Con Troll Ability of Linear Systems

8/14/2019 Con Troll Ability of Linear Systems

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Department of Mathematics 2005/06

Controllability of Linear Systems

Candidate Number: 16776

Submitted for the Master of ScienceLondon School of Economics

University of London


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CONTENTS

Table of contents

Abstract p 4

I) Introduction p 5

II) Control Theory p 6

Linear control systems p 6

Solution to ( ) ( )dx

Ax t Bu t dt

= + p 7

III) Controllability p 10

IV) Feedback and eigenvalues placement p 23

V) Algorithm and Program p 27

VI) exp( ), A where n n A and the Cayley-Hamilton theorem p 32

Bibliography p 46


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List of Figures

Figure I.1: A Control System p 5

Figure II.1: Algebraic Underdetermined Equation p 6

Figure IV.1: A Closed Loop System by State Feedback p 23


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Abstract

Control systems of the form

( ( ), ( )),dx

f x t u t dt

= 0,t ( ) n x t ( ) .mu t

are explained and their applications are described. Moreover, the special case of linear control systems of the form

( ) ( ) ( ),d

x t Ax t Bu t dt

= + 0,t ,n n A ,n m B ( ) ,n x t ( ) .mu t

are solved in terms of the matrices A and B.

Conditions on the matrices A and B are given to make the system controllable, reachable

spaces are introduced, Hautus test for controllability is proved and also the Pole

Placement Theorem.

Finally, a Matlab code for the Pole Placement Theorem is given and a numericalexample solved by the program.

The necessary ideas of the exponentiation of a matrix and the Cayley-Hamilton theorem

are introduced in the last chapter.


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Chapter I: Introduction

Control theory is a domain of applied mathematics that deals with the basic principles

behind the analysis of control systems. Roughly speaking, control systems are models

which are described by underdetermined differential equations of the type

( ( ), ( )),dx

f x t u t dt

= 0,t (*)

: ,n m n f and

where for each 0t we have ( ) n x t and ( ) .mu t The function u is called the

input, which can be frequently chosen, that is, it can be specified arbitrarily, and once

this is done, the differential equation (*) has a unique solution x (given an initial

condition (0), x and under some mild conditions on f ).

The models given by the differential equation (*) arise frequently in applications from

various sciences such as engineering, economics and biology.

One can think of a control system as a black box. This box, being given the input u,

manufactures the output x, according to the equation.

Schematically this can be represented as the following figure

Figure I.1: A control system

The aim in control theory is to choose the input u in such a manner that some desired

effect is produce on the output x.


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Chapter II: Control Theory

As it has been just said above, in control theory, one studies models which are described

by underdetermined differential equations. This means that some of the variables

satisfying the differential equations are free. An example of an underdeterminedalgebraic equation would be 2 2 1, x u+ = where both x and u are non-negative real

numbers. In this case we can see that this equation does not have an unique solution of

the form ( , ). x u Rather, there is freedom in choosing one of the variables, say u, and

once a particular u is chosen, then the variable x is uniquely determined by 21 . x u=

Figure I.2: Algebraic Underdetermined Equation

Linear Control Systems

In an analogous same manner, in linear control theory, one considers the differential

equation

( ) ( ) ( ),d

x t Ax t Bu t dt

= + 0,t 0(0) x x= (**)

where ,n n A ,n m B and for each 0 :t ( ) n x t and ( ) .mu t

The equation (**) is an underdetermined differential equation: indeed, u is free, and

once a particular ( [0, )) mu C has been chosen, this determines a unique solution x.

Later on, in the next section, we will show that this x is given by

( )0

0

( ) ( ) ,t

tA t A x x t e x e Bu d = = + 0.t

We shall also explain the meaning of the exponential of a matrix.


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It is important to note that the underdetermined differential equation (**) is a special

case of the much more general underdetermined differential equation (*), where

: m nn f is the linear function given by

( , ) , A B = + ( , ) .n n

That is the reason why (**) are called linear control systems.

Models described by equations of the type (**) occur frequently in applications (often

as linear approximation of more complicated non-linear models), and we give one

example below from the biological field.

Consider a lake with two species of fish whose populations 1 2, x x over a time period

[0, ]T are reasonably accurately modelled by1 11 12 1

2 21 22 2

( ) ( )

( ) ( )

x t a a x t d x t a a x t dt

=

for [0, ],t T

and with initial conditions

11

22

(0).

(0)i

i

x x

x x

=

This might very well be the model of the typical prey-predator system. Here, obviously

there is no input and this system can be solved rather easily.

Let us bring in the input: suppose we are interested in harvesting these fish (both types)

at harvesting rates 1 2,h h which are the inputs. In this case, the model describing the

evolution of the population becomes

1 11 12 1 1

2 21 22 2 2

( ) ( ) ( )

( ) ( ) ( )

x t a a x t h t d x t a a x t h t dt

=

for the same [0, ],t T and the same initial conditions.

We might desire several things, i.e. several different outputs. For instance suppose that

we want to harvest the species of fish over [0, ]T in such a way that by time t T = (theend of the fishing time) we are left with the desired final populations 1 2, f x x starting

from 1 2, .i i x x Perhaps by then one of the species is nearing extinction, because we have

wanted it to be this way.

So we would like to know if this is possible and if so, how we might achieve this.


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( )0

0

( ) ( ) .t

tA t A x t e x e Bu d = +

Proof.

First we show that this is indeed a solution. Formally, by working backwards we have

( ) ( )0 0

0 0

( ) ( )t t

tA t A tA tA t Ad d e x e Bu d e x e e Bu d dt dt

+ = +

( )0

0

( )t

tA tA t Ad Ae x e e Bu d dt

= +

( )0

0

( ) ( )t

tA tA t A tA tA Ae x Ae e Bu d e e Bu t = + +

( )0

0

( ) ( )t

tA tA t A tA tA A e x e e Bu d e Bu t

= + +

( )0

0

( ) ( )t

tA tA t A A e x e e Bu d Bu t

= + +

hence, this ( ) x does indeed satisfy (E.II.1) and the initial condition follows by setting t

to be zero0

0 ( )0 0 0

0

( ) 0 . A t Ae x e Bu d Ix x+ = + =

The uniqueness follows by considering two different solutions 1 x and 2 . x It follows that

1 2: x x x= satisfies the differential equation (E.II.2) with (0) 0 x = and so from the

previous Lemma we have that 0: ( ) 0t x t = which means that 1 2 . x x= And this ends

the proof.


