31 1 2014 1 DOI: 10.7641/CTA.2014.30591 Control Theory & Applications Vol. 31 No. 1 Jan. 2014,,, (, 210094) :,., :, ;,., ; Lyapunov ;,,. : ; ; ; ; : TP273 : A Fault diagnosis and fault tolerant control for the servo system driven by two motors synchronously CHEN Wei, WU Yi-fei, DU Ren-hui, WU Xiao-bei (School of Automation, Nanjing University of Sciense and Technology, Nanjing Jiangsu 210094, China) Abstract: In the servo system driven by two motors synchronously, actuator failures deteriorate the tracking performance or even result in system instability. To deal with this problem, we propose a fault diagnosis and fault tolerant control scheme based on adaptive sliding mode method. In this control scheme, the adaptive sliding mode observers for motor speed are designed, and the actuator failure factors are estimated online. When an actuator partly fails to work, the controller gains will be adjusted automatically. When an actuator fully breaks down, the control law of the system will be reconstructed. Considering that there are unmatched uncertainties in the system, we introduce compensations based on extended state observer (ESO) to the desired virtual signals. The stability of the servo system under normal and fault conditions are analyzed by Lyapunov theory, the convergence of the observers are also analyzed. It is shown by simulation results that the control strategy can assure the stable tracking of the command signals. The tracking performance almost remains unchanged when an actuator partly or fully fails to work. Key words: fault tolerant control; fault diagnosis; dual-motor; actuator failure; adaptive sliding mode control 1 (Introduction), [1 3].,,, [1 2],,. ( ),,.,,,. (fault tolerant control, FTC) [4].,,,. 3 [4] : [5 7],,,,, ; [8 10] : 2013 06 13; : 2013 08 02.. E-mail: supercw 86@163.com; Tel.: +86 13016959660. : (61074023); (CXLX11 0256); (GZKF 2012 03).
28 31,,., [11] ; [12 14],,,,.,,,..,,,,., :,,, (extended state observer, ESO),,.,,,,.,, ;,,., Lyapunov,. 2 (Problem description) 1,,. ( ) θ L (t) = ω L (t), θj (t) = ω j (t), ω L (t)= K L[θ 1 (t) θ L (t)]+k L [θ 2 (t) θ L (t)] J L + w 1 (t), ω j (t) = K Tjf j (t)u j (t) K L [θ j (t) θ L (t)] J mj + w 2j (t), (1) : θ L (t), ω L (t), J L, K L, w 1 (t),. θ j (t), ω j (t), J mj, K Tj, u j (t) (j = 1, 2( )), w 2j (t),. f j (t), 0 f j (t) 1, f j (t) = 0, 0 < f j (t) < 1, f j (t) = 1. x 1 = θ L (t), x 2 = ω L (t), x 3j = θ j (t), x 3 = x 31 + x 32, x 4j = ω j (t), x 4 = x 41 + x 42. a = K L J L, α 1j = K Tj J mj, α 2j = K L J mj, : a, α 1j, α 2j, : ẋ 1 = x 2, ẋ 2 = a(x 3 2x 1 ) + w 1, ẋ 3 = x 4, ẋ 4j = α 2j (x 1 x 3j ) + α 1j f j u j + w 2j, y = x 1. (2) y d : 1. (2) 2 w 1 w 2j w 2j, w 2j < ρ j. 3 y d. 4 f 1, f 2,. 1 Fig. 1 Structure of the servo system driven by two motors (2) 1 4,
