1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

546 CHAP'l'ER 12 • FOURIER SERIE.~ AND THE LAPLACE TRANSFORM

1
•EXAMPLE 12.10 Show that .C (t) = 2 ·
s

Solution Integration by parts yields

.C (t) = lim {R te-•tdt

R-+oo} 0

(

= 1 1m. -e - t -st - 2e^1 -st ) lt=R

R-+oo s S t=O

= hm. (-R -e -sR --^1 e -sR) +^0 +-^1 =0-0+ - =^1 -.^1

R-+oo s s^2 s2 52 52

For values of sin the right half-plane Re (s) > 0, an argument similar to that in

Example 12.8 shows that the limit approaches zero, establishing the result.

s

•EXAMPLE 12.11 Show that .C(cosbt) = --r--b 2 •

s +

Solution A direct approach using the definition is tedious. Instead, let's as-

sume that the complex constants ±ib are permitted and hence that the following

Laplace transforms exist:

.C (eibt) = 1 . and .C (e-ibt) = 1 ..

5 - ib s + ib

Using the linearity of the Laplace transform, we have

.C (cosbt) = ~.C (eibt) + ~.C (e-ibt)

1 1 1 1 s

=---+---=--

2 s - ib 2 5 + ib 52 + b2.

Inverting the Laplace transform is usually accomplished with the aid of a

table of known Laplace transforms and the technique of partial fraction expan-

sion. Table 12.2 gives the Laplace transforms of some well-known functions, and

Table 12.3 highlights some important properties of Laplace transforms.

• EXAMPLE 12. 12 Find .c-^1 (!:: ~).