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Chapter III: Controllability

As we said in the definition of linear systems, the interesting fact about

underdetermined equations is that we are allowed to choose the free variable in such a

way that we can produce a desired effect on the dependent variable.

Control theory seeks to do this in the differential equation

0( ( ), ( )), (0) , 0.dx

f x t u t x x t dt

= =

We will restrict our attention to linear systems

( ) ( ).dx

Ax t Bu t dt

= +

The state variable x are the variables to be controlled, which depend upon the free

variables u, the inputs. The main goal is then to solve the question: how do we choose

the control inputs to achieve regulation of the state variables?

We would like to define and characterize the property of controllability in terms of the

matrices A and B. To this end, we first define controllability and reachability of linear

systems.

Definition

The system

(E.III.1) ( ) ( )dx

Ax t Bu t dt

= +

for [ ]0,t T is said to be controllable at time T if for every pair of vectors 0 1, x x in ,n

there exists a control [ ]0,u C T such that the solution x of (E.III.1) with 0(0) x x=

satisfies 1( ) . x T x=

The system (E.III.1) is referred to as system ( , ). A B

Examples

Let 0 A = and 1, B = so that system ( , ) A B becomes

( ) 0 .dx

u t t T dt

=


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Given 0 1, , x x define [0, ]u C T to be the constant function

1 0

1( ) ( ) 0 .u t x x t T

T =

However, by the use of the fundamental theorem of calculus we can write

0 1 0 1 00 0 0

1( ) (0) ( ) '( ) ( ) ( )( 0)T T T dx x T x x T x d x d u d x x T x x

d T = = = = = =

and hence

1( ) . x T x=

Let us now study a system which is not controllable.

Consider the system1 0

0 1 A

=

and

1,

0 B

=

which gives:

11

22

( ) ( )

( )

dx x t u t dt

dx x t

dt

= +

=

The second equation implies that 2 2( ) (0) ,t x t x e= and so if 2 (0) 0, x > then for all 0t

we have 2 ( ) 0. x t > This in turn means that a final state with the 2 component x negative

is never reachable by any control.

Let us define more thoroughly what reachability means.

Definition

The reachable space of (E.III.1) at time T , denoted by ,T is defined as the set of all

n x for which there exists a control [ ]0,u C T such that

( )

0

( ) .T

T A x e Bu d =

Let us remark that this means that if we run the differential equation (E.III.1) with the

input u, and with initial condition 0(0) , x x= then x is the set of all points in the state

space that are reachable at time T starting from 0 by means of some input [ ]0, .u C T

Formally we have


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( )

0

| [0, ] s.t. ( ) .T

n T AT x u C T u e Bu d

= =

This means that T is made up of all the controls u that make the system ( , ) A B

controllable.

First we show that a reachable space is a subspace of . n

Lemma

The reachable space T is a subspace of .n

Proof

In order to prove this claim, we need to show that T is

- nonempty- closed under addition

- closed under scalar multiplication.

If we take 0,u = then

( )

0

( ) 0,T

T Ae Bu d =

and hence 0 .T

Next, suppose 1 2, T x x then there exist [ ]1 2, 0,u u C T such that( )

0

( )T

T Ai i x e Bu d

= for 1,2.i =

Thus [ ]1 2: 0,u u u C T = + and

( ) ( ) ( )1 2 1 2

0 0 0

( ) ( ) ( ) ,T T T

T A T A T Ae Bu d e Bu d e Bu d x x = + = +

and hence 1 2 .T x x+

Finally, if T x then there exists a [ ]0,u C T such that( )

0

( ) .T

T A x e Bu d =

If , then [ ]0,u C T and


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( ) ( )

0 0

( )( ) ( ) ,T T

T A T Ae B u d e Bu d x = =

thus .T x The proof is therefore finished.

So far, we have proved that , nT now we want to give a necessary and sufficient

condition for T to actually be .n As we will see, this condition depends on the rank

of the matrix 1[ | | | ].n B AB A B

Let us remember that the definition of rank is a bit technical:

Definition

The rank of a matrix A, written rank( ), is equal to the maximum number of linearly

independent rows of A or, equivalently, the dimension of the row space of A.

Essentially, the rank counts the number of genuinely independent rows in the matrix A.

Theorem

The following propositions are equivalent

(1) ,nT =

(2)1

rank[ | | | ] .n

B AB A B n

=

Proof

Let us do a contradiction to prove that (2) implies (1). To this end, suppose , nT

then there exists a 0 0 x belonging ton such that for all T r we have

T0 0. x r =

This is actually equivalent to the following equation

T ( )0

0

( ) 0T

T A x e Bu d = ([0, ]) .mu T

In particular, it will still be true if we chooseTT ( )

0( ) ,T Au u B e x= = with [0, ].t T As

a precaution it is worth reminding the reader that T and T have very different meanings

and the notation can become a bit tricky. This choice of u belongs to ([0, ]) .mT

So plugging this intro the integral gives


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TT ( ) T ( ) T0 0

0 0

0 ( ) ( ) ,T T

T A T A x e BB e x d y y d = =

where

T1

T ( ) 0

( )

( ) ,( )

T A

m

y

y B e x y

= =

see the Appendix to justify the change of the transpose in the exponential.

The product inside the integral can be written as

2 2

1 10 0

0 ( ) ( ) ,T T m m

k k k k

d y d = =

= =

the inversion of the summation and integration symbols is justified because the sum is

finite. However, each 20

( )T

k y d

is greater than or equal to zero, so the sum is only

zero if each 20

( )T

k d is actually zero, i.e.

