Optimal algorithm and application for point to point iterative learning control via updating reference trajectory

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1 DOI: /CTA Control Theory & Applications Vol. 33 No. 9 Sep. 2016,, (, ) :,.,,.,,,.. : ; ; ; ; : TP273 : A Optimal algorithm and application for point to point iterative learning control via updating reference trajectory TAO Hong-feng, DONG Xiao-qi, YANG Hui-zhong (Key laboratory of Advanced Process Control for Light Industry (Ministry of Education), Jiangnan University, Wuxi Jiangsu , China) Abstract: For the output tracking control problem of discrete linear system with non-repetitive disturbance, a point to point iterative learning control algorithm based on updating reference trajectory is proposed. Firstly, the iterative learning controller is optimized by constructing performance index with norm function, and the corresponding convergence conditions are given, then the system output can track with the desired points in updating reference trajectory. Furthermore, when the system output is affected by non-repetitive disturbance in some trials, a new multi-objective performance index function is constructed by Lagrange multiplier algorithm, and the robust iterative learning controller is optimized to improve the convergence speed and tracking accuracy. Finally, the simulation results of the motor driven single mechanical arm control system show the effectiveness and feasibility of the proposed algorithm. Key words: updating reference trajectory; iterative learning control; performance index; Lagrange multiplier; point to point 1 (Introduction) (iterative learning control), [1 3].,,,,.,,,,,. [4],,,. [5],,., [6],. [7] P : ; : taohongfeng@hotmail.com; Tel.: :. ( , ). Supported by National Natural Science Foundation of China ( , ).

2 CB , CAB CB 0 0, G = CA 2 B CAB CB 0,......, CA T 1 B CA T 2 B CA T 3 B CB. [8] d k =[CA CA 2 CA 3 CA T ] T x k (0),,,. u k y k u k =[u k (0) u k (1) u k (T 1)] T, y k =[y k (1) y k (2) y k (T )] T. [9], (1) k +1, ω, k+1,,., y k+1 = Gu k+1 + d k+1 + ω k+1, (3), [10], sup ω k+1 ξ, t [0,T ]. :,,. [11],. [12 13],. b x0 = sup t [0,T ] d k+1 d k.,,,.,,.,. 2 (Problem formulation) : { xk (t +1)=Ax k (t)+bu k (t), (1) y k (t) =Cx k (t), : x k (t) R m, u k (t) R r, y k (t) R n k, x k (0) k, t [0,T] T ; A, B, C, CB 0. (1), : : y k = Gu k + d k, (2),,, y k (t) y d (t), t {0, 1,,T}.,,, : r k (t i )=y d (t i ), i =1,,M, 0 t 1 <t 2 < <t M T., r k+1 (t) =r 0 (t)+h(t)f(t), (4) : h(t) =(t t 1 )(t t 2 ) (t t M ), f(t), r 0 (t). (4),,, r k (t i )=r 0 (t i )=y d (t i ). r k+1 =[r k+1 (0) r k+1 (1) r k+1 (T )] T, H =diag{h(0) h(1) h(t )}, f k =[f(0) f(1) f(t )] T. (4) : r k+1 = r 0 + Hf k. (5), r k (t i )=r 0 (t i )=y d (t i ), (5) : r k+1 = r k + Hf k. (6) (6) f k = F (r k y k ), F

