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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.  

 

2020 June 9

Error in video

Filed under: Circuits course — gasstationwithoutpumps @ 22:00
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After posting Resonance for non-linear impedance today, I realized that there was a small error in the most recent video (Tenth video for electronics book).  In particular, the estimation formula  was incorrect for the initial value for C2 based on the resonant frequency of the non-linear inductor-like component of the voice coil in parallel with the capacitor C2.  It used the average of the two roots of the quadratic equation for u in Resonance for non-linear impedance, rather than just the more positive root.

This error is unimportant for two reasons:

  • The initial value of C2 only needed to be close enough for the fitting algorithm to converge.
  • I didn’t use that estimate of C2 in any case, but used the simpler estimate based on a single data point on the downward slope that corresponded to where C2 dominated the impedance.

The corrected code is fairly simple:

cp(alpha,beta) = cos(pi*(alpha-beta)/2)
U(alpha,beta) = (-(alpha+beta)*cp(alpha,beta) + sqrt((alpha+beta)**2*cp(alpha,beta)**2 - 4*alpha*beta))/(2*alpha)                                              
N = (2*pi*f_peak)**(1+alpha)*M/U(alpha,-1)                                      
C2_a = 1./N   

This method produces an estimate that is closer to the final value after fitting than the simple estimator I used in the video.

Resonance for non-linear impedance

Filed under: Circuits course — gasstationwithoutpumps @ 08:51
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In modeling loudspeakers and electrodes, my book uses a non-linear impedance (j \omega\; 1\,s)^\alpha M, where -1 \le \alpha \le 1 is unitless, and M is in ohms.  This two-parameter component generalizes the standard passive components, as \alpha=0 gives resistors (with M=R), \alpha=1 gives inductors (with M= L/1\,s) and \alpha=-1 gives capacitors (with M= 1\,s/C).

This week I was thinking about resonance from a combination of an inductor-like component and a capacitor and more generally from two non-linear impedances with different exponents.  If we put two in series, we have an impedance Z = (j\omega\;1\,s)^\alpha M + (j\omega\;1\,s)^\beta N. To figure out where there are peaks or dips in the the magnitude of the impedance, it is easiest to look at |Z|^2 = Z Z^*, and let w = \omega \; 1s, to simplify the writing.

|Z|^2 = Z Z^*

= (w^\alpha M)^2 + (w^\beta N)^2 + (j w)^{\alpha+\beta}M N ( (-1)^\alpha + (-1)^\beta)

= (w^\alpha M)^2 + (w^\beta N)^2 + (j w)^{\alpha+\beta}M N ( (j^{-2\alpha} + j^{-2\beta})

= (w^\alpha M)^2 + (w^\beta N)^2 +w^{\alpha+\beta}M N ( (j^{\beta -\alpha} + j^{\alpha-\beta})

= (w^\alpha M)^2 + (w^\beta N)^2 +2 w^{\alpha+\beta}M N ( \cos(\frac{\pi(\alpha-\beta)}{2}))

We can first look for places where the magnitude goes to zero—are any of them for real values of \omega?  If we let u=w^{\alpha-\beta} M/N, so that \omega = w/1\,s = (u N/M)^{1/(\alpha-\beta)} / 1\,s, then

|Z|^2 = (w^\beta N)^2 \left( u^2 + 2 \cos(\frac{\pi(\alpha-\beta)}{2}) u +1 \right)

and the magnitude is zero if

u = - \cos(\frac{\pi(\alpha-\beta)}{2}) \pm j \sin(\frac{\pi(\alpha-\beta)}{2}),

which is real if \alpha-\beta is an even integer.  The only ones of interest are ±2, which correspond to the standard LC resonance with \alpha=1 and \beta=1, and give us the usual formula for resonance: \omega = (LC)^{-1/2}.

More generally, we can look for a minimum of the magnitude of the impedance that is not zero, by taking the derivative with respect to \omega or w and setting it to zero:

\frac{d (ZZ^*)}{d w} = \frac{1}{w}\left(2\alpha (w^\alpha M)^2 + 2\beta(w^\beta N)^2 + 2 (\alpha+\beta) w^{\alpha+\beta} M N \cos(\frac{\pi(\alpha-\beta)}{2}) \right)

Setting that to zero and changing variable again gives us 0 = \alpha u^2 + (\alpha+\beta)\cos(\frac{\pi(\alpha-\beta)}{2}) u + \beta, which gives us the solution

u = \frac{- (\alpha+\beta)\cos(\frac{\pi(\alpha-\beta)}{2}) \pm  \sqrt{(\alpha+\beta)^2 \cos^2(\frac{\pi(\alpha-\beta)}{2}) - 4\alpha\beta}}{2\alpha}.

The simplest case is the symmetric one, with \beta = - \alpha, which gives us u = \pm 1or \omega =(N/M)^{1/(2\alpha)}/ 1\,s.  This generalizes the standard LC resonance, with the standard result \omega = (LC)^{1/2}.

If \alpha and \beta have the same sign, then u is not real and so there is no resonance (except in the trivial case \alpha=\beta, in which case we have a simple non-linear impedance (j\omega)^\alpha (M+N) and still no resonance).

As the two exponents get further apart, the resonance gets more pronounced:

2020 May 13

Fifth video for electronics book

Filed under: Circuits course — gasstationwithoutpumps @ 20:05
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I’ve just published my fifth video for my Applied Analog Electronics book.  This video completes Chapter 28, giving the impedance for RLC circuits.

I filmed the video using OBS (Open Broadcaster Software), and this is the unedited first take.

2020 May 12

Fourth video for electronics book

Filed under: Circuits course — gasstationwithoutpumps @ 20:22
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I’ve just published my fourth video for my Applied Analog Electronics book.  This video is for §28.4 of the textbook (Bode plots for inductor combined with capacitors).

I filmed the video using OBS (Open Broadcaster Software), and this is the unedited second take. The first take was far too long and I messed up about 18 minutes in. I’ll have to do a separate video for the second half of that attempt.

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