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"glen herrmannsfeldt" wrote in
message
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.


The signal might be thought of as being quantized in the aplitude domain,
but it is clearly not quantized in the time domain. For a signal to be
quantized, it has to be fully quantized, that is quantized in both the time
domain and the amplitude domain.



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On Sun, 26 Aug 2007 21:37:37 -0800, (Floyd L.
Davidson) wrote:

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.


So why have you kept harping on the idea of discrete voltages?

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?).


Can you describe a real-world two-state, "digital" signal by your
definition that doesn't behave that way?

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.


You just keep drawing your circle of constraints smaller and smaller.
Yes, it is possible to keep reducing a circle of scope to vanishing
small radius, but it stops being relevant to anyone else long before
that. I'm guessing the folks left in this thread are here mostly for
amusement or with the hope of folks other than you learning something.
It's clear you're not going to get anything out of it.

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...


It's very basic obfuscation on your part. If you can't have a
meaningful dialogue with anyone without continually updating the
constraints and definitions to fit your view, you're not going to have
a lot of influence.

Eric Jacobsen
Minister of Algorithms
Abineau Communications
http://www.ericjacobsen.org
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Floyd L. Davidson wrote:
Actually, they are good definitions, but they are *not*
"valid standard definitions" for this discussion.


They are valid for any discussion in which I care to use them. But
thanks for your input.

jk

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"Arny Krueger" wrote:
"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.


That is another technically *absurd* statement. Can you
explain how one gets aliasing under the circumstances
stated?

"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.


The reason you state is incorrect, and that should be
obvious.

Incidentally, I took a look at "Telecommunications
System Engineering, Third Edition" by Roger L. Freeman
to see what definitions he uses. He cited another
reference, and quoted this as "the Nyquist sampling
theorem",

"If a band-limited signal is sampled at regular
intervals of time at a rate equal to or higher than
twice the highest significant signal frequency, then
the sample contains all the information of the
original signal. The original signal may then be
reconstructed by use of a low-pass filter."

His source is "/Reference/ /Data/ /for/ /Radio/ /Engineers/,
6th Ed., ITT/Howard W. Sams, Indianapolis, 1976.

That's two more absolutely credible sources who say you need
to learn more about this.

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)
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Jim Kelley wrote:
Floyd L. Davidson wrote:
Actually, they are good definitions, but they are *not*
"valid standard definitions" for this discussion.


They are valid for any discussion in which I care to use
them. But thanks for your input.


That is true. You can use any definition for any word
you like, in Alice's Wonderland.

And you won't be understood by anyone else, which seems to
be the point of many who post this sort of drivel.

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


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Floyd L. Davidson wrote:
And you won't be understood by anyone else, which seems to
be the point of many who post this sort of drivel.


Perhaps true insofar as the word might not be understood by someone
who also does not know how to use a dictionary. Admitedly, I had not
accounted for that possibility. 8-|

jk

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Jim Kelley wrote:
Floyd L. Davidson wrote:
And you won't be understood by anyone else, which seems to
be the point of many who post this sort of drivel.


Perhaps true insofar as the word might not be understood
by someone who also does not know how to use a
dictionary.


You, for example.

Admitedly, I had not accounted for that
possibility. 8-|


That was fairly obvious, and still is.

You are implying that any dictionary definition (hence
not your unique Alice in Wonderland definition) is
correct in any context. That is not the way a
dictionary is properly used.

"Digital", for example, has at least three different
definitions. You want to be able to pull any one of
them out of a hat, and say that it means what it
means... But that is back to Alice in Wonderland.

The word "Digital" is a Term of Art, and your common
language dictionary definition is not valid in a
technical discussion.

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

"Arny Krueger" wrote:
"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.


That is another technically *absurd* statement. Can you
explain how one gets aliasing under the circumstances
stated?


Real world signals have finite, not zero bandwidth. That means that they
have sidebands. As soon as any of the sidebands are at or exceed the nyquist
rate, there is distortion of the wave because of aliasing. Therefore the
sample rate has to be somewhat higher than the sample rate of the signal,
which is characterized by its carrier frequency.

"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.


The reason you state is incorrect, and that should be
obvious.


Incidentally, I took a look at "Telecommunications
System Engineering, Third Edition" by Roger L. Freeman
to see what definitions he uses. He cited another
reference, and quoted this as "the Nyquist sampling

theorem",

"If a band-limited signal is sampled at regular
intervals of time at a rate equal to or higher than
twice the highest significant signal frequency, then
the sample contains all the information of the
original signal. The original signal may then be
reconstructed by use of a low-pass filter."


Congrats to Freeman for avoiding making a different statement that avoids
the error made in the previous statement.



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Floyd L. Davidson wrote:
You are implying that any dictionary definition (hence
not your unique Alice in Wonderland definition) is
correct in any context.


I'm not implying that at all. I'm simply saying that the terms and
definitions I used are correct in the context in which I used them.

jk



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Arny Krueger wrote:


Incidentally, I took a look at "Telecommunications
System Engineering, Third Edition" by Roger L. Freeman
to see what definitions he uses. He cited another
reference, and quoted this as "the Nyquist sampling


theorem",


"If a band-limited signal is sampled at regular
intervals of time at a rate equal to or higher than
twice the highest significant signal frequency, then
the sample contains all the information of the
original signal. The original signal may then be
reconstructed by use of a low-pass filter."



Congrats to Freeman for avoiding making a different statement that avoids
the error made in the previous statement.


Although, unless the original signal is a squarewave, saying "the
sample contains all the information of the original signal" may be
overly optimistic.

jk



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Randy Yates wrote:
Jerry Avins writes:
Randy Yates wrote:
"Bob Myers" writes:
"Floyd L. Davidson" wrote:
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.


So you figure that posting invalid definitions to Usenet
or proving that you cannot understand standard
definitions won't be a problem for you?

It's amazing though, just who finds what with web
searches. Anyone who reads what you've had to say
here... well it could easily affect your work!

Whatever, an article with out of hand invalid statements
quoted from the two people who clearly post from an IEEE
host is a really good place to throw out something that
I've asked them for repeatedly, and they have weaseled
around the question in odd ways: IEEE definitions of
"digital" and "analog". I think it was Randy Yates who
claimed they had been posted but did admit that nobody
had ever cited IEEE as a source.

Well, it appears that the person who posted it was me.
IEEE apparently uses the standard definitions which I
have posted from other sources.

However, here is a very interesting discussion from an
IEEE dictionary:

An analog signal implies /continuity/,
as contrasted to a digital signal that
is concerned with /discrete/ states.
Often the means of carrying information
is the distinguishing feature between
analog and digital. The information
content of an analog signal is conveyed
by the value or magnitude of some
characteristics of the signal such as
phase, amplitude, frequency of the
voltage, the amplitude or duration of a
pulse, and so on. To extract the
information, it is necessary to compare
the value or magnitude of the signal to
a standard. The information content of
a digital signal is concerned with
discrete states of the signal, such as
the presence or absence of a voltage, a
contrast in the open or closed position,
or a hole or no hole in certain
positions on a card. The digital signal
is given meaning by assigning numerical
values or other information to the
various possible combinations of the
discrete states of the signal."

"The New IEEE Standard Dictionary of Electrical and
Electronic Terms", 5th ed., IEEE Std. 100-1992,
IEEE Press, New York, 1992.

I'm quoting it from Roger L. Freeman's "Telecommunications
System Engineering", 3rd ed., 1996.

For one, it clearly shows the FM-signal-through-a-limiter
example given by Jerry very clearly to be exactly as my
analysis indicated, and not at all what Jerry said.
Also they clearly state that the values assigned to a
digital symbol need not be "numerical" as someone argued
repeatedly in earlier posts.

Another example of credible references that support each
and every point that I've made. And it again highlights
that none of those saying it isn't so can find *anything*
credible to support their statements. (And that of course
is why it is not "fruitful" to argue with me. I don't buy
rotten fruit.)

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)
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Jim Kelley wrote:
Floyd L. Davidson wrote:
You are implying that any dictionary definition (hence
not your unique Alice in Wonderland definition) is
correct in any context.


I'm not implying that at all. I'm simply saying that
the terms and definitions I used are correct in the
context in which I used them.


The were easily demonstrated as *not* correct in the
context used, and your whole discussion since has been
to argue the implication as stated above.

Do you read what you write?

Lets review it:

Here are some valid standard defintions:

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

Your source is for common English, it is not defining
the Terms of Art used in technical discussions. This
*is* a discussion about those terms of art, not about
common English word usages. To whatever degree your
cited definition differs from the standard definitions
I've cited, you are wrong because of the context.

Here is the standardized definition for "quantization"

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.

And you looked at a invalid definition for this context and
stated:

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

Obviously the Term of Art used in technical discussion
does indeed mean there *must* be a value assigned. In
fact it makes no sense at all unless that step is
included.

Then you go on to produce other, equally invalid in this
context, definitions for other terms of art:

"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."

The first definition states that the form must be "of
numerical digits", and that is simply unnecessary. It
must be assigned a "value". The value can be numerical
digits, but it can be otherwise too. The point is that
it must be from a finite set of discrete values. (I
used the example of flags, where the value might be a
square flag, a round flag, or a triangular flag. No
numerical digits at all.)

As you can see, knowing how to use a dictionary is
vitally important...

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

Floyd L. Davidson wrote:

You are implying that any dictionary definition (hence
not your unique Alice in Wonderland definition) is
correct in any context.


I'm not implying that at all. I'm simply saying that
the terms and definitions I used are correct in the
context in which I used them.



The were easily demonstrated as *not* correct in the
context used, and your whole discussion since has been
to argue the implication as stated above.

Do you read what you write?

Lets review it:

Here are some valid standard defintions:

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

Your source is for common English, it is not defining
the Terms of Art used in technical discussions.


It perfectly defines the term as I use it.

This
*is* a discussion about those terms of art, not about
common English word usages. To whatever degree your
cited definition differs from the standard definitions
I've cited, you are wrong because of the context.


I don't use the term the way your reference defines it.

Here is the standardized definition for "quantization"

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.

And you looked at a invalid definition for this context and
stated:

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

Obviously the Term of Art used in technical discussion
does indeed mean there *must* be a value assigned. In
fact it makes no sense at all unless that step is
included.


It seems there are a great many things that don't make sense to you.
If a "term of art" were to be defined in such a way that it
contradicts the definition for the exact same term as it is published
in Websters dictionary, I would be inclined to disregard it.

Then you go on to produce other, equally invalid in this
context, definitions for other terms of art:

"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."

The first definition states that the form must be "of
numerical digits", and that is simply unnecessary. It
must be assigned a "value". The value can be numerical
digits, but it can be otherwise too. The point is that
it must be from a finite set of discrete values. (I
used the example of flags, where the value might be a
square flag, a round flag, or a triangular flag. No
numerical digits at all.)


As you can see, knowing how to use a dictionary is
vitally important...


Perhaps almost as important as realizing at which side of the analog
to digital convertor you're looking.

jk

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"Arny Krueger" wrote:
"Floyd L. Davidson" wrote in message

"Arny Krueger" wrote:
"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.


That is another technically *absurd* statement. Can you
explain how one gets aliasing under the circumstances
stated?


Real world signals have finite, not zero bandwidth. That means that they
have sidebands. As soon as any of the sidebands are at or exceed the nyquist


You meant 1/2 the Nyquist Rate. But the point is that
the statement above excludes all such signals. It
specifically says that the actual sampling rate will be
*above* the Nyquist Rate. There will not be any
aliasing at samplying rate which is higher than the
Nyquist Rate.

Therefore there *cannot* *be* any problem with aliasing.
There will, however, be quantization distortion.

rate, there is distortion of the wave because of aliasing. Therefore the
sample rate has to be somewhat higher than the sample rate of the signal,
which is characterized by its carrier frequency.


