Trial and error problem
5 次查看(过去 30 天)
显示 更早的评论
Dear Matlab Community Members
My objective is to find the values a, b, and c. of the following equation:
log(m)=log(a)+(b/(T-c))
I have three values of m & T
m=0.0701 & T=293.65
m=0.0262 & T=313.15
m=0.00433 & T=373.15
I want to know what is the best code or technique to find these values.
Thank you very much
8 个评论
the cyclist
2023-1-26
编辑:the cyclist
2023-1-26
How did you choose that function for the fit? It seems like a poor one, because of the singularity that will happen when T==c.
c = 280;
T = 260 : 380;
f = @(t) 1./(t -c);
plot(T,f(T))
Even if your values of T are strictly greater than c, (the "critical" temperature), this behavior will make it difficult for any algorithm to fit it.
One can try (e.g. with fitnlm), but MATLAB spits out warnings about overparameterization, and the fit is not good.
回答(1 个)
Torsten
2023-1-26
编辑:Torsten
2023-1-26
The results are not convincing, but that's the way to go with three data points.
fun1 = @(x)[0.0701-x(1)*exp(x(2)/(293.65-x(3)));0.0262-x(1)*exp(x(2)/(313.15-x(3)));0.00433-x(1)*exp(x(2)/(373.15-x(3)))];
fun2 = @(x)[log(0.0701)-(log(x(1))+x(2)/(293.65-x(3)));log(0.0262)-(log(x(1))+x(2)/(313.15-x(3)));log(0.00433)-(log(x(1))+x(2)/(373.15-x(3)))];
x0 = [2 1 450];
x = fsolve(fun1,x0)
fun1(x)
x = fsolve(fun2,x0)
fun2(x)
5 个评论
Torsten
2023-1-27
编辑:Torsten
2023-1-27
Isn't the form of the equation usually
m = 10^(A-B/(C+T))
where T is in degreeC ?
Thus
log10(m) = A - B/(T+C)
where log10 is not the natural, but the decadic logarithm and T is temperature in degreeC ?
m1=0.0701;
T1=293.65-273.15;
m2=0.0262;
T2=313.15-273.15;
m3=0.00433;
T3=373.15-273.15;
syms a b c
eqns = [a-b/(T1+c)== log10(m1), a-b/(T2+c)== log10(m2), a-b/(T3+c)==log10(m3)];
S = vpasolve(eqns,[a b c])
double(subs([a-b/(T1+c)- log10(m1), a-b/(T2+c)- log10(m2), a-b/(T3+c)-log10(m3)],[a,b,c],[S.a,S.b,S.c]))
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Symbolic Math Toolbox 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!