Imagine two dipole (or Yagi) antennas as might be used to create a circularly polarized combination antenna. The two are physically oriented (rotated) by 90 deg, and if in the same receiving plane, one signals is also delayed by 90 deg. The two signals can be combined with a duplexer and match to a common line (e.g. this: https://www.qsl.net/sv1bsx/antenna-pol/polarization.html ). My interest is in 2m (~ 144 MHz).

Now -- if used as a transmitter, it is easy to see how a circularly polarized signals is generated -- both antennas receive equal power with the correct phasing relationship. However as a receiver, if a linearly polarized signal is received, it may be picked up by only one of the two antennas, and the matching network will deliver some power to the feed line (as desired), but also some to the second (90 deg) antenna, which will re-radiate it thus degrading the overall antenna gain.

Can this be mitigated ? Is it possible to build a practical (at 144 MHz) coupler-type circuit that eliminates cross-transfer between the two antennas ?

  • $\begingroup$ Hello and welcome to ham.stackexchange.com! $\endgroup$
    – rclocher3
    Feb 10, 2021 at 22:23
  • 1
    $\begingroup$ A duplexer is used to combine energy from antennas meant to work with signals that are separated in frequency. The source you cite doesn't mention duplexers; it describes the use of transmission lines to create phase delay, which is the classical way to combine antennas for circular polarization. $\endgroup$
    – Brian K1LI
    Feb 11, 2021 at 15:15
  • $\begingroup$ I might suggest a new question that more specifically asks what your're after. It's a little vague here, you ask "it possible to build a...coupler-type circuit that eliminates cross-transfer between the two antennas?" and the answer is "yes, a Wilkinson power combiner for example", but then you also (not in the question) seem to want something that can avoid the 3dB loss associated with the Wilkinson when the input is from just one antenna. Now we're getting into the topic of network theory, and there's a lot that can be said about it, if you ask the right question. $\endgroup$ Feb 12, 2021 at 21:13
  • $\begingroup$ One way to go about it would be to specify precisely the scattering matrix you want, and then someone could tell you if it's possible or not, and with what kinds of networks. For example some kinds of networks can't be made from passive components, and some things aren't possible with isotropic materials. $\endgroup$ Feb 12, 2021 at 21:15

2 Answers 2


It may help to be familiar with the concept of scattering parameters or S-parameters. The S-parameters are the elements of a matrix which tell you the amplitude and phase coming out of every port, for the signal going into any other port. It assumes all the ports are terminated in matched loads, and it's usually valid for only one frequency.

For a 3-port device, there are 9 S-parameters. They are written in a matrix like this:

$$ [S] = \begin{bmatrix} S_{11} & S_{12} & S_{13} \\ S_{21} & S_{22} & S_{23} \\ S_{31} & S_{32} & S_{33} \end{bmatrix} $$

The subscripts are ordered output, then input. So, $S_{21}$ is related to how the input at port 1 contributes to what's observed at port 2.

In the case of feeding an array of two antennas, you'll probably use a Wilkinson power divider. Such a power divider with a 50/50 power split, where port 1 is the transceiver and ports 2 and 3 are the antennas, has a scattering matrix of:

$$ [S] = \begin{bmatrix} 0 & {-j \over \sqrt 2} & {-j \over \sqrt 2} \\ {-j \over \sqrt 2} & 0 & 0 \\ {-j \over \sqrt 2} & 0 & 0 \end{bmatrix} $$

Consider transmitting: we're really only concerned with the power in to port 1, so we only need to look at the first column of the matrix. $S_{11}$ is 0, meaning none of the power into port 1 is reflected back out port 1, meaning the SWR is 1:1.

$S_{21}$ and $S_{31}$ are each ${-j \over \sqrt 2}$. The $-j$ term just means the phase of the output is rotated 90 degrees from the input, which would make sense because the power divider is constructed from 1/4-wave transmission lines. The need for the $\sqrt{2}$ term should be obvious considering this is a passive network, and here we're also assuming ideal, lossless components. Thus, power out can't be more than power in. The S-parameters relate to amplitude and phase, so we must square them to get power. $ \left(1/\sqrt 2\right)^2 = {1 / 2} $, so half the power coming in port 1 goes out port 2, the other half goes out port 3.

