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090130Lecture10 - Power-symmetric leads to PR For the above...

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Power-symmetric leads to PR For the above choice of analysis and synthesis filters, ( ) 0 1 0 1 ( 1) 1 ( 1) 1 0 0 0 0 ( 1) 1 1 0 0 0 0 ( 1) det( ( )) ( ) ( ) ( ) ( ) ( )( ( )) ( )( ( )) [ ( ) ( ) ( ) ( )] m N N N N H z H z H z H z H z H z z H z H z z H z z H z H z H z H z z - - - - - - - - - - - - = - - - = - - ± - = + - - = ECE 700 WI 2009 The job then is to design H 0 ( z ) that satisfies the power-symmetric condition. To do that, we will take a diversion to look at what are called half-band filters. So the PR condition is satisfied.
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Half-Band Filters Consider a filter F ( z ) with real coefficients that satisfies F ( z ) + F (- z ) = 1. ( ) ( ) ( ) ( ) 1 If [ ] is real, ( ) ( ), So ( ) ( ) 1. j j j j j j F e F e f n F e F e F e F e ϖ ϖ ϖ π ϖ ϖ π ϖ - - + - = - = + = This makes the filter F ( z ) have a symmetric frequency response around w = ECE 700 WI 2009 π /2, hence the name half-band filter.
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Some more on half-band filters even odd ( ) ( ) 1 [ ] [ ]( ) 1 [ ] [ ] [ ] 1 1 [0] . 2 n n n n n n n n n n n n F z F z f n z f n z f n z f n z f n z f - - =-∞ =-∞ - - - =-∞ =-∞ =-∞ + - = + - = + - = = ECE 700 WI 2009 [ ] 0, even f n n = If F ( z ) is zero-phase, then f [ n ] = f *[- n ].
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