From my current knowledge, any metal whether grounded (and or earthed) or not will reflect radio waves as seen on dishes, Faraday cages, etc. On the other side the ground in transmission lines or the ground planes in PCB, I see more as a place where the fields are absorbed, as the energy travels between the two. It’s always constantly stated to have proper grounding for high speed signals.

All in all it seems in different scenarios, metal absorbs the fields, and other times reflects/guides the fields.

What is the main reason for this?

I am assuming it is to do with the impedance seen to the metal, for instance a Faraday cage or ground plane for antennas is certainly not seen as the impedance of the propagating waves, but that's only my limited knowledge understanding.

Which brings on to the actual question: Why would the wires acting as the ground plane on those monopole ground plane antennas be reflecting the signal, when they are directly connected to the coax shield?

  • \$\begingroup\$ Are you familiar with the concept of reflection coefficient? \$\endgroup\$ Jan 11 at 10:08
  • \$\begingroup\$ A decent coax shield does not absorb waves. A decent ground plane does not absorb waves. If you have evidence that they do (other than trivial amounts) then you should propose that evidence. \$\endgroup\$
    – Andy aka
    Jan 11 at 10:26
  • \$\begingroup\$ @Andyaka perhaps im viewing it wrong then. I came to that conclusion just based on the premise antennas do absorb. By absorb i mean more in the sense what a south pole is to a north pole in magnetism if that helps to see where im stuck. \$\endgroup\$ Jan 11 at 10:33
  • \$\begingroup\$ @TimWilliams cant say i am in terms of em waves. Im assuming this then depends on many facorts, including the different scenarios i gave? \$\endgroup\$ Jan 11 at 10:34
  • \$\begingroup\$ @Andyaka Both exhibit resistance (unless you've splurged for the superconducting variant, in which case just very much less) and therefore absorb waves. Put another way, the reflection is nearly perfect but not total, 97, 99, 99.9%, what have you. Or if you like, the Poynting vector is slightly into the coax walls. \$\endgroup\$ Jan 11 at 10:48

2 Answers 2


Taking example of the ground plane antenna, consider the case with a monopole, perpendicular to multiple ground spokes in a plane:

enter image description here

Suppose we have a feedpoint voltage Vs (and more generally a port, with feedline and incident and reflected waves, not shown).

Within a certain degree of symmetry, we can consider each ground spoke as its own monopole with respect to the rest.

Notice the geometric symmetry is not perfect, because the remaining spokes and the monopole element (considering Vs = 0 by superposition) do not form a plane perpendicular to the spoke in question, whereas they do for the monopole. This moderately affects the characteristic impedance of the element in question, but does not change the overall argument.

Considering a spoke as a driven element, like any other element, it exhibits a characteristic impedance, looking into its feedpoint. In normal use, that feedpoint is connected to an approximate short circuit -- Rg being a small value. But "approximate" is not "perfect". The element has some resistance along its length, which for reasons not important to go into here, can be represented as an equivalent total resistance at the base, and the rest of the element assumed ideal.

Given the symmetry (and limitations as noted), the characteristic impedance of the spoke as an element, will be comparable to the monopole element itself, typically 30-40Ω. We can understand the effect of Rg, when it is small in relation to this impedance, as the reflection coefficient at that port.

The reflection coefficient is given by,

$$ \Gamma = \frac{Z_L - Z_0}{Z_L + Z_0} $$

For \$Z_0 = 40\,\Omega\$ and say \$Z_L = R_g = 0.1\,\Omega\$, we have \$\Gamma = 0.995\ldots\$, quite high but not exactly one.

So, we have some energy absorbed in the element, but most of the incident energy is reflected in phase.

Important take-away: real situations are always both absorption and reflection simultaneously. (Or transmission, if applicable.) How much of each, is the question. Real in-practice ground planes, wires, shields, waveguides, etc. act as guides or mirrors to incident waves, usually to quite a good degree (a mirror below 90% reflectivity might be considered poor even at optical frequencies, and over 99% is typical at radio frequencies). But we must keep in mind there is always absorption in that small remainder below 100%.

