Adaptative high-gain extended Kalman filter and applications

tel-00559107, version 1 - 24 Jan 2011
therefore φu(τ,s) � d
dτ h(τ)� φ ′
u(τ,s)=F (τ), and
� τ
h(τ) =h(0) +
SinceΛ( τ) =φu(τ,s)h(τ)φ ′
u(τ,s), thus φu(s, 0)Λ(0)φ ′
0
B.1 Bounds on the Riccati Equation
φu(s, v)F (v)φ ′
u(s, v)dv.
Λ(τ) =
u(s, 0) = h(0). Therefore:
φu(τ,s)h(0)φ ′
�� τ
u(τ,s)+φu(τ,s) φu(s, v)F (v)φ
0
′
�
u(s, v)dv φ ′
=
u(t, s)
φu(τ,s)φu(s, 0)Λ(0)φ ′
u(s, 0)φ ′
� τ
u(τ,s)+ φu(τ,v)F (v)φ
0
′
=
u(τ,v)dv
φu(τ, 0)Λ0φ ′
� τ
u(τ, 0) + φu(τ,v)F (v)φ ′
u(τ,v)dv.
0
Let us take λ ∈ R ∗ , and define ˆ S = e λτ S. The differential equation associated to ˆ S is
d
dτ ˆ S(τ) = λeλτ S + eλτ dS
dτ
= λ ˆ S(τ) − A ′ S(τ) ˆ − S(τ)A ˆ − e−λτ S(τ)Q ˆ S(τ). ˆ
This equation is of the form (B.12), with F (τ) = λ ˆ S(τ) − e−λτ SQ ˆ S. ˆ According to the
computation above, we have:
ˆS(τ) =φu(τ, 0) ˆ S(0)φ ′
� τ �
u(τ, 0) + φu(τ,v) λ ˆ S(v) − e −λv �
SQ ˆ Sˆ
φ ′
u(τ,v)dv,
and consequently
S(τ) =φu(τ, 0)S0φ ′
u(τ, 0)e −λτ � τ
+ λ
Lemma 92
Let S : [0; e(S)[→ Sm be a maximal semi solution of
If S(0) = S0 is positive definite then
0
0
e −λ(τ−v) �
φu(τ,v) S − SQS
�
φu(τ,v)dv.
λ
d
S = −A′ S − SA − SQS.
dτ
e(S) =+∞ and S(τ) is positive definite for all τ ≥ 0.
Thus, for any arbitrary time subdivision {τi}i∈N∗ the solution to the continuous discrete
Riccati equation (B.1) is positive definite for all times provided that S0 is positive definite.
Proof.
Assume that S is not always positive definite. Let us define
θ = inf (τ|S(τ) /∈ Sm(+)).
In other words S(τ) ∈ Sm(+) for all τ ∈ [0; θ[. From Lemma 79:
d
dτ Tr(S) ≤−aTr(S) 2 +2bTr(S)
145