I am trying to fit some data using lsqcurvefit in MATLAB but I am fairly new to this area.
xdata1 = [0 60 660 1250];
ydata1 = [0 18 23 31];
In the image below, the red line is the fit I want to achieve. Sadly, Polyfit does not provide suitable results.
How can I achieve this fit? Thank you in advance!

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Matt J
Matt J 2018년 10월 11일

0 개 추천

I believe piece-wise linear fitting was in the scope of Bruno's free-knot spline fitting package,

추가 답변 (2개)

Chaoyu Zhang
Chaoyu Zhang 2018년 10월 11일
편집: Chaoyu Zhang 2018년 10월 15일

0 개 추천

You can use the method described below,
The target equation (3rd order or maybe higher) is
y = a*x.^3 + b*x.^2 + c*x + d;
A * p = y;
p is the parameters of the equation,
p = [a;b;c;d]
A is the matrix made of x.^3,x.^2,x,1,
A = [x(1).^3 x(1).^2 x(1) 1; ... ; x(n).^3 x(n).^2 x(n) 1]
y is the vector made of y,
y = [y(1); ... ;y(n)];
p = (A.'*A)^(-1)*A.'*y;
Now you get the parameters you need.

댓글 수: 5

Torsten
Torsten 2018년 10월 11일
This in principle is what polyfit does, but the OP was not satisfied with the fit curve.
Matt J
Matt J 2018년 10월 15일
편집: Matt J 2018년 10월 15일
Also, direct use of (A.'*A)^(-1)*A.'*y is generally not a good idea. Better numeric accuracy is obtained by using mldivide() or similar.
Image Analyst
Image Analyst 2018년 10월 15일
Matt, do you have a reference for the better accuracy claim? The closest I could find in the help said "If A is a square matrix, then A\B is roughly equal to inv(A)*B, but MATLAB processes A\B differently and more robustly." But I don't think that says it's more accurate.
Matt J
Matt J 2018년 10월 15일
편집: Matt J 2018년 10월 15일
Well, the poorer accuracy comes from the inversion of (A.'*A), since cond(A.'*A) is the square of cond(A).
>> A=rand(1000,100);
>> cond(A.'*A)
ans =
632.4462
>> cond(A)
ans =
25.1485
When solving with mldivide(), the QR decomposition is used, which avoids this inversion. With A=Q*R,
(A.'*A)^(-1)*A.'*y
=(R.'*R)^(-1)*R.'*Q.'*y
=R^(-1)*Q.'*y
So, the inversion involves only R^(-1) and cond( R )=cond(A).
Matt J
Matt J 2018년 10월 15일
편집: Matt J 2018년 10월 15일
Here is a test showing the increased error sensitivity of inv(A.'*A)*(A.'*y).
N=1000;
M=15;
A=vander(linspace(1,3.3,M)) + eye(M);
xt=rand(M,1);
yt=A*xt;
y=yt+randn(M,N)*1e-6;
x1=inv(A.'*A)*(A.'*y);
x2=A\y;
Error1=mean( sqrt(sum((x1-xt).^2)) )
Error2=mean( sqrt(sum((x2-xt).^2)) )
should give something like
Error1 =
5.0094
Error2 =
2.8313e-04

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Image Analyst
Image Analyst 2018년 10월 11일

0 개 추천

You cannot get that unless you put in a model curve for that shape. Otherwise functions are not going to know that it's a piecewise linear fit or some sharply kinked log function or whatever. And having more data points would help too. Then you can use fitnlm.
I'm attaching several examples for piecewise linear fit and non-linear fits.

댓글 수: 4

Julian
Julian 2018년 10월 11일
Thank you very much! yes I think I have to define the function first. sadly are we not able to generate more datapoints (physical wise)
Image Analyst
Image Analyst 2018년 10월 11일
Then can you just say that the curve is linear between the first 2 points, and linear from the second point onwards? So just have two formulas that define the "curve".
Julian
Julian 2018년 10월 11일
Yes that would greatly improve the fit. Can you describe me how it is possible to combine two linear fit into one function in Matlab ?
For three order of polyfit: y = p1+p2*x+p3*x^2+p4*x^3
Root of Mean Square Error (RMSE): 5.67944047963309E-15
Sum of Squared Residual: 1.2902417664678E-28
Correlation Coef. (R): 1
R-Square: 1
Parameter Best Estimate
---------- -------------
p1 3.92787554473843E-15
p2 0.340654276995852
p3 -0.000698994200659206
p4 3.57048623250019E-7
while, if taking function as: y = p1*x/(p2+x)^2-p3*x
Root of Mean Square Error (RMSE): 0
Sum of Squared Residual: 0
Correlation Coef. (R): 1
R-Square: 1
Parameter Best Estimate
---------- -------------
p1 9035.3274708592
p2 119.630504666011
p3 -0.0199834390844168
Obviously, the second function should be more reasonable than 3rd polyfit.

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