16-Bits was never enough

[psykostx]

Seriously though, how dare I say 16-bits wasn't enough? But "All of my CD's are 16-bit!" you say?

"I bet he doesn't even know what a bit is!!"

Do YOU know what a bit is? A better question, how does my computer process a 16-bit sample?!

Ok theres about to be a lot of numbers on your screen. Get ready to bookmark this page, or run away...

You already know a bit is a binary digit. It's either a 1 or a 0.
You also aready know that a cd contains waveform represented by 44100 16-bit samples per second.

So you know how many increments are in your sample rate, but how many increments are in your dynamic range?! I'm pretty sure most people say, obviously, 16 bits. CD resolution is 44100 by 16-bits. You probably like 24-bits because you know it has more dynamic range, and thus the potential to sound louder and clearer. So how many more increments do you have in 24-bit? You say 8 bits? Well, thats kinda like Nigel the guitarist saying that the volume knob on his Marshall TSL goes to 11. How much louder is it? One more. Of course.
NOT! (anyone know how much blacker a black album cover can be?)

So now we need to know how many whole number integers a computer can represent with 16-bit. For that, you need to know how to use the secret code of the binary number system. Wow, what a nerd I have become. Well heres a secret, nerds make better music than stoned out hippies!
dfegad
Stoned out nerds is another post altogether...

You will have to take my word for it that 16-bits is:

16384 8192 4096 2048 1024 512 256 128 64 32 16 8 4 2 1 0

16-bits = 16384 decimal integer increments.

So that means that a CD has a resolution of (44100 samples per second) x (16384 increments per sample). Yes you can simplify the math out farther, but that doesn't help out our case, because we want a static resolution, not dynamic. If you must know, its 722534400 bit increments a second, which is roughly 700Kbit per second per channel, or 1.4Mbit/s dynamic (streaming) resolution stereo.

So we have 16384 volume increments in our dynamic range. May seem like a lot to some people...and it would be, in a perfect world. Since I suck at logarithms (and the point of my post is to embarass you publicly, not me fuuck) we won't be proving that most of those increments are in the top 1/10 of the decibel range. I can however show you how many "volume increments" there are per decibel on AVERAGE.

Assume a 96db spl playback system is used.
16384 increments, divided by 96 decibels, is roughly 171 increments per decibel average...or almost exactly 6db per bit if you continue out the math. Already this sounds skimpy... you may not "hear" those 167 increments, but what is going on in the space between them? A ME's most loyal adversary and spiteful foe, DISTORTION!!! DUHN, DUHNNA DUHN!

But wait, 171 increments is plenty for each of thel 98 decibels. The wiki post regarding this matter even says I get 6db per bit! Its wrong! Remember, when I said average? Really, the decibel scale is logarithmic. That means that the number of increments within each decibel isn't always the same. In fact it gets smaller and smaller, as you go down in db. It gets so small that the last bit is either 1 or 0 (1 actually means silence, 0 means that another bit is in use). That means that the number of increments from 0db to -6db is twice that of -6 to -12 db. All the detail is in the peaks of your audio, the part you spend all your time and money compressing the detail out of........

16-bits isn't enough because really the bottom 8 bits of audio are distortion waiting to happen when you bring up the level....(again, I'll let someone else post the Log equation). A CD is just barely situated atop a sea of distortion...even more than a tape, but that shouldn't be news to a mastering engineer!

PS: Maybe this post could go in the mixing forum instead, seems like it could be useful here as well...

...also yes you can hire me to write articles/essays for you...not to mention software...

[mobius.media]

Originally Posted by psykostx

You will have to take my word for it that 16-bits is:

16384 8192 4096 2048 1024 512 256 128 64 32 16 8 4 2 1 0

16-bits = 16384 decimal integer increments.

So that means that a CD has a resolution of (44100 samples per second) x (16384 increments per sample). Yes you can simplify the math out farther, but that doesn't help out our case, because we want a static resolution, not dynamic. If you must know, its 722534400 bit increments a second, which is roughly 700Kbit per second per channel, or 1.4Mbit/s dynamic (streaming) resolution.

So we have 16384 volume increments in our dynamic range. May seem like a lot to some people...and it would be, in a perfect world. Since I suck at logarithms (and the point of my post is to embarass you publicly, not me fuuck) we won't be proving that most of those increments are in the top 1/10 of the decibel range. I can however show you how many "volume increments" there are per decibel on AVERAGE.

