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#1
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Digital Volume Control
I believe that a digital volume control (for example on a CD player)
sacrifices 1 bit of resolution for every 6dB of attenuation. This raises for me a couple of questions. If the original signal is 16/44 and it is upsampled to 24/96 before the digital volume control, would 6dB of attenuation result in 15-bit resolution or 23-bit resolution? Any explanation would be most helpful. Wadia claim to have a digital volume control on their CD player that does not reduce the bit resolution. The claim is that redundant information (where?) is used for the volume control and so there is no loss of resolution (as my workmate/Wadia-owner explained it). How does this work? |
#2
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nowater wrote:
I believe that a digital volume control (for example on a CD player) sacrifices 1 bit of resolution for every 6dB of attenuation. Most analog volume controls also lose 6 dB (=1 bit) of resolution for every 6 dB of attenuation, more or less. This raises for me a couple of questions. If the original signal is 16/44 and it is upsampled to 24/96 before the digital volume control, would 6dB of attenuation result in 15-bit resolution or 23-bit resolution? 16 bits or less. Upsampling doesn't increase the resolution of a signal, it just makes it take up more data space. If you take a 16 bit signal and upsample it to 24 bits and then attenuate it in the 24 bit domain, you can apply up to 8 bits of attenuation (about 46 dB) before you start losing resolution in the digital attenuator. Thing is that the signal doesn't stay in the digital domain, but instead looses additional amounts of resolution in the conversion to the analog domain, and other processes that follow. The whole discussion lacks practical relevance since the source material, prior to being put on the CD, has 75 or less dB dynamic range due to noise upstream in the live performance, initial recording process, and production processing. The commonly-stated bugabear in digital attenuation is low-level nonlinear distortion, which does not actually happen in a properly-dithered digital attenuator. Wadia claim to have a digital volume control on their CD player that does not reduce the bit resolution. IME, Wadia claim lots of things that are kinda strange, and/or have limited to nonexistent practical benefits. The claim is that redundant information (where?) is used for the volume control and so there is no loss of resolution (as my workmate/Wadia-owner explained it). You've got to be a true believer to believe a lot of what Wadia says. The very act of paying Wadia prices for an optical player that if perfect, would sonically indistinguishable from a good $60 DVD player seems to take a toll on the objectivity of Wadia equipment. How does this work? See above. It can work in theory, but just in the digital domain. |
#3
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François Yves Le Gal wrote:
On Tue, 7 Jun 2005 21:50:13 +0930, "nowater" wrote: I believe that a digital volume control (for example on a CD player) sacrifices 1 bit of resolution for every 6dB of attenuation. Most digital VC's use bit shifting between the LSB (least significant bit) and the MSB (most significant bit). The signal shifs towards the LSB for attenuation, thus reducing resolution. If the original signal is 16/44 and it is upsampled to 24/96 before the digital volume control, would 6dB of attenuation result in 15-bit resolution or 23-bit resolution? Any explanation would be most helpful. An upsampled signal doesn't contain more information than the original. Some manufacturers claim that their interpolation schemes boost resolution but what goes out isn't what goes in: information is *created* by the system. A 24/96 system with an attenuation of 6 dB results in the loss of one bit of resolution on the original signal. Wadia claim to have a digital volume control on their CD player that does not reduce the bit resolution. The claim is that redundant information (where?) is used for the volume control and so there is no loss of resolution (as my workmate/Wadia-owner explained it). How does this work? Wadia uses 22+ bits converters coupled with bit shifting and interpolation. According to them, a 36 dB attenuation doesn't result in the loss of information (22 - 6 = 6 x 6). This is pure BS: signal bits are thrown away. OTOH, you can design a sliding system based on a 32-bit processor coupled with 24-bit converters: a typical 16-bit signal can thus be slided down by 8 bits, or 48 dB, w/o any significant loss (monotonicity et al. kick in when it comes to the lower bits). This architecture is typical of pro gear or well designed consumer equipment. Hi François, I was looking at this very nice post of yours, wondering what happened to the days when you and I used to fight like cats and dogs? ;-) |
#4
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François Yves Le Gal wrote:
On Tue, 7 Jun 2005 08:50:18 -0400, "Arny Krueger" wrote: I was looking at this very nice post of yours, wondering what happened to the days when you and I used to fight like cats and dogs? Well Arny, when you behave like a schmuck, I still treat you like a schmuck, don't I? Likewise, I'm sure. BTW, despite what you may believe, I've never been on the wee-wee side of audiophilia. So you say. The difference between you and me is that you're still a Luddite somewhere, stuck into, say, 16/44 and refusing to admit - despite numerous objective tests - that this obsolete format doesn't allow for full signal resolution, either in bandwidth or dynamics. I never said that 16/44 allows for full signal resolution, either in bandwidth or dynamics. Obviously it doesn't. However, the question I like to deal with is practical sufficiency. Ditto for amps: if models of similar approaches and topologies used within their performance envelope essentially sound the same, there are some huge measurable as well as perceptible differences between different amps. And so on. It's true that most SETs sound like crap when playing complex sounds at goodly volumes. |
#5
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"nowater" wrote ...
