[Don Pearce[_3_]]

On Tue, 20 Oct 2009 09:24:39 -0400, "Arny Krueger"
wrote:

"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.



You are right, it is just noise, no distortion.

d

[Scott Dorsey]

Guys, please take this stupid thread elsewhere.
--scott
--
"C'est un Nagra. C'est suisse, et tres, tres precis."

[Anahata]

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.

I believe your original post asserted that the output of a CD player is
digital representation of analog information.

It occurs to me that it is in fact precisely the opposite.

By your logic, the output of a D-A converter is digital too.
I think we're having a bit of a definition-of-terms problem here.

The fact that an analog representation of a finite set of numbers can
only (in a perfect noise free world) take on a finite set of voltages
does not alter that fact that the voltage is an analog representation of
the number.

***

In addition, there is a separate arument which qestions the assumption
that the voltage even has a finite number of values when it's changing
with time. Actually it's a continuous waveform constructed from a number
of point values that characterize it, rather like the ways that smooth
bezier curves are generated from a small number of points, and it's
irrelevant that those points have a finite number of possible values.

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

[Don Pearce[_3_]]

On 20 Oct 2009 09:32:37 -0400, (Scott Dorsey) wrote:


Guys, please take this stupid thread elsewhere.
--scott

Don't read it if you don't like it. It is quite at home here.

d

[hank alrich]

William Sommerwerck wrote:

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.

Thank you. The signal from a DAC can vary continuously within the
bandwidth of the chosen sampling parameters.

--
ha
shut up and play your guitar

[hank alrich]

William Sommerwerck wrote:

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?

Yes. That's called conjecture. The next step is the attempt at proof, if
one honestly desires to validate or reject one's conjecture.

One who makes conjectures intended to be taken seriously will take that
next step. Otherwise a "thought experiment" can also be called "making
**** up".

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?

Apparently no one but you. I'm amazed the whole process has gotten this
far without your help.

Meanwhile, subject your conjecture to test in the physical world and let
us know how it went.

--
ha
shut up and play your guitar

[William Sommerwerck]

"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.

What modulation process? (Quantization could be considred a kind of
modulation, of course.)


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

Of course.


But that is all it is -- some distortion. There is nothing magical
about it that converts an analogue signal into a digital one.

You and Arny are missing the point. I'm reminded of the apocryphal story
that Winston Churchill once asked a society lady if she'd have sex with him
for £100. "I don't know. I suppose I might."

"Well, what about £1?"

"Mr. Churchill! What kind of a woman do you think I am?"

"I know what kind of a woman you are. I'm just trying to settle on a price."

Let's come back to the sticking point that started this argument. Once
you've quantized the samples (dithered or not), you've digitized the signal.
Assigning each level a number has major advantages in storage and
transmission, but it doesn't change the sampled/quantized signal. Both the
quantized samples and the numbers corresponding to them represent exactly
the same thing. One cannot be analog and the other digital.

If a signal contains quantization noise -- regardless of what the signal
"looks like" -- how can it not be digital?

[William Sommerwerck]

The fact that an analog representation of a finite set of numbers can
only (in a perfect noise free world) take on a finite set of voltages
does not alter that fact that the voltage is an analog representation of
the number.

You obviously don't understand the difference between analog and digital.
Analog signals do not have a finite set of voltages.

[William Sommerwerck]

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.

Thank you. The signal from a DAC can vary continuously within the
bandwidth of the chosen sampling parameters.

The output of a DAC has only a finite number of levels. It cannot vary
continuously.

[William Sommerwerck]

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

Apparently no one but you. I'm amazed the whole process has gotten this
far without your help.

Meanwhile, subject your conjecture to test in the physical world and let
us know how it went.

You still have explained/justified your confusion between time sampling and
amplitude quantization.

[Don Pearce[_3_]]

On Tue, 20 Oct 2009 06:53:18 -0700, "William Sommerwerck"
wrote:

"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.

What modulation process? (Quantization could be considred a kind of
modulation, of course.)


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

Of course.


But that is all it is -- some distortion. There is nothing magical
about it that converts an analogue signal into a digital one.

You and Arny are missing the point. I'm reminded of the apocryphal story
that Winston Churchill once asked a society lady if she'd have sex with him
for £100. "I don't know. I suppose I might."

"Well, what about £1?"

"Mr. Churchill! What kind of a woman do you think I am?"

"I know what kind of a woman you are. I'm just trying to settle on a price."

Let's come back to the sticking point that started this argument. Once
you've quantized the samples (dithered or not), you've digitized the signal.
Assigning each level a number has major advantages in storage and
transmission, but it doesn't change the sampled/quantized signal. Both the
quantized samples and the numbers corresponding to them represent exactly
the same thing. One cannot be analog and the other digital.

If a signal contains quantization noise -- regardless of what the signal
"looks like" -- how can it not be digital?


It is possible to generate precisely that same quantization noise
signal by other means, not involving quantization. The resulting
signal would look exactly the same. Would that also be a digital
signal? Presumably not. If it were the output from a black box, what
process - even just a theoretical one - would you propose to identify
the "digital" signal from the analogue one?

d

[Arny Krueger]

"William Sommerwerck" wrote in
message
"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.

What modulation process? (Quantization could be considered
a kind of modulation, of course.)

Exactly. Quantization is often treated as a modulation process.

But that is all it is -- some distortion. There is
nothing magical about it that converts an analogue
signal into a digital one.

You and Arny are missing the point. I'm reminded of the
apocryphal story that Winston Churchill once asked a
society lady if she'd have sex with him for £100. "I
don't know. I suppose I might."

"Well, what about £1?"

"Mr. Churchill! What kind of a woman do you think I am?"

"I know what kind of a woman you are. I'm just trying to
settle on a price."

The version of this story that I'm familiar with involved a famous writer,
name escapes me but not Churchill. Whatever.

Let's come back to the sticking point that started this
argument. Once you've quantized the samples (dithered or
not), you've digitized the signal. Assigning each level a
number has major advantages in storage and transmission,
but it doesn't change the sampled/quantized signal. Both
the quantized samples and the numbers corresponding to
them represent exactly the same thing. One cannot be
analog and the other digital.

If a signal contains quantization noise -- regardless of
what the signal "looks like" -- how can it not be digital?

I believe that the common generally agreed-upon usage is "Digital to Analog
Converter". In practice is literally true. Forget all of the pontification,
go into the lab and see what happens.

[Don Pearce[_3_]]

On Tue, 20 Oct 2009 06:55:10 -0700, "William Sommerwerck"
wrote:

The fact that an analog representation of a finite set of numbers can
only (in a perfect noise free world) take on a finite set of voltages
does not alter that fact that the voltage is an analog representation of
the number.

You obviously don't understand the difference between analog and digital.
Analog signals do not have a finite set of voltages.


Analogue square waves do. S/PDIF signals don't. You are simply wrong.

d

[Laurence Payne[_2_]]

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.

"Think, people, think!"

Think yourself!

Take the output. Record it on a wax cylinder. Still hear the
quantisation noise? So it's digital?

[Anahata]

On Tue, 20 Oct 2009 10:00:41 -0400, Arny Krueger wrote:

"I know what kind of a woman you are. I'm just trying to settle on a
price."

The version of this story that I'm familiar with involved a famous
writer,

George Bernard Shaw, and the version of the punch line I have is "we've
already established that: I'm just haggling over the price"

I believe that the common generally agreed-upon usage is "Digital to
Analog Converter". In practice is literally true. Forget all of the
pontification, go into the lab and see what happens.

Yes, apparently we are being told that the output of a D-A converter is
digital...

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

[Anahata]

On Tue, 20 Oct 2009 06:55:10 -0700, William Sommerwerck wrote:

You obviously don't understand the difference between analog and
digital. Analog signals do not have a finite set of voltages.

Like I said, we have a terminology problem at the root of this.

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

[William Sommerwerck]

It is possible to generate precisely that same quantization noise
signal by other means, not involving quantization. The resulting
signal would look exactly the same. Would that also be a digital
signal?

Yes, it would be. "If it looks like a duck", etc.


Presumably not. If it were the output from a black box, what
process -- even just a theoretical one -- would you propose to
identify the "digital" signal from the analogue one?

I'm not sure this is a fair question. Bear with me.

I'm one of those people who feel they're standing outside society, looking
in. This isn't necessarily a bad thing, as it encourages independent
thought. Whether such thinking inevitably results in anything useful is
another matter.