1 : 29 u j, x 3 = x 31 + x 32,, y ẋ 3j = x 4j, y d. S 2 : (4) ẋ 4j = α 2j (x 1 x 3j ) + α 1j f j u j + w 2j, y 2 = x 3. 3 (Fault tolerant controller design and stability analysis) w 1, S 1 S 2 : ẋ 1 = x 2, S 1 : ẋ 2 = a(y 2 2x 1 ) + w 1, (3) y = y 1 = x 1,, S 2 S 1, w 1, S 1 η 3, S 2, u j. 2. 2 Fig. 2 Fault tolerant control of the servo system driven by two motors 3. 3 Fig. 3 Flow chart of the decision block. : e 1 = y d x 1, e 2 = η 2 x 2, e 3 = η 3 ax 3, (5) η 2 x 2, η 3 ax 3, : 1, (5) ė 1 = ẏ d ẋ 1 = ẏ d η 2 + e 2. (6) e 1, η 2 η 2 = ẏ d + k 1 e 1, (7) : k 1, k 1 > 0. (7) (6) Lyapunov : ė 1 = k 1 e 1 + e 2. (8) V 1 = e2 1 2. (9) (9) (8) V 1 = k 1 e 2 1 + e 1 e 2. (10)
30 31 2, [15]. ESO w 1,. z 1 x 2, z 2, w 1, e 0 = z 1 x 2, ż 1 = a(x 3 2x 1 ) + z 2 β 1 e 0, ż 2 = β 2 fal(e 0, a 0, δ 0 ), (11) : β 1, β 2, fal(e 0, α 0, δ 0 ), ;,, [15] e 0 α0 sgn e 0, e 0 >δ 0 >0, fal(e 0, α 0, δ 0 ) = e 0 1 α δ 0, 0< e 0 δ 0, 0 0 < α 0 < 1, δ 0 > 0. (12) ESO w 1,, e w = w 1 z 2,, e w ê w = ẽ w, ê w e w, ẽ w, (5) ê w = ẽ w. (13) ė 2 = η 2 + 2ax 1 η 3 + e 3 w 1. (14) η 3 = e 1 + η 2 + 2ax 1 + k 2 e 2 z 2 ê w, (15) k 2, k 2 > 0. (15) (14) : ė 2 = e 1 k 2 e 2 + e 3 ẽ w. (16) ẽ w = r 0 e 2, (17) r 0, r 0 > 0, Lyapunov : V 2 = V 1 + e2 2 2 + ẽ2 w 2r 0. (18) (18) (16) (17) V 2 = k 1 e 2 1 k 2 e 2 2 + e 2 e 3. (19) t e 3, V 2 0, e 1, e 2, e 3. 3, S 1 = ė 3 + k 3 e 3, (20) k 3, k 3 > 0., : 1) f j 0.,,, e 3j = η 3 2 ax 3j. (21) S 1j = ė 3j + k 3 e 3j, (22) S 11 +S 12 =k 3 e 31 +ė 31 +k 3 e 32 +ė 32 =S 1. (23) ˆf j, f j, f j ˆf j = f t j, ˆfj = f j (0) f j dt. (24) 0 f j = r j aα 1j u j S 1, (25) r j, r j > 0, u j u j = F j + k 4j S 1 + aρ j sgn S 1 aα 1j ˆfj, (26) k 4j, k 4j > 0, sgn( ), F j F j = k 3 ( η 3 2 ax 4j) + η 3 2 aα 2j(x 1 x 3j ). (27) 2) f j = 0, f 1 = 0, f 2 0. 1, 1 2, f 1 = 0, (28) f 2 = r 2 aα 12 u 2 S 1. u 1 = 0, u 2 = F f + k 4 S 1 + a(ρ 1 + ρ 2 )sgn S 1, aα 12 ˆf2 : k 4, k 4 > 0, F f F f = k 3 [ η 3 a(x 41 + x 42 )] + η 3 + a[α 21 x 31 + (29) α 22 x 32 (α 21 + α 22 )x 1 ]. (30) 1 (2) 1 4,, (25) (26)(28) (29),. Lyapunov : V = V 2 + V S, V s = S 2 1 + f 2 1 2r 1 + f 2 2 2r 2. (31)
