32 Bit, does it make any sense?

[Master Zap]

Well it doesnt exactly take 5 minutes to answer the question "Do we get +127 to -128 or not?" does it? [img]http://www.musicplayer.com/ubb/biggrin.gif[/img] /Z

[rold]

[quote]Originally posted by PeeTee: [b]Mika...you work at Sweetwater? That explains a lot. Thank you for not letting me waste my time any further.[/b][/quote] Nika is a very knowledgeable guy and it is for all of our benefit that he is here. His affiliation with Sweetwater should not change that. First Sven and Alphajerk, now Nika. Are you on a mission to offend *everyone* here? Think twice about that...besides, it's MY job.. [img]http://www.musicplayer.com/ubb/biggrin.gif[/img]

[Nika]

[quote]Originally posted by Master Zap: [b]Well it doesnt exactly take 5 minutes to answer the question "Do we get +127 to -128 or not?" does it? [img]http://www.musicplayer.com/ubb/biggrin.gif[/img] /Z[/b][/quote] Zap, Sine wave with a noise floor down 293db, turned up to maximum gain going into an 8 bit converter: Yes. Of course. You get from -128 to 127. Now, please elaborate, because I fail to see what that proves, if that is TRULY what the point of the question was - merely to establish this minor specific sitaution in case you'll be recording sine waves at 8 bit this afternoon. Nika.

[Master Zap]

The question was for a 16 bit converter. You are mr "dodge the question" tonight (its night here). [img]http://www.musicplayer.com/ubb/biggrin.gif[/img] Reread the question. Or wait, I restate it: You have a sinewave that you plug into an 8 bit soundblaster. It will register as +127 to -128. You put the SAME sinewave into a 16 bit soundblaster card. Will it register as +127 to -128 sample steps, or not? /Z

[Nika]

[quote]Originally posted by Master Zap: [b]The question was for a 16 bit converter. You are mr "dodge the question" tonight (its night here). [img]http://www.musicplayer.com/ubb/biggrin.gif[/img] Reread the question. Or wait, I restate it: You have a sinewave that you plug into an 8 bit soundblaster. It will register as +127 to -128. You put the SAME sinewave into a 16 bit soundblaster card. Will it register as +127 to -128 sample steps, or not? /Z[/b][/quote] Ahh, sorry. Didn't mean to dodge. I forgot the question. No, if the reference voltages on the converters are the same then no, it will register from ~-32,000 to ~+32,000 Nika.

[Master Zap]

Exactly. I.e: More resolution. Adding bits yeilds more resolution. Q.E.D. I.e. I'm right, and you are a little bit less right. Ipso Facto. Pax Vobiscum. And whatnot [img]http://www.musicplayer.com/ubb/biggrin.gif[/img] [img]http://www.musicplayer.com/ubb/wink.gif[/img] [img]http://www.musicplayer.com/ubb/smile.gif[/img] Of course all your stuff is very interesting. I think you are a bit wrong at some things though. I do think resolving the noise is very very VERY important. Remember for instance that if you dither something, the dither is (most commonly) done in the lowest bit only (otherwise you would increase the S/N and you dont want that, do you?). Now multiply this signal by 1.51 times. Your former dither of 1 lsb becomes - due to rounding - 2 steps instead of one. Now you made the signal WORSE than it was before. If we had had the extra [i]resolution[/i] and dithered in 4 bits instead (which should be "enough" for most cases) this would not have happened. For a really good quality transfer [i]that is safe to put arithmetic operations on afterwards[/i] I would say, between thumb and forefinger, that resolving your noise to 3 or 4 bits is plenty, [i]and is very very very useful[/i]. Burying your noise in 1 bit I do not believe in, except for the FINAL STAGE in the digital chain. The mastering engineer may do 1 bit dithering, but nobobdy else should!! And then pray to god the CD player doesnt do digital volume knobs!! Also, as stated before, I find using the "analog" terms of S/N ratio *OR* dynamic range very unfitting for digital audio, which, in fact, is numbers, resolved to a certain number of bits. Nothing more, nothing less. /Z

[Master Zap]

Ok I did some tests. If the noise in the signal is pure white noise at a level that after truncation equals 1 LSB, truncating the bits is almost inaudible, coz you get free dithering by that very noise. If you add dithering at 1 LSB you get twice the noise and no other gain. However, as soon as you deviate the noise from white noise, the truncated one sounds very much different from the original. The dithered one sounds slightly better but with dither noise added (i.e. twice the noise). /Z

[Curve Dominant]

[quote][b]quote: ------------------------------------------------------------------------ Originally posted by Master Zap: The question was for a 16 bit converter. You are mr "dodge the question" tonight (its night here). Reread the question. Or wait, I restate it: You have a sinewave that you plug into an 8 bit soundblaster. It will register as +127 to -128. You put the SAME sinewave into a 16 bit soundblaster card. Will it register as +127 to -128 sample steps, or not? /Z ------------------------------------------------------------------------ Ahh, sorry. Didn't mean to dodge. I forgot the question. No, if the reference voltages on the converters are the same then no, it will register from ~-32,000 to ~+32,000 Nika. IP: Logged Master Zap Senior Member Posts: 310 Registered: Dec 2000 posted 09-09-2001 03:22 PM             ------------------------------------------------------------------------ Exactly. I.e: More resolution. Adding bits yeilds more resolution.[/b][/quote] No, actually. I think what he was saying...what he's BEEN saying ALL ALONG (Nika please correct me if I'm wrong and I'll get out the sauté pans), is that you are simply adding more numbers to your measurement. You are not actually increasing the resolution of the audio. That's why design engineers like Paul Frindle don't like the r-word - it's misleading. Sorry, Zap, but my money's on the guy who builds Oxfords. RE: Peetee...don't mind him. He's from the Digi board. [img]http://www.musicplayer.com/ubb/biggrin.gif[/img] E [img]http://www.musicplayer.com/ubb/smile.gif[/img]

