Lmi Slides

7/27/2019 Lmi Slides

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Linear Matrix Inequalities in Control

Guido Herrmann

University of Leicester

Linear Matrix Inequalities in Control p. 1/43

Presentation prepared for

Summerschool at the Universityof Leicester, September 2006

Affiliation now:

Department of MechanicalEngineering, University of Bristol


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Presentation

1. Introduction and some simple examples

2. Fundamental Properties and Basic Structure of Linear Matrix Inequalities (LMIs)

3. LMI-problems

4. Tricks in Matrix Inequalities - Approaches to create LMIs from Matrix Inequalities

(a) Congruence Transformation

(b) Change of Variables

(c) Projection Lemma

(d) S-procedure

(e) Schur Complement

5. Examples (L2-gain computation, non-linearities, etc)

6. Conclusions



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Introduction - A Simple Example

A linear system

x = Ax

is stable if and only if there is a positive definite P for

V(x) = xTPx (i.e. V(x) > 0 f or x = 0)

and

xTPAx +xTATPx < 0

x= 0



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A linear system

x = Ax


V(x) = xTPx (i.e. V(x) > 0 f or x = 0)

and

xTPAx +xTATPx < 0

x= 0

The two matrix inequalities involved here are

PA +ATP < 0

and

P > 0.



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A linear system

x = Ax


V(x) = xTPx (i.e. V(x) > 0 f or x = 0)

and

xTPAx +xTATPx < 0

x= 0

The two matrix inequalities involved here are

PA +ATP < 0

and

P > 0.

The matrix problem here is to find P so that these inequalities are satisfied. Theinequalities are linear in P.



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Introduction - LQR-optimal control

We would like to compute a state feedback controller u = Kx controlling

x = Ax +Bu

with an initial condition of x(0) = x0.

The cost function

J=

0

(xTQx + uTRu)dt

is to be minimized. We know that the solution to this problem is

K = R1BTP, ATP + PA + PBR1BTP + Q = 0

and J = minu

0

(xTQx + uTRu)dt = xT0 Px0.

How can we express this problem in terms of an LMI?



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Introduction

In control the requirements for controller design are usually

1. Closed Loop Stability

2. Robustness

3. Performance

4. Robust Performance

Control design requirements are usually best encoded in form of an optimizationcriterion

(e.g. robustness in terms of L2/small gain-requirements, performance via linear

quadratic control, H -requirements etc.)



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Introduction

We have seen that stability of a linear autonomous system can be easily expressed viaa linear matrix inequality

We will see that linear quadratic control problems can be expressed in terms of LMIs

L2 or H analysis/design problems can be expressed as LMI-problems

Some classes of nonlinearities are easily captured via matrix inequalities

This creates a synergy which allows to express a control design problem via differentseemingly contradictive requirements

For LMIs, very reliable numerical solution tools are available



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Fundamental LMI properties

A matrix Q is defined to be positive definite if it is symmetric and

xTQx > 0 x = 0

This is signified by

Q > 0



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xTQx > 0 x = 0


Q > 0

Likewise, Q is said to be positive semi-definite if it is symmetric and

xTQx 0 x thus Q 0



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xTQx > 0 x = 0


Q > 0

Likewise, Q is said to be positive semi-definite if it is symmetric and

xTQx 0 x thus Q 0

A matrixQ

is negative definite if it is symmetric and

xTQx < 0 x = 0 thus Q < 0

or negative semi-definite if it is symmetric and

xTQx 0 x = 0 thus Q 0Linear Matrix Inequalities in Control p. 7/43


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The Basic Structure of an LMI

Any linear matrix inequality (LMI) can be easily rewritten as

F(v) = F0 +m

i=1

viFi > 0

where v

R

m is a variable and F0,Fi are given constant symmetric matrices.

This matrix inequality is linear in the variablesvi.

For instance for the simple linear matrix inequality in the symmetric P

PA +ATP < 0

the variables v Rm are defined via P Rnn. Hence, m = n(n+1)2 in this case!



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The Basic Structure of an LMI

Another very generic way of writing down an LMI is

F(V1,V2, . . . ,Vn) = F0 + G1V1H1 + G2V2H2 + . . .

