Adaptative high-gain extended Kalman filter and applications

tel-00559107, version 1 - 24 Jan 2011
B.1 Bounds on the Riccati Equation
Proof.
Let M(t) be a primitive matrix of m(t), that is to say a matrix whose elements are the
primitives of the elements of m(t). We have the identity
� T
m(v)dv = M(T ) − M(0) =
0
k�
[M(τi) − M(τi−1)] .
We can apply the Taylor-Lagrange expansion on all elements Mkl:
i=1
Mkl(τi−1) =Mkl(τi)+(τi−1 − τi) mkl(τi)+ (τi−1 − τi) 2
where ξkl,i ∈ [τi−1, τi]. We have thus, the relation
k�
M(τi−1) =
i=1
k�
M(τi)+
i=1
k�
m(τi) ((τi−1 − τi)) +
where (Rkl)i = m ′
kl (ξkl,i), with ξkl,i ∈ [τi, τi−1]. Therefore
� T
m(v)dv −
0
i=1
k�
m(τi)(τi − τi−1) =
i=1
i=1
= � �
k (τi−1−τi)
i=1
2
2
Ri
2
k�
i=1
k�
[M(τi) − M(τi−1)] −
�
.
m ′
kl (ξkl,i)
�
(τi−1 − τi) 2
2
Ri
�
k�
((τi − τi−1)m(τi))
We now use the definition of the matrix inequality. Let x be a non zero element of Rn ,
x ′
�
k�
�
(τi−1 − τi)
i=1
2
��
Ri x
2
= 1
k� �
(τi−1 − τi)
2
i=1
2 x ′
�
Rix
≤ 1
⎛
⎞
≤
k�
�
⎝(τi−1
2
− τi) |xk||Rk,l|
2
i |xl| ⎠
i=1
k,l
1 � �
µ max |Rk,l|
2 k,l,i
i
� �
�k
τi−1 − τi
⎛
⎝ �
⎞
|xk||xl| ⎠
Thus giving the result.
≤ 1 �
µ max |Rk,l|
2 k,l,i
i
≤ 1
2
�
µ max |Rk,l|
k,l,i
i
i=1
� � k �
� T 1
2
i=1
τi−1 − τi
�
�
2n �x� 2�
.
i=1
1
2
k,l
⎛
⎝ �
|xk| 2 + |xl| 2
⎞
⎠
Remark 94
Suppose that m = ψ ′
(v, T )C ′
Cψ(v, T ) where ψ(v, T ) is a resolvent matrix as in Lemma
90. From the assumptions we put on our continuous discrete system, and the fact that the
147
k,l