THÈSE DE DOCTORAT Ecole Doctorale « Sciences et ...

the polynomial term on the left-hand side will ‘cross’ the
exponential on the right-hand side at a unique point.
c)Considering regularly persistent inputs, the positive
definition of the solution S of equations (3b)-(4b) can
only be ensured under condition on the sampling steps,
otherwise the solution of equation (3b) may become nonpositive
definite [21].
d) Notice that β 1 , β 2 implicitly depend on the maximum
allowable sampling interval ¯τ.
Lemma 7 Suppose the following system to be globally
exponentially stable,
ζ(t k+1 ) = F(t k )ζ(t k ),
and let u be a bounded and integrable function of time
that vanishes exponentially fast, then, system
z(t k+1 ) = F(t k )z(t k ) + u(t k ), k ≥ 0,
converges exponentially to 0.
Remark 8 This lemma is a straight extension of Lemma
III.1, in [18], for discrete-time systems.
4.3 Stability analysis
System (3a-d)-(4a-d) is said to be a global exponential
adaptive observer for system (1) if, along solutions to
(1), (3a-d)-(4a-d):
(i) for all (x 0 , ˆx 0 , λ 0 , ˆθ 0 ) ∈ R n ×R n ×R n ×R l and any
S 0 > 0 symmetric, there exist a 1 , a 2 , b 1 , b 2 ∈ R + ,
such that for all t ≥ t 0 , ‖e(t)‖ ≤ a 1 e −b1(t−t0) ‖e 0 ‖
and ||˜θ(t)|| ≤ a 2 e −b2(t−t0) ||˜θ 0 ||.
(ii) there exist ¯λ, ¯S ∈ R + , such that, for all t ≥ t 0 ,
‖λ(t)‖ ≤ ¯λ, ‖S(t)‖ ≤ ¯S.
Condition (i) ensure the exponential convergence of the
state and variable estimates to the corresponding true
value, whereas (ii) ensure that the other observer variables
do not explose in (in)finite time.
Theorem 9 Assuming Hypothesis 2 to hold, the input
to be regularly persistent, and choosing µ ≥ ξ if,
τ ≤ min{ (2ρ−1)β1
ρ 2 ‖C‖ 2 , ¯τ}, (9)
then system (3a-d)-(4a-d) is a global exponential adaptive
observer for system (1).
Proof. Like in [23,6], the variable η is introduced as
η = e − λ˜θ, where ˜θ = θ − ˆθ. Firstly, the global
exponential convergence of η to zero, along the solutions
to dynamical equations to (1) and (3a-d)-(4a-d), is
proved. After having noticed that variables λ and S remain
bounded, invoking Lemma 7, the exponential convergence
to the origin of errors e and ˜θ is deduced.
The candidate Lyapunov function is defined as, for t ≥
t 0 :
V (t) = η(t) T S(t)η(t).
Note that, according to Proposition 5 and since (9) is
satisfied, the symmetric matrix S is positive definite and
of bounded norm. Let k ≥ 0.
1. Let t ∈ [t k , t k+1 ). The variable η satisfies the following
dynamical equation, in view of (5a) and because,
here, ˙˜θ = 0 (3d):
˙η(t) = A(u)e(t) + φ(u)˜θ(t k ) − ˙λ(t)˜θ(t k )
(
˙λ(t))
= A(u)η(t) + A(u)λ(t) + φ(u) − ˜θ(tk ).
Thus, according to (3c),
˙η(t) = A(u)η(t). (10)
Consequently, differentiating the Lyapunov function
along the solutions of (10),
˙V (t) = η(t) T (
S(t)A(u) + A(u) T S(t) + Ṡ(t) )
η(t).
From (3b),
Integrating (11) over [t k , t],
and so,
˙V (t) = −µV (t). (11)
V (t) = e −µ(t−t k) V (t k ), (12)
V (t − k+1 ) = e−µτ k
V (t k ). (13)
2. Consider now t = t k+1 , it can be shown that, in view
of (5b),
η(t k+1 )=(M(τ k , t k+1 ) −λ(t k+1 )∆(t k+1 )C)e(t − k+1 )
−λ(t k+1 )˜θ(t k+1 )
)
= M(τ k , t k+1 )
(η(t − k+1 ) + λ(t− k+1 )˜θ(t − k+1
(
−λ(t k+1 ) ∆(t k+1 )Ce(t − k+1 ) + ˜θ(t
)
k+1 ) .
Remarking that ˜θ(t − k+1 ) = ˜θ(t k ) in view of (3d), and
using (4b-d),
η(t k+1 ) = M(τ k , t k+1 )η(t − k+1
). (14)
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