Master degree in Internet and Multimedia Engineering / Electronic Engineering
Antennas / Advanced Antenna Engineering
Slides of the lectures – Part 1
Course outline
❑ The course provides knowledge and skills concerning antennas, with reference to both their
working principles and their use in different applicative fields.
❑ The following main topics will be addressed:
o Fundamentals of electromagnetic radiation and antenna parameters.
o «Classical antennas», such as
• Linear antennas
• Aperture antennas
• Printed antennas
• Reflector antennas
o Reconfigurable and «smart» antennas, such as
• Antenna arrays
o Link between antennas
2
Course outline
❑ Notes for the course topics are available in the website
http://www.aem.diten.unige.it/
❑ under the section «For students» -> «Teaching material»
❑ Login/password are available on Aulaweb
❑ Some useful reference books are listed below:
o C. A. Balanis, Antenna theory: Analysis and design, 4th ed. Hoboken, NJ: John Wiley & Sons, 2016.
o S. J. Orfanidis, Electromagnetic Waves and Antennas, 2016. Available online at
www.ece.rutgers.edu/~orfanidi/ewa/
3
Course outline
❑ Lecture schedule for the common IME/EO part of the course (unless otherwise indicated):
o Wednesday 13 – 16
o Thursday 11 – 13
❑ For any need, please do not hesitate to contact the Teacher:
o Andrea Randazzo – andrea.randazzo@unige.it
4
Introduction to antennas
❑ The first question to answer is: What is an antenna?
❑ Basically, antennas are devices designed to transmit and/or receive
an electromagnetic radiation.
❑ Some standard definitions
Antenna
o Webster’s Dictionary: “a usually metallic device
(as a rod or wire) for radiating or receiving
radio waves.”
o IEEE Standard Definitions of Terms for Antennas
(IEEE Std 145–1983): “a means for radiating
or receiving radio waves”
o C.A. Balanis: “the transitional structure between
free-space and a guiding device”
Source
Guiding
structure
Radiated
wave
5
Introduction to antennas
❑ An important point: antennas usually may work both in transmission and receiving modes!
TX Antenna
𝑉, 𝐼 → 𝐄, 𝐇
RX Antenna
𝐄, 𝐇 → 𝑉, 𝐼
Antenna
Antenna
Reference
plane, (𝑉, 𝐼)
Reference
plane, (𝑉, 𝐼)
Guiding
structure
Guiding
structure
Radiated
wave 𝐄, 𝐇
Impinging
wave 𝐄, 𝐇
6
Introduction to antennas
❑ Frequency bands
❑ Wavelenght: 𝜆 =
𝑐
(𝑐 = 3 × 108 m/s is the speed of light)
𝑓
7
Introduction to antennas
ITU frequency bands
IEEE radar bands
Abbreviation
ITU band
number
ELF
1
3–30 Hz
100,000–10,000 km
SLF
2
30–300 Hz
10,000–1,000 km
Ultra low frequency
ULF
3
300–3,000 Hz 1,000–100 km
Very low frequency
VLF
4
3–30 kHz
100–10 km
Low frequency
LF
5
10–1 km
Medium frequency
High frequency
MF
HF
6
7
30–300 kHz
300–
3,000 kHz
3–30 MHz
Very high frequency
Ultra high
frequency
Super high
frequency
Extremely high
frequency
Terahertz or
Tremendously high
frequency
VHF
8
UHF
9
30–300 MHz 10–1 m
300–
3,000 MHz 1–0.1 m
SHF
10
3–30 GHz
EHF
11
30–300 GHz 10–1 mm
12
300–
3,000 GHz
Band name
Extremely low
frequency
Super low
frequency
THz or THF
Frequency
Wavelength
1,000–100 m
100–10 m
Band name
HF
VHF
UHF
L
S
C
X
Ku
K
Ka
V
W
mm or G
Frequency range
0.003 to 0.03 GHz
0.03 to 0.3 GHz
0.3 to 1 GHz
1 to 2 GHz
2 to 4 GHz
4 to 8 GHz
8 to 12 GHz
12 to 18 GHz
18 to 27 GHz
27 to 40 GHz
40 to 75 GHz
75 to 110 GHz
110 to 300 GHz​
100–10 mm
1–0.1 mm
8
Introduction to antennas
❑ Antennas are nowadays used in a wide range of applications.
❑ The most straightfoward ones are those related to communications:
o Wireless LAN
o Satellite communications
o Mobile phones (such as new 5G ones)
o …
❑ But many others requires antennas, for example (but many more may be envisioned)
o IOT devices and RFID tags
o Automotive sensors for assisted driving
o Civil and enviromental monitoring instrumentation
o …
❑ The aim of the antenna is always the same, i.e., to radiate and receive an electromagnetic
field, but the requirements are significantly different → Different antennas, specifically
tailored for the specific application, are often needed!
9
Introduction to antennas
❑ Some examples:
o Indoor wireless communications – Main requirement is a full coverage of building (or parts of
them). Consequently, it is necessary to have antennas able to radiate everywhere. In modern
systems, multiple input multiple output configurations are also needed to increase channel capacity.
Dipole antenna
10
Introduction to antennas
o Satellite communications – In this case, it is necessary to have antennas with very high gain and a
very narrow beam.
Parabolic dish antenna
11
Introduction to antennas
o Automotive applications – Nowadays, most cars are equipped with a lot of sensors for
communication and assisted driving. Different ones are based on radar systems, i.e., they use a
radiofrequency signal to probe the scenario nearby the car and consequently antennas are needed.
In several cases, printed antenna arrays are used, since they allow reconfigurability and the
extraction of additional infomartion (e.g., direction of arrival).
Rear collision
warning and lane
change assistant
Radio, WiFi,
GPS, bluetooth
Adaptive cruise
control and
collision avoidance
Example of 24 GHz
automotive radar
antenna
Blind spot
detection
Cross
traffic alert
12
Introduction to antennas
o Navigation support systems (e.g., GPS) – Circularly polarized antennas are required.
Printed or helical antennas
13
Introduction to antennas
o Industrial and civil monitoring applications
Industrial diagnostics
Subsurface monitoring
Antenna
By The Charles Machine Works - http://www.flickr.com/photos/officialditchwitch/5876764119, CC BY-SA 2.0,
https://commons.wikimedia.org/w/index.php?curid=15635167
14
Introduction to antennas
❑ Despite various type of antennas exist, from a theoretical point of view they can be modeled
(in TX mode) by considering a (eventually equivalent) current density 𝐉0 𝐫 located in a
volume 𝑉0 (e.g., the region of space occupied by the metallic part of the antenna) that
radiates in the considered propagation scenario.
❑ Usually free-space radiation conditions are assumed, in order to obtain a set of radiation
parameters describing the antenna behavior.
Real antenna
Electromagnetic model
𝐉0 𝐫
𝑉0
15
Introduction to antennas
❑ The problem of characterizing the antenna thus reduces to the problem of understanding
how a proper current density radiates!
❑ When working in RX mode, an electromagnetic field impinge on the antenna.
❑ Such a field induces a current density on the antenna, which in turn generate a current and
voltage at the output reference plane (e.g., the connector) of the antenna (we will better
discuss this topic later in the course).
❑ Such a behavior can be studied by exploiting the reciprocity principle → consequently most
of the analysis are performed considering the TX mode.
16
Review of radiation from EM sources
❑ In the following, we also assume to work in the frequency domain, i.e., an 𝑒 𝑗𝜔𝑡 time
dependence (with angular frequency 𝜔) is assumed.
❑ In order to study the radiation problem, it is convenient to use the potential theory.
❑ From Maxwell’s equations we have
∇∙𝐁 𝐫 =0
∇ × 𝐄 𝐫 = −𝑗𝜔𝐁 𝐫
❑ The following potential functions can be defined
𝐁 𝐫 =∇×𝐀 𝐫
𝐄 𝐫 = −∇𝜑 𝐫 − 𝑗𝜔𝐀 𝐫
17
Review of radiation from EM sources
❑ The potentials 𝐀 𝐫 e 𝜑 𝐫 are however not unique. To this end, we use the Lorenz gauge (or
radiation gauge), i.e.,
∇ ∙ 𝐀 𝐫 + 𝑗𝜔𝜖𝜇𝜑 𝐫 = 0 → 𝜑 𝐫 = 𝑗
∇∙𝐀 𝐫
𝜔𝜖𝜇
Only the vector
potential is needed!
❑ By inserting the vector potentials in the Maxwell equations, we obtain:
∇2 𝐀 𝐫 + 𝑘 2 𝐀 𝐫 = −𝜇𝐉 𝐫
❑ i.e., it satisfies an Helmoltz equation whose source is the current density distribution.
18
Review of radiation from EM sources
❑ The solution of the previous equation is:
′
−𝑗𝑘 𝐫−𝐫
𝑒
′
𝐀 𝐫 = −𝜇 න 𝐉 𝐫 ′ 𝑔 𝐫, 𝐫 ′ 𝑑𝐫 ′ = 𝜇 න 𝐉 𝐫 ′
𝑑𝐫
4𝜋 𝐫 − 𝐫 ′
𝑉
❑ where 𝑘 = 𝜔 𝜖𝜇 is the wavenumber
𝜖 is the dielectric permittivity, and
𝜇 is the magnetic permeability.
𝑉
Green’s function for free space
−𝑗𝑘 𝐫−𝐫 ′
1
𝑒
𝑔 𝐫, 𝐫 ′ = −
4𝜋 𝐫 − 𝐫 ′
❑ If the propagation medium is air, it
can be modeled as vacuum (at least for frequencies below some tens of GHz) and we have
𝜖 = 𝜖0 ≅ 8.85 × 10−12 F/m and 𝜇 = 𝜇0 ≅ 1.26 × 10−6 H/m.
19
Review of radiation from EM sources
❑ The electric and magnetic fields are then obtained as:
𝐄 𝐫 = −∇𝜑 𝐫 − 𝑗𝜔𝐀 𝐫
1
𝐇 𝐫 = ∇×𝐀 𝐫
𝜇
❑ By using the Lorentz’s gauge, 𝐄 can be finally expressed directly in terms of the vector
potential, i.e.,
1
𝐄 𝐫 =
∇ ∇∙𝐀 𝐫
𝑗𝜔𝜇𝜖
1
− 𝑗𝜔𝐀 𝐫 =
∇ ∇∙𝐀 𝐫
𝑗𝜔𝜇𝜖
+ 𝑘2𝐀 𝐫
20
Far-field radiation
❑ When dealing with antennas, usually we are interested in the fields at a great distance from
the antenna, i.e., for 𝑟 = 𝐫 ≫ 𝐫 ′ = 𝑟 ′ .
21
Far-field radiation
❑ In this case, the quantity 𝑅 = 𝐫 − 𝐫 ′ can be approximated as:
𝑅 = 𝐫 − 𝐫′ =
𝐫 − 𝐫′ ∙ 𝐫 − 𝐫′ =
′
′2
𝑟
𝑟
𝑟 2 − 2𝑟𝑟 ′ cos 𝜓 + 𝑟 ′ 2 = 𝑟 1 − 2 cos 𝜓 + 2
𝑟
𝑟
❑ Using the Taylor expansion:
𝑥 𝑥2
1+𝑥 ≅1+ − +⋯
2 8
Negligible since
(when 𝑛 > 1)
❑ The previous relationship can be written as:
2
𝑟′
𝑟
𝑛
≪
𝑟′
𝑟
2 2
𝑟′
𝑟′
1
𝑟′
𝑟′
𝑅 ≅ 𝑟 1 − cos 𝜓 + 2 −
2 − cos 𝜓 + 2
𝑟
2𝑟
8
𝑟
𝑟
22
Far-field radiation
❑ Finally, by neglecting higher-order terms, we can write:
𝑅 ≅ 𝑟 − 𝑟 ′ cos 𝜓 = 𝑟 − 𝐫 ′ 𝐫ො cos 𝜓 = 𝑟 − 𝐫ො ∙ 𝐫 ′
❑ Such approximation is used in the exponential term inside the integral.
❑ The term 1/𝑅 is instead approximated with 1/𝑟.
Sphericalwave factor
❑ The vector potential can thus be written as:
′
′
−𝑗𝑘 𝐫−𝐫
−𝑗𝑘 𝑟−𝐫ො∙𝐫
−𝑗𝑘𝑟
𝑒
𝑒
𝜇𝑒
ො ∙𝐫 ′
′
′
′
′
′
𝑗𝑘
𝐫
′
න
𝐀 𝐫 =𝜇න𝐉 𝐫
𝑑𝐫
≅
𝜇
න
𝐉
𝐫
𝑑𝐫
=
𝐉
𝐫
𝑒
𝑑𝐫
0
4𝜋 𝐫 − 𝐫 ′
4𝜋𝑟
4𝜋𝑟
𝑉0
𝑉0
𝑉0
Radiation vector
23
Far-field radiation
❑ In the previous relationship, the dependence from the distance 𝒓 is present only in the
multiplicative term outside the integral, which represents a spherical wave.
❑ The quantity:
′
𝐅 𝜃, 𝜙 = න 𝐉0 𝐫 ′ 𝑒 𝑗𝑘𝐫ො∙𝐫 𝑑𝐫 ′
𝑉0
❑ is called radiation vector and it depends only on the source type (i.e., on the antenna).
❑ The vector potential in the far-field region can be finally expressed as:
𝜇𝑒 −𝑗𝑘𝑟
𝐀𝐫 =
𝐅 𝜃, 𝜙
4𝜋𝑟
24
Far-field radiation
❑ The radiated fields are finally obtained by applying the potential-field transformations
1
𝐇 𝐫 = ∇×𝐀 𝐫
𝜇
1
𝐄 𝐫 =
∇ ∇∙𝐀 𝐫
𝑗𝜔𝜇𝜖
− 𝑗𝜔𝐀 𝐫
❑ In all cases, the differential operator ∇ is applied to terms of the type 𝐟 𝜃, 𝜙
approximated as ∇≅ −𝑗𝑘ො𝐫.
❑ As an example, let us consider the computation of ∇ ∙ 𝐟 𝜃, 𝜙
𝑒 −𝑗𝑘𝑟
𝑟
𝑒 −𝑗𝑘𝑟
and can be
𝑟
. Using known
differential identities we have:
𝑒 −𝑗𝑘𝑟
𝑒 −𝑗𝑘𝑟
𝑒 −𝑗𝑘𝑟
∇ ∙ 𝐟 𝜃, 𝜙
= 𝐟 𝜃, 𝜙 ∙ ∇
+
∇ ∙ 𝐟 𝜃, 𝜙
𝑟
𝑟
𝑟
25
Far-field radiation
❑ Since in spherical coordinates
𝜕𝜓
1 𝜕𝜓
1 𝜕𝜓
෡
෡
∇𝜓 =
𝐫ො +
𝜽+
𝝓
𝜕𝑟
𝑟 𝜕𝜃
𝑟 sin 𝜃 𝜕𝜙
❑ we obtain
𝑒 −𝑗𝑘𝑟
𝑒 −𝑗𝑘𝑟
∇
= −𝑗𝑘
𝐫ො + higher order terms
𝑟
𝑟
❑ For a similar reason, ∇ ∙ 𝐟 𝜃, 𝜙 only contains higher-order terms and can be neglected.
26
Far-field radiation
❑ Consequently,
≅∇
𝑒 −𝑗𝑘𝑟
𝑒 −𝑗𝑘𝑟
𝑒 −𝑗𝑘𝑟
∇ ∙ 𝐟 𝜃, 𝜙
≅ 𝐟 𝜃, 𝜙 ∙ −𝑗𝑘
𝐫ො = −𝑗𝑘ො𝐫 ∙ 𝐟 𝜃, 𝜙
𝑟
𝑟
𝑟
❑ Applying the previous property and using the potential-field transformations we obtain, for
the magnetic field:
1
1
𝑒 −𝑗𝑘𝑟
𝐇 𝐫 = ∇ × 𝐀 𝐫 = −𝑗𝑘ො𝐫 × 𝜇
𝐅 𝜃, 𝜙
𝜇
𝜇
4𝜋𝑟
𝑒 −𝑗𝑘𝑟
= −𝑗𝑘
𝐫ො × 𝐅 𝜃, 𝜙
4𝜋𝑟
𝑒 −𝑗𝑘𝑟
𝑒 −𝑗𝑘𝑟
෡ + 𝐹𝜙 𝜃, 𝜙 𝛟
෡ = −𝑗𝑘
෡ − 𝐹𝜙 𝜃, 𝜙 𝛉
෡
= −𝑗𝑘
𝐫ො × 𝐹𝑟 𝜃, 𝜙 𝐫ො + 𝐹𝜃 𝜃, 𝜙 𝛉
𝐹𝜃 𝜃, 𝜙 𝛟
4𝜋𝑟
4𝜋𝑟
27
Far-field radiation
❑ Similarly, for the electric field we have:
∇ ∇∙𝐀 𝐫
𝐄 𝐫 = −𝑗𝜔𝐀 𝐫 − 𝑗
𝜔𝜇𝜖
𝑒 −𝑗𝑘𝑟
𝑗
𝑒 −𝑗𝑘𝑟
= −𝑗𝜔𝜇
𝐅 𝜃, 𝜙 −
−𝑗𝑘ො𝐫 −𝑗𝑘ො𝐫 ∙ 𝜇
𝐅 𝜃, 𝜙
4𝜋𝑟
𝜔𝜇𝜖
4𝜋𝑟
𝑒 −𝑗𝑘𝑟
𝑘 2 𝑒 −𝑗𝑘𝑟
𝑒 −𝑗𝑘𝑟
= −𝑗𝜔𝜇
𝐅 𝜃, 𝜙 + 𝑗
𝜇
𝐅 𝜃, 𝜙 ∙ 𝐫ො 𝐫ො = −𝑗𝜔𝜇
𝐅 𝜃, 𝜙 − 𝐅 𝜃, 𝜙 ∙ 𝐫ො 𝐫ො
4𝜋𝑟
𝜔𝜇𝜖 4𝜋𝑟
4𝜋𝑟
𝑒 −𝑗𝑘𝑟
= −𝑗𝑘𝜂
𝐫ො × 𝐅 𝜃, 𝜙
4𝜋𝑟
𝑘𝜂 = 𝜔 𝜖𝜇
𝑒 −𝑗𝑘𝑟
෡ + 𝐹𝜙 𝜃, 𝜙 𝛟
෡
× 𝐫ො = −𝑗𝑘𝜂
𝐹𝜃 𝜃, 𝜙 𝛉
4𝜋𝑟
𝐅⊥ 𝜃,𝜙
𝐅⊥ 𝜃,𝜙
𝜇
= 𝜔𝜇
𝜖
Transverse component of 𝐅
28
Far-field radiation
❑ From these relationships we can devise the following property:
𝐄 𝐫 = 𝜂𝐇 𝐫 × 𝐫ො
❑ where 𝜂 =
→
1
𝐇 𝐫 = 𝐫ො × 𝐄 𝐫
𝜂
𝜇/𝜖 is the intrinsic impedance of the medium (in air/vacuum, it is equal to
~377 Ω).
❑ In the far-field region, 𝐄 and 𝐇 are orthogonal one to each other and both of them are
orthogonal to the direction of propagation.
❑ Moreover, they also satisfy the following property:
𝐄
=𝜂
𝐇
29
Far-field radiation
❑ Consequently, at a great distance from the source,
the field behaves locally (i.e., in the neighborhood
of a generic observation point 𝐫) as a plane wave.
Local planar wavefront
𝐄, 𝐇
𝐫ො
Actual spherical wavefront
30
Far-field radiation
❑ The considered far-field approximation is valid under the following conditions:
2𝐷 2
𝑟≥
𝜆
𝑟≫𝜆
𝑟≫𝐷
❑ where 𝐷 is the maximum linear dimension of the antenna. All the conditions must be
contemporarily satisfied in order to apply the far-field approximation.
𝐷
31
Far-field radiation – Field regions
𝑅1 = 0.62 𝐷3 /𝜆
𝑅2 = 2𝐷2 /𝜆
❑ Conventionally, three field regions are identified (*):
o Reactive near field region: «portion of the near-field
region very near to the antenna where the reactive field
predominates»
o Radiating near-field (Fresnel) region (**): «region
between the reactive near-field region and the far-field
region where the radiation fields predominate but the
angular field distribution is dependent from the distance»
Far-field
region
Radiating nearfield region
Reactive nearfield region
𝐷
𝑅1
o Far-field (Fraunhofer) region: «region where the angular
field distribution is independent from the distance»
𝑅2
(*) The boundaries between regions are not sharp. Moreover various criteria have been
proposed for defining them. A commonly adopted definition is shown here as an example.
(**) This region may not exist, depending on the dimension of the antenna.
32
Far-field radiation – Field regions
❑ Example: consider a half-wavelenght dipole (we will better define this antenna later). The
maximum size is equal to the lenght of the dipole, i.e., 𝐷 = 𝜆/2. Consequently we have:
2𝐷2
2𝜆2
𝜆
𝑅>
=
=
𝜆
4𝜆
2
𝑅≫𝜆
𝜆
𝑅≫𝐷=
→ 𝑅 ≫ 𝜆 𝑒. 𝑔. , 𝑅 > 10𝜆
𝐷 = 𝜆/2
2
❑ Considering a dipole for WiFI applications, we have 𝑓 = 2.45 GHz → 𝜆 =
𝑐
~12.24 cm.
𝑓
Consequently, the far field region is defined by 𝑅 ≳ 1.22 m
33
Far-field radiation – Field regions
❑ Example: consider a parabolic dish antenna of diameter 100 cm working at 12 GHz (we will
better define this antenna later). Consequently, since 𝜆 =
2𝐷2
2×1600𝜆2
𝑅>
=
= 3200𝜆
𝜆
𝜆
ቐ
𝑅≫𝜆
𝑅 ≫ 𝐷 = 40𝜆
𝑐
= 2.5 cm and 𝐷 = 40𝜆, we have:
𝑓
→ 𝑅 > 3200𝜆 = 80 m
❑ Moreover, the reactive near field region boundary is 𝑅1 = 0.62 𝐷 3 /𝜆 = 3.92 m.
34
Polarization of radiated fields
❑ An important information about the field radiated by an antenna is its polarization.
❑ By definition, the polarization of an electromagnetic field is the direction of the electric
field vector.
❑ An important note: the polarization is defined by considering the time-domain field
quantities!
❑ Depending on how this vector changes versus time during propagation we can have
different types of polarization, which in turn requires different type of antennas.
❑ The main polarization types are:
o Linear – The direction of the field vector does not change versus time
o Circular – The polarization vector rotate in time and trace a circle
o Elliptical – The polarization vector rotate in time and trace an ellipse
35
Polarization of radiated fields
❑ In order to better understand these concepts and find the conditions under which a certain
polarization is present, let us consider the following generic form of the radiated electric
field:
෡ 𝑒 −𝑗𝑘𝑟
෡ + 𝐵𝑒 𝑗𝜙𝑏 𝝓
𝐄 𝐫 = 𝐴𝑒 𝑗𝜙𝑎 𝛉
❑ Going back to the time domain, we have
෡ 𝑒 −𝑗𝑘𝑟 𝑒 𝑗𝜔𝑡 = ℜ 𝐴𝑒 𝑗 𝜔𝑡−𝑘𝑟+𝜙𝑎 𝛉
෡
෡ + 𝐵𝑒 𝑗𝜙𝑏 𝝓
෡ + 𝐵𝑒 𝑗 𝜔𝑡−𝑘𝑟+𝜙𝑏 𝝓
𝐄 𝐫, 𝑡 = ℜ 𝐴𝑒 𝑗𝜙𝑎 𝛉
෡
෡ + 𝐵 cos 𝜔𝑡 − 𝑘𝑟 + 𝜙𝑏 𝝓
= 𝐴 cos 𝜔𝑡 − 𝑘𝑟 + 𝜙𝑎 𝛉
𝐸𝜃 𝐫,𝑡
𝐸𝜙 𝐫,𝑡
Since 𝑒 𝑗𝑥 = cos 𝑥 + 𝑗 sin 𝑥 and
consequently ℜ 𝑒 𝑗𝑥 = cos 𝑥
36
Polarization of radiated fields
❑ If we are interested in the behavior versus time of the field vector, we can consider a generic
position → without loss of generally, we set 𝑟 = 0 in the previous formulas.
❑ If we also split the field vector into its 𝜃 and 𝜙 components, we have:
𝐸𝜃 𝑡 = 𝐴 cos 𝜔𝑡 + 𝜙𝑎
𝐸𝜙 𝑡 = 𝐵 cos 𝜔𝑡 + 𝜙𝑏
❑ From this relationship, we can see that the electric field vector, in general, moves in the plane
𝜃, 𝜙 with respect to time.
❑ Depending on the values of 𝐴, 𝐵, 𝜙𝑎 , and 𝜙𝑏 different rotation shapes can be identified.
37
Polarization of radiated fields
❑ Linear polarization: If 𝜙𝑎 = 𝜙𝑏 or 𝜙𝑏 = 𝜙𝑎 ± 𝜋 we have a linear polarization, i.e., the tip of
the polarization (E-field) vector moves along a line during time. The same happens if either
𝐴 or 𝐵 are null.
❑ Indeed, if for example we consider the case 𝜙𝑎 = 𝜙𝑏 , we have
𝐸𝜃 𝑡 = 𝐴 cos 𝜔𝑡 + 𝜙𝑎
𝐸𝜙 𝑡 = 𝐵 cos 𝜔𝑡 + 𝜙𝑎
❑ If we consider the case 𝜙𝑏 = 𝜙𝑎 − 𝜋, we have
𝐸𝜃 𝑡 = 𝐴 cos 𝜔𝑡 + 𝜙𝑎
𝐸𝜙 𝑡 = 𝐵 cos 𝜔𝑡 + 𝜙𝑎 − 𝜋 = −𝐵 cos 𝜔𝑡 + 𝜙𝑎
38
Polarization of radiated fields
Vertical component only
Both components
39
Polarization of radiated fields
❑ Circular polarization: If 𝐴 = 𝐵 and 𝜙𝑏 = 𝜙𝑎 ±
𝜋
we have a circular polarization, i.e., the tip of
2
the polarization (E-field) vector traces a circumference during time.
❑ Indeed, if for example we consider the case 𝜙𝑏 = 𝜙𝑎 −
𝜋
we obtain
2
𝐸𝜃 𝑡 = 𝐴 cos 𝜔𝑡 + 𝜙𝑎
𝜋
𝐸𝜙 𝑡 = 𝐴 cos 𝜔𝑡 + 𝜙𝑎 −
= 𝐴 sin 𝜔𝑡 + 𝜙𝑎
2
❑ which is exactly the parametric formula of a circle.
40
Polarization of radiated fields
𝜋
2
❑ Depending on the sign of the term , we may have two opposite directions of rotation, that
in turn produce two different cases of circular polarization.
❑ Following the IEEE convention [*], the two types of circular polarization can be defined as
follows
o Right circular polarization (RCP): curl the fingers of the right hand into a fist and point the thumb
towards the direction of propagation. If the fingers are curling in the direction of rotation of the
electric field, then we have a right circular polarization. In this case, the field rotate
counterclockwise.
o Left circular polarization (LCP): curl the fingers of the left hand into a fist and point the thumb
towards the direction of propagation. If the fingers are curling in the direction of rotation of the
electric field, then we have a left circular polarization. In this case, the field rotate clockwise.
[*] “IEEE Standard Definitions of Terms for Antennas,” IEEE Std 145-1983. Revised IEEE Std 145-1993.
41
Polarization of radiated fields
Left circular polarization
Right circular polarization
42
Polarization of radiated fields
❑ Which is the form of the radiated field in frequency domain for circularly polarized fields?
❑ If we assume 𝐴 = 𝐵 and 𝜙𝑏 = 𝜙𝑎 ±
𝜋
, and we return back from time domain to frequency
2
domain, we obtain
෡ 𝑒 −𝑗𝑘𝑟 (left circular polarization)
෡ + 𝑗𝛟
𝐄 𝐫 = 𝐴𝑒 𝑗𝜙𝑎 𝜽
෡ 𝑒 −𝑗𝑘𝑟 (right circular polarization)
෡ − 𝑗𝛟
𝐄 𝐫 = 𝐴𝑒 𝑗𝜙𝑎 𝜽
43
Polarization of radiated fields
❑ Elliptical polarization: In the general case, the equations
𝐸𝜃 𝑡 = 𝐴 cos 𝜔𝑡 + 𝜙𝑎
𝐸𝜙 𝑡 = 𝐵 cos 𝜔𝑡 + 𝜙𝑏
❑ describe an ellipse in the 𝜃, 𝜙 plane. Indeed, after some trigonometrical manipulations, they
can be written as
𝐸𝜃 𝐸𝜙
𝐸𝜃2 𝐸𝜙2
2 𝜙 −𝜙
+
−
2
cos
𝜙
−
𝜙
=
sin
𝑎
𝑏
𝑎
𝑏
𝐴2 𝐵 2
𝐴𝐵
44
Polarization of radiated fields
❑ We define axial ratio 𝐴𝑅 the ratio between the lengths of the major and minor axes of the
ellipse, i.e.,
𝑚𝑎𝑗𝑜𝑟 𝑎𝑥𝑖𝑠
𝐴𝑅 =
∈ 1, ∞
𝑚𝑖𝑛𝑜𝑟 𝑎𝑥𝑖𝑠
❑ If the axes are along the 𝜃 and 𝜙 directions, i.e., the ellipse is aligned to the referene system,
𝐴
𝐵
this parameter reduces to 𝐴𝑅 = . Otherwise a more complex formula is needed.
