Digital Signal Processing Reference
In-Depth Information
Example (1) If x(n) = (0.8)
n
u(n) and y(n) = (0.4)
n
u(n), then their convolution
is:
w
ð
n
Þ¼
x
ð
n
Þ
y
ð
n
Þ¼
sum
1
k
¼1
x
ð
k
Þ
y
ð
n
k
Þ
ð
2
:
5
Þ
Since the variable of summation is k, x and y should be re-expressed as functions
of k:
(
y
ð
k
Þ¼
ð
0
:
4
Þ
k
;
k
0
0
;
k\0;
(
x
ð
k
Þ¼
ð
0
:
8
Þ
k
;
k
0
0
;
k\0
(
) y
ð
n
k
Þ¼
ð
0
:
4
Þ
n
k
;
n
k
0
)
) k
n
0
;
n
k\0
)
) k [ n
Examination of Fig. (
2.7
) reveals that the product x(k)y(n - k), is non-zero
only when n C 0. Hence it follows that:
x(k)=(0.8)
k
u(k)
1
o
o
o
o
o
o
o
k
−2
−1
0
1
2
3
y(k)=(0.4)
k
u(k)
1
o
o
o
o
o
o
o
k
−2
−1
1
2
3
0
y(−k)
1
o
o
o
o
o
o
o
k
−2
−1
0
1
2
3
y(2−k)
1
o
o
o
o
o
o
o
k
−2
−1
0
1
2
3
Overlap
Convolution of x(n) = (0.8)
n
u(n) with y(n) = (0.84)
n
Fig. 2.7
u(n)
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