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I'm new to multi scale transformations. I was wondering if there is a special case where wave atom converts to curvelet transform? Can I use wave atom parameters to have curvelet properties?

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  • $\begingroup$ Thank you for this question. A natural interrogation would be: why would you want to use wave atoms instead of curvelets, and what curvelet properties are you interested in? $\endgroup$
    – Laurent Duval
    Jul 24, 2017 at 8:45
  • $\begingroup$ @LaurentDuval Because I could not handle 3D curvelet code to use under windows 64-bit and MATLAB environment. $\endgroup$
    – MJay
    Jul 25, 2017 at 4:54
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    $\begingroup$ Understood. Being closer to packets, and less elongated, they may behave a little differently. However, the reduced sparsity of wave atoms can be useful to tune in practice $\endgroup$
    – Laurent Duval
    Jul 25, 2017 at 8:18

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In their 2007 paper, Wave atoms and sparsity of oscillatory patterns, Demanet and Ying draw connections between wave atoms and other types of wavelets:

Wave atoms,

In plain words:

We introduce “wave atoms” as a variant of 2D wavelet packets obeying the parabolic scaling wavelength

with a better spatial localization, meant for texture analysis. Their choice ($\alpha=\beta=1/2$) was a compromise between two types of sparsification (preservation under warping, sparsity of oscillations). Yet, one could built different collections of such wave packets, with alternative choices of $\alpha$ and $\beta$, bridging the gap between the different transform choices.

Yet, their natural redundancy is 2, with an orthonormal variant, and a complex one with redundancy 4, different from that of the curvelets. So I do no think they can be used as a special case of curvelets.

A tutorial on 2D geometric transformation can be obtain in A Panorama on Multiscale Geometric Representations, Intertwining Spatial, Directional and Frequency Selectivity, 2011.

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  • $\begingroup$ Thank you for your response, I meant "is there a parameter in wave atom formulation to convert it to curvelet transform?" $\endgroup$
    – MJay
    Jul 25, 2017 at 5:05
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    $\begingroup$ So the global answer is no. From the paper, this is a fixed transformation (ie non-parametric) $\endgroup$
    – Laurent Duval
    Jul 25, 2017 at 8:03

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