Julio César García-Álvarez
Digital Electronic
Communications
Digital Electronic Communications
Julio César García-Álvarez
Digital Electronic
Communications
Julio César García-Álvarez
Universidad Nacional de Colombia
Manizales, Colombia
ISBN 978-3-031-53117-0
ISBN 978-3-031-53118-7
https://doi.org/10.1007/978-3-031-53118-7
(eBook)
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To my wife Monica, who dedicated her time
and energy to allow me to complete this
work.
To my mother Olga, who never hesitates on
my abilities and constantly asks about the
progress of this work.
To my children Isabella and Samuel, which is
why this work never has been left and
continues abiding until its completion.
To Professor Neil Guerrero González, as
evidence that this text can lead to the
formation of great engineers.
Preface
This book is based on practices carried out in the facilities of the Electricity
and Electronics Laboratory of the National University of Colombia. The book is
aimed at intermediate to postgraduate engineering students. The above implies
prior knowledge in integral calculus, differential calculus, circuit theory, and
statistics. However, a section is left to the reader to assess their knowledge, and
if reinforcement is required, previous sections are provided in the mentioned areas.
Each chapter is prepared in such a way that each item in the problem section is
left to student for a formulation of a laboratory report, only dedicating themselves
to filling in the concepts, results, and conclusions. This saves time for the student
to properly prepare a report, enhancing the progress for a clear report (Ballesteros
Horta and Hernández Maldonado, 2007). At the same time, as an example, it teaches
the student to formulate a technical report (Imbermón Muñoz et al. 2011). Electrical
element values, such as those for integrated circuits, capacitors, and resistances, are
calculated as a result of device electronic design and presented in the appropriate
tables to match design specifications, such as those for commercially available
electronic components. Moreover, the simulation must be done with the real values
obtained in the market. The implementation must be done on a PCB printed circuit,
including measurement access points (Jacks if possible). Universal or prefabricated
plates are not accepted. The breadboard will only be used for testing purposes.
Though the simulation can be carried out in professional software like Easy
EDA and the PCB in Proteus ARES, the design can be done with some other
simulation and design program. Table 1 illustrates the test points arranged in the
circuit and the measured variable.
The activities in this book suggest practical use in the laboratory, for proper
performance. It is important to clarify that this book is specific for the Telecommunications area, which is why it also requires specialized equipment. Therefore,
this book recommends the use of the following modes: (i) simulation using a
computational tool suitable for this purpose and (ii) assembly, using a laboratory
room with at least the following elements illustrated in Table 1 (Fig. 1).
vii
viii
Preface
Table 1 A model table for
system testing of Fig. 1
Test point
TP1
TP2
TP3
TP4
Variable
Fig. 1 Communication system with test points
Simulation Software This book illustrates an exploration of different software
environments that were useful in their time, but that gave way to others that
facilitate some processes. For this reason, some examples illustrate codes from
previous and other recent software environments. The list of languages or
applications is illustrated below:
– Easy Java Simulator (EJS) graphical interface can create laboratory test beds
by algorithm routines. Concepts from the Deitel et al. (2012) book can be used
as a starting point for learning about J ava programming, concepts taken from
the Deitel et al. (2012) book are taken as a basis.
– For electrical circuit analysis, the simulation is carried out in Proteus
ISIS and the PCB in Proteus ARES. However, the design can be done
with some other simulation and design program.
– For system modelling, we use MATLAB.
Signal Generator, used for arbitrary waveform generation and the carrier.
Measuring Instrument Two-channel oscilloscope and accessories (two BNC
(coaxial) cables, two oscilloscope probes, plastic screwdriver, and a bifilar cable).
It is possible to use some standard equipment that has a Baseband Transmission
System and the required transmission components, such as coaxial, optical
fiber or twisted pair wire. This book uses the PRO- 47 MAX EC-696 Analog
Communications Trainer for most of its experiments.
Case Study Methodology
The Case Study is a detailed analysis of a specific subject, in this case a communication system. It links laboratory work with technical reports. This allows the student
to generate a technical report (Cassani 2021; Garcia-Alvarez 2018). The following
are some cases used as practical applications:
Preface
ix
• Develop a communication system between instruments to control an irrigation
system.
• Implement a lighting control system that at the same time allows data transfer
through the light beam.
The Case Study Report
A general case study presents the following reports:
1. Technical report with the solution of telecommunications problems from a
systemic point of view.
2. Quote of the equipment, based on the specifications given in the technical report.
This document includes prices and must be raised from an established company,
including a quote for labor and cost of engineer(s), in addition to equipment and
space rentals.
3. Any proposal on the implementation of the network to be considered, in a real
region. Additionally, this must be proposed under a company name proposed by
the group or already established, if it exists.
In this book, the technical report will be specified, which contains the following
sections: (i) Introduction, (ii) Problem Statement/Theoretical Framework, and (iii)
Experimental Framework. Each case study is then linked to laboratory work. The
steps to follow are illustrated below.
The Introduction must be written in a continuous manner, that is, without
interruption or bullet points. This section consists of the following elements: (i)
search for information through keywords; (ii) list of at least three works published
in the last 5 years; these works must be published in databases such as IEEEXplore,
SCOPUS, or ScienceDirect; (iii) the problem statement, which is made up of one of
the following two aspects:
1. What the three published works have not been able to resolve.
2. An improvement or change of application of what was done by these three works.
This allows you to prepare the proposal (objective of the work), which is the final
part of the introduction.
The problem statement is distributed in the document as follows: (1) The problem
explained in a general and summarized way, in the introduction section; (2) the
problem and the proposal, expanded in the Background (Theoretical Framework)
and the proposal (Proposed method). For Part 1. When the problem is written in the
introduction, it must be written in a continuous manner, that is, without interruption
or bullet points. This section must relate to the three works published in the last 5
years. As before, these works must be published in databases such as IEEEXplore,
SCOPUS, or ScienceDirect. Thus, it is possible to prepare the proposal (objective
of the work), which is the final part of the introduction.
x
Preface
For Part 2. The problem statement is made up of one of the following two
aspects:
• What the three referenced works have not been able to resolve.
• An improvement or change of application of what was done by these three works.
As the problem has to do with communication systems, the following aspects
will be considered in this work:
• The characteristics of the source: Bandwidth in Baseband; Operating frequency (modulation) in its most common applications: Analog/Digital, Random/Deterministic.
• The characteristics of the channel: Channel Bandwidth (if the modulation is
analog) and Channel Capacity (if the modulation is digital). Modulation scheme
to use. Problems due to the channel.
• Noise model: Gaussian/Weibull/Uniform, type of interference. How do these
affect the modulation scheme?
• Expected issues: Signal dropout, SNR loss, low fidelity, transmission errors.
Up to this point, there are three sections: Introduction, Background, and Proposal.
Next, the Experimental Framework is added. In Communications Systems, this section identifies three groups: (1) Signal Variation due to Noise. Evidenced by Signal
Distortion, Aliasing, and statistical disparities. (2) Disparity in data. Evidenced by
Inter-Symbol Interference, Bit Error Rate, and Lost Packet measurement. (3) The
“Bottlenecks” in the Canal. Evidenced by the growth in bandwidth consumption
and the limitations given by the BER versus .Eb/N0 ratio. With the above, a brief
paragraph of each of these problems is described, and it is expanded in the case that
corresponds to the case study, justifying their relationship.
Two sections remain: (iv) the Results section illustrates the values found
through the Experimental Framework; (v) at the end, in the Conclusions section,
among other questions, the following question should be answered: Does the
proposed modulation scheme meet the conditions for the adequate transmission of a
signal?
For the assessment, the following guidelines are recommended (Eide et al.
2023):
1. Presentation. Most text editing software comes with templates. Although it does
not have to be in IEEE or APA format, it must meet the following conditions:
(i) Summary. (ii) Introduction section, with the definition of the system to
be implemented; (iii) Theoretical Framework, where they clearly define the
technology to which they applied the communication system; (iv) Experimental
Framework, where the variables to be measured, the software used, and the
analysis methodology of the communication system are defined; (v) Results,
where each figure and table of results obtained are explained; (vi) Conclusions,
with what was obtained from the results; (vii) References.
Preface
xi
2. Fluency about the definition. The introduction section should not be longer than
half a page. You will be graded if you do not explain the system coherently.
3. Clarity in the Theoretical Framework. This is a complete explanation of the technology to which they applied the communication system. It must be explained by
yourselves, so it’s essential that the redaction be original. If a figure or diagram
is used, it must also be explained, and the source from which it was extracted
should be named.
4. Clarity in the experimental methodology. This section is a compilation of all
the practices carried out during the semester, through the proposed system.
This is why it is the most important section. It is necessary to declare all the
variables associated with the system to be analyzed, including noise, attenuation,
sensitivity, bandwidth, and so on. Additionally, the transmitter, channel, and
receiver stages of the system where the variable is measured or analyzed, their
generation method (which need not involve source codes), and the expected
results of the analysis must be stated.
5. Results. A suitable presentation of the findings is included in this section.
Since given data are frequently shown in tables and figures, the text should
use connection terms like “Results of Table (...)” or “As it is shown in Fig.
(...)” when referencing given data. Any occurrence of “As illustrated in the
following.. . .” will be graded. Deliver, in respective tables, the calculation and
measurement procedures for the values of resistances, capacitances, amplifiers,
and other elements, in such a way that they coincide with the design parameters.
The simulation must be done with the real values obtained in the market.
Measurement access points must be described and listed.
6. Conclusions. The more consistent the conclusions are with each result, the better.
7. References. Web links are not accepted. All references cited must be in the
document.
Manizales, Colombia
December, 2023
Julio César García-Álvarez
References
Ballesteros Horta RUCDLV, Hernández Maldonado A (2007) La Plataforma Interactiva MOODLE para el Aprendizaje basado en Tareas. In: INFORMATICA
2007, p 32
Cassani D (2021) El arte de dar clases. Anagrama
Deitel P, Deitel H, Vidal REA, Martinez RR, Acosta ID (2012) Java: Cómo
Programar, 9th edn. Pearson Educación de México, Naucalpan de Juárez
Eide A, Mickelson S, Eide CL, Jenison R, Northup L (2023) Engineering Fundamentals and Problem Solving, 8th edn. McGraw Hill, New York
xii
Preface
Garcia-Alvarez JC (2018) Un acercamiento a laboratorios remotos embebidos en el
aula de clase. In: Quinto Congreso Internacional de Innovación Educativa CIIE,
Monterrey, pp 1459–1465
Imbermón Muñoz F, Silva García P, Guzmán Valenzuela C (2011) Competencias
en los procesos de enseñanza-aprendizaje virtual y semipresencial. Comunicar
18(36):107–114. https://doi.org/10.3916/C36-2011-03-01
Acknowledgments
This work is the result of the experience obtained from the research and extension
projects carried out in the research group Applied Electromagnetic Propagation
(PROPELA) (Minciencias COL0127629) of the Universidad Nacional de Colombia
Sede Manizales, between 2004 and 2023. This book is a product of the Laboratory
of Applied Electromagnetic Propagation, officially created by the Faculty of
Engineering and Architecture of Universidad Nacional de Colombia in December
of 2023.
I wish to thank Professor Néstor Peña Traslaviña, who bring me the basic
knowledge about signal theory and the guidelines about the topics for the first part
of this book. In the postgraduate degree (master’s-doctorate), I was a collaborator
in the work of Professor Nestor Peña Traslaviña in electromagnetic compatibility
(University of Los Andes), as well as Professor Hartmut Führ in algebra and wavelet
techniques (RWTH Aachen University). These observations as an assistant gave me
the initial push to the preparation of this book (2013). Thus, I greatly appreciate
those named above for their contribution in time, space, and energy.
xiii
Contents
Part I Single-Channel Communications
1
Introduction to Communication Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
2
Signal Analysis in Digital Communications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
3
Information Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
59
4
Performance of Digital Modulation Techniques . . . . . . . . . . . . . . . . . . . . . . . . . .
73
5
Elements of Digital Communication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
Part II Multipath Communications
6
Time–Frequency Channel Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
7
Multiple Input Multiple Output Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227
A Probability Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255
B The Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267
C Python Routines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275
Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281
xv
Acronyms
Although each acronym is defined a priori throughout each text, the main ones are
listed below.
AMI
ASK
FSK
PCM
PSK
PWM
RZ
TDM
WDM
Alternate Mark Inversion
Amplitude-Shift Keying
Frequency-Shift Keying
Pulse Code Modulation
Phase-Shift Keying
Pulse Width Modulation
Return to Zero
Time-Division Multiplexing
Wavelength Division Multiplexing
xvii
List of Symbols
The symbols used in this book are listed together with the proper units of
measurement, according to the International System of Units (SI).
Bit rate
Electric capacitance
Electric current
Electric potential, Electromagnetic force,
Amplitude
Electric power
Power ratio
Pulse width
Time
R
C
I
V
bit per second (bps)
farad (F)
ampere (A)
volt (V)
P
P
.τ
t
watt (W)
decibel (dB)
second (s)
second (s)
xix
International Standards
IEC 61280-4-1:2009. Fibre-optic communication subsystem test procedures –
Part 4–1: Installed cable plant – Multimode attenuation measurement.
IEC 80000-13:2008. Quantities and units Part 13: Information science and
technology.
G.711:1990. Pulse code modulation (PCM) of voice frequencies
xxi
Part I
Single-Channel Communications
The study of single-channel communication systems is the subject of this part.
One aspect of the study is an assessment of the efficiency constraints of the
system. Research turns its attention to understanding how channel capacity affects
information transmission. Finding the ideal conditions to maximize data transfer and
reduce errors is the aim. This involves testing the data transmission rates generated
by different modulation strategies. The considerations in electronic design will be
demonstrated by the use of the experiment findings reported in this part. This
section’s experiment findings will be used to highlight how crucial it is to choose the
best modulation method possible in order to achieve the best possible data transfer
and error.
Chapter 1
Introduction to Communication Systems
1.1 A Brief History of Communications
In mankind’s history there have always been big motivations for making technical
advances possible:
• Economics: Ease the trade process between nations.
• Military.
Communication Systems rely on those advances as a way to ease these motivations.
Throughout history, there are different communication mechanisms:
•
•
•
•
Face to face
Signals
Written word (letters)
Image/event description
In ancient times, the transmission of information often involved oral storytelling
(Religion, Tradition) and passing messages through messengers. Other mechanisms
include:
•
•
•
•
Writing knowledge (Papyrus, Library of Alexandria)
Printed editions (Guttenberg)
The network: commercial routes, food, wear, services, slaves
Storing services: accounting, logs
Communications are no exception, and its great progress has been made thanks
to them. Remote communications were born to facilitate trade between different
nations and empires.
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024
J. C. García-Álvarez, Digital Electronic Communications,
https://doi.org/10.1007/978-3-031-53118-7_1
3
4
1 Introduction to Communication Systems
Inadequate infrastructure and natural disasters frequently contributed to problems in the ancient communication networks, which resulted in the following
problems:
• Delay in transmission time (days, months)
• Inaccurate carriers (routes, addresses, receivers)
• Information message distortion (oral)
Paper was discovered in China in 105 AD. In the fifteenth century Johann
Gutenberg created the printing press: He favored rationalism, scientific research,
and literature. In the seventeenth century, the first newspapers appeared with the
objective of making current events available to the public. The word “postal service”
comes from the Latin positus (which means post), since in the beginning they were
postal posts located throughout the Roman Empire, and horse messengers carried
correspondence from position to position. The speed and efficiency of this type of
communication are better with the appearance of means of transportation such as:
Railways, Automobiles, and Planes.
1.2 Electronic Communications
With the discovery of electricity in the eighteenth century, the world enters to
a revolution in communications. From the studies of many scientists, compiled
by James Clerk Maxwell in 1873 in his Treatise on Electricity and Magnetism,
we began to contemplate the possibility of transmitting information by wireless
media. The telegraph appeared in the nineteenth century and used the Code Morse
International. In 1874, Thomas Alba Edison developed a method of quadruple
coding that allows transmitting twice as much information. The great advance
that the telegraph represented was improved widely with the development of the
telephone, since it allows the transmission of the human voice in the form of
electricity. Antonio Meucci was its inventor in 1871; however, he did not able
to patent it due to economic difficulties. There has been constant innovation and
progress in the field of telecommunications since Alexander Graham Bell received
a patent for this invention in 1876, particularly with regard to the devices that change
physical variables into electronic signals, such as:
•
•
•
•
•
Telegraph .⇒ Digital signals
Telephone .⇒ Voice, Data
Radio .⇒ Voice, Data
Television .⇒ Video, Data
Internet (computer) .⇒ Data
This revolution was possible with the use of physical transmission lines (cable pair,
coaxial, fiber optic) or air (propagation) as a propagation medium. It is possible
to emit or receive multiple signals that propagate through a medium, detecting
all signals propagated through that medium, in addition to identifying them by
1.3 Basic Elements of a Communication System
5
parameters; the most used is frequency. It can also be identified by the content of
the information (voice, data, images, among others).
The first radio broadcast in the USA happened in 1906. Television appeared at
the beginning of the twentieth century and had its peak during the Second World
War I:
1. Analog Television: It was broadcast over radio waves in the VHF and UHF
bands.
2. Cable Television: Allows signals to be delivered to the home of each subscriber
without the need for receivers and/or antennas. Additionally, it allows the return
of signals (loop of feedback) without any additional infrastructure.
3. Satellite Television: It began with the development of the career space during the
Cold War. Allows access to Television from remote and isolated areas.
4. IP Television (IPTV): Uses broadband connections over the IP protocol. Can be
used in televisions, computers, and mobile phones.
1.3 Basic Elements of a Communication System
To understand the function performed by communications systems, it is important
to know what type of variables are involved in the processing of information. The
important thing for a communication system is to maintain, regardless of the process
carried out, the greatest fidelity of a certain type of information that is to be sent
from one place to another. For this, certain techniques are used that help visualize
and process the information. In this chapter the theories necessary to determine
these techniques are explained. Fourier’s conjecture has a lot of truth: “Only” some
concepts need to be conveniently developed: What is meant by a function, how
to construct the integral of a function and establish the nature or convergence of
the series of functions in general and trigonometric in particular (Stremler 1989;
Tomasi 2003). It is the set of devices that constitute the link of information between
the source and destination. Its primary objective is to transmit information from one
point to another so that the message obtained from the receiver differs as much as
possible from the message that was originally sent (Couch II 1997; Lathi 1998; Stark
Henry B. Tuteur 1979; Carlson and Crilly 2009). The recovered message must be
fully reconstructed and usable at your destination point. In general, communication
systems are composed of a message, a transmitter, a means of transmission, and a
receiver. Figure 1.1 illustrates a setup for the basic elements required for an adequate
communication system:
Data and message. Message is the physical manifestation of the information that
is wanted to communicate: (1) Series of symbols (2) Voice (3) Music (4) Images
(5) Measurements of some physical variable
Transducer. Device capable of transforming or converting a certain type of input
energy into a different type of energy to the exit: (1) Microphone: Acoustic
6
1 Introduction to Communication Systems
Fig. 1.1 Example: wireless link
energy to electrical energy (2) Thermocouple: Transformation of thermal energy
into electrical energy (3) Speaker: Electrical Energy into Acoustic Energy
Signal. For a communications system to transmit appropriately, the message is
required to adopt some of the following ways: (1) Electrical, (2) Electromagnetics, (3) Optics. Consequently, a transducer is required that converts the message
into signal.
1.4 Aspects in the Design of Digital Communication Systems
The interference constitutes one of the most undesirable factors that modify the
content of information of a transmitted signal (Haykin 2001; Roddy and Coolen
1995; Miller 1999):
Signal Distortion. Manifested by flaws in the transmission medium, such as
cables or fiber optics.
Crosstalk. These are disturbances that the signal suffers due to the effect of
signals outside the system. As an example, radio stations interfering with a
walkie-talkie or Wi-Fi interfering with wireless devices making use of the same
allocated transmission frequency.
Random Noise. The initial signal may be partially or totally corrupted by erratic
and random processes. For instance, an increase in temperature has the potential
to modify the flow of electric current by influencing the motion of charged
particles within a conductor; this phenomenon is known as thermal noise.
1.6 Communication System Specifications
7
1.5 Measurements in a Communication System
These measurements are made on the different elements of a communication system
as follows:
• The source of information, .X, can be measured through the Entropy .H (X), due to
the statistical characteristics of the source ( mean, variance, and spectral density).
• In the transmitter, which is composed of the modulator, the filter, and the amplifier, it can be measured by the modulation index, the transmission bandwidth, and
the transmission power. The bandwidth contains the frequency response of the
signal. This means that for certain frequencies within the bandwidth, the signal
may or may not be transmitted properly. This frequency response is analogous to
the frequency response of a filter. This means that there are frequencies where the
message can be attenuated, called cutoff frequencies. Typically cutoff frequencies
are found when the channel gain is 3 dB below the maximum channel gain.
• The channel can be monitored through channel bandwidth and attenuation.
Bandwidth is the difference between the upper and lower frequencies where the
channel allows transmission of a message. For the message, its bandwidth is the
difference between the upper and lower frequencies that make up the signal that
represents the message. Normally the message bandwidth is reduced to the higher
frequency, called the fundamental frequency of the message. Thus, for effective
transmission, the channel bandwidth must exceed the value of the fundamental
frequency of the message. Otherwise, the message will be distorted.
• Measurements on the Receiver can be made using Sensitivity and Channel
Capacity.
• The received message .X' , which is interpreted as an estimate of the source, is
evaluated through the Error Rate.
1.6 Communication System Specifications
Assuming that you have been assigned a communication system, the first step
is to determine the technical characteristics of the system. The design for a
Communication System can be deployed by using the following features:
Modulation index. It defines the modulation scheme, given by:
– Amplitude
– Frequency
– Phase
– A combination of some of the above
Communication scheme:
– Continuous Wave Analog: Analog Baseband, Analog Carrier
– Pulsed Wave Analog: Analog Baseband, Pulsed Carrier
8
1 Introduction to Communication Systems
– Digital: Pulsed Baseband, Analog Carrier
– Line: Pulsed Baseband, Pulsed Carrier
Improvement Strategy. From easy to very complex:
– Increasing the carrier power
– Changing the modulation scheme
– Varying the modulation index
– Optimizing receiver sensitivity
1.7 Assessment
Exercise 1.1 Solve 3(x+1) = 81 for x.
Exercise 1.2 If in the equation 20002 − 19962 = 111ak 2 , where a and k are
integers, what is the maximum value of k − a?
Exercise 1.3 In a school, the ratio between the number of boys and the number of
girls is 2 : 3, and the ratio between the number of girls and the number of teachers
is 8 : 1. What is the ratio between the total number of students and the number of
teachers?
Exercise 1.4 An aircraft has a glide ratio of 12 : 1, meaning that the plane drops
1 meter in each 12 meter it travels horizontally. A building 45 m high lies directly
in the glide path to the runway. If the airplane is 12 m above the building, how far
from the building does the aircraft touch down on the runway?
Exercise 1.5 A perfect square is the square of an integer. How many positive
integers x are there such that both x and x + 99 are perfect squares?
Exercise 1.6 In a soccer tournament each of the eight teams plays each of the
other teams exactly once. Two points are credited for each game won, one point
for each game tied, and zero points for each game lost. How many points must a
team accumulate to ensure being in one of the first four places (that is, having more
points than at least four of the other teams) in the tournament?
Exercise 1.7 Two trains cover the route between two cities traveling on parallel
rails. The ratio between the speeds of the two trains is equal to the ratio between the
time they face each other when one passes the other traveling in the same direction
and the time they last facing each other when they cross each other traveling in
opposite directions. What is the ratio between the speeds of the two trains?
Exercise 1.8 If x, y, z are real numbers such that (x −3)2 +(y −4)2 +(z−5)2 = 0,
what is the result of x + y + z?
Exercise 1.9 If a is greater than c by 50%, and b is greater than c by 25%, by what
percentage is a greater than b?
1.7 Assessment
9
Exercise 1.10 The Gerrinsp publishing house designates the price of its latest book
titled “The Engineer in His Labyrinth” as follows:
C(n) =
.
⎧
⎪
⎪12n
⎨
1 ≤ n ≤ 24
11n
⎪
⎪
⎩10n
n ≥ 49
25 ≤ n ≤ 48
where n is the number of books ordered, and C(n) is the cost of n books. Noting that
the cost of 25 books is less than 24 books, for how many values of n is it cheaper to
buy more than n books than to buy exactly n books?
Exercise 1.11 The number of geese belonging to a certain group is increased such
that the difference between the population in year n + 2 and the population in year n
is directly proportional to the population in year n + 1. If the population in the years
1994, 1995, and 1997 was 39, 60, and 123, respectively, determine the population
in 1996.
Exercise 1.12 Augusta, Beatriz, Carlos, David, and Esteban go shopping. Each one
has a whole number of money bills to spend, and between them they have $56. The
absolute difference between the amounts that Augusta and Beatriz have is $19. The
absolute difference between the amounts that Beatriz and Carlos have is $7, between
Carlos and David it is $5, between David and Esteban it is $4, and between Esteban
and Augusta it is $11. How many money does Esteban have?
Exercise 1.13 Consider the functions f that satisfy the following equality:
f (x + 4) + f (x − 4) = f (x)
.
for every real number x. Each of these functions is periodic (f (x + p) = f (x) for
some whole number p), and there is a smallest common positive period p for all of
them. Calculate the value of period p.
Exercise 1.14 A shot takes a golf ball to a circular green of radius 12 m. Assuming
that all positions on the green are equally probable, what is the probability that the
ball stops less than 1 meter from the hole? Assume that the hole is at least 1 meter
from the edge of the green.
Exercise 1.15 Two cubic dice are fair in the sense that, when rolled, each of the
six faces has the same probability of coming up as the top face. However, on one of
the dice the number 4 has been replaced by the 3 and on the other the number 3 has
been replaced by the 4. When these two dice are thrown, what is the probability that
the sum obtained is odd?
Exercise 1.16 In the sixth, seventh, eighth, and ninth games of the basketball
season, a certain player scored 23, 14, 11, and 20 points, respectively. Her points
per game average are higher after nine games than it was after the first five games.
10
1 Introduction to Communication Systems
If her average after ten games is greater than 18 points, what is the fewest number
of points she could have scored in the tenth game?
References
Carlson B, Crilly PB (2009) Communication Systems: An Introduction to Signals and Noise in
Electrical Communication, 5th edn. McGraw Hill, New York
Couch II LW (1997) Digital and Analog Communication Systems. Prentice Hall, Hoboken
Haykin S (2001) Communication Systems. Wiley, Hoboken
Lathi BP (1998) Modern Digital and Analog Communications Systems, 3rd edn. Oxford Press,
Oxford
Miller GM (1999) Modern Electronic Communications. Prentice Hall, Hoboken
Roddy D, Coolen J (1995) Electronic Communications. Prentice Hall, Hoboken
Stark H, FB Tuteur (1979) Modern Electrical Communications Theory and Systems, 1st edn.
Prentice Hall, Hoboken
Stremler FG (1989) Sistemas de Comunicación. Alfa-Omega, Rock Rapids
Tomasi W (2003) Electronic Communications Systems, 5th edn. Prentice Hall, Phoenix
Chapter 2
Signal Analysis in Digital
Communications
2.1 The Baseband Signal
The signal that represents the message, when the signal is not modulated, is called
baseband. Baseband means that the characteristics of the message are not altered
at the time of transmission, that is, the signal is at the same frequency as its analog
representation (Garcia-Alvarez et al. 2014). Examples include the following: An
audio message is transmitted in baseband at a frequency range between 20 and
20 kHz; a video message is transmitted in baseband at the audio frequency of
.5.25 MHz, the luminance frequency of .1.25 MHz, and the chromatic frequency of
.4.83 MHz. Thus, it is intended to show the difficulties involved in sending a message
without any processing, giving rise to the need for modulation of the message in its
transmission.
2.1.1 Harmonic Signals
According to the Fourier transformation (Haykin 2001), any signal that describes the
above characteristics, such as those used in communications, is described by sinetype functions (Garcia-Alvarez and Gomez 2014). Indeed, they are modulated into
sine wave forms. A sinusoidal signal is described mathematically by the expression:
fi (t) = Ai sin (ωc t + ψi )
.
(2.1)
The quantities .Ai , ωc , and .ψ are called the amplitude, frequency, and phase of
the function, respectively, and you can describe this signal in two ways: One way
is to describe its evolution in time (time domain model), and the equation is a
mathematical representation of this evolution. The second way is to describe its
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024
J. C. García-Álvarez, Digital Electronic Communications,
https://doi.org/10.1007/978-3-031-53118-7_2
11
12
2 Signal Analysis in Digital Communications
frequency content (frequency domain model). The function .fi (t), in this case, has a
single frequency .ωi .
Example 2.1 The Time Domain sinusoidal function is a commonly used basis for
signal representation. The amplitude, frequency, and phase arguments, .Ai , .ωi , and
.φi , respectively, indicate the function’s evolution over time. As a result, this function
is defined as
fi (t) = Ai sin(ωi t + φi)
.
2.2 Modulation Principle
Both multiplexing and modulation are methods for moving the spectral components
of a baseband signal to a different frequency range (the latter is covered in part two
of this book). The purpose of moving a signal into a different frequency bandwidth
is to prevent interference, sometimes known as crosstalk, from occurring between
signals. These benefits will be explored in later chapters. These methods can also
be applied to effectively transfer information between locations. Modulation makes
the information signal more compatible with the medium. Multiplexing allows more
than one signal to be transmitted concurrently over a single medium. In the electrical
sense, they are variations of voltage (V) and current (I) with respect to time. In the
electromagnetic sense, they are variations of the electric field (.E) and magnetic field
(.H) with respect to time. In the optical sense, they are variations in the mode of light
propagation (polarization and wavelength).
The resulting signal from modulation is .s(t), representing the passband signal
(modulated):
⎞
⎛
g (t) =
A
''''
.
Amplitude
sin ⎝ ''''
ω
F requency
t + ''''
θ ⎠
P hase
• .A ⇒ Amplitude variation
• .ω ⇒ Frequency variation
• .θ ⇒ Phase wave variation
In that way, the information given by the baseband signal is carried by the variation
of parameters in the carrier wave.
Using phase velocity equation .Vp = x/t = λω/(2π ), carrier signal can be
converted into electromagnetic representation, traveling a distance .x :
⎞
⎛
g (x) =
.
⎟
⎜ 2π x
+ ''''
θ ⎟
sin ⎜
⎠
⎝ λ
''''
Amplitude
P hase
A
''''
Distance
2.3 Signal Representation
13
• .A ⇒ Amplitude variation
• .λ ⇒ Wavelength variation
• .θ ⇒ Polarization wave variation
Let .s(t) = cos ωs t the baseband (or original analog) signal and .fc (t) = cos ωc t
the carrier wave, with .ωs << ωc . A multiplier device generates the output signal
.g(t):
g(t) = s(t)fc (t) = (cos ωs t) (cos ωc t)
.
=
1
[cos (ωs + ωc ) t + cos (ωs − ωc ) t]
2
(2.2)
The bandwidth is obtained from Eq. (2.5), using .fc (t) as a periodic signal with
only the fundamental frequency .ω0 = ωc . The Fourier transform of generated signal
of Eq. (2.2) .G(ω) = F {g(t)} has bandwidth of .B = 4π
τ
τ=
.
2π
ωc
Therefore, modulation is a technique used to transport information on a high
frequency carrier signal, which is typically an electromagnetic wave, in order to
make it suitable for transmission over a communication channel, allowing more
transmission, information simultaneously, protecting it from possible interference
or noise. The modulation process involves two signals: (i) the modulator signal
and (ii) the carrier signal. The modulation procedure consists of using the carrier
wave as a transport for the message, represented by the modulating signal (audio,
video, data, etc.). These two signals result in the modulated signal. The idea of
using a carrier signal is to increase the transmission power, in such a way that
distortion phenomena, such as distance, propagation medium, and noise, do not
significantly affect the fidelity of the message (Arenas. and Garcia-Alvarez 2005).
The methodology for generating a carrier and modulating signal for a band-limited
communication channel uses two types of signals: analog and digital.
2.3 Signal Representation
The Fourier transform is a linear mapping between time space and frequency space
and vice versa.
f (t) ⇔ F (ω)
.
The spectral density function of .f (t) .F (ω) is obtained from the Fourier
transform of the signal .f (t). The Fourier transform is a linear mapping between
time space and frequency space and is defined as
14
2 Signal Analysis in Digital Communications
f (t) ⇔ F (ω)
.
f+∞
.F (ω) =
f (t) e−j ωt dt
−∞
The amplitude of the spectral components is proportional to .F (ω), so .F (ω)
constitutes the frequency spectrum of .f (t). It is also called spectral density function.
Unlike the Fourier series, the Fourier transform is applied to functions that are not
necessarily periodic and generates a continuous spectrum in .ω.
2.3.1 Hermitian
Hermite functions are also useful in mathematics, as they form an orthonormal
basis (complete orthonormal set) for the Hilbert space .L2 (R). For instance, Hermite
functions are applied for elementary quantum mechanics to solve the quantum
nonrelativistic harmonic oscillator problem (Celeghini et al. 2021). These functions
are obtained by multiplying Hermite polynomials and a Gaussian function and
have the property of being highly concentrated around the origin (Stremler 1989).
The Fourier transform is an operator that preserves the unitary structure of .L2 (R).
Therefore, the Hermite functions are the eigenfunctions of this operator and enable
a decomposition of .L2 (R) into four eigenspaces that correspond to the cyclic group
.C4 . This decomposition can have practical implications. The Fourier transform and
its inverse act as automorphisms in the Hilbert space .L2 (R), which means that
they keep the Hilbert space norm unchanged, according to the Plancherel theorem
(Celeghini et al. 2021). This theorem can be generalized to other spaces that are
relevant for physics, such as the space of smooth functions that vanish at infinity
faster than any polynomial inverse and the space of tempered distributions. In
both spaces, the Fourier transform and its inverse are also automorphisms that are
compatible with the usual topologies defined on them (Aakash et al. 2019).
2.4 Signal Classification
The names of some signals, classified by some properties, are shown as follows:
Signal
Baseband
Carrier
Noise
Attenuation
Interference
Features
Ergodic, nonperiodical
Deterministic, periodical
Random, nonperiodical
Ergodic, nonperiodical
Ergodic, nonperiodical
2.4 Signal Classification
15
Fig. 2.1 Analog (left) and discrete (right) signal
Periodic: They are those functions that satisfy the condition:
f (t + T ) = f (t)
T ∈R
.
(2.3)
where T is the period of the signal. That is, they are signals that are repeated
over time: For any .m = {1, 2, 3, . . .} ∈ Z+ , if T is the period of a signal, mT
are also periods of the signal, and T is considered the fundamental period of the
signal. Moreover, if a signal .f (t) is periodic with period T , the signal .f (mt) is
also periodic with period .T /m. Another property of periodic signals is that direct
current offset .ADC is the same when measured on the period interval:
1
.ADC =
T
f T
0
f T /2
1
f (x)dx =
b−a
−T /2
1
f (x)dx =
T
f b
f (x)dx
T = b−a
a
Aperiodic: They are those signals that do not repeat over time and therefore do
not meet the condition of Eq. (2.3). However, a constant function .f (t) = a has
infinite periods and no fundamental period.
Analog Signal: It is a signal that presents continuous values on both the
amplitude and time axes. Due to its precision, it is very susceptible to noise and
interference (Fig. 2.1).
Digital Signal: It is a signal that presents a finite number of possible values on
the amplitude and time axis. Therefore, the received signal is an approximation
of the sent signal, which is not convenient for sending precise signal values.
Example 2.2 The period of the signal .f = sin(mt) is .Tm = (2π/m)s, with its
fundamental period when .m = 1, i.e., .T1 = (2π ) s. Analogously, this signal can be
constructed from a given period T , leading to
(
f (t) = sin
.
2π
t
Tm
)
16
2 Signal Analysis in Digital Communications
2.5 Power Analysis
This modification is measurable and can be described by physical units. Instant
power is the work required to transport the signal at a specific instant of time. Thus,
the power is equal to the potential energy required for transport .v(t), multiplied
by the magnitude of the energy flow used for the same electrical current .i(t). In
electrical systems, using the Ohm’s law, power can be expressed as
P(t) = v(t)i(t) =
.
v 2 (t)
= i 2 (t)Z
Z
where Z is the impedance of the transmission medium or channel. Thus, the energy
used is the sum of the power used over time. It would be logical to bet that for a
long interval of time, the amount of energy needed would be considerable. This is
not the case, if we consider that the modified signals have a periodic behavior. Then,
we only need to know the average energy used in a short time interval.
2.5.1 Relationship Between Power and Energy
The energy of a signal is defined as
f+∞
|f (t)|2 dt
.E =
−∞
If the energy is finite and the power is zero, the signal is said to be energy. The
power of a signal is defined as
1
.P = lim
T0 →∞ T0
f
|f (t)|2 dt
T0
If the power is finite and the energy is infinite, the signal is said to be a power signal.
Since carrier is a periodic signal, the power measure is better given in terms of its
mean. Therefore, with T as the carrier time period, power and energy are measured
as
P=
1
T
E=
1
T
.
.
f T /2
−T /2
f T /2
−T /2
sc (t)dt
P(t)dt
2.5 Power Analysis
17
2.5.2 Statistical Parameters
The mean of a function is defined as the average of the value of the function over
the working domain. For time-dependent signals .f (t), the mean over the interval
.(a, b) is defined as
1
.f¯ =
b−a
fb
f (t) dt
a
The energy of a signal is defined as
f+∞
|f (t)|2 dt
.E =
−∞
If the energy is finite and the power is zero, the signal is said to be energy.
The power of a signal is defined as
1
T0 →∞ T0
f
|f (t)|2 dt
P = lim
.
T0
2.5.3 Power Measurement
A power analysis determines if there is enough power at the receiver to recover
the information. The time–frequency domain is used to locate the energy of the
transmitted signal. An energy signal is simple noticeable via time domain (with an
oscilloscope, for example). However, we cannot see the energy components of the
signal in time domain, so we require a frequency decomposition, as it is shown in
Fig. 2.2:
• The channel impulse response is a function of time and frequency.
• The function describes the channel in the time domain, but the space of these
functions is in the frequency domain.
• As a result, it is simple to see how a specific signal characteristic varies in
either domain. Frequency domain functions are characterized by the argument
.ω, measured in radians/s.
• The result of the function is the power density of the signal at frequency
.ω. To achieve this, we use convolution operator, translating the system into
multiplication in the frequency domain:
↔
y(t) = s(t) ∗ h(t) F Y (ω) = S(ω)H (ω)
.
18
2 Signal Analysis in Digital Communications
Normalize the maximal
(
)
power value
[dBm/Hz]
0dBm
Mean power is the sum
of main spectral density components
−3
Those below of -3dB
can be discarded
[Hz·2 ]
Fig. 2.2 Power measurement by spectral analysis
The problem is to find the coefficients .An and .Bn .
2
.An =
T
tf
0 +T
[
]
2π nt
f (t) cos
dt
T
t0
2
.Bn =
T
tf
0 +T
[
]
2π nt
f (t) sin
dt
T
t0
These coefficients are known as spectral components, and they quantify the
composition of the signal at each of the frequencies.
.A0
.A1 , .B1
..
..
.An , .Bn
Zero-frequency component
First harmonic coefficients
..
..
n-th harmonic coefficients
The fundamental frequency is the minimum frequency necessary to represent a
waveform. Since the Fourier series has infinite components, an exact representation
is only possible when all harmonics of the signal are used. Since the channel
bandwidth is finite, not all the series coefficients could be transmitted through.
2.5 Power Analysis
19
1
Amplitude
0.8
t
0.6
0.4
0.2
0
T
tiempo (s)
Fig. 2.3 Pulsed train signal .fp (t), normalized by its amplitude A
Moreover, none of the computer or digital system could handle all the coefficients
for processing the signal. Therefore, there must be a commitment between the
number of required harmonics and the quality of signal. The Gibbs effect appears
once the channel bandwidth is less than the signal bandwidth, eliminating spectral
harmonics in the signal. Let .fp (t) be the pulse train signal shown in Fig. 2.3. This
signal can be represented by Fourier series as
fp (t) =
∞
Σ
.
Cn ej nω0 t
(2.4)
−∞
where .ω0 is considered as the fundamental or natural frequency of the signal (in
Fig. 2.3, .ω0 = 2π/T ). The Fourier coefficients .Cn are calculated by definition as
1
T
Cn =
.
f T
2
− T2
fp (t)e−j nω0 t dt =
1
T
f −τ
2
− τ2
Ae−j nω0 t dt
]
A [ (j nω0 τ/2)
e
− e−(j nω0 τ/2)
j nω0 T
=
By trigonometrical properties:
Cn =
.
( nω τ )
A sin (nω0 τ/2)
A
0
= sinc
2 nω0 τ/2
2
2
where .sinc(•) stands for the unnormalized sinc function, which has the following
properties:
20
2 Signal Analysis in Digital Communications
• Maximum value at zero:
.
sinc(0) = lim
x→0
sin x
=1
x
• Zero crossing values
.
sinc(0) = 0 ⇒
sin x
= 0 ⇒ sin x = 0 ⇒ x = mπ
x
m = 0, 1, . . .
Taking these properties for the coefficients .Cn , we have
.
Cn = 0 ⇒
.
max {Cn } =
A
2
nω0 τ
2mπ
= mπ ⇒ nω0 =
2
τ
The first zero crossing .m = 1 leads to the following:
nω0 =
.
2π
τ
Thus, the first coefficient .C1 is the energy value of the component of the signal .fp (t)
located at the fundamental frequency .ω0 = 2π
τ . Moreover, the pulse width .τ gives
us the bandwidth relationship:
τ=
.
m 2π
n ω0
(2.5)
This ratio allows us to modify the bandwidth parameters by modulating the pulse
width. In effect .B = 2π
τ .
Example 2.3 The following example illustrates the effect of reducing the spectral
harmonics in the signal due to channel bandwidth restrictions. Consider the
transmission of the signal .v(t).
{
v(t) =
.
