630F04_hw4_soln

# 630F04_hw4_soln - ENEE 630 Fall2009 Homework Solution #4...

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ENEE 630 Fall’2009 Homework Solution #4 Solution to 1 Since the system is alias-free, H ( z ) f ( z ) = M T ( z ) 0 . . 0 In other words, H 0 ( z ) F 0 ( z ) + ··· + H M - 1 ( z ) F M - 1 ( z ) = M T ( z ) H 0 ( zW i ) F 0 ( z ) + ··· + H M - 1 ( zW i ) F M - 1 ( z ) = 0 , i = 1 , ··· ,M - 1 If replace z with zW M - i , it’ll continue to hold: H 0 ( z ) F 0 ( zW M - i ) + ··· + H M - 1 ( z ) F M - 1 ( zW W - i ) = 0 , i = 1 , ··· ,M - 1 Thus, F 0 ( z ) . . F M - 1 ( z ) . . . . F 0 ( zW M - 1 ) . . F M - 1 ( zW M - 1 ) H 0 ( z ) . . H M - 1 ( z ) = M T ( z ) 0 . 0 We can swap H k ( z )’s and F k ( z )’s and the system is still alias-free with the same T(z). Solution to 2 a)

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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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630F04_hw4_soln - ENEE 630 Fall2009 Homework Solution #4...

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