chap 4- laplace transform

UECM1653 Mathematics for Engineering I UECM1683 Mathematics for Physics I

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Chapter 4 : Laplace Transform

The Laplace transform is an important tool for solving certain kinds of initial values problem,

particularly those involving discontinuous forcing function, as occur frequently in areas such as

electrical engineering. It is also used to solve boundary value problems involving partial

differential equations to analyze wave and diffusion phenomena. The Laplace transform is an

example of a class called integral transforms, and it takes a function () of one variable t (which we shall refer to as time) into a function () of another variable s (the complex frequency).

Definition:

The Laplace transform of a function f is a function [()]() defined by

[()]() = ()0 = ()

The improper integration is with respect to t and defines a function of the new variables s for all

s such that this integral converges.

Note- we call it improper integral because the upper limit in the integral is infinite, so the domain

of integration is infinite. And because the lower limit in the integral is zero, it follows that the

negative values of t for () is ignored or suppressed that means () contains information of () only for 0. In general, however, unless the domain is clearly specified, a function () is normally intepreted as being defined for all real values, both positive and negative, of t.

Making use of the Heaviside unit step function (), where () = 0 < 0 1 0

We have

()() = 0 < 0 () 0 Thus the effect of multiplying f (t) by H(t) is to convert it into a causal function.

Graphically, the relationship between f (t) and f (t)H(t) is


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It follows that the corresponding Laplace transform () contains full information on the behavior of ()(). Consequently, strictly speaking one should refer to { ()(), ()} rather than { (), ()} as being a Laplace transform pair. However, it is common practice to drop the () and assume that we are dealing with causal functions.

Notation: Because the symbol [()]() may be awkward to write in computations, we will make the following convention. We will use lowercase letters for a function we put into the

transform and the corresponding uppercase letters for the transformed function.

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

Example 1:

Let c be any real number, and () = . The Laplace transform of f is the function defined by

[()] = [] = 0 = 0 = lim

0 = lim 1 0 = lim

1

+ 1 = 1

=


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Example 2:

Let a be any real number, and () = . The Laplace transform of f is the function defined by

Example 3:

Determine The Laplace transform of the ramp function () = .

Solution:

Exercise 1:

Determine The Laplace transform of the ramp function () = 3 .

Properties of Laplace Transform

Now we consider some of the properties of the Laplace transform that will enable us to find

further transform pairs {( ), ()} without having to compute them directly using the definition.

The Linear properties of Laplace Transform:

A fundamental property of the Laplace transform is its linearity, which may be stated as follows:

Let f and g be functions where Laplace transforms exist, and let and be constants. Then

[ () + ()] = [ ()] + [()] = () + ().


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Laplace Transform Table

)(tf )( fL

1 s1

t 2

1s

2t 3!2

s

,...1,0=nt n

1ns!n+

at 1as

)1a(+

+

tae as

1

atnet 1)(

!+ nas

n

btat ee ( )( )bsas

ba

)(tf )( fL

tcos 22 +s

s

tsin 22

+s

tacosh 22 as

s

tasinh 22 as

a

te ta cos ( ) 22 +

as

as

te ta sin ( ) 22

+ as

( )tt sin ( )222

2

+ss

( )tt cos ( )222

22

+

ss

Note that: ( + 1) = !

Example 4:

Determine The Laplace transform of the ramp function () = 3 + 24 .

Solution:


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Exercise 2:

Determine the Laplace transform of the following function

(i) )3sin(5)( 42 tettk t +=

(ii) )3cos(2)5cos()( ttttw =

(iii) tettttf 73 25)2sin(34)( ++=

(iv) )8sin()( 4 tetg t=

(v) )7sin(414)( 23 tetth t =

The Inverse Laplace Transform

If [()]() = (), then 1[()] = () denotes a function () whose Laplace transform is () and is called the inverse Laplace transform of () . That is [()] = () 1[()] = ()

Example 5:

1 1

= Since [ ] = 1

1 1 = 1 Since [1] = 1

Remark:

The inverse Laplace transform also has linearity property

1[ () + ()] = 1[ ()] + 1[()] = () + ().


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The most obvious way of finding the inverse transform of the function () is using a table of transforms. Sometimes it is possible to write down the inverse transform directly from the table,

but more often than not. It is first necessary to carry out some algebraic manipulation of (). In particular, we frequently need to determine the inverse transform of a rational function of the

form ()() , where () and () are polynomials in s. in such cases the procedure is first to

resolve the function into partial fractions and then to use the table of transforms.

Example 6:

Find the following inverse Laplace transforms.

(i) 1 36

(ii) 1 3(6)3 (iii) 1 132 + 429 (iv) 1 1

2+6 (v) 1 2

2+328


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Exercise 3:

Find the inverse Laplace transform of the following functions.

(i) 8

411

5)( 22 +

+=

ssssG

(ii) 5)4(1

351)(

+=

sssP

(iii) 222

)4(273)(

+

=sssF

(iv) 64

)( 2 +=

sssQ

The First Shifting Theorem (s shifting)

Exponential modulation theorem

The shifting theorems of this section will enable us to solve problems involving pulses and other

discontinuous forcing functions. We will show that the Laplace transform of () is the transform of (), shifted a units to the right. This shift is achieved by replacing s by s-a in () to obtain ( ).

Theorem:

If () is a function having Laplace transform (), then for any number a,

)()]([ asFtfeL ta =

or )()]([1 tfeasFL ta=

This conclusion is also called shifting in the s variable.

Proof:

)()()())](([0

)(

0asFdttfedttfeestfeL tasatstat ===


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Example 7:

Given )()][cos( 22 sFbssbtL =+

= . Find the Laplace Transform of )cos(bteat .

