How to calculate array factor in matlab

2021 12 23
0 个回答
更新时间：2021 12 24
24 次查看（30 天）

0 个投票

I m trying to write matlab code for this expression.
Can anyone help me with this?
Thanks!!!!

11 个评论
显示 9更早的评论 隐藏 9更早的评论

Rik
Rik 2021-12-23
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
FARHA KHAN 2021-12-23
编辑：Rik 2021-12-23
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
FARHA KHAN 2021-12-23
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
Rik 2021-12-23
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.
Rik
Rik 2021-12-23
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
FARHA KHAN 2021-12-23
alpha(m,n)=-k*r(m)*cos(phi0-phi(m,n)) @Rik
FARHA KHAN
FARHA KHAN 2021-12-23
编辑：Walter Roberson 2021-12-24
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
Rik
Rik 2021-12-23
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

请先登录，再进行评论。

请先登录，再回答此问题。

回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 MATLAB Support Package for Raspberry Pi Hardware 的更多信息

产品

版本

R2021a

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by