some examples on controllability and observability

ECE 602 Lumped Systems Theory December 08, 2008 1 ECE 602 Lecture Notes: Some Examples on Controllability and Observability Example 1. Suppose that A is 3 × 3. Is A 4 B linearly dependent on {A 3 B,A 2 B, AB}? First, suppose that A has full rank. Then A is invertible so the subspace spanned by {A 3 B,A 2 B, AB} is the same as that spanned by {A 2 B,AB,B}. Thus every column of A 4 B must be a in the span of {A 2 B,AB,B} and thus is linearly dependent on {A 2 B,AB,B}. Next, suppose that A does not have full rank. Then AB is in the range of A, which is not all of R 3 . We could rewrite A 4 B as a linear combination of columns of {A 2 B,AB,B}, but if B were in the null space of of A, then we could not recover it by premultiplying it again by A. Here’s an example: Suppose A is 1 0 0 0 1 0 0 0 0 (1) and B = [1 1 1] T . Then the controllability matrix C = 1 1 1 1 1 1 1 0 0 has rank 2 and has two linearly independent columns 1 1 1 and 1 1 0 . (2) No matter how many more times we multiply by A, we’ll never get anything but zero in the third column, hence although by Cayley-Hamilton there exist real scalars α 0 , α 1 , and α 2 , A 4 B = α 2 A 2 B + α 1 AB + α 0 B but α 0 must be zero. Of course this is a very simple example, but any n × n matrix A that has a nontrivial null space has a zero eigenvalue, hence its Jordan normal form can be chosen to have the zero eigenvalue in the ˜ a nn position, so the same argument holds. Example 2. Consider the system A = 2 0 -3 0 0 0 3/2 0 5/2 (3) B = -1 0 -2 (4) C = 1 1 -1 . (5)

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Suppose that A is 3 × 3. Is A4B linearly dependent on {A3B,A2B,AB}?First, suppose that A has full rank. Then A is invertible so the subspace spannedby {A3B,A2B,AB} is the same as that spanned by {A2B,AB,B}. Thus every columnof A4B must be a in the span of {A2B,AB,B} and thus is linearly dependent on{A2B,AB,B}.

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ECE 602 Lumped Systems Theory December 08, 2008 1

ECE 602 Lecture Notes:

Some Examples on Controllability and Observability

Example 1. Suppose that A is 3 3. Is A4B linearly dependent on {A3B,A2B,AB}?First, suppose that A has full rank. Then A is invertible so the subspace spannedby {A3B,A2B,AB} is the same as that spanned by {A2B,AB,B}. Thus every col-umn of A4B must be a in the span of {A2B,AB,B} and thus is linearly dependent on{A2B,AB,B}.Next, suppose that A does not have full rank. Then AB is in the range of A, which is notall of R3. We could rewrite A4B as a linear combination of columns of {A2B,AB,B},but if B were in the null space of of A, then we could not recover it by premultiplying itagain by A.

Heres an example: Suppose A is 1 0 00 1 00 0 0

(1)and B = [1 1 1]T . Then the controllability matrix

C = 1 1 11 1 1

1 0 0

has rank 2 and has two linearly independent columns 11

1

and 11

0

. (2)No matter how many more times we multiply by A, well never get anything but zero inthe third column, hence although by Cayley-Hamilton there exist real scalars 0, 1, and2, A

4B = 2A2B + 1AB + 0B but 0 must be zero. Of course this is a very simple

example, but any n n matrix A that has a nontrivial null space has a zero eigenvalue,hence its Jordan normal form can be chosen to have the zero eigenvalue in the ann position,so the same argument holds.

Example 2. Consider the system

A =

2 0 30 0 03/2 0 5/2

(3)B =

102

(4)C =

[1 1 1 ] . (5)
ECE 602 Lumped Systems Theory December 08, 2008 2

To obtain the controllability form, we first determine the controllability matrix C, thenbuild an equivalence transformation from its linearly independent columns augmented toa basis.

C = 1 4 55/20 0 02 13/2 41/4

(6)so, for example, we can choose

P1 = Q =

1 4 00 0 12 13/2 0

. (7)Then we find A = PAP1, B = PB and C = CP1 or

A =

0 19/2 01 9/2 00 0 0

(8)B =

100

(9)C =

[1 21/2 0

]. (10)

which is in controllability form with Ac 2 2 and Ac 1 1.To obtain the Kalman decomposition, we first determine whether the system is observable.The observability matrix is

O = 1 1 11/2 0 11/229/4 0 61/4

(11)which has full rank. Hence, the system is already in Kalman form: Ac = Aco, Ac = Acoand Bc = co, but there are no Aco nor Aco nor Bco.

Example 3. Consider the system

A =

2 0 30 0 03/2 0 5/2

(12)B =

102

(13)C =

[1 21/2 0

]. (14)
ECE 602 Lumped Systems Theory December 08, 2008 3

(Only the C matrix has changed.) To obtain the Kalman decomposition, we first deter-mine whether the system is observable. The observability matrix is

O = 1 0 11/2 0 26/229/4 0 61/4

(15)which does not have full rank. It has rank 2 so we expect to have one unobservable mode.It must either be controllable or not, so we first check the observability of the controllablesubsystem. We obtain

Oc =[

1 21/221/2 149/4

](16)

which has rank 2 so the uncontrollable mode must also be unobservable. This is consistentwith the fact that the third element of C is zero. Again the system in controllability formis already in Kalman form but this time, Ac = Aco, Ac = Aco and Bc = Bco. This timethere is no Aco nor Aco nor Bco.