Antenna Parameters - Lanka Education and Research...

Antenna Parameters

Ranga Rodrigo

University of Moratuwa

December 15, 2008

Ranga Rodrigo (University of Moratuwa) Antenna Parameters December 15, 2008 1 / 47

Summary of Last Week’s Lecture

Radiation Pattern for an Array with 10 Elements

0o

45o

90o

135o

180o

225o

270o

315o

φ0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.90.5 1

φHP

Figure 1: Half-Power Beam Width


Summary of Last Week’s Lecture Directivity

Directivity

We defined a parameter known as the directivityof the antenna, denoted by the symbol D, as theratio of the maximum power density radiated bythe antenna to the average power density.Thus the directivity of the Hertzian dipole isgiven by

D =[Pr]max

[Pr]av= 1.5 (1)


Summary of Last Week’s Lecture Directivity

D = 4π[f (θ, φ)]max∫ π

θ=0

∫ 2π

φ=0f (θ, φ) sinθdθdφ

(2)


Summary of Last Week’s Lecture Linear Antennas

:

Vrms, IrmsVmax

Vmin

I

Z0RF generator

forward wavereverse wave

Open circuit transmission line.


Summary of Last Week’s Lecture Linear Antennas

:

Vrms, IrmsVmax

Vmin

I

Z0RF generator

forward wavereverse wave

Radiation from a half-wave dipole.


Summary of Last Week’s Lecture Radiation Field Due to half-Wave Dipole

Now to find the radiation field due to the half-wavedipole, we divide it into a number of Hertzian dipoles,each of length dz′.


Summary of Last Week’s Lecture Radiation Field Due to half-Wave Dipole

y

x

z

P

L2

−L2

θ′

θ

r′

r

z′

dz′

iθ′

iφ′

z′ cosθ

Half-wave dipole as a number of Hertzian dipoles.


Summary of Last Week’s Lecture Field Equation for the Half-Wave Dipole

Evaluating the integral we obtain

Eθ = −ηI0

2πrcos [(π/2) cosθ]

sinθsin

(ωt −

πL

r). (3a)

SimilarlyH = Hφiφ.

where

Hφ = −I0

2πrcos [(π/2) cosθ]

sinθsin

(ωt −

πL

r). (3b)


Summary of Last Week’s Lecture Power Radiated for the Half-Wave Dipole

Power Radiated

Prad =0.609ηI2

0

πsin2

(ωt −

πL

r). (4)



Time-Average Radiated Power

〈Prad〉 =12

I20

(0.609ηπ

). (5)



Radiation Resistance of theHalf-Wave Dipole

Rrad =0.609ηπ

Ω. (6)

For free space, η = η0 = 120πΩ, and

Rrad = 0.609 × 120 = 73Ω. (7)


Summary of Last Week’s Lecture Radiation Pattern

Radiation Pattern of the Half-WaveDipoleRadiation pattern for the field is

cos [(π/2) cosθ]sinθ

.

Radiation pattern for the power desnsity is

cos2 [(π/2) cosθ]sin2 θ

.

These are slightly more directional that thecorresponding patterns for the Hertzian dipole.


Summary of Last Week’s Lecture Radiation Pattern

Directivity of the Half-Wave Dipole

From 2D = 1.642.


Summary of Last Week’s Lecture Center-Fed Linear Antennas of Arbitrary Length

Center-Fed Linear Antennas ofArbitrary Length

Eθ = −ηI0

2πrF(θ) sin(ωt − βr). (8a)

Hφ = −I0

2πrF(θ) sin(ωt − βr). (8b)

Rrad =η

π

∫ π/2

θ=0F2(θ) sinθdθ. (8c)

D =

[F2(θ)

]max∫ π/2

θ=0F2(θ) sinθdθ

. (8d)


Summary of Last Week’s Lecture Center-Fed Linear Antennas of Arbitrary Length

where

F(θ) =cos

(βL2 cosθ

)− cos βL

2

sinθ. (9)

F(θ) is the radiation pattern of the fields. For L = kλ,9 reduces to

F(θ) =cos(kπ cosθ) − cos(kπ)

sinθ. (10)


Antenna Parameters Directivity Continued

More on Directivity

We learned how to compute the directivitybased on the radiation pattern.Directivity has no units, and is usually given indB or dBi, where the reference directivity is 1,the directivity of an isotropic antenna.However, sometimes the reference directivity isthat of a half-wave dipole, 1.642. The directivitythen is in dBd.


Antenna Parameters Gain

Gain

The gain of a transmit antenna is proportional tothe directivity by a constant known as theantenna efficiency ot efficiency factor.This efficiency accounts for the power losseswithin the antenna.Gain, therefore, is not only a descriptor ofantenna’s power focusing ability, but also, takesinto account the power losses within theantenna.Therefore the gain is always less than thedirectivity.