1 : 31 (31) (19) V = k 1 e 2 1 k 2e 2 2 + e 2e 3 + V s, V s = S 1 Ṡ 1 + f 1 f1 r 1 + f 2 f2 r 2. (32) 1) f j 0, (22) (4)(21) Ṡ 1j = F j aα 1j f j u j aw 2j = (23) V s F j aα 1j ( ˆf j + f j )u j aw 2j. (33) V s = S 1 (S 11 + S 12 ) + f 1 f1 r 1 (33) + f 2 f2 r 2. (34) V s = S 1 [F 1 aα 11 ˆf1 u 1 aw 21 + F 2 aα 12 ˆf2 u 2 aw 22 ] aα 11 f1 u 1 S 1 aα 12 f2 u 2 S 1 + f 1 f1 r 1 + f 2 f2 r 2. (35) (25) (26) (35) V s = (k 41 + k 42 )S 2 1 a(ρ 1 S 1 + w 21 S 1 + ρ 2 S 1 + w 22 S 1 ) < (k 41 + k 42 )S 2 1 0. (36) 2) f 1 = 0, f 2 0, (5) ė 3 = η 3 aẋ 3 = η 3 a(x 41 + x 42 ). (37) (37) (2) ë 3 = η 3 a(α 21 x 1 α 21 x 31 + w 21 + α 22 x 1 α 22 x 32 + α 12 f 2 u 2 + w 22 ). (38) (37) (38) (20) Ṡ 1 = F f aα 12 ( ˆf 2 + f 2 )u 2 a(w 21 + w 22 ). (39) (28) (29) (39) V s V s = S 1 [F f aα 12 ˆf2 u 2 a(w 21 + w 22 )] = k 4 S 2 1 a(ρ 1 + ρ 2 ) S 1 a(w 21 + w 22 )S 1 < k 4 S 2 1 0. (40), V s 0( S 1 0, Vs <0), S 1, e 3, lim t e 3 = 0, (32) V = V 2 + V s < k 1 e 2 1 k 2 e 2 2 min(k 4, k 41 +k 42 )S 2 1 0, (41),.. 1, : 1), ẋ 2 = a(x 3 2x 1 ) + a(x 3 2x 1 ), ẋ 4j = α 2j (x 1 x 3j ) + α 1j f j u j + α 2j (x 1 x 3j ) + α 1j f j u j,,,, a(x 3 2x 1 ) α 2j (x 1 x 3j ) + α 1j f j u j, 2. w 1 = a(x 3 2x 1 ), w 2j = α 2j (x 1 x 3j ) + α 1j f j u j. 2),, w 1, w 2.,,,. 4 (Actuator fault diagnosis based on adaptive sliding mode observers),,,,. x 4j ˆx 4j = α 2j (x 1 x 3j ) + α 1j ˆf j u j + ls 2 + ρ j sgn S 2, (42) : ˆx 4j x 4j, ˆf j, f j, f j = f j ˆf j, l, l > 0, S 2, S 2 = x 4j ˆx 4j. f j = r jα 1j u j S 2, (43) r j, r j > 0, : 2 x 4j (42), (43), l > 0, w 2j < ρ j,. (2) (42) Ṡ 2 = ls 2 + α 1j f j u j + w 2j ρ j sgn S 2. (44) Lyapunov : V o = S2 f 2 2 2 + j. (45) 2r j
32 31 (45) (43) (44) V o = S 2 Ṡ 2 + f j r j f j = S 2 [ ls 2 + α 1j f j u j + w 2j ρ j sgn S 2 ] α 1j f j u j S 2 = ls 2 2 + S 2 w 2j ρ j S 2 < ls 2 2 0, (46) lim S 2 = 0,. t 5 (Simulation)., MATLAB 7.1. : J m1 = 4 10 3 kg m 2, K T1 = 1.1 N m/a, J m2 = 8 10 3 kg m 2, K T2 = 0.9 N m/a. J L = 1.6 10 2 kg m 2, K L = 1000 N m/rad., k 1 = 50, k 2 = 30, k 3 = k 41 = k 42 = 30, k 4 = 50, r 0 = 10, r 1 = r 2 = 5, r 1 = r = 8000, l = 90, 2 ρ 1 = ρ 2 = 0.5, β 1 = 50, β 2 = 5000, α 0 = 0.3, δ 0 = 0.1. 1 rad, 4 s, 4(a)., Stribeck, 4(b), 4(c) ESO,, 0.016 rad, 0.009 rad; ESO, 0.0036 rad, 4(d)..,, 4 s 1, 1 0.5, 5(a), 0.2 s, u 1, u 2, 5(b).,, 0.0052 rad, 5(c);,, 5(d). (a) (b) Stribeck (c) (d) 4 Fig. 4 System response with nonlinear friction
1 : 33 (a), 4 s 1, 0.25 s, 6(a), u 1, u 2,, 6(b). 1,, 0.0068 rad, 6(c);, 0.001 rad,, 6(d). 4 6,. (b) (a) (c) (b) (d) 5 Fig. 5 System response when an actuator partly fails to work (c)
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