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  #31  
Old 03.06.2014, 07:15 PM
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Originally Posted by TweakHead View Post
Yes, it's determined by that. But FFT is just a way of analysing a signal. The frequency resolution is how long (in time) that sample is - aka as block size. In other words, we agree on that.
GREAT! You are correct that FFT is a way of analyzing a signal, but it is also a way of constructing signals (remember, it perfectly captures the time domain signal), and that is relevant in the supersaw case.

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What I don't agree with is that there's no single complete cycle within that array. Because there is. And if you were to loop just one of them, the result would be exactly like that in the picture that I posted. So it's possible to produce a 30.5Hz sine wave within the constraints of that time span. If it was to end half a cycle earlier, the frequency of the signal would be the same, there would be no audio click, just a little silence before it starts over. And absolutely no sample leakage.
There are complete cycles of 30.5hz in that signal, but upon close look they are actually made from combination of 30 and 31hz sine waves (+maybe other). It is not possible to produce 30.5hz in the limitations of that time span. 30.5hz frequency simply does not exist in that time span. Look at the FFT analyzer in the youtube video I provided, the frequencies are points - 30hz, 31hz... Nothing exists inbetween the points, 30.5hz doesn't and can't exist in such signal, ever. Remember that the FFT construct signals from complete sine waves.


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The problem is that the samples collected for the FFT analysis are not synced to the frequency of the signal, it's dependant on the refresh rate you set for it and the number of samples it collects (block size again), so there's no perfect alignment between the two things - which need to be for producing accurate results instead of displaying waveforms cut at random points that will indeed be interpreted (calculated) as a different waveform altogether and hence show some other harmonics which should not be there. So if you were to fade in and out that array, discontinuity would not be a problem, no other waveform would be calculated instead of that present in the signal, but you'd need a fairly high refresh rate to compensate for the measurement of amplitude, presumably - not even pretending to be an expert here.
This is just way of saying that the block size (length of the signal) gives you the frequency resolution. We need not to be concerned with what lies outside the signal (Unless you are in the business of predicting - it's easy to say what's outside of our window when you know it beforehand by the way).

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But why does this matter so much to you? Even with all this in mind, let's assume we're on the same page as of now - that I finally made some sense of your words - how do you explain why note duration would be such an important thing for supersaws?

Let's say our frequency resolution is 1hz/1second sample and we want to use specific detune amount. Now I can determinate the exact amount of saw waves that are needed to create all the possible different combinations of the waveform. You see putting a saw wave at 30.5hz does NOTHING, that putting one at 30hz second at 31hz etc. doesn't already accomplish. Except we are dealing with saw waves here so there is actually a difference with the upper harmonics. That's why I used 16khz, and not 30hz...

Look at the FFT plot in the video I provided earlier. Once you have one sine wave at each of the FFT bins, generating more sine waves does nothing that can't be already done. That's the point...
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  #32  
Old 03.06.2014, 08:38 PM
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Correct me if I'm wrong here: but the number of points within a second is determined by our choice of sample rate, right?

so this would give us 44100 points a second, for example. If you divide this number by 30,5 you get 1445,9016393442623 points per cycle, right?

While for 30Hz you'd have 1470 points. For 31Hz it would be 1445,9016393442623 points. Of course the problem is there's nothing but whole numbers in there. So you're saying that we need more time so that there's enough points to get to a whole number and thus completing a perfect cycle, I think. And I'm saying that the mere phase difference should be enough to distinguish these pitches even within such a short time frame. That the time duration of the cycles is enough to determine pitch with precision, provided there's at least one complete cycle (???).
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  #33  
Old 03.06.2014, 10:03 PM
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Correct me if I'm wrong here: but the number of points within a second is determined by our choice of sample rate, right?
For digital time domain signal, yes. For frequency domain signal, the amount of points depend on the length of the signal, the usable bandwidth on the sample rate. Analog signal is however continuous in time domain, we can't ever hear a digital signal...

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so this would give us 44100 points a second, for example. If you divide this number by 30,5 you get 1445,9016393442623 points per cycle, right?
I didn't do the math, but I have no problems believing that what you are saying is correct. (but it is irrelevant).

