Plotting the function by the points that need to be determined

1 Ansicht (letzte 30 Tage)
Dmitry
Dmitry am 12 Jan. 2023
Bearbeitet: Torsten am 15 Jan. 2023
It is necessary to plot F(t) by points. The function F(t) is a sum from 0 to nD(t)(nD is the upper limit of the sum) which depends on 't', i.e. I have an array 't' and an array nD(t) is formed from it, it contains 20 values [496, 248, 165, ...], the first point will be the final sum of F(t) with an upper limit of 496, the first point will be the final sum of F(t) with an upper limit of 248, etc., it is necessary to plot F(t) at these 20 points.
My code:
%% initial conditions
global d k0 h_bar ksi m E;
Ef = 2.77*10^3;
Kb = physconst('boltzmann'); % 1.38*10^(-23)
T = 0.12:0.24:6.4;
m = 9.1093837*10^(-31);
Tc = 1.2;
%t = T./Tc;
t = 0.1:0.1:2;
nD = floor(375./(2.*pi.*t.*1.2) - 0.5);
D = 10^(-8); % толщина пленки
ksi = 10^(-9);
%d = D/ksi;
d = 1000;
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(3601+1i.*(2.*n+1))-((1+1i)./sqrt(2)).*sqrt(2.*n+1)); % константа
end
%% calculation
F = f_calc(t,nD);
plot(t,F, '-r');
%% F(t)
function F = f_calc(t,nD)
global d k0 h_bar ksi m;
F = 0;
for i = 1:20
n = nD(1,i);
F = F + 1/(2*n+1).*(k0.*real(((f_p1(n,t)-f_p2(n,t))./2))+(f_arg_2(n,t)-f_arg_1(n,t))./d);
end
F = -F;
%F = -(1/d).*F;
%F = F - C_2;
end
function p1 = f_p1(n,t)
p1 = ((1+1i)./sqrt(2)).*sqrt(t.*(2.*n+1));
end
function p2 = f_p2(n,t)
global E;
p2 = sqrt(3601+1i.*t.*(2.*n+1));
end
function n_lg = f_lg(n,t)
global d k0;
arg_of_lg = (1+exp(-1i*d*k0.*f_p1(n,t)))/(1+exp(-1i*d*k0.*f_p2(n,t)));
n_lg = log(abs(arg_of_lg));
end
function arg_1 = f_arg_1(n,t)
global d k0;
arg_1 = angle(1+exp(-1i*d*k0.*f_p1(n,t)));
end
function arg_2 = f_arg_2(n,t)
global d k0;
arg_2 = angle(1+exp(-1i*d*k0.*f_p2(n,t)));
end
  2 Kommentare
Torsten
Torsten am 12 Jan. 2023
So you expect F to be a matrix of size numel(t) x numel(nD) where element (i,j) is the sum for t(i), taken from 0 to nD(j) ?

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Antworten (2)

