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Author Topic: newbie has a question about bit depth!  (Read 4418 times)

David Barton

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newbie has a question about bit depth!
« on: June 05, 2006, 05:27:29 PM »

I have a question that troubles my analog mind:

As I understand it, bit depth governs s/n ratio. That’s great where head room is concerned, but what about sample resolution? I’ll explain:

Say I’m running at 16 bits - s:n is 96db. Then I switch to 24 bits and the s:n goes up. That might help me with respect to dynamic range, but what I’m looking for is higher resolution db-wise (as opposed to sample rate, which is time based). I want to use more bits-per-decibel, if that makes sense.

If I’m not making sense now, then think of a computer monitor. There is screen size and there is resolution. If you upgrade from a 19in. to a 23in., but the dots-per-inch on both screens are the same, you have increased the number of pixels on the screen, but the level of detail is the same. You can still see individual pixels, whereas raising the d.p.i. would put more pixels into the same space, smoothing the picture.

That is what I want to do with audio - use more bits per decibel and smooth the sound, so to speak. That is, use more bits to measure amplitude within a given range, so that increasing bit depth from 16 to 24 doesn't increase the range sampled, but more accurately samples the amplitude in that 96db range.

So my question is, how?


p.s. sorry for the wordiness - just trying to be precise Smile
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StudioRhythm

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Re: newbie has a question about bit depth!
« Reply #1 on: June 05, 2006, 05:40:59 PM »

As I understand it, there is really no such thing as "bits per decibel" or anything like it. It's counterintuitive, and I'm sure others will explain it to you in greater detail soon, but when a sound is sampled by an A/D and then reconstructed by a D/A, everything within the frequency and dynamic ranges captured is reconstructed perfectly (except for the coloration, noise and other by-products of the specific converters and filters used).

"Greater resolution" is a fiction as it pertains to digital audio. We don't hear samples, we hear a reconstructed waveform that has been filtered back into a continous wave, not a staircase of sample steps. Read some papers and you'll understand in more detail -- this is an extremely common and basic misconception that has taken on a life of its own.

Jon Hodgson

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Re: newbie has a question about bit depth!
« Reply #2 on: June 05, 2006, 06:26:49 PM »

The thing is to understand what more resolution in sample levels actually means in terms of what you hear.

Resolution and signal to noise are interchangable, if you're talking about the noise introduced by quantization rather than other components in the system (which in a 24 bit system are actually quite a bit larger than the quantization noise).

The obvious way to think about quantization error (the difference between an infinite resolution sample and the real resolution sample) is that the signal isn't being captured correctly. But another way to look at it is that once the signal passes through a DAC and filter, you get the original signal back (whatever your ADC resolution, be it 8, 16 or 24 bit) plus an error signal (which depends on the ADC resolution, becoming smaller for a higher resolution ADC).

That error signal is your noise, now the interesting thing is that for a given sample depth we can't reduce it, but we can change its "colour". If you don't do anything then the error is very dependent on the signal, with distinct frequency components depending on the input... your ear picks that up as distortion, not very pleasant.

You can randomize the noise, spreading it out, making it into white noise.. hiss basicaly, a lot less unpleasant. That's what dither is.

You can also get clever and shape it, so you have less noise in bands that your ear is sensitive to, and more where it is less sensitive.
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David Barton

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Re: newbie has a question about bit depth!
« Reply #3 on: June 06, 2006, 05:46:25 PM »

see i thought that increasing bit depth added dynamic range, and since you can't sample higher than 0db, you must increase range downwards. but you can only go so low until it doesn't make a difference (nobody can hear down to -144db).

but you're telling me that "everything is captured." so... what am i misunderstanding? if i hear you correctly, you're saying that both 16- and 24-bit systems capture the same range. this would mean that the 24-bit system samples with greater accuracy, right? if so, then my original question is moot since i'd already be sampling the same range in smaller incriments. is that it? sampling at 24bits captures the same range, but in smaller steps?

can you give me links to good papers? i've obviously looked over something quite fundamental.
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Jon Hodgson

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Re: newbie has a question about bit depth!
« Reply #4 on: June 06, 2006, 06:20:34 PM »

You're making distinctions where there are none.

Dynamic RANGE is not an absolute measurement, it is the difference between the loudest sound you can capture and the smallest.

The reason for that range is basically that the signal dissappears down into the noise.
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Graham Jordan

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Re: newbie has a question about bit depth!
« Reply #5 on: June 06, 2006, 07:26:16 PM »

Let me try to clarify a few things for you...

