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Adaptative high-gain extended Kalman filter and applications

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tel-00559107, version 1 - 24 Jan 2011<br />

5.2 Continuous-discrete Framework<br />

is, therefore, also symmetric positive definite. We consider the same function as before (Cf.<br />

Lemma 33):<br />

Λ : L∞ �<br />

[0,d] R n(n+1) �<br />

2 × L∞ [0,d] (Rnu ) −→ R + .<br />

With the same reasoning as in Lemma 33, we deduce the existence of a scalar λ0 d > 0 such that<br />

Gd ≥ λ0 d Id for any u <strong>and</strong> any matrix B(t) having the structure specified above. Therefore:<br />

�i=d<br />

�<br />

�yi i=0<br />

5.2.5 Preparation for the Proof<br />

� 0<br />

0,x � � 0<br />

− yi 0, ξ �� �2 = � x 0 − ξ 0�′ � 0 0<br />

Gd x − ξ �<br />

≥ λ 0 �<br />

� 0 0<br />

d x − ξ � �2 .<br />

In order to establish the preliminary inequalities needed in the proof, we first recall the<br />

matrix inversion lemma:<br />

Lemma 64 (matrix inversion lemma [67], Section 0.7.4)<br />

If M is symmetric positive definite, <strong>and</strong> λ > 0 then<br />

(M + λMC ′<br />

CM) −1 = M −1 − C ′<br />

(λ −1 + CMC ′<br />

) −1 C.<br />

The estimation error is denoted ɛ(t) =z(t) − x(t), <strong>and</strong> we consider the change of variables:<br />

− ˜x = ∆x, ˜z = ∆z <strong>and</strong> ˜ɛ = ∆ɛ,<br />

− ˜ b(., u) =∆b(∆ −1 ., u), <strong>and</strong> ˜ S = ∆ −1 S∆ −1 ,<br />

− ˜ b ∗ (., u) =∆b ∗ (∆ −1 ., u)∆ −1 .<br />

As before, the Lipschitz constant of the vector field remains the same in the new system of<br />

coordinates (Cf. Lemma 51).<br />

The error dynamics are given by<br />

in the continuous case, <strong>and</strong><br />

˙ɛ = A(u)ɛ +(b(z, u) − b(x, u)) (5.26)<br />

ɛk(+) = ɛk(−) − δtS −1 (+)C ′<br />

r −1<br />

θ Cɛk(−) (5.27)<br />

in the discrete case.<br />

We <strong>high</strong>light the relations below as they are useful for the following computations:<br />

where N = diag ({0, 1, . . . , n − 1}).<br />

∆A(u) =θA(u)∆,<br />

∆−1A ′<br />

(u) =θA ′<br />

(u)∆−1 ,<br />

Remark 65<br />

Notice that on intervals of the form [kδt, (k + 1)δt[, the derivative of θ is equal to zero.<br />

108<br />

(5.28)

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