Let us try to understand convolution by performing
spatial averaging on a matrix without using MATLAB built in function ‘conv2()’.
%CONVOLUTION IN MATLAB
%INPUT MATRIX
A = zeros(5);
A(:) = 1:25;
%KERNEL
avg3 = ones(3)/9;
%CONVOLUTION
Result = conv2(A,avg3,'same');
display(Result);
Steps to be performed:
NOTE :
To define a kernel for spatial averaging, fill the
kernel with ones and divide it by the number of elements in it.
For instance, consider kernel of size 4x4 , fill
the matrix with ones and divide it by 16. i.e the total number of elements in
the matrix.
MATLAB CODE:
%INPUT MATRIX
A = zeros(5);
A(:) = 1:25;
%KERNEL
avg3 = ones(3)/9;
%PAD THE MATRIX WITH ZEROS
B = padarray(A,[1 1]);
% PRE-ALLOCATE THE MATRIX
Output = zeros([size(A,1)
size(A,2)]);
%PERFORM COONVOLUTION
for i = 1:size(B,1)-2
for j = 1:size(B,2)-2
Temp = B(i:i+2,j:j+2).*avg3;
Output(i,j) = sum(Temp(:));
end
end
display(Output);
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10 comments:
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Congratulations very interesting and didactically, do you have a youtube channel? if no, think about it.
Man You are a life save, thanks for this easy explaining :) :)
Hey In understand your example very clearly. I have question. If i want to subtraction any pixel intensity from its 8 neighbors pixel intensity(this is for every pixels in the image) and then take the absolute sum of the subtraction. How should I define the kernel (avg3 in your example) and perform the task with conv2?
Thanks was very helpful for my assignment!
plz tell me how to use it , I cant understand...
Hello, I keep getting the following error "Subscript indices must either be real positive integers or logicals.
Error in test (line 16)
Output(i,j) = sum(Temp(:));"
do you have any advice? thank you
I tried with a kernel containing elements with different values (not a constant value as in your example) and the convolution with conv2 or with the for loop gave different results, why? I used the following code:
%CONVOLUTION IN MATLAB with conv2
clear
%INPUT MATRIX
A = zeros(5);
A(:) = 1:25;
%KERNEL
avg3 = rand(3);
%CONVOLUTION
Result = conv2(A,avg3,'same');
display(Result);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% EXPLICIT calculation using the same kernel
%PAD THE MATRIX WITH ZEROS
B = padarray(A,[1 1]);
% PRE-ALLOCATE THE MATRIX
Output = zeros([size(A,1) size(A,2)]);
%PERFORM COONVOLUTION
for i = 1:size(B,1)-2
for j = 1:size(B,2)-2
Temp = B(i:i+2,j:j+2).*avg3;
Output(i,j) = sum(Temp(:));
end
end
display(Output);
@Michele
Rotate the convolution mask by 180 degrees.
Example:
Temp = B(i:i+2,j:j+2).*rot90(avg3,2);
The mask in the given example is symmetric so rotating it by 180 degree yielded the same mask.
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