Let's go circular on 10!
Marko Cebokli S57UUU
ABSTRACT - The current polarisation standard on 5 & 10 GHz EME is Europe
vertical and America horizontal. A comparison of calculated signal losses in
this standard with maximum possible losses that could be specefic to circular
polarisation shows, that it would be good to change the standard to circular
polarisation, as used on 23 and 13cm. The only drawback is, that it is much
more difficult to build and tune a circular polarized antenna. This could be
overcome (at least on 3cm) by using standard parts for satellite DBS TV. Also,
the dielectric plate polariser might be a simple 'no-tune' solution.
1. How much signal do we loose with the current standard?
The USA are a big country, spanning several time zones. Also in Europe, the
difference in latitude and longitude between for example Moscow and Lisabon is
considerable. Besides that, the Moon orbit is inclined too. So it is quite
obvious that the polarisatin offset is far from a constant 90 degrees. I wrote
a simple program, using Moon ephemeris
from  to draw the loss caused by polarisation offset from 90 degrees.
Figures 1 and 2 represent the losses from S5 to NE USA, for both high and
low Moon declination. It can be seen
from fig 2 that not only there are losses, but that the polarisation
error can even go through 90 degrees (=total signal loss). Figs 3 and 4 show
how this can happen.
On average, the
loss is around 2dB. If one of the stations is south enough that the
Moon passes nearly overhead, we get a 'polarisation flip'. (figs 5,6,7)
Also, when stations become QRV on the other continents, many more problems
will arise. With a fixed waveguide instalation, it is impossible to rotate
polarisation, short of disassembling the whole dish!
2. Theoretically, how much better could linear be compared to circular?
In this section I'll talk only about fully polarised linear and circular
waves, because I want to show the maximum possible difference.
Fully polarised eliptical waves are somewhere in between, and I'll
talk about partly polarised waves in a later section.
It is well known that any linear polarised wave can be represented as a sum of
two circular polarised waves and vice versa - a circular polarised wave can be
represented as a sum of two linear polarised waves. This is not only
mathematics - you can physically generate a linear wave with two
circular antenas or a circular one using linear antennas.
In general, any eliptically polarised wave can be decomposed into a
sum of two arbitrary orthogonal eliptical waves, linear and circular being
only two extreme cases of eliptical polarisation.
Let's suppose now that there are Martians on the Moon and they want to play
tricks with Earthly hams. (We know that these little green guys are naughty
- they keep stealing our Mars probes).
First day, it's Martian boy's turn, and they're 'circular hostile'. They know
that Earthlings are transmitting right hand circular, so they cover the whole
surface of the Moon with right hand helices, terminated in 50ohm resistors.
Any right hand circular signal from Earth now gets totally absorbed in the
resistors. What about linear guys? Their signal contains both circular
components. The right hand component gets absorbed, but the other is reflected
totally. So 50% of the power gets back, but it is now left hand circular
polarised. A linear antena receives 50% of that, and total loss is 6dB for the
linear guys. But wait - a circular guy can easily change to left hand circular
by exchanging the cablees on his feed. Now he gets ALL of his signal back!
After finding that out, of course, Martians decide to scrap the helixes and
put up orthomode circular antenas with both ports terminated with resistors.
Now both right and left hand waves get absorbed. But that also means that
any linear wave will be absorbed, since it is a sum of two circular ones.
From this we can draw the following conclusions:
1-- if we selectively absorb one circular polarisation, we do more harm to
every linear polarisation than to the oposite hand circular.
2-- if we absorb both circular polarisations equally, we also absorb for
the same amount every linear polarisation.
Next day, it's the Martian girls who get their chance to do some practical
jokes, and they're 'linear friendly'. So they decide to replace the Lunar
surface with an array of parallel wires, to favour the linear Earthlings.
Now, if the linear guys align their polarisation with the wires, they'll get
all of their signal back. But they must align their polarisation - if they use
just any random linear polarisation, they will loose 6 dB on average, with the
possibility of total loss of signal, if they happen to use the wrong linear
On the other hand, for the circular guys of either hand, the loss will always
be 6 dB (half of the transmitted power gets reflected as a linear polarised
wave, of which half gets received by the circular antena).
From the above it is clear that
3-- any advantage for the linear polarisation can exist only for a
specific fixed polarisation
That means that even if there is any linear advantage, the current standard
cannot profit from it. To exploit any such linear advantage, everybody would
have to change to a polar mounted antenna, to keep the polarisation constant
at the lunar surface. It is easy to see that the relative polarisation between
local horizontal frame of reference and the moon changes: observe for example
the first quarter Moon. When it rises the "D" leans to the left (for abt 45
degrees, if you live at temperate latitudes), and when it sets, it is
tilted to the right. On the equator, the change is plus minus 90 deg, and on
the poles it doesn't change at all - every mount is a polar mount there, HI.
3. What physical mechanisms could cause a linear advantage, and how much?
At least since Apollo 11 we know that there are no parallel wires there.
It is also safe to assume that the rocks and mountains are random in angle.
The only mechanism that could be at work there is the 'Brewster' thing. If a
wave impinges on a boundary between two different dielectric materials at a
certain (Brewster) angle, it gets polarised. Typical values for Brewster
angles are around 60 degrees. According to the law of reflection, the polarised
wave is also going out at 60 degrees, that is away from us. Since EME works by
backscattering, this fenomenon is irelevant. The Earth as seen from the Moon
subtends an angle of abt 2 deg, so even for two stations out on the limbs, the
impact angle is never greater than one degree at the center of the Moon. At
such angles, the polarisation is totally negligible. At some radius away from
the center, the waves do impact at the Brewster angle, of course, but the
polarised reflected component is directed away from Earth. Even if there would
be any polarisation in the backscattered wave, it would get averaged away,
because the Moon is round.
