Discussion :: Exam Questions Paper
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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
A.
Sx(f) + Sy(f) + 2Sxy(f)
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B.
Sx(f) + Sy(f) + 2Sxy(f) + 2Syx(f)
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C.
Sx(f) + Sy(f)
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D.
Sx(f) + Sy(f) - 2Sxy(f) - 2Syx(f)
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Answer : Option C
Explanation :
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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