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I am trying to understand how to evaluate this equation in the context of acceleration data which contain engine orders

$a^{f_{e}^{crit}}(f)=\sum_{o}^{K}A^{o,f_{e}^{crit}}\mathscr{F}(cos(2\pi \cdot f_{e}^{crit} \cdot o \cdot t))$

$a^{f_{e}^{crit}}$ is the acceleration, $f_{e}^{crit}$ is the critical engine speed, $A^{f_{e}^{crit}}$ is the acceleration expressed as a complex number and $o=0.5,1,1.5,....$ are the engine orders

My confusion arises when I try to understand how it is possible to sum the time histories of the engine orders and then apply a fourier transform to frequency domain. I am not actually sure is it possible to have a time history of an engine order...

Any clarifications will be appreciated. Thanks!

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  • $\begingroup$ What do you mean by "engine order"? And is this acceleration signal the vehicle acceleration, or is it an accelerometer mounted to an engine block that's sensing vibration? $\endgroup$
    – TimWescott
    Jan 11, 2020 at 19:57
  • $\begingroup$ The engine orders are the harmonics of vibration described by N multiples of engine speed. It is an accelerometer mounted on the engine $\endgroup$
    – shoggananna
    Jan 12, 2020 at 17:56
  • $\begingroup$ That looks like a very peculiar way to express a fairly standard inverse Fourier transform, very possibly by an author who does not want to introduce the student to Dirac impulses. I'm also pretty sure that it's erroneous (because $\mathcal{F}^{-1} \left \lbrace i \mathcal{F} \left \lbrace \cos \omega t \right \rbrace \right \rbrace \ne \sin \omega t$). Can you reference the text out of which it comes? $\endgroup$
    – TimWescott
    Jan 12, 2020 at 18:22
  • $\begingroup$ You will find it on pg15 eqn 2.36. The description starts on pg14 via this link...liu.diva-portal.org/smash/… $\endgroup$
    – shoggananna
    Jan 12, 2020 at 18:51
  • $\begingroup$ It's a student thesis. Even given that it's a Master's thesis, it's by some guy who doesn't have to get all the signal-processing 'i's dotted and 't's crossed. I think you just need to accept that the guy is -- probably unknowingly -- playing fast and loose with the math. He goes off the rails at equation 2.37, but he makes complimentary mistakes going from 2.37 to 2.38, so the final answer is correct. $\endgroup$
    – TimWescott
    Jan 12, 2020 at 19:34

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