128Kbps channel on VOIP?

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

  1. Scott

    MM Guest

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

    MM, Oct 23, 2003
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  2. Any A/D (Analog->Digital) conversion is lossy regardless how high the signal
    is sampled.
    root/administrator, Oct 23, 2003
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  3. Scott

    chris Guest

    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, Oct 24, 2003
  4. Regardless what your nyquist rate is and your work in the acoustics field,
    the sampling is always an approximation and therefore it is lossy.
    root/administrator, Oct 24, 2003
  5. Scott

    Miguel Cruz Guest

    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 Cruz, Oct 24, 2003
  6. 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.
    root/administrator, Oct 24, 2003
  7. Scott

    Miguel Cruz Guest

    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 Cruz, Oct 24, 2003
  8. 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.
    root/administrator, Oct 24, 2003
  9. Scott

    Miguel Cruz Guest

    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 Cruz, Oct 24, 2003
  10. Scott

    Miguel Cruz Guest

    Sorry, INexplicable.

    Miguel Cruz, Oct 24, 2003
  11. Scott

    chris Guest

    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, Oct 25, 2003
  12. That's another subject that does not pertain to lossy/lossless discussion
    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.
    root/administrator, Oct 25, 2003
  13. Scott

    Miguel Cruz Guest

    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 Cruz, Oct 25, 2003
  14. 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 ... :D
    root/administrator, Oct 26, 2003
  15. Scott

    Miguel Cruz Guest

    There is no original analog signal. The person wanted to pass digital data
    across an analog-over-digital link. You remain the only person talking about
    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 Cruz, Oct 26, 2003
  16. 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

    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.
    root/administrator, Oct 26, 2003
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