Assume a 96db spl playback system is used.
16384 increments, divided by 96 decibels, is roughly 171 increments per decibel average...or almost exactly 6db per bit if you continue out the math. Already this sounds skimpy... you may not "hear" those 167 increments, but what is going on in the space between them? A ME's most loyal adversary and spiteful foe, DISTORTION!!! DUHN, DUHNNA DUHN!

Thanks. I've always wondered about that but didn't know how to calculate it as you have.

Is it possible to develop an alternative digital format with greater incremental resolution per dB?

[J.S.Vega III]

None... none more black.

Interesting and educational post, thanks!

[psykostx]

Originally Posted by mobius.media

Thanks. I've always wondered about that but didn't know how to calculate it as you have.

Is it possible to develop an alternative digital format with greater incremental resolution per dB?

Sure. Binary is easy. I should have labeled the columns. Here's a simple example of how to calculate decimal (base 10) numbers into binary (base 2).

4-bits can represent up to the number 7:
4 2 1 0
1 1 1 0
(That is, 1110 in binary is the number 7 in decimal)


The number 5 would be:
4 2 1 0
1 0 1 0
(That is to say, there is one 4, zero 2's, one 1, and zero 0's. 4 + 1 = 5 )

The number 8 (in 8-bits) would be:
64 32 16 8 4 2 1 0
0 0 0 1 0 0 0 0
(So 00010000 is the number 8, which brings me to the next point...)


Yes it is without a doubt possible to create another format, and it would be indeed be easier for a person to calculate (although it defintely wouldn't be real time) but computers work in binary, so every computer program is abstracted from binary operations. You would just be adding to the number of binary operations the computer would have to perform to represent and manipulate a waveform with decimals, thus reducing the performance of the sytem by a whole whole lot.
It would kinda be ridiculous as if everyone said "two plus two" instead of just saying "four." Computers speak binary, so its easier for them to do math that way, even though its absolutely mindnumbing for their carbon based creators....

[psykostx]

Originally Posted by J.S.Vega III

None... none more black.

Interesting and educational post, thanks!

Lol. Absolutely anytime my friend. Think for yourself, question authority, then realize the existence of authority itself is questionable!

[mobius.media]

Originally Posted by psykostx

Yes it is without a doubt possible to create another format, and it would be indeed be easier for a person to calculate (although it defintely wouldn't be real time) but computers work in binary, so every computer program is abstracted from binary operations. You would just be adding to the number of binary operations the computer would have to perform to represent and manipulate a waveform with decimals, thus reducing the performance of the sytem by a whole whole lot.
It would kinda be ridiculous as if everyone said "two plus two" instead of just saying "four." Computers speak binary, so its easier for them to do math that way, even though its absolutely mindnumbing for their carbon based creators....

Makes sense. That's what I was thinking.

So I guess Intel/AMD better get moving on making some much, much faster computers.

[psykostx]

Originally Posted by mobius.media

Makes sense. That's what I was thinking.

So I guess Intel/AMD better get moving on making some much, much faster computers.

I'm sure 64-bit audio is on its way. All it takes is a major software industry standards upgrade. Imagine a 32-bit wasteland underneath a blisteringly clear 32-bit peak range! Creative distortion may be a thing of the past...and physical hearing capability may start to matter logarithmically more than it does already! The future of audio is what photographs were to paintings...I promise. Start saving your money for holographic projection speakers...er panels, er paint??!!

[mobius.media]

Originally Posted by psykostx

I'm sure 64-bit audio is on its way. All it takes is a major software industry standards upgrade. Imagine a 32-bit wasteland underneath a blisteringly clear 32-bit peak range!

But isn't the point that as long as we're using current formats, 64 bit, like 24 vs 16 bit, will provide no greater dynamic resolution, just more largely pointless dynamic range added to the bottom?

Also, on another note, does anyone know how to calculate the frequency resolution of digital?

[Mihaly]

Originally Posted by psykostx

Seriously though, how dare I say 16-bits wasn't enough? But "All of my CD's are 16-bit!" you say?

"I bet he doesn't even know what a bit is!!"Stoned out nerds is another post altogether...

You will have to take my word for it that 16-bits is:

16384 8192 4096 2048 1024 512 256 128 64 32 16 8 4 2 1 0

16-bits = 16384 decimal integer increments.

..

Actually 16 bits = 65536 integer increments. So everything after that point is in error. And the 16th bit is worth 2*16384.

