1. Consider two random processes x(t) and y(t) have zero mean, and they are individually stationary. The random process is z(t) = x(t) + y(t). Now when stationary processes are uncorrelated then power spectral density of z(t) is given by





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  • By: guest on 02 Jun 2017 12.56 am
    The autocorrelation function of z(t) is given by Rz(t, u) = E[Z(t)Z(u)] = E[(x(t) + y(t)) (x(u) + y(u))] = E[x(t) x (u)] + E[x(t) y(u)] + E[y(t) x(u)] + E[y(t) y(u)] = Rx(E, u) + Rxy(t, u) + Ryx(t, u) + Ry(t, u) Defining t = t - u, we may therefore write Rz(t) = Rx(t) + Rxy(t) + Ryx(t) + Ry(t). When the random process x(t) and y(t) are also jointly stationary. Accordingly, taking the fourier transform of both sides of equation we get Sz(f = Sx(f) + Sxy(f) + Syx(f) + Sy(f) We thus see that the cross spectral densities Sxy(f) and Syx(f) represent the spectral components that must be added to the individual power spectral densities of a pair of correlated random processes in order to obtain the power spectral density of their sum. When the stationary process x(t) and y(t) are uncorrelated the cross-sectional densities Sxy(f) and Syz(f) are zero ∴ Sz(f) = Sx(f) + Sy(f).
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