ECE 5324/6324
NOTES
ANTENNA THEORY AND
DESIGN
2013
Om P. Gandhi
Text: Warren L. Stutzman and Gary A. Thiele, Antenna Theory and Design, Third
Edition (2013), John Wiley & Sons.
The identified page numbers and the equations with dashes (x-xxx) refer to the equations
of the text.
Example:
Solution:


 


− jβ ×V =
− j β rˆ × V for any vector 
Show that for far-field region ∇ × V =
V

 


A, H and E of the radiated fields from antennas.
such as 


 

A, H and E are of the form
The radiated fields 

 K(θ,φ) − jkr
ˆ ⇒ V(r, θ, φ)V
ˆ
V=
e
V
r


(1)
where K(θ, φ) would, in general, depend upon the current distribution on the
antenna.
In spherical coordinates
V(r, θ, φ)
  ∂ θˆ ∂
φˆ
∂   K(θ, φ) − jβr ˆ 
e
V
∇=
× V  rˆ +
+
 × 

 ∂r r ∂θ r sin θ ∂φ   r
 1

ˆ V)
ˆ V)
ˆ + 1 ∂V (θ×
ˆ + 1 ∂V (φ×
ˆ
= −
− jβ  V(rˆ × V)
r ∂θ
r sin θ ∂φ
 r

 

≅ − j β V (rˆ × Vˆ ) = − j β rˆ × V = − j β × V
(2)
(3)
since all terms other than the second term in Eq. 2 are a factor of 1/βr
smaller. For =
βr
2π r
λ
>> 1 , all of these terms can, therefore, be neglected.
This is a powerful relationship which can be applied for radiated fields
from any antenna.
 


 ∇×A
jβ× A β
H= =
−
− j rˆ × A
µ
µ
µ
 

 ∇×H


jβ× H
β
µ
E=
=
−
=
−
rˆ × H =
−
rˆ × H
jωε
jωε
ωε
ε
1
(4)
(5)
p. 44, 45, 50 Text
General Theory of Conduction Current Antennas
P
R

r′
r




J
( r ′) .
Formulate 
Steps
1.


Calculate 
A

 
µ
µ JS j(ωt −βR)
J(r′) j(ωt −βR)
=
e
dV′
dS′
∫
∫ e
4π V ′ R
4π S′ R

For surface current radiators
For volume current radiators
µ I j(ωt −βR)
=
d
∫ e
(2-101)
4π  R

=
A
For line current radiators

 


 ∇×A

β× A
β rˆ × A ∇ × E
1
H ==
−j
=
−j
= =
rˆ × E
µo
µo
µo
− jωµo η
2.







∇× H
β 2 ˆr
× ( ˆr × A) = jω ˆr × (ˆr × A)
=
E=
o µo

 jωε o  jωε

=− jω  A − A ⋅ rˆ rˆ  =− jω A θθˆ + A φφˆ


3.
(
(
)
)
(2-107)
(2-105)
  
  
  
A × B× C = A ⋅C B − A ⋅ B C



rˆ × rˆ × A = rˆ ⋅ A rˆ − ( rˆ ⋅ rˆ ) A
(
(
) ( ) (
) ( )
)
From Eq. (2-105) we can write




E
=
−
jω
A


(2-104)
which is transverse to direction of propagation ˆr .
4.
5.
 1
 
=
S
Re E × H*
2
  1 2π π
=
Re ∫ ∫ E θ H*φ − E φ H*θ r 2 sin θ dθdφ
Total Radiated Power
= ∫ S ⋅ ds
2
0 0
Calculate
(
)
(
2
)
(2-127)
(2-128)
Calculation of Magnetic Fields of Conduction Current Antennas

Definition of Magnetic Vector Potential A : A Simplifying Mathematical
Intermediate Step
I
F

R = R Rˆ
(x,y,z)
From Biot-Savart's law of electromagetism



I d × Rˆ
J dV′ × Rˆ
µo ∫
=
µo ∫
B=
2
2
 4πR
V 4π R

B = µo ∇ ×
1
−∇  
R

J dV′
∫ 4πR
V′
(1)
(2)
In going from Eq. 1 to Eq. 2, we have used the following steps
Rˆ
1
∇  =
−
R
R2
(3)

0

 
J
1 1
∇ × J(r′) − ∇ ×  
J ×∇ =
(4)

R R
R

The first term in Eq. 3 is zero, since the current density J is a function of source


coordinates r′ = (x ′, y′, z′) whereas the curl ∇ × J involves derivatives with respect to
field coordinates (x,y,z).
From Eq. 2

B = µo ∇ ×


J dV′
∫ 4πR ≡ ∇ × A
V′
(5)
Thus the magnetic field at the field point can be written as curl of magnetic vector


potential A where A is given by

 µo J dV′
A=
(6)
∫
4π V ′ R


Note that calculation of B is a lot simpler if the intermediate step of first calculating A is
undertaken since integral of Eq. 6 is much simpler than that of Eq. 5 or Eq. 1.
2a
Note that because of time retardation for propagating fields, Eq. 6 should be modified to

 µo J(r′)e j(ωt −βR)
A=
dV′
∫
4π
R
(7)
Same as Eq. 2-101
of the text
Once, the only complicated step that of integration for Eq. 7 has been done, the magnetic

field B from Eq. 5 can be simplified to

 


B =∇ × A =− jβ× A =− jβ Rˆ × A
 

jβ× A
H= −
µo
2b
(2-107) text
p. 45 Text
A Uniform Line Source
r̂
θ
R
φ̂
⊗
r
θ̂
Fig. 2.9 text. A uniform line source.
I( z′ ) = Io for −
L
L
< z′ <
2
2
(2-109)
R = r − z′ cos θ
(2-86)

µ − jβ r L / 2
A=
e
Io e jβz′ cos θdz′
∫
4πr
−L / 2
Az =
zˆ
 β L 

 cos θ 
 2 

β
L


cos θ 

2


− jβr sin 
µIo Le
4πr
 β L 

sin 
cos θ 


µI L
 2 

E = jω sin θ A z θˆ = jω o sin θ
4πr
 βL

cos θ 

 2

From Eq. (2-107)
(2-110)
θˆ
(2-111)

1
E
H = E θ ˆr × θˆ = θ φˆ
η

 η
Radiation pattern for a plot of normalized values of E(θ, φ) is given by
F(θ, φ) =
E(θ, φ)
E max
(2-112)
For an elemental or Hertzian dipole (βL/2 <<1)
F(θ) = sin θ
(2-113)
3
Otherwise
 β L 

sin 
 cos θ 
 2 

F(θ)= sin θ
 βL

cos θ 

 2

P(θ) = F 2 (θ)
Normalized Pattern factor
p. 34
(2-114)
(2-119)
An Infinitesimal (Hertzian) Current Dipole or An Ideal Dipole
L = ∆z
I = Io
βL π∆z
<< 1
=
λ
2
βL
cos θ << 1
2
In Equations 2-110 to 2-114;
sin x
≅1
x

 µI ∆z
E = jω o sin θ θˆ
4πr


(2-74a)

E
jβIo ∆z
H = θ φˆ =
sin θ
4πr

 η
 Io2  ∆z 2 sin 2 θ
=
S
η 
8  λ  ro2
rˆ
(2-74b)
(2-76)
 

1
Re E × H* ⋅ ds
∫∫
2
(
Radiated Power
P
=
=
φˆ
)
1 2π π
Re ∫ ∫ E θ H*φ − E φ H*θ r 2 sin θ dθ dφ
2
0 0
(
)
2
I2
1 2
 ∆z 
Io R r
= o πη  =
3
2
 λ 
4
(2-128)
Valid only for very short length
2
2 Hertzian dipoles ( ∆z ≤ 0.02λ )
2  ∆z 
2  ∆z 
R r= R a=
πη   = 80 π   Ω
(2-169) p. 57
3  λ 
 λ 
789.6
=
D
Smax
= 1.5
So
(2-148)
p. 54
A Linear Center-Fed Dipole
(See also pp. 152-160 of the text)
ẑ
z′ = +
L
2

A zˆ

H φˆ
Im

E θˆ
Ia
z′ cos
0
z′ = −
L
2
 L

′) Im sin β  − z′  
I(z=

 2
Note that
z′ <
L
2
 L
′) ′
Ia I(z=
I m sin  β 
=
z =0
 2
(6-1)
p. 152
(1)
As previously assumed on p. 3 of these Notes (from Eq. 2-86 of the text),
R = r − z′ cos θ
From Eq. 2-101

A
=
 0
µ
 L

Im  ∫ sin β  + z′   e jβz′ cos θdz′
4πr

 2
 − L / 2
L/2

L

+ ∫ sin β  − z′  e jβz′ cos θdz′ e− jβr zˆ
2


0
 j60 Im

e − jβ r
E = jω sin θ A z θˆ = jη
2Im F(θ) θ =
F(θ)e− jβr θˆ
4πr
r
(6-3)
(6-6) p. 154
where F(θ) is the function that gives the variation of radiated fields with angle θ.
Note that this expression for the radiated θ-directed E field can also be expressed
in terms of the feedpoint antenna current Ia using Eq. (1) on this page.
5
 βL

 βL 
cos 
cos θ  − cos 

2

 2 
F(θ) = 
sin θ
(2)
See p. 154 of the text, Fig. 6-4, for plots of F(θ) for several values of L/λ.
F(θ) is always zero for angle θ =0 i.e. no radiated fields along the length of the
dipole.
 1
E θ E*θ
15 I2m 2
*
ˆ
=
=
S
Re E=
H
r
F (θ) rˆ
θ φ
2
2η
πr 2
(
)
Ia2
30 Prad
F2 (θ)
2
=
⋅
F=
(θ)rˆ
πr 2 sin 2  β L 
πr 2 R a sin 2  β L 




 2
 2
  15 I2m 2π π 2
Radiated Power Prad= ∫∫ S ⋅ ds=
F (θ)r 2 sin θ dθ dφ
2 ∫ ∫
πr
0 0
1 2
1 2
Im R rm
Ia R a
= =
2
2
15
=
Ra
60 I2m π 2
∫ F (θ) sin θ dθ
Ia2 0
(3)
(4)
(5)
Where Ra is the antenna equivalent resistance at the feed point ( z′ = 0 ).
The antenna equivalent resistance Ra is given by Table l on p. 7.
Thus the directivity D of a linear center-fed antenna of end-to-end length L is
given by:
2
Smax 120 F (θ) max
D =
=
So
Ra
 βL 
sin 2 

