Questions about equivalents of audio/video and digital/analog.

[Floyd L. Davidson]

Floyd L. Davidson wrote:
..

Good grief man, it's the SAME DEVICE! The same cable,
the same modules, the same everything. All that changes
is that the user tweaks a pot.

You didn't accurately describe the device the first
time. Which is *exactly* why my answer was conditional
on there being *precisely* 12 voltages per octave. Then
you admit that there are not, that it can be any of an
infinite number of voltages because it is actually
continuously variable.

You just aren't ready to be honest at all, are you.

So I suppose we have to define "device" now as well as everything else.

Somebody else uses that game...

I gave you a picture, a micrograph, of a system that
~can~ produce a signal of precisely stepped
voltages. You promptly pronounce that as "digital". Then

No, I did not. I said *if* what you described was
accurate. It wasn't, and the difference negates
everything.

Why not be honest?

I zoom out, give you a broader picture of the same
system, and all of a sudden we discover sginals that can
morph seamlessly between stepped and non-stepped -

A, yes... a "continuous" set of values... which
clearly makes it analog. If it had actually been just
12 levels, as you initially said, it would be digital.

But you just had to be dishonest.

between "digital" and "analogue". Perhaps that simply
doesn't arise in your universe.

How anyone could miss the distinction is beyond me. But
worse yet, it is *obvious* that you have not missed that
distinction, and instead are merely trying to make a
point with deceitful and abject dishonesty.

Away now for a week, so you will have to figure the rest out by yourself!

How hard did you think it would be to figure you out?

 

[Floyd L. Davidson]

[email protected] (Floyd L. Davidson) writes:

Floyd: Have a nice day. Come visit me if you're ever in Fuquay, NC,
and I think you'll find me somewhat different than the usenet monster
you seem to imagine me to be.

I don't doubt that for a minute, and don't believe for a
second that you are a "usenet monster".

We could talk politics! (Or, maybe we shouldn't... ;-)

 

[Floyd L. Davidson]

Above is the definition of Nyquist's theorem. That is
not claimed to be the standard definition of the Nyquist
Rate, which I keep quoting and you continue to snip in
another dishonest attempt to make it appear other than
it is.

It is most definitely incorrect, and since you've now had
more than adequate time to identify and discuss the error,
I guess I'll have to point it out. The key item in question
from the definition you gave is:

That is not to be found in the standard defintion, which
I gave, for Nyquist Rate. Why is it you must be so
damned dishonest?

The standard definition of Nyquist Rate, as I have shown
several times now, is (from Federal Standard 1037C):

Nyquist rate:
The reciprocal of the Nyquist interval, i.e., the
minimum theoretical sampling rate that fully
describes a given signal, i.e., enables its
faithful reconstruction from the samples.

It says *nothing* that is not absolutely correct. We
might note though, that your discussion was not
absolutely correct either. You merely require the
sampling rate to be greater than 2 times the highest
frequency to be sampled. In fact, there is a measurable
amount greater that is required, which depends on the
quantum size. Just being greater is *not* enough.

I won't attempt to explain that to you, because it is
clearly too technical. ;-)

Here's another standard definition from FS-1037C:

Nyquist interval:
The maximum time interval between equally spaced
samples of a signal that will enable the signal
waveform to be completely determined. (188) Note 1:
The Nyquist interval is equal to the reciprocal of
twice the highest frequency component of the
sampled signal. Note 2: In practice, when analog
signals are sampled for the purpose of digital
transmission or other processing, the sampling rate
must be more frequent than that defined by
Nyquist's theorem, because of quantization error
introduced by the digitizing process. The required
sampling rate is determined by the accuracy of the
digitizing process.

Again, I'll point out that the Theorem is a mathematical
proof of all cases, including as the quantum size
approaches zero; practical examples of sampling must
have 1) greater than zero range of size for quantized levels,
and 2) will have errors in the range (sizes will actually
vary).

 

[Richard Dobson]

Floyd L. Davidson wrote:
..

You didn't accurately describe the device the first
time. Which is *exactly* why my answer was conditional
on there being *precisely* 12 voltages per octave. Then
you admit that there are not, that it can be any of an
infinite number of voltages because it is actually
continuously variable.

