Xenon02
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I don't have the guitar but if I had I don't have anything good to church the voltages etc.Averagesupernova said:Well it may be stupid to someone but I wouldn't say from my perspective or yours it's a stupid question. Do you have a guitar? If so, do some experiments.
The number comes from the link from the first post so I used them as a reference to my geogebra version of these graphs.Averagesupernova said:Where is your math for this and are you assuming peak amplitude or RMS?
What I took out of this lesson is that no matter the phase shift or the frequency. All signals with different frequencies while adding up will meet at some point where all peaks will add up resulting in high peak.
So I research alot and many sites shows the same graphs as mine which is the orange one. They did not use units so I interpreted them as an input signal. So input signal had peak of "4" what will the pickup do to this signal ? It won't cut it creating distortion so ??? Input is signal like my orange signal what is the output ? Reduced to 0.6V peak version ?
If yes then there is no logic in between input signal peaks with output voltage because the peak could have value "10" so what pickup will do ? What value he will give to this peak ?
BUT look at the table I gave in post #47 the higher notes we get from the guitar the bigger the amplitude or maybe the lower notes ?? I dunno the sound on the bridge. But yea, so there is some conclusion I can take from this picture BUT how does the input to the guitar looks like ? Why if I have 6 signals with 0.3V adding up did not exceed 1V ? Dunno. There is in the table that possibility and the chord gave only 0.3V output having individual strings with 0.3V peak. And we know signal with different frequencies all meet their peaks at some point so 0.3*6. Why am I confident they must meet ? My geogebra example.
Baluncore said:That is irrational. If two produce 300 mV peak, then one produces 150 mV peak and six produce 900 mV peak. That is less than one volt peak, not more.
Individual string have this much the table I've provided doesn't lie. Plus what if you pull the string harder then maybe the amplitude rises. But maybe not because the frequency of the sound is the same just how high the bouncing is increased. So it should have any impact. I think so reading what Averagesuperman has said in post #41
Baluncore said:This is an electric guitar. If you hit every string at once, you would complain if it did not clip and distort.
Clipping and distortion should the circuit do and not the pickup I assume.
And if randomly pulling the string makes distortion then playing on the guitar the same way will result randomly in distortion which doesn't make sense.
But you can agree that signal with different frequencies must meet at some point all their peaks must add up at some point like in my orange graph example right ? Resulting in high peak moment. So like how does the input looks like. Why the 6 string sound peak wasn't above 1V knowing from the table that strings have 0.3V or even 0.8V at some example.
So how does the input look like and why pickup with these signals amplitude didn't exceed 1V. Weird and interesting