Question about the 14-bit Paula trick

[8bitbubsy]

I don't fully understand how I convert a signed 16-bit PCM sample into an 8-bit pair for 14-bit Paula output...

I have played around with it a bit, and the only way I can make it sound good is by doing this:

move.w	d0,d1	; d1 = -32768..32767
lsr.b	#2,d1	; hi byte = normal sample, lo byte = low-vol (vol 1) sample

But this surely seems wrong?! Using logical bitshifting (not arithmetic) means that the lower byte is 0..63 instead of -32..31. If I do asr.b instead, I hear background noise that is not supposed to be there.
Can anyone explain why this is? And maybe I'm doing it all wrong?

[ross]

The sign bit is only on bit 15, on bit 7 there some lower significance part of the 16 bit signed integer.
So it is sure better a logical than an arithmetic shift for the lower byte.

[8bitbubsy]

Originally Posted by ross

The sign bit is on bit 15, on bit 7 there some lower significance part of the integer.
So it is sure better a logical than an arithmetic shift for the lower byte.

Ahh of course, that makes complete sense! Can't believe I didn't see that, lol.
So ideally I should insert the sign bit in bit 6 and 7 after the lsr?

EDIT: How about just subtracting 32 from d1.b? Is that correct?

[ross]

I don't think it makes a difference, just a little bias in the signal.

Test by ear if it's different for you

[8bitbubsy]

Seems like the 0..63 range is correct, or at least anything else I try makes it sound (and look) wrong. So maybe the "lower 6 bits" part just shouldn't be signed (-32..31).
This is a bit confusing to me, though!

[ross]

Well it actually makes sense, you are always adding-up at the most significant part of the sample, you surely generate all the valid 2^14 theoretical combinations, in a linear way.

For simplicity try to think how it work around the zero point and you will see that make sense.

[meynaf]

Following code converts 16-bit sample to the 8-bit pair :

; d3=16-bit sample -> d4=high byte, d5=low byte
 move.w d3,d4
 eori.w #$8000,d4
 bpl.s .pos1
 neg.w d4
.pos1
 move.b d4,d5
 lsr.w #8,d4
 lsr.b #2,d5
 tst.w d3
 bpl.s .pos2
 neg.b d4
 neg.b d5
.pos2

Not my own code, i ripped that somewhere so i can't really explain why it is done this way. It appears to be working fine. I just removed the "volume boost" part.
Also note that the 14-bit trick works best if calibrated (in that case, the code becomes a lot more complicated). In any case, better use a table.

[8bitbubsy]

I'll try it and see how it works out.

Regarding calibration, that requires unique calibration per Amiga machine, right?
If not, then no problem, the stuff I'm doing is already for generating a LUT. The reason I want it to be fast is because I am calculating a lot of LUT entries, but precision is more important.

[meynaf]

Yes it needs unique calibration per Amiga machine.

[8bitbubsy]

Alright, I'm not going to bother with the calibration then.

[ross]

Originally Posted by meynaf

Following code converts 16-bit sample to the 8-bit pair :

This is for unsigned 16-bit sample and extend the sign on the lower byte of the new signed couple.

[8bitbubsy]

But I thought you weren't supposed to store the lower 6 bits as signed -32..31?

[ross]

Originally Posted by 8bitbubsy

But I thought you weren't supposed to store the lower 6 bits as signed -32..31?

This code uses a reverse polarity output.
Why did the coder do such a thing? Maybe it sounded better in his equipment

And has been proven there are many ways to generate a '14-bit' signal on the Amiga and none of them bring perfect results yet.
So the method used in the first message is no more wrong than the others

[8bitbubsy]

Ah, so maybe it's stored as -63..0?

[ross]

Originally Posted by 8bitbubsy

Ah, so maybe it's stored as -63..0?

In this case, yes.

But, it seems better to me, for a 16-bit signed sample, to use values from 0 to 63 in the lowest byte.
In a perspective of pure additive mixing it seems fair that the low byte 'fills' the gap between the magnitude higher bytes, that have 64x_volume than the low.
But it is absolutely not certain that I'm right, indeed I would like to have the opinion of others.

[meynaf]

Originally Posted by ross

This is for unsigned 16-bit sample and extend the sign on the lower byte of the new signed couple.

Sorry, i didn't remove the trick that allows doing this once the table is filled :

 move.w (a0)+,d0		; our sample
 move.w (a3,d0.w*2),d0
 move.b d0,(a2)+		; low
 lsr.w #8,d0
 move.b d0,(a1)+		; high

[robinsonb5]

Just in case you're not already aware of this document, it might be useful:
http://bax.comlab.uni-rostock.de/dl/...mTheoretic.pdf