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498 CHAPTER 14 The Fourier Integral and Transforms

14.6 Sampled Fourier Series

We have just discussed the approximation of Fourier coefficients of a periodic function f (t).
This was done by approximating terms of an N-point DFT formed by sampling f (t) at N points
of [0, p].
We will now discuss the use of an inverse DFT to approximate sampled partial sums (partial
sums evaluated at selected points) of the Fourier series of a period function.
Consider the partial sum

M

S N (t) = d k e 2πikt/p .
k=−M
Subdivide [0, p] into N subintervals of equal length p/N and choose sample points t j = jp/N
for j = 0,1,··· , N − 1. Form the N-point sequence
N−1
u =[ f ( jp/N)] j=0
and approximate
1
d k ≈ U k
N
where
N−1

U k = f ( jp/N)e −2πijk/N .
j=0
In order to have |k|≤ N/8, we will require that M ≤ N/8 in forming the partial sum of the
Fourier series. We have
M
1
2πikt/p
S M (t) ≈ U k e .
N
k=−M
In particular, if we sample this partial sum at the partition point jp/N, then
M
1
S M ( jp/N) ≈ U k e 2πijk/N .
N
k=−M
We will show that the sum on the right is actually an N-point inverse DFT for a particular N-point
sequence which we will determine by exploiting the periodicity of the N-point DFT (U k+N =U k ).
Write
−1 M
1 2πijk/N 1 2πijk/N
S M ( jp/N) ≈ U k e + U k e
N N
k=−M k=0
M
M
1 1
= U −k e −2πijk/N + U k e 2πijk/N
N N
k=1 k=0
M M
1 2πij(−k+N)/N 1 2πijk/N
= U −k+N e + U k e
N N
k=1 k=0
M−1 M
1
= U k e 2πijk/N + U k e 2πijk/N .
N
k=N−M k=0

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October 14, 2010 16:43 THM/NEIL Page-498 27410_14_ch14_p465-504