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I'm sure there must be an easy way to do this, but given the Fourier transform of an isotropic filter kernel, $\hat{f}(\mathbf{u}) = \mathcal{F}f(\mathbf{z})$, can one calculate the value of the kernel at $\mathbf{z} = 0$?

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Since $$f(\mathbf{z})=\int_{\mathbf{R}^n}\hat{f}(\mathbf{u})e^{2\pi i\mathbf{z}\cdot\mathbf{u}}\;d\mathbf{u}$$

$$f(\mathbf{0})=\int_{\mathbf{R}^n}\hat{f}(\mathbf{u})\;d\mathbf{u}$$

So you simply integrate (or sum in the discrete case) over $\hat{f}(\mathbf{u})$.

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  • $\begingroup$ Thats what I thought... so why am I getting the wrong answer in MATLAB? FFT/IFFT scaling perhaps? Edit: yep! $\endgroup$ – geometrikal May 15 '13 at 8:08
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    $\begingroup$ In the other direction (from the "spatial" domain to the "frequency" domain) it is more intuitive: The component at f=0 frequency (DC level) is what you get when you average the signal at the "spatial" domain. $\endgroup$ – nimrodm May 15 '13 at 8:20
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    $\begingroup$ @nimrodm oh yes of course, great descriptive explanation. I forgot the inv FT is just the FT with a 180 deg phase change. $\endgroup$ – geometrikal May 15 '13 at 8:32

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