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502 CHAPTER 14 The Fourier Integral and Transforms

Subdivide [0,2π L] into N subintervals of length 2π L/N and choose partition points ξ j =
2π jL/N for j = 0,1,··· , N − 1. Using these, approximate
2π L
ˆ
f (ω) ≈ f (ξ)e −iωξ dξ
0
N−1
−2πij Lω/N 2π L
≈ f (2π jL/N)e
N
j=0
N−1
2π L
= f (2π jL/N)e −2πij Lω/N .
N
j=0
The sum on the right is nearly in the form of a DFT. If we put ω = k/L with k any integer, then
we have
N−1
2π L
ˆ
f (k/L) ≈ f (2π jL/N)e −2πijk/N . (14.19)
N
j=0
This approximates f (k/L), the Fourier transform of f evaluated at k/L, with the N-point DFT
ˆ
of the sequence
N−1

2π jL
f .
N j=0
ˆ
Again, we must restrict |k|≤ N/8 because the DFT is periodic of period N, while f (k/L) is not
periodic.

EXAMPLE 14.17
We will test the approximation (14.19) for the simple case that

e −t for t ≥ 0
f (t) =
0 for t < 0.
f has Fourier transform
∞

ˆ
f (ω) = f (ξ)e −iωξ dξ
−∞

∞ 1 − iω
−ξ −iωξ
= e e dξ = .
0 1 + ω 2
7
Choose L = 1, N = 2 = 128 and k = 3, so |k|≤ N/8. Now k/L = 3 and the approximation
(14.17) is
127
2π
ˆ
e
ˆ
f (k/L) = f (3) ≈ e −π j/64 −6πij/128
128
j=0
127
π
e
= e −π j/64 −3πij/64 = 0.12451 − 0.29884i.
64
j=0
For comparison, the exact value is
1 − 3i
ˆ
f (3) = = 0.1 − 0.3i.
10
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October 14, 2010 16:43 THM/NEIL Page-502 27410_14_ch14_p465-504