16.513 Control Systems
Controllability and Observability
(Chapter 6)
1
A General Framework in State-Space Approach
Given an LTI system:
x& = Ax + Bu; y = Cx
(*)
The system might be unstable or doesn’t meet the required
performance spec. How can we improve the situation?
The main approach: Let u= v- Kx (state feedback), then
x& = Ax + B(v - Kx); y = Cx + D(v - Kx)
= (A - BK)x + Bv;
= (C - DK)x - Dv
The performance of the system is changed by matrix K.
Questions:
• Is there a matrix K s.t. A-BK is stable?
• Can eig(A-BK) be moved to desired locations?
These issues are related to the controllability of (*) 2
1
Main Result 1: The eigenvalues of A-BK can be moved
to any desired locations iff the system (*) is controllable.
Another situation: the state x is not completely available.
Only a linear combination of x, e.g., y = Cx, can be
measured. How can we realize u = v-Kx?
A possible solution: build an observer to estimate x based
on measurement of y.
Main result 2: The observer error (difference between the
real x and estimated x) can be made arbitrarily small within
arbitrarily short time period iff (*) is observable.
We will arrive at these conclusions in Chapter 8. Before
that, we need to prepare some tools and go through these
fundamental problems: controllability and observability.3
Controllability and Observability
Consider an LTI system:
x& = Ax + Bu;
y = Cx + Du
Questions:
a) Can we move the state from one point in the state space to a
desired location: x0 → xd, by choosing u properly?
b) Can we build a controller to stabilize the system?
c) Suppose that y contains all the quantities that we
can measure. Can we evaluate the state x from y?
– The first two questions are related to the controllability of
the system
– The last one is related to observability.
– They describe, in some sense, the potential of a system that
can be explored through feedback design, e.g., stabilization
by output feedback.
4
2
An example: A circuit system
u
R1
R2
C1
C2
+ x −
1
+ x −
2
+
y
−
Input: u, a current source
State: x1,x2, voltages across the two capacitors
Output: y, voltage between two points
Observation:
The input u has no effect on x2
⇒ x2 cannot be altered by u ⇒ x2 is not controllable
The measurement y has no relation to x1, ⇒ any change on
x1 is not reflected on y ⇒ x1 is not observable.
5
Controllability: Definition
Consider the system
x& = Ax + Bu, x ∈ R n ; u ∈ R p .
Controllability is a relationship between state and input.
Definition: The system, or the pair (A,B), is said to be
controllable if for any initial state x(0)=x0 and any final
state xd, there exist a finite time T > 0 and an input u(t),
t∈[0,T] such that
T
x(T) = e AT x 0 + ∫ e A(T-τ) Bu(τ )dτ = x d
0
(1)
Comment: There may exist different T and u that satisfy (1).
As a result, there may be different trajectories starting from x0 and
end at xd. Controllability does not care about the difference. 6
3
Examples: uncontrollable networks.
+
u
−
R1= 1 Ω
R2= 1 Ω
+x −
R3= 1 Ω
+
u
−
R4= 1 Ω
Observation:
• If x(0)=0, then x(t)=0 for
all t > 0. The input u can
do nothing about it.
• If the resistance is changed
so that R1/R3 ≠ R2/R4, then
you can bring x to any desired
value.
+
x1
−
1F
+
x21 F
−
1Ω
1Ω
Observation:
• If x1(0)=x2(0)=0, then
x1(t)=x2(t) for all t > 0.
You cannot bring x(t) to any
point in the plane.
• This situation can be changed
by altering the parameters
of the components.
7
Theorem 1: The pair (A,B) is controllable if and only if
the n×n matrix,
t
t
0
0
Wc (t) = ∫ e Aτ BB' e A'τ dτ = ∫ e A(t − τ) BB' e A'(t − τ) dτ
is nonsingular for every t > 0.
Main idea of the proof: Suppose Wc(t1)-1 exists. Recall the solution
If we choose
x(t 1 ) = e At1 x 0 + ∫
t1
0
e A(t1 - τ) Bu( τ )dτ
−1
u(t) = − B' e A'(t1 − t) Wc (t1 )[e At1 x 0 − x d ]
Then, x(t 1 ) = e At x 0 − t e A(t - τ) BB' e A'(t −τ ) Wc −1 (t1 )[e At x 0 − x d ]dτ
∫
1
1
{∫
0
= e At1 x 0 −
t1
0
1
1
}
1
−1
e A(t1 - τ) BB' e A'(t1 −τ ) dτ Wc (t1 )[e At1 x 0 − x d ]
= e At1 x 0 − e At1 x 0 + x d = x d
The necessity of the condition is shown by contradiction. (p.146)
8
4
t
t
0
0
Wc (t) = ∫ e Aτ BB' e A'τ dτ = ∫ e A(t − τ) BB' e A'(t − τ) dτ
Equivalent conditions: The following are equivalent
conditions for the pair (A,B) to be controllable:
1)
2)
3)
4)
Wc(t) is nonsingular for every t > 0.
