herbie@watdcsu.UUCP (Herb Chong, Computing Services) (11/12/84)
When oversampling is used in CD players, how do they fill in the missing
samples? Do they hold the previous value (not likely)? Do they do
interpolation? If so, what kind do they do?
Herb Chong...
I'm user-friendly -- I don't byte, I nybble....
UUCP: {decvax|utzoo|ihnp4|allegra|clyde}!watmath!watdcsu!herbie
CSNET: herbie%watdcsu@waterloo.csnet
ARPA: herbie%watdcsu%waterloo.csnet@csnet-relay.arpa
NETNORTH, BITNET: herbie@watdcs, herbie@watdcsu
POST: Department of Computing Services
University of Waterloo
Waterloo, ON
N2L 3G1 (519)886-4733 x3524ron@brl-tgr.ARPA (Ron Natalie <ron>) (11/16/84)
> When oversampling is used in CD players, how do they fill in the missing > samples? Do they hold the previous value (not likely)? Do they do > interpolation? If so, what kind do they do? Yes. What they do is replicate the value the value four times and then take output the average of the four prior samples. So it is interpolated. -Ron
gregr@tekig1.UUCP (Greg Rogers) (11/16/84)
In the Phillips implementation of 4x oversampling, three samples of zero are inserted between each of the actual samples derived from the CD disc, after error correction has taken place. The actual sample followed by three zero samples is then fed to a digital filter and finally the resulting samples are sent to a 14 (!) bit D/A at a 176 khz rate. Greg Rogers Tektronix
sullivan@harvard.ARPA (John Sullivan) (11/21/84)
My understanding of oversampling is that they do a linear interpolation
between successive samples on the disk. The general idea is to thereby
have digital filtering instead of analog filtering, and get better phase
relationships at high frequencies.
As has been discussed before, at some point you have to filter out frequencies
above 22kHz. If this is done with a simple analog cut-off filter, the phases
of frequencies around 20kHz get changed a lot. If you put in a square wave
signal, you have to get some kind of ringing, but with this simple filter
it will be asymmetric--at the rising edge, the output is flat until the
input goes high, and then it rings quite a bit. This is basically because
the filter cannot anticipate the change that is about to occur. A perfect
22kHz cutoff filter that did not change phases would have symmetric output.
Look in any book on Fourier series for sample pictures of partial sums of
the series for a square wave. This waveform has the same frequency components
as the one from the analog filter, but the high frequency's phases haven't
been shifted.
The digital filter (or oversampling) essentially can anticipate changes,
because it looks at the n+1st input before giving you the interpolated values
between n and n+1. Given a square wave, it will output a square wave with
sloped edges. This will then still need analog filtering, but that is less
severe, and has less effect on the phases.
I have not read details of the methods actually used--this is just what
I guess they must be doing. If anyone has more information, please let me
know.
John M. Sullivan
{ihnp4,allegra,decvax}!harvard!sullivan