# One way propagation

The following discussion about one way propagation took place in the Moon-Net reflector. I found the subject so interesting that I decided to put together all the messages in this page.

On 19-Apr-2003 DM2BHG wrote:
Hi all Many thanks for all of the answers, I've got due to my question concerning "one way propagation". But to be honest: I'm not yet rally happy with the explanations. Perhaps it comes from my poor English and I couldn't formulate my question clearly. So I will do it once more: If I'm calling "CQ". Another station can copy me clearly and answers my call, but I can't hear these stn at all. What is the difference between these two signals? Why can he hear me, but I can't hear him? What happens to the signal, that it works only in one direction???

On 19-Apr-2003 GM4JJJ wrote:
Heinz, here is part of an article written by Paul N1BUG a number of years ago which attempts to explain this rather misunderstood one-way effect. Paul also wrote a DOS graphical program that demonstrated the effect most clearly for me (I am sorry I don't have a copy here, perhaps Paul could help). Here is part of Paul's article:

I suspect few people will argue these days that the polarization of an incoming signal with respect to the polarization of the receiving antenna makes a tremendous difference in how well the signal can be heard. A signal perfectly aligned in polarization with the receiving antenna system will suffer no loss of strength due to polarization mis-alignment. A signal arriving perpendicular to the polarization plane of the antenna (90 degrees to it) will suffer the greatest loss - How much depends on many factors, but for a typical modern yagi it is generally taken as between 20 and 30 dB. In between these extremes the loss varies logarithmically with the cosine of the angle (by angle I mean the difference in degrees between the polarization plane of the incoming signal and that of the antenna). A formula for this is:

```             Loss (dB) =  20 log(cos )

where  = the angle in degrees.```

This formula assumes a perfect yagi with no unwanted response to signals of the opposite polarization. It is VERY close up to about the 18 or 20 dB point, but beyond that needs a correction factor to keep the result down to a realistic level.

That's all well and good for two antennas of like polarization pointing toward each other along the Earth's surface. But what of two antennas of like polarization pointing off into space? The situation becomes a little more complicated. In my own crude way let me try to help you visualize what I mean. Suppose you had a large round ball sitting in front of you. If you took two small model yagis and planted one on the very top of the ball (be sure to align the elements so they are parallel to the surface of the ball directly below the antenna) so that it is pointing straight away from you toward a distant wall, and the other on the left side of the ball (elements parallel to the surface of the ball) pointing to the same distant wall, You will have the basis of this visualization. Both antennas are horizontally polarized with respect to the surface of the ball (simulated Earth), right? But stand back and sight along the boom of each yagi, toward that wall. If you can visualize a signal leaving each antenna and travelling to the the wall (simulated moon), you will see that the polarization planes of the two signals hitting the wall are actually 90 degrees out of sync with each other. This difference is what we call "Spatial" Polarization Offset! Gee Whiz!

The geometry involved on an EME path is a little more complex, what with elevation angles and so on. The point of reference used for such calculations is generally the Earth's polar axis, but a complete discussion of the mechanics involved is beyond the intent of this discourse (and is a fantastic way to get a headache). Here is the basic formula used to calculate a spatial polarization for an antenna with respect to the Earth's polar axis:

```                     (sin L * cos E - cos L * cos A * sin E)
P = ATN  ---------------------------------------
(cos L * sin A)

where L = Latitude of station           {footnote 1}
A = Azimuth of antenna
E = Elevation of antenna
P = Polarization angle```

The spatial polarization offset (read 'difference') between any two stations is simply P1 - P2, but keeping it within a range of -90 to +90 takes additional steps.

MOVING ON TO THE REAL WORLD

For some time we have been accustomed to seeing a "Polarization" or "Spatial Polarization Offset" figure in moon-tracking programs that track the moon for two stations. This figure is in degrees, with some programs keeping it within a range of -90 to +90 degrees while others allow it to approach 180. Along with this, many of the programs have also tried to include a "Polarization Loss" figure in dB, with 0 dB indicating a polarization offset of 0 (or 180) degrees and 25 dB or more indicating a polarization offset of 90 derees.

It doesn't take a rocket scientist to figure out that this just cannot be the case. We've all made some fine EME QSOs when software told us to expect 20+ dB loss of signal based on polarization! So what's the deal here? The problem is that programs like that don't consider the affects of Faraday rotation. The fact is that due to Faraday rotation in the ionosphere signals can arrive at the receive antenna at ANY polarization, regardless of the spatial polarization relationship between the two stations. Given that consideration, the next thing we're confronted with is the argument that ALL of this spatial polarization business should be disregarded since the unpredictable Faraday is going to modify the polarization anyway. In other words we "make our skeds blindly and take our chances".

BUT - and this is a B-I-G but - the ionosphere has another unique little property that should make us re-think the situation. Simply put, a signal polarization which rotates in a given direction when passing through the ionosphere on its way to the moon will rotate the same amount in the same direction (from the observer's viewpoint) on the return trip. You can visualize it this way: Suppose you were standing behind the reflector on your EME antenna and sighting down the boom to the moon. Further suppose for a moment that you could see the polarization plane of the signal leaving your antenna and traveling to the moon and back again to the antenna. If you sent out a burst of RF and observed that the polarization plane shifted 45 degrees to the right (clockwise) on the way out to the moon, your first assumption would probably be that the signal will rotate in the opposite direction (from your perspective) on its way back, thus "un-twisting" itself and arriving perfectly in line with the elements on your antenna. But the ionosphere doesn't see it in that way. In fact the polarization plane will rotate by the same amount in the same direction (clockwise) from your vantage point on the return trip - thus making it now 90 degrees out of alignment with your antenna.

