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Perhaps this is a silly question but the explanations mostly make perfect sense to me but in the case of sample rate, what happens when two waves with very diff

Perhaps this is a silly question but the explanations mostly make perfect sense to me but in the case of sample rate, what happens when two waves with very different frequencies overlap? Say I've got an 18kHz wave and a 9kHz wave and the 9kHz wave is ever so slightly out of phase. Then imagine there are 10 different waves under 20kHz all interfering with each other in different wages. Is it still possible to reproduce everything accurately? And on bit-depth and dynamic range: Given that much audio doesn't use the full range available to it, wouldn't higher bit depth increase the fidelity in the range the audio does fill? The article talks about the bit depth and range only in terms of the maximum volume but what about fidelity? What's the minimum difference in volume the human ear can hear?

TazeTSchnitzel 9y ago
IIRC, I think the article's point about maximum volume was about how you'd need to turn the volume up to deafening levels before the small differences would be noticeable.
- Veratyr 9y ago
  It doesn't seem to say anything like that. Looking at it again, I see this: > It is also worth mentioning that increasing the bit depth of the audio representation from 16 to 24 bits does not increase the perceptible resolution or 'fineness' of the audio. It only increases the dynamic range, the range between the softest possible and the loudest possible sound, by lowering the noise floor. However, a 16-bit noise floor is already below what we can hear. But I don't understand, why is this? Is the step in volume between two values fixed regardless of the bit depth? If so, why?
  - lgunsch 9y ago
    The video linked in the top-most comment will beautifully answer your questions. It's also on YouTube. You can think of it as in the sampling rate provides a mathimatical solution to the waveform - it's simply 100% accurate all the time. The bit depth adds extra "buckets" to divide the volumes into . If you have 2 bits, you then have volume levels 0, 1, 2, 3.
agency 9y ago
With regards to the first part of your question - as long as the signal is bandlimited the Nyquist-Shannon theorem applies. If all of the waves are under 20kHz, regardless of how they're interfering with each other, the signal can be reconstructed perfectly with a 40kHz sample rate.
- kuschku 9y ago
  Assuming that you have a perfect DAC (which you don’t), with perfect analog brickwall filters (which you don’t), with no compressed audio (which isn’t the case). With a perfect DAC, and perfect filters, and expensive headphones, there is no difference between 16bit/44.1kHz WAV and 24bit/192kHz WAV (until you add an EQ, then there is). With a usual home DAC, and its filters, and at best $100 headphones, there is a major difference between 16/44.1 AAC and 24/192 AAC. (And guess what the article attacked? The decision of Apple to sell lossy files in 24/192 for home usage. The discussion was never about lossless files, or professional usage)
- umaxumax 9y ago
  Incorrect. Nyquist-Shannon theorem does not guarantee phase! It only guarantees frequency reproduction. This paper illustrates the issue succinctly: http://www.wescottdesign.com/articles/Sampling/sampling.pdf http://www.wescottdesign.com/articles/Sampling/sampling.pdf
baddox 9y ago
I believe that recorded music is a ton of waves of different frequencies going at the same time. You still just need to sample at twice the highest frequency. Overlapping frequencies may have some audible effects that may be undesirable (like beating [0]), but those would be audible in the real (analog) signal as well as the sampled signal. [0] https://en.wikipedia.org/wiki/Beat_(acoustics) https://en.wikipedia.org/wiki/Beat_(acoustics)
gsteinb88 9y ago
So long as all frequencies you're trying to represent are below the Nyquist frequency, there's still only one band-limited reconstruction of the samples. So yes, it's still possible to reproduce everything accurately so long as all of the individual frequencies (no matter their relationships) fall below 1/2 the sample rate.
tomlu 9y ago
> Is it still possible to reproduce everything accurately? Sound is linear, so yes. Any signal can be represented correctly as a sum of frequencies. > wouldn't higher bit depth increase the fidelity in the range the audio does fill? When you do the maths quantisation turns into the original sound + quantization noise. The bit range determines the noise (dither helps here). So it's a question of whether the noise is loud enough to hear. There are samples on the internet you can try to determine yourself at which stage you can notice the noise. It's shocking how few bits are actually required. For a sloppy recording or repeated processing the quantisation noise goes up, so more bits are required.

TazeTSchnitzel 9y ago

IIRC, I think the article's point about maximum volume was about how you'd need to turn the volume up to deafening levels before the small differences would be noticeable.

Veratyr 9y ago
It doesn't seem to say anything like that. Looking at it again, I see this: > It is also worth mentioning that increasing the bit depth of the audio representation from 16 to 24 bits does not increase the perceptible resolution or 'fineness' of the audio. It only increases the dynamic range, the range between the softest possible and the loudest possible sound, by lowering the noise floor. However, a 16-bit noise floor is already below what we can hear. But I don't understand, why is this? Is the step in volume between two values fixed regardless of the bit depth? If so, why?
- lgunsch 9y ago
  The video linked in the top-most comment will beautifully answer your questions. It's also on YouTube. You can think of it as in the sampling rate provides a mathimatical solution to the waveform - it's simply 100% accurate all the time. The bit depth adds extra "buckets" to divide the volumes into . If you have 2 bits, you then have volume levels 0, 1, 2, 3.

agency 9y ago

With regards to the first part of your question - as long as the signal is bandlimited the Nyquist-Shannon theorem applies. If all of the waves are under 20kHz, regardless of how they're interfering with each other, the signal can be reconstructed perfectly with a 40kHz sample rate.

kuschku 9y ago
Assuming that you have a perfect DAC (which you don’t), with perfect analog brickwall filters (which you don’t), with no compressed audio (which isn’t the case). With a perfect DAC, and perfect filters, and expensive headphones, there is no difference between 16bit/44.1kHz WAV and 24bit/192kHz WAV (until you add an EQ, then there is). With a usual home DAC, and its filters, and at best $100 headphones, there is a major difference between 16/44.1 AAC and 24/192 AAC. (And guess what the article attacked? The decision of Apple to sell lossy files in 24/192 for home usage. The discussion was never about lossless files, or professional usage)
umaxumax 9y ago
Incorrect. Nyquist-Shannon theorem does not guarantee phase! It only guarantees frequency reproduction. This paper illustrates the issue succinctly: http://www.wescottdesign.com/articles/Sampling/sampling.pdf http://www.wescottdesign.com/articles/Sampling/sampling.pdf

baddox 9y ago

I believe that recorded music is a ton of waves of different frequencies going at the same time. You still just need to sample at twice the highest frequency. Overlapping frequencies may have some audible effects that may be undesirable (like beating [0]), but those would be audible in the real (analog) signal as well as the sampled signal. [0] https://en.wikipedia.org/wiki/Beat_(acoustics) https://en.wikipedia.org/wiki/Beat_(acoustics)

gsteinb88 9y ago

So long as all frequencies you're trying to represent are below the Nyquist frequency, there's still only one band-limited reconstruction of the samples. So yes, it's still possible to reproduce everything accurately so long as all of the individual frequencies (no matter their relationships) fall below 1/2 the sample rate.

tomlu 9y ago

> Is it still possible to reproduce everything accurately? Sound is linear, so yes. Any signal can be represented correctly as a sum of frequencies. > wouldn't higher bit depth increase the fidelity in the range the audio does fill? When you do the maths quantisation turns into the original sound + quantization noise. The bit range determines the noise (dither helps here). So it's a question of whether the noise is loud enough to hear. There are samples on the internet you can try to determine yourself at which stage you can notice the noise. It's shocking how few bits are actually required. For a sloppy recording or repeated processing the quantisation noise goes up, so more bits are required.