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Let's take the signal to be sampled to be made up of several sine waves close to one another in frequency, e.g. 1MHz, 1.001MHz, 1.002MHz,... 1.005MHz

I assume in the time domain we would see beats 1ms apart with little voltage in between. The minimum sampling frequency would follow the Nyquist theorem and be 2.01MHz. However, what would the minimum sampling duration (i.e. number of samples needed) be to make sure the signal is fully captured?

If we take just a few samples between the beats, there would be little signal to see. I assume it would be the beat period (1ms) because this is when the pattern in the time domain starts repeating itself.

Am I correct and is there a general rule?

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    \$\begingroup\$ If you are interested only in the beat notes, then undersampling would be an effective solution here. en.wikipedia.org/wiki/Undersampling \$\endgroup\$ Aug 24, 2021 at 9:13
  • \$\begingroup\$ Are you looking for a theoretical answer or a practical one? In theory, if your samples are perfect (no noise in either voltage or time), you can determine the amplitude and phase of a known-frequency sine wave from just two samples. Since you have a superposition of 6 sine waves (12 unknowns), you'll need a minimum of 12 samples. \$\endgroup\$
    – Dave Tweed
    Aug 24, 2021 at 11:40
  • \$\begingroup\$ I am looking for a theoretical answer with perfect samples. But if I sample at say 2.5MHz and take 12 samples, this would be over in 4.8us. What if this 4.8us sampling happens between the beats where there is hardly any signal, how would I reconstruct the signal then? I may be totally wrong but I imagine your example would only work if the 12 samples were spread out differently in time, perhaps not at regular intervals. \$\endgroup\$
    – Hyp
    Aug 24, 2021 at 12:22
  • \$\begingroup\$ "Hardly any" is not zero. There's still enough information to reconstruct the signal. And it's true that the samples don't have to be evenly spaced, although there are pathological cases that must be avoided. \$\endgroup\$
    – Dave Tweed
    Aug 24, 2021 at 12:54
  • \$\begingroup\$ Right, so what you are saying is that even within a few us of sampling I should be able to reconstruct the signal perfectly. In reality this would not be possible as the low signal would get drowned out in noise, I suppose. \$\endgroup\$
    – Hyp
    Aug 30, 2021 at 6:26

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The minimum sampling frequency would follow the Nyquist theorem and be 2.01MHz.

The bandwidth of the signal you've described is only 1.000 MHz to 1.005 MHz, or 5 kHz, so if you want to be really parsimonious, your sample rate only needs to exceed 10 kHz.

However, this assumes that your signal is in this known bandwidth. When you come to reconstruct your 10 kHz samples, if you plug into the reconstruction algorithm that your original signal occupies (for instance) the 5 kHz from 12.300 MHz to 12.305 MHz, then it will obediently reconstruct a signal in that bandwidth.

This process is known as undersampling, and if done correctly can save you a lot of samples and processing. If it's done incorrectly, for instance if your original '5 kHz' bandwidth signal also had some energy outside that range, the undersampling would alias it back into your samples, and corrupt the reconstruction.

In practice, the Nyquist criterion is an absolute minimum, and you will usually want to exceed it by some margin, so that the filters you are going to use to define your input bandwidth are actually realisable. Given your signal, I'd be inclined to sample it at a minimum of 5 MHz, using a low pass anti-alias filter that was flat to a bit more than 1 MHz, rolled off around 2 MHz, and was fully into the stopband by 4 MHz.

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  • \$\begingroup\$ dsp.stackexchange.com/questions/2857/… \$\endgroup\$
    – Hyp
    Aug 24, 2021 at 10:11
  • \$\begingroup\$ Thanks, it is a neat explanation. My focus was rather on the minimum number of samples / time needed. I have also found the link in the comment above. If an infinite number of samples is needed for a perfect reconstruction, what happens if I set the sampling window time to cover the longest repetition period (1ms derived from 1kHz beats in my example). And will the reconstruction be fundamentally different if I sample for less than 1ms? \$\endgroup\$
    – Hyp
    Aug 24, 2021 at 10:59

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