Multivariable Control Systems - ???????? - FUMcpanelprofsite.um.ac.ir/~karimpor/multi/Multivariable_lec4.pdfChapter 4 Chapter 4 SS b y o u v b e eedb c Co o Sys e stability of Multivariable

Multivariable Control Multivariable Control SystemsSystemsSystemsSystems

Ali KarimpourpAssistant Professor

Ferdowsi University of MashhadFerdowsi University of Mashhad

Chapter 4Chapter 4Stability of Multivariable Feedback Control SystemsS b y o u v b e eedb c Co o Sys e s

Topics to be covered include:

• Well - Posedness of Feedback Loop

• Internal Stability• Internal Stability

• The Nyquist Stability CriterionThe Generalized Nyquist Stability CriterionThe Generalized Nyquist Stability CriterionNyquist arrays and Gershgorin bands

C i F t i ti St bl T f F ti• Coprime Factorizations over Stable Transfer Functions

• Stabilizing Controllers

Ali Karimpour Feb 2011

2• Strong and Simultaneous Stabilization

Chapter 4

Stability of Multivariable Feedback Control Systems

W ll P d f F db k L• Well - Posedness of Feedback Loop

• Internal Stability

• The Nyquist Stability CriterionThe Generalized Nyquist Stability CriterionNyquist arrays and Gershgorin bands

• Coprime Factorizations over Stable Transfer Functionsp


St d Si lt St bili ti


3

• Strong and Simultaneous Stabilization

Chapter 4

Well - Posedness of Feedback Loop

Assume that the plant P and the controller K are fixed real rational t f t iproper transfer matrices.

The first question one would ask is whether the feedback interconnectionmakes sense or is physically realizablemakes sense or is physically realizable.

1,21Let =

+−

−= KssP

idsdnrsu3

1)(3

2 −+−−

+=


4

2+s 33

Hence, the feedback system is not physically realizable!

Chapter 4


Definition 4-1 A feedback system is said to be well-posed if all closed-loop transfer matrices are well-defined and propertransfer matrices are well-defined and proper.

Now suppose that all transfer matrices from the signals r n d and d toNow suppose that all transfer matrices from the signals r, n, d and di tou are respectively well-defined and proper.

Thus y and all other signals are also well-defined and the related transfery gmatrices are proper.

So the system is well-posed if and only if the transfer matrix from di


5

y p y iand d to u exists and is proper.

Chapter 4


So the system is well-posed if andonly if the transfer matrix from dionly if the transfer matrix from diand d to u exists and is proper.

Theorem 4-1 The feedback system in Figure is well-posed if and only if

invertible is )()( ∞∞+ PKI

( ) [ ] ⎥⎦

⎤⎢⎣

⎡+−= −

idd

KPKKPIu 1Proof

Thus well - posedness is equivalent to the condition that exist( ) 1−+ KPIThus well - posedness is equivalent to the condition that existand is proper.

( )+ KPI

And this is equivalent to the condition that the constant term of the


6

And this is equivalent to the condition that the constant term of the transfer matrix is invertible.)()( ∞∞+ PKI

Chapter 4


Transfer matrix is invertible.)()( ∞∞+ PKI

i i l i h f h f ll i di iis equivalent to either one of the following two conditions:

invertible is )(⎥⎤

⎢⎡ ∞KI

invertibleis)()( ∞∞+ KPIve b es)( ⎥

⎦⎢⎣ ∞− IP

invertible is )()(+ KPI

The well- posedness condition is simple to state in terms of state-space realizations. Introduce realizations of P and K:

⎥⎦

⎤⎢⎣

⎡≅

DCBA

P⎥⎥⎦

⎤

⎢⎢⎣

⎡≅

DC

BAK ))

ˆˆ

⎦⎣ DC ⎥⎦⎢⎣ DC

SO well- posedness is equivalent to the condition that


7invertible is ⎥⎦

⎤⎢⎣

⎡

− IDDI)

Chapter 4









8


Chapter 4

Internal Stability

Assume that the realizations for P(s) and K(s) are stabilizable and detectable. Let x and denote the state vectors for P and K, respectively.x̂

