How does output Voltage of an electric guitar work?

[Xenon02]

Distortion is non-linear. The old frequencies are still there, but more frequencies appear as harmonics of the inputs, plus sum and difference frequencies.

Linear means that the output follows the input, with a scale factor only.
Linear means there is no distortion and no new frequency terms created.

So it means that if Output is 0.6V then Input is "0.6" Wave vibration amplitude as I understand it cannot be "1.2" or any other value it must be "0.6" because it follows the output value in voltage. Although you've said about scale factor so.

About distortion good to know that old frequencies are still there, good to know.

PS.
Because the pickup does not reduce the amplitude of the input so there is no scale factor ? It must be 0.6V output = "0.6" Wave vibration input ????? Because when I showed the orange one with scaled version of the black signal or rather reduced. Then it was incorrect.

Did I understand it correctly ? Because It cannot be reduced as I know and that's how I interpreted the scale factor. Or rather pickups has the same scale factor or it changes everytime and it is not that easy as I thought. Dunno. I just gave an arbitrary example.

 

[Averagesupernova]

@Xenon02 concerning your diagram in post #69. What is that supposed to mean? You have sound waves going into the mic, sound waves coming out of the speaker, and electrical signals in between. So what? Where is the voltage on the input as well as the output of the transducers? There is none.
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You need to do as I said setting up an experiment with a guitar. View the signals in a spectral display using audacity.

 

[Baluncore]

It must be 0.6V output = "0.6" Wave vibration input ?????

The most difficult subjects can be explained to the most slow-witted man if he has not formed any idea of them already; but the simplest thing cannot be made clear to the most intelligent man if he is firmly persuaded that he knows already, without a shadow of doubt, what is laid before him.
– Leo Tolstoy. 1894.

 

[Xenon02]

@Xenon02 concerning your diagram in post #69. What is that supposed to mean? You have sound waves going into the mic, sound waves coming out of the speaker, and electrical signals in between. So what? Where is the voltage on the input as well as the output of the transducers? There is none.
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You need to do as I said setting up an experiment with a guitar. View the signals in a spectral display using audacity.

Dunno.

I imagined it like that :

The whole signal is scaled, like you have the high input and it's is just scaled by the pickup. The shape and everything is the same just scaled down by the pickup.

The scale factor is unknown if it's random or what but that's how I can somehow logically understand how the input is scaled in the output.

The most difficult subjects can be explained to the most slow-witted man if he has not formed any idea of them already; but the simplest thing cannot be made clear to the most intelligent man if he is firmly persuaded that he knows already, without a shadow of doubt, what is laid before him.
– Leo Tolstoy. 1894.

Again I apologize for me ;> If what I said is somehow correct then there is only one thing and I am done ... that's how lineary I see it in the picture or rather how the pickup scales the input if it indeed scales. It scales the final result of the input so.

 

[sophiecentaur]

. I'm a bit of a perfectionist

Which bit? Is it the bit that uses the wrong terminology? The effect that you are looking at is not 'interference'; it simple vector addition of signals at one location. To a first approximation, the output is the simple addition of the values of each of the signals involved.

To be honest, though. Perfect linearity of the pickup would depend on the magnetic field being uniform all over the flux gap so that it would be the same for all displacements of the string. Also, induced emf would probably be proportional to the string velocity in some pickups (as with some phono cartridges). I believe that the induced emf in the coil could be due to the change in the magnetic reluctance path as the string changes position. But now we're further down that rabbit hole and wasting quite a lot of our time down there. ;-)

First things first; learn about amplifiers first (and how to build them) .Learn about signal level handling and the problems of linearising amplifiers. There are a million and one IC audio amplifiers for small signals. Plenty of fun to be had there without bothering about the guitar pickup part of the exercise.

 

[Averagesupernova]

Dunno.

I imagined it like that :

Oh you can't be serious. Something is the way you think it is because you imagined it that way?

 

[Xenon02]

Oh you can't be serious. Something is the way you think it is because you imagined it that way?

