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sevny
USA
5 Posts |
 Posted - 02/22/2013 : 12:13:35 PM
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Origin Ver. and Service Release (Select Help-->About Origin): 8.0
Operating System:Win
Hello, I got another question about nonlinear curve fitting.
I am trying to do curve fitting with a set of equations with intermediate variables. Different from the problem before, these equations can be combined and expressed as one equation of y=f(x) form,however, the one equation will be super-complex to write down. Therefore I introduced several intermediates and seperated the equation into several equations (as follows).So in principle, the fitting and simulation should be straightforward. The problem is, when I tried to simulate the curve with a set of parameters, I got no curve appeared on the coordinates. And compiling was no problem. I am not sure if it is because the equations are still too complex for the computer to handle, as when I testing with simpler equations but similar format, it worked well. Please help me check the equations, if it is the reason, how can I solve the problem? Thank you very much for any comment.
void _nlsfTwosite_with_n(
// Fit Parameter(s):
double Vi, double V0, double R0, double M0, double L0, double fi, double K1, double K2,
double H1, double H2, double n1, double n2,
// Independent Variable(s):
double x,
// Dependent Variable(s):
double& y)
{
// Beginning of editable part
double ft1,ft2,p1,q1,r1,p2,q2,r2,L1,L2,x11,x12,x21,x22;
ft1=R0/(R0+x);
ft2=ft1/fi;
p1=1/K1+1/K2+(n1+n2)*M0*ft1-L0*(1-ft1);
q1=(n1/K2+n2/K1)*M0*ft1-(1/K1+1/K2)*L0*(1-ft1)+1/(K1*K2);
r1=-L0*(1-ft1)/(K1*K2);
p2=1/K1+1/K2+(n1+n2)*M0*ft2-L0*(1-ft2);
q2=(n1/K2+n2/K1)*M0*ft2-(1/K1+1/K2)*L0*(1-ft2)+1/(K1*K2);
r2=-L0*(1-ft2)/(K1*K2);
L1=-p1/3-2^(1/3)*(-p1^2+3*q1)/(3*(-2*p1^3+9*p1*q1-27*r1+3*3^0.5*(-(p1*q1)^2+4*q1^3+4*p1^3*r1-18*p1*q1*r1+27*r1^2)^0.5)^(1/3))+(-2*p1^3+9*p1*q1-27*r1+3*3^0.5*(-(p1*q1)^2+4*q1^3+4*p1^3*r1-18*p1*q1*r1+27*r1^2)^0.5)^(1/3)/(3*2^(1/3));
L2=-p2/3-2^(1/3)*(-p2^2+3*q2)/(3*(-2*p2^3+9*p2*q2-27*r2+3*3^0.5*(-(p2*q2)^2+4*q2^3+4*p2^3*r2-18*p2*q2*r2+27*r2^2)^0.5)^(1/3))+(-2*p2^3+9*p2*q2-27*r2+3*3^0.5*(-(p2*q2)^2+4*q2^3+4*p2^3*r1-18*p2*q2*r2+27*r2^2)^0.5)^(1/3)/(3*2^(1/3));
x11=K1*L1/(1+K1*L1);
x12=K1*L2/(1+K1*L2);
x21=K2*L1/(1+K2*L1);
x22=K2*L2/(1+K2*L2);
y=M0*V0*ft1*(n1*(x11-x12)*H1+n2*(x21-x22)*H2)/(Vi*L0);
// End of editable part
}
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Edited by - sevny on 02/22/2013 12:33:57 PM |
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Shirley_GZ
China
Posts |
Posted - 02/25/2013 : 03:02:43 AM
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Hi,
When the parameter Vi=1, V0=1, R0=1, M0=1, L0=1, fi=1, K1=1, K2=1, H1=1, H2=1, n1=1, n2=1, and X equals a value (such as 2 or 1.5), the parts of the equations (L1 and L2),
"-(p1*q1)^2+4*q1^3+4*p1^3*r1-18*p1*q1*r1+27*r1^2"
and
"-(p2*q2)^2+4*q2^3+4*p2^3*r2-18*p2*q2*r2+27*r2^2"
are negative. So the intermediate variables L1 and L2 are missing values.
So you got no curve appeared when you tried to simulate the curve.
Originlab Technical Service Team |
Edited by - Shirley_GZ on 02/25/2013 03:16:05 AM |
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sevny
USA
5 Posts |
Posted - 02/25/2013 : 10:35:53 AM
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Thanks a lot for your reply Shirley. I tried to use different values for the parameters, like Vi=6E-6, V0=0.00143, R0=20, M0=1E-5, L0=2E-4, fi=0.995813, K1=1E7, K2=1E14, H1=-10, H2=2, n1=0.5, n2=0.5, which were from the experiments (or close estimations), but still no curve; while reseasonable result was obtained when using Mathematica to solve a series of such equations with these parameter values.
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Edited by - sevny on 02/25/2013 10:39:30 AM |
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Curve fitting with intermediate variables