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14th August 2008
#1

Bit depth / dynamic range relationship?

Hi,
There’s something that I am wondering about these days, and I would be grateful if anyone can help me to find a right answer. I am wondering what is the exact relationship between bit depth and dynamic range? I know that every bit (in a fixed point system carries ~ 6dB; 6,02 to be precise), but what I would like to know is: what is the explanation for this? The answer that I came up to is that every digital system is dynamically limited by quantization distortion / noise (in the case that TPDF white noise was added during initial ADC) which is, on the other hand, directly proportional to the bit depth of the system (more bits = more discrete steps = less quantization distortion = greater dynamic range). Would this be the right explanation?
Also, is there a difference in this relationship (bit depth and dynamic range) between systems which are adding TPDF white noise during initial ADC and those that are not (how did those 6dB / bit came up, and with which of these two systems is that related with, with or without TPDF white noise)?
Please correct me if I am wrong and be so kind to provide me a right explanation.
P.S. I hope that this is the right forum to discuss this thread.
P.P.S. I did a search button and didn’t find any thread that discussed this question. I am sorry in advance if I am repeating a thread.
P.P.P.S. Sorry if my thread sounds confusing.
14th August 2008
#2
Lives for gear

The addition of each bit to the wordlength doubles the number of values that can be expressed. 20 log 2 = 6.02. Is that what you're getting at?

-Ben B
14th August 2008
#3

Quote:
Originally Posted by Ben B
The addition of each bit to the wordlength doubles the number of values that can be expressed. 20 log 2 = 6.02. Is that what you're getting at?

-Ben B
Thank you for your quick reply Ben B. I am aware that every additional bit doubles the number of discrete steps because number 2 is mantissa / fundament. I know that bit depth determines the number of discrete steps for example 24bit system has 2 ^ 24 = 16777216 discrete steps and dynamic range of app. 24 * 6 = 144dB, which means that every discrete step carries 144 / 16777216 = 0,00000858306dB and that maximum quantization distortion is half of that value 0,00000429153dB. But what I am looking for is the relation ship betwean bit depth and the dynamic range. Actualy, why every bit "carries" exactly 6dB of dynamic range?
14th August 2008
#4

Quote:
Originally Posted by uros
Thank you for your quick reply Ben B. I am aware that every additional bit doubles the number of discrete steps because number 2 is mantissa / fundament. I know that bit depth determines the number of discrete steps for example 24bit system has 2 ^ 24 = 16777216 discrete steps and dynamic range of app. 24 * 6 = 144dB, which means that every discrete step carries 144 / 16777216 = 0,00000858306dB and that maximum quantization distortion is half of that value 0,00000429153dB. But what I am looking for is the relation ship betwean bit depth and the dynamic range. Actualy, why every bit "carries" exactly 6dBs of dynamic range?
Uros, In PCM terms, bit depth directly pertains to the resolution of each sample in a set of digital audio data. The depth will limit binary data quantities such as dynamic range and signal-to-noise ratio. The bit depth will not limit frequency range, which is limited to sampling rate.

All you have to keep in mind is that on CDs, all 16 used bits equal a program at FULL headroom (-0 dBFS), which is the same as those squashed hip hop records on some CDs who have no more bits to fill up.
14th August 2008
#5
Lives for gear

It's probably a function of the maximum output voltage of the device (referenced to 0dBfs) compared to the lowest encodable level it is capable of. In other words, suppose you had two devices with the same maximum output levels post-D/A. Make one of them 16-bit and the other 24 bit. The 24 bit device (in theory) could resolve signals 144 dB below this maximum output voltage, while the 16 bit device could resolve signals only 96 dB below this point. 96/16=6, and 144/24=6. So, I think of this "downward" from the maximum output level (or any other reference point).

-Ben B
14th August 2008
#6
Gear Guru

Quote:
Originally Posted by uros
why every bit "carries" exactly 6dBs of dynamic range?
Simple - every bit has approximately twice the value of the sum of all bits below it. And when you double the voltage, the signal goes up 6 dB.

--Ethan
14th August 2008
#7

Quote:
Originally Posted by Ethan Winer
Simple - every bit has approximately twice the value of the sum of all bits below it. And when you double the voltage, the signal goes up 6 dB.

--Ethan
Thank you all, very much for your replies.
Ethan, that was the answer that I was looking for. Ben B also wrote that, but I didn't assume that it is "just" a logarithmic function that is responsible for the dynamic range that each bit carries, because every additional bit doubles the number of discrete steps; 20 log 2.

Thank you all very much.
14th August 2008
#8

Quote:
Originally Posted by Ethan Winer
Simple - every bit has approximately twice the value of the sum of all bits below it. And when you double the voltage, the signal goes up 6 dB.

