1635 inverse laplace transforms special functions

— (16/35)

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.