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

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) 

Answer 1

[Ayush]

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

• Read Further from here: https://www.mathworks.com/help/matlab/ref/polyfit.html

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

• Read further from here: https://www.mathworks.com/help/curvefit/fit.html

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]);

• Read further from here: https://www.mathworks.com/help/curvefit/fittype.html

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);

• Read Further from here: https://www.mathworks.com/help/optim/ug/lsqcurvefit.html

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

[Mushtaq Al-Jubbori]

Thanks @Ayush, but I need equation like (exp,tanh,...ect)

[Alex Sha]

Hi, try the fitting function below, suitable for all 4 set of data.

Data-1:

Sum Squared Error (SSE): 0.319897084641848
Root of Mean Square Error (RMSE): 0.141398613112419
Correlation Coef. (R): 0.995640062563947
R-Square: 0.991299134182341

Parameter	Best Estimate       
---------	-------------       
p1       	-6.82884032533297   
p2       	-0.00277440148921541
p3       	3.47264072105555    
p4       	-0.49209017455415

Data-2:

Sum Squared Error (SSE): 0.485351233701642
Root of Mean Square Error (RMSE): 0.148530871248071
Correlation Coef. (R): 0.999082387535941
R-Square: 0.998165617084517

Parameter	Best Estimate       
---------	-------------       
p1       	-10.8515771754327   
p2       	-6.43083370788535E-5
p3       	2.82220368533152    
p4       	-0.223614506000965

Data-3:

Sum Squared Error (SSE): 2.49436533045795
Root of Mean Square Error (RMSE): 0.288349633284429
Correlation Coef. (R): 0.998199004320542
R-Square: 0.996401252226522

Parameter	Best Estimate        
---------	-------------        
p1       	-15.151099984065     
p2       	-0.000749148428345008
p3       	1.39139482554941     
p4       	-0.129600115152763 

Data-4:

Sum Squared Error (SSE): 2.39172194030262
Root of Mean Square Error (RMSE): 0.235841981996747
Correlation Coef. (R): 0.999294509051636
R-Square: 0.998589515820749

Parameter	Best Estimate        
---------	-------------        
p1       	-19.2278440833496    
p2       	-0.000399076834625561
p3       	1.04194799212013     
p4       	-0.0824358135353087

[Mushtaq Al-Jubbori]

Very thanks Dear @Alex Sha but did you can modefied the equation to was more sharply at the beganning saturation Mushtaq

[Alex Sha]

Hi, it depends on your overall data。

Answer 2

[John D'Errico]

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.

[Mushtaq Al-Jubbori]

Very thanks, what the formula of equation

[John D'Errico]

You can see it right in the code. Read the answer. I chose two different simple models. The functional forms are written in Linmdl and in quadmdl.

[Mushtaq Al-Jubbori]

Very thanks

[Mushtaq Al-Jubbori]

@John D'Errico

your equation dont satisfied the condition