GOA COLLEGE OF ENGINEERING

“BHAUSAHEB BANDODKAR TECHNICAL EDUCATION COMPLEX"

FARMAGUDI, PONDA- GOA - INDIA

MATHEMATICS-III
UNIT 2: LAPLACE TRANSFORMS
LECTURE 1

DEVELOPED BY MATHEMATICS FACULTY

DEPARTMENT OF SCIENCE & HUMANITIES
GOA COLLEGE OF ENGINEERING
 Introduction
Laplace Transforms named after Pierre-Simon
Laplace is an integral transform that converts a
function of real variable t (often time) to a function of
frequency s. The transform has many application in
engineering. Specifically it transforms a differential
equation to algebraic equation, a convolution of
functions to multiplication of their transform etc. This
transformation enable finding solution and analysis of
the engineering problem dealt with.
 LAPLACE TRANSFORMS LECTURE 1

TOPICS TO BE COVERED:

➢Definition of Laplace Transforms.

➢Laplace transforms of exponential, trigonometric,
algrabiac functions.
➢Existence theorem for Laplace transforms.
➢Properties of Laplace transforms.
➢Computation of Laplace transforms using properties.
 Definition
• Definition: If f(t) a function defined for t > 0 is piece wise continuous and if
∞
‫׬‬0 𝑒 −𝑠𝑡 𝑓 𝑡 𝑑𝑡 is finite for some s > 0 then it is a function of s, called the Laplace
transform. It is denoted by
∞
L( f(t) )=𝑓 𝑠 = F(s)= ‫׬‬0 𝑒 −𝑠𝑡 𝑓 𝑡 𝑑𝑡
Laplace Transform of standard functions
∞ ∞ −𝑠𝑡 𝑒 −𝑠𝑡 𝑡 = ∞ 1 1
• f(t)=1, L(f(t)) =‫׬‬0 𝑒 −𝑠𝑡 𝑓𝑡 𝑑𝑡 = ‫׬‬0 𝑒 1 𝑑𝑡=
−𝑠 𝑡 = 0
=- 0 − 1 = ; s>0
𝑠 𝑠
∞ −𝑠𝑡 ∞
• f(t)=𝑡 α , L(f(t))=‫׬‬0 𝑒 𝑓 𝑡 𝑑𝑡= ‫׬‬0 𝑒 −𝑠𝑡 𝑡 α 𝑑𝑡 by substituting st=x
∞ −𝑥 𝑥 α 𝑑𝑥 1 ∞ −𝑥 α
= ‫׬‬0 𝑒 = α+1 ‫׬‬0 𝑒 𝑥 𝑑𝑥 where Γα is the gamma function
𝑠α 𝑠 𝑠
1 𝑛!
= α+1 Γ(α + 1) = 𝑛+1 , s > 0. when α=n is a positive integer.
𝑠 𝑠
 Laplace Transforms of Standard Functions
1Τ Γ(3Τ2) 1 π
• If f(t)= 𝑡 2 then L(f(t)) = = since Γ(1/2)= π
𝑠 1/2+1 2 𝑠 3ൗ2

3Τ Γ(5Τ2) 31 π 1 1
• f(t)= 𝑡 2 then L(f(t)) = = since Γ(3/2)= Γ(1/2)= π
𝑠 3/2+1 2 2 𝑠 5ൗ2 2 2