ˆˆ&

yDxCuDuCxy

yBxAxBuAxxˆˆˆ

ˆˆ

−=+=

−=+=&

Definition 4-2

ht ti iti lllft)ˆ(t tthit blllt tii (0,0))ˆ,(origin theif stable internally be tosaid is Figure of system The =xx


90 and0,0,0whenstatesinitialallfromzero togo),(states thei.e. stableally asymptotic is

==== nddrxx

i

Chapter 4

Internal Stability

⎤⎡⎤⎡⎤⎡⎤⎡− ˆˆ 1

e

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

−=⎥

⎦

⎤⎢⎣

⎡xx

CC

IDDI

yu

ˆ00

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

−⎥⎦

⎤⎢⎣

⎡

−+⎥

⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡−

xx

Axx

CC

IDDI

B

B

A

A

x

xˆ

~ˆ0

ˆ0ˆˆ0

0ˆ0

0

ˆ

1

&

&

⎠⎝

Theorem 4-2

The system of above Figure with given stabilizable and detectable

A~

The system of above Figure with given stabilizable and detectable

realizations for P and K is internally stable if and only if is a A~


10

Hurwitz matrix (All eigenvalues are in open left half plane).

Chapter 4

Internal Stability

⎥⎤

⎢⎡

⎥⎤

⎢⎡⎥⎤

⎢⎡ duKI ip

e

⎥⎦

⎢⎣−

=⎥⎦

⎢⎣−⎥⎦

⎢⎣− reIP

ip

Theorem 4-3The system in Figure is internally stable if and only if the transfer matrixThe system in Figure is internally stable if and only if the transfer matrix

⎥⎥⎦

⎤

⎢⎢⎣

⎡ +−+−=⎥

⎦

⎤⎢⎣

⎡ −−−

11

111

)()()()( PKIKPPKIKI

IPKI

⎥⎥⎦⎢

⎢⎣ ++⎥

⎦⎢⎣− −− 11 )()( PKIPPKIIP

from (di , -r) to (up , - e) be a proper and stable transfer matrix.


11

( i , ) ( p , ) p p

Chapter 4

Internal Stability

⎤⎡⎤⎡ −−− 111 )()( PKIKPPKIKIKI

⎥⎥⎦

⎤

⎢⎢⎣

⎡

++

+−+−=⎥

⎦

⎤⎢⎣

⎡− −− 11

11

)()()()(

PKIPPKIPKIKPPKIKI

IPKI

Note that to check internal stability it is necessary (and sufficient) to testNote that to check internal stability, it is necessary (and sufficient) to test whether each of the four transfer matrices is stable.

⎤⎡ ++ ss 11

11,

11Let

−=

+−

=s

KssP ⎥

⎦

⎤⎢⎣

⎡−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

++

+−

+−−

+=⎥

⎦

⎤⎢⎣

⎡

− rd

sssss

e

u ip

21

21

)2)(1(2


12

⎥⎦⎢⎣ ++ ss 22Stability cannot be concluded even if three of the four transfer matrices are stable.

