Environmental Engineering Reference
In-Depth Information
f
(
s
)
is called the
image function
of
f
(
t
)
.
The Laplace transformation of
f
(
t
)
with
t
∈
(
0
,
+
∞
)
as defined by Eq. (B.23) is
e
−
β
t
actually the Fourier transformation of
f
(
t
)
I
(
t
)
(for
β
>
0). Thus we have, by
the Fourier integral,
+
∞
e
iω
t
d
+
∞
1
2
e
−
β
t
e
−
βτ
e
−
iω
t
d
f
(
t
)
I
(
t
)
=
f
(
τ
)
I
(
τ
)
τ
ω
π
−
∞
−
∞
e
iω
t
⎡
⎤
+
∞
+
∞
1
2
⎣
e
−
(
β
+
iω
)
τ
d
⎦
d
=
f
(
τ
)
τ
ω
π
−
∞
0
+
∞
1
2
f
e
iω
t
d
=
(
s
)
ω
,
t
>
0
,
π
−
∞
where
s
=
β
+
i
ω
. Thus
+
∞
β
+
i
∞
1
2
1
2
f
e
(
iω
+
β
)
t
d
f
e
st
d
s
f
(
t
)=
(
s
)
ω
=
(
s
)
,
(B.24)
π
π
i
−
∞
β
−
i
∞
which is called the
inverse Laplace transformation
and denoted by
L
−
1
f
)
.
f
(
t
)=
(
s
f
The
f
(
t
)
is called the
inverse image function
of
(
s
)
.
f
Therefore we may find
(
s
)
by Eq. (B.23) and
f
(
t
)
by Eq. (B.24) from the
f
known
f
. The integral in Eq. (B.23) is a regular integral. The integral
in Eq. (B.24) is however an integral of a complex-valued function and is normally
obtained by using the residue theorem which will be given in Section B.2.3.
Example 1
. Find the Laplace transformation of the unit function
I
(
t
)
or
(
s
)
(
t
)
.
Solution.
By Eq. (B.23),
+
∞
e
−
st
+
∞
0
1
s
1
s
.
e
−
st
d
t
L
[
I
(
t
)] =
=
−
=
0
Its convergence domain is
Re
(
s
)
>
0. Thus
1
s
,
L
[
I
(
t
)] =
Re
(
s
)
>
0
.
Example 2.
Find the Laplace transformation of the exponential function e
α
t
,where
α
is a real constant.
Solution.
When
Re
(
)
>
s
0, by Eq. (B.23) we have
+
∞
+
∞
L
e
α
t
=
1
e
α
t
e
−
st
d
t
e
−
(
s
−
α
)
t
d
t
=
=
−
α
.
s
0
0
Example 3.
Find the Laplace transformation of
δ
(
t
)
.
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