Laplace Transform with Applications

Laplace Transform

And

Applications

EE 221

Building the Case…

Definition

Laplace

transform solution

in

s domain

inverse

Laplace

transform

solution in

time

domain

problem in

time

domain

- Other Transforms

• Fourier Transform

• Z-transform

• Wavelet-Transform

Laplace Transformation

linear

differential

equation

time

domain

solution

Laplace

transformed

equation

Laplace

solution

time domain

Laplace domain or

complex frequency domain

algebra

Laplace transform

inverse Laplace

transform

The Laplace Transform

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

0

)()()]([ dtetfsFtfLst

js where


Laplace Transform of the unit step.

|0

0

11)]([

stste

sdtetuL

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 t0 01)(

0

0

0

dttt

t

t



An important property of the unit impulse is a shifting

or sampling property. The following is an important.

2

12010

2010

0,0

)()()(

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 the

Pair relationships between approximately 10 of the

Laplace transform pairs we can go a long way.


Building transform pairs:

eL(

e

tasstatatdtedteetueL

0

)(

0

)]([

asas

etueL

stat

1

)()]([ |

0

astue at

1

)( A transform

pair



0

)]([ dttettuLst

0 0

0| vduuvudv

u = t

dv = e-stdt

2

1)(

sttu

A transform

pair



22

0

11

2

1

2

)()][cos(

ws

s

jwsjws

dteee

wtLst

jwtjwt

22)()cos(

s

stut 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([

as

asteL at


Time Integration:

The property is:

stst

t

st

t

es

vdtedv

and

dttfdudxxfuLet

partsbyIntegrate

dtedxxfdttfL

1,

)(,)(

:

)()(

0

0 00


Time Integration:

Making these substitutions and carrying out

The 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

anddtsedueustst



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

23

3

3

2

2

2

n

nnn

n

n

f

fsfssFsdt

tdfL

casegeneral

fsffssFsdt

tdfL

fsfsFsdt

tdfL


Transform Pairs:

____________________________________

)()( sFtf

f(t) F(s)

1

2

!

1

1

1)(

1)(

n

n

st

s

nt

st

ase

stu

t


Transform Pairs:

f(t) F(s)

22

22

1

2

)cos(

)sin(

)(

!

1

ws

swt

ws

wwt

as

net

aste

n

atn

at


Transform Pairs:

f(t) F(s)

22

22

22

22

sincos)cos(

cossin)sin(

)()cos(

)()sin(

ws

wswt

ws

wswt

was

aswte

was

wwte

at

at


Common Transform Properties:

f(t) F(s)

)(1

0

)(

)()(

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

)()(

)]0

([0),0

()(

)(00

),0

()0

(

sFs

t

df

ds

sdFttf

fnfsfnsfnssFnsndt

tfnd

asFtfate

ttfLsotetttutf

sFsotetttuttf


Theorem: 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 Value Theorem: Example:

Given;

225)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


Theorem: 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 ssFs

)()(lim)(lim ftfssF

0s t

Again, the utility of this theorem lies in not having to take the inverse

of 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 Value

Theorem

The Laplace Transform Final Value Theorem: Example:

Given:

ttesFnote

s

ssF

t3cos)(

3)2(

3)2()(

21

22

22

Find )(f .

0

3)2(

3)2(lim)(lim)(

22

22

s

ssssFf

0s0s

Apply Initial- and Final-Value Theorems to

this Example

• Laplace transform

of the function.

• Apply final-value

theorem

• Apply initial-value

theorem

)4()2(

2)(

ssssF

4

1

)40()20()0(

)0(2)(lim

tft

0)4()2()(

)(2)(lim 0

tft

The Inverse Laplace Transform

)(

)()(

sD

sNsF

31

Definition:

F(s) is generally a ratio of two polynomials:

Finding the inverse Laplace transform of F(s) involves two

steps:

1. Decompose F(s) into simple terms using partial fraction

expansion.

