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

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

Post by GuidoK » 15 Nov 2019 02:46

NOYB wrote:
14 Nov 2019 22:45


Zero is the absence of value and makes no sense for calculating dynamic range.
zero is not the absence of a value
The zero you describe is a digital zero but it's still signal, being the signal zero.
Just like that 0dB on a sound pressure scale is still signal.

So your reasoning that:
Try this with min being zero.
20Log(max/min) = db

Ratio of 1:0 is infinity.
would also mean that s/n calculations wouldnt be possible.

Can you please provide an example or instance of using zero as the minimum for calculating the dynamic range of something?
https://en.wikipedia.org/wiki/Dynamic_range#Audio

"Audio engineers use dynamic range to describe the ratio of the amplitude of the loudest possible undistorted signal to the noise floor, say of a microphone or loudspeaker.[18] Dynamic range is therefore the signal-to-noise ratio (SNR) for the case where the signal is the loudest possible for the system."
For redbook audio data. Signed 16-bit linear PCM...
The maximum magnitude is 32768 and is referenced to -0db.
The minimum magnitude is 1.
Minimum magnitude is −32768, not 1 (or 0)
https://en.wikipedia.org/wiki/Compact_D ... ifications
"Each audio sample is a signed 16-bit two's complement integer, with sample values ranging from −32768 to +32767"

Like I said, the way you define dynamic range determines whether its value for cd is 90dB or 96dB.
If you define the term "dynamic range" as "the difference between the smallest and largest non zero magnitude (where zero magnitude is determined as the noise floor)" then its 90dB for a CD, and if you define the term "dynamic range" as "the ratio of the amplitude of the loudest possible undistorted signal to the noise floor" then its 96dB for a CD.

I guess thats why every scientific paper has a section where terms / concepts of which are written about in that paper are defined. To prevent these kinds of misunderstandings/differences of opinion. A value or parameter only has meaning related to how it's defined.

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

Post by NOYB » 15 Nov 2019 04:37

GuidoK wrote:
15 Nov 2019 02:46
NOYB wrote:
14 Nov 2019 22:45

Zero is the absence of value and makes no sense for calculating dynamic range.
zero is not the absence of a value
The zero you describe is a digital zero but it's still signal, being the signal zero.
Just like that 0dB on a sound pressure scale is still signal.
"value" is an incorrect choice of words. But as you say zero is a signal of zero. Amplify it by anything and it is still zero. Thus using zero as the minimum means that dynamic range db is infinity 2Log(n/0).

Because 0db on a sound pressure scale is referenced to a non zero sound pressure (still signal). Right. It is signal. Not zero. 0db is whatever it is referenced to. If it is referenced to 0 then it has no meaning.
GuidoK wrote:
15 Nov 2019 02:46
So your reasoning that:
Try this with min being zero.
20Log(max/min) = db

Ratio of 1:0 is infinity.
would also mean that s/n calculations wouldnt be possible.
Incorrect. S/N ratio is not with respect to 0. Signal noise level is with respect to 0. Thus the S/N is the difference between signal level and noise level. Neither of which is 0. If the n in s/n was 0 then the ratio would be infinity.
GuidoK wrote:
15 Nov 2019 02:46
Can you please provide an example or instance of using zero as the minimum for calculating the dynamic range of something?
https://en.wikipedia.org/wiki/Dynamic_range#Audio

"Audio engineers use dynamic range to describe the ratio of the amplitude of the loudest possible undistorted signal to the noise floor, say of a microphone or loudspeaker.[18] Dynamic range is therefore the signal-to-noise ratio (SNR) for the case where the signal is the loudest possible for the system."
The noise floor is not 0. S/N ratio being ratio between the noise floor and loudest possible undistorted signal. That is analog signals. The reason noise level is used in analog is because that is the lowest possible signal (magnatude if you will). Also please note it is the ratio between them. Not the difference between them.

In digital the lowest possible magnitude is 1, not the noise floor. Noise floor is typically half the LSB (0.5)
Therefore:
SQNR = 20Log(max/0.5) = 20Log(32768/0.5) = 96.33 db.
DR = 20Log(max/1) = 20Log(32768/1) = 90.31 db.
GuidoK wrote:
15 Nov 2019 02:46
For redbook audio data. Signed 16-bit linear PCM...
The maximum magnitude is 32768 and is referenced to -0db.
The minimum magnitude is 1.
Minimum magnitude is −32768, not 1 (or 0)
https://en.wikipedia.org/wiki/Compact_D ... ifications
"Each audio sample is a signed 16-bit two's complement integer, with sample values ranging from −32768 to +32767"
The wiki quote is correct. But your application of it is incorrect. Range is not the same thing as magnitude.