{ } 20

1,..., : ( ) 0,T

k k m y d =

consequently 2( ) 0k y = or ( ) 0k y = for all [0, ]T and { }1,..., .k m

Hence T ( )0[0, ] : 0.T AT x e B = Differentiate this last equation k times with respect

to . This process gives T ( )0 0,k T A x A e B = and by suggestively setting T = we have

T0 0.

k x A B = This is the same as saying that T 10 [ | | | ] 0,n x B AB A B = but since we

have that 0 0, x we have to have1rank[ | | | ] ,n B AB A B n < which is a

contradiction since we know this rank to be exactly n, and thus .nT =

Let us now prove that (1) implies (2), set 1: rank[ | | | ] .n B AB A B n = < Then

there exists a nonzero 0n x such that

T 10 [ | | | ] 0,

n x B AB A B =

however, every matrix is a zero of its characteristic polynomial (by the Cayley-

Hamilton theorem) thus

2T T 1 T 10 0 0 1 0

0

( ) 0,n

n n n k n n k

k

x A B x I A A B x A A B

=

= + + + = + =

and by induction T0: 0.k k n x A B =


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These last two equations give: T00 : 0.k k x A B =

Now,

T T T 2 2 T0 0 0 0

0

1 10

2! !k k

k

x IB x AtB x A t B x A t Bk

== + + + =

since each term in the sum is zero, and we continue from here

T T0 0

0

10 .

!k k tA

k

x A t B x e Bk

=

= =

However, this implies that 0 ,T x since otherwise if some ([0, ])mu T then

( )0

0

( )T

T A x e Bu d =

T T ( )0 0 0

0 0

( ) 0 ( ) 0T T

T A x x x e Bu d Bu d = = =

which gives 0 0, x = a contradiction, so the rank has to be exactly n. And this ends the

proof.

Schematically we have the following results

( )

0

: | [0, ] s.t. ( )T

n T AT x u C T u e Bu d

= =

1rank[ | | | ]n nT B AB A B n

= = and hence (?)

1( , ) is controllable rank[ | | | ] .n A B B AB A B n =

Combining both results from the Lemma and from the Theorem we obtain:

Corollary

The following propositions are equivalent

(1) ( , ) B is controllable

(2) 1rank[ | | |] ,n B AB A B n = the dimension of the state space.

Before proceeding into the eigenvalue analysis for controllability let us return to the

examples we considered at the beginning and see why one of them is controllable and

the other is not using our new result: Corollary 3.5.

In the first example we had 1n = and


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1rank[ | | | ] rank[ ] rank[1] 1 ,n B AB A B B n = = = =

which is indeed the dimension of , the state space.

However, on the second example, the state space was 2 , so 2n = and we have

1 1 1rank[ | | | ] rank[ | |] rank 1 2 .0 0

n

B AB A B B AB n

= = = =

Lemma

Let ,n n A then the characteristic polynomial of A is the same as the characteristic

polynomial of 1 ,T AT where T is an invertible matrix in .n n

Proof

We have1 1det( ) det( ( ) ) I T AT T I A T =

1det( )det( )det( )T T I A=

det( ). I A=

so the characteristic polynomials are indeed the same.

Lemma

Let A be the following matrix

0 1 1

0 1 0

.0 0 1

n

A

a a a

=

Then its characteristic polynomial is1

0

det( ) .n

n k k

k

I A a

= = +

Proof

One has


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0 1 2 1

1 0 0

0 1 0

det( ) det

0 0 0 1n

n

I A

a a a a

=

+

1 2 1 0 2 1

1 0 0 1 0

det det0 0 1 0 0 1

n na a a a a a

= +

+ +

0

1 2 1

1 0

det0 0 1

n

a

a a a

= +

+

1 2 1 03 3

( ( ( ) ) )n nn n

a a a a

= + + + +

1 21 2 1 0

n n nn na a a a

= + + + + +

1

0

.n

n k k

k

a

== +

From now onwards, let ,n n A and 1.n B

Lemma

Let ( , ) A B be controllable and let A have characteristic polynomial1

0.

nn k k k

a

= +

Then there exists an invertible T in n n such that

1

0 1 1

0 1 0

0 0 1

n

T AT

a a a

=

and 1

1

0.

0

T B

=

Proof

Because ( , ) A B is controllable we have 1rank[ | | | ] .n B AB A B n = So if we define

as 1{ , , , },n B AB A B = then forms a basis of .n With respect to this basis ,

the matrix of the linear transformation A is


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0

1

1

0 0

1 0.

0 1 n

a

aQ

a

=

Let the characteristic polynomial of A be given by1

11 1 0

0

( ) ,n

n n n k n k

k

p z z a z a z a z a z

== + + + + = +

and define the auxiliary polynomials2

1 2 11 1 2 1 1

0

( )n

n n n k n k

k


+

== + + + + = +

32 3 2

2 1 3 2 20

( )n

n n n k n k

k


+

== + + + + = +

1 1( )n n p z z a = +

( ) 1.n p z =

The general polynomial being given by

1

( ) .n i

n i n i k i n k

k

p z z a z

== +

These auxiliary polynomials satisfy the recursion relation (for {1, , })i n

1 1( ) ( ) ( ).

i i i n zp z p z a p z

=

Now, also for {1, , }i n introduce the vectors ( ) .i ie p A B= The auxiliary polynomials

can be written for the matrix A as

1

( ) ,n i

n i n i k i n k

k

p A A a A

== +

and similarly for the recursion relation we have

1 1 .i i i n Ae e a e =

Using this recursion relation we can see that1

1 2span{ , , , } span{ , , , }.n

n B AB A B e e e =

Therefore, { {1, , }}ie i n is a basis for .n

In this basis, A has the matrix


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0 1 1

0 1 0

.0 0 1

na a a

The statement for 1

T B

1

1

0

0

T B

=

follows trivially.

Next we prove an if and only if lemma which says that if ( , ) A B is controllable, then it

will still controllable after the multiplication of an invertible matrix T .

Lemma

Let T be an invertible matrix in .n n Then the following propositions are equivalent

(1) ( , ) B is controllable

(2) 1 1( , )T AT T B is controllable.

