# 128Kbps channel on VOIP?

Discussion in 'VOIP' started by Scott, Oct 20, 2003.

1. ### MMGuest

So it's strictly quantization error. I know nothing of G.711 (and
little about VoIP for that matter). I was guessing.

MM

MM, Oct 23, 2003

Any A/D (Analog->Digital) conversion is lossy regardless how high the signal
is sampled.

3. ### chrisGuest

Depends entirely on the frequency content of the signal, and whether
you're sampling at a fast enough rate (nyquist rate). Trust me, I
work in the acoustics field.

-Chris

chris, Oct 24, 2003

Regardless what your nyquist rate is and your work in the acoustics field,
the sampling is always an approximation and therefore it is lossy.

5. ### Miguel CruzGuest

I think his point is that if, at the end of the day, you are trying to carry
digital data (which is after all the point of this thread) then you can go
to analog and back without necessarily losing data.

Given a sufficient sampling rate, the information that does get lost in the
A/D process was incidentally added as a side effect of the earlier D/A
process and is therefore spurious.

A simple proof of this is the fact that modems work.

miguel

Miguel Cruz, Oct 24, 2003

Modem works mainly because the origin is already a discrete data then it
gets converted to analog (D/A) -> stream through a media between two or
more points -> A/D (back to discrete data at the destination point). The
discrete data that gets converted to an analog form is only as an
approxination and will never (not almost) be faithfully identical to its
analog origination. Thus, it is a lossy conversion.

7. ### Miguel CruzGuest

Congratulations, you have successfully paraphrased what I said.
The OP wants to transmit digital data. It does not have an analog
origination, only intermediate analog representation.

miguel

Miguel Cruz, Oct 24, 2003

What you did say was the fact modems work and that was necessary as a proof.
What I wrote was to proof what you said does not necessary proof the
matter. So, your congratulation was not appropriate as well as not needed
in this matter and please save it for some other occasions.
That I understood clearly. The point I was trying to proof is that any A->D
conversion is lossy that some of you who has experience of being an
engineer working in the field of acoustic claimed A->D conversion is
lossless. Perhaps, that particular individual should refresh himself with
an A->D conversion knowledge.

9. ### Miguel CruzGuest

Due to an explicable myopic fixation on a single decontextualized point,
you're "proving" something that is not in dispute. If you like, you can go
on to prove that water is wet or that trees are made of wood.

The only thing that is lost in the A/D conversion the rest of us are talking
about, is the unwanted noise that was introduced by a prior D/A conversion.
There is no net loss.

miguel

Miguel Cruz, Oct 24, 2003
10. ### Miguel CruzGuest

Sorry, INexplicable.

miguel

Miguel Cruz, Oct 24, 2003
11. ### chrisGuest

I agree, A-D conversion does lose some of the original signal - namely
the frequency content above the nyquist frequency. To properly
digitize, you need to filter out those frequencies as they will
foldover when you do an fft. The quantization 'noise' depends on
your original signal dynamic range and sample size. D->A suffer the
same issues where you need some analog filtering to smooth the
resulting output.

As an example: Take a pure sine wave. You need to sample it fast
enough (nyquist criteria) to capture all of the frequency data.
Quantization noise depends on your sample size (and indirectly the
frequency content). For telephone voice data, we're typically
talking 8k sample rate, 8-bit.

Calling digitization an 'approximation' is overly general and not very
accurate. This original discussion was about whether the codec was
lossy, not the actual sampling. G711 does not alter or compress the
data like other codecs.

I don't need a refresher, but thank you for the offer. I do this
every day as I'm in charge of a data processing lab. On a typical
test, we acquire data from over 1000 sensors and process many
different outputs, ranging from autospectra, cross-correlation, hit
analysis, wavenumber processing, etc. Too often we get customers who
don't understand data acquisition or processing and frequently
misinterpret data (for example the fft window size/weighting does make
a difference!).

-Chris

-Chris

chris, Oct 25, 2003

That's another subject that does not pertain to lossy/lossless discussion
here.
Regardless there is an "unwanted noise" or not in a D/A reconstruction, the
conversion result is only an approximation, with a "no net loss" (as you
mentioned above). Just because an A/D->D/A conversion has a "no net loss"
or an average loss is zero, it does not mean the conversion is subjected no
loss (**). Thus, an A/D->D/A conversion is not lossless.

** A double mistake in an equation that cancels each other does not entitle
the equation is error free.

13. ### Miguel CruzGuest

If I take a pile of quarters, and temporarily throw in some pennies, then
later pick all the pennies back out, I haven't lost any quarters.

The quarters are the original digital signal. The pennies are the noise
introduced by the initial D/A conversion (and stop changing the rules by
switching the order around from D/A-A/D to A/D-D/A).
Of course it does, tautologically in fact. It's the essence of algebra.
That's what we have the phrase "cancels each other out" for. Or perhaps
you'd like to provide a counter-example?

miguel

Miguel Cruz, Oct 25, 2003

This is not a good example to compare the losses introduced in the A/D-D/A
conversion. In your explanation above with quarters and pennies, you
assumed the quarters represent the digital signals (where is the original
analog signal?). Anyway,in reality such an assumption is not true neither.
When you tossed in a bunch of pennies into a pile of quarters, they don't
homogenize even though mixed thoroughly (perhaps the piles need be
smeltered in zero gravity which is going to make things getting uglier).
Therefore, your explanation is not valid and can not be justified to
represent the losss in the A/D-D/A conversion.
Cancelling each other out is only valid if the two entities are equatable. A
double mistake in an equation, does not equate to a "cancels each other
out" phrase as you mentioned above. You were trying to compare an orange to
a mandarin citrus. Nice try ...

15. ### Miguel CruzGuest

There is no original analog signal. The person wanted to pass digital data
an A-D-A chain.
Good thing the discussion is not about an A-D/D-A conversion then, eh?
You're the one who came up with the phrasing of the double mistake
cancelling itself out.

I will leave you to talk to yourself since your main beefs seem to be with
(A) things you have said and (B) things nobody has said. Enjoy.

miguel

Miguel Cruz, Oct 26, 2003

The point in my continuant discussion is to show you that you can't recover
*completely* the original signal through a D/A conversion (that is based on
an approximation method) regardless of the Nyquist criteria. This does not
mean the Nyquist criteria is flawed. It just means when the original analog
signal gets through an A/D conversion, the signal has already been clipped
through the sampling process. And, because of Nyquist criteria, it is
possible to faithfully reconstruct or recover an approximate to the
original analog signals through a D/A conversion with a "zero net loss".
When one deals with an "approximation" in any conversion even such a
conversion yields a "zero net lost", it does not mean there is *NO* loss of
information at the end of conversion. There is your proof that A/D-D/A
conversion is not lossless.
My discussion is about the loss in A/D-D/A conversion and the quarter +
pennies pile is your ill-condition discussion in here.
I came out with a "double mistake that cancels *each other* ..." not
"itself" as you said above. There is a *BIG* different in *each other* and
*itself*, here.
You are definitely wrong to misinterpret that ended up with such bad
conclusions.

Anyway, I just merely tried to proof to you that an A/D-D/A conversion is
always lossy even with a *zero net loss*. However, it does not mean a "zero
net loss" is unacceptable, especially in the A/D-D/A conversion. After all,
an A/D-D/A conversion is based on an approximation approach.