What is the dynamic range and S/N ratio of CD format?

compact disc, dacs, mp3 players and streaming audio
NOYB
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Re: What is the dynamic range and S/N ratio of CD format?

Post by NOYB » 17 Nov 2019 05:46

GuidoK wrote:
17 Nov 2019 04:44
It's not what I want or am able to refute.
It is what you are not able to refute. The reason is because it is correct. Not because I says so. Though I do say so. But because it is the math that says so. It is the math that makes it correct.
GuidoK wrote:
17 Nov 2019 04:44
Its like I said; you're arguing with what Analog Devices says.
No. You are saying it based on their article and others such as Wikipedia and an O'Reilly book.
You do not speak for them. You speak for yourself. You are citing what they have said in general articles as basis for your attempted failed refutes.
GuidoK wrote:
17 Nov 2019 04:44
Your argument is like "yeah its not ment to be an engineering document", but all the math is right there in that document.
Really? What math in that document do you say I should be using for signed 16-bit two's compliment linear PCM dynamic range instead of what I'm using?
GuidoK wrote:
17 Nov 2019 04:44
Its not analog who says that its an incorrect document, its YOU who says it. So its your word against analog's word, not your word against mine....
You are the one here discussing it. Not them. You are using their document as basis for your claims. Yes I say their document has some incorrect statements. And I have provided basis for that claim. Where is your basis for what you are using from their document in support of your potion that shows it to be correct? Oh yeah. It usually starts with something like divide by zero. Almost forgot about. ;)
GuidoK wrote:
17 Nov 2019 04:44
As analog doesn't say that its a generalizing document with information that is not correct, you take it upon you to speak for them. They're your words, aren't they?
That you don't realize this is why this topic derails. As I'm writing this I wonder why I'm still here...
They don't have to say it is a generalizing document with information that is not correct. It is self evident by what is and is not in the document.
Yes the document has incomplete and incorrect information. It is so whether I say so or not. It is what the document contains and does not contain that makes it so. It is self evident.

It derails because of you using generalized information as though it is complete and accurate. And based on that post divide by zero nonsense. And then want to argue that the generalized documents you based it on says thus and so.

Possibly you are still here to keep the discussion deflected and derailed away from the case that has been laid out, and supported, that you have not been able to refute. Perhaps that is easier than accepting reality that the dynamic range is 90db instead of what generalized writings and marking propaganda has led people (maybe including yourself) to believe.

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Re: What is the dynamic range and S/N ratio of CD format?

Post by GuidoK » 17 Nov 2019 19:34

http://www.ti.com/lit/an/sbaa055/sbaa055.pdf

BB article on dynamic performance on their DA converters (yes, BB also makes DA converters)
(if confusing, BB is owned by TI)

"Theoretical dynamic range performance is limited by resolution, which determines the quantizationerror level:
Dynamic Range = 6.02n + 1.76, in dB
n = number of bits of resolution"

Same formula is listed in that link I posted form that other major DA converter manufacturer, Analog devices.
So cd: 16 bits, so actual theoretical dynamic range is even 98dB (if you relate it to quantizationerror level, but imho this is masked with dither noise in the end so that 1,76dB drouwns in the noise)
The two's complement signing doesnt reduce the numbers of bits of resolution
https://en.wikipedia.org/wiki/Two%27s_complement
(for the small example of a 3 bit table you can still see that there are 8 different values; 2³=8)
They both relate the maximum dynamic range to the noise floor/level of quantize error, and not the smallest signal above that (thats what teledyne does, therefore the 6dB difference)

But I guess this is of course also not true, and that the document is of course of general content and not an engineering document, as I have the feeling you also speak for BB.

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Re: What is the dynamic range and S/N ratio of CD format?

Post by NOYB » 17 Nov 2019 21:31

GuidoK wrote:
17 Nov 2019 19:34
http://www.ti.com/lit/an/sbaa055/sbaa055.pdf

BB article on dynamic performance on their DA converters (yes, BB also makes DA converters)
(if confusing, BB is owned by TI)

"Theoretical dynamic range performance is limited by resolution, which determines the quantizationerror level:
Dynamic Range = 6.02n + 1.76, in dB
n = number of bits of resolution"

Same formula is listed in that link I posted form that other major DA converter manufacturer, Analog devices.
So cd: 16 bits, so actual theoretical dynamic range is even 98dB (if you relate it to quantizationerror level, but imho this is masked with dither noise in the end so that 1,76dB drouwns in the noise)
The two's complement signing doesnt reduce the numbers of bits of resolution
https://en.wikipedia.org/wiki/Two%27s_complement
(for the small example of a 3 bit table you can still see that there are 8 different values; 2³=8)
They both relate the maximum dynamic range to the noise floor/level of quantize error, and not the smallest signal above that (thats what teledyne does, therefore the 6dB difference)

But I guess this is of course also not true, and that the document is of course of general content and not an engineering document, as I have the feeling you also speak for BB.
The audio data set of a redbook CD (signed 16-bit two's compliment linear PCM) is not "theoretical". It is actual.

