Process Control

CHAPTER III

LAPLACE TRANSFORM

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

In the application of Laplace transform variable time is eliminated and a new domain is introduced.

In the modeling of dynamic systems differential equations are solved by using Laplace transform.

dtetfsftfL st

0

)()()(

Properties of Laplace Transform

1. The Laplace transform contains no information about f(t) for t<0. (Since t represents time this is not a limitation)

2. Laplace transform is defined with an improper integral. Therefore the required conditions are

a. the function f(t) should be piecewise continuous

b. the integral should have a finite value; i.e., the function f(t) does not increase with time faster than e -st decreases with time.

3. Laplace transform operator transforms a function of variable t to a function of variable s. e.g., T(t) becomes T(s)

4. The Laplace transform is a linear operator.

5. Tables for Laplace transforms are available. In those tables inverse transforms are also given.

)()()()( 2121 tfbLtfaLtbftafL

)()(1 tfsfL

Transforms of Some Functions

1. The constant function

s

aaL

sL

ss

e

t

edtetfL

tf

t

t

stst

11

10)1()(

1)(0

00

2. The Step function

0

0

)()(

0,

0,0)(

1)1()(

0,1

0,0)(

s

adteatfL

ta

ttf

sdtetfL

t

ttf

st

st

3. The exponential function

as

AtfLAetfif

ase

as

dtedteetfL

te

ttf

at

t

t

tas

tasstat

at

)(,)(,

11

)(

0,

0,0)(

0

)(

0

)(

0

4. The Ramp function

svdtedv

dtdutu

duvvuudv

dttetfL

tt

ttf

stst

st

e,

,

,..

)(

0,

0,0)(

0

2

0

2

2

0

0

0

1)(

1

0

1

1

)(

sdttetfL

s

t

te

s

dtess

et

dttetfL

st

st

sttt

st

st

Laplace Transforms of Derivatives

)0()(

)().()(.0

))((0

)(

)(,

,eu

)(

0

00

st-

0

Fssf

sfsofe

dtestft

ttfedte

dt

df

tfvdtdt

dvdv

dtsedu

dtedt

dftf

dt

dL

ststst

st

st

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

)0(')0()(

)0()(

233

3

22

2

fsffssfsdt

fdL

fsfsfsdt

fdL

fssfdt

dfL

Solution of Differential equations by Laplace Transform

For the solution of linear, ordinary differential equations with constant coefficients Laplace Transforms are applied.

The procedure involves:1. Take the Laplace Transform of both sides of the

equation.2. Solve the resulting equation for the Laplace

transform of the unknown function. i.e., evaluate x(s).

3. Find the function x(t), which has the Laplace Transform obtained in step 2.

Example:

Solve the following equation.

Solution:

2)0(,,03 xwherexdt

dx

tetx

sssx

ssx

sxxssx

32)(

3

1.2

3

2)(

2)3)((

0)(3)0()(

Inversion by Partial FractionsExample Solve the following equation.

Solution:

Partial fractions is to be applied;

0)0(,,1 xwherexdt

dx

)1(

1)(

1)1)((

1)()0()(

sssx

sssx

ssxxssx

1,010

1

0,11

1

1..

1)1(

1)(

AA

sfors

BsA

s

s

Bs

s

As

s

B

s

A

sssx

1)s(s

1s.

s with sides both multiply

I METHOD

1,0

1,

).1(

BB

sfor

s

1-

1

1s

B

s

1)A(s

1)s(s

11).(s

1),(s with sides both multiply

1

0

1

1

1)s(s

Bs

1)s(s

1)A(s

1s

A

)1(

1

II METHOD

B

BA

A

BsAAs

s

B

ss

1)1(

1)(

s

B

s

A

sssx

tetx

sssx

1)(

1

11)(

(by using Laplace Transform table)

Example Solve the following equation.

2

14

)(2

)0()(

)0()0()(2

)0()0()0()(

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

422

2

23

22

2

3

3

ss

sx

xssx

xsxsxs

xxsxssxs

xxxwhere

exdt

dx

dt

xd

dt

xd t

Apply Laplace transform to both sides of the equation.

1212)(

)1)(2)(1)(2(

896)(

)22)(2(

896)(

)2(

89622

2

1422)(

24

23

24

24223

s

E

s

D

s

C

s

B

s

Asx

sssss

ssssx

sssss

ssssx

ss

sssss

ssssssx

to determine multiply by Set s to results

A S 0 -2

B S-2 2 1/12

C S+1 -1 11/3

D S+2 -2 -17/2

E S-1 1 2/3

particularshomogeneou

22 )3/2()2/17()3/11()12/1(2)(

1

1.

3

2

2

1.

2

17

1

1.

3

11

2

1.