When it is used as I described, in its "standard" arrangement, it ~is~
exactly quantised to 12 voltages per octave. Just like the frets on a
guitar. At other times, it may be quantized in some different way, or
all smoothed out. Same hardware, same cable, same interface. Only you
claimed this describes a digital signal.

Perhaps you were too quick to jump to conclusions, and not ask necessary
clarifying questions?
...

A, yes... a "continuous" set of values... which
clearly makes it analog. If it had actually been just
12 levels, as you initially said, it would be digital.

...

So we have at last reached a consensus, that that a signal can be
quantized, just as I have described, and nevertheless be analogue.
"Quantized" of itself is not a sufficient condition for a signal to be
classed as digital. You would require further information to make that
determination. As you have yourself now clearly indicated. The term
"digital" can at last be reserved for where it is truly appropriate.
"Quantized" is a subset, an aspect of, "digital", but it is manifestly
not the same as "digital".

QED. Isn't music wonderful, that it can demonstrate such things!

Richard Dobson

 

[Floyd L. Davidson]

Floyd L. Davidson wrote:
.
When it is used as I described, in its "standard"
arrangement, it ~is~ exactly quantised to 12 voltages

If it is quantized, it is digital. (I cannot see how
what you are describing is quantized though.)

All you have done is adjusted the range of an analog
signal to have 12 steps over a given voltage range.
That has nothing at all to do with quantization.

per octave. Just like the frets on a guitar. At other
times, it may be quantized in some different way, or all
smoothed out. Same hardware, same cable, same
interface. Only you claimed this describes a digital
signal.

I claimed that if you quantize something to a set of
only 12 voltages, that it is digitized. That is a true
fact. If you adjust the range of an analog signal to
have 12 steps, that is not quantizing it and it is not
digital.

Perhaps you were too quick to jump to conclusions, and
not ask necessary clarifying questions?

Why do you thing I put the "if" in my response?
Obviously I know you aren't likely to be honest or
clueful either one.

..

So we have at last reached a consensus, that that a
signal can be quantized, just as I have described, and
nevertheless be analogue.

If the signal is quantized, it is digital. That is a
fact, by the very definition of quantized.

"Quantized" of itself is not a sufficient condition
for a signal to be classed as digital. You would require

It absolutely is.

further information to make that determination. As you
have yourself now clearly indicated. The term "digital"
can at last be reserved for where it is truly
appropriate. "Quantized" is a subset, an aspect of,
"digital", but it is manifestly not the same as
"digital".

Look up any standard definition you like for "quantized",
and every one of them will indicate changing a
continuous range of values to a discrete value from a
finite set.

That of course defines digital too, as you find if you
look at *any* standard definition of the term.

Of course if you make up your own definitions, it can
mean anything you like. I won't know what it is, and
neither will anyone else. You won't be able to
communicate, and will be reduced to posting even more
nonsense.

 

[Bob Myers]

I gave, for Nyquist Rate. Why is it you must be so
damned dishonest?

I'M being dishonest? Floyd, I didn't type those
words - YOU did. If you don't agree with them
now, that's not my problem. Or if you intended
something else, simply say so. But if anyone is
being dishonest, or at the very least doing a rather
transparent job of back-pedaling like a madman,
it's clearly you.

It says *nothing* that is not absolutely correct. We
might note though, that your discussion was not
absolutely correct either. You merely require the
sampling rate to be greater than 2 times the highest
frequency to be sampled. In fact, there is a measurable
amount greater that is required, which depends on the
quantum size. Just being greater is *not* enough.

And you're doing it again. >I< am not the one who said
anything about it relating to the "highest frequency to be
sampled." It's quite plain that I related the minimum
sampling rate to the bandwidth of the signal to be sampled,
not its "highest frequency." And when this correct form
is used, all comes out correctly. Funny how that happens.

I won't attempt to explain that to you, because it is
clearly too technical. ;-)

For one who complains loudly and longly when others
use what you consider to be "insults," you are certainly
quick yourself with the snide and sarcastic comments.
Are you familiar with the story of the pot and the kettle
discussing their color?

Here's another standard definition from FS-1037C:

Nyquist interval:
The maximum time interval between equally spaced
samples of a signal that will enable the signal
waveform to be completely determined. (188) Note 1:
The Nyquist interval is equal to the reciprocal of
twice the highest frequency component of the
sampled signal.

And again, "highest frequency component" is at best
misleading, and at worst completely incorrect.

