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Unit -2
Syllabus:

Laplace transform, Existence theorem, Laplace transforms of derivatives and integrals, Initial and final value
theorems, Unit step function, Dirac- delta function, Laplace transform of periodic function, Inverse Laplace
transform, Convolution theorem, Application to solve simple linear and simultaneous differential equations
...
The basic knowledge of Laplace Transforms and its applications in solving differential equations
...
1
1
...
3
1
...
5
1
...
7
1
...
9
1
...
11
1
...
13
1
...
15
1
...
17
1
...
19
1
...
21

Inverse

Laplace 25

Transform
1
...
23

Linearity Property

27

1
...
25

Second shifting theorem

27-28

1
...
27

Inverse

32

Laplace

Transform

of

Derivatives
1
...
29

Multiplication by s:Heaviside
Formula

32

Expansion 33
for

Inverse

Laplace Transform
1
...
31
...


2

1
...

With the increasing complexity of engineering problems, Laplace transforms help in solving complex
problems with a very simple approach just like the applications of transfer functions to solve ordinary
differential equations
...


•

Analysis of Electrical Circuits
In electrical circuits, a Laplace transform is used for the analysis of linear time-invariant systems
...


•

Digital Signal Processing
One cannot imagine solving DSP (Digital Signal processing) problems without employing Laplace transform
...
It makes studying analytic part
of Nuclear Physics possible
...
It helps analyse the variables, which when altered,
produces desired manipulations in the result
...
This is an efficient and easier way to control processes that are
guided by differential equations
...
2 Introduction
The knowledge of Laplace Transformations has in recent years became an essential part of Mathematical background
required of engineers and scientists
...

The transform turns integral equations and differential equations to polynomial equations, which are much
easier to solve
...

This subject originated from the operational methods applied by the English engineer Oliver Heaviside (1850 – 1925)
to problems in electrical engineering
...


1
...


1
...
then
L{a1 F1(t) + a2 F2(t) } = a1 L{ F1(t) + a2 L{ F2(t) }
...
H
...


= L{a1 F1(t) + a2 F2(t) }
=

∞

∫e

− st

[ a 1 F1 (t) + a 2 F2 (t) ] dt

0

∞

∞

0

0

= a 1 ∫ e − st
...
H
...

∴ L{ a1 F1(t) + a2 F2(t) } = a1 L{ F1(t) + a2 L{ F2(t) }
This completes the proof of the theorem
...
Compute L {1}
Solution- Definition of Laplace Transform
∞

L {1}= f (s) = ∫0 𝑒𝑒 −𝑠𝑠𝑠𝑠
...
Compute L(3t)
∞

L {3t }= f (s) = ∫0 𝑒𝑒 −𝑠𝑠𝑠𝑠
...
𝑡𝑡𝑡𝑡𝑡𝑡

(Using Linearity Property)
=3

Example 3
...
(4 𝑡𝑡 + 9) 𝑑𝑑𝑑𝑑

=

∞

∞

4 ∫0 𝑒𝑒 −𝑠𝑠𝑠𝑠
...
𝑑𝑑𝑑𝑑

(Using Linearity Property)

=

𝟑𝟑𝟑𝟑 𝟗𝟗
+
𝒔𝒔𝟗𝟗 𝒔𝒔

Example 4
...
6 sin at 𝑑𝑑𝑑𝑑 using
𝑒𝑒 𝑎𝑎𝑎𝑎

∞

∫0 𝑒𝑒 𝑎𝑎𝑎𝑎 sin bx 𝑑𝑑𝑑𝑑 = 𝑎𝑎2 +𝑏𝑏2 (𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 − 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
∞

=6∫0 𝑒𝑒 −𝑠𝑠𝑠𝑠
...
Compute L {cos at}
Solution
...
cos at 𝑑𝑑𝑑𝑑
∞

𝑒𝑒 𝑎𝑎𝑎𝑎

We know ∫0 𝑒𝑒 𝑎𝑎𝑎𝑎 cos bx 𝑑𝑑𝑑𝑑 = 𝑎𝑎2 +𝑏𝑏2 (𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 + 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
=

