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curve fitting a custom equation

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Ayushi Sharma
Ayushi Sharma 2021 年 7 月 15 日
編集済み: Alex Sha 2021 年 7 月 18 日
a*exp(-x/T) +c*(T*(exp(-x/T)-1)+x)
This is the equation i'm trying to fit using the custom fit tool. Since the variable x is both in exponential and linear form, the data isn't really fitting, it's just a straight line. Is there way to fit the data to this equation?
As a side note: how do I view the numerical values instead of scientific ones in curve fitting tool?
Thank you
  1 件のコメント
Matt J
Matt J 2021 年 7 月 16 日
In the first line of your post, you show the model equation as,
a*exp(-x/T) +c*(T*(exp(-x/T)-1)+x)
However, in your screenshot of the cftool app, the model equation appears instead as,
a*exp(-x*T) +c*(T*(exp(-x/T)-1)+x)
You should edit your post so that we know which version of the equation is the intended one.

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Alex Sha
Alex Sha 2021 年 7 月 17 日
編集済み: Alex Sha 2021 年 7 月 18 日
Hi, all, as mentioned by Walter Roberson, the resule I provided above is obtained by using 1stOpt, the biggest advantage, for 1stOpt, is that the guessing of initial-start values are no longer needed for curve-fitting, equation-solving or any other optimization problem.
if the model function is: y=a*exp(-x*T) +c*(T*(exp(-x/T)-1)+x), the result is:
Root of Mean Square Error (RMSE): 15155741.0636478
Sum of Squared Residual: 4.59392974376683E15
Correlation Coef. (R): 0.996754359090101
R-Square: 0.993519252365118
Parameter Best Estimate
---------- -------------
a -33754000.9741343
t -1.00000000434611
c -33754006.8543357
Note the value of parameter t: t=-1.00000000434611, if taking t as -1.00, the chart will become:
The Sensitivity analysis of each parameter is as below:
while, if the model function is: y=a*exp(-x/T) +c*(T*(exp(-x/T)-1)+x), the result is:
Root of Mean Square Error (RMSE): 15137803.4043043
Sum of Squared Residual: 4.58306183814736E15
Correlation Coef. (R): 0.996383637725283
R-Square: 0.992780353526669
Parameter Best Estimate
---------- -------------
a -54592643.2783804
t -1.72718663083536
c -31608221.5261319
  2 件のコメント
Matt J
Matt J 2021 年 7 月 17 日
The solution I presented below constrained the search to T>=0, so it might be global subject to that constraint.
It's important to notice as well that whether T=-1 or T=0 is taken as the solution, it results in a cancelation of the exponential terms, leaving an essentially linear solution. It makes it seem like either the model was over-parametrized to begin with, or there isn't enough (x,y) data to accurately estimate the exponential terms.

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その他の回答 (1 件)

Matt J
Matt J 2021 年 7 月 15 日
You don't need a custom type. Your model is just exp2 but with modified data (x,y-x)
  11 件のコメント
Matt J
Matt J 2021 年 7 月 16 日
Thank you so much for your input. I have opted for this method.
@Ayushi Sharma You're welcome, but please Accept-click the answer if you consider the question addressed.
Alex uses a commercial program named 1stOpt from 7d-soft. ... I would say that if you do curve fitting, that it does a very nice job.
However, the fit Alex has shown us is based on an incorrect model equation, because @Ayushi Sharma had a typo in the first line of his post. If we substitute Alex's parameter estimates into the true model, it gives strong disagreement with the data:
load doubt
a=-54592639.7090172;
c=-31608221.6797308;
T=-1.72718650949557;
fun=@(x) a*exp(-x*T) +c*(T*(exp(-x/T)-1)+x);
plot(x,y,'o'); ax=axis;
hold on
plot( x,fun(x));
hold off
axis(ax)
legend('Data','Fit')

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