--Ethan
I thought that the bits used depended upon the signal's envelope, not upon a set of increments of 6 dBs per bit...IMO, bit 1=16777216, then 1 bit + 00000001 does not equal 2 Bits (12 dBs/-132 dBFS)...Or, does it?
14th August 2008
#9
Lives for gear

Verified Member
Quote:
Originally Posted by uros
dynamic range of app. 24 * 6 = 144dB
Came to think about the "1.76" that's missing in the equations above and wondered why exactly 1.76.. This one was spot on: SNR in ADCs: Where did all the bits go? - 6/7/2007 - EDN

The use of such equations implies lack of dither. Dither will make the figures slightly smaller than 6.02n+1.76 suggest, but the dynamic range will be infinite..

Quote:
Originally Posted by joerod
All you have to keep in mind is that on CDs, all 16 used bits equal a program at FULL headroom (-0 dBFS), which is the same as those squashed hip hop records on some CDs who have no more bits to fill up.
All signals approaching full scale needs all bits to be represented, not just at full scale.

If you have a number say 3(1*2^1+1*2^0 - binary "11") and wanted to use 4(1*2^2+0*2^1+0*2^0) - binary "100" ) as the next number, you'd need to add that other bit. That'll be good up to 7(1*2^2+1*2^1+1*2^0 - binary "111", where another bit needs to be added to write 8(1*2^3+0*2^1+0*2^0). And so on.
14th August 2008
#10

Quote:
Originally Posted by Lupo
Came to think about the "1.76" that's missing in the equations above and wondered why exactly 1.76.. This one was spot on: SNR in ADCs: Where did all the bits go? - 6/7/2007 - EDN

The use of such equations implies lack of dither. Dither will make the figures slightly smaller than 6.02n+1.76 suggest, but the dynamic range will be infinite..
Thank you for your reply, Lupo. Thanks for the article, too, it's great.
14th August 2008
#11

Quote:
Originally Posted by Lupo

All signals approaching full scale needs all bits to be represented, not just at full scale.
Thanks, that's not what I meant though...

Quote:
If you have a number say 3(1*2^1+1*2^0 - binary "11") and wanted to use 4(1*2^2+0*2^1+0*2^0) - binary "100" ) as the next number, you'd need to add that other bit. That'll be good up to 7(1*2^2+1*2^1+1*2^0 - binary "111", where another bit needs to be added to write 8(1*2^3+0*2^1+0*2^0). And so on.
That said, we might as well get the whole explanation on binary numeral system, so for those who want to know more, click here.

Thanks Lupo.
14th August 2008
#12
Gear Guru

Quote:
Originally Posted by joerod
I thought that the bits used depended upon the signal's envelope, not upon a set of increments of 6 dBs per bit...IMO, bit 1=16777216, then 1 bit + 00000001 does not equal 2 Bits (12 dBs/-132 dBFS)...Or, does it?
Actually, the highest bit sets the loudest possible level, and every bit added below that reduces the noise floor by 6 dB. The absolute output voltage level for full scale can be any arbitrary value.

--Ethan
14th August 2008
#13

Quote:
Originally Posted by Ethan Winer
The absolute output voltage level for full scale can be any arbitrary value.

--Ethan
The absolute output level of a 24 bit file can be greater than 144 dBs if each bit value is say, 8 dBs as opposed to 6, correct?
14th August 2008
#14
Lives for gear

Verified Member
Quote:
Originally Posted by joerod
Thanks, that's not what I meant though...
Quote:
Originally Posted by joerod
All you have to keep in mind is that on CDs, all 16 used bits equal a program at FULL headroom (-0 dBFS), which is the same as those squashed hip hop records on some CDs who have no more bits to fill up.
There's nothing inherently wrong in using all bits. It does not equate squashed loud sound. Squashed sound doesn't run out of bits, it runs out of numbers! The 16'th bit provides 32768 additional positions to the 32768(15 bit) already available. A signal at position 32769 uses 16 bits and is only a fraction of a dB less than -6dB compared to full scale.

The 6dB is not arbitary, it's exactly it's twice as much. Adding a bit doubles the available numbers and 20 log 2 is always 6.02 dB.
14th August 2008
#15

Quote:
The absolute output level of a 24 bit file can be greater than 144 dBs if each bit value is say, 8 dBs as opposed to 6, correct?
I think there's confusion as to dBFS, voltage, and what you actually hear. Let's disregard what comes out your speakers as far as maximum loudness is concerned - that's arbitrary and has nothing to do with bit-depth.