[videoeditor1]

Curve, Sorry man. If you're gonna join this cluster f***, you're gonna have to STATE your case - kinda like an hypothesis... [img]http://www.musicplayer.com/ubb/wink.gif[/img] At this point, I can't even remember who said what, but I've asked the principals [b]not[/b] to go back an do any editing, just so those unlearned amongst us (me, primarily) can follow. So, what's [b]your[/b] position on the whole shebang? I've been fasting all day, just in case I have to be the one to eat the crowfue. [img]http://www.musicplayer.com/ubb/smile.gif[/img] NYC Drew

[Uh Clem]

I'm beginning to think both Nika and Zap are correct - they are just arguing one in french, the other Japanese - "The river is to the east".... "you are wrong, that mountain was once a volcano". I've been reading about this all day. What I've found that makes sense: * a source sample at 24 bit will have 16777216 possible values that can be assigned to the level of the wave at any point in time. A 16 bit sampling will only have 64K. Half of the values will represent values above the zero crossing, the other half below. The technical term seems to be "number of quantization values", but I could see why "resolution" would seem a reasonable substitute. The difference is probably subtle, but isn't that the bitch of semantics? This directly relates to quantization errors (approximations of the actual level since the number of bits cannot correctly represent them), which shows up as random noise added to the original signal. * The 6dB of dynamic range per bit is widely accepted, but I don't think the concept is intuitive. [I suspect it goes something like this: adding 1 bit is essentially multiplying your QVs by 2, dB is a ratio, 6dB is double the power, (your new QV)/(your old QV) = 2x or a 6dB increase...but I could be wrong]. I have also found that even this is technically called Signal to Quantization Noise Ratio. It is 2^n, where n is the number of bits and is usually expressed in dB where each bit is roughly 6dB of SQNR. This is a ratio of maximum signal amplitude to max Q error. * I don't think Zap meant to get so deep into this. I think he really meant the obvious: more bits equal more possible discreet data values to assign to the amplitude measured at a given sample time. I don't think this can be disputed - if he mis-named it resolution, well OK, but I think we got the idea. And resolution is not so bad a term - at really low levels of amplitude, a lower bit depth cannot properly [i]resolve[/i] a signal - I think this is true. The quantization error becomes too high. * Lastly, I have not yet come up with what I feel is a reproducible test for this, but I'll work on it - I would like to be able to easily demonstrate the effect of bit depth, amplitude, and SQNR so that I can feel I have a handle on the whole thing. This message has been edited by stevepow on 09-09-2001 at 10:55 PM

[videoeditor1]

Stevepow, your 425 words make a helluva lot of sense. I'm agree with you 100%. NYC Drew

[Nika]

