I am doing some calculations where I determine power at the input of Rx given some field strength at the antenna. I have a question about the antenna aperture:
$A_e = D\frac{\lambda^2}{4\pi}$
$D$ is the antenna gain directivity (Rx in this case)
$\lambda$ is the wavelength
Normally, I think of the frequency of the incoming (or outgoing) signal to determine the wavelength. However, I am wondering if it may dependent on the resonance frequency of the antenna.
In-band the difference between resonance and desired signal is small, however for out-of-band this relation may not hold. I would generally think the aperture size is correlated to antenna size regardless of transmit frequency; i.e. it is not as if the Rx antenna gets bigger just because the transmit frequency decreases.
Alternatively, I can think of $\frac{\lambda^2}{4\pi}$ as "normalization" which means that lowered gain accounted for by D (which would be measured). We then convert the maximum directivity measured into a equivalent isotropic antenna using the factor $\frac{\lambda^2}{4\pi}$.
I know I am not accounting for inefficiencies, to me this is a separate matter. But may also help explain why the incoming power is less than incident field strength.
Edited to add why we don't need to include inefficiencies in this discussion. If we have 2 antennas (Tx and Rx) separated by a distance (the Directivity ).
$\frac{D_{Tx}}{A_{Tx}} = \frac{D_{Rx}}{A_{Rx}}$
If we assume that the Tx is an isotropic antenna then $(D_{Tx} = 1)$
$A_{Tx} = \frac{\lambda^2}{4\pi} = \frac{A_{Rx}}{D_{Rx}} \rightarrow A_{Rx} = D_{Rx} \frac{\lambda^2}{4\pi}$
However, I only want to be clear about how to use effective aperture equation. Therefore, we would assume the maximum amount that could be transferred would be some ideal - then apply inefficiencies due to loss (dielectric, conduction, impedance mismatch, polarization, some other loss).
Therefore, I think $\lambda$ is the transmit frequency and not the resonant frequency.
Can someone provide some clarification/verification?