This place has been great in helping me to understand my online image and signals class!

I've moved on to another question and wanted to see if I'm correctly grasping the subject. The next question is as follows:

If we sample a pure sine signal at "sam" samples per second, and there are 8 samples per cycle of the signal, and we have not undersampled the signal, what is the signal's frequency in cycles per second?

So using the Nyquist-Shannon theorem I know I have a Nyquist rate of $8\frac{samples}{cycle}$ so I believe to calculate the signals frequency in cycles per second I need to do the following:

$8\frac{sam}{cycle} = 2\,f_{max}$

$8\frac{sam}{cycle} = 2\left(\frac{sam}{sec}\right)$

$4\frac{sam}{cycle} = \frac{sam}{sec}$



1 Answer 1


There are sam samples per second and 8 times less pure sine cycles per second. This means that there are sam/8 pure sine cycles per second.

What shannon has to do with that? Probably you mean that Niquitz allows you to reduce sampling, make it 4 times more rarely, as low as sam/4 samples per second. That would sample twice per sine period.

enter image description here

  • $\begingroup$ Sorry, I'm a little confused. Are you saying shannon has nothing to do with this question or just being rhetorical? $\endgroup$
    – nullByteMe
    Commented Mar 20, 2016 at 17:30
  • $\begingroup$ There's something wrong with your statement that according to the sampling theorem one should sample at sam/16. Since sampling at rate sam means 8 samples per period, sampling at sam/16 would mean one sample every two periods, which is not enough. $\endgroup$
    – Matt L.
    Commented Mar 20, 2016 at 18:21
  • $\begingroup$ @MattL. so is it just sam/8? $\endgroup$
    – nullByteMe
    Commented Mar 20, 2016 at 18:32
  • $\begingroup$ @free_mind: No, that would mean just 1 sample per period. You need more than 2 samples per period to be able to reconstruct the sinusoid, so according to Shannon, your sampling rate must satisfy fs > sam/4. $\endgroup$
    – Matt L.
    Commented Mar 20, 2016 at 18:36
  • $\begingroup$ @MattL. this is all new to me, so sorry for my ignorance here, but am I on the right track in the statements I made in my question? Am I trying to calculate the signal's frequency in cycles per second correctly? $\endgroup$
    – nullByteMe
    Commented Mar 20, 2016 at 18:39

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