Wc(t) is nonsingular for at least one t > 0.
For every v∈Rn, v≠0, v’eAt B is not identically zero.
The matrix Gc = [B AB A2B … An-1B] has full
row rank, i.e., ρ(Gc) = n.
5) The matrix M(λ) = [A−λI B] has full row rank at all λ∈C.
6) M(λ) has full row rank at every eigenvalues of A.
Note: M(λ) has full row rank if λ is not an eigenvalue of A.
We only need to check the rank of M(λ) at eigenvalues of A.
Note : Of all the conditions, only 4) and 6) can be practically verified.
9
Everything starts from the matrix function
t
Wc (t) = ∫ e Aτ BB' e A'τ dτ
0
We first give some necessary background and tools.
Notice that Wc(t) is a symmetric matrix: Wc’(t)=Wc(t)
ƒ
ƒ
ƒ
ƒ
Actually it is positive semidefinite. Why?
All eigenvalues are non-negative.
Wc(t1) is nonsingular iff Wc(t1) > 0.
Controllability is related to the positive definiteness.
10
5
t
Wc (t) = ∫ e Aτ BB' e A'τ dτ
0
1) Wc(t) is nonsingular for any t > 0.
2) Wc(t) is nonsingular for at least one t > 0.
Explanations: 1) ⇔ 2) Only need to show 2) ⇒1)
Assume that Wc(t1) is nonsingular. Then Wc(t1)>0.
⇒ v’Wc(t1)v > 0 for every v∈Rn.
⇒ For a fixed v, v’Wc(t)v is analytic in t and
nondecreasing. Also, v’Wc(0)v=0, v’Wc(t1)v > 0
⇒ v’Wc(t)v > 0, for all t > 0.
⇒ For any t>0 and v∈Rn, v’Wc(t)v > 0.
⇒ Wc(t) > 0 for all t > 0.
⇒ Wc(t) nonsingular for all t > 0.
Note: If f(t) is analytic and f(t)=0 for t∈[a,b], then f(t)=0 for all 11t∈R.
The above relation can also be proven with the result
from last lecture
Theorem: Let M(t)∈Rn×n be a symmetric matrix
function. Suppose that every element of M(t) is an
analytic function, i.e., mij(t) can be differentiated
infinitely many times. Then the eigenvalues of M(t)
can be arranged as λ1(t),λ2(t),…,λn(t) where each λi(t)
is analytic.
If M(t) is nondecreasing, i.e., M(t1) ≤ M(t2), for any
t1< t2, then λ1(t), λ2(t) ,…, λn(t) are nondecreasing.
12
6
t
Wc (t) = ∫ e Aτ BB' e A'τ dτ
0
1) Wc(t) is nonsingular for any t > 0.
3) For every v∈Rn, v≠0, v’eAt B not identically zero.
Explanations: 1) ⇔ 3)
Suppose Wc(t1) is singular for certain t1.
Exist v∈Rn such that v’Wc(t1) v =0 .
Since v’eAτBB’eA’τ v≥0 for all τ, must have
⇒ v’eAτBB’eA’τ v=0, v’eAτB=0 for all τ∈[0,t1]
⇒ v’eAtB=0 for all t by analyticity.
On the other hand if v’eAtB=0 for all t,
⇒ v’Wc(t) v =0 for all t
⇒ Wc(t) must be singular since it is positive
semi-definite.
13
3) For every v∈Rn, v≠0, v’eAt B not identically zero.
4) The matrix Gc = [B AB A2B … An-1B] has full row
rank i.e., ρ(G) = n.
Explanation: 3) ⇔ 4)
The opposite of 3) and 4):
3a) Exists v∈Rn, v≠0, v’eAt B =0 for all t.
4a) ρ(G) < n
4a) ⇒ 3a) Exists v such that v’Gc=0 ⇒ v’B=v’AB=…=v’An-1B=0.
