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matlab代写-ELE7029
时间：2022-04-21
ELE7029: Optimal Digital Control System
Lecture III
Linear Matrix Inequalities
Goals of this lecture
· LMIs
· LMI Problems
· Solving LMIPs
S.-H. Lee (shlee@ieee.org) Elec. Engr., Hanyang University Spring 2022
Outline
1. Linear Matrix Inequalities
2. Solving LMIPs
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 1
Linear Matrix Inequalities
An inequality of the form
F(x) = F0 + x1F1 + · · ·+ xnFn = F0 +
n∑
i=1
xiFi ≥ 0, (3.1)
is called linear matrix inequality (LMI)
Here
x = (x1, · · · ,xn) ∈ Rn a vector of unknown scalars (decision,
optimization variable)
Fi = FTi ∈ Rm×m, i= 0, · · · ,n, are given symmetric matrices
F : Rn→ Rm×m is an affine fcn. of the variable x ∈ Rn
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 2
The canonical form is very inefficient from a storage point of view.
In most applications, LMIs arise rather in the form
L (X1, . . . ,Xm)where
L(·) and R(·) are affine functions of
some structured matrix variables X1, . . . ,Xm.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 3
e.g. An affine function of x ∈ R2
F(x) =
x1 + x2 x2 + 1
x2 + 1 x3
= F0 + x1F1 + x2F2 + x3F3 ≥ 0
where
F0 =
0 1
1 0
, F1 =
1 0
0 0
, F2 =
1 1
1 0
, F3 =
0 0
0 1
The LMI F(x)≥ 0 is equivalent to a set of nonlinear inequalities:
x1 + x2 ≥ 0,
x3 ≥ 0,
(x1 + x2)x3 − (x2 + 1)2 ≥ 0
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 4
e.g. Lyapunov inequality
Consider a system x˙ = Ax with A ∈ Rn×n.
Suppose the Lyapunov inequality ATY + YA+Q≤ 0 is affine inequality of
the variable Y.
Y satisfies the Lyapunov inequality iff the quadratic form V(x) = xTYx
satisfies V˙(x)≤ −xTQx for the system x˙ = Ax.
The inequality F(Y) = ATY + YA+Q≤ 0 is an LMI in Y.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 5
e.g. Bounded Real Lemma
Consider a system
x˙ = Ax+ Bw
z= Cx
with A ∈ Rn×n, B ∈ Rn×m, C ∈ Rp×n, and
γ > 0.
The BR lemma LMIs are
ATY + YA+ CTC YB
BTY −γI
≤ 0, Y ≥ 0
with a variable Y.
By this we mean if Y satisfies the LMIs, then the quadratic Lyapunov fcn.
V(x) = xTYx proves the RMS gain of the system is no more than γ.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 6
e.g. A time-varying linear dynamic system x˙(t) = A(t)x(t) with
A(t) ∈ {A1, . . . ,Aq}.
Want to find a quadratic Lyapunov fcn. V(x) = xTYx: Y > 0 and V˙(x)≤ 0
for all x and all possible values of A(t)
Y > 0, ATi Y + YAi ≤ 0, i= 1, . . . ,q
We can write as LMIs
Y ≥ I, ATi Y + YAi ≤ 0, i= 1, . . . ,q
If such Y exists, it proves boundedness of trajectories of the system
x˙ = A(t)x with
A(t) = θ1(t)A1 + · · ·+ θq(t)Aq
where θi(t)≥ 0
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 7
■ S-procedure:
Let F0 = FT0 , F1 = F
T
1 ∈ Rn×n
Assume xTF1x > 0
When is it true that for all x satisfying xTF1x ≥ 0 ⇒ xTF0x ≥ 0?
The implication
xTF1x ≥ 0 =⇒ xTF0x ≥ 0
holds iff there exists τ≥ 0 s.t. F0 ≥ τF1.
Note: For a given F1 ≥ 0, find Fo ≥ 0 and τ≥ 0 satisfying Fo ≥ τF1.
Note: The S-procedure provides conditions under which a particular
quadratic inequality is a consequence of another quadratic inequality.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 8
Let F1 = FT1 , F0 = F
T
0 ∈ Rn×n, v1, v0 ∈ Rn, and α1, α0 ∈ R.
Assume that there is some x s.t.
xTF1x+ 2v
T
1x+α1 > 0
holds.
Then the implication
xTF1x+ 2v
T
1x+α1 ≥ 0 =⇒ xTF0x+ 2vT0x+α0 ≥ 0
holds iff if there exists τ≥ 0 s.t.
F0 v0
vT0 α0
≥ τ
F1 v1
vT1 α1
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 9
e.g.
Let Li(·) be quadratic fcns
Li(ζ) = ζ
TTiζ, Ti = T
T
i ∈ Rn×n, i= 0, . . . ,p
When is it true that
ζTTiζ≥ 0, i= 1, . . . ,p =⇒ ζTToζ > 0
(hard to find)
Holds if there exist τi ≥ 0, i= 1, . . . ,p s.t.
To >
p∑
i=1
τiTi
(LMIs in To and τi’s)
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 10
e.g. Stability of a system x˙ = Ax+ d(x), ∥d(x)∥ ≤ γ∥x∥
Let us use a quadratic Lyapunov fcn V(x) = xTPx to prove stability of the
system
We need P> 0 and V˙(x)≤ −αV(x) for all x (α > 0 given):
V˙(x) +αV(x) = 2xTP(Ax+ d(x)) +αxTPx
= xT(ATP+ PA)x+ xTPd(x) + d(x)TPx+αxTPx
= xT(ATP+ PA+αP)x+ d(x)TPx+ xTPd(x)
=
x
d(x)
T
ATP+ PA+αP P
P 0
x
d(x)
≤ 0
whenever
d(x)Td(x)≤ γ2xTx
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 11
So we need P> 0 and
−
x
d(x)
ATP+ PA+αP P
P 0
x
d(x)
≥ 0
whenever
x
d(x)
T
γ2I 0
0 −I
x
d(x)
≥ 0
(hard to find)
By S-procedure, this happens iff
−
ATP+ PA+αP P
P 0
≥ τ
γ2I 0
0 −I
for some τ≥ 0
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 12
Thus, necessary and sufficient conditions for the existence of quadratic
Lyapunov fcn. can be expressed as LMI
P> 0,
ATP+ PA+αP+τγ2I P
P −τI
≤ 0
in variables P and τ.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 13
e.g. Find P> 0 satisfying the constraints on P:
For all ξ ̸= 0 and P satisfying
ξ
η
T
ATP+ PA PB
BTP 0
ξ
η
< 0,
whenever
ξ
η
T −CTC 0
0 I
ξ
η
≤ 0.