I am legitimately bothered, however, by the fact that, whenever anyone
suggests anything that conflicts with conventional viewpoints, they are
immediately attacked, usually on a personal level. Every time I start these
arguments (and they need to started occasionally), I rarely get any kind of
thoughtful response. Most are equivalent to knees slamming into the
responders' jaws, in a fit of violent antipathy.

It's obvious that most of you have no fundamental understanding of the
issues. This is (unfortunately) normal, because education tends to encourage
swallowing without digestion. If you've never ask yourself "Why is
thus-and-such so?" or "Do I really understand XYZ?" you need to start. Any
honest person knows that they don't understand everything they think they
know.

I'm sure most of you accept the idea that a Fourier analysis is the
frequency-domain representation of a time-domain signal. But the reason you
accept this (unless you went through the derivation) is because you were
told that it was so. If someone suggested that it might not be so, you would
be all a-dither, defending something you didn't actually understand.

Facts don't interest me. Principles do. Where possible, I want to understand
why something is, not simply know that it is.

In this case, the principle is that quantized samples are "digital"; numbers
are not needed, because quantized levels are direct representations of
numbers. (Duh...) This isn't open to debate, any more than that an "event"
can be represented in both the time and frequency domains.

Running quantized samples though a low-pass filter might produce a "smooth"
waveform, but that doesn't mean the waveform is an analog signal. You're
confusing waveform with information. Waveform is not information.

So my answer to your preceding question is this... If I add the appropriate
dithered quantization noise to an analog signal (that is, exactly the same
noise, in amplitude, phase, and spectral content that would have been
created by sampling/quantizing the noise)... then I have converted it into a
digital signal, because there is no way to distinguish them.

Quantized samples are digital (regardless of what you think). And
quantization errors are quantization errors, whether they are correlated or
noise-like. I don't see how randomizing them analog-izes" the information.
It's still quantized. Why is an error that looks like noise "analog" and an
error that looks like enharmonic distortion not?

I'm tired of arguing with people who fail to use their capacity for original
or independent thought.

[Don Pearce[_3_]]

On Tue, 20 Oct 2009 07:51:23 -0700, "William Sommerwerck"
wrote:

Facts don't interest me.

I think we now understand everything that has gone before.

Goodbye.

d

[Wecan do it]

"anahata" wrote in message
o.uk...
On Tue, 20 Oct 2009 06:55:10 -0700, William Sommerwerck
wrote:

You obviously don't understand the difference between
analog and
digital. Analog signals do not have a finite set of
voltages.

Like I said, we have a terminology problem at the root of
this.


Which is the definition of William Sommerwreck, redefining
terminology to suit his argument in spite of the facts.

To better understand William have a look on the thread about
the Beatles reissues/remasters and how he tries to redefine
the terminology for an industry (the record making industry)
that has been good enough for the recording industry until
William decided to jump into the conversations.

Everyone else is wrong and William is right.

That makes sense if you understand that William has confused
right and wrong as he does many concepts in the English
language.


peace
dawg

[Mark]

On Oct 19, 8:33*am, "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.

[hank alrich]

Don Pearce wrote:

On 20 Oct 2009 09:32:37 -0400, (Scott Dorsey) wrote:


Guys, please take this stupid thread elsewhere.
--scott

Don't read it if you don't like it. It is quite at home here.

d

Is this any fancier than just "show me the ones and zeros if it's
digital"?

--
ha
shut up and play your guitar

[hank alrich]

William Sommerwerck wrote:

"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.

What modulation process? (Quantization could be considred a kind of
modulation, of course.)


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

Of course.


But that is all it is -- some distortion. There is nothing magical
about it that converts an analogue signal into a digital one.

You and Arny are missing the point. I'm reminded of the apocryphal story
that Winston Churchill once asked a society lady if she'd have sex with him
for £100. "I don't know. I suppose I might."

"Well, what about £1?"

"Mr. Churchill! What kind of a woman do you think I am?"

"I know what kind of a woman you are. I'm just trying to settle on a price."

