Unformatted text preview: cos ω
P(ω) = 2 - C´∫
0 ( ω ( 1–
= 2 - C ´∫
(1 + 2 os ω )
) ω p-1 ) 2p-1 i.e. P(ω) = 2 – C ∫ sin p-1 p-1 sin ω dω
2 sin ω dω
7 Spectral Factorization
Recall the halfband condition for orthogonal filters:
H0(z) H0(z-1) + H0(-z) H0(-z-1) = 2
H0(ω) + H0(ω + π) =2
The product filter for the orthogonal case is
P(z) = H0(z) H0(z-1)
P(ω) = H0(ω)
⇒ P(ω) ≥ 0
p[n] = h0[n] ∗ h0[-n]
⇒ p[n] = p[-n]
The spectral factorization problem is the problem
of finding H0(z) once P(z) is known.
8 Consider the distribution of the zeros (roots) of P(z).
• Symmetry of p[n] ⇒ P(z) = P(z-1)
If z0 is a root then so is z0-1.
If p[n] are real, then the roots appear in complex,
(1 – z0 z-1)(1 – z0*z-1) = 1 – (z0 + z0*) z-1 + (z0z0*)z-2
)(1 123 123 real real 9 Im O1
z0 Im z0
1 Re *
z0 z0 1 1/z Re
Complex zeros Real zeros If the zero z0 is grouped into the spectral factor H0(z),
then the zero 1/z0 must be grouped into H0(z-1).
⇒ h0[n] cannot be symmetric.
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.
- Spring '03