Marty Albert wrote:
As I recall, at 23 cm and 80 Mbps we had an effective bandwidth of around
100 KHz on the "final" design... That design incorporated TDM, limited SS,
and WDM of the signals.

What exactly does this all mean?

Passing 80,000,000 bits/sec in 100,000Hz of bandwidth sounds
pretty fantastic - to the extent that makes me question the
validity of the measurements.

Today, we could use TDM, WDM, SDM, high-end SS, and a few other tricks and,
assuming a target data-rate of 100 Mbps, get the on air bandwidth down to
around 50-75 KHz, maybe even a little less.

Whoa. Hold on. Help me understand what units and methods
of measurement you're using. Right now, you're off by several
decimal places in even the most generous way.

Dana K6JQ

"Dana H. Myers" wrote in message
...
Marty Albert wrote:
As I recall, at 23 cm and 80 Mbps we had an effective bandwidth of around
100 KHz on the "final" design... That design incorporated TDM, limited
SS,
and WDM of the signals.

What exactly does this all mean?

Passing 80,000,000 bits/sec in 100,000Hz of bandwidth sounds
pretty fantastic - to the extent that makes me question the
validity of the measurements.

This in 100 Hz of bandwidth we can obtain 80 Kbps.
Shannon twirling in his grave.

Today, we could use TDM, WDM, SDM, high-end SS, and a few other tricks
and,
assuming a target data-rate of 100 Mbps, get the on air bandwidth down to
around 50-75 KHz, maybe even a little less.

Whoa. Hold on. Help me understand what units and methods
of measurement you're using. Right now, you're off by several
decimal places in even the most generous way.

Lost a decimal point for sure.

BTW ... why does everyone always mention 1200 baud?
Doesn't everyone use at least 9600 for local links, and
PACTOR II / III on HF?

Think I have a 1200 baud TNC around here ... yeah there it is
over in that cabinet. Big black box, says TAPR TNC-1 on it.

--

... Hank

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

TDM = Time Domain Multiplexing

WDM = Wide Dimensional Multiplexing

SDM = Statistical Domain Multiplexing.

TDM and, to a limited degree, WDM have been around and used for several
decades in high performance networking. SDM is relatively new, about 5
years.

There are far more ways to multiplex intelligence on a medium than PSK and
FSK, although both are used in high performance systems.

You are 100% correct... I may very well a few decimal places off... The
mathematical models show that 100 Mbps should be possible in as little as 10
KHz.

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


"Dana H. Myers" wrote in message
...
Marty Albert wrote:
As I recall, at 23 cm and 80 Mbps we had an effective bandwidth of
around
100 KHz on the "final" design... That design incorporated TDM, limited
SS,
and WDM of the signals.

What exactly does this all mean?

Passing 80,000,000 bits/sec in 100,000Hz of bandwidth sounds
pretty fantastic - to the extent that makes me question the
validity of the measurements.

Today, we could use TDM, WDM, SDM, high-end SS, and a few other tricks
and,
assuming a target data-rate of 100 Mbps, get the on air bandwidth down
to
around 50-75 KHz, maybe even a little less.

Whoa. Hold on. Help me understand what units and methods
of measurement you're using. Right now, you're off by several
decimal places in even the most generous way.

Dana K6JQ

"Marty Albert" writes:
You are 100% correct... I may very well a few decimal places off... The
mathematical models show that 100 Mbps should be possible in as little as 10
KHz.

Yes, you are a few decimal places off, but in the wrong direction ;-).

"Marty Albert" wrote in message
...
TDM = Time Domain Multiplexing

WDM = Wide Dimensional Multiplexing

SDM = Statistical Domain Multiplexing.

TDM and, to a limited degree, WDM have been around and used for several
decades in high performance networking. SDM is relatively new, about 5
years.

There are far more ways to multiplex intelligence on a medium than PSK and
FSK, although both are used in high performance systems.

You are 100% correct... I may very well a few decimal places off... The
mathematical models show that 100 Mbps should be possible in as little as
10
KHz.


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

--

... Hank

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

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
ink.net...

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

"The mathematical models show that 100 Mbps
should be possible in as little as 10 KHz."

If that is what it shows, there is an error.

The error is either with the model itself, or with the
assumptions fed into the model.

Basic Thermo 101 ... Shannon ... etc.

However, if you can get me 100 Mbps in 10 KHz,
I'll be glad to buy a whole bunch of 'em :-)

--

... Hank

http://home.earthlink.net/~horedson
http://home.earthlink.net/~w0rli
"Marty Albert" wrote in message
om...
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
ink.net...

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

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
link.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.


Alan
wa6azp

"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

I was wondering when someone would stumble across the Hartley-Shannon
"law"...

Like most laws of this nature, they break down at the extremes (i.e.
Relativity fails on the scale of the very small and Quantum Mechanics fails
on the scale of the very large... For my money, I'm betting on M Theory
(formerly String Theory) to be the true GUT).

Plug the numbers in for CAT6 over 100 m running a 10 Gbps data stream. You
will find that the equation fails in that it will tell that you can't do
that and, if set up right, it will send you the fact that it is impossible
OVER the 10 Gbps, 100 m run.

By the same token, if you plug in numbers for, say, 1.2 Kbps over CAT6 of
500 m length, that won't work either. The fact is, we (the telecomm/data
comm industry) do it every day.

Keep in mind that we all tend to "think inside the box" and say that things
(pick one) are impossible because of some theory or "law". As a general
rule, we end up being wrong.

Case in point, everyone knows that the velocity of light (c) is the "Cosmic
Speed Limit". Einstein said so and had the math to prove it. Quantum
Mechanics and M Theory both have solutions that allow for FTL... And some
have some rather unpleasant side effects.

But enough of the arcane math and other arenas...

Try to "think outside of the box"... Think about ways to do:

1) Ultra high speed data over RF media;
2) Make the system so cheap and easy that everyone will want one;
3) Perhaps get the attention of the commercial sector.

Everyone sitting around bemoaning the obstacles that need to be over come
are counter productive.

There is no such thing as "I can't do that."

There is only "I won't do that."

At the risk of sounding Republican (which I am) and of sounding like I agree
with President Bush (which I do not at least 50% of the time) you need to
decide how much, if any, effort, thinking, and support you are willing to
give.

If you are interested in digital modes, you will be willing to put in what
is needed.

If you are not interested in the digital modes, you should at least get out
of the way of those who do.

But, I wax philosophical... And even with 2 doctorate degrees, neither
qualifies me to be a philosopher.

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


"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
link.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.


Alan
wa6azp