Antenna Fundamentals - National Radio Astronomy ... · PDF fileAntenna Fundamentals An antenna...

Antenna FundamentalsAn antenna is a device for converting electromagnetic radiation in space into electrical currentsin conductors or vice-versa, depending on whether it is being used for receiving or fortransmitting, respectively. Passive radio telescopes are receiving antennas. It is usually easierto calculate the properties of transmitting antennas. Fortunately, most characteristics of atransmitting antenna (e.g., its radiation pattern) are unchanged when the antenna is used forreceiving, so we often use the analysis of a transmitting antenna to understand a receivingantenna used in radio astronomy.

Radiation from a Short Dipole Antenna (Hertz Dipole)

The coordinate system used to describe the radiation from a short ( ) dipole driven by acurrent source at frequency ·.

The simplest antenna is a short (length l much smaller than one wavelength Õ) dipoleantenna, which is shown above as two colinear conductors (e.g., wires). Since they aredriven at the small gap between them by a current source (a transmitter), the current in thebottom conductor is 180 deg out of phase with the current in the top conductor. The radiationfrom a dipole depends on frequency, so we consider a driving current I varying sinusoidallywith angular frequency ! Ù·:

I (!t) ;

where I is the peak current going into each half of the dipole. It is computationally convenientto replace the trigonometric function cos(!t) with its exponential equivalent, the real part of

l Ü Õ

= 2

= I0 cos

0

Antenna Fundamentals http://www.cv.nrao.edu/course/astr534/AntennaTh...

1 of 19 09/20/2010 11:34 AM

so the driving current can be rewritten as

with the implicit understanding that only the real part of I represents this current. The drivingcurrent accelerates charges in the antenna conductors, so we can use Larmor's formula tocalculate the radiation from the antenna by converting from the language of charges andaccelerations to time-varying currents.

Recall that electric current is defined as the time derivative of electric charge:

Along a wire, the current is the amount of charge flowing past any point per unit time. For awire on the z-axis

I v ;

where v is the instantaneous flow velocity of the charges.

It is a common misconception to believe that the velocities v of individual electrons in a wireare comparable with the speed of light c because electrical signals do travel down wires atnearly the speed of light. A wire filled with electrons is like a garden hose already filled with anincompressible fluid—water. When the faucet is turned on, water flows from the other end of afull hose almost immediately, even though individual water molecules are moving slowly alongthe hose. As a specific example, consider a current of 1 ampere flowing through a copper wireof cross section Û mm m . The number density of free electrons is about equal

to the number density of copper atoms in the wire, m . In mks units, the charge ofan electron is

One Ampere is one Coulomb per second, so the number N of electrons flowing past any pointalong the wire in one second is

N :25 0 s

The average electron velocity is only

v 0 m s

e (!t) (!t) Ài!t = cos À i sin

I e = I0Ài!t

I : Ñdt

dq

=dt

dq=

dz

dq

dt

dz=

dz

dq

= 1 0 2 = 1 À6 2

n 0 Ù 1 29 À3

Àe :80 0 statcoul :60 0 coul Ù 4 Â 1 À12 Â 1 coul3 0 statcoulÂ 1 9

Ù 1 Â 1 À19

=I

jej Ù1 coul sÀ1

1:60 0 coulÂ 1 À19 Ù 6 Â 1 18 À1

Ù N

ÛnÙ 6:25 0 sÂ 1 18 À1

10 m 0 mÀ6 2 Â 1 29 À3 Ù 6Â 1 À5 À1 Ü c


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Thus the nonrelativistic Larmor equation may be used directly to calculate the radiation from awire.

