Linear Matrix Inequalities in Control

Linear Matrix Inequalities (LMI) 

in Control 

Wojciech Kozinski PhD 

Warsaw University of Technology 

Faculty of Electrical Engineering 

Institute of Control and Industrial Electronics 

Scope of the lecture 

• Control – classical vs. modern 

• Optimisation approach 

• Computer tools support 

• LMI primer 

• Basic LMI tools for control 

• Analysis 

• Synthesis 

• Conclusions 

KSO materiały do wykładu 2008/09 2 

Control – classical vs. modern (1) 

• Classical control offers (almost) no guaranty on 

performance or robustness of the closed loop, 

• „Optimistic” design starting from desired closed loop 

behaviour, 

• Classical control techniques (only some possibilities): 

– pole placement, 

– LQR approach, 

– lead and lag units, 

– root locus, 

– algebraic approach (solving diophantine equation), 


1

Control – classical vs. modern (2) 

• Modern control techniques, 

– H analysis and synthesis, 

– H 2 

analysis and synthesis, 

– mixed criteria (+ pole placement), 

• They offer upper bounds estimates on performance and 

robustness of the closed loop, 

• They can tackle difficult problems, 

• Some disadvantages (high order controllers, a lot of 

mathematics, unable to solve some important problems). 








• Analysis 

• Synthesis 



Optimisation approach (general) 

w 

p 

u 

Δ 

P 

K 

q 

y 

z 

x 

q 

y 

z 

u 

p 

Ax B1 

p B2w 

B3u 

C x D p D w D u 

q 

C x 

y 

Czx 

K y 

q 

D 

D 

q1 

y1 

z1 

p 

p 

q2 

D 

y2 

w 

D w D u 

z2 

q3 

z3 

find 

min 

K 

T 

zw p 

forall 


2

Optimisation approach (2) 

scalar signal norms 

L 

L 2 

norm - amplitude 

f ( t) 

sup f ( t) 

f 

t 0 

norm - energy 

1 

( d 

2 

2 

2 

f t) 

dt fˆ( 

) 

2 

0 

2 



system norm H 

System description (causal, with L 2 gain γ>=0) 

x ( t) 

Ax( 

t) 

Bf ( t) 

y( 

t) 

Cx( 

t) 

Df ( t) 

H( 

s) 

C( 

sI 

Norm definition L 2 gain 

T 

T 

2 

2 

y( 

t) 

dt 

c 

0 

0 

H sup H( 

j 

R 

2 

f ( t) 

dt 

Norm interpretation 

) H 

sup max 

R 

( H( 

j 

A) 

)) 

1 

B 

D 



system norm H calculations 

Theorem 

Let A be a Hurwitz (stable) matrix. Then the L 2 gain of the 

system is less than γ if and only if max( 

D) 

and the matrix 

F 

2 

A B( 

I 

T T 

C C C D( 

T 1 T 

D D) 

D C 

2 T 1 T 

I D D) 

D C 

2 T 1 T 

B( 

I D D) 

B 

T T 2 T 

A C D( 

I D D) 

1 

B 

does not have eigenvalues on the imaginary axis. 

Hint : use bisection method. 


3

H 

2 

2 


– norm H 2 

If the input into system is white noise then output 

variance is called H 2 norm 

lim E y( 

t) 

t 

0 

2 

H 

If h(t) is impulse response matrix of the system, then 

T 

trace( 

h( 

t) 

h( 

t)) 

dt 

2 

2 

1 

2 

T 

trace( 

H( 

j ) H( 

j )) d 



system norm calculations 

H 2 

Theorem: 

Let A be a Hurwitz matrix. The H 2 norm of the system is 

infinite if D 0. 

