Mixing
Which Daubechies D6 wavelet to choose?
I was trying to calculate coefficients for D6 wavelet as an exercise. When I did factorization part (where you find polynomial $L(e^{iomega})$ such that $L(e^{iomega})L(e^{-iomega}) = Q(cos omega)$), I realized that this factorization is not unique. For example, I found this representation of the scaling function:
$h(x) = sqrt 2 [c_{0}h(2x) + c_{1}h(2x-1) + c_{2}h(2x-2) + c_{3}h(2x-3) + c_{4}h(2x-4) + c_{5}h(2x-5)]$.
Where
$c_0 approx -0.33267054$
$c_1 approx -0.80689156$
$c_2 approx -0.4598775$
$c_3 approx 0.13501099$
$c_4 approx 0.08544126$
$c_5 approx -0.03522629$
But Wikipedia suggests another coefficients for D6 wavelet. After reading wikipedia, I understand now that one factorization is somehow better than the others:
Among the 2A−1 possible solutions of the algebraic equations for the moment and orthogonality conditions, the one is chosen whose scaling filter has extremal phase.
Can you explain "extremal phase" concept to me? I also found this page:
https://www.mathworks.com/help/wavelet/gs/extremal-phase-wavelet.html
This page says this is about faster growth of "cumulative sums of the squared coefficients" which becomes 1 eventually. So why faster growth is preferred?
wavelets
asked Sep 20 '18 at 20:40
shamaz.mazumshamaz.mazum
133
add a comment |
I was trying to calculate coefficients for D6 wavelet as an exercise. When I did factorization part (where you find polynomial $L(e^{iomega})$ such that $L(e^{iomega})L(e^{-iomega}) = Q(cos omega)$), I realized that this factorization is not unique. For example, I found this representation of the scaling function:
$h(x) = sqrt 2 [c_{0}h(2x) + c_{1}h(2x-1) + c_{2}h(2x-2) + c_{3}h(2x-3) + c_{4}h(2x-4) + c_{5}h(2x-5)]$.
Where
$c_0 approx -0.33267054$
$c_1 approx -0.80689156$
$c_2 approx -0.4598775$
$c_3 approx 0.13501099$
$c_4 approx 0.08544126$
$c_5 approx -0.03522629$
But Wikipedia suggests another coefficients for D6 wavelet. After reading wikipedia, I understand now that one factorization is somehow better than the others:
Among the 2A−1 possible solutions of the algebraic equations for the moment and orthogonality conditions, the one is chosen whose scaling filter has extremal phase.
Can you explain "extremal phase" concept to me? I also found this page:
https://www.mathworks.com/help/wavelet/gs/extremal-phase-wavelet.html
This page says this is about faster growth of "cumulative sums of the squared coefficients" which becomes 1 eventually. So why faster growth is preferred?
wavelets
asked Sep 20 '18 at 20:40
shamaz.mazumshamaz.mazum
133
add a comment |
I was trying to calculate coefficients for D6 wavelet as an exercise. When I did factorization part (where you find polynomial $L(e^{iomega})$ such that $L(e^{iomega})L(e^{-iomega}) = Q(cos omega)$), I realized that this factorization is not unique. For example, I found this representation of the scaling function:
$h(x) = sqrt 2 [c_{0}h(2x) + c_{1}h(2x-1) + c_{2}h(2x-2) + c_{3}h(2x-3) + c_{4}h(2x-4) + c_{5}h(2x-5)]$.
Where
$c_0 approx -0.33267054$
$c_1 approx -0.80689156$
$c_2 approx -0.4598775$
$c_3 approx 0.13501099$
$c_4 approx 0.08544126$
$c_5 approx -0.03522629$
But Wikipedia suggests another coefficients for D6 wavelet. After reading wikipedia, I understand now that one factorization is somehow better than the others:
Among the 2A−1 possible solutions of the algebraic equations for the moment and orthogonality conditions, the one is chosen whose scaling filter has extremal phase.
Can you explain "extremal phase" concept to me? I also found this page:
https://www.mathworks.com/help/wavelet/gs/extremal-phase-wavelet.html
This page says this is about faster growth of "cumulative sums of the squared coefficients" which becomes 1 eventually. So why faster growth is preferred?
wavelets
asked Sep 20 '18 at 20:40
shamaz.mazumshamaz.mazum
133
I was trying to calculate coefficients for D6 wavelet as an exercise. When I did factorization part (where you find polynomial $L(e^{iomega})$ such that $L(e^{iomega})L(e^{-iomega}) = Q(cos omega)$), I realized that this factorization is not unique. For example, I found this representation of the scaling function:
$h(x) = sqrt 2 [c_{0}h(2x) + c_{1}h(2x-1) + c_{2}h(2x-2) + c_{3}h(2x-3) + c_{4}h(2x-4) + c_{5}h(2x-5)]$.
Where
$c_0 approx -0.33267054$
$c_1 approx -0.80689156$
$c_2 approx -0.4598775$
$c_3 approx 0.13501099$
$c_4 approx 0.08544126$
$c_5 approx -0.03522629$
But Wikipedia suggests another coefficients for D6 wavelet. After reading wikipedia, I understand now that one factorization is somehow better than the others:
Among the 2A−1 possible solutions of the algebraic equations for the moment and orthogonality conditions, the one is chosen whose scaling filter has extremal phase.
