Adaptative high-gain extended Kalman filter and applications

tel-00559107, version 1 - 24 Jan 2011
(a) if A ≥ B ≥ 0, and U is orthogonal, then U ′
AU ≥ U ′
BU,
B.1 Bounds on the Riccati Equation
(b) A, and B as in (a), and A is invertible, then ρ(BA −1 ) ≤ 1, where ρ denotes the
spectral radius,
(c) A, and B as in (a), then λi(A) ≥ λi(B), where λi(M) denote the i th eigenvalue of
the matrix M sorted in ascending order,
(d) A, and B as in (a), and both invertible then B −1 ≥ A −1 ≥ 0,
(e) A, and B as in (a), then det(A) ≥ det(B), and Tr(A) ≥ Tr(B),
(f) if A1 ≥ B1 ≥ 0, and if A2 ≥ B2 ≥ 0 then A1 + A2 ≥ B1 + B2 ≥ 0.
4. From facts 1.(c) and 3.(f) we deduce that:
B.1.1 Part One: the Upper Bound
(A ≥ B ≥ 0) ⇒ (|A| ≥| B| ≥ 0) .
This first part of the proof is decomposed into three steps. Let T ∗ be a fixed, positive scalar.
1. We prove that there is β1 such that for all τk ≤ T ∗ , k ∈ N, Sk(+) ≤ β1Id,
2. we show that there exists β2 such that for all T ∗ ≤ τk, k ∈ N, Sk(+) ≤ β2Id,
3. we deduce the result for all times.
The first fact can be directly proven as follows.
Lemma 78
Consider equation (B.1) and the assumptions of Lemma 77. Let T ∗ > 0 be fixed. There
exists β1 > 0 such that
Sk(+) ≤ β1Id,
for all τk ≤ T ∗ , k ∈ N, independently from the subdivision {τi}i∈N.
Proof.
For all τ ∈ [τk−1; τk], equation (B.1) gives:
� τ dS (v)
S (τ) = Sk−1(+) +
dτ dv,
= Sk−1(+)
� τ
+
τk−1
τk−1
⎡
⎣−
�
A(u)+ � b ∗ (z, u)
θ
�′
S − S
Since SQS is symmetric definite positive, we can write
⎡
� �
τ
S (τ) ≤ Sk−1(+) + ⎣− A(u)+ �b ∗ �′
(z, u)
S − S
θ
and fact 4 gives us
τk−1
� τ
|S(τ)| ≤| Sk−1(+)| +
133
τk−1
�
A(u)+ � b ∗ (z, u)
θ
�
�
A(u)+ � b ∗ (z, u)
θ
⎤
− SQS⎦
dv.
� ⎤
⎦ dv,
2s|S|dv, (B.2)