Multivariable Control SystemsMultivariable Zeros Element zeros Decoupling zeros (input and output) Transmission zeros or blocking zeros System zeros They are not very important in

Multivariable Control Multivariable Control SystemsSystems

Ali KarimpourAssociate Professor

Ferdowsi University of Mashhad

Lecture 4

References are appeared in the last slide.

Dr. Ali Karimpour Feb 2014

Lecture 4

2

Poles and Zeros in Multivariable Systems

Topics to be covered include:

Multivariable Poles and Zeros

Direction of Poles and Zeros

Smith-McMillan Forms

Matrix Fraction Description (MFD) and Smith-McMillan Forms

Transmission Zero Assignment



Lecture 4

3

Multivariable Poles (Through State Space Description )

Definition 4-1: The poles of a system with state-space description are eigenvalues of the matrix A.

The pole polynomial or characteristic polynomial is defined as niAi ,...,2,1,)( ),,( DandCBA

ip

AsIs )(

Thus the system’s poles are the roots of the characteristic polynomial

0)( AsIs

Note that if A does not correspond to a minimal realization then the poles by this definition will include the poles (eigenvalues) corresponding to uncontrollable and/or unobservable states.


Lecture 4

4

Multivariable Poles (Though Rosenbrock’s System Matrix)

Let

)()()()(

)(PsWsRsQsP

s

Thus the system’s poles are the roots of the following system

0)()( sPs

DuCxyBuAxx

DBCAsI

s)(P

0)( AsIs 0)()( AsIsPs

So they are compatible.


Lecture 4

5

Multivariable Poles (Thrugh Transfer Function Description )

Theorem 4-1: Finding pole polynomials through transfer function

The pole polynomial (s) corresponding to a minimal realization of a system with transfer function G(s) is the least common denominatorof all non-identically-zero minors of all orders of G(s).A minor of a matrix is the determinant of the square matrix obtained bydeleting certain rows and/or columns of the matrix.


Lecture 4

6

Example 4-1

)1)(1()1)(1()2)(1()1(0)2)(1(

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

2

sssssssss

ssssG

The non-identically-zero minors of order 1 are

21,

21,

11,

)2)(1(1,

11

ssssss

s

The non-identically-zero minor of order 2 are

2)2)(1()1(,

)2)(1(1,

)2)(1(2

sss

ssss

By considering all minors we find their least common denominator

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

Multivariable Poles (Thrugh Transfer Function Description )


Lecture 4

7

Multivariable Poles

Note that if A corresponds to a minimal realization then the poles by both definition will be the same.

DuCxyBuAxx

0)( AsIs

Theorem 4-1: Finding pole polynomials through transfer functionThe pole polynomial (s) corresponding to a minimal realization of a system with transfer function G(s) is the least common denominatorof all non-identically-zero minors of all orders of G(s).A minor of a matrix is the determinant of the square matrix obtained bydeleting certain rows and/or columns of the matrix.

Otherwise if A does not correspond to a minimal realization then the poles by |sI-A|=0 will include the poles (eigenvalues) corresponding to uncontrollable and/or unobservable states.


Lecture 4

8

Multivariable Poles

Exercise 4-1: Consider the state space realizationDuCxyBuAxx

000000

00111021

100111010321

0.108.06.700.36.18.0002.24.0004.08.2

DCBA

a- Find the poles of the system directly through state space form.

b- Find the transfer function of system ( note that the numerator and the denominator of each element must be prime ).

c- Find the poles of the system through its transfer function.

d- Compare poles from part “a” and “c” and explain the results.


Lecture 4

9

Multivariable Zeros

• Dynamic response.

• Stability of inverse system.

• Closed loop poles are on the open loop zeros at high gain.

• Blocking the inputs.

• Not affected by feedback.

Importance of multivariable zeros

• Stability analysis by inverse Nyquist


Lecture 4

10

Multivariable Zeros

Different definition of multivariable zeros

• Element zeros.

• Decoupling zeros (input and output).

• Transmission zeros or blocking zeros.

• System zeros.

• Invariant zeros.


Lecture 4

11

Multivariable Zeros

Element zeros

Decoupling zeros (input and output)

Transmission zeros or blocking zeros

System zeros

They are not very important in MIMO design.

