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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). H ( z ) = 1 1 - (1 / 2) z - 1 + (1 / 3) 3 z - 3 1 - (1 / 3) z - 1 = 1+(1 / 2) z - 1 1 - (1 / 4) z - 2 + (1 / 3) 3 z - 3 +(1 / 3) 4 z - 4 1 - (1 / 9) z - 2 . Hence, E 0 ( z ) = 1 1 - (1 / 4) z - 1 + (1 / 3) 4 z - 2 1 - (1 / 9) z - 1 , and E 1 ( z ) = 1 / 2 1 - (1 / 4) z - 1 + (1 / 3) 3 z - 1 1 - (1 / 9) z - 1 .
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