Fitting an exponential equation to this data with CFtool

Question

Stefano Russo 2023-11-1

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2041451-fitting-an-exponential-equation-to-this-data-with-cftool

评论： Sam Chak 2023-11-1

采纳的回答： Star Strider

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Hello,

I have the following points:

y=[10 80 120 180 280 415 680 972 1178 1322 1445];
t=[50 300 410 600 900 1190 1797 2400 3000 3600 4985];

and they should fit an equation which has the form of

I have problems with inputting this equation into cftool. My scope is to find a suitable range for all the coefficient, which, in theory, should all be positive values.

Any help is much appreciated.

Stefano

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Torsten 2023-11-1

What are the coefficients you want to fit in your equation ?

As t -> Inf, your equation tends to y(0). But this is not consistent with the trend of your data curve.

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Answer 1

Star Strider 2023-11-1

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https://ww2.mathworks.cn/matlabcentral/answers/2041451-fitting-an-exponential-equation-to-this-data-with-cftool#answer_1344806

在 MATLAB Online 中打开

I get this result with fitnlm —

y=[10 80 120 180 280 415 680 972 1178 1322 1445];

t=[50 300 410 600 900 1190 1797 2400 3000 3600 4985];

objfcn = @(b,x) b(1).*(1-exp(b(2).*x./b(1)))

objfcn = function_handle with value:

@(b,x)b(1).*(1-exp(b(2).*x./b(1)))

nlmdl = fitnlm(t, y, objfcn, [max(y);randn])

nlmdl =

Nonlinear regression model: y ~ b1*(1 - exp(b2*x/b1)) Estimated Coefficients: Estimate SE tStat pValue ________ ________ _______ __________ b1 2818.1 796.96 3.536 0.0063535 b2 -0.45067 0.044962 -10.023 3.5094e-06 Number of observations: 11, Error degrees of freedom: 9 Root Mean Squared Error: 83.5 R-Squared: 0.978, Adjusted R-Squared 0.976 F-statistic vs. zero model: 493, p-value = 6.35e-10

tv = linspace(min(t), max(t)).';

[yr,yci] = predict(nlmdl, tv);

figure

hp1 = plot(t, y, 'pb', 'MarkerFaceColor','b', 'DisplayName','Data');

hold on

hp2 = plot(tv, yr, '-r','DisplayName','Regression');

hp3 = plot(tv, yci, ':r','DisplayName','\pm95% Confidence Interval');

hold off

grid

legend([hp1 hp2 hp3(1)], 'Location','best')

tv = linspace(min(t), 10*max(t)).';

[yr,yci] = predict(nlmdl, tv);

figure

hp1 = plot(t, y, 'pb', 'MarkerFaceColor','b', 'DisplayName','Data');

hold on

hp2 = plot(tv, yr, '-r','DisplayName','Regression');

hp3 = plot(tv, yci, ':r','DisplayName','\pm95% Confidence Interval');

hold off

grid

legend([hp1 hp2 hp3(1)], 'Location','best')

.

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Sam Chak 2023-11-1

It's a nice prediction plot. With the ±95% confidence interval relatively wide, OP should consider obtaining more data.

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Answer 2

Sam Chak 2023-11-1

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Hi @Stefano Russo

The data doesn't show the steady-state value. Nonetheless, we can still try fitting the exponential model to the data. I believe that the exponential model may be relatively inaccurate for fitting only a portion of the transient response.

% Data

tdat = [50 300 410 600 900 1190 1797 2400 3000 3600 4985];

ydat = [10 80 120 180 280 415 680 972 1178 1322 1445];

% Proposed exponential model with coefficients a1 and a2

yfit = @(a, tdat) a(1)*(1 - exp(- a(2)/a(1)*tdat));

% The algorithm starts with the initial guess of coefficients a1 and a2

a0 = [2*max(ydat) 1];

% Call lsqcurvefit to fit the model

[asol, resnorm] = lsqcurvefit(yfit, a0, tdat, ydat)

Local minimum possible. lsqcurvefit stopped because the final change in the sum of squares relative to its initial value is less than the value of the function tolerance.

asol = 1×2

1.0e+03 * 2.8182 0.0005

resnorm = 6.2679e+04

% Computing the Coefficient of determination

ybar = mean(ydat);

dev = ydat - ybar;

Sdev = sum(dev.^2);

Rsq = (Sdev - resnorm)/Sdev % R-square

Rsq = 0.9782

% Plot fitting result

plot(tdat, ydat, 'bo', tdat, yfit(asol, tdat), 'r-'), grid on

legend('Data points', 'Fitted curve', 'location', 'best')

xlabel('t'), ylabel('y')

title({'$y(t) = a_{1} (1 - \exp(- \frac{a_{2}}{a_{1}} t))$'}, 'interpreter', 'latex', 'fontsize', 16)

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Torsten 2023-11-1

@Sam Chak

Do you think y(0) = 2818 according to the data given ?

Sam Chak 2023-11-1

Unfortunately, @Stefano Russo's data doesn't show the initial value y(0). If the response follows a 1st-order linear time-invariant system,

, then the initial value y(0) is most likely 0. However, I took a guess that OP intended to fit an exponential model. Technically, the general analytical solution for 1st-order LTI systems is given by:

If u = 1 and y(0) = 0, then the ideal solution becomes:

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