Laplace Transforms

1. Standard notation in dynamics and control (shorthand notation)

2. Converts mathematics to algebraic operations

3. Advantageous for block diagram analysis

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0

)( dtf (t)e=tf -stL

Laplace Transform

Example 1:Example 1:-

00

- - - -( ) ( )

00 0

-

0

( ) 0

1 1( ) -

( ) 0

st st

bt bt st b s t b s t

st

a a aa ae dt e

s s s

e e e dt e dt eb s s b

df dff e dt s (f) f( )

dt dt

L

L

L L L

Usually define f(0) = 0 (e.g., the error)

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?

00

00

0

where

2

2

2

=dt

fd

fsfsF= s

ffssF= s

ssΦ

dt

df φ

dt

dφ=

dt

fd

n

n

L

LL

22

22222

111

2

1

2cos

ωs

s=

ωs

jωs

ωs

jωs=

jωsjωs=

ee=t

tjt-j

LL

Note that

cos sin

cos sin

1

j t

jwt

e t j t

e t j t

j

2 2

sin2

j t j te et

j

s

L L

Unit Step Function:

0 0

1 0

1

tS t

t

S ts

L

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

Note that S(t-1) is the step starting at t = 1.

By Laplace transform

1( )

seF s

s s

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

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0

1 1( ) (1 )

h st hsF s e dt eh hs

Let h→0, f(t) = δ(t) (Dirac delta)

0

0 0

Impulse Function: t

11

1

1

lim

lim lim

hs

h

hs hs

h h

t ehs

e se

hs s

tL

Laplace transforms can be used in process control for:

1. Solution of differential equations (linear)

2. Analysis of linear control systems (frequency response)

3. Prediction of transient response for different inputs

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Please see Table 3.1 in Text

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Example 3.1

Solve the ODE,

5 4 2 0 1 (3-26)dy

y ydt

First, take L of both sides of (3-26),

25 1 4sY s Y s

s

Rearrange,

5 2

(3-34)5 4

sY s

s s

Take L-1,

1 5 2

5 4

sy t

s s

L

From Table 3.1,

0.80.5 0.5 (3-37)ty t e

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Example 2:Example 2:

system at rest (s.s.)

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

00

0000

2461162

2

3

3

at t=dt

du

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

udt

duy

dt

dy

dt

yd

dt

yd

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

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

1. Take Laplace Transform (L.T.)2. Factor, use partial fraction decomposition3. Take inverse L.T.

Rearranging,

Step 2a. Factor denominator of Y(s)

Step 2b. Use partial fraction decomposition

Multiply by s, set s = 0

input) (1

1

6116

2423

stepunits

U(s)

ssss

s+Y(s)=

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

321321

24 4321

s

α

s

α

s

α

s

α

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

s+

3

1

321

2

321321

24

1

0

4321

0

α

s

α

s

α

s

αsα

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

s+

ss

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For 2, multiply by (s+1), set s=-1 (same procedurefor 3, 4)

3

531 432 , α, αα

3

13

53

3

1 32

) y(tt

eeey(t)= ttt

Step 3. Take inverse of L.T.

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

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)3

3/5

2

3

1

1

3

1(

ssssY(s)=

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

the denominator can be factored into (s+2)(s+1).Using the partial fraction technique

The step response of the process will have exponential terms e-2t and e-t, which indicates y(t) approaches zero. 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.

23

12 ss

Y(s)=

1221

s

α

s

αY(s)=

))(s(sssY(s)=

21

1

2

12

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Properties of Laplace Transform

Linearity

where and are constants.

af t bg t

a f t g t

aF s bG s

a b

L

L +bL

Sifting Theorems

0 0

0 0

-

(a) Dead time

( )

(b) Multiplication by

t s t s

bt

bt

g t f t t S t t

g t e f t e F s

e

e f t F s b

L L

L

Final Value Theorem“offset”

Example 3: step responseExample 3: step response

offset (steady state error) is a.

sY(s))=y(s 0lim

aτs

a

τs

asY(s)

s

a

τsY(s)

s

1lim

1

1

1

0

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Initial Value Theorem

by initial value theorem

by final value theorem

sY(s)y(t)=st limlim

0

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

s+=For Y(s)

321

24

00 )=y(

3

1)=y(C

hap

ter

3

Transform of Time Integration

0

t F sf d

s L

Differentiation of F(s)

1n

n nn

d F st f t

ds L

Integration of F(s)

s

f tF d

t

L