[William Sommerwerck]

When Hemo (Marvin Miller) insists that Dr. Science (Dr. Frank Baxter)
respond with the two words that shows he understands the essence of blood,
Dr. Science comes back with "sea water".

So... Here's my equivalent of the "sea water" answer.


"Is the audio output of a CD player analog or digital?"

It's digital. Once you've converted analog to digital, the damage is done
and can't be undone. You can't go home again. Here's the explanation...

Consider the classic sampling of a band-limited signal. The sampling process
produces a string of pulses whose amplitudes are exactly the amplitude of
the signal at the instant of sampling. Because this amplitude can vary
continuously (ie, have any value, without restriction), the pulse chain is
an analog signal. (analog = data representation by continuously variable
values)

Fourier analysis shows that the pulse train contains the original signal,
unmodified. If we play the pulse train through a system free of
intermodulation distortion, we will hear the original signal, unchanged. *
This is true even if the system has "infinite" bandwidth, because the
original signal is a component of the pulse train; no filtering is required
to "recreate" it.

To convert this pulse train to a digital signal, we first have to quantize
its level. (I'm ignoring the use of dither, to clarify the point I'll be
making. We'll consider it later.) In a 16-bit system, the signal's original
amplitude range would have to be divided into 65536 equally spaced levels.
**

Once the quantization is performed, the signal is digital. (digital = data
representation by quantized values) Any sample can have an amplitude of only
one of 65536 values. Each of these represents a "number", as assuredly as
the bit settings in a two-byte register represent a "number". ("12345" is
not a number; it is the representation of a number.)

If you don't believe that (and of course, you don't), imagine that you had
hundreds of sheets of paper with lines whose lengths were directly
proportional to the quantized amplitudes, with the relative length printed
next to each line. If I said I would send you the number of my street
address, and you opened the envelope and found a paper with a long line on
it, would you have any trouble finding the line of the matching length on
the reference sheets, and determining that my condo number was 17610? I
don't think so.

Let's repeat the listening test. If the quantized samples are played, we'll
hear the original signal, with varying amounts of quantization noise. ***
Furthermore, if we converted the quantized levels to digital numbers, then
reversed the process, converting those numbers back to the corresponding
voltage levels, the quantization noise would still be present. In other
words, there's no way to audibly distinguish a quantized pulse train with
the same signal recreated from a string of "numbers" -- because there isn't
any difference. The level-quantized samples and the PCM bit sequences
representing them are /exactly the same numbers/.

"But wait!" you say. "If you run the signal through a low-pass
reconstruction filter" (which doesn't and never will exist, but we'll ignore
that), "all them sharp little edges will be rounded off, and the horizontal
lines will be tilted and curved, and we'll have a nice analog signal again.
Right?"

Wrong. The signal still contains the quantization errors, because they
appear in the "baseband" component of the signal, and cannot be removed by
filtering.

So... How can a signal with quantization noise can be analog? (I'm waiting,
Hemo.) It can't, of course.

Regardless of what the waveform /looks like/, the fact is that it comprises
only a finite number of signal levels. If it didn't, there wouldn't be any
quantization noise.

QED. The output of a CD player is digital, not analog.

As for dither... It randomizes the quantization noise to minimize its
audibility. But the noise is still present, because the signal does not vary
in a continuous fashion. It's just that the errors have been moved around to
reduce their correlation with the signal.


* Other than the frequency response variation caused by the finite sampling
width.

** One could have non-linear quantization. I'm ignoring that for simplicity,
and because it's not often done.

*** If you don't have a test CD with an undithered tone sweep, find one.
It's quite interesting to hear the bursts of quantization noise as the test
tone sweeps through frequencies that are submultiples of the sampling rate.
Of course, there's quantization noise at _all_ frequencies. It's just that,
when the test tone is not "too close" to a submultiple, the errors are
sufficiently "random", particularly with musical material, not to be
particularly noticeable at normal listening levels.

[Don Pearce[_3_]]

On Mon, 19 Oct 2009 05:33:17 -0700, "William Sommerwerck"
wrote:

When Hemo (Marvin Miller) insists that Dr. Science (Dr. Frank Baxter)
respond with the two words that shows he understands the essence of blood,
Dr. Science comes back with "sea water".

So... Here's my equivalent of the "sea water" answer.


"Is the audio output of a CD player analog or digital?"

It's digital. Once you've converted analog to digital, the damage is done
and can't be undone. You can't go home again. Here's the explanation...

Consider the classic sampling of a band-limited signal. The sampling process
produces a string of pulses whose amplitudes are exactly the amplitude of
the signal at the instant of sampling. Because this amplitude can vary
continuously (ie, have any value, without restriction), the pulse chain is
an analog signal. (analog = data representation by continuously variable
values)

Fourier analysis shows that the pulse train contains the original signal,
unmodified. If we play the pulse train through a system free of
intermodulation distortion, we will hear the original signal, unchanged. *
This is true even if the system has "infinite" bandwidth, because the
original signal is a component of the pulse train; no filtering is required
to "recreate" it.

To convert this pulse train to a digital signal, we first have to quantize
its level. (I'm ignoring the use of dither, to clarify the point I'll be
making. We'll consider it later.) In a 16-bit system, the signal's original
amplitude range would have to be divided into 65536 equally spaced levels.
**

Once the quantization is performed, the signal is digital. (digital = data
representation by quantized values) Any sample can have an amplitude of only
one of 65536 values. Each of these represents a "number", as assuredly as
the bit settings in a two-byte register represent a "number". ("12345" is
not a number; it is the representation of a number.)

If you don't believe that (and of course, you don't), imagine that you had
hundreds of sheets of paper with lines whose lengths were directly
proportional to the quantized amplitudes, with the relative length printed
next to each line. If I said I would send you the number of my street
address, and you opened the envelope and found a paper with a long line on
it, would you have any trouble finding the line of the matching length on
the reference sheets, and determining that my condo number was 17610? I
don't think so.

Let's repeat the listening test. If the quantized samples are played, we'll
hear the original signal, with varying amounts of quantization noise. ***
Furthermore, if we converted the quantized levels to digital numbers, then
reversed the process, converting those numbers back to the corresponding
voltage levels, the quantization noise would still be present. In other
words, there's no way to audibly distinguish a quantized pulse train with
the same signal recreated from a string of "numbers" -- because there isn't
any difference. The level-quantized samples and the PCM bit sequences
representing them are /exactly the same numbers/.

"But wait!" you say. "If you run the signal through a low-pass
reconstruction filter" (which doesn't and never will exist, but we'll ignore
that), "all them sharp little edges will be rounded off, and the horizontal
lines will be tilted and curved, and we'll have a nice analog signal again.
Right?"

Wrong. The signal still contains the quantization errors, because they
appear in the "baseband" component of the signal, and cannot be removed by
filtering.

So... How can a signal with quantization noise can be analog? (I'm waiting,
Hemo.) It can't, of course.

Regardless of what the waveform /looks like/, the fact is that it comprises
only a finite number of signal levels. If it didn't, there wouldn't be any
quantization noise.

QED. The output of a CD player is digital, not analog.

As for dither... It randomizes the quantization noise to minimize its
audibility. But the noise is still present, because the signal does not vary
in a continuous fashion. It's just that the errors have been moved around to
reduce their correlation with the signal.


* Other than the frequency response variation caused by the finite sampling
width.

** One could have non-linear quantization. I'm ignoring that for simplicity,
and because it's not often done.

*** If you don't have a test CD with an undithered tone sweep, find one.
It's quite interesting to hear the bursts of quantization noise as the test
tone sweeps through frequencies that are submultiples of the sampling rate.
Of course, there's quantization noise at _all_ frequencies. It's just that,
when the test tone is not "too close" to a submultiple, the errors are
sufficiently "random", particularly with musical material, not to be
particularly noticeable at normal listening levels.


I think we have heard it all now. Meltdown is complete.

Here's a little hint for you. The output of a CD player contains no
discernable quantization levels. The analogue anti-alias filter
following the oversampled DAC has removed every last trace of them.
The signal is pure analogue.

d

[William Sommerwerck]

I think we have heard it all now. Meltdown is complete.

Whose?


Here's a little hint for you. The output of a CD player contains
no discernable quantization levels. The analogue anti-alias filter
following the oversampled DAC has removed every last trace of
them. The signal is pure analogue.

Quantization and aliasing are unrelated effects. My explanation spelled out
the logic of this, which is plain to anyone who gives it a little thought.

Why don't you think this through? Take two brains and call me in the
morning.

I want an apology, too, by the way. Not for disagreeing with me, but for not
using your basic intelligence.

I won't discuss this any further.

PS: Before posting this, I asked An Internationally Famous Designer for his
views. He said "it's a fine point", but agreed that what is done cannot be
undone. Once you've quantized the signal, you cannot remove it the
quantization error. This, as I pointed out, is shown by the fact that the
quantization nose remains, even after the filtering. QED, as they say.

[Don Pearce[_3_]]

On Mon, 19 Oct 2009 06:03:50 -0700, "William Sommerwerck"
wrote:

I think we have heard it all now. Meltdown is complete.

Whose?


Here's a little hint for you. The output of a CD player contains
no discernable quantization levels. The analogue anti-alias filter
following the oversampled DAC has removed every last trace of
them. The signal is pure analogue.

Quantization and aliasing are unrelated effects. My explanation spelled out
the logic of this, which is plain to anyone who gives it a little thought.


The anti alias filter removes the quantization steps. This has nothing
whatever to do with any quantization distortion products that may
already have been generated during the initial A/D process. You are
yet again muddling all your terminologies.

Subject the output of a CD player to the closest examination you can
bring to bear, and you will find no flat regions corresponding to
quantization levels.

Why don't you think this through? Take two brains and call me in the
morning.


Every digital engineer in the world (apart from you, evidently) has
done all this long ago.

