This document discusses the Laplace transform, which is a mathematical technique that can be used to solve linear differential equations. It defines the Laplace transform of a function f(t) as the integral from 0 to infinity of e^-st f(t) dt, where s is a parameter. The document lists sufficient conditions for a function's Laplace transform to exist and provides several properties of the Laplace transform, including linearity, scaling properties, shifting properties, and transforms of derivatives and integrals. It gives examples to illustrate these concepts.

1. ENGINEERING MATHEMATICS -IV
SUBJECT CODE: BSH251
(Regulation 2016-17)
Common to all branches of B.E
UNIT-I
LAPLACE TRANSFORM
Engineering Maths - IV
Laplace Transform Page 1

2. Engineering Maths - IV
UNIT-V
LAPLACE TRANSFORM
1. INTRODUCTION
A transformation is an operation which converts a mathematical expression to a different but
equivalent form. The well known transformation logarithms reduce multiplication and division to a simpler
process of addition subtraction.
The Laplace transform is a powerful mathematical technique which solves linear equations
with given initial conditions by using algebra methods. The Laplace transform can also be used to solve
systems of differential equations, Partial differential equations and integral equations. In this chapter, we
will discuss about the definition, properties of Laplace transform and derive the transforms of some
functions which usually occur in the solution of linear differential equations.
2. LAPLACE TRANSFORM
Let 𝑓(𝑡) be a function of t defined for all 𝑡 ≥ 0 .then the Laplace transform of𝑓(𝑡), denoted by
𝐿[ 𝑓(𝑡)] is defined by
𝐿[ 𝑓(𝑡)] = ∫0
∞
𝑒−𝑠𝑡 𝑓(𝑡)𝑑𝑡
𝑡→∞
Provided that the integral exists, “s” is a parameter which may be real or complex. Clearly 𝐿[ 𝑓(𝑡)]is a
function of s and is briefly written as 𝐹(𝑠) (𝑖. 𝑒. ) 𝐿[ 𝑓(𝑡)] = 𝐹(𝑠)
Piecewise continuous function
A function 𝑓(𝑡) is said to be piecewise continuous is an interval 𝑎 ≤ 𝑡 ≤ 𝑏, if the interval can be sub
divided into a finite number of intervals in each of which the function is continuous and has finite right and
left hand limits.
Exponential order
A function 𝑓(𝑡) is said to be exponential order if lim 𝑒−𝑠𝑡𝑓(𝑡) is a finite quantity, where 𝑠 >
0(exists).
Example: 5. 1 Show that the function 𝒇(𝒕) = 𝒆𝒕𝟑
is not of exponential order.
Solution:
lim 𝑒−𝑠𝑡 𝑒𝑡3
=lim 𝑒−𝑠𝑡+𝑡3
= lim 𝑒𝑡3−𝑠𝑡
𝑡→∞ 𝑡→∞ 𝑡→∞
= 𝑒∞ = ∞, not a finite quantity.
Hence 𝑓(𝑡) = 𝑒𝑡3
is not of exponential order.
Sufficient conditions for the existence of the Laplace transform
The Laplace transform of 𝑓(𝑡) exists if
i)
ii)
𝑓(𝑡) is piecewise continuous in the interval 𝑎 ≤ 𝑡 ≤ 𝑏
𝑓(𝑡) is of exponential order.
Laplace Transform Page 2

3. Engineering Maths - IV
Note: The above conditions are only sufficient conditions and not a necessary condition.
Example: 5.2 Prove that Laplace transform of 𝒆𝒕𝟐
does not exist.
Solution:
lim 𝑒−𝑠𝑡 𝑒𝑡2
=lim 𝑒−𝑠𝑡+𝑡2
= lim 𝑒𝑡2−𝑠𝑡
𝑡→∞ 𝑡→∞ 𝑡→∞
= 𝑒∞ = ∞ ,not a finite quantity.
∴ 𝑒𝑡2
is not of exponential order.
Hence Laplace transform of 𝑒𝑡2
does not exist.
5.3 PROPERTIES OF LAPLACE TRANSFORM
Property: 1 Linear property
𝑳[𝒂𝒇(𝒕) ± 𝒃𝒈(𝒕)] = 𝒂𝑳[𝒇(𝒕)] ± 𝒃𝑳[𝒈(𝒕)] , where a and b are constants.
Proof:
𝐿[𝑎𝑓(𝑡) ± 𝑏𝑔(𝑡)] = ∫
∞
[𝑎𝑓(𝑡) ± 𝑏𝑔(𝑡)] 𝑒−𝑠𝑡 𝑑𝑡
0
=𝑎 ∫0
∞
𝑔(𝑡) 𝑒−𝑠𝑡𝑑𝑡
∞
𝑓(𝑡)𝑒−𝑠𝑡𝑑𝑡 ± 𝑏 ∫
0
𝐿[𝑎𝑓(𝑡) ± 𝑏𝑔(𝑡)] = 𝑎 𝐿[𝑓(𝑡)] ± 𝑏 𝐿[𝑔(𝑡)]
Property: 2 Change of scale property.
𝒂 𝒂
If 𝑳[𝒇(𝒕)] = 𝑭(𝒔), then 𝑳[𝒇(𝒂𝒕)] = 𝟏
𝑭 (𝒔
) ;𝒂 > 0
Proof:
Given 𝐿[𝑓(𝑡)] = 𝐹(𝑠)
∞
𝑒−𝑠𝑡 𝑓(𝑡) 𝑑𝑡 = 𝐹(𝑠) ⋯ ⋯ (1)
∴ ∫0
By the definition of Laplace transform, we have
𝐿[𝑓(𝑎𝑡)] = ∫0
∞
𝑒−𝑠𝑡 𝑓(𝑎𝑡) 𝑑𝑡 ⋯ ⋯ (2)
Put at= 𝑥 𝑖𝑒. , 𝑡 = 𝒙
⇒ 𝑑𝑡 = 𝑑𝑥
𝒂 𝑎
−𝒔𝒙
∞ 𝑑𝑥
(2) ⇒ 𝐿[𝑓(𝑎𝑡)] = ∫0 𝑒 𝒂 𝑓(𝑥) 𝑎
−𝒔𝒙
𝟏 ∞
= ∫ 𝑒 𝒂 𝑓(𝑥)𝑑𝑥
𝒂 0
Replace 𝑥 by t,
−𝑠
𝑡
1 ∞
𝐿[𝑓(𝑎𝑡)] = ∫ 𝑒 𝑎 𝑓(𝑡)𝑑𝑡
𝑎 0
𝟏 𝒔
𝒂 𝒂
𝑳[𝒇(𝒂𝒕)] = 𝑭 ( ) ;𝒂 > 0
Property: 3 First shifting property.
𝑳[𝒆−𝒂𝒕𝒇(𝒕)] = 𝑭(𝒔 + 𝒂)
𝑳[𝒆𝒂𝒕𝒇(𝒕)] = 𝑭(𝒔 − 𝒂)
If 𝑳[𝒇(𝒕)] = 𝑭(𝒔), then i)
ii)
Proof:
(i) 𝐿[𝑒−𝑎𝑡𝑓(𝑡)] = 𝐹(𝑠 + 𝑎)
Laplace Transform Page 3

4. Given 𝐿[𝑓(𝑡)] = 𝐹(𝑠)
∞
𝑒−𝑠𝑡 𝑓(𝑡) 𝑑𝑡 = 𝐹(𝑠) ⋯ (1)
∴ ∫0
By the definition of Laplace transform, we have
∞
𝑒−𝑠𝑡 𝑒−𝑎𝑡𝑓(𝑡) 𝑑𝑡
𝐿[𝑒−𝑎𝑡𝑓(𝑎𝑡)] = ∫
0
∞
𝑒−(𝑠+𝑎)𝑡 𝑓(𝑡) 𝑑𝑡
= ∫0
= 𝐹(𝑠 + 𝑎) by (1)
∞
𝑒−𝑠𝑡 𝑒𝑎𝑡𝑓(𝑡) 𝑑𝑡
(ii) 𝐿[𝑒𝑎𝑡𝑓(𝑎𝑡)] = ∫
0
∞
𝑒−(𝑠−𝑎)𝑡 𝑓(𝑡) 𝑑𝑡
= ∫0
= 𝐹(𝑠 − 𝑎) by (1)
Property: 4 Laplace transforms of derivatives 𝑳[𝒇′(𝒕)] = 𝒔𝑳[𝒇(𝒕)] − 𝒇(𝟎)
Proof:
∞ ∞
𝐿[𝑓′(𝑡)] = ∫ 𝑒−𝑠𝑡 𝑓′(𝑡) 𝑑𝑡 = ∫ 𝑢𝑑𝑣
0 0
0
= [𝑢𝑣]∞ − ∫ 𝑢𝑑𝑣
0
∞
𝑓(𝑡) (−𝑠)𝑒−𝑠𝑡 𝑑𝑡
= [𝑒−𝑠𝑡 𝑓(𝑡)]∞ − ∫
0
= 0 − 𝑓(0) + 𝑠𝐿[𝑓(𝑡)]
= 𝑠𝐿[𝑓(𝑡)] − 𝑓(0)
𝑳[𝒇′(𝒕)] = 𝒔𝑳[𝒇(𝒕)] − 𝒇(𝟎)
Property: 5 Laplace transform of derivative of order n
𝑳[𝒇𝒏(𝒕)] = 𝒔𝒏𝑳[𝒇(𝒕)] − 𝒔𝒏−𝟏 𝒇(𝟎) − 𝒔𝒏−𝟐𝒇′(𝟎)⋯ − 𝒔𝒏−𝟑 𝒇′′(𝟎) − ⋯ 𝒇𝒏−𝟏 (𝟎)
Proof:
We know that 𝐿[𝑓′(𝑡)] = 𝑠𝐿[𝑓(𝑡)] − 𝑓(0)⋯ ⋯ (1)
𝐿[𝑓𝑛(𝑡)] = 𝐿[[𝑓′(𝑡)]′]
= 𝑠𝐿[𝑓′(𝑡)] − 𝑓′(0)
= 𝑠[𝑠𝐿[𝑓(𝑡)] − 𝑓(0)] − 𝑓′(0)
= 𝑠2𝐿[𝑓(𝑡)] − 𝑠𝑓(0) − 𝑓′(0)
Similarly, 𝐿[𝑓′′′(𝑡)] = 𝑠3𝐿[𝑓(𝑡)] − 𝑠2𝑓(0) − 𝑠𝑓′(0) − 𝑓′′(0)
In general, 𝑳[𝒇𝒏(𝒕)] = 𝒔𝒏𝑳[𝒇(𝒕)] − 𝒔𝒏−𝟏 𝒇(𝟎) − 𝒔𝒏−𝟐𝒇′(𝟎) ⋯ − 𝒔𝒏−𝟑 𝒇′′(𝟎) − ⋯ 𝒇𝒏−𝟏 (𝟎)
Laplace transform of integrals
𝒕 𝑭(𝒔)
∫
𝟎 𝒇(𝒕)𝒅𝒕] =
𝒔
Theorem: 1 If𝑳[𝒇(𝒕)] = 𝑭(𝒔), then 𝑳 [
Proof:
𝑡
𝑓(𝑡)𝑑𝑡
Let 𝑔(𝑡) = ∫0
∴ 𝑔′(𝑡) = 𝑓(𝑡)
0
𝑓(𝑡)𝑑𝑡 = 0
And 𝑔(0) = ∫0
𝑢 = 𝑒−𝑠𝑡
∴ 𝑑𝑢 = −𝑠𝑒−𝑠𝑡𝑑𝑡
𝑑𝑣 = 𝑓′(𝑡)𝑑𝑡
∴ 𝑣 = ∫ 𝑓′(𝑡)𝑑𝑡
= 𝑓(𝑡)
Engineering Maths - IV
Laplace Transform Page 4

5. Engineering Maths - IV
Now𝐿[𝑔′(𝑡)] = 𝐿[𝑓(𝑡)]
𝑠𝐿[𝑔(𝑡)] − 𝑔(0) = 𝐿[𝑓(𝑡)]
𝑠𝐿[𝑔(𝑡)] = 𝐿[𝑓(𝑡)] ∴ 𝑔(0) = 0
𝐿[𝑔(𝑡)] = 𝐿[𝑓(𝑡)]
𝑠
𝒕 𝑭(𝒔)
∴ 𝑳 [∫
𝟎 𝒇(𝒕)𝒅𝒕] =
𝒔
𝒅𝒔
Theorem: 2 If𝑳[𝒇(𝒕)] = 𝑭(𝒔), then 𝑳[𝒕𝒇(𝒕)] = − 𝒅
𝑭(𝒔)
Proof:
Given 𝐿[𝑓(𝑡)] = 𝐹(𝑠)
∴ ∫
∞
𝑒−𝑠𝑡 𝑓(𝑡) 𝑑𝑡 = 𝐹(𝑠) ⋯ ⋯ (1)
0
Differentiating (1) with respect to s, we get
𝑑 ∞
𝑒−𝑠𝑡 𝑓(𝑡) 𝑑𝑡 = 𝑑
𝐹(𝑠)
∫
𝑑𝑠 0 𝑑𝑠
𝑑𝑠
∞ 𝜕
(𝑒−𝑠𝑡 )𝑓(𝑡) 𝑑𝑡 = 𝑑
𝐹(𝑠)
∫0 𝜕𝑠
𝑑𝑠
∞ 𝑑
(−𝑡)𝑒−𝑠𝑡 𝑓(𝑡) 𝑑𝑡 = 𝐹(𝑠)
∫0
𝒅𝒔
∞ 𝒅
𝑒−𝑠𝑡 𝑓(𝑡) 𝑑𝑡 = 𝐹(𝑠)
− ∫
0
𝑑𝑠
−𝐿[𝑡𝑓(𝑡)] = 𝑑
𝐹(𝑠)
𝒅𝒔
∴ 𝑳[𝒕𝒇(𝒕)] = − 𝒅
𝑭(𝒔)
𝒅𝒔𝒏
𝒏
Note: In general 𝑳[𝒕𝒏𝒇(𝒕)] = (−𝟏)𝒏 𝒅
𝑭(𝒔)
Example: 5.3 If 𝑳[𝒇(𝒕)] =
𝒔𝟐−𝒔+𝟏
(𝟐𝒔+𝟏)𝟐(𝒔−𝟏)
then find 𝑳[𝒇(𝟐𝒕)].
Solution:
Given 𝐿[𝑓(𝑡)] =
𝑠2−𝑠+1
(2𝑠+1)2(𝑠−1)
= 𝐹(𝑠)
2 2
𝐿[𝑓(2𝑡)] = 1
𝐹 (𝑠
)
=
1
2
𝑠 2 𝑠
( ) − +1
2
2 2
2
𝑠 𝑠
2
(2 +1) ( −1)
= 1
𝑠2 𝑠+4]
[ 4
− 2
𝑠−2
=
2 (𝑠+1)2 ( )
2
𝑠2−2𝑠+1
4(𝑠+1)2(𝑠−2)
Laplace transform of some Standard functions
Result: 1 Prove that 𝑳[𝒕𝒏] = 𝚪(𝒏+𝟏)
𝒔𝒏+𝟏
Proof:
Laplace Transform Page 5

6. ∞
We know that 𝐿[𝑓(𝑡)] = ∫ 𝑒−𝑠𝑡 𝑓(𝑡) 𝑑𝑡
0
∞
𝑒−𝑠𝑡 𝑡𝑛 𝑑𝑡
𝐿[𝑡𝑛] = ∫
0
[ 𝑛] −𝑢
∞ 𝑢 𝑛 𝑑𝑢
𝐿 𝑡 = ∫0 𝑒 (𝑠
)
𝑠
𝑠𝑛+1
∞ 𝑢𝑛
𝑒−𝑢
= ∫0 𝑑𝑢
=
1 ∞
𝑒−𝑢𝑢𝑛 𝑑𝑢
𝑠𝑛+1 ∫0
∴ 𝑳[𝒕𝒏] = 𝚪(𝒏+𝟏)
𝒔𝒏+𝟏
∞ −𝑢 𝑛
∵ ∫0 𝑒 𝑢 𝑑𝑢
Note: If n is an integer, then Γ(𝑛 + 1) = 𝑛!
n!
𝑠𝑛+1
∴ 𝐿[𝑡𝑛] = if n is an integer
If 𝑛 = 0 , then 𝐿[1] = 1
𝑠
If 𝑛 = 1 , then 𝐿[𝑡] = 1
𝑠2
Similarly 𝐿[𝑡2] = 2!
𝑠3
𝐿[𝑡3] = 3!
𝑠4
𝟏
𝒔−𝒂
Result: 2 Prove that 𝑳(𝒆𝒂𝒕) = , 𝒔 > 𝒂
Proof:
We know that 𝐿[𝑓(𝑡)] = ∫
∞
𝑒−𝑠𝑡 𝑓(𝑡) 𝑑𝑡
0
∞
𝑒−𝑠𝑡𝑒𝑎𝑡 𝑑𝑡
∴ 𝐿(𝑒𝑎𝑡) = ∫
0
∞
𝑒−𝑡(𝑠−𝑎) 𝑓(𝑡) 𝑑𝑡
= ∫0
= [ ]
−(𝑠−𝑎) 0
𝑒−𝑡(𝑠−𝑎) ∞
𝑠−𝑎
= − [0 − ( 1
)]
∴ 𝐿(𝑒𝑎𝑡) =
1
𝑠−𝑎
𝟏
𝒔+𝒂
Result: 3 Prove that 𝑳(𝒆−𝒂𝒕) = , 𝒔 > 𝒂
Proof:
We know that 𝐿[𝑓(𝑡)] = ∫
∞
𝑒−𝑠𝑡 𝑓(𝑡) 𝑑𝑡
0
∞
𝑒−𝑠𝑡𝑒−𝑎𝑡 𝑑𝑡
∴ 𝐿(𝑒−𝑎𝑡) = ∫
0
∞
𝑒−𝑡(𝑠+𝑎) 𝑓(𝑡) 𝑑𝑡
= ∫0
= [ ]
−(𝑠+𝑎) 0
𝑒−𝑡(𝑠+𝑎) ∞
𝑠+𝑎
= − [0 − ( 1
)]
∴ 𝐿(𝑒𝑎𝑡) =
1
𝑠+𝑎
Let 𝑠𝑡 = 𝑢 ⋯ ⋯ (1)
𝑡 = 𝑢
𝑠
𝑑𝑡 = 𝑑𝑢
𝑠
When 𝑡 → 0(1) => 𝑢 → 0
,
𝑡 → ∞,(1) => 𝑢 → ∞
Engineering Maths - IV
Laplace Transform Page 6

7. 𝒂
𝒔𝟐+𝒂𝟐
Result: 4 Prove that 𝑳[𝒔𝒊𝒏𝒂𝒕] =
Proof:
∞
𝑒−𝑠𝑡 𝑓(𝑡) 𝑑𝑡
We know that 𝐿[𝑓(𝑡)] = ∫0
∞
𝑒−𝑠𝑡 𝑠𝑖𝑛𝑎𝑡 𝑑𝑡
𝐿[𝑠𝑖𝑛𝑎𝑡] = ∫0
∴ 𝐿[𝑠𝑖𝑛𝑎𝑡] =
𝒂
𝒔𝟐+𝒂𝟐
, 𝑠 > |𝑎| [∵ ∫0
∞
𝑒−𝑎𝑡 𝑠𝑖𝑛𝑏𝑡 𝑑𝑡 =
𝒃
𝒂𝟐+𝒃𝟐
]
𝒔
𝒔𝟐+𝒂𝟐
Result: 5 Prove that 𝑳[𝒄𝒐𝒔𝒂𝒕] =
Proof:
∞
𝑒−𝑠𝑡 𝑓(𝑡) 𝑑𝑡
We know that 𝐿[𝑓(𝑡)] = ∫0
∞
𝑒−𝑠𝑡 𝑐𝑜𝑠𝑎𝑡 𝑑𝑡
𝐿[𝑐𝑜𝑠𝑎𝑡] = ∫0
∴ 𝐿[𝑐𝑜𝑠𝑎𝑡] =
𝑠
𝑠2+𝑎2
, 𝑠 > |𝑎| ∵ ∫0
∞
𝑒−𝑎𝑡 𝑐𝑜𝑠𝑏𝑡 𝑑𝑡 =
𝒂
𝒂𝟐+𝒃𝟐
𝒂
𝒔𝟐−𝒂𝟐
Result: 6 Prove that 𝑳[𝒔𝒊𝒏𝒉𝒂𝒕] =
Proof:
, 𝒔 > |𝒂|
2
𝑎𝑡 −𝑎𝑡
We have 𝐿[𝑠𝑖𝑛ℎ𝑎𝑡] = 𝐿 [𝑒 −𝑒
]
2
= 1
[𝐿(𝑒𝑎𝑡) − 𝐿(𝑒−𝑎𝑡)]
−
2 𝒔−𝒂 𝒔+𝒂
= 1
[ 𝟏 𝟏
]
= 1
[𝒔+𝒂−𝒔+𝒂
]
2 𝒔𝟐−𝒂𝟐
𝟐𝒂
2 𝒔𝟐−𝒂𝟐
= 1
[ ]
∴ 𝐿[𝑠𝑖𝑛ℎ𝑎𝑡] =
𝒂
𝒔𝟐−𝒂𝟐
, 𝑠 > |𝑎|
𝒔
𝒔𝟐−𝒂𝟐
Result: 7 Prove that 𝑳[𝒄𝒐𝒔𝒉𝒂𝒕] =
Proof:
, 𝒔 > |𝒂|
2
𝑎𝑡 −𝑎𝑡
We have 𝐿[𝑐𝑜𝑠ℎ𝑎𝑡] = 𝐿 [𝑒 +𝑒
]
2
= 1
[𝐿(𝑒𝑎𝑡) + 𝐿(𝑒−𝑎𝑡)]
+
2 𝒔−𝒂 𝒔+𝒂
= 1
[ 𝟏 𝟏
]
= 1
[𝒔+𝒂+𝒔−𝒂
]
2 𝒔𝟐−𝒂𝟐
𝟐𝒔
2 𝒔𝟐−𝒂𝟐
= 1
[ ]
∴ 𝐿[𝑐𝑜𝑠ℎ𝑎𝑡] =
𝒔
𝒔𝟐−𝒂𝟐
, 𝑠 > |𝑎|
𝟏
Example: 5.4 Find 𝑳 [𝒕𝟐]
Solution:
We have 𝐿[𝑡𝑛] = Γ(𝑛+1)
𝑠𝑛+1
Engineering Maths - IV
Laplace Transform Page 7

8. Put 𝑛 = 𝟏
𝟐
𝟏 Γ 𝟏
∴ 𝐿 [𝑡𝟐] = 𝟐
( +1)
𝟏
𝑠𝟐
+1
∵ Γ(𝑛 + 1) = 𝑛Γ𝑛
=
𝟏
Γ 𝟏
𝟐 𝟐
( )
𝟏
𝑠𝟐
+1 𝟐
∵ Γ (𝟏
) = √𝜋
= √𝜋
𝟑
2𝑠𝟐
𝟏
∴ 𝐿 [𝑡𝟐] =
√𝜋
2𝑠√𝑠
− 𝟏
Example: 5.5 Find the Laplace transform of 𝒕 𝟐 or
𝟏
√𝑡
Solution:
We have 𝐿[𝑡𝑛] = Γ(𝑛+1)
𝑠𝑛+1
Put 𝑛 = − 𝟏
𝟐
∴ 𝐿 [𝑡−
𝟏 𝟏
𝟐] = 𝟐
Γ(− +1)
𝟏
𝑠 𝟐
− +1
∵ Γ(𝑛 + 1) = 𝑛Γ𝑛
Γ 𝟏
= 𝟐
( )
𝟏
𝑠𝟐
𝟏
𝟐
∵ Γ ( ) = √𝜋
= √𝜋
√𝑠
∴ 𝐿 𝟏 𝜋
[ ] = √
√𝑡 𝑠
FORMULA
𝑳[𝒇(𝒕)] = 𝑭(𝒔) 𝑳[𝒇(𝒕)] = 𝑭(𝒔)
𝑳[𝟏] = 𝟏
𝒔
𝑳[𝒕] = 𝟏
𝒔𝟐
𝑳[𝒕𝒏] = 𝚪(𝒏+𝟏)
if n is not an integer
𝒔𝒏+𝟏
𝑳[𝒕𝒏] = 𝐧!
if n is an integer
𝒔𝒏+𝟏
𝑳(𝒆𝒂𝒕) = 𝟏
𝒔−𝒂
𝑳(𝒆𝒂𝒕) = 𝟏
𝒔+𝒂
𝒂
𝑳[𝒔𝒊𝒏𝒂𝒕] =
𝒔𝟐 + 𝒂𝟐
𝒔
𝑳[𝒄𝒐𝒔𝒂𝒕] =
𝒔𝟐 + 𝒂𝟐
𝒔
𝑳[𝒄𝒐𝒔𝒉𝒂𝒕] =
𝒔𝟐 − 𝒂𝟐
𝒂
𝑳[𝒔𝒊𝒏𝒉𝒂𝒕] =
𝒔𝟐 − 𝒂𝟐
Problems using Linear property
Example: 5.6 Find the Laplace transform for the following
i. 𝟑𝒕𝟐 + 𝟐𝒕 + 𝟏
ii. (𝒕 + 𝟐)𝟑
iii. 𝒂𝒕
v. 𝒔𝒊𝒏√𝟐 𝒕
vi. 𝒔𝒊𝒏(𝒂𝒕 + 𝒃)
vii. 𝒄𝒐𝒔𝟑𝟐𝒕
ix. 𝒔𝒊𝒏𝟐𝒕
x. 𝒄𝒐𝒔𝟐𝟐𝒕
xi. 𝒄𝒐𝒔𝟓𝒕𝒄𝒐𝒔𝟒𝒕
Engineering Maths - IV
Laplace Transform Page 8