Now what if a signal is applied to port 2? $S_{12} = {-j \over \sqrt 2}$, so half the power goes back to the transceiver. $S_{22} = S_{32} = 0$, meaning none of the power goes to the other ports. The remaining half of the power is lost in the resistor.

Port 3 is identical to port 2.

However as a receiver, if a linearly polarized signal is received, it may be picked up by only one of the two antennas, and the matching network will deliver some power to the feed line (as desired), but also some to the second (90 deg) antenna, which will re-radiate it thus degrading the overall antenna gain.

Not quite! Because $S_{23} = S_{32} = 0$, none of the power received at one antenna port makes it to the other antenna port. There is no re-radiation. But because $S_{12} = S_{13} = {-j \over \sqrt 2}$, only half the power received at one of the antennas (assuming for now that only one antenna is receiving a signal) makes it to the receiver, which does indeed reduce antenna gain by 3 dB.

Now what if both antennas are receiving a signal simultaneously? Let's say the signal at each antenna port is $1 / \sqrt 2$, so that combined they have power 1. In order to determine what appears at the receiver, we must calculate:

$$ \begin{align} &S_{12} {1 \over \sqrt 2} + S_{13} {1 \over \sqrt 2} \\ = &{-j \over \sqrt 2} {1 \over \sqrt 2} + {-j \over \sqrt 2} {1 \over \sqrt 2} \\ = &{-j \over 2} + {-j \over 2}\\ = &-j \end{align}$$

$-j$ is just 1 with a 90 degree phase shift, so all the power received by the antennas makes it to the receiver, just shifted in phase.

But this only works when the signals are in phase. What if port 2 is receiving $1 / \sqrt 2$ but port 3 is receiving the same thing, but 90 degrees shifted: $-j / \sqrt 2 $?

$$ \begin{align} &S_{12} {1 \over \sqrt 2} + S_{13} {-j \over \sqrt 2} \\ = &{-j \over \sqrt 2} {1 \over \sqrt 2} + {-j \over \sqrt 2} {-j \over \sqrt 2} \\ = &{-j \over 2} + {1 \over 2} \end{align}$$

Converting this to polar form we find the magnitude is $1/\sqrt 2$, so we lost half the power. So in the case of receiving, the Wilkinson power divider (or I guess combiner, in this direction) is lossless only when ports 2 and 3 are in phase.

Unsurprisingly, repeating the same thing with the antennas 180 degrees out of phase results in all the power lost.

When you want to generate circular polarization, the transmitter will feed the power divider, then one of the power divider's outputs will have an additional 90 degree delay in it. On receive, this delay has the effect of combining received signals of the same circular polarization to be in-phase at the power divider's (or now, combiner) antenna ports to be in-phase, and they will make it to the receiver without loss.

If the received signal is circularly polarized with opposite chirality, the signals are opposite phase at the power combiner, and the loss is theoretically infinite.

If the received signal is linearly polarized (at any angle, horizontal, vertical, or otherwise) then the phase shift is 90 degrees and half the power is lost.

There is no way around this with just receiver and one antenna. More generally, any antenna or antenna array you build must pick one point on the Poincaré sphere, and there will always be an opposite point on that sphere that's orthogonal to your antenna which you can't receive, with all other possible polarizations on the sphere having some loss between zero and infinity.

If you want to receive any polarization without loss, then you must build two antennas and two receivers, each occupying opposite points on the Poincaré sphere. There is then some combination of amplitude and phase coefficients with which those signals can be combined to receive an incoming signal at any polarization with zero loss.