12

12)(

SSSolution

eeeetx

ssssssx

tttt

tttt

t

ttt

t

t

t

t

BeAecececSolution

BeAS

ecececS

er

er

er

rrr

rrr

exdt

dx

dt

xd

dt

xd

23

221

2particular

32

21shomogeneou

2

23

rt

22

2

3

3

,1

,2

,1

0)1)(1)(2(

022

e xsolution, shomogeneoufor

422

Example

Solve the following equation;

)45(

25)(

52

)45)((

2)(4)0()(5

245

1)0(,245

ss

ssY

sssY

ssYYssY

ydt

dyL

yydt

dy

t

t

tbtb

ety

ety

bbb

bsbs

bse

bb

bbe

bb

bbline

8.0

8.0

312

21

3

21

23

12

13

5.05.0)(

)1(08.0

04.0

8.00

8.04.0)(

4.0,8.0,0

))((,,11 21

Example; Find the solution of the equation with given laplace

transform.

tt eety

sssY

B

sss

Bs

s

As

ss

s

A

sss

Bs

s

As

ss

s

II

BABA

BA

sBsAs

s

sB

s

sA

ss

s

I

s

B

s

A

ss

s

ss

ssY

4

2

.3

1.

3

4)(

4

1.

3

1

1

1.

3

4)(

3

1

4),4(4

)4(1

)4()4)(1(

53

4

1),1(4

)1(1

)1()4)(1(

5

.

3

1,

3

4

45

1

)1()4(5

4

)1(

1

)4(

)4)(1(

5

.

41)4)(1(

5

45

5)(

Example;

321)6116(

1)(

1

)(6

)0()(11

)0()0()(6

)0()0()0()(

0)0()0()0(,16116

23

2

23

2

2

3

3

s

D

s

C

s

B

s

A

sssssY

s

sY

yssY

ysysYs

yysyssYs

yyyydt

dy

dt

yd

dt

yd

Solve the following equation.

to determine

multiply by Set s to results

A S 0 1/6

B S+1 -1 -1/2

C S+2 -2 1/2

D S+3 -3 1/6

ttt eeety

sssssY

32 .6

1.

2

1.

2

1

6

1)(

3

1.

6

1

2

1.

2

1

1

1.

2

11.

6

1)(

Example (Repeated factors)

Solve the equation for which Laplace transform is given.

Heaviside expansion rule can not be used for A, because the denominator of coefficient B becomes infinity.

s

C

s

B

s

A

sss

ssY

22 )2(2)44(

1)(

4

1,..

)2(.

2.

)2(

1

2

1,)2()2.(

)2()2.(

2)2.(

)2(

1

22

222

222

Css

Cs

s

Bs

s

As

ss

s

Bss

Cs

s

Bs

s

As

ss

s

4

1-A-C,AC.2,A.20

2)C2(sB2)A(2s1

2)C(sBs2)As(s1s

)2(2)44(

1

2

22

s

C

s

B

s

A

sss

s

Equalizing the derivatives gives

Taking derivative w.r.t. s gives

Further differentiation gives

s

sfdttfL

ssYty

ssYty

t

st

st

)()(

)(lim)(lim

)(lim)(lim

0

0

0

Final Value Theorem

Initial Value Theorem

Transform of an integral

Example:

Solve the following equation for x(t)

tt

t

eetx

ssssx

CBA

ssCssBssAs

s

C

s

B

s

A

sss

s

ss

s

s

s

s

ssx

s

s

s

ssx

ssssx

ss

sxxssx

xtdttxdt

dx

1)(

1

1

1

11)(

1,1,1

)1)(()1)(()1)(1(13

11)1)(1(

13

)1(

13

1.

13)(

131)(

13

1)(

1)()0()(

3)0(,)(

2

2

2

2

22

2

2

22

2

2

0

Example:

For a mixing tank, evaluate the change of composition w.r.t. time due to a step change in composition.

w1

x1 w

x

(Overflow system with constant volume, having a component balance)

ttexextx

sx

ss

xsx

s

xxssx

s

xsxxssx

xxdt

dxxx

dt

dx

w

V

wxwxdt

dxV

xVdt

dwxwx

1)0()1()(

1

1)0(

)1()(

)0(1)(

)()0()(

,

)(

1

1

1

1

11

1

1

Example: Considering Ex.2.1 for a sudden change in one of the input flow rates, evaluate

concentration change w.r.t. time.

dt

xVdwxxwxw

dt

Vdwww

)(

)(

2211

21

Total mass balance;

Component balance

w1, x1

w2, x2

w, x

For constant V and ρ ;

tt

c

eectx

ssscsx

xs

cssx

s

csxxssx

xcxdt

dx

xw

xwxw

dt

dx

w

V

wxxwxwdt

dxV

15.0)1(*)(

1

15.0

)1(

1*)(

)0(*

)1)((

*)()0()(

5.0)0(*,

*

2211

2211

τ

Example: Consider a CSTR and evaluate concentration change with respect to time.

t

A

t

AA

AA

A

AAA

AAAA

AA

A

AAAA

AAAA

AAAA

eCeCVkq

qsC

sC

ss

C

Vkq

qsC

s

C

Vkq

qCssC

s

C

Vkq

qsCCssC

s

C

Vkq

qC

dt

dC

Vkq

V

qCCVkqdt

dCV

qCVkCqCdt

dCV

VkCCCqdt

dCV

i

i

i

i

i

i

i

i

1)0()1()(

1

1)0(

)1()(

)0()1)((

)()0()(

)(

)(