Digital Signal Processing Reference
In-Depth Information
8
X
N
1
k
¼
0
X
k
e
þ
j2
p
ki
=
8
x
i
¼
1
=
Since X
0
¼
8 and the rest are zero, we only need to evaluate the
summation for k
¼
0:
8
8
e
þ
j2
p
0i
=
8
x
i
¼
1
=
¼
1
This is true for all values of i (the 0 in the exponent means the
value of “i” is irrelevant). So we get an infinite sequence of 1s.
In general however, we would evaluate for i from 0 to N
1.
Due to the periodicity of the transform, there is no point in
evaluating when i
¼
N or greater. If we evaluate for i
¼
N, we will
find we get the same value as i
¼
0, and for i
¼
N
þ
1, we will get
the same value as i
¼
1.
12.1.2 Second DFT Example
Let's consider another simple example, with a time domain
signal {1,j,
1,
j,1,j,
1,
j}. This is actually the complex expo-
nential e
þ
j2
p
i
=
4
. This signal consists of a single frequency, and
corresponds to one of the frequency “bins” that the DFT will
measure. So we can expect a non-zero DFT output in this
frequency bin, but zero elsewhere. Let's see how this works out.
Starting with X
k
¼
P
N
1
x
i
e
j2
p
ki
=
N
and setting N
¼
8 and
i
¼
0
x
i
¼
{1,j,
1,
j,1,j,
1,
j}:
X
0
¼
X
7
as k
¼
0
ð
e
0
i
¼
0
x
i
1
;
¼
1
Þ
X
0
¼
1
þ
j
1
j
þ
1
þ
j
1
j
¼
0, so the signal has no DC
content, as expected. Notice that to calculate X
0
e
which is the DC
content of x
I
e
the DFT just sums (essentially averaging) the input
samples.
Next, evaluate for k
¼
1:
X
1
¼
X
7
i
¼
0
x
i
e
j2
p
i
=
8
¼
1
1
þ
j
e
j2
p=
8
1
e
j4
p=
8
j
e
j6
p=
8
þ
1
e
j8
p=
8
þ
j
e
j10
p=
8
1
e
j12
p=
8
j
e
j14
p=
8
X
1
¼
1
þ½
j
ð
0
:
7071
j0
:
7071
Þ þ
j
½
j
ð
0
:
7071
j0
:
7071
Þ
1
þ½
j
ð
0
:
7071
þ
j0
:
7071
Þ
j
½
j
ð
0
:
7071
þ
j0
:
7071
Þ
X
1
¼
0
If you take the time to work this out, you will see that all eight
terms of the summation cancel out. This also happens for X
3,
X
4,