They are clearly subset of poles.

They are the zeros that block the output in special condition.

Invariant zeros

Invariant zeros=System zeros (In the case of square plant)

zeros i.o.d.zeros o.d.zeros i.d.zeroson Transmissizeros System


Lecture 4

12

Multivariable Zeros (Through State Space Description )

DC

BAsIsP

yux

sP )(,0

)(DuCxyBuAxx

Laplace transform

The system zeros are then the values of s=z for which the Rosenbrock’s system matrix P(s) (if P(s) is square) loses rank resulting in zero outputfor some nonzero input.

000

,I

IDCBA

M g

Let

0)(

z

zg u

xMzI

Then the zeros are found as non trivial solutions of

This is solved as a generalized eigenvalue problem.

Rosenbrock system matrix


Lecture 4

13

DCBAsI

sP )(


Input decoupling zeros

System zero

rank. losses )( that of valueThe sPs

rank. losses that of valueThe A BsIs

Output decoupling zeros

rank. losses that of valueThe

C

AsIs

Invariant zeros

)(),(min)(result that s of values thoseare zerosInvariant

CnBnsP

)( of minorsgreatest all of gcdOr sP



Lecture 4

14

Multivariable Zeros

Element zeros


uxy

uxx

501010011

4000030000200001

1655

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

ss

ssssssssg

s= -2 and -4 are output decoupling zero (o.d.z)s= -3 and -4 are input decoupling zero (i.d.z)s= -4 is input-output decoupling zero (i.o.d.z)


s= -6/5 is transmission or blocking zeroSystem zeros and invariant zeros

s= -6/5, s=-2, s=-3 and s=-4 are system zero(invariant zeros)


Lecture 4

15

Let


uxy

uxx

1210112

1000

300020005

s= -2 is output decoupling zero (o.d.z)s= -5 is input decoupling zero (i.d.z)


s= -4 is transmission or blocking zeroInvariant zeros

s= -4, s=-5 are invariant zeros.

Element zeros

34

382)(

ss

sssG

s= -4, -4

System zeross= -4, s=-5, s=-2 are system zeros.

1210112300

1002000005

)(s

ss

sP

Multivariable Zeros


Lecture 4

16

Generally we have:

zeros i.o.d.zeros o.d.zeros i.d.zeroson Transmissizeros System

zeros SystemzerosInvariant zeroson Transmissi

System zerosInvariant zeros

Transmission zeros

Decoupling zeros

Multivariable Zeros


Lecture 4

17zerosmnExactlymCBrankandDzerosCBrankmnmostAtD

zerosDrankmnmostAtD

:)(0)(2:0

)(:0

For square systems with m=p inputs and outputs and n states, limits on the number of transmission zeros are:

Multivariable Zeros


Lecture 4

18

zerosmnExactlymCBrankandDzerosCBrankmnmostAtD

zerosDrankmnmostAtD

:)(0)(2:0

)(:0

Example 4-2: Consider the state space realizationDuCxyBuAxx

0014100

560100010

DCBA

0000010000100001

000

00141560

01000010

II

DCBA

M g

),( gIMeig

4z

10123 zerosofNumber



Lecture 4

19

Definition 4-2

zi is a zero (transmission zero) of G(s) if the rank of G(zi) is less than

the normal rank of G(s). The zero polynomial is defined as

)()( 1 ini zssz z

Where nz is the number of finite zeros (transmission zero) of G(s).

Multivariable Zeros (Through Transfer Function Description )


Lecture 4

20

The zero polynomial z(s) corresponding to a minimal realization of the

system is the greatest common divisor of all the numerators of all

order-r minors of G(s) where r is the normal rank of provided that

these minors have been adjusted in such a way as to have the pole

polynomial as their denominators.)(s

Theorem 4-2



Lecture 4

21

Example 4-3

)1)(1()1)(1()2)(1()1(0)2)(1(

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

2

sssssssss

ssssG

according to example 4-1 the pole polynomial is:

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

The minors of order 2 with as their denominators are)(s

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

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

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

2

2

22

ssss

sssss

sssss

The greatest common divisor of all the numerators of all order-2 minors is

1)( ssz



Lecture 4

22

Multivariable Zeros

Exercise 4-3: Consider the state space realizationDuCxyBuAxx

000000

00111021

100111010321

0.108.06.700.36.18.0002.24.0004.08.2

DCBA

a- Find the transmission zeros of the system directly through state space form.

b- Find the transfer function of system ( note that the numerator and the denominator of each element must be prime ).

c- Find the transmission zeros of the system through its transfer function.


