Advanced Feedback Control

8/3/2019 Advanced Feedback Control

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Advanced Feedback Control

. . . Models, Identification, Control, Robustness.

Monterrey, June 2007

Advanced Feedback Control – p.



Various structures of feedback

We will study various structures of feedback laws.Those structures depend on whether the feedback isapplied

• from the all state vector (if it can be measured),

• of from the ouput vector.

Besides, the feedback itself can be either• static (the control vector entries are linearcombinations of the measurements),

• or dynamic (the control vector becomes the ouputof dynamic system - the controller - from whichthe measurement vector is the input).




Various structures of feedback

Therefore, three kinds of structures will be considered• static state feedback (usually simply referred to

as state feedback),

• static ouput feedback,

• dynamic ouput feedback.

Indeed, dynamic state feedback is rarely used.

The system model to be considered is simply a

realization

x = Ax + Bu

y = Cx + Du

(1)

(x ∈ IRn, u ∈ IRm and y ∈ IR p) Advanced Feedback Control – p.



Static state feedback

This is a control law that might have been studied inthe case of SISO sytstems

A

C yu

D

B + +

+

+xxH yc +

+

K





The corresponding mathematical description is

u(t) = Hyc(t) + Kx(t). (2)

with

• K ∈ IRm×n: state feedback matrix;

• H ∈ IR p×n: feedforward matrix;• yc ∈ IR p: reference vector.





The induced feedback model is given by

x = (A + BK )x + BHyc

y = (C + DK )x + DHyc.(3)

• K is computed to ensure stability and either to

possibly reach transient performances (poleplacement) or to minimize some criterion (e.g.LQ control, optimal control).

• H is rather computed to reach staticperformances.


t t t t



tat c output ee ac

Assume that not all the entries of x are measured butonly the entries of y.

A

C yu

D

B + +

+

+xxH yc +

+

F




t t t t



tat c output ee ac

If D = O (no direct transmission to make simpler)then the closed-loop model is

x = (A + BF C )x + BHycy = Cx

(5)

If D = Othen the control vector u complies with

u = Hyc + F Cx + F Du

⇔ (II m − F D)u = Hyc + F Cx


tat c o tp t ee ac



tat c output ee ac

⇔ u = (II m − F D)−1H

yc + (II m − F D)−1F

C

⇔ u =ˆ

H yc +ˆ

F C (6)This leads to the following closed-loop model:

x = (A + BF C )x + BHyc

y = (C + DF C )x + DHyc.(7)

One can compute F and H for design purpose anddeduce F and H which are implemented in practice.

Advanced Feedback Control – p. 1

tat c output ee ac



tat c output ee ac

Another possibility is to modify the control law:

u = Hyc + F (y − Du) = Hyc + F y (8)

y = y − Du = Cx ∈ IR p is the new "measure" onehas to built (it’s part of the controller) so that one gets

A

Ê

C yu

D

B + +

+

+xxH yc +

+

F +

−y


tat c output ee ac



tat c output ee ac

Then, the closed-loop model directly depends on F and H :

x = (A + BF C )x + BHycy = (C + DF C )x + DHyc

(9)

F and H are computed to get satisfactoryperformances. Note however that, with this kind of structure, it might be preferable to measure u too.


ynam c ouput ee ac



ynam c ouput ee ac

A

Ê

C yu

D

B + +++xxH yc +

+

F 3

F 4

Ê

F 1

+

+ zF 2

+

+

z


ynam c ouput ee ac



ynam c ouput ee ac

The control law is given by

z = F 1z + F 2y

u = F 3z + F 4y + Hyc,(10)

where z ∈ IRlis the state vector of the feedback

system.

The transfer matrix of this controller is

GF (s) = F 3(sII l − F 1)−1F 2 + F 4. (11)


ynam c ouput ee ac



ynam c ouput ee ac

Linking controller and process realizations yields

(II m − F 4D)u = F 3z + F 4Cx + Hyc

u = (II m − F 4D)−1F 4 Cx+ (II m − F 4D)−1F 3 z+ (II m − F 4D)−1H yu = F 4 Cx+ F 3 z+ H y

Consider a concatenation of process and controllerstate vectors ξ′ = [x′ z′]′ to get:

ξ = A + Bˆ

F 4C Bˆ

F 3F 2C + F 2DF 4C F 1 + F 2DF 3

ξ + Bˆ

H F 2DH

yc

y =

C + DF 4C DF 3

ξ + DHyc.

(12)Advanced Feedback Control – p. 1

ynam c ouput ee ac



ynam c ouput ee ac

Let the next augmented dynamic model be defined:

ξ = Aξ + Bu

y = Cξ + Du,(13)

where ξ ∈ IRn+l and

A = A O

O Ol ; B = B O

O II l ;

C = C O

O II l ; D = D O

O Ol .


ynam c ouput ee ac



ynam c ouput ee ac

Also let some control law (static ouput feedback) beapplied on this model:

u = F y + Hyc, with (14)

F = F 4 F 3F 2 F 1

an H = H

Ol,m , (15)

After some few calculation, one gets the same

closed-loop model as the one obtained by applyingthe dynamic feedback on the original process model.

⇒ Applying a dynamic output feedback controller ona linear model is equivalent to applying a static feedback gain on an augmented s stem. Advanced Feedback Control – p. 1







Reference



Reference

The developments in this part are actually very easy toproduce whith quite simple calculation and matrixmanipulations. Only one reference might deserve to

be quoted, where the dynamic controller is formulatedas a static one applied on an augmented system:

P. Hippe et J. O’Reilly.Parametric compensator design.International Journal of Control, Vol 45(4), p.1455-1468, 1987.


oup ng etween c anne s



oup ng etween c anne s

The purpose in this part is to highlight the inherentdifficulty of controlling MIMO models due tocoupling bewteen the various inputs to outputs

channels.

For example, consider a process with two ouputs y1

and y2 and to control signals u1 and u2.It is interesting two control y1 that should track somereference yc1 as well as to control y2 that should track some reference yc1.

Unfortunately, in most cases, those control lawscannot be designed independently. An action on yc1,

an thus on u1 has an influence on y2 and the other wayaround.


n examp e: c em ca reactor




Mixer

Products (output chemicals)

Outgoing fluid

Heating/ Cooling

Reacting chemicals (input chemicals)

Entering fluid






The temperature of the reactor is highly influent onthe quality of the reaction which is itself influent onthe temperature of the environment.

The process includes two inputs:

• the rate (concentration) of entering chemicals,

• the temperature of the heating/cooling fluid,

and two ouputs

• the rate (concentration) of outgoing chemicals,• the temperature inside the reactor.






So two channels input/ouput:• one for the chemical rates;

•the other one for the temperature.

Why is there a couplig between the two channels?

• If the temperature of the outside fluid changes (in

order to control that of the reactor), then thequality of the reaction is modified and the rates of the products are changed.

• If the rates of the entering chemicals are changed(in order to control the rates of products), then thereaction is of course more or less important

inducing a change of temperature because thereaction either provides or absorbs heat.Advanced Feedback Control – p.

not er numer ca examp e




A = −1 0

0 0, 5

; B =

1 1

1 1

C = −1 0

0 1

; D =

0 0

0 0

.

It is an unstable square system (see the poles). Theemphasized entries are those responsible for thecoupling.