Otherwise, it equals: 

2 

T 

( ) trace T 

H trace CPC ( B QB) 

2 

Where P=P T and Q=Q T are the unique solutions of the 

Lapunov equations: 

AP 

T 

PA 

BB 

T 

T 

T 

0; 

A Q QA C C 

0 



Linear Fractional Transformation (LFT) 

z1 

y1 

M 

l 

w1 

u1 

z1 

y 

u 

T 

1 

1 

zw1 

w1 

M 

u 

l 

y 

F 

1 

M 

M 

11 

21 

M 

M 

12 

22 

w1 

u 

1 

1 

l 

( M, 

l 

) M11 

M12 

l 

( I M22 

l 

) M21 

1 

z2 

y2 

u 

M 

u2 

w 2 

z2 

y 

u 

T 

2 

2 

zw2 

w2 

M 

u 

u 

F 

y 

2 

M 

M 

11 

21 

M 

M 

12 

22 

w2 

u 

2 

1 

u( M, 

u) 

M22 

M21 

u( 

I M11 

u) 

M12 

2 


4


Analysis - example 

w 

u 

e 

F C Go 

q 

Δ 

+ 

p 

z 

- 

G 

0 

1 

; C 

2 

s 

Find the largest 

2.35(1 s) 

; 

(1 0.1s 

) 

F 

1.2 

; 

s 1.2 

and maintain stability 



Synthesis - example 

w 

- 

1 

s 

- 

10 

z 

Gi(s) G i 

( s) 

; 

2 

s as b 

x 

K a 1..3 ; b 8..12 ; 

For all plants G i (s) find stabilising 

static state space controller K. 



when the synthesis is well-posed? 

w 

u 

P(s) 

K(s) 

Closed loop description: 

cl 

y 

z 

x 

Ax B1w 

B2u 

z C1x 

D11w 

D12u 

y C x D w D u 

x f 

u 

2 

A x 

C x 

f 

f 


f 

f 

21 

D y 

22 

B y 

x A x B w x 

cl cl cl 

xcl 

z Cclxcl 

Dclw 

x 

f 

Minimize some measure ( T ) ( F ( P, 

K)) 

w z 

l 

f 

f 

5


when the synthesis is well-posed? 

A 

C 

cl 

cl 

A B2D f 

C2 

B C 

C 

1 

f 

12 

2 

D D C 

f 

2 

B2C 

f 

A 

f 

D C ; D 

12 

; B 

f 

cl 

cl 

B1 

Df 

D 

B D 

D 

11 

f 

21 

12 

21 

D D D 

1. Simultanouos stabilisability of the pair (A,B 2 ) and observability of the pair 

(C 2 ,A). 

2. D-matrix conditions: 

1. D 22 =0, 

2. D cl =0 (if H 2 norm is used), 

3. often D 11 =0 required (Robust Toolbox), 

4. D 12 full column rank matrix, 

5. D 21 full row rank matrix. 

f 

21 


Optimisation approach (warning) 

• Many (all ?) control analysis and synthesis problems can be formulated 

as optimisation problems. 

• Only some of them can be transformed into LMI (not BMI), solvable 

problems (e.g. there exists no general output controller synthesis for 

uncertain plant – like LFT framework). 

• As LMI – the price is paid in large number of slack (Lapunov) 

variables. 

• Many interesting papers on BMI solvers and non-smooth, non-convex 

optimisation have been published recently. P.Apkarian ONERA-CERT, 

D. Noll Uni Toulouse, M.Overton, Uni of New York. 








• Analysis 

• Synthesis 



6

Computer tools support 

• Matlab ® Robust Control Toolbox 

• Public domain SDP-solvers (Matlab ® based): 

– SP (Boyd & Vanderberghe) –old, 

– SeDuMi (Sturm, now – McMaster University), 

• Public domain preprocessors 

– YALMIP (Löfberg in ETH Zurich): 

• Many other possibilities (see YALMIP www 

page) 








• Analysis 

• Synthesis 



F( 

x) 

LMI Primer (1) - definition 

LMI is a matrix dependent on vector x 

F 

0 

m 

i 1 

T 

z Fz 

x i 

F i 

0, 

z 

0 

where m 

x R , Fi 

Positive definite matrix F is such that 

0, z 

n 

R . 

n n 

If LMI has a solution (is feasible) then the solution is a 

non empty set. The solution set is convex. 

x| 

F( 

x) 

0 


R 

7

LMI Primer (2) 

There are many LMIs which have the same solution: 

Let M R 

n n 

; det( M) 

0 and x M z 

then by congruence can be obtained: 

T 

T T 

T 

A 0 x Ax 0 z M AMz 0 M AM 

An example: 

A B 

0 

C D 

0 I A B 

I 0 C D 

0 I 

0 

I 0 

D C 

0 

B A 

0 



Multiple LMIs can be expressed as a single LMI. 