= F0 +n

i=1

GiViHi > 0

where the unstructured ViR

qipi are matrix variables, ni=1 qi

pi = m. We seek to find

Vi as they are variables.

The matrices F0,Gi,Hi are given.

From now on, we will mainly consider LMIs of this form.



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System of LMIs

A system of LMIs is

F1(V1, . . . ,Vn) > 0

.

.

. > 0

Fp(V1, . . . ,Vn) > 0

where

Fj(V1, . . . ,Vn) = F0j +n

i=1

Gi jViHi j

This can be easily changed into a single LMI ...



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System of LMIs

Lets define F0,Gi, Hi,Vi as

F0 =

F01 0 0 0

0 F02 0 0

0 0 . . . 0

0 0 0 F0p

= diag(F01, . . . ,F0p)

Gi = diag(Gi1, . . . ,Gip)

Hi = diag(Hi1, . . . ,Hip)

Vi = diag(

Vi, . . .V

i)



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System of LMIs

Lets define F0,Gi, Hi,Vi as

F0 =

F01 0 0 0

0 F02 0 0

0 0 . . . 0

0 0 0 F0p

= diag(F01, . . . ,F0p)

Gi = diag(Gi1, . . . ,Gip)

Hi = diag(Hi1, . . . ,Hip)

Vi =

diag(V

i, . . .V

i)

We then have the inequality

Fbig

(V1, . . . ,Vn) := F0 +n

i=1GiVi Hi > 0

which is just one single LMI.

But be aware that this time the new variable Vi is structured, i.e. not all elements of Vi are

free parameters!



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Different classes of LMI-problems: Feasibility Problem

We seek a feasible solution {V1, . . . ,Vn} such that

F(V1, . . . ,Vn) > 0

We are not interested in the optimality of the solution, only in finding a solution, whichsatisfies the LMI and may not be unique.



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Different classes of LMI-problems: Feasibility Problem

We seek a feasible solution {V1, . . . ,Vn} such that

F(V1, . . . ,Vn) > 0

We are not interested in the optimality of the solution, only in finding a solution, whichsatisfies the LMI and may not be unique.

Example: A linear system

x = Ax

is stable if and only if there is a matrix P satisfying

PA +AT

P < 0

and

P > 0.


bl b


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LMI-problems: Linear Objective Minimization

Minimization (or maximization) of a linear scalar function, (.), of the matrix variables Vi,

subject to LMI constraints:

min(V1, . . . ,Vn) s.t.such that, subject to

F(V1, . . . ,Vn) > 0

where we have used the abbreviation s.t. to mean such that.


LMI bl Li Obj i Mi i i i


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LMI-problems: Linear Objective Minimization

Minimization (or maximization) of a linear scalar function, (.), of the matrix variables Vi,

subject to LMI constraints:

min(V1, . . . ,Vn) s.t.such that, subject to

F(V1, . . . ,Vn) > 0

where we have used the abbreviation s.t. to mean such that.

Example: Calculating the H norm of a linear system.

x = Ax +Bw

z = Cx +Dw

the H norm of the transfer function matrix Tzw from w to z is computed by:

min s.t.

A

TP + PA PB C T

BTP I DTC D I

< 0, P > 0.

The LMI variables are P and ! The value of is unique, P is not.


LMI bl G li d i l bl


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LMI-problems: Generalized eigenvalue problem

min s.t. F1(V1, . . . ,Vn) +F2(V1, . . . ,Vn) < 0F2(V1, . . . ,Vn) < 0

F3(V1, . . . ,Vn) < 0

Note that in some cases, a GEVP problem can be reduced to a linear objectiveminimization problem, through an appropriate change of variables.


LMI bl G li d i l bl


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F3(V1, . . . ,Vn) < 0


Example: Bounding the decay rate of a linear system.

The decay rate is the largest such that

x(t) exp(t)x(0), >, x(t)

Lets choose the Lyapunov function V(x) = xTPx > 0 and ensure that V(x) 2V(x).The problem of finding the decay rate could be posed as

min s.t. ATP + PA + 2P < 0,P < 0,


LMI problems: Generali ed eigenvalue problem


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F3(V1, . . . ,Vn) < 0


Example: Bounding the decay rate of a linear system.

