If one has two signals (say, two acceleromters mounted perpendicularly) and a piece-wise resultant acceleration signal is determined, it appears that frequency content information cannot be determined on the resultant signal using an FFT.

  1. Can someone comment on why this is so?

It may be that the process of determining the resultant converts the negative content in the individual signals through the process $$r(i) = \sqrt{x(i)^2+y(i)^2}$$ which "loses" information (not sure if that is the best way to describe it). This same effect can be seen if one takes a single acceleration signal and processes the FFT of the absolute value of that signal.

  1. Is there any way of still being able to assess frequency content of a resultant vector?
  2. Is it even worth pursuing?
  • $\begingroup$ As a hack, try multiplying one channel by $\sqrt(-1)$ before adding, add the 2 channels and then take the complex DFT. $\endgroup$ – Stanley Pawlukiewicz Nov 17 '17 at 19:17
  • $\begingroup$ Interesting point. Am I correct in assuming that the point of multiplying one of the signals by $i$ is because the signals are orthogonal, and doing this multiplication rotates the signal through 90 degrees in the complex plane, so that the two vectors are now co-linear? Hence the ability to add the two signals rather than determining the conventional resultant, and, in so doing, preserve the signs. $\endgroup$ – Terrance Frangakis Nov 21 '17 at 12:07
  • $\begingroup$ Sometimes you try something, and if it works, one needs to explain it. seems rsinable $\endgroup$ – Stanley Pawlukiewicz Nov 21 '17 at 23:18
  • $\begingroup$ Unfortunately,if it is useful, complex varible don't generalize the same way to 3 dimensions. $\endgroup$ – Stanley Pawlukiewicz Nov 22 '17 at 0:09

First, you should note that the FFT refers to an algorithm for the fast computation of the DFT taking advantage of its symmetry.

The signals you have are:

enter image description here

where $\mathbf{a}_y$ and $\mathbf{a}_x$ can both have negative a positive values depending on the direction along the $x$ and $y$-axis. These signals are the output of your accelerometers.

What you call "a piece-wise resultant acceleration", calculated like $$r(i) = \sqrt{x(i)^2+y(i)^2},$$ is just $\rVert \mathbf a\lVert$, the magnitude of a vector quantity $\mathbf{a}$, which doesn't show the direction.

From NASA webpage, about vector components:


In conclusion, performing a DFT on your vector quantities $\mathbf{a}_x$ and $\mathbf{a}_y$ will give you frequency information both for about magnitude and the direction together, while performing a DFT on your scalar vector $\rVert \mathbf a\lVert$ will just give you the frequency domain of the magnitude, so yes, you are loosing information here.


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