9. iv. 𝒆𝟐𝒕+𝟑
viii. 𝒔𝒊𝒏𝟑𝒕
Engineering Maths - IV
Solution:
(i) Given 𝑓(𝑡) = 3𝑡2 + 2𝑡 + 1
𝐿[𝑓(𝑡)] = 𝐿[3𝑡2 + 2𝑡 + 1]
= 𝐿[3𝑡2] + 𝐿[2𝑡] + 𝐿[1]
= 𝐿[3𝑡2] + 𝐿[2𝑡] + 𝐿[1]
= 3𝐿[𝑡2] + 2𝐿[𝑡] + 𝐿[1]
𝑠3 𝑠2
= 3 2
+ 2 1
+ 1
𝑠
𝑠3 𝑠2
∴ 𝐿[3𝑡2 + 2𝑡 + 1] = 6
+ 2
+ 1
𝑠
𝑠4 𝑠3 𝑠2
(ii) Given 𝑓(𝑡) = (𝑡 + 2)3 = 𝑡3 + 3𝑡2(2) + 3𝑡22 + 23
𝐿[𝑓(𝑡)] = 𝐿[𝑡3 + 3𝑡2(2) + 3𝑡22 + 23]
= 𝐿[𝑡3] + 𝐿[6𝑡2] + 𝐿[12𝑡] + 𝐿[8]
= 𝐿[𝑡3] + 6𝐿[𝑡2] + 12𝐿[𝑡] + 8𝐿[1]
= 6
+ 12
+ 12
+ 12
𝑠
(𝑖𝑖𝑖) Given 𝑓(𝑡) = 𝑎𝑡
𝐿[𝑓(𝑡)] = 𝐿[𝑎𝑡 ] = 𝐿[𝑒𝑡𝑙𝑜𝑔𝑎 ]
𝐿[𝑎𝑡] =
1
𝑠−𝑙𝑜𝑔
𝑎
(iv)Given 𝑓(𝑡) = 𝑒2𝑡+3
𝐿[𝑓(𝑡)] = 𝐿[𝑒2𝑡+3] = 𝐿[𝑒2𝑡 . 𝑒3]
= 𝑒3𝐿[𝑒2𝑡]
𝑠−2
1
= 𝑒3 [ ]
𝑠−2
1
∴ 𝐿[𝑒2𝑡+3] = 𝑒3 [ ]
(v) 𝐿[𝑠𝑖𝑛√2𝑡] = √2
𝑠2+2
(vi)Given 𝑓(𝑡) = sin(𝑎𝑡 + 𝑏) = 𝑠𝑖𝑛𝑎𝑡𝑐𝑜𝑠𝑏 + 𝑐𝑜𝑠𝑎𝑡𝑠𝑖𝑛𝑏
𝐿[𝑓(𝑡)] = 𝐿[sin(𝑎𝑡 + 𝑏)]
= 𝐿[𝑠𝑖𝑛𝑎𝑡𝑐𝑜𝑠𝑏 + 𝑐𝑜𝑠𝑎𝑡𝑠𝑖𝑛𝑏]
= 𝑐𝑜𝑠𝑏 𝐿[sin𝑎𝑡] + 𝑠𝑖𝑛𝑏 𝐿[cos𝑎𝑡]
𝐿[sin(𝑎𝑡 + 𝑏)] = 𝑐𝑜𝑠𝑏 + 𝑠𝑖𝑛𝑏
𝑠 𝑠
𝑠2+𝑎2 𝑠2+𝑎2
4
(vii) Given 𝑓(𝑡) = cos3 2𝑡 = 1
[3𝑐𝑜𝑠2𝑡 + 𝑐𝑜𝑠6𝑡]
4
𝐿[𝑓(𝑡)] = 1
𝐿[3𝑐𝑜𝑠2𝑡 + 𝑐𝑜𝑠6𝑡]
4
= 1
[3𝐿(𝑐𝑜𝑠2𝑡) + 𝐿(𝑐𝑜𝑠6𝑡)]
= 1
[3 +
𝑠 𝑠
4 𝑠2+4 𝑠2+36
]
3𝑐𝑜𝑠𝜃 + 𝑐𝑜𝑠3𝜃
∵ cos3 𝜃 =
4
Laplace Transform Page 9

10. 𝐿[cos3 2𝑡] = 1
[3 +
𝑠 𝑠
4 𝑠2+4 𝑠2+36
]
4
(viii) Given 𝑓(𝑡) = sin3 𝑡 = 1
[3𝑠𝑖𝑛𝑡 − 𝑠𝑖𝑛3𝑡]
4
𝐿[𝑓(𝑡)] = 1
𝐿[3𝑠𝑖𝑛𝑡 − 𝑠𝑖𝑛3𝑡]
4
= 1
[3𝐿(𝑠𝑖𝑛𝑡) − 𝐿(𝑠𝑖𝑛3𝑡)]
= 1
[3 −
1 3
4 𝑠2+1 𝑠2+9
]
𝐿[sin3 𝑡] = 3
[ −
1 1
4 𝑠2+1 𝑠2+9
]
(ix) Given 𝑓(𝑡) = sin2 𝑡 = 1−𝑐𝑜𝑠2𝑡
2
2
𝐿[𝑓(𝑡)] = 𝐿 [1−𝑐𝑜𝑠2𝑡
]
2
= 1
[𝐿(1) − 𝐿(𝑐𝑜𝑠2𝑡)]
= 1
[1
−
2 𝑠 𝑠2+4
𝑠
]
𝐿[cos2 2𝑡] = 1
[1
−
2 𝑠 𝑠2+4
𝑠
]
(x) Given 𝑓(𝑡) = cos2 2𝑡 = 1+𝑐𝑜𝑠4𝑡
2
2
𝐿[𝑓(𝑡)] = 𝐿 [1+𝑐𝑜𝑠4𝑡
]
2
= 1
[𝐿(1) + 𝐿(𝑐𝑜𝑠4𝑡)]
= 1
[1
+
2 𝑠 𝑠2+16
𝑠
]
𝐿[cos2 2𝑡] = 1
[1
+
2 𝑠 𝑠2+16
𝑠
]
(xi) Given 𝑓(𝑡) = 𝑐𝑜𝑠5𝑡𝑐𝑜𝑠4𝑡
𝐿[𝑓(𝑡)] = 𝐿[𝑐𝑜𝑠5𝑡𝑐𝑜𝑠4𝑡]
2
= 1
[𝐿(𝑐𝑜𝑠9𝑡) + 𝐿(𝑐𝑜𝑠𝑡)]
= 1
[
𝑠 𝑠
2 𝑠2+81 𝑠2+1
+ ]
Problems using First Shifting theorem
𝑳[𝒆−𝒂𝒕𝒇(𝒕)] = 𝑳[𝒇(𝒕)]𝒔→𝒔+𝒂
𝑳[𝒆𝒂𝒕𝒇(𝒕)] = 𝑳[𝒇(𝒕)]𝒔→𝒔−𝒂
Example: 5.7 Find the Laplace transform for the following:
i.
ii.
𝒕𝒆−𝟑𝒕
𝒕𝟑𝒆𝟐𝒕
iii. 𝒆𝟒𝒕𝒔𝒊𝒏𝟐𝒕
iv. 𝒆−𝟓𝒕𝒄𝒐𝒔𝟑𝒕
v. 𝒔𝒊𝒏𝒉𝟐𝒕𝒄𝒐𝒔𝟑𝒕
vii. 𝒕𝟐𝟐𝒕
viii. 𝒕𝟑𝟐−𝒕
ix. 𝒆−𝟐𝒕𝒔𝒊𝒏𝟑𝒕𝒄𝒐𝒔𝟐𝒕
x. 𝒆−𝟑𝒕𝒄𝒐𝒔𝟒𝒕𝒄𝒐𝒔𝟐𝒕
xi. 𝒆𝟒𝒕𝒄𝒐𝒔𝟑𝒕𝒔𝒊𝒏𝟐𝒕
Engineering Maths - IV
Laplace Transform
Page
10

11. vi. cosh3tsin2t
(i) 𝒕𝒆−𝟑𝒕
𝐿[𝑡𝑒−3𝑡] = 𝐿[𝑡]𝑠→𝑠+3
= ( 1
)
𝑠2
𝑠→𝑠+3
∵ 𝐿(𝑡) = 1
𝑠2
∴ 𝐿[𝑡𝑒−3𝑡] =
1
(𝑠+3)2
(ii) 𝒕𝟑𝒆𝟐𝒕
𝐿[𝑡3𝑒2𝑡] = 𝐿[𝑡3]𝑠→𝑠−2
= (3!
)
𝑠4
𝑠→𝑠−2
∵ 𝐿(𝑡) =
3!
𝑠3+1
∴ 𝐿[𝑡3𝑒2𝑡] =
6
(𝑠−2)4
(iii) 𝒆𝟒𝒕𝒔𝒊𝒏𝟐𝒕
𝐿[𝑒4𝑡𝑠𝑖𝑛2𝑡] = 𝐿[𝑠𝑖𝑛2𝑡]𝑠→𝑠−4
2
= ( )
=
𝑠2+22
𝑠→𝑠−4
2
=
(𝑠−4)2+4
2
𝑠2−8𝑠+16+4
∴ 𝐿[𝑒4𝑡𝑠𝑖𝑛2𝑡] =
2
𝑠2−8𝑠+20
(iv) 𝑳[𝒆−𝟓𝒕𝒄𝒐𝒔𝟑𝒕]
𝐿[𝑒−5𝑡𝑐𝑜𝑠3𝑡] = 𝐿[𝑐𝑜𝑠3𝑡]𝑠→𝑠+5
𝑠
= ( )
=
𝑠2+32
𝑠→𝑠+5
𝑠+5
=
(𝑠+5)2+9
𝑠+5
𝑠2+10𝑠+25+9
𝑠+5
𝑠2+10𝑠+34
∴ 𝐿[𝑒−5𝑡𝑐𝑜𝑠3𝑡] =
(v) 𝑳[𝒔𝒊𝒏𝒉𝟐𝒕𝒄𝒐𝒔𝟑𝒕]
2
2𝑡 −2𝑡
𝐿[𝑠𝑖𝑛ℎ2𝑡𝑐𝑜𝑠3𝑡] = 𝐿 [(𝑒 −𝑒
) 𝑐𝑜𝑠3𝑡]
2
= 1
[𝐿(𝑒2𝑡𝑐𝑜𝑠3𝑡) − 𝐿(𝑒−2𝑡𝑐𝑜𝑠3𝑡)]
2 𝑠→𝑠−2
= 1
[𝐿(𝑐𝑜𝑠3𝑡) − 𝐿(𝑐𝑜𝑠3𝑡)𝑠→𝑠+2]
2
= 1
[( )
𝑠 𝑠
− ( )
𝑠2+32
𝑠→𝑠−2 𝑠2+32
𝑠→𝑠+2
]
∴ 𝐿[𝑠𝑖𝑛ℎ2𝑡𝑐𝑜𝑠3𝑡] = 1
[
𝑠−2 𝑠+2
2 (𝑠−2)2+9 (𝑠+2)2+9
− ]
(vi) 𝑳[𝒄𝒐𝒔𝒉𝟑𝒕𝒔𝒊𝒏𝟐𝒕]
2
3𝑡 −3𝑡
𝐿[𝑐𝑜𝑠ℎ3𝑡𝑠𝑖𝑛2𝑡] = 𝐿 [(𝑒 +𝑒
) 𝑠𝑖𝑛2𝑡]
Engineering Maths - IV
Laplace Transform
Page
11

12. 2
= 1
[𝐿(𝑒3𝑡𝑠𝑖𝑛2𝑡) + 𝐿(𝑒−3𝑡𝑠𝑖𝑛2𝑡)]
2 𝑠→𝑠−3
= 1
[𝐿(𝑠𝑖𝑛2𝑡) + 𝐿(𝑠𝑖𝑛2𝑡)𝑠→𝑠+3]
2
= 1
[( )
2 2
+ ( )
𝑠2+22
𝑠→𝑠−3 𝑠2+22
𝑠→𝑠+3
]
∴ 𝐿[𝑐𝑜𝑠ℎ3𝑡𝑠𝑖𝑛2𝑡] = 1
[
2 2
2 (𝑠−3)2+4 (𝑠+3)2+4
+ ]
(vii) 𝒕𝟐𝟐𝒕
𝐿[𝑡22𝑡] = 𝐿[𝑡2𝑒𝑙𝑜𝑔2𝑡
]
= 𝐿[𝑡2𝑒𝑡𝑙𝑜𝑔2] = 𝐿[𝑡2]𝑠→𝑠−𝑙𝑜𝑔2
= (2!
)
𝑠3
𝑠→𝑠−𝑙𝑜𝑔2
2
=
(𝑠−𝑙𝑜𝑔2)3
∴ 𝐿[𝑡22𝑡] =
2
(𝑠−𝑙𝑜𝑔2)3
(viii) 𝒕𝟑𝟐−𝒕
𝐿[𝑡32−𝑡] = 𝐿[𝑡3𝑒𝑙𝑜𝑔2−𝑡
]
= 𝐿[𝑡3𝑒−𝑡𝑙𝑜𝑔2] = 𝐿[𝑡3]𝑠→𝑠+𝑙𝑜𝑔2
= (3!
)
=
𝑠4
𝑠→𝑠+𝑙𝑜𝑔2
6
∴ 𝐿[𝑡32−𝑡] =
(𝑠+𝑙𝑜𝑔2)4
6
(𝑠+𝑙𝑜𝑔2)4
(ix) 𝑳[𝒆−𝟐𝒕𝒔𝒊𝒏𝟑𝒕𝒄𝒐𝒔𝟐𝒕]
𝐿[𝑒−2𝑡𝑠𝑖𝑛3𝑡𝑐𝑜𝑠2𝑡] = 𝐿[𝑠𝑖𝑛3𝑡𝑐𝑜𝑠2𝑡]𝑠→𝑠+2
2
= 1
𝐿[sin(3𝑡 + 2𝑡) + sin(3𝑡 − 2𝑡)]𝑠→𝑠+2
2
= 1
𝐿[sin 5𝑡 + sin 𝑡]𝑠→𝑠+2
2
= 1
[L(sin 5𝑡) + 𝐿(sin 𝑡)]𝑠→𝑠+2
= 1
[
2 𝑠2+52
5 1
+ ]
𝑠2+12
𝑠→𝑠+2
= 1
[ +
5 1
2 (𝑠+2)2+25 (𝑠+2)2+1
]
∴ 𝐿[𝑒−2𝑡𝑠𝑖𝑛3𝑡𝑐𝑜𝑠2𝑡] == 1
[ +
5 1
2 (𝑠+2)2+25 (𝑠+2)2+1
]
(x) 𝑳[𝒆−𝟑𝒕𝒄𝒐𝒔𝟒𝒕𝒄𝒐𝒔𝟐𝒕]
𝐿[𝑒−3𝑡𝑐𝑜𝑠4𝑡𝑐𝑜𝑠2𝑡] = 𝐿[𝑐𝑜𝑠4𝑡𝑐𝑜𝑠2𝑡]𝑠→𝑠+3
2
= 1
𝐿[cos(4𝑡 + 2𝑡) + cos(4𝑡 − 2𝑡)]𝑠→𝑠+3
2
= 1
𝐿[cos6t + cos2𝑡]𝑠→𝑠+3
Engineering Maths - IV
Laplace Transform
Page
12

13. Engineering Mathematics - II
2
= 1
[L(cos6𝑡) + 𝐿(cos2 𝑡)]𝑠→𝑠+3
= 1
[
2 𝑠2+62
𝑠 𝑠
+ ]
𝑠2+22
𝑠→𝑠+3
= 1
[ +
𝑠+3 𝑠+3
2 (𝑠+3)2+36 (𝑠+3)2+4
]
∴ 𝐿[𝑒−3𝑡𝑐𝑜𝑠4𝑡𝑐𝑜𝑠2𝑡] = 1
[
𝑠+3 𝑠+3
2 (𝑠+3)2+36 (𝑠+3)2+4
+ ]
(xi) 𝑳[𝒆𝟒𝒕𝒄𝒐𝒔𝟑𝒕𝒔𝒊𝒏𝟐𝒕]
𝐿[𝑒4𝑡𝑐𝑜𝑠3𝑡𝑠𝑖𝑛2𝑡] = 𝐿[𝑐𝑜𝑠3𝑡𝑠𝑖𝑛2𝑡]𝑠→𝑠−4
2
= 1
𝐿[sin(3𝑡 + 2𝑡) − sin(3𝑡 − 2𝑡)]𝑠→𝑠−4
2
= 1
𝐿[sin 5𝑡 − sin 𝑡]𝑠→𝑠−4
2
= 1
[L(sin 5𝑡) − 𝐿(sin 𝑡)]𝑠→𝑠−4
= 1
[
2 𝑠2+52
5 1
− ]
𝑠2+12
𝑠→𝑠−4
= 1
[ +
5 1
2 (𝑠−4)2+25 (𝑠−4)2+1
]
∴ 𝐿[𝑒4𝑡𝑐𝑜𝑠3𝑡𝑠𝑖𝑛2𝑡] = 1
[
5 1
2 (𝑠−4)2+25 (𝑠−4)2+1
+ ]
Exercise: 5.1
Find the Laplace transform for the following
1. cos2 3𝑡 Ans: 𝟏
[ 𝟑𝒔
+
𝒔
𝟒 𝒔𝟐+𝟗 𝒔𝟐+𝟖𝟏
]
2. 𝑠𝑖𝑛3𝑡𝑐𝑜𝑠4𝑡 Ans: 𝟏
[
𝟕 𝟏
− ]
3. 𝑡𝑒2𝑡 Ans:
𝟒 𝒔𝟐+𝟒𝟗 𝒔𝟐+𝟏
𝟏
4. 𝑡4𝑒−3𝑡
Ans:
(𝒔−𝟐)𝟐
𝟒!
5. 𝑒4𝑡𝑠𝑖𝑛2𝑡 Ans:
(𝒔−𝟑)𝟓
𝟐
6. 𝑒−5𝑡𝑐𝑜𝑠3𝑡 Ans:
(𝒔−𝟒)𝟐+𝟒
𝒔+𝟓
7. 𝑡33𝑡 Ans:
(𝒔+𝟓)𝟐+𝟗
𝟑!
8. 𝑡54−𝑡 Ans:
(𝒔−𝒍𝒐𝒈𝟑)𝟒
𝟓!
(𝒔+𝒍𝒐𝒈𝟒)𝟔
9. 𝑒−2𝑡𝑠𝑖𝑛3𝑡𝑐𝑜𝑠2𝑡 Ans: +
𝟓 𝟏
(𝒔+𝟐)𝟐+𝟐𝟓 (𝒔+𝟐)𝟐+𝟏
10. 𝑒−3𝑡𝑐𝑜𝑠4𝑡𝑐𝑜𝑠2𝑡 Ans: +
𝒔+𝟑 𝒔+𝟑
(𝒔+𝟑)𝟐+𝟑𝟔 (𝒔+𝟑)𝟐+𝟒
11. 𝑠𝑖𝑛ℎ𝑡𝑠𝑖𝑛4𝑡 Ans: −
𝟒 𝟒
(𝒔−𝟏)𝟐+𝟏𝟔 (𝒔+𝟏)𝟐+𝟏𝟔
12. 𝑐𝑜𝑠ℎ2𝑡𝑐𝑜𝑠2𝑡 Ans: 1
[ −
𝑠−2 𝑠+2
2 (𝑠−2)2+4 (𝑠+2)2+4
]
Laplace Transform
Page
13

14. Engineering Maths - IV
5.4 LAPLACE TRANSFORM OF DERIVATIVES AND INTEGRALS
Problems using the formula
𝑳[𝒕𝒇(𝒕)] = −𝒅
𝑳[𝒇(𝒕)]
𝒅𝒔
Example: 5.8 Find the Laplace transform for 𝒕𝒔𝒊𝒏𝟒𝒕
Solution:
𝑑𝑠
𝐿[𝑡𝑠𝑖𝑛4𝑡] = −𝑑
𝐿[𝑡𝑠𝑖𝑛4𝑡]
= −𝑑
[ 4
𝑑𝑠 𝑠2+4
]
2
= −[(𝑠 +16)0−4(2𝑠)]
(𝑠2+16)2
∴ 𝐿[𝑡𝑠𝑖𝑛4𝑡] =
8𝑠
(𝑠2+16)2
Example: 5.9 Find 𝑳[𝒕𝒔𝒊𝒏𝟐𝒕]
Solution:
𝐿[𝑡𝑠𝑖𝑛2𝑡] = −𝑑
𝐿[sin2 𝑡] = −𝑑
𝐿 [(1−𝑐𝑜𝑠2𝑡)
]
𝑑𝑠 𝑑𝑠 2
2 𝑑𝑠
= − 1 𝑑
[𝐿(1) − 𝐿(𝑐𝑜𝑠2𝑡)]
𝒔
2 𝑑𝑠 𝑠 𝒔𝟐+𝟒
= − 1 𝑑
[1
− ]
𝟐 𝟐
= − 1 𝑑
[𝒔 +𝟒−𝒔
]
2 𝑑𝑠 𝒔(𝒔𝟐+𝟒)
= − 1 𝑑
[
2 𝑑𝑠 𝒔(𝒔𝟐+𝟒)
𝟒
]
= − 4 𝑑
[
2 𝑑𝑠 𝒔(𝒔𝟐+𝟒)
𝟏
]
𝟐
= −2 [𝟎−(𝟑𝒔 +𝟒)
]
(𝒔𝟑+𝟒𝒔)
𝟐
𝟐
∴ 𝐿[𝑡𝑠𝑖𝑛2𝑡] = 𝟐(𝟑𝒔 +𝟒)
(𝒔𝟑+𝟒𝒔)
𝟐
Example: 5.10 Find 𝑳[𝒕𝒄𝒐𝒔𝟐𝟐𝒕]
Solution:
𝐿[𝑐𝑜𝑠22𝑡] = −𝑑
𝐿[cos2 2𝑡] = −𝑑
𝐿 [(1+𝑐𝑜𝑠4𝑡)
]
𝑑𝑠 𝑑𝑠 2
2 𝑑𝑠
= − 1 𝑑
[𝐿(1) + 𝐿(𝑐𝑜𝑠4𝑡)]
= − 1 𝑑
[1
+
2 𝑑𝑠 𝑠 𝒔𝟐+𝟏𝟔
𝒔
]
𝑠2
2 (𝒔𝟐+𝟏𝟔)𝟐
𝟐
= − 1
[− 1
+ (𝒔 +𝟏𝟔)𝟏−𝒔.𝟐𝒔
]
𝑠2
2 (𝒔𝟐+𝟏𝟔)𝟐
𝟐 𝟐
= − 1
[− 1
+ 𝒔 +𝟏𝟔−𝟐𝒔
]
∴ 𝐿[𝑐𝑜𝑠22𝑡] = 1
[ 1
−
2 𝑠2 (𝒔𝟐+𝟏𝟔)𝟐
𝟏𝟔−𝒔𝟐
]
Laplace Transform
Page
14

15. Engineering Maths - IV
Example: 5.11 Find the Laplace transform for 𝒕𝒔𝒊𝒏𝒉𝟐𝒕
Solution:
𝑑𝑠
𝐿[𝑠𝑖𝑛ℎ2𝑡] = −𝑑
𝐿[𝑠𝑖𝑛ℎ2𝑡]
= −𝑑
[ 2
𝑑𝑠 𝑠2−4
]
2
= −[(𝑠 −4)0−2(2𝑠)]
(𝑠2−4)2
∴ 𝐿[𝑡𝑠𝑖𝑛ℎ2𝑡] =
4𝑠
(𝑠2−4)2
Example: 5.12 Find the Laplace transform for 𝒇(𝒕) = 𝒔𝒊𝒏𝒂𝒕 − 𝒂𝒕𝒄𝒐𝒔𝒂𝒕
Solution:
𝐿[𝑠𝑖𝑛𝑎𝑡 − 𝑎𝑡𝑐𝑜𝑠𝑎𝑡] = L(sin 𝑎𝑡) − 𝑎 L(tcosat)
=
𝑎
𝑑𝑠
− 𝑎 (−𝑑
𝐿[𝑐𝑜𝑠𝑎𝑡])
=
𝑠2+𝑎2
𝑎 𝑠
𝑠2+𝑎2 𝑑𝑠 𝑠2+𝑎2
+ 𝑎 𝑑
[ ]
=
=
2 2
𝑎
+ 𝑎 [(𝑠 +𝑎 )1−𝑠(2𝑠)
]
𝑠2+𝑎2 (𝑠2+𝑎2)2
𝑎
+ 𝑎 [𝑠2+𝑎2−𝑠2
]
𝑠2+𝑎2 (𝑠2+𝑎2)2
=
𝑎
𝑠2+𝑎2 (𝑠2+𝑎2)2
2 2
+ 𝑎 [ 𝑎 −𝑠
]
2 2 2 2
= 𝑎(𝑠 +𝑎 )+𝑎(𝑎 −𝑠 )
(𝑠2+𝑎2)2
2 3 3 2
= 𝑎𝑠 +𝑎 +𝑎 −𝑎𝑠 )
(𝑠2+𝑎2)2
∴ 𝐿[𝑠𝑖𝑛𝑎𝑡 − 𝑎𝑡𝑐𝑜𝑠𝑎𝑡] =
2𝑎3
(𝑠2+𝑎2)2
Example: 5.13 Find the Laplace transform for the following
(i) 𝒕𝒆−𝟑𝒕𝒔𝒊𝒏𝟐𝒕 (ii) 𝒕𝒆−𝒕𝒄𝒐𝒔𝒂𝒕 9iii) 𝒕𝒔𝒊𝒏𝒉𝒕𝒄𝒐𝒔𝟐𝒕
Solution:
𝑑𝑠
(i) 𝐿[𝑡𝑒−3𝑡𝑠𝑖𝑛2𝑡] = 𝐿[𝑡𝑠𝑖𝑛2𝑡] = −𝑑
𝐿[𝑠𝑖𝑛2𝑡]
𝑠→𝑠+3 𝑠→𝑠+3
𝑑𝑠
= −𝑑
( 2
)
𝑠2+22
𝑠→𝑠+3
(𝑠2+4)2
2
= [(𝑠 +4)0−2(2𝑠)
]
𝑠→𝑠+3
= [
4𝑠
]
(𝑠2+4)2
𝑠→𝑠+3
∴ 𝐿[𝑡𝑒−3𝑡 𝑠𝑖𝑛2𝑡] =
4(𝑠+3)
((𝑠+3)2+4 )2
𝑠→𝑠+1 𝑑𝑠
(ii) 𝐿[𝑡𝑒−𝑡𝑐𝑜𝑠𝑎𝑡] = 𝐿[𝑡𝑐𝑜𝑠𝑎𝑡] = −𝑑
𝐿[𝑐𝑜𝑠𝑎𝑡]𝑠→𝑠+1
𝑑𝑠
= −𝑑
( 𝑠
)
𝑠2+𝑎2
𝑠→𝑠+1
Laplace Transform
Page
15

