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Also, does anyone know the formula? In 1-D, it is:

H(x) = 2/sqrt(pi) * G(x) * integral[0->x/ssqrt(2)] exp(s^2)ds

where the integral is Dawson's.

If this is OFF TOPIC then please, all you experts, point me to a forum where I might find my answer.

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Quoting - rogene

Also, does anyone know the formula? In 1-D, it is:

H(x) = 2/sqrt(pi) * G(x) * integral[0->x/ssqrt(2)] exp(s^2)ds

where the integral is Dawson's.

If this is OFF TOPIC then please, all you experts, point me to a forum where I might find my answer.

If you're looking for the Hilbert Transform there was working 1D versionof in the IPP 5.x

Presumably, based on that you could buildyour 2D version.

From implementation point of view you may also want to read something about the analytic signal http://en.wikipedia.org/wiki/Hilbert_transform#Analytic_representation

Hope it helps.

AndrewK

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