You can't have more 'bits per decibel', put simplisticly: 1bit = 6dB. Can't change that. For each extra bit of resolution, you get twice the peak signal compared to the noise floor, which is 6dB.

Also I think you're getting slightly confused with the specifics of 'dBs'. Remember dB is a measure of one signal compared to another, not an absolute measure in itself.

Remember, hearing/sound levels are in dB, relative to the threshold of hearing, i.e. 0dB is where you start hearing things, so 0dB in this case is the LOWEST signal level. 120dB is pretty much the limit before you blow out your ears. Dynamic range is the difference between the lowest and highest level signals. So in this case is 120dB (largest signal referenced to the smallest - remember dB always has a reference).

For digital signal levels, the usual reference is fullscale signal, so this is 0dB, more fully shown as 0dBFS (indicating that it is relative to full scale). Also, for pure digital signals the smallest signal is basically the same as the noise floor. The smallest signal is roughly a single bit in size. So, for 16-bit audio, the lowest signal level is about -96dBFS. This gives a dynamic range of 96dB. For 24-bit audio smallest signal is about -144dBFS, giving a dynamic range of 144dB.

Between 16- and 24-bit, dynamic range is not really simply 'extended up' or 'extended down'. For real signals that all depends on what digital level, in dBFS, you equate with normal signal level (in dBu, sound dB, or whatever). The main point is that 16-bit does not cover the dynamic range of the ear, but 24-bit does. Which is why in storage, media, converters, etc. you should never need more than 24-bits (internal digital processing is a separate issue).

With 16-bits you need to somehow fit the full ear range into just 96dB. This can be done by limiting the maximum signal level to equivalent of 96dB (sound), and keeping the noise level the same, or accepting a noise level of 24dB (sound, is 120dB - 96dB), or (more usually) a compromise between the two.

With 24-bit format, you don't need to make those compromises, when relating SPL to dBFS.

However, in all this remember that real analog electronics and converters, do not themselves have full 144dB dynamic range. You'll be paying a lot of money for A/D with 120dB dynamic range. I don't know the specifics of microphones, but I understand that they're nowhere near that level.

In your last post you talk about if 16-bit and 24-bit capture the 'same range'. I think you're a bit confused about the notion of the range. If you're referring to the ear's dynamic range (120dB -ish), then 16-bit doesn't capture this range, only 96dB of it - so you loose the highest level signals (either they get compressed, or clipped before A/D), and/or the lowest level signals (get lost in the noise floor of the A/D). With 24-bit you capture it all. This includes capturing the noise of your input devices, as they could be noisier than the A/D introduced noise. All of which are noisier than the inherent noise level with 24-bit.

Real A/Ds are at best equivalent to about 20bits, and real audio signals a lot less than this.

So say you have a signal in which the noise level, is -86dBFS (or equivalent in analog levels), and your A/D is accurate to -110dBFS levels, then the signal noise is 24dB more than the A/D noise. You basically have about 24dB (4 bits) more real 'accuracy' in the A/D than you need. These 4 bits are the sampled signal noise. Then there are still 5-6 bits of A/D generated noise (the remaining range from -110dBFS to -144dBFS).

Also, comparing sampling in digital audio to digitised pictures and monitor screens is not a good way to look at it. They are really too dissimilar. Let me just demonstrate with youre first post... On a screen, if you go to a larger monitor and increase the number of pixels, you're keeping the dpi the same. BUT, the way you actually look at a computer screen, what matters for 'smoothing' (so you can't see the pixels), is the angle subtended at your eye going from one pixel to the next. In other words, what are the 'dots per degree' as you look at the screen? You can get more dots per degree by, moving away from the screen. But then the image gets smaller, so you make the screen bigger, but keep the dpi the same. However, this has nothing to do with audio, and has no analogies with it.
Also, comparing image pixel smoothing with audio smoothing and detail, is way off base. If you think of pixels as being spatial 'samples' of the 'ideal' text font you're trying to read, then the sample 'rate' is far below the 'Nyquist' rate(spacing of samples in relation to the character feature detail size/scale). Fortunately on computer screens, they don't actually sample an ideal font, but generate a font out of colouring pixels so it kind of looks OK. To see what they would really look like if actually sampled a real font at that rate, think of scanning regular text at 100dpi, or printing at 100dpi. Yuk! The thing with audio is that is IS sampled at or above the Nyquist rate for the signals it is sampling (normally at least - otherwise you get that horrible metallic sound of aliasing), so it is already perfectly smoothed. And this has NOTHING to do with bit depth. Bit depth affects the noise level (at the risk of making an image comparison - how much 'fuzz' or 'snow' there is in the picture - more comparable to colour/grey level depth - this is equating image spatial dimensions to time in audio, and colour/grey level to voltage - but even this is not a direct comparison).

