Convolution The Laplace Transform

Chernet Argaw

Advisor: Ato Markos Fisseha

MiliyonNew Stamp

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Acknowledgments

First of all I would like to thank the almighty God for his endless blessing. Secondly, I want to

express my heartfelt gratitude that I have for my instructors; Dr. Tilahun Abebaw, you are not

an instructor but you are a teacher, whom I dont have a word to express with. Ato Ademe

Mekonnen, you are truly a father not only a teacher. I would also like to thank my friends for

being there for me in the time of misery and loneliness.

Finally, I want to thank my advisor Ato Markos Fisseha for his continuous follow up and

constructive comments that made working on this project much easier to me.

Thank You!

Chernet Argaw

Addis Ababa University

Department of Mathematics

13/10/2007

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Contents Page

Acknowledgments ......................................................................................................................................... 2

Chapter 1 Introduction ................................................................................................................................. 4

The Laplace transform .............................................................................................................................. 4

Inverse Transforms ................................................................................................................................... 5

Transforms of Derivatives ..................................................................................................................... 7

Translation on the ................................................................................................................. 7

Inverse form of Second Translation Theorem ...................................................................................... 9

Chapter 2 Convolution ................................................................................................................................ 10

Inverse of Convolution ............................................................................................................................ 12

Properties of Convolution ....................................................................................................................... 13

Application of Convolution ..................................................................................................................... 13

The Volterra integral equation ........................................................................................................... 13

Integro differential equation .............................................................................................................. 14

Transform of a periodic Function ........................................................................................................... 16

Conclusion ................................................................................................................................................... 18

Reference .................................................................................................................................................... 19

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Chapter 1 Introduction

The Laplace transform

In elementary calculus we have learned that differentiation and integration are transforms; this means,

these operations transform a function into another function. In this section we will examine a special

type of integral transform that transform a function into another function. This type of operation is

called the Laplace transform.

Definition (Laplace transform): Let be a function defined for 0. Then the integral

{()} = ()

(1)

is said to be the Laplace transform of , provided that the integral converges. When the defining integral

(1) converges, the result is a function of . In general we shall use the lower case letter to denote the

function being transformed and the corresponding capital letter to denote its Laplace transform.

For example: {()} = (), {()} = (), {()} = ()

Example: Evaluate {}

Solution: () = , then by using (1),

{} =

= ()

= lim

()

= lim

1

( + 3)()

= lim

1

( + 3)() ()

=1

( + 3)lim

() ()

=1

( + 3)lim

1 ()

=1

( + 3), > 3

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The Laplace transform is a linear transform

Let be a function defined for 0 and , . Then,

{() + ()} = {()} + {()} = ( + ) + () (2)

Because of this property given above, is said to be a linear transform.

For example, {1 + 5} = {1} + {5} = {1} + 5{}

=1

+

5

+ 3

Theorem: Transforms of some basic functions

()

{()}

1 1

!

, = 1,2,3,

1

sin

+

cos

+

sinh

cosh

Inverse Transforms If () represents the Laplace transform of a function (), {()} = (), we then say () is the

inverse Laplace transform of () and we write () = {()}.

Example: Transform Inverse transform

{1} =

= 1

{} =

=

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Theorem: Some important inverse transforms

()

{()}

1

1

!

1

+ sin

+ cos

sinh

cosh

In evaluating inverse transforms, it often happens that a function of under consideration does not

match exactly the form of the Laplace transform () given in the table above. It may be necessary to

fix up the function of by multiplying and dividing by an appropriate constant.

Example: Evaluate

1

+ 9

Solution:

To match the form given in a table of the above theorem, we identify = 9. So, = 3. We fix up the

expression by multiplying and dividing by 3.

1

+ 9 =

1

3

3

+ 9

=1

3sin3

Inverse Laplace transform is a linear transform

The inverse Laplace transform is also a linear transform. i.e. for constants ,

{() + ()} = {()} + {()}

= () + ()

Where and are the transforms of the function and .

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Example: Evaluate

2 + 6

+ 4

Solution:

2 + 6

+ 4 =

2

+ 4+

6

+ 4 = 2

+ 4 + 6

1

+ 4

= 2

+ 4 +

6

2

2

+ 4

= 2cos2 + 3sin2

Transforms of Derivatives

Theorem: - If ,, ,() are continuous on [0,) and are of exponential order and if ()() is

piecewise continuous on [0,), then

()() = () () () ()

() ()

Where () = {()}.

