INITIAL-VALUE AND FINAL-VALUE THEOREMS

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Taking the limit as s ! 1 (through real values) of the direct Laplace transform of the derivative, df t =dt, lim l 1 df t df t st e dt lim fsF s f 0 g lim s!1 0 s!1 dt dt

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THE LAPLACE TRANSFORM METHOD

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[CHAP. 16

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But e st in the integrand approaches zero as s ! 1. Thus,

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lim fsF s f 0 g 0

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Since f 0 is a constant, we may write f 0 lim fsF s g

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which is the statement of the initial-value theorem.

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EXAMPLE 16.3 In Example 16.1,

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lim fsI s g lim 10

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s!1

8s 10 8 2 s 2000

which is indeed the initial current, i 0 2 A. The nal-value theorem is also developed from the direct Laplace transform of the derivative, but now the limit is taken as s ! 0 (through real values). 1 df t df t st lim l e dt limfsF s f 0 g lim s!0 s!0 0 s!0 dt dt 1 1 df t st lim But df t f 1 f 0 e dt s!0 0 dt 0 and f 0 is a constant. Therefore, f 1 f 0 f 0 limfsF s g

f 1 limfsF s g

This is the statement of the nal-value theorem. The theorem may be applied only when all poles of sF s have negative real parts. This excludes the transforms of such functions as et and cos t, which become in nite or indeterminate as t ! 1.

PARTIAL-FRACTIONS EXPANSIONS

The unknown quantity in a problem in circuit analysis can be either a current i t or a voltage v t . In the s-domain, it is I s or V s ; for the circuits considered in this book, this will be a rational function of the form R s P s Q s

where the polynomial Q s is of higher degree than P s . Furthermore, R s is real for real values of s, so that any nonreal poles of R s , that is, nonreal roots of Q s 0, must occur in complex conjugate pairs. In a partial-fractions expansion, the function R s is broken down into a sum of simpler rational functions, its so-called principal parts, with each pole of R s contributing a principal part. Case 1: s a is a simple pole. principal part of R s is When s a is a nonrepeated root of Q s 0, the corresponding

A s a

where

A limf s a R s g

If a is real, so will be A; if a is complex, then a is also a simple pole and the numerator of its principal part is A . Notice that if a 0, A is the nal value of r t Case 2: s b is a double pole. When s b is a double root of Q s 0, the corresponding principal part of R s is

CHAP. 16]

THE LAPLACE TRANSFORM METHOD

B1 B2 s b s b 2 where the constants B2 and B1 may be found as B2 limf s b R s g

s!b 2

B1 lim s b R s

B2 s b 2

B1 may be zero. Similar to Case 1, B1 and B2 are real if b is real, and these constants for the double pole b are the conjugates of those for b. The principal part at a higher-order pole can be obtained by analogy to Case 2; we shall assume, however, that R s has no such poles. Once the partial-functions expansion of R s is known, Table 16-1 can be used to invert each term and thus to obtain the time function r t .

EXAMPLE 16.4 Find the time-domain current i t if its Laplace transform is s 10 s4 s2 s 10 I s 2 s s j s j I s

Factoring the denominator,

we see that the poles of I s are s 0 (double pole) and s j (simple poles). The principal part at s 0 is B1 B2 1 10 2 2 s s s s s 10 10 B2 lim s!0 s j s j s 10 10 10s 1 B1 lim s 2 2 1 lim 2 s!0 s!0 s 1 s s 1 s2 The principal part at s j is A 0:5 j5 s j s j s 10 A lim 2 0:5 j5 s!j s s j