16. (𝑠2+𝑎2)2
Engineering Maths - IV
2 2
= − [(𝑠 +𝑎 )1−𝑠(2𝑠)
]
𝑠→𝑠+1
2 2
= − [ 𝑎 −𝑠
]
(𝑠2+𝑎2)2
𝑠→𝑠+1
2 2
= [ 𝑠 −𝑎
]
(𝑠2+𝑎2)2
𝑠→𝑠+1
∴ 𝐿[𝑡𝑒−𝑡𝑐𝑜𝑠𝑎𝑡] =
(𝑠+1)2−𝑎2
((𝑠+1)2+𝑎2 )2
(iii) 𝐿[𝑡𝑠𝑖𝑛ℎ𝑡𝑐𝑜𝑠2𝑡]
2
𝑡 −𝑡
𝐿[𝑡𝑠𝑖𝑛ℎ𝑡𝑐𝑜𝑠2𝑡] = 𝐿 [𝑡 (𝑒 −𝑒
) 𝑐𝑜𝑠2𝑡]
2
= 1
[𝐿(𝑡𝑒𝑡𝑐𝑜𝑠2𝑡) − 𝐿(𝑡𝑒−𝑡𝑐𝑜𝑠2𝑡)]
2 𝑑𝑠 𝑑𝑠
= 1
[−𝑑
𝐿[𝑐𝑜𝑠2𝑡] + 𝑑
𝐿[𝑐𝑜𝑠2𝑡]
𝑠→𝑠−1 𝑠→𝑠+1]
2 𝑑𝑠
= 1
[−𝑑
( 𝑠
𝑠2+4 𝑠→𝑠−1 𝑑𝑠
𝑠
𝑠2+4 𝑠→𝑠+1
) + 𝑑
( ) ]
2 (𝑠2+4)2
2
= 1
[− [(𝑠 +4)1−𝑠(2𝑠)
]
𝑠→𝑠−1 (𝑠2+4)2
2
+ [(𝑠 +4)1−𝑠(2𝑠)
]
𝑠→𝑠+1
]
2
= 1
[− [ 4−𝑠
]
2
+ [ 4−𝑠
]
2 (𝑠2+4)2
𝑠→𝑠−1 (𝑠2+4)2
𝑠→𝑠+1
]
2
∴ 𝐿[𝑡𝑠𝑖𝑛ℎ𝑡𝑐𝑜𝑠2𝑡] = 1
[ (𝑠−1) −4
+
4−(𝑠+1)2
2 ((𝑠−1)2+4 )2 ((𝑠+1)2+4 )2
]
Problems using the formula
𝟐
𝑳[𝒕𝟐𝒇(𝒕)] = 𝒅
𝑳[𝒇(𝒕)]
𝒅𝒔𝟐
Example: 5.14 Find the Laplace transform for (i)𝒕𝟐𝒔𝒊𝒏𝒕 (ii) 𝒕𝟐𝒄𝒐𝒔𝟐𝒕
Solution:
𝑑𝑠2
2
(i) 𝐿[𝑡2𝑠𝑖𝑛𝑡] = 𝑑
𝐿[𝑠𝑖𝑛𝑡]
1
𝑑𝑠2 𝑠2+1
2
= 𝑑
[ ]
2
= 𝑑
([(𝑠 +1)0−1(2𝑠)]
)
𝑑𝑠 (𝑠2+1)2
= 𝑑
(
𝑑𝑠 (𝑠2+1)2
−2𝑠
)
= −2 𝑑
(
𝑑𝑠 (𝑠2+1)2
𝑠
)
=
2
2 2
−2[(𝑠 +1) (1)−𝑠(2)(𝑠 +1)(2𝑠)]
(𝑠2+1)4
2 2 2
= −2(𝑠 +1)[(𝑠 +1)−4𝑠 ]
(𝑠2+1)4
2
= −2[1−3𝑠 ]
(𝑠2+1)3
2
∴ 𝐿[𝑡2𝑠𝑖𝑛𝑡] = 6𝑠 −2
(𝑠2+1)3
Laplace Transform
Page
16

17. 𝑑𝑠2
2
(ii) 𝐿[𝑡2𝑐𝑜𝑠2𝑡] = 𝑑
𝐿[𝑐𝑜𝑠2𝑡]
𝑠
2
= 𝑑
[ ]
= (
𝑑𝑠2 𝑠2+4
2
𝑑 [(𝑠 +4)1−𝑠(2𝑠)]
𝑑𝑠 (𝑠2+4)2
)
2
= 𝑑
( 4−𝑠
)
𝑑𝑠 (𝑠2+4)2
=
2 2
( 2 2
[(𝑠 +4) −2𝑠)−(4−𝑠 )2(𝑠 +4)(2𝑠)]
(𝑠2+4)4
2 2 2
= 2𝑠(𝑠 +4)[(𝑠 +4)(−1)−(4−𝑠 )2]
(𝑠2+4)4
2
= 2𝑠[𝑠 −12]
(𝑠2+4)3
2
∴ 𝐿[𝑡2 cos2𝑡] = 2𝑠[𝑠 −12]
(𝑠2+4)3
Example: 5.15 Find the Laplace transform for (i)𝒕𝟐𝒆−𝟐𝒕𝒄𝒐𝒔𝒕 (ii) 𝒕𝟐𝒆𝟒𝒕𝒔𝒊𝒏𝟑𝒕
Solution:
(i) 𝐿[𝑡2𝑒−2𝑡𝑐𝑜𝑠𝑡] = 𝐿[𝑡2𝑐𝑜𝑠𝑡]𝑠→𝑠+2 𝑑𝑠2
2
= 𝑑
𝐿[𝑐𝑜𝑠𝑡]𝑠→𝑠+2
𝑑𝑠2
𝑠
2
= 𝑑
( )
𝑠2+1 𝑠→𝑠+2
𝑑𝑠 (𝑠2+1)2
2
= 𝑑
[(𝑠 +1)1−𝑠(2𝑠)
]
𝑠→𝑠+2
2
= 𝑑
[ 1−𝑠
]
𝑑𝑠 (𝑠2+1)2
𝑠→𝑠+2
= [
2 2
( 2 2
[(𝑠 +1) −2𝑠)−(1−𝑠 )2(𝑠 +1)(2𝑠)]
(𝑠2+1)4
]
𝑠→𝑠+2
(𝑠2+1)4
2 2
= (𝑠2 + 1)[[(𝑠 +1)(−2𝑠)−4𝑠(1−𝑠 )]
]
𝑠→𝑠+2
(𝑠2+1)3
3 3
= [−2𝑠 −2𝑠−4𝑠+4𝑠
]
𝑠→𝑠+2
(𝑠2+1)3
3
= [2𝑠 −6𝑠
]
𝑠→𝑠+2
∴
3
𝐿[𝑡2𝑒−2𝑡𝑐𝑜𝑠𝑡] = 2(𝑠+2) −6(𝑠+2)
((𝑠+2)2+1 )3
(ii)𝐿[𝑡2𝑒4𝑡𝑠𝑖𝑛3𝑡] = 𝐿[𝑡2𝑠𝑖𝑛3𝑡] 𝑠→𝑠−4 𝑑𝑠2
2
= 𝑑
𝐿[𝑠𝑖𝑛3𝑡]𝑠→𝑠−4
𝑑𝑠2
3
2
= 𝑑
( )
𝑠2+9 𝑠→𝑠−4
𝑑𝑠 (𝑠2+9)2
2
= 𝑑
[(𝑠 +9)0−3(2𝑠)
]
= 𝑑
[
−6𝑠
]
𝑑𝑠 (𝑠2+9)2
𝑠→𝑠−4
𝑠→𝑠−4
= −6 𝑑
[
𝑠
]
𝑑𝑠 (𝑠2+9)2
𝑠→𝑠−4
= −6 [
2
2
[(𝑠 +9) ( ) 2
1 −(𝑠)2(𝑠 +9)(2𝑠)]
(𝑠2+9)4
]
𝑠→𝑠−4
Engineering Maths - IV
Laplace Transform
Page
17

18. (𝑠2+9)4
Engineering Mathematics - II
2 2
= −6(𝑠2 + 9)[[(𝑠 +9)−4𝑠 ]
]
𝑠→𝑠−4
(𝑠2+9)3
2
= −6 [ 9−3𝑠
]
𝑠→𝑠−4
(𝑠2+9)3
2
= [18𝑠 −54
]
𝑠→𝑠−4
2
∴ 𝐿[𝑡2𝑒4𝑡𝑠𝑖𝑛3𝑡] = 18(𝑠−4) −54
((𝑠−4)2+9 )3
Exercise: 5.2
Find the Laplace transform for the following
2𝑎𝑠
(𝑠2+𝑎2)2
𝑠2−𝑎2
(𝑠2+𝑎2)2
6(𝑠+4)
(𝑠+4)2+9
8𝑠
(𝑠2+64)2
−
4𝑠
(𝑠2+16)2
(𝑠−2)2−4
((𝑠+4)2+4)2
1. 𝑡𝑠𝑖𝑛𝑎𝑡 Ans:
2. 𝑡𝑐𝑜𝑠𝑎𝑡 Ans:
3. 𝑡𝑒−4𝑡𝑠𝑖𝑛3𝑡 Ans:
4. 𝑡𝑐𝑜𝑠2𝑡𝑠𝑖𝑛6𝑡 Ans:
5. 𝑡𝑒−2𝑡𝑐𝑜𝑠2𝑡 Ans:
Problems using the formula
𝑳 [𝒇(𝒕)
] =
𝒕
∞
∫𝒔 𝑳[𝒇(𝒕)]𝒅𝒔
𝑡→0 𝑡
This formula is valid if lim 𝑓(𝑡)
is finite.
The following formula is very useful in this section
𝑠
𝑑𝑠
∫ = 𝑙𝑜𝑔𝑠
𝑠+𝑎
𝑑𝑠
∫ = log(𝑠 + 𝑎)
𝑠 𝑑𝑠 1
∫ = log(𝑠2 + 𝑎2)
𝑠2+𝑎2 2
𝑎 𝑑𝑠 𝑠
∫ = tan−1
𝑠2+𝑎2 𝑎
𝒕
Example: 5.16 Find 𝑳 [𝒄𝒐𝒔𝒂𝒕
]
𝑡→0 𝑡 0 0
Solution:
lim 𝑐𝑜𝑠𝑎𝑡
= 𝑐𝑜𝑠𝑎(0)
= 1
= ∞
∴ Laplace transform does not exists.
𝒕
Example: 5.17 Find 𝑳 [𝒔𝒊𝒏𝒂𝒕
]
𝑡→0
Solution:
lim 𝑠𝑖𝑛𝑎𝑡
= 𝑠𝑖𝑛𝑎(0)
= 0
𝑡 0 0
= lim𝑎𝑐𝑜𝑠𝑎𝑡
𝑡→0
(by applying L−Hospital rule)
Laplace Transform
Page
18

19. Engineering Maths - IV
lim𝑎𝑐𝑜𝑠𝑎𝑡 = 𝑎𝑐𝑜𝑠0 = 𝑎, finite quantity.
𝑡→0
Hence Laplace transform exists
𝑡
𝐿 [𝑠𝑖𝑛𝑎𝑡
] =
∞
𝐿[(𝑠𝑖𝑛𝑎𝑡)]𝑑𝑠
∫𝑠
∞ 𝑎
= ∫𝑠 𝑠2+𝑎2
𝑑𝑠
= [tan ]
𝑎 𝑠
−1 𝑠 ∞
𝑎
= [tan−1∞ − tan−1 𝑠
]
= [𝜋
− tan−1 𝑠
]
2 𝑎
∴ 𝐿 [𝑠𝑖𝑛𝑎𝑡
] = 𝑐𝑜t−1 𝑠
𝑡 𝑎
𝒕
𝟑
Example: 5.18 Find 𝑳 [𝒔𝒊𝒏 𝒕
]
Solution:
𝑡→0 𝑡→0
𝒔𝒊𝒏𝟑𝒕
= 𝟑𝒔𝒊𝒏𝒕−𝒔𝒊𝒏𝟑𝒕
𝒕 𝟒𝒕
𝒔𝒊𝒏𝟑𝒕 𝟑𝒔𝒊𝒏𝒕−𝒔𝒊𝒏𝟑𝒕
lim = lim
𝒕 𝟒𝒕
= 0−0
= 0
0 0
(by applying L−Hospital rule)
𝑡→0 4𝑡
= lim 3𝑠𝑖𝑛𝑡−𝑠𝑖𝑛3𝑡
= 0
𝑡
Hence Laplace transform exists
3
𝐿 [𝑠𝑖𝑛 𝑡
] = 𝐿 [3𝑠𝑖𝑛𝑡 −𝑠𝑖𝑛3𝑡
]
= 1
4𝑡
∞
𝐿[(3𝑠𝑖𝑛𝑡 − 𝑠𝑖𝑛3𝑡)]𝑑𝑠
∫
4 𝑠
𝑠2+1 𝑠2+9
= 1
∫
∞
(3 1
− 3
) 𝑑𝑠
4 𝑠
1
[ −1 −1 𝑠
= 3tan 𝑠 − tan ]
4 3 𝑠
∞
= 1
[3(tan−1 ∞ − tan−1 𝑠) − (tan−1 ∞ − tan−1 𝑠
)]
4 3
= 1
[(𝜋
− tan−1 𝑠) − (𝜋
− tan−1 𝑠
) ]
4 2 2 3
= 1
[cot−1 𝑠 − 𝑐𝑜t−1 𝑠
]
4 3
𝒕
Example: 5.19 Find 𝑳 [𝒆−𝟐𝒕 𝒔𝒊𝒏𝟐𝒕𝒄𝒐𝒔𝟑𝒕
]
𝒕 𝒕
Solution:
𝐿 [𝑒−2𝑡 𝒔𝒊𝒏𝟐𝒕𝒄𝒐𝒔𝟑𝒕
] = 𝐿 [𝒔𝒊𝒏𝟐𝒕𝒄𝒐𝒔𝟑𝒕
]
𝑠→𝑠+2
= 1
[∫
∞
𝐿(sin(3𝑡 + 2𝑡) − sin(3𝑡 − 2𝑡))𝑑𝑠]
2 𝑠 𝑠→𝑠+2
= 1
[∫
∞
𝐿((sin 5𝑡) − 𝐿(sin 𝑡))𝑑𝑠]
2 𝑠 𝑠→𝑠+2
Laplace Transform
Page
19

20. −
2 𝑠2+52 𝑠2+12
1
∞
[ 5
= 1
[ ∫
𝑠
]]
𝑠→𝑠+2
2 5
−1 𝑠 −1
𝑠
∞
= 1
[[tan −tan 𝑠] ]
𝑠→𝑠+2
= 1
[[(tan−1 ∞ − tan−1 𝑠
) − (tan−1 ∞ − tan−1 𝑠)] ]
2 5 𝑠→𝑠+2
2 2 5 2
= 1
[(𝜋
− tan−1 𝑠
) − (𝜋
− tan−1 𝑠) ]
𝑠→𝑠+2
= 1
[cot−1 𝑠
− 𝑐𝑜t−1 𝑠 ]
2 5 𝑠→𝑠+2
= 1
[cot−1 (𝑠+2)
− 𝑐𝑜t−1(𝑠 + 2)]
2 5
−𝒂𝒕 −𝒃𝒕
Example: 5.20 Find the Laplace transform for 𝒆 −𝒆
𝒕
Solution:
𝑡→0 𝑡 𝑡→0 0 0 0
−𝑎𝑡 −𝑏𝑡 0 0
lim𝑒 −𝑒
= lim 𝑒 −𝑒
= 1−1
= 0
(use L− Hospital rule)
𝑡→0
−𝑎𝑡 −𝑏𝑡
= lim −𝑎𝑒 +𝑏𝑒
1
= −𝑎 + 𝑏 = 𝑏 − 𝑎 = a finite quantity
Hence Laplace transform exists.
𝑡
−𝑎𝑡 −𝑏𝑡
𝐿 [𝑒 −𝑒
] =
∞ −𝑎𝑡 −𝑏𝑡
∫𝑠 𝐿[𝑒 − 𝑒 ]𝑑𝑠
∞ −𝑎𝑡 −𝑏𝑡
= ∫𝑠
[𝐿(𝑒 ) − 𝐿(𝑒 )]𝑑𝑠
𝑠+𝑎 𝑠+𝑏
∞ 1 1
= ∫
𝑠 ( − ) 𝑑𝑠
𝑠
= [log(𝑠 + 𝑎) − log(𝑠 + 𝑏)]∞
= [𝑙𝑜𝑔 ]
𝑠+𝑏 𝑠
𝑠+𝑎 ∞
= [𝑙𝑜𝑔
𝑎
𝑠
𝑠(1+ )
𝑏 ]
𝑠(1+𝑠
)
𝑠
∞
= 𝑙𝑜𝑔1 − 𝑙𝑜𝑔 𝑠+𝑎
= 0 − 𝑙𝑜𝑔 𝑠+𝑎
𝑠+𝑏 𝑠+𝑏
∵ 𝑙𝑜𝑔1 = 0
= 𝑙𝑜𝑔 𝑠+𝑎
𝑠+𝑏
Example: 5.21 Find the Laplace transform of
Solution:
𝟏−𝒄𝒐𝒔𝒕
𝒕
𝑡→0
lim 𝟏−𝒄𝒐𝒔𝒕
= 0
𝒕 0 𝑡→0 𝟏 1
lim 𝒔𝒊𝒏𝒕
= 0
= 0 (use L− Hospital rule)
𝑡
𝐿 [1−𝑐𝑜𝑠𝑡
] exists.
𝐿 [1−𝑐𝑜𝑠𝑡
] =
𝑡
∞
∫𝑠 𝐿[(1 − 𝑐𝑜𝑠𝑡)]𝑑𝑠
𝑠
𝑠 𝑠2+1
∞ 1
( − ) 𝑑𝑠
= ∫𝑠
Engineering Maths - IV
Laplace Transform
Page
20

21. 1
2
2
= [log𝑠 − log(𝑠 + 1)]
𝑠
∞
𝑠
∞
= [log𝑠 − log√𝑠2 + 1]
𝑠
= [𝑙𝑜𝑔 ]
√𝑠2+1 𝑠
∞
= 0 − 𝑙𝑜𝑔
𝑠
√𝑠2+1
= 𝑙𝑜𝑔 √𝑠2+1
𝑠
Example: 5.22 Find the Laplace transform for 𝒄𝒐𝒔𝒂𝒕−𝒄𝒐𝒔𝒃𝒕
𝒕
Solution:
𝑡→0
lim 𝑐𝑜𝑠𝑎𝑡−𝑐𝑜𝑠𝑏𝑡
= 1−1
=
0 𝑡 0 0
(use L− Hospital rule)
𝑡→0 1
= lim −𝑎𝑠𝑖𝑛𝑎𝑡+𝑏𝑠𝑖𝑛𝑏𝑡
= 0 = a finite quantity
Hence Laplace transform exists.
𝑡
𝐿 [𝑐𝑜𝑠𝑎𝑡−𝑐𝑜𝑠𝑏𝑡
]
=
∞
∫𝑠 𝐿[𝑐𝑜𝑠𝑎𝑡 − 𝑐𝑜𝑠𝑏𝑡]𝑑𝑠
∞
[𝐿(𝑐𝑜𝑠𝑎𝑡) − 𝐿(𝑐𝑜𝑠𝑏𝑡)]𝑑𝑠
= ∫𝑠
( −
𝑠2+𝑎2 𝑠2+𝑏2
𝑠
) 𝑑𝑠
∞ 𝑠
= ∫𝑠
[1
2
( 2 2) 1
2
2 2
= log 𝑠 + 𝑎 − log(𝑠 + 𝑏 )]
𝑠
∞
= [𝑙𝑜𝑔 ]
2 𝑠2+𝑏2
𝑠
1 𝑠2+𝑎2 ∞
2
2 𝑎2
𝑠2
𝑠2(1+𝑏2
)
𝑠 (1+ )
= 1
[𝑙𝑜𝑔 𝑠2
]
𝑠
∞
2
𝑎2
𝑏2
(1+ )
= 1
[𝑙𝑜𝑔 𝑠2
]
(1+ )
𝑠2
𝑠
∞
2 2 2 2
= 1
[𝑙𝑜𝑔1 − 𝑙𝑜𝑔 𝑠 +𝑎
] = − 1
[𝑙𝑜𝑔 𝑠 +𝑎
]
2 𝑠2+𝑏2 2 𝑠2+𝑏2
[∵ 𝑙𝑜𝑔1 = 0]
2 2
= 1
[𝑙𝑜𝑔 𝑠 +𝑏
]
2 𝑠2+𝑎2
Example: 5.23 Find the Laplace transform of
Solution:
𝒔𝒊𝒏𝟐𝒕
𝒕
𝑡→0
𝑠𝑖𝑛2𝑡
= 1−𝑐𝑜𝑠2𝑡
𝑡 2𝑡
lim 𝟏−𝒄𝒐𝒔𝟐𝒕
= 0
𝟐𝒕 0
𝑡→0 𝟐 1
lim 𝟐𝒔𝒊𝒏𝟐𝒕
= 0
= 0 (use L− Hospital rule)
Laplace transform exists.
Engineering Maths - IV
Laplace Transform
Page
21

22. 𝑡 2𝑡
2
𝐿 [𝑠𝑖𝑛 𝑡
] = 𝐿 [1−𝑐𝑜𝑠2𝑡
] = 1 ∞
∫ 𝐿[(1 − 𝑐𝑜𝑠2𝑡)]𝑑𝑠
2 𝑠
= 1 ∞
[𝐿(1) − 𝐿(𝑐𝑜𝑠2𝑡)]𝑑𝑠
∫
2 𝑠
𝑠 𝑠2+4
= 1 ∞
(1
− 𝑠
) 𝑑𝑠
∫
2 𝑠
1
[ 1
2 2
2
= log𝑠 − log(𝑠 + 4)]
𝑠
∞
2 𝑠
= 1
[log𝑠 − log√𝑠2 + 4]
∞
2
1
[
= 𝑙𝑜𝑔
𝑠
2
√𝑠 +4
∞
]
𝑠
2
= 1
[0 − 𝑙𝑜𝑔
𝑠
√𝑠2+4
]
= 1
𝑙𝑜𝑔 √𝑠2+4
2 𝑠
Example: 5.24 Find the Laplace transform for 𝒔𝒊𝒏𝟐𝒕𝒔𝒊𝒏𝟓𝒕
𝒕
Solution:
𝑡
𝐿 [𝑠𝑖𝑛2𝑡𝑠𝑖𝑛5𝑡
] =
∞
𝐿[𝑠𝑖𝑛2𝑡𝑠𝑖𝑛5𝑡]𝑑𝑠
∫𝑠
2
∞ 1
[𝐿(cos(−3𝑡) − 𝐿(𝑐𝑜𝑠7𝑡)]𝑑𝑠
= ∫𝑠
= 1 ∞
∫ [𝐿(cos(3𝑡) − 𝐿(𝑐𝑜𝑠7𝑡)]𝑑𝑠
2 𝑠
[∵ cos(−𝜃) = 𝑐𝑜𝑠𝜃]
𝑠
𝑠2+9 𝑠2+49
( − ) 𝑑𝑠
= 1 ∞ 𝑠
∫
2 𝑠
1
[1
2 2
( 2 )
1
2
2
= log 𝑠 + 9 − log(𝑠 + 49)]
𝑠
∞
= [𝑙𝑜𝑔 ]
4 𝑠2+49 𝑠
1 𝑠2+9 ∞
1
4
= [𝑙𝑜𝑔
2 9
𝑠2
𝑠 (1+ )
2 49
𝑠2
𝑠 (1+ )
]
𝑠
∞
9
𝑠2
4 (1+49
)
= 1
[𝑙𝑜𝑔
(1+𝑠2)
]
𝑠
∞
2 2
= 1
[𝑙𝑜𝑔1 − 𝑙𝑜𝑔 𝑠 +9
] = − 1
[𝑙𝑜𝑔 𝑠 +9
]
4 𝑠2+49 4 𝑠2+49
[∵ 𝑙𝑜𝑔1 = 0]
2
= 1
[𝑙𝑜𝑔 𝑠 +49
]
4 𝑠2+9
Problems using 𝑳 [
𝒕 1
𝑠
∫𝟎 𝒇(𝒕)𝒅𝒕] = 𝑳[𝒇(𝒕)]
Example: 5.25 Find the Laplace transform for (i)
𝒕 −𝟐𝒕
∫𝟎 𝒆 𝒅𝒕 (ii)
𝒕
∫𝟎𝒄𝒐𝒔𝟐𝒕𝒅𝒕
(iii) ∫𝟎
𝒕
𝒕𝒔𝒊𝒏𝟑𝒕𝒅𝒕 (iv) 𝒕
𝒕
∫
𝟎 𝒄𝒐𝒔𝒕𝒅𝒕
Solution:
(i) 𝐿 [∫
𝑡
𝑒−2𝑡𝑑𝑡] = 1
𝐿[𝑒−2𝑡] = 1
( 1
)
0 𝑠 𝑠 𝑠+2
Engineering Maths - IV
Laplace Transform
Page
22

