Adaptative high-gain extended Kalman filter and applications

tel-00559107, version 1 - 24 Jan 2011
− from the definition of r, D>0,
− (a − 2b
x0 ) is an increasing function of x0, so is A,
B.1 Bounds on the Riccati Equation
− B can be rewritten (a − 2b
x0 )2 � (4b 2 +2ar) 2 − 4r 2 a 2� , the right part is positive, and
the left one is an increasing function of x0, so is B.
X µ (x0) is therefore an increasing function of x0, so is µψ(x0) (i.e. P (X) was obtained
from the change of variables X = e 2bτ ).
Remark 85
1. Notice that since ψ(0) = x0, therefore for all τ ∈ [0,µψ(x0)], ψ(τ) ≤ ψ(0) = x0.
2. Let x∗ 0 > 0 be fixed, and denote ψx∗(τ) the associated ψ function. It is a decreasing
0
function over [0,µψ(x∗ 0 )]. The fact that µψ(x∗ 0 ) is an increasing function of x0 tells us
that for any x0 >x∗ 0 , ψx0 (τ) is a decreasing function of τ over the interval [0,µψ(x∗ 0 )] ⊂
[0,µψ(x0)].
In other words ψx0 (µψ(x∗ 0 )) 0 be fixed. There
exist two scalars β2 > 0 and µ>0 such that
Sk(+) ≤ β2Id,
for all T ∗ ≤ τk, k ∈ N, for all subdivision {τi}i∈N, τi − τi−1 0 be arbitrarily fixed. We define B(T ∗ )= 2b
a
+ 2b
a
1
e 2bT ∗ −1 + rT ∗ . From Lemma
82, x(T ∗ ) ≤ B(T ∗ ) independently from the subdivision {τi}i∈N.
Let us define µ as the maximum value such that the functions φ(τ), and ψ B(T ∗ )(τ) of Lemma
84 are both decreasing functions over the interval [0,µ]. Notice that µ depends on B(T ∗ ).
We claim that β2 = B(T ∗ )+rµ solves the problem.
Let us consider a time subdivision {τi}i∈N such that τi − τi−1 ≤ µ, for all i ∈ N.
We show first that if xk(+) ≤ β2, then xk+1(+) ≤ β2.
1. If xk(+) ≤ 2b
a then, from Lemma 80, xk+1(−) ≤ max � 2b
a ,xk(+) � = 2b
a . Therefore
xk+1(+) ≤ 2b
a + rµ ≤ β2,
2. if xk(+) ≥ 2b
a , and xk(+) ≤ B(T ∗ ), we use Lemma 80 again to deduce xk+1(−) ≤
max � 2b
a ,xk(+) � ≤ B(T ∗ ). Finally we have xk+1(+) ≤ B(T ∗ )+rµ = β2,
141