Since eAt is a linear combination of {I,A,A2,…An-1},
eAtB = k0 (t)B+k1(t) AB+ k2 (t) A2B+…+kn-1(t) An-1B
⇒ v’eAtB=0 for all t
3a) ⇒ 4a) v’eAt B =0 for all t ⇒ dj (v’eAtB)/dtj |t=0 = 0 for all j ≥ 0.
⇒ v’eAtAjB=0 for all j. ⇒ v’eAtGc=0 ⇒ ρ(Gc) < n.
For the relationship between 4) and 5), or 4) and 6), see page 147.
14
7
t
t
0
0
Wc (t) = ∫ e Aτ BB' e A'τ dτ = ∫ e A(t − τ) BB' e A'(t − τ) dτ
Equivalent conditions: The following are equivalent
conditions for the pair (A,B) to be controllable:
1)
2)
3)
4)
Wc(t) is nonsingular for any t > 0.
Wc(t) is nonsingular for at least one t > 0.
For every v∈Rn, v≠0, v’eAt B not identically zero.
The matrix Gc = [B AB A2B … An-1B] has full
row rank, i.e., ρ(Gc) = n.
5) The matrix M(λ) = [A−λI B] has full row rank at all λ∈C.
6) M(λ) has full row rank at every eigenvalues of A.
Note: M(λ) has full row rank if λ is not an eigenvalue of A.
We only need to check the rank of M(λ) at eigenvalues of A.
Note : Of all the conditions, only 4) and 6) can be practically verified.
15
t
Wc (t) = ∫ e Aτ BB' e A'τ dτ
0
Some observations: Recall from the proof of Theorem 1(slide 8).
To bring the state from x(0) = x0 to x(t1) = xd , a particular input is
u(t) = − B' e A'(t1 − t) W −1 (t1 )[e At1 x 0 − x d ]
ƒ This is actually the minimal energy control, i.e., if there is another
input w(t) to transfer x0 to xd within the same time interval, then
∫
t1
0
t1
w(τ )' w(τ )dτ ≥ ∫ u (τ )' u (τ )dτ
0
ƒ If (A,B) is controllable,
for all t > 0.
⇒ The transfer of the state can be accomplished in arbitrarily small
time interval
Wc(t)-1 exists
Note that as t1 decrease, both λmin[Wc(t1)] and λmax[Wc(t1)] decrease.
Then ||Wc(t1)-1|| increases. ⇒ larger magnitude of u is required.
As t1 → 0, ||Wc(t1)-1|| → ∞ , u(t) → ∞.
16
8
t
Wc (t) = ∫ e Aτ BB' e A'τ dτ
0
An example:
1
0⎤
⎡0
⎡1⎤
A=⎢ 0
0
1 ⎥, B = ⎢1⎥
⎣⎢− 1 − 2 − 3⎦⎥
⎣⎢1⎦⎥
x& = Ax + Bu,
Eigenvalues of Wc(t1)
10
10
10
10
10
10
Eigenvalues of Wc(t1)-1
2
10
0
10
-2
10
-4
10
-6
10
-8
0
1
2
3
t
1
u(t) = − B' e
A'(t1 − t)
4
10
5
8
6
4
2
0
-2
0
1
2
W (t1 )[e x 0 − x d ]
−1
At1
3
4
5
t
1
Magnitude of u increases as t1 is decreased.
17
Example: Determine the controllability for
A = ⎡− 1 0 ⎤, B = ⎡a ⎤
⎢⎣ 0 − 1⎥⎦
⎢⎣b ⎥⎦
Approach 1: G c = [B AB] = ⎡⎢ab -- ab⎤⎥
⎣
⎦
x& = Ax + Bu,
ρ (G ) < 2 = n for all possible a and b
The system not controllable whatever a and b are.
Approach 2: Check M(λ)=[A-λI B] at λ=-1
M( −1) = ⎡0 0 a ⎤
⎢⎣0 0 b ⎥⎦
ρ(M(-1)) < 2 for all possible a and b
Same conclusion on controllability
18
9
Example:
x& = Ax + Bu,
A = ⎡− 1 0 ⎤, B = ⎡a ⎤
⎢⎣ 0 − 2⎥⎦
⎢⎣b ⎥⎦
Approach 1: G c = [B AB] = ⎡a
⎣⎢b
-a ⎤
- 2b ⎥⎦
⎧ρ(G c ) = 2, if a ≠ 0 and b ≠ 0
detG c = −ab, ⎨
c
⎩ ρ(G ) < 2, if either a = 0 or b = 0
The system is controllable if a ≠ 0 and b ≠ 0.
