Scholarly article on topic 'Deterministic Kalman Filtering on Semi-Infinite Interval'

Deterministic Kalman Filtering on Semi-Infinite Interval Academic research paper on "Mathematics"

CC BY
0
0
Share paper
Keywords
{""}

Academic research paper on topic "Deterministic Kalman Filtering on Semi-Infinite Interval"

Hindawi Publishing Corporation

International Journal of Mathematics and Mathematical Sciences Volume 2012, Article ID 490139,11 pages doi:10.1155/2012/490139

Research Article

Deterministic Kalman Filtering on Semi-Infinite Interval

L. Faybusovich1 and T. Mouktonglang2,3

1 Mathematics Department, University of Notre Dame, Notre Dame, IN 46556, USA

2 Mathematics Department, Faculty of Science, Chiang Mai University, Chiang Mai 50220, Thailand

3 The Centre of Excellence in Mathematics, CHE, Sri Ayutthaya Road, Bangkok 10400, Thailand

Correspondence should be addressed to T. Mouktonglang, thanasak@chiangmai.ac.th Received 28 March 2012; Accepted 9 May 2012 Academic Editor: Aloys Krieg

Copyright © 2012 L. Faybusovich and T. Mouktonglang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We relate a deterministic Kalman filter on semi-infinite interval to linear-quadratic tracking control model with unfixed initial condition.

1. Introduction

In [1], 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 linear-quadratic stochastic control problem extension to semi-infinite interval leads to complications with the standard quadratic objective function (see, e.g., [2]). According to [1], the model which we are going to consider has the following form:

Here we assume that the pair (x,u) e a(xo) + Z, where Z is a vector subspace of the Hilbert space L"[0, +<x>) x L™[0, +<x>) (with L"[0, +<x>) a Hilbert space of R"-value square integrable

x = Ax + Bu,

x(0) = xo.

(1.2) (1.3)

functions) is defined as follows:

Z = {(x, u) e L£[0, +<x>) x +<x>) : x is absolutely continuous, x e Ln[0, +<x>),X = Ax + Bu,x(0) = 0}.

Here A is an n by n matrix; B is an n by m matrix; R = RT is an n by n and positive definite; Q = QT is an r by r and positive definite; C is an r by n matrix; y e Lrr [0, +<x>). Notice that in (1.1)-(1.3) x0 is not fixed and we minimize over all triple (x,u,x0) e Ln[0, +<x>) x Lm[0, +<x>) x Rn satisfying our assumption.

Notice also that we interpret (1.1)-(1.3) as an estimation problem of the form

where we try to estimate x with the help of observation y by minimizing perturbations u, v and choosing an appropriate initial condition x0.

2. Solution of the Deterministic Problem

Consider the algebraic Riccati equation

where L = BR-1BT. Assuming that the pair (A, B) is stabilizable and the pair (C, A) is detectable, there exists a negative definite symmetric solution Kst to (2.1) such that the matrix A + LKst is stable (see, e.g., Theorem 12.3 in [3]). According to [4], we have described a complete solution of the linear-quadratic control problem on a semi-infinite interval with the linear term in the objective function. The major motivation for this extension comes from [5] where we consider applications of primal-dual interior-point algorithms to the computational analysis of multicriteria linear-quadratic control problems in mini-max form. To compute a primal-dual direction it is required to solve linear-quadratic control problems with the same quadratic and different linear parts on each iteration. Using the results in [5], we can describe the optimal solution to (1.1)—(1.3) with fixed x0 as follows.

There exists a unique solution p0 e L£[0, to) satisfying the differential equation

x = Ax + Bu,

y = Cx + v,

KA + ATK + KLK - CTQC = 0,

p = -(A + LKst)Tp - CTQy.

Moreover, p0 can be explicitly described as follows:

The optimal solution (x,u) to (1.1)—(1.3) has the form

x = (A + LKst)x + Lpo, x(0) = xo, (2.4)

u = R-1BT (Kstx + p0). (2.5)

For details see [5].

