Bit-resolution decrease for internet

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.

Radium wrote:
> 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.


You are clearly misguided.

Ben
-- 
I'm not just a number. To many, I'm known as a String...

glucegen@excite.com (Radium) wrote in 
news:yvik75eq9k7.fsf@jpff.cs.bath.ac.uk:

> 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


-- 
Nothing beats the bandwidth of a station wagon filled with DLT tapes.

Wrong. 1 Hz sampling rate would equate to .5 Hz. Sampling rate must be
at least 2x the maximum frequency.

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 60 bits of disk space.

Rich Andrews <n0way@yahoo.com> wrote in message news:<yvi1xrcobq0.fsf@jpff.cs.bath.ac.uk>...
> 
> 1 bit per second?  Wouldn't that equate to .5 hz or did I miss something?

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 <n0way@yahoo.com> wrote in message news:<yvi1xrcobq0.fsf@jpff.cs.bath.ac.uk>...
> glucegen@excite.com (Radium) wrote in 
> news:yvik75eq9k7.fsf@jpff.cs.bath.ac.uk:
> 
> > 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.
�����������������������������������������������������������������������

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

44100 X 16 X 2 = 1,441,200

44100 X 1/88200 X 2 = 1

Jerry Avins <jya@ieee.org> wrote in message news:<3fdb3cb6$0$14947$61fed72c@news.rcn.com>...
> 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