𝑠𝑠
𝑠𝑠2 +𝑎𝑎 2

Example 6
...
cosh 5t 𝑑𝑑𝑑𝑑

( Using Linearity Property)

3𝑠𝑠
= 2 −52
𝑠𝑠

−

20
𝑠𝑠2 −52

5

5t 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑

Example 7
...
𝑡𝑡 4 𝑑𝑑𝑑𝑑 + ∫0 𝑒𝑒 −𝑠𝑠𝑠𝑠
...
𝑡𝑡 2 𝑑𝑑𝑑𝑑

1 4
𝑠𝑠 𝑠𝑠3

= 5+ +

EXERCISE
1
...


2
...


L { 6 sin 2t – 5 cos 2t }

4
...
𝑓𝑓(𝑠𝑠) =

12
6
− 𝑠𝑠2
𝑠𝑠4

+

− 5𝑠𝑠

Ans
...
L{ cosh at – cos at }

Ans 𝑓𝑓(𝑠𝑠)

𝑠𝑠+9
𝑠𝑠2 −9
2𝑎𝑎 2 𝑠𝑠
= � 4 −𝑎𝑎4 �
𝑠𝑠

24
𝑠𝑠5

Some basic formula of Laplace Transform:

S No
...


F(t)
L{f(t)} = f(s)

02
...


t

1
s
1
s2

t2

2!
s3

04
...
𝑓𝑓(𝑠𝑠) =

5
...
5

𝑓𝑓(𝑠𝑠) =

Laplace Transform
L-1{f(s)} = f(t)

n!
05
...


06
...

08

09
...


s n +1
1
s−a

at

1
s+a

e-at
(n -1)

at

(k -1)

at

t
e
(n − 1)!
t

e
Γ(k )

1
( s − a) n

n = 1,2,3,
...


a
s + a2
2

sin at

6

s
s + a2
2

11
...


13
...



...


cosh at

16
...


f(t – a)
...
6 First shifting theorem
Statement: If L ( F(t) ) = f(s) , then L ( eat F(t) ) = f(s-a)
∞

Proof: Definition of Laplace transformation
...


1
...

t
Proof:
a

−∞

∞

|------------------------ t < a ------------|0------|--- t > a -----|
∞

L { G(t) } =∫0 𝑒𝑒 −𝑠𝑠𝑠𝑠 𝐺𝐺(𝑡𝑡)𝑑𝑑𝑑𝑑
=

a

∞

0

a

−s t
∫ e
...


1
...


∫e

F ( x) dx

0

1
=
a

∞

1
=
a

∞

1
𝑎𝑎

x
−s  
a

∫e

s
−x  
a

F ( x) dx

0

∫e

s
−t  
a

F (t ) dt

[since

0

b

∫
a

𝑠𝑠
𝑎𝑎

= 𝑓𝑓( )

By Leibnitz principle for differentiation under integral sign,

8

f ( x) dx =

b

∫ f (t ) dt
a

]

1
...
r
...
s,

∞
 d
d  −s t
{f(s)}
∫ e F (t ) dt  =
ds  0
ds


∞

d
d −s t
∫0 ds (e ) F (t ) dt = ds { f(s) } ⇒
⇒

∞

∫ −t e

−s t

F (t ) dt =

0

∞

∫te

−s t

F (t ) dt = −

0

for n = m,

∞

∫e

−s t

[t

F (t ) ] dt = ( − 1)m

m

0

d
{ f(s) }
ds
d
{ f(s)}
ds

dm
{f(s) }
ds m

This completes the proof of multiplication by tn
...
10 Division by t

Statement: If L { F(t) } = f(s) then L {
∞

Proof: since = f(s) = ∫ e − s t F (t ) dt

1
∞
F(t) } = ∫𝑠𝑠 𝑓𝑓(𝑠𝑠)𝑑𝑑𝑑𝑑
t

0

On integrating both sides w
...
t
...


