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# E p 2 c sin p 1 p 1 sin d sin 2 sin d sin 2 d 7

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Unformatted text preview: cos ω P(ω) = 2 - C´∫ 2 0 ( ω ( 1– = 2 - C ´∫ 0 cos2 2 ω 0 cos c (1 + 2 os ω ) ) ω p-1 ) 2p-1 i.e. P(ω) = 2 – C ∫ sin p-1 p-1 sin ω dω sin 2 sin ω dω sin 2 ωdω 7 Spectral Factorization Recall the halfband condition for orthogonal filters: Recall halfband z domain: H0(z) H0(z-1) + H0(-z) H0(-z-1) = 2 Frequency domain: 2 2 H0(ω) + H0(ω + π) =2 The product filter for the orthogonal case is P(z) = H0(z) H0(z-1) 2 P(ω) = H0(ω) ⇒ P(ω) ≥ 0 p[n] = h0[n] ∗ h0[-n] ⇒ p[n] = p[-n] [n] n] The spectral factorization problem is the problem The of finding H0(z) once P(z) is known. of 8 Consider the distribution of the zeros (roots) of P(z). • • Symmetry of p[n] ⇒ P(z) = P(z-1) Symmetry If z0 is a root then so is z0-1. If p[n] are real, then the roots appear in complex, If conjugate pairs. conjugate (1 – z0 z-1)(1 – z0*z-1) = 1 – (z0 + z0*) z-1 + (z0z0*)z-2 (1 )(1 123 123 real real 9 Im O1 * z0 Im z0 1 Re * z0 z0 1 1/z Re 0 1 Oz 0 Complex zeros Real zeros If the zero z0 is grouped into the spectral factor H0(z), (z), then the zero 1/z0 must be grouped into H0(z-1). then must ⇒ h0[n] cannot be symmetric. 10 10 Daubechies’ choice: Choose H0(z) such that (i) all its zeros are inside or on the unit circle. (ii) it is causal...
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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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