Linear Control Systems
Lecture # 5
Stability and Matlab
– p. 1/2
Internal Stability
Internal stability deals with the boundedness and
asymptotic behavior (as t → ∞) of the solutions of
ẋ = Ax
The solution of ẋ = Ax is given by
x(t) = eAt x(0)
Definition: Thesystem
ẋ = Ax is stable if
eAt ≤ γ, ∀ t ≥ 0, γ > 0
and asymptotically (or exponentially) stable if
At e ≤ γe−λt , ∀ t ≥ 0, γ > 0, λ > 0
– p. 2/2
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. 3/2
In general
eAt =
mi
r X
X
i=1 k=1
Wik
tk−1
(k − 1)!
eλi t
Re [λi ] > 0 ⇒ tk−1 eλi t is unbounded
k−1 λ t Re [λi ] < 0 ⇒ t
e i ≤ γe−λt , γ > 0, λ > 0
Re [λi ] = 0 and k = 1 ⇒ eλi t is bounded
Re [λi ] = 0 and k ≥ 2 ⇒ tk−1 eλi t is unbounded
– p. 4/2
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
qi = nullity(A − λi I) = n − rank(A − λi I)
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. 5/2
If
nullity(A − λi I) = n − rank(A − λi I) < qi
for any eigenvalue with multiplicity higher than one, then A
is not diagonalizable
nullity(A − λi I) is equivalent to the geometric multiplicity
of the eigenvalue λi .
We need
qi − nullity(A − λi I)
number of generalized eigenvectors
– p. 6/2
Theorem: The system ẋ = 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. 7/2
Example: Study the stability of ẋ = 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. 8/2
Example: Study the stability of ẋ

−1 0
 0 1

A=
 0 0
0 0
Eigenvalues :
= Ax, where

3 1

1 0 

0 0 
0 0
− 1, 1, 0, 0
The system is unstable
– p. 9/2
Example: Consider the series and parallel connections of
two identical systems, each represented by
"
#
" #
h
i
0 1
0
ẋ =
x+
u, y = 1 0 x
−1 0
1
or
H(s) =



As = 

0
−1
0
1
1 0
0 0
0 0
0 −1
0
0
1
0
1
s2 + 1






,
A
=


p


Eigenvalues :
0
−1
0
0
± j, ±j
1 0
0 0
0 0
0 −1
0
0
1
0





– p. 10/2
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. 11/2
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. 12/2
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. 13/2
Input-Output Stability
Definition: The linear system ŷ(s) = H(s)û(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
y(t) =
Z
t
H(t − τ )u(τ ) dτ
0
where H(t) = L−1 {H(s)} is the impulse response matrix
– p. 14/2
Theorem: The system ŷ(s) = H(s)û(s) is BIBO stable if
and only if
Z ∞
kH(t)k dt < ∞
0
Proof of sufficiency:
ky(t)k =
≤
≤
≤
Z t
H(t
−
τ
)u(τ
)
dτ
0
Z t
kH(t − τ )k ku(τ )k dτ
0
Z ∞
kH(t − τ )k dτ ku
0
Z ∞
def
kH(σ)k dσ ku = ky
0
– p. 15/2
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
t1
|h(t1 − τ )| dτ >
0
ky
ku
– p. 16/2
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
y(t1 ) =
h(t1 −τ )u(τ ) dτ =
ku |h(t1 −τ )| dτ > ky
0
0
Contradcition
– p. 17/2
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. 18/2
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. 19/2
Since
−1
(sI − A)
=
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. 20/2
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
L−1
K
(s − p)α
=K
tα−1
(α − 1)!
ept
Theorem: H(s) is BIBO stable if and only if all poles of
every element of H(s) have negative real parts
– p. 21/2
What is the relationship between asymptotic and BIBO
stability?
The system ẋ = 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. 22/2
Example: Suppose A is diagonalizable
eAt =
n
X
eλi t vi wiT
i=1
n
At X
=L e
=
−1
(sI − A)
i=1
−1
H(s) = C(sI − A)
B+D =
n
X
i=1
1
s − λi
1
s − λi
vi wiT
Cvi wiT B + D
If Cvi = 0 or wiT B = 0, the eigenvalue λi cancels out of
H(s)
– p. 23/2
{A, B, C, D} → x = P z → {Λ, P −1 B, CP, D}
ẋ = Ax + Bu
→
ż = Λz + (P −1 B := B̃)u
y = Cx + Du
→
y = (CP := C̃)z + Du


Λ=
λ1
...


w1T


 .. 
−1
=:  . 
 , P = [v1 , · · · , vn ], P
T
λn
wn
CP = [Cv1 , · · · , cvn ] =: [c̃1 , · · · , c̃n ]

w1T B



b̃1
 .. 
 .. 
−1
P B =  .  =:  . 
TB
wn
b̃n
– p. 24/2
Uncontrollable Mode and Unobservable Mode
−1
H(s) = C(sI − A)
B=
n
X
i=1
1
s − λi
c̃i b̃i
– p. 25/2
Example: Suppose we want to stabilize an unstable
system described by
Gp (s) =
1
s−1
Consider a cascade compensator
Gc (s) =
so that
Gp (s)Gc (s) =
s−1
s+1
1
s−1
·
s−1
s+1
=
1
s+1
The system is BIBO stable. Is it asymptotically stable?
– p. 26/2
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
-
→ ẋ1 = x1 + v, y = x1
→ ẋ2 = −x2 + u, v = −2x2 + u
ẋ1 = x1 − 2x2 + u
"
#
" #
1 −2
1
ẋ =
x+
u
0 −1
1
– p. 27/2
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. 28/2