Approach 2: Check M(λ)=[A-λI B] at λ=-1
M(−1) = ⎡0 0 a ⎤, M(−2) = ⎡1 0 a ⎤,
⎢⎣0 - 1 b ⎥⎦
⎢⎣0 0 b ⎥⎦
ρ(M(-1))= ρ(M(-2))=2 iff a ≠ 0 and b ≠ 0
Same conclusion on controllability
19
A general SI system (diagonalizable)
x& = Ax + Bu,
⎡λ1 0
⎢
A = ⎢ 0 λ2
M M
⎢0 0
⎣
L
L
O
L
0⎤
⎡ b1 ⎤
0 ⎥, B = ⎢b2 ⎥
⎢M⎥
M⎥
⎥
⎢b ⎥
λn ⎦
⎣ n⎦
The above system is controllable if and only if
the eigenvalues are distinct and none of the bi’s is zero
20
10
Example:
α −β⎤
⎡ b1 ⎤
=
,
A = ⎡⎢
B
⎢⎣b2 ⎥⎦
⎣ β α ⎥⎦
x& = Ax + Bu,
⎡ b αb1 − βb 2 ⎤
G c = [B AB] = ⎢ 1
⎣b 2 βb1 + αb 2 ⎥⎦
detG c = β (b12 + b 22 ),
⎧ ρ(G c ) = 2, if β ≠ 0 and b12 + b22 ≠ 0
⎨
c
2
2
⎩ρ(G ) < 2, if either β = 0 or b1 + b2 = 0
The system is controllable if β≠ 0 and (b1,b2) ≠ (0,0)
21
Example:
⎡A 0 ⎤
⎡B ⎤
A=⎢ 1
, B = ⎢ 1⎥
⎥
0
A
2⎦
⎣ B2 ⎦
⎣
Suppose that the eigenvalues of A1 and those of A2 are disjoint.
⇒ (A,B) is controllable iff (A1,B1) is controllable and (A2,B2) is
controllable.
0
B1 ⎤
Proof: Consider M(λ ) = [λI − A B ] = ⎡λI - A1
⎢⎣ 0
λI - A 2 B2 ⎥⎦
Problem: Show that
⎡X 0 ⎤
rank ⎢
⎥ ≥ rank(X) + rank(Y)
⎣ Z Y⎦
If Y has full row rank, then rank ⎡X 0 ⎤ = rank(X) + rank(Y)
⎢ Z Y⎥
⎣
⎦
⎡X 0 ⎤
Give an example such that, rank ⎢
⎥ ≠ rank(X) + rank(Y)
⎣ Z Y⎦
22
11
Theorem: Consider the pair
⎡ A1 0 L 0 ⎤
⎡ B1 ⎤
⎢
⎥
⎢ ⎥
A = ⎢ 0 A2 L 0 ⎥ , B = ⎢ B2 ⎥
M M O 0
M
⎢0 0 L A ⎥
⎢B ⎥
m⎦
⎣
⎣ m⎦
Suppose that the eigenvalues of Ai and those of Aj are
disjoint for i ≠ j . Then (A,B) is controllable iff (Ai,Bi)
is controllable for all i.
23
Theorem: Let ρ(B) = p. The pair (A,B) is controllable iff
Gc n-p+1: = [B AB A2B … An-pB]
has full row rank. This is equivalent to Gcn-p+1Gc’n-p+1 being
nonsingular, and to Gcn-p+1Gc’n-p+1 > 0 (positive definite.)
Note: The following are equivalent
• X has full row rank;
• XX’ is nonsingular;
• XX’ is positive-definite.
• How to prove these?
24
12
Procedure to determine controllability:
Step 1: form the n×(p×(n-p+1)) matrix
Gcn-p+1=[b1 b2 … bp Ab1 Ab2 … Abp …… An-pb1 An-pb2 …An-pbp]
Step 2: From left to right, throw away columns
that are dependent on its LHS columns (or LI columns)
until n LI columns are detected or until the last column.
– Recall: y is dependent on the columns of Z iff ρ(Z) = ρ([Z y])
Step 3:
• If the maximal number of LI columns is n, controllable.