The decay rate is the largest such that

x(t) exp(t)x(0), >, x(t)

Lets choose the Lyapunov function V(x) = xTPx > 0 and ensure that V(x) 2V(x).The problem of finding the decay rate could be posed as

min s.t. ATP + PA + 2P < 0, i.e. F1(P) := ATP + PAP < 0, i.e. F2(P) := 2P

i.e. F3(P) := I


Tricks: Congruence transformation


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Tricks: Congruence transformation

We know that for Q RnnQ > 0

and a real W Rnn such that rank(W) = n, the following inequality holds

W QWT > 0

Definiteness of a matrix is invariant under pre and post-multiplication by a full rank real

matrix, and its transpose, respectively.

Often W is chosen to have a diagonal structure.


Tricks: Change of variables


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By defining new variables, it is sometimes possible to linearise nonlinear MIs

Example: State feedback control synthesis

FindF such that the eigenvalues ofA +BF are in the open left-half complex plane

This is equivalent to finding a matrix F and P > for

(A +BF)TP + P(A +BF) < 0 or ATP + PA + FTBTP + PBF< 0

! Terms with products of F and P are nonlinear or bilinear !




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By defining new variables, it is sometimes possible to linearise nonlinear MIs

Example: State feedback control synthesis

FindF such that the eigenvalues ofA +BF are in the open left-half complex plane

This is equivalent to finding a matrix F and P > for

(A +BF)TP + P(A +BF) < 0 or ATP + PA + FTBTP + PBF< 0

! Terms with products of F and P are nonlinear or bilinear !

Multiply with Q := P1 > 0 (A very simple case of congruence transformation):

QAT +AQ + QFTBT +BFQ < 0

This is a new matrix inequality in the variables Q > 0 and F (still non-linear).




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QAT +AQ + QFTBT +BFQ < 0

Define a second new variable L = FQ

QAT +AQ +LTBT +BL < 0

We now have an LMI feasibility problem in the new variables Q > 0 and L.

Recovery of F and P by

F = LQ1, P = Q1.


Tricks: Schur complement


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Schurs formula says that the following statements are equivalent:

i. =

11 12

T12 22

< 0

ii. 22 < 0

11 12122 T12 < 0

The main use is to transform quadratic matrix inequalities into linear matrix inequalities.




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i. =

11 12

T12 22

< 0

ii. 22 < 0

11 12122 T12 < 0


Example: Making a LQR-type quadratic inequality linear (Riccati inequality)

ATP + PA + PBR1BTP + Q < 0

where P > 0 is the matrix variable and Q,R > 0 are constant.




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i. =

11 12

T12 22

< 0

ii. 22 < 0

11 12122 T12 < 0



ATP + PA + PBR1BTP + Q < 0

where P > 0 is the matrix variable and Q,R > 0 are constant.

The Riccati inequality can be transformed into

ATP + PA + Q PB R

< 0




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p


ATP + PA + PBR1BTP + Q < 0 is equivalent to

ATP + PA + Q PB

R

< 0

where P > 0 is the matrix variable and Q,R > 0 are constant. This inequality can be usedto minimize the cost function

J=

0(xTQx + uTRu)dt

for the computation of the state feedback controller u = Kx controlling

x = Ax +Bu

with an initial condition of x(0) = x0.




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p


ATP + PA + PBR1BTP + Q < 0 is equivalent to

ATP + PA + Q PB

R

< 0

where P > 0 is the matrix variable and Q,R > 0 are constant. This inequality can be usedto minimize the cost function

J=

0(xTQx + uTRu)dt

for the computation of the state feedback controller u = Kx controlling

x = Ax +Bu

with an initial condition of x(0) = x0. We know that the solution to this problem is

K = R1BTP, ATP + PA + PBR1BTP + Q = 0

and J = minu

0

(xTQx + uTRu)dt = xT0 Px0.




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p

The alternative solution to the optimization problem is given by the following

LMI-problem:

minxT0 Px0 s.t.

ATP + PA + Q PB

R < 0P < 0

for which the optimal controller is given by K = R1BTP.


