0
$\begingroup$

Correct me please if I am wrong...

I have just started working in the domain of signal processing using the Fourier Transform ($\mathbf{FFT}$) and Power Spectral Density ($\mathbf{PSD}$). I am trying to adjust the Frequency Response Function ($\mathbf{FRF}$) so that response $Y$ computed using relation:

$$\mathbf{FFT}(Y)=\mathbf{FRF} \cdot \mathbf{FFT}(X)$$

match with actual response. $X$ is the input.

I was successful in determining $\mathbf{FRF}$ of system concerned. Then I realized that the input signal used ($X$) is a random one. Literature suggested that instead of above approach $\mathbf{FRF}$ should be obtained by using power spectral density approach, i.e.:

$$\mathbf{PSD}(Y)=|\mathbf{FRF}|^2 \cdot \mathbf{PSD}(X)$$

I adjusted the Frequency Response Function to match actual $\mathbf{PSD}(Y)$ with $\mathbf{PSD}(Y)$ computed by $|\mathbf{FRF}|^2 \cdot \mathbf{PSD}(X)$. FRF obtained by this approach is much different from FRF obtained by FFT approach.

PSD was determined by FFT of autocorrelation of a signal, as it was simple for me to code. Signal is collected once, one realization.

Thus my questions are:

  1. Do $$\mathbf{FFT}(Y)=\mathbf{FRF} \cdot \mathbf{FFT}(X)$$and $$\mathbf{PSD}(Y)=|\mathbf{FRF}|^2 \cdot \mathbf{PSD}(X)$$ give same FRF for same input $X$ and response $Y$?
  2. If yes then what could have gone wrong in my case?
  3. If no then what are reasons for difference?
  4. Is there any connection between these to relations (FFT and PSD).

Dinesh Zanwar,

Nagpur, India

$\endgroup$
  • $\begingroup$ Please note that FFT usually stands for Fast Fourier Transform which is an efficient algorithm implementing the DFT (Discrete Fourier Transform). So the FFT is something much more specific than the general Fourier transform. $\endgroup$ – Matt L. Aug 24 '14 at 10:44
  • $\begingroup$ Note that FRF is complex. Therefore for any complex number $x$, $|x|^2 = xx* \neq xx$. $\endgroup$ – learner Aug 25 '14 at 3:29
  • $\begingroup$ You write in your question that X is random and that you have one realization. For random inputs you have to use the PSD approach. However the PSD approach requires multiple, independent realizations. Otherwise it will not work, as you need to average over the realizations. $\endgroup$ – snowflake Feb 27 '16 at 0:43

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.