I believe that a digital volume control (for example on a CD player) sacrifices 1 bit of resolution for every 6dB of attenuation. This raises for me a couple of questions. If the original signal is 16/44 and it is upsampled to 24/96 before the digital volume control, would 6dB of attenuation result in 15-bit resolution or 23-bit resolution? Any explanation would be most helpful. Wadia claim to have a digital volume control on their CD player that does not reduce the bit resolution. The claim is that redundant information (where?) is used for the volume control and so there is no loss of resolution (as my workmate/Wadia-owner explained it). How does this work? When you "attenuate" the signal by 6dB, you aren't even *using* that bit of resolution anymore. So "sacrafice" seems like a mis- leading characterization. I reject the concept that attenuating a signal "sacrafices" the digital resolution. It is a silly semantic argument. Doesn't hold any water for me. |
#6
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"François Yves Le Gal" wrote in message ... On Tue, 7 Jun 2005 09:54:03 -0400, "Arny Krueger" wrote: I never said that 16/44 allows for full signal resolution, either in bandwidth or dynamics. Obviously it doesn't. Ahem. here's just an example: "we find that 16/44 is a pretty good match to the limits of the ears and typical recording (or even exceptional recording ) and playback circumstances. Its even overkill. That means that more resolution and higher sample rates are sonically moot." in RAHE Another one? OK: "The CD format makes it entirely possible to make a recording that has just about any content that can be recorded and heard". right here. And so on. Ahem. Perhaps you missed the part about "the question I like to deal with is practical sufficiency." But if you like to argue just for argument's sake, go for it. I'll go back to ignoring such silly time-wasters. |
#7
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François Yves Le Gal wrote:
On Tue, 7 Jun 2005 09:54:03 -0400, "Arny Krueger" wrote: I never said that 16/44 allows for full signal resolution, either in bandwidth or dynamics. Obviously it doesn't. However, the question I like to deal with is practical sufficiency. Ahem. here's just an example: "we find that 16/44 is a pretty good match to the limits of the ears and typical recording (or even exceptional recording ) and playback circumstances. Its even overkill. That means that more resolution and higher sample rates are sonically moot." in RAHE Another one? OK: "The CD format makes it entirely possible to make a recording that has just about any content that can be recorded and heard". right here. And so on. Easily explained by the sentence about practical sufficiency. Where's the beef? |
#8
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Richard Crowley wrote:
"François Yves Le Gal" wrote in message ... On Tue, 7 Jun 2005 09:54:03 -0400, "Arny Krueger" wrote: I never said that 16/44 allows for full signal resolution, either in bandwidth or dynamics. Obviously it doesn't. However, the question I like to deal with is practical sufficiency. Ahem. here's just an example: "we find that 16/44 is a pretty good match to the limits of the ears and typical recording (or even exceptional recording ) and playback circumstances. Its even overkill. That means that more resolution and higher sample rates are sonically moot." in RAHE Another one? OK: "The CD format makes it entirely possible to make a recording that has just about any content that can be recorded and heard". right here. And so on. Ahem. Perhaps you missed the part about "the question I like to deal with is practical sufficiency." Since it was in my OP, seems like putting that sentence back in to François' quote is reasonable. But if you like to argue just for argument's sake, go for it. I'll go back to ignoring such silly time-wasters. Agreed. |
#9
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Arny Krueger wrote:
nowater wrote: snip... Most analog volume controls also lose 6 dB (=1 bit) of resolution for every 6 dB of attenuation, more or less. OK I never thought about that ... kind of invalidates the argument against digital VC! This raises for me a couple of questions. If the original signal is 16/44 and it is upsampled to 24/96 before the digital volume control, would 6dB of attenuation result in 15-bit resolution or 23-bit resolution? 