Tricks: The S-procedure


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We would like to guarantee that a single quadratic function of x R

m

is such that

F0(x) 0 F0(x) := xTA0x + 2b0x + c0

whenever certain other quadratic functions are positive semi-definite

Fi(x) 0 Fi(x) := xTAix + 2b0x + c0, i {1,2, . . . ,q}




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We would like to guarantee that a single quadratic function of x R

m

is such that

F0(x) 0 F0(x) := xTA0x + 2b0x + c0

whenever certain other quadratic functions are positive semi-definite

Fi(x) 0 Fi(x) := xTAix + 2b0x + c0, i {1,2, . . . ,q}

Illustration:

Consider i = 1. We need to ensure F0(x) 0 for all x such that F1(x) 0.If there is a scalar, > 0, such that

Faug(x) := F0(x) + F1(x) 0 x s.t.F1(x) 0

then our goal is achieved.

Faug(x)

0 implies that F0(x)

0 if F1(x)

0 because F0(x)

Faug(x) if F1(x)

0.




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Faug(x) := F0(x) + F1(x) 0 x s.t.F1(x) 0Extending this idea to q inequality constraints:

F0(x)

0 whenever Fi(x)

0 (

)

holds if

F0(x) +q

i=1

iFi(x)

0, i

0 (

)




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Faug(x) := F0(x) + F1(x) 0 x s.t.F1(x) 0Extending this idea to q inequality constraints:

F0(x)

0 whenever Fi(x)

0 (

)

holds if

F0(x) +q

i=1

iFi(x) 0, i 0 ()

The S-procedure is conservative; inequality () implies inequality ()

Equivalence is only guaranteed for i = 1.

The is are usually variables in an LMI problem.


Tricks: The Projection Lemma


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We sometimes encounter inequalities of the form

(V) + G(V)HT(V) +H(V)TGT(V) < 0 ()

where V and are the matrix variables, is an unstructured matrix variable.

(.),G(.),H(.) are (normally affine) functions of V.




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(V) + G(V)HT(V) +H(V)TGT(V) < 0 ()



Inequality () is satisfied for some V if and only if

WT

G(V)(V)WG(V) < 0

WTH(V)(V)WH(V) < 0




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(V) + G(V)HT(V) +H(V)TGT(V) < 0 ()



Inequality () is satisfied for some V if and only if

WT

G(V)(V)WG(V) < 0

WTH(V)(V)WH(V) < 0

where WG(V) and WH(V) are the orthogonal complementsof G(V) and H(V), i.e.

WG(V)G(V) = 0 WH(V)H(V) = 0.

and [WTG(V)G(V)], [W

TH(V)H(V)] are both full rank.




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The main point is that we can transform a matrix inequality which is a function of two

variables, V and , into two inequalities which are functions of one variable:

(i) It can facilitate the derivation of an LMI.

(ii) There are less variables for computation.




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The main point is that we can transform a matrix inequality which is a function of two

variables, V and , into two inequalities which are functions of one variable:

(i) It can facilitate the derivation of an LMI.

(ii) There are less variables for computation.

It is often the approach is to solve for V using

WT

G(V)(V)WG(V) < 0

WT

H(V)(V)WH(V) < 0

and then for using

(V) + G(V)HT(V) +H(V)TGT(V) < 0

Note that this can be numerically unreliable!!


Examples: L2 gain- Continuous-time systems


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Linear systems: The H norm is equivalent to the maximum RMS (Root-Mean-Square)

energy gain, the H -gain of a linear system, the L2 gain.




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A system with input w(t) and output z(t) is said to have an L2 gain of if

z2 < w2 +, > 0

where w2 = t=0 w(t)w(t)dt.The L2 gain is a measure of the output relative to the size of its input.




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z2 < w2 +, > 0

where w2 = t=0 w(t)w(t)dt.The L2 gain is a measure of the output relative to the size of its input.

The H norm of x = Ax +Bw, z = Cx +Dw is given by:

min s.t.

ATP + PA PB C T

BTP I DTC D I

< 0, P > 0.




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ATP + PA PB C T

BTP

I DT

C D I < 0, P > 0.

The Schur complement gives

ATP + PA + 1CTC PB + 1CTDBTP + 1D

TC I+ 1DTD




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ATP + PA PB C T

BTP

I DT

C D I < 0, P > 0.



TC I+ 1DTD

In terms of xT wTT

, it follows that we need to find the minimum of so that

x

w

TATP + PA + 1C

TC PB + 1CTD

BTP + 1DTC I+ 1DTD

x

w

< 0, xT wTT = 0



T T


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ATP + PA PB C T

BTP

I DT

C D I < 0, P > 0.