1;
−1;
0≤t <π
π ≤ t < 2π
The corresponding Fourier series, for odd n, is
v(t) =
∞
Σ
4
.
n=1
nπ
sin(nt)
2.5 Power Analysis
21
The magnitudes of the harmonic coefficients are:
.B1
.B3
.B5
.B7
.· · ·
.Bn
4/.π
4/.3π
4/.5π
4/.7π
.· · ·
4/.nπ
The frequencies where the signal is allocated are the following:
.ω1
.ω3
.ω5
.ω7
.· · ·
.ωn
1 rad/s
3 rad/s
5 rad/s
7 rad/s
.· · ·
n rad/s
2.5.4 Power Measurement Units
The measurement of the power necessary for transmission is commonly done using
relative values, commonly 1 W or 1 mW. To measure relative power, the deciBel is
used, which uses the following expression:
.
P|dB = 10 log10
P
P0
(2.6)
where .P0 is the reference power. Variants of the decibel are the following:
P
.
1
P
P[dBm] =10 log10
.
0.001
V
P[dBμ] =20 log10
1
P[dBW] =10 log10
.
(2.7a)
(2.7b)
(2.7c)
In experimental test, a reference source is used, in order to give an accurate
measurement. In this case, the relative decibel (dBr) is used either voltage .Vs or
22
2 Signal Analysis in Digital Communications
power .Ps accordingly:
P [dBr] = 10 log
.
V
P
= 20 log
Ps
Vs
A pilot (test) signal gives the power reference value to the system. The power
measure for that signal is absolute, so the unit dBm0 is calculated by subtracting
the relative measure:
P [dBm0] = Pm [dBm] − Pr [dBr]
.
The neper (Np) is another unit based on a logarithmic scale, but using the Euler’s
number as a base:
.
P|Np =
P
1
ln
2 P0
(2.8)
The neper and the decibel have the following ratios:
1.0 Np = 20 log10 e ≈ 8.685889638 dB
.
1.0 dB =
.
1
ln 10 ≈ 0.115129255 dB
20
Example 2.4 Convert 3500 pW into dBm.
P ' [dBm] = 10 log
.
P [W]
0.001
With .P = 3 500 × 10−12 = 3.5 × 10−9 W, we have
P ' = 10 log
.
)
(
3.5 × 10−9
−6
= 10 (log 3.5 − 6 log 10)
=
10
log
3.5
×
10
10−3
= 5.44 − 60
= −54.56 dBm
Example 2.5 The power system of Fig. 2.4 has 0 dBr in the point B. Thus, the
estimated powers in the other test points are as follows:
• Test point A is located at the 1 dB-attenuated block input from B; thus A is 1 dBr
above B: .PA = 1 dBr.
• Test point C is located at the 3 dB-gain block output from B; thus C is 3 dBr
above B: .PC = 3 dBr.
• Test point D is located at the output of a 3 dB-gain-7 dB-attenuated cascode
system from B; thus .PD = 3 − 7 = −4 dBr.
2.5 Power Analysis
23
Fig. 2.4 A simplified power system
• Analogously to the above item, .PE = −7 dBr.
Now a source supplies power at test point A, giving a measure of .−5 dBm at test
point B. Using the relative power measures, we obtain
PA = −5[dBm] + 1[dBr] = −4 dBm
.
PC = −5[dBm] + 3[dBr] = −2 dBm
PD = −5[dBm] − 4[dBr] = −9 dBm
PE = −5[dBm] + 7[dBr] = −12 dBm
Accordingly, the absolute power measurement is given as follows:
PA = −4[dBm] − 1[dBr] = −5 dBm0
.
PB = −5[dBm] − 0[dBr] = −5 dBm0
PC = −2[dBm] − 3[dBr] = −5 dBm0
PD = −9[dBm] + 4[dBr] = −5 dBm0
PE = −12[dBm] + 7[dBr] = −5 dBm0
Therefore some additive noise or interference on some point of the system could be
detected, where an unlike absolute power value (dB0) was detected on.
Example 2.6 In the power system of Fig. 2.5, a load of impedance .75 o is coupled
at test point E, referenced at 21 dBr, giving a reading of .−8 dB.μ. Here, the power
gain (or loss) value of the device located between test points D and E is unknown.
Taking the power source at test point A, .PA = 4 mW, the following calculations can
be made:
PA = 10 log
.
4[mW]
= 6 dBm
1.0
PB = 6[dBm] − 3[dB] = 3 dBm
PC = 3[dBm] + 10[dB] = 13 dBm
24
2 Signal Analysis in Digital Communications
Fig. 2.5 A simplified power system
PD = 13[dBm] − 20[dB] = −7 dBm
PE = −8[dBμ] + 10 log
600
[dBZ] = 1 dBm
75
PF = 1[dBm] − 5[dB] = −4 dBm
Therefore, the power ratio of the device located between the test points D and E
is equal to .GDE = 1[dBm] − (−7)[dBm] = 8 dBm, confirming that the device is
an amplifier. The relative power measurements to the test point E are calculated as
follows:
• The gain between test points A and E is .GAE [dB] = GAB [dB] + GBC [dB] +
GCD [dB] + GDE [dB] = −3 + 10 − 20 + 8 = −5 dB. Since point A is at input,
the power measured at point A is 5 dBr (above of point E). With the reference
value .PE = 21 dBr, .PA = 21 + 5 = 26 dBr.
• Analogously to the above item, the gain between points B and E is .GBE [dB] =
GCD [dB] + GDE [dB] = 10 − 20 + 8 = −2 dB. With test point B at input, the
power measured at this point is 2 dBr (above of point E). Then .PB = 21 + 5 =
23 dBr.
• Using calculations of the two above items, power measured at point C is .−12 dBr
(below of point E). Then .PC = 21 + 12 = 33 dBr.
• Test point D is located at the input of a 8 dB-gain device from E; thus .PD =
21 − 8 = 13 dBr.
• Test point F is located at the output of a .−5 dB-gain device from E; thus .PD =
21 − 5 = 16 dBr.
Finally, just taking one point of the system, supposing no external interference or
noise, the absolute power value can be given for the remaining points. For instance,
taking points A (source), E (reference), and F (load), the absolute power would
remain the same:
PA = 6[dBm] − 26[dBr] = −20 dBm0
.
PE = 1[dBm] − 21[dBr] = −20 dBm0
PF = −4[dBm] − 16[dBr] = −20 dBm0
2.5 Power Analysis
25
2.5.5 Experiment: Baseband Signals
For this experiment we will use two signals that represent a message in a certain way.
The first is the signal SAudio .s(a, k) which is a three-frequency analog signal, and
the second SDigital .g(b, k) which is a discrete signal. These two functions are
described in the following equation:
f (k) =a ·
10
Σ
.
sin(2πρ(n)k).
(2.9)
n=1
g(a, k) = sign(f (a, k))
(2.10)
where k is a vector represented by .k = [0 : 1/Fs : T ]' . The value .Fs corresponds to
the sampling frequency and defines the number of elements of the generated signals.
Finally, .ρ(n) is a uniform random variable with parameters .(a, b) = (0, 300) Hz,
where a and b are the lower and upper limits, respectively. The generated algorithm
is encapsulated in the function SigGen(a,N), where N is the number of points
used to generate the signal. The amplitude value a is arbitrary, within the interval
.[0, 5]. The generation of the signals SAudio and SDigital follows the following
routine:
• Assign an arbitrary sample frequency as input .Fs .
• Time vector t is generated as follows:
[
]'
1
:T
.t = 0 :
Fs
where T is the elapsed time of the signal. A value .T = (2 ∗ Fs ) + 1 is suggested.
• Generate a vector of 10 random frequencies .F = [fi ]|10
n=1 with uniform
distribution .(a, b) = (0, 300).
• Discrete signals SAudio y SDigital are generated by Eq. (2.10).
• The definition cout has the following structure:
– cout.signal: Vector of signal in function of time
– cout.spectra: Vector of signal spectra
– cout.bandwidth: Float number of the signal bandwidth
– cout.tpower: Float number of the power calculated by time series
(cout.signal data)
– cout.spower: Float number of the power calculated by Fourier spectra
(cout.spectra data)
Example 2.7 Make a graph of SAudio as a function of time t (i) and as a function
of frequency (spectrum). Explain in text what characterizes it as a continuous signal.
26
2 Signal Analysis in Digital Communications
Example 2.8 Make a graph of SDigital as a function of time t (i) and as a
function of frequency (spectrum). Explain in text what characterizes it as a discrete
signal.
There are different methods to carry out these measurements on a communication
system: (i) process analytical systems; (ii) analytical perception systems; (iii)
experimental systems. In this text we will commonly use analytical process systems.
To do this, a mathematical model must be established initially, where its parameters
can be analyzed throughout the system. Figures 2.6 and 2.7 illustrate two examples
of calculating the cutoff points for the bandwidth according to the analyzed signal.
The transmission bandwidth is calculated as follows:
Step 1. From the spectral density plot, take the two frequency points where the
gain decreases rapidly. Use the two extreme points. The lower end point is called
the low cutoff frequency .ωLC , and the upper end point is called the high cutoff
frequency .ωH C . The bandwidth is .BW = ωH C − ωLC .
Step 2. Draw the spectrum of the SAudio and SDigital signals. Describe the
differences between these spectra in bandwidth and spectral efficiency.
Step 3. To calculate the PSD, we rely on the average power of a signal .P, using
the following routine:
Fig. 2.6 Calculation of bandwidth for multimode spectra
2.5 Power Analysis
27
Fig. 2.7 Calculation of bandwidth for multimode spectra
Calculate PSD
n = length(X)
% Number of elements
MeanPow = int(X^{2})/n
where X is the signal to evaluate. The RMS value of a signal is obtained using the
following routine:
Calculate RMS
n = length(X)
% Number of elements
rms = norm(X)/sqrt(n)
Step 4. Calculate the average power and RMS value of the SAudio and
SDigital signals using time series (integral of signal over signal period).
Convert these values into dBm and dBW.
Step 5. Decompose the SAudio and SDigital signal into an Euler representation using the ACarrier signal as a base as follows.
28
2 Signal Analysis in Digital Communications
Step 6. Assign the signal ACarrier as the base In Phase (I) and generate its
corresponding quadrature signal (Q).
Step 7. Because SAudio and SDigital are generated from a random
sequence of frequencies, generate 10 .X = {xn } with .n = 1, . . . , 10. This
will correspond to the coefficients I. Use a discrete uniform random variable
with interval .[1, 10].
Step 8. Generate a sequence of random numbers .Y = {yn } with .n = 1, . . . , 10.
This will correspond to the Q coefficients. Use a discrete uniform random
variable with interval .[−5, 5].
Step 9. Based on the signal ACarrier, generate .cos (ωc t), which we will call
ICarrier and its corresponding quadrature function .sin (ωc t) which we call
QCarrier.
Step 10. Perform the quadrature operation according to the following equation:
nτ
n − 1τ
<t <
N
N
√
Step 10a. Calculate the magnitude vector .M = (X2 )+ Y2 .
2
.
Step 10b. Calculate the phase vector .φ = arctan Y
X2
Step 10c. Draw the polar diagram .(M, φ) where .τ is the total interval of the
signal .s(t) and .N = 10 is the number of phase and quadrature coefficients, that
is, the elements of the respective vectors. We will call this signal .s(t) PassBand,
and it is used for modulation techniques, especially digital.
Step 11. Generate an estimate of the SDigital signal using only the first three
coefficients .φi , i = 1 · 3 from the Fourier Series representation (use Fig. 2.7 for
reference). Make an overlay plot between SDigital and its respective estimate
and calculate the mean square error between these two signals.
s(t) = xn cos (ωc t) + yn sin (ωc t) ,
.
2.6 Quantization
Signal quantization has the following advantages: (1) representing a value or
measurement in a finite number of symbols and (2) allowing numerical calculations
that would otherwise be extensive. However, precision is an aspect that must be
adjusted in order to reduce measurement error as much as possible. For this, tools
for analyzing the error in signal transmission are necessary. The first step in the
process of digitizing a signal is to establish a set of discrete time intervals at which
the input signal is sampled. Relevantly, digitization techniques are based on the use
of periodic signals .fT (t), so if a signal is sampled with sufficient frequency, this
signal can be recovered from its representative samples (with the use of an LPF, for
example). Quantization consists of approximating a value of a signal over time into
a finite value of a discrete set. It is not possible to assign an infinite set of values
that represent the infinite set of signal values taken in a period of time. On the other
2.6 Quantization
29
hand, our perception of reality, that is, of the signals we want to transmit, does not
allow us to distinguish small variations of the signals or a certain frequency range.
This is why it is possible to have an approximate representation of the original signal
through discrete amplitude levels.
2.6.1 Sampling and Hold
For the transmission of a digital system, it is required that the signal be sent in
the form of a finite number of symbols without losing relevant information. In the
sense of sending analog signals over a digital communication system, or processing
them in a digital computer, analog signals need to be converted into digital. This
process is carried out by an analog–digital converter (ADC), which consists of a
sample and hold circuit (Sample and Hold S/H), which samples the analog signal,
that is, measures its value in moments of time. .tj regularly spaced (Fig. 2.4), and
a quantizer, which quantifies the signal, that is, converts it into discrete numerical
values, as illustrated in Fig. 2.4. The reverse operation to the ADC is performed by
a digital analog converter (DAC) (Heinisuo and Vanhala 2006). These circuits fulfill
the function of taking a sample of the signal .x(t) at an instant of time k. This sample
.X[k] is a symbol of the infinite set which occurs in discrete spaces of time. That is
why the time expression loses validity now. The sampling and hold circuits must be
consistent with the sampling theorem, such that it does not present signal distortion
when detected and recovered. Sometimes it is advisable to filter the signal to be
sampled at its maximum frequency, so that spurious signals are eliminated.
2.6.2 A Measure of Digital Signals
Gottfried Wilhelm Leibnitz, German mathematician, developed the binary numeral
system in 1679. It paved the way to all modern computing and digital communications. This system uses two digits, i.e. 0 and 1, representing two states or actions (the
day can be represented as 1 and the night as 0, for instance). The unit of measure
of the information required to store or display this two-digit event was coined by
John W. Tukey in 1949 as binary information digit (binary means “consisting of
two things or parts”) or simply bit. Claude E. Shannon announced the word “bit” in
his paper “A Mathematical Theory of Communication”. The bit is considered as the
minimal information quantity, since only two events can be enclosed in. In fact, a bit
covers just two events: day–night, on–off, left–right, and so on. Therefore, two bits
can enclose the permutation of two times two events. Consider the weather event of
a rainy day. Thus, two bit could represent the two-day weather condition, leading
to following occurrences: rain–rain, no rain–rain. rain–no rain, and no rain–no rain.
The basic permutation operation states that, for a number K of times that D events
could occur, we have .D K possibilities. In the case of bits, .D = 2, therefore, the
30
2 Signal Analysis in Digital Communications
number of possible occurrences for K bits is equal to .2K . In the above mentioned
example, there would be four weather forecasts.
2.6.3 Dynamic Range
It is the relationship between the maximum and minimum expected quantization
values .(M0 ). This determines the number of bits that can be used to obtain a given
range. Using .3.8:
M0 =
.
max{f }
max(f )
Ac D K
=
=
= DK
min{f }
X0
Ac
Expressing it in decibels, the dynamic range becomes
M0 |dB = 20 log D K = 20K log D
.
[dB]
Example 2.9 If you want the dynamic range of a quantizer to be 30 dB, find: a. The
relationship between the maximum and minimum of quantization.
Using .4.38
M0 = 10
.
M0 |dB
20
30
= M0 = 10 20 = 31,6228
b. The number of bits required to quantize a signal and preserve the relationship of
values.
With .D = 2, and using .4.37:
K = log2 M0 = 4, 9829 ≈ 5 bits
.
2.6.4 The Information Sequence
A sequence is a series of finite symbols that follow a certain behavior and contain
a certain amount of information. The symbols belong to a finite set, which depends
on the quantization performed. According to(this,
) a quantized representation .{X}m ,
which is representing the sampled signal .s tj , can be decomposed into a series
.{Xk } , k = 1, 2, . . . , K, of finite symbols between the instants .tj and .tj +1 , which
are equivalent to a maximum number of quantization levels .M = D K .
The value K of said sequence is supplying the amount of information of the event
.{X}m , which can have a number of M possible combinations with probability
Pm = P {X = {X}m }
.
m = 1, . . . , M
2.6 Quantization
31
In fact, M is the number of quantization levels and can be performed with the
number of D symbols of the alphabet, so K can be expressed as
K = I {X = {X}m } = logD
.
1
Pm
Assuming that the probability of occurrence is equiprobable and equal to .1/M:
K = I {X = {X}m } = logD M
.
Note that then K becomes the length of the sequence representing .{X}m . Thus,
each symbol represents a range of discrete levels. Quantization noise can be
considered here because the minimum variation between sequences of information
is produced by the variation of one bit. Therefore, it is assumed that the quantization
noise is given by
}
{ ( )
η = s tj / {Xk } = X0
.
2.6.5 Quantization Stage
By having the signal sampled, this signal can be sent directly to the medium, using
the pulse train .xT (t) as a carrier, just as in PAM. Another method is to perform
a quantization of the signal .X[k] in order to be transmitted as a discrete source of
information. This quantification is performed by dividing each sample of continuous
value of .X[k] into a discrete value called level. This level, given by the discrete event
.X = Xm , m = 1, . . . , M, represents a discrete sample of the signal, so it is replaced
by the value of the level closest to it. In the quantification process, each sample is
approximated to the nearest level. Then, if a signal is required to be quantized into
M levels, each value of .X[k] will be represented by the quantized value .Xi . Based
on Fig. 2.8 in which it is assumed that the signal is unipolar, that is, its values are in
the interval .(0, ∞), we have that:
Xm = m
.
Ac
,
M
(2m − 1)
Ac
Ac
≤ X[k] ≤ (2m + 1)
2M
2M
where .Ac is the maximum value that the quantizer can reach, called the quantization
range. The set of quantification levels is given by
Xm,
.
And the level width is given by
m = 0, . . . , M
32
2 Signal Analysis in Digital Communications
Fig. 2.8 Unipolar quantization, .M = 4
Fig. 2.9 Bipolar quantization, .M = 11
X0 =
.
Ac
2M
According to Fig. 2.9, if it is assumed that the signal is bipolar, that is, its values are
in the range .(−∞, ∞), we have
Xm = m
.
2Ac
,
M
(2m − 1)
Ac
Ac
≤ X[k] ≤ (2m + 1)
M
M
2.6 Quantization
33
The set of quantization levels is given by
Xm ,
.
m=−
M
M
,...,
2
2
Note that the levels are distributed halfway between the two polarities; this means
that there is an amount of information used to represent the polarity or sign and an
amount of information used to represent the value of the quantized signal. In this
case, since there are two events to represent the sign, the amount of information
required, assuming that there is an equal probability that any of the events will
occur, is
I {X = +} = I {X = −} = log2 2 = 1bit
.
Therefore, one bit is extracted from the amount of information calculated for the
value of the quantized signal. The width of the levels is given by
X0 =
.
Ac
2M
For a bipolar quantifier, the polarity occurs in two events (polarity .+y−)
and is (represented
) by a symbol, and the magnitude is represented by a number
.K =
logD M − D2 of symbols. In the binary case, in effect, they would be
(
)
.K = log2 M − 1 bits.
Unless otherwise stated, binary alphabet sequences will be taken. In such a way,
each of the .M − 1 quantization levels is represented by a sequence of .K = log2 M
bits. For the bipolar case, the sign bit .{0} indicates positive polarity, while .{1}
expresses negative polarity.
Example 2.10 For the binary alphabet, in which the symbols 1,0 are tempted, one
can have a sequence of lengths .K = log2 M, which is called a binary information
sequence.
Example 2.11 To represent the minimum quantization level for 3 bits (1 sign, 2
magnitudes), the representation illustrated in Table 2.1 is taken. Note that:
M = 2K − 1 = 7 levels
.
X0 =
2Ac
= 4 mV
M
With which we have spacings of .4 mV.
Note that .X2 = MSB is used as the sign bit, while .X1 and .X0 = LSB represent
the magnitude of the value
Example 2.12 For a 7-bit digitizer (one sign bit, 6 information bits), whose
minimum LSB transition value is .X0 − 0.15 V, determine:
34
2 Signal Analysis in Digital Communications
Table 2.1 A 3 bit-length
sequence
m
3
2
1
0
.−0
.−1
.−2
.−3
( )
Range .s tj [mV]
.(10, 14)
.(6, 10)
.(2, 6)
.(η, 2)
.(−2, η)
.(−6, −2)
.(−10, −6)
.(−14, −10)
.{Xk }m [mV]
.{Xk }
12
8
4
0
0
.−4
.−8
.−12
011
010
001
000
100
101
110
111
a. The number of quantization levels.
M = 27 = 128
.
64 levels by polarity.
b. The maximum value that the quantizer can deliver and its corresponding
sequence.
The maximum value is
{
}
{Xk } = 0 1 1 1 1 1 1
.
which corresponds to the level .m = 6.3. Then
{X}63 = 63 × 0.15 = 9.45 V
.
Since the digitizer is bipolar, the maximum range it can have is .{X}63 −
{X}−63 = 2 × 9.45 = 18.9 V.
c. The value .{X}m that represents the sequence
{
}
{Xk } = 0 1 0 1 0 1 0
.
This sequence corresponds to the level .m = 42. Then
{X}42 = 42 × 0.15 = 6.3 V
.
d. The value .{X}m that represents the sequence
.
{
}
{fk } = 1 0 1 1 0 1 0
This sequence corresponds to the level .m = −26. Then
{X}−26 = −26 × 0.15 = −3.9 V
.
e. The sequence that represents
2.6 Quantization
35
{X}m = 4.75 V
.
This value represents the level
m=
.
4.75
= 31.7 ≈ 32,
0.15
so the sequence is
.
{
}
{Xk } = 1 1 0 0 0 0 0
f. The sequence that represents
{X}m = −2.72 V
.
This value represents the level
−2.72
= −18.13 ≈ 18,
0.15
m=
.
so the sequence is
.
{
}
{Xk } = 0 0 1 0 0 1 0
Example 2.13 For a 6-bit quantizer (one sign bit, 5 information bits), whose
maximum range is .±31.5 V, determine:
a. The number of quantization levels.
M = 26 = 64 The number of levels is divided by two once the sign bit
.
is taken, resulting in 32 levels for each polarity.
b. The value of the LSB and the maximum quantization error.
The LSB value is equivalent to the width of the quantization levels. Using 4.23,
31.5
2Ac
=
= 0.984 V
M
32
f0 =
.
By deduction, the maximum quantization error reaches half the amplitude of the
quantization levels:
ε=
.
Ac
= 0.492V
M
c. The value .{X}m that represents the sequence
36
2 Signal Analysis in Digital Communications
{
}
{Xk } = 0 0 1 1 0 1
.
This sequence represents the level .m = 13. Then:
{X}13 = 13 × 0.984 = 12.79 V
.
d. The value .{X}m that represents the sequence
.
{
}
{fk } = 1 1 1 0 0 1
This sequence represents the level .m = −25. Then:
{X}−25 = −25 × 0.984 = −24.6V
.
e. The sequence that represents .{X}m = 13.62 V
This value represents the level
m=
.
13.62
= 13.84 ≈ 14,
0.984
so the sequence is
.
{
}
{Xk } = 0 0 1 1 1 0
2.7 Orthogonal Representation of Modulated Signals
Measuring the performance of a communication system is done by representing it
in the phase diagram. For this, signal orthogonalization is used. Let the vector of
orthogonal signals .φ given by
{
}
φ = φ̂1 (t), φ̂2 (t), . . . , φ̂N (t)
.
which represents a signal according to Eq. (2.4) , leaving the set of symbols .{Cm }
that will represent the signal, where each projection is calculated:
Cm (t) =
.
√
Em φm (t) +
M
Σ
Cim φi (t) m = 1, . . . , M
i=1
And the set of symbols that will represent the baseband signal:
s = {s1 (t), s2 (t), . . . , sM (t)}
.
2.7 Orthogonal Representation of Modulated Signals
37
where each projection is calculated:
√
s (t) =
. m
Em φm (t) +
m
Σ
Cim φi (t) m = 1, . . . , M
i=1
This indicates that each coefficient value, which is the projection of the signal on
its orthogonal components, would represent a probable value of the presence of the
signal, and then:
Xm = Ŝm = {C1m , C2m , . . . , C(N)m} m = 1, . . . , M
√
.C( N)m =
Em
.
It is the set of statistically representative signals, thus leaving a bank of measurements on the passband signal. Some of these measurements are expanded upon
below.
Example 2.14 Let the sequence .U = (−2.3, 27) and unit vector .B = φ̂1 + φ̂2 + φ̂3 .
If the vector elements .φ̂i are orthogonal among them, these elements are linearly
independents, and the sequence X can be represented as
u = U · B = −2φ̂1 + 3φ̂2 + 27φ̂3
.
which is the unique form of U . Therefore, each element .φ̂i is a basis vector.
Example 2.15 Let the following sequences: .U = (2, −1, 3), .V = (4, 0, 6), and
W = (8, −2, −3), represented by the vector basis .B = φ̂1 + φ̂2 + φ̂3 . Therefore:
.
u = U · B =2φ̂1 − 1φ̂2 + 3φ̂3
.
v = V · B =4φ̂1 + 6φ̂3
w = W · B =8φ̂1 − 2φ̂2 − 3φ̂3
The generated vectors .u, .v, and .w are not orthogonal among them. However, a
prolongation of these vectors can generate another vector space. For example, the
sequence .κ = (−2, 3, 27) can be represented by .u, .v, and .w as basis vectors, from
their projections, that is .κ = αu + βv + γ w. First, the vector .κ generated by vector
basis .B is
κ = κ · B = −2φ̂1 + 3φ̂2 + 27φ̂3
.
Therefore, the projection leads to the following equation system:
2α + 4β + 8γ = −2
.
38
2 Signal Analysis in Digital Communications
−1α + 0β − 2γ = 3
3α + 6β − 3γ = 27
The solutions of these equation systems are .α = 1, .β = 3, and .γ = −2. Thus, the
sequence vector .κ can also be represented as .κ = u + 3v − 2w.
Example 2.16 The DC level of the signal is measured on its representation as
E {X} =
M
Σ
.
Xm Pm
m=1
Moreover, the quadratic value or average power can be calculated by
M
{ } Σ
2
E X2 =
Xm
Pm
.
m=1
2.7.1 The Phase Diagram
The analytical signal .Ŝ(t) = A(t)
{ exp(j θ}t) is the projection of the signal into two
orthogonal components .O(t) = φI , φQ {cos(ωc t), sin(ωc t)}. The phase diagram
is used to represent an analytical signal which is widely used in engineering for the
polar representation of the projections of the signal .xI (t) and .xQ (t), which are the
signals In Phase and Quadrature, respectively. Let the modulated signal be
x(t) = a(t) cos (ωc t + θ (t))
.
where, regardless of the modulation technique (amplitude, frequency, or phase), .a(t)
is the magnitude and .θ (t) is the phase of the modulated signal, both time variants.
In complex representation, this becomes the analytical signal:
X̂(t) = A(t) exp(j θ t) = xI (t) cos (ωc t) − xQ (t) sin (ωc t)
.
where .xI (t) and .xQ (t) are, respectively, the In Phase (I) and Quadrature (Q)
representation of the modulated signal. The Hilbert transform .{x(t)} allows the
passband signal to be decomposed into its in phase and quadrature components by
x̂(t) =
.
Then:
1
(x(t) + j H {x(t)})
2
(2.11)
2.8 Noise
39
/
(
)2 (
)2
1
1
x(t) +
H |x(t)| .
2
2
(
)
H |x(t)|
θ (t) = arctan
x(t)
A(t) =
.
(2.12)
(2.13)
The spectral representation looks like this:
X̂(ω) =
.
1
F̂ (ω) + sgn(ω)F̂ (ω)
2
(2.14)
2.8 Noise
In general, noise is defined as any type of unwanted energy present in the transmission and reception of a communication system. For example, in an audio recording,
any unwanted signal that falls in the hearing band (0–15 kHz) is perceptible to the
ear and will interfere with the information. Consequently, for audio circuits, this
unwanted signal within this band will be considered noise. Noise plays a crucial
role in communication systems. In theory, it determines the theoretical capacity
of the channel. In practice, it determines the number of errors that occur in a
digital communication. Now we consider how noise determines error rates in the
next section. In this section, we will provide a description of noise. Noise is a
random signal. In the transmission process, signals are always mixed with foreign
signals. When a signal carrying certain information is sent, it is susceptible to
being attenuated, deformed, or mixed with signals other than the transmission. Any
process imposed on a signal that introduces undesirable disturbances is called noise.
Noise is treated as a random process due to its unpredictable nature. The transmitter
and receiver of the overall system, together with the propagation medium, can be
treated as a random mapping from the infinite set of transmitted signals .{sm (t)} to
the received random process .n(t).
2.8.1 Channel Modeling with Noise
All sources of distortion, including fading, multipath, intersymbol interference
(ISI), nonlinear distortion, and additive noise, may be implicit in the signal due to
the propagation medium and electromagnetic implications before emerging to the
receiver:
• In general, noise is defined as any type of unwanted energy present in the transmission. Noise occurs in communication systems as unpredictable or random
40
2 Signal Analysis in Digital Communications
Fig. 2.10 Noiseless signal (left) and its respective spectra (right)
Fig. 2.11 Signal with band-limited noise (left) and its respective spectra (right)
behavior. When noise is introduced, both the signal form and the spectral density
change, as seen in Figs. 2.10 and 2.11.
• Noise is considered a phenomenon that in some way affects one or more
parameters of the signal. It is usually associated with the terms Disturbance and
Interference.
• When these events occur, the signal is said to be corrupted by by noise
(Fig. 2.12).
• In communication systems, noise affects the signal or message on the channel,
presenting erroneous information or total loss of the message to the receiver:
– Affects the carrier component, making the signal unable to be detected or
tuned.
– It affects the modulator component, so the signal does not behave in the
expected way, that is, the message arrives in error.
2.8 Noise
41
Fig. 2.12 Signal with white Gaussian noise (left) and its respective spectra (right)
2.8.2 Sources of Noise
There are different and very varied sources of noise. However, they can be classified
into different types: man-made (artificial) and natural. Examples include:
• The acoustic and electrical noise produced by the machines and the energy
distribution system (artificial).
• Electromagnetic pollution produced by wireless data networks (artificial).
• Atmospheric ionization and electric discharges (natural). The conductivity
(Puurtinen et al. 2006) changes according to the properties of the material
(air, Earth), presenting a variation of its reluctance to the propagation of the
wave.
• Mixer noise (artificial). Consider noise as a constant spectral value at all
frequencies, called white behavior (Garavaglia and Rosso 2007).
• Thermal noise produced by electronic components (artificial). Variable to temperature, called thermal behavior. It refers to the thermal movement of free
electrons (Karp 1961) through an energized conductor.
• Gunshot noise (artificial). It refers to the recombination of carrier charges, or
minority carriers (Li et al. 2006), in semiconductors. The value of the noise at a
certain instant cannot be exactly predicted, called random behavior. It is related
to Brownian noise, which has a property of self-similarity (Leporini 2001): If
2
.W (t) is a Brownian path over interval .[0, T ], then .W (c T )/c is a Brownian path
2
over .[0, c T ].
2.8.3 Noise Power Density
Taking these four behaviors into consideration, a general model for noise can be
established. Temperature is the first behavior to evaluate. The Boltzmann–Maxwell
equation law, combined with the work of Johnson–Nyquist, states that the power
42
2 Signal Analysis in Digital Communications
of the thermal noise generated within a source for a bandwidth of 1 Hz (in watt
per hertz) is the noise power density .η0 , measured in Watts per Hertz, which is
mathematically represented as
η0 = kT ◦
.
(2.15)
where k is the Boltzmann constant (.1.38 · 10−23 J/K), and .T ◦ is the absolute
temperature (Zhao et al. 2021). Taking into account that .0◦ K is equivalent to
◦
◦
.−273 C, if the ambient temperature has a value of .17 C, the absolute temperature
◦
will be .290 K.
2.8.4 Noise Variance and Power
Considering the second behavior, the simplest approach to define it is using a
Gaussian probability function (Allen et al. 2006). Being .σ 2 the variance of the
Gaussian probability, most of the values that represent the noise are within the
interval .[−2σ, 2σ ], giving as a result, the magnitude of the noise becomes .4σ (AjaFernández et al. 2006). The noise power then becomes equal to .16σ 2 . Because the
noise spectrum occurs at all frequencies, the noise is called Additive White Gaussian
(AWGN). The term Additive comes from the properties of the Gaussian random
variable; thus:
.
η0 |AW GN = 16σ 2
(2.16)
Considering the third behavior, the noise power .N0 can be calculated based on
Eq. (2.16), remaining
.
N0 |AW GN =
16σ 2
n0 |AW GN
=
R
R
(2.17)
where R is the resistance of the conductor or medium through which the signal
passes.
Example 2.17 The random walk is a form to model a noisy signal. Let .η[ti ] the
value generated at instant .ti by a Gaussian distribution .η[ti ] ∼ N(0, 1)R2 . The
signal .g[ti+1 ] at next time instant .ti+1 is generated as
σ
g[ti+1 ] = g[ti ] + √ η[ti ]
T
.
Example 2.18
• Brownian motion/Wiener process:
2.8 Noise
43
T →∞
g[ti ] −−−−→ Wt
.
ti /T →t
• Brownian bridge between .(a, b) ∈ C2 :
a + (b − a)
.
g(t) − g(0)
g(1) − g(0)
2.8.5 Wideband Properties
Considering the fourth behavior, the noise is commonly not white. On the contrary,
the presence of signal frequencies is examined. Given B as the bandwidth of the
signal or channel, the noise power .N0 can be computed, following Eq. (2.15), where
N0 =
.
kT ◦ B
R
(2.18)
P
N0
(2.19)
2.8.6 Signal-to-Noise Ratio
Single (Watt/Watt) form:
SNR ≡
.
Let us take the logarithm function of the signal-to-noise ratio:
.
log10 SNR = log10 P − log10 N0
dB
(2.20)
2.8.7 Sensitivity
It is the minimum input signal required at the receiver to produce a specific output
signal:
• Sensitivity describes the weakest signal power level that the receiver is able to
detect and decode.
• Sensitivity is determined by the lowest signal-to-noise ratio at which the signal
can be recovered.
• Different modulation and coding schemes have different minimum SNRs.
44
2 Signal Analysis in Digital Communications
• Range: .<0 to 60 dB.
• Sensitivity is determined by adding the required SNR to the noise present at the
receiver.
Noise limits the systems’ ability to detect low-power signals. Dynamic range is
the ability to detect low-power signals in the presence of high-amplitude signals.
The noise factor in a system is the ratio between the output power and the noise
power at the input. It is a measure of the degradation of the signal-to-noise ratio
(SNR) due to noise added by the system. It implies that the SNR worsens as the
signal is altered by noise in the amplifier. Using the definition, the noise factor is
equal to
kη =
.
•
•
•
•
SNRi
Pi (Ni G + Ns )
=
SNRo
Ni P o
P: Signal power
G: System gain
◦
.T : System temperature
.N: Noise power
.
.
Spot noise factor
kη' = 1 +
.
Ns
kT ◦
Using the noise equation, the noise factor can be approximated by
kη =
.
PZ
kT ◦ BG
where .P is the power of transmitted system and .G is the system gain. Z is the
impedance. .T ◦ is the system temperature and B is the bandwidth.
Friis Equation For L cascade stages, each with gain .Gi , we have
kη (L) =kη (1) +
.
kη (2) − 1
+
G(1)
kη (3) − 1
kη (L) − 1
+ · · · + | |L
G(1)G(2)
i=1 G(i)
The noise figure corresponds to the value in deciBell of the noise factor:
NF = 10 log10 kη
.
The power at the input is converted into dBm below:
2.8 Noise
45
Pin [dBm] = 10 log10 (Pin [mWatt]) = 16 dBm
.
For the transmitter (Tx), there is an output power equivalent to .PX [dBm] =
Pin [dBm] + GT x [dB] = 19 dBm. The signal-to-noise ratio (SNR) at the input of
the amplifier is equal to
SNRX [dB] = PX [dBm] − N0 [dBm] = 39 dB.
.
The amplifier has a noise figure equal to .NF = 3 dB. Thus, at the output of the
amplifier we have the following signal-to-noise ratio:
SNRY [dB] = SNRX [dB] − NF [dB] = 36 dB.
.
Using the amplifier gain .GAMP = 6 dB, the signal power at the amplifier output is
calculated as follows:
PY [dBm] = PX [dBm] + GAMP [dB] = 25 dBm.
.
Then, from the signal-to-noise ratio at the output of the amplifier SNR.Y , the noise
at the output of the amplifier is obtained as illustrated below:
NY [dBm] = PY [dBm] − SNRY [dB] = −9 dBm.
.
As it passes through the channel, the amplified signal power and noise are attenuated
by a factor of .L = 80 dB. Therefore:
PZ [dBm] = PY [dBm] − L[dB] = −55 dBm.
.
NZ [dBm] = NY [dBm] − L[dB] = −89 dBm.
.
At the receiving point (Z), the noise power of the channel .NC is added, so that,
with a channel bandwidth of B.= 40 MHz, .k = 1.3806488 × 10−23 J/K (Boltzmann
constant) and temperature of T.◦ = 303 K:
NC [watt] = kBT ◦ = 1.67 × 10−13 W
.
And the total noise power at the receiver is equal to
NT [watt] = NZ [watt] + NC [watt]
.
(
.
= 10
NZ [dBm]
10
)
+ NC [Watt] = 1.26 × 10−9 W.
46
2 Signal Analysis in Digital Communications
2.8.8 Quantization Noise
( )
The fact of “rounding” .fs tj in an approximate expression introduces an error
( )
with respect to the real value, thus inducing a noise called quantization noise, .s tj ,
and the closest quantization level .{X}m . This noise produces distortion of the signal
when it is reconstructed, also generating harmonics at frequencies higher than the
maximum signal frequency; therefore, one way to eliminate this noise can be by
filtering the signal when it is reconstructed. But in any case noise components occur
within the bandwidth of the signal, so this noise is decreased if the number of levels
M is increased. At the output of the receiver, the quantization level could represent
any value in the interval
(
.
)
Ac
Ac
(2m − 1),
(2m + 1)
2M
2M
For the unipolar case, that is, there is an uncertainty whose maximum value
occurs at each half of a quantization level. Assuming that all quantization noise
values are with same probability value in the minimum quantization range, i.e.,
.
−
Ac
Ac
≤η<
2M
2M
η follows a uniform random distribution on this interval, described by
.
{
PN (η) =
.
M
Ac
Ac
Ac
− 2M
≤ η < 2M
0
Other cases
Then the power of the quantization noise is (spectral density of a probabilistic
function)
N0 = W (η) =
.
f ∞
−∞
f Ac
η2 PN (η)dη =
2M
Ac
− 2M
M 2
η dη
Ac
| Ac
[ 3
]
Mη3 || 2M
M
A3c
Ac
A2c
=
=
+
=
|
3
3
3Ac − Ac
3Ac 8M
8M
12M 2
2M
With . AMc = {X}0 :
|
|
f02
|
.N0 =
⇒ W ε|
|
12
(
dB
X02
= 10 log
12
Likewise, for the bipolar case:
)
dB = 20 log X0 − 10.8 dB
2.8 Noise
47
{
PN (η) =
M
2Ac
− AMc ≤ η < AMc
0
other case.
.
N0
= ε2 =
f ∞
−∞
f Ac
ε P (ε)dε =
2
M
− AMc
M 2
ε dε
2Ac
| Ac
[
]
A2c
M A3c
A3c
Mε 3 || M
=
=
+
=
|
3
3
6Ac − Ac
6Ac M
M
3M 2
.
M
The minimum quantization value is .X0 , given by
.
X0 =
2Ac
M
N0 =
X02
12
(
X02
W ε|dB = 10 log
12
)
dB = 20 log X0 − 10.8 dB
Therefore, a large number of quantization levels imply a lower quantization error.
Likewise, the maximum value of the quantization noise is
.
max η =
√
X0
N0 = √
2 3
And its SNR would be expressed as
{ }
{ }
12E f 2
E f2
=
.SNR =
X0
X02
the SNR is also expressed in dB as:
⎛/ { }⎞
E f2
⎠
⎝
.SNRdB = 10.8 + 20 log
X0
[dB]
where .f¯2 is the expected quadratic value of the source .f (t). The quantization
process can then be modeled as
( )
{X}m = s tj + η
.
Being .{X}m a uniform random variable between the quantization intervals.
Taking into account that there are .M = D K levels, the value of the interval between
48
2 Signal Analysis in Digital Communications
levels .X0 is given by
|
Ac |
Ac
= K ||
M
D Unipolar
.
|
2Ac
2Ac |
X0 =
= K ||
M
D Bipolar
X0 =
So the quantization noise power is, for a unipolar quantizer,
(
Ac
.N0 [dB] = 20 log
DK
)
− 10.8 = 20 log Ac − 20K log D − 10.8
[dB]
and for a bipolar quantizer:
(
N0 [dB] = 20 log
.
2Ac
DK
)
− 10.8 = 20 log Ac − 20K log D − 4.8
[dB]
The signal-to-noise ratio for a unipolar quantizer is
⎛/ { }
⎞
E f 2 DK
⎠
.SNRdB = 10.8 + 20 log ⎝
Ac
⎛/ { }⎞
E f2
⎠
= 10.8 + 20K log D + 20 log ⎝
Ac
and for a bipolar quantizer:
⎞
⎛/ { }
E f 2 DK
⎠
.SNR[dB] = 10.8 + 20 log ⎝
2Ac
⎛/ { }⎞
E f2
⎠
= 4.8 + 20K log D + 20 log ⎝
Ac
Sometimes, to reduce quantization distortion (minimize quantization error), the
quantization levels are shifted in the form:
{
{X}m =
.
{X}m+1
(
)
fd tj , m
m odd
m even
2.8 Noise
49
Example 2.19 Calculate the SNR of the signal .f (t) = A cos ωt when quantized
at levels of .X0 = AV of amplitude, given that the maximum quantization range
.εSAc = A. Using Example 2.16:
/ { }
A
E f2 = √
2
.
