|
|
Post by JohnH on Aug 16, 2018 4:38:46 GMT -5
I was about to write something here, then saw that Id already written it above back in April. The 6db per octave issue is not complicated. Nothing to do witb exciter coil inductance, which is very small compared to tbe resistance that the guys are driving it through. So a constant voltage from the signal gen makes a constant current amplitude and a constant flux intensity. But induction is not based on flux but rate of change of flux. Higher frequency at which flux goes up and down gives higher rate of change. Hence +6db per octave increase in induction.
Real strings get lighter for the top strings so although they vibrate faster than the low strings, with about the same amplitude (they have about the same tension and you pluck them the same), this smaller mass provides a compensating slope so all strings give similar output. Integration by hardware, or post-processing maths, brings the results from exciter tests back into line with real (or theoretical real but perfect) strings.
|
|
abiscan
Rookie Solder Flinger
Posts: 17
Likes: 0
|
Post by abiscan on Aug 16, 2018 6:57:12 GMT -5
Please. Any literature backing this up? A clear hint to every induction increases by 6dB because Faraday? I understand your point. Mathematically this is what should happen while inducing. But I can't observe it. I clearly think we are missing something out here.
What I can observe is a high pass influence. Acting with +6dB only till 300Hz. After that the slope changes. Applying +6dB over the whole measurement would falsify the outcome.
|
|
|
|
Post by stratotarts on Aug 16, 2018 8:23:27 GMT -5
Please. Any literature backing this up? A clear hint to every induction increases by 6dB because Faraday? I understand your point. Mathematically this is what should happen while inducing. But I can't observe it. I clearly think we are missing something out here. What I can observe is a high pass influence. Acting with +6dB only till 300Hz. After that the slope changes. Applying +6dB over the whole measurement would falsify the outcome. You have a problem is in your experimental apparatus. The time constant of your exciter coil is much too long. Your -3dB cutoff is at about 2000 Hz, so the exciter coil is not operating in mainly constant current mode. That means that the pickup is not receiving a uniform excitation amplitude up to 15kHz or so, where it has to be measured. Either your coil inductance is too high, or the limiting resistor is too high. The resulting response is thus uneven and requires the correction that you have mentioned. However, your conclusion that this flaw affects everybody is wrong. I anticipated this problem a long time ago, and ensured that my exciter response was almost flat up to 20kHz. I have also measured the drive circuit independently and observed the Bode plot. My coil inductance is about 500uH, with a 100 ohm resistor this gives me a -3dB point around 32kHz. Thus my raw plots do exhibit a +6dB response up to resonance.
Here is a plot from my first experiments in 2014:
I can also show you plots of different pickups measured with the same test apparatus that show, one has a dip below resonance while the other does not. That also demonstrates that the "droop" or "dip" below resonance is actually some property of the pickup (most people think it's due to eddy current losses) and has nothing to do with the exciter.
|
|
abiscan
Rookie Solder Flinger
Posts: 17
Likes: 0
|
Post by abiscan on Aug 16, 2018 9:13:03 GMT -5
I just simulated some different driver coils.
Green mine. Blue yours ( stratotarts ). red hypothetical coil.
Imo i see a clear influence of the coils on the measured frequency response of the guitar PU. Wouldnt it be best to reach the linear part of the transmission as soon as possible? For example the red curve! like i said... driver coil and PU induction together is a bandpass. Please tell me ho you produced such coil. What wire did u use and what number of windings etc.? Do you know the input capacity of your soundcard? what length of cable did you use, cause your peaks appear quite low-end to me.
One more thing. I'd be very happy seeing more measurements. I measured single coils, pafs, JB, PB and P90 but never saw that dip (e: or better to say, the dip is exactly at the -3dB roll-off position connected to the driver coil)
|
|
|
|
Post by Yogi B on Aug 16, 2018 10:06:10 GMT -5
Either your coil inductance is too high, or the limiting resistor is too high. Surely you mean: too low?
Wouldn't it be best to reach the linear part of the transmission as soon as possible? For example the red curve! Theoretically, with a large enough inductance with low enough resistance such that the entire 'constant current' region was below the measured range then one could forgo the integrator. The issue being the values that would make that possible aren't practical -- I'd like to see someone try to make that 1000H inductor with only 50k series resistance and make it roughly pickup sized -- furthermore I believe the mutual inductance between driver and subject coils with such an inductive driver would significantly throw off the readings.