By S-procedure, this happens iff
ATP+ PA PB
BTP 0
< τ
−CTC 0
0 I
≤ 0
for some τ≥ 0.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 14
What if we have a matrix inequality such as
X22 > 0 and X11 − X12X−122 XT12 > 0
(nonlinear)
Find its linear equivalent form.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 15
Good to Note: Matrix partition
Let A=
A11 A12
A21 A22
, then
A=
A11 A12
A21 A22
=
I 0
A21A
−1
11 I
A11 0
0 ∆
I A−111A12
0 I
(3.3a)
where ∆= A22 −A21A−111A12, A11 nonsingular: (Schur complement of A11 in
A)
Or
A=
A11 A12
A21 A22
=
I A12A
−1
22
0 I
∆ˆ 0
0 A22
I 0
A−122A21 I
(3.3b)
where ∆ˆ= A11 −A12A−122A21, A22 nonsingular: (Schur complement of A22 in
A)
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 16
Let X = X∗ ≥ 0 be partitioned as
X =
X11 X12
X∗12 X22
If X+22 is the pseudo-inverse of X22,
X11 X12
X∗12 X22
=
I X12X
+
22
0 I
X11 − X12X+22X∗12 0
0 X22
I 0
X+22X
∗
12 I
Find
X11 X12
X∗12 X22
≥ 0 ⇐⇒
X11 − X12X+22X∗12 0
0 X22
≥ 0
⇐⇒
X11 − X12X+22X∗12 ≥ 0
X22 ≥ 0
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 17
■ Schur complement:
For matrices Q(x) = Q(x)T, R(x) = R(x)T, and S(x) depend affinely on x:
R(x)> 0, Q(x)− S(x)R(x)−1S(x)T > 0 ⇐⇒
Q(x) S(x)
S(x)T R(x)
> 0
Note:
▶
Q S
ST R
=
I S−1R
0 I
Q− SR−1ST 0
0 R
I 0
R−1S I
▶ If
Q S
ST R
> 0, then so is the Schur complement of R in
Q S
ST R
.
▶ Nonlinear (convex) inequalities can be converted to an LMI form using
the Schur complement.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 18
Schur complement conditions:
Let A> 0. Then
A B
BT C
> 0 ⇐⇒ A> 0, C− BTA−1B> 0.
Let C > 0. Then
A B
BT C
> 0 ⇐⇒ C > 0, A− BC−1BT > 0.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 19
Proof: Consider the quadratic program
inf
(u,v)
u
v
T
A B
BT C
u
v
= uTAu+ 2uTBv+ vTCv
The solution of the minimization problem satisfies:
u
v
∗
= 0 (and attained uniquely at 0) ⇐⇒
A B
BT C
> 0.
First, solve
min
u
u
v
T
A B
BT C
u
v
= min
u
uTAu+ 2uTBv
+ vTCv
The optimum u is
d
du
=⇒ 2Au+ 2Bv= 0 ⇐⇒ u∗ = −A−1Bv
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 20
Observe
uTAu= uTATu= (−A−1Bv)TAT(−A−1Bv) = vTBTA−1Bv
and
uTBv= vTBTu= −vTBTA−1Bv
which leads to
uTAu+ 2uTBv
+ vTCv= vT
C− BTA−1B v
Now if we minimize over v, then the minimum is achieved and attained
uniquely at 0 iff C− BTA−1B> 0. □
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 21
Good to Note:
A block matrix
A B
BT C
< 0
is equivalent to
C < 0, A− BC−1BT < 0 ⇐⇒
−A B
BT −C
> 0
or
A< 0, C− BTA−1B< 0 ⇐⇒
C BT
B A
< 0
⇐⇒
−C BT
B −A
> 0
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 22
Extensions to the Schur complement:
A− BC−1BT Q
QT R
< 0 ⇐⇒
A B QBT C 0
QT 0 R
< 0
⇐⇒
C BT 0B A Q
0 QT R
< 0
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 23
P Q
QT A− BC−1BT
< 0 ⇐⇒
P Q 0QT A B
0 BT C
< 0
⇐⇒
P 0 Q0 C BT
QT B A
< 0
Exercise: Proof
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 24
■ Redundant (Slack) Variables: Parametrization, Dilated LMI conditions
Assume there exists X > 0 with
I A
−X 0
0 X
I
AT
= AXAT − X < 0
Equivalent reformulation: There exist X and G with −X AG
GTAT X −G−GT
< 0
▶ New parametrization independent of the original variables
▶ Dilated synthesis approaches always encompass the corresponding
standard-LMI-based ones
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 25
Equivalent reformulation
I A
−X 0
0 X
I
AT
+
I A
A
−I
G
0 I
I
AT
+
I A
0
I
GT
AT −I I
AT
=
I A
−X AG
GTAT X −G−GT
I
AT
=
I A
−X 0
0 X
I
AT
= AXAT − X
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 26
Exercise: Equivalent reformulation
i) ∃ P> 0 s.t.