The corresponding transfer matrix is G(s) =

1

s2 + 0, 5s − 0.5 s

−0, 5 s

−0,5

s + 1 s + 1 = G11(s) G12(s)

G21(s) G22(s) .





a a p

Assume that one ignores (!!!) the coupling transfersG12(s) and G22(s) and that one designs somecontrollers only for diagonal transfers.

From u1 to y1:

G11(s) =s−

0, 5

s2 + 0, 5s − 0, 5 =1

s + 1 ,

From u2 to y2:

G22(s) =s + 1

s2 + 0, 5s − 0, 5=

1

s − 0, 5.





p

For each first order channel, ones applies

F i Gii(s)H iyci + yi

−

ui

WithH 1 = −1

,F 1 = −0.5

,H 2 = 0.5

andF 2 = 1

,one gets the two following closed-loop models:

G11 = G22 =

1

1 + 2s.







p

With such a simple (and false) reasoning, one shouldget the next step response:

−1

−0.5

0

0.5

1

T o : O u t ( 1 )

From: In(1)

0 2 4 6 8 10 12−1

−0.5

0

0.5

1

T o : O u t ( 2

)

From: In(2)

0 2 4 6 8 10 12

Step Response

Time (sec)

A m p l i t u d e





p

...whereas one actually getsStep Response

Time (sec)

A m p l i t u d e

−0.5

0

0.5

1From: U(1)

T o : Y ( 1 )

0 2 4 6 8 10 12−2

−1.5

−1

−0.5

0

0.5

T o : Y ( 2 )

From: U(2)

0 2 4 6 8 10 12


Hence...



... from these examples, one can conclude that:

• The coupling cannot always be neglected;

• The responses can be drastically distorted.

Indeed, some models can even be unstable due tocouplings...


Thus...



One can formulate several problems:

• Static decoupling (only for steady-stateresponse),

• Tansient decoupling (also for the transientresponse).

Those problems can be handled

• either from a frequency point of view (frequency

decoupling),• or from a time point of view (state-space

approach).


req. app. stat c ecoup ng



Some possibility is to use feedforward control:yc u y

H (s) G(s)

Y (s) = G(s)U (s) = G(s)H (s)Y c(s)

⇒ y∞ = limt→∞ y(t) = lim

s→0(sY (s)) = lim

s→0(sG(s)H (s)Y c(s)).

If ones considers that all the reference entries yci aresteps of magnitude αi, one has to satisfy:





y∞ = lims→0

sG(s)H (s)

1

s

α1

...

α p

=

α1

...

α p

⇔ G(0)H (0)α1

.

..

α p

= α1

.

..

α p

⇔ G(0)H (0) = II p. (18)

So H (s) = H (0) = H (constant feedforward matrix)has to check (18).





• If m = p (square model) then H = G(0)−1;• If m > p then H can be a pseudo-inverse of G(0)

(for example, the Moore-Penrose one);

• If m < p then no generic solution: not enoughactuators compared with the number of outputs.

So the limits are:• m ≥ p;

• G(0) must be of full rank;• The process must be stable or be stabilized firstbecause this is only a feedforward control.


xamp e



G(s) =

20(s + 1)

(s2 + 3s + 12)(s + 2)

−130(s − 0, 3)

s2 + 2s + 80

−10(s − 3)

(s2 + 3s + 12)(s + 8

15(s − 1)(s2 + 4s + 12)(s + 2)

43(s + 1(s2 + 2s + 32)(s + 2)

30(s + 1)s2 + 2s + 122

−9(s

−4)

s2 + 2s + 52

30(0, 5s + 4)

s2 + 2s + 412

3, 2

s + 2

This is a square stable model ⇒

H = G(0)−1 =

0, 833 −0, 572 −0, 075

0, 972 0, 927 −0, 332

−0, 537 0, 079 0, 718

.


req. app. ynam c ecoup ng



Just some idea that can sometimes be used !

The idea is to compute H (s) such that

Q(s) = G(s)H (s) checks

qii(s) = 0 ∀i ∈ {1,...,p}

qij(s) = 0 ∀{i, j = i} ∈ {1,...,p}2

But it is illusory to solve such constraints so one cansimply try to reach

|qij(iω)

|<< 1

∀{i, j

= i

} ∈ {1,...,p

}2


req. app. ynam c ecoup ng



There are several techniques in the literature based onthat simple idea (whose efficiency has still to beproved (author’s note)). With those techniques, one

has to check that the useful transfers qii(s)• have no instable zeros;

• are strictly proper;

• should be preferably of weak order.

In any case, one has to keep in mind that a decoupling

procedure does not ensure other performances andshould be accompanied by other control laws toguarantee stability, transient behaviour, and so son.




ecoup ng w t t me approac



These p independent linear 1st order differentialequations correspond, in Laplace’s domain, to:

Y i(s)Y ci(s) = −qiis − qii

∀i ∈ {1,...,p}.

that is to some transfers with unit static gain and one

pole qii.





If one looks for a state feedback such that thesetransfers are obtained, ones can write

x = Ax+Bu = Ax+B(Hyc+Kx) = (A+BK )x+BHyc

⇔ C x = y = (CA + CBK )x + CBHyc.

to be identified to

y = QCx − Qyc,

leading to (assuming that W = (CB)−1

exists)

K = W (QC − CA)

H = −W Q.(19)





Thus a very simple technique that is unfortunatelyonly useful when m = p = n.

Indeed,• It cannot be extended to static output feedback;

•it cannot be used when D

= O.

... so very restrictive !


ecoup ng: ome conc us on



• Frequential approach and feedforward sometimesefficient (not alone) for static decoupling.

• Time approach rarely used (except under drastic

constraints) but see the next part for some attemptto transient decoupling.

• There exist other methods of decoupling such as

the "relative gain" method whose efficiency hasnot convinced the author of these slides.

• Other techniques consists in tracking a reference

model which is usually chosen with no coupling...but it is not a decoupling approach in itself.

As a conclusion, decoupling is fundamental but sodifficult !Advanced Feedback Control – p.

References



• P. T. Tham

Notes - An introduction to Decoupling control.Department of Chemical and Process Engineering, University of

Newcastle upon Tyne, England... for the example of chemical reactor

•Course Notes, Chapter 6 : Analysis and Design of Multivariable

Control Systems.

Electrical Engineering Department, State University of Binghamton,

New-York, USA... for decoupling by time approach and other insights.

• E. H. Bristol

On a new measure of interaction in multivariable process control.

IEEE Transactions on Automatic Control, Vol 11, p. 133-134... for the

reader interested in "relative gain approach".• J. P. Corriou.

Commande des procédés.

Lavoisier Editions, TEC&DOC Collection, 1996 (in French, sorry !),

for connected information.Advanced Feedback Control – p.

genstructure ass gnment



Motivation: assigning the poles and possibly theassociated eigenvectors in order to try to shape thetransient response of the closed-loop system.

It can be way to obtain some transient input/ouputdecoupling.

Techniques based upon eigenstructure placement arealso called Modal Control.


atr x e genstructure



λ is an eigenvalue of A

∈lC n×n iff

P (λ) = det(λII n − A) = 0. (20)

A owns n eigenvalues λi

(which will be assumeddisctinct for the sake of conciseness). This set isreferred to as the spectrum λ(A).

A ∈ IRn×n ⇒ λ(A) is closed under conjugation.

There exists n non zeros vectors vi

∈lC n, called right

eigenvectors, such that

Avi = λivi ∀ i ∈ {1,...,n}. (21)

One should talk about eigendirections since they canbe multiplied by any non zero scalar. Advanced Feedback Control – p.



ee ac mo e e genstructure



Closed-loop model eigenstructure=eigenstructure of its state matrix

Ac = A + BFC.

⇒

Acvi = (A + BF C )vi = λivi ∀i ∈ {1,...,n}u′

iA = u′i(A + BF C ) = λiu

′i ∀i ∈ {1,...,n}

U ′V = II n

Ac = A + BF C = V ΛU ′.Advanced Feedback Control – p.

ee ac mo e e genstructure



Remark: In practice, the matrices are real meaningthat not only λ(Ac) but also the sets of eigenvectorsare closed underconjugation.

• Input directions:

wi = F CV i ∀i ∈ {1,...n}.• Output directions:

l′i = u′iBF ∀i ∈ {1,...n}.


n uence o t e e genva ues



It can be easily proved that the free response of amodel to an intial condition is

x(t) =

ni=1

αieλitvi. (26)

• Re(λi) < 0 ∀i otherwise there are non vanishingterms (instability).

•

|Re(λi)

| ց⇒the term (mode) reduces faster.

• |Im(λi)| ր⇒ the term induces strongeroscillation (none if λi is real).

So λ(A) has an influence on stability, settling time,oscillations, characterizing the transient behaviour.Advanced Feedback Control – p.

n uence o t e e genvectors



Consider the pertubed closed-loop model

x = (A + BF C )x + BHyc + Bd

y = Cx.

(27)

With the basis change x = V ξ, V being the modalmatrix of Ac = A + BF C :

ξ = Λξ + UBHyc + U Bd

y = C V ξ .(28)

Also consider the identity matrix:

II n = e1 . . . en , (29)Advanced Feedback Control – p.




• yc[i] has no effect on λ j iff

u′ jBHei = 0.

⇒ left eigenvectors distribute the effects of thereferences on the eignvalues

• λi has no effect on x[ j] (resp. y[ j]) iff

e′ jvi = 0.

⇒ right eigenvectors vi (resp. Cvi) distribute theeffects of the eigenvalues on the state entries(resp. outputs).





• d[i] has no effect on λ j iff

u′ jBei = 0.

⇒ left eigenvectors distribute the effects of somedisturbances on the eigenvalues

• λ j has no effect on u[ j] iff (less obvious)

e′ jwi = 0.

⇒ input directions distribute the effects of theeigenvalues on the control entries.


n uence o t e e genstructure



The effect of the environment on the system dynamicsis mainly described by the left eigenstructure whereasthe effect of these dynamics on the system outputs ismainly described by the right eigenstructure (IbrahimChouaib).

Remark: It can also be proved that eigenvectors have

an influence on the local sensitivity of eigenvalueswith respect to additive unstructured uncertaintyaffecting the state matrix (not detailed here).


tate ee ac ass gnment



Pole Placement Problem: find K ∈ IRm

×n

such that λ(Ac = A + BK ) equals some specified set.

The computation of feedforward matrix H will beconsidered later.

There is always some solution provided the pair(A, B) is controllable.







Characterisitc subspacesBecause for one λ, the associated v and w = Kvcomply with

(A − λII n)v + Bw = O,

then v

∈S (λ) where

S (λ) = {v ∈ lC n | ∃w ∈ lC m | (A−λII n)v +Bw = O}(A, B) controllable ⇒ dim(S (λ) = m.

⇒only (m

−1) should be exploited to assign

v ∈ S (λ), exactly what is offered by K .Advanced Feedback Control – p.




Define:

T λ =

A − λII n B

∈ lC n×(n+m),

Rλ = N λ

M λ

= Ker(T λ) with N λ ∈ lC n×m, M λ ∈ lC m

and with some parameter vector z ∈ lC m, it comes

π = v

w = N λz

M λz ,

leading to admissible eigenvector v ∈ S (λ) and

associated input direction w.Advanced Feedback Control – p.




Choice of z

Assume vd is ome desired eigenvector (with forinstance zero entries to try to reach some decouplingproperties). One has to assign an admissible v as closeas possible to vd. Solving a classical least squareproblem leads to

z = (N ′λN λ)−1N ′λvd. (30)

Remark: It is possible to rather give specifications onvarious ui and then to deduce suitable vdi.





Theorem 1 There exists K ∈ IRm

×n

solving the problem iff

(i) vectors vi are linearly independent;

(ii) vi = v j when λi = λ j;

(iii) vi

∈S (λi).

In this event the unique solution is given by

K = W V −1 (31)

where

W = w1 . . . wn .Advanced Feedback Control – p.



utput ee ac ass gnment



Pole Placement Problem: find F ∈ IRm

×n

such that λ(Ac = A + BF C ) equals some specified set.

Remark: it is possible but dangerous to assign only part of the spectrum following the same kind of reasoning as for state feedback.

Nessary condition for solving the problem : (A,B,C )minimal.





About the dof:• At first fight, m × p entries in F so problem can

be solved if mp

≥n but not so simple.

• In 1975, Kimura proved that m + p > n ⇒generic assignability.

• Later (1981), it was proved that condition ismp ≥ n but in the field of complex matrices...but no need for a complex F !

• In 1996, Wang proved that sufficient condition inthe field of real matrices is mp > n but theassociated design method is not very tractable.

• In practice, the tractable (or even non iterative)techniques require that Kimura’s condition holds.





What to do if Kimura’s condition does not hold?• Assign only part of the spectrum,

•or apply a dynamic feedback.

If it holds, there are several techniques available withdifferent restrictions:

• "Polynomial" design;

• Parametric approach;

• Geometric approach (very elegant);• Coupled Sylvester Equations (my favourite !),

•and many others I may not know or that still have

to be found.Advanced Feedback Control – p.




Principle of the "Sylvester approach":

Solve the system:

AV − V Λ = −BW (right eigenstructure) (32)

U ′A − ΛU ′ = −L′C (left eigenstructure) (33)

Ker(U ′) = Im(V ) (orthogonality) (34)

The main idea: assign {λi, i ∈ {1,...,p}} and the

associated vi as well as {λi, i ∈ { p + 1,...,n}} andthe associated ui, while respecting the three aboveequations.







Simplified algorithm (cont’d):• Compute

N λi

M λi

= Ker( N λi) ∈ lC (n+m)×ri)

(it generically exists when m + p > n);• Choose p parameter vectors zi ∈ lC ri such that

V p = [Cv1,...,Cv p] = [CN λ1z1, ... , CN λpz p] (37)

is a full rank matrix;







With direct transmission D:

Just find F by the above technique to assign thespectrum of

Ac = A + BF C

and then deduce

F = F (II p + DF )−1.





m + p ≤ nIt is possible to design a dynamic feedback or order lin order to assign n + l poles but one has to satisfy

l ≥ n − m − p + 1. (40)

Hence, Kimura’s condition holds for the "augmentedsystem" (see part of the various feedback structures)and then one computes a static gain for thisaugmented system that corresponds to a dynamic gainfor the original system.

