A signal on a noisy channel is input to a filter

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SUMMARY

The discussion centers on the analysis of a wide sense stationary (WSS) random signal {X(t)} transmitted over a noisy channel, where it is affected by additive zero-mean WSS random noise {N(t)}. The received signal Y(t) is processed through a linear filter with frequency response H(ω) = S_X(ω)/(S_X(ω) + S_N(ω)). The mean-square deviation E((Z(t)− X(t))^2) is expressed in terms of the power spectral densities (PSD) S_X and S_N, leading to the integral representation E((Z(t)− X(t))^2) = 1/2π ∫[|H(ω)−1|^2 S_X(ω) + |H(ω)|^2 S_N(ω)] dω.

PREREQUISITES
  • Understanding of Wide Sense Stationary (WSS) processes
  • Knowledge of Power Spectral Density (PSD)
  • Familiarity with linear systems and filters
  • Proficiency in convolution operations
NEXT STEPS
  • Study the properties of Wide Sense Stationary (WSS) processes
  • Learn about Power Spectral Density (PSD) and its applications
  • Explore linear filter design and frequency response analysis
  • Investigate convolution techniques in signal processing
USEFUL FOR

Signal processing engineers, communication system designers, and students studying random processes and filter theory will benefit from this discussion.

JohanL
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This is not homework. I am doing old exams so i have the full solution but need help understanding it.

1. Homework Statement

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 don't understand how you get to lines (i) and (ii) and i can't find any definitions that explains those steps.
 
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JohanL said:
This is not homework. I am doing old exams so i have the full solution but need help understanding it.

1. Homework Statement

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 don't understand how you get to lines (i) and (ii) and i can't find any definitions that explains those steps.

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.
 
Ray Vickson said:
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.

Sorry, WSS = Wide sense stationary and PSD = Power spectral density
 

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