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Given two functions $I, H$ we define the discrete convolution as

$$ I' (u,v) = I(u,v) \ast H(u,v) = \sum_{i = -\infty}^\infty \sum_{j = -\infty}^\infty I(u-i, v-j) H(i,j)$$

Now, I need to show that if $ H = H_1 \ast H_2 $ then $ I \ast H = (I \ast H_1) \ast H_2$. This is what I have done:

$$ I \ast H = \sum_{i = -\infty}^\infty\sum_{j = -\infty}^\infty I(u-i,v-j) \, \, H(i,j) = $$ $$ = \sum_{i = -\infty}^\infty\sum_{j = -\infty}^\infty I(u-i,v-j) \,\, \sum_{a = -\infty}^\infty\sum_{b = -\infty}^\infty H_1(i-a,j-b) H_2(a,b)$$

On the other hand:

$$ (I \ast H_1) \ast H_2 = \left( \sum_{i = -\infty}^\infty\sum_{j = -\infty}^\infty I(u-i,v-j) \, \, H_1(i,j) \right) \ast H_2(u,v) = $$ $$ = \sum_{a=-\infty}^\infty \sum_{b=-\infty}^\infty \sum_{i= -\infty}^\infty \sum_{j= -\infty}^\infty I(u-i-a,v-j-b) \, \, H_1(i,j) \, \, H_2(a,b)$$

Now, have I done any mistakes? How can I show that this two things are the same?

José D.
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  • Now you only need to "move" indexes $i$ and $j$ to get the same expression in both sides. – cjferes Aug 18 '14 at 15:50
  • I think it's correct but not in the most useful form. Going from your second-to-last line to the last line, you had the choice of whether to flip/translate $H_2$ or the expression in parentheses. You chose the expression in parentheses. I think it would be more profitable to choose $H_2$. –  Aug 18 '14 at 15:52
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    By the way, this problem is showing associativity of convolution. I think "separability" in 2-dimensional problems like this usually refers to whether you can separate, say, $I(u,v)$ into two factors, each depending on only one variable: $I_1(u)I_2(v)$. –  Aug 18 '14 at 15:54

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Note that $i,j,a,b$ are dummy variables and can be relabelled at will. In particular, we can replace $(i,j)\mapsto (i-a,j-b)$ in your last expression. This doesn't change the limits of summations since they're over all integers. How does this expression compare to the other desired form?

Semiclassical
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