group 2 bhadouria , arjun singh glave , theodore dean han, zhe

Group 2

Bhadouria, Arjun SinghGlave, Theodore Dean

Han, Zhe

Chapter 5. Laplace TransformChapter 19. Wave Equation

Laplace Transform

Chapter 5

5. Laplace Transform

• 5.1. Introduction & Definition• 5.2. Calculation of the Transform• 5.3. Properties of the Transform• 5.4. Application to the Solution of Differential

Equations• 5.5. Discontinuous Forcing Functions; Heavisid

e Step Function• 5.6. Impulsive Forcing Functions; Dirac Impuls

e Function• 5.7. Additional Properties

5.1. Introduction & Definition

• The Laplace transform is a widely used integral transform. Denoted , it is a linear operator of a function f(t) with a real argument t (t ≥ 0) that transforms it to a function F(s) with a complex argument s. The Laplace transform has the useful property that many relationships and operations over the originals f(t) correspond to simpler relationships and operations over the images F(s). The Laplace transform has many important applications throughout the sciences.

• The Laplace transform is used for solving differential and integral equations. In physics and engineering, it is used for analysis of linear time-invariant systems. In this analysis, the Laplace transform is often interpreted as a transformation from the time-domain, in which inputs and outputs are functions of time, to the frequency-domain, where the same inputs and outputs are functions of complex angular frequency, in radians per unit time. Given a simple mathematical or functional description of an input or output to a system, the Laplace transform provides an alternative functional description that often simplifies the process of analyzing the behavior of the system, or in synthesizing a new system based on a set of specifications.

http://en.wikipedia.org/wiki/Laplace_transform

Basic idea

Tim e d o m a inu n kn o w n f(t), d /d t, D iff E q s

F req u en c y d o m a inu n kn o w n F (s ), A lg E q s

L ap laceTra n s fo rm a tio n

S o lv eA lgeb ra icE q u a tio n s

F req u en c y d o m a inkn o w n F (s )

T im e d o m a inkn o w n f(t)

S o lveD iffe ren tia lE q u a tio n s

In ve rs eL ap laceTran s fo rm

http://faculty.mercer.edu/olivier_pd/documents/Ch2LaplaceTransforms.ppt

• Given a known function K(t,s), an integral transform of a function f is a relation of the form

• The Laplace Transform of f is defined as

where the kernel function is K(s,t) = e-st , a=0, b= . F (s) is the symbol for the Laplace transform, L is the Laplace transform operator, and f(t) is some function of time, t.

b

adttfstKsF )(),()(

0)()()( dttfesFtfL st

How to solve problems

tttt xyyy 12

tt eet 2

21

21y

s

t

sss

1X

231)H( 2

2311

2 sss

1x t

Time Domain Frequency Domain

Solve algebraic equation

Laplace transform

Inverse Laplace transform

Douglas Wilhelm Harder, University of Waterloo.

5.2. Calculation of the Transform

Since the Laplace Transform is defined by an improper integral, thus it must be checked whether the transform F(s) of a given function f(t) exists, that is

whether the integral converges.

0)()( dttfesF st

EXAMPLE 2.1. Consider the following improper integral.

We can evaluate this integral as follows:

Note that if s = 0, then est = 1. Thus the following two cases hold:

0dtest

1lim1limlim0

00

sb

b

bst

b

b st

b

st ess

edtedte

0. if,diverges and0; if,100

sdtess

dte stst

EXAMPLE 2.2.Consider the following improper integral.

We can evaluate this integral using integration by parts:

Since this limit diverges, so does the original integral.

0cos tdtst

1cossinlim

cossinlim

sinsinlim

coslimcos

00

00

00

bsbsb

tstst

tdtstst

tdtsttdtst

b

bb

b

bb

b

b

b

Exponential orderSuppose that f is a function for which the following hold:

(1) f is piecewise continuous on [0, b] for all b > 0. (2) | f(t) | Kect when t T, for constants c, K, M, with K, M > 0.

A function f that satisfies the conditions specified above is said to to have exponential order as t

EXAMPLE 2.3. Is f(t)=exp(4t)cost of exponential order?Solution: Yes.

ttt eetetf 444 10cos)(

Therefore, K=10, c=4, and T=10 for instance

EXAMPLE 2.4. Is f(t)=ln(t) of exponential order?Solution: Yes.