3 9 : 1209, (6) r k+1 = r k + HF(r k y k ). (7) (7) H F, λ k = HF, λ k,, (7), (8) r k+1 = r k + λ k (r k y k ). (8) r k+1 y k = (I + λ k )(r k y k ) I + λ k (r k y k ). I + λ k =1, r k+1 y k (r k y k ). f k.,,,. 3 (Point to point robust iterative learning control) : { uk+1 = u k + L(r k+1 y k ), (9) r k+1 = r k + λ k (r k y k ), : λ k, r k k 1, L. 1 (1), (9), : ρ(i GL) < 1, k, : 0 lim e k+1 1 k 1 ρ (ξ + b x 0 )., ξ =0, b x0 =0, lim e k+1 =0. k k +1, (1) e k+1 = r k+1 y k+1 = r k + λ k (r k y k ) Gu k d k GL(r k + λ k (r k y k ) y k )+d k d k+1 ω k+1 = (I GL)(I + λ k )(r k y k )+ d k d k+1 ω k+1. (10) (10) (I GL) = ρ, I + λ k =1, (10) e k+1 ρ e k + ξ + b x0, (11) k e k+1 = ρ k+1 e ρk+1 1 ρ (ξ + b x 0 ). (12) ρ <1, 0 lim k e k ρ (ξ + b x 0 )., ξ =0, b x0 =0, k.,,, (11) e k+1 ρ e k. 0 <ρ<1, e k+1 e k,. 3,, 4, 5,. 4 (Norm performance index optimal iterative learning without disturbance), u k+1,., : min J k+1 = e k u Q k+1 u k 2 + u R k+1 2, S (13) Q, R S. e k+1 = r k+1 Gu k+1 d k+1, e k+1 e k+1 = k (I + λ i )r 0 k (I + λ i )λ 0 Gu 0 i=0 i=1 k (I + λ i )λ 1 Gu 1 i=2 (I + λ k )λ k 1 Gu k 1 λ k Gu k Gu k+1 λ k d k d k+1 k (I + λ i )λ 1 d 1 i=2 (I + λ k )λ k 1 d k 1. (14) (14) (13), u k+1 G T Q(r k+1 Gu k+1 d k+1 )+ R(u k+1 u k )+Su k+1 =0. (15) (G T QG + R + S)u k+1 = Ru k + G T Qr k+1 G T Qd k+1. (16), I + λ k =1,

4 λ k 0, r k (t i )=r 0 (t i ), i =1, 2,,M. (16) : (G T QG + R + S)u k+1 = Ru k + G T Qr 0 G T Qd k+1. (17) (17) R S ri si, (r + s)i >0, G T QG > 0, (17) u k+1 =(G T QG + R + S) 1 Ru k + (G T QG + R + S) 1 G T Qr 0 (G T QG + R + S) 1 G T Qd k+1. (18), (17) (G T QG + R + S)u k+1 = (R + G T QG)u k + G T Q(r 0 y k ) G T Q(d k+1 d k ), (19) u k+1 =(G T QG + R + S) 1 (R + G T QG)u k (G T QG + R + S) 1 G T Q(d k+1 d k )+ (G T QG + R + S) 1 G T Qe k. (20) Z u =(G T QG + R + S) 1 (R + G T QG), Z e = (G T QG + R + S) 1 G T Q, (20) u k+1 = Z u u k + Z e e k Z e (d k+1 d k ). (21) 2 (21), (1) : 1) : ρ((g T QG + R + S) 1 R) < 1; 2) S ; 3) x k (0), k,. (1) u d, (21) u d u k+1 = u d Z u u k Z e e k +Z e (d k+1 d k ). (22) (22) G, e k+1 Z u Z e G e k + U, U = G Z u G u d + Z e d k+1 d k + Z u d k d k+1., ρ(z u Z e G)= ρ((g T QG + R + S) 1 R) < 1, (23)., S, x k (0) : Z u, G Z u G =0, d k+1 d k =0, Z u d k d k+1 =0, e k+1 ρ e k e k, (24) e k+1. 2 Amann [14 15], Owens [16 17],, (9), [15],.,, [16]. 3, 4,, (1). 5 (Norm index optimal iterative learning with non-repetitive disturbance) 1, k +1 e k+1 = r 0 Gu k+1 d k+1 ω k+1. (25) min J k+1 = r 0 Gu k+1 d k+1 ω k Q u k+1 u k 2 + u R k+1 2. S (26), min max J(u k+1,ω k+1 )= u k+1 ω k+1 r 0 Gu k+1 d k+1 ω k Q u k+1 u k 2 + u R k+1 2. S (27) ω k+1,, (27) ω k+1 r 0 Gu k+1 d k+1 ω k+1 2 Q. : max r 0 Gu k+1 d k+1 ω k+1 2. ω Q d(k+1) ω k+1 ξ, γ, max J 1 (ω k+1, γ), J 1 (ω k+1,γ)= r 0 Gu k+1 d k+1 ω k+1 2 Q γ( ω k+1 2 ξ 2 ). (28), γ 0, -, (28) ω k+1 γ

