Inverse Laplace Transforms.ppt

The Inverse Laplace Transform

The University of TennesseeElectrical and Computer Engineering Department

Knoxville, Tennessee

wlg

Inverse Laplace Transforms

Background:

To find the inverse Laplace transform we use transform pairsalong with partial fraction expansion:

F(s) can be written as;

)(

)()(

sQ

sPsF

Where P(s) & Q(s) are polynomials in the Laplace variable, s.We assume the order of Q(s) P(s), in order to be in properform. If F(s) is not in proper form we use long division and divide Q(s) into P(s) until we get a remaining ratio of polynomialsthat are in proper form.


Background:There are three cases to consider in doing the partial fraction expansion of F(s).

Case 1: F(s) has all non repeated simple roots.

n

n

ps

k

ps

k

ps

ksF

...)(

2

2

1

1

Case 2: F(s) has complex poles:

...)))()((

)()(

*11

1

1

js

k

js

k

jsjssQ

sPsF

Case 3: F(s) has repeated poles.

)(

)(...

)(...

)())((

)()(

1

1

1

12

1

12

1

11

11

1

sQ

sP

ps

k

ps

k

ps

k

pssQ

sPsF

rr

r

(expanded)

(expanded)


Case 1: Illustration:

Given:

)10()4()1()10)(4)(1(

)2(4)( 321

s

A

s

A

s

A

sss

ssF

274)10)(4)(1(

)2(4)1(| 11

ssss

ssA 94

)10)(4)(1(

)2(4)4(| 42

ssss

ssA

2716)10)(4)(1(

)2(4)10(| 103

ssss

ssA

)()2716()94()274()( 104 tueeetf ttt

Find A1, A2, A3 from Heavyside


Case 3: Repeated roots.

When we have repeated roots we find the coefficients of the terms as follows:

|111

)()(1 psr

sFpsds

dk r

|121

)()(!2 12

2

psrsFps

ds

dk r

|11

)()()!( 1 psj

sFpsdsjr

dk r

jr

jr


Case 3: Repeated roots. Example

2

1

1

2211

2 )3()3()3(

)1()(

K

K

A

s

K

s

K

s

A

ss

ssF

)(____________________)( 33 tuteetf tt ? ? ?


Case 2: Complex Roots:

...)))()((

)()(

*11

1

1

js

K

js

K

jsjssQ

sPsF

F(s) is of the form;

K1 is given by,

jeKKK

jsjssQ

sPjsK js

||||

))(()(

)()(

111

1

11

|



js

eK

js

eK

js

K

js

K jj

11

*11

|||

tj

etej

etj

etej

eKjs

eK

js

eKL

jj

1||

||||111

2

)()(|

1|2

1||

tje

tjeateK

tjete

je

tjete

jeK


)cos(||2|||

1111

teK

js

eK

js

eKL t

jj


Therefore:

You should put this in your memory:


Complex Roots: An Example.

For the given F(s) find f(t)

o

jj

j

jss

sK

ss

sA

js

K

js

K

s

AsF

jsjss

s

sss

ssF

js

s

10832.0)2)(2(

12

)2(

)1(

5

1

)54(

)1(

22)(

)2)(2(

)1(

)54(

)1()(

|

|

2|1

0|

11

2

2

*


Complex Roots: An Example. (continued)

We then have;

jsjsssF

oo

2

10832.0

2

10832.02.0)(

Recalling the form of the inverse for complex roots;

)(108cos(64.02.0)( 2 tutetf ot


Convolution Integral:

Consider that we have the following situation.

h(t)x(t) y(t)

x(t) is the input to the system.h(t) is the impulse response of the system.y(t) is the output of the system.

System

We will look at how the above is related in the time domainand in the Laplace transform.


Convolution Integral:

In the time domain we can write the following:

ttdxthdhtxthtxty

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

In this case x(t) and h(t) are said to be convolved and theintegral on the right is called the convolution integral.

It can be shown that,

sHsXsYthtxL )()()()(

This is very important

* note

Inverse Laplace TransformsConvolution Integral:

Through an example let us see how the convolution integral and the Laplace transform are related.

We now think of the following situation:

x(t) y(t)

X(s) Y(s)

te 4

h(t)

H(s)

)4(

1

s


From the previous diagram we note the following:

)()(;)()(;)()( thLsHtyLsYtxLsX

h(t) is called the system impulse response for the followingreason.

)()()( sHsXsY

If the input x(t) is a unit impulse, (t), the L(x(t)) = X(s) = 1.Since x(t) is an impulse, we say that y(t) is the impulseresponse. From Eq A, if X(s) = 1, then Y(s) = H(s). Since,

Eq A

.)(,

)()()()( 11

responseimpulsesystemthSo

thsHLresponseimpulsetysYL


A really important thing here is that anytime you are givena system diagram as follows,

H(s)X(s) Y(s)

the inverse Laplace transform of H(s) is the system’simpulse response.

This is important !!


Example using the convolution integral.

e-4t

x(t) y(t) = ?

t

tt

tt deededuety0

44

0

)(4)(4 )()(

)(4

1

4

1

4

1)( 444

0

44 | 0 tueeedeety ttt

t t


Same example but using Laplace.

x(t) = u(t) s

sX1

)(

h(t) = e-4tu(t) 4

1)(

ssH

)(14

1)(

4

4141

4)4(

1)(

4 tuety

sss

B

s

A

sssY

t


Practice problems:

?)(,)2(

3)(

2)()( thiswhat

ssYand

ssXIfa

).(),()()()()( 6 thfindtutetyandtutxIfb t

).(,)4(

2)()()()(

2tyfind

ssHandttutxIfc

Answers given on note page

)(2)(5.1)( 2 tuetth t


Circuit theory problem:

You are given the circuit shown below.

+_

t = 0 6 k

3 k 1 0 0 F

+

_v(t)1 2 V

Use Laplace transforms to find v(t) for t > 0.



We see from the circuit,

+_

t = 0 6 k

3 k 1 0 0 F

+

_v(t)1 2 V

voltsxv 49

312)0(



+

_v c(t) i(t)

3 k

1 0 0 F

6 k

05)(

0)(

0)()(

tvdt

tdv

RC

tv

dt

tdv

tvdt

tdvRC

c

c

cc

c

c

Take the Laplace transformof this equations includingthe initial conditions on vc(t)



)(4)(

5

4)(

0)(54)(

0)(5)(

5 tuetv

ssV

sVssV

tvdt

tdv

tc

c

cc

cc

Inverse Laplace Transforms.ppt

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