011)ln( limlim

ctt

ctt cete

t

Therefore, K=100, c=1, and T=10 for instance

Piecewise ContinuousIf an interval [a, b] can be partitioned by a finite number of

points a = t0 < t1 < … < tn = b such that

Then the function f is piecewise continuousOr we can say f is piecewise continuous on [a, b] if it is

continuous there except for a finite number of jump discontinuities.

nktf

nktf

k

k

tt

tt

,,1,)(lim)3(

1,,0,)(lim)2(

) t,(teach on continuous is f (1)

1

1kk

Picture from Paul's Online Math Notes

EXAMPLE 2.5. Piecewise continuous functionConsider the following piecewise-defined function f.

• (a)

From this definition of f, and from the graph of f below, we see that f is piecewise continuous on [0, 3].

32121,310,

)(

2

tttttt

tf

• (b)

From this definition of f, and from the graph of f below, we see that f is NOT piecewise continuous on [0, 3].

32,421,210,1

)( 1

2

ttttt

tf

http://ebookbrowse.com/ch06-laplace-transform-ppt-d116265990

THEOREM 2.1 Existence of the Laplace Transform

Let f(t) satisfy these conditions:(i) f(t) is piecewise continuous on for every A>0, (ii) f(t) is of exponential order as

That is to say(1) f is piecewise continuous on [0, b] for all b > 0. (2) | f(t) | Kect when t T, for constants c, K, M, with K, M > 0.

At0

t

Then the Laplace Transform of f exists for s > c. finite )()()(

0 dttfesFtfL st

Inverse Laplace transform operatorBy definition, the inverse Laplace transform operator, L-1, converts an s-domain function back to the

corresponding time domain function:

Importantly, both L and L-1 are linear operators.Thus,

1 1{ } ( )2

i sti

f t L F s F s e dsi

{ } { } { }L au t bv t aL u t bL v t aU s bV s

1{ }L aU s bV s au t bv t

Examples of Calculation2.6.

2.7

2.8

0 0

)( 1)(;)(as

dtedteesFetf tasstatat

attf sin)(

0,)()(1

sin/)sin(lim1

cos/)cos(lim

sinlimsin)sin()(

222

2

00

00

00

sas

asFsFas

a

ateasaate

as

a

ateasaate

atdteatdteatLsF

b stbst

b

b stbst

b

b st

b

st

;tan tconsaf(t) 0

0

0st sta a aF s ae dt es s s

Laplace transform table

5.3. Properties of the Transform

THEOREM 3.0The transform of an expression that is multiplied by a constant is the constant multiplied by the transform. That is:

1 1( ) ( ) and ( ) ( )L kf t kL f t L kF s kL F s

• A various types of problems that can be treated with the Laplace transform include ordinary and partial differential equations as well as integral equations.

• Suppose u and v are functions whose Laplace transforms exist for s > a1 and s > a2, respectively.

• Then, for s greater than the maximum of a1 and a2, the Laplace transform of au(t) + bv(t) exists. That is,

With

Therefore

finite is )()()()(0 dttbvtauetbvtauL st

)( )()( )()()(00

tvbLtuaLdttvebdttueatbvtauL stst

THEOREM 3.1 Linearity of the Transform

)( )()()( tvbLtuaLtbvtauL

EXAMPLE 3.1.f (t) = 5e-2t - 3sin(4t) for t 0.

by linearity of the Laplace transform, and using results of Laplace transform table, the Laplace transform F(s) of f is:

0,

1612

25)4sin(35

)4sin(35)}({)(

22

2

sss

tLeL

teLtfLsF

t

t

For any U(s) and V(s) such that the inverse transforms

For any constants a,b.

THEOREM 3.2 Linearity of the Inverse Transform

1{ }L aU s bV s au t bv t

1 1{ } { } .L U s u t and L V s v t exist

Basic idea : Consider a general expression,

Expand a complex expression for F(s) into simpler terms, each of which appears in the Laplace Transform table. Then you can take the L-1 of both sides of the equation to obtain f(t).

