solution_set6

# solution_set6 - EE 511 Solutions to Problem Set 6 1 The...

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Unformatted text preview: EE 511 Solutions to Problem Set 6 1. The power spectral density of the output noise process is equal to N / 2 for | f | ≤ B and equal to 0 otherwise. Therefore, the variance of the output noise process (zero-mean) is the area under the PSD = (2 B )( N / 2) = N B . 2. a) m Y ( t ) = E [ Y t ] = E [ X t + D ]- E [ X t ] = m ( t + D )- mt = mD . R Y ( t,s ) = E [( X t + D- X t )( X s + D- X s )] = R X ( t + D,s + D )- R X ( t + D,s )- R X ( t,s + D ) + R X ( t,s ) = σ 2 [min ( t + D,s + D )- min ( t + D,s )- min ( t,s + D ) + min ( t,s )] + m 2 D 2 For 0 ≤ t- s ≤ D , R Y ( t,s ) = m 2 D 2 + σ 2 ( s + D- s- t + s ) = m 2 D 2 + σ 2 ( s- t )+ σ 2 D . For t- s ≥ D , R Y ( t,s ) = m 2 D 2 + σ 2 ( s + D- s- ( s + D ) + s ) = m 2 D 2 . For- D ≤ t- s ≤ 0, R Y ( t,s ) = m 2 D 2 + σ 2 ( t + D- s- t + t ) = m 2 D 2 + σ 2 ( t- s )+ σ 2 D . For t- s ≤ - D , R Y ( t,s ) = m 2 D 2 + σ 2 ( t + D- ( t + D )- t + t ) = m 2 D 2 . Therefore, defining τ = t- s we have R Y ( τ ) =      m 2 D 2 + σ 2 ( D- | τ | ) | τ | ≤ D m 2 D 2 | τ | > D b) Since Y t is W.S.S. (from part (a)) and Y t is a Gaussian random process, Y t is also strictly stationary....
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## This note was uploaded on 10/26/2011 for the course ECE 171 taught by Professor Tu during the Spring '11 term at Aarhus Universitet.

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solution_set6 - EE 511 Solutions to Problem Set 6 1 The...

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