082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

24.7 The relationship between the Fourier transform and the Laplace transform 773
Solution (a) Either by integration, orfrom Table 21.1, we find
L{u(t)e 3t } = 1
s −3
provideds > 3
(b) F{u(t)e 3t } =
∫ ∞
0
e 3t e −jωt dt =
∫ ∞
0
[ e
(3−jω)t ∞
e (3−jω)t dt = .
3−jω]
0
Now, as t → ∞, e 3t → ∞, so that the integral fails to exist. Clearly, u(t)e 3t has a
Laplace transform butno Fourier transform.
Suppose f (t)isdefined tobe0fort < 0.Thenits Fourier transform becomes
F{f(t)} =
∫ ∞
0
f (t)e −jωt dt
and its Laplace transform is
L{f(t)} =
∫ ∞
0
f (t)e −st dt
ByreplacingsbyjωintheLaplacetransformweobtaintheFouriertransformof f (t)if
itexists.CaremustbetakenheresincewehaveseenthattheFouriertransformmaynot
exist forafunctionthatnevertheless has a Laplace transform.
Example24.15 Findthe Laplace transforms of
(a) u(t)e −2t
(b) u(t)e 2t
Lets = jω and comment uponthe result.
Solution (a) L{u(t)e−2t } = 1
s +2 .
(b) L{u(t)e 2t } = 1
s −2 .
Replacingsby jω in (a) gives
Now
F{u(t)e −2t } = 1
jω+2
1
jω+2 . Similarly, replacingsby jω in (b) gives 1
jω−2 .
sothatreplacingsbyjωintheLaplacetransformresultsintheFouriertransform.However,
F{u(t)e 2t } does not exist and even though we can lets = jω in the Laplace transformand
obtain , wecannot interpretthis asaFourier transform.
1
jω−2
TheFouriertransformdoespossesscertainadvantagesovertheLaplacetransform.While
the Laplace transform can only be applied to functions which are zero fort < 0, the