Digital Signal Processing Reference
In-Depth Information
discussed in Chapter 3. In the following, we highlight the main steps in calcu-
lating the convolution sum between two sequences
x
[
k
] and
h
[
k
].
Algorithm 10.1 Graphical procedure for computing the linear convolution
(1) Sketch the waveform for input
x
[
m
] by changing the independent variable
of
x
[
k
] from
k
to
m
and keep the waveform for
x
[
m
] fixed during steps
(2)-(7).
(2) Sketch the waveform for the impulse response
h
[
m
] by changing the inde-
pendent variable from
k
to
m
.
(3) Reflect
h
[
m
] about the vertical axis to obtain the time-inverted impulse
response
h
[
−
m
].
(4) Shift the sequence
h
[
−
m
] by a selected value of
k
. The resulting function
represents
h
[
k
−
m
].
(5) Multiply the input sequence
x
[
m
]by
h
[
k
−
m
] and plot the product function
x
[
m
]
h
[
k
−
m
].
(6) Calculate the summation
∞
m
=−∞
x
[
m
]
h
[
k
−
m
].
(7) Repeat steps (4)-(6) for
−∞ ≤
k
≤∞
to obtain the output response
y
[
k
]
over all time
k
.
The graphical approach for calculating the output response is illustrated through
a series of examples.
Example 10.6
Repeat Example 10.5 with input
x
[
k
]
=
0.8
k
u
[
k
] and impulse response
h
[
k
]
=
0
.
5
k
u
[
k
] to determine the output of the LTID system using the graphical con-
volution approach.
Solution
Following steps (1)-(3) of Algorithm 10.1, the DT sequences
x
[
m
]
=
0.8
m
u
[
m
],
h
[
m
]
=
0.5
m
u
[
m
] and its time reflection
h
[
−
m
]
=
0.5
−
m
u
[
−
m
] are plotted in
Fig. 10.6. Based on step (4), the sequence
h
[
k
−
m
]
=
h
[
−
(
m
−
k
)] is obtained
by shifting
h
[
−
m
]by
k
samples. To compute the output sequence, we consider
two cases based on the values of
k
.
Case 1
For
k
<
0, the waveform
h
[
k
−
m
] is on the left-hand side of the vertical
axis. As is apparent in Fig. 10.6, step (5a), waveforms for
h
[
k
−
m
] and
x
[
m
]do
not overlap. In other words, the product
x
[
m
]
h
[
k
−
m
]
=
0, for
−∞ ≤
m
≤∞
,
as long as
k
<
0. The output sequence
y
[
k
] is therefore zero for
k
<
0.
Case 2
For
k
≥
0, we see from Fig. 10.6, step (5b), that the non-zero parts of
h
[
k
−
m
] and
x
[
m
] overlap over the range
m
=
[0,
k
]. Therefore,
k
k
0
.
8
m
0
.
5
k
−
m
.
y
[
k
]
=
x
[
m
]
h
[
k
−
m
]
=
m
=
0
m
=
0
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