Using MATLAB to Integrate Data that has a Sinusoidal Signature

Hello,

I am currently working with datasets that yield angular rate in radians/sec. The plots from this rate data are sinusoidal, and I need to find the angles in radians. I attempted to use the cumtrapz MATLAB function, and it integrates (I can retrieve the original signal via delta_integrated_data/delta_time); however, I don't think it is integrating correctly based on the nature of the signal, i.e., a signal with amplitude, phase, and frequency. I was expecting something that resembled -cosine +C since the original signal resembled sine. I attached a figure of the angular rate and a figure of the integrated angular rate. Is there an alternate approach that could be taken? Would using the symbolic toolbox somehow be better? Or, am I overthinking this matter? Feedback is much appreciated. Thanks!

Answer 1

[Star Strider]

The ‘Angular Rate’ signal is offset from zero (the mean appears to be 0.06), and cumtrapz is integrating that constant offset (as a constant-of-integration) as well.

I would subtract the mean of your ‘Angular Rate’ signal from your ‘Angular Rate’ signal before integrating it.

[Walter Roberson]

The signal is approximately

 sin((pi*x)/1000)/20 + 3/50

which is going to integrate to

(3*x)/50 - (50*cos((pi*x)/1000))/pi

The 3/50*x is going to dominate the graph.

[C. Robinson]

Thanks! Removing the mean certainly worked. I am getting a signal that is much more expected; please see the attached plots. I plan on putting these signals (angular rate and angle) through FFTs. Would removing the mean affect that (i.e., the DC part of the signal)? Would I have to compensate for the removal of the mean?

[Star Strider]

My pleasure.

I usually recommend subtracting the mean from a signal (if there is a significant constant or d-c offset) before doing the fft (link) because it does not affect the other frequency components (it actually makes them more visible in the plot). You are not losing data, since you can easily recover (or store) the mean of your signal. So no compensation or other corrections are necessary.

[C. Robinson]

That makes sense. Thanks again for the help! Your feedback is very clear.

[Star Strider]

As always, my pleasure!