1
∞
F(t)} = ∫𝑠𝑠 𝑓𝑓(𝑠𝑠)𝑑𝑑𝑑𝑑
t
This completes the proof of theorem
...
Find L [3e2t sin t }
Solution:

1

L {sin t} = 𝑠𝑠2 +1

L [3e2t sin t }=

3
(𝑠𝑠−2)2 +1

( Use First shifting theorem)

Example 2
...
a
a
d
{
}
−
2
2
L { t sin at } =
= 2
ds s + a
s + a2

(

∴ L { t sin at } =

2𝑠𝑠
...


Example 3
...
a 
2  =−
= ds  2
 s + a 2 

)

(

(

)
(

)

 8
...
a s 2 + a 2 2 − 2as
...
2 s

4
s2 + a2


  8
...

∴L { t2 sin at } = = 
2
2 2

 s + a

(

)

Example 4
...
− 2 


 (s + 3)3 



d  2
3
= - ds 
 (s + 3)
 2
...


Example 5
...


2

 (1 − e t ) 

 t


Example 6
...


11

(𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶−𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶)
�
𝑡𝑡

Example 7
...


 = Log
∴L
(s 2 + b 2 )
t



∞

Example 8
...


−𝑠𝑠𝑠𝑠
∴ ∫0 𝑡𝑡𝑒𝑒 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆=(𝑠𝑠2 +1)2 and s = 2

4

∞

−2𝑡𝑡
∴ ∫0 𝑡𝑡𝑒𝑒 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆= 25

Example 9
...


Sin mt
dt
t

∫
0

 m 
Solution: L {Sin m t} =  s 2 + m 2 
∞

Now
∞

∴

∫
0

∫
0

∞

Sin mt
 m 
dt = ∫  2
ds = [ tan −1( s / m)]∞S
2 
t
Ss +m 

π
−1 s 
Sin mt
dt =  − tan ( ) as s → 0
m 
t
2
π

= 2

if m > 0

π

=− 2

if m < 0

12

∞

∴

∫
0

π
−1 s 
Sin mt
dt =  − tan ( ) is required solution
...
Find L ∫
t
0

Solution: since L {
∴ L { et

Sin t
∞
𝑑𝑑𝑑𝑑
𝜋𝜋
−1
−1
∫𝑆𝑆 �𝑠𝑠2 +12 � = [𝑡𝑡𝑡𝑡𝑡𝑡−1 ( 𝑠𝑠)]∞
𝑆𝑆 = 2 − 𝑡𝑡𝑡𝑡𝑛𝑛 𝑠𝑠 = 𝑐𝑐𝑐𝑐𝑡𝑡 𝑠𝑠
}
=
t

Sin t
t
-1
-1
t } =L {e Cot s} = cot (s – 1)

[ by shifting theorem ]

∞ e t Sin t  1
dt  =
...
Hence the solution
∴ L ∫
t
0
 s

EXERCISE
1
...
L F(t) if F(t) = 
0

0 < t <1
t >1

0
4

3
...


𝑓𝑓(𝑠𝑠) =

1
𝑠𝑠+2

Ans
...


𝑒𝑒 −2𝑠𝑠
𝑠𝑠

Ans
...
𝑓𝑓(𝑠𝑠) = 𝑠𝑠−1 �1 − 𝑒𝑒

Ans
...
L{  t −
t


6
...


∫t e

Sin t dt

Ans
...
25

0

3

∞

8
...


Ans
...
11
...
12
...


1
...
Existence theorem:
Statement: If F(t) is sectionally continuous for t ≥ 0 and exponential order b, then 𝐿𝐿{𝐹𝐹(𝑡𝑡)} = 𝑓𝑓(𝑠𝑠) exists for s >
b
...

continuous in every finite interval 0 ≤ t ≤ t0
I2

≤

∫

∞

t0

e − st F (t ) dt

≤

∫

∞

≤

∫

∞

≤

∫

∞

=

M
s−b

e − st F (t ) dt

0

0

0

e − st M ebt dt
e − ( s − b ) t
...


Lim

Solution 1:

e −bt
...
Hospital rule

= 𝟎𝟎

t n = 0(ebt ) , t → ∞ for any fixed positive value of b
...

