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I have an images $u \in R^{M \times N}$, a Gaussian kernel $G$ and the gradient operator $\nabla u_{i,j}=[\nabla_{x} u_{i,j},\nabla_{y} u_{i,j}] \in (R^{2})^{M \times N}$ and I need to perform 2d deconvlotion as $(G \otimes \nabla u)_{i,j}$. Since the gradient operator $\nabla u_{i,j}$ consists of the horizontal component $\nabla_{x} u_{i,j}$ and the vertical component $\nabla_{y} u_{i,j}$, should I perform 2d convolution separately on these two components and the result belongs to $(R^{2})^{M \times N}$? In this case, I have tried as $$ (G \otimes \nabla u)_{i,j} = [\sum_{(m,n)\in \Omega} G(i-m,j-n)\nabla_{x} u_{m,n}~~, \sum_{(m,n)\in \Omega} G(i-m,j-n)\nabla_{y} u_{m,n}]$$ where $\Omega = \{1,...,M\}\times \{1,...,N\}$ and in this case $(G \otimes \nabla u)_{i,j} \in (R^{2})^{M \times N}$ , or I should sum 2d deconvolution over the two components as $$ (G \otimes \nabla u)_{i,j} = \sum_{(m,n)\in \Omega} G(i-m,j-n)\nabla_{x} u_{m,n}~~+ \sum_{(m,n)\in \Omega} G(i-m,j-n)\nabla_{y} u_{m,n}$$ where in this case $(G \otimes \nabla u)_{i,j} \in R^{M \times N}$. I really appreiciate if someone could please help me to solve this problem.

eli
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  • Is this question answered by my answer over on Stack Overflow, or is there something that is still unclear? – Cris Luengo Dec 04 '20 at 22:17
  • @Cris Luengo Thank you so much. I had asked it here before and since I did not receive any response here I asked it in stack overflow and you kindly helped me. I learned how to do it using your answer. Many thanks. – eli Dec 07 '20 at 13:30

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