 2 
6
(6)
Table 1. Calculated values of the driving point resistance R a for end-fed monopoles of different
lengths
h/λ. (Multiply
2 to
obtain
the resistance
driving point
resistance
center-fed
Calculated
values ofbythe
driving
point
end fed for
monopoles
of
R a for
dipoles
of
length
L
=
2h.)
different h/λ. (Multiply by 2 to obtain the driving point resistance for dipoles.)
h/λ = L/2λ
h/λ = L/2λ
Ra
7
Ra
Fig. 1. The calculated resistance Ra and reactance Xa of an end-fed monopole antenna
of length h (in terms of wavelength λ). Multiply by 2 to obtain the driving
point resistance Ra for a center-fed dipole antenna of length L = 2h.
8
Example 1: Calculate and compare the directivities, gains, and power densities including E-fields created by dipole antennas of lengths L =
0.07 λ, 0.18 λ, 0.5 λ, and 1.1 λ. Power radiated by the antenna is 100 W and distance from the antenna to the field point ro = 10
km.
Note that the radiated power and the distance ro are needed to calculate the power density and maximum electric fields.
S max
θ =90
From p. 7
of Notes
L/2 λ
Ra ohms
I a* A
F (θ ) θ =90
0.035
0.09
0.25
0.55
0.994
6.68
73.12
1731.1
14.18
5.47
1.65
0.34
0.0241
1.844
1.0
1.951
i
= So D
E2
S=
2η
Including Ohmic losses for Prob. 5 of HW
From Eq. 2-153
antenna efficiency
D ii
 P 
=  rad 2  D μW/m2
 4π r 
Emax = 2η S max
mV/m
Rohmic iii Ω
1.47
1.516
1.64
2.76
0.117
0.1206
0.1305
0.2196
9.39
9.54
9.92
12.87
0.0388
0.1042
0.208
8.758
er =
Ra
Ra + Rohmic
0.9624
0.9846
0.997
0.995
From Eq.
2-155
G = erD
1.414
1.493
1.635
2.746
9
* Note that
2 Prad
1
Prad =   I a2 Ra from Eq. (4) on p. 6 of Class Notes; I a =
Ra
2
 βL

 βL
πL

πL 
cos 
cos θ  − cos 
cos θ  − cos 
 cos 

 2

 2 
 λ

 λ  . For L/λ < 1.38 – 1.4; F(θ) is max. for θ = 90°.
i
From Eq. (2)=
on p. 6 of Class Notes, F (θ )
=
sin θ
sin θ
F (θ ) max = 1 − cos (π L λ )
θ =90
ii
iii
2
120 F (θ ) max
From Eq. (6) on p. 6 of Notes, D =
Ra sin 2 (π L λ )
Rohmic is given by Eq. (9) on p. 13 of Class Notes;
RS = 1.988
f MHz
σ
; Take 2a = 3.264 mm (0.1285) ← 8 AWG wire (App. B on p. 783 of the Text).
For Aluminum, from App. B.1, σ = 3.5 × 107 S/m; take fMHz = 10 MHz.
9
Example 2:
Prad = 1 W;
f = 835 MHz;
L= 2h= 0.65λ= 0.65
r = 1 km
30
= 23.35 cm
0.835
h L
= = 0.325
λ 2λ
= 2×
Ra
Table1
Prad =
84.974 + 96.727
= 181.7Ω
2
1 2
Ia R a
2
Ia = 0.1049 A
=
Im
Ia
Ia
=
=
0.1177
 βL 
 πL 
sin 
 sin 

 2 
 λ 

 βL  
1 − cos 


60 Im
60 Im 
2  

=
E max=
F(θ)

= 10.265
r
r 
1
θ=90



E rms = 0.707 E peak = 7.26
mV
m
mV
m
E max 2 E rms 2
µW
Smax =
=
= 0.1398 2
2η
η
m
F2 (θ)
120
120
2.114
max =×
D=
=
1.759 (2.45 dBi)
Ra
2  β L  181.7 sin 2 (0.65π)
sin 

 2 
This is an improvement of only 1.073 times (or 0.3 dB) relative to a half wave dipole.
Smax = So D = 0.1398
10
µW
m2
Radiation patterns
xy plane
H-plane
y
z
2.45 dBi (0.3 dBd)
x
Fig. 2. The radiation pattern of a z-directed dipole antenna for the xy plane or H-plane
(normal to the orientation of the dipole).
See also p. 154 (Fig. 6-4)
yz plane
(E-plane)
θHP
HP
Fig. 3. The radiation pattern of the z-directed dipole antenna for the yz plane or the
E-plane.
HP = 2 × (90° - θHP)
(2-126)
p. 49
2

 βL

 βL  
2
 cos  2 cos θHP  − cos  2  
1
 βL  






=
1 − cos 

sin θHP
2 
 2 




11
(7)
pp. 57, 58 Text
Ohmic Losses for a Linear Dipole
dR
From Eq. 6-1, p. 152 Text
L
I( z′ ) = I m sin β  − z ′ 

 2
0 ≤ z′ ≤
 L

= Im sin β
+ z ′ 

 2
L
2
L
− ≤ z′ ≤ 0
2
 βL 
′)
Ia I(z
I m sin 
=
=

z′=0
 2 
(1)
(2)
Ohmic power lost in the antenna
Pohmic =
(
1 L/2 2
∫ I dR
2 − L/2
)
(3)
dR =
dz ′
R dz ′
= s
(2πa δ)σ
2πa
(4)
R=
s
1
=
σδ
ωµ
2σ
(5)
where
(2-171)
p. 58
is the surface resistance which depends on the conductivity σ of the material and
frequency ω (= 2πf). See App. B.1 of the Text for σ of various metals.
12
=
Pohmic
L/2
∫
0
0
R s L / 2 2
L

L
 
 ∫ Im sin 2 β  − z′  dz′ + ∫ I2m sin 2 β  + z′  dz′
4πa  0
2

2
 
−L / 2
βL / 2
βL / 2
1  sin (2ζ ) 
ζ
ζ−
∫ sin ζ d=
2β 
2  0
0
1  βL sin (β L) 
=
−

2β  2
2
L
 ′ 1
=
sin β  + z′  dz
β
2

2
(6)
2
(7)
L

β  + z′ 
where ζ = =
2

2
Pohmic = Im
=
R ohmic
R s L  sin (βL)  1 2
=
1−
IA R ohmic
8πa 
(βL)  2
Rs L
4πa
Antenna efficiency =
er
1
 βL 
sin 2 

 2 
 sin (βL) 
1 − (βL) 


Prad
Ra
P
=
=
Pin Prad + Pohmic R a + R ohmic
Gain G = er D
(8)
(9)
(2-177)
(2-155)
For a Short Dipole (L = Δz << λ)
2  ∆z 
R=
a 20 π 

 λ 
R=
ohmic
2
(2-172)
Rs ∆z Rs L
=
6π
6π a
(2-175)
For the general case of a linear dipole or a monopole Rohmic is calculated from the
general Eq. 9 given above.
Example 3:
For a Short Dipole
2 L
2
L
Ra =
20 π   ≅ 197.4  
λ
λ
2
See e.g. Table 1 on page 7 for L/λ =0.02, Ra = 2 × 0.0394 = 0.0788 Ω.
13
(2-172)
Using the conductivity of steel (see App. B.1 of the Text) σ = 2 × 106 S/m.
From Eq. 2-171 or Eq. 5
Rs =
1.4 ×10−3 f MHz
Ω
From Eq. 9, βL is small and we can expand sin x for small x
R ohmic=
Rs L
1
4πa  βL 2


 2 

(βL)3 
β
−
L


6 = R s L
1 −
βL

 6πa




Short dipole
(2-175)
p. 59
For L/λ = 0.02 dipole at f = 1 MHz; taking 2a = l/8"
λ = 300m
R ohmic =
=
er
Antenna Efficiency
;
L = 6m
1.4 × 10−3 × 6
= 0.2807Ω
1
−2
× 2.54 × 10
6π ×
16
Ra
0.0788
=
= 0.219
R a + R ohmic 0.0788 + 0.2807
(21.9%)
Gain G = er D = 0.219 ×1.5 = 0.3285.
Note that for short dipoles of thin wires, the ohmic resistance can be substantial
and even larger than Ra. Therefore, this leads to reduced efficiency of radiation.
Example 4:
For a Half Wave Dipole
L = 0.5λ; Ra = 73.12 Ω;
βL πL π
= ; βL = π; f = 10 MHz; L = 15m; 2a = 1/8"
=
2
2
λ
From Eq. 9
Rs L
R ohmic
=
=
4πa
=
er
(1.4 ×10−3 10 ) ×15=
1

4π×  × 2.54 ×10−2 
 16

73.12
= 0.9565
73.12 + 3.33
3.33Ω
(95.65%)
G = 0.9565 D = 0.9565 × 1.64 = 1.568
14
pp. 75-81 Text Dipoles Versus Monopoles Above a Perfect Ground or Reflector
I′a
Ia
Image antenna
I(z′) same as on page 5 of the Notes
0 ≤ θ ≤ 180
For
L
L

I ( z ′) I m′ sin β  − z ′  0 ≤ z ′ ≤
=
2
2

0 ≤ θ ≤ 90
For

j 60 I m′
E′ =
F (θ ) e − j β rθˆ
r

60 I m
E= j
F (θ ) e − j β rθˆ
r
(2)
 15 I m2 2
S=
F (θ ) rˆ
π r2
 15 I m′2 2
15
(4)
=
S′ =
F (θ ) rˆ
2
πr
π r2
Prad =
1 2
I a Ra
2
′ =
Prad
(6)
(1)
(3)
I a′2
F 2 (θ ) rˆ (5)
 βL
sin 2 

 2 
1 2
I a′ Ra′
2
(7)
F(θ) is given as Eq. (2) on p. 6 of the Notes.
Since a monopole radiates in the upper half space while a dipole radiates both in
the upper and lower half spaces,
S dipole =
1
′
for identical radiated powers
S monopole
2
′
Dmonopole
= 2 Ddipole
Ra′
X a′
=
monopole
1 Ra
2 Xa
(8)
(9)
(10)
dipole
For identical radiated powers
I a′ = 2 I a
(11)
I m′ = 2 I m
(12)
15
Example 4:
h
L
=
= 0.35 Monopole Antenna
λ 2λ
f = 1.5 MHz, λ = 200 m ; r = 1 km
h=
L
= 70 m ; Prad = 103 W (1 KW)
2
From Table 1 on page 7, R a′ = 127.1 Ω
(do not multiply by 2 for monopoles)
From Eq. 7, I′a = 3.967A
From Eq. 5,
S′
monopole
 βL

