The first assignment:
Fx(1:size(my_matrix,1)-1,i,j)=diff(my_matrix(:,i,j)/(a_scalar_dimension/1000));
simply seems to take the consecutive differences (using the diff funciton) over the rows of ‘my_matrix’ for the fixed second and third dimensions ‘i’ and ‘j’. We can assume this is in a nested loop of some design.
The second assignment:
Fx(size(my_matrix,1),i,j)=0;
sets the last element of ‘Fx’ for the same ‘i’ and ‘j’ coordinates to zero, apparently to equalise the row sizes of ‘Fx’ and ‘my_matrix’. In recent MATLAB versions, using end in place of the size call would have the same effect:
Fx(end+1,i,j)=0;
That’s my best guess, anyway.
