Multivariable Control Systems - ???????? - FUMcpanelprofsite.um.ac.ir/~karimpor/multi/Multivariable_lec2.pdfMultivariable Zeros Importance of multivariable zeros • Dynamic response

Multivariable Control Multivariable Control SystemsSystems

Ali KarimpourAssistant ProfessorAssistant Professor

Ferdowsi University of Mashhad

Chapter 2Chapter 2Introduction to Multivariable ControlIntroduction to Multivariable Control

Topics to be covered include:

Multivariable Connections

Multivariable Poles and zeros

Transmission zero assignment

Smith-McMillan Forms

M t i F ti D i ti (MFD) Matrix Fraction Description (MFD)

Scaling

Performance Specification

Bandwidth and Crossover Frequency

Ali Karimpour Feb 2012

2


Chapter 2

Introduction to Multivariable Control

Topics to be covered include: Multivariable Connections

M lti i bl P l d Multivariable Poles and zeros



Matrix Fraction Description (MFD) Matrix Fraction Description (MFD)

Scaling




3


Chapter 2


• Cascade (series) interconnection of transfer matrices( )

)(su )(sy )()()()()()( 12 susGsusGsGsy ==)(1 sG )(2 sG

)()()( 12 sGsGsG = )()( 21 sGsG≠

• Parallel interconnection of transfer matricesGenerally

)(1 sG)(su+

)(sy )()()())()(()( 21 susGsusGsGsy =+=

)(2 sG

++

)()()())()(()( 21y

)()()( 21 sGsGsG +=


4

21

Chapter 2


• Feedback interconnection of transfer matrices

)(1 sG )(2 sG)(sy)(su

+

− ( ))()()()()( 12 sysusGsGsy −= ( ))()()()()( 12 yy

)()()()()())()(()( 121

12 susGsusGsGsGsGIsy =+= −

)()())()(()( 121

12 sGsGsGsGIsG −+=

A useful relation in multivariable is push-through rule

12122

112 ))()()(()())()(( −− +=+ sGsGIsGsGsGsGI


5Exercise 1: Proof the push-through rule

Chapter 2


MIMO rule: To derive the output of a system, start from the output and write down the blocks as you meet them whenmoving backward (against the signal flow) towards the inputmoving backward (against the signal flow) towards the input.If you exit from a feedback loop then include a term ( ) 1−− LI

or according to the feedback sign where L is the transfer function around that loop (evaluated against the

( ) 1−+ LI

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

g

Chapter 2


Example 2-1: Derive the transfer function of the system shown in figurp y g

( )ω211

221211 )( PKPIKPPz −−+=


7

Chapter 2




l i i bl l d Multivariable Poles and zeros



Matri Fraction Description (MFD) Matrix Fraction Description (MFD)

Scaling




8


Chapter 2

Multivariable Poles ( State Space Description )

Definition 2 1: The poles of a system with state space descriptionipDefinition 2-1: The poles of a system with state-space description are eigenvalues of the matrix A.

i i i i i fi

niAi ,...,2,1,)( =λ),,,( DCBA

ip

The pole polynomial or characteristic polynomial is defined as

)det()( AsIs −=φ

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

0)det()( =−= AsIsφ

Note that if A does not correspond to a minimal realization then the poles by this definition will include the poles (eigenvalues)


9

p y p ( g )corresponding to uncontrollable and/or unobservable states.

Chapter 2

Multivariable Poles ( Transfer Function Description )

Theorem 2-1The pole polynomial corresponding to a minimal realization of a system with transfer function G(s) is the least common denominator

)(sφ

of all non-identically-zero minors of all orders of G(s).A minor of a matrix is the determinant of the square matrix obtained byq ydeleting certain rows and/or columns of the matrix.


10

Chapter 2

Multivariable Poles ( Transfer Function Description )

Example 2-4

⎥⎦

⎤⎢⎣

⎡

+−+−++−−+−

−++=

)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 are11111

21,

21,

11,

)2)(1(1,

11

++−−

++−

+ ssssss

s

Th id ti ll i f d 2The non-identically-zero minor of order 2 are

2)2)(1()1(,

)2)(1(1,

)2)(1(2

++−−

++++ sss

ssss )2)(1()2)(1()2)(1( ++++++ ssssss

By considering all minors we find their least common denominator


11)1()2)(1()( 2 −++= ssssϕ

Chapter 2

Multivariable Poles

Exercise 2: Consider the state space realization BuAxx +=&

DuCxy +=

⎤⎡⎤⎡⎥⎥⎤

⎢⎢⎡

⎥⎥⎤

⎢⎢⎡

−−

0001021010321

002.24.0004.08.2

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡=

⎥⎥⎥

⎦⎢⎢⎢

⎣

=

⎥⎥⎥

⎦⎢⎢⎢

⎣ −−−−

=000000

00111021

100111010

0.108.06.700.36.18.0002.24.0

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.


12

Chapter 2

Multivariable Zeros

Importance of 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.

p p p p g g

Different definition of multivariable zeros

• Element zeros. • Decoupling zeros (input and output).

• Transmission zero or blocking zero. • System zeros.


13• Invariant zeros.

Chapter 2

Multivariable Zeros

Element zeros

Decoupling zeros (input and output)

They are not very important in MIMO design.

Transmission ero or blocking eroThey are clearly subset of poles.

Transmission zero or blocking zero

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

System zero

zerosi.o.d.zeroso.d.zerosi.d.zerosonTransmissizerosSystem −++=

Invariant zero

zerosi.o.d.zeroso.d.zerosi.d.zeroson TransmissizerosSystem ++


14Invariant zeros=System zeros (In the case of square plant)

Chapter 2

Multivariable Zeros ( State Space Description )

⎥⎤

⎢⎡ −

⎥⎤

⎢⎡

⎥⎤

⎢⎡ BAsI

Px

P )(0

)(BuAxx +=& Laplace transform⎥⎦

⎢⎣ −

=⎥⎦

⎢⎣−

=⎥⎦

⎢⎣− DC

sPyu

sP )(,)(DuCxyBuAxx

+=+

Rosenbrock system matrix

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

⎥⎤

⎢⎡

=⎥⎤

⎢⎡ −

=0I

IBA

MLet

⎥⎦

⎢⎣

=⎥⎦

⎢⎣ −

=00

, IDC

M g

Then the zeros are found as non trivial solutions of

0)( =⎥⎦

⎤⎢⎣

⎡− z

g ux

MzI


15

⎦⎣ zu

This is solved as a generalized eigenvalue problem.

Chapter 2


⎥⎤

⎢⎡ −

=BAsI

sP )(System zero⎥⎦

⎢⎣ − DC

sP )(

Rosenbrock system matrix

I t d li

) P(s) squarefor (just rank. losses )( that of valueThe sPs

Input decoupling zeros

[ ] rank. losses that of valueThe A BsIs −

Output decoupling zeros⎤⎡ − AsI

rank. losses that of valueThe ⎥⎦

⎤⎢⎣

⎡−C

AsIs

In ariant erosInvariant zeros

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

CnBnsP ρρρ ++<


16

( ) ( ))(),(min)( CnBnsP ρρρ ++<

)( of minorsgreatest all of gcdOr sP

Chapter 2

Multivariable Zeros

Element zeros

Decoupling zeros (input and output)10001⎥⎤

⎢⎡

⎥⎤

⎢⎡− 655)4)(3)(2()( ++++ ssss

uxx

001

400003000020

⎥⎥⎥⎥

⎦⎢⎢⎢⎢

⎣

+

⎥⎥⎥⎥

⎦⎢⎢⎢⎢

⎣ −−

−=&

15

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

+=+

++++=

ssssssg

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

[ ] uxy 5010104000

+=⎦⎣⎦⎣ − s= -3 is input decoupling zero (i.d.z)

s= -4 is input-output decoupling zero (i.o.d.z)Transmission zero or blocking zeroTransmission zero or blocking zero

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


17

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

Chapter 2

Multivariable Zeros

Let00005 ⎤⎡⎤⎡− ⎥

⎤⎢⎡ + 00005s

[ ] [ ]

uxx

1210112

1000

300020005

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−=&

⎥⎦⎤

⎢⎣⎡

+−−

++

=34

382)(

ss

sssG

⎥⎥⎥⎥

⎦⎢⎢⎢⎢

⎣

−++

=

1210112300

10020)(

ss

sP

[ ] [ ]uxy 12101 −+=

Element zeross= -4, -4

⎦⎣ −12101

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

Transmission zero or blocking zero

s= -4 is transmission or blocking zeroInvariant zeros

s= -4 s=-5 are invariant zeros


18

s 4, s 5 are invariant zeros.System zeros

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

Chapter 2

Multivariable Zeros

Generally we have:

zeros i.o.d.zeros o.d.zeros i.d.zeroson Transmissizeros System −++=

S tI i tT i i zerosSystemzerosInvariant zeroson Transmissi ⊂⊂

System zerosInvariant zeros Decoupling zeros

Transmission zero


19

Chapter 2

Multivariable Zeros

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

zerosCBrankmnmostAtDzerosDrankmnmostAtD

+−=+−≠

)(2:0)(:0


20zerosmnExactlymCBrankandD −== :)(0)(

Chapter 2


Example 2-5: Consider the state space realization BuAxx +=&

DuCxy +=

[ ] 001400

100010

==⎥⎥⎤

⎢⎢⎡

=⎥⎥⎤

⎢⎢⎡

= DCBA [ ] 001410

560100 ==

⎥⎥⎦⎢

⎢⎣

=⎥⎥⎦⎢

⎢⎣ −−

= DCBA

⎤⎡⎤⎡ 00010010

⎥⎥⎥⎥⎤

⎢⎢⎢⎢⎡

=⎥⎦

⎤⎢⎣

⎡=

⎥⎥⎥⎥⎤

⎢⎢⎢⎢⎡

−−−=⎥

⎦

⎤⎢⎣

⎡−−

=010000100001

000

156001000010

II

DCBA

M g

),( gIMeig

4

zerosDrankmnmostAtD +−≠ )(:0

⎥⎦

⎢⎣

⎥⎦

⎢⎣ 00000014 4−=z

zerosmnExactlymCBrankandDzerosCBrankmnmostAtD−==

+−=:)(0

)(2:0


2110123 =−⋅−≤zerosofNumber

Chapter 2

Multivariable Zeros ( Transfer Function Description )

Definition 2-2

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

)()( n zssz z −Π= )()( 1 ii zssz −Π= =

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


22

Chapter 2


Theorem 2-2

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

Theorem 2-2

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

system is the greatest common divisor of all the numerators of allsystem 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.)(sφ


23

Chapter 2


Example 2-9

⎥⎦

⎤⎢⎣

⎡

+−+−++−−+−

−++=

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

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

2

sssssssss

ssssG

⎦⎣ ))(())(())(())()((

according to example 2-4 the pole polynomial is:

)1()2)(1()( 2 −++= ssssϕ

)(φThe minors of order 2 with as their denominators are)(sφ

)1()2)(1()2)(1(2 2−−+−+− sssss)1()2)(1(

,)1()2)(1(

,)1()2)(1( 222 −++−++−++ sssssssss

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


241)( −= ssz

Chapter 2

Multivariable Poles

Exercise 4: Consider the state space realization BuAxx +=&

DuCxy +=

⎤⎡⎤⎡⎥⎥⎤

⎢⎢⎡

⎥⎥⎤

⎢⎢⎡

−−

0001021010321

002.24.0004.08.2

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡=

⎥⎥⎥

⎦⎢⎢⎢

⎣

=

⎥⎥⎥

⎦⎢⎢⎢

⎣ −−−−

=000000

00111021

100111010

0.108.06.700.36.18.0002.24.0

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.


25

Chapter 2

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

is input zero direction and is output zero directionzu zy

We usually normalize the direction vectors to have unit length

1 1


26

12=zu 1

2=zy

Chapter 2


Pole directions:

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

somewhat crudely write

)()( GG H ∞=∞= )()( pGyupG Hpp

is input pole direction and is output pole directionpu py

HH Hp

Hppp pqAqptAt ==


27pH

ppp qBuCty ==

Chapter 2


Example:⎥⎤

⎢⎡ − 411)(s

G ⎥⎦

⎢⎣ −+

=)1(25.42

)(ss

sG

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

GsG ⎥⎦

⎤⎢⎣

⎡−⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −=⎥

⎦

⎤⎢⎣

⎡==

653.0757.0757.0653.0

271.000475.1

588.0809.0809.0588.0

45.442

51)3()( 0

⎦⎣⎦⎣⎦⎣⎦⎣ 653.0757.0271.00588.0809.045.45H

GzG ⎥⎦

⎤⎢⎣

⎡ −⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −=⎥

⎦

⎤⎢⎣

⎡==

60808.06.0

00050.1

55083083.055.0

65443

61)4()( ⎥

⎦⎢⎣⎥⎦

⎢⎣⎥⎦

⎢⎣

⎥⎦

⎢⎣ 6.08.00055.083.065.46

)()(

zuzyNo let the inp t as: Input zero directionOutput zero directionNow let the input as:

)()()( susGsy =tetu 4

6.08.0

)( ⎥⎦

⎤⎢⎣

⎡−= ⇒

41

608.0

)1(25441

21

−⎥⎦

⎤⎢⎣

⎡−⎥⎦

⎤⎢⎣

⎡−

−+

=ss

ss ⎥

⎦

⎤⎢⎣

⎡−+

=218.0

21

s


28

6.0 ⎦⎣ 46.0)1(25.42 ⎦⎣⎦⎣+ sss ⎦⎣+ 2.12s

⎥⎦

⎤⎢⎣

⎡−=

−

−

t

t

ee

ty 2

2

2.18.0

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

Chapter 2


Example:⎤⎡⎥⎦

⎤⎢⎣

⎡−

−+

=)1(25.4

412

1)(s

ss

sG

⎤⎡It has a zero at z = 4 and a pole at p = -2 . ⎥⎦

⎤⎢⎣

⎡∞∞∞∞

=−= )2()( GpG

??!!

⎥⎦

⎤⎢⎣

⎡ +−=+−=+

)3(254431)2()(

εεε GpG

??!!

001.0 examplefor Let =ε⎥⎦

⎢⎣ +− )3(25.4 εε

H⎤⎡⎤⎡⎤⎡− 8.06.00901083.055.0

p

uy

G ⎥⎦

⎤⎢⎣

⎡−⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=+−

6.08.08.06.0

00.0009010

55.083.083.055.0

)001.02(


29

pu

Input pole direction

py

Output pole direction

Chapter 2

Minimality

A state space system (A,B,C,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,D1) with fewer states having the y ( 1 1 1 1) gsame transfer function, the given system is not minimal.

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

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

Theorem. A system (A,B,C,D) is minimal iff it has no input decoupling zeros and no output decoupling zeros.


30

p g p p g

Chapter 2








Scaling




31


Chapter 2


Zeros position depends on the location of sensors and actuators.

Theorem: Transmission zeros cannot be assigned by state feedback.

Proof ⎤⎡ − BAsIProof ⎥⎦

⎤⎢⎣

⎡−

=DC

sP )(by definedisSystem

⎤⎡ + BBKAsI⎥⎦

⎤⎢⎣

⎡+−+−

=DDKCBBKAsI

sPf )( :isfeedback with System

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−−

=⎥⎦

⎤⎢⎣

⎡+−+−

=IK

IsP

IKI

DCBAsI

DDKCBBKAsI

sPf

0)(

0)(

( ) ( ))()( sPsPf ρρ =


32Theorem: Transmission zeros cannot be assigned by output feedback.