 βL 
cos θ  − cos 

2 cos 
15
(3.967)
 2

 2 
=
×
sin θ
π×106 sin 2 (0.7 π)
2
=
0.29 mW / m 2
θ=90
D′
monopole
=
S′
= 3.64= 2 × D
Prad
dipole
So =
4πr 2
From Eq. (6) on p. 6 of Class Notes
120
D=
Ra
Note that
F2 (θ)
max
2  βL 
sin 

 2 
L
= h which is the height of the monopole.
2
16
pp. 84-89 Small Diameter (<< λ) Loop Antennas
The loop antenna is a radiating (or receiving) coil of one or more turns of circular
or rectangular form. Ferrite or air core loops are used extensively in radio
receivers, direction finders, aircraft receivers, and UHF transmitters.
The theory of loop antennas is derived in a manner similar to the General Theory
of Conduction Current Antennas given on page 44 of Text and on page 2 of my
handout notes.
We start by assuming, as seen in Fig. 1, that the current I in the loop has the same
magnitude and phase. This is certainly possible for small diameter loops where
2πb < λ/10.
z
R
r
x
Loop in the xy plane
Fig. 1. A circular loop antenna of radius 'b'.
From General Theory of Conduction Current Antennas, from Eq. 2-101,

 µ 2π I φˆ ′e − jβR
bdφ ′
A=
∫
R
4π 0


R=
( x − x′ )2 + ( y − y′ )2 + z 2
x ′ = b cos φ ′; y′ = b sin φ′; z ′ = 0; x = r sin θ; y = 0; z = r cos θ
(1)
(2)
(3)
Note that we have defined the x-axis (the choice of which is arbitrary) such that
the field point lies in the xz plane. The field point F, therefore, has coordinates (x,
0, z) in Cartesian coordinate system and (r, θ, 0) is spherical coordinate system.
Substituting Eq. 3 into Eq. 2
1/ 2
R =  r 2 + b 2 − 2 br sin θ cos φ′


=−
r b sin θ cos φ′
17
 b

≅ r 1 − sin θ cos φ′
 r

(4)
since, for the far-field region, r >> b.
Using the far-field approximation for Eq. 1
 µI e− jβr 2π
φˆ ′ e jβb sin θ cos φ′bdφ′
A=
∫
2πr 0
For small radii βb =
(5)
2πb
<< 1 , we can write
λ
e jβb sin θ cos φ′ = 1 + jβb sin cos φ ′
(6)
We can also write (see Fig. 1(b))
φˆ ′ = − xˆ sin φ ′ + yˆ cos φ′
(7)
Also to be noted is that for the field point F
yˆ = φˆ
(8)
From Eq. 5 therefore, we can write

 jµI S − jβr
A=
βe
sin θ φˆ
4πr


(9)
(p. 86 Text Eq. 3-49)
where S = πb2 is the area of the loop.
On page 19, we compare the expressions for the radiated fields from a loop
antenna to those for an ideal (infinitesimal) dipole and show duality of the two
sets of fields. Ohmic resistance of a circular loop antenna can be written as
follows:
2πb
bR s
=
R ohmic = Rw =
(10)
(2πaδ)σ
a
(3-60)
p. 88 Text
where
=
Rs
f
1
= 1.988 MHz
σδs
σ
where "b" is the mean loop radius and "a" is the wire radius; Rs = 1/σδ is the
surface resistance at the frequency of interest previously defined on page 12 of
Class Notes.
The small loop antenna is inherently inductive. For a small circular loop of N
turns wound on a magnetic core
  8b 

L= N 2 b µeff µo n   − 2 
  a 

(11)
(Eq. 3-62 p. 88 Text)
18
Table 2. Field expressions for small diameter circular loop antennas and an ideal (infinitesimal) dipole antenna [see p. 4 of Class Notes].
Loop Antenna
Ideal (Infinitesimal) Dipole

H − θˆ

Magnetic Vector Potential ( A)
(9)
(3-48) Text
jIS
β sin θ e − j β rφˆ
4π r
Magnetic Field


 ∇× A
jβ
H= =
rˆ × A
−
µ
µ
Electric Field
19

 ∇× H

E=
=
−η rˆ × H
jωε
−
(3-50)
p. 86 Text
IS 2 − j β r
β e sin θθˆ
4π r
η IS 2 − j β r
β e sin θφˆ
4π r
(3-49)
p. 86 Text
µ I ∆z − j β r
e zˆ
4π r
(2-65)
p. 33 Text
j β I ∆z − j β r
sin θφˆ
e
4π r
since rˆ × zˆ =− sin θφˆ
(2-74b)
(2-70)
p. 33 Text
jη I ∆z − j β r
β e sin θθˆ
4π r
(2-74a)
p. 34 Text
Radiated Power Density
 
 1
 *
EE*
S
Re E × H =
rˆ
=
2
2η
(
)
Radiated Power
P
=
  1 2
S
∫ ds ≡ 2 I Rr
Rr (for single turn loop)
Directivity D =
S max
So
η I 2 ( ∆z )
η I 2S 2 4 2
β sin θ rˆ
32π 2 r 2
10I 2 ( β 2 S )
20 ( β S )
2
2
2 ( 4π r )
(3-52)
p. 86 Text
2
2
 S 
≅ 31, 200  2  Ω
λ 
N − turn loop
(3-54)
19
β 2 sin 2 θ rˆ
(2-76)
ωµβ
2
( I ∆z )
12π
 ∆z 
 ∆z 
80π   = 790  
 λ 
 λ 
2
1.5
2
Rr
2
2
(3-53)
1.5
NS 

31, 200  µeff 2  Ω
λ 

2
(2-77)
p. 33 Text
2
(2-169)
p. 57 Text
For an N-turn loop, Rohmic is also higher proportional to overall length of the wire
R ohmic
=N
N− turn
loop
bRs
a
(12)
The effective permeability μeff depends not only on the permeability μr of the
ferrite core material, but also on the core geometry, i.e., length to diameter ratio
R, given as follows:
µeff =
µr
1 + D ( µr − 1)
where 4D is the demagnetization factor approximately given by
D  0.37 R −1.44
(13)
D [4]
(14)
p. 87 Text
Example 5 (see also Ex. 3-1, p. 88, Text):
Calculate the input impedance, directivity, and gain for an N = 1000 turn loop
antenna wound with a AWG 22 copper wire on a ferrite rod of diameter 3/4".
This antenna is to be used at a frequency of 1.5 MHz. It is given that µeff = 50
for the ferrite that is used.
Solution: From p. 783 of the Text, for AWG 22 wire d = 2a = 0.644 mm ⇒
0.0253′′
From p. 58, Eq. 2-171
R=
s
1
=
σδ
f
ωµ
= 1.988 MHz ohms
σ
2σ
(15)
For copper σ = 5.7 × 107 S/m (p. 783, Text); Rs = 3.22 ×10-4 Ω at f = 1.5 MHz
Mean loop radius b =
3′′
+ a = 9.847 mm
8
From Eq. 12, for N = 1000-turn loop
Rohmic = 9.85Ω
From Eq. 3-54 Text (see also p. 19 of Class Notes)
𝑅𝑟 = 31,200 �𝜇𝑒𝑓𝑓
4
𝑁𝑆 2
𝜆2
� = 45Ω
(3-54) Text
R. Pettengill, H. Garland, and J. Mendl, “Receiving antennas for miniature receivers,” IEEE Transactions
on Antennas and Propagation, Vol AP-26, pp. 528-530, July 1977.
20
From Eq. 11 above
L = 0.232H
⇒
D = 1.5;
ωL = 2π×1.5 ×106 × 0.232 = 2.18 MΩ
𝑒𝑟 = 𝑅
𝑅𝑟
𝑜ℎ𝑚𝑖𝑐 +𝑅𝑟
= 0.82 (82%)
G = erD = 1.23
pp. 107-111 Antennas in Communication Systems
Pt
Pr
Gt
Gr
Fig. 4-4 (p. 107 Text). A communication link.
Voc = VA
Equivalent circuit for the receiving system.
Maximum available power to the receiver (for Z L = Z∗A )
2
i
2
2
VA2
1  VA 
1 E (∆z)
P
RA =
=

 =
Am
2  2R A 
8R A 8
RA
(1)
For an ideal (infinitesimal) dipole
VA = E i ∆z
PAm
3 2
2
Maximum effective aperture area = Ae,m = S = 8π λ = 0.119 λ
inc
21
(2)
(3)
=
Sinc Ei
2
2η
(4)
D = 1.5 for an ideal dipole
D=
4π
λ
2
4π
A e,m =
λ
2
×
3 2
λ = 1.5
8π
(5)
(4-22)
(6)
(4-23) Text
For a general antenna, therefore
4π
G = 2 Ae
λ
(7)
Ae = er Aem effective aperture area of an antenna
(4-27)
p. 108 Text
Available power including also the antenna losses
PA = S Ae
S = Gt
(4-26) Text
Pt
2
4πR
G t G rλ2
 G t Pt 
=
Pr SA
=
A er Pt
er 
=
(4πR)2
 4πR 2 
(4-31)
(4-33)
or
Pr = Pt
A et A er
R 2λ 2
Friis transmission formula
(4-33)
We can also write Eq. (4-33) in dB-form as follows:
Pr (dBm) = Pt (dBm) + G t (dB) + G r (dB) − 20 log R(km)
− 20 log f (MHz) − 32.44
(4-34)
Example 6:
For Ground Based TV Stations
Say, Channel 5
f ≅ 80 M Hz
f = 76 - 82 MHz
Prad ~ 5 - 10 kW
say, 10 kW = 104 W (40 dBW)
22
λ = 3.75 m
Gt ~ 20 - 50 (factor)
say, G t = 30 (factor) ⇒ (14.77 dB ~ 15 dB)
EIRP = Gt Prad → 55 dBW → 105.5 W (85 dBm)
Rmax ~ 20 - 30 miles ~ 50 km; since 1 mile = 1.6 km
say
G r ≅ 7 dB ≅ 5
λ2
2
G r = 5.6 m
A e,r =
4π
Using the logarithmic form of the Friis communication link formula Eq. (4-34)
Pr (dBm) = 70 + 15 + 7 − 34.0 − 38.06 − 32.44
=
−12.5 dBm =
10−12.5 mW =
56.2 µW
Example 7:
Calculate the open-circuit voltage developed across an antenna of resistance RA=
80 ohms for the above-calculated incident power density
Sinc
=
Pr 56.2
µW
=
= 10
A e 5.6
m2
Assume RA= 80Ω
2
Voc
V2
→ A
8R A
8R A
power picked
=
up and delivered to a matched load Sinc A e
Voc = 8R ASinc A e,r = 8 × 80 Pr = 640 × 56.2 ×10−6
= 188.65 mV  0.19V
23
Chapter 8 -- Antenna Arrays (see pp. 271..... Text)
For a uniformly excited (UE), equally-spaced linear array (ESLA)
F(x,y,z)
d cos φ
For N identical radiating elements (length, orientation, etc.) that are excited with
identical magnitudes but progressively phase-shifted currents i.e.
I , I e − jα , I e −2 jα ,  I e − j ( N −1)α
(1)