And while the dB scale is logarythmic the , the encoding for CD is linear PCM.

Mihaly.

[Nordenstam]

The first bit/first 6dB's is on or off. The next bit have two more choices added to that - four level options. The next one have 8 level options, and so on. Dividing total range by number of steps makes no sense.

The frequency resolution is set by the sample rate. A sampled system can preserve all frequencies below FS/2.

[Nordenstam]

PS: this discussion omits the fact that any digital audio system not using dithering is for all practical purposes a broken system. In a dithered system, dynamic range depends on the level of the dither noise - not on the available number of steps. A 16 bit signal with proper dithering gives about 120dB of dynamic range in the most sensitive frequency region. A 24 bit signal is only marginally better than that..

[mobius.media]

Originally Posted by Lupo

The frequency resolution is set by the sample rate. A sampled system can preserve all frequencies below FS/2.

Don't you mean the frequency bandwidth is set by the sample rate? Surely frequency has finite resolution as well.

ie. Can a digital system represent the difference between 11000 hz and 11000.000000000000000001 hz? Where does the rounding occur?

[psykostx]

Originally Posted by Mihaly

Actually 16 bits = 65536 integer increments. So everything after that point is in error. And the 16th bit is worth 2*16384.

And while the dB scale is logarythmic the , the encoding for CD is linear PCM.

Mihaly.

I was talking about a single PCM channel by the way.... and you are absolutely right, I used the wrong bit value. But the answers I got are still calculated correctly, so I would appreciate if you would please withdraw your comment about everything being wrong. I will fix the error... thanks for pointing it out! 16-bit unsigned int is 65535. Duh me. Heh, not a bad post off the top of my head tho!

BTW, I program but I don't do a lot of bitwise operations, I'm still on the higher level GUI/DLL stuff. Always learn everyday what a dumbass I am in some way or another!

ALSO, logarithmic scale represented by a linear value is still what I was talking about, just had it backwards...

PS: I did tell everyone to think for themselves didn't I!!! And look ... Thank you!

[psykostx]

Originally Posted by Lupo

PS: this discussion omits the fact that any digital audio system not using dithering is for all practical purposes a broken system. In a dithered system, dynamic range depends on the level of the dither noise - not on the available number of steps. A 16 bit signal with proper dithering gives about 120dB of dynamic range in the most sensitive frequency region. A 24 bit signal is only marginally better than that..

Dithering is filler for missing data... its a guassian blur effect, no? It was conceived to cover deficiences in dynamic resolution, correct?

Can audio be captured and played back with more detail so that it more closely resembles reality?

[Jon Hodgson]

It's a bit too easy to get caught up on the "stepped" image of the sample stream, you have to think of what comes out of the reconstruction filter after the DAC.

The best way to think of it is as the original signal plus an error (an analogue system would be the same so long as you're talking about signals within its bandwdith you get the orginal signal, plus noise, which is an error).

In a properly dithered sample stream that error is statistically random, i.e. noise, and the system maintains resolution below the LSB, you can hear below the digital noise floor just as you can below an analogue noise floor.

So I'm afraid the original post is rather misleading.

As a delivery system, 16 bits is probably adequate in most cases (it has a higher signal to noise ratio than vinyl), especially if we use higher resolutions in the preceding stages, but of course these days we can do better, it's a shame the shops are full of CDs and not DVD Audio.

[mobius.media]

Originally Posted by Jon Hodgson

It's a bit too easy to get caught up on the "stepped" image of the sample stream, you have to think of what comes out of the reconstruction filter after the DAC.

So I'm afraid the original post is rather misleading.

I don't think so. We're not talking about how it sounds, we're talking about how it exists in a computer, where there is definitely finite, quantifiable resolution.

[Jon Hodgson]

Originally Posted by mobius.media

Don't you mean the frequency bandwidth is set by the sample rate? Surely frequency has finite resolution as well.

ie. Can a digital system represent the difference between 11000 hz and 11000.000000000000000001 hz? Where does the rounding occur?

A sample stream can repreent any frequency below nyquist, with any phase relationship, if you change the frequency slightly or change the phase slightly then you change the instantaneous level at the sample instances.

[Jon Hodgson]

Originally Posted by mobius.media

I don't think so. We're not talking about how it sounds, we're talking about how it exists in a computer, where there is definitely finite, quantifiable resolution.

Yes, it has finite resolution, but what matters is what signal is represented and manipulated. It is most certainly about how it sounds.

