ECE535_10S_hw6soln

ECE535_10S_hw6soln - ECE 535 DISCRETE TIME SYSTEMS SPRING...

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Unformatted text preview: ECE 535 DISCRETE TIME SYSTEMS SPRING 2010 HOMEWORK ASSIGNMENT #6 Solutions 1. (a) We have e ( z ) = ( z- . 6 + j . 3)( z- . 6- j . 3) = z 2- 1 . 2 z + 0 . 45 e 1 1 0 1 = 1 1 0 1 2- 1 . 2 1 1 0 1 + 0 . 45 1 0 0 1 = . 25 . 8 . 25 . Also, O = 1 0 1 1 ; O- 1 = 1- 1 1 . Then L p = e 1 1 0 1 O- 1 1 = . 25 . 8 . 25 1- 1 1 1 = . 8 . 25 . The estimator is then x ( k + 1) = 1 1 0 1 x ( k ) + . 5 1 u ( k ) + . 8 . 25 ( y ( k )- 1 0 / x ( k ) ) . (b) We have L c = 1 1 0 1 - 1 L p = 1- 1 1 . 8 . 25 = . 55 . 25 . The estimator equations are then x ( k ) = 1 1 0 1 x ( k- 1) + . 5 1 u ( k- 1) x ( k ) = x ( k ) + . 55 . 25 ( y ( k )- 1 0 / x ( k ) ) . (c) We want A 22- L r A 12 = 0 . 6 so that L r = A 22- . 6 A 12 = 1- . 6 1 = 0 . 4 . We then have x 1 ( k ) = y ( k ) x 2 ( k ) = x 2 ( k- 1) + u ( k- 1)+ . 4 y ( k )- y ( k- 1)- 1 2 u (...
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ECE535_10S_hw6soln - ECE 535 DISCRETE TIME SYSTEMS SPRING...

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