kalman decomposition

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Lecture 4 and 5

Controllability and Observability: Kalman decompositions

Spring 2013 - EE 194, Advanced Control (Prof. Khan)

January 30 (Wed.) and Feb. 04 (Mon.), 2013

I. OBSERVABILITY OF DT LTI SYSTEMS

Consider the DT LTI dynamics again with the observation model. We are interested in deriving

the necessary conditions for optimal estimation and thus the additional noise terms can beremoved. The system to be considered is the following:

xk+1 = Axk + Buk, (1)

yk = C xk, (2)

as before we consider the worst case of one observation per k, i.e., p = 1.

Definition 1 (Observability). A DT LTI system is said to be observable in n time-steps when

the initial condition, x0 , can be recovered from a sequence of observations, y0, . . . , yn−1 , and

inputs, u0, . . . , un−1.

The observability problem is to find x0 from a sequence of observations, y0, . . . , yk−1, and

The lecture material is largely based on: Fundamentals of Linear State Space Systems, John S. Bay, McGraw-Hill Series in

Electrical Engineering, 1998.

February 6, 2013 DRAFT



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inputs, u0, . . . , uk−1. To this end, note that

y0 = C x0,

y1 = C x1,

= CAx0 + CBu0,

y2 = C x2,

= CA2x0 + CABu0 + CBu1,

...

yk−1 = CAk−1x0 + CAk−2Bu0 + . . . + CBuk−2.

In matrix form, the above is given by

y0

y1

y2...

yk−1

=

C

CA

CA2

...

CAk−1

x0 + Υ

u0

u1

u2

...

uk−1

.

The above is a linear system of equations compactly written as

Ψ0:k−1 = O0:k−1 ∈Rk×n

x0, (3)

and hence, from the known inputs and known observations, the initial condition, x0, can be

recovered if and only if

rank (O0:k−1) = n. (4)

Using the same arguments from controllability, we note that rank (O) < n for k ≤ n − 2 as the

observability matrix, O , has at most n − 1 rows. Similarly, adding another observation after the

nth observation does not help by Cayley-Hamilton arguments as

rank (O0:n−1) = rank (O0:n) = rank (O0:n+1) = . . . . (5)




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Hence, the observability matrix is defined as

O =

C

CA

CA2

...

CAn−1

, (6)

without subscripts.

Remark 1. In literature, rank (C ) = n is often referred to as the system being (A, B) –controllable.

Similarly, rank (O) = n is often referred to as the system being (A, C ) –observable.

Remark 2. Show that

rank (C ) = n ⇔ CC T 0 ⇔ (CC T )−1 exists, (7)

rank (O) = n ⇔ OT O 0 ⇔ (OT O)−1 exists. (8)

Remark 3. Show that

(A, C ) –observable ⇔ (AT , C T ) –controllable, (9)

(A, B) –controllable ⇔ (AT , BT ) –observable. (10)




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I I . KALMAN DECOMPOSITIONS

Suppose it is known that a SISO (single-input single-output) system has rank (C ) = nc < n,

and rank (O) = no < n. In other words, the SISO system, in question, is neither controllable nor

observable. We are interested in a transformation that rearranges the system modes (eigenvalues)into:

• modes that are both controllable and observable;

• modes that are neither controllable nor observable;

• modes that are controllable but not observable; and

• modes that are observable but not controllable.

Such a transformation is called Kalman decomposition.

A. Decomposing controllable modes

Firstly, consider separating only the controllable subspace from the rest of the state-space. To

this end, consider a state transformation matrix, V , whose first nc columns are a maximal set of

linearly independent columns from the controllability matrix, C . The remaining n − nc columns

are chosen to be linearly independent from the first nc columns (and linearly independent among

themselves). The transformation operator (matrix) is n × n, defined as

V = v1 . . . vnc | vnc+1 . . . vn V 1 V 2 . (11)

It is straightforward to show that the transformed system (xk = V xk) is

xk+1 = V −1AV xk + V −1Bu, (12)

yk = CV xk. (13)

Partition V −1 as

V −1 = V 1

V 2 , (14)

where V 1 is nc × n and V 2 is (n − nc) × n. Now note that

I n = V −1V =

V 1

V 2

V 1 V 2

=

V 1V 1 V 1V 2

V 2V 1 V 2V 2

=

I nc

I n−nc

. (15)




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The above leads to the following1 :

V 2V 1 = 0 ⇒ C lies in the null space of V 2, (16)

further leading to

V 2B = 0, and V 2AV 1 = 0. (17)

