curve fitting toolbox for y=y0.exp((x0/x)^v) where data is known, but y0,x0 and v are unknown constants

1 次查看（过去 30 天）
Em
Em 2020-5-2
评论： darova 2020-5-3
Hi everyone
I have some scatter data that should take the form of y=y0.exp((x0/x)^v) (where x is the independent variable).
I have tried using the cftool gui but one of 2 things happens; either no model appears on the plot and I get values that can't possibly be right, or I get a totally straight line appearing across the plot.
I should say that y0,x0 and v are unknown constants in a range that I have a rough idea about. I was wondering if a different MATLAB tool would be better for this even. I can't see whats wrong with my approach.
Thanks for your time.
  3 个评论
显示 1更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Alex Sha
Alex Sha 2020-5-3
Hi, Em, if fit as the function: y=y0.exp((x0/x)^v), then the result will be about：
Root of Mean Square Error (RMSE): 9.39279339716383E-8
Sum of Squared Residual: 2.20561419504511E-13
Correlation Coef. (R): 0.82258630535955
R-Square: 0.676648229765075
Parameter Best Estimate
---------- -------------
y0 5.04851528699078E-118
x0 8.3186904534098E-210
v -0.0113812101308709
While, if change your fit function from " y=y0.exp((x0/x)^v)" to " y=y0.exp(x0^v/x^v)", take x0^v as x1, the function become: y=y0.exp(x1/x^v), the result will be more better and stable as below, the biggest difference is: x0 in original function will never be negative, but x1 in new function can be any value:
Root of Mean Square Error (RMSE): 7.55037962863218E-8
Sum of Squared Residual: 1.42520581341159E-13
Correlation Coef. (R): 0.886423351349966
R-Square: 0.785746357818505
Parameter Best Estimate
---------- -------------
y0 3.84276923298628E-7
x1 -2.23082444717647E24
v 10.629499809373

类别

在 Help Center 和 File Exchange 中查找有关 Get Started with Curve Fitting Toolbox 的更多信息

标签

产品


版本

R2017a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by