Steve, Thanks for doing your homework. Good info! Curve, I think I understand what you're saying, and you're close about what I'm saying, but I should clarify. Drew, A dash of worcestershire sauce helps. Zap, OK. First we should clarify a couple of things. Noise=white noise. Any other form of noise is considered noise WITH SIGNAL, and yes, you're right, you'll get different results if you treat the [i]apparent[/i] noise floor as the [i]actual[/i] noise floor. The [i]actual[/i] noise floor is where the signal actually drops off into white noise only. If your signal drops off into pink noise, or other filtered noise then you have not actually hit the noise floor yet. "Signal to noise" ratio deals specifically with white, pure, natural noise. When discussions over signal to noise ratios are brought up it is implied that the noise being spoken of is the fundamental level of white noise. In our community noise=white noise. If you thought we were discussing a different, more variable type of noise then we should clear that up now. Any signal above the white noise floor is actually valid signal that DOES need to get captured in its entirety properly in order to give a complete representation of the signal. Next, your sine wave example. As I've stated, the bit depth tells us the maximum dynamic range that can be captured. If you want me to sample a signal that has a signal to noise ratio of 40db, this means that from the peak of the signal to the white noise floor there is 40db of dynamic range, then you will need to have 7 bits of dynamic range to handle this. Anything short of this will indeed not have enough "resolution", or "quantization steps" to accurately reproduce your material. Anything more than this does not give any additional resolution of the material, but it does give additional quantization steps. More on that to follow. Clearly the signal has the benefit of more quantization points, but if the majority of them aren't "valid", or do not provide useful information, then there is still the same "resolution" to the signal, even though the signal was measured more accurately and with more quantization points. Again, though, if you do not provide enough bit depth then I do agree that your "resolution" will suffer. One has to provide at least a certain amount of quantization points to accurately sample material. If you'd like to throw the following quote from me as a victory in your court that is fine, but I would never have argued against this, nor (do I believe) was this your point, "If you sample at a bit depth that yields a dynamic range that is LESS than the signal to noise ratio of the material then increasing the bit depth will increase the resolution of the signal." Now what you have provided for us as an example gets us nowhere near closer to "resolution" on this topic (pun was damn well intended!) You have given me a signal that has a signal to noise ratio of 293db. Obviously any bit depth under 48bits is not going to be capable of accurately sampling this waveform, and thus any number of bits you give me (8, 16, whatever) will not provide accurate resolution. Duh. I'm wondering if, by chance, the purpose of giving such a high number was specifically to get me to agree that it is "underresolved" to prove a point that doesn't need to be proven? I'm not sure. Now let's get back to real world scenarios. Let's talk about a sine wave with a signal to noise ratio of 42db. As I said back on day one in this, it will take 7 bits, or 128 quantization points to accurately capture and reproduce this sine wave. Now, to hopefully answer your next question, the signal, when maximized in an 8 bit converter, will indeed register from -128 to +127. When put into a 16 bit converter, will indeed register from ~-32,000 to ~+32,000. This is an example where more "quantization points" are used to capture this audio. Unfortunately, though, it is specifically NOT more resolution. Let me try to explain why: Even though we have essentially 65,000 points now to describe that sinewave, it is really divided up into 128 chunks of about 512 quantization steps. The signal is going to fall within that window, but exactly where within that window is not important because the resolution within that window of 512 quantization steps only helps to describe the white noise. But since this is white noise, it is unnecessary to describe it with precision, as Zap's experiment has essentially shown us. What this means is that the behaviour of the signal within that window of 512 quantization steps is totally random. * So the signal will pass through all of these 128 groups of 512 quantization steps, but the fact that it does that, and at what times it passes through these ranges is all we need to know. Exactly where within that window it passes is totally irrelevant. Now if we turn the signal down 6db so that we are only using 15 bits of our 16bit system then we still only have 128 quantization steps for the signal itself, each of which is now divided into 256 quantization steps for the noise. Now you can say, "what if the signal passes through that 512 or 256 quantization points in a very organized fashion?" Well then that would not be noise! That would be some sort of filtered noise, or not noise at all, and changes the situation entirely. We are no longer dealing with 42dB SNR. We're now dealing with some other signal to noise ratio. In other words, if the signal passes through that window of 512 quantization points [i]dead on center[/i] then you have changed the situation. This no longer is 42dB SNR. if it really passes through [i]dead on center[/i] then you actually have a sine wave that has more SNR than this sampling paradigm provides for (16 bits is what we're talking about), so this sine wave has a signal to noise ratio of greater than 96db, so we need to increase our bit depth until there IS a determinable resolution. But that 42db signal is only ITSELF ever quantized into 128 discreet increments, and thus its resolution does not change no matter how many quantization points we add. As I said before, all that does is give you better resolution of a totally random noise signal. So again, we need to be clear about using the word "resolution". The audio signal itself never has any better resolution than the minimum amount necessary to describe it, which can be ascertained by it's dynamic range (or signal to noise ratio). Increasing the bit depth does, in no way, benefit the "resolution" of the signal, and thus we try to avoid using that word to describe the effect of adding bit depth. All that adding bit depth does is allow us to accurately record material with wider dynamic range. Again, I don't care HOW we use the word "resolution". The SIGNAL itself does not have any additional "resolution", "quantization steps", "discreet measurement increments", or any other term to describe units of measurement when we increase the bit depth beyond what is necessary to accurately describe the signal. Any objection to this either shows that I'm really not good at describing things, or is a semantical ploy to give credence to a blatant validation for the subject's distaste for crow. I hope this helps. For those reading along, follow this sidebar* Nika. *Sidebar: This is exactly how dither is done. Every sample is put through a random number generator where it adds a random number of quantization steps to the sample. If we're dealing with 24 bits being dithered down to 16bits, a random number from -256 to +256 is added. Then the last eight bits are lumped off. Just like in the situation above, this has randomized the signal to within a window of 512 quantization steps before lowering the dynamic range by truncating off the last eight bits - which are now completely noise anyway. It is this dithering process at the end that negates the need to accurately sample the noise going IN to the system. This brings Paul Frindle's comments from page one into a little more light: [i]"Oh Guys, you seem to be making very hard work of this subject using the term 'resolution' and applying 'bits' to it as though they were hard boundaries. They are not at all hard boundaries if we dither the signal correctly.[/i] In other words, since we're going to be randomizing the signal anyway, it doesn't matter exactly HOW random it is coming in. So 16 bits or 24 bits - providing for better quality resolution and randomness of the noise is therefore unimportant.

[Curve Dominant]

I'll concur with that as well. That Stevepow is helluva tough! [img]http://www.musicplayer.com/ubb/smile.gif[/img] Drew, don't make the mistake of assuming that I know what I'm talking about. I'm rollin' the dice everytime I chime in on one of these tech-related threads. I'm gonna go back to the kitchen now, and cook myself a nice dish of Fettucini Crow-fedo. Ciao! E [img]http://www.musicplayer.com/ubb/smile.gif[/img]

[Guest]

oops This message has been edited by kBw on 09-10-2001 at 11:27 AM

[Nika]

For those looking for a test, I believe I've thought of one that can demonstrate this phenomenon. Take a microphone and split it and run it into both sides of a stereo mic pre. Not a "great" mic, nor a "great" mic preamp. We're trying to insure a failry low signal to noise ratio. Now run one side out of the pre into your 24 bit converters at full strength. Take the other side and turn it down 48db and run it into those converters. You've essentially reduced the bit depth by 8, so you're now sending a 16 bit signal into your system.* Now normalize both versions so that they are both at the same level. Now compare the two. What Zap is saying is that you'll hear more "resolution", or a better quality signal on the one that was recorded in at 24 bits originally. He's saying that the 16 bit one will sound grainier or something because it doesn't have all of that additional resolution there. What will actually happen is they will sound the same. This is because, even though we use less quantization points to record it, we have captured the signal itself at full necessary "resolution". Give it a try. Nika. *Alternative method - use an analog synthesizer or noise generator and run a demo sequence or play a patch once at a full 24 bits worth of level and the second time reduce the volume by 48db and do it again. This may not be easy without an active DI in the mix. Note: obviously there will be issues of the linearity of the gain structure in whatever device you're using to conduct this test.

[VerySoon]

"What will actually happen is they will sound the same." No. What will actually happen is that your normalized 16-bit signal will have it's noise floor raised. 24-bit is used so that the noise floor is "way down there" as "far away as it can be" from the actual signal. Record a signal as close as possible to full scale via a 16-bit DAC then a 24-bit DAC. The 24-bit recording will sound better! There are more numbers to represent it...more RESOLUTION. Or, you can call it HEADROOM. This is the sole reason why 24-bit is here today....the noise floor get's "pushed down." Does a recorded 8-bit signal when normalized to 24-bit sound the same as a recorded 24-bit signal? How about a 2-bit signal? Unequivocally NO! Zap is correct. Maybe your ears can't hear the difference between a signal recorded at 16-bit and one recorded at 24-bit. But, mine can! (I am already aware that you doubt the validity of 96kHz recordings. So, I am really not that surprised at your "theories" regarding 16-bit vs 24-bit.) And, I didn't even have to mention the effects on transient clarity (or...there's that word again: RESOLUTION) with a 16- vs 24-bit recording. Who normalizes their tracks anyway? That's a BIG no-no! A suggestion: Do all yourselves a favor and go to [url=http://www.digido.com]www.digido.com[/url] and read Bob Katz's articles titled "More Bits Please," "Dither" and "Level Practices in Digital Audio." This message has been edited by PeeTee on 09-10-2001 at 01:18 AM