E
we can write the total electric field 
T as follows


 

 − jβ⋅

− jβ⋅ rN −1
jβ⋅ r1
r
ET = Eoe
+ E1e
+  E N −1 e


r = xˆx + yyˆ + z ˆz


r
1

= (x − d)xˆ + yyˆ + zzˆ

β = β [sin θ cos φ xˆ + sin θ sin φ yˆ + cos θ zˆ ]
(1)
(2)
(3)
From Eq. 1, we can write


 − jβ⋅
ET =
E o e r 1 + e− jα e jβd sin θ cos φ + e−2 jα e2 jβd sin θ cos φ + 


  N −1
= E o e− jβ⋅ r
∑ e jnψ
(4)
n =0
(3-16)
since
  
  
β ⋅ ( r1 − r ) = −βd sin θ cos φ; β ⋅ ( r2 − r ) = −2βd sin θ cos φ
24
(6)
From Eq. 4

 

E
=
E
T
o ⋅ AF


where
N −1
e jnψ
=
∑
Array Factor
AF
=
1 − e jNψ
n =0
(7)
1 − e jψ
sin(Nψ / 2)
sin(ψ / 2)
AF = e j(N−1)ψ / 2
(8)
(8-19) p. 279 Text
Normalized AF
f(ψ) =
sin(N ψ / 2)
N sin(ψ / 2)
(9)
(8-22 Text)
UE, ESLA
where
ψ = ( βd x sin θ cos φ − α x ) for an xˆ − directed array
(
)
= βd y sin θ sin φ − α y for a yˆ − directed array
= ( βd z cos θ − α z ) for a zˆ − directed array (see Eq. 3-19 Text)





sin(N ψ / 2) *
E T = NEo f(ψ) = E o
sin(ψ / 2)







 ∇ × ET
jβ
ˆr × E T
=−
HT =
jωµ o
jωµ o


 
 1

*
E T ⋅ E*T
=
S
Re E T × H=
rˆ
T
2
2η
(
)
* From Eq. 11, for directions of max radiation
(10)
(11)
(12)
(13)
ψ
= 0, ± π, ± 2π,
2
p. 280 Text
A number of trends can be seen by examining the normalized array factor |f(ψ)|.
1. As N increases the main lobe narrows.
ψ = 0 where |f(ψ)| =1.
25
Peak for the main lobe occurs for
Plot of |f(ψ)| as a function of ψ.
Fig. (8-8) p. 280 Text.
For directions of zero (nulls of radiation)
Zero values of |f(ψ)| occur for
Nψ
= ± π, ± 2π,
2


i.e.
2π
ψ=±
,
N


±
4π
,
N
(14)
2. More than one major lobe will exist if it is possible to get values of ψ = ± 2π, ±
4π. The additional lobes are called Grating Lobes.
3. The minor lobes are of width 2π/N in the variable ψ and the major lobes (main
and grating) are twice this width i.e. 4π/N in the variable ψ.
4. The side lobe peaks decrease relative to the major lobe as
1
1
:
 3π 
 5π 
N sin 
N sin 


 2N 
 2N 
For large N, SLL decrease as
1:
(16)
2
2
:
: 
1:

 3π 5π
(17)
i.e.
0, 20 log
2
2
, 20 log
, 


5π
3π
or
0, -13.46, -17.90, 

 dB
(18)
5. As N increases, there are more side lobes in one period of f(ψ). See also the text,
Fig. 8-8, p. 280.
26
Case A. Broadside Arrays
All antennas excited in phase α = 0.
From Eq. 10, for an antenna stretched along the z-axis
ψ= βd cos θ = 0, ± 2π, ± 4π 

(19)
π
 λ
 2λ 
θ = ± , cos −1  ±  , cos −1  ±

2
 d
 d 
(20)
for major lobes
Subcase 1
d/λ < 1 i.e. interelement spacing less than λ
Two and only two major lobes of radiation for θ = ± π/2 i.e. in directions
broadside to the stretch of the array
For directions of first nulls ( θFN )
27
= θ FN left − θ FN right
BWFN
(24)
 λ 
−1  λ 
= cos −1  −
 − cos 

 Nd 
 Nd 
(25)
(8-31) p. 283 Text
λ
 λ  2λ
2sin −1 
radians =
114.6°
=
≅
Nd
 Nd  Nd
(26)
(8-33)
for
Nd >> λ
Example 6:
d/λ = 0.5 ,
N=8
From Eq. 23
π
1
− θFN = ± sin −1   = ± 14.5
2
4
(27)
BWFN = 29°
(28)
Angle for first-side lobe
 Nψ 
sin 
= ± 1
 2 
from Eq. 11
3π
Nψ
8πd
cos θ = ±
= 4(βd cos θ) =
2
λ
2
(29)
 3
θ = cos −1  ±  = ± 68 ; ± 112
 8
(30)
(-13.46 dB down relative to major lobe)
Subcase 2
1≤
d
≤2
λ
From Eq. 19, for major lobes
ψ = βd cos θ = 0, ± 2π, ± 4π , 


cos θ = 0, ±
λ
2λ
, ±
d
d
(31)
(31a)
This corresponds to six major lobes and a radiation pattern of the type shown on
the next page.
28
Example 7:
d/λ = 1.5
From Eq. 31, for major lobes
3π cos θ = 0, ± 2π, ± 4π
(32)
For angles of maximum radiation
cos θ = 0, ± 2 / 3, ± 4/ 3
θ = ± 90°; ± 48.2°,
(33)
(34)
± 131.80°
Six maxima of radiation
θ =90

=
θ 180 − 48.3
θ =48.3
= 131.8
−48.3

−131.8.3
d/λ > 1
−90
Angles for first nulls for each of these maxima are obtained from Eq. 21
3π cos θ FN = ±
Subcase 3
2π
2π
; ± 2π ±
N
N
(35)
d/λ = 1.0
For this case, there are four maxima of radiation (major lobes); the two fatter
lobes in the above figure coalesce into single modes with directions of
maximum radiation θ = 0°, 180°.
From Eq. 31a
(36)
29
θ= ± 90, 0, 180

(37)
Four maxima of radiation
From Eq. 21, directions of first nulls are:
ψ=
2π
2π
d cos θFN = ±
;
λ
N
cos θFN =±
λ
,
Nd
±1±
± 2π ±
2π
,
N
λ
,
Nd
(38)
(38a)
Example 8:
N = 8, d/λ = 1.0
From Eq. 38a
7
7
, −
8
8
(39)
− 28.96, 151.04; 208.96
(40)
1
1
1
cos θFN = ± , ± 1 ± = ± ,
8
8
8
θFN = 82.82, 97.18, 28.96,


BWFN = 14.36° for major lobes along ± 90°
= 57.92° for major lobes along 0°, 180°
30
p. 315
Case B.
Electronically-Scannable (Steerable) Antennas -- Phased Array
Antennas
The phase shift of currents (excitations) for adjacent antennas may be altered
α ≡ −βd cos θ o
(phase delay)
(41)
From Eq. 11 for directions of maximum radiation (major lobes)
ψ
= 0,
2

± π , ± 2π , 

(42)
ψ = 0 , ± 2π , ± 4π , 


(43)
βd ( cos θ − cos θo=
) 0 , ± 2π , ± 4π , 


(44)
λ
λ
cos θ = cos θ o , cos θo ± , cos θ o ± 2 , 


d
d
(45)
For two and only two major lobes for θ = ± θo, d/λ should be less than 0.5.
 α 
=
θo cos −1  − 
 βd 
θo
Example 9:
−1  π / 6 
N = 8 , d/λ = 0.3 ; α = -30° ; θo = cos 
 = ± 73.9°
 2π× 0.3 
For
α variable from -30° to -75°
θo varies from ± 73.9° to ± 46°
31
For directions of zero radiation, from Eq. (14) on p. 25 of Class Notes,
Nψ
= ±π
2
ψ= ±
2π
N
2π
β d ( cos θ − cos θ o ) =
±
N
θ FN cos θ o ±
cos=
2π
Nβd

2π 
=
θ FN cos −1 cos θ o ±
N β d 

λ 

= cos −1 cos θ o ±
Nd 

(45a)
d/λ = 0.3 , N = 8
Example 9 (continued):
α = -30° = -π/6
θo = ±73.9°
0.4167
1 

=
θ FN cos −1 0.2773 ±
2.4 

= 46.05° ; 98.01°
BWFN = 51.96°
Example 9, Part B: Let us compare the antenna array of N = 8, d = 0.3λ for the
following three conditions:
Direction of
max radiation
α
θo = ±90°
0
Broadside array
-30°
directions of max
radiation
from Ex. 9 on this
page
-108°
θo = ±73.9°
End fire antenna
array
α = -βd
from Eq. (20) on p. 27
of Class Notes
BWFN
from Eq. (26) on p. 28 of Class Notes
 λ 
BWFN = 2sin −1 

