I want to measure and compare the phase response of two identical microphones. To do this, I have a pressure chamber, which allows to apply the same pressure to both microphones.

I apply a sweeped sine wave to the chamber to measure the freq. range of interest. As Data I have the applied pressure and the measured pressure of the two microphones.

Right know I do the following:

fftChamber = fft(chamber_sig)
fftMic1 = fft(mic1_sig)
fftMic2 = fft(mic2_sig)
G1 = fftMic1./fftChamber
G2 = fftMic2./fftChamber
phaseResponse1 = angle(G1)
phaseResponse2 = angle(G2)

This method gives me the most reasonable result, but I'm not sure why dividing should be right or appropriate, as this is not really a transfer-function?

  • 1
    $\begingroup$ how is this not a transfer function? $\endgroup$ Jun 28 '16 at 10:40
  • $\begingroup$ well it somehow is. $\endgroup$ Jun 28 '16 at 12:20
  • 1
    $\begingroup$ it pretty much is the very definition of a transfer function $\endgroup$ Jun 28 '16 at 12:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.