Some special cases can also be handled with static

gain (since 2006 !)Advanced Feedback Control – p.




Pole placement and feedforward

It might be possible (depending on dimensions) to

computeˆ

H and H = (II m − F D)ˆ

H such that

(

−(C + DF C )(A + BF C )−1B + D)H = II p. (41)

to ensure a unit static gain otherwise add integratorsbefore to solve the problem... but with integrators,

Kimura’s condition is harder to satisfy.





Simple example with MATLAB

Model and desired spectrum:

>> A=[1 4 5;0 2 6;1 0 3];

>> B=[1 1;1 0;0 0];

>> C=[1 0 0;0 1 0];

>> D=eye(2);

>> n=3;m=2;p=2;>> lambda=[-1 -2 -3];



Choice of L solution to "left" Sylvester equation:



Choice of Ln− p, solution to left Sylvester equation:

>> Lam_nmoinsp=diag(lambda(p+1:n))

Lam_nmoinsp =

-3

>> L_nmoinsp=[1;1];

>> U_nmoinsp=sylv(A’,-Lam_nmoinsp’,-C’*L_nmoinsp)

U_nmoinsp =

-0.3025

0.0420

0.2101



Computation of 1 its kernel v1 and w1:



Computation of

N 1, its kernel, v1 and w1:

>> NN1=[A-lambda(1)*eye(3) B;U_nmoinsp’ zeros(n-p,m)]

NN1 =

2.0000 4.0000 5.0000 1.0000 1.0000

0 3.0000 6.0000 1.0000 0

1.0000 0 4.0000 0 0

-0.3025 0.0420 0.2101 0 0

>> R1=null(NN1)

R1 =

-0.0364

-0.3074

0.0091

0.8675

0.3892

>> v1=R1(1:3);w1=R1(4:5);



The same for λ2.



The same for λ2.

>> NN2=[A-lambda(2)*eye(3) B;U_nmoinsp’ zeros(n-p,m)]

NN2=

3.0000 4.0000 5.0000 1.0000 1.0000

0 4.0000 6.0000 1.0000 0

1.0000 0 5.0000 0 0

-0.3025 0.0420 0.2101 0 0

>> R2=null(NN2)

R2 =

-0.0305

-0.2499

0.0061

0.9629

0.0975

>> v2=R2(1:3);w2=R2(4:5);Advanced Feedback Control – p.


Computation of W V and F :



Computation of W p

, V p

and F :

>> Wp=[w1 w2];Vp=C*[v1 v2];

>> F_hat=Wp*inv(Vp)

F_hat =

-285.8000 31.0000

242.8000 -30.0000Verification of the cloded-loop spectrum:

>> eig(A+B*F_hat*C)

ans =

-3.0000

-2.0000

-1.0000



Deduction of F and computation of H :



p

>> F=F_hat*inv(eye(p)-D*F_hat)

F =

-0.9773 0.0227

0.1780 -0.7897

>> H_hat=inv(-(C+D*F_hat*C)*inv(A+B*F_hat*C)*B+D)

H_hat =

58.5529 -84.1059

-49.6706 71.3412

>> H=(eye(p)-F*D)*H_hat

H =

-0.2161 0.3106

-0.0962 0.1406Advanced Feedback Control – p.


Construction of the closed-loop model and verification of the static gain



p g

>> Ac=(A+B*inv(eye(m)-F*D)*F*C);

>> Bc=B*inv(eye(m)-F*D)*H;

>> Cc=(C+D*inv(eye(m)-F*D)*F*C);

>> Dc=D*inv(eye(m)-F*D)*H;

>> -Cc*inv(Ac)*Bc+Dc

ans =

1.0000 -0.0000

0.0000 1.0000


References• B. C. Moore

On the flexibility offered by state feedback in multivariable systems



On the flexibility offered by state feedback in multivariable systems

beyond closed-loop eigenvalue assignment .

IEEE Transactions on Automatic Control, Vol 21, p.689-692, 1976, for

the state feedback

• H. Kimura.

Pole assignement by gain output feedback . IEEE Transactions on

Automatic Control, Vol 20, p.509-516, 1975, for m + p > n

• J. Rosenthal et F. Sottile.Some remarks on real and complex output feedback . Systems and

Control Letters, Vol 33, p.73-80, 1997, for other conditions including

Wang’s one


References (2)• V. L. Syrmos et F. L. Lewis.

Output feedback eigenstructure assignment using two Sylvester



p f g g g y

equations.

IEEE Transactions on Automatic Control, Vol 38(3), p.495-499,

1993.for static output feedback (Sylvester approach)

• A. N. Andry, E. Y. Shapiro et J. C. Chung.

Eigenstructure assignment for linear systems

IEEE Transactions on Aerospace and Electronic Systems, Vol 19(5),

p.711-729, 1983, for the meaning of eigenstructure• O. Bachelier, J. Bosche et D. Mehdi

On pole placement via eigenstructure assignment approach.

IEEE Transactions on Automatic Control, Vol 51(9), p.1554-1558,

2006, for the case m + p = n

... and all the references therein.


Model reduction



The main idea: Approximate a high order (n) linearmodel by a reduced order (r) model to make thedesign simpler.

S

= x = Ax + Bu

y = Cx + Du →S

r = xr = Arxr + Bru

yr = C rxr + Dru(42)


ua ty o r

Several criteria can be considered:



• Preserve the dominant poles (i.e. neglect the fast(high frequency) dynamics);

• Approximate the input/ouput behaviour: for asame input vector, yr should be as close aspossible to y.


ome ex s tng met o s

According to these criteria, a non complete list of



g pexisting techniques is as follows:

• By modal approach;

• By "agregation";• By Schur decomposition;

•By minimization of norm (ex:

H∞-norm);

• By balancing transformation (the only onepresented here and maybe the most known!).


Balanced reduction

Only valid for asymptotically stable minimal models



y y p ybut there exists a (not well known) extension tounstable models.

The idea: neglect the dynamics of the state entries thatare the less controllable and observable in S.

But how to quantify controllability and observability?

Answer: through the grammians (or Gramm

matrices).


Balanced reduction

Controllability grammian



W c =

∞

0

eAτ BB′eA′τ dτ. (43)

which solves Lyapunov equation

AW c + W cA′ =

−BB′. (44)

Observability grammian

W o = ∞0

eA′

τ C ′

CeAτ dτ ⇒ (45)

A′W o + W oA =−

C ′C. (46)

(hence the stability assumption). Advanced Feedback Control – p.

Balanced reduction

The controllability grammian can be interpreted in



terms of energy.

There exists a basis in IRn in which W c is diagonal. In

this basis each diagonal entry wci of W c is thereciprocal of the minimum energy required to(asymptotically) bring the state vector to

[0, . . . , 0, 1, 0 . . . , 0]′. Thus, it can be seen as acontrollability index of xi.

For observability, the reasoning is based upon dualityto conclude that in the "diagonal" basis, woi is anobservability index of xi.


Balanced reduction

LS

b d ib d b h i l f i (A B C)



Let be decribed by the triplet of matrices (A,B,C )assuming D = 0 for the sake of conciseness.

Theorem: There exists a a full rank matrix T such that the realisationS = (T −1AT,T −1B,CT ) = (A, B, C ) is balancedi.e. both grammians equal to the same diagonal

matrixW o = W c = Σ.

