Environmental Engineering Reference
In-Depth Information
If
f
1
(
t
)=
t
and
f
2
(
t
)=
sin
t
, for example, we have
t
0
τ
t
∗
sin
t
=
sin
(
t
−
τ
)
d
τ
t
t
0
=
τ
cos
(
t
−
τ
)
|
−
cos
(
t
−
τ
)
d
τ
0
=
t
−
sin
t
.
Convolution Theorem
. Suppose that
f
1
(
t
)
and
f
2
(
t
)
satisfy the condition for the
existence of the Laplace transformation. Let
L
f
1
(
f
2
(
[
f
1
(
t
)] =
s
)
,
L
[
f
2
(
t
)] =
s
)
.
or
L
−
1
f
1
(
)
=
f
1
(
f
2
(
f
2
(
Thus
L
[
f
1
(
t
)
∗
f
2
(
t
)] =
s
)
s
)
s
)
s
f
1
(
t
)
∗
f
2
(
t
)
.
Proof.
Clearly,
f
1
(
t
)
∗
f
2
(
t
)
also satisfies the condition for the existence of the
Laplace transformation. Thus
+
∞
e
−
st
d
t
L
[
f
1
(
t
)
∗
f
2
(
t
)] =
[
f
1
(
t
)
∗
f
2
(
t
)]
0
t
e
−
st
d
t
+
∞
=
f
1
(
τ
)
f
2
(
t
−
τ
)
d
τ
0
0
+
∞
+
∞
e
−
st
d
t
=
f
1
(
τ
)
d
τ
f
2
(
t
−
τ
)
.
0
τ
−
τ
=
Let
t
u
. Thus
+
∞
+
∞
e
−
st
d
t
e
−
(
u
+
τ
)
s
d
u
e
−
st
f
2
(
f
2
(
t
−
τ
)
=
f
2
(
u
)
=
s
)
.
0
0
)] =
+
∞
0
f
2
(
f
1
(
f
2
(
e
−
s
τ
Finally,
L
[
f
1
(
t
)
∗
f
2
(
t
f
1
(
τ
)
s
)
d
τ
=
s
)
s
)
.
s
2
f
(
)
(
)=
Example 5.
Find
f
t
for
s
2
.
(
s
2
+
1
)
Solution.
By direct calculation or the table of Laplace transformations (AppendixC),
we have
L
−
1
s
=
cos
t
.
s
2
+
1
Thus
L
−
1
L
−
1
s
2
s
s
f
(
t
)=
=
1
·
s
2
2
s
2
s
2
(
+
1
)
+
+
1
t
=
cos
t
∗
cos
t
=
cos
τ
cos
(
t
−
τ
)
d
τ
0
t
1
2
[
1
2
(
=
+
(
τ
−
)]
τ
=
+
)
.
cos
t
cos
2
t
d
t
cos
t
sin
t
0
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