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I came across this equation while trying to process an image $z$,

$$ \mathcal{F}\left(\mathbf{D}^T \mathbf{D}\right) \mathcal{F}\left(z\right), $$

where $\mathcal{F}$ is the 2-D Fourier transform, and $\mathbf{D}$ is the gradient operator, which I think is in matrix form. However, I am not sure on how to implement this on code; it is simple if $\mathbf{D}$ is acting on the image, but the Fourier transform only acts on the operators themselves.

I came across these two posts that generates the horizontal and vertical gradients and transposes the gradient, but I am not sure on how to apply the horizontal and vertical gradients to acquire $\mathbf{D}$ since they have different sizes on a non-square image (I have thought of cutting the image into square sections, but the $\mathcal{F}\left(z\right)$ accounts the whole image), while the example given in the transposition of the gradient is acting on the image itself (although I could act the transpose of the gradient to the gradient itself, but that leads me to the first problem).

My initial thought is to use $\mathbf{D}$ as,

$$ \mathbf{D} = \begin{bmatrix} D_h \\ D_v \end{bmatrix}, $$

where $D_h$ and $D_v$ are the horizontal and vertical gradients, respectively, but I am not sure what their sizes should be if I were to append zeros, given for example that my image is $n \times m$.

This equation was found on the paper Optics Temperature Dependent Non Uniformity Correction Via $ {l}_{0} $ Regularized Prior for Airborne Infrared Imaging Systems, at equation (7).

So, my question is how to generate the gradient operator $\mathbf{D}$. Thank you in advance.

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  • $\begingroup$ Could you please cite the original material where you saw that equation? If it is available on the internet, and is not behind a paywall, please include a link. $\endgroup$
    – TimWescott
    Jun 6, 2021 at 21:45
  • $\begingroup$ Here is the original material--it is shown in Equation (7). Note that for simplicity I have set $\eta_0 = 0$, $\eta_1 = 1$ and $b = 0$. $\endgroup$
    – magus_e
    Jun 7, 2021 at 5:36

1 Answer 1

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There are 2 things to remember in the context you're asking:

  1. Image as Column Stacked Vector
    It seems what's confuses you is that you think the operator is working on the image in its 2D form (Matrix). Yet in practice you should reshape it into a column vector. In my answer you linked to, have a look how it operates on the image in mIx = reshape(mDh * mI(:), numRows, numCols - 1);.

  2. The Context of Your Paper
    In the paper above they use to apply the operators on the frequency domain to make the computation faster. So what you need to do is understand that $ {D}^{T} D $ is basically the Laplace Operator (See in Discrete Laplace Operator how it is implemented for 2D Discrete case). Then all needed is to use Replicate MATLAB's conv2() in Frequency Domain to build the operator in Fourier Transform and apply the equation.

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  • $\begingroup$ Let me confirm if I understand correctly: from the link, the $\mathcal{F}\left(\mathbf{D}^T \mathbf{D}\right)$ can be built? $\endgroup$
    – magus_e
    Jun 7, 2021 at 9:52
  • $\begingroup$ Of course it can be built. Moreover, it can be built directly without going through $ D $. $\endgroup$
    – Royi
    Jun 7, 2021 at 13:12
  • $\begingroup$ Thank you for the clarification! I wanted to clarify one more thing, which is the kernel to be used. What would be the kernel for the convolution, since they are divided into two the horizontal and vertical gradients? I am under the impression that the kernel that I should use implements both horizontal and vertical gradients. $\endgroup$
    – magus_e
    Jun 7, 2021 at 13:40
  • $\begingroup$ I just realized that $\mathbf{D}^T \mathbf{D}$ is the Laplace Operator. Thank you very much! $\endgroup$
    – magus_e
    Jun 8, 2021 at 7:19
  • $\begingroup$ @magus_e, Indeed, hence you can computer its Fourier Transform directly as I wrote. $\endgroup$
    – Royi
    Jun 9, 2021 at 4:03

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