I already posted on this, but yeah. Just saying that peak/valley/peak/valley is all the samples you get at 22KHz. Regardless of what the peak or valley values are, there's no gradiation along the waveform curve like there would be for lower frequencies - the information is not captured at that sample rate.
You are correct, I should have used 0-1-1-0-0-1-1-0 and 0-100-100-0-0-100-100-0.
I was only saying that wasn't possible to have this for 22KHz at a 44.1KHz sample rate. Which was what the thread was on. It might be more correct to call it a sawtooth wave.
Maybe we got confused on different interpretations of this:
There's a huge distinction between a value range of 0-1 and a value range of 0-65535, though, which affects all waveforms.
Which I took them to think meant that there were gradiating values between peak and trough when there can't be. And it seemed like you were agreeing with that.
You might be making a point of how much information is needed to convey an accurate signal. That is described by Nyquist's Theorem, which does state that 22KHz can be reproduced from a 44.1KHz sample rate. It's a lot more complicated of a mathematical proof than I have time to learn, but it's held up for a very long time.
Still, my point is mroe about what actually happens to 22KHz when quantized at 44.1KHz. Which was the point of this whole thread. You don't get a sine wave shape encoded. Playback is a different story, since I'm sure that playback equipment tries to fit a curve to it anyway.
but also completely irrelevant to the conversation that was being had.
Actually, you're the one that threw the thread off the original subject. The original subject was that a 22KHz waveform is represented the same as a square wave (technically sawtooth might be more accurate) at a 44.1KHz sample rate. Lower frequencies would have more sample points along the curve and a cleaner sine wave shape.
There's a huge distinction between a value range of 0-1 and a value range of 0-65535,
The value range is only the amplitude of the wave, not the shape of it, when you have only two samples per peak/trough. This specifically only applies to 22KHz and near it. At lower frequencies, the details of the amplitude helps actually define the waveform shape. At 22KHz specifically, the amplitude (sample value) only defines the volume and nothing more.
But by your wording, you claimed that you gain SNR by resampling from 96KHz down to 48KHz. Not that you gain it by starting from a higher sampling rate. Your SNR is at the same level while it's still at 96KHz.
Then you further obfuscate your posts by using outdated terminology (clock jitter is meaningless if you're doing your processing on a computer using DAW software. It is not a real-time process).
And my point is that the distinction is meaningless at 22KHz at a 44.1KHz sample rate. A sine wave, square wave, sawtooth wave - all get the same exact digital representation when quantized to digital at this rate.
Except you can't, because there aren't enough samples to represent 22KHz at 44,1KHz. It would indeed have to be 0-1-0-1 (or 0-65535-0-65535, depending on volume level). You only get two samples per wavelength at the upper limit (the nyquist limit).
Conservative "christians" have deluded themselves into believing they're actual Christians, even though Jesus spoke out against greed, materialism and the worship of money.
I agree without contest here - greed and materialism (and gluttony, while we're at it) are sins that are becoming troublingly common within the church.
I'll add that these specific people also aren't good at being fiscally conservative either (which is not the same thing as hoarding all your own money for yourself). It's an odd label to pick for yourself and then totally stray away from.
Well if they have pasted code into their question that they copied from elsewhere but don't understand it themselves, then you can bet they are a cut-rate overseas outsourced coder. Outside of that, I have no idea.
After purchase of very special audio equipment which allows precision tuning of channel delay on microsecond level I was frankly surprised that changes of 10 microseconds in delay between two channels are actually audible
Wouldn't that mostly be phase cancellation and not an actual audio difference? Try 22 microseconds and you'd probably hear no difference at all from the original.
And still "better quality" can be had for under $2 if you know where to look (e.g. Monoprice). Cheaper junk being sold for more money than that has nothing to do with the argument.
They have 50GB to fill. Even most DTS-MA discs are only 96KHz. It looks better for marketing and they already have the bits because mixing/mastering requires the higher sample rate.
A 22KHz signal captured at 44KHz would only have a 0 and a 65535 with nothing in between. Not much different than a 0 and a 1. Let's say it's perfectly aligned with the sample rate for simplicity. The first sample is at the peak (or 65535). Because it will be back to the peak again two samples later, the next sample will have to be at 0. And the third sample would be at 65535 again. Any waveform shape is imaginary at this limited resolution - there is only peak/valley/peak/valley.
And this is why amateur recordings sound so terrible. If you don't record at 96KHz / 24-bit or higher, you don't have any margin for mixing and smoothing.
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I already posted on this, but yeah. Just saying that peak/valley/peak/valley is all the samples you get at 22KHz. Regardless of what the peak or valley values are, there's no gradiation along the waveform curve like there would be for lower frequencies - the information is not captured at that sample rate.
We're talking about audio that's already in the digital domain. Oversampling may have happened at the capture stage, but that's over and done with.
When you said:
You are correct, I should have used 0-1-1-0-0-1-1-0 and 0-100-100-0-0-100-100-0.
