In electromagnetics and antenna theory, the aperture of an antenna is defined as "A surface, near or on an antenna, on which it is convenient to makeassumptions regarding the field values for the purpose of computing fields at external points. The aperture is often taken as that portion of a plane surface near the antenna, perpendicular to the direction of maximum radiation, through which the major part of the radiation passes."[1]
The effective area of an antenna is defined as "In a given direction, the ratio of the available power at the terminals of a receiving antenna to the power flux density of a plane wave incident on the antenna from that direction, the wave being polarization matched to the antenna." Of particular note in this definition is that both effective area and power flux density are functions of incident angle of a plane wave. Assume a plane wave from a particular direction
(\theta,\phi)
\|\vec{S}\|
By definition, if an antenna delivers
PO
|S(\theta,\phi)|
Ae
Ae(\theta,\phi)=
PO | |
\|\vec{S |
(\theta,\phi)\|}.
The power
PO
PR
η
PR
|S(\theta,\phi)|=|\vec{S} ⋅ \hat{a}|
\hat{a}
A
PO=ηPR=ηA|\vec{S} ⋅ \hat{a}|=ηA\|\vec{S}(\theta,\phi)\|\cos\theta\cos\phi,
Ae(\theta,\phi)=ηA\cos\theta\cos\phi.
The effective area of an antenna or aperture is based upon a receiving antenna. 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
Ae
Ae
See main article: Antenna factor.
See also: Effective height.
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. An alternate measure of antenna response that has a greater relationship to the physical length of such antennas is effective length
leff
leff=V0/Es,
V0
Es
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). For instance, a half-wave dipole has a much longer effective length than a short dipole. However the effective area of the short dipole is almost as great as it is for the half-wave antenna, since (ideally), given an ideal impedance-matching network, it can receive almost as much power from that wave. Note that for a given antenna feedpoint impedance, an antenna's gain or
Aeff
leff
In general, the aperture of an antenna cannot be directly inferred from its physical size.[4] However so-called aperture antennas such as parabolic dishes and horn antennas, have a large (relative to the wavelength) physical area
Aphys
AphysS
Ae
Aphys
ea
ea=
Ae | |
Aphys |
.
The aperture efficiency is a dimensionless parameter between 0 and 1 that measures how close the antenna comes to using all the radio wave power intersecting its physical aperture. 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:
Ae=Aphys
Note that when one simply speaks of an antenna's "efficiency", what is most often meant is the radiation efficiency, a measure which applies to all antennas (not just aperture antennas) and accounts only for the gain reduction due to losses. Outside of aperture antennas, most antennas consist of thin wires or rods with a small physical cross-sectional area (generally much smaller than
Ae
The directivity of an antenna, its ability to direct radio waves preferentially in one direction or receive preferentially from a given direction, is expressed by a parameter
G
Po
Piso
G=
Po | |
Piso |
=
Ae | |
Aiso |
.
As shown below, the aperture of a lossless isotropic antenna, which by this definition has unity gain, is
Aiso=
λ2 | |
4\pi |
,
λ
G=
Ae | |
Aiso |
=
4\piAe | |
λ2 |
.
See main article: Friis transmission equation. The fraction of the power delivered to a transmitting antenna that is received by a receiving antenna is proportional to the product of the apertures of both the antennas and inversely proportional to the squared values of the distance between the antennas and the wavelength. This is given by a form of the Friis transmission formula:[6]
Pr | |
Pt |
=
ArAt | |
d2λ2 |
,
Pt
Pr
Ar
At
d
d
d\gtrsim2a2/λ
a
λ
The aperture of an isotropic antenna, the basis of the definition of gain above, can be derived on the basis of consistency with thermodynamics.[7] [8] [9] 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
2} | |
\overline{v | |
n |
=4kBTRη(f),
η(f)
η(f)=1
η(f)=
hf/kBT | ||||||
|
.
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
P=
| |||||||
4R |
.
2} | |
\overline{v | |
n |
=
2 | |
v | |
rms |
PR=
| ||||||||||
4R |
=
4kBTRη(f1)B1 | |
4R |
=kBTη(f1)B1.
The same antenna, being bathed in black-body radiation of temperature T, receives a spectral radiance (power per unit area per unit frequency per unit solid angle) given by Planck's law:
Pf,A,\Omega(f)=
2hf3 | |
c2 |
1 | ||||||
|
=
2f2 | |
c2 |
kBTη(f),
η(f)
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:
PA=
infty | |
\int | |
0 |
\int4\pi
Pf,A,\Omega(f) | |
2 |
Aeff(\Omega,f)F\nu(f)d\Omegadf.
PA=2\piPf,A,\Omega(f)AeffB1=
4\pikBTη(f1) | ||||||
|
AeffB1,
λ1=c/f1
Since each system is in thermodynamic equilibrium at the same temperature, we expect no net transfer of power between the cavities. Otherwise one cavity would heat up and the other would cool down in violation of the second law of thermodynamics. Therefore, the power flows in both directions must be equal:
PA=PR.
4\pikBTη(f1) | ||||||
|
AeffB1 =kBTη(f1)B1,
Aeff=
| |||||||
4\pi |
.
We thus find that for a hypothetical isotropic antenna, thermodynamics demands that the effective cross-section of the receiving antenna to have an area of λ2/4π. This result could be further generalized if we allow the integral over frequency to be more general. Then we find that Aeff for the same antenna must vary with frequency according to that same formula, using λ = c/f. Moreover, the integral over solid angle can be generalized for an antenna that is not isotropic (that is, any real antenna). 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. The spectral density of the power flow in either direction will still be given by
kBTη(f)
2/(4\pi) | |
λ | |
1 |