From the derivation of Larmor's formula, recall that

E

so in the short dipole

For a sinusoidal driving current,

and

That is, the radiated electric field strength is proportional to the integral of the current

distribution along the antenna. The current at the center is just the driving current and the current must drop to zero at the ends of the antenna, where the conductivity goes tozero. For a short antenna, we can make the approximation that the current declines linearlyfrom the driving current at the center to zero at the ends:

Then

and

Substituting ! Ùc=Õ gives

? =rc2

qv_ sin Ò

E dz ? =

Z +l=2

z=Àl=2 dzdq

rc2v_ sin Ò

v i!v _ = À

E vdz ? =rc2

Ài! sin Ò Z

Àl=2

+l=2

dz

dq

E dz ? =rc2

Ài! sin Ò Z

Àl=2

+l=2

I

E ?I e = I0

Ài!t

I(z) e 1 = I0Ài!t

ÔÀ jzj

(l=2)

Õ

dz e Z

Àl=2

+l=2

I =2

I l0 Ài!t

E e ? =rc2

Ài! sin Ò2

I l0 Ài!t

= 2


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Since (cgs), the time-averaged Poynting flux (power per unit area) is

and

because . Note that the radiation from a short dipole has the samepolarization and the same doughnut-shaped power pattern (the power pattern is the angulardistribution of radiated power, often normalized to unity at the peak)

as Larmor radiation from an accelerated charge because all of the charges in the dipole arebeing accelerated along one line much shorter than one wavelength. From the observer'spoint of view, the power received depends only on the projected (perpendicular to the line ofsight) length (l ) of the dipole. Thus the electric field received is proportional to theapparent length of the dipole. The time-averaged total power emitted is obtained by integratingthe Poynting flux over the surface area of a sphere of any radius centered on theantenna:

Recall that , so the time-averaged power radiated by a short dipole is

where I (!t) is the driving current, l is the total length of the dipole, and Õ is thewavelength.

Most practical dipoles are half-wave dipoles ( ) because half-wave dipoles are resonant,

E e ? =Õrc2

Ài2Ùc sin Ò

2

I l0 Ài!t

E ? =c

ÀiÙ sin ÒÕ

I l0r

eÀi!t

jE j H j ?~ = j ?

~

hSi hE i; =c

4Ù2?

hSi =c

4Ù

Ò

Õ

I l0c

ÙÓ2

r2sin2 Ò

Ò

2

1Ó

hcos (!t)i =2 2 = 1

P / sin2 Ò (3A1)

sin Ò

r µ l

hP i hSidA r d¶rdÒ =

Z=

c

4Ù

Ò

Õ

I l0c

ÙÓ2Ò

2

1ÓZ 2Ù

¶=0

Z Ù

Ò=0 r2sin2 Ò

sin Ò

hP i 2Ù dÒ =c

4Ù

Ò

Õ

I l0c

ÙÓ2Ò

2

1Ó Z Ù

Ò=0

sin3 Ò

dÒ =3 R0

Ùsin3 Ò = 4

hP i ; =3c

Ù2Ò

Õ

I l0Ó2

(3A2)

0 cos

l =2 Ù Õ


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meaning that they provide a nearly resistive load to the transmitter. When each half of thedipole is Õ=4 long, the standing-wave current is highest at the center and naturally falls ascos(2Ùz=Õ) to almost zero at the ends of the conductors.

The ground-plane vertical shown below is very similar to the dipole. A ground-plane verticalis one half of a dipole above a conducting plane, which is called a "ground plane" becausehistorically the conducting plane for vertical antennas was the surface of the Earth. Thetransmitter is connected between the base of the vertical and the ground plane near the base. The conducting ground plane is a mirror that creates the lower half of the dipole as the mirrorimage of the upper half. Electric fields produced by the vertical must induce currents in theconducting plane that ensure that the horizontal component of the electric field goes to zeroon the conductor. This implies that the virtual electric fields from the image vertical must havethe same amplitude but be 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 inthe half space above the ground plane and zero below the ground plane.

The ground-plane vertical is just half of a dipole above a conducting plane. The lower half ofthe dipole is the reflection of the vertical in the mirror provided by the conducting "groundplane." The image vertical is 180 deg out of phase with the real vertical. Above the groundplane, the radiation from the ground-plane vertical is exactly the same as the radiation fromthe dipole.

