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Post by antigua on Nov 5, 2018 12:00:17 GMT -5
I've got a javascript based integration processors working just enough to see how the plots will come out, but it still needs a lot of added functionality in order to be used as a productive tool. This plot was generated without the use of a driver coil, it's measuring the difference across a resistor as described here guitarnuts2.proboards.com/thread/7775/pickups-resonant-peak-usb-oscilliscope%C2%A0 The result shows an increased impedance below 1kHz, which is to say, it's not flat. I've got this same sort of result when using the hardware integrator: My question: is the lack of flatness due to the real resistance of the pickup itself, or is it caused by the 1 meg resistance value that is between the pickup and the driving voltage?
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Post by ms on Nov 5, 2018 14:59:04 GMT -5
I am not sure what a "java based integration processor" is, but maybe that is what is causing the downward slope with increasing frequency since integrators cause a roll off with increasing frequency.
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Post by JohnH on Nov 5, 2018 15:48:41 GMT -5
Im also thinking that for this type of test, the integrator function is not needed. With the driver-coil tests, the integrator effectively converts from a signal derived from driving a coil with constant voltage at all frequencies (hence increasing flux with frequency), to one with constant flux at all frequencies. But this doesnt apply here.
Also, in theory, that 1M should be as high as possible so that the driver signal approaches a constant current source. Ifs hard (for me!) to predict how high it needs to be to optimise the accuracy before noise etc creeps in. May try x2 to see what happens?
Once you have this working as well as possible, how about a pair of such tests with/without an added 470pF, using one of the simpler alnico singles that also has a good set of driver-coil results?. Then, lets try to work the maths to see if we can derive one type of plot from the other. ie, based on these impedance tests, how well can we produce a plot similar to a driver coil or real guitar test?
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Post by antigua on Nov 5, 2018 20:40:37 GMT -5
I am not sure what a "java based integration processor" is, but maybe that is what is causing the downward slope with increasing frequency since integrators cause a roll off with increasing frequency. By integrated I mean a raw bode plot csv from the Velleman PSCU200 has it's amplitude data points run through this operation: amp = amp - 20 * Math.log10( f ) , the plot points are then feed into a Google line chart. I got the equation from here guitarnuts2.proboards.com/post/86877/thread I have observed this same curve with the physical integrator as well, so I don't think it's a math problem.
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Post by antigua on Nov 5, 2018 20:48:25 GMT -5
Im also thinking that for this type of test, the integrator function is not needed. With the driver-coil tests, the integrator effectively converts from a signal derived from driving a coil with constant voltage at all frequencies (hence increasing flux with frequency), to one with constant flux at all frequencies. But this doesnt apply here. The +20dB/decade effect happens with or without a driver coil. If I don't mathematically integrate the data, I get the familiar mountain shaped function, which is how it appeared when the bode plot was generated, sans driver coil. Even in the non-integrated plot above, you can see that the slope is not flat as it approaches zero. I'm thinking that 1meg resistor is to blame, and if it is, the next question becomes, what are the alternatives? The goal I have in mind is that anyone can buy a Velleman PCSU200, find a 1 meg resistor and create an integrated bode plot of their pickups by just transferring the plot CSV into this web page I'm working on.
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Post by Yogi B on Nov 5, 2018 22:19:44 GMT -5
I'm also thinking that for this type of test, the integrator function is not needed. With the driver-coil tests, the integrator effectively converts from a signal derived from driving a coil with constant voltage at all frequencies (hence increasing flux with frequency), to one with constant flux at all frequencies. But this doesn't apply here. Agreed. Integrating isn't what we want here, that "increased impedance" would continue increasing to infinity as we head towards 0Hz; yet we know the DC resistance of our pickups isn't infinite. I'm not sure that's an issue. Unless I have this wrong, we essentially have a voltage divider: V_\text{in} is measured by the channel 2 probe, V_\text{out} by channel 1, Z_\text{out} is the impedance of our pickup, Z_\text{in} is that plus the external resistance, R_\text{ext} = 1\,\mathrm{M\Omega} . Thus, rather than assuming the current is roughly constant, we can actually calculate the total impedance of the pickup directly, via: Z_\text{pickup} = Z_\text{out} = \cfrac{R_\text{ext}}{\cfrac{V_\text{in}}{V_\text{out}} - 1} I'm not sure we can. The issue I see is that here we have the entire impedance of the pickup lumped together in parallel with the voltage source, normally this isn't the case. Assuming the 3-part model, only the capacitance is in parallel -- the resistance and inductance are in series. I don't think we can use maths to truly 'untangle' those, as we'd only be as accurate as the model we picked in order to attempt it. For example, once we calculated the inductance (again using the 3-part model), the first step might be to divide through by the (subtle) HPF formed by the 1 Meg resistor, the pickup's inductance, and its resistance. Ultimately we may well 'fix' the plot, but we can only be as accurate as our chosen model allows. Luckily even in the 6-part model we only have the series resistance and inductance that need correcting -- the additional components are added in parallel. Overall I don't see that this would it be any better than throwing out the measured data, aside from calculated inductance and capacitance values, and plotting using only the model from scratch. That doesn't mean to say it can't be done, but I don't see the motivation in it. I still need to think more on this -- it hasn't yet clicked why, when the entire pickup's impedance is lumped together, resonant frequency should remain unaffected. Is just a happy coincidence that the changes don't overlap the critical frequency region?
P.S. Let me know if the embedded maths displays okay.
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Post by newey on Nov 5, 2018 22:50:35 GMT -5
Yogi- The technical argument here is way beyond my ken. But your maths display just fine. Good work setting up that Katek module for us . . .
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Post by ms on Nov 6, 2018 6:51:20 GMT -5
Concerning the measurement of the response from the impedance (to summarize and add a bit to what YB said):
1. What you are measuring here appears to something like the magnitude of the impedance in parallel with 1M resistor.
Note: You can in principle remove the effect of the 1M resistor by calculation if you measure the full impedance or use a model of the pickup.
2. If you have the impedance (amplitude and phase or, equivalently, real and imaginary parts) and a perfect model of the pickup you can get the response, with a caveat.
This is not easy. You need to separate out the coil resistance and the inductance, by which I mean that which you would measure at very low frequencies in order to avoid contamination with eddy current effects (on the impedance). This becomes the series part of the voltage divider. Then you are left with the capacitance, eddy current effects, and anything else (if there is anything) for the parallel part.
Even if you can do all this, there is still an eddy current effect that is not included. Remember, eddy currents can affect the impedance: this is included. But they also can cut down the response at high frequencies by intercepting some of the signal from the coil, and this is not included. So I think you need measurements with the coil to get that.
