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:
$$E[S(f_1)S(f_2)]$$
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How does the LPF factor into your question? – John May 28 '14 at 21:41
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The psd is limited to the range $[-B/2,B/2]$ instead of being infinite. – bigredx24x May 28 '14 at 21:44
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2$S(f_1)$ 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. – Dilip Sarwate May 29 '14 at 01:25
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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: $$ \begin{align} E[S_y(f_1)S_y(f_2)] &= \frac{1}{4} N_0^2\left|H(f_1)H(f_2)\right|^2 \end{align} $$
