The straight-line navigation control of an agricultural tractor subject to input saturation

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30 10 013 10 DOI: 10.7641/CTA.013.30034 Control Theory & Applications Vol. 30 No. 10 Oct. 013 1,, 1, 1, 3 (1., 1013;., 10096; 3., 10031 ) :,,,,,,,, :,, : ; ; ; : TP73 : A The straight-line navigation control of an agricultural tractor subject to input saturation DING Shi-hong 1,, JIANG Yue-xia 1, ZHAO De-an 1, LIN Xiang-ze 3 (1. Key Laboratory of Facility Agriculture Measurement and Control Technology and Equipment of Machinery Industry, Jiangsu University, Zhenjiang Jiangsu 1013, China;. Key Laboratory of Measurement and Control of Complex Systems of Engineering, Ministry of Education, Southeast University, Nanjing Jiangsu 10096, China; 3. College of Engineering, Nanjing Agricultural University, Nanjing Jiangsu 10031, China) Abstract: For an agricultural tractor, a linear path-tracking navigation algorithm is developed by using the saturation method, providing an effective solution to the ubiquitous input saturation control problem in agricultural tractor navigation. First, based upon the kinematics of the straight-linear navigation system, a linear system model for control design is obtained by linearization. Secondly, a non-saturated controller is designed to stabilize the system globally. Finally, by integrating a saturation function with the non-saturated controller, we obtain a saturated controller under which the closed-loop system is globally asymptotically stable. This controller is simple in structure without the need of sequential decrement of restriction on the saturation levels from outside to inside, thus improving the convergence of the closed-loop system. Simulation results validate the effectiveness of the proposed method. Key words: agricultural tractor; path tracking; controller design; stability 1 (Introduction),,,.,,, [1].,,.,, []., [3 4]. : : 013 01 11; : 013 04 09.. E-mail: dsh@mail.ujs.edu.cn; Tel.: +86 511-8879145. : (6103014, 6103054); (PAPD, (011)6); (011371000); (MCCSE01A01); (BK0183).

188 30,,.,, [5].,, [6],. [7] X 804, PID, (difference global positioning system, DGPS),. PID, [8], [9], [10],,., [11],.,,,,,,,,,,, [1].,,., ([13 15]), (3.1).,.,,,,,,,,,,,.,,. (System modelling and problem formulation),, 1 [8, 16]. 1 Fig. 1 The sketch map of agricultural tractor straight-line navigation 1: Ψ(rad), V x (m/s), L(m), δ(rad), x(m), y(m). 1, Ψ = V x tan δ, L ẋ = V x cos Ψ, (1) ẏ = V x sin Ψ, : Ψ, V x, L, δ, x, y1. (1),,,,,,, 1, y 0 Ψ 0, yψ., : ẏ 0 V x 0 Ψ = 0 0 V y 0 x L Ψ + 0 u, () δ 0 0 0 δ 1 : u, δ. : u max > 0, u u u max, y 0Ψ 0. 1, PID

