Signals and Systems
Fall 2012
Solution to Homework Assignment #1:
1.
)
(
)
(
s
X
s
Y
=
1
4
2
3
5
.
2
2
3
+
−
+
s
s
s
s
3
2
3
( )
2
( )
4
( )
( )
2.5
( )
s Y s
s Y s
sY s
Y s
sX s
⇒
+
−
+
=
By taking the inverse Laplace transform of the above equation, we obtain:
3
( )
2 ( )
4 ( )
( )
2.5 ( )
y t
y t
y t
y t
x t
+
−
+
=
2.
The input and output of the considered system are related by:
∫
−
=
+
t
d
x
t
x
t
y
t
y
0
)
(
)
(
)
(
3
)
(
τ
τ
By taking the Laplace transform, we have
0
( )
( )
(0)
3 ( )
( )
X s
sY s
y
Y s
X s
s
=
−
+
=
−
(
)
(
)
1
3
( )
( )
s
s
Y s
X s
s
−
⇒
+
=
It then follows that:
(
)
2
1
( )
( )
( )
3
s
Y s
H s
X s
s
s
−
=
=
+
3.
( )
2 ( )
( )
2 ( )
y t
y t
y t
u t
=
−
+
( )
2 ( )
( )
2 ( )
y t
y t
y t
u t
⇒
−
+
=
(*)
We first consider the following problem:
( )
2 ( )
( )
0
y t
y t
y t
−
+
=
(**)
Observe that the corresponding characteristic equation is
2
2
1
0
k
k
−
+
=
, giving the double root
1
k
=
. Hence, the solution of (**) has the following form:
1
2
t
t
H
y
C e
C te
=
+

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