Plotting the function by the points that need to be determined

Question

Dmitry 2023-1-12

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1892900-plotting-the-function-by-the-points-that-need-to-be-determined

编辑： Torsten 2023-1-15

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

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Torsten 2023-1-12

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) ?

Dmitry 2023-1-12

Yes, you are right.

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Answer 1

Torsten 2023-1-12

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1892900-plotting-the-function-by-the-points-that-need-to-be-determined#answer_1147455

编辑：Torsten 2023-1-15

在 MATLAB Online 中打开

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

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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Dmitry 2023-1-15

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 2023-1-15

编辑：Torsten 2023-1-15

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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Answer 2

Torsten 2023-1-15

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1892900-plotting-the-function-by-the-points-that-need-to-be-determined#answer_1149045

在 MATLAB Online 中打开

%% 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

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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