16 bits or less. Upsampling doesn't increase the resolution of a signal, it just makes it take up more data space. Whoops I knew that... should have worded my question more carefully (see below). If you take a 16 bit signal and upsample it to 24 bits and then attenuate it in the 24 bit domain, you can apply up to 8 bits of attenuation (about 46 dB) before you start losing resolution in the digital attenuator. Thanks FYLG and AK for helpful replies. I think this answers the question I really meant, namely: "If 16/44 is upsampled (say to 24 bits) prior to digital attenuation, can we then apply up to 48 dB of attenuation without losing any of the original 16 bits of resolution?" The answer seems to be Yes. This seems to be a neat solution to the "loss of bits" argument against digital volume control. It is also what I have implemented at home (16-bit CD = 24-bit upsampling = digital VC), one reason for my original question, and I now feel reassured. In fact, if I read AK correctly, it is a better solution than an analog attenuator - any analog attenuator. Thing is that the signal doesn't stay in the digital domain, but instead looses additional amounts of resolution in the conversion to the analog domain, and other processes that follow. Is this a caveat on the use of digital volume controls, or just a general observation about digital-analog hybrid systems, with and without digital volume control? The whole discussion lacks practical relevance since the source material, prior to being put on the CD, has 75 or less dB dynamic range due to noise upstream in the live performance, initial recording process, and production processing. I disagree about lack of practical relevance. If you have a CD player with internal digital volume control operating at 16 bits (like my brother in law's Denon), and you connect it directly to a sensitive high-power power amp, in order to get normal listening levels it may be necessary to dial back say 48dB of digital attenuation, taking 8 bits off your 16-bit signal and only having 8 bits of resolution (48dB) available. If I read AK correctly, the above scenario only allows about 20 dB of digital attenuation from the 16-bit (say 96dB resolution) before the attenuator results in less resolution than the 75dB of dynamic range in the recorded music signal prior to being put on CD. Wadia claim to have a digital volume control on their CD player that does not reduce the bit resolution. snip How does this work? See above. It can work in theory, but just in the digital domain. OK so I now see that the Wadia system works as claimed. They use 21-bit upsampling, so their digital attenuator would permit up to 30dB of attenuation with full 16-bit signal resolution. Thanks to posters for clarifying. Please correct my conclusions if I misunderstand you. |
#10
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"Arny Krueger" writes:
[...] If you take a 16 bit signal and upsample it to 24 bits and then attenuate it in the 24 bit domain, you can apply up to 8 bits of attenuation (about 46 dB) before you start losing resolution in the digital attenuator. Wrong. Even if the signal stays 24 bits all the way to the DAC, this is not correct. You lose resolution immediately. If you had a true 24-bit DAC, you'd have a usable SNR of about 144 dB relative to a full-scale level. When you attentuate X dB, you shave X dB off that SNR just as you would with a 16-bit DAC or a 9-bit DAC or whatever. The only advantage is that, with a high-resolution DAC, you can afford to throw away some of your SNR, so in that sense it does make a digital volume control more viable. What seems to be going on in your mind (and others) is that if you don't "lose any of the original bits" you don't lose anything. That is not correct since the noise floor of the DAC stays constant, and thus any attenuation of the maximum signal power degrades the SNR of the signal. -- % Randy Yates % "My Shangri-la has gone away, fading like %% Fuquay-Varina, NC % the Beatles on 'Hey Jude'" %%% 919-577-9882 % %%%% % 'Shangri-La', *A New World Record*, ELO http://home.earthlink.net/~yatescr |
#11
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nowater wrote:
Arny Krueger wrote: nowater wrote: Most analog volume controls also lose 6 dB (=1 bit) of resolution for every 6 dB of attenuation, more or less. OK I never thought about that ... kind of invalidates the argument against digital VC! This raises for me a couple of questions. If the original signal is 16/44 and it is upsampled to 24/96 before the digital volume control, would 6dB of attenuation result in 15-bit resolution or 23-bit resolution? 16 bits or less. Upsampling doesn't increase the resolution of a signal, it just makes it take up more data space. Whoops I knew that... should have worded my question more carefully (see below). If you take a 16 bit signal and upsample it to 24 bits and then attenuate it in the 24 bit domain, you can apply up to 8 bits of attenuation (about 46 dB) before you start losing resolution in the digital attenuator. Thanks FYLG and AK for helpful replies. I think this answers the question I really meant, namely: "If 16/44 is upsampled (say to 24 bits) prior to digital attenuation, can we then apply up to 48 dB of attenuation without losing any of the original 16 bits of resolution?" The answer seems to be Yes. Agreed. This seems to be a neat solution to the "loss of bits" argument against digital volume control. It is also what I have implemented at home (16-bit CD = 24-bit upsampling = digital VC), one reason for my original question, and I now feel reassured. In fact, if I read AK correctly, it is a better solution than an analog attenuator - any analog attenuator. ....other than the fact that