I want an apology, too, by the way. Not for disagreeing with me, but for not
using your basic intelligence.


Pompous as ever.

I won't discuss this any further.

PS: Before posting this, I asked An Internationally Famous Designer for his
views. He said "it's a fine point", but agreed that what is done cannot be
undone. Once you've quantized the signal, you cannot remove it the
quantization error. This, as I pointed out, is shown by the fact that the
quantization nose remains, even after the filtering. QED, as they say.


What have quantization errors and quantization noise to do with
anything? They are artefacts of the initial process, and cannot be
removed. It is the QUANTIZATION itself that is removed by the
anti-alias filter. Yet again, you are having trouble working out what
all the various terms mean.

And by the way, argument from authority is one of the logical
fallacies of the Debating Trade Tricks. You fail again.

d

[Arny Krueger]

"Don Pearce" wrote in message


I think we have heard it all now. Meltdown is complete.

Agreed. I have to admit I lacked the time or stomach to read the whole tome
word-for-word. The answer to the purported question said all I needed to
hear. I've heard the rest from the unwashed far too many times.

Here's a little hint for you. The output of a CD player
contains no discernable quantization levels.

True in general. I suspect that if you pay enough for one of those
audiophile specials with allegedly no reconstruction filtering, traces of
them might be there.

The analogue
anti-alias filter following the oversampled DAC has
removed every last trace of them.

That's how its supposed to work, and that is how it works out in almost
every reasonable case.

The signal is pure analogue.

Well, about as pure analog as it gets in ordinary life - even many special
places in life.

Not being a registered meter reader like yours truly, our esteemed colleague
William has missed out on some of the more instructive ironies of life.

Back in the day, I built up a highly modified Heath THD analyzer that
performed up to 100 times better than the best Benton Harbor chose to
provide. I had a correspondingly modified audio signal generator, and the
two together could get down well under 0.01% - actually 0.0025% under some
conditions. Both were about as analog as it gets.

I still remember the first time I hooked my lovingly-modded "pure analog"
signal generator and THD analyzer up to what I would call my first serious
digital audio interface - a Turtle Beach Pinnacle of Fiji if memory serves.

Right up front, spectrum analysis showed that the *digital* Fiji was in
general a better and purer signal source than my all-analog signal
generator, and also a better and more sensitive signal analyzer than my
all-analog THD analyzer.

And the rest is history.

A really good CD player can easily outperform "pure analog" hardware by
factors of from 3 to 10 or more. The no-name, no-price audio interfaces on
a typical PC motherboard can have residuals that are easily 80 to 90 dB
down. The world is full of "pure analog" hardware that they can do a good
job of measuring.

[Anahata]

On Mon, 19 Oct 2009 06:03:50 -0700, William Sommerwerck wrote:


Why don't you think this through? Take two brains and call me in the
morning.

This is the classic naive view about digital audio that has been argued
countless times. Information theory at least 40 years old (and still
valid) has left you way behind.

Summary:
Analog = signal + random noise
Digital = signal + random noise
(assuming it's done properly, which includes dithering)

End of story.

The earth is flat. It's obvious.

--
Anahata
==//== 01638 720444
http://www.treewind.co.uk ==//== http://www.myspace.com/maryanahata

[Richard Crowley]

"Since its inception, the many counter-intuitive results of quantum
mechanics have provoked strong philosophical debate and many
interpretations. Even fundamental issues such as Max Born's basic rules
concerning probability amplitudes and probability distributions took decades
to be appreciated...."
http://en.wikipedia.org/wiki/Quantum...l_consequences

[Neil Gould]

Arny Krueger wrote:
"Don Pearce" wrote in message
Here's a little hint for you. The output of a CD player
contains no discernable quantization levels.

True in general. I suspect that if you pay enough for one of those
audiophile specials with allegedly no reconstruction filtering,
traces of them might be there.

Well, there may be some ragged slopes, but that doesn't equate to quantized
levels (and I don't think it would sound very good, either).

The analogue
anti-alias filter following the oversampled DAC has
removed every last trace of them.

That's how its supposed to work, and that is how it works out in
almost every reasonable case.

The signal is pure analogue.

Well, about as pure analog as it gets in ordinary life - even many
special places in life.

The CD's output is an analogue of the post-DAC signal, regardless of how
analogous that may be of the original source material.

Best,

Neil

[Les Cargill[_2_]]

William Sommerwerck wrote:
I think we have heard it all now. Meltdown is complete.

Whose?


Here's a little hint for you. The output of a CD player contains
no discernable quantization levels. The analogue anti-alias filter
following the oversampled DAC has removed every last trace of
them. The signal is pure analogue.

Quantization and aliasing are unrelated effects. My explanation spelled out
the logic of this, which is plain to anyone who gives it a little thought.



No - bear with us. We're trying to implement a reconstruction
filter. That which follows a Nyqist-conformant D/A converter
will be analog, by the magic of the bandlimit part of the
Nyquist theorem.

snip
--
Les Cargill

[William Sommerwerck]

Quantization and aliasing are unrelated effects. My explanation spelled
out the logic of this, which is plain to anyone who gives it a little
thought.

