This is my "weekend" hobby problem.
I have some well-loved single-cycle waveforms from the ROMs of a classic synthesizer.
These are 8-bit samples (256 possible values).
Because they are only 8 bits, the noise floor is pretty high. This is due to quantization error. Quantization error is pretty weird. It messes up all frequencies a bit.
I'd like to take these cycles and make "clean" 16-bit versions of them. (Yes, I know people love the dirty versions, so I'll let the user interpolate between dirty and clean to whatever degree they like.)
It sounds impossible, right, because I've lost the low 8 bits forever, right? But this has been in the back of my head for a while, and I'm pretty sure I can do it.
Remember that these are single-cycle waveforms that just get repeated over and over for playback, so this is a special case. (Of course, the synth does all kinds of things to make the sound interesting, including envelopes, modulations, filters cross-fading, etc.)
For each individual byte sample, what I really know is that it's one of 256 values in the 16-bit version. (Imagine the reverse process, where the 16-bit value is truncated or rounded to 8 bits.)
My evaluation function is trying to get the minimum noise floor. I should be able to judge that with one or more FFTs.
Exhaustive testing would probably take forever, so I could take a lower-resolution first pass. Or do I just randomly push randomly chosen values around (within the known values that would keep the same 8-bit version) and do the evaluation and keep the cleaner version? Or is there something faster I can do? Am I in danger of falling into local minimums when there might be some better minimums elsewhere in the search space? I've had that happen in other similar situations.
Are there any initial guesses I can make, maybe by looking at neighboring values?
Edit: Several people have pointed out that the problem is easier if I remove the requirement that the new waveform would sample to the original. That's true. In fact, if I'm just looking for cleaner sounds, the solution is trivial.