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EE351kHW_15SolnFall2009

EE351kHW_15SolnFall2009 - Solution to HW 15 Due 30 November...

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Unformatted text preview: Solution to HW # 15, Due 30 November 2009 Problem 1 (a) E[ X (t )] = ∫ A cos(ω0t + θ ) −π π 1 A π [sin(ω0t + θ )]θ = −π = 0 dθ = 2π 2π A2 1 dθ = 2π 4π (b) E[ X 2 (t )] = ∫ A2 cos 2 (ω0t + θ ) −π π ∫−π (1 + cos(2ω0t + 2θ ))dθ = 1 dθ 2π π A2 A2 2π = 4π 2 (c) RX (τ ) = E[ X (t ) X (t + τ )] = ∫ A2 cos(ω0t + θ ) cos(ω0 (t + τ ) + θ ) −π π = A2 4π ∫−π (cos(−ω0τ ) + cos(2ω0t + ω0τ + 2θ ))dθ = π A2 cos(ω0τ ) 2 Problem 2 (a) < X (t ) >= 1 T0 ∫ T0 0 A cos(ω0t + θ )dt = A [sin(ω0t + θ )]T=0 0 = 0 t T0 A2 2T0 A ∫0 (1 + cos(2ω0t + 2θ ))dt = 2T0 T0 = T0 (b) < X 2 (t ) >= (c) 1 T0 ∫ T0 0 A2 cos 2 (ω0t + θ )dt = 2 A2 2 RX (τ ) =< X (t ) X (t + τ ) >= A2 = 2T0 1 T0 ∫ T0 0 A2 cos(ω0t + θ ) cos(ω0 (t + τ ) + θ )dt ∫ T0 0 A2 A2 (cos(−ω0τ ) + cos(2ω0t + ω0τ + 2θ ) )dt = T0 cos(ω0τ ) = cos(ω0τ ) 2T0 2 (d) From the results above E[ X (t )] =< X (t ) > E[ X 2 (t )] =< X 2 (t ) > E[ X (t ) X (t + τ )] =< X (t ) X (t + τ ) > ...
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