Chapter 2


What about following structures?

⎩⎨⎧

=−+=

CxyurBAxx

sG)(

)(&

⎩⎨⎧

+=+=

DyHwuGyFww

sK&

)(⎩ Cxy

Theorem: Transmission zeros cannot be assigned by dynamic feedback.

⎩ ywu

⎥⎤

⎢⎡ −− 0GCFsI

⎥⎤

⎢⎡

+−− 0

)( BBDCAIBHGCFsI

P

⎥⎥⎥

⎦⎢⎢⎢

⎣ −+−=

00)(

CBBDCAsIBHsPf

)()( sPFsIsP −=⎥⎥⎥

⎦⎢⎢⎢

⎣ −+−=

00)(

CBBDCAsIBHsPf

)()( sPFsIsPf =

⎥⎥⎤

⎢⎢⎡

−⎥⎥⎤

⎢⎢⎡ −

= 0000BAsI

IBDIBHGFsI


33⎥⎥⎦⎢

⎢⎣ −⎥⎥⎦⎢

⎢⎣ 0000 CI

Chapter 2


What about following structures?

⎩⎨⎧

=+=

CxyBuAxx

sG&

)(⎩⎨⎧

+=+=

uDHwuuGFww

sKˆˆ

)(&

⎩ Cxy

Theorem: Transmission zeros cannot be assigned by dynamic feedback.

⎩

⎥⎥⎤

⎢⎢⎡

−−−

=0

)( BDAsIBHGFsI

sP ⎥⎥⎤

⎢⎢⎡

−−−

= BDAsIBHGFsI

sP0

)(⎥⎥⎦⎢

⎢⎣ −

=00

)(C

BDAsIBHsPs

)()()( sPsPsP ′=⎤⎡⎤⎡

⎥⎥⎦⎢

⎢⎣ −

GFII

CBDAsIBHsPs

000

00)(

)()()( sPsPsPs =

⎥⎥⎥⎤

⎢⎢⎢⎡ −

⎥⎥⎥⎤

⎢⎢⎢⎡

−= IGFsI

BAsII

000

000


34⎥⎦⎢⎣ −⎥⎦⎢⎣ − DHC 000

Chapter 2



35

Chapter 2








Scaling




36


Chapter 2

Smith Form of a Polynomial Matrix

)(sΠSuppose that is a polynomial matrix.)(pp p y

Smith form of is denoted by , and it is a pseudo diagonali th f ll i f

)(sΠ )(ssΠ

in the following form

⎥⎤

⎢⎡Π

=Π0)(

)(s

s ds⎥⎦

⎢⎣

=Π00

)(ss

)(,,........)(,)()( Where 21 sssdiags rds εεε=Π

)(siε )(1 si+εis a factor of and1

)(−

=i

ii s

χχ

ε


37

Chapter 2


)(,,........)(,)()( Where 21 sssdiags rds εεε=Π

)(sε )(sεis a factor of and )( = isχ

ε)(siε )(1 si+εis a factor of and1

)(−

=i

i sχ

ε

10 =χ

=1χ gcd all monic minors of degree 1

=χ gcd all monic minors of degree 2=2χ gcd all monic minors of degree 2

………………………………………………………..

=rχ gcd all monic minors of degree r

………………………………………………………..


38

rχ g g

Chapter 2


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

• Multiplying a row or column by a constant;

• Interchanging two rows or two columns; andInterchanging two rows or two columns; and

• Adding a polynomial multiple of a row or column to another lrow or column.

)().....()()()()().......()( 121122 sRsRsRssLsLsLs nns Π=Π

)()()()( sRssLss Π=Π


39

Chapter 2


Example 2-11⎤⎡ +− )2(4 s

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

−+

+=Π

21)2(2

)2(4)(

s

ss

⎦⎣

10 =χ 11,2,2,1gcd1 =++= ssχ

)3)(1(34gcd 22 ++=++= ssssχ

χ χ1)(

0

11 ==

χχ

ε s )3)(1()(1

22 ++== sss

χχ

ε

⎥⎦

⎤⎢⎣

⎡++

=Π)3)(1(0

01)(

ssss


40

⎦⎣Exercise 5: Derive R(s) and L(s) that convert to)(sΠ )(ssΠ

Chapter 2


Theorem 2-3 (Smith-McMillan form)

Let be an m × p matrix transfer function, where

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

)]([)( sgsG ij= )(sgij

)()(

1)( sd

sG Π=

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

)(sd

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) .

)(~GTh i S ith M Mill f f G( ) d b d i d di tl b)(sGThen, is Smith McMillan form of G(s) and can be derived directly by

⎥⎤

⎢⎡

⎥⎤

⎢⎡Π

Π0)(0)(1)(1)(~ sMs

ssG ds


41⎥⎦

⎢⎣

=⎥⎦

⎢⎣

=Π=0000)(

)()(

)(sd

ssd

sG s

Chapter 2


Theorem 2-3 (Smith-McMillan form)