25
Example:
⎡0 1
A = ⎢3 0
⎢0 0
⎣⎢0 − 2
0
0
0
0
0⎤
⎡0
2⎥ , B = ⎢1
⎢0
1⎥
0⎦⎥
⎣⎢0
0⎤
0⎥
0⎥
1⎦⎥
n=4, p=2. ρ(B)=2= p.
G
c
n − p +1
⎡0
= [B AB A B] = ⎢1
⎢0
⎢⎣0
b1
2
0 1 0
0 0 2
0 0 1
1 −2 0
b 2 Ab1 Ab 2
0
2⎤
−1 0 ⎥
−2 0 ⎥
0 − 4⎥⎦
A 2 b1 A 2 b 2
The first 4 columns are LI. ⇒ ρ(Gcn-p+1)=4 = n
⇒ (A,B) controllable
26
13
Example:
⎡1
A = ⎢0
⎢0
⎢⎣0
G
c
n − p +1
0
2
0
0
0
1
2
0
0⎤
⎡1
0⎥ , B = ⎢1
⎢1
0⎥
⎢⎣0
3⎥⎦
⎡1
= [B AB A B] = ⎢1
⎢1
⎢⎣1
b1
2
0⎤
0⎥
0⎥
1⎥⎦
n=4, p=2. ρ(B)=2.
0
1
0
1
0⎤
0
3
0
8
0⎥
0
2
0
4
0⎥
1
3
3
9
9⎥⎦
b 2 Ab1 Ab 2 A 2 b1 A 2 b 2
The first 3 columns are LI.
The 4th is dependent on the first 3.
⎡1
[b1 b 2 Ab1 A b1 ] = ⎢⎢11
⎢
⎣1
2
0
0
0
1
1
3
2
3
1
8
4
9
⎤
⎥
⎥ has full row rank
⎥
⎦
Hence (A,B) is controllable.
27
Effect of equivalence transformation
Recall that equivalence transformation can make the structure
cleaner and simplify analysis.
Question:
Does similarity transformation retain the controllability property?
Theorem: The controllability property is invariant under any
equivalence transformation
Proof: Consider (A,B) with Gc=[B AB A2B …. An-1B].
Let the transformation matrix be P. Then (A,B) ⇔ (PAP-1, PB)
G c = [B AB L A n −1 B]
= [PB PAP -1PB L PA n -1P −1PB]
= [PB PAB L PA n -1B]
= P[B AB L A n -1B]
= PG c
Since P is nonsingular,
ρ(G c ) = ρ(G c )
28
14
Review: Definition of Controllability
Consider the system
x& = Ax + Bu, x ∈ R n ; u ∈ R p .
Controllability is a relationship between state and input.
Definition: The system, or the pair (A,B), is said to be
controllable if for any initial state x(0)=x0 and any final
state xd, there exist a finite time T > 0 and an input u(t),
t∈[0,T] such that
T
x(T) = e AT x 0 + ∫ e A(T-τ) Bu(τ )dτ = x d
0
(1)
29
Next Problem: Observability
30
15
Observability: A dual concept
Consider an n-dimensional, p-input, q-output system:
x& = Ax + Bu; y = Cx + Du
u
System
y
Assume that we know the input and can measure the
output, but has no access to the state.
Definition: The system, is said to be observable if for
any unknown initial state x(0), there exists a finite t1> 0
such that x(0) can be exactly evaluated over [0,t1] from
the input u and the output y. Otherwise the system is
said to be unobservable.
31
Examples: Two circuits
1H
1H
x1
+
+
1Ω
1Ω
1F
1Ω
1Ω
x2 = y
0 ⎤ ⎡ x1 ⎤ + ⎡1⎤ u,
− 1⎥⎦ ⎢⎣ x 2 ⎥⎦ ⎢⎣1⎥⎦
y = x2
The output is independent of x1;
Or any change on x1 will not be
reflected from the output.
⇒ Unobservable.
1F
x2 = y
−
−
⎡ x& 1 ⎤ ⎡− 1
⎢⎣ x& 2 ⎥⎦ = ⎢⎣ 0
x1
⎡ x& 1 ⎤ ⎡0
⎢⎣ x& 2 ⎥⎦ = ⎢⎣1
− 1⎤ ⎡ x1 ⎤ + ⎡1⎤ u,
− 2⎥⎦ ⎢⎣ x 2 ⎥⎦ ⎢⎣1⎥⎦
y = x2
x1 has some effect on x2= y.
We will see later that the
circuit is observable.
