フィルターのクリア

During calculations , a constant is obtained instead of a function

2 ビュー (過去 30 日間)
Dmitry
Dmitry 2022 年 12 月 28 日
コメント済み: Torsten 2022 年 12 月 28 日
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 件のコメント
Torsten
Torsten 2022 年 12 月 28 日
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).

サインインしてコメントする。

回答 (0 件)

カテゴリ

Help Center および File ExchangeInteractive Control and Callbacks についてさらに検索

製品

Community Treasure Hunt

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

Start Hunting!

Translated by