WEEK1 CHAPTER 1 MAT485/565

LAPLACE TRANSFORM

Definition

The Laplace transform of the function F(t) on [0 ,  ) is f(s), that is

 − st
ℒ {F(t)} = 0 e F( t ) dt = f(s)

provided the improper integral exist.

Note: Laplace transform is a special type of integral transform and it is very useful in solving
linear initial-value problems.

BUILDING THE BASIC FORMULA

a
(1) ℒ{a} =
s

Proof:
 − st m
ℒ {a} = 0 e a dt = lim a
m→ 0
e −st dt
m
e − st
= a lim
m→ − s
0
a
= − lim (e − ms − 1)
s m→
a
= − (0 − 1) if s > 0
s
a
= = f(s)
s

(2)
1
ℒ { e at } =
s−a

Proof:
 − st m − ( s − a )t
0 e m →  0
ℒ { e at } = e at dt = lim e dt

1 m
= lim − e − ( s − a )t
m→ s − a 0
1 m
= − lim e − ( s − a)t
s − a m→ 0
1
= − lim (e − ( s − a)m − 1)
s − a m→
1
= − (0 − 1) if s - a > 0
s−a
1
= = f(s)
s−a

Rusyah/Norma Laplace transform

WEEK1 CHAPTER 1 MAT485/565

 − st m
t e − st dt
0 e m →  0
(3) a) ℒ {t} = t dt = lim

u v
t + e-st
1
1 − e − st
- s
1 − st
0 e
s2
m
t 1 − st
= lim ( − e − st − e )
m→ s s2 0
1 m
= − lim (st + 1)e − st
2 m→ 0
s
= −
s
1

lim (sm + 1)e − ms − 1
2 m→

1
= − (0 − 1) if s > 0
s2
1
=
s2

1
ℒ {t} =
s2

 − st m 2
t e − st dt
0 e m →  0
b) ℒ { t2 } = t 2 dt = lim

u v
t2 + e-st
1
2t − e − st
- s
1 − st
2 + e
s2
1 − st
0 − e
s3

m
 t 2 2t 2  − st
= lim  − − − e
m →  s s 2 s 3 
 0
 m 2 2m 2  
= lim  − − − e − ms + 2 
m →   s s 2 s 3  s 3 

2
=0+
s3
2
=
s3

Rusyah/Norma Laplace transform

WEEK1 CHAPTER 1 MAT485/565

 − st m 3
t e − st dt
0 e m →  0
c) ℒ { t3 } = t 3 dt = lim

u v

t3 e-st
+
1
3t2 − e − st
- s
1 − st
6t e
+ s2
1 − st
6 - − e
s3
1 − st
0 e
s4
m
 t 3 3t 2 6t 6  − st
= lim  − − − − e
m →  s s 2 s 3 s 4 
 0
 m 3
3m 2
6m 6  − ms  6 
= lim  − − − − e −  − 
m →   2 3 4 4 

s s s s   s 
6
=0+
s4
6
=
s4
n!
Observe the pattern (behavior). Hence, ℒ{ tn } = , n = 1, 2, 3……..
n +1
s

 − st m − st
0 e m →  0
(4) ℒ { cos at } = cos at dt = lim e cos at dt

− st
Consider e cos at dt

u v

cos at e-st
+
1 − st
- a sin at − e
- s

1
e − st
+
- a2 cos at
2
s
1 − st a − st a2
 e − st cos at dt = − e cos at + e sin at − e − st cos at dt 
s 2 2
s s
s2 + a2 − st 1 a − st
e cos at dt = − e − st cos at + e sin at
2
s s s2

Rusyah/Norma Laplace transform

WEEK1 CHAPTER 1 MAT485/565

m
m s2  1 a  − st
 L{cos at } = lim  e − st
cos at dt = lim 2  − cos at + 2 sin at e
m→ s + a2
m→ 0
 s s  0

s2  1 a   1 
= lim  − cos ma + sin ma e − sm −  − 
s 2 + a 2 m →   s s2   s 
s2 1
= 0 + 
2 2 s
s +a
s
=
s2 + a2

s
ℒ {cos at } =
s + a2
2

 − st m − st
0 e m →  0
(5) ℒ { sin at } = sin at dt = lim e sin at dt

− st
Consider e sin at dt

u v

sin at + e-st

1 − st
a cos at − e
- s

1
e − st
+
- a2 sin at
2
s
1 − st a − st a2
 e − st sin at dt = − e sin at − e cos at − e − st sin at dt 
s 2 2
s s
s2 + a2 − st 1 a − st
e sin at dt = − e − st sin at − e cos at
2
s s s2

m
m  1 s2 a  −st

−st
 L{sin at } = lim e sin at dt = lim 2  − sin at − 2 cos at e
m → 0 m → s + a  s
2
s  0

s2  1 a   a 
= lim  − sin ma + cos ma e − sm −  − 
s 2 + a 2 m →   s s2   s2 
 s2 a 
=  0 + 
2 2
s +a  s2 
a
=
s2 + a2
a
ℒ {sin at } =
s + a2
2

Rusyah/Norma Laplace transform

WEEK1 CHAPTER 1 MAT485/565

e x − e −x e at − e −at
(6) By definition : sinh x =  sinh at =
2 2
 − st m − st  e at − e − at 
ℒ { sinh at } =  e sinh at dt = lim e   dt
0 m→ 0  2 
 
1 m − ( s − a )t
= lim e  − e − ( s + a)t dt
2 m→ 0
m
1  1 1 
= lim  − e − ( s − a )t + e − ( s + a )t 
2 m →  s − a s+a 0

1  1 1   1 1 
= lim  − e − ( s − a)m + e − ( s + a)m  −  − + 
2 m →  s − a s+a   s−a s+a

1 1 1 
= 0 + − 
2 s−a s+a
1  2a 
=  
2  s 2 − a 2 
a
=
2
s − a2

a
ℒ {sinh at } =
s − a2
2

e x + e− x eat + e−at
(7) By definition : cosh x =  cosh at =
2 2
 m  e at + e −at 
ℒ { cosh at } =  0
e −st cosh at dt = lim
m→ 0  e −st 
 2
dt


1 m
= lim
2 m→  0
e −( s−a )t + e −( s+a )t dt
m
1  1 1 
= lim  − e −( s−a )t − e −( s+a )t 
2 m →   s−a s+a 0

1  1 1   1 1 
= lim  − e −( s−a )m − e −( s+a )m  −  − − 
2 m →  s − a s+a   s − a s + a

1 1 1 
= 0 + + 
2 s−a s+a
1  2s 
=  
2  s2 − a2 
s
=
s − a2
2

s
ℒ {cosh at} =
s2 − a2

Rusyah/Norma Laplace transform

WEEK1 CHAPTER 1 MAT485/565

Example
Use the basic Laplace transform for the following:

5
a) ℒ {5} =
𝑠
−7
b) ℒ {-7} =
𝑠
3t 1
c) ℒ {e } =
𝑠−3
-4t 1
d) ℒ {e } =
𝑠+4
𝑠
e) ℒ {cos 6t} =
𝑠 2 +36
10
f) ℒ {sin 10t} =
𝑠 2 +100
1
1 12 12
g) ℒ {sinh12 𝑡 } = 1 =
𝑠2 − 144𝑠2 −1
144
𝑠
h) ℒ {cosh 2t} =
𝑠 2 −4
5!
i) ℒ {t5} = 𝑠6

LAPLACE TRANSFORM OF A PIECEWISE CONTINUOUS FUNCTION

Example
Use the definition to find ℒ{F(t)}.

0 0≤𝑡<3
a) F(t) = {
2 𝑡≥3

𝑡 0≤𝑡<1
b) F(t) = {
1 𝑡≥1

𝑠𝑖𝑛𝑡 0≤𝑡<𝜋
c) F(t) = {
0 𝑡≥𝜋

d)

F(t)

(2,2)
1

0 1
t

Rusyah/Norma Laplace transform

WEEK1 CHAPTER 1 MAT485/565

F(t)
e)

Solution
∞ 3 ∞
a) ℒ{F(t)} = ∫0 𝑒 −𝑠𝑡 𝐹(𝑡)𝑑𝑡 = ∫0 𝑒 −𝑠𝑡 (0)𝑑𝑡 + ∫3 𝑒 −𝑠𝑡 (2)𝑑𝑡
2
= lim 𝑒 −𝑠𝑡 |𝑚
3
𝑚→∞ −𝑠
−2
= lim (𝑒 −𝑠𝑚 − 𝑒 −3𝑠 )
𝑠 𝑚→∞
−2
= (0 − 𝑒 −3𝑠 ) , s > 0
𝑠
2
= 𝑠 𝑒 −3𝑠

∞ 1 ∞
b) ℒ{F(t)} = ∫0 𝑒 −𝑠𝑡 𝐹(𝑡)𝑑𝑡 = ∫0 𝑒 −𝑠𝑡 (𝑡)𝑑𝑡 + ∫1 𝑒 −𝑠𝑡 𝑑𝑡

u v
t 𝑒 −𝑠𝑡
+

−1
1 𝑒 −𝑠𝑡
𝑠
-
1
0 𝑒 −𝑠𝑡
𝑠2

1 𝑚
−𝑡 1 𝑒 −𝑠𝑡
= ( 𝑠 − 𝑠2 ) 𝑒 −𝑠𝑡 | + lim |
0 𝑚→∞ −𝑠 1

1 1 1 1
= − [(𝑠 + 𝑠2 ) 𝑒 −𝑠 ] + 𝑠2 + [− 𝑠 lim (𝑒 −𝑚𝑠 − 𝑒 −𝑠 )]
𝑚→∞

1 1 1 1
= − [(𝑠 + 𝑠2 ) 𝑒 −𝑠 ] + 𝑠2 + [− 𝑠 (0 − 𝑒 −𝑠 )] , s > 0

1
= 𝑠2 (1 − 𝑒 −𝑠 )