23. ∴ 𝐿 [∫
0
𝑡
𝑒−2𝑡𝑑𝑡] =
1
𝑠(𝑠+2)
(ii) 𝐿 [∫
𝑡
𝑐𝑜𝑠2𝑡𝑑𝑡] = 1
𝐿[𝑐𝑜𝑠2𝑡] = 1
(
0 𝑠 𝑠2+4
𝑠
)
∴ 𝐿 [∫
0
𝑠
𝑡
𝑐𝑜𝑠2𝑡𝑑𝑡] =
1
𝑠2+4
(iii) 𝐿 [
𝑠
𝑡 1
∫0 𝑡𝑠𝑖𝑛3𝑡𝑑𝑡] = 𝐿[𝑡𝑠𝑖𝑛3𝑡]
𝑠 𝑑𝑠
= 1
[−𝑑
[𝐿[𝑠𝑖𝑛3𝑡]]]
= −1
[ 𝑑
[ 3
𝑠 𝑑𝑠 𝑠2+9
]]
= −1
[ −6𝑠
𝑠 (𝑠2+9)2
]
𝑡
∴ 𝐿 [∫
0 𝑡𝑠𝑖𝑛3𝑡𝑑𝑡] =
6
(𝑠2+9)2
(iv) 𝐿 [𝑡
𝑑𝑠
𝑡 −𝑑 𝑡
∫0 𝑐𝑜𝑠𝑡𝑑𝑡] = 𝐿 [∫0𝑐𝑜𝑠𝑡𝑑𝑡]
= −𝑑
[1
( 𝑠
𝑑𝑠 𝑠 𝑠2+1
)]
1
𝑑𝑠 𝑠2+1
= − 𝑑
[ ]
−2𝑠
= − [ ]
∴ 𝐿 [∫
0
𝑡
𝑡𝑠𝑖𝑛3𝑡𝑑𝑡] =
(𝑠2+1)2
2𝑠
(𝑠2+1)2
𝒕
𝒕𝒄𝒐𝒔𝟒𝒕𝒅𝒕
Example: 5.26 Find the Laplace transform for 𝒆−𝒕 ∫
𝟎
Solution:
𝐿 [𝑒−𝑡 ∫
0
𝑡
𝑡𝑐𝑜𝑠4𝑡𝑑𝑡] = 𝐿 [
𝑠 𝑑𝑠
𝑠→𝑠+1 𝑠→𝑠+1
= − (1 𝑑
𝑡 −1 𝑑
∫0 𝑡𝑐𝑜𝑠4𝑡𝑑𝑡] = [ 𝐿(𝑐𝑜𝑠4𝑡)]
𝑠
)
𝑠 𝑑𝑠 𝑠2+16 𝑠→𝑠+1
𝑠 (𝑠2+16)2
2
= [−1 (𝑠 +16)1−𝑠(2𝑠)
]
𝑠→𝑠+1
𝑠 (𝑠2+16)2
2 2
= [−1 (𝑠 +16−2𝑠 )
]
𝑠→𝑠+1
2
= [−1 (−𝑠 +16)
]
𝑠 (𝑠2+16)2
𝑠→𝑠+1
2
= [1 (𝑠 −16)
]
𝑠 (𝑠2+16)2
𝑠→𝑠+1
∴ 𝐿 [𝑒−𝑡 ∫
0
𝑡
𝑡𝑐𝑜𝑠4𝑡𝑑𝑡] =
𝑠+1 ((𝑠+1)2+16 )2
2
1
[ (𝑠+1) −16
]
𝒕
𝒕 𝒔𝒊𝒏𝒕
𝒅𝒕
Example: 5.27 Find the Laplace transform of 𝒆−𝒕 ∫
𝟎
Solution:
𝐿 [𝑒−𝑡 ∫
0 𝑡 𝑡
∫0
𝑡 𝑠𝑖𝑛𝑡
𝑑𝑡] = 𝐿 [ 𝑡 𝑠𝑖𝑛𝑡
𝑑𝑡]
𝑠→𝑠+1
Engineering Maths - IV
Laplace Transform
Page
23

24. Engineering Mathematics - II
𝑠 𝑡
= [1
𝐿 (𝑠𝑖𝑛𝑡
)]
𝑠→𝑠+1
= [1 ∫
𝑠 𝑠
∞
𝐿(𝑠𝑖𝑛𝑡)𝑑𝑠]
𝑠→𝑠+1
∞
1
= [1
∫
𝑠 𝑠
]
𝑠2+1 𝑠→𝑠+1
𝑠 𝑠
= [1
[tan−1 𝑠 ]∞ ]
𝑠→𝑠+1
𝑠
= [1
(tan−1 ∞ − tan−1 𝑠) ]
𝑠→𝑠+1
𝑠 2
= [1
(𝜋
− tan−1 𝑠) ]
𝑠→𝑠+1
𝑠
= [1
cot−1 𝑠 ]
𝑠→𝑠+1
𝑡
∴ 𝐿 [𝑒−𝑡 ∫
0
𝑡 𝑠𝑖𝑛𝑡
𝑑𝑡] = 1
𝑠+1
𝑐𝑜t−1(𝑠 + 1)
Exercise: 5.3
Find the Laplace transform of
1. 𝑠𝑖𝑛
𝑡 𝑡
Ans: cot−1 𝑠
2
2. 𝑒−2𝑡 𝑠𝑖𝑛𝑡
𝑡
Ans: 𝑐𝑜t−1(𝑠 + 2)
3. 𝑠𝑖𝑛𝑎𝑡−𝑠𝑖𝑛𝑏𝑡
𝑡
Ans:
−𝑎𝑡
4. 𝑒 −𝑐𝑜𝑠𝑏𝑡
𝑡
Ans:
cot−1 𝑠
− cot−1 𝑠
𝑎 𝑏
𝑙𝑜𝑔 √𝑠2+𝑏2
𝑠+𝑎
−𝑡
5. 1−𝑒
𝑡
Ans: 𝑙𝑜𝑔 s+1
𝑠
𝑡
𝑡 𝑠𝑖𝑛𝑡
𝑑𝑡
6. 𝑒−𝑡 ∫
0
Ans:
1
𝑠+1
𝑐𝑜t−1(𝑠 + 1)
7. 𝑒−𝑡 ∫
𝑡
𝑡𝑐𝑜𝑠𝑡𝑑𝑡
0
Ans: [
1 𝑠2+2𝑠
𝑠+1 (𝑠2+2𝑠+2 )2
]
𝑡
𝑡𝑒−𝑡𝑠𝑖𝑛𝑡𝑑𝑡
8. 𝑒−𝑡 ∫
0 𝑠 𝑠2+2𝑠+2
Ans: 1
[ 2(𝑠+1)
]
∞
𝑓(𝑡)𝑒−𝑠𝑡𝑑𝑡 = 𝐿[𝑓(𝑡)]
Evaluation of integrals using Laplace transform
Note: (i) ∫
0
(ii) ∫0
∞
𝑓(𝑡)𝑒−𝑎𝑡 𝑑𝑡 = [𝐿[𝑓(𝑡)]]
𝑠=𝑎
(iii) ∫0
∞
𝑓(𝑡)𝑑𝑡 = [𝐿[𝑓(𝑡)]]
𝑠=0
𝒔𝟐+𝟒
Example: 5.28 If 𝑳[𝒇(𝒕)] = 𝒔+𝟐
, then find the value of
∞
∫𝟎 𝒇(𝒕)𝒅𝒕
Solution:
Given 𝐿[𝑓(𝑡)] = 𝑠+2
𝑠2+4
We know that ∫0
∞
𝑓(𝑡)𝑑𝑡 = [𝐿[𝑓(𝑡)]]
𝑠=0
𝑠+2
= [ ] = 2
𝑠2+4 𝑠=0 4
Laplace Transform
Page
24

25. 2
∞ 1
∫0 𝑓(𝑡)𝑑𝑡 =
𝒔𝟐−𝟗
∞
𝒆−𝟐𝒕𝒇(𝒕)𝒅𝒕
Example: 5.29 If 𝑳[𝒇(𝒕)] = 𝟓𝒔+𝟒
, then find the value of ∫
𝟎
Solution:
Given 𝐿[𝑓(𝑡)] = 5𝑠+4
𝑠2−9
We know that ∫0
∞
𝑒−2𝑡 𝑓(𝑡)𝑑𝑡 = [𝐿[𝑓(𝑡)]]
𝑠=2
= [5𝑠+4
] = 14
𝑠2−9 𝑠=2 −5
5
∞ −14
𝑒−2𝑡𝑓(𝑡)𝑑𝑡 =
∴ ∫0
Example: 5.30 Find the values of the following integrals using Laplace transforms:
(i) ∫𝟎
∞
𝒕𝒆−𝟐𝒕𝒄𝒐𝒔𝟐𝒕𝒅𝒕 (ii) ∫𝟎
∞
𝒕𝟐𝒆−𝒕𝒔𝒊𝒏𝒕𝒅𝒕 (iii)
𝒕
∞ 𝒆−𝒕−𝒆−𝟐𝒕
( ) 𝒅𝒕
∫𝟎
𝒕
(iv) ∫𝟎
∞ 𝟏−𝒄𝒐𝒔𝒕
( ) 𝒆−𝒕𝒅𝒕 (v)
𝒕
∞ 𝒆−𝒂𝒕−𝒄𝒐𝒔𝒃𝒕
∫
𝟎 ( ) 𝒅𝒕
Solution:
(i) ∫
∞
𝑡𝑒−2𝑡𝑐𝑜𝑠2𝑡𝑑𝑡 = 𝐿[𝑡𝑐𝑜𝑠2𝑡]
0 𝑠=2 𝑑𝑠
= [−𝑑
𝐿(𝑐𝑜𝑠2𝑡)]
𝑠=2
= −𝑑
( 𝑠
)
𝑑𝑠 𝑠2+4 𝑠=2
(𝑠2+4)2
2
= − [ (𝑠 +4)1−𝑠(2𝑠)
]
𝑠=2
2
= − [ (4−𝑠 )
]
(𝑠2+4)2
𝑠=2
= − (4−4)
= 0
(4+4)2
(ii) 𝑠=1 𝑑𝑠2
∞ 2 −𝑡 2 𝑑2
∫0 𝑡 𝑒 𝑠𝑖𝑛𝑡𝑑𝑡 = 𝐿[𝑡 𝑠𝑖𝑛𝑡] = 𝐿[𝑠𝑖𝑛𝑡]𝑠=1
1
2
= 𝑑
( )
𝑑𝑠2 𝑠2+1 𝑠=1
= 𝑑
[ −1(2𝑠)
]
𝑑𝑠 (𝑠2+1)2
𝑠=1
= −2 𝑑
[
𝑠
]
𝑑𝑠 (𝑠2+1)2
𝑠=1
2
[(𝑠 +1)
= −2 [
2
2
(1)−𝑠.2(𝑠 +1)(2𝑠))]
(𝑠2+1)4
]
𝑠=1
= −2 [
[(𝑠2+1)[(𝑠2+1)−4𝑠2)]]
(𝑠2+1)4
]
𝑠=1
(𝑠2+1)3
2
= −2 [(1−3𝑠 )
]
𝑠=1
(𝑠2+1)3
3
= [ 6𝑠 −2
]
𝑠=1
= 4
= 1
(iii) ∫0 𝑡 𝑡
∞ 𝑒−𝑡−𝑒−2𝑡 −𝑡 −2𝑡
( ) 𝑑𝑡 = 𝐿 [𝑒 −𝑒
]
𝑠=0
𝑠=0
8 2
= ∫
∞
[𝐿[𝑒−𝑡 − 𝑒−2𝑡]𝑑𝑠]
𝑠
Engineering Maths - IV
Laplace Transform
Page
25

26. 𝑠=0
∞
= ∫ [[𝐿(𝑒−𝑡) − 𝐿(𝑒−2𝑡)]𝑑𝑠]
𝑠
𝑠+1 𝑠+2
− 1
) 𝑑𝑠]
𝑠=0
∞
[( 1
= ∫𝑠
= {[log(𝑠 + 1) − log(𝑠 + 2)]∞}
𝑠 𝑠
=0
𝑠+2 𝑠
𝑠+1 ∞
= {[𝑙𝑜𝑔 ] }
𝑠=0
= {𝑙𝑜𝑔 𝑠
𝑠(1+ )
2
𝑠(1+𝑠
)
𝑠
1 ∞
}
𝑠=0
= [0 − 𝑙𝑜𝑔 𝑠+1
]
𝑠+2 𝑠=0
∵ 𝑙𝑜𝑔1 = 0
= [𝑙𝑜𝑔 𝑠+2
] = 𝑙𝑜𝑔2
𝑠+1 𝑠=0
(iv)
𝑡
∞ 1−𝑐𝑜𝑠𝑡 −𝑡
∫0 ( ) 𝑒 𝑑𝑡
∫0 𝑡 𝑡
∞ 1−𝑐𝑜𝑠𝑡 1−𝑐𝑜𝑠𝑡
( ) 𝑒−𝑡𝑑𝑡 = 𝐿 [ ]
𝑆=1
𝑆=1
∞
[𝐿[(1 − 𝑐𝑜𝑠𝑡)]𝑑𝑠]
= ∫𝑠
𝑆=1
∞
[[𝐿(1) − 𝐿(𝑐𝑜𝑠𝑡)]𝑑𝑠]
= ∫𝑠
𝑠 𝑠2+1 𝑆=1
∞ 1 𝑠
= ∫
𝑠 [( − ) 𝑑𝑠]
2 𝑠
∞
= {[log𝑠 − 1
log(𝑠2 + 1)] }
𝑆=1
𝑠
∞
= {[log𝑠 − log√𝑠2 + 1] }
𝑆=1
= {[𝑙𝑜𝑔
𝑠
√𝑠2+1 𝑠
∞
] }
𝑆=1
= [0 − 𝑙𝑜𝑔
𝑠
]
√𝑠2+1 𝑠=1
𝑠
= [𝑙𝑜𝑔 √𝑠2+1
]
𝑠=1
= 𝑙𝑜𝑔√2
𝑡
−𝑎𝑡
(v)
∞
(𝑒 −𝑐𝑜𝑠𝑏𝑡
) 𝑑𝑡
∫0
𝑡 𝑡
∞ 𝑒−𝑎𝑡−𝑐𝑜𝑠𝑏𝑡 𝑒−𝑎𝑡−𝑐𝑜𝑠𝑏𝑡
∫0 ( ) 𝑑𝑡 = 𝐿 [ ]
𝑆=0
𝑆=0
∞
[𝐿[(𝑒−𝑎𝑡 − 𝑐𝑜𝑠𝑏𝑡)]𝑑𝑠]
= ∫𝑠
𝑆=0
∞
[[𝐿(𝑒−𝑎𝑡) − 𝐿(𝑐𝑜𝑠𝑏𝑡)]𝑑𝑠]
= ∫𝑠
−
𝑠+𝑎 𝑠2+𝑏2
𝑠
) 𝑑𝑠]
𝑆=0
∞
[( 1
= ∫𝑠
1
2
2 2
= {[log(𝑠 + 𝑎) − log(𝑠 + 𝑏 )]
𝑠
∞
}
𝑆=0
2
𝑠
2
∞
= {[log(𝑠 + 𝑎) − log√𝑠 + 𝑏 ] }
𝑆=0
𝑠+𝑎
𝑠
∞
= {[𝑙𝑜𝑔 ] }
√𝑠2+𝑏2
𝑆=0
Engineering Maths - IV
Laplace Transform
Page
26

27. = [0 − 𝑙𝑜𝑔
𝑠+𝑎
]
√𝑠2+𝑏2
𝑠=0
𝑠+𝑎
= [𝑙𝑜𝑔 √𝑠2+𝑏2
]
𝑠=0
√𝑏2
= 𝑙𝑜𝑔
𝑎
= 𝑙𝑜𝑔 𝑏
𝑎
Exercise: 5.4
Find the values of the following integrals using Laplace transforms
1. ∫0
∞
𝑡𝑒−2𝑡𝑐𝑜𝑠𝑡𝑑𝑡 Ans:
3
25
13
250
2. ∫0
∞
𝑡𝑒−3𝑡𝑠𝑖𝑛𝑡𝑑𝑡 Ans:
𝑡
3. ∫0
∞ 𝑒−𝑎𝑡−𝑒−𝑏𝑡
( ) 𝑑𝑡 Ans: 𝑙𝑜𝑔 𝑏
𝑎
𝑡
4. ∫0
∞ sin2 𝑡
𝑒−2𝑡 𝑑𝑡 Ans:
4
1
𝑙𝑜𝑔2
5. ∫0
∞ 𝑐𝑜𝑠𝑎𝑡−𝑐𝑜𝑠𝑏𝑡
( ) 𝑑𝑡 Ans: 𝑙𝑜𝑔 𝑎
𝑡 𝑏
∞
𝒇(𝒕)𝒆−𝒔𝒕𝒅𝒕 = 𝑳[𝒇(𝒕)]
Laplace transform of Piecewise continuous functions
∫𝟎
Example: 5.31 Find the Laplace transform of 𝒇(𝒕) = {
𝒆−𝒕; 𝟎 < 𝑡 < 𝜋
𝟎 ; 𝒕 > 𝜋
Solution:
∞
𝑓(𝑡)𝑒−𝑠𝑡𝑑𝑡
𝐿[𝑓(𝑡)] = ∫0
= ∫0
𝜋
𝑒−𝑠𝑡𝑒−𝑡𝑑𝑡 +
∞
𝑒−𝑠𝑡 0𝑑𝑡
∫𝜋
𝜋
𝑒−(𝑠+1)𝑡𝑑𝑡
= ∫0
= [ ] =
𝑒−(𝑠+1)𝑡 𝜋 𝑒−(𝑠+1)𝜋−𝑒0
−(𝑠+1) 0 −(𝑠+1)
−(𝑠+1)𝜋
∴ 𝐿[𝑓(𝑡)] = 1−𝑒
−(𝑠+1)
Example: 5.32 Find the Laplace transform of 𝒇(𝒕) = {
𝒔𝒊𝒏𝒕; 𝟎 < 𝑡 < 𝜋
𝟎 ; 𝒕 > 𝜋
Solution:
𝐿[𝑓(𝑡)] =
∞ −𝑠𝑡
∫0 𝑓(𝑡)𝑒 𝑑𝑡
= ∫𝟎
∞
𝑒−𝑠𝑡0𝑑𝑡
𝝅
𝑒−𝑠𝑡𝑠𝑖𝑛𝑡𝑑𝑡 + ∫
𝜋
𝜋
𝑒−𝑠𝑡𝑠𝑖𝑛𝑡𝑑𝑡
= ∫0
𝒆−𝒔𝒕
𝟎
𝝅 𝑒−𝑠𝜋
[
= [ (−𝑠𝑠𝑖𝑛𝑡 − 𝑐𝑜𝑠𝑡)] = −𝑠𝑠𝑖𝑛𝜋 − 𝑐𝑜𝑠𝜋] −
(−𝒔)𝟐+𝟏 𝑠2+1
𝑒0
2
𝑠 +1
[−𝑠𝑠𝑖𝑛0 − 𝑐𝑜𝑠0]
1
−𝑠𝜋 −𝑠𝜋
= 𝑒
(0 + 1) − (−1) = 𝑒 +1
𝑠2+1 𝑠2+1 𝑠2+1
Engineering Maths - IV
Laplace Transform
Page
27

28. −𝒔𝝅
∴ 𝐿[𝑓(𝑡)] = 𝒆 +𝟏
𝒔𝟐+𝟏
( )
Example: 5.33 Find the Laplace transform of 𝒇 𝒕 = {
𝒕;𝟎 < 𝑡 < 1
𝟎 ; 𝒕 > 1
Solution:
∞
𝑓(𝑡)𝑒−𝑠𝑡𝑑𝑡
𝐿[𝑓(𝑡)] = ∫0
1 −𝑠𝑡
= ∫0 𝑒 𝑡𝑑𝑡 +
∞ −𝑠𝑡
∫1 𝑒 0𝑑𝑡
1
𝑡𝑒−𝑠𝑡𝑑𝑡
= ∫0
0
𝑒−𝑠𝑡 𝑒−𝑠𝑡 1 𝒆−𝒔 𝒆−𝒔 𝟏
= [𝑡 − (1) ] = − − 0 +
−𝑠 (−𝑠)2 −𝒔 𝒔𝟐 𝒔𝟐
𝒔
−𝒔 −𝒔
∴ 𝐿[𝑓(𝑡)] = − 𝒆
− 𝒆
+ 𝟏
𝒔𝟐 𝒔𝟐
Exercise: 5.5
( )
1. Find the Laplace transform of 𝑓 𝑡 = {
0; 0 < 𝑡 < 2
3 ; 𝑡 > 2
Ans:
𝟑𝒆−𝟐𝒔
𝒔
( )
2. Find the Laplace transform of 𝑓 𝑡 = {
𝑒𝑡 ; 0 < 𝑡 < 1
0 ; 𝑡 > 1
Ans:
𝟏−𝒆−(𝒔−𝟏)
𝒔−𝟏
( )
3. Find the Laplace transform of 𝑓 𝑡 = {
1; 0 < 𝑡 < 1
0 ; 𝑡 > 1
Ans:
𝟏−𝒆−𝒔
𝒔
Unit step function
The unit step function 𝑈(𝑡 − 𝑎) is defined as 𝑈(𝑡 − 𝑎) = {0; 𝑡 < 𝑎
0 ; 𝑡 > 𝑎
Example: 5.34 Find the Laplace transform of unit step functions.
Solution:
∞
𝑈(𝑡 − 𝑎)𝑒−𝑠𝑡𝑑𝑡
𝐿[𝑈(𝑡 − 𝑎)] = ∫0
∞
𝑒−𝑠𝑡𝑑𝑡
∞
(1)𝑒−𝑠𝑡𝑑𝑡 = ∫
𝑎
𝑎
= ∫00𝑑𝑡 + ∫𝑎
= [
−𝑠
𝑒−𝑠𝑡 ∞
] = 0 −
𝑎
𝑒−𝑠𝑎
−𝑠
=
𝑒−𝑠𝑎
𝑠
−𝑠𝑎
𝐿[𝑈(𝑡 − 𝑎)] = 𝑒
𝑠
Second Shifting theorem
Statement: If 𝐿[𝑓(𝑡)] = 𝐹(𝑠), then 𝐿[𝑓(𝑡 − 𝑎)𝑈(𝑡 − 𝑎)] = 𝑒−𝑎𝑠𝐹(𝑠)
Proof:
𝑈(𝑡 − 𝑎)𝑓(𝑡 − 𝑎) = {
0; 𝑡 < 𝑎
𝑓(𝑡 − 𝑎) ; 𝑡 > 𝑎
By the definition of Laplace transform,
∞
𝑈(𝑡 − 𝑎)𝑓(𝑡 − 𝑎)𝑒−𝑠𝑡𝑑𝑡
𝐿[𝑈(𝑡 − 𝑎)𝑓(𝑡 − 𝑎)] = ∫0
∞
𝑓(𝑡 − 𝑎)𝑒−𝑠𝑡𝑑𝑡
𝑎
= ∫00𝑑𝑡 + ∫𝑎
∞
𝑒−𝑠(𝑎+𝑥)𝑓(𝑥)𝑑𝑥
𝐿[𝑈(𝑡 − 𝑎)𝑓(𝑡 − 𝑎)] = ∫0
Engineering Maths - IV
Laplace Transform
Page
28

