Digital Signal Processing Reference
In-Depth Information
X
0
¼
x
0
e
j2
p
0
=
4
þ
x
1
e
j2
p
0
=
4
þ
x
2
e
j2
p
0
=
4
þ
x
3
e
j2
p
0
=
4
X
1
¼
x
0
e
j2
p
0
=
4
þ
x
1
e
j2
p
1
=
4
þ
x
2
e
j2
p
2
=
4
þ
x
3
e
j2
p
3
=
4
X
2
¼
x
0
e
j2
p
0
=
4
þ
x
1
e
j2
p
2
=
4
þ
x
2
e
j2
p
4
=
4
þ
x
3
e
j2
p
6
=
4
þ
x
3
e
j
þp
9
=
4
The term e
j2
p
k
=
4
repeats itself with a period of k
¼
4, as the
complex exponential makes a complete circle and begins
another. This periodicity means that e
j2
p
k
=
4
X
3
¼
x
0
e
j2
p
0
=
4
þ
x
1
e
j2
p
3
=
4
þ
x
2
e
j2
p
6
=
4
is equal when
evaluated for k
¼
0, 4, 8, 12
It is again equal for k
¼
1, 5, 9, 13
.
Because of this, we can simplify the last two terms of expressions
for X
2
and X
3
(shown in bold below). We can also remove the
exponential when it is to the power of zero.
.
X
0
¼
x
0
þ
x
1
þ
x
2
þ
x
3
X
1
¼
x
0
þ
x
1
e
j2
p
1
=
4
þ
x
2
e
j2
p
2
=
4
þ
x
3
e
j2
p
3
=
4
X
2
¼
x
0
þ
x
1
e
j2
p
2
=
4
þ
x
2
e
j2
p
0
=
4
þ
x
3
e
j2
p
2
=
4
X
3
¼
x
0
þ
x
1
e
j2
p
3
ftsg=
4
þ
x
2
e
j2
p
2
=
4
þ
x
3
e
j2
p
1
=
4
Now we are going to rearrange the terms of the four point
(N
¼
4) DFT, grouping the even and odd terms together.
X
0
¼½
x
0
þ
x
2
þ½
x
1
þ
x
3
X
1
¼½
x
0
þ
x
2
e
j2
p
2
=
4
þ½
x
1
e
j2
p
1
=
4
þ
x
3
e
j2
p
3
=
4
X
2
¼½
x
0
þ
x
2
þ½
x
1
e
j2
p
2
=
4
þ
x
3
e
j2
p
2
=
4
X
3
¼½
x
0
þ
x
2
e
j2
p
2
=
4
þ½
x
1
e
j2
p
3
=
4
þ
x
3
e
j2
p
1
=
4
Next we will factor x
1
and x
3
to get this particular form:
X
0
¼½
x
0
þ
x
2
þ½
x
1
þ
x
3
X
1
¼½
x
0
þ
x
2
e
j2
p
2
=
4
þ½
x
1
e
j2
p
1
=
4
þ
x
3
e
j2
p
3
=
4
¼½
x
0
þ
x
2
e
j2
p
2
=
4
þ½
x
1
þ
x
3
e
j2
p
2
=
4
e
j2
p=
4
X
2
¼½
x
0
þ
x
2
þ½
x
1
e
j2
p
2
=
4
þ
x
3
e
j2
p
2
=
4
¼½
x
0
þ
x
2
þ½
x
1
þ
x
3
e
j2
p
2
=
4
