Adaptative high-gain extended Kalman filter and applications

tel-00559107, version 1 - 24 Jan 2011
− ˜ b(., u) =∆b(∆ −1 ., u),
− ˜ b ∗ (·,u)=∆b ∗ � ∆ −1 ·,u � ∆ −1 .
3.5 Preparation for the Proof
Since C = � a1(u) 0 . . . 0 � , and ∆= diag �� 1, θ −1 , ...,θ −(n−1)�� then C∆ = C. We
have the following identity for the A(u) and∆:
⎛
⎞ ⎛
0 a2 (u) 0 · · · 0 1 0 0 · · · 0
⎜
.
⎜ 0 a3 (u) ..
⎟ ⎜
. ⎟ ⎜ 1
⎟ ⎜ 0 θ 0 .
A(u)∆= ⎜
.
⎜
.
.. . .. ⎟ ⎜
0 ⎟ ⎜
1
⎟ ⎜ 0 0 θ
⎝
0 an (u) ⎠ ⎜
⎝
0 · · · 0
2
. .. .
.
. .. . .. 0
1
0 · · · · · · 0 θn−1 ⎞
⎟
⎠
⎛ a2(u)
0 θ 0 · · · 0
⎜
a3(u)
⎜ 0 θ
= ⎜
⎝
2
. .. .
.
. .. . .. 0
an(u)
0 θn−1 ⎞
⎟ =
⎟
⎠
0 · · · 0
1
θ ∆A(u).
This relation leads to the set of equalities:
(a) ∆A = θA∆, (b) A ′
∆ = θ∆A ′
,
(c) A∆−1 = θ∆−1A, (d) ∆−1A ′
= θA ′
∆−1 .
Because we want to express the time derivative of ˜ε we need to know the time derivative of
∆, as θ is time dependent. We simply write
d∆
dt =
⎛
d(1)
dt 0 · · · 0
⎜ � �
⎜ d 1
⎜ 0 dt θ
.
⎜
⎝
.
.
..
�
0
d 1
0 · · · 0 dt θn−1 ⎞
⎟
⎠
�
=
⎛
0 0 · · · 0
⎜ 0 −
⎜
⎝
˙ θ
θ2 ⎞
⎟
. ⎟
.
. .. ⎟
0 ⎠ ,
0 · · · 0 − (n−1) ˙ θ
θ n
which can be rewritten as a multiplication of matrices with the use of
N = diag ({0, 1, 2, ..., n − 1}). We obtain the two identities10 :
d
F(θ,I)
(a) dt (∆) =− θ N∆ (b) d
dt
(3.7)
� ∆ −1 � = F(θ,I)
θ N∆−1 . (3.8)
The dynamics of the error are given by:
�
˙ε =˙z − ˙x = A (u) − S −1 C ′
R −1
θ C
�
ε + b (z, u) − b (x, u) ,
and the error dynamics after the change of variables are:
d˜ε
dt
d∆ = dt ε + ∆˙ε
= − ˙ θ
θ
10 remember that ˙ θ = F(θ, I).
N∆ε + ∆
�
A (u) − S−1C ′
R −1
θ C
�
ε + ∆( b (z, u) − b (x, u))
= − ˙ θ
θ N ˜ε + θA (u)˜ε − θ∆S−1∆∆−1C ′
R−1C∆−1∆ε �
= θ − F(θ,I)
θ2 N ˜ε + A˜ε − ˜ S−1C ′
R−1C ˜ε + 1
� ��
˜b
θ (˜z, u) − ˜b (˜x, u) .
45
+∆b � ∆ −1 ˜z, u � − ∆b � ∆ −1 ˜x, u �
(3.9)