1

n

ii

N s N sF s

D s s b

5

1 4sF s

s s

Perform a partial fraction expansion (PFE)

1 25( )

1 4 1 4sF s

s s s s

where coefficients and have to be determined.1 2

EXAMPLE 3.2.

To find : Multiply both sides by s + 1 and let s = -11

1 21 4

5 4 5 14 3 1 3s s

s ss s

To find : Multiply both sides by s + 4 and let s = -42

1 1

1 1 4

4 1 1 1( ) { ( )} { }3 1 3 4

4 1 1 1 4 1{ } { }3 1 3 4 3 3

t t

f t L F s Ls s

L L e es s

Therefore,

Let f(t) be continuous and f’(t) be piecewise continuous on 0≤t≤t。For every finite t。, and let f(t) be of exponential order as t so that there are constants K, c, T such that | f(t) | Kect when t T. Then L{f’(t)} exists for all s>c.

• This is a very important transform because derivatives appear in the ODEs we wish to solve.

• Similarly, for higher order derivatives:

where:

{ '( )} { ( )} 0L f t sL f t f

1 2 1( ) 1 2{ ( )} 0 0 ... 0 0n nn n n nL f t s F s s f s f sf f

0

0n

nn

t

d ffdt

THEOREM 3.3 Transform of the Derivative

Deriving the Laplace transform of f (t ) often requires integration by parts. However, this process can sometimes be avoided if the transform of the derivative is known:

For example, if f (t ) = t then f ‘ (t ) = 1 and f (0) = 0 so that, since:

That is:

( ) { ( )} (0) then {1} { } 0L f t sL f t f L sL t

2

1 1{ } therefore { }sL t L ts s

Proof

It has already been established that if:( ) { ( )} and ( ) { ( )}F s L f t G s L g t

then: { ( )} ( ) (0) and { ( )} ( ) (0)L f t F s f L g t G s g

Now let ( ) ( ) so (0) (0) and ( ) ( ) (0)g t f t g f G s sF s f

so that: { ( )} { ( )} ( ) (0) ( ) (0) (0)L g t L f t sG s g s sF s f f

Therefore: 2{ ( )} ( ) (0) (0)L f t s F s sf f

For example, if:

then:

and

Similarly:

And so the pattern continues.

( ) { ( )}F s L f t

{ ( )} ( ) (0)L f t F s f

2{ ( )} ( ) (0) (0)L f t s F s sf f

3 2{ ( )} ( ) (0) (0) (0)L f t s F s s f sf f

EXAMPLE 3.3

Then substituting in:

yields

So:

2( ) sin so that ( ) cos and ( ) sinf t kt f t k kt f t k kt

2{ ( )} ( ) (0) (0)L f t s F s sf f

2 2 2{ sin } {sin } {sin } .0L k kt k L kt s L kt s k

2 2{sin } kL kts k

)}sin({1

11

)1(11

)0(1

)}cos({ 22

22

2

2

2

2

tLss

sss

sfs

stdtdL

EXAMPLE 3.4

•COMMENT: Difference in )0(&)0(),0( fff

The values are only different if f(t) is not continuous at t=0

Example of discontinuous function: u(t)

1)0()0(

1)(lim)0(

0)(lim)0(

0

0

uf

tuf

tuf

t

t

*Additional Section

EXAMPLE 3.5Try to solve the differential equation:

( ) ( ) 1 where (0) 0f t f t f take the Laplace transform of both sides of the differential equation to yield:

( ) ( ) 1 so that ( )} { ( ) 1L f t f t L L f t L f t L

1 1( ) (0) ( ) which means that ( 1) ( )sF s f F s s F ss s

1( )( 1)

F ss s

Resulting in:

The right-hand side can be separated into its partial fractions to give:1 1( )

1F s

s s

From the table of transforms it is then seen that:

1 1 11 1 1 1( ) 11 1

tf t L L L es s s s

( ) 1 tf t e Thus,

THEOREM 3.4 Laplace Convolution Theorem

If both exist for s>c, then

{ } { }L f t F s and L g t G s

10

{ } ( ) ( )t

L F s G s f g t d

0{ ( ) ( ) } .

tor L f g t d F s G s for s c

0( )( ) ( ) ( )

tdefine f g t f g t d As Laplace convolution of f and g.