Chapter 4

Internal Stability

Theorem 4-4insystemThen thestable.isSuppose K

Stable

stable. is )( ifonly and if stable internally is figure the

in system Then the stable. is Suppose

1 PPKI

K

−+⎥⎥⎦

⎤

⎢⎢⎣

⎡

++

+−+−−−

−−

11

11

)()()()(

Remember PKIPPKI

PKIKPPKIKI

)(

Proof: The necessity is obvious. To prove the sufficiency let

PPKIQ 1)( −+=

KQIPPKIKI −=+− −1)( Stable

)()( 1 QKIKPKIK −−=+− −

QPPKI =+ −1)(

Stable

Stable


13QKIPKPKIIIPKIIPKI −=+−=−++=+ −−− 111 )()()(

QPPKI + )( Stable

Stable

Chapter 4

Internal Stability

Theorem 4-5insystemThen thestableisSuppose P

Stable

stableis)(ifonly and if stable internally is figure the

in system Then the stable. is Suppose

1−+ PKIK

P

⎥⎥⎦

⎤

⎢⎢⎣

⎡

++

+−+−−−

−−

11

11

)()()()(

Remember PKIPPKI

PKIKPPKIKI

stable. is )( + PKIK


1)( −+= PKIKQ

QPIPPKIKI −=+− −1)( Stable

QPKIK −=+− −1)(

PPQIPPKI )()( 1 −=+ −

Stable

Stable


14PQIPKIPKIIPKIIPKI −=+−=−++=+ −−− 111 )()()(

PPQIPPKI )()( + Stable

Stable

Chapter 4

Internal Stability

Theorem 4-6Thenstable.bothareandSuppose KP

Stable Stable

stable.is )( ifonly and if stable internally is figure in the system the

Then stable.both are and Suppose

1−+ PKI

KP

⎥⎥⎦

⎤

⎢⎢⎣

⎡

++

+−+−−−

−−

11

11

)()()()(

Remember PKIPPKI

PKIKPPKIKI

)(y


1)( −+= PKIQ

KQPIPPKIKI −=+− −1)( Stable

KQPKIK −=+− −1)(

QPPPKI =+ −1)(

Stable

Stable


15QIPKIIPKI =−++=+ −− 11 )()(

QPPPKI + )( Stable

Stable

Chapter 4

Internal StabilityN l ll ti

Theorem

No pole zero cancellation

)()(ofpolesRHPopenofnumberthei) and posed- wellisit ifonly and if

stableinternallyisfigure in the system he

+= nnsKsP

T

⎥⎥⎦

⎤

⎢⎢⎣

⎡

++

+−+−−−

−−

11

11

)()()()(

Remember PKIPPKI

PKIKPPKIKI

stable. is )( ii)

)()(ofpolesRHPopen ofnumber thei)1−+

+=

PKI

nnsKsP PK

Proof:

of polse RHP ofnumber theis of polse RHP ofnumber theis PnKn PK

See: “Essentials of Robust control” written by Kemin Zhou


16

Chapter 4









17


Chapter 4

The Nyquist Stability Criterion

f ibhfbilih k hiN

lRHPi thdlh)](d t[L t PPkGI

. of valuesreal sfor variousystemabove theofstability check the togoing are weNow

k

plane.RHPin the zerosandpoleshave )](det[Let co PPskGI +

tii ithi l))(det()( ofplot Nyquist thesystems, SISO asjust Then

PPskGIs +=φ

times. origin, theencircles oc PP −

However, we would have to draw the Nyquist locus of for each value of k in which we


18

were interested, whereas in the classical Nyquist criterion we draw a locus only once,and then infer stability properties for all values of k.

Chapter 4


∏ +=+i

i skskGI )](1[)](det[ λ

)(of eigenvaluean is)(Where sGsiλ )(g)(i

{ } ( )∑ +=+ i skskGI )(1)](det[ λϑϑ{ } ( )∑i

i )()]([

1- 0 →


19

Chapter 4


Theorem 4-7 (Generalized Nyquist theorem)Theorem 4 7 (Generalized Nyquist theorem)

If G(s) with no hidden unstable modes, has Po unstable (Smith-McMillan)

poles, then the closed-loop system with return ratio –kG(s) is stable if

and only if the characteristic loci of kG(s) taken together encircle theand only if the characteristic loci of kG(s), taken together, encircle the

point -1, Po times anticlockwise.


201)(1)(1 −++ =− NPZ skGskG 1)()(1 −+ =− NPZ sGskG⇒

Chapter 4


Example 4-1: Let

⎤⎡ 2

⎥⎥⎥⎥⎤

⎢⎢⎢⎢⎡

+−−

+++

=01

)1()1(5.0

)1(1)1(5.0

)(3

2

ss

ss

sG

⎥⎦⎢⎣ +−0

)1)(1( 3ss

Suppose that G(s) has no hidden modes, check the stability of system for different values of k.