❑ Another parameter is the tilt angle 𝜏, which indicates the the angle between the ellipse main
axis and the 𝜙 direction.
❑ Also for the elliptical case, we can have left and circular elliptically polarized fields.
45
Polarization of radiated fields
Left elliptical polarization
(null tilt angle)
Right elliptical polarization
(arbitrary tilt angle)
46
Parameters of antennas
❑ Antennas are typically characterized in terms of some parameters that describe, in synthetic form, their
radiation properties.
❑ Such parameters are defined in terms of the power, and in particular of the power density of the radiated
field.
❑ The Poynting vector, in far field, is given by:
−𝑗𝑘𝑟
1
1
𝑒
෡
෡ + 𝐹𝜙 𝜃, 𝜙 𝛟
𝐒 𝐫 = 𝐄 𝐫 × 𝐇∗ 𝐫 =
−𝑗𝑘𝜂
𝐹𝜃 𝜃, 𝜙 𝛉
2
2
4𝜋𝑟
𝑒 −𝑗𝑘𝑟
෡ − 𝐹𝜙 𝜃, 𝜙 𝛉
෡
× −𝑗𝑘
𝐹𝜃 𝜃, 𝜙 𝛟
4𝜋𝑟
∗
1
𝑒 −𝑗𝑘𝑟
𝑒 𝑗𝑘𝑟
෡ × 𝐹𝜃∗ 𝜃, 𝜙 𝛟
෡ − 𝐹𝜙∗ 𝜃, 𝜙 𝛉
෡ + 𝐹𝜙 𝜃, 𝜙 𝛟
෡
=
−𝑗𝑘𝜂
𝑗𝑘
𝐹𝜃 𝜃, 𝜙 𝛉
2
4𝜋𝑟
4𝜋𝑟
𝜂𝑘 2
෡ − 𝐹𝜃 𝜃, 𝜙 𝐹𝜙∗ 𝜃, 𝜙 𝛉
෡ ×𝛟
෡ − 𝐹𝜙 𝜃, 𝜙 𝐹𝜙∗ 𝜃, 𝜙 𝛟
෡ ×𝛉
෡×𝛟
෡×𝛉
෡ + 𝐹𝜙 𝜃, 𝜙 𝐹𝜃∗ 𝜃, 𝜙 𝛟
෡
=
𝐹𝜃 𝜃, 𝜙 𝐹𝜃∗ 𝜃, 𝜙 𝛉
2
2
32𝜋 𝑟
𝜂𝑘 2
𝜂𝑘 2
2
2ො
=
𝐹𝜃 𝜃, 𝜙 𝐫 + 𝐹𝜙 𝜃, 𝜙 𝐫ො =
𝐅⊥ 𝜃, 𝜙 2 𝐫ො = 𝑆𝑟 𝐫ො
2
2
2
2
32𝜋 𝑟
32𝜋 𝑟
47
Parameters of antennas
❑ The radiated power density (power per unit of surface) is thus defined as:
𝑑𝑃
𝑑𝑃
𝜂𝑘 2
𝑝𝑟 𝐫 =
=
=
𝐅 𝜃, 𝜙
𝑑𝑆 𝑟 2 𝑑Ω 32𝜋 2 𝑟 2 ⊥
2
❑ We define radiation intensity the quantity:
2
𝑑𝑃
𝜂𝑘
2
𝑈 𝜃, 𝜙 =
= 𝑟 2 𝑝𝑟 𝐫 =
𝐅
𝜃,
𝜙
⊥
𝑑Ω
32𝜋 2
❑ The radiation intensity represents the radiated power per unit of solid angle.
48
Parameters of antennas
❑ Some recalls on solid angles…
❑ Solid angles are the extension to 3D spaces of the concept of angles.
❑ We are all familiar with angles, and we know they are measured in radians [rad].
❑ One radian is defined as the angle subtended by an arc of length 𝑙 = 𝑟, 𝑟 being the radius of the
considered circumference.
❑ In general, the angle subtended by a generic arc of lenght 𝑙 is given by
𝑙
𝜙 = [𝑟𝑎𝑑]
𝑟
Angle 𝜙
𝑟
𝑙
❑ The conversion from radians to degrees is given by
𝜙 𝑑𝑒𝑔 = 𝜙 𝑟𝑎𝑑 × 180/𝜋
49
Parameters of antennas
❑ Solid angles extends the concept of angles.
❑ They are the «two-dimensional» angle subtended by a certain surface 𝑆 over a sphere.
❑ By definition, we have
𝑆
Ω= 2
𝑟
Surface 𝑆
𝑧
❑ Solid angles are measured in steradians.
❑ 1 sr is defined as the solid angle
subtended by a spherical surface
with area equal to that of a square
with each side of length 𝑟.
𝑟
𝑦
Solid angle Ω
𝑥
50
Parameters of antennas
❑ It is worth recalling that the infinitesimal area 𝑑𝑆 on the surface of a sphere of radius 𝑟 is given by
𝑑𝑆 = 𝑟 2 sin 𝜃 𝑑𝜃𝑑𝜙 [𝑚2 ]
❑ Consequently, the corresponding infinitesimal solid angle 𝑑Ω is given by
𝑑Ω = sin 𝜃 𝑑𝜃𝑑𝜙 [𝑠𝑟]
❑ The solid angle subtended by the whole sphere is thus
2𝜋
Ω𝑠𝑝ℎ𝑒𝑟𝑒 = න
𝑠𝑝ℎ𝑒𝑟𝑒
𝑑Ω = න
0
𝜋
2𝜋
න sin 𝜃 𝑑𝜃𝑑𝜙 = න
0
0
𝜋
𝜋
𝑑𝜙 න sin 𝜃 𝑑𝜃 = 2𝜋 − cos 𝜃 ቚ = 2𝜋 1 + 1 = 4𝜋
0
0
51
Parameters of antennas
❑ The total radiated power is thus obtained as
𝜋 2𝜋
𝑃𝑟𝑎𝑑 = න න 𝑈 𝜃, 𝜙 𝑑Ω
0 0
❑ We define radiation resistance 𝑅𝑟𝑎𝑑 the equivalent resistance that would dissipates the same
power 𝑃𝑟𝑎𝑑 , i.e.,
1
2𝑃𝑟𝑎𝑑
𝑃𝑟𝑎𝑑 = 𝑅𝑟𝑎𝑑 𝐼𝑖𝑛 2 → 𝑅𝑟𝑎𝑑 =
2
𝐼𝑖𝑛 2
𝐼𝑖𝑛
𝑃𝑟𝑎𝑑
❑ where 𝐼𝑖𝑛 is the current at the input terminals of the antenna.
Antenna
52
Parameters of antennas
❑ More generally, the antenna is characterized by an input impedance.
❑ Such quantity can be defined as «the ratio of the voltage to current at
antenna terminals» or «the ratio of the appropriate components of the
electric to magnetic fields at the reference input plane».
❑ The input impedance is in general a complex quantity, i.e.,
𝑍𝐴
Antenna
𝑍𝐴 = 𝑅𝐴 + 𝑗𝑋𝐴 [Ω]
Antenna resistance
Antenna reactance
❑ The resistive part can be though to be composed of two components:
𝑅𝐴 = 𝑅𝑟𝑎𝑑 + 𝑅𝑙𝑜𝑠𝑠
Radiation resistance
Loss resistance
53
Parameters of antennas
❑ We define isotropic antenna an ideal antenna that would radiate the power equally in every
direction.
❑ In this case, the radiation intensity is given by:
𝑑𝑃
𝑃𝑟𝑎𝑑
𝑃𝑟𝑎𝑑
𝑈𝐼 =
=
=
𝑑Ω 𝐼 Ω𝑠𝑝ℎ𝑒𝑟𝑒
4𝜋
❑ The corresponding radiated power density is given by:
𝑑𝑃
𝑈𝐼
𝑃𝑟𝑎𝑑
𝑝𝑟 =
=
=
𝑑S 𝐼 𝑟 2 4𝜋𝑟 2
54
Parameters of antennas
❑ The directivity function is defined as the ratio between the radiation intensity of the antenna
and the radiation intensity of an isotropic radiator that radiates the same power radiated by
the antenna, i.e.,
𝑈 𝜃, 𝜙
𝑈 𝜃, 𝜙
4𝜋
𝐷 𝜃, 𝜙 =
=
=
𝑈 𝜃, 𝜙
𝑈𝐼
𝑃𝑟𝑎𝑑 /4𝜋
𝑃𝑟𝑎𝑑
❑ The maximum value of the directivity function is called directivity and it is given by:
max 𝑈 𝜃, 𝜙
𝐷𝑚𝑎𝑥 = max 𝐷 𝜃, 𝜙 =
𝜃,𝜙
𝜃,𝜙
𝑈𝐼
=
𝑈𝑚𝑎𝑥
𝑈𝐼
55
Parameters of antennas
4𝜋
𝑈 𝜃, 𝜙
𝑃𝑟𝑎𝑑
↓
𝑃𝑟𝑎𝑑 𝐷 𝜃, 𝜙
𝑈 𝜃, 𝜙 =
4𝜋
𝐷 𝜃, 𝜙 =
❑ The radiated power density can be expressed
in terms of the directivity function as:
𝑝𝑟 𝐫 =
𝑑𝑃 𝑈 𝜃, 𝜙
𝑃𝑟𝑎𝑑 𝐷 𝜃, 𝜙
=
=
𝑑𝑆
𝑟2
4𝜋𝑟 2
❑ The quantity 𝑃𝑟𝑎𝑑 𝐷 𝜃, 𝜙 is called effective isotropic power and the corresponding
maximum value, 𝑃𝑟𝑎𝑑 𝐷𝑚𝑎𝑥 , is called effective isotropic radiated power (EIRP).
❑ EIRP represents the power that would have been radiated by an isotropic antenna to obtain
the same power as the actual source antenna in the direction of the main beam.
❑ EIRP is usually used in antenna national and international regulations. For example, the
ETSI EN 301 893 rule define a maximum EIRP of 20 dBm for the radiated WiFi signal in the
2.4 GHz band.
𝑃
𝑃𝑑𝐵𝑚 = 10 log10
1 mW
56
Parameters of antennas
❑ The gain function is defined as the ratio between the radiation intensity of the antenna and
the radiation intensity of an isotropic radiator that radiates the whole power in input to the
antenna, i.e.,
𝐺 𝜃, 𝜙 =
𝑈 𝜃, 𝜙
4𝜋
=
𝑈 𝜃, 𝜙
𝑃𝑇 /4𝜋
𝑃𝑇
❑ where 𝑃𝑇 is the power available at the input terminals of the antenna.
❑ The maximum value of the gain function is called gain and it is given by:
4𝜋
𝐺𝑚𝑎𝑥 = max 𝐺 𝜃, 𝜙 =
𝑈
𝜃,𝜙
𝑃𝑇 𝑚𝑎𝑥
57
Parameters of antennas
❑ If the antenna is ideal, all the power in input is radiated (and consequently 𝐺 = 𝐷).
❑ However, in practice there are always losses. Consequently the radiated power is lower than
the input power. Such behavior is described by the antenna efficiency, defined as:
𝑒=
𝑃𝑟𝑎𝑑
𝑃𝑇
TX
❑ Clearly, 0 ≤ 𝑒 ≤ 1.
❑ From the definitions of 𝐺 and 𝐷, it results:
𝑍𝑐
𝑃𝑇
𝑃𝑟𝑎𝑑
Antenna
𝐺 𝜃, 𝜙 = 𝑒𝐷 𝜃, 𝜙
58
Parameters of antennas
❑ It is worth noting that the antenna efficiency (and consequently the gain) does not take into
account the losses due to the eventually present load mismatch loss between the transmission line
and the antenna.
❑ In this case a reflection is present at the input terminals of the antenna, and thus not all the power
𝑃𝑇𝑋 provided by the TX is available at the antenna, i.e.,
𝑃𝑇 = 1 − Γ
2
Γ
𝑃𝑇𝑋
𝑍𝑐
TX
=𝑒𝑟𝑒
𝑃𝑇𝑋
Antenna
𝑃𝑇
❑ where Γ = (𝑍𝐴 − 𝑍𝑐 )/(𝑍𝐴 + 𝑍𝑐 ) is the reflection coefficient
at the antenna input, 𝑍𝐴 being the input impedance of the antenna.
❑ In order to take into account these losses, too, it is possible to introduce a realized gain function
𝐺𝑟𝑒 as
𝐺𝑟𝑒 𝜃, 𝜙 = 𝑒𝑟𝑒 𝐺 𝜃, 𝜙
59
Parameters of antennas
❑ A further parameter is the radiation pattern, defined as:
𝐺 𝜃, 𝜙
𝐷 𝜃, 𝜙
𝑈 𝜃, 𝜙
𝐅⊥ 𝜃, 𝜙 2
𝑔 𝜃, 𝜙 =
=
=
=
𝐺𝑚𝑎𝑥
𝐷𝑚𝑎𝑥
𝑈𝑚𝑎𝑥
𝐅⊥ 𝜃, 𝜙 2𝑚𝑎𝑥
❑ Such function represents a three-dimensional graph.
❑ Often, two-dimensional cuts of the radiation pattern are provided, since they allow an easier
visualization. In particular, the following cut planes are usually considered:
o “E” plane: section of the radiation pattern obtained by cutting the 3d graph with a plane containing the electric
field vector and the maximum value of the pattern;
o “H” plane: section of the radiation pattern obtained by cutting the 3d graph with a plane containing the
magnetic field vector and the maximum value of the pattern.
60
Parameters of antennas
❑ It is possible to identify three main classes of radiation patterns:
o Isotropic – the antenna radiates in the same way in every direction. The radiation pattern is a sphere.
«E» plane
3D pattern
𝜃
𝑔(𝜃, 𝜙)
«H» plane
𝜙
61
Parameters of antennas
o Directional – The antenna radiation is significantly higher in a given direction. The radiation pattern
is characterized by a main lobe (a portion of graph with a maxima delimited by nulls) in this
direction and usually several minor lobes.
3D pattern
«E» plane
𝜃 (𝜙 = 0°)
«H» plane
𝜃 (𝜙 = 90°)
Please note that the radiation pattern is often
provided in decibel, i.e., 𝑔𝑑𝐵 𝜃, 𝜙 = 10 log10 𝑔 𝜃, 𝜙
62
Parameters of antennas
Main lobe
Secondary
lobes
Null direction
Back lobe
63
Parameters of antennas
o Omnidirectional – The antenna has an isotropic behavior in one of the coordinate planes and a
directive behavior in the other plane.
3D pattern
«E» plane
𝜃
«H» plane
𝜙
64
Parameters of antennas
❑ One final additional note: gain (function), directivity (function), and radiation pattern are
related one to each other (as can be easily noted on the previous slides).
❑ Indeed, we have:
𝐷 𝜃, 𝜙 = 𝐷𝑚𝑎𝑥 𝑔 𝜃, 𝜙
𝐺 𝜃, 𝜙 = 𝐺𝑚𝑎𝑥 𝑔 𝜃, 𝜙 = 𝑒𝐷𝑚𝑎𝑥 𝑔 𝜃, 𝜙
❑ So, usually not all are reported in data sheets, but just the ones that allows to fully rebuild all
the others.
65
Parameters of antennas
❑ Example of data sheet (extract)
66
Parameters of antennas
❑ From the radiation pattern, it is possible to derive a further parameter, the beamwidth Δ𝜃𝑏 .
It is defined as the “angle between the directions for which the power density is half (-3 dB)
of the value in the direction of maximum radiation”.
❑ If the radiation pattern is not symmetrical, it is possible to define the apertures on the two
coordinate planes “E” and “H”.
Beamwidth, i.e.,
angle Δ𝜃𝑏 between
the two directions
at -3 dB
Point in which the
radiation pattern has a
value equal to -3 dB, i.e.,
half of the maximum
67
Parameters of antennas
❑ Another important parameter is the beam solid angle ΔΩ, i.e., the solid angle in which the
major part of the power is radiated.
ΔΩ
68
Parameters of antennas
❑ The beam solid angle of an antenna is defined as:
𝜋
2𝜋
𝛥𝛺 = න න 𝑔 𝜃, 𝜙 𝑑Ω
0
0
❑ By substituting the directivity inside the integral, we obtain:
𝜋 2𝜋
𝜋 2𝜋
𝜋 2𝜋
0 0
0 0
0 0
𝐷 𝜃, 𝜙
1 4𝜋
4𝜋
න න 𝑈 𝜃, 𝜙 𝑑Ω =
𝛥𝛺 = න න 𝑔 𝜃, 𝜙 𝑑Ω = න න
𝑑Ω =
𝐷𝑚𝑎𝑥
𝐷𝑚𝑎𝑥 𝑃𝑟𝑎𝑑
𝐷𝑚𝑎𝑥
69
Parameters of antennas
❑ For directive antennas, it is possible to devise a relationship between the beam solid angle
and the beamwidth.
❑ Let us suppose that the radiation pattern is approximated by the following function:
Δ𝜃𝐵
2
1,
𝑖𝑓 0 ≤ 𝜃 ≤
0,
Δ𝜃𝐵
𝑖𝑓
<𝜃≤𝜋
2
𝑔 𝜃, 𝜙 =
70
Parameters of antennas
❑ In this case we have:
𝜋 2𝜋
Δ𝜃𝐵 /2 2𝜋
𝛥𝛺 = න න 𝑔 𝜃, 𝜙 𝑑Ω = න
0 0
න sin 𝜃 𝑑𝜃 𝑑𝜙 = 2𝜋 1 − cos
0
0
Δ𝜃𝐵
2
❑ By using the approximation cos 𝑥 ≅ 1 − 𝑥 2 /2, we obtain:
1 Δ𝜃𝐵2
Δ𝜃𝐵2 𝜋
𝛥𝛺 ≅ 2𝜋 1 − 1 +
= 2𝜋
= Δ𝜃𝐵 2
2 4
8
4
❑ and finally:
𝐷𝑚𝑎𝑥 ≅ 𝜋
4
4𝜋
Δ𝜃𝐵 2
=
16
Δ𝜃𝐵 2
71
Parameters of antennas
❑ For asymmetric patterns, it is possible to obtain a similar relationship for the apertures in the
two planes.
❑ In particular, it results:
𝛥𝛺 = Δ𝜃1 Δ𝜃2
𝐷𝑚𝑎𝑥 =
4𝜋
Δ𝜃1 Δ𝜃2
72
Parameters of antennas
❑ Example: Let us consider a satellite in geostationary orbit (ℎ = 36000 𝑘𝑚). Find the gain and
the beamwidth required to completely illuminate the visible part of the earth.
❑ From the satellite, the earth approximatively appears as a disk with a surface equal to 𝜋𝑅 2 ,
where 𝑅 = 6400 𝑘𝑚 is the earth radius.
ℎ
ℎ
𝑅
𝑅
73
Parameters of antennas
❑ Consequently, the beam solid angle must be:
𝜋𝑅2
𝛥𝛺 ≅ 2 ≅ 0.1 steradians
ℎ
❑ The directivity results:
Please note that gain and
directivity
are
usually
provided in decibel, i.e.,
𝑑𝐵 = 10 log 𝐷
𝐷𝑚𝑎𝑥
10 𝑚𝑎𝑥
𝐷𝑚𝑎𝑥 =
4𝜋
4𝜋
=
= 126.56 → 21 𝑑𝐵
𝛥𝛺 𝜋𝑅2
ℎ2
❑ The beamwidth is:
𝐷𝑚𝑎𝑥 =
Δ𝜃𝐵 =
4
𝐷𝑚𝑎𝑥
16
Δ𝜃𝐵 2
= 0.36 𝑟𝑎𝑑 = 20.37°
74
Parameters of antennas
❑ Finally, the polarization of the antenna, in a given direction, is defined as the polarization of
the wave radiated by the antenna.
𝐄 𝐫, 𝑡
𝐫ො
Polarizarion
vector
❑ When the direction is not stated, the polarization is taken to be the polarization in the
direction of maximum gain.
75
Parameters of antennas
❑ Usually, the polarization is defined in terms of vertical and horizontal directions.
vertical
horizontal
𝐄 𝐫, 𝑡
𝐫ො
෡
𝜽
෡ 𝐄 𝐫, 𝑡
𝝓
X 𝐫ො
෡
𝜽
76
Master degree in Internet and Multimedia Engineering / Electronic Engineering
Antennas
Slides of the lectures – Part 2
Linear wire antennas
❑ Linear wire antennas represent a class of antennas which is widely used in practical
applications, thanks to their simplicity.
❑ They are typically constituted by a metallic wire (whose diameter 𝑑 is usually assumed to be
negligible with respect to the other dimensions).
❑ Le us consider the case in which the antenna is composed by a straight metallic wire of
length 𝑙 directed along the 𝑧 axis.
𝑧
𝑙
𝐉 𝐫
𝑦
𝑥
𝑑
2
Linear wire antennas
❑ By neglecting the diameter of the wire, the current density is expressed as:
𝐉 𝐫 = 𝐼 𝑧 𝛿 𝑥 𝛿 𝑦 𝐳ො
𝑧
𝑙
𝐉 𝐫
𝑦
𝑥
𝑑
3
Linear wire antennas
❑ The radiation vector has only component along z and it is given by:
′
𝐅 𝜃, 𝜙 = න 𝐉 𝐫 𝑒
𝑉
𝑗𝑘ො𝐫∙𝐫 ′
′
′
𝑑𝐫 = 𝐳ො න 𝐼 𝑧 𝛿 𝑥 𝛿 𝑦 𝑒
𝑉
𝑗𝑘ො𝐫∙𝐫 ′
′
′
𝑑𝐫 = 𝐳ො න 𝐼 𝑧 𝑒
𝑗𝑘𝑧 ′ 𝐫ො∙ො𝐳
𝑑𝑧 ′
𝑙
′
= 𝐳ො න 𝐼 𝑧 ′ 𝑒 𝑗𝑘𝑧 cos 𝜃 𝑑𝑧 ′ = 𝐹𝑧 𝜃 𝐳ො
𝑙
❑ The radiation vector depends only on 𝜃! → The radiation pattern has an ominidirectional
shape.
❑ In order to obtain the radiated fields, we need to rewrite the radiation vector in terms of the
spherical unit vectors.
4
Linear wire antennas
෡ sin 𝜃, we have
ො = 𝐫ො cos 𝜃 − 𝛉
❑ Since 𝐳
෡ + 𝐹𝑧 𝜃 cos 𝜃 𝐫ො
𝐅 𝜃, 𝜙 = 𝐹𝑧 𝜃 𝐳ො = −𝐹𝑧 𝜃 sin 𝜃 𝛉
෡
𝐅⊥ 𝜃,𝜙 =𝐹𝜃 𝜃 𝛉
❑ Consequently, the electric and magnetic field vectors are given by:
𝑒 −𝑗𝑘𝑟
𝑒 −𝑗𝑘𝑟
෡ = 𝐸𝜃 𝜃 𝛉
෡
𝐄 𝐫 = −𝑗𝑘𝜂
𝐅⊥ 𝜃, 𝜙 = 𝑗𝑘𝜂
𝐹𝑧 𝜃 sin 𝜃 𝛉
4𝜋𝑟
4𝜋𝑟
−𝑗𝑘𝑟
𝑒
𝑒 −𝑗𝑘𝑟
෡ = 𝐻𝜙 𝜃 𝛟
෡
𝐇 𝐫 = −𝑗𝑘
𝐫 × 𝐅⊥ 𝜃, 𝜙 = 𝑗𝑘
𝐹𝑧 𝜃 sin 𝜃 𝛟
4𝜋𝑟
4𝜋𝑟
5
Linear wire antennas
❑ The radiation intensity is finally:
𝜂𝑘 2
𝑈 𝜃 =
𝐅 𝜃, 𝜙
32𝜋 2 ⊥
2
𝜂𝑘
2 =
𝐹 𝜃
32𝜋 2 𝑧
2 sin2 𝜃
❑ Depending on 𝐼 𝑧 , we can have several types of antennas.
o 𝐼 𝑧 = 𝐼𝑙𝛿 𝑧
→ Hertzian dipole
o 𝐼 𝑧 = 𝐼(1 − 2
o 𝐼 𝑧 = 𝐼 sin 𝑘
𝑧
) → Small dipole
𝑙
𝑙
− 𝑧 → Standing-wave antenna (long dipole)
2
o 𝐼 𝑧 = 𝐼𝑒 −𝑗𝑘𝑧 → Traveling-wave antenna
6
Dipole antennas
❑ Dipoles are one of the most commonly adopted type of linear-wire antennas.
❑ We assume to have a wire of length 𝑙 and negligible diameter 𝑑 ≪ 𝜆, and that the feeding
point is in the middle of the wire.
𝑧
𝐼𝑖𝑛
𝑙
𝑦
𝑥
7
Dipole antennas
❑ If the length 𝑙 is comparable or greater than the wavelength 𝜆 (typically 𝑙 > 𝜆/10), the
current density over the dipole can be approximated as:
𝑙
𝐼 𝑧 = 𝐼 sin 𝑘
− 𝑧
2
𝑙
𝑙
− ≤𝑧≤
2
2
❑ Defining ℎ = 𝑙/2, the radiation vector becomes:
ℎ
𝐹𝑧 𝜃 = න 𝐼 sin 𝑘 ℎ −
𝑧′
−ℎ
∫ 𝑒 𝛼𝑥 sin 𝛽𝑥 + 𝛾 𝑑𝑥 =
′
𝑒 𝑗𝑘𝑧 cos 𝜃 𝑑𝑧 ′ =
2𝐼 cos 𝑘ℎ cos 𝜃 − cos 𝑘ℎ
𝑘
sin2 𝜃
𝑒 𝛼𝑥
𝛼 sin 𝛽𝑥 + 𝛾 − 𝛽 cos 𝛽𝑥 + 𝛾
𝛼 2 + 𝛽2
8
Dipole antennas
❑ The radiations intensity is thus given by
2
𝜂 𝐼 2 cos 𝑘ℎ cos 𝜃 − cos 𝑘ℎ
𝑈 𝜃 =
8𝜋 2
sin 𝜃
❑ Finally, the radiation pattern is obtained as
cos 𝑘ℎ cos 𝜃 − cos 𝑘ℎ
𝑔 𝜃 = 𝑐𝑛
sin 𝜃
2
❑ where 𝑐𝑛 is a normalization constant.
9
Dipole antennas
❑ The radiation pattern has the following shape (𝑙 = 0.5𝜆 is assumed as an example).
«H» plane
(horizontal)
𝜙
𝜓=
𝜋
−𝜃
2
«E» plane
(vertical)
10
Dipole antennas
❑ The radiations shape changes however with the length!
𝑙 = 0.5𝜆

0o

45o
45o
0.5
o
90
1 90o
o


45o
𝜃
0.5
135
1 90o
𝜃
45o
0.5
o
90
o
135o

45
45
o
0o
o
o
90
o
135o
𝑙=𝜆
𝑙 = 0.75𝜆
0

135
180o
180o
180o
𝑙 = 1.25𝜆
𝑙 = 1.5𝜆
𝑙 = 2𝜆

o
0


45o
45
0.5
o
1 90o
o
135o
135
180o
0


45o
o
90
o
𝜃
45
0.5
o
90
1 90o
o
135o
135
180o
0o

45o
o
𝜃
𝜃
o
135o
135
1 90o
45o
0.5
o
90
1 90o
𝜃
o
135o
135
180o
11
Dipole antennas
❑ Depending on the value 𝑙, the maximum value can be in directions different from 𝜃 = 90°.
❑ Moreover, also the maximum of the current density 𝐼 can be different from the value in the
feed point. Indeed we have:
𝑘𝑙
𝐼𝑖𝑛 = 𝐼 0 = 𝐼 sin
= 𝐼 sin 𝑘ℎ
2
𝐼 𝑧 = 𝐼 sin 𝑘
𝑙
− 𝑧
2
❑ This fact impacts on the radiation
resistance. Indeed, the radiation
resistance is usually defined in terms
of the input current. However, the
radiated power depends directly on the
maximum value of the current along the dipole 𝐼.