BTW: One of the best explainers on dither, uses image colour/grey-scale dithering (it's conceptually the same) to show it and how it relates to bit depth. Note that in this they used the highest resolution possible so that the pixel aspect of it is removed as far as possible (as it is in audio).
Here's a link to Wikipedia with something like what I saw:
http://en.wikipedia.org/wiki/Dithering
Look at the "Digital photography and image processing", NOT the audio section. The problem with the audio diagrams is that it can't really get across the notion of what you actually hear. But the images can.
Figure 1 = original (high bit depth, 'c.f 24-bit')
Figure 2 and 5 = truncated bit depth (c.f. 16-bit truncated) (note the lack of detail in areas where the image/signal change/level is close to the change by a single bit).
Figure 3 and 6 = dithered reduced bit depth (c.f. 16 bit from 24-bit with dither first) (note that there are no more 'sample levels' than in Fig2, but as you go across areas where the image is changing slowly, over a small areas of pixels the ratio of the numbers of pixels at the two sample levels changes. As your eye averages together dots that are very close together, you see much or overall detail than figure 2 - but not as much as Fig 1, but that's because the resolution/noise floor of our eyes is better than the level of dither).
If you screw up your eyes (effectively raising the noise floor on your vision), you get to a point where you cannot distinguish between Figs 1 and 3/5, but can still see the errors in Fig 2/4.

Hope this helps.
Graham
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David Barton

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Re: newbie has a question about bit depth!
« Reply #6 on: June 06, 2006, 11:36:02 PM »

Graham, thanks for taking the time to write all that out! i'm starting to grasp the range issue.

there is still an issue i don't understand, though, and it's the central point of my question, however poorly worded: why 6db per bit? you said i can't change. what is the reason? i realize that there may be no benefit, even if it were possible, due to the nature of D/A conversion being that a continuous wave is being reconstructed rather than a series of steps.

okay i just read what i wrote and i'll admit i feel kinda dumb right now... but that still leaves the 6db per bit question.
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tom eaton

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Re: newbie has a question about bit depth!
« Reply #7 on: June 07, 2006, 12:56:30 AM »

In a binary numerical system each time you add a digit you double the possible values that the binary number can represent.

1 bit can be either 1 or zero, so two values.
2 bits can be 00, 01, 10, or 11, so four values.
3 bits can be 000,001,010,011,100,101,110,111, eight possible values.
4 bits can represent 16 possible values
5 bits, 32
6 bits, 64
and so on

So if each additional bit on the binary number doubles the representable value, what does that mean when you represent an audio signal?  Well, for a signal to be twice as loud, it needs to be increased by 6dB.  If we have a signal that is x, x+6dB will be twice as loud.  So, if we have a two digit binary number, and we add an additional bit to the number, we increase by a factor of two the value that can be represented by that number.  An increase by a factor of two, of course, is doubling, and in audio, doubling is 6dB. So for each additional bit, the binary number can represent exactly twice as many values, exactly twice as much information, exactly twice as much signal... twice as much volume, which is to say: +6dB.

(well, almost...it's more like 6.02 if I recall correctly, excepting the LSB, but that's the theory behind it).

-tom

David Barton

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Re: newbie has a question about bit depth!
« Reply #8 on: June 07, 2006, 03:33:28 AM »

THAT is cool. you guys are awesome for putting in the time and effort to help me out. thank you so much!
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crm0922

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Re: newbie has a question about bit depth!
« Reply #9 on: June 07, 2006, 10:17:31 AM »

Don't forget that your computer's resolution, even on a very high performance system, is *far* below the the minimum resolution required hide "pixellation" from the human eye.  

Generally 300dpi is the resolution where the eye cannot easily discern the component pixels, and most screens run in the 72-96dpi range.  Thus 300dpi, and the common 600+dpi's seen in print is far more detailed.

The audio represented by 44.1/16 is very near the limits of the ear's ability to perceive frequency and dynamics.  Thus it is more like a high quality 300dpi+ print. Wink

In reality, the technologies are not very much alike, but I wanted to point out that video, due to speed/space/bandwidth requirements, is purposely limited in resolution.

Chris
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