Page 499 - Advanced engineering mathematics

14.3 The Fourier Transform 479

Frequency Differentiation The variable ω used in f (ω) is the frequency of f (t), since it occurs
ˆ
iωt
in the complex exponential e , which is cos(ωt) + i sin(ωt). In this context, the process of
ˆ
computing f (ω) is called frequency differentiation. We will show how derivatives of f (ω) relate
ˆ
to f (t).
Let n be a positive integer. Let f be piecewise continuous on [−L, L] for every positive L
∞ n
and assume that |t f (t)|dt converges. Then
−∞
d n −n n
f (ω) = i F[t f (t)](ω).
ˆ
dω n
This means that the nth derivative of the Fourier transform of f is i −n times the transform
n
of t f (t).
We will indicate a proof for the case n = 1. Write
d d ∞ −iωt ∞ ∂ −iωt
ˆ
f (ω) = f (t)e dt = [ f (t)e ]dt
dω dω −∞ ∂ω
−∞
∞ ∞

= f (t)(−it)e −iωt dt =−i [tf (t)]e −iωt dt
−∞ −∞
=−iF[tf (t)](ω).
As an example, using the result of Example 14.3, we can write
d 2 10 25 − 3ω 2
2 −5|t|
F[t e ](ω) = i 2 = 20 .
2 2
dω 2 25 + ω 2 (25 + ω )
The Fourier Transform of an Integral Let f be piecewise continuous on every interval
∞
[−L, L]. Suppose | f (t)|dt converges and that f (0) = 0. Then
−∞
t
1

ˆ
F f (τ)dτ (ω) = f (ω).
iω
−∞
t

To prove this, define g(t) = f (τ)dτ. Then g (t) = f (t) at each point at which f is
−∞
continuous. Further, g(t) → 0as t →−∞, and
∞

ˆ
lim g(t) = f (τ)dτ = f (0) = 0
t→∞
−∞
by assumption. Therefore, applying the operational formula,
ˆ

f (ω) = F[g (t)](ω)
t

= iω[g(t)](ω) = iωF f (τ)dτ (ω).
−∞
Convolution
Integral transforms usually have some kind of convolution operation. We have seen a convolution
for the Laplace transform. For the Fourier transform, we define the convolution of f with g to be
the function f ∗ g given by
∞

( f ∗ g)(t) = f (t − τ)g(τ)dτ.
−∞
b b
In making this definition, we assume that f (t)dt and g(t)dt exist for every interval [a,b]
∞ a a
and that, for every real number t, | f (t − τ)g(t)|dτ converges.
−∞
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October 14, 2010 16:43 THM/NEIL Page-479 27410_14_ch14_p465-504