Gain

Gain can be defined as

G =maximum radiation intensity

maximum radiation intensity from a ref-erence antenna with same power output

(11)



Gain

Often, the reference antenna is a linearhalf-wave antenna.If the reference antenna is assumed to be anisotropic antenna with 100 percent efficiency,the gain so defined for the subject antenna iscalled the gain with respect to an isotropicsource and is designated G0.

G0 =

maximum radiation intensity fromsubject antenna

radiation intensity from (lossless)isotropic source with same power input

(12)



Let the maximum radiation intensity of from thesubject antenna be U′m. Let this be related to thevalue of the maximum radiation intensity Um for a100 percent efficient subject antenna by a radiationefficiency factor k. Thus

U′m = kUm (13)

where 0 ≤ k ≤ 1. Therefore 12 may be written as

G0 =U′mU0

=kUm

U0, (14)

where U0 is the radiation intensity from a losslessisotropic source with the same power input. But Um

U0is

the directivity D in terms of intensities.Ranga Rodrigo (University of Moratuwa) Antenna Parameters December 15, 2008 21 / 47


Gain

G = kD (15)

where k is the efficiency factor.



Why Is Gain Always Less ThanDirectivity For a Practical Antenna?

The gain of a transmit antenna is proportional todirectivity by a constant known as the antennaefficiency. This efficiency accounts for the powerlosses within the antenna. Gain therefore is not onlya descriptor of the antenna’s power focusing abilitybut also takes into account power losses within theantenna. Therefore, gain is always less than thedirectivity.


Antenna Parameters Efficiency

Efficiency

1 If the efficiency of an antenna is 100%, all thepower delivered to the antenna is radiated.

2 Usually, however, some of this power is lost inconduction and dielectic losses.

3 Therefore, the efficiency is typically less than100%.

4 A useful antenna should have an efficiency thatis in the very least in the 80% range.


Antenna Parameters Efficiency

Efficiency

Pout = I2Rrad. (16)

Pin = I2(Rrad + Rloss). (17)

where Rloss is the loss resistance.

Efficiency =Pout

Pin=

Rrad

Rrad + Rloss. (18)


Antenna Parameters Antenna Aperture

The Aperture Concept

Consider a horn antenna as a receiving antenna.The power absorbed by the antenna is SA where Ais the aperture area and S is the magnitude of powerdensity of the incident electromagnetic wave.



Effective Area of an ApertureAn antenna has an effective aperture area as areceive antenna; it is the area which extracts thepower from the incident electromagnetic wave.As a transmit antenna, the effective area is thearea from which electromagnetic waves aregiven out.For aperture antennas such as horn antennas,the effective area is closely related to thephysical area.However, for wire antennas such as dipoles, theeffective area is much bigger than the physicalarea of the antenna.



Effective Aperture

Consider any type of collector or receivingantenna. The antenna collects power from thewave and delivers it to the terminating or loadimpedance ZT connected to its terminals.

The antenna may bereplaced by itsThevenin equivalenthaving an equivalentvoltage of V andinternal antennaimpedance ZA.

ZA

V

ZT



The voltage V is induced by the passing wave andproduces a current

I =V

ZT + ZA. (19)

In general, the antenna and terminating impedancesare complex, thus

ZT = RT + jXT (20a)

ZA = RA + jXA (20b)The antenna resistance comprises of two parts, aradiation resistance Rrad, and a loss resistance Rloss,that is

RA = Rrad + Rloss. (21)Ranga Rodrigo (University of Moratuwa) Antenna Parameters December 15, 2008 29 / 47


Let the power delivered by the antenna to theterminating impedance be W. Then

W = I2RT (22)

I =V√

(Rrad + Rloss + RT)2 + (XA + XT)2. (23)

W =V2RT

(Rrad + Rloss + RT)2 + (XA + XT)2 . (24)



Effective ApertureThe ratio of power W in the terminating impedanceto the power density of the incident wave will bedefined as the effective aperture Ae. Thus,

Effective aperture =WP

= Ae, (25)

where P = |P|. Thus,

Ae =V2RT

P[(Rrad + Rloss + RT)2 + (XA + XT)2

] . (26)

V is the induced voltage when the antenna isoriented for maximum response and the incidentwave has the same polarization as the antenna.


Antenna Parameters Maximum Effective Aperture

Maximum Effective ApertureWhen the terminating impedance is the complexconjugate of the antenna impedance, so that themaximum power is transferred, and if the antennalosses are zero (Rloss = 0 and therefore RA = Rrad),

XT = −XA (27)

RT = Rrad. (28)

So, the largest possible power is

W′ =V2RT

4R2T

=V2

4Rrad. (29)


Antenna Parameters Maximum Effective Aperture

The power W′ is delivered to the terminatingimpedance under the conditions of maximumpower transfer and zero antenna losses.The ration of this power to the power density ofthe incident wave is the maximum effectiveaperture Aem.