Proof: { }L f t F s

0)()( duufesF su

0

)()( dvvgesG svWe have,

})()({

])()([

)()(

)()(

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

0

0 0

0 0

0 0

)(

00

duutgufL

dtduutgufe

dudtutgufe

dudvvgufe

dvvgeduufesGsF

t

t

t

u

st

t

t

u

st

vus

svsu

10

{ } ( ) ( )t

L F s G s f g t d therefore

EXAMPLE 3.6 If f(t)=exp(t), g(t)=t, then

10

)(0

)()(*0

tet

eet

etdtetgf tt

EXAMPLE 3.7 ))((

1}*{bsas

eeL btat

)(*}))((

1{ )(

0

1 babaeedeeee

bsasL

btattbt abtat

Schiff. Joel L. The Laplace transform: theory and applications. P92

5.4. Application to the Solution of Differential Equations

• Laplace transforms play a key role in important engineering concepts and techniques.

Examples: • Transfer functions • Frequency response• Control system design• Stability analysis• …

Solution of ODEs by Laplace Transforms

Procedure:

1. Take the L of both sides of the ODE.

2. Rearrange the resulting algebraic equation in the s domain to solve for the L of the output variable, e.g., F(s).

3. Perform a partial fraction expansion.

4. Use the L-1 to find f(t) from the expression for F(s).

Equation with initial conditions

Laplace transform is linear

Apply derivative formula

Rearrange

Take the inverse

EXAMPLE 4.1.

'' 1, (0) '(0) 0f f f f

{ ''} { } {1}L f L f L

2 1( ) (0) '(0) ( )s F s sf f F ss

2 21 1( )

( 1) 1sF s

ss s s

( ) 1 cosf t t

'' 1, (0) '(0) 0f f f f Solve

resulted in the expression 3 21

6 11 6F s

s s s s

31 2 41

1 2 3 1 2 3

1/ 6 1/ 2 1/ 2 1/ 61 2 3

F ss s s s s s s s

s s s s

''' 6 '' 11 ' 6 1, (0) 0f f f f f

EXAMPLE 4.2.

Take L-1 of both sides:1 1 1 1 11/ 6 1/ 2 1/ 2 1/ 6{ ( )} { } { } { } { }

1 2 3L F s L L L L

s s s s

2 31 1 1 16 2 2 6

t t tf t e e e

EXAMPLE 4.3.3 ' 2 4 2, (0) 0xf f e f

Taking Laplace transforms of both sides of this equation gives:

4 2 6 23[ ( ) (0)] 2 ( )1 ( 1)

ssF s f F ss s s s

6 2 27 1 1 4 1( ) ( ) ( )( 1)(3 2) 5 3 2 5 1

27 1 1 4 1( ) ( )15 2 / 3 5 1

sF ss s s s s s

s s s

2 /39 4( ) 15 5

x xf t e e

K.A. Stroud. Engineering Mathematics. P1107

EXAMPLE 4.4.A mass m is suspended from the end of a vertical spring of constant k (force required to produce unit stretch). An external force f(t) acts on the mass as well as a resistive force proportional to the instantaneous velocity. Assuming that x is the displacement of the mass at time t and that the mass starts from rest at x=0,(a) Set up a differential equation for the motion(b) Find x at any time t

Solution:(a)The resistive force is given by –βdx/dt. The restoring force is given by –kx. Then by Newton’s law, 2

2 ( ) (1)d x dxm kx F tdtdt

2

2 ( ) (1)d x dxor m kx F tdtdt

0, '(0) 0 (2)where x x

(b) Taking the Laplace transform of (1), using we obtain

So that on using (2)

Where

There are three cases to be considered.

EXAMPLE 4.4.

{ } , ( )L f t F s L x X

2[ (0) '(0)] [ (0)] ( )m s X sx x sX x kx F s

2 2( ) ( ) (3)

[( / 2 ) ]F s F sX

ms s k m s m R

2

24kRm m

• Case 1, R>0. In this case letWe have

Then we find from (3)

• Case 2, R=0. In this case

thus

EXAMPLE 4.4.2R

1 /22 2

1 sin{ }( / 2 )

t m tL es m

( )/20

1 ( ) sin ( )t t u mx F u e t u du

m

1 /22

1{ }( / 2 )

t mL tes m

( )/0

1 ( )( )t t u mx F u t u e du

m

• Case 3. R<0, In this case letWe have

EXAMPLE 4.4.

2R

1 /22 2

1 sinh{ }( / 2 )

t m tL es m

( )/20

1 ( ) sin ( )t t u mx F u e h t u du

m

Schaum’s Outline of Theory and Problems of Advanced Mathematics for Engineers and Scientists. P115, Problem 4.46

5.5. Discontinuous Forcing Functions; Heaviside Step Function

0, 0( )

1, 0t

H tt

The unit step function is widely used. It is defined as:

0 0

1 1{ ( )} ( ) [ ]0

t tst st st

t t

tL H t H t e dt e dt e

ts s

Because the step function is a special case of a “constant”, it follows

More generally,0,

( )1,

t aH t a

t a

{ ( ) ( )} ( )asL H t a f t a e F s

It is possible to express various discontinuous functions in terms of the unit step function.

0{ ( )} (0) (1) 0 0

st asa st st

a

e eL H t a e dt e dt if sas s

{ ( )} 0aseL H t a s

s

1{ } ( ) 0aseL H t a s

s

Therefore, or

EXAMPLE 5.10

1 1( ) ( ( ) ( ))

0

a b

t a

H t H t a H t b a t bb a b a

t b

{ ( )}( )

as bse eL H ts b a

Schiff. Joel L. The Laplace transform: theory and applications. P92

EXAMPLE 5.2Determine L{g(t)} for

2

0, 0 1( )

( 1) , 1t

g tt t

2 23 3

2 2{ ( )} { ( 1)( 1) } { }s

s s eL g t L H t t e L t es s

Schiff. Joel L. The Laplace transform: theory and applications. P92

5.6. Impulsive Forcing Functions; Dirac Impulse Function

Pictorially, the unit impulse appears as follows:

Mathematically:

01)(0

0

0

dtttt

t

tt

tt0

0/1)( 0

tt

tt0

0);( 0

0lim

Picture from web.utk.edu

2

1 2010

20100 ,0

)()()(

t

t ttttttttg

dttttg

)0()()(0

gdtttg

We can prove that

Where 010 ttt

)()()(1)()( 0100

10

00

limlimlim tgtgtgdttttg

thus

)(t is known as Dirac delta function, or unit impulse function

further

The Laplace transform of a unit impulse:if we let f(t) = (t) and take the Laplace

1)()}({ 0

0

sst edtettL

* Rectangular Pulse Function

0 for 0

for 00 for

w

w

tf t h t t

t t

1 wt shF s es

*Additional Section

EXAMPLE 6.1Some useful properties of Dirac Impulse Function

1{ } ( ) 2 ats aL t aes a

1( ) ( ) duax dx u

a a

( )( ) xaxa

( ) ( ) ( )f t t T dt f T

5.7. Additional Properties

THEOREM 7.1 S-plane (frequency) shiftIf exists for s>s0, then for any real constant a,

for s+a>s0, or equivalently,)()}({ asFtfeL at

{ }L f t F s

)()}({1 tfeasFL at

Proof : )()()()}({0

)(

0

asFdtetfdtetfetfeL tasstatat

EXAMPLE 7.1.

22)()}sin({

asteL at

EXAMPLE 7.2.

22sin2cos2}

4)1(1{}

4)1()1({2

}4)1(1)1(2{}

4)1(12{}

5212{

21

21

21

21

21

tetes

Ls

sL

ssL

ssL

sssL

tt

THEOREM 7.2 Time ShiftIf exists for s>s0, then for any real constant

a>0,

for s>s0, or, equivalently,

{ }L f t F s

)()}()({ sFeatfatHL as

)()()}({1 atfatHsFeL as

)()()(

,

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

00

)(

0

sFedueufedueuf

autatuset

dteatfdteatfatHatfatHL

assuasa

aus

a

stst

Proof :

EXAMPLE 7.3.