2

Example 2:

show that et is not of exponential order as t → ∞
...


1
...
1

e − st F ' (t ) dt

]

∞

F (t ) 0 + s ∫

∞

0

e − st F (t )dt integrating by parts
15

= 𝑳𝑳𝑳𝑳𝑳𝑳 𝒆𝒆−𝒔𝒔𝒔𝒔 𝑭𝑭(𝒕𝒕) − 𝑭𝑭(𝟎𝟎) + 𝒔𝒔𝒔𝒔{𝑭𝑭(𝒕𝒕)}


...
14
...
𝐹𝐹𝑛𝑛−1 (0)

L{F " (t )} = s{L F ' (t )} − F ' (0) [Applying derivative Theorem]

L{F ' ' ' (t )}

=

s[s L{F (t )} − F (0)] − F ' (0)

=

s 2 L{F (t )} − sF (0) − F ' (0)

=

s L{F ' ' (t )} − F ' ' (0)

=

s s 2 L{F (t )} − s F (0) − F ' (0) − F ' ' (0)

=

s 3 L{F (t )} − s 2 F (0) − sF ' (o) − F ' ' (0)

[

]

Proceeding, we get

{

}

L F n (t ) = s n L{F (t )} – s n−1 F (0)
...
15 Initial Value Theorem:
If F(t) is continuous for all t ≥ 0 and is of exponential order as t → ∞ and if F'(t) is of Class A then

𝑡𝑡→0

s L{F (t )}

Lim

=

𝐿𝐿𝐿𝐿𝐿𝐿𝐹𝐹(𝑡𝑡)

s →∞

1
...

s →∞

4
s( s + a)

=0
By final value theorem

Lim F (t ) = Lim s
...

𝑠𝑠→0

s +1
s+2

4
𝑠𝑠(𝑠𝑠+𝑎𝑎)

=

4
𝑎𝑎

then find

Lim F (t )
t →0

Solution:

by initial value theorem

17

Lim F (t ) = Lim s
...

Note:

This theorem is applicable strictly if F(s) is proper fraction i
...
the numerator polynomial is of
lower order than the Denominator Polynomial
...
L{F (t )}

=

 2 s + 51 
Lim s
...


s →0
s  47 s + 67 

=

51
67

s →0

1
...
Find initial value of the transformed function
...
9(𝑠𝑠+1)
2
...
9
2
...
Find final value of the function
...
we can’t apply Final Value Theorem
9𝑠𝑠
𝑠𝑠(5𝑠𝑠+9)
𝑠𝑠
𝑠𝑠2 +2𝑠𝑠4

3
...


Find initial value function of 𝐹𝐹(𝑠𝑠) =

Ans
...
0
19

t

Cos at − cos bt
dt
5
...


∞

e − t − e −3 t
dt
∫0 t

6
...


Log (3)

1
...
Laplace Transform of unit step function

In engineering applications, we frequently encounter functions whose values change abruptly at specified values of
time t
...

The value of t = 0 is usually taken as a convenient time to switch on or off the given voltage
...


L{u (t − a)}

=
=
=
=

EXAMPLES

∞

∫0 𝑒𝑒 −𝑠𝑠𝑠𝑠 𝑢𝑢(𝑡𝑡 − 𝑎𝑎)𝑑𝑑𝑑𝑑
𝑎𝑎

∞

∫0 𝑒𝑒 −𝑠𝑠𝑠𝑠 𝑢𝑢(𝑡𝑡 − 𝑎𝑎)𝑑𝑑𝑑𝑑 + ∫𝑎𝑎 𝑒𝑒 −𝑠𝑠𝑠𝑠 𝑢𝑢(𝑡𝑡 − 𝑎𝑎)𝑑𝑑𝑑𝑑
∞

𝑒𝑒 −𝑠𝑠𝑠𝑠

∞

∫𝑎𝑎 𝑒𝑒 − 𝑠𝑠𝑠𝑠
...
Find unit step function of 𝐿𝐿{(𝑡𝑡 − 4)2 𝑢𝑢(𝑡𝑡 − 4)}
Solution: 𝐿𝐿{(𝑡𝑡 − 4)2 𝑢𝑢(𝑡𝑡 − 4)}

using the property of unit step function

= 𝑒𝑒 −4𝑠𝑠 L (𝑡𝑡)2
2𝑒𝑒 −4𝑠𝑠
𝑆𝑆 3

=

Example 2
...
19
...