Note 2: In practice, when analog
signals are sampled for the purpose of digital
transmission or other processing, the sampling rate
must be more frequent than that defined by
Nyquist's theorem, because of quantization error
introduced by the digitizing process. The required
sampling rate is determined by the accuracy of the
digitizing process.

You comments to date, though, have demonstrated
nothing but a complete lack of understanding of just
what the above actually means.

At this point, Floyd, it should be obvious to the few hardy
souls still following this thread that it is you against basically
everyone else. This is either due to your being the sole
person in the entire readership of this group who understands
these matters - which I find highly unlikely, especially given
your inability to actually explain anything - or, as most have
appear to have already agreed to be the case, that you are
simply an utterly unimaginative wretch who attempts to use
recitation of texts learned by rote to make up for a lack of
any real understanding or a willingness to even attempt to learn
something. In any event, you're simply no longer worth the
time, especially when others who might have picked up
something worthwhile from this thread have moved on. It
would seem that the best thing to do is to follow the lead of
other, no doubt wiser heads, and simply killfile you and move
on. Please understand the I bear you no ill will personally,
despite the ill will that you have demonstrated toward myself
and others. If anything, you strike me as a sad case. As has
already been said by another, were you to actually meet and
chat with any of the rest of us, you might find that we're hardly
the ignorant monsters of the internet that you seem to think -
but it seems you are very unlikely to ever know that, either.
In any case, this is the end of the line. I've wasted more than
enough time here, and it's ending now.

Bob M.

 

[Floyd L. Davidson]

I'M being dishonest? Floyd, I didn't type those
words - YOU did.

You stated the definition was for the Nyquist Rate,
which indeed something I had posted. But what you
quoted was not that definition but one for the Nyquist
Theorem, and claimed it gives the standard definition
for the Nyquist Rate. It doesn't, and that was very
clearly a less than honest attempt to make it appear to
be ambiguous. It isn't.

If you don't agree with them
now, that's not my problem.

It is is if you quote A and claim it is B.

Or if you intended
something else, simply say so. But if anyone is
being dishonest, or at the very least doing a rather
transparent job of back-pedaling like a madman,
it's clearly you.

You are not winning points for integrity there either.

And you're doing it again. >I< am not the one who said
anything about it relating to the "highest frequency to be
sampled." It's quite plain that I related the minimum
sampling rate to the bandwidth of the signal to be sampled,
not its "highest frequency." And when this correct form
is used, all comes out correctly. Funny how that happens.

Read what you said again. Regardless, it is not merely
twice the bandwidth either and that is just as wrong as
what you did say.

For one who complains loudly and longly when others
use what you consider to be "insults," you are certainly

I don't complain about insults. I do complain when they
are gratuitous. The above is not a gratuitous insult,
it is simply the truth based on what you have posted.
If you cannot get the basics right, we cannot go on to
anything more technical. I'm sorry if that insults you,
but that a valid statement based only on what you have
said here, and is not at all gratuitous.

quick yourself with the snide and sarcastic comments.
Are you familiar with the story of the pot and the kettle
discussing their color?

I'm sorry that you either do not read English or cannot
be honest. Take your pick, it has to be one or the
other. I have *never* complained just because someone
said something insulting.

And again, "highest frequency component" is at best
misleading, and at worst completely incorrect.

You'd think you would learn by now. A whole lot of
industry experts agreed to putting *that* definition
into the Federal Standard 1037C glossary.

I'm sure you know better than all of them... and that
is why I can find several references to cite that support
exactly what I've claimed, and you can't cite even a single
reference...

You comments to date, though, have demonstrated
nothing but a complete lack of understanding of just
what the above actually means.

So you say, but then you haven't demonstrated nearly the
understanding that I have.

At this point, Floyd, it should be obvious to the few hardy
souls still following this thread that it is you against basically
everyone else.

First, that isn't true at all. Second, I guess it
wouldn't be surprising that you might think a head count
of posters on Usenet has logical significance, given
your other illogical argumentation. I do grant that you
are supported by the loudest and most ignorant of the people
who posted. What does that say for you?

This is either due to your being the sole
person in the entire readership of this group who understands

Or that you can't count?

these matters - which I find highly unlikely, especially given
your inability to actually explain anything - or, as most have
appear to have already agreed to be the case, that you are
simply an utterly unimaginative wretch who attempts to use

Hmmm... gratuitousness is not a blessing when it comes to
insults.

recitation of texts learned by rote to make up for a lack of
any real understanding or a willingness to even attempt to learn
something.