29. ∞
= ∫ 𝑒−𝑠𝑎𝑒−𝑠𝑥𝑓(𝑥)𝑑𝑥
0
= 𝑒−𝑠𝑎
∞
𝑒−𝑠𝑥𝑓(𝑥)𝑑𝑥
∫0
∞
𝑒−𝑠𝑡𝑓(𝑡)𝑑𝑡
Replace x by t
𝐿[𝑈(𝑡 − 𝑎)𝑓(𝑡 − 𝑎)] = 𝑒−𝑠𝑎 ∫
0
= 𝑒−𝑠𝑎𝐿[𝑓(𝑡)] = 𝑒−𝑠𝑎𝐹(𝑠)
𝐿[𝑈(𝑡 − 𝑎)𝑓(𝑡 − 𝑎)] = 𝑒−𝑠𝑎𝐹(𝑠)
5.5 PERIODIC FUNCTIONS
Definition: A function 𝑓(𝑡) is said to be periodic if 𝑓(𝑡 + 𝑇) = 𝑓(𝑡) for all values of t and for certain values
of T. The smallest value of T for which 𝑓(𝑡 + 𝑇) = 𝑓(𝑡) for all t is called periodic function.
Example:
𝑠𝑖𝑛𝑡 = sin(𝑡 + 2𝜋) = sin(𝑡 + 4𝜋) ⋯
∴ 𝑠𝑖𝑛𝑡 is periodic function with period 2𝜋.
Let 𝒇(𝒕) be a periodic function with period T. Then
𝟏
𝑳[𝒇(𝒕)] =
𝟏 − 𝒆−𝒔𝑻
𝑻
∫ 𝒆−𝒔𝒕𝒇(𝒕)𝒅𝒕
𝟎
Problems on Laplace transform of Periodic function
Example: 5.35 Find the Laplace transform of 𝒇(𝒕) = {
𝒔𝒊𝒏𝑚𝒕; 𝟎 < 𝑡 < 𝝅
𝑚
𝟎 ; 𝝅
< 𝑡 < 𝟐𝝅
𝑚 𝑚
𝑚
𝒇(𝒕 + 𝟐𝝅
) = 𝒇(𝒕)
Solution:
The given function is a periodic function with period 𝑇 = 2𝜋
𝜔
𝐿[𝑓(𝑡)] =
1 𝑇
𝑒−𝑠𝑡𝑓(𝑡)𝑑𝑡
1−𝑒−𝑠𝑇 ∫0
=
1
−𝟐𝝅𝒔
1−𝑒 𝑚
𝝅
𝑚
0
𝟐𝝅
−𝑠𝑡 𝑚 −𝑠𝑡
𝝅
𝑚
[∫ 𝑠𝑖𝑛𝜔𝑡𝑒 𝑑𝑡 + ∫ 𝑒 (0)𝑑𝑡]
=
1
1−𝑒 𝜔
𝜋
∫𝜔 𝑠𝑖𝑛𝜔𝑡𝑒−𝑠𝑡𝑑𝑡
−2𝜋𝑠 0
=
1
−2𝜋 𝑠
1−𝑒 𝜔
𝑒−𝑠𝑡
[ (−𝑠𝑠𝑖𝑛𝜔𝑡 − 𝜔𝑐𝑜𝑠𝜔𝑡)]
(−𝑠)2+𝜔2
0
𝜋
𝜔
=
1−𝑒 𝜔
−𝑠𝜋
1 𝑒 𝜔
−2𝜋𝑠 { 𝑠2+𝜔2
[−𝑠𝑠𝑖𝑛𝜋 − 𝜔𝑐𝑜𝑠𝜋] +
𝜔
𝑠2+𝜔2
}
=
1−𝑒 𝜔
−𝑠𝜋
𝑠2+𝜔2
1 𝑒 𝜔 𝜔+𝜔
−2𝜋𝑠 [ ]
=
1
12−(𝑒 𝜔 )
−𝜋𝑠 2 [
−𝑠𝜋
𝜔(𝑒 𝜔 +1)
𝑠2+𝜔2
]
Let 𝑡 − 𝑎 = 𝑥 ⋯ (1)
𝑡 = 𝑎 + 𝑥
𝑑𝑡 = 𝑑𝑥
When 𝑡 = 𝑎, (1) => 𝑥 = 0
When 𝑡 = ∞, (1) => 𝑥 = ∞
Engineering Maths - IV
Laplace Transform
Page
29

30. =
1
−𝜋 𝑠
(1−𝑒 𝜔 )(1+𝑒 𝜔 )
−𝜋𝑠 [
−𝑠𝜋
𝜔(𝑒 𝜔 +1)
𝑠2+𝜔2
]
∴ 𝐿[𝑓(𝑡)] =
𝜔
−𝜋𝑠
(1−𝑒 𝜔 )(𝑠2+𝜔2)
Example: 5.36 Find the Laplace transform of 𝒇(𝒕) = {
𝑬;𝟎 ≤ 𝒕 ≤ 𝒂
−𝑬 ;𝒂 ≤ 𝒕 ≤ 𝟐𝒂
given that 𝒇(𝒕 + 𝟐𝒂) = 𝒇(𝒕).
Solution:
The given function is a periodic function with period 𝑇 = 2𝑎
𝐿[𝑓(𝑡)] =
1 𝑇
𝑒−𝑠𝑡𝑓(𝑡)𝑑𝑡
1−𝑒−𝑠𝑇 ∫0
=
1 2𝑎
𝑒−𝑠𝑡𝑓(𝑡)𝑑𝑡
1−𝑒−2𝑎𝑠 ∫0
=
1
1−𝑒−2𝑎𝑠 ∫0
[
𝑎
𝐸𝑒−𝑠𝑡 𝑑𝑡 + ∫
𝑎
2𝑎
−𝐸𝑒−𝑠𝑡𝑑𝑡]
=
1
1−𝑒−2𝑎𝑠 [𝐸 ∫0
𝑎
𝑒−𝑠𝑡𝑑𝑡 − 𝐸 ∫
𝑎
2𝑎
𝑒−𝑠𝑡𝑑𝑡]
𝐸
= [[ ]
1−𝑒−2𝑎𝑠 −𝑠 0 𝑎
𝑒−𝑠𝑡 𝑎 𝑒−𝑠𝑡 2𝑎
− [ ] ]
−𝑠
=
𝐸
=
1−𝑒−2𝑎𝑠
𝐸
1−𝑒−2𝑎𝑠 𝑠
−𝑎𝑠 −2𝑎𝑠 −𝑎𝑠
[𝑒
+ 1
− 𝑒
− 𝑒
]
−𝑠 𝑠 𝑠 𝑠
−𝑎𝑠 −2𝑎𝑠
[1−2𝑒 +𝑒
]
= 12−(𝑒−𝑎𝑠)2 𝑠
−𝑎𝑠 2
𝐸
[(1−𝑒 )
]
=
𝐸
(1−𝑒−𝑎𝑠)(1+𝑒−𝑎𝑠) 𝑠
−𝑎𝑠 2
[(1−𝑒 )
]
−𝑎𝑠
= 𝐸 (1−𝑒 )
𝑠 (1+𝑒−𝑎𝑠)
∴ 𝐿[𝑓(𝑡)] = 𝐸
tanh (𝑎𝑠
)
𝑠 2
Example: 5.37 Find the Laplace transform of 𝒇(𝒕) = {
𝟏;𝟎 ≤ 𝒕 ≤ 𝒂
𝟐
𝒂
𝟐
−𝟏 ; ≤ 𝒕 ≤ 𝒂
given that 𝒇(𝒕 + 𝒂) = 𝒇(𝒕).
Solution:
The given function is a periodic function with period 𝑇 = 𝑎
𝐿[𝑓(𝑡)] =
1 𝑇
𝑒−𝑠𝑡𝑓(𝑡)𝑑𝑡
1−𝑒−𝑠𝑇 ∫0
=
1 𝑎
𝑒−𝑠𝑡𝑓(𝑡)𝑑𝑡
1−𝑒−𝑎𝑠 ∫0
=
1
1−𝑒−𝑎𝑠
𝑎
0
𝑎
2
[∫2(1)𝑒−𝑠𝑡𝑑𝑡 + ∫𝑎(−1)𝑒−𝑠𝑡𝑑𝑡]
=
1
1−𝑒−𝑎𝑠
𝑎
0
𝑎
2
[∫2 𝑒−𝑠𝑡𝑑𝑡 − ∫𝑎 𝑒−𝑠𝑡𝑑𝑡]
=
1
1−𝑒−𝑎𝑠
0
𝑎
−𝑠𝑡
−𝑠 −𝑠 𝑎
2
−𝑠𝑡 2𝑎
[[𝑒
]
2
− [𝑒
] ]
Engineering Maths - IV
Laplace Transform
Page
30

31. = 1−𝑒−𝑎𝑠
−𝑠𝑎 −𝑠𝑎
1
[𝑒 2
+ 1
+ 𝑒−𝑎𝑠
− 𝑒 2
]
−𝑠 𝑠 𝑠 𝑠
= 1−𝑒−𝑎𝑠
−𝑠𝑎
𝑠
−𝑎𝑠
1
[1−2𝑒 2 +𝑒
]
=
1
12−(𝑒 2 )
−𝑠𝑎 2 [
−𝑠𝑎 2
(1−𝑒 2 )
𝑠
]
=
1
−𝑠𝑎
(1−𝑒 2 )(1+𝑒 2 )
−𝑠𝑎 [
−𝑠𝑎 2
(1−𝑒 2 )
𝑠
]
=
𝑠
−𝑠𝑎
1 (1−𝑒 2 )
−𝑠𝑎
(1+𝑒 2 )
(1−𝑒−2𝑥)
[∵ 𝑡𝑎𝑛ℎ𝑥 = ]
(1+𝑒−2𝑥)
∴ 𝐿[𝑓(𝑡)] = 1
tanh (𝑎𝑠
)
𝑠 4
Example: 5.38 Find the Laplace transform of 𝒇(𝒕) = {
𝒕;𝟎 ≤ 𝒕 ≤ 𝒂
𝟐𝒂 − 𝒕 ;𝒂 ≤ 𝒕 ≤ 𝟐𝒂
given that
𝒇(𝒕 + 𝟐𝒂) = 𝒇(𝒕).
Solution:
The given function is a periodic function with period 𝑇 = 2𝑎
𝐿[𝑓(𝑡)] =
1 𝑇
𝑒−𝑠𝑡𝑓(𝑡)𝑑𝑡
1−𝑒−𝑠𝑇 ∫0
=
1 2𝑎
𝑒−𝑠𝑡𝑓(𝑡)𝑑𝑡
1−𝑒−2𝑎𝑠 ∫0
=
1
1−𝑒−2𝑎𝑠 ∫0
[
𝑎
𝑡𝑒−𝑠𝑡𝑑𝑡 + ∫
𝑎
2𝑎
(2𝑎 − 𝑡)𝑒−𝑠𝑡𝑑𝑡]
1
𝑎
𝑒−𝑠𝑡 𝑒−𝑠𝑡 𝑒−𝑠𝑡 𝑒−𝑠𝑡
0 𝑎
2𝑎
= [[𝑡 ( ) − ( )] − [(2𝑎 − 𝑡) ( ) − (−1)( )] ]
1−𝑒−2𝑎𝑠 −𝑠 (−𝑠)2 −𝑠 (−𝑠)2
=
𝑠 𝑠
𝑠2 𝑠2 𝑠2 𝑠2
1
[−𝑎𝑒−𝑎𝑠
− 𝑒−𝑎𝑠
+ 1
+ 𝑒−2𝑎𝑠
+ 𝑎𝑒−𝑎𝑠
− 𝑒−𝑎𝑠
]
= 1−𝑒−2𝑎𝑠 𝑠2
1−𝑒−2𝑎𝑠
1
[1−2𝑒−𝑎𝑠+𝑒−2𝑎𝑠
]
= 12−(𝑒−𝑎𝑠)2 𝑠2
−𝑎𝑠 2
1
[(1−𝑒 )
]
=
1
(1−𝑒−𝑎𝑠)(1+𝑒−𝑎𝑠) 𝑠2
−𝑎𝑠 2
[(1−𝑒 )
]
−𝑎𝑠
= 1 (1−𝑒 )
𝑠2 (1+𝑒−𝑎 𝑠)
𝑠2 2
= 1
tanh (𝑎𝑠
)
Exercise: 5.6
1. Find the Laplace transform of
𝑓(𝑡) = {
1; 0 ≤ 𝑡 ≤ 𝑎
2
2
−1 ;𝑎
≤ 𝑡 ≤ 𝑎
given that 𝑓(𝑡 + 𝑎) = 𝑓(𝑡). Ans: 𝑘
tanh (𝑎𝑠
)
𝑠 2
Engineering Maths - IV
Laplace Transform
Page
31

32. Engineering Maths - IV
2. Find the Laplace transform of
𝑓(𝑡) = {
𝑡; 0 ≤ 𝑡 ≤ 𝑎
2𝑎 − 𝑡 ;𝑎 ≤ 𝑡 ≤ 2𝑎
given that 𝑓(𝑡 + 2𝑎) = 𝑓(𝑡). Ans: 𝑠2 2
1
tanh (𝑠𝜋
)
3. Find the Laplace transform of
𝑎
𝑡
; 0 ≤ 𝑡 ≤ 𝑎
2𝑎−𝑡
;𝑎 ≤ 𝑡 ≤ 2𝑎
𝑓(𝑡) = { given that 𝑓(𝑡 + 2𝑎) = 𝑓(𝑡).
𝑎
Find the Laplace transform of
Ans:
1
tanh (𝑠𝑎
)
𝑎𝑠2 2
4.
𝑓(𝑡)
𝑠𝑖𝑛𝑡;0 < 𝑡 < 𝜋
= { 0 ;𝜋 < 𝑡 < 2𝜋
𝑓(𝑡 + 2𝜋) = 𝑓(𝑡) Ans:
1
(1−𝑒−𝜋𝑠)(𝑠2+1)
5.6 INVERSE LAPLACE TRANSFORM
Definition
If the Laplace transform of a function 𝑓(𝑡) is 𝐹(𝑠)𝑖𝑒. , 𝐿[𝑓(𝑡)] = 𝐹(𝑠),then𝑓(𝑡) is called an
inverse Laplace transform of 𝐹(𝑠) and we write symbolically𝑓(𝑡) = 𝐿−1[𝐹(𝑠)], where 𝐿−1 is called the
inverse Laplace transform operator.
Inverse Laplace transform of elementary functions
𝑳[𝒇(𝒕)] = 𝑭(𝒔) 𝑳−𝟏[𝑭(𝒔)] = 𝒇(𝒕)
𝟏
𝑳[𝟏] =
𝒔
𝟏
𝐿−1 [ ] = 𝟏
𝒔
𝟏
𝑳[𝒕] =
𝒔𝟐
𝟏
𝐿−1 [ ] = 𝒕
𝒔𝟐
𝒏!
𝑳[𝒕𝒏] = 𝒊𝒇 𝒏 𝒊𝒔 𝒂𝒏 𝒊𝒏𝒕𝒆𝒈𝒆𝒓
𝒔𝒏+𝟏
𝒏!
𝐿−1 [ ] = 𝒕𝒏
𝒔𝒏+𝟏
𝟏 𝒕𝒏
𝐿−1 [ ] =
𝒔𝒏+𝟏 𝒏!
𝟏
𝑳[𝒆𝒂𝒕] =
𝒔 − 𝒂
𝟏
𝐿−1 [ ] = 𝒆𝒂𝒕
𝒔 − 𝒂
𝟏
𝑳[𝒆−𝒂𝒕] =
𝒔 + 𝒂
𝟏
𝐿−1 [ ] = 𝒆−𝒂𝒕
𝒔 + 𝒂
𝒂
𝑳[𝒔𝒊𝒏𝒂𝒕] =
𝒔𝟐 + 𝒂𝟐
𝟏 𝒔𝒊𝒏𝒂𝒕
𝐿−1 [ ] =
𝒔𝟐 + 𝒂𝟐 𝒂
𝒔
𝑳[𝒄𝒐𝒔𝒂𝒕] =
𝒔𝟐 + 𝒂𝟐
𝒔
𝐿−1 [ ] = 𝒄𝒐𝒔𝒂𝒕
𝒔𝟐 + 𝒂𝟐
Laplace Transform
Page
32

33. 𝒂
𝑳[𝒔𝒊𝒏𝒉𝒂𝒕] =
𝒔𝟐 − 𝒂𝟐
𝟏 𝒔𝒊𝒏𝒉𝒂𝒕
𝐿−1 [ ] =
𝒔𝟐 − 𝒂𝟐 𝒂
𝒔
𝑳[𝒄𝒐𝒔𝒂𝒕] =
𝒔𝟐 − 𝒂𝟐
𝒔
𝐿−1 [ ] = 𝒄𝒐𝒔𝒉𝒂𝒕
𝒔𝟐 − 𝒂𝟐
Result on inverse Laplace transform
Result: 1 Linear property
𝑳[𝒇(𝒕)] = 𝑭(𝒔) and 𝑳[𝒈(𝒕)] = 𝑮(𝒔) ,then 𝑳−𝟏[𝒂𝑭(𝒔) ± 𝒃𝑮(𝒔)] = 𝒂𝑳−𝟏[𝑭(𝒔)] ± 𝒃𝑳−𝟏[𝑮(𝒔)]
Where a and b are constants.
Proof:
We know that 𝐿[𝑎𝐹(𝑠) ± 𝑏𝐺(𝑠)] = 𝑎𝐿[𝐹(𝑠)] ± 𝑏𝐿[𝐺(𝑠)]
= 𝑎 𝐹(𝑠) ± 𝑏 𝐺(𝑠)
(𝑖.𝑒. )𝑎 𝐹(𝑠) ± 𝑏 𝐺(𝑠) = 𝐿[𝑎𝑓(𝑡) ± 𝑏𝑔(𝑡)]
Operating 𝐿−1 on both sides, we get
𝐿−1[𝑎𝐹(𝑠) ± 𝑏𝐺(𝑠)] = 𝑎𝑓(𝑡) ± 𝑏𝑔(𝑡)
𝐿−1[𝑎𝐹(𝑠) ± 𝑏𝐺(𝑠)] = 𝑎𝐿−1[𝐹(𝑠)] ± 𝑏𝐿−1[𝐺(𝑠)]
Result: 2 First shifting property
(i) 𝑳−𝟏 [𝑭(𝒔 + 𝒂) = 𝒆−𝒂𝒕𝑳−𝟏[𝑭(𝒔)]]
(ii) 𝑳−𝟏 [𝑭(𝒔 − 𝒂) = 𝒆𝒂𝒕𝑳−𝟏[𝑭(𝒔)]]
Proof:
Let 𝐿[𝑒−𝑎𝑡𝑓(𝑡)] = 𝐹[𝑠 + 𝑎]
Operating 𝐿−1 on both sides, we get
𝑒−𝑎𝑡𝑓(𝑡) = 𝐿−1[𝐹[𝑠 + 𝑎]]
𝐿−1[𝐹[𝑠 + 𝑎]] = 𝑒−𝑎𝑡𝐿−1[𝐹(𝑠)]
Result: 3 Multiplication by s.
𝒅𝒕
If 𝑳−𝟏[𝑭(𝒔)] = 𝒇(𝒕) and 𝒇(𝟎) = 𝟎, then 𝑳−𝟏[𝒔𝑭(𝒔)] = 𝒅
𝑳−𝟏[𝑭(𝒔)]
Proof:
We know that 𝐿[𝑓′(𝑡)] = 𝑠𝐿[𝑓(𝑡)] − 𝑓(0) = 𝑠𝐹(𝑠)
Operating 𝐿−1 on both sides, we get
𝑓′(𝑡) = 𝐿−1[𝑠𝐹(𝑠)]
𝑑𝑡
𝑑
𝑓(𝑡) = 𝐿−1[𝑠𝐹(𝑠)]
𝑑𝑡
𝑑
𝐿−1[𝐹(𝑠)] = 𝐿−1[𝑠𝐹(𝑠)]
𝑑𝑡
∴ 𝐿−1[𝑠𝐹(𝑠)] = 𝑑
𝐿−1[𝐹(𝑠)]
∵ 𝑓(𝑡) = 𝐿−1[𝐹(𝑠)]
∵ 𝑔(𝑡) = 𝐿−1[𝐺(𝑠)]
Engineering Maths - IV
Laplace Transform
Page
33

34. Engineering Maths - IV
Result: 4 Division by s.
𝑳−𝟏 [𝑭(𝒔)
] =
𝒔
𝒕
𝐿−1[𝐹(𝑠)]𝑑𝑡
∫𝟎
Proof:
We know that 𝐿 [∫
𝑡
𝑓 (𝑡)𝑑𝑡] = 1
𝐿[𝑓(𝑡)] = 1
𝐹(𝑠)
0 𝑠 𝑠
Operating 𝐿−1 on both sides ,we get
∫0 𝑠
𝑡 1
𝑓 (𝑡)𝑑𝑡 = 𝐿−1 [ 𝐹(𝑠)]
𝑠
𝑡 −1 −1 1
∫0 𝐿 [𝐹(𝑠)] 𝑑𝑡 = 𝐿 [ 𝐹(𝑠)]
𝑠
∴ 𝐿−1 [𝐹(𝑠)
] = ∫
0
𝑡
𝐿−1[𝐹(𝑠)] 𝑑𝑡
Result: 5 Inverse Laplace transform of derivative
𝑳−𝟏[𝑭(𝒔)] = −𝟏
𝑳−𝟏 [ 𝒅
𝑭(𝒔)]
𝒕 𝒅𝒔
Proof:
We know that 𝐿[𝑡𝑓(𝑡)] = −𝑑
𝐿[𝑓(𝑡)] = −𝑑
𝐹(𝑠)
𝑑𝑠 𝑑𝑠
Operating 𝐿−1 on both sides ,we get
𝑑𝑠
𝑑
𝑡𝑓(𝑡) = −𝐿−1 [ 𝐹(𝑠)]
𝐿−1[𝐹(𝑠)] = −1
𝐿−1 [ 𝑑
𝐹(𝑠)]
𝑡 𝑑𝑠
𝑓(𝑡) = −1
𝐿−1 [ 𝑑
𝐹(𝑠)]
𝑡 𝑑𝑠
𝐿−1[𝐹(𝑠)] = −1
𝐿−1 [ 𝑑
𝐹(𝑠)]
𝑡 𝑑𝑠
∞
𝐹(𝑠)𝑑𝑠]
Result: 6 Inverse Laplace transform of integral
𝑳−𝟏[𝑭(𝒔)] = 𝒕𝐿−1[∫
𝑠
Proof:
We know that 𝐿 [𝑓(𝑡)
] =
𝑡
∫𝑠
∞
𝐿(𝑓(𝑡)) 𝑑𝑠
= ∫𝑠
∞
𝐹(𝑠) 𝑑𝑠
Operating 𝐿−1 on both sides, we get
𝑡
𝑓(𝑡)
= 𝐿−1[∫𝑠
∞
𝐹(𝑠) 𝑑𝑠]
𝑓(𝑡) = 𝑡𝐿−1[∫
𝑠
∞
𝐹(𝑠) 𝑑𝑠]
𝐿−1[𝐹(𝑠)] = 𝑡𝐿−1[∫
𝑠
∞
𝐹(𝑠) 𝑑𝑠]
Problems under inverse Laplace transform of elementary functions
Example: 5.39 Find the inverse Laplace for the following
(i)
𝟏 𝟏
𝟐𝒔+𝟑 𝟒𝒔𝟐+𝟗
(ii) (iii)
𝒔𝟑−𝟑𝒔𝟐+𝟕
𝒔𝟒
(iv)
𝟑𝒔+𝟓
𝒔𝟐+𝟑𝟔
Solution:
Laplace Transform
Page
34

35. (i) 𝐿−1 𝟏
𝟐𝒔+𝟑
[ ] = 𝐿−1 [
𝟏
𝟑
𝟐
𝟐[𝒔+ ]
]
=
𝟏
𝟐
−𝟑𝒕
𝑒− 𝟐
(ii) 𝐿−1 [
𝟏
𝟒𝒔𝟐+𝟗
] = 𝐿−1 [
𝟏
𝟐 𝟗
𝟒[𝒔 +𝟒
]
]
𝟒
= 𝟏
𝐿−1 [
𝟏
𝟐 𝟗
𝟒
[𝒔 + ]
]
4 3⁄2 2
= 1 1
𝑠𝑖𝑛 3
𝑡
6 2
= 1
𝑠𝑖𝑛 3
𝑡
3 2 3 2
(iii) 𝐿−1 [𝑠 −3𝑠 +7
] = 𝐿−1 [𝑠
− 3𝑠
+ 7
]
𝑠4 𝑠4 𝑠4 𝑠4
𝑠 𝑠2 𝑠4
1 1
= 𝐿−1 [1
] − 3𝐿−1 [ ] + 7𝐿−1 [ ]
𝑠4
3 2 3
𝐿−1 [𝑠 −3𝑠 +7
] = 1 − 3𝑡 + 7𝑡
3!
3𝑠+5 𝑠 1
(iv) 𝐿−1 [ ] = 3𝐿−1 [ ] + 5𝐿−1 [ ]
𝑠2+36 𝑠2+36 𝑠2+36
𝐿−1 [ 3𝑠+5
] = 3𝑐𝑜𝑠6𝑡 + 5𝑠𝑖𝑛6𝑡
𝑠2+36 6
Inverse Laplace transform using First shifting theorem
𝑳−𝟏[𝑭(𝒔 + 𝒂)] = 𝒆−𝒂𝒕𝑳−𝟏[𝑭(𝒔)]
Example: 5.40 Find the inverse Laplace transform for the following:
(i)
𝟏 𝟏
(𝒔+𝟐)𝟐
𝟏
(ii) (iii)
𝟏
(iv)
(𝒔−𝟑)𝟒
𝒔+𝟐
𝟏
𝒔𝟐−𝟐𝒔+𝟐
𝒔
(v)
𝒔𝟐−𝟒𝒔+𝟏𝟑
𝒔
(vi) (vii)
(𝒔+𝟑)𝟐+𝟗
𝒔+𝟐
(viii)
(𝒔+𝟐)𝟐+𝟐𝟓
𝒔
𝒔𝟐+𝟒𝒔+𝟐𝟎
𝟐𝒔+𝟑
(𝒔+𝟑)𝟐
𝒔
(ix)
(𝒔−𝟒)𝟑
(x)
𝒔𝟐−𝟐𝒔+𝟐
(xi)
𝒔𝟐+𝟔𝒔+𝟐𝟓
(xii)
𝒔𝟐+𝟔𝒔−𝟕
Solution:
(i) 𝐿−1 [ 1
(𝑠+2)2 𝑠2
] = 𝑒−2𝑡 𝐿−1 [ 1
] = 𝑒−2𝑡 𝑡
(ii) 𝐿−1 [ 1
𝑠4
3
] = 𝑒3𝑡𝐿−1 [ 1
] = 𝑒−2𝑡 𝑡
3!
(iii) 𝐿−1 [
(𝑠−3)4
1
(𝑠+3)2+9
] = 𝑒−3𝑡𝐿−1 [
𝑠2+9
1
] = 𝑒−3𝑡 𝑠𝑖𝑛3𝑡
3
(iv) 𝐿−1 [ 1
] = 𝐿−1 [
1
(𝑠−1)2+1
1
𝑠2+1
] = 𝑒𝑡𝐿−1 [ ] = 𝑒𝑡𝑠𝑖𝑛𝑡
(v) 𝐿−1 [
𝑠2−2𝑠+2
1
] = 𝐿−1 [
1
(𝑠−2)2+9
] = 𝑒2𝑡𝐿−1 [
𝑠2+9
1
] = 𝑒2𝑡 𝑠𝑖𝑛3𝑡
3
(vi) 𝐿−1 [
𝑠2−4𝑠+13
𝑠+2
(𝑠+2)2+25
] = 𝑒−2𝑡𝐿−1 [ 𝑠
𝑠2+25
] = 𝑒−2𝑡𝑐𝑜𝑠5𝑡
(vii) 𝐿−1 [ 𝑠+2
𝑠2+4𝑠+20
] = 𝐿−1 [
𝑠+2
(𝑠+2)2+16
]
Engineering Maths - IV
Laplace Transform
Page
35

