solution_set8

# solution_set8 - EE 511 Solutions to Problem Set 8 1 E[Yt =...

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EE 511 Solutions to Problem Set 8 1. E [ Y t ] = E [ X t ] cos 2 πf c t . E [ Y t ] is periodic with period 1 /f c . R Y ( t,t + τ ) = E [ X t + τ cos 2 πf c ( t + τ ) X t cos 2 πf c t ] = R X ( τ ) 2 [cos 2 πf c τ + cos 2 πf c (2 t + τ )] R Y ( t,t + τ ) is periodic with period 1 / (2 f c ). Therefore, Y t is wide-sense cyclostationary with period 1 /f c . E [ Z t ] = E [ X t ] E [cos (2 πf c t + Θ)] = 0. R Z ( t,t + τ ) = E [ X t + τ cos (2 πf c ( t + τ ) + Θ) X t cos (2 πf c t + Θ)] = R X ( τ ) 2 [cos 2 πf c τ + E [cos (2 πf c (2 t + τ ) + 2Θ)]] = R X ( τ ) 2 cos 2 πf c τ Therefore, Z t is wide-sense stationary. 2. (a) As derived in class, we have R X ( τ ) = 1 T s k = -∞ R A ( k ) R p ( τ kT ) , where R p ( τ ) = i -∞ p ( t ) p ( t + τ ) dt. (b) R A ( k ) = 0 for all k n = 0. Therefore R X ( τ ) = 1 T R A (0) R p ( τ ) , where R A (0) = 1 and R p ( τ ) = b A 2 (2 T − | τ | ) | τ | < 2 T 0 else Therefore, we have (see Figure 1) R X ( τ ) = b A 2 (2 | τ | T ) | τ | < 2 T 0 else (c) R p ( τ ) = b A 2 ( T − | τ | ) | τ | < T 0 else R A (0) = E [ A 2 n ] = E [( B n + B n - 1 )( B n + B n - 1 )] = E [ B 2 n ] + E [ B 2 n - 1 ] = 2 . R A ( 1) = R A (1) = E [ A n A n - 1 ] = E [( B n + B n - 1 )( B n - 1 + B n - 2 )] = E [ B 2 n - 1 ] = 1

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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_set8 - EE 511 Solutions to Problem Set 8 1 E[Yt =...

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