What???? The rate of the signal is characterized by its carrier
frequency? 1) What if there is no carrier? 2) What if there is
a carrier and there is an upper sideband?

Clearly the carrier is not what characterizes the signal
frequency or the sample rate.

"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.


The reason you state is incorrect, and that should be
obvious.


Incidentally, I took a look at "Telecommunications
System Engineering, Third Edition" by Roger L. Freeman
to see what definitions he uses. He cited another
reference, and quoted this as "the Nyquist sampling
theorem",


"If a band-limited signal is sampled at regular
intervals of time at a rate equal to or higher than
twice the highest significant signal frequency, then
the sample contains all the information of the
original signal. The original signal may then be
reconstructed by use of a low-pass filter."


Congrats to Freeman for avoiding making a different statement that avoids
the error made in the previous statement.


He said "at a rate equal to or higher than twice the
highest significant signal frequency". That is
precisely the same as the one you claim is wrong:

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.

Did you forget what you were arguing previously?

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)
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Will you two just get a room. Please.




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Jim Kelley wrote:
Floyd L. Davidson wrote:
Jim Kelley wrote:

Floyd L. Davidson wrote:

You are implying that any dictionary definition (hence
not your unique Alice in Wonderland definition) is
correct in any context.

I'm not implying that at all. I'm simply saying that
the terms and definitions I used are correct in the
context in which I used them.

The were easily demonstrated as *not* correct in the
context used, and your whole discussion since has been
to argue the implication as stated above.
Do you read what you write?
Lets review it:
Here are some valid standard defintions:
"quantize - to subdivide into small but
measurable increments."
(Merriam Webster's Collegiate Dictionary, Tenth Edition)
Your source is for common English, it is not defining
the Terms of Art used in technical discussions.


It perfectly defines the term as I use it.


Yes. And in a technical discussion, you are off in Alice's
Wonderland.

This
*is* a discussion about those terms of art, not about
common English word usages. To whatever degree your
cited definition differs from the standard definitions
I've cited, you are wrong because of the context.


I don't use the term the way your reference defines it.


But nobody else, unless they are also in Wonderland,
knows what you mean if you don't use standard definitions.
And in a *technical* discussion, that means the technical
term of art, not the common English term.

....
Obviously the Term of Art used in technical discussion
does indeed mean there *must* be a value assigned. In
fact it makes no sense at all unless that step is
included.


It seems there are a great many things that don't make
sense to you.


Why people want to make absurd claims about terminology
is one of them. Maybe you can explain that...

If a "term of art" were to be defined in
such a way that it contradicts the definition for the
exact same term as it is published in Websters
dictionary, I would be inclined to disregard it.


It isn't a contradiction. The term of art is usually a
stricter definition. It is certainly going to be more
rigorous.

Regardless, to ignore it is abject foolishness. Can you
imagine a lawyer in court ignoring all of the term of
art definitions in favor of the standard common English
dictionary definitions? The other side would be
ecstatic... and the judge would probably throw the
lawyer out the door!

If you don't want to use the technical terminology, why
not just step out the door and avoid all technical
discussions? You won't make sense, it it does become
annoying.

As you can see, knowing how to use a dictionary is
vitally important...


Perhaps almost as important as realizing at which side
of the analog to digital convertor you're looking.


Exactly. Now, if you don't want to use terms in a what
that will be understood, how can you even tell which
side is which?

You'll end up making grossly trivial errors the way
Jerry Arvins and a couple of others here do...

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)
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Default Questions about equivalents of audio/video and digital/analog.

"Dan Coby" wrote:
Will you two just get a room. Please.


I believe that we already have one. Right here.

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)
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Default Questions about equivalents of audio/video and digital/analog.

Floyd L. Davidson wrote:
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?


Yes. How is it related to sampling rate? Why would sampling exactly at
the Nyquist rate be adequate if the quantization were fine enough? Don't
you read the crap you cut and paste?

Don't answer. I'm through.

Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
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Default Questions about equivalents of audio/video and digital/analog.


"Jim Kelley" wrote in message
...
"If a band-limited signal is sampled at regular
intervals of time at a rate equal to or higher than
twice the highest significant signal frequency, then
the sample contains all the information of the
original signal. The original signal may then be
reconstructed by use of a low-pass filter."


Although, unless the original signal is a squarewave, saying "the
sample contains all the information of the original signal" may be
overly optimistic.


Did you not understand the words "band-limited signal",
or the words "sampled at a rate equal to or higher than
twice the *highest significant signal frequency* " ?

And I guess you have no idea how Fourier analysis applies to square waves
either?

MrT.


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

Floyd L. Davidson wrote:
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.


There are only two possible values for the output of a hard limiter.
Make something of it. OTOH, bu using CMOS logic and varying the power
supply voltage randomly between the limits of 3 volts and 18 volts, I
can hive a digital signal an infinite number of values.

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


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

Jerry Avins wrote:
Floyd L. Davidson wrote:
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.


There are only two possible values for the output of a hard limiter.


There are an infinite number of errors in your
statement.

An FM signal, such as you specified earlier, does not
encode information as voltage levels.

The phase or frequency contains the information, and a
hard limiter does not restrain the values to only two.

There might in fact be only two, but we don't know that
and the limiter does not assure that.

For a typical FM signal modulated by audio we might well
have an infinite number of possible values after the
signal is passed through a hard limiter. It is also
true that we could modulate it with something else that
has only two values, and then it would be digital.

Make something of it. OTOH, bu using CMOS logic and varying the power
supply voltage randomly between the limits of 3 volts and 18 volts, I
can hive a digital signal an infinite number of values.


That would be funny if you were joking, but Jerry I
realize that you are dead serious.

Lets use, just for this example, an RS-232C interface
definition. We could use some other interface, and it
would always be the same result. But since your voltage
range fits within the range of voltages that are correct
for RS-232 (and few others do) we'll use that.