 Nd 
= 49.25°
θo = ±73.9°
BWFN = 51.96°
θo = 0°
λ 

=
BWFN 2 cos −1 1 −

 Nd 
= 108.6°
See Eq. 52 on p. 36 of Class Notes
32
For a one-dimensional antenna array
The array factor of a one-dimensional antenna array from Eq. (8) of Class Notes
p. 25 is as follows:
AF =
sin ( Nψ 2 )
sinψ 2
(1)
Where ψ is given by Eq. (1) on p. 25 of Class Notes.
From Eq. (1) here, for directions of max radiation
ψ = 0, ±2π, …
For directions of zero radiation or nulls of radiation
Nψ
=
±π , ±2π ,
2
or ψ = ±2π/N for first nulls of radiation.
Table of general relationships for one-dimensional z-directed
phased array antennas
ẑ
α
0
directions of max.
radiation principal lobe/s
ψ=0
θo = ±90°
broadside array
directions of first nulls
derived on p. 32
λ 

cos −1 cos θ o ±
Nd 

 λ 
= cos −1  ±
 Nd 
see Eq. (22) on p. 27 of
Class Notes
α
 α 
θ o cos −1  −
=

 βd 
see p. 31 of Class
Notes
α = -βd
θo = cos-1 (1)
= 0°
End fire array
λ 

cos −1 cos θ o ±
Nd 

 α
λ 
= cos −1  −
±

 β d Nd 
BWFN
 λ 
2sin −1 

 Nd 
see Eq. (26) on p. 28 of
Class Notes
calculate θFN1, θFN2
BWFN = θFN2 - θFN1
see Eq. 45a on p. 32
of Class Notes
λ 

cos −1 1 −
 Nd 
λ 

2 cos −1 1 −

 Nd 
 λ 
= 4sin −1 

 2Nd 
see Eq. 52 on p. 36
of Class Notes
33
Case C. End Fire Arrays
From the previous section, we can see that in order to get a single major lobe for
θo = 0° i.e. along the line or stretch of the array, we need
α = - βd
and
d/λ < 0.5
(47)
For this case, the two major lobes on the previous page coalesce into one major
lobe in the end fire direction.
Example 10:
N = 20 , d/λ = 0.4
2πd
α = −βd = −
= −144
λ


(48)
For directions of first nulls from Eq. 14
ψ = β d ( cos θFN − 1) = ±
2π
N
(49)
(49a)
7
θFN =
± cos −1   =
± 28.96
8
 
BWFN =θ
2 FN =57.92
(50)
(50a)
It is interesting to note that for a given stretch of the array (N-1)d or
approximately Nd, BWFN is smallest for broadside arrays, intermediate for
phased arrays and broadest (largest) for end fire arrays.
Example 11:
For N = 20, d = 0.4 broadside array ( α =0 )
1
−1  λ 
=
BWFN 2 sin
=
sin −1   14.36

 2=
 Nd 
8
as compared to 57.92° for an end fire array.
34
Example 10 (continued):
N = 20 ; d/λ = 0.4
For Hanson-Woodyard end fire array (p. 285 Text)


α=
− βd +
π 

N
(8-37)
d<
;
180° 

= − 144° +

20 

λ
1 
1 − 
2  20 
(8-38a)
d < 0.475 λ
= −153°
For directions of first nulls (from Eqs. 10, 14 on pp. 25, 26 of Class Notes)
2π
2π
ψ=
β d z cos θ − α z =
144° cos θ FN − 153° =
±
=
±
=
±18°
N
cos θ FN
=
153° ± 18° 171°
=
144°
144°
,
20
135°
= 0.9375
144° rather than 7/8 or 0.875 in Eq. 49a
θ FN = ± cos −1 ( 0.9375 )
= ±20.36°
BWFN = 40.72°
rather than 57.92° for an
ordinary end fire array
(see Eq. 50a on p. 34 of Class Notes)
35
Nd/λ
BWFN for antenna arrays
λ 

= 2 cos −1 1 −

 Nd 
 λ 
= 4sin −1 

 2Nd 
Example 11-1:
Nd
= 5.0
λ
BWFN = 23.07
for a broadside array
BWFN = 73.74 for an end fire array
p. 293 Directivity of a Uniformally Excited, Equally Spaced Antenna Array
From Eq. 13 we can write
2
AF





Nψ 2

 

2
sin
*

E ⋅E
E
2 ˆr
S = T T ˆr = o
ψ
2η
2η sin
2


2
2
Smax = So max AF max ⇒ N So max
Power radiated by the antenna array =
=
(54)
1 2
IA ( R A0 + R A1 + R N −1 )
2
N −1
1
2
IA ∑ R Ai
2
i =0
36
(53)
(55)
D array = N 2 Do
In general
R A0
N −1
(55a)
∑ R Ai
i =0
where Do and R A0 pertain to an isolated element of the antenna array.
Ignoring Mutual Impedance Effects
RA0 = RA1, 
= RN-1

Power radiated by the antenna array =
1
2
IA RA0 N = N Po
2
(56)
where Po is the power radiated by the zeroth element
2
AF max
Smax
=
=
D=
Do ⋅
Do N
N Po
N
(57)
4πr 2
where Do is the directivity of each of the antenna elements.
Example 12:
Calculate the directivity of an antenna array of 20 half wavelength (L = λ/2)
dipoles that are fed in phase and consequently radiate in broadside directions.
Neglect the mutual impedance effects for this problem.
Solution:
2
AF
D = max =NDo =20 ×1.64 =32.8
N
Example 13:
a. Calculate the directivity/gain of an array of 30 vertical monopoles above ground
each of length H = L/2 = 0.35 λ that are spaced a distance d = 0.2λ from each
other.
b. Calculate the relative phase difference between monopoles if the major lobe of
radiation is to be in the end fire direction assuming an ordinary end fire array.
c. Calculate the BWFN for this array.
Solution:
a. D = N Do = 30 × 3.636 = 109.08
b. From Eq. 47
2πd
α = −βd = −
= −72
λ


37
Each of the successive elements should be fed with a current that is lagging in
phase by 72° from the previous element.
c. From Eq. 49a,
cos θFN = 1 −
λ
1 5
λ
= 1− =
=1 −
6 6
6λ
Nd
BWFN = 2 cos-1 (5/6) = 67.11°
2-D and 3-D Uniformly Excited, Equally-Spaced Antenna Arrays
Nx: No. of antennas in x-direction
Ny: No. of antennas in y-direction
Nz: No. of antennas in z-direction
A 2-D array of identical elements
Neglecting phase terms

 

E T = E 1 AF x AF y AF z


38
N

sin  x ( βd x sin θ cos φ + α x ) 
 2

AF x =
1

sin  ( βd x sin θ cos φ + α x ) 
2

 Ny

sin 
βd y sin θ sin φ + α y 
 2

AF y =
1


sin  βd y sin θ sin φ + α y 
2

(
)
(
)
N

sin  z ( βd z cos θ + α z ) 
 2

AF z =
1


sin  ( βd z cos θ + α z ) 
2

As always







 ∇× E
− jβ ˆr × E T ˆr × E T
T
HT =
=
=
− jωµ o
η

 − jωµ o

 

S T = S 1 AF 2x AF 2y AF 2z



S1

where S 1 is the radiated power density due to one of the elements. These
arrays are also called mattress Arrays.
Example 14: A Unidirectional Broadside Array
In order to obtain a unidirectional broadside array, we can use a 2-D antenna array
of Nz = 1, Ny = 2, Nx which can be an arbitrary number. By using a back row of
antennas that are placed with dy = λ/4 and αy = 90°, we can obtain an antenna
pattern as shown.
ŷ
λ/4
x̂
39
Universal field-pattern chart for arrays of various numbers n of
isotropic point sources of equal amplitude and spacing
|Array factor|
or f (ψ )
40
41
∠90
∠0
An enlarged version of Fig. (b) from previous page.
α = 90°;
d = λ/4
Note a broad unidirectional (cardiod type) pattern possible
with this arrangement.
42
Reactance of Linear Dipoles
We have previously calculated Rin = Ra + Rohmic for linear dipole or monopole
antennas. We need to know the input reactance Xin or Xa in order to design
matching networks to match power in or out of the antenna.
Like the current distribution on a linear dipole, the input reactance can be written
as though a two-wire line of length L/2 had been opened up as shown in the
following:
E
d
D
E
L/2
S
D′
E′
z′
E
L/2
D
2
L/
a.
g/2
z = 0 2 z′
-g/2
D′
D
S(z)
− z′
D′
E′
b.
E′
c.
For a two-wire line of Fig. a, each of diameter d,
•
Zo =
120  2S 
n  
εr
 d 
(1)
For the opened-up line of Fig. b, we can define an average characteristic
impedance Zo
•
1
Zo =
L/2
L/2
∫
0
43
Zo (z) dz
(2)
For the completely opened-up transmission line of Fig. c, we can define
2
Zoa =
L
g / 2+ L / 2
120
 4z′  ′
n 
 dz
εr
 d 
∫
g/2
(3)
120   2L  
=
 n 
 − 1
εr   d  
The reactance Z D D′ of an open-circuited transmission line of length L/2 can be
written from Transmission Line Theory
 L
ZDD′ =
Zin =
− jZoa cot  β 
 2
(4)
Combining Eq. 3 and 4, we can write
 βL′ 
ZDD′ = jXin = − j Zoa cot 

 2 
(5)
where L′ ≅ (1.02 − 1.10)L is the effective "electrical" length of the antenna.
Reactance of Linear Monopoles Above Ground
We have previously shown that
R in
monopole
=
1
R in
2
dipole
(6)
Similarly,
Z=
oa
monopole
1
  2L  
=
Zoa
60 n 
 − 1
dipole
2
  d  
(7)
From Eq. 5
Xin
monopole
=
1
Xin
2
dipole
(8)
Example 15:
Calculate the feed point impedances Rin + jXin for linear dipoles of length (a) L =
0.5λ (half wave dipole) and (b) L = 0.3λ. Assume that the antenna wire is No. 19
AWG (d = 9.12 × 10-4 m from Table B.2, p. 623) and frequency f = 30 MHz.
Take copper as the material for the antenna.
44
a. From the table on driving point resistance, p. 7 of Class Notes
R ri
=
2 × 36.56 =
73.12Ω
(9)
 L sin (βL) 
1
 − 4β 
2π4

sin  
2
(10)
L / 2 λ=0.25
From p.13 of Class Notes, Eq. 9
=
R ohmic
Rs
πa
σcopper = 5.8 × 107 S/m
R=
s
1
= 2.61×10−4 f MHz
σδ
R s =2.6 ×10−4 30 =14.4 ×10−4 Ω
L=
for copper
@ f =30 MHz
(11)
λ
=5m
2
14.4 × 10−4 5
= 1.411Ω
R ohmic =
π × 4.06 × 10−4 4
(12)
R in = R ri + Rohmic = 74.53Ω
(13)
From Eq. 3 on p. 44 of Class Notes
 