36. 𝑠2+16
𝑠
= 𝑒−2𝑡𝐿−1 [ ] = 𝑒−2𝑡𝑐𝑜𝑠4𝑡
(viii) 𝐿−1 [ 𝑠
] = 𝐿−1 [𝑠+3−3
]
(𝑠+3)2 (𝑠+3)2
𝑠2
𝑠+3 3
= 𝐿−1 [ ] − 𝐿−1 [ ]
(𝑠+3)2 (𝑠+3)2
= 𝐿−1 1 1
[ ] − 3𝐿−1 [ ]
𝑠+3 (𝑠+3)2
= 𝑒−3𝑡 − 3𝑒−3𝑡𝐿−1 1
[ ]
= 𝑒−3𝑡 − 3𝑒−3𝑡𝑡
(ix) 𝐿−1 [ 𝑠
] = 𝐿−1 [𝑠−4+4
]
(𝑠−4)3 (𝑠−4)3
𝑠−4 4
= 𝐿−1 [ ] + 𝐿−1 [ ]
(𝑠−4)3 (𝑠−4)3
= 𝐿−1 1 1
[ ] + 4𝐿−1 [ ]
(𝑠−4)2 (𝑠−4)3
= 𝑒4𝑡𝐿−1 1 1
[ ] + 4𝑒4𝑡𝐿−1 [ ]
𝑠2 𝑠3
2
= 𝑒4𝑡𝑡 + 4𝑒4𝑡 𝑡
2!
= 𝑒4𝑡𝑡 + 2𝑒4𝑡𝑡2
(x) 𝐿−1 [ 𝑠
𝑠2−2𝑠+2
] = 𝐿−1 [
(𝑠−1)2+1 (𝑠−1)2+1
𝑠
] = 𝐿−1 [ 𝑠−1+1
]
𝑠−1 1
= 𝐿−1 [ ] + 𝐿−1 [ ]
(𝑠−1)2+1 (𝑠−1)2+1
𝐿−1 [ 𝑠
𝑠 1
= 𝑒𝑡𝐿−1 [ ] + 𝑒𝑡𝐿−1 [ ]
𝑠2+1 𝑠2+1
] = 𝑒𝑡𝑐𝑜𝑠𝑡 + 𝑒𝑡𝑠𝑖𝑛𝑡
(xi) 𝐿−1 [
𝑠2−2𝑠+2
2𝑠+3
𝑠2+6𝑠+25
] = 𝐿−1 [
2𝑠+3
] = 𝐿−1 [2(𝑠+3−3)+3
]
(𝑠+3)2+16 (𝑠+3)2+16
𝑠2+16
= 𝐿−1 [2(𝑠+3)−6+3
]
(𝑠+3)2+16
= 𝑒−3𝑡𝐿−1 2𝑠−3
[ ]
𝑠 1
= 𝑒−3𝑡 [2𝐿−1 [ ] − 3𝐿−1 [ ]]
𝑠2+16 𝑠2+16
𝐿−1 [ 2𝑠+3
𝑠2+6𝑠+25 4
] = 𝑒−3𝑡 (2𝑐𝑜𝑠4𝑡 − 3𝑠𝑖𝑛4𝑡
)
(xii) 𝐿−1 [ 𝑠
𝑠2+6𝑠−7
] = 𝐿−1 [
𝑠
(𝑠+3)2−16
] = 𝐿−1 [
𝑠+3−3
(𝑠+3)2−16
]
𝑠−3
= 𝑒−3𝑡𝐿−1 [ ]
𝑠2−16
𝑠2−16
𝑠
= 𝑒−3𝑡𝐿−1 [ ] − 3𝑒−3𝑡𝐿−1 [
1
𝑠2−16
]
𝐿−1 [ 𝑠
𝑠2+6𝑠−7 4
] = 𝑒−3𝑡 [𝑐𝑜𝑠ℎ4𝑡 − 3𝑠𝑖𝑛ℎ4𝑡
]
Exercise: 5.7
Find the inverse Laplace transform for the following:
Engineering Maths - IV
Laplace Transform
Page
36

37. 1. 2𝑠−3
𝑠2+52
Ans: 2𝑐𝑜𝑠5𝑡 − 3𝑠𝑖𝑛5𝑡
5
2. 3𝑠+5
Ans: 3𝑐𝑜𝑠4𝑡 + 5𝑠𝑖𝑛4𝑡
4
3.
𝑠2+16
1
6 2
Ans: 1
𝑠𝑖𝑛 3
𝑡
4.
4𝑠2+9
1
(𝑠+4)5
Ans:
4
𝑒−4𝑡 𝑡
4!
5.
1
𝑠2−4𝑠+13 3
𝟐𝒕
Ans: 𝒆
𝑠𝑖𝑛3𝑡
Inverse using the formula
𝑳−𝟏[𝑭(𝒔)] = −𝟏
𝑳−𝟏 [ 𝒅
𝑭(𝒔)]
𝒕 𝒅𝒔
Note: This formula is used when 𝐹(𝑠) is cot−1 ∅(𝑠) or tan−1 ∅(𝑠) or 𝑙𝑜𝑔∅(𝑠)
Example: 5.41 Find the inverse Laplace transform for the following
(i) 𝒄𝒐𝒕−𝟏 (𝒔
) (ii) 𝒕𝒂𝒏−𝟏 (𝒂
) (iii) 𝒄𝒐𝒕−𝟏𝒂𝒔
𝒂 𝒔
(iv) 𝒕𝒂𝒏−𝟏(𝒔 + 𝒂) (v) 𝒍𝒐𝒈(𝒔+𝒂
) (vi) 𝒄𝒐𝒕−𝟏 ( 𝟐
)
𝒔+𝒃 𝒔+𝟏 𝒔𝟐
(vii) 𝒕𝒂𝒏−𝟏 ( 𝟐
)
Solution:
(i) 𝐿−1 [𝑐𝑜𝑡−1 (𝑠
)] = −1
𝐿−1 [ 𝑑
(𝑐𝑜𝑡−1 (𝑠
))]
𝑎 𝑡 𝑑𝑠 𝑎
1+𝑠2
𝑎2
𝑡 𝑎 𝑡 𝑎2+𝑠2
𝑎2
𝑎
= −1
𝐿−1 [ −1
(1
)] = 1
𝐿−1 [ −1
(1
)]
𝑡
= 1
𝐿−1 [
𝑎
𝑠2+𝑎2
]
𝐿−1 [𝑐𝑜𝑡−1 (𝑠
)] = 1
𝑠𝑖𝑛𝑎𝑡
𝑎 𝑡
(ii) 𝐿−1 [𝑡𝑎𝑛−1 (𝑎
)] = −1
𝐿−1 [ 𝑑
(𝑡𝑎𝑛−1 (𝑎
))]
𝑠 𝑡 𝑑𝑠 𝑠
𝑡
= −1
𝐿−1 [
1
1+(𝑠
)
𝑎 2 𝑠2 𝑡
(−𝑎
)] = −1
𝐿−1 [
1
𝑠2+𝑎2 𝑠2
𝑠2
(−𝑎
)]
𝑡
= 1
𝐿−1 [
𝑎
𝑠2+𝑎2
]
𝐿−1 [𝑡𝑎𝑛−1 (𝑎
)] = 1
𝑠𝑖𝑛𝑎𝑡
𝑠 𝑡
(iii) 𝐿−1[𝑐𝑜𝑡−1𝑎𝑠] = −1
𝐿−1 [ 𝑑
(𝑐𝑜𝑡−1(𝑎𝑠))]
𝑡 𝑑𝑠
𝑡
= −1
𝐿−1 [
−1
1+𝑎2𝑠2 𝑡
(𝑎)] = 1
𝐿−1 [
𝑎
𝑎2
𝑎2(𝑠2+ 1
)
]
1
𝑎𝑡 𝑠2+ 1
𝑎2
1
𝑎𝑡
1
𝑎
𝑠𝑖𝑛 𝑡
1
𝑎
= 1
𝐿−1 [ ] = [ ]
𝐿−1[𝑐𝑜𝑡−1𝑎𝑠] = 1
𝑠𝑖𝑛 𝑡
𝑡 𝑎
(iv) 𝐿−1[𝑡𝑎𝑛−1(𝑠 + 𝑎)] = 𝑒−𝑎𝑡𝐿−1[𝑡𝑎𝑛−1𝑠]
Engineering Maths - IV
Laplace Transform
Page
37

38. Engineering Maths - IV
= 𝑒−𝑎𝑡 [−1
𝐿−1 [ 𝑑
(𝑡𝑎𝑛−1𝑠)]]
𝑡 𝑑𝑠
1
𝑡 1+𝑠2
= 𝑒−𝑎𝑡 (−1
) 𝐿−1 [ ]
𝑡
= −1
𝑒−𝑎𝑡𝐿−1 [ 1
1+𝑠2
]
−𝑎𝑡
𝐿−1 [𝑐𝑜𝑡−1 (𝑠
)] = −𝑒
𝑠𝑖𝑛𝑡
𝑎 𝑡
(v) 𝐿−1 [𝑙𝑜𝑔 (𝑠+𝑎
)] = −1
𝐿−1 [ 𝑑
(𝑙𝑜𝑔 (𝑠+𝑎
))]
𝑠+𝑏 𝑡 𝑑𝑠 𝑠+𝑏
= −1
𝐿−1 [ 𝑑
(log(𝑠 + 𝑎) − log(𝑠 + 𝑏))]
𝑡 𝑑𝑠
−
𝑡 𝑠+𝑎 𝑠+𝑏
= −1
𝐿−1 [ 1 1
]
= −1
[𝑒−𝑎𝑡 − 𝑒−𝑏𝑡]
𝑡
𝐿−1 [𝑙𝑜𝑔 (𝑠+𝑎
)] = −1
[𝑒−𝑎𝑡 − 𝑒−𝑏𝑡]
𝑠+𝑏 𝑡
(vi) 𝐿−1 [𝑐𝑜𝑡−1 ( 2
)] = 𝑒−𝑡𝐿−1 [𝑐𝑜𝑡−1 (2
)]
𝑠+1 𝑠
= 𝑒−𝑡 (−1
) 𝐿−1 [ 𝑑
(𝑐𝑜𝑡−1 (2
))]
𝑡 𝑑𝑠 𝑠
𝑡 1+ 4
𝑠2
𝑠2 𝑡
−𝑡
= 𝑒−𝑡 (−1
) 𝐿−1 [ −1
(−2
)] = − 𝑒
𝐿−1 [
1
𝑠2+4 𝑠2
𝑠2
( 2
)]
−𝑡
= − 𝑒
𝐿−1 [
2
𝑡 𝑠2+4
]
−𝑡
𝐿−1 [𝑐𝑜𝑡−1 ( 2
)] = − 𝑒
𝑠𝑖𝑛2𝑡
𝑠+1 𝑡
𝑡 𝑑𝑠
𝑠2 𝑠2
(vii) 𝐿−1 [𝑡𝑎𝑛−1 ( 2
)] = −1
𝐿−1 [ 𝑑
(𝑡𝑎𝑛−1 ( 2
))]
𝑡
= −1
𝐿−1 [
1
1+( )
𝑠2
2 2 𝑠3 𝑡
1
𝑠4+4 𝑠3
𝑠4
(−4
)] = 4
𝐿−1 [ ( 1
)]
𝑠
𝑡 𝑠4+4
= 4
𝐿−1 [ ]
𝑡
= 4
𝐿−1 [
𝑠
(𝑠2)2+22
]
𝑡
= 4
𝐿−1 [
𝑠
(𝑠2+2)2−(2𝑠)2
]
𝑡
= 4
𝐿−1 [
𝑠
(𝑠2+2+2𝑠)(𝑠2+2−2𝑠)
]
𝑡
= 4
𝐿−1 [ 𝑠
( −
1 1
−4𝑠 𝑠2+2+2𝑠 𝑠2+2−2𝑠
)] ∵ {
1
(𝑠2+𝑎𝑥+𝑏)(𝑠2+𝑎𝑥+𝑐)
= [ −
1 1 1
𝑐−𝑏 𝑠2+𝑎𝑥+𝑏 𝑠2+𝑎𝑥+𝑐
}
]
𝑡
= −1
𝐿−1 [( −
1 1
𝑠2+2𝑠+2 𝑠2−2𝑠+2
)]
𝑡
= −1
𝐿−1 [ −
1 1
(𝑠+1)2+1 (𝑠−1)2+1
]
∵ 𝑎2 + 𝑏2 = (𝑎 + 𝑏)2 − 2𝑎𝑏
Laplace Transform
Page
38

39. 1
𝑡 𝑠2+1
1
𝑠2+1
= −1
(𝑒−𝑡𝐿−1 [ ] − 𝑒𝑡𝐿−1 [ ])
𝑡
= −1
(𝑒−𝑡𝑠𝑖𝑛𝑡 − 𝑒𝑡𝑠𝑖𝑛𝑡)
𝑡
= 𝑠𝑖𝑛𝑡
(𝑒−𝑡 − 𝑒𝑡)
𝑡
= 𝑠𝑖𝑛𝑡
2𝑠𝑖𝑛ℎ𝑡
𝑠2
𝐿−1 [𝑡𝑎𝑛−1 ( 2
)] = 2𝑠𝑖𝑛𝑡𝑠𝑖𝑛ℎ𝑡
𝑡
Inverse using the formula
𝒅𝒕
𝑳−𝟏[𝒔𝑭(𝒔)] = 𝒅
𝑳−𝟏[𝑭(𝒔)]
𝒔𝟐+𝒃𝟐
𝟐 𝟐
Example: 5.42 Find 𝑳−𝟏 [𝒔𝒍𝒐𝒈 (𝒔 +𝒂
)]
Solution:
2 2 2 2
𝐿−1 [𝑠𝑙𝑜𝑔 (𝑠 +𝑎
)] = 𝑑
𝐿−1 [𝑠𝑙𝑜𝑔 (𝑠 +𝑎
)] ⋯ (1)
𝑠2+𝑏2 𝑑𝑡 𝑠2+𝑏2
2 2 2 2
𝐿−1 [𝑙𝑜𝑔 (𝑠 +𝑎
)] = 𝐿−1 𝑑
[𝑙𝑜𝑔 (𝑠 +𝑎
)]
𝑠2+𝑏2 𝑑𝑠 𝑠2+𝑏2
= −1
𝐿−1 [ 𝑑
(log(𝑠2 + 𝑎2) − log(𝑠2 + 𝑏2))]
𝑡 𝑑𝑠
𝑡
= −1
𝐿−1 [
𝑠2+𝑎2
1 1
𝑠2+𝑏2
2𝑠 − 2𝑠]
𝑡
= −2
𝐿−1 [ −
𝑠 𝑠
𝑠2+𝑎2 𝑠2+𝑏2
]
𝑡
= −2
[𝑐𝑜𝑠𝑎𝑡 − 𝑐𝑜𝑠𝑏𝑡]
= 2
[𝑐𝑜𝑠𝑏𝑡 − 𝑐𝑜𝑠𝑎𝑡]
𝑡
Substituting in (1), we get
𝑡2
2 2
𝐿−1 [𝑠𝑙𝑜𝑔 (𝑠 +𝑎
)] = 𝑑
[2
[𝑐𝑜𝑠𝑏𝑡 − 𝑐𝑜𝑠𝑎𝑡]]
𝑠2+𝑏2 𝑑𝑡 𝑡
= 2 [𝑡(−𝑏𝑠𝑖𝑛𝑏𝑡+𝑎𝑠𝑖𝑛𝑎𝑡)−(𝑐𝑜𝑠𝑏𝑡−𝑐𝑜𝑠𝑎𝑡
)
]
𝑡2
2 2
𝐿−1 [𝑠𝑙𝑜𝑔 (𝑠 +𝑎
)] = 2 [𝑡(−𝑏𝑠𝑖𝑛𝑏𝑡+𝑎𝑠𝑖𝑛𝑎𝑡)−(𝑐𝑜𝑠𝑏𝑡−𝑐𝑜𝑠𝑎𝑡)
]
𝑠2+𝑏2
Inverse using the formula
𝑳−𝟏 [𝑭(𝒔)
] =
𝒔
𝒕 −𝟏
∫𝟎 𝑳 [𝑭(𝒔)]𝒅𝒕
This formula is used when 𝐹(𝑠) =
𝑜𝑛𝑒 𝑡𝑒𝑟𝑚
𝑠(𝑎𝑛𝑜𝑡ℎ𝑒𝑟 𝑡𝑒𝑟𝑚)
𝒔(𝒔𝟐+𝒂𝟐)
Example: 5.43 Find 𝑳−𝟏 [ 𝟏
]
Solution:
𝐿−1 [ 1
𝑠(𝑠2+𝑎2)
1
(𝑠2+𝑎2)
] 𝑑𝑡
𝑡
𝐿−1 [
] = ∫0
𝑎
𝑡
[𝑠𝑖𝑛𝑎𝑡
] 𝑑𝑡
= ∫0
Engineering Maths - IV
Laplace Transform
Page
39

40. = [
1 −𝑐𝑜𝑠𝑎𝑡
𝑎 𝑎
𝑡
]
0
= −1
[𝑐𝑜𝑠𝑎𝑡]𝑡
𝑎2 0
= −1
(𝑐𝑜𝑠𝑎𝑡 − 𝑐𝑜𝑠0) = −1
(𝑐𝑜𝑠𝑎𝑡 − 1)
𝑎2 𝑎2
𝑠(𝑠2+𝑎2)
1 1−𝑐𝑜𝑠𝑎𝑡
∴ 𝐿−1 [ ] = 𝑎2
𝒔(𝒔𝟐−𝒂𝟐)
Example: 5.44 Find 𝑳−𝟏 [ 𝟏
]
Solution:
𝐿−1 [ 1
𝑠(𝑠2+𝑎2)
1
(𝑠2−𝑎2)
] 𝑑𝑡
𝑡
𝐿−1 [
] = ∫0
𝑎
𝑡
[𝑠𝑖𝑛ℎ𝑎𝑡
] 𝑑𝑡
= ∫0
= [
1 𝑐𝑜𝑠ℎ𝑎𝑡
𝑎 𝑎
𝑡
]
0
= 1
[𝑐𝑜𝑠ℎ𝑎𝑡]𝑡
𝑎2 0
= 1
(𝑐𝑜𝑠ℎ𝑎𝑡 − 𝑐𝑜𝑠ℎ0) = 1
(𝑐𝑜𝑠ℎ𝑎𝑡 − 1)
𝑎2 𝑎2
𝑠(𝑠2−𝑎2)
1 𝑐𝑜𝑠ℎ𝑎𝑡−1
∴ 𝐿−1 [ ] = 𝑎2
𝒔(𝒔+𝒂)
Example: 5.45 Find 𝑳−𝟏 [ 𝟏
]
Solution:
𝐿−1 [ 1
𝑠(𝑠+𝑎)
1
(𝑠+𝑎)
] 𝑑𝑡
𝑡
𝐿−1 [
] = ∫0
𝑡
𝑒−𝑎𝑡𝑑𝑡
= ∫0
−𝑎
= [ ]
0
𝑒−𝑎𝑡 𝑡
𝑎
= −1
(𝑒−𝑎𝑡 − 1)
∴ 𝐿−1 [
1
] = 1−𝑒−𝑎𝑡
𝑠(𝑠+𝑎) 𝑎
Inverse using Partial Fraction
Example: 5.46 Find
𝒔(𝒔+𝟐)(𝒔−𝟏)
𝒔−𝟐
𝑳−𝟏 [ ]
Solution:
= 𝐴
+ +
𝑠−2 𝐵 𝐶
𝑠(𝑠+2)(𝑠−1) 𝑠 𝑠+2 𝑠−1
= 𝐴(𝑠+2)(𝑠−1)+𝐵𝑠(𝑠−1)+𝐶𝑠(𝑠+2)
𝑠(𝑠+2)(𝑠−1)
𝐴(𝑠 + 2)(𝑠 − 1) + 𝐵𝑠(𝑠 − 1) + 𝐶𝑠(𝑠 + 2) = 𝑠 − 2 ⋯ (1)
Put 𝑠 = 0 in (1) Put 𝑠 = −2 in (1) Put 𝑠 = 1 in (1)
𝐴(2)(−1) = −2 𝐵(−2)(−3) = −4 3𝐶 = −1
Engineering Maths - IV
Laplace Transform
Page
40

41. ⇒ 𝐴 = 1 ⇒ 𝐵 = −4
= −2
6 3
⇒ 𝐶 = −1
3
∴
𝑠−2 2
= 1
− −
𝑠(𝑠+2)(𝑠−1) 𝑠 𝑠+2 3(𝑠−1)
1
𝑠−2 1 2 1 1 1
𝐿−1 [ ] = 𝐿−1 [ ] − 𝐿−1 [ ] − 𝐿−1 [ ]
𝑠(𝑠+2)(𝑠−1) 𝑠 3 𝑠+2 3 𝑠−1
𝐿−1 [ 𝑠−2
] = 1 − 2
𝑒−2𝑡 − 1
𝑒𝑡
𝑠(𝑠+2)(𝑠−1) 3 3
Example: 5.47 Find 𝐿−1 [ 2𝑠−3
(𝑠−1)(𝑠−2)2
]
Solution:
2𝑠−3 𝐴 𝐵
= + +
(𝑠−1)(𝑠−2)2 𝑠−1 𝑠−2 (𝑠−2)2
𝐶
2
= 𝐴(𝑠−2) +𝐵(𝑠−1)(𝑠−2)+𝐶(𝑠−1)
(𝑠−1)(𝑠−2)2
𝐴(𝑠 − 2)2 + 𝐵(𝑠 − 1)(𝑠 − 2) + 𝐶(𝑠 − 1) = 2𝑠 − 3 ⋯ (1)
Put 𝑠 = 1 in (1) Put 𝑠 = 2 in (1) Equating the coefficient of 𝑠2
𝐴 = −1 𝐶 = 1
𝐴 + 𝐵 = 0
𝐵 = −𝐴 ⇒ 𝐵 = 1
∴
2𝑠−3
= −1
+ 1
+
1
(𝑠−1)(𝑠−2)2 𝑠−1 𝑠−2 (𝑠−2)2
2𝑠−3 1 1 1
𝐿−1 [ ] = −𝐿−1 [ ] + 𝐿−1 [ ] + [ ]
(𝑠−1)(𝑠−2)2 𝑠−1 𝑠−2 (𝑠−2)2
𝑠2
= −𝑒𝑡 + 𝑒2𝑡 + 𝑒2𝑡𝐿−1 [ 1
]
(𝑠−1)(𝑠−2)2
2𝑠−3
∴ 𝐿−1 [ ] = −𝑒𝑡 + 𝑒2𝑡 + 𝑒2𝑡𝑡
𝟐
Example: 5.48 Find the inverse Laplace transform of 𝟓𝒔 −𝟏𝟓𝒔−𝟏𝟏
(𝒔+𝟏)(𝒔−𝟐)𝟑
Solution:
5𝑠2−15𝑠−11
= 𝐴 𝐵
+ + +
𝐶 𝐶
(𝑠+1)(𝑠−2)3 𝑠+1 𝑠−2 (𝑠−2)2 (𝑠−2)3
3 2
= 𝐴(𝑠−2) +𝐵(𝑠+1)(𝑠−2) +𝐶(𝑠+1)(𝑠−2)+𝐷(𝑠+1)
(𝑠−1)(𝑠−2)3
𝐴(𝑠 − 2)3 + 𝐵(𝑠 + 1)(𝑠 − 2)2 + 𝐶(𝑠 + 1)(𝑠 − 2) + 𝐷(𝑠 + 1) = 5𝑠2 − 15𝑠 − 11⋯ (1)
Put 𝑠 = −1 in (1)
𝐴(−27) = 9
Put 𝑠 = 2 in (1)
𝐷(3) = −21
Equating the coefficient of 𝑠3
𝐴 + 𝐵 = 0
𝐴 =
9
−27
⇒ 𝐴 = −1
3 3
𝐷 = −21
= −7 𝐵 = −𝐴 ⇒ 𝐵 = 1
3
Put 𝑠 = 0 in (1), we get
−8𝐴 + 4𝐵 − 2𝐶 + 𝐷 = −11
Engineering Maths - IV
Laplace Transform
Page
41