Notice that po does not depend on xo. To solve the original problem (1.1)—(1.3) we need to express the minimal value of the functional (1.1) in term of xo.

Theorem 2.1. Let (x,u) be an optimal solution of (1.1)—(1.3) with fixed x0 given by (2.2)-(2.5). Then

J(x,u,xo) = & - x^Kstxo - 2po(0)Txo + J \yTQy - PoLPo] dt. (2.6)

Remark 2.2. Notice that J(x, u, xo) is a strictly convex function of xo and hence minimum of J as a function of xo is attained at

x0opt = -K-/po(o). (2.7)

Hence (2.2)-(2.5) gives a complete solution of the original problem (1.1)—(1.3).

Proof. Let (y,w) e a(xo) + Z be feasible solution to (1.1)—(1.3), where xo is fixed. Consider

a(y,w) = [w - R-1BT(Ksty + po)]T ■ R ■ [w - R-1BT(Ksty + po)], (2.8)

where we suppressed an explicit dependence on time. Notice that by (2.5)

a(x,u) = o,

) (2.9)

^ y, w) = o,

for any feasible solution (y,w) implies that (y,w) = (x,u). Furthermore, let a(y,w) = ai + a2 + a3, where

a1 = wTRw,

a2 = -2(Ksty + po)TBw, (2.1o)

a3 = (Ksty + po) TL(Ksty + po).

Now Bw = y - Ay, and consequently

a2 = -2(yTKst + pD {y - Ay) = yT (KstA + ATKst)y - 2yTKsiy - 2pTy + 2pTAy,

(2.11)

a3 = yT KstLKsty + pTLpo + 2pJ}LKsty-

Consequently,

d / Tt. \ nd + KstA + A'Kstjy - " "

■ 2Po y + 2p1oLKsty + 2Po Ay.

a (y, to) = w^ Rw + pTLpo + yT(KstLKst + KstA + ATKst)y - -jt(yTKsty) - 2 -¡¡¡¿(piy)

(2.12)

Using (2.1) and (2.2), we obtain

a (y, w) = wTRw + pTLpo + yTCTQCy - 2 jt(pT0y) - di(yTKsty) - 2(CTQy)Ty

=wTRw+pTLpo- 2 dftipoy)- drt(yTKsty) + (y- Cy)TQ(y- Cy) - yTQy-

(2.13)

Hence, taking into account that p0(t) ^ 0, y(t) ^ 0, t ^ +<x> (see, for details [5]), we obtain

/«+œ f+œ

j A(y,w) dt = j |wTRw + (y - Cy)TQ(y - C^jdi

+ J [pTLpo - yTQy] dt + 2po(0)Txo + xoKstXo (2.14) = J(y,w,xo) + 2po(o)Txo + XoKstXo + c,

where c = Jo+œ [pTLpo - yTQy] dt.

Notice, that a(y,w) > o and A(x,u) = o. This shows that, indeed, (x,u) is an optimal solution to (1.1)—(1.3) (with fixed xo) and proves (2.6). □

Remark 2.3. By (2.14) and a(x,u) = o, we have J(y,w,xo) > J(x,u,xo) and the equality occurs if and only if (y, w) = (x,u) (see also (2.9)). Hence (x,u) is a unique solution to the problem (1.1)—(1.3). Similary reasoning works in discrete time case.

International Journal of Mathematics and Mathematical Sciences 3. Steady-State Deterministic Kalman Filtering

In light of (2.7), it is natural to consider the process

z(t) = - K-1po(t), t e [o, +œ) (3.1)

as a natural estimate for the optimal solution to problem (1.1)—(1.3). Let us find the differential equation for z.

