3.8 – 1
Solution to problem 3.8
a) The Laplace transform of
∫

=
⋅
+
⋅
+
t
d
e
x
x
x
0
2
2
3
τ
can be obtained as follows:
For the right hand side of the above expression, we know that
( )
( )
s
F
s
dt
t
f
L
t
1
0
*
*
=
∫
And thus we can write that:
( )
( )
1
1
1
0
+
=
=


∫
s
s
e
L
s
d
e
L
t
For the left hand side of the above expression, we apply the superposition principle (and we
assume that we have a system in steadystate, i.e. x(0)=x’(0)=0):
( )
( )
( )
)
(
2
)
(
3
2
3
2
3
2
2
2
s
X
s
X
s
s
X
s
x
L
dt
dx
L
dt
x
d
L
x
x
x
L
⋅
+
⋅
+
⋅
=
+
+
=
⋅
+
⋅
+
Combining the Laplace transforms for left and right hand side we now obtain:
( )
( )
1
2
)
(
2
)
(
3
2
+
=
⋅
+
⋅
+
⋅
s
s
s
X
s
X
s
s
X
s
( ) ( )
2
1
2
)
(
2
+
+
=
s
s
s
s
X
This expression can now be expanded in partial fractions:
( ) ( )
( )
2
1
1
2
1
2
)
(
4
2
3
2
1
2
+
+
+
+
+
+
=
+
+
=
s
s
s
s
s
s
s
s
X
α
Applying Heaviside expansion we obtain values for
α
1
,
α
3
and
α
4
.
Multiplication by s and putting s equal to 0 gives:
( ) ( )
1
2
1
2
0
2
1
=
+
+
=
=
s
s
s
Multiplication by (s+1)
2
and putting s equal to 1 gives:
( )
2
1
2
2
2
1
3

=

=
+
=

=
s
s
s
Multiplication by (s+2) and putting s equal to 2 gives:
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 Spring '09
 RafiqulGani
 Laplace, right hand

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