Chapter 3 Radio Telescopes and Radiometers

The simplest antenna is a short (total length $l$ much smaller than one wavelength $\lambda$) dipole antenna, which is shown in Figure 3.1 as two collinear conductors (e.g., wires or conducting rods). When they are driven at the small gap between them by an oscillating current source (a transmitter), the current going into the bottom conductor is 180 degrees out of phase with the current going into the top conductor. The radiation from a dipole depends on the transmitter frequency, so consider a sinusoidal driving current $I$ with angular frequency $\omega \equiv 2\pi \nu$:

$$I=I_0\cos (\omega t),$$

(3.1)

where $I_0$ is the peak current going into each half of the dipole. It is computationally convenient to replace the trigonometric function $\cos (\omega t)$ with its complex exponential equivalent (Appendix B.3), the real part of

$$\boxed{e^{-i\omega t}=\cos (\omega t)-i\sin (\omega t),}$$

(3.2)

so the driving current can be rewritten as

$$I=I_0{e}^{-i\omega t}$$

(3.3)

with the implicit understanding that only the real part of $I$ represents the actual current. The driving current accelerates charges in the antenna wires, so Larmor’s formula can be used to calculate the radiation from the antenna by converting from charges and accelerations to time-varying currents.

The electric current in a wire is defined as the flow rate of electric charge along the wire:

$$\boxed{I\equiv \frac{dq}{dt}.}$$

(3.4)

For a wire on the $z$-axis,

$$I=\frac{dq}{dt}=\frac{dq}{dz}\frac{dz}{dt}=\frac{dq}{dz}v,$$

(3.5)

where $v$ is the instantaneous flow velocity of the charges.

Many people incorrectly believe that the velocity $v$ of individual electrons in a wire is comparable with the speed of light $c$ because electrical signals do travel down wires at nearly the speed of light. However, a wire filled with electrons is like a garden hose already filled with water, a nearly incompressible fluid. When the faucet is turned on, water flows from the other end of a full hose almost immediately, even though individual water molecules have moved only a short distance along the hose. The example above shows that electrons move so slowly in a wire that Larmor’s nonrelativistic equation accurately predicts the radiation from antennas.

Example. Estimate the speed of electrons flowing through a copper wire of cross section $\sigma =1$ mm${}^2={10}^{-6}$ m${}^2$ and carrying a current of 1 ampere. The number density of free electrons is about equal to the number density of copper atoms in the wire, $n\approx {10}^{29}$ m${}^{-3}$. In MKS units, the charge of an electron is $$-e\approx 4.80\times {10}^{-10}\mathrm{statcoul}\times \frac{1\mathrm{coul}}{3\times {10}^9\mathrm{statcoul}}\approx 1.60\times {10}^{-19}\mathrm{coul}.$$ One ampere is defined as one coulomb per second, so the number of electrons flowing past any point along the wire in one second is $$\dot{N}=\frac{I}{|e|}\approx \frac{1\mathrm{coul}{\mathrm{s}}^{-1}}{1.60\times {10}^{-19}\mathrm{coul}}\approx 6.25\times {10}^{18}{\mathrm{s}}^{-1}.$$ The average electron velocity is only $$v\approx \frac{\dot{N}}{\sigma n}\approx \frac{6.25\times {10}^{18}{\mathrm{s}}^{-1}}{{10}^{-6}{\mathrm{m}}^2\cdot {10}^{29}{\mathrm{m}}^{-3}}\approx 6\times {10}^{-5}\mathrm{m}{\mathrm{s}}^{-1}\ll c.$$ Thus the nonrelativistic Larmor equation may safely be used to calculate the radiation from a wire.

Equation 2.136 from the derivation of Larmor’s formula

$$E_{\perp }=\frac{q\dot{v}\sin \theta }{r{c}^2}$$

can be applied to yield the $d{E}_{\perp }$ contributed by each infinitesimal dipole segment of length $dz$. If the dipole is short ($l\ll \lambda$), all of these electric fields are in phase and add directly to give the total $E_{\perp }$ produced by the dipole:

$$E_{\perp }={\int }_{z=-l/2}^{+l/2}\frac{dq}{dz}𝑑z\frac{\dot{v}\sin \theta }{r{c}^2}.$$

(3.6)

At distances $r\gg l$, ($1/r$) is nearly constant over the whole antenna and can be taken outside the integral. For a sinusoidal driving current, $\dot{v}=-i\omega v$ and

$$E_{\perp }=\frac{-i\omega \sin \theta }{r{c}^2}{\int }_{-l/2}^{+l/2}\frac{dq}{dz}v𝑑z=\frac{-i\omega \sin \theta }{r{c}^2}{\int }_{-l/2}^{+l/2}I𝑑z.$$

(3.7)

That is, the radiated electric field strength $E_{\perp }$ is proportional to the integral of the current distribution along the antenna. The current at the center is the driving current $I=I_0{e}^{-i\omega t}$, and the current must drop to zero at the ends of the antenna, where the conductivity goes to zero. The current distribution along a short dipole is the tail end of a standing-wave sinusoid, which declines almost linearly from the driving current at the center to zero at the ends:

$$I(z)\approx I_0{e}^{-i\omega t}\left[1-\frac{|z|}{(l/2)}\right].$$

(3.8)

Then

$${\int }_{-l/2}^{+l/2}I𝑑z\approx \frac{I_{0}l}{2}{e}^{-i\omega t}$$

(3.9)

and

$$E_{\perp }\approx \frac{-i\omega \sin \theta }{r{c}^2}\frac{I_{0}l}{2}{e}^{-i\omega t}.$$

(3.10)

Substituting $\omega =2\pi c/\lambda$ gives

$$E_{\perp }\approx \frac{-i2\pi c\sin \theta }{\lambda r{c}^2}\frac{I_{0}l}{2}{e}^{-i\omega t}=\frac{-i\pi \sin \theta }{c}\frac{I_{0}l}{\lambda }\frac{e^{-i\omega t}}{r}.$$

(3.11)

The time-averaged Poynting flux (power per unit area) follows from Equation 2.139; it is

$$⟨S⟩=\frac{c}{4\pi }⟨E_{\perp }^2⟩.$$

(3.12)

Thus

$$⟨S⟩=\frac{c}{4\pi }\left(\frac{1}{2}\right){\left(\frac{I_{0}l}{\lambda }\frac{\pi }{c}\right)}^2\frac{{\sin }^2\theta }{r^2},$$

(3.13)

where the factor $(1/2)$ reflects the fact that $⟨{\sin }^2(\omega t)⟩=⟨{\cos }^2(\omega t)⟩=1/2$. (This is a good relation to remember and an easy one to derive because ${\sin }^2(\omega t)+{\cos }^2(\omega t)=1$ and $⟨{\sin }^2(\omega t)⟩=⟨{\cos }^2(\omega t)⟩$.)

