setting record levels into ProTools with the intent of mixing in ProTools

[joy4u]

With the dithered mixer plugin, PT sounds great, apart that mixed tracks sounds worse that soloed tracks, maybe this is just my fault, because there is a BIG masking effect due to the wide band of each track due to a too perfect sampling, i will try to cut out some freq in a track that i really don't need... (yes, this is an old trick, but it should be used more in digital domain)

[metMayo]

Wow [img]images/icons/wink.gif[/img] over half way to the millenium (sp?)....I have to say though this thread is dragging on a bit~!!! [img]images/icons/tongue.gif[/img]
There's obviously some very interesting points/posts in there, but the vast majority is repetative, drifting more towards a different subject, or totally completely and utterly pointless alltogether!!!!

[img]images/icons/frown.gif[/img]

P.S. just like this one [img]images/icons/grin.gif[/img]

[Corey Shay]

I think the sonic degredations due to processing inputs and outputs being limited to 24 bits wide can be nearly entirely avoided with care if not avoided altogether. Keep your signal at a reasonable level. Don't clip. Don't go too high. Leave a little room up top. But don't record or mix your peaks below -20 dB full scale. There's a certain green area we should all shoot for probably.

Given I take this into condsideration while mixing, and use good plugins (oh, man I just got the Sony EQ, amazing), I am not worried about the sound of the TDM mixer anymore. I cannot say I can definitely hear a difference between the standard and dithered mixer. And I'm guessing if I really did, it's something you have to strain for. In my opinion, it seems this mixer really doesn't have a sound, and I myself prefer it that way, so I can control such things by hand.

[Baixo]

I've been following the other discussions about the disputed idea of Resolution vs. Dynamic Range that Nika has been participating in on other boards and I have to say I found evidence against this theory in something as simple as an Apogee digital audio primer on their website. I learned, like almost all of us, that resolution (defined as the accuracy of the measurements used to define amplitude) increases with the wordlength and that this resolution applies to the signal, not the noise. My own jury is still out so I don't yet feel like this dispute is resolved. Can we resurrect this for a minute? Sorry Nika and all, I just want to know the truth and can't find evidence on your behalf yet. I would love it to be true, as it solves a lot of problems of the practiceof recording.

from the Apogee pdf:

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:<HR>The number of bits in the digital word defines the dynamic range –
the difference in level between the loudest signal (full-scale)and the
quietest (when all the bits are set to zero except the smallest one).A 16-
bit system has a dynamic range of about 96 dB.A 24-bit system on the
other hand in theorywill give you 144 dB.Whether you can hear 144 dB,
and whether a real piece of audio equipment can actually deliver it,is
another question. <HR></BLOCKQUOTE>

Now we get to the good stuff:

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:<HR>The word-length defines something else,too:the size of the small-
est differencein level that you can capture,and that influences the per-
ceived quality of the recording.Let ’s look at the issue of quantization in
a little more detail.A “quantum ”in atomic physics is a single,tiny,per-
haps indivisible particle of matter/energy.Thus quantization in digital
audio refers to the tiny incremental step in digital level between a zero
and a one.Imagine an analog signal gradually and smoothly getting
louder.As it does so,the digital data output by an A/D converter fed by
that signal also increases,but instead of increasing smoothly like the
analog signal,it increases in a series of steps.The size of the steps is
determined by the number of bits,the smallest step being the difference
between the smallest,or “least significant ”bit being set to zero and set
to 1.(Don ’t worry:you can ’t really hear the steps.A filter in the D/A con-
verter smooths them all out.But you can hear the difference made by
the size of the steps!At least,you can most of the time.) <HR></BLOCKQUOTE>

Now I know this is a simplification of the issue, but it cannot be such an oversimplification that the information is simply wrong or misleading.

Thinking a lot about this idea, it brought up a problem that I'd like to figure out, which is, if every bit of wordlength doubles the number of steps of accuracy, and this accuracy is assumed to manifest itself across the entire dynamic range of the signal, and you've got 24 bits that get dithered to 16 bits, you've helped preserve some low-level detail with the dither, but how does a wordlength reduction like that possibly preserve resolution in the rest of the signal that is, well, above the 16th bits. This doesn't compute at all to me (not a math guy, but a logical mind). This problem would seem to corroborate Nika's claim because since the least significant bits represent low-level reproduction, how does stripping them off mess with the accuracy of the rest of the amplitude range of the signal?

I apologize for such stream-of-conciousness, it may not make sense.

How can each bit define the dynamic range of a system, with each bit representing roughly 6dB (a fixed range of volume) and yet the resolution of the whole of that dynamic range is somehow increased, making the steps of each bit's range grow in number?

[Nika]

[QUOTE]Originally posted by Baixo:
Now I know this is a simplification of the issue, but it cannot be such an oversimplification that the information is simply wrong or misleading.