The unit impulse function is consider as limiting form of the function
1

𝑎𝑎 ≤ 𝑡𝑡 ≤ 𝑎𝑎+∈

𝛿𝛿∈ (𝑡𝑡 − 𝑎𝑎) = � 𝜀𝜀
0,

where ∈> 0

𝑜𝑜𝑜𝑜ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒

Now Laplace Transform of unit Impulse function

L{δ∈ (t − a ) { =

∫

a +∈

a

1
e − st
...
e Laplace Transform of unit Impulse function is 1
...


F (t )

=

F (t )

=

F (t )

=

0 < t <1

 0

(t − 1)
 1


1< t < 2
t>2

(𝑡𝑡 − 1){𝑢𝑢(𝑡𝑡 − 1) − 𝑢𝑢(𝑡𝑡 − 2)} − 𝑢𝑢(𝑡𝑡 − 2)

(t − 1) u (t − 1) − (t − 2)u (t − 2)

By second shifting theorem
If

L{ F (t )

}

= f (s) then

L{F (t − a )u (t − a )} = e − as f ( s )
L{F (t )} = L(t ) =

Here

1
s2

1
...
Laplace Transform of Periodic Function:
Statement:

If F(t) is a periodic function with period T i
...

F (t + T) = F (t +2T) = F (t + 3T) = F (t + n T) = F (t) for Example if f(x)= sinx
f(x+2π)=sinx
𝐿𝐿{𝐹𝐹(𝑡𝑡)} =

𝑇𝑇
1
∫ 𝑒𝑒 −𝑠𝑠𝑠𝑠 𝐹𝐹(𝑡𝑡)𝑑𝑑𝑑𝑑
1−𝑒𝑒 −𝑠𝑠𝑠𝑠 0

Proof: by definition of Laplace Transform
∞

𝐿𝐿{𝐹𝐹(𝑡𝑡)} = ∫0 𝑒𝑒 −𝑠𝑠𝑠𝑠 𝐹𝐹(𝑡𝑡)𝑑𝑑𝑑𝑑

Since F (t) is periodic function with Periodic T, so that
𝑇𝑇

(𝑛𝑛+1)𝑇𝑇

2𝑇𝑇

𝐿𝐿{𝐹𝐹(𝑡𝑡)} = � 𝑒𝑒 −𝑠𝑠𝑠𝑠 𝐹𝐹(𝑡𝑡) 𝑑𝑑𝑑𝑑 + � 𝑒𝑒 −𝑠𝑠𝑠𝑠 𝐹𝐹(𝑡𝑡)𝑑𝑑𝑑𝑑 +
...


For Limit of x ;

If t = nT then x = 0

and if t = (n+1) T then x = T
=

𝑇𝑇

−𝑠𝑠(𝑥𝑥+𝑛𝑛𝑛𝑛)
∑∞
𝐹𝐹(𝑥𝑥 + 𝑛𝑛𝑛𝑛)𝑑𝑑𝑑𝑑
𝑛𝑛=0 ∫0 𝑒𝑒

F(n) is periodic function
...
∞

𝐿𝐿{𝐹𝐹(𝑡𝑡)} =

EXAMPLES
Example1:

Example 2:

a
= as G
...

