I read in a comment:

it doesn't matter which circular polarization you choose, because they are both present, and the whole point here is to select one of them instead of receiving both and getting mixing.

The premise of the point being made seems to be that an HF signal received after making a pass or passes through the ionosphere will be both left-hand and right-hand circularly polarized, and so a receiving antenna of either chirality will not be subject to fading as a linearly polarized antenna would be.

Is this possible, for a signal to be both left- and right-hand polarized?


Is this possible, for a signal to be both left- and right-hand polarized?

Yes, it's very much possible:

While the superposition of two orthogonal circular polarizations might¹ indeed look linear (just as the superposition of a horizontal and a vertical polarized wave at appropriate phasing is a circularly polarized wave), of course that means that a linearly polarized wave is at the same time circularly polarized in both directions.

Technically, this is widely exploited: Satellite receivers use polarization multiplex. That is awesome, because you get two totally independently useful "subchannels", as long all media the wave travels through is a largely a linear medium and isotropic. (And the microwave frequencies geostationary satellite downlink channel fulfills that pretty well.)

Even if that's not the case, you still get some isolation between RHCP and LHCP, and can use that for MIMO techniques to increase your data rate or robustness beyond what you can do on a single polarization.

¹ might because that's not necessarily the case. Remember the Poincaré sphere:

2128506 / CC BY-SA (https://creativecommons.org/licenses/by-sa/4.0

When you add waves of different polarizations, you wander on the surface of that sphere; only when you add RHCP and LHCP with the same magnitude, you end up with a linear polarization. The angle of that is then defined by the phase between the two constituent waves; every other combination, every attenuation that affects one rotational sense more than the other, will produce an elliptic polarization.

Let me rephrase your question:

Is this possible, for a wave to be both left- and right-hand polarized at the same time?

No, that's not possible, because any wave can only occupy one point in polarization space.

  • $\begingroup$ That's a good point about the distinction between wave and signal. I probably should have used wave in my question as it was closer to the point: the application under consideration was amateurs on HF, where almost certainly no one is doing any MIMO. $\endgroup$ – Phil Frost - W8II Jun 26 '20 at 15:45
  • $\begingroup$ @PhilFrost-W8II but people are using polarization multiplex, aren't they? $\endgroup$ – Marcus Müller Jun 26 '20 at 18:11
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    $\begingroup$ People yes. But hams? Not much that I know of, especially on HF. $\endgroup$ – Phil Frost - W8II Jun 26 '20 at 18:57
  • $\begingroup$ Interesting! Sure, a linearly polarized HF antenna is large enough to be cumbersome, adding a phased second element of orthogonal polarization doesn't sound tempting at all, and circularly polarized aperture antennas are mechanically infeasible alltogether. $\endgroup$ – Marcus Müller Jun 26 '20 at 19:08
  • $\begingroup$ There's also the problem of equipment. Very few radios a ham might have even have two antenna inputs. I had to build my own because the only options I know of are really expensive, like the USRP. And hams are notoriously cheap. $\endgroup$ – Phil Frost - W8II Jun 26 '20 at 19:57

No, it's not possible. The electric field vector can only point in one direction at a time, so there's no way it could simultaneously rotate in two directions.

However it is possible to consider any possible polarization as a superposition of left- and right-handed circular polarization.

When there are multiple radiation sources, either because there are multiple transmitters or because the same signal is being received through multiple paths, then the result at the receiver is the addition of each source. Adding two circular polarizations of opposite chirality together produces a linear polarization.

This can be shown graphically as a 3D parametric plot:

enter image description here
(editable source)

On the left in yellow, we have:

$$ \left\{ \begin{aligned} x(t) &= \cos(t) \\ y(t) &= \sin(t) \end{aligned} \right. $$

On the right in blue we have:

$$ \left\{ \begin{aligned} x(t) &= -\cos(t) \\ y(t) &= \sin(t) \end{aligned} \right. $$

And the middle in green is the addition of these two:

$$ \left\{ \begin{aligned} x(t) &= \cos(t) - \cos(t) \\ y(t) &= \sin(t) + \sin(t) \end{aligned} \right. $$

It's pretty plain to see this simplifies to $x(t) = 0$ as the opposite electric fields along the x axis cancel each other.

As the two sources change in relative phase, the plane of the resulting linear polarization rotates:

enter image description here
(editable source)

Here, green is showing

$$ \left\{ \begin{aligned} x(t) &= \cos(t+2) - \cos(t) \\ y(t) &= \sin(t+2) + \sin(t) \end{aligned} \right. $$

While it's not so immediately obvious to see the cancellation, it remains true that opposite helices like this will cancel each other in some plane, as long as they are equal in amplitude.

If the two sources are not equal in amplitude, then the result is elliptical polarization:

enter image description here
(editable source)

It is true that the ionosphere is time-variant, and so at one time the signal as received may be left-handed, and some time later right-handed. But is impossible for it to be both at the same time, although one could consider a linear polarization to be the superposition of both circular polarizations in equal amplitude.

The problem is an ionospheric channel does not guarantee equal amplitude. Thus, a circularly polarized receive antenna will still be subject to fading as the signal randomly wanders between left- and right-handed chirality, as well as linear and all the points between (elliptical polarizations).

  • 1
    $\begingroup$ ... so in other words yes, they can exist and their sum is equivalent to a linear polarization. $\endgroup$ – user253751 Jun 26 '20 at 9:46
  • $\begingroup$ @user253751 Any possible polarization can be represented by the superposition of two orthogonal polarizations of specified amplitude and phase. Vertical and horizontal polarization are one such orthogonal pair. LHCP and RHCP are another such pair. It is sometimes useful to use superposition to analyze a problem, but when you use superposition to say "ah ha! all polarizations are present so it doesn't matter what antenna I pick!" you are making an error. $\endgroup$ – Phil Frost - W8II Jun 26 '20 at 15:26
  • $\begingroup$ So if you simply say "they don't exist because they sum to a linear one" then you're really saying that no two polarizations can exist at the same time. $\endgroup$ – user253751 Jun 26 '20 at 15:30
  • $\begingroup$ @user253751 If you have a circularly polarized antenna that it could be convenient to consider everything as a superposition of RHCP and LHCP. Now if everything you received consisted of both RHCP and LHCP in equal amplitude, you could conclude your antenna would always receive half the power. But what you are missing is this is not the case on an ionospheric channel: sometimes all the power will be in the RHCP component, other times all in the LHCP component. This means sometimes your antenna will receive all the power, and other times none of it. It is no different than a linear antenna. $\endgroup$ – Phil Frost - W8II Jun 26 '20 at 15:31
  • $\begingroup$ @user253751 yes, that is what I'm saying, and you can also see it in Marcus Müller's answer below: "any wave can only occupy one point in polarization space." Now you can use superposition to divide that one point into two orthogonal components, but you must then consider both components throughout the analysis. This is the part you seem to be missing: if you erect a LHCP receive antenna then you must also consider what happens to the RHCP portion of the signal: you don't receive it at all. That is, you get polarization fading. $\endgroup$ – Phil Frost - W8II Jun 26 '20 at 15:41

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