MATLAB: Zero Crossing of Signal – Misunderstanding the attached matlab code.

signal processingzero crossing

Hi guys!
Im trying to understand the zero crossing points that uses interpolation approximation which I found the code here in the threads of matlab.
the code that I found it is this:
x=1:length(y);
zci = @(v) find(v(:).*circshift(v(:), [-1 0]) <= 0);
% Returns Approximate Zero-Crossing Indices Of Argument Vector
dy = zci(y);
% Indices of Approximate Zero-Crossings
for k1 = 1:size(dy,1)-1
    b = [[1;1] [x(dy(k1)); x(dy(k1)+1)]]\[y(dy(k1)); y(dy(k1)+1)];
    % Linear Fit Near Zero-Crossings
    x0(k1) = -b(1)/b(2);
    % Interpolate ‘Exact’ Zero Crossing
    mb(:,k1) = b;
    % Store Parameter Estimates (Optional)
end
I almost understand the code but I didn't understand the statement of b = [[1;1] [x(dy(k1)); x(dy(k1)+1)]]\[y(dy(k1)); y(dy(k1)+1)]; what does it stand for?
what does this statement mean? I want to understand what this statement does exactly , any help? thanks alot.
does it take a row of matrix and devide it by another row matrix and the result is stored in b? I really just missunderstanding that row among others rows in the code.
all what I need is a detalied explanation what this row b = [[1;1] [x(dy(k1)); x(dy(k1)+1)]]\[y(dy(k1)); is about ? and what it does functionality (takes two rows of matrix and devide between each other?) ?
thanks alot.

Best Answer

Thank you! I recognise my code!
The assignment: 
b = [[1;1] [x(dy(k1)); x(dy(k1)+1)]]\[y(dy(k1)); y(dy(k1)+1)];
loops through the zero-crossings returned in ‘dy’ as indexed with ‘k1’ and calculates the linear regression slope and intercept terms for that small segment of the original waveform.
The next line calculates the interpolated value of the exact zero-crossing, and returns it as ‘x0’. The ‘mb’ matrix stores the slope and intercept terms.
For the record, I now use: 
zci = @(v) find(diff(sign(v(:))));
since it is a bit more robust and does not have the wrap-around problems my original code for it does.
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