Adaptative high-gain extended Kalman filter and applications

tel-00559107, version 1 - 24 Jan 2011
For any k, we compute Sk(+) iteratively:
B.1 Bounds on the Riccati Equation
Sk(+) = e−λτkφu(τk, 0)S0φ ′
� u(τk, 0)
τk
+λ e
0
−λ(τk−v)
�
φu(τk,v) S(v) − S(v)QS(v)
�
φ
λ
′
u(τk,v)dv
k�
+
i=1
e −λ(τk−τi) φu(τk, τi)C ′
R −1 Cφ ′
u(τk, τi)(τi − τi−1) .
(B.13)
We consider this last equation for any k ∈ N ∗ such that τk ≥ T∗. It is of the form
Sk(+) = (I)+(II)+(III), in the same order as before.
− Since S0 is positive definite, (I) is positive definite,
�
�
− from the definition of λ,
is positive definite, and the same goes for
(II),
S(v) − S(v)QS(v)
λ
− let us define l < k as the maximal element of N such that τk − τl ≥ T∗. Since we
consider (B.13) for times τk ≥ T∗ such an element always exists 5 . Note that because
the subdivisions we use have a step less than µ, τk − τl ≤ T∗ + µ. All the terms of the
sum (III) are symmetric definite positive matrices and thus
(III) ≥
k�
i=l+1
e −λ(τk−τi) φu(τk, τi)C ′
R −1 Cφ ′
u(τk, τi)(τi − τi−1) .
From the properties of the resolvent, the right hand side expression can be rewritten,
with ū(τ) =u(τ + τl):
k�
i=l+1
e −λ(τk−τi) φū(τk − τl, τi − τl)C ′
R −1 Cφ ′
ū(τk − τl, τi − τl)(τi − τi−1) . (B.14)
For all i = {l +1, .., k} we have e −λ(τk−τi) ≥ e −λ(T∗+µ) , and R −1 is a fixed symmetric
positive definite matrix. Therefore in order to lower bound (B.14) we have to find a
lower bound for
k�
i=l+1
φū(τk − τl, τi − τl)C ′
Cφ ′
ū(τk − τl, τi − τl)(τi − τi−1) . (B.15)
Note that this sum is now computed for a subdivision of the interval [0, τk − τl] that
has the very specific property T∗ ≤ τk − τl ≤ T∗ + µ. Let us redefine this subdivision as
follows:
– let k∗ be equal to k − l, the subdivision has k∗ + 1 elements,
– we denote ˜τi = τi+l − τl –i.e. ˜τ0 = 0 and ˜τk∗ = τk − τl.
5 It can happen that k = 0. Recall that τ0 = 0 for all subdivisions.
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