>OK...so am I understanding correctly that you are transferring
>through your converter into your computer at 16bit, but then
>importing it into ProTools for 24bit?
>
>If this is the case, you're not "enhancing" the sound quality, but
>just using some kind of algorithm(s) in order to get a more complex
>file - information that was added, by the way, and has nothing to do
>with the original.
That's not entirely accurate. In fact, it's quite INaccurate.
Any adjustments made to the sound file will result in a mathematical
calculation which is never done as simple addition or subtraction.
The simple process of even just boosting the level by 3 db results in
a multiplication process that when fed 16 bits of data will result in
more than 16 bits of final calculation in the form of a decimal.
Some people would simply say truncate that result back to 16 bits and
you're accurate enough for government work.
The problem with that is that any additional work done on the file
will also result in a similar calculation. Let's say this time it's
a bit of hiss reduction, and maybe some EQ boost to the 1KHz to 3KHz
band. If each of those is working with the 16 bit PLUS decimal
value, then each calculation will result in a more accurate
representation of the final value. Only when all processing is done
should you drop back to 16 bits through whatever process, truncation,
dithering, noise shaping, etc. that you feel is appropriate.
If, on the other hand, you leave the file as 16-bit, each calculation
done for each adjustment you make will leave an error which the next
calculation will amplify until you start to get noticeable problems.
A totally made up example which gets the idea across could be as follows.
Initial value 179.
Your first calculation is your initial value * 0.5749.
Your second calculation is the result of your first value * 4.7768.
Your third calculation is the result of your second value * 2.48.
Your fourth calculation is the result of your third value * 1.17.
Your final value is the result of the fourth calculation brought back
to 16-bit world.
If you start at 16 bit and stay at 16 bit through this entire
process, then here's what you get.
First calculation (179 * 0.5749) results in 102.90710000 which is
immediately truncated to 102 so it will fit entirely in the 16-bits
which has no room for the decimal value. No, it won't round it up to 103.
Second calculation (102 * 4.7768) results in 487.23360000 which is
immediately truncated to 487 so it will fit entirely in the 16-bits
which has no room for the decimal value.
Third calculation (487 * 2.48) results in 1207.76000000 which is
immediately truncated to 1207 so it will fit entirely in the 16-bits
which has no room for the decimal value. No, it still won't round it
up to 1208.
Fourth calculation (1207 * 1.17) results in 1412.19000000 which is
immediately truncated to 1412 so it will fit entirely in the 16-bits
which has no room for the decimal value.
Your final value is 1412 after 4 simple adjustments to the audio data.
Now, if you take that 16-bit file and import it as 24-bit, do your
calculations and then truncate, dither, or otherwise produce a 16-bit
final file, you'll notice the value changes drastically.
First calculation (179 * 0.5749) results in 102.90710000 (the extra 8
bits holding the value to the right of the decimal place).
Second calculation (102.90710000 * 4.7768) results in 491.56663528.
Third calculation (491.56663528 * 2.48) results in 1219.08525549.
Fourth calculation (1219.08525549 * 1.17) results in 1426.32974893.
Let's assume you do a simple truncation at this point to go back to
16-bit. That value is now 1426, whereas when you worked strictly in
16-bit it turned out to be 1412.
Hmm, which one was more accurate?
You'll notice that in the "stay in 16-bit" example above, the
truncated results each time were small enough that they certainly
weren't using the full 16 bit capabilities of the digital
sample. (102, 487, 1207, 1412) The first number, 102, could be
represented in 6 bits, the second number, 487, could be represented
in 9 bits, the third number, 1207, could be represented in 11 bits,
and the fourth number, 1412, could be represented in 11 bits. So it's
obvious that |