tailieunhanh - Sabatier Agrawal Machado Advances in Fractional Calculus Episode 15

Tham khảo tài liệu 'sabatier agrawal machado advances in fractional calculus episode 15', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | FRACTIONAL-ORDER BOUNDARY CONTROL 545 The control objective is to stabilize u x t given the initial conditions 4 and 5 . We adopt the following Caputo definition for fractional derivative of order a of any function f t because the Laplace transform of the Caputo derivative allows utilization of initial values of classical integer-order derivatives with known physical interpretations 19 20 daf t 1 r f w t dT dta r a - n Jo t - T - 6 where n is an integer satisfying n 1 a n and r is the Euler s gamma function. In this paper we study the robustness of the controllers in the following format . dMu 1 t f t k z 0 p 1 7 where k is the controller gain ự is the order of fractional derivative of the displacement at the free end of the cable. Based on the definition 6 the Laplace transform of the fractional derivative is 19 20 saF s np f k 0 sa-1-k k 0 8 In the following the transfer function from the boundary controller f t to the tip end displacement will be derived for later use. Assuming zero initial conditions of u x 0 and ut x 0 take the Laplace transform of 1 2 and 3 with respect to t making use of 8 the original PDE of u x t with initial and boundary conditions can be transformed into the following ODE of U x s with boundary conditions. s U x s 0 9 U 0 s 0 10 Ux 1 s F s 11 where U x s is the Laplace transform of u x t and F s is the Laplace transform of f t . Solving the ODE 9 we have the following solution of U x s with two arbitrary constants C1 and C2 s can be treated as a constant in this step . U x s C1 exs2 C2e-xs2 . 12 Substitute 12 into 10 and 11 we have the following two equations. C1 C2 0 13 546 Liang Zhang Chen and Podlubny a a s2 C1 es 2 - C2e-s 2 F s . 14 Solving 13 and 14 simultaneously we can obtain the exact value of Cl and C2 a F s es2 Cl C2 a 15 s2 e2s 2 1 Now we have obtained the solution of U x s . Substituting x 1 into U x s and divide U x s by F s we obtain the following transfer function of the fractional wave equation P s P s U 1 s F s a 1 - .