According to the strict definition of an antenna as a device for converting betweenelectromagnetic waves in space and currents in conductors, the only antennas in most radiotelescopes are half-wave dipoles and their relatives, quarter-wave ground-plane verticals. Thelarge parabolic reflector of a radio telescope serves only to focus plane waves onto the feedantenna. [The term "feed" comes from radar antennas used for transmitting; the "feed"feeds transmitter power to the main reflector. Receiving antennas used in radio astronomywork the other way around, and the "feed" actually collects radiation from the reflector.]

Actual half-wave dipoles, backed by small reflectors about Õ=4 behind them to focus the dipolepattern in the direction of the main dish, are normally used as feeds at low frequencies (· < 1


5 of 19 09/20/2010 11:34 AM

GHz) or long wavelengths (Õ :3 m) because of their relatively small size. However, theradiation patterns of half-wave dipoles backed by small reflectors are not well matched tomost parabolic dishes, so their performance is less than optimum.

At shorter wavelengths, almost all radio-telescope feeds are quarter-wave ground-planeverticals inside waveguide horns. Radiation entering the relatively large (size > ) rectangularor circular aperture of the tapered horn is concentrated into a rectangular or circular waveguidewith parallel conducting walls. In the case of the rectangular waveguide shown below, the sidewalls are separated by slightly over Õ=2 so that vertical electric fields can travel down thewaveguide with low loss. The top and bottom walls are separated by somewhat less than Õ=2so only one mode with vertical electric fields can propagate. The Õ=4 vertical inserted througha small hole in the bottom wall collects most of this vertically polarized radiation and convertsit into an electric current that travels down the coaxial cable to the receiver. The backshortwall about Õ=4 behind the dipole ensures that the dipole sees only radiation coming from thehorn.

Most high-frequency feeds are quarter-wave ground-plane verticals inside waveguide horns.The only true antenna in this figure is the Õ=4 ground-plane vertical, which convertselectromagnetic waves in the waveguide to currents in the coaxial cable extending down fromthe waveguide.

Radiation Resistance

The power flowing through a circuit is , where V is the voltage (defined as energyper unit charge) and I is the current (defined as charge flow per unit time), so P hasdimensions of energy per unit time. The physicist George Simon Ohm observed that thecurrent flowing through most materials is proportional to the applied voltage, so many (but notall) objects have a well-defined resistance defined by R =I (Ohm's law). For them,

.

From Ohm's law for time-varying currents,

> 0

Õ

P = V Â I

= V

P R =R = V Â I = I2 = V 2


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If I (!t),

The radiation resistance of an antenna is defined by

R

For our short dipole, the radiation resistance is

Example: A "half-wave" dipole has length l =2. This is the length of a resonant dipoleantenna. Resonant antennas are used in most real applications because the impedance of aresonant antenna is resistive; nonresonant antennas have large capacitive or inductivereactances as well. A half-wave dipole doesn't strictly satisfy our criterion for being"short," so the current distribution along the dipole is actually closer to sinusoidal than linearand our calculated radiation resistance will not be exact. Proceeding nonetheless to estimatethe radiation resistance of a half-wave dipole,

Engineers and real test instruments use the mks "Ohm" (symbol Ê) as the unit of resistance.The conversion factor is 1 Ê = s cm , so

[The actual radiation resistance of a half-wave dipole in free space is about 73 Ê.]

For a given driving current, a ground-plane vertical of height l=2 emits exactly like a dipole oflength l above the ground plane and zero below the ground plane. Thus the total poweremitted by the vertical is half the power emitted by the dipole, and the radiation resistance ofthe vertical is half the radiation resistance of the dipole.