As a side note the second link in antigua's opening post (to syscomp design), is broken, but is available on the Internet Archive. In that they produce plots somewhat similar to ABisCan's using a driver coil with resistance of 129Ω and inductance 36.9mH. They (incorrectly) justified its use by noting that its self resonant frequency was above the frequency range to be measured. |
|
|
|
Post by antigua on Aug 16, 2018 12:30:57 GMT -5
What i am trying to say.. we need to get rid of this 6dB connection to faradays law. In my view it just isn't existing. I found this rising slope NOWHERE in any literature. You see the same curve even if you don't use an exciter coil, and instead measure the impedence of the pickup directly, as described here guitarnuts2.proboards.com/thread/7775/pickups-resonant-peak-usb-oscilliscope  As for the literature, AFAIK all of this conforms to your typical inductor with non-ideal leakages such as "inter winding capacitance" and series resistance:  |
|
abiscan
Rookie Solder Flinger
Posts: 17
Likes: 0
|
Post by abiscan on Aug 16, 2018 12:33:41 GMT -5
okay...  so... you use a integrator to get rid of the 6db/octave slope of the high pass that the driver coil produces. BUT! you can only assume it to be integrateable if the inductance is low enough for the low cut to be in the range of the PU use. so the -3dB roll-off has to be above roughly 20kHz. else you would affect the original data. OR you could (theoretically) produce a coil which would not affect the transmission (roll-off below 10Hz) OR you could mathematically post process the data by subtracting the driver-coil frequency response BUT... the slope of 6dB is coming from the driver coil and its high pass behaviour!  can we all agree on that? edit: i just saw antigua replied sooo.. lets make this clear.. we are talking about the frequency response here and NOT the impedance. This view u can only get if u put on a logarithmic scale on both axis AND in the first herz it is constant and NOT rising by 6dB because impedance is not measured in dB but in Ohm   |
|
|
|
Post by JohnH on Aug 16, 2018 16:47:01 GMT -5
It does seem that the next step is to measure the inductance of your exciter coil, and check you are running it through a large enough resistor so that the corner frequency where 2Pi.L.f = R is high, at least say 20kHz, prefferably higher. Then you have constant current through your coil at all frequencies. This gives the 6db slope, which being then a consistent gradient, can be taken out.
In theory, you could instead, try to wind a very low resistance coil and run it from a perfectly low impedance amp, and so try to achieve a constant voltage condition across the inductance of the coil, instead of constant current. Then youd get induced signal without the slope. But thats heavy wiring needed!
Going for the condition of constant current is what works best with the available apparatus. Anything in between just confuses the results needing difficult post-processing to extract what the test pickup is doing from in amongst the effects of the exciter coil.
|
|
|
|
Post by JohnH on Aug 16, 2018 17:05:21 GMT -5
Just further on achieving constant current, to all the guys doing testing: What exciter coil inductance are you using and through what series resistance?
I figure that if you want less than say 0.5db drop in slope linearity signal at say 10khz, then youd want the LR corner frequecy to be at least 30kHz. So then at 10khz, the coil reactance is 1/3 of R.
20log (1 + (1/3)^2) = 0.46db.
|
|
|
|
Post by antigua on Aug 16, 2018 19:51:33 GMT -5
BUT... the slope of 6dB is coming from the driver coil and its high pass behaviour! can we all agree on that? The exciter coils have a very low inductance, at least mine do. If you try to, for example, use another guitar pickup, or any coil that has a core in it, as an exciter, you'll have a bad time. So there's some relatively high resistance in series with the exciter coil, so looking at this depiction, it's flat where "R" is, and it the exciter + function generator should have a relatively high "R" so as to extend that flat portion out, beyond the resonance of the pickup being tested. |
|
|
|
Post by stratotarts on Aug 16, 2018 23:27:00 GMT -5
[...] BUT... the slope of 6dB is coming from the driver coil and its high pass behaviour! can we all agree on that? [...] The idea is to reference the response to a given time varying magnetic field of uniform intensity. That happens when the exciter coil is driven with constant current, because the strength of the exciter field is proportional to current through it (not the derivative dI/dt!). The slope is not any part of that. It is a purely linear response. The slope is only produced in the pickup coil because Faraday's law does impose a derivative function. The same +6dB slope is produced by the actual strings. It is the actual, true response because the typical guitar chain does not include any integrator. The purpose of the integrator is to isolate the LCR filter response of the pickup so that it can be more easily understood and modeled. The raw plot does not lend itself to easy visualization and analysis of resonant peaks, eddy losses, and so on. The use of post test processing to compensate for inductance in the exciter coil is feasible and interesting, but if it produces different results than what would normally be expected, it would be better to examine the processing, rather than the measurement methods which have been tested and confirmed, or the theory of the practical pickup device itself, which has been deeply explicated.
