7/27/2019 Lmi Slides
1/80
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
7/27/2019 Lmi Slides
2/80
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
Linear Matrix Inequalities in Control p. 2/43
7/27/2019 Lmi Slides
3/80
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
Linear Matrix Inequalities in Control p. 3/43
7/27/2019 Lmi Slides
4/80
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
The two matrix inequalities involved here are
PA +ATP < 0
and
P > 0.
Linear Matrix Inequalities in Control p. 3/43
7/27/2019 Lmi Slides
5/80
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
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.
Linear Matrix Inequalities in Control p. 3/43
7/27/2019 Lmi Slides
6/80
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?
Linear Matrix Inequalities in Control p. 4/43
7/27/2019 Lmi Slides
7/80
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.)
Linear Matrix Inequalities in Control p. 5/43
7/27/2019 Lmi Slides
8/80
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
Linear Matrix Inequalities in Control p. 6/43
7/27/2019 Lmi Slides
9/80
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
Linear Matrix Inequalities in Control p. 7/43
7/27/2019 Lmi Slides
10/80
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
Likewise, Q is said to be positive semi-definite if it is symmetric and
xTQx 0 x thus Q 0
Linear Matrix Inequalities in Control p. 7/43
7/27/2019 Lmi Slides
11/80
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
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
7/27/2019 Lmi Slides
12/80
7/27/2019 Lmi Slides
13/80
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!
Linear Matrix Inequalities in Control p. 8/43
7/27/2019 Lmi Slides
14/80
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.
Linear Matrix Inequalities in Control p. 9/43
7/27/2019 Lmi Slides
15/80
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 ...
Linear Matrix Inequalities in Control p. 10/43
7/27/2019 Lmi Slides
16/80
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)
Linear Matrix Inequalities in Control p. 11/43
7/27/2019 Lmi Slides
17/80
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!
Linear Matrix Inequalities in Control p. 11/43
7/27/2019 Lmi Slides
18/80
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.
Linear Matrix Inequalities in Control p. 12/43
7/27/2019 Lmi Slides
19/80
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.
Linear Matrix Inequalities in Control p. 12/43
bl b
7/27/2019 Lmi Slides
20/80
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.
Linear Matrix Inequalities in Control p. 13/43
LMI bl Li Obj i Mi i i i
7/27/2019 Lmi Slides
21/80
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.
Linear Matrix Inequalities in Control p. 13/43
LMI bl G li d i l bl
7/27/2019 Lmi Slides
22/80
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.
Linear Matrix Inequalities in Control p. 14/43
LMI bl G li d i l bl
7/27/2019 Lmi Slides
23/80
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.
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,
Linear Matrix Inequalities in Control p. 14/43
LMI problems: Generali ed eigenvalue problem
7/27/2019 Lmi Slides
24/80
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.
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
Linear Matrix Inequalities in Control p. 14/43
Tricks: Congruence transformation
7/27/2019 Lmi Slides
25/80
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.
Linear Matrix Inequalities in Control p. 15/43
Tricks: Change of variables
7/27/2019 Lmi Slides
26/80
Tricks: Change of variables
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 !
Linear Matrix Inequalities in Control p. 16/43
Tricks: Change of variables
7/27/2019 Lmi Slides
27/80
Tricks: Change of variables
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).
Linear Matrix Inequalities in Control p. 16/43
Tricks: Change of variables
7/27/2019 Lmi Slides
28/80
Tricks: Change of variables
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.
Linear Matrix Inequalities in Control p. 17/43
Tricks: Schur complement
7/27/2019 Lmi Slides
29/80
Tricks: Schur complement
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.
Linear Matrix Inequalities in Control p. 18/43
Tricks: Schur complement
7/27/2019 Lmi Slides
30/80
Tricks: Schur complement
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.
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.
Linear Matrix Inequalities in Control p. 18/43
Tricks: Schur complement
7/27/2019 Lmi Slides
31/80
Tricks: Schur complement
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.
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.
The Riccati inequality can be transformed into
ATP + PA + Q PB R
< 0
Linear Matrix Inequalities in Control p. 18/43
Tricks: Schur complement
7/27/2019 Lmi Slides
32/80
p
Example: Making a LQR-type quadratic inequality linear (Riccati inequality)
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.
Linear Matrix Inequalities in Control p. 19/43
Tricks: Schur complement
7/27/2019 Lmi Slides
33/80
p
Example: Making a LQR-type quadratic inequality linear (Riccati inequality)
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.
