# Deterministic Kalman Filtering on Semi-Infinite Interval

Authors: L. Faybusovich; T. Mouktonglang
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

---

## Abstract

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


---

## Body

## 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:
(1.1)J(x,u,x0)=∫0+∞‍[uTRu+(Cx-y¯)TQ(Cx-y¯)]dt,(1.2)x˙=Ax+Bu,(1.3)x(0)=x0.
Here we assume that the pair (x,u)∈a(x0)+Z, where Z is a vector subspace of the Hilbert space L2n[0,+∞)×L2m[0,+∞) (with L2n[0,+∞) a Hilbert space of Rn-value square integrable functions) is defined as follows:
(1.4)Z={(x,u)∈L2n[0,+∞)×L2m[0,+∞):xisabsolutelycontinuous,Z=x˙∈L2n[0,+∞),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¯∈L2r[0,+∞). Notice that in (1.1)–(1.3) x0 is not fixed and we minimize over all triple (x,u,x0)∈L2n[0,+∞)×L2m[0,+∞)×Rn satisfying our assumption.Notice also that we interpret (1.1)–(1.3) as an estimation problem of the form
(1.5)x˙=Ax+Bu,y¯=Cx+v,
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(2.1)KA+ATK+KLK-CTQC=0,
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 solutionρ0∈L2n[0,∞) satisfying the differential equation
(2.2)ρ˙=-(A+LKst)Tρ-CTQy¯.
Moreover, ρ0 can be explicitly described as follows:
(2.3)ρ0(t)=∫0+∞‍exp⁡[(A+LKst)Tτ]CTQy¯(t+τ)dτ.
The optimal solution (x,u) to (1.1)–(1.3) has the form
(2.4)x˙=(A+LKst)x+Lρ0,x(0)=x0,(2.5)u=R-1BT(Kstx+ρ0).
For details see [5].Notice thatρ0 does not depend on x0. To solve the original problem (1.1)–(1.3) we need to express the minimal value of the functional (1.1) in term of x0.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
(2.6)J(x,u,x0)=&-x0TKstx0-2ρ0(0)Tx0+∫0+∞‍[y¯TQy¯-ρ0TLρ0]dt.Remark 2.2.
Notice thatJ(x,u,x0) is a strictly convex function of x0 and hence minimum of J as a function of x0 is attained at
(2.7)x0opt=-Kst-1ρ0(0).
Hence (2.2)–(2.5) gives a complete solution of the original problem (1.1)–(1.3).Proof.
Let(y,w)∈a(x0)+Z be feasible solution to (1.1)–(1.3), where x0 is fixed. Consider
(2.8)Δ(y,w)=[w-R-1BT(Ksty+ρ0)]T⋅R⋅[w-R-1BT(Ksty+ρ0)],
where we suppressed an explicit dependence on time. Notice that by (2.5)
(2.9)Δ(x,u)≡0,Δ(y,w)≡0,
for any feasible solution (y,w) implies that (y,w)=(x,u). Furthermore, let Δ(y,w)=Δ1+Δ2+Δ3, where
(2.10)Δ1=wTRw,Δ2=-2(Ksty+ρ0)TBw,Δ3=(Ksty+ρ0)TL(Ksty+ρ0).
Now Bw=y˙-Ay, and consequently
(2.11)Δ2=-2(yTKst+ρ0T)(y˙-Ay)=yT(KstA+ATKst)y-2yTKsty˙-2ρ0Ty˙+2ρTAy,Δ3=yTKstLKsty+ρ0TLρ0+2ρ0TLKsty.
Consequently,
(2.12)Δ(y,w)=wTRw+ρ0TLρ0+yT(KstLKst+KstA+ATKst)y-ddt(yTKsty)-2ddt(ρ0Ty)+2ρ0Ty˙+2ρ0TLKsty+2ρ0TAy.
Using (2.1) and (2.2), we obtain
(2.13)Δ(y,w)=wTRw+ρ0TLρ0+yTCTQCy-2ddt(ρ0Ty)-ddt(yTKsty)-2(CTQy¯)Ty=wTRw+ρ0TLρ0-2ddt(ρ0Ty)-ddt(yTKsty)+(y¯-Cy)TQ(y¯-Cy)-y¯TQy¯.
Hence, taking into account that ρ0(t)→0,y(t)→0,t→+∞ (see, for details [5]), we obtain(2.14)∫0+∞‍Δ(y,w)dt=∫0+∞[wTRw+(y¯-Cy)TQ(y¯-Cy)]‍dt+∫0+∞‍[ρ0TLρ0-y¯TQy¯]dt+2ρ0(0)Tx0+x0Kstx0=J(y,w,x0)+2ρ0(0)Tx0+x0Kstx0+c,
where c=∫0+∞‍[ρ0TLρ0-y¯TQy¯]dt.
Notice, thatΔ(y,w)≥0 and Δ(x,u)≡0. This shows that, indeed, (x,u) is an optimal solution to (1.1)–(1.3) (with fixed x0) and proves (2.6).Remark 2.3.
By (2.14) and Δ(x,u)≡0, we have J(y,w,x0)≥J(x,u,x0) 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.


