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 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) − 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 xp(t) satisfying the conditions and substituting this in the previous step Continuing this process until we obtain x(t) Tóm tắt - Nội dung báo chứng minh rằng, nghiệm x(t) (hàm trạng thái) hệ dừng động học tuyến tính x (t ) = Bx (t ) + Du (t ) , dịch chuyển hệ từ trạng thái ban đầu đến trạng thái cuối tùy ý đồng thời thỏa mãn điều kiện cho trước hàm điều khiển u(t), tìm dạng đa thức bậc r + ( k + 2)( p + 1) − với hệ số vector Cơ sở lý thuyết phép chứng minh dựa phương pháp phân tách hệ phương trình ban đầu thành hệ tương đương, nghĩa sau số hữu hạn bước biến đổi, ta đưa hệ giai đoạn cuối p mà hệ hồn tồn điều khiển (xem [3], [4]) Tại đây, sau tìm hàm giả trạng thái xp(t) thỏa mãn điều kiện, ta tiến hành ngược trở lại vào giai đoạn trước Tiếp tục q trình nhận 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 Rationale For the calculation of the entire dynamic system and control: x(t ) = Bx(t ) + Du (t ), (1) 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 " where x(t )  Rn , u(t )  Rm ; 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 state x0 to terminal state xT through k checkpoints arbitrary and adds to the boundary conditions for the control function u (t ) and derivative at t = ti , i = 0,1, , k + 1, = t0  t1   tk +1 = T In [1], [2] the system (1) with conditions: x(0) = x , x(T ) = x , T (2) the function u (t ) is built in the form: T u(t ) = D*etB (  e− sB DD*esB ds)−1 (e−TB xT − x0 ), * * and in [3]: u (t ) = Dp+ e p Pr (t ), tB where u (t ) = Pr (t ) polynomial with variable t , matrix + p D , B p 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 The following example shows the result can be made more precise At the system:  x1 = x2  x2 = u1 x = x  x3 = u  (3) with conditions xi (0) = xi0 , xi (T ) = xiT , i = 1, 2,3, we have: n = 4, rankD = 2, M = 5, therefore existing u1 (t ) = P5 (t ), u3 (t ) = P5 (t ) ( Pi (t ) subsequent steps i polynomial coefficient vector) But the system is constituted by the two independent systems:  x j = x j +1 x = u , j  j +1 xi (0) = xi0 , xi (T ) = xiT , j = 1,3; i = 1, 2,3, rankD = 1, M = 3, system n = 2, u1 (t ) = u3 (t ) = P3 (t ) in each inferred In [5] the system (1) - (2) with the additional conditions x( ) = x ,    T , x(t ) = P3 p+2 (t ) , where p = q , q satisfy criterion Kalman’s u (t ) = Pk (t ), k  p + (see[1]) and THE UNIVERSITY OF DANANG, JOURNAL OF SCIENCE AND TECHNOLOGY, NO 6(79).2014, VOL In the case of leaving x( ) = x , the conditions corresponding polynomial of degree does not exceed p + Note that, M = p + if and only if each one contains a matrix control system linearly independent vector, ie:  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: Qx(t ) = x1 (t ), ( I − Q) x(t ) = y1 (t ), QBQ = B1 , rank (DBD B D) = i + 1, i = 0,1, , n −1 i In [3] proved that the system (1) - (2) is full controller if and only if q to Dq is subjective ( Qq = ) and the p smallest number In this article, we just examine the general control problem The requirement x(t ) is to get the value ti in any one given arbitrary values: x(ti ) = x00i , i = 0,1, , k +1, k , (4)  ji u(ti ) = ui , ji = 0,1, , ri , i = 0,1, , k + 1, (5) where uiji 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 (ti , x00i ) , checked by the driver at the time ti , for example, in