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Handout5 - Course 18.327 and 1.130 Wavelets and Filter...

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Course 18.327 and 1.130 Wavelets and Filter Banks Modulation and Polyphase Representations: Noble Identities; Block Toeplitz Matrices and Block z-transforms; Polyphase Examples 2 Modulation Matrix Matrix form of PR conditions: [F 0 (z) F 1 (z)] H 0 (z) H 0 (-z) = [ 2z œ ? 0 ] H 1 (z) H 1 (-z) So [ F 0 (z) F 1 (z)] = [2z œ ? 0] H œ1 (z) H m œ1 (z) = 1 Modulation matrix, H m (z) H 0 (z) H 1 (-z) - H 0 (-z) H 1 (z) (must be non-zero) H 1 (-z) -H 0 (-z) -H 1 (z) H 0 (z) 1
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Ω ? ? *+, ? Ω F 0 (z) = 2z - ? H 1 (-z) Require these F 1 (z) = 2z - ? H 0 (-z) to be FIR Suppose we choose = 2z - Then F 0 (z) = H 1 (-z) F 1 (z) = -H 0 (-z) 3 1 4 Synthesis modulation matrix: Complete the second row of matrix PR conditions by replacing z with œz: F 0 (z) F 1 (z) H 0 (z) H 0 (-z) z - ? 0 F 0 (-z) F 1 (-z) H 1 (z) H 1 (-z) 0 (-z) - ? Synthesis modulation matrix, F m (z) Note the transpose convention in F m (z). = 2 2
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5 Noble Identities 1. Consider U(z) = H(z 2 )X(z) Y(z) = ² {U(z ² ) + U(-z ² )} (downsampling) = ² { H (z) X (z ² ) + H(z) X (- z ² )} = H(z) • ² {X (z ² ) + X (-z ² )} Ω can downsample first First Noble identity: x [n] x [n] H(z 2 ) é 2 y[n] u[n] é 2 H(z) y[n] ô y[n] x[n] H(z 2 ) é 2 6 2. Consider x[n] u[n] y[n] U(z) = H(z) X(z) Y(z) = U(z 2 ) (upsampling) = H(z 2 ) X(z 2 ) Ω can upsample first Second Noble Identity: x[n] y[n] x[n] y[n] H(z) å 2 ô å 2 H(z) å 2 H(z 2 ) 3
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7 Derivation of Polyphase Form 1. Filtering and downsampling: x[n] y[n] é 2 H(z) H(z) = H even (z 2 ) + z -1 H odd (z 2 ); h even [n] = h[2n] h odd [n] = h[2n+1] x[n] H even (z 2 ) H odd (z 2 ) é 2 y[n] z -1 8 x[n] z -1 H even (z 2 ) H odd (z 2 ) é 2 é 2 y[n] + z -1 H even (z) H odd (z) é 2 é 2 y[n] + x[n] Polyphase Form x even [n] x odd [n-1] 4
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This note was uploaded on 12/04/2011 for the course ESD 18.327 taught by Professor Gilbertstrang during the Spring '03 term at MIT.

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Handout5 - Course 18.327 and 1.130 Wavelets and Filter...

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