The power pattern of a transmitting antenna is the angular distribution of its radiated power, often normalized to unity at the peak. From Equation 3.13 the normalized power pattern of a short dipole is

$$\boxed{P\propto {\sin }^2\theta .}$$

(3.14)

The radiation from a short dipole has the same polarization and the same doughnut-shaped power pattern as Larmor radiation from an accelerated charge because all of the charges in the short dipole are being accelerated along one line much shorter than one wavelength. From the observer’s point of view, the power received depends only on the projected (perpendicular to the line of sight) length $l\sin \theta$ of the dipole. The electric field strength received is proportional to the apparent length of the dipole, and the radiation from the dipole is linearly polarized parallel to the projected dipole. The time-averaged total power emitted is obtained by integrating the Poynting flux over the surface area of a sphere of any radius $r\gg l$ centered on the antenna:

	$⟨P⟩={\displaystyle \int ⟨S⟩𝑑A}$	$={\displaystyle \frac{c}{4\pi }}\left({\displaystyle \frac{1}{2}}\right){\left({\displaystyle \frac{I_{0}l}{\lambda }}{\displaystyle \frac{\pi }{c}}\right)}^2{\displaystyle {\int }_{\varphi =0}^{2\pi }}{\displaystyle {\int }_{\theta =0}^{\pi }}{\displaystyle \frac{{\sin }^2\theta }{r^2}}r\sin \theta d\varphi rd\theta$		(3.15)
		$={\displaystyle \frac{c}{4\pi }}\left({\displaystyle \frac{1}{2}}\right){\left({\displaystyle \frac{I_{0}l}{\lambda }}{\displaystyle \frac{\pi }{c}}\right)}^{2}2\pi {\displaystyle {\int }_{\theta =0}^{\pi }}{\sin }^3\theta d\theta .$		(3.16)

Recall that ${\int }_0^{\pi }{\sin }^3\theta d\theta =4/3$, so the time-averaged power radiated by a short dipole is

$$\boxed{⟨P⟩=\frac{{\pi }^2}{3c}{\left(\frac{I_{0}l}{\lambda }\right)}^2,}$$

(3.17)

where $I_0\cos (\omega t)$ is the driving current and $l/\lambda$ is the total length of the dipole in wavelengths.

Most practical dipoles are half-wave dipoles ($l\approx \lambda /2$) because half-wave dipoles are resonant, meaning that they provide a nearly resistive load to the transmitter. When each half of the dipole is $\lambda /4$ long, the standing-wave current is highest at the center and naturally falls as $I=I_0\cos (2\pi z/\lambda )$ to zero at the ends of the conductors.

The ground-plane vertical antenna shown in Figure 3.2 is very similar to the dipole. A ground-plane vertical is one half of a dipole above a conducting plane, which is called a “ground plane” because historically the conducting plane for vertical antennas was the surface of the Earth. The transmitter is connected between the base of the vertical, which is insulated from the ground, and the ground plane near the base. Many AM broadcast transmitting antennas are tall (at $\nu \sim 1$ MHz, $\lambda \sim 300$ m and a $\lambda /4$ vertical antenna is about 75 m high), insulated towers acting as quarter-wave verticals. The conducting ground plane is a mirror that creates the lower half of the dipole as the mirror image of the upper half. Electric fields produced by the vertical antenna induce currents in the conducting plane to make the horizontal component of the electric field go to zero on the conductor. The virtual electric fields from the image vertical have the same amplitude but are 180 degrees out of phase, exactly as in a half-wave dipole. Consequently the radiation field from a ground-plane vertical is identical to that of a dipole in the half space above the ground plane and zero below the ground plane.

According to the strict definition of an antenna as a device for converting between electromagnetic waves in space and currents in conductors, the only antennas in most radio telescopes are half-wave dipoles and their relatives, quarter-wave ground-plane verticals. The large parabolic reflector of a radio telescope serves only to focus plane waves onto the feed antenna. (The term “feed” comes from radar antennas used for transmitting; the “feed” antenna feeds transmitter power to the main reflector. Receiving antennas used in radio astronomy work the other way around, and the “feed” actually collects radiation from the reflector.)

Actual half-wave dipoles, backed by small reflectors about $\lambda /4$ behind them to focus the dipole pattern in the direction of the main dish, are normally used as feeds at low frequencies ( GHz) or long wavelengths ($\lambda >0.3$ m) because of their relatively small size. However, the radiation patterns of half-wave dipoles backed by small reflectors are not well matched to most parabolic dishes, so their performance is less than optimum.

For shorter wavelengths, almost all radio-telescope feeds are quarter-wave ground-plane verticals inside waveguide horns. Radiation entering the relatively large (size $>\lambda$) rectangular or circular aperture of the tapered horn is concentrated into a rectangular or circular waveguide with parallel conducting walls. In the case of the rectangular waveguide whose cross section is shown in Figure 3.3, the side walls are separated by slightly over $\lambda /2$ so that vertical electric fields can travel down the waveguide with low loss. The top and bottom walls are separated by somewhat less than $\lambda /2$ so only the mode with vertical electric fields can propagate (Section 3.4). The $\lambda /4$ vertical antenna inserted through a small hole in the bottom wall collects most of this vertically polarized radiation and converts it into an electric current that travels down the coaxial cable to the receiver. The backshort wall about 1/4 of the guide wavelength ${\lambda }_{\mathrm{w}}$ (Equation 3.144) behind the dipole ensures that the dipole sees only radiation coming from the direction of the horn opening.

Both dipoles and quarter-wave verticals are linearly polarized feeds. The voltage response of a linearly polarized feed to a linearly polarized source is proportional to $\cos \mathrm{\Delta }$, where $\mathrm{\Delta }$ is the angle between the feed and the source electric field, and the power response is proportional to ${\cos }^2\mathrm{\Delta }=[\cos (2\mathrm{\Delta })+1]/2$. Consequently the degree of polarization and the polarization position angle of a partially linearly polarized radio source can be measured by rotating the linearly polarized feed of a radio telescope while tracking the source. The degree of polarization $p$ of a partially linearly polarized source defined by Equation 2.58 is

$$p\equiv \frac{I_{\mathrm{p}}}{I}=\frac{I_{\mathrm{p}}}{I_{\mathrm{p}}+I_{\mathrm{u}}},$$

(3.18)

where $I_{\mathrm{p}}$ is the polarized flux density and $I=I_{\mathrm{p}}+I_{\mathrm{u}}$ is the total flux density of the source. The power response of the feed $R(\mathrm{\Delta })\propto I_{\mathrm{p}}{\cos }^2\mathrm{\Delta }+I_{\mathrm{u}}/2$ will be $R_{\parallel }\propto I_{\mathrm{p}}+I_{\mathrm{u}}/2$ when the feed and source polarizations are parallel ($\mathrm{\Delta }=0$) and $R_{⟂}=I_{\mathrm{u}}/2$ when they are perpendicular ($\mathrm{\Delta }=\pi /2$). In terms of the observables $R_{\parallel }$ and $R_{⟂}$, the degree of polarization is

$$p=\frac{I_{\mathrm{p}}}{I_{\mathrm{p}}+I_{\mathrm{u}}}=\frac{I_{\mathrm{p}}+I_{\mathrm{u}}/2-I_{\mathrm{u}}/2}{I_{\mathrm{p}}+I_{\mathrm{u}}/2+I_{\mathrm{u}}/2}=\frac{R_{\parallel }-R_{⟂}}{R_{\parallel }+R_{⟂}}.$$

(3.19)

Figure 3.4 shows how the relative power output $R(\mathrm{\Delta })$ of a linearly polarized feed varies as it is rotated through $\mathrm{\Delta }=\pi$ radians relative to the polarization position angle of sources with fractional polarizations $p=1.0$, 0.1, and 0.0.