You also have to think about what you mean by "the response of the pickup". The actual response includes the 6db/octave that the integrator takes out. It certainly looks better to have that taken out, but why should we not use the actual response? Also, taking out this rise with the integrator affects the location of the peak a bit since the peak is broad. This means that the peak measured measured with the "impedance" method is not exactly the same. Also, as had been pointed out, the peak measured from the magnitude of the impedance is not the same as the formal definition of resonance, which is where the phase crosses zero. So there are some difficulties in establishing an equivalence between different methods.
None of this prevents the useful measurement of the differences between different pickups, but you have to be careful how far you push the interpretations.
Edit: Taking another look at the setup in the link in the first post, I might be confusing whether the voltage is measured across the 1M resistor or the pickup. You can do either, but I guess the results are a bit different depending. If you measure the voltage across the pickup, you want to use a very large resistor so you inject a current. If you measure across the resistor, maybe you want to use the smallest possible resistor so that you measure the current through the pickup. In this case maybe applying the integrator gives funny results at low frequencies wit reasonable value resistors, I am not sure. I think if you want to measure impedance a more formal setup such as the one I have described should be used.
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Post by JohnH on Nov 6, 2018 16:05:23 GMT -5
Yes the phase angle between the applied signal and that across the pu is necessary info. Then it becomes possible to plot the impedances in real and imaginary parts.
In terms of visualising the arrangement as a voltage divider, we try to equate the ratio of measured pickup voltage and total applied voltage, to the ratio of pickup impedance and total impedance.
The total impedance is a simple sum when phase angle is 0 (fully resistive) and a pythagoras calc when phase angle is 90 degrees. But as the feed resistor becomes very high, it dominates the total impedance in both cases, simplifying the calc.
Once we back-figure impedance and phase, we can use it to directly reproduce an in-circuit response, except....
...eddy current effects are not included. I think it would be very interesting to compare a response calculated this way, with one derived fro. driver-coil tests, in order to try to see this eddy difference.
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Post by antigua on Nov 6, 2018 16:55:50 GMT -5
Even if you can do all this, there is still an eddy current effect that is not included. Remember, eddy currents can affect the impedance: this is included. But they also can cut down the response at high frequencies by intercepting some of the signal from the coil, and this is not included. So I think you need measurements with the coil to get that. In comparing the impedance curve using the 1 meg resistor method and a driver coil, it appears that the eddy current effects between the driver coil and pickup are trivial compared to the overall eddy current losses that occur between the conductive parts and the primary coil(s) of the pickup. For example, in the past I would have noted that the cover of a pickup is especially prone to cause eddy current losses because it's in between the coils and the guitar strings, but it is currently my understanding that if the cover were somehow turned upside down, and underneath the coils, the losses would be nearly the same, because it's the cover's proximity to the coil that causes the loss, not the proximity to the guitar strings. I've noted in the past that the brass base plates don't cause much attenuation relative to nickel silver, I don't believe that is because it is below the pickup, but because it's farther away from the coils, and doesn't envelop the coils, as a brass cover does. For this reason, I don't believe that a driver coil is especially necessary for the purpose of plotting a pickup's response, and another user on another forum has made an argument that the driver coil introduces its own form of interference www.tdpri.com/threads/physically-based-eddy-current-equivalent-circuit.846745/page-3 . In the demo above you can see I'm testing a Filter'tron, I am checking to see if it has the "dip" when integrated and measured without a driver coil. Even though the plot is not real clear, it nevertheless looks to me like there is no dip when the measurement is taken across a 1 meg resistor: 1 meg resistor: driver coil: Helmuth Lemme attributed the dip to eddy currents, but Helmuth Lemme also used a driver coil, so this might be a point on which we have all been mistaken. You also have to think about what you mean by "the response of the pickup". The actual response includes the 6db/octave that the integrator takes out. It certainly looks better to have that taken out, but why should we not use the actual response? Also, taking out this rise with the integrator affects the location of the peak a bit since the peak is broad. This means that the peak measured measured with the "impedance" method is not exactly the same. Also, as had been pointed out, the peak measured from the magnitude of the impedance is not the same as the formal definition of resonance, which is where the phase crosses zero. So there are some difficulties in establishing an equivalence between different methods. None of this prevents the useful measurement of the differences between different pickups, but you have to be careful how far you push the interpretations. Mostly because the Q factor is hard to observe when the overall plot is of a peak. Edit: Taking another look at the setup in the link in the first post, I might be confusing whether the voltage is measured across the 1M resistor or the pickup. You can do either, but I guess the results are a bit different depending. If you measure the voltage across the pickup, you want to use a very large resistor so you inject a current. If you measure across the resistor, maybe you want to use the smallest possible resistor so that you measure the current through the pickup. In this case maybe applying the integrator gives funny results at low frequencies wit reasonable value resistors, I am not sure. I think if you want to measure impedance a more formal setup such as the one I have described should be used. Someone scribbled out the schematic: <iframe width="13.340000000000032" height="11.519999999999982" style="position: absolute; width: 13.340000000000032px; height: 11.519999999999982px; z-index: -9999; border-style: none;left: 5px; top: 437px;" id="MoatPxIOPT0_48218284" scrolling="no"></iframe> <iframe width="13.340000000000032" height="11.519999999999982" style="position: absolute; width: 13.34px; height: 11.52px; z-index: -9999; border-style: none; left: 607px; top: 437px;" id="MoatPxIOPT0_40041606" scrolling="no"></iframe> <iframe width="13.340000000000032" height="11.519999999999982" style="position: absolute; width: 13.34px; height: 11.52px; z-index: -9999; border-style: none; left: 5px; top: 951px;" id="MoatPxIOPT0_44904703" scrolling="no"></iframe> <iframe width="13.340000000000032" height="11.519999999999982" style="position: absolute; width: 13.34px; height: 11.52px; z-index: -9999; border-style: none; left: 607px; top: 951px;" id="MoatPxIOPT0_37843601" scrolling="no"></iframe> The 1 meg resistance is in series with the pickup, relative to the voltage source and channel 2.
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Post by Yogi B on Nov 6, 2018 17:22:18 GMT -5
In terms of visualising the arrangement as a voltage divider, we try to equate the ratio of measured pickup voltage and total applied voltage, to the ratio of pickup impedance and total impedance. Exactly: \begin{aligned} \frac{V_\text{out}}{V_\text{in}} &= \frac{Z_\text{out}}{Z_\text{in}} \\\\ &= \frac{Z_\text{pickup}}{R_\text{ext} + Z_\text{pickup}} \\\\ \frac{V_\text{out}}{V_\text{in}} \cdot \left( R_\text{ext} + Z_\text{pickup} \right) &= Z_\text{pickup} \\\\ \frac{V_\text{out}}{V_\text{in}} \cdot R_\text{ext} &= Z_\text{pickup} \cdot \left( 1 - \frac{V_\text{out}}{V_\text{in}} \right) \\\\ \frac{\frac{V_\text{out}}{V_\text{in}} \cdot R_\text{ext}}{1 - \frac{V_\text{out}}{V_\text{in}}} &= Z_\text{pickup} \\ \text{dividing both numerator and }&\text{denominator of the LHS by }\frac{V_\text{out}}{V_\text{in}}\text{ yielding:} \\ \frac{R_\text{ext}}{\cfrac{V_\text{in}}{V_\text{out}} - 1} &= Z_\text{pickup} \end{aligned} Total impedance is always a simple sum, so long as you're using it's full complex value, and not it's magnitude. Except it's still the wrong circuit, we'd have the total impedance of the pickup, this could be used to see how that pickup would load another voltage source, but that isn't the same as what happens when the pickup is the voltage source. We need to separate out the parts of the impedance that are modelled in series with the pickups voltage source and those which are in parallel -- I don't know that there is a simple way of doing this.
