I am struggling trying to understand the 3rd line of the expression below. The second line can be simply visualized by letting the kernel size be 3 and $i=2$ for eg.

Then $ y_2 = \sum_a x_{2+a-1}k_a = x_2k_1 + x_3k_2 + x_4k_3$

In the 3rd line, there is a change of variable letting $u=i+a-1$. What happens then ? How do I compute $y_2$ now ?

$ y_2 = \sum_u x_uk_{u-i+1} = ?$

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  • $\begingroup$ Isn't it $u=i+a-1 \implies a=u-i+1$ ?? Then the sum is materialized by $u$ instead of $a$. What is the issue? $\endgroup$ – AlexTP Jun 15 '18 at 13:14
  • $\begingroup$ I dont know. I just dont know how to write down the expression for $y_2$. Can u help me write it down so I can draw some blocks and vizualise what has happened ? $\endgroup$ – Kong Jun 15 '18 at 13:32
  • 3
    $\begingroup$ Whichever variable changing is used, the expression of $y_2$ does not change. Your kernel is zero for index smaller than 1 and greater than 3, then it is non-zero for $1 \leq u - i + 1 \leq 3$. Just replace $i=2$ then calculate the $u$ yielding the desired index. $\endgroup$ – AlexTP Jun 15 '18 at 13:57

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