The radiation resistance R of free space can be obtained from the relations

hP i I iR = h 2

= I0 cos

hP i =2

I R02

ÑI02

2hP i(3A3)

R =3c

2Ù2Òl

Õ

Ó2

= Õ

l Ü Õ

R :5 0 s cm Ù 2Ù2

3 0 cm sÂ 3Â 1 10 À1

Ò

2

1Ó2

Ù 5 Â 1 À11 À1

(10 =9) À11 À1

R :5 0 Ê 0 Ê Ù 5 Â 1 À11 s

cm

9 cm10 sÀ11 Ù 5

0


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Since the electric field E is just the voltage per unit length V=l and the flux is the power per unitarea l ,

jSj

R

Converting to mks units yields the radiation resistance of space in Ohms:

R 20Ù Ê 77 Ê :

Since a black hole is a perfect absorber of radiation, its impedance must also be 120Ù Ê tomatch that of free space. A black hole spinning in an external magnetic field can generateelectrical power with a voltage/current ratio of 120Ù Ê, and this process may be important inpowering quasars (see Blandford & Znajek 1977, MNRAS, 179, 433).

The Power Gain of a Transmitting Antenna

The power gain G(Ò; ) of a transmitting antenna is defined as the power transmitted per unitsolid angle in direction (Ò; ) divided by the power transmitted per unit solid angle from anisotropic antenna driven by a transmitter supplying the same total power. Frequently the valueof G is expressed logarithmically in units of decibels (dB):

For a lossless antenna, energy conservation requires that the gain averaged over all directionsbe

or

jSj E nd P : ~ =c

4Ù2 a =

R

V 2

2

~ =cV 2

4Ùl2=

V 2

2R l02

0 = c

4Ù=

4Ù

3 0 cm sÂ 1 10 À1

0 =4Ù

3 0 cm s 0 sec cm ÊÂ 1 10 À1 Â 1=9Â 1 À11 À1 À1= 1 Ù 3

¶¶

G(dB) 0 (G) Ñ 1 Â log10

hGi ÑÊ

Rsphere

d

dÊRsphere

G=

4Ù

4Ù

hGi = 1 (3A4)


8 of 19 09/20/2010 11:34 AM

Consequently

for any lossless antenna. Different lossless antennas may radiate with different directionalpatterns, but they cannot alter the total amount of power radiated. Consequently, the gain ofa lossless antenna depends only on the angular distribution of radiation from that antenna. Ingeneral, an antenna having peak gain G must beam most of its power into a solid angleÁÊ such that

Thus the higher the gain, the narrower the beam or power pattern.

Example: What is the power gain of a short dipole? It is sufficient to recall only the angulardependence of the short-dipole power pattern

where Ò is the angle from the dipole axis. Thus

The maximum gain G is determined by energy conservation:

Recall that so

G

and

G(Ò; ) :

Expressed in dB, the maximum gain G of a short dipole is

dÊ dÊ Ù Z

sphere

G =

Z

sphere

1 = 4

max

ÁÊ : Ù 4Ù

Gmax

hSi ; / sin2 Ò

G : / sin2 Ò = G0 sin2 Ò

0

dÊ dÒ d¶ Ù : Z

sphere

G =

Z 2Ù

¶=0

Z Ù

Ò=0

G0 sin2 Ò sin Ò = 4

2ÙG dÒ Ù : 0

Z

0

Ù

sin3 Ò = 4

dÒ =3 R0

Ùsin3 Ò = 4

0 = 2Ù

4Ù

4

3=

2

3

¶ =2

3 sin2 Ò

0

G 0 (3=2) :76 dB: 0 = 1 log10 Ù 1


9 of 19 09/20/2010 11:34 AM

Note that G(Ò; ) is independent of the antenna length in wavelengths so long as

because the short-dipole power pattern is independent of l. Varying affects only theradiation resistance.

The Effective Area of a Receiving Antenna

How can we characterize antennas used for receiving, as in radio astronomy, rather than fortransmitting? The receiving counterpart of transmitting power gain is the effective area oreffective collecting area of an antenna.

