Memristors: how are memristors modeled in terms of impedance? How are memristors modeled in terms of impedance?
I have been searching the net for memristors, but could not get much of an idea. I'm not interested in how memristors are made (by combining materials), (yet), because I first want to understand what they do, compared to classical RCL components.
I had a look at https://en.wikipedia.org/w/index.php?title=Memristor&oldid=757282109 (intentionally linked to the version I have read), but this left me more than confused. There, I didn't even recognize some formulas for RCL components.
Then I saw Why does electrical impedance have as many parameters as it has? and the answer https://physics.stackexchange.com/a/187301 , which does not explain about how memristors are modeled in terms of impedance.
 A: A memristor is, by definition, a two-terminal element whose constitutive relation is of the type [1]
$$g(\varphi,q)=0,$$
where $\varphi = \int_{-\infty}^t v(s)\,\mathrm{d}s$ is the flux linkage and $q= \int_{-\infty}^t i(s)\,\mathrm{d}s$ is the charge.
If the relationship $g(\varphi,q)=0$ is linear, the memristor degenerates into a linear resistor and, thus, its impedance coincides with the resistance.
When the relationship $g(\varphi,q)=0$ is nonlinear, the memristor becomes a nonlinear element for which the concept of impedance is valid only in the so-called small signal approximation, that is, when the applied signal around a certain operating point is sufficiently small that the nonlinearity can be neglected. This kind of linearization is well known in circuit theory, and, for instance, it is commonly applied when analyzing diode circuits and transistor amplifiers. Let's see how this can be done in the case of the memristor.
If the relationship $g(\varphi,q)=0$ can be solved for $\varphi$, that is, we can write $\varphi = f_\mathrm{M}(q)$ (at least in a certain interval) the memristor is called charge-controlled and its $iv$ characteristic is given by [1]
$$v = R(q)i,\qquad\qquad(1)$$
where
$$R(q) = \frac{\mathrm{d}f_\mathrm{M}(q)}{\mathrm{d} q}$$
is a charge-dependent resistance. Since $q= \int_{-\infty}^t i(s)\,\mathrm{d}s$, the charge $q$ at any given instant depends on the past history of the current. But once you have reached a certain operating point $q$, the one of interest, you can think of applying a sine wave with infinitely small amplitude, so that the charge remains virtually constant. Around this operating point, you can see from (1) that the voltage across the memristor is proportional to the current, that is, for small signals the memristor behaves like a resistor with differential resistance $R(q)$. Hence, the small signal impedance will be $R(q)$, independent of frequency.
A similar conclusion can be obtained for the case in which the relationship $g(\varphi,q)=0$ can be solved for $q$.
To sum up, the small-signal (I cannot stress this enough) impedance of a memristor is resistive, independent of frequency, and, because of memristor nonlinearity, dependent on the operating point. 
A more detailed analysis of the circuit behaviour of the memristor can be found in Chua's papers [1-3]. In [3], in particular, it is discussed the impedance. For more general information on small-signal analysis you can have a look at [4] (Chua again!).
[1] L. O. Chua, "Memristor - The missing circuit element," IEEE Trans. Circuit Theory, CT-18, 507–519, 1971.
[2] L. O. Chua, "The fourth element", Proc. IEEE, 100, 1920-1927, 2012.
[3] L. O. Chua, "Nonlinear circuit foundations for nano devices, Part I: The Four-Element torus," Proc. IEEE, 91, 1830–1859, 2003.
[4] L. Chua, C. A. Desoer, and E. S. Kuh, Linear and nonlinear circuits, McGraw-Hill, 1987.