On the most basic level, your output voltage from a DAC is either on or off (speaker cone out/in, or at rest). As you increase your bit resolution in the digital domain, you define more points between 'on' and 'off,' refining the voltage from square to sine (assuming we're dealing with a sine wave and not a complex audio stream). This refinement in voltage means your speakers have a smoother movement between in/out and rest. The more places they have to move, the more detail you hear - but your LOUDNESS is NOT increased. In other words, a full scale 1-bit square wave is going to be equally as loud as a full scale square wave at 24-bits.
14th August 2008
#16

Quote:
Originally Posted by Lupo
There's nothing inherently wrong in using all bits. It does not equate squashed loud sound. Squashed sound doesn't run out of bits, it runs out of numbers!
IMO, this is an image of a 16 Bit squashed sound, Are all 16 bits used here or not?

[IMG]http://*************************/images/meter_hi.gif[/IMG]

Quote:
The 6dB is not arbitary, it's exactly it's twice as much. Adding a bit doubles the available numbers and 20 log 2 is always 6.02 dB.
Adding an extra bit adds 6 dB of headroom, that's why 16 bits can decode and deliver a maximum 96 dB SPL to an audio file or a CD file correct?
14th August 2008
#17
Gear Maniac

Quote:
Originally Posted by Ben B
The addition of each bit to the wordlength doubles the number of values that can be expressed. 20 log 2 = 6.02. Is that what you're getting at?

-Ben B

i can understand 1 bit = 6 dbs more or less
the cuestion for me is:

How many steps are on a 6 dbs to a right wave representation?
have it diferent nature for Fixed point and Floating point?
A full scale 48 bit double precision stem has 48 bit depth, isn't it?

And a full scale 32 bit floating point what about bit depth in footroom tems?
14th August 2008
#18
Lives for gear

Verified Member
Quote:
Originally Posted by joerod
IMO, this is an image of a 16 Bit squashed sound, Are all 16 bit used here or not?
16 bits are used at full scale, as they still would be if the level was turned down -6.00dB. Anything in the upper 50% (20log2) 6dB needs that last bit to be represented.

Quote:
Originally Posted by joerod
Adding an extra bit adds 6 dB of headroom, that's why 16 bits can decode and deliver a maximum 96 dB SPL to an audio file or a CD file correct?
Long story very short: Each bit doubles the range, plus the 1.76dB the article linked to above explains. The 6dB is a consequence of the change being a doubling and is actually 20*(log2) = 6.0205999132796239042747778944899 and so on. There is no specific dB number that matches a doublng exactly. The use of decibels is just a convenient way of thinking audio, while the actual change going on is a precise doubling. What I'm trying to get at is that the doubling is the cause and the 6dB is the effect. It's a one way route, converting from dB to bits doesn't work as the dB can't be specified to endless decimal places.

If you add 16 times 6.02etc to 1.76 you get 98,089dB, the theoretical range of 16 bit with a sine reference. That assumes there is no dither noise. When dither is added, the dynamic range is boundless - except for the slight issue of the actual noise contained in the dither. So in a well implemented CD system with dithering, there's around -85 to -93dB (+1.76) dB range as displayed in the level meter, while the actual perceivable dynamic range spans far further than that.

Hope this helps
14th August 2008
#19

Quote:
Originally Posted by Lupo
Hope this helps
Helped a lot! Thanks thumbsup
14th August 2008
#20
Lives for gear

Verified Member
Quote:
Originally Posted by Lupo
Hope this helps

In case noone ever told you, you're very good at explaining! :-)

Could you maybe also write on the following subjects:

- Are we alone in the universe?
- Is there a god?
- What is the meaning on Life?
- The middle east conflict - why won't they just give it up?
- How to get rich quickly with minimal effort.

Thanks
15th August 2008
#21
Gear Guru
Think of it like F vs C scales for temperature. Water boils and freezes at the same absolute temperature, either way, i.e. the absolute difference between them is always the same. The only difference is that you are laying two different grids over that same distance, of of which has more gradations than the other.

Since F squeezes more numbers between those two points, you can measure the change in temperature more finely. But it doesn't change the actual point at which the water boils.

So a 24 bit capture has more dynamic *resolution* between 0dB and, say, -50dB, because there are more gradiations in that interval than a 16 bit system does. But it doesn't make 0dB (the boling point) any louder.
15th August 2008
#22

Quote:
Originally Posted by Dean Roddey

So a 24 bit capture has more dynamic *resolution* between 0dB and, say, -50dB, because there are more gradiations in that interval than a 16 bit system does. But it doesn't make 0dB (the boling point) any louder.
I get it, I get it fine, thanks....Here is my silly analogy then:

Here is my picture at 178 X 138 pixels 300 dpi...Handsome ha? And that's a real cuban habano, BTW...

Now , below is my picture also at 178 X 138 pixels, but at 72 dpi!
Same size but bit rate is lower.