0)( =− sGIλ

)1(5.0

1 +=

sλ 32 )1(

1+

=s

λ


21

)( )(s

Chapter 4

The Nyquist Stability CriterionStability Criterion

)1(5.0

1 +=

sλ

)1( +s

32 )1(1+

=s

λ125.0−

Since G(s) has one unstable poles, we will have closed l bili if h l iloop stability if these loci give one net encirclementsof -1/k (ccw) when a negative feedback kI is appliedfeedback kI is applied.

pole. loop closed RHP one is theresont encircleme no 125.0-/1-- <<∞ kpoleloopclosedRHPthreeistheresontencirclemetwo0 5k/11250- <−<


22

pole.loopclosedRHP threeis theresont encircleme two0.5k/1125.0 <<pole. loop closed RHP twois theresont encircleme one 1k/10.5 <−<pole. loop closed RHP threeis theresont encircleme two 11/k- >

Chapter 4


Example 4-2: Let

⎥⎤

⎢⎡ −11)(

ssG ⎥

⎦

⎤⎢⎣

⎡−−++

=26)2)(1(25.1

)(sss

sG

Suppose that G(s) has no hidden modes, check the stability of system for different values of k.

0)( =− sGIλ

24132 −−− ssλ 24132 −+− ss)2)(1(5.2

241321 ++=

ssssλ

)2)(1(5.224132

2 ++−+−

=ss

ssλ


23

Chapter 4


)2)(1(5.224132

1 ++−−−

=ss

ssλ

)2)(1(5.224132

2 ++−+−

=ss

ssλ

Since G(s) has no unstable poles, we will have closed l bili if h l iloop stability if these loci give zero net encirclementsof -1/k when a negative feedback kI is appliedfeedback kI is applied.

nt.encircleme no /153.0 and 0/14.0 , 8.0/1 ∞<−<<−<−−<−<∞− kkk


24

system. loop closedin pole RHPonesont encirclemeone 4.0/18.0 −<−<− ksystem. loop closedin pole RHP twoso ntsencircleme two53.0/10 <−< k

Chapter 4

The Inverse Nyquist Stability Criterion

Theorem 4-7 (Generalized Nyquist theorem)

f ( ) i h hidd bl d h bl ( i h ill ) l h hIf G(s) with no hidden unstable modes, has Po unstable (Smith-McMillan) poles, then the closed-loop system with return ratio –kG(s) is stable if and only if the characteristic loci of kG(s), taken together, encircle the point -1, Po times anticlockwise.

NPZ 1)()(1 −+ =− NPZ sGskG

Theorem 4-7’ (Generalized Inverse Nyquist theorem)

1)(1)(1 −++ =− NPZ skGskG ⇒

If G(s) with no hidden unstable modes, has Zo unstable (Smith-McMillan) zeros, then the closed-loop system with return ratio –kG(s) is stable if and

l if th h t i ti l i f i G( )/k t k t th i l th i t

NPZ

only if the characteristic loci of invG(s)/k, taken together, encircle the point -1, Zo times anticlockwise.

= NZZ⇒


25

1/)(inv1/)(inv1 −++ =− NPZ ksGksG 1)()(1 −+ =− NZZ sGskG⇒

Chapter 4


Example 4-3: Let1)( =sG

)10)(5()(

++=

ssssG

Suppose that G(s) has no hidden modes, check the stability of system for different values of k.1

)10)(5(1

++=

sssλ0)( =− sGIλ )10)(5(1 ++=− sssλ

polesunstableTwo)750(750 >−<− kk poles.unstableTwo)750(750 >−<− kk

Stable.)7500(0750 <<<−<− kk

-750 pole. RHP One)0(0 <−< kk


26

Chapter 4

Nyquist arrays and Gershgorin bands

The Nyquist array of G(s) is an array of graphs (not necessarily closed) th ijth h b i th N i t l f ( )curve), the ijth graph being the Nyquist locus of gij(s) .

Theorem 4-8 (Gershgorin’s theorem)

:follows as elements diagonal thearound circles of unions in two containedare of seigenvalue The . dimensions ofmatrix complex a be Let ZmmZ iλ×

mizzm

ijj

ijiii ,....,2,1,1

=<− ∑≠=

λ mizzm

ijj

jiiii ,....,2,1,1

=<− ∑≠=

λ

The ‘bands’ obtained in this way are called Gershgorin bands, each is composed of Gershgorin circles.