If your signal varies between +3 and +18 volts, it has
exactly 1 value. That is a space signal. If you want
it to be mark, you can vary it between -3 and -18
volts...

Yes, you do have an infinite number of *voltages*
between 3 and 18, but only one signal value.

Tell me Jerry, do you actually get paid to do
engineering?

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)
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Default Questions about equivalents of audio/video and digital/analog.

On Mon, 27 Aug 2007 20:52:29 -0800, (Floyd L.
Davidson) wrote:

Jerry Avins wrote:
Floyd L. Davidson wrote:
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.


There are only two possible values for the output of a hard limiter.


There are an infinite number of errors in your
statement.

An FM signal, such as you specified earlier, does not
encode information as voltage levels.


We all know that. You seem to be trying to catch up, though.

The phase or frequency contains the information, and a
hard limiter does not restrain the values to only two.


The example is of the sort of hard limiting achieved by something like
a Class-C amplifier where the limiter does limit the output values to
two voltages. This is a specific function in an important stage of a
common FM demodulator architecture. Jerry spelled this out pretty
clearly, but you're having a real hard time understanding the folks
here who know what they're talking about. Pay attention.

There might in fact be only two, but we don't know that
and the limiter does not assure that.


A limiter in this sense absolutely does assure that. The idea is to
remove any AM modulation, which is fine in an FM signal since there's
no information in the amplitude. Limiting it to two voltages helps
clean up a lot of noise and distortion with a relatively simple
implementation.

For a typical FM signal modulated by audio we might well
have an infinite number of possible values after the
signal is passed through a hard limiter.


You're not following; the limiter limits it to two voltages. That's
what a limiter does in this case. It's a pretty common function,
with a very common symbol used frequently in block diagrams and
schematics. Jerry's question, which you've been dancing around
without any sort of coherent response that I can detect, is whether
the signal at the output of the limiter is digital or analog according
to you? The situations that you've been describing and explanations
for definitions that you claim to be absolute and universal don't seem
to work well for some of these cases.

It is also
true that we could modulate it with something else that
has only two values, and then it would be digital.


LOL!

Okay, so we can modulate it FM with audio, and limit it, and then it's
two voltages, but it's not digital. But if we modulate with only two
values, then it's two voltages and it is digital! Right?

Floyd, you must have a mind for detail to be able to keep track of all
the goofy, byzantine distinctions you try to make up.

Make something of it. OTOH, bu using CMOS logic and varying the power
supply voltage randomly between the limits of 3 volts and 18 volts, I
can hive a digital signal an infinite number of values.


That would be funny if you were joking, but Jerry I
realize that you are dead serious.

Lets use, just for this example, an RS-232C interface
definition.


Why not just stick with the example Jerry gave? You have a digital
CMOS circuit, the high voltage is one, the low voltage is zero (or,
just to keep you comfortable, mark and space). Vary the supply
voltage, over some range, so that the voltages for one and zero (oh,
wait, mark and space) change. You could even stretch the example a
little bit so that the receiving CMOS gate supply voltage stays at the
same level, because there's a lot of slop built into these things. So
does it stop being a digital circuit because there is an infinite
range of voltages being used?

We could use some other interface, and it
would always be the same result. But since your voltage
range fits within the range of voltages that are correct
for RS-232 (and few others do) we'll use that.

If your signal varies between +3 and +18 volts, it has
exactly 1 value. That is a space signal. If you want
it to be mark, you can vary it between -3 and -18
volts...

Yes, you do have an infinite number of *voltages*
between 3 and 18, but only one signal value.


So, in your mind, is it digital or analog?


Eric Jacobsen
Minister of Algorithms
Abineau Communications
http://www.ericjacobsen.org
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"Floyd L. Davidson" wrote in message
...
Yes, you do have an infinite number of *voltages*
between 3 and 18,


Only in theory, in practice nothing is infinite in this universe.
Noise will set the resolution limit.

MrT.


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"Mr.T" MrT@home wrote:
"Floyd L. Davidson" wrote in message
...
Yes, you do have an infinite number of *voltages*
between 3 and 18,


Only in theory, in practice nothing is infinite in this universe.
Noise will set the resolution limit.


The noise itself has an infinite number of possible
values, and therefore even if the signal itself is
supposed to be just 1 value, add the noise and there are
an infinite number of values.

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)
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Default Questions about equivalents of audio/video and digital/analog.

Floyd L. Davidson wrote:
Jerry Avins wrote:

(snip on analog vs. digital signals)

There are only two possible values for the output of a hard limiter.


There are an infinite number of errors in your
statement.


An FM signal, such as you specified earlier, does not
encode information as voltage levels.


The phase or frequency contains the information, and a
hard limiter does not restrain the values to only two.


Previously when I was in these discussions instead of
digital vs. analog it was modulated (and so in need of
a modem) vs. not modulated. Passing a digital signal
through an analog channel is said to require a modulated
carrier. In that case, it is an analog representation
of a digital signal.

There might in fact be only two, but we don't know that
and the limiter does not assure that.


For a typical FM signal modulated by audio we might well
have an infinite number of possible values after the
signal is passed through a hard limiter. It is also
true that we could modulate it with something else that
has only two values, and then it would be digital.


(snip)

OK, so when is it a direct representation of a digital
signal instead of an analog channel with a modulated
signal? I might believe it for NRZ, but just about
anything else I don't.

Lets use, just for this example, an RS-232C interface
definition. We could use some other interface, and it
would always be the same result. But since your voltage
range fits within the range of voltages that are correct
for RS-232 (and few others do) we'll use that.


If your signal varies between +3 and +18 volts, it has
exactly 1 value. That is a space signal. If you want
it to be mark, you can vary it between -3 and -18
volts...


Say you have a space alien trying to decode the signal.
In this case, as with other NRZ signals, there are two choices:
The higher voltage represents '1', or the lower voltage.
There also needs to be a way to know when the value is there,
usually an external clock signal.

