Path: utzoo!attcan!uunet!munnari.oz.au!sirius.ucs.adelaide.edu.au!hydra!francis From: francis@cs.ua.oz.au (Francis Vaughan) Newsgroups: sci.electronics Subject: Re: Digital/Analog converter help needed Message-ID: <1176@sirius.ucs.adelaide.edu.au> Date: 23 Jul 90 00:54:23 GMT References: <3550.26a1d2d7@ccvax.ucd.ie> <1839@island.uu.net> <1168@sirius.ucs.adelaide.edu.au> <1847@island.uu.net> Sender: news@ucs.adelaide.edu.au Reply-To: francis@cs.ua.oz.au Organization: Adelaide Univerity, Computer Science Lines: 60 In article <1847@island.uu.net>, rich@island.uu.net (Rich Fanning) writes: |>In article <1168@sirius.ucs.adelaide.edu.au> francis@cs.ua.oz.au writes: |>>Uh No. If you take the top 16 bits off a 32 bit result you will only get |>>the full 16 bit resolution of the source material at full volume. |> |>>The Denon Audio Test disk demonstrates this very well. They have a sequence |>>of tracks of orchestral music recorded on the disk at differing levels of |>>attenuation. If you normalise the volume on playback for the different tracks |>>the difference in sound quality is remarkable. Tracks that are recorded 60db |>>down sound worse than telephone lines. |> |>I can only reason, not having the disk, that the recording was not made using |>dither. This will cause obvious distortions in the kind of tests you describe. |>A properly dithered recording will simply have random noise added. Of course, |>it won't sound better as the attenuation increases, but it ought to sound |>much like the original with just noise added. No. Dither only helps remove some scratchy sounding stuff that corresponds to one bit of rounding. Dither cannot help with more than a one bit rounding. The noise added to cover the lost bits as you describe would require that if 45 db of attenuation was needed (not a lot by the way) then the signal would occupy 8 of the 16 bits and the dither noise would also need 8 bits. A signal to noise of exactly 0db. Not exactly High Fidelity. The whole point is that at about -45db the signal is only 8 bits wide (if it was all the way up to FFFF on peaks, even fewer bits otherwise). This is getting down to telephone dynamic range (you still have the frequency response of CD) and the distortion is comensurate. |>Professional digital mixers, from what I hear, can sound excellent. |>And they only store 16 bits after the mix is complete. No doubt internally |>they use 32 bit accumulators to help prevent rounding errors, but the signal |>as it's recorded is still only 16 bits. True, but they are fading with respect to your current listening levels. That is FFFF is still the loudest you want to listen to, not the loudest that your system can ever reproduce, and they attenuate wrt to that. The digital fader board has attenuated the signal but because you are listening at an analogue attenuated level the distortion products are also attenuated by the same amount, and not as noticable. |>From this, I reason that a turning down a digital volume control which |>emits 16-bit codes to the D/A is the same as a fade on a CD. If you cannot |>hear artifacts when a song fades out on a CD, then a digital volume control |>driving a 16-bit D/A ought to be adequate. Again wrong. Sanity check time. What you imply is that an arbitrarily truncated 16 bit wide signal sounds as good as the origonal signal. This is patently false. You are assuming you cannot hear the distortion artifacts on a fade out. That was the whole point of the example with the test disk. If they are normalised you can. This is what would happen in your proposed system. Francis Vaughan.