Proposition 3.1. One has

z = Az + K-1CTQ(y - Cz). (3.2)

Remark 3.2. Notice that K-1 is a solution to the algebraic equation

L - PCTQCP + AP + PAT = o. (3.3)

In other words, the differential equation (3.2) is a precise deterministic analogue for the stochastic differential equation describing the optimal (steady-state) estimation in Kalman filtering problem. See, for example, [2] .

Proof. Using (2.2) and (3.1), we obtain

z = K-1 (A + LKst)Tpo + K-1CTQy = - (.K-A + L) (Kstz) + K-1CTQy.

Since Kst is a solution to (2.1), we have

-K-jATKst - LKst = A - K-1CTQC. (3.5)

Hence,

z = Az - K-jCTQCz + K-jCTQy. (3.6)

Hence, we obtain (3.2). □

Remark 3.3. Notice that due to (3.1) a(z,o) = o and consequently (z, o) would be an optimal solution to (1.1)—(1.3) if it were feasible for this problem.

4. The Solution of the Discrete Deterministic Problem

It is natural to consider the discrete version for the problem (1.1)—(1.3). In this case, the problem can be reformulated as follows:

1 ^ r 1

J(x,u,x0) = [uTkRuk + (Cxk - yk)TQ(Cxk - yk)\ —> min, (4.1)

xk+i = Axk + Buk, (4.2)

x0 = xo. (4.3)

Here we let x denote a sequence {xk} c rn for k = 0,...,<x>. We say that x e I" (n) if X ¿=111 xf H2 < where || ■ || is a norm induced by an inner product (,) in r". Let (x,u) e l" (n) x (n).

Like in the continuous case, we assume that the pair (x,u) e a(x0) + Z, where Z is a vector subspace of the Hilbert space l" (n) x lmm (n).

Observe now the inner product in H has the following form:

{(x,y), (u,v))h = £{(Xk,uk) + <yk,vk)}. (4.4) k=0

The vector subspace Z now takes the following form:

Z = {(x,u) e H : xk+1 = Axk + Buk, k = 0,1,..., x0 = 0}. (4.5)

Here A is an n by n matrix. B is an n by m matrix. R = RT is an n by n and positive definite. Q = QT is an r by r and positive definite. C is an r by n matrix and y e l22 (n).

As in the continuous case, we interpret (4.1)-(4.3) as an estimation problem of the

xk+1 = Axk + Buk,

yk = Cxk + vk,

where we try to estimate x with the help of observation y by minimizing perturbations u, v and choosing an appropriate initial condition x0.

According to [4], a general cost function for a discrete linear-quadratic control problem with linear term on the cost function has the following form:

J(x,u,x0) = jr1 [xTkQxk + uj^Ru^ + xlfk + uTkfyk —> min, (4.7)

where f e V^ (n) and $ e IVf (n). The solution to the particular class of problems can be completely described by solving several system of recurrence relations and the following discrete algebraic Riccati equation (DARE):

K = ATKA - ('ATKB) (r + BTKB) ^AtKBY + Q.

We assume that this equation has a positive definite stabilizing solution Kst. For sufficient conditions, see [6] .

In our situation, we have

J(x,u,xo) = \uTkRuk + (Cxk - yk)TQ(Cxk - yj] min.

It is easy to see that = -CTQyk and = o,k = o, 1,.... By [4], there is a unique solution p = {pk} e ¡n (n) of the following recurrence relations

AT - (,ATKstB) (R + BTKstB) '

pk+i + cTQVk.

(4.10)

For details on an explicit solution of the above recurrence relation, see [4]. For simplicity, we let

R = (R + BTKstB),

(4.11)

and we also let

L = BR BT.

(4.12)

So our recurrence relation for p now takes the form

pk = [AT - ATKstL]pk+i + CTQyk

(4.13)

with the corresponding DARE

K = ATKA - (ATKB)R (ATKB)T + CTQC, K = ATKA - ATKLKA + CTQC.

