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
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