I'm sure you can find hundreds of documents that will say things like theoretically 6.02n ... is 96db, 98db, etc. i.e. 6.02 x 16. It is absolutely true that a 16-bit data set has a dynamic range of 96.33db. I said that earlier. And that is essentially what these documents you refer to are saying. And I'm sure you can find hundreds of them.

However. That does not directly translate to signed 16-bit two's compliment linear PCM audio data. The reason is that there are 2 data sets contained in that 16 bit range. The positive phase (0 to 32767) AND the negative phase (32768 to 65535). Therefore. There are only 15 factor of 2 changes available for the positive phase. And likewise there are only 15 factor of 2 changes available for the negative phase.

Since a 6.0206db change requires BOTH the positive phase AND the negative phase to be changed by a factor of 2. The N for signed 16-bit two's compliment linear PCM (redbook CD) is 15. Not 16. Thus 6.0206N = 6.0206 x 15 = 90.31db.

You keep taking these documents that are about other things than signed 16-bit two's compliment linear PCM audio data and take their formulas and try to apply them to signed 16-bit two's compliment linear PCM audio data. Doing so yields incorrect results because it is a misapplication of the formulas.

As an aside. The Teldyn document I linked to earlier uses the the LSB value (1). Not the noise floor.
http://www.fte.com/WebHelpII/AES/Conten ... cRange.htm

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Re: What is the dynamic range and S/N ratio of CD format?

Post by GuidoK » 17 Nov 2019 21:57

NOYB wrote:
17 Nov 2019 21:31

As an aside. The Teldyn document I linked to earlier uses the the LSB value (1). Not the noise floor.
http://www.fte.com/WebHelpII/AES/Conten ... cRange.htm
No, they're literally saying:
"The lowest level that can be represented is the minimum possible non-zero number that can be represented, and that of course in all cases is the value 1, or -1."
zero is the lsb value, aka 0000 0000 0000 0000. And two's complement signing changes nothing to that.

Thats something different. They relate it to 1 or -1 where as Analog and BB/TI relate it to 0 as minimum possible value, being either the noise floor (after dither) or the quantization error level (pre dither) depending on where you pick the point. They literally say that in their documents. Do I need to quote that?
That's the difference and from that comes that difference of 6dB. So its a difference in definition of the terminology.
Something I understood right away when I posted my first question "how do you define dynamic range"
You still don't seem to understand that and disregard the documents from analog (oldest manufacturer of DA converters) and TI/BB as being general, non engineering documents not really ment as an engineering document and therefore incorrect. But I didnt read any of that in those documents that analog and ti/bb say that the content of it is of general nature and therefore technically incorrect. So apparently you speak for both analog and ti/bb now.

If that's your point, fine. Maybe add that to your list why you started this topic:
1) To learn. I know more now than I did when I started this thread.
2) To expose people to what the dynamic range really is.
3) To show that I speak for Analog Devices and TI/BB to say that their documents are of general nature and therefore techically not correct.
If it's so obvious to you that both Analog and TI/BB are wrong, why don't you contact them and tell them?
I'm sure they'd be grateful that someone points them to the stupid mistakes they've made in their papers, misinforming the public/customers....
Oh and correct that wikipedia page on dynamic range in the process too will ya? It's apparantly full of nonsense.

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Re: What is the dynamic range and S/N ratio of CD format?

Post by NOYB » 18 Nov 2019 00:16

GuidoK wrote:
17 Nov 2019 21:57
NOYB wrote:
17 Nov 2019 21:31

As an aside. The Teldyn document I linked to earlier uses the the LSB value (1). Not the noise floor.
http://www.fte.com/WebHelpII/AES/Conten ... cRange.htm
No, they're literally saying:
"The lowest level that can be represented is the minimum possible non-zero number that can be represented, and that of course in all cases is the value 1, or -1."
zero is the lsb value, aka 0000 0000 0000 0000. And two's complement signing changes nothing to that.
That is not noise floor. And 0 (0000 0000 0000 0000) is not the LSB value. LSB value is 1 (0000 0000 0000 0001). Claiming that 0 is the LSB value demonstrates a lack of very basic math knowledge. Please stop claiming that documents say things that they do not say.