ΦTPΦ− P< 0
ii) ∃ P> 0 s.t. −P ΦTP
PΦ −P
< 0
iii) ∃ P> 0 and G s.t. −P ΦTGT
GΦ P− (G+GT)
< 0
iv) ∃ P> 0 and G and F s.t. −P−ΦTF− FTΦ FT +ΦTGT
F+GΦ P− (G+GT)
< 0
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 27
Proof:
ii) → iii) Apply Schur complement by choosing G= GT = P> 0
ΦTPΦ−P< 0 =⇒
−P ΦTP
PΦ P
< 0 =⇒
−P ΦTGT
GΦ P− (G+GT)
< 0
iii) → ii)
I
Φ
T −P ΦTGT
GΦ P− (G+GT)
I
Φ
= ΦTPΦ− P< 0
ii) → iv) ∃ P> 0 s.t. ΦTPΦ− P< 0 =⇒ ∃ F s.t.
ΦTPΦ− P+ FP−1F < 0.
ΦTPΦ− P+ FP−1F = −P−ΦTF− FTΦ+ (F+ PΦ)T P−1 (F+ PΦ)
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 28
Apply Schur complements by choosing G= GT = P> 0 as in the proof
of ii) → iii)
− P−ΦTF− FTΦ+ (F+ PΦ)T P−1 (F+ PΦ)< 0 =⇒ −P−ΦTF− FTΦ FT +ΦTGT
F+GΦ P− G+GT
< 0
iv) → ii)
I
Φ
T −P−ΦTF− FTΦ FT +ΦTGT
F+GΦ P− (G+GT)
I
Φ
= ΦTPΦ− P< 0
□
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 29
■ Elimination of matrix variables:
Lemma 3.1 (Projection Lemma)
Given a symmetric matrix Ψ and two matrices U, V, there exist M such that
Ψ +UMVT + VMTUT < 0 (3.4)
iff the projection inequalities
U˜TΨU˜ < 0, V˜TΨV˜ < 0 (3.5)
hold, where U˜ and V˜ are orthogonal complements (bases of the null spaces) of
U and V, respectively, i.e., U˜TU = 0 and V˜TV = 0.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 30
Proof: Using U˜TU = 0 and V˜TV = 0, we find
U˜T
Ψ +UMVT + VMTUT
U˜ = U˜TΨU˜ + U˜TUMVTU˜ + U˜TVMTUTU˜
= U˜TΨU˜ < 0
and
V˜T
Ψ +UMVT + VMTUT
V˜ = V˜TΨV˜ + V˜TUMVTV˜ + V˜TVMTUTV˜
= V˜TΨV˜ < 0
□
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 31
■ Generic LMI Problems:
▶ Feasibility problem (FEASP):
Given an LMI F(x)> 0, find xfeas s.t. F(xfeas)> 0 or determine that the
LMI is feasible.
e.g.
▶ Given Ai ∈ Rn×n, i= 1, . . . ,n, find X satisfying the LMI
X > 0, ATi X + XAi < 0, i= 1, . . . ,n
or determine that no such X exists.
▶ Given Ai ∈ Rn×n, i= 1, . . . ,n, and γ > 0, find X satisfying
ATi X + XAi + C
TC XB
BTX −γI
< 0, i= 1, . . . ,n, X > 0
or determine that no such X exists.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 32
▶ Find X satisfying the LMI:
AXAT − X < 0, X > 0
▶ Find a soln. X to N TL(X)N▶ This feasibility problem is solved by the auxiliary problem:
minimize t
subject to: L(X)−R(X)≤ t I
Feasible iff tmin ≤ 0.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 33
▶ Linear objective minimization problem (MINCX):
Given an LMI F(x)> 0, minimize a linear objective with F(xmincx)> 0
being satisfied.
e.g.
▶ minimize cTx subject to F(x)> 0
▶ minimize tr (X) subject to F(X)> 0
▶ minimize t subject to F(X)< t I (with t as a DV)
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 34
▶ Generalized eigenvalue problem (GEVP):
Given an LMI F(x,λ)> 0, minimize λ with F(xgevp,λ)> 0 being
satisfied.
e.g.
▶ minimize λ subject to λB(x)− A(x)> 0, B(x)> 0, C(x)> 0
▶ maximize α subject to ATX + XA+ 2αX < 0, X > 0
▶ minimize γ subject to
ATX + XA+ CTC XB
BTX −γI
< 0, X > 0
▶ minimize γ subject to L(X)< γR(X), L(X)> 0, R(X)> 0
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 35
Solving LMIPs
Develop an LMI system:
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 36
▶ Assign a name to the LMI system
e.g. lmisys
▶ Describe the matrix variables and LMIs
▶ Write the commands as an m file.
e.g. lmimodel.m
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 37
% Load an LMI model:
lmimodel % in LMIEDIT, write commands as an m file
lminbr(lmisys) % # LMIs
matnbr(lmisys) % # matrix variables
lmiinfo(lmisys) % info. about variables and term content
% Choose a solver for generic LMIP: feasp/mincx/gevp
% <==== Call feasp/mincx/gevp here
% Solution matrix from the decision vector(s)
Xopt=dec2mat(lmisys,xopt,X)
% Evaluate LMIs to check if the DV(s) indeed feasible
evlmi=evallmi(lmisys,xopt);
[lhs,rhs]=showlmi(evlmi,1) % LHS and RHS
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 38
e.g. Choosing a solver
▶ Feasibility problem: feasp
[tmin,xfeas]=feasp(lmisys); % feasible iff tmin<=0
Note: feasp solves minimize tmin for L(·)−R(·)≤ tminI
▶ Linear objective problem: mincx
ndc=decnbr(lmisys); % total # decision variables
c=zeros(ndc,1); c(1:q)=nu; % for nu*x(1:q)
c=zeros(ndc,1); c(diag(sX))=1; % for trace(X)
[cxopt,xopt]=mincx(lmisys,c);
▶ Generalized eigenvalue problem: gevp
[mu,xopt]=gevp(lmisys,nlfc); % last nlfc LMIs involving obj.
Note: gevp solves minimize γ for L(·)< γR(·) with R(·)> 0.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 39
Note: “Solve” means that we can find DVs that satisfy the LMIP, or
determine that no solns. exist.
Note: The dimension of the variable Y = YT ∈ Rn×n is n(n+ 1)/2:
Y11
Y12
Y22
Y13
...
