Linear PCM audio: 44.1 KHz, monaural, 1-bit-per-second

Radium <glucegen1@gmail.com> writes:

> On Jul 21, 9:15 pm, Jerry Avins <j...@ieee.org> wrote:
>
>> Radium wrote:
>
>> > On Jul 20, 8:42 pm, Jerry Avins <j...@ieee.org> wrote:
>
>> >> Bits don't come in fractional parts.
>
>> > I already said the bit-resolution is not fractional.
>
>> If 44,100 samples are represented by 1 bit, than each sample represents
>> 1/44,100th of a bit. I see that as fractional.
>
> No. 1 bit only results after 44,100 cycles are completed. It's like an
> extremely-low-geared processing unit. It takes 44,100 cycles to gain 1
> bit.

Go build one and show us.
--
% Randy Yates % "Bird, on the wing,
%% Fuquay-Varina, NC % goes floating by
%%% 919-577-9882 % but there's a teardrop in his eye..."
%%%% <yates@ieee.org> % 'One Summer Dream', *Face The Music*, ELO
http://home.earthlink.net/~yatescr

Jerry Avins wrote:

> glen herrmannsfeldt wrote:

(snip)

>> A decimal digit is worth about 3.32 bits. In any
>> digital system where the number of possible levels
>> isn't a power of two you have fractional bits.

> A decimal digit may be *worth* about 3.32 binary bits, but because bits
> don't come in fractional parts, *representing* a decimal digit requires
> rounding up to 4.

It does if it is represented in a binary system.
You could have a system with 10 different voltage levels allowing for
3.32 bits to be represented. It is often most convenient to
use two voltage levels, but there is no requirement on that.

-- glen

Randy Yates wrote:
> glen herrmannsfeldt <gah@ugcs.caltech.edu> writes:
>
>> Jerry Avins wrote:
>> (snip)
>>
>>> Bits don't come in fractional parts.
>> a little out of context, but mostly I don't agree.
>>
>> A decimal digit is worth about 3.32 bits. In any
>> digital system where the number of possible levels
>> isn't a power of two you have fractional bits.
>
> No system, binary or not, can have less than two possible levels.
> -

Image N-dimensional vector quantization using a codebook where the
number of entries differs from 2^(k*N) (k = 1,2,3,.....). This will
result in a fractional number of bits per sample ...


Regards,
Laurent

Laurent Schmalen wrote:
> Randy Yates wrote:
>> glen herrmannsfeldt <gah@ugcs.caltech.edu> writes:
>>
>>> Jerry Avins wrote:
>>> (snip)
>>>
>>>> Bits don't come in fractional parts.
>>> a little out of context, but mostly I don't agree.
>>>
>>> A decimal digit is worth about 3.32 bits. In any
>>> digital system where the number of possible levels
>>> isn't a power of two you have fractional bits.
>>
>> No system, binary or not, can have less than two possible levels.
>> -
>
> Image N-dimensional vector quantization using a codebook where the
> number of entries differs from 2^(k*N) (k = 1,2,3,.....). This will
> result in a fractional number of bits per sample ...

True, but it seems unrelated to what Randy said.

Steve

Jerry Avins wrote:
>

>
> Assume you are a salesman. You sell 44,100 items per second, and after
> one second you have earned one cent. How much did each customer pay you?
> What coin did each customer use?
>

Well if they are all the same customer which is the case presented then
the answer is obvious. Why are you saying this isn't possible? It is not
only possible but quite simple to implement. Just sum all the 41000
samples and depending on if the answer is positive or negative set the
bit accordingly. I think you mis-read his question. He didn't ask "would
you do this?" he asked "could you do this?"

-jim

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Laurent Schmalen <loron@gmx.de> writes:

> Randy Yates wrote:
>> glen herrmannsfeldt <gah@ugcs.caltech.edu> writes:
>>
>>> Jerry Avins wrote:
>>> (snip)
>>>
>>>> Bits don't come in fractional parts.
>>> a little out of context, but mostly I don't agree.
>>>
>>> A decimal digit is worth about 3.32 bits. In any
>>> digital system where the number of possible levels
>>> isn't a power of two you have fractional bits.
>> No system, binary or not, can have less than two possible levels.
>> -
>
> Image N-dimensional vector quantization using a codebook where the
> number of entries differs from 2^(k*N) (k = 1,2,3,.....). This will
> result in a fractional number of bits per sample ...

Hi Laurent,

This is essentially the same as what Glen said, and I have the
same response.

