What's the meaning of "sample rate" in MP3/Vorbis/AAC?, Considering the audio was transformed Options

[sheh]

What's the meaning of "sample rate" in MP3/Vorbis/AAC? Wouldn't it be meaningless after transform, same as bit depth?

[saratoga]

No, the transform occurs on sampled data and so the sample rate determines which samples correspond to which frequencies.

[sheh]

I guess the way I put it does seem like a retread. The thing is, some statements above make it sound like it could be an implementation problem rather than a complete inherent impossibility.

JAZ says "the compressed data is specific for one max frequency". Makes it sound like he refers to decoding to higher frequencies. Then the mention of Opus using 48kHz internally for all sample rates (though, afterwards I found an old discussion that suggest 44.1kHz *is* a problem).

Benski says "If you didn't know the sample rate, you wouldn't be able to decode the audio", makes it sound like just needing the rate for the math. And the "easily" in "can decode easily to sample rates that are multiples" like a problem of complexity.

...sample those sinusoids at any intervals...

Could you explain what you mean more precisely? Its difficult to understand what your idea is.

This is what I was wondering about. Perhaps it's just my oversimplified idea of what the transformed data represents. Taken to an extreme, let's say you have 10 seconds of 1kHz sine wave at a fixed level. If encoding ultimately just means storing the frequency, level, and phase, it seems you should be able to recreate that 1kHz sine wave with as many or as few samples as you'd like. What would actually be stored in such a case?

the sample rate determines which samples correspond to which frequencies.

What does "which samples correspond to which frequencies" mean?

This post has been edited by sheh: Oct 9 2012, 05:22

[saratoga]

Benski says "If you didn't know the sample rate, you wouldn't be able to decode the audio", makes it sound like just needing the rate for the math. And the "easily" in "can decode easily to sample rates that are multiples" like a problem of complexity.

I'm interpreting your question to actually ask "Is there some way to start with (M)DCT transformed data and then transform to time domain at an arbitrary sample rate that is algorithmically more efficient than a combination of inverse (M)DCT and resampling.

My answer is no. The choice of domain makes no difference with respect to sampling and starting or ending in one domain or the other is orthogonal to questions about sampling. I can see no way to exploit one to make the other easier.

If encoding ultimately just means storing the frequency, level, and phase, it seems you should be able to recreate that 1kHz sine wave with as many or as few samples as you'd like. What would actually be stored in such a case?

The process of fitting a function to data for purposes of changing the sampling rate is called resampling. So your idea here is probably something like "resampling then fourier transforming". This will work but its no different than the usual process once you actually implement it. You've just expressed the same process in a different way.

the sample rate determines which samples correspond to which frequencies.

What does "which samples correspond to which frequencies" mean?

Are you familiar with the discrete fourier transform (or fast fourier transform)? In these (and all) discrete fourier transforms, the choice of sampling rate determines which samples in the Fourier domain contain which frequencies.

This post has been edited by saratoga: Oct 9 2012, 19:04