contsys1 l3 laplace trans

8/12/2019 ContSys1 L3 Laplace Trans

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Mechanical Engineering Science 8

Dr. Daniil Yurchenko

Laplace Transform


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Block Diagram

+

Y

Y

X

X

Closed-Loop

Open Loop

?

?

?

Here we consider systems described by ODEs

ya xadt

xd a

dt xd

a nn

nn

n

n

011

1

1 .......


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Definition of the Laplace

Transform For the given real f(t)

0)()( dt et f s F st

F(s) Laplace transform of f(t)

Capital letters Laplace transform f(t)

)]([)( t f L s F


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Examples of Laplace Transform

Unit Step Function

f(t)

t0 t,00 t,1)()( t ut f

s se

sdt e s F st st 1)1(011)(

00

st u L /1)]([


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Exponential Function 0,)( t t t f

20

200

00

10

1

1)(

s s

edt e

s s

te

tde s

dt te s F

st st

st

st st

1

!][ n

n

s

nt L


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Exponential Function 0,)( t et f at

a sa se

a sdt ee s F t sa st at

1)

1(0

1)(

0

)(

0

a se L at

1][


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Properties of L - Transform

Theorem 1if k is a constant, then

)()]([ skF t kf L

)()()]()([ sG s F t g t f L

Theorem 2if F(s) and G(s) are L-transform of f(t) and g(t), then


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Theorem 3 (Initial Value of f(t))

)(lim)0()(lim0

s sF f t f st

Theorem 4 (Final Value of f(t))

)(lim)()(lim 0 s sF f t f st


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Theorem 5 (L-transform of differentiation)

)0()()(

f s sF dt

t df L

For higher order derivatives

)0(')0()()(

)0()()( 2

0

2

2

2

f sf s F sdt

t df sf s F s

dt

t f d L

t

)0(.. .)0()0()()( )1()1(21 nnnn

n

n

f f s f s s F sdt

t f d L


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Theorem 6 (L-transform of integration)

s s F

d f L

t )()(

0

)()( a s F t f e L at

Theorem 7

Table of Laplace transform is attached


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ExampleVibration of a Single-Degree-of-Freedom System

b xa x

t y Kx xC x M

)0(,)0(

),(

Y(s) X(s)?

)]([][ t y L Kx xC x M L

)()()]0()([)]0()0()([][][][ 2

sY s KX x s sX C x sx s X s M x KL xCL x ML


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Example

)()()}0()({)}0()0()({ 2 sY s KX x s sX C x sx s X s M

)()(})({})({ 2 sY s KX a s sX C b sa s X s M

)(}{)(}{ 2 sY b sa M Ca s X K Cs Ms

K Cs Ms sY s X

sY s X K Cs Ms

when

2

2

1)()(

)()(}{

0, ba


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Example

K Cs Ms sY s X

2

1)()(

Y(s) X(s)

By definition Transfer Function is the ratioof the L-transform of the output variable tothe L-transform of the input variable (with

zero ICs)

K Cs Ms 21

How to find x(t) ? Use inverse L-transform


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Inverse Laplace Transform

const cdse s F i

t f ic

ic

st -,)(2

1)(

)]([)( t f L s F )]([)( 1 s F Lt f

Instead of using this formula, wewill use Partial-Fraction expansion

or the table of L-transform


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Example: Response to a unit step

K Cs Ms sY

s X 2)(

)(

Y(s) X(s)

K Cs Ms 21

s sY t U t y

1)( ),()(

23

,2,3,122 s s K Cs Ms

K C M Let


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0232 s s

To find the roots of this quadratic equation we equateit to zero

Roots of a quadratic equation

02 cbsasare found as:

a

acbb s

a

acbb s

2

4 ;

2

4 2

2

2

1

In our case a=1,b=3,c=2, thus

22

833 ;1

2

833 22

2

1 s s


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s s s s X

1)1)(2(

1)(

Look in the table of L-transform

t t ee s s s

Lt x s X L 2121

)1)(2(1

)()( 211

The response of the system to a unit step excitation

0)0(,0)0(

),(23

x x

t u x x x t t eet x 212

1)( 2


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Partial-Fraction Expansion

)1)(2(1

)( s s s

s X

What to do if we do not have L-transform tables?

mmmm

nnnn

m

n

b sb sb sba sa sa sa

s B s A

s F 1

110

11

10

...

