Linear Control Systems
Lecture # 8
Observability
&
Discrete-Time Systems
– p. 1/2
Definition: The system
ẋ = Ax,
y = Cx
or the pair (A, C), is said to be observable on [t0 , tf ] if any
initial state x(t0 ) = x0 can be uniquely determined from
y(t) on [t0 , tf ]. It is said to be (re)constructible on [t0 , tf ] if
any final state x(tf ) = xf can be uniquely determined from
y(t) on [t0 , tf ]
Without loss of generality, take t0 = 0
The observability Gramian of (A, C) is defined by
Wo (0, tf ) =
Z
tf
e
AT t
C T CeAt dt
0
– p. 2/2
Lemma: Wo (0, tf ) is positive definite if and only if there is
no vector xa 6= 0 such that
CeAt xa ≡ 0,
∀ t ∈ [0, tf ]
Proof:
xTa Wo (0, tf )xa
=
Z
tf
0
T AT t T
xa e
C CeAt xa
dt
xTa Wo (0, tf )xa = 0 ⇔ CeAt xa ≡ 0, ∀ t ∈ [0, tf ]
– p. 3/2
Theorem: The pair (A, C) is observable (constructible) on
[0, tf ] if and only if the observability Gramian Wo (0, tf ) is
positive definite
Proof of sufficiency: Suppose Wo (0, tf ) is positive definite
and let x0 be the initial state
y(t) = Cx(t) = CeAt x0
Multiply from the left by
e
AT t
T
A
e tC T
T
C y(t) = e
AT t
C T CeAt x0
Integrate from 0 to tf
Z
tf
e
0
T
A t
T
C y(t) dt =
Z
tf
e
AT t
C T CeAt dt x0
0
– p. 4/2
Z
tf
e
AT t
C T y(t) dt = Wo (0, tf )x0
0
This equation has a unique solution
Z tf
−1
AT t T
x0 = Wo (0, tf )
e
C y(t) dt
0
Since
x(tf ) = eAtf x0
and eAtf is nonsingular, x(tf ) is determined uniquely by
x(tf ) = e
Atf
Wo−1 (0, tf )
Z
tf
e
AT t
C T y(t) dt
0
– p. 5/2
Proof of Necessity: We want to show that positive
definiteness of Wo (0, tf ) is a necessary condition for
observability (constructibility) over [0, tf ]
Suppose Wo (0, tf ) is not positive definite. Then, there is a
vector xa 6= 0 such that
CeAt xa ≡ 0,
∀ t ∈ [0, tf ]
For the case of observability, The output due to x(0) = x0
is
CeAt x0
and the output due to x(0) = x0 + xa is
CeAt (x0 + xa ) = CeAt x0 + CeAt xa = CeAt x0
– p. 6/2
The initial states x0 and x0 + xa produce the same output.
Hence, x0 cannot be uniquely determined from the output
For the case of constructibility
y(t) = CeAt x(0) = CeAt e−Atf x(tf )
x(tf ) = xf and x(tf ) = xf + eAtf xa correspond to the
same output because
³
´
CeAt e−Atf xf + eAtf xa = CeAt e−Atf xf
+ CeAt e−Atf eAtf xa
= CeAt e−Atf xf + CeAt xa
= CeAt e−Atf xf
– p. 7/2
Lemma: The observability Gramian Wo (0, tf ) is positive
definite if and only if rank O = n, where


C


 CA 

O=
.


.
.


CAn−1
is the observability matrix (O is np × n when A is n × n
and C is p × n)
– p. 8/2
Proof:
Wo (0, tf ) is singular if and only if
there is xa 6= 0 such that CeAt xa ≡ 0, ∀ t ∈ [0, tf ]
⇔ C
∞ k
X
t
k=0
k!
Ak xa ≡ 0, ∀ t ∈ [0, tf ]
⇔ CAk xa = 0, ∀ k ≥ 0
⇔ CAk xa = 0, ∀ k = 0, 1, . . . , n − 1
– p. 9/2



⇔ 



C
CA
..
.
CAn−1



⇔ rank 




 xa = 0


C
CA
..
.
CAn−1



<n


Theorem: The pair (A, C) is observable (reconstructible)
if and only if rank O = n
– p. 10/2
Duality
(A, B) controllable ⇔ rank [B, AB, . . . , An−1 B] = n


BT


T
T


B A
=n

⇔ rank 
..

.