  • $\begingroup$ Thank you -- I was also coming to see this is an implementation of a Wilkinson power divider. Small typo: "So, 𝑆21 S21 is related to how the input at port 2 contributes to what's observed at port 1" $\endgroup$
    – jp314
    Feb 11, 2021 at 17:42
  • $\begingroup$ I half-understood the topic before reading this excellent answer, and now my understanding has moved ahead quite a bit. Thanks! Learning more about radio is one of the things that keep me coming back to this site. $\endgroup$
    – rclocher3
    Feb 11, 2021 at 18:43
  • $\begingroup$ @jp314 thanks, fixed. Though the error was in the previous sentence, I think. $\endgroup$ Feb 11, 2021 at 20:04
  • $\begingroup$ fixed, thank you! $\endgroup$ Feb 11, 2021 at 20:53
  • $\begingroup$ ""Fortunately this isn't too much of a problem, since both signal and noise are attenuated, and so signal to noise ratio isn't affected". Not quite true since the non-receiving antenna will still receive NOISE and send it back to the output. Thus SNR is degraded by 3 dB. $\endgroup$
    – jp314
    Feb 12, 2021 at 6:10

There is no passive solution which can magically choose the matching polarisation solution for the combiner.

There are two types of splitters - a simple impedance match (say transforming 50 to 25 Ohms, with the two output cables in parallel), or a Wilkinson-type splitter that does this but includes a resistor. The resistor provides isolation between the ports - so if a signal arrives at port 2, half of it leaves from port 1, and half is absorbed, none leaves from port 3.

The "splitter" used in the transmit case, splits the signal in half without loss, feeding both antennas equally. This works as you expect.

In the receive case the two types of splitter end up giving the same result:

  1. The simple impedance matching splitter will reflect some power and radiate some of the power out of the other polarisation antenna.
  2. The Wilkinson splitter will absorb half of the power from the receiving antenna, and not re-radiate it on the other polarisation. Half the power leaves at the common port.

More general rambling
I like to look it like this: whatever is going on inside the antenna, in the end it is a 1 port device with some radiation pattern and polarisation. If its a circularly polarised antenna, then it will behave like one, whatever is going on inside.

Remember even when matched, all antennas scatter a fair fraction of the received power, they're not blackbodies that capture all the incident energy. Of course the two types will differ. But the scattering of incident fields is not (directly) part of how we measure an antenna.

The QSL site mentions using a splitter for stacking yagis for higher gain. This shows the fallacy even more clearly - it's like asking the array, on receive, to have the same high gain as the peak of the stacked array, but everywhere, which is also not possible from a passive antenna.

  • $\begingroup$ My question isn't particular to circular polarization, although it is (to me) a good example. The question is directed to "can you isolate 2 antennas that are combined with a splitter -- so that if one receives a signal, the other doesn't re-transmit some of it (and waste received power) ?". The splitter will work nearly ideally when used as a transmitter, but is non-ideal in the receive mode. $\endgroup$
    – jp314
    Feb 10, 2021 at 23:18
  • $\begingroup$ Yes, you use a Wilkinson splitter which is well matched, and doesn't send anything back out of the other antenna port. But it has a 3 dB loss in that direction. $\endgroup$
    – tomnexus
    Feb 11, 2021 at 2:36
  • $\begingroup$ Staying with the polarisation example, you can make a combiner with two outputs for orthogonal modes. V and H are trivial, LHCP and RHCP also possible. Then attach two receivers and choose the strongest signal. $\endgroup$
    – tomnexus
    Feb 11, 2021 at 2:39
  • $\begingroup$ I am interested in the receiver case mostly. With the answers, I understand that Wilkinson splitter has 3 dB loss from either port if properly matched. I actually don't care about re-radiation from the other antenna -- so is performance better by NOT using the 2*Z resistor in a Wilkinson ? $\endgroup$
    – jp314
    Feb 12, 2021 at 6:05
  • $\begingroup$ You can remove the resistor, but then it reflects some power back (not matched) and reradiates some from the other port, case 1 as described above. I think what you're looking for is beamforming where you have N receivers and an algorithm combines the signals digitally, in the best phase for optimum performance. As done in a MIMO wireless lan or 4G phone. $\endgroup$
    – tomnexus
    Feb 12, 2021 at 9:10

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