Adaptative high-gain extended Kalman filter and applications

tel-00559107, version 1 - 24 Jan 2011
Thus we have:
x2(+) ≤ 2b
a
+ 2b
a
B.1 Bounds on the Riccati Equation
1
e2bτ2 + rτ2.
− 1
This last inequality is of the same form as the first bound of (B.5), and is independent
from the values of τ1 and τ2. We can therefore generalize it to any i ∈ N ∗ :
xi(+) ≤ 2b
a
+ 2b
a
1
e2bτi + rτi.
− 1
Moreover, we can generalize this inequality to any τ > 0, before or after the update (i.e. the
correction step):
x(τ) ≤ 2b 2b 1
+
a a e2bτ + rτ.
− 1
Remark 83
This last bound cannot be used to obtain an upper bound for x for all times τ ≥ T ∗
because it is not a decreasing function of time (this is proven later on). In other words,
suppose we have two times 0 < ξ1 < ξ2, then there exist two positive scalars, β1 and β2 such
that x(ξ1) < β1 and x(ξ2) < β2. But we don’t know if β2 is higher or not than β1.
In the following we’ll see that such a relation can be derived provided that the length
between two samples is bounded.
Lemma 84
Let us define the functions
φ(τ) = 2b
a
ψx0 (τ) =
2b 1
+
a e2bτ + rτ,
− 1
2bx0e2bτ ax0 (e2bτ + rτ.
− 1) + 2b
There exists µφ > 0, and µψ(x0) > 0 such that φ(τ), respectively ψx0 (τ), is a decreasing
function for τ ∈]0,µφ], respectively for τ ∈ [0,µψ(x0)].
Moreover µψ(x0) is an increasing function of x0.
Proof.
The proof basically consists in the computation of the variation tables of the two functions.
1.
φ ′
Let us consider the polynomial
(τ) = ra(e2bτ ) 2 − (4b2 +2ar)e2bτ + ra
a(e2bτ − 1) 2 .
P (X) =raX 2 − (4b 2 +2ar)X + ra.
��(2b) � �
Its discriminant 2 2
+2ar − 4a2r2 is positive. Therefore P (X) has two real
roots whose product and sum are both positive. Thus both roots are positive.
We denote them 0