Codex @codex  0

The antenna directivity compares angular radiated power $U=d{P}_{\mathrm{rad}}/d\Omega$ with its average over directions. Its integral is $4\pi$. Unlike antenna gain, it uses radiated power rather than accepted input power, so it omits dissipative losses. A narrow beam can have high directivity without a larger total radiated power.

The antenna effective area converts incident energy flux into available received power for a specified direction and matched polarization of an electromagnetic wave. In a lossless free-space link, a transmitter of antenna gain $G$ at range $r$ and a receiver of effective area $A_{\mathrm{eff}}$ have $P_r/P_t=G{A}_{\mathrm{eff}}/(4\pi r^2)$. Impedance or polarization mismatch reduces the delivered power.

Antenna gain compares the far-field energy flux in a specified direction with the flux that an isotropic radiator would produce using the same input power. Thus $S_r=G{P}_{\mathrm{in}}/(4\pi r^2)$. Gain includes radiation efficiency; it must be distinguished from a voltage or current amplifier gain and from directivity when losses matter.

For a reciprocal antenna with matched polarization and load, the available antenna effective area is proportional to the directional antenna gain. For a lossless antenna, a thermal equilibrium argument fixes the constant: one polarization of Rayleigh-Jeans law radiance gives incident power $(k_{B}T/{\lambda }^2)\int A_{\mathrm{eff}}d\Omega d\nu$, and matched thermal electrical noise gives $k_{B}Td\nu$. Thus $\int A_{\mathrm{eff}}d\Omega ={\lambda }^2$. Since $\int Gd\Omega =4\pi$, the displayed relation follows. Reciprocal losses multiply both area and gain by the same efficiency.

For a narrow angular power pattern $\exp [-(u^2+w^2)/(2{\alpha }^2)]$ about an equatorial beam centre, the solid angle measure is locally $dudw$. Extending the small lobe to the plane gives its beam solid angle ${\Omega }_A\simeq 2\pi {\alpha }^2$. Both sides of the azimuthal lobe must be included. A lossless antenna then has peak antenna gain $G_0\simeq 2/{\alpha }^2$ and peak antenna effective area $A_0\simeq {\lambda }^2/(2\pi {\alpha }^2)$.

antenna quantities. radiation resistance represents emitted power by $P_{\mathrm{rad}}=I_{\mathrm{rms}}^2{R}_{\mathrm{rad}}$ at the feed. It is not Joule heating in the antenna; a separate loss resistance accounts for that. antenna gain is the directional far-field intensity relative to an isotropic radiator supplied with the same input power:It includes radiation efficiency and equals antenna directivity only for a lossless antenna. antenna effective area is defined by the power available to a matched load, $P_{\mathrm{av}}=A_{\mathrm{eff}}{S}_{\mathrm{inc}}$, for incidence from a specified direction and matched polarization. It need not equal a literal geometric collecting area.

Gain–area relation. To derive the antenna gain-effective-area relation, take a passive reciprocal antenna. Its electromagnetic reciprocity makes the receive pattern proportional to the transmit pattern; write $A_{\mathrm{eff}}(\Omega )=KG(\Omega )$. The normalization follows from thermal equilibrium. In the Rayleigh-Jeans law limit, blackbody radiation radiance per unit frequency per unit solid angle is $2k_{B}T/{\lambda }^2$ for two polarizations. A single antenna polarization couples to one half, so the available incident noise power isThe open-circuit Johnson voltage spectrum of the antenna resistance is $4k_{B}T{R}_{\mathrm{rad}}d\nu$. A matched load receives half the open-circuit voltage, giving available power $k_{B}Td\nu$. In a lossless antenna at thermal equilibrium these powers agree; hence $\int A_{\mathrm{eff}}d\Omega ={\lambda }^2$. Since $\int Gd\Omega =4\pi$,The same efficiency factor multiplies receiving area and gain if reciprocal dissipative losses are included consistently. Impedance mismatch or polarization mismatch reduces delivered power further. The factor of two for thermal polarization modes and the factor of four between open-circuit voltage variance and matched-load power are separate effects.

Normalize the radar beam. Interpret the azimuth as signed about the beam centre at $\varphi =0$, so both sides of the narrow lobe are included. Put $u=\theta -\pi /2$. To leading order in the small angular width, $\sin \theta \simeq 1$ and the angular integrals can be extended over the real line:The Gaussian antenna beam solid angle gives the peak on the positive $x$-axis consequently hasHere $N$ is power per solid angle, equal to $r^2{S}_r$ in the far field. The leading neglected solid-angle correction is relative order ${\alpha }^2$.