[Nika]

[QUOTE]Originally posted by PeeTee: [b]"What will actually happen is they will sound the same." No. What will actually happen is that your normalized 16-bit signal will have it's noise floor raised.[/b] If both of them have a very low dynamic range (lets say 42db) then the noise floors of both recordings will be at -42dbFS. What you're talking about (I believe) is the added quantization noise, but because that is so much farther below the noise floor itself, it will not make the two recordings sound different. [b]This is the sole reason why 24-bit is here today....the noise floor get's "pushed down."[/b] Right. We need the quantization noise to be lower than the noise floor. With 96dB of dynamic range we were not obtaining this still. This is why we've gone to 24 bits: to give us more dynamic range so that the quantization noise can always be below the actual noise floor of the recording. [b]Does a recorded 8-bit signal when normalized to 24-bit sound the same as a recorded 24-bit signal? How about a 2-bit signal? Unequivocally NO![/b] That depends on the signal to noise ratio of the material you're dealing with. If the signal is only 6dB then yes, it DOES sound the same. [b]Maybe your ears can't hear the difference between a signal recorded at 16-bit and one recorded at 24-bit. But, mine can![/b] I can to, but only when dealing with material with decent SNR. Remember that the best A/D converters at 16 bit still only have a dynamic range of around 85 dB? So any material that has greater than 85 dB of range between it's peak and the white noise floor will sound better in 24 bit. And of course mixing and effects sound better in 24 bit, but that's not what we're discussing here. [b]And, I didn't even have to mention the effects on transient clarity (or...there's that word again: RESOLUTION) with a 16- vs 24-bit recording.[/b] Go ahead and mention them. I'm very curious to hear what you have to say about them and how you believe they apply. [b]Who normalizes their tracks anyway? That's a BIG no-no![/b] It's for a TEST, PeeTee. It's to bring the levels to the same amount for the sake of answering some questions that some people had. [b]A suggestion: Do all yourselves a favor and go to [url=http://www.digido.com]www.digido.com[/url] and read Bob Katz's articles titled "More Bits Please," "Dither" and "Level Practices in Digital Audio."[/b] I read every word, (I've read them before also) yet I don't see how any of this applies, other than possibly sending us on a goose chase so that you don't have to explain in your own words what you believe to be the case on this topic: "umm....umm...umm..........well go read what THAT guy wrote and then come back to me." In fact, in one point he specifically mentioned that you were better off giving additional headroom and that you would have no signal loss from doing so. This falls completely in line with what I was saying.

[Master Zap]

Nika: With that definition of "noise" (which, incidentally, does not agree at all with the definition of noise we were [url=http://www.sweetwater.com/insync/word.tpl?cart=&find=S/NRatio]supposed to agree on earlier[/url] [img]http://www.musicplayer.com/ubb/biggrin.gif[/img] ) you are right. If you actually READ what I write (I have read everyting you write) you will find that we are in complete agreement, except for the meaning of that word "resolution". I use the technical term, which is number of quantization steps. The audio being quantized is [i]irrelevant[/i]. You are using a completely different definition of "resolution" which I do not think is properly labeled "resolution" at all. But on the other hand, stevepow thinks I "mislabeled" it "resolution", so it all boils down to what we mean with that one word. You are talkign about useful, resolvable data. Kinda like heisenbergs uncertainity principle for audio [img]http://www.musicplayer.com/ubb/wink.gif[/img] I completely agree with this. That was actually what my very first post was about (I do not have the strength to scroll but the gist of my post was something like "why on earth resolve the noise to 16 bits, thats dumb") I look at this as pure numerics, and from that standpoint statements such as: [quote] The SIGNAL itself does not have any additional "resolution", "quantization steps", "discreet measurement increments", or any other term to describe units of measurement when we increase the bit depth beyond what is necessary to accurately describe the signal. [/quote] Semantics. You cannot deny that there exists more discrete sample levels. They might be "meaningless" from the signals standpoint, nevertheless they are there. [quote] As I said before, all that does is give you better resolution of a totally random noise signal. [/quote] Yes this was what I said in my first post [img]http://www.musicplayer.com/ubb/biggrin.gif[/img] and It's still "resolution". You even said so yourself, right there. [img]http://www.musicplayer.com/ubb/wink.gif[/img] [img]http://www.musicplayer.com/ubb/smile.gif[/img] Now, lets get this straight, once and for all. When I use the word "resolution", I am talking about number of sample steps. Nothing else. Never talked about something else, and never am I going to use the word "resolution" to mean something else. With me so far? The main point of DISAGREEMENT here, apart from your stubborn misuse of "resolution" (or at least, stubborn use of a *different meaning* to the word "resolution" than I do [img]http://www.musicplayer.com/ubb/smile.gif[/img] ) is this statement [quote] Adding bits does not increase resolution. It increases headroom [/quote] Lets break down this into two parts to really cut it up [quote] Adding bits does not increase resolutoin [/quote] Using my definition of resolultion it does, of course, since I am talking about sample steps and not a single thing else. Using your definition of "resolution" the answer it "well it depends on what you put in". I can agree to that [i]if[/i] we consider for a second your usage of "resolution" to be "correct". (Can we PLEASE come up with a better term than "resolution" for this, which clearly signifies that it's "useful resolution needed to accurately portray the signal"?) The second statement [quote] Adding bits increases headroom [/quote] This, however, is my MAIN objection. [i]No it doesn't[/i]. Not unless you do something. Since, on any given A/D converter, well every one I've used, samples are done from a reference voltage. Adding bits to such a device is done in the "bottom" end, i.e. it (sorry to sound repetetive) increases reso..uh...the number of quantization steps [img]http://www.musicplayer.com/ubb/wink.gif[/img] a given voltage is divided into. Hence, adding a bit gives us an extra "line in between the lines" for describing the range of voltages up to ref voltage*. IT STILL DOES NOT GIVE US THE ABILITY TO DESCRIBE A VALUE ABOVE THE REF VOLTAGE*!!!! It does not "give" us HEADROOM at all! It may give us the *potential* to *utilize* headroom, but that aint headroom in itself. Nika's statement may have been well-meaning in his own way, from his very own little viewpoint, but he is only confusing the hell out of people by saying them. By saying things like [quote] Adding bits does not increase resolution, it increases headroom[/quote] or things like (from the football field analogy) [quote] adding a bit does not make extra lines between the lines, it makes the field twice as large[/quote] will give people the completely skewed sense that the bits go on the top, not the bottom, and that their 24 bit converter is still recording in 16 bits unless they crank their signal to +48 dB !!! /Z (* = sidebar: Some converters actually convert signals to ref voltage * 2 but lets not go there ... *grin* )

