Convolution using convn does not give expected result
显示 更早的评论
Dear all,
I am trying to use a kernel with values which should decrease the values in 3D array that you use. However, when I use convn and 'same', this leads to a plateau with a much higher value as the maximum value of the kernel. Am I using the wrong way to calculate the convolution, or what is happening?
load('s.mat')
A=ones(110,110,110);
C_con=convn(A,s_full,'same');
figure; surf(C_con(:,:,50));
采纳的回答
There's no reason to think convolution in general won't increase the values. As a trivial example:
A=ones(5);
B=convn(A,2,'same')
4 个评论
Lieke Pullen
2022-1-14
Not at alll. Here's a revised example with even lower kernel values
A=ones(10);
B=convn(A,[1,1,1,1,1],'same')
Now, if you had normalized the sum of the kernel, that would be a different story:
B=convn(A,[1,1,1,1,1]/5,'same')
Lieke Pullen
2022-1-14
Ah okay, then the values of the convolution are probably correct. Then the other question, why do I get a ''plateau''? I would expect, since the kernel itself has a sort of Gaussian shape, I would the convolution to also have this kind of shape. If I use a sphere, it looks like this:
I would expect a more smoother shape, with just one peak and not a plateau. How does this work?
You seem to be under the impression that convn(A,B) should always have the shape of B, but convolution is commutative, i.e., convn(A,B)= conv(B,A), so how can the shape of the result have more to do with B than with A? I could just as easily view A as the kernel here, so shouldn't the shape follow A rather than B?
更多回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Logical 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