I was only saying that wasn't possible to have this for 22KHz at a 44.1KHz sample rate. Which was what the thread was on. It might be more correct to call it a sawtooth wave.
Maybe we got confused on different interpretations of this:
There's a huge distinction between a value range of 0-1 and a value range of 0-65535, though, which affects all waveforms.
Which I took them to think meant that there were gradiating values between peak and trough when there can't be. And it seemed like you were agreeing with that.
You might be making a point of how much information is needed to convey an accurate signal. That is described by Nyquist's Theorem, which does state that 22KHz can be reproduced from a 44.1KHz sample rate. It's a lot more complicated of a mathematical proof than I have time to learn, but it's held up for a very long time.
Still, my point is mroe about what actually happens to 22KHz when quantized at 44.1KHz. Which was the point of this whole thread. You don't get a sine wave shape encoded. Playback is a different story, since I'm sure that playback equipment tries to fit a curve to it anyway.
but also completely irrelevant to the conversation that was being had.
Actually, you're the one that threw the thread off the original subject. The original subject was that a 22KHz waveform is represented the same as a square wave (technically sawtooth might be more accurate) at a 44.1KHz sample rate. Lower frequencies would have more sample points along the curve and a cleaner sine wave shape.
There's a huge distinction between a value range of 0-1 and a value range of 0-65535,
The value range is only the amplitude of the wave, not the shape of it, when you have only two samples per peak/trough. This specifically only applies to 22KHz and near it. At lower frequencies, the details of the amplitude helps actually define the waveform shape. At 22KHz specifically, the amplitude (sample value) only defines the volume and nothing more.
Regardless, that is still the claimed limit of the Nyquist theorem to accurately reproduce 22KHz sound.
But by your wording, you claimed that you gain SNR by resampling from 96KHz down to 48KHz. Not that you gain it by starting from a higher sampling rate. Your SNR is at the same level while it's still at 96KHz.
Then you further obfuscate your posts by using outdated terminology (clock jitter is meaningless if you're doing your processing on a computer using DAW software. It is not a real-time process).
And my point is that the distinction is meaningless at 22KHz at a 44.1KHz sample rate. A sine wave, square wave, sawtooth wave - all get the same exact digital representation when quantized to digital at this rate.
Except you can't, because there aren't enough samples to represent 22KHz at 44,1KHz. It would indeed have to be 0-1-0-1 (or 0-65535-0-65535, depending on volume level). You only get two samples per wavelength at the upper limit (the nyquist limit).
but AmiMoJo (according to his other posts) apparently makes assumptions about a persons race and gender from usernames
And so, apparently, do you.
Dynamic range is not the same thing as SNR.
Conservative "christians" have deluded themselves into believing they're actual Christians, even though Jesus spoke out against greed, materialism and the worship of money.
I agree without contest here - greed and materialism (and gluttony, while we're at it) are sins that are becoming troublingly common within the church.
I'll add that these specific people also aren't good at being fiscally conservative either (which is not the same thing as hoarding all your own money for yourself). It's an odd label to pick for yourself and then totally stray away from.
many of my programming question google searches end up with a SO link and I will continue to use the service.
Despite having an account, a Google search is really the only way I use the service.
In this example of the housing market,
What example? I don't see any housing market example.
Well if they have pasted code into their question that they copied from elsewhere but don't understand it themselves, then you can bet they are a cut-rate overseas outsourced coder. Outside of that, I have no idea.
*ouch*
The vast majority of K-12 education is funded through state and local taxes. Exactly 0 of this is sent to the DoD.
"Now, that's not because most Stack Overflow contributors are hostile jerks.
But they are.
Have you heard of 44.1KHz/16-bit CDs?
After purchase of very special audio equipment which allows precision tuning of channel delay on microsecond level I was frankly surprised that changes of 10 microseconds in delay between two channels are actually audible
Wouldn't that mostly be phase cancellation and not an actual audio difference? Try 22 microseconds and you'd probably hear no difference at all from the original.
And still "better quality" can be had for under $2 if you know where to look (e.g. Monoprice). Cheaper junk being sold for more money than that has nothing to do with the argument.
Resampling does nothing to the dynamic range (and nothing to dB either). The dynamic range is solely defined by the number of bits per sample.
They have 50GB to fill. Even most DTS-MA discs are only 96KHz. It looks better for marketing and they already have the bits because mixing/mastering requires the higher sample rate.
A 22KHz signal captured at 44KHz would only have a 0 and a 65535 with nothing in between. Not much different than a 0 and a 1. Let's say it's perfectly aligned with the sample rate for simplicity. The first sample is at the peak (or 65535). Because it will be back to the peak again two samples later, the next sample will have to be at 0. And the third sample would be at 65535 again. Any waveform shape is imaginary at this limited resolution - there is only peak/valley/peak/valley.
And this is why amateur recordings sound so terrible. If you don't record at 96KHz / 24-bit or higher, you don't have any margin for mixing and smoothing.