Advanced Mechatronics Engineering

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1 Advanced Mechatronics Engineering German University in Cairo 21 December, 2013

2 Outline Necessary conditions for optimal input Example Linear regulator problem Example

3 Necessary conditions for optimal input The problem is to find an optimal input (u ) that causes the system ẋ(t) = f(x(t), u(t), t), (1) to follow a trajectory (x ) that minimizes the performance measure tf J(x(t), t) = h(x(t f ), t f ) + g(x(τ), u(τ), τ)dτ. (2) t 0

4 Necessary conditions for optimal input The Hamiltonian is given by H(x(t), u(t), p(t), t) g(x(t), u(t), t) + p T (t) [f(x(t), u(t), t)]. (3) The necessary conditions of optimality are ẋ (t) = H p (x (t), u (t), p (t), t), (4) ṗ (t) = H x (x (t), u (t), p (t), t), (5) 0 = H u (x (t), u (t), p (t), t). (6)

5 Example The system ẋ 1 (t) = x 2 (t) (7) ẋ 2 (t) = x 2 (t) + u(t) (8) is to be controlled so that its input is conserved. Therefore, the performance measure is given by J(u) = tf The Hamiltonian (36) is given by t u2 (t)dt. (9) H(x(t), u(t), p(t), t) = 1 2 u2 (t)+p 1 (t)x 2 (t) p 2 (t)x 2 (t)+p 2 (t)u(t). (10)

6 Example H(x(t), u(t), p(t), t) = 1 2 u2 (t)+p 1 (t)x 2 (t) p 2 (t)x 2 (t)+p 2 (t)u(t). The necessary conditions for optimality are (11) ṗ 1(t) = H x 1 = 0 (12) ṗ 2(t) = H x 2 = p 1(t) + p 2(t), (13) and 0 = H u = u (t) + p 2(t). (14)

7 Linear Regulator Problems The plant is described by the linear state equations ẋ(t) = A(t)x(t) + B(t)u(t) (15) which may have time-varying coefficients. The performance measure to be minimized is J = 1 2 xt (t f )Hx(t f ) + tf t [ x T Qx + u T Ru ] dt, (16) where the final time t f is fixed. Further, H and Q are real symmetric positive semi-definite matrices. Finally, R is a real symmetric positive definite matrix. The Hamiltonian is H(x(t), u(t), p(t), t) g(x(t), u(t), t) + p T (t) [f(x(t), u(t), t)]. (17)

8 Linear Regulator Problems H(x(t), u(t), p(t), t) = 1 2 xt Qx+ 1 2 ut Ru+p T A(t)x(t)+p T B(t)u(t), (18) and the necessary conditions for optimality are ẋ (t) = A(t)x (t) + B(t)u (t), (19) ṗ (t) = Q(t)x (t) A T (t)p (t), (20) 0 = R(t)u (t) + B T (t)p (t). (21) Solving (54) for u (t) yields u (t) = R 1 (t)b T (t)p (t). (22) Substitution of (55) into (52) yields ẋ (t) = A(t)x (t) B(t)R 1 (t)b T (t)p (t). (23)

9 Linear Regulator Problems Putting (53) and (56) into the following matrix format ẋ (t) A(t) B(t)R 1 (t)b T (t) x (t) =. (24) ṗ (t) Q(t) p (t) p (t) The solution of these equations has the following form x (t f ) x (t) = ϕ(t f, t), (25) p (t f ) p (t) where ϕ(t f, t) is the state-transition matrix of the system (57). x (t f ) ϕ 11 (t f, t) ϕ 12 (t f, t) x (t) =, (26) p (t f ) ϕ 21 (t f, t) ϕ 22 (t f, t) p (t)

10 Linear Regulator Problems From the boundary condition, the final co-states are related to the final states using p (t f ) = Hx (t f ). (27) Solving for p (t f ), we obtain p (t) = [ϕ 22 (t f, t) Hϕ 12 (t f, t)] 1 [Hϕ 11 (t f, t) ϕ 21 (t f, t)] x (t). (28) p (t) = K(t)x (t). (29) The optimal input is given by u (t) = R 1 (t)b T (t)k(t)x (t). (30)