𝑎𝑡 ∞ −𝑠𝑡 ∞ −𝑠𝑡 𝑎𝑡 ∞ −(𝑠−𝑎)𝑡

• If f(t)=𝑒 , L(f(t))= ‫׬‬0 𝑒 𝑓 𝑡 𝑑𝑡=‫׬‬0 𝑒 𝑒 𝑑𝑡= ‫׬‬0 𝑒 𝑑𝑡

1 −(𝑠−𝑎)𝑡 𝑡=∞ 1 1
=- 𝑒 =- 0−1 = for s-a > 0
(𝑠−𝑎) 𝑡 = 0 (𝑠−𝑎) (𝑠−𝑎)
 Laplace Transforms of standard functions
2𝑡 1
• If f(t)= 𝑒 ; then L(f(t)) = for s > 2
𝑠−2
−3𝑡 1
• If f(t)= 𝑒 ; then L(f(t)) = for s >-3
(𝑠+3)
∞ −𝑠𝑡
• If f(t)=Cos(at) ; L(f(t)) = ‫׬‬0 𝑒 𝑓 𝑡 𝑑𝑡
∞ −𝑠𝑡 𝑒 −𝑠𝑡 𝑡=∞
• =‫׬‬0 𝑒 Cos(at)𝑑𝑡= 𝑠2+𝑎2 −𝑠𝐶𝑜𝑠𝑎𝑡 + 𝑎𝑆𝑖𝑛𝑎𝑡
𝑡=0
1 𝑠
= 2 2 0 − (−𝑠) = 2 2 for s >0
𝑠 +𝑎 𝑠 +𝑎
𝑒 𝑎𝑥
න 𝑒 𝑎𝑥 Cos(bx)𝑑𝑥 = 2 𝑎𝑐𝑜𝑠𝑏𝑥 + 𝑏𝑠𝑖𝑛𝑏𝑥
𝑎 + 𝑏2
 Laplace Transforms of standard functions
𝑎𝑥
𝑒 𝑎𝑥
න𝑒 𝑆𝑖𝑛𝑏𝑥 𝑑𝑥 = 2 𝑎𝑆𝑖𝑛𝑏𝑥 − 𝑏𝐶𝑜𝑠𝑏𝑥
𝑎 + 𝑏2

• If f(t)=Sin(at);
∞ −𝑠𝑡 ∞ −𝑠𝑡
L(f(t)) = ‫׬‬0 𝑒 𝑓 𝑡 𝑑𝑡= ‫׬‬0 𝑒 Sin(at)𝑑𝑡
𝑒 −𝑠𝑡 𝑡=∞
= 2 2 −𝑠𝑆𝑖𝑛 𝑎𝑡 − aCos(𝑎𝑡)
𝑠 +𝑎 𝑡=0
1 𝑎
= 2 2 0 − (−𝑎) = 2 2
𝑠 +𝑎 𝑠 +𝑎

𝑠 3
If f(t)= Cos(2t), L(f(t))= If f(t)= Sin(3t), L(f(t))=
𝑠 2 +4 𝑠 2 +9
 Laplace Transforms of Standard Functions
The list of functions and their Laplace transforms
Functions Laplace Transforms
α 1
𝑡 Γ(α + 1)
𝑠 α+1
𝑛 𝑛!
𝑡
𝑠 𝑛+1
1
𝑒 𝑎𝑡
𝑠−𝑎
𝑠
Cos at
𝑠 2 +𝑎2
𝑎
Sin at
𝑠 2 +𝑎2
 Existence Theorem
• Theorem(Existence): If f(t) is a piece-wise continuous function and
there exist constants M>0,𝛼>0 and such that
𝑓(𝑡) ≤ 𝑀𝑒 𝛼𝑡 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0.
Then Laplace transform of f(t) exists.
• Proof: Since f(t) is piece wise continuous 𝑒 −𝑠𝑡 𝑓(𝑡) is also piece wise
continuous and therefore integrable. We only have to ensure that
the integral is bounded.
∞ −𝑠𝑡 ∞ −𝑠𝑡 ∞ −𝑠𝑡 𝛼𝑡
. ‫׬‬0 𝑒 𝑓 𝑡 𝑑𝑡 ≤ ‫׬‬0 𝑒 𝑓(𝑡) 𝑑𝑡 ≤ 𝑀 ‫׬‬0 𝑒 𝑒 𝑑𝑡
∞ −(𝑠−𝛼)𝑡 𝑒 −(𝑠−𝛼)𝑡 𝑡 = ∞ 𝑀
≤M ‫׬‬0 𝑒 𝑑𝑡=M = for s>𝛼
−(𝑠−𝛼) 𝑡 = 0 𝑠−𝛼
 Existence Theorem
The function 𝑡 𝑛 , 𝑒 𝑎𝑡 , Cosat, Sinat are all continuous and
𝑡𝑛 ≤ 𝑒𝑡, 𝑒 𝑎𝑡 ≤ 2𝑒 𝑎𝑡
Cosat ≤ 1 ≤ 𝑒 0𝑡 , Sinat ≤ 1 ≤ 𝑒 0𝑡
Which Implies that they fulfil the existence condition.
𝑡 2
However the function 𝑒
is one function that does not fulfil the existence
condition. To prove that, lets assume that for some M and 𝛼
𝑡2
𝑒 ≤ 𝑀𝑒 𝛼𝑡 for all t >0. Then for t > 𝑙𝑜𝑔𝑒 𝑀 + 𝛼; 𝑡 2 > (𝑙𝑜𝑔𝑒 𝑀 + 𝛼)t
𝑡2
𝑒 > 𝑒 (𝑙𝑜𝑔𝑒 𝑀+𝛼)𝑡 = 𝑒 𝑙𝑜𝑔𝑒 𝑀𝑡+𝛼𝑡 > 𝑒 𝑙𝑜𝑔𝑒 𝑀+𝛼𝑡 =M 𝑒 𝛼𝑡 for t>1
𝑡 2
Which is a contradiction. Therefore 𝑒 does not satisfy the condition.
 Properties of Laplace Transforms
• Suppose functions f(t), g(t) have a Laplace transforms F(s),G(s)
Sr Property Function in time t Function in s
no
1 Linearity 𝛼𝑓 𝑡 + 𝛽𝑔(𝑡) 𝛼𝐹(𝑠) + 𝛽𝐺(𝑠)