All audio systems have errors, analogue and digital, what matters is how those errors sound.

[psykostx]

Originally Posted by mobius.media

I don't think so. We're not talking about how it sounds, we're talking about how it exists in a computer, where there is definitely finite, quantifiable resolution.

Thank you! I know I made a rather large error in the original post, it was purely and simply haste. I practice drums, not binary, ya know! I'm learning programming and I thought I'd share some information. All these robots just came out of the woodwork all "invalid though, invalid thought" My opinion is still valid Mihal and Lupo, as was my description of digital media. I got my binary conversion wrong, and I'm sorry! I thank you for correcting me!

[mobius.media]

Originally Posted by Jon Hodgson

A sample stream can repreent any frequency below nyquist, with any phase relationship, if you change the frequency slightly or change the phase slightly then you change the instantaneous level at the sample instances.

So you're saying there is infinite frequency resolution? How can a finite data stream represent infinity?

That's ridiculous.

[dcollins]

Originally Posted by psykostx

You will have to take my word for it that 16-bits is:

16384 8192 4096 2048 1024 512 256 128 64 32 16 8 4 2 1 0

16-bits = 16384 decimal integer increments.

Oh dear.


DC

[dcollins]

Originally Posted by psykostx

Dithering is filler for missing data... its a guassian blur effect, no? It was conceived to cover deficiences in dynamic resolution, correct?

Oh dear.

DC

[psykostx]

Originally Posted by psykostx

and the point of my post is to embarass you publicly, not me fuuck

makes this post timeless!!!!!! Classic example of why you shouldn't post off the top of your head!!!! Lololololol

[psykostx]

Originally Posted by dcollins

Oh dear.

DC

Don't "oh dear" me, it rounds off the square wave which is composed of harmonic distortion in the analog world to a more suitable sinewave...it is pretty much a guassian blur that is generated by noise, as far as I can tell....

ALSO, I ALREADY ADMITTED TO THE BINARY MISTAKE! ENOUGH! WE GET IT! ITS OBVIOUS NOW THAT I LOOKED IT UP! LOL

[Jon Hodgson]

Originally Posted by mobius.media

So you're saying there is infinite frequency resolution? How can a finite data stream represent infinity?

That's ridiculous.

Well Shannon's proof has stood up to the scrutiny of mathemeticians for a few decades and pretty much all modern communications relies on it, so I'd say you're a bit quick labelling it as ridiculous.

The finite sample rate is not an issue with frequency and phase accuracy, the finite quantization error is, but in a properly implemented system it's no different from the limitation caused by noise in an analogue system.

[Nordenstam]

http://www.lavryengineering.com/docu...ing_Theory.pdf

[mobius.media]

Originally Posted by Jon Hodgson

Well Shannon's proof has stood up to the scrutiny of mathemeticians for a few decades and pretty much all modern communications relies on it, so I'd say you're a bit quick labelling it as ridiculous.

The finite sample rate is not an issue with frequency and phase accuracy, the finite quantization error is, but in a properly implemented system it's no different from the limitation caused by noise in an analogue system.

You're sidestepping.

Frequency in reality has infinite resolution within a given bandwidth.
Frequency as stored in a digital file does not.

[Jon Hodgson]

Originally Posted by mobius.media

You're sidestepping.

Frequency in reality has infinite resolution within a given bandwidth.
Frequency as stored in a digital file does not.

True, but neither does freqency stored on a piece of analog tape, for the same reason, when you get below a certain level of difference, you get into the random error caused by noise, given an infinitely long piece of tape or an infinite digital sample stream, you would on average have the infinite frequency resolution, in the real world you don't, but nobody's noticed with tape yet.

[mobius.media]

Originally Posted by Jon Hodgson

True, but neither does freqency stored on a piece of analog tape, for the same reason, when you get below a certain level of difference, you get into the random error caused by noise, given an infinitely long piece of tape or an infinite digital sample stream, you would on average have the infinite frequency resolution, in the real world you don't, but nobody's noticed with tape yet.

No one's talking about tape.

This is a discussion about quantifying digital resolution, which I think is valuable, since it's rarely considered.

[dcollins]

Originally Posted by psykostx

Don't "oh dear" me, it rounds off the square wave which is composed of harmonic distortion in the analog world to a more suitable sinewave...it is pretty much a guassian blur that is generated by noise, as far as I can tell....

This thread is making me sad.

DC