Set defined by by q LMIs: 

F( 

x) 

1 

F ( x) 

2 

0; F ( x) 

An equivalent single LMI: 

where 

q 

0;...; F ( x) 

m 

0 

xiFi 

x 

i 1 

F 

1 2 

q 

diag F ( x) 

F ( x) 

... F ( ) 0 

1 2 

q 

Fi diagFi 

Fi 

... Fi 

; i 0,1 ,..., m 

0 



Linear constraints can be expressed as an LMI: 

General linear constraint: 

Ax 

b 

where 

n 

b R , 

A 

R 

, 

n m 

x 

m 

R . 

can be converted to n scalar inequalities in LMI form 

m 

bi Aijx 

j 

0; i 1,..., 

n 

j 1 

and then into single LMI. 


8


The Schur complement lemma: 

Converts convex nonlinear inequality 

R( 

x) 

where 

0, 

Q( 

x) 

1 T 

S( 

x) 

R( 

x) 

S( 

x) 

T 

T 

Q( x) 

Q( 

x) 

, R( 

x) 

R( 

x) 

0 and S( 

x) 

into the equivalent LMI 

Q( 

x) 

T 

S( 

x) 

S( 

x) 

R( 

x) 

Very important lemma! 

0 

0 

depend affinely on x 


The condition 

( A( 

x)) 

1 

I 

A( 

x) 

T 


Maximum singular value: is defined as: 

T 

( A( 

x)) 

max( 

A( 

x) 

A( 

x) 

) 

where matrix A(x) depends affinely on x 

( A( 

x)) 

1 can be expressed as a LMI 

A( 

x) 

A( 

x) 

A( 

x) 

0 

I 

T 

I I 

A( 

x) 

A( 

x) 

T 

0 


whereP 


Algebraic Riccati inequality: 

T 

1 T 

A P PA PBR B P Q 0 

P 

T 

T 

0; Q Q ; R 0; 

can be expressed as a LMI 

T 

A P PA 

T 

B P 

Q 

PB 

R 

0; 

P 

0. 


9


• Three types of LMI problems 

– feasibility problem LMIP, 

A(x) 0 

– eigenvalue problem EVP, 

minimum c T x subject to A(x) 0 

– generalised eigenvalue problem GEVP, 

minimum λ subject to A( x) 

B( 

x); 

B( 

x) 

0; C( 

x) 

where 

T 

T 

B ( x) 

B( 

x) 

0; A( 

x) 

A( 

x) 

0. 








• Analysis 

• Synthesis 



Basic LMI tools for control (1) 

Stability 

Autonomous linear system 

x 

Ax 

T 

Lapunov function V( x) 

x Px 0 except x 

Condition on it’s derivative: 

dV( 

x) 

dt 

x Px 

T 

x Px 

T T 

x ( A P 

LMI for stability A T P PA 

0 

PA) 

x 

0 

P 

0 

T 

A P 

0 

PA 

0 


10


Stability 

Equivalence between LMI definition and matrix formula 

p1 

p2 

Example for n=2 P 

0 

p p 

A T P PA 

2a 

p1 

a 

11 

12 

a12 

0 

a 

a 

11 

12 

a 

a 

21 

22 

11 

p1 

p 

2 

2 

2a21 

p2 

a a 

22 

p2 

p 

3 

3 

a11 

a 

2a 

12 

p1 

p 

2 

22 

p2 

p 

3 

a 

a 

11 

21 

0 

p3 

a 

21 

a 

a 

12 

22 

a 

2a 

21 

22 



Stability 

Additional conditions on s-plane regions 

Guaranteed damping 

C stab 

{ s C | re( 

s) 

} s s 2 0 

Maximal damping and oscillation 

r s q 

C stab 

{ s C | s q r} 

0 

s q r 

Vertical strip 

( s s) 

2 

2 

0 

C stab 

{ s C | 

1 

re( 

s) 

2} 

0 

0 ( s s) 

2 

1 



Stability 

Theorem: 

