630_rec_3_sol - ENEE630 ADSP 1. RECITATION 3 w/ solution...

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ENEE630 ADSP RECITATION 3 w/ solution September 24/25, 2009 1. In this problem, the term ‘polyphase components’ stands for the Type 1 components with M = 2. (a) Let H ( z ) represent an FIR filter of length 10 with impulse response coefficients h ( n ) = (1 / 2) n for 0 n 9 and zero otherwise. Find the polyphase components E 0 ( z ) and E 1 ( z ). (b) Let H ( z ) be IIR with h ( n ) = (1 / 2) n u ( n ) + (1 / 3) n u ( n - 3). Find the polyphase components E 0 ( z ) and E 1 ( z ). Give simplified, closed form expressions. (Hint: 1 1+ x = 1 - x 1 - x 2 ) (c) Let H ( z ) = 1 / (1 - 2 R cos θz - 1 + R 2 z - 2 ), with R > 0 and θ real. This is a system with a pair of complex conjugate poles at Re ± . Find the polyphase components E 0 ( z ) and E 1 ( z ). Solution: (a). E 0 ( z ) = 1 + (1 / 2) 2 z - 1 + (1 / 2) 4 z - 2 + (1 / 2) 6 z - 3 + (1 / 2) 8 z - 4 , E 1 ( z ) = 1 / 2 + (1 / 2) 3 z - 1 + (1 / 2) 5 z - 2 + (1 / 2) 7 z - 3 + (1 / 2) 9 z - 4 . (b).
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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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630_rec_3_sol - ENEE630 ADSP 1. RECITATION 3 w/ solution...

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