[Richard Dobson]

Floyd L. Davidson wrote:
...
...
Indeed, if we do look at a "valid standard definition"
for the term of art,

quantization:

A process in which the continuous range of values
of an analog signal is sampled and divided into
nonoverlapping (but not necessarily equal)
subranges, and a discrete, unique value is assigned
to each subrange.

From Federal Standard 1037C.

We can see that it *clearly* does mean to make it digital.
That is the *only* purpose for quantization.



This only addresses values, it does not address time. So how would you
classify this signal:

the output of a standard 1V/oct (voltage control) music keyboard - a
monophonic (= non-overlapping) series of stepped voltages corresponding
precisely to the 12-tone equal-termperament subdivisions of the octave.
This control signal is typically applied to the frequency control input
of an analogue voltage-controlled oscillator (VCO; think MiniMoog), in
order to synthesise tones at the specified frequency. Thus, values are
quantized. There is no time quantization (no clock) - the notes can be
played at any time, and changed at any speed (presumably within the
limits of the human player).

I would call this an analogue signal; it meets exactly the definition
above, it is only you who extrapolates from it the notion of "digital".
And manifestly, making a digital signal is ~not~ the only purpose for
quantization!

And if the Federal Standard had meant to make it mean "digital" surely,
given its importance, they would have said so.

Richard Dobson

[Randy Yates]

(Floyd L. Davidson) writes:
Randy Yates wrote:
(Floyd L. Davidson) writes:
[...]
Nyquist's theorem:
A theorem, developed by H. Nyquist, which states
that an analog signal waveform may be uniquely
reconstructed, without error, from samples taken at
equal time intervals. The sampling rate must be
equal to, or greater than, twice the highest
frequency component in the analog signal. Synonym
sampling theorem.
[...]
You are looking at the definition of the Theorem, not
the definition of the rate, and then saying the
definition of "Nyquist Rate" should not have the words
"equal to".

I agree this is a definition of the theorem, but in the definition of
the theorem it states the definition of the rate, and that statement
is wrong.

I can see, though, that you and others in this thread have become very
unbalanced. It's not good for your mental health. You're in Alaska,
right? Get outside and go bear hunting or something! Pull your chair
away from the computer!
--
% 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 %
%%%% % 'Yours Truly, 2095', *Time*, ELO
http://home.earthlink.net/~yatescr

[Floyd L. Davidson]

(Don Pearce) wrote:
On Fri, 24 Aug 2007 17:15:40 -0800, (Floyd L.
Davidson) wrote:

(Don Pearce) wrote:
On Fri, 24 Aug 2007 12:38:04 -0800, (Floyd L.
Davidson) wrote:

I'm still unable to comprehend how you think that was a
"howler". Once again have yet to explicitly state what
you thought was wrong with the definiton provided and
you do not give an alternate.

Right let me spell it out for you. That glossary explained the Nyquist
frequency. As part of that definition it explicitly gave the
requirement that the Nyquist frequency be EQUAL to twice the highest
frequency being reproduced.

Here is the definition it has of the *rate* (you incorrectly call it
the Nyquist "frequency"):

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.

Here is the theorem:

Nyquist's theorem:
A theorem, developed by H. Nyquist, which states
that an analog signal waveform may be uniquely
reconstructed, without error, from samples taken at
equal time intervals. The sampling rate must be
equal to, or greater than, twice the highest
frequency component in the analog signal. Synonym
sampling theorem.

It appears that you are somewhat confused as to what is
being defined. The definition for the Nyquist rate says
absolutely nothing about being equal to anything.
Instead it says it is the minimum rate that will "fully
describe" the signal.

That is 100%, definitively incorrect.

What is not correct about it. What do you claim is
correct instead?

Explain *your* definition. (Oh, and do so for all values
of sampling rate as the size of the quantum steps approach
zero.)

Whatever, I can't tell what you are disagreeing with. You
read one definition and claim it is something else, you don't
say what you think is wrong with it or what would be right.

Maybe you disagree with the way the words are spelled,
with the use of the term "analog" or you just can't
understand what it says...

It is a howler made by many
people who don't understand sampling. To find it in a list that you
regard as definitive must give you cause to consider the quality of
the rest of the list.

You are the howler.

