Adaptative high-gain extended Kalman filter and applications

tel-00559107, version 1 - 24 Jan 2011
Proof.
d
dτ Tr(S(τ)) = −Tr(SQS) − Tr(A ′
B.1 Bounds on the Riccati Equation
S) − Tr(SA)
≤ −Tr(SQS)+2|Tr(A ′
S)|.
From the matrix facts: Tr(SQS) > q
n (Tr(S)) 2 , with q = min
�x�=1 x′ Qx, and
|Tr(A ′
S)| ≤ � Tr(A ′ A) � Tr(S ′ S)
≤ Tr(A ′
A) 1 � 1
2 Tr(S) 2� 2
�
≤ sup Tr(A
t
′
A) 1 �
2 Tr(S).
Therefore
d
dτ Tr(S) ≤−aTr(S) 2 +2bTr(S)
where a and b are as in the lemma.
Useful bounds for Tr(S(τ)) can be derived from Lemma 79. In the following, x stands for
Tr(S).
Lemma 80
Let a, b be two positive constants. Let x : [0,T[→ R + (possibly T =+∞) be an absolutely
continuous function satisfying for almost all 0 < τ 2b
a
Proof.
− If x(τ) ≤ 2b
a
− Otherwise, we define
˙x(τ) ≤−ax 2 (τ)+2bx(τ).
and 0. The solution x(τ) is such that:
x(τ) ≤ max � (x(0), 2b
a )� for all τ ∈ [0,T[.
then for all τ > 0 ∈ [0,T[ we have the two inequalities:
⎧
⎪⎨
⎪⎩
x(τ) ≤ 2b 2b 1
+
a a e2bτ − 1 ,
2bx0e
x(τ) ≤
2bτ
ax0 (e2bτ − 1) + 2b .
for all τ ∈ [0,T[, trivially we have x(τ) ≤ max (x(0), 2b
a ).
E =
�
τ ∈ [0,T[: x(τ) > 2b
�
a
(B.5)
which is a non empty set. Take any connected component ]α,β [⊂ E such that α > 0.
There are two situations:
1. β = T , and x(β) ≥ 2b
2b
a , x(α) = a ,
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