I don't know, I read it is linear so I tried to show an example.
Linear means that the output follows the input, with a scale factor only.
Linear means there is no distortion and no new frequency terms created.

So no distortions, no new frequencies, Input has the same shape as the output only the scale is different because maybe pickup has it's own scale factor ... I don't know. Just making conclusion from what I have.
I don't know how to tell you that I just thought of it like that or pictured it like that. That's why I said if it's something like that ... Did I say "I imagined it like that so it is that and there is no way it is wrong". I just followed what was written and made an example ...

I also followed that the pickup has some max value on the output because someone wrote about it I didn't get it out of nowhere. So probably (I don't know for sure), the input isn't limitless and has it's max value like pickup that cannot exceed, and the pickup factor just reduces the whole signal and that's it ... I don't know if there is something like pickup factor that is scaling the input value. I just pictured it like that from the quote :

Linear means that the output follows the input, with a scale factor only.

So Output = Input with a factor so like Output = Input * Factor.

 

[Averagesupernova]

I don't know, I read it is linear so I tried to show an example.
Linear means that the output follows the input, with a scale factor only.
Linear means there is no distortion and no new frequency terms created.

So no distortions, no new frequencies, Input has the same shape as the output only the scale is different because maybe pickup has it's own scale factor ... I don't know. Just making conclusion from what I have.
I don't know how to tell you that I just thought of it like that or pictured it like that. That's why I said if it's something like that ... Did I say "I imagined it like that so it is that and there is no way it is wrong". I just followed what was written and made an example ...

I also followed that the pickup has some max value on the output because someone wrote about it I didn't get it out of nowhere. So probably (I don't know for sure), the input isn't limitless and has it's max value like pickup that cannot exceed, and the pickup factor just reduces the whole signal and that's it ... I don't know if there is something like pickup factor that is scaling the input value. I just pictured it like that from the quote :

Linear means that the output follows the input, with a scale factor only.

So Output = Input with a factor so like Output = Input * Factor.

That really doesn't get to the crux of the issue.
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The issue by my observations are:
You cannot measure input voltage on a guitar pickup. You seem to imply that the input is a known constant and that is not something that is easily done.
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The other issue is you seem to believe it is impossible for a guitar pickup to output more than a volt. This is also false. I don't know the standards and it may be common practice attempt to keep the signal output below a volt.

 

[Xenon02]

The issue by my observations are:
You cannot measure input voltage on a guitar pickup. You seem to imply that the input is a known constant and that is not something that is easily done.

I implied that the Input is a vibration signal not an input voltage. The pickup picks the vibrations so it must be something physical, the strings goes up and down in a specific frequency like a sinewave, so I thought the sinwave or waveform is the input of the pickup not voltage. That's why I was using "2" and not 2V as an input value, while output I clearly used Volts as a unit.

So it's something random ? Pickup will make it into voltage no matter the input amplitude value ?

Or perhaps indeed the input signal is unmeasurable because this is only vibration, and there is no input signal in the pickup. So the videos I saw where just a myth a bit, or I didn't understand something.
If that's what you've tried to tell me then okey the vibration isn't the signal input or rather it does not create sinus vibrating signal, just only the change in magnetic field, and the pickup just makes an output. So no input sinus I guess. That's what I was supposed to understand ?

The other issue is you seem to believe it is impossible for a guitar pickup to output more than a volt. This is also false. I don't know the standards and it may be common practice attempt to keep the signal output below a volt.

If it's false ok, I'll try to find more info about it.

 

[Averagesupernova]

So it's something random ? Pickup will make it into voltage no matter the input amplitude value ?

You are putting words in my mouth. I've never said that.

 

[Averagesupernova]

Or perhaps indeed the input signal is unmeasurable because this is only vibration, and there is no input signal in the pickup.

I never said it was unmeasurable. I said it is difficult to measure.
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A true accurate definition of what the input is needs to include the distance the string is moving at the pickup. Needs to be uniform in distance to the pickup compared to all strings. Also, several tests need to be done with different tensions with the same string. In other words, different frequencies using the same thickness string. Also I would like to see comparisons between different parts of the pickup with the same string and pitch. Without these things nailed down you really only know that a vibrating string makes a signal come out of the pickup.

