8.2 Memristor-Based Chaotic Circuit (MCC)
323
L
di L (t)
dt
=v C 2 (t)
(8.2c)
dϕ M (t)
dt
=v C 1 (t)
(8.2d)
for all t ≥ t 0 , with ICs w(t 0 ) = (v C 1 (t 0 ), v C 2 (t 0 ), i L (t 0 ), ϕ M (t 0 )) T . Here, we have
taken into account that ϕ C 1 (t; t 0 ) + ϕ M (t 0 ) = ϕ M (t; t 0 ) + ϕ M (t 0 ) = ϕ M (t). The
SEs (8.2) are an IVP for a fourth-order system of ODEs in the state variables w(t).
For a better understanding of the order reduction of the MCC dynamics in the
(ϕ, q)-domain, with respect to the (v, i)-domain, let us verify that the state space in
the (v, i)-domain can be foliated in a continuum of invariant manifolds. Consider
the function Q : R 4 → R of the state variables in the (v, i)-domain
Q(w) = f (ϕ M ) +
1
R
ϕ M + C 1 v C 1 −
L
R
i L
(8.3)
and, for any Q 0 ∈ R, let
M(Q 0 ) = {w ∈ R
4
: Q(w) = Q 0 }
(8.4)
which is a three-dimensional manifold in R 4 that coincides with the Q 0 -level set of
function Q(·). Note that, for any w ∈ R 4 , we have w ∈ M(Q(w)).
Property 8.1 The state space R 4 of the MCC in the (v, i)-domain can be foliated
in ∞ 1 3D manifolds M(Q 0 ) by varying Q 0 ∈ R. Manifolds are nonintersecting
and they span the whole state space R 4 . Each manifold is positively invariant for the
dynamics of the MCC in the (v, i)-domain, i.e., if the ICs w(t 0 ) ∈ M(Q 0 ), where
Q 0 = Q(w(t 0 )), then the solution (v C 1 (t), v C 2 (t), i L (t), ϕ M (t)) of the IVP (8.2)
belongs to M(Q 0 ) for any t ≥ t 0 . On each manifold M(Q 0 ) the dynamics of the
MCC is described in the (ϕ, q)-domain by the third-order system of ODEs (8.1).
Proof Suppose for purposes of contradiction that there exists x ∈ R 4 such that
w ∈ M(Q 1 ) ∩ M(Q 2 ), with Q 1 = Q 2 . This implies Q(w) = Q 1 and Q(w) =
Q 2 , which is a contradiction. Then, manifolds are nonintersecting. To see that they
cover the whole space R 4 it is enough to recall that, for any w ∈ R 4 , we have
w ∈ M(Q(w)).
Let us now show that each manifold is positively invariant by analyzing the
associated circuit of the MCC in the (ϕ, q)-domain. We have from KqL at the cutset made of nodes A and B (see Fig. 8.2)
q M (t; t 0 ) + q C 1 (t; t 0 ) + q C 2 (t; t 0 ) + q L (t; t 0 ) = 0
for any t ≥ t 0 . From KqL at node B
q R (t; t 0 ) + q C 2 (t; t 0 ) + q L (t; t 0 ) = 0
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