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

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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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...

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...

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...

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

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

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

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

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

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

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

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

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

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...

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...

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...

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

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

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

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...

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...

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...

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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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...

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...

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...

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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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...

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...

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...