Admissibility condition for wavelets
$begingroup$
The admissibility condition for a wavelet $psi$ is:
$int frac{|hatpsi(x)|^2}{|x|} dx < infty$, with $hatpsi$ the Fourier transform of $psi$.
A necessary and sufficient condition should be:
$$int psi(x) dx = 0quad mathrm{(or} ;hatpsi(0) = 0mathrm{)}$$
But I can't see how you should get from one to the other...
Does someone have some suggestions?
numerical-methods wavelets fourier-transform
edited Apr 30 '16 at 16:48
Laurent Duval
5,34311240
asked Apr 30 '16 at 16:04
Robbe MotmansRobbe Motmans
180212
$endgroup$
add a comment |
$begingroup$
The admissibility condition for a wavelet $psi$ is:
$int frac{|hatpsi(x)|^2}{|x|} dx < infty$, with $hatpsi$ the Fourier transform of $psi$.
A necessary and sufficient condition should be:
$$int psi(x) dx = 0quad mathrm{(or} ;hatpsi(0) = 0mathrm{)}$$
But I can't see how you should get from one to the other...
Does someone have some suggestions?
numerical-methods wavelets fourier-transform
edited Apr 30 '16 at 16:48
Laurent Duval
5,34311240
asked Apr 30 '16 at 16:04
Robbe MotmansRobbe Motmans
180212
$endgroup$
add a comment |
$begingroup$
The admissibility condition for a wavelet $psi$ is:
$int frac{|hatpsi(x)|^2}{|x|} dx < infty$, with $hatpsi$ the Fourier transform of $psi$.
A necessary and sufficient condition should be:
$$int psi(x) dx = 0quad mathrm{(or} ;hatpsi(0) = 0mathrm{)}$$
But I can't see how you should get from one to the other...
Does someone have some suggestions?
numerical-methods wavelets fourier-transform
edited Apr 30 '16 at 16:48
Laurent Duval
5,34311240
asked Apr 30 '16 at 16:04
Robbe MotmansRobbe Motmans
180212
$endgroup$
The admissibility condition for a wavelet $psi$ is:
$int frac{|hatpsi(x)|^2}{|x|} dx < infty$, with $hatpsi$ the Fourier transform of $psi$.
A necessary and sufficient condition should be:
$$int psi(x) dx = 0quad mathrm{(or} ;hatpsi(0) = 0mathrm{)}$$
But I can't see how you should get from one to the other...
Does someone have some suggestions?
numerical-methods wavelets fourier-transform
numerical-methods wavelets fourier-transform
edited Apr 30 '16 at 16:48
Laurent Duval
5,34311240
asked Apr 30 '16 at 16:04
Robbe MotmansRobbe Motmans
180212
edited Apr 30 '16 at 16:48
Laurent Duval
5,34311240
asked Apr 30 '16 at 16:04
Robbe MotmansRobbe Motmans
180212
edited Apr 30 '16 at 16:48
Laurent Duval
5,34311240
edited Apr 30 '16 at 16:48
Laurent Duval
5,34311240
edited Apr 30 '16 at 16:48
Laurent Duval
5,34311240
5,34311240
asked Apr 30 '16 at 16:04
Robbe MotmansRobbe Motmans
180212
asked Apr 30 '16 at 16:04
Robbe MotmansRobbe Motmans
180212
asked Apr 30 '16 at 16:04
Robbe MotmansRobbe Motmans
180212
180212
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
The admissibility condition (at least in the most common versions in 1D) can be stated as follows: function $phi$ is a synthesis (or reconstruction) wavelet for analysis wavelet $psi$ if the two constants:
$$c^{pm}_{phi,psi} =
2pi int_{0}^{+infty}tfrac{hat{overline{psi}}(pmomega)hatphi(pmomega)}{omega} domega
$$
are finite, non null and equal (see M. Holschneider, Wavelets, an analysis tool, 1995, p. 65 sq.).
This can be satisfied by a lot of couples of functions. If you now want the synthesis wavelet to be the same as the analysis wavelet, ie $phi = psi$, then a necessary condition is that $$int psi = 0$$ to avoid the singularity at $0$. This is the root for using discrete orthogonal wavelets (otherwise, you have for instance biorthogonal wavelets, used in JPEG 2000 image compression). In other words, the wavelet has zero-mean, which is equivalent to say (when integrals exists) that its spectrum vanished at zero (on the frequency axis).
But it is not sufficient in theory, and requires additional smoothed/boundedness/localization on $psi$. So my suggestion is to refer to accurate sources for sufficient conditions.
answered Apr 30 '16 at 16:47
Laurent DuvalLaurent Duval
5,34311240
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The admissibility condition (at least in the most common versions in 1D) can be stated as follows: function $phi$ is a synthesis (or reconstruction) wavelet for analysis wavelet $psi$ if the two constants:
$$c^{pm}_{phi,psi} =
2pi int_{0}^{+infty}tfrac{hat{overline{psi}}(pmomega)hatphi(pmomega)}{omega} domega
$$
are finite, non null and equal (see M. Holschneider, Wavelets, an analysis tool, 1995, p. 65 sq.).