How about this for a digital signal: I take an analog telephone
and say into it either "one" or "zero". That is, the english words.

The information content is digital, but the representation is
pretty definitely analog.

-- glen



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

glen herrmannsfeldt wrote:
Floyd L. Davidson wrote:

Previously when I was in these discussions instead of
digital vs. analog it was modulated (and so in need of
a modem) vs. not modulated.


Okay...

Passing a digital signal
through an analog channel is said to require a modulated
carrier.


That is not necessarily true. The problem has nothing
to with analog vs digital. It instead a question of
whether DC can be amplified or only AC, and over what
bandwidth, on that analog channel.

In that case, it is an analog representation
of a digital signal.


I'm not sure that makes sense for an electrical signal.
A digital representation of an analog signal does (the
result when restored is called quasi-analog though).

However, I do think that your last example, discussed
below, appears to be a valid example of an analog
representation of digital signal, except it is not an
electronic representation...

There might in fact be only two, but we don't know that
and the limiter does not assure that.


For a typical FM signal modulated by audio we might well
have an infinite number of possible values after the
signal is passed through a hard limiter. It is also
true that we could modulate it with something else that
has only two values, and then it would be digital.


(snip)

OK, so when is it a direct representation of a digital


Any time you encode discrete values from a finite set,
that is digital. Period. Whether it can be passed over
an inherently analog channel or not is fairly
meaningless. The voltage, phase, or whatever that is
encoded with the information may have only a discrete
set of values for the information, but they obviously
take on an infinite number of possible values for the
characteristic itself.

Hence the voltage on a binary system carries only two
values, on and off, but that is the value of the
information. The voltage that is encoded might be +3 to
+15 for an on and -3 to -15 for an off. And when the state
switch from on or off to the opposite value, those
voltages do not change instantly, and they do cover an
infinite number of voltages during that change.

That is an infinite number of possible voltages, but
they have a value of either on or off.

The information values are what makes it a digital signal.

But indeed, you can pass that signal through an analog
amplifier. Depending on the characteristics, it may or
may not destroy the information. Obviously the
amplifier would need to pass DC voltages unless we
encode the information in a way that guarantees some set
minimum time interval between state changes (T1 digital
carrier systems typically do that, for example).

signal instead of an analog channel with a modulated
signal? I might believe it for NRZ, but just about
anything else I don't.


If the digital signal has DC components it can be
modulated onto an analog carrier to pass through an AC
coupled analog channel. It could also be re-encoded
in a manner that will pass through an AC only channel,
and then be transmitted over the same AC coupled analog
channel.

Most physical channels are inherently analog! Wire
cables and fiber optic cables are two examples. Digital
signals are commonly sent via either of them.

Lets use, just for this example, an RS-232C interface
definition. We could use some other interface, and it
would always be the same result. But since your voltage
range fits within the range of voltages that are correct
for RS-232 (and few others do) we'll use that.


If your signal varies between +3 and +18 volts, it has
exactly 1 value. That is a space signal. If you want
it to be mark, you can vary it between -3 and -18
volts...


Say you have a space alien trying to decode the signal.
In this case, as with other NRZ signals, there are two choices:
The higher voltage represents '1', or the lower voltage.


Yes, but you aren't saying anything of significane to the
rest of the discussion.

There also needs to be a way to know when the value is there,
usually an external clock signal.


This is even farther off track. (Not that it isn't
interesting! clock synchronization is a really
fascinating topic in my opinion.)

How about this for a digital signal: I take an analog telephone
and say into it either "one" or "zero". That is, the english words.

The information content is digital, but the representation is
pretty definitely analog.


The information content encoded and sent electronically
is not digital. There is nothing digital about the
sound of your voice. And while "one" and "two" might be
digital, those are not the values that are encoded and
set over the telephone wires. It is the difference
between "one" and "uno" that is encoded, and that is an
infinite number of variations.

But, I do think you've stated a good case for the analog
representation of a digital signal! The "one/two" code
(like Paul Revere's message) is in fact digital, and it
is being sent over an analog channel, which is the
channel from your mouth to your ear!

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)
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Default Questions about equivalents of audio/video and digital/analog.

Floyd L. Davidson wrote:

(I wrote)
Floyd L. Davidson wrote:


Previously when I was in these discussions instead of
digital vs. analog it was modulated (and so in need of
a modem) vs. not modulated.


Okay...


Passing a digital signal
through an analog channel is said to require a modulated
carrier.


That is not necessarily true. The problem has nothing
to with analog vs digital. It instead a question of
whether DC can be amplified or only AC, and over what
bandwidth, on that analog channel.

(snip)

When DSL started to become popular, there was discussion
that a DSL modem wasn't a modem because the DSL signal
was digital. Everyone (just about) knows that v.90 needs
a modem because it goes through the voice telephone system.

But DSL does use a modulated carrier, and the box is
a modem.

However, I do think that your last example, discussed
below, appears to be a valid example of an analog
representation of digital signal, except it is not an
electronic representation...


(snip)

Any time you encode discrete values from a finite set,
that is digital. Period. Whether it can be passed over
an inherently analog channel or not is fairly
meaningless. The voltage, phase, or whatever that is
encoded with the information may have only a discrete
set of values for the information, but they obviously
take on an infinite number of possible values for the
characteristic itself.


But why phase modulation, for example? Two reasons
that I see. One is that the channel does not have
the appropriate frequency response, and the other is
the need for a clock. Consider 10baseT ethernet.
Phase modulation allows for transformer coupling that
is needed to avoid ground loops and ensure a balanced
signal to avoid EMI problems. Using synchronous
phase modulation allows for the recovery of the clock
from the signal, which is also important.

If, for example, the signal was not modulated and one
decided to send 1000000 zeros in a row, there would be
no way to recover the clock to know how many zeros
were sent. If you can't separate the bits, you are
losing an important part of a digital signal.

