Why is the Nyquist frequncy on the bode diagrams the same as the sampling frequency?
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
I notice that when I create a discrete dynamical system with a given sampling frequency, and I plot the bode diagrams, the phase and amplitude responses are plotted up to the samping frequency. On the documentation page for the bode command, it calls this upper limit on the frequency the Nyquist frequency. Even in their example, this Nyquist frequency is exactly the same as the sampling frequency. They create a system with a sampling time of 0.5 seconds (2 Hz, or 6.28 rad/s) and the resulting bode diagrams go up to 6.28 rad/s. So why is this the case, when I know that the Nyquist frequency is defined as half of the sampling frequency?
2 Kommentare
Akzeptierte Antwort
Voss
am 13 Nov. 2024
"sampling time of 0.5 seconds (2 Hz, or 6.28 rad/s)"
There's the problem. 2 Hz is 2*2*pi rad/s = 4*pi rad/s, which is approx 12.56 rad/s (not 6.28 rad/s).
Thus the corresponding Nyquist frequency is half of that or approx 6.28 rad/s.
2 Kommentare
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Time and Frequency Domain Analysis finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!