Inverse Laplace Transform-Signals And Systems-Lecture Slides, Slides for Signals and Systems Theory. Fakir Mohan University, Balasore

Signals and Systems Theory

Description: Prof. Sahendra Agneya delivered this lecture at Fakir Mohan University, Balasore for Signals and Systems course. Inverse, Laplace, Transform, Function, Defined, Linear, Operator, Initial, Value, Problem
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1

• Consider a given function F(s), is it possible to find a function f(t) defined on [0,  ), such that

• If this is possible, we say f(t) is the inverse Laplace transform of F(s), and we write

Inverse Laplace Transform

? )()}({ sFtfL

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• First we note that the Inverse Laplace Transform is a “ Linear Operator”.

Some Examples.

, ss

s . F(s)

, b(s-a)

b . F(s)

, b(s-a)

s-a . F(s)

102 233

2

1

2

22

22

 

 

 

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• Consider the Initial Value Problem:

• We shall use Laplace Transform and Inverse Laplace Transform to solve this I.V.P.

Applications

.12)0(' ,2)0( ;85'2" 

 

yy eyyy t

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• Given the following I.V.P: (#36, P. 406)

Next let us consider a D.E. with variable coefficents

. 1)0(' 2)0(

2'"

 



y y

ytyty

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• Discontinuous functions play a very important role in Engineering, for example:

• This is known as the unit step function.

Laplace Transform of Discontinuous and Periodic Functions

. t 0 1, 0, t 0, :)(  tu

t 0

1

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Shifts and scalar multiples of the Unit Step Functions

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• For example: … • What is the Laplace Transform of u(t- a), a > 0?

Unit step functions can be used to represent any piecewise continuous function.

).(} s 1{ is

s 1 of Transform Laplace inverse The

thatfollowsIt . s 1

1 s 1 lim

|1lim

)())}(({

1

0

atueL

e

e

e s

e

e s

dte

dtatuesatuL

sa

sa

sa

sNsa

N

N a

st

Na

-st

st



  

   

 











 

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Next, what is: ?))}(()({  satuatfL

))}(({))}(()({ have then we,let weif ,particularIn

).()()}}({{ :Transform Laplace Inverse have we

moreover, , )}({)(

, ,let , )(

)()())}(()({ Since

1

0

)(

0

satgLesatutgL f(t - a)g(t)

atuatfsfLeL

sfLedvvfe

dtdvatvdtatfe

dtatuatfesatuatfL

sa

sa

saavs

a

st

st



 









 

 

 

 

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• Let us consider the following :

Some Examples

?)}({ .3

?)} (){(cos .2 ?))}(1({ .1

2

3 1

2

 

  t

s eL

tutL stutL

s

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• Periodic functions play a very important role in the study of dynamical systems

• Definition: A function f (t) is said to be periodic of period T, if for all t D(f) , we have

f (t + T) = f (t). • For examples, sine waves, cosine waves and

square waves are periodic functions. • What can we say about the transforms of

periodic functions?

Laplace Transform of Periodic Functions

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• It is not difficult to see that f (t) can be written as the sum of translates of f T (t). Namely,

Let f T (t) be the part of f over the basic period [0, T]. This is known as the Windowed

version of the periodic function f .

. 1

)}({)}({

thenT, periodwith function periodic continuous piecewise a is f If :Theorem

have weHence

. )()()( 0

sT T

n T

e sfLsfL

nTtunTtftf

 

 

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• Note that in this case f T (t) is given by :

• f T (t) = 1 - u(t - 1) . Hence

Example: The square wave with period 2

t 1 2

1

. )1(

1 1

)}({)}({

:Therefore

. )1(11)}({

2 ss T

s s

T

ese sfLsfL

e ss

e s

sfL



 

 

 



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• #10, 15, 31,

Some problems in the exercise

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• Definition: Given two functions f (t) and g(t) piecewise continuous on [0,). The convolution of f and g , denoted by

• Covolution is 1.commutative, 2. distributive, 3. associative and with 4. existence of zero.

Convolution Operator “ * “.

. 12

t tf,f1 ,11 example,For

. )()())((

:bydefinedis,

4 2

0

tt

dssgstftgf

gf

t





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• An important property of convolution is the • Theorem:

Laplace Transform of convolution

. ly,consequent , ))((

then, and If . order lexponentia of and , ) [0,on continuous

piecewise and function any twoFor

1 g)(t) (fG(s)}(t){F(s)L G(s) F(s) }(s) t gfL{

G(s)L{g}(s) F(s)L{f}(s)

g(t)f(t)

- 

 

 

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• 1. Writing

• 2. Then apply the Fubini’s theorem on interchanging the order of integration.

