What precisely is conv2(A,B,'same')?

Hi All,

I am aware of the following:

• Given two matrices A and B, the matrix is a submatrix of their convolution , and I know the definition of .
• and A have the same size.

My question is:

If A is a matrix and B is , then what are the values of i and j such that ? That is, where does start inside the matrix ?

Many thanks

[Guillaume]

Isn't the doc clear enough?

• 'same' — Return the central part of the convolution, which is the same size as A.

[Anonymous]

It wasn't clear to me where the "central part" is chosen to be though. I guessed it was something like the answer that Bruno gave below. It's not the most natural looking formula in the world, so I wanted confirmation.

[Richard Lockwood]

I believe it is the middle half, from 1/4 to 3/4 of the total interval

[Image Analyst]

RIchard, no, not correct. It's not the middle 50%.

'Same' is the part of the output signal where the moving kernel's central element overlaps some part of the signal. So it includes signal where the central element of the moving kernel overlaps an element of the input signal and includes locations where the kernel "falls off" the end of the signal as long as the central element is still over the input signal.

'Full" is the entire output signal, even when the kernel is at the end and only a single element of the kernel overlaps the signal.

'Valid' is where the kernel fully overlaps the signal. So it would not include the ends of the signal where some parts of the moving kernel "fall off" the end of the input signal (they do not overlap).

Answer 1

[Bruno Luong]

For vectors A and B

C = conv(A,B,'same') 

is equivalent to

Cfull = conv(A,B,'full')
C = Cfull(floor(length(B)/2) + (1:length(A)))

For 2D or ND, repeat the above indexing for each dimension d, using size(.,d) instead of length(.).

[Anonymous]

Thanks Bruno, that's really helpful.

[Bruno Luong]

Yeap you are welcome, and you'll note that the MATLAB document is poorly written in term of accuracy and clarity

conv(u,v,'same') "Central part of the convolution of the same size as u."

[Anonymous]

This is exactly my point, thanks for being understanding.

[Bruno Luong]

This is ta reply to

Nunzio Russo 's deleted question

"Hello Bruno can you pls explain it again for 2dimensional arrays ? I don't get it right now. I don't know what you mean with variable d ? Perhaps you can explain it again and write it down as a formula. Thank you ! "

In 2D

C = conv2(A,B,'same') 

is equivalent to

Cfull = conv2(A,B,'full')
C = Cfull(floor(size(B,1)/2) + (1:size(A,1)), ...
          floor(size(B,2)/2) + (1:size(A,2)))

Fot nd case

C = convn(A,B,'same') % is eqivalent to

Cfull = convn(A,B,'full');
idx = arrayfun(@(B,A) floor(B/2)+(1:A), size(B), size(A), 'unif', 0); % NOTE: traling singleton needs to be handled
C = Cfull(idx{:})

[Nunzio Russo]

Bruno Thank you again !

I deleted my question yesterday after coming to the same conclusion. :)

Answer 2

[Image Analyst]

Sounds like homework. This is very easy. Just make up arrays on paper, like cut out gridded paper or draw something. Then just physically move the paper along and look where you are. Really, it's trivial.