 

[Xenon02]

You are putting words in my mouth. I've never said that.

I also have not said that the input is in voltage ;>
I was deducing but I did it incorrectly my bad here sorry.

A true accurate definition of what the input is needs to include the distance the string is moving at the pickup. Needs to be uniform in distance to the pickup compared to all strings. Also, several tests need to be done with different tensions with the same string. In other words, different frequencies using the same thickness string. Also I would like to see comparisons between different parts of the pickup with the same string and pitch. Without these things nailed down you really only know that a vibrating string makes a signal come out of the pickup.

So it's better to just assume that I have some output voltage and be happy with it then yes ? And not think about input signal ? It just works I guess. Although these vibrations are interesting in which I'll just pass I guess

Okey then last thing then, usually the electric signal of a chord/single string etc do not exceed 1V, I do not mean now that pickup max voltage output is 1V I just say now that usually the output voltage is usually 1V, so how 6 strings with amplitude 0.8V it's peak is equal also 0.8V and not higher ? It is now linear adding all sinewaves right all sinewaves from all 6 strings together. And the sum is 0.8V like single string amplitude ?
I used superposition and I just saw that different frequencies will always result in one moment where all peaks add up. So it is weird though that it stays only in 0.8V the sum of all 6 strings which individually are 0.8V as well.

 

[Averagesupernova]

By deducing or whatever, you are implying that all tones are picked up the same. We have covered this. It is my suspicion that such is not the case. The higher pitched strings will not swing as far in their oscillation. They swing faster but cover less distance. So, we don't really know. However, in reading one of your links I did see that it was common for guitar amps to have a better high end response which implies the low tones generally come out of the guitar with a higher amplitude. This response of the amp tends to flatten the response. More than once I have wished for the opportunity to delve into the guts of a guitar amp. So far, no one has come to me with an amp problem.

 

[Baluncore]

The higher pitched strings will not swing as far in their oscillation. They swing faster but cover less distance.

Yes. The induced voltage is proportional to the rate the magnetic field cuts the PU coil, which is greatest at the high-frequency end of the octave, so compensates.

The rate that the field cuts the PU coil, is greater for the lower-amplitude higher-harmonics, in the three or so octaves above the string's fundamental at that time.

 

[Drakkith]

Okey then last thing then, usually the electric signal of a chord/single string etc do not exceed 1V, I do not mean now that pickup max voltage output is 1V I just say now that usually the output voltage is usually 1V, so how 6 strings with amplitude 0.8V it's peak is equal also 0.8V and not higher ?

Is this actual data or are you making it up?

 

[Xenon02]

By deducing or whatever, you are implying that all tones are picked up the same. We have covered this. It is my suspicion that such is not the case. The higher pitched strings will not swing as far in their oscillation. They swing faster but cover less distance. So, we don't really know. However, in reading one of your links I did see that it was common for guitar amps to have a better high end response which implies the low tones generally come out of the guitar with a higher amplitude. This response of the amp tends to flatten the response. More than once I have wished for the opportunity to delve into the guts of a guitar amp. So far, no one has come to me with an amp problem.

Wait a sec so in the link from post#1 the output voltage is not measure from pickup ?

But if still the pickup makes higher amplitudę for the bridge part then again high amplitude adds up, the rule of superposition and so applied and all sounds from 6 strings have different frequencies so still it contradicts with the fact that there must be high peak and there is not. What is the flaw of superposition here ? In fact it was confirmed here that output signal I can use superposition on output signal. Because adding Individual signals from E1 and E2 will look the same as if E1+E2 already where played at the same time. So the sum of individual is equal to the voltage output.

Is this actual data or are you making it up?

From first link in post#1 I took as a reference E1 and E2 but the chor consists of 6 strings while here I had E1 and E2. But seeing that the output is high because 0.8V or a bit smaller like 0.6V then I deduced that E3 ... E6 of that chor is also high so adding all 6 up for chor should result in high peak although in the article it says it is max 0.8V which doesn't make sense to me, knowing the superposition or the fact that peak values add up at some point because all signals have different frequencies

 

[Averagesupernova]

I don't know where the voltage is taken in the link in post #1. However, the quote below comes from that link.

