How to modify the code to get my desired array?

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If we keep 10 antennas at equal distance in a circle on XY-Plane, then their coordinates are given by the code given below:
function r=circularArray(N)
radius = 0.5/sind(180/N);
rx = radius*cosd(360*(0:N-1).'/N);
ry = radius*sind(360*(0:N-1).'/N);
r = [rx, ry, zeros(N,1)];
end
Here you just put value of N where N is the number of antennas. So if N=10, then the above code gives:
r =
1.6180 0 0
1.3090 0.9511 0
0.5000 1.5388 0
-0.5000 1.5388 0
-1.3090 0.9511 0
-1.6180 0 0
-1.3090 -0.9511 0
-0.5000 -1.5388 0
0.5000 -1.5388 0
1.3090 -0.9511 0
Now if I want to keep 9 antennas at same equal distance on x and y -axes separatelyi.e., 4 antennas along x-axis and and 4 antennas along y-axis where the inter-antenna distance is same between any two consective antennas and 1 antenna is placed on the origin. Then what change is to be done in the above code to get the desired geometry i.e., L-type geometry instead of circile?

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Torsten
Torsten 2022-7-14
编辑:Torsten 2022-7-14
N = 9;
d = 0.25;
rx = [(N-1)/2*d:-d:d,zeros(1,(N+1)/2)].';
ry = [zeros(1,(N+1)/2),d:d:(N-1)/2*d].';
r = [rx,ry,zeros(N,1)]
r = 9×3
1.0000 0 0 0.7500 0 0 0.5000 0 0 0.2500 0 0 0 0 0 0 0.2500 0 0 0.5000 0 0 0.7500 0 0 1.0000 0
  8 个评论
Torsten
Torsten 2022-7-16
编辑:Torsten 2022-7-16
N = 4;
d = 0.25;
N_on_each_axis = (N-1)/3;
r = zeros(3*N_on_each_axis+1,3);
v = (N_on_each_axis*d:-d:d).';
r(2:N_on_each_axis+1,1) = v;
r(N_on_each_axis+2:2*N_on_each_axis+1,2) = v;
r(2*N_on_each_axis+2:3*N_on_each_axis+1,3) = v;
r
r = 4×3
0 0 0 0.2500 0 0 0 0.2500 0 0 0 0.2500
But you have to make sure that N-1 is divisible by 3 !

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