So what? Consider this: If you add a second station whose QTH is far enough away from yours so that the spatial polarization offset between the two of you is 45 degrees, something interesting begins to take shape. Suppose you transmit and the Faraday causes the polarization plane of your signal to rotate 45 degrees clockwise by the time it reaches the other station. It started out at 0 degrees (the spatial polarization of your antenna) and was twisted 45 degrees clockwise, so it arrives at his antenna perfectly aligned with it (his spatial polarization is 45 degrees greater than yours, don't forget). Sounds like the Faraday is doing us a big favor, right? WRONG! Now suppose the other station transmits a string of O's back to you (having heard your signal very well by virtue of the perfect polarization alignment). The polarization plane of the signal starts out at 45 degrees (his spatial polarization) and then gets rotated 45 degrees clockwise on its way to you, making the polarization of the signal 90 degrees in "spatial" terms. Yikes! Your antenna is still at 0 degrees "spatial" polarization, and his signal is coming in at 90 degrees! Unless you're a BIG GUN, you just won't be able to hear him. Good grief, one-way propagation!

WHEREAS THE ABOVE SITUATION IS AVOIDABLE...

Fortunately, there is a way to figure out what times are most likely to yield frustrating one-way conditions in advance. Going back to the example above... If the spatial offset between the two stations were 0 degrees, then Station A transmitting to station B would result in a clockwise twisting of 45 degrees, arriving 45 degrees out of alignment with his antenna. On the return trip, the exact same relationship holds. If the Spatial offset were 90 degrees, then station A transmitting to station B would result in a signal at 45 degrees "spatial polarization" being received by an antenna with a polarization of 90 degrees, for a difference of 45 degrees. On the return trip, 90 degrees starting from his antenna gets twisted clockwise to 135 degrees (135 degrees is equivalent to -45 degrees. It should always be kept in a range of +/- 90 degrees by adding or subtracting 180 degrees as necessary. 180 degrees is physically the same as 0 degrees), which is an offset of 45 degrees with regard to your antenna. In either of these cases the signal will be mis-aligned by the same amount (45 degrees) at both stations. There will be a resulting loss of 3 dB, but at least it's reciprocal (and 3 dB isn't as bad as 20+ dB).

I could (and probably should) sit here and go through various possible combinations of "spatial offset" and Faraday rotation for a L-O-N-G time. I'm not going to, but I would like to bring up one other point before going on. If one works away at this problem long enough (I've been fretting away at it for 5 years now), one then begins to think in terms of "windows of opportunity". Considering the original 45 degree spatial offset again... The absolute best that can be achieved (under normal circumstances) in this case is a Faraday of 0 degrees or 90 degrees. This would indeed result in a reciprocal path with 3 dB loss of signal at both ends. BUT, if the Faraday changes by only 15 degrees in either direction, the signal misalingment at one station will become 30 degrees (1.24 dB loss) and at the other station 60 degrees (6 dB). Notice how quickly the signal falls off at one end of the path for very small changes in prevailing Faraday - thus a small "window of opportunity" for reciprocal conditions, what with Faraday being such a changeable entity. On the other hand, consider either of the perfect cases - that is either spatial 0, Faraday 0 or spatial 90, Faraday 90. Under these circumstances the path is reciprocal with 0 dB loss at either end. If the Faraday changes 15 degrees in either direction the reciprocity will remain, and the signal will drop by only 0.3 dB at both stations. Indeed, the Faraday must shift 45 degrees one way or the other before the signal drops to -3 dB and even then it is still reciprocal. Hence a greater "window of opportunity".

The bottom line is that one-way conditions are much more likely to occur when the spatial offset is in the vicinity of 45 degrees and drops to nearly 0 when the offset is 0 or 90 degrees. Data collected over the past few years clearly shows that statistically the chances of completing an EME QSO drop off sharply when the spatial offset approaches 45 degrees, especially for the smaller stations who just don't have the signal to spare. Many have clearly benefited in the past few years by careful consideration of these polarization issues.

On 19-Apr-2003 N1BUG wrote:
It is still available for download from my web site as part of the Z-Track package (either version 3.08 or 3.21). Within that package you will find AN.EXE -- this is the utility that demonstrates the one way condition. Choose 'Spatial Polarity/Faraday Demo from the main AN menu and experiment with the various graphical demonstrations to get an idea how one way propagation happens. The following link will aid in finding it for those who might wish to try this. http://www.n1bug.net/software/ztrack.html

On 19-Apr-2003 WB9UWA  wrote:
Hi Heinz,

You have seen the complicated explanations, but since English is not your primary language, perhaps another German could give you the long version. The short explanation is simple. The polarity of the signal changes in the Ionoshere. This can change in a manner that results in a good polarity match in one direction, but not the other. In any case, Xpol can make faraday a non-issue.

(Note: I'm not showing any E-Mail address here in order to avoid them from being collected by SpamBots. You can possibly find the E-Mail addresses of the above OM at QRZ.COM.)

This discussion is still not closed. If you have any opinions or additional related information you would like to be published here, just send me an E-Mail

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