2 Frequency shift 𝑒 𝛼𝑡 f(t) F(s-𝛼)

3 Time scale f(at) 1 𝑠

F
𝑎 𝑎
4 Time shifting f(t-a)H(t-a) 𝑒 −𝑎𝑠 𝐹(𝑠)
H(t)-Heavyside’s unit step function
 Properties of Laplace Transforms
• Suppose functions f(t), g(t) have a Laplace transforms F(s),G(s)
Sr Property Function in Function in frequency s
no time t
5 Time domain 𝑑𝑓 sF(s)-f(0)
derivative 𝑑𝑡
6 Second derivative 𝑑2 𝑓 𝑠 2 F(s)-sf(0)-𝑓 / (0)
𝑑𝑡 2
𝑛
7 General derivative 𝑑𝑛 𝑓
𝑛 𝑛−𝑘 𝑘−1
𝑛 𝑠 𝐹 𝑠 − ෍ 𝑠 𝑓 (0)
𝑑𝑡
𝑘=1
8 Integral of function 𝑡 1
F(s)
න 𝑓 𝑢 𝑑𝑢 𝑠
 Properties of Laplace Transforms
• Suppose functions f(t), g(t) have a Laplace transforms F(s),G(s)
Sr Property Function in time t Function in frequency s
no
9 Frequency domain tf(t) 𝑑𝐹
-
𝑑𝑠
derivative
𝑛𝐹
10 Frequency domain 𝑡 𝑛 f(t) 𝑑
(−1)𝑛 𝑛
Higher derivative 𝑑𝑠
𝑡
11 Convolution න 𝑓 𝑡 − 𝑢 𝑔(𝑢)𝑑𝑢 F(s)G(s)
0
𝑝
12 Periodic function f(t+p)=f(t) 1 −𝑠𝑡 𝑓 𝑡 𝑑𝑡
න 𝑒
1 − 𝑒 −𝑠𝑝 0
 Properties of Laplace Transforms
• Suppose functions f(t), g(t) have a Laplace transforms F(s),G(s)
Sr Property Function in Function in frequency s
No time t
𝑓(𝑡) ∞
13 Frequency domain
integral න 𝐹 𝑢 𝑑𝑢
𝑡 𝑠
 Properties of Laplace Transforms
Proof of properties of Laplace Transforms:
1. Linearity:
∞ −𝑠𝑡
L(𝛼𝑓 𝑡 + 𝛽𝑔(𝑡))=‫׬‬0 𝑒 𝛼𝑓 𝑡 + 𝛽𝑔(𝑡) dt
∞ −𝑠𝑡 ∞ −𝑠𝑡
= 𝛼 ‫׬‬0 𝑒 𝛼𝑓 𝑡 𝑑𝑡+ 𝛽 ‫׬‬0 𝑒 𝑔(𝑡)dt
= 𝛼F(s)+𝛽G(s)
𝛼𝑡 ∞ −𝑠𝑡 𝛼𝑡
2.Frequency Shift: L(𝑒 f(t))= ‫׬‬0 𝑒 𝑒 𝑓 𝑡 dt
∞ −(𝑠−𝛼)𝑡
= ‫׬‬0 𝑒 𝑓 𝑡 dt = F(s- 𝛼) for s- 𝛼 > 𝑎
 Properties of Laplace Transforms
∞ −𝑠𝑡
• 3. Time scale . L (f(at))= ‫׬‬0 𝑒 𝑓 𝑎𝑡 dt sub at=x
1 ∞ −𝑠𝑥
=
𝑎
‫׬‬0 𝑒 𝑎 𝑓 𝑥 dx = 𝑎1 F(𝑎𝑠 ) a>0