Ynn
∈ R
n(n+1)/2 or
Y11
Y21
Y22
Y31
...
Ynn
∈ R
n(n+1)/2
Note: The dimension of the variable Y ∈ Rn×n is n2.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 40
e.g.
>> n=3; setlmis([])
>> [X,nX,sX]=lmivar(1,[n 1])
X =
1
nX =
6
sX =
1 2 4
2 3 5
4 5 6
>> [Y,nY,sY]=lmivar(2,[n n])
Y =
2
nY =
15
sY =
7 8 9
10 11 12
13 14 15
>>
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 41
e.g. Linear objective
Given
A1 =
−1 2
1 −3
, A2 =
−0.8 1.5
1.3 −2.7
Find Y > I to minimize trY satisfying
I < Y
AT1Y + YA1 < Y
AT2Y + YA2 < Y
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 42
A1=[-1 2; 1 -3]; A2=[-1.4 0.9; 0.7 -2];
setlmis([]);
[Y,nY,sY]=lmivar(1,[2 1]);
lmiterm([1 1 1 0],1); % LMI #1: I
lmiterm([-1 1 1 Y],1,1); % LMI #1: Y
lmiterm([2 1 1 Y],A1,1); % LMI #2: A1*Y
lmiterm([2 1 1 Y],1,A1’); % LMI #2: Y*A1’
lmiterm([-2 1 1 Y],1,1); % LMI #2: Y
lmiterm([3 1 1 Y],A2,1,’s’); % LMI #3: A2*Y+Y*A2’ (with FLAG=’s’)
lmiterm([-3 1 1 Y],1,1); % LMI #3: Y
lmis=getlmis;
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 43
>> ndc=decnbr(lmis);
>> c=zeros(ndc,1); c(diag(sY))=1;
>> [cyopt,yopt]=mincx(lmis,c);
>> cyopt
cyopt =
2.0012
>> Yopt=dec2mat(lmis,yopt,Y)
Yopt =
1.0005 0.0000
0.0000 1.0007
>> evlmi=evallmi(lmis,yopt);
>> [lhs,rhs]=showlmi(evlmi,3)
lhs =
-2.8013 1.6008
1.6008 -4.0027
rhs =
1.0005 0.0000
0.0000 1.0007
>> eig(lhs-rhs)
ans =
-6.1123
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 44
-2.6928
>> eig(A2*Yopt+Yopt*A2’-Yopt)
ans =
-6.1123
-2.6928
>>
>> [lhs,rhs]=showlmi(evlmi,2)
lhs =
-2.0008 3.0017
3.0017 -6.0040
rhs =
1.0005 0.0000
0.0000 1.0007
>> eig(lhs-rhs)
ans =
-8.6108
-1.3951
>> eig(A1*Yopt+Yopt*A1’-Yopt)
ans =
-8.6108
-1.3951
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 45
Note: To solve a FEASP, we need to write LMI conditions
L(X)The FEASP solver finds tmin ≤ 0 to guarantee
L(X)−R(X)< tminI ≤ 0
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 46
e.g. Feasibility
Given
A1 =
−1 2
1 −3
, A2 =
−0.8 1.5
1.3 −2.7
Find Y > I satisfying
I < Y
AT1Y + YA1 < Y
AT2Y + YA2 < Y
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 47
>> [tmin,yopt]=feasp(lmis);
>> tmin
tmin =
-2.9180
>> Yopt=dec2mat(lmis,yopt,Y)
Yopt =
99.9899 27.8674
27.8674 54.9808
>> evlmi=evallmi(lmis,yopt);
>> [lhs,rhs]=showlmi(evlmi,3)
lhs =
-229.8104 24.7265
24.7265 -180.9089
rhs =
99.9899 27.8674
27.8674 54.9808
>> eig(lhs-rhs)
ans =
-329.9052
-235.7848
>> eig(A2*Yopt+Yopt*A2’-Yopt)
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 48
ans =
-329.9052
-235.7848
>>
>> [lhs,rhs]=showlmi(evlmi,2)
lhs =
-88.5102 98.4820
98.4820 -274.1501
rhs =
99.9899 27.8674
27.8674 54.9808
>> eig(lhs-rhs)
ans =
-358.4683
-159.1628
>> eig(A1*Yopt+Yopt*A1’-Yopt)
ans =
-358.4683
-159.1628
>>
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 49
Note: To solve a GEVP, we need to write LMI conditions
L(X) 0.
The GEVP solver finds λmin to guaaraantee
L(X)< λminR(X).
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 50
e.g. Generalized eigenvalue problem
Given
A1 =
−1 2
1 −3
, A2 =
−0.8 1.5
1.3 −2.7
Find Y > I to minimize λ:
I < Y
AT1Y + YA1 < λY
AT2Y + YA2 < λY
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 51
>> [lambda,yopt]=gevp(lmis,2); % the last 2 lmis are LFTs
>> lambda
lambda =
-0.5359
>> Yopt=dec2mat(lmis,yopt,Y)
Yopt =
493.1196 -671.1353
-671.1353 917.6951
>> evlmi=evallmi(lmis,yopt);
>> [lhs,rhs]=showlmi(evlmi,3)
lhs =
1.0e+03 *
-2.5888 3.4530
3.4530 -4.6104
rhs =
493.1196 -671.1353
-671.1353 917.6951
>> eig(lhs-lambda*rhs)
ans =
1.0e+03 *
-6.4423
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 52
-0.0008
>> eig(A2*Yopt+Yopt*A2’-lambda*Yopt)
ans =
1.0e+03 *
-6.4423
-0.0008
>> [lhs,rhs]=showlmi(evlmi,2)
lhs =
1.0e+03 *
-3.6708 5.0131
5.0131 -6.8484
rhs =
493.1196 -671.1353
-671.1353 917.6951
>> eig(lhs-lambda*rhs)
ans =
1.0e+03 *
-9.7632
-0.0000
>> eig(A1*Yopt+Yopt*A1’-lambda*Yopt)
ans =
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 53
1.0e+03 *
-9.7632
-0.0000
>>
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 54
Lecture III
Linear Matrix Inequalities
Goals of this lecture
· LMIs
· LMI Problems
· Solving LMIPs
S.-H. Lee (shlee@ieee.org) Elec. Engr., Hanyang University Spring 2022
Outline
1. Linear Matrix Inequalities
2. Solving LMIPs
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 1
Linear Matrix Inequalities
An inequality of the form
F(x) = F0 + x1F1 + · · ·+ xnFn = F0 +
n∑
i=1
xiFi ≥ 0, (3.1)
is called linear matrix inequality (LMI)
Here
x = (x1, · · · ,xn) ∈ Rn a vector of unknown scalars (decision,
optimization variable)
Fi = FTi ∈ Rm×m, i= 0, · · · ,n, are given symmetric matrices
F : Rn→ Rm×m is an affine fcn. of the variable x ∈ Rn
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 2
The canonical form is very inefficient from a storage point of view.
