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Unformatted text preview: Course 18.327 and 1.130 Course Wavelets and Filter Banks Maxflat Filters: Daubechies and Filters: Daubechies and Meyer Formulas. Meyer Spectral Factorization Formulas for the Product Filter Halfband condition: P(ω) + P( ω + π) = 2 Also want P(ω) to be lowpass to lowpass and p[n] to be symmetric. P (ω + π) P (ω ) 2 ω π 0 Daubechies’ Approach ~ D~ esign Design a polynomial, P(y), of degree 2p - 1, such that P(0) = 2 ~ ~ P(l)(0) = 0; l = 1, 2, …, p - 1 (0) 1, ~ P(l)(1) = 0; l = 0, 1, …, p - 1 (1) 0, P(y) 2 0 1 y 2 Can achieve required flatness at y = 1 by including a Can term of the form (1 – y)p i.e. ∼ P(y) = 2(1 – y)p Bp(y) P(y) Where Bp(y) is a polynomial of degree p – 1. (y) How to choose Bp(y)? Let Bp(y) be the binomial series expansion for (y) (1 – y)-p, truncated after p terms: p(p + 1) 2 2p Bp(y) = 1 + py + y + … + 2p - 2 yp-1 (y) py 2 p–1 = (1 – y)-p + O(yp) (1 O(y ( ) < Higher order terms 3 ∞ (1 – y)-1 = ∑ yk (1 k=0 (1 – (1 y)-p ∞ =∑ k=0 (p + k –1 ) y k k | y| < 1 Then ∼ P(y) = 2(1 – y)p[(1-y)-p + O (yp)] P(y) = 2 + O(yp) O(y 4 Thus P(l)(0) = 0 ; l = 1, 2, … , p-1 (0) 1, So we have p-1 ∼ P(y) = 2 (1-y)p ∑ p + k - 1 k k=0 ( Now let y= Thus Thus 1 - eiω 2 1 – cos ω = 2 ( )( 1 – e-iω 2 )y ) k maintains symmetry ∼ P(ω) = P (1 – cos ω ) 2 1 + cos ω p + k + 1 1 – cos ω cos = 2( ∑( ) )( 2 ) k 2 p p-1 k k=0 5 z domain: domain: ( P(z) = 2 1 + z P(z) 2 p )( 1 + z-1 2 p p-1 ) ∑ k=0 ( k )( ) ( p+k-1 k 1-z 2 1 – z-1 2 k ) 6 Meyer’s Approach ∼ Work with derivative of P(y): ∼ P´(y) = - C´ yp-1 (1 –y)p-1 (y) (1 So y ∼ P(y) = 2 - C´∫ yp-1 (1-y)p-1 d y P(y) ∼ (P(0) = 2) 0 Then ω 1 –...
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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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