Need help to get the minimum absolute value of intersection point

below are my two vector L1(976x1) and L2(976x101) . i am calculatin the minimum absolute value where they intersecting. the figure below is for the command of

figure, plot(Tspan,y(:,1)), hold on, plot(tspan,SigV(:,81))

so if i run Tspan(rowOfMinValue) and tspan(rowOfMinValue) i am getting different values at column no 81 than the graph for both case.i need the value exactly where they are intersecting.

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tspan =[0.0976:-0.0001:0]    % Span for L1
Tspan = [0 : 0.0001: 0.0976] % Span for L2
L1 = y( : ,1);
L2 = SigV;
 
 
dL =L2(:,1:end)- L1;
 
 
columns = length(dL(1,:));
rowOfMinValue = [];
for col = 1 : columns
    % Find min value in this column, and it' s row number (index).
    [minValue(col), rowOfMinValue(col)] = min(abs(dL(:, col)));
end
% Report to command window:
rowOfMinValue

[Matt J]

We can't run your code, because we don't have y or SigV.

[md mayeen uddin]

Hi mat, i have attached both y and SigV here. since SigV is (976x101) i just shared you the SigV(: , 81) which is represented in Figure 1

Answer 1

[KSSV]

Use this file exchange: https://in.mathworks.com/matlabcentral/fileexchange/22441-curve-intersections

[md mayeen uddin]

This is giving me a precise answer but only for the last column of SigV. i want to Store values for the intersection of everz column. can it be possible?

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L1 = y( : ,1);
 L2 = SigV;
 Xf = zeros(length(SigV(1,:)),1);
 Yxf =zeros(length(SigV(1,:)),1);
 for i = length(SigV(1,:))
     [yi(i),xi(i)] = polyxpoly(y(:,1),Tspan,SigV(:,i),tspan);
     Xf(i,1) = xi(i);
     Yxf(i,1) = yi(i);
 end 

[md mayeen uddin]

it works , thanks a lot KSSV.

for i =1: length(SigV(1,:))