antennas2.v3

1

ANTENNAS FOR COMMUNICATIONS

Radiation

DEFINITIONS AND UNITS

: electric field intensity [V/m]

: magnetic field intensity [A/m]

: electric flux density [C/m2] :

1 A/m = 4 10-3 Oe E

H

D

ED

: electric flux density [C/m2] :

: magnetic flux density [Vs/m2=Wb/m2=T] :

: electric current density [A/m2]

: magnetic current density [V/m2]

: electric charge density [C/m3]

: magnetic charge density [Wb/m3]mF101 9

in free space:

1 T = 104 G

DB

J

M

veqvmq

HB

ED

2ANTENNAS FOR COMMUNICATIONS

: electric permittivity [F/m] : = r 0 : magnetic permeability [H/m] : = r 0

mF 10360

mH 10 4 70

2

ELECTROMAGNETIC POTENTIALS

Obtention of the EM fields radiated or scattered by electric and/or magnetic sources.

Sources

J

M

veq vmq Radiated or Scattered Fields

Maxwells Equations+

Boundary Conditions


E H

J

nearE

nearH

scatteredEapertureE

veq

farE

farH

scatteredE

0

inH

E


scatteredH

0inH0

inE,

3

apertureS EnM

nearE

nearH

scatteredE

apertureS HnJ

farE

farH

scatteredE

S surfaceJ n H


scatteredHS surfaceM n E ,

For an homogeneus medium and time-harmonic fields and functions, the EM fields should satisfy:

HjME

F d

vm

ve

qB

qD

EjJH

Faraday

Ampre

Gauss

0 veqjJ

Continuity equation

Maxwells equations


or

vm

ve

qMjJHH

qJjMEE

22

22

22with

4

A simplified and straightforward procedure needs additional functions:

Maxwells Equations+SourcesJ

M

veq vmq Radiated FieldsE

H

+Boundary Conditions

AB

JAA 22


PotentialsA

F

e m FjHFD

AjE

m

e

vmmm

veee

qMFF

q

22

22

22

WAVE EQUATION SOLUTIONS

EM fields can be found by the superposition of the fields generated by the electric sources and the magnetic sources:

me

me

HHH

EEE

contribution of

contribution of magnetic sourcesSourcesJ

M

veq vmq


contribution of electrical sources

5

Dual set of equations:

mm HjME

for magnetic sources

ee HjE

for electric sources

vmm

m

mm

mm

qH

E

EjH

0

mE F

0

e

vee

ee

H

qE

EjJH

1eH A

Lorentz condition


2 2

2 2

m m

m m vm

H j F

F F Mq

2 2

2 2

e e

e e ve

E j A

A A Jq

0 mjF

0 ejA

To solve these four (two scalar and two vector) wave equations the Greens function technique is used.

The solution of an scalar wave equation is found when a unit source (Dirac delta) is used as a driving function. This solution is the Greens function.

fieldpoint

rR

'r'rrR

sourcepoint z

'',', 22 rrrrGrrG

ee Rjrrj '


'rrRR

x

y

R

err

errGj

4'4',

- outwardly traveling wave+ inwardly traveling wave

6

By a superposition of the impulse response solutions with the Dirac delta source at different locations as the actual driving function, the solution of the equation is found. In the limit this is an integral.

harmonic solution

''

rrtrq

general solution

' '4

' ' ' ','

'

'dv

rrerqdvrqrrGr

V

rrjve

Vve

e

V : volume enclosing the sources


' '4

, ,

'dv

rrv

trqtr

V

ve

e

1

v

v : speed of propagation

Dual set of solutions and quantities:for electric sources for magnetic sources

1 eRj

1 eRj

From these potentials we can get the electric and magnetic fields and flux densities:

' ' 4

''4

1

'

'

dvR

erMF

dvR

erq

Rj

V

V vmm

' ' 4

' ' 41

'

'

dvR

erJA

dvR

erq

Rj

V

j

V vee


FjHFD

m

AjE

AB

e

7

However, by solving Maxwells equations only for electric sources we can get the fields for these electric sources and also for an equivalent problem with magnetic sources:

1AJHE ee

1FMEH mm

h : intrinsic impedance


Example: Fields radiated by a very thin linear wire of a very short length (infinitesimal dipole) with a constant current distribution (uniform current distribution).

