Isolate horizonal part of curve
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I have the data in the graph below (blue dotted line). I have fitted a curve (red line). How can I isolate the flat/horizontal part?

Code example:
close all;
clear all;
clc;
load('ExampleData');
ft = fittype('(a.*x.^b)',...
'dependent',{'y'},'independent',{'x'},...
'coefficients',{'a','b'});
f = fit(dataX,dataY,ft,'StartPoint',[600 -1]);
coeffvals = coeffvalues(f);
figure
plot(dataX,dataY,'-ob')
hold on
plot(dataX,f(dataX),'-r')
legend('Actucal data','Fitted')
xlabel('dataX')
ylabel('dataY')
The ExampleData file is also attached.
3 comentarios
Star Strider
el 31 de En. de 2023
Plotting it on a loglog scale produces a straight line (as would be expected from a power relation) —
LD = load(websave('ExampleData','https://www.mathworks.com/matlabcentral/answers/uploaded_files/1279075/ExampleData.mat'));
dataX = LD.dataX;
dataY = LD.dataY;
ft = fittype('(a.*x.^b)',...
'dependent',{'y'},'independent',{'x'},...
'coefficients',{'a','b'});
f = fit(dataX,dataY,ft,'StartPoint',[600 -1])
coeffvals = coeffvalues(f);
figure
loglog(dataX,dataY,'-ob')
hold on
plot(dataX,f(dataX),'-r')
legend('Actual data','Fitted')
xlabel('dataX')
ylabel('dataY')
.
Respuestas (4)
John D'Errico
el 31 de En. de 2023
What part of this curve is horizontal?
syms x
F = exp(-10*x)
fplot(F,[0,3])
Well, clearly, ithe horizontal part lies above x==0.5.
fplot(F,[0.5,3.5])
Oh wait. It must start above x==1.
fplot(F,[1,4])
Wow. That is strange. It must start above x==1.5.
fplot(F,[1.5,4.5])
I think I'm gonna get it right soon. It DEFINITELY starts at x==2.
fplot(F,[2,5])
This is really, really strange.
Or, maybe, just maybe, there is NO horizontal part of the curve. Wherever you look, the curve has EXACTLY the same shape.
0 comentarios
Walter Roberson
el 31 de En. de 2023
Declare your horizontal cutoff to be the place where abs() of the gradient is less than some threshold. If necessary, low-pass filter the data first (to remove experimental noise)
0 comentarios
Image Analyst
el 31 de En. de 2023
Maybe adjust the axis to start and end wherever you want, like
xlim([0.5e-6, 5e-6]);
but like John said, there is no flat part so you just have to make some judgment about where you think the flat part starts.
0 comentarios
MichailM
el 31 de En. de 2023
1 comentario
Image Analyst
el 31 de En. de 2023
You can try my piecewise linear demo, attached.

My attached demo does the "Novice Method" above. It looks like you're using the "Expert Method" above.
Or you can use the triangle method, also attached.
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