EE3210 Tutorial 5 (Solution)
1.
Since the input is
δ
[
n
] and the system is initially at rest, we can conclude that
h
[
n
]
= 0 for
n
< 0. Now,
]
1
[
2
]
[
]
[
−
−
=
n
y
n
x
n
y
.
Therefore,
8
]
2
[
2
]
3
[
]
3
[
4
]
1
[
2
]
2
[
]
2
[
2
]
0
[
2
]
1
[
]
1
[
1
]
1
[
2
]
0
[
]
0
[
−
=
−
=
=
−
=
−
=
−
=
=
−
−
=
h
x
h
h
x
h
h
x
h
h
x
h
and so on. Combining the two cases, we can write the impulse response in closed
form as follows:
]
[
)
2
(
]
[
n
u
n
h
n
−
=
.
2.
Since the system is initially at rest, we must have
h
[
n
] = 0 for
n
< 0.
Next, we consider the following polynomial equation:
0
16
.
0
6
.
0
2
=
−
−
r
r
.
Factorize it,
0
)
8
.
0
)(
2
.
0
(
=
−
+
r
r
.
The two roots are -0.2 and 0.8. Hence, for
n
≥
0, we have
(
)
n
n
C
C
n
h
)
8
.
0
(
2
.
0
]
[
+
−
=
2
1
.
We determine two initial conditions recursively as follows:
3
]
1
[
16
.
0
]
0
[
6
.
0
]
1
[
5
]
1
[
5
]
2
[
16
.
0
]
1
[
6
.
0
]
0
[
5
]
0
[
=
−
+
+
=
=
−
+
−
+
=
h
h
h
h
h
h
δ
δ
Now we solve for
C
1
and
C
2
:
2
1
2
1
8
.
0
2
.
0
3
5
C
C
C
C
+
−
=
+
=
Therefore,
C
1
= 1 and
C
2
= 4.
Hence,
(
)
]
[
]
)
8
.
0
(
4
2
.
0
[
]
[
n
u
n
h
n
n
+
−
=
.

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