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LF: Re: Jason: 65 channels?

To: [email protected]
Subject: LF: Re: Jason: 65 channels?
From: "Alberto di Bene" <[email protected]>
Date: Mon, 11 Feb 2002 13:41:16 +0100
References: <[email protected]>
Reply-to: [email protected]
Sender: <[email protected]>
[email protected] wrote:

G'day Alberto, Steve and all,

 its great to see the concept of FDK and its friendly implementation in
Jason. But, one question came to my mind: Instead of using a 4-bit pair for
each 6 bit character, why not take it all the way and transmit 6 bits at
once? Of course, this would mean 65 frequencies instead of 17, four times the
bandwidth at first glance.

But then you could double the symbol duration and halve the FFT bin bandwidth
to 42 mHz, picking up some SNR on the way. Additionally, even with a Hamming
or Bartlett window, the noise in channels only two bins apart should have no
significant correlation. Thus the total occupied  bandwidth would increase
from the current 4.12 Hz to no more than 5.42 Hz.

What do you think?
Hello Markus,
   I have been out of town for the weekend, so am able to answer only today.
If I do correctly understand you, what you propose is to multiply by 4 the 
number
of the signalling elements, halving at the same time their spacing and the baud 
rate,
for an unchanged throughput in terms of bit/sec. This would led to a doubling 
of the
bandwidth. Then you implicitly suggest a reduction of the number of bins between
different elements from 3 to 2, thus gaining another 0.666... factor of 
reduction of
the band, which would go from 4.12 to 4.12 * 2 * 0.6666...  =  5.48 (roughly) 
Hz.
Please correct me if I misunderstood you.

I have one remark on this. Placing the signalling elements on alternate bins 
(spacing
of two instead of three as in the present scheme), can cause uncertainty due 
both
to the leakage intrinsic in the FFT process, and to possible instabilities of 
the Tx
and/or the Rx.  In other words, let's make a simple example. Let's number the
bins 1, 2, 3, 4, 5, 6  etc.  Now, suppose I use the even numbered bins only.
In the case I find a signal in bin number 5, should I consider it belonging 
actually
to the signalling element of bin 4 or of bin 6 ??  There is no way to tell.
Using three bins as spacing, as I do now, and keeping the previous example,
the 'correct' bins are, suppose, the Nr. 1, 4, 7, 10, etc. If I find a signal 
in bin 5,
I attribute it to bin 4, as I do for bin 3. So, bin 6, 7, 8 are all 'belonging' 
to bin 7,
and so on. In this way I have more chances to keep the signalling elements
orthogonal, as is required for a correct MFSK uncoherent decoding.

So, to sum up, I would tend to discard the proposal of reducing the bin spacing.
What remains is the proposal of doubling the bandwidth in exchange for the
possibility of keeping unchanged the throughput while at the same time 
increasing
the S/N ratio. It can be done, but I would like to know also what the general
consensus is on this, and any possible comments from the other LFers.   TNX.

73  Alberto  I2PHD





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