LAPLACE TRANSFORMS.
Introduction
Let f(t) be a function defined in [0 , ∞).
Definition:
The integral


0
e st f (t )dt
…..
(1) , is called Laplace Transforms
for f(x), if that integral convergent.
Notation: £{f(x)} where £ is an operator.


0
e st f (t )dt is improper integral.
Then


0
T

e st f (t )dt = lim
0
T→∞
e st f (t )dt .
(1) depends on parameter S, then

£{f(t)} =

0
e st f (t )dt = F(S).
Generally: £{f(t)} = F(S)
£{g(t)} = G(S)
£{y(t)} = Y(S)
Example: Show that £{1} =
1
.
S

Solution. : £{1} =  e st 1dt =
0
=
T
lim
T 
lim
T 
 st
 e dt =
0
lim
T 
e  st
]
S
T
0
1
1
[- e  sT  ] = ∞ … (2)
s
s
a) If S < 0, then 2) → £{1} = ∞
b) If S = 0, then 2) → £{1} = ∞
1
1
1
c) If S > 0, then 2) → £(1) = - e-sT +
=0+
S
S
S
1
Ł(1) =
.
S
Example: Using the definition, determine the Laplace Transforms for the following
functions:
a) f(t) = a. b) f(t) = t. c) f(t) = tn. d) f(t) = eat.
Let a be constant and n be a positive integer.
1


0
0
Solution: a) £{a} =  e  st .a dt = a  e st dt = a[
£{a} =
e  st
]
s

0
1
a
= a[ ] =
s
s
a
, s>0
S
1
S
5
a = 5 → £{5} =
S
1
1
1
a=
→ £{ } =
3
3
3S
Substitute: a = 1 → £{1} =

b) £{t} =  e  st t dt = t∫e-stdt - ∫{∫e-stdt}
0
d (t )
dt
dt

e  st
e  st
]
-∫
0
s
s
 st

1 e
1 1
1
= 0+ [
]
= ( )= 2
0
s s
s s
s
= t.
£{t} =

c) £{t } =  e
n
 st
0
d (t n )
dt
t dt = t ∫e dt - ∫{ ∫e dt}
dt
n
= tn .
=
→ £{tn} =
1
, S>0
S2
n
-st
-st

e  st
e  st
]
-∫
.n.tn-1dt
0
s
s
n
n
0 + ∫e-st tn-1dt = £{tn-1}.
S
S
n
£{tn-1}.
S

£{t } =  e st .t n1dt
n-1
0
= tn-1 ∫e-stdt - ∫{∫e-stdt}
d (t n 1 )
dt.
dt

t n1e  st
e  st
]
-∫
(n-1)tn-2dt.
0
s
s
n 1
= 0 +
£{tn-2}
s
=
2
n 1
£{tn-2}
s
n

2
Thus: £{tn-2} =
£{tn-3}
s
.
.
2
£{t2} =
£{t}
s
1
£{t} =
£{1}
s
1
£{1} =
s
n
→ £{tn} = ( ) £{tn-1}
s
n n 1
= ( )(
) £{tn-2}
s
s
n n 1 n  2
= ( )(
)(
) £{tn-3}
s
s
s
.
.
.
n n 1 n  2
1
= ( )(
)(
) …( ) £{1}
s
s
s
s
n! 1
= ( n )( )
s
s
Then: £{tn-1} =
£{tn} =
n!
s n 1
n = 0 → £{t0} = £{1} =
n = 0, 1, 2, …
1
;
S

d) £{eat} =
 st at
 e e dt = ∫e-t(s-a)dt =
0
£{eat} =
If:
n = 3 → £{t3} =
a = 0, then £{1} =
e  ( s a )t
]
 ( s  a)
1
,
sa

0
=
6
s4
1
, s>0.
S a
s>0
1
.
S
3
1
.
S 3
1
a = -2, then £{e-2t} =
.
S 2
a = 3, then £{e3t} =
e) Let
f(t) = cos at. Then:

£{cos at} =  e st kos at dt
0
= cos at ∫e-stdt - ∫{∫e-stdt}
d
(cos at)dt.
dt

kos at e  st
e  st
]
-∫
(-asin at)dt
0
s
s
1 a
d
=
- [sin at ∫e-st dt - ∫{∫e-stdt} (sin at)dt]
s s
dt
 st
 st

1 a
e
e
=
- {sin at.
]
-∫
(a cos at)dt}
0
s s
s
s
1 a
a
=
- { 0 + ∫e-stcos at dt}
s s
s
2
1
a
=
- 2 £{cos at}
s
s
2
1
a
(1+ 2 )Ł{cos at} =
s
s
=
Ł{cos at} =
s
s
= 2
;
2
s 2
s 4
Ł{cos 2t} =
2
s
,
s  a2
2
s>0
Ł{cos ½ t} =
4s
4s 2  1

f) £{sin at) =

e-stsin at dt
0
= sin at ∫e-stdt - ∫{∫e-stdt}
d
(sin at) dt
dt

sin at.e  st
e  st
]
-∫
.a cos at dt
0
s
s

a
e  st
e  st
= 0 +
{cos at.
]
-∫
(-a sin at)dt}
0
s
s
s
a 1 a
= { - ∫e-stsin at dt
s s s
=
4
=
a
a2
£{sin at}
s2
s2
→
£{sin at} =
a
s  a2
2
s >0
d
d
(sinh t) = cosh t and
(cosh t) = sinh t.
dt
dt
By the same calculation we get:
Notice:
a
,
s  a2
s
,
Ł{kosh at} = 2
s  a2
Ł{sinh at} =
2
s0
s0
Example:
Using the definition of the Laplace transformation, determine £{f(t)}, if:
1
t,
5
0≤t<5
1,
t≥5
f(t) =
Solution:
5
£{f(t)} =
=
=
=
=
=

 st 1
 st
0 e . 5 t dt + 5 e .1dt
1
d
e  st
-st
-st
[t ∫e dt - ∫{∫e dt} (t)dt] +
]
5
dt
s
1
e  st 5
e  st
e 5 s
{ t.
]
- ∫
dt } +
0
5
s
s
s
5 s
 st
5 s
5
1 5e
1 e
e
] +
2
0
5 s
5 s
s
5 s
5 s
1 e
1
e
e 5 s
- { 2 - 2 }+
5
s
s
s
s
1
( 1 - e-5s ).
5s 2

5
5
Theorem: If £{f1(t)} and £{f2(t)} exist, α and β are constants, then:
£{αf1(t) + βf2(t)} = α £{f1(t)}+ β £{f2(t)}
Theorem: £{a1f1(t) + a2f2(t) + … + anfn(t)} = a1£{f1(t)} + a2£{f2(t)} + … + an£{fn(t)},
where f1(t), f2(t), …, fn(t) exist
and a1, a2, …, an are constants.
Example: Determine £{f(t)} if f(t) = 2t4 – e- 4t.
Solution : £{f(t)} = £{2t4 – e- 4t}
= 2 £{t4} – £{e- 4t}
4!
1
=2( 5 )s
s  (4)
48
1
=
5
s4
s
1 at
(e + e-at), determine £{cosh at}.
2
1
Solution: £{cosh at} = £{ (eat + e-at)}
2
1
1
= £{eat} + £{e-at}
2
2
1
1
1
1
= (
)+ (
)
2 sa
2 sa
s
= 2
.
s  a2
Example: If cosh at =
1 at -at
(e – e ) shows that
2
a
₤{sinh at} = 2
.
s  a2
Exercise: If sinh at =
Example: Find the Laplace transform of f(t) = sin 3t.kos 5t.
Solution: f(t) = sin 3t.kos 5t
1
= {sin(3t + 5t) + sin(3t – 5t)}
2
1
= {sin 8t – sin(-2t)}
2
1
= {sin 8t – sin 2t}
2
6
1
(sin 8t – sin 2t)}
2
1
1
= ₤{sin 8t} - ₤{sin 2t}
2
2
1
8
1
2
= ( 2
)- ( 2
)
2 s  64
2 s 4
3s 2  48
=
.
( s 2  64)( s 2  4)
₤{f(t)} = ₤{
Properties of Laplace Transform
First Shift Theorem
Theorem: If ₤{f(t)} = F(s) and a constant, then ₤{eat.f(t)} = F(s – a).