(4.14)

The optimal solution to (4.1)-(4.3) has the following form:

Xk+1 = (AT - ATKstLj Xk + Lpk+1,

uk = -R ^BTKstAxk + R "1BTpk+1.

(4.15)

(4.16)

For details, see [4]. To solve the original problem (4.1)-(4.3) we need to express the minimal value of the functional (4.1) in terms of x0.

Theorem 4.1. Let (x,u) be an optimal solution of (4.1)-(4.3) with fixed x0 given by (4.15)-(4.16). Then

J(x,u,x0) = -4Kstx0 - pT0x0 + [2yTkQyk - pTk+1LPk+i\. (4.17)

2 2 k=0

Proof. For simplicity of notation, we use K for Kst. Let

A(yfc,wfc) = [wk + R"1BT(KAyk - pfc+1)]T ■ R ■ \wk + R^B1 (KAyk - pfc+1)]

(4.18)

= ai + a2 + a

(4.19)

A1 = wTRwk, A2 = 2(KAyk - pk+1)TBwk

= 2(KAyk - pk+1) T(yk+1 - Ayk) = 2yi^Kyk+1 - 2y[ATKAyk - 2pTk+1yk+1 + 2pTk+1 Ay^ A3 = (KAyk - Pk+1)TL(KAyk - pk+1)

= yTATKLKAyk + PT+1LPk+1 - 2yTATKLpk+v

We assume that (y,w) e a(xo) + Z. Since ATKA - ATKLKA = K - CTQC and [AT -

ATKL]Pk+1 = Pk - cTQyk, we have

a (yk, Wk) = wTkRwk - 2yT [ATKA - ATKLKA]yk - yTkÀfKLKAyk

T t T _ T T — T T

- 2Pk+1yk+1 + 2Pk+1Ayk + pk+1LPk+1- 2yiA KLPk+1 + 2ykAT Kyk+1 = wTRwk - 2ykk [k - CTQC] yk - ykkATKLKAyk - 2p\+xyk+1

+ 2yTATPk+1 + PT+1LPk+1 - 2yTATKLPk+1 + 2yiAKyk+1

= wTRwk - 2yT [K - CTQC\ yk - y[ATKLKAyk - 2p\+xyk+1

+ 2yT AT - ATKL]pk+1 + PT+1 LPk+1 + 2yTATKyk+1

International Journal of Mathematics and Mathematical Sciences

= wTRwk - 2yT [k - CTQC\ yk - ylATKLKAyk - 2pl+1yk+l

+ 2yfcT [Pk - cTQVk] + Pk+1LPk+1 + 2yTATKyk+1

= wTRwk - 2yT [K - CTQC\ yk - yTATKLKAyk - 2pk+1yk+1

+ 2yTkPk - 2yTkCTQVk + PTk+1LPk+1 + 2ylATKyk+1-

By recalling now the definition of R, we have

wkRwk = wT(R + BTKBS)wk

= wTRwk + wTBTKBwk = wlRwk + (yk+1 - Ay^TKyk+1 - Ayk) = w\Rwk + yk+Kyk+1 + yTATKAyk - 2yTATKyk+1-

(4.20)

(4.21)

Therefore,

a (Уk, wk) = wTRwk - ykKyk - ylCTQCyk - 2pl+1yk+1 + 2ylpk

T T t t

- 2ykC Qyk + Pk+1LPk+1 + yk+1Kyk+1.

