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#41
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CD Level Variations
Dave Platt wrote:
I'm sure many of you have experienced the frustration in different playback levels when playing CDs in a CD carousel or a CD jukebox, especially in shuffle or random mode. The reason for this may be "technological advances" as most of the seemingly "low- level" CDs are early issues. Anyway, large differences in levels tend to make variations in music programming less than optimum. In my experience, the "technological advances" you cite are precisely backwards. Most modern CDs (popular music in particular) are being mixed/mastered at a very high level, with a very great deal of compression being applied to the signal. The goal of this appears to be a deliberate desire to have the music sound "loud" or "punchy" - that is, attention-getting. It's purely market-and-marketing driven. The compression is often so great that there are two very negative side effects: - The music often has very little dynamic range left - the distance between loudest and softest passages is often only a few dB. see, for example, the march of progress on display he http://forums.lukpac.org/viewtopic.php?t=643 -- -S. "They've got God on their side. All we've got is science and reason." -- Dawn Hulsey, Talent Director |
#42
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CD Level Variations
"Laurence Payne" wrote in
message On Tue, 16 Mar 2004 07:12:14 -0000, (Dave Platt) wrote: Leaving a small fraction of a dB of digital headroom seems like good practice, What fraction of a dB would correspond to one bit under maximum? 6 dB, way too much. I leave 1 dB headroom. This number is based on testing a ton of digital audio gear. |
#43
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CD Level Variations
"Laurence Payne" wrote in
message On Tue, 16 Mar 2004 07:12:14 -0000, (Dave Platt) wrote: Leaving a small fraction of a dB of digital headroom seems like good practice, What fraction of a dB would correspond to one bit under maximum? 6 dB, way too much. I leave 1 dB headroom. This number is based on testing a ton of digital audio gear. |
#44
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CD Level Variations
"Laurence Payne" wrote in
message On Tue, 16 Mar 2004 07:12:14 -0000, (Dave Platt) wrote: Leaving a small fraction of a dB of digital headroom seems like good practice, What fraction of a dB would correspond to one bit under maximum? 6 dB, way too much. I leave 1 dB headroom. This number is based on testing a ton of digital audio gear. |
#45
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CD Level Variations
"Laurence Payne" wrote in
message On Tue, 16 Mar 2004 07:12:14 -0000, (Dave Platt) wrote: Leaving a small fraction of a dB of digital headroom seems like good practice, What fraction of a dB would correspond to one bit under maximum? 6 dB, way too much. I leave 1 dB headroom. This number is based on testing a ton of digital audio gear. |
#46
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CD Level Variations
Laurence Payne wrote in message . ..
On Tue, 16 Mar 2004 07:12:14 -0000, (Dave Platt) wrote: Leaving a small fraction of a dB of digital headroom seems like good practice, What fraction of a dB would correspond to one bit under maximum? Well if you mean you go from the value FFFF (hexadecimal) (=65536) to FFFE (65535) then the decibel change will be (20*lg(65535/65536)) where lg is the logarithm using 10 as base. A rather small value indeed. Note that one step close to the maximum level (FFFF) is very small in decibel compared to a step at the minumum level. A step between 0001 to 0002 is aprox 6 dB. You have doubbled to voltage value. The quantization is linear but the ear's sensitivity is logarithmic. |
#47
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CD Level Variations
Laurence Payne wrote in message . ..
On Tue, 16 Mar 2004 07:12:14 -0000, (Dave Platt) wrote: Leaving a small fraction of a dB of digital headroom seems like good practice, What fraction of a dB would correspond to one bit under maximum? Well if you mean you go from the value FFFF (hexadecimal) (=65536) to FFFE (65535) then the decibel change will be (20*lg(65535/65536)) where lg is the logarithm using 10 as base. A rather small value indeed. Note that one step close to the maximum level (FFFF) is very small in decibel compared to a step at the minumum level. A step between 0001 to 0002 is aprox 6 dB. You have doubbled to voltage value. The quantization is linear but the ear's sensitivity is logarithmic. |
#48
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CD Level Variations
Laurence Payne wrote in message . ..
On Tue, 16 Mar 2004 07:12:14 -0000, (Dave Platt) wrote: Leaving a small fraction of a dB of digital headroom seems like good practice, What fraction of a dB would correspond to one bit under maximum? Well if you mean you go from the value FFFF (hexadecimal) (=65536) to FFFE (65535) then the decibel change will be (20*lg(65535/65536)) where lg is the logarithm using 10 as base. A rather small value indeed. Note that one step close to the maximum level (FFFF) is very small in decibel compared to a step at the minumum level. A step between 0001 to 0002 is aprox 6 dB. You have doubbled to voltage value. The quantization is linear but the ear's sensitivity is logarithmic. |
#49
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CD Level Variations
Laurence Payne wrote in message . ..
