introduction to modern control...

Introduction to Modern Control SystemsConvex Optimization, Duality and Linear Matrix Inequalities

Kostas Margellos

University of Oxford

AIMS CDT 2016-17 Introduction to Modern Control Systems November 23, 2016 1 / 28

Agenda for today’s lectures1

1 09.30 - 10.30

Convex Optimization

2 10.30 - 10.45

Break

3 10.45 - 11.30

Duality Theory

4 11.30 - 12.15

Linear Matrix Inequalities (LMIs)

1Thanks to Prof. Paul Goulart and Dr. James Anderson for slides input!AIMS CDT 2016-17 Introduction to Modern Control Systems November 23, 2016 2 / 28

So far we have discussed about ...

Standard form optimization problems:

Convexity

Convex optimization programs

Optimality conditions

Duality

What is left to study?

Linear Matrix Inequalities, LMIs for short


Linear Matrix Inequalities

What are LMIs ?

Non-obvious LMIs

LMIs in state feedback control

LMIs in optimization

Semidefinite programmingPrimal and dual formulation


What are LMIs?

A Linear Matrix Inequality (LMI) is a constraint of the form:

x1A1 + x2A2 + · · · xnAn � B

where the matrices (A1, . . . ,An,B) are all symmetric.

This is a constraint that imposes matrix

B −n∑i

xiAi

to be positive semidefinite.

It is equivalent to imposing n polynomial inequalities

Not element-wise constraints.

All leading principle minors of this matrix are positive.


What are LMIs?

A Linear Matrix Inequality (LMI) is a constraint of the form:

x1A1 + x2A2 + · · · xnAn � B


Consider the constraint

Q =

[Q11 Q12

Q21 Q22

]� 0

This is equivalent to 2 inequalities:

Q11 ≥ 0

det(Q) ≥ 0⇔ Q11Q22 − Q12Q21 ≥ 0


LMIs are convex constraints

Theorem

The following LMI constraint is convex.

F (x) = B −n∑i

xiAi � 0

Proof: Let x , y such that F (x),F (y) � 0, and λ ∈ (0, 1).

F (λx + (1− λ)y) = B −∑i

(λxi + (1− λ)yi

)Ai

= λB + (1− λ)B − λ∑i

xiAi − (1− λ)∑i

yiAi

= λF (x) + (1− λ)F (y)

� 0


LMIs are convex constraints

Theorem

The following LMI constraint is convex.

F (x) = B −n∑i

xiAi � 0

Alternative proof: We want to show that the set {x : F (x) � 0} isconvex. We have that ...

{x : F (x) � 0} = {x : z>F (x)z ≥ 0, for all z}

=⋂z

{x : z>F (x)z ≥ 0}

... but this is an infinite intersection of sets affine in x ... so it is convex!

LMI much harder than linear constraints – an infinite number of them!

Result can be piecewise affine – LMIs nonlinear!


Why are LMIs interesting?

Linear Matrix Inequalities:

Appear in many common control design problems (more later on)

Most of the problems presented so far can be written using LMIconstraints

Linear constraints

Ax ≤ b ⇔ diag(Ax) � diag(b)

Quadratic constraints (It will be clear in a couple of slides)

x>Qx + b>x + c ≥ 0, Q � 0 ⇔(c + b>x x>

x −Q−1)� 0


Non-obvious LMIs

Some cases (like the QP) are harder to write as LMIs.

There are two useful “tricks”:

Schur complement

The S-procedure

Schur complement: Turns a nonlinear constraint into an LMI

Assume that Q(x) = Q(x)>,R(x) = R(x)> : affine functions of x

We then have that

R(x) � 0 and Q(x)− S(x)R(x)−1S(x)> � 0

⇔[Q(x) S(x)S(x)> R(x)

]� 0


Schur complement

Schur complement: Turns a nonlinear constraint into an LMI

Assume that Q(x) = Q(x)>,Q(x) = Q(x)> : affine functions of x

We then have that

R(x) � 0 and Q(x)− S(x)R(x)−1S(x)> � 0

⇔[Q(x) S(x)S(x)> R(x)

]� 0

Example 1:

‖A‖2 ≤ t ⇔ A>A � t2I , t ≥ 0 ⇔(tI A>

A tI

)� 0

Example 2: The QP

x>Qx + b>x + c ≥ 0, Q � 0 ⇔(c + b>x x>

x −Q−1)� 0


S-procedure

S-procedure: Turns implications to LMIs

When is it true that one quadratic inequality implies another? In otherwords, when does

x>F x ≥ 0, x 6= 0⇒ x>Fx ≥ 0

It is not difficult to see that the condition is satisfied if there exists

τ ∈ R such that τ ≥ 0 and F � τ F

The only if part also holds, but less obvious and difficult to prove.


S-procedure: Example

Example: Use a quadratic Lyapunov function to show that

x = Ax + g(x)

is stable if ‖g(x)‖2 ≤ γ‖x‖2, γ > 0.

