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Author Topic: Digital Resoloution and bit depth  (Read 7155 times)

Wiz.

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Digital Resoloution and bit depth
« on: February 10, 2005, 05:26:41 pm »

I have been having a long discussion with someone regards bit depth and George, if you have the time I would like your input.


You have a 16 bit and 24 bit file.

in the area of 0db to -6db of each file, how many steps of resoloution are available.

Is there more available steps of resoloution in the 24 bit file vs the 16 bit file, or are they the same.

I understand how dynamic range equates.


Its just that I think there are more steps of resoloution in a 24 bit file, between 0 and -6db, compared to 16 bit, and he thinks the are the same, only the dynamic range is increased, and the extra resoloution available all lies below -96db.



thanks in advance

Peter Knight
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Nika Aldrich

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Re: Digital Resoloution and bit depth
« Reply #1 on: February 10, 2005, 05:33:35 pm »

Dear god.


Smile


I have to go to a meeting.  I'll chime in when I get back later tonight.

Nika
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Wiz.

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Re: Digital Resoloution and bit depth
« Reply #2 on: February 10, 2005, 05:50:57 pm »

will do

thanks


Peter
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howlback

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Re: Digital Resoloution and bit depth
« Reply #3 on: February 10, 2005, 09:25:10 pm »

Wiz. wrote on Thu, 10 February 2005 17:26

I have been having a long discussion with someone regards bit depth and George, if you have the time I would like your input.


You have a 16 bit and 24 bit file.

in the area of 0db to -6db of each file, how many steps of resoloution are available.

Is there more available steps of resoloution in the 24 bit file vs the 16 bit file, or are they the same.

I understand how dynamic range equates.


Its just that I think there are more steps of resoloution in a 24 bit file, between 0 and -6db, compared to 16 bit, and he thinks the are the same, only the dynamic range is increased, and the extra resoloution available all lies below -96db.



thanks in advance

Peter Knight


Hi Peter,

Each bit in a Linear PCM system is EVENLY spaced throughout the amplitude range.  This is probably the kind of digital system you are talking about (although your converters may do something else...).  So, there are more steps available in a 24 bit system, regardless of the amplitude of the signal.  The error resolution of a system is equal to one half the least significant bit (LSB).  More bits = smaller error.

The important question to ask is whether that matters.  There are at least 2 major concepts to keep in mind.  Quantization error, and quantization distortion.  

Error occurs when the noise caused by quantization is not correlated with the input signal.  Perceptually, this is like low level white noise, noise that you probably can't hear for high amplitude signals.  

As signal level decreases, the quantization error increases, eventually it becomes audible because it is correlated with the input signal; this is quantization distortion.  Adding dither (random noise) to the INPUT signal randomizes the quantization distortion (and error for that matter), practically eliminating its perceptual effect.  In this respect it doesn't matter whether you are using 16 bits properly dithered, or 24 bits.  

I have a feeling your question is actually more complicated.  You probably are wondering what bit depth to use when recording.  There is no simple answer to that question.  But if you like 24 bit recording, its probably not only due to its lower quantization error.

There are other issues too regarding the conversion process, but maybe that gets you on your way to answering the question.
Nika can spank me if he doesn't like what I say.  I'm certain he's sick of writing about this, and that's why he is praying to God for somebody else to post something.  

Best,

KW
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George Massenburg

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Re: Digital Resoloution and bit depth
« Reply #4 on: February 10, 2005, 09:50:05 pm »

Wiz. wrote on Thu, 10 February 2005 16:26

I have been having a long discussion with someone regards bit depth and George, if you have the time I would like your input.


You have a 16 bit and 24 bit file.

in the area of 0db to -6db of each file, how many steps of resoloution are available.

Is there more available steps of resoloution in the 24 bit file vs the 16 bit file, or are they the same.

I understand how dynamic range equates.

Its just that I think there are more steps of resoloution in a 24 bit file, between 0 and -6db, compared to 16 bit, and he thinks the are the same, only the dynamic range is increased, and the extra resoloution available all lies below -96db.

thanks in advance

Peter Knight


Peter,

The exasperated reponsees are because of the very basic nature of your question.  More or less like walking into a college Literature classroom and asking, "What kind of stuff did Shakespeare write, anyway?"

Theoretically (and that word is important because it's not directly measurable) there are 2^24 = 16777216 steps between 0FS (full scale) and the threshold of the first step (which turns out to be kind of a philosophical question in and of itself).  so, because -6.0206dB is half of the maximum amplitude, there are 8388608 steps between 0 and -6.0206dB.  In a 16 bit system (all of the previous caveats apply) there are 32768 steps.