## 3. Steady-State Deterministic Kalman Filtering

In light of (2.7), it is natural to consider the process
(3.1)z(t)=-Kst-1ρ0(t),t∈[0,+∞)
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(3.2)z˙=Az+Kst-1CTQ(y¯-Cz).Remark 3.2.
Notice thatKst-1 is a solution to the algebraic equation
(3.3)L-PCTQCP+AP+PAT=0.
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
(3.4)z˙=Kst-1(A+LKst)Tρ0+Kst-1CTQy¯=-(Kst-1AT+L)(Kstz)+Kst-1CTQy¯.
Since Kst is a solution to (2.1), we have
(3.5)-Kst-1ATKst-LKst=A-Kst-1CTQC.
Hence,
(3.6)z˙=Az-Kst-1CTQCz+Kst-1CTQy¯.
Hence, we obtain (3.2).Remark 3.3.
Notice that due to (3.1) Δ(z,0)≡0 and consequently (z,0) 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:
(4.1)J(x,u,x0)=12∑k=1∞‍[ukTRuk+(Cxk-y¯k)TQ(Cxk-y¯k)]⟶min⁡,(4.2)xk+1=Axk+Buk,(4.3)x0=xo.Here we letx denote a sequence {xk}⊂ℝn for k=0,…,∞. We say that x∈l2n(ℕ) if ∑i=1∞‍∥xi∥2<∞, where ∥·∥ is a norm induced by an inner product 〈,〉 in ℝn. Let (x,u)∈l2n(ℕ)×l2m(ℕ).Like in the continuous case, we assume that the pair(x,u)∈a(x0)+Z, where Z is a vector subspace of the Hilbert space l2n(ℕ)×l2m(ℕ).Observe now the inner product inH has the following form:
(4.4)〈(x,y),(u,v)〉H=∑k=0∞‍{〈xk,uk〉+〈yk,vk〉}.The vector subspaceZ now takes the following form:
(4.5)Z={(x,u)∈H:xk+1=Axk+Buk,k=0,1,…,x0=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 and y¯∈l2r(ℕ).As in the continuous case, we interpret (4.1)–(4.3) as an estimation problem of the form
(4.6)xk+1=Axk+Buk,y¯k=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:
(4.7)J(x,u,x0)=∑k=1∞‍12[xkTQxk+ukTRuk]+xkTψk+ukTϕk⟶min⁡,
where ψ∈l2n(ℕ) and ϕ∈l2m(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):
(4.8)K=ATKA-(ATKB)(R+BTKB)-1(ATKB)T+Q.
We assume that this equation has a positive definite stabilizing solution Kst. For sufficient conditions, see [6].In our situation, we have(4.9)J(x,u,x0)=12∑k=1∞‍[ukTRuk+(Cxk-y¯k)TQ(Cxk-y¯k)]⟶min⁡.
It is easy to see that ψk=-CTQy¯k and ϕk=0,k=0,1,…. By [4], there is a unique solution ρ¯={ρ¯k}∈l2n(ℕ) of the following recurrence relations
(4.10)ρk=[AT-(ATKstB)(R+BTKstB)-1BT]ρk+1+CTQy¯k.
For details on an explicit solution of the above recurrence relation, see [4]. For simplicity, we let
(4.11)R¯=(R+BTKstB),
and we also let
(4.12)L=BR¯-1BT.
So our recurrence relation for ρ now takes the form
(4.13)ρk=[AT-ATKstL]ρk+1+CTQy¯k
with the corresponding DARE
(4.14)K=ATKA-(ATKB)R¯-1(ATKB)T+CTQC,K=ATKA-ATKLKA+CTQC.
The optimal solution to (4.1)–(4.3) has the following form:
(4.15)xk+1=(AT-ATKstL)Txk+Lρ¯k+1,(4.16)uk=-R¯-1BTKstAxk+R¯-1BTρ¯k+1.
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