Torsten
Torsten am 12 Jan. 2023
Bearbeitet: Torsten am 15 Jan. 2023
Maybe something like this ?
%% initial conditions
global d k0 h_bar ksi m E;
Ef = 2.77*10^3;
Kb = physconst('boltzmann'); % 1.38*10^(-23)
T = 0.12:0.24:6.4;
m = 9.1093837*10^(-31);
Tc = 1.2;
%t = T./Tc;
t = 0.1:0.1:2;
nD = floor(375./(2.*pi.*t.*1.2) - 0.5);
D = 10^(-8); % толщина пленки
ksi = 10^(-9);
%d = D/ksi;
d = 1000;
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(3601+1i.*(2.*n+1))-((1+1i)./sqrt(2)).*sqrt(2.*n+1)); % константа
end
%% calculation
F = f_calc(t,nD);
hold on
plot(t,F(:,1),"Color","red");
plot(t,F(:,numel(nD)),"Color","blue");
hold off
grid on
function F = f_calc(t,nD)
global d k0 h_bar ksi m;
F = zeros(numel(t),numel(nD));
for i = 1:numel(t)
for j = 1:numel(nD)
n = nD(j);
for k = 0:n
F(i,j) = F(i,j) + 1/(2*k+1).*(k0.*real(((f_p1(k,t(i))-f_p2(k,t(i)))./2))+(f_arg_2(k,t(i))-f_arg_1(k,t(i)))./d);
end
end
end
F = -F;
%F = -(1/d).*F;
%F = F - C_2;
end
function p1 = f_p1(n,t)
p1 = ((1+1i)./sqrt(2)).*sqrt(t.*(2.*n+1));
end
function p2 = f_p2(n,t)
global E;
p2 = sqrt(3601+1i.*t.*(2.*n+1));
end
function n_lg = f_lg(n,t)
global d k0;
arg_of_lg = (1+exp(-1i*d*k0.*f_p1(n,t)))/(1+exp(-1i*d*k0.*f_p2(n,t)));
n_lg = log(abs(arg_of_lg));
end
function arg_1 = f_arg_1(n,t)
global d k0;
arg_1 = angle(1+exp(-1i*d*k0.*f_p1(n,t)));
end
function arg_2 = f_arg_2(n,t)
global d k0;
arg_2 = angle(1+exp(-1i*d*k0.*f_p2(n,t)));
end
  7 Kommentare
Dmitry
Dmitry am 15 Jan. 2023
I think we misunderstood each other:
We have:
Each of these sums is finite, there are 20 such sums in total, since there are twenty values in the array t[]. Therefore, our final graph will consist of these 20 points.
Torsten
Torsten am 15 Jan. 2023
Bearbeitet: Torsten am 15 Jan. 2023
And what's the sense that you take a different number of elements for the sum depending on t ?
Do you know in advance how fast the infinite series converges depending on t ?
See the answer below.

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Torsten
Torsten am 15 Jan. 2023
%% initial conditions
global d k0 h_bar ksi m E;
Ef = 2.77*10^3;
Kb = physconst('boltzmann'); % 1.38*10^(-23)
T = 0.12:0.24:6.4;
m = 9.1093837*10^(-31);
Tc = 1.2;
%t = T./Tc;
t = 0.1:0.1:2;
nD = floor(375./(2.*pi.*t.*1.2) - 0.5);
D = 10^(-8); % толщина пленки
ksi = 10^(-9);
%d = D/ksi;
d = 1000;
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(3601+1i.*(2.*n+1))-((1+1i)./sqrt(2)).*sqrt(2.*n+1)); % константа
end
%% calculation
F = f_calc(t,nD);
plot(t,F)
grid on
function F = f_calc(t,nD)
global d k0 h_bar ksi m;
F = zeros(1,numel(t));
for i = 1:numel(t)
for k = 0:nD(i)
F(i) = F(i) + 1/(2*k+1).*(k0.*real(((f_p1(k,t(i))-f_p2(k,t(i)))./2))+(f_arg_2(k,t(i))-f_arg_1(k,t(i)))./d);
end
end
F = -F;
%F = -(1/d).*F;
%F = F - C_2;
end
function p1 = f_p1(n,t)
p1 = ((1+1i)./sqrt(2)).*sqrt(t.*(2.*n+1));
end
function p2 = f_p2(n,t)
global E;
p2 = sqrt(3601+1i.*t.*(2.*n+1));
end
function n_lg = f_lg(n,t)
global d k0;
arg_of_lg = (1+exp(-1i*d*k0.*f_p1(n,t)))/(1+exp(-1i*d*k0.*f_p2(n,t)));
n_lg = log(abs(arg_of_lg));
end
function arg_1 = f_arg_1(n,t)
global d k0;
arg_1 = angle(1+exp(-1i*d*k0.*f_p1(n,t)));
end
function arg_2 = f_arg_2(n,t)
global d k0;
arg_2 = angle(1+exp(-1i*d*k0.*f_p2(n,t)));
end

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