eventually the output of any practical digital volume control will have to be converted to analog, which tends to wipe out the superiority of the digital volume control over the analog volume control. Thing is that the signal doesn't stay in the digital domain, but instead looses additional amounts of resolution in the conversion to the analog domain, and other processes that follow. Is this a caveat on the use of digital volume controls, or just a general observation about digital-analog hybrid systems, with and without digital volume control? It's just a general observation about digital-analog hybrid systems. The whole discussion lacks practical relevance since the source material, prior to being put on the CD, has 75 or less dB dynamic range due to noise upstream in the live performance, initial recording process, and production processing. I disagree about lack of practical relevance. If you have a CD player with internal digital volume control operating at 16 bits (like my brother in law's Denon), and you connect it directly to a sensitive high-power power amp, in order to get normal listening levels it may be necessary to dial back say 48dB of digital attenuation, taking 8 bits off your 16-bit signal and only having 8 bits of resolution (48dB) available. If you connect a volume control to a highly sensitive power amp (e.g., one whose input voltage spec is significantly less than the output voltage spec of the DAC) both good practice and the specifics of this application suggest that you first attenuate the output of the DAC with an appropriate analog attenuator. The input gain control of the power amp may be a readily availalable tool for the purpose of matching its sensitivity to the maximum output of the DAC. Thus, the full resolution of the DAC is available for the purpose at hand, which is providing a controllable attenuator that exploits all of the hardware at hand. If I read AK correctly, the above scenario only allows about 20 dB of digital attenuation from the 16-bit (say 96dB resolution) before the attenuator results in less resolution than the 75dB of dynamic range in the recorded music signal prior to being put on CD. As long as this 20 dB of attenuation by the digital attenuator results in 20 dB attenuation of the final SPL, there is no problem. This will happen if any excess gain of the power amp is essentually absorbed by an analog attenuator between the DAC and the power amp. The noise floor of most power amps is around 80-95 dB below their maximum output. If the full output of a 16 bit DAC is just barely sufficient to drive the power amp to full output, then any residual noise at the output of the DAC will be at, near, or below the residual noise of the power amp. There are exceptional power amps with residual noise on the order of 110 dB below full output. To fully exploit such a power amp, a digital attenuator with 16 bit resolution will be required. However, room noise etc, may allow this requirement to be relaxed. For example, a typical listening room will have residual acoustical noise on the order of 35-45 dB. 96 dB over this is 131-151 dB SPL which is at or beyond the threshold of pain. A real-world audio system will probably need to develop no more than 120 dB SPL, with 115 dB being a more practical maximum. A 16 bit (96 dB) digital attenuator will have residual noise on the order of 19 to 24 dB SPL which will only be noticable in a typical residential listening room if you put your head close to the speakers and don't actually play a recording. |
#12
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Randy Yates wrote:
"Arny Krueger" writes: [...] If you take a 16 bit signal and upsample it to 24 bits and then attenuate it in the 24 bit domain, you can apply up to 8 bits of attenuation (about 46 dB) before you start losing resolution in the digital attenuator. Wrong. Even if the signal stays 24 bits all the way to the DAC, this is not correct. You lose resolution immediately. If you had a true 24-bit DAC, you'd have a usable SNR of about 144 dB relative to a full-scale level. No you wouldn't. The dynamic range of the input signal, which comes from a 16 bit CD, limits the total usable dynamic range of the system to about 96 dB, assuming a perfectly noiseless power amp. |
#13
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"Arny Krueger" writes:
Randy Yates wrote: "Arny Krueger" writes: [...] If you take a 16 bit signal and upsample it to 24 bits and then attenuate it in the 24 bit domain, you can apply up to 8 bits of attenuation (about 46 dB) before you start losing resolution in the digital attenuator. Wrong. Even if the signal stays 24 bits all the way to the DAC, this is not correct. You lose resolution immediately. If you had a true 24-bit DAC, you'd have a usable SNR of about 144 dB relative to a full-scale level. No you wouldn't. The dynamic range of the input signal, which comes from a 16 bit CD, limits the total usable dynamic range of the system to about 96 dB, assuming a perfectly noiseless power amp. If the dynamic range of the system was 96 dB, then the attenuator would degrade it to 96 dB. I see your view. It's wrong. The 16-bit CD signal is represented by the four-digit hexadecimal number 0xklmn. This is converted to a 24-bit number "left justified" so that it becomes 0xklmn00. The attenuator operates on that input and produces a value 0xopqrst. That value is free to range over almost 144 dB of a 24-bit system (except for the 8 LSBs, which amounts to a loss of 0.000133 dB from a true 24-bit system). Another way to look at it is this. The attenuation procedure computes the following product: a*x, where "x" is the 16-bit input signal and "a" is a positive, real number between 0 and 1. Since "a" has infinite precision, then so does a*x. The final "noise" in the system is related to the number of bits to which the product a*x is quantized. -- % Randy Yates % "So now it's getting late, %% Fuquay-Varina, NC % and those who hesitate %%% 919-577-9882 % got no one..." %%%% % 'Waterfall', *Face The Music*, ELO http://home.earthlink.net/~yatescr |
#14
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François Yves Le Gal writes:
The difference between you and me is that you're still a Luddite somewhere, stuck into, say, 16/44 and refusing to admit - despite numerous objective tests - that this obsolete format doesn't allow for full signal resolution, either in bandwidth or dynamics. Can you cite or comment on these "numerous objective tests"? -- % Randy Yates % "My Shangri-la has gone away, fading like %% Fuquay-Varina, NC % the Beatles on 'Hey Jude'" %%% 919-577-9882 % %%%% % 'Shangri-La', *A New World Record*, ELO http://home.earthlink.net/~yatescr |
#15
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"Randy Yates" wrote ...
The 16-bit CD signal is represented by the four-digit hexadecimal number 0xklmn. This is converted to a 24-bit number "left justified" so that it becomes 0xklmn00. The attenuator operates on that input and produces a value 0xopqrst. That value is free to range over almost 144 dB of a 24-bit system (except for the 8 LSBs, which amounts to a loss of 0.000133 dB from a true 24-bit system). What's the point in the real world? You can't add dynamic range merely by shifting a 16-bit sample into a 24-bit word. You act as if there were some magic imputed by converting to 24-bit before attenuating. |
#16
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A point that folks seem to be missing:
The point of 24-bit sampling is HEADROOM. It allows you to turn up the input and get a reliable 16-or-so bits without having the signal clip unexpectedly if your input levels were too high. It allows you to mix multiple 16-bit-or-so samples without the cumulative signal becoming too loud to handle, and without having to attenuate them prematurely and lose part of the original signal before you have to. More bits can be useful IF you have a specific use for them. But "go faster stripes" on a car don't really make it go faster, and extra bits for their own sake don't necessarily buy you anything. |
#17
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"Richard Crowley" wrote in message ... "Randy Yates" wrote ... The 16-bit CD signal is represented by the four-digit hexadecimal number 0xklmn. This is converted to a 24-bit number "left justified" so that it becomes 0xklmn00. The attenuator operates on that input and produces a value 0xopqrst. That value is free to range over almost 144 dB of a 24-bit system (except for the 8 LSBs, which amounts to a loss of 0.000133 dB from a true 24-bit system). What's the point in the real world? You can't add dynamic range merely by shifting a 16-bit sample into a 24-bit word. You know that, I know that but somehow this has escaped the attention of Mr. yates. You act as if there were some magic imputed by converting to 24-bit before attenuating. Indeed. |
#18
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"Richard Crowley" writes:
"Randy Yates" wrote ... The 16-bit CD signal is represented by the four-digit hexadecimal number 0xklmn. This is converted to a 24-bit number "left justified" so that it becomes 0xklmn00. The attenuator operates on that input and produces a value 0xopqrst. That value is free to range over almost 144 dB of a 24-bit system (except for the 8 LSBs, which amounts to a loss of 0.000133 dB from a true 24-bit system). What's the point in the real world? You can't add dynamic range merely by shifting a 16-bit sample into a 24-bit word. That much is true. However, the operation you describe is only the first of several that occur in a digital attenuator, and it is the subsequent operations that provide the extra dynamic range. You act as if there were some magic imputed by converting to 24-bit before attenuating. If you consider a little arithmetic magic, then I guess so. -- % Randy Yates % "My Shangri-la has gone away, fading like %% Fuquay-Varina, NC % the Beatles on 'Hey Jude'" %%% 919-577-9882 % %%%% % 'Shangri-La', *A New World Record*, ELO http://home.earthlink.net/~yatescr |
#19
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"Joe Kesselman" wrote ...
A point that folks seem to be missing: The point of 24-bit sampling is HEADROOM. Only if the original sample was 24-bits. If the original was 16-bit, then none of what follows makes any sense.... It allows you to turn up the input and get a reliable 16-or-so bits without having the signal clip unexpectedly if your input levels were too high. It allows you to mix multiple 16-bit-or-so samples without the cumulative signal becoming too loud to handle, and without having to attenuate them prematurely and lose part of the original signal before you have to. More bits can be useful IF you have a specific use for them. But "go faster stripes" on a car don't really make it go faster, and extra bits for their own sake don't necessarily buy you anything. |
#20
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"Randy Yates" wrote in message ... "Richard Crowley" writes: "Randy Yates" wrote ... The 16-bit CD signal is represented by the four-digit hexadecimal number 0xklmn. This is converted to a 24-bit number "left justified" so that it becomes 0xklmn00. The attenuator operates on that input and produces a value 0xopqrst. That value is free to range over almost 144 dB of a 24-bit system (except for the 8 LSBs, which amounts to a loss of 0.000133 dB from a true 24-bit system). What's the point in the real world? You can't add dynamic range merely by shifting a 16-bit sample into a 24-bit word. That much is true. However, the operation you describe is only the first of several that occur in a digital attenuator, and it is the subsequent operations that provide the extra dynamic range. You act as if there were some magic imputed by converting to 24-bit before attenuating. If you consider a little arithmetic magic, then I guess so. Unless you can actually show how you think this works, sounds like you're just relying on magic that you don't really understand. |
#21
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"Richard Crowley" wrote in message ... "Joe Kesselman" wrote ... The point of 24-bit sampling is HEADROOM. Only if the original sample was 24-bits. If the original was 16-bit, then none of what follows makes any sense.... I suggest you look up the meaning of "headroom". MrT. |
#22
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"Mr.T" wrote ...
"Richard Crowley" wrote ... "Joe Kesselman" wrote ... The point of 24-bit sampling is HEADROOM. Only if the original sample was 24-bits. If the original was 16-bit, then none of what follows makes any sense.... I suggest you look up the meaning of "headroom". And I suggest you look up the meaning of "magic". Unless you have devised some method of recreating the missing 8 bits of data it doesn't matter WHERE you put the 16-bits of real data within the 24-bit word. You've got only 16-bits of dynamic range. Period. |
#23
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"Richard Crowley" wrote in message ... "Mr.T" wrote ... "Richard Crowley" wrote ... "Joe Kesselman" wrote ... The point of 24-bit sampling is HEADROOM. Only if the original sample was 24-bits. If the original was 16-bit, then none of what follows makes any sense.... I suggest you look up the meaning of "headroom". And I suggest you look up the meaning of "magic". I see no mention of that in *my* post? Unless you have devised some method of recreating the missing 8 bits of data it doesn't matter WHERE you put the 16-bits of real data within the 24-bit word. You've got only 16-bits of dynamic range. Period. Exactly, and 24 bits gives you HEADROOM to avoid clipping. I assume you have never done any *real* recording. MrT. |
#24
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"Mr.T" wrote ...