Thus Signal to Noise Ratio given by quantization is equal to:
SNR = 10.8 + 20 log
.
A
A
− 10 log 2 = 7, 79 + 20 log
A
A
[dB]
Example 2.20 Repeat the previous example, for the signal .f (t) =
using quantization levels of .X0 = 0.1 V.
√
2 cos ωt,
/ { }
E f2 = 1
.
SNR = 10.8 + 20 log
1
= 30, 8 dB
0, 1
Example 2.21 By quantizing the signal .f (t) = cos ωt, an SNR of .30 dB is
obtained. Find:
a. The width of the quantization intervals.
/ { }
1
E f2 = √
2
.
⎛/ { }⎞
E f2
⎠ = 30 dB
SN RdB = 10.8 + 20 log ⎝
X0
⇒ f0 =
(
(
)
)
/ { }
SNRdB −10,8
SNRdB −10,8
1
−
−
20
20
E f 2 10
= √ 10
= 0.078 V
2
b. The number of levels required to completely cover the signal .f (t).
Because it is a bipolar signal, the quantizer reaches
of .2Ac . Therefore
/ { a range
}
.M = (2Ac ) /0.078. If it is assumed that .Ac =
E f 2 = √1 , we have .M ≈
2
18, leaving 9 for each polarity.
c. Number of bits used in quantization.
Since .M = D K quantization levels are used, given the binary base .D = 2,
then .K = log2 M = 5 bits, of which one is sign and the remaining 4 are used for
magnitude.
Example 2.22 Calculate the minimum SNR value of a 3-bit quantizer (one sign bit,
two information bits), in the worst case.
With .K = 3, the worst case happens at the minimum quantization value. This is:
50
2 Signal Analysis in Digital Communications
/ { }
2Ac
2Ac
Ac
= K =
E f 2 = X0 =
M
2
4
.
Therefore:
⎛/ { }⎞
E f2
⎠ = 10.8 dB
.SNRdB = 10.8 + 20 log ⎝
X0
Example 2.23 Calculate the maximum SNR value of a 3-bit quantizer (one sign
bit, two information bits), at its maximum amplitude.
The maximum amplitude occurs in
(
)
/ { }
3Ac
2Ac M
2
−1 =
=
. E f
M
2
4
Therefore:
SNRdB = 10, 8 + 20 log
.
( √
M
E {f 2 }
2Ac
)
= 10, 8 + 20 log
(
3Ac M
8Ac
)
= 10, 8 +
20 log 3 = 20, 34 dB
Example 2.24 Find the SNR for a unipolar signal quantized to 8 bits.
We have that .D = 2 and .K = 8. The Signal to Noise Ratio given by quantization
is equal to:
⎛/ { }⎞
⎛/ { }⎞
E f2
E f2
⎠ = 59 + 20 log · ⎝
⎠
.SNRdB = 10.8 + 20K log D + 20 log ⎝
Ac
Ac
Example 2.25 Show that a bipolar binary quantizer has a quarter of the SNR of a
unipolar quantizer. With .D = 2, we have, for a unipolar quantizer, the Signal to
Noise Ratio:
⎛/ { }⎞
E f2
⎠
.SNRdB = 10.8 + 20K log 2 + 20 log ⎝
Ac
Then for a bipolar quantizer
⎛/ { }⎞
E f2
⎠
.SNRdB = 4.8 + 20K log 2 + 20 log ⎝
Ac
Then, the difference between the quantizers is expressed as
.
SNRdB |Unipolar − SNRdB |Bipolar = 6 dB
which represents a quarter of the SNR of the unipolar quantizer.
2.8 Noise
51
Example 2.26 Show that a bipolar binary quantizer can have the same SNR as
a unipolar quantizer by removing the sign bit. Taking the SNR for the bipolar
quantizer, taking into account that one of the K bits is used as a sign:
⎛/ { }⎞
E f2
⎠
SNRdB = 4.8 + 20(K − 1) log 2|Magnitude + 20 log 2|Sign + 20 log ⎝
Ac
⎛/ { }⎞
E f2
⎠
= 4.8 + 20(K − 1) log 2|Magnitude + 20 log ⎝
Ac
.
Note that to equal the SNR value of a unipolar quantizer, the power given by the
sign is added. This shows that the SNR is the same in both, as long as the sign bit
is considered in bipolar quantization and the signal is quantized with the remaining
.K − 1 bits.
Example 2.27 Find the SNR resulting from the quantization process of the signal
f (t) = . A2c cos ωt in 8 bits.
/ { }
With . E f 2 = A√c , D = 2 and .K = 8, and using the result of example 4.14:
.
2 2
⎛/ { }⎞
E f2
⎝
⎠
.SN RdB = 10.8 + 20(K − 1) log D + 20 log
Ac
(
1
= 53 + 20 log √
2 2
)
= 44 dB
Example 2.28 Given a failure in the quantizer, the amplitude of the quantization
levels
varies from 10 to .20 mV. Determine, when the signal is quantized .f (t) =
√
2 cos ωt:
a. The variation in SNR.
/ { } √
2
. E f2 = √ = 1
2
Using equation 4.28, and taking the zero truncated interval .X0 = 0.01,
SNR0,01 = 10.8 + 20 log 100 = 50.8 dB
.
For .X0 = 0, 02
SNR0,02 = 10, 8 + 20 log 50 = 44.8 dB
.
The Signal to Noise Ratio variation can be calculated as:
52
2 Signal Analysis in Digital Communications
SNR0.01 − SNR0.02 = 50.8 − 44.8 = 6 dB
.
b. The noise power due to the variation of values of quantization levels.
Given the result above:
{ }
E f2
= 50.8 dB
SNR0.01 = 10 log
N0 (1)
.
{ }
E f2
SNR0.02 = 10 log
= 44.8 dB
N0 (2)
{ }
Given the energy normalized value .E f 2 = 1, the noise values are .N0 (1) =
8.3176 × 10−6 W and .N0 (2) = 33.1131 × 10−6 W. The ratio of variation of noise
power can be calculated as:
.
N0 (1)
= 3.98 = 6 dB
N0 (2)
Review Questions
•
•
•
•
•
•
•
What are carrier charges?
What are free electrons?
What is the relationship between conductivity and resistance?
Where does the term white come from?
What is Brownian noise?
What is a Gaussian function?
Could the telephone copper pair be used to propagate television signals
with a bandwidth of .6 MHz?
2.9 Problems
Problem 2.1 Construct a sine wave function using period T =
√
5 s.
Problem 2.2 Generate a quantized signal database by the following procedure:
Step 1. Given a selected arbitrary value of carrier frequency ωc , generate a
bank of signals from the signals ACarrier and DCarrier, increasing the
frequencies in octaves, specifically ωc = [ωc , 2ωc , 4ωc , 8ωc ].
Step 2. Calculate the power of each of the analog and digital bank signals, and
assign them as PT x (ωc )[ACarrier] and PT x (ωc )[DCarrier], respectively.
These values are those that appear in Table 2.2.
2.9 Problems
53
Table 2.2 Initial data
Frequency
ωc
2ωc
4ωc
8ωc
PT x
Table 2.3 Properties of the sine cardinal (sinc) function
Function
sinc (a t)
Fourier(transform
)
f
1
|a| rect a
Fourier transform
angular frequency
)
(
√ 1 rect ω
2π a
2
2π a
Problem 2.3 Calculate the frequency (in hertz) and period (in hertz) of minimum
sampling, and transmission rate for:
• The signal x (t) = cos(100π t), which is quantized into 512 equally spaced levels
• The signal x (t) = sin (200π t), which is quantized into 1024 equally spaced
levels
• A train of equally spaced unitary unipolar square pulses at T = 1/(20π ), which
is quantized into 4 equally spaced levels
• A train of unitary unipolar triangular pulses of equal slope and spaced at T =
1/(5π ), which is quantized into 256 equally spaced levels
With the result, what would be the conditions to interleave both quantized signals in
a single channel?
Problem 2.4 Specify the minimum sampling frequency and period for signal
acquisition
g (t) = sinc (320t)
.
Table 2.3 and Eq. (2.21) may be useful.
( )
(
(
t
τ)
τ)
.rect
−u t −
=u t+
2
2
τ
(2.21)
Problem 2.5 A signal with a bandwidth of 32×103 Hz is quantized into 24 equally
spaced levels in a unipolar range Ac = 1 V:
a. Determine a table showing the resulting quantization levels using PCM.
b. Determine the channel bandwidth required for effective transmission with a
signal-to-noise ratio of 20 dB.
Problem 2.6 A signal is sampled at 32 Hz and quantized into 24 equally spaced
levels in a unipolar range Ac = 1 V:
54
2 Signal Analysis in Digital Communications
1. Determine a table showing the resulting quantization levels using PCM.
2. Determine the power of quantization error, assuming that the probability of
placing a value in some interval follows a uniform distribution.
Problem 2.7 The signal x(t) = cos (3π t) is quantized into 16 equally spaced
levels. If the sampling rate is 100 Hz, determine:
a. A table indicating the resulting quantization levels using PCM, if the minimum
quantization level is at the minimum signal value
b. A table indicating the resulting quantization levels using PCM, if the minimum
quantization level is at the value zero and one bit is used for the sign
Problem 2.8 For a PCM system with quantization interval q = 0.02 V, and
quantization range of Ac = ±3 V, find the number of quantization levels and the
quantization noise power.
Problem 2.9 Table 2.4 illustrates the data, quantized by PCM, of two signals A and
B. If the range of the quantizer is Ac = ±1:
a. Determine the value of the quantization interval and the amount of information
required for these samples to have a quantization noise power of 10−4 W.
b. Find the sequences that represent the signals A and B in PCM.
c. Determine the values of Table 2.4, if the quantizer is DPCM, using the
estimator:
a. Ax(tj ) = x(tj ) − x(tj −1 ).
b. x(tj ) = 0.7x(tj ) − 0.3x(tj −1 ).
c. x(tj ) = 0.3x(tj −1 ) − 0.4x(tj −2 ) + 0.3x(tj −3 ).
d. x(tj ) = 0.2x(tj −1 ) − 0.4x(tj −2 ) + 0.4x(tj −3 ).
d. Using the same PCM quantization interval, determine the information reduction
when using each DPCM quantizer (a through d).
Problem 2.10 A signal is sampled at a frequency of 8 kHz, and then it is quantized
by some method. However, it is required that the signal to quantization noise is at
least SNR = 50 dB. Assuming that for the method you have to
SNR [dB] = 1.8 + 6Hb {X}
.
[dB]
where Hb {X} is the number of representation bits. Calculate:
a. Hb {X}.
Table 2.4 Quantized data of two signals
Signal A 0.13 0.19
0.22
0.3
0.25
0.35
0.42
0.52
0.65
0.75 0.86
Signal B 0.13 0.09 −0.02 −0.2 −0.45 −0.65 −0.72 −0.85 −0.65 −0.45 0.26
2.9 Problems
55
b. The minimum storage capacity required in the channel to overcome quantization
noise, if the bandwidth consumed using the quantizer is twice the transmission
rate R.
Problem 2.11 A Delta-Modulation system is tested with a signal described by
m (t) = 5 sin (20t)
.
a. Determine the minimum sample period Ts .
b. Determine the step size A required to avoid slope overloading and to minimize
granular noise.
Equation (2.22) may be useful.
|
|
| dm (t) |
A
|
|
≥ max |
.
dt |
Ts
(2.22)
Problem 2.12 Consider a White Gaussian noise process of zero mean and power
spectral density N0 /2 that is applied to the input of the high-pass RL filter shown in
Fig. 2.9:
a. Find the autocorrelation function and power spectral density of the random
process at the output of the filter.
b. What are the mean and variance of this output?
Problem 2.13 White Gaussian noise of zero mean and power spectral density N0 /2
is applied to the filtering scheme shown in Figure (a). The frequency responses of
these two filters are shown in Figure (b). The noise at the low-pass filter output is
denoted by n(t):
a. Find the power spectral density and the autocorrelation function of n(t).
b. Find the mean and variance of n(t).
c. What is the rate at which n(t) can be sampled so that the resulting samples are
essentially uncorrelated?
56
2 Signal Analysis in Digital Communications
Problem 2.14 The power spectral density of a narrowband noise n(t) is as shown
in figure. The carrier frequency is 5 Hz:
a. Find the power spectral densities of the in phase and quadrature components of
n(t).
b. Find their cross-spectral densities.
Problem 2.15 Consider a Gaussian noise n(t) with zero mean and the power
spectral density sN (ω) shown in figure:
a. Find the probability density function of the envelope of n(t).
b. What are the mean and variance of this envelope?
Problem 2.16 A pair of noise processes n1 (t) and n2 (t) are related by
n2 (t) = n1 (t) cos (ωc t + θ ) − n1 (t) sin (ωc t + θ )
.
where ωc is fixed, and θ is the value of a random variable O whose probability
density function is defined by
{
fO (θ ) =
.
1
2π
0 ≤ θ ≤ 2π
0
otherwise
The noise process n1 (t) is stationary, and its power spectral density is as shown in
Figure. Find and plot the corresponding power spectral density of n2 (t).
References
57
Fig. 2.13 Communication system
Table 2.5 Parameters of the
system in Fig. 2.13
Parameter
Input power
Transmitter gain
Amplifier gain
Amplifier noise figure
Inserted noise in amplifier
Channel bandwidth
Channel temperature
Link loss
Symbol
Pin
GT x
GAMP
NFAMP
N0
Value
40 mW
3 dB
6 dB
3 dB
−20 dBm
B
T◦
L
40 MHz
30 ◦ C
80 dB
Problem 2.17 Figure 2.13 illustrates a communication system with the parameters
of Table 2.5. Determine the required sensitivity at the receiver (Z point).
Problem 2.18 Plot the superimposed transmit and receive waveforms for messages
transmitted in Baseband, using the values of Wl, Wh, and SNRval defined above.
These figures are drawn with adequate tension and time scales. According to this
result, determine the most efficient technique.
Problem 2.19 Give some examples about the effect on communication systems
due to the addition of noise, and channel limitations in bandwidth. Sketch some
differences between the signals measured at sender and receiver points for a
baseband transmission.
Problem 2.20 In a baseband transmission, the fundamental frequency is altered by
noise. Describe the amplitude variations in signal envelope it would happen. Would
the increase of the signal amplitude do reduce this alteration?
Problem 2.21 The saturation effect of the electronic devices leads to the distortion
of the shape of the signal. Give special considerations to reduce this effect at receiver
stage.
References
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for electromagnetics. In: 2019 URSI Asia-Pacific Radio Science Conference, AP-RASC 2019.
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Aja-Fernández S, Estépar RSJ, Alberola-López C, Westin CF, Aja-fern S, Alberola-l C,
Westin CF (2006) Image Quality Assessment Based on Local Variance, New York, vol 1,
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Garcia-Alvarez JC, Gomez S (2014) Performance of an Improved LS3/5A Speaker. NOOS 4
Garcia-Alvarez JC, Aguirre-Ordonez S, Marino-Duque S (2014) Perceptual Sound Quality
Assessment. NOOS 4
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7–33. https://doi.org/10.1007/978-981-16-6452-6_2
Chapter 3
Information Theory
3.1 Basics on Information Theory
For every transmission system, the medium or channel is the fundamental basis
for the establishment of reliable information (Ash 1965; Gallager 1968; Abramson
1963). The main objective of channel coding is the protection of digital information
to transmit or store ensuring greater immunity against noise, so that it does not
reach the receiver, regardless of the goodness of the channel. This information, in a
transmission, or storage system, noncodified digital (Fig. 6.1), is created in a binary
source (data, voice, image, .. . .) normally with a treatment that, by conversion or
compression, reduces the cost of transmission. After a series–parallel conversion,
these bits become digital symbols belonging to the same alphabet. Through the
mapping, these symbols translate into points of the constellation that the modulator
transforms into a signal that is transmitted through the channel, or, in the case of
information storage, the symbols are passed to a unit of writing that records them in
a digital support. The receiver captures the signal that has been modified in the space
that must have been traveling as a channel and that the demodulator transforms
into points of the constellation and the plaintiff into digital symbols, or, where
appropriate, recovers the symbols thanks to the unit of reading, which have been
modified during the time they have been stored in the support. In any case, these
digital symbols, which are an estimate since they do not usually coincide with the
originals for the modification in the channel, suffer a parallel–series conversion that
delivers to the destination some bits estimate of those delivered by the binary source.
What is intended in any transmission is that the information experiences the least
possible modification and that those bits that are delivered to the destination are as
similar as possible to those originated in the source. The way to quantify this similar
could be to quantify the reliability of the channel, but it is an arduous task. The most
advisable thing is to quantify the reliability of the transmission by means of the
residual error rate or BER (bit error). In this way you can ask if the error rate of the
transmission system is less or not than the user considers acceptable. The problem
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024
J. C. García-Álvarez, Digital Electronic Communications,
https://doi.org/10.1007/978-3-031-53118-7_3
59
60
3 Information Theory
Fig. 3.1 Digital channel model
appears when this maximum error rate is exceeded and the information becomes
unintelligible. The first solution in which it is thought is the increase in transmission
power, to increase the Signal to Noise Ratio (SNR). But it is not always possible,
since many times the transmission power is already limited by other aspects. The
solution that will be discussed in this project is that of the channel coding, since, as
Shannon demonstrated in 1948, with an appropriate coding of the information, the
errors introduced in it by a noisy channel or storage can be reduced to a arbitrary
level without sacrificing the transmission rate or storage. Information transmission
can be represented as the channel coding of Fig. 3.1, where the source (reading from
a data storage system, for example) emits a series of digital symbols. Appropriate
modulated symbols are produced throughout the coding process to be sent across
the electromagnetic medium. After processing the signals impacted by any physical
disturbance, the decoder provides the sink with the expected symbols. Therefore, it
is a block that, initially, transforms words of digital symbols into words of digital
symbols belonging to the same alphabet, in order to obtain greater immunity to
noise. This provides certain advantages with respect to the .S(tJ ) signal: Most of
the processes involving symbols are without memory (Memoryless Digital Process),
that is, noise effects or transmission errors are not cumulative and can be completely
regenerated. Bits represent unique signal levels that can be easily rebuilt by a level
detector, isolating additive noise. An efficient code that reduces the unnecessary
repetition of data (compression) can be used. The signal power can be improved
using the bandwidth of the transmission channel.
For this it is not enough to send the message directly on the channel. Specifically,
the channel has parameters such as the bandwidth, the frequency response, and
the signal-to-noise ratio. These parameters determine how efficient the information
can be sent by means of a transmission channel. The present work presents the
complications that are presented when passing a baseband message through a
transmission medium with band restrictions and frequency. This chapter aims to
illustrate the effects of message baseband transmission, illustrating how channel
restrictions affect the shape of the message, reaching the point of modifying it
completely.
3.1 Basics on Information Theory
61
3.1.1 Information Rate
It is a measure of the information that can be transferred through a system per unit
of time. The first approach was proposed in 1920 by R. Hartley, in proportion to the
channel bandwidth B, which corresponds to the
R α B · t,
.
where the information rate is a measure of the speed with which information is
transferred. The information rate or speed of the information source .R is defined as
the relationship given by
R=
.
H {X}
T
(3.1)
where .H {X} is the entropy of the source and T is the time interval over which the
source transmits the event or a series of events. For a single event .X = i, for which
we have a measure of its amount of information, we have
R=
.
I {X = i}
Ti
(3.2)
Considering M equiprobable, statistically independent symbols of equal duration
T , the maximum transmission rate is
.
1 Σ
H {X}
=
I {X = m}P {X = m}
T
T
M
R=
.
m=1
1
1 Σ
log2 M = log2 M bits /s
MT
T
M
.
=
m=1
Example 3.1 The information rate of a binary symmetric channel is
R=−
.
1
1
p log2 (p) −
(1 − p) log2 (1 − p)
TA
TB
(3.3)
which .TA TB determines the duration of existence given the event zero. If the
duration of each event is determined by time entity, then:
R=−
.
1
1
p log2 (p) −
(1 − p) log2 (1 − p) 0 ≤ k ≤ 1
kT
T (1 − k)
where k is called the duty cycle of the source.
62
3 Information Theory
3.1.2 Shannon’s Theorem and Channel Capacity
The most important question associated with a communication channel is the
maximum rate at which information can be transferred. Side information can be
transferred by serial if the serial is allowed to change. Analog signals transmitted
over physical channels cannot rapidly change arbitrarily. The rate at which a signal
can change is determined by the bandwidth. In fact, it is governed by the same
Nyquist–Shannon law as in the case of sampling, a signal of bandwidth B can
change at a maximum rate of 2B. If each change is used to represent a symbol,
the maximum information rate is 2B. The Nyquist–Shannon theorem makes no
observation about the magnitude of the change. If changes of different magnitudes
are each associated with a separate symbol, the information rate can be increased.
So, if at each instant the signal changes, one of the M symbols of equal probability
that can be transmitted can be taken. So, the information rate R is
R=
.
1
logD M = 2B logD M
max {Ti }
This equation states that as M tends to infinity, so will the information rate. Is there a
limit to the number of levels? The limit is established by the presence of noise. If we
continue to subdivide the magnitude of the changes into their decreasing intervals,
we reach a point where the levels cannot be distinguished separately due to the
presence of noise. Noise therefore provides a limit on the maximum rate at which
information can be transmitted. Therefore, what really matters is the ratio (SNR).
There is a theoretical maximum for the rate at which information passes error-free
over the channel. This maximum is called the channel capacity C. The Hartley–
Shannon law establishes that the capacity of channel C is given by the maximum
amount of information of a given event .Y = n that has been transmitted as an event
.X = m:
C = max {I {Y = n|X = m}} .
.
= max {I {Y = n ∩ X = m}} |I {X = m}
(3.4)
(3.5)
As demonstrated with entropy, the maximum value is found when M equiprobable symbols are transmitted. Then:
C = max {I {Y = n ∪ X = m}} | logD M = B logD (SN R + 1)
.
(3.6)
Note that the signal-to-noise ratio is linear in this expression.
Example 3.2 A 16 kHz channel operating with an SNR of 15 dB has a maximum
information rate of .C = 10000 log2 (31.623) = 49.28 kbps
The theorem does not state how the channel capacity is fulfilled. In fact, channels
only come close to this limit. The task of providing optimal spectral efficiency
3.1 Basics on Information Theory
63
is the goal of coding techniques, which will be seen later. Failure to find perfect
performance is measured by the bit error rate (BER). Typically the BER is of the
order of .10−6 . For a given channel there is a technique that allows the transmission
of errors through the channel at a speed R, as long as .R < C, the channel
capacity. The identity holds only when the SNR is infinite. As we saw previously,
the important part of this theory is the existence of channel capacity. This offers
lower error transmission, but it does not define what code is required. Indeed, what
can be deduced from the proof of the theorem is that it must be long enough. No
one has finally found a code that allows the use of a channel due to its capacity.
However, Shannon proved that the code exists. The Hartley–Shannon law tells us
that the maximum information rate of a zero bandwidth channel is zero. Thus,
sources with zero bandwidth contain no information. To allow the channel new
information, we must introduce capacity for nonperiodic change. The consequence
of a nonperiodic change is to introduce an expansion of frequencies in the signal.
A detailed theoretical consideration leads us to the Nyquist–Shannon law where a
bandwidth signal B can change in a nonperiodic manner at a maximum rate of 2B.
Let us consider the case of a square signal, which we can apply to the rectum of
newspaper serials. If a square signal wave is replaced with a nonperiodic binary
sequences, the spectrum changes substantially. The discrete harmonic components
are replaced by a continuous range of frequencies whose form is the sampling
function .sinc(w). There are several features to note:
• The bandwidth of the signal can only be approximated to a finite value. Most
of the energy is contained in a limited region called the main lobe. Nonetheless,
some energy remains at all frequencies.
• If there are two changes in the period T, the width of the main lobe is .4/T .
The Nyquist–Shannon law establishes that if a serial has a bandwidth of .4/T .
Therefore the maximum number of changes that can take place is .8/T per second
or changes in the period T . If each change is taken to represent 1 tilt, this will
mean that this signal does not carry as much information as the bandwidth allows.
• The spectrum has positive and negative frequencies. These are symmetrical about
the origin.
The bandwidth of a communication channel is limited by the physical construction
of the channel. The Nyquist–Shannon law places a limit on the maximum number of
changes, events per unit of time that the signal can make. It is the modulator’s task
to make best use of the available band width. However, the modulator is restrained
by its physical design and construction, and it can and cannot approach the Nyquist
limit in its operation, for a given bandwidth. This Hartley–Shannon law establishes
that if the SNR tends to infinity, so will the capacity. This claim that the methods
that have now been described does not comply with this law. This inability to satisfy
the Hartley–Shannon equation is often explained short of determining that it is a
theoretical relationship and that should be treated for the practical environment, this
is neither a good nor correct explanation. The law is valid for systems whose inputs
and outputs can take some value. As this is not true for real systems, the equation is
not satisfied, but there is no reason why it should be (deriving an alternative formula
64
3 Information Theory
for binary channels). The notable feature of the Hartley–Shannon equation is not that
the channels used do not obey it, but a message can be transmitted without errors.
If the previous results are considered, it can be seen that, for a finite SNR, there is
always a finite probability of error. This is the existence of a channel capacity, not its
functional form, which makes the equation an important relationship. It is important
to understand the difference between the maximum bit rate and channel capacity.
The maximum data rate is determined by system design and assigned operating
frequency. The channel capacity is the maximum rate that the channel can transmit
error-free data. Clearly, the channel capacity can not exceed the maximum data rate.
However, there is not any technique describing how error-free transmission can be
made.
3.1.3 Shannon’s First and Second Theorem
If the information rate R is equal to or less than the capacity of the channel .C,
there will be a coding technique that allows transmission over the channel with a
negligible BER; therefore, it must be true that .R ≤ C. This restriction is valid
even in the presence of channel noise. This condition is necessary, but if .R > C,
it is not possible to transmit messages without errors. Channel capacity is then
defined as the maximum amount of reliable information through the channel. For
each communication system, the channel capacity is then expressed with respect
to the bandwidth B and the SNR. Although the relationship between the channel
capacity affected by a band-limited Gaussian noise is introduced by the HartleyShannon theorem, there are terms here that have not been clarified yet, such as Data
Rate. These and other related terms will be explained below.
Example 3.3 (Binary Symmetric Channels) Returning to the equation, the binary
channel (BC) is a channel with a binary input and output. Associated with each
output is the probability p that the output is correct and the probability .1 − p that it
is not correct. For each channel, the channel capacity becomes
C = 1 + H {X} = 1 + p log2 p + (1 − log2 (1 − p))
.
(3.7)
Here, p is the probability of bit error. If .p = 0, then .C = 1. If .p = 0.5, .C = 0.
So if there is an equal probability of receiving a 1 or a 0, regardless of what is
sent, the channel is completely unfeasible and you cannot send messages without
errors. As defined, the channel capacity is a dimensionless number. Normally if the
capacity is at an arbitrary rate of R bps, to do this, we name each output to change
in the channel. For a channel bandwidth B, we can transmit at least 2B changes per
second. So the capacity in bps is 2B. For the binary channel we have
Rb = 2B(1 + p) = log2 p(1 − p) log 2(1 − p)
.
(3.8)
3.2 Signal-to-Noise Ratio
65
For the binary channel the maximum bit rate Rb is 2B. I know that .C < Rb ; in
effect, the capacity is always less than the bit rate. The data rate .R, or information
rate describes the rate of transfer of data bits over the channel, in theory. We have
Rb ≥ C ≥ R
.
Shannon’s channel coding theorem applies to the channel, but not to the source. If
the source is optimally encoded, we can restate the coding theorem: A source of
information with entropy .H {X} can be transmitted seamlessly over a channel as
long as: .H {X} < C. The capacity of the binary channel is much smaller than that
calculated from the Hartley–Shannon Law (Eq. 3.6). This is because the equation
applies to systems whose outputs can take certain values. Systems that obey this law
are used because they are technically convenient and not because they are desirable.
The noise-free source coding theorem (also called Shannon’s first theorem) states
that a symbol with an average number of bits per symbol .Rb , which represents a
source of entropy .H {X}, can be found such that:
RKc ≥ H {X}
.
(3.9)
3.1.4 Receiver Power
The receiver power is equal to the modulation power .Pm minus the loss powers:
Pr = Pm − Ploss = Pm − P RL1 − P RL2 − PM
.
(3.10)
The losses .P RL1 and .P RL2 are the losses of the windings, and the loss .PM is
due to magnetic dispersion.
3.2 Signal-to-Noise Ratio
Signal-to-noise ratio SNR is defined as the proportion between the power of the
signal that is transmitted and the power of the noise that corrupts it. This margin is
measured in decibels (dB).
In watt/watt and dB form, respectively:
SN R =
.
PRx
Pη
SNR|dB = 10 log10 PRx − 10 log10) Pη
66
3 Information Theory
3.2.1 Sensitivity and Channel Capacity
Sensitivity describes the power level of the weakest signal that the receiver is
capable of detecting and decoding, i.e., sensitivity is determined by adding the SNR
required to the noise present in the receiver and is in a range from 0 to 60 dB. They
are connected by the Shannon–Hartley theorem, which refers to the useful part of
the signal and the false or irrelevant data.
Sensitivity determines the ability of a receiver to respond to weak signals that
exceed the noise floor. It is defined as the minimum signal power that the receiver
can detect, such that a useful information signal is recovered. Therefore, sensitivity
is associated with the signal-to-noise ratio at the receiver. It also usually specifies
.μV of the received signal. Also called the receiver threshold. The sensitivity
depends on the noise power present at the input to the receiver. Sensitivity is part of
the restriction that the channel determines for the transmission of the signal without
distortion. As a general rule, a communications channel cannot propagate signals
with a bandwidth greater than the passband of the channel itself.
Therefore, the fundamental problem consists of the efficient use of bandwidth
to propagate information through an electronic communications system. In 1948
C. E. Shannon, a scientist at Bell Laboratories, related the information capacity
of a channel (in bps), with its bandwidth (in Hz) and its signal-to-noise ratio in a
mathematical expression called Shannon limit information capacity:
)
(
P
.C = B log2 1 +
(3.11)
N0
[
]
P
.C = 3.32 B log10 1 +
N0
• C: Information capacity in the channel [bps]
• B: Bandwidth [Hz]
• . NP0 : Signal-to-noise ratio [Watt/Watt]
It is a measure that is used to compare the desired level of a signal with the level of
the background noise, it is defined as the ratio of the power of the signal to that of
the noise commonly expressed in dB, and a ratio greater than 1 will then indicate a
noise component larger than the signal itself, both the SNR and the bandwidth and
capacity of a communication channel.
Example 3.4 A radio station wants to transmit the music that a user programs. A
preliminary analysis considers that there is no loss of information if the music is
sent via MODEM through a cable whose bandwidth is 64 kHz, as long as there is
an SNR of 20 dB. Determine the channel capacity.
C = B log2 (1 + SNR) = 64[kHz] · log2 (1 + 100.2 )
.
.
= 426.12 kbps
3.2 Signal-to-Noise Ratio
67
Assuming that from the preliminary analysis, the listener has a good musical quality,
if the signal is quantized in 12 bits, at 24 kHz:
R = fs · H {X} = 24 · 103 · 12 = 288 kbps
.
Which would be a good quality transmission (at least for the listener), since in this
case .R < C?
Example 3.5 For a voice band communications channel, with a signal-to-noise
ratio of 30 dB and a bandwidth of .2.7 kHz, the channel capacity is given by
C = 3.32 (2700 Hz) log10 [1 + 30 dB]
.
C = ±26.9 kbps
Therefore, a communications channel of this type will be able to transmit a
maximum of .26.9 kbps
Exercise 3.1 Using Eq. (3.11), determine the bandwidth necessary to transmit data
at 10 Mb/s over a channel with a signal-to-noise ratio of 28 dB.
3.2.2 Shannon Theoretical Limit
Because the BER increases according to the number of bits per symbol, a limit is
chosen for which the BER can be small independent of the number of bits. A given
value of the signal-to-noise ratio .Eb /No is sought such that
.
Eb
> 2 loge 2 = 1.39(1.42 dB)
N0
(3.12)
Using the limit .k → ∞ where k represents the number of bits per symbol, the
minimum value is for .−3 dB of the Shannon limit
.
Eb
> 1n2 = 0, 693(−1, 58 dB)
N0
This is the limit value for which there can be a small probability of error.
The Levinson–Durbin algorithm is an error detection mechanism that calculates
the predictor coefficients. Another error detection mechanism is through the Shannon model, which is a source encoder with a fidelity criterion (Perlmutter et al.
1998). The source to be encoded, .{x[n]; n ∈ Z}, is considered a random process,
where .Z are the integers. .x[n] takes values in the k-dimensional Euclidean space
with marginal distribution .pX . This distribution can be parametric, but in practice it
is considered an empirical distribution .pL estimated from a training or learning set
.L = {xl ; l = 1, . . . , |L|} as
68
3 Information Theory
PL (G) =
.
1 Σ
1(x ∈ G)
|L|
x∈L
for any event G, where .1 (x ∈ G) is the indicator function.
{
1 (x ∈ G) =
.
1 x∈G
0 otherwise
(3.13)
The dimension k is the parameter of the particular application. In information
theory, Shannon indicates that performance can be improved by using a vector of
dimensions as large as the expense of added complexity in terms of computational
memory.
An encoder for a source .x[n] consists of a pair .(α, β) of encoder and decoder. An
encoder .α: .A → {0, 1} is a mapping of the input alphabet A (typically a subset of
k
.R ) into a set of all binary sequences of finite length. The space range .W ≡ α(A),
which refers to the alphabet, is the set of sequences required for the representation of
the model. To ensure that the sequence of symbols (variable-length binary vectors)
carried from the channel codebook can be decoded only if the starting point is
known, it is required that W be freely prefixed or satisfies the prefixing condition;
thus, not finding a sequence in the alphabet is evidence of an error, the root of which
is the prefix of any other word in the alphabet.
The decoder .β : {0, 1}∗ → L is a mapping of the space of binary sequences
of finite length into a set .L = {β(w); w ∈ W } called codebook reproduction, with
members called codewords from the codebook or templates. The members of L are
- in which typically, but not always, is simply the
chosen from a playback alphabet .A
input alphabet, A. For a given encoder care must be taken about the definition of
.β(w) only for .w ∈ W . This can be arbitrarily defined outside of this set.
Any data compressor can be decomposed into two independent parts: (1) a model
of the source to be compressed used as a predictor and (2) an encoder. The function
of the predictor is to determine the correct probability for each source symbol .ai
of the source alphabet base r, and the encoder assigns a code of variable length
to each symbol based on said probability, thus obtaining the alphabet base code s.
The probability of the symbol or code to be compressed is obtained by calculating
the probability of the source symbol .s = − log2 P (r), generated by the prediction
model. For encoding to be possible, and therefore compression, it must be true that:
.0 < p(r) < 1.
If .P (r) = 0, the source symbol would be encoded with a codeword of infinite
length. Also, r would not belong to the source alphabet. If .P (r) = 1, the generated
codeword would have a length of 0 bits and the decoder would be unable to detect
it. On the other hand, the font alphabet would only contain one font symbol. For
a lossless or reversible compression system, the decompressor would be made up
of the same elements of the compressor acting in reverse: (1) a decoder and (2) a
composition stage (Gonzalez and Wintz 1980). The model proposed by Shannon for
3.3 Problems
69
lossless data compression and decompression is presented, observing that both the
receiver and the transmitter can be divided into two fundamental stages.
The numerator refers to the amount of information. The denominator refers to
the binary alphabet .{1.0}. When .p /= 0.5, .a information rate of the BSS channel
drops. Now, when .p = 0.5, .H {X} = 0.47 bit per symbol. It means that, in average,
each binary symbol (1 or 0) of the source is providing .C = 0.47 bits of information.
For a BSS, the entropy is maximal when both events become equiprobable. This
property is generalized for a source whose alphabet has M equiprobable symbols:
0 ≤ H {X} ≤ log2 M
.
Example 3.6 Suppose you have a binary alphabet. Of which they have two symbols. If a message is sent, which is represented by a succession of K independent
and equiprobable symbols, each of them will have a probability of occurrence of .0.5
and .I {X = i} = log2 ; 2 = 1 bit of information. Then the source information is
I {X} =
M
Σ
.
I {X = i} = Kbits
(3.14)
i=1
Now suppose an sth source represented in M binary symbols. If we assume that
these sources are with same probability, it has to be .I {Xs } = log2 M of information.
Hence, there is a fundamental identity in compute systems and processors, which is
I {Xs } = K = log2 M
.
Example 3.7 A source can be represented by 13 bits, so the alphabet can have
M = 2k = 213 = 8192 possible symbols.
.
3.3 Problems
Problem 3.1 A digital source emits levels of −1.0, 0.0, and 1.0 V, with a
probability of 0.2 each, and levels of 3.0, 4.0, and 5.0 V with a probability of
0.3 each. Determine the entropy of the source and the minimum time required to
transmit one of the levels over a channel whose bandwidth is 4.4 kHz and a signalto-noise ratio of 20 dB.
Problem 3.2 Find the amount of information required to transmit the values of a
thermometer with range [20, 80] ◦ C and a precision of: (i) 0.5◦ , (ii) 1◦ , and (iii)
2 ◦ C. Determine the time required to transmit the information from each of the
thermometers over a telephone channel with a bandwidth of 3 kHz and a signalto-noise ratio of 30 dB.
70
3 Information Theory
Problem 3.3 There is the following alphabet for the letters of the English language,
with its corresponding probability:
Xi
E
T
A
O
R
N
F
Pi
0.14878570
0.09354149
0.08833733
0.07245769
0.06872164
0.06498532
0.024552972
Xi
I
S
D
L
U
P
V
Pi
0.05644515
0.05537763
0.04376834
0.04123298
0.02762209
0.0257539
0.01160928
Xi
M
C
W
G
Y
B
–
Pi
0.02361889
0.02081665
0.01868161
0.01521216
0.01521216
0.01267680
–
Xi
K
X
J
Q
Z
H
–
Pi
0.00867360
0.00146784
0.00080064
0.00080064
0.00053376
0.05831331
–
a. Find the amount of information in each letter and the entropy of the complete
alphabet.
b. Determine whether, by sending the word MAD every 10 more, it can be
transmitted over a channel with bandwidth equal to 200 Hz and a signal-to-noise
ratio of 5 dB.
c. Repeat the previous item, if the words MAD, END, and WAR are sent alternately,
each one every 10 ms.
Problem 3.4 For a ternary system, add noise to the received signal and determine
the signal-to-noise ratio (SNR) at the receiver (sensitivity) and its channel capacity
C, using two methods:
a. Through a simulation program, generating the noise through a Gaussian random
variable with mean 0 and variance varying between 0.1 and 0.9, in intervals of
0.1. Fill Table 3.1 with these values.
b. By setting up a communication system, using the self-induced coaxial cable as
a transmission medium. The noise to be measured is that generated in a defined
bandwidth, above the empty channel noise floor. For both cases, the calculation
of the channel capacity will use the bandwidth B of the transmission medium in
the assembly as bandwidth B.
Table 3.1 Measured and simulated data
Frequency
ωc
2ωc
4ωc
8ωc
σN
[0.1, . . . 0.9]
[0.1, . . . 0.9]
[0.1, . . . 0.9]
[0.1, . . . 0.9]
Sensitivity
Measured
Simulated
C
Measured
Simulated
3.3 Problems
71
Problem 3.5 An alphabet consists of 48 symbols, of which 15 occur with probability 15/48 and the remaining have probability 33/48. Determine the average
required transmission time of each symbol, passing through a channel of width band
B = 5.3 MHz and signal-to-noise ratio SNR = 6 dB.
Problem 3.6 An alphabet consists of 271 symbols, of which 15 occur with
probability 1/16 and the remaining 256 each with probability 1/4096. Determine
the minimum time required to transmit a word over a line whose bandwidth is
5.3 kHz and an SNR of 10 dB.
Problem 3.7 A connection is designed to obtain a maximum transmission rate of
R = 4800 bps. However, the available bandwidth is 3 kHz for a signal-to-noise ratio
of 30 dB. Calculate the available channel capacity and the new signal-to-noise ratio
value needed to obtain the rate R.
Problem 3.8 Draw the curve of the channel capacity C as a function of the
bandwidth B for a constant signal-to-noise ratio of a channel affected by a white
Gaussian noise with spectral density N0 /2. Based on the curve, determine the
channel capacity for bandwidth B = 0.1 MHz and signal-to-noise ratio SN R =
1 dB.
Problem 3.9 Determine the minimum time required (seconds) to transmit a digit
over a line whose bandwidth is 6.4 kHz with a signal-to-noise ratio of 20 dB.
Problem 3.10 Calculate the pulse width (pulse transmission time) required to pass
through a line whose bandwidth is 4.3 kHz under a signal-to-noise ratio SN R of
20 dB.
Problem 3.11 For the following cases, determine the amount of information
required to represent x in the quantizer:
a. A signal x(t) = cos (3π t), quantized into 16 equally spaced levels at a sampling
rate of 100 Hz
b. A signal of bandwidth 32 Hz, and quantized into 24 equally spaced levels in a
unipolar range of Ac = 1 V
c. A signal with a bandwidth of 32 · 103 Hz, quantized into 24 equally spaced levels
in a unipolar range of Ac = 1 V
Problem 3.12 Some source of information sends in 12-bit PCM, the next binary
sequence: {fk } = 101101000001111000111101.
With the following parameters: Ac = 4096, fs = 1200 Hz, determine:
a. The entropy of the binary source
b. If it is possible for the sequences transmitted by the source on a channel of
capacity C = XX kbps, without losing information
Problem 3.13 Using Eq. (3.10), calculate the reception power of the signals, and
assign them as PRx [ACarrier] and PRx [DCarrier]. The random values of the
attenuation models are:
72
3 Information Theory
a. Lf [dB] Gaussian with mean 80 dB and variance 1.
b. Ls [dB] Gaussian with mean 0 dB and variance 1 (Normal).
c. Lp [dB] Weibull with mean 20 log(d) dB and variance β 2 , where β is the Weibull
scale parameter.