[Nika]

[QUOTE]Originally posted by Master Zap: [b]Nika: With that definition of "noise" (which, incidentally, does not agree at all with the definition of noise we were [url=http://www.sweetwater.com/insync/word.tpl?cart=&find=S/NRatio]supposed to agree on earlier[/url] [img]http://www.musicplayer.com/ubb/biggrin.gif[/img] ) you are right.[/b] Zap, First, in neither THAT definition of noise, nor in the subsequent linked version of [url=http://www.sweetwater.com/insync/word.tpl?cart=&find=NoiseFloor]noise floor[/url] was there anything that disagreed with the notion that noise had to be au natural - white noise. I think this is by definition, though I've yet to find a definition that accompanies that. Maybe it's accepted enough that the definition assumes it also? I've always heard of and accepted noise to be white noise. Anything else is not "noise", but rather "filtered noise". [b]...you are right.[/b] And was that 100% right, or just "mostly right?" [img]http://www.musicplayer.com/ubb/smile.gif[/img] Careful about agreeing with me on anything! PeeTee might get on your case now! [b]If you actually READ what I write (I have read everyting you write) you will find that we are in complete agreement, except for the meaning of that word "resolution". I use the technical term, which is number of quantization steps. The audio being quantized is [i]irrelevant[/i]. [/b] I've read everything you've written. At least once. And now I think that we're also in agreement that the term "resolution" is just a bad word to use altogether (as I pointed out in the beginning, designers don't like to use it. Ref. Paul Frindle's quote for that). Certainly, by definition there are more quantization steps, but the signal itself has no further resolution as we concur. [b]You are talking about useful, resolvable data.[/b] Hmm. Is this particularly odd? To talk about the programme material we're recording? [b]I completely agree with this. That was actually what my very first post was about (I do not have the strength to scroll but the gist of my post was something like "why on earth resolve the noise to 16 bits, thats dumb")[/b] Your first post didn't much mention "noise", so it's no surprise that our communication has been weak. You thought you said something that I certainly didn't read into! I'm not saying that that's not what you intended, but communication requires the sender and the receiver to have the same message. That clearly didn't happen - no fault to either of us. I also agree that I used the term "headroom" improperly when I meant that more bits gives you better dynamic range. And the football analogy was, as I said in the beginning, flawed for various reasons. What I was INTENDING to get across was that there were two different ways of [i]looking[/i] at the scenario, and one way might leave people with a mental picture or a mental image that was not indicative of what was actually happening. For the future I've come up with an improvement to the football analogy that I've been working on for the past day or so. I think it works better than the last one. I certainly recognize that it was flawed. I had some improvements for it I was going to bring in to fix it, but we dismissed the analogies per your request and I never went back to it. I can fill you in on my improvements later in case you would like to borrow this from "Nika's Patented World of Football Analogies". [b]Nika's statement may have been well-meaning in his own way, from his very own little viewpoint, but he is only confusing the hell out of people by saying them. By saying things like, " Adding bits does not increase resolution, it increases headroom", or things like (from the football field analogy) "adding a bit does not make extra lines between the lines, it makes the field twice as large" will give people the completely skewed sense that the bits go on the top, not the bottom, and that their 24 bit converter is still recording in 16 bits unless they crank their signal to +48 dB !!![/b] Fair. Just the same, I believe that your insisting that you'll get more "resolution" when you go to different bit depths will give people the wrong impression that low SNR audio will actually sound better at higher bit depths. I have to say that I wasn't the only one confused about whether or not this was your point. Drew believed that it was your point according to his text, and you've now convinced PeeTee that this is what you think also. If you're concerned that my communication was less than perfect and may confuse people you're correct, but I think it's apparent that we EACH were less than ideal in our printed text. I suppose there's a question about whether or not we actually agreed in the beginning, or if we just agree now, but IF we agreed in the beginning I think it a sign that EACH of our communications were bad if it took us 50 posts to realize this!! [img]http://www.musicplayer.com/ubb/smile.gif[/img] [b](* = sidebar: Some converters actually convert signals to ref voltage * 2 but lets not go there ... *grin* )[/b] Roger Nichols said most are referenced to 10V. All of mine are referenced to 4V. In the hi-fi world they've been making it a buzz word for marketing: "this one is a 6V converter!" "Yeah, well this one is 8V, so it's better!". I'm not kidding! Anyway, thank you for providing some entertainment for my weekend. I'm sure we'll meet up elsewhere. It's been a pleasure. Then again, you now know how I learn best. Under those auspices, I learn best when there's a conversation to be had, so I'll always welcome a good one. (So..........what are your thoughts on abortion??) Cheers! Nika.

