630F09_hw3_soln - ENEE 630 Fall’2009 Homework Solution #3...

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Unformatted text preview: ENEE 630 Fall’2009 Homework Solution #3 Solution to 1 It can be found that ˆ X ( z ) = 1 2 z- 1 [ H ( z ) F ( z ) + H 1 ( z ) F 1 ( z )] X ( z ) + + 1 2 z- 1 [ H (- z ) F ( z )- H 1 (- z ) F 1 ( z )] X (- z ) Or, call it simply ˆ X ( z ) = T ( z ) X ( z ) + A ( z ) X (- z ) Since H 1 ( z ) = H (- z ) ,F ( z ) = H ( z ) ,F 1 ( z ) = H 1 ( z ), A(z)=0 and T ( z ) = 1 2 z- 1 [ H 2 ( z ) + H 2 (- z )] a) Now, H ( w ) = e jwN 2 R ( w ). So, T ( w ) = 1 2 e- jw ( N +1) ( | H ( w ) | 2 + (- 1) N | H ( π- w ) | 2 ) Thus, N must be even to avoid T ( e j π 2 ) = 0. b) If N is even, then both polyphase components are linear phase. Use of (iii) decimators and (i) H 1 ( z ) = H (- z ) and (ii) results in the simplification shown in Figure 1. It can’t be reduced further. Since both E ( z ) and E 1 ( z ) are linear phase, each of A and B in the Figure requires N 2 + 1 multipliers. Thus A+B needs N+2. MPU= N +2 2 ....
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630F09_hw3_soln - ENEE 630 Fall’2009 Homework Solution #3...

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