42. 8
+ 4
− 2𝐶 − 7 = −11
3 3
4 − 2𝐶 = 7 − 11
−2𝐶 = −8 ⇒ 𝐶 = 4
2
∴ 5𝑠 −15𝑠−11
= −1
+ + −
1 4 7
(𝑠+1)(𝑠−2)3 3(𝑠+1) 3(𝑠−2) (𝑠−2)2 (𝑠−2)3
2
𝐿−1 [5𝑠 −15𝑠−11
] = −1
𝐿−1 [ 1
] + 1
𝐿−1 [ 1
] + 4𝐿−1 [
(𝑠+1)(𝑠−2)3 3 𝑠+1 3 𝑠−2
1
(𝑠−2)2
] − 7𝐿−1 [
1
(𝑠−2)3
]
3 3 𝑠2 𝑠3
= −1
𝑒−𝑡 + 1
𝑒2𝑡 + 4𝑒2𝑡 𝐿−1 [ 1
] − 7𝑒2𝑡 𝐿−1 [ 1
]
𝑠2
2 2
𝐿−1 [5𝑠 −15𝑠−11
] = −1
𝑒−𝑡 + 1
𝑒2𝑡 + 4𝑒2𝑡 𝐿−1 [ 1
] − 7𝑒2𝑡 𝑡
(𝑠+1)(𝑠−2)3 3 3 2
Example: 5.49 Find the inverse Laplace transform of
𝟒𝒔+𝟓
(𝒔+𝟏)(𝒔𝟐+𝟒)
Solution:
4𝑠+5
=
𝐴
+ 𝐵𝑠+𝑐
(𝑠+1)(𝑠2+4) 𝑠+1 𝑠2+4
2
= 𝐴(𝑠 +4)+(𝐵𝑠+𝑐)(𝑠+1)
(𝑠+1)(𝑠2+4)
𝐴(𝑠2 + 4) + (𝐵𝑠 + 𝑐)(𝑠 + 1) = 4𝑠 + 5 ⋯ ⋯ (1)
Put 𝑠 = −1in (1)
𝐴(1 + 4) + 0 = 4(−1) + 5
Equating coefficients of 𝑠2term in (1)
𝐴 + 𝐵 = 0
Put 𝑠 = 0in (1)
𝐴(4) + 𝐶 = 5
𝐴(5) = 1 ⇒ 𝐴 = 1
5
𝐵 = −𝐴 ⇒ 𝐵 = −1
5
𝐶 = 5 − 4𝐴 = 5 − 4
5
= 25−4
= 21
5 5
∴
4𝑠+5
=
(𝑠+1)(𝑠2+4) 𝑠+1
−1 21
1
𝑠+
5
+ 5 5
𝑠2+4
= −
1 𝑠
+ 21 1
5(𝑠+1) 5(𝑠2+4) 5 (𝑠2+4)
𝐿−1 [ 4𝑠+5 𝑠
(𝑠+1)(𝑠2+4) 5 𝑠+1 5 𝑠2+4 5
1
𝑠2+4
] = 1
𝐿−1 [ 1
] − 1
𝐿−1 [ ] + 21
𝐿−1 [ ]
= 1
𝑒−𝑡 − 1
𝑐𝑜𝑠2𝑡 + 21 𝑠𝑖𝑛2𝑡
5 5 5 2
(𝑠+1)(𝑠2+4) 5 5 10
4𝑠+5 1 1 21
𝐿−1 [ ] = 𝑒−𝑡 − 𝑐𝑜𝑠2𝑡 + 𝑠𝑖𝑛2𝑡
Exercise: 5.8
Find the Inverse Laplace transforms using partial fraction for the following
1.
1
(𝑠+1)(𝑠+3)
Ans:
2
1
(𝑒−𝑡 − 𝑒−3𝑡)
2.
1
𝑠(𝑠+1)(𝑠+2)
Ans:
2
1
(𝑒−2𝑡 − 2𝑒−𝑡 + 1)
3.
54−3𝑠−5
(𝑠+1)(𝑠2−3𝑠+2)
Ans: 2𝑒−𝑡
3𝑡 3𝑡
𝑡 𝑡
+ 2 𝑒 2 𝑐𝑜𝑠ℎ 2 +8𝑒 2 𝑠𝑖𝑛ℎ
2 2
5. 7 INITIALAND FINAL VALUE THEOREMS
Initial value theorem
Engineering Maths - IV
Laplace Transform
Page
42

43. 𝑡→0 𝑠→∞
Proof:
We know that 𝐿[𝑓′(𝑡)] = 𝑠 𝐿[𝑓(𝑡)] − 𝑓(0)
= 𝑠𝐹(𝑠) − 𝑓(0)
∴ 𝑠𝐹(𝑠) = 𝐿[𝑓′(𝑡)] + 𝑓(0)
= ∫
∞
𝑒−𝑠𝑡𝑓′(𝑡)𝑑𝑡 + 𝑓(0)
0
Taking limit as 𝑠 → ∞ on both sides, we have
Engineering Maths - IV
Statement: If𝐿[𝑓(𝑡)] = 𝐹(𝑠), then lim 𝑓(𝑡) = lim𝑠𝐹(𝑠)
𝑠→∞ 𝑠→∞
lim𝑠𝐹(𝑠) = lim[∫0
∞
𝑒−𝑠𝑡𝑓′(𝑡)𝑑𝑡 + 𝑓(0)]
𝑠→∞ 0
∞
= lim[∫ 𝑒−𝑠𝑡 𝑓′ (𝑡)𝑑𝑡] + 𝑓(0)
𝑠→∞
∞
lim[𝑒−𝑠𝑡𝑓′(𝑡)]𝑑𝑡 + 𝑓(0)
= ∫0
∵ 𝑒−∞ = 0
𝑠→∞ 𝑡→0
= 0 + 𝑓(0)
= 𝑓(0)
= lim𝑓(𝑡)
𝑡→0
∴ lim𝑠𝐹(𝑠) = lim𝑓(𝑡)
𝑡→∞ 𝑠→0
Final value theorem
Statement: If the Laplace transforms of 𝑓(𝑡) and 𝑓′(𝑡) exist and 𝐿[𝑓(𝑡)] = 𝐹(𝑠),then lim𝑓(𝑡) = lim𝑠𝐹(𝑠)
Proof:
We know that 𝐿[𝑓′(𝑡)] = 𝑠 𝐿[𝑓(𝑡)] − 𝑓(0)
= 𝑠𝐹(𝑠) − 𝑓(0)
∴ 𝑠𝐹(𝑠) = 𝐿[𝑓′(𝑡)] + 𝑓(0)
= ∫
∞
𝑒−𝑠𝑡𝑓′(𝑡)𝑑𝑡 + 𝑓(0)
0
Taking limit as 𝑠 → 0 on both sides, we have
𝑠→0 𝑠→0 0
lim𝑠𝐹(𝑠) = lim[∫
∞
𝑒−𝑠𝑡𝑓′(𝑡)𝑑𝑡 + 𝑓(0)]
= lim[
𝑠→0 0
∞
∫ 𝑒−𝑠𝑡𝑓′(𝑡)𝑑𝑡] + 𝑓(0)
𝑠→0
∞
lim[𝑒−𝑠𝑡𝑓′(𝑡)]𝑑𝑡 + 𝑓(0)
= ∫0
∞
𝑓′(𝑡)𝑑𝑡 + 𝑓(0)
= ∫0
= [𝑓(𝑡)]∞ + 𝑓(0)
𝑡→∞ 𝑠→0
0
= 𝑓(∞) − 𝑓(0) + 𝑓(0)
= 𝑓(∞)
= lim𝑓(𝑡)
𝑡→∞
∴ lim𝑓(𝑡) = lim𝑠𝐹(𝑠)
Example: 5.50 Verify the initial value theorem for the function 𝒇(𝒕) = 𝒂𝒆−𝒃𝒕
Laplace Transform
Page43

44. Engineering Maths - IV
Solution:
Given 𝑓(𝑡) = 𝑎𝑒−𝑏𝑡
𝐹(𝑠) = 𝐿[𝑓(𝑡)]
= 𝐿[𝑎𝑒−𝑏𝑡]
= 𝑎
1
𝑠+𝑏
𝑠𝐹(𝑠) = 𝑎𝑠
𝑠+𝑏
Initial value theorem is lim𝑓(𝑡) = lim𝑠𝐹(𝑠)
𝑡→0 𝑠→∞
𝑡→0
lim𝑓(𝑡) = lim 𝑎𝑒−𝑏𝑡
𝑡→0
= 𝑎 ⋯ ⋯ ⋯ (1)
𝑠→∞ 𝑠→∞ 𝑠+𝑏
𝑎𝑠
lim𝑠𝐹(𝑠) = lim [ ]
= lim [
𝑎𝑠
𝑏
𝑠→∞ 𝑠(1+𝑠
)
𝑎
𝑏
𝑠→∞ (1+𝑠
)
] == lim [ ]
𝑡→0 𝑠→∞
= 𝑎 ⋯ ⋯ ⋯ (2)
From (1) and (2), lim𝑓(𝑡) = lim𝑠𝐹(𝑠)
∴ Initial value theorem is verified
Example: 5.51Verify the initial value theorem and Final value theorem for the function
𝒇(𝒕) = 𝟏 + 𝒆−𝒕[𝒔𝒊𝒏𝒕 + 𝒄𝒐𝒔𝒕].
Solution:
Given 𝑓(𝑡) = 1 + 𝑒−𝑡[𝑠𝑖𝑛𝑡 + 𝑐𝑜𝑠𝑡]
𝐹(𝑠) = 𝐿[𝑓(𝑡)]
= 𝐿[1 + 𝑒−𝑡[𝑠𝑖𝑛𝑡 + 𝑐𝑜𝑠𝑡]]
= 𝐿[1] + 𝐿[𝑒−𝑡[𝑠𝑖𝑛𝑡 + 𝑐𝑜𝑠𝑡]]
= 𝐿[1] + 𝐿[𝑠𝑖𝑛𝑡 + 𝑐𝑜𝑠𝑡]𝑠→𝑠+1
𝑠 𝑠2+1
1 𝑠
= 1
+ [ + ]
𝑠2+1 𝑠→𝑠+1
= 1
+ +
1 𝑠+1
𝑠 (𝑠+1)2+1 (𝑠+1)2+1
𝐹(𝑠) = 1
+ +
1 𝑠+1
𝑠 𝑠2+2𝑠+2 𝑠2+2𝑠+2
𝑠𝐹(𝑠) = 1 + +
𝑠 𝑠2+𝑠
𝑠2+2𝑠+2 𝑠2+2𝑠+2
Initial value theorem is lim𝑓(𝑡) = lim𝑠𝐹(𝑠)
𝑡→0 𝑠→∞
lim𝑓(𝑡) = lim[1 + 𝑒−𝑡[𝑠𝑖𝑛𝑡 + 𝑐𝑜𝑠𝑡]]
𝑡→0 𝑡→0
= 1 + 0 + 1 = 2 ⋯ ⋯ ⋯ (1)
lim𝑠𝐹(𝑠) = lim [1 +
𝑠→∞ 𝑠→∞
+
𝑠 𝑠2+𝑠
𝑠2+2𝑠+2 𝑠2+2𝑠+2
]
Laplace Transform
Page44

45. = 1 + lim [
1
1
(1+𝑠
)
2 2 2 2
𝑠→∞ 𝑠(1+𝑠+𝑠2) (1+𝑠+𝑠2)
+ ]
𝑡→0 𝑠→∞
= 1 + 0 + 1 = 2 ⋯ ⋯ ⋯ (2)
From (1) and (2), lim𝑓(𝑡) = lim𝑠𝐹(𝑠)
𝑡→∞ 𝑠→0
∴ Initial value theorem is verified
Final value theorem is lim𝑓(𝑡) = lim𝑠𝐹(𝑠)
lim𝑓(𝑡) = lim(1 + 𝑒−𝑡[𝑠𝑖𝑛𝑡 + 𝑐𝑜𝑠𝑡])
𝑡→∞ 𝑡→∞
= 1 + 0 = 1 ⋯ ⋯ ⋯ (3)
𝑠→0 𝑠→0
lim𝑠𝐹(𝑠) = lim [1 + +
𝑠 𝑠2+𝑠
𝑠2+2𝑠+2 𝑠2+2𝑠+2
]
𝑡→∞ 𝑠→0
= 1 + 0 + 0 = 1 ⋯ ⋯ ⋯ (4)
From (3) and (4), lim𝑓(𝑡) = lim𝑠𝐹(𝑠)
∴ Final value theorem is verified.
Example: 5.52 Verify the initial value theorem and Final value theorem for the function
𝒔(𝒔+𝟐)𝟐
𝟏
𝒇(𝒕) = 𝑳−𝟏 [ ]
Solution:
𝑠(𝑠+2)2
1
Given 𝑓(𝑡) = 𝐿−1 [ ] ⋯ (1)
(𝑠+2)2
𝑡 −1 1
= ∫
0 𝐿 [ ] 𝑑𝑡 = 𝑠2
𝑡 −2𝑡 −1 1
∫
0 𝑒 𝐿 [ ] 𝑑𝑡
𝑡 −2𝑡
= ∫
0 𝑒 𝑡 𝑑𝑡
= ∫
𝑡
𝑡𝑒−2𝑡 𝑑𝑡
0
𝑒−2𝑡
= [𝑡 ( ) − ]
−2 (−2)2
0
(1)𝑒−2𝑡 𝑡
−2𝑡 −2𝑡
= −𝑡 𝑒
− 𝑒
− 0 + 1
2 4 4
−2𝑡 −2𝑡
∴ 𝑓(𝑡) = 1
− 𝑡𝑒
− 𝑒
4 2 4
From (1), 𝐹(𝑠) =
1
𝑠(𝑠+2)2
1
𝑠𝐹(𝑠) =
(𝑠+2)2
Initial value theorem is lim𝑓(𝑡) = lim𝑠𝐹(𝑠)
𝑡→0 𝑠→∞
𝑡→0 𝑡→0 4 2 4
−2𝑡 −2𝑡
lim𝑓(𝑡) = lim [1
− 𝑡𝑒
− 𝑒
]
= 1
− 0 − 1
= 0
4 4
∴ lim𝑓(𝑡) = 0 ⋯ (2)
𝑡→0
𝑠→∞
lim𝑠𝐹(𝑠) = lim
𝑠→∞ (𝑠+2)2
1
= 0
Engineering Maths - IV
Laplace Transform
Page45

46. 𝑡→0 𝑠→∞
𝑡→∞ 𝑠→0
∴ Initial value theorem is verified
Final value theorem is lim𝑓(𝑡) = lim𝑠𝐹(𝑠)
Engineering Maths - IV
∴ lim𝑠𝐹(𝑠) = 0 ⋯ (3)
𝑠→∞
From (2) and (3), lim𝑓(𝑡) = lim𝑠𝐹(𝑠)
𝑡→∞ 𝑡→∞ 4 2 4
−2𝑡 −2𝑡
lim𝑓(𝑡) = lim [1
− 𝑡𝑒
− 𝑒
]
= 1
− 0 − 0 = 1
⋯ (4)
4 4
𝑠→0 𝑠→0 (𝑠+2)2
lim𝑠𝐹(𝑠) = lim [ 1
]
= 1
⋯ (5)
𝑡→∞ 𝑠→0
4
From (4) and (5), lim𝑓(𝑡) = lim𝑠𝐹(𝑠)
∴ Final value theorem is verified
Example: 5.53 Verify the initial value theorem and Final value theorem for the function
𝒇(𝒕) = 𝒆−𝒕(𝒕 + 𝟐)𝟐
Solution:
Given 𝑓(𝑡) = 𝑒−𝑡(𝑡 + 2)2
= 𝑒−𝑡(𝑡2 + 4𝑡 + 4)
𝐹(𝑠) = 𝐿[𝑓(𝑡)]
= 𝐿[𝑒−𝑡(𝑡2 + 4𝑡 + 4)]
= 𝐿[𝑡2 + 4𝑡 + 4]𝑠→𝑠+1
= [𝐿(𝑡2) + 4𝐿(𝑡) + 4𝐿(1)]𝑠→𝑠+1
𝑠3 𝑠2
= [2!
+ 4 1
+ 4 1
]
𝑠 𝑠→𝑠+1
=
2 1
+ 4 + 4
(𝑠+1)3 (𝑠+1)2 𝑠+1
1
𝑠𝐹(𝑠) = +
2𝑠 4𝑠
+ 4𝑠
(𝑠+1)3 (𝑠+1)2 𝑠+1
Initial value theorem is lim𝑓(𝑡) = lim𝑠𝐹(𝑠)
𝑡→0
𝑡→0 𝑠→∞
lim𝑓(𝑡) = lim[𝑒−𝑡(𝑡2 + 4𝑡 + 4)]
𝑡→0
= 4 ⋯ (1)
𝑠→∞
2𝑠
lim𝑠𝐹(𝑠) = lim [ +
𝑠→∞ (𝑠+1)3 (𝑠+1)2 𝑠+1
4𝑠
+ 4𝑠
]
𝑠→∞ 3
𝑠
𝑠 (1+ )
1 3
2𝑠 4𝑠
= lim [ +
2
𝑠 (1+𝑠
)
1 2 +
4𝑠
1
𝑠(1+𝑠
)
]
𝑠→∞ 2
𝑠
𝑠 (1+ )
1 3
2
= lim [ +
4
𝑠(1+𝑠
)
1 2 +
4
1
(1+𝑠
)
]
= 0 + 0 + 4
Laplace Transform
Page46

47. 𝑡→0 𝑠→∞
𝑡→∞ 𝑠→0
∴ Initial value theorem is verified
Final value theorem is lim𝑓(𝑡) = lim𝑠𝐹(𝑠)
Engineering Maths - IV
= 4 ⋯ (2)
From (1) and (2), lim𝑓(𝑡) = lim𝑠𝐹(𝑠)
lim𝑓(𝑡) = lim[𝑒−𝑡(𝑡2 + 4𝑡 + 4)]
𝑡→∞ 𝑡→∞
𝑠→0 𝑠→0
= 0 ⋯ (3)
2𝑠
lim𝑠𝐹(𝑠) = lim [ +
(𝑠+1)3 (𝑠+1)2 𝑠+1
4𝑠
+ 4𝑠
]
= 0 ⋯ (4)
From (3) and (4), lim𝑓(𝑡) = lim𝑠𝐹(𝑠)
𝑡→∞ 𝑠→0
∴ Final value theorem is verified.
Example: 5.54 If 𝑳[𝒇(𝒕)] =
𝟏
𝒔(𝒔+𝟏)
, find the 𝐥𝐢𝐦𝒇(𝒕) and 𝐥𝐢𝐦𝒇(𝒕) using initial and final value
𝒕→𝟎 𝒕→∞
theorems.
Solution:
Given 𝐿[𝑓(𝑡)] =
1
𝑠(𝑠+1)
⋯ (1)
𝑖𝑒. , 𝐹(𝑠) =
1
=> 𝑠𝐹(𝑠) =
𝑠(𝑠+1) (𝑠+1)
1
Initial value theorem is lim𝑓(𝑡) = lim𝑠𝐹(𝑠)
𝑡→0 𝑠→∞
= lim
1
𝑠→∞ (𝑠+1)
= 0
Final value theorem is lim𝑓(𝑡) = lim𝑠𝐹(𝑠)
𝑡→∞ 𝑠→0
= lim
1
𝑠→0 (𝑠+1)
= 1
Exercise: 5.9
1. Verify the initial value theorem for the function
2. Verify the initial value theorem for the function
3. Verify the initial value theorem for the function
𝑓(𝑡) = 𝑒−𝑡𝑠𝑖𝑛𝑡
𝑓(𝑡) = sin2 𝑡
𝑓(𝑡) = 1 + 𝑒−𝑡 + 𝑡2
4. Verify the Final value theorem for the function 𝑓(𝑡) = 1 − 𝑒−𝑎𝑡
5. Verify the Final value theorem for the function 𝑓(𝑡) = 𝑡2𝑒−3𝑡
5.8 CONVOLUTION THEOREM
Definition: Convolution of two functions
The convolution of two functions 𝑓(𝑡) and 𝑔(𝑡) is denoted by 𝑓(𝑡) ∗ 𝑔(𝑡) and defined by
𝑡
𝑓(𝑢)𝑔(𝑡 − 𝑢)𝑑𝑢.
𝑓(𝑡) ∗ 𝑔(𝑡) = ∫0
State and prove Convolution theorem
Statement: If 𝐿[𝑓(𝑡)] = 𝐹(𝑠) and 𝐿[𝑔(𝑡)] = 𝐺(𝑠), then 𝐿[𝑓(𝑡)] ∗ 𝐿[𝑔(𝑡)] = 𝐹(𝑠)𝐺(𝑠)
Laplace Transform
Page47

48. Engineering Maths - IV
Proof:
𝑡
𝑓(𝑢)𝑔(𝑡 − 𝑢)𝑑𝑢
We have 𝑓(𝑡) ∗ 𝑔(𝑡) = ∫0
𝐿[𝑓(𝑡) ∗ 𝑔(𝑡)] = ∫
∞
[𝑓(𝑡) ∗ 𝑔(𝑡)] 𝑒−𝑠𝑡𝑑𝑡
0
∞ 𝑡
𝑓(𝑢)𝑔(𝑡 − 𝑢)𝑑𝑢𝑒−𝑠𝑡𝑑𝑡
= ∫0 ∫0
= ∫0 ∫0
∞ 𝑡
𝑓(𝑢)𝑔(𝑡 − 𝑢)𝑒−𝑠𝑡 𝑑𝑢𝑑𝑡 ⋯ (1)
Now we have no change the order of integration.
𝑢 = 0, 𝑢 = 𝑡;𝑡 = 0, 𝑡 = ∞
Change of order is . Draw horizontal strip PQ
At P, 𝑡 = 𝑢, At A 𝑢 = ∞
∞ ∞
𝑓(𝑢)𝑔(𝑡 − 𝑢)𝑒−𝑠𝑡𝑑𝑡𝑑𝑢
𝐿[𝑓(𝑡) ∗ 𝑔(𝑡)] = ∫0 ∫𝑢
∞
= ∫
0 𝑓(𝑢) [∫
𝑢
∞
𝑔(𝑡 − 𝑢)𝑒−𝑠𝑡 𝑑𝑡]𝑑𝑢 ⋯ (2)
Put 𝑡 − 𝑢 = 𝑥 ⋯ (3)
𝑡 = 𝑢 + 𝑥 ⇒ 𝑑𝑡 = 𝑑𝑥
When 𝑡 = 𝑢;(3) ⇒ 𝑥 = 0
When 𝑡 = ∞; (3) ⇒ 𝑥 = ∞
(2) ⇒ 𝐿[𝑓(𝑡) ∗ 𝑔(𝑡)] = ∫0
∞
𝑓(𝑢) [∫0
∞
𝑔(𝑥)𝑒−𝑠(𝑢+𝑥)𝑑𝑥]𝑑𝑢
∞
= ∫
0 𝑓(𝑢) [∫
0
∞
𝑔(𝑥)𝑒−𝑠𝑢𝑒−𝑠𝑥𝑑𝑥]𝑑𝑢
∞
𝑔(𝑥)𝑒−𝑠𝑥𝑑𝑥
∞ −𝑠𝑢
= ∫
0 𝑓(𝑢)𝑒 𝑑𝑢 ∫
0
= 𝐿[𝑓(𝑢)]𝐿[𝑔(𝑥)]
∴ 𝐿[𝑓(𝑡) ∗ 𝑔(𝑡)] = 𝐹(𝑠)𝐺(𝑠)
Note: Convolution theorem is very useful to compute inverse Laplace transform of product of two terms
Convolution theorem is 𝐿[𝑓(𝑡) ∗ 𝑔(𝑡)] = 𝐹(𝑠)𝐺(𝑠)
𝐿−1[𝐹(𝑠)𝐺(𝑠)] = 𝑓(𝑡) ∗ 𝑔(𝑡)
𝐿−1[𝐹(𝑠)𝐺(𝑠)] = 𝐿−1[𝐹(𝑠)] ∗ 𝐿−1[𝐺(𝑠)]
Problems under Convolution theorem
Example: 5.55 Find 𝑳−𝟏 [
𝟏
(𝒔+𝒂)(𝒔+𝒃)
] using convolution theorem.
Solution:
𝐿−1 1 1 1
[ ] = 𝐿−1 [ ] ∗ 𝐿−1 [ ]
(𝑠+𝑎)(𝑠+𝑏) (𝑠+𝑎) (𝑠+𝑏)
= 𝑒−𝑎𝑡 ∗ 𝑒−𝑏𝑡
= ∫0
𝑡
𝑒−𝑎𝑢𝑒−𝑏(𝑡−𝑢) 𝑑𝑢
∫0
= 𝑒−𝑏𝑡 𝑡
𝑒−𝑎𝑢 𝑒𝑏𝑢𝑑𝑢
∫0
= 𝑒−𝑏𝑡 𝑡
𝑒(𝑏−𝑎)𝑢 𝑑𝑢
Laplace Transform
Page48

