At 10:32 PM 1/28/06, you wrote:
Respectfully, you are talking about compressing the content. That won't help
with cramming a 5600 baud circuit into a 2400hz bandwidth. It might help send
more content faster - making a slower circuit look like a 5600 baud
circuit, but it won't help put a 5600 baud
No, the claim I am investigating is concerning a statement that there is a
need in amateur radio for a new mode with a symbol rate of 5600baud and a
bandwidth of 2400hz.
I'm very interested in how such a feat can be accomplished.
I'm with you. My first thought is that such a thing would be
Yes, 16QAM or 8PSK, if possible. QPSK with its
sidebands would be broader than 2400 Hz.
Jose, CO2JA
--- Tim Gorman [EMAIL PROTECTED] wrote:
Respectfully, you are talking about compressing the
content. That won't help
with cramming a 5600 baud circuit into a 2400hz
bandwidth. It might
If Nyquist is the governing limit, how do computer modem designers
get 56 kbps through a 3 khz telphone line? That's a C/B of more than
18.
73,
Dave, AA6YQ
--- In digitalradio@yahoogroups.com, Tim Gorman [EMAIL PROTECTED]
wrote:
No, the claim I am investigating is concerning a
Most telephone circuits have a bandwidth of about 2400hz which will support a
symbol rate of 2400baud. They just use a modulation scheme that allows
multiple bits per baud to be be sent, eg 32QAM or higher.
They can't shove more than 2400 symbol changes per second down a pipe of
2400hz
I suspect they mean 5600 bits per second.
73,
Dave, AA6YQ
--- In digitalradio@yahoogroups.com, Tim Gorman [EMAIL PROTECTED]
wrote:
Most telephone circuits have a bandwidth of about 2400hz which
will support a
symbol rate of 2400baud. They just use a modulation scheme that
This is a commonly confused area. And you asked the right question.
Nyquist is not the governing limit. The Nyquist limit applies to
symbols/sec, not bits/sec. A symbol can carry more than one bit.
(Interestingly, one of the first such applications was early wire
telegraphy systems that
As a non-engineer I am struggling to keep up here. ;-\
... it is the Shannon limit, which is s/n based, that
governs bits, and the Nyquist limit that governs symbols
(baud).
Can you explain why the symbol (symbol rate) construct
allows more efficient communications than a bit?
I
A more complete context of the proposal would include the following:
- - - - -
10. The real catalyst for change, however, is the need to permit higher
speed data in the Amateur bands from 1.8 MHz to 450 MHz, above which
there are no limits except to contain the transmitted signal within the
Again, this is a way to get 5600 bits per second into a 2400hz bandwidth by
sending 2.4 bits per symbol.
It won't help get 5600 symbols per second into a 2400 hz bandwidth.
tim ab0wr
On Sunday 29 January 2006 09:00, Jose Amador wrote:
Yes, 16QAM or 8PSK, if possible. QPSK with its
This was my opinion also. In order to get 5600baud, even in a 6khz bandwidth,
significant power levels will be needed to reach a signal to noise ratio
sufficient to allow the baud rate to be realized.
Even a 5600bps rate in a 2400hz ratio is going to require a HUGE signal to
noise ratio be
If there is redundancy in the data -- e.g. images, or natural language
text -- loss-less compression before transmission and decompression
after reception might get you there.
73,
Dave, AA6YQ
--- In digitalradio@yahoogroups.com, Tim Gorman [EMAIL PROTECTED] wrote:
Question
Respectfully, you are talking about compressing the content. That won't help
with cramming a 5600 baud circuit into a 2400hz bandwidth. It might help send
more content faster - making a slower circuit look like a 5600 baud
circuit, but it won't help put a 5600 baud circuit into a 2400hz
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