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Unformatted text preview: -1-1)z-L+1 + (1 +2z-1 ) (1/2)(3-4z-1 )z-L+1 = (1/2z-L+1 )(12z-1-8z-1 ) = 2z-L The result shows that the output is a delayed version of the input and therefore achieves perfect reconstruction. Problem 3: The filter bank diagram can be simplified as shown below. The filters for each branch are H 1 (z) = 1 +2z-1 H 01 (z) = H (z) H 1 (z 2 ) = (3+4z-1 )( 1 +2z-2 ) = 3+4z-1 +6 z-2 +8z-3 H 001 (z) = H (z) H (z 2 ) H 1 (z 4 ) = 9+12z-1 +12z-2 +16z-3 +18z-4 +24z-5 +24z-6 +32z-7 H 000 (z) = H (z) H (z 2 ) H (z 4 ) = 27+36z-1 +36z-2 +48z-3 +36z-4 +48z-5 +48z-6 +64z-7 Therefore if the transfer function of branch(i) = T i (z), we then have T 1 (z) = T 2 (z) = T 3 (z) = T 1 (z) =...
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This note was uploaded on 12/28/2011 for the course ECE 158 taught by Professor Staff during the Fall '08 term at UCSB.
- Fall '08