Imagine an ideal antenna that collects all of the radiation falling on it from a distant pointsource and converts it to electrical power—a "rain gauge" for collecting photons. The totalspectral power that it collects will be the product of its geometric area A and the incidentspectral power per unit area, or flux density S. By analogy, if any real antenna collectsspectral power P , its effective area A is defined by

A ;

where S is the flux density in the "matched" polarization.

What does matched polarization mean? Any electromagnetic wave can be decomposed intotwo orthogonal polarized components. For example, the transverse electric field can beresolved into horizontal and vertical components, or horizontal and vertical linearpolarizations. If the horizontal and vertical electric fields are equal in amplitude and out ofphase, the radiation is circularly polarized. Any radio wave can also be decomposed into left-and right-circular polarizations. If the wave is essentially random (noise generated byblackbody radiation for example), the two orthogonal components will vary rapidly in intensitybut have equal powers when averaged over long times. Such radiation is called unpolarized.Thus for an unpolarized source,

S :

Blackbody radiation is unpolarized. Most radio astronomical sources are unpolarized or nearlyso.

Any antenna with a single output collects only one of the two polarizations from anelectromagnetic wave. For example, a linear dipole antenna collects radiation only from thelinear polarization whose electric field is parallel to the antenna wires. Electric fieldsperpendicular to the dipole antenna do not produce currents in the antenna, so the linear dipoleis completely insensitive to the linear polarization perpendicular to its wires. A pair of crosseddipoles is needed to collect power from both orthogonal polarizations simultaneously.

¶ l Ü Õ

l Ü Õ

· e

e Ñ P·

S(matched)

(3A5)

(matched)

90 Î

(matched) = 2

S


10 of 19 09/20/2010 11:34 AM

Just as energy conservation implies that all lossless transmitting antennas have the sameaverage power gain, all lossless receiving antennas have the same average collecting area. This average collecting area can be calculated via another thermodynamic thought experiment.

A cavity in thermodynamic equilibrium at temperature T containing a resistor R is coupled toan antenna, also at temperature T , through a filter passing frequencies in the range · to· ·.

Imagine an antenna inside a cavity in full thermodynamic equilibrium at temperature Tconnected through a transmission line to a matched resistor in a second cavity at the sametemperature. Suppose further that the connection contains a filter that passes only a narrowrange of frequencies between · and · ·. Since this entire system is in thermodynamicequilibrium, no net power can flow between the antenna and the resistor. Otherwise, onecavity would heat up and the other would cool down, in violation of the second law ofthermodynamics. Thus the total spectral power from all directions collected by the antenna

must equal the Nyquist spectral power P T produced by the resistor. Using theRayleigh-Jeans approximation

B

gives

The average collecting area is defined by

+ d

+ d

P S (Ò; ) dÊ · = Ae (matched) = Ae2

S=

Z

4Ù

Ae ¶2

B·

· = k

· =Õ22kT

P dÊ T · =2Õ22kT

Z

4Ù

Ae = k

dÊ Z

4Ù

Ae = Õ2


11 of 19 09/20/2010 11:34 AM

The effective collecting area of a receiving antenna is independent of its radiation environment,so this result applies for any type of radiation, not just blackbody radiation. Without usingMaxwell's equations we have obtained the remarkable result true for all lossless antennas:

hA i

Any antenna, from a short dipole to the 100-m diameter Green Bank Telescope, has thesame average collecting area that depends only on wavelength.

In the case of an isotropic antenna, the effective collecting area in any direction equals theaverage collecting area:

Consequently, a nondirectional antenna operating at a very short wavelength Õ will have avery small effective collecting area and poor sensitivity for reception. For this reason, mostsatellite broadcast services, GPS or satellite FM radio for example, transmit at relatively longwavelengths (10 to 20 cm). Likewise, practical radio telescopes can be constructed fromarrays of dipoles at long wavelengths (Õ m) but not at short wavelengths because thenumber of dipoles needed to produce useful collecting areas would be too large.