We also have high reflectance for \$\Gamma \rightarrow -1\$. What would that do?

We could connect the element(s) open-circuit, which gives this condition. The meaning of negative Γ is that waves reflect in opposing phase. What does this do to the monopole-with-ground-plane antenna? Well, the reflected waves would oppose the driven element's waves, and... nothing happens. Which is as we expect: Vs would be essentially open-circuit on one side (all Rg's infinite), so feedline current cannot flow.

  • \$\begingroup\$ Really great response that helped me a lot! Many thanks. Ill be refering back to this as i learn. But one small question, you said the characterstic impedance of the spoke is around 30-40 simlar to the monopole, and when finding the reflection coefficent you said the feedpoint is an approcxmate short curcuit, i understand (and from you answer) this is causing the reflections. But why is the feedpoint short curcuit here yet id storngly assume its 50 for the center spoke since its not reflecting. If its a long answer i dont mind a short and simple one and ill try reseach myself too. \$\endgroup\$ Jan 11 at 12:42
  • \$\begingroup\$ Correct, I glossed over the simplification from Vs as a feedpoint, to Vs = 0 (shorted). In practice, it will be terminated with line and amplifier impedance. This has the effect of raising the impedance in a similar way as the geometry note, so effectively I've rolled it into that. \$\endgroup\$ Jan 11 at 12:50

Ground plane nor the shield of the coaxial plane don't absorb electromagnetic waves. Metal forms only a low resistance surface where the electric field is forced to be perpendicular against the metal. The electromagnetic wave always adapts itself so that its electric field is perpedicular against metal surfaces. That (depending on the geometric form, dimensions, frequency and also on other materials like insulators nearby) sometimes forces a new reflected wave to be generated, sometimes only the propagation direction of the wave is forced - like in transmission lines it happens along the line; in coaxial cables the wave exists only in the space between the conductors only.

Electromagnetic waves do not penetrate into metals assuming there's no resistance in the metal. That assumption is quite useful for short enough distances, but practical metals, of course have some resistance, too and the wave energy gradually is dissipated due the induced current on the metal surface. But the the parts are designed in practice so short that most of the wave energyis not absorbed. It's reflected or directed along the metal surface.

Why? That's only a mathematical consequence of Maxwell's equations that the waves obey. Mathematicians say it "boundary conditions define which differential equation solutions can exist!". Metal is such boundary.

Why do electromagnetic phenomenons obey such equations? Nobody knows. Physicists have found those basic equations valid by making careful measurements. That happened in the about 50 years long period 100...150 years ago.

  • \$\begingroup\$ If ground plane does not absorb waves, then I can angle two together at an acute angle and get a perfect corner reflector still, right? Or since coax shield doesn't absorb, I can take an infinite length of it and still get signal out the end? \$\endgroup\$ Jan 11 at 10:57
  • \$\begingroup\$ Makes sense thanks! I see what you mean how the coax sort of retransmits, keeping the energy between the center and sheild. However, you said "sometimes forces a new reflected wave", surley in the event it dosent, i would classify that as absorbtion? \$\endgroup\$ Jan 11 at 10:58
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    \$\begingroup\$ @TimWilliams is it annoying that you didn't be fast enough to write something informative at first? be faster next time. \$\endgroup\$
    – unawriter
    Jan 11 at 11:05
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    \$\begingroup\$ @Georgekirby I'm writing an answer as we speak, actually. \$\endgroup\$ Jan 11 at 11:12
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    \$\begingroup\$ @Georgekirby the fields of the wave still exist in the line. The energy is in the fields. The voltage is one of the extremely reduced property of the electric field. The current is induced to the metal. It's often practical to follow the reduced quantitities voltage and current, as we electricians do, but the propagating wave is the one which contains the energy which flows from the transmitter towards the antenna of from antenna towards the receiver. The reduced quantities voltage and current are quite good tools for following the actual wave when the line has 2 parallel or twisted conductors \$\endgroup\$
    – unawriter
    Jan 11 at 13:20

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