Proof: ₤{f(t)} =  e  st . f (t )dt = F(s). [definition].
0

₤{eat.f(t)} =

e-st.eatf(t) dt
0

=

e-(s-a)t.f(t) dt.
[suppose p = s – a]
0

=

e-p.f(t) dt = F(p) = F(s – a).
0
→ ₤{eat f(t)} = F(s – a).
Examples.
a) Find the Laplace transform for f(t) = t4e3t.
4!
24
= 5 = F(s)
4 1
s
s
24
₤{t4e3t} = F(s – 3) =
.
( s  3) 5
₤{t4}
=
b) Find the Laplace transform for f(t) = 2e4tsin 4t.
4
= F(s).
s  16
₤{2e4tsin 4t}= 2 ₤{e4tsin 4t} = 2 F(s – 2)
₤{sin 4t} =
2
7
4
]
( s  4) 2  16
8
= 2
.
s  8s  32
= 2[
Theorem. If ₤{f(t)} = F(s), then for n = 1, 2, 3, …
dn
₤{tn.f(t)} = (-1)n n [F(s)].
ds
Example: Find the Laplace transform for f(t) = t2sin 2t .
2
Solution : ₤{sin 2t) = 2
= F(s).
s 4
2
d2
d2
₤{t2sin 2t} = (-1)2 2 [ 2
] = 2 [2(s2 + 4)-1]
ds s  4
ds
d
=
[-2(s2 + 4)-2(2s)]
ds
= -4(s2 + 4)-2 – 4s(-2)(s2 + 4)-3(2s)
 4( s 2  4)  16s 2
=
( s 2  4) 3
=
12s 2  16
( s 2  4) 3
Inverse Laplace Transforms (ILT)
Definition: If ₤{f(t)} = F(s), then Inverse Laplace Transforms for F(s) as written as:
₤-1{F(s)} = f(t).
₤-1 is known as operator for inverse Laplace transforms.
Notice:
₤{f(t)} = F(s)
If f(t) = a, then ₤{a} =
a
s
a
→ ₤-1{ } = a.
s
Examples:
4
4
a) ₤-1{ } = 4, because ₤{4} = .
s
s
1
1
b) ₤-1{
} = e4t, because ₤{e4t} =
.
s4
s4
8
2
2
}= ₤-1{ 2
} = sin 2t.
s 4
s  22
3
1
d) ₤-1{
} = ₤-1{
} = e5/3 t
3s  5
s  5/3
4s
7
s
e) ₤-1{ 2
} = ₤-1{
} = cos t.
49
2
4s  49
s2 
4
6
6
/
4
3
f) ₤-1{ 2
} = ₤-1{ 2
} = sin t.
2
4s  9
s  9/ 4
c) ₤-1{
2
Properties of Inverse Laplace Transform.
Theorem: If ₤-1{F(s)} = f(t) and ₤-1{G(s)} = g(t) and if
α and β are constants then:
₤-1{α.F(s) + β.G(s)} = α ₤-1{F(s)} + β ₤-1{G(s)}.
Examples:
a) ₤-1{
12
2
2!
} = ₤-1{6( 3 )} = 6 ₤-1{ 3 } = 6 t2.
3
s
s
s
b) ₤-1{
2
1
1
}= ₤-1{2(
} = 2 ₤-1{
} = 2 e-3t.
s3
s3
s3
c) ₤-1{
4
4
3
4 -1
3
4
} = ₤-1{ ( 2
)} =
₤ { 2
} = sin 3t.
3 s 9
3
3
s 9
s 9
d) ₤-1{
2s
2 -1
s
1
3
}=
₤ { 2