(4.22)

We then rearrange the terms and complete the square to obtain a useful expression for a:

a(y^ wk) = wlRwk - ylKyk + yl+1Kyk+1 - 2Pl+1yk+1 + 2plyk +Pl+1LPk+1 + ylClQCyk- 2ylClQyk

lll _T _T

= wkRwk - ykKyk + yk+1Kyk+1 - 2Pk+1yk+1 + 2Pkyk + pl+1Lpk+1 + (Cyk - yk)TQ(Cyk - yk) - 2ylQyk-

Notice, since we fixed x0, we let y0 = x0 and take summation of both sides:

EMyk,wk) = - xlKx0 + 2Plx0 + ^ \wlRwk + {Cyk - yk)lQ(Cyk - yk)l k=0 k=0

(4.23)

+ X lpk+1Lpk+1- 2ylQyJ •

(4.24)

By the definition of A(yk,wk), A(xk,uk) = 0. Therefore,

0 = -xlKx0 + 2plx0 + 2J(x, u,x0) + ^j [pfc+1LPk+1 - 2ylQyJ • (4.25)

As a result,

1 1 °°

J(x, u, x0) = 2xlKx0 - plx0 + ^ [2ykQyk - pl+1LPk+J • (4.26) 2 2 k=0

Then the proof is completed. □

As in continuous case, for the discrete case, J(x,u,x0) is a strictly convex function of x0 and hence minimum of J as a function of x0 is attained at

x0°pt = K-J-P0, (4.27)

where p is the unique l2 solution to (4.13).

Since we have (4.27), it is natural to consider the process

zk = K-Jpk, (4.28)

as an estimate for the optimal solution to problem (4.1)-(4.3). Let us find the recurrence relation for zk.

Proposition 4.2. Assuming that the closed loop matrix A-LKA is invertible, one has

Zk+1 = Azk - K-1 [A1 - AlKstL] (yk - Czk) • (4.29) Proof. We can rewrite (4.13) in the form

KstZk = [at - AlKstL] KstZk+1 + CTQyk. (4.30)

Using the algebraic Riccati equation

Kst = AlKstA - AlKstLKstA + ClQC, (4.31) we can rewrite (4.30) in the form

ClQCzk + (.AlKst - AlKstLKst)Azk = [at - AlKst^KstZk+1 + CTQyk, (4.32)

which is equivalent to

(AT - ATKstL) KstZk+1 = (AT - ATKstL) KstAzk - CTQ {yk - Czk). (4.33)

The result follows. □

Remark 4.3. Notice that (4.29) is the analogue of the "limiting" discrete Kalman filter [6, Page 384, (17.6.1)].

5. Concluding Remarks

In this paper, we relate a deterministic Kalman filter on semi-infinite interval to linear-quadratic tracking control model with unfixed initial condition. Solutions of the deterministic problems both continuous and discrete cases are described. This extends the result of Sontag to semi-infinite interval.

Acknowledgments

The research of L. Faybusovich was partially supported by National science foundation. Grant DMS07-12809. The research of T. Mouktonglang was partially supported by the Centre of Excellence in Mathematics and the Commission for Higher Education (CHE), Sri Ayut-thaya Road, Bangkok, Thailand.

References

[1] E. D. Sontag, Mathematical Control Theory, Springer, 1990.

[2] M. H. A. Davis, Linear Estimation and Stochastic Control, Chapman and Hall, London, UK, 1977.

[3] W. M. Wonham, Linear Multivariable Control, a Geometric Approach, Springer, 1974.

[4] L. Faybusovich, T. Mouktonglang, and T. Tsuchiya, "Implementation of infinite-dimensional interior-point method for solving multi-criteria linear-quadratic control problem," Optimization Methods & Software, vol. 21, no. 2, pp. 315-341, 2006.

[5] L. Faybusovich and T. Mouktonglang, "Linear-quadratic control problem with a linear term on semiinfinite interval: theory and applications," Tech. Rep., University of Notre Dame, 2003.

[6] P. Lancaster and L. Rodman, Algebraic Riccati Equations, The Clarendon Press Oxford University Press, Oxford, UK, 1995.

Copyright of International Journal of Mathematics & Mathematical Sciences is the property of Hindawi Publishing Corporation and its content may not be copied or emailed to multiple sites or posted to a listserv without the copyright holder's express written permission. However, users may print, download, or email articles for individual use.