In fact I've probably been working with digital systems
for significantly longer than most of those who have
demonstrated their lack of understanding, including you.

But I am not dumb enough to cite my own opinion as the
sole source of information. You are. I'm not so silly
as to state my opinion without providing references to
authoritative sources that support it. You are.

Now you have the audacity to say that because I can and
do cite authoritative sources to support my opinions,
that it is *I* who has a lack of understanding. You on
the other hand have yet to cite *anything*, supporting
or otherwise. You can't, we all know it; because there
are no authoritative sources that agree with you.

Try being at least a little bit rational in your
comments, please. It is embarrassing for me to have
people who do understand this topic read a thread where
I actually continue stubbornly to argue with someone who
comes up with the above sort of comment and actually
thinks it is valid.

....

In any case, this is the end of the line. I've wasted more than
enough time here, and it's ending now.

You wasted a lot of *everyone's* time. I hope you are
finally being honest, and do cease posting nonsense.

 

[Bill's News]

Toes, my friend, toes. OT. Words to a Mozart horn concerto:

Moses suppose his toeses are roses
But Moses supposes erroneously ...

From "Singin' in the Rain" (1952) soundtrack:
"Moses"
(1952) (uncredited)
(Also known as "Moses Supposes")
Music by Roger Edens
Lyrics by Betty Comden and Adolph Green
Sung and Danced by Gene Kelly and Donald O'Connor

 

[Jerry Avins]

Floyd L. Davidson wrote:

...

It isn't "telephone system thinking", it's Information
Theory. That applies to a great deal more than high
fidelity audio.

So Information Theory tells us that a quantized signal is digital?
Consider the output of the limiters in an FM IF driving a Foster-Seely
discriminator. It has two states -- saturated and zero -- before the
tank that smooths the edges. I guess Information theory says that FM
radio is digital (maybe unless you use an Avins-Seely ratio detector,
but even those work better with at least one limiter).

Jerry

 

[Jerry Avins]

Not that I necessarily agree or disagree with Floyd's original point,
but citing a written reference holds more water than a post from an
individual on a usenet newsgroup, in my opinion.

Floyd maintains that any signal whose values are restricted to a finite
set -- IOW, "quantized" -- is digital. I cited a two-level analog signal
and I can demonstrate a digital signal with a relatively large
continuous range of values. His definitions are simply too restrictive
to accommodate those, and he seems to be having a fit.

Jerry

 

[Randy Yates]

Floyd maintains that any signal whose values are restricted to a
finite set -- IOW, "quantized" -- is digital. I cited a two-level
analog signal and I can demonstrate a digital signal with a relatively
large continuous range of values. His definitions are simply too
restrictive to accommodate those, and he seems to be having a fit.

I've decided that it's not fruitful to continue this discussion since
the knowledge I work with admits anough understanding to get a lot of
real work done. These sorts of discussions take too much time and
produce little or no fruit.

My ability to do work does not depend on others' judgement of the
correctness of my definitions.
--
% Randy Yates % "She has an IQ of 1001, she has a jumpsuit
%% Fuquay-Varina, NC % on, and she's also a telephone."
%%% 919-577-9882 %
%%%% <[email protected]> % 'Yours Truly, 2095', *Time*, ELO
http://home.earthlink.net/~yatescr

 

[Jerry Avins]

Floyd L. Davidson wrote:

...

Nyquist rate:
The reciprocal of the Nyquist interval, i.e., the
minimum theoretical sampling rate that fully
describes a given signal, i.e., enables its
faithful reconstruction from the samples. Note: The
actual sampling rate required to reconstruct the
original signal will be somewhat higher than the
Nyquist rate, because of quantization errors
introduced by the sampling process.

It does not say what you claimed it does.

Do you buy the "because clause? I don't.

"The actual sampling rate required to reconstruct the original signal
will be somewhat higher than the Nyquist rate, because of quantization
errors introduced by the sampling process."

All qualified practitioners will recognize that as wrong. Are you qualified?

Jerry

Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

 

[Floyd L. Davidson]

Floyd L. Davidson wrote:

...