Proof

Using the relation 1 1( ) ,n nT AT T A T = we have that1 1 1 1 1 1 1rank[ | | | ] rank[ | | | ] ,n n B AB A B n T B T ATT B T A TT B n = =

and the claim follows.

Recall that a subspace nV is said to be A-invariant if , AV V which is equivalent

to saying that if x V then . Ax V This means that V is A-invariant when the whole

trajectory remains in V whenever the initial state belongs to V .

Lemma

One has that 1ran[ | | | ]n B AB A B is the smallest A-invariant subspace containing

ran . B


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Proof

Let us first prove that 1ran[ | | | ]n B AB A B is a subspace containing ran . B

Suppose that ran , x B this means that there exists mu such that , x Bu= and this can

be seen my computing

1 0[ | | | ] .

0

n

u

B AB A B Bu x

= =

Next, let us show that 1ran[ | | | ]n B AB A B is an A-invariant subspace. To

accomplish this, let 1ran[ | | | ]n x B AB A B which means that

1

21[ | | | ]n

n

u

u x B AB A B

u

=

11 2 1

0

nn k

n k k

Bu ABu A Bu A Bu +=

= + + + =

and operating A on x we have

2 12 1

1

.n

n n k n n k

k

Ax ABu A Bu A Bu A Bu A Bu =

= + + + + =

If we let p denote the characteristic polynomial of A then by Cayley-Hamilton we have( ) 0. p A = So, if

11

1 1 00

( ) ,n

n n n k n k

k


== + + + + = +

then1

11 1 0

0

.n

n n k n k

k

A a A a A a I a A

== =

If we now substitute this new equation for n A in the equation for Ax we have

2 1 11 2 1 1 1 0( )n nn n n Ax ABu A Bu A Bu a A a A a I Bu = + + + +

2 11 1 2 2 1 10 ( ) ( ) ( )

nn n n n n B AB u a u A B u a u A B u a u

= + + + +

1 1

1 0

n nk k

k k nk k

A Bu a A Bu

= =

= +


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1

1

0 ( )n

k k k n

k

B A B u a u

== +

1 11

1 1

0

[ | | | ] nn

n n n

u a u B AB A B

u a u

=

1ran[ | | | ],n B AB A B

so 1ran[ | | | ]n B AB A B is indeed A-invariant.

Finally we need to prove that this subspace is the smallest of all the subspaces with this

property. In order to do so, let V be a subspace of n such that

- ran B V

- , AV V

then the claim is that 1ran[ | | | ] .n B AB A B V Let us proceed as in the previous

parts of the proof. Let 1ran[ | | | ]n x B AB A B then

1

21 1

1

ran[ | | | ] .n

n k k

k

n

u

u x B AB A B A Bu

u

=

= =

Since { }1, , : ran ,n Bu B and since ran , B V it follows that , Bu V for all

in the set { }1, , .n And, as we have shown, V is A-invariant, and so 0 : .k k A Bu V

As V is a subspace, it follows that

1 11

1

.n

n k n k

k

x Bu A Bu A Bu V =

= + + =

This ends the proof of the lemma.

Hautus test for controllability

The following propositions are equivalent(1) ( , ) B is controllable

(2) : rank[ ] . B I A n =


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Proof

Let us first prove that (1) implies (2). To achieve such goal, let ( , ) A B be controllable.

We now consider two cases for . If is not an eigenvalue of A, then we have

rank[ ] . B I A n = However, if happens to be an eigenvalue of A then there exists

v such that T ( ) 0,v I A = which can also be written as T T .v A v=

If T 0v B = then we haveT 1 T T 1 T 1[ | | | ] [ | | | ] 0n n nv B AB A B v B v B v A B = =

a contradiction to ( , ) A B being controllable.

To show that (2) implies (1) let ( , ) B be non controllable. Choose a basis for

1ran[ | | | ]n B AB A B and extend it to a basis B for .n

Then by previous lemma there exists n nT such that

11 121

22

,0

A AT AT

A =

11 .

0

BT B

=

Now Let 2 be an eigenvalue of 22 A and let 2 0 x be such thatT T2 22 2 2 . x A x=

Also, let T[ ] 0.v B I A = Hence : rank[ ] . B I A n =

Now we define the state feedback which has the control u in terms of the variable x by

means of an appropriate matrix. It is called in such a way because it feeds the u into the

system in a closed loop manner.

State Feedback

Let ,m n F then ( ) ( ).u t Fx t =

Theorem

If ( , ) A B is controllable then ( , ) A BF B+ is still controllable.

Proof

We have

1rank[ ] rank [ ]

0 1

F B I A BF B I A

=

rank[ ]. B I A=


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Chapter IV: The eigenvalue placement theorem

Let us do a summary of the linear system ( , ) A B

( ) ( ) 0dx

Ax t Bu t t dt

= +

with the following matricesn x mu n n A n m B

and the state feedback being defined as

( ) ( ) .m nu t Fx t F =

The state feedback has been defined in such a way so that we can create a closed loop

system, that is, a system that is fed ( )u t in terms of ( ). x t

Closed loop system

( ) ( ) ( ) ( ) ( ) ( ).dx

Ax t Bu t Ax t BFx t A BF x t dt

= + = + = +

Figure IV.1: A Closed Loop System by State Feedback

A sensible question to ask now is the following : what are the conditions on A and B

such that for every monic, real polynomial p of degree n there exists m n F such that

the characteristic polynomial of A BF + is p?

In order to answer this question we first state and prove some technical lemmas which

will eventually lead to a major proposition called the Pole Placement Theorem.


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Lemma

If ( , ) A B is controllable and ranb B with 0,b then there exist 1 1, , nu u such that

the sequence defined by

{ }1

1

:

: 1, , 1k k k

x b

x Ax Bu k n+

=

= +

is independent.