Let's come back to the sticking point that started this argument. Once
you've quantized the samples (dithered or not), you've digitized the signal.
Assigning each level a number has major advantages in storage and
transmission, but it doesn't change the sampled/quantized signal. Both the
quantized samples and the numbers corresponding to them represent exactly
the same thing. One cannot be analog and the other digital.

If a signal contains quantization noise -- regardless of what the signal
"looks like" -- how can it not be digital?

Show me the ones and zeros.

--
ha
shut up and play your guitar

[Don Pearce[_3_]]

On Tue, 20 Oct 2009 08:23:52 -0700, (hank alrich)
wrote:

Don Pearce wrote:

On 20 Oct 2009 09:32:37 -0400, (Scott Dorsey) wrote:


Guys, please take this stupid thread elsewhere.
--scott

Don't read it if you don't like it. It is quite at home here.

d

Is this any fancier than just "show me the ones and zeros if it's
digital"?

Not really. It is simply a matter of the man not knowing the
difference between quantized and digitized.

d

[William Sommerwerck]

A properly reconstructed dithered digital system
IS THE SAME AS an analog system.

So the dithering just magically causes the quantization errors to disappear?
Where do they go?

[Scott Dorsey]

In article ,
hank alrich wrote:
Don Pearce wrote:

On 20 Oct 2009 09:32:37 -0400, (Scott Dorsey) wrote:


Guys, please take this stupid thread elsewhere.

Don't read it if you don't like it. It is quite at home here.


Is this any fancier than just "show me the ones and zeros if it's
digital"?

No, not really. It's the same stuff we all went through here in the
late 1980s, and it's the reason there's such a good discussion of the
issues in the FAQ.

In general, if it's in the FAQ, it's not worth rehashing on the group.
That's what the FAQ is for, after all.
--scott
--
"C'est un Nagra. C'est suisse, et tres, tres precis."

[Scott Dorsey]

William Sommerwerck wrote:
A properly reconstructed dithered digital system
IS THE SAME AS an analog system.

So the dithering just magically causes the quantization errors to disappear?

Yes.

Where do they go?

They become noise.
--scott
--
"C'est un Nagra. C'est suisse, et tres, tres precis."

[Don Pearce[_3_]]

On Tue, 20 Oct 2009 08:29:15 -0700, "William Sommerwerck"
wrote:

A properly reconstructed dithered digital system
IS THE SAME AS an analog system.

So the dithering just magically causes the quantization errors to disappear?
Where do they go?


Just remembered a couple of old papers I did years ago on aliasing and
dither

http://www.donepearce.plus.com/papers/index.htm

The other two are dead at the moment. I may revive them.

d

[jwvm]

On Oct 19, 4:51*pm, "Neil Gould" wrote:

snip

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).


Actually not if reconstruction has been performed properly. The output
signal must also be bandwidth limited to less than half the sampling
frequency so the highest frequencies permissible can only consist of
sinusoidal waveforms.


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


There is nothing digital about the output. With modern D/A converters
and dithering, real-world noise will completely mask any quantization
effects.

[jwvm]

On Oct 20, 11:33*am, (Scott Dorsey) wrote:
William Sommerwerck wrote:


snip


Where do they go?

They become noise.
--scott


This added noise is not even audible in properly-made real-world
recordings since the background noise will be much greater.

[William Sommerwerck]

So the dithering just magically causes the quantization errors to
disappear?

Yes.

Where do they go?

They become noise.

So how is an error that's not sonically obnoxious no longer an error?

[William Sommerwerck]

Not really. It is simply a matter of the man not knowing
the difference between quantized and digitized.

You da man.

[William Sommerwerck]

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

There is nothing digital about the output. With modern D/A
and dithering, real-world noise will completely mask any
quantization effects.

I see exactly what you're saying, though I don't agree.

So we're down to where the rubber meets the road. Tell me...

If I fed a complex, band-limited signal into an ADC, then properly converted
it back from numbers -- and I had some magic way to overlap the waveforms
for close examination -- what, if any differences would I see, and what
would be the cause of the differences?

This is a roundabout way of stating the following... If the "restored" (I
refuse to say "reconstructed") signal is truly "analog", then any amplitude
change, no matter how small, in the original must be reproduced in the
output. Otherwise, the output signal isn't analog.

[Mark]

On Oct 20, 12:09*pm, "William Sommerwerck"
wrote:
So the dithering just magically causes the quantization errors to

disappear?

