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Chapter 7 Antennas
ECE 3317Dr. Stuart Long
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7-1
What is an “antenna”?
An antenna is a structure that provides electrical current that thenproduces an electromagnetic wave. They can also be thought ofas transducers that transfer electromagnetic energy between atransmission line and free space. They are the usual sources ofEM waves.
In this chapter we will only focus on three types of antennas:
1. Infinitesimal Current Sources.
2. Linear Wire Antennas (Dipoles).
3. Uniform Linear Arrays.
7-2
Examples of antennas
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7-3
Vector Potential
The vector potential is defined somewhat indirectly as:
vector potenti
N
al
:ot
e
= ∇×
∇ ⋅ =
A
A
B A
B ( )
( )is uniquely defined. If we let
then
and give same which means that w
not
e need to specify
0
0
ψψ
′
∇ ⋅ ∇× ≡
′ = +′× = × +
′× = × +
B
A A B
A
A AA A
A A
∇
∇ ∇ ∇ ∇
∇
also.
⋅ A ∇
7.1
7-4
Scalar Potential
( )
From Ma
xwell's equations
reca l l:
j
j
ω
ω
= ∇ ×
∇ × = −
∇ × = − ∇ ×
B A
E B
E A
( )
( )but:
So let
0
Φ r
0
o
ψ
j
j
ω
ω
∇ × ∇ =
∇ × + =
+ = −
E A
E A ∇
Φwhere is the scalar poten ial
Φt
jω= − −E A ∇
7.2
7.3
7-5
Lorentz Condition
Thus far we have defined
vector potential
calar potential
To uniquely define we also have to specify
Φ Φ s
jω
∇ ⋅
= ∇×
= − − ∇
A
A
A
B A
E A
Φ Lorentz condition (gauge) 0 jω µ ε∇ ⋅ + =A 7.4
7.1
7.3
7-6
2 2
2 2 v
Using vector identities and the Lorentz condition we obtain:
Differential equation for
Di eΦ ffΦ
k
k
µ
ρε
∇ + =
∇ + =
A
A A - J
- rential equation for Φ
Vector and Scalar Potential
7.6
7.5
7-7
The solution to the previous equations are given by:
Solution(
)( ) 4
jk
V
e dVµπ
′− −
′
′′=
′−∫∫∫
JAr rrr
r r
v
for
Solution for
Φ
1 ( )
( )
4
jk
V
e dVρπε
′− −
′
′′=
′− Φ∫∫∫
A
r rrrr r
field point source point k ω µε
′→
→
= rr
7.7
7.8
Vector and Scalar Potential
7-8
Spherical Coordinates
θr
z
x
φ
θ̂
r̂
φ̂
Note: unit vectors depend on the location of the vector.
y
(fig.7.2)
( , , )r θ φ
7-9
y
′−r r
′r
r
field pointsource point→′
→
r
r
z
x
Spherical Coordinates( , , )r θ φ
7-10
Infinitesimal Hertzian Dipole
I
z∆
2a → ←
y
′r
r
z
x
Electrically short
Physically th in a
z
z
λ∆ <<
∆<<
7-11
Fields Due to an Infinitesimal Hertzian Dipole
I
z∆
2a → ←
x
y
z
′r
r
dV’
dV’r’
r’
r’r’
7-12
To find :
use to obtain ˆˆˆ cos sin
we first in spherical coordinates
I ˆ4
jkrz er
θ θ
µ
µπ
−
= −
= = ∇ ×
∆=
B H A
A
A
z r
z
θ
