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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
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Lyapunov Stability
Lyapunov Stability
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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
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Lyapunov Stability
Lyapunov Stability
Example. Frictionless pendulum:
x1 = θ
x2 = θ=⇒
[x1x2
]=
[x2
−gml sin(x1)
]What are the equilibria?[
00
]: stable
[π0
]: unstable
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(Globally) Asymptotic Stability
(Globally) Asymptotic Stability
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(Globally) Asymptotic Stability
(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.”
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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
Stability of LTI Systems I
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Stability of LTI Systems I
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.
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Stability of LTI Systems I
Stability of LTI Systems I
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.
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Stability of LTI Systems I
Stability of LTI Systems I
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.
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Lyapunov’s Direct Method
Lyapunov’s Direct Method
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Lyapunov’s Direct Method
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.
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Lyapunov’s Direct Method
Lyapunov’s Direct Method
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.
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Region of Asymptotic Stability
Region of Asymptotic Stability
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Region of Asymptotic Stability
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
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Region of Asymptotic Stability
Region of Asymptotic Stability
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.
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Stability of LTI Systems II
Stability of LTI Systems II
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Stability of LTI Systems II
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.
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Stability of LTI Systems II
Stability of LTI Systems II
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.
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Stable Subspace
Stable Subspace
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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.
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Stable Subspace
Stable Subspace
x = Ax, x(0) = x0
Stable Subspace: Σs := x0 ∈ Rn | limt→∞
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.
S. Bolouki (UIUC) 27 / 32
Stable Subspace
Stable Subspace
x = Ax, x(0) = x0
Stable Subspace: Σs := x0 ∈ Rn | limt→∞
x(t) = 0
What if the eigenvalues are complex?
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Stability via Linearization
Stability via Linearization
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Stability via Linearization
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!
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Input-Output Stability
Input-Output Stability
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Input-Output Stability
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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