Download Module 6: Antennas - MSU College of Engineering

Module 6:
Antennas
6.0 Introduction
In chapter 2, the fundamental concepts associated with electromagnetic radiation were
examined. In this chapter, basic antenna concepts will be reviewed, and several types of antennas
will be examined. In particular, the antennas commonly used in making EMC-related
measurements will be emphasized.
6.1 The Radiation Mechanism
Antennas produce fields which add in phase at certain points of space. Consider a loop of
wire that carries a current.
R1
dI1
R2
D
dI2
Here two elements of current d I 1 and d I 2 are separated by a distance D. The current elements
are located at distances R1 and R2, respectively from a distant observation point. If
R2-R1 0.1λ
D 0.1λ
Then the fields produced by the current elements add out of phase, and the amount of radiation is
small. However, if
R2-R1 0.1λ
D 0.1λ
Then the fields produced by the current elements add in phase, and the amount of radiation is
large.
-Reception Mechanism
Electromagnetic fields which are incident upon an antenna induce currents on the surface
of the antenna which deliver power to the antenna load.
6-1
induced
current
incident
field
ZL
load
impedance
transmission
line
antenna
6.2 Radiated Power
The power radiated by a distribution of sources is that power which passes through a
sphere of infinite radius. This, therefore, is the power which leaves the vicinity of the source
system, and never returns.
In chapter 2 the time-average Poynting vector was presented
P =
1
R e { E × H ∗}
2
At points far from the antenna (the radiation zone)
( )
[
µ e − jkr
θNθ + φNφ
E r ≈ − jω
4π r
( ) ( )
(θ φ ) ∫ ( ) H r ≈
where
N
,
]
r×E r
η
J S r ′ e jk (r ⋅r ′ )d v ′
=
v
is known as the “radiation vector.” The radiation vector is related to the vector potential by
(
µ e − jkr
A r ,θ , φ) =
N (θ , φ )
4π r
with
6-2
( )
∫ ( ) η
) ( ) ( )
(
) ( ) }
(
{
η
η
ωµ [
]
η π
φ ] [
φ ] [θ
η
µ e − jkr
A r ≈
4π r
Now in the radiation zone
J S r ′ e jk (r ⋅r ′ )d v ′
v
 r × E ∗  
1 

P ≈ ReE × 
2

 

A × B × C = B A ⋅C − C A ⋅ B
Using the vector identity
P ≈
1
Re E ⋅ E ∗ r − E ⋅r E ∗
2
E ⋅E∗
≈r
2
......because E ⋅ r = 0
2
......because
[
1 

N θ N θ∗ + N φ N φ∗
≈r


2 4 r 
θ N θ + N φ ⋅ N θ∗ + N φ∗ = N θ N θ∗ + N φ N φ∗
Finally
P ≈r
]
1
 N 2 + N 2
φ 

r 2 8 λ2  θ
This represents the average power flow density and lies in the direction of wave propagation.
The power radiated through a sphere of infinite radius is given by
W = lim ∫ n ⋅ P ds
r →∞
s
Applying the expression for the time average Poynting vector leads to
π 2π
W = lim
r →∞
∫
0
2 
2
 1 η 
 2
r
r
N
N
⋅
+
φ   r sin θd θd φ
∫0  r 2 8 λ2  θ

6-3
π 2π
=
∫
0
Let K =
2
2
η 


N
N
+
φ  sin θd θd φ
∫0 8 λ 2  θ
2
2
dW
η
= 2  N θ + N φ  ......”radiation intensity”
dΩ 8λ
= power radiated per unit solid angle. ......where
d Ω = sin θd θd φ .
The total radiated power is then
π 2π
W =
∫ ∫ K (θ , φ )d Ω .
0 0
6.3 Antenna Terminology
Antenna Patterns
Radiation pattern - A plot of the radiation characteristics of an antenna. There are two
types of radiation patterns:
1. Power pattern - A plot of the radiated power at a constant radius.
2. Field pattern - A plot of the electric or magnetic field magnitude at a constant radius.
An antenna pattern consists of a number of lobes. The largest lobe is usually called the
main lobe, while the other smaller lobes are called side lobes. The minima between lobes are
called nulls.
main
lobe
side lobe
null
Radiation patterns are three-dimensional, but are usually measured and displayed as two6-4
dimensional patterns, which are sometimes called cuts. For most antennas, two cuts give a good
representation of the three-dimensional pattern.
The radiation patterns of linearly polarized antennas are often specified in terms of Eplane and H-plane patterns. The E-plane contains the direction of maximum radiation and the
electric field vector. The H-plane contains the direction of maximum radiation and the magnetic
field vector.
E
Epl
an
e
H-
pla
ne
H
No antenna has a truly isotropic pattern (one which is the same in all directions). Rather
antennas (real ones anyway) tend to radiate more effectively in some directions rather than others.
Directive gain - The ratio of the radiation intensity K(θ,φ) to the uniform radiation intensity for
an isotropic radiator with the same total radiation power W.
g d (θ , φ ) =
K (θ , φ ) 4 π
K (θ , φ )
=
W
W 
 
