magnitude of acceleration FRFs

[MariaMV]

Hi all,
I have got the magnitude of acceleration FRFs of a real test and no complex numbers. i need to have displacement FRFs to calculate PSD of the response displacement. in order to obtain the disp. FRFs, the acc. FRFs need to be divided by (w^2), w is the related circular frequency.
the question is that is it possible to get disp. FRFs of acc. FRFs without having complex numbers of acc FRFs? and also after that disp. PSD? because displacement power spectral density (PSD) is calculated by disp. FRFs times force PSD.

PSDd=FRFd*PSDf

Cheers

 

[Skogsgurra]

Don't see any need for real and imaginary components if all you need to do is calculate speed or displacement from acceleration. Just remember to stay with the same representation (RMS all the way) and to use consistent dimensions, SI, preferably.

Of course, if your vibrations are composed of many different frequencies (not sinoid vibration) you may need another approach.

Gunnar Englund

http://www.gke.org

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Half full - Half empty? I don't mind. It's what in it that counts.

 

[MariaMV]

Thank you for your response, actually i just found out that the formula i was using for converting acc. FRFs to disp. FRFs might be wrong. do you have any idea what is the right relationship between these two??

thank you very much

 

[IRstuff]

Acceleration is the 2nd derivative of displacement. Therefore, their magnitudes are related by (2*pi*f)²

TTFN
faq731-376

 

[MariaMV]

Thank you IRstuff, my question was about converting from FRF acc to FRF disp.
(2*pi*f)^2 can be still applied in this domain. as far as i know this relationship is for frequency domain of the acc and disp. so i was wondering if i still could use that for FRFs???

Best regards,

 

[amanuensis]

there is another error.
To shift from PSDd to PSDf you have to multiply by the square of the FRF. Indeed the unit is g2/Hz for PSD