Saving Data for Each for loop

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Sohel Rana
Sohel Rana am 22 Jan. 2021
Beantwortet: Gaurav Garg am 25 Jan. 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 Kommentare
Sohel Rana
Sohel Rana am 22 Jan. 2021
Thanks for mentioning it.

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Gaurav Garg
Gaurav Garg am 25 Jan. 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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