12
Dipole antennas
❑ Indeed, the overall radiated power is
2𝜋 𝜋
𝜋
𝑃𝑟𝑎𝑑 = න 𝑈 𝜃 𝑑Ω = න න 𝑈 𝜃 sin 𝜃 𝑑𝜃 𝑑𝜙 = 2𝜋 න 𝑈 𝜃 sin 𝜃 𝑑𝜃
𝜋
0
0
𝜂𝐼2
cos 𝑘ℎ cos 𝜃 − cos 𝑘ℎ
=
න
4𝜋
sin 𝜃
0
0
2
𝜂𝑝ℎ 2
𝑑𝜃 =
𝐼
4𝜋
𝑝ℎ
❑ The radiation resistance is thus given by
𝑃𝑟𝑎𝑑 =
𝑅𝑝𝑒𝑎𝑘
1
1
𝑅𝑟𝑎𝑑 𝐼𝑖𝑛 2 = 𝑅𝑝𝑒𝑎𝑘 𝐼 2 =→ 𝑅𝑟𝑎𝑑 =
2
2
sin2 𝑘ℎ
13
Dipole antennas
❑ 𝑅𝑝𝑒𝑎𝑘 is a radiation resistance referred to the peak current, which is given by
𝑅𝑝𝑒𝑎𝑘 =
𝜂𝑝ℎ
2𝜋
→
𝑅𝑟𝑎𝑑 =
𝜂𝑝ℎ
2𝜋sin2 𝑘ℎ
❑ Some examples:
o 𝑙 = 0.5𝜆 → 𝑅𝑝𝑒𝑎𝑘 = 𝑅𝑟𝑎𝑑 = 73.08 Ω
o 𝑙 = 0.75𝜆 → 𝑅𝑝𝑒𝑎𝑘 = 185.68 Ω, 𝑅𝑟𝑎𝑑 = 371.36 Ω
o 𝑙 = 𝜆 → 𝑅𝑝𝑒𝑎𝑘 = 198.95 Ω, 𝑅𝑟𝑎𝑑 → ∞
o 𝑙 = 1.25𝜆 → 𝑅𝑝𝑒𝑎𝑘 = 106.46 Ω, 𝑅𝑟𝑎𝑑 = 212.92 Ω
o 𝑙 = 1.5𝜆 → 𝑅𝑝𝑒𝑎𝑘 = 𝑅𝑟𝑎𝑑 = 105.42 Ω
14
Dipole antennas
❑ The beam solid angle can be then expressed as
2𝜋 𝜋
𝜋
cos 𝑘ℎ cos 𝜃 − cos 𝑘ℎ
ΔΩ = න න 𝑔 𝜃 sin 𝜃 𝑑𝜃 𝑑𝜙 = 2𝜋𝑐𝑛 න
sin 𝜃
0
0
2
2𝜋𝑅𝑝𝑒𝑎𝑘
𝑑𝜃 = 2𝜋𝑐𝑛
𝜂
0
❑ Consequently the directivity is directly obtained as
𝐷𝑚𝑎𝑥 =
4𝜋
ΔΩ
=
1
𝜂
𝑐𝑛 𝜋𝑅𝑝𝑒𝑎𝑘
15
Dipole antennas
❑ In the general case, the constant 𝑐𝑛 must be computed numerically.
❑ Some examples:
o 𝑙 = 0.5𝜆 → 𝑐𝑛 = 1, 𝐷𝑚𝑎𝑥 = 1.64 = 2.15 𝑑𝐵
o 𝑙 = 0.75𝜆 → 𝑐𝑛 = 0.34, 𝐷𝑚𝑎𝑥 = 1.88 = 2.74 𝑑𝐵
o 𝑙 = 𝜆 → 𝑐𝑛 = 0.25, 𝐷𝑚𝑎𝑥 = 2.41 = 3.82 𝑑𝐵
o 𝑙 = 1.25𝜆 → 𝑐𝑛 = 0.34, 𝐷𝑚𝑎𝑥 = 3.28 = 5.16 𝑑𝐵
o 𝑙 = 1.5𝜆 → 𝑐𝑛 = 0.51, 𝐷𝑚𝑎𝑥 = 2.23 = 3.48 𝑑𝐵
o 𝑙 = 2𝜆 → 𝑐𝑛 = 0.18, 𝐷𝑚𝑎𝑥 = 2.53 = 4.03 𝑑𝐵
Highest directivity!
16
Half-wavelength dipole antenna
❑ An antenna often employed in practice is the half-wavelength dipole, for which 𝑙 = 𝜆/2.
❑ In this case, the current over the antenna reduces to
𝐼 𝑧 = 𝐼 cos 𝑘𝑧
❑ and the radiation pattern becomes
𝜋
cos 2 2 cos 𝜃
𝑔 𝜃 =
sin2 𝜃
❑ The radiation resistance is equal to 𝑅𝑟𝑎𝑑 = 73.08 Ω.
❑ Consequently, the directivity is 𝐷𝑚𝑎𝑥 =
𝜂
𝜋𝑅𝑝𝑒𝑎𝑘
= 1.64 = 2.15 𝑑𝐵
17
Half-wavelength dipole antenna
❑ The radiation pattern is the following one.
«E» plane (vertical)

0o

45o
45o
0.5
o
90
Beamwidth:
Δ𝜃𝑏 = 78.08°
1 90o
o
135o
135
180o
18
Half-wavelength dipole antenna
❑ An example of a real dipole antenna implementation for WiFi routers.
Arms of the
dipole
𝑙 ≅ 5 𝑐𝑚
Feeding point
19
Half-wavelength dipole antenna
❑ The dipole is realized by folding back the external conductor of a coaxial cable.
❑ The length of each arm is equal to a quarter of the wavelenght, in order to obtain a half-
wavelength dipole.
❑ It is worth noting that in this case, the antenna is built by using a coaxial cable, and not just a
simple ideal wire.
❑ Assuming a LMR400 coaxial cable, we have the following parameters
LMR400 coaxial cable
Velocity of Propagation
85%
Dielectric Constant
1.38
Impedance
50 𝛺
Attenuation dB/100 m @ 2.5 GHz 0.33
20
Half-wavelength dipole antenna
❑ In the ideal case (metallic wires with negligble diameter), the dipole length should be
𝜆0
𝑐
3 108
𝑙=
=
=
m = 0.0615 m = 6.15 cm
2
2𝑓0 2 2.441 109
❑ where 𝑓0 = 2.441 GHz is the central frequency for the 2.4 GHz band.
❑ However, in the coaxial cable the wavelength is higher, since the nominal velocity of
propagation is 𝑣𝑐𝑜𝑎𝑥 = 0.85𝑐. Consequently the length of the dipole becomes
𝜆𝑐𝑜𝑎𝑥 𝑣𝑐𝑜𝑎𝑥
0.85 3 108
𝑙=
=
=
m = 0.0522 m = 5.22 cm
2
2𝑓0
2 2.441 109
21
Hertzian dipole antenna
❑ A particular case is represented by dipole antennas with length very small with respect to
the wavelength, i.e., 𝑙 ≪ 𝜆.
❑ In this case, we have 𝐼 𝑧 = 𝐼𝑙𝛿 𝑧 and the radiation vector becomes
𝑙
2
′
𝐹𝑧 𝜃 = න 𝐼𝑙𝛿 𝑧 ′ 𝑒 𝑗𝑘𝑧 cos 𝜃 𝑑𝑧 ′ = 𝐼𝑙
𝑙
−
2
❑ Consequently the radiated field reduces to
𝑒 −𝑗𝑘𝑟
𝐄 = 𝐸𝜃 ෡
𝛉 = 𝑗𝑘𝜂𝐼𝑙
sin 𝜃 ෡
𝛉
4𝜋𝑟
22
Hertzian dipole antenna
❑ Finally, the radiation pattern is
𝜂𝑘 2
2
2
𝑈(𝜃) 32𝜋 2 𝐼𝑙 sin 𝜃
𝑔 𝜃 =
=
= sin2 𝜃
2
𝜂𝑘
𝑈𝑚𝑎𝑥
𝐼𝑙 2
2
32𝜋
Beamwidth
: Δ𝜃𝑏 = 90°
«E» plane (vertical)
23
Monopole antennas
❑ A monopole antenna is obtained by putting a half dipole on a ground plane, as schematically
shown in the figure.
𝑧
ℎ
𝑧
𝑦
Ground
plane
𝑥
𝐼𝑖𝑛
Coaxial
feed
24
Monopole antennas
❑ By assuming an infinite PEC plane, the monopole antenna is equivalent to a dipole antenna,
whose lower part is the image of the monopole.
𝑧
𝑧
Equivalent
antenna
ℎ
ℎ
𝑙 = 2ℎ
ℎ
❑ Consequently, the radiation pattern, over the ground plane, has the same shape as that of the
dipole.
25
Monopole antennas
❑ Example: 𝜆/4 monopole over ideal
infinite ground plane
The radiation pattern
is limited to the
upper half plane!
𝜓=
𝜋
−𝜃
2
«E» plane (vertical)
26
Monopole antennas
❑ However, since the field is radiated only on the upper half space, the radiated power (once 𝐼𝑖𝑛 is
fixed) is half that of the equivalent dipole, i.e.,
1
𝑃𝑚𝑜𝑛𝑜𝑝𝑜𝑙𝑒 = 𝑃𝑑𝑖𝑝𝑜𝑙𝑒
2
1
𝑅𝑚𝑜𝑛𝑜𝑝𝑜𝑙𝑒 = 𝑅𝑑𝑖𝑝𝑜𝑙𝑒
2
❑ Consequently:
𝐷𝑚𝑜𝑛𝑜𝑝𝑜𝑙𝑒 = 2𝐷𝑑𝑖𝑝𝑜𝑙𝑒
❑ Monopole antennas used in practice have usually a length equal to 𝝀/𝟒.
❑ When a higher directivity is needed, a monopole of length 5𝜆/8 (𝟓/𝟖-wave monopole), can be
used (indeed, the equivalent dipole, which has length 𝑙 = 1.25𝜆 = 5𝜆/4, is the one providing the
maximum directivity).
27
Monopole antennas
❑ In practical applications the ground plane has finite extent. This affects the radiation pattern.
Some examples:
Square ground
plane of side
𝑊 = 0.5𝜆
Square ground
plane of side
𝑊 = 2.5𝜆
28
Loop antennas
❑ Up to now, we considered only linear antennas. However, wire antenna in which the wire is
curved also exists.
❑ The simplest case of curved wire structure is the loop antenna. Although different shapes
can be adopted, the main common ones are circular and square loops.
𝑧
𝑧
𝑙
𝐼𝑖𝑛
𝐼𝑖𝑛
𝑎
𝑥
𝑦
𝑦
𝑥
29
Loop antennas
❑ We concentrate on the case of circular loops.
❑ In such a case, the antenna is composed by a wire with circular shape, which we assume is
located in the x-y plane, as shown in the figure.
❑ If the diameter of the wire is small (ideally negligible), the current density over the antenna
can be modeled as
𝑧
෡
𝐉 𝐫 =𝐼 𝜙 𝛿 𝜌−𝑎 𝛿 𝑧 𝛟
𝐫
𝐼 𝜙′
❑ The current 𝐼 𝜙 may be constant,
sinusoidal, or expressed as a Fourier
summation.
𝜃
𝑎
𝜙′ 𝐫
𝑥
ෝ′
𝝆
′
𝑦
෡
𝛟
෡′
𝛟
ෝ
𝝆
𝜙
30
Loop antennas
❑ We consider explicitly the case of constant current (the other cases are quite cumbersome
from amathematical point of view).
❑ Assuming 𝐼 𝜙 = 𝐼, the radiation vector becomes
𝐅 𝜃, 𝜙 = න 𝐉 𝐫 𝑒
𝑉
𝑗𝑘ො𝐫∙𝐫 ′
෡
𝑑𝐫 ′ = න 𝐼 𝜙 ′ 𝛿 𝜌′ − 𝑎 𝛿 𝑧 ′ 𝛟𝑒
𝑗𝑘ො𝐫∙𝐫 ′
2𝜋
෡ ′ 𝑒 𝑗𝑘ො𝐫∙𝐫 ′ 𝑑𝜙 ′
𝜌′ 𝑑𝜌′ 𝑑𝜙 ′ 𝑑𝑧 ′ = 𝑎𝐼 න 𝛟
𝑉
0
෡ ′ depends upon 𝜙 and 𝜙 ′ . With some geometrical reasonments, it can be
❑ However, 𝛟
expressed as
෡ ′ = cos 𝜙 ′ − 𝜙 𝛟
෡ − sin 𝜙 ′ − 𝜙 𝝆
ෝ
𝛟
31
Loop antennas
❑ Since the integral is performed on a circumference lying in the x-y plane we have 𝑧 ′ = 0 and
𝜌′ = 𝑎 and consequently
𝐫ො ∙ 𝐫 ′ = 𝑎 sin 𝜃 cos 𝜙 ′ − 𝜙
❑ Consequently, the radiation vector is
2𝜋
෡ ′ 𝑒 𝑗𝑘𝑎 sin 𝜃 cos 𝜙′ −𝜙 𝑑𝜙 ′
𝐅 𝜃, 𝜙 = 𝑎𝐼 න 𝛟
0
32
Loop antennas
❑ Consequently, we have
2𝜋
𝐅 𝜃, 𝜙 = 𝑎𝐼 න
0
2𝜋
cos 𝜙 ′ − 𝜙
= 𝑎𝐼 න
0
′
෡ − sin 𝜙 ′ − 𝜙 𝝆
ෝ 𝑒 𝑗𝑘𝑎 sin 𝜃 cos 𝜙 −𝜙 𝑑𝜙 ′
cos 𝜙 ′ − 𝜙 𝛟
෡ 𝑗𝑘𝑎 sin 𝜃 cos 𝜙′ −𝜙
𝛟𝑒
2𝜋
𝑑𝜙 ′ − 𝑎𝐼 න
′
ෝ𝑒 𝑗𝑘𝑎 sin 𝜃 cos 𝜙 −𝜙 𝑑𝜙 ′
sin 𝜙 ′ − 𝜙 𝝆
0
2𝜋
෡ න cos 𝜙 ′ − 𝜙 𝑒 𝑗𝑘𝑎 sin 𝜃 cos 𝜙′ −𝜙 𝑑𝜙 ′ = 2𝜋𝑗𝐼𝑎𝐽1 𝑘𝑎 sin 𝜃 𝛟
෡
= 𝛟𝑎𝐼
0
Since by definition the Bessel function 𝐽1
2𝜋
1
is given by 𝐽1 =
∫0 cos 𝜓 𝑒 𝑗𝜒 cos 𝜓 𝑑𝜓
2𝜋𝑗
This term vanisheses because the
integrand is an odd function and
consequently its integral over the
period is zero.
33
Loop antennas
❑ Finally, the radiation intensity becomes
𝜂𝑘 2
𝑈 𝜃, 𝜙 =
𝐅 𝜃, 𝜙
32𝜋 2 ⊥
2 2 2
𝜂𝑘
𝐼 𝑎
2 =
𝐽1 𝑘𝑎 sin 𝜃
8
2
❑ and consequently the radiation pattern is given by
𝑔 𝜃, 𝜙 = 𝑐𝑛 𝐽1 𝑘𝑎 sin 𝜃
2
❑ where 𝑐𝑛 is a normalization constant.
34
Loop antennas
❑ Some examples.
𝑎 = 0.1𝜆
𝜃
𝑎 = 0.8𝜆
𝜃
35
Elementary loop antenna
❑ If the radius of the loop is very small (typically 𝑎 < 𝜆/6𝜋), it is possible to approximate the
Bessel function with
𝑘𝑎 sin 𝜃
𝐽1 𝑘𝑎 sin 𝜃 ≅
2
❑ Consequently, the radiation vector reduces to
෡ ≅ 𝑗𝜋𝐼𝑎2 𝑘 sin 𝜃 𝛟
෡
𝐅 𝜃, 𝜙 = 2𝜋𝑗𝐼𝑎𝐽1 𝑘𝑎 sin 𝜃 𝛟
❑ Comparing this result with the corresponding formula for the elementary dipole (i.e.,
෡ we note that they have basically the same form and are perpedicular one
𝐅 𝜃, 𝜙 = 𝐼𝑙 sin 𝜃 𝜽,
to each other.
36
Elementary loop antenna
❑ Indeed, by exploiting the duality theorem, the radiation from a small loop of radius 𝑎 ≪ 𝜆
and current 𝐼 is equivalent to the one of an elementary magnetic dipole of length 𝑙 ≪ 𝜆
perpendicular to the loop and with magnetic current 𝐼𝑚 such that
𝐼𝑚 𝑙 = 𝑗𝜂𝑘 𝜋𝑎2 𝐼
𝑧
𝑧
𝐼
𝑎
𝐼𝑚
𝑎
𝑥
𝑦
𝑦
𝑥
37
Helical antenna
❑ Helical antennas are often used in global positioning systems (GPS), since they allow the
radiation of circularly polarized fields.
❑ These antennas are built with a wire that has the form of a helix with 𝑁 loops at a distance 𝑆,
as schematically shown in the figure.
❑ Usually, a ground plane is also present (as in monopole antennas).
𝐷
𝑠
Ground
plane
38
Helical antenna
❑ The height of the antenna is 𝐿 = 𝑁𝑆 and the length of the wire is
𝐿𝑢 = 𝑁𝐿0 = 𝑁 𝑆 2 + 𝐶 2 = 𝑁 𝑆 2 + 𝜋 2 𝐷 2
𝑆 2 + 𝐶 2 is the length of the wire between each turn, 𝐶 = 𝜋𝐷 being the
circumference of the helix and 𝐷 the diameter of the loops.
❑ where 𝐿0 =
❑ The helical antenna is characterized by its pitch angle, defined as
𝛼 = tan−1
𝑆
𝑆
−1
= tan
𝐶
𝜋𝐷
39
Helical antenna
❑ When 𝛼 = 0 we have 𝑆 = 0, and the antenna becomes a loop composed by a wire winded N
times on itself.
❑ When 𝛼 = 90° we have 𝐷 = 0, and thus the antennas becomes a monopole of length ℎ = 𝑁𝑆.
❑ For 0° < 𝛼 < 90°, the behavior of the antenna is a combination of the ones of monopoles and
loops and the helical antenna has generally an elliptical polarization.
❑ In particular, the helical antenna is characterized by two possible operation modalities,
depending on the geometrical parameters of the wire:
o Normal mode
o Axial mode
40
Helical antenna
•
Normal mode: The normal mode is achieved when 𝐶 ≪ 𝜆 and 𝐿𝑢 ≪ 𝜆 (i.e., the helix is
“small” compared to the wavelength).
•
In the normal mode, the antenna can be thought as composed by N elementary dipoles and
N elementary loops (as schematically shown in the figure).
𝐷
𝐷
𝑠
Equivalent
model
𝑠
41
Helical antenna
❑ The radiated field is a combination of the fields due to elementary dipoles and loops.
❑ In particular, since the dimensions are small, the current can be assumed constant, and the
far field can be supposed to be independent from the number of loops and dipoles.
❑ Consequently, the field can be written as (see previous lectures):
𝑒 −𝑗𝑘𝑟
𝐷
2
෡
𝐄 = 𝑗𝑘𝜂𝐼𝑆
sin 𝜃 𝛉 + 𝜂𝑘 𝜋
4𝜋𝑟
2
Hertzian dipole field
2
𝑒 −𝑗𝑘𝑟
෡
𝐼
sin 𝜃 𝛟
4𝜋𝑟
Elementary loop field
❑ The radiation pattern is consequently omnidirectional, with maximum radiation direction
perpendicular to the helical axis.
42
Helical antenna
❑ Example: Helical antenna composed by
Horizontal
plane
𝑁 = 5 turns with 𝐷 = 0.02𝜆 and 𝑠 = 0.01𝜆.
𝜙
Vertical
plane
𝜃
43
Helical antenna
❑ Effect of the ground plane radius.
𝑅𝑔 = 0.1𝜆
𝑅𝑔 = 0.5𝜆
44
Helical antenna
❑ The field has generally an elliptical polarization with axial ratio given by
AR =
𝐸𝜃
𝐸𝜙
=
1
𝑘𝜂 𝐼 𝑆 4𝜋𝑟 sin 𝜃
2
𝐷
1
𝜂𝑘 2 𝜋 2
𝐼 4𝜋𝑟 sin 𝜃
4𝑆
2𝜆𝑆
=
=
𝜋𝑘𝐷2
𝜋𝐷 2
❑ When 𝐴𝑅 → 0 (i.e., 𝐸𝜃 = 0), we obtain a horizontal linear polarization (helix becomes a loop).
❑ When 𝐴𝑅 → +∞ (i.e., 𝐸𝜙 = 0 ), we obtain a vertical linear polarization (helix becomes a
monopole).
❑ When 𝐴𝑅 = 1 a circular polarization is obtained. Such condition is obtained when
2𝜆𝑆
= 1 → 𝐶 = 𝜋𝐷 = 2𝑆𝜆
𝜋𝐷 2
45
Helical antenna
❑ Axial mode: the axial mode is obtained when the size of the antenna is comparable or
greater than 𝜆.
❑ In particular, when
3 𝐶 4
< <
4 𝜆 3
𝜆
𝑆≅
4
𝐶
≅ 1 optimal value
𝜆
or
12° ≤ 𝛼 ≤ 14°
❑ the radiation pattern has a maximum in the direction of the helical axis (directive antenna
with end-fire pattern) and the polarization is almost circular.
46
Helical antenna
❑ In such conditions, the normalized field pattern can be modeled as a combination (array) of loops
with non-uniform current, resulting in
𝑁
sin 2 𝜓
𝜋
𝐸𝑛 = sin
cos 𝜃
𝜓
2𝑁
sin 2
❑ where
𝐿0
𝐿0
𝜓 = 𝑘 𝑆 cos 𝜃 −
= 𝑘 𝑆 cos 𝜃 −
𝐿0 /𝜆0
𝑝
𝑆/𝜆0 + 1
❑ with 𝑝 =
= 𝑘 𝑆 cos 𝜃 − 𝑆 + 𝜆0
𝐿0 /𝜆0
(ordinary end-fire radiation).
𝑆/𝜆0 +1
47
Helical antenna
❑ Example: Helical antenna with
𝑁 = 10 turns, C = 𝜆 and 𝑠 = 𝜆/4.
Vertical
plane
𝜃
48
Helical antenna
❑ The antenna parameters in the axial mode can be obtained by some empirical formulas:
𝐶
𝑅𝑖𝑛 = 140
𝜆
Δ𝜃𝑏 =
3
52𝜆2
𝐶 𝑁𝑆
𝐶2𝑆
𝐷𝑚𝑎𝑥 = 15𝑁 3
𝜆
❑ Moreover, the axial ratio given by
2𝑁 + 1
𝐴𝑅 =
2𝑁
Almost circular, for 𝑁 ≫ 1
49
Helical antenna
❑ A design example – Consider a 10-turn helix antenna. Determine the circumference, pitch
angle, and separation between turn to operate in axial mode at 1 GHz. Calculate the
resulting input resistance, beamwidth, directivity, and axial ratio.
𝑐
3×108 m/s
❑ The wavelength is 𝜆 = =
= 0.3 𝑚.
𝑓
1×109 Hz
❑ For an optimal design, we have 𝐶 = 𝜆 = 0.3 m and 𝑆 =
𝜆
−1 4
❑ The corresponding pitch angle is 𝛼 = tan
𝜆
= 0.075 m.
4
1
𝜆
= tan−1 = 14°.
4
❑ The resulting directivity is
𝐷𝑚𝑎𝑥 = 15𝑁
𝐶2𝑆
𝜆3
𝜆2
= 150
𝜆
4
𝜆3
150
=
≅ 37.5 → 15.74 dB
4
50
Helical antenna
❑ The beamwidth is
Δ𝜃𝑏 = 52
3
𝜆2
𝐶 𝑁𝑆
=
52
1 × 10/4
≅ 32.9°
❑ The axial ratio is equal to
2𝑁 + 1 21
𝐴𝑅 =
=
= 1.05
2𝑁
20
51
Master degree in Internet and Multimedia Engineering / Electronic Engineering
Antennas / Advanced Antenna Engineering
Slides of the lectures – Part 3
Aperture antennas
❑ Several antennas used in practical applications can be modeled as “aperture antennas”.
❑ Some examples are the following:
o Open waveguides
o Horn antennas
o Slot antennas
o Dish antennas
Guided field
(𝐄𝐠 , 𝐇𝐠 )
Aperture
❑ As the name said, in aperture
antennas it is possible to define
an aperture (real or virtual) from
which the field is radiated.
Radiated field
(𝐄, 𝐇)
2
Aperture antennas
An example of an antenna
with a real aperture
An example of an antenna
with a virtual aperture
3
Introduction to apertures
❑ In order to characterize the radiation from such antennas it is necessary to understand the
radiation from a generic aperture, which is characterized by a region 𝐴 (the aperture)
surrounded by a “screen”, which in general could be made by a dielectric material, a perfect
electric conductor (PEC) or perfect magnetic conductor (PMC).
4
Introduction to apertures
❑ The radiation from this structure can be expressed in terms of the field on the aperture.
❑ Let us denote with 𝐄𝑎 and 𝐇𝑎 the electric and magnetic field vectors on the aperture 𝐴.
❑ Such vectors are assumed to be
known and we want to determine
the radiated fields 𝐄 𝐫 e 𝐇 𝐫 .
❑ To this end, we can use the
Fields on the
aperture
equivalence principle, i.e., the
field on the aperture is substituted
with a set of currents that radiates
the same field.
5
Introduction to apertures
❑ In particular, when dealing with a dielectric screen, we can define the following equivalent
current densities:
electric current density
magnetic current density
normal to the
aperture
ෝ 𝐫 × 𝐇𝑎 𝐫
𝐉𝑠 𝐫 = 𝐧
ቊ
,𝐫 ∈ 𝐴
ෝ 𝐫 × 𝐄𝑎 𝐫
𝐉𝑚𝑠 𝐫 = −𝐧
Aperture
fields
❑ The screen is assumed to be an infinite surface over which the tangential field is approximated as
null.
❑ Such assumption is not always true → In order to compute correctly the radiated fields we should
know the fields also on the screen.
❑ However, this is usually not possible (since it depends itself on the field over the aperture) → we
neglect the contribution of the fields on the screen.
❑ Such approximation provides good results when the aperture is big (especially for what concerns
the main lobe).
6
Introduction to apertures
❑ Similar equivalent problems can be defined when considering PEC or PMC screens, i.e.,
𝐉𝑠 𝐫 = 0
൝
ෝ 𝐫 × 𝐄𝑎 𝐫
𝐉𝑚𝑠 𝐫 = −2 𝐧
ෝ 𝐫 × 𝐇𝑎 𝐫
𝐉 𝐫 =2 𝐧
൝𝑠
𝐉𝑚𝑠 𝐫 = 0
,
,
𝐫∈𝐴
𝐫∈𝐴
→
→
PEC screen
PMC screen
7
Radiation from magnetic sources
❑ In order to characterize the radiation from aperture antennas, it is necessary first to
understand how magnetic sources radiate.
❑ Maxwell equations with magnetic sources become
1
∇∙𝐄 𝐫 = 𝜌 𝐫
𝜖
1
∇ ∙ 𝐇 𝐫 = 𝜌𝑚 𝐫
𝜇
∇ × 𝐇 𝐫 = 𝐉 𝐫 + 𝑗𝜔𝜖𝐄 𝐫
∇ × 𝐄 𝐫 = −𝐉𝑚 𝐫 − 𝑗𝜔𝜇𝐇 𝐫
magnetic charge density
magnetic current density
8
Radiation from magnetic sources
❑ In particular, since the Maxwell equations are linear with respect to the sources, we can split
the problem into two sub-problems
1
∇∙𝐄 𝐫 = 𝜌 𝐫
𝜖
∇∙𝐇 𝐫 = 0
∇ × 𝐇 𝐫 = 𝐉 𝐫 + 𝑗𝜔𝜖𝐄 𝐫
∇ × 𝐄 𝐫 = −𝑗𝜔𝜇𝐇 𝐫
∇∙𝐄 𝐫 =0
1
∇ ∙ 𝐇 𝐫 = 𝜌𝑚 𝐫
𝜇
∇ × 𝐇 𝐫 = 𝑗𝜔𝜖𝐄 𝐫
∇ × 𝐄 𝐫 = −𝐉𝑚 𝐫 − 𝑗𝜔𝜇𝐇 𝐫
𝜑 𝐫 , 𝐀 𝐫 → 𝐄𝑒 𝐫
𝜑𝑚 𝐫 , 𝐀𝑚 𝐫 → 𝐄𝑚 𝐫
Full EM problem
𝐄 𝐫 = 𝐄𝑒 𝐫 + 𝐄𝑚 𝐫
❑ The solution of the two problems can be then expressed in terms of a linear combination of
the standard potentials 𝜑 𝐫 e 𝐀 𝐫 and of two magnetic potentials 𝜑𝑚 𝐫 e 𝐀𝑚 𝐫 .
