Deterministic Kalman Filtering on Semi-infinite Interval

Transcription

1 Deterministic Kalman Filtering on Semi-infinite Interval L. Faybusovich and T. Mouktonglang Abstract We relate a deterministic Kalman filter on semi-infinite interval to linear-quadratic tracking control model with unfixed initial condition. 1 Introduction In [4], E. Sontag considered the deterministic analogue of Kalman filtering problem on finite interval. The deterministic model allows a natural extension to semi-infinite interval. It is of a special interest because for the standard linearquadratic stochastic control problem extension to semi-infinite interval leads to complications with the standard quadratic objective function (see e.g. [1]). According to [4], the model which we are going to consider has the following form: J(x, u, x ) = [u T Ru + (Cx ȳ) T Q(Cx ȳ)] dt, (1) ẋ = Ax + Bu, (2) x() = x. (3) Here we assume that the pair (x, u) a(x ) + Z, where Z is a vector subspace of the Hilbert space L n 2 [, + ) L m 2 [, + ) (with L n 2 [, + ) a Hilbert space of R n value square integrable functions) defined as follows: Z = {(x, u) L n 2 [, + ) L m 2 [, + ) : x is absolutely continuous, ẋ L n 2 [, + ), ẋ = Ax + Bu, x() = } Here A is an n by n matrix; B is an n by m matrix; R = R T is an n by n and positive definite; Q = Q T is an r by r and positive definite; C is an r Mathematics Department, University of Notre Dame, USA This paper is based upon work supported in part by National Science Foundation Grant No. DMS Mathematics Department, Faculty of Science, Chiang Mai University, Thailand 522. This paper is based upon work supported in part by the Center of Excellence in Mathematics, Sri Ayutthaya Rd., Bangkok 14, Thailand 1

2 by n matrix; ȳ L r 2[, + ). Notice that in (1) - (3) x is not fixed and we minimize over all triple (x, u, x ) L n 2 [, + ) L m 2 [, + ) R n satisfying our assumption. Notice also that we interpret (1) - (3) as an estimation problem of the form ẋ = Ax + Bu, ȳ = Cx + v, where we try to estimate x with the help of observation ȳ by minimizing perturbations u, v and choosing an appropriate initial condition x. 2 Solution of the deterministic problem Consider the algebraic Riccati equation KA + A T K + KLK C T QC =, (4) where L = BR 1 B T. Assuming that the pair (A, B) is stabilizable and the pair (C, A) is detectable, there exists a negative definite symmetric solution K st to (4) such that the matrix A+LK st is stable (see e.g. Theorem 12.3 in [5]). Using the result of [2], we can describe the optimal solution to (1) - (3) with fixed x as follows. There exists a unique solution ρ L n 2 [, ) satisfying the differential equation ρ = (A + LK st ) T ρ C T Qȳ. (5) Moreover, ρ can be explicitly described as follows: ρ (t) = The optimal solution (x, u) to (1) - (3) has the form exp[(a + LK st ) T τ]c T Qȳ(t + τ) dτ. (6) ẋ = (A + LK st )x + Lρ, x() = x, (7) u = R 1 B T (K st x + ρ ). (8) For details see [2]. Notice that ρ does not depend on x. To solve the original problem (1) - (3) we need to express the minimal value of the functional (1) in term of x. Theorem 1. Let (x, u) be an optimal solution of (1) - (3) with fixed x given by (5) - (8). Then J(x, u, x ) = x T K st x 2ρ () T x + [ȳ T Qȳ ρ T Lρ ] dt. (9) 2