 4π 
 4π 
 is the total power radiated by an isotropic radiator per unit solid angle.
W 
.....where 
Directivity - The maximum value of directive gain.
Gain - Directivity expressed in dB.
G = 10 log 10 (directivity) = gain in dB
Beamwidth - The beamwidth of a radiation pattern is the angle between the half-power points of
the pattern.
6-5
3 dB (half power)
points
φ
side lobe
Radiation efficiency - The radiation efficiency of an antenna is the ratio of the power radiated by
the antenna to the total power supplied to the antenna. The total power supplied to the antenna
consists of the power radiated and the power given up to resistive losses.
E =
W
W + WL
......where
E = radiation efficiency
W = power radiated
WL = power lost
Radiation resistance - The radiation resistance of an antenna is the equivalent resistance through
which its input current must flow in order that the power dissipated in the resistance is equal to
the total radiated power.
1
I 0 I 0∗ R r = W
2
or R r =
2W
......radiation resistance
I 0 I 0∗
......where I0 is the input current to the antenna.
From the stand point of the source that drives an antenna, radiation resistance is
indistinguishable from Ohmic resistance. In both cases, the source must continuously supply
energy to the antenna in order to keep the current amplitude constant with time. In the case of
Ohmic resistance, this resistance converts energy into propagating electromagnetic waves.
Input impedance - An arbitrary antenna with a pair of input terminals ‘a’ and ‘b’ is shown below.
6-6
Z in
I in
a
V
Antenna
b
When the antenna is not receiving power from waves generated by other sources the
Thevenin equivalent circuit looking into the terminals of the antenna consists only of an
impedance
Z in =
V
= R in + jX in
I in
where Rin is the input resistance and X in is the input reactance. The input resistance is the sum of
two components
R in = R ri + R L
where Rri is the input radiation resistance and R L is the input loss resistance. R L accounts for that
portion of the input power that is dissipated as heat, while the input radiation resistance R ri
accounts for power that is radiated by the antenna. R ri is related to Rr by
2
I 
R ri =  m ax  R r .
 I0 
Radiation efficiency can be expressed
ηr =
Pra d
R ri
.
=
Pin
R ri + R L
6.4 Hertzian Dipole
The simplest radiation source consists of a short segment of current
6-7
z
dz
2
A, B, E
r
I
dz
y
−
dz
2
x
A Hertzian dipole consists of a uniform current I flowing in a short wire dz terminated by point
charges.
Here
( ) δ ( )δ ( )
 Iz
J r =

It is seen that
I =
x
0
∫ J ⋅ds =
−dz
dz
≤z≤
2
2
...... e lse w h ere
y ...... fo r
∫ ∫ I δ ( x )δ ( y )d xd y = I
x y
The charge associated with the current is found using the continuity equation
∇ ⋅ J = − j ωρ ⇒ ρ = −
1 dJ z
jω d z
The current density may be expressed
 
dz 
dz  

J = Iz δ ( x )δ ( y ) u  z +  − u  z −  

2
2 
 
where u(t) represents the unit step function.
6-8
J
z
dz
−
2
ρ=
=
dz
2
1
d
I δ ( x )δ ( y )
jω
dy
dz 
dz  
 

+
−
−
u
z
u
z




 

2
2  

I
dz 
dz  
 

δ ( x )δ ( y ) − δ  z +  + δ  z −  

2
2 
jω
 
( ) µ ∫ ( )
π
( ) µ ∫ (
A r =
where
0
4
v
e − jkR
J r′
dv ′
R
R = r − r ′ ≈ r = r for r >> d z
so,
e − jkr
A r ≈
4π r
J r ′ )d v ′
0
− jkr
≈
or
µ0 e
4π r
d u (t )
= δ (t )
dt
dz
dz
, (-) point charge at z = −
2
2
= (+) point charge at z =
Vector potential
......because
v
dz
2
∫ ∫ δ ( x ′ )δ ( y ′ )d x ′d y ′ ∫ Iz d z ′
x y
( ) µ
z ′=−
e − jkr
A r ≈z
Idz
4π
r
0
6-9
dz
2
θ
( ) (
now use z = r co s θ − θ sin θ to get
µ0
e − jkr
A r ≈ r c os θ − sin θ
Id z
4π
r
E-M fields
)
φ
∂Ar 
∂
−
rA
(
)
θ
∂θ 
r  ∂ r
B =∇× A =
φ
∂
µ
∂
= 0 Idz  (− sin θe − jkr ) −
∂θ
4π
r  ∂r
=
φ
µ0
sin θ  − jkr

Id z  jk sin θ +
e
4π
r 
r 
µ
so
B =φ
E =−
0
4π
Id z
1 + jkr
sin θ e − jkr
2
r
j
∇ × B at all points where J = 0
ωµ 0 ε 0
j
=−
ωµ 0 ε 0
!

e − jkr  
 co s θ

r 

!
!
 r

∂
θ ∂
B φ sin θ −
rB φ 

r ∂r
 r sin θ ∂θ

(
)
!
( )

j Id z  r
∂  1 + jkr
θ ∂  1 + jkr
2
− jkr 
− jkr 
e
e
sin
sin
=−
θ
−
θ





 r ∂r  r
 
ωε 0 4 π  r sin θ ∂θ  r 2
"
#
j Idz  r 1 + jkr
E =−
2 sin θ cos θ e − jkr
2

ωε 0 4 π  r sin θ r
#
θ
jkr − 1 − jkr  − jkr 
 k
− sin θ  − j [1 + jkr ] +
e 
 r

r
r2

6-10
!
!
 1 + jkr − k 2 r 2   − jkr
j Id z  1 + jkr
2 co s θ + θ sin θ 
=−
 e
r
r3
r3
ωε 0 4 π 


now use
$ !
µ 0 η0
.....where η0 =
=
k
ε0
1
1
=
ωε 0 ω µ 0 ε 0
!
Then
Id z e − jkr
E =r
4π r
µ0
ε0
 2 η0
η0
2 
Id z e − jkr 
1 
co
s
j
θ
θ
ωµ
+
+
+
+