References
Abramson N (1963) Information Theory and Coding. McGraw-Hill, New York
Ash RB (1965) Information Theory. Interscience Publishers, Geneva
Gallager RG (1968) Information Theory and Reliable Communication. Wiley, Hoboken
Gonzalez RC, Wintz P (1980) Digital Image Processing. Addison-Wesley, Boston
Perlmutter SM, Cosman PC, Tseng Cw, Olshen RA, Gray RM, Li KC, Bergin CJ (1998) Medical
image compression and vector quantization. Statist. Sci. 13(1):30–53
Chapter 4
Performance of Digital Modulation
Techniques
4.1 Spectral Analysis of the Baseband Signal
Errors are interpreted as the difference between the received baseband signal and the
transmitted baseband signal. In communication systems it is not possible to perform
this operation because only the received signal is available. However, the effects that
the error produces on the received signal can be seen. This is possible through signal
representation models, which serve as a tool for error analysis in communication
systems. When obtaining an Alphabet of the possible coefficients that represent
the signal, the dissimilar symbols will indicate the presence of the error. Suppose
that the message .X[k] is composed of an alphabet of k elements, and due to the
noise effect, the order of its elements is inverted; thus, there is a probability that the
inversion will result in .k ' < k elements that do not belong to the alphabet:
• The analysis of signal attenuation due to losses in transmission lines is difficult
to analyze in the time domain, i.e., transient analysis.
• Some properties of transmission media are frequency dependent, such as skin
effect, dielectric losses, dispersion, resonance, complex permittivity, among
others.
For above commented reasons, frequency domain analysis allows the characterization of a linear network at discrete frequencies. Such type or characterization
lessens the computational calculations. Moreover, frequency analysis is beneficial
for following reasons: (1) Ease and accuracy of measurement at high frequencies;
(2) Simplified mathematics; (3) Allows separation of electrical phenomena (loss,
resonance, among others).
Example 4.1 An audio signal .s(t) has a limited bandwidth B. When it is modulated, the spectra of the generated signal .g(t) are, by Fourier properties,
G(ω) = S(ω + B) + S(ω − B)
.
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024
J. C. García-Álvarez, Digital Electronic Communications,
https://doi.org/10.1007/978-3-031-53118-7_4
73
74
4 Performance of Digital Modulation Techniques
+
Filter 1
S
wc
Y
Filter 2 R
H
S'
N
Fig. 4.1 Communication system with channel restricted by noise
So the modulated signal has a bandwidth of 2B, divided into lower sideband .S(ω −
B) and higher sideband .S(ω + B). This duplication of the bandwidth is important in
signal reconstruction but can be deprecated in the transmission stage. For instance,
in single sideband (SSB) analog modulation, just the upper or the lower sideband is
transmitted (Haykin 2001).
Basically, modulation consists of varying a parameter of the carrier wave
according to variations in the modulating signal, which is the information we want
to transmit, and is manifested in sounds and images. If the amplitude of the carrier
is varied as a function of the modulating signal, an amplitude modulation (AM) is
being performed, and if the frequency of the carrier is varied as a function of the
modulating signal, a frequency modulation (FM) is being performed. Figure 4.1
illustrates a communication system, which aims to establish a link between two
points, with a channel affected by noise, for the adequate transmission of a
signal. It is required that the signal can be transmitted over a channel of B in
bandwidth. Modulation and multiplexing are electronic techniques for transmitting
information efficiently from one place to another:
• Modulation makes the information signal more compatible with the medium.
• Multiplexing allows more than one signal to be transmitted concurrently over a
single medium.
4.2 Modulation Schemes
A modulation scheme is the way in which modulation is carried out, according to the
characteristics of the modulating signal and the carrier signal. Thus, the modulated
signal can be classified as follows:
4.2.1 By Signal Type
Continuous Wave Modulation (CW). It uses a continuous signal as a carrier, such
as a sinusoid. It is useful for sending analog messages, such as the signal coming
from a transducer. Examples include AM--FC, AM--SSB, AM--VSB, FM, and
PM.
Pulsed Wave Modulation (PW). Uses a digital signal as a carrier, such as a train
of periodic pulses. It is also called sampling modulation as it involves a digital
4.3 The Principle of Modulation
75
transformation over time through which the baseband signal is changed into a
symbolic language. Examples include: PAM, PPM, PWM.
Line Modulation When the carrier is digital, it is called coded modulation, quantization, or digital line modulation. This scheme requires a digital transformation
through which the baseband signal is changed into a symbolic language of levels
or edges. Examples include: DPCM, Delta, NRZ, AMI, Manchester.
Digital Modulation. When the carrier is analog, it is called digital passband
(digital passband modulation). It involves an approach by which the baseband
signal is modulated by analog symbols. Examples of these are ASK--OOK,
FSK--SS, PSK, 16--QPSK, 16--QAM, OFDM.
4.2.2 By Variation of the Carrier
Amplitude. One of the simplest methods of signal transmission is amplitude
modulation. This method consists of varying the amplitude of the carrier signal,
through the amplitude of the modulator. This process implies that there are
several periods of the carrier signal in each period of the modulator.
Frequency.
Phase.
Code.
Example 4.2 In an amplitude modulator with full carrier (AM-DSB-FC), the
carrier fc (t) = cos(ωc t) is modified in amplitude by a low frequency signal, called
modulator, which is the signal that contains the information to be transmitted.
4.3 The Principle of Modulation
In the electrical sense, they are variations of voltage (V) and current (I) with respect
to time. In the electromagnetic sense, they are variations of the electric field (.E) and
magnetic field (.H) with respect to time. In the optical sense, they are variations in
the mode of light propagation (polarization and wavelength). The resulting signal
from modulation is .g(t), representing the passband signal (modulated):
⎛
g (t) =
.
A
''''
Amplitude
⎞
sin ⎝ ''''
ω
F requency
t + ''''
θ ⎠
P hase
where .A ⇒ amplitude variation, .ω ⇒ frequency variation, and .θ ⇒ phase wave
variation. In that way, the information given by the baseband signal is carried by
the variation of parameters in the carrier wave.
76
4 Performance of Digital Modulation Techniques
Let there be two processes .xQ (t) and .xI (t) with mean zero, multiplied by a carrier
signal with constant frequency .ωc . The process is then formed:
x(t) = xQ (t) cos ωc t − xQ (t) sin ωc t
.
= a(t) cos (ωc t + θ(t))
This process is called amplitude modulated with respect to .a(t) and phase modulated with respect to .θ (t). Calculating the autocorrelation function of the modulated
signal, we have
Rxx (τ ) = E {x(t + τ )x(t)}
.
= RxI xI (τ ) cos ω0 τ + RxQ xQ (τ ) sin ω0 τ
which means that the modulation preserves the correlation characteristics of the
baseband signals, transporting them appropriately through the carrier frequency .ω0 .
4.4 Modulation Parameters
Thus, for a message to be sent properly, it must conform to the parameters of the
channel through which it is transmitted. These parameters are presented below:
• Bandwidth is the difference between the upper and lower frequencies where the
channel allows the transmission of a message. For the message, its bandwidth is
the difference between the upper and lower frequencies that make up the signal
that represents the message. Normally the message bandwidth is reduced to the
higher frequency, called the fundamental frequency of the message. Thus, for
effective transmission, the bandwidth of the channel must exceed the value of the
fundamental frequency of the message. Otherwise, the message will be distorted.
• The frequency response within the channel bandwidth means that for certain
frequencies within the bandwidth, the signal may or may not be transmitted
properly. This frequency response is analogous to the frequency response of a
filter. This means that there are frequencies where the message can be attenuated,
called cutoff frequencies. Normally the cutoff frequencies are found when the
channel gain is 3 dB below the maximum channel gain.
• Noise is modeled as a random signal that is added to the message in the channel.
Thus, if the power of the noise is much greater than that of the message, the
message will be corrupted and cannot be easily reconstructed. To counteract
noise, modulation is used, which allows the signal to be reconstructed despite
the noise presented in the channel.
Digital information can be represented in several ways that differ fundamentally
in the levels and shape of the waves they use. Codification is the choice of a particular form of representation. It must allow the system’s objectives to be achieved
4.5 Line Encoding
77
in the most economical way. Technical feasibility, communications reliability,
and equipment complexity must be considered. To choose a code, the following
technical factors must be taken into account: The significant DC components are
due, since the terminals and repeaters must be coupled by means of transformers.
Low frequencies must be avoided since cable compensation requires high values
that cannot be easily accommodated by the equipment.
By obtaining the maximum spectral efficiency as a percentage of the data rate,
the majority of the transmitted signal’s frequency components can be recovered.
Nevertheless, some of these signal components are cut off by the transmission
media’s bandwidth restrictions, and the signal can be affected by other transmitted
signals due to electromagnetic susceptibility (crosstalk). Therefore, by reducing the
sequence data length to the minimum, it is possible to increase the signal-to-noise
ratio. It also prevents oscillation dampening in the timing recovery circuits.
4.5 Line Encoding
The characteristics of the digital signal determine the efficiency of data transmission. To do this, it is not enough to send the train of pulses corresponding to
the quantized information signal. Such characteristics are the effective bandwidth
over the channel and the level of synchronism and the bit error rate (BER), which
determine the degree of efficiency of information transmitted through a physical
link, independent of the parameters that the transmission medium provided.
The theoretical basis for implementing baseband coding algorithms is presented
below, in which it is intended to show the differences between coding methods,
in addition to familiarizing the student with the applications that these coding
techniques can have.
When a sequence of information is generated, which represents a particular
signal or symbol, the aim is that this sequence can be sent through the channel
without any distortion. This is why there are techniques that allow this sequence
to be converted into electrical levels that can pass through a specific medium.
Despite this, it is not enough to send electrical representations of the sequence to
have effective communication. It is necessary that it can be transmitted with the
lowest possible probability of error and free of the influence of noise on the channel.
Regardless of the protocol used, the transmission speed of an application (throughput) always depends on whether the data being sent reaches its final destination.
Transmission failure, which appears as attenuated or corrupted signal levels, or as
erroneous detection, results in data retransmission. These repeated retransmission
attempts are what increase network traffic and seriously decrease system efficiency.
The theoretical basis for implementing baseband coding algorithms is presented
below, in which it is intended to show the differences between coding methods,
in addition to familiarizing the student with the applications that these coding
techniques can have. The physical level in a computer network and, in general, for
any digital transmission system, is the fundamental basis for the establishment of
78
4 Performance of Digital Modulation Techniques
reliable information communication. To do this, it is not enough to send the train of
pulses corresponding to the quantized signal of the information. Parameters such as
the effective bandwidth over the channel and the bit error rate (BER) determine
how efficiently information can be sent through a physical link, independent of
the parameters that the transmission medium provides. Among other desirable
properties, a line code must have the following:
• Adequate Timing Content: It must be possible to extract timing or clock
information from the signal.
• Efficiency: For a given bandwidth and data transmission power, the code must
have minimum probability of detection error, that is, maximum immunity to
channel noise and intersymbol interference (ISI).
• Ability to Detect and Correct Errors: It must be possible to detect, and preferably
correct, the error in the detection. In the bipolar case, for example, a single error
will cause bipolar violation and can be easily detected.
• Favorable power spectral density: The signal spectrum must match the frequency
response of the channel. For example, if a channel has high attenuation at the
lowest frequencies, the signal spectrum must have a small PDS within this range
to avoid excessive signal distortion. It is also desirable to have zero PDS when
.ω = 0 (DC), since high frequency coupling is used in repeaters. Significant
power in the low frequency components causes the dc to wander within the pulse
current when high frequency coupling is used.
• Transparency: It must be possible to correctly transmit a digital signal independent of the pattern of 1’s and 0’s. We see that a long succession of zeros could
cause errors in the timing extraction. If the data is encoded so that every possible
sequence of decoded data is received faithfully, the code will be transparent.
Let the bit duration .Tb and the pulse amplitude A be conditional on normalizing
the average power of the line code to unity. Assuming that the data stream is
generated randomly, and the binary symbols 0 and 1 are similar and equiprobable,
in general, a line code can be represented as
N
Σ
s(t) =
.
an g(t − nTb )
n=−N
With .G(ω) the Fourier transform of the pulse .g(t), we may then define the Fourier
transform of .s(t) as
S(ω) =
N
Σ
.
an G(ω) exp(−j ωnTb ).
(4.1)
n=−N
= G(ω)
N
Σ
n=−N
an exp(−j ωnTb )
(4.2)
4.5 Line Encoding
79
The power spectral density of .s(t) is
| N
|2 ⎤
| Σ
|
1
|
|
|G(ω)|2 E |
an exp(−j ωnTb )| ⎦ .
.Ss (ω) = lim ⎣
|
|
T →∞ T
⎡
(4.3)
n=−N
[
= |G(ω)|
2
lim
T →∞
N
1 Σ
T
N
Σ
]
E [an am ] exp(j (m − n)ωTb )
(4.4)
n=−N m=−N
where T is the duration of the binary data sequence, and .E denotes the statistical
expectation operator. Define the autocorrelation of the binary data sequence as
R(k) = E [an an+k ]
.
By letting .m = n + k and .T = (2N + 1)Tb , we may write
[
Ss (ω) = |G(ω)|
.
2
lim
N →∞
n=N
Σ
1
(2N + 1)Tb
k=N
Σ−n
]
R(k) exp(j kωTb )
n=−N k=−N −n
Replacing the outer sum over the index n by .2N + 1, we get
]
[
k=N −n
|G(ω)|2
2N + 1 Σ
lim
R(k) exp(j kωTb ) .
.Ss (ω) =
Tb N →∞ 2N + 1
|G(ω)|2
=
Tb
[ k=∞
Σ
k=−N −n
(4.5)
]
R(k) exp(j kωTb )
(4.6)
k=−∞
Also
L
Σ
(an an+k )pl
R(k) =
.
l=1
where .pi is the probability of getting the product.(an an+k )l , and there are L possible
values for the .an an+k product. .G(ω) is the spectrum of the pulse-shaping signal for
representing a digital symbol. The bandwidth efficiency (bps/Hz) for the encoded
signal is given by
η=
.
Tb
T
where .Tb is the bit rate and T is the minimum period of the digital pulse train.
(4.7)
80
4 Performance of Digital Modulation Techniques
4.5.1 Speed vs. Bandwidth
In the methods described in this section, you will find that choosing the modulation
technique places a limit on the data rate. This limit is imposed by the number of
levels in each symbol duration interval. For a binary system, there are two levels
for the symbol duration interval, and the maximum data rate is equal to the symbol
transmission rate. It is also noted that the addition of frames and synchronization
information degrades the data rate, as in general. For a binary system, the data rate
drops below the symbol lifetime rate. These limitations exist by virtue of practice,
not principle; by introducing more complex circuitry and signal processing, the data
rate can be increased. The sequence of bits to be transmitted is represented by a train
of square pulses whose duration .τ is given by the duration of the bit. The frequency
of this pulse train is called the bit rate .Rb , given in bits per second (bit/s).
Rb =
.
1
ωb
=
Tb
2π
where .Tb is the duration of a bit and .ωb is the transmission frequency (Hz). The
different methods for sending the pulse train have characteristics with respect to
the bit rate. Consider the two types of modulation in Fig. 5.38, in which a change
in bit duration occurs: In one a bit is transmitted for every 2 level changes, while
the second is transmitted for each level change, for which for the first system is
transmitted at 2 bps for each Hz, while the second does so at 1 bps for each Hz. For
.S1 , we have .T = 2Tb = 2 bps Hz, and for .S2 , we have: .T = Tb = 1 bps./Hz. In the
same way, we observe the spectra in Fig. 4.2, where
∞
Σ
PT (t) =
P nej nωb t
.
n=−∞
where
Pn =
.
1
T
f T /2
−T /2
PT (t)e−j nωb t dt =
1
T
f τ/2
−τ/2
e−j nωb t dt =
( nω τ )
τ
b
sinc
T
2
n /= 0
We have for L’hopital:
.
(
)
( nω τ )
τ ω2b τ cos nω2b τ
τ sen 2b
τ
=
= lim
.P0 = lim
ωb τ
nωb τ
n→0
n→0 T
T
2
2
Then the different methods for sending the pulse train have a bandwidth
characteristic given by
BBB = n|Pn =0
.
4.5 Line Encoding
81
Fig. 4.2 A modulation scheme with different bit rate (1/Tb). Left: signal in time domain. Right:
signal frequency spectrum
The value of n where .Pn = 0 is given by
.
( nω τ )
b
Sinc
2
=0⇒
nωb τ
= mπ
2
The minimum point is .m = 1. At this value, we have the first cutoff point
of the signal’s spectral density. The frequency interval from .ω = 0 to the cutoff
point is called the null bandwidth .B0 , where the greatest amount of signal power is
concentrated.
BBB = n|Pn =0 =
.
2π
ωb τ
At maximum bit toggle .ωb = 2π
Tb = 2π Rb , where .τ is the pulse width. Taking
T
.τ =
(the
pulse
in
bits
is
at
least
half the period)
2
2π
Tb
.BBB =
=2
2π T
T
T 2
b
[
1
bit
]
If a sequence is sent at .Rb bps, the required bandwidth is: .2Rb TTb (the higher the
bit rate, the greater the bandwidth). In this way, the null bandwidth can be defined
as the frequency interval in which the greatest amount of energy in the spectrum is
82
4 Performance of Digital Modulation Techniques
concentrated. For Fig. 4.2, there would be a zero bandwidth value for .S1 of .BBB =
1 Hz bit and for .S2 of .BBB = 2 .Hz/bit
Tb está dado en (bps)−1 , T está dado en Hz−1
.
In the laboratory environment, digital information is any type of signal that
represents both defined states of a system; that is, current equal to a closed switch,
the absence of current equal to an open switch. Thus for a base system B with
n digits or word size, it is possible to represent Z combinations or a number of
different values, like this:
Z = BK
.
K=
.
ln(Z)
ln(B)
In a base two or binary system, several bits are necessary to represent complex
information.
K = log2 (Z)
.
Digital data can be transmitted using different transmission or line codes, such as
nonreturn to zero code (NRZ), return to zero code (RZ), Manchester or biphase
modulation, alternate mark modulation (AMI), brand investment (MIC), and others.
An example for the sequence .{1 0 0 1 1 0 1 0 1 0}.
Example 4.3 We wish to transmit with a modem at 1200 bit per second through a
telephone channel with a bandwidth between 300 and 3400 Hz:
• Find the bandwidth of the generated signal. In the case of a binary alternate
signal, .T = 2τ . Using Eq. (2.5) the bandwidth becomes .B = 4π/τ . Given the
bit length .T = 1/1200 s, .B = 8π/T = 9600π Hz.
• Find the bit rate required to fit the telephone channel. Using .B ' = 3400 − 300 =
1100 Hz, the bit length is adjusted into .T ' = 8π/B ' ≈ π/137.5; thus the bit rate
is approximately 137 bit per second.
4.6 Line Edge Transition Modulation
Figure 4.4 presents the advantages and disadvantages of each type of modulation.
4.6.1 Reversible Polarity Code (RB)
Three levels are used (0, 1, and a polarization level). This bias level can be chosen
below or between the other two levels (it can be a zero reference). The wave returns
to the bias level during half of each bit interval.
4.6 Line Edge Transition Modulation
83
4.6.2 AMI (Altern Mark Inversion)
In order to avoid the disadvantages of the P CM signal, a pseudo-ternary alphabet
is used, as illustrated in Fig. 4.3a and b where three signal values are distinguished.
Logical zeros are transmitted as zero volts and binary ones as positive or negative
pulses in alternating form of RZ symbols (Lathi 1998; Stremler 1989; Haykin 2001).
The first binary one is represented by .+A, the second by .−A, the third by .+A, and
Fig. 4.3 (a) AMINRZ. (b) AMIRZ. (c) BRZ. (d) PNRZ. (e) UNRZ. (f) URZ
84
4 Performance of Digital Modulation Techniques
so on. A predecessor of this technique, named Mark Inversion Code (MI), uses two
levels .+D and .−D, and its coding principle is described as follows: A binary zero
is encoded by the amplitude .−D and .+D, each with a time equal to half the unit
interval or bit time equal to .(T b/2). Binary ones are encoded as AMI .100%, that is,
alternating levels D with a duration equal to the unit interval or T b. AMI is derived
from RZ and vice versa by inverting the ones alternately. It is also known as bipolar
RZ (BRZ). .{1} is represented by alternations of .s1 (t), and .s2 (t) · {0} is represented
by .s3 (t). In general, the characteristics of AMI Modulation are: Alphabet: Bipolar.
.U = {−A, 0, A} Representative orthogonal functions:
f Tb
A
2
dt = A,
Tb
−A
2
dt = −A,
Tb
2
C11 =
.
0
f Tb
2
C21 =
.
0
C21 = 0
C22 = 0
C31 = C32 = 0
Symbols: .X1 = (A, 0), X2 = (−A, 0), X3 = (0, 0) Probability set: .P1 , P2 , P3 The
projection process yields the following symbols:
⎡
⎤ ⎡
⎤
]
s1 (t)
A 0 [
φ̂1 (t)
. ⎣ s2 (t) ⎦ = ⎣ −A 0 ⎦
φ̂2 (t)
0 0
s3 (t)
(4.8)
The justification for determining the AMI sequence as pseudo-ternary is that
alternating pulses are considered probable events. Accordingly, the entropy of the
source is not .log2 3 = 1.58 bits, as in a complete ternary signal, but .log2 2 = 1 bit.
Likewise, only the information quantities .I {X1 } , I {X2 } are evaluated (Fig. 4.4).
Fig. 4.4 Baseband signals modulated by phase alternation
4.7 Phase Alternation Techniques
85
Example 4.4 Determine the DC value and average power of the AMI symbol
sequence.
E{X} =
M
Σ
.
Xm Pm = (A, 0)P1 + (−A, 0)P2 + (0, 0)P3 = (A (P1 − P2 ) , 0)
m=1
M
{ } Σ
2
E X2 =
Xm
Pm = |(A, 0)|2 P1 + |(−A, 0)|2 P2 = A2 (P1 + P2 )
m=1
SNR =
A2 (P1 + P2 )
N0
Replacing on .PE :
(/
PE = Erfc
.
SNR
2
)
4.7 Phase Alternation Techniques
The bipolar code uses additional levels of modulation to achieve certain specifications or characteristics such as timing transitions, the absence of an erroneous
DC value, and ease of monitoring the signal so that errors can be detected. Many
varieties of baseband modulation techniques achieve an appreciable amount of level
variations that favor synchronism and eliminate the erroneous DC component, with
a corresponding increase in the bandwidth of the transmitted signal. The different
techniques are illustrated in Fig. 4.5.
Fig. 4.5 (a) Phase diagram for AMI modulation. (b) Phase diagram for baseband phase-based
modulation
86
4 Performance of Digital Modulation Techniques
4.7.1 Manchester or Dual-Phased
One of the most used techniques, as it provides high pulse density, giving an efficient
synchronization signal, and has an expected value equal to zero, is the biphase
digital code, also called Manchester Modulator. It is a bipolar modulator, composed
of two representative signals .s1 (t) and .s2 (t). Its main application is in local LAN
networks (Tadokoro et al. 2013; Fensel et al. 2013; García-Álvarez 2007)
{
s (t) =
. 1
−A T2b ≤ t ≤ Tb
+A 0 ≤ t ≤ T2b
A two-phase digital modulator uses one cycle of a square wave with a given
phase to encode a one and one cycle of an opposite phase to encode a zero. In this
split-phase binary representation, the variation in the mean value is eliminated by
symmetry. In this method the .{1} is represented by the A level for the first half of
the bit interval and then shifted to .−A for the remaining half: The .{0} indicates
inversely, as shown in Fig. 4.6. The two-phase modulator is used primarily in short
distance links where the cost of terminal equipment is more significant than the
use of bandwidth. One of Manchester’s applications is the one for Ethernet-LAN
networks developed by Xerox Digital and Intel Corp.
The Manchester or biphase code has two levels .+V and .−V , encoding the signals
.f1 (t) and the marks for .f2 (t) as follows, for the NRZ(L) code (Fig. 4.7a). Its main
application is in local area networks (LANs).
−V
+V
f1 (t) =
.
si
si
t > T2b
0 ≤ t ≥ T2b
f2 (t) = −f1 (t)
.
Fig. 4.6 Signals generated by different phase alternation techniques
4.7 Phase Alternation Techniques
87
Fig. 4.7 (a) Signal generated using Manchester technique. (b) Signal generated using Miller
technique
Using Eq., the spectra of the Manchester encoded signal are
S(ω) = A Tb sinc
.
2
2 ωTb
4π
(
2
sin
ωTb
4
)
4.7.2 Differential Manchester
It is another split-phase binary representation. It uses the same symbols as in
Manchester except that the phase inversion in relation to the previous symbol is done
with the .{1}. At .{0} there is no phase change (Haykin 2001; Couch II 1997; Stremler
1989). In Manchester, .{1} is represented by the symbol .s1 (t) and .{0} by the symbol
.s2 (t). In CMI, .{1} represents alternations of .s1 (t) and .s2 (t) (phase alternation), and
.{0} represents maintaining .s1 (t) or .s2 (t) (no phase alteration). For both Manchester
and Differential Manchester you have the following characteristics: Alphabet:
Bipolar. Label .= {−A, A}. Representative orthogonal functions:
s (t) = Aφ̂1 (t) + −Aφ̂2 (t),
. 1
s2 (t) = −Aφ̂2 (t) + Aφ̂2 (t)
The generated symbols are: .X1 = (A, −A), X2 = (−A, A), with respective
probabilities: .P1 , P2 = 1 − P1 .
(
)
( )
1
.
Information measure: .I {X = X1 } = log2 P11 , I {X = X2 } = log2 1−P
1
Example 4.5 Determine the DC value and the average power of a sequence of
Manchester Differential symbols.
E{X} =
M
Σ
.
m=1
Xm Pm = (A, −A)P1 + (−A, A)P2 = (A (P1 − P2 ) , A (P2 − P1 ))
88
4 Performance of Digital Modulation Techniques
M
{ } Σ
2
E X2 =
Xm
Pm = |(A, −A)|2 P1 + |(−A, A)|2 P2 = 2A2
m=1
SNR =
2A2
N0
Replacing in .PE :
(/
PE = Erfc
.
SNR
2
)
4.7.3 Delay Modulation or Miller Coding
This technique is a variant of NRZI and Manchester Differential. .{1} produces halfbit transitions, and .{0} produces voltage levels. In this code a 1 is represented by
a signal transition in the center of a bit interval (Fig. 4.7b). A 0 is represented
by no transition, unless it is followed by another zero, in which case the signal
transition occurs at the end of the bit interval. Therefore, alternations (both phase
and level) occur when consecutive values are presented, whether .{1} or .{0} (HenaoRíos et al. 2021; Maniktala 2013; Rajagopal et al. 2000). For example, .{0} would
be represented by null level transition, unless it is preceded by another .{0}, in which
case a level transition occurs. .{1} is represented by the symbols .s1 (t) and .s2 (t) · {0}
by .s3 (t) and .s4 (t). Phase and level alternations occur in consecutive blocks of .{1}
s or .{0} s. For Miller it has the following characteristics: Bipolar alphabet set:
.A = {−A, A}
Representative orthogonal functions:
f Tb
2
C11 =
.
0
2
A dt = A,
Tb
f Tb
2
C21 =
.
−A
0
2
dt = −A,
Tb
f Tb
2
C31 =
f Tb
2
0
−A
Tb
2
C22 =
−A
2
dt = −A
Tb
f Tb
Tb
2
A
2
dt = A
Tb
2
dt = A,
Tb
C32 = A
2
dt = −A,
Tb
C42 = −A
A
0
C41 =
C12 =
f Tb
s1 (t) = Aφ̂1 (t) + −Aφ̂2 (t),
s2 (t) = −Aφ̂1 (t) + Aφ̂2 (t)
4.7 Phase Alternation Techniques
89
s3 (t) = Aφ̂1 (t) + Aφ̂2 (t),
s4 (t) = −Aφ̂1 (t) − Aφ̂2 (t)
Symbol set: .X1 = (A, −A), X2 = (−A, A), X3 = (A, A), X4 = (−A, −A)
Probability set: .P1 , P2 , P3 , P4
Information quantity: .I {X1 } , I {X2 } , I {X3 } , I {X4 }
Example 4.6 Determine the DC value and the average power of the Miller symbol
sequence.
E{X} =
M
Σ
.
Xm Pm = (A, −A)P1 + (−A, A)P2 + (A, A)P3 + (−A, −A)P4
m=1
= (A (P1 + P3 − P2 − P4 ) , A (P2 + P3 − P1 − P4 ))
M
{ } Σ
2
Xm
Pm
E X2 =
m=1
= |(A, −A)| P1 + |(−A, A)| P2 + |(A, A)|2 P3 + |(−A, −A)|2 P4 = 2A2
2
2
SNR =
2A2
N0
Replacing in .PE :
(/
PE = Erfc
.
SNR
2
)
Example 4.7 An example of Manchester modulation corresponding to an information sequence is shown in Fig. 4.8. The voltage level is .A = 5 V. Determine, taking
the sequence itself as probabilities:
1. The amount of information of each symbol
5
4
P1 = = 0, 56, P2 = = 0, 44
9
9
( )
( )
9
9
= 1, 17 bits
= 0, 8480 bits, I {X1 } = log2
I {X1 } = log2
4
5
.
Fig. 4.8 Sample of a signal generated by Miller coding
90
4 Performance of Digital Modulation Techniques
2. The entropy of the modulated signal
H {X} = P1 I {X1 } + P2 I {X2 } = 0, 99 bits/sequence
.
3. The expected value of the modulated signal
E{X} = (A (P1 − P2 ) , A (P2 − P1 )) = (0, 12A, −0, 12A) = (0, 6, −0, 6)V
.
4. The average power of the modulated signal
{ }
E X2 = 2A2 = 50 W
.
4.8 Techniques with Memory
The major limitation of bipolar modulation is its dependence on the density of .{1}
values per sequence to maintain the timing of repeater stations. The different types
of digital signals present drawbacks in their operation when there are sequences of
only zeros (for example, some systems tolerate up to sequences of 14 consecutive
zeros before determining a synchronism correction, even if the system does not
consider actions for sequences of more of 14 consecutive zeros, the low density of
pulses on the channel could cause loss of synchronism in the receiver). Zero-padding
techniques try to compensate for the problem by inserting ones into long strings of
zeros. This results in a better arrangement of power in the channel, increases the
amount of information, and provides greater synchronization in the channel. As will
be seen below, these techniques only apply to transition amplitude modulators (RZ
and AMI) (Henao-Ríos et al. 2021; Couch II 1997).
For these techniques, the following is available. There is a sequence of symbols
.{sm (t)} modulated in AMI, from which it follows that there is alternation of symbols
.s1 (t), s2 (t). Let SV be one of those two symbols that does not comply with the
alternation (if the preceding symbol was .s1 (t), {V} = s1 (t) and viceversa). This
symbol is called alternation violation. On the contrary, let .{R} be one of those two
symbols that fulfills the alternation (if the preceding symbol was .s1 (t), {R} = s2 (t)
and vice versa). This symbol is called alternation fill. This sometimes involves
changing the alternation of the original symbols. The necessary precautions must
be taken to eliminate the symbols inserted in the data reception, and to do so, it
is necessary to differentiate it from normal pulses. Note that, since the violation
symbol implies a parity violation, this pulse is transmitted with identical polarity to
the pulse that precedes it. Therefore, the receiver recognizes that there is a parity
violation, since there are two consecutive impulses of the same polarity.
These two new symbols are inserted into the sequence, according to the
techniques described below.
4.8 Techniques with Memory
91
4.8.1 BNZS Modulation
BNZS modulation performs substitutions that increase the number of ones by
substituting consecutive strings of N zeros by means of a special sequence of
length N, which contains alternation violation pulses. Thus, the pulse intensity is
increased, while the original data is obtained by recognizing the alternation violation
pulses. These violations are recognized by the receiver, which takes the sequence
and proceeds to replace it with a sequence of .N zeros.
This type of modulation guarantees the presence of synchronization signals
continuously without imposing any restriction on the source of data information.
The BNZS code supports variations of any encoding transparently.
4.8.2 B3ZS Modulation
This algorithm is specified by the DS-3 standard. It is also used in Canadian LD-4
coaxial cable systems.
In the B3ZS format, every sequence of three consecutive zeros in the data source
is processed with one of the following sequences: .00 V or .R∩V, where .V represents
a violation of alternation and R represents padding corresponding to the alternation.
In either situation, the violation occurs in the bit that occupies the last position of
the sequence of three consecutive zeros. In this way the position of the replacement
is easily detectable.
The decision to determine which of the two sequences is used is made such
that the number of pulses R is odd. So, if an odd number of values .{1} have been
transmitted, the sequence to replace the three-zero one is .00 V; if an even number
of .{1} values have been transmitted, the sequence to replace the three zeros is R0V.
In this way, all violations contain an odd number of alternating bipolar pulses. The
violations alternate in polarity to prevent the existence of a nonzero expected value,
with its corresponding erroneous detection of a DC signal. An even number of
bipolar pulses between violation pulses occur as a consequence of an error on the
transmission channel. Additionally, each violation pulse is immediately preceded by
a zero. Therefore, the baseband modulated signal consists of a considerable amount
of redundancy, whose systematic structure facilitates its monitoring and verification.
The substitution algorithm is presented in Table 4.1.
Table 4.1 Algorithm for B3ZS coding
.{X}m
Value of previous pulse
.+A
.−A
Odd number of pulses between
sequences .{0 0 0}
or it is the first sequence block
.{0 0 A}
.{0 0 − A}
Even name of pulses between
sequences .{0 0 0}
.{−A 0 − A}
.{A 0 A}
92
4 Performance of Digital Modulation Techniques
Table 4.2 Example 4.8 (A)
.{Xk }
101
000
11
000
000
001
000
1
.{X}m
.A0 − A
.00 − A
.A − A
.A0A
.−A0−
.00A
.00A
.−A
1
A
A
Table 4.3 Example 4.8 (B)
.{Xk }
101
000
11
000
000
001
000
.{X}m
.A0 − A
.A0A
.−AA
.−A0−
.A0A
.00 − A
.00 − A
Table 4.4 Algorithm for B6ZS coding
Value of the previous pulse
.{fT (t)}
.+A
.
.−A
.
{
{
0 A −A 0 −A A
0 −A A 0 A −A
}
}
Example 4.8 Determine the result of the following sequence:
.
{Xk } = . . . 101000110000000010001
When passing through a B3ZS modulator, and with the following considerations,
the above sequence presented an odd number of pulses after a block of only zeros.
Table 4.2 shows the segmentation of the sequence given the detection of sequences
of 3 zeros.
The sequence above presented an even number of pulses after a block of all zeros.
Table 4.3 shows the segmentation of the sequence given the detection of
sequences of 3 zeros.
The previous examples indicate that with the process of replacing sequences of
consecutive zeros with bipolar pulses, the pulse density is considerably increased
per sequence. It is left to the reader to demonstrate that the minimum pulse density
is .0.33 per sequence. In this way, this technique supplies a high-density baseband
modulated signal of continuous alternating pulses.
4.8.3 B6ZS Modulation
This modulation technique replaces sequences of six consecutive zeros and is shown
in the Table 4.4. Note that this technique produces alternating violations in the
second and fifth positions of the sequence that has been replaced (0VR0VR).
4.8 Techniques with Memory
93
Table 4.5 Example 4.9 (A)
.{Xk }
1
000000
1011
000000
000000
0001
.{X}m
.−A
.0 − AA0A − A
.A0 − AA
.0A − A0 − AA
.0A − A0 − AA
.000 − A
Table 4.6 Example 4.9 (B)
.{fk }
.{fT (t)}
1
A
000000
1011
000000
000000
0001
.0A − A0 − AA
.−AOA − A
.0 − AAOA − A
.0 − AA0A − A
.000 − A
Table 4.7 Algorithm for B8ZS coding
Value of previous pulse
.{sm (t)}
.+A
.
.−A
.
{
{
0 0 0 A −A 0 −A A
0 0 0 −A A 0 A −A
}
}
Example 4.9 Determine the result of the following sequence, .{fk } =
. . . 100000010110000000000000001, when passing through a B6ZS modulator.
Table 4.5 shows the data output sequence, given the preceding pulse either
positive or negative. The output sequence is given at detection of sequences of 6
zeros.
4.8.4 B8ZS Modulation
This technique replaces any string of eight zeros with a string that has two code
violations. The violation–fill sequence is .{000VROVR}.
4.8.5 High-Density Bipolar n
This modulator recommended by the ITT -T is called High-Density Bipolar Code
(HDB). In general terms, it is expressed as HDBn, where n indicates the length of
the sequence of consecutive zeros that can be allowed without being replaced. The
HDBn technique eliminates sequences of .n + 1 zeros by inserting symbols given
by the sequences .{000 · · · 0 V} or .{R00 · · · 0 V}, depending on the characteristics of
the previous sequence. .{000 · · · 0 V} is used when there are an odd number of ones
between sequences of .{000 · · · 0} and for the first sequence of .{000 · · · 0} (Tables
4.6 and 4.7).
.{R00 · · · 0V } is used when there are an even number of ones between sequences
of .{000 · · · 0}.
94
4 Performance of Digital Modulation Techniques
4.8.6 HDB3
This type of modulation is used to eliminate sequences of 4 consecutive zeros.
This modulation starts from
{ AMI, }with the difference that the sequence of four
zeros given by .{sm (t)} = 0 0 0 0 is replaced by the violation and fill symbols
(V and .R, respectively). The HDB3 code allows, according to the above, up to 3
consecutive zeros. In the case that the binary signal contains four consecutive zeros,
this sequence was replaced by a string that contains a bipolar violation pulse in the
last bit. The following Table 4.8 shows the algorithm used by the HDB3 code. Note
that the replacement is carried out with the sequences .{000 V} or .{R00 V}.
Example 4.10 Two PCM sequences are illustrated below, modulated in AMI, and
then processed in HDB3. Note then that the HDB3 modulation table is similar to the
B3ZS code, except that the number of zeros to replace is four. The HDB3 technique
is used to eliminate sequences of consecutive zeros greater than three elements. To
clarify the understanding of the table, the following cases apply (Tables 4.9 and
4.10):
Case 1. In the case of sequences of 4 zeros, the last element 0 of the sequence
must be replaced by a pulse A. This value A represents the value .{1}. The
conversion of the four zeros is then .000 V, causing a violation of the alternation.
In addition to this, the pulses V must alternate in polarity.
Table 4.8 Algorithm for the HDB3 coding
.{X}m
Value of the previous pulse
.+A
.−A
Odd number of pulses between
sequences .{0 0 0 0}
or it is the first sequence block
.{0 0 0 A}
.{0 0 0 − A}
Even number of pulses between
sequences .{0 0 0 0}
.{−A 0 0 − A}
.{A 0 0 A}
Table 4.9 Example A
1
.A
.A
0
0
0
1
.−A
.−A
0
0
0
0
0
0
0
0
0
0
0
.V
.(−)
0
0
0
0
0
.V
.(−)
0
0
0
1
.A
.A
0
0
0
0
0
0
0
0
0
0
0
.V
.(A)
0
0
.R
.(A)
0
0
0
0
0
0
1
1
.−A
.A
.−A
.A
0
0
0
PCM
.AMI
.HDB3
Table 4.10 Example B
1
.A
.A
0
0
0
1
.−A
.−A
0
0
0
0
0
0
1
1
.A
.−A
.A
.−A
0
0
.V
.(A)
1
1
.A
.−A
.−A
.A
0
0
0
PCM
.AMI
.HDB3
4.9 Eye Diagram
95
Table 4.11 Process of sequence in Example 4.11
.{X}
101
0000
11
0000
0000
10
.{Sm }
.A0 − A
.000 − A
.A − A
.A00A
.−A00 − A
.A0
Fig. 4.9 Signals generated by NRZ, AMI, and HDB3 modulators. B: valance symbol, V: alternate
violation symbol
Case 2. If there is an even number of signal elements 1 among the signal elements
.V inserted according to rule 1 and the signal element .V presents, it must be
replaced by a signal element .R the first of the four 0’s of the sequence. The
signal .R agrees with the alternation of the original signal. The last of the four
zeros in the sequence is again changed by a signal element V, this is R00V. The
signal elements .R and .V have the same polarity.
To maintain a zero DC component, as many positive and negative violations must
be transmitted alternately, this condition of alternating violations, to maintain a zero
DC component, forces a filler pulse to be placed when the pulse that precedes the
violation does not have the same polarity.
Example 4.11 Determine the result of the sequence .{X} = 1010000110000000010
when passing through an HDB3 modulator, when the previous sequence presented a
positive violation pulse and an odd number of pulses after a block of only zeros. The
Table 4.11 shows the segmentation of the sequence given the detection of sequences
of 4 zeros. Figure 4.9 shows the signal over time, compared to the NRZ and AMI
schemes.
4.9 Eye Diagram
The eye diagram is a method for the performance evaluation of a communications
system using signal distortion when it passes through the transmission medium. This
method allows to know the imperfections of the channel, through the evidence of
alterations in the form of the signal detected in each period, preventing the detected
signal from having a form identical to the transmitted.
96
4 Performance of Digital Modulation Techniques
Fig. 4.10 Eye-diagram for different scenarios. (a) Ideal signal,without noise; (b) Signal with
bandwidth limitations, without noise; (c) Signal with bandwidth limitations and added noise
The result is that the strokes become damaged, as Fig. 4.10 shows, due to
phenomena such as distortions in synchronism, influence of interference and
interference-Símbolos (ISI).
4.9.1 Building of Eye Diagram
The eye diagram is a parametric plot between the transmitted and received signals.
The signal .{xT x (t)} is used as the time base signal. The eye diagram is constructed
from twice the carrier period .2(ωc−1 ), so a vertical deflection is provided in the
signal. As the time base is usually limited to a width of two symbol periods (bit
period if a binary signal has been displayed), the signal is confined to the center of
the peaks (the sampling points), synchronizing the signal to a frequency multiple
of the carrier frequency. The result is that the signal from the source will be
permanently displayed and superimposed on each detection. Therefore, the eye
diagram is generated by superimposing all possible sequences of m symbols, as
shown in Fig. 4.13. The image we get is similar to an eye, and the further apart
the levels are, that is, the more open the eye is, the better performance the system
observes.
