630F09_hw6_sol

# 630F09_hw6_sol - ENEE 630 Homework#6 Solution to 1(a R =...

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ENEE 630 Homework #6 Solution to 1 (a) R = E [ u ( n ) u H ( n )] where u (n) = α (n) s (n) + v (n) with α (n) uncorrelated with v (n). We have R = E [ | α ( n ) | 2 ] s ( n ) s H ( n ) + E [ v ( n ) v H ( n )] = σ 2 α s ( n ) s H ( n ) + R v (b) P = E [ u ( n ) d * ( n )] = 0 Thus, w 0 = R - 1 P = 0 . (c) If σ 2 α = 0, then R= R v . Since d(n)=v(n-k), P = E [( α ( n ) s ( n ) + v ( n )) v * ( n - k )] = E [ v ( n ) v ( n - 1) . . . v ( n - M + 1) v * ( n - k )] = r v ( k ) r v ( k - 1) . . . r v ( k - M + 1) , 0 k M - 1 Thus, w 0 = R - 1 v P . (d) P = E [ u ( n ) d * ( n )] = σ 2 α s ( n ) e jwτ = σ 2 α e jwτ e jw ( τ - 1) . . . e jw ( τ - M +1) and w 0 = R - 1 P . 1

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Solution to 2 (a) R = E [ u ( n ) u H ( n )] = E [ | A 1 | 2 ] s ( w 1 ) s H ( w 1 ) + I E [ | v ( n ) | 2 ] = σ 2 1 s ( w 1 ) s H ( w 1 ) + σ 2 v I (b) Knowing p = σ 2 0 s ( w 0 ), w 0 = R - 1 p . Now what is R - 1 ? Recall the matrix inversion lemma : A = B - 1 + CD - 1 C H A - 1 = B - BC ( D + C H BC ) - 1 C H B Choose A=R, B - 1 = σ 2 v I , D - 1 = σ 2 1 , C= s ( w 1 ). Then, R - 1 = σ - 2 v I - σ - 2 v s ( w 1 ) s H ( w 1 ) σ 2 v σ 2 1 + s H ( w 1 ) s ( w 1 ) Since s H ( w 1 ) s ( w 1 ) = M , w 0 = σ 2 0 σ 2 v s ( w 0 ) - ( σ 2 0 σ 2 v s ( w 1 ) s H ( w 1 ) σ 2 0 σ 2 1 + M ) s ( w 0 ) Solution to 3 e ( n ) = w H u ( n ) = M X k =1 w * k u k ( n ) where w = ( w 1 w 2 ... w M ) u T ( n ) = ( u 1 ( n ) u 2 ( n ) ... u M ( n )) E [ | e ( n ) | 2 ] = w H R w . The problem is: min w ( w H R w ) subject to w H s = 1. 2
We apply the method of Lagrange multipliers: min w ( w
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## This note was uploaded on 10/31/2010 for the course EE 630 taught by Professor Wu during the Spring '10 term at Aarhus Universitet, Aarhus.

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630F09_hw6_sol - ENEE 630 Homework#6 Solution to 1(a R =...

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