On Tue, 16 Mar 2004 07:12:14 -0000, (Dave Platt) wrote: Leaving a small fraction of a dB of digital headroom seems like good practice, What fraction of a dB would correspond to one bit under maximum? Well if you mean you go from the value FFFF (hexadecimal) (=65536) to FFFE (65535) then the decibel change will be (20*lg(65535/65536)) where lg is the logarithm using 10 as base. A rather small value indeed. Note that one step close to the maximum level (FFFF) is very small in decibel compared to a step at the minumum level. A step between 0001 to 0002 is aprox 6 dB. You have doubbled to voltage value. The quantization is linear but the ear's sensitivity is logarithmic. |
#51
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CD Level Variations
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#52
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CD Level Variations
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#53
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CD Level Variations
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#54
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CD Level Variations
On Tue, 16 Mar 2004 23:33:58 +0000, Laurence Payne
What fraction of a dB would correspond to one bit under maximum? When you ask about "1 bit" are you asking about .. THE most significant bit (msb) (perhaps referenced to a "signed" 32767) or THE least significant bit (lsb) or some other bit or some particular bit based upon and arbitary level. There's quite a difference to the way the question should be answered. |
#55
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CD Level Variations
On Tue, 16 Mar 2004 23:33:58 +0000, Laurence Payne
What fraction of a dB would correspond to one bit under maximum? When you ask about "1 bit" are you asking about .. THE most significant bit (msb) (perhaps referenced to a "signed" 32767) or THE least significant bit (lsb) or some other bit or some particular bit based upon and arbitary level. There's quite a difference to the way the question should be answered. |
#56
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CD Level Variations
On Tue, 16 Mar 2004 23:33:58 +0000, Laurence Payne
What fraction of a dB would correspond to one bit under maximum? When you ask about "1 bit" are you asking about .. THE most significant bit (msb) (perhaps referenced to a "signed" 32767) or THE least significant bit (lsb) or some other bit or some particular bit based upon and arbitary level. There's quite a difference to the way the question should be answered. |
#57
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CD Level Variations
On Tue, 16 Mar 2004 23:33:58 +0000, Laurence Payne
What fraction of a dB would correspond to one bit under maximum? When you ask about "1 bit" are you asking about .. THE most significant bit (msb) (perhaps referenced to a "signed" 32767) or THE least significant bit (lsb) or some other bit or some particular bit based upon and arbitary level. There's quite a difference to the way the question should be answered. |
#59
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CD Level Variations
(Dave Platt) writes:
In article , Barry Mann wrote: - The sound is often pushed to such a high volume level that it's hitting the maximum-positive/negative number boundaries of the 16-bit digital number system used by CDs. In other words, it's clipping - quite hard, often very harshly, and often for a prolonged period. I don't have a problem with the CD touching the top level at one or more spots. This is good management of dynamic range. It's just a simple computer trick to do this. Redbook CD's have a 16 bit range. "Clipping" implies a loss of information because the signal would have gone beyond the limit if it could have. "Touching" means the signal reached the limit and didn't need to go past it. Touching the limit, for a single sample, usually doesn't cause technical problems. Two or more samples, touching the limit, usually means that the original signal tried to go past the limit, and has in fact been clipped. Agreed. If one were directly outputing the data from an A/D, you could point the finger to this stage, but usually there is some (a crapload?) digital processing to the signal before D/Aing. It is hard to ensure that all DSP algorithms result in an equivalent analog waveform that doesn't exceed the D/A limits. In fact, it can be shown (although it is a pathological case) that a digital signal can be generated which would result in an infinite output level if a theoretical brick-wall reconstruction filter is applied. Of course such a filter is not physically realizable, but long filter are and the potential for overflow is real. I believe that the usual working definition of a "digital over" is two or more consecutive samples at the positive or negative limit. Leaving a small fraction of a dB of digital headroom seems like good practice, because (as you note) some CD players will show clipping artifacts at the digital limit, and I understand that some can actually show clipping problems even if the limit isn't reached. CD players with digital oversampling filters (many/most of 'em) can have problems with clipping during the oversampling process - even if two adjacent samples in the original 16-bit data stream are below the limit, the reconstructed/oversampled waveform calculated by the filters can go over the limit and be clipped. As long as the digital signal is maintained below the "equivalent analog limit," oversampling shouldn't do this. If it does, the designer screwed up a gain somewhere. -- % Randy Yates % "With time with what you've learned, %% Fuquay-Varina, NC % they'll kiss the ground you walk %%% 919-577-9882 % upon." %%%% % '21st Century Man', *Time*, ELO http://home.earthlink.net/~yatescr |
#60
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CD Level Variations
(Dave Platt) writes:
In article , Barry Mann wrote: - The sound is often pushed to such a high volume level that it's hitting the maximum-positive/negative number boundaries of the 16-bit digital number system used by CDs. In other words, it's clipping - quite hard, often very harshly, and often for a prolonged period. I don't have a problem with the CD touching the top level at one or more spots. This is good management of dynamic range. It's just a simple computer trick to do this. Redbook CD's have a 16 bit range. "Clipping" implies a loss of information because the signal would have gone beyond the limit if it could have. "Touching" means the signal reached the limit and didn't need to go past it. Touching the limit, for a single sample, usually doesn't cause technical problems. Two or more samples, touching the limit, usually means that the original signal tried to go past the limit, and has in fact been clipped. Agreed. If one were directly outputing the data from an A/D, you could point the finger to this stage, but usually there is some (a crapload?) digital processing to the signal before D/Aing. It is hard to ensure that all DSP algorithms result in an equivalent analog waveform that doesn't exceed the D/A limits. In fact, it can be shown (although it is a pathological case) that a digital signal can be generated which would result in an infinite output level if a theoretical brick-wall reconstruction filter is applied. Of course such a filter is not physically realizable, but long filter are and the potential for overflow is real. I believe that the usual working definition of a "digital over" is two or more consecutive samples at the positive or negative limit. Leaving a small fraction of a dB of digital headroom seems like good practice, because (as you note) some CD players will show clipping artifacts at the digital limit, and I understand that some can actually show clipping problems even if the limit isn't reached. CD players with digital oversampling filters (many/most of 'em) can have problems with clipping during the oversampling process - even if two adjacent samples in the original 16-bit data stream are below the limit, the reconstructed/oversampled waveform calculated by the filters can go over the limit and be clipped. As long as the digital signal is maintained below the "equivalent analog limit," oversampling shouldn't do this. If it does, the designer screwed up a gain somewhere. -- % Randy Yates % "With time with what you've learned, %% Fuquay-Varina, NC % they'll kiss the ground you walk %%% 919-577-9882 % upon." %%%% % '21st Century Man', *Time*, ELO http://home.earthlink.net/~yatescr |
#61
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CD Level Variations
(Dave Platt) writes:
In article , Barry Mann wrote: - The sound is often pushed to such a high volume level that it's hitting the maximum-positive/negative number boundaries of the 16-bit digital number system used by CDs. In other words, it's clipping - quite hard, often very harshly, and often for a prolonged period. I don't have a problem with the CD touching the top level at one or more spots. This is good management of dynamic range. It's just a simple computer trick to do this. Redbook CD's have a 16 bit range. "Clipping" implies a loss of information because the signal would have gone beyond the limit if it could have. "Touching" means the signal reached the limit and didn't need to go past it. Touching the limit, for a single sample, usually doesn't cause technical problems. Two or more samples, touching the limit, usually means that the original signal tried to go past the limit, and has in fact been clipped. Agreed. If one were directly outputing the data from an A/D, you could point the finger to this stage, but usually there is some (a crapload?) digital processing to the signal before D/Aing. It is hard to ensure that all DSP algorithms result in an equivalent analog waveform that doesn't exceed the D/A limits. In fact, it can be shown (although it is a pathological case) that a digital signal can be generated which would result in an infinite output level if a theoretical brick-wall reconstruction filter is applied. Of course such a filter is not physically realizable, but long filter are and the potential for overflow is real. I believe that the usual working definition of a "digital over" is two or more consecutive samples at the positive or negative limit. Leaving a small fraction of a dB of digital headroom seems like good practice, because (as you note) some CD players will show clipping artifacts at the digital limit, and I understand that some can actually show clipping problems even if the limit isn't reached. CD players with digital oversampling filters (many/most of 'em) can have problems with clipping during the oversampling process - even if two adjacent samples in the original 16-bit data stream are below the limit, the reconstructed/oversampled waveform calculated by the filters can go over the limit and be clipped. As long as the digital signal is maintained below the "equivalent analog limit," oversampling shouldn't do this. If it does, the designer screwed up a gain somewhere. -- % Randy Yates % "With time with what you've learned, %% Fuquay-Varina, NC % they'll kiss the ground you walk %%% 919-577-9882 % upon." %%%% % '21st Century Man', *Time*, ELO http://home.earthlink.net/~yatescr |
#62
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CD Level Variations
Randy Yates writes:
(Dave Platt) writes: In article , Barry Mann wrote: - The sound is often pushed to such a high volume level that it's hitting the maximum-positive/negative number boundaries of the 16-bit digital number system used by CDs. In other words, it's clipping - quite hard, often very harshly, and often for a prolonged period. I don't have a problem with the CD touching the top level at one or more spots. This is good management of dynamic range. It's just a simple computer trick to do this. Redbook CD's have a 16 bit range. "Clipping" implies a loss of information because the signal would have gone beyond the limit if it could have. "Touching" means the signal reached the limit and didn't need to go past it. Touching the limit, for a single sample, usually doesn't cause technical problems. Two or more samples, touching the limit, usually means that the original signal tried to go past the limit, and has in fact been clipped. Agreed. Actually, let me correct myself here. It is possible for an analog waveform to exceed the reference voltage of an A/D and still not clip digitally. It *will* be clipped when it is reproduced over a D/A, but it *hasn't* been clipped by the A/D. -- % Randy Yates % "Rollin' and riding and slippin' and %% Fuquay-Varina, NC % sliding, it's magic." %%% 919-577-9882 % %%%% % 'Living' Thing', *A New World Record*, ELO http://home.earthlink.net/~yatescr |
#63
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CD Level Variations
Randy Yates writes:
(Dave Platt) writes: In article , Barry Mann wrote: - The sound is often pushed to such a high volume level that it's hitting the maximum-positive/negative number boundaries of the 16-bit digital number system used by CDs. In other words, it's clipping - quite hard, often very harshly, and often for a prolonged period. I don't have a problem with the CD touching the top level at one or more spots. This is good management of dynamic range. It's just a simple computer trick to do this. Redbook CD's have a 16 bit range. "Clipping" implies a loss of information because the signal would have gone beyond the limit if it could have. "Touching" means the signal reached the limit and didn't need to go past it. Touching the limit, for a single sample, usually doesn't cause technical problems. Two or more samples, touching the limit, usually means that the original signal tried to go past the limit, and has in fact been clipped. Agreed. Actually, let me correct myself here. It is possible for an analog waveform to exceed the reference voltage of an A/D and still not clip digitally. It *will* be clipped when it is reproduced over a D/A, but it *hasn't* been clipped by the A/D. -- % Randy Yates % "Rollin' and riding and slippin' and %% Fuquay-Varina, NC % sliding, it's magic." %%% 919-577-9882 % %%%% % 'Living' Thing', *A New World Record*, ELO http://home.earthlink.net/~yatescr |
#64
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CD Level Variations
Randy Yates writes:
(Dave Platt) writes: In article , Barry Mann wrote: - The sound is often pushed to such a high volume level that it's hitting the maximum-positive/negative number boundaries of the 16-bit digital number system used by CDs. In other words, it's clipping - quite hard, often very harshly, and often for a prolonged period. I don't have a problem with the CD touching the top level at one or more spots. This is good management of dynamic range. It's just a simple computer trick to do this. Redbook CD's have a 16 bit range. "Clipping" implies a loss of information because the signal would have gone beyond the limit if it could have. "Touching" means the signal reached the limit and didn't need to go past it. Touching the limit, for a single sample, usually doesn't cause technical problems. Two or more samples, touching the limit, usually means that the original signal tried to go past the limit, and has in fact been clipped. Agreed. Actually, let me correct myself here. It is possible for an analog waveform to exceed the reference voltage of an A/D and still not clip digitally. It *will* be clipped when it is reproduced over a D/A, but it *hasn't* been clipped by the A/D. -- % Randy Yates % "Rollin' and riding and slippin' and %% Fuquay-Varina, NC % sliding, it's magic." %%% 919-577-9882 % %%%% % 'Living' Thing', *A New World Record*, ELO http://home.earthlink.net/~yatescr |
#65
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CD Level Variations
Randy Yates writes:
(Dave Platt) writes: In article , Barry Mann wrote: - The sound is often pushed to such a high volume level that it's hitting the maximum-positive/negative number boundaries of the 16-bit digital number system used by CDs. In other words, it's clipping - quite hard, often very harshly, and often for a prolonged period. I don't have a problem with the CD touching the top level at one or more spots. This is good management of dynamic range. It's just a simple computer trick to do this. Redbook CD's have a 16 bit range. "Clipping" implies a loss of information because the signal would have gone beyond the limit if it could have. "Touching" means the signal reached the limit and didn't need to go past it. Touching the limit, for a single sample, usually doesn't cause technical problems. Two or more samples, touching the limit, usually means that the original signal tried to go past the limit, and has in fact been clipped. Agreed. Actually, let me correct myself here. It is possible for an analog waveform to exceed the reference voltage of an A/D and still not clip digitally. It *will* be clipped when it is reproduced over a D/A, but it *hasn't* been clipped by the A/D. -- % Randy Yates % "Rollin' and riding and slippin' and %% Fuquay-Varina, NC % sliding, it's magic." %%% 919-577-9882 % %%%% % 'Living' Thing', *A New World Record*, ELO http://home.earthlink.net/~yatescr |
#66
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CD Level Variations
Norbert Hahn writes:
Your zero value is at 32768, thus going from 32768 to 32769 doubles the output voltage, but the reference level remains 65536. Or am I grossly mistaken? Almost all digital systems use "signed two's complement" representation of the data. For a 16-bit system, this results in integers from -32768 (full-scale negative) to +32767 (full-scale positive), with 0 corresponding to 0. -- % Randy Yates % "Rollin' and riding and slippin' and %% Fuquay-Varina, NC % sliding, it's magic." %%% 919-577-9882 % %%%% % 'Living' Thing', *A New World Record*, ELO http://home.earthlink.net/~yatescr |
#67
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CD Level Variations
Norbert Hahn writes:
Your zero value is at 32768, thus going from 32768 to 32769 doubles the output voltage, but the reference level remains 65536. Or am I grossly mistaken? Almost all digital systems use "signed two's complement" representation of the data. For a 16-bit system, this results in integers from -32768 (full-scale negative) to +32767 (full-scale positive), with 0 corresponding to 0. -- % Randy Yates % "Rollin' and riding and slippin' and %% Fuquay-Varina, NC % sliding, it's magic." %%% 919-577-9882 % %%%% % 'Living' Thing', *A New World Record*, ELO http://home.earthlink.net/~yatescr |
#68
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CD Level Variations
Norbert Hahn writes:
Your zero value is at 32768, thus going from 32768 to 32769 doubles the output voltage, but the reference level remains 65536. Or am I grossly mistaken? Almost all digital systems use "signed two's complement" representation of the data. For a 16-bit system, this results in integers from -32768 (full-scale negative) to +32767 (full-scale positive), with 0 corresponding to 0. -- % Randy Yates % "Rollin' and riding and slippin' and %% Fuquay-Varina, NC % sliding, it's magic." %%% 919-577-9882 % %%%% % 'Living' Thing', *A New World Record*, ELO http://home.earthlink.net/~yatescr |
#69
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CD Level Variations
Norbert Hahn writes:
Your zero value is at 32768, thus going from 32768 to 32769 doubles the output voltage, but the reference level remains 65536. Or am I grossly mistaken? Almost all digital systems use "signed two's complement" representation of the data. For a 16-bit system, this results in integers from -32768 (full-scale negative) to +32767 (full-scale positive), with 0 corresponding to 0. -- % Randy Yates % "Rollin' and riding and slippin' and %% Fuquay-Varina, NC % sliding, it's magic." %%% 919-577-9882 % %%%% % 'Living' Thing', *A New World Record*, ELO http://home.earthlink.net/~yatescr |
#70
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CD Level Variations
Norbert Hahn wrote in message . ..