George

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Nika Aldrich

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Re: Digital Resoloution and bit depth
« Reply #5 on: February 10, 2005, 10:42:05 pm »

Wiz. wrote on Thu, 10 February 2005 17:26

I understand how dynamic range equates.


I'm not so sure this is the case.  I think it might be best if you tell us what you know, or how you see it first and then we'll probably be more effective at finding an effective way of responding - a way that gives you all the information you need to know if order to settle this dispute with your cohort.

Let's try these questions on for size.  See what you can do about answering them as comprehensively as possible and then we'll know more about how to proceed:

1.  Why does each bit give us 6dB of dynamic range?

2.  What kind of effect occurs to the waveform when we have to round to the closest quantization step?

3.  Are Signal to Noise Ratio and Dynamic Range the same?  Why or why not?

4.  In a 16 bit system what is the SNR?  If SNR and DR are different, what is the DR of a 16 bit system?

5.  If I have a 16 bit system that covers the range from 0V to 1V and I make it a 17 bit system, what happened to the size of the range that this converter covers?

I hope this will help.

Nika
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Wiz.

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Re: Digital Resoloution and bit depth
« Reply #6 on: February 11, 2005, 12:19:12 am »

Thanks to everyone for replying...


Oh goody a test...Cool



1. Why does each bit give us 6dB of dynamic range?


dB = 20 x Log (V1/V2)

( i didnt pull that equation from memory, i knew of its existence)



2. What kind of effect occurs to the waveform when we have to round to the closest quantization step?


I dont really know.


3. Are Signal to Noise Ratio and Dynamic Range the same? Why or why not?


I dont think so, Dynamic range i think is theoretical, SNR is actual.

4. In a 16 bit system what is the SNR? If SNR and DR are different, what is the DR of a 16 bit system?

Well, the dynamic range is 96db, i dont know what the SNR is.


5. If I have a 16 bit system that covers the range from 0V to 1V and I make it a 17 bit system, what happened to the size of the range that this converter covers?

well, it still overs  0 to 1V. Just with more steps.




I hope this will help.

Nika


Me too


Look, i probably got everything above wrong...Cool

And i guess, I might be asking this question, in a forum, where this stuff is probably, well, to put it badly, beneath you all, if thats the case i am sorry...I was just trying to get a clearer answer. I have been reading and reading and reading and reading.



This is how I understand it to work.

16 bit, gives a dynamic range of 96 db, for the reason I explained above.( i know its not EXACTLY 6db, close enough)

24 bit gives a dynamic range of 144db.


11111111  in binary, 8 bit word, equals in Decimal 256

this is where the bit depth/resoloution comes from, 8 bits gives 256 possible steps, 16 bit gives 65, 535 steps etc etc....


See the way I look at it, and its probably wrong, is that at the time of sample (say 44.1khz) the ADC, takes a snapshot, and determines the level of what it sees, to the resoloution of the bit depth created.

So, it looks and sees a signal , between 0 and -6db, and uses 16 bits to represent that level, in a 16 bit file, and 24 bits to represent that level in a 24 bit file.

In my mind there are more steps available, between 0 and -6db, in a 24 bit file, than there are in a 16 bit file, but my friend disagrees, to his mind, all the extra steps are below -96 db, and there is exactly the same amount of resoloution available to that 0 to -6db window, whether it is 16 bit or 24 bit.

Its almost like we see resoloution the exact opposite of each other. That I see more steps available at 0db than i do at -144db, where he sees the opposite.

I apologise , If i cant be clear enough.

But my basic question remains.


with a 24 bit file, is there more resoloution (steps available,possible values)  to document a signal, between 0 and -6db, than there are in a 16 bit file


thanks for all your time and effort


Peter Knight
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Nika Aldrich

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Re: Digital Resoloution and bit depth
« Reply #7 on: February 11, 2005, 12:35:54 am »

Wiz,

Sorting this out between you and your friend is not as important in the long run as making sure that you understand what is going on with this stuff.  So we're going to take it slow and easy.  For starters:

Quote:

5. If I have a 16 bit system that covers the range from 0V to 1V and I make it a 17 bit system, what happened to the size of the range that this converter covers?

well, it still overs  0 to 1V. Just with more steps.


That's the one you got right.  Now think for a second - why does the dynamic range improve if I just take the region and cut it into more and more pieces?  How does that help our dynamic range?  Why do these smaller quantization steps manifest themselves as lowering the noise?

I'm going to have to answer it from there in the morning, but think on that one for a bit and see what you can come up with.

Cheers!
Nika
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Duardo

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Re: Digital Resoloution and bit depth
« Reply #8 on: February 11, 2005, 12:45:23 am »

Quote:

Its just that I think there are more steps of resoloution in a 24 bit file, between 0 and -6db, compared to 16 bit, and he thinks the are the same, only the dynamic range is increased, and the extra resoloution available all lies below -96db.