Proof of Convolution theorem can be done by

 

 0

)()()())(( dssgstfstutgf

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• Solve the initial value problem

Applications

 



 

t dssgsttgtty

stgLstLsyL

yytgyy

0 .)()sin()(sin)(

have wen theorem,convolutio by the Thus . ))}(({)}({sin)}({

that Transform Laplace usingafter found We. order lexponentia of

and )[0,on continuous pieceewise is g(t) Where .0)0(' and , 0)0( ; )("

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• Consider the following equation:

Example 2: Integral-Differential Equation

.2)( this, 1

12)}({ find We

fraction). partial using Y(s)for (solving n theorem.Convolutio and Transform Laplace

using solved becan which ,)(1)(' as; written becan equation This growth). population of

study on the Volterra V.by introducedEquation An (

10 , )(1)('

2

0

2

t

t

t s

ety ss

syL

etyty

. )y(dsestyty

 





 

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• Consider the linear system governed by the I.V.P:

• Thus given g(t) we wish to find the solution y(t). g(t) is called the input function and y(t) the output. The ratio of their Laplace Transforms,

Transfer function and Impulse response function

. 0)0(' and 0)0( . 0for , )('"

 

yy ttgcybyay

)( )()( i.e. system.linear the

offunction transfer thecalled is )( )}({ )}({

sG sYsH

sH sgL syL

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• we get

• The inverse Laplace Transform of H(s), written h(t) = L-1{H(s)}(t) is called the Impulse response function for the system. Graph!!

For our example, take the Laplace transform of the I.V.P

. 1 )( )()(

isfunction transfer the thus, becomes eq. the,conditions initial trivialhave weSince

).()()}0()({)}0(')0()({

2

2

2

cbsassG sYsH

G(s)cY(s)bY(s)Y(s)as

sGscYyssYbysysYsa

 





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• Namely:

• This can be checked easily (using Laplace transform). Now to solve a general I.V.P. such as

• This is a non-homogeneous eq with non-trivial initial values.

This function h(t) is the unique solution to the homogeneous problem

. /1)0(' and 0)0( with ; 0'" ahhchbhah 

1

0

0 0

with , )('"

y)y'( , y) y(

tgcybyay

 



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• They are the equivalent to the original I.V.P. Namely:

We shall split the given I.V.P into two problems

easily. checked becan This . form theof is I.V.P. original theofsolution

Then the . say, ,by given is (**) ofsolution the and be tofound is (*) ofsolution The

.000 0000 (*)

10

(t) y g)(t) (hy(t)

(t)y g)(t) (hy(t)

y) and y'(y) , y( cyby'(**) ay" ,) and y'()y( g(t) , cyby' ay"

k

k



  

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• Theorem: Let I be an interval containing the origin. The unique solution to the initial value problem

Theorem on solution using Impulse Response Function

. 00 ,0 :to solution unique theis )( and system the

for function response impulse theis where , : (S)

by given is I,on continuous is and constants are and where

00 :(P)

10

10

k

k

y), y'( y)y( cyby'ay" ty

h (t)yg)(t)(hy(t)

gca, b, ,y), y'( y) g ; y(cy by'ay"







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Example • #24, P.428 • Let a linear system be governed by the given

initial value problem.

• Find the transfer function H(s), the impulse response function h(t) and solve the I.V.P.

• Recall: y(t) = (h*g)(t) + yk(t)

0)0(' ,2)0( );(9''  yytgyy

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Dirac Delta Function • Paul A. M. Dirac, one of the great physicists

from England invented the following function: • Definition: A function (t) having the following

properties:

• is called the Dirac delta function. It follows from (2) that for any function f(t) continuous in an open interval containing t = 0, we have

 

 



- 1, dt (t) (2)

and , 0 tallfor , 0(t) )1(

 

  . )()( )( afdtattf

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Remarks on Theory of Distribution.

• Symbolic function, generalized function, and distribution function.

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Heuristic argument on the existence of -function.

• When a hammer strikes an object, it transfer momentum to the object. If the striking force is F(t) over a short time interval [t0, t1], then the total impulse due to F is the integral

momentum".in Change Impulse " that means This

motion, of law 2nd sNewton'By .dt F(t)Impulse

1

0

1

0

1

0

01

t

t





t

t

t

t ) . )-mv(t mv(t dt

dt dvm F(t) dt

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This heuristic leads to conditions 1 and 2.

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What is the Laplace Transform of -function?

• By definition, we have

).(') :F.T.C.) (by theget we(*), of sidesboth ateDifferenti

(*)

thatfollowsIt .1)( ,

and , 0)( ,

that whenNote

. )( )( ))}(( {

t

0

atuaδ (t

a). u(ta)dxδ (x

dxaxat

dxax a t

edtatedtatesatL

t

-

-

t

asstst











  









 



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Application: • Consider the symbolic Initial Value Problem:

Transform. Laplace of method by theit solve shall We t.at timeposition mequilibriu thefromnt displaceme

theis x(t) wheremass., on the impulsean exertinghammer aby struck is mass thelater, seconds but vibrate, tobeginsIt system.

spring-mass for theposition mequilibriu thebelowmeter 1 rest from released is spring a toattached mass a represents This

0010 ; ) ( 39"

 . ) , x'( ) x(txx 

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Linear Systems can be solved by Laplace Transform.(7.9)

• For two equations in two unknowns, steps are: • 1. Take the Laplace Transform of both equations

in x(t) and y(t), • 2. Solve for X(s) and Y(s), then • 3. Take the inverse Laplace Transform of X(s)

and Y(s), respectively. • 4. Work out some examples.

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