A length of coaxial cable radiates negligible amount of RF.
Same length, if cut into pieces of length (λ/2 * VF), and these pieces joined again cross-connected (i.e. core of a piece to shield of adjacent piece & vice versa), then this length of coax becomes a very good radiator of RF. Why & how?

EDIT on Feb 05, 2017:
Added sketch below to improve the question. enter image description here

  • 1
    $\begingroup$ Why? Physics? How? Application of the laws of physics. This is a very broad question. Could you please explain what you've understood so far, and where you came across this very particular antenna? That would make this easier to answer – also, as far as I can tell you didn't tell us whether these pieces are connected in series or parallel. Could you maybe add a drawing or similar to your question? $\endgroup$ – Marcus Müller Jan 15 '17 at 22:05
  • $\begingroup$ When the net current (braid + central conductor) is not 0, then you have common mode current and the coax does emmit. So a coax certainly can emmit. $\endgroup$ – Juancho Jan 16 '17 at 2:25
  • $\begingroup$ @MarcusMüller: The pieces are connected in series, but connections are crossed, i.e. core of a piece is connected to shield of adjacent pieces, and shield of a piece is connected to cores of adjacent pieces. $\endgroup$ – abcd567 Jan 16 '17 at 4:35
  • $\begingroup$ @MarcusMüller: I came across this antenna here 1-rason.org, 2-arundale.com, 3-repeater-builder.com, 4-nodomainname.co.uk, 5-rberger.com, 6-balarad.net $\endgroup$ – abcd567 Jan 16 '17 at 4:38
  • $\begingroup$ @Juancho You are right that a coax with common-mode current will radiate, but my reference was a coax with net current (braid + central conductor) Zero, like Case 1 in this diagram. $\endgroup$ – abcd567 Jan 18 '17 at 19:43

If you look at this, reduced to two segments:


you'll notice that the voltage on the outer conductor right of the crossover is exactly the opposite (assuming infinetly small crossover and perfect impedance matching) than on the left side.

So let's assume we simply feed the two outer conductors at that crossover point with exactly such a voltage:


Looking familiar? Yep, that is a classical dipole!

  • $\begingroup$ Except... The feed point is not in the middle but at the end. Or does this not really matter? $\endgroup$ – Dieter Vansteenwegen ON4DD Jan 16 '17 at 15:12
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    $\begingroup$ @DieterVansteenwegenON4DD I'd argue what counts is only the current distribution on the "visible" conductor / the E-field between points on that Dipole, so I don't really see the difference (might be confused, brain feels mushy today) $\endgroup$ – Marcus Müller Jan 16 '17 at 15:35
  • $\begingroup$ @MarcusMüller I think it is a collinear array of end-fed monopoles, like this diagram. $\endgroup$ – abcd567 Jan 18 '17 at 19:47
  • $\begingroup$ @DieterVansteenwegenON4DD I think it is a collinear array of end-fed monopoles, like this diagram. $\endgroup$ – abcd567 Jan 18 '17 at 19:47

I am not sure, but I feel that the attached diagram explains how a Coaxial Collinear Antenna works. Suggestions/comments are welcomed.

Click on image to see full size image how coaxial collinear antenna works


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