2. Find the inverse of each term by matching entries in

Laplace Transform Table.

j

j

tsdsesF

jtfsFL

)(2

1)()]([1

Partial Fraction Expansions

32)3()2(

1

s

B

s

A

ss

s• Expand into a term for each

factor in the denominator.

• Recombine RHS

• Equate terms in s and

constant terms. Solve.

• Each term is in a form so

that inverse Laplace

transforms can be applied.

)3()2(

2)3(

)3()2(

1

ss

sBsA

ss

s

3

2

2

1

)3()2(

1

ssss

s

1BA 123 BA

The Inverse Laplace Transform

4

6

1

53)(

2

ssssF

0 t),())2sin(353(

4

6

1

53)(

2

111

tute

sL

sL

sLtf

t

33

Example:

Find the inverse Laplace transform of

Solution:

)1)(4(

18892)(

2

23

sss

ssssF

Example of Solution of an ODE

0)0(')0(2862

2

yyydt

dy

dt

yd • ODE w/initial conditions

• Apply Laplace transform

to each term

• Solve for Y(s)

• Apply partial fraction

expansion

• Apply inverse Laplace

transform to each term

ssYsYssYs /2)(8)(6)(2

)4()2(

2)(

ssssY

)4(4

1

)2(2

1

4

1)(

ssssY

424

1)(

42 tt eety

APPLICATION LAPLACE

TRANSFORM TO CIRCUIT

ANALYSIS

40

Example:

Find v0(t) in the circuit shown below, assuming zero

initial conditions.

41

Solution:

Transform the circuit from the time domain to the s-

domain, we have

s

L

tu

3

sC

1 F

3

1

ss H 1

s

1 )(

42

Solution:

Apply mesh analysis, on solving for V0(s)

Taking the inverse

transform give

220)2()4(

2

2

3)(V

ss

0 V, )2sin(2

3)( 4

0 ttetv t

43

Example:

Determine v0(t) in the circuit shown below, assuming zero

initial conditions.

V )()21(8 :Ans 22 tutee tt

44

Example:

Find v0(t) in the circuit shown below. Assume v0(0)=5V .

V )()1510()( v:Ans 2

0 tueet tt

45

Example:

The switch shown below has been in position b for a long

time. It is moved to position a at t=0. Determine v(t) for t

> 0.

where0, t ,I)IV()( v:Ans 0

/

00 RCReRt t

46

Example:

Consider the circuit below. Find

the value of the voltage across the

capacitor assuming that the value

of vs(t)=10u(t) V and assume that

at t=0, -1A flows through the

inductor and +5 is across the

capacitor.

47

Solution:

Transform the circuit from time-domain (a) into s-

domain (b) using Laplace Transform. On rearranging the

terms, we have

By taking the inverse transform, we get

2

30

1

35V1

ss

V )()3035()(v 2

1 tueet tt

48

Example:

The initial energy in the circuit below is zero at t=0. Assume that vs=5u(t)

V. (a) Find V0(s) using the thevenin theorem. (b) Apply the initial- and

final-value theorem to find v0(0) and v0(∞). (c) Obtain v0(t).

Ans: (a) V0(s) = 4(s+0.25)/(s(s+0.3)) (b) 4,3.33V, (c) (3.33+0.667e-0.3t)u(t) V.

49

Example:

The output of a linear system is y(t)=10e-tcos4t when the input is equal to

x(t)=e-tu(t). Find the transfer function of the system and its impulse

response.

Solution:

Transform y(t) and x(t) into s-domain and apply H(s)=Y(s)/X(s), we get

Apply inverse transform for H(s), we get

16)1(

44010

16)1(

)1(10

)(

)()(

22

2

ss

s

sX

sYsH

)()4sin(40)(10)( tutetth t

50

Example :

The transfer function of a linear system is

Find the output y(t) due to the input e-3tu(t) and its impulse

response.

6

2)(

s

ssH

)(12e-(t)2 0; t,42 :Ans -6t63 tuee tt

Documents

Laplace Transform with Applications