The transformed system in Eq. (12) is thus given by

A V −1AV =

V 1

V 2

A

V 1 V 2

=

V 1AV 1 V 1AV 2

V 2AV 1 V 2AV 2

Ac Acc

0 Ac

, (18)

B V −1B =

V 1

V 2

B =

V 1B

V 2B

Bc

0

. (19)

Now recall that a similarity transformation does not change the rank of the controllability matrix,

C , of the transformed system. Mathematically,

rank (C ) = rank

V −1B (V −1AV )V −1B . . . (V −1AV )n−1V −1B

,

= rank

V −1B V −1AB . . . V −1An−1B

,

= rank

V −1

B AB . . . An−1B

,

= rank B AB . . . An−1B ,

= rank (C ) ,

= nc.

Furthermore,

rank (C ) = rank

B AB . . . Anc−1

B. . . A

n−1B

,

= rank

Bc AcBc . . . Anc−1c Bc

0 0 . . . 0

. . .

An−1c Bc

0

,

= rank

Bc AcBc . . . Anc−1c Bc

0 0 . . . 0

,

= nc,

1This is because all of the freedom in the linear independence of C is already captured in V 1.




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where to go from second equation to the third, we use Cayley-Hamilton theorem. All of this to

show that the subsystem (Ac, Bc) is controllable.

Conclusions: Using the transformation matrix, V , defined in Eq. (11), we have transformed

the original system, xk+1 = Axk + Buk into the following system:

xk+1 = Axk + Buk, (20)

xc(k + 1)

xc(k + 1)

=

Ac Acc

0 Ac

xc(k)

xc(k)

+

Bc

0

uk, (21)

such that the transformed state-space is separated into a controllable subspace and an uncontrollable

subspace.

Consider the transformed system in Eq. (20) and partition the (transformed) state-vector, x(k),

into an nc × 1 component, xc(k), and an n − nc × 1, xc(k). We get

xc(k + 1) = Acxc(k) + Accxc(k) + Bcuk, (22)

xc(k + 1) = Acxc(k). (23)

Clearly, the lower subsystem, xc(k), is not controllable. In the transformed coordinate space, xk,

we can only steer a portion, xc(k), of the transformed states. The subspace of Rn that can be

steered is denoted as R(C ), defined as the row space of C , where R(C ) is called the controllable

subspace.

Finally, recall that the eigenvalues of similar matrices, A and V −1AV , are the same. Recall

the structure of V −1AV from Eq. (18); since V −1AV is block upper-triangular, the eigenvalues

of A are given by the eigenvalues of Ac and Ac, called the controllable eigenvalues (modes)

and uncontrollable eigenvalues (modes), respectively.

Example 1. Consider the CT LTI system:

x = 2 1 1

5 3 6−5 −1 −4

x + 1

00

u,

C =

1 2 −4

0 5 −5

0 5 −5

,




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with rank (C ) = 2 < 3. Defines V as

V =

1 2

0 5

0 5

0

0

1

,

and note that the requirement for choosing V are satisfied. Finally, verify that the transformed

system matrices are

V −1AV =

0 6 1

1 −1 0

0 0 0

, V −1B =

1

0

0

.

Verify that the top-left subsystem is controllable. From this structure, we see that the two-

dimensional controllable system is grouped into the first two state variables and that the third

state variable has neither an input signal nor a coupling from the other subsystem. It is therefore

not controllable.

B. Decomposing observable modes

Now we can easily follow the same drill and come up with another transformation, xk = W xk,

where W contains no linearly independent rows of O and the remaining rows are chosen to be

linearly independent of the first no rows (and linearly independent among themselves). Finally,

the transformed system,

xk+1 = Axk + Buk,

=

Ao 0

Aoo Ao

xk +

Bo

Bo

uk, (24)

yk =

C o 0

xk (25)

is separated into an observable subspace and an unobservable subspace. In fact, we can now

write this result as a theorem.

Theorem 1. Consider the DT LTI system,

xk+1 = Axk,

yk = C xk,




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where A ∈ Rn×n and C ∈ R

p×n. Let O be the observability matrix as defined in Eq. (6). If

rank (O) = no < n, there there exists a non-singular matrix, W ∈ Rn×n such that

W −1AW = Ao 0

Aoo Ao , CW = C o 0 ,

where Ao ∈ Rno×no , C o ∈ R

no×n , and the pair (Ao, C o) is observable.

