How does this quarter-wave stub balun work?

Question

I was searching for 1:1 balun designs that would be efficient at VHF yet could easily be made from commonly available materials when I came across these two:

While the design on the right seems pretty straightforward, the workings of the left one are a bit of a mystery to me. A quarter-wave shorted stub will, as far as I understand, look like an open circuit at the resonant frequency, so it seems like it shouldn't have any effect, at least if treated as a perfect transmission line...
So how (if at all) does it function as a balun, and what are the necessary conditions?

Is it important for the coax from which the stub is made to have the same impedance as the antenna (75 ohms)?
Is the crossover (i.e. connecting the shield of the stub to the feedline's center conductor and vice versa) important, or will the balun also work without it?
Does the stub have to be oriented in a particular manner relative to the main feedline in order for it to work?

Phil Frost - W8II · Accepted Answer

I've never seen a balun design like this and and I'm not entirely certain it works. But if it does, here's how I think it could work:
As you say, a shorted quarter-wave stub appears like an open circuit. So the stub is a parallel open circuit to the differential mode: it has no effect.
To the common-mode, the stub and the parallel section of feedline make a quarter-wave transmission line.
Hm, a quarter-wave transmission line at the feedpoint sounds like a sleeve or bazooka balun. These baluns are made from coaxial, not balanced feedlines. And they can be open or short on the feedline end: if they are open, the sleeve radiates. If they are closed, the sleeve does not radiate.
I suspect (though I'm not entirely certain) this is the same idea. If you try to make a sleeve balun that's short on the feedline end and thus does not radiate but with balanced rather than coaxial transmission line, you get a folded balun. If you make it open on the feedline end, you get this design.
It works because the common-mode sees, from the perspective of the feedpoint, a short (because the transmission line is open 1/4 wavelength away). This ensures the voltage at each leg of the dipole is symmetrical, which is good. But what's making them symmetrical is common-mode current on that 1/4 wave section of balanced transmission line, which is probably not the best since that means the final 1/4 wavelength of the feedline is radiating and that's probably not what you want. There won't be any common-mode currents on the rest of the feedline, at least.
Again, not entirely certain this is correct. But maybe someone can show that I'm right or wrong in the comments.

sm5bsz · Answer

I think it does not work. In the first place, the shorted quarter wave will present a very high impedance between screen and center conductor so there will be almost no current in the center conductor. Just removing the center conductor and in effect place a metal rod or metal tube with the same outer diameter would give exactly the same function.
What you then have is two quarter wave radiators connected to the center conductor with a single radiator as a counterweight. You would get a tilted polarization plane and a significant current on the feedline towards the radio.
The proper way is to place a tube alongside the feeder and connect it to the screen of the feeder at a distance of a quarter wave. Look here: https://diydrones.com/profiles/blogs/some-more-antennae-and-when-is-a-balun-not-a-balun

How does this quarter-wave stub balun work?

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