we then evaluate with vector in spherical coordinates
as seen i (eqn. 2n 7.1 )
I ˆˆ cos sin
4
jkrzer
µ θ θπ
−∆ = −
∇×
A
A A
r θ
I
z∆
2a → ←
x
y
z
′r
r
Fields Due to an Infinitesimal Hertzian Dipole
7-13
( ) 1sinr
Ar φθ θ
∂∇ × =
∂A ( )sin Aθθ
φ∂
−∂
( ) 1 1sin
rArθ θ φ
∂∇ × =
∂A rA
r φ∂
−∂ ( )
( ) ( )1
1 I 1ˆ 1 sin4
r
jkr
ArAr r
jk zer jkr
θφ θ
θµ π
−
∂ ∂ ∇ × = − ∂ ∂
∆= ∇ × = +
A
H A φ
7.12
with vector in spherical
co
ordi
nat
s
e∇ × A A
7.13
I
z∆
2a → ←
x
y
z
′r
r
Fields Due to an Infinitesimal Hertzian Dipole
7-14
( ) ( )
2
2 2
,1 1
r r
The electric field outside the dipole is given by :
: has only a component with two terNote ms ˆ
1
I 1 1 1 1 ˆˆ 2cos 1 sin4
jkr
j
jk zer jkr jkrjkr jkr
ωε
µθ θ
ε π
−
= ∇ ×
∆= + + + +
H
E H
E r
φ
θ
2 3 2 3
1 1 1 1 1
r r,
r,
r,
rhas an component and ˆˆ a com ponent
E r θ
7.14
I
z∆
2a → ←
x
y
z
′r
r
Fields Due to an Infinitesimal Hertzian Dipole
7-15
7.17a
7.17b
I
z∆
2a → ←
x
y
z
′r
r
Radiation Field (Far Field)
7-16
: field , field , and the direction of propagation are all perpendicular to each other.
and
field , and fiel
d are
sin proportional t
Note
o
vk
r
µ ωηε
θ
= = =
E H
EH
E H
The wave propagates in the radial direction.
The surfaces of constant phase are spherical.
I
z∆
2a → ←
x
y
z
′r
r
Radiation Field (Far Field)
7-17
Hertzian Dipole (Ideal Current Source)
I(z)
zz∆
I0
7-18
Real Electrically Short Antenna (I→0 at ends)
I(z)
z∆ z
I0
7-19
Practical Approximation to Hertzian Dipole
I(z)
z∆
z∆
z∆
Capacitive Plate Antenna
7-20
Radiation Patterns
|E| vs. angle θ at a constant r
|E|
θ0
(rectangular plot)
ππ−
0
IE sin4
|r
|k zθ
µ θε π
∆=
7-21
x
Field Patterns
z
θ
IE sin4
|r
|k zθ
µ θε π
∆=
(fig.7.4)
θE
(polar plot)
7-22
Power Pattern
r
2r
S versus angle plot
Not : S ne si
θ
θ
0 dB
θ -30°30°
60°
120°
150° -150°
-120°
-60°
(polar plot)
7-23
3 dB Beamwidth
Angle in degrees between points 3 dB down from
maximum.
HPBW
7-24
max
max
Electric Field
Power
E20log 20log 0.707 3 dBE
P 1 10log 10lo
g 3 dBP 2
= = −
= = −
max0.707← Emax0.5 →P
HPBW
3 dB Beamwidth
7-25
2
Surface Element sindS r d dθ θ φ=
Coordinate System
rdθ
sinr θ
sinr dθ φ
dr
dφ
7-26
Directive Gain
( )2
rad
22
rad0 0
Directive Gain
Total Radiated P
ower
4,
P
P
sin
r
r
S rD
S r d dπ π
φ θ
πθ φ
θ θ φ= =
=
= ∫ ∫
7.19
7.19
rdθ
sinr θ
sinr dθ φ
dr
dφ
7-27
Directivity of Hertzian Dipole
22
2
rad
2
0Directivity (maxi1.5 mun directive gain s i in
I z sin2 4
4 I z P3 4
3 ( ,
) si
n2
r
D
kSr
k
D
η θπ
π ηπ
θ φ θ
= =
∆ =
∆ =
=
= 90 direction)θ
7.19
7-28
Linear Antenna
z
z∆
0z =
2z h=