10 : 189., PID., (1),,, (1), ().,.,,,,.,,. 3 (Controller design) x 1 = y, x = V x Ψ, x 3 = (Vx /L)δ,, () 1 = x, = x 3, 3 = V (3) x L u = v, v. I) v(3). (3), : v = k 3 x 3 k 3 k x k 3 k k 1 x 1, (4) s 3 + k 3 s + k 3 k s + k 3 k k 1 = 0 Hurwitz, (3) (4). II) (4), u.,,,.,,. σ εi (x) = { (sgn x)ε i, x > ε i, x, x ε i, (4), : v = k 3 σ ε3 (x 3 + k σ ε (x + k 1 σ ε1 (x 1 ))), (5) k 3 ε 3 k [ε 3 + (k 1 + k )ε + k1ε 1 ] > 0, k ε > ε 3 + k 1 (ε + k 1 ε 1 ), k 1 ε 1 > ε. S 3 (t) = x 3 + k σ ε (x + k 1 σ ε1 (x 1 )), S (t) = x + k 1 σ ε1 (x 1 ), S 1 (t) = x 1. (6), (5), (3). 3. 1 T 1, t T 1, S 3 (t) = x 3 + k σ ε (S ) ε 3.,, T 1 S 3 (T 1 ) ε 3., t 0, S 3 (t) > ε 3 /, S 3 (t) > ε 3, t > 0, (7) S 3 (t) < ε 3, t > 0. (8) S 3 (t) > ε 3 /, t > 0, (3)(5) 3 = k 3 σ ε3 (S 3 ) < k 3ε 3. (9) (9)x 3 (t) x 3 (0) < k 3 ε 3 t/., t > 0, x 3 (t) < x 3 (0) k 3 ε 3 t/., S 3 S S 3 = x 3 + k σ ε (S ) x 3 (0) k 3ε 3 t + k ε., t, S 3, (7). (7). (8)., T 1 S 3 (T 1 ) ε 3 /. t T 1 S 3 (t) = x 3 + k σ ε (S ) ε 3. Ω 1 = {x : S 3 (t) = x 3 + k σ ε (S ) ε 3 }. Ω 1,,,, P 1 P P 3 P 4 Ω 1, Q 1 Q Q 3 Q 4 Ω 1.. S x 3 Fig. The phase plot of S x 3 x P 1 P, S ε. Ṡ 3 = ẋ 3 k ε = k 3 σ ε3 (S 3 ) = k 3 ε 3 < 0, x P 1 P. (10)

190 30 Ṡ 3 = ẋ 3 + k ε = k 3 σ ε3 (S 3 ) = k 3 ε 3 < 0, x P 3 P 4. (11) (10) (11), xp 1 P P 3 P 4 Ω 1., x Q 1 Q Q 3 Q 4 Ω 1. xp P 3 Q Q 3 Ω 1. x P P 3, S 3 = ε 3, S ε., Ṡ 3 = ẋ 3 + k Ṡ = k 3 S 3 + k Ṡ = k 3 ε 3 + k Ṡ, x P P 3. (1) x {x : S 3 ε 3, S ε }, Ṡ x 3 + k 1 σ ε1 (x 1 ) x 3 + k 1 σ ε1 (x 1 ). σ ε1 (x 1 ), σ ε1 (x 1 ) = 0, x 1 > ε 1, σ 1 (x 1 ) = x, x 1 ε 1., x {x : S 3 ε 3, S ε }, σ 1 (x 1 ) ε + k 1 ε 1. (13) x {x : S 3 ε 3, S ε }, Ṡ x 3 + k 1 σ ε1 (x 1 ) ε 3 + k ε + k 1 (ε + k 1 ε 1 ) = ε 3 + (k 1 + k )ε + k 1ε 1. (14) x P P 3 x {x : S 3 ε 3, S ε }. (14)(1) S 3 k 3 ε 3 +k [ε 3 +(k 1 +k )ε +k 1ε 1 ], x P P 3. (6)k 3 ε 3 k [ε 3 + (k 1 + k )ε + k 1ε 1 ] > 0., (6) S 3 < 0, x P P 3. (15) S 3 > 0, x Q Q 3. (16) (15) (16), xp P 3 Q Q 3., xω 1. : t T 1, S 3 (t) ε 3. T > T 1 t T, S 3 (t) ε 3, S (t) ε. T > T 1, t T S 3 (t) ε 3.,, T > T 1 t T, S (t) ε., T > T 1 S (T ) < ε., t > T 1, S (t) ε., S (t) ε, t > T 1 S (t) ε, t > T 1. 1. 1 t > T 1, S 3 (t) ε 3. t > T 1, ẋ = x 3 = k σ ε (S ) + x 3 + k σ ε (S ) = k σ ε (S ) + S 3 k σ ε (S ) + ε 3. (17) S (t) ε 3, t > T 1, (17) ẋ k σ ε (S ) + ε 3 k ε + ε 3, x (t) x (0) (k ε ε 3 )t. (18) (6), k ε > ε 3., (18)S (t) S (t) = x (t)+k 1 σ ε1 (x 1 ) x (0) (k ε ε 3 )t + k 1 ε 1. t, S <, S ε, t > T 1., S (t) ε, t > T 1., T > T 1 S (T ) < ε. t > T, S (t) ε. Ω = {x : S (t) = x + k 1 σ ε1 (x 1 ) ε }. Ω,, 3,, H 1 H H 3 H 4 Ω, L 1 L L 3 L 4 Ω. 3 S 1 x Fig. 3 The phase plot of S 1 x x H 1 H, S 1 = x 1 < ε 1, S = x + k 1 σ ε1 (x 1 ) = x k 1 ε 1., x H 1 H, Ṡ = ẋ = x 3. x(t ) Ω. H 1 H Ω, t > T > T 1, x(t ) H 1 H., t = t, S 3 (t ) ε 3. x 3 (t ) = S 3 (t ) k σ ε (S (t )) ε 3 k ε. k ε ε 3 > 0., x H 1 H, Ṡ = ẋ = x 3 < ε 3 k ε < 0, xh 1 H Ω. xh 3 H 4, L 1 L, L 3 L 4 Ω. x H H 3, x 1 ε 1, Ṡ = ẋ + k 1 σ ε (x 1 ) = x 3 + k 1 x = k σ ε (S ) + S 3 + k 1 x.

