Jerry Avins wrote:
> Andor wrote:
<Snip>
> > I don't see any problem once you have accepted the fact that the
> > product of message time with the bandwdith has to stay constant
> > (assuming equal SNR of all channels involved).
>
> Yes, of course, but the message must have finite length, so it is not a
> continuous data stream in a strict sense. I call it a batch job,
> "intermittently continuous". For two to one, one begins recording at
> full speed and immediately transmitting at half speed*. That way, a
> buffer only large enough to hold half the message is needed and the
> delay is least.
>
> Jerry
> ___________________________
> * Possible with digital storage, but that means sampling, again, not
> truly continuous.

Or a big bin for the tape :)
However considering that most quality magnetic tape units used AC
supersonic bias, maybe even audio tape recordings could be considered
to be sampled.
Regards
Rocky

Vladimir Vassilevsky wrote:
> Andor wrote:
>
>
> > Let's assume you wanted to send a message of some given length M
> > through a channel with some SNR and fixed bandwdith B, and the same
> > message through a channel with equal SNR but bandwidth B/2. I think
> > you'll agree that the time required needs to be 2M.
>
> We have an analog signal of the bandwidth B. We have a perfect channel
> with the bandwidth less then B. Is it possible to transmit the signal
> through this channel without any loss and at the same time scale?

After sending any signal through a perfect channel with bandwidth B, it
will be infinite in length - changing the time scale won't matter
anymore.

> Looks like any procedure which can accomplish that requires quantization
> both in time and in amplitude, i.e. has to be digital and lossy. Is
> there any proof?

Well, if you have a perfect channel you have to assume non-quantized
sampling. In that case, you don't lose information.

Regards,
Andor

Vladimir Vassilevsky wrote:
> Given an analog bandlimited signal, is it possible to create a
> continuous time reversible procedure to reduce the bandwidth without
> losses? In the other words, is it possible to create a bandwidth
> reducing analog modulation?
>
> Looks like the answer is no, however I can't make a mathematical proof
> of this.

A bandlimited signal can not be time limited, and when dealing
with infinite anything (including infinite time signals), the results
can look strange.

An infinite number of filters with infinitely narrow bandwidths
around center frequency f but driving oscillator outputs at f/2
might do the job, but each filter will have an infinite delay.
The process is perfectly reversable.  Does that long a start-up
delay still count as continuous time?

More realistically, one might try scaling the filter width by
the inverse S/N ratio of the channel in exchange for imperfect
reconstruction and delays slightly less than infinite.


IMHO. YMMV.
-- 
rhn A.T nicholson d.O.t C-o-M

Vladimir Vassilevsky wrote:
> 
> 
> Jerry Avins wrote:
> 
> 
>>> OK, let's consider the noisy channels with the different bandwidth 
>>> but with the same Shannon's capacity. What fundamental limitation 
>>> does not allow for the existence of the operator which directly 
>>> converts one channel into another?
>>
>>
>>
>> Such an operator exists in certain cases. If each channel has the same 
>> capacity and they are used to the same fraction of that capacity, they 
>> can convey the same amount of information in the same time. Provided, 
>> of course, that a recoding exists which will allow it.
> 
> 
> So, back to the original problem: it is possible to design such an 
> operator which does NOT use the quantization in time and amplitude?

I don't see how, but someone clever might. One can in theory sample the 
signal with the LSB in the noise and encode in a way that suits the 
channel. At the far end, the signal can be reconstructed. If it can be 
done digitally, it can be probably done analog. I described analog 
signal in, channel you describe, and analog signal out. Are the means 
relevant?

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

Jerry Avins wrote:


>> OK, let's consider the noisy channels with the different bandwidth but 
>> with the same Shannon's capacity. What fundamental limitation does not 
>> allow for the existence of the operator which directly converts one 
>> channel into another?
> 
> 
> Such an operator exists in certain cases. If each channel has the same 
> capacity and they are used to the same fraction of that capacity, they 
> can convey the same amount of information in the same time. Provided, of 
> course, that a recoding exists which will allow it.

So, back to the original problem: it is possible to design such an 
operator which does NOT use the quantization in time and amplitude?

VLV

Vladimir Vassilevsky wrote:
> 
> 
> Jerry Avins wrote:
> 
> 
>>>
>>> What you are saying translates to "the perfect channels of the 
>>> bandwidth B0 and B1 are not equivalent if B0 != B1". This does not 
>>> seem to be correct. Any ideas?
>>
>>
>>
>> Do you maintain that "Noiseless channels of different bandwidth are 
>> not equivalent" is incorrect? I think there's a difference, but since 
>> the carrying capacity of a noiseless channel is infinite, your 
>> aleph-nul is as big as mine.
> 
> 
> OK, let's consider the noisy channels with the different bandwidth but 
> with the same Shannon's capacity. What fundamental limitation does not 
> allow for the existence of the operator which directly converts one 
> channel into another?

