magnitude of acceleration FRFs
magnitude of acceleration FRFs
(OP)
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
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






RE: magnitude of acceleration FRFs
Of course, if your vibrations are composed of many different frequencies (not sinoid vibration) you may need another approach.
Gunnar Englund
www.gke.org
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RE: magnitude of acceleration FRFs
thank you very much
RE: magnitude of acceleration FRFs
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RE: magnitude of acceleration FRFs
(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,
RE: magnitude of acceleration FRFs
To shift from PSDd to PSDf you have to multiply by the square of the FRF. Indeed the unit is g2/Hz for PSD