TC I+ 1DTD

In terms of xT wTT

, it follows that we need to find the minimum of so that

x

w

TATP + PA + 1C

TC PB + 1CTD

BTP + 1DTC I+ 1DTD

x

w

< 0, xT wTT = 0

or

xTATPx +xTPAx +1

xTCTCx +xT(PB +

1

CTD)w + wT(BTP +

1

DTC)x + wT

1

DTDw wTw

= xTATPx +xTPATx + 2xTPBw + 1

zTz wTw < 0Linear Matrix Inequalities in Control p. 26/43



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Defining V = xTPx

V = xTATPx +xTPAx + 2xTPBw

Thus, we require:

xTATPx +xTPATx + 2xTPBw +1

zTz wTw = V+ 1

zTz wTw < 0




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Defining V = xTPx

V = xTATPx +xTPAx + 2xTPBw

Thus, we require:

xTATPx +xTPATx + 2xTPBw +1

zTz wTw = V+ 1

zTz wTw < 0

and integration in the interval [0,) implies

V(t = ) V(t = 0) +

t=0

1

zT(s)z(s)ds

t=0

wT(s)w(s)ds < 0

t=0

zT(s)z(s)ds <

t=02wT(s)w(s)ds + (V(t = 0)

V(t = ))

t=0zT(s)z(s)ds 0.

and

V(x(k+ 1)) V(x(k)) = V(x(k+ 1)) = xT(k)ATPAx(k) xT(k)Px(k) < 0 x(k) = 0

or

ATPA P < 0.


Examples: l2 gain- Discrete-time systems



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z2 < w2 +, > 0

where w2 =

k=0 w

T(k)w(k) .





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z2 < w2 +, > 0

where w2 =

k=0 w

T(k)w(k) .

For linear systems

x(k+ 1) = Ax(k) +Bw(k)

y = Cx(k) +Dw(k)

the value of the finite l2-gain, , (H norm; the maximum RMS energy gain) is:

min s.t.ATPA P + 1CTC ATPB + 1CTD

1D

TC I+BTPB + 1DTD

< 0

P < 0

for P > 0.Linear Matrix Inequalities in Control p. 30/43


The l2 gain relationship readily follows for V = xTPx from:


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The l2 gain relationship readily follows for V x Px from:

V(x(k+ 1)) +1

yT(k)y(k) wT(k)w(k) < 0

k=0

V(x(k+ 1))

V(x())V(x(0))

+1

k=0

yT(k)y(k)

k=0

wT(k)w(k) < 0





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2 ga a s p a y s

V(x(k+ 1)) +1


k=0

V(x(k+ 1))

V(x())V(x(0))

+1

k=0

yT(k)y(k)

k=0

wT(k)w(k) < 0

Problem : The matrix inequality

ATPA P + 1CTC ATPB + 1CTD

1D

TC I+BTPB + 1DTD

< 0

is not linear. P > 0 and > 0 are variables.





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2 g p y

V(x(k+ 1)) +1


k=0

V(x(k+ 1))

V(x())V(x(0))

+1

k=0

yT(k)y(k)

k=0

wT(k)w(k) < 0

Problem : The matrix inequality

ATPA P + 1CTC ATPB + 1CTD

1D

TC I+BTPB + 1DTD

< 0

is not linear. P > 0 and > 0 are variables.

The Schur Complement implies

ATPA P ATPB CT

BT

PA I+BT

PB DT

C D I< 0


Examples: l2 gain- Discrete-time systemsATPA P + 1 CTC ATPB + 1 CTD

0


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1D

TC

I+BTPB + 1D

TD

< 0

Congruence transformation & Change of variable approach:

ATPA P +CTC ATPB +CTD

DTC 2I+ BTPB +DTD < 0Defining Q = P and = 2:

min s.t.ATQA Q +CTC ATQB +CTD

DTC I+BTQB +DTD

< 0

Q < 0

Q > 0 and the scalar > 0 are variables.

The l2-gain is readily computed with =.