This can be satisfied by a lot of couples of functions. If you now want the synthesis wavelet to be the same as the analysis wavelet, ie $phi = psi$, then a necessary condition is that $$int psi = 0$$ to avoid the singularity at $0$. This is the root for using discrete orthogonal wavelets (otherwise, you have for instance biorthogonal wavelets, used in JPEG 2000 image compression). In other words, the wavelet has zero-mean, which is equivalent to say (when integrals exists) that its spectrum vanished at zero (on the frequency axis).
But it is not sufficient in theory, and requires additional smoothed/boundedness/localization on $psi$. So my suggestion is to refer to accurate sources for sufficient conditions.
answered Apr 30 '16 at 16:47
Laurent DuvalLaurent Duval
5,34311240
$endgroup$
add a comment |
$begingroup$
The admissibility condition (at least in the most common versions in 1D) can be stated as follows: function $phi$ is a synthesis (or reconstruction) wavelet for analysis wavelet $psi$ if the two constants:
$$c^{pm}_{phi,psi} =
2pi int_{0}^{+infty}tfrac{hat{overline{psi}}(pmomega)hatphi(pmomega)}{omega} domega
$$
are finite, non null and equal (see M. Holschneider, Wavelets, an analysis tool, 1995, p. 65 sq.).
This can be satisfied by a lot of couples of functions. If you now want the synthesis wavelet to be the same as the analysis wavelet, ie $phi = psi$, then a necessary condition is that $$int psi = 0$$ to avoid the singularity at $0$. This is the root for using discrete orthogonal wavelets (otherwise, you have for instance biorthogonal wavelets, used in JPEG 2000 image compression). In other words, the wavelet has zero-mean, which is equivalent to say (when integrals exists) that its spectrum vanished at zero (on the frequency axis).
But it is not sufficient in theory, and requires additional smoothed/boundedness/localization on $psi$. So my suggestion is to refer to accurate sources for sufficient conditions.
answered Apr 30 '16 at 16:47
Laurent DuvalLaurent Duval
5,34311240
$endgroup$
add a comment |
$begingroup$
The admissibility condition (at least in the most common versions in 1D) can be stated as follows: function $phi$ is a synthesis (or reconstruction) wavelet for analysis wavelet $psi$ if the two constants:
$$c^{pm}_{phi,psi} =
2pi int_{0}^{+infty}tfrac{hat{overline{psi}}(pmomega)hatphi(pmomega)}{omega} domega
$$
are finite, non null and equal (see M. Holschneider, Wavelets, an analysis tool, 1995, p. 65 sq.).
This can be satisfied by a lot of couples of functions. If you now want the synthesis wavelet to be the same as the analysis wavelet, ie $phi = psi$, then a necessary condition is that $$int psi = 0$$ to avoid the singularity at $0$. This is the root for using discrete orthogonal wavelets (otherwise, you have for instance biorthogonal wavelets, used in JPEG 2000 image compression). In other words, the wavelet has zero-mean, which is equivalent to say (when integrals exists) that its spectrum vanished at zero (on the frequency axis).
But it is not sufficient in theory, and requires additional smoothed/boundedness/localization on $psi$. So my suggestion is to refer to accurate sources for sufficient conditions.
answered Apr 30 '16 at 16:47
Laurent DuvalLaurent Duval
5,34311240
$endgroup$
The admissibility condition (at least in the most common versions in 1D) can be stated as follows: function $phi$ is a synthesis (or reconstruction) wavelet for analysis wavelet $psi$ if the two constants:
$$c^{pm}_{phi,psi} =
2pi int_{0}^{+infty}tfrac{hat{overline{psi}}(pmomega)hatphi(pmomega)}{omega} domega
$$
are finite, non null and equal (see M. Holschneider, Wavelets, an analysis tool, 1995, p. 65 sq.).
This can be satisfied by a lot of couples of functions. If you now want the synthesis wavelet to be the same as the analysis wavelet, ie $phi = psi$, then a necessary condition is that $$int psi = 0$$ to avoid the singularity at $0$. This is the root for using discrete orthogonal wavelets (otherwise, you have for instance biorthogonal wavelets, used in JPEG 2000 image compression). In other words, the wavelet has zero-mean, which is equivalent to say (when integrals exists) that its spectrum vanished at zero (on the frequency axis).
But it is not sufficient in theory, and requires additional smoothed/boundedness/localization on $psi$. So my suggestion is to refer to accurate sources for sufficient conditions.
answered Apr 30 '16 at 16:47
Laurent DuvalLaurent Duval
5,34311240
edited Jan 11 at 21:43
answered Apr 30 '16 at 16:47
Laurent DuvalLaurent Duval
5,34311240
answered Apr 30 '16 at 16:47
Laurent DuvalLaurent Duval
5,34311240
answered Apr 30 '16 at 16:47
Laurent DuvalLaurent Duval
5,34311240
5,34311240
add a comment |
add a comment |
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