9
Radiation from magnetic sources
It is possible to «convert» the solution for
the electric currents into a solution for
magnetic currents with the substitutions:
❑ Duality principle
Electric sources (A)
∇2 𝐀 𝐫 + 𝑘 2 𝐀 𝐫 = −𝜇𝐉 𝐫
′
Magnetic sources (B)
∇2 𝐀𝑚 𝐫 + 𝑘 2 𝐀𝑚 𝐫 = −𝜖𝐉𝑚 𝐫
′
𝑗𝑘 𝐫−𝐫
𝜇
𝑒
𝐀 𝐫 =
න 𝐉 𝐫′
𝑑𝐫 ′
′
4𝜋 𝑉0
𝐫−𝐫
𝑗𝑘 𝐫−𝐫
𝜖
𝑒
𝐀𝑚 𝐫 =
න 𝐉𝑚 𝐫 ′
𝑑𝐫 ′
′
4𝜋 𝑉0
𝐫−𝐫
𝜇𝑒 −𝑗𝑘𝑟
𝜇𝑒 −𝑗𝑘𝑟
′
′
𝑗𝑘ො
𝐫
∙𝐫
′
≅
න𝐉 𝐫 𝑒
𝑑𝐫 =
𝐅 𝜃, 𝜙
4𝜋𝑟
4𝜋𝑟
𝜖𝑒 −𝑗𝑘𝑟
𝜖𝑒 −𝑗𝑘𝑟
ො ′
′
𝑗𝑘
𝐫∙𝐫
′
න 𝐉𝑚 𝐫 𝑒
≅
𝑑𝐫 =
𝐅 𝜃, 𝜙
4𝜋𝑟
4𝜋𝑟 𝑚
1
𝐇 𝐫 = ∇×𝐀 𝐫
𝜇
1
𝐄 𝐫 = − ∇ × 𝐀𝑚 𝐫
𝜖
𝑉
𝐄 𝐫 = −𝑗𝜔𝐀 𝐫 − 𝑗
∇ ∇∙𝐀 𝐫
𝜔𝜇𝜖
𝑉
𝐇 𝐫 = −𝑗𝜔𝐀𝑚 𝐫 − 𝑗
∇ ∇ ∙ 𝐀𝑚 𝐫
𝜔𝜇𝜖
(A)
𝐄
→
(B)
𝐇
𝐇
→
−𝐄
𝐉
→
𝐉𝑚
𝐀
→
𝐀𝑚
𝜖
→
𝜇
𝜇
→
𝜖
𝑘
→
𝑘
𝜂
→
1
𝜂
1
𝜂
→
𝜂
10
Radiation from magnetic sources
❑ Considering that electric currents produce the following radiated fields:
𝑒 −𝑗𝑘𝑟
𝑒 −𝑗𝑘𝑟
෡ + 𝐹𝜙 𝜃, 𝜙 𝛟
෡ = −𝑗𝑘𝜂
𝐄𝑒 𝐫 = −𝑗𝑘𝜂
𝐹𝜃 𝜃, 𝜙 𝛉
𝐫ො × 𝐅 𝜃, 𝜙 × 𝐫ො
4𝜋𝑟
4𝜋𝑟
𝑒 −𝑗𝑘𝑟
𝑒 −𝑗𝑘𝑟
෡ + 𝐹𝜃 𝜃, 𝜙 𝛟
෡ = −𝑗𝑘
𝐇𝑒 𝐫 = −𝑗𝑘
−𝐹𝜙 𝜃, 𝜙 𝛉
𝐫ො × 𝐅 𝜃, 𝜙
4𝜋𝑟
4𝜋𝑟
❑ By means of the duality principle, we can directly find that the contribution of the magnetic
currents is
𝑒 −𝑗𝑘𝑟
𝑒 −𝑗𝑘𝑟
෡ + 𝐹𝑚 𝜃, 𝜙 𝛟
෡ = 𝑗𝑘
𝐄𝑚 𝐫 = 𝑗𝑘
−𝐹𝑚𝜙 𝜃, 𝜙 𝛉
𝐫ො × 𝐅𝑚 𝜃, 𝜙
𝜃
4𝜋𝑟
4𝜋𝑟
𝑘 𝑒 −𝑗𝑘𝑟
𝑘 𝑒 −𝑗𝑘𝑟
෡ + 𝐹𝑚 𝜃, 𝜙 𝛟
෡ = −𝑗
𝐇𝑚 𝐫 = −𝑗
𝐹𝑚𝜃 𝜃, 𝜙 𝛉
𝐫ො × 𝐅𝑚 𝜃, 𝜙 × 𝐫ො
𝜙
𝜂 4𝜋𝑟
𝜂 4𝜋𝑟
11
Radiation from apertures
❑ Let us consider now to the specific case of planar
apertures that lie on the x-y plane with a PEC screen.
❑ Only the magnetic radiation vector is required (since
𝐉𝑠 𝐫 = 0 and consequently 𝐅 𝜃, 𝜙 = 0), and it can be
obtained by integrating the equivalent source on the
aperture 𝐴, i.e.,
𝐅𝑚 𝜃, 𝜙 = න 𝐉𝑚𝑠
′
′
𝑗𝑘ො
𝐫
∙𝐫
𝐫 𝑒
𝑑𝐫 ′
𝐴
′
= −2ො𝐳 × න 𝐄𝑎 𝐫 ′ 𝑒 𝑗𝑘ො𝐫∙𝐫 𝑑𝑥 ′ 𝑑𝑦 ′
𝐴
12
Radiation from apertures
❑ Since
𝑘ො𝐫 ∙ 𝐫 ′ = 𝑘 𝐱ො sin 𝜃 cos 𝜙 + 𝐲ො sin 𝜃 sin 𝜙 + 𝐳ො cos 𝜃 ∙ 𝑥 ′ 𝐱ො + 𝑦 ′ 𝐲ො + 0ො𝐳
= 𝑘 sin 𝜃 cos 𝜙 𝑥 ′ + 𝑘 sin 𝜃 sin 𝜙 𝑦 ′ = 𝑘𝑥 𝑥 ′ + 𝑘𝑦 𝑦 ′
𝑘𝑥
𝑘𝑦
2D Fourier transform
of the electric field
❑ we obtain:
𝐅𝑚 𝜃, 𝜙 = −2ො𝐳 × න 𝐄𝑎
𝑗 𝑘𝑥 𝑥 ′ +𝑘𝑦 𝑦 ′
′
𝐫 𝑒
𝑑𝑥 ′ 𝑑𝑦 ′ = −2ො𝐳 × 𝐟 𝜃, 𝜙
𝐴
❑ where 𝑘𝑥 = 𝑘 cos 𝜙 sin 𝜃 and 𝑘𝑦 = 𝑘 sin 𝜙 sin 𝜃.
13
Radiation from apertures
ො , we can just consider the tangential
❑ Since we have a cross product with the unit vector 𝐳
components of aperture fields (because 𝐳ො × 𝐳ො = 𝟎) → the z component of 𝐄𝑎 and 𝐟 is
neglected in the following.
❑ Consequently, the magnetic radiation vector becomes :
𝐅𝑚 𝜃, 𝜙 = −2ො𝐳 × 𝐟 𝜃, 𝜙 = −2ො𝐳 × 𝑓 𝜃, 𝜙 𝐱ො + 𝑓𝑦 𝜃, 𝜙 𝐲ො = −2 𝑓𝑥 𝜃, 𝜙 𝐲ො − 𝑓𝑦 𝜃, 𝜙 𝐱ො
෡ cos 𝜙
෡ cos 𝜃 sin 𝜙 + 𝝓
= −2𝑓𝑥 𝜃, 𝜙 𝐫ො sin 𝜃 sin 𝜙 + 𝜽
𝐲ො
+ 2𝑓𝑦 𝜃, 𝜙
෡ sin 𝜙
෡ cos 𝜃 cos 𝜙 − 𝝓
𝐫ො sin 𝜃 cos 𝜙 + 𝜽
𝐱ො
෡
෡ − 2 𝑓𝑥 𝜃, 𝜙 cos 𝜙 + 𝑓𝑦 𝜃, 𝜙 sin 𝜙 𝝓
= … 𝐫ො − 2 𝑓𝑥 𝜃, 𝜙 cos 𝜃 sin 𝜙 − 𝑓𝑦 𝜃, 𝜙 cos 𝜃 cos 𝜙 𝜽
෡
෡ − 2 𝑓𝑥 𝜃, 𝜙 cos 𝜙 + 𝑓𝑦 𝜃, 𝜙 sin 𝜙 𝝓
= … 𝐫ො + 2 cos 𝜃 𝑓𝑦 𝜃, 𝜙 cos 𝜙 − 𝑓𝑥 𝜃, 𝜙 sin 𝜙 𝜽
14
Radiation from apertures
❑ Since
𝑒 −𝑗𝑘𝑟
𝑒 −𝑗𝑘𝑟
෡ + 𝐹𝑚 𝜃, 𝜙 𝛟
෡
𝐄 𝐫 = 𝐄𝑚 𝐫 = 𝑗𝑘
𝐫ො × 𝐅𝑚 𝜃, 𝜙 = 𝑗𝑘
−𝐹𝑚𝜙 𝜃, 𝜙 𝛉
𝜃
4𝜋𝑟
4𝜋𝑟
❑ the 𝜃 and 𝜙 components of the electric field are finally equal to
𝑒 −𝑗𝑘𝑟
𝐸𝜃 𝐫 = 𝑗𝑘
𝑓𝑥 𝜃, 𝜙 cos 𝜙 + 𝑓𝑦 𝜃, 𝜙 sin 𝜙
2𝜋𝑟
𝑒 −𝑗𝑘𝑟
𝐸𝜙 𝐫 = 𝑗𝑘
cos 𝜃 𝑓𝑦 𝜃, 𝜙 cos 𝜙 − 𝑓𝑥 𝜃, 𝜙 sin 𝜙
2𝜋𝑟
15
Radiation from apertures
❑ The previous relationship can be extended to the cases in which we have a dielectric screen.
❑ In particular, let us consider that the electric and magnetic fields forms a so-called Huygens
source, i.e., they satisfy the condition:
1
𝐇𝑎 = 𝒛ො × 𝐄𝑎
𝜂
❑ where 𝜂 is the intrinsic impedance of the propagation medium.
❑ In this case it results
𝐅 𝜃, 𝜙 = න 𝒛ො × 𝐇𝑎
𝐴
′
𝐫 ′ 𝑒 𝑗𝑘𝐫ො∙𝐫 𝑑𝐫 ′ = න 𝒛ො ×
𝐴
1
𝒛ො × 𝐄𝑎 𝐫 ′
𝜂
′
𝑒 𝑗𝑘𝐫ො∙𝐫 𝑑𝐫 ′ = −
1
𝒛ො × 𝐅𝑚 𝜃, 𝜙
𝜂
16
Radiation from apertures
❑ Consequently, only 𝐅𝑚 is required (as in the PEC screen case).
❑ After some mathematical passages, it can be found that the 𝜃 and 𝜙 components of the
radiated field are given by:
𝑒 −𝑗𝑘𝑟 1 + cos 𝜃
𝐸𝜃 𝐫 = 𝑗𝑘
𝑓𝑥 𝜃, 𝜙 cos 𝜙 + 𝑓𝑦 𝜃, 𝜙 sin 𝜙
2𝜋𝑟
2
𝑒 −𝑗𝑘𝑟 1 + cos 𝜃
𝐸𝜙 𝐫 = 𝑗𝑘
𝑓𝑦 𝜃, 𝜙 cos 𝜙 − 𝑓𝑥 𝜃, 𝜙 sin 𝜙
2𝜋𝑟
2
Same expressions as in
the PEC screen case.
17
Radiation from apertures
❑ If the fields on the aperture are not exactly a Huygens source, but they satisfy the following
conditions (generalized Huygens sources):
1
ෝ × 𝐄𝑎 𝐫
𝐇𝑎 𝐫 = 𝐧
𝜂𝑇
❑ with 𝜂 𝑇 ≠ 𝜂, the previous relationships become (setting 𝐾𝑐 =
𝜂
)
𝜂𝑇
𝑒 −𝑗𝑘𝑟 1 + 𝐾𝑐 cos 𝜃
𝐸𝜃 𝐫 = 𝑗𝑘
𝑓𝑥 𝜃, 𝜙 cos 𝜙 + 𝑓𝑦 𝜃, 𝜙 sin 𝜙
2𝜋𝑟
2
𝑒 −𝑗𝑘𝑟 𝐾𝑐 + cos 𝜃
𝐸𝜙 𝐫 = 𝑗𝑘
𝑓𝑦 𝜃, 𝜙 cos 𝜙 − 𝑓𝑥 𝜃, 𝜙 sin 𝜙
2𝜋𝑟
2
18
Radiation from apertures
❑ In general, for every considered case, the radiated field ican be expressed as
𝑒 −𝑗𝑘𝑟
𝐸𝜃 𝐫 = 𝑗𝑘
𝑐 𝜃 𝑓𝑥 𝜃, 𝜙 cos 𝜙 + 𝑓𝑦 𝜃, 𝜙 sin 𝜙
2𝜋𝑟 𝜃
𝑒 −𝑗𝑘𝑟
𝐸𝜙 𝐫 = 𝑗𝑘
𝑐 𝜃 𝑓𝑦 𝜃, 𝜙 cos 𝜙 − 𝑓𝑥 𝜃, 𝜙 sin 𝜙
2𝜋𝑟 𝜙
❑ where
o
𝑐𝜃 𝜃
𝑐𝜙 𝜃
=
o
𝑐𝜃 𝜃
𝑐𝜙 𝜃
= 2 1 + cos 𝜃 (dielectric screen with Huygens source)
1 + cos 𝜃
o
𝑐𝜃 𝜃
𝑐𝜙 𝜃
=2
1
(PEC screen)
cos 𝜃
1
1
1 + 𝐾𝑐 cos 𝜃
𝜂
, 𝐾𝑐 = 𝜂 (dielectric screen with generalized Huygens source)
𝐾𝑐 + cos 𝜃
𝑇
19
Parameters of apertures
❑ The radiation intensity is then given by:
1
1
2
2
2
𝑈 𝜃, 𝜙 = 𝑟
𝐄 𝐫 =𝑟
𝐸𝜃 𝐫 2 + 𝐸𝜙 𝐫
2𝜂
2𝜂
2
−𝑗𝑘𝑟
1
𝑒
= 𝑟 2 ൥ 𝑗𝑘
𝑐𝜃 𝑓𝑥 𝜃, 𝜙 cos 𝜙 + 𝑓𝑦 𝜃, 𝜙 sin 𝜙
2𝜂
2𝜋𝑟
2
−𝑗𝑘𝑟
+ 𝑗𝑘
𝑒
𝑐𝜙 𝑓𝑦 𝜃, 𝜙 cos 𝜙 − 𝑓𝑥 𝜃, 𝜙 sin 𝜙
2𝜋𝑟
2
൩
𝑘2
2
2
= 2 𝑐𝜃2 𝑓𝑥 𝜃, 𝜙 cos 𝜙 + 𝑓𝑦 𝜃, 𝜙 sin 𝜙 + 𝑐𝜙2 𝑓𝑦 𝜃, 𝜙 cos 𝜙 − 𝑓𝑥 𝜃, 𝜙 sin 𝜙
8𝜋 𝜂
1
2
2
= 2 𝑐𝜃2 𝑓𝑥 𝜃, 𝜙 cos 𝜙 + 𝑓𝑦 𝜃, 𝜙 sin 𝜙 + 𝑐𝜙2 𝑓𝑦 𝜃, 𝜙 cos 𝜙 − 𝑓𝑥 𝜃, 𝜙 sin 𝜙
2𝜆 𝜂
20
Parameters of apertures
❑ Let us consider as an example the case of a Huygens source (but similar considerations can
be made also for the other cases). The radiation intensity becomes
1
2
2
𝑈 𝜃, 𝜙 = 2 𝑐𝜃2 𝑓𝑥 𝜃, 𝜙 cos 𝜙 + 𝑓𝑦 𝜃, 𝜙 sin 𝜙 + 𝑐𝜙2 𝑓𝑦 𝜃, 𝜙 cos 𝜙 − 𝑓𝑥 𝜃, 𝜙 sin 𝜙
2𝜆 𝜂
2
1 1 + cos 𝜃
2
= 2
ቀ 𝑓𝑥 𝜃, 𝜙 2 cos 2 𝜙 + 𝑓𝑦 𝜃, 𝜙 sin2 𝜙 + 𝑓𝑥 𝜃, 𝜙 𝑓𝑦∗ 𝜃, 𝜙 sin 𝜙 cos 𝜙
2𝜆 𝜂
2
+ 𝑓𝑥∗ 𝜃, 𝜙 𝑓𝑦 𝜃, 𝜙 sin 𝜙 cos 𝜙 + 𝑓𝑦 𝜃, 𝜙
2
cos 2 𝜙 + 𝑓𝑥 𝜃, 𝜙
2 sin2 𝜙 − 𝑓 𝜃, 𝜙 𝑓 ∗ 𝜃, 𝜙
𝑦
𝑥
sin 𝜙 cos 𝜙
− 𝑓𝑦∗ 𝜃, 𝜙 𝑓𝑥 𝜃, 𝜙 sin 𝜙 cos 𝜙ቁ
1 1 + cos 𝜃
= 2
2𝜆 𝜂
2
1 1 + cos 𝜃
= 2
2𝜆 𝜂
2
2
𝑓𝑥 𝜃, 𝜙
2 cos 2 𝜙 + sin2 𝜙
2
𝑓𝑥 𝜃, 𝜙
2+
𝑓𝑦 𝜃, 𝜙
2
+ 𝑓𝑦 𝜃, 𝜙
2
cos 2 𝜙 + sin2 𝜙
1
1 + cos 𝜃
= 2
2𝜆 𝜂
2
2
𝐟 𝜃, 𝜙
2
21
Parameters of apertures
❑ Since usually, for aperture antennas, the maximum radiation is in the direction
perpendicular to the aperture (i.e., 𝜃 = 0), the maximum value of the radiation intensity is
then:
1
1
2
𝑈𝑚𝑎𝑥 = 2 𝐟 𝑚𝑎𝑥 = 2 𝑓𝑥 𝜃, 𝜙
2𝜆 𝜂
2𝜆 𝜂
2 + 𝑓 𝜃, 𝜙 2
𝑦
𝜃=0
❑ The radiation pattern for an aperture with Huygens source is then:
1 + cos 𝜃 2 𝐟 𝜃, 𝜙 2
𝑔 𝜃, 𝜙 =
2
𝐟 2𝑚𝑎𝑥
22
Parameters of apertures
❑ The radiated power is given by:
𝑈 𝜃, 𝜙
1
2
= 2 ቂ𝑐𝜃2 𝑓𝑥 𝜃, 𝜙 cos 𝜙 + 𝑓𝑦 𝜃, 𝜙 sin 𝜙
2𝜆 𝜂
1
1 + cos 𝜃
+ 𝑐𝜙2 𝑓𝑦 𝜃, 𝜙 cos 𝜙 − 𝑓𝑥 𝜃, 𝜙 sin 𝜙 ቃ →
2𝜆2 𝜂
2
𝜋
2 2𝜋
2
𝑃𝑟𝑎𝑑 = න න 𝑈 𝜃, 𝜙 𝑑Ω
0 0
2
𝐟 𝜃, 𝜙
2
For Huygens sources!
❑ where the integral along 𝜃 is limited to 0, 𝜋/2 since the radiation is assumed to be only in half
space 𝒛 ≥ 𝟎.
❑ The radiated power can be computed using the energy conservation principle. By assuming that
all the power on the aperture is radiated (ideal aperture antenna) and considering Huygens
sources we have:
1
1
∗
′
𝑃𝑟𝑎𝑑 = 𝑃𝑎 = න ℘𝑧 𝑑𝑆 = න 𝐳ො ∙ Re 𝐄𝑎 × 𝐇𝑎 𝑑𝑆 =
න 𝐄𝑎 2 𝑑𝑆 ′
2
2𝜂
′
𝐴
𝐴
𝐴
23
Parameters of apertures
❑ Moreover, considering that the maximum radiation is obtained for 𝜃 = 0 (and thus 𝑘𝑥 =
𝑘𝑦 = 0), the quantity 𝐟 2𝑚𝑎𝑥 can be expressed as:
2
′
2
′
𝐟 2𝑚𝑎𝑥 = න 𝐄𝑎 𝐫 ′ 𝑒 𝑗 𝑘𝑥𝑥 +𝑘𝑦 𝑦 𝑑𝑥 ′ 𝑑𝑦 ′
𝐴
= න 𝐄𝑎 𝐫 ′ 𝑑𝑆 ′
𝑘𝑥 =0,𝑘𝑦 =0
𝐴
❑ The maximum value of the radiation intensity is thus given by:
2
1
1
2
𝑈𝑚𝑎𝑥 = 2 𝐟 𝑚𝑎𝑥 = 2 න 𝐄𝑎 𝐫 ′ 𝑑𝑆 ′
2𝜆 𝜂
2𝜆 𝜂
𝐴
24
Parameters of apertures
❑ Consequently, it directly derives that the directivity is:
2
2
1
′
′
′
′
‫𝑆𝑑 𝐫 𝐄 ׬‬
𝑈𝑚𝑎𝑥
4𝜋 ‫𝑆𝑑 𝐫 𝑎𝐄 𝐴׬‬
4𝜋 𝑚
2𝜆2 𝜂 𝐴 𝑎
𝐷𝑚𝑎𝑥 = 4𝜋
= 4𝜋
= 2
= 2 𝐴𝑒𝑓𝑓
1
𝑃𝑟𝑎𝑑
𝜆
𝜆
2 𝑑𝑆 ′
2 𝑑𝑆 ′
𝐄
‫׬‬
𝐄
‫׬‬
𝑎
𝑎
𝐴
2𝜂 𝐴
❑ which, for an ideal aperture, is also equal to the gain 𝐺𝑎 = 4𝜋
𝑈𝑚𝑎𝑥
.
𝑃𝑎
❑ We define maximum effective area of the aperture the following quantity:
𝐴𝑚
𝑒𝑓𝑓 =
‫𝑎𝐄 𝐴׬‬
𝐫 ′ 𝑑𝑆 ′
‫ 𝑎𝐄 𝐴׬‬2 𝑑𝑆 ′
2
≤𝐴
25
Parameters of apertures
❑ The aperture efficiency is thus defined as:
𝐫 ′ 𝑑𝑆 ′
‫𝑎𝐄 𝐴׬‬
𝐴𝑚
𝑒𝑓𝑓
𝑒𝑎 =
=
𝐴
𝐴‫𝐄 ׬‬
𝐴
𝑎
2 𝑑𝑆 ′
2
=
𝐫′
𝑑𝑆 ′
𝐴 ‫𝑎𝐄 𝐴׬‬
2 𝑑𝑆 ′
‫𝑎𝐄 𝐴׬‬
2
𝑎𝑝𝑒𝑟𝑡𝑢𝑟𝑒 𝑡𝑎𝑝𝑒𝑟 𝑙𝑜𝑠𝑠
𝑒𝑎𝑡𝑙
‫𝑎𝐄 𝐴׬‬
𝐫 ′ 𝑑𝑆 ′
2
‫ 𝐫 𝑎𝐄 𝐴׬‬′ 𝑑𝑆 ′
2 ≤1
𝑝ℎ𝑎𝑠𝑒 𝑒𝑟𝑟𝑜𝑟 𝑙𝑜𝑠𝑠
𝑒𝑝𝑒𝑙
❑ The maximum efficiency (𝑒𝑎 = 1) is obtained for uniform apertures (for which 𝐄𝑎 is constant
on the aperture). In this case, the effective area is equal to the real area of the aperture.
26
Parameters of apertures
❑ Aperture field tapering
No tapering →
Higher efficiency
|𝐄𝑎 |
Low tapering
Maximum tapering
→ lower efficiency
𝑥
27
Open-waveguide antenna
Input connector
❑ Open waveguide antennas are one of the simplest aperture antennas.
❑ The tangential field on the aperture is given by the transverse
components of the field that propagate inside a waveguide.
𝑦
Metallic walls
Aperture
Aperture
𝑥
Guided field
(𝐄𝐠 , 𝐇𝐠 )
𝑏
𝑧
𝐄𝑎 , 𝐇𝑎
𝑎
28
Open-waveguide antenna
❑ By assuming that only the fundamental mode is present, we can model the aperture fields
(transverse component) as:
𝐄 𝐫 = 𝐸𝑦 𝑥 𝐲ො = 𝐸0 cos
𝜋𝑥
𝐲ො
𝑎
❑ where
𝜂 𝑇𝐸 =
𝐇 𝐫 = 𝐻𝑥 𝑥 𝐱ො = −
𝜂
𝜆 2
1 − 2𝑎
𝐸0
𝜋𝑥
cos
𝐱ො
𝜂 𝑇𝐸
𝑎
Generalized
Huygens source
❑ This condition can be achieved if we work between the first and second cutoff frequencies, i.e.,
when the frequency is
𝑐
𝑐 𝑐
𝑓∈
, min ,
2𝑎
𝑎 2𝑏
29
Open-waveguide antenna
❑ The radiated field is then (since 𝑓𝑥 = 0):
𝑒 −𝑗𝑘𝑟
𝑒 −𝑗𝑘𝑟
𝐸𝜃 = 𝑗𝑘
𝑐 𝜃 𝑓𝑦 𝜃, 𝜙 sin 𝜙 = 𝑗
𝑐 𝜃 𝑓𝑦 𝜃, 𝜙 sin 𝜙
2𝜋𝑟 𝜃
𝜆𝑟 𝜃
𝑒 −𝑗𝑘𝑟
𝑒 −𝑗𝑘𝑟
𝐸𝜙 = 𝑗𝑘
𝑐 𝜃 𝑓𝑦 𝜃, 𝜙 cos 𝜙 = 𝑗
𝑐 𝜃 𝑓𝑦 𝜃, 𝜙 cos 𝜙
2𝜋𝑟 𝜙
𝜆𝑟 𝜙
❑ where
Windowed cosine cos
𝑎
2
𝑏
2
′
𝑓𝑦 𝜃, 𝜙 = න න 𝐸𝑦 𝑥 𝑒
−
𝑎
𝑏
−
2
2
𝑗 𝑘𝑥 𝑥 ′ +𝑘𝑦 𝑦 ′
𝑘𝑦 𝑏
𝑘𝑥 𝑎
cos
sin
2𝑎
2
2
= 𝐸0
𝑏
2
𝑘𝑦 𝑏
𝜋
𝑘𝑥 𝑎
1− 𝜋
2
𝑎
2
𝜋𝑥
𝑑𝑥 𝑑𝑦 = 𝐸0 න cos
𝑎
𝑎
−
′
′
2
′
𝑒
𝜋𝑥 ′
𝑎
𝑥
𝑟𝑒𝑐𝑡 𝑎
𝑗𝑘𝑥 𝑥 ′
′
𝑏
2
𝑑𝑥 න 𝑒
−
𝑏
2
𝑗𝑘𝑦 𝑦 ′
𝑑𝑦 ′
𝑦
Rectangle function 𝑟𝑒𝑐𝑡 𝑏
30
Open-waveguide antenna
❑ In order to obtain a compact and more general representation, the following auxiliary
quantities are introduced
2𝜋
𝑘𝑥 𝑎 𝑎
𝑘𝑥 = 𝑘 sin 𝜃 cos 𝜙 =
sin 𝜃 cos 𝜙 → 𝑣𝑥 =
= sin 𝜃 cos 𝜙
𝜆
2𝜋
𝜆
𝑘𝑦 𝑏 𝑏
2𝜋
𝑘𝑦 = 𝑘 sin 𝜃 sin 𝜙 =
sin 𝜃 sin 𝜙 → 𝑣𝑦 =
= sin 𝜃 sin 𝜙
𝜆
2𝜋
𝜆
❑ Consequently, the Fourier transform of the electric field becomes
𝑘𝑦 𝑏
𝑘𝑥 𝑎
sin 𝜋
2𝑎 cos 𝜋 2𝜋
2𝑎𝑏 cos 𝜋𝜈𝑥
2𝜋
𝑓𝑦 𝜃, 𝜙 = 𝐸0
= 𝐸0
2𝑏
2 sinc 𝜈𝑦
𝑘
𝑏
𝜋
𝜋
1
−
4𝜈
𝑦
𝑘 𝑎
𝑥
𝜋
1−4 𝑥
2𝜋
2𝜋
31
Open-waveguide antenna
❑ From the previous field it is possible to calculate the radiation pattern. However, it is simpler to
calculate separately the graphs on the cut planes. In particular, the two most significant cut planes
for this antenna are the x-z (for which 𝜙 = 0) and y-z (for which 𝜙 = 90°) planes.
❑ Let us consider the x-z plane, which in this case is the H plane.
