Subject: RE: modems

From: TOMSON ERIC <[email-masked]>
Date: Mon, 17 Jun 2002 12:21:12 +0200

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:[email-masked]]

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

---

From: Vernon Schryver <[email-masked]>
Date: Mon, 17 Jun 2002 07:36:31 -0600 (MDT)

> From: TOMSON ERIC <[email-masked]>

> ...
> That's why V.90 could only exceed this limit (56.000bps in
> downstream) by using compression techniques.
> ...

Doesn't v.90 exceed 30 Kbit/sec because in one direction it is not
analog for a significant part of the path and plays games with the
digital to analog convertors on the single local loop, and so has more
than 3 KHz at 30 db to play with?  Isn't v.42bis the modem compression
standard?  Hasn't anyone who has actually read the standards or used
v.90 modems with v.32bis and done measurements seen better than 56
kbit/sec or at least noticed that v.42 and v.42bis can be turned on
and off independently of each other and of the raw bit rate that v.90
negotiates?

This whole thread is not only of limited relevance to this mailing
list or much of anything in this century, but not as technically
accurate as one might hope.

Vernon Schryver    [email-masked]