How to implement Array factor in MATLAB?

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FARHA KHAN
FARHA KHAN 2021-12-5
回答： ATHULRAJ 2023-2-27
Ran in:
clc
clear all
close all
%arf of cca 4,16 ele
c=3*(10^8);
f=30*(10^9);
lambda=c/f;
l=[1 3] %mode
l = 1×2
1 3
r=[0.5 1.5]*lambda %radii of each ring
r = 1×2
0.0050 0.0150
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]
phi = 2×16
90.0000 180.0000 270.0000 360.0000 0 0 0 0 0 0 0 0 0 0 0 0 22.5000 45.0000 67.5000 90.0000 112.5000 135.0000 157.5000 180.0000 202.5000 225.0000 247.5000 270.0000 292.5000 315.0000 337.5000 360.0000
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
Unable to perform assignment because the size of the left side is 1-by-1 and the size of the right side is 2-by-16.
Please help.

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回答（3 个）

Walter Roberson
Walter Roberson 2021-12-5
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.
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Walter Roberson
Walter Roberson 2021-12-5
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.

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FARHA KHAN
FARHA KHAN 2021-12-5
编辑：Walter Roberson 2021-12-5
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
is this right?
I am getting 16 ele but im supposed to get 20 @Walter Roberson
  2 个评论
Walter Roberson
Walter Roberson 2021-12-5
What leads you to expect to get 20 output elements when you are working with 2 x 16 arrays?
FARHA KHAN
FARHA KHAN 2021-12-5
编辑：FARHA KHAN 2021-12-5
yeah
but 12 elements are zeros so i said 20 .
Now I am gettig right 2*16
clc
clear all
close all
%arf of cca 4,16 ele
c=3*(10^8);
f=30*(10^9);
lambda=c/f;
l=[1 3] %mode
r=[0.5 2.5]*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= 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
By this code
Just can you check if it is right? @Walter Roberson

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ATHULRAJ
ATHULRAJ 2023-2-27
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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