Want to insert a matrix in a for loop and change it with a variable
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Ashi Khajotia
il 19 Apr 2023
Modificato: chicken vector
il 19 Apr 2023
Hello,
So, I have a variable phi(1x41) changing with lambda(1x41) and there is matrix (2x2) inside a for loop that needs to be changed with phi but the dimension of matrix should be (2x2x41) instead of (2x42) as it shown in workspace. Can anyone help me in this?
d1 = 0.1077;
lam = 3:0.1:7;
th = 0;
n1 = 2.32;
for j = 1:length(lam)
phi1 =2*pi.*(d1./lam).*sqrt((n1).^2 - sind(th).^2);
P1 = [exp(1i.*phi1) 0; 0 exp(-1i.*phi1)];
end
3 Commenti
chicken vector
il 19 Apr 2023
Modificato: chicken vector
il 19 Apr 2023
d1 = 0.1077;
lam = 3:0.1:7;
th = 0;
n1 = 2.32;
phi1 = 2*pi.*(d1./lam).*sqrt((n1).^2 - sind(th).^2);
P1 = zeros(2,2,length(lam));
P1(1,1,:) = exp(1i.*phi1);
P1(2,2,:) = exp(-1i.*phi1)];
This is just one of the many ways you could build your matrix.
Note that the for loop changes the values of the variable j, which your are not using so it is completely useless.
At each iteration you overwrite the variable P1 with the same values since:
x = 1:3;
M = [x 0; 0 x];
Produces:
M = [1 2 3 0 ; 0 1 2 3];
To effectively use the loop I suggest you to initialise the variable outside the loop and use j to change indexing inside P1.
P1 = zeros(2,2,length(lam));
phi1 =2*pi.*(d1./lam).*sqrt((n1).^2 - sind(th).^2);
for j = 1 : length(lam)
P1(:,:,j) = [exp(1i.*phi1(j)) 0 ; 0 exp(1i.*phi1(j))];
end
Anyway I don't reccomend you doing this because vectorisation is much faster than looping and you would notice a great difference with large datasets.
Risposta accettata
Dyuman Joshi
il 19 Apr 2023
d1 = 0.1077;
lam = 3:0.1:7;
th = 0;
n1 = 2.32;
%Define phi1 outside the loop as it is not varying with the loop
phi1 = 2*pi.*(d1./lam).*sqrt((n1).^2 - sind(th).^2);
numlam = numel(lam);
%Preallocation
P1 = zeros(2,2,numlam);
for j = 1 : numlam
P1(:,:,j) = [exp(1i.*phi1(j)) 0; 0 exp(1i.*phi1(j))];
end
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