aseetHL

sta

10)10( })10({

THEOREM 7.3 Multiplication by 1/s (Integrals)If exists for s>s0, then

for s>max{0,s0}, or, equivalently,

{ }L f t F s

ssFtfDLdfL

t )()}({})({ 100

dfssFL

t

0

1 )(})({

Proof : ststte

svdtedvdttfdudftfDtguset

1,,)(,)()()(0

10

ssFdtetf

setg

s

tgdes

etgs

des

tgdtetgtgLdfL

stst

ststststt

)()(1])(1[

))((1])(1[)1)(()()}({})({

0

000

0

0,)()(,)()(for 0,g(t) 0, tIf00

stst ethanslowertgdttfsodtetft

Notice:

EXAMPLE 7.4. )}{sin(

11)

1)(1(})cos({})({ 2200

tLss

ss

dLdfLtt

COMMENT:Another way to prove Time Integration Property

ssF

dτeτfs

tdτdeτf

tdτdeτf

dtedττfdττfL

st

τ

st

τ

st

sttt

0

0

0

0 00

1

From Douglas Wilhelm Harder, University of Waterloo.

THEOREM 7.4 Differentiation with Respect to s (Multiplication by tn)If exists for s>s0, then

for s>s0, or, equivalently,

{ }L f t F s

)()1(})({1 tftds

sFdL nnn

n

Proof :

EXAMPLE 7.5.

)()1()}({ sFdsdtftL n

nnn

)()1()()1(

)()1()()()}({

0

000

sFs

dtetfs

dtes

tfdtettfdtetfttftL

n

nnst

n

nn

stn

nnstnstnn

1

!)1()1(}1{)}({ nn

nnnn

sn

sdsdtLtutL

00

000

cossin1

cossin1sin1sin

dtettdtets

dtettts

etts

dtett

stst

ststst

00

000

sincos1

sincos1cos1cos

dtettdtets

dtettts

etts

dtett

stst

ststst

EXAMPLE 7.6.

tttf sin)(

0000

sincos1sin1sin dtettdtets

dtets

dtett stststst

Firstly

where

substitute

Douglas Wilhelm Harder, MMathdwharder@alumni.uwaterloo.ca

EXAMPLE 7.6.

tttf sin)(

0000

sincos1sin1sin dtettdtets

dtets

dtett stststst

222

22

sin1

11

1

sin1

11

11sin

sttL

ssss

ttLs

ssss

ttL

substitute

1

11

2sin

121sin

222

22

2

sdsd

ssttL

ssssttL

therefore

THEOREM 7.5 Integration with Respect to sIf there is a real number s0 such that exists for s>s0, and exists, then

for s>s0, or, equivalently,

{ }L f t F s

EXAMPLE 7.7. To evaluate

')'(})({ dssFttfL

s

0lim /t f t t

ttfdssFL

s

)(}')'({1

}{ln1

sasL

sassas

dsd 11ln

')'

1'1(lnln

11

1 dssass

ass

as s

s

')'

1'1(ln ds

sassas

s

1}11{)( 1

ate

sasLtf

te

sasL

at

1}{ln1

THEOREM 7.6 Large s Behavior of F(s)Let f(t) be piecewise continuous on for each finite and of exponential order as , then(i) (ii)

00 t t

Proof : Since f(t) is of exponential order as then there exist real constants K and c, with K 0 such that | f(t) | Kect for all t T. Since f(t) is piecewise continuous onThere must be a finite constant M such that | f(t) | M, onFor all s>c:

0tt

( ) 0F s as s ( )sF s is bounded as s

t

00 t t

00 t t

csK

sM

tcseK

seMdtKedtMe

dtetfdtetfdtetfdtetfsF

tcsst

t

tcst st

t

stt ststst

0

)()(

0

000

)(1

)()()()()(

0

0

0

0

0

THEOREM 7.7 Initial-Value TheoremLet f be continuous and f’ be piecewise continuous on for each finite , and let f and f’ be of exponential order as then

00 t t

Proof : with the stated assumptions on f and f’, it follows that

0t t [ ( )] (0)lim

ssF s f

NOTE:The utility of this theorem lies in not having to take the inverse of F(s) in order to find out the initial condition in the time domain. This is particularly useful in circuits and systems.

