Adaptative high-gain extended Kalman filter and applications

tel-00559107, version 1 - 24 Jan 2011
for almost all t>0, for k1, k2 > 0. Then, if x (0) < k2 1
4k2 , we have
2
x(t) ≤ 4 x (0) e −k1t .
3.7 Technical Lemmas
Lemma 41 ([38])
Consider �b (˜z) − �b (˜x) − �b ∗ (˜z)˜ε as in the inequality (3.12) (omitting to write u in ˜ �
b) and
�
suppose θ ≥ 1. Then ��b (˜z) − �b (˜x) − �b ∗ �
�
(˜z)˜ε � ≤ Kθn−1 �˜ε� 2 , for some K>0.
Lemma 42 (adaptation function)
For any ∆T >0, there exists a positive constant M(∆T ) such that:
− for any θ1 > 1,and
− any γ1 > γ0 > 0,
there is a function F (θ, I) such that the equation
˙θ = F (θ, I (t)) , (3.16)
for any initial value 1 ≤ θ (0) < 2θ1, and any measurable positive function I (t), has the
properties:
1. that there is a unique solution θ (t) defined for all t ≥ 0, and this solution satisfies
1 ≤ θ (t) < 2θ1,
�
�
2. � F(θ,I)
�
�
� ≤ M,
θ 2
3. if I (t) ≥ γ1 for t ∈ [τ,τ + ∆T ] then θ (τ + ∆T ) ≥ θ1,
4. while I (t) ≤ γ0, θ (t) decreases to 1.
Remark 43
The main property is that if I (t) ≥ γ1, θ (t) can reach any arbitrarily large θ1 in an
arbitrary small time ∆T , and that this property can be achieved by a function satisfying
F (θ, I) ≤ Mθ 2 with M independent from θ1 (but dependant from ∆T ).
Proof.
Let F0 (θ) be defined as follows:
� 1
F0 (θ) = ∆T θ2 if θ ≤ θ1
2
(θ − 2θ1) if θ >θ 1
(the choice 2θ1 is more or less arbitrary) and let us consider the system
�
˙θ
θ (0)
=
=
F0 (θ)
1
.
Simple computations give the solution:
⎧
⎪⎨
∆T
θ (t) =
∆T − t
⎪⎩
θ1∆T
2θ1 −
θ1t + (2 − θ1)∆T
while θ ≤ θ1
when θ >θ 1.
1
∆T
50