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Linear Control Systems Lecture # 5 Stability

– p. 1/26

Internal Stability Internal stability deals with the boundedness and asymptotic behavior (as t → ∞) of the solutions of x˙ = Ax

The solution of x˙ = Ax is given by x(t) = eAt x(0)

Definition: Thesystem x˙ = Ax is stable if  eAt  ≤ γ, ∀ t ≥ 0, γ > 0

and asymptotically (or exponentially) stable if   eAt ≤ γe−λt ,

∀ t ≥ 0, γ > 0, λ > 0

– p. 2/26

If A is diagonalizable eAt =

n X

eλi t vi wiT

i=1

The behavior depends on Re [λi ]  λ t  Re [λi ] < 0 ⇒ e i  ≤ e−λt , λ > 0  λ t  Re [λi ] = 0 ⇒ e i  = 1

Re [λi ] > 0 ⇒ eλi t is unbounded

– p. 3/26

In general eAt =

r X mi X

i=1 k=1

tk−1 eλ i t Wik (k − 1)!

Re [λi ] > 0 ⇒ tk−1 eλi t is unbounded  k−1 λ t   e i  ≤ γe−λt , γ > 0, λ > 0 Re [λi ] < 0 ⇒ t Re [λi ] = 0 and k = 1 ⇒ eλi t is bounded

Re [λi ] = 0 and k ≥ 2 ⇒ tk−1 eλi t is unbounded

– p. 4/26

When will the dimension of the Jordan block Ji be higher than one? (A − λi I )vi = 0 Let qi be the algebraic multiplicity of λi . If rank(A − λi I) = n − qi

then there are qi linearly independent eigenvectors associated with the eigenvalue λi If this is true for every eigenvalue that has multiplicity higher than one, then we can find n linearly independent eigenvectors and A is diagonalizable

– p. 5/26

If rank(A − λi I) > n − qi

for any eigenvalue with multiplicity higher than one, then A is not diagonalizable

– p. 6/26

Theorem: The system x˙ = Ax is Stable if and only if Re [λi ] ≤ 0,

for i = 1, 2, . . . , n

and for every eigenvalue with Re [λi ] = 0 and algebraic multiplicity qi ≥ 2, rank(A − λi I) = n − qi

Asymptotically (or exponentially) stable if Re [λi ] < 0,

for i = 1, 2, . . . , n

– p. 7/26

Example: Study the stability of x˙ = Ax, where   −1 0 3 1  0 −2 1 0      A= 0 0 −3 1  0 0 0 −3 Eigenvalues :

− 1, − 2, − 3, − 3

The system is asymptotically stable

– p. 8/26

Example: Study the stability of x˙ = Ax, where   −1 0 3 1  0 1 1 0    A= 0 0 0 0  0 0 0 0 Eigenvalues :

− 1, 1, 0, 0

The system is unstable

– p. 9/26

Example: Consider the series and parallel connections of two identical systems, each represented by " # " # h i 0 1 0 x˙ = x+ u, y = 1 0 x 1 −1 0 or 

  As =  

0 −1 0 1

1 0 0 0

1 H(s) = 2 s +1   0 0   0 0   , Ap =   0 1 −1 0

Eigenvalues :

0 −1 0 0

± j, ±j

1 0 0 0 0 0 0 −1

0 0 1 0

    

– p. 10/26

Series Case: λ1 = j, q1 = 2, n − q1 = 2



  As − λ1 I =  

−j 1 0 0 −1 −j 0 0 0 0 −j 1 1 0 −1 −j





     →   

rank(As − λ1 I) = 3

0 1 0 0 0 −j 0 0 0 0 0 1 1 0 0 −j

    

The system is unstable

– p. 11/26

Parallel Case: λ1 = j, q1 = 2, n − q1 = 2



  Ap − λ1 I =  

−j 1 0 0 −1 −j 0 0 0 0 −j 1 0 0 −1 −j





     →   

rank(Ap − λ1 I) = 2

0 1 0 0 0 −j 0 0 0 0 0 1 0 0 0 −j

    

The system is stable

– p. 12/26

What is the effect of state transformations on stability? A → x = P z → P −1 AP Av = λv P −1 Av = λP −1 v (P −1 AP )(P −1 v) = λP −1 v A and P −1 AP have the same eigenvalues. If vi is an eigenvector of A, then P −1 vi is an eigenvector of P −1 AP     rank P −1 AP − λI = rank P −1 (A − λI )P = rank (A − λI )

Internal stability is invariant under state transformations

– p. 13/26

Input-Output Stability Definition: The linear system yˆ(s) = H(s)ˆ u(s) is Bounded-Input Bounded-Output (BIBO) stable if for every bounded input u(t), the output y(t) is bounded Equivalently, for every ku > 0 there is ky > 0 such that ku(t)k ≤ ku , ∀ t ≥ 0 ⇒ ky(t)k ≤ ky , ∀ t ≥ 0

