Pelajari tentang Multivariable Control Systems

Multivariable Control Systems

Ali Karimpour Professor Ferdowsi University of Mashhad

Lecture 3

References are appeared in the last slide.

2

Topics to be covered include:

v

Multivariable Connections

v

Multivariable System Representation

v Polynomial Matrix Description & Rosenbrock’s System Matrix

v General Control Problem Formulation

v Matrix Fraction Description (MFD)

v

Multivariable Connections

3

• Cascade (series) interconnection of transfer matrices

• Parallel interconnection of transfer matrices

)

1(s

G G₂(s) )

(s

U Y(s)

)

1(s G

)

2(s G )

(s U

+ +

) (s Y

) ( ) ( )

( ) ( ) ( )

(s G₂ s G₁ s U s G s U s

Y = =

) ( ) ( )

( )) ( )

( ( )

(s G₁ s G₂ s U s G s U s

Y = + =

) ( ) ( )

(s G₂ s G₁ s

G =  G₁(s)G₂(s)

Generally

) ( )

( )

(s G₁ s G₂ s

G = +

4

• Feedback interconnection of transfer matrices

)

1(s

G G₂(s)

) (s ) Y

(s

U ₊

− ^{Y(s) = G}₂^(s)G₁^(s)

(

^U(s)−Y(s)

)

) ( ) ( )

( ) ( ) ( ))

( ) ( (

)

(s I G₂ s G₁ s ¹G₂ s G₁ s U s G s U s

Y = + ⁻ =

) ( ) ( ))

( ) ( (

)

(s I G₂ s G₁ s ¹G₂ s G₁ s

G = + ⁻

A useful relation in multivariable is push-through rule

1 2

1 2

2 1 1

2( ) ( )) ( ) ( )( ( ) ( ))

(I +G s G s ⁻ G s = G s I +G s G s ⁻ Exercise 3-1: Proof the push-through rule

5

MIMO rule: To derive the output of a MIMO system,

Start from the output and write down the blocks as you meet them when moving backward (against the signal flow) towards the input.

(

^{I − L}

)

⁻¹

(

^{I + L}

)

⁻¹

If you exit from a feedback loop then include a term

or according to the feedback sign where L is the transfer function around that loop (evaluated against the signal flow starting at the point of exit from the loop).

Parallel branches should be treated independently and their

contributions added together.

6

Example 3-1: Derive the transfer function of the system shown in figure

(

21

)

^

1 22

12

11 P K(I P K) P P

z = + − ⁻

11 P z =

21

1 22

12K(I P K) P P

z = − ⁻

7

Topics to be covered include:

v

Multivariable Connections

v

Multivariable System Representation

v Polynomial Matrix Description & Rosenbrock’s System Matrix

v General Control Problem Formulation

v Matrix Fraction Description (MFD)

8

General form of a polynomial matrix description

System Variables System Inputs

System Outputs

Rosenbrock’s system matrix

9

Example 3-2: A position control system.

  _



 



= 



 



= 



 





















+

− +

) (

) 0 (

1 ) (

) 1 (

0 )

( ) (

2 2

t i t t

y

t t e

i t dt R

L d dt

K d

dt K f d dt

J d

a

a a

a a

b





)

(t

) (t u



 



 dt P d

10

  _



 



= 



 



= 



 





















+

− +

) (

) 0 (

1 ) (

) 1 (

0 )

( ) (

2 2

t i t t

y

t t e

i t dt R

L d dt

K d

dt K f d dt

J d

a

a a

a a

b





)

(t

) (t u

Polynomial matrix description

















−

=

















−



















−

+

− +

) ( 0 0

) (

) (

) (

0 0

1

1

2 0

s Y s

E s I

s R

s L s

K

K fs

Js

a a a

a b

Rosenbrock’s system matrix

11

Rosenbrock’s system matrix

( ( ) ( ) ( ) ( ) ) ( )

)

( s R s P

¹

s Q s W s U s

Y =

⁻

+

) ( )

( ) ( )

( )

( s R s P

¹

s Q s W s

G =

⁻

+

Suppose P is nonsingular.