To measure all four Stokes parameters of an arbitrarily polarized source, it is necessary to combine the voltage outputs of two orthogonally polarized feeds. For example, two orthogonal quarter-wave verticals can be inserted into a square waveguide to receive both the horizontally and the vertically polarized components simultaneously. If their output voltages are added in phase (phase difference $\delta ={\varphi }_x-{\varphi }_y=0$ in Figure 2.15), the feed combination will respond to radiation linearly polarized in position angle $\pi /4={45}^{\circ }$. If a phase difference $\delta =\pi /2$ is inserted either mechanically (by moving one feed $\lambda /4$ behind the other) or electrically (by inserting a $\lambda /4$ longer cable between one feed and the point where the two outputs are added), then $\delta =\pi /2$ and the feed combination will respond to circular polarization.

The power flowing through a circuit is

$$\boxed{P=VI,}$$

(3.20)

where $V$ is the voltage (defined as energy per unit charge) and $I$ is the current (defined as the charge flowing through the circuit per unit time), so $P$ has dimensions of energy per unit time. The physicist George Simon Ohm observed that the current flowing through most (but not all) materials is proportional to the applied voltage, so most objects have a well-defined resistance $R$ defined by Ohm’s law,

$$\boxed{R\equiv \frac{V}{I}.}$$

(3.21)

When Ohm’s law holds,

$$\boxed{P=I^{2}R=\frac{V^2}{R}.}$$

(3.22)

The average power in a resistive circuit with time-varying currents is

$$⟨P⟩=⟨I^2⟩R.$$

(3.23)

In the particular case of sinusoidal currents $I=I_0\cos (\omega t)$ and $⟨I^2⟩=I_0^2/2$, so

$$⟨P⟩=\frac{I_0^{2}R}{2}.$$

(3.24)

Thus the (frequency-dependent) radiation resistance of an antenna is defined by

$$\boxed{R\equiv \frac{2⟨P⟩}{I_0^2}.}$$

(3.25)

For a short dipole, the power emitted is given by Equation 3.17 and the radiation resistance is

$$R=\frac{2{\pi }^2}{3c}{\left(\frac{l}{\lambda }\right)}^2.$$

(3.26)

A ground-plane vertical of height $l/2$ emits exactly like a dipole of length $l$ above the ground plane and nothing below the ground plane. Thus the total power emitted by the vertical is half the power emitted by the dipole, and the radiation resistance of the vertical is half the radiation resistance of the dipole.

The radiation resistance ${𝑹}_{\mathrm{𝟎}}$ of free space (sometimes called the impedance $Z_0$ of free space) can be obtained from the relations

$$|\overrightarrow{S}|=\frac{c}{4\pi }{E}^2\mathit{\hspace{1em}\hspace{0.25em}}\mathrm{and}\mathit{\hspace{1em}\hspace{0.25em}}P=\frac{V^2}{R}.$$

(3.27)

The electric field $E$ is just the voltage per unit length $V/l$ and the flux is the power per unit area $l^2$, so

$$|\overrightarrow{S}|=\frac{c{V}^2}{4\pi l^2}=\frac{V^2}{R_0{l}^2}$$

(3.28)

and

$$R_0=\frac{4\pi }{c}=\frac{4\pi }{3\times {10}^{10}\mathrm{cm}{\mathrm{s}}^{-1}}=4.19\times {10}^{-10}\mathrm{s}{\mathrm{cm}}^{-1}.$$

(3.29)

Converting from CGS to MKS units yields the radiation resistance of space in ohms:

$$R_0=\frac{4\pi }{3\times {10}^{10}\mathrm{cm}{\mathrm{s}}^{-1}\cdot 1/9\times {10}^{-11}\mathrm{s}{\mathrm{cm}}^{-1}{\mathrm{\Omega }}^{-1}}=120\pi \mathrm{\Omega }\approx 377\mathrm{\Omega }.$$

(3.30)

The tapered opening of a waveguide horn feed (Figure 3.3) acts as an impedance transformer to match the impedance of the waveguide to the impedance of free space to minimize standing waves and couple power efficiently between the waveguide and space, just as the bell of a trombone is an acoustic transformer matching sound vibrations of air in the trombone to the outside environment.

A black hole is a perfect absorber of radiation, so its resistance must also be $120\pi \mathrm{\Omega }$ to match that of free space. A black hole spinning in an external magnetic field can generate electrical power with a voltage/current ratio of $120\pi \mathrm{\Omega }$, and this process may be important in powering quasar jets [12].

The power gain $G(\theta ,\varphi )$ of a transmitting antenna is defined as the power transmitted per unit solid angle in direction $(\theta ,\varphi )$ relative to an isotropic antenna, which has the same gain in all directions. Frequently, the value of $G$ is expressed logarithmically in units of decibels (dB):

$$\boxed{G(\mathrm{dB})\equiv 10{\log }_{10}(G).}$$

(3.31)

For any lossless antenna, energy conservation requires that the gain averaged over all directions be

$$\boxed{⟨G⟩=1.}$$

(3.32)

Consequently, all lossless antennas obey

$$\boxed{{\int }_{\mathrm{sphere}}Gd\mathrm{\Omega }=4\pi .}$$

(3.33)

Different lossless antennas may radiate with different directional patterns, but they do not alter the total amount of power radiated. Consequently, the gain of a lossless antenna depends only on the angular distribution of radiation from that antenna. In general, an antenna having peak gain $G_0$ must beam most of its power into a solid angle $\mathrm{\Delta }\mathrm{\Omega }$ such that $\mathrm{\Delta }\mathrm{\Omega }\approx 4\pi /G_0$. This motivates the definition of the beam solid angle ${\mathrm{\Omega }}_{\mathrm{A}}$:

$${\mathrm{\Omega }}_{\mathrm{A}}\equiv \frac{4\pi }{G_0}.$$

(3.34)

Thus the higher the gain, the smaller the beam solid angle.

The antenna efficiency $\eta$ is defined as the ratio of radiated power to input power. If ohmic losses reduce $\eta$, then the gain $G$ in Equations 3.31 through 3.34 should be replaced by the directivity defined by $D\equiv G/\eta$.