In order to observe the chirp frequency dependence of the diagram, the number
of periods must be greater than the memory counter per channel for all possible
ISI. A modulation technique is assumed such that low frequency components
are eliminated. A number .kM T is considered as the signal sweep period, so the
development of each .KM bits is considered, and the central bits are analyzed. A
typical value of .KM is 3, so the center symbol is evaluated.
Example 4.12 Generate the eye diagram of the Manchester signal explained in
Sect. 4.7.1, through the following procedure:
Step 1. Part of the digital carrier signal .xT (t) passes through a band-limited
channel (Fig. 4.10). Note that the duration of the oscilloscope’s detection interval
4.9 Eye Diagram
97
Fig. 4.11 Test for measurement of eye diagram
Fig. 4.12 Setup for reference eye diagram using signals .xT (t) and .s(t)
is 2T , but this interval is in .τ/2, .5τ/2, that is, in half the duration of the signal
period (Fig. 4.11). Thus, the synchronization signal .xT x (t) is divided into N
equal intervals, leaving N signals .xn (t).
Step 2. There are two critical points: the extreme level of the signal .|A| and the
value zero (Fig. 4.12). The values .tk indicate the detection points of the signal
.xT (t).
Step 3. The eye diagram shown in Fig. 4.13 illustrates the ideal condition where
the waveforms corresponding to the superimposed intervals are identical such
that a shift in the trace does not occur.
Step 4. Starting at .k = 0, plot the two periods covered by .k = [0, τ ]. At the point
.k = τ the oscilloscope trigger is reactivated, so that the next pattern .k = τ + 1
begins at the same point of .k = 1. At .k = τ + 2 a detection occurs again and
98
4 Performance of Digital Modulation Techniques
Fig. 4.13 (a) Superposition of orthogonal signals in the phase constellation diagram. (b) General
measurements
Fig. 4.14 Detailed distortion measurements in an eye diagram
the next two periods are plotted. Over many periods .τ many traces coincide, and
this along with a normal persistence of a phosphor tube creates the appearance of
a stationary pattern similar to that shown in Fig. 4.15. Thus, the received signal
n−1τ nτ
.xRx (t) is superimposed on the same interval .[
N , N ].
4.9 Eye Diagram
99
Fig. 4.15 Algebraic nomenclature for some measurements illustrated in Fig. 4.14
4.9.2 Eye-Diagram Parameters
There are then the following evaluation parameters, according to Fig. 4.14:
• The timing error allowed in sampling at the receiver side is given by the width
inside the eye.
• The sensitivity to timing error is given by the slope of the open eye.
• The margin of error or noise immunity of the system is given by the height of the
eye opening.
• The sync detection range (T1) is the time interval between which the signal can
be detected. The time interval separates two successive zero crossings. When the
signal crosses the zero amplitude level, or swings from positive to negative or
vice versa, it is known to as a zero crossing.
• Amplitude margin .Aη . Height (in Volts) from zero to the beginning of the
variation band .A1 .
• Zero crossing distortion .(T2 ). Width of the area where the signals usually cross
zero.
• The slope .(Ahv ) is the sensitivity to timing errors. Downward (or upward) slope
of the signal when it changes level.
• The maximum distortion .(A1 ). Height (in volts) of the amplitude variation band.
• Variation attenuation
( )
Aη
.
.AA = 20 log
A1
100
4 Performance of Digital Modulation Techniques
• There is vertical and horizontal degradation of the eye. This vertical degradation
is defined by
EDV =
.
ED1 + ED0
A+ − A−
where A + and A- are the level values defined for the modulation (A and -A)
in this case, and ED1 and ED0 are the level deviation of the acquired signal
compared to the defined level.
• The complete horizontal opening of the eye is equal to Tb. The horizontal
degradation of the eye is then expressed as
EDH =
.
T − EOH
T
where EOH is the horizontal opening of the eye.
• Vertical power degradation comes from
LH = 20 log EDV
.
Note that the effect of noise Fig. 4.14 and ISI is the same as closing the eye,
reducing the noise margin. The noise margin is the difference between the zero
level and the lowest limit on the peak, where the signal could normally be detected.
The tilt of the waveform around the edge of the eye’s edges indicates the sensitivity
of the system to timing errors. If the sampling errors move the sampled point away
from the center, but the decline is long, the small time error can result in a large level
change, then the sampled value may be in error. Pulse distortion also distorts the zero
crossing point. This reduces the range over which sampling can occur. Additionally,
zero crossing detection is used to generate the receiver synchronization signal,
increasing the phase shift. Power degradation is a measure of the additional power
in dB required to achieve an ideal eye opening. Horizontal eye degradation increases
time jitter, which affects synchronism, increasing the variation of sampling times at
the receiver. These effects can translate as a degradation of the signal integrity, as it
is displayed in Fig. 4.15. However, the exact calculation of system degradation due
to drift depends on the synchronizing circuit used in the receiver.
4.10 Ethernet
Probably the most common form of Medium Access Control (MAC) in LAN
networks is Carrier Sense Multiple Access/Collision Detection (CSMA/CD), which
is based on the use of communications between computers with a shared physical
4.10 Ethernet
101
Table 4.12 Collected data
for collision detection
through a 2.5 km CSMA/CD
bus (10 kbit per frame)
Bit rate .[Mb/s]
10
100
200
.Tt [μs]
.2A
1000
100
50
.0.05
.0.5
.1.0
Bit length
500
5 000
10 000
medium (Garcia-Alvarez et al. 2006). Initially, the case is considered in which there
is a fast transmission of this type such that a higher throughput is achieved. We have
the following variables:
Let the propagation delay be .TZ , and the time it takes for a MAC layer frame to
be transmitted .TL and the time it takes to transmit a MAC layer frame .TF .
TZ =
z
vp
L
R
TZ
. A =
TL
TL =
TC = 2TZ
.
where z is the distance between transmitter and receiver [m], .vp is the propagation velocity .[m/s], L is the number of symbols by frame, R is the transmission rate
of the system [bps], and .TC is the maximal time required to detect a collision .[s].
Computer networks using CSMA/CD work well as long as the propagation delay
.A is small compared to the time required to transmit a frame. This is required in the
sense of detecting collisions during transmission. The maximum time that a node in
a network using CSMA/CD must wait to discover a collision is .TC = 2TZ , and also,
note that .2A is a fraction of the frame that will be transmitted before a collision
is detected. Therefore, if .TZ is much smaller than .TL not much time is wasted if a
collision is detected.
Example 4.13 If a .2.5 km cable is considered and we take a propagation speed
of .vp = 2 × 108 m/s, then .TC = 25 μs. Then, considering collision detection, an
average frame size of 10 000 symbols is taken. Table 4.12 summarizes information
about collision detection as the link bit rate increases.
Table 4.12 shows us that it is possible for an entire frame to have been transmitted
before we can expect to detect any collisions. This is clearly unacceptable. One
solution to this problem could be to tolerate a minimum frame size that increases
as the transmission rate and cable length increase, but this would waste network
capacity. It is for these reasons that high-speed LAN and MAN networks must use
other MAC methods.
102
4 Performance of Digital Modulation Techniques
4.11 Problems
Problem 4.1 Describe the algorithm for generating SDigital and the corresponding
transmission using NRZ, RZ, AMI, Manchester, and Miller modulation.
Problem 4.2 Generate the signals AM-DSB-FC, FM-WB, PWM, and 8-QPSK from
the signals SAudio, SDigital, ACarrier, and DCarrier generated in
Chap. 4.
Problem 4.3 Implement the baseband modulation systems of the theoretical framework using a development system or simulation software. Given the data stream
1110010100, sketch the transmitted sequence of pulses for each of the following
line codes:
• Unipolar nonreturn-to-zero (UNRZ)
• Polar nonreturn-to-zero (PNRZ)
• Manchester
Problem 4.4 Design a platform where different types of coding can be selected; in
addition, the program must allow adding a noise source to the channel.
Problem 4.5 Verify that the output signal corresponds to the encoded input, in each
of the modulation systems. Make the corresponding theoretical coding with the
word that is entered through the port to make the comparison (this is only for an
example that will be delivered in the report).
Problem 4.6 Determine the power spectral density (PDS) of each of the digital
modulation systems in B.B. Measure the respective bandwidth efficiency η on the
spectral density graph.
Problem 4.7 The signal sx (t) = a0 + a1 s(t) + a2 s 2 (t), where s(t) = cos ω1 t +
cos ω2 t, passes through the circuit in Fig. 4.16. By trigonometric identities, the
quadratic term corresponds to a signal at twice the frequency, as illustrated in
Eq. (4.9). Using ω1 /2π = 10 kHz and ω2 /2π = 40 kHz, which passes through
the transmission system in Fig. 4.16. Assuming that there is no effect due to internal
noise of the devices, determine the frequencies ωi that remain of the signal at point
Z, taking into account that ω3 /2π = 100 kHz, ωX /2π = 40 kHz, BX = 30 kHz,
ωZ /2π = 130 kHz, BZ = 40 kHz. Equations (4.9) and (4.10) may be useful.
1
(1 − cos 2x) .
2
1
cos x1 cos x2 = (cos(x1 − x2 ) + cos(x1 + x2 ))
2
cos 2 x =
.
Problem 4.8 An oscillator generates the described carrier signal by
X(t) = A cos (ωc t − O)
.
(4.9)
(4.10)
4.11 Problems
103
Fig. 4.16 Heterodyne
amplifier
Table 4.13 PCM symbols
for a set of events
Event
A
B
C
D
Probability
0.40
0.30
0.20
0.10
PCM symbol
01
11
10
00
with constant amplitude A and frequency ωc . The phase O is random with
probability density function defined by
{
pO (θ ) =
.
1
2π ,
0 ≤ θ ≤ 2π
0,
otherwise
Find the power spectral density of X(t) in terms of the probability density function
of the frequency ωc . What happens to the power spectral density when the frequency
ωc takes a constant value?
Problem 4.9 Table 4.13 illustrates an alphabet of events with their corresponding
PCM symbol and probability of occurrence. If these events are sent at 4800 baud
(symbols/s):
a. Determine the amount of information for each event.
b. Assuming that the sequence ABACAD is sent periodically as a test signal,
determine the binary entropy in the PCM source.
c. Draw the waveform of the Manchester modulated signal representing the
sequence ABACAD.
d. Assuming that in Manchester modulation, the relationship between bandwidth
consumption and transmission rate is B/R = 2 bit−1 , determine the channel
capacity required for transmission, if there is a signal-to-noise ratio of 20 dB.
Problem 4.10 There is the following sequence of information:
{fk } = 101010010101000000000000110111
a. Find the entropy of the binary source.
b. Find the entropy of the modulated source in AMI.
c. Find the power of the modulated sequence in AMI, with A = 300 mV.
d. Find the power of the modulated sequence in B6ZS, with A = 300 mV.
e. Determine if the AMI modulator, transmitting at 9600 bps, is compatible with a
channel of bandwidth B = XX00 Hz, at a signal-to-noise ratio of 4 dB.
104
4 Performance of Digital Modulation Techniques
Problem 4.11 There is the following sequence of information:
{fk } = 100000000101000000000000110111
a. Find the entropy of the binary source.
b. Find the entropy of the modulated source in AMI.
c. Find the power of the modulated sequence in AMI, with A = 500 mV.
d. Find the power of the modulated sequence in HDB5, with A = 300 mV.
e. Determine if this AMI system, transmitting at 9200 bps, is compatible with a
channel of bandwidth of B = XX00 Hz, at a signal-to-noise ratio of 4 dB.
Problem 4.12 There is the following sequence of information:
{fk } = 100101000101000010000000110111
a. Find the entropy of the binary source.
b. Find the entropy of the modulated source in RB.
c. Find the expected value of the modulated sequence in RB, with A = 750 mV,
V = 200 mV.
d. Find the power of the sequence modulated in RB, with A = 300 mV, V =
200 mV.
e. Determine if this RB system, transmitting at 9600 bps, is compatible with a
channel of bandwidth of B(x) = 100x [Hz], at a signal-to-noise ratio of 4 dB.
Problem 4.13 Analyze the changes produced by the added noise in each system,
use the necessary graphs and/or measurements to add them to the report, and
conclude. Calculate the reception error, bandwidth efficiency of the modulation
schemes explained. From these measurements the student must conclude, and these
conclusions will be included in a report that will be sent with the code of the
functions (only the baseband coding law functions) duly commented. Based on the
results, determine the best coding method.
Problem 4.14 Draw the eye diagram for a signal modulated by AM − F C
technique, using ACarrier as carrier, and SAudio as baseband. Determine the
modulation index by simulation.
Problem 4.15 Draw the eye diagram for the signal modulated by 8 − QP SK
technique, using ACarrier as carrier and SDigital as modulating. Calculate
the modulation index by simulation.
Problem 4.16 Add Gaussian noise to the signal PassBand generated from
Example 2.8, and make the respective eye-diagram measurements as shown in
Fig. 4.11. Parameters of the Gaussian noise are zero mean and variance between
0.1 and 0.9, by 0.1 intervals.
Problem 4.17 Plot the eye diagram for messages transmitted online (NRZ, RZ,
AMI Manchester, and Miller), using 10 uniform type noise variance values. Confirm
whether, for a pulsed transmission, the noise power increases compared to an analog
transmission. Similarly, check that the channel bandwidth decreases.
For a baseband transmission, what happens when the noise power increases?
What happens if the channel bandwidth decreases? From these measurements the
References
105
student must conclude, and these conclusions will be included in a report that will
be sent with the code of the functions (only the baseband coding law functions) duly
commented.
Problem 4.18 For the system in Fig. 4.1, draw the signal spectrum at the receiver
(point R) and determine the signal-to-noise ratio (SNR) at the receiver (sensitivity)
and its channel capacity. The noise is generated by a Gaussian random variable with
mean 0 and variance σN2 varying between 0.1 and 0.9, in intervals of 0.1. Although
the signal is required to be transmitted over a channel of BHz in bandwidth, this
time it is assumed that the channel width is large enough. Therefore, the carrier
bandwidth is assumed for the calculations.
References
Couch II LW (1997) Digital and Analog Communication Systems. Prentice Hall, Hoboken
Fensel A, Tomic S, Kumar V, Stefanovic M, Aleshin SV, Novikov DO (2013) SESAME-S: semantic smart home system for energy efficiency. Informatik-Spektrum 36(1):46–57. https://doi.org/
10.1007/s00287-012-0665-9. https://link.springer.com/article/10.1007/s00287-012-0665-9
Garcia-Alvarez JC, Ibarra-Ortiz OD, Rodriguez A (2006) Descripción de la Calidad del Servicio
en una Cabecera LAN integrada. In: XII Congreso Internacional de Telecomunicaciones
SENACITEL, Ministerio de Telecomunicacionesde Cuba, Valdivia, Chile, pp 1–5
García-Álvarez JC (2007) Telemedicine integrated-services perspectives. Revista del Congreso
Internacional de Ingeniería Electrónica 1(1):1–4
Haykin S (2001) Communication Systems. Wiley, Hoboken
Henao-Ríos JL, Guerrero-González N, García-Álvarez JC (2021) Experimental validation
of inverse M-PPM modulation for dimming control and data transmission in visible
light communications. IEEE Latin Amer. Trans. 19(2):280–287. https://latamt.ieeer9.org/
index.php/transactions/article/download/3174/634/59632. https://latamt.ieeer9.org/index.php/
transactions/article/view/3174/634
Lathi BP (1998) Modern Digital and Analog Communications Systems, 3rd edn. Oxford Press,
Oxford
Maniktala S (2013) Power Over Ethernet Interoperability, 1st edn. McGraw-Hill Education, New
York. https://doi.org/10.26537/neutroaterra.v0i12.388. https://www.accessengineeringlibrary.
com/content/book/9780071798259. https://www.accessengineeringlibrary.com/content/book/
9780071798259.abstract. http://parc.ipp.pt/index.php/neutroaterra/article/view/388
Rajagopal S, Bashyam S, Cavallaro JR, Aazhang B (2000) Efficient VLSI architectures for baseband signal processing in wireless base station receivers. In: Proceedings IEEE International
Conference on Application-Specific Systems, Architectures, and Processors. Rice University.
Nokia Corporation. Texas Instruments Inc
Stremler FG (1989) Sistemas de Comunicación. Alfa-Omega, Rock Rapids
Tadokoro M, Kubo T, Murayama D, Yamada T, Suzuki KI, Yoshimoto N, Kubo R (2013) Hybrid
TDma and contention-based access techniques in ethernet passive optical networks for smart
sensor-actuator networks. In: Proceedings of the 2013 IEEE International Communications
Quality and Reliability Workshop, CQR 2013, pp 1–6
Chapter 5
Elements of Digital Communication
5.1 Design of a Digital Communication System
What is intended in any transmission is that the information undergoes the least
possible modification and that those bits that are delivered to the destination are
as similar as possible to those originating at the source. The way to quantify this
similarity may be to quantify the reliability of the channel, but this is an arduous
task. The most advisable thing is to quantify the reliability of the transmission by
the residual error rate or bit error rate (BER). In this way, one can ask whether or
not the error rate of the transmission system is lower than what the user considers
acceptable. The problem appears when this maximum error rate is exceeded and the
information becomes unintelligible. Figure 5.1 illustrates the general model of an
electronic system. The first solution that is thought of is to increase the transmission
power, to increase the signal-to-noise ratio (SNR). But it is not always possible since
many times the transmission power is already limited by other aspects. The solution
discussed in this project is that of channel coding, since, as Shannon demonstrated
in 1948, with appropriate coding of the information, the errors introduced into it by
a noisy channel or storage can be reduced to an arbitrary level without sacrificing
transmission or storage rate. The signal that carries the message is called carrier.
The carrier is a signal that has a defined frequency and power. A signal cannot be
transmitted over long distances without changing its parameters (Tomasi, 2003).
The main characteristic of a carrier signal is its periodicity. A periodic signal .xT (t)
is one that satisfies (Stremler, 1989; Couch II, 1997; Schwartz, 1990; Lathi, 1998)
x(t + T ) = x(t)∀t
.
(5.1)
where T is a fixed time interval, called period. In this case, .xT (t) would have infinite
energy in an infinite time interval, because this signal is measured throughout the
time domain. Indeed .E → ∞. However, the energy is finite and equal within any
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024
J. C. García-Álvarez, Digital Electronic Communications,
https://doi.org/10.1007/978-3-031-53118-7_5
107
108
5 Elements of Digital Communication
Fig. 5.1 Communication system diagram
interval T , so we can have an expression for the average energy of the signal,
dividing by the period T :
ET =
.
1
T
t0 +T
|xT (t)|2 dt
(5.2)
t0
With this, the representation of a periodic signal is treated in the same way as
a finite energy signal, by finding the value of T and replacing it with the interval
.[t2 , t1 ]. The process of changing signal parameters is called modulation. Therefore,
the carrier alone would not be of much use if it has no information to carry. The
frequency of the carrier signal must be higher than the frequency of the modulating
one. A carrier signal is defined by the following parameters: (i) the Amplitude, (ii)
the Frequency, and (iii) the Phase.
Definition 5.1 Defining system as a set of objects that are configured in a certain
way to accomplish a certain task or objective, one can supose that for each
modulation stage, the system can interact as part of another system or a higher
hierarchy.
Theoretically, information generated by source X would reach source Y without
errors, under the right conditions. However, in order for it to be able to do so over
long distances and with efficient power consumption, it needs to be transported
by another signal, at an assigned frequency. To complete the transmission of
information, the following must be applied:
5.2 Electronic Devices in a Digital Communication System
109
• The existence of the information signal, called baseband or modulating signal
• The existence of a periodic and harmonic signal that physically transmits
information, called carrier
• The process of interaction between the two elements, resulting in the bandpass
or modulated signal, called modulation
5.2 Electronic Devices in a Digital Communication System
• The transmitter (Sect. 5.2.1) a collection of electronic components and circuits
that converts the electrical signal into a signal suitable for transmission over a
given medium. Transmitters are made up of oscillators, amplifiers, tuned circuits
and filters, modulators, frequency mixers, frequency synthesizers, and other
circuits.
• The communication channel is the medium by which the electronic signal is
sent from one place to another. Types of media include:
– Electrical conductors
– Optical media
– Free space
– System-specific media (e.g., water is the medium for sonar)
• A receiver is a collection of electronic components and circuits that accepts
the transmitted message from the channel and converts it back into a form
understandable by humans. Receivers contain amplifiers, oscillators, mixers,
tuned circuits and filters, and a demodulator or a detector that recovers the
original intelligence signal from the modulated carrier.
5.2.1 Transmitter
It is responsible for adapting the signal to the transmission medium with the
objective of making more efficient transmissions. It consists of (1) modulator, (2)
amplifiers, (3) filters, and (4) antennas. The signal whose information is transmitted
is called the source. The goal of transmission is to preserve as much information as
possible from the source signal. To do this, adequate power is required to:
• That the signal is detectable
• That the detected signal is intelligible
• That the information extracted from the signal does not contain errors
The source itself can be transmitted. However, in order for it to be able to do so
over long distances and with efficient power consumption, it needs to be transported
by another signal, at an assigned frequency. To accomplish the transmission of
information, there must apply the following:
110
5 Elements of Digital Communication
• The existence of the information signal, namely baseband
• A wave that physically transmits the information, namely carrier
• An interaction between the above elements, resulting in the passband signal
The interaction is called modulation.
5.2.2 Mixer
Modulation is a technique used to transport information on a high-frequency
carrier signal, which is typically a sine wave, in order to make it suitable for
transmission over a communication channel, allowing more information to be
transmitted simultaneously, protecting it from possible interference or noise.
Information, as well as sounds and images, when converted into electrical audio
and video signals, cannot be radiated due to the fact that they are baseband signals.
On the other hand, the carrier wave alone would not be of much use if it does not
have information to transport. Modulation procedures consist of using the carrier
wave as transport for electrical information (audio, video, data, etc.). Modulation is
necessary basically for two reasons:
• It is difficult to make a baseband electrical signal, such as audio (between 3 Hz
and 20 kHz), and transform into an efficient electromagnetic wave.
• If it were possible to radiate a baseband wave, it would be unfeasible to radiate
it, since the transmitting devices would be of non implementable dimensions.
Basically, modulation consists of varying a parameter of the carrier wave according
to variations in the modulating signal, which is the information we want to transmit,
and is manifested in sounds and images. If the amplitude of the carrier is varied
as a function of the modulating signal, an amplitude modulation (AM) is being
performed, and if the frequency of the carrier is varied as a function of the
modulating signal, a frequency modulation (FM) is being performed. Figure 5.2
illustrates a communication system, which aims to establish a link between two
points, for the proper transmission of a signal.
The carrier is the direct representation of the physical transmitted wave. The
carrier transmits the information. The following are the carrier propagation modes:
1. Transverse electromagnetic wave, through a transmission line.
2. Radiated electromagnetic wave, through air.
+
Filter 1
S
wc
Y
Filter 2 R
H
N
Fig. 5.2 Communication system with noiseless restricted channel
S'
5.2 Electronic Devices in a Digital Communication System
111
3. Waveguide propagation, as in Fiber Optics.
4. Acoustic propagation.
The Baseband Signal (commonly expressed as modulating signal) has the information that can be sent directly and unmodified over the medium or can be used
to modulate a carrier for transmission over the medium. In telephone or intercom
systems, the voice is placed on the wires and transmitted via a baseband signal. In
some computer networks, the data are applied directly to coaxial or twisted pair
cables for transmission using baseband signals.
5.2.3 Oscillators
Emphasis will be placed on so-called linear oscillators, which use positive feedback
loops, through passive or active arrangements. By using positive feedback, the poles
of the system are located toward the critical stability value. In this sense, only a
transient signal state is required (e.g., a pulse signal) for the oscillation to occur.
Example 5.1 (Varicap Diode as Resonator) The circuit shown in Fig. 5.3 acts as
a variable tuner. In this case, input impedance .Z11 is given by
Z11 =
.
R
(1 + (L/R)s)
vi
(1/Cs)(R + Ls)
=
=
2
R + Ls + (1/Cs)
LC s + (R/L)s + (1/LC)
ii
When denominator is .s 2 + (ω0 /Q)s + ω02 and .ω0 is the cutout frequency given by
the pole, the Quality Factor Q is calculated as
1
LC
1 L
Q=
R C
ω0 =
.
(5.3)
a. With the calculated values of capacitance and inductance at the previous item,
how much the quality factor changes when the varicap diode is connected?
Figure 5.3 illustrates the circuit with the connected varicap diode. Figure 5.3
−γ
shows a model of the varicap diode, where .CD = 35×10−12 1 + Vvi0
+0.1×
10−12 Farads, .vi = 4 + 0.001 sin ωt V, .Rs = ηVT /IDQ , .η = 1.5, .VT = 23 mV,
.V0 = 0.6 V, and .γ = 0.5. The oscillator circuit with a feedback gain, given by
the operational amplifier of the first stage, is described in Figure 5.5.
b. In the circuit of Fig. 5.4, with .R = 50 Ω, calculate the values for inductance L
and capacitance C, such that the quality factor .Q(vi /ii ) is equal to 30, and the
resonance frequency is equal to 20 kHz.
112
5 Elements of Digital Communication
Fig. 5.3 (a) Varicap diode model. (b) Compensated resonant circuit. (Source: Author creation)
Fig. 5.4 (a) Oscillator. (b) Wien’s oscillator. (Source: Author creation)
5.3 Passband Modulation
Modulation is a set of techniques used to transport information on a high-frequency
carrier signal, which is typically a sine wave, in order to make it suitable for
transmission over a communication channel. It allows more information to be
transmitted simultaneously, protecting it from possible interference or noise.
Information, knowing as a digital set of symbols that represent the physical
signal, as sounds and images, when converted into electrical audio and video signals,
cannot be radiated due to the fact that they are baseband signals. On the other hand,
5.4 Frequency Modulation
113
Fig. 5.5 Triangle oscillator. (Source: Author creation)
the carrier wave alone would not be of much use if it does not have information to
transport. Modulation procedures consist of using the carrier wave as transport for
electrical information (audio, video, data, etc.). Modulation is necessary basically
for two reasons:
• It is difficult to make a baseband electrical signal, such as audio (between 3 Hz
and 20 kHz), and transform into an efficient electromagnetic wave.
• If it were possible to transmit a baseband signal by audio waves, it would be
unfeasible to radiate it, since the transmitting devices would be of gigantic
dimensions.
5.4 Frequency Modulation
Amplitude modulation has two drawbacks in practice: On the one hand, information
is not always transmitted with sufficient quality, since noise greatly affects the
amplitude of the transmitted signal. On the other hand, amplitude affectation by
noise leads to the receiver to hardly to eliminate interference. This is the reason,
among others, that it is adequate to transmit a message using Frequency Modulation
(FM). When we refer to Continuous Wave (CW), it means that the signal that
represents the message continuously alters some parameter of the carrier signal at
the time of transmission. Thus, the aim is to show the difficulties involved in sending
a message using this method.
For every transmission system, the medium or channel is the fundamental
basis for establishing reliable communication of information. To do this, it is
not enough to send the message directly through the channel. Specifically, the
channel has parameters such as Bandwidth, frequency response, and Signal-to-Noise
Ratio. These parameters determine how efficiently information can be sent over a
transmission channel. The present work presents the advantages and disadvantages
of transmitting a frequency modulated (FM) message through a transmission
114
5 Elements of Digital Communication
Fig. 5.6 Frequency
modulated signal. (Source:
Author creation)
medium with band, frequency, and noise restrictions. To verify the above, a message
is modulated, transmitted through a channel, detected, and demodulated through FM
(WB/NB) techniques, determining the parameters at each stage. The elements that
make up an FM communication system are analyzed, determining how it affects the
modulation index and the carrier signal in the transmission. Finally, it is verified
how the transmission power is modified by varying the parameters.
Frequency modulation consists of varying the frequency of the carrier wave
according to the intensity of the information wave. The amplitude of the modulated
wave is constant. This means that the carrier oscillates with greater or lesser
frequency, depending on the amplitude of the modulating wave. For example, if
we apply a .Δc Hz modulator, the modulated signal moves .Δc times in one second
with respect to the frequency of the carrier signal, called Central Frequency, as
illustrated in Fig. 5.6. Thus, the degree of this variation will depend on the volume
with which the carrier is modulated, which is called Modulation Index. Then, as
the noise or interference significantly alters the amplitude of the wave, but not its
frequency, the information transmitted in FM is preserved, since the information
is extracted from the frequency variation and not from the amplitude, which is
constant. As a consequence of these modulation characteristics, it can be observed
how the sound or image quality is improved when it is modulated in frequency than
when it is modulated in amplitude. In addition, additional forms of data or signals
requiring more baseband bandwidth, which include audio streaming or video, can
be transmitted without affecting the transmission rate. This is the reason why radio
broadcasts use modulated frequency, or in other words, the birth of the stations
that in the mid-60s chose this system to broadcast their programs with higher
sound quality, giving rise to musical radio broadcasting. Other uses of modulated
frequency are mobile telephony, television, and civil band communication services.
The frequency or phase of the sinusoidal carrier varies proportionally with the
modulating signal .x(t) in two ways: (1) Frequency Modulation (FM) and (2) Phase
Modulation (PM).
Modulation comprises FM and PM and refers to the process by which the
phase angle of a carrier sine wave is varied according to the message signal.
Frequency modulation is a type of angular modulation that transmits information
through a carrier wave by varying its frequency. The instantaneous phase .θ (t) of
the modulated signal is proportional to the instantaneous value of the modulating
signal. The carrier is represented by
y (t) = A cos [θ (t)] .
(5.4)
θ (t) = ωc t + φ(t)
(5.5)
.
5.4 Frequency Modulation
115
In which the frequency and/or phase are linearly related to the message signal .x(t).
Depending on the nature of the spectral relationship between the modulation index
.x(t) and .(wc , φ), there are different types of frequency or phase modulation. The
angle .θ (t) of the carrier varies depending on the message signal.
1. If the variation is of frequency, .θ (t) = ωc t + kf x(t)dt.
2. If the variation is phase, .θ (t) = ωc t + x(t).
An angle modulated signal, also referred to as an exponentially modulated signal,
has the form
j φt
.sm (t) = A cos [ωt + θ (t)] = Re Ae
(5.6)
where the instantaneous phase .φi (t) is defined as
φi (t) = ωt + θ (t)
.
and the instantaneous frequency of the modulated signal is defined as
ωi (t) =
.
d
d
[ωt + θ (t)] = ω + θ (t)
dt
dt
The functions .θ (t) and .dθ (t)/dt are referred to as the instantaneous phase and
frequency deviations, respectively. The phase deviation of the carrier .φ(t) is related
to the baseband message signal .s(t). Depending on the nature of the relationship
between .ψ(t) and .s(t), we have different forms of angle modulation.
d
θ (t) = kf s(t)
dt
t
.φ(t) = kf
s(λ)dλ + ωt
.
(5.7)
(5.8)
t0
where .kf is a frequency deviation constant [radian/s/V]. It is assumed that .t0 →
−∞ and .φ(−∞) = 0. Combining equations (5.7) and (5.8) with Eq. (5.6), we can
express the frequency modulated signal as
s (t) = A cos ωt + kf
. m
t
−∞
s(τ )dτ
5.4.1 Spectrum of Frequency Modulated Signal
Since frequency modulation is a nonlinear process, an exact description of the
spectrum of a frequency modulated signal for an arbitrary message signal is
116
5 Elements of Digital Communication
more complicated than linear process. However if .s(t) is sinusoidal, then the
instantaneous frequency deviation of the angle modulated signal is sinusoidal and
the spectrum can be relatively easy to obtain. If we assume .s(t) to be sinusoidal:
s(t) = Am cos ωm t
.
then the instantaneous phase deviation on the modulated signal is
φ(t) =
.
kf Am
sin ωm t
ωm
(5.9)
The modulated signal, for the FM signal, is given by
Sm (t) = A cos (ωt + β sin ωt)
.
where the parameter .β is called the modulation index defined, for FM, as
kf Am
ωm
β=
.
The parameter .β is defined only for sine wave modulation, and it represents
the maximum phase deviation produced by the modulating signal. If we want to
compute the spectrum of .Sm (t) given in Eq. (5.9), we can express .Sm (t) as
Sm (t) = Re Aej ωt ejβ sin ωm t
.
In the preceding expression, the exponential function of .jβ sin ωm t is periodic with
a period .Tm = 2π/ωm . Thus, we can represent it in a Fourier series of the form:
∞
e(jβ sin ωm t) =
Cx (nfm )e(j 2π nf )
.
n=−∞
where
Cx (nfm ) =
ωm
2π
.
1
=
2π
τ
ωM
− ωτ
M
π
−π
e(jβ sin ωm t) e(−j ωm t) dt
ej (β sin θ−nθ) dθ
This is the .jn (β), as it is shown in Sect. B.3. Therefore, we can obtain the following
expression for the FM signal with tone modulation:
∞
Sm (t) = A
jn (β) cos [(ω + nωm )t]
.
n=−∞
5.4 Frequency Modulation
117
Fig. 5.7 Narrow band .β = 0.2 (left) and wideband .β = 5 (right) FM spectrum. (Source: Author
creation)
The signal .Sm (t) is thus composed by a series of sinusoidal signals multiplied by a
constant (Bessel function); The spectrum of .Sm (t) consists of an infinite number of
Dirac delta functions. The number of significant (energy contained) spectral lines is
limited, so we can use the FM modulation, with finite bandwidth. An example of
narrow band and wideband FM spectrum is shown in Fig. 5.7.
Among others, the spectrum of an FM signal has the following properties:
• The FM spectrum consists of a carrier component plus an infinite number of
sideband components at frequencies .f ± nfm (.n = 1, 2, 3 . . .). But the number
of significant sidebands depends primarily on the value of .β. In comparison,
the spectrum of an AM signal with tone modulation has only three spectral
components (at frequencies .f, f + fm , and .f − fm ).
• The relative amplitude of the spectral components of an FM signal depends on
the values of .jn (β). The relative amplitude of the carrier depends on .j0 (β), and
its value depends on the modulating signal (unlike AM modulation where the
amplitude of the carrier does not depend on the value of the modulating signal).
• The phase relationship between the sideband components is such that the oddorder lower sidebands are reversed in phase.
• The number of significant spectral components is a function of argument .β.
When .β ⪡ 1, only .J0 and .J1 are significant (in Table 5.1, .j0 (β) approaches unity,
while .j1 (β) to .jn (β) approach zero) so that the spectrum will consist of carrier
plus two sideband components, just like an AM spectrum with the exception of
the phase reversal of the lower sideband component.
• A large value of .β implies a large bandwidth since there will be many significant
sideband components.
• Transmission bandwidth of 98% of power always occurs after .n = β + 1
(underlined values in Table 5.1).
• Carrier and sidebands null often at special values of .β are the zeros of Bessel
function .jn (β) = 0.
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5 Elements of Digital Communication
Table 5.1 Bessel function .jn (β)
.β
n
0
1
2
3
4
5
6
7
8
9
10
11
0
.0.2
.0.5
1
2
5
8
10
.1.00
.0.99
.0.938
.0.765
.0.224
.−0.178
.−0.246
.0.1
.0.242
.0.440
.0.577
.−0.328
.0.005
.0.031
.0.115
.0.353
.
0.172
. 0.235
.−0.113
.−0.291
.−0.105
. 0.186
. 0.338
. 0.321
. 0.223
. 0.126
. 0.061
. 0.026
0
.0.02
.0.129
.
.0.002
.0.034
.
.0.007
.
.
.
.
0.047
0.365
0.391
0.261
0.131
0.053
0.018
.
0.043
0.255
. 0.058
.−0.22
.−0.234
.−0.14
. 0.217
. 0.318
. 0.292
. 0.208
. 0.123
.
.
5.4.2 FM Power
The AM power depends on the amplitude of the modulated signal and is defined as
PAM =
.
x 2 (t) A2
+
2
2
From the previous section, we saw that a single tone modulated FM signal has
an infinite number of sideband components, and hence the FM spectrum seems to
have infinite. However, for any .β, a large portion of transmitted power is contained
in finite bandwidth. Hence the determination of FM bandwidth depends on the
number of significant sidebands required for an adequate transmission. The limits
of bandwidth allow the transmission of signal without significant distortion. To
determine transmission bandwidth, let us analyze the power ratio .Sn , which is the
fraction of the power contained in the carrier plus n sidebands to the total power of
FM signal.
Sn =
.
PF M
PF M (n)
Since an FM modulated signal has constant amplitude, it can be shown that its power
is
PF M =
.
A2
1
= A
2
2
∞
|jk (β)|2
k=−∞
5.4 Frequency Modulation
119
PF M (n) =
.
1
A
2
n
|jk (β)|2
k=−∞
Since .x 2 (t) ≥ 0, in general, less power is required to send an FM modulated signal
than an AM modulated one.
Example 5.2 We search the number of sidebands n such that the power ratio
achieves 98%. Thus,
1
2A
Sn =
.
1
2A
n
k=−∞
∞
|jk (β)|2
≥ 0.98
|jk (β)|2
k=−∞
Using Table 5.1, we can show that the bandwidth of FM signal .BT depends on the
number of sidebands and FM modulation index .β, which can be expressed as
BT ≈ 2(β + 1)fm
.
It is known as Carson’s law.
5.4.3 Narrow Band Modulation
An FM modulation system is called narrow band if the frequency deviation meets
the condition:
.kf
f (t) dt ⪡ 1
The shape of a narrowband FM signal is
y (t) = A cos (ωc t) − Akf
.
f (t) dt sin (ωc t)
Using a harmonic signal,
Sm (t) = A cos [ωt + φ(t)] = A cos(ωt) cos φ(t) − A sin(ωt) sin φ(t)
.
≈ A cos ωt − Aφ(t) sin ωt
As .φ becomes very small, .cos φ = 1 and .sin φ ≈ φ. With this approximation,
the name narrow band refers to the fact that such a modulated signal has the same
bandwidth as a DSB-AM signal, i.e., 2.Bbb .
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5 Elements of Digital Communication
Similarly, a PM modulation system is called narrowband if the frequency
deviation meets the condition:
kp f (t) ⪡ 1
.
The shape of a narrowband PM signal is
y (t) = A cos (ωc t) − Akp f (t) sin (ωc t)
.
Based on Carson’s law, the condition for a modulation system to be wideband or
narrowband can be redefined:
1. .mf ⪢ 1 .→ Wideband (WB) modulation
2. .mf ⪡ 1 .→ Narrowband (NB) modulation
5.5 Frequency Modulation Methods
5.5.1 Indirect Modulation Method
This method is also called Armstrong Method. In this method, a narrowband
modulated signal is first produced and then converted into a wideband modulated
signal using frequency multipliers (Fig. 5.8). The frequency multiplier makes such
operation on the argument of a sine input by a value n, to bring the signal to the
desired frequency range.
y (t) = A cos ωc t + kf
.
x (t) dt
Step 1. Increase the bandwidth of the narrowband FM signal. This is achieved
through a multiplier device. For example, using a square law device,
Fig. 5.8 Indirect FM
modulator scheme
5.5 Frequency Modulation Methods
121
⎛
⎞
2
1
A
2
+ cos ⎝ 2θ (t) ⎠
.x (t) =
2
2
doubleBW
Step 2. In general, using an nth law multiplier, it is possible to obtain any desired
bandwidth.
Step 3. Shift the signal spectrum to a higher frequency without modifying its
content. For this, a multiplier is used for a cosine signal. That is, AM modulation
is applied with a carrier frequency .ω2 .
Angle modulation is obtained by the orthogonal addition of an unmodulated
carrier and a DSB modulated carrier. Orthogonal addition requires that these two
carriers are at exactly the same frequency and .90◦ out of phase. The input stage of
the generator consists of a signal input, an integrator, an amplifier, and a selection of
FM and PM modes. In FM mode, the output of the integrator is sent to the multiplier
of an Armstrong modulator. In this case, the combination of the integrator and the
Armstrong modulator produces a signal that is frequency modulated by the input
signal. In PM mode, the Armstrong modulator directly produces a signal phase
modulated by the input signal; the buffer only provides insulation.
5.5.2 Direct FM Generation Method
A Voltage Controlled Oscillator (VCO) is used as a frequency modulator, which
generates a square signal whose instantaneous frequency is proportional to the
input signal. The modulating signal is the one coming from the function generator.
Figure 5.9 shows a VCO using a triangular generator. In demodulation, a digital
Phase Locked Loop (PLL) and a low-pass filter are used which eliminates carrier
residues. The modulating signal directly controls the carrier frequency. A common
method used to generate FM directly is by varying the inductance or capacitance
of an oscillator. The main advantage of direct FM is that frequency deviations are
possible and fewer frequency multipliers are then required.
It consists of varying the instantaneous frequency of the carrier directly with the
information signal. One of the devices that is capable of performing this task is a
Voltage Controlled Oscillator (VCO). A VCO varies linearly the frequency of its
internal oscillator with respect to a voltage.
5.5.3 Frequency-Shift Keying
When a digital signal is modulated by FM, it is called frequency-shift keying (FSK).
In this case, the frequency of the carrier signal changes for a reduced set of values.
122
5 Elements of Digital Communication
Fig. 5.9 General scheme of a direct FM circuit. (Source: Author creation. Schematic generated in
EasyEDA)
A cos(2πf0 t + ϕp ) f0 = fp + Δfp 1
.UF SK (t) =
A cos(2πf1 t + ϕp ) f1 = fp − Δfp 0
(5.10)
The space between the two frequencies must be reduced to avoid increasing
bandwidth. The minimum space is twice the bandwidth of the baseband signal. The
frequency of the carrier signal is varied to represent two states in binary 1 or 0,
where the peak amplitude and phase remain constant for each bit interval. The FSK
demodulator must be able to determine which of the two possible frequencies is
present at any given time, the advantage of this type of modulator is that it is less
susceptible to errors, and the receiver is able to look for specific frequency changes
of more than a number of intervals, so voltage spikes (noise) can be ignored.