For a gross oversimplification...you're both right.  There are more "steps", evenly spaced, in a 24-bit system...but even though they're evenly spaced, they only manifest themselves as more detailed "resolution" below -96 dBFS.

-Duardo
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jfrigo

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Re: Digital Resoloution and bit depth
« Reply #9 on: February 11, 2005, 12:50:22 am »

George et al,

I think he's asking in a practical sense if the top 96dB sound better in the 24 bit system than the 16 bit system; in other words, do you gain more than just increased dynamic range and lower noise when recording?

The simplest answer in a practical rather than theoretical sense is that the top 96dB (a few dB of dither notwithstanding) is not somehow better in 24 bit compared to a properly implimeneted 16 bit system. The audible way the error in a 16 bit system manifests itself is as noise. A 16 bit recording is noisier than a 24 bit recording and you lose the ability to capture some of the quieter signals, but 16 bits is perfectly capable of accurate reproduction within its dynamic limits.

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steve parker

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Re: Digital Resoloution and bit depth
« Reply #10 on: February 11, 2005, 07:03:45 am »

Peter,

you are about to learn some surprising and satisfying things. . .
and you are absolutely in the right place.
some of the people here will give the clearest explanations you will find anywhere.

happy learnin'

steve parker.
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Nika Aldrich

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Re: Digital Resoloution and bit depth
« Reply #11 on: February 11, 2005, 11:12:14 am »

OK, I'll see what I can do, here.  

Let's start simple.  Let's talk about a sampling system with a range from 0V to 4V and four quantization steps.  Each step is 1V, yes?  That means that if I put a signal into it, each time I take a sample I am going to have an error, and that error is going to be +/-.5V, yes?  Each time I take a sample it is going to have to round up or round down, and the amount it is going to round is going to be no greater than half the amplitude of a quantization step.

What I have just done is ADDED a signal to my original waveform.  By rounding the signal I have, by default, added some harmonic content to the waveform.  I could subtract my original signal from the sampled version of it and this would yield the specific waveform that I added.  You can do this on graph paper - overlay the original waveform on top of the sampled version of it and look at the areas where you have error.  That error is a waveform unto itself.  We can extract the error from this and analyze it separately, as a separate waveform.  It IS, after all, a separate waveform.  It is the waveform that you added to the original waveform by means of sampling.

Now, let's look at that error waveform.  What is it?  What is its amplitude?  Well since we didn't round any more than .5V in either direction we know that the error waveform can't have an amplitude of greater than -.5 to +.5, so it has a total amplitude of 1V, yes?  As for its frequency content, that gets a bit tricky, but for the sake of this discussion we're going to say that its frequency content is noise.  We can think that the likelihood of any particular sample in this error-waveform being at any particular given amplitude is pretty random, since it is just the error remaining from the sampling process.  In truth that has some qualifiers, but run with me.  So since all of the sample points on this error-waveform are random we'll say that this waveform is simple noise.

What we have determined, and what we can now say is that the result of sampling is that we have added random noise to our waveform at an amplitude of 1V.  Since 1V is 1/4 the amplitude of the maximum signal you can put into this system (4V) we could compare the amplitude of this noise to full scale and realize that its amplitude is 12dB lower than any full-scale waveform you could put into the system.  Therefore, 4 quantization steps gives you 12dB of dynamic range - noise is added 12dB below the maximum peak.  

Now, what happens when we double the number of quantization steps?  You cut each "region" in your sampling system in half, yes?  As you correctly stated above, the overall range of 0V to 4V is still valid, but the quantization steps are no longer at 0V, 1V, 2V, etc.  They are now at 0V, .5V, 1V, 1.5V, etc.  Yes?  With the cutting in half of the quantization steps we also cut the error-signal's maximum amplitude in half, yes?  The error-signal can no longer have an amplitude of +/-.5V.  It is now restricted to +/-.25V.  That means its overall amplitude is .5V, which is half of what it was before.  Since reducing a signal by 50% is equivalent to a reduction by 6dB, we have just lowered the error signal (the signal that gets added to our original signal by means of quantizing) by 6dB.  Therefore the dynamic range is no longer 12dB but is instead 18dB.  

And what would happen if we double the number of quantization steps again?  And again?  Each time double the number of quantization steps we lower the noise that gets added to the system by 6dB.  Since adding a bit in a binary system is equivalent to doubling the number of quantization steps, each bit added gives us 6dB of dynamic range.

STOP ME NOW IF YOU'RE NOT FOLLOWING.