|
|
abiscan
Rookie Solder Flinger
Posts: 17
Likes: 0
|
Post by abiscan on Aug 17, 2018 3:54:49 GMT -5
i just can ask once again... i can find NO literature backing this 6dB slope in connection to faraday. please. if you could hand me over this calculation it would be great. but till then the connection to me is a false interpretation. (type in faraday and 6dB into google)
consider the calculation of a tranfer function. and by this u have to consider the high pass behaviour of the driver coil. a simple RL high pass got a rising slope of 20/decade which is the same as 6dB/octave and gets to a point where the transmission is not touched (0dB). a straight forwards implementation of your faraday interpretation would superimpose a +6dB slope all over the transfer funtion and not just till the cutoff frequency. driver coil and PU is a new system. it may be closer to the string PU model but you cant explain the 6dB slope with faraday but with the high pass behavior because the slope extends to RC systems at well, where there is no inductance therefore no induction. in EVERY filter book you look, you will find the 20dB/decade. how come in none of the induction books there is any info about it?
"quote:
Ratios and slopes
A frequency ratio expressed in octaves is the base-2 logarithm (binary logarithm) of the ratio:
{\displaystyle {\text{number of octaves}}=\log _{2}\left({\frac {f_{2}}{f_{1}}}\right)} {\displaystyle {\text{number of octaves}}=\log _{2}\left({\frac {f_{2}}{f_{1}}}\right)}
An amplifier or filter may be stated to have a frequency response of ±6 dB per octave over a particular frequency range, which signifies that the power gain changes by ±6 decibels (a factor of 4 in power), when the frequency changes by a factor of 2. This slope, or more precisely 10 log10(4) ≈ 6.0206 decibels per octave, corresponds to an amplitude gain proportional to frequency, which is equivalent to ±20 dB per decade (factor of 10 amplitude gain change for a factor of 10 frequency change). This would be a first-order filter."
i will update my paper for you guys so you will get my point a little better, i really dont know if you get my point, cause in my head all is set but i am not to good in english and probably didacticly bad
best greetings
|
|
|
|
Post by Yogi B on Aug 17, 2018 7:35:23 GMT -5
i just can ask once again... i can find NO literature backing this 6dB slope in connection to faraday. please. if you could hand me over this calculation it would be great. but till then the connection to me is a false interpretation. (type in faraday and 6dB into google) It might be one of those things that seems so obvious to some that they never explicitly point it out. But it is always implicitly there whenever Faraday's law is stated. The induced EMF in a closed loop is equal to the negative time rate of change in linked magnetic flux. Thus if the frequency at which the flux is oscillating increases by a factor of 2 (one octave), the rate of change doubles, therefore double the EMF (+6dB) is induced. Also, Googling "Faraday's Law in the Frequency Domain" leads here (see the first point under Eq. 4).
Most of the time when people discuss or plot filter responses they are doing so in terms of voltage, not current. Here's the standard (voltage) plot of a high pass filter showing the voltage across the inductor, followed by a plot of the current through the same inductor. As I said before, the magnetic field created by the driver coil and responsible for the voltage induced into the pickup under test is dependent on the current through the driver coil. Thus the 6dB/octave slope in voltage has no direct influence on the measurement other than the fact it is coincident with the region where the current is constant, and it is this region of constant current which allows the 6dB/octave slope of Faraday's law to be clearly observed. |
|
|
|
Post by JohnH on Aug 17, 2018 7:48:08 GMT -5
The electrical theory is just that induction is proportional to rate of change of flux. The 6db per octave slope is just an outcome of the associated differential maths. No further electrical theory is involved.
We are saying that with constznt current amplitude, flux amplitude is constant.
So lets assume flux = a.sin(2Pi.f.t) where f is frequency and t is time and a is a constant.
