Unit VI Laplace Transforms
Definition:- If f(t) is a real valued function for all ≥
, then the laplace transform of f(t),
denoted by L{f(t)} and is defined as
=
= ( )
Where ‘s’ is parameter, real or complex. Note: L af t ± b g(t) = a L f t
± b L{g(t)} where ‘a’ and ‘b’ are constants.
Laplace transform of standard functions:1. Prove that,
= , where ‘a’ is constant.
Proof: By definition
Lf t
=
La =
=
e
e
= − [
If a=1, then
=
f t dt
a dt
−
= − [0 − 1]
=
]
2. Prove that
=
( >
Proof: By definition L e
=
) (
=
3. Prove that Proof:
ℎ
= {
= =
= ℎ
=
=
[ [
4. Prove that Proof:
ℎ
= {
= = = ℎ
=
[
[ [
}
−
−
(
(
=
[
(
+
] )
+
(
)
)
=
]
}
(
)
e dt =
]
)
]
e
)
]
]
(
e
)
(
=
) (
dt )
=
5. Prove that
=
Proof: By definition
=
{
=
(
= −
6. Prove that Proof: By definition
=
=
{
= −
Proof: By definition
=
(
= −
=
}
+
[0 − 1( . 0 + . 1)]
}
=
7. Prove that
{
=
=
)
= −
)
{−
}
+
[0 − 1( . 1 + . 0)]
}
=
−
= 0 − 0 + =
=
=
⋮
=
=
=
= 8. Prove that L{ } =
{
Proof: By definition
=
( + 1)/
=
!
⋮
}
{
)
}
{
}
!
{ }
1 =
!
…
.
{ }
where n is a positive integer
, where ‘n’ is constant
when t=0, x=0 and when t=∞, x=∞
{ } =
(
{ }
put st=x , s dt= dx, dt=dx/s
therefore L{ } =
( )
Definition:
Examples:
L{ } = ( + 1)/
( )=
( +
)=
1. Find L{f(t)} where f(t)= Solution: By definition,
replacing n by n+1
, < < 1 , > 1
L{f(t)} = =
= (
=
)
=
+
+
+ 0
(0)
=
2. Find the Laplace transform of f(t) = Solution : By definition,
, < < 4 , > 4
L{f(t)} = = = = = - {4 =-
+
+
− 1. (
-0}- {
- {
)
-1}+
5
+5 -1}- {0-
}
=
- {
-1}
, < < , >
3. Find the Laplace transform of f(t)=
Solution: By definition, L{f(t)} = = {
= =
(
sin 2 )
+
+
(0)
}
{
}
= =
{
}
4. Find the Laplace transform of f(t)=
Solution:
,
<
,
<
>
L{f(t)} =
5. Find the Laplace transform of f(t)=
,
, < < >
First shifting property: If L{f(t)} = f(s) then, L{ Proof: By definition, L{
f(t)} = = = f(s-a)
(
)
f(t)} = f(s-a)
Standard results: 1. L{ 2. L{ 3. L{ 4. L{ 5. L{
}= (
sin
cos
!
}= }=
) (
sin ℎ }=
)
(
cos ℎ }=
)
(
(
)
)
because, L{ }=
!
because, L{sin
}=
because, L{sin
}=
because, L{sinh
}=
because, L{cosh
}=
Examples: 1. Find L{
}
We know that, sin sin
=
Therefore L{sin 2 sin 3 } = =
2. Find L{
cos( −
) − cos( +
cos( ) − cos(5 ) −
t}
We know that, cos
=
(put =2t)
L{ cos 2t} = L{
}
= [L{1}+L{cos 4 }]
= { +
}
)
3. Find L{
t}
We know that sin3 = 3sin - 4sin θ or sin θ = L{sin 2t} = L{ =
{sin2t} - {sin6t} } - {
}
t cos2t cos3t}
We know that cosA cosB = {cos
+
+ cos
L{cost cos2t cos3t} =
(
=
5. Find L{
+4 +4
6. Find L{ (
1 +
+
cos 3t = 3
) + (cos 4 +
6 + cos 4 + +
+
-2sin3t +3cos3t}
-2sin3t +3cos3t} = −
!