Maximum effective aperture =W′

P= Aem. (30)

Aem =V2

4PRrad. (31)


Antenna Parameters Maximum Effective Aperture of a Hertzian Dipole

Maximum Effective Aperture of aHertzian DipoleConsider the Hertzian dipole with length dl λ, witha uniform current distribution. The terminatingresistance RT is assumed to be equal to Rrad, andthe antenna loss resistance Rloss is assumed to beequal to zero.

V = Edl. (32)

We found that

Rrad = 80π2

(dlλ

)2

Ω (33)


Antenna Parameters Maximum Effective Aperture of a Hertzian Dipole

The (magnitude) of the power density in free space is

P =E2

η=

E2

120π(34)

The maximum effective aperture of the Hertziandipole is

Aem =120πE2(dl)2λ2

320π2E2(dl)2 =3λ2

8π= 0.119λ2. (35)


Antenna Parameters Maximum Effective Aperture of a Half-Wave Dipole

Maximum Effective Aperture of aHalf-Wave Dipole

y

x

z

P

L2

−L2

θ′

θ

r′

r

z′

dz′

iθ′

iφ′

z′ cosθ

Half-wave dipole as a number of Hertzian dipoles.



Maximum Effective Aperture of aHalf-Wave DipoleWe noticed that when we considered the Hertziandipole at a distance z′ from the origin of a half-wavedipole, the current in this Hertzian dipole isI0 cos(πz′/L) cosωt. For the half-wave dipole L = λ/2.The infinitesimal voltage dV due to the voltageinduced by the incident wave is this Hertzian dipoleof length dz′ is

dV = E(dz′) cos(2πz′/λ) (36)

We can find the total induced voltage by integratingover z′.



Radiator Aem D D (dB)

Isotropicλ2

4π1 0

Short dipole1 3λ2

8π1.5 1.76

Half-wave dipole30λ2

73π1.64 2.14

1A short dipole is always of finite length even though it may be veryshort. The current along a short dipole is uniform.


Received Power in a Communication System

Received Power in a CommunicationSystem

The received power in relation to a radio link isgiven by the Friis transmission formula, whichwas presented by harold T. Friis while at BellLabs in 1946.



Suppose the transmit antenna and the receiveantenna are aligned for maximum reception.If the transmit antenna is isotropic, the powerdensity of the electromagnetic wave at thereceiver is

Sr =Pt

4πR2 (37)

where Pt is the transmit power, and R is thedistance from the transmitter.However, if the transmitter has a gain Gt, thepower density at the receiver is

Sr =PtGt

4πR2 (38)



If the effective area of the receive antenna is Aer,the power collected at the receiver, Pr = SrAer, is

Pr =PtGtAer

4πR2 . (39)

Directivity and gain are related to the aperturearea. Thus, the gain of the transmitter Gt isrelated to its aperture Aet by

Gt =4πλ2 Aet. (40)

Therefore, from Equations 39 and 40 we get

Pr =PtAetAer

R2λ2 . (41)



Power given by Equation 41 can also beexpressed in terms of antenna gains:

Pr = PtGtGr

(λ

4πR

)2

. (42)



Example 3.1What is the maximum effective aperture of amicrowave antenna with a directivity of 800?Assume that the antenna is designed for 10 GHz.

Solution 3.2

D =4πλ2 Aem.

λ = 3 cm = .03 m.

Aem = 800.032

4π= 0.057 m2.



Example 3.3What is the maximum power received at a distanceof 2 km over a 7 GHz free-space link consisting of atransmit antenna with a 30 dB gain and a receiveantenna with a 25 dB gain? The gains areexpressed with respect to a lossless isotropicsource. The transmit antenna input power is 1 kW.



Solution 3.4

Pr = PtGtGr

(λ

4πR

)2

,

Pr (dBW) = Pt (dBW) + Gt (dB) + Gr (dB) + 10 log(λ

4πR

)2

,

= 30 + 30 + 25 − 97.7,= −12.7 (dB) ,= 0.2 pW .


Variation of Directivity with Length


As the length of the antenna becomes longer,the antenna beamwidth becomes narrower.However, as the antenna becomes longer that awavelength, side-lobes appear in the radiationpattern.This means that the radiation is now not focusedin one main-lobe but is focused in multipledirections.



Reference

Indra J. Dayawansa.Antennas and propagation.Lecture notes, University of Moratuwa, 2003.

John D. Kraus.Antennas.McGraw-Hill, 1950.

Nannapaneni Narayana Rao.Elements of Engineering Electromaganetics.Prentice Hall, 4th edition, 1994.


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