I have an acceleration signal
I can integrate it twice using (Orgin 2021 and Matlab) in order to find out displacement signals.
but need to know if what I did is true or wrong and give me a better explanation.
The result of the displacement was as shown in figure (1).
I think it look like a wave but had an anon-zero mean line. the mean line tends according to that equation (y = -520.77x - 0.5844) ,i think that comes from the integrations.
when I remove these terms from the signal result to get (zero-mean line) the wave looks as shown in figure (2).
Hi Mohamed rezk Omar ,
the displacement in fig. 2 and the acceleration in fig. 1 fit well.
The general problem is that, from the mathematical point of view, the result of an integration includes an arbitrary constant. From the physical viewpoint, this is correct because exclusively measuring the acceleration provides neither information about the initial velocity nor on the initial position. So, any constant velocity can be added to the result of the first integration, and any constant displacement can be added to the result of the second integration which produces an outcome similar to fig. 1 b.
In other words, if your measurement on a real object is limited to the acceleration, its displacement might be indeed as shown in fig. 1 b. In order to avoid this uncertainty, you had to measure the position, too, under ideal circumstances only twice, under real conditions (measument errors) in certain time intervals.
To me, the only surprise in your data is that mathematical programs usually set the (generally unknown) integration constant to zero but, if I understand rightly, your fig. 1 b is the unmodified result.
I hope I understood your question correctly.
It all depends at which point on the cycle you start your integration. If it starts at zero acceleration you get a large residual velocity. If you start at something like 100 degrees the residual velocity gets small, but it starts at the peak of displacement so there is a fixed displacement term.
This is because of your start conditions. If the starting acceleration is zero, it is at the point where the sine crosses the axis. The integration of this is a cosine, and should be at a peak, but starts at zero (because the total so far is zero), so the cosine wave for the velocity starts offset by its initial amplitude, and the centre-line of the velocity wave-form keeps that offset.
Your best action is just to subtract the velocity offset.
Erraneously, I had symbolic integration in mind in my first answer (hence my surprise) but in real applications the acceleration is not given by an equation but as a stream of measured values, of course. The attached program demonstrates the difference between the results of symbolic and numerical integration (here in its simplest form). Malcolm White 's point is illustrated by the penultimate figure, while the last figure looks similar to your figure 1 b.
The equations predict it too, depending on the boundary values (starting velocity, for instance).
The equation from integrating the acceleration might be
V=1/(aw) cos(wt) exp(-at) +Vo
a=attenuation, w =frequency, t = time
If this is set to zero at t=0, then it means Vo=-1/(aw).
By starting the velocity from a value of zero at t=0 this accidentally puts in an offset velocity Vo.
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