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For many papers which talks about using wavelet transforms for feature detection in image processing, it is stated that it is advantageous for the wavelet to have zero mean? Why is this so? Thanks in advance.

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Please consider that the wavelet transform coefficients represent the local details of the signal. In (incomplete) DWT or a Gaussian pyramid, there is also a trend signal after the transform that contains the global behavior.

The first condition on this local-global separation is that constant or near constant parts of the signal get mapped to zero in the local details. The scalar product of a (wavelet) function with a constant function gives its DC contents. So DC of wavelet equal 0 means that the computed coefficients are high at regions of interest like edges.

The same goes for higher moments. If one wants to regard slow linear or quadratic "gradients" in the signal as not essential for the contents, or as part of the trend signal, then the first resp. second moment of the wavelet must vanish.


Or to consider it from a slightly different angle: Each scale of the wavelet transform corresponds to the application of a high-pass filter. One defining property of a high-pass filter is that the amplitude response around the frequency zero is small, ideally zero. Vanishing moments of the wavelet function correspond to a flat graph of the Fourier transform of the wavelet at frequency zero.

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    $\begingroup$ Very nice theoretical analysis, but can you add some formula, and/or figures to better illustrate it? $\endgroup$ – lennon310 Feb 10 '14 at 15:46
  • $\begingroup$ The nice images are in the link in the comment to the related question, especially the cat image. The question was for a quality, the zero DC of the wavelet function, I'm at a loss to add quantitative formulas (without replicating wikipedia articles or text books). $\endgroup$ – LutzL Feb 10 '14 at 15:56
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    $\begingroup$ @LutzL: For image processing applications, what detrimental effect does it have if the DC of the wavelet function is not zero? It's best if you can show some equations to make it mathematically sound. $\endgroup$ – user1734134 Feb 11 '14 at 14:24
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It is advantageous for a simple reason:

  • The DC component carries little information for signal processing.

(well, you can find counterexamples - but from my experience, they would weight 0.01%)

Indeed, the mean of wavelet coefficient, when convolved with an image, would transfer to this mean times the DC component. Having this information in your wavelet coefficients vector would just add some redundant, mostly uninformative bias.

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    $\begingroup$ Hi @meduz: Thanks for your answer. Can you show your reasoning mathematically? $\endgroup$ – user1734134 Feb 14 '14 at 10:25
  • $\begingroup$ You can find counter examples (like if you want to use the average value in your vector of coefficients) so there is litlle proof to do mathematically. What you can do is to look for the prior distribution of coefficients in the signal versus noise. $\endgroup$ – meduz Mar 3 '14 at 13:47

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