L10 - Chapter 3

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CCB3013 - Chemical Process Dynamics, Instrumentation

and Control 16/16/2014

Chapter 3

Laplace Transform

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Chapter Objectives

End of this chapter, you should be able to:

• Use Laplace Transform of representative functions

• Solve differential equations by Laplace Transform

techniques

• Explain other Laplace Transform properties

– Final value theorem

– Initial value theorem

– Time delay (translation in time)

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3.1 Laplace Transform of Representative

Functions

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

0

)()]([)( dtetf =tfLsF -st

(3.1)

where F(s) is the symbol for the Laplace transform

s is a complex independent variable

f(t) is some function of time

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Transforms of some important functions

Constant function: f(t) = a

s

a

s

ae

s

adtae=aL st-st

0][0

0

(3.2)

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Step function:

The unit step function is defined as

0 1

0 0)(

t

ttS (3.3)

The Laplace transform of the unit step function is

obtained with a = 1

stSL

1)]([ (3.4)

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Exponential Functions

bteb > 0

bs

esb

dtedtee=eL tsbtsb-stbtbt

11

][0

)(

0

)(

0

(3.5)

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Derivatives

)0()()0(][

)()(

][

00

0

fssFffsL

etfsdtetf

dtedtdf=dtdfL

stst

-st

(3.6)

Usually define f(0) = 0

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Second-order derivatives

dt

dL

dt

fdL

2

2

wheredt

df

)0()0()(

)0()0()(

)0()(

2

2

2

fsfsFs

ffssFs

ss

dt

dL

dt

fdL

(3.7)

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Higher-order derivatives

)0()0(

...)0()0()(

)1()2(

21

nn

nnn

n

n

fsf

fsfssFsdt

fdL

(3.8)

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Trigonometric functions

Euler identity:2

costjtj ee

t

cos sin

cos sin

1

j t

jwt

e t j t

e t j t

j

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Cosine function

22

222

1

11

2

1

2cos

s

s

s

jsjs

jsjs

eeLtL

tjtj

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Sine function

22

222

1

11

2

1

2sin

s

s

jsjs

j

jsjsj

j

eeLtL

tjtj

(3.9)

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Rectangular pulse function

0

1 1( ) (1 )

hst hsF s e dt e

h hs

ht

htht

tf

0

0 /10 0

)( (3.10)

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Rectangular pulse function

If h = 1, unit rectangular pulse input.

Difference of two step inputs S(t) – S(t-1). (S(t-1) is a

step starting at t = h = 1)

By Laplace transform

1( )

seF s

s s

Can be generalized to steps of different magnitudes (a1, a2).

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Impulse function

Let h→0, f(t) = δ(t) (Dirac delta) L(δ) = 1

Laplace transforms can be used in process control for:

• Solution of differential equations (linear)

• Analysis of linear control systems (frequency

response

• Prediction of transient response for different inputs

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General procedure for solving differential

equations

repeated factors

and complex

factors may arise.

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Example 3.1 Solve the ODE,

5 4 2 0 1 (3-26)dy

y ydt

(3.11)

First, take L of both sides of (3-11),

2

5 1 4sY s Y ss

Rearrange,

5 2

(3-34)5 4

sY s

s s

(3.12)

Take L-1,

1 5 2

5 4

sy t

s s

L

From L Table

0.80.5 0.5 (3-37)ty t e (3.13)

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Example 3.2

00

0000

2461162

2

3

3

at t=dt

du

)=(y)=(y)=y(

udt

duy

dt

dy

dt

yd

dt

yd

To find transient response for u(t) = unit step at t > 0 :

1. Take Laplace Transform (L.T.)

2. Factor, use partial fraction decomposition

3. Take inverse L.T.

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Step 1 Take L.T. (note zero initial conditions)

3 26 11 6 ( ) 4 2s Y(s)+ s Y(s)+ sY(s) Y s = sU(s) + U(s)

Rearranging,

input) (1

1

6116

2423

stepunits

U(s)

ssss

s+Y(s)=

Step 2a. Factor denominator of Y(s)

))(s+)(s+)=s(s+s++s+s(s 3216116 23

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Step 2b. Use partial fraction decomposition

321321

24 4321

s

α

s

α

s

α

s

α

))(s+)(s+s(s+

s+

Multiply by s, set s = 0

3

1

321

2

321321

24

1

0

4321

0

α

s

α

s

α

s

αsα

))(s+)(s+(s+

s+

ss

For a2, multiply by (s+1), set s = -1 (same procedure for a3, a4)

3

531 432 , α, αα

3

3/5

2

3

1

1

3

1

ssssY(s)=

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Step 3. Take inverse of L.T.

3

1

3

53

3

1 32

) y(tt

eeey(t)= ttt

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Dynamic Analysis

2

2

3 4 1s s

Example 1:2 16

1.333 14 12

b

a

2 13 4 1 (3 1)( 1) 3( )( 1)3

s s s s s s

Transforms to e-t/3, e-t (real roots) - (no oscillation)

2

2

1

s

s s

Example 2:

2 11

4 4

b

a

2 3 31 ( 0.5 )( 0.5 )

2 2s s s j s j

Transforms to 0.5 0.53 3

sin , cos2 2

t te t e t

(oscillation)

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Remarks

• You can use this method on any order of ODE, limited only

by factoring of denominator polynomial (characteristic

equation)

• Must use modified procedure for repeated roots,

imaginary roots

One other useful feature of the Laplace transform is that one

can analyze the denominator of the transform to determine its

dynamic behavior. For example, if

23

12 ss

Y(s)=

the denominator can be factored into (s+2)(s+1).

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Remarks

Using the partial fraction technique

12

21

s

α

s

αY(s)=

The step response of the process will have exponential

terms e-2t and e-t, which indicates y(t) approaches zero.

))(s(sssY(s)=

21

1

2

12

However, if

• We know that the system is unstable and has a transient

response involving e2t and e-t. e2t is unbounded for large

time.

• We shall use this concept later in the analysis of

feedback system stability.

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Other properties of L( ):

A. Final value theorem

sY(s))=y(s 0lim

“offset”

Example 3: Step response

aτs

a

τs

asY(s)

s

a

τsY(s)

s

1lim

1

1

1

0

offset (steady state error) is a.

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Other properties of L( ):

Time-shift theorem

y(t)=0 t < Y(s)=et-yL s-

Initial value theorem

sY(s)y(t)=st limlim

0

Example 4.

)3)(2)(1(

24)(

s+s+s+s

s+=sFor Y

by initial value theorem: 0)0( y

by final value theorem:3

1)( y

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Conclusion!

• Laplace transform definition

• Laplace transform of some typical functions

• Inverse Laplace transforms

• Examples

• Dynamic Analysis

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L10 - Chapter 3

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