Laplace Transforms – Translation Theorems - Sinclair Community ...

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Oct 8, 2013 (3 years and 10 months ago)

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Laplace Transforms – Translation Theorems
Department of Mathematics, Sinclair Community College, Dayton, OH 1
By definition, the Laplace transform of  is
L  =





= 
Some of the most common transforms are:
L


1

=
1

L





=

!




L






=
1




L


sin


=



+




L


sinh


=







L


cos


=



+


L


cosh


=








These are obtained from the definition by integrating by parts. In each of these transforms, the
expression in the bracket is  and the result on the right side is .

First Translation Theorem
If we seek the Laplace transform of






, an exponential multiple of 



, then
L






=

 
 =  −



This result is known as the First Translation Theorem, so called because 

 −

translates the graph of
 “” units to the right (if  > 0), or “||” units to the left (if  < 0). The notation for the theorem is
L








= L




|

→ 
*
= 


|

→ 
*
= 

 −


where “ → −” means “s” in 



is replaced by “ −”.

Example
: Consider L



+

,

. Here, a = 5 and  = 
,
. Then,
L 

+

,
 =
*L 
,
|
→ +
=
*
6!

/
0
→ +
=
720

 −5

/


Example
: Consider L




cos 3

. Here, a = -1 and 



= cos 3. Then,
L 


cos 3 = L cos 3|
→ 
*
=
*



+9
6
→ 
=
 +1

 +1


+9

Department of Mathematics, Sinclair Community College, Dayton, OH 2
Taking the inverse of a Laplace Transform, that is, L




, will return the original function 



.
For example:
L



7
1

8
=
1
L



7

!




8
=


L



7
1



8
=




L



7



+


8
=
sin



L



7






8
=
sinh


L



9



+


:
=
cos

L



9






:
=
cosh



Example
: Find L

9


;
:
.
L

7
1

,
8
=
1
5!
L

7
5!

,
8
=
1
5!

+
=

+
5!

Example
: Find L

9
,

<
+
:.
L

7
2 −6


−5
8
= L

7
2


−5
8
−L

7
6


−5
8
= 2 L

9



−5
:

6
√5
L

>
√5


−5
?
= 2cosh√5
 −
6
√5
sinh√5


Inverse Form of the First Translation Theorem
Since L 


 = L |
→ 
= |
→ 
=  −,
**
taking the inverse we get
L

 − = L

|
→
=


.
*
Here “ − →” means that “ −” is replaced by “”.

Example
: Find L

9
A



<
A
<
:.
L

7

 +2

+

8 = L

*7



+

80
 →
=


sin
Example
: Find L

9

 B
C
:.
L

7
1
 −3
D
8 =
1
3!
L

7
3!
 −3
D
8 =
1
3!
L

*7
3!

D
80
B→
=


B

B
3!

Department of Mathematics, Sinclair Community College, Dayton, OH 3
Second Translation Theorem
If  − is multiplied by the unit step function U −, the transform is changed to


.
Alternatively, if E is multiplied by the unit step function U −, the transform is changed to



L E +. That is,
L  −U − =



L EU − =


L E +

This is known as the Second Translation Theorem.
Example
: Evaluate L

 −3

U −3

.
 = 3, = 

, = L




=
2!

B
→ L

 −3

U −3

=
2

B

B


Example
: Evaluate L  −3

U −3 using the alternative method.
 = 3,E =  −3

,E +3 = 

→ L  −3

U −3 =

B
L 

 =
2

B

B


Inverse Form of the Second Translation Theorem
The inverse form of the Second Translation Theorem is
L







=  −U −


Example
: Find L

9
F
G<H

I
:.
12
L

7
2

B
8 =
12



=
12
 −2

→ L

>




B
?=
12
 −2

U −2

Department of Mathematics, Sinclair Community College, Dayton, OH 4
Exercises:
1.
L





B

sin
2



5.
L





2

B





U



2



2.
L








3

B


6.
L



7







+
1

8

3. L





2

U



2


7. L



J



K




+
1
L

4.
L




B


U



3





Answers:
1.
2


+
3


+
4
5.
6







1

D

2.
3
!



1

D

12



1

B
+
12



1



8



1

6.
U



1










U



1


3.








7.


cos


U
O


π
2
Q

4.



B


+
1