abt some problem with nyquist's and shannon theory

[Joseph2k]

I take "duplex" to mean full-duplex so single cable comm. Then
conceptually label the routers 1,2,...,10, and the cabling would go as
(1,2),(1,3),...,(1,10),(2,3),(2,4),....,(9,10) where (m,n) denotes the


This is easily answered by reference to this piece written by a
pseudo-intellectual pedant:
http://en.wikipedia.org/wiki/Shannon-Hartley_theorem

C= BW x Log2(1+S/N)= 4e3*Log(1+1000)/Log(2)=40Kbps for Shannon

Nyquist's Theorem applies to the equivalent bit rates of noiseless
channels:

C = 2*W log base2( L ) bits/sec
o where 2W is 2 times the highest frequency contained in the
noiseless channel, and
o where L = number of discrete levels (e.g., binary = two levels, 0
and 1)

C=8e3*log2(2)= 8Kbps

Despite the claims of others, real life is just this simple.

Two issues, 10 fully interconnected routers takes only 9 cables in a correct
tree configuration; standard V.90 modems regularly push 50 kbps down a POTS
line with a bandwidth of 3.4kHz with no better S/N ratio.