"Richard Crowley" wrote ... "Mr.T" wrote ... "Richard Crowley" wrote ... "Joe Kesselman" wrote ... The point of 24-bit sampling is HEADROOM. Only if the original sample was 24-bits. If the original was 16-bit, then none of what follows makes any sense.... I suggest you look up the meaning of "headroom". And I suggest you look up the meaning of "magic". I see no mention of that in *my* post? Your methodology of thinking that putting 16-bits worth of data into a 24-bit word somehow gives you extra "headroom" can only be described as "magic". There is cerainly no technical explanation for it. At least you and your magical 24-bit friends have offered none. Unless you have devised some method of recreating the missing 8 bits of data it doesn't matter WHERE you put the 16-bits of real data within the 24-bit word. You've got only 16-bits of dynamic range. Period. Exactly, and 24 bits gives you HEADROOM to avoid clipping. I assume you have never done any *real* recording. I've been recording long before digital was invented. I have recorded hundreds (thousands?) of hours back in the analog era, and hundreds more in 16-bit. I now record mostly 24-bit and also produce digital video. I know not only the analog implications of headroom, but also practice both digital and analog circuit design as well as writing code for everything from micro- controllers to supercomputers. I have possibly been recording since before you were born and have been involved in the digital world for many years before it was ever practical to apply to audio. Unless you can explain how these empty 8 bits expand your "headroom" you appear to be talking through your hat. |
#25
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"Richard Crowley" wrote in message ... I suggest you look up the meaning of "headroom". And I suggest you look up the meaning of "magic". I see no mention of that in *my* post? Your methodology of thinking that putting 16-bits worth of data into a 24-bit word somehow gives you extra "headroom" can only be described as "magic". Not, that's exactly what headroom is for when doing REAL recording. One doesn't know exactly what the peak levels will be ahead of time. A couple of bits of headroom is a nice thing to have. (assumes the hardware can achieve better than 16 bits of course, but that's common these days) There is cerainly no technical explanation for it. Yes there is. Look up recording headroom like I told you. At least you and your magical 24-bit friends have offered none. You simply refuse to look. Exactly, and 24 bits gives you HEADROOM to avoid clipping. I assume you have never done any *real* recording. I've been recording long before digital was invented. I have recorded hundreds (thousands?) of hours back in the analog era, and hundreds more in 16-bit. I now record mostly 24-bit and also produce digital video. I know not only the analog implications of headroom, but also practice both digital and analog circuit design as well as writing code for everything from micro- controllers to supercomputers. I have possibly been recording since before you were born and have been involved in the digital world for many years before it was ever practical to apply to audio. All that so called experience and you still have no idea what headroom is. Hint: its the *unused* part of the recorders dynamic range *above* the actual signal peak. Simply allows you to avoid clipping when you don't know ahead of time what the peak level will be. It's obvious you are too busy trying to big-note yourself to see what *I* have actually said. Unless you can explain how these empty 8 bits expand your "headroom" you appear to be talking through your hat. It's quite simple, IF the recorder has less than 16 bits resolution, then you achieve nothing by going to 24 bits. *IF* your recorder *CAN* achieve better than 16 bits (17, 18 bits etc) then the extra bits can be used for recording HEADROOM. IF *you* don't require the extra headroom, fine. MrT. |
#26
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"Mr.T" MrT@home wrote in message u... "Richard Crowley" wrote in message ... "Joe Kesselman" wrote ... The point of 24-bit sampling is HEADROOM. Only if the original sample was 24-bits. If the original was 16-bit, then none of what follows makes any sense.... I suggest you look up the meaning of "headroom". Your typical 16 - 24 bit conversion adds no headroom. |
#27
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"Mr.T" MrT@home wrote in message u... "Richard Crowley" wrote in message ... "Mr.T" wrote ... "Richard Crowley" wrote ... "Joe Kesselman" wrote ... The point of 24-bit sampling is HEADROOM. Only if the original sample was 24-bits. If the original was 16-bit, then none of what follows makes any sense.... I suggest you look up the meaning of "headroom". And I suggest you look up the meaning of "magic". I see no mention of that in *my* post? Unless you have devised some method of recreating the missing 8 bits of data it doesn't matter WHERE you put the 16-bits of real data within the 24-bit word. You've got only 16-bits of dynamic range. Period. Exactly, and 24 bits gives you HEADROOM to avoid clipping. I assume you have never done any *real* recording. Repeat: Your typical 16- 24 bit conversion adds no headroom. In a digital volume control, what matters is more like "foot room". |
#28
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"Mr.T" MrT@home wrote in message u... It's quite simple, IF the recorder has less than 16 bits resolution, then you achieve nothing by going to 24 bits. Not necessarily. You can achieve what I called "foot room" which is relevant to digital attenuators. *IF* your recorder *CAN* achieve better than 16 bits (17, 18 bits etc) then the extra bits can be used for recording HEADROOM. Only if you turn the levels down. IF *you* don't require the extra headroom, fine. What you always get with a 16-24 bit conversion is the potential for better dynamic range in follow-on processing. |
#29
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Richard Crowley wrote:
Your methodology of thinking that putting 16-bits worth of data into a 24-bit word somehow gives you extra "headroom" can only be described as "magic". I think we're still talking at cross purposes and applying different assumptions about what's going to be done with the data. If all you're doing is record-and-playback, then of course you can't play back more information than you originally recorded. BUT: even assuming that you have 16-bit starting samples, as soon as you start mixing them together you either need more bits or you lose information. Add two waves that peak at the top of the 16-bit range, and you either need another bit or you lose information off the bottom. This *is* a headroom consideration -- without that headroom, your mixer becomes lossy. No magic, just engineering. (And as I pointed out, originally recording at 24 increases the odds of your getting a good reliable 16-meaningful. Actually, I really like Sony's (?) patent for floating-point digitization, but I haven't seen any affordable hardware or software which uses it yet. I'm still kicking myself for not having filed that patent when I first came up with the idea!) |
#30
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*IF* your recorder *CAN* achieve better than 16 bits (17, 18 bits etc) then
the extra bits can be used for recording HEADROOM. Bingo. And generally that's a better use for the extra bits -- in the recording stage -- than turning up the preamp and putting the extra bits on the bottom would be. I really think we've got two separate arguments going on -- whether more than 16 bits buys you anything significant in finished output, and whether it has practical uses before that point. I think the answers are "probably not" and "hell yes". The latter's unquestionable and easily demonstrable. The former's an audiophile argument and probably impossible to separate from placebo effect and opinion. |
#31
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Unless you can actually show how you think this works, sounds
like you're just relying on magic that you don't really understand. See other thread. No magic, just leaving room for accumulation and error. |
#32
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"Mr.T" MrT@home wrote in message u... "Richard Crowley" wrote in message ... I suggest you look up the meaning of "headroom". And I suggest you look up the meaning of "magic". I see no mention of that in *my* post? Your methodology of thinking that putting 16-bits worth of data into a 24-bit word somehow gives you extra "headroom" can only be described as "magic". Not, that's exactly what headroom is for when doing REAL recording. One doesn't know exactly what the peak levels will be ahead of time. A couple of bits of headroom is a nice thing to have. (assumes the hardware can achieve better than 16 bits of course, but that's common these days) We agree completely that recording in 24 bits is highly desirable and will most certainly increase the dynamic range i.e. provide significant headroom. What you fail to realize is that adding 8 more bits to a pre-recorded 16 bit sample buys you NOTHING. |
#33
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"Joe Kesselman" wrote in message ... Unless you can actually show how you think this works, sounds like you're just relying on magic that you don't really understand. See other thread. No magic, just leaving room for accumulation and error. In an attenuator? (Refer to the subject line if you have lost track of the discussion.) |
#34
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"Arny Krueger" wrote in message ... Your typical 16 - 24 bit conversion adds no headroom. Of course not, why would you think it would? MrT. |
#35
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"Arny Krueger" wrote in message ... Repeat: Your typical 16- 24 bit conversion adds no headroom. Of course not, where did I say it did? MrT. |
#36
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"Arny Krueger" wrote in message ... "Mr.T" MrT@home wrote in message u... It's quite simple, IF the recorder has less than 16 bits resolution, then you achieve nothing by going to 24 bits. Not necessarily. You can achieve what I called "foot room" which is relevant to digital attenuators. *IF* your recorder *CAN* achieve better than 16 bits (17, 18 bits etc) then the extra bits can be used for recording HEADROOM. Only if you turn the levels down. Duh! IF *you* don't require the extra headroom, fine. What you always get with a 16-24 bit conversion is the potential for better dynamic range in follow-on processing. Yes, but if you wish to reply to something other than what I said, why tack it onto my post? Why not the person who said it? MrT. |
#37
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"Joe Kesselman" wrote in message ... I really think we've got two separate arguments going on -- whether more than 16 bits buys you anything significant in finished output, and whether it has practical uses before that point. I think the answers are "probably not" and "hell yes". The latter's unquestionable and easily demonstrable. The former's an audiophile argument and probably impossible to separate from placebo effect and opinion. Agreed. MrT. |
#38
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"Mr.T" MrT@home wrote in message u... "Arny Krueger" wrote in message ... Your typical 16 - 24 bit conversion adds no headroom. Of course not, why would you think it would? That was the (mis)statement that started this whole discussion. |
#39
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"Richard Crowley" wrote in message ... We agree completely that recording in 24 bits is highly desirable and will most certainly increase the dynamic range i.e. provide significant headroom. What you fail to realize is that adding 8 more bits to a pre-recorded 16 bit sample buys you NOTHING. Why do you think I fail to realise that, since I never mentioned it? As others have pointed out though, it *does* buy you less computational error when mathematically manipulating the data in any way that causes rounding errors. MrT. |
#40
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"Richard Crowley" wrote in message ... "Mr.T" MrT@home wrote in message u... "Arny Krueger" wrote in message ... Your typical 16 - 24 bit conversion adds no headroom. Of course not, why would you think it would? That was the (mis)statement that started this whole discussion. Maybe, but not mine. MrT. |
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