Phased Array Math

Thinned arrays

The emitters send energy to a ground antenna, which is a very distant point. Except for the square meter sized receive antenna, every watt going in some other direction is wasted power and potential interference. There may be many emitters on a thinsat, but for now we will model with one. The thinsats are widely spaced, so sunlight can permeate the entire array - there will be many grating lobes.

In fact, because the spacing is much larger than the wavelength, the array will be sparse and we encounter the thinned array curse. That means that most of the radio power is splattered at angles away from the main beam. For an array that is 99.9999% sparse, approximately 99.9999% of the power is splattered away from the main lobe.

Is this a problem? Not if we can spread out the splatter and eliminate concentrated sidelobes. With 64,000 dipole emitters, the main lobe voltage will be 64K times the field from one emitter, and the power density will be 4.1E9 times the power (intensity squared) from one emitter. If the emitters are widely spaced, the ground spot where this occurs will be small compared to a uniform beam, and the rest of the energy will spray into 2π steradians, the dipole emission pattern.

The power for the array is P_T = N P_O , where N is the number of emitters and P_O is the power of each one. The area of the transmitter is A_T = N x^2 where x is the average spacing. If the transmitting antennas are arranged in a disk with diameter D_T then A_T = ( { \pi \over 4 } ) {D_T}^2 . Therefore D_T = x \sqrt{ { 4 \over \pi } N } .

The ground spot diameter will be D_G = 1.22 \lambda L / D_T = 1.22 \left( { \lambda \over x } \right) L / \sqrt{ { 4 \over \pi } N } , where \lambda is the wavelength and L is the distance from transmitter to ground spot. The power intensity at the ground spot will be approximately I_G = N P_T / ( 2 \pi L^2 ) for a dipole antenna. So the total power received will be approximately:

P_G \approx { \pi \over 4 } {D_G}^2 I_G = { \pi \over 4 } \left( 1.22^2 \left( { \lambda \over x } \right)^2 { { L^2 } \over { { 4 \over \pi } N } } \right) ~ \left( { { N P_T } \over {2 \pi L^2} } \right) = { { 1.22^2 \pi } \over 32 } \left( { \lambda \over x } \right)^2 P_T \approx 0.146 \left( { \lambda \over x } \right)^2 P_T

Assuming a wavelength of 8mm, a distance of 10,000 km, and an array width of 1km, the ground spot will be 1.22 × 8E-3 m × 1E7 m / 1e3m or about 100 meters. If each of 64K emitters is radiating 100 mW, then the average isotropic power is 64K×1E-1/(2π × 1E7²) = 1E-11 W/m² and the main lobe power will be 64K times that, or 6.5E-7 W/m². The total power into the main lobe will be 5.1E-3W, a tiny fraction of the entire 6.4 kW transmitted by the whole array. However, with a noise factor of 20, a bit at 300K needs 4E-20 joules, so a meter-square antenna collecting 6.5e-7W can receive 1.6E13 bits per second. That is what the entire 6.4 kW transmitter array accomplishes, sending 1E-10 of its power to the intended single receive antenna. We can probably dial down the power a lot for mere 10 gigabit transmission.

Wasteful, huh? Radio is like that. A television set might receive 2E7 bits of information a second. A billion television sets receive 1E16 bits per second, or 8E-4 joules per second, or 400 microwatts receive energy for all those television sets. A typical digital UHF television transmitter emits 1MW, and there are about 20,000 transmitters around the world, so they total about 20 GW. The ratio of usable received energy to transmitted energy is 2E-14. Pretty lousy ...

If the 64K emitters are spaced half a wavelength apart, in a disk, then the transmit array must be only 1.1 meters across. That will put all the power of the main lobe into a 89km diameter ground spot receiving 6.5e-7 W/m² or about 4kW. Close enough to 6.4kW - if we want to pick up the dregs of the signal, we will need a wider receive antenna with low power around the fringes.

A big, sparse array is probably good. People live in bunches, and if we want to send separate signals to nearby people, we will need small ground spots. With all but 1E-10 of our power going someplace besides the receive antenna, we might as well send it out as isotropically as we can, pestering the whole world a tiny bit rather than nearby neighbors a whole lot.

Power Sky

When we use server sky technology for transmitting grid power ("Power Sky") we want to eliminate all the sidelobe and grating power that we can - it should all go into the main beam. Given the sparse array curse, that means spreading out the receive power over a wide antenna, by purposely reducing the power density in the central lobe. If we can do that, while simultaneously scattering "zeros" in all other directions, perhaps by adding semi-passive lens antennas in the beam path, then we can use large semi-sparse arrays in orbit to simulate a denser array.

Assume a 1 terawatt array in a Lagrange-like halo orbit at lunar distances (approximately 400k km, 40 times farther from the rectenna than M288 ). At 7% conversion and transmission efficiency from satellite solar to grid power, that requires 1E10m² of collector, an area 100km on a side. With 2cm 15GHz wavelengths, our transmitter could theoretically hit a ground antenna 100 meters on a side. Obviously, we will need to spread the power a lot more than that. If our rectenna produces 100kW per square meter, the 1E7m² antenna will be 3.6 km in diameter. The world might need 200 of these, sited in low biological productivity regions near the equatorial ocean, and distributing the power with power loop technology.

Phased Array Definitions and Equations

This will get pictures later.

A thinsat array is three dimensional. The emitters are in a cartesian grid:

For now, ignore polarization and dipole radiation patterns - assume each emitter produces an isotropic, single frequency sinusoidal electric field with a phase. Assume "x" and "y" angles, rotated eastward and northward.

Choose the (x, y, z) of the array as zero phase. If we are aiming in a direction