 
10
− 1= 996.3Ω
Zoa= 120 n 
−4  
  9.25 ×10  
(14)
Taking L ′ ≅ 1.02 L = 0.51λ
 2π 0.510λ 
cot  ×
−0.0314
=
2 
 λ
From Eq. 5 on p. 44 of Class Notes
 βL′ 
jXin =
− j 996.3 cot 
+ j 31.3Ω
=
 2 
(15)
Z in = R in + jX in = 74.53 + j31.3Ω
(16)
45
Note that if we had constructed a slightly shorter, say L = 0.49λ dipole
L ′ = 0.49 × 1.02λ = 0.5λ
jX in = −j 996.3 cot β L′ ⇒ 0
Z in = R ri + R ohmic + j0 = 69.46 + 1.41+ j0 ≅ 71 + j0Ω
(17)
b. You can solve for the numbers for part b of the problem following the procedure
indicated above.
Example 16:
Feedpoint impedance for a linear monopole of length L/2 = 0.25λ.
Solution:
From Eq. 16
Z in
monopole
=
1
Z in
2
dipole
= 37.27 + j15.65Ω
Examples on Calculation of Im (ZA) or Reactance of Antennas
Example 17: (See also Fig. 6-6, p. 157 Text)
L/λ = 0.4; wire radius a = 0.0005λ (same as in Fig. 6-6, p. 157 Text). Assume
L ′ = 1.04 L .
From Eq. 3 on p. 44 of Class Notes
  0.8λ  
Z=
1 682.15Ω
oa 120 n 
 −=
  0.001λ  
From Eq. 5 on p. 44 of Class Notes
 πL′ 
ZDD′ = − j Zoa cot 
 = − j682.15 cot (0.4π×1.04) = − j184.3Ω
 λ 
taking L ′ = 1.04L (from Table 6-2 on p. 159 Text). From the graph in Fig. 6-6, p.
157 Text
Im ( ZA ) =
− j180Ω
Example 18:
L/λ = 0.3; wire radius a = 0.0014λ (one of the wire radii on p. 8 of Class Notes).
46
From Eq. 3, p. 44 of Class Notes
  0.6λ  
Z=
1 524.08Ω
oa 120 n 
 −=
  0.0028λ  
L′ 
π 

ZDD′ = − j524.08 cot  L′  = − j524.08 cot  0.3π×  = − j351.4Ω
L
λ 

taking L′ / L = 1.04 .
From graph on p. 8 of Class Notes
Reactance X a = −2 × 150 = −300Ω
which is close.
Examples on Mutual Impedance Effects
Example 19: A half-wave dipole above ground
x
Distance to ground d g = λ / 4
1
2d g = distance to image antenna2 = λ / 2
dg
dg
From Fig. 8-25a, b, for d/λ = 0.5
z
2
Image
antenna
I2 =
I1∠180 =
−I1
Z12 =
−12.5 − j30;
V1 =I1Z11 + I2 Z12 =I1 [ Z11 − Z12 ]
Z1=
V1
= (73 + j42.5) − (−12.5 − j30)
I1
= 85.5 + j72.5
Feedpoint impedance of the half-wave dipole placed at a distance of λ / 4 from
the ground = 85.5 + j72.5Ω rather than 73 + j42.5Ω .
Radiation Pattern
We can consider the above situation as a 2-element (N = 2) antenna array in the x
direction and write


E T = E 1 AF
47
Figure 8-25 The mutual impedance between two resonant parallel dipoles as a
function of their spacing relative to a wavelength. (a) The real part. (b) The
imaginary part.
48
Figure 8-26 The mutual impedance between two resonant collinear dipoles as a
function of spacing relative to a wavelength. (a) The real part. (b) The imaginary
part.
49
AF =
sin ( Nψ / 2 )
sin ( ψ / 2 )
where
ψ = βd x sin θ cos φ + α x = 2βd g sin θ cos φ + π
= π sin θ cos φ + π
From pp. 36-37 of Class Notes, Eqs. (54)-(57)
R A,isolated
73
2
D=
G=
Do AF max
=
1.64 × N 2 4 ×
= 5.60
R A,with ground effect
85.5
Without ground effect
D = G = 1.64
λ
Example 20: A broadside array of five
monopoles (α = 0)
/4
d = λ/2
Ant.
#1
#2
#3
#4
#5
d12 = λ / 2
d13 = λ
d15 = 2λ
d14 = 3λ / 2
I=
1 I=
2 I=
3 I=
4 I5 because it is a broadside array
V
Z1 = 1 = Z11 + Z12 + Z13 + Z14 + Z15
I1
=
1
63.7 + j27.5
j9) ]
[(73 + j42.5) + (−12.5 − j30) + (4 + j18) + (−1.8 − j12) + (1 + =
2
2
= 31.85 + j13.75Ω
monopoles
Z5 = Z1 by symmetry
50
d 25 = 3λ / 2
Same as Z12
Z2 = Z12 + Z22 + Z23 + Z24 + Z25
d 24 = λ
1
=
[ 2(−12.5 − j30) + (73 + j42.5) + (4 + j18) + (−1.8 − j12)]
2
monopoles
1
= [50.2 − j11.5] = 25.1 − j5.75Ω
2
Z2 = Z4 by symmetry
Same as Z23
Z3 = Z13 + Z23 + Z33 + Z34 + Z35
= 2Z13 + 2Z23 + Z33
Same as Z13
1
[ 2(4 + j18) + 2(−12.5 − j30) + (73 + j42.5)]
2
monopoles
56 + j18.5
=
= 28 + j9.25
2
=
Note that for each of the antennas, the input impedances are slightly different and
each of these values are different than
73 + j42.5
or 36.5 + j21.25Ω
2
for an isolated λ/4 monopole.
Directivity
From Eq. (55a) on p. 37 of the Class Notes, including mutual impedance effects
2
D = Do AF max
R A isolated
5
∑ R Ai
i =1
=3.28 × N 2 25 ×
36.5
= 21.1
2 × 31.85
2 × 25.1
28
+
+
Re ( Z1 + Z5 ) Re ( Z2 + Z4 ) Re ( Z3 )
Do= 3.28 ⇒ 2 × 1.64 for a single isolated λ/4 monopole above ground
Note that a directivity of 21.1 is higher than NDo of 5 × 3.28 = 16.4 which would
be obtained for this antenna array neglecting mutual impedance effects.
51
Inclusion of mutual impedance effects can often lead to an increased gain relative
to the value had the mutual impedance effects been neglected.
λ
Example 21: Two monopole antennas separated by
. (Note that the second
/4
antenna is grounded.)
=
V1 Z11I1 + Z12 I2
λ/4
=
0 Z12 I1 + Z22 I2
I1
λ/4
I2
Z
(36 − j25) / 2
I2 =
− 12 I1 =
−
I1
Z22
(73 + j42) / 2



43.8 e− j34.87
=
−
I =
0.52 e− j64.78 e j180 I1
 1
84.22 e+ j29.91

= 0.52 e+ j115.2 I1
For the driven antenna 1
Z=
1
V1
I
= Z11 + Z12 2
I1
I1
(1)
From p. 307, Fig. 8-25 of the Text (see also p. 48 of the Class Notes)
Z12=
d =λ / 4

36 − j25
= 21.91 e− j34.8
2
monopole
From Eq. (1)


73 + j42
+ 0.52 e j115.2  21.91 e− j34.87 
2


= 38.4 + j32.2Ω
Z1
=
Calculate current I1 for a transmitter power of 100 W
Antenna 1 is the only antenna that is driven and is to be fed (current in antenna 2
is created by induction)
1 2
1 2
P=
=
I1 R=
I1 × 38.4
rad 100W
A1
2
2
I1 = 2.28A
Because of induced current (by mutual impedance effect)
52


I2 =×
0.52 I1 e j115.2 =
1.19 e j115.2 A
Example 22: Calculate the feedpoint impedances of two parallel antennae separated by a
distance of λ/4 and fed with a phase shift α = -90º. Each of the antennas is a λ/2
dipole.
Rad. pattern
1
2
From Fig. 8-25, p. 307 Text (p. 48 of Class Notes)
Z11 =
73 + j42.5 ; Z12 =
36 − j25
73 + j42.5
V
V1 = I1Z11 + I2 Z12 ⇒ Z1 = 1 = Z11 − j(36 − j25)
I1
Z1 =48 + j6.5Ω
V2
= j(36 − j25) + (73 + j42.5)
I2
V2 = I1Z12 + I2 Z22 ⇒ Z2 =
Z2 =
98 + j78.5Ω
1 2
I1 × 48
→ 3.29 KW
2
1 2
Power fed to Ant. 2 = I1 × 98
→ 6.71 KW
2
1
Total power = I12 ( R A1 + R A2 ) =
10 KW
2
Power fed to Ant. 1 =
I1 = 11.7 A
G=
N 2G1
R A,isolated
2
4 ×1.64 ×
=
∑ R Ai
i =1
= 3.28
53
73
146
Methods of Matching Power to the Antennas
A. Transmission Line Matching Method
Example 23:
Match an antenna of impedance Za = 10 - j300Ω to a twin-wire line of
characteristic impedance Zo = 300Ω using (a) series elements and (b) a shunt
element. Take f = 30 MHz
za =
Z a 10 − j300
=
= 0.033 − j1
Zo
300
This normalized impedance is shown as point A on the Smith Chart on page 55 of
Class Notes. If the antenna is not matched
Z − Zo
Voltage reflection coefficient ρ = a
= 0.9672∠270

 Z a − Zo
VSWR
=
1+ ρ
= 60.0
1− ρ
Power reflection coefficient =
Pr
2
= ρ = 0.9354
Pinc
(1)
(2)
(3)
i.e. 93.54% of the input power Pinc is reflected and only 6.46% of the transmitter
power is radiated -- a truly poor situation!
Approach A: Use of series elements to match the antenna
From point A, we move on the transmission line circle C to point B on p. 55 -Smith Chart, which corresponds to the intersection with real part zB of 1.0 circle.
Length AB = (0.231 + 0.125)λ = 0.356λ ⇒ 3.56m
(4)
Z BB′ = z BZ o = (1 + j8)300 = 300 + j2400Ω
(5)
As shown in Fig. 2, we can compensate for j2400Ω by using two capacitors as
shown each of reactance
jXse = − j
2400
= − j1200
2
This gives the values of series capacitances
Cse = 4.42 pF
54
(6)
Example of the Transmission Line Matching Method
f = 30 MHz ;
λo = 10 m ;
55
Za = 10 - j300Ω
Fig. 2.
Approach B:
An alternative design using a shunt element to match the antenna
An undesirable feature of the above design Approach A is that it takes a fairly
long length 

