ECE511_F10_exam2_solution

ECE511_F10_exam2_solution - 1. 1 - - - e = e = Ae - - 2(1 -...

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1. a) 1 ee A e , 2(1 e ) 2(1 e ) f() 0 ,otherwise −λθ −λπ Θ λ = λ π < θ < π −− θ= Where 1 A 2(1 e ) = ; Then X E(X) xf (x)dx cos f ( )d cos A e d 0 ∞π π Θ −∞ −π = = θ θ θ= θ⋅ λ ⋅ θ≠ ∫∫ In the same way, E(Y) sin A e d 0 π λ θ = ( f( ) sin θ ⋅ λ is odd function ) 1 E(XY) cos sin A e d sin2 A e d 0 2 ππ θ λ θ = θ λ θ = Since E(XY) E(X)E(Y) = , X and Y are uncorrelated. b) 22 0 00 E(X ) cos A e d (1 cos2 ) A e d Ae c o s 2 ed A 1 e c o s 2 0 = θ ⋅λ θ = + θ θ ⎡π =− + θ λ θ = + θ λ θ ⎢⎥ ⎣⎦ Since, 0 2 1e cos2 e d 4 1 π θ⋅λ + λ So, 2 2 2 1 1 2 E(X ) A 1 e 1 44 24 11 ⎡⎤ −λ + =−+ =+ = λ + ++ λλ
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22 0 2 2 2 1 E(Y ) sin A e d (1 cos2 ) A e d 2 1e (1 cos 2 ) A e d A 1 e 4 1 11 2 1 4 24 1 ππ −λθ −π −λπ π λ θ =− θ λ θ ⎡⎤ ⎢⎥ =−θ λθ =−− + λ ⎣⎦ = λ+ + λ ∫∫ 2 2 0 1 E ( X Y ) c o ss i n A ed ( 1 c o s 4 ) A 8 1 cos 4 ) A e d 4 π = θ
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ECE511_F10_exam2_solution - 1. 1 - - - e = e = Ae - - 2(1 -...

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