𝐄 𝜃, 0 2
𝑔𝐻 𝜃 = 𝑔 𝜃, 0 =
=
𝐄 𝜃, 0 2𝑚𝑎𝑥
2
+ 𝐸𝜙 𝜃, 0
2+
𝐸𝜙 𝜃, 0
𝐸𝜃 𝜃, 0
𝐸𝜃 𝜃, 0
2
2
𝑚𝑎𝑥
❑ Since 𝜙 = 0, we directly have
𝑒 −𝑗𝑘𝑟
𝐸𝜃 𝐫 = 𝑗
𝑐 𝜃 𝑓𝑦 𝜃, 𝜙 sin 𝜙 = 0
𝜆𝑟 𝜃
𝑒 −𝑗𝑘𝑟
𝑒 −𝑗𝑘𝑟
𝐸𝜙 𝐫 = 𝑗
𝑐 𝜃 𝑓𝑦 𝜃, 𝜙 cos 𝜙 = 𝑗
𝑐 𝜃 𝑓𝑦 𝜃, 𝜙
𝜆𝑟 𝜙
𝜆𝑟 𝜙
32
Open-waveguide antenna
❑ Consequently
𝑔𝐻 𝜃 =
2
+ 𝐸𝜙 𝜃, 0
2+
2
𝐸𝜃 𝜃, 0
𝐸𝜃 𝜃, 0
𝐾𝑐 + cos 𝜃
=
𝐾𝑐 + 1
2
𝐸0
𝐸𝜙 𝜃, 0
2
=
𝑚𝑎𝑥
2𝑎𝑏 cos 𝜋𝜈𝑥
sinc 𝜈𝑦
𝜋 1 − 4𝜈𝑥2
2𝑎𝑏 2
𝐸0
𝜋
❑ where we used that 𝑣𝑦 =
2
𝐸𝜙 𝜃, 0
𝐸𝜙 𝜃, 0
2
2
𝑐𝜙 𝜃 𝑓𝑦 𝜃, 0
=
𝑐𝜙 𝜃 𝑓𝑦 𝜃, 0
𝑚𝑎𝑥
2
𝑚𝑎𝑥
2
𝑐𝜙2 𝜃 𝑓𝑦 𝜃, 0
= 2
𝑐𝜙 0 𝑓𝑦 0,0 2
2
𝐾𝑐 + cos 𝜃
=
𝐾𝑐 + 1
2
cos 𝜋𝜈𝑥
1 − 4𝜈𝑥2
2
= 𝑐𝐻2 𝜃
cos 𝜋𝜈𝑥
1 − 4𝜈𝑥2
2
𝑏
𝑎
𝑎
sin 𝜃 sin 𝜙 = 0 and 𝑣𝑥 = sin 𝜃 cos 𝜙 = sin 𝜃.
𝜆
𝜆
𝜆
33
Open-waveguide antenna
❑ Similarly, for the E plane, which correspond to the y-z plane (𝜙 = 90°), we obtain
𝐸𝜃 𝜃 2
1 + 𝐾𝑐 cos 𝜃
𝑔𝐸 𝜃 =
=
1 + 𝐾𝑐
𝐸𝜃 𝜃 2𝑚𝑎𝑥
❑ with 𝑣𝑥 = 0 and 𝑣𝑦 =
2
sinc 2 𝜈𝑦 = 𝑐𝐸2 𝜃 sinc 2 𝜈𝑦
𝑏
sin 𝜃.
𝜆
❑ The beamwidths can be obtained from the previous relationships and are given by:
𝜆
𝜆
Δ𝜃𝐻 = 1.189 = 68.12°
𝑎
𝑎
𝜆
𝜆
Δ𝜃𝐸 = 0.886 = 50.76°
𝑏
𝑏
❑ where we exploited that sinc 2 𝑣𝑦 = 1/2 when 𝑣𝑦 = 0.443.
34
Open-waveguide antenna
❑ The gain of the antenna is:
4𝜋
4𝜋
𝐺𝑎 = 2 𝐴𝑒𝑓𝑓 ≅ 2 𝑎𝑏𝑒𝑎
𝜆
𝜆
❑ For this antenna, the aperture efficiency can be analytically computed and is equal to
8
𝑒𝑎 = 2 ≅ 0.81
𝜋
❑ Consequently
4𝜋
𝐺𝑎 = 2 0.81𝑎𝑏
𝜆
35
Open-waveguide antenna
❑ Indeed, the amplitude of the field exhibits a significant tapering effect along the x direction,
although it is constant along the y direction.
❑ The phase is instead constant, and consequently for this case 𝑒𝑝𝑒𝑙 = 1 and 𝑒𝑎 = 𝑒𝑎𝑡𝑙 .
36
Open-waveguide antenna
❑ Example: Calculate the gain and beamwidth of a WR-90 open waveguide (𝑎 = 2.282 𝑐𝑚 e 𝑏
= 1.016 𝑐𝑚) working at 10 GHz.
❑ The gain is:
4𝜋
4𝜋
𝐺𝑎 = 0.81 2 𝑎𝑏 = 0.81
2.282 10−2 1.016 10−2 = 2.6 → 4.2 𝑑𝐵
−2
2
𝜆
3 10
❑ The beamwidths in the relevant planes are
𝜆
Δ𝜃𝐻 = 68.12° ≅ 68°
𝑎
𝜆
Δ𝜃𝐸 = 50.76° ≅ 114°
𝑏
❑ The working frequency band is 𝑓 ∈ 6.6, min 13.1,14.8
GHz = 6.6,13.1 GHz
37
Open-waveguide antenna
❑ The radiation pattern has the following shape.
3D pattern
E plane
(y-z)
𝜃
H plane
(x-z)
𝜃
38
Horn antennas
❑ Horn antennas are obtained by flaring out the end of a waveguide
❑ In this way, since the area of the aperture is higher, the gain can be significantly increased.
39
Horn antennas
❑ There are various types of horn antennas. The main ones are exemplified in the following
figure.
Pyramidal Horn antenna
Sectorial (E plane) Horn antenna
Sectorial (H plane) Horn antenna
Conical Horn antenna
40
Horn antennas
❑ In the following we will consider the case of the pyramidal horn antenna.
𝐿𝑎
𝑦
𝛼
𝑎
𝑥
𝐴
𝑅𝑎
𝑎
𝑏
𝑅𝐴
𝑧
𝐵
𝐿𝑏
𝐴
𝛽
𝑏
𝑅𝑏
𝑅𝐵
𝐵
41
Horn antennas
❑ Let us consider the horizontal section. We can derive some relationship among the
geometrical characteristics of the antenna.
❑ Let us first consider the two triangles in the figure. Since they share the lower angle, we have
𝑅𝑎
𝑅𝐴
𝐴
=
→ 𝑅𝑎 =
𝑅
𝐴
𝐴−𝑎
𝐴−𝑎 𝐴
2
2
𝐿𝑎
𝛼
𝑎
❑ Moreover,
𝐴
𝑅𝑎
𝐴
2
2
𝐿𝑎 = 𝑅𝑎 +
2
2
2
𝐴
= 𝑅𝑎2 +
4
𝑅𝐴
42
Horn antennas
❑ An important parameter is the length Δ𝑎 , which can be obtained as
Δ𝑎 = 𝐿𝑎 − 𝑅𝑎 =
𝐴2
𝐴2
2
𝑅𝑎 +
− 𝑅𝑎 = 𝑅𝑎 1 + 2 − 𝑅𝑎 ≅
ณ
4
4𝑅𝑎
𝐴2
4𝑅𝑎
𝑅𝑎
≪1
𝐴2
𝐴2
1 + 2 − 𝑅𝑎 =
8𝑅𝑎
8𝑅𝑎
𝑥
❑ More in general, the excess length at an arbitrary location 𝑥 along
𝐿𝑎
the aperture is expressed as
2
Δ𝑎 𝑥 =
𝑥
2𝑅𝑎
❑ Clearly, we have Δ𝑎 = Δ𝑎 𝐴/2 .
𝛼
𝑎
𝐴
𝑅𝑎
Δ𝑎 𝑥
Δ𝑎
43
Horn antennas
❑ Similarly, for the vertical plane we have:
𝐵
𝑅𝑏 =
𝑅
𝐵−𝑏 𝐵
𝑦
𝐿𝑏
2
𝐵
𝐿2𝑏 = 𝑅𝑏2 +
4
𝛽
𝑏
𝐵2
Δ𝑏 =
8𝑅𝑏
𝐵
𝑅𝐵
𝑅𝑏
Δ𝑏
𝑦2
Δ𝑏 𝑦 =
2𝑅𝑏
44
Horn antennas
❑ The fields on the aperture can be modeled by assuming that the field originating from the
waveguide spread radially up to the new aperture.
Δ𝑎 𝑥
ฑ
𝑥2
Δ𝑏 𝑦
ฑ
𝑦2
𝜋𝑥 −𝑗𝑘 2𝑅 −𝑗𝑘 2𝑅
𝑎 𝑒
𝑏
𝐸𝑦 𝑥, 𝑦 = 𝐸0 cos
𝑒
𝐴
𝑎
Waveguide transverse
field type term
𝐴
Spherical phase term
2
2
𝑦
𝑥
𝐸0
𝜋𝑥 −𝑗𝑘2𝑅
−𝑗𝑘2𝑅
𝑎𝑒
𝑏
𝐻𝑥 𝑥, 𝑦 = − cos
𝑒
𝜂
𝐴
45
Horn antennas
❑ The previous relationship can be rewritten in a more useful form by rearranging the
exponents of the exponential term. Let us for example consider the term in x. We can write
𝑥2
2
2𝑅𝑎 𝜋 2 2𝑥
= 𝜎𝑎
𝐴2
2
𝐴
8𝑅𝑎
𝑥2
2𝜋 Δ𝑎 𝑥
𝜋 Δ𝑎
𝑘
=
Δ𝑎 =
4
2𝑅
𝜆 Δ𝑎
2
𝜆
ต𝑎
Δ𝑎 𝑥
2
Δ
𝐴
𝑎
𝜎𝑎2 = 4
=
𝜆
2𝜆𝑅𝑎
❑ A similar relationship is obtained for the 𝑦 term, i.e.,
𝑦2
𝜋
2𝑦
𝑘
= 𝜎𝑏2
2𝑅
2
𝐵
ต𝑏
Δ𝑏 𝑦
2
2
Δ
𝐵
𝑏
𝜎𝑏2 = 4
=
𝜆
2𝜆𝑅𝑏
46
Horn antennas
❑ The field over the aperture can be thus rewritten as
𝜋𝑥 −𝑗𝜋𝜎𝑎2 2𝑥 2 −𝑗𝜋𝜎𝑏2 2𝑦 2
𝐸𝑦 𝑥, 𝑦 = 𝐸0 cos
𝑒 2 𝐴 𝑒 2 𝐵
𝐴
2
𝜋 2 2𝑦 2
𝐸0
𝜋𝑥 −𝑗𝜋2𝜎𝑎2 2𝑥
−𝑗
𝐴 𝑒 2 𝜎𝑏 𝐵
𝐻𝑥 𝑥, 𝑦 = − cos
𝑒
𝜂
𝐴
❑ Some notes:
o The field at the end of the waveguide is 𝐸𝑦 = 𝐸0 cos
𝜋𝑥
𝑎
and it changes gradually to the previous form.
o The field on the aperture is a Huygens source
o The quantities 𝜎𝑎 and 𝜎𝑏 defines how much the waveguide is flared out. Typical values are 𝜎𝑎 ≅ 1.22 and 𝜎𝑏 ≅ 1.
47
Horn antennas
❑ Similarly to the open waveguide case, the Fourier transform of the aperture fields has only y
component, which is given by:
𝐴
2
𝐵
2
𝑓𝑦 𝜃, 𝜙 = න න 𝐸𝑦 𝑥, 𝑦 𝑒 𝑗 𝑘𝑥𝑥+𝑘𝑦 𝑦 𝑑𝑥𝑑𝑦
𝐴
2
𝐵
2
𝐴 𝐵
−2 − 2
2
2
𝜋 2 2𝑦
𝜋𝑥 −𝑗𝜋2𝜎𝑎2 2𝑥
−𝑗
𝐴 𝑒 2 𝜎𝑏 𝐵 𝑒 𝑗 𝑘𝑥 𝑥+𝑘𝑦 𝑦 𝑑𝑥𝑑𝑦
= 𝐸0 න න cos
𝑒
𝐴
𝐴 𝐵
−2 − 2
𝐴
2
2
𝐵
2
2
𝜋 2 2𝑦
𝜋𝑥 −𝑗𝜋2𝜎𝑎2 2𝑥
−𝑗
𝑗𝑘
𝑥
𝐴 𝑒 𝑥 𝑑𝑥 න 𝑒 2 𝜎𝑏 𝐵 𝑒 𝑗𝑘𝑦 𝑦 𝑑𝑦
= 𝐸0 න cos
𝑒
𝐴
𝐴
−2
𝐵
−2
Known 1D Fourier transforms, that can be expressed in terms of combinations of Fresnel integrals.
48
Horn antennas
❑ We can thus write:
Combination of Fresnel integrals → known special function!
𝑓𝑦 𝜃, 𝜙 = 𝐸0
𝐴𝐵
𝐹 𝜈 ,𝜎 𝐹 𝜈 ,𝜎
4 1 𝑥 𝑎 0 𝑦 𝑏
❑ with 𝐹0 and 𝐹1 known functions and
𝐴
sin 𝜃 cos 𝜙
𝜆
𝐵
𝜈𝑦 = sin 𝜃 sin 𝜙
𝜆
𝜈𝑥 =
49
Horn antennas
❑ Using the formula for the radiation pattern of an aperture with a Huygens source we directly
find:
𝑔 𝜃, 𝜙 =
1 + cos 𝜃
2
2
𝑓𝑦 𝜃, 𝜙
𝑓𝑦
2
𝑚𝑎𝑥
2
=
1 + cos 𝜃
2
2
𝑓𝑦 𝜃, 𝜙
𝑓𝑦 0, 𝜙
2
2
2
𝐴𝐵
2
2 𝐸
2
0 4 𝐹1 𝜈𝑥 , 𝜎𝑎 𝐹0 𝜈𝑦 , 𝜎𝑏
𝐹1 𝜈𝑥 , 𝜎𝑎 𝐹0 𝜈𝑦 , 𝜎𝑏
1 + cos 𝜃
1 + cos 𝜃
=
2 =
2
2
𝐹1 0, 𝜎𝑎 𝐹0 0, 𝜎𝑏 2
𝐴𝐵
𝐸0
𝐹 0, 𝜎𝑎 𝐹0 0, 𝜎𝑏
4 1
50
Horn antennas
❑ On the H (x-z) plane, since 𝜙 = 0, 𝑣𝑥 =
𝐴
𝐴
𝐵
sin 𝜃 cos 𝜙 = sin 𝜃 and 𝜈𝑦 = sin 𝜃 sin 𝜙 = 0.
𝜆
𝜆
𝜆
Consequently, the radiation pattern reduces to :
1 + cos 𝜃
𝑔𝐻 𝜃, 𝜙 = 𝑔 𝜃, 0 =
2
2
𝐹1 𝜈𝑥 , 𝜎𝑎 𝐹0 0, 𝜎𝑏 2
1 + cos 𝜃
=
𝐹1 0, 𝜎𝑎 𝐹0 0, 𝜎𝑏 2
2
2
𝐹1 𝜈𝑥 , 𝜎𝑎 2
𝐹1 0, 𝜎𝑎 2
❑ Similarly on the E (y-z) plane, since 𝜙 = 90° and thus 𝑣𝑥 = 0, we have:
1 + cos 𝜃
𝑔𝐸 𝜃, 𝜙 = 𝑔 𝜃, 90° =
2
2
2
𝐹0 𝜈𝑦 , 𝜎𝑏
𝐹0 0, 𝜎𝑏 2
51
Horn antennas
❑ But which are the behaviors of the terms containing the Fresnel integrals?
1
a=1.22
1.2
a=0
1
a=1.5
|F0(:y,b)/F0(0,b)|
|F1(x,a)/F1(0,a)|
1.4
a=2.0
0.8
0.6
0.4
b=0.0
0.8
b=0.5
0.6
b=1.5
b=1.0
0.4
0.2
0.2
0
0
1
2
x
3
4
0
0
1
2
y
3
4
❑ There is a strong dependance on the parameters 𝜎𝑎 and 𝜎𝑏 , i.e., on the geometrical shape of
the flaring section!
52
Horn antennas
❑ The beamwidths in the E and H planes are (with 𝜎𝑎 = 1.22 and 𝜎𝑏 = 1) are given by
𝜆
𝜆
Δ𝜃𝑎 = 1.36 = 77.90°
𝐴
𝐴
𝜆
𝜆
Δ𝜃𝑏 = 0.94 = 53.88°
𝐵
𝐵
❑ The radiated power can be approximated with the power of the field on the aperture. Since
the phase terms (𝑒 𝑗… ) do not influence the computation of the Poynting vector, it results
(similarly to the case of a waveguide):
𝑃𝑎 =
1
𝐸0 2 𝐴𝐵
4𝜂
53
Horn antennas
❑ The gain is thus equal to:
2
1
𝐴𝐵
𝑓
2
𝐸
0 4 𝐹1 0, 𝜎𝑎 𝐹0 0, 𝜎𝑏
𝑈𝑚𝑎𝑥
2𝜆2 𝜂 𝑦 𝑚𝑎𝑥 4𝜋
𝐺𝑎 = 4𝜋
= 4𝜋
= 2
2 𝐴𝐵
1
𝑃𝑎
𝜆
𝐸
2
0
𝐸 𝐴𝐵
4𝜂 0
2
4𝜋
1
= 2 𝐴𝐵 𝐹1 0, 𝜎𝑎 𝐹0 0, 𝜎𝑏
𝜆
8
2
𝑒𝑎
❑ In the typical case 𝜎𝑎 = 1.22 and 𝜎𝑏 = 1 the efficiency is equal to 𝑒𝑎 = 0.51 and consequently:
4𝜋
𝐺𝑚𝑎𝑥 = 0.51 2 𝐴𝐵
𝜆
54
Horn antennas
❑ Example. Let us consider a Horn antenna working in the X band at 10 GHz. The antenna is
fed by a WR-90 waveguide. Calculate the aperture sides when 𝑅𝐴 = 4𝜆 (assuming that 𝜎𝑎
= 1.22 and 𝜎𝑏 = 1), the beamwidths, the gain and visualize the radiation pattern.
❑ The dimensions of the feeding waveguide are 𝑎 = 2.282 cm
wavelength is equal to 𝜆 =
𝑐
= 3 cm and consequently 𝑅𝐴 = 12 cm.
𝑓
and 𝑏 = 1.016 cm . The
❑ From the definition of 𝜎𝑎 and the geometrical relationships of the Horn antennas, we have
that
2
2
𝐴
𝐴
𝐴 𝐴−𝑎
𝜎𝑎2 =
=
=
→ 𝐴2 − 𝑎𝐴 − 2𝜆𝜎𝑎2 𝑅𝐴 = 0
2𝜆𝑅𝑎 2𝜆 𝐴 𝑅
2𝜆𝑅𝐴
𝐴
𝐴−𝑎
⇓
1
𝐴=
𝑎 + 𝑎2 + 8𝜆𝑅𝐴 𝜎𝑎2 = 11.56 cm
2
55
Horn antennas
❑ Moreover, in order for the antenna to be physically realizable, we also need 𝑅𝐴 = 𝑅𝐵 .
Consequently, we have
1
𝐵=
𝑏 + 𝑏2 + 8𝜆𝑅𝐴 𝜎𝑏2 = 9.01 cm
2
❑ The beamwidths are then
Δ𝜃𝑎 = 77.90°𝜆/𝐴 = 20.22°
Δ𝜃𝑏 = 53.88°𝜆/𝐵 = 17.94°
❑ and the gain is
4𝜋
𝐺𝑚𝑎𝑥 = 0.51 2 𝐴𝐵 = 74.13 → 18.7 dB
𝜆
56
Horn antennas
Theoretical
Numerical
❑ The radiation pattern is shown below.
H (x-z)
plane
𝜃
𝜃
E (y-z)
plane
57
Horn antennas
❑ An example of
matlab script
using the
Antenna toolbox
58
Horn antennas
❑ Optimal values of 𝜎𝑎 and 𝜎𝑏
❑ The optimal values of 𝜎𝑎 and 𝜎𝑏 are those that maximize the gain 𝐺𝑚𝑎𝑥 when 𝑅𝐴 = 𝑅𝐵 are
fixed.
❑ If 𝐴 and 𝐵 are large (i.e., 𝐴 ≫ 𝑎 and 𝐵 ≫ 𝑏), then it results 𝑅𝐴 =
𝐴2
2
this case, since 𝜎𝑎 =
2𝜆𝑅
𝑎
𝐵2
2
and 𝜎𝑏 =
, we have
2𝜆𝑅
𝐴−𝑎
𝑅𝑎 ≅ 𝑅𝑎 and 𝑅𝐵 ≅ 𝑅𝑏 . In
𝐴
𝑏
4𝜋
1
𝜎𝑎 2𝜆𝑅𝑎 𝜎𝑏 2𝜆𝑅𝑏 1
2
𝐺𝑚𝑎𝑥 = 2 𝐴𝐵 𝐹1 0, 𝜎𝑎 𝐹0 0, 𝜎𝑏
= 4𝜋
𝐹1 0, 𝜎𝑎 𝐹0 0, 𝜎𝑏
2
𝜆
8
𝜆
8
𝜋 𝑅𝑎 𝑅𝑏
=
𝜎𝑎 𝐹1 0, 𝜎𝑎 2 𝜎𝑏 𝐹0 0, 𝜎𝑏 2
𝜆
𝑓𝑎 𝜎𝑎
2
𝑓𝑏 𝜎𝑏
59
Horn antennas
❑ The behaviors of the functions 𝑓𝑎 and 𝑓𝑏 are shown in the figure.
𝜎𝑏 = 1.02
𝜎𝑎 = 1.26
❑ The functions 𝑓𝑎 and 𝑓𝑏 are maximum for 𝜎𝑎 = 1.26 and 𝜎𝑏 = 1.02. Often, such values are
approximated as 𝜎𝑎 = 1.5 = 1.22 and 𝜎𝑏 = 1.
60
Horn antennas
❑ A particular case is when the aperture has the same form factor as the waveguide, i.e.,
𝑏 𝐵
= =𝑟
𝑎 𝐴
❑ In this case, we have 𝜎𝑏 =
𝐵
2𝜆𝑅𝑏
=𝑟
𝐴
2𝜆𝑅𝑏
= 𝑟𝜎𝑎 (since 𝑅𝑏 =
𝐵
𝑟𝐴
𝑅𝐵 =
𝑅 = 𝑅𝑎 )
𝐵−𝑏
𝑟𝐴−𝑟𝑎 𝐴
and
consequently:
𝜋 𝑅𝑎 𝑅𝑏 2
𝐺𝑚𝑎𝑥 =
𝜎𝑎 𝑟 𝐹1 0, 𝜎𝑎 𝐹0 0, 𝑟𝜎𝑎
𝜆
2
𝑓𝑟 𝜎𝑎
61
Horn antennas
❑ Depending on the value of 𝑟, the maximum value of 𝑓𝑟 is obtained for different values of 𝜎𝑎 ,
e.g.,
1
𝑟 = → 𝜎𝑎 = 1.4749 → 𝑒𝑎 = 0.4743
2
4
𝑟 = → 𝜎𝑎 = 1.4982 𝑊𝑅90
9
2
𝑟 = → 𝜎𝑎 = 1.5127 (𝑊𝑅42)
5
Optimal 𝜎𝑎
62
Horn antennas
❑ Design of a horn antenna
❑ When designing a horn antenna, it is necessary to satisfy not only the constrains related to
the radiation properties of the antenna, but also the geometrical and practical constraints.
❑ In particular, for a horn antenna, the size of the waveguide is fixed (in order to support the
fundamental mode at the given frequency) and we have that 𝑅𝐴 = 𝑅𝐵 .
❑ Since 𝑅𝐴 =
𝐴−𝑎
𝐴 𝐴−𝑎
𝑅𝑎 =
𝐴
2𝜆𝜎𝑎2
and 𝑅𝐵 =
𝐵−𝑏
𝐵 𝐵−𝑏
𝑅𝑏 =
, the last condition leads to:
𝐵
2𝜆𝜎𝑏2
𝐵 𝐵−𝑏
𝜎𝑏2
= 2
𝐴 𝐴−𝑎
𝜎𝑎
❑ It is thus necessary to define the values of 𝐴 and 𝐵 that guarantee the desired gain and, at the
same time, satisfy the previous relationship.
63
Horn antennas
❑ If the form factor 𝑟 =
𝑏
𝐵
𝜎
= = 𝑏 is fixed, the problem is simple, since the previous
𝑎
𝐴
𝜎𝑎
geometrical constrain is always satisfied, and consequently we have:
4𝜋
𝐺𝑚𝑎𝑥
𝐺𝑚𝑎𝑥 = 𝑒𝑎 2 𝐴𝑟𝐴 → 𝐴 = 𝜆
𝜆
4𝜋𝑒𝑎 𝑟
❑ where 𝑒𝑎 is the optimal for the considered value of 𝑟.
64
Horn antennas
❑ In the general case, we need to solve a non-linear system of equation given by:
𝑓1 𝐴, 𝐵 = 𝐵 𝐵 − 𝑏 𝜎𝑎2 − 𝐴 𝐴 − 𝑎 𝜎𝑏2 = 0
൞
4𝜋
𝑓2 𝐴, 𝐵 = 𝐺𝑚𝑎𝑥 − 𝑒𝑎 2 𝐴𝐵 = 0
𝜆
❑ Such problem must be solved numerically, for example by using an iterative Newton-type
approach.
65
Horn antennas
❑ Example: Design a horn antenna working in the X band (at 10 GHz, i.e., 𝜆 = 3 cm) with a gain of
20 dB.
❑ The dimensions of the feeding waveguide are 𝑎 = 2.282 cm and 𝑏 = 1.016 cm. The form factor is
equal to 𝑟 = 𝑏/𝑎 = 4/9.
❑ If the size ratio of the aperture is kept constant, we have
𝐴=𝜆
𝐺𝑚𝑎𝑥
100
= 3 × 10−2 ×
≅ 18.5 cm
4
4𝜋𝑒𝑎 𝑟
4𝜋 0.4713
9
→
𝐵 = 𝑟𝐴 = 8.23 cm
❑ Moreover
𝐴 𝐴−𝑎
𝑅𝐴 =
= 22.2 cm
2𝜆𝜎𝑎2
66
Horn antennas
❑ If a generic ratio is assumed, the problem can only be solved numerically.
❑ In this case the optimal values that can be obtained are
𝐴 = 13.7 cm
𝐵 = 10.7 cm
𝑅 = 16.4 cm
❑ As can be seen, for obtaining the same gain, the dimensions of the antenna are significantly
lower!
❑ This is due to the reduced constraints on the design.
67
Reflector antennas
❑ Reflector antennas are constituted by a metallic structure (with a properly designed shape)
that “reflect” the field produced by another radiating element (one of the antennas that we
considered up to now, e.g., a dipole, a open-waveguide, or a horn).
❑ In particular, the aim is to direct the radiation toward a given direction, in order to obtain a
highly directive antenna.
❑ The reflector can have different shape, e.g.,
o Plane
o Corner
o Curved (e.g., parabolic)
68
Reflector antennas
Planar reflector
corner reflector
Parabolic reflector
Cassegrain antenna (dual reflector)
69
Parabolic dish antenna
❑ Dish antennas are reflector antennas characterized
by a high gain (usually > 30 dB) and narrow beam.
❑ They are constituted by a metallic reflector with
parabolic shape.
❑ The reflector “reflects” the field generated by
another antenna, the feed.
❑ The feed is located in the focus of the parabola.
❑ In this way, all the rays originating from
the feed are reflected along the axis of the
parabola (in the following we assume that
it is the z axis).
70
Parabolic dish antenna
𝐹
❑ From the properties of the parabolic surfaces, we have:
𝑅 + ℎ = 2𝐹
ℎ
ෝ
𝐧
❑ where 𝐹 is the focal length.
❑ The field originates from the feed with the same
initial phase.
❑ Since the traveled distance is the same for every
𝑅
𝐷
𝜌
𝜓
𝜓0
𝑧
“ray”, we have that on a plane perpendicular to
the z axis (i.e., the x-y plane) the field has the
same (constant) phase.
71
Parabolic dish antenna
❑ The spherical wave generated by the feed is thus
𝑦′
𝑦
transformed in a plane wave on a plane passing
through the focus and perpendicular to the
parabola axis!
❑ Such plane can be considered as a virtual
aperture from which the radiated fields
originate.
ℎ
𝑥′
𝜒
𝑅
𝜒
𝑥
𝜓
❑ Consequently, the parabolic dish antenna
can be studied by using the concepts we
already discussed for aperture antennas.
𝜌
𝑧
𝐹
Virtual
aperture
72
Parabolic dish antenna
❑ Before addressing the radiation properties, it is
𝑦′
𝑦
useful to introduce some geometrical parameters
that characterized the antenna.
ℎ
❑ By using the constant length property, we can write:
𝑥′
2𝐹
𝐹
𝑅 + ℎ = 𝑅 + 𝑅 cos 𝜓 = 2𝐹 → 𝑅 =
=
1 + cos 𝜓 cos2 𝜓
2
𝜒
𝑅
𝜌
𝜒
𝑥
𝜓
𝑧
❑ A generic point on the aperture can be expressed
𝐹
in cylindrical coordinates 𝜌, 𝜒 with:
sin 𝜓
𝜓
𝜌 = 𝑅 sin 𝜓 = 2𝐹
= 2𝐹 tan
1 + cos 𝜓
2
Virtual
aperture
73
Parabolic dish antenna
❑ An important parameter for the dish antennas is
𝑦′
𝑦
the maximum angle 𝜓0 , corresponding to 𝜌 = 𝐷/2,
which is related the efficiency and is given by
𝐷
𝜓0 = 2 atan
4𝐹
ℎ
𝑥′
𝜒
𝑅
𝜌
𝜒
𝑥
𝜓
❑ The maximum angle 𝜓0 is thus determined by the
ratio 𝐹/𝐷. Some typical values are:
𝑧
𝐹
𝜓0
o 𝐹/𝐷 = 0.25 → 𝜓0 = 90°
o 𝐹/𝐷 = 0.35 → 𝜓0 = 71°
o 𝐹/𝐷 = 0.5 → 𝜓0 = 53°
Virtual
aperture
74
Parabolic dish antenna
❑ In order to fully characterize the radiated field, two possible ways can be followed:
o Calculate the current density produced by the feed on the reflector and compute the field radiated
by this source
o Calculate the fields on the virtual aperture and use aperture radiation concepts.