...)()(

)(

In general, if:

We consider the case n


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Partial-Fraction ExpansionIf F(s) involves distinct real poles

m

m

p sq

p sq

p sq

s F ...)(2

2

1

1

Then

There are two ways of calculating unknown coefficients

mnn

nn

mmmm

nnnn

p s p s p sa sa sa sa


......

...

...

21

11

10

11

10

11

10


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Partial-Fraction Expansion1) Get F(s) to the common denominator

)...(...))...(())...((

)(21

1221

m

mm

p s p s p s p s p sq p s p sq

s F

And compare the numerator to the original one, sothat

Equating the coefficients with the same power we obtain aset of linear equations with respect to q 1,q2, q3,, qm

nnnn

mm

a sa sa sa

p s p sq p s p sq

11

10

1221

...

...))...(())...((


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)1)(2(1

)( s s s

s X If

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

12)1)(2(1

321

321

s s s s sq s sq s sq

sq

sq

sq

s s sThen

12:

023:

0:

10

3211

3212

q s

qqq s

qqq s

1

2/1,2/1

3

2

1

q

qq


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11

22/1

21

)1)(2(1

)( s s s s s s

s X Thus

Then

t t t t eeee s s s

L s X Lt x

2121

21

21

11

22/1

21

)()(

22

11


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Partial-Fraction Expansion2) To multiply the original F(s) ))((lim k p sk p s s F q k

In our case of

1)2(

1lim)1(

)1)(2(

1lim

,21

)1(1

lim)2()1)(2(

1lim

,21

)1)(2(1

lim)1)(2(

1lim

113

222

001

s s s

s s sq

s s s

s s sq

s s s

s s sq

s s

s s

s s

12)1)(2(

1)( 321

s

q

s

q

s

q

s s s s X


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Partial-Fraction Expansion If the second order polynomial has two complexroots, we DO NOT EXPAND it. LEAVE it as is.

1)1(1)( 2

3212 s s

q sq sq

s s s s X

The polynomial in the numerator has to be of the

first order. CLASS! GET UNKNOWN COEFFICIENTS If the denominators roots are multiple, the process offinding the coefficients is slightly more complicated.


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Partial-Fraction Expansion formultiple poles

na s s X

)(1

)(If

nn

n a sq

a sq

a sq

a sq

a s ...

)(1

33

221

Then

For instance

3

2

)1(32

)( s

s s s X


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Then

3:

22:

1:

3210

211

12

qqq s

qq s

q sThus

33

221

3

2

111)1(32

)( s

q s

q s

q s

s s s X

3

322

13

2

1)1(1

)1(32

sq sq sq

s s s

201

3

2

1

qqq


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n

nnn

bas s

p sq

bas s

p sqbas s

p sqbas s 222

222

112 ...)(

1

Then

nbas s s X

)(1

)( 2If

and0

2

bas shas complex-conjugate roots


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Partial-Fraction Expansion Guide

Y X


s B s A

s F

mm

mmnn

nn

m

n

1

1

10

11

10

...

...)()(

)(

1. Equate the denominator to zero and find roots ofthe equation.

2. You will write your denominator as multiplicationof first and second order terms.

3. If the second order polynomial has two complexroots, we DO NOT EXPAND it! LEAVE it as is!

0... 11

10 mmmm b sb sb sb

)...()( 222

21 k c sc sb sac s


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Partial-Fraction Expansion Guide

has

4. Use the Laplace transform table or partialfraction expansion to get inverse Laplace transform

To recover the system response )(t x

042

acb

0222

2 c sb sa

This happens when

contsys1 l3 laplace trans

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