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Post by JohnH on Nov 7, 2018 2:27:14 GMT -5
My thought was to take an RLC model arrangement that we know can work, and see how its values and results vary comparing calculated passive impedance to measured impedance, or derive values using tbe impedance mesurements and see if it can match driver coil tests. This can be with 3, 4 or 6 part models. The locating of the signal source would be an assumption, and the whole thing is just an exploation to see where it might lead. We have the thinking in place for relating models to tests in terms of matching output with load, but it would be interesting to see if similar matches can occur with impedance.
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Post by Yogi B on Nov 8, 2018 12:51:56 GMT -5
Also, as had been pointed out, the peak measured from the magnitude of the impedance is not the same as the formal definition of resonance, which is where the phase crosses zero. True, but that just means we need a new formula. See aquin43's post below. In the following I forgot that \frac{d|x|}{dx} = \frac{x}{|x|} is only applicable to x \in \Reals , and in fact |x| isn't (directly) complex differentiable. Starting with Z_\text{pickup} = ( Z_R + Z_L ) \parallel Z_C , the extrema of the magnitude of impedance occur when:\gdef\dd#1{\frac{d}{d#1}}
\begin{aligned} \left. \dd{f} \middle| ( Z_R + Z_L ) \parallel Z_C \right| &= 0 \\\\ \dd{f} \left| ( R + 2 \pi j f L ) \parallel \frac{1}{2 \pi j f C} \right| &= 0 \\\\ \dd{f} \frac{1}{\left| \frac{1}{ R + 2 \pi j f L } + 2 \pi j f C \right|} &= 0 \\\\ 2 \pi j \cdot \left( C - \frac{L}{(R+ 2 \pi j fL)^2} \right) \cdot \frac{ 2 \pi j f C + \frac{1}{R + 2 \pi j f L} }{ \left| 2 \pi j f C + \frac{1}{R + 2 \pi j f L} \right|^3 } &= 0 \end{aligned}
This has four roots, the first pair when:\begin{aligned} C - \frac{L}{(R + 2 \pi j f L)^2} &= 0 \\\\ C &= \frac{L}{(R + 2 \pi j f L)^2} \\\\ R + 2 \pi j f L &= \pm\sqrt{\frac{L}{C}} \\\\ f &= \frac{-R \pm \sqrt{\frac{L}{C}}}{2 \pi j L} \\\\ f &= \frac{R \mp \sqrt{\frac{L}{C}}}{2 \pi L} \cdot j \end{aligned} The second pair when:\begin{aligned} 2 \pi j f C + \frac{1}{R + 2 \pi j f L} &= 0 \\\\ 2 \pi j f C &= -\frac{1}{R + 2 \pi j f L} \\\\ 2 \pi j f C \cdot (R + 2 \pi j f L) &= -1 \\\\ 2 \pi j f R C + 4 \pi^2 j^2 f^2 L C &= -1 \\\\ \frac{2 \pi j R C}{4 \pi^2 j^2 L C} \cdot f + f^2 &= \frac{\mathllap{-}1}{4 \pi^2 j^2 L C} \\\\ \frac{R}{2 \pi j L} \cdot f + f^2 &= \frac{1}{4 \pi^2 L C} \\\\ \left( \frac{R}{4 \pi j L} + f \right)^2 &= \frac{1}{4 \pi^2 L C} + \frac{R^2}{16 \pi^2 j^2 L^2} \\\\ \frac{R}{4 \pi j L} + f &= \pm\sqrt{\frac{1}{4 \pi^2 L C} - \frac{R^2}{16 \pi^2 L^2}} \\\\ f &= - \frac{R}{4 \pi j L} \pm \sqrt{\frac{1}{4 \pi^2 L C} - \frac{R^2}{16 \pi^2 L^2}} \\\\ f &= \pm \sqrt{\frac{4 L - R^2 C}{16 \pi^2 L^2 C}} + \frac{R}{4 \pi L} \cdot j \end{aligned} Now, my question is how to interpret the calculated frequency being complex. I'm not concerned with the first pair of roots (the purely imaginary pair), as those relate to minima -- it is the second pair which is relevant to the maximum impedance.
Plotting out the impedance of the 3-part Seth Lover model used in GF (R=7k39, L=4.16, C=154p) the maximum impedance is at roughly 6.288kHz.
Plugging the same values into the above equation gives f \approx \pm 6286.42 + 141.365 j , the magnitudes of which are the 6288 value that we're after.
Now we just have to work out how to go backwards.
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Post by aquin43 on Nov 9, 2018 8:11:04 GMT -5
Hello,
The continuous frequency we measure can't be complex. If you work out the absolute value of the impedance and differentiate that, (differentiating its square is easier), there is a real solution at Omega_peak = sqrt(sqrt(L) * sqrt(2 * C * R^2 + L) - C * R^2 )/(L * sqrt(C)) This, of course, differs from Omega_0 = 1/sqrt(L * C) The ratio is: Omega_peak/Omega_0 = sqrt(sqrt(L) * sqrt(2 * C * R^2 + L) - C * R^2 )/sqrt(L) In the limit, as R goes to zero, this ratio becomes 1, so for high Q circuits, it is almost true that Omega_peak = 1/sqrt(L * C)
Arthur
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Post by ms on Nov 9, 2018 11:25:45 GMT -5
Antigua:
"In comparing the impedance curve using the 1 meg resistor method and a driver coil, it appears that the eddy current effects between the driver coil and pickup are trivial compared to the overall eddy current losses that occur between the conductive parts and the primary coil(s) of the pickup."
That was what i thought until making some measurements with very high eddy current loss in metal between the coil and the pickup. I think those measurements are somewhere on this forum, and I think they showed that there is an effect in addition to the effect on the inductance.
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Post by antigua on Nov 9, 2018 15:34:17 GMT -5
Here is the same "voltage across resistor in series with pickup" measurement, with various resistances: Note that the 1meg resistor yields the highest Q factor, but also shows that broad curve ahead of the resonance, meanwhile the lower value resistors 100k and 390k have a flatter slope ahead of the resonance, but the Q factor greatly reduced. I need the flatness of the lower value resistors, but the Q factor of the higher resistance.
