Isolate horizonal part of curve
4 ビュー (過去 30 日間)
古いコメントを表示
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 件のコメント
Star Strider
2023 年 1 月 31 日
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')
.
回答 (4 件)
John D'Errico
2023 年 1 月 31 日
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 件のコメント
Walter Roberson
2023 年 1 月 31 日
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 件のコメント
Image Analyst
2023 年 1 月 31 日
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 件のコメント
MichailM
2023 年 1 月 31 日
1 件のコメント
Image Analyst
2023 年 1 月 31 日
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.
参考
カテゴリ
Help Center および File Exchange で Linear and Nonlinear Regression についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!