Linear Matrix Inequalities in Control p. 19/43
Tricks: Schur complement
7/27/2019 Lmi Slides
34/80
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.
Linear Matrix Inequalities in Control p. 20/43
Tricks: The S-procedure
7/27/2019 Lmi Slides
35/80
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}
Linear Matrix Inequalities in Control p. 21/43
Tricks: The S-procedure
7/27/2019 Lmi Slides
36/80
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.
Linear Matrix Inequalities in Control p. 21/43
Tricks: The S-procedure
7/27/2019 Lmi Slides
37/80
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 (
)
Linear Matrix Inequalities in Control p. 22/43
Tricks: The S-procedure
7/27/2019 Lmi Slides
38/80
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.
Linear Matrix Inequalities in Control p. 22/43
Tricks: The Projection Lemma
7/27/2019 Lmi Slides
39/80
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.
Linear Matrix Inequalities in Control p. 23/43
Tricks: The Projection Lemma
7/27/2019 Lmi Slides
40/80
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.
Inequality () is satisfied for some V if and only if
WT
G(V)(V)WG(V) < 0
WTH(V)(V)WH(V) < 0
Linear Matrix Inequalities in Control p. 23/43
Tricks: The Projection Lemma
7/27/2019 Lmi Slides
41/80
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.
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.
Linear Matrix Inequalities in Control p. 23/43
Tricks: The Projection Lemma
7/27/2019 Lmi Slides
42/80
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.
Linear Matrix Inequalities in Control p. 24/43
Tricks: The Projection Lemma
7/27/2019 Lmi Slides
43/80
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!!
Linear Matrix Inequalities in Control p. 24/43
Examples: L2 gain- Continuous-time systems
7/27/2019 Lmi Slides
44/80
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.
Linear Matrix Inequalities in Control p. 25/43
Examples: L2 gain- Continuous-time systems
7/27/2019 Lmi Slides
45/80
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.
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.
Linear Matrix Inequalities in Control p. 25/43
Examples: L2 gain- Continuous-time systems
7/27/2019 Lmi Slides
46/80
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.
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.
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.
Linear Matrix Inequalities in Control p. 25/43
Examples: L2 gain- Continuous-time systems
7/27/2019 Lmi Slides
47/80
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
Linear Matrix Inequalities in Control p. 26/43
Examples: L2 gain- Continuous-time systems
7/27/2019 Lmi Slides
48/80
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
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
Linear Matrix Inequalities in Control p. 26/43
Examples: L2 gain- Continuous-time systems
T T
7/27/2019 Lmi Slides
49/80
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
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
Examples: L2 gain- Continuous-time systems
7/27/2019 Lmi Slides
50/80
Defining V = xTPx
V = xTATPx +xTPAx + 2xTPBw
Thus, we require:
xTATPx +xTPATx + 2xTPBw +1
zTz wTw = V+ 1
zTz wTw < 0
Linear Matrix Inequalities in Control p. 27/43
Examples: L2 gain- Continuous-time systems
7/27/2019 Lmi Slides
51/80
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.
Linear Matrix Inequalities in Control p. 29/43
Examples: l2 gain- Discrete-time systems
A system with input w(t) and output z(t) is said to have an L2 gain of if
7/27/2019 Lmi Slides
54/80
A system with input w(t) and output z(t) is said to have an L2 gain of if
z2 < w2 +, > 0
where w2 =
k=0 w
T(k)w(k) .
Linear Matrix Inequalities in Control p. 30/43
Examples: l2 gain- Discrete-time systems
A system with input w(t) and output z(t) is said to have an L2 gain of if
7/27/2019 Lmi Slides
55/80
A system with input w(t) and output z(t) is said to have an L2 gain of if
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
Examples: l2 gain- Discrete-time systems
The l2 gain relationship readily follows for V = xTPx from:
7/27/2019 Lmi Slides
56/80
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
Linear Matrix Inequalities in Control p. 31/43
Examples: l2 gain- Discrete-time systems
The l2 gain relationship readily follows for V = xTPx from:
7/27/2019 Lmi Slides
57/80
2 ga a s p a y s
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
Problem : The matrix inequality
ATPA P + 1CTC ATPB + 1CTD
1D
TC I+BTPB + 1DTD
< 0
is not linear. P > 0 and > 0 are variables.