Proof

We proceed by steps, 1 0 x , and therefore it is independent. Now suppose we have

constructed the sequence i x for 1, , .i k = Let { }1span , , ,k x x = that is, the linear

space generated by { }1 , , .k x x We need to choose k u such that 1 .k k k x Ax Bu+ = +

Should this not be possible then : k k u Ax Bu + (*). In particular choose 0u = so we

have k Ax (**). Combining (*) and (**) we have

:u Bu or the same in other words: ran . B Finally for all i k < we have that

1 ,i i i Ax x Bu+= since both 1i x + and i Bu satisfy 1i x + and .i Bu

This gives now gives that indeed i Ax for all { }1, ,i k and so it follows that

. A This in turn implies that we have , n = and .k n=

Lemma

If ( , ) B is controllable and ranb B with 0,b then there exists F such that

( , ) BF b+ is controllable.

Proof

It follows from the previous lemma. Define F by ,k k Fx u= for { }1, , 1k n and

0.n Fx = Then1( ) ,n k A BF b x

+ = and so ( , ) A BF b+ is controllable.

Note: we could have used any other arbitrary vector in n to define .n Fx


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25

We are now in a position to finally give a necessary and sufficient condition for the

system ( , ) A B to be controllable in terms of characteristic polynomials, this result is

called the:

Eigenvalue Placement TheoremThe following propositions are equivalent

(1) ( , ) A B is controllable

(2) for every monic polynomial p with real coefficients of degree n, there existsm n F such that the characteristic polynomial of A BF + is p

Proof

Let us prove the only if part first. Suppose that ( , ) A B is not controllable. We want to

reach the conclusion that the characteristic polynomial of A BF + is never equal to p.

So, if ( , ) A B is not controllable then there exist and nv such that T Tv A v=

and that T 0.v B = Thus, T T: ( ) ,m n F v A BF v + = which is the same as saying that

,m n F is an eigenvalue of . A BF +

Having this result now we take any monic polynomial p of degree n with real

coefficients such that ( ) 0. p This gives the conclusion that we were looking for,

which is that the characteristic polynomial of A BF + will never be equal to p.

[Note: take for instance the natural choice ( ) , n p x x= if 0 and ( ) 1n p x x= + if

0. = ]

The proof of the if is considerably longer and needs to be divided into two cases, first

one with 1m = and then a more general case for larger and arbitrary m.

For 1m = find T such that

1

0 1 1

0 1 0

0 0 1

n

T AT

a a a

=

and 1

1

0

0

T b

=

Find [ ]1 n f f f = such that


26/47

26

1 1

0 1 1

0 1 0

,0 0 1

' ' 'n

T AT T bf

a a a

+ =

(this can be accomplish indeed with 0 1 0 1 1', , ')n n na f a a f a + = + = where

1

0

'n

n k k

k

a

= +

is the desired characteristic polynomial.

It just remains to realize that

1 1

0 1 1

0 1 0

( )0 0 1

' ' 'n

T A bfT T

a a a

+ =

and so

1 1

0 1 1

0 1 0

0 0 1

' ' 'n

A bfT T T

a a a

+ =

has a characteristic polynomial which is1

0'.

nn k k k

a

= +

The general case is now done as follows. First, suppose ( , ) A B is controllable. Set

ranb B with 0,b then 0 F such that 0( , ) A BF b+ is controllable.

We can also write ' f such that the characteristic polynomial of ' A BF bf + + is the

desired characteristic polynomial. However, there exists u such that b Bu= (because

we have that ran ).b B

Finally, define 0 ', F F uf = + then characteristic polynomial of A BF + is the desired

polynomial.

We will do a numerical example using this theorem (about the linearized motion of a

satellite where the inputs are the thrust in the radial direction and the thrust in the

tangential direction) in the following Chapter.

We will also give the a code which was successfully impletemented in Matlab.


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27

Chapter V: Algorithm and Program

The following is a Matlab code of the eigenvalue placement theorem which is

commented and described.

% The program is Pole Assignment

% Given (A, B) and poles p, this function will find out if the given pair is controllable

and

% if so it will relocate the poles. The outputs are relocated poles and the feedback

matrix {z}.

% Checking for controllability

[n,n]=size(A);

C=[];

for i=1:n,

C=[C A^(i-1)*B];

end

if rank(C)==n,

disp('(A,B) is controllable')

else

disp('(A,B) is uncontrollable'), RETURN

end

[n,m]=size(B);

r=n-1;

while r~=n,

% form a random matrix of size m x n

F=rand(m,n);

Ao=A+B*F;

% find the eigvenvalues of Ao

x=eig(Ao);

% to check whether they are distinct


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28

V=[];

for i=1:n,

v=[];

for j=1:n,

v=[v x(i)^(j-1)];end

V=[V;v];

end

r=rank(V);

end

r=n-1;

while r~=n,

% form a random matrix u of size m x 1

u=rand(m,1);

b=B*u;

% check for controllability

C=[];

for i=1:n,

C=[C Ao^(i-1)*b];

end

r=rank(C);

end

% given poles p(1), ... , p(n), to find coefficients of the polynomials

x=[1 -p(1)];

for i=2:n,

x = conv(x, [1 -p(i)]);

end

for i=1:n,

d(i)=x(i+1);

end


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29

% to find the characteristic polynomial of A

x=eig(Ao);

y=[1 -x(1)];

for i=2:n,

y=conv(y,[1 -x(i)]);end

for i=1:n,

ao(i)=y(i+1);

end

% Transformation matrix

P=[zeros(1,n); eye(n-1) zeros(n-1,1)];

x=[1];

for i=1:(n-1),

x=[ao(i) x];

end

U=x;

t=x;

for i=1:(n-1),

t=t*P;

U=[U;t];

end

T=C*U;

fo=[];

for i=1:n,

fo=[fo (d(n+1-i)-ao(n+1-i)) ];

end

fo=fo*inv(T);

x=eig(Ao-b*fo);

disp('poles for controllable pair have been relocated at');

disp(x);

disp('feedback matrix is');

z=(F-u*fo);


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30

disp(z);

Now, consider the system

1 1

2 2 1

3 3 2

4 4

( ) ( )0 1 0 0 0 0

( ) ( ) ( )3 0 0 2 1 0 .( ) ( ) ( )0 0 0 1 0 0

( ) ( )0 2 0 0 0 1

x t x t

x t x t u t d x t x t u t dt x t x t

= +

This system is the linearized equation of motion of a satellite where the input 1u is the

thrust in the radial direction and 2u the thrust in the tangential direction.

