> This is not so surprising to me, and I think you should be careful with > your conclusions:
Indeed...and I welcome discussion about my findings... > *) The modulation is a phase modulation, that is primarily not working > with amplitude shifts. That makes it more robust against amplitude > noise. It's quite the same with the difference between AM an FM, where > AM is very sensitive to atmospheric noise. The downside of FM is > bandwidth usage, with a phase modulation it is kind of the minimum > sampling rate. Hmm, I would agree with you provided it was ONE carrier, but now we have 15, summed, and with 13.5 periods up to 34.5 / symbol. > *) You probably tested with a perfect signal. Would be interesting what > is happening if you add white noise to your samples. Sox can do that for > you for example. Yes, I was testing with perfect signal, i.e. the binary 16-bit -> 6bit data. > *) You would not want to implement a real-world system with less than 8 > bits, and 8 bits still is not really that much. You cannot sample a > signal perfectly: You have more amplitude variations than just from your > signal, like background noise, QRM, other signals. Especially other > strong nearby signals will result in big amplitudes at your ADC, which > in turn saturates at its highest and lowest sample value, and you start > to loose information. In German that's 'Grossignalfestigkeit', in > English I think it's called large signal immunity or large signal > rejection. You can get along with less bits, but you have to apply very > narrow and exact filtering before ADC. Indeed, and a real imlementation should take some latitude for this. > *) SNR to bits resolution has a know relation: > SNR = (1.76 + n*6.02) dB > where n is the number of bits. For 8 bits you should get 749.92 dB SNR. > ADCs are never perfect and add some noise, which lowers your SNR. Thus > you get a certain effective SNR, which corresponds to certain effective > (non-integer) bits resolution. Even good ADCs loose some 1 bit resolution. That sounds a bit big, I get 20*log(1/256) = -48 dB or 20 * log(256) = 48 dB. Your formula gives 49.92 i guess a keyboard slip. Where does the 1.76 come from? Anyway, the experiment was simple, and sets the limits for a noisefree channel, and *seems* to indicate that even with 8 bits there is a small margin. Anyway, my first real attempt at this will be a test generator, and in the "lab" with control over noise and other simulations. Tnx for the tip about noise insertion with sox, also easy to do.... Regards, Gullik > Regards > > Patrick ------------------------------------------------------------------------------ Live Security Virtual Conference Exclusive live event will cover all the ways today's security and threat landscape has changed and how IT managers can respond. Discussions will include endpoint security, mobile security and the latest in malware threats. http://www.accelacomm.com/jaw/sfrnl04242012/114/50122263/ _______________________________________________ Freetel-codec2 mailing list [email protected] https://lists.sourceforge.net/lists/listinfo/freetel-codec2
