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 α e ( z ) = ( z- . 6 + j . 3)( z- . 6- j . 3) = z 2- 1 . 2 z + 0 . 45 α 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 = α 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) ˆ 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 ˆ x 1 ( k ) = y ( k ) ˆ x 2 ( k ) = ˆ 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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