You probably should look up Shannon's "Communication in
the Presence of Noise" from 1949.

Over to you - your turn to explain to me how that was in fact correct.

It appears to me that the definition they gave is
precisely correct, and again *you* are abjectly
clueless.

I guessed you would think it was correct. You can't sample at a rate
equal to twice the frequency you are sampling. The wanted signal has
collided with its image and you can't disambiguate them. Thank you for
showing us that you are clueless.

The definition they have for Nyquist Rate does not suggest
anything different.

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.

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)

[Floyd L. Davidson]

(Don Pearce) wrote:
On Fri, 24 Aug 2007 20:54:55 -0800, (Floyd L.
Davidson) wrote:

"Bob Myers" wrote:
"Randy Yates" wrote in message
...
One thing that's incorrect and has been discussed many times here
before is that the inequality must be strict. That is, the wording
should have omitted "equal to".

That's one error. It's not the only one.

So demonstrate where there is another!

No need. You claim your definitions to be correct because they appear
to be borne out by a list you claim to be definitive. The list has
been shown to be errored, so your authority has vanished. Deal with
it.

All you would need to show the definitions I posted are
not valid is provide a conflicting definition of each
from an authoritative source.

You haven't, because there are none.

You claim that the list itself is errored, but you
cannot show an error in it other than one of your own
imagination, where you think one definition implies that
of another... except it does not and the specific
"error" that you claim exists is not part of the
definition given for the term in question.

The point still remains that even if you can find some
error some place in the list, the agrument from
authority is valid for the definitions of the terms
"analog" and "digital" until you can show some other
expert source that disagrees.

You can't.

You apparently can't learn the rules of logic either,
as that has all been explained previously.

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)

[Don Pearce]

On Sat, 25 Aug 2007 06:16:10 -0800, (Floyd L.
Davidson) wrote:


I guessed you would think it was correct. You can't sample at a rate
equal to twice the frequency you are sampling. The wanted signal has
collided with its image and you can't disambiguate them. Thank you for
showing us that you are clueless.

The definition they have for Nyquist Rate does not suggest
anything different.

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.

--

Floyd, be a good boy and **** off you lying little toad. Don't bother
replying any more because you are now in my killfile along with Phil
Alison. Why is it always the six-fingered inbreds from the outback
that cause the most grief around here?

d

--
Pearce Consulting
http://www.pearce.uk.com

[Floyd L. Davidson]

Richard Dobson wrote:
Floyd L. Davidson wrote:
..
..
Indeed, if we do look at a "valid standard definition"
for the term of art,
quantization:
A process in which the continuous range of values
of an analog signal is sampled and divided into
nonoverlapping (but not necessarily equal)
subranges, and a discrete, unique value is assigned
to each subrange.
From Federal Standard 1037C.
We can see that it *clearly* does mean to make it
digital.
That is the *only* purpose for quantization.


This only addresses values, it does not address time. So
how would you classify this signal:

the output of a standard 1V/oct (voltage control) music
keyboard - a monophonic (= non-overlapping) series of

Monophonic measn one channel. The output could be
monophonic and still be overlapping.

stepped voltages corresponding precisely to the 12-tone
equal-termperament subdivisions of the octave. This

It is a digital output if there are precisely 12 voltages
per octave.

control signal is typically applied to the frequency
control input of an analogue voltage-controlled
oscillator (VCO; think MiniMoog), in order to synthesise
tones at the specified frequency.

An analog output device, that has a digital control circuit.

Thus, values are
quantized.

The DC control voltage is quantized. It is digital.
The tone output from the VCO is not quantized and is
analog.

from it the notion of "digital". And manifestly, making
a digital signal is ~not~ the only purpose for
quantization!

And if the Federal Standard had meant to make it mean
"digital" surely, given its importance, they would have
said so.

I believe that what you had to say there demonstrates
why you are so utterly confused on the topic of analog
and digital.

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)

[Floyd L. Davidson]

Randy Yates wrote:
(Floyd L. Davidson) writes:
Randy Yates wrote:
(Floyd L. Davidson) writes:
[...]
Nyquist's theorem:
A theorem, developed by H. Nyquist, which states
that an analog signal waveform may be uniquely
reconstructed, without error, from samples taken at
equal time intervals. The sampling rate must be
equal to, or greater than, twice the highest
frequency component in the analog signal. Synonym
sampling theorem.
[...]
You are looking at the definition of the Theorem, not
the definition of the rate, and then saying the
definition of "Nyquist Rate" should not have the words
"equal to".