All eigenvalues of the real matrix A are in the region 

described by C 

stab 

{ s C| 

P Qs 

T 

Q s 0} 

with P 

T 

P 

if and only if there exists symmetric matrix X=X T >0 

p X 

p 

11 

k1 

X 

q11AX 

 

q AX 

k1 

T 

q11XA 

 

 

T 

q XA 

1k 

T 

p1 

k 

X q1 

k 

AX qk1XA 

 

T 

p X q AX q XA 

kk 

kk 

kk 

0 

Equivalent LMI using Kronecker multiplication: 

P X Q AX 

Q 

T 

XA 

T 

0 


11


Stability – example - feasibility 

For given real matrix A check 

it its eigenvalues are inside 

the selected region: 

• usage of previously presented 

theorem, 

• feasibility type problem, 

r 

-4 -2 

q=-3 

Im 

1 

Re 

0 

-1 


Optimiser: SeDuMi 

Demo no.1 

Preprocesor: YALMIP 

Given is matrix A with eigenvalues -1 j1. If radius of the 

circle with origin (-3,0) is greater then sqrt(5) the problem 

is feasible, otherwise unfeasible. 



Stability of non-linear/time-varying 

Two admissible descriptions of autonomous non-linear 

or time varying system in form of convex hull of matrices: 

x t) 

A( 

t) 

x( 

t), 

A( 

t) 

Co{ 

A, 

A ,... A } 

( 

1 2 L 

L 

x ( t) 

A( 

t) 

x( 

t), 

A( 

t) 

LMIs to prove stability: 

i 

A i 

i 1 

T 

P P 0 

T 

Ai P PAi 

0, i 1,..., 

L 

, 

i 

0, 

L 

i 

i 1 

1. 


12


Bounded real lemma 

The system definition 

x Ax Bu 

G( 

s) 

sup ( G( 

s)) 

sup ( G( 

j )) 

Re( s) 

0 

R 

y Cx Du 

x(0) 

0 

G( 

s) 

1 

C( 

sI A) 

B 

Minimize γ 2 subject to: 

T 

T 

T 

A P PA C C PB C D 

T T 

T 2 

B P D C D D I 

D 

Find (the upper bound) 

0; P 

P 

T 

0 



S-procedure (simplified) 

If there exist symmetric matrices T 0 ,...,T p and nonnegative 

numbers 

1 

0,..., 

p 

0 such that 

T 

0 

i 

p 

1 

iT i 

0 

T 

T 

then x T0 x 0; x 0if 

x Ti 

x 0; i 1, , 

p. 

For p=1 the reverse is true. 



S-procedure example 

There are two condtions given for P 

P 

T 

0, 

0 and any 

T 

T 

A P PA 

T 

B P 

PB 

0 

0 

T C T C 

T 

A P 

PA 

T 

B P 

T 

C C 

PB 

I 

0; P 

P 

T 

T . 

Using S-procedure one can obtain LMIs: 

0; 

0 


13


H inf norm calculations 

For system: 

x 

y 

Ax 

Cx 

Minimize γ subject to: 

T 

A X XA 

T 

B X 

C 

XB 

I 

D 

T 

C 

T 

D 

I 

Bu 

Du 

0; X 

Variable γ is the upper bound of H inf 

0; whereX 

T 

X . 



Let 

G( s) 

2 

s 

10 

2s 

10 

Demo no.2 

The state space representation is 

A 

2 

8 

1.25 

; B 

0 


1 

; C 

0 

0 1.25; D 0. 

Find norm H inf and compare it with the result of Matlab ® 

function. 



H 2 norm calculations 

For system: 

x 

y 

Ax 

Cx 

Bu 

Minimize trace(Q) subject to: 

T T 

AP PA BB 0 

Q 

PC 

T 

whereP 

CP 

P 

T 

P , Q 

0; P 

0 

T 

Q . 


14


Demo no.3 


Let 

G( s) 

2 

s 

10 

2s 

10 

The state space representation is 

A 

2 

8 

1.25 

; B 

0 

1 

; C 

0 

0 

1.25; D 

0. 

Find norm H 2 and compare it with the result of Matlab ® 

function. 