1z h= −
7-29
2
1
1 2
cos
If total length is comparable to a wavele ngth
( ) I I( )
ˆ sin U( )4
ˆ
U( ) Ι ( z)
jkr
h jkzh
h h z
ejkr
e dzθ
λ
µ θ θε π
η
θ
−
−
+ ≈ ⇒ =
=
=
= ∫
E
EH
θ
φ
z
z∆
0z =
2z h=
1z h= −
7.23a
7.23b
7.24
Linear Antenna
7-30
z
z∆
0z =
2z h=
1z h= −
Same field as Hertzian dipole with ½ current
Electrically short dipole
Example
( )01 2
ˆ sin4 2
jkr Ijke h hr
µθ θε π
− = + E
I(z)
7-31
Example
Half-Wave Dipole
1 2
0
0 2
h h h4
I( ) cos
cos cos2U( ) 2
sin
z I kz
Ik
λ
π θθ
θ
= = =
=
=
7.25
7.26
0II(z)
z∆
4z λ
=
4z λ
= −
7-32
Example
-9 -3-6
0 dB
infin. dipole 2λ
θ
Note: This is a normalized plot of the radiated fields. Equal
driving currents result in much higher fields for dipoles 2λ
-30°30°
60°
120°
150° -150°
-120°
-60°
7-33
Example 7-34
Monopole Over a Ground Plane
Ground plane
7-35
Monopole Over a Ground Plane
z
hz∆
2h
z
monopole dipole
same radiation pattern in upper half plane
same current at driving point
the voltage1 1Z Z2 2
•
•
⇒ =
7-36
Total of NElements
z
y
x d← →
rγθ
φ
Uniform Linear Array
7-37
1) Identical radiators oriented in direction and centers along the -axis.
2) All elements are equally spaced a distance .
3) All elements are driven by currents with equal magnitude and with
ˆˆ
d
p
z y-
20 1 2
rogressive phase shifts .
I I( ) I I( ) ; ; I ( I )j j
ψ
z z e z eψ ψ= = =
z
y
x d← →
rγθ
φ
Uniform Linear Array
7-38
( ) ( )
( )
The total field of the antenna array can be found by adding the fields from each antenna (principle of superposition).
Total field of an arrayF ,
.
t e
e
θ θ φ
θ
=
⇒
E
E E E
E
( )
Element factor field due to a single element located at origin
Array factor.
F ,
θ φ ⇒
7.35
z
y
x d← →
rγθ
φ
Uniform Linear Array
7-39
total number of elements
phase shift in driving current between adjacent elements
spacing between elements
cos sin
N
d
ψ
γ θ
=
=
=
= sinφ
( )( cos )
( cos ) Array Factor 1F , 1
jN kd
j kd
ee
ψ γ
ψ γθ φ+
+
−=
−7.36
z
y
x d← →
rγθ
φ
Uniform Linear Array
7-40
7.37
z
y
x d← →
rγθ
φ
Uniform Linear Array
( )
( )
( )
Array Factor
not
1ing that
Arr
( cos )1F , ( cos
=2 sin 2
)1
cossin2F , ay Factor M
cossiagnitu de
2
n
j
jN kdej kde
kdN
kd
x xe
ψ γθ φ ψ γ
γ ψ
θ φγ ψ
+−=
+−
+ =
+
−
total number of elements
phase shift in driving current between adjacent elements
spacing between elements
cos sin
N
d
ψ
γ θ
=
=
=
= sinφ
7-41
2-Element Array
( ) ( )
with
in pl
2
x-yane ( plane)
sin sin sinF , 2cos
2sin
2
sinF , 2cos
2
cos
co
s2
2
N
kd
kd
kdkd
θ
θ φ ψθ φ
π
φ ψφ
γ ψγ ψ
π
=
=
+= =
+=
+ +
( )
cossin