I agree this is a definition of the theorem, but in the definition of
the theorem it states the definition of the rate, and that statement
is wrong.

It does not define the rate. There *is* a formal
definition of the rate provided, and it is absolutely
correct.

If *you* read something into it that is clearly in
conflict with what they say, it is time to question your
interpretation of what you read into it.

I can see, though, that you and others in this thread have become very
unbalanced.

Sorry sonny, but gratuitous insults are almost always a
reflection of the mental state of the people who make
them. They indicate self fears by those who make them.

One of the things that should be obvious from the length
of this thread is that *I* am not the one wallowing all
over creation with a variety of different and
conflicting attempts, all of which fail, to prove
something that obviously isn't true. I have simply been able
to followup on the initial statements that I made with
*logical* and rational continuations of exactly the same
thing, without contradictions, without variations, and
without wearing a tin foil hat.

It's not good for your mental health. You're in Alaska,
right? Get outside and go bear hunting or something! Pull your chair
away from the computer!

You might be right about mental health problems. Given
the amount of therapeutic noise that you and others have
generated as you slam back and forth trying to imagine a
hole in the basic wall you've butted up against... you
probably should seek a professional evaluation.


--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)

[Floyd L. Davidson]

(Don Pearce) wrote:
On Sat, 25 Aug 2007 06:16:10 -0800, (Floyd L.
Davidson) wrote:


I guessed you would think it was correct. You can't sample at a rate
equal to twice the frequency you are sampling. The wanted signal has
collided with its image and you can't disambiguate them. Thank you for
showing us that you are clueless.

The definition they have for Nyquist Rate does not suggest
anything different.

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.

--

Floyd, be a good boy and **** off you lying little toad. Don't bother
replying any more because you are now in my killfile along with Phil
Alison. Why is it always the six-fingered inbreds from the outback
that cause the most grief around here?

I see that you find it difficult to handle facts and
logic when you meet up with someone who can sort them
out at the drop of a hat.

Gratuitous insults are virtually always a fair
indication of the reflection the writer sees in a
mirror, so my only comment on your statements above is
that you seem to have a truly horrifying mental image of
yourself to use as an example when you want to insult
someone.

Whatever, if you had had your facts straight to begin
with, you wouldn't feel so crushed now.

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)

[Don Bowey]

On 8/25/07 7:30 AM, in article , "Don
Pearce" wrote:

On Sat, 25 Aug 2007 06:16:10 -0800, (Floyd L.
Davidson) wrote:


I guessed you would think it was correct. You can't sample at a rate
equal to twice the frequency you are sampling. The wanted signal has
collided with its image and you can't disambiguate them. Thank you for
showing us that you are clueless.

The definition they have for Nyquist Rate does not suggest
anything different.

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.

--

Floyd, be a good boy and **** off you lying little toad. Don't bother
replying any more because you are now in my killfile along with Phil
Alison. Why is it always the six-fingered inbreds from the outback
that cause the most grief around here?

d

The problem with people like Floyd is that, when you get frustrated from his
moronic misleading replies and lies, he will internalize that he has "won,"
and will feel empowered to continue in kind. Someone else posted that his
goal is to win at any cost (including his veracity) and facts will be
twisted or ignored to meet that goal. It would be unfortunate if he posts
his views to Wikapedia.

Phil, on the other hand, seems to be technically correct, though sometimes
vague. The latter is, I believe, to leave something for the OP to finish.
I always read his posts.

[Al in Dallas]

On Fri, 24 Aug 2007 13:55:54 -0700, Jim Kelley
wrote:

"Floyd L. Davidson" wrote:

So when will any of you be able to cite credible support
for your claims that the standard definitions of analog
and digital signals/data are not valid.

Here are some valid standard defintions:

"quantize - to subdivide into small but measurable increments."
(Merriam Webster's Collegiate Dictionary, Tenth Edition)

Note that in the definition, there appears no mention of assigning a
value. Assigning a value would then be considered a part of a
separate and distinct process of converting to digital form, as in

"digital - of, or relating to data in the form of numerical digits",

and as opposed to

"analog - of, relating to, or being a mechanism in which data is
represented by continuously variable physical quantities."