No - bear with us. We're trying to implement a reconstruction
filter.

There is no such thing. To imagine that there is only throws everything into
confusion.


That which follows a Nyqist-conformant D/A converter
will be analog, by the magic of the bandlimit part of the
Nyquist theorem.

No, because the quantization errors remain within the audio band. (I tried
to make this clear by "stepping through" the developmental sequence in the
original explanation.) The filter cannot remove them.

The other point, of course is... how can you correctly recreate discarded
data?

[Meindert Sprang]

"Don Pearce" wrote in message
...
The anti alias filter removes the quantization steps.

Of course it does not!!
Let's simplify the system William describes to a one bit system, sampling at
2 kHz. If you put a 1kHz sinewave in, you get a square wave in the digital
domain, which will be a sinewave again after the D/A conversion and
filtering, simply because the filter removes anything above 2kHz. So indeed,
you won't be able to measure any quantisation on the output, *as long as you
put this sine wave in*. Now, lower the input fequency to, say, 1Hz. Et
voila, on the output appears a square wave with rounded edges... Now you can
clearly observe that this system can only represent two stable levels.

The only thing an anti alias filter does is smooth the transitions between
quantization steps.

Meindert

[Don Pearce[_3_]]

On Tue, 20 Oct 2009 09:13:27 +0200, "Meindert Sprang"
wrote:

"Don Pearce" wrote in message
...
The anti alias filter removes the quantization steps.

Of course it does not!!
Let's simplify the system William describes to a one bit system, sampling at
2 kHz. If you put a 1kHz sinewave in, you get a square wave in the digital
domain, which will be a sinewave again after the D/A conversion and
filtering, simply because the filter removes anything above 2kHz. So indeed,
you won't be able to measure any quantisation on the output, *as long as you
put this sine wave in*. Now, lower the input fequency to, say, 1Hz. Et
voila, on the output appears a square wave with rounded edges... Now you can
clearly observe that this system can only represent two stable levels.

The only thing an anti alias filter does is smooth the transitions between
quantization steps.

Meindert


No. Simple as that. Do you seriously think that at 1Hz, you would get
out a level corresponding to a quantization step? You don't. You get
the right level, intermediate between two quantization steps. That is
what dither does, and that is why it is an integral and necessary part
of the encoding process.

I'll ignore the first part of your comment, where you have failed to
understand how the Nyquist condition works. Anti-alias filtering too
for that matter.

In fact, why did you post this at all? The whole thing is just
gibberish.

d

[Meindert Sprang]

"Don Pearce" wrote in message
...
No. Simple as that. Do you seriously think that at 1Hz, you would get
out a level corresponding to a quantization step? You don't. You get
the right level, intermediate between two quantization steps. That is
what dither does, and that is why it is an integral and necessary part
of the encoding process.

The system William described, had no dithering. Neither had my
simplification of that system.

Your remark that an anti alias filter removes the quantization steps is only
partially true. It only does when the digital signal was dithered.

I'll ignore the first part of your comment, where you have failed to
understand how the Nyquist condition works. Anti-alias filtering too
for that matter.

That is probably why my customers are so happy with the DSP systems I
designed for them...

Meindert

[Don Pearce[_3_]]

On Tue, 20 Oct 2009 09:49:37 +0200, "Meindert Sprang"
wrote:

"Don Pearce" wrote in message
...
No. Simple as that. Do you seriously think that at 1Hz, you would get
out a level corresponding to a quantization step? You don't. You get
the right level, intermediate between two quantization steps. That is
what dither does, and that is why it is an integral and necessary part
of the encoding process.

The system William described, had no dithering. Neither had my
simplification of that system.


It is not a simplified system, it is a broken system.

And even without dither you still won't find any quantization steps.
They will only be evident in the sloping parts of the waveform, but
you won't find them. They will have been filtered away.

Your remark that an anti alias filter removes the quantization steps is only
partially true. It only does when the digital signal was dithered.

I'll ignore the first part of your comment, where you have failed to
understand how the Nyquist condition works. Anti-alias filtering too
for that matter.

That is probably why my customers are so happy with the DSP systems I
designed for them...

Meindert


Oh god, not another one. Do read your own post, will you? Review your
numbers and claims.

d

[Meindert Sprang]

"Don Pearce" wrote in message
...
On Tue, 20 Oct 2009 09:49:37 +0200, "Meindert Sprang"
wrote:

"Don Pearce" wrote in message
...
No. Simple as that. Do you seriously think that at 1Hz, you would get
out a level corresponding to a quantization step? You don't. You get
the right level, intermediate between two quantization steps. That is
what dither does, and that is why it is an integral and necessary part
of the encoding process.

The system William described, had no dithering. Neither had my
simplification of that system.


It is not a simplified system, it is a broken system.

Why? Because it proves you wrong?