)()(

1)( ssd

sG Π= ⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡Π=Π=

000)(

000)(

)(1)(

)(1)(~ sMs

sds

sdsG ds

s)( ⎦⎣⎦⎣)()(

⎭⎬⎫

⎩⎨⎧

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

)()(,

)()()( 21 sssdiagsM r

δε

δε

δε

⎭⎬

⎩⎨ )(

,,)(

,)(

)(21 sss

grδδδ

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

11 )()(~)()(~ RRLL

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


42

11 )()(,)()( −− == sRsRsLsL

Chapter 2


Example 2-13⎤⎡ 14

⎥⎥⎥⎤

⎢⎢⎢⎡

−+−

++= 12)1(

1)2)(1(

4

)( ssssG

⎥⎥⎦⎢

⎢⎣ +++ )2)(1(2)1( sss

))(()()2(4

)()(1)(⎤⎡ +−

ds

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

)()(,)(

)(1)( ++=⎥

⎦

⎤⎢⎣

⎡−+

=ΠΠ= sssds

sssd

sG

⎤⎡ 01⎥⎦

⎤⎢⎣

⎡++

=Π)3)(1(0

01)(

ssss

⎥⎥⎤

⎢⎢⎡

++=Π=

0)2)(1(

1

)(1)(~ ssssG


43⎥⎥⎥

⎦⎢⎢⎢

⎣ ++

Π

230

)()(

)(

ss

ssd

sG s

Chapter 2


Example 2-14⎥⎤

⎢⎡ −

Π 82411

1)(1)( 22G⎥⎥⎥

⎦⎢⎢⎢

⎣ −−−−−+

++=Π=

824824

)2)(1()(

)()(

22

22

ssssss

sss

sdsG

⎤⎡ ⎤⎡ ⎤⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−+

−=Π

824824

11)(

22

22

sssssss

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−=Π

)4(30)4(30

11)(

2

21

sss

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−=Π

00)4(30

11)( 2

2 ss

⎦⎣ ⎦⎣ ⎦⎣

⎥⎤

⎢⎡

Π )4(3001

)( 2 ⎥⎤

⎢⎡

ΠΠ )4(001

)()( 2

⎥⎥⎥

⎦⎢⎢⎢

⎣

−=Π00

)4(30)( 23 ss

⎥⎥⎥

⎦⎢⎢⎢

⎣

−=Π=Π00

)4(0)()( 24 sss s )()()()()()( 4312 sRsRssLsLss Π=Π

⎤⎡ 01

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

1)()(

1)( sRsGsLsRssLsd

ssd

sG s =Π=Π=⎥⎥⎥⎥⎥⎤

⎢⎢⎢⎢⎢⎡

+−

++

=120

0)2)(1(

)(~ss

ss

sG


44

)()( sdsd

⎥⎥

⎦⎢⎢

⎣ 00

Chapter 2







Scaling




45


Chapter 2

Matrix Fraction Description (MFD)

function transferSISOan is 1)(Let 2

+=

ssg65

)( 2 ++ ssg

( )( ) 12 651)(can writeWe −+++= ssssg ( )( )651)( can write We +++= ssssg

polynomial polynomialp y p y

This is a Right Matrix Fraction Description (RMFD)

( ) ( )165)( writealsocan We 12 +++=− ssssg

polynomial polynomial


46This is a Left Matrix Fraction Description (LMFD)

Chapter 2


A model structure that is related to the Smith-McMillan form isA model structure that is related to the Smith-McMillan form is matrix fraction description (MFD).

• Right matrix fraction description (RMFD)

• Left matrix fraction description (LMFD)Left matrix fraction description (LMFD)

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

( )0,...,0,)(,...,)()( 1 ssdiagsN rεε∆

= ( )1,...,1,)(,...,)()( 1 ssdiagsD rδδ∆

=Let define:

or)()()(~ 1−= sDsNsG )()()(~ 1 sNsDsG −=


47

Chapter 2


)()()(~ 1−DNG )()()(~ 1 NDG −or)()()( 1= sDsNsG )()()( 1 sNsDsG =

We know thatWe know that

)(~)(~)(~)( sRsGsLsG = )(~)()()(~ 1 sRsDsNsL −= 1)()( −= sGsG DN( ) 1)()()()(~ −= sDsRsNsL

This is known as a right matrix fraction description (RMFD) whereg p ( )

~ )()()(,)()(~)( sDsRsGsNsLsG DN ==


48

Chapter 2


)()()(~ 1−DNG )()()(~ 1 NDG −or)()()( 1= sDsNsG )()()( 1 sNsDsG =

We know thatWe know that

)(~)(~)(~)( sRsGsLsG = )(~)()()(~ 1 sRsNsDsL −= )()( 1 sGsG ND−=( ) )(~)()()( 1 sRsNsLsD −=

This is known as a left matrix fraction description (LMFD) wherep ( )

)(~)()(,)()()( sRsNsGsLsDsG ND ==


49

Chapter 2


We can show that the MFD is not unique, because, for any nonsingularq y g

mm× )(sΩmatrix we can write G(s) as:

( ) ( )( ) 111( ) ( )( ) 111 )()()()()()()()()( −−− ΩΩ=ΩΩ= ssGssGsGsssGsG DNDN

Where is said to be a right common factor. When the only right )(sΩ g y gcommon factors of and is unimodular matrix, then, weSay that the RMFD is irreducible

)(

)(sGN )(sGD

( ))()( sGsGSay that the RMFD is irreducible. ( ))(),( sGsG DN

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

zssGsGzs N ==∗ at rank losses )( ifonly and if )( of zero a is

zssGsGps atsingularis)(ifonlyandif)(ofpoleais ==∗


50( )(s)Gφ(s)zssGsGps

D

D

det is G(s) of polynomial pole that themeans This at singular is )( ifonly and if )(ofpolea is