Translation on the

A special function that is the number 0 (off) up to a certain time = and then the number 1 (on) after

that time. This function is called the unit step function or the Heaviside function.

Definition (Unit step function): The unit step function ( ) is defined to be

( ) = 0,0 < 1,

Notice that we define ( ) only on the non negative . In a broader sense ( ) = 0 for

< . The graph of ( ) is

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The unit step function can also be used to write piecewise defined function in a compact form. A general

piecewise defined function of the type

() = (), 0 <

(),

is the same as

() = () ()( ) + ()( ) (4)

Similarly, a function of the type

() = 0, 0 <

(), < 0,

Can be written as

() = ()[( ) ( )] (5)

Example: Find the compact form of () = 2,0 < 32, 3

Solution:

By using (4) above

() = () ()( ) + ()( )

In this case = 3,() = 2,() = 2. Then

() = 2 2( 3) 2( 3)

= 2 4( 3)

Theorem (Second translation Theorem): If () = {()} and > 0, then

{( )( )} = () (6)

Proof: By the additive interval property of integrals,

( )( )

Can be written as two integrals

{( )( )} = ( )( )

+ ( )( )

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= ( )

Now if we let = , = , then

{( )( )} = ()()

= ()

= {()}

By using the above theorem {( )} =

Inverse form of Second Translation Theorem

If () = {()}, the inverse form is

{()} = ( )( ) (7)

Example: By using (6) of the above theorem find the Laplace transform of the function given by

() = 2, 0 < 32, 3

() = 2 4( 3)

Solution:

{()} = {2 4( 3)}

= 2{1} 4{( 3)}

=2

4

In a similar manner to the above, we have

{()( )} = {( + )}

Theorem (Derivative Transforms): If () = {()} and = 1,2,3, , then

{()} = (1)

() (8)

Example:

{} =

{} =

1

3 = ( 3) =

1

( 3)

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Chapter 2 Convolution

Addition of transforms provides no problem, we know that

{ + } = {} + {}

But the transform of a product is generally different from the the product of transforms of the factors.

i.e.

{} {}{}

For example, take = and = 1. Then

=

{} = {} =1

1

But {} =

and {1} =

gives

{}{} =1

11

=

1

Thus

{} {}{}

Therefore, we use another method to solve the multiplication of transforms. This method of solving the

multiplication of transforms is called convolution.

Definition (Convolution): If the functions and are piecewise continuous on the interval [0,), then

a special product, denoted by is defined by the integral

= ()( )

(1)

and is called the convolution of and . The convolution is a function of .

Example: Let () = cos and () = sin. Find () ()

Solution:

Using (1) above we get

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() () = cos sin = cos sin( )

Since we know that

cos sin =1

2[sin( + ) sin( )]

Our integral becomes

cos sin =1

2 [sin() sin(2 )]

=1

2 sin() +

1

2cos(2 )

=1

2 sin() +

1

2(cos cos())

=

sin(), since cosine is an even function.

Theorem (Convolution Theorem): If () and () are piecewise continuous on [0,) and of

exponential order, then

() = {()} = {( )()} = {()}{()} = ()() (2)

Proof:

Let () =

() and () = ()

We now set = + , where is at first constant. Then = , and varies from to . Thus

() = ()( )

= ( )

Then, in and in vary independently. Hence we can insert the -integral into the -integral.

()() = ()

( )

= ()

( )

Here we integrate for fixed over from to and then over from 0 to . Under the assumption on

and the order of integration can be reversed. We then integrate first over from 0 to and then

over from 0 to , i.e.

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()() =

()( )

=

( )(), (1)

=

()

= {()} = (),

() = {()} = {( )()} = {()}{()} = ()()

Example: Evaluate {1 }

Solution:

With () = 1 and () = , the convolution theorem states that the Laplace transform of the

convolution of and is the product of their Laplace transform, then

{1 } =

= 1

4

= 1

4 =

1

44!

=6

Inverse of Convolution The convolution theorem is sometimes useful in finding the inverse Laplace transform of the product of

two Laplace transforms. i.e.

{()()} = ( )() (3)

Example: Evaluate

1

( )

Solution:

Using equation (3) above,

() =

has the inverse () = , and () =

has the inverse () = 1 with () = and

( ) = 1. Then we get,

1

( ) = () ()

= ()( )

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= 1

=1

[]

=1

[ 1]

Properties of Convolution

Application of Convolution

The Convolution theorem and the result by setting () = 1 are useful in solving other types of

equations in which an unknown function appears under an integral sign. In our case, we look the

Volterra integral equation and Integro differential equation.

The Volterra integral equation

It is the equation of the form

() = () + ()( )

In this case, () and ()are known.

Example: Solve () = cos +

( )

1 = Commutative law

2 ( + ) = + Distributive law

3 ( ) = ( ) Associative law

4 0 = 0 = 0 Multiplication by zero

5 1 = Identity

6 ( )() 0 may not hold.

7 When () = 1, ()

=

()

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Solution:

In the integral we identify () = , so that () = . We take the Laplace transform of each term.

In particular, the transform of the integral is the product of {()} = () and {} =

.

() = cos +

( )

{()} = {cos} +

( )

() =

+ 1+ {()}{}

() =

+ 1+ ()

1

+ 1

By rearranging we solve for ()

() =

+ 1+

1

+ 1

Then we take the inverse transform

{()} =

+ 1 +

1

+ 1

() = cos + sin

Integro differential equation

In a single-loop or series circuit Kirchhoffs second law states that the sum of the voltage drops across an

inductor, resistor and capacitor is equal to the impressed voltage (). Now, it is now that the voltage

drops across an inductor, resistor and capacitor are respectively

,() and

()

Where () is the current and , and are constants. It follows that the current in a circuit is governed

by the Integro differential equation

+ () +

1

()

= () (4)

Example: Determine the current () in a single loop circuit where = 1, = 2, = 0.2 with

(0) = 0. And the impressed voltage is () = 1.

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Solution: First, we find the Laplace transform of each term in the equation

+ 2 + 5

= 1, (0) = 0

Then,

() (0) + 2() + 5 ()

=1

We multiply throughout by and use the fact that (0) = 0 to obtain

() + 2() + 5() = 1

Solving for () and completing the square on the denominator

() =1

+ 2 + 5=

1

( + 1) + 4=1

2

2

( + 1) + 4

Taking the inverse Laplace transform to give us the current as a function of time

() =1

2 sin2

Example: Evaluate

+ + = 0

Such that (0) = 1 and (0) = 0.

Solution: Then {()} = () 1 and {()} = () ,

and because and are multiplied by ,

[() ] + [() 1]

[()] = 0

The result of differentiation and simplification is the differential equation

( + 1)() + () = 0

This equation is separable

()

()=

+ 1

Its general solution is

() =

, where C is a constant.

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Transform of a periodic Function

If a periodic function has period > 0, then ( + ) = ().

Theorem (Transform of a periodic function): If () is piecewise continuous on [0,) of exponential

order, and periodic with period , then

{()} =1

1 ()

(5)

Proof: We write the Laplace transform of as two integrals

{()} = ()

+ ()

Then we let = + , then the last integral becomes

()

= ()( + )

= ()

= {()}

Therefore,

{()} = ()

+ {()}

By rearranging we solve for {()}

{()} =1

1 ()

Example: Find the Laplace transform of the periodic function of

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Solution: The function () is called meander function and has period = 2. For 0 < 2, () can be

defined by

() = 1, 0 < 11, 1 < 2

and outside the interval by ( + ) = ( + 2) = ()

Now, by the above theorem (5)

{()} =1

1 ()

=1

1

+

=1

1

1

|

+1

|

=1

1 1

+

=1

1 2 + 1

=1

1 + 1

Example: Solve the following IVP

+ = cos , (0) = 5

Solution: Using Laplace transform and the fact that (0) = 5, we get

{} + {} = {cos}

() (0) + () =

+ 1

() + () 5 =

+ 1

() =5

+ 1+

( + 1)( + 1)=

5 + + 5

( + 1)( + 1)

By applying partial fraction

() =9

2

1

+ 1 +

1

2

+ 1+1

2

1

+ 1

Taking the Laplace inverse both sides, we get

() =9

2 +

1

2cos +

1

2sin

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Conclusion

Even though, it is not true that the integral of a product of functions is the product of the

integrals. Convolution theorem provides us a way to evaluate the Laplace transform of the

special product (1) is the product of the Laplace transform of and . This means that it is

possible to find the Laplace transform of the convolution of two functions without actually

evaluating the integral. Therefore, the Convolution theorem is the better way of finding the

Laplace transform of a product of functions.

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Reference

[1] [Dennis G. Zill] A First Course in Differential Equations with modeling applications, 10th ed. 2013.

[2] [Edwards and Penny] Elementary Differential equations with boundary value problems, 5th ed. 2004.

[3][Erwin Kreyszig] Advanced Engineering Mathematics, 10th ed. 2011.