Yes.
Where do they go?
They become noise.

So how is an error that's not sonically obnoxious no longer an error?

its still an error but it is noise the same as the noise in an anlog
system...

no one ever said a digital system is PERFECT, it still has a dynamic
range limitation like and anlaog system and a BW limitation like an
analog system.. Using dither, the low end of the dynamic range of
the digital system is bounded by random noise instead of quantization
errors, just like an analog system.

So yes there are "errors" but the errors are random noise, no
different then analog.

And if implemented correctly, these random noise errors are
indistinguishable from random noise in an analog system, and i don't
just mean that you can't HEAR the difference, there IS NO difference.



Mark

[Mark]

On Oct 20, 12:25*pm, Mark wrote:
On Oct 20, 12:09*pm, "William Sommerwerck"

wrote:
So the dithering just magically causes the quantization errors to

disappear?

Yes.
Where do they go?
They become noise.

So how is an error that's not sonically obnoxious no longer an error?

its still an error but it is noise the same as the noise in an anlog
system...

no one ever said a digital system is PERFECT, it still has a dynamic
range limitation like and anlaog system and a BW limitation like an
analog system.. * Using dither, the low end of the dynamic range of
the digital system is bounded by random noise instead of quantization
errors, just like an analog system.

So yes there are "errors" but the errors are random noise, no
different then analog.

And if implemented correctly, these random noise errors are
indistinguishable from random noise in an analog system, and i don't
just mean that you can't HEAR the difference, there IS NO difference.

Mark

I forgot to add this

http://www.national.com/an/AN/AN-804.pdf

Mark

[Mark]

On Oct 20, 12:30*pm, Mark wrote:
On Oct 20, 12:25*pm, Mark wrote:





On Oct 20, 12:09*pm, "William Sommerwerck"

wrote:
So the dithering just magically causes the quantization errors to

disappear?

Yes.
Where do they go?
They become noise.

So how is an error that's not sonically obnoxious no longer an error?

its still an error but it is noise the same as the noise in an anlog
system...

no one ever said a digital system is PERFECT, it still has a dynamic
range limitation like and anlaog system and a BW limitation like an
analog system.. * Using dither, the low end of the dynamic range of
the digital system is bounded by random noise instead of quantization
errors, just like an analog system.

So yes there are "errors" but the errors are random noise, no
different then analog.

And if implemented correctly, these random noise errors are
indistinguishable from random noise in an analog system, and i don't
just mean that you can't HEAR the difference, there IS NO difference.

Mark

I forgot to add this

http://www.national.com/an/AN/AN-804.pdf

Mark- Hide quoted text -

- Show quoted text -

and one final thing ...really this is the last...

when I say a dithered digital system is the same as anlog, i really
mean the same..

Take for example, an anlog system with a noise floor say -96 dB, you
know you can have a tone burried BELOW the noise say at -106 dB and
you can HEAR it and you can SEE it on a spectrum analyzer.

Well the same is true for a dithered digital system!!!!

You can have a -106 dB tone below the noise on a digital system with
dither and you can HEAR it and see it on a spectrum analyzer...
Imagine that, the tone is BELOW one MSB but yet it is encoded and
comes out the output!!!!.

That is the power of dither. These are not just my opinions, these
are facts.. Think about how that works, how a tone BELOW one MSB can
get through, and you will begin to understand dither.

Mark

[Arny Krueger]

"William Sommerwerck" wrote in
message
The fact that an analog representation of a finite set
of numbers can only (in a perfect noise free world) take
on a finite set of voltages does not alter that fact
that the voltage is an analog representation of the
number.

You obviously don't understand the difference between
analog and digital. Analog signals do not have a finite
set of voltages.

Right, real-world analog signals have probabilities. Real world digital
signals have exact values.

[jwvm]

On Oct 20, 12:19*pm, "William Sommerwerck"
wrote:
The CD's output is an analogue of the post-DAC signal,
regardless of how analogous that may be of the original
source material.
There is nothing digital about the output. With modern D/A
and dithering, real-world noise will completely mask any
quantization effects.

I see exactly what you're saying, though I don't agree.

So we're down to where the rubber meets the road. Tell me...

