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
2 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
Jeff
el 18 de Nov. de 2017
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.
Respuesta aceptada
John D'Errico
el 18 de Nov. de 2017
Editada: John D'Errico
el 18 de Nov. de 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!
4 comentarios
Más respuestas (0)
Ver también
Categorías
Más información sobre Descriptive Statistics en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!