12

System order is the number of independent initial condition that is necessary to describe the system.

System order in Rosenbrock’s system matrix is equal to degree of det(P(s)).

















−

=

















−



















−

+

− +

) ( 0 0

) (

) (

) (

0 0

1

1

2 0

s Y s

E s I

s R

s L s

K

K fs

Js

a a a

a b

For previous example:

System order is 3 and P(s) is 2×2

s KK R

s L fs Js

s

P( ) = ( ² + )( _a + _a)+ _b

13

Remark2: G(s) is strictly proper if D=0 otherwise it is proper.



 





= −



 





 −



 





−

−

) ( 0 )

( ) (

s Y s

U s X D

C

B A

sI Du

Cx y

Bu Ax

x

+

=

+

=



D B

A sI

C s

G( ) = ( − )⁻¹ +

Remark1: (sI-A) is n×n and also system order is n. But generally dimension of P(s) in Rosenbrok’s system matrix is not the same as system order.(See previous example)

14

Example(Two important remarks)

) ( ) 2

( ) ( )

(

) ( )

( ) 1

( ^{2 3}

s U s s

s Y

s U s s

s

− +

=

= +





2 2

3 1

) 1 (

2 2 3

) 1 ) (

( )

( ) ( )

( )

( +

= +

− + +

= +

= ⁻

s s s

s s s

W s

Q s P

s R s

G

Remark1: G(s) is strictly proper but W(s)=2-s !



 





= −



 





 −



 





−

− +

) ( 0 )

( ) ( 2

1 ) 1

( ^{2 3}

s Y s

U s s

s

s 

 





 





−

 =







 





 −

 





 







−

− +

) ( 0 0 )

( ) (

) ( 2

1 0

) 1 (

0

0 0

1

3 2

s Y s

U s s s

s

s 



Remark 2: Another form of Resenbrock’s system matrix.

15

Topics to be covered include:

v

Multivariable Connections

v

Multivariable System Representation

v Polynomial Matrix Description & Rosenbrock’s System Matrix

v General Control Problem Formulation

v Matrix Fraction Description (MFD)

16

𝑤 exogenous inputs: Inputs that are not used to control the system. i.e. references, disturbances, noises

𝑧 exogenous outputs: Outputs that we want to control them and push them to zero.

𝑢 control signals: Signals that produced by controller to control the system.

𝑣 sensed outputs: Signals that used by controller.

Problem description: Derive K(s) such that closed loop system be stable and 𝑧 be as small as possible for bounded 𝑤.

17

closed loop system be stable and 𝑧 be as small as possible for bounded 𝑤.



 







 



= 



 





u w P

P

P z P

22 21

12 11



 K u =

(

^{P P K I P K P}

)

^{w Nw}

z = ₁₁ + ₁₂ ( − ₂₂ )⁻¹ ₂₁ =

Problem description: Make N stable and as small as possible.

) , (P K F

N = _l

18

𝑤 = 𝑟 𝑑 𝑛

𝑧 = 𝑟 − 𝑦

𝑣 = 𝑟 − 𝑦 − 𝑛 𝑢

𝐼 − 𝐺_𝑑 𝑠 0 − 𝐺(𝑠)

control problem formulation. ^𝑟 _{𝐾(𝑠) 𝐺(𝑠)} ^𝑦

𝑛

+ + +

+ +

−

𝑢

𝐼 − 𝐺_𝑑 𝑠 − 𝐼 − 𝐺(𝑠) ^{z =}

(

^{P + P K I −P K − P}21

)

^{w = Nw}

1 22

12

11 ( )