You don't get it. They have not defined that the values they are applying are for signed 16-bit two's compliment linear PCM audio data. Which contains 2 sets of data within the 16-bit range. Yet you still try to apply the math as though it is a single 16-bit data set. For all your harping on definitions you sure cite a lot of articles that don't define the data set they are applying formulas to beyond just the number of bits. The type of data, what it is used for, the structure, etc. matters.

Why would I correct that Wikipeda page. It's not the problem. Your misapplication of it is the problem. Continuing the lack of knowledge exhibit.

But I guess that is all you can do since you can't refute the case that has been laid out and actually supported with actual math and illustrations that are specific to the data, it's format, and it's use. And all you do is continue to come back with incorrectly applied formulas. Possibly just to muddy things up because you can't refute it. It's a typical behavior of the losing side of a debate when they can't make a solid case or refute the other side.

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Re: What is the dynamic range and S/N ratio of CD format?

Post by GuidoK » 18 Nov 2019 00:54

Your teledyne document also doesn't speak of signed 16-bit two's compliment linear PCM audio data, so I guess thats wrong too. That's your reasoning.
Both my links point to that they're talking about 16 bit digital pcm audio, and the document from analog states it as two's complement signed. And again two's complement doesnt take away any resolution.
And 0000 0000 0000 0000 is definately a bit combination that can come from the cd data. Saying not is just silly. Look at the wikipedia link, look at the little table the constructed and see that all 0 is actually there.

Both my links clearly state that the theoretic dynamic range for a digital audio signal is presented as
6.02n + 1.76, in dB
And wikipedia also uses those exact formula's:
https://en.wikipedia.org/wiki/Signal-to ... al_signals

But you clearly have a different theory on that, thus telling that analog and ti/bb are wrong with their articles.
I don't mind. I'll wait until they've corrected those documents.
Last edited by GuidoK on 18 Nov 2019 01:08, edited 1 time in total.

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Re: What is the dynamic range and S/N ratio of CD format?

Post by NOYB » 18 Nov 2019 01:08

Show the math to be wrong.

A maximum amplitude sine wave:
positive phase peak value: +32767 two's compliment: +32767
negative phase trough value: -32768 two's compliment: +32768

A minimum amplitude sine wave:
positive phase peak value: +1 two's compliment: +1
negative phase trough value: -1 two's compliment: +65535

It is difficult to paint the picture textually but try to visualize this. The two's compliment moves the negative phase of the sine wave up to the top half of the 16-bit range (32768 to 65535). Thus the maximum negative phase magnitude of -32768 is now represented as +32768 and the minimum negative phase magnitude (-1) is represented as +65535. A range of 32767 (65535 - 32768 = 32767).

Each phase of the sine wave is still contained in a range of 32767. 0 to 32767 for the positive phase and 32768 to 65535 for the negative phase.

To change the level by 6.0206db, BOTH the positive phase AND the negative phase has to be changed by a factor of 2.
So when a -0dbFS sine wave is reduced by 6.0206db, it's positive phase peak value becomes 16384, and it's negative phase trough value becomes 49152 (32768 + 16384).

There are 15 such 6.0206db steps available.
15 x 6.0206db = 90.31db = 20Log(32768/1) = 20Log(2^15/1) = 20Log(max/min) = db

To illustrate:
sine wave at:
-0db: positive phase value: +32767 and negative phase value: -32768 (+32768)
-6db: positive phase value: +16384 and negative phase value: -16384 (+49152)
-12db: positive phase value: +8192 and negative phase value: -8192 (+57344)
-18db: positive phase value: +4096 and negative phase value: -4096 (+61440)
-24db: positive phase value: +2048 and negative phase value: -2048 (+63488)
-30db: positive phase value: +1024 and negative phase value: -1024 (+64512)
-36db: positive phase value: +512 and negative phase value: -512 (+65024)
-42db: positive phase value: +256 and negative phase value: -256 (+65280)
-48db: positive phase value: +128 and negative phase value: -128 (+65408)
-54db: positive phase value: +64 and negative phase value: -64 (+65472)
-60db: positive phase value: +32 and negative phase value: -32 (+65504)
-66db: positive phase value: +16 and negative phase value: -16 (+65520)
-72db: positive phase value: +8 and negative phase value: -8 (+65528)
-78db: positive phase value: +4 and negative phase value: -4 (+65532)
-84db: positive phase value: +2 and negative phase value: -2 (+65534)
-90db: positive phase value: +1 and negative phase value: -1 (+65535)

-oodb: positive phase value: 0 and negative phase value: 0 (0)

That is 15 steps of 6db. 6db x 15 steps = 90db dynamic range (rounded)

The values in parenthesis are the two's compliment of the negative phase values. The two's compliment does not change the number of 6db steps that exist in each phase. It simply makes the negative values positive for more efficient arithmetic processing.

oo represents infinity

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Re: What is the dynamic range and S/N ratio of CD format?

