Lecture 9: Controllability and Observability - Illinois Institute of ...

Modern Control Systems
Matthew M. Peet
Illinois Institute of Technology
Lecture 9: Controllability and Observability

Controllability
We would like to prove that
To do this, we will prove that
• Im(W t ) ⊂ R t
• R t ⊂ Im(C(A, B))
• Im(C(A, B)) ⊂ Im(W t )
So far, we have show that
R t = Im(W t ) = Im(C(A, B))
Next, we will show that
R t ⊂ C AB
Im(W t ) ⊂ R t
M. Peet Lecture 9: Controllability 2 / 15

This proof M. Peet is especially useful, as it gives theLecture control 9: Controllability input which will get us to x3 / 15
Controllability: Im(W t ) ⊂ R t
Theorem 1.
Im(W t ) ⊂ R t
Proof.
First, suppose that x ∈ Im(W t ) for some t > 0. Then x = W t z for some z.
• Now let u(s) = B T e AT (t−s) z. Then
Γ t u =
=
∫ t
0
∫ t
= W t z = x
• Thus x ∈ Im(Γ t ) = R t .
We conclude that Im(W t ) ⊂ R t
0
e A(t−s) Bu(s)ds
e A(t−s) BB T e AT (t−s) zds

Proof by Contradiction: Im(C(A, B)) ⊂ Im(W t )
The last proof is a proof by contradiction.
Once you eliminate the impossible, whatever remains, no matter how
improbable, must be the truth.
There are two mutually exclusive possibilities
1. x ∈ Im(C(A, B)) ⊂ Im(W t )
2. There exists some x ∈ Im(C(A, B)) such that x ∉ Im(W t ) .
We eliminate the second possibility by showing that:
• If x ∉ Im(W t ) then x ∉ Im(C(A, B)).
In shorthand:
(¬2 ⇒ ¬1) ⇔ (1 → 2)
An alternative would be to find an x which disproves the second possibility.
M. Peet Lecture 9: Controllability 4 / 15

Proof by Contradiction: Im(C(A, B)) ⊂ Im(W t )
Theorem 2.
Im(C(A, B)) ⊂ Im(W t ).
Proof.
Suppose x ∉ Im(W t ). Then x ∈ Im(W t ) ⊥ .
• As we have shown, this means x ∈ ker(W t ), so W t x = 0.
• Thus
x T W t x =
=
∫ t
0
∫ t
0
x T e A(t−s) BB T e AT (t−s) xds
u(s) T u(s)ds = 0
where u(s) = B T e AT (t−s) x.
• This implies u(s) = B T e AT (t−s) x = 0 for all s ∈ [0, t].
M. Peet Lecture 9: Controllability 5 / 15

Proof by Contradiction: Im(C(A, B)) ⊂ Im(W t )
Proof.
• This means that for all s ∈ [0, t],
d k
ds k BT e AT s x = B T ( A T ) k
e
A T s x = 0
• At s = 0, this implies B T ( A T ) k
x = 0 for all k.
• We conclude that
x T [ B AB · · · A n−1 B ] = 0
• Thus C(A, B) T x = 0, so x ∈ ker C(A, B) T . As before, this means
x ∈ Im(C(A, B)) ⊥ .
• We conclude that x ∉ Im(C(A, B)). This proves by contradiction that
Im(C(A, B)) ⊂ Im(W t ).
M. Peet Lecture 9: Controllability 6 / 15

Summary: R t = Im(W t ) = Im(C(A, B))
We have shown that
R t = Im(W t ) = Im(C(A, B))
Moreover, we have shown that for any x d ∈ R Tf , we can find a controller
• Choose any z such that x d = W Tf z. (z = W −1
T f
x if W Tf is invertible)
• Let u(t) = B T e AT (T f −t) z.
• Then the system ẋ(t) = Ax(t) + B(t) with x(0) = 0 has solution with
x(T f ) = x d .
• x d = Γ Tf u.
M. Peet Lecture 9: Controllability 7 / 15

Representation and Controllability
Question: Is the representation (A, B, C, D) of the system y = Gu,
unique
ẋ(t) = Ax(t) + Bu(t)
y(t) = Cx(t) + Du(t)
Question: Do there exist (Â, ˆB, Ĉ, ˆD) such that y and u also satisfy,
ẋ(t) = Âx(t) + ˆBu(t)
y(t) = Ĉx(t) + ˆDu(t)
Answer: Of Course! Recall the similarity transform: z(t) = T x(t) for any
invertible T . Then y and u also satisfy,
ż(t) = T ẋ(t) = T Ax(t) + T Bu(t)
= T AT −1 z(t) + T Bu(t)
y(t) = Cx(t) + Du(t)
= CT −1 x(t) + Du(t)
M. Peet Lecture 9: Controllability 8 / 15