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Post by aquin43 on Nov 9, 2018 18:45:47 GMT -5
Hello,
Here is my approach to pickup impedance measurement: 1) Use the highest source voltage you can. 2) Use a feed resistor that will give a few uA to ground. 3) Use the highest available probe resistance. Ground one end of the pickup and connect the other to the source via the feed resistor. connect the probe between pickup live and ground. Plot the voltage curve (not dB). Do not apply any frequency scaling. Calculation: Rf = feed resistor, Rp = probe resistance, Vt = source voltage, Vm = measured voltage. Measured Impedance = Rm = Vm * Rf/Vt The pickup, probe and feed resistor are effectively in parallel because a voltage on the pickup terminal will drive a current through all three. So, convert them to admittances (1/R), subtract the unwanted ones and convert back. Actual Impedance Ra = 1/(1/Rm - 1/Rf - 1/Rp) This is true at resonance and is a good approximation very close to resonance. If the test set can measure phase as well and it is desired to plot the true impedance curve, then a more refined calculation is possible. For each data point: r = Vm * Rf/Vt # read voltage and convert to Ohms phi = measured phase * pi/180 # read phase - convert to radians yr = cos(phi)/r # real part of admittance in parallel with: yi = -sin(phi)/r # imaginary part of admittance yr_corr = (yr - yt - yp) # subtract the feed resistor and probe admittance phi_corr = -atan2(yi, yr_corr) # phase of the corrected impedance r_corr = 1/sqrt(yi^2 + yr_corr^2) # magnitude of the corrected impedance Plot: True R = r_corr True Phase = phi_corr * 180/pi
I use a 10M resistor with 10V drive and a buffer amplifier with several G input resistance made from an OPA1642, This gives a voltage around 300mV - 400Mv at a typical pickup peak impedance, so the correction is under 10%
I have also tried the method with a 1M feed resistor and a 10M probe and the results are comparable.
Arthur
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Post by Yogi B on Nov 9, 2018 21:45:04 GMT -5
Once more unto the breach... In order to use the definition |z| = \sqrt{\Re(z)^2 + \Im(z)^2} , we must separate out the real and imaginary parts of our impedance: \begin{aligned} Z_\text{pickup} &= ( Z_R + Z_L ) \parallel Z_C \\\\ &= (R + 2 \pi j f L) \parallel \frac{1}{2 \pi j f C} \\\\ &= \frac{1}{\frac{1}{R + 2 \pi j f L} + 2 \pi j f C} \\\\ &= \frac{1}{\frac{R - 2 \pi j f L}{R^2 + 4 \pi^2 f^2 L^2} + 2 \pi j f C} \\\\ &= \frac{R^2 + 4 \pi^2 f^2 L^2}{R - 2 \pi j f L + 2 \pi j f C \cdot (R^2 + 4 \pi^2 f^2 L^2)} \\\\ &= \frac{R^2 + 4 \pi^2 f^2 L^2}{R + 2 \pi j f \cdot (R^2 C + 4 \pi^2 f^2 L^2 C - L)} \\\\ &= \frac{ (R^2 + 4 \pi^2 f^2 L^2) (R - 2 \pi j f \cdot [R^2 C + 4 \pi^2 f^2 L^2 C - L]) }{ R^2 + 4 \pi^2 f^2 \cdot (R^2 C + 4 \pi^2 f^2 L^2 C - L)^2 } \\\\ &= \frac{ (R^2 + 4 \pi^2 f^2 L^2) (R - 2 \pi j f \cdot [R^2 C + 4 \pi^2 f^2 L^2 C - L]) }{ R^2 + 4 \pi^2 f^2 \cdot ( [R^2 C]^2 + 2 \cdot R^2 C \cdot 4 \pi^2 f^2 L^2 C - 2 \cdot R^2 C \cdot L + [4 \pi^2 f^2 L^2 C]^2 - 2 \cdot 4 \pi^2 f^2 L^2 C \cdot L + L^2 ) } \\\\ &= \frac{ (R^2 + 4 \pi^2 f^2 L^2) (R - 2 \pi j f \cdot [R^2 C + 4 \pi^2 f^2 L^2 C - L]) }{ R^2 \cdot ( 1 + 4 \pi^2 f^2 R^2 C^2 + 4 \pi^2 f^2 \cdot 4 \pi^2 f^2 L^2 C^2 - 8 \pi^2 f^2 L C ) + 4 \pi^2 f^2 L^2 \cdot ( 4 \pi^2 f^2 R^2 C^2 + [4 \pi^2 f^2 L C]^2 - 8 \pi^2 f^2 L C + 1 ) } \\\\ &= \frac{ (R^2 + 4 \pi^2 f^2 L^2) (R - 2 \pi j f \cdot [R^2 C + 4 \pi^2 f^2 L^2 C - L]) }{ R^2 \cdot ( [1 - 4 \pi^2 f^2 L C]^2 + 4 \pi^2 f^2 R^2 C^2 ) + 4 \pi^2 f^2 L^2 \cdot ( [1 - 4 \pi^2 f^2 L C]^2 + 4 \pi^2 f^2 R^2 C^2 ) } \\\\ &= \frac{ \cancel{(R^2 + 4 \pi^2 f^2 L^2)} \cdot (R - 2 \pi j f \cdot [R^2 C + 4 \pi^2 f^2 L^2 C - L]) }{ \cancel{(R^2 + 4 \pi^2 f^2 L^2)} \cdot ([1 - 4 \pi^2 f^2 L C]^2 + 4 \pi^2 f^2 R^2 C^2) } \\\\ &= \frac{ R }{ (1 - 4 \pi^2 f^2 L C)^2 + 4 \pi^2 f^2 R^2 C^2 } - \frac{ 2 \pi f \cdot (R^2 C + 4 \pi^2 f^2 L^2 C - L) }{ (1 - 4 \pi^2 f^2 L C)^2 + 4 \pi^2 f^2 R^2 C^2 } \cdot j \end{aligned} That would've been much clearer without the 2 \pi 's, okay \omega it is then: Z_\text{pickup} = \frac{ R }{ (1 - \omega^2 L C)^2 + \omega^2 R^2 C^2 } - \frac{ \omega \cdot (R^2 C + \omega^2 L^2 C - L) }{ (1 - \omega^2 L C)^2 + \omega^2 R^2 C^2 } \cdot j Differentiating we get: \begin{aligned} \frac{d}{d\omega} |Z_\text{pickup}| &= \frac{d}{d\omega} \sqrt{\Re(Z_\text{pickup})^2 + \Im(Z_\text{pickup})^2} \\ \\ &= \frac{d}{d\omega} \frac{\sqrt{R^2 + (\omega \cdot [R^2 C + \omega^2 L^2 C - L])^2}}{(1 - \omega^2 L C)^2 + \omega^2 R^2 C^2} \\ \\ &= \frac{d}{d\omega} \frac{\sqrt{R^2 + (\omega R^2 C + \omega^3 L^2 C - \omega L)^2}}{(1 - \omega^2 L C)^2 + \omega^2 R^2 C^2} \\ \\ &= \frac{d}{d\omega} \frac{ \sqrt{R^2 + \omega^2 R^4 C^2 + 2 \omega^4 R^2 L^2 C^2 - 2 \omega^2 R^2 L C + \omega^6 l^4 c^2 - 2 \omega^4 L^3 C + \omega^2 L^2} }{ (1 - \omega^2 L C)^2 + \omega^2 R^2 C^2 } \\ \\ &= \frac{d}{d\omega} \frac{ \sqrt{ R^2 \cdot (1 + \omega^2 R^2 C^2 + \omega^4 L^2 C^2 - 2 \omega^2 L C) + \omega^2 L^2 \cdot (\omega^2 R^2 C^2 + \omega^4 L^2 C^2 - 2 \omega^2 L C + 1) } }{ (1 - \omega^2 L C)^2 + \omega^2 R^2 C^2 } \\ \\ &= \frac{d}{d\omega} \frac{ \sqrt{ (R^2 + \omega^2 L^2) \cdot (1 + \omega^2 R^2 C^2 + \omega^4 L^2 C^2 - 2 \omega^2 L C) } }{ (1 - \omega^2 L C)^2 + \omega^2 R^2 C^2 } \\ \\ &= \frac{d}{d\omega} \frac{ \sqrt{ (R^2 + \omega^2 L^2)} \cdot \sqrt{(1 - \omega^2 L C)^2 + \omega^2 R^2 C^2 } }{ (1 - \omega^2 L C)^2 + \omega^2 R^2 C^2 } \\ \\ &= \frac{d}{d\omega} \sqrt{ \frac{ R^2 + \omega^2 L^2 }{ (1 - \omega^2 L C)^2 + \omega^2 R^2 C^2 } } \\ \\ &= \frac{1}{2} \sqrt{ \frac{ (1 - \omega^2 L C)^2 + \omega^2 R^2 C^2 }{ R^2 + \omega^2 L^2 } } \cdot \frac{d}{d\omega} \!\left( \frac{ R^2 + \omega^2 L^2 }{ (1 - \omega^2 L C)^2 + \omega^2 R^2 C^2 } \right) \\ \\ &= \frac{1}{2} \sqrt{ \frac{ (1 - \omega^2 L C)^2 + \omega^2 R^2 C^2 }{ R^2 + \omega^2 L^2 } } \cdot \frac{ \frac{d}{d\omega} \! \left( R^2 + \omega^2 L^2 \right) \cdot \left( (1 - \omega^2 L C)^2 + \omega^2 R^2 C^2 \right) - \left( R^2 + \omega^2 L^2 \right) \cdot \frac{d}{d\omega} \! \left( (1 - \omega^2 L C)^2 + \omega^2 R^2 C^2 \right) }{ \left( (1 - \omega^2 L C)^2 + \omega^2 R^2 C^2 \right)^2 } \\ \\ &= \frac{ 2 \omega L^2 \cdot \left( (1 - \omega^2 L C)^2 + \omega^2 R^2 C^2 \right) - \left( R^2 + \omega^2 L^2 \right) \cdot \left( 2 \cdot (1 - \omega^2 L C) \cdot ( - 2 \omega L C) + 2 \omega R^2 C^2 \right) }{ 2 \cdot \sqrt{ R^2 + \omega^2 L^2 } \cdot \left( (1 - \omega^2 L C)^2 + \omega^2 R^2 C^2 \right)^{3/2} } \\ \\ &= \frac{ \cancel{2} \omega \cdot \left( (L - \omega^2 L^2 C)^2 + \omega^2 R^2 L^2 C^2 - \left( R^2 + \omega^2 L^2 \right) \cdot \left( R^2 C^2 - 2 L C \cdot (1 - \omega^2 L C) \right) \right) }{ \cancel{2} \cdot \sqrt{ R^2 + \omega^2 L^2 } \cdot \left( (1 - \omega^2 L C)^2 + \omega^2 R^2 C^2 \right)^{3/2} } \end{aligned} The extrema occur at the roots of \frac{d}{d\omega} |Z_\text{pickup}| = 0 , which include the trivial case of \omega = 0 . The other four roots are given by: \def\ccancel#1#2{\mathrlap{#2}\textcolor{#1}{\cancel{\phantom{#2}}}}
\begin{aligned} (L - \omega^2 L^2 C)^2 + \omega^2 R^2 L^2 C^2 - \left( R^2 + \omega^2 L^2 \right) \cdot \left(R^2 C^2 - 2 L C \cdot (1 - \omega^2 L C) \right) &= 0 \\ \\ L^2 - 2 \omega^2 L^3 C + \omega^4 L^4 C^2 + \omega^2 R^2 L^2 C^2 - \left( R^2 + \omega^2 L^2 \right) \cdot \left(R^2 C^2 - 2 L C + 2 \omega^2 L^2 C^2 \right) &= 0 \\ \\ L^2 - 2 \omega^2 L^3 C + \omega^4 L^4 C^2 + \omega^2 R^2 L^2 C^2 - \left( R^4 C^2 - 2 R^2 L C + 2 \omega^2 R^2 L^2 C^2 + \omega^2 R^2 L^2 C^2 - 2 \omega^2 L^3 C + 2 \omega^4 L^4 C^2 \right) &= 0 \\ \\ L^2 - \ccancel{purple}{2 \omega^2 L^3 C} + \ccancel{red}{\omega^4 L^4 C^2} + \cancel{\omega^2 R^2 L^2 C^2} - R^4 C^2 + 2 R^2 L C - 2 \omega^2 R^2 L^2 C^2 - \cancel{\omega^2 R^2 L^2 C^2} + \ccancel{purple}{2 \omega^2 L^3 C} - \ccancel{red}{2} \omega^4 L^4 C^2 &= 0 \\ \\ \omega^4 L^4 C^2 + 2 \omega^2 R^2 L^2 C^2 &= L^2 + 2 R^2 L C - R^4 C^2 \\\\ \omega^4 + \frac{2 \omega^2 R^2}{L^2} &= \frac{L^2 + 2 R^2 L C - R^4 C^2}{L^4 C^2} \\\\ \left( \omega^2 + \frac{R^2}{L^2} \right)^2 &= \frac{L^2 + 2 R^2 L C - R^4 C^2}{L^4 C^2} + \frac{R^4}{L^4} \\\\ \omega^2 + \frac{R^2}{L^2} &= \pm \sqrt{ \frac{L^2 + 2 R^2 L C - \cancel{R^4 C^2} + \cancel{R^4 C^2}}{L^4 C^2} } \\\\ \omega^2 &= \frac{ - R^2 C \pm \sqrt{ L^2 + 2 R^2 L C }}{L^2 C} \\\\ \omega &= \pm \sqrt{ \frac{ - R^2 C \pm \sqrt{ L^2 + 2 R^2 L C }}{L^2 C} }
\end{aligned} Of these, the positive real root is given by: \omega = \sqrt{ \frac{ - R^2 C + \sqrt{ L^2 + 2 R^2 L C }}{L^2 C} } (Thankfully) this is equivalent to what Arthur posted above: \omega_\text{peak} = \frac{\sqrt{\sqrt{L} \cdot \sqrt{2 \cdot C \cdot R^2 + L} - C \cdot R^2 }}{L \cdot \sqrt{C} } Hurrah!