5.6 Pulse Width Modulation
The pulse width modulation scheme is a pulsed scheme used to directly send energy
quantity information to a switched transmission system. The circuit described in
Fig. 5.10 gives us a description on PWM technique:
5.6 Pulse Width Modulation
123
Fig. 5.10 General scheme of PWM circuit. (Source: Author creation. Schematic generated in
EasyEDA)
To generate SAudio, SPWM, it is recommended to use the following routine in
MATLAB:
PWM Modulation
Fs = 50;
AParameter = 6;
FParameter = 5;
Fc = Fs/FParameter;
t = [0:2*Fs+1]’/Fs;
t2 = [0:(2*FParameter*(Fs+1))-1]’/Fs;
% SAudio signal
SAudio = sin(2*pi*t)+...
sin(4*pi*t)+sin(8*pi*t)+3;
% SPAM signal
h = modem.pammod(’M’, AParameter);
SPAM = modulate(h, floor(SAudio));
SPAM = interp(SPWM,FParameter);
SPWM = modulate(SAudio./AParameter,Fc,Fs,’pwm’);
SAudio = interp(SAudio,FParameter);
To transmit the PWM modulation signals through the channel, it is recommended
to use the following routine in MATLAB:
PWM Modulation
Wl = 0.001;
Wh = 0.999;
SNRval = 10;
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5 Elements of Digital Communication
[b,a] = cheby1(4,1,[Wl Wh]);
% Bandpass Chebyshev Type I
YPAM = filter(b,a,SPAM);
YPWM = filter(b,a,SPWM);
%Transmission of SPAM and SPWM
YPAMNoi = awgn(YPAM,...
SNRval,’measured’);
YPWMNoi = awgn(YPWM,...
SNRval,’measured’);
% Add white Gaussian noise.
hdem = modem.pamdemod(’M’, AParameter);
ZPAM = demodulate(hdem, floor(YPAMNoi));
ZPWM = demod(YPWMNoi,Fc,Fs,’pwm’);
% Create a Welch spectral estimator.
% Calculate and plot the PSD
% for each signal
h2 = spectrum.welch;
plot(t2,SPAM,’-’,t2,YPAM,’x-’),
figure,
plot(t2,SPAM,’-’,t2,YPAMNoi,’x-’),
figure,
plot(t2,SAudio,’-’,t2,ZPAM,’x-’),
figure,
subplot(2,2,1), psd(h2,SAudio,’Fs’,Fs);
subplot(2,2,2), psd(h2,SPAM,’Fs’,Fs);
subplot(2,2,3), psd(h2,YPAM,’Fs’,Fs);
subplot(2,2,4), psd(h2,YPAMNoi,’Fs’,Fs);
figure,
plot(t2,SPWM,’-’,t2,YPWM,’x-’),
figure,
plot(t2,SPWM,’-’,t2,YPWMNoi,’x-’)
figure,
plot(t2,SAudio,’-’,t2,interp(ZPWM,FParameter),’x-’),
figure,
subplot(2,2,1), psd(h2,SAudio,’Fs’,Fs);
subplot(2,2,2), psd(h2,SPWM,’Fs’,Fs);
subplot(2,2,3), psd(h2,YPWM,’Fs’,Fs);
subplot(2,2,4), psd(h2,YPWMNoi,’Fs’,Fs);
In this way, Wl and Wh define the bandwidth values of the channel, and SNRval
defines the signal-to-noise ratio of the channel (Garcia-Alvarez, 2002). The PWM
modulator is made up of a 100 kHz triangular wave generator, a signal that
5.6 Pulse Width Modulation
125
when compared with the input signal produces a train of pulses whose width is
proportional to the input signal and the frequency is the same as the triangular signal
generated. The average value of the PWM signal is proportional to the modulating
signal, so it can be easily recovered by low-pass filtering the modulated signal.
Generate a PPM signal
//////////////////////CONFIGURATION///////////////////////////////
#define CHANNEL_NUMBER 12
//set the number of channels
#define CHANNEL_DEFAULT_VALUE 1500
//set the default servo value
#define FRAME_LENGTH 22500
//set the PPM frame length in microseconds (1ms = 1000us)
#define PULSE_LENGTH 300 //set the pulse length
#define onState 1
//set polarity of the pulses: 1 is positive, 0 is negative
#define sigPin 10
//set PPM signal output pin
/*this array holds the values for the ppm signal. These values can
be changed.
(usually servo values move between 1000 and 2000)*/
int ppm[CHANNEL_NUMBER];
Generate a PPM signal: Step 2
void setup(){
//initialize default ppm values
for(int i=0; i<CHANNEL_NUMBER; i++){
ppm[i]= CHANNEL_DEFAULT_VALUE;
}
pinMode(sigPin, OUTPUT);
digitalWrite(sigPin, !onState);
//set the PPM signal pin to the default state (off)
cli();
TCCR1A = 0; // set entire TCCR1 register to 0
TCCR1B = 0;
OCR1A = 100; // compare match register, change this
TCCR1B |= (1 << WGM12); // turn on CTC mode
TCCR1B |= (1 << CS11);
// 8 prescaler: 0.5 microseconds at 16MHz
TIMSK1 |= (1 << OCIE1A); // enable timer compare interrupt
sei();
}
Generate a PPM signal: Step 3
void loop(){
/*
Here modify ppm array and set any channel to value between
1000 and 2000.
Timer running in the background will take care of the rest
126
5 Elements of Digital Communication
and automatically
generate PPM signal on output pin using values in ppm array
*/
}
ISR(TIMER1_COMPA_vect){ //leave this alone
static boolean state = true;
TCNT1 = 0;
if (state) { //start pulse
digitalWrite(sigPin, onState);
OCR1A = PULSE_LENGTH * 2;
state = false;
} else{ //end pulse and calculate when to start the next pulse
static byte cur_chan_numb;
static unsigned int calc_rest;
digitalWrite(sigPin, !onState);
state = true;
if(cur_chan_numb >= CHANNEL_NUMBER){
cur_chan_numb = 0;
calc_rest = calc_rest + PULSE_LENGTH;//
OCR1A = (FRAME_LENGTH - calc_rest) * 2;
calc_rest = 0;
}
else{
OCR1A = (ppm[cur_chan_numb] - PULSE_LENGTH) * 2;
calc_rest = calc_rest + ppm[cur_chan_numb];
cur_chan_numb++;
}
}
}
Exercise 5.1 Using the routine 5.6, we obtain from the output port 12 of Arduino
UNO the PPM generated pulse, of 300 ms width, sent each .22.5 ms. The following
steps are followed for PWM transmission:
• Connect the function generator to the input of the COAX1 transmitter.
• Select with the INPUTS 1 button on this input, and provide the function
generator with a sine wave signal of 1 kHz.
• Observe the signal at point A of the emitter with the oscilloscope. Set the signal
amplitude to 3 V peak-peak.
• Activate the transmission in pulse modulation (PWM) using the Modulation
button, and with the FDM button leave it in the Direct option.
• Observe the signal at point C of the emitter (modulated signal).
• Activate coaxial line transmission on the transmitter using the Transmission
button.
• Put the receiver into operation, and select coaxial line reception, using the
Reception button.
• Observe the signal at test points A and C in the receiver stage.
For the SDigital message, the same procedure mentioned above is carried out,
and the following parameters are adjusted: square signal of 2 V peak-peak and
5.7 Phase Passband Modulation
127
Table 5.2 Calculated values for the pulsed transmission channel
Wl
0.001
0.001
0.001
0.1
0.2
0.5
0.001
0.001
0.001
Wh
0.999
0.999
0.999
0.999
0.999
0.999
0.5
0.7
0.9
SNRval RMS(YPWM)/ RMS(SAudio) RMS(YPWMNoi)/RMS(SAudio)
10
5
1
10
10
10
10
10
10
fundamental frequency of 1 kHz. On the other hand, the channel consists of a coaxial
cable accessory to the baseband transmission system, and the effect of noise is not
considered. It should be noted that a prior recognition of the transmission system
must be done to carry out this experiment.
Exercise 5.2 Check the relationship between the variation of channel capacity and
the modulation index. Determine the parameters of each of the filters for the proper
transmission of the PWM signal.
Exercise 5.3 Using the result from Table 5.20, determine the changes that occur in
the transmission PWM.
Exercise 5.4 Vary the frequency of the SAudio message, while observing on the
oscilloscope how its waveform changes in the receiver. Using this, determine the
frequency at which the received message is very similar to the pure carrier signal,
when SAudio is transmitted. Fill the data with respective results in the Table 5.2.
5.7 Phase Passband Modulation
The phase of the carrier signal is shifted to represent digital data. In this system,
a binary zero is represented by a signal with the same phase as the phase of the
previously sent signal, and a binary 1 is represented by the transmission of a signal
whose phase is in opposition to the previous one. The phase modulated signal is
then
y (t) = A cos ωc t + kp x (t)
.
Use the symbols given by .s(t) = cos ψm cos ωc t − sin ψm sin ωc t, where the phase
shift is given by .ψm = 2π
M (m − 1), where m is the index of the M symbols
representing the signal. The energy is then equal to
128
5 Elements of Digital Communication
T
E = 0T
.
0
s 2 (t)dt = T2
baseband signal
A2 dt = A2 T
carrier signal
(5.11)
In this case, a phase variation occurs, and this signal can be represented as the linear
combination of two orthonormal functions.
2
cos ωc t
.φ1 (t) =
0 ≤ t < T /2.
(5.12)
T
2
sin ωc t
φ2 (t) = −
T /2 ≤ t ≤ T
(5.13)
T
Because they are in quadrature, they are represented in the complex plane as .s(t) =
Cm1 φ1 (t) + Cm2 φ2 (t), where their respective projections are
Cm1 (t) =
.
Cm2 (t) =
T
cos ψm t.
2
(5.14)
T
sin ψm t
2
(5.15)
Phase shift keying (PSK) is a type of angular modulation that consists of varying
the phase of the carrier signal between a number of discrete values. The difference
with conventional phase modulation (PM) is that while in this the phase variation is
continuous, depending on the modulating signal, in PSK the modulating signal is a
digital signal and, therefore, with a limited number of states (Linn, 2006).
5.7.1 Quadrature Phase Modulation
Quadrature shift keying (QPSK) refers to a technique for varying the phase of a
carrier wave .φ a wave of amplitude, and frequency .ω by applying a carrier signal
digital. This digital modulation is represented in the constellation diagram by four
points equidistant from the origin of the coordinates. With four phases, QPSK can
encode two bits per symbol. The assignment of bits to each symbol is usually done
using the gray code, which means that between two adjacent symbols, the symbols
only differ by 1 bit, thus minimizing the bit error rate (Wegmueller et al., 2006).
Definition 5.2 To observe the location in the complex plane of each of the signals
sent, the emph(constellation diagram) is used. Such diagram is represented by a set
of symbols, assigning them a specific amplitude and phase; in this case it can be
seen that the symbols sent by each channel are clearly differentiated.
5.8 Channel
129
5.8 Channel
It serves as a physical link between the transmitter and the receiver of the signal and
is the medium utilized to transfer data signals. The physical characteristics of the
medium determine the channel design since the signal needs to be transformed into
an electromagnetic wave in order to be sufficiently transmitted over the channel.
Attenuation, or the gradual drop in signal strength with the distance between the
transmitter and receiver, is one of the consequences of the physical restriction
imposed by the channel (Castillo et al., 2001). Additional considerations influencing
transmission via the channel include: (1) reflection loss resulting from coupling
problems between the medium’s characteristic impedance and the receiver’s input
impedance; (2) any electromagnetic phenomenon in the case of wireless links; and
(3), in the case of fiber optics, microscopical imperfections in the fabrication of this
medium.
5.8.1 Spectral Assignment
Bandwidth (BW) is that portion of the electromagnetic spectrum occupied by a
signal. Channel bandwidth refers to the range of frequencies required to transmit
the desired information. The Signal Bandwidth comprises the minimal and maximal
frequencies where a signal .s(t) can be represented.
Example 5.3 A voice signal contains frequencies between 300 and 3000 Hz, that
is, a bandwidth of 2700 Hz.
Example 5.4 AM radio spectrum is between 450 and 500 kHz. Commercial FM
radio spectra is assigned between 88 and 108 MHz.
5.8.2 Channel Bandwidth
It is the difference between the maximum and the minimum frequency that a channel
can transmit. For example, a copper pair has a bandwidth of 2 MHz; therefore
it is suitable for transmitting voice signals. Today, virtually the entire frequency
spectrum between approximately 30 kHz and 300 MHz has been served. There is
tremendous competition for these frequencies, between companies, individuals, and
government services in individual carriers and between the different nations of the
world. The electromagnetic spectrum is one of our most precious natural resources.
An important characteristic of the application of filters in telecommunications is
the understanding that there is always a frequency band limitation at the time of
transmitting. The Electromagnetic Spectrum is the resource by which permission for
transmission is given. This resource is managed by an entity, whether governmental
130
Table 5.3 Spectral
allocation in the United States
5 Elements of Digital Communication
Classification
Extremely low
Infra low
Very low
Low
Medium
High
Very high
Ultrahigh
Super high
Extremely high
Tremendously high
Acronym
ELF
ILF
VLF
LF
MF
HF
VHF
UHF
SHF
EHF
THF
Freq. range
.<300 Hz
300 Hz–3 kHz
3–30 kHz
30–300 kHz
300 kHz–3 MHz
3–30 MHz
30–300 MHz
300 MHz–3 GHz
3–30 GHz
30–300 GHz
300 GHz–3 THz
or private, and that permission is granted through a license. This permit specifies,
among others, the power limit at which a signal is allowed to be transmitted and
the Spectral Assignment. The latter has two parts: (1) the Assigned Frequency over
which the carrier signal can be transported, and (2) the Bandwidth within which the
signal is allowed to transmit, with right not to be interfered with (García-Álvarez
et al., 2003). In Table 5.3 there are some frequency ranges where the spectral
assignment for radio waves is made.
5.8.3 Bandwidth Optimization
Communication engineering is devoted to making the best use of the spectrum.
Great effort goes into developing communication techniques that minimize the
bandwidth required to transmit given information by preserving spectrum space.
This provides more room for additional communication channels and gives other
services or users an opportunity to have a chance to communicate.
To preserve the greatest fidelity of the information that is to be sent from one
place to another, certain techniques are used that allow the source signal to be
decomposed. Thus, the source signal is analyzed through the power distribution
and frequency composition of the signals used. Spectral Analysis of signals is
the technique that analyzes the source as a function. In this case, the signal is
transformed through integration processes and convergent series of representative
functions. Spectrum management is provided by agencies set up by Colombia and
other countries to control spectrum use. The Agencia Nacional del Espectro (ANE)
and the Comision Reguladora de Telecomunicaciones (CRT) are two agencies that
deal in spectrum management. Standards are specifications and guidelines necessary
to ensure compatibility between transmitting and receiving equipment.
5.8 Channel
131
5.8.4 Media Frequency Assignation
• Wired: Multiple channels, local assignment
– Coaxial Cable
– Fiber Optic
• Guided: Single channel, local assignment
– Waveguides
– Infrared
• Wireless: Single channel, restricted assignment
– Antennas
– Satellites
The design of devices within a communication system is based on the parameters
given by the following tables:
Medium
Coaxial
Optical fiber
Microphone
Waveguide
Twisted pair
Bandwidth
1 to 3 GHz (RG11)
102 GHz
11.4 kHz
1 to 30 GHz
250 to 500 MHz (Cat.6)
Attenuation (typical)
Low
Low
High
Low, for short distances
High
Medium
Coaxial
Fiber Optics
Audio
Waveguide
Twisted pair
Cable budget (per 50 ft)
Around $10
Around $8
Around $10
Around $100
Around $9
Implementation cost
Low
High
Low
High
Low
• Operating Frequency f
• Wavelength λ
These parameters are related to Phase Velocity: vp = f λ.
Exercise 5.5 Fill the following table with the results of wavelengths required for
transmission at a certain phase speed.
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5 Elements of Digital Communication
f (kHz)
0.22
220
2 · 105
λ
vp = 343.2 m/s
vp = 3 · 108 m/s
Which frequencies can an acoustic transmission system (vp = 343.2 m/s) can be
implemented for? Which ones for an electromagnetic radiation transmission system
(vp = 3 · 108 m/s)? As a hint, following table illustrates some examples of communication applications, which frequency/wavelenght allocation (radio spectrum)
is based on vp = 3 · 108 m/s.
Source
AC current
FM broadband
Mobile
Satellite Ka band
Ultraviolet band
f
60 Hz
100 MHz
800 MHz
20 GHz
1 PHz
λ
5000 km
3m
37.5 cm
15 mm
100 nm
5.8.5 Attenuation Analysis
The phenomenon of Attenuation is implicit in the frequency response of the channel.
Therefore, one way to measure attenuation is by measuring the power value of
the messages both in transmission and in reception. The general link loss model
is presented by the equation:
PRx =
.
PT x
L
(5.16)
which represents the estimated value of the received power, given a propagation
model, where L is a random variable. Thus, for a message to be sent properly, it
must conform to the parameters of the channel through which it is transmitted. The
Path Loss (PL) model adds the loss term .(LM ) in Eq. (5.16).
PR =
.
PT GT GR
PT GT GR
=
LP L
LF S LM
(5.17)
The value .LM depends on the model used to characterize the channel link. In this
case, the frequency response of the channel is analyzed. The related parameters are
as follows:
• Bandwidth. It is the difference between the upper and lower frequencies where
the channel allows the transmission of a message. For the message, its bandwidth
5.9 Receivers
133
is the difference between the upper and lower frequencies that make up the
signal that represents the message. Normally the message bandwidth is reduced
to a higher frequency, called the fundamental frequency of the message. Thus,
for effective transmission, the channel bandwidth must exceed the value of the
fundamental frequency of the message. Otherwise, the message will be distorted.
• Frequency Response. Within the bandwidth of the channel, it contains a frequency response. This means that for certain frequencies within the bandwidth,
the signal may or may not be transmitted properly. This frequency response
is analogous to the frequency response of a filter. This means that there are
frequencies where the message can be attenuated, called cutoff frequencies.
Typically cutoff frequencies are found when the channel gain is 3 dB below the
maximum channel gain.
Each link has an associated propagation model, according to the transmission
medium and the conditions around the link. In addition to the quadratic law of
attenuation, the Friis equation defines the most general propagation model. As
illustrated in the equation, the losses are dependent on the frequency, the gain at the
transmitter (including the antenna in the case of a wireless link), and the transmit
power. The procedure to generate the data necessary for the analysis is illustrated
below.
In wired channels, the attenuation loss is linear. In wireless channels (Radio,
Mobile, and Satellite) attenuation loss is calculated by Path Loss models:
• Path loss is a reduction in the signal’s power, which is a direct result of the
distance between the transmitter and the receiver in the communication path.
• It is proportional to the square of the distance between the transmitter and the
receiver.
• There are many models used in the industry today to estimate the path loss.
• Each model has its own requirements that need to be met in order to be used
correctly.
5.9 Receivers
It is responsible for extracting the signal from the transmission medium and carry
out the opposite operations to the transmitter to return the signal to its initial
condition (baseband). It is composed of (1) antennas, (2) filters, (3) amplifiers, and
(4) demodulator.
If the power is finite and the energy is infinite, the signal is said to be a power
signal.
Exercise 5.6 Equation (5.18) is used to calculate the bit-to-noise ratio in an 8-PSK
system.
C
B
Eb .
=
N0 8−P SK
N
Rb
(5.18)
134
5 Elements of Digital Communication
5.9.1 Coherent Detector
Coherent detection consists of multiplying a replica of the carrier signal .x̂c (t) on the
signal received after the channel .s(t). Taking the received signal .r(t) = s(t) + n(t),
the product is realized by the signal .x̂c (t) = cosωc t. By multiplying the received
signal .g(t) by an in phase carrier .f ' (ωc t), we obtain a zero-frequency component
equivalent to the amplitude of baseband signal. It is important to verify how the
amplitude and frequency at the signal output .g ' (t) depend on the phase variation.
Example 5.5 Two signals are used: One is varied in amplitude using a modulating
signal, and the second is a signal with frequency variation. The envelope detector
is expected to be stable under different frequency variations. The procedure is as
follows:
• The SAudio and SDigital signals are used as modulator signals.
• The signal with amplitude variation will be an AM-DSB signal, which we will
call AMS (Amplitude-Modified Signal).
• The frequency-shifted signal will be a wideband FM signal, which we will call
FMS (Frequency-Shifted Signal)
5.9.2 Envelope Detector
One of the simplest methods of signal transmission is amplitude modulation. This
method consists of varying the amplitude of the carrier signal, through the amplitude
of the modulator. This process implies that there are several periods of the carrier
signal in each period of the modulator. Therefore, we have an envelope signal, built
with the shape of the modulating signal.
Therefore, we have an envelope signal, built with the shape of the modulating
signal. This work illustrates the performance of an envelope detector in its use with
AM receivers. Here we implement simple circuits that detect with good fidelity the
envelope of the modulated signal.
The envelope detector circuit (Fig. 5.1) is built through the parallel between a
capacitor and a resistor, in series with a switch-type diode. The stability analysis
of this system is important because it allows studying its performance. Specifically,
the circuit is expected to behave as a linear low-pass filter, whose cutoff frequency
is higher than the maximum frequency of the envelope, guaranteeing the stability
of the filter when there are phase variations due to channel distortion. This section
illustrates the performance of an envelope detector in its use with AM receivers.
Here we implement simple circuits that detect with good fidelity the envelope of
the modulated signal. In this practice, a practical approach to diodes is given, using
them within rectifier circuits.
In electronics, a rectifier is the element or a circuit that allows alternating current
to be converted into direct current. Rectifiers have several uses, including power
5.9 Receivers
135
supply components and radio signal detectors. The core element of a rectifier is a set
of switching diodes, whether solid-state semiconductors, vacuum valves, or gaseous
valves such as mercury vapor valves. The first important aspect in rectification is
the loss between the peak input voltage and the peak output voltage, due to the
bias voltage of the diodes (around .0.7 V for silicon P-N diodes and .0.3 V for
Schottky diodes). Half-wave rectification circuits, by using two separate transformer
secondary points, obtain a voltage loss of a diode. On the other hand, a diode
bridge rectifier circuit obtains a voltage loss of two diodes, representing a significant
power loss for low voltage sources. Because the diodes do not conduct below the
driving voltage, the circuit will only deliver power for a portion of each half cycle,
which is defined as a clipping in the duty cycle. The second important aspect in
rectification is the stability of the DC voltage at the output, because the rectifier
circuit alone does not deliver a constant value. To produce a DC voltage from an
AC source, a smoothing circuit is required. This circuit simply requires a capacitor
called smoothing or reserve, placed in parallel to the output of the rectifier. However,
an amount of oscillation AC called ripple still persists, which means that the output
signal is not yet constant. Thus, the design of the capacitor represents a problem.
For a given load, for example, a high value capacitor will reduce ripple but will be
more expensive and would create high peak currents in the secondary transformer
and load. In extreme cases where many rectifiers are loaded on a power distribution
system, it is difficult for the system operator to maintain a correct sine curve
in magnitude and phase. This work illustrates the behavior of diodes for various
rectifier circuits and determines guidelines for the proper design of a rectifier circuit
using diodes. In this work, a Short-Circuit simulation will also be carried out, which
consists of short-circuiting the output of the rectifier and thus checking if the rectifier
was designed according to the specifications, specifically, if the diodes can withstand
an extreme value of current.
Example 5.6 (Detection in a Simple Sideband System)
Step 1. Since the continuous wave (CW) amplitude modulation (AM) mode is
selected, the SAudio signal is used as the modulator and ACarrier as the
carrier. The generated signal Passband-AM-SSB-SC .y(t) is
y(t) = A(t) cos (ωc t) + I m {H {A(t)}} sin (ωc t)
.
Visualize the correct modulation at the emitting point. A sine wave with carrier
frequency must be observed, moving in amplitude according to the modulation
index, which is used as the carrier of the modulated signal in AM.
The bandwidth, quality factor, order, and transfer function parameters must be
defined for the filters.
To adjust the modulation index, the amplitude variation of the SAudio signal is
used. Knowing the amplitude of the carrier signal, make measurements using
modulation index values of .0.1–.0.9 in steps of .0.1.
136
5 Elements of Digital Communication
For each modified modulation index, different carrier frequency values are used:
ωc , .2ωc , .4ωc , and .8ωc . The designed filters must vary with central frequency .ωc ,
.2ωc , .4ωc , and .8ωc .
The channel attenuation values .Lp , .Ls , and .Lf are the same as those generated
for Eq. (3.10).
Use the noise variance values .σ .
Calculate the sensitivity and capacity of the channel.
.
5.9.3 Half-wave Rectification Envelope Detector
Figure 5.11 corresponds to a half-wave rectifier circuit. This circuit is built through
the parallel between a capacitor and a resistor, in series with a switch-type diode.
The stability analysis of this system is important because it allows studying its
performance. Specifically, the circuit is expected to behave as a linear low-pass filter,
whose cutoff frequency is higher than the maximum frequency of the envelope,
guaranteeing the stability of the filter when there are phase variations due to channel
distortion. For the Short-Circuit simulation, change the value of the load resistance
to .RL = 0.5 Ω, and measure the value of the current in one of the diodes in each
circuit. The materials in the table are required.
• Generator AC/AC
• At least four 1N4001 diodes. As alternatives there are 1N4004 or 1N4003 diodes.
Refrain from bringing the integrated diode bridge.
• Resistors from .100 Ω to 5 W
• Resistors from .1000 Ω to 5 W (minimum of 1 W)
• 10 k.Ω resistors
• Capacitors of 470, 1000, and .1470 μF
Diodes have been obtained with characteristics similar to those illustrated in
Table 5.4.
Example 5.7 (Carrier Frequency Variation) Two signals are used: One is varied
in amplitude using a modulating signal, and the second is a signal with frequency
Fig. 5.11 Half-wave rectifier. (Source: Author creation)
5.9 Receivers
137
Table 5.4 Parameters of the
diode(s) used in circuit
depicted in Fig. 5.11
Parameter
Operating current (at 20 .◦ C)
Bias voltage
Maximal allowed current
Internal resistance .rd
Value
Unit
ampere
volt
ampere
ohm
Fig. 5.12 ASK modulation
technique. (Source: Author
creation)
variation. The envelope detector is expected to be stable under different frequency
variations. The procedure is as follows:
• The SAudio and SDigital signals are used as modulator signals.
• The signal with amplitude variation will be an AM-DSB signal, which we will
call AMS (Amplitude-Modified Signal).
• The frequency-shifted signal will be a wideband FM signal, which we will call
FMS (Frequency-Shifted Signal)
5.9.4 FM Detection Measurement Parameters
Figure 5.12 illustrates the frequency detector circuit using a resonant diode bridge.
Thus, for a message to be sent properly, it must conform to the parameters of the
channel through which it is transmitted. The modulation index is given by
mf =
.
Δω
ωbb
where .Δω is the maximum frequency deviation. Thus, the bandwidth is calculated
using Carson’s rule:
BW = 2ωbb 1 + mf
.
(5.19)
Finally, the channel will always be susceptible to interference. This interference
is modeled as Noise and is presented as a random signal that is added to the message
on the channel. Thus, if the power of the noise is much greater than that of the
message, the message will be corrupted and cannot be easily reconstructed. To
counteract noise, Modulation is used, which allows the signal to be reconstructed
despite the noise presented in the channel. The test bench to be used is a two-signal
138
5 Elements of Digital Communication
FM transmission system in baseband. To do this, there are two test signals: One is
a signal composed of several sine functions, named SAudio, which represents an
audio message, and the other is a square signal, named SDigital, which represents a
digital message.
Exercise 5.7 With the signals SAudio and SDigital, generate the FM transmission
with the input parameters: modulation index .ηf . It is recommended to use the
following program, which will be called GenFM.(ηf , Ac , ωc ).
Program Code
% Frequency deviation in modulated signal
YAudio = fmmod(SAudio,Fc,Fs,dev);
% Modulate channel.
YDigital = fmmod(SDigital,Fc,Fs,dev);
SAudioDem =...
fmdemod(ZAudioNoi,Fc,Fs,dev);
% Demodulate channel.
%SAudio demodulation.
SDigitalDem =...
fmdemod(ZDigitalNoi,Fc,Fs,dev);
%SDigital demodulation.
h = spectrum.welch;
% Create a Welch spectral estimator.
% Calculate and plot the PSD
% for each signal
subplot(4,1,1), plot(t,YAudio,’-’),
subplot(4,1,2), plot(t,ZAudio,’-’),
subplot(4,1,3), plot(t,ZAudioNoi,’-’),
subplot(4,1,4),
plot(t,SAudio,’b-’,t,SAudioDem,’r-’),
figure,
subplot(4,1,1), psd(h,YAudio,’Fs’,Fs);
subplot(4,1,2), psd(h,ZAudio,’Fs’,Fs);
subplot(4,1,3), psd(h,ZAudioNoi,’Fs’,Fs);
subplot(4,1,4), psd(h,SAudio,’Fs’,Fs),
hold on, psd(h,SAudioDem,’Fs’,Fs),
hold off,
figure,
subplot(4,1,1), plot(t,YDigital,’-’),
subplot(4,1,2), plot(t,ZDigital,’-’),
subplot(4,1,3), plot(t,ZDigitalNoi,’-’),
subplot(4,1,4),
plot(t,SDigital,’b-’,t,...
SDigitalDem,’r-’),
figure,
subplot(4,1,1), psd(h,YDigital,’Fs’,Fs);
5.9 Receivers
139
subplot(4,1,2), psd(h,ZDigital,’Fs’,Fs);
subplot(4,1,3),
psd(h,ZDigitalNoi,’Fs’,Fs);
subplot(4,1,4), psd(h,SDigital,’Fs’,Fs),
hold on, psd(h,SDigitalDem,’Fs’,Fs),
hold off
Example 5.8 An FM receiver has .B = 200 kHz bandwidth, noise figure NF .=
8 dB, and noise temperature at the input of .T = 100 ◦ C. Determine the minimum
carrier power at the input to achieve SNR .= 23 dB at detection. To exceed the noise
figure, an input signal-to-noise ratio of SNR.i = 23 + 8 = 31 dB would be required.
Then, the minimum signal power at the receiver to overcome the noise figure is
Pi (dB) = SNRi + 10 log N = 31 + 10 log
.
KT B
= −245 dBm
0.001
where .K = 1.38 · 10−23 is the Boltzmann constant.
Example 5.9 Determine for a modulation coefficient .m = 0.2 and an unmodulated
carrier power .Pc = 2000 W:
a. The total sideband power
PT SB =
.
m 2 Pc
= 40 W
2
b. Upper and lower sideband power
PU SB = PLSB =
.
m 2 Pc
= 20 W
4
c. Modulated carrier power
Pmc = Pc = 2000 W
.
d. The total power transmitted
m2
P T = Pc 1 +
= 2040 W
2
.
Example 5.10 For an AM receiver, with the following gains and losses:
Gains: RF amplification of 30 dB, an IF amplifier of 44 dB, and an audio
amplifier of 24 dB gain
Losses: Pre-selector loss of 2 dB, mixer loss of 6 dB, and detector loss of 8 dB
the net gain of the receiver in dB is .
Gains .−
Losses .= 98 − 16 = 82 dB.
140
5 Elements of Digital Communication
Example 5.11 For an FM modulator with unmodulated carrier amplitude .Vc =
20 V, modulation index .m = 1, and load resistance .RL = 10 Ω, calculate the power
on the modulated carrier and at each side frequency, and plot the power spectrum of
the modulated wave. The power on the modulated carrier is
Pc =
.
Vc2
= 20 W
2R
Using the Bessel table, for a modulation index .m = 1, 3, significant sidebands are presented. The amplitude of the modulated carrier and the sidebands
are calculated from the Bessel components .Jn .n = 0, 1, 2, 3. Being .{Jn } =
{0.77, 0.44, 0.11, 0.02}, the amplitude of the modulated carrier is equal to
Vmc = J0 Vc = 15.4 V
.
In the same way, the amplitudes of the sidebands are calculated
.
V1 = J1 Vc = 8.8 V.
(5.20a)
V2 = J2 Vc = 2.2 V.
(5.20b)
V3 = J3 Vc = 0.4 V
(5.20c)
The total power is
3
P T = Pc +
.
i=1
V2
Pi = mc +
2R
3
i=1
Vi2
= 20.1 W
R
5.10 Filter
In a communication system, the filter is a spectral efficiency optimization device.
At the transmitter, the filter provides gain to the passband signal in the assigned
bandwidth and cancels the residual modulation sidelobes. At the receiver, the filter
cuts off the noise allocated out of the passband bandwidth.
5.10.1 Filter Design
In communication systems, the filter fulfills different functions. For the transmitter,
the filter is a device for optimizing spectral efficiency, since it adds gain to the
modulating signal in the assigned bandwidth and cancels the residual harmonics
5.10 Filter
141
of the modulation. At the receiver, the filter eliminates noise that appears outside
the bandwidth of the modulating signal.
However, for the channel, the filter is the representation of the frequency
limitations of the transmission medium and the attenuation of signal power over
distance and other related phenomena. This causes signal degradation evident in all
transmission media. There are different sources of attenuation:
• Frequency response of devices
• Change of propagation mode
• Imperfections in the transmission medium. Example: Defects in fiber optics or
cables
This decrease can occur for the entire signal or for some of its parameters. The
following work describes the considerations for parameterizing the filters at each
stage (transmitter, channel, and receiver) to optimize the spectral efficiency of
the system. For validation, we will consider Channel Sensitivity and Capacity as
measurements. The filter commonly used for communication systems is of the first
order low-pass type, which can be modified to a high-pass filter or add others in
series forming a bandpass or band-eliminate filter and higher order filters, depending
on the restriction of bandwidth. The frequency response of the filter is extremely flat
(with minimal ripples) in the passband. Viewed on a Bode plot with a logarithmic
scale, the response decays linearly from the cutoff frequency toward minus infinity.
For a first order filter, it is .−20 dB per decade (approx. .−6 dB per octave). Although
there are other types of filters used for communication systems, the Butterworth
filter is the only filter that maintains its shape for higher orders (only with a steeper
drop from the cutoff frequency).
5.10.2 General Filter Parameters
Each filter must be designed to allow the frequency components that belong to the
signal and eliminate interference. The filter has the following characteristics:
• Bandwidth is the difference between the upper and lower frequencies where the
channel allows the transmission of a message. For the message, its bandwidth
is the difference between the upper and lower frequencies that make up the
signal that represents the message. Normally the message bandwidth is reduced
to a higher frequency, called the fundamental frequency of the message. Thus,
for effective transmission, the channel bandwidth must exceed the value of the
fundamental frequency of the message. Otherwise, the message will be distorted.
• Within the bandwidth of the channel, the Fig. 5.13 displays the frequency
response of a filter. This means that for certain frequencies within the bandwidth,
the signal may or may not be transmitted properly. This frequency response
is analogous to the frequency response of a filter. This means that there are
frequencies where the message can be attenuated, called cutoff frequencies.
Typically cutoff frequencies are found when the channel gain is 3 dB below
142
5 Elements of Digital Communication
Fig. 5.13 Frequency
response for filter output
the maximum channel gain. There are commonly used shapes for the frequency
response (Elliptical, Chebyshev, and Butterworth).
• Filter order
• Filtering Factor Q. Use the graph line of Fig. 5.13 for calculation.
Filter design parameters are adjusted by the role the filter plays in the system
(transmitter, receiver). Subtract (in dB) losses due to passive components in the
transmit chain after the amplifier.
•
•
•
•
Filter loss
Feedline loss
Jumpers loss
Antenna/laser/wire amplifier gain
Definition 5.3 The Transfer Function is defined as the ratio of the Laplace
transform of the output to the input with all initial conditions equal to zero. Transfer
functions are defined only for linear time invariant systems. Transfer functions can
usually be expressed as the ratio of two polynomials in the complex variable, s.
The filter in its general form is presented as the following transfer function:
H (s) =
.
K(s + z1 )(s + z2 ) . . . (s + zM )
(s + p1 )(s + p2 ) . . . (s + pN )
(5.21)
The roots of the numerator polynomial .zm are called zeros. The roots of the
denominator polynomial .pn are called poles. For calculation purposes in the Bode
diagram, we have Eq. (5.21) with .s = j ω:
K'
.H (ω) =
. . . zjMω + 1
jω
jω
jω
p2 + 1 . . . pN + 1
p1 + 1
jω
z1 + 1
jω
z2 + 1
5.10 Filter
143
Example 5.12 Unlike AM-SSB, the AM Residual Sideband (AM-BLR) modulation system can be used in practice, since it considers the use of physically
feasible filters. What is the condition that AM-BLR filters must meet for correct
modulation and demodulation? First, the total transmission power for Full Carrier
Double Sideband Amplitude Modulation (AM-DSB-FC) is defined by the equation:
m2
1+
.PAM−DSB−F C = Pc
2
where .Pc is the carrier power and m is the modulation index. Show that when
the modulation index is equal to 1 in single sideband amplifier modulation with
suppressed carrier (AM-SSB-SC), whose power is defined by the equation
PAM−SSB−SC = Pc
.
m2
4
there is a reduction of at least 8 dB in the total transmit power compared to AMDSBFC.
5.10.3 The Filter on the Transmitter and Receiver
This section presents some configurations of the filters, for example, the bandpass
filter shown as BPF in Fig. 5.14. Knowing the function that filter plays into
the system is useful for the student to become familiar with the important use
of this device in electronic communications. The design of electronic circuits
through Operational Amplifiers (OAs) allows solving problems involved with
linear operations such as addition, multiplication, differentiation, and integration.
Understanding the importance of this linear element within electronic systems, it is
important to recognize the operation of the operational amplifier as a derivative and
integrator operator in analog electronics. This work illustrates two simple examples
of existing integrators and derivators. The materials required are the following:
• Resistors and capacitors according to designs
• Amplifier TL084 or LM324
Fig. 5.14 Example of filter
located in the transmitter
144
5 Elements of Digital Communication
5.10.4 Emphasis/De-emphasis Filters
A signal is applied to the left and right inputs. Each signal goes through a preemphasis circuit to increase the signal-to-noise ratio at reception. The signals then
go to two audio amplifiers: one is an adder and the other is a differential amplifier.
At the output of the adder, the signal L+R (left + right) is obtained, while at the
output of the differentiator, we have L-R. The L+R signal goes directly to an adder,
but the L–R signal first goes to a multiplier where it is combined with a 38 kHz
subcarrier. It is then sent to an adder. The output signal of this summation is sent to
the input of the FM modulator, whose frequency deviation can be adjusted between
5 and 100 kHz. The modulated signal is then passed through the variable gain RF
amplifier.
5.10.5 The Forward Filter
The lead filter is used to increase the sensitivity in phase detection of the carrier
signal. However, it is very susceptible to noise. Figure 5.15 illustrates an Operational
Amplifier configured as a shunt.
Once the cutoff frequency is found, for a shunt with high frequency gain (.ω →
∞) equal to 10 dB, and a cutoff frequency of 1 kHz, the values of the electronic
components are illustrated in Table 5.5, taking into account business values.
Fig. 5.15 High-pass filter
using operational amplifier
Table 5.5 Values of the
electronic components for the
circuit in Fig. 5.15.
Component
.R1
.Rf
C
Value
Units
k.Ω
k.Ω
nF
5.10 Filter
145
Fig. 5.16 FSK receiver
5.10.6 Phase Locking
It requires a local oscillator. Let .g ' (t) be the detected signal at the receiver stage:
'
g (t) =
.
A cos ω1 t
0 ≤ t < t1
A sin ω2 t
t 1 ≤ t < t2
with .ω1 > ω2 . After coherent detection, the measured signals are, according to the
test points in Fig. 5.16:
• At .0 ≤ t < t1 , the signal gets reconstructed when the filter bandwidth is .ωLP F <
ωm = min {ω1 − ω2 , ω2 }. In fact,
A A
+ cos (2ω1 t)
2
2
A
A
gB (t) = A cos (ω1 t) cos (ω2 t) = cos (ω1 + ω2 ) t + cos (ω1 − ω2 ) t
2
2
A
, ωLP F < ω1
gC (t) = A2
2 , ωLP F < ωm
A
cos (ω1 − ω2 ) t, ωLP F < ω1
gD (t) = 2
0,
ωLP F < ωm
A
[1 − cos (ω1 − ω2 ) t] (distortion), ωLP F < ω1
gE (t) = A2
ωLP F < ωm
2,
gA (t) = A cos2 (ω1 t) =
.
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5 Elements of Digital Communication
• At .t1 ≤ t < t2 , the signal gets reconstructed when the filter bandwidth is .ωLP F <
ωm = min {ω1 − ω2 , ω2 }. In fact,
.gA (t) = A cos (ω1 t) cos (ω2 t) =
gB (t) = A cos2 (ω1 t) =
gC (t) =
gD (t) =
gE (t) =
0,
0,
A
A
cos (ω1 + ω2 ) t + cos (ω1 − ω2 ) t
2
2
A
[1 + cos (2ω2 t)]
2
(ωLP F < ω1 ) ∧ (ωLP F < ω1 − ω2 )
(ωLP F < ω1 ) ∧ (ωLP F < ωm )
A
2 [1 + cos (2ω2 ) t] ,
A
2,
(ωLP F < ω1 ) ∧ (ωLP F < ω1 − ω2 )
(ωLP F < ω1 ) ∧ (ωLP F < ωm )
− A2 [1 + cos (2ω2 ) t] (distortion), (ωLP F < ω1 ) ∧ (ωLP F < ω1 − ω2 )
− A2 ,
(ωLP F < ω1 ) ∧ (ωLP F < ωm )
5.10.7 Integrating or Delay Amplifier
The integrating amplifier performs the functions of accumulator and memory. This
means that the device fulfills the function of shifting a signal, useful in quadrature
modulators. The second important function is to serve as a low-pass filter in the
receivers.
Exercise 5.8 (Frequency Detector Using Filter Bank) This project provides a
complete analysis of an electronic device based on class knowledge. We will work
with discrete devices, calculating their values in each case. The circuit in Fig. 5.17
is proposed as an efficient detector of selected frequencies.