2 
2  sin θ .
0
r
j
r
4
r
r
j
r
ωε
π
ωε
0
0




$ !µ
Thus the fields may be expressed:
B =φ
Er =
Eθ =
Id z
 jk 1 
sin θ  + 2  e − jkr
4π
r 
 r
0
Idz
j 
1
2 η0 co s θ  2 − 3  e − jkr
4π
kr 
r
Id z
j 
 jk 1
η0 sin θ  + 2 − 3  e − jkr
kr 
4π
 r r
It is seen that these expressions contain terms having three different rates of decay: 1/r,
1/r2, and 1/r3.
Near-zone fields (induction zone)
%
&
The near zone fields are those which are strongest near r = 0 , or when r <<
these are the terms of B which vary as 1/r 2 and the terms of E which vary as 1/r 3:
'
[(
(
Id z e − jkr η0
2 r co s θ + θ sin θ
E ≈−j
4π r 3 k
' φ( µ
B≈
e − jkr
Id z sin θ 2
4π
r
0
6-11
]
λ
. Thus
2π
&
The near zone E -field looks like the field of an electrostatic dipole.
The near zone fields do not contribute to radiated power. Instead they result in reactive
power (time changing energy stored in the fields near the antenna). Only the 1/r terms contribute
to radiated power.
Far-zone fields (radiation zone fields)
& )
The far-zone fields are those which are strongest as r → ∞ . Thus, these are the terms
of E and B which vary as 1/r:
' φ( µ
π
* +
B≈
0
e − jkr
jk sin θ
r
Id z
4
jk Idz
e − jkr
η sin θ
E ≈θ
4π 0
r
The radiation zone fields are those that contribute to radiated power. Note that fields form a
transverse wave.
Radiated Power
-
P =
/
, . .
W = lim ∫ n ⋅ P ds
- -
r →∞
s
1
R e { E × H ∗ } where n = r and d s = r 2 d Ω
2
0
B has terms that go as 1/r and 1/r 2 and E has terms that go as 1/r, 1/r 2 and 1/r3.
2
Since d s ≈ r as r → ∞ only the 1/r terms contribute to W.
1
0
where E r and H
- - ,
0 1
1
R e { E r × H r ∗} ⋅ r r 2 d Ω
∫
r →∞ 2
W = lim
r
are the 1/r terms from E and H called the “radiation zone” fields.
π 2π
W =
1 r r∗ 2
∫0 ∫0 2 E θ H φ r d Ω
6-12
1  Id z 
= 
2  4 π 
2 π 2π
 ωµ 0
 k

sin θ   sin θ  r 2 sin θd θd φ
r
 r

∫ ∫ 
0 0
1
=
2
2π
2
π
 Id z 
3
 4 π  ωµ 0 k ∫ d φ ∫ sin θd θ
0
0
2 53 6 4 76
=2 π
1 I 2dz2
 8π 
=

2 ωµ 0 k 
 3 
2 1 6π
use:
=4 3
µ0
= k η0
ε0
ωµ 0 = ω µ 0 ε 0
1 I 2dz 2 2
=
k η0
12 π
use: k =
2π
λ
π  dz 
= η0 I 2   Watts
3 λ
2
 dz 2 
W = 4 0π I  2 
 λ 
2
2
Radiation resistance
π  dz 
1 ∗
II R r = W = η0 I 2  
2
3 λ
2π
or R r = η0
3
2
2
 dz 
 dz 
  = 8 0π 2  
 λ
λ
2
Example: Calculate the radiation resistance of a 1 cm length of uniform current if the frequency is
900 MHz and the host medium is air:
3 × 108 m s
c
λ=
=
= 3 3.3 cm
f
900 × 106 H z
2
 1 × 1 0 −2 m 
R r = 8 0π 
 = 0 .7 1 1Ω
−2
 3 3.3 × 1 0 m 
2
It is seen that the radiation resistance of a short current segment is only on the order of a fraction
6-13
of an Ohm, making it a relatively inefficient radiator.
Directive gain
K (θ )
g d (θ ) =
W / 4π
2
π 2π
W =
∫ ∫ K (θ )d Ω
0 0
1  Idz 
ωµ 0 sin θ [ k sin θ ]
4π
2  4 π 
=
g d (θ ) =
sin 2 θ
2
4
1  1  Idz 
 4   ( 2 π ) 
(
)
 
k
2
ωµ
π
 
0
 3
 3
4 π  2  4 π 
[
]
g d (θ ) =
3 2
sin θ
2
Directivity
π
D = m ax g d (θ ) = g d   = 1.5
2
[
]
G = g a in = 1 0 lo g 1 0 (1.5 ) = 1.7 6 d B
6.5 Radiation from a cylindrical dipole
The cylindrical dipole antenna is one of the most commonly used antenna in the VHF/UHF
frequency range.
The cylindrical dipole may be viewed as an open-ended transmission line which has been flared
out.
I(z)
Rg
l
+
Vg
-
z=0
Open circuited transmission line
6-14
z=l
I(z)
Rg
+
Vg
+
V0
-
-
z=0
z=-l
Dipole antenna
Far zone fields
88 88
z
8
8
z=l
I (z )
R
9
9
r
r′
I0
B ( r ), E ( r )
+
V0
_
y
x
z = -l
It is noted that I ( z = ± l ) = 0 (current at the tips of the antenna is zero).
6-15
: (: ) <
The current density present on the dipole is given by
 z I ( z )δ ( x )δ ( y )
J r =