Linear Matrix Inequalities in Control p. 31/43
Examples: l2 gain- Discrete-time systems
The l2 gain relationship readily follows for V = xTPx from:
7/27/2019 Lmi Slides
58/80
2 g p y
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
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
Linear Matrix Inequalities in Control p. 31/43
Examples: l2 gain- Discrete-time systemsATPA P + 1 CTC ATPB + 1 CTD
0
7/27/2019 Lmi Slides
59/80
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 =.
Linear Matrix Inequalities in Control p. 32/43
Examples: Sector boundedness
The saturation function is defined as
7/27/2019 Lmi Slides
60/80
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
Linear Matrix Inequalities in Control p. 33/43
Examples: Sector boundedness
The saturation function is defined as
7/27/2019 Lmi Slides
61/80
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)
Linear Matrix Inequalities in Control p. 33/43
Examples: Sector boundedness
7/27/2019 Lmi Slides
62/80
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.
Linear Matrix Inequalities in Control p. 34/43
7/27/2019 Lmi Slides
63/80
Examples: A slightly more detailed example
Consider the L2-gain for the SISO-system with saturated input signal u:
7/27/2019 Lmi Slides
64/80
x = Ax + bsat(u), x Rn
and a limited measurement range of the output y:
y = sat(cx + dsat(u)).
Linear Matrix Inequalities in Control p. 35/43
Examples: A slightly more detailed example
Consider the L2-gain for the SISO-system with saturated input signal u:
7/27/2019 Lmi Slides
65/80
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.
Linear Matrix Inequalities in Control p. 35/43
Examples: A slightly more detailed example
7/27/2019 Lmi Slides
66/80
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
Linear Matrix Inequalities in Control p. 36/43
Examples: A slightly more detailed example
7/27/2019 Lmi Slides
67/80
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
Examples: A slightly more detailed example
With the S-procedure
7/27/2019 Lmi Slides
68/80
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
Linear Matrix Inequalities in Control p. 37/43
Examples: A slightly more detailed example
With the S-procedure
7/27/2019 Lmi Slides
69/80
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.
Linear Matrix Inequalities in Control p. 37/43
Examples: A slightly more detailed example
This is equivalent to
7/27/2019 Lmi Slides
70/80
ATP + PA Pb 0 cTw2
bTP 2w1 w1 dw20 w1 0
w2c w2d 0 2w2 + 1
< 0.
We would like to minimize , while P, w1, w2 are variables. Not an LMI!
Linear Matrix Inequalities in Control p. 38/43
Examples: A slightly more detailed example
This is equivalent to
7/27/2019 Lmi Slides
71/80
ATP + PA Pb 0 cTw2
bTP 2w1 w1 dw20 w1 0
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
Examples: A slightly more detailed example
First Step:
7/27/2019 Lmi Slides
72/80
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.
Linear Matrix Inequalities in Control p. 39/43
Examples: A slightly more detailed example
0 0 ATP + PA Pb 0 cTw2
7/27/2019 Lmi Slides
73/80
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
Linear Matrix Inequalities in Control p. 40/43
Examples: A slightly more detailed example
0 0 ATP + PA Pb 0 cTw2
7/27/2019 Lmi Slides
74/80
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
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
Linear Matrix Inequalities in Control p. 40/43
Examples: A slightly more detailed example
0 0 ATP + PA Pb 0 cTw2
7/27/2019 Lmi Slides
75/80
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
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
Linear Matrix Inequalities in Control p. 40/43
Examples: A slightly more detailed example
T
ATP + PA 0 cTw2T
ATP + PA Pb cTw2T
7/27/2019 Lmi Slides
76/80
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
Linear Matrix Inequalities in Control p. 41/43
Examples: A slightly more detailed example
T
ATP + PA 0 cTw2T
ATP + PA Pb cTw2T
7/27/2019 Lmi Slides
77/80
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
Linear Matrix Inequalities in Control p. 41/43
Examples: A slightly more detailed example
I 0 0 I 0 cTATP + PA Pb 0
T
7/27/2019 Lmi Slides
78/80
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
.
Linear Matrix Inequalities in Control p. 42/43
Examples: A slightly more detailed example
I 0 0 I 0 cTATP + PA Pb 0
T
7/27/2019 Lmi Slides
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.
Linear Matrix Inequalities in Control p. 42/43
Summary
Matrix inequalities have shown to be versatile tool to
7/27/2019 Lmi Slides
80/80
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
Linear Matrix Inequalities in Control p. 43/43