Such an operator exists in certain cases. If each channel has the same 
capacity and they are used to the same fraction of that capacity, they 
can convey the same amount of information in the same time. Provided, of 
course, that a recoding exists which will allow it.

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

Jerry Avins wrote:


>>
>> What you are saying translates to "the perfect channels of the 
>> bandwidth B0 and B1 are not equivalent if B0 != B1". This does not 
>> seem to be correct. Any ideas?
> 
> 
> Do you maintain that "Noiseless channels of different bandwidth are not 
> equivalent" is incorrect? I think there's a difference, but since the 
> carrying capacity of a noiseless channel is infinite, your aleph-nul is 
> as big as mine.

OK, let's consider the noisy channels with the different bandwidth but 
with the same Shannon's capacity. What fundamental limitation does not 
allow for the existence of the operator which directly converts one 
channel into another?

VLV

Andor wrote:
> Jerry Avins wrote:

   ...

>>Yes, of course, but the message must have finite length, so it is not a
>>continuous data stream in a strict sense.
> 
> 
> No, the data can be a continuous stream. Re-read what I wrote.

Do you mean that two channels of B/2 can replace one of B? Sure. Anyway, 
signals sent in binary form on low-noise channels rarely use the 
channel's full capacity. We get 56K bits/second worth of information on 
a telephone line, but we that is achieved with a lower rate of symbols 
each packed with more information. There is no magic, but there is 
inspiring sleight of hand.

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

Vladimir Vassilevsky wrote:
> 
> 
> Jerry Avins wrote:
> 
>>> We have an analog signal of the bandwidth B. We have a perfect 
>>> channel with the bandwidth less then B. Is it possible to transmit 
>>> the signal through this channel without any loss and at the same time 
>>> scale?
>>> Looks like any procedure which can accomplish that requires 
>>> quantization both in time and in amplitude, i.e. has to be digital 
>>> and lossy. Is there any proof?
> 
> 
>> A proof probably lies in the meaning of bandwidth.
>>
>> We have a pile of oranges to big to fit into a particular box. We want 
>> to transport those oranges, all at the same time and all in that box, 
>> without making orange juice. Can it be done? I think not. Is there a 
>> proof? I think the same proof applies your question and mine.
> 
> 
> What you are saying translates to "the perfect channels of the bandwidth 
> B0 and B1 are not equivalent if B0 != B1". This does not seem to be 
> correct. Any ideas?

Do you maintain that "Noiseless channels of different bandwidth are not 
equivalent" is incorrect? I think there's a difference, but since the 
carrying capacity of a noiseless channel is infinite, your aleph-nul is 
as big as mine.

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

Jerry Avins wrote:

> Andor wrote:
> > Jerry Avins wrote:
> >
> >
> >>Andor wrote:
> >>
> >>>Andor wrote:
> >>
> >>   ...
> >>
> >>
> >>>Is recording onto a recorder and playing back at a different speed a
> >>>"continuous process"? How does it make a difference if the recorder is
> >>>analog or digital?
> >>
> >>A device that plays back faster than it records will shortly predict
> >>future inputs. One that plays back more slowly must eventually choke.
> >>Such a device can at best be intermittently continuous. :-)
> >
> >
> > Let's assume you wanted to send a message of some given length M
> > through a channel with some SNR and fixed bandwdith B, and the same
> > message through a channel with equal SNR but bandwidth B/2. I think
> > you'll agree that the time required needs to be 2M.
> >
> > In other words, if doubling the transmission time for any message is
> > not an option, you need at least two channels (assuming the SNR stays
> > constant) with half the bandwidth. One can think of a scheme with
> > several recorders alternately recording and playing back two continuous
> > signals on the two channels.
> >
> > For the general case, the number of recorders required depends only on
> > the ratio of the two bandwidths. Such a transmission scheme can be
> > implemented with arbitrary small, but finite, delay.
> >
> > A second option would be to separate the signal with bandwidth B into
> > two signals with bandwidth B/2 - this would be even simpler using
> > analog technology (filters and mixers) - and recombine the original
> > signal at the receiving end.
> >
> > I don't see any problem once you have accepted the fact that the
> > product of message time with the bandwdith has to stay constant
> > (assuming equal SNR of all channels involved).
>
> Yes, of course, but the message must have finite length, so it is not a
> continuous data stream in a strict sense.

No, the data can be a continuous stream. Re-read what I wrote.