0


....where I ( z ) = I 0
@
N
sin k (l − z )
sin kl
J r ′ e jk (r ⋅r ′ )d v ′
=
,
v
where r ′ = z ′z
: (θ φ )
N
;
for z ≤ l
elsew h ere
: (θ φ ) ∫ : ( : ) > =
A C ( ABA )
<
The radiation vector is
?
;
,
<
r ⋅ r ′ = z ′ z ⋅ r = z ′ co s θ
I z
= 0
sin kl
l
∫ ∫ δ ( x ′ )δ ( y ′ )dx ′d y ′ ∫ sin k (l − z ′ )e
x y
<
I z
= 0
sin kl
jkz ′ co s θ
dz ′
z ′=− l
l
∫ sin k (l − z ′ )e
<
z ′=− l
l
jkz ′ co s θ
dz ′
0
I z
I z
= 0 ∫ sin k (l − z ′ )e jkz ′ co s θ dz ′ + 0 ∫ sin k ( l + z ′ )e jkz ′ co s θ d z ′
sin kl 0
sin kl − l
Using the relationship
e ax
∫ e sin (b x + c )d x = a 2 + b 2 [a sin (b x + c ) − b co s(b x + c )]
ax
: (θ φ ) <
gives
N
Let
F 0 (θ , kl ) =
then
:
=z
,
zI 0 co s ( kl co s θ ) − co s kl
k sin kl
sin 2 θ
co s( kl co s θ ) − co s kl
sin θ
N (θ , φ ) =
(D <
“radiation function”
< )
GF F E > IF F H
2 I 0 F 0 (θ , kl )
r co s θ − θ sin θ
k sin kl sin θ
6-16
z
: (: )
<[
<
]
µ 0 e − jkr
θNθ + φN φ
E r = − jω
4π r
j ωµ 0 I e − jkr
=
F (θ , kl )θ ......far zone E -field
2πk
r
co s( kl co s θ ) − co s ( kl )
where F (θ , kl ) =
sin ( kl ) sin θ
<
J
: (: ) < : (: )
<
r×E r
j ωµ 0 I 0 e − jkr
=
H r =
F (θ , k l )φ
η
2 πk η r
: ( : ) θ<
: ( : ) φ<
Use ωµ 0 = k η0 then
E r =
H r =
j η0 I 0 e − jkr
F (θ , kl )
2π
r
jI 0 e − jkr
F (θ , kl )
2π r
Radiation zone fields
produced by a dipole
Radiation pattern
K (θ , φ ) =
2
2
η0 


+
N
N
2 
θ
φ 
8λ
so
2
η 4 I0
η
2
K (θ ) = 02 N θ = 02
F 2 (θ , kl )
2
8λ
8λ k
2
I0
2
= η0
2 F (θ , kl )
8π
Often times
far-zone field:
K (θ , φ ) is plotted as opposed to K (θ , φ ) since K describes the pattern of the
co s( kl co s θ ) − co s kl
sin θ
2
4
x
x
+
+
co s x = 1 −
2! 4!
K (θ ) ≈ F (θ , kl ) =
6-17
K
Special case: kl << 1 → l << λ
F (θ , kl << 1) ≈
1−
1
1
( kl co s θ ) 2 − 1 + ( kl ) 2
2
2
sin θ
1
( kl ) 2 [1 − co s 2 θ ]
≈ 2
sin θ
≈
1
( kl ) 2 sin θ → Same pattern as Hertzian dipole.
2
z
90
L
1

0.707  pow er point 
2

θ = 90
1
b eam w id th = 90
Other cases:
6-18
L
broadside
direction
L
120
90
1
60
120
0. 5
150
30
180
210
330
240
2l= lam bda/2
120
90
120
30
180
1. 1688
30
0
210
300
270
2. 3376
60
180
330
240
90
150
0
210
300
270
2l= lam bda
1. 3985
60
0. 69923
150
30
0
300
270
60
180
330
240
2
1
150
0
210
90
330
240
2l= (3/ 2)lam bda
270
300
2l= 2lam bda
Most often the length l=λ/4 (half-wave dipole) is used, since if it is a nearly resonant structure
with its current maximum at the driving point (z=0). For a dipole with a non-zero wire radius, the
length must be slightly shorter than λ/4 to produce resonance.
Note if we define the input impedance Z in as
Z in =
in p u t vo lta g e
in p u t cu rren t
V0
I0
then resonance occurs when Z0 is purely real (just as in circuit theory).
Radiated power
π
2 2π
2π
I0
W = ∫ ∫ K (θ , φ )d Ω = η0
8π 2
θ =0 φ =0
π
∫ d φ ∫ F (θ , kl ) sin θd θ
2
0
0
2 π
I0
= η0
4π
∫ F (θ , kl ) sin θd θ
2
0
This expression must be integrated numerically.
Special case #1:
kl<<1
(Short dipole)
1

F (θ , kl ) ≈  ( kl ) sin θ 
2

2
6-19
2
MOP N QP
2
π
I0 1
2
( kl ) ∫ sin 3 θd θ
W ≈ η0
4π 4
0
4
η0 ≈ 1 2 0 π Ω
3
2
I 0 1  2 lπ  2 4


