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Problem (1)
Solution:
(a) The system described by
3
[ ]
[ ]
y n
n x n
=
is linear, causal, unstable and timevariant.
For the input
1
[ ]
x n
, the output is
3
1
1
[ ]
[ ]
y n
n x n
=
;
For the input
2
[ ]
x n
, the output is
3
2
2
[ ]
[ ]
y n
n x n
=
;
For the input
3
1
2
[ ]
[ ]
[ ]
x n
ax n
bx n
=
+
, the output is
{
}
3
3
3
3
3
3
1
2
1
2
1
2
[ ]
[ ]
[ ]
[ ]
[ ]
[ ]
[ ]
[ ]
y n
n x n
n
ax n
bx n
n ax n
n bx n
ay n
by n
=
=
+
=
+
=
+
;
So the system is linear.
Since the current output does not depend on the future inputs, the system is causal.
If
[ ]
x n
is bounded, the
[ ]
y n
is not bounded as
n
→ ∞
. So the system is unstable.
For
the
input
1
0
[ ]
[
]
x n
x n
n
=

,
the
output
is
3
3
1
1
0
[ ]
[ ]
[
]
y n
n x n
n x n
n
=
=

.
While
3
0
0
0
1
[
]
(
) [
]
[ ]
y n
n
n
n
x n
n
y n

=


≠
, the system is timevariant.
(b) The system described by
3
0
[ ]
[
]
k
y n
b
x n
k
=
=
+

∑
is nonlinear, causal, stable and timeinvariant.
For the input
1
[ ]
x n
, the output is
3
1
0
[ ]
[
]
k
y n
b
x n
k
=
=
+

∑
;
For the input
2
[ ]
x n
, the output is
3
2
0
[ ]
[
]
k
y n
b
x n
k
=
=
+

∑
;
For the input
3
1
2
[ ]
[ ]
[ ]
x n
ax n
cx n
=
+
, the output is
3
3
3
3
3
1
2
1
2
0
0
0
[ ]
[
]
[
]
[
]
[ ]
[ ]
k
k
k
y n
b
x n
k
b
ax n
k
cx n
k
ay n
cy n
=
=
=
=
+

= +

+

≠
+
∑
∑
∑
;
So the system is nonlinear. (Note: the superposition holds only when
a
+
c
=1)
Since the current output does not depend on the future inputs, the system is causal.
If
[ ]
x n
< ∞
, then the output
[ ]
y n
< ∞
. So the system is stable.
For the input
1
0
[ ]
[
]
x n
x n
n
=

, the output is
3
3
1
1
0
0
0
0
[ ]
[
]
[
]
[
]
k
k
y n
b
x n
k
b
x n
k
n
y n
n
=
=
= +

=
+
 
=

∑
∑
.
So the system is timeinvariant.
(c) The system described by
[ ]
[
2]
y n
x
n
=
 +
is linear, noncausal, stable and timeinvariant.
For the input
1
[ ]
x n
, the output is
1
1
[ ]
[
2]
y n
x
n
=
 +
;
For the input
2
[ ]
x n
, the output is
2
2
[ ]
[
2]
y n
x
n
=
 +
;
For the input
3
1
2
[ ]
[ ]
[ ]
x n
ax n
bx n
=
+
, the output is
3
3
1
2
1
2
[ ]
[
2]
[
2]
[
2]
[ ]
[ ]
y n
x
n
ax
n
bx
n
ay n
by n
=
 +
=
 +
+
 +
=
+
;
So the system is linear.
The outputs depend on the future value of input. For example,
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 Fall '10
 ShuHungLeung
 Digital Signal Processing, Signal Processing

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