11 Example It is desired to determine the input (using the principle of optimality and the Hamilton-Jacobi-Bellman equation) that causes the plant ẋ 1 = x 2 (t) (31) ẋ 2 = x 1 (t) 2x 2 (t) + u(t) (32) to minimize the performance measure J = 10x 2 1 (T ) T 0 [ x 2 1 (t) + 2x 2 2 (t) + u 2 (t) ]. (33)

12 State Transition Matrix Consider the scalar case ẋ(t) = ax(t). (34) Taking the Laplace transform of (34), we obtain sx (s) x(0) = ax (s), (35) X (s) = x(0) s a = (s a) 1 x(0). (36) Finally, inverse Laplace transform of (36) yields x(t) = e at x(0). (37)

13 State Transition Matrix Now consider the following homogenous state equation ẋ(t) = Ax(t). (38) sx(s) x(0) = AX(s), (39) X(s) = (si A) 1 x(0). (40) The inverse Laplace transform yields x(t) = L 1 [ (si A) 1] x(0) = e At x(0). (41) Therefore, the state transition matrix (e At ) is given by e At = L 1 [ (si A) 1]. (42)

14 State Transition Matrix Calculate the state transition matrix of the following system ] [ẋ1 = ẋ 2 [si A] = [si A] 1 = [ ] [ ] x1 x 2 [ ] (s + 1) 0 2 (s + 3) (43) (44) [ (s+3) (s+1)(s+3) 0 2 (s+1) (s+1)(s+3) (s+1)(s+3) ] = [ 1 (s+1) 0 ( 1 (s+1) 1 (s+1) ) 1 (s+3) e At = L 1 [ (si A) 1], (45) [ e At e = t ] 0 (e t e 3t ) e 3t. ]

15 State Transition Matrix Calculate the state transition matrix of the following system ] [ ] [ ] [ẋ1 0 1 x1 = 2 3 ẋ 2 [si A] = x 2 [ ] s 1 2 (s + 3) (46) (47) e At = L 1 [ (si A) 1], (48) [ 2e = t e 2t e t e 2t ] 2e t + 2e 2t e t + 2e 2t. [si A] 1 = [ (s+3) (s+1)(s+2) 2 (s+1)(s+2) 1 (s+1)(s+2) s (s+1)(s+2) ]

16 State Transition Matrix If the matrix A can be transformed into a diagonal form, then the state transition matrix e At is given by e λ 1t e At = Pe Dt P 1 0 e λ 2t... 0 = P P 1, (49) e λnt where P is a digonalizing matrix for A. Further, λ i is the ith eigenvalue of the matrix A, for i = 1,..., n.

17 State Transition Matrix Derivation: Consider the following homogenous state equation ẋ = Ax, (50) and the following similarity transformation: x = Pξ, ẋ = P ξ. (51) Substituting (51) in (50) yields ξ = P 1 APξ = Dξ. (52) Solution of (52) is using (51) ξ(t) = e Dt ξ(0), (53) x(t) = Pξ(t) = Pe Dt ξ(0), x(0) = Pξ(0). (54) Therefore x(t) = Pe Dt P 1 x(0) = e At x(0). (55)

18 State Transition Matrix Calculate the state transition matrix of the following system ] [ẋ1 = ẋ 2 [ ] [ ] x1 x 2 (56) The eigenvalues of A are λ 1 = 0 and λ 2 = 2. A similarity transformation matrix P is [ ] 1 1 P =. (57) 0 2 Using (49) to calculate the state transition matrix e At = Pe Dt P 1 (58) [ ] [ ] [ ] 1 1 e = e 2t [ e At 1 1 = 2 (1 ] e 2t ) 0 e 2t. (59)

19 Thanks Questions please

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