How to implement Array factor in MATLAB?

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.

Answer 1

[Walter Roberson]

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.

[FARHA KHAN]

CAN YOU HELP ME OUT WHAT CODE TO BE WRITTEN

[FARHA KHAN]

I want somthething like

when m=1

i=1:4

and when m=2

i=1:16

i understand the point you have explained @Walter Roberson

could you help me out with the approach

[Walter Roberson]

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]

yeah but its not doing double summation

Answer 2

[FARHA KHAN]

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

[Walter Roberson]

What leads you to expect to get 20 output elements when you are working with 2 x 16 arrays?

[FARHA KHAN]

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

Answer 3

[ATHULRAJ]

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