We want to show that there exists a Lyapunov functionV (x) = x>Px , P � 0 such that

V (x) ≤ −αV (x), for all x , [α > 0 given]

if ‖g(x)‖2 ≤ γ‖x‖2

Or, in other words, if ‖g(x)‖2 ≤ γ‖x‖2, we want

V (x) + αV (x) ≤ 0


S-procedure: Example (cont’d)

We want to show that there exists a Lyapunov functionV (x) = x>Px , P � 0 such that

V (x) ≤ −αV (x), for all x , [α > 0 given]

if ‖g(x)‖2 ≤ γ‖x‖2

Letting z = g(x), we have

V (x) + αV (x) = x>(A>P + PA + αP

)x + 2z>Pz

=

[xz

]> [A>P + PA + αP P

P 0

] [xz

]

Hence, if ‖g(x)‖2 ≤ γ‖x‖2, we want

−[A>P + PA + αP P

P 0

]� 0



The condition

‖g(x)‖2 ≤ γ‖x‖2 ⇒ z>z ≤ γ2x>x

This can be written as [xz

]> [−γ2 00 I

] [xz

]≤ 0

We thus have


P 0

]� 0 if

[γ2 00 −I

]� 0



We thus have


P 0

]� 0 if

[γ2 00 −I

]� 0

Applying S-procedure (finally!) we have


P 0

]� τ

[γ2 00 −I

]� 0

Overall,

−[A>P + PA + αP + τγ2I P

P −τ I

]� 0, P � 0, τ > 0


LMIs in state feedback control - By means of example

Consider a system G : x = Ax + Bu

Convex optimization

5

State feedback

Control Theory Using LMIs

Gux

Assume we are given the system G : x = Ax + Bu

The state feedback problem is to find a gain matrix K such that the controlpolicy u = Kx renders the closed loop stable.

Gux

The closed loop dynamics are:

x = (A + BK)x.

Thus the goal is to find K such that A + BK is Hurwitz.

K

Determine a feedback gain matrix K such that u = Kx renders the closedloop system stable.

Convex optimization

5

State feedback

Control Theory Using LMIs

Gux

Assume we are given the system G : x = Ax + Bu

The state feedback problem is to find a gain matrix K such that the controlpolicy u = Kx renders the closed loop stable.

Gux

The closed loop dynamics are:

x = (A + BK)x.

Thus the goal is to find K such that A + BK is Hurwitz.

K

Closed loop system: x =(A + BK

)x .

Goal: Determine K such that A + BK is Hurwitz.




)x .


Lyapunov stability: A matrix A is stable if and only if there existsP = P> � 0 such that

A>P + PA � 0

Equivalent representation: Multiply by P−1 from the left and right:

P−1A>PP−1 + P−1PAP−1 � 0

and set Q = P−1. We then have

QA> + AQ � 0




)x .


Lyapunov stability: A matrix A is stable if and only if there existsQ = Q> � 0 such that

QA> + AQ � 0

Enforce this condition with A + BK in place of A and determine K and Q:

Q(A + BK

)>+(A + BK

)Q � 0

which leads to

QA> + (QK>)B> + AQ + B(QK ) � 0




)x .


Lyapunov stability: A matrix A is stable if and only if there existsQ = Q> � 0 such that

QA> + AQ � 0

We are left with this condition which is not nice!

QA> + (QK>)B> + AQ + B(QK ) � 0

Setting Z = KQ we have

QA> + Z>B> + AQ + BZ � 0

Solve this LMI to determine Q and Z and then compute K = ZQ−1


LMIs in optimization

Consider the following optimization program

(SDP) :

min c>x

subject to: x1A1 + x2A2 + · · · xnAn � B


We could also have equality constraints

Optimization over LMI constraints

Why is this class of optimization programs interesting?

Semidefinite programming (SDP)

Many control analysis and synthesis problems can be written as SDPs

Most of the problems presented so far can be written as SDPs


Semidefinite programming

Primal SDP problem:

min c>x



Lagrangian:

L(x ,Λ) = c>x +∑i

〈Λ,Ai 〉xi − 〈Λ,B〉,

where 〈X ,Y 〉 = trace(X>Y ) =∑

i ,j XijYij .

This fact relies on the following property (dual cone for semidefinitematrices):{

x ∈ Sn∣∣∣ trace(X>Y ) ≥ 0, ∀Y � 0

}= {X ∈ Sn | X � 0}



Primal SDP problem:

min c>x



Lagrangian:

L(x ,Λ) = c>x +∑i

〈Λ,Ai 〉xi − 〈Λ,B〉

=∑i

(ci + 〈Λ,Ai 〉

)xi − 〈Λ,B〉

Dual function:

g(λ) =

{−〈Λ,B〉 if ci + 〈Λ,Ai 〉 = 0 for i = 1 . . . n

−∞ otherwise



Primal SDP problem:

min c>x



Dual function:

g(λ) =

{−〈Λ,B〉 if ci + 〈Λ,Ai 〉 = 0 for i = 1 . . . n

−∞ otherwise

The dual problem:max −〈B,Λ〉

subject to: 〈Ai ,Λ〉 = −ci , for all i

Λ � 0



Primal SDP problem:

min c>x




Λ � 0

Weak duality:p∗ − d∗ = c>x + 〈B,Λ〉 [primal feasibility]

≥ c>x +∑i

〈Ai ,Λ〉xi [dual feasibility]

=∑i

cixi −∑i

cixi = 0



Primal SDP problem:

min c>x




Λ � 0

Weak duality: p∗ − d∗ ≥ 0

Strong duality:Also true under Slater’s condition (constraint qualification). Constraints inthe primal need to be satisfied with ≺ instead of �.


What was this lecture about?

All about optimization ...

Convex optimization programs are “nice”Local solution = Global solution

Recognizing convex functions and setsIf you do this life is easier; you can recast seemingly difficult problemsto convex ones

Formulating dual optimization programsSometimes much easier than the primal ones (think of MILPs, SDPs)

Linear matrix inequalitiesMore difficult but appear in many stability and control design problems

Don’t forget the “big picture” ...

Determine controllers via optimization programs

Allows trading between different objectives

Allows enforcing constraints

Efficient alternative to PID and LQR - price to pay is computation!


Thank you! Questions?

Contact at:[email protected]


introduction to modern control...

Documents