Gas station without pumps

2021 May 4

Resonance with nonlinear impedances

Filed under: Circuits course — gasstationwithoutpumps @ 08:24
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My book uses nonlinear impedance Z=(j\omega\;1\,s)^\alpha M for modelling loudspeakers and electrodes.  The loudspeaker models include an inductor-like component with \alpha>0 and the electrode models include a capacitor-like component with \alpha<0. Standard linear components are special cases of the nonlinear impedance: inductors have \alpha=1, resistors have \alpha=0, and capacitors have \alpha=-1.

This past week I was thinking about the resonance that I see in some loudspeakers around 4MHz, which can be modeled with a capacitor in parallel with the main nonlinear impedance.  How can I estimate the capacitance to get initial values for model fitting?  In general, when do I get resonance with two nonlinear impedances?

Although I initially looked at the capacitor case, I realized that the general case of two nonlinear impedances in series gives me simple math that I can easily generalize to special cases and to parallel connections, so let’s look at Z=(j\omega\;1\,s)^\alpha M + (j\omega\;1s)^\beta N.  The first thing to do is to replace j by the polar form e^{j\pi/2}, getting Z=e^{j \alpha\pi/2}(\omega\;1\,s)^\alpha M + e^{j \beta\pi/2}(\omega\;1s)^\beta N.  The we can apply Euler’s formula to get the real and imaginary parts:

\Re(Z) = \cos(\alpha\pi/2)(\omega\;1\,s)^\alpha M + \cos(\beta\pi/2)(\omega\;1s)^\beta N.

\Im(Z) = \sin(\alpha\pi/2)(\omega\;1\,s)^\alpha M +\sin(\beta\pi/2)(\omega\;1s)^\beta N.

We will get resonance whenever the imaginary part goes to zero:

0= (\omega\;1\,s)^{\alpha-\beta} M\sin(\alpha\pi/2) +N \sin(\beta\pi/2), or

\omega = \left(\frac{ -N \sin(\beta\pi/2)}{M\sin(\alpha\pi/2)} \right)^{1/(\alpha-\beta)} s^{-1}.

The special case of an inductor and a capacitor sets \alpha=1, M = L / 1\,s, \beta = -1, and N= 1\,s/C,  yielding \omega= \left(\frac{N}{M}\right)^{1/2} s^{-1} = \sqrt{\frac{1}{LC}}, which is the standard result.

We get a resonance whenever \alpha and \beta have opposite signs.

We can deal with parallel rather than series impedances by looking at the sum of admittances instead of the sum of impedances.  To get the admittances, the exponents \alpha and \beta get negated and the coefficients M and N inverted, giving us 

\omega = \left(\frac{ -M \sin(\beta\pi/2)}{N\sin(\alpha\pi/2)}\right)^{1/(\beta-\alpha)} s^{-1}.

Note: this post is a much simpler analysis than last year’s in Resonance for non-linear impedance, because here I am just looking for where the phase goes to zero, rather than where the magnitude of impedance is minimized.

Update 2021 May 4: The two definitions of resonance I’ve used (minimum |Z| and \Im(Z)=0) are not the same—I tried doing a parametric plot of the magnitude vs. the phase for one asymmetric example (\alpha=0.6 and \beta=-0.2) and saw that the minimum magnitude did not occur at 0°.  So I’ll need to think some more about what I want “resonance” to mean for nonlinear impedances.  

 

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