Examples: Sector boundedness

The saturation function is defined as


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sat(u) = [sat1(u1), . . . ,sat2(um)]T

and sati(ui) = sign(ui) min{|ui|, ui} , ui > 0 i {1, . . . ,m}

ui is the ith saturation limit



The saturation function is defined as


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sat(u) = [sat1(u1), . . . ,sat2(um)]T

and sati(ui) = sign(ui) min{|ui|, ui} , ui > 0 i {1, . . . ,m}

ui is the ith saturation limit

It is easy to verify that the saturation function, sati(ui) satisfies the following inequality

uisati(ui) sat2i (ui)




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It is easy to verify that the saturation function, sati(ui) satisfies the following inequality

uisati(ui) sat2i (ui)

or

sati(ui)[ui sati(ui)]wi 0

for some wi > 0.



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Examples: A slightly more detailed example

Consider the L2-gain for the SISO-system with saturated input signal u:


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x = Ax + bsat(u), x Rn

and a limited measurement range of the output y:

y = sat(cx + dsat(u)).



Consider the L2-gain for the SISO-system with saturated input signal u:


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x = Ax + bsat(u), x Rn

and a limited measurement range of the output y:

y = sat(cx + dsat(u)).

The limits at the actuator inputs u can be due to mechanical limits (e.g. valves) or due to

digital-to-analogue converter voltage signal limits.

Output signals can be constrained due to sensor voltage range limits or simply by

analogue-to-digital converter limits.

The analysis of such systems is vital to practical control systems and will be pursued in

greater detail later. We may consider here an L2 gain analysis.




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We may define

s = sat(u).

Hence, it follows

sw1(u s) 0, w1 > 0

For the output signal y:

yw2(cx + ds y) 0, w2 > 0




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We may define

s = sat(u).

Hence, it follows

sw1(u s) 0, w1 > 0

For the output signal y:

yw2(cx + ds y) 0, w2 > 0

We know that from

V+1

y2 u2 0

follows that our system has the L2-gain .

We have to consider the two saturation nonlinearities! Linear Matrix Inequalities in Control p. 36/43


With the S-procedure


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V+ 1

y2 u2 + 2sw1(u s) + 2yw2(cx + ds y) < 0 f or xT u y = 0the system has also an L2-gain of .

The expression V implies:

xTATPx +xPAx +xTPbs + sbTPx

+1

y2 u2 + 2sw1(u s) + 2yw2(cx + ds y) 0



With the S-procedure


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V+ 1

y2 u2 + 2sw1(u s) + 2yw2(cx + ds y) < 0 f or xT u y = 0the system has also an L2-gain of .

The expression V implies:

xTATPx +xPAx +xTPbs + sbTPx

+1

y2 u2 + 2sw1(u s) + 2yw2(cx + ds y) 0

Rewriting:

x

s

u

y

T

ATP + PA Pb 0 cTw2

bTP 2w1 w1 dw20 w1 0

w2c w2d 0 2w2 + 1

x

s

u

y

< 0

for

xT s u y

= 0.



This is equivalent to


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ATP + PA Pb 0 cTw2


w2c w2d 0 2w2 + 1

< 0.

We would like to minimize , while P, w1, w2 are variables. Not an LMI!



This is equivalent to


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ATP + PA Pb 0 cTw2


w2c w2d 0 2w2 + 1

< 0.

We would like to minimize , while P, w1, w2 are variables. Not an LMI!

Using Projection Lemma twice, we can derive a significantly simpler matrix inequality

which delivers the L2-gain.

First Step:

ATP + PA Pb 0 cTw2

bTP 0 0 dw2

0 0 0w2c w2d 0 2w2 + 1

+0

1

0

0

w1 0 1 1 0 +0

1

1

0

w1 0 1 0 0 < 0.Linear Matrix Inequalities in Control p. 38/43


First Step:


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ATP + PA Pb 0 cTw2

bTP 0 0 dw2

0 0 0w2c w2d 0

2w2 +

1

+

0

1

0

0

w1

0 1 1 0

+

0

11

0

w1

0 1 0 0

< 0.

Defining the matrices

g1 =

0

1

0

0

, h1 = 0

11

0

, 1 = ATP + PA Pb 0 cTw2

bTP 0 0 dw2

0 0 0w2c w2d 0 2w2 + 1

,

allows us to write 1 + g1w1hT1 + h1w

T1 g

T1 < 0.