1− r

𝑇𝑇 −𝑠𝑠𝑠𝑠
1
𝑒𝑒 𝐹𝐹(𝑡𝑡)𝑑𝑑𝑑𝑑
∫
−𝑠𝑠𝑠𝑠
0
1−𝑒𝑒

L{F (t )}

We know

L{F (t )}

Now

=

If F(t) = t2, 0 < t < 2 and F (t + 2) = F (t) then
find

Solution:

𝑇𝑇

(1 + 𝑒𝑒 −𝑠𝑠𝑠𝑠 + 𝑒𝑒 −2𝑠𝑠𝑠𝑠 +
...
t 2 dt
2

 −st t 2
 − t 1 
e
...
3 1 − e −2 s (1 + 2 s + 2 s 2 )
−2 s
s
1− e

[

]

Find the Laplace transform of a square wave Periodic function, if

 E 0F (t ) = 
− E − E < t < ω
and

Solution:

F (t + w) = F (t)

Since F(t) is a periodic function with period w
...
0dt 
∫π / ω
 ∫0


𝜋𝜋/𝜔𝜔
𝑒𝑒 −𝑠𝑠𝑠𝑠 (−𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠 𝜔𝜔𝑡𝑡−𝜔𝜔 𝑐𝑐𝑐𝑐𝑐𝑐 𝜔𝜔𝑡𝑡)
�
𝑠𝑠2 +𝜔𝜔2
0

−𝜋𝜋𝜋𝜋
(1−𝑒𝑒 𝜔𝜔 )(𝑠𝑠2 +𝜔𝜔2 )

Find Laplace Transform of the square-wave function of period a defined as

f (t )
Solution:

ω

e − st F (t )dt + ∫

Here f (t) is a Periodic function with period

L {F (t )} =

Example 4:

e −st F (t ) dt

Find Laplace Transform of the rectified semi-wave function defined by
𝑠𝑠𝑠𝑠𝑠𝑠 𝜔𝜔 𝑡𝑡,
𝑓𝑓(𝑡𝑡) = �
0

Solution:

0

𝐸𝐸
1−𝑒𝑒 −𝑠𝑠𝑠𝑠/2 +𝑒𝑒 −𝜔𝜔/2 +𝑒𝑒 −𝑠𝑠𝑠𝑠
�
�
𝑠𝑠
1−𝑒𝑒 −𝑠𝑠𝑠𝑠

=

Example 3:

ω

∫

=

L{ f (t )} =

 1 0≤t < a/2

− 1 a / 2 < t < a
1
1 − e −as

∫

=

1  a / 2 −st
e
...
Find the Laplace Transform of t 3 u (t − 3)
Ans

e −3s
(2 + 6 s + 9 s 2 )
s3

8
6

t < 2

t > 2
8 2e
Ans
...
Find Laplace Transform of unit step function f (t ) = 

for t > 5 
10
2
Ans
...
Find the Laplace Transform of unit step function of f (t ) = 

4
...
2 −
s T s (1 − e − st )
5
...


t 0 < t < 1
where

0 1 < t < 2

Ans
...
21
...
22
...
23
...
24
...
25 Second shifting Theorem:
Theorem

L−1 { f ( s )} = F (t ), then

{

}

L−1 e − as f ( s ) = G (t ) where
𝐹𝐹(𝑡𝑡 − 𝑎𝑎) 𝑡𝑡 > 𝑎𝑎
𝐺𝐺(𝑡𝑡) = �
0,
𝑡𝑡 < 𝑎𝑎
∞

Proof: f ( s ) = ∫ e −st F (t ) dt Definition
0

e − as f (s )

=

∫

∞

=

∫

∞

0

0

e − as
...
0𝑑𝑑𝑑𝑑 + ∫𝑎𝑎 𝑒𝑒 −𝑠𝑠𝑠𝑠 𝐹𝐹(𝑡𝑡 − 𝑎𝑎)𝑑𝑑𝑑𝑑 =

0

0

𝑎𝑎

e − su F (u − a )du
e − st F (t − a )dt
∞

27

∫

∞

0

e − st G (t )dt = L{G (t )}

1
...
F  du
a ∫0
a

=

1 ∞ −st  t 
e
...
s + 3 / 2 

=

3

1 − 2 t −1  1 
e L 
=
2
 s

 e −2 s
L−1  2
 s

Example 4:

Evaluate

Solution:

𝑳𝑳−𝟏𝟏 � 𝟐𝟐� = 𝒕𝒕 = 𝑭𝑭(𝒕𝒕)

=

Example 5:

Solution:

𝟏𝟏
𝒔𝒔

𝟏𝟏
𝒔𝒔

3

− t
1
e 2
2π t





𝒕𝒕 − 𝟐𝟐 𝒕𝒕 > 2
𝟎𝟎
𝒕𝒕 < 2

𝑳𝑳−𝟏𝟏 �𝒆𝒆−𝟐𝟐𝟐𝟐
...
a > 0



 s 
L−1  2
= cos h wt = F (t )
2 
 s −ω 
 s
 cos h ω (t − a) t ≥ a
by second shifting
L−1  2

...

