— (16/35)
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Inverse Laplace Transforms
Special Functions
Heaviside or Step function
Periodic functions
Impulse or
δ
Function
Laplace Transform of Step Function
Theorem
If
F
(
s
) =
L
[
f
(
t
)]
exists for
s > a
≥
0
, and if
c
is a nonnegative
constant, then
L
[
u
c
(
t
)
f
(
t

c
)] =
e

cs
L
[
f
(
t
)] =
e

cs
F
(
s
)
,
s > a.
Conversely, if
f
(
t
) =
L

1
[
F
(
s
)]
, then
u
c
(
t
)
f
(
t

c
) =
L

1
[
e

cs
F
(
s
)]
.
This theorem states that the translation of
f
(
t
) a distance
c
in the
positive
t
direction corresponds to the multiplication of
F
(
s
) by
e

cs
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Laplace Transforms: Part B
— (17/35)
Inverse Laplace Transforms
Special Functions
Heaviside or Step function
Periodic functions
Impulse or
δ
Function
Example with Step Function
1
Example with Step Function:
Consider the following
initial
value problem
:
y
00
+ 2
y
0
+ 5
y
=
u
2
(
t
)

u
5
(
t
)
,
y
(0) = 0
,
y
0
(0) = 0
.
Take
Laplace transforms
and obtain
(
s
2
+ 2
s
+ 5)
Y
(
s
) =
e

2
s
s

e

5
s
s
This rearranges to
Y
(
s
) =
e

2
s

e

5
s
s
(
s
2
+ 2
s
+ 5)
Partial fraction decomposition gives
1
s
(
s
2
+ 2
s
+ 5)
=
A
s
+
B
(
s
+ 1) + 2
C
(
s
+ 1)
2
+ 4
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Laplace Transforms: Part B
— (18/35)
Inverse Laplace Transforms
Special Functions
Heaviside or Step function
Periodic functions
Impulse or
δ
Function
Example with Step Function
2
Example with Step Function:
Partial fraction decomposition gives
1 =
A
(
(
s
+ 1)
2
+ 4
)
+ (
B
(
s
+ 1) + 2
C
)
s
With
s
= 0, 1 = 5
A
or
A
=
1
5
The
s
2
coefficient gives 0 =
A
+
B
, so
B
=

1
5
The
s
1
coefficient gives 0 = 2
A
+
B
+ 2
C
, so
C
=

1
10
Thus,
Y
(
s
) =
1
5
s

1
5
(
s
+ 1) +
2
10
(
s
+ 1)
2
+ 4
(
e

2
s

e

5
s
)
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Laplace Transforms: Part B
— (19/35)
Inverse Laplace Transforms
Special Functions
Heaviside or Step function
Periodic functions
Impulse or
δ
Function
Example with Step Function
3
Example with Step Function:
With the
Laplace transform
Y
(
s
) =
1
5
s

1
5
(
s
+ 1) +
2
10
(
s
+ 1)
2
+ 4
!
(
e

2
s

e

5
s
)
The theorem for step functions allows the
inverse Laplace transform
yielding
y
(
t
)
=
u
2
(
t
)
10
2

2
e

(
t

2)
cos(2(
t

2))

e

(
t

2)
sin(2(
t

2))

u
5
(
t
)
10
2

2
e

(
t

5)
cos(2(
t

5))

e

(
t

5)
sin(2(
t

5))
0
1
2
3
4
5
6
7
8
9
10
0.2
0.15
0.1
0.05
0
0.05
0.1
0.15
0.2
0.25
0.3
t
y
(
t
)
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Laplace Transforms: Part B
— (20/35)
Inverse Laplace Transforms
Special Functions
Heaviside or Step function
Periodic functions
Impulse or
δ
Function
Periodic Functions
Definition
A function
f
is said to be
periodic with period
T >
0
if
f
(
t
+
T
) =
f
(
t
)
for all
t
in the domain of
f
.
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
0
0.2
0.4
0.6
0.8
1
t
y
(
t
)
A
sawtooth
waveform
f
(
t
) =
t,
0
≤
t <
1
,
0
,
1
≤
t <
2
.
and
f
(
t
) has period 2
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Laplace Transforms: Part B
— (21/35)
Inverse Laplace Transforms
Special Functions
Heaviside or Step function
Periodic functions
Impulse or
δ
Function
Periodic and Window functions
Consider a
periodic function
f
(
t
). Define the
window function
,
f
T
(
t
), as follows:
f
T
(
t
) =
f
(
t
) [1

u
T
(
t
)] =
f
(
t
)
,
0
≤
t
≤
T
0
,
otherwise
The
Laplace transform
F
T
(
s
) satisfies:
F
T
(
s
) =
Z
∞
0
e

st
f
T
(
t
)
dt
=
Z
T
0
e

st
f
T
(
t
)
dt.
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