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Unformatted text preview: 11)zL+1 + (1 +2z1 ) (1/2)(34z1 )zL+1 = (1/2zL+1 )(12z18z1 ) = 2zL 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 +2z1 H 01 (z) = H (z) H 1 (z 2 ) = (3+4z1 )( 1 +2z2 ) = 3+4z1 +6 z2 +8z3 H 001 (z) = H (z) H (z 2 ) H 1 (z 4 ) = 9+12z1 +12z2 +16z3 +18z4 +24z5 +24z6 +32z7 H 000 (z) = H (z) H (z 2 ) H (z 4 ) = 27+36z1 +36z2 +48z3 +36z4 +48z5 +48z6 +64z7 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
 Staff

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