There's a huge difference between the jargon of experts and the
language of common people. Doctors and surgeons don't restrict
themselves to the definitions found in ordinary dictionaries. Neither
do experts in communications theory. If that's the best cite the
audiophiles have, then they're admitting they're hobbiests.

--
Al in St. Lou

[Don Pearce]

On Sat, 25 Aug 2007 08:37:40 -0700, Don Bowey
wrote:

On 8/25/07 7:30 AM, in article , "Don
Pearce" wrote:

On Sat, 25 Aug 2007 06:16:10 -0800, (Floyd L.
Davidson) wrote:


I guessed you would think it was correct. You can't sample at a rate
equal to twice the frequency you are sampling. The wanted signal has
collided with its image and you can't disambiguate them. Thank you for
showing us that you are clueless.

The definition they have for Nyquist Rate does not suggest
anything different.

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.

--

Floyd, be a good boy and **** off you lying little toad. Don't bother
replying any more because you are now in my killfile along with Phil
Alison. Why is it always the six-fingered inbreds from the outback
that cause the most grief around here?

d

The problem with people like Floyd is that, when you get frustrated from his
moronic misleading replies and lies, he will internalize that he has "won,"
and will feel empowered to continue in kind. Someone else posted that his
goal is to win at any cost (including his veracity) and facts will be
twisted or ignored to meet that goal. It would be unfortunate if he posts
his views to Wikapedia.

Phil, on the other hand, seems to be technically correct, though sometimes
vague. The latter is, I believe, to leave something for the OP to finish.
I always read his posts.


Indeed Phil is usually technically correct, but his posts are simply
so strewn with the filthiest invective that I am prepared to forego
the occasional nugget. Floyd, unfortunately, doesn't have even that
redeeming feature.

d

--
Pearce Consulting
http://www.pearce.uk.com

[Al in Dallas]

On Sat, 25 Aug 2007 06:30:52 -0800, (Floyd L.
Davidson) wrote:

Richard Dobson wrote:
Floyd L. Davidson wrote:
..
..
Indeed, if we do look at a "valid standard definition"
for the term of art,
quantization:
A process in which the continuous range of values
of an analog signal is sampled and divided into
nonoverlapping (but not necessarily equal)
subranges, and a discrete, unique value is assigned
to each subrange.
From Federal Standard 1037C.
We can see that it *clearly* does mean to make it
digital.
That is the *only* purpose for quantization.


This only addresses values, it does not address time. So
how would you classify this signal:

the output of a standard 1V/oct (voltage control) music
keyboard - a monophonic (= non-overlapping) series of

Monophonic measn one channel. The output could be
monophonic and still be overlapping.

"Monophonic" is a term of art in the music world. In that world, it
means one note at a time, in contrast with polyphony, which was the
revolution that gave the world Western Music as we know it today.

stepped voltages corresponding precisely to the 12-tone
equal-termperament subdivisions of the octave. This

It is a digital output if there are precisely 12 voltages
per octave.

control signal is typically applied to the frequency
control input of an analogue voltage-controlled
oscillator (VCO; think MiniMoog), in order to synthesise
tones at the specified frequency.

An analog output device, that has a digital control circuit.

Thus, values are
quantized.

The DC control voltage is quantized. It is digital.
The tone output from the VCO is not quantized and is
analog.

from it the notion of "digital". And manifestly, making
a digital signal is ~not~ the only purpose for
quantization!

And if the Federal Standard had meant to make it mean
"digital" surely, given its importance, they would have
said so.

I believe that what you had to say there demonstrates
why you are so utterly confused on the topic of analog
and digital.

They seem to be experts in music, not communications theory. Perhpas
their definition of "digital" is extremely useful for them. However, I
don't understand why they've come into an electronics group and
started an argument about electronics terms of art.

--
Al in St. Lou

[Richard Dobson]

Floyd L. Davidson wrote:
...

It is a digital output if there are precisely 12 voltages
per octave.


