Digital Signal Processing Reference
In-Depth Information
Therefore, in its complex Fourier series form, we have
N
1
X p (k)e j( 2 π/N)kn
x p (n)
=
(3.70)
k
=
0
To find these coefficients, let us multiply both sides by e jmω 0 k and sum over n
from n
= 0to (N
1 ) :
N
1
N
1
N
1
x p (n)e jmω 0 k
X p (k)e j( 2 π/N)kn e jmω 0 k
=
(3.71)
n
=
0
n
=
0
k
=
0
By interchanging the order of summation on the right side, we get
X p (k) N 1
e j( 2 π/N)k(n m)
N
1
(3.72)
k
= 0
n
= 0
It is next shown that [ N 1
0 e j( 2 π/N)k(n m) ] is equal to N when n
=
m and zero
n
=
m , the summation reduces to [ N 1
n
e j 0 ] =
for all values of n
=
m .When n
=
= 0
N ,andwhen n
m , we apply (3.52) and find that the summation yields zero.
Hence there is only one nonzero term X p (k)N in (3.72). The final result is
=
N
1
1
N
x p (n)e jnω 0 k
X p (k)
=
(3.73)
n
=
0
Now we notice that
N
1
N
1
1
N
1
N
x p (n)e jnω 0 k
x(n)e j( 2 π/N)nk
=
n
=
0
n
=
0
1
N
X p (e ) ω k = ( 2 π/N)k =
=
X p (k)
(3.74)
In other words, when the DTFT of the finite length sequence x(n) is evaluated
at the discrete frequency ω k
( 2 π/N)k ,(whichisthe k th sample when the
frequency range [0 , 2 π ] is divided into N equally spaced points) and dividing
by N , we get the value of the Fourier series coefficient X p (k) .
The expression in (3.70) is known as the discrete-time Fourier series
(DTFS) representation for the discrete-time, periodic function x p (n) and (3.73),
which gives the complex-valued coefficients of the DTFS is the inverse DTFS
(IDTFS). Because both x p (n) and X p (k) are periodic, with period N , we observe
=
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