Prepare for your exams
Get points
Guidelines and tips

Prepare for your exams

Study with the several resources on Docsity

Earn points to download

Earn points by helping other students or get them with a premium plan

Guidelines and tips

Sell on Docsity

Log in Sign up

Prepare for your exams

Study with the several resources on Docsity

Find documents

Prepare for your exams with the study notes shared by other students like you on Docsity

Search Store documents

The best documents sold by students who completed their studies

Search through all study resources

Docsity AINEW

Summarize your documents, ask them questions, convert them into quizzes and concept maps

Explore questions

Clear up your doubts by reading the answers to questions asked by your fellow students

Earn points to download

Earn points by helping other students or get them with a premium plan

Share documents

20 Points

For each uploaded document

Answer questions

5 Points

For each given answer (max 1 per day)

All the ways to get free points

Get points immediately

Choose a premium plan with all the points you need

Study Opportunities

Choose your next study program

Get in touch with the best universities in the world. Search through thousands of universities and official partners

Community

Ask the community

Ask the community for help and clear up your study doubts

University Rankings

Discover the best universities in your country according to Docsity users

Free resources

Our save-the-student-ebooks!

Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors

From our blog

Exams and Study

Go to the blog

Laplace Transform: Existence and Properties, Summaries of Applied Mathematics

Dr. Babasaheb Ambedkar Technological University Applied Mathematics

The existence and properties of Laplace transforms. It explains the conditions for the existence of Laplace transforms and provides a theorem and proof for the same. It also discusses the first translation property and provides an example to find Laplace transforms of different functions. relevant for students studying mathematics, engineering, and physics.

Typology: Summaries

2020/2021

Available from 01/19/2022

sid_ortal 🇮🇳

3 documents

1 / 10

This page cannot be seen from the preview

Don't miss anything!

Notes-II

Unit-I LAPLACE TRANSFORM

1.4: Existence of Laplace Transforms

The Laplace transform of any function 𝑓(𝑡) will exist if the integral in Section

(1.1) is convergent. For the purpose, the whole integrand 𝑒−𝑠𝑡𝑓(𝑡) goes to zero

fast enough as 𝑡→∞, at least like an exponential function with a negative

exponent. This implies that 𝑓(𝑡) itself should not grow faster than, 𝑒𝑘𝑡(𝑠𝑎𝑦).

This motivates the inequality (1.2) given in the following theorem:

Theorem 1.1: Let 𝑓(𝑡) be a function that is piecewise continuous on every finite

interval in the range 𝑡≥0 and satisfies

|𝑓(𝑡)|≦ 𝑀 𝑒𝑘𝑡

for all 𝑡 ≧0, … …… …… …… …… …(1.2)

and for some constants 𝑘 and 𝑀. Then the Laplace transform of 𝑓(𝑡) exists

for all 𝑠>𝑘.

Proof: Since 𝑓(𝑡) is continuous, the integrand 𝑒−𝑠𝑡𝑓(𝑡) is integrable over any

finite interval on the t-axis.

From (1.2), assuming that 𝑠>𝑘, we obtain

|𝐿{𝑓(𝑡)}|=|∫ 𝑒−𝑠𝑡

∞

0𝑓(𝑡)𝑑𝑡|≦∫ |𝑓(𝑡)|

∞

0𝑒−𝑠𝑡𝑑𝑡

≦∫ 𝑀

∞

0𝑒𝑘𝑡 𝑒−𝑠𝑡𝑑𝑡=𝑀

𝑠−𝑘 ,

where the condition 𝑠>𝑘 was needed for the existence of the last integral.

This completes the proof of the theorem.

1.5: Properties of Laplace Transform

1.5.1: First Translation Property or First Shifting Property

Statement: If 𝐿{𝑓(𝑡)}=𝑓(𝑠), then 𝐿{𝑒−𝑎𝑡𝑓(𝑡)}=𝑓(𝑠+𝑎).

Proof: 𝐿{𝑒−𝑎𝑡𝑓(𝑡)}=∫𝑒−𝑠𝑡

∞

0{𝑒−𝑎𝑡𝑓(𝑡)}𝑑𝑡

=∫ 𝑒−(𝑠+𝑎)

∞

0𝑓(𝑡)𝑑𝑡

=𝑓(𝑠+𝑎)

Partial preview of the text

Download Laplace Transform: Existence and Properties and more Summaries Applied Mathematics in PDF only on Docsity!

Notes-II

Unit-I LAPLACE TRANSFORM

1.4: Existence of Laplace Transforms

The Laplace transform of any function 𝑓(𝑡) will exist if the integral in Section

(1.1) is convergent. For the purpose, the whole integrand 𝑒

−𝑠𝑡

𝑓(𝑡) goes to zero

fast enough as 𝑡 → ∞, at least like an exponential function with a negative

exponent. This implies that 𝑓(𝑡) itself should not grow faster than, 𝑒

𝑘𝑡

This motivates the inequality (1.2) given in the following theorem:

Theorem 1.1: Let 𝑓(𝑡) be a function that is piecewise continuous on every finite

interval in the range 𝑡 ≥ 0 and satisfies

𝑘𝑡

for all 𝑡 ≧ 0 , … … … … … … … … … … ( 1. 2 )

and for some constants 𝑘 and 𝑀. Then the Laplace transform of 𝑓(𝑡) exists

for all 𝑠 > 𝑘.

