Signals and Systems
Fall 2012
Solution to Homework Assignment #15
1.
(a)
(
)
(
)
(
)
2
2
2
2
2
3
3
2
7
4
( )
3
6
25
4
3
4
3
4
s
s
X s
s
s
s
s
+
+
=
= ×
−
×
+
+
+
+
+
+
We know that:
(
)
2
2
sin
( )
L
at
s
b
e
bt u
t
s
a
b
−
←→
⋅
+
+
(
)
2
2
cos
( )
L
at
s
s
a
e
bt u
t
s
a
b
−
+
←→
⋅
+
+
Therefore, we obtain the following signal in time domain:
( )
4
4
7
3
cos3
( )
sin3
( )
4
t
t
s
s
x t
e
t u
t
e
t u
t
−
−
= ×
⋅
−
×
⋅
(b)
4
4
1
1
1
( )
2
2
2
s
s
e
X s
e
s
s
s
−
−
−
=
=
−
×
+
+
+
We know that
( )
1
1
L
at
s
e
u
t
a
−
←→
+
and
(
)
( )
(
)
,
0
L
as
s
x t
a u
t
a
e
X
s
a
−
−
−
←→
>
Therefore, we obtain the following signal in time domain:
( )
( )
(
)
(
)
2
4
2
4
t
t
s
s
x t
e
u
t
e
u
t
−
−
−
=
−
−
2.
)
4
(
2
)
(
)
(
)
(
2
.
0
)
(
−
+
=
+
+
t
x
t
x
t
y
t
y
t
y
Taking the Laplace transform of the above equation:
( )
(
)
( )
(
)
3
2
4
0.2
2
s
Y
s
s
s
s
X
s
s
e
−
+
+
=
+
We can write the input as below:
≥
<
≤
<
=
−
−
2
2
0
0
0
)
(
2
t
e
t
e
t
t
x
t
.

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