How to fit a step function?

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Mr M.
Mr M. 2015-9-16
评论： Jörn Froböse 2021-6-10
The problem is only the first parameter (p(1)) is fitted, but p1(2) = p0(2), unchanged! Why?
X = [-1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0];
Y = [0.1 -0.15 0.05 0.0 -0.05 10.2 10.5 11.5 8.2 9.2 10.0 10.5 10.2 10.9 10.5 11.5];
plot(X,Y)
step = @(x)x > 0;
fun = @(p,x) p(1)*step(x-p(2))
p0 = [9,-0.5];
p1 = lsqcurvefit(fun,p0,X,Y);
Z = fun(p1,linspace(min(X),max(X)));
hold on
plot(linspace(min(X),max(X)),Z,'r')
  1 个评论
Adam Danz
Adam Danz 2021-4-13
Ran in:
It's not clear to me what the fit should look like or the resolution of the steps. For example,
X = [-1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0];
Y = [0.1 -0.15 0.05 0.0 -0.05 10.2 10.5 11.5 8.2 9.2 10.0 10.5 10.2 10.9 10.5 11.5];
plot(X,Y, 'b-o')
xr = [X(1),repelem(X(2:end),1,2)]; % X is row vec
yr = [repelem(Y(1:end-1),1,2),Y(end)]; % Y is row vec
hold on
plot(xr, yr, 'r-','LineWidth',2)

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回答（1 个）

Figen Ece Demirer
Figen Ece Demirer 2021-4-13
X = [-1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0];
Y = [0.1 -0.15 0.05 0.0 -0.05 10.2 10.5 11.5 8.2 9.2 10.0 10.5 10.2 10.9 10.5 11.5];
cut=0;
plot(X,Y,'g--o')
step = @(x)x > cut;
step2 = @(x)x < cut;
fun = @(p,x) p(1)*step(x)+p(2)*step2(x);
p0 = [9,-0.5];
p1 = lsqcurvefit(fun,p0,X,Y);
Z = fun(p1,linspace(min(X),max(X)));
hold on
plot(linspace(min(X),max(X)),Z,'r')

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