+
=
s ( s − 3)( s + 2) 2 s 5 ( s − 3) 10 s + 2
 s 2 + 2s − 3 
L−1 

 s ( s − 3)( s + 2) 

Example 7:

Solution:

=

1 −1  1  4 −1  1  3 −1  1 
L  + L 
− L 

2  s  5  s − 3  10  s + 2 

=

1 4 3
3
+ e t − e −2t
2 5
10

If

 e −1/ s
L−1 
 s

 cos 2 t
,
 =
πt


 e−a / s
L−1 
 s





R
0

∞

t

0

0

∫ ∫e

− st


...
dt on checking the order of integration, we get

𝐿𝐿{𝜑𝜑(𝑡𝑡)}=
=
=
=

∫

∞

0

∞

∞

∞

0

u

∫ ∫

e − st F (u ) G (t − u ) dt
...
f ( s )
t

L−1 { f ( s )
...


2
( s + 3) 2 + 4

=



1
− 2 L−1 
∟
2
 ( s + 3) + 4 

=

 1 
− 2e −3t L−1  2

 s + 4

=

− 2e −3t
sin 2t
2

34

Example 3: Find inverse Laplace Transform of
Solution:

Since 𝐿𝐿−1 �

1
�
(𝑠𝑠+1)3

 1 t t 2e −t
 1
dt
L−1 
= ∫
3
 s ( s + 1)  2 0 2

=

𝑒𝑒 −𝑡𝑡 𝑡𝑡 2
∠2

1
s ( s + 1) 3

By shifting property

Division by s

[

=

1 2
t (−e −t ) − 2t (e −t ) + 2(−e t )
2

=


t2 

1 − e 1 + t + 
2


]

t
0

−t

𝑠𝑠

Example 4: Find inverse Laplace Transform of (𝑠𝑠2+𝑎𝑎2)2
Solution:

 a 
L−1  2
= Sin at
2 
s +a 

 d  a 
L − 1  2
= −t Sin at
2 
 ds  s + a 
(Inverse Laplace Transform of Derivation)
=

=

−𝟐𝟐𝟐𝟐𝟐𝟐
�
(𝒔𝒔𝟐𝟐 +𝒂𝒂𝟐𝟐 )𝟐𝟐

𝑳𝑳−𝟏𝟏 �

= −𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕


 t
s
Sin at
L−1  2
=
2 2
 ( s + a )  2a

Example 5:

Apply Heaviside Expansion theorem to obtain

Solution:

F (s) = 2s2 + 5s – 4

 2 s 2 + 5s − 4 

L−1  3
2
 s + s − 2s 

G (s) = s3 + s2 – 2s
G (s) = s (s – 1) (s + 2)
G (s) = 0 gives s = 0, 1, – 2
𝛼𝛼1 =0,

F (α1 ) = −4,

𝛼𝛼2 = 1,

α3 = – 2
F (α 2 ) = 3,

F (α 3 ) = −6
35

G '(s) = 3s2 + 2s – 2

G ' (α 2 ) = 3,

G ' (α1 ) = −2,

𝐿𝐿 − 1 �

Example 6
...
g ( s )} = F * G = ∫ F (u ) G (t − u ) du
0

Here

f (s) =

F (t ) =

F (u ) =

1
𝑠𝑠
and 𝑔𝑔(𝑠𝑠) = 2 +1
𝑠𝑠
s +4
2

1
Sin 2t and G (t ) = cos t
2
1
sin 2 u and 𝐺𝐺(𝑡𝑡 − 𝑢𝑢) = 𝑐𝑐𝑐𝑐𝑐𝑐( 𝑡𝑡 − 𝑢𝑢)
2

1

𝑡𝑡

𝐿𝐿−1 �

1 𝑡𝑡
𝑠𝑠
=
� 𝑆𝑆𝑆𝑆𝑆𝑆2𝑢𝑢
...


