can any one help me to find empirical equation with 3 or 4 parameters for this data, And of course with different parameters for each curve

8 次查看（过去 30 天）
Mushtaq Al-Jubbori
编辑： Mushtaq Al-Jubbori 2024-1-9
x1=0.75:0.25:4.5;
x2=0.75:0.25:6;
x3=1.25:.25:8.5;
x4=1.75:0.25:12.25;
y1=[2.1 3.045 4.2 4.515 6.09 6.72 6.8145 6.825 6.825 6.825 6.72 6.825 6.825 6.93 6.825 6.825];
y2=[1.89 2.31 2.625 2.94 3.675 3.99 4.41 5.04 6.09 6.51 8.190 9.345 10.5 10.92 10.92 10.92 10.92 10.71 10.92 10.5 10.92 10.92];
y3=[2.415 2.625 3.045 3.675 3.99 4.2 4.83 5.565 5.88 6.51 6.825 7.875 8.61 9.975 10.5 11.34 11.55 13.965 14.9625 15.12 15.12 15.33 15.12 14.805 15.12 14.91 15.12 15.12 15.225 15.12];
y4=[2.94 3.15 3.255 3.57 3.885 4.305 4.41 5.04 5.25 5.67 6.09 6.3 6.615 7.77 7.98 8.631 9.24 9.975 10.605 11.34 12.075 13.335 14.175 14.7 16.17 17.325 18.585 19.215 19.2465 19.215 19.215 19.215 19.215 19.425 19.215 19.215 18.9 19.215 19.215 18.69 19.215 19.215 19.2150];
plot(x1,y1)
hold
plot(x2,y2)
plot(x3,y3)
plot(x4,y4)

请先登录，再回答此问题。

回答（2 个）

Ayush
Ayush 2024-1-8
Hi @Mushtaq Al-Jubbori
Depending on the complexity of the model you want to fit, you can use simple polynomial fitting or more advanced non-linear curve fitting. Here are some MATLAB functions you can use for different types of fits:
Polynomial Fitting: For polynomial fitting, you can use the polyfit function, which finds the coefficients of a polynomial that fits the data in a least-squares sense.
p = polyfit(x, y, n); % 'n' is the degree of the polynomial
fitted_values = polyval(p, x); % Evaluate the polynomial at the points x
Non-linear Curve Fitting: For non-linear curve fitting, you can use the fit function from the Curve Fitting Toolbox, which allows you to specify a custom model or choose from a variety of built-in library models.
f = fit(x', y', 'modelname', 'StartPoint', [a0, b0, c0, ...]);
% 'modelname' could be 'exp1', 'exp2', 'gauss1', 'gauss2', 'rat01', etc.
% 'StartPoint' is an array with initial guesses for the parameters
If you need a custom equation, you can define it using a fit type object:
ft = fittype('a*x^2 + b*x + c', 'independent', 'x', 'dependent', 'y');
f = fit(x', y', ft, 'StartPoint', [a0, b0, c0]);
Non-linear Least Squares Fitting: For more control over the fitting process, you can use the lsqcurvefit function, which solves non-linear curve-fitting (data-fitting) problems in least-squares sense
fun = @(p,x) p(1)*x.^2 + p(2)*x + p(3); % Example of a quadratic model
p0 = [a0, b0, c0]; % Initial guess for the parameters
[p,resnorm] = lsqcurvefit(fun, p0, x, y);
Remember to replace 'modelname' with the appropriate model name or custom equation, and [a0, b0, c0, ...] with your initial guesses for the parameters.
Thanks,
Ayush
  4 个评论
显示 2更早的评论
Mushtaq Al-Jubbori
Mushtaq Al-Jubbori 2024-1-9
编辑：Mushtaq Al-Jubbori 2024-1-9
Very thanks Dear @Alex Sha but did you can modefied the equation to was more sharply at the beganning saturation Mushtaq