32
16
Some observations:
Recall the state-space solution,
t
y(t) = Ce At x 0 + C ∫ e A(t - τ) Bu( τ )dτ + Du(t)
0
Since both u and y are known, define
t
y(t) = y(t) − C ∫ e A(t - τ) Bu( τ )dτ − Du(t)
0
Then y(t) is known. Also we have
y(t) = Ce At x 0
The problem is whether x 0 can be detected given y(t)
over an interval [0, t1 ]
Clearly, the observability depends only on matrices A
and C. We can assume that B=0 and D=0.
33
Theorem: The system (A,B,C,D) is observable if and
only if the n×n matrix
t
Wo (t ) = ∫ e A'τ C ' Ce Aτ dτ
0
is nonsingular for any t > 0.
Proof. Consider the case where B = 0 and D = 0. Then
y(t) = Ce At x 0
∫
t1
0
e A't C' y(t) = e A't C' Ce At x 0
t1
e A't C' y(t)dt = ∫ e A't C' Ce At x 0 dt
0
t1
= ∫ e A't C' Ce At dt x 0 = Wo (t1 )x 0
0
If Wo(t1) is nonsingular, then
t1
x 0 = Wo (t1 ) −1 ∫ e A't C' y(t)dt
0
and observability is ensured. (Sufficiency proved)
34
17
Proof continued. (necessity)
Take a comparison between Wo and Wc
t
Wo (t) = ∫ e A'τ C' Ce Aτ dτ
0
t
Wc (t) = ∫ e Aτ BB' e A'τ dτ
0
Mathematically they have the same structure. So we
have the following equivalent conditions:
1) Wo(t) is nonsingular for any t > 0.
2) Wo(t) is nonsingular for at least one t > 0.
3) For every v∈Rn, v≠0, CeAt v not identically zero.
If W0(t) is singular for any t > 0, then there exists v∈Rn,
v≠0 such that CeAtv=0 for all t.
If we have y(t)= CeAtx0 for some x0, then
Ce At (x 0 + kv) = Ce At x 0 + kCe At v = y(t)
This means that different initial conditions will yield the same
output. And the difference can never to told. ⇒ unobservable.35
Duality between controllability and observability
By comparing
t
t
0
0
Wo (t) = ∫ e A'τ C' Ce Aτ dτ Wc (t) = ∫ e Aτ BB' e A'τ dτ
It is straightforward to conclude the following
Theorem of duality: The pair (A,B) is controllable if
and only if (A1, C1) = (A’,B’) is observable.
x& = Ax + Bu
Dual systems
z& = A1 z = A' z
y = C1 z = B' z
36
18
Equivalent conditions for observability:
1) The pair (A,C) is observable.
2) Wo(t) is nonsingular for some t > 0.
3) The observability matrix
⎡ C ⎤
G = ⎢⎢ CA ⎥⎥
M
⎢⎣CA n −1 ⎥⎦
o
has full column rank, i.e., ρ(Go) = n.
4) The matrix
M o ( λ) = ⎡A − λI⎤
⎢⎣ C ⎥⎦
has full column rank at every eigenvalue of A.
37
Theorem: The pair (A,C) is observable if and only if
G
o
n -q +1
⎡ C ⎤
= ⎢⎢ CA ⎥⎥
M
⎢⎣CA n −q ⎥⎦
has full column rank, where q=ρ(C).
Theorem: The obserbability property is invariant
under any equivalence transformation;
38
19
Examples: Two circuits
1H
1H
x1
+
+
1Ω
1Ω
1Ω
1Ω
x2 = y
1F
x1
1F
−
−
⎡ x& 1 ⎤ ⎡− 1
⎢⎣ x& 2 ⎥⎦ = ⎢⎣ 0
y = [0
0 ⎤ ⎡ x1 ⎤ + ⎡1⎤ u,
− 1⎥⎦ ⎢⎣ x 2 ⎥⎦ ⎢⎣1⎥⎦
1] x
G o = ⎡ C ⎤ = ⎡0
⎢⎣CA ⎥⎦ ⎢⎣0
⎡ x& 1 ⎤ ⎡0
⎢⎣ x& 2 ⎥⎦ = ⎢⎣1
y = [0
1⎤
− 1⎥⎦
ρ(Go) = 1 < 2
Unobservable
x2 = y
− 1⎤ ⎡ x1 ⎤ + ⎡1⎤ u,
− 2⎥⎦ ⎢⎣ x 2 ⎥⎦ ⎢⎣1⎥⎦
1] x
G o = ⎡ C ⎤ = ⎡0
⎢⎣CA ⎥⎦ ⎢⎣1
1⎤
− 2⎥⎦
ρ(Go) = 2
Observable
39
Theorem: Consider the pair
⎡ A1 0 L 0 ⎤
⎢
⎥
A = ⎢ 0 A2 L 0 ⎥ , C = [C1 C2 L Cm ]
M M O 0
⎢0 0 L A ⎥
m⎦
⎣
Suppose that the eigenvalues of Ai and those of Aj are
disjoint for .i ≠ j . Then (A,C) is observable iff (Ai,Ci)
is observable for all i.
