Need help to get the minimum absolute value of intersection point

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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.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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
  2 个评论
md mayeen uddin
md mayeen uddin 2021-9-4
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

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KSSV
KSSV 2021-9-4
  2 个评论
md mayeen uddin
md mayeen uddin 2021-9-4
编辑:md mayeen uddin 2021-9-4
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?
__________________________________________________________________________
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

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