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#1
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Internet Stream Audio [Was Bit-resolution decrease for internet]
1 Hz sampling rate would equate to .5 Hz. 1-bit/sec, however would
not. Bit/time is the bit-rate. Sample rate is different from bit-rate. It is also important to know the difference between *bit-resolution* and *bit-rate*. 1 byte = 8 bits If in a wave file, the bit-resolution is made to equal 1 /(sampling rate X number of channels), then the bit-rate will definitely be 1-bit/second. If the sample rate is 44,100 Hz in a stereo (2-channel) wave file of this type, the bit-resolution would be 1/(44100 x 2)-bit or 1/88200-bit. Bit-rate = sample-rate X bit-resolution X numbers of channels Multiply the 44100 X 2 X 1/88200 and you get 1! 44100 Hz X 1/88200-bit X 2 channels = 1 bit per second 1 minute of this file would comsume only 60 bits of disk space. It would definitely work for the internet. Unlike conventional MP3s and WMAs, the high-frequency content of the PCM music will be restored due to the high sample rate. 60 bits = 60/8 bytes Rich Andrews wrote in message ... (Radium) wrote in : I would like to use an audio codec based on WAVE PCM. It should be a little different though. The bit-resolution should be set to equal 1/(sampling rate X # of channels). The bit-rate should be set to equal 1 bit per second. I would like to use this codec to transport audio files though the internet via email. I am looking for frequency response. In digital audio the sampling rate must be at least twice the highest frequency in the signal. It would like a highest frequency of at least 200 KHz. This would require a sample rate of at least 400 KHz. In this codec the bit-resolution is decreased to maintain a low bit rate of 1 bit/sec. The bit-resolution is divided by the sampling rate and the # of channels to acheive this. 1 bit per second? Wouldn't that equate to .5 hz or did I miss something? r |
#2
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Radium wrote:
1 Hz sampling rate would equate to .5 Hz. 1-bit/sec, however would not. Bit/time is the bit-rate. Sample rate is different from bit-rate. It is also important to know the difference between *bit-resolution* and *bit-rate*. 1 byte = 8 bits If in a wave file, the bit-resolution is made to equal 1 /(sampling rate X number of channels), then the bit-rate will definitely be 1-bit/second. If the sample rate is 44,100 Hz in a stereo (2-channel) wave file of this type, the bit-resolution would be 1/(44100 x 2)-bit or 1/88200-bit. Bit-rate = sample-rate X bit-resolution X numbers of channels Multiply the 44100 X 2 X 1/88200 and you get 1! 44100 Hz X 1/88200-bit X 2 channels = 1 bit per second 1 minute of this file would comsume only 60 bits of disk space. It would definitely work for the internet. Unlike conventional MP3s and WMAs, the high-frequency content of the PCM music will be restored due to the high sample rate. 60 bits = 60/8 bytes The saddest thing about this drivel is that your ability to express it is so good. What a waste! Take "Bit-rate = sample-rate X bit-resolution X numbers of channels", which you so disastrously misinterpret. Consider a two-channel CD. The sample rate is 44,100 samples per second. The number of channels is 2. The bit resolution is 16 (The system encodes sound levels as 16-bit signed integers.) When I multiply those numbers, I get 1,441,200 bits/second. You get 1 because you don't know what bit resolution is. Jerry -- Engineering is the art of making what you want from things you can get. ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
#3
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Radium wrote:
1 Hz sampling rate would equate to .5 Hz. 1-bit/sec, however would not. Bit/time is the bit-rate. Sample rate is different from bit-rate. It is also important to know the difference between *bit-resolution* and *bit-rate*. 1 byte = 8 bits If in a wave file, the bit-resolution is made to equal 1 /(sampling rate X number of channels), then the bit-rate will definitely be 1-bit/second. If the sample rate is 44,100 Hz in a stereo (2-channel) wave file of this type, the bit-resolution would be 1/(44100 x 2)-bit or 1/88200-bit. Bit-rate = sample-rate X bit-resolution X numbers of channels Multiply the 44100 X 2 X 1/88200 and you get 1! 44100 Hz X 1/88200-bit X 2 channels = 1 bit per second 1 minute of this file would comsume only 60 bits of disk space. It would definitely work for the internet. Unlike conventional MP3s and WMAs, the high-frequency content of the PCM music will be restored due to the high sample rate. 60 bits = 60/8 bytes The saddest thing about this drivel is that your ability to express it is so good. What a waste! Take "Bit-rate = sample-rate X bit-resolution X numbers of channels", which you so disastrously misinterpret. Consider a two-channel CD. The sample rate is 44,100 samples per second. The number of channels is 2. The bit resolution is 16 (The system encodes sound levels as 16-bit signed integers.) When I multiply those numbers, I get 1,441,200 bits/second. You get 1 because you don't know what bit resolution is. Jerry -- Engineering is the art of making what you want from things you can get. ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
#4
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Radium wrote:
1 Hz sampling rate would equate to .5 Hz. 1-bit/sec, however would not. Bit/time is the bit-rate. Sample rate is different from bit-rate. It is also important to know the difference between *bit-resolution* and *bit-rate*. 1 byte = 8 bits If in a wave file, the bit-resolution is made to equal 1 /(sampling rate X number of channels), then the bit-rate will definitely be 1-bit/second. If the sample rate is 44,100 Hz in a stereo (2-channel) wave file of this type, the bit-resolution would be 1/(44100 x 2)-bit or 1/88200-bit. Bit-rate = sample-rate X bit-resolution X numbers of channels Multiply the 44100 X 2 X 1/88200 and you get 1! 44100 Hz X 1/88200-bit X 2 channels = 1 bit per second 1 minute of this file would comsume only 60 bits of disk space. It would definitely work for the internet. Unlike conventional MP3s and WMAs, the high-frequency content of the PCM music will be restored due to the high sample rate. 60 bits = 60/8 bytes The saddest thing about this drivel is that your ability to express it is so good. What a waste! Take "Bit-rate = sample-rate X bit-resolution X numbers of channels", which you so disastrously misinterpret. Consider a two-channel CD. The sample rate is 44,100 samples per second. The number of channels is 2. The bit resolution is 16 (The system encodes sound levels as 16-bit signed integers.) When I multiply those numbers, I get 1,441,200 bits/second. You get 1 because you don't know what bit resolution is. Jerry -- Engineering is the art of making what you want from things you can get. ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
#5
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Jerry Avins wrote:
Radium wrote: 1 Hz sampling rate would equate to .5 Hz. 1-bit/sec, however would not. Bit/time is the bit-rate. Sample rate is different from bit-rate. It is also important to know the difference between *bit-resolution* and *bit-rate*. 1 byte = 8 bits If in a wave file, the bit-resolution is made to equal 1 /(sampling rate X number of channels), then the bit-rate will definitely be 1-bit/second. If the sample rate is 44,100 Hz in a stereo (2-channel) wave file of this type, the bit-resolution would be 1/(44100 x 2)-bit or 1/88200-bit. Bit-rate = sample-rate X bit-resolution X numbers of channels Multiply the 44100 X 2 X 1/88200 and you get 1! 44100 Hz X 1/88200-bit X 2 channels = 1 bit per second 1 minute of this file would comsume only 60 bits of disk space. It would definitely work for the internet. Unlike conventional MP3s and WMAs, the high-frequency content of the PCM music will be restored due to the high sample rate. 60 bits = 60/8 bytes The saddest thing about this drivel is that your ability to express it is so good. What a waste! Take "Bit-rate = sample-rate X bit-resolution X numbers of channels", which you so disastrously misinterpret. Consider a two-channel CD. The sample rate is 44,100 samples per second. The number of channels is 2. The bit resolution is 16 (The system encodes sound levels as 16-bit signed integers.) When I multiply those numbers, I get 1,441,200 bits/second. You get 1 because you don't know what bit resolution is. He gets one becuse he thinks that you can represent the sound with 1/88200 bits per sample, instead of 16 bits pers sample. Whereas 16bits pers sample gives you 65536 levels, 1/88200 bits would give you approximately 1 level (to 5 significant figures) or if you prefer, the sort of sound reproduction of Beethoven's 5th that you'd get from plugging your speakers into a 9V PP3 battery (or just leaving them unplugged from the amp) Radium, how to you intend to represent a whole second of audio, with one bit? Thats two states - great for classifyng your audio into "this" or "that", and nothing else. Ben -- A7N8X FAQ: www.ben.pope.name/a7n8x_faq.html Questions by email will likely be ignored, please use the newsgroups. I'm not just a number. To many, I'm known as a String... |
#6
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Jerry Avins wrote:
Radium wrote: 1 Hz sampling rate would equate to .5 Hz. 1-bit/sec, however would not. Bit/time is the bit-rate. Sample rate is different from bit-rate. It is also important to know the difference between *bit-resolution* and *bit-rate*. 1 byte = 8 bits If in a wave file, the bit-resolution is made to equal 1 /(sampling rate X number of channels), then the bit-rate will definitely be 1-bit/second. If the sample rate is 44,100 Hz in a stereo (2-channel) wave file of this type, the bit-resolution would be 1/(44100 x 2)-bit or 1/88200-bit. Bit-rate = sample-rate X bit-resolution X numbers of channels Multiply the 44100 X 2 X 1/88200 and you get 1! 44100 Hz X 1/88200-bit X 2 channels = 1 bit per second 1 minute of this file would comsume only 60 bits of disk space. It would definitely work for the internet. Unlike conventional MP3s and WMAs, the high-frequency content of the PCM music will be restored due to the high sample rate. 60 bits = 60/8 bytes The saddest thing about this drivel is that your ability to express it is so good. What a waste! Take "Bit-rate = sample-rate X bit-resolution X numbers of channels", which you so disastrously misinterpret. Consider a two-channel CD. The sample rate is 44,100 samples per second. The number of channels is 2. The bit resolution is 16 (The system encodes sound levels as 16-bit signed integers.) When I multiply those numbers, I get 1,441,200 bits/second. You get 1 because you don't know what bit resolution is. He gets one becuse he thinks that you can represent the sound with 1/88200 bits per sample, instead of 16 bits pers sample. Whereas 16bits pers sample gives you 65536 levels, 1/88200 bits would give you approximately 1 level (to 5 significant figures) or if you prefer, the sort of sound reproduction of Beethoven's 5th that you'd get from plugging your speakers into a 9V PP3 battery (or just leaving them unplugged from the amp) Radium, how to you intend to represent a whole second of audio, with one bit? Thats two states - great for classifyng your audio into "this" or "that", and nothing else. Ben -- A7N8X FAQ: www.ben.pope.name/a7n8x_faq.html Questions by email will likely be ignored, please use the newsgroups. I'm not just a number. To many, I'm known as a String... |
#7
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Jerry Avins wrote:
Radium wrote: 1 Hz sampling rate would equate to .5 Hz. 1-bit/sec, however would not. Bit/time is the bit-rate. Sample rate is different from bit-rate. It is also important to know the difference between *bit-resolution* and *bit-rate*. 1 byte = 8 bits If in a wave file, the bit-resolution is made to equal 1 /(sampling rate X number of channels), then the bit-rate will definitely be 1-bit/second. If the sample rate is 44,100 Hz in a stereo (2-channel) wave file of this type, the bit-resolution would be 1/(44100 x 2)-bit or 1/88200-bit. Bit-rate = sample-rate X bit-resolution X numbers of channels Multiply the 44100 X 2 X 1/88200 and you get 1! 44100 Hz X 1/88200-bit X 2 channels = 1 bit per second 1 minute of this file would comsume only 60 bits of disk space. It would definitely work for the internet. Unlike conventional MP3s and WMAs, the high-frequency content of the PCM music will be restored due to the high sample rate. 60 bits = 60/8 bytes The saddest thing about this drivel is that your ability to express it is so good. What a waste! Take "Bit-rate = sample-rate X bit-resolution X numbers of channels", which you so disastrously misinterpret. Consider a two-channel CD. The sample rate is 44,100 samples per second. The number of channels is 2. The bit resolution is 16 (The system encodes sound levels as 16-bit signed integers.) When I multiply those numbers, I get 1,441,200 bits/second. You get 1 because you don't know what bit resolution is. He gets one becuse he thinks that you can represent the sound with 1/88200 bits per sample, instead of 16 bits pers sample. Whereas 16bits pers sample gives you 65536 levels, 1/88200 bits would give you approximately 1 level (to 5 significant figures) or if you prefer, the sort of sound reproduction of Beethoven's 5th that you'd get from plugging your speakers into a 9V PP3 battery (or just leaving them unplugged from the amp) Radium, how to you intend to represent a whole second of audio, with one bit? Thats two states - great for classifyng your audio into "this" or "that", and nothing else. Ben -- A7N8X FAQ: www.ben.pope.name/a7n8x_faq.html Questions by email will likely be ignored, please use the newsgroups. I'm not just a number. To many, I'm known as a String... |
#8
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Jerry Avins wrote:
(snip) The saddest thing about this drivel is that your ability to express it is so good. What a waste! Take "Bit-rate = sample-rate X bit-resolution X numbers of channels", which you so disastrously misinterpret. Consider a two-channel CD. The sample rate is 44,100 samples per second. The number of channels is 2. The bit resolution is 16 (The system encodes sound levels as 16-bit signed integers.) When I multiply those numbers, I get 1,441,200 bits/second. You get 1 because you don't know what bit resolution is. Somehow this reminds me of people who memorize physics formulas and then figure out which one applies to a given problem. I had thought once that one should give problems in unusual variables, so that people who memorize the formula, but don't understand the application of the formula will get the wrong answer. Maybe use f for velocity, m for distance, and ask them to find the time, a. Oh well. -- glen |
#9
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Jerry Avins wrote:
(snip) The saddest thing about this drivel is that your ability to express it is so good. What a waste! Take "Bit-rate = sample-rate X bit-resolution X numbers of channels", which you so disastrously misinterpret. Consider a two-channel CD. The sample rate is 44,100 samples per second. The number of channels is 2. The bit resolution is 16 (The system encodes sound levels as 16-bit signed integers.) When I multiply those numbers, I get 1,441,200 bits/second. You get 1 because you don't know what bit resolution is. Somehow this reminds me of people who memorize physics formulas and then figure out which one applies to a given problem. I had thought once that one should give problems in unusual variables, so that people who memorize the formula, but don't understand the application of the formula will get the wrong answer. Maybe use f for velocity, m for distance, and ask them to find the time, a. Oh well. -- glen |
#10
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Jerry Avins wrote:
(snip) The saddest thing about this drivel is that your ability to express it is so good. What a waste! Take "Bit-rate = sample-rate X bit-resolution X numbers of channels", which you so disastrously misinterpret. Consider a two-channel CD. The sample rate is 44,100 samples per second. The number of channels is 2. The bit resolution is 16 (The system encodes sound levels as 16-bit signed integers.) When I multiply those numbers, I get 1,441,200 bits/second. You get 1 because you don't know what bit resolution is. Somehow this reminds me of people who memorize physics formulas and then figure out which one applies to a given problem. I had thought once that one should give problems in unusual variables, so that people who memorize the formula, but don't understand the application of the formula will get the wrong answer. Maybe use f for velocity, m for distance, and ask them to find the time, a. Oh well. -- glen |
#11
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Internet Stream Audio [Was Bit-resolution decrease for internet]
44100 X 16 X 2 = 1,441,200
44100 X 1/88200 X 2 = 1 Jerry Avins wrote in message ... Radium wrote: 1 Hz sampling rate would equate to .5 Hz. 1-bit/sec, however would not. Bit/time is the bit-rate. Sample rate is different from bit-rate. It is also important to know the difference between *bit-resolution* and *bit-rate*. 1 byte = 8 bits If in a wave file, the bit-resolution is made to equal 1 /(sampling rate X number of channels), then the bit-rate will definitely be 1-bit/second. If the sample rate is 44,100 Hz in a stereo (2-channel) wave file of this type, the bit-resolution would be 1/(44100 x 2)-bit or 1/88200-bit. Bit-rate = sample-rate X bit-resolution X numbers of channels Multiply the 44100 X 2 X 1/88200 and you get 1! 44100 Hz X 1/88200-bit X 2 channels = 1 bit per second 1 minute of this file would comsume only 60 bits of disk space. It would definitely work for the internet. Unlike conventional MP3s and WMAs, the high-frequency content of the PCM music will be restored due to the high sample rate. 60 bits = 60/8 bytes The saddest thing about this drivel is that your ability to express it is so good. What a waste! Take "Bit-rate = sample-rate X bit-resolution X numbers of channels", which you so disastrously misinterpret. Consider a two-channel CD. The sample rate is 44,100 samples per second. The number of channels is 2. The bit resolution is 16 (The system encodes sound levels as 16-bit signed integers.) When I multiply those numbers, I get 1,441,200 bits/second. You get 1 because you don't know what bit resolution is. Jerry |
#12
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Internet Stream Audio [Was Bit-resolution decrease for internet]
44100 X 16 X 2 = 1,441,200
44100 X 1/88200 X 2 = 1 Jerry Avins wrote in message ... Radium wrote: 1 Hz sampling rate would equate to .5 Hz. 1-bit/sec, however would not. Bit/time is the bit-rate. Sample rate is different from bit-rate. It is also important to know the difference between *bit-resolution* and *bit-rate*. 1 byte = 8 bits If in a wave file, the bit-resolution is made to equal 1 /(sampling rate X number of channels), then the bit-rate will definitely be 1-bit/second. If the sample rate is 44,100 Hz in a stereo (2-channel) wave file of this type, the bit-resolution would be 1/(44100 x 2)-bit or 1/88200-bit. Bit-rate = sample-rate X bit-resolution X numbers of channels Multiply the 44100 X 2 X 1/88200 and you get 1! 44100 Hz X 1/88200-bit X 2 channels = 1 bit per second 1 minute of this file would comsume only 60 bits of disk space. It would definitely work for the internet. Unlike conventional MP3s and WMAs, the high-frequency content of the PCM music will be restored due to the high sample rate. 60 bits = 60/8 bytes The saddest thing about this drivel is that your ability to express it is so good. What a waste! Take "Bit-rate = sample-rate X bit-resolution X numbers of channels", which you so disastrously misinterpret. Consider a two-channel CD. The sample rate is 44,100 samples per second. The number of channels is 2. The bit resolution is 16 (The system encodes sound levels as 16-bit signed integers.) When I multiply those numbers, I get 1,441,200 bits/second. You get 1 because you don't know what bit resolution is. Jerry |
#13
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Internet Stream Audio [Was Bit-resolution decrease for internet]
44100 X 16 X 2 = 1,441,200
44100 X 1/88200 X 2 = 1 Jerry Avins wrote in message ... Radium wrote: 1 Hz sampling rate would equate to .5 Hz. 1-bit/sec, however would not. Bit/time is the bit-rate. Sample rate is different from bit-rate. It is also important to know the difference between *bit-resolution* and *bit-rate*. 1 byte = 8 bits If in a wave file, the bit-resolution is made to equal 1 /(sampling rate X number of channels), then the bit-rate will definitely be 1-bit/second. If the sample rate is 44,100 Hz in a stereo (2-channel) wave file of this type, the bit-resolution would be 1/(44100 x 2)-bit or 1/88200-bit. Bit-rate = sample-rate X bit-resolution X numbers of channels Multiply the 44100 X 2 X 1/88200 and you get 1! 44100 Hz X 1/88200-bit X 2 channels = 1 bit per second 1 minute of this file would comsume only 60 bits of disk space. It would definitely work for the internet. Unlike conventional MP3s and WMAs, the high-frequency content of the PCM music will be restored due to the high sample rate. 60 bits = 60/8 bytes The saddest thing about this drivel is that your ability to express it is so good. What a waste! Take "Bit-rate = sample-rate X bit-resolution X numbers of channels", which you so disastrously misinterpret. Consider a two-channel CD. The sample rate is 44,100 samples per second. The number of channels is 2. The bit resolution is 16 (The system encodes sound levels as 16-bit signed integers.) When I multiply those numbers, I get 1,441,200 bits/second. You get 1 because you don't know what bit resolution is. Jerry |
#14
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Radium wrote:
44100 X 16 X 2 = 1,441,200 44100 X 1/88200 X 2 = 1 a = b + c ... (1) 5a = 5b + 5c ... (2) 4b + 4c = 4a ... (3) Add (2) and (3): 5a + 4b + 4c = 4a + 5b + 5c ... (4) Subtract 9a: -4a + 4b + 4c = -5a + 5b + 5c ... (5) Simplify: 4(b + c - a) = 5(b + c - a) ... (6) Divide by (b + c - a): 4 = 5 ... (7) Paul |
#15
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Radium wrote:
44100 X 16 X 2 = 1,441,200 44100 X 1/88200 X 2 = 1 a = b + c ... (1) 5a = 5b + 5c ... (2) 4b + 4c = 4a ... (3) Add (2) and (3): 5a + 4b + 4c = 4a + 5b + 5c ... (4) Subtract 9a: -4a + 4b + 4c = -5a + 5b + 5c ... (5) Simplify: 4(b + c - a) = 5(b + c - a) ... (6) Divide by (b + c - a): 4 = 5 ... (7) Paul |
#16
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Radium wrote:
44100 X 16 X 2 = 1,441,200 44100 X 1/88200 X 2 = 1 a = b + c ... (1) 5a = 5b + 5c ... (2) 4b + 4c = 4a ... (3) Add (2) and (3): 5a + 4b + 4c = 4a + 5b + 5c ... (4) Subtract 9a: -4a + 4b + 4c = -5a + 5b + 5c ... (5) Simplify: 4(b + c - a) = 5(b + c - a) ... (6) Divide by (b + c - a): 4 = 5 ... (7) Paul |
#17
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Radium wrote:
44100 X 16 X 2 = 1,441,200 44100 X 1/88200 X 2 = 1 Why won't you respond to my postings asking you how you intend to represent and use fractional bits? Ben -- A7N8X FAQ: www.ben.pope.name/a7n8x_faq.html Questions by email will likely be ignored, please use the newsgroups. I'm not just a number. To many, I'm known as a String... |
#18
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Radium wrote:
44100 X 16 X 2 = 1,441,200 44100 X 1/88200 X 2 = 1 Why won't you respond to my postings asking you how you intend to represent and use fractional bits? Ben -- A7N8X FAQ: www.ben.pope.name/a7n8x_faq.html Questions by email will likely be ignored, please use the newsgroups. I'm not just a number. To many, I'm known as a String... |
#19
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Radium wrote:
44100 X 16 X 2 = 1,441,200 44100 X 1/88200 X 2 = 1 Why won't you respond to my postings asking you how you intend to represent and use fractional bits? Ben -- A7N8X FAQ: www.ben.pope.name/a7n8x_faq.html Questions by email will likely be ignored, please use the newsgroups. I'm not just a number. To many, I'm known as a String... |
#20
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Ben Pope wrote:
Radium wrote: 44100 X 16 X 2 = 1,441,200 44100 X 1/88200 X 2 = 1 Why won't you respond to my postings asking you how you intend to represent and use fractional bits? What is wrong with fractional bits? A decimal digit is worth 3.3 bits. There are many computations that use fractional bits. Sampling isn't usually one, though. -- glen |
#21
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Ben Pope wrote:
Radium wrote: 44100 X 16 X 2 = 1,441,200 44100 X 1/88200 X 2 = 1 Why won't you respond to my postings asking you how you intend to represent and use fractional bits? What is wrong with fractional bits? A decimal digit is worth 3.3 bits. There are many computations that use fractional bits. Sampling isn't usually one, though. -- glen |
#22
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Ben Pope wrote:
Radium wrote: 44100 X 16 X 2 = 1,441,200 44100 X 1/88200 X 2 = 1 Why won't you respond to my postings asking you how you intend to represent and use fractional bits? What is wrong with fractional bits? A decimal digit is worth 3.3 bits. There are many computations that use fractional bits. Sampling isn't usually one, though. -- glen |
#23
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Internet Stream Audio [Was Bit-resolution decrease for internet]
glen herrmannsfeldt wrote:
Ben Pope wrote: Radium wrote: 44100 X 16 X 2 = 1,441,200 44100 X 1/88200 X 2 = 1 Why won't you respond to my postings asking you how you intend to represent and use fractional bits? What is wrong with fractional bits? A decimal digit is worth 3.3 bits. Implementation is a fairly large problem. I'm fairly sure a transistor is either logically "off" or logically "on". If you had 1/88200 of a transistor, I doubt it would work, so physically you cannot have fractional bits. If you had 1/88200 of a logical bit, how do you determine the state of that bit? it has 1.0000something states. What is a fraction of a state? Does it even mean anything if it could be represented? The point is you cannot have a fraction of a bit. A bit is the atomic unit of digital computation, there is nothing smaller as it cannot represent state, without state you have nothing. You can have fractional values represented in binary but thats not the same thing. It still uses an integer number of bits. There are many computations that use fractional bits. Not really. You can't measure something with 3.3 bits, can you? You can't create a machine that can represent 2^3.3 (~9.85) states. Sampling isn't usually one, though. Nothing can use fractional bits, they just don't exist in a physical world (somebody will pull out an anology in quantum physics now, but I don't see its relevance in todays computing environment, besides, you can still can't have a fraction of a qubit) Ben -- A7N8X FAQ: www.ben.pope.name/a7n8x_faq.html Questions by email will likely be ignored, please use the newsgroups. I'm not just a number. To many, I'm known as a String... |
#24
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Internet Stream Audio [Was Bit-resolution decrease for internet]
glen herrmannsfeldt wrote:
Ben Pope wrote: Radium wrote: 44100 X 16 X 2 = 1,441,200 44100 X 1/88200 X 2 = 1 Why won't you respond to my postings asking you how you intend to represent and use fractional bits? What is wrong with fractional bits? A decimal digit is worth 3.3 bits. Implementation is a fairly large problem. I'm fairly sure a transistor is either logically "off" or logically "on". If you had 1/88200 of a transistor, I doubt it would work, so physically you cannot have fractional bits. If you had 1/88200 of a logical bit, how do you determine the state of that bit? it has 1.0000something states. What is a fraction of a state? Does it even mean anything if it could be represented? The point is you cannot have a fraction of a bit. A bit is the atomic unit of digital computation, there is nothing smaller as it cannot represent state, without state you have nothing. You can have fractional values represented in binary but thats not the same thing. It still uses an integer number of bits. There are many computations that use fractional bits. Not really. You can't measure something with 3.3 bits, can you? You can't create a machine that can represent 2^3.3 (~9.85) states. Sampling isn't usually one, though. Nothing can use fractional bits, they just don't exist in a physical world (somebody will pull out an anology in quantum physics now, but I don't see its relevance in todays computing environment, besides, you can still can't have a fraction of a qubit) Ben -- A7N8X FAQ: www.ben.pope.name/a7n8x_faq.html Questions by email will likely be ignored, please use the newsgroups. I'm not just a number. To many, I'm known as a String... |
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Internet Stream Audio [Was Bit-resolution decrease for internet]
glen herrmannsfeldt wrote:
Ben Pope wrote: Radium wrote: 44100 X 16 X 2 = 1,441,200 44100 X 1/88200 X 2 = 1 Why won't you respond to my postings asking you how you intend to represent and use fractional bits? What is wrong with fractional bits? A decimal digit is worth 3.3 bits. Implementation is a fairly large problem. I'm fairly sure a transistor is either logically "off" or logically "on". If you had 1/88200 of a transistor, I doubt it would work, so physically you cannot have fractional bits. If you had 1/88200 of a logical bit, how do you determine the state of that bit? it has 1.0000something states. What is a fraction of a state? Does it even mean anything if it could be represented? The point is you cannot have a fraction of a bit. A bit is the atomic unit of digital computation, there is nothing smaller as it cannot represent state, without state you have nothing. You can have fractional values represented in binary but thats not the same thing. It still uses an integer number of bits. There are many computations that use fractional bits. Not really. You can't measure something with 3.3 bits, can you? You can't create a machine that can represent 2^3.3 (~9.85) states. Sampling isn't usually one, though. Nothing can use fractional bits, they just don't exist in a physical world (somebody will pull out an anology in quantum physics now, but I don't see its relevance in todays computing environment, besides, you can still can't have a fraction of a qubit) Ben -- A7N8X FAQ: www.ben.pope.name/a7n8x_faq.html Questions by email will likely be ignored, please use the newsgroups. I'm not just a number. To many, I'm known as a String... |
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Radium wrote:
44100 X 16 X 2 = 1,441,200 44100 X 1/88200 X 2 = 1 I assume that you mean 1/88200 to represent "bit resolution". You misunderstand. The formula calls for resolution measured in bits. For CDs, that is 16 bits. The step size represented by one count out of 16 bits is 1/65,536 of the maximum peak-to-peak level that the converters can handle. Step size is not directly related to bit rate. Everyone is entitled to an opinion. Some opinions are worth more than others. Yours seem to be worth very little. Jerry -- Engineering is the art of making what you want from things you can get. ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Radium wrote:
44100 X 16 X 2 = 1,441,200 44100 X 1/88200 X 2 = 1 I assume that you mean 1/88200 to represent "bit resolution". You misunderstand. The formula calls for resolution measured in bits. For CDs, that is 16 bits. The step size represented by one count out of 16 bits is 1/65,536 of the maximum peak-to-peak level that the converters can handle. Step size is not directly related to bit rate. Everyone is entitled to an opinion. Some opinions are worth more than others. Yours seem to be worth very little. Jerry -- Engineering is the art of making what you want from things you can get. ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
#28
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Radium wrote:
44100 X 16 X 2 = 1,441,200 44100 X 1/88200 X 2 = 1 I assume that you mean 1/88200 to represent "bit resolution". You misunderstand. The formula calls for resolution measured in bits. For CDs, that is 16 bits. The step size represented by one count out of 16 bits is 1/65,536 of the maximum peak-to-peak level that the converters can handle. Step size is not directly related to bit rate. Everyone is entitled to an opinion. Some opinions are worth more than others. Yours seem to be worth very little. Jerry -- Engineering is the art of making what you want from things you can get. ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
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Ben Pope wrote:
... \ Nothing can use fractional bits, they just don't exist in a physical world (somebody will pull out an anology in quantum physics now, but I don't see its relevance in todays computing environment, besides, you can still can't have a fraction of a qubit) Ben Not only bits, but digits. You can buy a 3 1/2 digit DVM. Digits and bits are measures of precision. When the actual precision is not an integer times an integer power of the number base, fractions come into the representation. Physical realization isn't needed to give the representation meaning. Or were you just pulling my leg? Jerry -- Engineering is the art of making what you want from things you can get. ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
#30
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Ben Pope wrote:
... \ Nothing can use fractional bits, they just don't exist in a physical world (somebody will pull out an anology in quantum physics now, but I don't see its relevance in todays computing environment, besides, you can still can't have a fraction of a qubit) Ben Not only bits, but digits. You can buy a 3 1/2 digit DVM. Digits and bits are measures of precision. When the actual precision is not an integer times an integer power of the number base, fractions come into the representation. Physical realization isn't needed to give the representation meaning. Or were you just pulling my leg? Jerry -- Engineering is the art of making what you want from things you can get. ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
#31
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Ben Pope wrote:
... \ Nothing can use fractional bits, they just don't exist in a physical world (somebody will pull out an anology in quantum physics now, but I don't see its relevance in todays computing environment, besides, you can still can't have a fraction of a qubit) Ben Not only bits, but digits. You can buy a 3 1/2 digit DVM. Digits and bits are measures of precision. When the actual precision is not an integer times an integer power of the number base, fractions come into the representation. Physical realization isn't needed to give the representation meaning. Or were you just pulling my leg? Jerry -- Engineering is the art of making what you want from things you can get. ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
#32
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Jerry Avins wrote:
Ben Pope wrote: ... \ Nothing can use fractional bits, they just don't exist in a physical world (somebody will pull out an anology in quantum physics now, but I don't see its relevance in todays computing environment, besides, you can still can't have a fraction of a qubit) Ben Not only bits, but digits. You can buy a 3 1/2 digit DVM. Digits and bits are measures of precision. When the actual precision is not an integer times an integer power of the number base, fractions come into the representation. Physical realization isn't needed to give the representation meaning. Or were you just pulling my leg? A bit is not a measure of precision. It is a state machine with 2 states. The "value" of the bit is completely irrelevant in this discussion. From Radiums calculations, he appears to want to store 88200 samples in one bit, you can't. That would require splitting the bit into 88200 "chunks", you can't. You can't implement a unit of storage with non-integer number of states in the digital domain. And you simply can't have a unit of storage with less than 2 states, otherwise it will contain no information (if you have one level you know what it will be and is therefore completely deterministic, if you have zero levels you don't have anything) and will therefore be completely useless to you. I realise that you can bunch a load of bits together (a byte, two bytes, a word, whatever), then order them - together the (ordered) set of states will provide an integer (since you are either in one state or another, no half-states) index which you could multiply by some pre-determined fractional value that each bit represents, to give an overall value that can represent a fraction. That I'm happy with. A 3½ digit DVM or display is not a good example here. Ben -- A7N8X FAQ: www.ben.pope.name/a7n8x_faq.html Questions by email will likely be ignored, please use the newsgroups. I'm not just a number. To many, I'm known as a String... |
#33
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Jerry Avins wrote:
Ben Pope wrote: ... \ Nothing can use fractional bits, they just don't exist in a physical world (somebody will pull out an anology in quantum physics now, but I don't see its relevance in todays computing environment, besides, you can still can't have a fraction of a qubit) Ben Not only bits, but digits. You can buy a 3 1/2 digit DVM. Digits and bits are measures of precision. When the actual precision is not an integer times an integer power of the number base, fractions come into the representation. Physical realization isn't needed to give the representation meaning. Or were you just pulling my leg? A bit is not a measure of precision. It is a state machine with 2 states. The "value" of the bit is completely irrelevant in this discussion. From Radiums calculations, he appears to want to store 88200 samples in one bit, you can't. That would require splitting the bit into 88200 "chunks", you can't. You can't implement a unit of storage with non-integer number of states in the digital domain. And you simply can't have a unit of storage with less than 2 states, otherwise it will contain no information (if you have one level you know what it will be and is therefore completely deterministic, if you have zero levels you don't have anything) and will therefore be completely useless to you. I realise that you can bunch a load of bits together (a byte, two bytes, a word, whatever), then order them - together the (ordered) set of states will provide an integer (since you are either in one state or another, no half-states) index which you could multiply by some pre-determined fractional value that each bit represents, to give an overall value that can represent a fraction. That I'm happy with. A 3½ digit DVM or display is not a good example here. Ben -- A7N8X FAQ: www.ben.pope.name/a7n8x_faq.html Questions by email will likely be ignored, please use the newsgroups. I'm not just a number. To many, I'm known as a String... |
#34
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Jerry Avins wrote:
Ben Pope wrote: ... \ Nothing can use fractional bits, they just don't exist in a physical world (somebody will pull out an anology in quantum physics now, but I don't see its relevance in todays computing environment, besides, you can still can't have a fraction of a qubit) Ben Not only bits, but digits. You can buy a 3 1/2 digit DVM. Digits and bits are measures of precision. When the actual precision is not an integer times an integer power of the number base, fractions come into the representation. Physical realization isn't needed to give the representation meaning. Or were you just pulling my leg? A bit is not a measure of precision. It is a state machine with 2 states. The "value" of the bit is completely irrelevant in this discussion. From Radiums calculations, he appears to want to store 88200 samples in one bit, you can't. That would require splitting the bit into 88200 "chunks", you can't. You can't implement a unit of storage with non-integer number of states in the digital domain. And you simply can't have a unit of storage with less than 2 states, otherwise it will contain no information (if you have one level you know what it will be and is therefore completely deterministic, if you have zero levels you don't have anything) and will therefore be completely useless to you. I realise that you can bunch a load of bits together (a byte, two bytes, a word, whatever), then order them - together the (ordered) set of states will provide an integer (since you are either in one state or another, no half-states) index which you could multiply by some pre-determined fractional value that each bit represents, to give an overall value that can represent a fraction. That I'm happy with. A 3½ digit DVM or display is not a good example here. Ben -- A7N8X FAQ: www.ben.pope.name/a7n8x_faq.html Questions by email will likely be ignored, please use the newsgroups. I'm not just a number. To many, I'm known as a String... |
#35
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Ben Pope wrote:
Jerry Avins wrote: Ben Pope wrote: ... \ Nothing can use fractional bits, they just don't exist in a physical world (somebody will pull out an anology in quantum physics now, but I don't see its relevance in todays computing environment, besides, you can still can't have a fraction of a qubit) Ben Not only bits, but digits. You can buy a 3 1/2 digit DVM. Digits and bits are measures of precision. When the actual precision is not an integer times an integer power of the number base, fractions come into the representation. Physical realization isn't needed to give the representation meaning. Or were you just pulling my leg? A bit is not a measure of precision. It is a state machine with 2 states. The "value" of the bit is completely irrelevant in this discussion. I think your range of allowed use is entirely too restrictive. I use a bit in my lathe, and after one broke in two, I used half a bit. Howard Hughes made a fortune selling bits to oil-well drillers, and repairing them. [retracting tongue from cheek] Even taking the restricted meaning of binary digit, digits are parts of numbers. Orders if magnitude have their place, but it is sometimes important to use in-between values. Hence 3.5 digit meters. From Radiums calculations, he appears to want to store 88200 samples in one bit, you can't. That would require splitting the bit into 88200 "chunks", you can't. Radium is an opinionated ass. That he's cock sure doesn't make hid drivel worth considering. Don't expect rationality. If he were capable of hearing other people, this discussion would be long over. You can't implement a unit of storage with non-integer number of states in the digital domain. And you simply can't have a unit of storage with less than 2 states, otherwise it will contain no information (if you have one level you know what it will be and is therefore completely deterministic, if you have zero levels you don't have anything) and will therefore be completely useless to you. a system capable of distinguishing 16 states is said to be a 4-bit system. One that can have 32 states is a 5-bit system. How would you characterize the information capacity in bits of a system that can have 12 states? I get 3.585 bits. That's log2(12). I realise that you can bunch a load of bits together (a byte, two bytes, a word, whatever), then order them - together the (ordered) set of states will provide an integer (since you are either in one state or another, no half-states) index which you could multiply by some pre-determined fractional value that each bit represents, to give an overall value that can represent a fraction. That I'm happy with. A 3½ digit DVM or display is not a good example here. Why not? Jerry -- Engineering is the art of making what you want from things you can get. ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
#36
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Ben Pope wrote:
Jerry Avins wrote: Ben Pope wrote: ... \ Nothing can use fractional bits, they just don't exist in a physical world (somebody will pull out an anology in quantum physics now, but I don't see its relevance in todays computing environment, besides, you can still can't have a fraction of a qubit) Ben Not only bits, but digits. You can buy a 3 1/2 digit DVM. Digits and bits are measures of precision. When the actual precision is not an integer times an integer power of the number base, fractions come into the representation. Physical realization isn't needed to give the representation meaning. Or were you just pulling my leg? A bit is not a measure of precision. It is a state machine with 2 states. The "value" of the bit is completely irrelevant in this discussion. I think your range of allowed use is entirely too restrictive. I use a bit in my lathe, and after one broke in two, I used half a bit. Howard Hughes made a fortune selling bits to oil-well drillers, and repairing them. [retracting tongue from cheek] Even taking the restricted meaning of binary digit, digits are parts of numbers. Orders if magnitude have their place, but it is sometimes important to use in-between values. Hence 3.5 digit meters. From Radiums calculations, he appears to want to store 88200 samples in one bit, you can't. That would require splitting the bit into 88200 "chunks", you can't. Radium is an opinionated ass. That he's cock sure doesn't make hid drivel worth considering. Don't expect rationality. If he were capable of hearing other people, this discussion would be long over. You can't implement a unit of storage with non-integer number of states in the digital domain. And you simply can't have a unit of storage with less than 2 states, otherwise it will contain no information (if you have one level you know what it will be and is therefore completely deterministic, if you have zero levels you don't have anything) and will therefore be completely useless to you. a system capable of distinguishing 16 states is said to be a 4-bit system. One that can have 32 states is a 5-bit system. How would you characterize the information capacity in bits of a system that can have 12 states? I get 3.585 bits. That's log2(12). I realise that you can bunch a load of bits together (a byte, two bytes, a word, whatever), then order them - together the (ordered) set of states will provide an integer (since you are either in one state or another, no half-states) index which you could multiply by some pre-determined fractional value that each bit represents, to give an overall value that can represent a fraction. That I'm happy with. A 3½ digit DVM or display is not a good example here. Why not? Jerry -- Engineering is the art of making what you want from things you can get. ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
#37
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Ben Pope wrote:
Jerry Avins wrote: Ben Pope wrote: ... \ Nothing can use fractional bits, they just don't exist in a physical world (somebody will pull out an anology in quantum physics now, but I don't see its relevance in todays computing environment, besides, you can still can't have a fraction of a qubit) Ben Not only bits, but digits. You can buy a 3 1/2 digit DVM. Digits and bits are measures of precision. When the actual precision is not an integer times an integer power of the number base, fractions come into the representation. Physical realization isn't needed to give the representation meaning. Or were you just pulling my leg? A bit is not a measure of precision. It is a state machine with 2 states. The "value" of the bit is completely irrelevant in this discussion. I think your range of allowed use is entirely too restrictive. I use a bit in my lathe, and after one broke in two, I used half a bit. Howard Hughes made a fortune selling bits to oil-well drillers, and repairing them. [retracting tongue from cheek] Even taking the restricted meaning of binary digit, digits are parts of numbers. Orders if magnitude have their place, but it is sometimes important to use in-between values. Hence 3.5 digit meters. From Radiums calculations, he appears to want to store 88200 samples in one bit, you can't. That would require splitting the bit into 88200 "chunks", you can't. Radium is an opinionated ass. That he's cock sure doesn't make hid drivel worth considering. Don't expect rationality. If he were capable of hearing other people, this discussion would be long over. You can't implement a unit of storage with non-integer number of states in the digital domain. And you simply can't have a unit of storage with less than 2 states, otherwise it will contain no information (if you have one level you know what it will be and is therefore completely deterministic, if you have zero levels you don't have anything) and will therefore be completely useless to you. a system capable of distinguishing 16 states is said to be a 4-bit system. One that can have 32 states is a 5-bit system. How would you characterize the information capacity in bits of a system that can have 12 states? I get 3.585 bits. That's log2(12). I realise that you can bunch a load of bits together (a byte, two bytes, a word, whatever), then order them - together the (ordered) set of states will provide an integer (since you are either in one state or another, no half-states) index which you could multiply by some pre-determined fractional value that each bit represents, to give an overall value that can represent a fraction. That I'm happy with. A 3½ digit DVM or display is not a good example here. Why not? Jerry -- Engineering is the art of making what you want from things you can get. ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
#38
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Jerry Avins wrote:
Ben Pope wrote: A bit is not a measure of precision. It is a state machine with 2 states. The "value" of the bit is completely irrelevant in this discussion. I think your range of allowed use is entirely too restrictive. Not given the context. I use a bit in my lathe, and after one broke in two, I used half a bit. Howard Hughes made a fortune selling bits to oil-well drillers, and repairing them. [retracting tongue from cheek] Yes, well done. So anyway, back to the discussion: Even taking the restricted meaning of binary digit, digits are parts of numbers. Thats nice. A bit is still only capable of 2 states. A binary digit is still capable of 2 states (0 and 1) Orders if magnitude have their place, but it is sometimes important to use in-between values. Hence 3.5 digit meters. A 3.5 digit meter is designed to display fractions. It is used to display a "1" (or nothing) in the most significant digit, this also must be an integer since x.5 digit meters do not have the possiblity of point (decimal or otherwise) to the left of the ½ digit. From Radiums calculations, he appears to want to store 88200 samples in one bit, you can't. That would require splitting the bit into 88200 "chunks", you can't. Radium is an opinionated ass. That he's cock sure doesn't make hid drivel worth considering. Don't expect rationality. If he were capable of hearing other people, this discussion would be long over. But he's not alone in being unable to grasp the fact that a bit has a finite amount of storage, namely two states. You seem to be struggling also, mostly becuase you seem to have forgotten what we are talking about. You can't implement a unit of storage with non-integer number of states in the digital domain. And you simply can't have a unit of storage with less than 2 states, otherwise it will contain no information (if you have one level you know what it will be and is therefore completely deterministic, if you have zero levels you don't have anything) and will therefore be completely useless to you. a system capable of distinguishing 16 states is said to be a 4-bit system. One that can have 32 states is a 5-bit system. How would you characterize the information capacity in bits of a system that can have 12 states? I get 3.585 bits. That's log2(12). Yes, but you cannot implement it with a state register containing 3.585 bits can you? You'd need 4. So what you have to say still doesn't demonstrate the possiblilty of fractional bits. Merely your inability to distinguish the mathematical domain from real life. I realise that you can bunch a load of bits together (a byte, two bytes, a word, whatever), then order them - together the (ordered) set of states will provide an integer (since you are either in one state or another, no half-states) index which you could multiply by some pre-determined fractional value that each bit represents, to give an overall value that can represent a fraction. That I'm happy with. A 3½ digit DVM or display is not a good example here. Why not? You're not even talking about the same thing as me, thats why. It's an example of something completely different. Ben -- I'm not just a number. To many, I'm known as a String... |