We input this in the program Controllability.m

0 1 0 0

3 0 0 2,

0 0 0 1

0 2 0 0

A

=

0 0

1 0,

0 0

0 1

B

=

and we define the points (1) 1, p = (2) 2, p = (3) 3, p = (4) 4, p = which are the

eigenvalues we wish to have in this system. We also try the eigenvalues (1) 1, p =

(2) 1, p = (3) 5, p = and (4) 2. p =

The output of the program is as follows

>> A = [0 1 0 0 ; 3 0 0 2 ; 0 0 0 1 ; 0 -2 0 0]

A =

0 1 0 03 0 0 20 0 0 10 -2 0 0

>> B = [0 0 ; 1 0 ; 0 0 ; 0 1]

B =

0 01 0


31/47

31

0 00 1

>> p(1)=1

p =

1>> p(2)=2

p =

1 2

>> p(3)=3

p =

1 2 3

>> p(4)=4

p =

1 2 3 4

>> controllability(A,B) is controllablepoles for controllable pair have been relocated at

4.00001.00003.00002.0000

feedback matrix is118.8359 -24.5533 -5.2740 64.245364.0536 -13.1355 -2.5757 34.5533

>>

>> A = [0 1 0 0 ; 3 0 0 2 ; 0 0 0 1 ; 0 -2 0 0]

A =

0 1 0 03 0 0 20 0 0 10 -2 0 0

>> B = [0 0 ; 1 0 ; 0 0 ; 0 1]

B =

0 01 00 00 1

>> p(1) = 1


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32

p =

1

>> p(2) = -1

p =1 -1

>> p(3) = 5

p =

1 -1 5

>> p(4) = -2

p =

1 -1 5 -2

>> controllability(A,B) is controllablepoles for controllable pair have been relocated at

5.0000-2.0000-1.00001.0000

feedback matrix is3.6791 3.4798 4.2638 -0.46391.7086 2.0414 2.5879 -0.4798

>>

Therefore, wit the first set of eigenvalues the matrix F is

118.8359 24.5533 5.2740 64.2453,

64.0536 13.1355 2.5757 34.5533 F

=

and for the second set we have

3.6791 3.4798 4.2638 0.4639.

1.7086 2.0414 2.5879 0.4798 F

=

Since F was defined such that ( ) ( ),u t Fx t = then we have our control u for both cases

with the eigenvalues we wanted.


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33

Chapter VI: exp( ), A where n n A and the Cayley-Hamilton theorem

The objective of this chapter is to set the necessary tools required in the process of

constructing a consistent control theory. The aim is to understand these tools now

instead of having to digress during the theory because this would take our attention

away. The only required knowledge required is single variable calculus and some linear

algebra: namely vector spaces and matrix theory, in particular expansion of

determinants, digitalisation of matrices, eigenvalues and eigenvectors. The main

objective is not to go into the details of linear algebra, but to explain the exponentiation

of a matrix and the Cayley-Hamilton theorem.

The ( , )i j -entry of a generic matrix A will be denoted by ( ),ij

A or by ( )ij

a once the

cofactors and minors are introduced.

Definition A.1 Let denote the following norm

{ }max such that 1 and .ij A A i j n=

Clearly we have

(*) |ij A A

for all i and j.

Although there are several possible choices for a norm, this one is convenient because it

allows a relatively straightforward proof of the convergence of the exponential of a

matrix.

Lemma A.2

Suppose that A and B are two real n-square matrices, then . AB n A B

Proof

This follows by using the inequality that we have pointed out previously, we first

estimate the absolute value of the general entry of the matrix AB, denoted by ( ) ,ij AB


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35

Proof

By the convention of our notation, the ( , )i j -entry of e is

0

0

1( ) ( ) , with ( ) .

! A k

ij ij ij ijk

e A A I k

=

= =

To show that this exponential of a matrix converges, we show that the series

2

0

1 1( ) ( )

2! !k

ij ij ij ijk

I A A Ak

=+ + + =

is absolutely convergent, and hence convergent. To this end, let a n A= then

1ij I I = for 0,k = and11 1 1( )

! ! !k k k k

ij A A n Ak k k for 1.k (By Lemma A.3)

Hence

2 32 3 21 1 1 1| | | |( ) | |( ) | 12! 3! 2! 3!ij ij ij ij I A A A A n A n A+ + + + + + + +

2 31 1 112! 3!

a a an

= + + + +

11 .

aen= +

The same in abbreviated notation

1

0 0 1

1 1 1 1 1|( ) | 1 1 .

! ! !

ak k k k

ijk k k

e A n A a

k k n k n

= = =

= + = +

And the result follows since the series converges absolutely.

Lemma A.5

The series

1 1! !

r s

k r s k ij

A Br s+ =

converges for all i and j.

Proof

As in the proof above, let a n A= and b n B= and estimate double sum

1 1 1( )

! ! ! !r s r s

ijk r s k k r s k ij

A B A Br s r s+ = + =

=


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36

1! !

r s

k r s k

A Br s+ =

(by the (*) inequality)

1! !

r s

k r s k

n A Br s+ =

(by Lemma A.2)

1 11 ( )( )! !

r sr s

k r s k n n A n B

r s

+ = (by Lemma A.3)

21! !

r sr s

k r s k

n A Br s

+

+ ==

21! !

r s

k r s k

n a br s

+ ==

1! !

r s

k r s k

a br s+ =

a b

e+

= .<

Lemma A.6

Suppose that A and B are two real n-square matrices that commute, i.e. , AB BA= then

we have . A B A Be e e+ =

Proof

Formally we have0

1( )

!a b k

k

e a bk

+

== + and

0 0

1 1.

! !a b k k

k k

e e a bk k

= =

=

The terms of degree k in these expansions are respectively

( )1 1( )! !

k k r sr

r s k

A B A Bk k + =

+ = and 1 1 .! !

r s

r s k

A Br s+ =

Because of the binomial coefficients we have

( )1 1 ! 1 1 1 ,! ! !( )! !( )! ! !

k r

k k k r k r r k r r s

= = =

and hence both terms are equal for all k and all r and s such that .k r s= +

Define

0

1( ) : .

