7.7 Examples on Memristor Circuit Programming
299
N for any t ≥ t 0 is governed by the non-autonomous SEs (7.27) with input u x (t)
and u y (t) given by (7.44). Hence, any solution w(t; t 0 , w 0 ) of N is embedded into
the (continuous) family of manifolds M(k 0 + H 12 H
−1
22 u y (t) − u x (t)) parametrized
by time t ∈ [t 0 , +∞) (see (7.35) and Theorem 7.4).
Although the solution of N is instant by instant on a known manifold, complex
dynamic behavior and bifurcation phenomena can emerge due to the effect of the
external sources with time-varying momentum in the (ϕ, q)-domain.
The next section presents selected examples for a thorough illustration of
the explicit form of the SEs of memristor circuits N = N D
N R and their
programming by tuning invariant manifolds via suitable external sources.
7.7 Examples on Memristor Circuit Programming
In this section we discuss some fundamental examples of memristor circuits in LM
where we apply the technique developed in the chapter for programming via pulses
different manifolds and reduced-order dynamics. First, we reconsider the Memristor
Chaotic Circuit (MCC) studied in Chap. 6, then we study a circuit with a flux- and a
charge-controlled memristor, and, finally, a class of memristor neural networks with
a star topology.
7.7.1 Memristor Chaotic Circuit
In Sect. 6.3 of Chap. 6 we have considered a chaotic circuit named MCC obtained
by replacing the nonlinear resistor of a Chua’s oscillator with a flux-controlled
memristor. We have seen that for the MCC there coexist different regimes, i.e.,
convergent, periodic, and chaotic regimes for the same set of circuit parameters
and fixed nonlinearities. Here, for programming purposes we add a charge source
q a (t; t 0 ) and a flux source ϕ e (t; t 0 ) in the MCC as shown in Fig. 7.7.
For MCC we have γ M = 1, γ C = 1 and λ L = 1, hence n C = 2, n L = 1 and then
n M = n x = 1, n y = 2. MCC satisfies (A2) and (A3); moreover, by inspection it is
seen that it also satisfies (A1).
A simple circuit analysis based on Fig. 7.7 permits to derive H and B for the
hybrid representation in (7.16) (being G = 0 since there are no negative resistors,
and n E = n A = 1). Namely, we have
H 11 =
1
R
, H 12 =
−
1
R 0
= H
T
21
(7.45a)
H 21 =
−
1
R
0
, H 22 =
1
R −1
1 r
(7.45b)
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