Digital Signal Processing Reference
In-Depth Information
X
5,
X
6,
and X
7
. Let's look at X
2
now. We will also express x
i
using
complex exponential format of e
þ
j2
p
i
=
4
.
X
k
¼
X
7
i
¼
0
x
i
e
j2
p
ki
=
8
X
2
¼
X
7
i
¼
0
¼
X
7
x
i
e
j4
p
i
=
8
i
¼
0
e
þ
j2
p
i
=
4
e
j4
p
i
=
8
¼
X
7
i
¼
0
e
þ
j2
p
i
=
4
e
j2
p
i
=
4
Remember that when exponentials are multiplied the expo-
nents are added (x
2
¼
x
5
). Here the exponents are identical,
except of opposite sign, so they add to zero.
X
2
¼
X
7
x
3
¼
X
7
¼
X
7
i
¼
0
e
þ
j2
p
i
=
4
e
j2
p
i
=
4
i
¼
0
e
0
i
¼
0
1
¼
8
The sole frequency component of the input signal is X
2
because our input is a complex exponential frequency at the exact
frequency that X
2
represents.
12.1.3 Third DFT Example
Next, we can try modifying x
i
such that we introduce a phase
shift, or delay (like substituting a sine wave for a cosine wave). If
we introduce a delay, so x
i
starts at j instead of 1, but is still the
same frequency, the input x
i
is still rotating around the complex
plane at the same rate, but starts at j (angle of
/ 2) rather than 1
(angle of 0). Now the sequence x
i
¼
{j,
1,
j,1,j,
1,
j,1} or
e
þ
j
ð
2
pð
i
þ
1
Þ=
4
Þ
.
The DFT output will result in X
0,
X
1,
X
3,
X
4,
X
5,
X
6,
and X
7
¼
0, as
before. Changing the phase cannot cause any new frequency to
appear in the other bins.
Next, evaluate for k
¼
2:
X
k
¼
X
7
i
¼
0
p
x
i
e
j2
p
ki
=
8
X
2
¼
X
7
¼
X
7
i
¼
0
x
i
e
j4
p
i
=
8
i
¼
0
e
þ
j
ð
2
pð
i
þ
1
Þ=
4
Þþ
1
Þ
e
j4
p
i
=
8
We need to sum the two values of the two exponents:
þ
j
ð
2
pð
i
þ
1
Þ=
4
Þþ
j4
p
i
=
8
¼þ
j2
p
i
=
4
þ
j2
p=
4
j2
p
i
=
4
¼
j
p=
2
Substituting back this exponent value:
X
2
¼
X
7
i
¼
0
¼
X
7
¼
X
7
e
þ
j
ðð
2
p
i
=
4
Þþ
1
Þ
e
j4
p
i
=
8
i
¼
0
e
þ
j
p=
2
i
¼
0
j
¼
j8