Representation and Controllability
Thus the pair (T AT −1 , T B, CT −1 , D) is also a representation of the map
y = Gu.
• Furthermore x(t) → 0 if and only if z(t) → 0.
• So internal stability is unaffected.
Controllability is Unaffected:
C(T AT −1 , T B) = [ T B T AT −1 T B T AT −1 T AT −1 T B · · · T A n−1 B ]
= T C(A, B)
M. Peet Lecture 9: Controllability 9 / 15

Invariant Subspaces
Definition 3.
A subspace, W ⊂ X, is Invariant under the operator A : X → X if x ∈ W
implies Ax ∈ W .
For a linear operator, only subspaces can be invariant.
Proposition 1.
If W if A-invariant, then there exists an invertible T , such that
]
[ [Ā11
T AT −1 Ā
=
12
I
and T W = Im
Ā 21 Ā 22 0]
That is, for any x ∈ W , T x =
] [¯x1
, which is clearly
0
Ā-invariant.
M. Peet Lecture 9: Controllability 10 / 15

Invariant Subspaces
Proposition 2.
C AB is A-invariant.
Proof.
The proof is direct
A · C(A, B) = A [ B AB · · · A n−1 B ] = [ AB A 2 B · · · A n B ]
But, by Cayley-Hamilton,
A n =
n−1
∑
a i A i
so if x ∈ C AB , there exists a z such that
Ax = A [ B AB · · · A n−1 B ] z
⎡
⎤
z n a 0
= [ B · · · A n−1 B ] z 1 + z n a 1
⎢
⎥
⎣ . ⎦ ∈ C AB
z n+1 + z n a n−1
i=0
M. Peet Lecture 9: Controllability 11 / 15

Controllability Form
Since C AB is an invariant subspace of A, there exists an invertible T such that
]
[Ā11
T AT −1 Ā
=
12
0 Ā 22
[¯x
and T x = for any x ∈ C
0]
AB .
• Clearly B ∈ C AB .
[ ] ¯B1
• Thus T B = .
0
Definition 4.
The pair (A, B) is in Controllability Form when
[ ]
A11 A
A =
12
and B =
0 A 22
and the pair (A 11 , B 1 ) is controllable.
[ ]
B1
,
0
M. Peet Lecture 9: Controllability 12 / 15

Controllability Form
When a system is in controllability form, the dynamics have special structure
ẋ 1 (t) = A 11 x 1 (t) + A 12 x 2 (t) + B 1 u(t)
ẋ 2 (t) = A 22 x 2 (t)
The uncontrolled dynamics are autonomous.
• Cannot be stabilized or controlled.
We can formulate a procedure for putting a system in Controllability Form
1. Find an orthonormal basis, [ ]
v 1 · · · v r for CAB .
2. Complete the basis in R n : [ ]
v r+1 · · · v n .
3. Define T = [ ]
v 1 · · · v n .
4. Construct Ā = T AT −1 and ¯B = T B
◮ Works for ANY invariant subspace.
M. Peet Lecture 9: Controllability 13 / 15

Controllability Form
Example
Let
⎡
1 −1
⎤
0
⎡ ⎤
1
A = ⎣−1 1 0 ⎦ and B = ⎣0⎦
0 0 −1
0
Construct C(A, B) = [ B AB A 2 B ] .
⎡
1 −1
⎤ ⎡ ⎤
0 1
⎡ ⎤
1
AB = ⎣−1 1 0 ⎦ ⎣0⎦ = ⎣−1⎦ ,
0 0 −1 0 0
⎡
⎤ ⎡ ⎤ ⎡ ⎤
1 −1 0 1 0
A 2 B = A(AB) = ⎣−1 1 0 ⎦ ⎣−1⎦ = ⎣0⎦
0 0 −1 0 0
Thus
⎡
1 1
⎤
0
C(A, B) = ⎣0 −1 0⎦
0 0 0
Thus rank C(A, B) = 2 < n = 3 which means not controllable.
M. Peet Lecture 9: Controllability 14 / 15

Controllability Form
Example Continued
Using Gramm-Schmidt, we can construct an orthonormal basis for C AB
⎧⎡
⎤ ⎡ ⎤⎫
⎧⎡
⎤ ⎡ ⎤⎫
⎨ 1 1 ⎬ ⎨ 1 0 ⎬
C AB = span ⎣0⎦ , ⎣−1⎦
⎩
⎭ = span ⎣0⎦ , ⎣1⎦
⎩ ⎭ = span {v 1, v 2 }
0 0
0 0
Let v 3 = [ 0 0 1 ] T
. Then
⎡
T −1 = ⎣ 1 0 0
⎤
0 1 0⎦ = I
0 0 1
So T = I, which is because the system is already in controllability form. We
could also have used
⎡
0 1
⎤
0
⎡
1 −1
⎤
0
T −1 = ⎣1 0 0⎦ to get T AT −1 = ⎣−1 1 0 ⎦ = A
0 0 1
0 0 −1
M. Peet Lecture 9: Controllability 15 / 15