27

composed of Gershgorin circles.

Chapter 4

Nyquist arrays and Gershgorin bandsNyquist array, with Gershgorin bands for a sample systemNyquist array, with Gershgorin bands for a sample system

If all the Gershgorin bands l d th i t 1 thexclude the point -1, then

we can assess closed-loop stability by counting the

encirclements of -1 by theencirclements of -1 by the Gershgorin bands, since this

tells us the number of encirclements made by the y

characteristic loci.

f h h i b d f ( ) l d h i i h h ( ) iIf the Gershgorin bands of G(s) exclude the origin, then we say that G(s) is diagonally dominant (row dominant or column dominant).

The greater the degree of dominant ( of G(s) or I+G(s) ) – that is, the narrower


28

the Gershgorin bands- the more closely does G(s) resembles m non-interacting SISO transfer function.

Chapter 4


And in general:


29

Chapter 4


Diagonal dominance of a matrix G(s): mm×

Row Diagonal dominance: If for all s on the Nyquist contour,

misgsgm

ijj

ijii ...,,2,1)()(1

=>∑≠=

Column Diagonal dominance: If for all s on the Nyquist contour,

ij≠

misgsgm

jiii ...,,2,1)()( =>∑Ali Karimpour Feb 2011

30ijj

jiii1∑≠=

Chapter 4









31


Chapter 4

Coprime Factorizations over Stable Transfer Functions

Two polynomials m(s) and n(s), with real coefficients, are said to be coprime ifp• their greatest common divisor is 1 or

• they have no common zeros or• there exist polynomials x(s) and y(s) such that 1)()()()( =+ snsysmsx

Exercise1 : Let n(s)=s2+5s+6 and m(s)=s find x(s) and y(s) if n and m are coprime

Si il l t t f f ti ( ) d ( ) i th t f t bl t f

Exercise1 : Let n(s) s +5s+6 and m(s) s find x(s) and y(s) if n and m are coprime. Exercise 2 : Let n(s)=s2+5s+6 and m(s)=s+2 show that one cannot find x(s) and y(s) in

x(s)m(s)+y(s)n(s)=1

Similarly, two transfer functions m(s) and n(s) in the set of stable transfer functions are said to be coprime over stable transfer functions if there exists x(s) and y(s) in the set of stable transfer functions such that


321)()()()( =+ snsysmsx Bezout identities

Chapter 4


Definition 4-3 Two matrices M and N in the set of stable transfer matrices are right coprime over the set of stable transfer matricesmatrices are right coprime over the set of stable transfer matrices if they have the same number of columns and if there exist matrices X and Y in the set of stable transfer matrices s tXr and Yr in the set of stable transfer matrices s.t.

[ ] INYMXNM

YX rrrr =+=⎥⎦

⎤⎢⎣

⎡N ⎦⎣

Similarly, two matrices and in the set of stable transfer matrices are left coprime over the set of stable transfer matrices if they have the

N~M~

are left coprime over the set of stable transfer matrices if they have the same number of rows and if there exist two matrices Xl and Yl in the setof stable transfer matrices such that


33[ ] IYNXMYX

NM lll

l =+=⎥⎦

⎤⎢⎣

⎡ ~~~~

Chapter 4


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

1−= NMP

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

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

NMP ~~ 1−=

stableofsetin thecoprimeareand~~lcfandrcfIn NMNM


34matricesfunction ansfer tr stable ofset in the coprimeare and,, lcf and rcfIn NMNM

Chapter 4


Remember: 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 −=


35

In Matrix Fraction Description D(s) and N(s) are polynomial matrices

Chapter 4


Theorem 4-9 Suppose P(s) is a proper real-rational matrix and

⎤⎡⎥⎦

⎤⎢⎣

⎡≅

DCBA

sP )(

is a stabilizable and detectable realization. Let F and L be such that A+BF and A+LCare both stable, and define

⎤⎡ −+⎤⎡

LBBFAYM ⎥

⎤⎢⎡ +−+

⎤⎡LLDBLCA

YX)(

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+≅⎥

⎦

⎤⎢⎣

⎡ −

IDDFCIF

XNYM

l

l 0⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−≅⎥

⎦

⎤⎢⎣

⎡

− IDCIF

MN

YX rr 0)(

~~

Then rcf and lcf of P are:

NMP ~~ 1−=1−= NMP


36

NMP= NMP

Exercise: Let P(s)=(s+1)/(s+2) find two different rcf for P.