By carefully applying a sequence of such controllability and observability transformations,

the complete Kalman decomposition results in a system of the following form:

xco(k + 1)

xco(k + 1)xco(k + 1)

xco(k + 1)

= Aco 0 A13 0

A21 Aco A23 A24

0 0 Aco 0

0 0 A43 Aco

xco(k)

xco(k)xco(k)

xco(k)

+ Bco

Bco

0

0

u(k), (26)

y(k) =

C co 0 C co 0

xco(k)

xco(k)

xco(k)

xco(k)

. (27)




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C. Kalman decomposition theorem

Theorem 2. Given a DT (or CT) LTI system that is neither controllable nor observable, there

exists a non-singular transformation, x = T

x , such that the system dynamics can be transformed

into the structure given by Eqs. (26) – (27).

Proof: The proof is left as an exercise and constitutes Homework 1.




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APPENDIX A

SIMILARITY TRANSFORMS

Consider any invertible matrix, V ∈ Rn×n, and a state transformation, xk = V zk. The

transformed DT-LTI system is given by

xk+1 = Axk + Buk,

⇒ V zk+1 = AV zk + Buk,

⇒ zk+1 = V −1AV A

zk + V −1B B

uk,

yk = CV C

zk.

The new (transformed) system is now defined by the system matrices, A, B, and C . The two

systems, xk and zk, are called similar systems.

Remark 4. Each of the following can be proved.

• Given any DT-LTI systems, there are infinite possible transformations.

• Similar systems have the same eigenvalues; the eigenvectors of the corresponding system

matrices are different.

• Similar systems have the same transfer functions; in other words, the transfer function of

the DT-LTI dynamics is unique with infinite possible state-space realizations.

• The rank of the controllability and observability are invariant to a similar transformation.

APPENDIX B

TRIVIA: NULL SPACES

Consider a set of m vectors, xi ∈ Rn, i = 1, . . . , m.

Definition 2 (Span). The span of m vectors, xi ∈ Rn, i = 1, . . . , m is defined as the set of all

vectors that can be written as a linear combination of {xi}s , i.e.,

span(x1, . . . , xm) =

y

y =mi=1

αixi

, (28)

for αi ∈ R.




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Definition 3. The null space, N (A) ⊆ Rn , of an n × n matrix, A , is defined as the set of all

vectors such that

Ay = 0, (29)

for y ∈ N (A).

Consider a real-valued matrix, A ∈ Rn×n and let x1 = 0 and x2 = 0 be the maximal set of

linearly independent vectors2 such that

Ax1 = 0, Ax2 = 0, (30)

which leads to

A

2

i=1 αixi =

2

i=1 αiAxi = 0. (31)

Following the definition of span and null space, we can say that span(x1, x2) lies in the null

space of A, i.e.,

N (A) = span(x1, x2). (32)

In this particular case, the dimension of N (A) is 2 and3 rank (A) = n − 2.

2In other words, there does not exist any other vector, x3, that is linearly independent of x1 and x2 such that Ax3 = 0.

3This further leads to an alternate definition of rank as n− nc, where nc dim( N (A)).




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Lets recall the uncontrollable case with rank (C ) = nc. Clearly, C has nc linearly independent

(l.i.) columns (vectors in Rn); let the matrix V 1 ∈ R

n×nc consists of these nc l.i. columns. On

the other hand, let the matrix

V 1 ∈ R

n×n−nc consists of the remaining columns of C and note

that

every column of V 1 ∈ span(columns of V 1),

or, span(columns of V 1) contains every column of V 1.

The above leads to the following:

Remark 5. If there exists a matrix, V 2 , such that V 2V 1 = 0 , then

every column of V 1 ∈ N (V 2),

⇒ span(columns of V 1) ⊆ N (V 2),

⇒ every column of V 1 ∈ N (V 2).




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A. Kalman decomposition, revisited

Construct a transformation matrix, V ∈ Rn×n, whose whose first n × nc block is V 1. Choose

the remaining n − nc columns to be linearly independent from the first nc columns (and linearly

independent among themselves); call the second n × (n − nc) block as V 2:

V

V 1 V 2

, V −1

V 1

V 2

,

I n = V −1V =

V 1

V 2

V 1 V 2

=

V 1V 1 V 1V 2

V 2V 1 V 2V 2

=

I nc 0

0 I n−nc

.

From the above, note that V 2V 1 = 0. Now consider the transformed system matrices, A and B:

A V −1AV = V 1

V 2

A V 1 V 2 = V 1AV 1 V 1AV 2

V 2AV 1 V 2AV 2

,

B V −1B =

V 1

V 2

B =

V 1B

V 2B

.

Note that

AV 1 ∈ span(columns of V 1) ⊆ N (V 2),

B ∈ N (V 2).


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