Proof: Since 𝑓(𝑡) is continuous, the integrand 𝑒

−𝑠𝑡

𝑓(𝑡) is integrable over any

finite interval on the t-axis.

From (1.2), assuming that 𝑠 > 𝑘, we obtain

−𝑠𝑡

∞

0

∞

0

−𝑠𝑡

∞

0

𝑘𝑡

−𝑠𝑡

where the condition 𝑠 > 𝑘 was needed for the existence of the last integral.

This completes the proof of the theorem.

1.5: Properties of Laplace Transform

1.5.1: First Translation Property or First Shifting Property

Statement: If 𝐿

(𝑠), then 𝐿

−𝑎𝑡

Proof: 𝐿{𝑒

−𝑎𝑡

−𝑠𝑡

∞

0

−𝑎𝑡

−

( 𝑠+𝑎

)

∞

0

Corollary: If 𝐿{𝑓(𝑡)} = 𝑓

(𝑠), then 𝐿{𝑒

𝑎𝑡

Example 1.2: Find the Laplace Transforms of

(i) (𝟏 + 𝒕 𝒆

−𝒕

𝟑

(ii) 𝒆

−𝟑𝒕

(𝐜𝐨𝐬 𝟒𝒕 + 𝟑 𝐬𝐢𝐧 𝟒𝒕) (iii) 𝒆

−𝒕

𝟐

(iv) 𝒆

𝟐𝒕

𝟒

𝒕 (v) 𝒆

−𝟐𝒕

𝟑

Solution: (i) Let 𝑓(𝑡) = ( 1 + 𝑡 𝑒

−𝑡

3

− 3 𝑡

3

−𝑡

− 2 𝑡

2

−𝑡

3

− 3 𝑡

3

−𝑡

− 2 𝑡

2

− 3 𝑡

3

−𝑡

− 2 𝑡

2

4

2

3

4

2

3

(ii) Let 𝑓

− 3 𝑡

cos 4 𝑡 + 3 sin 4 𝑡

− 3 𝑡

cos 4 𝑡 + 3 𝑒

− 3 𝑡

sin 4 𝑡

− 3 𝑡

cos 4 𝑡 + 3 𝑒

− 3 𝑡

sin 4 𝑡}

− 3 𝑡

cos 4 𝑡

− 3 𝑡

sin 4 𝑡}

2

(iii) Let 𝑓

−𝑡

sin

2

𝑒

−𝑡

2

1 − cos 2 𝑡

[𝑒

−𝑡

cos 2 𝑡]

−𝑎𝑡

[𝐿{𝑒

−𝑡

cos 2 𝑡}]

[

2

]

(iv) Let 𝑓

= sin

4

3

8

1

8

cos 4 𝑡 −

1

2

cos 2 𝑡

Proof: 𝐿

−𝑠𝑡

∞

0

−𝑠𝑡

𝑎

0

−𝑠𝑡

∞

𝑎

−𝑠𝑡

𝑎

0

−𝑠𝑡

∞

𝑎

−𝑠𝑡

∞

𝑎

Let 𝑡 − 𝑎 = 𝑇, giving 𝑑𝑡 = 𝑑𝑇

at 𝑡 = 𝑎, 𝑇 = 0 & 𝑡 = ∞, 𝑇 = ∞

∴ From (1), we obtain

−𝑠

( 𝑎+𝑇

)

∞

0

−𝑎𝑠

−𝑠𝑇

∞

0

−𝑎𝑠

Example 1.3: (i) Find 𝑳{𝑭(𝒕)} where 𝑭(𝒕) = {

𝝅

𝟑

𝝅

𝟑

𝝅

𝟑

(ii) Find 𝑳{𝑭(𝒕)} where 𝑭(𝒕) = {

𝟐

(iii) Find 𝑳{𝑭(𝒕)} where 𝑭(𝒕) = {

𝒕−𝒂

(iv) Find 𝑳

where 𝑭

𝟐𝝅

𝟑

𝟐𝝅

𝟑

𝟐𝝅

𝟑

Solution: (i) Here 𝑓(𝑡 − 𝑎) = sin (𝑡 −

𝜋

3

) and 𝑎 =

𝜋

3

) = sin (𝑡 −

= sin 𝑡

∴ 𝐿{𝑓(𝑡)} = 𝐿{sin 𝑡} =

2

Hence 𝐿{𝐹(𝑡)} = 𝑒

−𝑎𝑠

−

𝜋

3

𝑠

2

−

𝜋

3

𝑠

2

(ii) Here 𝑓

2

3

[∵ 𝐿

𝑛

𝑛+ 1

]

