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A signal on a noisy channel is input to a filter

  1. Dec 9, 2015 #1
    This is not homework. Im doing old exams so i have the full solution but need help understanding it.

    1. The problem statement, all variables and given/known data

    A WSS random signal {X(t)}t∈R with 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##.


    3. 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.
     
    Last edited: Dec 9, 2015
  2. jcsd
  3. Dec 9, 2015 #2

    Ray Vickson

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    What are WSS and PSD? I can guess the first one, but why should I need to? I cannot even guess about the second one.
     
  4. Dec 9, 2015 #3
    Sorry, WSS = Wide sense stationary and PSD = Power spectral density
     
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