STABILITY - WordPress.com 1 Lyapunov Stability 2 (Globally) Asymptotic Stability 3 Stability of LTI...

STABILITY

Sadegh Bolouki

Lecture slides for ECE 515

University of Illinois, Urbana-Champaign

Fall 2016

S. Bolouki (UIUC) 1 / 32

Overview

1 Lyapunov Stability

2 (Globally) Asymptotic Stability

3 Stability of LTI Systems I

4 Lyapunov’s Direct Method

5 Region of Asymptotic Stability

6 Stability of LTI Systems II

7 Stable Subspace

8 Stability via Linearization

9 Input-Output Stability


Lyapunov Stability

Lyapunov Stability


Lyapunov Stability

Lyapunov Stability

x = f (x), x(0) = x0

Assumption: f ∈ C2(Rn) (has continuous second order derivative)

xe is called an equilibrium if f (xe) = 0.

DefinitionAn equilibrium xe is stable in the sense of Lyapunov if for any ε > 0, there existsδ > 0 such that

‖x0 − xe‖ < δ ⇒ ‖x(t)− xe‖ < ε


Lyapunov Stability

Lyapunov Stability

Example. Frictionless pendulum:

x1 = θ

x2 = θ=⇒

[x1x2

]=

[x2

−gml sin(x1)

]What are the equilibria?[

00

]: stable

[π0

]: unstable


(Globally) Asymptotic Stability





x = f (x), x(0) = x0

DefinitionAn equilibrium xe is asymptotically stable if (i) it is stable in the sense of Lyapunovand (ii) for some δ0:

‖x0 − xe‖ < δ0 ⇒ limt→∞‖x(t)− xe‖ = 0

DefinitionAn equilibrium xe is globally asymptotically stable if it is asymptotically stable withδ0 =∞.

If there is a globally asymptotically stable equilibrium, it is the only equilibriumof the system.

→We associate global asymptotic stability with the system and simply say

“the system is globally asymptotically stable.”


Globally Asymptotic Stability

⇓Asymptotic Stability

⇓Lyapunov Stability

Is the reverse also true?

Example. Let x = Ax, x(0) = x0, where

A =

[0 00 0

].

Then, x(t) = x0, ∀t. Thus, entire R2 forms the equilibrium set.

All equilibria are stable in the sense of Lyapunov. (Why?)

No equilibrium is (globally) asymptotically stable. (Why?)

Stability of LTI Systems I





x = Ax, x(0) = x0

(Globally) asymptotic stability

The vector 0 is an equilibrium.

Additivity⇒ If any other equilibrium is asymptotically stable, then 0 isasymptotically stable, too.

Homogeneity⇒ If 0 is asymptotically stable, it is also globally asymptoticallystable.

There cannot be an asymptotically stable equilibrium other than 0.




TheoremAn LTI system x = Ax is globally asymptotically stable if and only if for everyeigenvalue λ of A:

Re(λ) < 0,

i.e., if A is a Hurwitz matrix.




x = Ax, x(0) = x0

Lyapunov stability

The vector 0 is an equilibrium.

Additivity All or none of the equilibriums are stable in the sense of Lyapunov.

→ Again, we associate stability with the system.

TheoremAn LTI system x = Ax is stable in the sense of Lyapunov if and only if

(i) Re(λ) ≤ 0 for every eigenvalue λ of A;

(ii) for every eigenvalue λ of A that Re(λ) = 0,

algebraic multiplicity of λ = geometric multiplicity of λ.


Lyapunov’s Direct Method





Theorem

Let xe be an equilibrium of the system x = f (x). Then,

(i) xe is stable in the sense of Lyapunov if for some positive definite functionV : Rn → R and an open set Ω ⊂ Rn containing xe we have

ddt

V(x(t))≤ 0 whenever x(t) ∈ Ω;

(ii) xe is asymptotically stable if for some positive definite function V : Rn → R andan open set Ω ⊂ Rn containing xe we have

ddt

V(x(t))< 0 whenever x(t) ∈ Ω, x(t) 6= xe;

(iii) xe is globally asymptotically stable if for some positive definite functionV : Rn → R we have limx→∞ V(x) =∞ and

ddt

V(x(t))< 0 for all x(t), x(t) 6= xe.




