Linear Control SystemsLecture # 5

Stability and Matlab

– p. 1/28

Internal StabilityInternal stability deals with the boundedness andasymptotic behavior (as t → ∞) of the solutions of

x = Ax

The solution of x = Ax is given by

x(t) = eAtx(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. 2/28

If A is diagonalizable

eAt =

n∑

i=1

eλitviwTi

The behavior depends on Re[λi]

Re[λi] < 0 ⇒∣

∣eλit∣

∣ ≤ e−λt, λ > 0

Re[λi] = 0 ⇒∣

∣eλit∣

∣ = 1

Re[λi] > 0 ⇒ eλit is unbounded

– p. 3/28

In general

eAt =

r∑

i=1

mi∑

k=1

Wik

tk−1

(k − 1)!eλit

Re[λi] > 0 ⇒ tk−1eλit is unbounded

Re[λi] < 0 ⇒∣

∣tk−1eλit∣

∣ ≤ γe−λt, γ > 0, λ > 0

Re[λi] = 0 and k = 1 ⇒ eλit is bounded

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

– p. 4/28

When will the dimension of the Jordan block Ji be higherthan one?

(A − λiI)vi = 0

Let qi be the algebraic multiplicity of λi. If

qi = nullity(A − λiI) = n − rank(A − λiI)

then there are qi linearly independent eigenvectorsassociated with the eigenvalue λi

If this is true for every eigenvalue that has multiplicity higherthan one, then we can find n linearly independenteigenvectors and A is diagonalizable

– p. 5/28

Ifnullity(A − λiI) = n − rank(A − λiI) < qi

for any eigenvalue with multiplicity higher than one, then Ais not diagonalizablenullity(A − λiI) is equivalent to the geometric multiplicityof the eigenvalue λi.We need

qi − nullity(A − λiI)

number of generalized eigenvectors

– p. 6/28

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 algebraicmultiplicity qi ≥ 2,

rank(A − λiI) = n − qi

Asymptotically (or exponentially) stable if

Re[λi] < 0, for i = 1, 2, . . . , n

– p. 7/28

Example: Study the stability of x = Ax, where

A =

−1 0 3 1

0 −2 1 0

0 0 −3 1

0 0 0 −3

Eigenvalues : − 1,−2,−3,−3

The system is asymptotically stable

– p. 8/28

Example: Study the stability of x = Ax, where

A =

−1 0 3 1

0 1 1 0

0 0 0 0

0 0 0 0

Eigenvalues : − 1, 1, 0, 0

The system is unstable

– p. 9/28

Example: Consider the series and parallel connections oftwo identical systems, each represented by

x =

[

0 1

−1 0

]

x +

[

0

1

]

u, y =[

1 0]

x

orH(s) =

1

s2 + 1

As =

0 1 0 0

−1 0 0 0

0 0 0 1

1 0 −1 0

, Ap =

0 1 0 0

−1 0 0 0

0 0 0 1

0 0 −1 0

Eigenvalues : ± j, ±j

– p. 10/28

Series Case:

λ1 = j, q1 = 2, n − q1 = 2

As−λ1I =

−j 1 0 0

−1 −j 0 0

0 0 −j 1

1 0 −1 −j

→

0 1 0 0

0 −j 0 0

0 0 0 1

1 0 0 −j

rank(As − λ1I) = 3

The system is unstable

– p. 11/28

Parallel Case:

λ1 = j, q1 = 2, n − q1 = 2

Ap−λ1I =

−j 1 0 0

−1 −j 0 0

0 0 −j 1

0 0 −1 −j

→

0 1 0 0

0 −j 0 0

0 0 0 1

0 0 0 −j

rank(Ap − λ1I) = 2

The system is stable

– p. 12/28

What is the effect of state transformations on stability?

A → x = Pz → P−1AP

Av = λv

P−1Av = λP−1v

(P−1AP )(P−1v) = λP−1v

A and P−1AP have the same eigenvalues. If vi is aneigenvector of A, then P−1vi is an eigenvector of P−1AP

rank(

P−1AP − λI)

= rank[

P−1(A − λI)P]

= rank (A − λI)

Internal stability is invariant under state transformations

– p. 13/28

Input-Output Stability

Definition: The linear system y(s) = H(s)u(s) isBounded-Input Bounded-Output (BIBO) stable if for everybounded input u(t), the output y(t) is boundedEquivalently, for every ku > 0 there is ky > 0 such that

‖u(t)‖ ≤ ku, ∀ t ≥ 0 ⇒ ‖y(t)‖ ≤ ky, ∀ t ≥ 0

By taking the inverse Laplace transform

y(t) =

∫ t

0

H(t − τ )u(τ ) dτ

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

– p. 14/28

Theorem: The system y(s) = H(s)u(s) is BIBO stable ifand only if

∫

∞

0

‖H(t)‖ dt < ∞

Proof of sufficiency:

‖y(t)‖ =

∥

∥

∥

∥

∫ t

0

H(t − τ )u(τ ) dτ

∥

∥

∥

∥

≤

∫ t

0

‖H(t − τ )‖ ‖u(τ )‖ dτ

≤

∫

∞

0

‖H(t − τ )‖ dτ ku

≤

∫

∞

0

‖H(σ)‖ dσ kudef= ky

– p. 15/28

Proof on Necessity: Use a contradiction argument (in theSISO case). Given ku > 0, suppose there is ky > 0 suchthat

|u(t)| ≤ ku, ∀ t ≥ 0 ⇒ |y(t)| ≤ ky, ∀ t ≥ 0

but∫

∞

0|h(t)| dt is not finite

There is t1 (dependent on ky/ku) such that

∫ t1

0

|h(t1 − τ )| dτ >ky

ku

– p. 16/28

Let

u(t) =

ku, when h(t1 − t) > 0

0 when h(t1 − t) = 0

−ku when h(t1 − t) < 0

|u(t)| ≤ ku, for 0 ≤ t ≤ t1

y(t1) =

∫ t1

0

h(t1−τ )u(τ ) dτ =

∫ t1

0

ku|h(t1−τ )| dτ > ky

Contradcition

– p. 17/28

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∫

∞

0

δ(t − T ) dt = 1

– p. 18/28

WhenH(s) = C(sI − A)−1B + D

the elements hij(s) of H(s) are proper rational functions ofs

hij(s) =nij(s)

dij(s)

where nij(s) are dij(s) are polynomials with

deg(nij) ≤ deg(dij)

– p. 19/28

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 someelements of H(s) because some eigenvalues could becancelled

– p. 20/28

Given a strictly proper rational function H(s), leth(t) = L−1{H(s)}. When will

∫

∞

0

|h(t)| dt < ∞

H(s) can be expressed as the sum of terms of the form

K

(s − p)α

L−1

{

K

(s − p)α

}

= Ktα−1

(α − 1)!ept

Theorem: H(s) is BIBO stable if and only if all poles ofevery element of H(s) have negative real parts

– p. 21/28

What is the relationship between asymptotic and BIBOstability?The system x = Ax is asymptotically stable if all theeigenvalues of A have negative real parts.The system H(s) = C(sI − A)−1B + D is BIBO stable ifall poles of all elements of H(s) have negative real partsThe 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 wecould have a situation where the system is BIBO stable butnot asymptotically stable

– p. 22/28

Example: Suppose A is diagonalizable

eAt =

n∑

i=1

eλitviwTi

(sI − A)−1 = L{

eAt}

=

n∑

i=1

1

s − λi

viwTi

H(s) = C(sI − A)−1B + D =

n∑

i=1

1

s − λi

CviwTi B + D

If Cvi = 0 or wTi B = 0, the eigenvalue λi cancels out of

H(s)

– p. 23/28

{A,B,C,D} → x = Pz → {Λ, P−1B,CP,D}

x = Ax + Bu → z = Λz + (P−1B := B)u

y = Cx + Du → y = (CP := C)z + Du

Λ =

λ1

. . .

λn

, P = [v1, · · · , vn], P

−1 =:

wT1...

wTn

CP = [Cv1, · · · , cvn] =: [c1, · · · , cn]

P−1B =

wT1 B...

wTnB

=:

b1...bn

– p. 24/28

Uncontrollable Mode and Unobservable Mode

H(s) = C(sI − A)−1B =

n∑

i=1

1

s − λi

cibi

– p. 25/28

Example: Suppose we want to stabilize an unstablesystem described by

Gp(s) =1

s − 1

Consider a cascade compensator

Gc(s) =s − 1

s + 1

so that

Gp(s)Gc(s) =1

s − 1·s − 1

s + 1=

1

s + 1

The system is BIBO stable. Is it asymptotically stable?

– p. 26/28

Find a state model of the system

-s−1

s+1-

1

s−1-

u v y

Gp(s) =1

s − 1→ x1 = x1 + v, y = x1

Gc(s) =s − 1

s + 1→ x2 = −x2 + u, v = −2x2 + u

x1 = x1 − 2x2 + u

x =

[

1 −2

0 −1

]

x +

[

1

1

]

u

– p. 27/28

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 arenot acceptable because they can be excited by initialconditions or disturbancesFor example

x1(0) = α, x2(0) = 0, u(t) ≡ 0 ⇒ x2(t) ≡ 0

⇒ x1(t) = αet

– p. 28/28