-
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
3ANTENNAS FOR COMMUNICATIONS
E H
J
nearE
nearH
scatteredEapertureE
veq
farE
farH
scatteredE
0
inH
E
4ANTENNAS FOR COMMUNICATIONS
scatteredH
0inH0
inE,
-
3
apertureS EnM
nearE
nearH
scatteredE
apertureS HnJ
farE
farH
scatteredE
S surfaceJ n H
5ANTENNAS FOR COMMUNICATIONS
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
6ANTENNAS FOR COMMUNICATIONS
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
7ANTENNAS FOR COMMUNICATIONS
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
8ANTENNAS FOR COMMUNICATIONS
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
9ANTENNAS FOR COMMUNICATIONS
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 '
10ANTENNAS FOR COMMUNICATIONS
'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
11ANTENNAS FOR COMMUNICATIONS
' '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
12ANTENNAS FOR COMMUNICATIONS
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
13ANTENNAS FOR COMMUNICATIONS
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
15ANTENNAS FOR COMMUNICATIONS
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
16ANTENNAS FOR COMMUNICATIONS
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 , , , '''
17ANTENNAS FOR COMMUNICATIONS
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
18ANTENNAS FOR COMMUNICATIONS
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
19ANTENNAS FOR COMMUNICATIONS
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
20ANTENNAS FOR COMMUNICATIONS
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
21ANTENNAS FOR COMMUNICATIONS
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
22ANTENNAS FOR COMMUNICATIONS
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
23ANTENNAS FOR COMMUNICATIONS
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
24ANTENNAS FOR COMMUNICATIONS
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
25ANTENNAS FOR COMMUNICATIONS
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
,
26ANTENNAS FOR COMMUNICATIONS
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
27ANTENNAS FOR COMMUNICATIONS
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
28ANTENNAS FOR COMMUNICATIONS
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
29ANTENNAS FOR COMMUNICATIONS
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
30ANTENNAS FOR COMMUNICATIONS
[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
31ANTENNAS FOR COMMUNICATIONS
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
32ANTENNAS FOR COMMUNICATIONS
[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
33ANTENNAS FOR COMMUNICATIONS
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
34ANTENNAS FOR COMMUNICATIONS
[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
35ANTENNAS FOR COMMUNICATIONS
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'''
'
'
'' ' ' '
36ANTENNAS FOR COMMUNICATIONS
-
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
37ANTENNAS FOR COMMUNICATIONS
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
38ANTENNAS FOR COMMUNICATIONS
[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
39ANTENNAS FOR COMMUNICATIONS
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
40ANTENNAS FOR COMMUNICATIONS
440D
For a directivity of 40 dB the beamwidth is approximately 1