W ≈ (1 2 0 π )
4π 4  λ  3
2
≈ 4 0π I 0
2
Special case #2:
π

 kl = 

2
2
l
  Watts
λ
(Half-wave dipole)
π

π
π

co s  co s θ  − co s   co s  co s θ 
2

2
2

 π
=
F θ ,  =
 2
sin θ
π
sin   sin θ
2
π

co s 2  co s θ 
I0
2

W = η0
dθ
4 π ∫0
sin θ
2 π
M PIPRP N PIPSP Q
1 . 2 2 b y nu m erical integ ratio n
thus
W = 3 6 .6 I 0
2
Directivity
D=
Special case #1: kl<<1
m ax {K (θ )}
W
4π
(Short dipole)
2
2
I0
I0
2
(
)
θ
≈
η
K (θ ) = η0
F
,
kl
0
8π 2
8π 2
6-20
1

 2 ( kl ) sin θ 
2
2
⇒ K (θ ) m ax
I0 1
π

( kl ) 2
= K  θ =  = η0
2

2
8π 4
so
2
I0 1
2
η0 2 ( kl )
8π 4
D=
2
1  I 0 1
2
(
)
η
kl
0
4 π  4 π 4
4 
3 
=
3
2
( )
G = 1 0 lo g 1 0 ( D ) = 1 0 lo g 1 0 3 2 = 1.7 6 d B
Special case #2:
kl =
π
2
(half-wave dipole)
 π
 π  
2
  
co
s
co
s

 2 
I0   2


= η0

8π 2 
π


sin


2


π

K (θ ) m ax = K  θ = 

2
2
2
I0
= η0
8π 2
2
I0
η0 2
8π
D=
1
4π
 I 2

 η 0 (1.2 2 )
 0 4π



=
2
= 1.6 4
1.2 2
more directive than short dipole
G = 1 0 lo g 1 0 ( 1.6 4 ) = 2 .1 5 d B
Radiation resistance
Rr =
2W
2W
=
I 0 I 0∗ I 2
0
6-21
π
= 6 0 ∫ F 2 (θ , kl ) sin θd θ
0
Special case #1:
kl<<1
(short dipole)
π
2
1

R r = 6 0 ∫  ( kl ) sin θ  sin θd θ
2

0 
2
l
= 8 0π   Ω
λ
2
Special case #2:
kl = π 2
(half-wave dipole)
Rr =
(
2 3 6 .6 I 0
I0
2
2
) = 7 3.2 Ω
Note that a resonant half-wave dipole (Zin = Rin + jXin = 73.2 + j0) is matched quite well by a 75Ω
coaxial cable transmission line.
Example
1
l
and to a half-wave
=
 λ  10
What input currents are needed to a short dipole of length 
dipole to radiate a power of 1kW?
I0
Rr
2
 1
short dipole: R r = 8 0 π   = 7 .9 Ω → I 0 = 1 5 .9 A
 10 
half-wave dipole: R r = 7 3.2 Ω → I 0 = 5.2 3 A
2
6.6 Radiation from a small loop antenna
6-22
UU UU
z
E ( r ), B ( r )
T
V
U
r
I
R
a
T
y
r′
x
The fields associated with a small loop antenna can be shown to be
W ( W ) φX η ( )
W ( W ) X W ( W ) θX (
E r =
0
I ka
4
r×E r
=−
H r =
η
Radiated Power
K (θ , φ ) = K (θ ) =
=
using
η0 = 1 2 0 π
λ=
2
e − jkr
sin θ
r
I k a ) e − jkr
sin θ
4
r
2
2
η0 
2
 η0 N
+
N
N
2 
θ
φ  =
 8 λ2 φ
8λ 
η0 2 2
2 2
2
2 I k (π a ) sin θ
8λ
2π
k
K (θ , φ ) =
1 5π 2
4
I ( ka ) sin 2 θ
4
6-23
2
Note that the radiation pattern of the small loop is identical to the radiation pattern of the
Hertzian dipole. The small loop is the magnetic dipole analog of the electric (Hertzian) dipole
W =
∫ ∫ K (θ )d Ω =
2π
∫
φ =0
π
dφ
1 5π 2
4
∫θ =0 4 I ( ka ) sin 3 θd θ
π
1 5π 2 2
4
=
I ( ka ) ∫ sin 3 θd θ
2
θ =0
Y [ Z [\
4
3
= 1 0 π 2 I 2 ( ka ) W a tts
4
Radiation resistance
Rr =
example: a
2W
2
( ka ) 4 Ω
2 = 2 0π
I
2
λ = 0 .5 → R r = 2 0 π [2 π ( 0 .0 5 )] = 1.9 2 Ω
4
Directivity
Since the radiation pattern of the small loop is the same as that of the Hertzian dipole, the
directive gain, the directivity, and the gain are the same:
D = 3/2
G = 1.76dB
6.7 Currents above a perfectly conducting ground
Antennas are often placed above conducting surfaces, for purposes of measurements in the
lab, or in practical situations as when an antenna is placed on a car roof. Often the earth itself is
modeled as a perfect conductor (although this not always a good approximation because the
conductivity of the earth is fairly low).
Fields produced by currents by antennas above a ground plane can be calculated using the
method of images.
Method of images
Consider a current carrying element above a perfectly conducting plane:
6-24
V V
z
I
E, B = ?
] ]
^
x
p erfect con d u cto r, E = B = 0
The current element will produce an E -field which will induce currents to flow on the surface of
the conductor. These currents will produce an additional “scattered field”. The total field must
obey the boundary condition E tang ential = 0 on the conductor surface at z = 0.
We may replace the problem by an equivalent problem. The ground plane is removed and
replaced by an “image” current.
z
Free space
x
Free space
"image" current
Here the fields produced by both the current and its image will be identical to the fields produced
by the current above the ground plane as long as the boundary condition on the total
field E tan = 0 at z = 0 is obeyed.
Question: What image current will result in the boundary condition being satisfied?
Since any current distribution can be viewed as a superposition of Hertzian dipoles, we only need
to identify the image of a Hertzian dipole.
The field due to a Hertzian dipole on the z-axis is given by
6-25
X
W X
η0
2 
1 
Id z e − jkr  2 η0
Id z e − jkr 
+
E =r

