"[1] Of particular note in this definition is that both effective area and power flux density are functions of incident angle of a plane wave.
, which are the azimuth and elevation angles relative to the array normal, has a power flux density
; this is the amount of power passing through a unit area normal to the direction of the plane wave of one square meter.
watts to the transmission line connected to its output terminals when irradiated by a uniform field of power density
However, due to reciprocity, an antenna's directivity in receiving and transmitting are identical, so the power transmitted by an antenna in different directions (the radiation pattern) is also proportional to the effective area
[1] Most antenna designs are not defined by a physical area but consist of wires or thin rods; then the effective aperture bears no clear relation to the size or area of the antenna.
measured in metres, which is defined for a receiving antenna as[2] where The longer the effective length, the greater is the voltage appearing at its terminals.
However, the actual power implied by that voltage depends on the antenna's feedpoint impedance, so this cannot be directly related to antenna gain, which is a measure of received power (but does not directly specify voltage or current).
But since changing the physical size of an antenna inevitably changes the impedance (often by a great factor), the effective length is not by itself a useful figure of merit for describing an antenna's peak directivity and is more of theoretical importance.
[3] In general, the aperture of an antenna cannot be directly inferred from its physical size.
which is opaque to such radiation, essentially casting a shadow from a plane wave and thus removing an amount of power
If the aperture efficiency were 100%, then all the wave's power falling on its physical aperture would be converted to electrical power delivered to the load attached to its output terminals, so these two areas would be equal:
But due to nonuniform illumination by a parabolic dish's feed, as well as other scattering or loss mechanisms, this is not achieved in practice.
Since a parabolic antenna's cost and wind load increase with the physical aperture size, there may be a strong motivation to reduce these (while achieving a specified antenna gain) by maximizing the aperture efficiency.
[6][7][8] Suppose that an ideal isotropic antenna A with a driving-point impedance of R sits within a closed system CA in thermodynamic equilibrium at temperature T. We connect the antenna terminals to a resistor also of resistance R inside a second closed system CR, also at temperature T. In between may be inserted an arbitrary lossless electronic filter Fν passing only some frequency components.
Each cavity is in thermal equilibrium and thus filled with black-body radiation due to temperature T. The resistor, due to that temperature, will generate Johnson–Nyquist noise with an open-circuit voltage whose mean-squared spectral density is given by where
, but in general is given by The amount of power supplied by an electrical source of impedance R into a matched load (that is, something with an impedance of R, such as the antenna in CA) whose rms open-circuit voltage is vrms is given by The mean-squared voltage
can be found by integrating the above equation for the spectral density of mean-squared noise voltage over frequencies passed by the filter Fν.
For simplicity, let us just consider Fν as a narrowband filter of bandwidth B1 around central frequency f1, in which case that integral simplifies as follows: This power due to Johnson noise from the resistor is received by the antenna, which radiates it into the closed system CA.
However, that radiation is unpolarized, whereas the antenna is only sensitive to one polarization, reducing it by a factor of 2.
To find the total power from black-body radiation accepted by the antenna, we must integrate that quantity times the assumed cross-sectional area Aeff of the antenna over all solid angles Ω and over all frequencies f: Since we have assumed an isotropic radiator, Aeff is independent of angle, so the integration over solid angles is trivial, introducing a factor of 4π.
And again we can take the simple case of a narrowband electronic filter function Fν which only passes power of bandwidth B1 around frequency f1.
Since each system is in thermodynamic equilibrium at the same temperature, we expect no net transfer of power between the cavities.
Then we find that Aeff for the same antenna must vary with frequency according to that same formula, using λ = c/f.
Since the angle of arriving electromagnetic radiation only enters into Aeff in the above integral, we arrive at the simple but powerful result that the average of the effective cross-section Aeff over all angles at wavelength λ must also be given by
Although the above is sufficient proof, we can note that the condition of the antenna's impedance being R, the same as the resistor, can also be relaxed.
In principle, any antenna impedance (that isn't totally reactive) can be impedance-matched to the resistor R by inserting a suitable (lossless) matching network.
Since that network is lossless, the powers PA and PR will still flow in opposite directions, even though the voltage and currents seen at the antenna and resistor's terminals will differ.
, and in fact this is the very thermal-noise power spectral density associated with one electromagnetic mode, be it in free-space or transmitted electrically.
And an antenna, also having a single electrical connection, couples to one mode of the electromagnetic field according to its average effective cross-section of