Rate of change of flux = 2Pi.a.f.cos(2Pi.f.t)
So rate of change has amplitude depedent on f. Double f, doubles the rate of change of flux. This doubling per frequency octave is 6db per octave (strictly, 6.02db per octave= 20log(2) )
done!
|
|
|
|
Post by antigua on Aug 17, 2018 13:45:12 GMT -5
i just can ask once again... i can find NO literature backing this 6dB slope in connection to faraday. please. if you could hand me over this calculation it would be great. but till then the connection to me is a false interpretation. (type in faraday and 6dB into google)
I was in the same boat a while back, and found that Yogi's explanation of it not being mentioned due to being obvious is closest to the truth. For example, in this article about band pass filters www.electronics-tutorials.ws/filter/filter_2.html you see reference to -20dB/decade, which is close to -6dB/octave mentioned in passing as what the slope happens to be. |
|
|
|
Post by stratotarts on Aug 17, 2018 23:08:44 GMT -5
I just simulated some different driver coils. [...] Please tell me ho you produced such coil. What wire did u use and what number of windings etc.? Do you know the input capacity of your soundcard? what length of cable did you use, cause your peaks appear quite low-end to me. I use about 100 turns on a Strat coil, I believe the wire gauge is #32 but I can't remember exactly:  |
|
abiscan
Rookie Solder Flinger
Posts: 17
Likes: 0
|
Post by abiscan on Aug 22, 2018 7:58:40 GMT -5
Hey thanks a lot!
I am getting clooser to it. So u got a coil with a 100 turns? (when i put on a AWG 18 on a humbucker coil it isnt even close to full as yours) did u maybe mean 1000 turns?
then u put on a 100Ohm resistor in series to get it into constant mode?
i will investigate it furthermore...
|
|
|
|
Post by stratotarts on Aug 22, 2018 16:47:58 GMT -5
Hey thanks a lot!
I am getting clooser to it. So u got a coil with a 100 turns? (when i put on a AWG 18 on a humbucker coil it isnt even close to full as yours) did u maybe mean 1000 turns?
then u put on a 100Ohm resistor in series to get it into constant mode?
i will investigate it furthermore...
Yes, it is actually 50-100 turns of AWG #26. I think you must have a different gauge than 18. 100 turns of #18 would be quite thick. Electrical house wiring is AWG #14. |
|
abiscan
Rookie Solder Flinger
Posts: 17
Likes: 0
|
Post by abiscan on Aug 23, 2018 7:15:53 GMT -5
u are right... it wasn the AWG it was a european measurement and it had 0.22mm.. so actually almost a awg 32. i am gonna try to find another wire
|
|
abiscan
Rookie Solder Flinger
Posts: 17
Likes: 0
|
Post by abiscan on Aug 28, 2018 10:25:32 GMT -5
i've produced a new driver coil and still getting to some problems here, hoping you could help me out here.
i simulated your coil again and plotted the current as well...
To me the current aint constant as well. So an integrator with a -6dB/octave slope would still falsify the outcome, wouldnt it? imo a post measurement subtraction of the driver coil response still seems like the best solution to me, or am i somehow wrong about the integrator influences?
with one tenth of your inductance (0.5uH) i would be d'accord with calling the current constant.
here you can see the picture of the ken wilmot integrator simulation.
imo this produces a totally different frequency resonse of the measured pu. this can't be, what we all are looking for. you get the perfect RCL frequency response only by subracting the high pass FR of the driver coil. this integration in my eyes does something different. soon i will update the paper... if anyone got additional information regarding the integrator model please help me out here.
BR
|
|
|
|
Post by stratotarts on Aug 29, 2018 9:59:46 GMT -5
The integrator circuit has been validated by simulation and testing. The actual circuit is faithful to the 20dB/decade slope by less than 0.5dB between 200Hz and 20kHz. Here is an actual plot from a calibration run (integrator input connected directly to exciter voltage output).
The simulation produces identical results:
The link you provided to illustrate real vs. ideal integrator performance concerns only the low frequency performance. The integrator circuit revision 5.7 included an improvement to the low frequency cutoff that lowers the -3dB frequency to 8.85Hz. In that case, the discrepancy is reduced to fractional dB at 100Hz. That was an improvement from 5.6 which was only "flat" down to 200Hz or so. The LT1058 has enough gain to provide very good linearity up to at least 15kHz. Beyond that, there is not usually much importance to measurement accuracy as it is well outside the pickup's operating frequency range.