} = cos t.
2
16
8
4
16 s  9
s  9 / 16
e) ₤-1{
f) ₤-1{
2
2s  5
2s
5
} = ₤-1{ 2
} + ₤-1{ 2
}
2
s  25
s  25
s  25
s
5
= 2 ₤-1{ 2
} + ₤-1{ 2
}
s  25
s  25
= 2 cosh 5t + sinh 5t.
3s  5
3s
5
} = ₤-1{
} + ₤-1{
}
2
2
16 s  9
16 s  9
16 s 2  9
3 -1
s
5 -1
1
=
₤ { 2
}+
₤ { 2
}
16
16
s  9 / 16
s  9 / 16
9
3
3
5 4 -1
3/ 4
kosh t +
.
₤ { 2
}
16
4
16 3
s  9 / 16
3
3
5
3
=
cosh t +
sinh t.
16
4
12
4
=
First-Shift Theorem (Inverse).
If ₤-1{F(s)} = f(t) and a is constant, then:
₤-1{F(s – a)} = eat f(t) 0r
₤-1{F(s – a)} = eat ₤-1{F(s)}.
Examples:
a) ₤-1{
b) ₤-1{
1
1
} = e4t ₤-1{ } = e4t.1 = e4t.
s
s4
3s
3( s  1)  3
} = ₤-1{
}
4
( s  1)
( s  1) 4
3
3
= ₤-1{
} - ₤-1{
}
3
( s  1) 4
( s  1)
6
3 -1
1
2
₤ {
} - ₤-1{
}
3
2
2
( s  1) 4
( s  1)
3
2
1
6
= e-t₤-1{ 2 } - e-t₤-1{ 3 }
2
2
s
s
3 -t 2 1 -t 3
= e t - e t
2
2
1
= e-t(3t2 – t3).
2
=
c) ₤-1{
8( s  2)  3
8s  13
} = ₤-1{
}
s  4s  5
( s  2) 2  9
8( s  2)
3
= ₤-1{
} - ₤-1{
}
2
( s  2)  9
( s  3) 2  9
s
3
= 8e-2t₤-1{ 2
} - e-2t ₤-1{ 2
}
s 9
s 9
= 8e-2tcosh 3t – e-2tsinh 3t
e 3t  e 3t
e 3t  e 3t
= 8e-2t(
) – e-2t(
)
2
2
1
= (7et + 9e-5t).
2
2
10
d) ₤-1{
e) ₤-1{
1
1
1
} = ₤-1{- } + ₤-1{
}
2s
s ( s  2)
2( s  2)
1
1
1
1
= - ₤-1{ } + ₤-1{
}
2
s
2
s2
1
1
= - + e2t
2
2
1 2t
= (e – 1).
2
1
s3
3s  1
} = ₤-1{ }+ ₤-1{ 2
}
2
s
s 1
s ( s  1)
1
s
1
= ₤-1{ } - ₤-1{ 2
}+ 3₤-1{ 2
}
s
s 1
s 1
= 1 – cos t + 3sin t.
Applications of Laplace transforms.
Theorem:
If ₤{y(t)} = Y(s), then:
₤{y`(t)} = sY(s) – y(0)
₤{y``(t)} = s2Y(s) – sy(0) – y`(0)
₤{y```(t)}= s3Y(s) – s2y(0) – sy`(0) – y``(0)
:
.
₤{y(n)(t)
= snY(s) – sn-1y(0) – sn-2y`(0) – …- y(n-1)(0).
Exercises:
By using Laplace transform determine the following equations.
1.
2.
3.
4.
5.
y` + y = kos t, if y(0) = 0
y` + 3y = 13 sin 2t , y(0) = 6.
y` + y = te-2t , y(0) = 0
y`` - 4y = 4e2t , y(0) = 0 and y`(0) = 5.
y`` + 2y` - 3y = t , y(0) = 2 and y`(0) = 1.
11