Content-Transfer-Encoding: 8bit
Message-ID: <5.1.0.14.0.20020116135224.00af0e08@gemini.herts.ac.uk>
Date: Wed, 16 Jan 2002 15:13:55 +0000
To: rsgb_lf_group@blacksheep.org
From: "James Moritz" <j.r.moritz@herts.ac.uk>
Subject: Re: LF: Re: G3YXM's Jason test sigs
In-reply-to: <3C458059.44CF7601@usa.net>
References: <5.1.0.14.0.20020116102434.00a7a7f8@gemini.herts.ac.uk>
MIME-Version: 1.0
Content-Type: text/plain; charset=us-ascii; format=flowed
Precedence: bulk
Reply-To: rsgb_lf_group@blacksheep.org
Sender: <majordom@post.thorcom.com>

Dear Alberto, LF Group

At 14:30 16/01/2002 +0100, you wrote:
>So, what would all you judge more important, sideband independence,
>or bandwidth used ? Your vote will decide the final format for Jason.
Provided the software is able to cope with receiving the signal as either 
USB or LSB (for example. by having a user option to change the sign of the 
measured frequency differential), I don't think having to select the 
sideband would be any problem - I think reduced BW would be useful; it 
would make it easier to avoid QRM like Loran and make the signal slightly 
easier to generate.

Also, from your previous e-mail:

>This [eliminating the bit which defines the nibble as 1st or 2nd in the 
>character] 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.
I was thinking more that this would mean 2 x 3 bits for each character 
instead of 2 x 4, reducing the number of tones required by 1/2, and so 
further halving the bandwidth. Or the "spare" bit could be used to include 
some sort of error correction, perhaps.

And, regarding the 16 tone encoding scheme:
>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 ?

That would be no problem for 16 codes - you only need to re-define possible 
values of delta as 0 to 15 instead of 1 to 16. If it is important that two 
successive tones never have the same frequency, ie. there can be no delta = 
0, you would need 17 tones - then in the example it would be 13+16 mod 17 = 
12; perhaps this is something like what Stuart is suggesting?

Cheers, Jim Moritz
73 de M0BMU