2F ,cos
sin2
kdN
kd
γ ψ
θ φγ ψ
+
=+
z
y
x d← →
rγθ
φ
7-42
2-Element Array
special case in plane ( plane) with phase shift x-y 0
sinF , 2cos
2
no
2
2
kd
θ ψ
π φφ
π= =
=
maxima always to line of array for Broadsid =0 eψ ⇒⊥
z
y
x d← →
rγθ
φ
sinF , 2cos
22kd φ ψ
φπ +
=
7-43
F 2cos sin2
20
2
2 =2
d
kd
πφ
πθ
ψλ
π λ πλ
=
=
=
=
=
x
maxima always to line of array for Broadsid =0 eψ ⇒⊥
z
y
x d← →
rγθ
φ
x
y
2-Element Array || Example
φ
7-44
maxima always to line of array for Broadsid =0 eψ ⇒⊥
z
y
x d← →
rγθ
φ
x
y
F 2cos sin4
20
4
2 =4 2
d
kd
πφ
πθ
ψλ
π λ πλ
=
=
=
=
=
φ
2-Element Array || Example
7-45
maxima always to line of array for Broadsid =0 eψ ⇒⊥
z
y
x d← →
rγθ
φ
x
y
( )F 2cos sin
20
2 =2
d
kd
π φ
πθ
ψλ
π λ πλ
=
=
=
=
=
2-Element Array || Example
φ
7-46
sin2 2F 2cos
2
2
2
4
2 =4 2
d
kd
π πφ
πθ
πψ
λ
π λ πλ
=
=
=
=
=
+
z
y
x d← →
rγθ
φ
x
y
2-Element Array || Example
Endfire
φ
(with phase shift)
7-47
z
y
x d← →
rγθ
φ
x
y
sinF 2cos
2
2
2
2 =2
d
kd
π π φ
πθ
ψ πλ
π λ πλ
=
=
=
=
=
+
2-Element Array || Example
φ
(with phase shift)
7-48
sinF 2cos
2
2
2
2 =2
22
d
kd
φ
πθ
πψ
λ
π λ πλ
π π
=
=
=
=
=
+
z
y
x d← →
rγθ
φ
x
y
2-Element Array || Example
φ
(with phase shift)
7-49
Phased Arrays
-1
= sin = 02
in plane max field at where
By adjusting can change direction of mai n
b
= sin
e am
m m
m
kd
kd
πθ φ φ ψ
ψφ
ψ
+
−
electronically
z
y
x d← →
rγθ
φ
sinF , 2cos
22kd φ ψ
φπ +
=
7-50
Uniform Linear Array (Handy Facts)
( )
1) | | is symmetric about line of array ( - axis)
2) Principal maximun occurs when (magnitude of this max is )
3) Secondary maxima
ˆF
cos 0
cos = 2m +1 m = 1,2,3...ne2
ar2
w
kd N
kd
N
γ ψ
γ ψ π
+ =
+±
y
-1
ith magnitude at of
4) Nulls when
3 3 sin 2 2
cos = m = 1,2,3... 2
N
kd N m
π π
γ ψ π
+±
z
y
x d← →
rγθ
φ( )
cossin
2F ,cos
sin2
kdN
kd
γ ψ
θ φγ ψ
+
=+
7-51
0.75
0.75
20
2 3=2
3cos = sin2
d
kd
kd
πθ
ψλ
π λ πλ
γ ψ π φ
=
=
=
=
+x
y
4-Element Array || Example
z
y
x d← →
rγθ
φ
φ
7-52
( ) -1
o o
o o
1) Symmetric about - axis
2) Principal max at of
3) Secondary max of a
ˆ
0 4
3 sin3 2 sin = 1.08 4 = 2m +1 = 3t ,
4) Nulls at
0 56.4 2 2 2
3 sin24 = = 19.5 44.
2, , 8
m
m
n
n
N
n
φ
π φπ π φ
π φπ φ
⇒
⇒
=
±
y
o90
4-Element Array || Example
x
y
φ
7-53
=at 0
down from max.
F =4.0
3db F =(0.707)(4.0)=2.83
sinFsin
2 . 83
4
φ
αα
=
⇒
=
max0.707← Emax0.5 →P
HPBW
3 dB Beamwidth
( )sinF
sin
3 sin3 sin4
π φ
π φ=
7-54
sinF
sin2.83
4
αα
=
=
max0.707← Emax0.5 →P
3 dB Beamwidth
HPBW
HPBW
=1.43
3 sin =1.43
=8.73
for rad
3db
ians
HPBW=17.5
α
π φ
φ
HPBW
HPBW=2
φ
HPBWφ
7-55