1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

574 CHAPTER 12 • FOURI ER SERIES AND T HE LAPLACE TRANSFORM

Solution Letting F(s) = .C(f (t)) and using .C(t) =

1

s^2 in the convolution

t heorem, we obtain

2s 1

F(s)= s2+1 - szF(s).

Solving for F (s), we get

F (s) _ 2s

3

_ 2s 2s

• (s2 + 1)^2 - s^2 + 1 - (s2 + 1)^2 '

and t he solution is

f (t) = 2cost- tsint.

Engineers and physicists sometimes consider forces t hat produce large effects

but that are applied over a very short time interval. The force acting at the time

an earthquake starts is an example. This phenomenon leads to the idea of a

unit impulse funct ion, o (t). Let's consider the small positive constant a. The

function Oo. ( t) is defined by

oo. (t) = { 5:

for 0 < t <a;

otherwise.

The unit impulse function is obtained by letting the interval width go to zero,

or

o (t) = lim oo. (t).

o.-o

Figure 12.29 shows the graph of Oo. (t) for a= 10, 40, and 100. Alt hough o (t) is

called t he Dirac delta function, it is not an ordinary function. To be precise

it is a distribution, and the theory of distributions permits manipulation of o (t)

as t hough it were a function. Here, we treat o (t) as a function and investigate

its properties.

• EXAMPLE 12 .3 1 Show that .C (o (t)) = 1.
  Solution By definition, t he Laplace transform of Oa (t) is

.C (o,. (t)) = ["° Oo. (t) e-•tdt = t ~e-•tdt =^1 - e- •a.

k k a sa

Letting a ..... 0 in equation and using L' Hopital's rule, we obtain

1 -e-•a 0 + se-•o.

.c (o (t )) = lim .c (oo. (t)) = lim = lim = 1.

a-o a~o sa a.-o s