In most applications, LMIs arise rather in the form
L (X1, . . . ,Xm)
L(·) and R(·) are affine functions of
some structured matrix variables X1, . . . ,Xm.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 3
e.g. An affine function of x ∈ R2
F(x) =
x1 + x2 x2 + 1
x2 + 1 x3
= F0 + x1F1 + x2F2 + x3F3 ≥ 0
where
F0 =
0 1
1 0
, F1 =
1 0
0 0
, F2 =
1 1
1 0
, F3 =
0 0
0 1
The LMI F(x)≥ 0 is equivalent to a set of nonlinear inequalities:
x1 + x2 ≥ 0,
x3 ≥ 0,
(x1 + x2)x3 − (x2 + 1)2 ≥ 0
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 4
e.g. Lyapunov inequality
Consider a system x˙ = Ax with A ∈ Rn×n.
Suppose the Lyapunov inequality ATY + YA+Q≤ 0 is affine inequality of
the variable Y.
Y satisfies the Lyapunov inequality iff the quadratic form V(x) = xTYx
satisfies V˙(x)≤ −xTQx for the system x˙ = Ax.
The inequality F(Y) = ATY + YA+Q≤ 0 is an LMI in Y.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 5
e.g. Bounded Real Lemma
Consider a system
x˙ = Ax+ Bw
z= Cx
with A ∈ Rn×n, B ∈ Rn×m, C ∈ Rp×n, and
γ > 0.
The BR lemma LMIs are
ATY + YA+ CTC YB
BTY −γI
≤ 0, Y ≥ 0
with a variable Y.
By this we mean if Y satisfies the LMIs, then the quadratic Lyapunov fcn.
V(x) = xTYx proves the RMS gain of the system is no more than γ.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 6
e.g. A time-varying linear dynamic system x˙(t) = A(t)x(t) with
A(t) ∈ {A1, . . . ,Aq}.
Want to find a quadratic Lyapunov fcn. V(x) = xTYx: Y > 0 and V˙(x)≤ 0
for all x and all possible values of A(t)
Y > 0, ATi Y + YAi ≤ 0, i= 1, . . . ,q
We can write as LMIs
Y ≥ I, ATi Y + YAi ≤ 0, i= 1, . . . ,q
If such Y exists, it proves boundedness of trajectories of the system
x˙ = A(t)x with
A(t) = θ1(t)A1 + · · ·+ θq(t)Aq
where θi(t)≥ 0
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 7
■ S-procedure:
Let F0 = FT0 , F1 = F
T
1 ∈ Rn×n
Assume xTF1x > 0
When is it true that for all x satisfying xTF1x ≥ 0 ⇒ xTF0x ≥ 0?
The implication
xTF1x ≥ 0 =⇒ xTF0x ≥ 0
holds iff there exists τ≥ 0 s.t. F0 ≥ τF1.
Note: For a given F1 ≥ 0, find Fo ≥ 0 and τ≥ 0 satisfying Fo ≥ τF1.
Note: The S-procedure provides conditions under which a particular
quadratic inequality is a consequence of another quadratic inequality.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 8
Let F1 = FT1 , F0 = F
T
0 ∈ Rn×n, v1, v0 ∈ Rn, and α1, α0 ∈ R.
Assume that there is some x s.t.
xTF1x+ 2v
T
1x+α1 > 0
holds.
Then the implication
xTF1x+ 2v
T
1x+α1 ≥ 0 =⇒ xTF0x+ 2vT0x+α0 ≥ 0
holds iff if there exists τ≥ 0 s.t.
F0 v0
vT0 α0
≥ τ
F1 v1
vT1 α1
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 9
e.g.
Let Li(·) be quadratic fcns
Li(ζ) = ζ
TTiζ, Ti = T
T
i ∈ Rn×n, i= 0, . . . ,p
When is it true that
ζTTiζ≥ 0, i= 1, . . . ,p =⇒ ζTToζ > 0
(hard to find)
Holds if there exist τi ≥ 0, i= 1, . . . ,p s.t.
To >
p∑
i=1
τiTi
(LMIs in To and τi’s)
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 10
e.g. Stability of a system x˙ = Ax+ d(x), ∥d(x)∥ ≤ γ∥x∥
Let us use a quadratic Lyapunov fcn V(x) = xTPx to prove stability of the
system
We need P> 0 and V˙(x)≤ −αV(x) for all x (α > 0 given):
V˙(x) +αV(x) = 2xTP(Ax+ d(x)) +αxTPx
= xT(ATP+ PA)x+ xTPd(x) + d(x)TPx+αxTPx
= xT(ATP+ PA+αP)x+ d(x)TPx+ xTPd(x)
=
x
d(x)
T
ATP+ PA+αP P
P 0
x
d(x)
≤ 0
whenever
d(x)Td(x)≤ γ2xTx
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 11
So we need P> 0 and
−
x
d(x)
ATP+ PA+αP P
P 0
x
d(x)
≥ 0
whenever
x
d(x)
T
γ2I 0
0 −I
x
d(x)
≥ 0
(hard to find)
By S-procedure, this happens iff
−
ATP+ PA+αP P
P 0
≥ τ
γ2I 0
0 −I
for some τ≥ 0
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 12
Thus, necessary and sufficient conditions for the existence of quadratic
Lyapunov fcn. can be expressed as LMI
P> 0,
ATP+ PA+αP+τγ2I P
P −τI
≤ 0
in variables P and τ.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 13
e.g. Find P> 0 satisfying the constraints on P:
For all ξ ̸= 0 and P satisfying
ξ
η
T
ATP+ PA PB
BTP 0
ξ
η
< 0,
whenever
ξ
η
T −CTC 0
0 I
ξ
η
≤ 0.