!

nk

nk

S A Ak =

=

Then


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37

0 0 0 0

1 1 1 1( ) ( ) ,

! ! ! !

n n n nr s r s

n nr s r s

S A S B A B A Br s r s= = = =

= =

but similarly

( )0 0

1 1 1( ) .

! ! !

n nk r s r s

n r k r s k k r s k

S A B A B A Bk r s= + = = + =

+ = =

When we compare terms, we realize that the expansion of the partial sum ( )nS A B+

consists of the same terms as that of ( ) ( )n nS A S B provided that .r s n+

The subtlety here is showing that the sum of the remaining terms (the ones )r s n+ >

tend to zero as k grows large without bound.

By Lemma A.5 we have that the i, j-entry of ( ) ( ) ( )k k k S A S B S A B + satisfies

1 1( ( ) ( ) ( )) .

! !

r sk k k ij

r s k ij

S A S B S A B A Ar s+ >

+

However, by Lemma A.5, this sum tends to zero as k grows, but we also have that

( ) ( ) ( ) , A B A Bk k k S A S B S A B e e e+ +

hence this proves the lemma.

Lemma A.7

Suppose I is the identity matrix, then0

1lim ( ) .hAh

e I Ah

=

Proof

Expand the exponential (we have shown this to be a valid operation in Lemma A.4)

1

2

1 1( )

!hA k k

k

e I A h Ah k

= =

and use the usual substitution a n h A= to estimate the series

1 1

2 2

1 1( )

! !k k k k

ijk k ij

h A h Ak k

= =

(by the triangle inequality)

1

2

1| | |( ) |

!k k

ijk

h Ak

==

1

2

1| |

!k k

k

h Ak

= (by the (*) inequality)


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38

1 12

1| |

!k k k

k

h n Ak

= (by Lemma A.3)

1 1

2

1| | | |

!k k k k k

k

h n a n hk

==

2

1 1| | !

k

k

an h k

==

( 1 )a A

e aa

=

11 .

ae A

a

=

At this point it is important to note that 0a as 0.h However, the key here relies

on real analysis, which tells us that the derivative of the exponential function is the

exponential function itself,

0

00

1lim 1,

aa

aa

e d e e

a da =

= = =

(by LHopitals rule) so the series tends to zero with h.

Lemma 1.8

The function tAe is a differentiable matrix-valued function of t , and its derivative is

.tA Ae

Proof

Apply Newtons quotient definition of derivative

( )

0

1lim ( )tA t h A tAh

d e e e

dt h+

=

because of the commutability of tA and hA (both t and h are scalars), we have

( )1 1( ) ( )t h A tA hA tAe e e I eh h

+ =

and by Lemma A.7, the term is parenthesis tends to A as h goes to zero, and this proves

the result.

The exponential of a matrix, although well-defined and convergent requires further

explanations, in particular it requires an intelligent way of computing it, and certainly

exponentiating the entries of A is not a good way of doing so.


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39

Nonetheless, there is straightforward case which can be computed easily, and that is

when the matrix A is diagonal, with diagonal entries . i

It is important to note that inspection of the series shows that Ae is also diagonal in this

case, and that its diagonal entries are .ie This can easily be seen by the definition, if A

is a diagonal matrix with entries , i then each term in the expansion of e will be a

diagonal matrix with entries , k i each matrix being divided by !,k and hence Ae is a

diagonal matrix (example later on).

Definition A.9

A matrix A is diagonalizable if there exists a matrix P such that 1: D P AP = is a

diagonal matrix.

The following two lemmas are part of the details of linear algebra which we have

chosen to omit, so the proofs will not be given.

Lemma A.10

If A is diagonalizable, then 1. A PDP =

Lemma A.11

If A is diagonalizable, then 1 1( ) .k k PDP PD P =

Lemma A.12

If A is diagonalizable then

1

1

0

0 n

A

e

e P P

e

=

where i are the eigenvalues of A.

Proof

One has 00

1, with

! A k

k

e A A I k

== =


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40

11

1( )

!k

k

I PDP k

== +

1 1

1

1( )

!k

k

PIP PDP k

== +

1

1

1!

k

k

P I D P k

=

= +

1

1

1!

k

k

P I D P k

=

= +

1 D Pe P =

1

1

0

.

0 n

e

P P

e

=

Example

Compute Ae for 0

.0

A

=

Solution

The key step here is to realize that2

2 22

2

0

0

k k k

k k

J A J

= where {2

1 is odd.1 is evenk

k J k

= +

2 12 1 2 1

2 12 1

0

0

k k k

k k

K A

K

++ +

++

=

where {2 1 1 is odd .1 is evenk k K k + = +

Therefore

2 3

0

1 1 1! 2! 3!

A k

k

e A I A A Ak

== = + + + +

2 3

2 3

1 0 0 0 01 1

0 1 0 2! 3!0 0

= + + + +

2 4 3 5

3 5 2 4

1 1 1 11

2! 4! 3! 5!1 1 1 1

13! 5! 2! 4!

+ + +

= + + +


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41

2 2 1

2 1 2

( 1) ( 1)(2 )! (2 1)!

( 1) ( 1)(2 1)! (2 )!

k k k k

k k k k

x xk k

x xk k

+

+

+ =

+

cos sin .sin cos

=

Example (diagonal 2 2 real matrix)

Compute Ae for 1

2

0.

0 A

=

Solution

We already know the solution to this one, we know that 1

2

0 ,0

nnn A =

and hence

1

2

1

0 0 2

0 01 1.

! ! 0 0

n A n

nn n

ee A

n n e

= =

= = =

Example

Compute Ae for 0 1

.0 0

A

=

Solution

The crucial step is detecting that this is a nilpotent matrix, i.e. 2 0, A = and so on, so we

only need to sum up a finite number of terms.

1 0 0 1 1 1.

