Page 513 - Advanced engineering mathematics

14.5 The Discrete Fourier Transform 493

N−1

U k = u j e −2πijk/N
j=0
for k = 0,±1,±2,···. We will also abbreviate the phrase “discrete Fourier transform” to DFT.

EXAMPLE 14.14
N−1
Let u =[c] j=0 , a constant sequence, with c a given complex number. The N-point DFT of u is
given by
N−1 N−1
−2πijk/N −2πijk/N
U k = ce = c e .
j=0 j=0
Observe that this is
N−1
j

U k = c e −2πik/N ,
j=0
a finite geometric series. In general, for |r| < 1,
N−1 N
1 −r
j
r = .
1 −r
j=0
Then
1 − (e )
−2πik/N N
U k = c
1 − e −2πik/N
−2πik
1 − e
= c = 0
1 − e −2πik/N
for k = 0,±1,±2,··· because, for any integer k,
e −2πik = cos(2πk) − i sin(2πk) = 1.
The N-point DFT of a constant sequence is an infinite sequence of zeros.

EXAMPLE 14.15
N−1
We will find the N-point DFT of u =[sin( ja)] , in which N is a positive integer and a is a
j=0
given complex number. To avoid trivialities, suppose a is not an integer multiple of π.Wehave
N−1

U k = sin( ja)e −2πijk/N .
j=0
Use the fact that

1 ija −ija
sin( ja) = e − e .
2i
Then
N−1 N−1
1 1
ija −2πijk/N
e
U k = e e − e −ija −2πijk/N
2i 2i
j=0 k=0
N−1
N−1
1 1
) .
) −
= (e ia−2πik/N j (e −ia−2πik/N j
2i 2i
j=0 j=0
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October 14, 2010 16:43 THM/NEIL Page-493 27410_14_ch14_p465-504