Michael,
You're right. Now, it was not my intention to be as exhaustive as a LAN/WAN
course.
However, if the following information can satisfy your legitimate need for
exhaustivity, let me give some examples of those V.xx standards from CCITT/ITU :
Spec Datarate Baudrate Bit/baud Modulation
V.21 300 300 1/1 FM
V.22 1200 600 2/1 PM
V.26 2400 600 4/1 PM
V.26 bis 2400 600 4/1 PM
V.27 bis 4800 1600 4/1 PM
V.29 9600 2400 4/1 QAM
V.32 bis 14400 2400 6/1 QAM
V.33 14400 2400 6/1 QAM
V.34 28.8k/33.6k 2400 12/1 QAM
V.90 56k/28.8k 2400 23.3/1 & 12/1 QAM
As we can see, by using modulation techniques using more and more
levels/signals, this allows to transport more and more bits per baud, then more
and more bits per second. Now, this is not unlimited : Shannon explained us
why...
Regarding Shannon :
taking his Theorem 17's formula (I just re-read it to be sure) into account : C
= W x Log (P+N / N),
applying it to an ordinary telephone line (supporting 3000Hz-4000Hz with an
average signal-to-noise ratio of 20-30dB),
we can calculate an average maximum datarate of around 30.000bps.
That's why V.90 could only exceed this limit (56.000bps in downstream) by using
compression techniques.
Well, may I now suggest to close the "modem" case here?
-----Original Message-----
From: Michael Hammer [mailto:mhammer(_at_)cisco(_dot_)com]
What, no mention of constellations?
When you talk of baud or symbols per second, it helps to keep in mind that
a symbol is a shift from one analog 'waveform' to another. These analog
waves have three characteristics: frequency, phase, and amplitude. The
earlier modems simply shifted between two frequencies (e.g. Bell 103, if
memory serves). The later V.xx standards typically used a combination of
amplitudes and phases. If you plotted these on an 2-D surface it appears
like a grid of dots. Depending on the number of dots, you get many
different possible states to jump between. For example, two amplitudes and
4 phases would produce 8 states, which are numbered 000-111. Thus a shift
to a new state represents an output of three bits.
You can only take this so far as it becomes harder and harder to "squeeze"
more dots in and still distinguish one state from another. To understand
where this limit exists, study Shannon's Theorem. After that
pre-compression algorithms come into play.
Mike