Hence the voltage on a binary system carries only two
values, on and off, but that is the value of the
information. The voltage that is encoded might be +3 to
+15 for an on and -3 to -15 for an off. And when the state
switch from on or off to the opposite value, those
voltages do not change instantly, and they do cover an
infinite number of voltages during that change.


That is an infinite number of possible voltages, but
they have a value of either on or off.


The information values are what makes it a digital signal.


But indeed, you can pass that signal through an analog
amplifier. Depending on the characteristics, it may or
may not destroy the information.


If the information is destroyed, then the signal
didn't get through.

Obviously the
amplifier would need to pass DC voltages unless we
encode the information in a way that guarantees some set
minimum time interval between state changes (T1 digital
carrier systems typically do that, for example).


signal instead of an analog channel with a modulated
signal? I might believe it for NRZ, but just about
anything else I don't.


If the digital signal has DC components it can be
modulated onto an analog carrier to pass through an AC
coupled analog channel. It could also be re-encoded
in a manner that will pass through an AC only channel,
and then be transmitted over the same AC coupled analog
channel.


Most physical channels are inherently analog! Wire
cables and fiber optic cables are two examples. Digital
signals are commonly sent via either of them.


Some can pass a digital signal without modulation,
others can't.

(snip)

-- glen

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

On Mon, 27 Aug 2007 23:54:40 -0800, (Floyd L.
Davidson) wrote:


Any time you encode discrete values from a finite set,
that is digital. Period.


In that case, the quantum nature of matter means that *everything*
*everywhere* is digital. So no need to talk about this analog stuff
anymore.

Eric Jacobsen
Minister of Algorithms
Abineau Communications
http://www.ericjacobsen.org
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Default Questions about equivalents of audio/video and digital/analog.

In article ,
(Floyd L. Davidson) wrote:

Most physical channels are inherently analog! Wire
cables and fiber optic cables are two examples. Digital
signals are commonly sent via either of them.


I'll probably regret jumping in here, but:

The *message* may be digital, but the *signals* are most definitely
analog.

Isaac
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"isw" wrote in message
...
In article ,
(Floyd L. Davidson) wrote:

Most physical channels are inherently analog! Wire
cables and fiber optic cables are two examples. Digital
signals are commonly sent via either of them.


I'll probably regret jumping in here, but:

The *message* may be digital, but the *signals* are most definitely
analog.


The SIGNALS are electrical or optical.

Bob M.




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

glen herrmannsfeldt wrote:
Floyd L. Davidson wrote:
glen herrmannsfeldt wrote:
Floyd L. Davidson wrote:


Previously when I was in these discussions instead of
digital vs. analog it was modulated (and so in need of
a modem) vs. not modulated.


Okay...


Passing a digital signal
through an analog channel is said to require a modulated
carrier.


That is not necessarily true. The problem has nothing
to with analog vs digital. It instead a question of
whether DC can be amplified or only AC, and over what
bandwidth, on that analog channel.

(snip)

When DSL started to become popular, there was discussion
that a DSL modem wasn't a modem because the DSL signal
was digital. Everyone (just about) knows that v.90 needs
a modem because it goes through the voice telephone system.


Everyone who thinks that is wrong. A v.90 modem is
digital on both sides, and will not pass through a
"voice telephone system". It requires a *digital*
switching system to work. V.90 is not a D/A-A/D
protocol, it is a digital level encoding scheme.

(Indeed, a lot equipment originally used for 56Kbps
digital services was often called a "modem" by not just
customers, but also by telecommunications people. None
of them were technically modems, they were all level
changers, with digital signals in one side and out the
other in a different, but equally digital, format.)

But DSL does use a modulated carrier, and the box is
a modem.


But is that because it has a digital side, or is that
because the bandwidth restrictions of the channel cannot
pass the input signal due to the low frequency
components?

It is a bandwidth problem, and it has nothing at all to
do with digital or analog.

In fact, a v.90 modem sends an entirely digital signal
down that very same line. Of course in the process it
necessarily uses the same bandwidth that on a DSL is
allocated for a normal voice channel, and therefore
while DSL can co-exist separately with POTS the v.90
modem cannot.

However, I do think that your last example, discussed
below, appears to be a valid example of an analog
representation of digital signal, except it is not an
electronic representation...


(snip)

Any time you encode discrete values from a finite set,
that is digital. Period. Whether it can be passed over
an inherently analog channel or not is fairly
meaningless. The voltage, phase, or whatever that is
encoded with the information may have only a discrete
set of values for the information, but they obviously
take on an infinite number of possible values for the
characteristic itself.


But why phase modulation, for example?


It has *nothing* do to with whether it is digital or
analog...

Two reasons
that I see. One is that the channel does not have
the appropriate frequency response, and the other is
the need for a clock.


Phase modulation does not uniquely conserve bandwidth
(Manchester encoding uses twice the bandwidth of NRZ,
for example) nor is it unique in the ability to recover
a clocking rate from the data.

Consider 10baseT ethernet.
Phase modulation allows for transformer coupling that
is needed to avoid ground loops and ensure a balanced
signal to avoid EMI problems.


Alternate mark inversion (AMI) provides the same
characteristics.

But while all of this is indeed very interesting
stuff... it has *nothing* to do with analog vs. digital
or the definitions of either. I don't see any point to
your discussion.

Using synchronous
phase modulation allows for the recovery of the clock
from the signal, which is also important.


That is one way to do it, but there are others.

If, for example, the signal was not modulated and one
decided to send 1000000 zeros in a row, there would be
no way to recover the clock to know how many zeros


There would be no way *only* if you select an encoding
scheme such as NRZ. Manchester encoding provides for
easy clock recover, but so do other encoding schemes.

were sent. If you can't separate the bits, you are
losing an important part of a digital signal.


That is not true. Consider AMI using B8ZS encoding...


Hence the voltage on a binary system carries only two
values, on and off, but that is the value of the
information. The voltage that is encoded might be +3 to
+15 for an on and -3 to -15 for an off. And when the state
switch from on or off to the opposite value, those
voltages do not change instantly, and they do cover an
infinite number of voltages during that change.