About the exciter coil - I get a value of 350uH from the online calculator here link . That gives you a -3dB cutoff at about 45kHz with a 100 ohm series resistor. Of course, the impedance of the inductor rises with frequency. At 15kHz it is about 30 ohms, but keep in mind that the 100 ohm resistor is usually in series with the generator impedance which adds another 100-200 ohms. Disregarding vector math to arrive at a quick estimate, the error in that case would be about (100+30)/100 = 2.3dB.
However, considering a 50 turn coil, which should have an inductance of about 10uH, the inductive reactance at 15kHz is only about an ohm, and so is unlikely to disturb the constant current effect of the 100 ohm drive coil series resistor. I've always considered it important to use the minimum number of turns necessary to provide a working signal, for this reason.
|
|
|
|
Post by straylight on Sept 9, 2018 13:07:06 GMT -5
EDIT: I completely changed what I wrote here
Whether the electronic integrator is any good or not is not an issue here. It's entrely possible to handle the process programatically. I'm using R as it innately applies operators to an entire list rhater than needing the programmer to construct a loop each time.
Given the dataframe bodeData that looks somehting like where Mag is the magnetude of the response in dB
Freq Mag IntMag
1 40.00000 -22.109387885 1.635407e-01
2 40.80000 -21.846253948 2.546712e-01
3 41.61600 -21.600581582 3.283401e-01
4 42.44832 -21.532305589 2.246127e-01
5 43.29729 -21.570279567 1.463526e-02
6 44.16323 -21.295177938 1.177335e-01
7 45.04650 -21.158338444 8.256951e-02
8 45.94743 -20.944981667 1.239229e-01
9 46.86638 -20.822339990 7.456110e-02We can apply -20dB/decade as programatic integration with respect to frequency and store it as IntMag
bodeData$IntMag <- (bodeData$Mag - 20 * log10(bodeData$Freq) )
And it should be obvious that if the induced electrical field is proportional to the rate of change of magnetic flux then the response will be dependent on frequency.
The source of 20dB/decade (6dB/octave) is function of measuring gain in dB since voltage gain in dB = 20 × log (V2 / V1). I hope i'm not pointing out the wrong bit of the obvious here.
Note carefully that to get a nice straight line on our plots we're using a logarithmic value in Y and a logarithmic scale in X
|
|
|
|
Post by antigua on Sept 10, 2018 21:16:18 GMT -5
EDIT: I completely changed what I wrote here
Whether the electronic integrator is any good or not is not an issue here. It's entrely possible to handle the process programatically. I'm using R as it innately applies operators to an entire list rhater than needing the programmer to construct a loop each time.
Given the dataframe bodeData that looks somehting like where Mag is the magnetude of the response in dB
Freq Mag IntMag
1 40.00000 -22.109387885 1.635407e-01
2 40.80000 -21.846253948 2.546712e-01
3 41.61600 -21.600581582 3.283401e-01
4 42.44832 -21.532305589 2.246127e-01
5 43.29729 -21.570279567 1.463526e-02
6 44.16323 -21.295177938 1.177335e-01
7 45.04650 -21.158338444 8.256951e-02
8 45.94743 -20.944981667 1.239229e-01
9 46.86638 -20.822339990 7.456110e-02We can apply -20dB/decade as programatic integration with respect to frequency and store it as IntMag
bodeData$IntMag <- (bodeData$Mag - 20 * log10(bodeData$Freq) )
And it should be obvious that if the induced electrical field is proportional to the rate of change of magnetic flux then the response will be dependent on frequency.
The source of 20dB/decade (6dB/octave) is function of measuring gain in dB since voltage gain in dB = 20 × log (V2 / V1). I hope i'm not pointing out the wrong bit of the obvious here.
Note carefully that to get a nice straight line on our plots we're using a logarithmic value in Y and a logarithmic scale in X
stratotarts had mentioned doing the integration programatically a while back. The CGR drivers and software are open source, but I preferred to use the Velleman oscilloscope for a variety of reasons, and it would have to be submitted as a feature request with them. Outputting the data csv and processing it separately is a good idea, and I can possibly put something together with javascript, so that anyway can drop their data into the thing and get an integrated plot back out, so that all anyone needs to buy is either a Velleman or a CGR and they're good to go. I'm also of the opinion that an external exciter coil is not especially necessary, as the majority of the damping effects exist as a result of the coil's physical reaction to the metals, and not the string's. So all anyone should need is the $100 oscilloscope and a few resistors. I think this would be a worthwhile project, so I will try to bust it out soon. |
|
|
|
Post by antigua on Sept 18, 2018 0:08:28 GMT -5
EDIT: I completely changed what I wrote here
Whether the electronic integrator is any good or not is not an issue here. It's entrely possible to handle the process programatically. I'm using R as it innately applies operators to an entire list rhater than needing the programmer to construct a loop each time.
