, several functions are given with corresponding values of s for the expression Aest .

9. (a) 16; (b) 16; (c) 10. (a) 21; (b) 42; (c) 3

Table 8-1 f t 10e 5 cos 500t 308 2e 3t cos 100t 458 100:0

1 +1

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A 10 5 2 100.0

_ 10

When Fig. 8-10 is examined for various values of s, the three cases are evident. If 0, there is no damping and the result is a cosine function with maximum values of Vm (not shown). If ! 0, the function is an exponential decay with an initial value Vm . And nally, with both ! and nonzero, the damped cosine is the result.

15. Only (b) is not a proportion since 3(12)

_ 20

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A driving voltage of the form v Vm est applied to a passive network will result in branch currents and voltages across the elements, each having the same time dependence est ; e.g., Ia e j est , and Vb e j est . Consequently, only the magnitudes of currents and voltages and the phase angles need be determined (this will also be the case in sinusoidal circuit analysis in 9). We are thus led to consider the network in the s-domain (see Fig. 8-11).

5(7); that is, 36

Fig. 8-11

16. (a)

_ 45

A series RL circuit with an applied voltage v Vm e j est will result in a current i Im e j est Im est , which, substituted in the nodal equation Ri L will result in RIm est sLIm est Vm e j est from which Im Vm e j R sL The inductance theredi Vm e j est dt

6; (b)

_ 90

Note that in the s-domain the impedance of the series RL circuit is R sL. fore has an s-domain impedance sL.

17. (a) d; (b) 35; (c) 5 (d) 4 18. (a) 21; (b) 3; (c) 5 2 19. (a) 16; (b) 62; (c) 10 3 20. (a) yes, since 15 10

and L 2 H, has an applied voltage v 10 e 2t cos 10t 308 . Obtain the current i by an s-domain analysis. v 10 Since i Iest , 10 308 est 10Iest 2sIest Substituting s 2 j10, I 10 308 10 308 0:48 43:38 10 2 2 j10 6 j20 or I 10 308 10 2s 308 est Ri L di di 10i 2 dt dt

Fig. 8-2 Example 8.2. A simple rst-order network has Laplace-domain transfer function and frequency transfer function A s s 1 and A j! 1 j! Determine the network phase angle and the amplitude ratio Mdb , and discuss the nature of the Bode plot. After A j! is converted to polar form, it becomes apparent that tan 1 ! q 20 log jA j! j 20 log 1 ! 2 10 log 1 ! 2 8:7 (8.8)

18 12 ;