solution_set7 - EE 511 Solutions to Problem Set 7 1. (i) SX...

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EE 511 Solutions to Problem Set 7 1. (i) S ˆ X ( f ) = | H ( f ) | 2 S X ( f ) where H ( f ) = - j sgn( f ). Therefore, S ˆ X ( f ) = S X ( f ). (ii) Suppose h ( τ ) is the impulse response of the Hilbert transformer, - h ( τ ) = h ( - τ ) is the impulse response of the inverse Hilbert transformer. Therefore, we have R ˆ XX ( τ ) = R X ( τ ) ⋆ h ( τ ) and R X ˆ X ( τ ) = R ˆ X ( τ ) ( - h ( τ )) = R X ( τ ) ( - h ( τ )) = - R ˆ XX ( τ ) , where denotes convolution. From the above result, we have S ˆ XX ( f ) = - S X ˆ X ( f ) . (iii) E [ X t ˆ X t ] = R X ˆ X (0) and E [ X t ˆ X t ] = R ˆ XX (0). Therefore, we have E [ X t ˆ X t ] = 0. (iv) R Z ( τ ) = E [( X t + τ + j ˆ X t + τ )( X t - j ˆ X t )] = R X ( τ ) - jR X ˆ X ( τ ) + jR ˆ XX ( τ ) + R ˆ X ( τ ) = 2 R X ( τ ) + 2 jR ˆ XX ( τ ) Therefore S Z ( f ) = 2 S X ( f ) + 2 jS ˆ XX ( f ) = 2 S X ( f ) + 2 jS X ( f )[ - j sgn( f )] = 2 S X ( f )[1 + sgn( f )] = b 4 S X ( f ) f > 0 0 else
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solution_set7 - EE 511 Solutions to Problem Set 7 1. (i) SX...

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