(Kega) wrote: Well if you mean you go from the value FFFF (hexadecimal) (=65536) to FFFE (65535) then the decibel change will be (20*lg(65535/65536)) where lg is the logarithm using 10 as base. A rather small value indeed. Note that one step close to the maximum level (FFFF) is very small in decibel compared to a step at the minumum level. A step between 0001 to 0002 is aprox 6 dB. You have doubbled to voltage value. The quantization is linear but the ear's sensitivity is logarithmic. We're talking about 16 bit systems, don't we? So a step from 0001 to 0002 will again be 20*lg(1/65536) as it is when going from 65536 to 65535. All steps are of the same size as linear PCM is used. Your zero value is at 32768, thus going from 32768 to 32769 doubles the output voltage, but the reference level remains 65536. Or am I grossly mistaken? Norbert Yes I think you are. A double of output voltage is always approx 6 dB. When counting with dB you always count with the ratio between output and input (or between to levels). Decibel is a logaritmic scale and is a tenth of a Bel. A bel (B) is: B = lg (P_out/P_in) where P stands for Power. And since P(ower) is proportinal to the square of V(oltage) then B = lg(V_out**2/V_in**2) = lg((V_out/V_in)**2) = 2 * lg(V_out/V_in) and expressed in dB gives: 20*lg(V_out/V_in) A double in voltage gives: 20*lg(2/1) is approx 20*0.3=6 and a halfing: 20*lg(1/2)= -6 (approx) Therefor you have smaller dB-steps at higher voltage than you have in lower voltage in a linear PCM system (as the CD format is). This is a (kind of)drawback in audio PCM since you use more bits than you necessarily must use to obtain a suitable HiFi-level. I guess they (the constructor of CD) choosed linear PCM when designing CD because at that time it was easier to construct a reasonable good sounding linear D/A-converter than a logarithmic one. In the analog days you more or less learned that high volumes produced more distortion than low volumes. In PCM you have the opposite. High volumes that does not exceeds the headroom limit have lesser distortion than low volumes. But since you have a lots of bits the overall distortion at normal levels are very low. O boy. It is a bit difficult to explain in a language that is not my mother tounge. Hope you all understand my points. It was also very long time ago a studied PCM (back in 1977/78). BTW. Just recently I read how dts was coded and boy that was complicated (but beautiful) using Adaptive Differentiell PCM with prediction and splitting up the signal into sub frequency bands. But I am impressed. Back in 77 we (at school) just touched the technique of ADPCM. Best Regards from Sweden /Kent |
#71
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CD Level Variations
Norbert Hahn wrote in message . ..
