If the DTFT of a finite sequence is taken, the result is a continuous function
of
ω
ω
.
If the DFT of the same sequence is taken, the results are
N
N
evenly spaced samples of the DTFT. In other words, the DTFT of a finite signal
can be evaluated at
N
N
points with the DFT.
X
(
ω
)
=
D
T
F
T
{
x
(
n
)
}
=
∑
n
=
−
∞
∞
x
(
n
)
e
−
j
ω
n
X
(
ω
)
=
D
T
F
T
{
x
(
n
)
}
=
∑
n
=
−
∞
∞
x
(
n
)
e
−
j
ω
n
(1)
and because of the finite length
X
(
ω
)
=
∑
n
=
0
N
−
1
x
(
n
)
e
−
j
ω
n
.
X
(
ω
)
=
∑
n
=
0
N
−
1
x
(
n
)
e
−
j
ω
n
.
(2)
If we evaluate
ω
ω
at
N
N
equally space points, this becomes
X
(
2
π
N
k
)
=
∑
n
=
0
N
−
1
x
(
n
)
e
−
j
2
π
N
k
n
X
(
2
π
N
k
)
=
∑
n
=
0
N
−
1
x
(
n
)
e
−
j
2
π
N
k
n
(3)
which is the DFT of
x
(
n
)
x
(
n
)
.
By adding zeros to the end of
x
(
n
)
x
(
n
)
and taking a longer DFT, any density of points can be evaluated. This is
useful in interpolation and in plotting the spectrum of a finite length
signal. This is discussed further in Chapter
(Reference).
There is an interesting variation of the Parseval's theorem for the DTFT of a
finite
length-
N
N
signal. If
x
(
n
)
≠
0
x
(
n
)
≠
0
for
0
≥
n
≥
N
−
1
0
≥
n
≥
N
−
1
,
and if
L
≥
N
L
≥
N
,
then
∑
n
=
0
N
−
1
|
x
(
n
)
|
2
=
1
L
∑
k
=
0
L
−
1
|
X
(
2
π
k
/
L
)
|
2
=
1
π
∫
0
π
|
X
(
ω
)
|
2
ⅆ
ω
.
∑
n
=
0
N
−
1
|
x
(
n
)
|
2
=
1
L
∑
k
=
0
L
−
1
|
X
(
2
π
k
/
L
)
|
2
=
1
π
∫
0
π
|
X
(
ω
)
|
2
ⅆ
ω
.
(4)
The second term in
(
Equation 4) says the
Riemann sum is equal to its limit in this case.
