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mean image filter

开发者 https://www.devze.com 2022-12-29 01:44 出处:网络
Starting to learn image filtering and stumped on a question found on website: Applying a 3×3 mean filter twice does not produce quite the same result as applying a 5×5 mean filter once. However, a 5

Starting to learn image filtering and stumped on a question found on website: Applying a 3×3 mean filter twice does not produce quite the same result as applying a 5×5 mean filter once. However, a 5×5 convolution ker开发者_C百科nel can be constructed which is equivalent. What does this kernel look like?

Would appreciate help so that I can understand the subject better. Thanks.


Marcelo's answer is right. Another way of seeing it (more easy to think it first in one dimension) : we know that the mean filter is equivalent to a convolution with a rectangular window. And we know that the convolution is a linear operation, which is also associative.

Now, applying a mean filter M to a signal X can be written as

Y = M * X

where * denotes convolution. Appying the filter twice would then give

Y = M * (M * X) = (M * M) * X  = M2 * X

This says that filtering twice a signal with a mean filter is the same as filtering it once with an equivalent filter given by M2 = M * M. Now, this consists of applying the mean filter to itself, what gives a "smoother" filter (a triangular filter in this case).

The process can be repeated, (see first graph here) and it can be shown that the equivalent filter for many repetitions of a mean filter (N convolutions of the rectangular filter with itself) tends to a gaussian filter. Further, it can be shown that the gaussian filter has that property you didn't found in the rectangular (mean) filter: two passes of a gaussian filter are equivalent to another gaussian filter.


3x3 mean:

[1 1 1]
[1 1 1] * 1/9
[1 1 1]

3x3 mean twice:

[1 2 3 2 1]
[2 4 6 4 2]
[3 6 9 6 3] * 1/81
[2 4 6 4 2]
[1 2 3 2 1]

How? Each cell contributes indirectly via one or more intermediate 3x3 windows. Consider the set of stage 1 windows that contribute to a given stage 2 computation. The number of such 3x3 windows that contain a given source cell determines the contribution by that cell. The middle cell, for instance, is contained in all nine windows, so its contribution is 9 * 1/9 * 1/9. I don't know if I've explained it that well, so I hope it makes sense to you.


Actually I believe that 3x3 twice should give:

[1 2 3 2 1]
[2 4 6 4 2]
[3 6 9 6 3] * 1/81
[2 4 6 4 2]
[1 2 3 2 1]

The reason is because the sum of all values must be equal to 1.

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