Getting Naan error while integrating with multiple parameters

1 Ansicht (letzte 30 Tage)
Aakash Munnangi
Aakash Munnangi am 16 Mär. 2023
Bearbeitet: VBBV am 16 Mär. 2023
Hi, I am evaluating spectral radiance of a black body. It is giving me NaaN errors. The following is th
B =
h=6.63*(10^-34);
k=1.381*(10^-23);
c=3*(10^8);
T=6000;
fun = @(x) (2*h*(x.^3))./(c*c*((exp((h*x)./(k*T))-1)));
x=integral(fun,380*(10^-9),750*(10^-9));
Warning: Inf or NaN value encountered.

Antworten (1)

VBBV
VBBV am 16 Mär. 2023
Bearbeitet: VBBV am 16 Mär. 2023
h=6.63*(10^-34);
k=1.381*(10^-23);
c=3*(10^8);
T=6000;
fun = @(x) (2*h*(c.^2)).*(x.^(-5)).*(exp((h.*c)./(k*x*T))-1).^(-1);
x=integral(fun,380*(10^-9),750*(10^-9))
  2 Kommentare
Aakash Munnangi
Aakash Munnangi am 16 Mär. 2023
I was able to correct it. here is the code. However, the energy version of Planks law definite integral giving valuee of 0. I think I am missing some precision. need help.
working version with wave lengths
h=6.63*(10^-34);
k=1.381*(10^-23);
c=3*(10^8);
T=6000;
lambda1 = 380*10^-9;
lambda2= 750*10^-9;
M_e = @(x) (2 * pi * h * c^2) ./ (x.^5 .* (exp((h * c)./(x * k * T)) - 1));
M_e_int1 = integral(M_e,lambda1,lambda2)
lambda1 = 0;
lambda2= Inf;
M_e = @(x) (2 * pi * h * c^2) ./ (x.^5 .* (exp((h * c)./(x * k * T)) - 1));
M_e_int2 = integral(M_e,lambda1,lambda2)
fraction_of_energy = M_e_int1/M_e_int2;
//Not working or giving value of 0 , for energy version of planc law integration.
h=6.63*(10^-34);
k=1.381*(10^-23);
c=3*(10^8);
T=6000;
Energy_lower_limit = 100;
Energy_upper_limit = 10000;
M_e = @(x) (x.^3.*2)./((h^3 * c^2)*((exp(x/(k * T)) - 1)));
M_e_int1 = integral(M_e, Energy_lower_limit, Energy_upper_limit)
VBBV
VBBV am 16 Mär. 2023
Bearbeitet: VBBV am 16 Mär. 2023
The energy version of planks law you are using may be incorrect. Check the values of input limits for energy being considered. Are they in same units ? Can you open a new question ?

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Particle & Nuclear Physics finden Sie in Help Center und File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by