40
20
Comments on controllability and observability
Controllability: Recall from the proof of the theorem on slide 8.
To bring the state from x(0) = x0 to x(t1) = xd , a particular input is
u(t) = −B' e A'(t1 − t) Wc−1 (t1 )[e At1 x 0 − x d ]
ƒ Usually we don’t apply such an input to move the state between
points in the state space. (open-loop control, very sensitive).
ƒ A more practical problem is to use a simple linear feedback
control, such as u = Fx + u0, to bring the state asymptotically
toward a certain point.
– Such a problem is related to stabilizability.
A future result:
ƒ If a system is controllable, then it is stabilizable.
41
Observability: Theoretically, the initial state x(0) can be computed
from the output y(t) by the following formula
t1
x 0 = Wo (t1 ) −1 ∫ e A't C' y(t)dt
0
And the time interval [0,t1] can be arbitrarily small.
ƒ Again, this approach is complicated and sensitive.
ƒ Actually, we are not only interested in computing x(0) with
other available information. Most often, we would like to
detect the state for all time.
ƒ What we will be doing later is to build a LTI system with
u and y as input and z as state so that z(t)→x(t) as t → 0.
u
LTI
System
State estimator
y
xe
If the system is observable,
then the estimated state can
be made to approach the real
42
state asymptotically.
21
So far, we have learned
• Controllability
• Observability
Next, we will study
• Canonical decomposition: to divide the state space
into controllable/uncontrollable,
observable/unobservable subspaces
43
Canonical Decomposition
Consider an LTI system,
x& = Ax + Bu, y = Cx + Du
Let z = Px, where P is nonsingular, then
z& = Ax + Bu, y = C z + Du
where A = PAP -1 , B = PB, C = CP −1 , D = D
Recall that under an equivalence transformation, all properties,
such as stability, controllability and observability are preserved.
We also have G c = PG c ,
G o = G o P −1
Next we are going to use equivalence transformation to obtain
certain specific structures which reflect controllability and
observability.
44
22
Controllability decomposition
Recall Gc=[B AB … An-1B]. Suppose that ρ(Gc) = n1 < n.
Then Gc has at most n1 LI columns.
They form a basis for the range space of Gc.
Theorem: Suppose that ρ(Gc) = n1 < n. Let Q be a
nonsingular matrix whose first n1 columns are LI
columns of Gc. Let P=Q-1. Then
A12 ⎤
⎡B ⎤
, B = PB = ⎢ c ⎥, A c ∈ R n1 ×n1 , Bc ∈ R n1 ×p
A c ⎥⎦
⎣0⎦
⎡A
A = PAP −1 = ⎢ c
⎣0
C = Cc
Cc
[
]
Moreover, the pair (A c , Bc ) is controllab le and
Cc ( sI − A c ) −1 Bc + D = C(sI − A) −1 B + D
See page 159 for the proof.
45
Discussion:
After state transformation, the equivalent system is
z& 1 = A c z1 + A12 z 2 + Bc u
z& 2 =
Acz2
The input u has no effect on z2. This part of state is uncontrollable.
The first sub-system is controllable if z2=0. If z2≠0, then
t
t
z1 (t1 ) = e A t z10 + ∫ e A (t - τ) Bc u( τ )dτ + ∫ e A (t - τ) A12 z 2 ( τ )dτ
1
c 1
z 2 (τ ) = e A cτ z 20
c
1
1
0
c
1
0
Given a desired value for z1, say z1d. If we let
t1
t1
v(t1 ) = ∫ e A c (t1 - τ) A12 e A cτ z 20 dτ , Wc (t1 ) = ∫0 e A τ Bc Bc ' e A 'τ dτ
c
c
0
−1
and u(t) = − Bc ' e A c '(t1 − t) Wc (t1 )[e A c t1 z10 + v(t1 ) − z1d ]
Then you can verify that z1(t1)=z1d.
