Regional Solution of Constrained LQ Optimal Control

Transcription

1 Regional Solution of Constrained LQ Optimal Control José DeDoná September 2004

2 Outline 1 Recap on the Solution for N = 2 2 Regional Explicit Solution Comparison with the Maximal Output Admissible Set 3 Example 4 Conclusions

3 Motivation In the previous part, we provided a global characterisation of receding horizon constrained optimal control for horizon N = 2. Later in the course, we will extend the solution to the case of arbitrary horizon. This gives practically valuable insights into the form of the control law. Indeed, for many problems it is feasible to compute, store and retrieve the function u = N (x), thus eliminating the need to solve the associated QP on line. In this part we address a different question: Given that we only ever apply the first move from the optimal control sequence of length N, is it possible that the first element of the control law might not change as the horizon increases beyond some modest size, at least locally in the state space?

4 Recap on the Solution for N = 2 Theorem (RHC Characterisation for N = 2) The RHC law has the form sat (Gx + h) if x Z, 2 (x) = sat (x) if x Z, sat (Gx h) if x Z +, (1) where the gains, G R 1 n and the constant h R are = R 1 B t PA, G = + BA 1 + (B) 2, h = B 1 + (B) 2, and the sets (Z, Z, Z + ) are defined by Z = {x : (A B)x < }, Z = {x : (A B)x }, Z + = {x : (A B)x > }.

5 Recap on the Solution for N = 2 Thus, the solution for N = 2 is given by: sat (Gx + h) if x Z, 2 (x) = sat (x) if x Z, sat (Gx h) if x Z +, We now ask the following question: Under what conditions is the control law u k = sat (x k ) optimal for the receding horizon control problem, and for arbitrary horizon N?

6 Regional Explicit Solution Consider the fixed-horizon optimisation problem for arbitrary horizon N: P N (x) : V opt N (x) = min V N({x k }, {u k }), subject to: x 0 = x, x k+1 = Ax k + Bu k for k = 0, 1,...,N 1, u k U = [, ] for k = 0, 1,...,N 1, with objective function: V N ({x k }, {u k }) = 1 2 xt N Px N N 1 ( ) x t k Qx k + u t k Ru k. k=0

7 Regional Explicit Solution We have already shown, for N = 2, that for all x Z, where u opt k = sat (x k ), k = 0, 1, Z = {x : (A B)x }. We can see that this characterisation is valid locally, i.e., inside a region Z, defined by linear inequalities.

8 Regional Explicit Solution To extend the regional solution u opt = sat k (x k ) to larger horizons, we need to define a number of additional sets in the state space. This is no surprise, since as we increase the horizon we would expect additional sets of inequalities to be required in order for the solution u opt = sat k (x k ) to still be valid. Define the gains i = A i 1 (A B), i = 1, 2,...,N 1. Also, define the sets X i by the linear inequalities: X i = { x : i x i, i x i }, i = 1, 2,...,N 1. (2) where the saturation bounds i are defined as i 2 i 1 + A k B, i = 1, 2,...,N. (3) k=0

9 Regional Explicit Solution Also define the nonlinear mapping φ nl (x) = Ax Bsat (x), and the sets: Y i = i 1 j=1 X j, i = 2, 3,...,N, (4) Z = { x : φ k nl (x) Y N k, k = 0, 1,...,N 2 }, for N 2. (5)

10 Regional Explicit Solution Theorem Consider the fixed horizon optimal control problem P N (x). Then for all x Z the optimal control sequence {u opt 0,...,uopt N 1 } is u opt k = sat (x k ), for k = 0, 1,...,N 1, where x k = φ k nl (x). The proof follows the same arguments as in the proof of the case N = 2 and uses induction. (More involved.)

11 Regional Explicit Solution We now consider the receding horizon control (RHC) implementation of the previous (fixed horizon) control law. That is, given the control sequence {u opt 0,...,uopt } that achieves the N 1 minimum in P N (x), the control applied to the plant is N (x) = u opt 0. We need to introduce the concept of an invariant set for the closed loop system x k+1 = φ nl (x k ) = Ax k Bsat (x k ). To this end, define the set Z = { x : φ k nl (x) Y N, k = 0, 1, 2,... }, N 2. The set Z is the maximal positively invariant set contained in Z and Y N for the closed loop system x k+1 = φ nl (x k ) = Ax k Bsat (x k ).

12 Regional Explicit Solution We then have the following result: Theorem For all x Z the RHC law N is given by N (x) = sat (x). The proof follows from the fact that Z Z and that Z is positively invariant under the control N (x) = sat (x).

13 Comparison with the Maximal Output Admissible Set Consider the maximal output admissible set O, defined as the set where constraints are always satisfied with the linear controller u = x: O { x : (A B) i x for i = 0, 1,... }. (6) It can be shown that the set Z, where the control law N (x) = sat (x) is optimal, is bigger than the maximal output admissible set O, as we will prove next. In fact, Z can be considerably bigger than O, as we will illustrate with an example.

14 Comparison with the Maximal Output Admissible Set Proposition Proof. O Z. Since Z is the maximal positively invariant set in Y N, it suffices to show that O Y N N 1 i=1 X i (see (4)). Assume, therefore, that x O, so that (see (6)) A j x, j = 0, 1,..., (7) where A = A B.

15 Comparison with the Maximal Output Admissible Set Proof (Ctd.) For any i {1, 2,...,N 1}, A i = (A B)A i 1 = A(A B)A i 2 = A 2 (A B)A i 3 = A 3 A i 3 = AA i 1 A 2 BA i 3 BA i 1 BA i 1 ABA i 2 = A 2 A i 2 ABA i 2. i 2 = A i 1 A A j BA i 1 j, j=0 BA i 1 ABA i 2 BA i 1 BA i 1

16 Comparison with the Maximal Output Admissible Set Proof (Ctd.) From the previous expression, we obtain i 2 A i 1 A x = A i x + j=0 From (7) and (8), we obtain the inequality i 2 A i 1 A x A i x + j=0 A j BA i 1 j x. (8) A j B A i 1 j x (9) i A j B = i (see (3)). (10) j=0 This implies x X i for all i {1, 2,...,N 1} (see (2)), yielding the desired result.

17 Example Consider the system x k+1 = Ax k + Bu k with [ ] [ ] A =, B =, =1, N = 10, Q = I and R = Y N = N 1 i=1 X i x 2 k O Y N E O = maximal output admissible set E = positively invariant ellipsoid O E Z Y N. x 1 k

18 Example Simulation with initial condition x 0 = [ ] t. Notice that the control remains saturated during the initial three steps x 2 k 0 uk x 1 k Shown in the figures are the sequences obtained with numerical (QP) RHC and with the control law u k = sat (x k ). (Both sequences coincide.) k

19 Example Simulation with initial condition x 0 = [ ] t. Notice that the control remains saturated during the initial five steps x 2 k 0 uk x 1 k Shown in the figures are the sequences obtained with numerical (QP) RHC and with the control law u k = sat (x k ). (Both sequences coincide.) k

20 Conclusions State Space MPC = - sat(x) Unconstrained Locally, the tactical MPC design has a closed-form solution N (x) = sat (x). The control law N (x) = sat (x) can be thought of as a particular case of an Anti-windup design. Simple implementation of optimisation based strategies can lead to new ideas in robustness, stability, etc.