Adaptative high-gain extended Kalman filter and applications

tel-00559107, version 1 - 24 Jan 2011
2.5
2
1.5
1
0.5
Load torque estimate
Overshoots
10 10.5 11 11.5 12 12.5 13 13.5 14 14.5
Figure 4.9: Choice of a value for 2θ1.
4.2 Simulation
Real values
! 0 = 5
! 0 = 10
! 0 = 8
! 0 = 15
We chose a sigmoid function whose equation appears in the definition of the observer
(4.5). A graphical representation is displayed in Figure 4.10. The parameters
β and m play the same role as the bounding parameters, γ0 and γ1, in the
properties of the innovation shown above. The first parameter, β, controls the
duration of the transition part of the sigmoid. The higher β is, the smaller is the
transition. In practice, the best results are obtained for a small transition time
(i.e. a large value of β) 7 .
When m = 0, µ(0) = 0.5. The role of the parameter m is to pull the sigmoid to
the right. m is actually divided into two components, i.e., m = m1 +m2. We want
m1 is to be such that µ(Id) ≈ 0 when Id is around 0. Because of the presence of
noise in the measured signal, we need to add m2 to this value. The procedure is
explained below. The choice of m1 can be made either via a graphical method or
by solving a nonlinear equation 8 .
− λ: Contrary to the observer of [38] “λ small enough” is no longer required, since
θ increases when estimation error becomes too large. However, the situation may
arise when the innovation oscillates up and down after some large disturbance.
7
Note that the need for a small transition time is consistent with the requirements expressed in Theorem
36 of Chapter 3, i.e., when neither θ is high-gain nor the estimation error is sufficiently small, the state of the
system is unclear. We want to reach one of those two domains.
8
Let us choose 0 < ε1 < 1, small and l, a transition length. We keep m = 0 and solve the equation
µ(l/2) − µ(−l/2) = (1 − ε1) − ε1 in order to find the corresponding β.
Now choose ε2 > 0, sufficiently small, is used as the zero value (the function µ(x) = 0 has no solution). Set β
to the value found previously and solve for m using the equation µ(0) = ε2.
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