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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 a

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

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Title: Laplace transform mathes
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