Example. What is the power gain of a lossless short dipole? It is sufficient to recall only the angular dependence of the short-dipole power pattern (Equation 3.14) $$P\propto {\sin }^2\theta ,$$ where $\theta$ is the angle from the dipole axis. Thus $$G\propto {\sin }^2\theta =G_0{\sin }^2\theta .$$ The maximum gain $G_0$ is determined by energy conservation: $${\int }_{\mathrm{sphere}}G𝑑\mathrm{\Omega }={\int }_{\varphi =0}^{2\pi }{\int }_{\theta =0}^{\pi }{G}_0{\sin }^2\theta d\theta \sin \theta d\varphi =4\pi ,$$ $$2\pi G_0{\int }_0^{\pi }{\sin }^3\theta d\theta =4\pi .$$ Recall that ${\int }_0^{\pi }{\sin }^3\theta d\theta =4/3$ so $$G_0=\frac{4\pi }{2\pi }\frac{3}{4}=\frac{3}{2}$$ and $$G(\theta ,\varphi )=\frac{3{\sin }^2\theta }{2}.$$ Expressed in dB, the maximum gain $G_0$ of a short dipole is $$G_0=10{\log }_{10}(3/2)\approx 1.76\mathrm{dB}.$$ Note that $G(\theta ,\varphi )$ is nearly independent of the antenna length so long as $l\ll \lambda$ because the power pattern of a short dipole is nearly independent of $l$. Varying $l\ll \lambda$ affects only the radiation resistance.

The receiving counterpart of transmitting power gain is the effective area or effective collecting area of a receiving antenna. Imagine an ideal antenna of geometric area $A$ that could collect all of the radiation falling on it from a distant point source and convert it to electrical power—a “rain gauge” for collecting photons. The total spectral power incident on the antenna is the product of its geometric area and the incident spectral power per unit area, or flux density $S_{\nu }$. However, any single antenna can respond to only one polarization, so its output $P_{\nu }$ can equal all of the input spectral power ($P_{\nu }=A{S}_{\nu }$) from a fully polarized source whose polarization matches that of the antenna, but only half of the incident power ($P_{\nu }=A{S}_{\nu }/2$) from an unpolarized source and nothing at all from an orthogonally polarized source. The output of a real antenna is always smaller than this and most radio sources are nearly unpolarized, so radio astronomers find it useful to define the effective collecting area $A_{\mathrm{e}}$ of an antenna whose output spectral power is $P_{\nu }$ in response to an unpolarized point source of total flux density $S_{\nu }$ by

$$\boxed{A_{\mathrm{e}}\equiv \frac{2P_{\nu }}{S_{\nu }}.}$$

(3.35)

The average collecting area

$$⟨A_{\mathrm{e}}⟩\equiv \frac{{\int }_{4\pi }{A}_{\mathrm{e}}𝑑\mathrm{\Omega }}{{\int }_{4\pi }𝑑\mathrm{\Omega }}=\frac{1}{4\pi }{\int }_{4\pi }{A}_{\mathrm{e}}𝑑\mathrm{\Omega }$$

(3.36)

of any lossless antenna can be calculated via another thermodynamic thought experiment.

Imagine an antenna inside a cavity in full thermodynamic equilibrium at temperature $T$ connected through a transmission line to a matched resistor (whose resistance equals the radiation resistance of the antenna) in a second cavity at the same temperature (Figure 3.5). A filter between the cavities passes only currents in a narrow range of frequencies between $\nu$ and $\nu +d\nu$. Because this entire system is in thermodynamic equilibrium, no net power can flow through the wires connecting the antenna and the resistor. Otherwise, one cavity would heat up and the other would cool down, in violation of the second law of thermodynamics. The total spectral power $P_{\nu }$ from all directions collected in one polarization is half the total spectral power in the unpolarized blackbody radiation, so

$$P_{\nu }=\frac{1}{2}{\int }_{4\pi }{A}_{\mathrm{e}}(\theta ,\varphi )B_{\nu }𝑑\mathrm{\Omega }$$

(3.37)

must equal the Nyquist spectral power $P_{\nu }$ produced by the resistor. Inserting the Nyquist formula from Equation 2.119 and Planck’s law from Equation 2.85,

$$P_{\nu }=kT\left[\frac{{\displaystyle \frac{h\nu }{kT}}}{\exp \left({\displaystyle \frac{h\nu }{kT}}\right)-1}\right]\mathit{\hspace{1em}\hspace{1em}}\mathrm{and}\mathit{\hspace{1em}\hspace{1em}}{B}_{\nu }=\frac{2kT}{{\lambda }^2}\left[\frac{{\displaystyle \frac{h\nu }{kT}}}{\exp \left({\displaystyle \frac{h\nu }{kT}}\right)-1}\right]$$

(3.38)

leads to

$$kT=\frac{2kT}{2{\lambda }^2}{\int }_{4\pi }{A}_{\mathrm{e}}(\theta ,\varphi )𝑑\mathrm{\Omega },$$

(3.39)

and finally,

$${\int }_{4\pi }{A}_{\mathrm{e}}(\theta ,\varphi )𝑑\mathrm{\Omega }=4\pi ⟨A_{\mathrm{e}}⟩={\lambda }^2.$$

(3.40)

Without using Maxwell’s equations we have obtained the remarkable result

$$\boxed{⟨A_{\mathrm{e}}⟩=\frac{{\lambda }^2}{4\pi }}$$

(3.41)

which implies that all lossless antennas, from tiny dipoles to the 100-m diameter Green Bank Telescope (GBT), have the same average collecting area. $⟨A_{\mathrm{e}}⟩$ is proportional to ${\lambda }^2$ because space has two more dimensions than a transmission line has.

The collecting area of an isotropic receiving antenna is proportional to ${\lambda }^2$, so most satellite broadcast services, GPS (Global Positioning System) or satellite FM radio for example, operate at relatively long wavelengths (10 to 20 cm). Likewise, practical radio telescopes constructed from arrays of dipoles are reasonably sensitive only at long wavelengths.

By analogy with Equation 3.34, the beam solid angle of a lossless receiving antenna is defined as

$${\mathrm{\Omega }}_{\mathrm{A}}\equiv {\int }_{4\pi }\frac{A_{\mathrm{e}}(\theta ,\varphi )}{A_0}𝑑\mathrm{\Omega },$$

(3.42)

where $A_0$ is the maximum effective collecting area, so

$$\boxed{A_0{\mathrm{\Omega }}_{\mathrm{A}}={\lambda }^2.}$$

(3.43)

The much larger peak collecting area of the GBT implies it has a much smaller beam solid angle ${\mathrm{\Omega }}_{\mathrm{A}}$.

Many antenna properties are the same for both transmitting and receiving. It is often easier to calculate the gain of a transmitting antenna than the collecting area of a receiving antenna, and it is often easier to measure the receiving power pattern of a large radio telescope than to measure its transmitting power pattern. Thus this receiving/transmitting “reciprocity” greatly simplifies antenna calculations and measurements. Reciprocity can be understood via Maxwell’s equations or by thermodynamic arguments.

