During calculations , a constant is obtained instead of a function

1 Ansicht (letzte 30 Tage)
Dmitry
Dmitry am 28 Dez. 2022
Kommentiert: Torsten am 28 Dez. 2022
I have a complex function that depends on 'd' and 'n' F(n,d) I need to sum a series over 'n' and plot F(d).
I know that after summing by 'n' I should get a function F(d) which is rapidly oscillating and decaying. But after summing, I get a constant that does not depend on d.
My cod:
%% initial conditions
global t k0 h_bar ksi m E;
Ef = 2.77*10^3;
Kb = physconst('boltzmann'); % 1.38*10^(-23)
D = 5:5.1:50;
m = 9.1093837*10^(-31);
Tc = 1.2;
t = 1;
ksi = 10^(-9);
d = D./ksi;
E = Ef/(pi*Kb*Tc);
h_bar = (1.0545726*10^(-34));
k0 = (ksi/h_bar)*sqrt(2.*m.*pi.*Kb.*Tc);
C_2 = 0;
for n = 0:49
C_2 = C_2 + (1/(2.*n+1)).*k0.*real(sqrt(E+1i.*(2.*n+1))-((1+1i)./sqrt(2)).*sqrt(2.*n+1)); % константа
end
%% calculation
F = f_calc(d);
plot(d,F,'o');
%% F(d)
function F = f_calc(d)
global t k0 h_bar ksi m;
F = 0;
for n = 0:49
F = F + 1/(2*n+1).*imag(f_lg(n,t)+1i*d.*k0.*((f_p1(n)-f_p2(n))./2)+1i*f_arg_1(n,d)-1i*f_arg_2(n,d));
end
F = -(1./d).*F;
plot(d,F,'o');
end
function p1 = f_p1(n)
global t;
p1 = ((1+1i)./sqrt(2)).*sqrt(t.*(2.*n+1));
end
function p2 = f_p2(n)
global t E;
p2 = sqrt(E+1i.*t.*(2.*n+1));
end
function n_lg = f_lg(n,d)
global t k0;
arg_of_lg = (1+exp(-1i*d*k0.*f_p1(n)))/(1+exp(-1i*d*k0.*f_p2(n)));
n_lg = log(abs(arg_of_lg));
end
function arg_1 = f_arg_1(n,d)
global t k0;
arg_1 = angle(1+exp(-1i*d*k0.*f_p1(n)));
end
function arg_2 = f_arg_2(n,d)
global t k0;
arg_2 = angle(1+exp(-1i*d*k0.*f_p2(n)));
end
  1 Kommentar
Torsten
Torsten am 28 Dez. 2022
But after summing, I get a constant that does not depend on d.
You get a vector of values that depends on d. But the difference between its elements is very small compared to their absolute value (1e11).

Melden Sie sich an, um zu kommentieren.

Antworten (0)

Kategorien

Mehr zu 2-D and 3-D Plots finden Sie in Help Center und File Exchange

Produkte

Community Treasure Hunt

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

Start Hunting!

Translated by