Non-linear system solver

2 visualizaciones (últimos 30 días)
JuiChun Lin
JuiChun Lin el 11 de Jun. de 2018
Editada: Stephan el 11 de Jun. de 2018
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?

Respuesta aceptada

Stephan
Stephan el 11 de Jun. de 2018
Editada: Stephan el 11 de Jun. de 2018
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

Más respuestas (0)

Categorías

Más información sobre Systems of Nonlinear Equations en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

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

Start Hunting!

Translated by