Burke and Graham-Smith [20] state the electromagnetic case for reciprocity clearly: “An antenna can be treated either as a receiving device, gathering the incoming radiation field and conducting electrical signals to the output terminals, or as a transmitting system, launching electromagnetic waves outward. These two cases are equivalent because of time reversibility: the solutions of Maxwell’s equations are valid when time is reversed.”

The strong reciprocity theorem states,

If a voltage is applied to the terminals of an antenna A and the current is measured at the terminals of another antenna B, then an equal current (in both amplitude and phase) will appear at the terminals of A if the same voltage is applied to B. (Figure 3.6)

It can be formally derived from Maxwell’s equations (see a partial derivation in Wilson et al. [116, Appendix D]) or by network analysis (see Kraus et al. [63, “Antennas”, p. 252]).

Most radio astronomical applications do not depend on the detailed phase relationships of voltages and currents, so it is sufficient to use a weak reciprocity theorem that relates the angular dependences of the transmitting power pattern and the receiving collecting area of any antenna: “The power pattern of an antenna is the same for transmitting and receiving”; that is,

$$\boxed{G(\theta ,\varphi )\propto A_{\mathrm{e}}(\theta ,\varphi ).}$$

(3.44)

The weak reciprocity theorem can be proven by another simple thermodynamic thought experiment: An antenna is connected to a matched load inside a cavity initially in equilibrium at temperature $T$. The antenna simultaneously receives power from the cavity walls and transmits power generated by the resistor. The total power transmitted in all directions must equal the total power received from all directions because no net power can be transferred between the antenna and the resistor; otherwise the resistor would not remain at temperature $T$. Moreover, in any direction, the power received and transmitted by the antenna must be the same, else the cavity wall in directions where the transmitted power was greater than the received power would rise in temperature and the cavity wall in directions of lower transmitted/received power ratio would cool, leading to a violation of the second law of thermodynamics.

The constant of proportionality relating $G$ and $A_{\mathrm{e}}$ can be derived from Equations 3.32 and 3.41:

$$⟨A_{\mathrm{e}}⟩=\frac{{\lambda }^2}{4\pi }\mathit{\hspace{1em}\hspace{1em}}\mathrm{and}\mathit{\hspace{1em}\hspace{1em}}⟨G⟩=1.$$

(3.45)

Thus energy conservation and the weak reciprocity theorem imply

$$\boxed{A_{\mathrm{e}}(\theta ,\varphi )=\frac{{\lambda }^{2}G(\theta ,\varphi )}{4\pi }}$$

(3.46)

for any antenna. This extremely useful equation shows how to compute the receiving power pattern from the transmitting power pattern and vice versa.

Example. Use the transmitting power pattern of a short dipole (Equation 3.14) to calculate the effective collecting area of a short dipole used as a receiving antenna: $A_{\mathrm{e}}(\theta ,\varphi )=$ $\mathrm{ }{\displaystyle \frac{{\lambda }^{2}G(\theta ,\varphi )}{4\pi }}={\displaystyle \frac{{\lambda }^2}{4\pi }}{\displaystyle \frac{3{\sin }^2\theta }{2}},$ $A_{\mathrm{e}}=$ $\mathrm{ }{\displaystyle \frac{3{\lambda }^2{\sin }^2\theta }{8\pi }}.$ The effective collecting area of a short receiving dipole does not depend on the length $l$ of the dipole itself because the transmitting power pattern of a short ($l\ll \lambda$) dipole is independent of $l$.

A convenient practical unit for the power output per unit frequency from a receiving antenna is the antenna temperature $T_{\mathrm{A}}$. Antenna temperature has nothing to do with the physical temperature of the antenna as measured by a thermometer; it is only the temperature of a matched resistor whose thermally generated power per unit frequency in the low-frequency Nyquist approximation (Equation 2.117) equals that produced by the antenna:

$$\boxed{T_{\mathrm{A}}\equiv \frac{P_{\nu }}{k}.}$$

(3.47)

It is widely used for the following reasons:

1. 1.

   1 K of antenna temperature is a conveniently small power per unit bandwidth. $T_{\mathrm{A}}=1$ K corresponds to $P_{\nu }=k{T}_{\mathrm{A}}=1.38\times {10}^{-23}\mathrm{J}{\mathrm{K}}^{-1}\cdot 1\mathrm{K}=1.38\times {10}^{-23}\mathrm{W}{\mathrm{Hz}}^{-1}$.

2. 2.

   It can be calibrated by a direct comparison with hot and cold loads (another word for matched resistors) connected to the receiver input.

3. 3.

   The units of receiver noise are also K, so comparing the signal in K with the receiver noise in K makes it easy to compare the signal and noise powers.

Combining Equations 3.35 and 3.47 shows that an unpolarized point source of flux density $S$ increases the antenna temperature by

$$\boxed{T_{\mathrm{A}}=\frac{P_{\nu }}{k}=\frac{A_{\mathrm{e}}{S}_{\nu }}{2k},}$$

(3.48)

where $A_{\mathrm{e}}$ is the effective collecting area. It is often convenient to express the point-source sensitivity of a radio telescope in units of “kelvins per jansky” rather than in units of effective collecting area (m${}^2$). The effective collecting area corresponding to a sensitivity of $1\mathrm{K}{\mathrm{Jy}}^{-1}$ is

$$A_{\mathrm{e}}=\frac{2k{T}_{\mathrm{A}}}{S_{\nu }}=\frac{2\cdot 1.38065\times {10}^{-23}\mathrm{J}{\mathrm{K}}^{-1}\cdot 1\mathrm{K}}{{10}^{-26}\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{Hz}}^{-1}}=2761{\mathrm{m}}^2.$$

(3.49)

In an arbitrary radiation field $I_{\nu }(\theta ,\varphi )$, Equation 3.37 becomes

$$P_{\nu }=\frac{1}{2}{\int }_{4\pi }{A}_{\mathrm{e}}(\theta ,\varphi )I_{\nu }(\theta ,\varphi )𝑑\mathrm{\Omega }.$$

(3.50)

Replacing $P_{\nu }$ by antenna temperature $T_{\mathrm{A}}$ using Equation 3.47 and inserting $I_{\nu }(\theta ,\varphi )=2k{T}_{\mathrm{b}}(\theta ,\varphi )/{\lambda }^2$ (Equation 2.33) gives

	$k{T}_{\mathrm{A}}$	$={\displaystyle \frac{1}{2}}{\displaystyle {\int }_{4\pi }}{A}_{\mathrm{e}}(\theta ,\varphi ){\displaystyle \frac{2k}{{\lambda }^2}}{T}_{\mathrm{b}}(\theta ,\varphi )𝑑\mathrm{\Omega },$		(3.51)
	$T_{\mathrm{A}}$	$={\displaystyle \frac{1}{{\lambda }^2}}{\displaystyle {\int }_{4\pi }}{A}_{\mathrm{e}}(\theta ,\varphi )T_{\mathrm{b}}(\theta ,\varphi )𝑑\mathrm{\Omega }.$		(3.52)

In the limit of a very extended source having nearly constant $T_{\mathrm{b}}$ across the entire beam,

$$T_{\mathrm{A}}=\frac{T_{\mathrm{b}}}{{\lambda }^2}{\int }_{4\pi }{A}_{\mathrm{e}}(\theta ,\varphi )𝑑\mathrm{\Omega }$$

(3.53)

so

$$\boxed{T_{\mathrm{A}}=T_{\mathrm{b}}.}$$

(3.54)

In words, the antenna temperature produced by a smooth source much larger than the antenna beam equals the source brightness temperature.

