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EE630F09hw9_soln - ENEE 630 Fall 2009 Homework Solution#9...

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ENEE 630 Fall 2009 Homework Solution #9 Solution to 1 E r xx [ k ]] = r xx [ k ] = σ 2 x δ [ k ] (for white Gaussian process) E [ r xx [ k ]] = 1 ( N - k ) 2 N - 1 - k n =0 N - 1 - k m =0 E [ x [ n ] x [ n + k ] x [ m ] x [ m + k ]] = 1 ( N - k ) 2 N - 1 - k n =0 N - 1 - k m =0 ( r 2 xx [ k ] + r 2 xx [ m n ] + r xx [ m n + k ] r xx [ m n k ]) Noting that L m =0 L n =0 g [ m n ] = L j = - L ( L + 1 − | j | ) g ( j ), we have E [ r 2 xx [ k ]] = r 2 xx [ k ] + 1 ( N - k ) 2 N - 1 - k j = - ( N - 1 - k ) { ( N k − | j | )( r 2 xx [ j ] + r xx [ j + k ] r xx [ j k ]) } V ar [ r 2 xx [ k ]] = 1 N - k N - 1 - k j = - ( N - 1 - k ) (1 | j | N - k )( r 2 xx [ j ] + r xx [ j + k ] r xx [ j k ]) = 1 N - k N - 1 - k j = - ( N - 1 - k ) (1 | j | N - k )( σ 4 x δ [ j ] + σ 4 x δ [ j + k ] δ [ j k ]) = 1 N - k ( σ 4 x + σ 4 x δ [ k ]) = 2 σ 4 x N k = 0 σ 4 x N - k 1 k N 1 For k 1, as the lag increases the variance also increases since fewer lag products are averaged. For the biased ACF estimation, ˆ r xx [ k ] = 1 N N - 1 - k summationdisplay n =0 x [ n ] x [ n + k ] E r 2 xx [ k ]] = 1 N 2 N - 1 - k m =0 N - 1 - k n =0 ( r 2 xx [ k ] + r 2 xx [ m n ] + r xx [ m n + k ] r xx [ m n k ]) = ( N - k N ) 2 r 2 xx [ k ] + 1 N 2 N
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