∞ 𝜋 ∞
c) ℒ{F(t)} = ∫0 𝑒 −𝑠𝑡 𝐹(𝑡)𝑑𝑡 = ∫0 𝑒 −𝑠𝑡 (sin 𝑡)𝑑𝑡 + ∫𝜋 𝑒 −𝑠𝑡 (0)𝑑𝑡
𝜋
= ∫0 𝑒 −𝑠𝑡 (sin 𝑡)𝑑𝑡

Rusyah/Norma Laplace transform

WEEK1 CHAPTER 1 MAT485/565

u v
−𝑠𝑡
sin t 𝑒
+

−1
cos t - 𝑠
𝑒 −𝑠𝑡

+ 1
- sin t 𝑒 −𝑠𝑡
𝑠2

𝜋 sin 𝑡 cos 𝑡 1
∫0 𝑒 −𝑠𝑡 (sin 𝑡)𝑑𝑡 = − ( 𝑠
+ 𝑠2
) 𝑒 −𝑠𝑡 − 𝑠2
∫ 𝑒 −𝑠𝑡 𝑠𝑖𝑛𝑡 𝑑𝑡

𝑠2 +1 sin 𝑡 cos 𝑡
𝑠2
∫ 𝑒 −𝑠𝑡 sin 𝑡 𝑑𝑡 = − ( 𝑠
+ 𝑠2
) 𝑒 −𝑠𝑡 + 𝐶

−𝑠2 sin 𝑡 cos 𝑡 𝜋

∴ ℒ{𝐹(𝑡)} = ( + ) 𝑒 −𝑠𝑡 |
𝑠2 +1 𝑠 𝑠2 0
𝑠2 1 −𝑠𝜋 1
= − 𝑠2 +1 [(0 − 𝑠2 ) 𝑒 − (0 + )]
𝑠2
𝑠2 1 𝑒 −𝑠𝜋 +1
= 𝑠2 +1 (𝑒 −𝑠𝜋 + 1) =
𝑠2 𝑠2 +1

0 , 0≤𝑡 ≤1
d) F(t) = {
𝑡 , 𝑡 ≥1
∞ 1 ∞
ℒ{F(t)} = ∫0 𝑒 −𝑠𝑡 𝐹(𝑡)𝑑𝑡 = ∫0 𝑒 −𝑠𝑡 (0)𝑑𝑡 + ∫1 𝑒 −𝑠𝑡 (𝑡)𝑑𝑡

u v
−𝑠𝑡
t 𝑒
+

−1
1 𝑒 −𝑠𝑡
𝑠
-
1
0 𝑒 −𝑠𝑡
𝑠2

𝑡 1 𝑚
= lim − (𝑠 + 𝑠2) 𝑒 −𝑠𝑡 |
𝑚→∞ 1

𝑚 1 1 1
= lim − [( 𝑠 + 𝑠2) 𝑒 −𝑠𝑚 − (𝑠 + 𝑠2) 𝑒 −𝑠 ]
𝑚→∞

1 1 𝑠+1 −𝑠
= − [(0) − (𝑠 + 𝑠2 ) 𝑒 −𝑠 ] = 𝑠2
𝑒 , s>0

1−𝑡 , 0≤𝑡 ≤ 1
e) F(t) = { 0 , 𝑡 ≥ 1

∞ 1 ∞
ℒ{F(t)} = ∫0 𝑒 −𝑠𝑡 𝐹(𝑡)𝑑𝑡 = ∫0 𝑒 −𝑠𝑡 (1 − 𝑡)𝑑𝑡 + ∫1 𝑒 −𝑠𝑡 (0)𝑑𝑡

Rusyah/Norma Laplace transform

WEEK1 CHAPTER 1 MAT485/565

u v
−𝑠𝑡
1-t 𝑒
+

−1
-1 𝑒 −𝑠𝑡
𝑠
-
1
0 𝑒 −𝑠𝑡
𝑠2

1 1 1
= − (1 − 𝑡) 𝑠 𝑒 −𝑠𝑡 + 𝑠2 𝑒 −𝑠𝑡 |
0

1 1 1
= 𝑠2 𝑒 −𝑠 + 𝑠 − 𝑠2

TRY THIS
Find ℒ{𝐹(𝑡)} using the definition of Laplace transformation, given that:
3𝑡 2 , 0 < 𝑡 < 𝜋
𝐹(𝑡) = {
cos 𝑡 , 𝑡≥ 𝜋

6 3𝜋2 6𝜋 6 𝑠
(Ans: − 𝑒 −𝜋𝑠 ( + + + ))
𝑠3 𝑠 𝑠2 𝑠3 𝑠 2 +1

Rusyah/Norma Laplace transform