Meindert

[Don Pearce[_3_]]

On Tue, 20 Oct 2009 10:16:02 +0200, "Meindert Sprang"
wrote:

"Don Pearce" wrote in message
...
On Tue, 20 Oct 2009 09:49:37 +0200, "Meindert Sprang"
wrote:

"Don Pearce" wrote in message
...
No. Simple as that. Do you seriously think that at 1Hz, you would get
out a level corresponding to a quantization step? You don't. You get
the right level, intermediate between two quantization steps. That is
what dither does, and that is why it is an integral and necessary part
of the encoding process.

The system William described, had no dithering. Neither had my
simplification of that system.


It is not a simplified system, it is a broken system.

Why? Because it proves you wrong?

Meindert


You snip the reasoning. That marks the end of this conversation as I
do not intend to repeat it to a dishonest man.

d

[Meindert Sprang]

"Don Pearce" wrote in message
...
You snip the reasoning. That marks the end of this conversation as I
do not intend to repeat it to a dishonest man.

Don,

I am a self-taught professional who makes his living from designing
hardware, software and emdedded systems and I have a strong interest in
music. I may not be up to speed with everything I need to know but I am
always willing to learn new things. But I also recognize my knowlegde based
on my experience. For instance, if someone throws a fourier transform at me
in pure mathematical form, I bail out. But if I see an FFT algorithm
explained in diagams with bit reversals and twiddle factors, I am perfectly
capable of understanding what happens and I can implement the algorithm.
But one thing I am very, very sure of: I am not dishonest!

Meindert

[William Sommerwerck]

No, because the quantization errors remain within the audio band.
(I tried to make this clear by "stepping through" the developmental
sequence in the original explanation.) The filter cannot remove them.

Demonstrate those errors. Show them "stepping through". Do it on the
test bench. Your conjecture is testable. That should be inarguable.

Ever heard of a thought experiment?


The other point, of course is... how can you correctly recreate
discarded data?

You're asking this in the context of a world where 192 KHz AD's are
becoming much more common? You hear much over 90+ KHz or so?

Digital recording and playback have been commonplace for a quarter century.
Yet people -- recording engineers! -- still don't understand the difference
between time sampling and quantization!

This is beyond my comprehension.

Isn't there anyone out there with the least understanding of how digital
recording and playback actually work?

[William Sommerwerck]

The anti alias filter removes the quantization steps.

Of course it does not!!
Let's simplify the system William describes to a one bit system, sampling
at
2 kHz. If you put a 1kHz sinewave in, you get a square wave in the digital
domain, which will be a sinewave again after the D/A conversion and
filtering, simply because the filter removes anything above 2kHz. So
indeed,
you won't be able to measure any quantisation on the output, *as long as
you
put this sine wave in*. Now, lower the input fequency to, say, 1Hz. Et
voila, on the output appears a square wave with rounded edges... Now you
can
clearly observe that this system can only represent two stable levels.

The only thing an anti alias filter does is smooth the transitions between
quantization steps.

Thank you. I'm not surprised that support come from a fellow German. We're
very logical people.

[Meindert Sprang]

"William Sommerwerck" wrote in message
...
The anti alias filter removes the quantization steps.

Of course it does not!!
Let's simplify the system William describes to a one bit system,
sampling
at
2 kHz. If you put a 1kHz sinewave in, you get a square wave in the
digital
domain, which will be a sinewave again after the D/A conversion and
filtering, simply because the filter removes anything above 2kHz. So
indeed,
you won't be able to measure any quantisation on the output, *as long as
you
put this sine wave in*. Now, lower the input fequency to, say, 1Hz. Et
voila, on the output appears a square wave with rounded edges... Now you
can
clearly observe that this system can only represent two stable levels.

The only thing an anti alias filter does is smooth the transitions
between
quantization steps.

Thank you. I'm not surprised that support come from a fellow German. We're
very logical people.

Close. I'm Dutch :-)

Meindert

[Don Pearce[_3_]]

On Tue, 20 Oct 2009 04:45:24 -0700, "William Sommerwerck"
wrote:

The anti alias filter removes the quantization steps.

Of course it does not!!
Let's simplify the system William describes to a one bit system, sampling
at
2 kHz. If you put a 1kHz sinewave in, you get a square wave in the digital
domain, which will be a sinewave again after the D/A conversion and
filtering, simply because the filter removes anything above 2kHz. So
indeed,
you won't be able to measure any quantisation on the output, *as long as
you
put this sine wave in*. Now, lower the input fequency to, say, 1Hz. Et
voila, on the output appears a square wave with rounded edges... Now you
can
clearly observe that this system can only represent two stable levels.

The only thing an anti alias filter does is smooth the transitions between
quantization steps.

Thank you. I'm not surprised that support come from a fellow German. We're
very logical people.


So that is two Germans happy with the idea that you can not only
sample a 1kHz sine wave at 2kHz, but also perform the anti alias
filtering at 2kHz. And to think they had ambitions to rule the world.

d

[Meindert Sprang]

"Don Pearce" wrote in message
...
So that is two Germans happy with the idea that you can not only
sample a 1kHz sine wave at 2kHz, but also perform the anti alias
filtering at 2kHz. And to think they had ambitions to rule the world.