It means that in this basis, for each entry xi, thecontrollability and observability indices are the same

⇒one has to neglect the dynamics of the less

controllable and observable xi.Advanced Feedback Control – p.

Balanced reduction

In the balanced basis, the model is (here, D is kept)



˙x1 = A11x1 + A12x2 + B1u

˙x2 = A21x1 + A22x2 + B2u

y = C 1x1 + C 2x2 + Du.

(47)

The limit between the preserved dynamics (that of x1

and the neglected ones (that of x2) depends on apossible gap in the diagonal entries of Σ.


Balanced reduction

The idea is then to neglect the dynamics of x2

, the lessll bl d b bl f ( h h l



2controllable and observable part of x (thus the lessinfluent on the input/ouput behaviour) by imposing˙x2 = 0. This technique is sometimes called the"singular perturbations approximation"

It is also possible to simply truncate x by imposing

x2 = 0 but it does not preserve the static gain so it israther rough as a reduction.


Balanced reduction

So, the reduced model S is given by



r

Sr

= xr = Arxr +Bru

yr = C rxr + Dru.

(48)

where xr = x1 and

Ar = A11 − A12A−122 A21

Br = B1

−A12A−1

22 B2

C r = C 1 − C 2A−122 A21

Dr = D − C 2A−122 B2.

(49)


Balanced reduction

MATLAB corresponding functions



• minreal: Compute the minimal realization of original system S;

• balreal: Compute the balanced realization

(A, B, C, D) of S;

• modred: Compute the final realization(Ar, Br, C r, Dr) of reduced system Sr.


References• E. J. Davison.

A method for simplifying linear dynamic systems.

IEEE T ti A t ti C t l V l 11(1) 93 101 1966



IEEE Transactions on Automatic Control, Vol. 11(1), p. 93-101, 1966,

for a first glimpse to (non presented) modal approach

• B. Moore.

Principal component analysis in linear systems: controllability,

observability and model reduction

IEEE Transactions on Automatic Control, Vol. 26, p. 17-31, 1981, for

the bases of balanced reduction• L. Fortuna et G. Muscato

Model reduction via singular perturbation approximation of

normalized right coprime factorizations.

Proceedings of "European Control Conference ECC’95", Roma, Italy,

September 1995. for the balanced reduction of unstable models.


Norms

Definition and properties



A norm enables ones to compare an element withanother in a set which does not necessarily own a

relation of order. It is usually denoted by ||u||• whereu is the concerned element and • stands for theconsidered norm.

(i) ||u||• ≥ 0

(ii) ||u||• = 0 ⇔ u = 0

(iii) ||au||• = |a|.||u||•, ∀a ∈ lC(iv) ||u + v||• ≤ ||u||• + ||v||• (triangular inequality)

(50)


Vector norms

Euclidean norm

I d t f l { } ∈ { lC n}2



Inner product of a couple {x; y} ∈ { lC n}2:

< x, y > =

ni=1

x∗i yi = x∗y (51)

From this inner product, one can define the Euclidean

norm of 2-norm (the most natural):

||x||2 = √< x, x > =

ni=1

x2i = √x∗x. (52)

There are many other vector norms not detailed here.Advanced Feedback Control – p.

Vector function norms

Now the vectors depend on some real or complexvariable (t or s)



variable (t or s).

L2 and

H2-norms

Let Ln2 be the set of vector functions X (s) ∈ lC n, with

s ∈ lC whose square can be "summed" along theimaginary axis:

||X

||2 = 1

2π ∞

−∞X (iω)∗X (iω)dω

1/2

<∞

.

(53)

||X

||2 is called the

L2-norm of X (

Lfor Lebesgue).

It can be shown that Ln2 is an Hilbert space.Advanced Feedback Control – p.





Vector function norms

L2-gain



LLet a mathematical operator R be defined over thefollowing sets:

R : Lnw

2 → Lnw

2

w(t)

→e(t)

Then,

GL2(R) = supw∈Hnw

2

||e||2||w||2(55)

is the

L2-gain of

Rwhich corresponds to the highest

energy gain associated with R.Advanced Feedback Control – p.



Matrix norm

and Σ, if q = min

{m, n

}, complies with



{ }

Σ =

σ1 0 · · · 0 0

0 σ2 · · · 0 0

... ... . . . ... ...

0 0 · · · σq 0

si q = m,

Σ =

σ1 0 · · · 0

0 σ2 · · · 0...

.... . .

...

0 0 · · · σq

0 0 · · · 0

si q = n,

Σ = diag{σ1, · · · , σq} si q = m = n.

(58)


Matrix norm

σi are the singular values of M :

¯(M ) ≥ ≥ ≥ (M ) ≥ 0



σ(M ) = σ1 ≥ σ2 ≥ · · · ≥ σq = σ(M ) ≥ 0.(59)

• rank (M ) = number of non-zero singular values= number of linearly independant rows orcolumns.

• M such that rank (M ) < min{m; n} is rank deficient, otherwise full rank .

•M square and rank deficient cannot be inverted

and owns n − r zero singular values.

• σi = eigenvalues of M M ∗ (si m ≤ n) or M ∗M (si n

≤m).

• M Hermitian ⇒ σi = |λi|. Advanced Feedback Control – p. 1

Matrix norm

σ(M ) is a norm called 2-norm because it is inducedby the Euclidean vector norm in the following way:



by the Euclidean vector norm in the following way:

σ(M ) = ||M ||2 = max

x = 0 ∈ lC n

||M x

||2

||x||2 = max

x = 0 ∈ lC n

x∗M

∗

M xx∗x

(60)

Besides

σ(M )

≤

||M x||2

||x||2 ≤σ(M ). (61)

shows that the gain from x to (M x) lies in the range

[σ(M );σ(M )]




L∞ − H∞-norm of a transfer

LetRL

ne×nw

∞(resp.

RHne×nw

∞), be the set of proper

transfer matrices G(s) ∈ lC ne×nw (i e with finite



RL RHtransfer matrices G(s) ∈ lC (i.e. with finitedirect transmission) and with no pole on the

imaginary axis I

(resp. with no pole over lC +

∪ I ).

Then the L∞ (resp. H∞-norm) simply corresponds tothe frequency for which the transfer is the highest inthe sense of the 2-norm. Hence:

||G||∞ = supω ||G(iω)||2 = sup σ(G(iω)).ω

(63)


Transfer matrix norm



20log(σ(G(iω)))

20log(σ(G(iω)))

Pulsation ω (rad/s)

G a i n ( d B

)

20log(||G||∞)

Figure 1: Gain Bode diagramm in the MIMO caseAdvanced Feedback Control – p. 1







H2-norm

Stochastic interpretation



If the wi are white noises scaled such thatW (iω)W (iω)∗ = II nw

, then the expectation of the

inner product of the induced input vector checks

ne

i=1 E (e∗i (t)ei(t)) = ||G||

2

2.(68)

Remark: For this reason the so-called

H2-problem can

be related to the celebrated LQG-problem


2 2-norm

H2-norms in terms of gain

R i di h h H i h h L i h



Reminding that the H∞-norm is the the L2-gain thenthe

H2-norm checks

||G||2 = supW (s)

∈Hnw∞

||E ||2

||W

||∞. (69)

which, unlike for the H∞-norm, is not verymeaningful.


