1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

566 CHAPTER 12 • FOURIER SERIES AND THE LAPLACE TRANSFORM

of Q (s) occur at 0 ± 2i and 0 ± 3i. Computing the residues yields

Res (Y 2i] = p (^2 i) =^5 (^2 i) = ~ and

' Q' (2i) 4 (2i)^3 - 26 (2i) 2'

Res [Y 3i] = p (3i) = 5 (3i) = -21.

' Q' (3i) 4 (3i)^3 - 26 (3i)

We find that A 1 + iB 1 =! + Oi and A2 + iB2 = -! + Oi, which correspond to

a 1 + ib 1 = 0 + 2i and a 2 + ib 2 = 0 + 3i, respectively. Thus we obtain

2(!)(s- 0)-2(0)2 2(-!)(s-0)-2(0)3 s s

Y (s) - + - --- --

• s^2 + 4 s^2 + 9 - s^2 + 4 s^2 + 9 '

and the desired solution is

.c-^1 (Y (s)) = .c-^1 ( 52 :

4

) - .c-^1 (s 2 :

9

) = cos2t -cos3t.

• EXAMPLE 12. 26 Find .c-^1 (Y (s)) if Y (s) =

83

+

352

• 5

+I.

s(s+ 1)^2 (s2 +I)

Solution The partial fractional expression for Y (s) has the form

Y ( )

D C1 C2 2A (s -0) - 2B (1)

s = - + - - + + - -'--- '-:---"-'-

8 s+l (s+l)^2 (s-0)^2 +12

The linear factor s is nonrepeated, so we have

. s^3 + 3s^2 - s + 1
D = Res [Y (s), OJ= bm 2 = 1.
•-o (s + 1) (s^2 + 1)

The factor s + 1 is repeated, so we have

. d s^3 + 3s^2 - s + 1
C 1 =Res[Y(s),-l]=hm-d ( 2 l)
·--1 s 8 8 +

I

. -3s^4 + 4s^3 - 1
2
d
= 1m = - an

• -- 1 s2 (s + 1)2
  . s^3 + 3s^2 - s + 1

C2 = Res[(s+l)Y(s),- 1] = lim ( 2 l) =-2.

·-- 1 s s +

The term s^2 + 1 is an irreducible quadratic, with roots ±i, so that

A + i -'B - Res (Y i-1m ·1 - 1· s3 + 3s2 2 - s + 1 -- --1 -i

' •-'s(s+l) (s+i) 2 '