Oops, typo. The filter was supposed to be at 1kHz....

Meindert

[Don Pearce[_3_]]

On Tue, 20 Oct 2009 13:56:13 +0200, "Meindert Sprang"
wrote:

"Don Pearce" wrote in message
...
So that is two Germans happy with the idea that you can not only
sample a 1kHz sine wave at 2kHz, but also perform the anti alias
filtering at 2kHz. And to think they had ambitions to rule the world.

Oops, typo. The filter was supposed to be at 1kHz....

Meindert


And the sampling?

d

[Meindert Sprang]

"Don Pearce" wrote in message
...
On Tue, 20 Oct 2009 13:56:13 +0200, "Meindert Sprang"
wrote:

"Don Pearce" wrote in message
...
So that is two Germans happy with the idea that you can not only
sample a 1kHz sine wave at 2kHz, but also perform the anti alias
filtering at 2kHz. And to think they had ambitions to rule the world.

Oops, typo. The filter was supposed to be at 1kHz....

Meindert


And the sampling?

2kHz.

Meindert

[Don Pearce[_3_]]

On Tue, 20 Oct 2009 13:59:00 +0200, "Meindert Sprang"
wrote:

"Don Pearce" wrote in message
...
On Tue, 20 Oct 2009 13:56:13 +0200, "Meindert Sprang"
wrote:

"Don Pearce" wrote in message
...
So that is two Germans happy with the idea that you can not only
sample a 1kHz sine wave at 2kHz, but also perform the anti alias
filtering at 2kHz. And to think they had ambitions to rule the world.

Oops, typo. The filter was supposed to be at 1kHz....

Meindert


And the sampling?

2kHz.


Ok, so you don't understand one of the most basic rules of sampling.

d

[Meindert Sprang]

"Don Pearce" wrote in message
...
On Tue, 20 Oct 2009 13:59:00 +0200, "Meindert Sprang"
wrote:

"Don Pearce" wrote in message
...
On Tue, 20 Oct 2009 13:56:13 +0200, "Meindert Sprang"
wrote:

"Don Pearce" wrote in message
...
So that is two Germans happy with the idea that you can not only
sample a 1kHz sine wave at 2kHz, but also perform the anti alias
filtering at 2kHz. And to think they had ambitions to rule the
world.

Oops, typo. The filter was supposed to be at 1kHz....

Meindert


And the sampling?

2kHz.


Ok, so you don't understand one of the most basic rules of sampling.

d

Which is, according to you?

M.

[Don Pearce[_3_]]

On Tue, 20 Oct 2009 14:03:48 +0200, "Meindert Sprang"
wrote:

"Don Pearce" wrote in message
...
On Tue, 20 Oct 2009 13:59:00 +0200, "Meindert Sprang"
wrote:

"Don Pearce" wrote in message
...
On Tue, 20 Oct 2009 13:56:13 +0200, "Meindert Sprang"
wrote:

"Don Pearce" wrote in message
...
So that is two Germans happy with the idea that you can not only
sample a 1kHz sine wave at 2kHz, but also perform the anti alias
filtering at 2kHz. And to think they had ambitions to rule the
world.

Oops, typo. The filter was supposed to be at 1kHz....

Meindert


And the sampling?

2kHz.


Ok, so you don't understand one of the most basic rules of sampling.

d

Which is, according to you?

M.


You can not sample a 1kHz signal at 2kHz. According to me, of course.
You clearly know better.

d

[Meindert Sprang]

"Don Pearce" wrote in message
...
You can not sample a 1kHz signal at 2kHz. According to me, of course.
You clearly know better.

According to Nyquist you can. You even mentioned that yourself earlier....

M.

[Don Pearce[_3_]]

On Tue, 20 Oct 2009 14:12:39 +0200, "Meindert Sprang"
wrote:

"Don Pearce" wrote in message
...
You can not sample a 1kHz signal at 2kHz. According to me, of course.
You clearly know better.

According to Nyquist you can. You even mentioned that yourself earlier....


No, Nyquist never said any such thing. Go back to the books. Have you
truly done A/D design? It is sounding less likely with your every
post.

d

[William Sommerwerck]

That which follows a Nyquist-conformant D/A converter
will be analog, by the magic of the bandlimit part of the
Nyquist theorem.

No, because the quantization errors remain within the
audio band.

Not according to all the constraints of the Nyquist Theorem.
They're *out* of band, except for least-[significant-]bit errors
which are masked by dithering. That's the point...

Dithering doesn't mask quantization errors. It simply moves them around to
make them more like random noise.

Furthermore, what makes digital "digital" is quantization error. If there
were no quantization error (an impossibility), signal values would be able
to vary continuously, and you would have an analog signal.

The direct output of a DAC (prior to any filtering) is always representable
by a finite number of voltage levels. It is therefore digitial. It
reproduces the quantization errors (dithered or not) of the original
sampling process. No amount of filtering can change this.