H2-norm

H2-norms and grammians

R i d f h ll bili d b bili



Remind of the controllability and observabilitygrammians:

W c =

∞0

eAtBBT eAT tdt ; W o =

∞0

eAT tC T CeAtd

(70)that satisfy

AW c + W cA

T

= −BB

T

,AT W o + W oA = −C T C. (71)


H2-norm

H2-norms and grammians (cont’d)

Th



Then

||G||2 = trace(BT W oB) = trace(CW cC T ).(72)

This provides an analytical expression of the

H2-norm

which is valid for calculation.

Unlike the

H∞-norm, the

H2-norm can be computed

directly through its analytic expression.


approac

For simplicity, only real matrices are considered.

M t i i liti



Matrix inequalities

This is related to the notion of sign definition (partialorder of Löwner).

M ∈ IRn×

n

is positive definite (M > 0) (resp.semi-positive definite (M ≥ 0)) iff

xT M x > 0 (resp.≥ 0) ∀x = 0 ∈ IRn. (73)

M is negative definite (M < 0) (resp. semi-negative

definite (M ≤ 0)) iff (−M ) > 0 (resp. (−M ) ≥ 0).Advanced Feedback Control – p. 11

atr x nequa t es

In practice mostly symmetric matrices are handled so

sign definition is now considered only for thosematrices



matrices.With this assumption,

M < (≤) 0 ⇔ λmax(M ) < (≤) 0

M > (≥) 0 ⇔ λmin(M ) > (≥) 0

(74)

A straightforward notation isM > (≥) N ⇔ M − N > (≥) 0

M < (≤) N ⇔ M − N < (≤)0.(75)


LMI bases

Example of MI:

M = M T = AX3 + (X3)T AT + eBY Y T (eB)T < 0



M = M = AX + (X ) A + e Y Y (e ) < 0

with, for instance, X and Y that are unknown.

Among all possible MI only two will be consideredbecause they are often encountered:

• LMI : Linear matrix inequalities, that can besolved,

• BMI : Bilinear matrix inequalities.


ropert es o

•If M 1 < 0 and M 2 < 0 then one can stack these

properties in one single MI:



M 1 O

O M 2 < 0. (76)

• If M = M T is such that

M =

M 1 M 2M T

2 M 3

< 0, (77)

then M 1 < 0 and M 2 < 0.


LMI

LMI are interesting because they can be solved !

The most famous LMI come from



The most famous LMI come from...

Theorem 2 Let the autonomous continuous (resp.

discrete) model

x = Ax (resp. xk+1 = Axk)

This model is asymptotically stable iff ∃P = P T > 0such that

AT P + P A < 0, (resp. − P + AT P A < 0).

(Lyapunov’s inequality and its discrete counterpartdue to Stein). Advanced Feedback Control – p. 11

BMI

Consider the next MI with respect to X and Y :

AX + XT AT + XBY + Y T BT XT > 0



AX + X A + XBY + Y B X > 0

Because of the two last terms, it is bilinear.

Unfortunately those BMI are very difficult to solve inspite of some existing software.

Some crucial control problems are unfortunately veryeasily formulated as BMI, not as LMI (ex: static

output feedback stabilization).


A useful tool !

Schur’s lemma

Let S Q = QT and R = RT be matrices



Let S , Q = Q and R = R be matrices.

Q S

S T R < 0 ⇔

R < 0

Q − SR−1

S T

< 0(78)

This lemma enables to handle Stein’s inequality as an

LMI wrt A:

−P + AT P A O

O −P < 0 ⇔ −P AT P

P A −P < 0.Advanced Feedback Control – p. 11



e process

• u: control vector issued from control law;

• w: disturbance to be rejected (in practice, notalways actual exogeneous signals);



always actual exogeneous signals);

• y: measured ouput for the purpose of control;• e: vector of signals to be controlled.

E (s)

Y (s)

= P (s)

W (s)

U (s)

, with P (s) =

P ew(s) P eu(s)

P yw(s) P yu(s)

.

The idea is to reduce the transfer from w to e.


e process

P (s) = D + C (sII − A)−1

B, with (79)



B = Bw Bu ; C = C e

C y ; D = Dew Deu

Dyw Dyu

.

(80)

Or in other words

x(t) = Ax(t) + Bww(t) + Buu(t)

e(t) = C ex(t) + Deww(t) + Deuu(t)y(t) = C yx(t) + Dyww(t) + Dyuu(t)

x(t) ∈ IRn

, w(t) ∈ IRnw

, u(t) ∈ IRnu

, e(t) ∈ IRne

et y(t) ∈ IRny

.Advanced Feedback Control – p. 1

e process

Assumptions

• A1 : (A; Bu) and (A; C y) are respectively



A1 : (A; Bu) and (A; Cy) are respectively

stabilisable and detectable;

• A2 : Dyu = Ony,nu;

• A3 : Dew = One,nw(only for H2-problem).

A1 is rather classical and completely compulsory.A2 is just technical and induces no loss of generality.A3 is necessary for the

H2-norm to be defined.


Static controller

One considers a static state feedback controller

u = Kx. (81)



u Kx. (81)

In such a case, since y = x, the process model reducesto

x(t) = Ax(t) + Bww(t) + Buu(t)e(t) = C ex(t) + Deww(t) + Deuu(t).

(82)


ynam c contro er

One considers a dynamic output feedabck controller

xc(t) = Acxc(t) + Bcy(t)(83)



c( ) c c( ) + cy( )

u(t) = C cxc(t) + Dcy(t)(83)

where xc(t) ∈ IRn. Thus,

K (s) = Dc + C c(sII n − Ac)−1

Bc. (84)In the H2-case (not studied in these slides), one has toconsider a strictly proper controller i.e. Dc = Onu,ny

.


ose - oop mo e

What is the state-space model of F(P (s), K (s)) ?

With static controller



Under Assumption A2:

x

e = A

f B

f C f Df

x

w = A + B

uK B

wC e + DeuK Dew

In the

H2-case, Df = 0.


ose - oop mo e

With dynamic controller

Still under Assumption A2:



p

xxc

e

=

Af Bf

C f Df

xxc

w

=

A + BuDcC y BuC c Bw + BuDcDyw

BcC y Ac BcDyw

C e

+ Deu

Dc

C y

Deu

C c

Dew

+ Deu

Dc

Dyw

x

xc

w

.

In the H2-case, Df = 0.


•-pro em

Problem 1 Let P (s) and γ •

> 0 be given. Also let

assumptions A1 to A3 hold. Find a stabilizing (staticor dynamic) feedback such that



y ) f

||F(P (s), K (s))

||•< γ

•.

If • = ∞, then Assumption A3 can be omitted.

With no additional constraints (such as weightingmatrices), the problem is referred to as standard.


•-pro em

H2 or

H∞?

... not exactly the same philosophy.



In the H∞-case, one looks after the L2-gain i.e. thehighest possible energy transfer or, from the frequencyviewpoint, the energy transfer at the worst frequency.

In the H2-case, one considers energy transfer over thewhole frequency range, not focusing on the worst.


