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From: "mac.com" <bresh@...>
Date: Thu Feb 14, 2002 4:21pm
Subject: Re: [LUG] Re: [GEN] 24 bits vs. 96K

In a digital system, each voltage reading is expressed as a number that the
computer can interpret. Just as 2:00 p.m. can be expressed as 1400 hrs, so
any value can be expressed using binary digits -- 1s and 0s. Also, 1/3 can
be expressed as 0.3, or more accurately 0.33, or better yet as 0.333 and so
on: the greater the number of decimal places, the more precise the
expression of the translation. Hence, the greater the number of digital bits
used to express a voltage reading, the more accurate is the translation (not
the louder). For 16 bits, the number of digital bits used to translate or
"digitize" a voltage reading is 16 which gives 65,536 possible voltage
values.  24 bits gives 16,777,217 possible voltage values.


on 2/14/02 9:45 AM, dedric_terry at dterry@k... wrote:

>> I disagree with this. Bit depth has nothing to do with dynamic
> range. Bit
>> depth has to do with resolution. A 16 bit converter and 24 bit
>
> I can see why one would assume, but bit depth does determine
> dynamic range in the form of power level, and the level of the
> noise floor.  16 bits give you 96dB of range (0dB to -96dB), and
> 24 give you 144dB (0 to -144dB).  Since waveforms are positive
> and negative going, this is often represented as +/- 48dB (16
> bits), or +/-72dB (24 bits), but power is an absolute valueIf the
> two were not relative to a fixed value, then you would have to
> resample to change bit depth, instead of dither/truncate/round or
> pad (in the case of increased bit depth).  More processing would
> be involved to redefine, say 1 bit per 2dB instead of the usual
> 3dB.  The only thing that usually doesn't change is 0dBFS, which
> is usually represented by all 1's in each bit position (but in
> internal processing that could represent another number too to
> handle overs and carry-over summing for example, but the fixed
> bit output would be limited to 0dBFS for power level purposes).
> The extra bits (down to 1 bit) simply represent a quiter range.  If
> this were not so, then 16 bit converters and 24 bit converters
> would clip at different power levels.  The misconception is that
> the more bits the louder the signal, but that isn't accurate, and
> isn't what dynamic range is referring to (in this case, the change
> in the noise floor is the change in dynamic range).
>
> Sampling frequency determines actual resolution.  Put bit depth
> on the x axis of a graph (+/- [dynamic range]/2) and sampling on
> the y axis (time in seconds, ms, or us) and you will see how
> resolution changes with sampling frequency.
>
> Sorry for the long OT reply.  Now related to Logic, anyone with
> Logic 5 have a report on the effect of the new POW-r dithering?
>
> Regards,
> Dedric
>
> --- In logic-users@y..., Jonathan Christensen <jon@s...> wrote:
>>> Bit Rate determines dynamic range.  You will definitely hear
> mor of the soft
>>> sounds, and really loud sounds without compression when
> you use 24 bits
>>> instead of 16.  It is certainly a much more audible difference.
>>
>> I disagree with this. Bit depth has nothing to do with dynamic
> range. Bit
>> depth has to do with resolution. A 16 bit converter and 24 bit
> converter
>> will clip at exactly the same point (provided they are accurately
>> calibrated). The difference is the quality of the quiter bits,
> providing
>> a more accurate depiction of the instrument being recorded.
>>
>> best,
>> Jon
>>