The magnitude of +5 is the same as the magnitude of -5. Both have a magnitude of 5.
Test it out. From the middle rung of a stairway, take one step up. Typically about +7 inches. Now from the same middle rung as before, take one step down. Typically about -7 inches. Each step was 7 inches. Not 14 inches. That is the magnitude of each step was 7 inches. Whether it is up or down has no bearing on the magnitude of the step.

Place +5 volts across a 5 ohm resistor. 1 amp of current flows producing 5 watts of power.
Place -5 volts across a 5 ohm resistor. 1 amp of current flows producing 5 watts of power.
The voltage magnitude is 5. Not 10. The current magnitude is 1. Not 2. The power magnitude is 5. Not 10.
GuidoK wrote:
15 Nov 2019 02:46
Like I said, the way you define dynamic range determines whether its value for cd is 90dB or 96dB.
In a way sure. It can be defined incorrectly in order to come up with an incorrect value. i.e. 96 db.

But when defined correctly the answer is correct. i.e. 90 db.
GuidoK wrote:
15 Nov 2019 02:46
If you define the term "dynamic range" as "the difference between the smallest and largest non zero magnitude (where zero magnitude is determined as the noise floor)" then its 90dB for a CD, and if you define the term "dynamic range" as "the ratio of the amplitude of the loudest possible undistorted signal to the noise floor" then its 96dB for a CD.
Zero magnitude is not the noise floor. If we had equipment with 0 magnitude noise floor that would be infinite SNR.

The ratio of the amplitude of the loudest possible undistorted signal to the noise floor. For analog signal. But even then the noise floor is not 0.

The ratio of the amplitude of the loudest possible undistorted signal to the softest signal (not noise floor). For digital. For signed 16-bit data that is 32768 and 1.
GuidoK wrote:
15 Nov 2019 02:46
I guess thats why every scientific paper has a section where terms / concepts of which are written about in that paper are defined. To prevent these kinds of misunderstandings/differences of opinion. A value or parameter only has meaning related to how it's defined.
What scientific paper are you referring too that is giving you definitions like noise level being 0?
Besides. Just because they define a value as xyz doesn't make it correct or the application and usage of it correct.

I'll take what these guys say over a Wikipeda article or an O'Riley beginner primer book author.
Teledyne Lecroy
Precision and Dynamic Range
http://www.fte.com/WebHelpII/AES/Conten ... cRange.htm

http://fte.com/company/aboutus.aspx

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

Post by GuidoK » 15 Nov 2019 05:05

NOYB wrote:
15 Nov 2019 04:37


What scientific paper are you referring too that is giving you definitions like noise level being 0?
I dont refer to any scientific paper. I'm saying that a scientific paper defines the concepts it talks about.
Thats why I asked you how you defined dynamic range.
Thats not so hard to understand is it?
First the definition has to be set before you can give meaning to values talked about.
I'll take what these guys say over a Wikipeda article or an O'Riley beginner primer book author.
Teledyne Lecroy
Precision and Dynamic Range
http://www.fte.com/WebHelpII/AES/Conten ... cRange.htm

http://fte.com/company/aboutus.aspx
And they define dynamic range as
"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.

The difference between the maximum and minimum representable values, expressed in decibels is defined as the “dynamic range”."

so yes, within the definition of that guide the dynamic range of a CD is 90dB

You make it a right or wrong discussion, but its really not.
Its meaning only stretches as far as how its defined in the scope of the paper.
So if there is any question about its value, the first question should be: "how do you define it?" instead of bickering about its value because that just leads to a 'is so / na ah' discussion.

For its meaning there are probably as much examples to be found where its minimum value is defined as the noise floor as there are the lowest signal above the noise floor.
Like this document from keysight where its defined as the noise floor:
http://na.support.keysight.com/pna/help ... yn_Rge.htm
So there is no correct one and incorrect one. They're both correct as they define it (and adjust their value accordingly to it).

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

Post by NOYB » 15 Nov 2019 06:51

GuidoK wrote:
15 Nov 2019 05:05
NOYB wrote:
15 Nov 2019 04:37
What scientific paper are you referring too that is giving you definitions like noise level being 0?
I dont refer to any scientific paper. I'm saying that a scientific paper defines the concepts it talks about.
Thats why I asked you how you defined dynamic range.
Thats not so hard to understand is it?
First the definition has to be set before you can give meaning to values talked about.
Now I know why you are coming up with things like noise floor of zero that don't make sense.

There is only one correct definition for the dynamic range of a signed 16-bit integer data set expressed as db. It is not a matter of opinion. Either the correct definition, formulas and values are used or they are not. When they are the correct answer is the result. i.e. 90 db.
GuidoK wrote:
15 Nov 2019 05:05
NOYB wrote:
15 Nov 2019 04:37
I'll take what these guys say over a Wikipeda article or an O'Riley beginner primer book author.
Teledyne Lecroy
Precision and Dynamic Range
http://www.fte.com/WebHelpII/AES/Conten ... cRange.htm

http://fte.com/company/aboutus.aspx
And they define dynamic range as
"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.

The difference between the maximum and minimum representable values, expressed in decibels is defined as the “dynamic range”."

so yes, within the definition of that guide the dynamic range of a CD is 90dB
They define it that way because that it is the correct definition. And when calculated with the correct values in the correct formula, as they do, results in the correct answer. 90 db.
GuidoK wrote:
15 Nov 2019 05:05
You make it a right or wrong discussion, but its really not.
Its meaning only stretches as far as how its defined in the scope of the paper.
So if there is any question about its value, the first question should be: "how do you define it?" instead of bickering about its value because that just leads to a 'is so / na ah' discussion.
It is right or wrong. The dynamic range of a signed integer data set expressed as db is exact. There are not two correct answers.

It only leads to a such as that discussion when those who don't know the correct definition, formula and values try to debate using incorrect definitions and apply incorrect formulas and/or values.

You now know what the correct definition is. The correct formula. And what the correct values are. But yet you still seem to want to say there is some other correct answer. So show me the definition you chose, the formula, and the values used in the calculation and show why they are correct.
GuidoK wrote:
15 Nov 2019 05:05
For its meaning there are probably as much examples to be found where its minimum value is defined as the noise floor as there are the lowest signal above the noise floor.
Like this document from keysight where its defined as the noise floor:
http://na.support.keysight.com/pna/help ... yn_Rge.htm
So there is no correct one and incorrect one. They're both correct as they define it (and adjust their value accordingly to it).
Again as I said earlier. That is analog. We are talking about digital. A signed 16-bit integer data set. Ones and zeros. The smallest signal that can be represented corresponds to a magnitude of 1. Not the noise floor.

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

Post by GuidoK » 15 Nov 2019 23:42

Interesting piece about dynamic range and data word size in digital audio according to analog devices (intergrated circuit manufacturer):
https://www.analog.com/en/education/edu ... range.html

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

Post by NOYB » 16 Nov 2019 02:01

GuidoK wrote:
15 Nov 2019 23:42
Interesting piece about dynamic range and data word size in digital audio according to analog devices (intergrated circuit manufacturer):
https://www.analog.com/en/education/edu ... range.html
Just like the O'Reilly beginners primmer book the author completely disregards that redbook audio CD data is signed.

Never once is signed 16-bit mentioned in relation to CD (redbook audio data).
Never once is redbook mentioned.

Then in table 1. Dynamic Ranges they give ranges like 90 to 95 db for 16-bit audio converters. So that could cover both singed and unsinged, or dithering shaping, or who knows what.

Don't see any support for claim that dynamic range of signed 16-bit integer data set (redbook audio CD data) is 96 db.

Using the incorrect maximum magnitude of 2^16 (65536) instead of the correct maximum magnitude of 2^15 (32768) is the cause of the discrepancy here. Which stems from ignoring the fact that it is signed data.

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

Post by GuidoK » 16 Nov 2019 13:47

The aspect of signed data is covered in that article.
That there is no mentioning of the word redbook is no meaningful argument. The word redbook alone is not relavant (it is also not mentioned in your link from teledyne....)