True, you can use oversampling and other techniques to move part of the
quantization error into the ultrasonic region, thus gaining resolution in
the audible band. But you can never move all of it. And as long as you have
quantization error, you have a digital signal.

QED.

What is so difficult to understand about this? Don't you understand doors?

Ready your gum wrappers, people. The secrets of the universe are written
thereon.

[William Sommerwerck]

Thank you. I'm not surprised that support come from
a fellow German. We're very logical people.

Close. I'm Dutch :-)

Close enough. I've been told I must be of Dutch extraction because my name
is spelled with a "c". But my family came from Hildesheim.

[William Sommerwerck]

You can not sample a 1kHz signal at 2kHz. According to me,
of course. You clearly know better.

I hate taking Mr. Pearce's side in this, but he's correct. A 1kHz signal
sampled at 2kHz would "alias" to a DC signal.

It's interesting that the Wikipedia article points out that the original
statement of the Nyquist theorem is incorrect, as it states that sampling at
twice the rate of the highest frequency is sufficient.

[Meindert Sprang]

"Don Pearce" wrote in message
...
On Tue, 20 Oct 2009 14:12:39 +0200, "Meindert Sprang"
wrote:

"Don Pearce" wrote in message
...
You can not sample a 1kHz signal at 2kHz. According to me, of course.
You clearly know better.

According to Nyquist you can. You even mentioned that yourself
earlier....


No, Nyquist never said any such thing. Go back to the books.

Sigh.... you're splitting hair now. Does Nyquist-Shannon Sampling Theorem
sound better?

Have you truly done A/D design? It is sounding less likely with your every
post.

Yes I have. Three major players in the anti-shoplifting industry used my
A/D-DSP designs, I've designed a few image sampling and processing systems,
a few RF signal generators using DDS, done some FFT processing and digital
filtering on various DSP's (Analog Devices, Motorola). Want some schematics,
PCB layouts or source code too?

Meindert

[William Sommerwerck]

A 1kHz signal sampled at 2kHz would "alias" to a DC signal.

Whoops. I goofed. You'd get a kind of "sawtooth" whose amplitude was not
properly defined, varying from 0 to the peak signal level, according to the
relative phase of the sampling.

[Meindert Sprang]

"William Sommerwerck" wrote in message
...
You can not sample a 1kHz signal at 2kHz. According to me,
of course. You clearly know better.

I hate taking Mr. Pearce's side in this, but he's correct. A 1kHz signal
sampled at 2kHz would "alias" to a DC signal.

No it doesn't. Only when the samping occurs at the zero crossings, you end
up with DC, 0V to be precisely. At any other point in time, you end up with
samples of exactly the same voltage, anywhere between 0V and the peak
voltage of the signal, but with alternating signs, aka a digital squarewave.

If you look at it mathematically, sampling 1kHz with 2kHz is the same as
adding and subtracting both frequencies, resulting in 1kHz and 3kHz. (2+1
and 2-1). Sampling 1kHz with 1kHz does indeed result in DC and 2kHz.

Meindert

[Arny Krueger]

"Meindert Sprang" wrote in
message
"Don Pearce" wrote in message
...

The anti alias filter removes the quantization steps.

Of course it does not!!
Let's simplify the system William describes to a one bit
system, sampling at 2 kHz. If you put a 1kHz sinewave in,
you get a square wave in the digital domain, which will
be a sinewave again after the D/A conversion and
filtering, simply because the filter removes anything
above 2kHz. So indeed, you won't be able to measure any
quantisation on the output, *as long as you put this sine
wave in*. Now, lower the input fequency to, say, 1Hz. Et
voila, on the output appears a square wave with rounded
edges... Now you can clearly observe that this system can
only represent two stable levels.

I see we have a purported factual claim from someone who has obviously never
actually done the experiment whose results they purport to present.

If you actually look at (with an oscilliscope using an expanded scale) or
measure a 1 Hz wave coming out of a CD player with enough LF bandpass to
actually produce a signficantly large 1 Hz wave, you will find no such
square wave, or any evidence of stairstepping.

I've done both the examination with a scope and also a detailed spectral
analysis. No significant stair steps, no square waves, nada. Just tiny
residuals if anything at all.

The reason for this is that if the stairstepping were present, its frequency
would be at half the sampling frequency (Nyquist) and/or harmonics thereof
the first of which is the sampling frequency. However, frequencies of
Nyquist or above are attenuated very many dB by the reconstruction filter.
Only tiny traces of them might be found, if they are measurable at all.

[William Sommerwerck]

Of course it does not!!
Let's simplify the system William describes to a one bit
system, sampling at 2 kHz. If you put a 1kHz sinewave in,
you get a square wave in the digital domain, which will
be a sinewave again after the D/A conversion and
filtering, simply because the filter removes anything
above 2kHz. So indeed, you won't be able to measure any
quantisation on the output, *as long as you put this sine
wave in*. Now, lower the input fequency to, say, 1Hz. Et
voila, on the output appears a square wave with rounded
edges... Now you can clearly observe that this system can
only represent two stable levels.

