"Alan" wrote in message
...
In article "Marty
Albert" writes:
I will see if the University that I am using to develop the model will
allow
that at this point... It is actually their intellectual property.

I doubt, however that you will find any major errors in the algorithms....
There have been many professors, PhDs, and grad students looking at it to
find those errors as well as engineers from Motorola, Maxim, and TI.

Take Care & 73
--
From The Desk Of
Marty Albert, KC6UFM


"Hank Oredson" wrote in message
hlink.net...

The model is wrong.
Post it and I'll be glad to explain why.


A few things come to mind:

1) Multiplexing does not increase the bandwidth capability of a
channel.
You mention various forms of multiplexing, but these will not increase
the channel capacity. They are just different ways of utilizing what
is available.

2) The Hartley-Shannon Law gives the maximum bandwidth of a channel as
C = B log2(1+(s/n)) bits/second; where B is bandwidth (Hz) and s/n is
expressed as a value, not in dB.

Given this, to get 80 megabits of signal in a 100 kilobit channel, you
will need a signal/noise ratio of about 2408 dB. Since you were only
starting with a 10 watt signal, with about 100 dB path loss (after
including
the two j-poles), and a terrrestrial noise floor of about -124 dBm for the
100 kHz wide channel, you get only about 60 - 64 dB s/n in your receiver
(assuming things like lossless coax, etc.).
Thus you are about 2340 dB short on signal to accomplish the task as
described. See http://encyclopedia.laborlawtalk.com/Shannon_limit for
more
discussion of this.


Your numbers are a bit too far from what can reasonably be believed.


Thanks for doing the math :-)
I thought he claimed a 10 KHz channel ... but might be wrong.

You could cool the receiver and heat the transmitter (1500W,
larger dish) and get perhaps 20 - 40 more db ... that would help :-)

Then you might be only 2300 db short.

--

... Hank

http://home.earthlink.net/~horedson
http://home.earthlink.net/~w0rli