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scanned exercice

What is the advantage of using this equation? I guess we may use Taylor series, but I tried my best and could not get the equation.

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  • $\begingroup$ Welcome to SE.DSP. Generally, it is well-appreciated that one rewrites questions instead of doing a copy-paste of a scan. And personal thoughts are a plus $\endgroup$ Dec 18, 2016 at 21:13
  • $\begingroup$ I see,it is the first time for me to do so. and I want to ask,what if the drawback of this way? $\endgroup$
    – jenn
    Dec 19, 2016 at 2:07
  • $\begingroup$ K is floor(N/2) $\endgroup$
    – jenn
    Dec 19, 2016 at 2:08
  • $\begingroup$ Homogeneity of the notations and searchability are better when typing: the search engine cannot find words or equations easily in images $\endgroup$ Dec 19, 2016 at 3:23
  • $\begingroup$ how did you define the leftmost point and the rightmost one?a little confused $\endgroup$
    – jenn
    Dec 20, 2016 at 2:48

1 Answer 1

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An image is 2D, the kernel seems 1D. So let us double check: from (b), the output image is determined for each fixed $i$, and then only depends in $j$: Equation (1) only uses the second coordinate as the moving index.

So the kernel is effectively a $1\times N $ matrix. Its sum is $1$, and as $N$ is odd, you can interpret it as an equally-weighted moving average filter, applied along lines only. So it can behave like an horizontal motion blur (Question (a)).

The direct convolution requires $N-1$ adds, and one multiply (by $1/N$). Equation (1) tries to make it more efficient, both in terms of operations and calls to samples. It simply uses a recursive formulation for 1D moving average filters (a line of the image, here $x_j = I(.,j)$:

$$ \sum_{j=1}^N x_j = \sum_{j=0}^{N-1} x_j - x_0 + x_N $$

I suspect that in your formula, $K$ should be defined as $2K+1=N$. There is consistent with $K=\mathrm{floor}(N/2)$ for $N$ odd.

In other words: instead of doing sums of pixels each time you go to the right, you can reuse previous computations. The novel mean $I_0(i,j)$ is the last mean $I_0(i,j-1)$ minus $1/N$ the leftmost sample (the one that just left the previous frame) plus $1/N$ the rightmost sample (the one that just arrived).

Almost nothing to do with Taylor. The advantages:

  • a running buffer of $3$ values only is needed,
  • its is independent of the kernel size,
  • some computations can be saved if $N$ is relatively large (say $N\ge 5$).
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  • $\begingroup$ K is floor(N/2) $\endgroup$
    – jenn
    Dec 19, 2016 at 1:12
  • $\begingroup$ what the disadvantage of this way? $\endgroup$
    – jenn
    Dec 19, 2016 at 1:34
  • $\begingroup$ how did you define the leftmost point and the rightmost one?a little confused $\endgroup$
    – jenn
    Dec 20, 2016 at 2:48

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