How to implement Array factor in MATLAB?
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clc
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%arf of cca 4,16 ele
c=3*(10^8);
f=30*(10^9);
lambda=c/f;
l=[1 3] %mode
r=[0.5 1.5]*lambda %radii of each ring
k=(2*pi)/(lambda);
phio=0;
phi=0;
theta=0;
n=[4 16]; %no of ele
phi=[90:90:360 zeros(1,12); 22.5:22.5:360]
AF(1,1) = 0;
for m=1:2
for i=1:n(m)
AF(m,i)=exp((1j)*(((k)*(r(m))*(sin(deg2rad(theta)))*(cos(deg2rad(phi-phi(m,i))))+((l(m))*((-k)*r(m)*cos(deg2rad(phio-phi(m,i))))))));
end
end
Please help.
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Réponses (3)
Walter Roberson
le 5 Déc 2021
phi=[90:90:360 zeros(1,12); 22.5:22.5:360] is a 2 x 16 array.
AF(m,i)=exp((1j)*(((k)*(r(m))*(sin(deg2rad(theta)))*(cos(deg2rad(phi-phi(m,i))))+((l(m))*((-k)*r(m)*cos(deg2rad(phio-phi(m,i))))))));
The subtraction phi-phi(m,i) is taking all of phi and subtracting a scalar from it. That is going ot give you a 2 x 16 result because phi is 2 x 16. So the right hand side of the expression is going to be 2 x 16
Your formula defines AF(phi,I) where phi and I appear to be both M x N matrices. With your phi being 2 x 16 you should expect that the dimensions of AF should be 2 x 16 x (size of I) -- and that is after the double summation. Your code is not even doing the double summation.
4 commentaires
Walter Roberson
le 5 Déc 2021
Your existing code
for m=1:2
for i=1:n(m)
already does that, if m(1)=4 and m(2)=16 .
Note: the equation you posted is low resolution and difficult to read.
FARHA KHAN
le 5 Déc 2021
Modifié(e) : Walter Roberson
le 5 Déc 2021
2 commentaires
Walter Roberson
le 5 Déc 2021
What leads you to expect to get 20 output elements when you are working with 2 x 16 arrays?
ATHULRAJ
le 27 Fév 2023
AF= zeros(2,16);
for m=1:2
for i=1:n(m)
AF=AF+exp((1j)*(((k)*(r(m))*(sin(deg2rad(theta)))*(cos(deg2rad(phi-phi(m,i))))+ ((l(m))*((-k)*r(m)*cos(deg2rad(phio-phi(m,i))))))));
end
end
AFK=AF
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