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Unformatted text preview: 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 ( z ) Y 1 ( z ) Y 2 ( z ) = 1 1 1 1 W W 2 1 W 2 W Y ( z ) Y 1 ( z ) Y 2 ( z ) where W = 1 1 1 1 W W 2 1 W 2 W To get F ( z ) , set Y 1 ( z ) and Y 2 ( z ) to zero, then Y ( z ) Y 1 ( z ) Y 2 ( z ) = Y ( z ) Y ( z ) Y ( z ) Y ( z ) = [ z 2 R ( z 3 ) z 1 R 1 ( z 3 ) R 2 ( z 3 ) ] Y ( z ) Y 1 ( z ) Y 2 ( z ) = Y ( z )[ z 2 R ( 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 ( z ) so F ( 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 ( Wz ) F 2 ( z ) = WF ( W 2 z ) (b) The magnitude response is roughly given below: 1 1 π /3 2 π /3 π 2 π ϖ  F 1 (e j ϖ )  5 π /3  F (e j ϖ )   F (e j ϖ )   F (e j ϖ )  2 Problem 2...
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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.
 Spring '10
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