How to modify the code to get my desired array?

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?

Answer 1

[Torsten]

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

[Torsten]

Because you wanted to place your antennas on x- and y-axis with distance "d" of consecutive pairs (L-type geometry).

[Torsten]

Further, normally distance is kept as d=0.5 but you have kept it 0.25, why?

Because I don't know what "normal" is.

I think in your circular arrangement, the antennas are also not 0.5 units apart from each other. Or is there a different rule for what is "normal" ?

[Torsten]

I think with what I provided as code for the L-shape, it should be no problem for you to do it alone.

[Torsten]

What result do you want to get ?

0              0             0
1.0000         0             0
0.7500         0             0
0.5000         0             0
0.2500         0             0
0             1.0000         0         
0             0.7500         0         
0             0.5000         0         
0             0.2500         0                
0             0              1.0000         
0             0              0.7500        
0             0              0.5000         
0             0              0.2500  

?

[Torsten]

N = 4;
d = 0.25;
r = zeros(3*N+1,3);
v = (N*d:-d:d).';
r(2:N+1,1) = v;
r(N+2:2*N+1,2) = v;
r(2*N+2:3*N+1,3) = v;
r

r = 13×3

0 0 0 1.0000 0 0 0.7500 0 0 0.5000 0 0 0.2500 0 0 0 1.0000 0 0 0.7500 0 0 0.5000 0 0 0.2500 0 0 0 1.0000

[Torsten]

Strange, because you said in your previous post

I think its correct because when we look at x-coordinate, then both y and z must be zero.Likewise for y both x and z must be zero and for z, both x and y mus be zero. But where is the code?

But doesn't matter: If you had understood the code, you would have changed N = 4 to N = 1 and everything would have been fine.

[Torsten]

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 !

[Sadiq Akbar]

Thank you very much dear Torsten for your help.