I see we have a purported factual claim from someone who has obviously
never
actually done the experiment whose results they purport to present.

If you actually look at (with an oscilliscope using an expanded scale) or
measure a 1 Hz wave coming out of a CD player with enough LF bandpass to
actually produce a signficantly large 1 Hz wave, you will find no such
square wave, or any evidence of stairstepping.

I've done both the examination with a scope and also a detailed spectral
analysis. No significant stair steps, no square waves, nada. Just tiny
residuals if anything at all.

The reason for this is that if the stairstepping were present, its
frequency
would be at half the sampling frequency (Nyquist) and/or harmonics thereof
the first of which is the sampling frequency. However, frequencies of
Nyquist or above are attenuated very many dB by the reconstruction filter.
Only tiny traces of them might be found, if they are measurable at all.

I won't gainsay your experiment or observations. However, as long as
quantization noise is present in the output, the signal is digital. Please
see my other post.

"Think, people, think!"

[Don Pearce[_3_]]

On Tue, 20 Oct 2009 06:02:05 -0700, "William Sommerwerck"
wrote:

I won't gainsay your experiment or observations. However, as long as
quantization noise is present in the output, the signal is digital. Please
see my other post.

Quantization noise is a distortion produced by a modulation process.
Once the distortion is there, there is no way to remove it. Although
that energy can not be prevented, it can be spread out into wideband
noise by dithering, which is audibly preferable.

But that is all it is - some distortion. There is nothing magical
about it that converts an analogue signal into a digital one. Once the
signal is out of the DAC, and filed smooth by oversampling and
filtering you have an analogue signal that is just a tiny bit
distorted. There are no steps, no digits, Just signal plus noise and
distortion.

d

[Arny Krueger]

"William Sommerwerck" wrote in
message
Of course it does not!!
Let's simplify the system William describes to a one bit
system, sampling at 2 kHz. If you put a 1kHz sinewave
in, you get a square wave in the digital domain, which
will be a sinewave again after the D/A conversion and
filtering, simply because the filter removes anything
above 2kHz. So indeed, you won't be able to measure any
quantisation on the output, *as long as you put this
sine wave in*. Now, lower the input fequency to, say,
1Hz. Et voila, on the output appears a square wave with
rounded edges... Now you can clearly observe that this
system can only represent two stable levels.

I see we have a purported factual claim from someone who
has obviously never actually done the experiment whose
results they purport to present.

If you actually look at (with an oscilliscope using an
expanded scale) or measure a 1 Hz wave coming out of a
CD player with enough LF bandpass to actually produce a
signficantly large 1 Hz wave, you will find no such
square wave, or any evidence of stairstepping.

I've done both the examination with a scope and also a
detailed spectral analysis. No significant stair steps,
no square waves, nada. Just tiny residuals if anything
at all.

The reason for this is that if the stairstepping were
present, its frequency would be at half the sampling
frequency (Nyquist) and/or harmonics thereof the first
of which is the sampling frequency. However, frequencies
of Nyquist or above are attenuated very many dB by the
reconstruction filter. Only tiny traces of them might be
found, if they are measurable at all.

I won't gainsay your experiment or observations. However,
as long as quantization noise is present in the output,
the signal is digital.

No, its just a signal with some noise just like any real world analog
signal.

Any proper digital system is dithered, and the purpose of dither is to
modify quantization noise so that it has a pseudo-randomized spectrum.
Therefore, the actual quantization noise that one finds at the output of a
well-designed CD player (and almost all are well-enough designed to pass
this test) will be indistingushable from the usual residual noise you find
in any real world analog system.

The actual responsibility for dithering CDs is with the people who produce
the CD recording. The player is just the messenger. The good news is that
this means that any enhancments to the dithering process can benefit the
users of all properly-designed players.

[Arny Krueger]

"Don Pearce" wrote in message

On Tue, 20 Oct 2009 06:02:05 -0700, "William Sommerwerck"
wrote:

I won't gainsay your experiment or observations.
However, as long as quantization noise is present in the
output, the signal is digital. Please see my other post.

Quantization noise is a distortion produced by a
modulation process.

Right, and since any proper digital process is dithered, the actual
quantization noise at its output will strongly resemble the usual noise that
we find in real-world analog systems.

Once the distortion is there, there
is no way to remove it. Although that energy can not be
prevented, it can be spread out into wideband noise by
dithering, which is audibly preferable.

Not only is dither preferable, it is a *requirement* for proper
digitization.

But that is all it is - some distortion. There is nothing
magical about it that converts an analogue signal into a
digital one. Once the signal is out of the DAC, and filed
smooth by oversampling and filtering you have an analogue
signal that is just a tiny bit distorted. There are no
steps, no digits, Just signal plus noise and distortion.

Actually, properly dithered signals just have what appears to be a noise
floor. The noise may be pseudo-random and in some sense predictable, but if
you dither with a truely random noise source, the noise floor at the output
will be truely random.