Non linear multiple curve fitting

2022 3 14
1 个回答
回答已采纳
更新时间：2022 3 15
5 次查看（30 天）

0 个投票

Dear all,
I'm trying to do the multiple nonlinear regression of some rheology curves (rheology.png). In particular, viscisty (visc.txt) changes with both temperature (T.txt) and shear rate (shear_rate.txt).
I would find a nonlinear regression of the form given in the attached 'regression_model.txt', where eta is the viscosity, T the temperature and gamma the shear rate.
but I'm having problems... I tried with nonlinfit, but this worked on only if a fix a given value of temperature (so, for a single variable the regression was ok).
How can I perform this regression in a simple way?
I would thanks in advance anyone who will contribute to help me
Best regards,
AP

请先登录，再回答此问题。

 采纳的回答

Torsten
Torsten 2022-3-14
编辑：Torsten 2022-3-14

0 个投票

Form a big matrix A:
First column: a vector of 1's
Second column: log(gamma) or log10(gamma) (depending on what "lg" means)
Third column: (log(gamma)).^2 or (log10(gamma)).^2 (depending on what "lg" means)
Fourth column: T*log(gamma) or T*log10(gamma) (depending on what "lg" means)
Fifth column: T
Sixth column: T.^2
and a big vector b, consisting of one column with the corresponding values for log(eta) or log10(eta) ( (depending on what lg means)
Then you can solve for the vector v =[ A0; A1; A11; A12; A2; A22] of constants in your model by typing
v = A\b.

5 个评论
显示 3更早的评论 隐藏 3更早的评论

Alessio Pricci
Alessio Pricci 2022-3-14
Hi Torsten,
I tired to implement your suggestion, but it seems that it does not work. Here I report the implementation of the code:
%% Parameters used for the viscosity curves
D1=7.40688e+12;
D2=373.15;
A1=30.62;
A2=51.6;
n=0.2411;
tau=72297.9;
%% shear rate vector (equal for all 3 temperatures)
shear_rate=linspace(1e-1,1e4,1e5);
%% temperature and corresponding viscosity vector
T=210+273.15;
v1=D1*exp(-(A1*(T-373.15))./(A2+(T-373.15)))./((1+((D1*exp(-(A1*(T-373.15))./(A2+(T-373.15)))).*shear_rate./tau)).^(1-n));
T=245+273.15;
v2=D1*exp(-(A1*(T-373.15))./(A2+(T-373.15)))./((1+((D1*exp(-(A1*(T-373.15))./(A2+(T-373.15)))).*shear_rate./tau)).^(1-n));
T=280+273.15;
v3=D1*exp(-(A1*(T-373.15))./(A2+(T-373.15)))./((1+((D1*exp(-(A1*(T-373.15))./(A2+(T-373.15)))).*shear_rate./tau)).^(1-n));
%% Implementation of your suggestion
T1=ones(1,length(shear_rate))*(210+273.15);
T2=ones(1,length(shear_rate))*(245+273.15);
T3=ones(1,length(shear_rate))*(280+273.15);
A1=[ones(1,length(shear_rate)); log(shear_rate); (log(shear_rate)).^2; T1.*log(shear_rate); T1; T1.^2]';
A2=[ones(1,length(shear_rate)); log(shear_rate); (log(shear_rate)).^2; T2.*log(shear_rate); T2; T2.^2]';
A3=[ones(1,length(shear_rate)); log(shear_rate); (log(shear_rate)).^2; T3.*log(shear_rate); T3; T3.^2]';
A=[A1; A2; A3];
b=[log(v1) log(v2) log(v3)]';
v=A\b;
%% Verify at the higher temperature (280[°C])
T=280+273.15;
aaaa=v(1)+v(2)*log(shear_rate)+v(3).*(log(shear_rate).^2)+v(4)*T.*log(shear_rate)+v(5)*T+v(6)*T^2;
aaaa=exp(aaaa);
%% Graphical representation
loglog(shear_rate,aaaa,'b-',shear_rate,v3,'r-');
Torsten
Torsten 2022-3-14
I was talking about columns, not rows.
ones(1,length(shear_rate)), e.g., is a row vector, not a column vector.
Alessio Pricci
Alessio Pricci 2022-3-14
I modified, by transposing the vectors, but the results does not change:
T1=ones(1,length(shear_rate))'*(210+273.15);
T2=ones(1,length(shear_rate))'*(245+273.15);
T3=ones(1,length(shear_rate))'*(280+273.15);
A1=[ones(1,length(shear_rate))', log(shear_rate)', (log(shear_rate)).^2', T1.*log(shear_rate)', T1, T1.^2];
A2=[ones(1,length(shear_rate))', log(shear_rate)', (log(shear_rate)).^2', T2.*log(shear_rate)', T2, T2.^2];
A3=[ones(1,length(shear_rate))', log(shear_rate)', (log(shear_rate)).^2', T3.*log(shear_rate)', T3, T3.^2];
A=[A1; A2; A3];
b=[log(v1) log(v2) log(v3)]';
v=A\b;
T=280+273.15;
aaaa=v(1)+v(2)*log(shear_rate)+v(3).*(log(shear_rate).^2)+v(4)*T.*log(shear_rate)+v(5)*(T)+v(6)*T^2;
aaaa=exp(aaaa);
loglog(shear_rate,aaaa,'b-',shear_rate,v3,'r-');
Torsten
Torsten 2022-3-14
编辑：Torsten 2022-3-15
D1 = 7.40688e+12;
D2 = 373.15;
A1 = 30.62;
A2 = 51.6;
n = 0.2411;
tau = 72297.9;
T1 = 210+273.15;
T2 = 245+273.15;
T3 = 280+273.15;
shear_rate=(linspace(1e-1,1e4,1e5)).';
V = @(T) D1*exp(-(A1*(T-373.15))./(A2+(T-373.15)))./((1+((D1*exp(-(A1*(T-373.15))./(A2+(T-373.15)))).*shear_rate./tau)).^(1-n));
T = T1;
v1 = V(T1);
T = T2;
v2 = V(T2);
T = T3;
v3 = V(T3);
A = [ones(3*numel(shear_rate),1),...
[log10(shear_rate);log10(shear_rate);log10(shear_rate)],...
[(log10(shear_rate)).^2;(log10(shear_rate)).^2;(log10(shear_rate)).^2],...
[T1*log10(shear_rate);T2*log10(shear_rate);T3*log10(shear_rate)],...
[T1*ones(numel(shear_rate),1);T2*ones(numel(shear_rate),1);T3*ones(numel(shear_rate),1)],...
[T1^2*ones(numel(shear_rate),1);T2^2*ones(numel(shear_rate),1);T3^2*ones(numel(shear_rate),1)]];
b = [log10(v1);log10(v2);log10(v3)];
v0 = A\b
v = lsqnonlin(@(x)fun(x,A,b),v0)
T = T3;
aaaa = v(1) + v(2)*log10(shear_rate) + v(3)*(log10(shear_rate)).^2 + v(4)*T.*log10(shear_rate) + v(5)*T + v(6)*T^2;
aaaa = 10.^(aaaa);
%% Graphical representation
loglog(shear_rate,aaaa,'b-',shear_rate,v3,'r-');
function res = fun(x,A,b)
res = 10.^(A*x) - 10.^b;
end

请先登录，再进行评论。

更多回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 Support Vector Machine Regression 的更多信息

产品

版本

R2020b

标签

Community Treasure Hunt

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

Start Hunting!

Translated by