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 [1] 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.

Fig. 1



Fig. 2



Fig. 3



Fig. 4

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)

Fig. 5



Fig. 6



Fig. 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 polarisation. 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. [2]



4. But NASA found out that a circular wave gets 40% depolarised.... [3]


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 non-polarised component. 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' [4]. 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 [5]. (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.



4. Conclusion:


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!



5. References:



[1] Manfred Belter DJ2ZF EME-Positionberechnungen des Mondes CQ/DL 1/86,3/86

[2] N.S. Soboleva: Izmerenie poljarizacii radioizluchenia Luni na volne 3.2cm pri pomoshi bolshogo Pulkovskogo radioteleskopa Actronomicheski zurnal XXXiX 1962 pp 1124..1126

[3] Zisk, Pettengill, Catuna: High-resolution maps of the lunar surface at 3.8cm wavelength The Moon 10 (1974) pp17..50

[4] 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 conference)

[5] 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 Earth:



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 waveguide.
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 precisely enough.

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