Exercise 5.9 Explain the operation of the circuit. Design the circuit in such a way
that it can detect signals whose fundamental frequencies are at 10 kHz, 100 kHz,
and 1 MHz. The filter gains are expected to be 20 dB at these frequencies, with
Fig. 5.17 Proposed assembly
5.10 Filter
147
Fig. 5.18 Linearly sparse
channel
a bandwidth of .1 kHz,10 kHz, and 100 kHz, respectively. The gain of the input
amplifier (which uses the integrated .U0 ) must be 20 dB for the entire bandwidth
of .U0 . The green bulb lights up if the signal is 10 kHz, the yellow bulb lights up if
the signal is 100 kHz, and the red bulb lights up if the signal is 1 MHz. It cannot be
turned on, more than one bulb at a time.
Exercise 5.10 Determine the procedure for calculating the values of resistances,
capacitances, amplifiers, and other circuit elements. Then run the simulation on
Proteus ISIS and the PCB on Proteus ARES. However, the design can be
done with some other simulation and design program.
5.10.8 Analysis with Noise
Figure 5.18 illustrates the effect of the filter frequency response:
r(t) = (c(t) ∗ s(t)) + n(t)
.
The following work describes the parameter considerations given to these two
filters (transmitter and receiver) to optimize the system spectral efficiency. For
the sake of validation, we consider the Sensitivity and the Channel Capacity as
measures. As a result, we found that the following filter parameters are suggested
for best performance of our communication system: at the transmitter, Quality
Factor of XXt, Order Filter of YYt, Filter function ZZt, and Bandwidth AAt, and
at the receiver, Quality Factor of XXr, Order Filter of YYr, Filter function ZZr,
and Bandwidth AAr. For instance, at eight times the carrier frequency, the channel
capacity improved from CC to DD with respect to the carrier frequency.
Example 5.13 The equation (5.22) is used to determine the bandwidth of a signal
transmitted by FSK modulation.
f0 − f1
.B = 2
+R
2
(5.22)
5.10.9 Filter Design
We consider the channel cost U in a linear proportion to the assigned bandwidth. In
other words, the transmission via the bandwidth B has a cost of .U = B units, as it
is displayed in the next graph.
148
5 Elements of Digital Communication
Taking the cheap option: the main harmonic or the first lobe,
v(t) =
.
4
sin(t)
π
• The bandwidth is .B = 1 rad/s.
• The cost is .U = 1 unit, but the receiver cannot easily reconstruct the signal. The
difference using .U = 3 is appreciated in the following diagrams:
5.10 Filter
149
3
v(t) =
.
n=1
4
sin(nt)
nπ
Taking the three first harmonics, the bandwidth cost increases.
• The bandwidth is .B = 3 rad/seg.
• The cost is .U = 3 units.
5
v(t) =
.
n=1
4
sin(nt)
nπ
Taking the first five harmonics,
• The bandwidth is .B = 5 rad/s.
• The cost is .U = 5 units.
Taking 10 and 100 first harmonics is equivalent to bandwidth increasing from
B = 10 to 100 rad/s. Accordingly, the cost proportionally increases from .U = 10
to 100 units. So, a more expensive option could be .B = 1000 rad/s. We achieve
a higher cost of .U = 1000 units, but the receiver can completely reconstruct the
signal at a higher cost, as it appears in the following graphs.
.
150
5 Elements of Digital Communication
Selecting the cost function .(U ) is the basic aspect of the filter design. At
their fundamental level, filters function by resonating at a specific frequencies.
Consequently, a well constructed resonant circuit will dampen out unwanted signals
while permitting the proper frequency of signal to pass through. The filter quality
factor, often known as Q factor, is a quantitative indicator of how much signal is
allowed through in relation to how effectively undesired frequencies are dampened
out. Our goal is to achieve the highest feasible Q-factor, as it is shown in the next
diagram.
• Each filter must be designed to allow the frequency components that belong to
the signal and eliminate interference (5.33).
• The filter has the following characteristics.
– Cutoff frequencies
– Filter order
– Filtering factor Q
– Shape of the frequency response (Elliptical, Chebyshev, and Butterworth)
For this experiment we will use two signals that represent a message in a certain
way: the signal ACarrier .ca (k) which is an analog signal and DCarrier .cd (k)
which is a discrete signal. These two functions are described in the following
equation:
ca (a, k) =a cos(ωc k).
.
(5.23a)
5.10 Filter
151
cd (a, k) =a · sign(cos(ωc k))
(5.23b)
where k is a vector represented by .k = [0 : 1/Fs : T ]' . The value .Fs corresponds
to the sampling frequency and defines the number of elements of the generated
signals. The generated algorithm is encapsulated in the function SigGen(a,N),
where N is the number of points used to generate the signal. The amplitude value a
is arbitrary, within the interval .[0, 5].
Exercise 5.11 Make a graph of ACarrier as a function of time t (i) and as
a function of frequency (spectrum). Explain in text the advantages of using a
continuous signal as a carrier.
Exercise 5.12 Make a graph of DCarrier as a function of time t (i) and as a
function of frequency (spectrum). Describe the advantages of using a discrete signal
as a carrier.
Exercise 5.13 Using the spectral density of ACarrier and DCarrier illustrated
in the figures generated in Exercises 5.11 and 5.12, describe the differences between
these spectra in bandwidth and spectral efficiency.
Exercise 5.14 Based on the spectral density plot, find the amplitude, frequency,
bandwidth, and phase of the ACarrier and DCarrier carriers. Explain the
windowing error that occurs when plotting the ACarrier carrier and the effect
it has on bandwidth.
Exercise 5.15 Based on the results of Exercise 5.14, explain the randomness
phenomenon presented in the phase of the spectrum of the signal DCarrier.
Experimental Guidelines
• According to Activity 5.6, if you want to implement a microphone on
the transmission side, with the objective of transmitting voice messages,
determine the specifications that this element must meet.
• According to Activity 5.6, if you want to implement a hearing aid on the
receiving side, with the objective of amplifying voice messages, determine
the specifications that this element must meet.
• Determine the differences between the signals between the transmitter and
receiver points of a continuous signal and a discrete one in baseband.
• Determine the effects that determine the variation of the power at the
receiving point due to the variation of the fundamental frequency.
• If a signal is increased in amplitude, determine if at the transmitter point
the increase is linear. Determine which electronic device causes this effect
and what its function is in the transmitter.
• Based on the results of Activities 2.7 and 2.8:
1. Using the spectral density, determine the bandwidth for SAudio and
SDigital.
(continued)
152
5 Elements of Digital Communication
2. Describe the differences between a continuous signal and a discrete one
in baseband.
3. In conclusion, determine some special considerations observed between
the two signals.
Review Questions
• For the following communication systems:
– AM-FC;
– AM-SSB;
– AM-VSB;
– FM;
– PSK, 4-PSK, 8-PSK;
– ASK-OOK;
– FSK-SS;
– PWM;
– 16-QPSK;
– 16-QAM (OFDM);
– Manchester;
– DPCM-Delta;
– PWM;
– 64-QAM;
Define:
– The modulation index, as the relationship between the modulating and
carrier signal
– The type of modulator/carrier signal used
– The parameters of the bandpass filter
– Applications
• What is the predominant advantage of crystal oscillators over LC tank
circuit oscillators? (a) cost, (b) simplicity in connection, (c) physical size,
(d) precision and stability, and (e) weight of the elements.
• For an active sinusoidal oscillator, where are the both poles of the transfer
function located on? (a) as conjugate complex on the right half-plane, (b)
as conjugate complex on the real axis, (c) as conjugate complex on the
imaginary axis, (d) as conjugate complex on the left half-plane, and (e)
one pole on the imaginary axis and the other on the real axis.
• What are the adjustments that should be made for noise filtering?
• What does the Hartley oscillator differ from other types?
5.11 Problems
153
5.11 Problems
Problem 5.1 A cable communication system requires the transmission of a bipolar
signal of function:
v(t) =
.
1;
−1;
0 ≤ t < π2
π
2 ≤t <π
How many Bandwidth would you recommend for the transmission of this signal,
regarding that the implementation cost of transmission line is the square of the
bandwidth (U = B 2 )?
Problem 5.2 With respect to the circuit in Fig. 5.15, explain how it was possible to
verify that the operational amplifier behaves as a shunt. Specify some improvements
that can be made.
Problem 5.3 Explain, with respect to the circuit in Fig. 5.34, how it was possible
to verify that the operational amplifier behaves as an integrator.
Problem 5.4 Determine other solutions for the problem of the DC level at the
output of the filters.
Problem 5.5 From Activity 5.66, are the parameter values between the filter on the
transmitter and the filter on the receiver the same? Justify.
Problem 5.6 Determine the settings required for proper signal tuning of a QPSK
system. According to these settings, determine the critical points of the circuit.
Problem 5.7 A sinusoidal information signal modulates a 10 MHz carrier such
that the maximum frequency deviation is 50 kHz. Determine the bandwidth of the
modulated signal if the bandwidth of the baseband signal is (a) 500 kHz, (b) 500 Hz,
and (c) 10 kHz.
Problem 5.8 According to the FCC, the maximum frequency deviation allowed
on commercial FM radio stations is 75 kHz. Consider as an information signal a
sinusoidal signal of frequency 15 kHz. Calculate the bandwidth of the modulated
signal allowed by the FCC. Note that the FM bandwidth depends not only on the
modulation index used and but also on the bandwidth of the baseband signal, as in
the case of Amplitude Modulation.
Problem 5.9 A sinusoidal signal with amplitude Am = A and frequency fm =
10 kHz is used as a test signal when it is frequency modulated with fc = 100 MHz
carrier frequency, obtaining the modulated signal xF M (t) of Eq. (5.24), which
consumes a bandwidth BF M described by Eq. (5.25).
xF M (t) xm (t)=Am cos(2πfm t)
.
kf Am
sin (2πfm t)
= Ac cos 2πfc t +
fm
(5.24)
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5 Elements of Digital Communication
Fig. 5.19 Armstrong
oscillator
BF M ≈ 2 kf Am + fm
.
(5.25)
If the frequency deviation is kf = 75 kHz/V:
a. Calculate the limit value, Am = A0 , so that the frequency modulation is
considered narrowband (modulation index less than 0.01).
b. Calculate the increase in the modulation index and bandwidth BF M when Am =
2A0 .
Problem 5.10 The Armstrong modulator illustrated in Fig. 5.19 has two stages:
a narrow band FM modulator (NB–FM) and a frequency multiplier H1 . The
modulator has a carrier frequency fc = ωc /2π = 300 kHz, with a frequency
deviation of Δf = 100 Hz. The multiplier shifts the frequency components of the
modulated signal by a factor N. Find the value N = N0 such that the modulation
index is four times that sent in narrow band and the value of the frequency deviation
using N = 6N0 . Equations (5.26) and (5.27) may be useful.
Δf
sin (2πfm t)
xF M (t) xm (t)=Am cos(2πfm t) = Ac cos 2πfc t +
fm
.
BF M ≈ 2 (Δf + fm )
.
(5.26)
(5.27)
Problem 5.11 Consider a phase modulation (PM) system, with the modulated wave
defined by
s(t) = Ac cos ωc t + kp m(t)
.
where kp is a constant and m(t) is the message signal. The additive noise n(t) at the
phase detector input is
n(t) = nI cos(ωc t) − nQ (t) sin ωc t
.
Assuming that the carrier-to-noise ratio at the detector input is high compared with
unity, determine (a) the output signal-to-noise ratio and (b) the figure of merit of
the system. Compare your results with the FM system for the case of sinusoidal
modulation.
5.11 Problems
155
Fig. 5.20 Pre-emphasis filter
Problem 5.12 By using the pre-emphasis filter shown in Fig. 5.20, and with a
voice signal as the modulating wave, a frequency modulated transmitter produces a
signal that is essentially frequency modulated by the lower audio frequencies and
phase modulated by the higher audio frequencies. Explain some reasons for this
phenomenon.
Problem 5.13 A phase modulation system uses a pair of pre-emphasis and deemphasis filters defined by the transfer functions
ω
.
ω0
(5.28)
1
1 + j ωω0
(5.29)
Hpe (ω) = 1 + j
.
Hde (ω) =
Show that the improvement (QoS) in output signal-to-noise ratio produced by using
this pair of filters is
QoS =
.
B/ω0
tan−1 (B/ω0 )
where B is the baseband bandwidth. Evaluate this improvement for the case when
B = 15 kHz and ω0 /2π = 2.1 kHz, and compare your result with the corresponding
value for a frequency modulated system.
Problem 5.14 Determine the maximum bit rate at which data can be transmitted
over a binary FSK signal using mark frequency of 99 kHz and space frequency of
101 kHz, such that a bandwidth of 22 kHz is occupied.
Problem 5.15 FSK Audio One type of FSK modulation uses an audible tone
(30–8000 Hz) as a carrier signal. In this way it is possible to establish, for binary
transmission, a frequency mark f1 for the symbol {1}, and a frequency space f0 for
156
5 Elements of Digital Communication
Fig. 5.21 Digrama de fase del módem 16-QAM
the symbol {0}. Determine the maximum bit rate at which data can be transmitted
over a binary FSK signal using a mark frequency of 4 kHz and a space frequency of
6 kHz, such that a bandwidth of 6 kHz is occupied.
Problem 5.16 An 8-PSK system transmits at Rb = 20 Mb/s with a carrier-tonoise ratio of C/N = 11 dB. Determine the bandwidth required to obtain a Binary
Energy-to-Noise ratio of Eb /N0 = 14 dB.
Problem 5.17 Through a modem 16-QAM (Fig. 5.21), which transmits at arbitrary
rate of R kbaud, with Baseband Modulation = NZR, the binary sequence is sent:
{fk } = 10010110010100001001000010011110
through a channel that is affected by Gaussian White Noise of N0 = 10 mW/Hz.
Find:
a. The entropy of the binary source
b. The transmission power of the sequence modulated by the modem
c. The average BER. Do it with respect to symbols closest.
d. The bandwidth consumed by the modem
Problem 5.18 Design a BPSK Signal Detection and Modulation device. Encode,
transmit, receive, and decode a test signal using BPSK passband modulation
techniques.
a. Determine for this type of digital modulation the spectral efficiency—
relationship between the sampling frequency and the transmission bandwidth—
and its dependence on the modulation index.
5.11 Problems
157
Fig. 5.22 PWM circuit
b. Determine the behavior of the power at the inputs and outputs of the different
modules.
c. Determine how the sampling frequency and the carrier signal affect digital
transmission.
d. Check the operation of digital links through a monitoring signal.
e. Determine how transmission speed affects channel capacity.
Problem 5.19 Figure 5.22 illustrates a device for transmitting a PWM signal, for
an audio signal inserted at the input.
a. Describe the operation of the circuit in Fig. 5.22
b. Write an algorithm for generating a PWM signal spectra. Draw the signal
spectrum, and analytically determine the bandwidth in each of the stages of the
system, by modulating with an index of m = 3.
c. Using the RMS of the input and output signals as measurement variables,
determine the PWM modulation index.
Problem 5.20 Through a 16-QAM modem (Fig. 5.23), which transmits at R kbaud,
with NZR Baseband Modulation, the binary sequence is sent:
{fk } = 10110100011100001000000000011110
through a channel that is affected by Gaussian White Noise of N0 = 10 mW/Hz.
Find:
a. The entropy of the binary source
b. The transmission power of the sequence modulated by the modem
c. The average BER. Do it with respect to symbols closest.
d. The bandwidth consumed by the modem
Problem 5.21 Through a modem 16-QAM (Fig. 5.24), which transmits at R kbaud,
with NZR Baseband Modulation, the binary sequence is sent:
{fk } = 00110100010100001010000011011111
through a channel that is affected by a Gaussian White Noise of N0 = 10 mW/Hz.
Find:
a. The entropy of the binary source
b. The transmission power of the sequence modulated by the modem
158
5 Elements of Digital Communication
Fig. 5.23 Phase diagram for a 16-QAM modem
Fig. 5.24 Phase constellation diagram for a 16-QAM modem
5.11 Problems
159
c. The average BER. Do it with respect to symbols closest.
d. The bandwidth consumed by the modem
Problem 5.22 Add noise to the constellation diagram of Fig. 5.24, using a Weibull
model with ten different variance values. Set the Bit Error Rate and Channel
Capacity.
Problem 5.23 Through a 32-QAM modem, which transmits at R = 10∗XX Mbps,
with Baseband Modulation = NZR, the binary sequence is sent:
{fk } = 101010010101000000000000110111
Find:
a. The entropy of the binary sequence
b. The transmission power of the sequence modulated by the modem
c. The average BER. Do it with respect to the nearest symbols.
d. Support for an Ethernet channel of bandwidth B MHz and that is affected by
Gaussian White Noise of N0 = 20 mW/Hz
Problem 5.24 Figure 5.25 illustrates an alphabet of QAM symbols, and Table 5.6
illustrates their corresponding probabilities. If these symbols are sent at 4800 baud
(symbols/s):
Fig. 5.25 32-QAM phase diagram
160
Table 5.6 QAM symbols,
top–bottom, left–right order,
according to Fig. 5.25
5 Elements of Digital Communication
Symbol number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
Data
I
−0.1
0.1
−0.2
0.0
0.2
−0.3
−0.1
0.1
0.3
−0.4
−0.2
0.0
0.2
0.4
−0.3
−0.1
0.1
0.3
−0.4
−0.2
0.0
0.2
0.4
−0.3
−0.1
0.1
0.3
−0.2
0.0
0.2
−0.1
0.1
Q
0.4
0.4
0.3
0.3
0.3
0.2
0.2
0.2
0.2
0.1
0.1
0.1
0.1
0.1
0.0
0.0
0.0
0.0
−0.1
−0.1
−0.1
−0.1
−0.1
−0.2
−0.2
−0.2
−0.2
−0.3
−0.3
−0.3
−0.4
−0.4
Probability
0.01
0.01
0.02
0.02
0.02
0.02
0.04
0.04
0.02
0.02
0.03
0.05
0.03
0.02
0.05
0.1
0.1
0.05
0.02
0.03
0.05
0.03
0.02
0.02
0.04
0.04
0.02
0.02
0.02
0.02
0.01
0.01
a. Determine the amount of information for each event.
b. Assuming that, as a test signal, the sequence 1-2-4-8-16-32 is sent periodically,
draw the phase diagram of the modulated sequence signal.
c. Assuming that the relationship between bandwidth consumption and transmission rate is B/R = 2 bit−1 , determine the capacity of the channel required for
transmission, if there is a signal-to-noise ratio of 20 dB.
5.11 Problems
161
Problem 5.25 There is the following sequence of information:
{fk } = 100000000101000000000000110111
a. Find the expected value of the modulated sequence in HDB5. A = 500 mV.
b. Determine if the HDB5 system, transmitting at 9200 bps, is compatible with a
channel of bandwidth Bxx = XX00 Hz.
c. Find the average power of the modulated sequence in 32-QAM (Fig. 5.25).
Symbols are NOT equiprobable.
d. Determine if the 32-QAM system, when adding AMI modulation, transmitting
at 9200bps, is compatible with a bandwidth channel Bxx = 18.4 kHz.
Problem 5.26 There is the following sequence of information:
{fk } = 100101000101000010000000110111
a. Find the expected value of the modulated sequence in RB.
b. Determine if the RB system, transmitting at 9600 bps, is compatible with a
channel of Bandwidth B = 19.2 kHz.
c. Find the average power of the modulated sequence in 32-QAM (Fig. 5.25).
Symbols are NOT equiprobable.
d. Determine if the 32-QAM system, when adding RB modulation, transmitting at
6200 bps, is compatible with a Bandwidth channel B = 12.4 kHz.
Problem 5.27 A filter has a transfer function:
H (s) =
.
ks 2
s 2 + αs + 1
With k > 0, 0 < a < 2. Determine the type of filter (low-pass, high-pass, all-pass,
and bandpass).
Problem 5.28 Design the filter shown in Fig. 5.26 by calculating the adequate
inductance L1 for the elements R1 = 1 kΩ, R2 = 10 kΩ, R3 = 90 kΩ, RL = 10
kΩ, C = 1 nF. The operating frequency is ω = 106 rad/s.
VCC
+
RL
–
nin
R1
C
R2
Fig. 5.26 Filter for oscillator stage
R3
L1
n0
162
5 Elements of Digital Communication
Problem 5.29 Consider a White Gaussian noise process of zero mean and power
spectral density equal to N0 /2, applied to the input of the RL high-pass filter shown
in Fig. 5.27.
a. Find the autocorrelation function and power spectral density of the random
process at the filter output.
b. What is the mean and variance of this output?
Problem 5.30 A White Gaussian noise of zero mean and power spectral density
N0 /2 passes through the filter system of Fig. 5.28a. The frequency responses of
these filters are illustrated in Fig. 5.28b. The noise at the output of the low-pass filter
is illustrated as n(t).
a. Find the power spectral density and autocorrelation function of n(t).
b. Find the mean and variance of n(t).
c. What is the rate at which n(t) can be sampled such that the resulting samples are
essentially uncorrelated?
Problem 5.31 A stationary Gaussian process X(t) with zero mean and power
spectral density SX (ω) is applied to a linear filter whose impulse response h(t) is
presented in Fig. 5.29. A sample Y is taken from the random process at the filter
output at time T .
Fig. 5.27 RL filter
Fig. 5.28 Noise filtering process
Fig. 5.29 Filter impulsive
response
5.11 Problems
163
Fig. 5.30 Two-cascade filter system
a. Determine the mean and variance of Y.
b. What is the probability density function of Y ?
Problem 5.32 For the following filters with open loop transfer function,
a.
.
(s + 3)(s + 4)
(s + 1)(s + 2)
H (s) =
(s + 3)
s(s + 1)(s + 2)(s + 4)
H (s) =
b.
.
Determine the closed-loop root locus and its corresponding Bode diagram.
Problem 5.33 Consider two linear filters connected in cascade as in the Fig. 5.30.
Let X(t) be a stationary process with autocorrelation function RX (τ ). The random
process appearing at the first filter output is Y1 (t) and that at the second filter output
is Y2 (t).
a. Find the autocorrelation function of Y2 (t).
b. Find the cross-correlation function RY1 Y2 (τ ) of Y1 (t) and Y2 (t).
Problem 5.34 Complete the data in below table with numerical parameters of the
referenced transmission media.
Medium
Coaxial
Optical fiber
Microphone
Waveguide
Twisted pair
Bandwidth
Attenuation
5.5 dB/100 m
Cable budget
Implementation cost
Problem 5.35 Explain the reason for the variation of the link loss value calculated
by the Friis equation, compared to that calculated by the simulation program model.
164
5 Elements of Digital Communication
Fig. 5.31 Communication system with noise
Fig. 5.32 (a) Passive resonant circuit. (b) Active resonant circuit, with varicap diode
Explain the contribution that the inclusion of the random variables made to the
equation.
Problem 5.36 Calculate the voltage gain v0 /vi of the circuit in Fig. 5.15.
Problem 5.37 Figure 5.31 shows a general communication system with noisy
channel.
a. Generate a signal DSB − F C, with arbitrary carrier frequency ωc and modulator
frequency ωm , with ωc = 2ωm , 8ωm , 10ωm , 100ωm . Calculate the spectrum of
each generated signal. Establish a relationship between the carrier component
and the side components of the spectrum.
b. Analyze the link in Fig. 5.31 using an AM-DSB system, according to the noise,
filters, and channel attenuation parameters carried out in the filter practices.
Problem 5.38 Oscillator with Varactor Diode Figure 5.32b illustrates the implementation of a variable tuner circuit. It is intended to enhance the design of the
circuit, which has as parameters C1 = C2 = C3 = C4 = 10μF, R1 = 47 kΩ,
L = 1mH, using the circuit in Fig. 5.32, in such a way that it has a high quality
factor (Q) between 0.679 and 1.22 MHz.
a. Calculate the voltage gain V0 (ω)/Vi (ω) and the input impedance Z11 (ω) as a
function of the frequency ω for the circuit in Fig. 5.32. Calculate (analytically
and under simulation) the quality factor Q.
b. Since the capacitance of the Varactor Diode varies between 17 and 55 pF for
corresponding voltage values between 2.9 and 29 V, calculate the value R4 so
that the circuit in Fig. 5.32 works under the design parameters set out above,
5.11 Problems
Table 5.7 Diode current
values for circuit in Fig. 5.34
165
CF , F
470
1000
1470
ID−sc , A-RMS
using VBI AS = 60 V and R2 = 2.2 MΩ. This means that the extreme values of
the potentiometer at R4 must correspond to the extreme values of the varactor
capacitance (17 and 55 pF).
c. Calculate for the circuit in Fig. 5.32 the voltage gain V0 (ω)/Vi (ω) and the input
impedance Z11 (ω) as a function of the frequency ω for the two extreme values
of R4 . Calculate (analytically and under simulation) the quality factor Q.
Problem 5.39 Using simulation, draw the superimposed input and output waveforms for the circuit in Fig. 5.11 with a 1000 μF capacitor and a 1 kΩ load resistor.
Fill Table 5.7 with the results on the Short-Circuit simulation.
Problem 5.40 Using simulation, illustrate the superimposed input and output
waveforms for the envelope detector, without the capacitor, and a 10 kΩ load
resistor. From this result, determine the bias voltage required to activate the diodes
in the rectifier circuits for the half-wave, for the full wave one, and for the diode
bridge one.
Problem 5.41 An AM generator with output impedance of Zs = 50 Ω, carrier
voltage of Vc = 15 V, and modulation index m = 0.5 is connected to a load of
ZL = 60 − j 40 Ω through a transmission line of characteristic impedance Z0 =
75 Ω and length l = 0.7λ.
a. Determine the total modulated power that the generator can transmit.
b. Determine the reflected power.
c. Determine the transmitted power.
d. Determine the distance and length of a stub coupling required for maximum
power transfer.
Problem 5.42 Determine the noise factor and noise figure of the AM receiver in
Problem 5.41, when the parameters are as follows:
•
•
•
•
•
Input signal power Pi = 2 · 10−10 W
Noise power at the input Ni = 2 · 10−18 W
Internal noise power in the RF amplification stage NRF = 6 · 10−15 W
Internal noise power in the IF amplification stage NI F = 6 · 10−11 W
Internal noise power in the audio amplification stage NA = 6 · 10−9 W
Problem 5.43 Draw the signal spectrum, and analytically determine the bandwidth
in each of the stages of the system, if the modulation scheme is:
a. AM-FC with modulation index m = 0.5
b. AM-SSB with modulation index m = 1
166
5 Elements of Digital Communication
Table 5.8 Ripple and offset values in rectifiers, for a load of RL = 0.1 kΩ
CF (F)
No
470
1000
1470
Isec (A. RMS)
VDC (V)
Vripple (mV p–p)
c. AM-VSB
d. ASK-OOK with modulation index m = 2
Problem 5.44 From the results obtained from the experiment with the full-wave
rectifier circuit, determine the carrier voltage necessary for the output of the circuit
to be equal to the output of the bridge or half-wave rectifier circuits.
a. Determine and justify which rectifier obtained the best performance compared to
the other circuits considered. Determine what special considerations should be
taken into account when assembling an envelope detector circuit.
b. Determine and justify the optimal values of capacitance and charge.
Explain the reason for using switching diodes in envelope detectors.
c. Determine special considerations that should have been taken into account when
designing an AM–SSB system. Fill Table 5.8 with the results obtained from the
simulation of circuit shown in Fig. 5.34.
Problem 5.45 Consider a square law detector, using a nonlinear device whose
transfer characteristic is defined by
v0 (t) = a1 vi (t) + a2 vi2 (t)
.
where a1 and a2 are constants, vi (t) is the input, and v2 (t) is the output. The input
consists of the amplitude modulated wave
vi (t) = Ac [1 + ηa m(t)] cos(ωc t)
.
a. Evaluate the output v0 (t).
b. Find the conditions for which the message signal m(t) may be recovered from
v0 (t).
Problem 5.46 Power efficiency η is measured by the ratio between the power of the
transmitted modulating signal and the total transmitted power PT . An AM-DSB-FC
station transmits with an efficiency of η1 = 0.16. Using Eq. (5.30), determine the
modulation index m that is required for this efficiency to double η2 = 0.32.
m2
.PT = Pc
1+
2
(5.30)
5.11 Problems
167
Fig. 5.33 Power spectral density of noise
Problem 5.47 A DSB-SC modulated signal is transmitted over a noisy channel,
with the power spectral density of the noise being as shown in Fig. 5.33. The
message bandwidth is 4 kHz, and the carrier frequency is 200 kHz. Assuming that
the average power of the modulated wave is 10 W, determine the output signal-tonoise ratio of the receiver.
Problem 5.48 Determine the noise analysis to an SSB receiver, specifically evaluating the following:
• The output signal-to-noise ratio
• The channel signal-to-noise ratio
Hence, show that the figure of merit for the SSB receiver is exactly the same as that
for the DSB-SC receiver. Note that unlike the DSB-SC receiver, the middle-band
frequency of the spectral density function of the narrowband-filtered noise at the
front end of the SSB receiver is offset from the carrier frequency ωc by an amount
equal to B/2, where B is the message bandwidth.
Problem 5.49 Evaluate the autocorrelation functions and cross-correlation functions of the in phase and quadrature components of the narrowband noise at the
coherent detector input for (a) the DSB-SC system, (b) an SSB system using the
lower sideband, and (c) an SSB system using the upper sideband.
Problem 5.50 An unmodulated carrier of amplitude Ac , frequency ωc , and bandlimited white noise are summed and then passed through an ideal envelope detector.
Assume the noise spectral density to be of height N0 /2 and bandwidth 2B, centered
about the carrier frequency ωc . Determine the output signal-to-noise ratio for the
case when the carrier-to-noise ratio is high.
168
5 Elements of Digital Communication
Problem 5.51 Evaluate the autocorrelation functions and cross-correlation functions of the in phase and quadrature components of the narrowband noise at the
coherent detector input for the DSB-SC system.
Problem 5.52 In a receiver using coherent detection, the sinusoidal wave generated
by the local oscillator suffers from a phase error θ (t) with respect to the carrier
wave cos(ωc t). Assuming that θ (t) is a sample function of a zero-mean Gaussian
process of variance σθ2 and that most of the time the maximum value of θ (t) is small
compared with unity, find the mean-square error of the receiver output for DSB-SC
modulation. The mean-square error is defined as the expected value of the squared
difference between the receiver output and the message signal component of the
receiver output.
Problem 5.53 Show that the figure of merit for the SSB receiver is exactly the same
as that for the DSB-SC receiver. Note that unlike the DSB receiver, the midband
frequency of the spectral density function of the narrowband-filtered noise at the
front end of the SSB receiver is offset from the carrier frequency ωc by an amount
equal to B/2, where B is the message bandwidth.
Problem 5.54 Let a message signal m(t) be transmitted using single-sideband
modulation. The power spectral density of m(t) is
SM (ω) =
.
a |ω|
B , |ω| ≤ B
0,
otherwise
where a and B are constants. White Gaussian noise of zero mean and power spectral
density N0 /2 is added to the SSB modulated wave at the receiver input. Find an
expression for the output signal-to-noise ratio of the receiver.
Problem 5.55 A sinusoidal signal with amplitude Ac and frequency wc is used
as carrier. This signal is corrupted by a band limited noise through channel. The
received signal is processed by an ideal envelope detector.
Assume the noise spectral density to be in height N0 /2 and bandwidth 2B,
centered about the carrier frequency ωc . Determine the output signal-to-noise-ratio
for the case when the carrier-to-noise ratio is high.
Problem 5.56 Consider a signal encoded by PAM technique, transmitted through
a channel with Gaussian white noise and minimal bandwidth Bo, where Ts is the
sample period. Noise has zero mean and spectral power density equal to N0/2. The
PAM signal is composed of standard pulse g(t), so its Fourier transform H (ω) is
defined as:
1
2BT , |ω| < BT
.H (ω) =
|ω| > BT
0,
5.11 Problems
Table 5.9 Data for Problem
5.57
169
Frequency
ωc
2ωc
4ωc
8ωc
P T × P Rx
Eq. (5.17)
P Rx
Friis model
Fig. 5.34 Integrating
amplifier
Considering a sinusoidal modulating wave at maximum load, show that the PAM
system and baseband transmission have equal signal-to-noise ratios for the same
average transmitted power.
Problem 5.57 Using the results of Table 2.2 (Problem 2.2), assign a value L for a
given type of transmission medium (Coaxial Cable, Fiber Optic, Antenna), decrease
the ACarrier and DCarrier signals by that factor, attenuation, using Eq. (5.16).
Calculate the power PRx of the attenuated signals using the spectrum of the signals.
Fill the Table 5.9 with the given results. Convert those values into dBm and dBW.
Problem 5.58 Calculate the transfer function of the circuit in Fig. 5.34, with Rf =
10 kΩ, R1 = 10 kΩ, and C = 100 nF, using simulation mode of Frequency
Response (AC-sweep). Keep the voltage value Vin at 10 V, and measure the output
voltage vout of the amplifier. What is the cutoff frequency?
Problem 5.59 Determine the frequency response of v0 (ω)/vi (ω) of the circuit of
Fig. 5.15, with Rf = 10 kΩ, R1 = 10 kΩ, C = 100 nF, and a bias voltage of the
±12 V Operational Amplifier.
Problem 5.60 Apply a sinusoidal input signal of 20 V peak–peak to the terminal
Vin . With the help of the oscilloscope, compare the input signal vin against
the output signal vout , varying the frequency of the sinusoidal signal by
1, 10, 100, 103 , 104 , 105 , 106 Hz.
Problem 5.61 Apply a square input signal of 20 V peak–peak to the terminal
Vin . With the help of the oscilloscope, compare the input signal vin against
the output signal vout , varying the frequency of the sinusoidal signal by
1, 10, 100, 103 , 104 , 105 , 106 Hz.
170
Table 5.10 Values of the
electronic components for the
circuit in Fig. 5.34.
5 Elements of Digital Communication
Component
R1
Rf
C
Value
Units
kΩ
kΩ
nF
NOTE: The amplifier is working in mode inverter, so the output is negative
compared to the input
The following materials are required:
• Breadboard
• Resistors and capacitors according to designs
• Amplifier TL084 or LM324
Problem 5.62 Figure 5.34 illustrates an Operational Amplifier configured as an
integrator. This circuit is called Miller Integrator. Calculate the voltage gain
function v0 /vi . Once the cutoff frequency is given, for an integrator with gain DC
equal to 10 dB, cutoff frequency of 10 kHz, and input impedance Z11 of 10 kΩ, the
values of the electronic components are illustrated in Table 5.10, taking into account
commercial values.
Problem 5.63 Using a simulation software, draw the frequency response of the
circuit in Fig. 5.34, either using the Bode plot or using the AC sweep option of the
simulator. Next, we proceed with the assembly of the device. Draw the behavior of
the output signal, when applying a sinusoidal input signal of 10±Vpeak − −peak,
5±kHz to the circuit in Fig. 5.34.
a. Describe the magnitude and phase characteristics of the output signal compared
to the input, detailing the frequency range where a phase delay of 90◦ occurs,
that is, where the integrator is optimal.
b. Suppose the TL084 amplifier of Fig. 5.34 presents an internal offset value
voff set . Fill the Table 5.11 with the values of the gain at different frequency
values in the setup, where the voltage measurements v0 are in Volts peak–peak
and the frequency measurements f are in kHz. The amplitude of the input signal
vi is constant and equal to 10 V peak–peak.
Problem 5.64 Draw the response of the output signal when a square pulse signal
whose voltage value is equal to 10 V peak–peak is applied to Vin to the Miller
integrator. The period of the signal T is the inverse of the frequency, for example, if
the square pulse signal has a frequency of 106 H, the period is equal to T = 1/106 =
10−6 seconds.
Problem 5.65
a. Explain why the value DC that appears at the output of the circuit in Fig. 5.34 can
represent a problem.
5.11 Problems
171
Table 5.11 Assembly results
f (kH z)
0.010
0.050
0.100
0.500
1.0
5.0
10.00
50.00
500.0
v0
20 log10 (v0 /vi )
+
Filter 1
S
Y
Filter 2 R
H
wc
S'
N
Fig. 5.35 Communication system with noisy channel
b. The solution to this problem can be by connecting a 500 kΩ potentiometer in
parallel with the capacitor. This can be checked for a frequency of 100 Hz, by
varying the value of the potentiometer. Determine the behavior of the DC voltage
at that moment.
Problem 5.66 Figure 5.35 illustrates a communication system, which aims to
establish a link between two points, for the proper transmission of a signal.
a. Generate the signals ACarrier and DCarrier, using frequencies in octave
increments, specifically ω0 , 2ω0 , 4ω0 , and 8ω0 being ω0 the fundamental frequency of the signals to be generated.
b. Build two filters: the first to eliminate harmonics in the modulation, and the
second to eliminate out-of-band noise in the receiver. Fill the Table 5.12 with
the results obtained after making the following procedure:
•
•
•
•
Adjust the filter bandwidth values Wl and Wh Wl = 0.001; Wh = 0.999.
Calculate the IIR filter coefficients [b, a] = cheby1(4, 1, [W lW h]).
ZAudio = filter(b,a,YAudio).
ZDigital = filter(b,a,YDigital).
c. For the system in Fig. 5.35, and in each of the signals at their defined frequencies,
determine the following parameters for each of the filters: Quality Factor (Q),
Order of Filter (n), Transfer Function (H (ω)), Root Location in Closed Loop,
and Bandwidth (B).
d. Calculate the power of the signals ACarrier and DCarrier, and assign them
as PT x [ACarrier] and PT x [DCarrier], respectively.
172
5 Elements of Digital Communication
Table 5.12 Calculated values for the audio transmission channel
Wl
0.001
0.001
0.001
0.1
0.2
0.5
0.001
0.001
0.001
Wh
0.999
0.999
0.999
0.999
0.999
0.999
0.5
0.7
0.9
SNRval
10
5
1
10
10
10
10
10
10
RMS(YAudio)/RMS(SAudio)
RMS(YAudioNoi)/RMS(SAudio)
Table 5.13 Calculated values for the digital transmission channel
Wl
Wh
SNRval RMS(YDigital)/ RMS(SDigital) RMS(YDigitalNoi)/RMS(SDigital)
0.001 0.999 10
0.001 0.999 5
0.001 0.999 1
0.1
0.999 10
0.2
0.999 10
0.5
0.999 10
0.001 0.5
10
0.001 0.7
10
0.001 0.9
10
Problem 5.67
a. According to the result obtained in experiment, determine the spectral characteristics of SAudio.
b. According to the result obtained experimentally, determine the characteristics of
the message SDigital.
c. Write in Tables 5.19 and 5.20 the results obtained by varying the parameters.
Problem 5.68 According to the result obtained in Problem 5.67, determine the
variations of SAudio when increasing the noise power.
Problem 5.69 According to the result obtained in Problem 5.67, determine the
variations of SAudio when reducing the bandwidth. Fill the Table 5.13 with the
results of given experiment for this purpose.
Problem 5.70 According to the result obtained in Problem 5.67, determine the
variations of SDigital when increasing the noise power.
Problem 5.71 According to the result obtained in the Problem 5.67, determine the
variations of SDigital when reducing the bandwidth.
5.11 Problems
173
Table 5.14 Assembly results
f (kH z)
0.010
0.050
0.100
0.500
1.00
5.00
10.00
50.00
500.0
v0
20 log10 (v0 /vi )
Table 5.15 Assembly results
f
0.010
0.050
0.100
0.500
1, 000
5, 000
10.00
50.00
500.0
v0
20 log10 (v0 /vi )
Problem 5.72 From results you fill in Table 5.14, describe the magnitude and phase
characteristics of the output signal compared to the input.
Problem 5.73 Table 5.15 illustrates the value of the gain at different frequency
values in the assembly, where the voltage measurements v0 are in Volts peak–peak
and the frequency measurements are in kHz. The amplitude of the input signal vi is
constant and equal to 10 V peak–peak.
Problem 5.74 Determine the values of the components of the circuit in Fig. 5.15
where the operational amplifier behaves as a shunt.
Problem 5.75 Explain if a DC value is presented at the output of the circuit in the
figure and if this represents a problem in the transmission.
Problem 5.76 Thus, the solution to this problem can be by connecting a 500 kΩ
potentiometer in parallel with the capacitor. This can be checked for a frequency of
100 Hz, by varying the value of the potentiometer. Explain the behavior of the DC
voltage in this case. Determine another solution.
Problem 5.77 Describe the magnitude and phase characteristics of the output
signal compared to the input.
Problem 5.78 Table 5.15 illustrates the value of the gain at different frequency
values in the assembly, where the voltage measurements v0 are in Volts peak–peak
174
5 Elements of Digital Communication
and the frequency measurements are in kHertz. The amplitude of the input signal vi
is constant and equal to 10 V peak–peak.
Problem 5.79 A phase modulation system uses a pair of pre-emphasis and deemphasis filters defined by the transfer functions.
Hpe (ω) = 1 + (j ω/ω0 ).
(5.31a)
1
1 + (j ω/ω0 )
(5.31b)
.
Hde (ω) =
Show that the improvement (QoS) in output signal-to-noise ratio produced by using
this pair of filters is
QoS =
.
(B/ω0 )
tan−1 (B/ω0 )
where B is the baseband bandwidth. Evaluate this improvement for the case
when B = 15 kHz and ω0 /(2π ) = 2.1 kHz, and compare your result with the
corresponding value for a frequency modulated system.
Problem 5.80 Show that the figure of merit for the SSB receiver is exactly the same
as that for the DSB-SC receiver. Note that unlike the DSB receiver, the frequency
range of the spectral density function of the narrowband-filtered noise at the front
end of the SSB receiver is offset from the carrier frequency ωc by an amount equal
to B/2, where B is the message bandwidth.
Problem 5.81 Consider a phase modulation (PM) system, with the modulated wave
defined by
s(t) = Ac cos[ωc t + kp m(t)]
.
where kp is a constant and m(t) is the message signal. The additive noise n(t) at the
phase detector input is
n(t) = nI (t) cos(ωc t) − nQ (t) sin ωc t
.
Assuming that the carrier-to-noise ratio at the detector input is high compared with
unity, determine (a) the output signal-to-noise ratio and (b) the figure of merit of
the system. Compare your results with the FM system for the case of sinusoidal
modulation.