Given the dataframe bodeData that looks somehting like where Mag is the magnetude of the response in dB
..
Here are a couple data files produced by the VElleman PCSGU200, the files are named for the pickups in question. These plots were made by driving the pickups with a voltage and measuring the difference across a 1 meg resistor, as opposed to using a driver coil. echoesofmars.com/misc/velleman_pcsu200_dimarzio_super_dist.txtechoesofmars.com/misc/velleman_pcsu200_fender_fidelitron_neck.txtEach file has two plot lines, the first is unloaded, the second is loaded with 470pF and 200k. I haven't had a chance to start on a javascript based data intake and plotter, but one thing I'm curious to see is whether or not the Fideli'tron shows any pre-resonances dips when integrated mathematically. I can also create plots with an exciter coil so that the results can be compared, but that will have to wait until later in the week because among other things that requires changing around my test rig. |
|
|
|
Post by aquin43 on Oct 5, 2018 10:28:33 GMT -5
Hello, thanks for this interesting article. I have couple of points that might be useful in addition:
Resonance
If you have a basic oscilloscope and a signal generator, there is another way to measure the true resonant frequency without the need for a frequency response plot. This is by finding the point at which the phase response is 90 degrees. A simple second order system like a guitar pickup, with a resonance at omega_n = 2*pi*f_n has a response of the form 1/(1-(omega/omega_n)^2 + j*(omega/omega_n)/Q) At resonance (which is not the frequency of peak response, especially when Q is low), omega = omega_n so the real term vanishes and the response is Q/j = -jQ. This has a phase lag of 90 degrees. As usual, excite the coil with an external driving coil and buffer resistor, where the current is in phase with the driving voltage because you have kept the driving coil reactance low compared with the coil buffer resistor. The voltage induced in the pickup coil will depend on the rate of change of the driving current. The rate of change of sine is cosine implying a phase lead of 90 degrees. So, fortuitously, the overall phase shift of this system at resonance is zero. In addition, phase varies rapidly about the resonant frequency, even when the Q is low. Most oscilloscopes can measure channel 1 + channel 2 and also invert channel 2. All that is necessary is to difference the input and output on the scope and adjust the source frequency and relative channel gains until the difference is zero. It can only be zero when the phase difference is 180 degrees. This is quick and easy to do in practice. The frequency is now at the true resonance where omega=omega_n. If you load the pickup with, say, a 10n capacitor (C) to swamp its own and stray capacitance, you can calculate the inductance from L = 1/(C*(2*pi*fn)^2) Step ResponseIf you have a function generator that can output ramp or triangle waves, it is possible to measure the step response of the pickup. When excited by a triangle wave, the voltage induced in the pickup coil is a square wave. Typically, the response will overshoot. The percentage overshoot can be measured on the scope screen, usually automatically. The Q can be calculated from the percentage overshoot as follows: lnos = ln(oshoot%/100) Q = 0.5/sqrt(pi^2 + lnos^2)
It is interesting that the measured Q is usually quite a bit lower than what would be calculated from the inductance and dc resistance, presumably because of hysteresis and eddy current losses. It is possible to calculate an equivalent parallel load resistor to represent the losses but the formula is a bit messy.
Arthur
|
|
|
|
Post by ms on Oct 5, 2018 13:17:20 GMT -5
At resonance (which is not the frequency of peak response, especially when Q is low), omega = omega_n so the real term vanishes and the response is Q/j = -jQ. This has a phase lag of 90 degrees.
If you measure impedance instead of response, the value is real at resonance, and so the imaginary part crosses zero (as does the phase, of course) and makes a convenient point to measure, as on the attached plots.  |
|
|
|
Post by aquin43 on Oct 6, 2018 6:39:23 GMT -5
At resonance (which is not the frequency of peak response, especially when Q is low), omega = omega_n so the real term vanishes and the response is Q/j = -jQ. This has a phase lag of 90 degrees.
If you measure impedance instead of response, the value is real at resonance, and so the imaginary part crosses zero (as does the phase, of course) and makes a convenient point to measure, ... What you say is true, but at the moment I only have an oscillator and scope so I would have to make a test rig to measure impedance. The only test rig I have at the moment is 10 turns of hook-up wire in series with 50 ohms connected to the 50 ohm generator output.
