Considerations on model predictive control of McKibben pneumatic artificial muscle

1. May 22 Friday, 2015 @ 2 From 15:30-15:45 McKibben 58 SCI15 2. 2 3. 3 [1] B. Tondu et al., 2000. [1] m D L0 . P0 P [HUR ][ ] McKibben () () 4. 4 [2] T. V. Minh et al., 2010. [3] A. P-Arrese et al., 2010. , . [4] G. Andrikopoulos et al., 2014. () [2] [3] [4] [5,6] [5] Kogiso et al., 2013. [6] , 2015. 5. 5 x(t) = f (x(t), u(t)) y(t) = h(x(t)) x(t) 2 S x := [ P]T if 12[5] proportional directional control valve L M M L l0 air compressor Fig. 1. Illustration of considered PAM system. equilibrium where neither players want to deviate from respective utility values. The resulting parameters are a l (or might be a global) optimum to the optimization prob and then the model of the PAM system can be automatic identied. II. PHYSICAL MODELING OF THE PAM SYSTEM This paper considers a PAM system with a vertic suspended weight and a proportional directional con valve, as illustrated in Fig. 1. A model of the considered P system is in a switched system with 32 nonlinear subsyst f : 3 3 , x(t) = f(x(t), u(t)) if x(t) X, y(t) = 1 0 0 0 0 1 x(t), where the state variable x 3 and output vari (measured by sensors) y 2 are respectively dene m(t) = (t) mi(t) (1 (t)) mo(t) mi(t) = 8 >>>>>>>>>>< >>>>>>>>>>: A0 Ptankp T r k R 2 k+1 k+1 k 1 if P(t) 2 k+1 k k 1 Ptank, A0 Ptankp T q 2k R(k 1) P (t) Ptank 1 k r 1 P (t) Ptank k 1 k if P(t) > 2 k+1 k k 1 Ptank, () Ff (t) = 8 >>>< >>>: cvL(t) + ccsgn((t)), if (t) 6= 0, cc, if (t) = 0 and Fo(t) > cc, Fo(t), if (t) = 0 and Fo(t) 2 [ cc, cc], cc, if (t) = 0 and Fo(t) < cc, ML(t) = 8 >>< >>: F(P, , t) Mg Ff (t) K(L0 L(1 (t)))3 , if (t) L L0 L , F(P, , t) Mg Ff (t), otherwise, [Kikuue, IEEE TRO 06] V (t) = D1(t)2 + D2(t) + D3 [Kagawa, CEP 97], [Minh, Mechatronics 10] P(t) = k1 RT V (t) m(t) k2 V (t) V (t) P(t) [Richer, JDSMC 00] F(P, , t) = APg(t) at n a Cq1 1 + eCq2 Pg(t) (t) o2 as [Tondu, IEEE CSM 00], [Kang, ICRA 09] [Itto, IEEE ICIT11] y := [ P]T P 6. 6 control valve inner gas V + + + - PAM with weight 1/s 1/ML Ffric elasticity yPu tank Pout P PAM L 1/s 1/s 1/s 7. 7 J(xt|t, uN t ) = NX j=1 ||t|t+j ||2 2,Q + ||Pt|t+j P||2 2,QP + ||t|t+j||2 2,Q contraction velocityerror of contraction ratio error of pressure Voltage of minimum aperture x(t + 1) = f (x(t), u(t)) if x(t) 2 X y(t) = 1 0 0 0 0 1 x(t) u u(t) u Voltage of max aperture () () 8. 8 (MPC) v.s. PID(PID) MPC Matlab 2014b, fmincon, iMac Core i5 3.2GHz 16GB 50 103 sin (t/20) + 200 [kPa] 150 103 sin (2t/20) + 450 [kPa] 0.04 sin (2.5t/20) + 0.20 [ ] 0.02 sin (t/20) + 0.10 [ ] N = 1, QP = 1, Q = 1 N = 5, Q = 108 , Q = 6.0 104 () 10 [ms] air compressor pressure tank PAM pressure sensor control valve LDM flow meter 9. 9 MPC v.s. PID 0 5 10 15 20 25 30 35 40 100 200 300 400 500 600 700 Time [s] Pressure[kPa] Reference MPC PID 0 5 10 15 20 25 30 35 40 100 200 300 400 500 600 700 Time [s] Pressure[kPa] Reference MPC PID 50 kPa /20 150 kPa 2 /20 MPC 1313 MPC 1306 10. 