Regarding the maximum magnitude of 65536, that is related to if the the absolute minimum is 0 (so the difference between the maximum value and the minimum value)
If the 'correct' maximum value is set at 32768, then the 'correct' minimum value is -32767, and the difference between those two is still 65536. The maximum value is 65536 higher than the minimum value.
The signed data doesn't result in that the maximum amount of musical information in the data becomes less. This is a result from that the signing method used is two's complement signing. The signing doesn't take up any information that can't be used for audio in the datastream. It's a very easy conversion which is usually used to make adding/subtraction/etc easier in computers and in PCM audio to virtualize a sinus oscillating around a zero.
This was already covered in this topic in page 2 by Erin1
Thats how I see it too.
If you have a different opinion on that then you have a different interpretation of what two's complement signing is than how its defined everywhere else.

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

Post by NOYB » 16 Nov 2019 22:31

GuidoK wrote:
16 Nov 2019 13:47
The aspect of signed data is covered in that article.
Not in reference to CD data. Only in reference to DSP etc.
GuidoK wrote:
16 Nov 2019 13:47
That there is no mentioning of the word redbook is no meaningful argument. The word redbook alone is not relavant (it is also not mentioned in your link from teledyne....)
The article from teldyne... actualy described the data. Singed 16-bit. It didn't cover two's complement. Though that does not change the dynamic range.
GuidoK wrote:
16 Nov 2019 13:47
Regarding the maximum magnitude of 65536, that is related to if the the absolute minimum is 0 (so the difference between the maximum value and the minimum value)

If the 'correct' maximum value is set at 32768, then the 'correct' minimum value is -32767, and the difference between those two is still 65536. The maximum value is 65536 higher than the minimum value.
Show the dynamic range of a sine wave expressed as db for that.
db = 20Log(max/min)
GuidoK wrote:
16 Nov 2019 13:47
The signed data doesn't result in that the maximum amount of musical information in the data becomes less. This is a result from that the signing method used is two's complement signing. The signing doesn't take up any information that can't be used for audio in the datastream. It's a very easy conversion which is usually used to make adding/subtraction/etc easier in computers and in PCM audio to virtualize a sinus oscillating around a zero.
This was already covered in this topic in page 2 by Erin1
Thats how I see it too.
If you have a different opinion on that then you have a different interpretation of what two's complement signing is than how its defined everywhere else.
There is so much wrong there. Started to go over it piece by piece but it's just to much wrong. So I'll just try to go through the numbers instead.

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 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 » 17 Nov 2019 00:23

-oodb: positive phase value: 0 and negative phase value: 0 (0)
..
oo represents infinity
That's the 16th step, the LSB thus 96dB (where LSB is matched by dither noise to prevent quantize errors)

Interesting that you think you know better than _arguably_ the oldest digital IC manufacturer (analog was the first to launch a DA converter on the market).
The link I presented was from their education library. But as everything in this topic, of course they don't know their stuff.
But go ahead, tell Analog Devices they're wrong.
Tell them that (quote from the link):
In digital terms, SNR and dynamic range are used synonymously to describe the ratio between the largest representable number to the quantization error
is not true and that you know best.
You're not arguing me, you're arguing Analog Devices.

If you already have all the wisdom, why start this topic?

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

Post by NOYB » 17 Nov 2019 01:38

GuidoK wrote:
17 Nov 2019 00:23
-oodb: positive phase value: 0 and negative phase value: 0 (0)
..
oo represents infinity
That's the 16th step, the LSB thus 96dB (where LSB is matched by dither noise to prevent quantize errors)
-90 dbFS to -infinity dbFS is how many db? My math says that is infinity db.
That is not the 16th step. Making that step does not result in -96dbFS. It results in -infinity dbFS.
There are only 15 6db steps. Again for the bazillionth time try putting 0 into the db formula. db = 20Log(max/min).

Dithering changes values not the range the values exist in. The range is the same regardless of dithering. The ranges are still 0 to 32767 for the positive phase and 32768 to 65535 for the negative phase.
Also for a refresher please go back and read the first paragraph of this threads opening post with regard to dithering.
GuidoK wrote:
17 Nov 2019 00:23
Interesting that you think you know better than _arguably_ the oldest digital IC manufacturer (analog was the first to launch a DA converter on the market).
The link I presented was from their education library. But as everything in this topic, of course they don't know their stuff.
But go ahead, tell Analog Devices they're wrong.
Tell them that (quote from the link):
In digital terms, SNR and dynamic range are used synonymously to describe the ratio between the largest representable number to the quantization error
is not true and that you know best.
You're not arguing me, you're arguing Analog Devices.
Like the O'Reilly book and Wikipedia, this article is for a more casual audience. It is not technically complete and in some cases not technically accurate. For it's purpose it doesn't need to be. You are taking it's contents as though it is an engineering document.

