Digital Signal Processing Reference
In-Depth Information
Table 10.2. Convolution of
x
[
k
] and
h
[
k
] using the sliding tape method for Example 10.9
m
...
−
5
−
4
−
3
−
2
−
1
...
y
[
k
]
h
[
m
]
−
2
−
2
−
1
x
[
m
]
x
[
−
m
]
−
1
x
[
−
3
−
m
]
−
1
−
3
x
[
−
2
−
m
]
−
1
−
2
−
3
x
[
−
1
−
m
]
−
1
−
1
x
[0
−
m
]
−
1
x
[1
−
m
]
−
1
−
3
x
[2
−
m
]
−
1
x
[3
−
m
]
−
1
x
[4
−
m
]
−
1
−
4
x
[5
−
m
]
−
1
and is stored in the last column of Table 10.1. We repeat the process for increas-
ing values of
k
until the overlap between
x
[
m
] and
h
[
k
−
m
] is eliminated.
In Table 10.1, this occurs for
k
>
7, beyond which the output response
y
[
k
]is
zero.
By comparison with the result obtained in Example 10.7, we note that the
output response
y
[
k
] obtained using the sliding tape method is identical to the
one obtained using the graphical approach.
Example 10.9
For the following pair of the input sequence
x
[
k
] and impulse response
h
[
k
]:
−
1
k
=−
1
3
k
=−
1
,
2
1
k
=
0
−
2
k
=
1
,
3
0 otherwise,
calculate the output response using the sliding tape method.
1
k
=
0
x
[
k
]
=
and
h
[
k
]
=
2
k
=
1
0
otherwise
Solution
The output
y
[
k
] can be calculated by convolving the input sequence
x
[
k
] with
the impulse response
h
[
k
]. Since convolution satisfies the distributive property,
i.e.
y
[
k
]
=
x
[
k
]
∗
h
[
k
]
=
h
[
k
]
∗
x
[
k
]
,
9
y
[
k
]
4
≈
2
Table 10.2 reverses the role of the input sequence
x
[
k
] with that of the impulse
response
h
[
k
] and computes the following summation:
1
−2
k
−5
−4 −3
−1
023
5
∞
−3
−3
y
[
k
]
=
h
[
m
]
x
[
k
−
m
]
,
−4
m
=−∞
Fig. 10.8. Output response
calculated using the sliding tape
method in Example 10.9.
implying that the input sequence is time-reversed and time-shifted, while the
impulse response is kept fixed. The results of Table 10.2 are plotted in Fig. 10.8.
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