• Analysis 

• Synthesis 



Analysis (1) 

Robustness problems 

1. Robust stability, 

2. Robust performance. 

Question: what is the biggest admissible Δ ? 

G G G 0 

( I ) 

G 0 

Δ 

Δ 

G0 

G0 


15

Analysis (2) 

Uncertainity descriptions 

1. Norm-bounded uncertainity without structure 

1 

( s) 

W( 

s) 

2. Norm-bounded uncertainity with structure 

Δ partitioned into blocks with their own bounds 

3. Sector-bounded uncertainity (dynamic or static) 

mapping : u L y L 

0 

( y( 

t) 

2 

T 

au( 

t)) 

( y( 

t) 

2 

bu( 

t)) 

dt 

1 

0 


Analysis (3) 

Robust stability – problem statement 

w 

p 

Δ 

P 

q 

z 

x 

q 

z 

p 

Ax B1 

p B2w 

C1x 

D11p 

D12w 

C2x 

D21p 

D22w 

q 

1 

Is the system stable for all admissible Δ? 


M 

1 

Analysis (3) 

Robust stability – LMIs 

T 

A P PA 

T 

B P 

PB 

0 

B 

B 1 

B 2 

M 

2 

C2 

0 

D 

0 

21 

T 

22 

D 

I 

I 

0 

0 

2 

I 

C2 

0 

D 

0 

21 

D 

I 

22 

M 

3 

C1 

D 

0 I 

11 

T 

D12 

I 0 

0 0 I 

C1 

D11 

D 

0 I 0 

min s. 

t.( 

M1 M2 

M3 

0; P 0; 

12 

0) 


16

Analysis (4) 

Robust stability – example 

q Δ 

k 

w 

u 

e 

F C G o 

k 

p 

z 

2 

k 

q 

z 

2 

k G0C( 

I G0C) 

1 

k( 

I G C) 

0 

1 

kG0CF( 

I 

G CF( 

I 

0 

G0C) 

G C) 

0 

1 

1 

p 

w 


A 

C 

A 

Analysis (5) 

Robust stability – example 

G 

0 

C1 

C 

k max 

2 

0 

1 

; C 

2 

s 

B0Dr 

C 

BrC0 

0 

kC0 

C 

0 

0 

0.748 

B0Cr 

Ar 

0 

2.35(1 s) 

; F 

(1 0.1s 

) 

0 0 

; D 

0 0 

B0Dr 

C 

f 

BrC 

f 

A 

f 

D 

D 

11 

21 

; B 

D 

D 

12 

22 

1.2 

; 

s 1.2 

B0Dr 

k 

Brk 

0 

0 

k 

0 

0 


; 

B0Dr 

Df 

Br 

Df 

B 

f 

; 

Demo no.4 



Run the presented example with some simulations to 

confirm the result. 


17

Analysis (6) 

Integral Quadratic Constraints (IQC) 

Powerful method for investigating stability of uncertain systems 

. 

Let pˆ and qˆ 

be Fourier Transforms of p and q signals. 

. 

IQC is defined by ( j ) where 

pˆ( 

j 

qˆ( 

j 

) 

) 

* 

( j 

pˆ( 

j 

) 

qˆ( 

j 

) 

d 

) 

0 


Analysis (7) 


For autonomous system 

x 

( t) 

Ax( 

t) 

B p( 

t) 

q( 

t) 

p( 

t) 

C x( 

t) 

q 

q( 

t); 

p 

D p( 

t) 

qp 

01 

For small ε>0 it is true 

H( 

j ) 

I 

* 

H( 

j ) 

( j ) 

I 

H( 

s) 

C ( sI 

1 

A) 

B 


q 

2 I for all 

Kalman-Yakubovich-Popov (KYP) lemma converts this 

condition into LMI. 

R 

p 

D 

qp 

Analysis (8) 


Kalman-Yakubovich-Popov (KYP) lemma: 

Let the pair of matrices (A,B) is stabilisable and matrix A has no 

eigenvalues on imaginary axis. Then the statements are equivalent: 

( j 

1 

I A) 

B 

I 

T 

PA A P 

T 

B P 

* 

PB 

0 

( j 

1 

I A) 

B 

I 

0, P 

There is powerful Matlab ® toolbox called iqc-beta (from MIT). 