By S-procedure, this happens iff
ATP+ PA PB
BTP 0
< τ
−CTC 0
0 I
≤ 0
for some τ≥ 0.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 14
What if we have a matrix inequality such as
X22 > 0 and X11 − X12X−122 XT12 > 0
(nonlinear)
Find its linear equivalent form.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 15
Good to Note: Matrix partition
Let A=
A11 A12
A21 A22
, then
A=
A11 A12
A21 A22
=
I 0
A21A
−1
11 I
A11 0
0 ∆
I A−111A12
0 I
(3.3a)
where ∆= A22 −A21A−111A12, A11 nonsingular: (Schur complement of A11 in
A)
Or
A=
A11 A12
A21 A22
=
I A12A
−1
22
0 I
∆ˆ 0
0 A22
I 0
A−122A21 I
(3.3b)
where ∆ˆ= A11 −A12A−122A21, A22 nonsingular: (Schur complement of A22 in
A)
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 16
Let X = X∗ ≥ 0 be partitioned as
X =
X11 X12
X∗12 X22
If X+22 is the pseudo-inverse of X22,
X11 X12
X∗12 X22
=
I X12X
+
22
0 I
X11 − X12X+22X∗12 0
0 X22
I 0
X+22X
∗
12 I
Find
X11 X12
X∗12 X22
≥ 0 ⇐⇒
X11 − X12X+22X∗12 0
0 X22
≥ 0
⇐⇒
X11 − X12X+22X∗12 ≥ 0
X22 ≥ 0
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 17
■ Schur complement:
For matrices Q(x) = Q(x)T, R(x) = R(x)T, and S(x) depend affinely on x:
R(x)> 0, Q(x)− S(x)R(x)−1S(x)T > 0 ⇐⇒
Q(x) S(x)
S(x)T R(x)
> 0
Note:
▶
Q S
ST R
=
I S−1R
0 I
Q− SR−1ST 0
0 R
I 0
R−1S I
▶ If
Q S
ST R
> 0, then so is the Schur complement of R in
Q S
ST R
.
▶ Nonlinear (convex) inequalities can be converted to an LMI form using
the Schur complement.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 18
Schur complement conditions:
Let A> 0. Then
A B
BT C
> 0 ⇐⇒ A> 0, C− BTA−1B> 0.
Let C > 0. Then
A B
BT C
> 0 ⇐⇒ C > 0, A− BC−1BT > 0.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 19
Proof: Consider the quadratic program
inf
(u,v)
u
v
T
A B
BT C
u
v
= uTAu+ 2uTBv+ vTCv
The solution of the minimization problem satisfies:
u
v
∗
= 0 (and attained uniquely at 0) ⇐⇒
A B
BT C
> 0.
First, solve
min
u
u
v
T
A B
BT C
u
v
= min
u
uTAu+ 2uTBv
+ vTCv
The optimum u is
d
du
=⇒ 2Au+ 2Bv= 0 ⇐⇒ u∗ = −A−1Bv
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 20
Observe
uTAu= uTATu= (−A−1Bv)TAT(−A−1Bv) = vTBTA−1Bv
and
uTBv= vTBTu= −vTBTA−1Bv
which leads to
uTAu+ 2uTBv
+ vTCv= vT
C− BTA−1B v
Now if we minimize over v, then the minimum is achieved and attained
uniquely at 0 iff C− BTA−1B> 0. □
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 21
Good to Note:
A block matrix
A B
BT C
< 0
is equivalent to
C < 0, A− BC−1BT < 0 ⇐⇒
−A B
BT −C
> 0
or
A< 0, C− BTA−1B< 0 ⇐⇒
C BT
B A
< 0
⇐⇒
−C BT
B −A
> 0
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 22
Extensions to the Schur complement:
A− BC−1BT Q
QT R
< 0 ⇐⇒
A B QBT C 0
QT 0 R
< 0
⇐⇒
C BT 0B A Q
0 QT R
< 0
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 23
P Q
QT A− BC−1BT
< 0 ⇐⇒
P Q 0QT A B
0 BT C
< 0
⇐⇒
P 0 Q0 C BT
QT B A
< 0
Exercise: Proof
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 24
■ Redundant (Slack) Variables: Parametrization, Dilated LMI conditions
Assume there exists X > 0 with
I A
−X 0
0 X
I
AT
= AXAT − X < 0
Equivalent reformulation: There exist X and G with −X AG
GTAT X −G−GT
< 0
▶ New parametrization independent of the original variables
▶ Dilated synthesis approaches always encompass the corresponding
standard-LMI-based ones
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 25
Equivalent reformulation
I A
−X 0
0 X
I
AT
+
I A
A
−I
G
0 I
I
AT
+
I A
0
I
GT
AT −I I
AT
=
I A
−X AG
GTAT X −G−GT
I
AT
=
I A
−X 0
0 X
I
AT
= AXAT − X
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 26
Exercise: Equivalent reformulation
i) ∃ P> 0 s.t.
