How to calculate the duty cycle of a time signal in a for loop?

15 vues (au cours des 30 derniers jours)
Susan le 14 Mar 2023
Commenté : Susan le 16 Mar 2023
Hi All,
I have multiple time signals and like to compute their duty cycles automatically within a "for loop". For example the duty cycle of the following signal is 74.29%. The signal is attached and the sample rate is 15360000.
Assume I have a bunch of these signals. How can I calculate the ratio of the pulse width (duration of the on state) to the pulse period (the total duration of an on-and-off state) for each signal in a for loop?
Thanks in advance!

Réponse acceptée

Walter Roberson
Walter Roberson le 14 Mar 2023
load waveform
wr = real(waveform);
wi = imag(waveform);
subplot(2,1,1); plot(wr); ylabel('real');
subplot(2,1,2); plot(wi); ylabel('imag');
mr = movmedian(wr, 5);
mi = movmedian(wi, 5);
mean(abs(mr)<0.01) * 100
ans = 74.2301
mean(abs(mi)<0.01) * 100
ans = 74.4255
  9 commentaires
Walter Roberson
Walter Roberson le 15 Mar 2023
We can tell from the graphs that the only part of the signal this is "repeatable" is the zeros.
You can do things like
%newer file
load waveform
wr = real(waveform(:).');
wi = imag(waveform(:).');
maskr = abs(wr) < 1e-3;
stopsr = strfind([maskr 0], [1 1 0]) + 2;
mean(~maskr(stopsr(2):stopsr(end-1)-1)) * 100
ans = 77.2091
maski = abs(wr) < 1e-3;
stopsi = strfind([maski 0], [1 1 0]) + 2;
mean(~maski(stopsi(2):stopsi(end-1)-1)) * 100
ans = 77.2091
This measures for the signal starting from end of the second stretch of zeros (so skipping the first pulse). But you can see it made little difference (gave a lower duty cycle in fact.)
To go much beyond that you would need a more explicit definition of what should be skipped.
Susan le 16 Mar 2023
@Walter Roberson I see. Thanks again for your help. Appreciate it.

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