Users may insert a simple control (may be called "Key Follow" but is
basically just an analogue level control) that can reduce/expand the
size of the steps so that 24 notes, say, cover one octave; or 11 notes
cover an octave and a fifth. This is of course an analogue control, as
the amount of key follow (and hence the division of the octave achieved
by each step) is continuously variable. Which is another way of saying
that the quantization itself is infinitely variable. The VCO is
calibrated such that a change of one octave results in a pitch change of
one octave. Users may well subvert that calibration (and the whole
12-note octave convention) for creative purposes.


control signal is typically applied to the frequency
control input of an analogue voltage-controlled
oscillator (VCO; think MiniMoog), in order to synthesise
tones at the specified frequency.


An analog output device, that has a digital control circuit.



The key aspect of Voltage Control (as designed by Robert Moog) is that
any module can control (and be controlled by) any other. Inputs are
content-agnostic - the VCO does not have a "digital control circuit" -
just a control circuit to which can be connected any analogue input. In
short - the VCO's control inputs are all analogue; so that, for example,
by inserting a filter (slew-rate limiter) between the keyboard output
and the VCO input, you get a portamento from one note to the next, not a
straight jump. You can equally connect the output of one VCO to the
frequency input of another one, to do FM (Vibrato etc). These are ~all~
analogue signals, being handled by analogue electronics. The
electronics, indeed, on many early synths was notorious for being
somewhat unstable, so that oscillator frequencies adn voltage ranges
could drift as the machine warmed up or cooled down. Later technology
brought in the DCO - the Digitally-Controlled analogue Oscillator, to
eliminate such instabilities.

You might find this company's products interesting:

http://www.analoguesystems.co.uk/modules.htm

See for example the "Voltage Quantiser" and "Voltage controlled slew
limiter" modules.

Richard Dobson

[Floyd L. Davidson]

Richard Dobson wrote:
Floyd L. Davidson wrote:
..
It is a digital output if there are precisely 12
voltages
per octave.


Users may insert a simple control (may be called "Key
Follow" but is basically just an analogue level control)
that can reduce/expand the size of the steps so that 24

So there are not precisely 12 voltages per octave, but
rather there are now ever many you choose, and the
adjustment is continuously variable.

You described one device before, and now describe
a different device...

How do you expect a valid answer if you purposely
distort the question with false information?

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)

[Eric Jacobsen]

On Fri, 24 Aug 2007 23:32:21 GMT, Richard Dobson
wrote:

Don Pearce wrote:
..

And I guess someone will ask "how does one decide this or that source
is 'valid'? ". Wikipedia? Opportunities for more fruitless bipolar
arguments there, I fancy!


There is another way. Work it out for yourself, from first principles.


Well, I seem to have spent my life doing that, as much as I am able. But
agreeing upon terminology, the core vocabulary of the subject, is by
definition a group exercise. Otherwise, people take a term and
arbitrarily make it mean what they want it to mean, which seems to be
the issue here. Converging to an agreement would be great, but after an
avalanche of posts on this thread, people seem no closer now that at the
start. Calling each other "delusional"! First principles? Which ones?! :-)

Richard Dobson

Floyd's been beaten up about this before, but he just keeps coming
back for more.

"Standards" and associated definitions are created for their own use
in a specific context within the scope of the standard, and no
further. Stating that there is "a standard" definition that should
apply to everyone, everywhere, belies a substantial misunderstanding
of what "standards" are and how they work.

Get enough experienced, competent, expert comm people in the same room
and pretty soon you'll have enough different points of view that
you'll have some pretty substantial disagreements on the meaning of
things as fundamental as SNR.

All that means is that it's smart to have a short dialogue to synch up
contexts and definitions before you proceed, and if you do sense that
communication is breaking down due to different definitions, you stop
for just long enough to synch up and then move on. Pounding one's
fist on the table and demanding that one definition is superior to
another is not productive, IMHO. Clearly it's important to
understand what one means when using a term, but there's certainly no
central global clearinghouse that magically decides what terms mean.
If there was it'd be obsolete in a week because the technology and the
areas where it's used is constantly changing.

If one can't just express what they mean or manage to synch up somehow
with the folks with whom they're communicating, then that individual
is just going to have a harder time making progress with people. I
think the current thread is a pretty good example of that. For those
sick *******s among us that like this sort of thing it's been pretty
doggone amusing, too.

