The name Nyquist–Shannon sampling theorem honors Harry Nyquist (1889-1976) and Claude Shannon (1916-2001). Shannon built on the earlier work of Nyquist to develop what became known as the Nyquist-Shannon theorem.
The theorem was also discovered independently by E. T. Whittaker, by Vladimir Kotelnikov, and by others. and so is sometimes known as Nyquist–Shannon–Kotelnikov, Whittaker–Shannon–Kotelnikov, Whittaker–Nyquist–Kotelnikov–Shannon and also the cardinal theorem of interpolation.
Nyquist theory is generally understood, but this understanding usually relates to time sampling and the conversion to the frequency domain. Rotational order analysis and the effect of the Nyquist frequency are, however, less well understood.
Most engineers will understand the basic principle that Harry Nyquist defined and that was further developed by Claude Shannon. To re-create the frequency of interest you need at least two samples. As engineers experienced in real-world engineering know, more may well be required than that minimum.
However, when analysing in the order domain, how many samples are required? For time domain analysis and for synchronous analysis?
Order analysis can be computed from data in the time domain or data in the synchronous domain.
So what is the relationship between the synchronous sample rate and the highest order that can be analysed? And what is the relationship between the time sample rate and the highest order that can be analysed?
What is the maximum order I can analyse with synchronous data?
Where the highest order is Omax and where N is the synchronous sample rate, in samples per revolution
So Omax = N/2
So the highest order that can be analysed is the synchronous sample rate divided by two.
What is the maximum order I can analyse with time based data?
Where the highest order is Omax and where S is the time sample rate, that is samples per second and where R is the shaft speed in revolutions per second, effectively RPM/60
So Omax = S/(2*R)
So the highest order that can be analysed is the time sample rate divided by twice the revolutions per second.
Note the appearance of the twice factor, in both equations. It is not a coincidence that this factor is the same as the Nyquist value as discussed earlier. It is in fact the same relationship.
James Wren
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