(Kega) wrote: Well if you mean you go from the value FFFF (hexadecimal) (=65536) to FFFE (65535) then the decibel change will be (20*lg(65535/65536)) where lg is the logarithm using 10 as base. A rather small value indeed. Note that one step close to the maximum level (FFFF) is very small in decibel compared to a step at the minumum level. A step between 0001 to 0002 is aprox 6 dB. You have doubbled to voltage value. The quantization is linear but the ear's sensitivity is logarithmic. We're talking about 16 bit systems, don't we? So a step from 0001 to 0002 will again be 20*lg(1/65536) as it is when going from 65536 to 65535. All steps are of the same size as linear PCM is used. Your zero value is at 32768, thus going from 32768 to 32769 doubles the output voltage, but the reference level remains 65536. Or am I grossly mistaken? Norbert Yes I think you are. A double of output voltage is always approx 6 dB. When counting with dB you always count with the ratio between output and input (or between to levels). Decibel is a logaritmic scale and is a tenth of a Bel. A bel (B) is: B = lg (P_out/P_in) where P stands for Power. And since P(ower) is proportinal to the square of V(oltage) then B = lg(V_out**2/V_in**2) = lg((V_out/V_in)**2) = 2 * lg(V_out/V_in) and expressed in dB gives: 20*lg(V_out/V_in) A double in voltage gives: 20*lg(2/1) is approx 20*0.3=6 and a halfing: 20*lg(1/2)= -6 (approx) Therefor you have smaller dB-steps at higher voltage than you have in lower voltage in a linear PCM system (as the CD format is). This is a (kind of)drawback in audio PCM since you use more bits than you necessarily must use to obtain a suitable HiFi-level. I guess they (the constructor of CD) choosed linear PCM when designing CD because at that time it was easier to construct a reasonable good sounding linear D/A-converter than a logarithmic one. In the analog days you more or less learned that high volumes produced more distortion than low volumes. In PCM you have the opposite. High volumes that does not exceeds the headroom limit have lesser distortion than low volumes. But since you have a lots of bits the overall distortion at normal levels are very low. O boy. It is a bit difficult to explain in a language that is not my mother tounge. Hope you all understand my points. It was also very long time ago a studied PCM (back in 1977/78). BTW. Just recently I read how dts was coded and boy that was complicated (but beautiful) using Adaptive Differentiell PCM with prediction and splitting up the signal into sub frequency bands. But I am impressed. Back in 77 we (at school) just touched the technique of ADPCM. Best Regards from Sweden /Kent |
#72
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CD Level Variations
Norbert Hahn wrote in message . ..
(Kega) wrote: Well if you mean you go from the value FFFF (hexadecimal) (=65536) to FFFE (65535) then the decibel change will be (20*lg(65535/65536)) where lg is the logarithm using 10 as base. A rather small value indeed. Note that one step close to the maximum level (FFFF) is very small in decibel compared to a step at the minumum level. A step between 0001 to 0002 is aprox 6 dB. You have doubbled to voltage value. The quantization is linear but the ear's sensitivity is logarithmic. We're talking about 16 bit systems, don't we? So a step from 0001 to 0002 will again be 20*lg(1/65536) as it is when going from 65536 to 65535. All steps are of the same size as linear PCM is used. Your zero value is at 32768, thus going from 32768 to 32769 doubles the output voltage, but the reference level remains 65536. Or am I grossly mistaken? Norbert Yes I think you are. A double of output voltage is always approx 6 dB. When counting with dB you always count with the ratio between output and input (or between to levels). Decibel is a logaritmic scale and is a tenth of a Bel. A bel (B) is: B = lg (P_out/P_in) where P stands for Power. And since P(ower) is proportinal to the square of V(oltage) then B = lg(V_out**2/V_in**2) = lg((V_out/V_in)**2) = 2 * lg(V_out/V_in) and expressed in dB gives: 20*lg(V_out/V_in) A double in voltage gives: 20*lg(2/1) is approx 20*0.3=6 and a halfing: 20*lg(1/2)= -6 (approx) Therefor you have smaller dB-steps at higher voltage than you have in lower voltage in a linear PCM system (as the CD format is). This is a (kind of)drawback in audio PCM since you use more bits than you necessarily must use to obtain a suitable HiFi-level. I guess they (the constructor of CD) choosed linear PCM when designing CD because at that time it was easier to construct a reasonable good sounding linear D/A-converter than a logarithmic one. In the analog days you more or less learned that high volumes produced more distortion than low volumes. In PCM you have the opposite. High volumes that does not exceeds the headroom limit have lesser distortion than low volumes. But since you have a lots of bits the overall distortion at normal levels are very low. O boy. It is a bit difficult to explain in a language that is not my mother tounge. Hope you all understand my points. It was also very long time ago a studied PCM (back in 1977/78). BTW. Just recently I read how dts was coded and boy that was complicated (but beautiful) using Adaptive Differentiell PCM with prediction and splitting up the signal into sub frequency bands. But I am impressed. Back in 77 we (at school) just touched the technique of ADPCM. Best Regards from Sweden /Kent |
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CD Level Variations
Norbert Hahn wrote in message . ..