 j ωµ 0 + +
 sin θ
2  co s θ + θ
4π r  r
4π r 
j ωε 0 r 
r
j ωε 0 r 2 
Case I: Horizontal dipole
I
I
θ
θ
__
r
d
__
r
E, E i
ri
di
E ri
E, E i
Er
ri
Eθ
θi
θi
Ii
choose:
di = d ,
Ii = I ,
thus:
ri = r ,
θi = θ
and:
(E )
So,
= − ( E r )tan ,
a( a )
ri tan
E + Ei
tan
` `
Ii
Ii = − I
(E )
θi tan
( )
= − Eθ
tan
= 0 → then boundary condition is satisfied.
6-26
E θi
Case II: Vertical dipole
θ
θ
I
I
bb
r
d
bb
r
E, E i
di
E ri
ri
θi
ri
Eθ
Ii
choose:
di = d ,
thus:
ri = r ,
and:
(E )
Summary:
Er
θi
Ii
So,
E, E i
Ii = I ,
θi = π − θ → cos θi = − cos θ
= − ( E r )tan ,
d( d )
ri tan
E + Ei
c c
Ii = I
tan
(E )
θi tan
( )
= − Eθ
tan
= 0 → then boundary condition is satisfied.
1. Horizontal currents image in opposite directions.
2. Vertical currents image in the same direction.
6-27
E θi
e e
6.8 Monopole above a ground plane
free
space
E, B = ?
I
l
I0
coaxial
cable
The equivalent problem is a dipole.
l
I
l
Ii = I
Thus the fields radiated by a monopole above a ground plane are identical to those of a dipole in
free space. However, the radiated power, and thus the radiation resistance are half that of the
dipole, since the monopole only radiates into half the space that the dipole does.
W m o nop ole
1
= W dip ole
2
2 π
I0
= η0
4π
2
∫ F (θ , kl ) sin θ d θ
2
0
π
R rm on op loe
2
1
= R rdip ole = 6 0 ∫ F 2 (θ , kl ) sin θ d θ
2
0
6.9 Broadband antennas
Although dipole antennas possess many attractive characteristics for measurement of
6-28
radiated emissions, they are not ideal for gathering data over a wide range of frequencies. The
radiated emissions range typically extends from 30 MHz to 16 GHz, and the length of a dipole
must be physically adjusted to provide a length of ½ λ at each measurement frequency.
A more practical technique is to employ broadband measurement antennas. A broadband
antenna has the following characteristics:
1. The input/output impedance is fairly constant over the frequency band.
2. The antenna pattern is fairly constant over the frequency band.
Two types of broadband antennas will be examined: The biconical antenna, and the log-periodic
antenna.
The biconical antenna is typically used in the frequency range of 30 MHz to 200 MHz.
The log-periodic antenna is typically used in the band from 200 MHz to 1 GHz.
Biconical Antennas
An infinite biconical antenna consists of two cones of half angleθ h with a small gap at the
feed point.
z
Eθ
θh
free
space
Hφ
θ
voltage
source
θh
6-29
r
g h
f
i j
In the space surrounding the cones J = 0 . Symmetry suggests that the fields are H = φ H φ and
E = θ E θ . Maxwell’s equations can be solved to give the form of the field as
H 0 e − jk 0 r
Hφ =
sin θ r
and
k 0 H 0 e − jk 0 r
Eθ =
= η0 H φ
ωε 0 sin θ r
Where H0 is constant.
It is noted that these fields form transverse electromagnetic (TEM) waves (the electric and
magnetic fields are orthogonal and transverse to the direction of propagation). Therefore, a
unique voltage between two points on the cones may be defined.
The voltage produced between two points on opposite cones that are both a distance ‘r’
from the feed point is
θh
k k
∫ E ⋅ dl
V (r ) = −
θ =π −θh
1 

= 2 η0 H e − jk 0 r ln  c ot θ h 

2 
The current on the surface of the cones is given by
I (r ) =
2π
∫H
φ =0
φ
r sin θ d φ
= 2 πH 0 e − jk 0 r
and the input impedance is then
Z in =
1 
V (r )
η 
= 0 ln  co t θh 
2 
I ( r ) r =0 π 
1 

= 1 2 0 ln  c ot θ h 

2 
which is purely resistive.
6-30
Usually, the cone half-angle is chosen to provide a match to the feed line characteristic
impedance.
The total time average radiated power is given by
W =
∫
P ⋅ds
s
2 π π −θh
=
∫ ∫
φ = 0 θ =θh
2
Eθ 2
r sin θ d θ d φ
2 η0
θh
= πη0 H 0
2
dθ
sin θ
θ =0
∫
2
1 

