Stability of linear systems Daniele Carnevale Dipartimento di Ing. - - PowerPoint PPT Presentation

Matrix properties Stability

Stability of linear systems

Daniele Carnevale Dipartimento di Ing. Civile ed Ing. Informatica (DICII), University of Rome “Tor Vergata”

Fondamenti di Automatica e Controlli Automatici

A.A. 2014-2015

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Matrix properties Stability Cayley-Hamilton Minimal characteristic polynomial

Cayley-Hamilton

A generic square matrix A renders zero its characteristic polynomial pA(λ) such as pA(A) = An + αn−1An−1 + · · · + α1A + α0I = 0, this also implies that the matrix An can be written as a linear combination of lower

• rder power of A such as

−An = αn−1An−1 + · · · + α1A + α0I, (1) that is named the Cayley-Hamilton Theorem. Consequently An+h for any h ≥ 0 can be rewritten as a linear combination of Aj with j ∈ {0, 1, . . . , n − 1}. . The matrix A nullify its characteristic polynomial by definition, in fact pA(λ) = det(λI − A) ⇒ pA(A) = det(A − A) = 0. (2)

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Matrix properties Stability Cayley-Hamilton Minimal characteristic polynomial

Minimal characteristic polynomial

The minimal characteristic polynomial qA(λ) is defined as the minimum order polynomial such that qA(A) = 0. Note that in general the degree of qA(λ) is smaller than pA(λ). The polynomial qA(λ) can be obtained as the least common multiple (LCM) denominator of the rational matrix (sI − A)−1. Note that the roots of qA(λ) are the roots of pA(λ), i.e. the eigenvalues of the matrix

• A. As an example, if ν are the distinct eigenvalues of the matrix A, then

qA(λ) =

ν

• i=1

(λ − λi)mi. (3) Note that mi = ni − µi + 1, (4) than mi > 1 iif the matrix A is not diagonalizable. mi is the dimension of the Jordan block associated to the eigenvalue λi. Then, the free state response of a linear systems has polynomial terms only if mi > 1.

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Matrix properties Stability Cayley-Hamilton Minimal characteristic polynomial

Minimal characteristic polynomial: examples

Consider A =   1 −1 2 1   → pA(s) = (s − 1)2(s + 1). In this case the minimal polynomial qA(s) is the LCM denominator of (sI − A)−1 that is (sI − A)−1 =   

1 s−1 1 s+1 2 (s−1)(s+1) 1 s−1

   → qA(s) = (s − 1)(s + 1), in fact defining λ1 = −1, λ2 = 1, then µ1 = 2 = n1 → m1 = 1, µ2 = 1 = n2 → m2 = 1.

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Matrix properties Stability Cayley-Hamilton Minimal characteristic polynomial

Minimal characteristic polynomial: examples cont’d

Consider now A =   1 1 −1 2 1   → pA(s) = (s − 1)2(s + 1). yielding (sI − A)−1 =   

1 s−1 1 (s−1)2 1 s+1 2 (s−1)(s+1) 1 s−1

   → qA(s) = (s − 1)2(s + 1), since λ1 = −1, λ2 = 1, then µ1 = 1 < n1 → m1 = 2, µ2 = 1 = n2 → m2 = 1. In this case the degree of pA(s) is equal to qA(s), i.e. pA(s) = qA(s).

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Matrix properties Stability LTI stability condition Examples

The stability property do not depend on the system input

To retrieve if the equilibrium (set) xe associated to ue is stable, than consider the solution of a linear continuous time (in discrete time is similar) such as x(t) = ϕ(t, t0, x0, ue) = eA(t−t0)x0 + t

t0

eA(t−τ)B dτue (5) = eA(t−t0)x0 + L−1 (sI − A)−1B ue s

• (t − t0).

(6) By linearity, the stability property of xe can be equivalently derived by considering directly the stability property of the origin. Then ||x(t)|| = ||eA(t−t0)x0|| ≤ ||eA(t−t0)||||x0|| ≤ ||eA(t−t0)||δǫ, (7) that if Re{λi} < 0 for all λi ∈ σ(A), implies that ||eA(t−t0)|| ≤ m and then selecting δǫ < ǫ/m yield ||x(t)|| < ǫ, and lim

t→+∞ ||x(t)|| = 0,

so that the origin is Globally Asimptotically Stable (GAS).

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Matrix properties Stability LTI stability condition Examples

Stabiity conditions of LTI

A continuous time LTI system (all the equilibria have the same properties) is stable iif Re{λi} ≤ 0 for all λi ∈ σ(A) and µi = ni for all λi such that Re{λi} = 0 globally asymptotically stable iif Re{λi} < 0 for all λi ∈ σ(A) unstable otherwise. A discrete time LTI system (all the equilibria have the same properties) is stable iif |λi| ≤ 1 for all λi ∈ σ(A) and µi = ni for all λi such that |λi| = 1 globally asymptotically stable iif |λi| < 1 for all λi ∈ σ(A) unstable otherwise.

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Matrix properties Stability LTI stability condition Examples

Economic linear model

Let d(k) be the demand of a given product at month k − th, s(k) the production offer and p(k) the product cost. At market equilibrium it has to hold that d = s. We are interested in understanding if such equilibrium can be reached in the market. Assume that the dynamic relation between s, p and d is given by d(k + 1) = D − αp(k + 1) (8) s(k + 1) = S + βp(k), (9) with D > 0, S > 0, α > 0 and β > 0. Then p(k + 1) = D − d(k + 1) α =

• equilibriums=d

D − s(k + 1) α = − β α p(k) + D − S α . Does exist a price pe that is an equilibrium and is it globally stable, unstable or globally asymptotically stable? Υ. There is a bracket missing in the use of Υ function up to Definition 5.

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