b X t is a Gaussian rp Y t and V t are jointly Gaussian rps Thus Y t and V t

# B x t is a gaussian rp y t and v t are jointly

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b) X t is a Gaussian r.p. Y t and V t are jointly Gaussian r.p.’s. Thus, Y t and V t are independent when they are uncorrelated: R Y V ( τ ) = 0 , τ R S Y V ( ω ) = N 0 2 H ( ω ) G * ( ω ) = 0 ω R . If H ( ω ) G * ( ω ) = 0 ω R Y t and V t are independent. Page 2 of 3

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ELEC 533 (Professor Behnaam Aazhang): Homework 8 Problem 4 Problem 5 Y 0 t + αY t = X t m . s . with Y 0 = 0 a) take expectation of the differential equation: E[ Y 0 t ] + α E[ Y t ] = E[ X t ] m . s . with Y 0 = 0 d dt μ Y ( t ) + αμ Y ( t ) = μ X ( t ) = 0 the initial condition Y 0 = 0 E[ Y 0 ] = μ y (0) = 0 b) multiply the differential equation with the conjugate of X t : Y 0 t X * s + αY t X * s = X t X * s take expectation : E[ Y 0 t X * s ] + E[ αY t X * s ] = E[ X t X * s ] R Y 0 X ( t, s ) + αR Y X ( t, s ) = R XX ( t, s ) d dt R Y X ( t, s ) + αR Y X ( t, s ) = R XX ( t, s ) = a 2 δ ( t - s ) the initial condition Y 0 = 0 R Y X (0 , s ) = E[ Y 0 X s ] = E[0 X s ] = 0 R Y X (0 , s ) = 0 s R c) multiply the differential equation by Y * s Y 0 t Y * s + αY t Y * s = X t Y * s take expectation : E[ Y 0 t Y * s ] + E[ αY t Y * s ] = E[ X t Y * s ] R Y 0 Y ( t, s ) + αR Y Y ( t, s ) = R XY ( t, s ) d dt R Y Y ( t, s ) + αR Y Y ( t, s ) = R XY ( t, s ) the initial condition Y 0 = 0 R Y Y (0 , s ) = E[ Y 0 Y s ] = E[0 Y s ] = 0 R Y Y (0 , s ) = 0 s R d) Y t is WSS since its mean is constant and the autocorrelation depends only on ( t - s ). Y t is Gaussian,
• Spring '14
• Aazhang,Behnaam
• Derivative, Trigraph, Professor Behnaam Aazhang, T T0

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