Antenna aperture

In general, the aperture of an antenna is not directly related to its physical size.[3] However some types of antennas, for example parabolic dishes and horn antennas, have a physical aperture (opening) which collects the radio waves. In these aperture antennas, the effective aperture ${\displaystyle A_{e}}$ is always less than the area of the antenna's physical aperture ${\displaystyle A_{\text{phys}}}$, otherwise the antenna could produce more power from its terminals than the radio power entering its aperture, violating conservation of energy. An antenna's aperture efficiency, ${\displaystyle e_{a}}$ is defined as the ratio of these two areas:

${\displaystyle e_{a}={A_{e} \over A_{\text{phys}}}\,}$

The aperture efficiency is a dimensionless parameter between 0 and 1.0 that measures how close the antenna comes to using all the radio wave power entering its physical aperture. If the antenna were perfectly efficient, all the radio power falling within 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 ${\displaystyle A_{e}=A_{\text{phys}}}$ and the aperture efficiency would be 1.0. But all antennas have losses, such as power dissipated as heat in the resistance of its elements, nonuniform illumination by its feed, and radio waves scattered by structural supports and diffraction at the aperture edge, which reduce the power output. Aperture efficiencies of typical antennas vary from 0.35 to 0.70 but can range up to 0.90.

The directivity of an antenna, its ability to direct radio waves in one direction or receive from a single direction, is measured by a parameter called its isotropic gain ${\displaystyle G}$, which is the ratio of the power ${\displaystyle P_{o}}$ received by the antenna to the power ${\displaystyle P_{\text{iso}}}$ that would be received by a hypothetical isotropic antenna, which receives power equally well from all directions. It can be seen that gain is also equal to the ratio of the apertures of these antennas

${\displaystyle G={P_{o} \over P_{\text{iso}}}={A_{e} \over A_{\text{iso}}}}$

As shown in the section at bottom, the aperture of a lossless isotropic antenna, which by definition has unity gain, is

${\displaystyle A_{\text{iso}}={\lambda ^{2} \over 4\pi }\,}$

where ${\displaystyle \lambda }$ is the wavelength of the radio waves. So

${\displaystyle G={A_{e} \over A_{\text{iso}}}={4\pi A_{e} \over \lambda ^{2}}\,}$

and for an antenna with a physical aperture of area ${\displaystyle A_{\text{phys}}}$

${\displaystyle G={4\pi A_{\text{phys}}e_{a} \over \lambda ^{2}}\,}$

So antennas with large effective apertures are high gain antennas, which have small angular beam widths. As receiving antennas, they are most sensitive to radio waves coming from one direction, and are much less sensitive to waves coming from other directions. As transmitting antennas, most of their power is radiated in a narrow beam in one direction, and little in other directions. Although these terms can be used as a function of direction, when no direction is specified, the gain and aperture are understood to refer to the antenna's axis of maximum gain, or boresight.

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 distance between the antennas and wavelength. This is given by a form of the Friis transmission formula:.[1]

${\displaystyle {\frac {P_{r}}{P_{t}}}={\frac {A_{r}A_{t}}{d^{2}\lambda ^{2}}}\,}$

where:

• ${\displaystyle P_{t}}$ is the power fed into the transmitting antenna input terminals;[1]
• ${\displaystyle P_{r}}$ is the power available at receiving antenna output terminals;[1]
• ${\displaystyle A_{r}}$ is the effective area of the receiving antenna;[1]
• ${\displaystyle A_{t}}$ is the effective area of the transmitting antenna;[1]
• ${\displaystyle d}$ is the distance between antennas.[1] The formula is only valid for ${\displaystyle d}$ large enough to ensure a plane wave front at the receive antenna, sufficiently approximated by ${\displaystyle d\geqq 2a^{2}/\lambda }$ where ${\displaystyle a}$ is the largest linear dimension of either of the antennas.[1]
• ${\displaystyle \lambda }$ is the wavelength of the radio frequency;[1]

The variables ${\displaystyle A_{r}}$, ${\displaystyle A_{t}}$, ${\displaystyle d}$, and ${\displaystyle \lambda }$ must be expressed in the same units of length, such as meters, and variables ${\displaystyle P_{t}}$ and ${\displaystyle P_{r}}$ must be in the same units of power, such as watts.

In the case of thin element antennas such as monopoles and dipoles, there is no simple relationship between physical area and effective area. However, the effective areas can be calculated from their power gain figures:[4]

Wire antenna	Power gain	Effective area
Short dipole (Hertzian dipole)	1.5	0.1194 ${\displaystyle \lambda }$2
Half-wave dipole	1.64	0.1305 ${\displaystyle \lambda }$2
Quarter-wave monopole	3.28	0.2610 ${\displaystyle \lambda }$2[5]

This assumes that the monopole antenna is mounted above an infinite ground plane and that the antennas are lossless. When resistive losses are taken into account, particularly with small antennas, the antenna gain might be substantially less than the directivity, and the effective area is less by the same factor.[6]

For antennas which are not defined by a physical area, such as monopoles and dipoles consisting of thin rod conductors, the aperture bears no obvious relation to the size or area of the antenna. An alternate measure of antenna gain that has a greater relationship to the physical structure of such antennas is effective length l_eff measured in metres, which is defined for a receiving antenna as:[7]

${\displaystyle l_{\text{eff}}=V_{0}/E_{s}\,}$

where

V₀ is the open circuit voltage appearing across the antenna's terminals

E_s is the electric field strength of the radio signal, in volts per metre, at the antenna.