The information from Apogee is correct: The bit depth does indeed tell us the smallest size of measurement step. In 24 bit on many converters it is around 125 nanovolts. In 16 bit on the same converters it is around 32,000 nanovolts. There is nothing in this that contradicts what I have said. I'm not sure how much of the thread you've read, and there are certainly a lot of analogies that can be applied (all of which break down at a certain point) but let's revisit this one really quickly:

A football field is divided into yards. Would the game be more precise if it were divided into feet? What about inches? Would it be less precise if it were divided into 5 yard increments? What about 10 yard increments?

In order to determine this you have to be able to tell what the error in the game inherently is already. In football we have the eyesight of the referees, the size of the ball, the size of a player's foot or knee when he's down, and many other areas that make the game imprecise. There MIGHT be an advantage to measuring the field in feet, but to go with inches or even finer would probably not actually help the accuracy of the game at all. I would say that it would help the "statistical resolution of the field", but would not help the "resolution (accuracy) of the game," or where the ball get's placed on the field.

Audio is akin to this in that there is a certain amount of error (noise) in any signal. The quantization steps don't really have to be any more smaller than the rate of error in your signal. When noise gets added to your signal it will in effect "randomize" the voltage of the signal itself.

So if you have, let's say, 1 V of noise in your system, the location of any sample point is going to be +/- 1 V from where it originally was. Now, how accurate do you think we need to make the "quanta"? We certainly don't need it to be on the 125 nanovolt level.

So, yes, Apogee was correct. More bits does give us smaller quanta, or measurement or quantization steps, but if the noise level is high enough (dynamic range is small enough) then these steps are not utilized and it becomes like measuring a football field off in millimeters.


Thinking a lot about this idea, it brought up a problem that I'd like to figure out, which is, if every bit of wordlength doubles the number of steps of accuracy, and this accuracy is assumed to manifest itself across the entire dynamic range of the signal, and you've got 24 bits that get dithered to 16 bits, you've helped preserve some low-level detail with the dither, but how does a wordlength reduction like that possibly preserve resolution in the rest of the signal that is, well, above the 16th bits.

Dither is fascinating, but tricky to comprehend. Let's make sure we have all of the other ducks in a row and we'll address that one.

How can each bit define the dynamic range of a system, with each bit representing roughly 6dB (a fixed range of volume) and yet the resolution of the whole of that dynamic range is somehow increased, making the steps of each bit's range grow in number?

You lost me on that one. I can't see what angle you're coming at it from.


Cheers!
Nika.

[Dr. J]

I love the football anology. I think we should get Dennis Miller, and a drunk John Madden to commentate with Nika, then this discussion would really get interesting.

[img]images/icons/smile.gif[/img]

[Baixo]

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:<HR>
How can each bit define the dynamic range of a system, with each bit representing roughly 6dB (a fixed range of volume) and yet the resolution of the whole of that dynamic range is somehow increased, making the steps of each bit's range grow in number?

You lost me on that one. I can't see what angle you're coming at it from.


Cheers!
Nika.[/QB]<HR></BLOCKQUOTE>

Let me try to clarify. I envision bit depth as happening from top to bottom, the top end (MSB) being fixed and the range of it extends downward with every bit, like you were unfurling a scroll to the floor.

The difficulty I'm having is in seeing the relationship between bits-as-dynamic-range, e.g., each bit giving you a fixed range (6dB each time, imagine the further unfurling downward) and bits as an increase of resolution for the entire DR. One adds range to an increasingly low-level domain, the other adds possible steps to the entire show.

For example, take the MSB. The MSB can encode 6dB of the highest-level signal. in any wordlength system, how does tacking on another bit, which happens AT THE BOTTOM OF THE DYNAMIC RANGE, make the resolution of the MSB (and it's higher-up brothers) incur an increase in steps within that bit range, since it's at the top? Do you see what I'm driving at?

RE: top of your last post, are you saying that noise is the factor that makes the bits-as-resolution concept tricky? Essentially that the accuracy is wasted on noise?

[Nika]

[QUOTE]Originally posted by Baixo:
The difficulty I'm having is in seeing the relationship between bits-as-dynamic-range, e.g., each bit giving you a fixed range (6dB each time, imagine the further unfurling downward) and bits as an increase of resolution for the entire DR. One adds range to an increasingly low-level domain, the other adds possible steps to the entire show.

For example, take the MSB. The MSB can encode 6dB of the highest-level signal. in any wordlength system, how does tacking on another bit, which happens AT THE BOTTOM OF THE DYNAMIC RANGE, make the resolution of the MSB (and it's higher-up brothers) incur an increase in steps within that bit range, since it's at the top? Do you see what I'm driving at?

We should be clear. Tacking on another bit does NOT happen at the bottom of the dynamic range. Think of it this way:

A converter is a 4V converter, meaning that the swing from full scale to full scale (+ and -) is 4 Volts. This means that in a one bit system the quantization steps are 4 Volts away from each other. In a 2 bit system they are 1.33 Volts away from each other. In a 16 bit system the steps are .00006 V. Etc. Adding more bits adds more quantization steps all the way through the system - not just on top or bottom.

Did I answer that OK at all?

RE: top of your last post, are you saying that noise is the factor that makes the bits-as-resolution concept tricky?

Yes

Essentially that the accuracy is wasted on noise?

Yes.

Nika.