OK, so I need more significant figures? How do I get the Curve Fit Tool to output the results with more sig figs?
Fourier series expansion using the Curve Fitting Tool
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I have some very perplexing results I obtained using the Curve Fitting Tool which I am having difficulty interpreting and reconciling the results. This is the 1st time I used this tool so I hope someone with some experience can assist me with the results I obtained. I am using MATLAB 2012b.
I used the code below to generate a pressure profile vs time. The time vector & the P results I have attached in the Pdata.mat file.
A=1; p=A*(1-time/0.01).*exp(-time/0.01);
I used the code generation wizard within the Curve Fitting Tool to generate the m-file that is attached. Within that m-file the following initial guess at the Fourier series coefficients:
opts.StartPoint = [7.1054e-15 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 12.5664];
I then executed the FIT operation & the results for Fourier series coefficients are in the .txt file. I then copied those results & pasted them into MATLABcurvefit.m to check the results. I got nothing that resembled what is in Pdata.mat. Despite this discrepancy the Curve Fitting Tool generated fitted curve depicted in curve fit results.jpg. The fit depicted here is EXCELLENT!
So I am completely BAFFLED. How does the Curve Fitting Tool show excellent results, yet when I used the results from the .txt file in MATLABcurvefit.m the results are dreadful?
Really would appreciate an explanation.
Réponse acceptée
John D'Errico
le 18 Nov 2017
Modifié(e) : John D'Errico
le 18 Nov 2017
Almost always when someone complains of exactly this problem, it is because they did not in fact use the correct coefficients. Instead, they copied 4 significant digits, pasted them in. But MATLAB did not estimate 5 digits. So what did you do?
Here are the coefficients from the text file, exactly as you said.
a0 = 2.589e+07 (1.61e+07, 3.567e+07)
a1 = -2.918e+07 (-4.097e+07, -1.739e+07)
b1 = -3.622e+07 (-4.928e+07, -2.317e+07)
a2 = -7.159e+06 (-8.454e+06, -5.865e+06)
b2 = 3.284e+07 (2.019e+07, 4.55e+07)
a3 = 1.73e+07 (1.139e+07, 2.321e+07)
b3 = -8.643e+06 (-1.298e+07, -4.311e+06)
a4 = -7.895e+06 (-1.112e+07, -4.667e+06)
b4 = -3.616e+06 (-4.282e+06, -2.951e+06)
a5 = 7.232e+05 (1.553e+05, 1.291e+06)
b5 = 2.853e+06 (1.888e+06, 3.819e+06)
a6 = 4.273e+05 (3.462e+05, 5.083e+05)
b6 = -5.679e+05 (-8.266e+05, -3.092e+05)
a7 = -1.092e+05 (-1.471e+05, -7.122e+04)
b7 = 3528 (-1.4e+04, 2.105e+04)
a8 = 5249 (2399, 8099)
b8 = 6116 (4905, 7326)
w = 3.406 (3.325, 3.488)
Then look at what you did...
a0=2.589e+07;
a1=-2.918e+07;
b1=-3.622e+07;
a2=-7.159e+06;
b2=3.284e+07;
...
And, exactly as I said, you used 4 digit approximations to a list of numbers. That is a BAD idea. It is assured to result in crapola. What did you get? Yep. Crapola.
The point is, you did not in fact use the results. You used a poor approximation to the results, and you got a poor result. SURPRISE!
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