Can someone in simple terms why you would use Nyquist frequency limits when processing a signal? What benefit does it provide, and how does it affect the results? And how does it relate to the Nyquist rate?
-
1$\begingroup$ Can you give some explanation how your question relates to physics? Are you doing signal processing in the context of some specific physics experiment? Otherwise, this question would make more sense on Signal Processing or Electrical Engineering. $\endgroup$– The PhotonCommented May 4, 2019 at 16:22
-
$\begingroup$ @ThePhoton Yes you are right, Signal Processing is more fitting. $\endgroup$– NiklasCommented May 4, 2019 at 16:33
-
$\begingroup$ However on signal processing you might be told this question is so basic that they're not going to answer. Have you read the Wikipedia article on the Nyquist-Shannon Sampling Theorem? $\endgroup$– The PhotonCommented May 4, 2019 at 16:36
2 Answers
The Nyquist sampling theorem (sometimes called the Nyquist-Shannon sampling theorem) says, if you have a signal that is bandlimited with bandwidth $B$, then if you sample it with a sampling period $T_s$ strictly less than $1/2B$, then the original signal can be perfectly reconstructed from the samples.
We call the minimum sampling rate for ideal reconstruction, $f_N = 2B$ ($f_N$ being in samples per second and $B$ in hertz), the Nyquist limit.
If you sample a signal with a sample rate greater than the Nyquist limit, it is (in principle) possible to perfectly reconstruct the original continuous-time signal.
If you sample a signal with a sample rate below the Nyquist limit, you can not perfectly reconstruct the signal due to aliasing.
So if you want to retain "complete" information about the signal you are sampling, you must sample above the Nyquist limit.
As suggested by a comment, you should give us some context. There a several ways to answer the question. I'll provide a partial answer that is particularly relevant to physics, esp discrete periodic systems such as atoms in a solid. But there are other aspects, especially as concerns continuous systems and time-domain questions.
For a discrete periodic system, the Nyquist frequency is the highest frequency possible. There are no higher frequencies.
It's easier to visualize space rather than time :-) so let's consider spatial frequencies. Imagine a system of balls threaded on a string, with the distance between neighboring balls the same for all neighboring pairs. A low spatial frequency / long wavelength looks like a wave. As you increase the spatial frequency, the wavelength gets shorter. Continue increasing the spatial frequency until the wavelength is two times the ball spacing. At that frequency one ball is displaced "left", the next "right", the third "left" again, etc. Every other ball is displaced maximally to the left or right.
There is no higher spatial frequency. Any arbitrary waveform on the system will consist of a sum of frequencies up the maximum. There simply are no other frequencies to consider.
From a physics point of view, there is no consequence. From a math point of view, summations are limited, and algorithms to speed the calculation ("Fast Fourier Transform") can be used.
-
$\begingroup$ It's probably worth mentioning the term Brillouin zone in this answer. $\endgroup$ Commented May 4, 2019 at 16:38