If a lossless antenna is pointed at a compact source covering a solid angle ${\mathrm{\Omega }}_{\mathrm{s}}$ much smaller than the beam and having uniform brightness temperature $T_{\mathrm{b}}$, then

$$T_{\mathrm{A}}=\frac{A_0{T}_{\mathrm{b}}{\mathrm{\Omega }}_{\mathrm{s}}}{{\lambda }^2},$$

(3.55)

where $A_0$ is the on-axis effective collecting area. Substituting $A_0{\mathrm{\Omega }}_{\mathrm{A}}={\lambda }^2$ (Equation 3.43) gives the result:

$$\boxed{\frac{T_{\mathrm{A}}}{T_{\mathrm{b}}}=\frac{{\mathrm{\Omega }}_{\mathrm{s}}}{{\mathrm{\Omega }}_{\mathrm{A}}}.}$$

(3.56)

Stated in words, the antenna temperature equals the source brightness temperature multiplied by the fraction of the beam solid angle filled by the source. A $T_{\mathrm{b}}={10}^4$ K source covering 1% of the beam solid angle will add 100 K to the antenna temperature. The ratio $({\mathrm{\Omega }}_{\mathrm{s}}/{\mathrm{\Omega }}_{\mathrm{A}})$ is called the beam filling factor.

The main beam of an antenna is defined as the region containing the principal response out to the first zero; responses outside this region are called sidelobes or, very far from the main beam, stray radiation. The main beam solid angle ${\mathrm{\Omega }}_{\mathrm{MB}}$ is defined as

$$\boxed{{\mathrm{\Omega }}_{\mathrm{MB}}\equiv \frac{1}{G_0}{\int }_{\mathrm{MB}}G(\theta ,\varphi )d\mathrm{\Omega }.}$$

(3.57)

The fraction of the total beam solid angle lying inside the main beam is called the main beam efficiency or, loosely, the beam efficiency:

$$\boxed{{\eta }_{\mathrm{B}}\equiv \frac{{\mathrm{\Omega }}_{\mathrm{MB}}}{{\mathrm{\Omega }}_{\mathrm{A}}}.}$$

(3.58)

Antennas useful for radio astronomy at short wavelengths must have collecting areas much larger than the ${\lambda }^2/(4\pi )$ collecting area of an isotropic antenna and much higher angular resolution than a short dipole provides. Because arrays of dipoles are impractical at wavelengths  m or so, most radio telescopes use large reflectors to collect and focus power onto their small feed antennas, such as waveguide horns or dipoles backed by small reflectors, that are connected to receivers. The most common reflector shape is a paraboloid of revolution because it can focus the plane wave from a distant point source onto a single focal point.

To focus plane waves onto a single point, the reflector must keep all parts of an on-axis plane wavefront in phase at its focal point. Thus the total path lengths to the focus must all be the same, and this requirement is sufficient to determine the shape of the desired reflecting surface. Clearly the surface must be rotationally symmetric about its axis. In any plane containing the axis, the surface looks like the curve in Figure 3.7.

The requirement of constant path length can be written by equating the on-axis path length $(f+h)$ from any height $h$ to the reflector and then back to the prime focus at height $f$ with the off-axis path length:

$$(f+h)=\sqrt{r^2+{(f-z)}^2}+(h-z).$$

(3.59)

This yields the reflector height $z$ as a function of radius $r$:

	$$\sqrt{r^2+{(f-z)}^2}=f+z,$$
	$$r^2+f^2+z^2-2fz=f^2+z^2+2fz;$$

the result is

$$\boxed{z=\frac{r^2}{4f}.}$$

(3.60)

This is the equation of a paraboloid with focal length $f$.

The ratio of the focal length $f$ to the diameter $D$ of the reflector is called the $𝒇\mathbf{/}𝑫$ ratio or focal ratio. Note that the gain, collecting area, and beamwidth of a reflector antenna depend only weakly and indirectly on $f/D$, via the effect of $f/D$ on illumination taper. In principle, $f/D$ is a free parameter for the telescope designer, but in practice it is constrained. If the reflector $f/D$ is too high, the support structure needed to hold the feed or subreflector at the focus of a large radio telescope becomes very long and unwieldy. Consequently large radio telescopes usually have $f/D\approx 0.4$, an order-of-magnitude lower than the typical focal ratio of an optical telescope. The drawback of a low $f/D$ is a small field of view. The focal ellipsoid is the volume around the exact focal point that remains in reasonably good focus, and the focal circle is defined by the intersection of the focal ellipse and the transverse plane at $z=f$. Ruze [96] showed that the angular radius of the focal circle is proportional to ${(f/D)}^2$, and only a small number (about seven) of discrete feeds can fit inside the focal circle of an $f/D\approx 0.4$ paraboloid. Large arrays of feeds or imaging cameras require larger $f/D$ ratios, obtained either by using a shallower paraboloid or by using magnifying subreflectors to increase the effective focal length.

The primary mirrors of most radio telescopes are circular paraboloids or sections thereof for the following reasons:

1. 1.

   The effective collecting area $A_{\mathrm{e}}$ of a reflector antenna can approach its projected geometric area $A=\pi D^2/4$.

2. 2.

   They are electrically simple (compared with a phased array of dipoles, for example).

3. 3.

   A single reflector can work over a wide range of frequencies. Changing frequencies only requires changing the feed antenna and receiver located at the focal point, not building a whole new radio telescope.

How far away must a point source be for the received waves to satisfy the assumption that they are nearly planar across the reflector? The answer depends on both the wavelength $\lambda$ and the reflector diameter $D$. Figure 3.8 shows the spherical wave emitted by a point source a finite distance $R$ from a flat aperture, an imaginary circular hole that covers the reflector. It could be located at the plane $z=h$ shown in Figure 3.7, for example.