49. −𝑏𝑡
= 𝑒 [
𝑏−𝑎
]
0
𝑒(𝑏−𝑎)𝑢 𝑡
−𝑏𝑡
= 𝑒
[𝑒(𝑏−𝑎)𝑡 − 1 ]
𝑏−𝑎
𝑏−𝑎
−𝑏𝑡
= 𝑒
[𝑒𝑏𝑡−𝑎𝑡 − 1 ]
=
1
𝑏−𝑎
[𝑒−𝑏𝑡+𝑏𝑡−𝑎𝑡 − 𝑒−𝑏𝑡 ]
∴ 𝐿−1 [ ] =
1 1
(𝑠+𝑎)(𝑠+𝑏) 𝑏−𝑎
[𝑒−𝑎𝑡 − 𝑒−𝑏𝑡 ]
Example: 5.56 Find the inverse Laplace transform
𝒔𝟐
(𝒔𝟐+𝒂𝟐)(𝒔𝟐+𝒃𝟐)
by using convolution theorem.
Solution:
𝐿−1 [
𝑠2
(𝑠2+𝑎2)(𝑠2+𝑏2)
𝑠 𝑠
] = 𝐿−1 [
(𝑠2+𝑎2) (𝑠2+𝑏2)
]
𝑠 𝑠
= 𝐿−1 [ ] ∗ 𝐿−1 [ ]
(𝑠2+𝑎2) (𝑠2+𝑏2)
= 𝑐𝑜𝑠𝑎𝑡 ∗ 𝑐𝑜𝑠𝑏𝑡
𝑡
= ∫0𝑐𝑜𝑠𝑎𝑢 𝑐𝑜𝑠𝑏(𝑡 − 𝑢)𝑑𝑢
= ∫0
𝑡 cos(𝑎𝑢+𝑏𝑡−𝑏𝑢)+cos(𝑎𝑢−𝑏𝑡+𝑏𝑢)
𝑑𝑢
= 1 ∫
2 0
2
𝑡
(cos(𝑎𝑢 + 𝑏𝑡 − 𝑏𝑢) + cos(𝑎𝑢 − 𝑏𝑡 + 𝑏𝑢)) 𝑑𝑢
= 1 𝑡
[cos(𝑎 − 𝑏) 𝑢 + 𝑏𝑡 + cos(𝑎 + 𝑏) 𝑢 − 𝑏𝑡]𝑑𝑢
∫
2 0
=
2 𝑎 −𝑏 𝑎 +𝑏
+ ]
0
𝑡
1
[𝑠𝑖𝑛[(𝑎−𝑏)𝑢+𝑏𝑡] 𝑠𝑖𝑛[(𝑎+𝑏)𝑢+𝑏𝑡]
= 1
[sin(𝑎𝑡−𝑏𝑡+𝑏𝑡)
+ sin(𝑎𝑡−𝑏𝑡+𝑏𝑡)
− 𝑠𝑖𝑛𝑏𝑡
+ 𝑠𝑖𝑛𝑏𝑡
]
2 𝑎−𝑏 𝑎+𝑏 𝑎−𝑏 𝑎+𝑏
= 1
[sin 𝑎𝑡
+ sin 𝑎𝑡
− 𝑠𝑖𝑛𝑏𝑡
+ 𝑠𝑖𝑛𝑏𝑡
]
2 𝑎−𝑏 𝑎+𝑏 𝑎−𝑏 𝑎+𝑏
= 1
[(𝑎+𝑏)𝑠𝑖𝑛𝑎𝑡+(𝑎−𝑏)𝑠𝑖𝑛𝑎𝑡−(𝑎+𝑏)𝑠𝑖𝑛𝑏𝑡+(𝑎−𝑏)𝑠𝑖𝑛𝑏𝑡
]
2 𝑎2−𝑏2
= 1
[2𝑎𝑠𝑖𝑛𝑎𝑡−2𝑏𝑠𝑖𝑛𝑏𝑡
]
2 𝑎2−𝑏2
= 1
[2(𝑎𝑠𝑖𝑛𝑎𝑡−𝑏𝑠𝑖𝑛𝑏𝑡)
]
2 𝑎2−𝑏2
∴ 𝐿−1 [
𝑠2
] = 𝑎𝑠𝑖𝑛𝑎𝑡−𝑏𝑠𝑖𝑛𝑏𝑡
(𝑠2+𝑎2)(𝑠2+𝑏2) 𝑎2−𝑏2
Example: 5.57 Find the inverse Laplace transform
𝟏
(𝒔𝟐+𝒂𝟐)(𝒔𝟐+𝒃𝟐)
by using convolution theorem.
Solution:
1 1
𝐿−1 [ 1
] = 𝐿−1 [
(𝑠2+𝑎2)(𝑠2+𝑏2) (𝑠2+𝑎2) (𝑠2+𝑏2)
]
1 1
= 𝐿−1 [ ] ∗ 𝐿−1 [ ]
(𝑠2+𝑎2) (𝑠2+𝑏2)
= 1
𝑠𝑖𝑛𝑎𝑡 ∗ 1
𝑠𝑖𝑛𝑏𝑡
𝑎 𝑏
Engineering Maths - IV
Laplace Transform
Page49

50. = 1 𝑡
∫ 𝑠𝑖𝑛𝑎𝑢 𝑠𝑖𝑛𝑏(𝑡 − 𝑢)𝑑𝑢
𝑎𝑏 0
∫
𝑎𝑏 0
= 1 𝑡 cos(𝑎𝑢−𝑏𝑡+𝑏𝑢)−cos(𝑎𝑢+𝑏𝑡−𝑏𝑢)
𝑑𝑢
=
1
∫
2𝑎𝑏 0
2
𝑡
(cos(𝑎𝑢 − 𝑏𝑡 + 𝑏𝑢) − cos(𝑎𝑢 + 𝑏𝑡 − 𝑏𝑢)) 𝑑𝑢
= 1 𝑡
∫ [cos[(a + b)u − bt] − cos[(a − b)u + bt]]𝑑𝑢
2 0
=
2𝑎𝑏 𝑎+𝑏 𝑎−𝑏
− ]
0
𝑡
1
[𝑠𝑖𝑛[(𝑎+𝑏)𝑢−𝑏𝑡] 𝑠𝑖𝑛[(𝑎−𝑏)𝑢+𝑏𝑡]
=
=
2𝑎𝑏 𝑎+𝑏 𝑎−𝑏 𝑎 +𝑏 𝑎 −𝑏
1
[sin(𝑎𝑡+𝑏𝑡−𝑏𝑡)
− sin(𝑎𝑡−𝑏𝑡+𝑏𝑡)
+ 𝑠𝑖𝑛𝑏𝑡
+ 𝑠𝑖𝑛𝑏𝑡
]
2𝑎𝑏 𝑎+𝑏 𝑎−𝑏 𝑎+𝑏 𝑎−𝑏
1
[sin 𝑎𝑡
− sin 𝑎𝑡
− 𝑠𝑖𝑛𝑏𝑡
+ 𝑠𝑖𝑛𝑏𝑡
]
=
1
[(𝑎−𝑏)𝑠𝑖𝑛𝑎𝑡−(𝑎+𝑏)𝑠𝑖𝑛𝑎𝑡+(𝑎−𝑏)𝑠𝑖𝑛𝑏𝑡+(𝑎+𝑏)𝑠𝑖𝑛𝑏𝑡
]
2𝑎𝑏 𝑎2−𝑏2
=
1
[−2𝑏𝑠𝑖𝑛𝑎𝑡+2𝑎𝑠𝑖𝑛𝑏𝑡
]
2𝑎𝑏 𝑎2−𝑏2
=
1
[2(𝑎𝑠𝑖𝑛𝑏𝑡−𝑏𝑠𝑖𝑛𝑎𝑡)
]
2𝑎𝑏 𝑎2−𝑏2
1 𝑎𝑠𝑖𝑛𝑏𝑡−𝑏𝑠𝑖𝑛𝑎𝑡
∴ 𝐿−1 [ ] =
(𝑠2+𝑎2)(𝑠2+𝑏2) 𝑎𝑏(𝑎2−𝑏2)
Example: 5.58 Find the inverse Laplace transform
𝒔
(𝒔𝟐+𝟒)(𝒔𝟐+𝟗)
by using convolution theorem.
Solution:
𝑠 1 𝑠
𝐿−1 [ ] = 𝐿−1 [
(𝑠2+4)(𝑠2+9) (𝑠2+4) (𝑠2+9)
]
1 𝑠
= 𝐿−1 [ ] ∗ 𝐿−1 [ ]
(𝑠2+4) (𝑠2+9)
2
= 1
𝑠𝑖𝑛2𝑡 ∗ 𝑐𝑜𝑠3𝑡
= 1 𝑡
∫ 𝑠𝑖𝑛2𝑢 𝑐𝑜𝑠3(𝑡 − 𝑢)𝑑𝑢
2 0
2
∫
2 0
= 1 𝑡 sin(2𝑢+3𝑡−3𝑢)+sin(2𝑢−3𝑡+3𝑢)
𝑑𝑢
= 1 𝑡
[sin(3t − 𝑢) + sin(5u − 3t)]𝑑𝑢
∫
4 0
=
4 −1 5
− ]
0
𝑡
1
[−cos(3𝑡−𝑢) cos(5𝑢−3𝑡)
= 1
[cos(3t−t)
− cos(5t−3𝑡)
− 𝑐𝑜𝑠3𝑡
+ 𝑐𝑜𝑠3𝑡
]
4 1 5 1 5
= 1
[𝑐𝑜𝑠2𝑡 − cos2𝑡
− 𝑐𝑜𝑠3𝑡 + 𝑐𝑜𝑠3𝑡
]
4 5 5
= 1
[5𝑐𝑜𝑠2𝑡−𝑐𝑜𝑠2𝑡−5𝑐𝑜𝑠3𝑡+𝑐𝑜𝑠3𝑡
]
4 5
20
= 1
[4𝑐𝑜𝑠2𝑡 − 4𝑐𝑜𝑠3𝑡]
∴ 𝐿−1 [
𝑠
] = 𝑐𝑜𝑠2𝑡−𝑐𝑜𝑠3𝑡
(𝑠2+4)(𝑠2+9) 5
Example: 5.59 Find 𝑳−𝟏 [ 𝒔
] by using convolution theorem.
(𝒔𝟐+𝒂𝟐)
𝟐
Engineering Maths - IV
Laplace Transform
Page50

51. Engineering Maths - IV
Solution:
𝑠 1
𝐿−1 [ ] = 𝐿−1 [
(𝑠2+𝑎2)2 (𝑠2+𝑎2) (𝑠2+𝑎2)
𝑠
]
1 𝑠
= 𝐿−1 [ ] ∗ 𝐿−1 [ ]
(𝑠2+𝑎2) (𝑠2+𝑎2)
𝑎
= 1
𝑠𝑖𝑛𝑎𝑡 ∗ 𝑐𝑜𝑠𝑎𝑡
= 1 𝑡
∫ 𝑠𝑖𝑛𝑎𝑢 𝑐𝑜𝑠𝑎(𝑡 − 𝑢)𝑑𝑢
𝑎 0
∫
𝑎 0
= 1 𝑡 sin(𝑎𝑢+𝑎𝑡−𝑎𝑢)+sin(𝑎𝑢−𝑎𝑡+𝑎𝑢)
𝑑𝑢
= 1
2
𝑡
∫ [𝑠𝑖𝑛𝑎𝑡 + sin(2𝑎𝑢 − 𝑎𝑡)]𝑑𝑢
2𝑎 0
= 1
[∫
𝑡
𝑠𝑖𝑛𝑎𝑡𝑑𝑢 +
2𝑎 0
𝑡
∫0 sin(2𝑎𝑢 − 𝑎𝑡) 𝑑𝑢]
2𝑎
= 1
[𝑠𝑖𝑛𝑎𝑡 ∫
0
𝑡
𝑑𝑢 + ∫0
𝑡
sin(2𝑎𝑢 − 𝑎𝑡) 𝑑𝑢]
1
0
𝑡
= [𝑠𝑖𝑛𝑎𝑡(𝑢) − (
cos(2𝑎𝑢−𝑎𝑡)
2𝑎 2𝑎
𝑡
) ]
0
= 1
[𝑡𝑠𝑖𝑛𝑎𝑡 − cos(2𝑎𝑡−𝑎𝑡)
+ 𝑐𝑜𝑠𝑎𝑡
] 2
𝑎 2𝑎 2𝑎
= 1
[𝑡𝑠𝑖𝑛𝑎𝑡 − cos 𝑎𝑡
+ 𝑐𝑜𝑠𝑎𝑡
] 2𝑎 2𝑎 2𝑎
= 1
2𝑎
𝑡𝑠𝑖𝑛𝑎𝑡
𝑠 𝑡𝑠𝑖𝑛𝑎𝑡
∴ 𝐿−1 [ ] =
(𝑠2+𝑎2)2 2𝑎
Example: 5.60 Find 𝑳−𝟏 [ 𝟏
] by using convolution theorem.
(𝒔𝟐+𝒂𝟐)
𝟐
Solution:
1 1
𝐿−1 [ ] = 𝐿−1 [
1
(𝑠2+𝑎2)2 (𝑠2+𝑎2) (𝑠2+𝑎2)
]
1 1
= 𝐿−1 [ ] ∗ 𝐿−1 [ ]
(𝑠2+𝑎2) (𝑠2+𝑎2)
𝑎
= 1
𝑠𝑖𝑛𝑎𝑡 ∗ 1
𝑠𝑖𝑛𝑎𝑡
𝑎
= 1 𝑡
𝑠𝑖𝑛𝑎𝑢 𝑠𝑖𝑛𝑎(𝑡 − 𝑢)𝑑𝑢
𝑎2 ∫0
𝑎2 ∫0
= 1 𝑡 cos(𝑎𝑢−𝑎𝑡+𝑎𝑢)−cos(𝑎𝑢+𝑎𝑡−𝑎𝑢)
𝑑𝑢
=
1
2
𝑡
[cos(2𝑎𝑢 − 𝑎𝑡) − 𝑐𝑜𝑠𝑎𝑡]𝑑𝑢
2𝑎2 ∫0
1
2𝑎2 ∫0
𝑡
= [ cos(2𝑎𝑢 − 𝑎𝑡)𝑑𝑢 − ∫0
𝑡
𝑐𝑜𝑠𝑎𝑡 𝑑𝑢]
1
2𝑎2 ∫0
𝑡
= [ cos(2𝑎𝑢 − 𝑎𝑡)𝑑𝑢 − 𝑐𝑜𝑠𝑎𝑡 ∫0
𝑡
𝑑𝑢]
=
2𝑎2 2𝑎 0
𝑡
1
[(sin(2𝑎𝑢−𝑎𝑡)
0
)𝑡
) − 𝑐𝑜𝑠𝑎𝑡(𝑢 ]
=
1
[sin(2𝑎𝑡−𝑎𝑡)
− sin(−𝑎𝑡)
− 𝑡𝑐𝑜𝑠𝑎𝑡]
2𝑎2 2𝑎 2𝑎
Laplace Transform
Page51

52. =
1
[sin𝑎𝑡
+ 𝑠𝑖𝑛𝑎𝑡
− 𝑡𝑐𝑜𝑠𝑎𝑡]
2𝑎2 2𝑎 2𝑎
1
2𝑎2 2𝑎
= [2sin𝑎𝑡
− 𝑡𝑐𝑜𝑠𝑎𝑡]
(𝑠2+𝑎2)2 2𝑎2 𝑎
1 1 sin𝑎𝑡
∴ 𝐿−1 [ ] = [ − 𝑡𝑐𝑜𝑠𝑎𝑡]
Example: 5.61 Find 𝑳−𝟏 [
𝒔𝟐
(𝒔𝟐+𝒂𝟐)
𝟐] by using convolution theorem.
Solution:
𝐿−1 [
𝑠2
(𝑠2+𝑎2)2
𝑠 𝑠
] = 𝐿−1 [
(𝑠2+𝑎2) (𝑠2+𝑎2)
]
𝑠 𝑠
= 𝐿−1 [ ] ∗ 𝐿−1 [ ]
(𝑠2+𝑎2) (𝑠2+𝑎2)
= 𝑐𝑜𝑠𝑎𝑡 ∗ 𝑐𝑜𝑠𝑎𝑡
𝑡
𝑐𝑜𝑠𝑎𝑢 𝑐𝑜𝑠𝑎(𝑡 − 𝑢)𝑑𝑢
= ∫0
= ∫0
𝑡 cos(𝑎𝑢+𝑎𝑡−𝑎𝑢)+cos(𝑎𝑢−𝑎𝑡+𝑎𝑢)
𝑑𝑢
= 1
2
𝑡
∫ [𝑐𝑜𝑠𝑎𝑡 + cos(2𝑎𝑢 − 𝑎𝑡)]𝑑𝑢
2 0
∫
2 0
= 1
[
𝑡
𝑐𝑜𝑠𝑎𝑡𝑑𝑢 + ∫0
𝑡
cos(2𝑎𝑢 − 𝑎𝑡) 𝑑𝑢]
2
= 1
[𝑐𝑜𝑠𝑎𝑡 ∫
0
𝑡 𝑡
𝑑𝑢 + ∫0 cos(2𝑎𝑢 − 𝑎𝑡) 𝑑𝑢]
1
0
𝑡
= [𝑐𝑜𝑠𝑎𝑡(𝑢) + (
sin(2𝑎𝑢−𝑎𝑡)
2 2𝑎
𝑡
) ]
0
= 1
[𝑡𝑐𝑜𝑠𝑎𝑡 + sin(2𝑎𝑡−𝑎𝑡)
+ 𝑠𝑖𝑛𝑎𝑡
]
2 2𝑎 2𝑎
= 1
[𝑡𝑐𝑜𝑠𝑎𝑡 + sin𝑎𝑡
+ 𝑠𝑖𝑛𝑎𝑡
]
2 2𝑎 2𝑎
= 1
[𝑡𝑐𝑜𝑠𝑎𝑡 + 2sin𝑎𝑡
]
2 2𝑎
∴ 𝐿−1 [
𝑠2
(𝑠2+𝑎2)2 2 𝑎
] = 1
[𝑡𝑐𝑜𝑠𝑎𝑡 + sin𝑎𝑡
]
Example: 5.62 Find 𝑳−𝟏 [
𝒔𝟐
(𝒔𝟐+𝟒)
𝟐] by using convolution theorem.
Solution:
𝐿−1 [
𝑠2
] = 𝐿−1 [
𝑠 𝑠
(𝑠2+22)2 (𝑠2+22) (𝑠2+22)
]
𝑠 𝑠
= 𝐿−1 [ ] ∗ 𝐿−1 [ ]
(𝑠2+22) (𝑠2+22)
= 𝑐𝑜𝑠2𝑡 ∗ 𝑐𝑜𝑠2𝑡
𝑡
= ∫0𝑐𝑜𝑠2𝑢 𝑐𝑜𝑠2(𝑡 − 𝑢)𝑑𝑢
= ∫0
𝑡 cos(2𝑢+2𝑡−2𝑢)+cos(2𝑢−2𝑡+2𝑢)
𝑑𝑢
= 1
2
𝑡
[𝑐𝑜𝑠2𝑡 + cos(4𝑢 − 2𝑡)]𝑑𝑢
∫
2 0
∫
2 0
= 1
[
𝑡
𝑐𝑜𝑠2𝑡𝑑𝑢 + ∫0
𝑡
cos(4𝑢 − 2𝑡) 𝑑𝑢]
Engineering Maths - IV
Laplace Transform
Page52

53. 2
= 1
[𝑐𝑜𝑠2𝑡 ∫
0
𝑡 𝑡
𝑑𝑢 + ∫0 cos(4𝑢 − 2𝑡) 𝑑𝑢]
1
0
𝑡
= [𝑐𝑜𝑠2𝑡(𝑢) + (
sin(4𝑢−2𝑡)
2 4
𝑡
) ]
0
= 1
[𝑡𝑐𝑜𝑠2𝑡 + sin(4𝑡−2𝑡)
− sin(−2𝑡)
]
2 4 4
= 1
[𝑡𝑐𝑜𝑠2𝑡 + sin2𝑡
+ 𝑠𝑖𝑛2𝑡
]
2 4 4
= 1
[𝑡𝑐𝑜𝑠2𝑡 + 2sin2𝑡
]
2 4
∴ 𝐿−1 [
𝑠2
(𝑠2+𝑎2)2 2 2
] = 1
[𝑡𝑐𝑜𝑠2𝑡 + sin2𝑡
]
Example: 5.63 Find 𝑳−𝟏 [
𝟏
𝒔(𝒔𝟐+𝟒)
] by using convolution theorem.
Solution:
1 1 1
𝐿−1 [ ] = 𝐿−1 [
𝑠(𝑠2+4) 𝑠 𝑠2+4
]
= 𝐿−1 [1
] ∗ 𝐿−1 [ 1
]
𝑠 𝑠2+4
= 1 ∗ sin2𝑡
2
= sin2𝑡
∗ 1
2
2
= ∫0
𝑡 sin2𝑢 (1)
𝑑𝑢
= [
−𝑐𝑜𝑠2
𝑢 4 0
𝑡 1
4
] = (−𝑐𝑜𝑠2𝑡 + 1)
4
= 1
(1 − 𝑐𝑜𝑠2𝑡)
Example: 5.64 Find the inverse Laplace transform
𝒔+𝟐
(𝒔𝟐+𝟒𝒔+𝟏𝟑)
𝟐 by using convolution theorem.
Solution:
𝑠+2 1
𝐿−1 [ 𝑠+2
] = 𝐿−1 [
(𝑠2+4𝑠+13)2 𝑠2+4𝑠+13 𝑠2+4𝑠+13
]
𝑠+2 1
= 𝐿−1 [ ] ∗ 𝐿−1 [ ]
𝑠2+4𝑠+13 𝑠2+4𝑠+13
= 𝐿−1 𝑠+2 1
[ ] ∗ 𝐿−1 [ ]
(𝑠+2)2+9 (𝑠+2)2+9
= 𝑒−2𝑡𝐿−1 𝑠 −2𝑡 −1 1
[ ] ∗ 𝑒 𝐿 [ ]
𝑠2+9 𝑠2+9
−2𝑡
= 𝑒−2𝑡𝑐𝑜𝑠3𝑡 ∗ 𝑒 𝑠𝑖𝑛3𝑡
3
= ∫0 3
𝑡 𝑠𝑖𝑛3(𝑡−𝑢)
𝑒−2𝑢𝑐𝑜𝑠3𝑢 𝑒−2(𝑡−𝑢) 𝑑𝑢
= ∫0
𝑡 𝑠𝑖𝑛(3𝑡−3𝑢)
𝑒−2𝑢𝑐𝑜𝑠3𝑢 𝑒−2𝑡+2𝑢 𝑑𝑢
= 1
3
𝑡
∫ 𝑒−2𝑢 −2𝑡+2𝑢 𝑐𝑜𝑠3𝑢 sin(3𝑡 − 3𝑢) 𝑑𝑢
3 0
3 2
∫0
−2𝑡
= 𝑒 𝑡 sin(3𝑢+3𝑡−3𝑢)−sin(3𝑢−3𝑡+3𝑢)
𝑑𝑢
Engineering Maths - IV
Laplace Transform
Page53

54. −2𝑡
= 𝑒
6
𝑡
∫
0
[𝑠𝑖𝑛3𝑡 − sin(6𝑢 − 3𝑡)]𝑑𝑢
6
−2𝑡
= 𝑒
[∫
𝑡
𝑠𝑖𝑛3𝑡𝑑𝑢 − ∫
0 0
𝑡
sin(6𝑢 − 3𝑡) 𝑑𝑢]
6
−2𝑡
= 𝑒
[𝑠𝑖𝑛3𝑡 ∫
0
𝑡 𝑡
𝑑𝑢 − ∫
0 sin(6𝑢 − 3𝑡) 𝑑𝑢]
𝑒−2𝑡
0
𝑡
= [𝑠𝑖𝑛3𝑡(𝑢) + (
cos(6𝑢 −3𝑡)
6 6
𝑡
) ]
0
6
−2𝑡
= 𝑒
[𝑡𝑠𝑖𝑛3𝑡 + cos(6𝑡−3𝑡)
− cos(−3𝑡)
]
6 6
−2𝑡
= 𝑒
[𝑡𝑠𝑖𝑛3𝑡 + cos3𝑡
− 𝑐𝑜𝑠3𝑡
]
6 6 6
6
−2𝑡
= 𝑒
𝑡𝑠𝑖𝑛3𝑡
𝑠+2 𝑒−2𝑡
∴ 𝐿−1 [ ] = 𝑡𝑠𝑖𝑛3𝑡
(𝑠2+4𝑠+13)2 6
Example: 5.65 Find the inverse Laplace transform
𝟏
(𝒔+𝟏)(𝒔𝟐+𝟒)
by using convolution theorem.
Solution:
𝐿−1 [ 1
(𝑠2+4)(𝑠+1)
] = 𝐿−1 [ 1
𝑠+1 𝑠2+4
1
]
1 1
= 𝐿−1 [ ] ∗ 𝐿−1 [ ]
𝑠+1 𝑠2+4
= 𝑒−𝑡 ∗ 𝑐𝑜𝑠2𝑡
𝑡
𝑒−(𝑡−𝑢)𝑐𝑜𝑠2𝑢 𝑑𝑢
= ∫0
−𝑡 𝑢
𝑡
= 𝑒 ∫0𝑒 𝑐𝑜𝑠2𝑢 𝑑𝑢
−𝑡 [
= 𝑒 (𝑐𝑜𝑠2𝑢 + 2𝑠𝑖𝑛2𝑢)]
12+22
0
𝑒𝑢 𝑡
5
−𝑡
= 𝑒
[𝑒𝑡(𝑐𝑜𝑠2𝑡 + 2𝑠𝑖𝑛2𝑡) − 𝑒0(𝑐𝑜𝑠0 − 0)]
5
−𝑡
= 𝑒
[𝑒𝑡(𝑐𝑜𝑠2𝑡 + 2𝑠𝑖𝑛2𝑡) − 1]
∴ 𝐿−1 [
1
(𝑠2+4)(𝑠+1) 5
−𝑡
] = 𝑒
[𝑒𝑡(𝑐𝑜𝑠2𝑡 + 2𝑠𝑖𝑛2𝑡) − 1]
Exercise: 5.10
Find the inverse Laplace transforms using convolution theorem for the following
1.
1
Ans: 1 − 𝑐𝑜𝑠𝑡
2.
𝑠(𝑠2+1)
𝑠
8 2
Ans: 1
[sin2𝑡
− 𝑡𝑐𝑜𝑠2𝑡]
3.
(𝑠2+4)2
𝑠2
Ans: 1
[𝑡𝑐𝑜𝑠2𝑡 + sin2𝑡
]
2 2
4.
(𝑠2+4)2
1
2
Ans: 1
[𝑒−𝑡 + 𝑠𝑖𝑛𝑡 − 𝑐𝑜𝑠𝑡]
5.
(𝑠+1)(𝑠2+1)
1
(𝑠+1)(𝑠2+4)
Ans: − 1
𝑒−𝑡 + 1
𝑐𝑜𝑠2𝑡 − 1
𝑠𝑖𝑛2𝑡
5 5 10
∵ ∫ 𝑒𝑎𝑡 𝑐𝑜𝑠𝑏𝑡𝑑𝑡 =
𝑒𝑎𝑡
𝑎2 + 𝑏2
(𝑎𝑐𝑜𝑠𝑏𝑡 + 𝑏𝑠𝑖𝑛𝑏𝑡)
Engineering Maths - IV
Laplace Transform
Page54