0
$\begingroup$

If I have a white noise fed into a filter what would be the output of the filter - what would you expect to see? I know that white noise has a unit power spectral density (PSD), how these thing relate to each other?

$\endgroup$
  • $\begingroup$ The definition of a white noise process is that when applied to the input of a LTI filter with transfer function $H(f)$, the output process has PSD proportional to $|H(f)|^2$. $\endgroup$ – Dilip Sarwate Aug 9 '14 at 9:51
1
$\begingroup$

For a linear time-invariant filter, the output of the filter would be coloured noise.

Again assuming linear time-invariant filter, the effect of the filter on the output can be further characterized in terms of its power spectrum density (PSD). Specifically, for an input $X(f)$ with power spectrum density $S_x(f)$ to a linear time-invariant filter with frequency response $H(f)$, the power-spectrum density $S_y(f)$ of the output of the filter is given by:

$$ S_y(f) = |H(f)|^2 S_x(f) $$

So, for a white noise input with unit power spectrum density $S_x(f)=1$, the corresponding PSD of the output would be $$ S_y(f) = |H(f)|^2 $$

$\endgroup$
  • $\begingroup$ Do you also know how these thing relates to auto correlation . $\endgroup$ – Lakshmi Aug 8 '14 at 21:09
  • $\begingroup$ @Lakshmi See this answer $\endgroup$ – Dilip Sarwate Aug 9 '14 at 9:51
  • $\begingroup$ Given the autocorrelation $R_h(\tau)$ of the (again linear-time-invariant) filter's impulse response, it can be shown that the Fourier transform $\mathcal{F}\{R_h(\tau)\} = |H(f)|^2$. Also, for WSS process (see @dilip answer link) $\mathcal{F}\{R_y(\tau)\} = S_y(f)$. So in this case, the autocorrelation of the filtered noise corresponds to the autocorrelation of the filter's impulse response. $\endgroup$ – SleuthEye Aug 9 '14 at 15:44

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.