AB = 0.356λ over which the transmission line is not matched. For

the alternative Approach B, we work in terms of admittances.
YA =
1
;
10 − j300
yA =
1
≅ 0.033 + j1
0.033 − j1
(7)
This is shown by point a on the Smith Chart on page 55.
Now, we need to move only a distance

ab = (0.231− 0.125)λ = 0.106λ = 1.06 m


(8)
and use (as sketched in Fig. 3) a shunt element to match the line.
Fig. 3.
− j Ysh = − j
L sh =
8
j
mho ⇒ −
300
ω L sh
300
6 = 0.2 µH
8 × 2π × 30 × 10
56
(9)
(10)
USE OF LUMPED ELEMENTS FOR MATCHING AN ANTENNA
Example 24: A Matching Circuit for an Antenna of a Cellular Telephone
Topology 1
The antenna impedance is given to be 50 - j20Ω. The solid-state source to which
this impedance is to be matched has an internal impedance, say 15 + j130Ω. A
possible matching circuit is sketched as follows:
Vs
Fig. 1
For maximum power transfer to the antenna
∗
Z AB = Z S = 15 − j130Ω
(1)
jXsh (50 − j20)
+ jXse
50 + j ( Xsh − 20 )
=
ZAB
( 20 Xsh + j50 Xsh ) 50 − j ( Xsh − 20 )
+ j Xse
2
(50)2 + ( Xsh − 20 )
=
(2)
=
15 − j130Ω
Equating real parts on both sides of Eq. 2
2
=
− 40 Xsh 
1000 Xsh + 50 Xsh ( Xsh − 20
) 15  2900 + Xsh

35 X2sh + 600 Xsh − 43,500 = 0
(3)
−600 ± (600)2 + 4 × 35 × 43,500 −600 ± 2540
=
Xsh =
70
70
=
−44.86 ; + 27.71Ω
Taking the capacitive shunt reactance -j44.86Ω and equating the imaginary parts
on both sides of Eq. 2, we get
57
Xse = − j104.6Ω =
X
=
sh
1
jωCse
1
= 44.86Ω
ωCsh
For f = 900 MHz
Csh = 3.94 pF
X
=
se
1
= 104.6Ω
ωCse
Cse = 1.69 pF
The matching circuit for Topology 1 is as follows:
A
B
Fig. 2
Topology 2
A
B
Fig. 3
This problem may be easier to solve in terms of admittances
58
=
YAB
1
15 + j130
1
1
=
=
+
2
2
15 − j130 (15) + (130)
50 + j(Xse − 20) jXsh
(4)
Equating real parts
15
50
=
2
17,125 (50) + (Xse − 20)2
(5)
50 ×17125
= 57, 083
15
2
2500 + (Xse − 20)
=
=
X e 253.6;
− 213.6Ω
Taking the sines inductance
X
=
Lse 44.85 nH
se 253.6Ω ⇒ =
Xsh = −85.59Ω ⇒ Csh = 2.06 pF
Implications for Power Transfer
a. Without conjugate matching, for an oscillator voltage Vs = 2V RMS power
Power transferred to =
the load = I2rms R A
(2)2
=
× 50 12.25 mW
(15 + 50)2 + (130 − 20)2
b. With conjugate matching
Power transferred to the load = I′rms 2 Re
=
Z∗s
=
×15 66.7 mW
(15 + 15)2
Needed for 600 mW power transferred to the load
Vs = 6V RMS
59
(2)2
A REACTIVE THREE-ELEMENT CIRCUIT FOR ANTENNA MATCHING
Example 25: A reactive three-element network is a versatile circuit for matching power
onto the antenna. To illustrate the procedure, let us look at the circuit of Fig. 1. The
antenna equivalent impedance is RA + jXA.
RA
jX A
D′
A 3-reactance matching network.
Figure 1
In order to match power into the antenna, it is necessary that the impedance of the
network between points A and B be purely resistive and have the same value as Zo, the
characteristic impedance of the transmission line.
From Fig. 1, the expression for the impedance ZAB can be written as:
=
ZAB
( R A + jX A + jX1 ) jX 2 + jX
R A + jX A + jX1 + jX 2
3
(1)
We select X1 and X2 such that the reactance in the denominator of the first term is zero,
i.e.,
X A + X1 + X 2 ≡ 0
(2)
Equation 1 can then be rewritten as:
=
ZAB
( R A − jX 2 ) jX 2 + jX
RA
3
(3)
We select X2 such that
X 22
= Zo
RA
60
(4)
and X3 such that
X3 = -X2
(5)
This would then give
ZAB = Zo + j0
and the antenna would then be matched onto the transmission line.
To illustrate the procedure by a numerical example, let us say that the antenna is a
monopole and its impedance ZA has been calculated and found to be 1.5 - j460Ω.
Let us take Zo = 300 ohms (we must, of course, make sure that the diameter of the
feeder line is not overly thin for the current-carrying requirement). From Eq. 4,
X2 =
± 1.5 × 300 =
±21.2Ω
(6)
The upper sign corresponds to an inductance L = 21.1/ω and the lower sign corresponds
to a capacitance
C=
1
.
ω × 21.2
We can use either type.
Case 1: For inductive element X2
jX 2 =ω
j L 2 =j21.2Ω
If ω is prescribed, L 2 can be calculated. From Eq. 2,
X1 = - X2 - XA = -21.2 + 460
= 438.8Ω
This implies an inductor for jX1 . From Eq. 5,
X3 = - 21.2Ω
One possible 3-reactance matching network is, therefore, shown in Fig. 2.
61
Ω
Ω
Ω
Figure 2
Case 2: For capacitive element X2
1
jX 2 = =
− j21.2Ω
j ωC 2
From Eq. 2 on p. 60 of Class Notes,
X1 = -X2 - XA = +21.2 + 460
= 481.2Ω
(an inductor)
jX1 =ω
j L1 =j481.2Ω
From Eq. 5 on p. 60 of Class Notes,
jX3 = - jX2 = + j21.2Ω
(also an inductor)
and a second possible 3-reactance matching network is shown in Fig. 3.
j21.2 Ω
Z o = 300Ω feeder line
j481.2 Ω
-j21.2 Ω
Figure 3
62
63
H. Jasik, Antenna Engineering Handbook, McGraw Hill & Co.
Gain of a dipole antenna placed in a corner reflector of corner angles α = 180°
(flat reflector) and α = 90° (90° corner reflector).
α = 90°
α = 180°
α = 30°
α = 30°
From: K. F. Lee, Principles of Antenna Theory, John Wiley & Sons, 1984
64
H. V. Cottany and A. C. Wilson, "Gains of Finite Size Corner Reflector Antennas," IEEE
Transactions on Antennas and Propagation, Vol. AP-6, 1958, pp. 366-369.
Contours of constant gain for a 90º corner reflector
25.12 × 1.64 = 41.2
Width of the reflecting planes in wavelengths, W/λ
Maximum for infinite reflectors = 12.9 dB (19.5)
65
Some Commonly Used Feeder Lines for Antennas
1. Twin Wire Transmission line
Zo =
120  2S 
n  
εr
 d 
(Replace ε r by εeff for relatively thin dielectric sheathing)
Example:
2S
= 12.2 for Zo = 300Ω (air-filled line)
d
2. Wire Above Ground Transmission Line
=
Zo
60
 2S 
n  
=
εr
 d 
60
 4h 
n  
εr
 d 
Example:
h
= 3.05 for Zo = 150Ω (air-filled line)
d
3. Coaxial Line
Zo =


60
b
n  
εr
a
Example:
66
b
= 3.345 for εr = 2.1 (Teflon) coaxial line of Zo = 50Ω
a
Some of the other transmission lines useful for printed antennas are:
a.
Miscrostripline
b.
Slot line etc.
Ground Effect on Radiation Pattern of an Antenna
We have previously considered the effect of ground for the radiation from a
vertical monopole antenna. The net effect was that the monopole antenna of
length L/2 radiates electromagnetic fields much like a dipole of length L albeit for
the upper half plane i.e. for field points above ground.
For a horizontal dipole antenna placed at a distance h from the ground as sketched
in Fig. 1, an image antenna 1′ is created, which has a current excitation that is
equal in magnitude (for high conductivity ground) but 180º out of phase with that
in the installed antenna #1.
90 − φ
Fig. 1. A horizontal dipole antenna above ground.
From Eq. 10 on p. 24 of the Class Notes, this can be considered as a two-element
array (Ny = 2) with a phase difference αy = π or 180º.
67
 ψ
sin
2  



 2  E 2 cos  1 2β h sin θ sin φ + π 
E T E=
=
=
)
 (
1 AF y E1
1
2
ψ


sin  
2
 

= E1 2 sin ( β h sin θ sin φ )
(1)


E
neglecting the phase factors both in writing |AF|y and 
1 . Note that Eq. 1 could
also have been written by following a procedure similar to that for Eq. 4 on page
24 of the Class Notes.



− jβ r ′ − r
− jβ 2h sin θ sin φ ) 
E T= E1 + Ei= E1 1 − e ( 1 1 ) = E1 1 − e (





φ
= E1 e− jβ h sin θ sin φ − e− jβ h sin θ sin=
2 E1 sin ( β h sin θ sin φ )


(2)
ignoring the phase factors, as also done in writing Eq. 1. From Eqs. 1 and 2
AF = 2 sin ( βh sin θ sin φ ) ⇒ 2 sin ( βh sin φ )
(3)
for θ = π/2 i.e. xy plane.
For maxima of radiation
π
π
βh sin φ o = ± , ± 3 , 
2
2


(4)
For first nulls of radiation
βh

sin φ FN = 0,
± π, ± 2π, 
(5)
Example 26
a. Calculate the spacing h to ground for a half-wave dipole antenna if the maximum
of radiation is desired for angle φo = 30º off the horizon.
b. Calculate the directions of maximum and zero radiation for the selected h.
c. Calculate the gain of the antenna, without and with mutual impedance effects.
Solution: From Eq. 4 for φo = 30º, sin φo = 0.5
a.
h=
λ
,
4 sin φ o
λ
= ,

 2
3λ
,
2
3λ
, 
4 sin φ o
5λ
, 
2
(6)
68
h = λ/2
2h sin φo
φο
In order to keep the number of principal maxima to a minimum number, we select
the smallest spacing to the ground plane i.e.
h = λ/2
(7)
b. For this spacing itself, we note from Eq. 4 that the directions of maximum
radiation are:
βh sin φ o = π sin φ o = +
π
2
(8)


φ
o = 30 (wanted), φ o = 150 (unwanted)