Ah, well, this is the sort of thing that'll just always be an issue as
long as people are involved.

Eric Jacobsen
Minister of Algorithms
Abineau Communications
http://www.ericjacobsen.org

[Richard Dobson]

Floyd L. Davidson wrote:
...

Users may insert a simple control (may be called "Key
Follow" but is basically just an analogue level control)
that can reduce/expand the size of the steps so that 24


So there are not precisely 12 voltages per octave, but
rather there are now ever many you choose, and the
adjustment is continuously variable.

You described one device before, and now describe
a different device...

How do you expect a valid answer if you purposely
distort the question with false information?


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.

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

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

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

Richard Dobson

[Randy Yates]

(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.
--
% Randy Yates % "...the answer lies within your soul
%% Fuquay-Varina, NC % 'cause no one knows which side
%%% 919-577-9882 % the coin will fall."
%%%% % 'Big Wheels', *Out of the Blue*, ELO
http://home.earthlink.net/~yatescr

[Eric Jacobsen]

On Fri, 24 Aug 2007 20:54:15 -0800, (Floyd L.
Davidson) wrote:

I don't believe you understand the theorem.
Incidentally, Nyquist didn't come up with the theorem,
hence you really don't want to look at what Nyquist
wrote much as at Shannon's mathematical proof of what
Nyquist proposed.

Floyd, you're out of your league here.

Nyquist's and Shannon's careers overlapped a little bit at Bell Labs
and they collaborated a bit on some things...Nyquist reviewed a lot of
Shannon's early work, IIRC.

In any case, Nyquist clearly created the defining early work in this
area, and the correctness of that work has given it a lot of staying
power. It's not really been superceded by anything. Saying "you
really don't want to look at what Nyquist wrote" belies a deficiency
in your own understanding and makes me suspect that your motivation is
primarily to poison that well for anyone else who might want to
reference it. You seem to like to define your own playing field
smaller and smaller and claim "I'm absolutely correct in this tiny
little circle" and somehow try to make that relevant to everybody
else. At least, that's my take on it.

Shannon's work was primarily in laying the foundation for Information
Theory, and his sampling theorem was pretty much just re-working
Nyquist's theorem from an Information Theory perspective. That's
important partly because context changes interpretation. e.g., you
can't expect a single definition of a term to apply universally to all
possible cases when the contexts and technologies are constantly
changing.

Eric Jacobsen
Minister of Algorithms
Abineau Communications
http://www.ericjacobsen.org

[John E. Hadstate]

"Eric Jacobsen" wrote in message
...

"Pearls before swine". However, I understand that the
reason you do it is to keep people like Floyd from confusing
the less-informed readers who will come along later.

I have a rule that changes the color of messages from
certain posters to Hot Pink, and marks them Read and
Ignored. Floyd is a charter member of that group and he is
really very pretty in pink ;-)

[Bob Myers]

"Floyd L. Davidson" wrote in message
...
"Bob Myers" wrote:
"Floyd L. Davidson" wrote in message
...
Here is the theorem:

Nyquist's theorem:
A theorem, developed by H. Nyquist, which states
that an analog signal waveform may be uniquely
reconstructed, without error, from samples taken at
equal time intervals. The sampling rate must be
equal to, or greater than, twice the highest
frequency component in the analog signal. Synonym
sampling theorem.

The standard definition of Nyquist Rate from the
glossary is not incorrect.

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:

The sampling rate must be
equal to, or greater than, twice the highest
frequency component in the analog signal.

....and it contains a very common misunderstanding of
Nyquist's theorem. The sampling rate is NOT required
to be "equal to or greater than twice the highest frequency
component in the analog signal," even ignoring the
problematic "equal to" case in the above. Rather, the
sampling rate must be twice the BANDWIDTH of the
signal in question. For example, if one is sampling an
AM signal which comprises a 10 MHz carrier modulated
by an audio signal of 0 - 5 kHz, the highest frequency
component would be expected to be at 10.005 MHz -
yet sampling at 20.010 MHz or higher is NOT required
to fully recover the information contained within this
signal. The carrier itself carries no information, so
that's all there is to it. The AM signal in question could
be sampled at a MUCH lower rate (in this case, a bit
greater than 10 kHz would suffice), and still be fully
captured. This in fact forms the basis for what's often
referred to as "digital downcoversion" in receivers, and
also is the basis for the "equivalent-time sampling" operation
of digital sampling oscilloscopes.