OK, you didn't stop me.  Plowing forth:

Let's take a 16 bit system.  That gives us 96dB of dynamic range.  That means that the amount of error that gets added is 96dB down from full scale.  Let's go back to our 4V converter.  That means that the amount of error is about .00006V in total amplitude.  Let us talk about putting a signal into this system for quantization that has a SNR of, say, 12dB.  That means that the signal itself has error - random noise - and the amplitude of the random noise on the signal itself, before we ever put it into the sampling system, is 12dB lower than the peak value.  It's like taking a pure sine wave generator and a noise generator.  Set the pure sine generator to put out a sine wave at about 4V and set the noise generator to generate white noise at 12dB down from that, or 1V.  Sum the two together.  There's your signal.

Now we're going to run this 4V, 12dB SNR signal into our 16 bit sampling system.  As we know, the sampling system is going to add noise to our signal at an amplitude of .00006V.  This little bit of noise, however, is absolutely DWARFED by the fact that there is already 1V of noise present!  Sure, we added noise by quantizing, but how are you ever going to notice that you added .00006V of noise when you already had 1V of noise there?  There is no way that you, as a human, would ever hear the addition of this noise.  You, as a human, DO have the ability to hear things even if noise is present at a higher amplitude.  If I play noise at full scale and a sine wave at 4kHz 25dB lower than full scale you would hear it.  Our ears are sensitive to things below the noise floor - but only to a certain degree - 25dB is the maximum for this.  Even if I decreased the noise coming out of my signal generator, thereby increasing the SNR of the signal going in, such that the SNR was now 60dB - the amplitude of the noise is now .004V - you still wouldn't hear the added noise from the sampling process.  That noise is too far below the amplitude of the already-present noise in the system.  Depending on the signal, etc., I could lower the amplitude of the noise generator much lower still and you STILL wouldn't detect a difference.

Now, if this is the case - if I have a signal whereby the noise added because of sampling at 16 bits was so low that it could not be detected, what would be the benefit of adding more quantization steps?  Again, remember, the adding of quantization steps only lowers the error-signal (noise) in the system.  If I can't detect that noise anyway, what is the benefit of reducing it?

Now, how are we doing on answering your question?

Nika
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PookyNMR

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Re: Digital Resoloution and bit depth
« Reply #12 on: February 11, 2005, 01:59:21 pm »

Nika Aldrich wrote on Fri, 11 February 2005 09:12


if I have a signal whereby the noise added because of sampling at 16 bits was so low that it could not be detected, what would be the benefit of adding more quantization steps?  



So bats can enjoy the music too.  Wink

I think the original question arose from the idea that increased bit depth (to 24 bit) would give more increments (numercial values) for each sample which would prvide for more "resolution" in those last 6 dB (if I even interpret the original question properly).

Nika, do you know of any books that talk about this kind of stuff?  I'd prefer one with lots of pictures...  Smile

Nathan
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David Schober

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Re: Digital Resoloution and bit depth
« Reply #13 on: February 11, 2005, 03:01:19 pm »

Nika,

I just wanna drop in on this dialog and say thanks for your time and patience with all of us in explaining these things.  (sometimes no doubt, over and over)  This thread is a pefect example of the time and effort you take.  We really appreciate it and it's very helpful to all of us.  And it's great to see a forum where people can feel free to ask questions from the simple to complex and esoteric without feeling foolish.

Additionally you rarely, if ever, tout your book or say, "Go read my book and get back to me."  You slog on thru these threads, taking time to say what probably is in the book.

I know there are many of us here who would say, "Thanks very much!"

btw....I'm heading over to your site right now and ordering a copy!

Now...back to today's lesson....
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David Schober

Wiz.

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Re: Digital Resoloution and bit depth
« Reply #14 on: February 11, 2005, 07:06:44 pm »

Firstly

I wish to say a hearty thank you, to everyone who has replied to my, somewhat basic question, for the peeps on THIS board. I am sure there are a heap of lurkers, out there, who like me have a hard time wrapping our heads around this issue, and given the fact there is so much white noise on the net, on this very topic.

so again thank you.

To Nika in particular.

What a fantastic reply. I had to read thru it very slowly, and re read paragraphs over and over, but as I did, the clouds parted and the sun began to shine...Cool


A truly wonderfull explanation.

I particularly like the fact that you didnt just provide the answer, you provided the tools for me to reach the answer for myself, the mark of a GREAT teacher.

I would have taken you ages, to put it in, and I VERY much appreciate it.

I also notice, you have written a book on the subject, i also admire the fact, that you didnt plug your book in anyway, when it would have been very easy to do so, i admire ethics, again the mark of a person of integrity, and maturity.

If there is anyway i could ever help, just ask.

I will continue to lurk here and post if I ever think I can be of assistance.

Once again

thankyou


Peter Knight
aka Wiz
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