❑ In both cases, the mathematical treatment is quite involving. Moreover, the actual field
depends upon the particular feed used to illuminate the reflector.
❑ However, for computing the efficiency and the gain it is possible to operate in a simplified
way.
75
Parabolic dish antenna
❑ Let us consider the power radiated by the feed
𝑑Ω = sin 𝜓 𝑑𝜓𝑑𝜒
in a small solid angle 𝑑Ω = sin 𝜓 𝑑𝜓𝑑𝜒 directed
along the radial (with respect to the feed) direction
෡ , which depends upon the angle 𝜓 and 𝜒.
𝐑
𝑦
𝑑𝐴 = 𝜌𝑑𝜌𝑑𝜒
❑ The cone described by this solid angle is
“transformed” by the reflector in a small
cylinder with cross section of diameter
𝑑𝐴 = 𝜌𝑑𝜌𝑑𝜒.
𝑥′
𝑦′
❑ If an ideal reflector is considered (i.e., made
by PEC material), the power in the cone and
in the cylinder need to be the same (for the
energy conservation principle).
෡
𝐑
𝜒
𝜌
𝜒
𝑥
𝜓
𝑧
𝐹
76
Parabolic dish antenna
❑ Assuming that the fields on the aperture are
𝑑Ω = sin 𝜓 𝑑𝜓𝑑𝜒
Huygens sources (which is reasonable since the
aperture fields form a plane wave), the power
density on the aperture is equal to 𝐄𝑎 2 /2𝜂.
𝑦
𝑑𝐴 = 𝜌𝑑𝜌𝑑𝜒
❑ Consequently, we have that
෡
𝐑
𝜒
2
𝑈𝑓𝑒𝑒𝑑 𝜓, 𝜒 𝑑Ω =
𝐄𝑎
𝑑𝐴,
2𝜂
𝑥′
𝑦′
𝜌
𝜒
𝑥
𝜓
𝑧
𝐹
❑ with
𝑑Ω = sin 𝜓 𝑑𝜓𝑑𝜒
𝑑𝐴 = 𝜌𝑑𝜌𝑑𝜒
77
Parabolic dish antenna
❑ By using the geometrical relationship that we derived previously, we can rewrite such power
balance as
𝐄𝑎 𝜌, 𝜒
𝑈𝑓𝑒𝑒𝑑 𝜓, 𝜒 =
2𝜂
2
𝜌 𝑑𝜌
𝐄𝑎 𝜌, 𝜒
=
sin 𝜓 𝑑𝜓
2𝜂
2 𝑅 sin 𝜓
𝑅2
=
𝐄 𝜌, 𝜒
sin 𝜓 cos 2 𝜓 2𝜂 𝑎
2
𝐹
2
𝑅
𝑑𝜌
𝑑
𝜓
1 1
𝐹
=
2𝐹 tan
= 2𝐹
=
𝜓
𝜓
𝑑𝜓 𝑑𝜓
2
cos2 2 cos2
2
2
❑ From this formula we can write the amplitude of the field over the aperture as
1
4𝐹
𝐄𝑎 𝜌, 𝜒 =
2𝜂𝑈𝑓𝑒𝑒𝑑 𝜓, 𝜒 = 2
2𝜂𝑈𝑓𝑒𝑒𝑑 𝜓, 𝜒
𝑅
𝜌 + 4𝐹 2
78
Parabolic dish antenna
❑ The field on the aperture is tapered, i.e., its value is lower on the edges.
❑ A measure of how much the field is reduced is the edge illumination, defined as:
𝐄𝑎 𝑎, 𝜒
1 + cos 𝜓0 𝑈𝑓𝑒𝑒𝑑 𝜓0 , 𝜒
𝐴𝑒𝑑𝑔𝑒 =
=
𝐄𝑎 0, 𝜒
2
𝑈𝑓𝑒𝑒𝑑 0, 𝜒
❑ As we already discussed for the apertures, this fact results in the presence of a non-
negligible aperture taper loss!
❑ To evaluate this effect, let us now obtain the gain.
79
Parabolic dish antenna
❑ The gain of the aperture is defined as:
4𝜋𝑈𝑚𝑎𝑥
𝐺𝑎 =
𝑃𝑎
❑ where the power on the aperture can be written as
𝜓0 2𝜋
1
2
න 𝐄 𝑑𝑆 = න න 𝑈𝑓𝑒𝑒𝑑 𝜓, 𝜒 sin 𝜓 𝑑𝜓𝑑𝜒
𝑃𝑎 =
2𝜂 𝐴 𝑎
0
0
❑ Moreover, we now that the aperture gain can be written as:
𝐺𝑎 =
4𝜋𝐴
𝑒 𝑒
𝜆2 𝑎𝑡𝑙 𝑝𝑒𝑙
80
Parabolic dish antenna
❑ However, when dealing with reflector antennas, the gain is not that of the aperture, but it
must be referred to the power in input to the feed, i.e.,
Lost power
4𝜋𝑈𝑚𝑎𝑥 4𝜋𝑈𝑚𝑎𝑥 𝑃𝑎
𝐺𝑚𝑎𝑥 =
=
= 𝐺𝑎 𝑒𝑠𝑝𝑙
𝑃𝑓𝑒𝑒𝑑
𝑃𝑎 𝑃𝑓𝑒𝑒𝑑
Available
power
❑ where the term 𝑒𝑠𝑝𝑙 = 𝑃𝑎 /𝑃𝑓𝑒𝑒𝑑 is the
spillover efficiency, which the represent
the fraction of power provided by the feed
that is effectively reflector by the reflector.
The remaining part is lost since it since it
flows out of the borders.
Lost power
81
Parabolic dish antenna
❑ The total gain is then:
𝐺𝑚𝑎𝑥 = 𝐺𝑎 𝑒𝑠𝑝𝑙 =
4𝜋𝐴
𝑒𝑎𝑡𝑙 𝑒𝑝𝑒𝑙 𝑒𝑠𝑝𝑙
2
𝜆
❑ The total theoretical efficiency is
𝑒𝑎 = 𝑒𝑎𝑡𝑙 𝑒𝑝𝑒𝑙 𝑒𝑠𝑝𝑙
❑ In practice there are also other effects that can further reduce the efficiency, i.e., feed
blocking, non-idealities of the materials, etc.
82
Parabolic dish antenna
❑ The term PEL is moreover usually negligible, since the phase over the aperture is almost constant.
❑ Consequently, the ATL and SPL terms are the most significant ones and their product is called
illumination efficiency 𝑒𝑖𝑙𝑙 = 𝑒𝑎𝑡𝑙 𝑒𝑠𝑝𝑙 .
❑ From the previous relationships, the efficiency can be explicitly written as
2
𝜓
2𝐹 ‫ 𝐴׬‬2𝜂𝑈𝑓𝑒𝑒𝑑 tan 2 𝑑𝜓𝑑𝜒
‫𝐴𝑑 𝑎𝐄 𝐴׬‬
𝑒𝑎𝑡𝑙 =
=
2
𝐴 ‫ 𝜋 𝐴𝑑 𝑎𝐄 𝐴׬‬2𝐹 2 tan2 𝜓 2𝜂𝑈
𝑓𝑒𝑒𝑑 sin 𝜓 𝑑𝜓𝑑𝜒
2 ‫𝐴׬‬
𝜓0 2𝜋
‫׬‬0 ‫׬‬0 𝑈𝑓𝑒𝑒𝑑 sin 𝜓 𝑑𝜓𝑑𝜒
𝑃𝑎
𝑒𝑠𝑝𝑙 =
= 𝜋 2𝜋
𝑃𝑓𝑒𝑒𝑑
‫׬‬0 ‫׬‬0 𝑈𝑓𝑒𝑒𝑑 sin 𝜓 𝑑𝜓𝑑𝜒
2
1
𝜓
𝑒𝑖𝑙𝑙 = cot 2 0
𝜋
2
2
𝜓0 2𝜋
‫׬‬0 ‫׬‬0
𝜋 2𝜋
𝜓
𝑈𝑓𝑒𝑒𝑑 tan 𝑑𝜓𝑑𝜒
2
‫׬‬0 ‫׬‬0 𝑈𝑓𝑒𝑒𝑑 sin 𝜓 𝑑𝜓𝑑𝜒
2
No need to remeber
exactly these long
formulas for the exam….
83
Parabolic dish antenna
❑ An example of behavior for a feed with a symmetric radiation pattern is shown below.
Best efficiency
84
Parabolic dish antenna
❑ The best efficiency is 0.82 and it is obtained for 𝜓0 = 53.31°. The corresponding 𝐹/𝐷 ratio is
1
cot 𝜓0 /2
4
= 0.5.
❑ The corresponding edge illumination is 0.285 (i.e., −10.9 dB). In general, the best compromise
between the ATL and SPL efficiencies is obtained when the edge efficiency is −11 dB.
❑ The value of the efficiency of 0.82 is however overestimated, since it does not take into
account other losses that are present in real antennas (e.g., non ideality of the metallic
surface). In practice, the efficiency is usually in the range 0.55–0.65.
85
Parabolic dish antenna
❑ Finally, we have:
4𝜋𝐴
𝜋𝐷
𝐺𝑚𝑎𝑥 = 𝑒𝑎 2 = 𝑒𝑎
𝜆
𝜆
2
𝑒𝑎 = 0.55 − 0.65
❑ In this case, the beamwidth is given by:
Δ𝜃3𝑑𝐵 = 70°
𝜆
𝐷
𝐄𝑎 𝑎, 𝜒
𝐴𝑒𝑑𝑔𝑒 = −20 log10
𝐄𝑎 0, 𝜒
❑ More in general
Δ𝜃3𝑑𝐵 =
1.05°𝐴𝑑𝐵
𝑒𝑑𝑔𝑒 + 55.95°
𝜆
𝐷
86
Parabolic dish antenna
❑ Example. Calculate the focal distance, gain and beamwidth of a parabolic dish antenna of
diameter 80 cm and working at 12 GHz (under optimal conditions).
❑ In order to work in optimal conditions, we need
𝐹 = 0.5𝐷 = 40 cm.
𝐹
= 0.5. Consequently we directly obtain that
𝐷
𝑐
3×108
❑ Since the wavelength is 𝜆 = =
m = 2.5 cm, the gain is equal to (assuming 𝑒𝑎 = 0.6)
𝑓
12×109
𝐺𝑚𝑎𝑥 = 𝑒𝑎
𝜋𝐷
𝜆
2
𝜋0.8
= 0.6
0.025
2
≅ 6064 → 37.8 dB
❑ The corresponding beamwidth is given by:
Δ𝜃3𝑑𝐵 = 70°
𝜆
0.025
= 70°
≅ 2.2°
𝐷
0.8
87
Parabolic dish antenna
88
Printed antennas
❑ Printed (or patch) antennas are composed by a metallic patch (with height 𝑡 ≪ 𝜆0 , being 𝜆0
the wavelength in air) located over a dielectric substrate (with height typically in the range
0.003𝜆0 < ℎ < 0.05𝜆0 and relative dielectric permittivity usually between 2 < 𝜖𝑟 < 12).
❑ Below the substrate a ground plane is also present.
❑ The patch can be of arbitrary shape.
Metallic patch
Substrate
𝜖𝑟
ℎ
Ground plane
89
Printed antennas
❑ Clearly, if we consider a completely arbitrary patch, in order to characterize the radiation, it
is necessary to use numerical simulators.
❑ For patches of simple shape (e.g., rectangular or circular), it is possible to derive the
radiation parameters in a closed form.
Rectangular
Triangular
Circular
Elliptical
Disc sector
Circular ring
90
Printed antennas
❑ Printed antennas can be fed in different ways. The most common are exemplified below.
Microstrip
line
Coaxial
connector
Microstrip feed
Probe feed
Slot
Microstrip line
Proximity-coupled feed
Microstrip line
Aperture-coupled feed
91
Rectangular patch antennas
❑ One of the main adopted printed antennas is the rectangular (or microstrip) patch antenna.
❑ In this case, the metallic patch over the substrate has a rectangular shape with sides 𝐿 and 𝑊.
𝑧
𝐿
𝑦
𝑊
𝑥
ℎ
92
Rectangular patch antennas
❑ At the working frequency, the antenna can be modeled as a resonant cavity. Consequently,
the electric field below the patch can be modeled as:
𝜋𝑥
𝐸𝑧 𝑥 = −𝐸0 sin
𝐿
𝑗𝐸0
𝜋𝑥
𝐻𝑦 𝑥 =
cos
𝜂
𝐿
❑ The resonance frequency, which is the working frequency of the antenna, is given by:
𝜔0 =
𝜋𝑣
𝜋𝑐
=
𝐿
𝐿 𝜖𝑟
→
𝑓𝑟 =
𝑐
2𝐿 𝜖𝑟
Velocity of propagation in the substrate
93
Rectangular patch antennas
❑ It should however be considered that, differently from ideal cavities that are bounded by
metallic walls, here the field at the boundaries of the patch is free to exit, as shown below.
𝑧
Fringe fields
Δ𝐿
Fringe fields
𝐿
Δ𝐿
𝑥
ℎ
94
Rectangular patch antennas
❑ Due to the presence of the fringing fields, the patch looks electrically larger than it really is.
❑ It is then possible to define an effective length as
𝐿𝑒𝑓𝑓 = 𝐿 + 2Δ𝐿
❑ where the fringe length Δ𝐿 can be approximated as
𝑊
+ 0.264
ℎ
Δ𝐿 = 0.412
ℎ
𝑊
𝜖𝑟,𝑒𝑓𝑓 − 0.258
+ 0.8
ℎ
𝜖𝑟,𝑒𝑓𝑓 + 0.3
95
Rectangular patch antennas
❑ In the previous relationship, in order to take into account the fact that the fringing fields are
partly in air and partly in the substrate, an effective dielectric permittivity 𝜖𝑟,𝑒𝑓𝑓 is
introduced.
❑ The effective dielectric permittivity is essentially constant at low frequency. Then when the
frequency increases, it start to increase and approach the values of the actual dielectric
permittivity of the substrate.
𝜖𝑟,𝑒𝑓𝑓
𝜖𝑟
96
Rectangular patch antennas
❑ If the working frequency is not too high, the effective dielectric permittivity can be
approximated as (low-frequency value)
𝜖𝑟 + 1 𝜖𝑟 − 1
ℎ
𝜖𝑟,𝑒𝑓𝑓 =
+
1 + 12
2
2
𝑊
1
−2
❑ It is worth noting that, considering the presence of the fringing fields, the actual operating
(resonance) frequency needs to be corrected and becomes
𝑓𝑟 =
𝑐
2𝐿𝑒𝑓𝑓 𝜖𝑟,𝑒𝑓𝑓
97
Rectangular patch antennas
❑ Example. Design a rectangular patch antenna working at 10 GHz using a RT/duroid 5880
substrate.
❑ The RT/duroid 5880 has a relative dielectric permittivity 𝜖𝑟 = 2.2 and a thickness ℎ ≅ 1.6 mm.
❑ For an efficient design, it has been empirically found that
𝑊=
𝑐
2𝑓𝑟
2
≅ 1.19 cm
𝜖𝑟 + 1
❑ The effective relative dielectric permittivity is then
𝜖𝑟,𝑒𝑓𝑓 =
𝜖𝑟 + 1 𝜖𝑟 − 1
ℎ
+
1 + 12
2
2
𝑊
−
1
2
= 1.97
98
Rectangular patch antennas
❑ The effective length is
Wavelength in the
𝑐
substrate, 𝜆 = 𝑓 𝜖
𝑟,𝑒𝑓𝑓
𝜆
𝑐
𝐿𝑒𝑓𝑓 = =
= 1.07 cm
2 2𝑓𝑟 𝜖𝑟,𝑒𝑓𝑓
❑ Consequently, the actual patch length is
𝑊
+ 0.264
ℎ
𝐿 = 𝐿𝑒𝑓𝑓 − 2Δ𝐿 = 𝐿𝑒𝑓𝑓 − 2 × 0.412
ℎ = 1.07 cm − 2 × 0.08 cm
𝑊
𝜖𝑟,𝑒𝑓𝑓 − 0.258
+ 0.8
ℎ
= 0.91 cm
𝜖𝑟,𝑒𝑓𝑓 + 0.3
99
Rectangular patch antennas
❑ As we observed in the previous lecture, the fringing fields extend for a certain region around
the patch. Basically, these are the regions from which the radiated fields originate.
Consequently, they can be though as virtual radiating apertures.
❑ In particular, the effects of the region 2 and 4 are negligible, and it is thus sufficient to
consider the terms due to the regions 1 and 3.
2
Fringe fields 𝑧
Δ𝐿
𝑦
Fringe fields
𝐿
Δ𝐿
𝑥
1
𝐿
3
𝑥
ℎ
𝑊
4
100
Rectangular patch antennas
❑ The aperture field 𝐸𝑎 can be expressed in terms of 𝐸𝑧 by exploiting the Faraday law, i.e.,
Since the field is zero outside
the fringing region
න
𝐴𝐵𝐶𝐷
Since the component of the electric
field tangential to a PEC is zero
𝐄 ∙ 𝑑𝐥 = 𝐸𝑎 Δ𝐿 + ณ
0 + ณ
0 + 𝐸𝑧 ቚ
𝐵𝐶
𝐴𝐵
𝑥=
𝐶𝐷
𝐿ℎ=0
2
𝐷𝐴
↓
ℎ
ℎ
𝜋𝑥
ℎ
𝐸𝑎 = − 𝐸𝑧 ቚ 𝐿 =
𝐸0 sin ฬ 𝐿 =
𝐸0
Δ𝐿
Δ𝐿
𝐿
Δ𝐿
𝑥=
𝑥=
2
The magnetic field is
approximated as zero
in the fringing region
2
𝑧
𝐿
Δ𝐿
B
A
ℎ
𝑥
𝐸𝑎
𝐸𝑧
D
C
101
Rectangular patch antennas
❑ Consequently, for the virtual apertures 1 and 3 we have:
ℎ
𝐄𝑎 =
𝐸 𝐱ො
Δ𝐿 0
𝑦
❑ The radiated fields can be then computed
as the field due to two uniform apertures
located in positions ±
𝐿𝑒𝑓𝑓
2
along the x axis.
❑ In particular, since a ground plane is present
𝐿𝑒𝑓𝑓 /2
1
𝐿𝑒𝑓𝑓 /2
3
𝑥
below the substrate, we can assume a PEC screen.
102
Rectangular patch antennas
❑ The Fourier transforms of the aperture fields are given by:
Phase shift due to the aperture displacement along the x axis
𝐟1,3 = න
Δ𝐿
2
−
න
𝑊
2
Δ𝐿
𝑊
−
2
2
′
′
𝐄𝑎 𝑒 ±𝑗𝑘𝑥 𝐿𝑒𝑓𝑓 /2 𝑒 𝑗 𝑘𝑥𝑥 +𝑘𝑦 𝑦 𝑑𝑥 ′ 𝑑𝑦 ′ = 𝐄𝑎 𝑒 ±𝑗𝑘𝑥𝐿𝑒𝑓𝑓 /2 න
−
𝐿𝑒𝑓𝑓
𝐿
ℎ
±𝑗 2𝑘𝑥
±𝑗 2 𝑘𝑥
= 𝑊Δ𝐿 𝐸0 𝐱ො sinc 𝑣𝑎𝑥 sinc 𝑣𝑦 𝑒
≅ 𝑊ℎ𝐸0 𝐱ො sinc 𝑣𝑦 𝑒
Δ𝐿
❑ where
Δ𝐿
2
sinc 𝑣𝑎𝑥 = sinc
Δ𝐿
sin 𝜃 cos 𝜙
𝜆
Δ𝐿
2
′
𝑒 𝑗𝑘𝑥 𝑥 𝑑𝑥 ′ න
𝑊
2
−
𝑊
2
′
𝑒 𝑗𝑘𝑦 𝑦 𝑑𝑦 ′
= 1 sinceΔ𝐿 ≪ 𝜆
Δ𝐿
𝑣𝑎𝑥 =
sin 𝜃 cos 𝜙
𝜆0
𝑊
𝑣𝑦 = sin 𝜃 sin 𝜙
𝜆0
103
Rectangular patch antennas
❑ The resulting radiation vector is given by:
𝐿𝑒𝑓𝑓
𝐿𝑒𝑓𝑓
𝑘𝑥
−𝑗
𝑘
2 𝑥 = 𝑊ℎ𝐸0 𝐱
ො sinc 𝑣𝑦
𝐟 = 𝐟1 + 𝐟3 = 𝑊ℎ𝐸0 𝐱ො sinc 𝑣𝑦 𝑒 2
+ 𝑊ℎ𝐸0 𝐱ො sinc 𝑣𝑦 𝑒
𝑗
𝐿𝑒𝑓𝑓
𝐿𝑒𝑓𝑓
𝑘𝑥
−𝑗
𝑘
2 𝑥
𝑒 2
+𝑒
𝑗
𝐿𝑒𝑓𝑓
= 𝑊ℎ𝐸0 𝐱ො sinc 𝑣𝑦 2 cos
𝑘 = 2𝑊ℎ𝐸0 sinc 𝑣𝑦 cos 𝜋𝑣𝑥 𝐱ො
2 𝑥
𝐿𝑒𝑓𝑓
𝑊
❑ where 𝑣𝑥 =
sin 𝜃 cos 𝜙 and 𝑣𝑦 = 𝜆 sin 𝜃 sin 𝜙.
𝜆
❑ Finally, the radiated field is obtained as:
𝑒 −𝑗𝑘𝑟
𝑒 −𝑗𝑘𝑟
𝐸𝜃 𝜃, 𝜙 = 𝑗𝑘
𝑓𝑥 cos 𝜙 = 𝑗𝑘𝑊ℎ𝐸0
sinc 𝑣𝑦 cos 𝜋𝑣𝑥 cos 𝜙
2𝜋𝑟
𝜋𝑟
−𝑗𝑘𝑟
𝑒
𝑒 −𝑗𝑘𝑟
𝐸𝜙 𝜃, 𝜙 = −𝑗
cos 𝜃 𝑓𝑥 sin 𝜙 = −𝑗𝑘𝑊ℎ𝐸0
cos 𝜃 sinc 𝑣𝑦 cos 𝜋𝑣𝑥 sin 𝜙
2𝜋𝑟
𝜋𝑟
104
Rectangular patch antennas
❑ The radiation pattern is then given by:
𝐄 𝜃, 𝜙 2
𝑔 𝜃, 𝜙 =
= cos 2 𝜙 + cos 2 𝜃 sin2 𝜙 sinc 2 𝑣𝑦 cos 2 𝜋𝑣𝑥
2
𝐄 𝜃, 𝜙 𝑚𝑎𝑥
❑ which, on the E and H planes, reduces to:
𝑔𝐸 𝜃 = cos 2 𝜋𝑣𝑥
𝑔𝐻 𝜃 = cos 2 𝜃 sinc 2 𝑣𝑦
𝜙=0
𝜙=
𝜋
2
105
Rectangular patch antennas
❑ Example. Let us consider the rectangular patch designed in the previous example (i.e., with
sides 𝑊 = 1.19 cm and 𝐿 = 0.91 cm at 10 GHz using a RT/duroid 5880 substrate). The
resulting radiation pattern is shown below.
Analytical
Numerical
E (x-z)
plane
𝜃
Analytical
Numerical
H (y-z)
plane
𝜃
106
Rectangular patch antennas
𝑓𝑟 = 10 GHz, 𝐿𝑒𝑓𝑓 = 1.07 cm, ℎ = 1.6 mm
❑ Using the formulas for
the radiated field, it can
be found that the
directivity is
Values in dB
2𝜋𝑊
𝐷𝑚𝑎𝑥 =
𝜆0
𝜋
𝜋
𝐼2 = න න
0
0
2
𝜋
𝐼2
𝑘 𝑊
sin 0 cos 𝜓
2
cos 𝜓
2
3
sin 𝜓 cos
2
𝑘0 𝐿𝑒𝑓𝑓
sin 𝜓 sin 𝜁 𝑑𝜓𝑑𝜁
2
107
Rectangular patch antennas
❑ Similarly, the radiation resistance (i.e., approximatively the resistive part of the input
impendace) can be obtained as
15𝐷𝑚𝑎𝑥 𝜆𝑜
𝑅𝑟𝑎𝑑 =
2
𝑊
2
30𝜋
=
𝐼2
3
𝑓𝑟 = 10 GHz, 𝐿𝑒𝑓𝑓 = 1.07 cm, ℎ = 1.6 mm
❑ As can be seen, the
input resistance depends
upon 𝑊, which also
impact on the directivity,
and has quite high values.
108
Rectangular patch antennas
❑ It is however possible to change the input resistance by adding an inset feed. This allows to
easily match the impedance of the feeding microstrip.
𝑅𝑖𝑛 𝑥0 = 𝑅𝑟𝑎𝑑 cos
2
𝜋
𝑥0
𝐿
Feeding
microstrip
𝑥0
𝑦
𝑥
Inset
109
Rectangular patch antennas
❑ Example. Let us consider the rectangular patch designed in the previous example (i.e.,
working at 10 GHz using a RT/duroid 5880 substrate. Design the inset feed length necessary
to obtain a input resistance of 50 Ω.
❑ According to the previous relationships, the radiation resistance is equal to 𝑅𝑟𝑎𝑑 ≅ 260 Ω.
Consequently, we have
𝐿
𝑅𝑖𝑛
−1
𝑥0 = cos
≅ 0.3 cm
𝜋
𝑅𝑟𝑎𝑑
110
Master degree in Internet and Multimedia Engineering / Electronic Engineering
Antennas / Advanced Antenna Engineering
Slides of the lectures – Part 4
Arrays of antennas
❑ In several cases, it is useful to use antennas able to adapt to the framework in which they
operate. For example, they should:
o Change the direction and number of radiation lobes;
o Add nulls in some directions;
o Estimate the directions of incidence of impinging waves;
o …
❑ Such tasks can be accomplished by using arrays of antennas.
2
Arrays of antennas
❑ An antenna array is constituted by a set of radiating elements all of the same type and with
the same orientation.
❑ Moreover, the feeds of the various antennas differ only for a complex multiplicative factor.
❑ Arrays can be of different types, mainly for what concerns the positions of the antennas.
❑ The most common layouts are:
o Linear
o Planar
o Circular
3
Arrays of antennas
𝑧
𝑧
Linear arrays
𝑦
𝑦
𝑥
𝑥
𝑧
𝑧
𝑦
𝑦
𝑥
Planar array
𝑥
Circular array
4
Arrays of antennas
❑ The basic property of the antenna arrays is that the change of the position of the elements
produce a phase shift in the radiated field.
❑ Let us consider a radiating element located in a volume 𝑉0 centered in the origin of the
reference system and characterized by a current density 𝐉0 𝐫 .
❑ The radiation properties are then defined in terms of a radiation vector
𝐅0 𝜃, 𝜙 = ‫𝐉 𝑉׬‬0
0
𝑧
′
𝑗𝐤∙
𝐫′
𝐫 𝑒
𝑑𝐫 ′
𝐉𝑑 𝐫 = 𝐉0 𝐫 − 𝐝
𝐝
❑ If we move the antenna in a position 𝐝, the current
density of the new elements can be written as
𝐉𝑑 (𝐫) = 𝐉0 (𝐫 − 𝐝)
𝐉0 𝐫
𝑦
𝑥
5
Arrays of antennas
❑ The new radiation vector is given by:
𝐅𝑑 𝜃, 𝜙 = න 𝐉𝑑
′
′
𝑗𝐤∙𝐫
𝐫 𝑒
𝑑𝐫 ′ = න
′
′′ +𝐝
′
𝑗𝐤∙𝐫
′
′′
𝑗𝐤∙
𝐫
𝐉𝟎 𝐫 − 𝐝 𝑒
𝑑𝐫 = න 𝐉𝟎 𝐫 𝑒
𝑑𝐫 ′′
𝑉𝑑
𝑉0
𝑉𝑑
=𝑒
𝑗𝐤∙𝐝
න 𝐉𝟎 𝐫
′′
𝑉0
𝑒
𝑗𝐤∙𝐫 ′′
𝑑𝐫 ′′ = 𝑒 𝑗𝐤∙𝐝 𝐅0 𝜃, 𝜙
Change of variables:
𝐫 ′′ = 𝐫 ′ − 𝐝
𝑉𝑑 → 𝑉0
❑ where we set 𝐤 = 𝑘ො
𝐫.
if we also multiply the current with a feed coefficient 𝑎 ∈ ℂ , i.e.,
𝐉𝑑 𝐫 = 𝑎𝐉0 𝐫 − 𝐝 , for the linearity of the problem we have
❑ Moreover,
𝐅𝑑 𝜃, 𝜙 = 𝑎𝑒 𝑗𝐤∙𝐝 𝐅0 𝜃, 𝜙
6
Arrays of antennas
❑ More generally, considering 𝑁 elements located in positions 𝐝1 , 𝐝2 ,…, 𝐝𝑁
with feed
coefficients equal to 𝑎1 , 𝑎2 ,…,𝑎𝑁 , it results that the current density of the 𝑛th element is
𝐉𝑛 𝐫 = 𝑎𝑛 𝐉0 𝐫 − 𝐝𝑛 ,
𝑛 = 1, … , 𝑁
❑ and consequently the radiation vector of each element is given by
𝐅𝑛 𝜃, 𝜙 = 𝑎𝑛 𝑒 𝑗𝐤∙𝐝𝑛 𝐅0 𝜃, 𝜙 ,
𝑛 = 1, … , 𝑁
❑ where 𝐉0 and 𝐅0 are the current density and the radiation vector of the elements as if they
were in the origin of the reference system.
