Digital Signal Processing Reference
In-Depth Information
1
1
0
0
−1
−1
0
1
2
0
2
4
6
8
10
(a) Sig[0]
δ
[n]
(b) Sig[0](Imp)
1
1
0
0
−1
−1
0
1
2
0
2
4
6
8
10
(c) Sig[1]
δ
[n−1]
(d) Sig[1](Imp)
1
1
0
0
−1
−1
0
1
2
0
2
4
6
8
10
(e) Sig[2]
δ
[n−2]
(f) Sig[2](Imp)
1
1
0
0
−1
−1
0
1
2
0
2
4
6
8
10
(g) Superpos. (a),(c),& (e)
(h) Superposition of (b),(d),& (f)
Figure 2.20:
(a) First sample of signal, multiplied by
δ(n)
; (b) Impulse response, weighted by first signal
sample; (c) Second signal sample; (d) Impulse response, scaled by second signal sample, and delayed
by one sample; (e) Third signal sample; (f ) Impulse response scaled by third signal sample, delayed by
two samples; (g) Input signal, the superposition of its components shown in (a), (c), and (e); (h) The
convolution, i.e., the superposition of responses shown in (b), (d), and (f ).
= 0
.
7
n
Example 2.4.
Let
x
[
n
]
=
[
1
,
−
0
.
5
,
0
.
75
]
and
h
[
n
]
Compute the first 3 output values of
y
as
shown in subplot (h) of Fig. 2.20, using Eqn. (2.6).
= 0 for values of
n<
0. We set
the range of
k
as 0:1:2 to compute the first three output samples, and as a result,
n
= 0:1:2, which may be
explained as follows: since the maximum value of
k
we will compute is 2, we need not exceed
n
= 2 since
if
n
exceeds 2,
k
-
n
is less than zero and as a result,
h
We note that
x
[
n
]
= 0 for
n<
0 and
n>
2. We also note that
h
[
n
]
[
n
]
= 0. To compute
y
[
k
]
for higher values of
k
,a
correspondingly larger range for
n
is needed.
We get
2
y
[
0
]=
x
[
n
]
h
[
0
−
n
]
n
=
0
The sum above is
y
[
0
]=
x
[
0
]
h
[
0
]+
x
[
1
]
h
[−
1
]+
x
[
2
]
h
[−
2
]=
x
[
0
]
h
[
0
]=
1