0 𝑡<0
• 4. Time Shifting. For function H(t)=ቊ called Heavyside’s
1 𝑡≥0
unit step function. L(f(t)H(t-a))= 𝑒 −𝑠𝑎 L(f(t+a))
∞ −𝑠𝑡 ∞ −𝑠𝑡
L(f(t-a)H(t-a))= ‫׬‬0 𝑒 f(t-a)H(t-a) dt= ‫𝑒 𝑎׬‬ f(t-a) dt
∞ −𝑠(𝑥+𝑎) −𝑠𝑎 ∞ −𝑠𝑥 −𝑠𝑎
= ‫׬‬0 𝑒 f(x) dx= 𝑒 ‫׬‬0 𝑒 f(x) dx= 𝑒 F(s) sub x=t-a
 Properties of Laplace Transforms
𝑑𝑓 ∞ −𝑠𝑡 𝑑𝑓
• 5 Time domain derivative: L = ‫׬‬0 𝑒 dt
𝑑𝑡 𝑑𝑡
𝑡=∞ ∞ −𝑠𝑡
= 𝑒 −𝑠𝑡 𝑓(𝑡)ඌ-
𝑡=0 0
‫׬‬ (−𝑠)𝑒 f(t)dt integration by parts
∞ −𝑠𝑡
= 0-f(0)+s‫׬‬0 𝑒 f(t)dt= -f(0)+sF(s)= sF(s)-f(0)
• 6 Time domain second derivative:
L( 𝑓 // (𝑡))= sL(𝑓 / (𝑡))- 𝑓 / (0)=s(sF(s)-f(0))- 𝑓 / (0)
=𝑠 2 𝐹 𝑠 − 𝑠𝑓 0 − 𝑓 / (0)
 Properties of Laplace Transforms
• 7. Time domain general derivative: Continuing from 6, we have
L( 𝑓 /// (𝑡))= 𝑠 3 𝐹 𝑠 − 𝑠 2 𝑓 0 − 𝑠𝑓 / (0)- 𝑓 // (0)
L( 𝑓 𝑛 (𝑡))=𝑠 𝑛 𝐹 𝑠 − σ𝑛𝑘=1 𝑠 𝑛−𝑘 𝑓 𝑘−1 (0) by generalizing.
• 8. Integral of time domain function:
𝑡
define g(t)= ‫׬‬0 𝑓 𝑢 𝑑𝑢 then g(0)=0 and 𝑔/ (t)=f(t)
/ 𝑡
L(𝑔 (t)) = L(f(t))= sL(g(t))-g(0)= sL(‫׬‬0 𝑓 𝑢 𝑑𝑢)-0
𝑡 𝑡 1
sL(‫׬‬0 𝑓 𝑢 𝑑𝑢)=L(f(t))=F(s) therefore L(‫׬‬0 𝑓 𝑢 𝑑𝑢)= 𝐹(𝑠)
𝑠
 Properties of Laplace Transforms
• 9 Frequency domain derivative
𝑑𝐹 𝑑 ∞ −𝑠𝑡 ∞ 𝑑
= ‫׬‬0
𝑒 𝑓 𝑡 dt = ‫׬‬0 (𝑒 −𝑠𝑡 ) 𝑓 𝑡 dt
𝑑𝑠 𝑑𝑠 𝑑𝑠
∞ ∞ −𝑠𝑡
= ‫׬‬0 −𝑡𝑒 −𝑠𝑡 𝑓 𝑡 dt = − ‫׬‬0 𝑒 t 𝑓 𝑡 dt =- L(tf(t))
10 Frequency domain higher order derivative
𝑑𝑛𝐹 𝑑𝑛 ∞ −𝑠𝑡 ∞ 𝑑𝑛 −𝑠𝑡
𝑛 = 𝑛 ‫׬‬0
𝑒 𝑓 𝑡 dt = ‫׬‬0 𝑛 (𝑒 ) 𝑓 𝑡 dt
𝑑𝑠 𝑑𝑠 𝑑𝑠
∞ 𝑛 −𝑠𝑡 𝑛 ∞ −𝑠𝑡 𝑛
= ‫׬‬0 (−𝑡) 𝑒 𝑓 𝑡 dt =(−1) ‫׬‬0 𝑒 𝑡 𝑓 𝑡 dt
𝑛 𝑛
= (−1) 𝐿(𝑡 f(t))
 Properties of Laplace Transforms
• 11 Convolution Theorem.
𝑡 ∞ −𝑠𝑡 𝑡
L(‫׬‬0 𝑓 𝑡 − 𝑢 𝑔(𝑢)𝑑𝑢)= ‫׬‬0 𝑒 ‫׬‬0 𝑓 𝑡 − 𝑢 𝑔(𝑢)𝑑𝑢 dt
∞ 𝑡 −𝑠𝑡
=‫׬‬0 ‫׬‬0 𝑒 𝑓 𝑡 − 𝑢 𝑔(𝑢)𝑑𝑢 dt changing the order of integration
∞ ∞ −𝑠𝑡
= ‫׬‬0 ‫ 𝑡 𝑓 𝑒 𝑢׬‬− 𝑢 𝑑𝑡 𝑔(𝑢) du sub x=t-u
∞ ∞ −𝑠(𝑥+𝑢)
= ‫׬‬0 ‫׬‬0 𝑒 𝑓 𝑥 𝑑𝑥 g(u)du
∞ −𝑠𝑥 ∞ −𝑠𝑢
= ‫׬‬0 𝑒 𝑓 𝑥 𝑑𝑥 ‫׬‬0 𝑒 𝑔(𝑢) du = F(s)G(s)
 Properties of Laplace Transforms
• 12 Periodic function: f(t+p)=f(t) for all t>0.
∞ −𝑠𝑡 𝑝 −𝑠𝑡 ∞ −𝑠𝑡
L(f(t))=‫׬‬0 𝑒 𝑓 𝑡 dt= ‫׬‬0 𝑒 𝑓 𝑡 dt+ ‫𝑒 𝑝׬‬ 𝑓 𝑡 dt
𝑝 −𝑠𝑡 ∞ −𝑠𝑡
= ‫׬‬0 𝑒 𝑓 𝑡 dt+ ‫𝑒 𝑝׬‬ 𝑓 𝑡 dt by sub t-p=x in second integral
𝑝 −𝑠𝑡 ∞ −𝑠(𝑥+𝑝)
=‫𝑒 ׬‬ 𝑓 𝑡 dt+ ‫׬‬ 𝑒 𝑓 𝑥 + 𝑝 dx
0 0
𝑝 −𝑠𝑡 −𝑠𝑝 ∞ −𝑠𝑥
= ‫׬‬0 𝑒 𝑓 𝑡 dt+ 𝑒 ‫׬‬0 𝑒 𝑓 𝑥 dx; since f(x+p)=f(x)
−𝑠𝑝 ∞ −𝑠𝑡 𝑝 −𝑠𝑡
(1- 𝑒 ) ‫׬‬0 𝑒 𝑓 𝑡 dt= ‫׬‬0 𝑒 𝑓 𝑡 dt
∞ −𝑠𝑡 1 𝑝 −𝑠𝑡
Hence ‫׬‬0 𝑒 𝑓 𝑡 dt = ‫׬‬0 𝑒 𝑓 𝑡 dt
(1− 𝑒 )−𝑠𝑝
 Properties of Laplace Transforms
• 13 Integration in the frequency domain:
∞ ∞ ∞ −𝑢𝑡 ∞ ∞ −𝑢𝑡
• ‫𝐹 𝑠׬‬ 𝑢 𝑑𝑢= ‫𝑠׬‬ ‫׬‬0 𝑒 𝑓 𝑡 𝑑𝑡 𝑑𝑢=‫׬ 𝑠׬‬0 𝑒 𝑓 𝑡 𝑑𝑡𝑑𝑢
∞ ∞ −𝑢𝑡
= ‫׬‬0 ‫𝑓 𝑒 𝑠׬‬ 𝑡 𝑑𝑢𝑑𝑡 by changing the order of integration

