Page 497 - Advanced engineering mathematics

14.3 The Fourier Transform 477

Symmetry

ˆ
F[ f (t)](ω) = 2π f (−ω).
If we replace ω by t in the transformed function f , and then take the transform of this function
ˆ
of t, we obtain 2π times the original function f with t replaced by −ω.

Modulation If ω 0 is a real number, then
1
ˆ
ˆ
F[ f (t)cos(ω 0 t)](ω) = f (ω + ω 0 ) + f (ω − ω 0 )
2
and
i
ˆ
ˆ
F[ f (t)sin(ω 0 t)](ω) = f (ω + ω 0 ) − f (ω − ω 0 )
2
To prove the first expression, put
1 iωt −iωt
cos(ωt) = e + e ,
2
then use the linearity of F and the frequency-shifting theorem to write

1 1
F[ f (t)cos(ω 0 t)](ω) = F e iω 0 t f (t) + e −iω 0 t f (t) (ω)
2 2
1 1
= F[e iω 0 t f (t)](ω) + F[e −iω 0 t f (t)](ω)
2 2
1 1
ˆ
ˆ
= f (ω − ω 0 ) + f (ω + ω 0 ).
2 2
The second conclusion is proved by a similar calculation.

Operational Formula To apply the Fourier transform to a differential equation we must be able
to transform a derivative. This is called an operational rule. Recall that the kth derivative of f
is denoted f (k) . As a convenience, we let f (0) = f - the zero-order derivative of a function is just
the function.
Now let n be any positive integer and suppose that f (n−1) is continuous and f (n) is piecewise
∞ (n−1)
continuous on each interval [−L, L]. Suppose also that | f |dt converges and that
−∞
lim f (k) (t) = lim f (k) (t) = 0
t→∞ t→−∞
for k = 0,1,2,··· ,n − 1. Then
n
ˆ
F[ f (n) (t)](ω) = (iω) f (ω).
That is, under the given conditions, the Fourier transform of the nth derivative of f is the
nth power of iω times the Fourier transform of f .
Proof Since

d
f (n) (t) = f (n−1) (t),
dt

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October 14, 2010 16:43 THM/NEIL Page-477 27410_14_ch14_p465-504