To be sure, if you were trying to accurately capture
the form of a single cycle (or even a few cycles) of
a "10 MHz signal," you'd need a sampling rate
far in excess of 20 MHz - but that is also per the
theorem, since such signals' complete spectrums are
very, very wide.

As you seem to accept such things only if "authoritative
sources" are cited, I'd suggest you check the application
notes provided by either Agilent Technologies or
Tektronix on their web sites, re their digital oscilloscopes.
Hopefully, you will consider either of these companies as
knowing a bit about what they're talking about in this area.


Bob M.

[Bob Myers]

"Eric Jacobsen" wrote in message
...

Floyd's been beaten up about this before, but he just keeps coming
back for more.

"Standards" and associated definitions are created for their own use
in a specific context within the scope of the standard, and no
further. Stating that there is "a standard" definition that should
apply to everyone, everywhere, belies a substantial misunderstanding
of what "standards" are and how they work.

Yes, if there's one thing that's become adequately clear from
this discussion, it's the fact that Floyd has never ever seen any
actual standards-setting body at work.

It puts one in mind of what is traditionally said about not wanting
to watch either sausage or law being made...

Bob M.

[Floyd L. Davidson]

Richard Dobson wrote:
Floyd L. Davidson wrote:
..

Users may insert a simple control (may be called "Key
Follow" but is basically just an analogue level control)
that can reduce/expand the size of the steps so that 24
So there are not precisely 12 voltages per octave, but
rather there are now ever many you choose, and the
adjustment is continuously variable.
You described one device before, and now describe
a different device...
How do you expect a valid answer if you purposely
distort the question with false information?


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 http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)

[Floyd L. Davidson]

Randy Yates wrote:
(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 http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)

[Floyd L. Davidson]

"Bob Myers" wrote:
"Floyd L. Davidson" wrote in message
...
"Bob Myers" wrote:
"Floyd L. Davidson" wrote in message
...
Here is the theorem:

Nyquist's theorem:
A theorem, developed by H. Nyquist, which states
that an analog signal waveform may be uniquely
reconstructed, without error, from samples taken at
equal time intervals. The sampling rate must be
equal to, or greater than, twice the highest
frequency component in the analog signal. Synonym
sampling theorem.

The standard definition of Nyquist Rate from the
glossary is not incorrect.

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:

The sampling rate must be
equal to, or greater than, twice the highest
frequency component in the analog signal.

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).

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)

[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]

Richard Dobson wrote:
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

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.

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.

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.

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)

[Bob Myers]

"Floyd L. Davidson" wrote in message
...
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?

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]

"Bob Myers" wrote:
"Floyd L. Davidson" wrote in message
...
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?

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.

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.

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.

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

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.

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.

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...

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.

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.

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)

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

[Jerry Avins]

Randy Yates wrote:
"Bob Myers" writes:

"Floyd L. Davidson" wrote in message
...

Lets be clear... The definitions I cited are standard.
I posted 5 or 6 varied references to the same definitions.
And everyone knows, ""standards" are holy.

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

[Randy Yates]

Jerry Avins writes:

Randy Yates wrote:
"Bob Myers" writes:

"Floyd L. Davidson" wrote in message
...

Lets be clear... The definitions I cited are standard.
I posted 5 or 6 varied references to the same definitions.
And everyone knows, ""standards" are holy.
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.

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 %
%%%% % '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]

Jerry Avins wrote:
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).

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?

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)

[Jerry Avins]

Floyd L. Davidson wrote:
Jerry Avins wrote:
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).

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

[Floyd L. Davidson]

Jerry Avins wrote:

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 http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)

[Floyd L. Davidson]

Jerry Avins wrote:
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?

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 http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)

[Floyd L. Davidson]

Jerry Avins wrote:
Floyd L. Davidson wrote:
Jerry Avins 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?

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 http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)

[Floyd L. Davidson]

glen herrmannsfeldt wrote:
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.

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



--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)

[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]

"Jerry Avins" wrote in message

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.

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.