Mathematical Models of Control Systems

8/18/2019 Mathematical Models of Control Systems

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EMT 360/3

Control EngineeringMathematical Models ofControl Systems


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System Modelling

• To model a system, we have todetermine dierential eqation thatre!resents the system and how it

change with res!ect to time• "etermine dierential eqations sing

#a!lace o!erator

• "etermine transfer fnction• Standard form of transfer fnction$

% &irst order

% Second order


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Differential equations of physical systems

• #iqid level tan'

• (C Circit

• Car ss!ension


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Example 1: Liquid Level

• &low in ) *ow ot + rate ofaccmlation of liqid in the tan'

khdt

dh AQin

dt

dh AkhQin

dt dh AQout Qin

+=

=−

=−


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Example 2: ! !ircuit

• "ierential eqation that relates ot to in

Ca!acitorcrrent

dt

dVout CRVout Vin

R

dt

dVout C Vout Vin

iRVout Vin

dt

dvC i

+=

=−

=−

=


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Example ": !ar Suspension

• Mass/s!ring/dam!er system#e$ton%s Second La$:

Damping:

Spring:

&pplying #e$ton%s Second La$:

-oth ' and " o!!ose the motion, hence the)ve sign

kxdt

dx D

dt

xd m Fin

Findt

dx Dkxdt

xd m

kx F dt

dx D Dv F

dt

xd M ma F

out

out

++=

+−−=

=

==

==

2

2

2

2

2

2

dt

dva

dt

dxv

=

=


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#a!lace Transform (eview

• . system re!resented y a dierential eqationis diclt to model as a loc' diagram1

• Ths, #a!lace Transform is sed to re!resent thein!t, ot!t, and the system as se!arateentities1

• . transformation from time 2t) domain tocom!le freqency 2s) domain


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#a!lace Transform4ro!erties

• #inearity

• "ierentiation

• &inal vale theorem• 5nitial vale theorem

• (egion of convergence


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Laplace Transform

#a!lace Transform is dened as

where s is a #a!lace o!erator in a form of com!le freqency given y

where σ and jω are the real and imaginary freqency com!onentsres!ectively

[ ]

∫

∞−==

0

st dt et f t f L s F ).()()(

ω σ j s +=


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Soltion$

7tained a #a!lace Transform for a nit ste!

f 2t 8

9

t

am!le 9$

s

1

s

e

s

e 0 s s

=

−

−

−

=

−∞− ..

0,1)( ≥= t t f

[ ]

∞

−

∞−

−=

= ∫

0

0

.1)(

se

dt et f L

st

st


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7tained a #a!lace Transform for the ram! fnction,

Soltion$

Eam!le :$

0t t t f ≥= ,)([ ]

2

00

0

1

1)(

1

)(

s

dt e s s

te s F

s

evdu

edvt u

vduuvudv

dt tet f L

st st

st

st

st

=

+

−=

−==

==−=

=

∫

∫ ∫

∫

∞ −∞−

−

−

∞−


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&ind the #a!lace Transform of

Soltion$

am!le 3$

0t t u Aet f at ≥= − ),()(

[ ]

a s

A

ea s

A

dt e A

tdt e Aedt et f L

0t

t a s

0

a)t -(s

0

st at st

+=

+−=

=

=

∞

=+−

∞+

∞

−−−

∫

∫

)(

|)(


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• 5ntegration is hard, tales areeasier;

am!le


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5nverse #a!lace

• &ind inverse #a!lace transform of

• Soltion$

f(t)u(t)

dse s F j2

1 s F L j

j

st 1

=

= ∫ ∞+

∞−

−

)()]([σ

σ π

a s

1 s F

−=)(

at et f =)(


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7tain the inverse la!lace transform of the following

32

1)( .5

4

3)( .4

2

3)( .3

5

1)( .2

1)( .1

−=

−=

+=

−

=

−=

s s F

s

s F

s s F

s

s F

s s F

1

12)( .7

4

1)( .6

2

2

+

+−=

+

+=

s s

s F

s

s s F


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!haracteristic of Laplace 'ransform

(1) Linear

5f a9 and a: are constant, and &92s8 and &:2s8 are #a!lace Transform

(2) Diferential Theorem

#et

and

[ ] )()()()( 22112211 s F a s F at f at f a L +=+

∫ ∞

−

=

0

)()(dt e

dt

t df

dt

t df L st

∫ ∫ −= duvvudvu st eu −=

( )dt t df dv )(=dt e sdu st −−= .