That is an infinite number of possible voltages, but
they have a value of either on or off.


The information values are what makes it a digital signal.


But indeed, you can pass that signal through an analog
amplifier. Depending on the characteristics, it may or
may not destroy the information.


If the information is destroyed, then the signal
didn't get through.


Of course. But it has *nothing* to do with the amplifier
being analog. There *are* analog amplifiers that will not
destroy it. (And that is no different for analog data either,
which will also be destroyed if the amplifier does not have
suitable characteristics to pass it.)

Obviously the
amplifier would need to pass DC voltages unless we
encode the information in a way that guarantees some set
minimum time interval between state changes (T1 digital
carrier systems typically do that, for example).


Don't ignore what has already been made clear!

signal instead of an analog channel with a modulated
signal? I might believe it for NRZ, but just about
anything else I don't.


If the digital signal has DC components it can be
modulated onto an analog carrier to pass through an AC
coupled analog channel. It could also be re-encoded
in a manner that will pass through an AC only channel,
and then be transmitted over the same AC coupled analog
channel.


Most physical channels are inherently analog! Wire
cables and fiber optic cables are two examples. Digital
signals are commonly sent via either of them.


Some can pass a digital signal without modulation,
others can't.


But is has nothing to do with analog vs. digital. If the
amplifier cannot handle the signal's bandwidth, it makes no
difference if the signal is analog or digital, it will not
"pass" the data from input to output.

You are trying to impute something to digital that is
actually common to analog as well.

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)
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Default Questions about equivalents of audio/video and digital/analog.

isw wrote:
In article ,
(Floyd L. Davidson) wrote:

Most physical channels are inherently analog! Wire
cables and fiber optic cables are two examples. Digital
signals are commonly sent via either of them.


I'll probably regret jumping in here, but:

The *message* may be digital, but the *signals* are most definitely
analog.


That is not correct. Whether a message is or not
digital is entirely unrelated to whether the signal used
to transmit it is analog or digital (and it can indeed
be either, without regard to the message).

"Message" specifically means a complete set of data
formatted for transmission, and is not related to
analog/digital data signals.

For the term "signal", you can choose from several
definitions (FS-1037C):

signal:

1. Detectable transmitted energy that can
be used to carry information.

2. A time-dependent variation of a
characteristic of a physical phenomenon,
used to convey information.

3. As applied to electronics, any transmitted
electrical impulse.

4. Operationally, a type of message, the text
of which consists of one or more letters,
words, characters, signal flags, visual
displays, or special sounds, with prearranged
meaning and which is conveyed or transmitted
by visual, acoustical, or electrical means.

Hence you can see that using "message" and "signal" in
the same sentence is bound to cause confusion in the
context of this particular discussion. It simply does
not mean what you were thinking of. When used properly
the terms "signal" and "message" would be something like
this, "Our actions are intended to send Congress a
message, and we wish to signal our intense displeasure
with corruption."

But we've been discussing signals that meet either
the 1st or 2nd definition above, and specifically not
numbers 3 or 4.

In context, the signals are either digital or analog,
and which they are depends mostly on whether the data,
or individual parts of the information (message) that
the signal carries, is digital or analog.

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

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


"Floyd L. Davidson" wrote in message
...
Yes, you do have an infinite number of *voltages*
between 3 and 18,


Only in theory, in practice nothing is infinite in this universe.
Noise will set the resolution limit.


The noise itself has an infinite number of possible
values,


OK, prove it.

and therefore even if the signal itself is
supposed to be just 1 value, add the noise and there are
an infinite number of values.


Only for those who failed mathematics at high school.

MrT.


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

In article ,
(Floyd L. Davidson) wrote:

isw wrote:
In article ,
(Floyd L. Davidson) wrote:

Most physical channels are inherently analog! Wire
cables and fiber optic cables are two examples. Digital
signals are commonly sent via either of them.


I'll probably regret jumping in here, but:

The *message* may be digital, but the *signals* are most definitely
analog.


That is not correct. Whether a message is or not
digital is entirely unrelated to whether the signal used
to transmit it is analog or digital (and it can indeed
be either, without regard to the message).


You specifically mentioned "wire cables and fiber optic cables", so lets
talk about those and ignore other possible transmission media.

In both of those (as they are actually used in the real world),
communication is accomplished by the propagation along them of
electromagnetic fields; never anything else.

Doesn't matter one whit whether you turn the field on and off, or vary
the amplitude or any other characteristic of it continuously, as a means
of sending a message from one end to the other, those fields can take on
*any value* from the maximum level injected into the cable by the
transmitter down to far below the ambient noise level, depending (for
example) on the length of cable being used. IOW, those signals are,
without exception, *analog*.

Isaac


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

"Mr.T" MrT@home wrote:
"Floyd L. Davidson" wrote in message
...
Yes, you do have an infinite number of *voltages*
between 3 and 18,

Only in theory, in practice nothing is infinite in this universe.
Noise will set the resolution limit.


The noise itself has an infinite number of possible
values,


OK, prove it.


Eh? You seem to have misunderstood what was said. It
has *nothing* to do with resolution.

It has to do with the fact that no matter what level the
noise is, it could be either a little bit more or a
little bit less. That means whatever value you think
you have resolved, could in fact actually have had two
other possible values.

Which of course means that the number of voltages
between 3 and 18 is indeed infinite, with or without
noise.

Your ability to resolve those values is an entirely
different topic.

and therefore even if the signal itself is
supposed to be just 1 value, add the noise and there are
an infinite number of values.


Only for those who failed mathematics at high school.


Add a random number with an infinite range of possible
values to *anything*, and you have a result with an
infinite range of possible values. Pretty simple math.

I'm sorry to hear that you didn't do well with math in
highschool.

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)
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Default Questions about equivalents of audio/video and digital/analog.

isw wrote:
(Floyd L. Davidson) wrote:
In both of those (as they are actually used in the real world),
communication is accomplished by the propagation along them of
electromagnetic fields; never anything else.

Doesn't matter one whit whether you turn the field on and off, or vary
the amplitude or any other characteristic of it continuously, as a means
of sending a message from one end to the other, those fields can take on
*any value* from the maximum level injected into the cable by the


Yes, the electrical fields can take any value. It is
inherently an analog medium. But that has no
relationship to the signal which is used to send a
message.

The *signal* does or does not have the ability to take
on various values. If the signal uses discrete symbols,
it is a digital signal. If the symbols have a
continuous range of values, it is an analog signal.

This is not an insignificant distinction. It precisely
the reason that at Bell Labs Claude Shannon studied the
theory of how the two differ. As a result of his Theory
of Information the telecommunications industry began to
develop the technology required to implement digital
system to replace the existing analog systems. They did
that based on what Shannon had shown to be theoretically
the most effective for telecommunications.

Digital systems typically trade SNR (which can be very
low) for bandwidth (which will be very high) compared to
using analog signaling.

The inherent noise immunity of a digital system is
great, and because the analog noise in the medium is
*not* directly proportional to the signal value, a
digital signal can be transmitted with zero errors (due
to noise) if the SNR on the analog medium is above a
minimal level. It happens that that SNR is so low that
a system using analog signals would be unusable at the
same SNR. (Fiber optic cables are an example, where
they are virtually useless with analog signaling for the
typical long circuit lengths that are provided when
digital signaling is used.)

The actual minimum SNR depends on the type of digital
encoding used. But some typical values for various
communications purposes are interesting to look at. A
dialup telephone connection is supposed to have at least
a 24 dB SNR. That is relatively useful for voice
communications, but a typical dialup modem won't work
very well unless the SNR of the connection is above 30
dB above random noise (because it has been converted to
a bandwidth limited analog signal).

On the other hand a binary polar signal (such as the
RS-232C digital interface to that dialup modem) will
have an error rate of less than 1 in 10^5 with an SNR of
only 9.5 dB.

But that isn't even the most significant benefit! Noise
is additive on an analog system, but not on a digital
system, which is specifically the difference between
digital and analog that has revolutionized all
telecommunications in the years since Shannon showed
that digital was superior.

What that means is if we use 5 analog channels tandem
linked to get our message from one location to another,
the end to end noise must be added to determine the SNR,
and that total SNR must meet the above criteria for a
higher SNR than is needed by a digital system.

But if 5 digital channels are tandem linked, only the
errors are additive and not the noise.

That is, with analog both the noise and the errors in
the first link are sent to all succeeding links, and
that noise causes errors in each link on analog system.
On a digital system only the errors are inputted to the
succeeding links but not the noise, so noise in the
first link does not cause errors in the succeeding links
as it does with an analog system.

transmitter down to far below the ambient noise level, depending (for
example) on the length of cable being used. IOW, those signals are,
without exception, *analog*.


No, those signals are digital if the symbols they use
are discrete. The fact that the voltage, for example
can range from 0 to 1 volt has no significance in terms
of the signal *if* that signal uses exactly two symbols,
one of which is represented by any voltage less than 0.4
volts and one of which is represented by any voltage
greater than 0.6 volts. That describes a digital signal
(which indeed is being sent through an inherently analog
channel).


encoder medium decoder
+------+ +------+
| | | |
input ----+ +---------------+ +---- output
| | | |
+------+ +------+

| |
|---Analog ---|
| Channel |

| |
|--------------- Digital --------------|
| Channel |

Typical examples of the above are where the input to the
"encoder" is a DS1 and the "medium" is a twisted pair
cable, or where that input is an OC3 and the "medium" is
a fiber optic cable, or where the encoder is a satellite
modem and the medium is a "bent-pipe" geosynchronous
satellite.

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

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

isw wrote:
(Floyd L. Davidson) wrote:

The SIGNALS are electrical or optical.


Really? Nothing could be acoustical? Are you deaf?


I should have put a smiley on that, sorry for missing it.

Regardless, that does not address the incorrectness of
stating that all signals are analog. Morse code is not
analog.


The carrier or tone that is keyed on and off to send the code starts out


Morse code does not necesarily have either a carrier or
a tone involved, but we can ignore that for this
discussion without changing the validity of our
conclusions.

at a certain strength at the transmitter and grows weaker in a
continuous fashion as the receiver moves further and further away, until
at some point it becomes impossible to understand the *message* it is
carrying.


Well, lets take exactly that as an example, because it
is a good one. We could use a tone as the carrier if
you like, and send it down a regular twisted pair cable.
I'm going to describe this for Morse Code signaling, but
I'd like to point out that virtually any FSK modem does
exactly the same thing with exactly the dynamic range
I'm describing here. Instead of on/off though, it uses
two tones. Everything else is the same, except the
modem is many times faster than a human can decode Morse
Code.

If we put it on the cable at 0 dBm, we'll likely have an
SNR of roughly 50 dB or so, plus or minus a few.

The message is sent using on/off keying of a tone, so at
the cable head we have a 50 dB range which is used to
determine on vs. off. If we head down the road several
miles and get to a point where the signal level has
dropped 10 dB (about the maximum that can be used by a
POTS line), we now have a 40 dB SNR range to deal with.
We could go twice that distance again (losing 10 dB of
signal each time) and get to a point where our signal is
-30 dB and we have only a 20 dB SNR.

At 20 dB SNR there is no reason at all that you won't
get perfect copy, with no errors. Clearly the *signal*
has not changed, even though it has dropped 30 dB in
power. That is because the symbols used are discrete.

From perhaps -40 dBm to 0 dBm there is *no* *change* *in*
*the* *value* *of* *the* *symbols*!

That is, the signal is analog. How can it be digital if it can take on
*any* value?


Obviously it does *not* take on any value. The value
for Morse code is either on or off. There is no "on at
-22.4 dB" value, just on.

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

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