Digital Signal Processing Reference
In-Depth Information
and ancillary condition
y
[
−
1]
=
2.
(a) Determine the response
y
[
k
] by iterating the difference equation for
0
≤
k
≤
5.
(b) Determine the zero-state response
y
zi
[
k
] for 0
≤
k
≤
5.
(c) Calculate the zero-input response
y
zs
[
k
] for 0
≤
k
≤
5.
(d) Verify that
y
[
k
]
=
y
zi
[
k
]+
y
zs
[
k
].
10.2
Repeat Problem 10.1 for the applied input
x
[
k
]
=
0.5
k
u
[
k
] and the input-
output relationship
y
[
k
+
2]
−
y
[
k
+
1]
+
0
.
5
y
[
k
]
=
x
[
k
]
,
with ancillary conditions
y
[
−
1] = 0 and
y
[
−
2]
=
1.
10.3
Repeat Problem 10.1 for the applied input
x
[
k
]
=
(
−
1)
k
u
[
k
] and the
input-output relationship
y
[
k
+
2]
−
0
.
75
y
[
k
+
1]
+
0
.
125
y
[
k
]
=
x
[
k
]
,
with ancillary conditions
y
[
−
1]
=
1 and
y
[
−
2]
=−
1
.
10.4
Show that the convolution of two sequences
a
k
u
[
k
] and
b
k
u
[
k
]isgiven
by
(
k
+
1)
a
k
u
[
k
]
a
=
b
(
a
k
u
[
k
])
∗
(
b
k
u
[
k
])
=
1
a
−
b
(
a
k
+
1
−
b
k
+
1
)
u
[
k
]
a
=
b
.
10.5
Calculate the convolution (
x
1
[
k
]
∗
x
2
[
k
]) for the following pairs of
sequences:
(a)
x
1
[
k
]
=
u
[
k
+
2]
−
u
[
k
−
3]
,
x
2
[
k
]
=
u
[
k
+
4]
−
u
[
k
−
5];
(b)
x
1
[
k
]
=
0
.
5
k
u
[
k
]
,
x
2
[
k
]
=
0
.
8
k
u
[
k
−
5];
(c)
x
1
[
k
]
=
7
k
u
[
−
k
+
2]
,
x
2
[
k
]
=
0
.
4
k
u
[
k
−
4];
(d)
x
1
[
k
]
=
0
.
6
k
u
[
k
]
,
x
2
[
k
]
=
sin(
π
k
/
2)
u
[
−
k
];
(e)
x
1
[
k
]
=
0
.
5
k
,
x
2
[
k
]
=
0
.
8
k
.
10.6
For the following pairs of sequences:
k
0
≤
k
≤
3
2
−
1
≤
k
≤
2
(a)
x
[
k
]
=
and
h
[
k
]
=
0
otherwise
0
otherwise;
2
−
k
0
≤
k
≤
3
0 otherwise,
calculate the DT convolution
y
[
k
]
=
x
[
k
]
∗
h
[
k
] using (i) the graphical
approach and (ii) the sliding tape method.
k
k
≤
2
0
(b)
x
[
k
]
=
and
h
[
k
]
=
otherwise
10.7
Using the sliding tape method and the following equation:
∞
y
[
k
]
=
h
[
m
]
x
[
k
−
m
]
,
m
=−∞
calculate the convolution of the sequences in Example 10.8 and show that
the convolution output is identical to that obtained in Example 10.8.
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