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Post by Yogi B on Nov 10, 2018 23:28:33 GMT -5
This, of course, differs from Omega_0 = 1/sqrt(L * C) Yep, but that isn't 100% accurate with the additional resistance, thankfully having the phase information makes things simpler. Recall the equation for the pickups impedance: Z_\text{pickup} = \frac{ R }{ (1 - \omega^2 L C)^2 + \omega^2 R^2 C^2 } - \frac{ \omega \cdot (R^2 C + \omega^2 L^2 C - L) }{ (1 - \omega^2 L C)^2 + \omega^2 R^2 C^2 } \cdot j The impedance's phase is zero when its imaginary part is zero, i.e. when \omega = 0 or: \begin{aligned} R^2 C + \omega^2 L^2 C - L &= 0 \\ \omega^2 L^2 C &= L - R^2 C \\ \omega^2 &= \frac{\cancel{L}}{L^{\cancel{2}} C} - \frac{R^2}{L^2 C} \\[3ex] \omega &= \sqrt{\frac{1}{LC} - \frac{R^2}{L^2}} \end{aligned} For the example values I used before (R=7k39, L=4.16, C=154p), we get \frac{1}{2 \pi} \cdot \omega = \frac{1}{2 \pi} \cdot \sqrt{\frac{1}{LC} - \frac{R^2}{L^2}} \approx \frac{ \sqrt{1.561 \cdot 10^9 - 3.417 \cdot 10^6} }{ 2 \pi } \approx 6281\,\mathrm{Hz} Compare to: \frac{1}{2 \pi} \cdot \omega_0 = \frac{1}{2 \pi} \cdot \sqrt{\frac{1}{LC}} \approx \frac{ \sqrt{1.561 \cdot 10^9} }{ 2 \pi } \approx 6288\,\mathrm{Hz} Admittedly it appears close enough, but in reality is isn't (see spoiler at the end of this post). We still need to actually calculate L and C, to do that we must again use 'capacitance swamping'. That is take another measurement with the addition of a sufficiently large capacitor in parallel with the pickup and record the new resonant frequency. The equation for the resonant angular frequency when loaded then becomes: \omega_\text{load} = \sqrt{\frac{1}{L(C + C_\text{load})} - \frac{R^2}{L^2}} Thus we now have a pair of equations we can solve simultaneously for L and C.
The equation lacking resistance just won't cut it for this purpose. Dividing the equations (to eliminate L), gives:
\begin{aligned} \frac{\omega_\text{load}}{\omega} &= \frac{ \sqrt{\frac{1}{L (C + C_\text{load})}} }{ \sqrt{\frac{1}{LC}} } \\ \\ \frac{\omega^2_\text{load}}{\omega^2} &= \frac{ \frac{1}{L (C + C_\text{load})} }{ \frac{1}{LC} } \\ \\ \frac{\omega^2_\text{load}}{\omega^2} &= \frac{ \cancel{L} C } { \cancel{L} (C + C_\text{load}) } \\ \\ \omega^2_\text{load} \cdot (C + C_\text{load}) &= C \omega^2 \\ \\ \omega^2_\text{load} C_\text{load} &= C \cdot (\omega^2 - \omega^2_\text{load}) \\ \\ \frac{\omega^2_\text{load} C_\text{load}}{\omega^2 - \omega^2_\text{load}} &= C \end{aligned} Using LTspice to simulate the 3-part Seth Lover model in parallel with a 10nF capacitor, gives a loaded resonant frequency of approx 721 Hz, plugging that in (along with the 6281 Hz unloaded frequency) results in the following:
C = \frac{\omega^2_\text{load} C_\text{load}}{\omega^2 - \omega^2_\text{load}} = \frac{\cancel{4 \pi^2} f^2_\text{load} C_\text{load}}{\cancel{4 \pi^2} \cdot (f^2 - f^2_\text{load})} = \frac{(721)^2 \cdot (10 \cdot 10^{-9})}{(6281)^2 - (721)^2} \approx 134\,\mathrm{pF} This get's worse the larger the external capacitor, using a 33nF cap results in a resonant frequency of approx 322 Hz:
C = \frac{(322)^2 \cdot (33 \cdot 10^{-9})}{(6281)^2 - (322)^2} \approx 87.0\,\mathrm{pF}
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Post by aquin43 on Nov 12, 2018 7:07:50 GMT -5
Hello,
The problem with all of these theoretical calculations is that the information at the pickup terminals doesn't reflect the simple LCR model. With pickups that have eddy current losses inductively coupled to the coil (all but a few pure strat types, as far as I can see) , the effective coil resistance rises with frequency while the inductance falls to a lesser extent. This is predicted by the mutually coupled loss model and is easily measured experimentally. It becomes evident at quite low frequencies in some pickups.There is also a low pass step introduced by the eddy current losses intercepting part of the magnetic drive from the strings. The phase relationships observed at the terminals can vary widely from what one would expect from a simple LCR circuit.
I am new to this pickup modelling and am not familiar with the common pickup models, but I am led to believe that they are behavioural models that incorporate ad hoc filters and loads that have no physical representation.
Arthur
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Post by JohnH on Nov 12, 2018 14:28:13 GMT -5
Hello, The problem with all of these theoretical calculations is that the information at the pickup terminals doesn't reflect the simple LCR model. With pickups that have eddy current losses inductively coupled to the coil (all but a few pure strat types, as far as I can see) , the effective coil resistance rises with frequency while the inductance falls to a lesser extent. This is predicted by the mutually coupled loss model and is easily measured experimentally. It becomes evident at quite low frequencies in some pickups.There is also a low pass step introduced by the eddy current losses intercepting part of the magnetic drive from the strings. The phase relationships observed at the terminals can vary widely from what one would expect from a simple LCR circuit. I am new to this pickup modelling and am not familiar with the common pickup models, but I am led to believe that they are behavioural models that incorporate ad hoc filters and loads that have no physical representation. Arthur True enough, it depends how accurately you want a model to represent reality. With three LRC parts, you can capture the resonant frequency, with a credible inductance but not the eddy and other damping effects. Still useful though, for characterising some key aspects of pickups. But with more parts on the model, you can match resonant peak height and frequency, and also same when in circuit, and start to capture the dips in between. The GuitarFreak spreadsheet in the reference section uses a 6 part model for each pickup. If you can run excel or Libeoffice, you might like to try it out. Here is a model for an SSL1 single coil, derived from tests by Antigua: There are 6 parts, so 6 facts are used to derive it, being frequency and height of loaded and unloaded resonant peaks (that's 4), DCR, and one more point on the curve to capture the shape. But there's too much maths (for me) to derive such a model explicitly, so creating it is an iterative process. GuiatrFreak has a collection of such models.
