HW6_sol

HW6_sol - -1-1)z-L+1 + (1 +2z-1 ) (1/2)(3-4z-1 )z-L+1 =...

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ECE 158 Fall 2010 Gibson Homework #6 Solutions Problem 2: a) With H 0 (z) = 3+4z -1 and H 1 (z) = 1 +2z -1 , G 0 (z) and G 1 (z) can be found as H 0 (z) G 0 (z) + H 1 (z) G 1 (z) = 2z -L (perfect reconstruction condition) H 0 (-z) G 0 (z) + H 1 (-z) G 1 (z) = 0 (alias cancellation condition) Therefore G 1 (z) = - G 0 (z) = -((3-4z -1 )/(1-2z -1 )) G 0 (z) Substituting G 1 (z) into the first equation gives H 0 (z) G 0 (z) - H 1 (z) G 0 (z) = (-4z -1 /(1-2z -1 )) G 0 (z) = 2z -L+1 G 0 (z) = (1/2)(2z -1 -1)z -L+1 We can then find G 1 (z) as G 1 (z) = (1/2)(3-4z -1 )z -L+1 b) Using H 0 (z), H 1 (z), G 0 (z), and G 1 (z) from the previous section, we need to show again that
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H 0 (z) G 0 (z) + H 1 (z) G 1 (z) = 2z -L (perfect reconstruction condition) H 0 (-z) G 0 (z) + H 1 (-z) G 1 (z) = 0 (alias cancellation condition) By substituting the filters into the first equation we have (3+4z -1 ) (1/2)(2z
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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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HW6_sol - -1-1)z-L+1 + (1 +2z-1 ) (1/2)(3-4z-1 )z-L+1 =...

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