ΦTPΦ− P< 0
ii) ∃ P> 0 s.t. −P ΦTP
PΦ −P
< 0
iii) ∃ P> 0 and G s.t. −P ΦTGT
GΦ P− (G+GT)
< 0
iv) ∃ P> 0 and G and F s.t. −P−ΦTF− FTΦ FT +ΦTGT
F+GΦ P− (G+GT)
< 0
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 27
Proof:
ii) → iii) Apply Schur complement by choosing G= GT = P> 0
ΦTPΦ−P< 0 =⇒
−P ΦTP
PΦ P
< 0 =⇒
−P ΦTGT
GΦ P− (G+GT)
< 0
iii) → ii)
I
Φ
T −P ΦTGT
GΦ P− (G+GT)
I
Φ
= ΦTPΦ− P< 0
ii) → iv) ∃ P> 0 s.t. ΦTPΦ− P< 0 =⇒ ∃ F s.t.
ΦTPΦ− P+ FP−1F < 0.
ΦTPΦ− P+ FP−1F = −P−ΦTF− FTΦ+ (F+ PΦ)T P−1 (F+ PΦ)
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 28
Apply Schur complements by choosing G= GT = P> 0 as in the proof
of ii) → iii)
− P−ΦTF− FTΦ+ (F+ PΦ)T P−1 (F+ PΦ)< 0 =⇒ −P−ΦTF− FTΦ FT +ΦTGT
F+GΦ P− G+GT
< 0
iv) → ii)
I
Φ
T −P−ΦTF− FTΦ FT +ΦTGT
F+GΦ P− (G+GT)
I
Φ
= ΦTPΦ− P< 0
□
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 29
■ Elimination of matrix variables:
Lemma 3.1 (Projection Lemma)
Given a symmetric matrix Ψ and two matrices U, V, there exist M such that
Ψ +UMVT + VMTUT < 0 (3.4)
iff the projection inequalities
U˜TΨU˜ < 0, V˜TΨV˜ < 0 (3.5)
hold, where U˜ and V˜ are orthogonal complements (bases of the null spaces) of
U and V, respectively, i.e., U˜TU = 0 and V˜TV = 0.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 30
Proof: Using U˜TU = 0 and V˜TV = 0, we find
U˜T
Ψ +UMVT + VMTUT
U˜ = U˜TΨU˜ + U˜TUMVTU˜ + U˜TVMTUTU˜
= U˜TΨU˜ < 0
and
V˜T
Ψ +UMVT + VMTUT
V˜ = V˜TΨV˜ + V˜TUMVTV˜ + V˜TVMTUTV˜
= V˜TΨV˜ < 0
□
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 31
■ Generic LMI Problems:
▶ Feasibility problem (FEASP):
Given an LMI F(x)> 0, find xfeas s.t. F(xfeas)> 0 or determine that the
LMI is feasible.
e.g.
▶ Given Ai ∈ Rn×n, i= 1, . . . ,n, find X satisfying the LMI
X > 0, ATi X + XAi < 0, i= 1, . . . ,n
or determine that no such X exists.
▶ Given Ai ∈ Rn×n, i= 1, . . . ,n, and γ > 0, find X satisfying
ATi X + XAi + C
TC XB
BTX −γI
< 0, i= 1, . . . ,n, X > 0
or determine that no such X exists.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 32
▶ Find X satisfying the LMI:
AXAT − X < 0, X > 0
▶ Find a soln. X to N TL(X)N
minimize t
subject to: L(X)−R(X)≤ t I
Feasible iff tmin ≤ 0.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 33
▶ Linear objective minimization problem (MINCX):
Given an LMI F(x)> 0, minimize a linear objective with F(xmincx)> 0
being satisfied.
e.g.
▶ minimize cTx subject to F(x)> 0
▶ minimize tr (X) subject to F(X)> 0
▶ minimize t subject to F(X)< t I (with t as a DV)
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 34
▶ Generalized eigenvalue problem (GEVP):
Given an LMI F(x,λ)> 0, minimize λ with F(xgevp,λ)> 0 being
satisfied.
e.g.
▶ minimize λ subject to λB(x)− A(x)> 0, B(x)> 0, C(x)> 0
▶ maximize α subject to ATX + XA+ 2αX < 0, X > 0
▶ minimize γ subject to
ATX + XA+ CTC XB
BTX −γI
< 0, X > 0
▶ minimize γ subject to L(X)< γR(X), L(X)> 0, R(X)> 0
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 35
Solving LMIPs
Develop an LMI system:
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 36
▶ Assign a name to the LMI system
e.g. lmisys
▶ Describe the matrix variables and LMIs
▶ Write the commands as an m file.
e.g. lmimodel.m
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 37
% Load an LMI model:
lmimodel % in LMIEDIT, write commands as an m file
lminbr(lmisys) % # LMIs
matnbr(lmisys) % # matrix variables
lmiinfo(lmisys) % info. about variables and term content
% Choose a solver for generic LMIP: feasp/mincx/gevp
% <==== Call feasp/mincx/gevp here
% Solution matrix from the decision vector(s)
Xopt=dec2mat(lmisys,xopt,X)
% Evaluate LMIs to check if the DV(s) indeed feasible
evlmi=evallmi(lmisys,xopt);
[lhs,rhs]=showlmi(evlmi,1) % LHS and RHS
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 38
e.g. Choosing a solver
▶ Feasibility problem: feasp
[tmin,xfeas]=feasp(lmisys); % feasible iff tmin<=0
Note: feasp solves minimize tmin for L(·)−R(·)≤ tminI
▶ Linear objective problem: mincx
ndc=decnbr(lmisys); % total # decision variables
c=zeros(ndc,1); c(1:q)=nu; % for nu*x(1:q)
c=zeros(ndc,1); c(diag(sX))=1; % for trace(X)
[cxopt,xopt]=mincx(lmisys,c);
▶ Generalized eigenvalue problem: gevp
[mu,xopt]=gevp(lmisys,nlfc); % last nlfc LMIs involving obj.
Note: gevp solves minimize γ for L(·)< γR(·) with R(·)> 0.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 39
Note: “Solve” means that we can find DVs that satisfy the LMIP, or
determine that no solns. exist.
Note: The dimension of the variable Y = YT ∈ Rn×n is n(n+ 1)/2:
Y11
Y12
Y22
Y13
...