Hence 𝐿{𝐹(𝑡)} = 𝑒

−𝑎𝑠

−( 1 )𝑠

3

−𝑠

3

(iii) Here 𝑓(𝑡 − 𝑎) = 𝑒

𝑡−𝑎

𝑡

Hence 𝐿{𝑓(𝑡)} = 𝐿{𝑒

𝑡

1

𝑠− 1

−𝑎𝑠

(iv) Here 𝑓

= cos (𝑡 −

2 𝜋

3

) and 𝑎 =

2 𝜋

3

2 𝜋

3

) = cos (𝑡 −

2 𝜋

3

) or, 𝑓

= cos 𝑡

cos 𝑡

2

Hence, 𝐿{𝐹(𝑡)} = 𝑒

−𝑎𝑠

−

2 𝜋

3

𝑠

2

1.5.3 Change of Scale Property:

Statement: If 𝐿

(𝑠), then 𝐿

1

𝑎

𝑠

𝑎

Proof:

−𝑠𝑡

∞

0

Let 𝑎𝑡 = 𝑇, then 𝑎 𝑑𝑡 = 𝑑𝑇, giving 𝑑𝑡 =

1

𝑎

At 𝑡 = 0 : 𝑇 = 0 & 𝑡 = ∞: 𝑇 = ∞

From (1.3), we obtain

−𝑠(

𝑇

𝑎

)

∞

0

−(

𝑠

𝑎

)𝑇

∞

0

∞

0

[∫ 𝑒

−𝑠𝑡

∞

𝑠

𝑑𝑠] 𝑑𝑡 ,

after changing the order of integration.

∞

0

[

−𝑠𝑡

]

∞

0

[ 0 +

−𝑠𝑡

] 𝑑𝑡

−𝑠𝑡

∞

0

Example 1.5: Find the Laplace Transform of 𝑭(𝒕) , where (i) 𝑭(𝒕) =

𝒆

𝒂𝒕

−𝐜𝐨𝐬 𝒃𝒕

𝒕

(ii) 𝑭

𝒆

−𝒂𝒕

−𝒆

−𝒃𝒕

𝒕

(iii) 𝑭

𝐬𝐢𝐧 𝟐𝒕

𝒕

(iv) 𝑭

𝐬𝐢𝐧

𝟐

𝒕

(v) 𝑭

𝒆

𝒕

−𝐜𝐨𝐬 𝒕

𝒕

(vi) 𝑭

𝐬𝐢𝐧 𝟑𝒕 𝐜𝐨𝐬 𝒕

𝒕

Solution: (i) Let 𝑓(𝑡) = 𝑒

𝑎𝑡

− cos 𝑏𝑡, then

𝑎𝑡

− cos 𝑏𝑡

𝑎𝑡

cos 𝑏𝑡

1

𝑠−𝑎

𝑠

2

+𝑏

2

𝑎𝑡

− cos 𝑏𝑡

∞

𝑠

2

∞

𝑠

= [log

log

2

]

= [log

2

]

= − log

2

= log

2

(ii) Let 𝑓

−𝑎𝑡

−𝑏𝑡

, then

−𝑎𝑡

−𝑏𝑡

−𝑎𝑡

−𝑏𝑡

−𝑎𝑡

−𝑏𝑡

∞

𝑠

∞

𝑠

[

log

− log

)]

= [log (

)]

= − log (

) = log (

(iii) Let 𝑓

= sin 2 𝑡, then

𝐿{𝑓(𝑡)} = 𝐿{sin 2 𝑡} =

2

sin 2 𝑡

∞

𝑠

2

∞

𝑠

= [tan

− 1

)]

= tan

− 1

(∞) − tan

− 1

− tan

− 1

(vi) Let 𝑓

= sin 3 𝑡 cos 𝑡 =

1

2

sin 4 𝑡 + sin 2 𝑡

, then

= 𝐿 [

sin 4 𝑡 + sin 2 𝑡

]

[

sin 4 𝑡

sin 2 𝑡

}]

[

2

]

sin 3 𝑡 cos 𝑡

∞

𝑠

∞

𝑠

[

2

] 𝑑𝑠

[tan

− 1

) + tan

− 1

)]

[

tan

− 1

tan

− 1

− {tan

− 1

) + tan

− 1

)}]

[{

} − {tan

− 1

) + tan

− 1

)}]

[𝜋 − tan

− 1

) − tan

− 1

Laplace Transform: Existence and Properties, Summaries of Applied Mathematics

Related documents

Partial preview of the text

Download Laplace Transform: Existence and Properties and more Summaries Applied Mathematics in PDF only on Docsity!

Notes-II

Unit-I LAPLACE TRANSFORM

[𝑒

[𝐿{𝑒

[

]

[∵ 𝐿

]

[∫ 𝑒

𝑑𝑠] 𝑑𝑡 ,

[

]

[ 0 +

] 𝑑𝑡

]

]

[

)]

)]

)]

= 𝐿 [

]

[

}]

[

]

[

] 𝑑𝑠

)]

[

)}]

[{

)}]

)].