Puzzled by Nyquist frequency

Question

Say I have a 1kHz sine, so no higher harmonics, then I need to sample it at least at 2kHz to be able to reconstruct it.
But if I sample at 2kHz, but all my samples are on the zero-crossing, then my sampled signal doesn't show a sine at all, rather the ECG of a deceased patient. How can that be explained?

This can be expanded to higher sampling frequencies too. If I sample a more complex waveform at 10kHz, I should at least get the first 5 harmonics, but if the waveform is such that the samples are each time zero, then again we get nothing. This isn't far-fetched, it's perfectly possible for a rectangle wave with a duty cycle < 10%.

So why is it that the Nyquist-Shannon criterion seems to be invalid here?

Answer 1

You actually need just over 2 kHz sampling rate to sample 1 kHz sine waves properly. It's $$ f_N < f_S / 2 $$ not $$ f_N \le f_S / 2 $$

P.S. If you took your signal into complex space, where a sinusoid is of the form $$v(t) = Ae^{j(2 \pi f t - \theta)} = A(\cos(2 \pi f t - \theta) + j \sin(2 \pi f t - \theta))$$ where t is time, A is amplitude, f is frequency, and θ is phase offset, $$ f_N = f_S / 2 $$ is the point where the frequency "folds over", i.e. you cannot distinguish f from -f. Further increases in frequency will appear, after sampling, to have the sampling frequency subtracted from them, in the case of a pure sinusoid.

Non-Sinusoids

For the case of a square wave at 1 kHz with a duty cycle less than or equal to 10% which is sampled at 10 kHz, you are misunderstanding the input.

First you would need to decompose your waveform into a Fourier series to figure out what the amplitudes of the component harmonics are. You will probably be surprised that the harmonics for this signal are quite large past 5 kHz! (The rule of thumb of third harmonic being 1/3 as strong as the fundamental, and 5th being 1/5 of fundamental, only applies to 50% duty cycle square waves.)

The rule of thumb for a communications signal is that your complex bandwidth is the same as the inverse of the time of your smallest pulse, so in this case you're looking at a 10 kHz bandwidth minimum (-5 kHz to 5 kHz) for a 10% duty cycle with the fundamental at 1 kHz (i.e. 10 kbps).

So what will ruin you is that these strong higher-order harmonics will fold over and interfere (constructively or destructively) with your in-band harmonics, so it's perfectly expected that you might not get a good sampling because so much information is outside the Nyquist band.

Answer 2

Mike explains it well: it's the aliasing which makes the harmonics disappear in the sampled signal, the folding of the higher frequencies from \$F_S + f\$ to \$F_S - f\$.
When working with sampled signals you always have to make sure to filter out anything above \$F_S / 2\$.

In this spectrum the blue part is your base band signal's spectrum from \$-F_S / 2\$ to \$F_S / 2\$. (See this question about negative frequencies).
Note that this spectrum is repeated around every multiple of \$F_S\$. In this example there's no problem; the original signal is separated from the images, and can be reconstructed.

In this example (only positive frequencies shown) we can see that the base band signal extends to beyond \$F_S / 2\$. Due to the folding aliases overlap with our base signal, and there's no way we can filter them out again. That's why you need a (sharp) low-pass filter.

Now you may say that the pulse will look completely different after low-pass filtering, and that's right, but if you don't want that you've chosen your sample frequency too low. (For a discontinuous signal like the pulse, which has an infinite spectrum, you will always have distortion, whatever your \$F_S\$). Remember that you can reconstruct the signal only for frequencies smaller than \$F_S / 2\$.

Answer 3

Theorem is ok. Your signal should NOT contain frequencies equal or higher than half of sampling rate, accordint to Nyquist. Shannon probably allowes it, but it is his version of theorem, which probably causes ambiguity at critical frequency.

Edit (Re: downvoting for short answer ?): I dont see necessity to explain the sampling method itself. The question is about confusion "is critical frequency included into band or it is not", and if wording of theorem by Shannon contains fault. It actually does (as I see it in the world wiki). Or most likely the wiki authors cited his word inprecisely. And by the way, there are 4 independent authors in 20th century of this very theorem, so the confusion of anyone learning the idea from random sources can get worse.

Answer 4

If you have 2 samples on a sine wave of \$N Hz\$, and they occur at the zero-crossings, at \$\frac{1}{2}N\$ and \$1N\$ then you can determine the frequency of the signal by the time between the two samples.

\$f=\dfrac{1}{2t}\$

Where \$f\$ is the frequency and \$t\$ is the time between two zero-crossing samples.

But according to Wikipedia:

In essence, the theorem shows that a bandlimited analog signal that has been sampled can be perfectly reconstructed from an infinite sequence of samples if the sampling rate exceeds 2B samples per second, where B is the highest frequency in the original signal.

So a sampling frequency of twice the frequency is wrong - it should be just over twice the frequency. That way successive samples capture slightly different portions of the waveform.

Answer 5

When sampling at a particular rate F, every frequency component f will generate aliases of the form kF+f and kF-f for all integer values of k. In common usage, there are no frequency components above F/2 when the signal is sampled, so the only components in the range 0 to F/2 will be those that were present in the original signal. After sampling, there will be signal components above F/2 (generated as aliases of those below). The most troublesome of these for any frequency f in the original signal will be the one at frequency F-f.

Note that as frequency f approaches F/2 from below, the first alias frequency will approach F/2 from above. If the input contains a signal at frequency F/2-0.01Hz, there will be an alias at frequency F/2+0.01Hz--just 0.02Hz above it. Separating the original and alias signals will be theoretically possible, but in practice difficult. The sampled waveform will appear as the sum of two equal-strength waves of nearly equal frequency. As such, its amplitude will appear to change with the relative phase of the higher-frequency waves. In the case where the input frequency is exactly F/2, the alias frequency will also be exactly F/2. Since there will be no frequency separation at all between the original and alias, separation will be impossible. The phase relationship between the original and aliased signals will determine the amplitude of the resulting signal. If the original and aliased signals are 180 degrees out of phase, the original and aliased signals will precisely cancel.