# Is the effective aperture of an antenna a constant?

1. Mar 11, 2013

### yungman

When finding the effective aperture of an antenna we use:
$$A_{eff}\;=\;D_0\frac{\lambda^2}{4\pi}\;\hbox{ where }\;\frac{\lambda^2}{4\pi}\;\hbox { is the effective aperture of a isotropic antenna.}$$

I read from Wikipedia and other sources that in order to use this equation, you assume the polarization of the incident wave matches the receiving antenna. AND the output impedance of the receiving antenna is conjugate match to the load for maximum power transfer.

But in the Antenna Theory book by Balanis:
$$A_{eff}\;=\;e_0|\hat{\rho_w}\cdot\hat{\rho_a}|^2 D_0\frac{\lambda^2}{4\pi}\;\hbox { where }\; e_0=e_{cd}(1-|\Gamma|^2)$$

$e_{cd}$ is the ohmic loss, $(1-|\Gamma|^2)$ is the load matching loss, and $|\hat{\rho_w}\cdot\hat{\rho_a}|^2$ is the polarization loss factor.

Balanis includes all the loss factors to calculate the effective area. This implies the effective area of the receiving antenna depends on the polarization of the incident TEM wave AND the load matching. This means the effective area varies with other outside parameters. I don't think that is correct. Please give me your thoughts.

Thanks

2. Mar 11, 2013

### the_emi_guy

First equation (assuming D0 is directivity) assumes no losses.

Second equation includes all of the losses which are often combined into "radiation efficiency".

Another way this is commonly written:

$$A_{eff}\;=\;G\frac{\lambda^2}{4\pi};$$

Where G is the antenna gain.

G = ηD Where:
D is the directivity.
η is the radiation efficiency and includes all of the losses (resistive, mismatch etc).