# Why doesn't the sampling theorem work in this case?

In this example, under "Discussion", we can see that the Nyquist sample rate will result in samples with the value zero only. But why is this? I thought the sampling theorem said that if you sample with the Nyquist sample rate or higher, then the original signal can be exactly recovered? But in this case it can't? So does that mean the sample theorem is not actually valid?

• Welcome to DSP.SE. Please register for the site, as you should find it quite helpful :) – geometrikal Sep 5 '15 at 12:12
• Nyquist sampling applies to bandlimited signals. Any finite length signal containing a finite amplitude sinusoid at Fs/2 is not bandlmited to below Fs/2, and thus not close enough to being bandlimited for practical purposes. – hotpaw2 Sep 6 '15 at 19:45

The formula given is

$$F_s > 2F_{\mathrm{max}}$$

not

$$F_s \ge 2F_{\mathrm{max}}$$

otherwise it is possible that you could be sampling exactly on the zero crossing points. In practice you would choose sometime a bit higher. For example, CD audio uses 44100 Hz to represent signals up to 20000 Hz.

• I tried to sample a 1Hz sin wave with Fs = 3 Hz. That didn't work quiet as expected. Fs=2Hz was just ok if I had my sin wave shifted in order to match its max and min. I am still trying to understand this theorem ... – Eduardo Reis Oct 4 '18 at 14:31
• @EduardoReis In what way did it not work out? – geometrikal Oct 4 '18 at 14:35
• I just added this notebook with my experiment. github.com/eduardo4jesus/notebooks/blob/master/… – Eduardo Reis Oct 4 '18 at 14:56