Question:
A Wide-sense stationary (weakly stationary, (WSS)) random signal {X(t)}t∈R with Power spectral density(PSD) S_X(ω) is transmitted on a noisy channel where it is disturbed by an additive zero-mean WSS random noise {N(t)}t∈R that is independent of the signal X and has PSD S_N (ω).
The recived signal Y (t) = X(t)+N(t) is input to a linear system (/filter) with output signal Z(t) that has frequency response
$$H(ω) = S_X (ω)/(S_X (ω) + S_N (ω))$$
Express the mean-square deviation
$$E((Z(t)− X(t))^2)$$
in terms of
$$S_X , S_N$$
The solution:
Writing h for the impulse response of the filter and ⋆ for convolution the fact that h ⋆ N is zero-mean and independent of X (as N is) readily gives that
$$E((Z(t) − X(t))^2) = E(((h ⋆X)(t) + (h ⋆N)(t) − X(t))^2) =$$
$$(i) = E((((h−δ)⋆X)(t) + (h ⋆N)(t))^2) =$$
$$= E(((h−δ)⋆X)(t)^2 + (h⋆N)(t)^2) = $$
$$ (ii) = 1/2π \int^{+\infty}_{-\infty} [|H(ω)−1|^2 S_X(ω) + |H(ω)|^2 S_N(ω)] dω = $$
$$. . . = 1/2π \int^{+\infty}_{-\infty} S_X(ω)S_N(ω)/(S_X (ω) +S_N(ω)) dω $$
I dont understand how you get to lines (i) and (ii) and i cant find any definitions that explains those steps. I would appreciate some help understanding this.
Ok, i think i figured this out now.
(i) delta is the identity for convolution.
(ii) For general input X and output Y.
$$ R_Y(\tau) = \frac{1}{2\pi} \int^{+\infty}_{-\infty} S_Y(ω) exp^{jω\tau} dω = $$
$$ S_Y(ω) = \int^{+\infty}_{-\infty} R_Y(\tau) exp^{-jω\tau}d\tau = (iii ) = |H(ω)|^2 S_X(ω)$$
$$ E[Y(t)^2] = R_Y(0) = \frac{1}{2\pi} \int^{+\infty}_{-\infty} S_Y(ω) dω =\frac{1}{2\pi} \int^{+\infty}_{-\infty} |H(ω)|^2 S_X(ω) dω = $$
Now all that remains, i think, is to derive step (iii)
edit: i found the derivation of (iii) but its a long one, ill post it later.