= 2 πη0 H 0 ln  c ot θn 

2 
Radiation resistance is given by
Rr =
2W
I (0 )
2
=
2
1 

4 πη0 H 0 ln  c ot θh 

2 
4π 2 H 0
2
or
1 

R r = 1 2 0 ln  c ot θ h 

2 
m
which is the same as the input impedance Z in.
the
i
l
It is noted that the radiated fields are spherical waves with E in the θ direction and H in
direction. For linearly polarized waves incident on the antenna from the broadside
direction ( θ = 9 0 o ), the antenna responds to the field component that is parallel to its axis. Also
the input impedance and pattern are theoretically constant over an infinite range of frequencies.
Infinite length cones are obviously impossible to construct, therefore real biconical
antennas consist of truncated cones. The finite length of the cones causes reflections as the waves
travel outward along the cones. This produces standing waves that result in the input impedance
having an imaginary (reactive) component, rather than being purely real.
Often wires are used to approximate the cone surfaces:
6-31
truncated biconical antenna
composed of wire elements
Other variations:
cone
ground
plane
Discone
Bowtie
The fields of the discone antenna above the ground plane are the same as those of the
biconical antenna by the method of images. The radiation resistance of the discone is ½ that of
the biconical antenna.
The bow-tie antenna consists of flat triangular plates or a wire which outlines the same
area as the plates. The bow-tie antenna is frequently used for reception of the UHF television
signals. Using wires instead of solid metal triangles tends to reduce the bandwidth of the bow-tie
antenna.
Log-periodic antennas
The log-periodic antenna achieves a large operational bandwidth through repetitive
dimensioning of structures. The structural dimensions increase in proportion to the distance from
the origin of the structure. As a result, the input impedance and radiation properties repeat
periodically and are functions of the logarithm of frequency.
The log-periodic dipole array is a common log-periodic measurement antenna. This
antenna shares the properties of all log-periodic antennas in that element distances, lengths, and
separations are related by a constant such that
τ =
ln
l n −1
=
dn
R
= n
d n −1 R n −1
6-32
l n +1
α
l n −1
ln
dn
Rn
log-periodic array
The most efficient way of operating a log-periodic array is such that the currents on
adjacent elements are reversed in phase.
criss-cross feed method
In this way the shorter elements will not interfere with elements to the right.
The bandwidth of the log-periodic antenna is approximated by determining the frequency
at which the shortest element is, and the frequency at which the longest element is ½ λ.
At a particular frequency only the elements which are at or near a resonance are active.
Thus, the active region of the antenna adjusts depending on frequency.
6-33
low frequency
measurement
high frequency
measurement
emitter
emitter
active
region
active
region
6.10 Aperture antennas
Aperture antennas are characterized by an aperture or opening from which radiated fields
are emitted. These include horns, slots and microstrip patch antennas. The operation of such
antennas is best explained by Huygen’s principle:
Each point in an advancing wavefront acts as a source of spherical, secondary wavelets,
that propagate outward.
Incident
field
The secondary wavelets cause the wave to spread as it travels away from the aperture.
6-34
This is known as diffraction. The antenna pattern of an aperture antenna is actually a diffraction
pattern.
In the far zone of a simple aperture the magnitude of the electric field is given by
n
( )
jk
E r ≈
2π
o
n
( ) n n
e − jk r −r ′
∫S E a r ′ r − r ′ d s ′
where S is the aperture surface, E a ( r ′ ) is the magnitude of the electric field in the aperture and
r ′ is a position vector that sweeps over points in the aperture.
p p
At points far from the aperture
r − r ′ ≈ r − z ′ co s θ
Assuming that the field in the aperture is uniform and that the aperture width is small
− jkr
jk ∆ xe
2 πr
E (r , θ , φ ) ≈
a/2
∫ E a e jk z ′ co sθ d z ′ = jE a ∆ ×
e
− jkr
r
−a / 2
 a

