4 ms·
Since the MDCT is discrete, I assume it operates on power-of-2-sized batches of samples. So (like you, without looking at the code) I would have assumed that m
by richard_todd 6y ago
Since the MDCT is discrete, I assume it operates on power-of-2-sized batches of samples. So (like you, without looking at the code) I would have assumed that more samples/s mean you need more transform blocks, which means you have to allocate fewer output bits per output block to hit your target rate.
- neltnerb 6y agoYou are probably right. I forgot about the whole power of two thing for ffts. That would definitely irritate the same part of my brain that would be put off by interpolating discrete samples even if they're inaudible. Same vein as how 7 is more random than 6.