Remember that ddt V(x(t))

= ∇V(x(t))

f (x(t)).

Example. Van der Pol oscillator:[x1x2

]=

[x2

−(1− x21)x2 − x1

]There is a unique equilibrium: 0.By Lyapunov’s direct method, we can show that the equilibrium 0 is stable in the sense of Lyapunov if we choose

V(x) =12

(x21 + x2

2), Ω = x ∈ R2 : |x1| < 1;

asymptotically stable if we choose

V(x) =12

(x21 + x2

2) +16

x1x2, Ω = x : x21 < 1/2.


Region of Asymptotic Stability





x = f (x), x(0) = x0Let:

- xe be an equilibrium point, that is f (xe) = 0;

- V : Rn → R be positive definite, that is

V(x) > 0 x 6= xe

V(x) = 0 x = xe; and

- V have continuous first partial derivatives.

Define Ω` := x : V(x) < `. Ω` is an open set that contains xe. (Why?)

Theorem

Assume that Ω` is bounded and ∇V(x)f (x) < 0 for any x ∈ Ω`. Then

x0 ∈ Ω` =⇒ limt→∞

x(t) = xe.

Notice that under assumptions of the theorem, xe becomes asymptotically stable.

region of AS




Example. Consider the following nonlinear system:[x1x2

]=

[−x1 − x2

x1 − x2 + x32

].

The unique equilibrium point is xe = 0. Choosing the Lyapunov function

V(x) = ‖x‖2,

∇V(x)f (x) < 0 for every x 6= 0 that satisfies x22 < 1.


Stability of LTI Systems II





x = Ax, x(0) = x0

TheoremThe LTI system above is asymptotically stable if and only if for some positive definiteQ, the Lyapunov equation

ATP + PA = −Q

has a symmetric, positive definite solution for P.

sketch of proof:

(⇒): P =∫∞

0 eAT tQeAtdt solves the Lyapunov equation.

(⇐): V(x) = xTPx makes a “valid” Lyapunov function.




Example. Consider the following LTI system:

x = Ax =

[0 1−2 2

]x.

Let Q = I. Then, the Lyapunov equation ATP + PA = −I has the solution

P =

[5/4 1/41/4 3/8

].

Since P is positive definite, the systems is globally asymptotically stable.


Stable Subspace

Stable Subspace


Stable Subspace

Stable Subspace

x = Ax, x(0) = x0

Stable Subspace: Σs := x0 ∈ Rn | limt→∞

x(t) = 0

If the eigenvalues are real and distinct:

Σs = Span

v |Av = λv, Re(λ) < 0.

In other words:(i) Find the eigenvectors associated with the negative eigenvalues;(ii) Find the subspace generated by them.


Stable Subspace

Stable Subspace

x = Ax, x(0) = x0


x(t) = 0

If the eigenvalues are real but not necessarily distinct:

Σs = Span

v | (A− λI)kv = 0, Re(λ) < 0, k = 1, 2, . . ..

In other words:(i) Find the generalized eigenvectors associated with the negativeeigenvalues;(ii) Find the subspace generated by them.


Stable Subspace

Stable Subspace

x = Ax, x(0) = x0


x(t) = 0

What if the eigenvalues are complex?


Stability via Linearization





x = f (x) (1)

Assume xe is an equilibrium point. Linearize the system around xe:

δx = Aδx,

where A is the Jacobian of f at xe.

TheoremGiven the nonlinear model (1),

(i) xe is asymptotically stable for (1) if all eigenvalues of A have negative real parts.

(ii) xe is not asymptotically stable for (1) if A has an eigenvalue with positive realpart.

(iii) That’s all!


Input-Output Stability





x = Ax + Buy = Cx + Du

DefinitionA system is said to be bounded-input bounded output stable (BIBO stable) if, givenx0 = 0, any bounded input results in a bounded output.

Theorem

An LTI system is BIBO stable if each pole of every entry of G(s) has negative realpart.

Remember thatG(s) = C(sI − A)−1B + D

BIBO stability is weaker than asymptotic stability. (Why?)


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