The Russians have measured the polarisation of the thermal radiation to be
less than 3 percent. 
4. But NASA found out that a circular wave gets 40% depolarised.... 
Sure. (Well, actually it was MIT not NASA) But they never said that linear
would fare any better - because they
knew it wouldn't! What happens here is depolarisation. In section 2
I was talking about fully polarised waves. Many people mistakenly assume that
an eliptically polarised wave is not fully polarised. But it is - because a
single polarised antena can be made, that receives all its power, and another
'orthogonal' antena can be made, that receives no power from this wave.
With a partially polarised or unpolarised wave, that's not possible.
With a totally non-polarised wave, any antena (of any polarisation) will
receive exactly half of its power. To receive all it's power, you need an
antena that receives two orthogonal polarisations (of any kind).
A partly polarised wave can be represented as a sum of a polarised and a
In a fully polarised wave, the tip of the E-field vector, as seen from
'behind', always traces out an elipse (the line and circle being the extreme
cases of a very thin and fat elipse). In a non-polarised wave, it dances
randomly. (It is also sometimes called a random-polarised wave).
A monochromatic (single frequency) wave is always fully polarised. Because of
the relative movement between Earth and Moon, the signal gets spread over
frequency, and can therefore also get depolarised. With increasing frequency
the spreadnig and depolarisation get worse.
By representing a circular wave as a sum of two linear ones (as was done in
section 2), it can be shown that you cannot depolarize a circular wave without
also depolarising linear ones.
5. How difficult is it to make a circular feed?
The only real drawback of circular polarisation is that it is much
more difficult to make a good circular antena than a linear one.
Adjusting a screw type polariser is quite demanding on the instrumentation:
one must simultaneously check return loss and axial ratio. Besides that, the
measurement of axial ratio is very sensitive to reflections, so an anechoic
chamber is a nice thing to have, HI. This sensitivity can be easily
demonstrated by walking around the antennas while they are measured. Because
dish feeds usually have very broad directional patterns, especially the
ground reflection is very bothersome. One solution is to make the height big
compared to the distance between the antennas, which must of course satisfy
the far field criterion. On 10Ghz, with 1m disance and two tripods 1.5m high,
I still had to place microwave absorbers on the floor. One can check for the
ground problem by placing a reflective plate on the floor, then raising it for
half a wavelength or so and checking if the indication flutters.
Considering this, a 'no tune' solution would be very desirable.
One possibility would be to use circular orthomode feeds as used for satellite
TV. Most of them are made for the band around 12GHz, so their return loss and
axial ratio may not be so good at 10.4 GHz.
Another possibility is the 'septum polarizer' .
I think the best solution would be a linear orthomode coupler and a dielectric
plate polariser. Linear orthomode couplers for sat TV are mostly designed
from 10.7GHz up, and would probably function at 10.4 GHz. The clasical
HAM-style 'dual probe' design with two coax connectors on a round waveguide
as used on 23 and 13 cm is also an option. A waveguide variant was described
by W3IMU . (W3IMU proposed a 'squeezed tube' circular polarizer - that is
definitely the simplest solution, but it requires tuning.)
The dielectric plate polarizer is a simple piece of 5mm thick PTFE plate
cut to a kind of 'X' shape, a few cm long (for 10GHz), that is fitted inside
the wavegude at an angle of 45 degrees. In the (-: near :-) future, I'll try
if this can be done 'no-tune' style.
With the current standard we do loose decibels. It halfway works for EU-USA,
but when other continents enter, it becomes a mess.
There are no reasons to believe that circular is any worse than linear.
Even if there would be a linear advantage, the current standard couldn't
profit from it.
It is time that we go circular!
 Manfred Belter DJ2ZF EME-Positionberechnungen des Mondes CQ/DL 1/86,3/86
 N.S. Soboleva: Izmerenie poljarizacii radioizluchenia Luni na volne
3.2cm pri pomoshi bolshogo Pulkovskogo radioteleskopa Actronomicheski zurnal
XXXiX 1962 pp 1124..1126
 Zisk, Pettengill, Catuna: High-resolution maps of the lunar surface at
3.8cm wavelength The Moon 10 (1974) pp17..50
 OK1DAI: OK1KIR 5.7 and 10.3 GHz EME station HYPER Bulletin d'informations
des radioamateurs actifs en hyperfrequences NUMERO SPECIAL jul 88 ('98 EME
 Dick Turrin W2IMU:A circular waveguide x band ortho-mode coupler
W2IMU EME notes may 89, Reprinted in the '96 EME conference proceedings
Some additional drawings
While presenting the paper in Rio, I drew some sketches at the board,
here I try to redraw them:
First, the one that shows that the polarised component goes away from
A is the specular component, that doesn't change polarisation.
B is the
polarised Brewster component, that goes away from Earth.
C is the scattered
component, which is depolarised.
D is the Brewster angle.
The next one shows that even if there was any polarisation in the
returned signal, it would cancel out:
And the last one shows that the monostatic (own echo) and bistatic (QSO)
modes are not really different in terms of reflection angles:
The sizes and the distance are drawn to scale here.
A few notes about the losses in the polariser
The W2IMU squeezed tube polariser has probably less than 0.01 dB of
loss (one hundredth of a dB!) - since it's just a short piece of
The dielectric plate polariser is around 0.1dB
The linear ortho with external 90 degree hybrid and cables is NOT suitable
at 10 GHz, because of losses and the difficulty of cutting the cables
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