Adaptative high-gain extended Kalman filter and applications

tel-00559107, version 1 - 24 Jan 2011
Since x(τ) > 2b
a , equation (B.6) implies:
Which gives the first inequality:
therefore
x
(x − 2b
a ) ≥ e2bτ .
x ≥
x ≤ 2b
a
�
x − 2b
�
e
a
2bτ
+ 2b
a
We obtain the second inequality simply from (B.6):
Remark 81
x(τ) ≤
B.1 Bounds on the Riccati Equation
1
e2bτ − 1 .
2bx0e 2bτ
ax0 (e 2bτ − 1) + 2b .
1. The first inequality of (B.5) can be used for any initial value S(0), since it tends toward
+∞ when τ → 0,
2. the second one requires some knowledge on S(0) in order to be useful,
3. the two bounds obtained are higher than 2b
a , therefore they are true for any τ ∈ [0,T].
�
Let us denote r = sup Tr(C ′
R−1 �
C) . According to equation (B.1), the problem turns
into proving that xk(+), the solution of:
�
dx
dτ = −ax2 xk(+) ≤
+2bx
xk(−)+(τk − τk−1) r,
is upper bounded for all T ∗ ≤ τk, k ∈ N, independently from the subdivision {τi},i∈ N.
(B.7)
The bounds of Lemma 80 are valid on intervals of the form 2 ]τi−1, τi]. Let us find an
expression that we can use in order to upper bound x at any time.
Lemma 82
The solution of (B.7) is such that:
x(τ) ≤ 2b
a
for any τ > 0, before or after an update.
+ 2b
a
1
e2bτ + rτ,
− 1
2 Or of the form [τi−1, τi], it depends on which bound one considers.
137