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630F09_hw2_sol - ENEE 630 F2009 Homework Solution#2 Problem...

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ENEE 630 F’2009 Homework Solution #2 Problem 1 solution: (a) Since W is a constant matrix, the output of W can be written as Y 0 0 ( z ) Y 0 1 ( z ) Y 0 2 ( z ) = 1 1 1 1 W W 2 1 W 2 W Y 0 ( z ) Y 1 ( z ) Y 2 ( z ) where W = 1 1 1 1 W W 2 1 W 2 W To get F 0 ( z ) , set Y 1 ( z ) and Y 2 ( z ) to zero, then Y 0 0 ( z ) Y 0 1 ( z ) Y 0 2 ( z ) = Y 0 ( z ) Y 0 ( z ) Y 0 ( z ) Y ( z ) = [ z - 2 R 0 ( z 3 ) z - 1 R 1 ( z 3 ) R 2 ( z 3 ) ] Y 0 0 ( z ) Y 0 1 ( z ) Y 0 2 ( z ) = Y 0 ( z )[ z - 2 R 0 ( z 3 ) + z - 1 R 1 ( z 3 ) + R 2 ( z 3 )] = (2 + z - 1 + z - 2 + 3 z - 3 + z - 5 - z - 7 ) Y 0 ( z ) so F 0 ( z ) = 2 + z - 1 + z - 2 + 3 z - 3 + z - 5 - z - 7 similarly, F 1 ( z ) = 2 W 2 + Wz - 1 + z - 2 + 3 W 2 z - 3 + z - 5 - Wz - 7 F 2 ( z ) = 2 W + W 2 z - 1 + z - 2 + 3 Wz - 3 + z - 5 - W 2 z - 7 Note that, F 1 ( z ) = W 2 F 0 ( Wz ) F 2 ( z ) = WF 0 ( W 2 z ) (b) The magnitude response is roughly given below: 1
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1 π /3 2 π /3 π 2 π ϖ | F 1 (e j ϖ ) | 0 5 π /3 | F (e j ϖ ) | | F (e j ϖ ) | | F (e j ϖ ) | 2 0 0 Problem 2 solution: (a) Since h k [ n ] = h 0 [ n ] e j 2 πkn 5 , the coefficients are in general complex.
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