[ILUG] Irishwan.......

[Niall O Broin]

On Thu, Feb 07, 2002 at 05:31:18AM +0000, Paul Jakma wrote:

> (from a book):
> 
> nyquist's theorem states that for a noiseless channel of bandwidth H 
> with V discrete levels:
> 
> 	maximum data rate = 2H . log2(V) bits/sec.

I presume that the above is from a book (can't help thinking of Manuel
learning English, somehow) but the statement 

> so for a noiseless binary channel:
> 
> 	max data rate = 2H . 1

is only true if you define a binary channel as one on which you transmit at
two levels, ON and OFF, leading to this incorrect assumption

> so an x Hz channel has can carry a max of 2x bits/sec under perfect
> conditions.

If 'twere so, modems would never have passed 6.2 kbits/second because the
capacity of a standard telephone channel is 3.1KHz. Of course the wires to
your house can carry a lot more, witness ISDN and ADSL but the 300Hz -
3.4KHz range is assumed in a lot of the backbone coax. links. I don't know
how much bandwidth is allocated for one voice channel in a fibre link, but
it won't be much more.

The reason that a modem can transmit many more bits per second than this is
that it doesn't just transmit at two levels but rather many different levels
(the V discrete levels of Nyquist's theorem). So if you can transmit at say
four different levels, you double the bit rate from two, eight different
levels triples your bit rate and so on. [As a little aside, baud rate on a
transmission channel refers to the rate of transmission of these multivalued
symbols and NOT bits. With modern modems, baud rate is much lower than bit
rate]. So, you can transmit as many bits down a channel of given bandwidth
as you like with a sufficiently large V.


But TANSTAAFL which in this case means that there is a limit on how large we
can make V in a real channel because eventually we can't distinguish between
levels because the difference between them is less than the noise on the
channel which is where Shannon came in

> Shannon adds the concept of signal to noise ratio:
> 
> 	max bits/sec = H . log2 (1 + s/n)
> 
> true for any channel subject to gaussian noise, irrespective of 
> sample rate.

and as you can see from that, in a noise free channel (unfortunately there
is no such thing) of arbitrary bandwidth you can transmit an infinite number
of bits per sec.

> > Anyway, I hope this makes things clearer. If I were not riding an 802.11
> > deficient train, I might be able to research the answer on google.

We can't get ADSL into our houses and he's looking for 802.11 on a train ?



Niall