I thought the operative thingy here was Shannon? anyway....
I don't what the question actually is here, but I offer a few observations:-
1. The maximum instantanious full duplex signalling rate of communications on
a telephone circuit has been calculated as 2400 baud.
2. a 'baud' is a symbol/sec, that symbol may be complex.
3. The signaling rate achievable is dependant on the circuit signal to noise
ratio.
4. a "56K" modem is an inherently asymetric device in that it relies on the
fact that modern POTS exchanges are digital, thus have essentially only the
quantization noise to throw into the S/N equations.
5. Because, at the generating end, your ISP can synthesis a "perfect"
waveform and guarantee that it will be delivered unblemished to the bit of
copper that separates your house from the exchange, the lower aggregate S/N
allows much higher information rates _in that direction_ only.
6. But because you have a bit of copper and an D/A converter, you still have
to go thru all the malarky necessary to optimise your modem for, amongst
other things, phase noise and distortion, group delay, echo cancelling and so
on which limit the total thruput in the other direction to about 30Kb/s as
predicted by Shannon (as well for the little bit coming in your direction).
7. All this takes time. On a modern modem up to 30 seconds can be spent
optimising the filter, delay, echo and frequency response characteristics for
the circuit that you have been given.
8. On a radio, there is no "circuit" (especially not in a packet radio
environment with more than two radios talking and listening). You only have a
very short "training" time, typically no more than 64 bits, you can't do much
with that and that is what is limiting speeds.
If you want to contribute then start looking at ways to add bits to the
symbol rate without increasing training time (bit length). This is an area of
real research and you could make a name for yourself...
Dirk G1TLH
On 24-Nov-1999 Charles Suprin wrote:
> Quick tutorial on information theory and nyquist and the difference.
>
> The Nyquist rate deals with perfect reconstruction of a sampled signal.
> Detection and demodulation do not require perfect reconstruction of the
> signal.
>
> Channel capacity is defined as the maximum rate for reliable
> communication through a channel. This is a function of many things,
> channel bandwidth, noise spectrum, SNR and a whole bunch of other
> things. How to achieve channel capacity in the general case is a
> challenging problem. It has been solved in some cases.
>
> A clear example of this is the 56k telephone modem. The channel is 3-4
> kHz wide. Perfect reconstruction of the received signal can be
> accomplished with an 8kHz sampling rate. However we are getting almost
> 7 times that in data trasmission. (Yes a 56k modem does closer to 53k
> in the telephone system.)
>
> For the person looking for the SNR plots for the different bands such a
> beast does not exist. The best that you can do is find a chart that
> describes the noise in each of the bands. Everybody transmits with
> different power to receivers at different range. This changes the
> received signal power and therefore the SNR at the receiver.
>
> Demodulations can be computationally intensive. 56k modems usually have
> an ASIC or a dedicated processor to deal with the load. My question
> is: Where are the computational bottlenecks and can anything be done
> about them?
>
> Charles Suprin
> Not for work, not works opinions.
--
Dirk-Jan Koopman, Tobit Computer Co Ltd
At the source of every error which is blamed on the computer you will find
at least two human errors, including the error of blaming it on the computer.