+4( ) -2(
) }
) +3(
L{ (sint − cost) } = L{sin t + cos t − 2sintcost} = L{ 1-sin2t} = -
(put
} cos θ =
cos 3t + cos 3
=
=
−
(cos3t + cos )
=
L{
= 2 )
}
= { 4. Find L{
(put
)
2
2 )
= 3 )
7. Find L{sin2t cos3t} We know that, cosA sinB = {sin(A+B)-sin(A-B)} L{sin2t cos3t} =
L{sin5t-sint}
=
8. Find L{
{
-
}
}
We know that cos3 = 4cos θ - 3cos , cos θ = L{ cos 2t } = L{ = =
}
−
{
9. Find L{ sin5t cos2t } 10. L { 11. L {(
}
−
} +
) +
12. L {
}
13. Find L{
(
14. Find L{ L{ e
}
+
)} }
sinh4t} = L{ e = =
(
L{ e − e −
)}
(put
= 2 )
15. Find L{
(
L{ e
−
)}
(2cos5t − 3sin5t)} = 2L {e = 2 {( = (
16. Find L{
)
)
cos5t} − 3L { e } - 3 {(
}
because, sin θ =
= L{ e − e cos2t} = L{
17. Find L{
−
(
}
)
}
We know that, sinA cosB =
{sin(A+B)+sin(A-B)}
L{ e sin2t cost} = L{ e (sin3t + sint) }
= L{ e sin3t + e sint) } = {(
L{
19. L{
}
}
L{ e sin t} = L{ e (
18. Find L{
)
sin5t}
}
} = {(
20. L{sinhat sinat}
}
!
)
}
)
+(
)
}
Properties of Laplace transforms:
Property 1: If L{f(t)}= f(s) then L{ ( )} = (− ) Proof: By definition ( ) =
[
]
( )
Differentiating both sides with respect to s [f(s)] =
f(t) dt
According to Leibnitz rule for differentiation under integral sign [f(s)]
=
[f(s)]
= -
[f(s)]
f(t) dt
=
(− ) f(t) dt f(t) dt
(-1) [f(s)] = L{t f(t)}
Similarly
(− 1) (− 1) (− 1)
⋮
=
{
=
{
=
{
f(t)} f(t)}
}
Property 2: If Proof: By definition
=
( ) then prove that
=
=
f(t) dt
Integrating both sides between s to ∞ with respect to s =
f t dt ds
= = =
=
[changing the order of integration]
f t ds dt
[0 −
= L {
( )
}
]
( )
Examples: (1) Find {
}
We know that
{
{
} =
} = (− 1)1 = − [
(
[
( )
+
)
] ] = (
)
(2) Find
{
We know that
}
{
{
} =
} = (− 1)1
= − [
[
(
]
(
)
)
] = (
)
(3) Find L{t2 sin at} We know that L {sin at} =
Therefore L t sin at = (− 1) =
[(
= (
}
t cos ht = =
{t (
[ t
=
[
[
)
)}
+ t
( (
]
)
=
(4) Find {
]
=
]
) ) (
(
(
) ]
)
)
!
= [(
5. Find { −
We know that Therefore
= 3 [(
}
sin 4 4
7. Find
{
cos ℎ3
}
We know that
Therefore
1−
+
)
= (
(
= −
}
= {
= {
= [
!
=
−
=
=
)
]
)
]
)
= (− 1)
= ( 6. Find {
)
!
+ (
[(
[
(
(
+
[ −
)
]
)
+
−
]
)
)}
!
]
} ]
( + )
=
=
= 0 −
8. Find {
=
}
We know that sin
=
Therefore
9. Find {
{
∴L{
=
−
{1 −
∴L
}
} = } =
= [
(
)
= −
}
{
(
=
We know that
10. Find
= −
=
} = − =
[ −
]ds
= [logs − log (s − 1)] = [
] =
−
− +
−
ds
+
]
)
= [ 11. Find {
] = 0 −
}
We know that L e sint =
12. Find { 13. Find
{
14. Show that
}
[
(
sin4t] dt
= (− 1)
= (
Put s=2
15. P.T.
sin4t] dt = (
We have, e
+ 1 =
] =
=
[t e
ds = [tan (
)
= −
[t e
=
}
e
=
=
)
=
f t dt = L f t
[
(
) )
=
tcos2tdt = L tcos2t
.
]
)]
( + 1)
= (− 1)
Put s=3, e
16. Evaluate
= (
tcos2tdt = (
[
] = − [
)
)
=
using Laplace transform
We have,
e
(
e
f t dt = L f t ) dt = L{ =
= [
= [
( + )−
= 0 −
Put s=0 e
(
−
}
=
)dt = log
( + )]
] = [
= −
]
(
)
.
]
Property 3: If Proof: Let
=
L F t
= ( ) then [
f t dt and hence
=
e
=
F t dt
∞ − 0
−
= 0 − 0 + =
17. Find {
Let
=
} =
= {
}
= − (1) [ = −[
[
] =
(
=
] =
t cos at dt =
(
(
)
( )
)
)
.
]
e
( )
]
=
( )
( ) and
e F′ t dt −s
F′ t dt
0 = 0
18. Find {
}
Let f t = e = { =
(
)
= [
L[ e
f t dt] =
=
}
(
= −
=
( )
(
ds
∞ )] )
= ( )