46
23
Example:
G c2 = [B
x& = Ax + Bu
⎡1 1
A = ⎢0 1
⎢⎣0 1
⎡0
AB] = ⎢1
⎢⎣0
0⎤
0⎥ ,
1⎥⎦
1
0
1
⎡0
Let Q = [b1 b 2 q ] = ⎢1
⎢⎣0
⎡0
B = ⎢1
⎢⎣0
n=3, p=2, n-p+1=2.
Only need to check Gc2
1⎤
ρ (G2c ) = 2 < 3, uncontrollable
0⎥
1⎥⎦
1⎤
q is picked to make
1⎥, P = Q −1
Q nonsingular
0⎥⎦
1
1
1
1
0
1
1 ⎤ ⎡1
⎡−1 1
A = PAQ= ⎢ 0
0
1 ⎥ ⎢0
⎢⎣ 1
0 −1⎥⎦ ⎢⎣0
⎡1 0⎤ ⎡B ⎤
B = PB= ⎢0 1⎥ = ⎢ c ⎥
⎢⎣0 0⎥⎦ ⎣ 0 ⎦
1⎤
0⎥
1⎥⎦
1
1
1
0⎤ ⎡0
0⎥ ⎢1
1⎥⎦ ⎢⎣0
1
0
1
1⎤ ⎡1
1⎥ = ⎢1
0⎥⎦ ⎢⎣0
0
1
0
0⎤ ⎡A
1⎥ = ⎢ c
1⎥⎦ ⎣ 0
A12⎤
,
Ac ⎥⎦
Note: the last column of Q is different from the book (page 161).
As a result, Ā12 is different from that in the book, which is 0.
47
Observability decomposition (follows from duality)
⎡ C ⎤
Recall G = ⎢⎢ CA ⎥⎥
M
⎢⎣CA n −1 ⎥⎦
o
Theorem: Suppose that ρ(Go) = n1 < n. Let P be a nonsingular
matrix whose first n1 rows are LI rows of Go. Then
⎡A
A = PAP −1 = ⎢ o
⎣ A 21
C = Co 0 ,
[
]
⎡B ⎤
0⎤
, B = PB = ⎢ o ⎥, A o ∈ R n1×n1 , Bo ∈ R n1×p
A o ⎥⎦
⎣ Bo ⎦
Co ∈ R q × n 1
Moreover, the pair (A o , Co ) is observable and
Co ( sI − A o ) −1 Bo + D = C(sI − A) −1 B + D
Discussion: After state transformation, the equivalent system is
z& 1 = A o z1 + Bo u
z& 2 = A 21z1 + A o z 2 + Bo u,
y = Co z1 + Du
z2 may be affected by z1
but has no effect on y or z1
48
24
Summary for today:
• Controllability
• Observability
• Canonical decomposition
– Controllable/uncontrollable
– Observable/unobservable
Next Time:
• Controllability and observability continued
–
–
–
–
–
Controllability/observability decomposition
Minimal realization
Conditions for Jordan form conditions
Parallel results for discrete-time systems
Controllability after sampling
• State feedback design (introduction)
49
Problem Set #9
1. Is the following state equation controllable? observable?
⎡ 0 1 − 1⎤
⎡0 ⎤
x& = ⎢⎢− 1 − 1 1 ⎥⎥ x + ⎢⎢1⎥⎥ u,
⎢⎣0⎦⎥
⎣⎢− 1 − 1 0 ⎥⎦
y = [1 0 1]
If not controllable, reduce it to a controllable one;
If not observable, reduce it to an observable one.
2. Is the following state equation controllable? observable?
⎡1 0 − 2 ⎤
⎡0 1 ⎤
x& = ⎢⎢0 1 − 2⎥⎥ x + ⎢⎢0 1⎥⎥ u,
⎢⎣1 0⎥⎦
⎣⎢0 0 2 ⎦⎥
y = [1 0 1]
If not controllable, reduce it to a controllable one;
If not observable, reduce it to an observable one.
50
25
Bonus Problem: Show that
⎡X 0 ⎤
rank ⎢
⎥ ≥ rank(X) + rank(Y)
⎣ Z Y⎦
If Y has full row rank, then rank ⎡X 0 ⎤ = rank(X) + rank(Y)
⎢ Z Y⎥
⎣
⎦
⎡X 0 ⎤
Give an example such that, rank ⎢
⎥ ≠ rank(X) + rank(Y)
⎣ Z Y⎦
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26