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Post by Yogi B on Nov 12, 2018 19:15:52 GMT -5
True enough, it depends how accurately you want a model to represent reality. With three LRC parts, you can capture the resonant frequency, with a credible inductance but not the eddy and other damping effects. Still useful though, for characterising some key aspects of pickups. For now I'm focusing on the three part model simply as a proof of concept, and to outline the general procedure of translating the pickup's measured impedance into a plot comparable to those made with a driver coil & integrator. The three part model keeps the maths less verbose, with less variables to keep track of.
Anyway, taking the equation for the resonant frequency from above, and solving for C we get: C =\frac{L}{\omega_0^2 L^2 + R^2}
Substituting \omega = \omega_\text{load} and C = C + C_\text{load} for the capacitively swamped condition, again solving for C, we get: C = \frac{L}{\omega_\text{load}^2 L^2 + R^2} - C_\text{load}
Thus we have: \def\w{\omega_0} \def\wl{\omega_\text{load}} \def\Cl{C_\text{load}}
\begin{aligned} C = \frac{L}{\w^2 L^2 + R^2} &= \frac{L}{\wl^2 L^2 + R^2} - \Cl \\ \\ \frac{L}{\w^2 L^2 + R^2} &= \frac{L - \Cl \cdot (\wl^2 L^2 + R^2)}{\wl^2 L^2 + R^2} \\ \\ L \cdot ( \wl^2 L^2 + R^2 ) &= ( L - \Cl \cdot (\wl^2 L^2 + R^2) ) \cdot ( \w^2 L^2 + R^2 ) \\ \\ \wl^2 L^3 + \cancel{R^2 L} &= \w^2 L^3 - \w^2 \wl^2 L^4 \Cl - \w^2 R^2 L^2 \Cl + \cancel{R^2 L} - \wl^2 R^2 L^2 \Cl - R^4 \Cl \\ \\ \w^2 \wl^2 \Cl \cdot L^4 - (\w^2 - \wl^2) \cdot L^3 + R^2 \Cl \cdot (\w^2 + \wl^2) \cdot L^2 + R^4 \Cl &= 0 \end{aligned}
Yep, we have a quartic equation in L, I'm not going to go through solving it, or indeed writing out its roots, as both would probably take far too much space, but I will note that for R = 7390 , \omega_0 = 2 \pi \cdot 6282 , \omega_\text{load} = 2 \pi \cdot 721 & C_\text{load} = 10 \cdot 10^{-9} the roots are: L \approx -0.0266 \pm 0.1767 j, \quad 0.7027, \quad 4.159 I find it intriguing that we have two positive real roots, but that's something to look at another time, in this case the value we want is the last of the four, 4.159 \approx 4.16 . We could at this point solve to find C, which again results in a quartic, but for my purposes that isn't needed.
Now recall the equations for our original measurement of voltage across the pickup: V_\text{out} = V_\text{in} \cdot \frac{Z_\text{pickup}}{Z_\text{pickup} + R_\text{ext}}
And the pickups impedance: \begin{aligned} Z_\text{pickup} &= (Z_R + Z_L) \parallel Z_C \\\\&= \frac{(Z_R + Z_L) \cdot Z_C}{Z_R + Z_L + Z_C} \end{aligned}
Now, lets consider the equation for the voltage output of the three part pickup model: \begin{aligned} V_\text{pickup} &= V_\text{in} \cdot \frac{Z_C}{Z_R + Z_L + Z_C} \end{aligned} Edit: Note here I'm assuming that V_\text{in} is equal in both cases, obviously that is unlikely to be the case, but I'm unsure how we can use this measuring technique to give an indication of the amplitude of pickups voltage under normal operation. As such: \def\Zp{Z_\text{pickup}} \def\ccancel#1#2{\mathrlap{#2}\textcolor{#1}{\cancel{\phantom{#2}}}}
\begin{aligned} \frac{V_\text{pickup}}{V_\text{out}} &= \frac{Z_C}{Z_R + Z_L + Z_C} \cdot \frac{\Zp + R_\text{ext}}{\Zp} \\\\ &= \frac{\cancel{Z_C}}{\ccancel{red}{Z_R + Z_L + Z_C}} \cdot \frac{\Zp + R_\text{ext}}{\frac{(Z_R + Z_L) \cdot \cancel{Z_C}}{\ccancel{red}{Z_R + Z_L + Z_C}}} \\\\ &= \frac{\Zp + R_\text{ext}}{Z_R + Z_L} \\\\ V_\text{pickup} &= V_\text{out} \cdot \frac{\Zp + R_\text{ext}}{Z_R + Z_L}
\end{aligned}
Substituting Z_\text{pickup} = \dfrac{V_\text{out} R_\text{ext}}{V_\text{in} - V_\text{out}} : \begin{aligned} V_\text{pickup} &= V_\text{out} \cdot \frac{\frac{V_\text{out} R_\text{ext}}{V_\text{in} - V_\text{out}} + R_\text{ext}}{Z_R + Z_L} \\\\ &= V_\text{out} R_\text{ext} \cdot \frac{\frac{V_\text{out}}{V_\text{in} - V_\text{out}} + 1}{Z_R + Z_L} \\\\ &= V_\text{out} R_\text{ext} \cdot \frac{\frac{\cancel{V_\text{out}} + V_\text{in} - \cancel{V_\text{out}}}{V_\text{in} - V_\text{out}}}{Z_R + Z_L} \\\\ &= \frac{V_\text{in} V_\text{out} R_\text{ext}}{(V_\text{in} - V_\text{out})(Z_R + Z_L)} \end{aligned}
Finally we can substitute Z_R for the measured DC resistance and Z_L for 2 \pi j f L , where L is the impedance as calculated above. Resulting in: V_\text{pickup} = \frac{V_\text{in} V_\text{out} R_\text{ext}}{(V_\text{in} - V_\text{out})(R + 2 \pi j f L)}
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Post by Yogi B on Nov 14, 2018 0:01:23 GMT -5
While I don't have a pickup testing setup, I think it might be useful to include some graphs of the results so far. So, what I've done is used LTspice to simulate a couple of the 6-part models. Firstly here's the circuit I'm using: The first of the two models is the (uncovered) '57 Classic: R1 = 8.36k, L1 = 4.87, C1 = 100p, Rd = 290k, Rl = 600k, Ld = 9.0, and R_DC = 8.02k The first step being to plot the impedance of the pickup itself and with the 470pF capacitor added in parallel, measuring the resonant frequency of each: Plugging those, the pickups DC resistance and the value of the load cap (I've switched to the standard 470pF) into the quartic we have for L we get an inductance of approximately 4.3669 (see this wolfram|alpha query, note I've used ω for the unloaded case, w for the loaded case, and X for the additional capacitance) Finally substituting that value of L into last equation of my previous post and plotting it along side the regular output from the 6-part model, gives: While not perfect I think that's pretty good myself, so now for something a bit more challenging. I wanted to try something that had a bit more of an interesting response, so what better than a Filtertron, since that's where this thread started. (That won't be a surprise to anyone who looked at the subcircuit definition above, additionally those were the values used). First the resonant frequency measurements: Then comparison between calculated and actual response: Whilst still being pretty accurate at frequencies above roughly 500-600Hz, below that point not so much. I guess that's due to the particularly low damping impedance of the Filtertrons 6-part model. Also the difference in amplitude between the 'actual' and calculated responses is greater, this I put down to damping and load resistances of the 6-part model being significantly smaller than those used other pickups.