10 MPC v.s. PID 0 5 10 15 20 25 30 35 40 0 0.05 0.1 0.15 0.2 0.25 0.3 Time [s] ContractionRatio[-] Reference MPC PID 0 5 10 15 20 25 30 35 40 0 0.05 0.10 0.15 0.20 0.25 0.30 Time [s] ContractionRatio[-] Reference MPC PID 0.02 /20 0.04 2.5 /20 MPC 150447 MPC 172652 11. 11 0 5 10 15 20 25 30 35 40 100 200 300 400 500 600 700 Time [s] Pressure[kPa] Simulation Experiment 0 5 10 15 20 25 30 35 40 0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 Time [s] ContractionRatio[-] Reference Simulation Experiment 0 5 10 15 20 25 30 35 40 0 1 2 3 4 5 6 Time [s] InputVoltage[V] Input sequences 0 5 10 15 20 25 30 35 40 100 200 300 400 500 600 700 Time [s] Pressure[kPa] Reference Simulation Experiment 0 5 10 15 20 25 30 35 40 0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 Time [s] ContractionRatio[-] Simulation Experiment 0 5 10 15 20 25 30 35 40 0 1 2 3 4 5 6 Time [s] InputVoltage[V] Input sequences (:15 %) (:500 kPa) 12. 12 PID () M 2M torqueend-effect M M M M 2M 13. 13 14. 14 x(t) = f (x(t), u(t)) y(t) = h(x(t)) x(t) 2 S x := [ P]T y := [ F]T if L0 D0 D1 D2 D3 M K R T cc A0 Cq1 Cq2 k1 k2 cv Ptank Pout k : natural diameter of PAM [m] : natural length of PAM [m] : source absolute pressure [Pa] : coecients for PAM volume [m^3] : atmospheric pressure [Pa] : specic heat ratio for air [-] : ideal gas constant [J/kg K] : absolute temperature [K] : coecient of elasticity [N/m^3] : initial angle btw braided thread & cylinder long axis [deg] : correction coecient [-] : correction coecient [1/Pa] : Coulomb friction [N] : viscous friction coecient [Ns/m] : orice area of PDC valve [m^2] : polytropic indexes [-] : mass of weight [kg] , : McKibbne, , Vol. 26, No. 7, 2013. 12 : For the PAM system model, its steady-state behavior is characterized by and its transient behavior is characterized by A0 k1 k2 cv parameters: parameters: Cq2(M)K(M) (M) Cq1(M) cc(M) 15. 15 4 ActiveLink TAA10 FESTO 20-200 FESTO10-250N AirMuscle ActiveLink, TAA10 250 ActiveLink, AMT-S 250 Festo, DMSP-10-200N-RM-CM 200 , : McKibbne, , Vol. 51, No. 4, 2015. 16. 16 0 0.05 0.10 0.15 0.20 0.25 contractionratio 100 200 300 400 500 600 700 pressure [kPa] 100 200 300 400 500 600 700 pressure [kPa] 0 0.05 0.10 0.15 0.20 0.25 contractionratio 100 200 300 400 500 600 700 pressure [kPa] 0 0.05 0.10 0.15 0.20 0.25 contractionratio 0 0.05 0.10 0.15 0.20 0.25 contractionratio 100 200 300 400 500 600 700 pressure [kPa] ActiveLink TAA10 FESTO 20-200 AirMuscle -1709 -3980 -3650-3550 FESTO 10-250 [, 15] 17. 17 ActiveLink TAA10 FESTO 20-200 AirMuscle FESTO 10-250 0 0.1 0.2 0.3 2 3 4 5 6 7 x10 -5 contraction ratio PAMvolume[m3 ] 0 0.05 0.1 0.15 contraction ratio x10 -4 1.8 1.9 2.0 2.1 2.2 2.3 PAMvolume[m3 ] 0 0.05 0.1 0.15 0.2 0.25 0.5 0.7 0.9 1.1 1.3 1.5 x10 -4 contraction ratio PAMvolume[m3 ] 0 0.05 0.1 0.15 0.2 0.25 1 2 3 4 5 6 x10 -5 contraction ratio PAMvolume[m3 ]