Regression of a vector in a optimization problem
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Hello everyone,
I need to fit experimental data to an analytical solution. The analytical solution has the form:
- C(z,t) = C_eq*f(z,t,D)
where f(z,t,D) is a known function of time (t) and position (z), and D and C_eq are parameters to regress.
I have already determined D and C_eq using the routine fminsearch. However, I would like to consider that C_eq does not necessarily have to be constant and can change over time.
My question is whether it is possible to regress C_eq as a vector instead of a constant? In this case, which routine is the most appropriate?
P.D: parameter D could also be considered as a vector if necessary.
Thanks in advance.
5 comentarios
Torsten
el 31 de Jul. de 2019
Do you have experimental data at different z-positions ?
Ronny Rives
el 31 de Jul. de 2019
Torsten
el 31 de Jul. de 2019
You shouldn't work with too many degrees of freedom for C_eq.
E.g. if you consider it as a function of time and position, you could choose it as
C_eq = C_experimental/f(z,t,D)
which gives a 100% fit between experimental and theoretical values.
If you consider it as a function of time and you have experimental data at times
t1 < t2 < ... < tn,
you can choose C_eq between 0 and t1 such that it fits best your data at t1, you can choose C_eq between t1 and t2 such that it fits best your data at t2 and so on.
But I think this only gives you more degrees of freedom in the fitting process, but doesn't make much sense from the physics point of view.
Ronny Rives
el 31 de Jul. de 2019
Torsten
el 1 de Ag. de 2019
Use "lsqcurvefit" with the parameter vector x = (C_eq(1),C_eq(2),...,C_eq(n)).
Respuesta aceptada
Más respuestas (1)
Sai Bhargav Avula
el 13 de Ag. de 2019
0 votos
As mentioned by Torsten, lsqcurvefit can be used to obtain a vector as a result of the regression. But those values use the entire data for getting the output values . For you particular case you should segment the data based on time stamps and perform lsqcurvefit in a for loop.
1 comentario
Matt J
el 13 de Ag. de 2019
I don't think a loop would be appropriate here, actually, because as I understand it, the parameter D is shared by all time blocks.
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