[Master Zap]

Remember I'm Swedish. English aint a native language to me. If I do not express myself "on the dot" clearly at least I have an excuse [img]http://www.musicplayer.com/ubb/wink.gif[/img] For me though, resolution is a fine word. It simply means count of quantization steps, nothing else. You wont call the resolution of a 1024x768 monitor anything else just because you show a blurry picture on it, would you? I sure don't. Whatever the word that describes what you want to describe is, I dont know, but I dont think "resolution" is it. Maybe my usage of the word "resolution" is colored by its Swedish translation... for me the meaning is completely clear. -- Btw Nika I am going through the 96k thread with great enjoyment. One thing I always wondered about higher sample rates (although I never use them since my target is 44.1K and my system doesnt do 88.2k and I feel anything else would get destroyed in the converseion anyway) are these two things: ONE: 44k may resolve a 22khz tone... but as a [i]squarewave[/i], not a sinewave. True, theoretically the first harmonic of that squarewave is way up at 44khz and hence "inaudible"... but just maybe the ear doesnt quite do it that way, and actually is able to tell the difference between a square 22khz and a sine 22khz? Just hypo-thetically? TWO: 44100hz can represent 22050z tone. Nyquist and guys. But can it represent 22051hz? Accurately? Try it you'll like it.... Can it represent a 22khz tone exactly in phase with the sampling frequency (every zero-crossing is sampled)? These are clear flaws of 44k... but are they audible or not? Who knows. [img]http://www.musicplayer.com/ubb/wink.gif[/img] /Z This message has been edited by Master Zap on 09-10-2001 at 03:14 AM

[Uh Clem]

You cannot improve the noise floor of an incoming signal regardless of the bit depth you use to record it. True, the quantization error and resulting noise will be lower with more bits, but in Nika's test, he is going in with a poor S/N inherent in the original analog audio. It can only later get better if you break out SF's NR plug-in [img]http://www.musicplayer.com/ubb/wink.gif[/img]. So as I re-read some of the posts, it is not clear that everyone is on the same page with regards to noise and I think the terms quantization error and source S/N should be used to keep straight the two. E.g, it seems Nika's Mic Test post referred to the source S/N and PeeTee's to QE - the river/mountain thing all over again. I don't know if I believe that test makes any sense either - I need to sleep on it and then try it first. As far as the normalize and who does it - its a test procedure - think of it as simulating rasing the fader on a low level signal - that is done all the time in the real world and is essentially the same thing. I don't like the test so far as proposed because you have to use your ears to make the comparison. I've seen too many examples of where even the "golden ear" types will fail at those sorts of comparisons when the effects become subtle. Sorry Zap, I didn't really take issue with your use of resolution - I actually felt it made sense. I only mentioned that there may be a more precise and less abiguous term that you could adopt without changing your position that would bring you points across more clearly. If your posistion is only to hold to that word and your intended use of it, that becomes less intersting to me. I think all of us believe 24 bits is better for the general case of recording - understanding why and demonstrating it even to ourselves seems valid to me - builds character and all that. So I'm trying to sort this out and I think Nika is getting his explanations down better as he goes. That last bit (no pun intended) seemed pretty on target. The whole dynamic range thing makes sense - we can distinguish with good ears about 120dB of dynamic range from a very quiet to a very loud, but hon-health risk sound. So as long as my SQNR is -120dB or more, my ears will never hear it. 24 bits or a SQNR (theoretical) of -144dB seems like a safe goal. Even a solid 20 bit system that did deliver true -120dB SQNR would probably be fine for most ears. On headroom - sure you set that yourself. Take your nominal level at -12dBFS and now you've given yourself 12dB of headroom (2 bits) still need 20 bits to resolve (oooh) a signal that can range from -12dBFS to -132 (all the dynamic range my ears can hear). So I think that is why Nika says you have more headroom - maybe he should say you can allow more headroom without pushing your low level sources into the quantization noise. In this example to get all my dynamic range on that signal with a 20bit converter, I'd have no headroom - I'd have to set my nominal levels at 0dBFS and somewhere some guy would crack the snare harder than he did during sound check ('cause they always do) and then you clip. Off to read Bob's stuff....

[Master Zap]

Well perhaps I am overly stubborn in insisting on resolution meaning resolution, and as I said it might be colored by the Swedish word for resolution which means....uh.... resolution? [img]http://www.musicplayer.com/ubb/smile.gif[/img] [img]http://www.musicplayer.com/ubb/eek.gif[/img] [img]http://www.musicplayer.com/ubb/rolleyes.gif[/img] (Did I just translate myself into a dead end there? [img]http://www.musicplayer.com/ubb/biggrin.gif[/img] But see the 1024x768 monitor mentioned in my prev. post) As a matter of fact I wasnt as much opposed to Nikas concept of usable resolution as his matter-of-fact stating that "Adding bits wont get your more resolution" as in "it never does" and "Adding bits gives you headroom" as in "automatically". Reading these interpretations into the sentences he wrote, both are naturally completely wrong. Had he written "Adding bits wont automatically give you more resolution if the input audio is poorly resolved to begin with" and "...by taking the above into accound and opimizing the bit utilization nu turning stuff down on input, you actually gain more headroom instead." I wouldn't have flinched and this thread wouldn't even have a page 2 [img]http://www.musicplayer.com/ubb/wink.gif[/img] Actually, had he just stopped on that usable bit thing I probably wouldn't have flinched much either... but he started that headroom thing... and he wrote it in such a way that it REALLY seemed as he thought new bits go on the high end of the word, not on the low end. Thats how I interpreted it, which of course is completely wrong. Oh Nika, oh Nika, had you just not mentioned "headroom" this would never have happened [img]http://www.musicplayer.com/ubb/wink.gif[/img] /Z This message has been edited by Master Zap on 09-10-2001 at 04:17 AM

[Nika]

Yeah, yeah, that "headroom" thing. Bad idea. "Allows more headroom"... And as for the test, Steve, I wasn't looking for something for the audiophiles. There are several potential flaws in it. I was more looking for a really quick "wow! They DON'T sound different" sort of quick and dirty test. If somebody really wanted to test for the smallest details they would flip one fader out of phase and put the two together and see if the sound nulls. It wouldn't, though, because your gain staging between the two different inputs would be inherently non linear. Also, since you think my explanations are getting better, here is the repair to the football analogy. Have you ever seen a football game? You know how they tackle the guy and the ref comes over and "spots the ball" where he was down? What, is it a laser guided system that tells EXACTLY where he was down? Is it like in tennis when the court has sensors under it to tell EXACTLY where the ball was down? No. The ref just looks and calls it somewhere. Whereever he calls it it is "down". The ball itself is a foot long. The yard hashes are 3" thick. A player's knee is about 4" long. A player's stride can be over a yard! There's no way that we're going to bring this game down to a precision of inches! The construction of the turn itself isn't even that accurate! So the difference between where the MIDDLE of the ball touches the field, and where the ref spots the ball is considered "noise". It is a random variation from where the middle of the ball ACTUALLY hit the turf and where the ref spots the ball. Now let's say that the rules of football stipulated that the ball had to be placed on a hash mark. I would agree that 1 yard hashmarks are probably too wide. We could probably get more precise than that in a football game. So we need to add additional hashmarks. So do we go to inches? Feet? Millimeters? Tough call! At a certain point we're going to be getting so silly stupid in this matter that the field is WAY more accurately marked out than the level of precision that the game can be played with. Sure, the FIELD has more resolution, but does the GAME? I think that that analogy essentially works in one respect, but it doesn't address the increase in SNR. For that we go back to the first one. Ugh, sometimes analogies break down too quickly. No bother, live and learn. Or just learn. Nika.

[Master Zap]

Nika, do you know why the Football analogy doesnt work? Because I am from Sweden. In Sweden the word "football" is used for the proper game "football", the one you guys mislabel "soccer". What you guys mislabel "football" is actually "rugby". [img]http://www.musicplayer.com/ubb/biggrin.gif[/img] /* Just being naughty */ /Z

[Uh Clem]

Yes, you were doing better without the analogy [img]http://www.musicplayer.com/ubb/wink.gif[/img]. So here's one for you, Nika: can you explain why 1 bit represents 6dB or dyanmic range, or as it was called, Signal to Quantization Noise Ratio? I think I get it, but I can't explain it even to myself. Actually I think it is more like adding 1 bit increases the SQNR by 6dB and I think to account for positive and negative values you need a minimum of 2bits to get the 1st 6dB. Then you add one bit and double the SQNR and every doubling of signal is a 6dB increase! Maybe that's it - sounds kinda good. So say we start with 2Bits and record the very clean sine wave at a 0dBFS and don't sacrifice any bits to headroom. What do we end up getting? My wave is a constant volume and has no self-noise. I get a sqarewave because I only have value choices of 0, -1, 1. Boy, you have to hand it to [b]Zap[/b], now, when you see that and say to yourself, "I don't have enough bits to express this freakin' wave!, I've got to get better..uh...uh..uh..resolution" [img]http://www.musicplayer.com/ubb/eek.gif[/img]! I can approximate some of this in Wavelab in that it lets me generate sine waves at various levels at different bit depths - with no source noise. An 80Hz wave at 8 bit at -40dBFS is so close to the limits of the SQNR range that it no longer is even identifiable as 80 by my ear because it is so squared off. It realy on has 1 or 2 levels either side of zero - so that seems about right for a little over 6dB of range for it to live in, so it gets 1bit for sign and 1 for either on or off. The problem with using Wavelab is that the quana error does not appear to be random - make sense, since it is generating the wave with perfectly alligned sampling points. That also agrees with my prediction above with my perfect sine wave and 44.1/2bit sample. As I said before, I think you guys have at least nearly agreed (+/- a bit or two) all along. You just got different names for the same sports. The poor fellow who started this topic is probably thinking what the heck happened? But thanks for digging into this - I've been needing to get a handle on it for a while. I feel pretty close, if not there. Now I have to go convince myself why in the hell Nyquist threorem makes any sense for complex wave forms. Intuitively, I'd think I needed more samling points to get a better picture of the actual Wave, but 2x the frequency range is fine? Off to the 96K topic, I guess. This message has been edited by stevepow on 09-10-2001 at 02:08 PM

[Nika]

[QUOTE]Originally posted by stevepow: [b]So here's one for you, Nika: can you explain why 1 bit represents 6dB or dyanmic range, or as it was called, Signal to Quantization Noise Ratio? I think I get it, but I can't explain it even to myself. [/b] I'm with you there! That's a tough one. I can mentally "see" it, but I can't seem to describe it. First one has to have a pretty good grounding in what a "decibel" is and understand the principles of waves. I'm not going to be able to explain it without totally mangling it, though. [b]Actually I think it is more like adding 1 bit increases the SQNR by 6dB and I think to account for positive and negative values you need a minimum of 2bits to get the 1st 6dB. Then you add one bit and double the SQNR and every doubling of signal is a 6dB increase! Maybe that's it - sounds kinda good.[/b] That is getting there, but the question that remains is, if we start with a 6db sine wave (1bit), how do we add 2db to it? [b]My wave is a constant volume and has no self-noise. I get a sqarewave because I only have value choices of 0, -1, 1. [/b] OK, now I can deal with this one. You DON'T end up with a square wave. You only end up with that if you take your sampling points and draw a line between them in a very "dot to dot" fashion. Remember that those are merely points that represent where the signal was at that time. Before those three points actually hit a D/A converter, though, they will be put through a filter that will essentially look at the points and draw a sine wave through them, and then send THAT data on to the converters. So even though we only sample 3 points, those three points represent a lot more than just what you see there. For this reason, one bit can represent a lot more than just off and on square waves. You have to get that picture out of your head that "these points that once came from a sine wave now look like a square wave" because what you SEE is only half way through a complete process. That process includes: dither, then filter, then convert, then downsample and filter, then [processing] then dither, then upsample, then filter, then convert, then filter. While Nyquist made it sound pretty easy with "just sample at greater than two points in a cycle", for accurate reproduction it is dependant upon a complete cycle of processes akin to the above. By only looking at what a computer's representation of that signal is half way the through the complete system you are teasing yourself that there is a whole lot that isn't there that actually is still implied. [b]Now I have to go convince myself why in the hell Nyquist threorem makes any sense for complex wave forms. Intuitively, I'd think I needed more sampling points to get a better picture of the actual wave, but 2x the frequency range is fine? Off to the 96K topic, I guess. [/b] Steve, reference my paragraphs above about the "picture" of the waveform vs. it's representation in samples. >2x is fine for sampling complex waveforms in short because of this: By the time they're filtered properly, they're not complex anymore. When a complex waveform that has components in it that are >20k is filtered for sampling, all of the components above 20k are filtered out and all that is left is a waveform that CAN be represented as though it is under 20k. A sine wave at 15k that has a whole bunch of noise in it above 30k that is run through a brickwall 20k filter will lop off everything above 20k, leaving just the pure sine wave. If you follow this out, as waveforms approach 20k and are put through a filter, the results become more and more sinusoidal, so that at 20k all you CAN have is a sine wave. Thus, there's no complex waveform to sample, and 44.1k will do the trick just fine. The bottom line is that complex waveforms get filtered down to sinusoidal waveforms under 20k before they are sampled, so sampling at 44.1k will do just fine. I hope this helps! Let me know if it doesn't. Thanx! Nika. This message has been edited by Nika on 09-10-2001 at 10:07 PM

[Master Zap]

[quote]So here's one for you, Nika: can you explain why 1 bit represents 6dB or dyanmic range, or as it was called, Signal to Quantization Noise Ratio? I think I get it, but I can't explain it even to myself[/quoute] Well, 1 extra bit added to a digital word doubles it's possible numeric range. 1 bit = 2 values (0 and 1) 2 bits = 4 values (00, 01, 10, and 11) 3 bits = 8 values (000, 001, 010, 011, 100, 101, 110, 111) 4 bits = 16 values (you figure 'em out [img]http://www.musicplayer.com/ubb/wink.gif[/img] ) 5 bits = 32 values 6 bits = 64 values..... e.t.c. 16 bits = 65536 values On the logarithmic dB scale 6 dB means doubling. So it's really kinda obvious... it's just doubling some thing, one way or the other... it's still a multiplication by 2.... nothing magic about it [img]http://www.musicplayer.com/ubb/biggrin.gif[/img] (Actually, isnä't it 6.23 or something like that which TRULY is doubling? And the old "6 doubles" is just a convenient but fairly accurate rule of thumb thingy? Please correct me if I'm wrong!) The quantization "noise" comes from the fact that you cant resolve the "actual" curve to more than a quantization point and whatever DIFFERNCE you end up with is "precieved" as "noise" by the ear... and since the accuracy equals one sample step (cant get closer than that) you get a semi-random noise with an amplitude of +/- 0.5 sample steps... i.e. a total amplitude of 1 sample step... i.e. a "S/N ratio" equaling 6 times the bitcount.... Hmm. I dunno if that description helps at all ..... the hardest thing in the world, is to explain the obvious. [img]http://www.musicplayer.com/ubb/smile.gif[/img] As Nika said, there is only two ways to learn: By Trial and Error, or Teaching. I dunno if I learned something from the above. Wonder what that means? [img]http://www.musicplayer.com/ubb/eek.gif[/img] /Z

[Nika]

[quote]Originally posted by Master Zap: [B](Actually, isnä't it 6.23 or something like that which TRULY is doubling? And the old "6 doubles" is just a convenient but fairly accurate rule of thumb thingy? Please correct me if I'm wrong!)[/quote] You're correct. It's not quite 6, but we use that out of convenience. And I can understand why adding a bit adds 6db. That makes sense to me. But how does 1 bit represent 6 db to start with? And how do we raise a 6db signal by 2db digitally? That's my own stumbling block for the morning. Thanx! Nika.

[Master Zap]

Well, it doesn't. There is no such thing as "6 dB". There is only 6 dB MORE or LESS than something else. dB is relative and you always need a reference. So the first bit, the lowest bit, the sample level +000000000000001 has to have *some* number of dB. You can connect that bit to a speaker and measure the SPL and declar "this bit has -39.2 dB SPL". Now, if everything is linear, multiplying the number by two you get +000000000000010 and this is 6dB more... since we doubled it! Now, in digital though the reference aint the bottom, coz 00000000 is still "minus infinity" dB because it is ZERO/nada/nothing/zippo, so one starts from the one fixed level we have - the max, +111111111111111 and label this 0dBFS. This is the digital reference level. You cant get higher* ever. Dividing by 2 in digital - or turning down 6dB - is the same as shifting the bits one step right, so we get +011111111111111 which is 6dB lower. This holds all the way down to +000000000000001 which is 93 dB lower. Come that last bit... that last number... out of convenience one likes to say that that last division by 2 also is 6dB... but thats actually not the case. Divide 1 by 2 in integer math and you get.... zero. So the last little quantize step is not actually 6dB in size, its infinite. Which is why all talk about "dynamic range" in digital is actually complete bollox, since it is always infinite, even in a 1 bit system. Dynamic RESOLUTION - however - is not. And "max-signal-to-smallest-error-amplitude" (commonly named "signal-to-noise" ratio or mislabeled "dynamic range") is also not infinite. /Z (* = if you are TRULY in to digital, play spot-the-error in this sentence)

[Guest]

From the book Advanced Digital Audio P.44-47. It give you the eplanation of 1 bit for 6dB. It's kind of statistic which can be explain in plain words. BTW, this link is very interesting