On Sun, 5 Mar 2000, Kenn Humborg wrote:
On Wed, Mar 01, 2000 at 12:58:29PM +0000, Niall wrote:
> In any event, this cable is of much higher quality than any installed
> local loop telephone cable. A fundamental law of information transmission
> tells us that sending 45 Mbit/s requires a signal frequency of at least
> 90MHz
Only if you insist on carrying the signal on a single carrier.
And, more correctly, it's the bandwidth, not the frequency that
correlates with information content. The 10MHz of bandwidth
between 0 and 10MHz has the same information carrying capacity as
the 10MHz between 1GHz and 1.01GHz. The difference is in the
practicalities of modulation/demodulation and transmission.
which law is this again? (remember reading it in Tannenbaum's network
book). Also, this law only applies to "direct"/"binary" (what is the
fancy computer term for this?) signalling methods, doesn't it? ie
where there are 2 states for the signal such as rs232, old old
9600baud modems (i believe), 10baseT ethernet.. etc..
No disrespect to Owen and his friend in Genesis, but I'm still
skeptical. I'll be impressed when I see it, but, until then, it's
vapourware...
but note that the most recent comms technologies don't use a direct
wire state <==> bit stream model anymore. Eg 56k modems and
gigabit. According to (shannons??) law gigabit ethernet would require
a bandwidth of 2GHz, but it actually uses only 800Mhz (4x
200MHz). The wire bit stream no longer represents the higher level
data bit stream, it represents symbols. And the symbol stream can
encode a much higher bitrate datastream.
So if gigabit is possible over 800MHz, might 45MBit/s not be possible
over telephone lines?
Later,
Kenn
regards,
--
Paul Jakma paul at clubi.ie
PGP5 key: http://www.clubi.ie/jakma/publickey.txt
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