7
Arrays of antennas
❑ By neglecting the mutual coupling, the total current density and radiation vector can be
written as
𝐉𝑡𝑜𝑡 𝐫 = 𝑎1 𝐉0 𝐫 − 𝐝1 + 𝑎2 𝐉0 𝐫 − 𝐝2 + ⋯ + 𝑎𝑁 𝐉0 𝐫 − 𝐝𝑁
𝐅𝑡𝑜𝑡 𝜃, 𝜙 = 𝐅1 𝜃, 𝜙 + 𝐅2 𝜃, 𝜙 + ⋯ + 𝐅𝑁 𝜃, 𝜙
= 𝑎1 𝑒 𝑗𝐤∙𝐝1 𝐅0 𝜃, 𝜙 + 𝑎2 𝑒 𝑗𝐤∙𝐝2 𝐅0 𝜃, 𝜙 + ⋯ + 𝑎𝑁 𝑒 𝑗𝐤∙𝐝𝑁 𝐅0 𝜃, 𝜙 = 𝐴 𝜃, 𝜙 𝐅0 𝜃, 𝜙
❑ where 𝐴 𝜃, 𝜙 is the array factor, which is defined as:
𝑁
𝐴 𝜃, 𝜙 = 𝑎1 𝑒 𝑗𝐤∙𝐝1 + 𝑎2 𝑒 𝑗𝐤∙𝐝2 + ⋯ + 𝑎𝑁 𝑒 𝑗𝐤∙𝐝𝑁 = ෍ 𝑎𝑛 𝑒 𝑗𝐤∙𝐝𝑛
𝑛=1
8
Arrays of antennas
❑ The radiation intensity and the gain function of the array can be then expressed as:
𝑈𝑡𝑜𝑡 𝜃, 𝜙 = 𝐴 𝜃, 𝜙
2 𝑈 𝜃, 𝜙
0
𝐺𝑡𝑜𝑡 𝜃, 𝜙 = 𝐴 𝜃, 𝜙
2
𝐺0 𝜃, 𝜙
❑ where 𝑈0 𝜃, 𝜙 and 𝐺0 𝜃, 𝜙 are the radiation intensity and the gain function of the single
elements.
❑ The array factor can thus modify significantly the radiation properties of the single elements.
❑ This can be done by properly positioning and feeding the array elements!
9
Linear arrays
ො, 𝑛 =
❑ Let us consider the case of a one-dimensional array with elements located in 𝐝𝑛 = 𝑧𝑛 𝐳
0,1, … , 𝑁 − 1.
❑ A zero-based indexing is used for simplicity, thus assuming
𝑧
that the first element is located in the origin.
𝑧𝑁−1
❑ The array factor is given by:
𝑁−1
𝑁−1
⋮
𝑛=0
𝐫
𝑧𝑛
𝑁−1
𝐴 𝜃, 𝜙 = ෍ 𝑎𝑛 𝑒 𝑗𝐤∙𝐝𝑛 = ෍ 𝑎𝑛 𝑒 𝑗𝑘𝑧𝑛 𝐫ො∙ො𝐳 = ෍ 𝑎𝑛 𝑒 𝑗𝑘𝑧𝑛 cos 𝜃
𝑛=0
𝜃
⋮
𝑛=0
𝑧1
𝑧0 = 0
𝑦
𝑥
10
Linear arrays
❑ Moreover, if the elements are equispaced, i.e., 𝑧𝑛 = 𝑛𝑑 (being 𝑑 the inter-element distance),
the previous relationship reduces to
𝑧
𝑁−1
𝐴 𝜃, 𝜙 = ෍ 𝑎𝑛 𝑒 𝑗𝑛𝑘𝑑 cos 𝜃
⋮
𝑛=0
❑ The angular dependency is in the factor 𝜓 = 𝑘𝑧 𝑑 = 𝑘𝑑 cos 𝜃.
❑ We can express the array factor in terms of 𝜓 as:
𝑁−1
𝑑
𝑧2
𝑑
𝑑
𝐴 𝜓 = ෍ 𝑎𝑛 𝑒 𝑗𝑛𝜓
𝑛=0
𝑧3
𝑧1
𝑧0 = 0
𝑦
𝑥
11
Linear arrays
❑ If the array is located along the 𝑥 or 𝑦 axes, the array factor has the same form, except that 𝜓
becomes:
𝑁−1
𝜓 = 𝑘𝑑 sin 𝜃 cos 𝜙
→
𝐴 𝜃, 𝜙 = ෍ 𝑎𝑛 𝑒 𝑗𝑘𝑥𝑛 sin 𝜃 cos 𝜙
array along 𝑥
𝑛=0
𝑁−1
𝜓 = 𝑘𝑑 sin 𝜃 sin 𝜙
→
𝐴 𝜃, 𝜙 = ෍ 𝑎𝑛 𝑒 𝑗𝑘𝑦𝑛 sin 𝜃 sin 𝜙
array along 𝑦
𝑛=0
𝑧
𝐫
𝛾
❑ In general, it is possible to define 𝜓 = 𝑘𝑑 cos 𝛾, where 𝛾 is the angle between
the line on which the elements are located and the considered direction.
❑ Consequently, without loss of generality, in the following we consider
an array directed along the z axis.
𝑥
𝑦
12
Linear arrays
❑ By using the Euler formula, the array factor can be expressed as
𝑁−1
𝑁−1
𝐴 𝜓 = ෍ 𝑎𝑛 𝑒 𝑗𝑛𝜓 = ෍ 𝑎𝑛 cos 𝑛𝜓 + 𝑗 sin 𝑛𝜓
𝑛=0
𝑛=0
❑ Consequently, it is periodic with respect to 𝜓 with period −𝜋, 𝜋 .
❑ Since 𝜓 = 𝑘𝑑 cos 𝜃, the real range of variation of 𝜓 is −𝑘𝑑, 𝑘𝑑 . Such range is called visible
region, and its “length” is 𝜓𝑣𝑖𝑠 = 2𝑘𝑑.
❑ Depending on the overlapping of these two regions, we can have different cases.
13
Linear arrays
❑ Case 1: 𝑘𝑑 < 𝜋
𝑑<
𝜆
2
→
𝜓𝑣𝑖𝑠 < 2𝜋
❑ Only a limited part of the periodicity range is used;
Visible region
−𝑘𝑑
𝑘𝑑
14
Linear arrays
❑ Case 2: 𝑘𝑑 = 𝜋
𝑑=
𝜆
2
→
𝜓𝑣𝑖𝑠 = 2𝜋
❑ All the virtual period is replicated onto the real angles.
Visible region
−𝑘𝑑
𝑘𝑑
15
Linear arrays
❑ Case 3: 𝑘𝑑 > 𝜋
𝑑>
𝜆
2
→
𝜓𝑣𝑖𝑠 > 2𝜋
❑ The values of 𝐴 are repeated in the visible region → an aliasing effect is present!
Visible region
−𝑘𝑑
𝑘𝑑
16
Linear arrays
❑ Case 4: 𝑘𝑑 ≥ 2𝜋 𝑑 > 𝜆
→
𝜓𝑣𝑖𝑠 > 4𝜋
❑ Replicas of the main lobe, called grating lobes, appears in the visible region!
Visible region
Grating lobes
−𝑘𝑑
𝑘𝑑
17
Linear arrays
𝑑 = 𝜆/4
𝑑 = 𝜆/2
𝑑=𝜆
18
Uniform linear arrays
❑ A uniform linear array is a special case of linear array for which the elements are equally-
spaced and the feed coefficients are all equal, i.e.,
𝑧𝑛 = 𝑛𝑑,
❑ In this case we have:
1
𝑎𝑛 =
𝑁
Geometric series
Setting 𝑧 = 𝑒 𝑗𝜓
𝑁−1
𝑁−1
1 𝑧𝑁 − 1
1 𝑒 𝑗𝑁𝜓 − 1
𝑁𝜓
𝑗 2
𝑁𝜓
𝑗 2
𝑁𝜓
−𝑗 2
1 𝑗𝑛𝜓
1
1𝑒
𝑒
−𝑒
𝑒
= ෍ 𝑧𝑛 =
=
=
𝜓
𝜓
𝑁
𝑁
𝑁 𝑧−1
𝑁 𝑒 𝑗𝜓 − 1
𝑁 𝑗𝜓
𝑗
−𝑗
𝑛=0
𝑛=0
𝑒 2 𝑒 2 −𝑒 2
𝑁𝜓
sin
𝜓
1 𝑗 𝑁−1
2
2
= 𝑒
𝜓
𝑁
sin
2
𝐴 𝜓 =෍
19
Uniform linear arrays
❑ Assuming that the elements are isotropic antennas, the radiation pattern is then given by:
𝑁𝜓 2
𝑁𝑘𝑑 cos 𝜃 2
sin
sin
2
2
2
𝑔 𝜃 = 𝐴 𝜃
=
=
𝜓
𝑘𝑑 cos 𝜃
𝑁 sin
𝑁
sin
2
2
20
Uniform linear arrays
❑ The array factor 𝐴 𝜓 is characterized by 𝑁 − 1 zeros inside the period located in:
𝑁𝜓𝑛
2𝜋
= 𝜋𝑛 → 𝜓𝑛 =
𝑛
2
𝑁
❑ Such zeros are nulls of the radiation pattern only if they are contained in the visible region,
i.e., if 𝜓𝑛 ≤ 𝑘𝑑.
❑ Since 𝜓 = 𝑘𝑑 cos 𝜃, the nulls of the radiation pattern are given by:
𝜃𝑛
= cos−1
2𝜋𝑛
𝑘𝑁𝑑
21
Uniform linear arrays
❑ The beamwidth (with respect to 𝜓) is obtained as:
−3 db
𝜓≪1
𝑁𝜓 2
𝑁𝜓 2
2
sin
sin
𝑁𝜓
1
2
2
≅
= sinc
=
𝜓
𝑁𝜓
2𝜋
2
𝑁 sin
2
2
Δ𝜓𝑏
𝑁𝜓1/2
2𝜋
2𝜋
= 0.443 → 𝜓1/2 =
0.443 → Δ𝜓𝑏 = 2𝜓1/2 = 0.886
2𝜋
𝑁
𝑁
−𝜓1/2
+𝜓1/2
❑ In order to derive the aperture with respect to the real angles, we can note that (for 𝜃 sufficiently
small):
𝑑𝜓
Δ𝜓
𝑑𝜓
≅
→ Δ𝜓𝑏 ≅
Δ𝜃
𝑑𝜃
Δ𝜃
𝑑𝜃 𝜃=𝜋 𝑏
2
22
Uniform linear arrays
❑ Since
𝑑𝜓
𝑑𝜃
= 𝑘𝑑 sin 𝜃 we have:
Δ𝜓𝑏
2𝜋
𝜆
𝜆
Δ𝜓𝑏 = 𝑘𝑑 sin 𝜃 𝜃=𝜋 Δ𝜃𝑏 = 𝑘𝑑Δ𝜃𝑏 → Δ𝜃𝑏 =
= 0.886
= 0.886
≅ 0.886
𝑘𝑑
𝑘𝑁𝑑
𝑁𝑑
𝐿
2
When 𝑁 ≫ 1
When 𝑁 ≫ 1
𝐿 = 𝑁 − 1 𝑑 ≈ 𝑁𝑑
𝑑
⋮
𝑧
23
Uniform linear arrays
❑ An important parameter for antenna arrays is the secondary side-lobe level.
❑ For uniform arrays, the maximum value of the secondary lobe is approximatively located in
the middle between the first two zeros 𝜓1 =
2𝜋
4𝜋
3𝜋
and 𝜓2 = , i.e., for 𝜓 = .
𝑁
𝑁
𝑁
❑ Consequently, we obtain:
𝑁 3𝜋 2
3𝜋 2
sin
sin
1 2
2
𝑁
2
2
𝐴 𝜓
≅
ณ
=
≅ 0.2 2 → −13 𝑑𝐵
3𝜋 =
1 3𝜋
3𝜋
3𝜋
𝜓= 𝑁
𝑁≫1
𝑁 sin
2𝑁
2
2
−13 db
𝜓1
𝜓2
24
Uniform linear arrays
❑ Example. Array factor of a uniform linear array with 𝑁 = 11 and 𝑑 = 𝜆/2.
Δ𝜃𝑏 = 9.2°
3 dB
𝑅 = −13 dB
Null directions 𝜃𝑛 =
⋯ , 56.9°, 68.7°, 79.5°, …
25
Uniform linear arrays
❑ The directivity can be computed by using the relationship 𝐷𝑚𝑎𝑥 =
❑ The beam solid angle is given by
2𝜋
ΔΩ = න
𝜋
−𝑘𝑑
0
=
ณ
𝑑𝜓=−𝑘𝑑 sin 𝜃𝑑𝜃
𝑁−1
2𝜋 𝑘𝑑
=
න 𝑐0 ෍ 𝑎𝑛 𝑒 𝑗𝑛𝜓
𝑘𝑑 −𝑘𝑑
𝑛=0
𝑁−1
=
Normalization constant
න 𝑔 𝜃 sin 𝜃 𝑑𝜃 𝑑𝜙
0
4𝜋
.
ΔΩ
𝑁−1
∗ −𝑗𝑚𝜓
෍ 𝑎𝑚
𝑒
𝑚=0
𝑘𝑑
2𝜋 න
𝑘𝑑
2
𝑁−1
𝑐0 ෍ 𝑎𝑛 𝑒 𝑗𝑛𝜓
𝑛=0
sin 𝜃
𝑑𝜓
−𝑘𝑑 sin 𝜃
𝑁−1
2𝜋 𝑘𝑑
∗ 𝑗𝜓 𝑛−𝑚
𝑑𝜓 =
න 𝑐0 ෍ 𝑎𝑛 𝑎𝑚
𝑒
𝑑𝜓
𝑘𝑑 −𝑘𝑑
𝑛,𝑚=0
𝑁−1
2𝜋
2𝜋
sin 𝑘𝑑 𝑛 − 𝑚
∗
∗
𝑐0 ෍ 𝑎𝑛 𝑎𝑚
න 𝑒 𝑗𝜓 𝑛−𝑚 𝑑𝜓 =
𝑐0 ෍ 𝑎𝑛 𝑎𝑚
2𝑘𝑑
𝑘𝑑
𝑘𝑑
𝑘𝑑
𝑛
−
𝑚
−𝑘𝑑
𝑛,𝑚=0
𝑁−1
= 4𝜋𝑐0 ෍
𝑛,𝑚=0
𝑛,𝑚=0
∗
𝑎𝑛 𝑎𝑚
sin 𝑘𝑑 𝑛 − 𝑚
𝑘𝑑 𝑛 − 𝑚
26
Uniform linear arrays
❑ The pattern normalization constant is equal to
1
𝑐0 =
𝐴 0
2
=
1
2
σ𝑁−1
𝑛=0 𝑎𝑛
❑ Thus we have:
2
σ𝑁−1
𝑛=0 𝑎𝑛
∗ sin 𝑘𝑑 𝑛 − 𝑚
𝑎
𝑎
𝑛,𝑚=0 𝑛 𝑚 𝑘𝑑 𝑛 − 𝑚
4𝜋
𝐷𝑚𝑎𝑥 =
=
ΔΩ σ𝑁−1
❑ Such a relationship is valid for any set of feed coefficients 𝑎𝑛 .
27
Uniform linear arrays
❑ For a uniform array it reduces to:
1
σ𝑁−1
𝑛=0 𝑁
2
𝑁2
𝐷𝑚𝑎𝑥 =
=
1
1
sin
𝑘𝑑
𝑛
−
𝑚
𝑁−1 sin 𝑘𝑑 𝑛 − 𝑚
σ𝑁−1
σ
𝑛,𝑚=0 𝑁 𝑁 𝑘𝑑 𝑛 − 𝑚
𝑛,𝑚=0 𝑘𝑑 𝑛 − 𝑚
❑ If moreover, we have 𝑑 = 𝜆/2 the previous relationship can be further simplified to:
𝑁2
𝑁2
𝑁2
𝑁2
𝐷𝑚𝑎𝑥 =
= 𝑁−1
= 𝑁−1
=
=𝑁
sin
𝜋
𝑛
−
𝑚
σ
σ
𝑁
sinc
𝑛
−
𝑚
sinc
0
𝑛,𝑚=0
𝑛=0
σ𝑁−1
𝑛,𝑚=0 𝜋 𝑛 − 𝑚
28
Planar arrays
❑ In two-dimensional arrays, the elements are located on a planar grid.
❑ Let us consider a rectangular grid with elements equally-spaced along the two direction
(𝑀 × 𝑁 elements with inter-element distances equal to 𝑑𝑥 and 𝑑𝑦 ).
❑ The array factor can be written as
𝑁 elements
𝑧
𝑀−1 𝑁−1
𝐴 𝜃, 𝜙 = ෍ ෍ 𝑎𝑚𝑛 𝑒 𝑗𝐤∙𝐝𝑚𝑛
𝑑𝑦
𝑚=0 𝑛=0
𝑑𝑥
𝑦
❑ where
𝐝𝑚𝑛 = 𝑚𝑑𝑥 𝐱ො + 𝑛𝑑𝑦 𝐲ො
𝑥
29
Planar arrays
❑ Let us assume that
𝑦
𝑥
𝑎𝑚𝑛 = 𝑎𝑚
𝑎𝑛
𝑦
𝑥
❑ where 𝑎𝑚
(𝑚 =, … , 𝑀 − 1) and 𝑎𝑛 (𝑛 = 1, … , 𝑁) are the sets of feed coefficients of two linear arrays
along the 𝑥 and 𝑦 axes. In this case we obtain
𝑀−1 𝑁−1
𝑀−1
𝑁−1
𝑦
𝑥
𝑥 𝑗𝑚𝑑𝑥 𝐤∙ො𝐱
𝐴 𝜃, 𝜙 = ෍ ෍ 𝑎𝑚
𝑎𝑛 𝑒 𝑗𝐤∙ 𝑚𝑑𝑥 𝐱ො+𝑛𝑑𝑦 𝐲ො = ෍ 𝑎𝑚
𝑒
𝑀−1
𝑚=0 𝑛=0
𝑁−1
𝑀−1
𝑚=0
𝑛=0
𝑁−1
𝑦
𝑦
෍ 𝑎𝑛 𝑒 𝑗𝑛𝑑𝑦𝐤∙𝐲ො
𝑦
𝑥 𝑗𝑚𝑘𝑥 𝑑𝑥
𝑥 𝑗𝑚𝜓𝑥
= ෍ 𝑎𝑚
𝑒
෍ 𝑎𝑛 𝑒 𝑗𝑛𝑘𝑦𝑑𝑦 = ෍ 𝑎𝑚
𝑒
෍ 𝑎𝑛 𝑒 𝑗𝑛𝜓𝑦 = 𝐴𝑥 𝜓𝑥 𝐴𝑦 𝜓𝑦
𝑚=0
𝑛=0
𝑚=0
❑ with 𝜓𝑥 = 𝑘𝑑𝑥 sin 𝜃 cos 𝜙 and 𝜓𝑦 = 𝑘𝑑𝑦 sin 𝜃 sin 𝜙
𝑛=0
Array factors of linear arrays
directed along the 𝑥 and 𝑦 axes.
30
Planar arrays
❑ The array factor of a planar array is consequently just equal to the product of the array factor
of two linear arrays!
𝑥 =
❑ In the special case of a uniform planar array, since 𝑎𝑚
1
1
1
𝑦
and 𝑎𝑛 = , we have 𝑎𝑚𝑛 =
𝑀
𝑁
𝑀𝑁
and it results
𝐴 𝜓𝑥 , 𝜓𝑦
𝑀𝜓𝑥
sin
2
=
𝜓
𝑀 sin 𝑥
2
𝑁𝜓𝑦
sin
2
𝜓𝑦
𝑁 sin
2
❑ The radiation parameters can be then obtained similarly to the case of linear arrays.
31
Planar arrays
❑ Example. Visualize the array factor of a uniform rectangular array with 𝑀 = 11 and 𝑁 = 15
elements along the x and y axes, with interelement distances 𝑑𝑥 = 𝑑𝑦 = 0.5𝜆.
32
Phased arrays
❑ In several applications, there is the need of changing, during operation, the direction of the
main lobe (e.g., in smart antennas that adapt to the environment).
❑ For sake of simplicity, in the following we will consider the case of a linear array (the
extension to planar structures is however straightforward).
❑ Let us consider to have an array whose main lobe is perpendicular to the line on which the
elements are located (broadside array), e.g., an uniform linear array.
❑ We want to electronically steer the main lobe, so that it is directed in a direction 𝜃0 . The
corresponding value of 𝜓 is
𝜓0 = 𝑘𝑑 cos 𝜃0
33
Phased arrays
❑ Let 𝐴′ 𝜓 be the desired array factor.
❑ 𝐴′ can be obtained by translating (with respect to 𝜓) an array factor 𝐴 with main lobe
centered in 𝜓 = 0 (as in the case of uniform arrays).
34
Phased arrays
❑ In particular, we have
𝑁−1
𝐴′ 𝜓
= 𝐴 𝜓 − 𝜓0
= 𝐴 𝜓′
𝑁−1
′
𝑗𝑛𝜓
= ෍ 𝑎𝑛 𝑒
= ෍ 𝑎𝑛 𝑒 𝑗𝑛 𝜓−𝜓0
𝑛=0
𝑛=0
𝑁−1
𝑁−1
= ෍ 𝑎𝑛 𝑒 −𝑗𝑛𝜓0 𝑒 𝑗𝜓𝑛 = ෍ 𝑎𝑛′ 𝑒 𝑗𝜓𝑛
𝑛=0
𝑛=0
𝜓 ′ = 𝜓 − 𝜓0 = 𝑘𝑑 cos 𝜃 − 𝑘𝑑 cos 𝜃0
❑ Consequently, such translation can be obtained by changing the feed coefficients with
𝑎𝑛′ = 𝑎𝑛 𝑒 −𝑗𝑛𝜓0
35
Phased arrays
❑ It is worth noting that it is only needed that the feed coefficients 𝑎𝑛 must provide a
broadside array factor 𝐴. It is not necessary that they are constant!
❑ Moreover, it is just sufficient to add a progressive phase shift to the original feed coefficients!
❑ Beam steering however changes some of the parameters previously discussed.
❑ Since 𝜓 = 𝜓 ′ + 𝑘𝑑 cos 𝜃0 and the visible region is still defined by the condition −𝑘𝑑 ≤ 𝜓 ≤
𝑘𝑑, we have:
−𝑘𝑑 1 + cos 𝜃0 ≤ 𝜓 ′ ≤ 𝑘𝑑 1 − cos 𝜃0
❑ The amplitude of the visible region is still 𝜓𝑣𝑖𝑠 = 2𝑘𝑑.
36
Phased arrays
Visible region
Periodicity range
𝜓0
𝜓
𝜓′
37
Phased arrays
❑ In order to avoid grating lobes, we need that 𝜓 ′ ≤ 2𝜋.
❑ Consequently, from the previous condition, we can directly observe that we need:
𝜓 ′ ≤ 𝑘𝑑 1 ± cos 𝜃0
≤ 𝑘𝑑 1 + cos 𝜃0
≤ 2𝜋
𝜆
⇒ 𝑑≤
1 + cos 𝜃0
❑ For broadside arrays (i.e., 𝜃0 = 90°) such conditions reduces to that obtained in the previous
sections (i.e., 𝑑 ≤ 𝜆).
❑ If 𝜃0 = 0,180° (endfire array), the condition becomes 𝑑 ≤ 𝜆/2.
38
Phased arrays
❑ Similarly, it can be found that the beamwidth of a steered “uniform” array becomes:
𝜆
𝑁𝑑
𝜃0 = 0, 𝜋
𝜆
sin 𝜃0 𝑁𝑑
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
2 0.886
Δ𝜃𝑏 =
0.886
❑ In fact, (at least for 𝜃0 ≠ 0, 𝜋), the aperture Δ𝜓𝑏 does not change.
❑ Consequently, since 𝜓 = 𝑘𝑑 cos 𝜃 − 𝑘𝑑 cos 𝜃0 , we have
Δ𝜓𝑏 =
𝑑𝜓
= −𝑘𝑑 sin 𝜃 and consequently:
𝑑𝜃
𝑑𝜓
Δ𝜃𝑏 = 𝑘𝑑 sin 𝜃0 Δ𝜃𝑏
𝑑𝜃 𝜃=𝜃
0
39
Phased arrays
❑ Example: Let us steer the beam steering for a uniform linear array composed by 𝑁 = 11
elements with inter-element spacing 𝑑 = 𝜆/2.
40
Phased arrays
Phased array architectures
Series feed
𝐼0
∠𝜓0
∠𝜓0
…
∠𝜓0
∠𝜓0
Controller
Parallel feed
Controller
∠𝜓0
∠2𝜓0
∠ 𝑁 − 1 𝜓0
𝐼0
41
Multi-beam arrays
❑ By using antenna arrays it is also possible to simply obtain multi-beam radiation patterns,
i.e., with 𝐿 (main) lobes directed in different directions 𝜓𝑙 .
❑ To this end, it is sufficient to sum 𝐿 array factors with the main lobe steered in the wanted
direction.
❑ These array factor can be obtained starting from a set of broadside array factors 𝐴𝑙 by
exploiting the phase variation concept seen in the previous lecture.
❑ Indeed, we have
𝐿
𝐿
𝑁−1
𝑁−1
𝐿
𝑙
𝑁−1
𝑙
𝐴 𝜓 = ෍ 𝑤𝑙 𝐴𝑙 𝜓 − 𝜓𝑙 = ෍ 𝑤𝑙 ෍ 𝑎𝑛 𝑒 𝑗𝑛 𝜓−𝜓𝑙 = ෍ ෍ 𝑤𝑙 𝑎𝑛 𝑒 −𝑗𝑛𝜓𝑙 𝑒 𝑗𝑛𝜓 = ෍ 𝑎𝑛 𝑒 𝑗𝑛𝜓
𝑙=1
𝑙=1
𝑛=0
𝑛=0
𝑙=1
𝑛=0
42
Multi-beam arrays
❑ The set of coefficients needed to obtain the multi-beam pattern is thus given by:
𝐿
𝑙
𝑎𝑛 = ෍ 𝑤𝑙 𝑎𝑛 𝑒 −𝑗𝑚𝜓𝑙
𝑙=1
❑ The weights 𝑤𝑙 (in general complex numbers) define the “importance” of the different lobes.
𝑙
❑ The coefficients 𝑎𝑛 are the feed coefficients that produces the 𝑙th array factor 𝐴𝑙 (usually of
broadside type).
43
Multi-beam arrays
❑ Example. Create a multi-beam
array with 𝐿 = 3 beams in the
directions 60°, 90°, and 135°.
44
Multi-beam arrays
Multi-beam array architecture
Controller
∠𝑎0
∠𝑎𝑁−2
∠𝑎𝑁−2
𝑎𝑁−2
𝑎𝑁−2
…
𝑎′
𝐼0
45
Narrow-beam and low-sidelobe arrays
❑ In several cases, there is the need of having a radiation pattern with a narrow main beam and with
lower side-lobe level than the simple uniform linear array.
❑ Such an aim can be obtained with the following “standard” arrays:
o Binomial array;
o Dolph-Chebyshev array;
o Taylor’s array.
46
Binomial arrays
❑ The binomial array has the property that it does not have secondary lobes. However, it has a
relatively large main lobe.
❑ The feed coefficients and the corresponding array factor are:
𝑎𝑛 =
𝑁−1 !
, 𝑛 = 0,1, … , 𝑁 − 1
𝑛! 𝑁 − 1 − 𝑛 !
𝐴 𝜓 =
𝑗𝜓
𝑗𝜓 𝑁−1
𝑒 2 + 𝑒− 2
=
𝜓
2 cos
2
𝑁−1
❑ The beamwidth (in the 𝜓 variable) is:
−
Δ𝜓3𝑑𝐵 = 4 acos 2
0.5
𝑁−1
47
Binomial arrays
❑ Example. A binomial array with 𝑁 = 11 elements with distance 𝑑 = 𝜆/2. in this case, the
beamwidth is Δ𝜃𝑏 = 20.3° (the corresponding value for a uniform array is Δ𝜃𝑏 = 9.2°).