∞ 𝑒 −𝑢𝑡 𝑢 = ∞ ∞ 𝑒 −𝑢𝑡 𝑢 = ∞
=‫׬‬ ඌ f(t)dt= ‫׬‬0 ඌ f(t)dt
0 −𝑡 𝑢=𝑠 −𝑡 𝑢=𝑠
∞ 0−𝑒 −𝑠𝑡 ∞ 𝑒 −𝑠𝑡 ∞ −𝑠𝑡 f(t) f(t)
= ‫׬‬0 −𝑡 f(t)dt= ‫׬‬0 𝑡 f(t)dt= ‫׬‬0 𝑒 𝑡
dt= L
𝑡
 Computation of Laplace transform
• Find the Laplace transform of L(𝑡 𝑛 ) =
𝑛!
𝑠 𝑛+1
a) 2Sin3t+4𝑒 −𝑡 -2 t3 b) 2Cos 2 t-3e3𝑡 c) 2e2𝑡 Cos3tSin5t-3 t3 L(Cosat)= 2 2
𝑠
𝑠 +𝑎
Solution: L(𝑒 𝑎𝑡 ) =
1
By Property of Linearity 𝑠−𝑎
a)L(2Sin3t+4𝑒 −𝑡 -2 t3 ) L(Sinat)= 2 2
𝑎
𝑠 +𝑎
−𝑡 3 3 1 3!
=2L(Sin3t)+4L(𝑒 )-2L( t )=2 2 +4 − 2 4
𝑠 +9 𝑠+1 𝑠
6 4 12
= 2 + − 4
𝑠 +9 𝑠+1 𝑠

b)L(2Cos 2 t−2e3𝑡 ) =L( 1+Cos2t-2e3𝑡 )= L(1)+L(Cos2t)-2L(e3𝑡 )