P 

T 

0, 

0 

0, 


18







• Analysis 

• Synthesis 



Synthesis (1) 

What can be done within LMI framework? 

• Design of static, state controller for uncertain 

plants with H 2 /H inf norm minimization, pole 

placement, 

• Design of dynamic, output controller for nominal 


placement, 

• Design of dynamic, output controller for uncertain 


placement, if uncertainity measurment is available 

(so called gain-scheduling). 


Synthesis (2) 

Typical strategy 

Pole placement and H inf +H 2 norms minimisation, 

• Select the pole placement region and run the test to 

check, if it is feasible, 

• Minimise H inf norm to find the smallest possible one 

and controller, 

• For several, greater but fixed values of H inf norm 

minimise H 2 norms, 

• Plot H 2 norms versus H inf fixed norms (Pareto like 

curve), 

• Select trade-off point or discard the design. 


19

Synthesis (3) 

Design of static, state controller 

w 

x 

z 

+ 

u 

A x 

p 

C x 

p 

u Kx 

G 

K 

x 

B ( w 

p 

D ( w 

p 

z 

u) 

u) 

x 

z 

z 

z 

2 

w 

u 


P 

K 

x 

z2 

zinf 

z 

Ax Buu 

Bww 

C x D 

uu 

D 

ww 

C2x 

D2uu 

D2 

ww 

C x D u D w 

z 

zu 

zw 

2 

Synthesis (4) 


Norms definitions (in signal terms) 

Matrix definitions for general setup 

B 

C 

C 

C 

u 

z 

B ; 

p 

C ; 

Cp 

; 

0 

C ; 

p 

p 

B 

D 

w 

D 

D 

zu 

B ; 

u 

2u 

p 

D ; 

p 

D ; 

p 

Dp 

; 

1 

D 

zw 

D 

D 

w 

2w 

D ; 

p 

z w z; 

z2 

1 D ; 

p 

Dp 

; 

0 

z 

u 



G( 

s) 

1 

2 

s 

Demo no.5 


a) Pole placement and H inf minimisation to obtain lower 

bound, 

b) Several runs – pole placement and H 2 minimisation for 

fixed values H inf ,; generate Pareto-like curve, 

c) Pole placement and H inf + H 2 minimisation. 


20

Synthesis (5) 


Analysis of the design (closed loop) 

x A x B w; x x; A A B K B B ; 

u 

z 

z 

z 

cl 

2 

cl 

cl 

cl 

K xcl; 

( C D 

uK) 

x 

( C2 

D2u 

K) 

xcl 

( C D K) 

x 

z 

zu 

cl 

If norm H 2 is used then 

cl 

cl 

D 

ww; 

D2 

ww; 

D w; 

D 2w 

zw 

0. 

cl 

; 

cl w 


u 

Synthesis (6) 


Another structure for uncertain plants. 

w 

- 

e 

1/s 

v 

+ 

u 

G 

K 

x 

z 

No problems with static gain. 



Demo no.6 

10 

s) 

; 

s as b 

1..3 ; b 8..12 ; 

G i 

( 

2 

a 


Design state space controller for all „corner” plants 

using pole-placement. 


21

Synthesis (7) 

Design of dynamic, output controller 

w 

- 

u 

x Ax 

z Cx 

k 

x 

u 

Ak 

x 

C x 

k 

z 

G 

y 

Gk 

B( 

w u) 

D( 

w u) 

k 

k 

Bk 

y 

D y 

k 

w 

u 

P 

Gk 

z w z; 

z2 

y 

z 

u 

z2 


z 

Design norms as an example 


Synthesis (8) 


General design setup 

x 

Ax Buu 

Bww 

z C x D 

uu 

D 

ww 

z2 

C2x 

D2uu 

D2 

ww 

y C x D u D w 

z 

C x 

z 

y 

- H 2 norm 

yu 

D u 

D 

zu 

D D 

D 

yw 

zw 

w 

D 

2u 

k yw 2w 

D-matrix considerations: 

- well-posed feedback 

det( I D k 

Dyu) 

0 

- plant (usually) strictly proper 

D D yu 

0; D 0. 

0 

yw 

- well-posed optimization 

D w 

0; 

D2u 

0. 

