plotting Intensity function which consists of an exponential sum

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Elinor Ginzburg
Elinor Ginzburg 2020-3-25
评论： Sriram Tadavarty 2020-3-26
hello,
Im trying to plot the followin function in matlab:
where k=10,d=1,A=1,0.1,0.01, epsilon is a random number between [0,1], the summation n is from 1 to 20 and theta goes from [-pi/2,pi/2]
I've written this code:
% param defenition %
L=100; N=20; d=1; k=10; Ar=1;
A=0.1;
theta=linspace(-pi/2,pi/2,200);
phase=(d+A*epsilon)*sin(theta);
epsilon=rand(0,1);
function psi = Psi(x)
psi=1;
for n=1:N
psi=psi+(exp(i*k*phase).^n);
end
end
I=([abs(psi)].^2);
fplot(I,[-pi/2,pi/2])
but when I try to run it, I get nothing printed in the figure. what did I do wrong?
Thank you very much for your time and attention.
  3 个评论
显示 1更早的评论
Elinor Ginzburg
Elinor Ginzburg 2020-3-26
thank you for your reply, but it still doesn't work. perhaps you have another idea? now it looks like this:
%% Question 03 %%
% param defenition %
L=100; N=20; d=1; k=10; Ar=1;
A=0.1;
theta=linspace(-pi/2,pi/2,200);
epsilon=rand(0,1);
phase=(d+A*epsilon)*sin(theta);
psi=1;
for n=1:N
psi=psi+(exp(i*k*phase).^n);
end
I=((abs(psi)).^2);
plot(I)

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回答（1 个）

Birdman
Birdman 2020-3-26
Try this:
L=100; N=20; d=1; k=10; Ar=1;
A=0.1;
theta=linspace(-pi/2,pi/2,200);
epsilon=rand;
phase=(d+A*epsilon)*sin(theta);
psi=1;
for n=1:N
psi=psi+(exp(1i*k*phase).^n);
end
I=((abs(psi)).^2);
plot(I)
  2 个评论
Elinor Ginzburg
Elinor Ginzburg 2020-3-26
it worked! thank you so much! I had no idea I need to use rand instead of rand(0,1). how can you keep track between all the changes in different versions? thanks once again!

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