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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.

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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.

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  • $\begingroup$ Thanks @ArtBrown - I am also cautious about that $\cos \psi$ because of the same reason. $\endgroup$ – Avijit Nov 12 '13 at 3:14
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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.

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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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