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It follows at once that the principal part at s j is 0:5 j5 s j
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The partial-fractions expansion of I s is therefore I s 1 1 1 1 10 2 0:5 j5 0:5 j5 s s j s j s
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and term-by-term inversion using Table 16-1 gives i t 1 10t 0:5 j5 e jt 0:5 j5 e jt 1 10t cos t 10 sin t
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Heaviside Expansion Formula If all poles of R s are simple, the partial-fractions expansion and termwise inversion can be accomplished in a single step:   X n P s P ak ak t l 1 e 4 Q s Q 0 ak k 1 where a1 ; a2 ; . . . ; an are the poles and Q 0 ak is dQ s =ds evaluated at s ak .
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[CHAP. 16
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In 8 we introduced and utilized the concept of generalized impedance, admittance, and transfer functions as functions of the complex frequency s. In this section, we extend the use of the complex frequency to transform an RLC circuit, containing sources and initial conditions, from the time domain to the s-domain.
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Table 16-2 Time Domain s-Domain s-Domain Voltage Term RI s
sLI s Li 0
sLI s Li 0
I s V0 sC s
I s V0 sc s
Table 16-2 exhibits the elements needed to construct the s-domain image of a given time-domain circuit. The rst three lines of the table were in e ect developed in Example 16.1. As for the capacitor, we have, for t > 0, vC t V0 so that, from Table 16-1, VC s V0 I s Cs s 1 C t i  d
EXAMPLE 16.5 In the circuit shown in Fig. 16-4(a) an initial current i1 is established while the switch is in position 1. At t 0, it is moved to position 2, introducing both a capacitor with initial charge Q0 and a constant-voltage source V2 . The s-domain circuit is shown in Fig. 16-4(b). The s-domain equation is
RI s sLI s Li 0 in which V0 Q0 =C and i 0 i1 V1 =R.
I s V0 V2 sC sC s
CHAP. 16]
THE LAPLACE TRANSFORM METHOD
Fig. 16-4
THE NETWORK FUNCTION AND LAPLACE TRANSFORMS
In 8 we obtained responses of circuit elements to exponentials est , based on which we introduced the concept of complex frequency and generalized impedance. We then developed the network function H s as the ratio of input-output amplitudes, or equivalently, the input-output di erential equation, natural and forced responses, and the frequency response. In the present chapter we used the Laplace transform as an alternative method for solving di erential equations. More importantly, we introduce Laplace transform models of R, L, and C elements which, contrary to generalized impedances, incorporate initial conditions. The input-output relationship is therefore derived directly in the transform domain. What is the relationship between the complex frequency and the Laplace transform models A short answer is that the generalized impedance is the special case of the Laplace transform model (i.e., restricted to zero state), and the network function is the Laplace transform of the unit-impulse response.
EXAMPLE 16.17 Find the current developed in a series RLC circuit in response to the following two voltage sources applied to it at t 0: (a) a unit-step, (b) a unit-impulse. The inductor and capacitor contain zero energy at t 0 . Therefore, the Laplace transform of the current is I s V s Y s . (a) V s 1=s and the unit-step response is I s 1 Cs 1 1 s LCs2 RCs 1 L s  2 !2 d 1 t i t e sin !d t u t L!d s  R 2 1 !d 2L LC
where R ;  2L (b) V s 1 and the unit-impulse response is I s 1 s L s  2 !2 d 1 t i t e !d cos !d t  sin !d t u t L!d and
The unit-impulse response may also be found by taking the time-derivative of the unit-step response. EXAMPLE 16.18 Find the voltage across terminals of a parallel RLC circuit in response to the following two current sources applied at t 0: (a) a unit-step, (b) a unit-impulse. Again, the inductor and capacitor contain zero energy at t 0 . Therefore, the Laplace transform of the current is V s I s Z s .
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