sin  π co s θ 
 λ

π co s θ
The size of the main lobe in the resulting pattern is inversely proportional to the aperture
width ‘a’. Thus in order for the main lobes to be narrow, the aperture dimensions must be on the
order of a wavelength or greater.
z
S
x
θ
r
N o rm aliz ed far-z on e E -field vs . O bs ervat io n a ngle
1
0. 9
r −r′
0. 8
a
ds ′
r − z ′ c os θ
|E | (norm aliz ed)
0. 7
0. 6
0. 5
0. 4
0. 3
0. 2
∆x
0. 1
0
Radiation from a narrow aperture
0
Horn antennas
6-35
20
40
60
80
100
The ta (d egre es )
120
140
160
180
Horn antennas are flared waveguides. There are three basic types of horn antennas.
oq
E-plane horn - flare is in the plane that contains the E -field vector.
H-plane horn - flare is in the plane that contains the H -field vector.
Pyramidal horn - flare is in both planes.
The aperture distribution of a horn is typically the same as the mode of the feeding waveguide,
with a phase taper across the aperture.
13
r
E
2 .5 7 λ
z
H
0 .5 λ
Geometry of a typical horn antenna
The antenna as a receiving element
Any transmitting antenna can also be used for the purpose of “receiving” – intercepting a
portion of the power radiated by some source. Instead of being driven by transmission line, a
receiving antenna delivers power to a load connected at its terminals.
Consider a spherical wave radiated by some distant source, and incident on a receiving
antenna. Over the local region of the antenna, the spherical wave can be approximated as a plane
wave. The plane wave induces a current in the antenna, which in turn produces an additional
scattered field. But the induced current also causes a voltage to appear across the load
impedance. This voltage then acts like a driving voltage causing additional currents to flow, just
as in a transmitting antenna, which produce still another scattered field.
Thus, by superposition, the total current flowing on the antenna may be viewed as that due
to a scatterer interacting with a plane wave plus that of a transmitting antenna.
Transmitting/receiving equivalent circuits
6-36
V1 = Z 11 I 1 + Z 12 I 2 
 network description of transmitting/receiving system
V 2 = Z 21 I 1 + Z 22 I 2 
Assume: Z 1 2 << Z 1 1 for r → large (little coupling from receiving to transmitting antenna)
t
≈0
V1 = Z 11 I 1 + Z 12 I 2 ≈ Z 11 I 1
(Z )
1 in
V1 +
-
=
V1
≈ Z 11
I1
s
linear
medium
I1
Et
I2
I2
antenna 1
(transmitting)
ZL
antenna 2
(receiving)
r
I1
Z 11
Equivalent circuit for
transmitting antenna
+
V1
-
V 2 = − I 2 Z L = Z 21 I 1 + Z 22 I 2
I2 =
− Z 21 I 1
Z 22 + Z L
6-37
I2
Z 22
+
Z 21 I 1
-
ZL
Equivalent circuit for
receiving antenna
Note: When antenna 2 is in receiving, we can not assume Z 2 1 ≈ 0 since this term describes the
coupling effect between transmitter and receiver.
Note: When antenna 2 is transmitting, then Z 2 1 ≈ 0 so that
u
≈0
V 2 = Z 21 I 1 + Z 22 I 2 ≈ Z 22 I 2
(Z )
2 in
=
V2
≈ Z 22
I2
Z 2 2 ≈ (Z 2 )in
Receiving/transmitting reciprocity
There are three basic reciprocity relations between receiving and transmitting antennas:
1)
The antenna pattern for reception is identical to that for transmission.
2)
The equivalent impedance in the receiving antenna equivalent circuit is identical to the
input impedance of the antenna when it is transmitting.
3)
The effective receiving cross-section area of an antenna is proportianal to its diective gain
as a transmitting element.
We have already considered (2) above. The others require the use of the Lorentz reciprocity
theorem. From this theorem we can show that Z 1 2 = Z 2 1 .
Relationship between gain and effective receiving cross-sectional area
Definition:
Wr
effective receiving cross-section area (m2).
Pa v
where: W r = received power (power delivered to the load)
Pa v = average Poynting vector (power density) maintained by transmitting
A er =
6-38
antenna (Watts/m2)
A er is a function of: 1) Load impedance
2) Aspect of antenna to oncoming wave
3) Polarization of oncoming wave
Note: W r = A er Pav = total power intercepted by receiving antenna.
d Wt
dΩ
Recall: g dt (θ , φ ) =
 Wt 
 
 4π 
directive gain of transmitting antenna, W t = transmitted power.
4 π r 2 Pav
Wg
=
⇒ Pav = t dt2
Wt
4πr
So:
Wr = Wt
A er g dt
4 πr 2
received power in terms of transmitted power.
Maximum power transfer relation:
Assume load impedance is conjugate matched ⇒ maximum power transferred to load
I2
Z 22
+
Z 21 I 1
-
Z L = Z 2∗2
receiving antenna
Z L = Z 2∗2 = R 2 − jX 2
2
Z 21 I 1
1
2
1 − Z 21 I 1
Wr = I 2 R2 =
R2 =
2
2 2 R2
8 R2
6-39
2
I1
Z 11
+
V1
transmitting antenna
Z 1 1 = R 1 + jX 1
1 2
I R
2 1 1
Wt =
2
Z 21 I 1
2
Z 21
Wr
8 R2
=
=
1 2
4 R1 R 2
Wt
I 1 R1
2
but:
2
Z 21
Wr
A g
= er 2dt =
Wt
4πr
4 R1 R 2
so:
2
R 1 R 2 A er2 g dt 1
2
πr 2
R 2 R 1 A er1 g dt 2
Z 21 =
Z 12
now use Z 1 2 = Z 2 1
=
πr 2
antenna 1 transmitting, antenna 2 receiving
antenna 1 receiving, antenna 2 transmitting
(Reciprocity of network)
g dt 1 A er2 = g dt 2 A er1
or
A er1
g dt 1
=
A er2
g dt 2
6-40
Now since antennas 1 and 2 were arbitrary,
A er
= constant
g dt
For polarization matched conditions (receiving antenna oriented to intercept maximum amount of
power), the constant can be shown to be
Ae λ2
Thus,
=
g d 4π
4π
(Derivation is kind of messy).
λ2
Universal relationship between gain of an antenna acting as a transmitter
and effective area of same antenna acting as a receiver.
 λ 
Wr = Wt 
 D D
 4πr  r t
2
So,
“Friis” equation
Reciprocity between transmission and reception patterns
Case 1)
Antenna 1 transmitting, antenna 2 receiving
(2b)
θ
(2a)
(1)
Measure transmitting pattern of antenna 1 by varying θ.
2
1 

= Z 21 
W t1
 4 R1 R 2 
W r2
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W r2 b
W r2 a
Case 2)
2
=
Z 21 b
Z 21
2
a
Antenna 2 transmitting, antenna 1 receiving
θ
(1b)
(2)
(1a)
Measure receiving pattern of antenna 1 by varying θ.
2
1 

= Z 12 
 4 R1 R 2 
W r1
Wt2
W r1 b
W r1 a
By reciprocity:
Z 12 = Z 21 ⇒
W r2 b
W r2 a
=
=
Z12
Z12
W r1 b
W r1 a
2
b
2
a
power patterns are equal.
Thus, the reception pattern of an antenna is identical to its transmission pattern.
References
1. Demarast, K. Engineering Electromagnetics, Prentice Hall Inc., 1998
2. Paul, C. Introduction to Electromagnetic Compatibility, John Wiley & Sons, 1992
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