10 : 191 x H H 3, S = ε., x H H 3, x ε + k 1 ε 1., x H H 3, Ṡ k ε + ε 3 + k 1 x (6) k ε + ε 3 + k 1 (ε + k 1 ε 1 ). k ε > ε 3 + k 1 (ε + k 1 ε 1 ), S < 0, x H H 3., H H 3 Ω. L L 3., xh 1 H H 3 H 4 L 1 L L 3 L 4 Ω, t T, 3 S 3 (t) ε 3, S (t) ε. T 3 > T t T 3, S 3 (t) ε 3, S (t) ε, x 1 (t) ε 1. 1, t T, S (t) ε, S 3 (t) ε 3, t T, 1 = x = k 1 σ ε1 (x 1 ) + x + k 1 σ ε1 (x 1 ) = k 1 σ ε1 (x 1 ) + S. (19) (19)(6) ẋ 1 < k 1 ε 1 + ε < 0, x 1 > ε 1, ẋ 1 > k 1 ε 1 ε > 0, x 1 < ε 1, T 3, t T 3, x 1 (t) ε 1, S (t) ε, S 3 (t) ε 3, t > T 3, (5) (4)., t > T 3 (3)(5) 1 = x, = x 3, 3 = k 3 σ ε3 (x 3 + k σ ε (x + k 1 σ ε1 (x 1 ))) = k 3 x 3 k 3 k x k 3 k k 1 x 1. I) s 3 + k 3 s + k 3 k s + k 3 k k 1 = 0 (0) Hurwitz.,, (5), (3). u = k 3L V x σ ε3 ( V x L δ + k σ ε (V x Ψ + k 1 σ ε1 (y))) (1), (), 1 (1), ().,, ()., Teel [13], : u = L Vx σ ε3 (y 3 + σ ε (y + σ ε1 (y 1 ))), () : ε 3 > ε, ε > ε 1, y 1 =y + V xψ + V xδ, y = V xψ + V xδ, y 3 = V xδ. ε i, i = 1,, 3ε 3 > ε > ε 1., y (0), y 3 (0), y 3 (0), (), Lε 1 /V x. ε 1,, 1,. 4 (Simulation), MATLAB,. (1), (1)(),..4 m, V x = 1 m/s., u rad/s. (6), (1). k 1 = 1 5, k = 1, k 3 = 5 3, () ε 1 = 3, ε = 1, ε 3 = 1 10. ε 3 = 5 6, ε = 1 3, ε = 1 1. (x(0), y(0), Φ(0), δ(0)) = (4, 3, π 4, π 4 ), 5%()., 4 9, : (1), (). 4 6 X = (y, Φ, δ), X = y + Φ + δ 7. 8

19 30. 9 4 9,, (), (1)., 4 85%,. 9,., 7 X Fig. 7 The response curves of X 4 Fig. 4 The response curves of lateral deviation 8 y-x Fig. 8 The tractor tracking phase plot of y-x 5 Fig. 5 The response curves of course angle 9 Fig. 9 The response curves of controller 6 Fig. 6 The response curves of steering angle 5 (Conclusions),,,,,.,

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