0 0 ATP + PA Pb 0 cTw2


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g1 = 10

0

, h1 = 110

, 1 = bT

P 0 0 dw20 0 0

w2c w2d 0 2w2 + 1

,

The null space matrices Wg1

and Wh1

satisfy

WTg1 g1

&

WTh1 h1

full rank; Wg1 g1 = 0, Wh1 h1 = 0





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g1 = 10

0

, h1 = 110

, 1 = bT

P 0 0 dw20 0 0

w2c w2d 0 2w2 + 1

,


and Wh1

satisfy

WTg1 g1

&

WTh1 h1


Hence,

Wg1 =

I 0 0 0

0 0 1 0

0 0 0 1

, Wh1 =

I 0 0 0

0 1 1 0

0 0 0 1





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g1 = 10

0

, h1 = 110

, 1 = bT

P 0 0 dw20 0 0

w2c w2d 0 2w2 + 1

,


and Wh1

satisfy

WTg1 g1

&

WTh1 h1


Hence,

Wg1 =

I 0 0 0

0 0 1 0

0 0 0 1

, Wh1 =

I 0 0 0

0 1 1 0

0 0 0 1

Hence, it follows

Wg11WT

g1 =ATP + PA 0 cTw2

0 0w2c 0 2w2 + 1

, Wh11WTh1 =ATP + PA Pb cTw2

bTP dw2w2c w2d 2w2 + 1



T

ATP + PA 0 cTw2T

ATP + PA Pb cTw2T


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Wg11WT

g1 = 0 0w2c 0 2w2 + 1

, Wh11WTh1 = bTP dw2w2c w2d 2w2 + 1

If Wh11WT

h1< 0 then also Wg11W

Tg1< 0 (easily seen from a further analysis using the

Projection lemma).We may carry on investigating Wh11W

Th1

only



T

ATP + PA 0 cTw2T

ATP + PA Pb cTw2T


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Wg11WT

g1 = 0 0w2c 0 2w2 + 1

, Wh11WTh1 = bTP dw2w2c w2d 2w2 + 1

If Wh11WT

h1< 0 then also Wg11W

Tg1< 0 (easily seen from a further analysis using the

Projection lemma).We may carry on investigating Wh11W

Th1

only

Wh11WT

h1=2 + g2w2h

T2 + h2w2g

T2

where

g2 =

0

0

1

, h2 =

cT

d

1

, 2 =

ATP + PA Pb 0

bTP 00 0 1

This allows us to derive the null space matrices Wg2 and Wh2 for g2 and h2

Wg2 = I 0 0

0 1 0

, Wh2 =

I 0 cT0 1 d



I 0 0 I 0 cTATP + PA Pb 0

T


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Wg2 = I 0 00 1 0 , Wh2 = I 0 c0 1 d , 2 = bTP 00 0 1

Thus,

Wg22WT

g2=

ATP + PA Pb

bTP

, Wh22W

Th2

=

ATP + PA + c

Tc Pb +

cTd

bTP + dc + d2

.



I 0 0 I 0 cTATP + PA Pb 0

T


79/80

Wg2 = 0 1 0 , Wh2 = 0 1 d , 2 = bTP 00 0 1

Thus,

Wg22WT

g2=

ATP + PA Pb

bTP

, Wh22W

Th2

=

ATP + PA + c

Tc Pb +

cTd

bTP + dc + d2

.

Wg2

2WT

g2< 0 is always satisfied if Wh

2

2WT

h2.

Hence, the L2 gain is computed using

ATP + PA + cTc Pb +

cTd

bTP + dc + d2 < 0, P > 0The L2-gain of the linear system (A,b,c,d) is an upper bound of the non-linear operator.

The L2-gain of the non-linear and linear operator are identical.


Summary

Matrix inequalities have shown to be versatile tool to


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1. represent L2, H , linear quadratic performance constraints, H2 etc.

2. analyze linear parameter varying systems, mild non-linear systems

3. combine several analysis problems in one frame work

Matrix inequalities can often be transformed into linear matrix inequalitiesby congruence transformation, change of variable approach, etc.

Existence of a large variety of powerful tools for solving LMIs (semi-definiteprogramming)

LMIs have become a standard tool in the analysis and controller design of linear andnon-linear control systems


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