fit a curve to data without using curve fitting toolbox

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Anqi Li
Anqi Li am 12 Dez. 2014
Kommentiert: Andreas Goser am 12 Dez. 2014
I have a set of data=[x1 x2] which looks periodical. I want to fit them into this Fourier transform equation:
x2 = A1 + A2.*sin(x1) + A3.*cos(x1) + A4.*sin(2*x1) + A5.*cos(2*x1)
by using least squares optimisation to know the optimum A = [A1;A2;A3;A4;A5]
I don't have curve fitting toolbox. please let me know how to do it without using toolbox.
  1 Kommentar
Andreas Goser
Andreas Goser am 12 Dez. 2014
I see you already have answers, but I wonder why you do not have the Curve Fitting Toolbox. Is the information about the university you work at current (your profile)?

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Mohammad Abouali
Mohammad Abouali am 12 Dez. 2014
Bearbeitet: Mohammad Abouali am 12 Dez. 2014
Assuming your X1 and X2 are vector, i.e. size(x1)=size(x2)=[N 1] and N>=5 (since there are 5 coefficients, A1 to A5); then
C=[ones(numel(x1),1) sin(x1(:)) cos(x1(:)) sin(2*x1(:)) cos(2*x1(:))];
A=(C'*C)\(C'*x2(:));
or even:
A=C\x2(:);
  6 Kommentare
Mohammad Abouali
Mohammad Abouali am 12 Dez. 2014
so x2(:) makes sure that your data is a column vector data, i.e. size(x2(:))=[N 1]. I just wanted to be sure that it is not a row data but a vertical column vector. If it is already a column vector you can just replace it with X2 and drop (:).
A=C\x2 conceptually is pretty much solving C*A=x2; Your system of equations can be written as matrix C (the one with sine and cosine functions) multiplied by the unknown column vector, and x2 is your known variable vectors. once you do A=C\x2 pretty much you are solving the system of linear-equations. This operation is known as mldivide. If you want to know more about it go to mldivide help section, there you can find more information on how to use this.
Anqi Li
Anqi Li am 12 Dez. 2014
It's very useful! thank you very much!

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