Digital Signal Processing Reference
In-Depth Information
100
100
50
50
0
0
−1
−0.5
0
0.5
1
−1
−0.5
0
0.5
1
(a) Norm Freq, x
1
[n]
(e) Norm Freq, x
2
[n]
2
2
0
0
−2
−2
−1
−0.5
0
0.5
1
−1
−0.5
0
0.5
1
(b) Norm Freq, x
1
[n]
(f) Norm Freq, x
2
[n]
100
100
0
0
−100
−100
−1
−0.5
0
0.5
1
−1
−0.5
0
0.5
1
(c) Norm Freq, x
1
[n]
(g) Norm Freq, x
2
[n]
100
100
0
0
−100
−100
−1
−0.5
0
0.5
1
−1
−0.5
0
0.5
1
(d) Norm Freq, x
1
[n]
(h) Norm Freq, x
2
[n]
Figure 1.5:
DTFT of first sequence, with its magnitude and phase and real and imaginary parts being
shown, respectively, in plots (a)-(d); DTFT of second sequence, which is the first sequence offset in
frequency by 2
π
radians, with its magnitude, phase, and real and imaginary parts being shown, respectively,
in plots (e)-(h). All frequencies are normalized, i.e., in units of
π
radians.
which results in Fig. 1.6. The reader should be able to verify by visual comparison of plots (a) and (e) that
the magnitude of frequency response has in fact been shifted by
π/
8 radian.
1.6.5 CONVOLUTION
The DTFT of the time domain convolution of two sequences is equal to the product of the DTFTs of
the two sequences.
X
1
(e
jω
)X
2
(e
jω
)
DT FT (x
1
[
]∗
x
1
[
]
=
n
n
)
Example 1.6.
. Obtain the time domain convo-
lution by taking the inverse DTFT of the product of the DTFTs of each sequence, and confirm the result
using time domain convolution.
Consider the two sequences
[
1
,
0
,
1
]
and
[
1
,
0
,
0
,
−
1
]