Can you explain "extremal phase" concept to me? I also found this page:
https://www.mathworks.com/help/wavelet/gs/extremal-phase-wavelet.html
This page says this is about faster growth of "cumulative sums of the squared coefficients" which becomes 1 eventually. So why faster growth is preferred?
wavelets
wavelets
asked Sep 20 '18 at 20:40
shamaz.mazumshamaz.mazum
133
asked Sep 20 '18 at 20:40
shamaz.mazumshamaz.mazum
133
asked Sep 20 '18 at 20:40
shamaz.mazumshamaz.mazum
133
asked Sep 20 '18 at 20:40
shamaz.mazumshamaz.mazum
133
asked Sep 20 '18 at 20:40
shamaz.mazumshamaz.mazum
133
133
add a comment |
add a comment |
1 Answer
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For each root $z$ of $Q$, $1/z$ is also a root. Daubechies original construction chose among the pair ${z, 1/z}$ the root with smaller absolute value. Later, Daubechies realized this was probably not a good choice. She discusses this in "Ten Lectures on Wavelets", section 8.1.1. The choice of smallest magnitude roots (equivalent to extremal phase) makes the Daubechies wavelet "bunch up" at the start of their support. In order to fix this, Daubechies proposed the "symlets", which have smallest deviation from linear phase. In order to find symlets, you go through every combination of roots and compute the L2 norm of the phase with the linear regression subtracted out. The set of roots resulting in the smallest norm produces a more symmetric wavelet which is more spread out over its support.
So the premise of your question is not correct: The choice of roots that results in extremal phase is not better than other choices, and is arguably worse in many respects.
answered Nov 21 '18 at 19:44
user14717user14717
3,8281020
add a comment |
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1 Answer
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For each root $z$ of $Q$, $1/z$ is also a root. Daubechies original construction chose among the pair ${z, 1/z}$ the root with smaller absolute value. Later, Daubechies realized this was probably not a good choice. She discusses this in "Ten Lectures on Wavelets", section 8.1.1. The choice of smallest magnitude roots (equivalent to extremal phase) makes the Daubechies wavelet "bunch up" at the start of their support. In order to fix this, Daubechies proposed the "symlets", which have smallest deviation from linear phase. In order to find symlets, you go through every combination of roots and compute the L2 norm of the phase with the linear regression subtracted out. The set of roots resulting in the smallest norm produces a more symmetric wavelet which is more spread out over its support.
So the premise of your question is not correct: The choice of roots that results in extremal phase is not better than other choices, and is arguably worse in many respects.
answered Nov 21 '18 at 19:44
user14717user14717
3,8281020
add a comment |
For each root $z$ of $Q$, $1/z$ is also a root. Daubechies original construction chose among the pair ${z, 1/z}$ the root with smaller absolute value. Later, Daubechies realized this was probably not a good choice. She discusses this in "Ten Lectures on Wavelets", section 8.1.1. The choice of smallest magnitude roots (equivalent to extremal phase) makes the Daubechies wavelet "bunch up" at the start of their support. In order to fix this, Daubechies proposed the "symlets", which have smallest deviation from linear phase. In order to find symlets, you go through every combination of roots and compute the L2 norm of the phase with the linear regression subtracted out. The set of roots resulting in the smallest norm produces a more symmetric wavelet which is more spread out over its support.
So the premise of your question is not correct: The choice of roots that results in extremal phase is not better than other choices, and is arguably worse in many respects.
answered Nov 21 '18 at 19:44
user14717user14717
3,8281020
add a comment |
For each root $z$ of $Q$, $1/z$ is also a root. Daubechies original construction chose among the pair ${z, 1/z}$ the root with smaller absolute value. Later, Daubechies realized this was probably not a good choice. She discusses this in "Ten Lectures on Wavelets", section 8.1.1. The choice of smallest magnitude roots (equivalent to extremal phase) makes the Daubechies wavelet "bunch up" at the start of their support. In order to fix this, Daubechies proposed the "symlets", which have smallest deviation from linear phase. In order to find symlets, you go through every combination of roots and compute the L2 norm of the phase with the linear regression subtracted out. The set of roots resulting in the smallest norm produces a more symmetric wavelet which is more spread out over its support.
So the premise of your question is not correct: The choice of roots that results in extremal phase is not better than other choices, and is arguably worse in many respects.
answered Nov 21 '18 at 19:44
user14717user14717
3,8281020
For each root $z$ of $Q$, $1/z$ is also a root. Daubechies original construction chose among the pair ${z, 1/z}$ the root with smaller absolute value. Later, Daubechies realized this was probably not a good choice. She discusses this in "Ten Lectures on Wavelets", section 8.1.1. The choice of smallest magnitude roots (equivalent to extremal phase) makes the Daubechies wavelet "bunch up" at the start of their support. In order to fix this, Daubechies proposed the "symlets", which have smallest deviation from linear phase. In order to find symlets, you go through every combination of roots and compute the L2 norm of the phase with the linear regression subtracted out. The set of roots resulting in the smallest norm produces a more symmetric wavelet which is more spread out over its support.
So the premise of your question is not correct: The choice of roots that results in extremal phase is not better than other choices, and is arguably worse in many respects.
answered Nov 21 '18 at 19:44
user14717user14717
3,8281020
answered Nov 21 '18 at 19:44
user14717user14717
3,8281020
answered Nov 21 '18 at 19:44
user14717user14717
3,8281020
answered Nov 21 '18 at 19:44
user14717user14717
3,8281020
3,8281020
add a comment |
add a comment |
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