To be specific to your context, we are considering a (2^(R),k) rate
distortion code, where k is integer (the number of samples per vector)
and there is no integer N such that R = k*N. The bit rate of the code
is R bits/second. Therefore to achieve a rate of < 1 bit/second, we
require R < 1. Since the minimum k is 1, this implies Q < 1, where
Q is the sample size.

And now we are back to Glen's statement.
--
% Randy Yates % "Remember the good old 1980's, when
%% Fuquay-Varina, NC % things were so uncomplicated?"
%%% 919-577-9882 % 'Ticket To The Moon'
%%%% <yates@ieee.org> % *Time*, Electric Light Orchestra
http://home.earthlink.net/~yatescr

glen herrmannsfeldt wrote:
> Jerry Avins wrote:
>
>> glen herrmannsfeldt wrote:
>
> (snip)
>
>>> A decimal digit is worth about 3.32 bits. In any
>>> digital system where the number of possible levels
>>> isn't a power of two you have fractional bits.
>
>> A decimal digit may be *worth* about 3.32 binary bits, but because
>> bits don't come in fractional parts, *representing* a decimal digit
>> requires rounding up to 4.
>
> It does if it is represented in a binary system.
> You could have a system with 10 different voltage levels allowing for
> 3.32 bits to be represented. It is often most convenient to
> use two voltage levels, but there is no requirement on that.

We're not tuned into the same station. Bits are binary digits, just as
the roman numerals we use are decimal digits. I claim that one decimal
digit or four bits can represent one particular level out of ten. Do you
think I miss a point? If so, what point?

Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Laurent Schmalen wrote:
> Randy Yates wrote:
>> glen herrmannsfeldt <gah@ugcs.caltech.edu> writes:
>>
>>> Jerry Avins wrote:
>>> (snip)
>>>
>>>> Bits don't come in fractional parts.
>>> a little out of context, but mostly I don't agree.
>>>
>>> A decimal digit is worth about 3.32 bits. In any
>>> digital system where the number of possible levels
>>> isn't a power of two you have fractional bits.
>>
>> No system, binary or not, can have less than two possible levels.
>> -
>
> Image N-dimensional vector quantization using a codebook where the
> number of entries differs from 2^(k*N) (k = 1,2,3,.....). This will
> result in a fractional number of bits per sample ...

On the whole, by a particular accounting. How could you represent part
of a single bit and no part of any other?

Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

jim wrote:
>
> Jerry Avins wrote:
>
>> Assume you are a salesman. You sell 44,100 items per second, and after
>> one second you have earned one cent. How much did each customer pay you?
>> What coin did each customer use?
>>
>
> Well if they are all the same customer which is the case presented then
> the answer is obvious. Why are you saying this isn't possible? It is not
> only possible but quite simple to implement. Just sum all the 41000
> samples and depending on if the answer is positive or negative set the
> bit accordingly. I think you mis-read his question. He didn't ask "would
> you do this?" he asked "could you do this?"

I wasn't clear. Each transaction if paid for when it happens: no credit.
What coin is used for the payments?

Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Jerry Avins wrote:
>
> jim wrote:
> >
> > Jerry Avins wrote:
> >
> >> Assume you are a salesman. You sell 44,100 items per second, and after
> >> one second you have earned one cent. How much did each customer pay you?
> >> What coin did each customer use?
> >>
> >
> > Well if they are all the same customer which is the case presented then
> > the answer is obvious. Why are you saying this isn't possible? It is not
> > only possible but quite simple to implement. Just sum all the 41000
> > samples and depending on if the answer is positive or negative set the
> > bit accordingly. I think you mis-read his question. He didn't ask "would
> > you do this?" he asked "could you do this?"
>
> I wasn't clear. Each transaction if paid for when it happens: no credit.
> What coin is used for the payments?
>

I don't see the relevance of that question to the original problem.
There is nothing I saw that indicated he wanted to determine the value
of the 1 bit per second before each second had elapsed.

Actually this thread caught my eye because there seems to be some sort
of inverse law of silliness operating here. If the OP had asked can he
take 41KHz 16 bit PCM and convert it to 8kHz 8bit he would have gotten
some pretty serious answers. If he had asked could he convert to 2kHz
with 4 bits the answers would have been a bit less sober. It appears
that if the sample and bit rate of the question drops low enough the
answers appear to approach lunacy. That's a very interesting tidbit of
DSP trivia I wasn't aware of.

-jim


> Jerry
> --
> Engineering is the art of making what you want from things you can get.
> ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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