Adaptative high-gain extended Kalman filter and applications

tel-00559107, version 1 - 24 Jan 2011
Those two roots are:
and
B.1 Bounds on the Riccati Equation
X∗ = (4b2 +2ar) − � (4b2 +2a) 2 − 4a2r2 � 2ar √ �
(4b2 +2ar) 2 − 4a2r2 + 4a2r2 − (4b2 +2a) 2 − 4a2r2 ≤
,
2ar
≤ 1,
X µ = (4b2 +2ar)+ � (4b2 +2a) 2 − 4a2r2 > 1.
2ar
We conclude that there exists µφ > 0 such that φ(τ) is a decreasing function over the
interval [0,µφ].
2. Let us remark first that ψx0 (0) = x0. A few computations give us:
ψ ′
x0 (τ) =ra2 x2 0 (e2bτ ) 2 − x0(ax0 − 2b)(4b2 +2ra)e2bτ + r(ax0 − 2b) 2
[ax0(e2bτ − 1) + 2b] 2
.
As before we consider the polynomial
Its discriminant is
P (X) =ra 2 x 2 0X 2 − x0(ax0 − 2b)(4b 2 +2ra)X + r(ax0 − 2b) 2 .
x 2 0(ax0 − 2b) 2 ((2b) 2 +2ra) 2 − 4r 2 a 2 x 2 0(ax0 − 2b) 2 .
It is positive. Again both roots are positive: 0 0, a function of x0, such that ψ(τ) is a decreasing function over the
interval ]0,µψ(x0)].
In order to check that µψ(x0) increases with x0, we show that X µ is an increasing
function of x0.
X µ = x0(ax0 − 2b)(4b2 +2ar)
2ra2x2 � 0
(ax0 − 2b)
+
2 (4b2 +2ar) 2x2 0 − 4r2a2x2 0 (ax0 − 2b) 2
2ra2x2 =
0
(a − 2b
x0 )(4b2 +2ar)
2ra2 �
+
(a − 2b
x0 )2 (4b2 +2ar) 2 − 4r2a2 (a − 2b
x0 )2
2ra2 = A+√B D .
3 Since the coefficient of the second order term of P (X) is positive, then the polynomial has negative sign
for X ∈]X ∗ ,X µ [. We only to check that P (1) ≤ 0.
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