convolution of the corner pixel

Hi,

I am using 2-D image here and I used this command

conv2(Img, [1,1,1; 1,0,1; 1,1,1] / 8, 'same')

Example:

image1 = [1 2 3 3 0;3 4 1 0 1 ;2 3 0 7 5;4 0 2 5 1;0 3 1 6 4];

image2=conv2(image1, [1,1,1;1,0,1;1,1,1] / 8, 'same');

I got result like this:

1.125000 1.5000 1.25000 0.625000 0.5000 1.5000 1.875000 2.75000 2.5000 1.875000 1.75000 2 2.75000 1.875000 1.75000 1 1.875000 3.125000 3.25000 3.375000 0.87500 0.875000 2 1.62500 1.5000

By using conv2 I think it will give the average of the surrounding pixels

In image1 3rd row 3rd column 0 is replaced by the average of the surrounding pixels like (4+1+0+3+7+0+2+5)/8=2.75000

What about the corner pixel i.e 1st row and 1st column? 1 is replaced by 1.25000

How this 1.25000 came

How it is taking the average of the corner pixel

Can anyone suggest something about this

Thanks in advance

Answer 1

[Andrei Bobrov]

>> image2_11  = [0 0 0;0 1 2;0 3 4]
image2_11 =
     0     0     0
     0     1     2
     0     3     4
>>L = true(3);L(2,2)=false
L =
   1     1     1
   1     0     1
   1     1     1
>> image2_11 = image2_11.*L
ans =
   0     0     0
   0     0     2
   0     3     4
>>image2_11 = sum(image2_11(:))/8 
image2_11 = 1.125

Answer 2

[Image Analyst]

When the lower right corner of the kernel just barely overlaps your image, only the "1" in the lower right of the kernel overlaps only the "1" of your image. (1*1) / 8 = 0.125. Not sure what to suggest. That's how convolution works. You can crop it away if you want using the 'same' option, or you can program up various other ways to handle the edge effects.

Answer 3

[Bjorn Gustavsson]

If you have access to the image processing toolbox you can use imfilter (that also does linear filtering). Imfilter have some extra luxurious handling of edges - 'replicate', 'symmetric' (maybe even 'circular'?). These might do what you want for edges and corners. If you haven't got the IP-toolbox, I have to say as Image Analyst: you'll have to write it yourself.

HTH