It's also easy to see why most guitar amps have a significant amount of treble boost - it's necessary because the output of the higher strings is almost always lower than expected.

 

[Xenon02]

I don't know where the voltage is taken in the link in post #1. However, the quote below comes from that link.

Here it says at the top pickup voltage and not amp voltage so. E1 In Samick TV is 800mV peak, E2 is 200mV peak so the rest of the strings have smaller amplitude ?

Still the sum of E1 and E1 should give at some point 1V looking at this graph I've made

Why it must ? No matter what phase shift I've applied it always had this one peak the sum of all peaks.

So why is it that the chord is 0.85 V ? Why the sum is different from what I add here or is my superposition incorrect if so there is my geogebra link in first post, I would be glad to see how it should look like

 

[Drakkith]

From first link in post#1

Stop assuming the data in that article is accurate. It isn't, the author of the article said so themselves, more than once, and I already touched on this in a previous post. You can see that the data is wildly different between two different guitars and doesn't even following a similar pattern, making anything except the broadest of statements about the data unreliable.

Still the sum of E1 and E1 should give at some point 1V looking at this graph I've made

The voltages listed are RMS voltages, That means it's an average, so any peaks have been averaged out.

 

[Xenon02]

The voltages listed are RMS voltages, That means it's an average, so any peaks have been averaged out.

RMS and peak voltage is listed. In the bracket (). It's even written Average RMS (Peak)

The author was repeating these measure and they were similar, of course not accurate but very similar. Good point of reference.

So my calculations where conflicting, peaks should add up at some point so E1 + E2 = 1V, the chord is 0.8V which doesn't make sense because the chord consist of E1 = 0.8V and E2 of 0.2V

Their peaks must add up because of different frequencies so peaks must meet and because of superposition.

Any answer why superposition didn't work ? Logically E1+E2 = 1V peak, and there are 4 more strings so the peak should be bigger than >1V but it is 0.8V nonsense. Superposition didn 't work, logic behind it is illogical

 

[Drakkith]

RMS and peak voltage is listed. In the bracket (). It's even written Average RMS (Peak)

Ah, so it is. My mistake.

So my calculations where conflicting, peaks should add up at some point so E1 + E2 = 1V, the chord is 0.8V which doesn't make sense because the chord consist of E1 = 0.8V and E2 of 0.2V

How hard did the author strike the strings for each note and then the chord?

 

[Xenon02]

How hard did the author strike the strings for each note and then the chord?

That's a good point. Maybe you are right about that part that striking the chord results in smaller amplitude of each string. The question is, is it that radically reduced ? So that E1 from 0.8V will be smaller like 0.4V or 0.5V so that the rest strings will have like 0.1V

Although how much the amplitude changes. Author said he was striking them the same way repeatedly. So I assumed that E1 is equal 0.8V. But isn't the amplitude more dependent on the frequency like it was said before ? Like bridge E1 was 0.8V and E1 on neck is 0.45 V because of different frequencies than how hard it was stroke.

So let's say the chord that consist of E1 is 0.8V then the chord peak doesn't have any sense. If E1 on the bridge changes drastically from 0.8V to 0.6 V then it might be possible to have output of 0.8V when the rest of the strings have less than 0.1V if it makes any sense.

But yea did the E1 amplitude changed then the author played chord ? Single E1 is 0.8V then what is the value of E1 in chord of the bridge part ?

Interesting

 

[Drakkith]

That's a good point. Maybe you are right about that part that striking the chord results in smaller amplitude of each string.

My point is that the results in the article are from an almost entirely uncontrolled experiment. You would need to reliably strike the strings the same way each time to get an accurate comparison between the notes and chords. That's just not going to happen by hand. Amplitude, strike location, and strike direction will all vary with each strike by hand. Sometimes insignificantly, sometimes significantly. So trying to draw any conclusions from these numbers is problematic.