- freedom to select 

0; D or D 0 

k 

0 

k 


Synthesis (9) 


Structure of the closed loop and conversion rules: 

B B; 

B B; 

w 

z u 

w 

G 

C C; 

D D; 

D 

- 

u 

w 

u 

y C D 

Gk C2 

; D2u 

; D2 

w 

0 1 

Norms defined by: 

z w z; 

z2 

z 

u 

C 

C 

y 

z 

C; 

C; 

D 

D 

D; 

D; 

D; 

D; 


yu 

zu 

D 

D 

yw 

zw 

1; 

0 

0 

; 

22

Demo no.7 



1 

G( 

s) 

2 

s 

Design output, feedback controller using pole placement, 

and H inf + H 2 minimisation. 


Synthesis (10) 


Another closed loop structure and conversion rules 

w 

- 

y 

Gk 

u 

G 

Norms defined by: 

z w z; 

z2 

z 

u 

z 

Bu 

C 

C 

C 

C 

2 

y 

z 

B; 

Bw 

0; 

C; 

D 

C 

; D 

0 

C; 

D 

C; 

D 

zu 

u 

2u 

yu 

D; 

D 

D; 

D 

D; 

D 

zw 

2w 

yw 

w 

D 

; D 

1 

0; 

1; 

1; 

0 

; 

0 


Demo no.8 



G( 

s) 

1 

2 

s 

Design output, feedforward controller using pole placement, 

and H inf + H 2 minimisation. 


23


H-inf loop shaping 

z1 

W1 

w 

- 

y 

K 

u 

G 

d 

+ 

W3 

z3 

z 

Basic idea: 

W1 

( s) 

S( 

s) 

W ( s) 

T( 

s) 

3 

1 

The sensitivity functions: 

L( 

s) 

S( 

s) 

T( 

s) 

G( 

s) 

K( 

s) 

( I 

( I 

G( 

s) 

K( 

s)) 

1 

( I 

L( 

s)) 

1 

G( 

s) 

K( 

s)) 

G( 

s) 

K( 

s) 

I 

1 

S( 

s) 

( I 

1 

L( 

s)) 

L( 

s) 


Optimisation setup: 

w 

min 

K 

u 

T 

P 

K 

w z 


H-inf loop shaping 

y 

( s) 

z1 

z3 

z 

„Clever” selection of the 


z1 

z3 

y 

x 

z 

y 

z 

w 

P 

u 

C x 

W1 

0 

I 

D u 

WG 

1 

W3G 

G 

Ax Buu 

Bww 

C x D 

uu 

D 

ww 

C x D u D w 

z 

y 

yu 

zu 

yw 

D w 

zw 

w 

u 


Demo no.9 



G ( s) 

1 

G ( s) 

2 

G ( s) 

3 

1 a 

2 

s ( s a) 

G0 

( I ); 

1 

2 2 

2 

s ( s 2s 

) 

G0 

( I ); 

1 s 1 b s 

e ; 

2 

2 

s s b s 

G0 

( I ); 

filters W 1 and W 3 

1 

2 

2 

3 


( s) 

1 

( s) 

( s) 

3 

2 

s 

; 

s a 

2 

s 2s 

2 

s 2s 

2s 

; 

b s 

2 

; 

Design controller by the loop shaping method taking into 

account all possible neglected dynamics. 

24


Model predictive control (MPC) 

• For actual plant state x(k), find next M control u(k) 

steps (by optimisation), knowing next N (future) 

reference values. M

Demo no.10 



1. LQR control with control signal saturation 

(overshoot), 

2. MPC optimal control – no input signal saturation, 

3. MPC control with input signal saturation (LMI) 



Gain Scheduled H-inf Controller 

w 

p 

u 

? 

P(s) 

K(s) 

? 

y 

q 

z 

The plant description 

x 

A( 

) x B1 

( ) w B2 

( ) u 

z C1 

( ) x D11( 

) w D12( 

) u 

y C ( ) x D ( ) w 

A 

: 

R 

n 

2 

, D 

n 

1 


21 

p1 

m2 

12 

R , 

 

T 

L 

D 

The unknown parametres form 

a hypercube 

( t) 

, t 

i 

i 

i 

0 

21 

R 

p2 

m1 



w 

p 

u 

? 