Notice that I stole Darius's Admiral's hat, please don't tell. I figured I give myself a brief rank promotion at GS...heh

Oops, I did promise to be serious in this thread...Oh well...
15th August 2008
#23
Lives for gear

Verified Member
Quote:
Originally Posted by Dean Roddey

So a 24 bit capture has more dynamic *resolution* between 0dB and, say, -50dB, because there are more gradiations in that interval than a 16 bit system does.
Yes--the term "resolution" is a tricky one in this case.
15th August 2008
#24

Quote:
Originally Posted by Sunbreak Music
Yes--the term "resolution" is a tricky one in this case.

That's right Cass, that's why even when the capture has more dynamic *resolution* between -0dBFS and, say, -50dB, like with 32 Bit files, it doesn't make any difference to the sound captured if it came from lower resolution.
15th August 2008
#25
Lives for gear

Quote:
Originally Posted by Ethan Winer
Simple - every bit has approximately twice the value of the sum of all bits below it. And when you double the voltage, the signal goes up 6 dB.

--Ethan
Indeed, that's it.
15th August 2008
#26
Lives for gear

Quote:
Originally Posted by jordanstoner
On the most basic level, your output voltage from a DAC is either on or off (speaker cone out/in, or at rest). As you increase your bit resolution in the digital domain, you define more points between 'on' and 'off,' refining the voltage from square to sine (assuming we're dealing with a sine wave and not a complex audio stream). This refinement in voltage means your speakers have a smoother movement between in/out and rest. The more places they have to move, the more detail you hear - but your LOUDNESS is NOT increased. In other words, a full scale 1-bit square wave is going to be equally as loud as a full scale square wave at 24-bits.
That is the best explanation I have ever heard! thumbsup

"The more places they have to move, the more detail you hear - but your LOUDNESS is NOT increased."

The top level is the same, but the grid is finer, and therefore quieter sounds can be represented (greater dynamic range).
15th August 2008
#27
Lives for gear

Verified Member
Quote:
Originally Posted by joerod
I get it, I get it fine, thanks....Here is my silly analogy then:

Here is my picture at 178 X 138 pixels 300 dpi...Handsome ha? And that's a real cuban habano, BTW...

Now , below is my picture also at 178 X 138 pixels, but at 72 dpi!
Same size but bit rate is lower.

Notice that I stole Darius's Admiral's hat, please don't tell. I figured I give myself a brief rank promotion at GS...heh

Oops, I did promise to be serious in this thread...Oh well...
Your analogy is completely unrelated to the point -- 178x138 @ *any* dpi is still 178x138. The dpi is a display or reproduction resolution. The bit DEPTH (completely unrelated to the BIT RATE, which is a measure of transfer speed and has little to do with anything about PCM audio - or images for that matter) is the same, the pixel information is the same. The only time it would make a difference is if you're working in something that is displaying it via that information (a graphic/layout program perhaps). Still, the image is identical.

If you want a graphic comparison, compare a 256-color GIF to a 16.7million color jpg. Such as here: JPG vs. GIF

Although it's more extreme (24-bit vs. 8-bit), it's a reasonably related analogy.
15th August 2008
#28
Lives for gear

Verified Member
Quote:
Originally Posted by 24-96 Mastering
...
- Is there a god?
I know that one.

No, there is not.

... sorry God.
15th August 2008
#29

Quote:
Originally Posted by jordanstoner
a full scale 1-bit square wave is going to be equally as loud as a full scale square wave at 24-bits.
thumbsup

Quote:
Originally Posted by MASSIVE Master
Your analogy is completely unrelated to the point -- 178x138 @ *any* dpi is still 178x138. The dpi is a display or reproduction resolution. The bit DEPTH (completely unrelated to the BIT RATE, which is a measure of transfer speed and has little to do with anything about PCM audio - or images for that matter) is the same, the pixel information is the same. The only time it would make a difference is if you're working in something that is displaying it via that information (a graphic/layout program perhaps). Still, the image is identical.

If you want a graphic comparison, compare a 256-color GIF to a 16.7million color jpg. Such as here: JPG vs. GIF

Although it's more extreme (24-bit vs. 8-bit), it's a reasonably related analogy.
In my example, I meant that 178X138 = it's the loudness level (figuratively)
Thus,
Pic 1 / 288 DPI = Bit rate i.e. 48 Bit
Pic 2 / 72 DPI = Bit rate i.e.16 Bit

So, it didn't matter how many more bits are added (DPI), the loudness is the same (178X138). But, you knew that.

It was just another way to look at it...
15th August 2008
#30
Moderator

Verified Member
Couple quick geeky points... the plural of the abbreviation for decibel, dB, is not dBs; it is simply dB. Also, the d is lower case and the B is upper case because it is derived from a proper name. Next up, more bits does not equate to more headroom; it's better described as more footroom. It allows the recording of smaller and smaller signals, not larger and larger signals. The top of the scale is fixed. And last, digital graphics analogies are rather poor when it comes to digital audio. The two really don't work the same way.
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