Chapter 4


The right coprime factorization of a transfer matrix can be given a feedback control interpretation.

BA&

P

DuCxyBuAxx

+=++&

Fxvu +=BvxBFAx ++= )(& the transfer matrix from v to u is

⎤⎡ BBFAand that from v to y is

⎤⎡ BBFADvxDFCy

vFxu++=

+=)( ⎥

⎦

⎤⎢⎣

⎡ +≅

IFBBFA

sM )( ⎥⎦

⎤⎢⎣

⎡++

≅DDFCBBFA

sN )(

)()()( N )()()( 1MNP)()()( 1MN )()(P


37

)()()( svsNsy = )()()( 1 sMsNsP −=⇒)()()( 1 susMsN −= )()( susP=

Exercise3 : Derive a similar interpretation for left coprime factorization.

Chapter 4









38


Chapter 4

Stabilizing Controllers

Theorem 4-10 Suppose P is stable.Then the set of all stabilizing controllersThen the set of all stabilizing controllersin Figure can be described as

1)( −= PQIQK )( −= PQIQK

for any Q in the set of stable transfer matrices and non singular. )()( ∞∞− QPI

Proof:11 )()()( −− +=⇒=−⇒−= PKIKQQPQIKPQIQK System is stable.

Now suppose the system is stable, so define then stable, is )( 1−+ PKIK

KKPIPKIKQ 11 )()( −− +=+= KKPQQ =+⇒


39

KKPIPKIKQ )()( +=+= KKPQQ +⇒

1)( −−= PQIQKsor nonsingula is )()( ∞∞− QPI

Chapter 4


Example 4-4 For the plant

)2)(1(1)(

++=

sssP

Suppose that it is desired to find an internally stabilizing controller so that ySuppose that it is desired to find an internally stabilizing controller so that yasymptotically tracks a ramp input.

Solution: Since the plant is stable the set of all stabilizing controller is derived from

let so r,nonsingula is )()( such that Q stableany for )( 1 ∞∞−−= − QPIPQIQK

+ bas3++

=s

basQ

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

)3)(2)(1(11)(11 1

++++−+++

=+++

+−=−=+−=−= −

sssbassss

sssbasPQPKIPKTS


40

)3)(2)(1()3)(2)(1( ++++++ ssssss

3611

++

=ssQ

Chapter 4


NMNMP

P11 .~~ andmatrix rational

-realproper a be Let :11-4 Theorem−− ==

rrll YXXYK 11

controller gstabilizin a exists Then there−− ==

IXNYM

MN

YX

l

lrr =⎥⎦

⎤⎢⎣

⎡ −⎥⎦

⎤⎢⎣

⎡

− ~~ Where

Proof. See “Multivariable Feedback Design By Maciejowski”


41

Chapter 4

Remember

Theorem 4-9 Suppose P(s) is a proper real-rational matrix and

⎤⎡⎥⎦

⎤⎢⎣

⎡≅

DCBA

sP )(

is a stabilizable and detectable realization. Let F and L be such that A+BF

and A+LC are both stable, and defineand A LC are both stable, and define

⎤⎡ −+⎤⎡

LBBFA ⎤⎡ +−+⎤⎡

LLDBLCAYX

)(

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+≅⎥

⎦

⎤⎢⎣

⎡ −

IDDFCIF

XNYM

l

l 0⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−≅⎥

⎦

⎤⎢⎣

⎡

− IDCIF

MN

YX rr 0)(

~~


42

Chapter 4



43Proof. See “Multivariable Feedback Design By Maciejowski”

Chapter 4


Example 4-5 For the plant

1)2)(1(

1)(−−

=ss

sP

The problem is to find a controller that

1. The feedback system is internally stable.2. The final value of y equals 1 when r is a unit step and d=0.3. The final value of y equals zero when d is a sinusoid of 10 rad/s and r=0.

