The
window function
specifies values of
f
(
t
) over a single period
This can be replicated
k
periods to the right as
f
T
(
t

kT
)
u
kT
(
t
) =
f
(
t

kT
)
,
kT
≤
t
≤
(
k
+ 1)
T
0
,
otherwise
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Laplace Transforms: Part B
— (22/35)
Inverse Laplace Transforms
Special Functions
Heaviside or Step function
Periodic functions
Impulse or
δ
Function
Laplace for Periodic Functions
By summing
n
time shifted replications of the
window function
,
f
T
(
t

kT
)
u
kT
(
t
),
k
= 0,...,
n

1, gives
f
nT
(
t
), the periodic extension
of
f
T
(
t
) to the interval [0
, nT
],
f
nT
(
t
) =
n

1
X
k
=0
f
T
(
t

kT
)
u
kT
(
t
)
Theorem
If
f
is periodic with period
T
and is piecewise continuous on
[0
, T
]
,
then
L
[
f
(
t
)] =
F
T
(
s
)
1

e

sT
=
R
T
0
e

st
f
(
t
)
dt
1

e

sT
.
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Laplace Transforms: Part B
— (23/35)
Inverse Laplace Transforms
Special Functions
Heaviside or Step function
Periodic functions
Impulse or
δ
Function
Laplace for Periodic Functions
Proof:
From our earlier theorem, we have for each
k
≥
0,
L
[
f
T
(
t

kT
)
u
kT
(
t
)] =
e

kT s
L
[
f
T
(
t
)] =
e

kT s
F
T
(
s
)
.
By linearity of
L
, the
Laplace transform
of
f
nT
is
F
nT
(
s
)
=
Z
nT
0
e

st
f
(
t
)
dt
=
n

1
X
k
=0
L
[
f
T
(
t

kT
)
u
kT
(
t
)]
=
n

1
X
k
=0
e

kT s
F
T
(
s
) =
F
T
(
s
)
n

1
X
k
=0
(
e

T s
)
k
=
F
T
(
s
)
1

(
e

T s
)
n
1

e

sT
.
The last term comes from summing a geometric series. With
e

sT
<
1,
F
(
s
) =
lim
n

>
∞
Z
nT
0
e

st
dt
=
lim
n

>
∞
F
T
(
s
)
1

(
e

T s
)
n
1

e

sT
=
F
T
(
s
)
1

e

sT
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Laplace Transforms: Part B
— (24/35)
Subscribe to view the full document.
Inverse Laplace Transforms
Special Functions
Heaviside or Step function
Periodic functions
Impulse or
δ
Function
Sawtooth Function
Return to
sawtooth
waveform
f
(
t
) =
t,
0
≤
t <
1
,
0
,
1
≤
t <
2
.
and
f
(
t
) has period 2
The
theorem
for the
Laplace transform
of
periodic function
gives
L
[
f
(
t
)] =
R
2
0
e

st
f
(
t
)
dt
1

e

2
s
But
Z
2
0
e

st
f
(
t
)
dt
=
Z
1
0
te

st
dt
=
1

se

s

e

s
s
2
,
so
L
[
f
(
t
)] =
1

se

s

e

s
s
2
(1

e

2
s
)
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Laplace Transforms: Part B
— (25/35)
Inverse Laplace Transforms
Special Functions
Heaviside or Step function
Periodic functions
Impulse or
δ
Function
IVP with Periodic Forcing Function
1
Example:
Consider the following initial value problem:
y
00
+ 4
y
=
f
(
t
)
,
y
(0) = 0
,
y
0
(0) = 0
,
with the
square
waveform as the
periodic forcing function
:
f
(
t
) =
1
,
0
≤
t <
1
,
0
,
1
≤
t <
2
.
and
f
(
t
) has period 2
The
theorem
for the
Laplace transform
of
square
waveform gives
L
[
f
(
t
)] =
R
2
0
e

st
f
(
t
)
dt
1

e

2
s
But
Z
2
0
e

st
f
(
t
)
dt
=
Z
1
0
e

st
dt
=
1

e

s
s
,
so
L
[
f
(
t
)] =
1

e

s
s
(1

e

2
s
)
=
1
s
(1 +
e

s
)
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Laplace Transforms: Part B
— (26/35)
Inverse Laplace Transforms
Special Functions
Heaviside or Step function
Periodic functions
Impulse or
δ
Function
IVP with Periodic Forcing Function
2
Example:
Taking the
Laplace transform
of the
IVP
with
L
[
y
(
t
)] =
Y
(
s
), we have:
s
2
Y
(
s
)

sy
(0)

y
0
(0) + 4
Y
(
s
) =
1
s
(1 +
e

s
)
Thus,
Y
(
s
) =
1
s
(
s
2
+ 4) (1 +
e

s
)
Partial fractions decomposition gives
1
s
(
s
2
+ 4)
=
1
/
4
s

s/
4
s
2
+ 4
,
while
1
1 +
e

s
=
1
1

(

e

s
)
= 1

e

s
+
e

2
s

...
+ (

1)
n
e

ns
+
Joseph M. Mahaffy,
h
[email protected]
i
 Fall '08
 staff