Laplace transform

The Laplace Transform

The University of TennesseeElectrical and Computer Engineering Department

Knoxville, Tennessee

wlg


The Laplace Transform of a function, f(t), is defined as;

0

)()()]([ dtetfsFtfL st

The Inverse Laplace Transform is defined by

j

j

tsdsesFj

tfsFL

)(

2

1)()]([1

*notes

Eq A

Eq B


We generally do not use Eq B to take the inverse Laplace. However,

this is the formal way that one would take the inverse. To use

Eq B requires a background in the use of complex variables and

the theory of residues. Fortunately, we can accomplish the same

goal (that of taking the inverse Laplace) by using partial fraction

expansion and recognizing transform pairs.

*notes


Laplace Transform of the unit step.

*notes

|0

0

11)]([

stst es

dtetuL

stuL

1)]([

The Laplace Transform of a unit step is:

s

1


The Laplace transform of a unit impulse:

Pictorially, the unit impulse appears as follows:

0 t0

f(t) (t – t0)

Mathematically:

(t – t0) = 0 t 0

*note

01)(0

0

0

dtttt

t



An important property of the unit impulse is a sifting or sampling property. The following is an important.

2

1 2010

20100 ,0

)()()(

t

ttttt

ttttfdttttf



In particular, if we let f(t) = (t) and take the Laplace

1)()]([ 0

0

sst edtettL


An important point to remember:

)()( sFtf

The above is a statement that f(t) and F(s) are transform pairs. What this means is that for each f(t) there is a unique F(s) and for each F(s)there is a unique f(t). If we can remember thePair relationships between approximately 10 of the Laplace transform pairs we can go a long way.


Building transform pairs:

eL(

e

tasstatat dtedteetueL0

)(

0

)]([

asas

etueL

stat

1

)()]([ |

0

astue at

1

)( A transform pair



0

)]([ dttettuL st

0 00| vduuvudv

u = tdv = e-stdt

2

1)(

sttu

A transform pair



22

0

11

2

1

2

)()][cos(

ws

s

jwsjws

dteee

wtL stjwtjwt

22)()cos(

ws

stuwt

A transform

pair


Time Shift

0 0

)( )()(

,.,0,

,

)()]()([

dxexfedxexf

SoxtasandxatAs

axtanddtdxthenatxLet

eatfatuatfL

sxasaxs

a

st

)()]()([ sFeatuatfL as


Frequency Shift

0

)(

0

)()(

)]([)]([

asFdtetf

dtetfetfeL

tas

statat

)()]([ asFtfeL at


Example: Using Frequency Shift

Find the L[e-atcos(wt)]

In this case, f(t) = cos(wt) so,

22

22

)(

)()(

)(

was

asasFand

ws

ssF

22 )()(

)()]cos([

was

aswteL at


Time Integration:

The property is:

stst

t

stt

es

vdtedv

and

dttfdudxxfuLet

partsbyIntegrate

dtedxxfdttfL

1,

)(,)(

:

)()(

0

0 00


Time Integration:

Making these substitutions and carrying outThe integration shows that

)(1

)(1

)(00

sFs

dtetfs

dttfL st


Time Differentiation:

If the L[f(t)] = F(s), we want to show:

)0()(])(

[ fssFdt

tdfL

Integrate by parts:

)(),()(

,

tfvsotdfdtdt

tdfdv

anddtsedueu stst

*note



Making the previous substitutions gives,

0

00

)()0(0

)()( |

dtetfsf

dtsetfetfdt

dfL

st

stst

So we have shown:

)0()()(

fssFdt

tdfL



We can extend the previous to show;

)0(...

)0(')0()()(

)0('')0(')0()()(

)0(')0()()(

)1(

21

233

3

22

2

n

nnnn

n

f

fsfssFsdt

tdfL

casegeneral

fsffssFsdt

tdfL

fsfsFsdt

tdfL

The Laplace TransformTransform Pairs:

____________________________________

)()( sFtf

f(t) F(s)

1

2

!

1

1

1)(

1)(

nn

st

s

nt

st

ase

stu

t


f(t) F(s)

22

22

1

2

)cos(

)sin(

)(

!

1

ws

swt

ws

wwt

as

net

aste

natn

at


f(t) F(s)

22

22

22

22

sincos)cos(

cossin)sin(

)()cos(

)()sin(

ws

wswt

ws

wswt

was

aswte

was

wwte

at

at

Yes !

The Laplace TransformCommon Transform Properties:

f(t) F(s)

)(1

)(

)()(

)0(...)0(')0()()(

)()(

)([0),()(

)(0),()(

0

1021

00

000

sFs

df

ds

sdFttf

ffsfsfssFsdt

tfd

asFtfe

ttfLetttutf

sFetttuttf

t

nnnnn

n

at

sot

sot

The Laplace TransformUsing Matlab with Laplace transform:

Example Use Matlab to find the transform of tte 4

The following is written in italic to indicate Matlab code

syms t,slaplace(t*exp(-4*t),t,s)ans = 1/(s+4)^2

The Laplace TransformUsing Matlab with Laplace transform:

Example Use Matlab to find the inverse transform of

19.12.)186)(3(

)6()(

2prob

sss

sssF

syms s t

ilaplace(s*(s+6)/((s+3)*(s^2+6*s+18)))

ans = -exp(-3*t)+2*exp(-3*t)*cos(3*t)

The Laplace TransformTheorem: Initial Value

If the function f(t) and its first derivative are Laplace transformable and f(t)Has the Laplace transform F(s), and the exists, then)(lim ssF

0

)0()(lim)(lim

ts

ftfssF

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.

Theorem:

s

Initial Value Theorem


Initial ValueTheorem:Example:

Given;

22 5)1(

)2()(

s

ssF

Find f(0)

1)26(2

2lim

2512

2lim

5)1(

)2(lim)(lim)0(

2222

222

2

2

22

sssss

ssss

ss

ss

s

ssssFf

ss s

s

The Laplace TransformTheorem: Final Value Theorem:

If the function f(t) and its first derivative are Laplace transformable and f(t)has the Laplace transform F(s), and the exists, then)(lim ssF

s

)()(lim)(lim ftfssF0s t

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.

Final ValueTheorem


Final Value Theorem:Example:

Given:

ttesFnotes

ssF t 3cos)(

3)2(

3)2()( 21

22

22

Find )(f .

03)2(

3)2(lim)(lim)(

22

22

s

ssssFf

0s0s

Devices & Hardware

Laplace transform