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Jerry Avins wrote:
Ben Pope wrote: A bit is not a measure of precision. It is a state machine with 2 states. The "value" of the bit is completely irrelevant in this discussion. I think your range of allowed use is entirely too restrictive. Not given the context. I use a bit in my lathe, and after one broke in two, I used half a bit. Howard Hughes made a fortune selling bits to oil-well drillers, and repairing them. [retracting tongue from cheek] Yes, well done. So anyway, back to the discussion: Even taking the restricted meaning of binary digit, digits are parts of numbers. Thats nice. A bit is still only capable of 2 states. A binary digit is still capable of 2 states (0 and 1) Orders if magnitude have their place, but it is sometimes important to use in-between values. Hence 3.5 digit meters. A 3.5 digit meter is designed to display fractions. It is used to display a "1" (or nothing) in the most significant digit, this also must be an integer since x.5 digit meters do not have the possiblity of point (decimal or otherwise) to the left of the ½ digit. From Radiums calculations, he appears to want to store 88200 samples in one bit, you can't. That would require splitting the bit into 88200 "chunks", you can't. Radium is an opinionated ass. That he's cock sure doesn't make hid drivel worth considering. Don't expect rationality. If he were capable of hearing other people, this discussion would be long over. But he's not alone in being unable to grasp the fact that a bit has a finite amount of storage, namely two states. You seem to be struggling also, mostly becuase you seem to have forgotten what we are talking about. You can't implement a unit of storage with non-integer number of states in the digital domain. And you simply can't have a unit of storage with less than 2 states, otherwise it will contain no information (if you have one level you know what it will be and is therefore completely deterministic, if you have zero levels you don't have anything) and will therefore be completely useless to you. a system capable of distinguishing 16 states is said to be a 4-bit system. One that can have 32 states is a 5-bit system. How would you characterize the information capacity in bits of a system that can have 12 states? I get 3.585 bits. That's log2(12). Yes, but you cannot implement it with a state register containing 3.585 bits can you? You'd need 4. So what you have to say still doesn't demonstrate the possiblilty of fractional bits. Merely your inability to distinguish the mathematical domain from real life. I realise that you can bunch a load of bits together (a byte, two bytes, a word, whatever), then order them - together the (ordered) set of states will provide an integer (since you are either in one state or another, no half-states) index which you could multiply by some pre-determined fractional value that each bit represents, to give an overall value that can represent a fraction. That I'm happy with. A 3½ digit DVM or display is not a good example here. Why not? You're not even talking about the same thing as me, thats why. It's an example of something completely different. Ben -- I'm not just a number. To many, I'm known as a String... |
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Internet Stream Audio [Was Bit-resolution decrease for internet]
Jerry Avins wrote:
Ben Pope wrote: A bit is not a measure of precision. It is a state machine with 2 states. The "value" of the bit is completely irrelevant in this discussion. I think your range of allowed use is entirely too restrictive. Not given the context. I use a bit in my lathe, and after one broke in two, I used half a bit. Howard Hughes made a fortune selling bits to oil-well drillers, and repairing them. [retracting tongue from cheek] Yes, well done. So anyway, back to the discussion: Even taking the restricted meaning of binary digit, digits are parts of numbers. Thats nice. A bit is still only capable of 2 states. A binary digit is still capable of 2 states (0 and 1) Orders if magnitude have their place, but it is sometimes important to use in-between values. Hence 3.5 digit meters. A 3.5 digit meter is designed to display fractions. It is used to display a "1" (or nothing) in the most significant digit, this also must be an integer since x.5 digit meters do not have the possiblity of point (decimal or otherwise) to the left of the ½ digit. From Radiums calculations, he appears to want to store 88200 samples in one bit, you can't. That would require splitting the bit into 88200 "chunks", you can't. Radium is an opinionated ass. That he's cock sure doesn't make hid drivel worth considering. Don't expect rationality. If he were capable of hearing other people, this discussion would be long over. But he's not alone in being unable to grasp the fact that a bit has a finite amount of storage, namely two states. You seem to be struggling also, mostly becuase you seem to have forgotten what we are talking about. You can't implement a unit of storage with non-integer number of states in the digital domain. And you simply can't have a unit of storage with less than 2 states, otherwise it will contain no information (if you have one level you know what it will be and is therefore completely deterministic, if you have zero levels you don't have anything) and will therefore be completely useless to you. a system capable of distinguishing 16 states is said to be a 4-bit system. One that can have 32 states is a 5-bit system. How would you characterize the information capacity in bits of a system that can have 12 states? I get 3.585 bits. That's log2(12). Yes, but you cannot implement it with a state register containing 3.585 bits can you? You'd need 4. So what you have to say still doesn't demonstrate the possiblilty of fractional bits. Merely your inability to distinguish the mathematical domain from real life. I realise that you can bunch a load of bits together (a byte, two bytes, a word, whatever), then order them - together the (ordered) set of states will provide an integer (since you are either in one state or another, no half-states) index which you could multiply by some pre-determined fractional value that each bit represents, to give an overall value that can represent a fraction. That I'm happy with. A 3½ digit DVM or display is not a good example here. Why not? You're not even talking about the same thing as me, thats why. It's an example of something completely different. Ben -- I'm not just a number. To many, I'm known as a String... |
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