This Wikipedia entry tells me that the Thevenin equivalent circuit for an arbitrary receiving antenna on which an electric field $E_b$ is incident is a voltage source $V_a$ in series with an impedance $R_a + j X_a$ where (I have re-arranged the terms a bit to frame my question...) $$ V_a = E_b\; \frac{\cos {\psi}}{\sqrt{\pi Z_0}} \; \left( \lambda \sqrt{R_a G_a} \right) $$ ... given that $G_a$ is the directive gain of the antenna in the directive gain of the antenna in the direction of incidence, and $\psi$ is the angle by which the electric field is 'misaligned' with the antenna.

The article does mention that this is derived from reciprocity, from which I assume that there should be some reasoning beginning with the Rayleigh-Carson theorem: $$ \iiint_{V_1} \vec{J}_1 \cdot \vec{E}_2 \;dV = \iiint_{V_2} \vec{J}_2 \cdot \vec{E}_1 \;dV $$ I am trying to understand how I can apply this, and in fact how I can approach an arbitrary antenna structure in general (I do understand how a dipole and a loop can be analyzed)

Unfortunately the article itself doesn't point out any sources where this relationship is derived, so I was wondering if anyone could point me to any textbook or paper where this derivation may be found?

My motivation is something like this -- the relation mentioned in the Wikipedia article is actually for a sinusoidal input -- and the frequency determines $\lambda$, $R_a$, and $G_a$ in the expression (and $X_a$ in the equivalent circuit). I am trying to understand if any insight can be obtained about the equivalent voltage source $V\left(t\right)$ given an arbitrary $E\left(t\right)$ -- maybe, for example, as a differential or integral equation? The $X_a$ can be replaced by frequency independent $C_a$ and $L_a$ in series -- and for the voltage source I would integrate over $\lambda$ -- but I don't know how to deal with $R_a\left(\lambda\right)$ (which is ideally only the radiation resistance) and $G_a\left(\lambda\right)$ for arbitrary antenna geometries. So I was hoping that the derivation would offer me some clues...

Update OK, so it seems that I went on the wrong track here -- it is actually quite easy. I am answering my own question below.


3 Answers 3


A slight variant on your fine answer...

A reference is Ramo et al, Fields and Waves in Communication Electronics, chapter 12.

First, reciprocity: $Z_{21}=Z_{12}$ tells you that (assuming a conjugate-matched load):

$$ g_{dt} A_{er} = g_{dr} A_{et}$$

For both transmitting (subscript t) and receiving (r) antennas, $g_d$ is the antenna directional gain.

$A_{er}$ is the effective area of the receiving antenna, defined as the ratio of useful power removed from the receiving antenna $W_r$ to average power density $P_{av}$ in the incoming radiation.

Thus the ratio $g_d/A_e$ is the same for both transmitting and receiving antennas.

For large aperture antennas, it can be shown that the maximum possible gain satisfies: $$ \frac{(g_d)_{max}}{A_e} = \frac{4 \pi}{\lambda^2} $$

For other geometries, $A_e$ is defined to give the same result. For example, for a Hertzian dipole, with a maximum directivity of 1.5: $$ (A_e)_{max} = \frac{\lambda^2}{4 \pi} (g_d)_{max} = \frac{3}{8 \pi} \lambda^2 $$

Anyway, for the problem at hand, as you deduced, the useful power removed from the receiving antenna is:

$$ W_r = P_{av} A_{er} \text{, with the power density } P_{av} = \frac{E_b^2}{2 Z_o} , Z_o=377 \text{ ohms} $$

(Here, electric field and voltage are sinusoids measured as peak values.)

With a conjugate-matched load with real part $R_L$, equating load power dissipated with power delivered gives for the receiving antenna's Thevenin equivalent source voltage $V_a$:

$$\frac{(V_a/2)^2}{2 R_L} = \frac{E_b^2}{2 Z_o} A_{er} $$

$$ V_a = 2 \sqrt{A_{er}} \sqrt{\frac{R_L}{Z_o}} \, E_b $$

Substituting for $A_{er}$ from the reciprocity relation, the maximum voltage $V_{a,max}$ is:

$$ V_{a,max} = \sqrt{\frac{(g_{dr})_{max}}{\pi }} \sqrt{\frac{R_L}{Z_o}} \,\, \lambda E_b $$

I'm cautious about the $\cos \psi$ factor because beam patterns differ for different antennas.

  • $\begingroup$ Thanks @ArtBrown - I am also cautious about that $\cos \psi$ because of the same reason. $\endgroup$
    – Avijit
    Commented Nov 12, 2013 at 3:14

Using reciprocity, we can relate the gain $G$ (a transmission characteristic) and effective aperture $A_e$ (a reception characteristic) in any given direction $\left(\theta,\phi\right)$ as: $$ A_e\left(\theta, \phi\right) = G\left(\theta, \phi\right) \frac {\lambda^2}{4\pi} $$

So squaring both sides of the expression for $V_a$, I get: $$ {V_a}^2 = {E_b}^2 \frac {R_a \; \lambda^2\;G_a \cos^2 \psi} {\pi Z_0} $$ I assume that $G_a$ is actually $G_a\left(\theta,\phi\right)$, where $\left(\theta,\phi\right)$ is the direction of the source with respect to the antenna. Then we have: $$ \frac {{V_a}^2}{R_a} = \frac {{E_b}^2}{Z_0} \frac {\lambda^2 }{\pi}G_a\left(\theta,\phi\right) \cos^2\psi = 4 \frac {{E_b}^2}{Z_0} A_e\left(\theta,\phi\right) \cos^2 \psi $$ ...or... $$ \frac {{V_a}^2}{4R_a} = \frac {{E_b}^2}{Z_0} A_e\left(\theta,\phi\right) \cos^2 \psi $$ Now $\dfrac {{E_b}^2}{Z_0}$ is the magnitude of the Poynting vector in in the far field, and $\dfrac {{V_a}^2}{4R_a}$ is the power dissipated across a matched load $R_a-jX_a$, so for $\psi=0$ the equation simply says 'the power dissipated in a matched load equals the Poynting vector times the effective aperture', which is actually the definition of effective aperture!

So all that is left is $\psi$ -- and now we can see that it is actually the difference of polarization between the incoming wave and the antenna.


This seems too simple so I'm not sure this is the answer you are looking for. An antenna system is a part of the circuit it is attached to .It is an inductive(for the most part) unit.Most common circuits with antenna units are designed around a 50 ohm impediance model.


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