•-synt es s

In the following slides, some solutions are given for

• the H2-design by state static feedback,



• the

H∞-design by state static feedback,

• the H∞-design by output dynamic feedback,




2 stat c es gn

The idea is that X 2 is a matrix upper bound of either

the observability grammian (X 2 > W o: primalversion) or of the controllability grammian (X 2 > W c:d l i ) S h L i d



dual version). So the Lypaunov equations used to

calculate the grammians are here replaced by LMIs.

The dual version enables ones to derive some K .


2 stat c es gn

Theorem 3 There exists u = Kx , K

∈IRnu×n such

that the H2-norm of F(P (s), K (s)) is less than γ 2 iff there exist two symmetric positive definite matrices



{X 2; T

} ∈ {IRn×n

}2 , and a matrix L

∈IRnu×n such

that

AX 2 + BuL + X 2AT

+ LT

BT u Bw

BT w −II nw

< 0,

−T C eX 2 + DeuLX 2C T

e + LT DT eu −X 2

< 0,

trace(T ) = γ 22 . Advanced Feedback Control – p. 1

2 stat c es gn

In this event, K is given by

K = LX −12 .



• Notice that with various LMI solvers, it ispossible to minimize γ 2 while satisfying the LMI

constraints.• Also notice that X 2 can be inverted since it is

positive definite.






∞ stat c es gn

Theorem 4 There exists u = Kx , K

∈IRnu×n such

that the H∞-norm of F(P (s), K (s)) is less thanγ ∞ > 0 iff there exists a symmetric positive definite



matrix X ∞ ∈

IRn×n , and a matrix L

∈IRnu×n such

that

AX ∞ + B

uKX ∞ + X ∞AT + X ∞K T BT

u(•

) (•

)

BT w −γ ∞II nw (•)

C eX ∞ + DeuKX ∞ Dew −γ ∞II ne

< 0.

(87)


∞ stat c es gn

In this event, K is given by

K = LX −1∞ .



• Notice that with various LMI solvers, it ispossible to minimize γ ∞ while satisfying the LMI

constraints.• Also notice that X ∞ can be inverted since it is

positive definite.


∞ ynam c es gnCondition for solvability

Under assumptionsA

1-A

2, the H∞ dynamic problemcan be solved iff there exist R = RT and S = S T suchthat



that

N R O

O IInw

T

AR + RAT RC T e Bw

C eR −γ ∞II ne Dew

BT w DT ew −γ ∞II nw

N R O

O IInw

<

N S O

OII ne

T

AT S + SA SBw C T e

BT wS −γ ∞II nw DT

ew

C e Dew −γ ∞II ne

N S O

OII ne

< 0

R II n

II n S

≥ 0

(88)Advanced Feedback Control – p. 1

∞ ynam c es gn

where Span(N R) = Ker([BT u DT

eu]) and

Span(N R) = Ker([C y Dyw]).

M d t ll i t iff



Moreover, a n-order controller exists iff

rang(II n − RS ) = n. (89)

• It is also possible to achieve

min γ ∞ under the LMI constraints.R=RT ;S =S T


∞ ynam c es gn

How to recover K (s) ?

Achieve the singular value decomposition of

(II RS) in order to obtain M ; N IRn×n 2



(II n

−RS ) in order to obtain

{M ; N

} ∈ {IR

}such that

M N T = II n

−RS.

Then X ∞ = S N

N T

−M −1RN

is solution to the condition of the bounded real lemmawhich therefore becomes an LMI (thus solvable) wrt

(Ac, Bc, C c, Dc). Advanced Feedback Control – p. 1

∞ ynam c es gn

Example

e =

¾

z

u

¿

x 1 ( )+ x

w =

¾

b

v

¿



K (s)

u y

u

x

+

+

1

s

v

+

z P (s)

K (s)

+b x

y

Find the state-space model, write the LMI system,deduce the minimum value of γ ∞ and explain how to

recover K (s).


∞ ynam c es gn

State-space model:

b



x = [0]x + 1 0 b

v w

+ [1]u

z

u

e

=

1

0

x +

0 0

0 0

b

v

+

0

1

u

y = [1]x +

0 1

b

v

+ [0]u.

(90)


∞ ynam c es gn

which correspond to these matrices:

A = 0 Bw =

1 0

Bu = 1



C e = 1

0

Dew =

0 00 0

Deu =

01

C y = 1 Dyw = 0 1 Dyu = 0.

Assumptions A1 and A2 are easily verified.


c ∞ ynam c es gn

[BT u DT

eu] = [1

|0 1] = [C y Dyw]

⇒N R = N S

1 00 1

.



−1 0 A is scalar then so are R and S . The first LMI to besolved is

1 0 0 0

0 1 0 0

−1 0 0 0

0 0 1 0

0 0 0 1

T

0 R 0 1 0

R −γ ∞ 0 0 0

0 0 −γ ∞ 0 0

1 0 0 −γ ∞ 0

0 0 0 0

−γ ∞

1 0 0 0

0 1 0 0

−1 0 0 0

0 0 1 0

0 0 0 1


∞ ynam c es gn

which becomes

⇔

γ ∞ −R −1 0

−R γ ∞ 0 0

> 0



⇔ −1 0 γ ∞

00 0 0 γ ∞

> 0.

and, by Schur’s lemma, is equivalent to

γ ∞ > 0,

γ 2∞ − 1 − R2 > 0.


∞ynam c es gn

In a totally similar way, the 2nd inequality reduces to

γ 2∞ − 1 − S 2 > 0.

R 1



whereas the 3rd one , i.e. R 1

1 S ≥ 0 leads to

R ≥ 0

RS − 1 ≥ 0

⇔

S ≥ 0

RS − 1 ≥ 0.


∞ynam c es gn

The whole of constraints yields

γ 2∞ − 1 > min (max{R2; S 2}).R,S



which shows that the optimum is reached for

R = S = 1 ⇒ γ ∞ =

√2.

But in this case, the optimal controller is not of ordern = 1. Anyway, for a suboptimal case RS = 1, one

gets

M = −N =√

RS − 1,




o e p acement

LMI region

Any set D ⊂ lC defined by

T



D = {z ∈ lC | α + βz + β T

z < 0} (91)where α = αT ∈ IRl×l and β ∈ IRl×l is an openLMI-region of order l.

These regions are always convex and, if α and β arereal (as in the above definition), symmetric wrt the

real axis.




o e p acement

LMI formulation of a disc

... of center ρ and radius r.

|z − ρ| < r ⇔ (z − ρ)(z − ρ) − r2 < 0



⇔ −r + (z − ρ)1r

(z − ρ) < 0.

Applying Schur’s lemma, it comes

−r z − ρ

z − ρ −r = α + βz + β T z < 0

⇔α =

−r −ρ

−ρ

−r < 0 ; β =

0 1

0 0 .


-sta ty

A matrix (or by extension a model) is said

D-stable

when its eigenvalues lie inside D.

Theorem 5 Let be an LMI-region. A matrix A is



DD-stable iif ∃X D = X T D > 0 such that

M D

(A, X D

) = α⊗

X D

+ β ⊗

(AX D

) + β T

⊗(X

DAT ) < 0.

This is an LMI wrt X D or xrt A.


D-stabilizationOne considers a static state feedback control law.

