Ground Antenna Array Zeros
Summary: The zeros of the ground antenna array can be lined up to overlap nonselected orbiting arrays, nulling them out if they are not selected.
Note: this is incomplete. It is only for an array at zenith, from 0° latitude. The ground array will probably need to be larger to select arrays near the horizon. I also computed \alpha incorrectly, big errors far from zenith.
from Haupt pages 55 to 57:
\large { \psi ~=~ { \LARGE \left( { 2 \pi d } \over \lambda \right) } \sin( \phi ) } ~~~~~ where \phi is the angle from zenith
For an equatorial antenna array pointed at zenith:
\large { Power ~=~ } \LARGE \left( { \sin( N \psi / 2 ) } \over { N \sin( \psi / 2 ) } \right)^2 ~ ~ ~
So \psi where the power is zero are given by:
\large { \psi_0 ~=~ \pm { \LARGE \left( { 2 m \pi } \over N \right) } ~~~~ m = 1, 2, 3, ... }
So we can find \sin( \phi_m ) from:
{ \LARGE \left( { 2 \pi d } \over \lambda \right) } { \large \sin( \phi_m) ~=~ } { \LARGE \left( { 2 \pi m } \over N \right) } or { \large \sin( \phi_m ) ~=~ } { \LARGE { { m \lambda } \over { N d } } }
The ground antenna is at radius R_E . What is the angle \large \gamma of the thinsat array( at radius R_{288} ) from zenith? We can compute this from the SSA triangle:
{ \large \sin( \gamma ) ~=~ { \LARGE \left( { R_E } \over { R_{288} } \right) } \sin( \phi_m ) ~=~ { \LARGE \left( { R_E \lambda } \over { R_{288} N d } \right) } m }
But that is the wrong angle! What we want is \large \alpha ~=~ \phi_m  \gamma . The following is approximately right near zenith because R_{288} ~ \approx ~ 2 \times R_E in this particular instance, but not always!
From the law of sines, what we actually want is:
{ \large \sin( \alpha ) ~=~ { \LARGE \left( {  \vec{R_{288}}  \vec{R_E}  } \over { R_{288} } \right) } \sin( \phi_m ) ~=~ { \LARGE \left( {  \vec{R_{288}}  \vec{R_E}  ~ \lambda } \over { R_{288} N d } \right) } m }
W ~ \approx ~ N d is approximately the length of the antenna  and since we can leave some elements off, we can adjust N until the angular spacing is exactly the spacing of the arrays. In other words, all the arrays besides the one in the center of the beam are in nulls. At least for the nearby arrays at small angles; for the ones farther out, they will move into the edges of powered lobes  except that those lobes will be at much lower power levels.
This is cool!
So, what is the orbiting array angle versus antenna size, assuming a small angle approximation of \sin( \gamma ) \approx \gamma ? That angle is 2 { \large \pi} / P given P arrays in orbit .
{ \LARGE { { 2 \pi } \over P } } { \large ~ \approx ~ } { \LARGE { {  \vec{R_{288}}  \vec{R_E}  ~ \lambda } \over { R_{288} W } } } ~~~ or ~~~ { \large L ~ \approx ~ }{ \LARGE { { { \lambda P } \over { 2 \pi } } { {  \vec{R_{288}}  \vec{R_E}  } \over R_{288} } } }
\lambda 
4.28 mm 
R_E 
6378 km 
R_{288} 
12789 km 


P 
W 
100 
3.4 cm 
3600 
1.22 m 
10000 
3.4 m 
This is the selectivity of the antenna, assuming many other orbiting arrays are broadcasting simultaneously to the same region. In the distant future, we will have thousands of arrays near the same angular address  all addressing different ground antenna regions. With 10,000 angular addresses, the angular addresses are 8.04 kilometers apart; the arrays might be arranged in a rotating tilted disk around orbit center, radius 30 km, spaced 1 km apart, about 2800 arrays, or 28 million arrays in the entire orbit. At 7842 thinsats each, that is 220 billion total thinsats.