0 1 0 0 0 1 Ae I A

= + = + =

Definition A.13

Suppose A is an n-square matrix. Let ijM denote the ( 1) squaren submatrix of A

obtained by deleting its ith row and jth column. The determinant det( )ijM is called the

minor of the element ija of A, and we define the cofactor of , ija denoted by ,ij A to be

the signed minor: ( 1) det( ).i jij ij A M +=

Note that ijM denotes a matrix, whereas ij A denotes a scalar.


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42

Definition A.14

Consider a n-square matrix A over a field K . The classical adjoint, also called the

adjoint, of A, denoted by adj A is the transpose of the matrix of cofactors of A, i.e.

11 21 1

12 22 2

1 2

adj .

n

n

n n nn

A A A

A A A A

A A A

=

Lemma A.15 (Laplace Expansions)

One has1 1

det .n n

ij ij ij iji j

a A a A= =

= =

Proof

Omitted.

Lemma A.16

Suppose A is an n-square matrix, and suppose B is the matrix obtained from A by

replacing the ith row of A by the row vector ( )1 2 ... .i i inb b b One has 1 .n

ij ij jb A

= Additionally, if i j then

1 1

0.n n

jk ik kj kik k

a A a A= =

= =

Proof

As usual with our convention ( ), ij B b= and by the Laplace expansions in Lemma A.15

we have

1

det .n

ij ij j

B b B=

=

Now, we can safely say that ij ij B A= for 1,..., j n= since ij B does not depend upon the

ith row of B. Therefore the equation above becomes

1

det .n

ij ij j

B b A=

=


43/47


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44

Proposition A.18 (Cayley-Hamilton theorem)

Every matrix is a root of its characteristic polynomial.

Proof

Suppose that A is an arbitrary real n-square matrix, and let ( ) t denote its characteristic polynomial, say

11

1 1 00

( ) det( ) .n

n n n k n k

k

t tI A t a t a t a t a t

= = = + + + + = +

Now let ( ) B t denote the classical adjoint of the matrix .tI A The elements of ( ) B t are

cofactors of the matrix tI A and hence are polynomials in t of degree not exceeding

1,n thus1

11 1 0

0( ) .

nn k

n k k B t B t B t B B t

== + + + =

By Lemma A.17 we have ( ) ( ) det( ) ,tI A B t tI A I = or

1 1

0 0

( ) .n n

k n k k k

k k

tI A B t t a t I

= =

= +

The key step is now to equate the coefficients of corresponding powers of t , this can be

done since both sides are of order n,

1n B I =

2 1 1n n n B AB a I =

3 2 2n n n B AB a I = in general 1 1n k n k n k B AB a I + + = and so on

0 1 1 B AB a I =

0 0 . AB a I =

Next, multiply the matrices above by 1, ,..., , ,n n A A A I respectively, this yields

1n

n A B I =

1 12 1 1

n n nn n n A B A B a A

=

2 1 23 2 2

n n nn n n A B A B a A

= and so on

20 1 1 AB A B a A =

0 0 . AB a I =

Finally, add the above matrix equations


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45

11 2

1 2 00

0 ( ),n

n n n k n n n k

k

A a A a A a I A a A A

== + + + + = + =

which is the Cayley-Hamilton theorem.

Lemma A.19Let A and B be two real n-square matrices, then T T T( ) , AB B A= where the T denotes the

transpose of a matrix.

Proof

According to our notation, ( ) ij A a= and ( )ij B b= so the ij-entry of AB is 1 ,n

ik kjk a b

= which is also the ji-entry of T( ) . AB

However, column j of B now becomes row j of T

, B and row i of A becomes column i of T . A This means that the ji-entry of T T B A is

1

22

1

( ) .

i

ni

ij j nj kj ik k

in

a

ab b b b a

a=

=

Thus, T T T( ) AB B A= because the corresponding entries are equal.

Lemma A.20

For any real n-square matrix A we have: T T( ) ( ) .k k A A=

Proof (By induction)

By setting A B= in Lemma A.19 we have 2 T 2 T( ) ( ) . A A= And now we proceed by

induction. Let ( ) P n be the statement T T( ) ( ) .n n A A= We know it to be true for 2.n =

Now assume ( ) P n is true for some integer n, then T T( ) ( ) ,n n A A= multiply both sides

by T , A this gives T T T T( ) ( ) .n n A A A A= The left hand side is simply T 1( ) ,n A + whereas

the right hand side becomes T T T 1 T( ) ( ) ( ) ,n n n A A AA A+= = by Lemma A.19. So if ( ) P n

is true for some integer n, then so is ( 1) P n + and thus ( ) P n is true for all 2.n


46/47

46

We finish with two lemmas concerning the determinants and transpose of the

exponential of a square matrix.

Lemma A.21

Suppose A is a real n-square matrix, then Tr det . Ae e=

Proof

Let us start by a reduction of A of the form 1 , A P TP = where T is upper triangular.

Note that since k T will still be triangular, with diagonal entries equal to ( ), k T jjt e will be

triangular as well, with diagonal entries equal to ( ). jjt e

Hence

Tr det exp , jjt T T jj j j

e e t e= = =

and this is the result we wanted since 1 . A T e P e P =

Lemma A.22Suppose A is a real n-square matrix, then T T( ) exp( ), Ae A= where exp denotes

exponential of a square matrix as usual.

Proof By definition and Lemma A.4 we have

T TT T T T

0 0 0 0

1 1 1 1( ) ( ) ( ) exp( ),

! ! ! ! A k k k k

k k k k

e A A A A Ak k k k

= = = =

= = = = =

we have used Lemma A.20 to invert the transpose and the exponentiated power.


47/47

Bibliography

Linear Multivariable Control: A Geometric Approach

W. Murray Wonham Springer Verlag 1978

Mathematical Control Theory: An Introduction (Systems & Control: Foundations and

Applications)

Jerzy Zabczyk Springer Verlag 1982

An Introduction to Geometric State Theory in Linear Multivariable Control (A

Summary with Focus in the Disturbance Decoupling Problem)

Marvi Teixeira 1993

Advanced Linear Algebra

S. Roman Springer Verlag 1992

Con Troll Ability of Linear Systems

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