48
Dolph-Chebyshev arrays
❑ The Dolph-Chebyshev arrays are characterized by secondary lobes of equal amplitude.
❑ Given a desired secondary sidelobe level, they provide the smallest beamwidth.
❑ The array factor is given by:
𝐴 𝜓 = 𝑇𝑁−1 𝑥 ,
𝜓
𝑥 = 𝑥0 cos
2
❑ where 𝑇𝑁−1 is a Chebyshev polynomial of degree 𝑁– 1 and 𝑥0 depends of the desired
sidelobe level.
❑ The feed coefficients are computed numerically.
49
Dolph-Chebyshev arrays
❑ Example. A Dolph-Chebyshev array with 𝑁 = 11 elements, distance 𝑑 = 𝜆/2, and secondary
side-lobe level -20 dB. The beamwidth is Δ𝜃𝑏 ≅ 10° (the corresponding value for a uniform
array is Δ𝜃𝑏 = 9.2°).
50
Narrow-beam and low-sidelobe arrays
❑ Taylor arrays provide an array factor with secondary sidelobe level lower than a predefined
threshold.
❑ The amplitudes of the secondary lobes decrease when the angle increase.
❑ The beamwidth is however larger than
Dolph-Chebyshev arrays (with the same
secondary sidelobe level).
❑ The feed coefficients are given by:
𝑎𝑛 = 𝐼0
2𝑛
𝛼 1−
−1
𝑁−1
2
𝑛 = 0, … , 𝑁 − 1
❑ where 𝐼0 is the modified Bessel function of first kind and zeroth order and N is an odd number (a
similar formula can be written for the even case).
51
Taylor arrays
❑ The parameter 𝛼 depends on the desired sidelobe level 𝑅 (in dB), i.e.,
0
𝛼 = ቐ0.76609 𝑅 − 13.26 0.4 + 0.09834 𝑅 − 13.26
0.12468 𝑅 + 6.3
𝑅 ≤ 13.26
13.126 < 𝑅 ≤ 60
60 < 𝑅 < 120
❑ The beamwidth (in the 𝜓 space) is given by:
Broadening factor
Δ𝜓3𝑑𝐵 = 0.866
2𝜋𝑏
𝑁
❑ where 𝑏 = 0.01330𝑅 + 0.9761 (valid if 20 < 𝑅 < 100 𝑑𝐵)
52
Taylor arrays
❑ Example. A Taylor array with 𝑁 = 11 elements, distance 𝑑 = 𝜆/2, and secondary side-lobe
level -20 dB. The beamwidth is Δ𝜃𝑏 ≅ 11° (the corresponding value for a uniform array is
Δ𝜃𝑏 = 9.2°).
53
Narrow-beam and low-sidelobe arrays
Uniform
Binomial
Dolph-Chebyshev
Taylor
54
Multi-beam array with Taylor coefficients
❑ Example. Create a multi-beam array with 𝐿 = 3 beams in the directions 60°, 90°, and 135°.
Compare the results provided by considering a uniform broadside pattern and a Taylor one.
Uniform
Taylor
55
Master degree in Internet and Multimedia Engineering / Electronic Engineering
Antennas / Advanced Antenna Engineering
Slides of the lectures – Part 5
Circuit representation of antennas
❑ The antenna, when transmitting, is connected to a generator.
❑ It is possible to identify an equivalent circuit that describe this condition.
TX mode
Reference
plane, (𝑉, 𝐼)
Antenna
𝑍𝑔
𝑍𝐴
TX
𝑉
Guiding
structure
Transmitter
Radiated
wave 𝐄, 𝐇
Transmitter
TX antenna
2
Circuit representation of antennas
❑ The total power provided by the generator is given by:
1
1
∗
𝑃𝑡𝑜𝑡 = 𝑅𝑒 𝑉𝐼 = 𝑅𝑒
2
2
𝑉𝑉 ∗
1 2
= 𝑉 𝑅𝑒
𝑍𝐺 + 𝑍𝐴 ∗
2
𝑍𝐺 + 𝑍𝐴
1 2 𝑅𝐺 + 𝑅𝐴
= 𝑉
𝑍𝐺 + 𝑍𝐴 2
2
𝑍𝐺 + 𝑍𝐴 2
❑ Such a power is divided between the generator and antenna impedances, i.e., the power
available on 𝑍𝐺 and 𝑍𝐴 are
1 2
𝑅𝐺
1
𝑉
=
𝑅𝑔 𝐼 2
2
2
𝑍𝐺 + 𝑍𝐴
2
1 2
𝑅𝐴
1
𝑃𝑇 = 𝑉
= 𝑅𝐴 𝐼 2
2
2
𝑍𝐺 + 𝑍𝐴
2
𝑃𝐺 =
𝑍𝑔
𝑍𝐴
𝑉
Transmitter
TX antenna
3
Circuit representation of antennas
❑ In particular, the power 𝑃𝑇 available on the antenna can be divided into two terms, the radiated power
𝑃𝑖𝑟𝑟 and the dissipated power due to losses, i.e.,
1
1
1
𝑃𝑇 = 𝑅𝐴 𝐼 2 = 𝑃𝑟𝑎𝑑 + 𝑃𝑑𝑖𝑠𝑠 = 𝐼 2 𝑅𝑟𝑎𝑑 + 𝐼 2 𝑅𝑑𝑖𝑠𝑠
2
2
2
❑ Consequently, as we already introduced in the initial part of the course, in TX mode the antenna can be
modeled as an impedance 𝑍𝐴 = 𝑅𝐴 + 𝑗𝑋𝐴 , where:
o The reactive part 𝑋𝐴 is related to the energy stored in the near-field region.
o The real part 𝑅𝐴 = 𝑅𝑟𝑎𝑑 + 𝑅𝑙𝑜𝑠𝑠 accounts for the radiated power and the eventual losses (e.g., due to
non-idealities of the materials). The quantity 𝑅𝑟𝑎𝑑 is the radiation resistance.
4
Circuit representation of antennas
❑ The radiation efficiency of the antenna is thus expressed as:
𝑃𝑟𝑎𝑑 𝑅𝑟𝑎𝑑
𝑒=
=
𝑃𝑇
𝑅𝐴
Neglecting losses, 𝑒 = 1 and
consequently 𝑅𝐴 = 𝑅𝑟𝑎𝑑
❑ In order to provide the maximum power to the antenna, it is necessary to have power matching
between the generator and the antenna, i.e.,
𝑍𝐴 = 𝑍𝐺∗
𝑅𝐴 = 𝑅𝐺 ,
𝑋𝐴 = −𝑋𝐺
❑ In this case, the power available to the antenna is given by:
1
1 𝑉2
𝑃𝑇𝑚𝑎𝑥 = 𝑃𝑡𝑜𝑡 =
2
8 𝑅𝐺
5
Receiving antennas
❑ When used in reception, the antenna capture part of the power of the electromagnetic wave
that impinges on it.
❑ In particular, the impinging field is “transformed” into a voltage/current at the output of the
antenna, i.e., 𝐄, 𝐇 → 𝑉, 𝐼 .
RX mode
Receiver
Antenna
𝑍𝐴
RX
𝑍𝐿
Reference
plane, (𝑉, 𝐼)
𝑉
Impinging
wave 𝐄, 𝐇
RX antenna
Receiver
6
Receiving antennas
❑ From a circuital point of view, the power available to the load can be obtained as we did for
the TX mode, i.e.,
1 2
𝑅𝐿
1
2
𝑃𝑟𝑖𝑐 = 𝑉
=
𝑅
𝐼
𝐿
2
𝑍𝐿 + 𝑍𝐴 2 2
❑ In particular, in order to have the maximum power on the load, it is now required to have
𝑍𝐿 = 𝑍𝐴∗ (matching condition).
❑ In this case, the power available on the load is given by:
1 𝑉2
𝑃𝑟𝑖𝑐𝑚𝑎𝑥 =
8 𝑅𝐴
7
Receiving antennas
❑ The power available on the load can be also written as
𝑃𝑟𝑖𝑐 = 𝐴𝑒𝑓𝑓 𝜃𝑖𝑛𝑐 , 𝜙𝑖𝑛𝑐 𝑝𝑖𝑛𝑐 (𝐫𝑎𝑛𝑡 )
❑ where 𝑝𝑖𝑛𝑐 is the power density of the EM wave in correspondence to the antenna and
𝐴𝑒𝑓𝑓 𝜃, 𝜙 is the effective area of the antenna.
❑ Such a quantity is usually a fraction of the physical area of the antenna and depends upon
the direction of incidence of the imping wave.
❑ The power available to the load can also be obtained by considering the value of the
(eventually equivalent) voltage at the antenna terminals.
8
Receiving antennas
❑ By using the reciprocity principle, the voltage at the output of the RX antenna can be written as:
𝑉 = 𝐄𝑖𝑛𝑐 𝐫𝑎𝑛𝑡 ∙ 𝐡 𝜃𝑖𝑛𝑐 , 𝜙𝑖𝑛𝑐
❑ where 𝐄𝑖𝑛𝑐 is the electric field impinging on the antenna (we assume that it is generated in the far-
field region) and 𝐡 𝜃, 𝜙 is the effective length.
❑ The effective length is defined by considering the antenna in TX mode as:
𝐅⊥ 𝜃, 𝜙
𝐡 𝜃, 𝜙 = −
𝐼𝑖𝑛
෡ is the
෡ + 𝐹𝜙 𝜃, 𝜙 𝛟
❑ where 𝐼𝑖𝑛 is the current in input to the antenna and 𝐅⊥ 𝜃, 𝜙 = 𝐹𝜃 𝜃, 𝜙 𝛉
transverse component of the radiation vector.
9
Receiving antennas
❑ It is worth noting that 𝐡 has the same direction of the radiated field, since
𝑒 −𝑗𝑘𝑟
𝐄 r = 𝑗𝑘𝜂
𝐼 𝐡 𝜽, 𝝓
4𝜋𝑟 𝑖𝑛
❑ Consequently, the effective length define the polarization of the antenna (both in TX and RX
modes).
❑ If the receiving antenna has a polarization which is different from that of the imping EM
wave, a polarization mismatch is present.
❑ Such mismatch is quantified by the following polarization efficiency:
𝐄𝑖𝑛𝑐 ∙ 𝐡 2
𝑒𝑝𝑜𝑙 =
𝐄𝑖𝑛𝑐 2 𝐡 2
10
Receiving antennas
❑ Clearly, the maximum value of the polarization efficiency, i.e., 𝑒𝑝𝑜𝑙 = 1, is obtained when the
effective length vector is parallel to the polarization vector of the impinging field.
❑ If the antenna and the load are not perfectly matched, load mismatch losses are also present.
❑ If 𝑍𝐿 is the impedance of the load and 𝑍𝐴 the impedance of the antenna, it is possible to
define a load mismatch efficiency as:
𝑒𝑙𝑜𝑎𝑑 =
4𝑅𝐿 𝑅𝐴
𝑍𝐿 + 𝑍𝐴 2
❑ In this case, the maximum value is achieved when the antenna is matched to the load, i.e.,
𝑍𝐴 = 𝑍𝐿∗ .
11
Receiving antennas
❑ The effective area can be expressed in terms of the effective length as
1
𝑅𝐿 𝐼 2
𝑃𝑟𝑖𝑐
𝜂𝑅𝐿 𝑉 2
𝜂𝑅𝐿 𝐄𝑖𝑛𝑐 ∙ 𝐡 𝜃, 𝜙 2
2
𝐴𝑒𝑓𝑓 𝜃, 𝜙 =
=
=
2 𝐄
2=
1
𝑝𝑖𝑛𝑐
𝑍
+
𝑍
𝑍𝐴 + 𝑍𝐿 2 𝐄𝑖𝑛𝑐 2
𝐴
𝐿
𝑖𝑛𝑐
𝐄𝑖𝑛𝑐 2
2𝜂
𝜂 4𝑅𝐴 𝑅𝐿 𝐄𝑖𝑛𝑐 ∙ 𝐡 𝜃, 𝜙 2
𝜂
2
2
=
𝐡
𝜃,
𝜙
=
𝑒
𝑒
𝐡
𝜃,
𝜙
𝑝𝑜𝑙
𝑙𝑜𝑎𝑑
4𝑅𝐴 𝑍𝐴 + 𝑍𝐿 2 𝐄𝑖𝑛𝑐 2 𝐡 𝜃, 𝜙 2
4𝑅𝐴
❑ Similarly, the gain function can be expressed as:
2
1
𝐼
𝐡
𝜃,
𝜙
2 2
𝑖𝑛
𝐄 𝐫 𝑟
𝑟2
𝑈 𝜃, 𝜙
4𝜋 𝑘𝜂 4𝜋𝑟
𝜋𝜂
2𝜂
2
𝐺 𝜃, 𝜙 = 4𝜋
= 4𝜋
=
=
𝐡
𝜃,
𝜙
1
𝑃𝑇
𝜂
𝑅𝐴 𝐼𝑖𝑛 2
𝜆2 𝑅𝐴
𝑅𝐴 𝐼𝑖𝑛 2
2
12
Receiving antennas
❑ Consequently, the following relationship between the gain function and the effective area is
obtained:
𝜂 𝐡 𝜃, 𝜙
𝐴𝑒𝑓𝑓 𝜃, 𝜙 = 𝑒𝑝𝑜𝑙 𝑒𝑙𝑜𝑎𝑑
4𝑅𝐴
2
𝜂 𝜆2 𝑅𝐴
𝜆2
= 𝑒𝑝𝑜𝑙 𝑒𝑙𝑜𝑎𝑑
𝐺 𝜃, 𝜙 = 𝑒𝑝𝑜𝑙 𝑒𝑙𝑜𝑎𝑑
𝐺 𝜃, 𝜙
4𝑅𝐴 𝜋𝜂
4𝜋
❑ In case of polarization and load match (i.e., 𝑒𝑝𝑜𝑙 = 𝑒𝑙𝑜𝑎𝑑 = 1), the previous relationship
reduces to
𝜆2
𝐴𝑒𝑓𝑓 𝜃, 𝜙 =
𝐺 𝜃, 𝜙
4𝜋
13
Receiving antennas
❑ Consequently, in this case we have
𝐺 𝜃, 𝜙 = 4𝜋
𝐴𝑒𝑓𝑓 𝜃, 𝜙
𝜆2
❑ In the direction of maximum radiation, it results:
𝐴𝑒𝑓𝑓
𝐺𝑚𝑎𝑥 = 4𝜋 2
𝜆
❑ where 𝐴𝑒𝑓𝑓 is the effective area in the direction of maximum radiation / reception of the
antenna under the hypothesis of polarization and load matching.
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Link between antennas
❑ Let us now consider a wireless system composed by two communicating antennas, as shown
below.
TX Antenna
RX Antenna
𝑃𝑇
𝑃𝑟𝑖𝑐
𝑇𝑋 , 𝐴𝑇𝑋
𝐺𝑚𝑎𝑥
𝑒𝑓𝑓
𝑅
𝑅𝑋 , 𝐴𝑅𝑋
𝐺𝑚𝑎𝑥
𝑒𝑓𝑓
❑ Using the definition of the gain, the power density of the wave impinging on the RX
antennas can be expressed as
𝑇𝑋
𝑇𝑋
𝑃𝑇 𝐺𝑚𝑎𝑥
𝑃𝐸𝐼𝑅𝑃
𝑝𝑅 =
=
4𝜋𝑅2
4𝜋𝑅2
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Link between antennas
❑ The power at the output of the antenna is thus given by:
𝑇𝑋 𝐴𝑅𝑋
𝑃
𝐺
𝑇
𝑚𝑎𝑥
𝑒𝑓𝑓
𝑃𝑟𝑖𝑐 = 𝑝𝑅 𝐴𝑅𝑋
=
𝑒𝑓𝑓
4𝜋𝑅2
𝐹𝑟𝑖𝑖𝑠 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
❑ By using the expression of the effective area with respect to the gain, such a relationship can
be rewritten as:
𝑇𝑋 𝐺 𝑅𝑋 𝜆2
𝑃𝑇 𝐺𝑚𝑎𝑥
𝑚𝑎𝑥
𝑃𝑟𝑖𝑐 =
𝑒𝑝𝑜𝑙 𝑒𝑙𝑜𝑎𝑑
(4𝜋𝑅)2
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Link between antennas
❑ The quantity
4𝜋𝑅
𝐿𝑓 =
𝜆
❑ is the free-space attenuation. 𝐺𝑓 =
2
1
is the corresponding free-space gain.
𝐿𝑓
❑ By using a decibel notation, the Friis formula can be also written as:
𝑇𝑋 𝐺 𝑅𝑋
𝑃𝑇 𝐺𝑚𝑎𝑥
𝑚𝑎𝑥
𝑇𝑋
𝑅𝑋
𝑃𝑟𝑖𝑐 =
𝑒𝑝𝑜𝑙 𝑒𝑙𝑜𝑎𝑑 → 𝑃𝑟𝑖𝑐𝑑𝐵𝑊 = 𝑃𝑇𝑑𝐵𝑊 + 𝐺𝑚𝑎𝑥
+
𝐺
− 𝐿𝑓𝑑𝐵 + 𝑒𝑑𝐵
𝑚𝑎𝑥
𝑑𝐵
𝑑𝐵
𝐿𝑓
17
Link between antennas
❑ Example: Let us consider a TV satellite at height 𝐻 ≅ 36000 𝑘𝑚. The EIRP of the satellite is
53 𝑑𝐵𝑊. The receiver on earth is equipped with a parabolic dish of diameter 𝐷𝑅𝑋 = 80 𝑐𝑚
and with efficiency 𝑒𝑎 = 65%. Assuming a working frequency of 12 GHz, compute the gain
of the antennas and the power received on earth.
❑ The RX antenna gain is:
2
𝐴
𝜋𝐷
𝜋0.8
𝑅𝑋
𝑅𝑋
𝑅𝑋
𝐺𝑚𝑎𝑥 = 4𝜋 2 𝑒𝑎 = 𝑒𝑎
= 0.65
𝜆
𝜆
0.025
2
= 6569 → 38.2 dB
18
Link between antennas
❑ The free-space attenuation is given by:
4𝜋𝑅 2
4𝜋36000 × 103
𝐿𝑓 =
=
𝜆
0.025
2
= 3.3 1020 → 205 dB
❑ Consequently, it results (assuming 𝑒𝑝𝑜𝑙 = 𝑒𝑙𝑜𝑎𝑑 = 1):
𝑅𝑋
𝑃𝑒𝑖𝑟𝑝 𝐺𝑚𝑎𝑥
𝑃𝑟𝑖𝑐 =
= 4 10−12 W = 4 pW → 𝑃𝑅 𝑑𝐵𝑚 = −84 dBm
𝐿𝑓
19
Link between antennas
❑ When dealing with a wireless link, however, the knowledge of the received power is usually
not sufficient.
❑ Indeed, the antenna also captures part of the environmental electromagnetic noise. → A
more useful information is the signal-to-noise ratio at the receiver!
❑ If 𝐵 is the noise bandwidth (as a first approximation, the antenna bandwidth can be
considered), the noise power at the terminals of the antenna can be expressed as:
𝑁𝑎𝑛𝑡 = 𝑘𝐵𝑇𝑎𝑛𝑡
❑ where 𝑘 = 1.3803 10−23 W/Hz is the Boltzmann constant and 𝑇𝑎𝑛𝑡 is the noise temperature
of the antenna.
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Link between antennas
❑ The antenna noise temperature depends on the power density of the radiated noise that
impinge on the antenna.
❑ Generally, a noise source can be characterized by its brightness temperature 𝑇 𝜃, 𝜙 .
❑ The power of the noise radiated in the direction 𝜃, 𝜙 is then equal to 𝑁 𝜃, 𝜙 = 𝑘𝑇 𝜃, 𝜙 𝐵.
❑ The noise temperature of the antenna is thus defined as:
1 𝜋 2𝜋
𝑇𝑎𝑛𝑡 =
න න 𝑇 𝜃, 𝜙 𝑔 𝜃, 𝜙 𝑑Ω
ΔΩ 0 0
21
Link between antennas
❑ Let us first consider an antenna located on the earth and pointing towards the sky.
❑ In this case, we have an extended noise source (the sky), for which we can assume that
𝑇 𝜃, 𝜙 is about constant and equal to a value 𝑇𝑠𝑘𝑦 .
❑ Consequently, we have:
1
𝑇𝑎𝑛𝑡 = 𝑇𝑠𝑘𝑦
න 𝑔 𝜃, 𝜙 𝑑Ω
ΔΩ Ω
❑ where Ω is the solid angle towards the extended noise source.
Ω
22
Link between antennas
❑ Let us now consider an antenna located on the earth that points horizontally.
❑ Consequently, it receives part of the noise generated by the earth and part of the noise due to
the sky.
❑ In this case we have:
ΔΩ𝑔𝑟𝑜𝑢𝑛𝑑
ΔΩ𝑠𝑘𝑦
𝑇𝑎𝑛𝑡 = 𝑇𝑠𝑘𝑦
+ 𝑇𝑔𝑟𝑜𝑢𝑛𝑑
= 𝑒𝑠𝑘𝑦 𝑇𝑠𝑘𝑦 + 𝑒𝑔𝑟𝑜𝑢𝑛𝑑 𝑇𝑔𝑟𝑜𝑢𝑛𝑑
ΔΩ
ΔΩ
❑ where ΔΩ𝑠𝑘𝑦 and ΔΩ𝑔𝑟𝑜𝑢𝑛𝑑 are the solid angle regions
containing the earth and sky noise sources.
Ω𝑠𝑘𝑦
Ω𝑔𝑟𝑜𝑢𝑛𝑑
23
Link between antennas
❑ Example: Let us consider an antenna located on earth. The fraction of the beam that “sees”
the sky is 𝑒𝑠𝑘𝑦 = 80%. Calculate the antenna noise temperature.
❑ Typical values of the noise temperature of sky and earth are
𝑇𝑠𝑘𝑦 = 4 °𝐾,
𝑇𝑔𝑟𝑜𝑢𝑛𝑑 = 290 °𝐾
❑ Consequently, we have
𝑇𝑎𝑛𝑡 = 4°𝐾 0.8 + 290 °𝐾 0.2 = 32.6 °𝐾
24
Link between antennas
❑ The receiver can be thought as composed by the block shown below.
𝑃𝑟𝑖𝑐
𝑁𝑎𝑛𝑡
Transmission line
/ guiding structure
Low-noise
amplifier (LNA)
Receiver (downmix,
IF filter, …)
𝑃𝑜𝑢𝑡
𝑁𝑜𝑢𝑡
❑ Each block introduce its own noise and can be modeled as shown below.
Gain (𝐺)
𝑃𝑖 , 𝑁𝑖
+
Noise (𝑇𝑒 )
𝑃𝑜 , 𝑁𝑜
25
Link between antennas
❑ We can thus model the receiving chain as shown below.
Transmission line
𝑃𝑟𝑖𝑐
𝑁𝑎𝑛𝑡
Receiver
LNA
𝐺𝑙𝑖𝑛𝑒
𝐺𝑙𝑛𝑎
𝐺𝑟𝑒𝑐
+
+
+
𝑇𝑙𝑖𝑛𝑒
𝑇𝑙𝑛𝑎
𝑇𝑟𝑒𝑐
𝑃𝑟𝑖𝑐
𝑁𝑎𝑛𝑡
𝐺
𝑃𝑜𝑢𝑡
𝑁𝑜𝑢𝑡
𝑃𝑜𝑢𝑡
𝑁𝑜𝑢𝑡
+
𝑇𝑒
26
Link between antennas
❑ The overall system can be modeled using a single block with
𝐺 = 𝐺𝑙𝑖𝑛𝑒 𝐺𝑙𝑛𝑎 𝐺𝑟𝑒𝑐
1
1
𝑇𝑒 = 𝑇𝑙𝑖𝑛𝑒 +
𝑇 +
𝑇
𝐺𝑙𝑖𝑛𝑒 𝑙𝑛𝑎 𝐺𝑙𝑖𝑛𝑒 𝐺𝑙𝑛𝑎 𝑟𝑒𝑐
❑ Concerning the transmission line, the following models can be used :
𝐺𝑙𝑖𝑛𝑒 = 𝑒 −2𝛼𝑙
1
𝑇𝑙𝑖𝑛𝑒 =
1 − 𝐺𝑙𝑖𝑛𝑒 𝑇𝑝ℎ𝑦𝑠
𝐺𝑙𝑖𝑛𝑒
❑ where 𝛼 is the attenuation, 𝑙 is the length, and 𝑇𝑝ℎ𝑦𝑠 is the physical temperature of the line (in °K).
27
Link between antennas
❑ Consequently, we have that the overall noise temperature of the system antenna + receiver is:
1
1
1
𝑇𝑠𝑦𝑠 = 𝑇𝑎𝑛𝑡 + 𝑇𝑒 = 𝑇𝑎𝑛𝑡 +
1 − 𝐺𝑙𝑖𝑛𝑒 𝑇𝑝ℎ𝑦𝑠 +
𝑇 +
𝑇
𝐺𝑙𝑖𝑛𝑒
𝐺𝑙𝑖𝑛𝑒 𝑙𝑛𝑎 𝐺𝑙𝑖𝑛𝑒 𝐺𝑙𝑛𝑎 𝑟𝑖𝑐
❑ From the previous relationship, we can note that it is important to use a LNA in input to the
receiver.
❑ Indeed, high values of 𝐺𝑙𝑛𝑎 allow to reduce the contributions of the noise temperature of the
receiver 𝑇𝑟𝑖𝑐 . Moreover, since such amplifiers are designed in order to have low values of
𝑇𝑙𝑛𝑎 (between 20 °K for a cooled system and 100 °K at ambient temperature), their
contribution on the overall noise temperature is low.
❑ Clearly, the LNA should be positioned as near as possible to the antenna, in order to limit the
effects of 𝐺𝑙𝑖𝑛𝑒 .
28
Link between antennas
❑ Finally, the signal-to-noise ratio at the receiver output can be expressed as:
𝑇𝑋
𝑅𝑋
𝑃𝑜𝑢𝑡
𝐺𝑃𝑟𝑖𝑐
𝑃𝑟𝑖𝑐
𝑃𝑇 𝐺𝑚𝑎𝑥
𝐺𝑚𝑎𝑥
𝑆𝑁𝑅𝑜𝑢𝑡 =
=
=
=
𝑒 𝑒
𝑁𝑜𝑢𝑡 𝐺𝑘 𝑇𝑒 + 𝑇𝑎𝑛𝑡 𝐵 𝑘𝑇𝑠𝑦𝑠 𝐵
𝐿𝑓 𝑘𝑇𝑠𝑦𝑠 𝐵 𝑝𝑜𝑙 𝑙𝑜𝑎𝑑
𝑅𝑋
1 𝐺𝑚𝑎𝑥
1
𝑇𝑋
= 𝑃𝑇 𝐺𝑚𝑎𝑥
𝑒𝑝𝑜𝑙 𝑒𝑙𝑜𝑎𝑑
𝐿𝑓 𝑇𝑠𝑦𝑠 𝑘𝐵
𝑃𝐸𝐼𝑅𝑃
𝑅𝑋
𝐺
𝑚𝑎𝑥
𝑇𝑋
𝑆𝑁𝑅𝑜𝑢𝑡𝑑𝐵 = 𝑃𝑇 𝐺𝑚𝑎𝑥
𝑑𝐵 − 𝐿𝑓𝑑𝐵 +
𝑇𝑠𝑦𝑠
𝑃𝐸𝐼𝑅𝑃𝑑𝐵
❑ The quantity
𝐺𝑅
𝑇𝑠𝑦𝑠
− 𝑘𝑑𝐵 − 𝐵𝑑𝑏 + 𝑒𝑑𝐵
𝑑𝐵
is the figure of merit of the receiver and takes into account both the
receiver itself and the antenna.
29
Link between antennas
❑ Example: Let us consider a wireless link with : 𝑇𝑎𝑛𝑡 = 40 °𝐾, 𝐿𝑙𝑖𝑛𝑒𝑑𝐵 = −𝐺𝑙𝑖𝑛𝑒𝑑𝐵 = 0.1 𝑑𝐵,
𝑅𝑋
𝑇𝑝ℎ𝑦𝑠 = 290 °𝐾, 𝐺𝑙𝑛𝑎 = 50 𝑑𝐵, 𝑇𝑙𝑛𝑎 = 80 °𝐾, 𝑇𝑟𝑖𝑐 = 2000 °𝐾, 𝐺𝑟𝑖𝑐
= 45 𝑑𝐵. Calculate the noise
temperature and the figure of merit.
❑ The system noise temeprature is equal to
1
1
1
𝑇𝑠𝑦𝑠 = 40 +
− 1 290 + −0.1/10 80 + 5 −0.1/10 2000 = 128.62 °𝐾 ≅ 21 𝑑𝐵𝐾
10−0.1/10
10
10 10
❑ The figure of merit of the receiver is thus equal to
𝐺𝑅
𝑇𝑠𝑦𝑠
= 45 − 21 = 24 𝑑𝐵/𝐾
𝑑𝐵
30