Solutions of polynomials for linear stationary system with conditions to state function and controllibility function

NGHIỆM ĐA THỨC CỦA HỆ DỪNG TUYẾN TÍNH 110 Le Hai Trung SOLUTIONS OF POLYNOMIALS FOR LINEAR STATIONARY SYSTEM WITH CONDITIONS TO STATE FUNCTION AND CONTROLLIBILITY FUNCTION NGHIỆM ĐA THỨC CỦA HỆ DỪNG T[.]

110 Le Hai Trung

SOLUTIONS OF POLYNOMIALS FOR LINEAR STATIONARY

SYSTEM WITH CONDITIONS TO STATE FUNCTION AND

CONTROLLIBILITY FUNCTION

NGHIỆM ĐA THỨC CỦA HỆ DỪNG TUYẾN TÍNH VỚI ĐIỀU KIỆN

LÊN HÀM TRẠNG THÁI VÀ HÀM ĐIỀU KHIỂN

Le Hai Trung

1 The University of Danang, University of Education; Email: trungybvnvr@yahoo.com

Abstract - The aim of this article is to prove that, solution x(t) (state

function) of the linear stationary dynamic system

x t =Bx t +Du t , which transfers the system from any initial

conditions in to any final conditions and at the same time satisfies

conditions given for the controllability function u t( ) is possible to

find in the type of polynomials of degree r+ (k+ 2)(p+ − 1) 1 with

vector coefficients The basis of the theory is a method to prove the

cascade splitting to transform the original system into an equivalent

system, which means that after the final step of conversion routines

p, we get a system that is completely controllability (see [3], [4]) In

the final step, we obtain a pseudo-state function x p (t) satisfying the

conditions and substituting this in the previous step Continuing this

process until we obtain x(t)

Tóm tắt - Nội dung bài báo chứng minh được rằng, nghiệm x(t)

(hàm trạng thái) của hệ dừng động học tuyến tính

x t =Bx t +Du t , dịch chuyển hệ từ trạng thái ban đầu bất kỳ đến trạng thái cuối tùy ý và đồng thời thỏa mãn các điều kiện cho trước đối với hàm điều khiển u(t), có thể tìm được dưới dạng đa thức bậc r+ (k+ 2)(p+ − 1) 1 với các hệ số vector Cơ sở lý thuyết của phép chứng minh dựa trên phương pháp phân tách hệ phương trình ban đầu thành các hệ tương đương, nghĩa là sau một số hữu hạn bước biến đổi, ta đưa được hệ về giai đoạn cuối p mà tại đó

hệ là hoàn toàn điều khiển được (xem [3], [4]) Tại đây, sau khi tìm được hàm giả trạng thái x p (t) thỏa mãn các điều kiện, ta tiến hành thế ngược trở lại vào giai đoạn trước đó Tiếp tục quá trình trên cho đến khi nhận được x(t)

Key words - state function; controllability function; linear stationary

dynamic system; polynomial solutions; method cascade splitting

Từ khóa - hàm trạng thái; hàm điều khiển; hệ dừng động học tuyến

tính; nghiệm đa thức; phương pháp phân tách

1 Rationale

For the calculation of the entire dynamic system and

control:

where ( ) n,

x t R ( ) m;

u t R B D, - is the matrix with

corresponding size, t[0, ],T the requirement is to find the

vector-controllability function u t , which transforms ( )

system (1) from the initial statex to terminal state 0 T

x

through kcheckpoints arbitrary and adds to the boundary

conditions for the control function ( )u t and derivative at

,

i

t=t i=0,1, ,k+1, 0=   t0 t1 t k+1=T

In [1], [2] the system (1) with conditions:

0

(0) , ( ) T,

the function ( )u t is built in the form:

0

T

tB sB sB TB T

u t =D e e− DD e ds − e− x −x

and in [3]: ( ) tB p ( ),

p r

u t =D e P t+

where ( )u t =P t r( ) polynomial with variable t , matrix

,

D B+ described later

But with the emergence of the hat matrix brings ( )u t

many difficulties in the study and the nature of the survey,

( )

u t and ( ) x t in practice, so the article will introduce how

to define functions ( )u t , ( ) x t variables as a polynomial

according to the time t To accomplish that, we building

functions ( )u t , ( ) x t in polynomials form of a high enough

level and then surveyed them as additional parameters

A Ailon (see [5]) in 1986 for the system (1) with conditions (2) has demonstrated that existed in the form of

a polynomial function ( )u t with smaller steps 2n In [4]

demonstrated that "control function can be represented as

a polynomial M =2r+1, r= −n rankD."

The following example shows the result can be made more precise At the system:

1 2

2 1

3 4

4 2

=



 =

 =

 =



(3)

with conditions 0

i i

x =x ( ) T,

x T =x i =1, 2,3, 4 we have:n =4, rankD =2, M =5, therefore existing

1( ) 5( ), 3( ) 5( )

u t =P t u t =P t (P t subsequent steps i i( ) polynomial coefficient vector) But the system is constituted by the two independent systems:

1

1 ,

j j

j j

+ +

=





0

(0) , ( ) T, 1,3; 1, 2,3, 4

systemn =2, rankD =1, M =3, inferred

u t =u t =P t

In [5] the system (1) - (2) with the additional conditions

x =x 0  T, x t( )=P3p+2( )t , where p=minq , q

satisfy criterion Kalman’s (see[1]) and ( ) k( ), 3 2

u t =P t k p+

THE UNIVERSITY OF DANANG, JOURNAL OF SCIENCE AND TECHNOLOGY, NO 6(79).2014, VOL 1 111

In the case of leaving x( ) =x, the conditions

corresponding polynomial of degree does not exceed

2p +1 Note that, M =2p+1 if and only if each one

contains a matrix control system linearly independent

vector, ie:

rank DBD B D = +i i= n−

In [3] proved that the system (1) - (2) is full controller

if and only if q to D q is subjective (Q = q 0) and the p

smallest number

In this article, we just examine the general control

problem The requirement ( )x t is to get the value t i in any

one given arbitrary values:

0 0

( )i i, 0,1, , 1, ,

x t =x i= k+ k (4)

and for ( )u t :

( ) i, 0,1, , , 0,1, , 1,

i

j

j u t i u i j i r i i k

where j i

i

u is the given vector, s is the derivative t

level s Requirements set out to build ( ) x t and ( ) u t in

polynomials with conditions (1), (4), (5)

The above problem occurs when the determination of

control system dynamics to the conditions experienced

"orbit" ( )x t and check points 0

0

( ,t x i i), checked by the driver at the time t , for example, in the problem of motion i

of the object with the soft landing v.v…

Note that the condition (4), (5) is transferred through

the following conditions (1):

i

des

0,1, , 1, i 0,1, , i 1

2 Results and Survey Research

To construct ( )x t and ( ) u t , we use the split method of

equation (1) into the equation in the subspace, is described

in [3], to make a few changes, use:

CL R R then:

ker ,

R =imC + C R =imC+kerC T (7)

and solve equations Cv = corresponding to w v if and

only if Qw =0

When v=C w+ +Pv, where ,Q P is the projections on

kerC T, ker C in formula (7), 1 1

C+=C− I−Q C− is

invertible matrix of the matrix C in imC T,I- identity

matrix, Pv is an element in ker C

Using the results with C=D for (1) we get: equation

(1) corresponds solved ( ) u t if and only if conditions occur:

( ) ( )

With the above conditions which are met, we have:

u t =D x t+ −D Bx t+ +Pu t

where Pu t( )is a function-vector in ker D

Expression (8) is transformed into:

Qx t =QBQ Qx t +QB I−Q I−Q x t (9) Notation:

1

( ) ( ),

Qx t =x t (I−Q x t) ( )= y t1( ),QBQ=B1,

1

QB I−Q =D

Then (9) has the form:

where B D1, 1in subspace

Conditions (6) and expression (8) move to:

i

des

j j

j x t i Qx i x i i k j i r i

In that way, the equation (1) with conditions (4), (5) is equivalent to the following relationship:

u t =D x t+ −D Bx t+ +Pu t (12)

1( ) 1( ) 1( ),

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

with conditions (11) Where Pu t is function-vector in( )

ker D and satisfy conditions:

( ( )) i, 0,1, , 1, 0,1, ,

i

j

j Pu t i Pu i i k j i r i

Element Pu t can be found in the form ( ) P t , r( )

0

k i i

=

= , so we have the following proposition:

Lemma 2 There exists a polynomial P t satisfying s( )

the following conditions:

1

0

i

k j

i

=

(16) Indeed, for

0

( )

s i

i

=

= condition (16) with t =0 0 we obtain:

0

0

1 , 0,1, ,

!

j i

Review

0 1

s i

i



= +

0

0

1

!

i

j



=

To determine the remaining coefficientsb , from i

conditions (16) we obtain the system:

0 0

0 0

1 0

1 0

s

s

s

s













− +

+

+













(17)

0,1, , 1

i= k + With each i generation the ratio of the

formation of thei+ determinant line Wronxki for the 1 function 0 1

, , s

t+ t at the point t and the determinant of i

112 Le Hai Trung system (17) has non-zero value (see [6])

Consider the expression (14), where

DL imD D Lemma 1 gives us:

1T ker 1, ker T 1 ker 1T

1, 1

Q P is the projections on kerD1T, kerD1 in (18),

1 1 ( 1), 1

D+ =D− I−Q D− is invertible matrix of the matrix

1

Donto imD1T

Equation (14) solved y t1( )if and only correspond to

satisfy conditions:

when:

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

y t =D x t+ −D B x t+ +P y t P y t1 1( )kerD1

Equation (19) is transformed to the form:

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

Q x t =Q B Q Q x t +Q B I−Q I−Q x t (20)

Notation:Q x t1 1( )=x t2( ),(I−Q x t1) ( )1 =y t2( ),

1 1 1 2,

Q B Q =B Q B I1 1( −Q1)=D2, when (20) is equation:

2( ) 2 2( ) 2 2( ),

similar to (1) and (14), but in the space "narrow" than

From conditions (11) and expression (14) we move on:

i

des

Note that the functions x t required to satisfy more 2( )

2

k + conditions thanx t 1( )

Now equation (1) with conditions (4), (5) is equivalent to:

u t =D x t+ −D Bx t+ +Pu t

2 2 1

( ) ( ) ( ) ( ),

x t =x t +y t +y t

y t =D x t+ −D B x t+ +P y t

1

with condition (22) and any function P y t1 1( )kerD1,

satisfying the following conditions:

1 1( ) 1( ) ( ) 1( ) 0i,

j

j P y t i j P I Q x t i P I Q x i

0,1, , 1, i 0,1, , i 1

To the extent that P y t allows us to grab1 1( ) P r k+ +1,

satisfying conditions (25) It exists by Lemma 2

Thus (28) is modified by using the Lemma 1 when

2

C=D , we do so with the value q By that statement:

Lemma 3 Equation (1) with conditions (4), (5) is

equivalent to system:

u t =D x t+ −D Bx t+ +Pu t (26)

( ) ( ) ( ),

s s s s s s s s

y t =D x t+ −D B x t+ +P y t (28)

1 1

x t =B x t +D y t s= p− (30)

with conditions:

1 1

des

j j

j x t s i j Q x s− s−i x s

0,1, , 1, i 0,1, , i ,

and P y t s s( )kerD s :

j

j P y t s s i j P I s Q s− x s− t i P I s Q s− x s− i

0,1, , 1, i 0,1, , i 1

Where,

Q B− − I−Q− =D DL imD− D− Q P is s, s

the projections on ker T

s

D and:

imD− =imD + D D− =imD + D

D+ =D− I−Q D− - is the inverse matrix of the matrix

s

Don T

s

imD

To build x t and ( ) u t , we build enough ( ) x t p( ) polynomial form P r+ +(k 2)(p+ −1) 1( )t , satisfying conditions (32) and polynomial construction P y t p p( ) satisfies (32),

s= In (29) with p s= −p 1 received x s−1( )t To the extent that P y s−1 s−1( )t we get the polynomial, satisfying (32) whens= −p 1 Find the formula (27) with s= − p 1 polynomial y s−1( )t In (29) with s= −p 1 received x s−2( )t

v.v….Summary ( )x t and ( ) u t are determined by formulas

(27) to getPu t in Lemma 1 ( ) Thus to prove:

Theorem The existence of the state function ( ) x t of system (1) with conditions (4), (5) as a polynomial according t to coefficient vector of degree

r+ k+ p + − and the control functions ( ) u t as a polynomial of degree not exceeding r+(k+2)(p + − 1) 1

Remark 1 With no math test scores ( k =0) and limited

to the control function ( )u t ( r =0), the functions ( )x t and

( )

u t can be built-order as polynomials of degree not exceeding 2 p+ 1 M These results, as seen, can not do better

Remark 2 When addressing the problem of control does

not necessary have to build the projectionsP Q Maybe i, i from the equation (1) representing the amount of ingredients

as much as possible through function ( )u t composition of

functions ( )x t A portion ( ) u t of the remaining Pu t ( ) Conditions of solving equation (1) is written as

1 (.) 1 ( ) 1

x = x + y , inferred x1=B x1 1+D y1 1 v.v… Received

1, 1

B D may not coincide with B D in section 1, because 1, 1

we have different representations of space:

THE UNIVERSITY OF DANANG, JOURNAL OF SCIENCE AND TECHNOLOGY, NO 6(79).2014, VOL 1 113

and the corresponding projection by the way is not the

only determining

However, p ( Q = p 0) is fixed for any space division,

which testifies to the Kalman’s criterion (see [1])

REFERENCES

[1] Красовский Н Н Теория управления движением Издательство

«Наука», Москва 1968 475 с

[2] Уонэм М Линейные многомерные системы управлегия

Издательство «Наука», Москва 1980 375 с

[3] Раецкая Е В Критерий полной условной управляемости

сингулярно возмущенной системы Оценки функции состояния

и управлющей функции Кибернетика и технологии XXI века:

Воронеж, 2004 с 28 – 34

[4] Раецкая Е В Условная управляемость и наблюдаемость

[5] линейных систем: Дисс канд физ – мат наук Воронеж, 2004 [6] Ailon A., Langholz G More on the cintrollability of linear

time-invariant systems Int J Contr 1986 44 № 4 P 1161 – 1176

[7] Зубова С П, Ле Хай Чунг Об полиномиальных управлениях

линейной стационарной системы с контрольной точкой

Современные проблеммы механики и прикладной математики Сборник трудов международной школы – семинара Воронеж

2007 – с 133-136

[8] Каган В Ф Теория определителей Одесса, Гос Из – во

Украины, 1922 – 521 с

[9] S P Zubova, LeHaiTrung Construction of polynomial controls for

linear stationary system with control points and additional constrains Automation and Remote Control No: Volume 71,

Number 5 Pages: 971-975 2010

[10] Lê Hải Trung Về hàm trạng thái đa thức cho hệ dừng động học

tuyến tính Tạp chí Khoa học và Công Nghệ Đại học Đà Nẵng Số:

11[60] Trang: 53 - 57.2012

(The Board of Editors received the paper on 01/06/2014, its review was completed on 18/06/2014)