If I fed a complex, band-limited signal into an ADC, then properly converted
it back from numbers -- and I had some magic way to overlap the waveforms
for close examination -- what, if any differences would I see, and what
would be the cause of the differences?

This is a roundabout way of stating the following... If the "restored" (I
refuse to say "reconstructed") signal is truly "analog", then any amplitude
change, no matter how small, in the original must be reproduced in the
output. Otherwise, the output signal isn't analog.

OK. I understand what you are saying. If the reproduced signal is
different from the original signal, then it is digital. So by that
definition, any analog recording that does not reproduce the input
signal exactly is also digital. Since no analog recording technology
ever reproduces the input signal exactly, all analog recording
techniques are also digital. So we have had digital recordings ever
since Edison invented the phonograph. It is time to queue up the
Twilight Zone theme......

[Mark]

On Oct 20, 12:19*pm, "William Sommerwerck"
wrote:
The CD's output is an analogue of the post-DAC signal,
regardless of how analogous that may be of the original
source material.
There is nothing digital about the output. With modern D/A
and dithering, real-world noise will completely mask any
quantization effects.

I see exactly what you're saying, though I don't agree.

So we're down to where the rubber meets the road. Tell me...

If I fed a complex, band-limited signal into an ADC, then properly converted
it back from numbers -- and I had some magic way to overlap the waveforms
for close examination -- what, if any differences would I see, and what
would be the cause of the differences?

This is a roundabout way of stating the following... If the "restored" (I
refuse to say "reconstructed") signal is truly "analog", then any amplitude
change, no matter how small, in the original must be reproduced in the
output. Otherwise, the output signal isn't analog.

Correct, if you overlaped the input and output waveforms the
difference would be only random noise, just like an analog system.

and yes a properly dithered digital system DOES ENCODE DATA BELOW 1
MSB.

And the dither does not just MASK the quantization erros, it elimiates
them..

Think of it the same as the way as in tape recording high frequency
bias eliminates the tape tranfer function distortion,,, it "smears"
them out.. its the very same principal in play..


Mark

[Arny Krueger]

"William Sommerwerck" wrote in
message
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.

Thank you. The signal from a DAC can vary continuously
within the bandwidth of the chosen sampling parameters.

The output of a DAC has only a finite number of levels.

Being an analog signal, the output of a DAC can have any of an infinte
number of levels. Furthermore the levels are not exact but can only be
specified as probabilities.

It cannot vary continuously.

Being an analog signal, the output of a DAC includes noise. If the noise has
non-zero bandwidth, and it always does, then it does vary continuously, even
if the data word being converted remains the same. The continual variations
are due to the noise.

I understand that this is a different meaning of "vary continuously" than
you probably meant, but it is the truth.

Practically speaking, the output voltage of a good DAC is a more adjustable
and continously-variable than almost any other voltage you might commonly
encounter. For example, some of us have had the experience of adjusting the
output of a DC power supply to obtain a certain voltage, as measured with a
4 digit DVM. If you want an exact voltage as indicated by the low-order
digit of this DVM, well lots of luck! The all-analog power supply generally
can't do work that precise. Too much fiddle in the adjusting knob.

The output of a good 16 bit DAC has almost 5 digits of resolution, while the
DVM digit we can't generally match the analog DC source to anyway, has only
4 digits.

Bottom line is that the output of a good 16 bit DAC is more precise and
stable and thus more continuous than just about any analog signal source.
Furthermore, DACs with 20 or more bit resolution are becoming more common
and economical to obtain.

[Arny Krueger]

"William Sommerwerck" wrote in
message

If I fed a complex, band-limited signal into an ADC,

Easy enough to do. I presume that the band limiting would be such that
there is nothing in this signal above the Nyquist frequency of the ADC's
conversion process.

then properly converted it back from numbers -- and I had some
magic way to overlap the waveforms for close examination
-- what, if any differences would I see, and what would
be the cause of the differences?

Been there, done that.

Any additional band limiting due to the normal operation of the ADC is
usually the biggest cause of differences. If there is none, then it might
be hard to find a way to overlap the two signals that would show a visible
difference that was not due to the display device itself.

The obvious tool for displaying the two waves would be an oscilloscope. The
vertical channels of oscilloscopes are usually not all that linear compared
to a good ADC.