I just did a test, recorded my bass guitar through a high gain distortion pedal and then direct to my sound card at 96kHz/24bit, and then using a spectrum analyser, I looked at the frequency content, and saw it for certain notes extending the whole range (upto 48kHz). Now these high frequency overtones were at quite low volume, but still noticeable visibly (of course I can't hear them directly).

How high do these overtones extend, would 192kHz be enough to capture all of them or not (my sound card does not go that high)? I also assume that for electric guitars it does go a bit higher still, but just to get some idea either is fine.

And yes I did DI and not use speakers and microphones which would have put in a much higher noise floor, so please do the same. Since I'm talking about tones that we can't hear directly (possibly their interaction in chords might be noticeable if they were at a high enough volume), the requirement for the highest overtone is basically whether it's measurable in the sense of being above the noise floor.

Edit (see comment below): When looking at the overtones generated by distortion, how high do the overtones extend measurably different from the noise floor?

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    As the answers suggest, the upper limit worth recording is related to our aural capability, not the actual spectral output of the guitar+synth boxes. A really good spectrum analyzer, applied to the electronic signal rather than the acoustic output of a speaker, might well indicate power into the MHz range, since the distorted signal has all sorts of "junk" in it. – Carl Witthoft Sep 29 '15 at 20:22
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    Some acoustic instruments generate significant amounts of sound energy up to 100 kHz and beyond - for example cymbals, harpsichords, and some organ pipes - and not only for notes with high pitched fundamental frequencies. There is no big deal about generating electrical waveforms that go up to the MHz or even GHz region. A well-designed distortion pedal should be filtering out the frequency content that is irrelevant to what it is meant to do - but there are plenty of badly designed ones out there. – user19146 Sep 29 '15 at 21:01
  • All the answers and comments so far contain lots of good information, but so far a real answer to my question is missing. So I'll try to restate the question in a way that it's not buried in between descriptions of how I got to the question. – God Of Atheism Sep 30 '15 at 9:43

Depending on the amount of clipping, the waveform of a very distorted signal can approach a square wave. Mathematically, a square wave is an infinite sum of overtones, with each higher overtone at a lower intensity than the previous.

Near the bottom of this page is the infinite sum of a mathematically exact square wave: http://mathworld.wolfram.com/FourierSeriesSquareWave.html Each term in the sum represent an overtone, and only the odd ones are present. (notice the n=1,3,5... under the sigma - the summation symbol, and the infinity sign above) So theoretically the answer to your question is that the frequencies extend infinitely.

However, no real-world electronics can actually produce a mathematically perfect square wave, so the overtones aren't infinite in real life. We can get very close to a square wave with electronics, so in reality the highest overtone present could be very, very high frequency indeed.

Can we generate a frequency using distortion so high that we can't record it? Probably. Digital signals used for clock and recording purposes are intended to be square waves. Again, in reality they cannot be made to be perfect square waves, but think about the fact that we can approximate a square wave at 192 kHz. That means the overtone series for that approximate square wave starts at 192 kHz and goes up a long way from there. At the very least, we need to be able to keep several of the lowest overtones intact in the wiring and signalling used for recording or we would have sync and encoding issues.

If you're more than just curious about this, then I'm not sure how useful this information will be in the real world. Hardly ever can humans hear above 20 kHz, and above 48 kHz is unheard of except through bone conduction (suggesting the ossicles in the middle ear are part of the upper limit). More limiting is the abilities of 99% of the sound storage and reproduction devices in the world, which can top out as low as 16 kHz and are rarely designed to go about 20 kHz. Even if we could put a firm number on the frequency of the highest overtone one could generate with a distortion pedal, that number would likely be so far beyond anything with could use that it's not worth spending too much time on.

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    digital clocking is done in the megahertz range, way beyond the audio range. Thus we do get very good square wave approximations for 192k clock signals. – dwoz Sep 29 '15 at 19:47
  • also...there are a number of higher-end production desks that will pass 80k or even 90k audio signal. The Amek 9098i being one. Certainly you don't find that capability on pro-sumer gear. – dwoz Sep 29 '15 at 19:48
  • @dwoz There are definitely devices that can operate well above 48 kHz, but the real issue, IMHO, is having tweeters/drivers can can effectively extend that far. As far as I've seen, only esoteric and expensive professional or audiophile equipment even aspires to ultrasonic acoustics. – Todd Wilcox Sep 29 '15 at 19:55
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    since the Amek console I mentioned costs in the $500k range, I think it could be called "esoteric!" :) – dwoz Sep 29 '15 at 20:03
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    There are good reasons to record digitized audio at higher sample rates, irrespective of human hearing. One important issue is that any artefacts created by signal processing (equalization, filtering, etc) can be pushed into the inaudible frequency range and then deleted, rather than irreversibly polluting the frequency range that you can hear. For the same reason, it's important to filter out any frequencies that are too high to sample from the analog signal, before you digitize it, and that is much easier with a wider range (e.g. 20-48kHz, not 20-24) to roll off the filter response – user19146 Sep 29 '15 at 20:50

The short answer is that recording distorted bass guitar at higher sampling rates than 96k/24bit is not going to result in any kind of perceivable difference, other than using up lots of disk storage much more quickly.

The long answer is that current music recording technology ignores supersonic frequencies and there's still some debate about whether anything is lost.

As Wilcox alludes to, the higher the frequency capture, the better the square waves look. What he really is trying to say there is that the IMPULSE RESPONSE is better when we toss in the high frequencies. Those high frequencies don't become perceived as notes, but they CAN influence the spatialization cues we perceive.

In countless discussions with professional audio engineers and mixers, they express a preference for higher bit rates not because they're yearning for some "high-frequency shimmer" but rather because the audio image "seems to be more defined in space." This is explainable not because they have golden ears, but because the source material has more coherent transients. It's a very subtle thing though...most low-mid level gear renders the question moot for almost all listeners.

Most professional audio engineers that I've spoken with notice that 96k is easier to get to "lay in," but that going to 192k only introduces more computer management issues without a real sonic benefit.

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    When I think about Sgt. Pepper's being recorded on two synchronized four-tracks, and then Supposed Former Infatuation Junkie (by Alanis Morisette) being recorded on 20-bit, 48 kHz ADAT, I stop worrying about sample rates and bit depths and track counts. None of it matters without a good song, and none of it will hold back a great song. – Todd Wilcox Sep 29 '15 at 20:14
  • The 4-track tape machines used by Ken Scott to record Sgt. Pepper were arguably better machines than any ADAT. :) In that era, high-end analog was definitely better than digital, though that is not necessarily true today. – dwoz Sep 29 '15 at 21:00

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