You're adding in bits, and this is not how things are done. Let's say during the scan process, it detects that the gain has to be raised =+3.01 dB...then how many dB does each bit represent? Db is a logarithmic function and in most cases bit resolution is a linear function...not all cases though. Calculations are done with Floating point calcualtions, which basically means they're expodential function calculations. It's like a E^123 + E^123456= E^123456789. The more bits you have of the total resolution of the original wave, the more accurrate the answer can be represented in binary. In the most case if your wave is at a good level, then you're working with the majority of the bits, so the calculation and quantization error will almost be exact. Well in the case of a fade out, this is not the case because the original wave bit resolution is very finite at those points. This is the primary reason to work in 24bit over 16bit and to have all calculations done in 32bit, to reduce the amount of quantization error. Bottom line, there IS quantization errors during normalization, thus the reason why most mastering engineers will try to avoid this process if there processed .Wav is within -1.0 dB of 0dB. Although, the argument is, if I have a -1dB peaking .Wav and If I normalize, will I hear a difference? Some will say, why do the extra calculation, because 1dB really isn't a noticable change so, why introduce the errors due to quantization. The other side, could say, well the bit resolution is very high so the quantization error will not be that drastic and I won't be able to hear it anyways, so I'ld might as well take the level to the maximum. Both are right, but at this point it's really an argument not worth fighting over, because in both cases you probably can't tell the difference with 24bit and 32bit files.

The only true way to avoid calculation quantization errors in digital audio is to go back to analog processing, where you have a bit resolution of infinite. 32bit and next 64 bit resolutions, is just narrowing the gap between infinite, to where it no longer becomes noticable.

Red sed:
"You're adding in bits, and this is not how things are done. "

What I was suggesting that it could be an optional way for thing to be done.

You say "during scan it find the peak level needs to up by 3.01dB" I'm saying "stuff the dBs - scanning has found the peak levels (plus all the others) need to go up the binary equiv of 535 decimal". That gives me a track with a know full-scale peak level. How many dBs that took to get it there does not interest me in the slightest.

Just pure addition - no fractions. Not talking about mucking about with bit-depths or anything, just straight normal track peak normalisation. A simplistic suggestion sure, but in the context I am talking about, for a simple task in it's simplest form.

Straight binary addition = *no* quantisation errors.


geoff

This is almost amusing....

I'm not trying to slam you Geoff, but you do not appear to have the understanding of digital audio that you think you do. I hate to say something like this but do not know how to put it otherwise since you will not accept the facts.

Red posted earlier that if you are raising the audio level (by normalizing for example) you are doing multiplication with the DSP and not just straight addition. Your proposed method of normalizing without quantization error is not correct and will not work.

Think about this. PCM digital audio represents finite voltages levels with binary values. For 16 bit audio 0 dB FS would be 65,535 (actually, this is not exactly the way it works because 2's compliment is used to represent positive and negative voltage digital audio - I am simplifying for this explanation). A 3 dB increase will multiply your voltage or binary value by 1.413 and a 3 dB decrease will divide by 1.413 - from the textbooks!! If you have a sample value of 32,450 and raise that by 3 dB, your new value for the sample will be 45,851.85. Note the fractional part -this will be rounded and thus a QUANTIZING ERROR. Also, note that the 3 dB increase raised the sample value of 32,450 by 13,401.85. Now, let's look at another sample in the file we have just raised by +3 dB. Lets say it's value is 10,125 - so multiplying by 1.413 gives you a new value of 14,306.625. A difference of 4,181.625 versus the increase of 13,401.85 for the other sample in the same file. See something here? You cannot just arbitrarily add a fixed number to the sample values to increase the volume as you will end up with complete trash.

You can't say just forget about the *dB's stuff* and add a whole number to each sample. A level increase requires a logarithmic increase in voltage\sample value, which means that different sample values require different increases (with different fractional parts) for the SAME change in level (whether it is up or down).

I suggest you pick up the book "Principles of Digital Audio" by Ken Pohlman. It is very informative though possibly a bit deep for some - not a "Dummies" book.

3.01/6.02 if you want to split hairs. Guess I'm just looking at the peak bit then. I'll just need the dummies book, or to invent logarithmic bits.

I suppose I should give up doing this late at night ....

geoff

Great indebt explanation fishtank!!!!

Another reason Jeff, to show you that you can't just do simple addition of bits is, the point that I made to you earlier. "0+2"=2...the wrong answer that you want...but when adding +6dB to the level the real equation becomes 0x2=0, thus the correct answer. Fishtanks, explanation was the total indebt technical explanation. Hopefully by me giving you the "digital audio for dummies" explanation, somewhere inbetween you can figure out what is going on.