Negative sign is ignored in Eq. 8 since that gives angles φo = -30º, -150º (both
into the ground).
We will see later how to eliminate the unwanted radiation for φo = 150º. If we
had taken a larger h of say 3 λ/2 from Eq. 6, we would have had many more
directions of maximum radiation.
For directions of first null, from Eq. 5, φFN = 0 and sin-1(1) or 0 and 90º for the
principal maximum of φo = 30º, and φFN = 180º and sin-1(1) or 180º and 90º for
the principal maximum at φo = 150º. The radiation pattern is sketched in Fig. 2.
φo = 150º
φo = 30º
Fig. 2.
69
c. Ignoring mutual impedance effects Ra1 = 73Ω (same as for an isolated half wave
dipole). From Eq. 1
Emax = 2E1;
Smax =
E 2max
= 4S1
2η
Gain = 4G1 = 4 × 1.64 = 6.56
Height above ground, h/λ
Fig. 3.
Angles φ of maximum and zero radiation for a horizontal dipole
antenna above ground (From Eq. 1, 2, or 3).
Elimination of Unwanted Principal Lobes of Radiation
As seen in Example 24, there is an unwanted principal lobe of radiation for
φo = 150º that we would like to eliminate leaving thereby one and only one
principal lobe of radiation for the desired direction φo = 30º. A possible solution
for this problem is as sketched in Fig. 4.
e jo
Fig. 4. An arrangement of two horizontal dipoles above ground.
70
We take two horizontal dipoles 1, 2 above ground. Distance to ground h is the
same for both dipoles 1 and 2. Shown in Fig. 4 also are the two image antennas
1′, 2′ . Assuming that antenna #1 is leading in phase by α (i.e. antenna 2 is
lagging in phase by α).
α - β d1 cos φo = 0
(9)
for addition of signals along φo = 30º principal lobe.
α + β d1 cos φo = π
(10)
for complete cancellation of radiation in the back direction.
Note in both Eqs. 9 and 10, φo = 30º and d1 cos φo = 0.866 d1. From Eqs. 9 and
10, both the unknown α and d1 can now be found:
π
α = = 90 (phase lead angle for antenna #1)

 2
βd1 cos φ o =
π
λ
⇒ d1 =
= 0.289λ
2
4 cos φo
(11)
(12)
This arrangement would cancel the principal lobe for φo = 150º (in Fig. 2) while
reinforcing the principal lobe for the φo = 30º angle of radiation.
71
p. 349 Text
General Theory of Aperture Antennas (or Displacement Current Antennas). For comparison, see also the General Theory of
Conduction Current Antennas on p. 44 of the Text or p. 2 of Class Notes.
R
aperture

Source of Fields H a


Js= nˆ × H a
µ
4π
∫
Sa


M=
s E a × nˆ
(9-10)
 
J(r′) j(ωt −βR)
e
dS′
R
µ e − jβ r
n̂ ×
=
4πr
Define Equivalent surface currents
∫

ε
F= −
4π
 
R= r − r′


ˆ
H a e jβr ⋅ r ′dS′
(9-12)
Sa
 
Ms (r′) j(ωt −βR)
e
dS′
R
ε e − jβ r
n̂ ×
=
4πr
72
Q

 

∇×A
jβ× A
H1 =
= −
µ
µ
 



∇ × H1
jβ× H1
E1 =
=−
=− jωA
jωεo
jωεo
∫
Sa
(9-11)
∫
(9-13)


ˆ
E a e jβr ⋅ r ′dS′
Sa
P




∇ × F jβrˆ × F
jβF × rˆ
−
= =
−
E2 =
ε
ε
ε

= − jωη F × rˆ
(1)
(3)
(2)


 

E T = E1 + E 2 = − jωA − jωη F × rˆ


 


jβ e − jβ r
E T = E1 + E 2 = −
rˆ × ∫  nˆ × E a − η rˆ × nˆ × H a  e− jβrˆ⋅ r ′dS′


4πr
Sa




∇ × ET
jβ rˆ × E T
rˆ × E T
H T ==
−
=
−
− jωµo
jωµo
η
 ∗
 1
 
ET ⋅ ET
S
Re E × H∗=
rˆ
=
2
2η
 
Total radiated power = ∫ S ⋅ dS
(
(
)
)
sphere
72
(9-16)
(9-17)
Chapter 9 – Aperture Antennas

A=

Source of Fields E a
p. 466 Text
A Rectangular Microstrip Patch Antenna
λ/2
Notes
W
y
1. Note that the E-fields have no
variation in the y direction;
hence are identical for the front
and back edges separated by
width W.
E2
E1
2. Because of a separation
distance of λ/2, the E-fields at
edges E1 and E 2 are 180º out
of phase; also no variation in ydirection.
Ez
Fig. 1. Distributions and variations of electric fields at the four edges of the patch antenna.
(9-2b)
p. 346 Text
z
y
x
uniform magnitude; - -directed
half cycle cosinusoidal variation
Fig. 2. Equivalent surface currents at the four edges.
Maximum radiation in z-direction (normal to the patch) due to two uniform magnitude
"dipoles" corresponding to edges
and
; radiation from edges F and B i.e. the front
and back edges of Fig. 1 cancels out.
73
The array factor for the equivalent current dipoles at edges E1, E 2 can be written from
Eqs. 8 and 9 on page 25 of the Class Notes. For an x-directed array of two elements
N = 2; α x =;
0
ψ = β L sin θ cos φ + α x
0
ψ
ψ
 Nψ 
sin 
 2 sin   cos  
2 
2 =
 2  cos  βL sin θ cos φ 
=
Normalized AF =


ψ
ψ
 2

2 sin
2 sin  
2
2
(1)
For a two-element (two-edge "currents") antenna array


E T = E o AF
For a uniformly-excited rectangular aperture
 of length W (e.g. edges E1, E 2 ), from
Eqs. 9-36a, b (note that equivalent current Ms || − yˆ here rather than parallel to x̂ on
page 354 of the text)
=
E θ E o cos φ f (θ, φ)
(11-5a)
E φ =−E o cos θ sin φ f (θ, φ)
(11-5b)
where
Ly
 βW

sin 
sin θ sin φ 
2
 AF
f (θ, φ) = 
βW
sin θ sin φ
2
 βW

sin 
sin θ sin φ 
 2
 cos  βL sin θ cos φ) 
=


βW
 2

sin θ sin φ
2
In Eqs. 9-36a, b
thickness t
 βL 
sin  z u 
 2  →1
βLz
u
2
74
since t << λ
(11-5c)
p. 470 Text
Microstrip Patch Antenna
y
Max. rad.
.
L
W
x
For x-z plane ( φ =0 ) or E-plane

E field is θ̂ -directed with components in x- and z-directions
E=
θ E o f (θ, φ) ;
Eφ = 0

For direction of maximum radiation, E || xˆ
 βL

=
FE (θ) cos 
sin θ 
 2

Maximum for θ = 0 i.e. along z-direction.
BWFN:
βL
π
sin=
θ
2
2
For L 
λ
, for xz or E-plane
2
 λ 
→ θFN
= sin −1 

 2L 
 λ 
BWFN = 2 sin −1 

 2L 
= 180º
HPBW:
βL
π
sin=
θ
2
4
 λ 
→ θHP
= sin −1 

 4L 
 λ 
−1  1 

HPBW = 2 sin −1 
 → 2 sin   → 60
4L
2


 
75
(11-6a)
For y-z plane ( φ =90 )
 βW

sin 
sin θ 
 2

FH (θ=
) cos θ
βW
sin θ
2
(11-6b)
Maximum for θ = 0

E =−E o cos θ F(θ, φ)
FH (θ)
For yz or H-plane
βW
sin θFN =
π;
2
 λ 
sin −1  
θFN =
W
 λ 
BWFN = 2sin −1  
W
ZA = 90
ε 2r  L 
 
εr − 1  W 
2
for a half-wave rectangular patch antenna.
W
For Duroid
( ε r 2.2)
and
2.7
=
=
L
ZA= 50Ω
 1 

θFN yz plane= sin −1 
= 47.8
1.35


 λ 
BWFN = 2 sin −1   = 95.6º
W
76
(11-7)
Table 7.1. Radiation characteristics of commonly-used horn antennas.
Type of Horn
Pyramidal
Property that
is Optimized
for a Given
Length
Gain
Optimum
Properties
A = 3Lλ
B = 0.81A
Gain = 15.3 L/λ
(optimum)
Sectoral Hplane horn
Beam width
in H-plane
A=
Sectoral Eplane horn
Beam width
in E-plane
B=
Conical
Gain
Notation:
D=
Half-power Beam Widths in
Degrees
H (or xz)
E (or yz)
Plane
Plane
80
53
(A / λ)
(B / λ)
3Lλ
78
(A / λ)
(9-124)
51
(B / λ)
2Lλ
68
(A / λ)
54
(B / λ)
(9-138)
2.8Lλ
70
(D / λ)
60
(D / λ)
A is the horn dimension in x direction
B is the horn dimension in y direction
D is the horn diameter
L is the length of the horn from the throat to the aperture
77
Directive
Gain
0.51
4πAB
λ2
(9-96)
0.63
0.65
0.52
4πAB
λ2
4πAB
λ2
4π(area)
λ2
Table 7.2*. Comparative characteristics of parabolic reflectors with different illuminations.
(See also Table 9-2, p. 389 Text).
3 dB Beam
Width in
Degrees
Peak Side
Lobe
Level
(dB)
Relative
Gain
First Null
Position in
Degrees
m = 0 (uniform)
50.8 λ / L
-13.2
1.00
57.3 λ / L
m=1
68.8 λ / L
-23
0.81
85.9 λ / L
m=2
83.1 λ / L
-32
0.667
114.6 λ / L
m=3
95.1 λ / L
-40
0.575
143.2 λ / L
m=4
111.2 λ / L
-48
0.515
171.9 λ / L
m = 0 (uniform)
58.4 λ / D
-17.6
1.00
69.9 λ / D
m=1
72.8 λ / D
-24.6
0.75
92.2 λ / D
m=2
84.2 λ / D
-30.7
0.55
116.3 λ / D
m=3
94.5 λ / D
-36.1
0.45
138.7 λ / D
Illumination
A. Rectangular Aperture of Length L
L
 πx 
=
G(x) cos m   for | x | <
2
 L 
B. Circular Aperture of Diameter D
  2ρ 2 
G(ρ) = 1 −   
  D  
*
m
M. I. Skolnik, Radar Handbook, Chapter 9, McGraw-Hill Book Company, Inc., New York,
1970.
78