1 𝑠 2
= + 2 - 2𝐶𝑜𝑠 2 𝑡 = (1 + 𝐶𝑜𝑠2𝑡)
𝑠 𝑠 +4 𝑠−3
 Computation of Laplace transform
• c) L(2 e2𝑡 Cos3tSin5t-3 t3 )=
2SinACosB=Sin(A+B)+Sin(A-B)
=L(2e2𝑡 Cos3tSin5t)-3L( t3 ) 2SinASinB=Cos(A-B)+Cos(A+B)
2CosACosB=Cos(A-B)-Cos(A+B)
=L(e2𝑡 (Sin8t+Sin(-2t))-3L( t3 )
=L(e2𝑡 Sin8t- e2𝑡 Sin2t)-3L(t3 )=L(e2𝑡 Sin8t)-L(e2𝑡 Sin2t)-3L(t3 )
8 2 3!
= 2 − -3 If L(f(t))= F(s) then L(eα𝑡 𝑓(𝑡)) = 𝐹(𝑠 − α)
(𝑠−2) +64 (𝑠−2)2 +4 𝑠4 𝑎
L(Sinat)= 2 2 , L(eα𝑡 Sinat)=
𝑎
𝑠 +𝑎 2 2
(𝑠−α) +𝑎
8 2 18 𝑠
L(Cosat)= 2 2 , L(eα𝑡 Cosat)=
𝑠−α
= 2 − 2 - 𝑠 +𝑎 (𝑠−α)2 +𝑎2
𝑠 −4𝑠+68 𝑠 −4𝑠+8 𝑠4
 Computation of Laplace Transforms
• Find the Laplace transform of
𝑆𝑖𝑛3𝑡 𝑡
a) tCos2t b) c)‫׬‬0 𝑆𝑖𝑛𝑥 + 2𝑥 2 𝑑𝑥
𝑡
Solution:
𝑑 𝑑 𝑠 If L(f(t))=F(s) then
a)L( tCos2t)= - 𝐿(Cos2t ) = - 𝑑𝐹
𝑑𝑠 𝑑𝑠 𝑠 2 +4 L(tf(t))= -
𝑑𝑠
𝑓(𝑡) ∞
(𝑠 2 +4).1−2𝑠.𝑠 𝑠 2 −4 L =‫𝐹 𝑠׬‬ 𝑠 𝑑𝑠
=- 2 2 = 2 2 𝑡
𝑡 1
(𝑠 +4) (𝑠 +4) L ‫׬‬0 𝑓 𝑥 𝑑𝑥 = F(s)
𝑠
 Computation of Laplace Transforms
𝑆𝑖𝑛3𝑡 ∞ ∞ 3
b) L = ‫𝑛𝑖𝑆(𝐿 𝑠׬‬3𝑡) 𝑑𝑠= ‫ 𝑠׬‬2 𝑑𝑥
𝑡 𝑥 +9
1 −1 𝑥 𝑥 = ∞ 𝜋 𝑠
=3 Tan ฬ = − Tan−1
3 3 𝑥=𝑠 2 3
𝑡 2 1 2 1 1 2!
c) L ‫׬‬0 𝑆𝑖𝑛𝑥 + 2𝑥 𝑑𝑥 = L 𝑆𝑖𝑛𝑥 + 2𝑥 = 2 +2 3
𝑠 𝑠 𝑠 +1 𝑠
1 4
= 2 + 4
𝑠(𝑠 +1) 𝑠
 Computation of Laplace transforms

• Find the Laplace transform of the following

(a,1) (2a,1)
a)
y-axix

. . . . . .

a x-axis

b)
(1,1) (3,1)
. . . . . .

