# Meaning of sampling frequency?

Say there is a communication system, which can manage sampling frequency up to 4 MHz (given by software). Can the system be able to process signals with maximum frequency more than 2 MHz? Because according to sampling theorem sampling frequency > 2*maximum frequency.

• You'll have clearer understanding, and an easier time with the units, if you use the term "sampling rate" with units of 2M samples per second. The Hz designation has the implication of an implicit conversion factor of one sample per cycle. – Cedron Dawg Jun 30 '20 at 0:11

Probably no, but... maybe. The classical sampling theorem is often misstated: if a continuous signal is band-limited, it can be recovered exactly (in theory) if the sampling frequency is twice its maximum frequency.

However, there are cases where signals can be recovered above the Nyquist limit, provided that we have more information, because ambiguities may happen in the recovery:

• finite-time signal cannot be strictly band-limited, but processing can afford a small amount of distorsion,
• some band-limited signals can be recovered with about twice their bandwidth (and not just twice the max frequency).
• there are more generic sampling theorems (like: do you sample the derivative of the signal at the same time), but more involved as well.

Generally, the above possibilities are not implemented in traditional hardware or software.

Nyquist didn't say that. The Nyquist-Shannon sampling theorem says that the sampling rate asymptotically approaches twice the bandwidth of the signal, more or less no matter where the signal's center frequency is. So you care about the width of the signal, not its maximum frequency.

But where you got the expression wrong in one direction pertaining to bandwidth, you got it wrong in the other pertaining to the sampling frequency. $$f_s \ge 2 BW$$ is only true when your collection time is infinite. If you want to get your results before the universe dies, then not only is $$f_s > 2 BW$$, but for some desired do-not-exceed delay $$\tau_d$$, $$f_s > f_{min}$$, where $$f_{min} = BW + \frac{A}{\tau_d}$$.

The $$A$$ in the expression above is a bit loose, but it's going to be above 1, and it depends on other factors. These factors mostly have to do with how much aliasing you can stand, but they also have to do with how much complexity you can put into your reconstruction filter.

This is why audiophiles want music sampled at 192kHz even though the official upper range of human hearing is 20kHz. While for the most part I don't have patience with people who only want giant speaker cables that have been dipped in liquid nitrogen, they're right about the sampling rate part.

• You mean $a = \sin 2 \pi t / f$, and if you don't change the rules on me, yes. Because (in part) a continuous sine wave has a bandwidth of zero, which is much less than $2MHz$. With the correct circuitry ahead of the ADC, you could correctly sample at signal at exactly 2MHz, but not with a straight ADC. – TimWescott Jun 30 '20 at 3:50

If you have a pure tone signal that has a frequency higher than the Nyquist frequency, it doesn't "disappear", it looks like it has a different frequency (mirror image around the Nyquist). This is called aliasing. If it happens to be exactly at Nyquist, it could disappear if your sampling happens to hit the zero crossings. The same is true for a tone with a frequency the same as your sampling rate, or 1 1/2, etc. So like the other answers said, with a little extra knowledge about your signal, it may still be possible to accurately interpret your signal. This is for pure tones.