James Moritz wrote:

Dear Alberto, Steve, Dave, LF group,

The fastest computer I have is 300MHz, so whether this is adequate remains
to be seen. So I would certainly be interested in some tests from G3YXM
this evening.

Hi Jim and the group,
 probably 300 MHz will not be enough... at least until the slowed down
version is ready.

Perhaps silly questions, but why 2 different tone frequencies for each
symbol? Would it not be possible to have 16 different tones with 1 fixed
"carrier" tone, to generate the same 16 difference frequencies and roughly
halve the BW?

If you use a fixed "carrier" tone, you have to alternate it with the tones sent.
Don't forget that at any one time, a single tone is transmitted.
So, you transmit the carrier, then a tone. Their difference conveys information,
but when you retransmit again the carrier, this time no information is sent...

Also, is it necessary to have different symbols for the first
and second 3 bits of each carrier? I realize this removes the ambiguity,
but since a particular tone is either the beginning or the end, there are
only 2 possible decodes, one of which will be garbage - both decodes could
be displayed side by side, and it could be left to the operator to decide
which was correct.

This would gain one bit each nibble, increasing the alphabet from 64 to 256
symbols. For keyboard-to-keyboard communication, I think 64 symbols
are enough.

Another thought is a differential encoding scheme - suppose you have 16
tones and 16 symbols, then the next tone would be the modulo 16 sum of the
current tone frequency and the next symbol. For example, if tone 9 is
currently being transmitted, and the next symbol is 5, the next tone will
be 14. If the next symbol after that is 8, the next tone will be 6 (+16),
and so on.

Hmmm, suppose I have just sent tone 13. Next, I have to send a delta
of 16 (the deltas range from 1 to 16). 13 + 16 mod 16 = 13, so I would
end up sending the same tone, with a delta of 0.  Or maybe I misunderstood
what you said ?

The receiving software would still be looking for the difference
frequency between successive tones - however, sending would be twice as
fast because only 1 tone would be sent for each symbol, with the previous
tone acting as the reference. I think I'm missing something, but I can't
think what at the moment...

No, sending would be at the same speed. Each tone sent is acting as a
reference for the next one, already in the present scheme.

Cheers, Jim Moritz
73 de M0BMU

Thanks for your remarks Jim

73  Alberto  I2PHD