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RE: 20-bit processing

No, I don't think that's right. Imagine doing a sequence of arithmetic operations and rounding to a resolution of 0.001 every time:
  1. X = 4.001 (the original value)
  2. X = X / 4
  3. X = X - 1
  4. X = X * 1000
  5. X = X + 1
The right answer, rounded to the nearest 0.001, is 1.250. But if you round to the nearest 0.001 at each step then you'll end up with a value of 1.000 -- an error 250 times greater than your resolution of 0.001.

But if instead you rounded to the nearest 0.00001 at every intermediate step, and then rounded the final answer to the nearest 0.001, you'll get the right answer of 1.250.

This, apparently, is what the 20 bits were for, to provide an extra 4 bits of resolution for the intermediate values of the arithmetic computations. The starting value (the quantized analog sample) and the final value (the number put on the CD) may only be 16 bits, just like in the example above the starting and final values are rounded to the nearest 0.001. But you must have more resolution for internal arithmetic manipulation or else there could be a severe degree of error in the final numbers put on CD.


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