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How to calculate array factor in matlab

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FARHA KHAN 2021-12-23

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编辑： Walter Roberson 2021-12-24

Untitled.png

I m trying to write matlab code for this expression.

Can anyone help me with this?

Thanks!!!!

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Rik 2021-12-23

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What have you tried so far? This doesn't look like it should be too complicated to write.

Note that the subscripts are very hard to read in your image.

FARHA KHAN 2021-12-23

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编辑：Rik 2021-12-23

在 MATLAB Online 中打开

I have tried this: Here there are two rings 1st ring 4ele and 2nd ring ring 16ele. subscripts for first summation:m=1:M (here M=1,2) subscripts for second summation:Nm (here Nm=4,16) i.e when m=1 Nm=4 and when m=2 Nm=16.

I hope you undersood this .

I am not getting the result. help me @Rik

AF(1,1)= 0;n=[4 16]; %no of ele
phi1=[90:90:360 zeros(1,12); 22.5:22.5:360]
l=[1 4] %mode
r=[0.5 2.5]*lambda %radii of each ring
k=(2*pi)/(lambda);
phio=0;
phi=0;
theta=0;
for m=1:2  
    for i=1:n(m)
        AF= AF + (exp((1j).*(((k)*(r(m))*(sin(deg2rad(theta)))*(cos((deg2rad(phi))-(deg2rad(phi1(m,i))))) + ((l(m))*((-k)*r(m)*cos((deg2rad(phio))-(deg2rad(phi1(m,i))))))))));
    end
    AFk=abs(AF);
end

FARHA KHAN 2021-12-23

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I am getting the output @Rik. as 9.741 .

I want to know this program is right?As I am not getting required output which should be zero.

Rik 2021-12-23

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I can't properly read you function, so I can't absolutely confirm you did it correctly.

The main advice I have for you is to split your calculation into several parts you can more easily read. One main thing would be to move the calls to deg2rad outside the loop.

If you post a clearer version of your mathematical notation I can try to help you check if this is correct.

FARHA KHAN 2021-12-23

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Can you see this @Rik

Rik 2021-12-23

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在 MATLAB Online 中打开

Yes, this I can read.

I would convert this to Matlab as follows:

lambda=__;
M=2;
N=[4 16];
K=(2*pi)/lambda;
r=[0.5 2.5]*lambda;
theta=0;
phi=[90:90:360 zeros(1,12); ...
    22.5:22.5:360];
phi_=0;
alpha=__;
s=0;
for m=1:M
    for n=1:N(m)
        tmp=K * r(m) * sin(theta) * cos(phi_-phi(m,n)) + alpha(m,n);
        s=s + W(m) * exp(1j*tmp);
    end
end

FARHA KHAN 2021-12-23

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alpha(m,n)=-k*r(m)*cos(phi0-phi(m,n)) @Rik

FARHA KHAN 2021-12-23

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编辑：Walter Roberson 2021-12-24

在 MATLAB Online 中打开

intializing s as?

M=2
for m=1:M
    for n=1:n(m)
        tmp=(k * r(m) * sin(theta) * cos(phio-phi(m,n))) - (k * r(m) * cos(phio-phi(m,n)));
        s=s + W(m) * exp(1j*tmp);
    end
end

@rik

FARHA KHAN 2021-12-23

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tq it was bit helpful @Rik

Rik 2021-12-23

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You could also calculate alpha in your loop:

s=0;
for m=1:M
    for n=1:N(m)
        alpha=-K*r(m)*cos(phi0-phi(m,n));
        tmp=K * r(m) * sin(theta) * cos(phi0-phi(m,n)) + alpha;
        s=s + W(m) * exp(1j*tmp);
    end
end