(Kega) wrote: Well if you mean you go from the value FFFF (hexadecimal) (=65536) to FFFE (65535) then the decibel change will be (20*lg(65535/65536)) where lg is the logarithm using 10 as base. A rather small value indeed. Note that one step close to the maximum level (FFFF) is very small in decibel compared to a step at the minumum level. A step between 0001 to 0002 is aprox 6 dB. You have doubbled to voltage value. The quantization is linear but the ear's sensitivity is logarithmic. We're talking about 16 bit systems, don't we? So a step from 0001 to 0002 will again be 20*lg(1/65536) as it is when going from 65536 to 65535. All steps are of the same size as linear PCM is used. Your zero value is at 32768, thus going from 32768 to 32769 doubles the output voltage, but the reference level remains 65536. Or am I grossly mistaken? Norbert Yes I think you are. A double of output voltage is always approx 6 dB. When counting with dB you always count with the ratio between output and input (or between to levels). Decibel is a logaritmic scale and is a tenth of a Bel. A bel (B) is: B = lg (P_out/P_in) where P stands for Power. And since P(ower) is proportinal to the square of V(oltage) then B = lg(V_out**2/V_in**2) = lg((V_out/V_in)**2) = 2 * lg(V_out/V_in) and expressed in dB gives: 20*lg(V_out/V_in) A double in voltage gives: 20*lg(2/1) is approx 20*0.3=6 and a halfing: 20*lg(1/2)= -6 (approx) Therefor you have smaller dB-steps at higher voltage than you have in lower voltage in a linear PCM system (as the CD format is). This is a (kind of)drawback in audio PCM since you use more bits than you necessarily must use to obtain a suitable HiFi-level. I guess they (the constructor of CD) choosed linear PCM when designing CD because at that time it was easier to construct a reasonable good sounding linear D/A-converter than a logarithmic one. In the analog days you more or less learned that high volumes produced more distortion than low volumes. In PCM you have the opposite. High volumes that does not exceeds the headroom limit have lesser distortion than low volumes. But since you have a lots of bits the overall distortion at normal levels are very low. O boy. It is a bit difficult to explain in a language that is not my mother tounge. Hope you all understand my points. It was also very long time ago a studied PCM (back in 1977/78). BTW. Just recently I read how dts was coded and boy that was complicated (but beautiful) using Adaptive Differentiell PCM with prediction and splitting up the signal into sub frequency bands. But I am impressed. Back in 77 we (at school) just touched the technique of ADPCM. Best Regards from Sweden /Kent |
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CD Level Variations
Randy Yates wrote:
Norbert Hahn writes: Your zero value is at 32768, thus going from 32768 to 32769 doubles the output voltage, but the reference level remains 65536. Or am I grossly mistaken? Almost all digital systems use "signed two's complement" representation of the data. For a 16-bit system, this results in integers from -32768 (full-scale negative) to +32767 (full-scale positive), with 0 corresponding to 0. The number you use here is what we see on a PC, it is not what is on a CD. The previous poster, however, ... Laurence Payne wrote in message ... Well if you mean you go from the value FFFF (hexadecimal) (=65536) to FFFE (65535) then the decibel change will be (20*lg(65535/65536)) where lg is the logarithm using 10 as base. A rather small value indeed. Note that one step close to the maximum level (FFFF) is very small in decibel compared to a step at the minumum level. A step between 0001 to 0002 is aprox 6 dB. You have doubbled to voltage value. The quantization is linear but the ear's sensitivity is logarithmic. using the representation of samples on the CD. The numbers there are unsigned long integers using the Motorola order of bits. Thus I made my example using the representation on the CD. Norbert |
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CD Level Variations
Randy Yates wrote:
Norbert Hahn writes: Your zero value is at 32768, thus going from 32768 to 32769 doubles the output voltage, but the reference level remains 65536. Or am I grossly mistaken? Almost all digital systems use "signed two's complement" representation of the data. For a 16-bit system, this results in integers from -32768 (full-scale negative) to +32767 (full-scale positive), with 0 corresponding to 0. The number you use here is what we see on a PC, it is not what is on a CD. The previous poster, however, ... Laurence Payne wrote in message ... Well if you mean you go from the value FFFF (hexadecimal) (=65536) to FFFE (65535) then the decibel change will be (20*lg(65535/65536)) where lg is the logarithm using 10 as base. A rather small value indeed. Note that one step close to the maximum level (FFFF) is very small in decibel compared to a step at the minumum level. A step between 0001 to 0002 is aprox 6 dB. You have doubbled to voltage value. The quantization is linear but the ear's sensitivity is logarithmic. using the representation of samples on the CD. The numbers there are unsigned long integers using the Motorola order of bits. Thus I made my example using the representation on the CD. Norbert |
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CD Level Variations
Randy Yates wrote:
Norbert Hahn writes: Your zero value is at 32768, thus going from 32768 to 32769 doubles the output voltage, but the reference level remains 65536. Or am I grossly mistaken? Almost all digital systems use "signed two's complement" representation of the data. For a 16-bit system, this results in integers from -32768 (full-scale negative) to +32767 (full-scale positive), with 0 corresponding to 0. The number you use here is what we see on a PC, it is not what is on a CD. The previous poster, however, ... Laurence Payne wrote in message ... Well if you mean you go from the value FFFF (hexadecimal) (=65536) to FFFE (65535) then the decibel change will be (20*lg(65535/65536)) where lg is the logarithm using 10 as base. A rather small value indeed. Note that one step close to the maximum level (FFFF) is very small in decibel compared to a step at the minumum level. A step between 0001 to 0002 is aprox 6 dB. You have doubbled to voltage value. The quantization is linear but the ear's sensitivity is logarithmic. using the representation of samples on the CD. The numbers there are unsigned long integers using the Motorola order of bits. Thus I made my example using the representation on the CD. Norbert |
