A first approximation is taking the difference between two successive values and dividing by the difference of times between them. There are other expressions taking more observations into account, see nearly any book on numerical analysis. Of course, their validity depends on the difference of times. If it is too big, the approximation will be bad.
https://blog.prosig.com/2010/12/16/methods-of-conversion-between-acceleration-velocity-and-displacement/
https://dsp.stackexchange.com/questions/51621/acceleration-time-series-to-velocity-conversion-using-dfts
I think it is quite simple: Analogous to phyics, the velocity is the difference of a variable from one period to the next one, and the acceleraion is the difference of the difference (Example: If xt is the series of original absolute data, dxt=xt-xt-1 is the velocity and ddxt=dxt-dxt-1 the acceleration series. If the time periods are different - e.g. because of missing data -, you may proceed according to Guy's proposal.
https://dsp.stackexchange.com/questions/51621/acceleration-time-series-to-velocity-conversion-using-dfts
https://dsp.stackexchange.com/questions/51621/acceleration-time-series-to-velocity-conversion-using-dfts
https://dsp.stackexchange.com/questions/51621/acceleration-time-series-to-velocity-conversion-using-dfts
Anton Rainer Thank you for the answer.
What is the formula for higher order differentiation?
As time series are not continuous, it is not a case for differentiation, but one can, of course, dalculate differences of any order. If e.g. we say dx and ddx in my first answer are d1 (first difference) and d2 (second difference), then d3xt = d2xt-d2xt-1 etc. One could do that for any order, but I doubt if differences of order above 3 are sensible (for interpretation).
d2=d1-d1-1=(xt-xt-1)-(xt-1-xt-2)=xt-2xt-1+xt-2. If you continue this calculations for d3, d4 etc., you will arrive at results that could remind you on algebra in school.
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