By taking the inverse Laplace transform Z t H(t − τ )u(τ ) dτ y(t) = 0

where H (t) = L−1 {H (s)} is the impulse response matrix

– p. 14/26

Theorem: The system yˆ(s) = H(s)ˆ u(s) is BIBO stable if and only if Z ∞ kH(t)k dt < ∞ 0

Proof of sufficiency:  Z t    H(t − τ )u(τ ) dτ  ky(t)k =   0 Z t kH(t − τ )k ku(τ )k dτ ≤ 0

≤

Z

t

kH(t − τ )k dτ ku

0

≤ ku

Z

t

kH(σ)k dσ ≤ ku 0

Z

∞

def

kH(σ)k dσ = ky 0 – p. 15/26

Proof on Necessity: Use a contradiction argument (in the SISO case). Given ku > 0, suppose there is ky > 0 such that |u(t)| ≤ ku , ∀ t ≥ 0 ⇒ |y(t)| ≤ ky , ∀ t ≥ 0

R∞

but 0 |h(t)| dt is not finite There is t1 (dependent on ky /ku ) such that Z

0

t1

|h(t1 − τ )| dτ >

ky ku

– p. 16/26

Let    ku , u(t) =  0  −k u

when h(t1 − t) > 0 when h(t1 − t) = 0 when h(t1 − t) < 0

|u(t)| ≤ ku ,

for 0 ≤ t ≤ t1 Z t1 Z t1 ku |h(t1 −τ )| dτ > ky y(t1 ) = h(t1 −τ )u(τ ) dτ = 0

0

Contradiction

– p. 17/26

Example: Time delay element y(t) = u(t − T ) H(s) = e−sT h(t) = L−1 {H(s)} = δ(t − T ) |u(t)| ≤ ku ⇒ |y(t)| ≤ ku

Or Z

∞

δ(t − T ) dt = 1

0

– p. 18/26

When H(s) = C(sI − A)−1 B + D

the elements hij (s) of H(s) are proper rational functions of s nij (s) hij (s) = dij (s) where nij (s) are dij (s) are polynomials with deg(nij ) ≤ deg(dij )

– p. 19/26

Since (sI − A)−1 =

1 det(sI − A)

Adjoint(sI − A)

the poles of hij (s) are roots of det(sI − A) = 0

that is, eigenvalues of A Not all eigenvalues of A will appear as poles of some elements of H(s) because some eigenvalues could be cancelled

– p. 20/26

Given a strictly proper rational function H(s), let h(t) = L−1 {H(s)}. When will Z ∞ |h(t)| dt < ∞ 0

H(s) can be expressed as the sum of terms of the form K (s − p)α −1

L



K (s − p)α



tα−1 ept =K (α − 1)!

Theorem: H(s) is BIBO stable if and only if all poles of every element of H(s) have negative real parts – p. 21/26

What is the relationship between asymptotic and BIBO stability? The system x˙ = Ax is asymptotically stable if all the eigenvalues of A have negative real parts. The system H(s) = C(sI − A)−1 B + D is BIBO stable if all poles of all elements of H(s) have negative real parts The poles of H(s) are eigenvalues of A Asymptotic stability ⇒ BIBO stability

What about the opposite implication? Some eigenvalues of A may not appear as poles of H(s). If such eigenvalues have nonnegative real parts, then we could have a situation where the system is BIBO stable but not asymptotically stable – p. 22/26

Example: Suppose A is diagonalizable eAt =

n X

eλi t vi wiT

i=1



−1

(sI − A)

=L e

−1

H(s) = C(sI − A)

 At

=

B+D =

n X i=1

n X i=1

1 vi wiT s − λi

1 Cvi wiT B + D s − λi

If Cvi = 0 or wiT B = 0, the eigenvalue λi cancels out of H(s)

– p. 23/26

Example: Suppose we want to stabilize an unstable system described by Gp (s) =

1 s−1

Consider a cascade compensator Gc (s) =

so that

s−1 s+1

1 s−1 1 Gp (s)Gc (s) = · = s−1 s+1 s+1

The system is BIBO stable. Is it asymptotically stable? – p. 24/26

Find a state model of the system u

✲ s−1

s+1

Gp (s) = Gc (s) =

1 s−1

s−1 s+1

v ✲

1 s−1

y ✲

→ x˙ 1 = x1 + v, y = x1

→ x˙ 2 = −x2 + u, v = −2x2 + u

x˙ 1 = x1 − 2x2 + u " # " # 1 −2 1 x˙ = x+ u 0 −1 1

– p. 25/26

A=

"

1 −2 0 −1

#

Eigenvalues : 1, −1

The system is not asymptotically stable. The eigenvalue +1 is called a hidden mode. Unstable hidden modes are not acceptable because they can be excited by initial conditions or disturbances For example x1 (0) = α, x2 (0) = 0, u(t) ≡ 0 ⇒ x2 (t) ≡ 0 ⇒ x1 (t) = αet – p. 26/26...