the problem of motion of the object with the soft landing v.v… Note that the condition (4), (5) is transferred through the following conditions (1):  ji x(ti ) = Bx0jii −1 + Duiji −1 = x0jii −1 , (6) x1 (t ) = B1 x1 (t ) + D1 y1 (t ), 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: Lemma (see [4]) If C  L( Rs , R ) then: where B1 , D1 in subspace Conditions (6) and expression (8) move to: des  j x1 (ti ) = Qx0i = x1i , i = 0,1, , k + 1, ji = 0,1, , ri + (11) ji ji 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) x1 (t ) = x1 (t ) + y1 (t ), (13) x1 (t ) = B1 x1 (t ) + D1 y1 (t ), (14) with conditions (11) Where Pu (t ) is function-vector in ker D and satisfy conditions:  ji ( Pu(ti )) = Puiji , i = 0,1, , k + 1, ji = 0,1, , ri (15) Element Pu (t ) can be found in the form Pr (t ) , k r =  ri , so we have the following proposition: i =0 k +1  ji Ps (ti ) = aiji , i = 0,1, , k + 1, ji = 0,1, , i , s = k + +  i (16) s Indeed, for Ps (t ) =  bi t i condition (16) with t0 = we i =0 obtain: j0 a0 , i = 0,1, , j0 ! bi = (7) and solve equations Cv = w corresponding to v if and only if Qw = Review Ps (t ) = F (t ) + When v = C + w + Pv, where Q, P is the projections on ker C T , ker C in formula (7), C + = C −1 ( I − Q), C −1 is invertible matrix of the matrix C in imCT , 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: Qx(t ) = QBx(t ) (8) s bt ,   i i= +1 i 0 F (t ) =  i =0 j0 i a0 t j0 ! To determine the remaining coefficients bi , from conditions (16) we obtain the system:  b t + b t + + b t = a    ( + 1)b t + + sb t = a   s!  ( + 1)!   + + bt  ( − + 1)! b t ( s −  )!  v0 +1 +1 +2 +2 i s i s i s −1 0 +1 0 i i i s i i − i +1 0 +1 With the above conditions which are met, we have: u(t ) = D+ x(t ) − D+ Bx(t ) + Pu(t ), (10) i =0 Results and Survey Research R s = imC T + ker C , R = imC + ker CT Then (9) has the form: Lemma There exists a polynomial Ps (t ) satisfying the following conditions: des i = 0,1, , k + 1, ji = 0,1, , ri + QB( I − Q) = D1 i and for u (t ) : ji 111 i (17) s − i s i = i i i = 0,1, , k + With each i generation the ratio of the formation of the  i + determinant line Wronxki for the function t0 +1 , , t s at the point ti and the determinant of 112 Le Hai Trung xp (t ) = Bp xp (t ) + Dp y p (t ), s = 1, , p −1 system (17) has non-zero value (see [6]) Consider the expression (14), D1  L(imD, ker D) Lemma gives us: where T with conditions: des imD = imD + ker D1 ,ker D = imD1 + ker D T (30) T  ji xs (ti ) =  ji Qs−1 xsj−i i = xsji , (18) Q1 , P1 is the projections on ker D1T , ker D1 in (18), D1+ = D1−1 ( I − Q1 ), D1−1 is invertible matrix of the matrix i = 0,1, , k + 1, ji = 0,1, , ri + s, (31) and Ps ys (t )  ker Ds : D1 onto imD1T  ji Ps ys (ti ) =  ji Ps ( I − Qs −1 ) xs −1 (ti ) = Ps ( I − Qs −1 ) xsj−i i Equation (14) solved y1 (t ) if and only correspond to satisfy conditions: i = 0,1, , k + 1, ji = 0,1, , ri + s − Q1 x(t ) = Q1 B1 x1 (t ) (19) (32) Where, Qs −1 xs −1 (t ) = xs (t ), ( I − Qs −1 ) xs −1 (t ) = ys (t ), Qs −1 Bs −1Qs −1 = Bs , Qs −1 Bs −1 ( I − Qs −1 ) = Bs , when: y1 (t ) = D x (t ) − D B1 x1 (t ) + P1 y1 (t ), P1 y1 (t )  ker D1 + 1 + Qs −1Bs −1 (I − Qs −1 ) = Ds , D  L(imDs −1 , ker DsT−1 ), Qs , Ps is the projections on ker DsT and: Equation (19) is transformed to the form: Q1 x1 (t ) = Q1 B1Q1 (Q1 x1 (t )) + Q1 B1 ( I − Q1 )( I − Q1 ) x1 (t ) (20) Notation: Q1 x1 (t ) = x2 (t ), ( I − Q1 ) x1 (t ) = y2 (t ), Q1 B1Q1 = B2 , Q1 B1 ( I − Q1 ) = D2 , when (20) is equation: x2 (t ) = B2 x2 (t ) + D2 y2 (t ), (21) similar to (1) and (14), but in the space "narrow" than From conditions (11) and expression (14) we move on: des  j x2 (ti ) = Q1 x1i = x2 i , i = 0,1, , k + 1, ji = 0,1, , ri + ji ji (22) i Note that the functions x2 (t ) required to satisfy more k + conditions than x1 (t ) Now equation (1) with conditions (4), (5) is equivalent to: u(t ) = D+ x(t ) − D+ Bx(t ) + Pu(t ), x(t ) = x2 (t ) + y2 (t ) + y1 (t ), imDs −1 = imDsT + ker Ds , ker DsT−1 = imDs + ker DsT , Ds + = Ds −1 ( I − Qs ), Ds −1 - is the inverse matrix of the matrix Ds on imDsT To build x(t ) and u (t ) , we build enough x p (t ) polynomial form Pr +( k + 2)( p +1)−1 (t ) , satisfying conditions (32) and polynomial construction Pp y p (t ) satisfies (32), s = p In (29) with s = p − received xs −1 (t ) To the extent that Ps−1 ys−1 (t ) we get the polynomial, satisfying (32) when s = p − Find the formula (27) with s = p − polynomial ys−1 (t ) In (29) with s = p − received xs−2 (t ) v.v… Summary x(t ) and u (t ) are determined by formulas (27) to get Pu (t ) in Lemma Thus to prove: y1 (t ) = D1+ x1 (t ) − D1+ B1 x1 (t ) + P1 y1 (t ), x1 (t ) = B1 x1 (t ) + D1 y1 (t ), (23) s =1 (24) with condition (22) and any function P1 y1 (t )  ker D1 , satisfying the following conditions:  ji P1 y1 (ti ) =  ji P1 ( I − Q) x(ti ) = P1 ( I − Q) x0jii , i = 0,1, , k + 1, ji = 0,1, , ri + To the extent that P1 y1 (t ) allows us to grab Pr +k +1 , satisfying conditions (25) It exists by Lemma Thus (28) is modified by using the Lemma when C = D2 , we so with the value q By that statement: Lemma Equation (1) with conditions (4), (5) is equivalent to system: u(t ) = D+ x(t ) − D+ Bx(t ) + Pu(t ), (26) x(t ) = x1 (t ) + y1 (t ), (27) ys (t ) = Ds + xs (t ) − Ds + Bs xs (t ) + Ps ys (t ), (28) xs (t ) = xs+1 (t ) + ys+1 (t ), (29) 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 + 2)( p + 1) − and the control functions u (t ) as a polynomial of degree not exceeding r + (k + 2)( p + 1) − Remark With no math test scores ( k = ) and limited to the control function u (t ) ( r = ), the functions x(t ) and u (t ) can be built-order as polynomials of degree not exceeding p +  M These results, as seen, can not better Remark When addressing the problem of control does not necessary have to build the projections Pi , Qi Maybe 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 x1 = (.) x1 + ( ) y1 , inferred x1 = B1 x1 + D1 y1 v.v… Received B1 , D1 may not coincide with B1 , D1 in section 1, because we have different representations of space: THE UNIVERSITY OF DANANG, JOURNAL OF SCIENCE AND TECHNOLOGY, NO 6(79).2014, VOL R = coimD + ker D, R = imD + co ker D, n m and the corresponding projection by the way is not the only determining However, p ( Qp = ) 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 113 [5] линейных систем: Дисс канд физ – мат наук Воронеж, 2004 [6] Ailon A., Langholz G More on the cintrollability of linear timeinvariant systems Int J Contr 1986 44 № 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 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 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) ... 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 + 2)( p + 1) − and the control functions u (t... 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. .. 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