3 Remark. Notice that J(x, u, x ) is a strictly convex function of x and hence minimum of J as a function of x is attained at x opt = K 1 st ρ (). (1) Hence (5) - (8) gives a complete solution of the original problem (1) - (3). Proof. Let (y, w) a(x ) + Z be feasible solution to (1) - (3), where x is fixed. Consider (y, w) = [w R 1 B T (K st y + ρ )] T R[w R 1 B T (K st y + ρ )]. where we suppressed an explicit dependence on time. Notice that (x, u). Furthermore, (y, w) = , where 1 = w T Rw, 2 = 2(K st y + ρ ) T Bw, 3 = (K st y + ρ ) T L(K st y + ρ ). Now Bw = ẏ Ay and consequently Consequently, 2 = 2(y T K st + ρ T )(ẏ Ay) = y T (K st A + A T K st )y 2y T K st ẏ 2ρ T ẏ + 2ρ T Ay, 3 = y T K st LK st y + ρ T Lρ + 2ρ T LK st y. (y, w) = w T Rw + ρ T Lρ + y T (K st LK st + K st A + A T K st )y Using (4) and (5), we obtain d dt (yt K st y) 2 d dt (ρt y) + 2 ρ T y + 2ρT LK st y + 2ρ T Ay. (y, w) = w T Rw + ρ T Lρ + y T C T QCy 2 d dt (ρt y) d dt (yt K st y) 2(C T Qȳ) T y = w T Rw + ρ T Lρ 2 d dt (ρt y) d dt (yt K st y) +(ȳ Cy) T Q(ȳ Cy) ȳ T Qȳ. Hence, taking into account that ρ (t), y(t), t + (see for details [2]), we obtain 3

4 (y, w) dt = + [w T Rw + +(ȳ Cy) T Q(ȳ Cy)] dt [ρ T Lρ ȳ T Qȳ] dt + 2ρ () T x + x K st x = J(y, w, x ) + 2ρ () T x + x K st x + c where c = [ρ T Lρ ȳ T Qȳ] dt. Notice, that (y, w) and (x, u). This shows that, indeed, (x, u) is an optimal solution to (1) - (3) (with fixed x ) and proves (9). 3 Steady state deterministic Kalman filtering In light of (1), it is natural to consider the process z(t) = K 1 st ρ (t), t [, + ) (11) as a natural estimate for the optimal solution to problem (1) - (3). Let us find the differential equation for z. Proposition 1. Remark: Notice that K 1 st ż = Az + K 1 st C T Q(ȳ Cz). (12) is a solution to the algebraic equation L P C T QCP + AP + P A T = (13) In other words, the differential equation (12) is a precise deterministic analogue for the stochastic differential equation describing the optimal (steady-state) estimation in Kalman filtering problem. See e.g. [1]. Proof. Using (5) and (11), we obtain: ż = Kst 1 (A + LK st ) T ρ + Kst 1 C T Qȳ = (Kst 1 A T + L)(K st z) + Kst 1 C T Qȳ. Since K st is a solution to (4), we have Hence, Hence, we obtain (12). K 1 st A T K st LK st = A K 1 st C T QC. ż = Az K 1 st C T QCz + K 1 st C T Qȳ. Remark: Notice that due to (11) (z, ) and consequently (z, ) would be an optimal solution to (1) - (3) if it were feasible for this problem. 4

5 4 Concluding remarks The differential equation (12) allows one to recursively estimate of z based on observation ȳ provided we know z() = x opt. Notice that according to (6) z() = K 1 st exp[(a + LK st ) T τ]c T Qȳ(τ) dτ, requires the knowledge of the whole ȳ. However, due to the integral nature of this relationship the infinite integral can be very well approximated by the integral over the finite interval [, T ] (since the matrix A + LK st ) T is stable) and hence the recursive nature of (12) is recovered for t T for some finite T. Notice, further, that similar problem with discrete time can be solved using the formalism developed in [3]. Acknowledgments: The research of L. Faybusovich was partially supported by NSF grant DMS The research of T. Mouktonglang was partially supported by the Center of Excellence in Mathematics, Sri Ayutthaya Rd., Bangkok 14, Thailand. References [1] M.H.A Davis, Linear Estimation and Stochastic Control, Chapman and Hall, London, [2] L. Faybusovich and T. Mouktonglang, Linear-quadratic control problem with a linear term on semiinfinite interval: theory and applications, Technical Report, University of Notre Dame, December 23. [3] L. Faybusovich, T. Mouktonglang and T. Tsuchiya, Implementation of infinite - dimensional interior - point method for solving multi - criteria linear - quadratic control problem. Optim. Methods Softw. 21(26) no. 2, [4] E. D. Sontag, Mathematical Control Theory, Springer, 199. [5] W. M. Wonham, Linear Multivariable Control, a Geometric Approach, Springer,