Signals and Systems
Fall 2013
Solution to Homework Assignment #1
1.
Find the differential equation relating
y
(
t
) and
x
(
t
):
3
2
2
2
( )
( )
4
( )
3 ( )
( )
2
( )
s Y s
s Y s
sY s
Y s
s X s
s X s
⇒
−
+
+
=
+
Taking inverse Laplace transform of the above equation, leads to:
( )
( )
4 ( )
3 ( )
( )
2 ( )
y t
y t
y t
y t
x t
x t
−
+
+
=
+
.
2.
With the given equation:
∫
−
=
+
t
d
x
t
x
t
y
t
y
0
)
(
)
(
)
(
3
)
(
τ
τ
Taking Laplace transform:
0
( )
( )
(0)
3 ( )
( )
X s
sY s
y
Y s
X s
s
=
−
+
=
−
(
)
(
)
1
3
( )
( )
s
s
Y s
X s
s
−
⇒
+
=
(
)
2
1
( )
( )
( )
3
s
Y s
H s
X s
s
s
−
⇒
=
=
+
.
3.
Solve the differential equation
)
(
)
(
9
)
(
t
f
t
y
t
y
+
−
=
with initial conditions
y
(0)
=
0,
)
0
(
y
=
6, and with
f
(
t
)
=
0.
)
(
)
(
9
)
(
t
f
t
y
t
y
+
−
=
( )
9 ( )
0
y t
y t
⇒
+
=
(*)
Solving
( )
9 ( )
0
y t
y t
+
=
(**)
Characteristic equation:
2
9
0
k
+
=
, gives the complex roots are
3
k
j
= ±
.
Hence, the solution of (**) has the form of:
1
2
cos3
sin3
H
y
C
t
C
t
=
+

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*