Post by GuidoK » 18 Nov 2019 01:23

00 doesnt represent infinity, it represents 0, the equilibrium position of a sine.
But that equilibrium position doesn't mean no noise because of its quatize errors (thus after dither: noise on that level).
Those quatize errors or dither noise is the noise floor.
A sine wobbeling between -1 an 1 is 6dB up from that. So there are 16 steps:

step 1: from 0 to 1/-1 (+6dB above noise floor level)
step 2: from 0 to 2/-2 (+12dB above noise floor level)
step 3: from 0 to 4/-4 (+18dB above noise floor level)
step 4: from 0 to 8/-8 (+24dB above noise floor level)
step 5: from 0 to 16/-16 (+30dB above noise floor level)
step 6: from 0 to 32/-32 (+36dB above noise floor level)
step 7: from 0 to 64/-64 (+42dB above noise floor level)
step 8: from 0 to 128/-128 (+48dB above noise floor level)
step 9: from 0 to 256/-256 (+54dB above noise floor level)
step 10: from 0 to 512/-512 (+60dB above noise floor level)
step 11: from 0 to 1024/-1024 (+66dB above noise floor level)
step 12: from 0 to 2048/-2048 (+72dB above noise floor level)
step 13: from 0 to 4096/-4096 (+78dB above noise floor level)
step 14: from 0 to 8192 /-8192 (+84dB above noise floor level)
step 15: from 0 to 16384 /-16384 (+90dB above noise floor level)
step 16: from 0 to 32767 /-32767 (+96dB above noise floor level)

So 96dB s/n ratio and I quoted this before from the site from Analog devices:
In digital terms, SNR and dynamic range are used synonymously to describe the ratio between the largest representable number to the quantization error
Your model describes the ratio between the largest representable number and the first number up from the quantization error, as how TDY describes it.

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Re: What is the dynamic range and S/N ratio of CD format?

Post by NOYB » 18 Nov 2019 04:02

GuidoK wrote:
18 Nov 2019 01:23
00 doesnt represent infinity, it represents 0, the equilibrium position of a sine.
But that equilibrium position doesn't mean no noise because of its quatize errors (thus after dither: noise on that level).
Those quatize errors or dither noise is the noise floor.
A sine wobbeling between -1 an 1 is 6dB up from that. So there are 16 steps:

step 1: from 0 to 1/-1 (+6dB above noise floor level)
step 2: from 0 to 2/-2 (+12dB above noise floor level)
step 3: from 0 to 4/-4 (+18dB above noise floor level)
step 4: from 0 to 8/-8 (+24dB above noise floor level)
step 5: from 0 to 16/-16 (+30dB above noise floor level)
step 6: from 0 to 32/-32 (+36dB above noise floor level)
step 7: from 0 to 64/-64 (+42dB above noise floor level)
step 8: from 0 to 128/-128 (+48dB above noise floor level)
step 9: from 0 to 256/-256 (+54dB above noise floor level)
step 10: from 0 to 512/-512 (+60dB above noise floor level)
step 11: from 0 to 1024/-1024 (+66dB above noise floor level)
step 12: from 0 to 2048/-2048 (+72dB above noise floor level)
step 13: from 0 to 4096/-4096 (+78dB above noise floor level)
step 14: from 0 to 8192 /-8192 (+84dB above noise floor level)
step 15: from 0 to 16384 /-16384 (+90dB above noise floor level)
step 16: from 0 to 32767 /-32767 (+96dB above noise floor level)
From 0 to 1/-1, or to any non zero value, is infinite db. For the bazillion-th time. db = 20Log(max/min).
Though a mathematician would probably say undefined. But for this purpose infinite should be okay.

The problem you are having is that you are trying to start at zero instead of full scale. This is typical of a person who does not have the math knowledge and understanding of db and how to correctly apply it. A correct way, though not the only way, but probably the most typical, in a case such as this is to start with -0dbFS. That is reference -0db to the full scale value. The full scale value here is 32767. Thus 32767 = -0dbFS. Work it from there and you'll be more likely to have better results.

