284
7 Pulse Programming of Memristor Circuits
C ˙
ϕ C (t; t 0 ) = −h
−1 (ϕ C (t; t 0 ) − ϕ e (t; t 0 ) + h(q M (t 0 )))
+ q C (t 0 ) + q M (t 0 ).
(7.14)
On the contrary, if R = −R 1 but ϕ M = h(q M ) is not invertible, then the DAEs
cannot be cast in the form of a SE. It can be seen that in this case the memristor
circuit in Fig. 7.5 describes a nonphysical situation because it does not have a
globally defined SE and it exhibits impasse points (cf. Chap. 4).
2. If R = −R 1 the DAEs reduce to the next SE if and only if H (q M (t)) =
h(q M (t)) + (R 1 + R)q M (t) is strictly increasing
C ˙
ϕ C (t; t 0 ) = −H(ϕ C (t; t 0 ), ϕ e (t; t 0 ), q M (t 0 ))
+q C (t 0 ) + q M (t 0 )
(7.15)
where we have let H(ϕ C (t; t 0 ), ϕ e (t; t 0 ), q M (t 0 )) = H −1 (ϕ C (t; t 0 ) − ϕ e (t; t 0 ) +
h(q M (t 0 )) + (R 1 + R)q M (t 0 )). Note that, even if the memristor is passive, there
are values of R < 0 for which the required invertibility condition on H (q M (t))
fails, so that once more the SE does not exist.
It is worth noting that in both cases R = −R 1 , or R = −R 1 , as shown
in Chap. 5, a technique that guarantees the explicit derivation of the SE for the
memristor circuit in Fig. 7.5 is the insertion of an inductor in series with the
charge-controlled memristor and the negative resistor.
7.4 State Equations in the Flux-Charge Domain
We consider henceforth memristor circuits N ∈ LM satisfying (A1). Consider also
the following assumptions—denoted as Assumption 2, i.e., (A2), and Assumption
3, i.e., (A3)—enabling to specify all possible configurations of N = N D
N A
according to the structure of N A :
(A2) The subnetwork N A of N has no memristors.
(A3) The subnetwork N A of N has no negative resistors.
Clearly, N admits of only these four configurations:
(i) if both (A2) and (A3) are fulfilled, then μ F = μ Q = 0 and ρ G = ρ R = 0, that
is N A = N R and N = N D
N R
(ii) if only (A2) is fulfilled, then μ F = μ Q = 0, ρ G = 0 and ρ R = 0, that is
N = N D
N R
A
ϕ
G
A
q
R
(iii) if only (A3) is fulfilled, then μ F = 0, μ Q = 0 and ρ G = ρ R = 0, that is
N = N D
N R
A
ϕ
M
A
q
M
7 Pulse Programming of Memristor Circuits
C ˙
ϕ C (t; t 0 ) = −h
−1 (ϕ C (t; t 0 ) − ϕ e (t; t 0 ) + h(q M (t 0 )))
+ q C (t 0 ) + q M (t 0 ).
(7.14)
On the contrary, if R = −R 1 but ϕ M = h(q M ) is not invertible, then the DAEs
cannot be cast in the form of a SE. It can be seen that in this case the memristor
circuit in Fig. 7.5 describes a nonphysical situation because it does not have a
globally defined SE and it exhibits impasse points (cf. Chap. 4).
2. If R = −R 1 the DAEs reduce to the next SE if and only if H (q M (t)) =
h(q M (t)) + (R 1 + R)q M (t) is strictly increasing
C ˙
ϕ C (t; t 0 ) = −H(ϕ C (t; t 0 ), ϕ e (t; t 0 ), q M (t 0 ))
+q C (t 0 ) + q M (t 0 )
(7.15)
where we have let H(ϕ C (t; t 0 ), ϕ e (t; t 0 ), q M (t 0 )) = H −1 (ϕ C (t; t 0 ) − ϕ e (t; t 0 ) +
h(q M (t 0 )) + (R 1 + R)q M (t 0 )). Note that, even if the memristor is passive, there
are values of R < 0 for which the required invertibility condition on H (q M (t))
fails, so that once more the SE does not exist.
It is worth noting that in both cases R = −R 1 , or R = −R 1 , as shown
in Chap. 5, a technique that guarantees the explicit derivation of the SE for the
memristor circuit in Fig. 7.5 is the insertion of an inductor in series with the
charge-controlled memristor and the negative resistor.
7.4 State Equations in the Flux-Charge Domain
We consider henceforth memristor circuits N ∈ LM satisfying (A1). Consider also
the following assumptions—denoted as Assumption 2, i.e., (A2), and Assumption
3, i.e., (A3)—enabling to specify all possible configurations of N = N D
N A
according to the structure of N A :
(A2) The subnetwork N A of N has no memristors.
(A3) The subnetwork N A of N has no negative resistors.
Clearly, N admits of only these four configurations:
(i) if both (A2) and (A3) are fulfilled, then μ F = μ Q = 0 and ρ G = ρ R = 0, that
is N A = N R and N = N D
N R
(ii) if only (A2) is fulfilled, then μ F = μ Q = 0, ρ G = 0 and ρ R = 0, that is
N = N D
N R
A
ϕ
G
A
q
R
(iii) if only (A3) is fulfilled, then μ F = 0, μ Q = 0 and ρ G = ρ R = 0, that is
N = N D
N R
A
ϕ
M
A
q
M