请先登录，再进行评论。

John D'Errico
John D'Errico 2024-1-8
编辑：John D'Errico 2024-1-8
Ran in:
Sadly, while you want some magical form for each of these curves, that is not so easy to find. Essentially, it looks like each of those curves is well modeled by a simple low order polynomial model, coupled with a maximum, a saturation point. Nothing more sophisticated is necessry to do the above. And there is not sufficient information content in the data to build somethign more sophisticateed anyway.
x1=0.75:0.25:4.5;
x2=0.75:0.25:6;
x3=1.25:.25:8.5;
x4=1.75:0.25:12.25;
y1=[2.1 3.045 4.2 4.515 6.09 6.72 6.8145 6.825 6.825 6.825 6.72 6.825 6.825 6.93 6.825 6.825];
y2=[1.89 2.31 2.625 2.94 3.675 3.99 4.41 5.04 6.09 6.51 8.190 9.345 10.5 10.92 10.92 10.92 10.92 10.71 10.92 10.5 10.92 10.92];
y3=[2.415 2.625 3.045 3.675 3.99 4.2 4.83 5.565 5.88 6.51 6.825 7.875 8.61 9.975 10.5 11.34 11.55 13.965 14.9625 15.12 15.12 15.33 15.12 14.805 15.12 14.91 15.12 15.12 15.225 15.12];
y4=[2.94 3.15 3.255 3.57 3.885 4.305 4.41 5.04 5.25 5.67 6.09 6.3 6.615 7.77 7.98 8.631 9.24 9.975 10.605 11.34 12.075 13.335 14.175 14.7 16.17 17.325 18.585 19.215 19.2465 19.215 19.215 19.215 19.215 19.425 19.215 19.215 18.9 19.215 19.215 18.69 19.215 19.215 19.2150];
plot(x1,y1,'o')
If we look at the above curve, we can see a straight line, coupled with a saturation level. We might do that iusing a simple model, such as...
Linmdl = fittype('a + b*min(x,xsat)','indep','x')
Linmdl =
General model: Linmdl(a,b,xsat,x) = a + b*min(x,xsat)
So xsat is the point at which the curve saturates. All that really matters is I choose at least a decent estimate of the saturation point location.
fittedmdl1 = fit(x1',y1',Linmdl,'start',[1 1 2])
fittedmdl1 =
General model: fittedmdl1(x) = a + b*min(x,xsat) Coefficients (with 95% confidence bounds): a = -0.67 (-1.116, -0.2243) b = 3.72 (3.41, 4.03) xsat = 2.015 (1.945, 2.084)
plot(fittedmdl1,x1,y1)
There simply is nothing more in that data. No sophisticated model using tan or exp. Your data justifies nothing more. The y value at the saturation point is simple to find too.
ysat = fittedmdl1(fittedmdl1.xsat)
ysat = 6.8240
Similarly, we can model each of the other curves the same way. There, I could use a quadratic polynomial, since there is some amount of curvature in the curve below the break point.
quadmdl = fittype('a + b*min(x,xsat) + c*min(x,xsat).^2','indep','x')
quadmdl =
General model: quadmdl(a,b,c,xsat,x) = a + b*min(x,xsat) + c*min(x,xsat).^2
fittedmdl2 = fit(x2',y2',quadmdl,'start',[1 1 1 4])
fittedmdl2 =
General model: fittedmdl2(x) = a + b*min(x,xsat) + c*min(x,xsat).^2 Coefficients (with 95% confidence bounds): a = 2.271 (1.557, 2.984) b = -0.776 (-1.478, -0.07389) c = 0.7863 (0.6328, 0.9397) xsat = 3.833 (3.763, 3.904)
plot(fittedmdl2,x2,y2)
ysat = fittedmdl2(fittedmdl2.xsat)
ysat = 10.8500
Again, we see a very nice fit, entirely reasonable. I'll do a fit for the third curve, just to show how easy it is. Again, I'll use the quadmdl form.
fittedmdl3 = fit(x3',y3',quadmdl,'start',[1 1 1 6])
fittedmdl3 =
General model: fittedmdl3(x) = a + b*min(x,xsat) + c*min(x,xsat).^2 Coefficients (with 95% confidence bounds): a = 2.205 (1.336, 3.074) b = -0.2222 (-0.7667, 0.3223) c = 0.4144 (0.3378, 0.4911) xsat = 5.853 (5.763, 5.942)
plot(fittedmdl3,x3,y3)
You can fit the last curve. A quadratic model will surely suffice there too for the lower region. Again, if you think some complicated model with exp and tan functions is necessary, you are simply wrong. The data does not justify it, nor could you estimate such a model. And since you provide no physical reason why some specific form should exist, we cannot possibly guess what it might be.
DON'T OVERFIT YOUR DATA! Apply basic common sense to what you have, and don't look for something, more complicated. While that more complex model might exist in the shadows, you won't be able to know what it is. And if you did try to find something complex, you would get complete garbage as a result of that exploration on such limited information.
  4 个评论
显示 2更早的评论

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Interpolation 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by