𝑧 = 𝐼 − 𝐺_𝑑 𝑠 0 − 𝐺 𝑠 𝐾(𝑠) 𝐼 + 𝐺 𝑠 𝐾(𝑠) ⁻¹ 𝐼 − 𝐺_𝑑 𝑠 − 𝐼

𝑟 𝑑 𝑛

𝑧 = 𝐼 + 𝐺 𝑠 𝐾(𝑠) ⁻¹ −𝐺_𝑑(𝑠) 𝐼 + 𝐺 𝑠 𝐾(𝑠) ⁻¹ 𝐺 𝑠 𝐾(𝑠) 𝐼 + 𝐺 𝑠 𝐾(𝑠) ⁻¹ 𝑟 𝑑 𝑛

19

Topics to be covered include:

v

Multivariable Connections

v

Multivariable System Representation

v Polynomial Matrix Description & Rosenbrock’s System Matrix

v General Control Problem Formulation

v Matrix Fraction Description (MFD)

20

function transfer

SISO

a is 6 ) 5

(

Let

₂

+

= +

s s s

g

( 1 ) ( 2 5 6 ) 1

) ( can write

We g s = s + s + s +

⁻

polynomial polynomial

This is a Right Matrix Fraction Description (RMFD)

( 5 6 ) ( 1 )

) ( write also

can

We g s = s

²

+ s +

⁻¹

s +

polynomial polynomial

This is a Left Matrix Fraction Description (LMFD)

Remark: Are LMFD and RMFD the same for any system?

21

function transfer

SISO an

is 6 ) 5

(

Let

₂

+

= +

s s s

g

( 1 ) ( 2 5 6 ) 1

) ( can write

We g s = s + s + s +

⁻

polynomial polynomial

This is a Right Matrix Fraction Description (RMFD)

polynomial polynomial

This is a also a Right Matrix Fraction Description (RMFD)

Remark: Uniqueness of MFD?

( ( 1 )( ) ) ( ( )(

²

5 6 ) ) 1

) ( write also

can

We g s = s + s + a s + a s + s +

⁻

22

Matrix Fraction Description for Transfer Function Matrix

polynomial matrix

polynomial matrix

Left Matrix Fraction Description (LMFD)

) ) (

( ) 1

(

G N s

s

s = d

G is a pq matrix

( ( ) ) ( ) ( ) ( )

) (

G s = d s I

_p ⁻¹

N s = D

_L⁻¹

s N

_L

s

polynomial matrix

polynomial matrix

( ( ) ) ( ) ( )

) ( )

(

G s = N s d s I

_q ⁻¹

= N

_R

s D

_R⁻¹

s

Right Matrix Fraction Description (RMFD)

23

Matrix Fraction Description for Transfer Matrix

) ) (

( ) 1

(

G N s

s

s = d

Suppose G is a pq matrix so

Left Matrix Fraction Description (LMFD)

( ( ) ) ( ) ( ) ( )

) (

G s = d s I

_p ⁻¹

N s = D

_L⁻¹

s N

_L

s

( ( ) ) ( ) ( )

) ( )

(

G s = N s d s I

_q ⁻¹

= N

_R

s D

_R⁻¹

s

Right Matrix Fraction Description (RMFD)

Degree of denominator matrix is defined as: deg D_L(s) = deg det D_L(s) = rp

Degree of denominator matrix is defined as: deg D_R(s) = deg det D_R (s) = rq

24

We can show that the MFD is not unique, because, for any nonsingular m

m matrix we can write G(s) as:(s)

(

^{( ) ( ) 1}

)

^{1( )}

(

^{( ) ( )}

)(

^{( ) ( )}

)

¹

) ( )

(s = N s  s  s ⁻ D⁻ s = N s  s D s  s ⁻

G _{R R R R}

is said to be a right common factor.

)

(s

When the only right common factors of and is unimodular matrix, then, we say that the RMFD is irreducible.

) (s

N_R D_R(s)

(N_R(s),D_R(s))

25

If D_L(s) and N_L(s) are not irreducible, find irreducible one.

