How to plot a best fit curve to my data points?

39 vues (au cours des 30 derniers jours)
Nisar Ahmed
Nisar Ahmed le 2 Juin 2022
Commenté : Sam Chak le 3 Juin 2022
Hi,
I have attached a data (inverted_qp.mat Inv_qpAVO freq;) and a plot between fre and Inv_qpAVO is shown below in red color.
circled points on left hand side are the issue and I want to a best-fit curves passes through these points just like shown on the right hand side figure (in blue color a solid line).
The best fit curve should start from the fist point to the end point as given a solid blue line. Thanks
  7 commentaires
John D'Errico
John D'Errico le 3 Juin 2022
Modifié(e) : John D'Errico le 3 Juin 2022
@Nisar Ahmed Please stop using a new answer every time you want to make a comment.
Sam Chak
Sam Chak le 3 Juin 2022
@Image Analyst, thanks for the insights into the analysis of the data that provides a basis for understanding the curve fitting. 👍

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Réponses (2)

Kevin Holly
Kevin Holly le 2 Juin 2022
If you have the Curve Fitting Toolbox, you can use the Curve Fitting app after you import your data into the workspace. Within the app, there are many curve fits you could select from. Here is a video showing the process.
  1 commentaire
John D'Errico
John D'Errico le 3 Juin 2022
<Moved to a comment from @Nisar Ahmed
@Kevin Holly I tried all the options Curve Fitting Toolbox have but not got best fit results

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Nisar Ahmed
Nisar Ahmed le 3 Juin 2022
@Alex Sha thank you for you answer,
I am sorry I did not understand, do you mean, I need to compute
f(x) = p1 + p2.*x + p3*x.^1.5 + p4.*log(x) + p5.*exp(p6.*x)
while using
p1 = -0.137921750658333;
p2 = 4.83156544429652E-5;
p3 = -3.87422699831525E-7;
p4 = 0.000904269701487609;
p5 = 0.138139976684668;
p6 = -0.000327312583052385;
and x = Inv_qpAVO;
I am trying as I wrote above and having this error
Array indices must be positive integers or logical values.
Error in InQP_f (line 29)
f(x) = p1 + p2.*x + p3*x.^1.5 + p4.*log(x) + p5.*exp(p6.*x)
  1 commentaire
Torsten
Torsten le 3 Juin 2022
Modifié(e) : Torsten le 3 Juin 2022
p(1) = -0.137921750658333;
p(2) = 4.83156544429652E-5;
p(3) = -3.87422699831525E-7;
p(4) = 0.000904269701487609;
p(5) = 0.138139976684668;
p(6) = -0.000327312583052385;
f = @(x)p(1) + p(2)*x + p(3)*x.^1.5 + p(4).*log(x) + p(5).*exp(p(6).*x);
plot(x,f(x))

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