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I don't understand why the resonance of an antenna is not affected by the transmission line to the receiver. For example, given a dipole antenna like the following:

dipole antenna

The antenna length is carefully chosen to create a resonance at the target frequency, however, isn't this ruined by the transmission line essentially lengthening the effective length of the antenna?

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    $\begingroup$ If the transmission line is not matched to the antenna, than yes it will mess things up. $\endgroup$
    – Jon Custer
    Commented Mar 11, 2016 at 19:09
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    $\begingroup$ Also, the dipole shouldn't be directly connected to the coax cable like this, you need to use a balun to transform the balanced signal from the antenna to an unbalanced signal. $\endgroup$ Commented Mar 11, 2016 at 19:37
  • $\begingroup$ The resonance of the antenna is affected, the radiation pattern is not. Just think of any LC resonator, if you place a load on it, the resonance changes. $\endgroup$
    – hyportnex
    Commented Mar 11, 2016 at 22:01
  • $\begingroup$ Depends on your definition of "mess up". One can use coaxial cables as very efficient impedance matching devices. Pick the correct length and cable impedance and it will match your antenna near perfectly (at one frequency) to the receiver or transmitter. That's being used in e.g. nuclear magnetic resonance experiments all the time. As a student I built several cable boxes for that purposes. $\endgroup$
    – CuriousOne
    Commented Mar 11, 2016 at 22:07
  • $\begingroup$ @JonCuster "If the transmission line is not matched to the antenna, than yes it will mess things up." <-- Assuming by "matched" you mean impedance matched, that is not true. $\endgroup$
    – DanielSank
    Commented Jan 13, 2017 at 22:41

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A dipole is fed by a frequency varying voltage [or current] source at the center between the two halves. In theory, you send a sinusoidal signal down the transmission line which will see the dipole two dipole leads as an impedance $Z_{ant}$. For a half-wave dipole, $Z_{ant} \approx 72 + j42.5$ $\Omega$.

The classical textbook analysis of the radiation from a dipole begins by ignoring the feed and assuming the sinusoidal voltage source as above to calculate the theoretical radiating fields and radiation resistance $R_{rad}$. Then, texts will introduce the concept of balanced impedance lines and explain that a coaxial feed is an unbalanced line, i.e. current flows on the surface of center conductor. (Recall that the modal propagation down the line is a TEM wave). Since a dipole requires a balanced feed, you can use a balun to change the unbalanced line (uneven currents) to balanced line (with equal currents) exciting both halves of the dipole. This also serves the purpose of "choking" (or blocking/grounding) any currents that could be induced down the shield of the coaxial feed.

Textbooks often do not discuss the impact of the feed structure itself on the radiation. Yes, this metallic line will scatter fields incident on it, and thus interact with the radiation pattern. When you measure the gain in an anechoic chamber, you should see this variation in the measured omni pattern and change front-to-back ratio (if you define such a thing for a dipole...). However, It shouldn't be much since the line is usually assumed to be orthogonal to the polarization of the radiating fields.

To answer your question directly, the transmission line doesn't lengthen the dipole feed. Like I said above, the fields propagating down the line are TEM, so basically it's just a voltage source exciting the dipole leads; with a properly attached balun, you will excite the half-wave sinusoidal current distribution on the dipole.

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  • $\begingroup$ A pretty good answer with some further material to build the reader's knowledge, but the explanation of the balun is a bit mysterious. IIRC the co-ax's outside shield's radiating is pretty much logically equivalent to the imbalance of the shield and core currents - the one implies the other and contrariwise, so that to say the balun prevents radiation because it converts unbalanced to balanced currents is tautologous - like saying it prevents radiation by preventing radiation! I've never liked that explanation. I don't think there's a simple explanation: what you're really doing is .... $\endgroup$ Commented Jan 14, 2017 at 3:26
  • $\begingroup$ .... shifting the boundary conditions on both the transmission line and antenna through the use of the isolation device so that the core and shield of the transmission co-ax are forced to carry the same, but opposite sense, mutually shielding currents, or at least they get near enough to this condition that the antenna's efficiency isn't too degraded. You can really only delve into this issue by writing down detailed solutions to the MEs for the two subsystems and thinking carefully about what boundary conditions model which physical devices and how the two subsystems are linked, or .... $\endgroup$ Commented Jan 14, 2017 at 3:32
  • $\begingroup$ ... at least that's my recollection of what I had to do to understand that the balun does and why RS-422 gives the improvements it does, amongst other things. $\endgroup$ Commented Jan 14, 2017 at 3:36
  • $\begingroup$ @WetSavannaAnimalakaRodVance I don't think you necessarily have to go back to Maxwell's Equations and talk about the interface between a coaxial TEM mode and the dipole bare wires. Doing so would be rather complex. $\endgroup$
    – AntennaGuy
    Commented Jan 15, 2017 at 0:55
  • $\begingroup$ I'm just saying that in my experience, you really do have to look carefully at why the isolation device forces symmetrical, self shielding currents, and careful consideration of current / voltage boundary conditions is the only clear, rigorous way to do it. Otherwise, you end up with circular explanations such that it prevents radiation by balancing currents, which to my mind is a tautology because one can't happen without the other (or, contrapositively, each one implies the other). It's probably too much detail for a broad answer like this, but I found the balun pretty mysterious and my .... $\endgroup$ Commented Jan 15, 2017 at 1:01

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