Lecture 4

23







Transmission zero assignment



Lecture 4

24

Directions of Poles and Zeros

Zero directions:

Let G(s) have a zero at s = z, Then G(s) losses rank at s = z and

there will exist nonzero vectors and such that zu zy

0)(0)( zGyuzG Hzz

uz is input zero direction and yz is output zero direction

We usually normalize the direction vectors to have unit length

12zu 1

2zy


Lecture 4

25

Pole directions:

up is input pole direction and yp is output pole direction

Let G(s) have a pole at s = p. Then G(p) is infinite and we may

somewhat crudely write

)()( pGyupG Hpp

Hp

Hppp pqAqptAt

pH

ppp qBuCty



Lecture 4

26

Example4-4:

)1(25.4

412

1)(s

ss

sG

It has a zero at z = 4 and a pole at p = -2 . H

GsG

653.0757.0757.0653.0

271.000475.1

588.0809.0809.0588.0

45.442

51)3()( 0

H

GzG

6.08.08.06.0

00050.1

55.083.083.055.0

65.443

61)4()(

zuInput zero direction

zyOutput zero directionNow let the input as:

)()()( susGsy tetu 4

6.08.0

)(

41

6.08.0

)1(25.441

21

ss

ss

2.18.0

21

s

t

t

ee

ty 2

2

2.18.0

)( which does not contain any component of the input signal e4t



Lecture 4

27

Example4-5:

)1(25.4

412

1)(s

ss

sG

It has a zero at z = 4 and a pole at p = -2 .

)2()( GpG

)3(25.4431)2()(

GpG

pu

Input pole direction

py

Output pole direction

??!!

H

G

6.08.08.06.0

00.0009010

55.083.083.055.0

)001.02(

001.0 examplefor Let



Lecture 4

28

One Property of Zero Direction

DCBAsI

sP )(


System zero

rank. losses )( that of valueThe sPs

Thus we have a zero at s=z:

0

z

z

ux

DCBAzI

0

z

z

ux

DuCxyBuAxx

Now show that the output of following system is zero for all t

ztzz eutuxx )(,)0(

0,0)( tallforty


Lecture 4

29

Minimality

A state space system (A, B, C and D) is minimal if it is a system with the least number of states giving its transfer function.

If there is another system (A1, B1, C1 and D1) with fewer states having the same transfer function, the given system is not minimal.

For SISO systems, a system is minimal if and only if the transfer function numerator polynomial and denominator polynomial have no common roots.

For MV systems one must use the definition of MV zeros.

Theorem4-3 . A system (A, B, C and D) is minimal if and only if it has no input decoupling zeros and no output decoupling zeros.


Lecture 4

30










Lecture 4

31

Smith Form of a Polynomial Matrix

)(sSuppose that is a polynomial matrix.

Smith form of is denoted by , and it is a pseudo diagonalin the following form

)(s )(ss

000)(

)(s

s dss

)(,,........)(,)()( Where 21 sssdiags rds

)(si )(1 siis a factor of and1

)(

i

ii s


Lecture 4

32

)(,,........)(,)()( Where 21 sssdiags rds

)(si )(1 siis a factor of and1

)(

i

ii s

10

1 gcd {all monic minors of degree 1}

2 gcd {all monic minors of degree 2}

r gcd {all monic minors of degree r}

………………………………………………………..

………………………………………………………..



Lecture 4

33

The three elementary operations for a polynomial matrix are used to find Smith form.

• Multiplying a row or column by a non-zero constant;

• Interchanging two rows or two columns; and

• Adding a non-zero polynomial multiple of a row or column to another row or column.