B T (AT )n−1
³
´
⇔ AT , B T observable
Similarly, observability of (A, C) is equivalent to
¡ T
¢
T
controllability of A , C
– p. 11/2
Controllability and observability are invariant to state
transformations
{A, B, C, D} → x = P z → {P −1 AP, P −1 B, CP, D}
h
¢n−1
rank P −1 B, P −1 AP P −1 B, . . . , P −1 AP
P −1 B
£ −1
¤
−1
−1 n−1
= rank P B, P AB, . . . , P A
B
© −1
ª
n−1
= rank P [B, AB, . . . , A
B]
¡
i
= rank [B, AB, . . . , An−1 B]
Matlab:
O = obsv(A, C); rank(O)
– p. 12/2
Mathematical Preliminaries: Consider the equation
y = Mx
where M is an m × n matrix. The equation has a solution
if and only if y can be expressed as a linear combination of
the columns of M
rank M = rank [M, y]
The range space of M , denoted by R(M ), is the set of all
linear combinations of the columns of M . It is a subspace
of Rm of dimension equal to rank M . The equation
y = M x has a solution if and only if y ∈ R(M ). The
equation y = M x has a solution for every y ∈ Rm if and
only if
rank(M ) = m ⇔ R(M ) = Rm
– p. 13/2
Given that y = M x has a solution. Is the solution unique?
Suppose there are two solutions xa and xb such that
y = M xa ,
y = M xb
M (xa − xb ) = 0
The null space of M , denoted by N (M ), is the set of all
vectors x ∈ Rn such that M x = 0. It is a subspace of Rn
of dimension (n − rank M )
The solution of y = M x is unique if and only if there is no
vector x 6= 0 such that M x = 0
Equivalently, rank M = n
– p. 14/2
Discrete-Time Systems
The definitions of controllability (reachability) and
observability (constructibility) are the same as in the
continuous-time case
Let us study steering the state of the system
x(k + 1) = Ax(k) + Bu(k)
Find u(k) to steer the state of the system from x0 to xf in
finite time
x(k) = Ak x(0) +
k−1
X
Ak−j−1 Bu(j)
j=0
– p. 15/2



k
k−1
x(k) − A x0 = [B, AB, . . . , A
B] 


u(k − 1)
u(k − 2)
..
.
u(0)






Can we choose u(0), . . . , u(k − 1) such that x(k) = xf for
some k?
The answer is Yes if for some k,
³
´
xf − Ak x0 ∈ R [B, AB, . . . , Ak−1 B]
– p. 16/2
Due to Cayley-Hamilton theorem
³
´
¡
¢
k−1
n−1
R [B, AB, . . . , A
B] = R [B, AB, . . . , A
B]
for all k ≥ n
Thus, we take k = n and rewrite the equation as


u(n − 1)


 u(n − 2)  def
n−1
 = CU
xf − A
x0 = C 
..


.


u(0)
– p. 17/2
Reachability: x0 = 0 and xf is any vector in Rn
xf = C U
There is a solution for every xf if and only if
R(C) = Rn ⇔ rank C = n
Controllability: xf = 0 and x0 is any vector in Rn
−An x0 = C U
There is a solution if and only if −An x0 ∈ R(C)
rank C = n is a sufficient condition for the existence of a
solution. Is it necessary?
– p. 18/2
If A is nonsingular, then the vector An x0 can be any vector
in Rn . In this case, the condition rank C = n is necessary
If A is singular, the equation
−An x0 = C U
may be solvable even if rank C < n
– p. 19/2
Example:
A=
"
1 1
0 0
#
, B=
C = [B, AB] =
"
"
1 1
0 0
#
1
0
#
rank C = 1
– p. 20/2
−
"
α+β
0
#
=
"
1 1
0 0
#"
u(1)
u(0)
#
−(α + β) = u(0) + u(1)
There are infinitely many solutions. Two examples are
u(0) = −α − β, u(1) = 0
u(0) = −α,
u(1) = −β
– p. 21/2
Summary:
rank C = n is a necessary and sufficient condition for
reachability of x(k + 1) = Ax(k) + Bu(k)
rank C = n is a sufficient condition for controllability of
x(k + 1) = Ax(k) + Bu(k)
If A is nonsingular, then rank C = n is a necessary and
sufficient condition for controllability of
x(k + 1) = Ax(k) + Bu(k)
Remark: Strictly speaking, we should say that the pair
(A, B) is reachable if and only rank C = n. However, by
abuse of notation, we will continue to call the pair (A, B)
controllable if and only if rank C = n. This will keep the
discussion of controllability the same for both
continuous-time and discrete-time systems
– p. 22/2
Observability: Consider the system
x(k + 1) = Ax(k),
y(k) = Cx(k)
and study the question of uniquely determining x(0) from
the output y(k)






y(0)
Cx(0)
C






 y(1) 
 CAx(0) 
 CA 

 x(0)
=
=
.
.
.






.
.
.
.
.
.






y(k − 1)
CAk−1 x(0)
CAk−1
– p. 23/2
The equation will have a unique solution x(0) if and only if


C


 CA 
=n
rank 
..


.


CAk−1
By Cayley-Hamilton theorem


C


 CA 
 = rank
rank 
..


.


CAk−1






C
CA
..
.
CAn−1



, ∀ k ≥ n


– p. 24/2
Take k = n and write the equation as
Y = O x(0)
where



Y =


y(0)
y(1)
..
.
y(n − 1)






The equation has a unique solution if and only if
rank O = n
– p. 25/2