Nothing in that article, the O'Reilly book, or the Wikipeda article you have link to changes what the actual audio data or it's format on a redbook CD. That is signed 16-bit two's compliment linear PCM. That being integer values in the range of 0 to 65535 representing BOTH the positive phase AND the negative phase of audio. There are 16 steps of 6db over an entire 16 bit range. However. That entire range contains BOTH the positive phase AND the negative phase. Since a 6db change requires BOTH the positive phase AND the negative phase to be changed by a factor of 2. There are only 15 6db steps available.
GuidoK wrote:
17 Nov 2019 00:23
If you already have all the wisdom, why start 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.

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

Post by GuidoK » 17 Nov 2019 02:09

NOYB wrote:
17 Nov 2019 01:38
It is not technically complete and in some cases not technically accurate. For it's purpose it doesn't need to be. You are taking it's contents as though it is an engineering document.
yeah right. thats a very good argument. Be sure you tell them that.
Generalizing something without actually proving the point. Nothing in that link you prove wrong and still you say its wrong because.....it's not really meant to be accurate =D>

But yeah, you really proved that you know better than Analog Devices do.....
Be sure to tell them that, and in the process, explain to them how it really works.
You can probably reach them here
https://form.analog.com/form_pages/supp ... pport.aspx
I'm sure they will take up all your wisdom and will be really greatful for giving them the insight on how it really works.
They really need to be exposed to what dynamic range really is

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

Post by NOYB » 17 Nov 2019 04:07

GuidoK wrote:
17 Nov 2019 02:09
NOYB wrote:
17 Nov 2019 01:38
It is not technically complete and in some cases not technically accurate. For it's purpose it doesn't need to be. You are taking it's contents as though it is an engineering document.
yeah right. thats a very good argument. Be sure you tell them that.
Generalizing something without actually proving the point. Nothing in that link you prove wrong and still you say its wrong because.....it's not really meant to be accurate =D>

But yeah, you really proved that you know better than Analog Devices do.....
Be sure to tell them that, and in the process, explain to them how it really works.
You can probably reach them here
https://form.analog.com/form_pages/supp ... pport.aspx
I'm sure they will take up all your wisdom and will be really greatful for giving them the insight on how it really works.
They really need to be exposed to what dynamic range really is
There is no reason to tell them. They already know that it is not an engineering document. They already know they are generalizing. That is not the issue. The issue is in you taking it as an engineering document that is complete and accurate. It obviously is not. If it were it would detail out the claim of 96db dynamic range for CD. Also as you like to harp on they would also define the definition of dynamic range they are using.

Criticizing my critic of that document does not refute the case I have laid out for 90db dynamic range.

But please. If you do decide to attempt to refute. Please use meaningful values and formulas (calculations). No more of this divide by zero stuff.

Here it is again.

Signal to Quantization Noise Ratio:
SQNR = 2Log(max/half the LSB) = 96.33 db
max = 32767
half the LSB = 0.5

Dynamic Range:
DR = 2Log(max/min) = 90.31 db
max = 32767
min = 1
Last edited by NOYB on 17 Nov 2019 04:14, 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 » 17 Nov 2019 04:13

NOYB wrote:
17 Nov 2019 04:07
They already know they are generalizing. That is not the issue. The issue is in you taking it as an engineering document that is complete and correct. It obviously is not.
This is all speculation to make ends meet in your theory.
Apparently you now also speak for analog devices. =D>

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

Post by NOYB » 17 Nov 2019 04:25

GuidoK wrote:
17 Nov 2019 04:13
NOYB wrote:
17 Nov 2019 04:07
They already know they are generalizing. That is not the issue. The issue is in you taking it as an engineering document that is complete and correct. It obviously is not.
This is all speculation to make ends meet in your theory.
Apparently you now also speak for analog devices. =D>
No it is not speculation. It is evident by what is and is not in the document.
What I have laid out is not theory. It is actual values and actual math.
No need to speak for them. The document speaks for them. It's content reveals what it is and is not.

Is this what you've wanted all along? To deflect (derail) the discussion away from the actual case that has be laid out, that you have not been able to refute.

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

Post by GuidoK » 17 Nov 2019 04:44

It's not what I want or am able to refute.
Its like I said; you're arguing with what Analog Devices says.
Your argument is like "yeah its not ment to be an engineering document", but all the math is right there in that document. 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....
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...