The maximum departure $\mathrm{\Delta }$ from a plane wave occurs at the edge of the aperture. The far-field distance $R_{\mathrm{ff}}$ is somewhat arbitrarily defined by requiring that . At the aperture edge, the Pythagorean theorem gives

$$R^2={(R-\mathrm{\Delta })}^2+{\left(\frac{D}{2}\right)}^2.$$

(3.61)

Thus

$$R=\frac{\mathrm{\Delta }}{2}+\frac{D^2}{8\mathrm{\Delta }}.$$

(3.62)

In the limit $\mathrm{\Delta }\ll D$, we have $\mathrm{\Delta }/2\ll D^2/(8\mathrm{\Delta })$ and

$$R\approx \frac{D^2}{8\mathrm{\Delta }}.$$

(3.63)

Given the $\mathrm{\Delta }=\lambda /16$ criterion, the far-field distance is

$$\boxed{R_{\mathrm{ff}}\approx \frac{2D^2}{\lambda }.}$$

(3.64)

If , the path-length errors will introduce significant phase errors in the waves coming from the off-axis portions of the reflector, reducing the effective collecting area and degrading the antenna pattern.

Example. What is the far-field distance of the Green Bank Telescope ($D=100$ m) observing at $\lambda =1$ cm? Equation 3.64 gives $$R_{\mathrm{ff}}=\frac{2{(100\mathrm{m})}^2}{1\mathrm{cm}}=\frac{2\times {10}^4{\mathrm{m}}^2}{0.01\mathrm{m}}=2\times {10}^6\mathrm{m}=2000\mathrm{km}.$$ Such a large far-field distance makes ground-based measurements of the GBT antenna pattern impractical. To measure small errors in the GBT reflector surface using radio holography, it is necessary to observe a geostationary satellite having an orbital altitude $R\sim$ 36,000 $\gg 2000$ km. Similarly, the easiest way to determine the transmitting power pattern for a large radar antenna such as the $D=305$-m Arecibo reflector is to scan across a celestial point source in the far field and use the reciprocity theorem to equate the transmitting and receiving patterns.

In optics, the term aperture refers to the opening through which all rays pass. For example, the aperture of a paraboloidal reflector antenna would be the plane circle, normal to the rays from a distant point source, that just covers the paraboloid (Figure 3.9). The phase of the plane wave from a distant point source would be constant across the aperture plane when the aperture is perpendicular to the line of sight.

Another example of an aperture is the mouth of a waveguide horn antenna (Figure 3.10).

How can the beam pattern, or power gain as a function of direction, of an aperture antenna be calculated? For simplicity, first consider a one-dimensional aperture of width $D$ (Figure 3.11) and calculate the electric field pattern at a distant ($R\gg R_{\mathrm{ff}}$) point.

When used in a transmitting antenna, the feed can illuminate the aperture antenna with a sine wave of fixed frequency $\nu =\omega /(2\pi )$ and electric field strength $g(x)$ that varies across the aperture. The illumination induces currents in the reflector. The currents will vary with both position and time:

$$I\propto g(x)\exp (-i\omega t).$$

(3.65)

The constant of proportionality doesn’t matter yet; it can be calculated later from energy conservation. Huygens’s principle asserts that the aperture can be treated as a collection of small elements which act individually as small antennas. Huygens’s principle actually applies to waves of any type, sound waves for example. The electric field produced by the whole aperture at large distances is just the vector sum of the elemental electric fields from these small antennas. The field from each element extending from $x$ to $x+dx$ is

$$df\propto g(x)\frac{\exp (-i2\pi r(x)/\lambda )}{r(x)}dx,$$

(3.66)

where $r(x)$ is the distance between the source and the aperture element at position $x$ (Figure 3.11). In the far field (Equation 3.64), the Fraunhofer approximation

$$r\approx R+x\sin \theta$$

(3.67)

is valid. This equation is usually written in the form

$$r\approx R+xl,$$

(3.68)

where

$$\boxed{l\equiv \sin \theta .}$$

(3.69)

For the small angles $\theta \ll 1$ rad relevant to large $D\gg \lambda$ apertures, $l=\sin \theta \approx \theta$.

At large distances, the quantity

$$\frac{1}{r}\approx \frac{1}{R}$$

(3.70)

is nearly constant across the aperture and can be absorbed by the constant of proportionality in Equation 3.66. Although $\exp (-i2\pi R/\lambda )$ is a constant because $R$ is fixed, the variable part $xl$ of $r=R+xl$ in the numerator of Equation 3.66 cannot be ignored at any distance:

$$df\propto g(x)\exp (-i2\pi xl/\lambda )dx.$$

(3.71)

When $\theta \ne 0$ the phase $2\pi xl/\lambda \approx 2\pi x\sin \theta /\lambda$ varies linearly across the aperture, and different parts of the aperture add constructively or destructively to the total electric field $f(l)$. Defining

$$\boxed{u\equiv \frac{x}{\lambda }}$$

(3.72)

to express position along the aperture in units of wavelength yields

$$\boxed{f(l)={\int }_{\mathrm{aperture}}g(u)e^{-i2\pi lu}du.}$$

(3.73)

In words, this very important equation says that in the far field, the electric-field pattern $f\mathit{}\mathrm{(}l\mathrm{)}$ of an aperture antenna is the Fourier transform (Appendix A.1) of the electric field distribution $g\mathit{}\mathrm{(}u\mathrm{)}$ illuminating that aperture.

What is the electric-field pattern of a uniformly illuminated one-dimensional aperture of width $D$ at wavelength $\lambda$? Uniform illumination means that the strength of the illumination is constant over the aperture:

This question is best answered in two steps: first find the far-field pattern of a unit aperture ($D=\lambda$) and then use the similarity theorem (Equation A.11) for Fourier transforms to scale the first result to an aperture of any size.

The unit rectangle function is defined as

(3.74)

and $\mathrm{\Pi }(u)=0$ otherwise. The function symbol (an uppercase pi) is easy to remember because it looks like the function graph shown in the top panel of Figure 3.12.

Inserting $\mathrm{\Pi }(u)$ into Equation 3.73 gives the field pattern $f(l)$ of the uniformly illuminated unit aperture:

$$f(l)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{\Pi }(u)e^{-i2\pi lu}𝑑u.$$

(3.75)

Thus

$$f(l)={\int }_{-1/2}^{+1/2}{e}^{-i2\pi lu}du=\frac{e^{-i2\pi lu}}{-i2\pi l}{|}_{-1/2}^{+1/2}=\frac{e^{-i\pi l}-e^{i\pi l}}{-i2\pi l}.$$

(3.76)

Next, difference the mathematical identities (Appendix B.3)

	$e^{i\pi l}=$	$\mathrm{ }\cos (\pi l)+i\sin (\pi l),$
	$e^{-i\pi l}=$	$\mathrm{ }\cos (\pi l)-i\sin (\pi l)$

to derive

$$e^{i\pi l}-e^{-i\pi l}=2i\sin (\pi l).$$

Inserting this result into Equation 3.76 gives

$$f(l)=\frac{-2i\sin (\pi l)}{-2i\pi l}=\frac{\sin (\pi l)}{(\pi l)}\equiv \mathrm{sinc}(l).$$

(3.77)

The useful sinc function defined in Equation 3.77 is plotted in the middle panel of Figure 3.12.