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CD Level Variations
Randy Yates wrote:
Norbert Hahn writes: Your zero value is at 32768, thus going from 32768 to 32769 doubles the output voltage, but the reference level remains 65536. Or am I grossly mistaken? Almost all digital systems use "signed two's complement" representation of the data. For a 16-bit system, this results in integers from -32768 (full-scale negative) to +32767 (full-scale positive), with 0 corresponding to 0. The number you use here is what we see on a PC, it is not what is on a CD. The previous poster, however, ... Laurence Payne wrote in message ... Well if you mean you go from the value FFFF (hexadecimal) (=65536) to FFFE (65535) then the decibel change will be (20*lg(65535/65536)) where lg is the logarithm using 10 as base. A rather small value indeed. Note that one step close to the maximum level (FFFF) is very small in decibel compared to a step at the minumum level. A step between 0001 to 0002 is aprox 6 dB. You have doubbled to voltage value. The quantization is linear but the ear's sensitivity is logarithmic. using the representation of samples on the CD. The numbers there are unsigned long integers using the Motorola order of bits. Thus I made my example using the representation on the CD. Norbert |
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CD Level Variations
(Kega) wrote:
Norbert Hahn wrote in message . .. (Kega) wrote: Or am I grossly mistaken? Yes I think you are. A double of output voltage is always approx 6 dB. When counting with dB you always count with the ratio between output and input (or between to levels). Well, it all depends on the selection of a reference value. IHMO the reference value for a 16 bit audio signal should be either 32768 or 65536. I don't think it will help for the discussion to base your example 1 ... Well if you mean you go from the value FFFF (hexadecimal) (=65536) to FFFE (65535) then the decibel change will be (20*lg(65535/65536)) where lg is the logarithm using 10 as base. A rather small value indeed. on 65536 and example 2 ... .. A step between 0001 to 0002 is aprox 6 dB. You have doubbled to voltage value. The quantization is linear but the ear's sensitivity is logarithmic. on quite a differenct level. When those numbers are translated into voltages by the D/A process - using 5,0000 V as reference level and a symmetric output voltage, hence 10 V voltage swing 65535 - 5,000000 V 65534 - 4,9999237 while 00000 - -5,000000 V 00001 - -4,9999237 clearly showing that a step from 00000 to 00001 is not 6 dB. And what is going on he 32768 - 0,000000 V 32769 - 0,000305 V Again, those numbers are taken from the representation we find on the CD. I hope I clarified my view somewhat. Norbert |
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CD Level Variations
(Kega) wrote:
Norbert Hahn wrote in message . .. (Kega) wrote: Or am I grossly mistaken? Yes I think you are. A double of output voltage is always approx 6 dB. When counting with dB you always count with the ratio between output and input (or between to levels). Well, it all depends on the selection of a reference value. IHMO the reference value for a 16 bit audio signal should be either 32768 or 65536. I don't think it will help for the discussion to base your example 1 ... Well if you mean you go from the value FFFF (hexadecimal) (=65536) to FFFE (65535) then the decibel change will be (20*lg(65535/65536)) where lg is the logarithm using 10 as base. A rather small value indeed. on 65536 and example 2 ... .. A step between 0001 to 0002 is aprox 6 dB. You have doubbled to voltage value. The quantization is linear but the ear's sensitivity is logarithmic. on quite a differenct level. When those numbers are translated into voltages by the D/A process - using 5,0000 V as reference level and a symmetric output voltage, hence 10 V voltage swing 65535 - 5,000000 V 65534 - 4,9999237 while 00000 - -5,000000 V 00001 - -4,9999237 clearly showing that a step from 00000 to 00001 is not 6 dB. And what is going on he 32768 - 0,000000 V 32769 - 0,000305 V Again, those numbers are taken from the representation we find on the CD. I hope I clarified my view somewhat. Norbert |
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CD Level Variations
(Kega) wrote:
Norbert Hahn wrote in message . .. (Kega) wrote: Or am I grossly mistaken? Yes I think you are. A double of output voltage is always approx 6 dB. When counting with dB you always count with the ratio between output and input (or between to levels). Well, it all depends on the selection of a reference value. IHMO the reference value for a 16 bit audio signal should be either 32768 or 65536. I don't think it will help for the discussion to base your example 1 ... Well if you mean you go from the value FFFF (hexadecimal) (=65536) to FFFE (65535) then the decibel change will be (20*lg(65535/65536)) where lg is the logarithm using 10 as base. A rather small value indeed. on 65536 and example 2 ... .. A step between 0001 to 0002 is aprox 6 dB. You have doubbled to voltage value. The quantization is linear but the ear's sensitivity is logarithmic. on quite a differenct level. When those numbers are translated into voltages by the D/A process - using 5,0000 V as reference level and a symmetric output voltage, hence 10 V voltage swing 65535 - 5,000000 V 65534 - 4,9999237 while 00000 - -5,000000 V 00001 - -4,9999237 clearly showing that a step from 00000 to 00001 is not 6 dB. And what is going on he 32768 - 0,000000 V 32769 - 0,000305 V Again, those numbers are taken from the representation we find on the CD. I hope I clarified my view somewhat. Norbert |
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