Assume I have a random process $X(t)$ that is white gaussian noise with psd $S(f)$. Now, let's filter that noise through a LPF of bandwidth $B$. How would I evaluate the following expression:


  • $\begingroup$ How does the LPF factor into your question? $\endgroup$
    – John
    Commented May 28, 2014 at 21:41
  • $\begingroup$ The psd is limited to the range $[-B/2,B/2]$ instead of being infinite. $\endgroup$
    – bigredx24x
    Commented May 28, 2014 at 21:44
  • 2
    $\begingroup$ $S(f_1)$ and $S(f_2)$ are not random quantities and so taking their expected values gives just $Sf_1)S(f_2)$. Since the expectation is pretty meaningless here, I am inclined to believe that what you want to know is something quite different from what you have written. $\endgroup$ Commented May 29, 2014 at 1:25

1 Answer 1


In general, for an input signal $x$ with power-spectrum density $S_x(f)$ passed through a linear-time-invariant (LTI) system with transfer function $H(f)$, the power-spectrum density of the output, $S_y(f)$ is given by:

$$ \begin{align} S_y(f) &= S_x(f)\left|H(f)\right|^2 \end{align} $$ A complete derivation can be found in various textbooks (such as "Communication Systems Engineering", by John G. Proakis and Masoud Salehi) and lecture notes (such as these). Note that a consequence of this result for Gaussian noise is also quoted here.

Thus, if evaluating the output PSD at $f_1$ and $f_2$, we have $$ \begin{align} S_y(f_1) &= S_x(f_1)\left|H(f_1)\right|^2 \\ S_y(f_2) &= S_x(f_2)\left|H(f_2)\right|^2 \end{align} $$ And consequently, $$ \begin{align} E[S_y(f_1)S_y(f_2)] &= S_y(f_1)S_y(f_2) \\ &= S_x(f_1)S_x(f_2)\left|H(f_1)H(f_2)\right|^2 \end{align} $$

Note that for an input consisting of white (or at least white over some finite observable bandwidth much greater than B) Gaussian noise with power-spectral density $N_0/2$, we get: $$ \begin{align} E[S_y(f_1)S_y(f_2)] &= \frac{1}{4} N_0^2\left|H(f_1)H(f_2)\right|^2 \end{align} $$


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