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CHAP. 16]



B1 B2 þ s  b ðs  bÞ2 where the constants B2 and B1 may be found as 2

B2 ¼ limfðs  bÞ RðsÞg s!b

 B1 ¼ lim ðs  bÞ RðsÞ 



B2 ðs  bÞ2

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 RðsÞ has no such poles. Once the partial-functions expansion of RðsÞ is known, Table 16-1 can be used to invert each term and thus to obtain the time function rðtÞ. EXAMPLE 16.4

Find the time-domain current iðtÞ if its Laplace transform is s  10 s4 þ s2 s  10 IðsÞ ¼ 2 s ðs  jÞðs þ jÞ IðsÞ ¼

Factoring the denominator,

we see that the poles of IðsÞ are s ¼ 0 (double pole) and s ¼ j (simple poles). The principal part at s ¼ 0 is


B1 B2 1 10 þ 2 ¼  2 s s s s   s  10 ¼ 10 B2 ¼ lim s!0 ðs  jÞðs þ jÞ     s  10 10 10s þ 1 B1 ¼ lim s 2 2 þ ¼1 ¼ lim 2 s!0 s!0 s þ 1 s ðs þ 1Þ s2 The principal part at s ¼ þj is A 0:5 þ j5 ¼ sj sj   s  10 A ¼ lim 2 ¼ ð0:5 þ j5Þ s!j s ðs þ jÞ


It follows at once that the principal part at s ¼ j is 

0:5  j5 sþj

The partial-fractions expansion of IðsÞ is therefore IðsÞ ¼

1 1 1 1  10 2  ð0:5 þ j5Þ  ð0:5  j5Þ s sj sþj s

and term-by-term inversion using Table 16-1 gives iðtÞ ¼ 1  10t  ð0:5 þ j5Þe jt  ð0:5  j5Þejt ¼ 1  10t  ðcos t  10 sin tÞ

Heaviside Expansion Formula If all poles of RðsÞ are simple, the partial-fractions expansion and termwise inversion can be accomplished in a single step:   X n PðsÞ Pðak Þ ak t l1 e ð4Þ ¼ 0 QðsÞ Q ðak Þ k¼1 where a1 ; a2 ; . . . ; an are the poles and Q 0 ðak Þ is dQðsÞ=ds evaluated at s ¼ ak .