5 9 : 1211 Q(r 0 Gu k+1 d k+1 )=(Q γi)ω k+1, (29) ξ 2 = ω k+1 2. (30) Q γi 0, (26). (29) ω k+1 =(Q γi) + Q(r 0 Gu k+1 d k+1 ), (31) (Q γi) + Q γi. (31) (30) ξ 2 =(r 0 Gu k+1 d k+1 ) T Q(Q γi) + (Q γi) + Q(r 0 Gu k+1 d k+1 ). (32) (31) (32), ω k+1, γ u k+1,,, Θ(γ): Θ(γ) =(r 0 Gu k+1 d k+1 ) T Q(I+ (Q γi) + Q) (r 0 Gu k+1 d k+1 ) γξ 2. (33) (33) γ (32). (28) (33) : min Θ(γ) = 0 γ Q min (r 0 Gu k+1 d k+1 ) T Q(I+ 0 γ Q (Q γi) + Q)(r 0 Gu k+1 d k+1 )+γξ 2. (34),, max J 1 (ω k+1,γ)= min Θ(γ). (35) 0 γ Q (27) min min J(u k+1,ω k+1 )= 0 γ Q u k+1 Θ(γ)+ u k u S k+1 u k 2. R (36), (36) u k+1 (G T ΣG + R + S)u k+1 = Ru k + G T Σr 0 G T Σd k+1. (37) = Q[I +(Q γi) + Q], (37) u k+1 =(G T ΣG + R + S) 1 Ru k + (G T ΣG + R + S) 1 G T Σr 0 (G T ΣG + R + S) 1 G T Σd k+1. (38) ρ((g T ΣG + R + S) 1 R) < 1. (39)., (33) γ ξ 2 =(r 0 Gu k+1 d k+1 ) T Q(Q γi) + (Q γi) + Q(r 0 Gu k+1 d k+1 ). (40), (38) (40) (36), k, u k+1 =(G T ΣG + S) 1 G T Σr 0 (G T ΣG + S) 1 G T Σd k+1. (41) e k+1 = r 0 G(G T ΣG + S) 1 G T Σr 0 + G(G T ΣG + S) 1 G T Σd k+1 d k + ω k+1. (42) 4,, S,,. 6 (Simulation study), [18], ẋ 1 = x 2, ẋ 2 = N Ψ t x 1 B c Ψ t x 2 + K t Ψ t x 3, ẋ 3 = K b Γ x 2 R r Γ x Γ u, y = x 1, (43) : N = m 2 gl + m 1 gl, Ψ t = Ξ +(1/3)m 2 l 2 + (1/10)m 1 l 2 D c, x 1 = θ, x 2 = θ, x 3 = i, g, θ, i, K t, K b, B c, D c, l, m 1, m 2, R r, u, Ξ, Γ. K t =1N m, K b =0.085 V s/rad, R r =0.075 Ω, B c =0.015 kg m 2 /s, D c =0.05, l =0.6m,m 1 =0.05 kg,m 2 =0.01 kg, Ξ =0.05 kg m 2,Γ= Ω, g=9.8m/s 2. t =2s, t s =0.1s, A= ,B= , C =[100], D=0.

6 T {0, 1, 2,, 20},, M =7, t i {0, 2, 6, 10, 14, 18, 20}. x k (0) = k 1 [ ] T. λ k, 0. Q, R, S Q = I, R =0.005I, S =0., P : u k+1 = u k + k p e k, (9), (21), 1 3. P.,,, 2,., 3,,, P. 3 Fig. 3 The curves of system performance index without disturbance 1 Fig. 1 The curves of system actual output and reference point tracking (43) k(sin t +cost), 5, γ 0.5, Fig. 2 The curves of system output error without disturbance 1 (43) (9), π 2 π, 2. 2,,, 4 Fig. 4 The curves of system output error with disturbance 4, 10,,,,, P.

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