Reciprocity Theorems

Many antenna properties are the same for both transmitting and receiving. It is often easier tocalculate the gain of a transmitting antenna than the collecting area of a receiving antenna, andit is often easier to measure the receiving power pattern than to measure the transmittingpower pattern of a large radio telescope. Thus this receiving/transmitting "reciprocity" greatlysimplifies antenna calculations and measurements. Reciprocity can be understood via Maxwell'sequations or by thermodynamic arguments.

Burke & Smith (1997) state the electromagnetic case for reciprocity clearly: "An antenna canbe treated either as a receiving device, gathering the incoming radiation field and conductingelectrical signals to the output terminals, or as a transmitting system, launchingelectromagnetic 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:

hA i : e ÑÊ

R4Ùd

dÊR4ÙAe

e =Õ2

4Ù(3A6)

hA i e

A (Ò; ) A i e ¶ = h e =Õ2

4Ù

> 1


12 of 19 09/20/2010 11:34 AM

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

can be formally derived from Maxwell's equations [see a partial derivation in Rohlfs & WilsonSection 5.4] or by network analysis [see Kraus "Antennas", p. 252].

The strong reciprocity theorem implies that the transmitter voltages V and V are related tothe receiver currents I and I by

for any pair of antennas A and B.

For most radio astronomical applications, we are not concerned with the detailed phaserelationships of voltages and currents, and we can use a weak reciprocity theorem thatrelates the angular dependences of the transmitting power pattern and the receiving collectingarea of any antenna:

The power pattern of an antenna is the same for transmitting and receiving.

That is:

A B

A B

IB

VA =IA

VB

G(Ò; ) (Ò; ) ¶ / Ae ¶ (3A7)


13 of 19 09/20/2010 11:34 AM

The weak reciprocity theorem can be proven by another simple thermodynamic thoughtexperiment: An antenna is connected to a matched load inside a cavity initially in equilibrium attemperature T . The antenna simultaneously receives power from the cavity walls andtransmits power generated by the resistor. The total power transmitted in all directions mustequal the total power received from all directions since no net power can be transferredbetween the antenna and the resistor; otherwise the resistor would not remain at temperatureT . Moreover, in any direction, the power received and transmitted by the antenna must be thesame, else the cavity wall in directions where the transmitted power was greater than thereceived power would rise in temperature and the cavity wall in directions of lowertransmitted/received power ratio would cool, leading to a violation of the second law ofthermodynamics.

The weak reciprocity theorem states that the transmitting and receiving power patterns of anantenna cannot differ as shown here, without violating the second law of thermodynamics.

The constant of proportionality relating G and A can be derived from our earlier results for anisotropic antenna

Thus energy conservation and the weak reciprocity theorem imply

A (Ò; )

for any antenna. This extremely useful equation lets us compute the receiving power patternfrom the transmitting power pattern and vice versa.

Example: We can use our calculation of the transmitting power pattern of a short dipole tocalculate its effective collecting area when used as a receiving antenna:

e

hA i nd hGi e =Õ2

4Ùa = 1

e ¶ =4Ù

Õ G(Ò; )2 ¶(3A8)


14 of 19 09/20/2010 11:34 AM

A (Ò; )

A

Example: What is the power per unit bandwidth P collected by a short dipole at · 0 GHz

broadside to ( ) the Sun, a thermal source whose flux density is Jy?

The broadside (sin ) collecting area of the short dipole is

A

so

Clearly, dipoles or other nearly isotropic antennas have very small collecting areas at shortwavelengths. For an unpolarized source,

P S

Arrays of dipoles do make sense at long wavelengths. For example, the Long WavelengthArray (LWA) for · 20 to 80 MHz ( 4 to 15 m) will consist of 53 stations, each a 100 m

100 m array of crossed dipoles. The effective collecting area of the LWA will be up to m at Õ 5 m.

