Radar Equation: the derivation
Joachim Köppen DF3GJ ... Kiel Oct 2022
(One)
The transmitting antenna sends out radio waves via its radiation pattern
ΦTX(x) where x shall be the offset angle from the beam centre
(with Φ(x=0) = 1). If it is supplied by power PTX, a ray in direction x
contains the power
P(x) = PTX * GTX * ΦTX(x)
On the way to the Moon at a distance r, the radio waves spread out in a spherical way,
so that the power per unit surface area decreases like the distance squared:
power_density = P(x) / (4 π r²)
This basically is the free-space loss. But it simply describes how the amount of power
spreads out over the surface of the sphere whose radius is the distance from the
transmitter. Thus a patch of the lunar surface with an area A receives this amount
of power
deposited power = power_density * A = PTX * GTX * ΦTX(x) / (4 π r²) * A
We note that this is only from the ray sent off from the transmitter into direction x.
If we wanted to know the total amount of power deposited, we would have to integrate
over x, i.e. the antenna pattern.
Thus the transmitter antenna produces a spot on the lunar surface, according to its
radiation pattern: the centre is brighter than the outer areas.
(Two)
Now we have a bright spot on the Moon, which will re-radiate some of the incident
power. How much each part of the spot radiates, depends on the material of the lunar
soil, its dielectric constant, the size and shape of the 'pebbles', and the structure
of the surface.
In the article on Radar Studies of the Moon by J.V.Evans
(J.Res. National Bureau of Standards, Section D, p.1637 (1969)
available at https://nvlpubs.nist.gov/nistpubs/jres/69D/jresv69Dn12p1637_A1b.pdf)
he writes on page 1640 ... 'If the surface of the Moon were solid rock having a
dielectric constant k = 5 (the lowest value likely), then the reflection coefficient
would be 14 percent.'
This means that the remaining 86% of the power is absorbed,
and thus warms up the lunar soil! Since the soil is composed of sand, pebbles, and
irregularly shaped rocks, the reflection is even less ... thus the effective cross
section of the Moon is only 7 percent of its geometrical section (see his fig.2).
In radar technology a target is characterised by its 'radar cross section', i.e. the
area which would be responsible for the amount of re-radiated power.
(Three)
What happens to the power emitted by the spot? Let's think of a transmitter on
the Moon that emits the power Preflected.
The radio waves propagate in all directions like spherical shells, and - as before -
the power density diminishes with the distance r covered:
power_density = Preflected / (4 π r²)
When the waves reach the Earth and hit the receiving antenna (which shall have an
effective area Aeff) it picks up the power
Preceived = Preflected / (4 π r²) * Aeff
The effective area (also capture or collecting area) of an antenna is smaller
than the geometrical area of the dish. It is the area which is responsible for
catching the incoming radio waves, and it is tightly linked with the 'solid angle
of the antenna' ΩA - which is the solid angle of the cone through
which all the power transmitted or received by the antenna would flow with a
constant gain, and it is quite close
to the square of the HPBW (in radians) - by this fundamental relation
Aeff = λ² / ΩA
The gain G of the antenna is the ratio of the solid angles of the full sphere to that
of the antenna
G = 4 π / ΩA
Thus we have
Aeff = λ² / (4 π) * G
for the centre of the receiving beam. For any offset angle the antenna pattern comes
in (note that ΦRX(x=0) = 1):
Aeff(x) = λ² / (4 π) * G * ΦRX(x)
The effective antenna area is a measure of the gain!
Finally, let us put things together: the power received by a receiving antenna
(at offset x_RX) from the lunar echo of the power P_TX transmitted by the
transmitting antennna (at offset x_TX) is
Preceived = PTX
* GTX * ΦTX(xTX) * 1/(4 π rTX²) * moon_cross-section
* moon_reflection_efficiency
* 1/(4 π rRX²) * λ²/(4 π) * GRX * ΦRX(xRX)
Integrating over all offset angles, one gets equation 2 in Evans' article.
Note that the radar equation looks quite symmetric: