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

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Susan 2023-3-14

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评论： Susan 2023-3-16

采纳的回答： Walter Roberson

waveform.mat

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!

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采纳的回答

Walter Roberson 2023-3-14

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https://ww2.mathworks.cn/matlabcentral/answers/1928875-how-to-calculate-the-duty-cycle-of-a-time-signal-in-a-for-loop#answer_1193225

在 MATLAB Online 中打开

waveform.mat

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

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Walter Roberson 2023-3-15

在 MATLAB Online 中打开

waveform.mat

To be honest, I guessed.

I tried without the movmedian() first, but got values around 54%, which I realized was because the data just happens to have a lot of zero-ish values.

Oh wait, I calculate the wrong thing, don't I? You should not be interested in the portion that is near 0, you should be interested in the portion that is not near 0 -- abs(mr)>0.01 rather than abs(mr)<0.01

The basic idea was that the signal is changing rapidly, and that putting it through a filter should remove a bunch of values that happen to be near 0 in the rapidly changing regions. But I realize now that can be flawed depending on how you happen to do the filtering, especially if the data just happens to be fairly symmetrical.

So maybe movmax() would be better... no, that is still only about 54%

load waveform
wr = real(waveform);
wi = imag(waveform);
mr = abs(wr(3:end) - wr(1:end-2)) > 0.001;
mi = abs(wi(3:end) - wi(1:end-2)) > 0.001;
mean(mr) * 100
ans = 72.0809
mean(mi) * 100
ans = 72.0698

Ah, that works with your original file. I will try your newer file in a moment.

Walter Roberson 2023-3-15

在 MATLAB Online 中打开

waveform.mat

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 2023-3-16

@Walter Roberson I see. Thanks again for your help. Appreciate it.

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