'rrR

z

l/2

fieldpoint

rR

zzr ''

very short dipole: l

8

z fieldpoint

R ' '

4 'dv

RerJA

Rj

V

R

vector potential:

xy

r

zzr ''

xr cossinsincossin

AArrAA

far from the wire: rR

sincos4

4

'4

2

2

rr

elIA

zr

elIdzr

eIzA

rje

rje

l

l

rje

radiated fields at the field points ( ):0

J


zy

0cossinsinsincoscoscos

jH

jJHE

AH

1

sin cos cos cos sin sin sin cos sin cos cos sin 0

rxyz

11cos

sin114

2

elIE

re

rjlIjH

rjer

rje

0

111 4sin

2

2

2

E

errjr

lIjE

rjr

rje

r

in the limit: rRl 0 and consequently:the equations for the electric and magnetic fields are valid everywhere but in the source itself

2

r

radiansphere

rE E r E E far-field border

(radiated field region) for

small antennas


are valid everywhere but in the source itself

variations with r2 and r3 are neglected when:

21 rr

out of the radiansphere these variations can be neglected

2

9

FAR-FIELD RADIATION

Antennas of finite dimensions radiate spherical waves. In spherical waves the vector potential takes the general form:

,, ,, ,, rArArrAA r

Far from the sources the expression for the vector potential can be reduced to:

rr

eAArAArj

r , , , '''


Neglecting the terms of higher-order than 1/r the far-fields can be expressed as...

AH

1 AjAjE

AHe

jAjEe

ee

ee

er

EAjH

EAjH

H

0 0erE

FEm

1

FjFjHm

electricsources eE j A

eE j A


FjHFjH

H

m

m

mr

0

mm

mm

mr

HFjEHFjE

E

0

magneticsources

10

... or in a compact form:

ArrjEe

l t i

Right Hand Rule

EH

EH

r

r

z TEMr

ee ErArjH

1

FrrjHm

HrFrjE

electric sources

magnetic sources

r

H r


yxmm HrFrjE

me

me

HHH

EEE

total radiated fields

equations valid whereas:

22Dr

D: largest dimension of the source (D>l)

RADIATION AND SCATTERING EQUATIONS

In the vast majority of problems the main difficulty is solving the integral that yields the vector potentials:

' ' 4 '

dvR

erMFRj

V

' ' 4 '

dvR

erJARj

V

By using the expressions of the fields derived from the curl of the vector potentials

and the vector identities

4 ' RV 4 ' RV

AH

1

FE

1


AHe FEm

JfJfJf MfMfMf

11

we come up with expressions suitable for being calculated with the help of a computer.

'11' 1

41

2

2'

dveRjME

dveR

RjJH

Rj

Rj

Ve

The corresponding fields can also be written in integral form as a function of the electric and magnetic sources.

These expressions are valid everywhere. Until this point no approximation has been done.

4 2'

dveR

MEVm

ee HjE

1

mm EjH

1

near

fiel

ds


n

According to the ratio r'/r the distance between the source points and the field (or observation) points, can be approximated with

212 ''

and the vector potential is:

ar fi

elds ...sin

2'cos' 22

r

rrrR

2

222 cos'2'1cos'2'

rr

rrrrrrrR field point

xy

z

rR

'r


fa

' ' 4

' ' 4

22

sin2'

cos'

''dveerJ

redv

RerJA r

r-jrj

V

r-jRj

V

12

Constraining the phase errors in the integral to /8, i.e.

