Page 493 - Advanced engineering mathematics

14.3 The Fourier Transform 473

This is the pulse
f (t) = k[H(t + a) − H(t − a)].

Then
∞

ˆ
f (t) = f (t)e −iωt dt
−∞
a −k a
= ke −iωt dt = e −iωt
iω
−a −a
k 2k
=− [e −iωa − e iωa ]= sin(aω).
iω ω
These examples were done by integration. Usually the Fourier transform of a function is
computed using tables or software. In MAPLE, use

fourier(f(t),t,ω);
This is in the inttrans set of subroutines, for integral transforms. The Laplace transform is
also in this set.

Now suppose that f is continuous and f is piecewise smooth on every interval [−L, L].
Because f (ω) is the coefficient in the complex Fourier integral representation of f ,
ˆ
1 ∞
ˆ
f (t) = f (ω)e iωt dω. (14.11)
2π
−∞
Equation (14.11) defines the inverse Fourier transform.Given f satisfying certain conditions,
ˆ
we can compute its Fourier transform f , and, conversely, given this transform, we can recover f
from equation (14.11). For this reason we call the equations
∞ 1 ∞

ˆ
ˆ
f (ω) = f (t)e −iωt dt and f (t) = f (ω)e iωt dω
2π
−∞ −∞
−1
a transform pair. We also denote the inverse Fourier transform as F :
−1
ˆ
ˆ
F [ f ]= f exactly when F[ f ]= f .
−1
In MAPLE, F [ f ] can be computed using
invfourier[F,ω,t];
EXAMPLE 14.6
Let

1 −|t| for −1 ≤ t ≤ 1
f (t) =
0 for |t| > 1.
Then f is continuous and absolutely integrable, and f is piecewise continuous. A routine

integral gives us the Fourier transform of f :
∞

ˆ
f (ω) = f (t)e −iωt dt
−∞
1 2(1 − cos(ω))
= (1 −|t|)e −iωt = .
−1 ω 2

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October 14, 2010 16:43 THM/NEIL Page-473 27410_14_ch14_p465-504