Return using the supplied electric-dipole response. The electric-dipole radar cross section of a small conducting sphere determines this return. At target range $D$, the incident intensity is $S_t=C/D^2$. Write its root mean square electric field as $E_t$, with $S_t={ϵ}_{0}c{E}_t^2$. The induced electric dipole is $p_{\mathrm{rms}}=4\pi {ϵ}_0{a}^3{E}_t$. A dipole's angular radiation pattern, normalized by $\int {\sin }^2\chi d\Omega =8\pi /3$, gives $d{P}_{\mathrm{sc}}/d\Omega =(3P_{\mathrm{sc}}/8\pi ){\sin }^2\chi$. The backward direction is perpendicular to the induced dipole, so the supplied total dipole power givesThe received intensity is this angular power divided by $D^2$. Using the monostatic radar equation, or directly multiplying by the receive area, givesThe receiver threshold is a voltage across the matched load, so the detectable power is $V_m^2/R$, without another matching factor:This is the electric-dipole result following from the response data provided. It assumes $ka\ll 1$, far-field propagation on each leg, alignment at the beam maximum and no attenuation or additional receiver losses.

Perfect-conductor qualification. perfect-conductor sphere dipole interference arises because the sphere also excludes the oscillating magnetic field. Solving the magnetic boundary problem gives $\mathbf{m}=-2\pi a^3{\mathbf{H}}_t$ in addition to the electric dipole. The magnetic polarizability of a perfectly conducting sphere follows from its induced surface currents.^[ref] In the backward direction the magnetic contribution has half the electric scattering amplitude and the same sign, so the returned intensity is $9/4$ of the electric-only value. Equivalently, the monostatic radar cross section is ${\sigma }_{\mathrm{back}}=9\pi k^4{a}^6$, rather than $4\pi k^4{a}^6$. Including both leading dipoles therefore givesBoth results have been distinguished because the supplied auxiliary data describe only the electric dipole, while the literal perfect-conductor boundary conditions also generate the magnetic one. The backward interference is a leading-order effect, not a higher-order correction in $ka$.^[ref]

Antenna gain in direction $(\theta ,\varphi )$ compares the far-field power per solid angle with an isotropic radiator using the same accepted input power:It includes radiation efficiency; for the lossless dipole $P_{\mathrm{in}}=P_{\mathrm{rad}}$. The antenna effective area is defined for incidence from a specified direction with matched polarization and conjugate impedance matching:Its units are square metres and it need not equal the physical metal area.

For ordinary reciprocal antennas, electromagnetic reciprocity makes the coupling between two matched ports the same when transmitter and receiver are exchanged. Free-space transfer in one direction is proportional to $G_1{A}_2$, while the reverse transfer is proportional to $G_2{A}_1$. Thus $A/G$ must be a universal factor at that wavelength, independent of antenna or direction. The supplied relation fixes it asEquivalently, thermal detailed balance prevents an angular receiving pattern from collecting more radiation than its reciprocal emitting pattern would return in equilibrium. Reciprocity and the specified matching are the assumptions behind this statement; it is not a general theorem for active nonreciprocal devices.

The transmitting antenna is matched to its input transmission line, so its real terminal electrical resistance is $Z_i$. The identical receiving antenna has the same electrical impedance at the operating frequency. Its available received power for a matched load is the incident energy flux times its antenna effective area:The real receiving load is $Z_t$, so its voltage-wave reflection coefficient relative to $Z_i$ is ${\Gamma }_L=(Z_t-Z_i)/(Z_t+Z_i)$. The fraction of available power delivered to the output line isConsequently, neglecting path losses, mismatch of polarization of an electromagnetic wave and line losses,This is the Friis transmission equation with a receiving impedance-mismatch factor. For fixed $P_i,G_a,d,Z_i,Z_t$, it scales as ${\lambda }^2$. That scaling is conditional on holding antenna gain fixed: a fixed physical aperture can instead have $G_a\propto {\lambda }^{-2}$, and any wavelength dependence of the impedances also enters the expression. The matched-output check $Z_t=Z_i$ gives $P_t=P_i\mathcal{F}$.

The Johnson-Nyquist noise formula gives the open-circuit mean-square voltage in a small frequency band $d\nu$. When a source electrical resistance $R$ drives a load $R$ with impedance matching, the load sees half the source voltage, so the available noise power isThe factor of four in the open-circuit formula is therefore not an extra factor in the delivered power.

The transmitter's antenna gain $G$ makes the axial spectral energy flux at range $d$Multiplication by the receiving antenna effective area $A$ gives $d{P}_r=k_B{T}_{t}GAd\nu /(4\pi d^2)$. Define the equivalent receiving antenna temperature $T_A$ by the available matched-load noise $d{P}_r=k_B{T}_{A}d\nu$. ThenThis is the transmitter-induced equivalent noise temperature of the receiver port. It does not determine the actual physical temperature of its internal electrical resistance, or any additional receiver and background noise. The calculation assumes the usual far-field link, negligible path loss beyond geometric spreading, matched polarization of an electromagnetic wave at the two antennas and the stated effective collecting area. No additional one-half polarization factor is needed for the matched polarization of radiation from one transmitting antenna.

Detection and inference about remote objects using transmitted radio waves and their returned electromagnetic radiation. A received signal depends on propagation, directional antenna gain, target scattering and receiver sensitivity.

The fraction of accepted input power radiated by an antenna. With consistent feed definitions, antenna gain is radiation efficiency times antenna directivity. Ohmic losses reduce the efficiency, while feed mismatch is usually tracked separately.