(4.17)J(x,u,x0)=12x0TKstx0-ρ¯0Tx0+12∑k=0∞‍[2y¯kTQy¯k-ρ¯k+1TLρ¯k+1].Proof.
For simplicity of notation, we useK for Kst. Let
(4.18)Δ(yk,wk)=[wk+R¯-1BT(KAyk-ρ¯k+1)]T⋅R¯⋅[wk+R¯-1BT(KAyk-ρ¯k+1)]=Δ1+Δ2+Δ3,
where
(4.19)Δ1=wkTR¯wk,Δ2=2(KAyk-ρ¯k+1)TBwk=2(KAyk-ρ¯k+1)T(yk+1-Ayk)=2ykTATKyk+1-2ykTATKAyk-2ρ¯k+1Tyk+1+2ρ¯k+1TAyk,Δ3=(KAyk-ρ¯k+1)TL(KAyk-ρ¯k+1)=ykTATKLKAyk+ρ¯k+1TLρ¯k+1-2ykTATKLρ¯k+1.
We assume that (y,w)∈a(x0)+Z. Since ATKA-ATKLKA=K-CTQC and [AT-ATKL]ρ¯k+1=ρ¯k-CTQy¯k, we have
(4.20)Δ(yk,wk)=wkTR¯wk-2ykT[ATKA-ATKLKA]yk-ykTATKLKAyk-2ρ¯k+1Tyk+1+2ρ¯k+1TAyk+ρ¯k+1TLρ¯k+1-2ykTATKLρ¯k+1+2ykTATKyk+1=wkTR¯wk-2ykT[K-CTQC]yk-ykTATKLKAyk-2ρ¯k+1Tyk+1+2ykTATρ¯k+1+ρ¯k+1TLρ¯k+1-2ykTATKLρ¯k+1+2ykTATKyk+1=wkTR¯wk-2ykT[K-CTQC]yk-ykTATKLKAyk-2ρ¯k+1Tyk+1+2ykT[AT-ATKL]ρ¯k+1+ρ¯k+1TLρ¯k+1+2ykTATKyk+1=wkTR¯wk-2ykT[K-CTQC]yk-ykTATKLKAyk-2ρk+1Tyk+1+2ykT[ρ¯k-CTQy¯k]+ρ¯k+1TLρ¯k+1+2ykTATKyk+1=wkTR¯wk-2ykT[K-CTQC]yk-ykTATKLKAyk-2ρ¯k+1Tyk+1+2ykTρ¯k-2ykTCTQy¯k+ρ¯k+1TLρ¯k+1+2ykTATKyk+1.
By recalling now the definition of R¯, we have
(4.21)wkTR¯wk=wkT(R+BTKB)wk=wkTRwk+wkTBTKBwk=wkTRwk+(yk+1-Ayk)TK(yk+1-Ayk)=wkTRwk+yk+1TKyk+1+ykTATKAyk-2ykTATKyk+1.
Therefore,
(4.22)Δ(yk,wk)=wkTRwk-ykTKyk-ykTCTQCyk-2ρ¯k+1Tyk+1+2ykTρ¯k-2ykTCTQy¯k+ρ¯k+1TLρ¯k+1+yk+1TKyk+1.
We then rearrange the terms and complete the square to obtain a useful expression for Δ:
(4.23)Δ(yk,wk)=wkTRwk-ykTKyk+yk+1TKyk+1-2ρ¯k+1Tyk+1+2ρ¯kTyk+ρ¯k+1TLρ¯k+1+ykTCTQCyk-2ykTCTQy¯k=wkTRwk-ykTKyk+yk+1TKyk+1-2ρ¯k+1Tyk+1+2ρ¯kTyk+ρ¯k+1TLρ¯k+1+(Cyk-y¯k)TQ(Cyk-y¯k)-2y¯kTQy¯k.
Notice, since we fixed x0, we let y0=x0 and take summation of both sides:
(4.24)∑k=0∞‍Δ(yk,wk)=-x0TKx0+2ρ¯0Tx0+∑k=0∞‍[wkTRwk+(Cyk-y¯k)TQ(Cyk-y¯k)]+∑k=0∞‍[ρ¯k+1TLρ¯k+1-2y¯kTQy¯k].
By the definition of Δ(yk,wk), Δ(xk,uk)=0. Therefore,
(4.25)0=-x0TKx0+2ρ¯0Tx0+2J(x,u,x0)+∑k=0∞‍[ρ¯k+1TLρ¯k+1-2y¯kTQy¯k].
As a result,
(4.26)J(x,u,x0)=12x0TKx0-ρ¯0Tx0+12∑k=0∞‍[2y¯kTQy¯k-ρ¯k+1TLρ¯k+1].
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
(4.27)x0opt=Kst-1ρ¯0,
where ρ¯ is the unique l2 solution to (4.13).Since we have (4.27), it is natural to consider the process
(4.28)zk=Kst-1ρ¯k,
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 matrixA-LKA is invertible, one has
(4.29)zk+1=Azk-Kst-1[AT-ATKstL]-1(y¯k-Czk).Proof.
We can rewrite (4.13) in the form
(4.30)Kstzk=[AT-ATKstL]Kstzk+1+CTQy¯k.
Using the algebraic Riccati equation(4.31)Kst=ATKstA-ATKstLKstA+CTQC,
we can rewrite (4.30) in the form
(4.32)CTQCzk+(ATKst-ATKstLKst)Azk=(AT-ATKstL)Kstzk+1+CTQy¯k,
which is equivalent to
(4.33)(AT-ATKstL)Kstzk+1=(AT-ATKstL)KstAzk-CTQ(y¯k-Czk).
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.



---

*Source: 490139-2012-06-25.xml*