The longer the effective length the more voltage and therefore the more power the antenna will receive. An antenna's gain or A_eff increases according to the square of l_eff, and that this proportionality also involves the antenna's radiation resistance. Therefore, this measure is of more theoretical than practical value and is not, by itself, a useful figure of merit relating to an antenna's directivity.

Diagram of antenna A and resistor R in thermal cavities, connected by filter F_ν. If both cavities are at the same temperature ${\displaystyle T}$, ${\displaystyle P_{\text{A}}=P_{\text{R}}}$

The aperture of an isotropic antenna, the basis of the definition of gain above, can be derived by a thermodynamic argument.[8][9][10] Suppose an ideal (lossless) isotropic antenna A located within a thermal cavity CA, is connected via a lossless transmission line through a band-pass filter F_ν to a matched resistor R in another thermal cavity CR (the characteristic impedance of the antenna, line and filter are all matched). Both cavities are at the same temperature ${\displaystyle T}$. The filter F_ν only allows through a narrow band of frequencies from ${\displaystyle \nu }$ to ${\displaystyle \nu +\Delta \nu }$. Both cavities are filled with blackbody radiation in equilibrium with the antenna and resistor. Some of this radiation is received by the antenna. The amount of this power ${\displaystyle P_{\text{A}}}$ within the band of frequencies ${\displaystyle \Delta \nu }$ passes through the transmission line and filter F_ν and is dissipated as heat in the resistor. The rest is reflected by the filter back to the antenna and is reradiated into the cavity. The resistor also produces Johnson–Nyquist noise current due to the random motion of its molecules at the temperature ${\displaystyle T}$. The amount of this power ${\displaystyle P_{\text{R}}}$ within the frequency band ${\displaystyle \nu }$ to ${\displaystyle \nu +\Delta \nu }$ passes through the filter and is radiated by the antenna. Since the system is at a common temperature it is in thermodynamic equilibrium; there can be 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

${\displaystyle P_{\text{A}}=P_{\text{R}}}$

The radio noise in the cavity is unpolarized, containing an equal mixture of polarization states. However any antenna with a single output is polarized, and can only receive one of two orthogonal polarization states. For example, a linearly polarized antenna cannot receive components of radio waves with electric field perpendicular to the antenna's linear elements; similarly a right circularly polarized antenna cannot receive left circularly polarized waves. Therefore, the antenna only receives the component of power density S in the cavity matched to its polarization, which is half of the total power density

${\displaystyle S_{\text{matched}}={1 \over 2}S}$

Suppose ${\displaystyle B_{\nu }}$ is the spectral radiance per hertz in the cavity; the power of black-body radiation per unit area (meter²) per unit solid angle (steradian) per unit frequency (hertz) at frequency ${\displaystyle \nu }$ and temperature ${\displaystyle T}$ in the cavity. If ${\displaystyle A_{\text{e}}(\theta ,\phi )}$ is the antenna's aperture, the amount of power in the frequency range ${\displaystyle \Delta \nu }$ the antenna receives from an increment of solid angle ${\displaystyle d\Omega =d\theta d\phi }$ in the direction ${\displaystyle \theta ,\phi }$ is

${\displaystyle dP_{\text{A}}(\theta ,\phi )=A_{\text{e}}(\theta ,\phi )S_{\text{matched}}\Delta \nu d\Omega ={1 \over 2}A_{\text{e}}(\theta ,\phi )B_{\nu }\Delta \nu d\Omega }$

To find the total power in the frequency range ${\displaystyle \Delta \nu }$ the antenna receives, this is integrated over all directions (a solid angle of ${\displaystyle 4\pi }$)

${\displaystyle P_{\text{A}}={1 \over 2}\int \limits _{4\pi }A_{\text{e}}(\theta ,\phi )B_{\nu }\Delta \nu d\Omega }$

Since the antenna is isotropic, it has the same aperture ${\displaystyle A_{\text{e}}(\theta ,\phi )=A_{\text{e}}}$ in any direction. So the aperture can be moved outside the integral. Similarly the radiance ${\displaystyle B_{\nu }}$ in the cavity is the same in any direction

${\displaystyle P_{\text{A}}={1 \over 2}A_{\text{e}}B_{\nu }\Delta \nu \int \limits _{4\pi }d\Omega }$

${\displaystyle P_{\text{A}}=2\pi A_{\text{e}}B_{\nu }\Delta \nu }$

Radio waves are low enough in frequency so the Rayleigh–Jeans formula gives a very close approximation of the blackbody spectral radiance[11]

${\displaystyle B_{\nu }={2\nu ^{2}kT \over c^{2}}={2kT \over \lambda ^{2}}}$

Therefore,

${\displaystyle P_{\text{A}}={4\pi A_{\text{e}}kT \over \lambda ^{2}}\Delta \nu }$

The Johnson–Nyquist noise power produced by a resistor at temperature ${\displaystyle T}$ over a frequency range ${\displaystyle \Delta \nu }$ is

${\displaystyle P_{\text{R}}=kT\Delta \nu }$

Since the cavities are in thermodynamic equilibrium ${\displaystyle P_{\text{A}}=P_{\text{R}}}$, so

${\displaystyle {4\pi A_{\text{e}}kT \over \lambda ^{2}}\Delta \nu =kT\Delta \nu }$

• Antenna (radio)