Ynn
∈ R
n(n+1)/2 or
Y11
Y21
Y22
Y31
...
Ynn
∈ R
n(n+1)/2
Note: The dimension of the variable Y ∈ Rn×n is n2.
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 40
e.g.
>> n=3; setlmis([])
>> [X,nX,sX]=lmivar(1,[n 1])
X =
1
nX =
6
sX =
1 2 4
2 3 5
4 5 6
>> [Y,nY,sY]=lmivar(2,[n n])
Y =
2
nY =
15
sY =
7 8 9
10 11 12
13 14 15
>>
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 41
e.g. Linear objective
Given
A1 =
−1 2
1 −3
, A2 =
−0.8 1.5
1.3 −2.7
Find Y > I to minimize trY satisfying
I < Y
AT1Y + YA1 < Y
AT2Y + YA2 < Y
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 42
A1=[-1 2; 1 -3]; A2=[-1.4 0.9; 0.7 -2];
setlmis([]);
[Y,nY,sY]=lmivar(1,[2 1]);
lmiterm([1 1 1 0],1); % LMI #1: I
lmiterm([-1 1 1 Y],1,1); % LMI #1: Y
lmiterm([2 1 1 Y],A1,1); % LMI #2: A1*Y
lmiterm([2 1 1 Y],1,A1’); % LMI #2: Y*A1’
lmiterm([-2 1 1 Y],1,1); % LMI #2: Y
lmiterm([3 1 1 Y],A2,1,’s’); % LMI #3: A2*Y+Y*A2’ (with FLAG=’s’)
lmiterm([-3 1 1 Y],1,1); % LMI #3: Y
lmis=getlmis;
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 43
>> ndc=decnbr(lmis);
>> c=zeros(ndc,1); c(diag(sY))=1;
>> [cyopt,yopt]=mincx(lmis,c);
>> cyopt
cyopt =
2.0012
>> Yopt=dec2mat(lmis,yopt,Y)
Yopt =
1.0005 0.0000
0.0000 1.0007
>> evlmi=evallmi(lmis,yopt);
>> [lhs,rhs]=showlmi(evlmi,3)
lhs =
-2.8013 1.6008
1.6008 -4.0027
rhs =
1.0005 0.0000
0.0000 1.0007
>> eig(lhs-rhs)
ans =
-6.1123
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 44
-2.6928
>> eig(A2*Yopt+Yopt*A2’-Yopt)
ans =
-6.1123
-2.6928
>>
>> [lhs,rhs]=showlmi(evlmi,2)
lhs =
-2.0008 3.0017
3.0017 -6.0040
rhs =
1.0005 0.0000
0.0000 1.0007
>> eig(lhs-rhs)
ans =
-8.6108
-1.3951
>> eig(A1*Yopt+Yopt*A1’-Yopt)
ans =
-8.6108
-1.3951
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 45
Note: To solve a FEASP, we need to write LMI conditions
L(X)
L(X)−R(X)< tminI ≤ 0
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 46
e.g. Feasibility
Given
A1 =
−1 2
1 −3
, A2 =
−0.8 1.5
1.3 −2.7
Find Y > I satisfying
I < Y
AT1Y + YA1 < Y
AT2Y + YA2 < Y
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 47
>> [tmin,yopt]=feasp(lmis);
>> tmin
tmin =
-2.9180
>> Yopt=dec2mat(lmis,yopt,Y)
Yopt =
99.9899 27.8674
27.8674 54.9808
>> evlmi=evallmi(lmis,yopt);
>> [lhs,rhs]=showlmi(evlmi,3)
lhs =
-229.8104 24.7265
24.7265 -180.9089
rhs =
99.9899 27.8674
27.8674 54.9808
>> eig(lhs-rhs)
ans =
-329.9052
-235.7848
>> eig(A2*Yopt+Yopt*A2’-Yopt)
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 48
ans =
-329.9052
-235.7848
>>
>> [lhs,rhs]=showlmi(evlmi,2)
lhs =
-88.5102 98.4820
98.4820 -274.1501
rhs =
99.9899 27.8674
27.8674 54.9808
>> eig(lhs-rhs)
ans =
-358.4683
-159.1628
>> eig(A1*Yopt+Yopt*A1’-Yopt)
ans =
-358.4683
-159.1628
>>
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 49
Note: To solve a GEVP, we need to write LMI conditions
L(X)
The GEVP solver finds λmin to guaaraantee
L(X)< λminR(X).
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 50
e.g. Generalized eigenvalue problem
Given
A1 =
−1 2
1 −3
, A2 =
−0.8 1.5
1.3 −2.7
Find Y > I to minimize λ:
I < Y
AT1Y + YA1 < λY
AT2Y + YA2 < λY
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 51
>> [lambda,yopt]=gevp(lmis,2); % the last 2 lmis are LFTs
>> lambda
lambda =
-0.5359
>> Yopt=dec2mat(lmis,yopt,Y)
Yopt =
493.1196 -671.1353
-671.1353 917.6951
>> evlmi=evallmi(lmis,yopt);
>> [lhs,rhs]=showlmi(evlmi,3)
lhs =
1.0e+03 *
-2.5888 3.4530
3.4530 -4.6104
rhs =
493.1196 -671.1353
-671.1353 917.6951
>> eig(lhs-lambda*rhs)
ans =
1.0e+03 *
-6.4423
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 52
-0.0008
>> eig(A2*Yopt+Yopt*A2’-lambda*Yopt)
ans =
1.0e+03 *
-6.4423
-0.0008
>> [lhs,rhs]=showlmi(evlmi,2)
lhs =
1.0e+03 *
-3.6708 5.0131
5.0131 -6.8484
rhs =
493.1196 -671.1353
-671.1353 917.6951
>> eig(lhs-lambda*rhs)
ans =
1.0e+03 *
-9.7632
-0.0000
>> eig(A1*Yopt+Yopt*A1’-lambda*Yopt)
ans =
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 53
1.0e+03 *
-9.7632
-0.0000
>>
S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 54