===∗

Chapter 2


Example 2-16 ⎥⎤

⎢⎡ −

)2)(1(1

)2)(1(1

⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢

++−−

++−+

++++

=)2)(1(

82)2)(1(

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

)(22

ssss

ssss

ssss

sG

⎥⎥

⎦⎢⎢

⎣ +−

+−

)1(42

)1(2

ss

ss

⎥⎤

⎢⎡ 01

=⎥⎦

⎤⎢⎣

⎡ −

⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢

+−

++

⎥⎥⎥⎤

⎢⎢⎢⎡

−+==3011

120

0)2)(1(

114014001

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

2

ss

ss

sssRsGsLsG⎦⎣

⎥⎥⎥

⎦⎢⎢⎢

⎣

⎥⎦⎢⎣ −

00

1142s

11

2

21

2

2

10

31)1)(2(

2424

01

3011

100)1)(2(

0020

01

114014001

−

−

⎥⎥⎥⎥⎤

⎢⎢⎢⎢⎡

+

+++

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−+=⎥⎦

⎤⎢⎣

⎡ −⎥⎦

⎤⎢⎣

⎡+

++

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+s

sss

sssss

sss

ss

ss


513

02400114 ⎥⎦⎢⎣⎥⎦⎢⎣ −−⎥⎦⎢⎣⎥⎦⎢⎣ − sss

)(sGN )(sGDRMFD:

Chapter 2


Example 2-16 ⎥⎤

⎢⎡ −

)2)(1(1

)2)(1(1

⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢

++−−

++−+

++++

=)2)(1(

82)2)(1(

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

)(22

ssss

ssss

ssss

sG

⎥⎥

⎦⎢⎢

⎣ +−

+−

)1(42

)1(2

ss

ss

⎥⎤

⎢⎡ 01

=⎥⎦

⎤⎢⎣

⎡ −

⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢

+−

++

⎥⎥⎥⎤

⎢⎢⎢⎡

−+==3011

120

0)2)(1(

114014001

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

2

ss

ss

sssRsGsLsG⎦⎣

⎥⎥⎥

⎦⎢⎢⎢

⎣

⎥⎦⎢⎣ −

00

1142s

1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+−−+

++

=⎥⎦

⎤⎢⎣

⎡ −

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+

++

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−+

−−

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


52)(sGN)( sG DLMFD:

⎦⎣⎦⎣⎦⎣⎦⎣⎦⎣ 114s

Chapter 2


Example 2-17 Consider⎥⎤

⎢⎡ +

⎥⎤

⎢⎡ +−

−

− )()()(212

1 sssssssGsGsG ⎥

⎦⎢⎣ ++⎥

⎦⎢⎣ +−

==2222

)()()(ssss

sGsGsG ND

? eirreducibl )( and )( Are sGsG ND

⎤⎡⎤⎡⎥⎦

⎤⎢⎣

⎡

+++

=⎥⎦

⎤⎢⎣

⎡+

+⎥⎦

⎤⎢⎣

⎡=

221111

110

)(2

sssss

sss

sGN⎥⎦

⎤⎢⎣

⎡

+−+−

=⎥⎦

⎤⎢⎣

⎡+−

−⎥⎦

⎤⎢⎣

⎡=

221111

110

)(2

sssss

sss

sGD

Now let:

⎥⎦

⎤⎢⎣

⎡+−

−=

1111

)(~s

ssGD ⎥

⎦

⎤⎢⎣

⎡+

+=

1111

)(~s

ssGN

Now let:

? eirreducibl )(~ and )(~ Are sGsG ND

⎥⎤

⎢⎡ −

=⎥⎤

⎢⎡ −⎥⎤

⎢⎡

=11111

)(~ sssGD ⎥

⎤⎢⎡ +

=⎥⎤

⎢⎡ −⎥⎤

⎢⎡

=11111

)(~ sssGN⎥

⎦⎢⎣ +−⎥

⎦⎢⎣ +−⎥⎦

⎢⎣ 111110

)(ss

sGD ⎥⎦

⎢⎣ +⎥

⎦⎢⎣ +⎥⎦

⎢⎣ 111110

)(ss

sGN

Now let:⎥⎤

⎢⎡ −11

)(G ⎥⎤

⎢⎡ −11

)(ˆ sG


53? eirreducibl )(ˆ and )(ˆ Are sGsG ND

but why? Yes

⎥⎦

⎢⎣ +−

=11

)(s

sGD ⎥⎦

⎢⎣ +

=11

)(s

sGN

Chapter 2


Checking irreducibility:

Form:[ ] ⎥

⎦

⎤⎢⎣

⎡+−+−−

=1111

1111)(ˆ)(ˆ

sssGsG DN

Do preliminary transformation to make right part zero:

1) Add C on C ⎤⎡ − 0111 2) Add -C1 on C3 ⎤⎡ − 00111) Add -C2 on C4 ⎥⎦

⎤⎢⎣

⎡−+ 0111

0111s

2) Add C1 on C3

3) Add C1 on C2

⎥⎦

⎤⎢⎣

⎡−+ 0211 s

⎥⎤

⎢⎡ 00013) Add C1 on C2

4) Add 0.5(s+2)C3 on C2

⎥⎦

⎢⎣ −+ 0221 s

⎥⎤

⎢⎡ 0001) ( ) 3 2

5) Change C3 and C2 ⎥⎤

⎢⎡

=Ω01

)(:isfactorCommon s

⎥⎦

⎢⎣ − 0201

⎥⎤

⎢⎡ 0001


54

⎥⎦

⎢⎣ −

Ω21

)( :isfactor Common s

e.irreducibl are )(ˆ and )(ˆ so unimodular is (s) Since sGsG NDΩ

⎥⎦

⎢⎣ − 0021

Chapter 2


Deriving common factor:

Form:[ ] ⎥

⎦

⎤⎢⎣

⎡

+−+++−+

=2222

)()(22

ssssssssss

sGsG DN

Do preliminary transformation to make right part zero:

1) Add C on C ⎤⎡ −+ 02 ssss 2) Add C1 on C3 ⎤⎡ + 002 sss1) Add -C2 on C4 ⎥⎦

⎤⎢⎣

⎡

−+++

02220

sssssss 2) Add C1 on C3

⎥⎦

⎤⎢⎣

⎡

+++

022200

ssssss

3) Add –(s+1)C1 on C2 ⎥⎤

⎢⎡ 000s3) Add (s+1)C1 on C2 ⎥

⎦⎢⎣ +−+ 02)2(2 ssss

4) Add 0.5(s+2)C3 on C2 ⎥⎤

⎢⎡ 000s) ( ) 3 2 ⎥

⎦⎢⎣ + 0202 ss

5) Change C3 and C2 ⎥⎤

⎢⎡ 000s

⎥⎤

⎢⎡

=Ωs

s0

)(:isfactorCommon


55

⎥⎦

⎢⎣ + 0022 ss ⎥

⎦⎢⎣ +

=Ωss

s22

)( :isfactor Common

reducible. are )( and )( so unimodularnot is (s) Since sGsG NDΩ

Chapter 2







Scaling




56


Chapter 2

Scaling

ryedGuGy d ˆˆˆ;ˆˆˆˆˆ −=+= ryedGuGy d ;+

rDreDeyDyuDudDd eeeud ˆ,ˆ,ˆ,ˆ,ˆ 11111 −−−−− =====

rDyDeDdDGuDGyD eeeddee −=+= ;))

yy eeeddee

ddedue DGDGDGDG ˆ,ˆ 11 −− == ryedGGuy d −=+= ;ddedue , yy d ;


57

Chapter 2







Scaling




58


Chapter 2


• Nominal stability NS: The system is stable with no model• Nominal stability NS: The system is stable with no model uncertainty.

• Nominal Performance NP: The system satisfies the performance specifications with no model uncertainty.

• Robust stability RS: The system is stable for all perturbed plants about the nominal model up to the worst case model uncertaintyabout the nominal model up to the worst case model uncertainty.

• Robust performance RP: The system satisfies the performance p y pspecifications for all perturbed plants about the nominal model up to the worst case model uncertainty.


59

Chapter 2


• Time Domain PerformanceTime Domain Performance

• Frequency Domain Performance


60

Chapter 2

Time Domain Performance

Although closed loop stability is an important issue,Although closed loop stability is an important issue,

the real objective of control is to improve performance, that is, to make the output y(t) behave in a more desirable manner.

Actually, the possibility of inducing instability is one of the disadvantages of feedback control which has to be traded off against performance improvement.

The objective of this section is to discuss ways of evaluatingclosed loop performance.


61

closed loop performance.

Chapter 2


Step response of a system

• Rise time, tr , the time it takes for the output to first reach 90% of i fi l l hi h i ll i d b llits final value, which is usually required to be small.

• Settling time, ts , the time after which the output remains withing , s , p5% ( or 2%) of its final value, which is usually required to be small.

O ershoot P O th k l di id d b th fi l l hi h


62

• Overshoot, P.O, the peak value divided by the final value, whichshould typically be less than 20% or less.

Chapter 2



• Decay ratio, the ratio of the second and first peaks, which should typically be 0.3 or less.

• Steady state offset, ess, the difference between the final value and the desired final value, which is usually required to be small.


63

y q

Chapter 2



E i ti th t t l i ti (TV) di id d b th ll• Excess variation, the total variation (TV) divided by the overall change at steady state, which should be as close to 1 as possible.

The total variation is the total movement of the output as shown.

/i tiE TVTV ∑Ali Karimpour Feb 2012

640/variationExcess vTVvTV

ii ==∑

Chapter 2


• ISE : Integral squared error ττ deISE 2

0)(∫

∞=

• IAE : Integral absolute error ττ deIAE ∫∞

=0

)(

• ITSE : Integral time weighted squared error τττ deITSE 2)(∫∞

=• ITSE : Integral time weighted squared error τττ deITSE0

)(∫=

• ITAE : Integral time weighted absolute error τττ deITSE )(0∫∞

=


65

Chapter 2

Frequency Domain Performance

Let L(s) denote the loop transfer function of a system which isLet L(s) denote the loop transfer function of a system which is closed-loop stable under negative feedback.