2

max

22 2

8sin

2' Drr

r

(Fraunhofer)

'r DD: diameter of the minimum sphere that encloses the antenna

In the far zone of the sources the integration still needs to be performed to compute the vector potentials although now it is quite simple

Nr

edverJr

eAr-j

rj

V

r-j

4' '

4cos'

'

Lr

edverMr

eFr-j

rj

V

r-j

4' '

4cos'

'

ar fi

elds

, : radiation vectorsN

L


vector potentials, although now it is quite simple.fa

The previous development for the far fields can also be explained by taking into consideration that and are parallel. In that case,

The radiation vectors can be expressed as:

cos'' rrrrrR R

r

R

The radiation vectors can be expressed as:

These vectors do not depend with the distance to the field point and are only valid at far field

' ' ' ' ' '

'

'dverJdverJN rj

V

rrj

V

' ' ' ' ' '

'

'dverMdverML rj

V

rrj

V

ar fi

elds

x y

z

r

'rr'r


These vectors do not depend with the distance to the field point, and are only valid at far-field.fa

13

The radiation vectors are easily computed using the Fourier Transform: source points: field point direction:

zzyyxxr ''''

zyxr cossinsincossin wavenumber vector:

phase shift: and finally:

yzyxr cos sinsin cossin

zyx zyx

zzyxrrr zyx '' '' '

JFTdveeerJdverJN DzjyjxjVrrjV zyx

3'''

'

'

'' ' ' ' ar

fiel

ds


The radiation vectors are the Fourier Transform of the current distributions. MFTdveeerMdverML DzjyjxjVrrjV zyx

3

'''

'

'

'' ' ' '

fa

Procedure for solving a problem using the radiation and scattering integrals:

Identify the volumes V' (or surfaces S') where theactual current J or equivalent currents J and/or Mactual current J or equivalent currents JS and/or MSexist.Specify the actual current density J and/or theequivalent currents JS and/or MS formed by using,

aS

aS

EnM

HnJ

Determine N and L.Determine the potentials A and/or F

: unit normal vector to the surface S'

: total electric and magnetic fields over the surface S'

n

aa HE

,


Determine the total radiated E andH fields using the approximations.

far fieldsCombine them to find the totalfields.

or determine the fields E and H dueto the electric and magneticcurrents, respectively.

near and far fields

14

The total radiated fields by the electric and the magnetic sources are then:

Er 0 Hr 0

The electric and magnetic field components are orthogonal to each other and form a TEMr mode field.

NLr

ejE

NLr

ejE

rj

rj

4

4

LN

rejH

LNr

ejH

rj

rj

4

4

ar fi

elds


The radial component of the fields is not zero, but is very small when compared with the and F components.

fa

Maximum Directivity

Electrically small antennas have a directivity of 1.5

z

We had already seen that for an electrically small antenna with z directed current and arbitrary current distribution

'' '

' ' 'jkrr oV VN J r e dv J r N z

The radiated Electric field is

sin( )2

jkr

oeE j N

r

x

y

z

0 J J z


xS

0

15

Radiation intensity:

Directivity:

0 1530

45

60

75285

300

315

330345

-10dB

-20dB

-30dB

E-plane

222022

sin32

lIkU

Maximum Directivity

Directivity:

zz

90

105

120

135

150165180195

210

225

240

255

270

0 1530

45

60300

315

330345

-10dB

-20dB

30dB

beamwidth: 90

Radiation pattern cuts

2sin 5.1D dBD 76.1max


Image from:http://www.elliskaiser.com/doughnuts/tips.html

75

90

105

120

135

150165180195

210

225

240

255

270

285 -30dB

H-plane

Longer dipoles: D AND Rrad

2002

(Bar

celo

na)]

/2 dipole

Maximum Directivity

a et

al.,

Ant

enas

,Ed.

