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Are you a digital expert? Please help!

Discussion in 'Recording' started by bluemusic, Dec 13, 2007.

  1. bluemusic

    bluemusic Active Member

    Jan 29, 2005
    Up for a difficult question about bit and how it works?

    We know that sample rate sets the bandwith of the system by mesuring differences in the voltage level over time. In order for the desired frequency bandwith to be faithfully encoded in the digital domain, the selected sample rate must be twice as high as the higeste frequency to be recorded. We also know that higher bit the more detail mesurement.

    Since 16 bit represent 6 dB x bit + 1,8dB we can say that the dynamic range of a 16 bit signal is 97.8 and for 24 bit it is 145.8.

    We also know that the differences in the samples mesured would be in the LSB ( least significant bit). Because of the binural code a code can only be 0 or 1, therefore there will be problems when a samples is between what the converts can't handle.

    So far so good. What we have a problem understanding here is:

    Is the amplitude scale for the mesuring constant, meaning 16 bit and 24 bit has the same roof, but more posibilities for 24 between this scale. Ergo a more precise mesurament. Why doesn't 24 bit signal distort at the same volume as 16 bit when thay have the same roof?

    Or does the scale expands upward to have more headroom for 24 bit? But then there will also be few changes when we talk about possible sample values.

    Another way to see it:( this is not a correct way mathwise but ok example)

    16 bit is 1 cm with 10 possibilities inside
    24 bit is 1 cm with 20 possibilities inside
    = more possibilites, same roof.


    16 bit is 1 cm with 10 possibilities inside
    24 bit is 2 cm with 20 possibilities inside
    = higher roof, same possibilities.

    Am I confusing you? So am I!!!!

    All help are greatly appreaciated.

    Best regards Bluemusic
  2. RemyRAD

    RemyRAD Guest

    Sep 26, 2005
    Digital 0 DBfs is the same for 8, 16, 24 & 32. What changes is its lower range. But a broader noise floor to overload signal means a comfort pad between those two points for processing of an inexperienced manner. That is to say, if you're just processing stuff willy-nilly and am really not sure what you are doing, you are safer with 24/32-bit. Whereas I know what I'm doing and don't require any more than the 96 DB basic specification of 16 bit. Because I know what I'm doing, I don't need 15 foot wide lanes to drive in. I can drive in 8 foot wide lanes. My recording is the same.

    Out-of-control in a controlled manner
    Ms. Remy Ann David
  3. cfaalm

    cfaalm Active Member

    Feb 21, 2005
    Home Page:
    Your rule of thumb is OK. The formula is:

    16 bits = 20 log(2^16) + 1,76 = 98,09 dB
    24 bits = 20 log(2^24) + 1,76 = 146,25 dB

    This is the way I see it.

    At the top of a digital scale is 0dB Full Scale. The bits determine how much dynamic range fits under that. So with 24 bits it is 146dB. So you could say that it reaches down to 146 dB beneath 0 dBFS. It is more a matter of being able to capture real quiet sounds than real loud sounds.

    Now people tend to relate the dBFS scale to dB SPL. That is not a 1:1 story, but not too complicated either.

    When you record stuff that has a higher dynamic range than 98dB SPL, where there are actually worthwile details outside that range, you will need a 24bit system to capture it all. Let's say the top of the range is a deafening 138dB SPL, that value would translate to be close to 0dB FS on the meter of your digital system. Because you record in 24 bits, the passage at 38dB SPL i.e. -100 dB FS on the meter is also captured. On a 16 bit system that would already be out of reach.
  4. dterry

    dterry Active Member

    Apr 14, 2006
    Where I think most of the confusion about bit depth comes in is how bit depth applies to voltage/amplitude representation/resolution vs. power ratio (dynamic range).

    dB is of course a power ratio, and you gain more dynamic range with more bits (e.g. signal power range), and as Remy pointed out - it's on the lower end of the range (so we hear more detail in reverb tails, ambient spaces, and general depth of a recording or mix, if there is information there to be heard or perceived at least). D

    Don't confuse that with representing the same amplitude/voltage range with more bits - there you get tighter resolution (more steps for the same amplitude/voltage range in the converter at quantization (not external analog voltage range), but in terms of power ratio, we extend the dynamic range downward.
  5. DrGonz

    DrGonz Active Member

    Jun 24, 2007
    Phoenix, AZ
    So this basically creates more Headroom or less???
    This is really bugging me now? :lol:
  6. bluemusic

    bluemusic Active Member

    Jan 29, 2005
    Great guys. Acctually I know all of this, it just got lost when I went into koma this morning:) Now it's back!
  7. dterry

    dterry Active Member

    Apr 14, 2006
    bluemusic - tech theory has a way of doing that. Writing or recording music is usually a good cure though :)

    DrGonz - more bits (as in 32-bit float or 48 bit fixed) only gives you more "headroom" when used for internal software processing, summing, etc, but you are still limited to 0dBFS when rendering/exporting to a final mix in 16 or 24 bit. When representing audio signals, greater bit depth (recorded data) extends the dynamic range down.

    Basically for internal calculations one can use additional bits either for math overflows (headroom), or to represent much quieter signals, but when output to a fix bit depth (i.e. 16 or 24 bit) the 0dBFS limit is imposed and the additional bits (24 vs. 16) can only extend the dynamic range down, not above 0.

    The same applies to recorded data - since it must be represented within a fixed system there must be a 0dBFS limit, so recording at 24 bits only extends the range down, not up.

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