e ¶ =4Ù

Õ G(Ò; )2 ¶=

Õ2

4Ù 2

3 sin2 Ò

e =8Ù

3Õ2 sin2 Ò

· = 1

Ò 0 = 9 Î S :2 0 Ù 1 Â 1 6

Ò = 1

e =8Ù

3Õ2

Õ :03 m =c

·=

10 Hz10

3 0 m sÂ 1 8 À1= 0

A :07 0 m cm : e =8Ù

3 0:03 m)Â ( 2

= 1 Â 1 À4 2 Ù 1 2

· = Ae (matched) = 2

A Se

P :07 0 m · = 1 Â 1 À4 2 Â2

1:2 0 JyÂ 1 6

Â1 Jy

10 W m HzÀ26 À2 À1

P :4 0 W Hz · = 6 Â 1 À25 À1

= Õ ÙÂ A 0 e = 4Â 1 6 2 = 1


15 of 19 09/20/2010 11:34 AM

The proposed Long-Wavelength Array of crossed dipoles. Image credit


16 of 19 09/20/2010 11:34 AM

Eight test dipoles of the LWDA (Long-Wavelength Development Array) on the VLA site. Imagecredit

Antenna Temperature

A convenient practical unit for the power output per unit frequency from a receiving antenna isthe antenna temperature T . Antenna temperature has nothing to do with the physicaltemperature of the antenna as measured by a thermometer; it is only the temperature of amatched resistor whose thermally generated power per unit frequency equals that producedby the antenna. It is widely used because:

1 K of antenna temperature is a conveniently small power. T K corresponds to

.

1.

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

2.

The units of receiver noise are also K, so comparing the signal in K with the receiver noisein K makes it easy to decide if a signal will be detectable.

3.

T

Example: What is the increase in the antenna temperature of our short dipole produced bythermal emission from the Sun at · 0 GHz?

A

A = 1

P T :38 0 J K K :38 0 W Hz · = k A = 1 Â 1 À23 À1 Â 1 = 1 Â 1 À23 À1

A Ñ k

P·(3A9)

= 1


17 of 19 09/20/2010 11:34 AM

so

T :046 K

Beam Solid Angle

The beam solid angle Ê of a lossless antenna is defined as

where P (Ò; ) is the power pattern normalized to unity maximum:

P :

The beam solid angle is a useful parameter for estimating the antenna temperature producedby a compact source covering solid angle Ê and having uniform brightness temperature T . "Compact" means that the source is much smaller than the beam so that the variation of P issmall across the source. The power per unit bandwidth received from the source by theantenna pointing at it is

P T :

Thus the antenna temperature T =k is

T :

Stated in words, the antenna temperature equals the source brightness temperature multipliedby the fraction of the beam solid angle filled by the source. A T 0 K source covering 1%

P :4 0 W Hz · = 6 Â 1 À25 À1

A =1:38 0 J KÂ 1 À23 À16:4 0 W HzÂ 1 À25 À1

Ù 0

A

Ê (Ò; )dÊ A ÑZ

4Ù

Pn ¶ (3A10)

n ¶

n =Gmax

G(Ò; )¶

Ê dÊ A =1

Gmax

Z

4Ù

G =4Ù

Gmax

s B

n

P (Ò; ) dÊ · =

Z

4Ù

Ae ¶2

B·

· Ù4ÙÕ2

Õ G kT Ê2(max) B s

= k B

Ês

ÊA

A = P·

A Ù TBÊs

ÊA

B = 1 4


18 of 19 09/20/2010 11:34 AM

of the beam solid angle will add 100 K to the antenna temperature.

Main Beam Solid Angle

The main beam of an antenna is defined as the region containing the principal response out tothe first zero; responses outside this region are called sidelobes or, very far from the mainbeam, stray radiation. The main beam solid angle Ê is defined as

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

Ñ

MB

Ê (Ò; )dÊ MB ÑZ

MB

Pn ¶ (3A11)

B Ñ ÊA

ÊMB (3A12)


19 of 19 09/20/2010 11:34 AM

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