As I pointed out earlier in this thread. The value zero (0) is meaningless to db. This is because zero (0) multiplied or divided by anything is zero (0). Since db is a log function it does not work with zero (0). You have been making this same very basic mistake throughout this thread. Beginning even with your very first post. Your very first post clued me in that either you don't understand db or you are playing dumb. So I just ignored it. Perhaps I should have continued ignoring. But thought maybe some discussion could help you improve your understanding of db. So far though you just keep repeatedly committing the same mistakes with zero. Resulting in repeatedly posting of mathematical nonsense.

BTW. Going from 1/-1 (-90 db) to 0 is -infinity db.

And oh please after all your harping on defining things don't tell me that I'm not allowed to define a symbol "oo" to represent infinity.

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Re: What is the dynamic range and S/N ratio of CD format?

Post by GuidoK » 18 Nov 2019 04:23

0 is the noise floor. It's not an actual 0 in numeric value, only in digital.
Say the noise floor is 1Vpp noise, then +1/-1 is a 2Vpp sine wave, +2/-2 a 4Vpp sine wave etc etc.
That's how dB's work. Look at dB's measured for loudness to the ear. They're not related to 0 soundpressure, but to a base line (20 μPa).
Therefore it's not 'infinity'.
So the digital zero sets the base line. It's the equivalent of 20 μPa on the sound pressure scale, in level being 0dB, but with cd audio its eventually the dither noise. How many Vpp that is depends on what the output level of the device is set to.
So that first 6dB step from 0 to +1/-1 is definately there.
Thats why they call it signal to noise ratio
If 0, being the noise floor would be infinate, a cd player (and any digital device) would have an infinate SNR.
Last edited by GuidoK on 18 Nov 2019 04:31, edited 1 time in total.

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Re: What is the dynamic range and S/N ratio of CD format?

Post by NOYB » 18 Nov 2019 04:30

GuidoK wrote:
18 Nov 2019 04:23
0 is the noise floor. It's not an actual 0 in numeric value.
Say the noise floor is 1Vpp noise, then +1/-1 is a 2Vpp sine wave, +2/-2 a 4Vpp sine wave etc etc.
That's how dB's work. Look at dB's measured for loudness to the ear. They're not related to 0 soundpressure, but to a base line.
Therefore it's not 'infinity'.
So the digital zero sets the base line. It's the equivalent of 20 μPa on the sound pressure scale, in level 0dB, but with cd audio its eventually the dither noise. How many Vpp that is depends on what the output level of the device is set to.
We are talking about actual numeric values here though. The data set values. 0 is not the noise floor. It is 0.
Regardless of noise, dithering etc. The data is range limited to 0 to 32767 and 32768 to 65535 (-1 to -32768). The data has a dynamic range that corresponds to the difference between the maximum and minimum non zero value of the range.
Last edited by NOYB on 18 Nov 2019 04:38, edited 1 time in total.

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Re: What is the dynamic range and S/N ratio of CD format?

Post by GuidoK » 18 Nov 2019 04:32

0 (equilibrium position of the waveform signal) is the noise floor because of the dithering. Otherwise you get quantize errors. Those are horrible for the ear and very easy to pick out in music.

If quantize errors wouldnt exist, then 0 would be infininate silent, and SNR would also be infinity as no dithering would be needed. But the dither level added sits 6dB below +1/-1.

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Re: What is the dynamic range and S/N ratio of CD format?

Post by NOYB » 18 Nov 2019 05:25

Go ahead. Put that 0 into the db formula. You can't calculate db from zero. Stop with the nonsense math. Asked you before to stop the divide by zero nonsense.

db = 20Log(max/min)

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Re: What is the dynamic range and S/N ratio of CD format?

Post by GuidoK » 18 Nov 2019 06:40

0 is the noise floor. The equilibrium position of the quantized sine wave isn't 'dead silence' but a set amount of noise (or undithered it would be quantize errors). Every value calculated is related to that. So the +1/-1 sine is +1 from the noise floor and -1 from the noise floor.
see
https://en.wikipedia.org/wiki/Signal-to ... al_signals

They way you put it a SNR would not exist. Maybe add that one too to your list.

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Re: What is the dynamic range and S/N ratio of CD format?

Post by NOYB » 18 Nov 2019 06:48

If you say 0 is the noise floor. Then you cannot calculate db from that.

No. The way I put it is, db = 20Log(max/min). That is not non exist. The way you put it is infinite.