Bode plot of )( ωjL

1G

180)( 180 =∠ ωjL

)(1

180ωjLGM =

1)( =cjL ω

180)( +∠= cjLPM ω


66cPM ωθ /max =

Chapter 2


Let L(s) denote the loop transfer function of a system which isLet L(s) denote the loop transfer function of a system which is closed-loop stable under negative feedback.

1G

180)( 180 =∠ ωjLNyquist plot of )( ωjL

)(1

180ωjLGM =

1)( =cjL ω

180)( +∠= cjLPM ω


67cPM ωθ /max =

Chapter 2


Stability margins are measures of how close a stable closed-loop system is to instability.

From the above arguments we see that the GM and PM provide stabilityFrom the above arguments we see that the GM and PM provide stabilitymargins for gain and delay uncertainty.

More generally, to maintain closed-loop stability, the Nyquist stability condition tells us that the number of encirclements of the critical point-1 by must not change. )( ωjL

Thus the actual closest distance to -1 is a measure of stability


68

Chapter 2


One degree-of-freedomconfiguration

nGKIGKdGGKIrGKIGKsy d111 )()()()( −−− +−+++=

Complementary sensitivity function )(sT

Sensitivity function )(sS

The maximum peaks of the sensitivity and complementary sensitivity functions are defined as


69)(max)(max ωωωω

jTMjSM Ts ==

Chapter 2


( ) ( ) 11 )()()()( −− +=+= ωωωω jLIjKjGIjS ( ) ( )

sM1

)(max ωω

jSM s =

Thus one may also view Ms as a robustness measure.


70

y s

Chapter 2


There is a close relationship between MS and MT and the GM and PM.

][12

1sin2;1

1 radMM

PMM

MGM s ≥⎟

⎠

⎞⎜⎜⎝

⎛≥≥ −

21 MMM sss ⎠⎜⎝−

For example, with MS = 2 we are guaranteed GM > 2 and PM >29˚.

⎞⎛

For example, with MS 2 we are guaranteed GM 2 and PM 29 .

][12

1sin2;11 1 radMM

PMM

GMTTT

≥⎟⎟⎠

⎞⎜⎜⎝

⎛≥+≥ −

For example, with MT = 2 we are guaranteed GM > 1.5 and PM >29˚.


71

Chapter 2

Trade-offs in Frequency Domain

One degree-of-freedomconfiguration

nGKIGKdGGKIrGKIGKsy d111 )()()()( −−− +−+++=

Complementary sensitivity function )(sT

Sensitivity function )(sS

)(sT

IsSsT =+ )()(

TdSGS + KSdKSGKSAli Karimpour Feb 2012

72

TndSGSrrye d −+−=−= KSndKSGKSru d −−=

Chapter 2

Trade-offs in Frequency DomainTndSGSrrye + KSndKSGKSru =

IsSsT =+ )()(TndSGSrrye d −+−=−= KSndKSGKSru d −−=

( ) 1)()( −+= sLIsS

• Performance, good disturbance rejection ∞→→→ LITS or or 0

• Performance, good command following ∞→→→ LITS or or 0

• Mitigation of measurement noise on output 0oror0 →→→ LIST• Mitigation of measurement noise on output 0or or 0 →→→ LIST

• Small magnitude of input signals 0Tor0or0 →→→ LKSmall magnitude of input signals 0Tor 0or 0 →→→ LK

• Physical controller must be strictly proper 0Tor 0or 0 →→→ LK

• Nominal stability (stable plant) small be L


73• Stabilization of unstable plant ITL →or large be

Chapter 2





Transmission zero assignment g


M i F i D i i (MFD) Matrix Fraction Description (MFD)

Scaling




74


Chapter 2


Definition 2-3

below. from db 3- crossesfirst )S(j wherefrequency theis , ,bandwidth The B

ωω

40

60g

L

Definition 2-4 0

20

L

TDefinition 2 4

bfdb3)T(jt hi hfrequency highest theis , ,bandwidth The BTω

-40

-20S

above. from db3-crosses )T(jat which ω

Definition 2-510

010

110

210

3-60

Frequency (rad/sec)

bfdb0)L(jh frequency theis , ,frequency crossover gain The cω


75above. from db0crosses )L(j where ω

Chapter 2


Specifically, for systems with PM < 90˚ (most practical systems) we have

BTcB ωωω ≤≤60

g

20

40

60

L

-20

0T

S

100

101

102

103

-60

-40

S

In conclusion ωB ( which is defined in terms of S ) and also ωc

10 10 10 10Frequency (rad/sec)


76( in terms of L ) are good indicators of closed-loop performance, while ωBT ( in terms of T ) may be misleading in some cases.

Chapter 2


Example 2-18: Comparison of and as indicators of performance.BTωBω

1,1.0;1

1;)2(

Let ==++

+−=

+++−

= ττττ

zszs

zsTzss

zsL1)2( ++++ τττ szszss

036.0=Bω054.0=cω 1=BTω 1.0 zero RHPan is There =zc BT


77

Chapter 2


Example 2-17: Comparison of and as indicators of performance.BTωBω

1,1.0;1

1;)2(

Let ==++

+−=

+++−

= ττττ

zszs

zsTzss

zsL1)2( ++++ τττ szszss

036.0=Bω054.0=cω 1=BTω 1.0 zero RHPan is There =zc BT


78

Chapter 2

Exercises


79

Chapter 2

Exercises(Continue)( )


80

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