Fitting an exponential equation to this data with CFtool

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

[Torsten]

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.

Answer 1

[Star Strider]

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')

.

[Sam Chak]

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

Answer 2

[Sam Chak]

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)

[Torsten]

@Sam Chak

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

[Sam Chak]

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: