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

)}. ( { ) (1

s F L t f

2

• 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

3

• 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

4

• 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

5

• 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

t0

1

6

Shifts and scalar multiples of the Unit Step Functions

7

• 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

|1

lim

)())}(({

1

0

atueL

e

e

es

e

es

dte

dtatuesatuL

sa

sa

sa

sNsa

N

Na

st

Na

-st

st

8

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

9

• Let us consider the following :

Some Examples

?)}({ .3

?)} (){(cos .2

?))}(1({ .1

2

31

2

t

s

eL

tutL

stutL

s

10

• 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

11

• 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

sTT

nT

e

sfLsfL

nTtunTtftf

12

• 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

t1 2

1

. )1(

1

1

)}({)}({

:Therefore

. )1(11

)}({

2 ssT

ss

T

ese

sfLsfL

ess

e

ssfL

13

• #10, 15, 31,

Some problems in the exercise

14

• 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

. )()())((

:by defined is ,

42

0

tt

dssgstftgf

gf

t

15

• 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)

-

16

• 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

17

• Solve the initial value problem

Applications

tdssgsttgtty

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( ; )("

18

• 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

etyss

syL

etyty

. )y(dsestyty

19

• 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

sHsgL

syL

20

• 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

21

• 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

22

• 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

23

• 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"

24

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

25

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

26

Remarks on Theory of Distribution.

• Symbolic function, generalized function, and distribution function.

27

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

28

This heuristic leads to conditions 1 and 2.

29

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

30

Application:• Consider the symbolic Initial Value Problem:

x +9x=3δ \( t - π \) ; ital x \( 0 \) = 1 ital , x ' \( 0 \) = 0 . } {} # {} # size 12{ Thisrepresents a mass attached ¿a spring is released¿

rest } {} # size 12{1meter below the equilibrium position for the mass−spring } {} # size 12{ system . begins ¿vibrate , but π secondslater , the mass is struck } {} # size 12{ by a hammer exerting an impulse on the mass . , where x \( t \) is the} {} # size 12{ displacement ¿ the equilibrium position at time t . } {} # size 12{We shall solveby the method of ℒ Transform . } {} } } {

31

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.