The point is that phase is the most sensitive measure of resonance and is the only true indicator of when omega_n = 1/sqrt(L*C).
If the oscilloscope can do X-Y display, then it is possible to use the Lissajous method where the plot of output against input is an ellipse that closes to a line when the phase difference is zero. This makes the measurement very quick and easy.
Arthur
|
|
|
|
Post by aquin43 on Oct 7, 2018 10:00:19 GMT -5
On reflection, the impedance of a pickup doesn't become resistive at omega=1/sqrt(LC). This is only an approximation that works when Q is high.
Representing omega = 2*pi*frequency by w, the expression for the impedance (Z) of an inductor with resistance in parallel with a capacitor is:
Z = (jwL + R)/(1 - (wLC)^2 + jwCR)
so, when w=w0 or wLC = 1 we are left with:
Zw0 = (jw0L + R)/jw0CR
Zw0 = L/CR - j/(w0C)
There is a lagging imaginary component that becomes less significant as R becomes smaller, Q increases, and the real part L/CR becomes dominant.
I think that if you wish to measure the true inductance of the pickup then any method that relies on amplitude should be avoided. The Lissajous phase method using an exciter coil seems to be the most sensitive method.
Note that either the bandwidth of the exciter coil must be very large to keep the phase shift of the drive current low or a small sensing resistor can be placed between the return side of the coil and ground to provide the current drive phase reference.
Arthur
|
|
|
|
Post by ms on Oct 7, 2018 16:15:42 GMT -5
On reflection, the impedance of a pickup doesn't become resistive at omega=1/sqrt(LC). This is only an approximation that works when Q is high. Representing omega = 2*pi*frequency by w, the expression for the impedance (Z) of an inductor with resistance in parallel with a capacitor is: Z = (jwL + R)/(1 - (wLC)^2 + jwCR) so, when w=w0 or wLC = 1 we are left with: Zw0 = (jw0L + R)/jw0CR Zw0 = L/CR - j/(w0C) There is a lagging imaginary component that becomes less significant as R becomes smaller, Q increases, and the real part L/CR becomes dominant. I think that if you wish to measure the true inductance of the pickup then any method that relies on amplitude should be avoided. The Lissajous phase method using an exciter coil seems to be the most sensitive method. Note that either the bandwidth of the exciter coil must be very large to keep the phase shift of the drive current low or a small sensing resistor can be placed between the return side of the coil and ground to provide the current drive phase reference. Arthur On the other hand, if the only loss is in parallel, say eddy current loss, then the peak in amplitude and the 0 phase are the same. In a real pickup, the dominant loss at resonance can be the eddy current loss, and so it is almost true. That is why it is so close on why plots, I think. |
|
|
|
Post by straylight on Jan 23, 2019 19:51:40 GMT -5
Here are a couple data files produced by the VElleman PCSGU200, the files are named for the pickups in question. These plots were made by driving the pickups with a voltage and measuring the difference across a 1 meg resistor, as opposed to using a driver coil. echoesofmars.com/misc/velleman_pcsu200_dimarzio_super_dist.txtechoesofmars.com/misc/velleman_pcsu200_fender_fidelitron_neck.txtEach file has two plot lines, the first is unloaded, the second is loaded with 470pF and 200k. I haven't had a chance to start on a javascript based data intake and plotter, but one thing I'm curious to see is whether or not the Fideli'tron shows any pre-resonances dips when integrated mathematically. I can also create plots with an exciter coil so that the results can be compared, but that will have to wait until later in the week because among other things that requires changing around my test rig. I have been fighting with these files for a while. Some rewrites of the way i handled the data internally were required to be able to load from a generic source. My initial data model made too many assumptions based on my own process. I digress, I'm having two problems with the data: The first is I don't think it's high enough resolution (enough data points across the frequency domain) for my statistical smoothing to work. No big deal, it's sparse enough to get peak readings without denoising. I'll see if I can determine the lower limit of datapoints for loess() to work without falling back on simpleLoess() and throwing up errors. The second is the data isn't behaving as expected. I'm expecting the response between 100 and 500 Hz to be linear (and thus flat when integrated) It's not. Is this a quirk of measuring with a resistor? It's kind of inconvenient as I'm using the height of the flat section to do some calculations.
 |
|
can we all agree on that?
