| intervalkid wrote on Wed, 23 August 2006 05:39 |
I will admit that I am pretty ignorant and am hoping to learn.
I can only discuss according to my understanding as of now so here we go.
To my understanding (or so I thought, but am now in doubt)
for each sample the bit depth is applied. For 16/44.1 you would have 16 bits 44.1 thousand times per second. If this is so then 24 bits 192 thousands times per second seems like it would be closer to perfect representation. Now I notice that a wav. file of 16 bit 44.1khz is listed as having a bit rate or 1411kbps which is twice what I would expect from the above understanding.
Nonetheless, for whatever reason this is, I would assue that a 24 bit 192Khz wav file would have a bit rate of 9216kbps. If this is true, again I assume that the more information the higher fidelity, at least if the file has been band passed within audible frequencies.
Please tell me where I have erred.
Thanks
|
Ok, the first thing to understand is that more samples equals more data, but it does NOT equal more information.
What do I mean by this? Information is something that tells you something you would not otherwise know. How much data is required to represent all the information required depends on what you are trying to say, and also on the contect in which the data is presented. Take the following two messages
"Everton 2, Liverpool 1"
and
"Everton football team scored two goals in the match today against liverpool who score 1 goal in that match"
Which one has more data? The second one obviously, which one has more information? Well that depends on what you already know, if the message comes out of the blue then the second one does, but if you know you are listening to todays football scores, then they both contain exactly the same information.
Now let's take something a little different. You have a circle on a piece of paper, that circle is made up of an infinite number of points. How many samples of that circle do you need to take to be able to reproduce that circle exactly? An infinite number? No, actually it is three. Any three points on that circly allow you to reproduce the complete circle, so long as you know what you are trying to reproduce is a circle.
So, hopefully you can understand and accept that more data is not equal to more information.
Right now people are reading this and thinking (or possibly screaming at the screen) "But an audio signal is a hell of a lot more complex than a bloody circle!!". Well that's quite true, but it does have a mathematical representation. If we start with the simple case of a periodic signal, then ANY periodic signal can be composed of a combination of harmonics of the fundamental, with varying level and phase, up to the highest frequency contained in that signal.
With non periodic signals the maths gets more complicated, but the maths still exists. Not only that, but we can be about as certain as it is possible to be certain about anything that the maths is correct. The formulae have been prodded, twisted and tested by hundreds of mathemeticians, thousands of engineers and billions of people over the past century in many fields, including communications.
So unless the time space continuum somehow gets warped as soon pass a Keith Richards riff up to 20kHz or so through a wire (as opposed to passing a signal with a bandwidth of GHz and containing a few TV channels), then I think that Fourier et al can be taken as written.
So back to that mathematical representation of the audio signal.
We can break it down into two elements. The first is a filter, it is a lowpass filter and for the purposes of simplifying the discussion and avoiding reams of maths working out what the consequences of filter imperfections are we'll assume it is perfect, with signal components being passed perfectly below the cutoff and not being passed at all above it.
Then we need a signal to drive that filter, now what that signal looks like in its totality does not actually matter so long as EVERYTHING IN THE PASSBAND IS IDENTICAL TO THE ORIGINAL SIGNAL.
Let's say your original signal was a 0.1v magnitude sine wave at 20 Hz. Now you add a 1v square wave at 100kHz. What you see looks nothing like a sine wave, but put it through a brick wall filter with a 20kHz cutoff and what you will get back is that original sine wave.
So, what signal can we feed into the filter that has the same content below the cutoff point as the original signal? Well it turns out that a series of pulses at slightly more than twice the frequency of the cutoff actually contains all the same components below the cutoff as the original signal. It also contains a load of stuff above the cutoff which is completely different to the original signal (which had absolutely nothing above the cutoff, because our first stage was to bandlimit that to the audible range), so it looks very different, but our filter will remove all those additional components, leaving us with only what we want.
So, we need to generate a series of pulses at slightly more than twice the frequency of the filter cutoff, what information do we need to generate that series of pulses? The height of each pulse, which luckily happens to be the level of the original signal at a particular point in time, so we can get those levels by sampling the input at the same rate as we want to generate the pulses.
We now have ALL the INFORMATION required to reproduce the original signal. Capturing more DATA does not give us more information.
Ok, hopefully what I've written so far is understandable. But it is based purely on sample rate, where does sample bit-depth come into this? Well it affects how accurate each sample is.
People tend to think of finite sample resolution as meaning that something is missed out from the original signal, but it is easier to understand if you turn that around and think of it as something being ADDED to the original signal.
Imagine you have a sample based system which is perfect, except that the sampling has a finite number of quantization steps, what come out of the end?
Your original signal, PLUS an error signal.
Not that different from feeding audio through just about anything, put it through a channel on the best analogue desk in existance and what come out of the end is your original signal, plus and error (noise and distortion), though the error may well be so tiny you can't hear it.
And actually that's what that error is, noise and distortion. If the error is correlated to the signal, then you hear it as distortion, if it isn't (i.e. it is random), then you hear it as noise. But since in a 24bit system the error resulting from the quantization steps is not only below audibility (assuming you haven't set full scale such that it drives your eardrums through your brain), but way below the other errors in the system (i.e. the analogue noise).
