SM 5 BSZ - Selecting Polarisation for EME
(April 25 1997)

#### Polarisation in the EME path

When an electromagnetic wave is transmitted from the earth, reflected on the moon surface, and received again on the earth, polarisation is affected by four different phenomena.

1. Spatial polarisation. Imagine you are standing on the moon, looking at the earth, and two antennas on it. If the antennas are far apart, they will not be parallel due to the curvature of the earth's surface. This phenomenon is easily calculated and it affects linear polarisation only, and it is independent of the frequency.

2. Reflection properties of the moon. At low frequencies, the moon behaves as a perfect sphere, metallic or dielectric. As a result, linear polarisation is conserved while circular polarisation changes rotation. At high frequencies, the moon behaves as a chaotic surface, on which the wave is reflected many times before it leaves the surface - like light on a sheet of paper. This leads to a more or less complete loss of polarisation and the energy in the reflected signal is divided into two orthogonal components. When one is received, the other is necessarily lost. Two complete receivers are required to receive both of the signals into which the wave is divided in order to avoid loosing 3dB due to this phenomenon. On 144MHz, the moon behaves like a perfect sphere, the amount of energy lost into the "wrong" polarisation is in the order of 1%. On 10GHz the effect is large. There is an ongoing discussion among amateurs active on that band and I do not know the outcome.

3. Faraday rotation. The free electrons in the ionosphere are controlled by the earth's magnetic field. They move in spiral trajectories. A circularly polarised wave that travels through the ionosphere will have slightly different speed depending on whether it rotates the same way as the electrons or opposite. This small difference in speed will not cause any noticeable effect on circularly polarised waves - the time difference will be in the nanosecond range. For linearly polarised waves, faraday rotation is very important. A horizontal wave can equally well be described as two simultaneous circular waves, both with half the amplitude, rotating in opposite directions. They will add in one plane and cancel in the other. If one of these circular waves is delayed relatively to the other, they will no longer add in the same plane - they will add in a twisted plane. The important aspect of faraday rotation for EME is that it may change the polarisation plane of linearly polarised waves, and that the polarisation change is independent of the direction in which the wave travels. If faraday rotation is 45 degrees, a horizontally polarised antenna will produce a wave that is twisted by 45 degrees when it passed the ionosphere. After reflection at the lunar surface it will come back to the same point above the ionosphere, still twisted 45 degrees. After passing the ionosphere a second time, the signal is twisted 45 degrees again in the same direction, so it will be vertically polarised and not audible by a horizontal antenna. Faraday rotation is proportional to 1/f^2. A typical daytime value on 144MHz is 180 degrees at zenith and four times more at 10 degrees elevation.

4. Ionospheric absorbtion like faraday rotation varies like 1/f^2. A typical daytime value at 144MHz is 0.1dB at 10 degrees elevation. Only during very high solar activity this phenomenon is strong enough to affect the polarisation by attenuating different polarisations by different amounts. During such circumstances, faraday rotation is very strong and may vary rapidly, and it is hard to draw any conclusions with information deduced from narrow band signals. With a radio amateur bandwidth, the time resolution is in the millisecond range, and polarisation may fluctuate quicker than that during strong auroras. The effect is that the radio wave is split into two independent signals with orthogonal polarisation.

#### What to do depends on frequency

At low frequencies, 50 to 432 MHz, where yagi antennas are attractive, linear polarisation is used by most stations because it is much simpler than circular. Obviously EME communication would be much easier if everybody was using circular polarisation instead, but that will never happen because many operators who work EME are more interested in other propagation modes where linear polarisation is already a standard for which there is no technical reason to make a change. So if you want to make a good EME station for 50 to 432 MHz it is a good idea to design a system that allows linear polarisation with selectable polarisation plane. Since situations occur when optimum transmit and receive are 90 degrees from each other, an electronically switchable system is to prefer over a mechanical rotation of the whole antenna. But if the time to turn 90 degrees can be made short, a second or a few, a mechanical polarisation rotation will work equally well.

At very high frequencies, maybe 10GHz and above, where the moon is no longer a perfect reflector, the polarisation of the transmitted wave gets lost during the reflection. The reflected energy is spread out over a considerable frequency range and it is spread out over all polarisations. In order to achieve best receiver sensitivity, a two channel receiver is needed. The two channels should be connected to a pair of orthogonal feed elements. The two signals can be fed to the two ears of the operator, or even better, filtered through matched filters, square law detected, added to produce a single voltage representing the total power within the passband for the total wave. The rectified signal corresponding to voltage should then be low pass filtered and used to control a tone oscillator to restore the morse code. At frequencies where the reflection at the moon surface completely destroys the polarisation, it makes no difference what polarisation is used on the transmit side - anything will do perfectly well.

#### How to use switchable linear polarisation

It is obvious what to do on the receive side - select the polarisation that gives the strongest signal! On a very weak signal this may take quite some time, and it is much easier and faster is a two channel receive system is used as described here: What is Polarisation ? Signals vs Noise (detecting weak signals)

What to do on the transmit side is less obvious. When calling CQ I have found it much more efficient to alternate between horizontal and vertical than using circular polarisation. Circular is a certain loss of 3dB while alternating gives a good signal now and then for everybody. A really strong signal makes the other station much more willing to answer!!

If you want to be heard in a particular part of the world, and if you can hear an EME signal from someone in that area (within say 1000 kilometres) you can calculate what transmit polarisation to use in order for your signal to arrive with correct polarisation at the other end. In order to do this, you first have to calculate the spatial polarisation. You can do it with moon tracking software like SKYMOON by W5UN. You may also use equations published by KL7WE, Tim Pettis in Central States proceedings. He also gives a rule on what to do in words, which can be expressed as follows:

1. Calculate the polarisation at which the signal would arrive if there was no faraday rotation.

2. Find the optimum receive polarisation, which will usually be different.

3. When changing to Tx, rotate from the optimum Rx angle towards and through the angle from step 1, and go an equal distance beyond.

This rule comes from simple algebra:
Polarisation for the other stn = P(0=Horisontal, 90=Vertical)
Received polarisation angle = R (0=Horisontal, 90=Vertical)
Spatial polarisation offset = S
Optimum transmit polarisation = T

First assume that F=0 (would be true on 1296 and above)
Then, we can set up the following equations:

R=P+S (Wave emitted with angle P twisted by S before arriving at angle=R)
P=T-S (Wave emitted with angle T twisted by -S before arriving at angle=P)

These equations combine to give T=R, which is not surprising!!!

If we also include faraday rotation, (theory says that faraday rotation is the same regardless of the direction of the wave) the following equations are obtained:

R=P+S+F
P=T-S+F

Eliminating F gives:

T=2P+2S-R

We may safely assume that 2P is 0 or 180, and therefore:

T=2S-R

Which is another way of describing the KL7WE rule which translates into a formula like: T=R-2*(R-S).