(1,0) (2,0) (3,0) x-axis

 Computation of Laplace transforms
a)The saw tooth function is periodic with period a the function is defined by
𝑥
f(x)= ; 0≤ 𝑥 ≤ 𝑎 𝑎𝑛𝑑 𝑓 𝑥 + 𝑎 = 𝑓(𝑥).
𝑎
1 𝑎 −𝑠𝑥 1 𝑎 −𝑠𝑥 𝑥
L(f(x)) = −𝑎𝑠 ‫׬‬0 𝑒 𝑓 𝑥 𝑑𝑥= ‫𝑒 ׬‬ 𝑑𝑥
1−𝑒 1−𝑒 −𝑎𝑠 0 𝑎
1 1 𝑒 −𝑠𝑥 𝑥 = 𝑎 𝑎 𝑒 −𝑠𝑥
= 𝑥 ฬ − ‫׬‬0 𝑑𝑥 1 𝑝
1−𝑒 −𝑎𝑠 𝑎 −𝑠 𝑥 = 0 −𝑠 𝐿 𝑓 𝑡 = න 𝑒 −𝑠𝑡 𝑓 𝑡 𝑑𝑡
1−𝑒 −𝑠𝑝
1 1 𝑒 −𝑠𝑥 𝑥 = 𝑎 1 𝑒 −𝑠𝑥 𝑥 = 𝑎 0
= 𝑥 ฬ + ฬ
1−𝑒 −𝑎𝑠 𝑎 −𝑠 𝑥 = 0 𝑠 −𝑠 𝑥 = 0
1 1 𝑒 −𝑠𝑎 1 −𝑠𝑎
= 𝑎 − 0 − (𝑒 −1
1−𝑒 −𝑎𝑠 𝑎 −𝑠 𝑠2
(1−𝑎𝑠𝑒 −𝑠𝑎 −𝑒 −𝑠𝑎 )
=
𝑎 1−𝑒 −𝑎𝑠 𝑠2
 Computation of Laplace transforms
𝑡 ;0 ≤ 𝑡 ≤ 1
b)The function is defined as f(t)=൜ , f(t+2)=f(t)
2 − 𝑡;1 < 𝑡 ≤ 2
1 1 2
L(f(t))= −2𝑠
1−𝑒
‫׬‬0 𝑡𝑒 −𝑠𝑡 𝑑𝑡 + ‫׬‬1 (2 − 𝑡)𝑒 −𝑠𝑡 𝑑𝑡
1 𝑒 −𝑠𝑡 𝑒 −𝑠𝑡 𝑡 =1 𝑒 −𝑠𝑡 𝑒 −𝑠𝑡 𝑡 =2
= 𝑡 − ቤ + 2−𝑡 + ቤ
1−𝑒 −2𝑠 −𝑠 𝑠2 𝑡 =0 −𝑠 𝑠2 𝑡 =1
1 𝑒 −𝑠 𝑒 −𝑠 −1 𝑒 −𝑠 𝑒 −2𝑠 −𝑒 −𝑠
= −2𝑠 − 2 −0+ +
1−𝑒 −𝑠 𝑠 𝑠 𝑠2
1 𝑒 −𝑠 𝑒 −𝑠 −1−𝑒 −2𝑠 +𝑒 −𝑠 1 1−2𝑠𝑒 −𝑠 +𝑒 −2𝑠
= −2𝑠 2 − 2 =
1−𝑒 𝑠 𝑠 1−𝑒 −2𝑠 𝑠2
 Computation of Laplace transforms
• Find the Laplace transform of
a) H(t-1)(𝑡 2 +2𝑡 − 4) b) H(t-𝜋2)Sint L(H(t-a)f(t-a))=𝑒 −𝑎𝑠 𝐿(𝑓 𝑡 )

Solution: L(f(t)H(t-a))= 𝑒 −𝑎𝑠 L(f(t+a))

L(u(t-1)(𝑡 2 +2𝑡 − 4) )=𝑒 −𝑠 L((𝑡 + 1)2 +2(t+1)-4)

2 1 1
=𝑒 −𝑠 𝐿 𝑡2 + 4𝑡 − 1 = 𝑒 −𝑠 + 4 2 −
𝑠3 𝑠 𝑠
𝜋 𝜋 𝜋
𝜋 −2𝑠 𝜋 −2𝑠 −2𝑠 𝑠
b)L(H(t- )Sint)=𝑒 𝐿 𝑆𝑖𝑛 𝑡 + =𝑒 𝐿 𝐶𝑜𝑠𝑡 =𝑒
2 2 𝑠2 +1
 Computation of Laplace transform
Verify convolution theorem for Laplace transform for the functions
f(t)=t; g(t)= 𝑒 2𝑡
𝑡 𝑡
(f*g)(t)=‫׬‬0 𝑓 𝑡 − 𝑢 𝑔 𝑢 𝑑𝑢 = ‫׬‬0 𝑡 − 𝑢 𝑒 2𝑢 𝑑𝑢
𝑒 2𝑢 𝑢 =𝑡 𝑡 𝑒 2𝑢 𝑡 𝑒 2𝑢 𝑢 = 𝑡 𝑡 𝑒 2𝑡 1
= 𝑡−𝑢 ฬ + ‫׬‬0 𝑑𝑢 =0 - + ฬ =- + −
2 𝑢 =0 2 2 4 𝑢 =0 2 4 4
𝑡 𝑒 2𝑡 1 1 1 2𝑡 1 1 1 1
L(- + − )= - 𝐿 𝑡 + 𝐿(𝑒 )- 𝐿 1 =− 2 + −
2 4 4 2 4 4 2𝑠 4(𝑠−2) 4𝑠
−2 𝑠−2 +𝑠2 −𝑠(𝑠−2) −2𝑠+4+𝑠 2 −𝑠2 +2𝑠 −2𝑠+4+𝑠 2 −𝑠2 +2𝑠 1
= 2 = 2 = 2 = 2
4𝑠 (𝑠−2) 4𝑠 (𝑠−2) 4𝑠 (𝑠−2) 𝑠 (𝑠−2)
 Computation of Laplace transform
1
• L((f*g)(t))=
𝑠 2 (𝑠−2)
1 2𝑡 1
• L(f(t))=L(t)= L(g(t))=L(𝑒 ) =
𝑠2 (𝑠−2)
1 1 1
• L(f(t))L(g(t))= 2 = 2 = L((f*g)(t))
𝑠 (𝑠−2) 𝑠 (𝑠−2)
• Which is as per convolution theorem. Hence it is verified.
 Use of Laplace transforms to compute
Integrals
Using Laplace transforms compute the integrals
∞ −2𝑡 ∞ −3𝑡 𝑆𝑖𝑛2𝑡
a)‫׬‬0 𝑒 𝑡𝐶𝑜𝑠3𝑡𝑑𝑡 b) ‫׬‬0 𝑒 𝑑𝑡
𝑡
𝑑 𝑑 𝑠 𝑠2 −9
a)L(𝑡𝐶𝑜𝑠3𝑡)= - 𝐿(𝐶𝑜𝑠3𝑡) = − = 2 2
𝑑𝑠 𝑑𝑠 𝑠 2 +9 (𝑠 +9)
∞ −𝑠𝑡 𝑠2 −9 ∞ −2𝑡 22 −9 −5
‫׬‬0 𝑒 𝑡𝐶𝑜𝑠3𝑡𝑑𝑡= (𝑠2 +9)2 therefore ‫׬‬0 𝑒 𝑡𝐶𝑜𝑠3𝑡𝑑𝑡= (22 +9)2= 169
𝑆𝑖𝑛2𝑡 ∞ 2 −1 𝑢 𝑢=∞ 𝜋 −1 𝑠
b)L =‫ 𝑠׬‬2 𝑑𝑢=𝑇𝑎𝑛 ฬ = - 𝑇𝑎𝑛
𝑡 𝑢 +4 2 𝑢=𝑠 2 2
∞ −𝑠𝑡 𝑆𝑖𝑛2𝑡 𝜋 −1 𝑠 ∞ −3𝑡 𝑆𝑖𝑛2𝑡 𝜋 −1 3
‫׬‬0 𝑒 𝑡
𝑑𝑡=
2
- 𝑇𝑎𝑛
2
therefore ‫׬‬0 𝑒 𝑡
𝑑𝑡=
2
- 𝑇𝑎𝑛
2
 Problems on Laplace transforms
Compute the Laplace transform of
1)3𝑒 2𝑡 − 2𝐶𝑜𝑠𝑡 + 5𝑆𝑖𝑛3𝑡 2) 3𝑆𝑖𝑛2 3𝑡 − 2𝑡 2 𝑒 −𝑡
2𝑡 𝑡
3)3Cos2tCos4t+5𝑒 𝐶𝑜𝑠𝑡 4)‫׬‬0 2 𝑥 2 +4Sin3xdx
𝐶𝑜𝑠2𝑡−𝐶𝑜𝑠3𝑡 2
5) 6)3𝑡 𝑆𝑖𝑛2𝑡
𝑡
𝑒 −𝑡 −𝑒 −2𝑡
7) 8) H(t-2)Sin3t
𝑡
𝑆𝑖𝑛𝑡 0 < 𝑡 < 𝜋
9) f(t)=ቊ f(t+2𝜋)=f(t)
0 1 ≤ 𝑡 ≤ 2𝜋
 Problems on Laplace transforms
Verify Convolution theorem for the functions f(t)=𝑒 2𝑡 , g(t)=Cost .

Using Laplace transforms evaluate the integrals

∞ −𝑡 1−𝐶𝑜𝑠𝑡 ∞ −2𝑡
a)‫׬‬0 𝑒 𝑑𝑡 b) ‫׬‬0 𝑒 𝑡𝑆𝑖𝑛3𝑡 dt
𝑡
THANK YOU!