tube CD player (was Re: sizes (was Re: vacuum tube computer))

From: Philip.Belben_at_pgen.com <(Philip.Belben_at_pgen.com)>
Date: Thu Feb 4 03:52:56 1999

Eric has replied to my post with a lot of good points, but I don't
understand this one:

> After you do the D/A, you have a huge amount of sampling noise at Fs/2
(and
> its harmonics). For a CD player, Fs is 44.1 KHz, so the sampling noise
is at
> 22.05 KHz. Since the desired audio frequency response extends to 20 KHz,
a
> non-oversampled player needs "brick wall" analog filters, with a pass
band to
> 20 KHz, and a stop band starting below 22.05 KHz.
>
> Digital oversampling is used to move the sampling noise to a higher
> frequency. 4x oversampling moves it to 88.2 KHz. The oversampling
filter
> is a low-pass filter, so the audio content still has a frequency response
> to somewhere above 20 KHz. But now your analog filter can have a much
> shallower slope, which is easier to implement and introduces much less
> phase distortion.
>
> Your analog oversampling scheme does not eliminate the noise at Fs/2,
> and in fact introduces more sampling noise at higher frequencies. It
does
> not help reduce the requirements of the analog filter.
Don't confuse sampling noise with quantisation noise. Quantisation noise
is due to the replacement of an analogue value by a digital representation
not exactly the same; it is reduced simply by having more bits of precision
(there are other tricks, including things involving oversampling, but they
only cloud the issue).

Sampling noise is present in a sampled waveform whether it is a sequence of
digital numbers or of analogue values. It arises from the deviation
between the original waveform and the sampler output (with no filtration at
all this is a series of steps) at the intermediate points between samples.
Oversampling interpolates intermediate samples, and uses a filter to reduce
the sampling noise due to not having known what those intermediate values
should have been.

While I admit that the linear interpolation I described is a rotten
oversampling filter - it was only mentioned by way of an example - the
mathematics are the same whether you use a digital computer to generate the
digital numbers at your oversampled points or a switched analogue circuit
such as I described to generate analogue values at the intermediate points.
This analogue circuit can also (theoretically) reduce quantisation noise
even compared with digital oversampling, since the intermediate values are
not limited to the precision of the DAC. In practice, quantisation noise
probably is probably slightly higher, since the intermediate values are
limited to the precision of the resistor networks that derive them.

So why does this sort of analogue filter not reduce the sampling noise at
Fs/2? If it introduces noise at higher frequencies (which I agree it
does), this is because the noise arises from conversion of samples to a
continuous function, not from conversion of digital to analogue.


> The logical extreme of oversampling is to use such a high factor that you
only
> need one bit of data. This is called a delta-sigma D/A converter. For
16-bit
> source data this would require at least 65536x oversampling, but there
are
> some tricks to reduce this to e.g. 128x. This is the basis of the
so-called
> "1-bit" D/A converters. The advantage is that you only need a
single-pole
> analog filter (and in practice can get by without even that), and that
the
> tradeoff from quantization to time domain has been maximized, which is
good
> because it is much easier to get high precision in the time domain as
> described above.

I have met delta-sigma encoding, but I never managed to work out how you
get the sampling rate down to a manageable value. Since we're getting a
bit off topic here, could you post not an explanation but a reference to
somewhere I can look up this particular aspect of the problem.

Philip.
Received on Thu Feb 04 1999 - 03:52:56 GMT