{ '( )} ( ( )) (0) ( ) (0)L f t sL f t f sF s f

Since f’ satisfies the conditions of THEOREM 4.7, it follows that

{ '( )} 0L f t as s

[ ( )] (0)lims

sF s f

thus

Given22 51

2

)(s)(sF(s)

Find f(0)

1262

2

25122]

512[][0

2222

222

2

2

22

lim

limlimlim

)s(ssssssss

ssss

)(s)(sssF(s))f(

s

sss

EXAMPLE 7.8.

THEOREM 7.8 Final Value TheoremLet f be continuous and f’ be piecewise continuous on, and let f and f’ be of exponential order as then

0 t

EXAMPLE 7.9.

Suppose

then

t

0[ ( )] ( )lim lim

s tsF s f t

NOTE:It can be used to find the steady-state value of a closed loop system (providing that a steady-state value exists.Again, the utility of this theorem lies in not having to take the inverseof F(s) in order to find out the final value of f(t) in the time domain. This is particularly useful in circuits and systems.

5 25 4sF s

s s

0

5 2lim 0.5lim5 4t s

sf f ts

*Additional Section

Given:tte(s)Fnote

)(s

)(sF(s) t 3cos12322

2322 2

)(ffind

.0]

232)2(

232)2([)]([)( limlim00

s

ssssFfss

EXAMPLE 7.10.

THEOREM 7.9 Transform of Periodic FunctionIf f is periodic with period T on and piecewise continuous one period, then

0 t

EXAMPLE 7.11.

0for s 0

1{ ( )} ( )1

T stsTL f t f t e dt

e

(a)cos(t) is repeated with period 2 (b)cos(t) is repeated with period tcos

tL cos

tf

tfL

πs

πs

sπ

πst

es

sse

e

dtettfL

11

1

cos2

0From Douglas Wilhelm Harder, University of Waterloo.

Consider f(t) below:(a) (b)

s

s

s

s

s

st

ese

es

e

e

dtetfL 222

1

0

11

1

1

1

tf tu

s

tuL 1 tfL

EXAMPLE 7.12.

ss

dtetuL st 110

From Douglas Wilhelm Harder, University of Waterloo.

THEOREM 7.10 Scaling in TimeIf exists for s>s0, then for any real constant a>0,

for s>s0

Proof :

EXAMPLE 7.13.

)(1)}({asF

aatfL

{ }L f t F s

)(1)(1)()}({

1,,

0

)(

0 asF

adueuf

adteatfatfL

dua

dtautatulet

a uas

st

2222

2

2)(1)1

)(1(1)}{sin(

ssstL

222 1)/(/1cos

ωss

ss

ωωtL

*Additional Section

THEOREM 7.11 Time delayTime delays occur due to fluid flow, time required to do an analysis (e.g., gas chromatograph). The delayed signal can be represented asthen

EXAMPLE 7.14.

θ θ time delayf t

θ{ θ } sL f t e F s

θ2 2{ θ } {cos ( θ)} s sL f t L t e

s

*Additional Section

Common Transform Properties Tablef(t) F(s)

)()(

)(1)(

)(1

1)(

')'()(

)0(10...)0('2)0(1)()(

)()1()(

)(

0)(

)(0),()(

)()(

0

sFetfasF

aatf

dtetfe

tf

dssFttf

fnfsfnsfnssFnsndt

tfnd

dssFdtft

ssFt

df

sFaseaatfatH

asFtfate

s

T stsT

s

n

nnn

EXAMPLE 7.15Solve the ODE,

5 4 2 0 1df f fdt

First, take L of both sides of the equation,

25 1 4sF s F ss

Rearrange,

5 25 4sF s

s s

Take L-1, 1 5 2{ }

5 4sf t L

s s

From Table, 0.80.5 0.5 tf t e

References• Schiff. Joel L. The Laplace transform: theory and applications.• Murray. R. SPIEGEL. Schaum’s Outline of Theory and

Problems of Advanced Mathematics for Engineers and Scientists.

• K.A. Stroud. Engineering Mathematics. • http://en.wikipedia.org/wiki/Laplace_transform• Douglas Wilhelm Harder, Math. University of Waterloo• http://faculty.mercer.edu/olivier_pd/documents/Ch2Laplace

Transforms.ppt• http://ebookbrowse.com/ch06-laplace-transform-ppt-d1162

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