# Sampling at subnyqvist frequencies

I went through this material on Sampling theorem. http://www.wescottdesign.com/articles/Sampling/sampling.html

It uses a powerline signal example and concludes that if a signal is cyclic and repeating over a period N then you could sample the signal at lesser than N samples per second.In the example he shows for a 60Hz power signal he can sample and recover the signal by sampling at 19 Hz. He uses an expression

Ts = (M + P/N )*1/F where M>0 and N>0 where p is not equal to 0. I haven't understood it. Can anybody here tell me how it is done.

The point in the article is that you can sample the signal at lower frequencies than 60Hz by subsampling. Consider the following two scenarios of sampling the 60Hz signal:

• You sample like Nyquist wanted, at freq $\geq 120\mbox{Hz}$. In one period of your 60Hz signal, or $T=16.67\mbox{ms}$, you will have a correct representation of your signal via sampling.

• You sample at a frequency even lower than the $60\mbox{Hz}$ signal freq. The trick is to sample at frequencies that do not divide the main frequency, which means you cannot use $2, 3, 6, 10, 12, 15, 20, \mbox{ or } 30 \mbox{Hz}$ frequencies, because then you'd always get the same point of the signal.

Let's say you sample at $19\mbox{Hz}$ like in the article. The whole point of this is to save the points on your graph, and get the desired signal point-by-point. So you start sampling your signal, get one point, then whole 3 periods of the signal will pass, and you get a second point, which you put on the graph, and so on until you have enough.

How many points do we need? The Nyquist theorem you stated tells us the number of points we need to be sure we got 100% of the signal. $$T_s = (M + \frac{P}{N})\frac{1}{F},$$ where $F$ is our signal frequency of $60\mbox{Hz}$, $T_s = 1/F_s = 1/19\mbox{s} = 52.63\mbox{ms}$ is our sampling period, and $M, N, P$ are integers where $N$ and $P$ must have no common factors. If we try to get such integers, we can see that for $N = 19, M = 3, \mbox{ and } P = 3$ our equation holds. So, we need 19 points for our graph, in order to have the whole signal. This means we need to wait 19 times our sampling period to get our total time of sampling, so we get $$T_{tot} = 19\cdot 52.63\mbox{ms} = 999.97\mbox{ms}$$

So we see that sampling is possible at even lower frequencies, but at much longer time :) Hope this clarified your question! :)

• ok. That helped me understand it well.Thank you. I have a doubt. why don't we use this method normally in DSP. Is it because synchronisation is an issue and you will lose data if you do this because of the time taken. Any good references on this method and usage. That would be very helpful. – Karan Talasila Jun 2 '13 at 17:46
• Well I could name two reasons, the first being what you said, loss of data for real-time sampling, and the second the huge amount of time you need. In this example, we need 1s to sample the signal instead of the regular 16.67ms, so 60 times longer sampling time. In any practical system this is really unacceptable :) – Vidak Jun 2 '13 at 18:58
• How do i see on simulation about this subsampling. do i need to superimpose a 19Hz wave on a 60Hz wave and find the intersection points and reconstruct the wave. I want to see it in a figure.Any suggestions? – Karan Talasila Jun 3 '13 at 13:50

You only need to sample at twice the signal bandwidth to satisfy Nyquist, not twice the maximum frequency. So if the powerline signal has a spectrum which only extends a few Hz either size of 60 Hz then the sampling rate can be significantly < 120 Hz. Of course if the signal contains any significant energy at harmonics of 60 Hz (120 Hz, 180 Hz, etc) then the sampling rate will need to be much higher.

• I am aware of the bandwidth thing. But the wave he shows has harmonics of the order 5. There is no mention of bandwidth. – Karan Talasila May 31 '13 at 17:53