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The explicit four-tap Daubechies pair satisfies the orthogonal filter and two-moment conditions

Statement

The pair $(h, g)$ in Explicit four-tap Daubechies filter and its high-pass partner is admissible in Periodic finite orthogonal compact-filter wavelet transform. In addition, $\sum_{j = 0}^{3} j^{r} g_{j} = 0$ for $r = 0, 1$ , so the high-pass filter annihilates constant and affine data.

Facts & Assumptions

Given: The exact four entries of $h$ and the definition $g_{j} = (- 1)^{j} h_{3 - j}$ .

[F1]

Explicit four-tap Daubechies filter and its high-pass partner: $h = (1 + \sqrt{3}, 3 + \sqrt{3}, 3 - \sqrt{3}, 1 - \sqrt{3}) / (4 \sqrt{2})$ and $g = (h_{3}, - h_{2}, h_{1}, - h_{0})$ .

[F2]

Periodic finite orthogonal compact-filter wavelet transform: admissibility requires unit even-shift autocorrelation for each filter, zero cross-correlation at every even shift, $\sum_{j} h_{j} = \sqrt{2}$ , and $\sum_{j} g_{j} = 0$ .

Proof

1.1givenF1algebra

Write $r = \sqrt{3}$ . Direct addition gives $\sum h_{j} = 8 / (4 \sqrt{2}) = \sqrt{2}$ . The numerator of $\sum g_{j}$ is $(1 - r) - (3 - r) + (3 + r) - (1 + r) = 0$ , and the numerator of $\sum j g_{j}$ is $- (3 - r) + 2 (3 + r) - 3 (1 + r) = 0$ . Thus $g$ has the required zero sum and two vanishing moments.

1.2givenF1algebra

With denominator $32$ , the numerator of $\sum h_{j}^{2}$ is $(4 + 2 r) + (12 + 6 r) + (12 - 6 r) + (4 - 2 r) = 32$ , so the zero-shift autocorrelation is $1$ . The only possible nonzero even-shift autocorrelation of a four-tap filter is at shift $2$ or $- 2$ ; at shift $2$ its numerator is $(3 - r) (1 + r) + (1 - r) (3 + r) = 2 r - 2 r = 0$ , and the shift $- 2$ sum is the same by reindexing. Reversal with alternating signs gives $\sum g_{j}^{2} = 1$ and $\sum g_{j} g_{j - 2} = h_{3} h_{1} + h_{2} h_{0} = 0$ .

2.1step 1.1step 1.2givenF1F2algebra∎

For the cross-correlation $\sum_{j} h_{j} g_{j - 2 s}$ , only $s = - 1, 0, 1$ can have overlapping taps. These sums are respectively $h_{0} g_{2} + h_{1} g_{3} = h_{0} h_{1} - h_{1} h_{0} = 0$ , $h_{0} g_{0} + h_{1} g_{1} + h_{2} g_{2} + h_{3} g_{3} = 0$ by pairwise cancellation, and $h_{2} g_{0} + h_{3} g_{1} = h_{2} h_{3} - h_{3} h_{2} = 0$ . For all other integers $s$ the supports do not overlap, so the sum is zero. Together with steps 1.1 and 1.2 this verifies every condition in [F2].

Depends on

Used by

Cited to discharge well-definedness by Explicit four-tap Daubechies filter and its high-pass partner.

Dependency tree · two levels

3 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Gilbert Strang and Kevin Amaratunga, Wavelets and Filter Banks, MIT OpenCourseWare 18.327, Handout 8 (standard reference, not scraped)
Jelena Kovačević, Vivek K. Goyal, and Martin Vetterli, Fourier and Wavelet Signal Processing (standard reference, not scraped)