The power pattern $p(l)$ is the square of the field pattern $f(l)$. The power pattern $p(l)={\mathrm{sinc}}^2(l)$ of a uniformly illuminated unit aperture is graphed in the bottom panel of Figure 3.12. The central peak of the power pattern between the first zeros at $l=\pm 1$ is called the main beam. The smaller peaks are called sidelobes. They are separated by zeros or nulls in the power pattern at $l=\pm 1,\pm 2,\mathrm{\dots }.$

Next apply the powerful similarity theorem for Fourier transforms: if $f(l)$ is the Fourier transform of $g(u)$, then

$$\frac{1}{|a|}f\left(\frac{l}{a}\right)$$

is the Fourier transform of $g(au)$, where $a\ne 0$ is a dimensionless scaling factor. According to the similarity theorem, making a function $g$ wider () or narrower ($a>1$) makes its Fourier transform $f$ narrower and taller, or wider and shorter, respectively, always conserving the area under the transform. Consequently the beamwidth of an aperture antenna is inversely proportional to the aperture size in wavelengths and the on-axis field strength is directly proportional to the aperture size in wavelengths.

The scale factor for a uniformly illuminated one-dimensional aperture of width $D$ operating at wavelength $\lambda$ is $a=\lambda /D$, so the electric field pattern becomes

$$f(l)=\left(\frac{D}{\lambda }\right)\frac{\sin (\pi lD/\lambda )}{(\pi lD/\lambda )}\propto \frac{D}{\lambda }\mathrm{sinc}\left(\frac{lD}{\lambda }\right).$$

If the aperture is large ($D/\lambda \gg 1$), the relevant angles $\theta$ are so small ($\theta \ll 1$ radian) that $l=\sin \theta \approx \theta$ and

$$\boxed{f(\theta )=\frac{D}{\lambda }\mathrm{sinc}\left(\frac{\theta D}{\lambda }\right).}$$

(3.78)

The power pattern is proportional to the square of the electric field pattern, so

$$P(l)\propto {\left(\frac{D}{\lambda }\right)}^2{\mathrm{sinc}}^2\left(\frac{lD}{\lambda }\right).$$

If $\theta \ll 1$ radian, then

$$\boxed{P(\theta )={\left(\frac{D}{\lambda }\right)}^2{\mathrm{sinc}}^2\left(\frac{\theta D}{\lambda }\right).}$$

(3.79)

Radio astronomers use the angle between the half-power points to specify the angular width of the main beam, calling it the half-power beamwidth (HPBW) or the full width between half-maximum points (FWHM). The narrow beamwidth ${\theta }_{\mathrm{HPBW}}\ll 1\mathrm{rad}$ of a large ($D\gg \lambda$) one-dimensional uniformly illuminated aperture satisfies

	$$P({\theta }_{\mathrm{HPBW}}/2)=\frac{1}{2}={\mathrm{sinc}}^2\left(\frac{{\theta }_{\mathrm{HPBW}}D}{2\lambda }\right),$$		(3.80)
	$$0.443\approx \frac{{\theta }_{\mathrm{HPBW}}D}{2\lambda },$$		(3.81)
	$${\theta }_{\mathrm{HPBW}}\approx 0.89\frac{\lambda }{D}.$$		(3.82)

The similarity theorem implies the general scaling relation

$${\theta }_{\mathrm{HPBW}}\propto \frac{\lambda }{D}.$$

(3.83)

The constant of proportionality varies slightly with the illumination taper. Even an ideal aperture antenna of finite size has a finite resolving power that is limited by diffraction, the spreading of rays passing through a finite aperture, and Equation 3.82 specifies the diffraction-limited resolution of a uniformly illuminated aperture antenna.

The weak reciprocity theorem (Section 3.1.5) says that the preceding analysis of the transmitting power pattern of an aperture antenna also yields its receiving power pattern, or the variation of $A_{\mathrm{e}}$ with orientation. In receiving terms, the analog of the power pattern is called the point-source response. For a uniformly illuminated aperture, scanning a radio telescope beam in angle $\theta$ across a point source will cause the antenna temperature to vary as ${\mathrm{sinc}}^2(\theta )$, and the width of the half-power response will equal the transmitting HPBW. The receiving HPBW is sometimes called the resolving power of a telescope because two equal point sources separated by the HPBW are just resolved by the Rayleigh criterion that the total response has a slight minimum midway between the point sources.

Practical feeds such as small waveguide horns or half-wave dipoles backed by small subreflectors cannot illuminate a large aperture uniformly. A better approximation to their illumination is the cosine-tapered field pattern (cosine-squared tapered power pattern)

(3.84)

and $g(u)=0$ otherwise (Figure 3.13). The ($\pi /2$) normalization factor in Equation 3.84 ensures that

$${\int }_{-1/2}^{+1/2}g(u)𝑑u=1.$$

(3.85)

The corresponding field pattern of a one-dimensional unit aperture is given by

$$f(l)={\int }_{-1/2}^{+1/2}\frac{\pi }{2}\cos (\pi u)e^{-i2\pi lu}𝑑u.$$

(3.86)

This Fourier transform can be evaluated as follows:

	$f(l)$	$={\displaystyle \frac{\pi }{4}}{\displaystyle {\int }_{-1/2}^{+1/2}}(e^{i\pi u}+e^{-i\pi u})e^{-i2\pi lu}𝑑u$		(3.87)
		$={\displaystyle \frac{\pi }{4}}[{\displaystyle \frac{e^{i\pi (1-2l)u}}{i\pi (1-2l)}}{|}_{-1/2}^{+1/2}+{\displaystyle \frac{e^{-i\pi (1+2l)u}}{-i\pi (1+2l)}}{|}_{-1/2}^{+1/2}]$		(3.88)
		$={\displaystyle \frac{\pi }{4}}\left[{\displaystyle \frac{e^{i\pi (1/2-l)}-e^{-i\pi (1/2-l)}}{i2\pi (1/2-l)}}+{\displaystyle \frac{e^{i\pi (1/2+l)}-e^{-i\pi (1/2+l)}}{i2\pi (1/2+l)}}\right]$		(3.89)
		$={\displaystyle \frac{\pi }{4}}\left[{\displaystyle \frac{2i\sin [\pi (1/2-l)]}{i2\pi (1/2-l)}}+{\displaystyle \frac{2i\sin [\pi (1/2+l)]}{i2\pi (1/2+l)}}\right]$		(3.90)
		$={\displaystyle \frac{\pi }{4}}\left[{\displaystyle \frac{\cos (\pi l)}{\pi (1/2-l)}}+{\displaystyle \frac{\cos (\pi l)}{\pi (1/2+l)}}\right]={\displaystyle \frac{\pi }{4}}\cos (\pi l)\left({\displaystyle \frac{\pi }{{\pi }^2/4-{\pi }^2{l}^2}}\right)$		(3.91)