0]01[10

3210

Clearly ==⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡−

= DCBA

To derive coprime factorization let F = [1 -5] and L = [-7 -23]T

clearly A+BF and A+LC are stable.

⎤⎡ −+⎤⎡

LBBFA ⎤⎡ +−+⎤⎡

LLDBLCA )(


44⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+

+≅⎥

⎦

⎤⎢⎣

⎡ −

IDDFCIF

LBBFA

XNYM

l

l 0⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

++≅⎥

⎦

⎤⎢⎣

⎡

− IDCIF

LLDBLCA

MN

YX rr 0)(

~~

Chapter 4


Solution: The set of all stabilizing controller is:Solution: The set of all stabilizing controller is:

1))(( −−+= llll NQXMQYK

⎥⎥⎥⎤

⎢⎢⎢⎡ −+

≅⎥⎦

⎤⎢⎣

⎡ −IF

LBBFA

XNYM

l

l 0⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ +−+≅⎥

⎦

⎤⎢⎣

⎡

− IDCIF

LLDBLCA

MN

YX rr 0)(

~~⎥⎦⎢⎣ +⎦⎣ IDDFCl ⎥⎦⎢⎣ −⎦⎣ IDCN

2 38972108)1)(2(1 ++=

−=

−−==

ssXsYssMN 2222 )1(,

)1(,

)1(,

)1( +=

+=

+=

+=

sX

sY

sM

sN ll

2 38972108)1)(2(~1~ ++−−− sssss


452222 )2(389,

)2(72108,

)2()1)(2(,

)2(1

+++

=+

=+

=+

=s

ssXs

sYs

ssMs

N rr

Chapter 4


Solution: The set of all stabilizing controller is:

1))(( −−+= llll NQXMQYK

Clearly for any stable Q the condition 1 satisfied

To met condition 2 the transfer function from r to y must satisfyTo met condition 2 the transfer function from r to y must satisfy

To met condition 3 the transfer function from di to y must satisfy

5.36)0(1))0(~)0()0()(0( =⇒=+ rrr QMQYN→+= )()~()( srMQYNsy rr

To met condition 3 the transfer function from di to y must satisfy

jjQjNjQjXjM rrr 9062)10(0))10(~)10()10()(10( +−=⇒=−Now define 11)( ++= xxxsQ

→−= )()~()( sdNQXNsy irr


46

2321 )1(1)(

++

++=

sx

sxxsQr

Exercise 6: Derive QrExercise 5: Derive transfer function from di to y.Exercise 4: Derive transfer function from r to y.

Chapter 4









47


Chapter 4

Strong and Simultaneous

Practical control engineers are reluctant to use unstable controllers, especially when

the plant itself is stable.p

If the plant itself is unstable, the argument against using an unstable controller is

less compelling.p g

However, knowledge of when a plant is or is not stabilizable with a stable controller

is useful for another problem namely, simultaneous stabilization, meaning stabilizationis useful for another problem namely, simultaneous stabilization, meaning stabilization

of several plants by the same controller.

Simultaneous stabilization of two plants can also be viewed as an example of aSimultaneous stabilization of two plants can also be viewed as an example of a

problem involving highly structured uncertainty.

A plant is strongly stabilizable if internal stabilization can be achieved with a controller


48

A plant is strongly stabilizable if internal stabilization can be achieved with a controller

itself is a stable transfer matrix.

Chapter 4

Strong and Simultaneous

Theorem 4-13: P is strongly stabilizable if and only if it has an even number

of real poles between every pairs of real RHP zeros( including zeros at infinity)of real poles between every pairs of real RHP zeros( including zeros at infinity).

Proof. See “Linear feedback control By Doyle”.


49

Chapter 4

Exercises


50

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