Using Convolution theorem evaluate:



s2
L−1  2
2
2
2 
 ( s + a )( s + b ) 

s2
s
s
= 2

...
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶(𝑡𝑡 − 𝑢𝑢)𝑑𝑑𝑑𝑑

1 𝑡𝑡
∫ 𝐶𝐶𝐶𝐶𝐶𝐶{(𝑎𝑎
2 0

− 𝑏𝑏)𝑢𝑢 + 𝑏𝑏𝑏𝑏} + 𝐶𝐶𝐶𝐶𝐶𝐶{(𝑎𝑎 + 𝑏𝑏)𝑢𝑢 − 𝑏𝑏𝑏𝑏}𝑑𝑑𝑑𝑑

1 [𝑆𝑆𝑆𝑆𝑆𝑆{(𝑎𝑎−𝑏𝑏)𝑢𝑢+𝑏𝑏𝑏𝑏
2
𝑎𝑎−𝑏𝑏

=

+

a sin at − b sin bt
a 2 − b2

=

𝑆𝑆𝑆𝑆𝑆𝑆{(𝑎𝑎+𝑏𝑏)𝑢𝑢−𝑏𝑏𝑏𝑏 𝑡𝑡
�
𝑎𝑎+𝑏𝑏
0

𝟐𝟐

Example 8
...
Find Inverse Laplace Transform of
(i)

2
tan −1  
s

(ii)

s+a
log 

 s+b

e
Sin 2t
(ii)
Ans
...
Apply Heaviside's expansion formula to
Evaluate

(i)



2s − 1
L−1 

 s ( s − 1)( s + 1) 

(ii)



1 −t 4 2 t 7 3 t
2( s 2 − 2)
L−1 
 Ans
...


1+

1 t 3 −t
e − e
2
2

3
...
Evaluate

(i)



1
L 

 ( s + a ) ( s + b) 

e − at − e −bt
Ans
...


1 t
5
e − e 2 t + e 3t
2
2

(i)

∫

∞

Cosx 2 dx

Ans
...


π
2

−1

0

0

1
...
Application to solve simple linear and simultaneous differential equations
...
31
...
The
advantage of this method is that it yield the particular solution directly without the necessity of first finding the
general solution and then evaluating the arbitrary constant
...
Take Laplace transform of both sides of the given differential equation, using initial condition
...


38

2
...


3
...
This gives y as a function of t which is the desired solution
...
𝑛𝑛−1 𝐹𝐹(0)

EXAMPLES:

Example 1:

Solve the equation

d3y
d 2 y dy
+ 2 2 − − 2 y = 0 where y = 1
dx 3
dx dx
dy
=2
dx

Solution:

𝑑𝑑 2 𝑦𝑦
𝑑𝑑𝑥𝑥 2

= 2 at t = 0

The given equation is y"' + 2y" – y' – 2y = 0

Taking Laplace transform of both sides, we get
L ( y"' ) + 2 L (y") - L ( y ' ) – 2 L (y) = 0
[s3 y– s2y (0) - s y '(0) – y'' (0)] + 2 [s2 y – sy(0) – y'(0)] – [ s y – y (0)] –2y=0
Using the given conditions

y (0) = 1, y ' (0) = 2
y ' ' (0) = 2, in equation (1)

(s3 + s2 – s – 2)

y

= s2 + 4s + 5

y

=

s 2 + 4s + 5
( s − 1) ( s + 1) ( s + 2)

y

=

5
1
1
−
+
3 ( s − 1) ( s + 1) 3 ( s + 2)

[using Partial Fractions]

Taking the inverse Laplace Transform of both sides, we get

y=

5 −1  1  −1  1  1  1

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Title: Laplace transform mathes
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