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While for 30Hz you'd have 1470 points. For 31Hz it would be 1445,9016393442623 points. Of course the problem is there's nothing but whole numbers in there. So you're saying that we need more time so that there's enough points to get to a whole number and thus completing a perfect cycle, I think. And I'm saying that the mere phase difference should be enough to distinguish these pitches even within such a short time frame. That the time duration of the cycles is enough to determine pitch with precision, provided there's at least one complete cycle (???).
No, this is not at all what I am saying. The limitations I am describing applies to analog domain and infinite sample rate digital signals as well. You can't have a 30.5hz sine wave in one second time window, be it analog, digital, infinite sample rate or whatever. The sine just won't fit in the window! A signal can only consist of sine waves that fit in the window (because by definition those are sine waves, other waves are, not sine waves - meaning that they are composed of many different sine waves)...

What you are describing is simply that periodic signals in analog domain may not be periodic in the digital domain in the sense that the sample points would repeat themselves periodically as the analog wave does. To get around this some of the digital cycles are going to have more points, this is unimportant, because they will still fit into the one second window perfectly. However, this does mean we can't take a fourier transform of a 1.0000001 second signal, because the FFT size of 44100.1 samples does not exist. So there is that type of restriction with digital, which can be overcome with oversampling though.
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  #34  
Old 03.06.2014, 10:38 PM
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I thought that interpolation had become integrated with most audio applications by now, but I'm merely presenting a guess here - as I'm not and don't pretend to be an expert on such matters, at the time being we're sailing outside the waters that are most familiar to me, so keep that in mind.

Your last sentence kind of confirms what I was trying to say with the point numbers, I can picture a scenario where a perfectly contoured waveform in analogue wouldn't translate well in digital due to it not being aligned with the points and where that signal would happen to translate into that grid - this is interesting and honestly haven't thought much about it just yet.

No developer here either, but it feels like some algorithm could be implemented to make a good guess based on some results. Meaning, that the estimative could be almost spot on based on the behaviour of the wave at certain points. Always thought that was what interpolation means and does to the waveform. But I'm guessing efforts in that are made to work within a minor error margin, but we're never talking about absolute precision here, just not so good guesses and better guesses - and we still have to factor further latency introduced by the processing of this somehow, I guess.
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  #35  
Old 04.06.2014, 02:59 AM
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Originally Posted by TweakHead View Post

Your last sentence kind of confirms what I was trying to say with the point numbers, I can picture a scenario where a perfectly contoured waveform in analogue wouldn't translate well in digital due to it not being aligned with the points and where that signal would happen to translate into that grid - this is interesting and honestly haven't thought much about it just yet.
This is very simple thankfully. Any analog waveform can be captured and almost perfectly reproduced with digital sampling as long as the sample rate is 2x the lowest frequency. But like was pointed out, there is a small catch. If you were to loop the digital signal it's not the same thing as looping the analog signal. So if the original analog signal is 2.5 sample points long, looping the digital signal will produce something different than looping the analog signal. The FFT would also be different because in digtal you could not take 2.5 sample point length FFT. Once you oversample though, there is no problem. This is also why wavetable oscillators of most frequencies need to be longer than one analog cycle and possibly one reason why extreme pitch resolution requires a lot of processing power - the digital oscillators can't be periodic even if the resulting analog waveform is.

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No developer here either, but it feels like some algorithm could be implemented to make a good guess based on some results. Meaning, that the estimative could be almost spot on based on the behaviour of the wave at certain points. Always thought that was what interpolation means and does to the waveform. But I'm guessing efforts in that are made to work within a minor error margin, but we're never talking about absolute precision here, just not so good guesses and better guesses - and we still have to factor further latency introduced by the processing of this somehow, I guess.
I am not a developer (far from it, lol) either. There certainly seem to be a lot of unexplored waters in audio. In the end, I guess it's not that big of a market.

Anyway, oversampling or in other words interpolation with sinc function (also goes by the name of low pass filter) gives you more points along the analog signal - well extremely close anyway (analog filters are usually worse than digital ones, and won't be linear phase). But once again, oversampling affects the frequency domain so that higher frequencies can be produced, the frequency resolution remains the same. Only way to increase frequency resolution is to have a longer signal.

If you want to know how the analog signal is constructed from digital check: http://lavryengineering.com/pdfs/lav...ing-theory.pdf
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  #36  
Old 04.06.2014, 08:34 AM
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While we're at it, check this out:

http://www.gearslutz.com/board/maste...hoes-past.html

This is a very interesting thread where developers are going about minimum versus linear phase EQ and go about many interesting things, certainly worth a read.

Last edited by TweakHead : 04.06.2014 at 10:26 AM.
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