Theorem 6 Let D be an LMI-region. P (s) is

-stabilizable by state feedback iif X = X T > 0



Dy f f

∃ D Dand L such that

M D(A, Bu, X D, L) = α ⊗ X D + β ⊗ (AX D) + β T

⊗ (X DAT

)+

β ⊗ (BuL) + β T ⊗ (LT BT u ) < 0.

In this event, K is given by K = LX −1

D .


xt synt es s

One still considers a static state feedback control law.

Theorem 7 Let D be an LMI-region. P (s) is-stabilizable by state feedback that ensures



D||F(P (s), K )||• < γ • ∀ • ∈ {∞; 2} if ∃{X = X T > 0; T = T T ; L} such that

Z ∞(P (s), X , L , γ ∞) =

AX + BuL + XAT + LT BT u (•) (•)

BT w −γ ∞II nw (•)

C eX + DeuL Dew

−γ ∞II ne

Z 21(P (s), X , L) =

AX 2 + BuL + X 2A′ + L′B′u Bw

BT w

−II nw

< 0


xt synt es s

Z 22(P (s), X , L , T ) = −T C eX 2 + DeuL

X 2C T e + LT DT eu −II n

< 0

2



trace(T ) < γ 2

2

M D(A, Bu, X , L) = α⊗X +β ⊗(AX )+β T ⊗(XAT )+β ⊗(BuL)+β T ⊗(LT BT u ) <

In this event, K is given by K = LX −1.

•A great interest in LMI is that one can stack several

LMI constraints with preserving the LMI nature of theproblem but...


xt synt es s

• ...The condition is only sufficient because one

imposes the constraint

X = X 2 = X ∞ = X D.



This is referred to as the Lyapunov ShapingParadigm.

•In such a mixt synthesis only one criterion is

minimized (either γ 2 or γ ∞) and the other one isarbitrarily chosen. Another possibility is tominimize a weighted objective function.

• The design of dynamic controllers is alsopossible but harder.


Introduction to robust controlPolytopic uncertainty

(...still considering state feedback)



The process model is assumed to be uncertain (notprecisely known) but also assumed to belong to afamily of models defined by

M = M (τ ) = A(τ ) Bw(τ ) Bu(τ )

C e(τ ) Dew(τ ) Deu(τ ) =N

j=1

(τ jM j


o ytop c uncerta nty

in which τ = [τ 1,...,τ N ]T contains the coefficients of

a convex combination (i.e. τ j ≥ 0 andN

j=1

(τ j) = 1)



and M j are the vertices of a so-called polytope of matrices:

M j = A j B jw B ju

C je D jew D jeu

. (92)

Remarque 1 There are many possible descriptions of uncertainties either in the frequency domain (i.e.affecting the transfer matrix) or in the time domain(i.e. affecting the state-space model). Here only few

time-domain uncertainties are introduced. Advanced Feedback Control – p. 1


Why polytopic uncertainty?

It has a very interesting special case i.e the affineparametric uncertainty:



M = M 0 +

p

i=1

(δiN i), (93)

M 0 is the nominal part, the matrices N i are known andthe δi are unknown parameters obeying

δimin≤ δi ≤ δimax

∀i ∈ {1,...,p}. (94)

In this case, the polytope has N = 2 p vertices.



Example:

M =

−1 + δ1 δ2 3

1 −2 + 2δ1 0

where

|δ1| ≤ 0, 5

|δ2| ≤ 0, 2.⇔



M 0 =

−1 0 3

1 −2 0

; N 1 =

1 0 0

0 2 0

; N 2 =

0 1 0

0 0 0

which corresponds to the polytopic description:

[M 1

|M 2

|M 3

|M 4] =

−0, 5 0, 2 3 −0, 5 −0, 2 3 −1, 5 0, 2 3 −1, 5 −0, 2 3

1 −1 0 1 −1 0 1 −3 0 1 −3 0



Yes, but why polytopic uncertainty?

Simply because it is easily handled through LMImachinery.



For example, assume one wants to analyze the robuststability of a polytopic matrix A with two vertices A1

and A2. This matrix is robustly stable if

∃P = P T > 0 such that

AT

1

P + P A1 < 0 & AT

2

P + P A2 < 0.

The condition is only sufficient since only one uniqueP is considered and it is not a function of the

ucnertainty (one talks about quadratic stability).Advanced Feedback Control – p. 1

o ust m xt synt es s

Theorem 8 Consider an uncertain process model

P (s, τ ) and some LMI-region D. There exists aD-stabilizing state feedback K that ensures||F(P (s), K )||• < γ • ∀ • ∈ {∞; 2} if



∃{X = X T > 0; T = T T ; L} such that

Z ∞j

=Z

(P j

(s), X , L , , γ ∞

) < 0

Z 21j= Z (F(P j(s), X , L)) < 0

Z 22j= Z (F(P j(s), X , L , T )) < 0 ∀ j ∈ {1,...,N },

trace(T ) < γ 22M jD = M D(A j + B ju, X , L) < 0

In this event, K is given by K = LX −1. Advanced Feedback Control – p. 1

o ust m xt synt es s

There are two reasons why the condition is

conservative:

• Lyapunov shaping paradigm,



• Quadratic stability.

Avoiding the shaping paradigm is quite difficult butthere exist techniques that consider a matrix X (τ )(i.e. which is dependent on the uncertainty).


Robustness andH∞

Many authors consider that

H∞ approach belongs to

the realm of robust control whereas it is simplydisturbance rejection.



The reason is as follows:Consider the uncertain matrix (Linear FractionalTransform (LFT)-based uncertainty):

A = A + B∆C (95)

where ∆ = ∆(II − D∆)−1 and ||∆|| ≤ ρ, with ∆

complex (this special LFT is called norm-boundeduncertainty).Problem: find the largest value of ρ such that A is

Hurwitz for all ∆ .This value is the so-called complexstabilit radius. Advanced Feedback Control – p. 1



ReferencesThere are so many (I used quite many in French but...)

• S. Boyd, L. El Ghaoui, E. Feron and V. Balakrishnan

Linear Matrix Inequalities in System and Control.

Volume 15 of SIAM Studies in Applied Mathematics, USA, 1994, still



the ultimate reference for LMI in control• P. Gahinet and P. Apkarian

A LMI approach To H∞ control.

International Journal of Robust and Nonlinear Control, Vol 4,

p.421-448, 1994 for the solution to dynamic H∞-problem

• M. Chilali and P. Gahinet

H∞ design with pole placement constraints.

IEEE Transactions on Automatic Control, Vol 41(3), p.358-367, 1996.,

for the pole placement in LMI-regions


References (2)

• P. Khargonekar, I. R. Petersen and K. Zhou,.

Robust stabilization of uncertain linear systems: Quadtric

stabilizability and H∞ control.IEEE Transactions on Automatic Control, Vol 35, p.356-361, 1990, for

the robust stability against norm-bounded uncertainty)

C. Scherer and S. Weiland.



• Lectures Notes DISC Course on Linear Matrix Inequalities in Control

available for downloading from the web, very general and elegant

approach covering many of the aspects of these slides and even much

more.

... and all the references therein.

See also the very good slides proposed by D. Henrion on his web page:

http://www.laas.fr/ henrion


Hoping you enjoyed theseslides, the control communitynow needs you to investigate



many of the problems that arestill unsolved !


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Advanced Feedback Control