So Information Theory tells us that a quantized signal
is digital? Consider the output of the limiters in an FM
IF driving a Foster-Seely discriminator. It has two
states -- saturated and zero -- before the tank that
smooths the edges. I guess Information theory says that
FM radio is digital (maybe unless you use an Avins-Seely
ratio detector, but even those work better with at least
one limiter).

You aren't making a lick of sense Jerry. That suggests
you don't have even a foggy notion of what you are
talking about.

Tell us exactly what information is encoded in those
"saturated and zero" states?

 

[Jerry Avins]

You aren't making a lick of sense Jerry. That suggests
you don't have even a foggy notion of what you are
talking about.

Tell us exactly what information is encoded in those
"saturated and zero" states?

Very little; the information is in the zero crossings. The signal is
quantized in amplitude. Is it digital or not? If not, does your
definition hold?

Jerry

 

[Floyd L. Davidson]

Floyd maintains that any signal whose values are
restricted to a finite set -- IOW, "quantized" -- is
digital. I cited a two-level analog signal and I can
demonstrate a digital signal with a relatively large
continuous range of values. His definitions are simply
too restrictive to accommodate those, and he seems to be
having a fit.

I'll admit to a really great fit of laughter at that one!

You are so thoroughly confused that it is hilarious.

The recognized standard definitions say that a quantized
signal is digital. You can indeed have a two-level
analog signal, but the fact is that the *possible*
values are infinite (all values between your two listed
ones, for example). You cannot possibly have a digital
signal with a continuous range of values (large or
small, relative or otherwise).

I've cited multiple credible sources that agree with
what I say. You can't cite even one. There are none.

 

[Floyd L. Davidson]

Floyd L. Davidson wrote:

...


Do you buy the "because clause? I don't.

"The actual sampling rate required to reconstruct the
original signal will be somewhat higher than the Nyquist
rate, because of quantization errors introduced by the
sampling process."

All qualified practitioners will recognize that as wrong. Are you qualified?

All qualified practitioners will recognize that you are wrong, and
obviously unqualified.

It is in fact a correct statement. Do you know what quantization
distortion is?

 

[Floyd L. Davidson]

Very little; the information is in the zero
crossings.

The voltage amplitude has nothing to do with whether
the signal is digital or analog. It can be anything,
with any characteristics you'd like to imagine.
That is because it carries no information.

The signal is quantized in amplitude.

First, it is not. It varies between two voltages, and
does so continuously (and apparently too quickly for a
slow person to follow, eh?). But since the variations
contain no information and therefore do not represent
symbols of any kind, the amplitude does not determine
whether the signal is digital or analog.

Is it
digital or not? If not, does your definition hold?

We can't tell Jerry. You have not stated anything that
describes the symbols set. The information is carried
by some other characteristic of that signal (e.g., phase
or frequency). Not knowing if it carries only discrete
values from a finite set, or if the symbols are
continuously variable, we just don't know what it is.

This is *very* basic...

 

[Floyd L. Davidson]

Jerry Avins wrote:

(snip)


This sounds like what I previously tried to describe as quantized
but not sampled. The signal has two states, but the transition
can happen at any time.

Jerry's signal does not have two states. Voltage
amplitude is *not* what determines signal "state"
(value) with an FM signal.

 

[glen herrmannsfeldt]

Jerry Avins wrote:

(snip)

So Information Theory tells us that a quantized signal is digital?
Consider the output of the limiters in an FM IF driving a Foster-Seely
discriminator. It has two states -- saturated and zero -- before the
tank that smooths the edges. I guess Information theory says that FM
radio is digital (maybe unless you use an Avins-Seely ratio detector,
but even those work better with at least one limiter).

This sounds like what I previously tried to describe as quantized
but not sampled. The signal has two states, but the transition
can happen at any time.

-- glen

 

[Arny Krueger]

Floyd L. Davidson wrote:

...


Do you buy the "because clause? I don't.

You do well to disagree with it. It is false. The errors that are introduced
are aliasing, not quantization errors. You could change the size of the
quantization steps any which way you want, and the aliasing would still be
there.

"The actual sampling rate required to reconstruct the
original signal will be somewhat higher than the Nyquist rate, because of
quantization errors introduced by the sampling process."

All qualified practitioners will recognize that as wrong.

Agreed. It's an incorrect statement for the reason I stated above.