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Isolate horizonal part of curve

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MichailM
MichailM el 31 de En. de 2023
Comentada: Image Analyst el 31 de En. de 2023
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
MichailM
MichailM el 31 de En. de 2023
Correct. I rephrased.
Star Strider
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])
f =
General model: f(x) = (a.*x.^b) Coefficients (with 95% confidence bounds): a = 600.4 (561.2, 639.5) b = -1.038 (-1.041, -1.035)
coeffvals = coeffvalues(f);
figure
loglog(dataX,dataY,'-ob')
hold on
plot(dataX,f(dataX),'-r')
legend('Actual data','Fitted')
xlabel('dataX')
ylabel('dataY')
.

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Respuestas (4)

John D'Errico
John D'Errico el 31 de En. de 2023
What part of this curve is horizontal?
syms x
F = exp(-10*x)
F = 
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.

Walter Roberson
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)

Image Analyst
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.

MichailM
MichailM el 31 de En. de 2023
It seems that my description of the flat/horizontal part caused some sort of confusion with regards to the strict definition of it. dataY shows an abrupt decrease for the the very early values of dataX which fits well to the behavior of the power function. From a point an after, that decrease becomes sort of linear with superimposed experimental noise as Walter mentioned.
I will need to detect the part where the curve in the first post becomes sort of linear, after it has experienced the abrupt decrease. I hope that explains things somehow better. Apologies for the confusion.
  1 comentario
Image Analyst
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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