Non-linear system solver

Hello I have an equation

i=a*exp(-b*x*38.94)+c*exp(d*x*38.94)

and b and d have know range of 0 to 1. There are 101 sets of data points which are

i=[1.60E+00
1.52E+00
1.44E+00
1.36E+00
1.28E+00
1.20E+00
1.12E+00
1.04E+00
9.54E-01
8.73E-01
7.92E-01
7.11E-01
6.30E-01
5.48E-01
4.67E-01
3.86E-01
3.05E-01
2.23E-01
1.42E-01
6.00E-02
-2.18E-02
-1.04E-01
-1.87E-01
-2.70E-01
-3.54E-01
-4.40E-01
-5.29E-01
-6.20E-01
-7.17E-01
-8.23E-01
-9.42E-01
-1.09E+00
-1.28E+00
-1.61E+00
-2.59E+00
-1.52E+00
-1.23E+00
-1.05E+00
-9.04E-01
-7.85E-01
-6.78E-01
-5.78E-01
-4.84E-01
-3.93E-01
-3.04E-01
-2.16E-01
-1.30E-01
-4.40E-02
4.15E-02
1.27E-01
2.12E-01
2.97E-01
3.81E-01
4.66E-01
5.51E-01
6.35E-01
7.20E-01
8.05E-01
8.89E-01
9.74E-01
1.06E+00
1.14E+00
1.23E+00
1.31E+00
1.40E+00
1.48E+00
1.57E+00
1.65E+00
1.73E+00
1.82E+00
1.90E+00
1.99E+00
2.07E+00
2.16E+00
2.24E+00
2.33E+00
2.41E+00
2.50E+00
2.58E+00
2.67E+00
2.75E+00
2.83E+00
2.92E+00
3.00E+00
3.09E+00
3.17E+00
3.26E+00
3.34E+00
3.43E+00
3.51E+00
3.60E+00
3.68E+00
3.76E+00
3.85E+00
3.93E+00
4.02E+00
4.10E+00
4.19E+00
4.27E+00
4.36E+00
4.44E+00]
x=[0:-0.01:-1]

how do I solve a, b, c and d using these 101 sets of data points?

Answer 1

[Stephan]

Hi,

you could use

x = lsqcurvefit(fun,x0,xdata,ydata,lb,ub)

to solve this with the known bounds for b and d. When you define your problem this way:

fun = @(x,xdata)(x(1)*exp(-x(2)*xdata*38.94)+x(3)*exp(x(4)*xdata*38.94));
xdata = [0:-0.01:-1];
xdata = xdata';
lb = [-Inf 0 -Inf 0];
ub = [Inf 1 Inf 1];
x0 = [0 0 0 0];
ydata = [1.60E+00
1.52E+00
1.44E+00
1.36E+00
1.28E+00
1.20E+00
1.12E+00
1.04E+00
9.54E-01
8.73E-01
7.92E-01
7.11E-01
6.30E-01
5.48E-01
4.67E-01
3.86E-01
3.05E-01
2.23E-01
1.42E-01
6.00E-02
-2.18E-02
-1.04E-01
-1.87E-01
-2.70E-01
-3.54E-01
-4.40E-01
-5.29E-01
-6.20E-01
-7.17E-01
-8.23E-01
-9.42E-01
-1.09E+00
-1.28E+00
-1.61E+00
-2.59E+00
-1.52E+00
-1.23E+00
-1.05E+00
-9.04E-01
-7.85E-01
-6.78E-01
-5.78E-01
-4.84E-01
-3.93E-01
-3.04E-01
-2.16E-01
-1.30E-01
-4.40E-02
4.15E-02
1.27E-01
2.12E-01
2.97E-01
3.81E-01
4.66E-01
5.51E-01
6.35E-01
7.20E-01
8.05E-01
8.89E-01
9.74E-01
1.06E+00
1.14E+00
1.23E+00
1.31E+00
1.40E+00
1.48E+00
1.57E+00
1.65E+00
1.73E+00
1.82E+00
1.90E+00
1.99E+00
2.07E+00
2.16E+00
2.24E+00
2.33E+00
2.41E+00
2.50E+00
2.58E+00
2.67E+00
2.75E+00
2.83E+00
2.92E+00
3.00E+00
3.09E+00
3.17E+00
3.26E+00
3.34E+00
3.43E+00
3.51E+00
3.60E+00
3.68E+00
3.76E+00
3.85E+00
3.93E+00
4.02E+00
4.10E+00
4.19E+00
4.27E+00
4.36E+00
4.44E+00];
x = lsqcurvefit(fun,x0,xdata,ydata,lb,ub)

you wil get the vector x with:

x =

      0.1152    0.0991   -0.2550    0.0000

which represents the values for a,b,c and d.

but consider that:

your xdata runs from

x=[0:-0.01:-1]

this is a direction which maybe correct in your case - but not the usual direction! if you execute the same code with

x=[-1:0.01:0]

which is the "natural" direction you get:

x =

     -0.2550    0.0000    5.4618    0.0991

So you should try to find out in which direction your measured values run, to geht the correct result.

Best regards

Stephan