For fun I also tried this method simulating using a 3-part model, not much to report there -- the 'actual' and calculated traces were within rounding error of each other.
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Post by antigua on Nov 14, 2018 3:32:02 GMT -5
Finally we can substitute Z_R for the measured DC resistance and Z_L for 2 \pi j f L , where L is the impedance as calculated above. Resulting in: V_\text{pickup} = \frac{V_\text{in} V_\text{out} R_\text{ext}}{(V_\text{in} - V_\text{out})(R + 2 \pi j f L)} I can try this out with my plotting script, but I'm not sure what the j term is in 2πjfL, what is that exactly?
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Post by Yogi B on Nov 14, 2018 16:33:33 GMT -5
I can try this out with my plotting script, but I'm not sure what the j term is in 2πjfL, what is that exactly? The imaginary unit, sqrt(-1), more commonly denoted as i in broader areas of mathematics.
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Post by aquin43 on Nov 15, 2018 17:21:14 GMT -5
Having made a few impedance plots, I can see that both the inductance and series resistance of many pickups varies with frequency so measuring the inductance by swamping the coil capacitance can't give accurate results. It is not difficult, given the right equipment, to measure the magnitude and phase of the pickup impedance. Two resonant frequency measurements, one with the addition of a small capacitor to lower the frequency by, say 1.5 kHz will allow both the inductance and capacitance to be calculated. With a very high Q, this is easy, but with Q around 2 or so, it is necessary to take the losses into account as well. An impedance measurement as outlined earlier allows this.
Current pickup models represent the losses as being in parallel with the coil. While it is true that any series combination has a parallel equivalent, it is hard to see how eddy current losses can actually appear in parallel with the coil; they are magnetically coupled and so must be in series in the same way as the induced signal.
An eddy current loss represented by a current loop with inductance and resistance coupled magnetically to the coil by a mutual inductance introduces an extra series resistance and negative inductance, both modified by a frequency dependent factor of the form f^2 / (f^2 + fc^2). If one measures the coil impedance below 1kHz or so, where the coil capacitance least interferes with the measurement and decomposes it into a series resistance and inductance, this behaviour can be seen quite readily, particularly with pickups that have a low Q due to high eddy current losses. The inductance falls with frequency and the resistance rises.
In order to measure the inductance and capacitance near the coil's natural resonance, I have developed a suite of programs (for Linux) that allow a set of impedance measurements to be broken into three parts covering the peak and the rising and falling sections. Algebraic curves are fitted to these sections, quadratic in the case of the rising and falling sections and quartic in the case of the peak. Curve fitting averages out the measurement errors and allows the peak and the 3dB frequencies to be determined. From these, we can obtain the peak frequency and the Q. The free program gnuplot, in addition to graph plotting can do curve fitting. Another scripting program, gawk, is ideal for processing lists of results and also can carry out double-precision arithmetic.
Basically, we have W0^2 = 1/(L*C) + (R/L)^2 ---- cf Yogi B
but Q = W0*L/R, so R/L + W0/Q
therefore W^2*(1 - 1/Q^2) = 1/(L * C)
this leads to: L * C = a, say
measure again with added Ct: L * (C +Ct) = b
b/a = (C + Ct)/C = 1 + Ct/C
Ct/C = b/a - 1, from which we get C and then L and also R from W0, L and Q
Arthur
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Post by Yogi B on Nov 16, 2018 8:30:32 GMT -5
Current pickup models represent the losses as being in parallel with the coil. While it is true that any series combination has a parallel equivalent, it is hard to see how eddy current losses can actually appear in parallel with the coil; they are magnetically coupled and so must be in series in the same way as the induced signal. I have kinda got this in the back of my mind, but I want to look at the more familiar models first. From this it sounds like you are determining the "3dB" frequencies and Q from the magnitude of impedance, wouldn't it be easier and potentially more accurate to use the phase? (The half-power or -3dB frequencies being when the phase is \pm 45\degree i.e. when the resistive and reactive parts of impedance are equal in magnitude.
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Post by aquin43 on Nov 16, 2018 10:40:20 GMT -5
Yes, phase varies rapidly at the peak, but not so much at the 45 degree points. The zero phase frequency isn't the same as the peak frequency, though, depending on the Q. I might try a version of the curve fitter to work off the phase and compare the results. The best way to measure the zero phase point is using Lissajous an an oscilloscope. The sensitivity of this measurement can be high - one part in 500 or better being possible.
As an aside, I have been thinking that a table driven simulator might be the best possible for the user, rather than the designer. Two tables, one of the complex transmission function and the other of the complex output admittance would completely characterise the pickup without the need for any modelling at all.
Arthur
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Post by ms on Nov 17, 2018 7:38:11 GMT -5
Current pickup models represent the losses as being in parallel with the coil. While it is true that any series combination has a parallel equivalent, it is hard to see how eddy current losses can actually appear in parallel with the coil; they are magnetically coupled and so must be in series in the same way as the induced signal.
It is like a transformer. The load on the secondary is in parallel with the inductance of the primary. But in this case it is like a very poorly coupled transformer, and so other is an inductor in series with the resistance due to eddy currents.
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Post by aquin43 on Nov 17, 2018 13:42:02 GMT -5
If you look into the secondary of a tightly coupled 1:1 transformer with winding resistances R and shorted primary, what you see is 2 x R, i.e. the primary and secondary resistances in series. Loosen the coupling and you add a series leakage inductance to that.
Arthur
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