Using under-sampled data, what determinations can be made about the Nyquist rate (correct sampling rate)? If my instrumentation limits me to sampling data at a fixed but insufficient rate what, if anything, can be determined from that data about the sampling rate that would be required to properly represent the signal? Are there signal analysis techniques that would provide any indicators?
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$\begingroup$ One thought I've recently had after thinking about this more is what may be displayed with an FFT. Presumably high counts in the bins nearest the acquisition frequency would indicate there would be more that could be captured "beyond" or above the current frequency limit/acquisition rate. $\endgroup$– EnemyBagJonesFeb 24, 2014 at 4:17
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$\begingroup$ Not necessarily. If you sample at, say, 100 Hz, you can correctly sample data with a frequency up to 50 Hz. If you instead sample data with a frequency of, say, 90 Hz, it will appear at 100-90 = 10 Hz in the FFT due to the folding, so will data with a frequency of 110 Hz, 190 Hz, and so on. $\endgroup$– OscarFeb 25, 2014 at 21:52
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2 Answers
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This may be a trivial answer, but you can of course use the aliasing information to determine the correct sampling rate, given that you know that the signal is undersampled. In the simplest case, given that the signal appears to be at $k f_s,\ k < 0.5$ one can devise that the frequency of the signal is $(1-k)f_s$. In general it may be at $(N \pm k)f_s$ for some integer $N$.
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$\begingroup$ It may be trivial to you, but I haven't tried that yet. Thank you. $\endgroup$ Feb 24, 2014 at 4:15
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$\begingroup$ OK, good! The first suggested answer was so complicated (I do not in fact understand it), which made me think I might not even have understood the question fully. $\endgroup$– OscarFeb 24, 2014 at 6:08
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If the modulation is asymmetric (as in SSB/USB) and appears to be "upside-down", then that may be evidence that a better sample rate is higher.
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$\begingroup$ Right, if you have such assumptions about the signal, you might be able to recognize what happened. $\endgroup$– user7358Feb 22, 2014 at 9:16
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$\begingroup$ Ok thanks, I'll look into that. I don't believe the signals are asymmetric, but it is certainly a good point. $\endgroup$ Feb 23, 2014 at 16:12