86

1

−1

0

0

2( ) = (3 − 0.2

−2 + 2

−2 ) , 2 = (1),

−0.2

2

−2

1

where function

z(t),

t I is a solution of differential equation

z = A2 ( )z B1 ( ),

= S z. Characteristic polynomial of the matrix

A ( ) is equal to

2

( ) = | −

1

2

3

( )| = 3 + (1 + 0.1 ) 2 + (1 + 0.7 ) + 0.5 .

Hence it

follows

that matrix

2

A2 ( )

is a Hurwitz matrix for any > 0.

Coefficients

0 = 0.5 , 1 = 1 +

that matrices R,

R* , R* 1

are the same as in example 1.

Based on ratio A1 = R* A2*R* 1, B1

= R* B2 , S1 = S1R* 1 from (1.74) we get

̇ = , ̇ = , ̇ = − − − − ( ), =

1

2

2

3

3

0

1

1

2

3

3

= 1 = 0.5 1 + 0.7 2

+ 0.1 3,

(1.75)

where

0

1

0

0

1( ) = (

0

0

1 ),

= ( 0

), 1 = (0.5 0.7 0.1).

− 0

− 1

2

−1

From (1.75) it follows that identities are true:

( ( )) = − ( ) − 2( ) − 3( ), ( ) = 0.5 1( ) + 0.7 2( ) + 0.1 3( ),

,

∞

0

∞

= ∫ [0.1 2 + 0.5 2

+ 0.1 2] + ∫

[0.6 2( ) + 0.4 2

] ≤

2

3

2

3

+

0

∞

0

≤ ∫

2 ( ) =

, = ∫

[0.6 2( ) + 0.4 2( ))] , | |

< ∞, , ∞,

1

2

3

1

−

0

2

∞

∞

∞

∫ 0.1 2( ) < ∞, ∫

0.5 2( ) < ∞, ∫

0.1 2

( ) < ∞.

2

3

0

0

0

Then like in example 1 from condition lim y3 (t) = 0, we get

y1 = 0.

t

The value of

0

we define from condition of Hurwitz for the matrix A2 ( ). As

2

( )

= |

−

2

( )|

= 3

+ (1 + 0.1 ) 2 + (1 + 0.7 ) + 0.5 , then matrix A ( )

3

2

is a Hurwitz matrix for any

, 0 < < . Consequently,

0 = . So, in the sector

(0, 0

= ) Iserman’s problem has a solution for the regulated system (1.74).

For example 2, we

can easily find all sets of feedback vector S1 = (s1, s2 , s3 ),

where

= s x s x

s ,

for which Iserman’s problem has a solution. Notice that

1 1

2 2

3

* 1

(t) =

y (t) b y

(t)

y (t), t I ,

S1 = S1R

= ( 0 , 1

, 2 ),

where

2

2

0 1

1

3

Consequently, for any vector S1

= (

, ,

) satisfying condition

0

> 0,

2

2

> 0,

0

1

2

1

2 > 0

Iserman’s problem

has

a

solution. For

example

2, values

0 = 0.5, 1 =

= 0.7, 2 = 0.1, ( ) = 0.5 2 + 0.7 2 + 0.1 3, where

0

> 0,

1 0 2 = 0,1 > 0, 2 > 0.

(t) = ( 2 0 1 2 )x1(t)

Vector

S = S1R* = ( 2

0

,

0

,

),

1

1

2

2

0

( 0 2 )x2 (t) 0 x3 (t). For example 2, ( ) = −0.2 1 + 0.4 2 − 0.5 , .

So, all sets of vector S = ( 2

0

2

,

0

2

,

0

) R3 ,

where

0

> 0,

1

1

1 0 2 > 0, 2 > 0.

Using properties of boundedness of the improper integral I1

by choosing feed-

( ) = { ( ) C(R

, R ) | 0 ( ) <

,

(0) = 0, | ( ) |

,

1

1

2

1

0

*

(1.78)

0

< , , | |

,

0 <

< }

*

*

*

As the value

* , 0 < * < , > 0

is an arbitrarily small number, then inclu-

sions (1.77), (1.78) contain all nonlinearities from sector

[0, 0 ].

Practical systems of

automatic control refer

to the systems with limited

resources, for

such systems

function

( )

satisfies the conditions (1.77), (1.78).

The equilibrium states of the system (1.76), (1.77) are determined from the solu-

tion of algebraic equations

Ax B (

*

) = 0, = 0, (

*

) = 0,

*

= Sx

*

*

*

1 *

2 *

As matrix

A is a Hurwitz matrix, detA 0,

( * ) = 0

only when * = 0, then

z = A z B ( ), = S z, ( )

,

(1.79)

1

1

1

0

where

A

0

0

B

x

A1

=

0

0

1

, B1

=

2

, z =

,

0

0

0

1

arbitrarily small number. In other words, when | |< ,

where > 0 is an arbitrarily

small number, the function ( ) = , , > 0.

Then trivial solution of the

system (1.78) equal

to z = 0 is asymptotically stable in small,

if matrices A,

A1( ) = A1 B1 S1,

0 < < 0 are Hurwitz matrices, where

0

is Hurwitz

limit value of matrix

A1( ). So when matrix A1( ), 0 < < 0

is a Hurwitz

Definition

7.