)().....()()()()().......()( 121122 sRsRsRssLsLsLs nns

)()()()( sRssLss



Lecture 4

34

Example 4-6

21)2(2

)2(4)(

s

ss

10 1}1,2,2,1gcd{1 ss

)3)(1(}34gcd{ 22 ssss

1)(0

11

s )3)(1()(1

22 sss

)3)(1(0

01)(

ssss

Exercise 4-4: Derive R(s) and L(s) that convert Π(s)to Πs(s)



Lecture 4

35

Theorem 4-4 (Smith-McMillan form)

)()(

1)( ssd

sG

Let be an m × p matrix transfer function, where

are rational scalar transfer functions, G(s) can be represented by:

)]([)( sgsG ij )(sg ij

Where Π(s) is an m× p polynomial matrix of rank r and d(s) is the least

common multiple of the denominators of all elements of G(s) .

)(~ sGThen, is Smith McMillan form of G(s) and can be derived directly by

000)(

000)(

)(1)(

)(1)(~ sMs

sds

sdsG ds

s



Lecture 4

36

Theorem 4-4 (Smith-McMillan form)

)()(

1)( ssd

sG

000)(

000)(

)(1)(

)(1)(~ sMs

sds

sdsG ds

s

)()(,,........

)()(,

)()()(

2

2

1

1

ss

ss

ssdiagsM

r

r

)()()()(~ sRsGsLsG )(~)(~)(~)( sRsGsLsG

11 )()(~,)()(~ sRsRsLsL

and unimodular are )( and )(~,)(,)(~ matrices The sRsRsLsL



Lecture 4

37

Example 4-7

)2)(1(21

)1(2

)1(1

)2)(1(4

)(

sss

ssssG

)2)(1()(,5.0)2(2

)2(4)(,)(

)(1)(

sssds

sss

sdsG

)3)(1(0

01)(

ssss

230

0)2)(1(

1

)()(

1)(~

ss

ssssd

sG s



Lecture 4

38

Example 4-7

824824

11

)2)(1(1)(

)(1)(

22

22

ssssss

sss

sdsG

824824

11)(

22

22

sssssss

)4(30)4(30

11)(

2

21

sss

00)4(30

11)( 2

2 ss

00)4(30

01)( 2

3 ss

00

)4(001

)()( 24 sss s )()()()()()( 4312 sRsRssLsLss

)(~)(~)(~)(~)()(~)(

1)()(

1)( sRsGsLsRssLsd

ssd

sG s

00

120

0)2)(1(

1

)(~ss

ss

sG



Lecture 4

39










Lecture 4

40

Matrix Fraction Description (MFD)

Matrix Fraction Description for Transfer Matrix

)()(

1)(G sNsd

s Suppose G is an pq matrix so

Left Matrix Fraction Description (LMFD) )()()()()(G 11 sNsDsNIsds LLp

)()()()()(G 11 sDsNIsdsNs RRq Right Matrix Fraction Description

(RMFD)

But this forms are not irreducible.

Irreducible RMFD and LMFD can be derived directly through SMM form.


Lecture 4

41

Matrix Fraction Description&

Smith-McMillan form

Let is a matrix and its the Smith McMillan is )(sG mm )(~ sG

0,...,0,)(,...,)()( 1 ssdiagsN r

1,...,1,)(,...,)()( 1 ssdiagsD r

Let define:

or)()()(~ 1 sDsNsG )()()(~ 1 sNsDsG

We know that

)(~)(~)(~)( sRsGsLsG )(~)()()(~ 1 sRsDsNsL 1)()( sDsN RR 1)()()()(~ sDsRsNsL

)(~)(~)(~)( sRsGsLsG )(~)()()(~ 1 sRsNsDsL )()( 1 sNsD LL )(~)()()( 1 sRsNsLsD

It is easy to see that when a RMFD is irreducible, then

zsssGzs R at rank losses (s)Nor)(N ifonly and if )( of zeroion transmissa is L

(s)Gφ(s)pssDsDsGps

Ddet is G(s) of polynomial pole that themeans This at singular is )(or )( ifonly and if )( of pole a is RL


Lecture 4

42

Example 4-8

)1(42

)1(2

)2)(1(82

)2)(1(4

)2)(1(1

)2)(1(1

)(22

ss

ss

ssss

ssss

ssss

sG

3011

00

120

0)2)(1(

1

114014001

)(~)(~)(~)(2

2

ss

ss

ssssRsGsLsG

1

2

21

2

2

310

31)1)(2(

2424

01

3011

100)1)(2(

0020

01

114014001

s

sss

sssss

sss

ss

ss

)(sNR )(sDRRMFD:


Smith-McMillan form


Lecture 4

43

Example 4-8

)1(42

)1(2

)2)(1(82

)2)(1(4

)2)(1(1

)2)(1(1

)(22

ss

ss

ssss

ssss

ssss

sG

3011

00

120

0)2)(1(

1

114014001

)(~)(~)(~)(2

2

ss

ss

ssssRsGsLsG

)(sN L)( sD L

LMFD:

00)2(30

11

1101)4)(1(00)1)(2(

3011

0020

01

10001000)1)(2(

114014001 1

2

1

2

2 ss

ssssss

ssss

sss


Smith-McMillan form


Lecture 4

44







Transmission zero assignment Transmission zero assignment


Lecture 4

45


Zeros position depends on the location of sensors and actuators.

Theorem4-5: Transmission zeros cannot be assigned by state feedback.

Proof

DCBAsI

sP )(by defined is System

DDKCBBKAsI

sPf )( :isfeedback with System

IK

IsP

IKI

DCBAsI

DDKCBBKAsI

sPf

0)(

0)(

)()( sPsPf

Theorem4-6: Transmission zeros cannot be assigned by output feedback.

Changing zero position can affect the output response.


Lecture 4

46

What about following structures?

CxyBuAxx

sG

)(

Theorem4-7: Transmission zeros cannot be assigned by series compensation.

00

0)(

CBDAsIBHGFsI

sPs

uDHwuuGFww

sKˆˆ

)(

)()()( sPsPsPs

DHI

GFsI

CBAsI

I

CBDAsIBHGFsI

sPs

000

0

000

00

00

0)(


controllersystemnew zzz


Lecture 4

47


CxyurBAxx

sG)(

)(

Theorem4-8: Transmission zeros cannot be assigned by dynamic feedback.

00

0)(

CBBDCAsIBH

GCFsIsPf

DyHwuGyFww

sK

)(

)()( sPFsIsPf

000

00

00

0

00

0)(

CBAsI

I

IBDIBHGFsI

CBBDCAsIBH

GCFsIsPf


systemcontrollernew zpz


Lecture 4

48



Example 4-9: Suppose

11

11

32

11

)(

ss

sssG zeroion transmissa is 1s

Now let:

1110

)(sK

12

12

31

11

)()(

ss

ss

ss

ssKsG

Feed forward controller

zeroion transmissanot is 1s


Lecture 4

49


For zero assignment procedure see following papers.

R.V Patel and P. Misra “Transmission zero assignment in linear multivariable systems” ACC/WM5 1992.

A. Khaki Sedigh “Transmission zero assignment for linear Multivariable plans” 10th IASTED International Symposium, 1991.


Lecture 4

Exercises

50

Exercise 4-5: Consider following system.

xy

uxx

001001

213102131

a) Find the SMM form of the system system.b) Find the pole and zero polynomial of the system.c) Find the RMFD and LMFD of the system.

Exercise 4-6: Consider following transfer function.

2

422

1

)(sss

sGa) Find the SMM form of the system system.b) Find the pole and zero polynomial of the system.c) Find the RMFD and LMFD of the system.


Lecture 4

Exercises(Continue)

51

Exercise 4-7: Consider following transfer function.

124

23

31

42

42

21

)(

ss

ss

ss

ss

ssG

a) Find the SMM form of the system.b) Find the pole and zero polynomial of the system.c) Find the RMFD and LMFD of the system.


Lecture 4

52

References

• Control Configuration Selection in Multivariable Plants, A. Khaki-Sedigh, B. Moaveni, Springer Verlag, 2009.

References

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

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

• http://saba.kntu.ac.ir/eecd/khakisedigh/Courses/mv/

Web References

• http://www.um.ac.ir/~karimpor

• تحلیل و طراحی سیستم هاي چند متغیره، دکتر علی خاکی صدیق

Documents

Multivariable Control SystemsMultivariable Zeros Element zeros Decoupling zeros (input and output) Transmission zeros or blocking zeros System zeros They are not very important in