)(t f v =

)()0(

)()()(

00

s sF f

dt t f seet f dt

t df L st st

+−=

−−=

∫ ∞

−∞−


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Mlti!lication y time

Mlti!lication y tn

7ther !ro!erties of #a!lace Transform

n

nn

ds

sd

nt ! t L

)(

)1()}({ −=

ds

sdF t tf L

)()}({ −=


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• To nd inverse #a!lace transform of a com!licated fnction, wecan convert the fnction into a sm of sim!ler terms sing !artialfraction e!ansion1

• 5f , where the order of N(s) is less than D(s), then a!artial fraction

e!ansion can e made1

Example 1:

• =2s8 mst e divided y "2s8 sccessively ntil the reslt has aremainder whose nmerator is of order less than its denominator1

• Ta'ing the inverse #a!lace transform, we otain

4artial &raction E!ansion

)()()(

s D s " s F =

][)()(

# s s

2 Lt

dt

t d t f

2

11

++

++= −δ δ

# s s$ s% s2 s s F

2

2&

1++

+++=)(

# s s

2

1 s s F 21 ++++=)(


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Example 1a:

7tain the inverse la!lace transform of the following

34

16176)(

2

23

+++++

= s s

s s s s F


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• Three cases of !artial fraction e!ansion$

2a8(oots of the "enominator of &2s8 are (eal and "istinct

28(oots of the "enominator of &2s8 are (eal and (e!eated

2c8 (oots of the "enominator of &2s8 are Com!le and 5maginary

Example 2:

(•) >e write !artial fraction e!ansion as a sm of terms where eachfactor of the original denominator forms the denominator of eachterm, and constants, called resides form the nmerators1

(•) Mlti!ly eqation aove y 2s?98,

4artial &raction E!ansion 2Cont@d8

( ) ( )2 s1 s

2 s F

++

=)(

( ) ( ) 2121

2)(1

++

+=

++=

s

'

s

A

s s s F

( )( )

2

1

2

2)(1 +

++=

+=

s

s ' A

s s F


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• #etting s a!!roach %9, we get .+:1

• &ind - with the similar a!!roach, letting s a!!roach %:, we get -+%:1

• Ths,

(•) Ta'ing inverse #a!lace transform, we get


2 s

2

1 s

2 s F 1 +

−+

=)(

)()()( t u22t f t 2t ee

1

−−

−=


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• Example 3

(•) .dd additional term consisting of denominator factor of redced mlti!licity,

(•) .+: can e otained as !reviosly descried1(•) 5solate - y mlti!lying y 2s?:8:,

2:1:


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• Ta'e inverse #a!lace transform,


( )( ) ( ) ( )2 s2

2 s

2

1 s

2

2 s1 s

2

s F 221 +−+−+=++= )()(

)()( t 21 e2te2e2t f t 2t −−−=

−−


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Example 4

• The fnction can e e!anded in

• Eq 2:1398

Asing the !revios way, can e fond1

Mlti!ly Eq 2:1398 y the lowest common denominator,

-alancing coecients,


( )# s2 s s& s F

21 ++=)(

( ) ( )52523

22 +++

+=++ s sC 's

s

A

s s s

5

3= A

sC s '

++

+=

5

6

5

30

2

5

6 ,

5

3 −=

−= C '


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&ind the inverse #a!lace transforms of the following fnctions14erform !artial%fraction e!ansion on B2s8 rst, then se the#a!lace transform tale1

)2(

)1(2)( .4

)1)(4(

)2(100)( .3

)3()1(

10)(G .2

)3)(2(

1)(G .1

2

2

2

++

+=

+++

=++

=

++=

−

s s s

s s

e s s s

s s

s s

s

s s s s

s


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Eam!le 9Biven the following dierential eqation, solve for (t) if all

initial conditions are ero1 Ase the #a!lace transform1

Laplace transform solution of a differential equation

)(3232122

2

t u dt

d

dt

d =++


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Soltion

Sstitte the corres!onding &2s8 for each term in theeqation

)8)(4(

32)(

)3212(

32)(

32

)()3212(

32)(32)(12)(

2

2

2

++=

++=

=++

=++

s s s s

s s s s

s s s s

s s s s s s

l i 2 d8


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Soltion 2Cont@d8

g !artial%fraction e!ansion

ence

g inverse #a!lace transforms

)()21()(

)8(

1

)4(

21)(

1)4(

32

2)8(

32

1)8)(4(

32

)8()4()8)(4(

32)(

84

8

3

42

01

321

t ueet

s s s s

s s *

s s *

s s *

s

*

s

*

s

*

s s s s

t t

s

s

s

−−

−→

−→

→

+−=

+

+

+

−=

−=

+

=

−=

+

=

=

++

=

+

+

+

+=

++

=


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Eam!le : $ =on%ero initial condition

Biven the following dierential eqation, determine theres!onse of the system , (t)

Ta'ing #a!lace Transform

0)0(,1)0( ==dt

d

)3)(1(

24)(

34

24)(

34

42)(

42

]34)[(

2)(34)(4)(

2)(3)]0()([4)]0(')0()([

;234

2

23

2

2

2

2

2

2

2

2

++

++=

++

++=

++

++=

++=++

=+−+−

=+−+−−

=++

s s s

s s s)

s s s

s s s)

s s

s s s)

s s

s s s)

s s) s s) s s) s

s s) ( s s) ( s) s) s

(dt d(

dt (d


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t t eet (

s

C

s

'

s

A

s s s

s s s)

3

2

6

1

2

1

3

2)(

31)3)(1(

24)(

−− −+=

++

++=

++

++=

Solve the following dierential


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Solve the following dierentialeqations

91

:1

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Mathematical Models of Control Systems

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