HW_11(483)S09_Solns

HW_11(483)S09_Solns - 1 EE 483 – Digital Signal Processing Spring 2009 Homework#11 Solutions 11.1 Analysis yields w 1 n = β 1 α 1 w 2 n 1 α 2

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Unformatted text preview: 1 EE 483 – Digital Signal Processing Spring 2009 Homework #11 Solutions 11.1 Analysis yields w 1 [ n ] = β 1 α 1 w 2 [ n- 1] + α 2 w 4 [ n- 1] + α 3 w 6 [ n- 1] { } , w 2 [ n ] = w 1 [ n ] + x [ n ], w 3 [ n ] = β 2 α 2 w 4 [ n- 1] + α 3 w 6 [ n- 1] { } , w 4 [ n ] = w 2 [ n ] + w 3 [ n ], w 5 [ n ] = β 3 α 3 w 6 [ n- 1] { } , w 6 [ n ] = w 4 [ n ] + w 5 [ n ], y [ n ] = α 1 w 2 [ n ] + α 2 w 4 [ n ] + α 3 w 6 [ n ] + α x [ n ]. In matrix form the above set of equations are given by 1 1 1 1 1 2 2 2 3 3 4 4 5 5 6 6 1 2 3 [ ] [ ] 1 [ ] [ ] [ ] [ ] 1 1 [ ] [ ] [ ] [ ] 1 1 [ ] [ ] [ ] [ ] w n w n w n w n w n w n w n w n w n w n w n w n a a a y n y n β α β α β = + 1 3 1 2 2 2 2 3 3 4 3 3 5 6 [ 1] [ ] [ 1] [ 1] [ 1] [ 1] [ 1] [ ] [ 1] w n x n w n w n w n w n w n x n y n α β α β α β α α- -...
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This note was uploaded on 02/02/2010 for the course EE 483 taught by Professor Mitra during the Fall '08 term at USC.

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HW_11(483)S09_Solns - 1 EE 483 – Digital Signal Processing Spring 2009 Homework#11 Solutions 11.1 Analysis yields w 1 n = β 1 α 1 w 2 n 1 α 2

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