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Let's say there is a reception antenna connected to a detection system with a coax, and a transmitter sends a plane wave. What happens is that the antenna gets a certain power, following the [Schelkunoff and Friis. Antenna theory and practice 1952] or [Papas. Theory of electromagnetic wave propagation], the absorbed power will be \begin{equation} P_{abs} = \frac{\lambda^2} {4\pi} g(\theta_0, \phi_0) S^{inc}(\theta_0, \phi_0) |\vec{p}_{rec}\cdot \vec{p}_{inc}|^2 \end{equation}

Where $\theta_0$ and $\phi_0$ are angles from where the plane wave was emitted. $S^{inc}$ is incident pointing vector and $\vec{p}$ - are polarization vectors of incident wave and a receiver, and g is the gain of the receiver.

So the point is: what is the amplitude of a received signal (=travelling wave) at a coax connected to the antenna (current or voltage)? Is it as following:

$$\frac{I_0^2Z} {2} = P_{abs} $$

Where Z is 50 Ohms (if the coax has 50 Ohms impedance), and the current is $$ I =I_0 sin(\omega t)$$

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  • $\begingroup$ Taking a sinusoidal current, one have the power equal to $I^2R/2$ where I is the amplitude of the current. This value is average. basically it comes from the integration of square of current. $\endgroup$
    – Pierre Polovodov
    Commented Oct 22, 2021 at 15:10
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    $\begingroup$ @petr it's an old source of confusion, between Amplitude (peak value) and magnitude (RMS). You need to define terms carefully. But if you know $P_{abs}$ then you can work out RMS or peak current as normal. Your equation is correct if you mean $I = I{sin}(\omega t)$ Engineers more commonly use RMS as it's the standard and a bit simpler for calculating power, voltage drop, etc. $\endgroup$
    – tomnexus
    Commented Oct 22, 2021 at 18:07
  • $\begingroup$ Yup! Peak = 1.414 * RMS; RMS = Peak * .707. But I know that you knew that. :-) $\endgroup$
    – Mike Waters
    Commented Oct 22, 2021 at 18:59
  • $\begingroup$ Thanks for the precisions. I come from the interface between engineering and physics. So, that's why the choice was $I =I_0 sin(\omega t)$ $\endgroup$
    – Pierre Polovodov
    Commented Oct 23, 2021 at 8:10
  • $\begingroup$ I have edited a post a bit in order to better reflect the problem $\endgroup$
    – Pierre Polovodov
    Commented Oct 23, 2021 at 8:12

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The voltage at the terminals of the antenna is related to the captured power by the radiation resistance of the antenna. One way to think of the radiation resistance is the load impedance terminating the antenna that will result in maximum power transfer to the load.

The radiation resistance of some popular antennas (in the linked article) are:

  • 73.1 $\Omega$ - Half-wave dipole
  • 36.5 $\Omega$ - Quarter-wave monopole over a ground plane
  • 100 $\Omega$ - Resonant 1 $\lambda$ loop

Ideally, a balun should be used to transform the input impedance seen by the antenna to its radiation resistance. Connecting a 50$\Omega$ cable directly to a half-wave dipole will result in a slight mismatch. (When I built dipoles for low power applications I always used RG-59 instead of RG-58 coax and matched at the transmitter.)

The origin of 50$\Omega$ and 75$\Omega$ systems in RF engineering and CATV is discussed in A. S. Gilmour's book Microwave Tubes (quoted here) and in Conrad Young's paper to the Piedmont Chapter of the Society of Cable Television Engineers.

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  • $\begingroup$ ok I understand that the received power can be obtained from Friis formula. And then one should take into account a mismatch between coaxial line and antenna resistance. Thank you for the references btw $\endgroup$
    – Pierre Polovodov
    Commented Mar 2, 2022 at 8:35
  • $\begingroup$ Great! Thanks for the positive feedback. $\endgroup$
    – AG5CI
    Commented Mar 2, 2022 at 12:43

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