Checking irreducibility(left common factor):

Form:

Do preliminary transformation(on columns) to make right part zero:

1) Add -C₂ on C_{4 }



 





− +

−

0 1 1

1

0 1 1

1

s 2) Add -C₁ on C₃

3) Add C₁ on C₂ 4) Add 0.5(s+2)C₃ on C₂ 5) Change C₃ and C₂



 





= −

2 1

0 ) 1

( : is factor

Common Q s



 





− +

−

0 2 1

1

0 0 1

1 s



 





−

+2 2 0 1

0 0 0

1

s 



 





−2 0 0

1

0 0 0 1



 





−2 0 0 1

0 0 0 1

Since Q(s) is unimodular so D_L(s) and N_L(s) are irreducible.

26

If D_L(s) and N_L(s) are not

irreducible, find irreducible one.

Checking irreducibility(left common factor):

Form:

Do preliminary transformation(on columns) to make right part zero:



 + +

 +

 

 − +

+

= −

= ⁻

2 2

2 ) 2

( ) ( )

( ¹

s s

s s s

s s

s s s s

N s D s

G _{L L}

  _



 





+

− +

+

+

−

= +

2 2

2 ) 2

( )

(

2 2

s s

s s

s s s s

s s s

D s

N_{L L}

1) Add -C₂ on C₄



 





− +

+

− +

0 2 2

2

2 0

s s

s

s s

s

s 2) Add C₁ on C₃



 





+ +

+

0 2

2 2

0

2 0

s s

s

s s s

3) Add –(s+1)C₁ on C_{2 }



 





+

−

+ 2 ( 2) 2 0 0 0 0

s s

s s

s

4) Add 0.5(s+2)C₃ on C₂



 





+2 0 2 0 0 0 0

s s

s

Dr. Ali Karimpour Feb 2022

27

If D_L(s) and N_L(s) are not

irreducible, find irreducible one.

Checking irreducibility (left common factor):



 + +

 +

 

 − +

+

= −

= ⁻

2 2

2 ) 2

( ) ( )

( ¹

s s

s s s

s s

s s s s

N s D s

G _{L L}

  _



 





+

− +

+

+

−

= +

2 2

2 ) 2

( )

(

2 2

s s

s s

s s s s

s s s

D s

N_{L L}

1) Add -C₂ on C₄ 2) Add C₁ on C₃ 3) Add –(s+1)C₁ on C₂ 4) Add 0.5(s+2)C₃ on C₂



 





+2 0 2 0 0 0 0

s s

s

5) Change C₃ and C₂



 





+2 2 0 0 0 0 0

s s

s 



 





= +

s s

s s

Q 2 2

) 0 ( : is factor Common

Why this procedure leads to ….?

28

Let:

By suitable preliminary transformation :

 N

_L

( s ) D

_L

( s )  →  Q ( s ) 0 

q p is G s

N s D s

G( ) = _L⁻¹( ) _L( ) 

We must show that Q(s) is the greatest left common devisor.



^NL^{(s) D}L^(s)



^U(s) =



^{Q(s) 0}





 N

_L

( s ) D

_L

( s )   = Q ( s ) 0  U

⁻¹

( s )



Since U is unimodular so its inverse is also unimodular thus



 



= 

− =

) ( )

(

) ( )

) ( ( )

(

22 21

12 1 11

s V s

V

s V s

s V V s

U

So we have

) ( )

( )

( )

( )

( )

( s Q s V

₁₁

s D s Q s V

₁₂

s

N

_L

=

_L

=

29

) ( )

( )

( )

( )

( )

( s Q s V

₁₁

s D s Q s V

₁₂

s

N

_L

=

_L

=

Clearly Q(s) is a left common divisor but why greatest common divisor?

) ( )

( )

( )

( )

(s U₁₁ s D s U₂₁ s Q s

N_L + _L =

Let W(s) is another left common divisor so:

) ( )

( )

( )

( )

( )

( )

( s N s U

₁₁

s W s D s U

₂₁

s Q s

W

_L

+

_L

=

So W(s) is a left common divisor of Q(s), → Q(s) is gcd.



^N_L^{(s) D}_L^(s)



^{U(s) =}



^{Q(s) 0}



^

Now let: _



 



= 

) ( )

(

) ( )

) ( (

22 21

12 11

s U s

U

s U s

s U U

30

Theorem 1:

^D_L^{(s) and N}_L(s) are left coprime (irreducible) if and only if there exist two polynomial matrix X_L(s) and Y_L(s) such that following

equation satisfied.

I s

Y s D s

X s

N

_L

( )

_L

( ) +

_L

( )

_L

( ) =

This equation is called simple Bezout identity.

Example 3-6: Let N_L(s)=s+2 and D_L(s)=s²+5s+6, are they left coprime?

One cannot derive X(s) and Y(s) s.t. N_L(s)X(s)+D_L(s)Y(s)=1 Example 3-7: Let N_L(s)=s+1 and D_L(s)=s²+5s+6, are they left coprime?

Let X(s)=-s-2 and Y(s)=0.5 s.t. N (s)X(s)+D (s)Y(s)=1

Example 3-8: Let N(s)=2s and D_L(s)=2s²+10s+2, are they left coprime?

31

Are they coprime?

 N

_L

( s ) D

_L

( s )  →  Q ( s ) 0 

q p is G s

N s D

s

G( ) = _L⁻¹( ) _L( ) 

Bezout identity:

q p p

p 





I s

Y s D s

X s

N

_L

( )

_L

( ) +

_L

( )

_L

( ) =

 N

_L

( s ) D

_L

( s )  U ( s ) =  Q ( s ) 0 

Are they coprime?



 



→ 



 





0 ) ( )

( )

( Q s

s D

s N

R R

q p is G s

D s N s

G( ) = _R( ) _R⁻¹( ) 

Bezout identity:

q q q

p 





I s

D s Y s

N s

X

_R

( )

_R

( ) +

_R

( )

_R

( ) =



 



=



 





0 ) ( )

( ) ) (

( Q s

s D

s s N

V

R R



^N_L^{(s) D}_L^(s)



^U_l^(S)=



^{I 0}





 



= 



 





0 )

( ) ) (

( I

s D

s s N

V

R R r

If Q(s) is unimodular

If Q(s) is unimodular

32

Now let P be a proper real-rational matrix. A right-coprime factorization (rcf) of P is a factorization of the form

Similarly, a left-coprime factorization (lcf) of P has the form N

M

P ~ ₋₁ ~

=

where N and M are right-coprime in the set of stable transfer matrices.

−1

= NM P

Coprime Factorizations over Stable Transfer Functions

2 2

1 1

2 3 3

1 2 2

2 2

2 2 1

3 2 3

2 3 1





 

 



 





=

=

+

=

+

=

+

= +

+

y y

dt u d

dt u d

dt d dt

d dt

d

33

Exercise 3-2: Derive Rosenbrock’s system matrix for following system. What is the order of system?

1 1 2

3 1 2 3 1



 

 



=

+

−

=

−

= +

y dt u d

dt d dt

d

Exercise 3-3: Derive Rosenbrock’s system matrix for following system. What is the order of system?

34

following system.

b) Check the irreducibility of derived MFD in part “a”

c) Derive an irreducible MFD for the system. _^















+ +

−+ +

=

2 2 2

) 2 (

2

) 2 (

) 1 ) (

(

s s s

s

s s s

G

Exercise 3-6: Derive an irreducible RMFD for following system.















 +

=

2 1 2 0 )

(

s s s s

G

Exercise 3-7: Derive general control problem formulation for following system and derive N.(we need y track r)

𝐺(𝑠) 𝑟 𝐾(𝑠)

𝑑 𝑦

𝑛

+ + +

+ +

−

ℎ(𝑠) 𝑢

35

Web References

• http://karimpor.profcms.um.ac.ir/index.php/courses/9319

• Multivariable Feedback Control, S. Skogestad, I. Postlethwaite, Wiley,2005.

• Multivariable Feedback Design, J M Maciejowski, Wesley,1989.

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