Trial and error problem
5 ビュー (過去 30 日間)
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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 件のコメント
the cyclist
2023 年 1 月 26 日
編集済み: the cyclist
2023 年 1 月 26 日
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 件)
Torsten
2023 年 1 月 26 日
編集済み: Torsten
2023 年 1 月 26 日
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 件のコメント
Torsten
2023 年 1 月 27 日
編集済み: Torsten
2023 年 1 月 27 日
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]))
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