Solution:

Use First shifting theorem,

)()]cos([ asFbteL at = And )()][cos( 22 sFbssbtL =+

=

So

22)()()()]cos([

basasasFbteL at+

==

Example 8:

Given 54 24][

stL = . Find the Laplace Transform of tet 54 .

Solution:

Example 9:

Find 1[1 ( + 2)2 ]

Solution:

Exercise 4:

Determine

1. [3 sin 2] 2. 1 1(+1)2(2+4)


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Derivative-of-transform property:

This property relates operations in the time domain to those in the transformed s domain, but

initially we shall simply look upon it as a method of increasing our repertoire of Laplace

transform pairs. The property is also sometimes referred to as the multiplication-by-t property. A

statement of the property is contained in the following theorem.

Derivative of transform:

Theorem:

If () is a function having Laplace transform of [()] = ()

then the functions () ( = 1, 2, . . . ) also have Laplace transforms, given by

[()] = (1) ()

, = 1, 2,

Example 10:

Find [ sinh(2)].

Solution:

Using derivative of transform

Example 11:

Find [2 sin(3)]

Solution:


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Exercise 5:

1. Find [32] 2. Find [ cosh ]

Heaviside function (Unit step function)

Functions having jump discontinuities are efficiently treated by using the unit step function or

Heaviside function H, defined by

0 and j = 1,2, , n-1. Then

)0(f)0(sf...)0(fs)0(fs)s(Fs)s](f[L )1n()2n(2n1nn)n( =

For second derivative case

[ ] )0()0()()( 2 fsfsFssfL =

Example 20:

Solve 1)0(;14 == yyy

Solution:


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Example 21:

Solve 2)0(,0)0(;34 ===++ yyeyyy t

Solution:

Example 22:

Solve the differential equation

22 + 5 + 6 = 2

subject to the initial conditions = 1 and

= 0 at = 0.

Exercise 8:

1. Solve + 4 = cos ; (0) = 0 2. Solve + 6 + 9 = sin ; (0) = 0 ,(0) = 0 3. Solve

22 + 2 + 5 = 1 ; = 1 and = 0 at = 0

4. Solve 2

2 3 + 2 = 24 ; = 1 and = 1 at = 0


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Laplace Transform

1. Solve the initial value problem 2)0(;2 ==+ yeyy t by

(i) the method of integrating factor

(ii) the Laplace transform Sept 2011

2. (a) Find a particular solution of .323 2xeyyy =++

(b) Find the inverse Laplace transform of the following function using the Laplace

transform table provided.

(i) 53

22 ++ ss

(ii) )2(

2

sse s

Sept 2011

3. Use Laplace transform to solve the following initial value problem

0)0(,0)0(,65 ===+ yyeyyy t

Jan 2012

4. Obtain the inverse Laplace transform of

+++

13432

21

sssL .

Jan 2012

5. Use Laplace transform to solve the following system

0)0()0()0(0324232

====+=++

yxxyxyyyxx

Chapter 4 : Laplace TransformDefinition:The Laplace transform of a function f is a function ,,..() defined by,,..()=,0--,-.,..=()Notation: Because the symbol ,,..() may be awkward to write in computations, we will make the following convention. We will use lowercase letters for a function we put into the transform and the corresponding uppercase letters for the transform...Example 1:Example 2:Let a be any real number, and ,.= ,-. . The Laplace transform of f is the function defined byExample 3:Exercise 1:Laplace Transform TableNote that: ,+1.= !Example 4:Determine the Laplace transform of the following functionThe Inverse Laplace TransformIf ,,..()=,., then ,-1.,,..=,. denotes a function () whose Laplace transform is () and is called the inverse Laplace transform of () . That is,,..=,. ,-1.,,..=,.Remark:The inverse Laplace transform also has linearity property,-1., ,.+ ,..= ,-1., ,..+ ,-1.,,..= ()+ ().The First Shifting Theorem (s shifting)Theorem:Example 7:Given . Find the Laplace Transform of .Use First shifting theorem,SoExample 8:Given . Find the Laplace Transform of .Derivative-of-transform property:This property relates operations in the time domain to those in the transformed s domain, but initially we shall simply look upon it as a method of increasing our repertoire of Laplace transform pairs. The property is also sometimes referred to as the...Heaviside function (Unit step function)Second Shifting Theorem (shifting in the t variable)If () is a function having Laplace transform (), then for a positive constant a,Example 13:Find the Laplace transformExample 14:Find the Laplace transformExample 15:Find the Laplace transform , whereExample 16:FindConvolutionIf f (t) and g(t) are defined for , then the convolution of f with g is a function defined byExample 17:The Convolution Theorem:If () and () have Laplace transforms () and () respectively, then,And the inverse convolution theorem isExample 18:Find by using the convolution theoremExample 19:FindSolution of Initial value problemTransform of a derivativeTheorem:Transform of a Higher derivativeTheorem:For s > 0 and j = 1,2, , n-1. ThenFor second derivative caseExample 20:SolveExample 21:SolveExample 22:Solve the differential equation,,-2.-,-2..+ 5,-.+ 6=2,-.subject to the initial conditions = 1 and ,-.=0 at =0.Exercise 8:Solve ,-.+ 4=,cos-. ; ,0.=0Solve ,-.+ 6,-.+ 9=,sin-. ; ,0.=0 , ,-.,0.=0Solve ,,-2.-,-2..+ 2,-.+ 5=1 ; = 1 and ,-.=0 at =0Solve ,,-2.-,-2.. 3,-.+ 2=2,-4. ; = 1 and ,-.=1 at =0Laplace Transform

chap 4- laplace transform

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