P(s) 

K(s) 

? 

y 

q 

z 

Find a dynamic controller 

x 

u 

K 

AK 

( ) x 

C ( ) x 

K 

K 

K 

BK 

( ) y 

D ( ) y 

which ensures internal stability 

and a guaranteed L 2 

-gain 

T 

0 

T 

z zd 

T 

2 T 

0 

K 

w wd , 

T 

0 


26



• Gain scheduled H-inf controller can be 

calculated in three steps: 

1. After the application of elimination lemma one can 

obtain Lapunov matrices X and Y for all vertices of 

hypercube. 

2. For each vertex of hypercube calculate the vertex 

controller. 

3. For each value of parameter calculate the 

corresponding controller and use it in the control. 



Elimination Lemma 

• Definition. An orthogonal complement of given 

matrix N is any maximal rank matrix, such that 

T 

N N 0 and N N 0. 

n k n m 

T n n 

• Lemma. Let M R , N R , H H R 

The following statements are equivalent: 

m k 

1. There exists X R such that 

T T T 

NXM MX N H 0 

2. The following two conditions hold 

N HN 

T 

M HM 

T 

0 

0 




Step 1. Solve 2*2 L +1 LMIs for X and Y Lapunov matrices 

NX 

0 

0 

I 

T 

T 

XA A X 

T 

B1 

X 

C 

1 

XB1 

I 

D 

11 

T 

C1 

T 

D11 

I 

NX 

0 

0 

I 

0 

Where: 

N X 

null( C 2 

D21 

) 

) 

NY 

0 

0 

I 

T 

T 

YA AY 

C Y 

B 

1 

T 

1 

T 

YC1 

I 

D 

T 

11 

B1 

D11 

I 

NY 

0 

0 

I 

0 

N 

Y 

null 

T T 

( B2 

D12 

X 

I 

I 

Y 

0 


27



Step 2. Calculate controller for each vertex of hypercube 

by solving LMI. Some constraints can be added. 

XCl 

0 

T 

T T 

XCl 

0 

P KQ Q K P 

0 I 

0 I 

0 

where: 

AK 

B I X Y I 

K 

K 

XCl 

T T 

C D 0 M N 0 

MN T I YX 

0 

K 

A0 

X 

Cl 

X 

Cl 

A0 

T 

B0 

X 

Cl 

C 

K 

XClB0 

I 

D 

NB: Not all symbols are defined (see next slide). 

11 

T 

C0 

T 

T 

D11 

; P B 0 D12 

; Q C D21 

I 


0 



Step 2. (continued) 

Extended plant parameters 

A 0 B1 

0 

A0 

; B0 

; B 

0 0 0 I 

D 

12 

0 

k 

D ; D 

12 

k 

21 

0 

D 

21 

Closed loop description 

A 

C 

Cl 

Cl 

B 

D 

Cl 

Cl 

A0 

C 

0 

B0 

D 

11 

; 

k 

B 

D 

12 

k 

B2 

; C0 

0 

A 

C 

K 

K 

BK 

D 

K 

C 

C 

1 

0 ; C 

D 

21 

0 

C 

2 

Ik 

k 

; 

0 




Step 3. Controller for given value of parameter. 

Case of one parameter: 

( t) 

, t 

There are two controllers 

0 

K and K 

For given value of parameter θ 

; 

0 

1 

K (1 )K 

K 

Extension for multiparameter case. Possible simplifications. 


28


Demo no.11 


For the plant with measurable, time-varying parameter a 

design the gain scheduling controller. 

1 

G( s) 

; a 1.0 

s( 

s a) 

1.0 








• Analysis 

• Synthesis 



Conclusions 

• New (~15 years old) control analysis and control synthesis 

tool, 

• Active field of research with many unsolved problems: 

– BMI as natural formulation of control problems, 

but not convex, 

– suboptimal, low order controllers, controllers with structure (nonsmooth, 

non-convex algorithms). 

– output controllers for uncertain plants, 

– synthesis for systems with delays, saturations, 


29

Thank you for your attention! 


30

Linear Matrix Inequalities in Control

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