Trivial solution

z(t) 0,

t I

of

the system

(1.79)

is

called

absolutely stable, if: 1) matrices

A,

A ( ), [ ,

0

]

are Hurwitz matrices; 2) for

1

all

( )

0

a

solution

of the

differential

equation (1.78) has

the

property

lim z(t;0, z

0

, ) = 0, z

0

= z(0),

| z

0

|< .

t

Notice that: 1) exploring properties of solutions of a system

with differential

inclusion

z A z B ( ),

= S z, ( )

0

;

2) As

( )

0

,

then equation

1

1

1

(1.78) has a non-unique solution, outgoing from the starting point

z

0

= (x

,

0

,

0

);

0

from

any

starting

point

z

R

n 2

, | z

|<

tend

to

equilibrium

0

0

z* = (x* = 0, * = 0, * = 0); 4) Equilibrium state

z* = 0

is asymptotically stable by

Lyapunov when t .

Definition 8. Conditions of absolute stability of the system (1.76), (1.77) are cal-

led relations, binding constructive parameters of the system

(A, B, 1, 1, S, 1, 2 )

under which the equilibrium state z* = 0

(trivial solution

z(t)

0,

t I ) is abso-

lutely stable.

Problem 5. Find the condition of absolute stability of equilibrium states of the

system (1.76), (1.77).

Function (t) = Sx 1

2 , t I

is a control formed by the principle of

feedback, and the row vector S1

= (S, 2 ,

1 ) R

n 2

is called the feedback vector.

Iserman’s problem consists in choosing a vector of feedback

S Rn 2 , such that

1

from asymptotic stability of

trivial solution z(t) 0, t I

of

linear system

absolute stability of

z = A1z B1 S1z = A1( )z, 0 < 0 = 0 we get

trivial solution z(t) 0, t I

of the system (1.76), (1.77), where

0

is the limiting

2) for any ( ) = 0 0 solution of

the system

(1.78)

is

asymptotically stable; 3) for any

0 trivial solution

of the

system

(1.78)

is

absolutely stable.

Problem 6. Find a sector

[ ,

0

],

where Iserman’s problem has a solution.

Note that as follows from work [11] Iserman’s problem does not always have a

solution. The problem consists of: finding such a vector of feedback

S R

n 2

,

that in

1

the sector

[ ,

0

],

=

0

Iserman’s problem had a solution.

0

From (1.79) when

( ) =

( ),

we get

matrix

A1 ( )

is a Hurwitz matrix for any small enough > 0.

contained in the inclusion

Therefore,

it

is necessary

to

transform the

( ) 1.

equation (1.80) so, that the function

( (t)),

t I

is explicitly presented through

phase coordinates of the system, taking into account Hurwitz of matrices

A,

A1 ( .)

Moreover, nonsingular

transformation should

be such that

improper integral

It should be noted that the existing conversion of a linear system from [12;

§

5,

Theorem 2] based on the controllability of a pair

( A, B)

does not satisfy the above

where

In 2 is a unit matrix of order

(n 2) (n 2),

1 ( ) =

n

an 1

n 1

a3 a2

is a characteristic polynomial of the matrix

A,

matrix

A1

has a double zero eigen-

value.

Lemma 11. Let the row vector

= ( , ,

) R

n 2

such that

n 2

1

n

n 1

B

0.

B = 0, A B = 0, , A B = 0,

A

(1.81)

1

1

1

1

1

1

1

Then equation (1.79) can be represented as

y1 = y2 , y2 = y3 , , yn 1 = yn 2 ,

y

= a

y

a y

a

y

A

n 1

B ( ),

n 2

4

n 1

n 2

2

3

3

1

1

y1 = z, y2

= A1z, , yn 1

n

n 1

yi = yi (t), i = 1, n 2,

where

= A1 z, yn 2 = A1

z,

z = z(t),

t I.

Proof. Consider the first equation from (1.79). Multiplying it on the left by

,

we

In a similar way, differentiating by

t,

we get the following system of differential

equations:

y

= y

,

y

2

z

, , y

= y

,

y

= A

n 1

z

= A

n 1

[ A z

B ( )] =

(0) = A

n 1

n 2

n

2

3

4

3

1

0

1

1

1

1

= [ a

An 1 a An

a A3

a A2

]z An 1B ( ) =

n 1

1

n

1

3

1

2

1

1

1

= a

n 1

y

n 2

a

n

y

n 1

a y

4

a

y An 1B ( ),

3

2 3

1

1

where y

n 2

(0) = An 1z

. The lemma is proved.

1

0

Lemma 12. Let the conditions of lemma 1 be satisfied, and let, moreover, the

rank of the matrix

R =

* ,

A* * , , A*n 1 *

(1.83)

1

1

(t) = 1 y1 (t) 2 y2 (t) n 2 yn 2 (t),

t I;

(1.84)

2) if

y = z = 0, y

2

= A z = 0, , y

n 2

= An 1z = 0, then

z = 0.

1

1

1

Proof.

Notice

that rank

R = n 2

then

and

only

then,

when vectors

* , A* *

, , A*n 1 *

are linearly independent. As vectors * , A* * , , A*n 1 * form

1

1

1

1

a basis in Rn 2 ,

then vector

S* Rn 2

can be represented unambiguously in the form

1

*

= 1

*

*

*

*n 1

*

.

Then

S1

2 A1

n 2 A1

= S z = z A z

A

n 1

z =

y

y

y

.

n

2

2

n

2

n 2

1

1

1

1

2

1

1

Now the second equation from (1.79) can be written as (1.84). On the other hand,

(

*

*

from (1.83) it follows that the pair

, A )

is controllable. From controllability of

1

(

*

*

n 1

z = 0,

the

pair

, A )

it

follows that equalities

z = 0, A z = 0, , A

which

1

1

1

entail z = 0.

Consequently, from

yi

= 0, i = 1, n 2

it follows that z = 0, t I. The

lemma is proved.

Introducing the notation

0

0

1

0

0

0

0

1

0

0

0

A1

=

B1

=

S1 = (

, ,

),