UPC

, 2da

. Edi

cin

,

maximum directivity dipole


[Imag

es fr

om: A

. Car

dam

a

16

Resistencia de Radiacin []Directividad

3.33 (=5.22 dB)n

()

002

(Bar

celo

na)]

Maximum Directivity

2.4 (=3.80 dB)

73

1.64 (=2.16 dB)199

Res

iste

ncia

de

Rad

iaci

et a

l., A

nten

as,E

d. U

PC, 2

da. E

dici

n, 2


Full wave dipoleHalf wave dipoleH=5/8

[Imag

e fro

m: A

. Car

dam

a

arm length: H = larm radius: aconductivity:

Z R jX

Maximum Directivity

Resistance

ant ant antZ R jX

radR tlX


[Images from: A. Cardama et al., Antenas, Ed. UPC, 2da. Edicin, 2002 (Barcelona)]

Reactance a BW

17

HALF-WAVELENGTH DIPOLE A fundamental form of antenna is a wire whose length is half the

transmitting wavelength. It is a standing-wave antenna whose length is l= /2 or L=0.5.

It is the unit from which many more complex forms of antennas are fabricated.

Maximum Directivity

Directivity is D =1 64=2 15 dB

0 cos( )I z I kz 2

2

sincos5.0cos64.1D


Directivity is Dmax=1.64=2.15 dB Maximum effective area is 0.13 2 Radiation Resistance is 73 Most dipoles require a little pruning to reach the desired resonant frequency. The feed-point impedance is low at the resonant frequency and odd

harmonics thereof. The impedance is high near even harmonics.

HALF-WAVELENGTH DIPOLE: RESONANCE

Dipole2

Zant=73 + j 43

02 (B

arce

lona

)]

Resonant Dipole

Zant= 73 295.0

Maximum Directivityet

al.,

Ant

enas

,Ed.

UPC

, 2da

. Edi

cin

, 20


[Imag

e fro

m: A

. Car

dam

a e

18

/2 DIPOLE: RADIATION CHARACTERISTICS The radiation pattern of a dipole antenna in free space is strongest at right

angles to the wire.

, max tDD 2

2 cos5.0cos, t

Maximum Directivity

linear scale

dB scale

-3dB=78

, max tDD 2sin,t


Beamwidth (-3 dB) is 78.

E-plane radiation pattern

Maximum Directivity

Wire antennas have a limited directivity. At most D=3.3, and for a half-wave dipole it is 1.64.

Resonant antennas will have dimensions of the order of half a wavelenght and they will have broad patterns and small directivities.

This is the result of the relationship between current distribution and radiation vector.

JFTdveeerJdverJN DzjyjxjVrrjV zyx

3'''

'

'

'' ' ' '

MFTdveeerMdverML DzjyjxjVrrjV zyx

3'''

'

'

'' ' ' '


19

Consider a current distribution on a LxL square. The current distribution has a triangular distribution. The radiation vector has a in the transformed domain a duration that is inversely proportional to L.

Maximum Directivity

L 2L


In consequence, the larger the antenna dimension the larger the directivity

For aperture type antennas

Maximum Directivity

E

E

H-plane sectoral horn

E-plane sectoral horn

pyramidal horn

0E 0E0E


[Image courtesy of Universidad Politcnica de Valencia]

20

Maximum Directivity

eA

The aperture efficiency of an antenna is the ratio of the effective antenna aperture and its physical area.

ap A

02

4

me

DA

Since


2

4o apD A

We obtain

Directivity is proportional to the antenna surface in terms of

Maximum Directivity

Example

Typically for a reflector antenna the aperture efficiency is of the order of 0.6 and for a horn antenna 0.5. Estimate the dimensions of an antenna aperture to have a directivity of 40 dB at a frequency of 10 GHz.

At 10 GHz, = 3 cm 24 210 3 104 ap

A

Assuming an aperture efficiency of 0.5, A=0.36 m2, and an aperture side of 60 cm.Note that to have the same directivity at a frequency 10 times smaller (1 GHz) would require an aperture side 10 times larger (6 m).Note also that by specifying the directivity, and assuming a pencil beam type of radiation pattern the beamwidth is also specified


440D

For a directivity of 40 dB the beamwidth is approximately 1

antennas2.v3

Documents

Transcript of antennas2.v3