Saving Data for Each for loop

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Sohel Rana
Sohel Rana el 22 de En. de 2021
Respondida: Gaurav Garg el 25 de En. de 2021
Hi,
I tried this task several times but don't know why I'm not getting the expected results. Here in the code, I used a signle value for n1 and n2. As a result, I got single graph or single value for wave. If I have five values for n1 and n2 each (length of n1 and n2 will be always same), then I should get five graphs or five values for wave. I assume I have to use two for loops: one for n1 and n2, and another one for lam. Then for each n1 & n2 value, I will get a range for lam. In the code, I used B=n2*k:(n1*k-n2*k)./((length(lam))-1):n1*k; so that I can get B values same length (not values) to lam. If lam has 100 values for example, B will have 100 values. You are free to change the expression in B between n2*k and n1*k. B will have a range and the min and max value will be n2*k and n1*k, respectively. Thanks in advance.
clc; clear all; close all;
format long
m=1; p=0.5296; n3=1; a1=4.1*m; a2=62.5*m; dn=10^-3;
n1=1.44507;
n2=1.4444;
lam=1.51*m:0.0001*m:1.55*m;
NA=sqrt(n1^2-n2^2);
k=2*pi./lam;
B=n2*k:(n1*k-n2*k)./((length(lam))-1):n1*k;
neff=B./k;
V=(2*pi*a1*sqrt(n1^2-n2^2))./lam;
b=(neff.^2-n2^2)./(n1^2-n2^2);
%Bessels function
aa=V.*sqrt(1-b);
bb=V.*sqrt(b);
LHS=aa.*(besselj(1,aa)./besselj(0,aa));
RHS=bb.*(besselk(1,bb)./besselk(0,bb));
yc=LHS-RHS;
% core effective index
difference=abs(yc);
min_diff=min(difference);
intersect=find(difference==min_diff);
nc=neff(intersect);
% coupling coefficient
lam_fbg=1.530;
n=(pi^2*(2*a1)^2*NA^2)./(lam.^2+pi^2*(2*a1)^2*NA^2); % overlaping factor
kk=(pi.*n.*dn)./lam_fbg; % coupling coefficient
% detuning factor
L=3000;
d=(2*pi*nc./lam)-pi/p;
dk=d./kk; % detuning ratio
% Bragg normalized reflectivity
kl=kk.*L;
x1=sqrt(1-dk.^2);
x2=(sinh(kl.*x1)).^2;
x3=1-dk.^2;
x4=(cosh(kl.*x1)).^2;
x5=dk.^2;
R=x2./(x3.*x4+x5.*x2);
plot(lam,R);
[ymax, inx]=max(R);
wave=lam(inx);
  4 comentarios
Sohel Rana
Sohel Rana el 22 de En. de 2021
Thanks for mentioning it.

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Respuestas (1)

Gaurav Garg
Gaurav Garg el 25 de En. de 2021
Hi Sohel,
The above function throws an error when we try to run it with n1 and n2 having 5 values.
The error displayed is -
Incorrect dimensions for raising a matrix to a power

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