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F09_630_rec_5a_sol

# F09_630_rec_5a_sol - ENEE630 ADSP 1 RECITATION 5 w solution...

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ENEE630 ADSP RECITATION 5 w/ solution October 8/9, 2009 1. Consider the QMF bank in Fig. R5.1, where H 0 ( z ) = 1 + 2 z - 1 + 3 z - 2 + z - 3 , and H 1 ( z ) = H 0 ( - z ). X ( z ) Y ( z ) - H 0 ( z ) - 2 - 2 - F 0 ( z ) - H 1 ( z ) - 2 - 2 - F 1 ( z ) - Figure R5.1: (a) Design an FIR synthesis bank F 0 ( z ) , F 1 ( z ) such that the system is alias-free. (b) Design an IIR synthesis bank F 0 ( z ) , F 1 ( z ) such that the system is perfect reconstruction. Is this bank stable? (By default, we assume it is causal) (c) Design a stable IIR synthesis bank F 0 ( z ) , F 1 ( z ) such that the system is alias-free and amplitude-distortion-free. (Hint: a * + z - 1 1+ az - 1 is an all-pass filter.) Solution: E 0 ( z ) = 1 + 3 z - 1 , E 1 ( z ) = 2 + z - 1 . F 0 ( z ) = z - 1 R 0 ( z 2 ) + R 1 ( z 2 ), F 1 ( z ) = - F 0 ( - z ) = z - 1 R 0 ( z 2 ) - R 1 ( z 2 ). (a) R 0 ( z ) = E 1 ( z ), R 1 ( z ) = E 0 ( z ). T ( z ) = 2 z - 1 E 0 ( z 2 ) E 1 ( z 2 ). (b) R 0 ( z ) = 1 /E 0 ( z ), R 1 ( z ) = 1 /E 1 ( z ). T ( z ) = 2 z - 1 . However, check that R 0 ( z ) is not stable. (c) Let e E 0 ( z ) = 3 + z - 1 , then E 0 ( z ) e E 0 ( z ) is an all-pass filter. R 0 ( z ) = 1 e E 0 ( z ) , R 1 ( z ) = E 0 ( z ) e E 0 ( z ) E 1 ( z ) . T ( z ) = 2 z - 1 E 0 ( z 2 ) e E 0 ( z 2 ) . Note: 1. When designing a QMF bank, we usually use the polyphase decomposition as in this problem.

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F09_630_rec_5a_sol - ENEE630 ADSP 1 RECITATION 5 w solution...

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