Problem 5.82 Complete Table 5.16 with the results obtained, both for the signal
ACarrier and for the signal DCarrier, for each of the generated frequencies.
The loss values Lf , Ls , and Lp are random, so variants should appear in each row
of Table 5.16.
5.11 Problems
Table 5.16 Results on the
inclusion of filters in the
system in Fig. 5.35
175
σN2
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Lf
Ls
Lp
PRx
Sensitivity
Capacity
Fig. 5.36 Baseband transmission
Problem 5.83 From results of Problem 5.82, write a discussion on the convenience
of having an analog and a digital carrier when designing filters.
Problem 5.84 Using the results of Problem 5.82, write a discussion on the variation
of Channel Capacity as the carrier frequency varies (e.g., from the change to two
octaves of frequency ω0 to4ω0 ).
Problem 5.85 Figure 5.36 describes the test bench to be used to evaluate the
transmission of the carrier signal. For this, the two signals ACarrier and
DCarrier are used, both generated by signal generating equipment, adjusted to
a frequency greater than 1 MHz.
Problem 5.86 Using a laboratory trainer, the assembly of Fig. 5.37 is carried out,
which uses the signals ACarrier and DCarrier. The transmission means are
the following:
• A coaxial cable wound with another in a self-induced manner
• A Fiber Optic cable in single-mode configuration
In this case, the gain of the transduction stages of the system and the effect of noise
are not considered. It should be noted that a prior recognition of the transmission
system must be done to carry out this experiment.
176
5 Elements of Digital Communication
Fig. 5.37 Different tests of baseband over optical fiber baseband transmission
Table 5.17 Assembly data
Signal
ACarrier
DCarrier
Coaxial armature
P (A) P (B)
Fiber optic
P (A) P (B)
Problem 5.87 Draw the graph that the oscilloscope records the signal spectrum at
points A and D defined in Fig. 5.37. Fill in the power data in Table 5.17, and calculate
the channel attenuation.
Problem 5.88 Figure 5.38 illustrates the devices that transmit a signal in ASK, for
a pulsed audio signal (encoded of an audio signal) to the input.
a. Describe the operation of the circuit in Fig. 5.38.
b. Write the procedure required for generating an ASK signal.
c. Describe the procedures for calculating and measuring the values of resistances,
capacitances, amplifiers, and other elements, so that they coincide with the design
parameters of an ASK modulator. The simulation must be done with the real
values obtained in the market. Measurement access points must be described
and listed. The simulation can be done with some other simulation and design
program.
Problem 5.89 Complete Table 5.18 with the test points arranged in the circuit and
the measured variable.
5.11 Problems
177
Fig. 5.38 BPSK modulation circuit
Table 5.18 Values measured
at test points
Test Point
TP1
TP2
TP3
TP4
Variable
Problem 5.90 Experiment: FM Modulation The following are the steps to carry
out FM transmission (Acevedo VIllada and García-Álvarez, 2004):
Step 1. Put the transmitter into operation. Select frequency modulation (FM)
using the MODULATION button on the transmitter. Visualize the signal at test
point C of the transmitter on the oscilloscope. A square wave of about 100 kHz
should be observed, which is used as the FM carrier.
Step 2. Now provide the signal generator with a sinusoidal signal of a few kHz
and an amplitude of 3 V peak-peak. Note that when viewing the test point C
of the emitter on the oscilloscope, a stable image cannot be obtained, because the
signal is not of constant frequency. Reduce the frequency to 1 kHz, and observe
the continuous variation of the carrier frequency.
Step 3. Selecting frequency demodulation (FM) using the DEMODULATION
button on the receiver and connecting the oscilloscope to the S1 output of the
receiver, we can see the already demodulated signal.
a. Make a time and spectral plot of the superimposed transmit and receive signal
for messages transmitted in Frequency Modulation, using the values of Wl,
Wh, and SNRval defined above. The figures must have tension and time
scales. Based on this result, determine the differences in using the message
SAudioDem versus SAudio.
b. Draw the signal spectrum, and analytically determine the bandwidth in each of
the stages of the system, if the modulation scheme is:
– FM with modulation index m = 3
– PSK with modulation index m = 1/3
– With modulation index
178
Table 5.19 Calculated
values for the transmission of
audio signal
Table 5.20 Calculated
values for the transmission of
data stream
5 Elements of Digital Communication
Wl
0.001
0.001
0.001
0.001
0.001
0.001
0.1
0.5
0.5
0.001
0.001
0.001
Wh
0.999
0.999
0.999
0.999
0.999
0.999
0.999
0.999
0.999
0.5
0.9
0.9
SNRval
20
20
20
5
5
5
20
20
20
20
20
20
dev
10
50
80
10
50
80
80
10
80
80
10
80
BW
P(SAudioDec)/
P(SAudio)
Wl
0.001
0.001
0.001
0.001
0.001
0.001
0.1
0.5
0.5
0.001
0.001
0.001
Wh
0.999
0.999
0.999
0.999
0.999
0.999
0.999
0.999
0.999
0.5
0.9
0.9
SNRval
20
20
20
5
5
5
20
20
20
20
20
20
dev
10
50
80
10
50
80
80
10
80
80
10
80
BW
P(SAudioDec)/
P(SAudio)
c. Determine the changes that the message SDigitalDem undergoes, against
SDigital.
d. Check that the bandwidth of the modulated signals ZAudio and ZDigital is
greater with respect to the bandwidth in baseband. Determine this proportion.
e. Establish the differences between FM–NB and FM–WB techniques. Determine
the limits of the frequency deviation dev so that ZAudio and ZDigital can be
considered wide band (FM-WB).
f. Determine the procedure to calculate the modulation index, specifying the
parameter that modifies it.
g. Complete Tables 5.19 and 5.20 with the results obtained by varying the channel
parameters.
h. Determine the behavior of the analog signal if the noise power increases in an
FM broadcast. Likewise, when the channel bandwidth decreases. Determine the
bandwidth limit by increasing the frequency deviation.
5.11 Problems
179
i. Determine the changes evidenced by the digital frequency modulated signal:
1. With the increase in noise power
2. With the decrease in bandwidth
Determine the deviation value dev required for a wide band modulation (FM–
WB).
j. When setting up with the Trainer, plot the time and spectrum diagrams of the
frequency modulated signal under the considered parameters, using SAudio as
the modulating signal. The same is illustrated for SDigital.
k. From what is observed in the spectral component, determine the variations in
the SAudio message at the receiver, with respect to the source.
l. From what is observed in the spectral component, determine the variations in
the SDigital message at the receiver, with respect to the source.
m. Increase the amplitude of the SAudio source, and determine the rate of change
in frequency deviation (Hertz/Volt). Explain the behavior.
o. Check whether Carson’s rule of Eq. (5.19) when reviewing the spectral components of the signals passing through the channel, both for SAudio and for
SDigital. Determine the case in which digital transmission requires only the
fundamental frequency of the modulator.
p. Make a complete description of the laboratory procedure with the respective
analysis.
Problem 5.91 The effects of message frequency modulation transmission illustrate
how channel restrictions affect the form of the message, reaching the point of
completely modifying it. First, the effects that were illustrated were due to the
frequency response of the channel and the addition of noise.
a. Using the results, determine the signal differences between the sender and the
receiver points of a Frequency Modulated transmission.
b. Establish the fundamental differences between FM and AM, making a comparison in Bandwidths and Transmission Power.
c. Establish the fundamental characteristics of a PLL and a DPLL.
d. Determine an advantage of upward frequency translation in modulation.
e. Suggest some applications for the use of FM, comparing the modulating signals
SAudio and SDigital.
f. Set the method to calculate the modulation index in FM.
g. Establish the differences between FM–NB and FM–WB.
h. Determine the relationship between the signal-to-noise ratio and the variation of
the modulation index. What variation trend (Increases/Decreases) is the one that
allows an improvement in the SNR?
i. Determine the cases of the parameter dev so that Carson’s rule of Eq. (5.19) is
satisfied. Establish what additional considerations must be made.
Problem 5.92 Consult the effect of the Hartley oscillator on the VCO.
180
5 Elements of Digital Communication
Problem 5.93 Generate the PassBand signal, using the Amplitude Modulated
signal:
s(t) = x(t) cos(ωc t)
.
Decompose the signal SAudio xQ and SDigital into an Euler representation
using the signal ACarrier as the basis vector of functions.
a. Based on the ACarrier signal, generate the signals
.
φI (t) = cos (ωc t) .
(5.32)
φQ (t) = sin (ωc t)
(5.33)
which we will call ICarrier and QCarrier, respectively. The signal φQ (t)
is in quadrature with φI (t).
b. Calculate the Hilbert Transforms of SAudio xQ (t) = H {x(t)}, which will
correspond to the respective signals in quadrature xQ . The analytical signal s(t)
is as follows:
s(t) = x(t)φI (t) + xQ (t)φQ (t),
.
nτ
n − 1τ
<t <
N
N
c. Perform the quadrature operation according to the following equation:
.
⟨x, φ⟩ =
∞
−∞
x(t)φI (t)dt
√
2
d. Calculate the amplitude vector M = X2 +
Y .
Y2
e. Calculate the phase vector Ф = arctan X2 .
f. Draw the polar diagram (M, Ф).
Where τ is the full interval of signal s(t). Declare this signal s(t) as CWPassBand
and DPassBand for next problems.
Problem 5.94 Draw the phase diagram of the in phase xI (t) and quadrature xQ (t)
coefficients, generated from Example 5.93, using SDigital as modulating and
DCarrier as carrier signals.
References
Acevedo VIllada JH, García-Álvarez JC (2004) Laboratorio remoto de comunicaciones. B. Eng.
degree work. Universidad Nacional de Colombia, Manizales
Castillo J, Rodriguez O, Garcia-Alvarez JC (2001) Estudio de modelos de propagación para
comunicación móvil. B. Eng. degree work. Universidad Nacional de Colombia, Manizales
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Schwartz M (1990) Information Transmission, Modulation and Noise. McGraw-Hill, New York
Stremler FG (1989) Sistemas de Comunicación. Alfa-Omega, Rock Rapids
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org/10.1109/IEMBS.2006.260789. http://www.ncbi.nlm.nih.gov/pubmed/17946088
Part II
Multipath Communications
This part is dedicated to the analysis of communication systems that have multiple
channel paths. The analysis includes examining the effects of multipath fading and
interference on signal quality. It also involves designing techniques to mitigate
these effects and improve overall system performance. The multiplexing technique
is used in this type of communication system to ensure reliable and efficient
communication in multiple user environments. This technique allows for the
simultaneous transmission of multiple data streams over a single communication
channel.
Chapter 6
Time–Frequency Channel Assignment
“In theory, all telephone communications from France,
Germany and the United Kingdom could be sent at the same
time using only one optical fiber”
Department of Photonics Engineering. Technological University
of Denmark
6.1 The Multiplexing Model
In Fig. 6.1 the three basic elements of multiplexing are found: (1) the Multiplexer,
which combines the n channels into a single medium; (2) the Transmission Medium,
by which the channels are assigned, and (3) the Demultiplexer, which redistributes
the channels to their respective recipients.
6.2 Time Division Multiplexing
It is the type of multiplexing used especially for digital transmission systems,
especially line schemes (Al-Jamal et al. 2003). In it, the total bandwidth of the
transmission medium is allocated to each channel for a fraction of the total time
(time slot), shown in Fig. 6.2 at TDM-row, by multiplexing low-speed channels
into high-speed channels (Lopez Vasquez and García Álvarez 2004). It can be
synchronous or asynchronous (statistical). The most common is PCM-TDM, where
each channel time-spacing is assigned as .Δt, as it is shown in Figure 6.2.
For digital transmission, each channel can be multiplexed by the following
devices:
• Pulse amplitude modulator (Sect. 2.6.1)
• Quantifier, explained in Sect. 2.6.5.
• Baseband modulator, some cases are illustrated in Sect. 4.5.
Exercise 6.1 A signal has bandwidth .B0 = 8 kHz. The allocated time slot for a
lossless channel becomes .T0 = 1/B0 = 125 μs. Now suppose that you have 10
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024
J. C. García-Álvarez, Digital Electronic Communications,
https://doi.org/10.1007/978-3-031-53118-7_6
185
186
6 Time–Frequency Channel Assignment
Fig. 6.1 General multiplexing model
Fig. 6.2 Time and frequency diagrams of three multiplexing techniques
channels to transmit signals with the same bandwidth. Therefore, each time slot is
divided by the number of channels. Thus, each total time slot is .Ta = T0 /10 =
12.5 μs, and an overall bandwidth of .Ba = 10B0 = 80 kHz.
6.2.1 PAM/PCM
Developed in 1937 in the AT&T laboratories in Paris, it was the most used digital
modulation method in telephone networks. Strictly, it is an encoding method
and not a modulation. It is a digital process that is applied when the average
data transmission capacity of the channel is much greater than what each of
the information sources individually requires to transmit. The transmission of
information from different sources is carried out by the same transmission medium,
but not at the same time. Transmissions from different sources are interleaved in the
time domain.
6.2 Time Division Multiplexing
187
Synchronous Digital Hierarchy It gives each channel a time slot in the highspeed channel. Some characteristics are: constant delays, fast processing, low
bandwidth efficiency. All channels of the multiplexed frame have the same time
duration and the same number of bits, as illustrated in Fig. 6.3. However, if
one of the sources generates data with a higher rate than the others, then the
multiplexer must either subsample the information from that source or more
quickly acquire the information from the other sources to ensure that all rates are
equal. Additionally, if one of the sources stops transmitting, then the multiplexer
must insert padding data to ensure that the frame length is preserved.
Asynchronous Transfer Mode The bandwidth is shared based on demand, when
there is a sample it is put on the output channel and when there is no sample
it goes to the next channel, indicating which channel it belongs to so that the
demultiplexer understands it, it requires more processing and has variable delays
(García-Álvarez and Mejia 2007; Mejia et al. 2005) (Fig. 6.4).
Fig. 6.3 Time division multiplexing
A1
A2
A3
Time, t
B2
| 1| | 1|
| 2| 2| |
| 3| 3| 3|
SDH
B3
| 1| 1| 2|
| 2| 3| 3|
| 3| | |
Time, t
ATM
C1
C3
Time, t
Fig. 6.4 SDH vs. ATM systems
188
6 Time–Frequency Channel Assignment
6.2.2 Overhead
They are unreplicated signals, that is, uniquely encoded so that they are not confused
with information signals. These signals are assigned by the system and should not
be confused with addressing signals (such as the Internet IP address). One of the
functions of these fill bits is to set the clock signal. Thus, by assigning a bit or
character each time period, the clock in the receiver is synchronized to adjust the
data. Compared to sending the clock signal on a different channel, synchronizing
with filler characters only takes a small fraction of the time for each channel. The
G.711 standard gives the guidelines of this type of characters.
6.2.3 Frames
The information data for all channels, synchronization, and overhead compose the
frame. This sequence includes:
•
•
•
•
•
User information and data control
Representation of each signal in channel
Channel bit-rate information
Intersymbol interference and latency alerts
Unique control character for frame start
Example 6.1 A TDM service is composed by three channels of 1200 bit per
second, and one channel of 2400 bit per second. Each channels is provided with
a 10-bit slot. Therefore, a basic bit assignment in the frame would be:
.
|
|
| C1 (10) C4 (10) C2 (10) C3 (10) C4 (10) |
Where .Ci (10) means an assignment of 10 bits for the i-th channel. The reason of
the duplicity of assignment in channel 4 relies on that channel has the double bit
rate of the channels 1 to 3. Each control bit .δi (1) is added one for each channel. The
frame’s setup leads to:
.
|
|
| δ1 (1)C1 (10) δ4 (1)C4 (10) δ2 (1)C2 (10) δ3 (1)C3 (10) δ4 (1)C4 (10) |
Then, the frame length is 55 bits (50 for channels plus 5 for control). The channel
bit rate is .Rch = 1200 × 3 (channels 1, 2, 3) .+ 2400 × 1 (channel 4) .= 6000 bit per
second. The overall bit rate is calculated as follows:
(
)
5
1+
= 6600 bps
.Rb = Rc
50
6.2 Time Division Multiplexing
189
Example 6.2 The following structure is a frame of 1152 bits:
|
|
| M0 (48) CA (48) F0 (48) CA (48) CA (48) F1 (48) |
|
|
| M1 (48) CB (48) F0 (48) CB (48) CB (48) F1 (48) |
|
|
.
| M1 (48) CC (48) F0 (48) CC (48) CC (48) F1 (48) |
|
|
| M (48) C (48) F (48) C (48) C (48) F (48) |
1
D
0
D
D
1
This structure is arranged by adding 24 control bits, leading to 1176 bits. Thus, the
rate efficiency is equal to:
η=
.
1152
= 0.98
1176
The F sequence stands for synchronization purposes .F0 = {01010101}. The M
sequence is taken by the receiver for control after synchronization.
Example 6.3 In a three-port TDM synchronous multiplexer whose bit rates are
1200, 3200 and 4800 bps, data enters and is stored in an elastic memory in the
interval .Tj whose duration is 1 s. The multiplexed signal has a speed of 14.4 k bps
and does not contain synchronization or frame storage characters. In the time
interval .Tj +1 , the sequence generated at the previous instant is analyzed. The
structure of the frame is as follows:
|
|
| |CA CB CC |1 · · · |CA CB CC |1200 |
|
|
. | |∗ CB CC |1201 · · · |∗ CB CC |2400 |
|
|
| |∗ ∗ C |
|
C 2401 · · · |∗ ∗ CC |4800
The nomenclature .|CA |1 stands for the sequence 1 assigned to channel A. Also, .∗
means that this channel does not transmit.
Example 6.4 There is a frame of 38.4 k bps, made up of tributaries as follows:
• A tributary .CA of 9600 bps
• Two tributaries .CB ,.CC of 4800 bps
• m tributaries of 2400 bps
Each character in the frame consists of 10 bits. Each tributary is assigned 20 data
characters and one synchronization character. 2% of the data characters in the
multiplexed signal are used for padding. Tributary characters are also composed
of 10 bits. To calculate the value of tributaries m, it is necessary to determine how
many characters are directly assigned to them. Note that the character rate of the
multiplexed signal is equivalent to dividing the total frame rate by the length of
each character .R = 38 400/10 = 3840 characters per second. Then, the number
of synchronization characters results from the relationship between the character
rate of the multiplexed signal between the number of characters per tributary (20
data plus 1 synchronism), thus .nF = 3840/21 = 182 characters. Therefore, the
synchronization subframe is .F (182). With this result, it is possible to extract the
190
6 Time–Frequency Channel Assignment
number of synchronization characters from that of tributary channels, in effect:
nC = 3840 − 182 = 3658 characters, that is, all tributaries occupy .C(3658).
However, the filler bits must be extracted from the tributaries. With the input data of
2%, we have .nδ = 3658 × 0.02 = 73 characters. In this way, the filler subplot
remains .δ(73). These filler characters are extracted from the tributaries, leaving
.3658 − 73 = 3585 characters. Now, a 9600 bps tributary of 10 bits per character is
assigned .R = 9600/10 = 960 characters per s. Similarly, the other two tributaries
of 4800 bps will take .4800 × 2/10 = 960 characters per second. Extracting these
three tributaries, .3585 − 960(CA ) − 960(CB + CC ) = 1665 characters are available.
Since the transmission rate of the remaining tributaries is 2400 bps, each of these
is assigned, with 10 characters per channel, .2400/10 = 240 characters. Thus, the
number of tributaries m is obtained through the following relationship:
.
m=
.
1665
≈ 6 tributaries
240
The frame is, taking into account only the tributary channels, as follows:
.
|
|
| CA (960) CB (960) CC (960) C1 (240) · · · C6 (240) |
6.3 Frequency Division Multiplexing (FDM)
It is the first technique developed in electronic communications, initially applied
to continuous wave (CW) modulation schemes. Each input channel is assigned a
carrier and thus each source is converted from several that originally occupied the
same frequency spectrum, to a different band of frequencies (different carriers),
and transmitted simultaneously over a single transmission medium. Thus many
relatively narrow band channels can be transmitted over a single broadband
transmission system (Parrado et al. 2007). The main advantage of this multiplexing
technique is its efficiency, because the spectrum of each of the modulated signals
did not present crossover between them (overlapping), thus guaranteeing successful
transmission.
It is used when the bandwidth of the transmission medium is much greater than
the bandwidth of the individual signals to be transmitted. Each of the signals is
modulated with a carrier of a different frequency. To avoid overlapping, a protective
strip is included that separates the signals from each other. Channels are always
assigned, even if there is no data to transmit (Oppenheim 1998). If there are several
signals which can be shifted in frequency in such a way that the spectra do not
overlap, these signals can be transmitted over the same channel in order to make
more use of their bandwidth. The resulting concept is known as frequency division
multiplexing (FDM). In Fig. 6.5 the multiplexing performance can be seen.
6.3 Frequency Division Multiplexing (FDM)
191
Ch 1
1
-
Ch 2
1
2
-
3
-
2
3
FDM Group
Ch 3
Fig. 6.5 Frequency division multiplexing
To transmit messages, carrier waves with different frequencies must be used.
These modulated signals are added and transmitted simultaneously on the same
channel. After they are sent, bandpass filters are used at the receiver to separate
the signals again.
Example 6.5 Assume a commercial AM broadcast system, where the baseband of
the transmitted signals is 5 kHz. The spectrum available for these transmissions is
from 535 to 1605 kHz. How many radio stations could operate?
In AM, the bandwidth of the modulated signal corresponds to twice the
bandwidth of the baseband signal. Therefore, it is possible to multiplex 107
different radio stations using FDM, each with a bandwidth of 10 kHz. Note that this
calculation corresponds to the ideal case in which protection strips are not required.
6.3.1 Asymmetrical Digital Subscriber Line
It is a local communication link between a subscriber and a public network and
uses the available telephone network (copper pair) for data transmission/reception.
It supports broadband signals with frequencies up to 2 MHz (Giraudo et al. 2001;
Céspedes et al. 2004).
192
6 Time–Frequency Channel Assignment
The Asynchronous Digital Subscription Line (ADSL) allows access to digital
data through the cables of a telephone network, which are commonly a twisted pair
of copper wires (Leon 2005).
The growing user demand creates the need to improve the communication system
through the strategy of changing the modulation scheme (Garcia-Alvarez 2002). In
the late 80s Joseph Lechleider proposed a new technology to support data traffic
asynchronously, which would exploit the characteristic of most applications where
much of the information flows from the provider to the subscriber and the least
amount from the subscriber to the provider. Thus, Asymmetric Digital Subscriber
Line (ADSL) technology arises, based on DSL, where the term asymmetric refers to
the characteristic that the channel has of assigning a higher data transmission speed
for downloading data than for uploading, thus making better use of the bandwidth
of each user.
ADSL is a broadband access technique that allows you to have and run highbandwidth multimedia applications (video conferencing, digital television, real-time
video) (Rivera et al. 2003; Ibarra et al. 2007). It operates at frequencies above
the operating frequencies of standard telephony over the same twisted pair, so the
signals resulting from the data transmission process do not interfere with the voice
signals of the telephone channel, thus allowing the transmission of voice and data,
simultaneously on the same channel (Londono Giraldo and García Álvarez 2004).
For data transmission, a multi-carrier modulation technique known as Discrete
Multitone (DMT) is used, which is based on the use of a number of subcarriers in the
same frequency band, each subcarrier has a frequency and it carries information that
is modulated either in QAM or in PSK. Frequency Division Multiplexing (FDM) is
then performed to accommodate the different modulated waves of each subcarrier
in the transmission channel. There is another technique to assign the frequency band
called echo cancellation; this is not widely used since it allows an increase in the
downstream transmission rate but reduces the upload rate and its implementation is
much more complex, which is why it is preferred to use FDM.
Discrete Multitone (DMT) was implemented in the 80s by IBM; it is a multicarrier modulation technique. It is used for ADSL transmission technologies
because great efficiency and great performance are required since in these transmission systems transmission there are many factors that can affect the quality of
the information sent and received such as channel attenuation, noise, or distorted
symbols due to the aforementioned.
The fundamental idea of DMT modulation is to divide the available bandwidth
into a number of orthogonal sub-channels; in other words, they send the necessary
information distributed over a number of subcarriers of different frequencies
sufficiently separated to be distinguished and in each one of which the information
is modulated in QAM (Riche 2012).
Each subcarrier can transmit a certain number of bits, the assignment of the
number of bits per subcarrier is carried out proportionally to the signal-to-noise
ratio (SNR) presented by each sub-channel, thus to the sub-channels with higher
SNR will be assigned a greater number of bits and those with a lower SNR will be
assigned a smaller number of bits.
6.3 Frequency Division Multiplexing (FDM)
193
Voice, POTS
Voice, POTS
1
-
Uplink
Downlink
Uplink
4kHz 25kHz 138kHz
2
-
3
-
1 1MHz
ADSL link
Downlink
Fig. 6.6 ADSL channel distribution
It is important to remember how the available bandwidth in the twisted pair of the
line is divided for ADSL operation; the available bandwidth of .1104 MHz is divided
into 256 sub-channels ranging from 0 Hz to 1104 MHz in baseband. Therefore, each
tone occupies .4.3125 kHz of the total bandwidth. The first tone or sub-channel is
used for traditional voice telephony services (POTS), the tones between .4.3125 kHz
and .25.875 kHz (tones of the 2 to 6) are used to avoid interference between POTS
and ADSL, tones 7 to 32 located between .25.768 kHz and 138 kHz are used for
data transfer in the upstream direction, while tones 33 to 256 located between
138 kHz and 1104 kHz are used for data transfer in the downstream direction.
Fig. 6.6 illustrates the bandwidth distribution of each sub-channel in the ADSL
system. At this point we must clarify that although the frequency range for data
transfer in the up and down direction varies quite a bit, the bandwidth of each subchannel for both processes is not affected, so it will always be 4.3125 kHz.
The QAM modulator is a type of digital modulation that allows 2 waves to
occupy the same bandwidth and therefore can be modulated with the same carrier;
both signals have the same frequency but are out of phase by .90◦ . Therefore, the
carrier signal will be out of phase between one message signal and another, which
finally results in 2 orthogonal channels over the same bandwidth. Because the
resultant contains the amplitude of the cosine and sine waves, then the graph of
a QAM modulation is given by means of a constellation map locating coordinates
(x,y) in the plane. The size of the constellation depends on the number of bits carried
by each subcarrier, b bits results in a constellation with .2b points.
194
6 Time–Frequency Channel Assignment
Fig. 6.7 16-QAM constellation
The maximum size of a constellation for ADSL transmission is 32 768, which
corresponds to sending a maximum of 15 bits on a subcarrier. It is asymmetric
because the frequency band used to upload information is narrower than that
assigned to download. There are three FDM bands for ADSL:
* POTS: Between 0 kHz and 20 kHz it is used for telephony (“Plain Old Telephone
Service”).
* Upstream: Between 25 kHz and 200 kHz it is used to upload data.
* Downstream: Frequencies above 250 KHz are used to download data.
In Fig. 6.7 we can see the example of a 16-QAM constellation diagram that
corresponds to transmitting 4 bits per subcarrier, and in this figure we can see that
the constellation is made up of the amplitudes of the messages Q and I.
Once the concepts of QAM and DMT have been reviewed, we now have an idea
of how the modulation process is carried out for the transmission of information
through ADSL. Figure 6.7 illustrates how both techniques are integrated to allow
transmission data through the telephone channel.
Once the previous concepts are clear, we move on to explain the process of
sending the subcarriers through the same channel called multiplexing. The FDM
technique is used so that the signals obtained from the DMT are combined into one
and sent over a link that has sufficient bandwidth to accommodate and transmit it.
It is possible thanks to the center frequencies of the carriers assigned in the process.
6.3 Frequency Division Multiplexing (FDM)
195
In DMT, the modulated signals are sufficiently spaced in frequency to be able to
combine them without having excessive bandwidth.
The main objective of using FDM is to maximize the use of bandwidth by
ensuring that several signals share the same transmission channel. Additionally, in
the FDM process, guard bands are assigned. These bands are frequencies not used
for transmitting and their function is to separate the channels or signals from each
other, which avoids overlaps and errors when demultiplexing the signal.
The process of demultiplexing the signal is carried out by means of filters that
detect the center frequency of each subcarrier that was assigned to the signals to be
transmitted, which allows the signal transmitted by the channel to be decomposed
and thus obtain the original signals.
Next, the simulation of an ADSL transmission will be done applying DMT,
QAM, and FDM techniques. The channels that need to be managed will be purely
data, allocating 2 channels to upload transmission and 3 channels to download
transmission.
It is required that the upstream and downstream transmission be simultaneous,
that is, full duplex communication over the channel without generating interference
between the signals.
The mathematical modeling of the processes necessary for data transmission in
ADSL is presented below.
As explained above, QAM modulation sends 2 message signals Q and I on the
same carrier; the form of representation of both signals would be the following:
Q(t) = A1 cos(2πfc t + φ)
.
I (t) = A2 sin(2πfc t + φ)
.
The Q and I messages are represented as sine and cosine because these signals
have a phase difference of .90◦ so they meet the condition. The result of modulating
these signals in QAM is the following:
s(t) = Q(t) + I (t)
.
Where we can replace the term .2πfc by .ωc . If .ωc is the bandwidth of Q and I
then the bandwidth of s(t) is .2ωc . From the resulting signal we can take .A1 Q(t) as
the quadrature component of the signal and .−A2 I (t) as the in phase component. In
the frequency domain the modulated signal is expressed as follows:
S(f ) =
.
1
i
[MQ (f − fc ) + MQ (f + fc )] + [MI (f − fc ) − MI (f + fc )]
2
2
Bandwidth efficiency can be calculated by the following equation.
γ =
.
RT
BB
196
6 Time–Frequency Channel Assignment
Where .RT is the Transmission Speed in bits per second, and .BB is the bandwidth
in Hertz. On the other hand, the narrowband noise of the signal is given by:
n(t) = nQ (t)Cos(2πfc t) − nI (t)Sin(2πfc t)
.
Where .nQ is the quadrature component and .nI is the phase component of the
noise. After the QAM modulation process is carried out and sent via subcarriers,
multiplexing is carried out. In this case, all the resulting DMT signals are sent on the
same channel in an adjacent manner. Finally, FDM multiplexing can be expressed
through summation since the signal resulting from this process is the sum of the
modulated signals on each subcarrier.
SM (t) =
n
Σ
.
Xn (t)
i=1
Where x(t) represents a sub-channel of the system.
6.3.2 Experimental Framework
It is proposed to create an interface in Easy Java Simulation that can duplicate the
data transmission process via ADSL. For the experimental development of this
work, it was decided to transmit a 2-bit message, that is, for each transmission
period for each channel there are only 3 states in the message which are described
in Table 6.1. Likewise, it is determined to work with 5 subscriber lines, of which 3
will be Downstream and 2 Upstream. The subcarrier frequency values for each of
the channels are shown in Table 6.2.
To set up the modulation system, it was necessary to generate a coding that was
robust enough to decipher not only the message sent by each user and also include
the information about which user sent said message. A 16QAM modulation scheme
was used that contains enough of symbols to meet the conditions presented; the
coding scheme implemented for this work is described in Table 6.3.
Thereafter, each of the subscriber lines containing useful data is generated
through the main channel, through the discrete multitone (DMT) modulation
scheme making use of the previously mentioned subcarrier frequencies, in this
process each message It is encoded and modulated in a 16-QAM scheme using
phase and quadrature information in such a way that it generates a sinusoidal carrier
Table 6.1 States of digital
input
Digital input
1
0
x
Link
Yes
Yes
No
Sent
1
0
Undetermined
6.3 Frequency Division Multiplexing (FDM)
Table 6.2 Assigned
subcarrier frequencies
197
Channel
Upstream channel 1
Upstream channel 2
Downstream channel 1
Downstream channel 2
Downstream channel 3
Frequency
40 kHz
60 kHz
200 kHz
300 kHz
400 kHz
Table 6.3 ADSL coding scheme
Channel
Downstream channel 1
Downstream channel 2
Downstream channel 3
Upstream channel 1
Upstream channel 2
Downstream channel 1
Downstream channel 2
Downstream channel 3
Upstream channel 1
Upstream channel 2
Data input
1
1
1
1
1
0
0
0
0
0
16-QAM code In phase amplitude Quadrature amplitude
0000
0.5
0.5
0100
.−0.5
0.5
1000
.−0.5
.−0.5
1100
0.5
.−0.5
1110
1
.−0.5
0011
1
1
0111
.−1
1
1011
.−1
.−1
1101
0.5
.−1
1111
1
.−1
in phase and another in quadrature, which when superimposed corresponds to the
signal sent by the user, finally within the channel this signal will be superimposed
with those generated by other users.
Finally, after the transmission process, the detection part is carried out, which
consists of finding the components of the message for each of the subcarriers. This
is achieved by multiplying in parallel the signal received by the carrier in phase
and in quadrature to, the resulting ones are passed through a gain amplifier 2 that
doubles the amplitude of the resulting one and passed through an average filter,
thus obtaining the in phase and quadrature component with which the constellation
diagram is generated.
Differentiated, it is shown in Fig. 6.8 each one of the 5 channels. The difference
relies on the central frequency of each one presented in the experimental framework section. Note that the spectrum is concentrated at the corresponding carrier
frequency. In this way, it can be said that the transmitted information is wrapped
in the carrier frequency. A channel has .4.3125 kHz of bandwidth; this channel is
centered on the fundamental frequency of each subcarrier so the sidebands will be
.fc + 2.15625 and .fc − 2.15625 for the upper and lower subcarriers, respectively.
In this case it can be seen that these do not present any overlap, thus incurring the
successful transmission of data through this technique.
In the constellation diagram shown in Fig. 6.9 you can clearly see the difference
between the own channel symbols and the other surrounding channels. This
guarantees optimal data detection because no type of interference occurs between
the channels.
198
6 Time–Frequency Channel Assignment
Fig. 6.8 Spectral allocation for the possible data inputs
Fig. 6.9 Constellation diagram
6.4 Wavelength Division Multiplexing
Methodology similar to FDM, since multiple light rays of different frequencies
are sent over the same optical fiber. Each color (wavelength) carries different
information. It can be applied to both digital and analog signals, as it is shown
in Fig. 6.10.
Example 6.6 At Bell Laboratories in 1997, a WDM system was developed consisting of 100 light rays traveling on the same optical fiber, each at 10 Gbps, that is, a
total capacity of 1 Tbps. Currently, there are commercial WDM systems with 160
channels each at 10 Gbps.
6.4 Wavelength Division Multiplexing
199
Fig. 6.10 Two signals accessing to a noisy channel
Example 6.7 Near 2000, Alcatel has developed 256-channel WDM systems with a
transmission capacity of 39.8 Gbps per channel, that is, a total capacity of 10.1 Tbps
(Stallings 2000).
r(t) =
K
Σ
.
sk (t) + n(t)
(6.1)
k=1
Exercise 6.2 A sinusoidal signal of frequency .ωc = 104 radian/s is sampled at the
rate of .8 k Hz and then applied to a sample-and-hold circuit to produce a flat-topped
time-multiplexed signal .s(t) with pulse duration .T = 500 μs.
a. Compute the used channel allocation .|S(ω)| , denoting the magnitude spectrum
of the time-multiplexed signal .s(t).
b. Compute the envelope of .|S(ω)| . Hence, confirm that the frequency at which this
envelope goes through zero for the first time is equal to .(1/T ) = 20 kHz.
6.4.1 Fiber Optics
Exercise 6.3 Four data nodes numbered A, B, C, D are linked by an FDDI network
as shown in Fig. 6.11, each by a dual access station (DAS), which are distanced as
follows:
•
•
•
•
Link A–B: 6 km
Link A–C: 4 km
Link B–D: 3 km
Link C–D: 1 km
The characteristics of the fiber used and the transmission requirements are the
following:
• Wavelength: 1300 nm
200
6 Time–Frequency Channel Assignment
Fig. 6.11 Logical connection
of a FDDI network
X
X
X
X
X
X
X
X
•
•
•
•
•
•
•
62.5/125 μm multimode fiber
Numerical aperture 0.275
Attenuation NOT greater than 2.5 dB/km
Maximum distance of cable sections: 2 km
Power required at the output (Tx): [−19, −14] dBm
Power required at the input (Rx): [−31, −14] dBm
Sensitivity −31 dBm
(a) Determine the amount of fiber and the number of repeaters required (number of
splices, if necessary).
(b) Make a quote (price is not required) for the cables, connectors, repeaters, and
DAS necessary to build the FDDI network. (Connections for the redundant
ring are marked with X.) Specifically, the attenuation values of the cables and
connectors, the transmission and reception power of the DAS, and the gain of
the repeaters are required.
(c) What is the difference between a DAS and a dual access concentrator (DAC)?
If a DAC is included in the FDDI network, where would it be located for the
network?
(d) Using the quote data, determine the power transmitted and received at each
station and repeater, and the network architecture.
Exercise 6.4 Four data nodes numbered A, B, C, D are linked by a WDM network
as shown in Fig. 6.12, each by an optical add-drop multiplexer (OADM), which are
distanced as follows:
•
•
•
•
Link A–B: 125 km
Link A–C: 100 km
Link B–D: 60 km
Link C–D: 30 km
The characteristics of the fiber used and the transmission requirements are the
following:
6.5 Optical Frequency Division Multiplexing
201
Fig. 6.12 Logical connection
of a WDM network
•
•
•
•
•
•
•
Wavelengths: 1310 and 1550 nm
62.5/125 μm multimode fiber
Attenuation NOT greater than 1 dB/km @1310 nm and 0.3 dB/km @1550 nm
Maximum distance of cable sections: 80 km
Power required at the output (Tx): [−19, −14] dBm
Power required at the input (Rx): [−31, −14] dBm
Sensitivity −31 dBm
(a) Determine the amount of fiber, the number of repeaters (Erbium Doped Fiber
Amplifiers –EDFA–) and splices (if necessary) required.
(b) Provide a quote (price is not required) for the cables, connectors, repeaters,
and OADM needed to build the WDM network. Specifically, the attenuation
values of the cables and connectors, the transmission and reception power of
the OADMs, and the gain of the repeaters are required.
(c) What is the difference between an OADM and an optical switch (OXC)? If
an OXC is included in the WDM network, where would it be located for the
network?
(d) Using the quote data, determine the power transmitted and received at each
station and repeater, for the two wavelengths (1310, 1550 nm), and the network
architecture.
6.5 Optical Frequency Division Multiplexing
This multiplexing technique has the following steps, illustrated in Fig. 6.13: the
burst of continuous input binary bits are mapped, these are modulated with QPSK
having symbols, each of these symbols are made up of an integer number of bits.
Having a number N of symbols that are in parallel, an inverse Fourier transform
(IFFT) is performed. In order to send them, they need to be in series again as at
the beginning (Álvarez Álvarez and Gómez Muñoz 2014). The information passed
through the channel is received and the serial to parallel conversion is performed,
then the cyclic prefix is eliminated taking into account the size assigned for each
channel, subsequently the fast Fourier transform (FFT) is performed, a conversion
202
6 Time–Frequency Channel Assignment
{ }
-
101010
QPSK
IFFT
Serial/Paralel
Orthogonal MUX
Fig. 6.13 OFDM system
is performed from parallel to serial and finally demodulated in QPSK. Finally you
get the sent message.
6.5.1 Mathematical Modeling
In QPSK modulation, the four different input conditions require more than a single
input bit. With two bits there are four possible conditions: 00, 01, 10, and 11.
Consequently, the input binary data is combined into groups of two bits, called
dibits. Each code dibit generates one of four possible output phases.
VQPSK (t) = Acos(ωc t + φ)
.
(6.2)
Thus for each dibit synchronized in the modulator, a single change in the output is
obtained.
In QPSK transmission, two bits (one dibit) are synchronized in the bit divider.
After both bits have entered in series, they exit in parallel. One bit goes to channel I
and the other to channel Q. The I bit modulates a carrier in phase with the reference
oscillator (“InPhase”), and the Q bit modulates a carrier that is .90◦ out of phase, or
in quadrature, with respect to the reference carrier (Tomasi 2003).
When the linear adder combines the two quadrature signals, there are four
possible resulting phasors.
VQPSK (t) = I cos(ωc t) + Qsin(ωc t)
.
(6.3)
s (t) = sin(ωc t) + cos(ωc t)s2 (t) = sin(ωc t) − cos(ωc t).
(6.4)
s3 (t) = −sin(ωc t) + cos(ωc t)s4 (t) = −sin(ωc t) − cos(ωc t)
(6.5)
. 1
Table 6.4 illustrates the signal output of the QPSK process given two input bits.
The corresponding phase constellation is shown in Fig. 6.14. Finally, the Fig. 6.15
shows the signal generated from a series sequence of the two-bit combinations,
given in time domain.
6.6 Code Division Multiplexing
203
Table 6.4 Signals generated
from pair of bits
Bits
.VQP SK (t)
00
.Asin
01
10
11
)
ωc t − 3π
4 )
(
3π
.Asin ωc t +
4
)
(
π
.Asin ωc t −
4)
(
π
.Asin ωc t +
4
(
Q
Fig. 6.14 Constelation
diagram for QPSK technique
01
11
1
00
10
Fig. 6.15 Output signal after QPSK processing. (Tomasi 2003), Cap 12, pag 485
6.6 Code Division Multiplexing
This multiplexing technique applies the spread spectrum and each input signal is
assigned a unique code. The receiver captures the transmitted signals at the same
time, but thanks to the code it can select the signal of interest (Fig. 6.16).
204
6 Time–Frequency Channel Assignment
Fig. 6.16 CDM system
Fig. 6.17 DSL communication system with channel restricted by noise
6.6.1 2B1Q Modulation
High Speed Digital Subscriber Line (HDSL) technology uses line modulation
scheme. In particular, it uses the 2-Binary-1-Quaternary (2B1Q) technique for
data transmission. Some works illustrated the effects of baseband transmission of
messages (Gagnaire 1997). Specifically, they illustrate how channel restrictions
affect the form of the message, reaching the point of completely modifying it.
Figure 6.17 illustrates a communication system, which aims to establish a link
between two points using 2B1Q, for the proper transmission of a signal. The
2B1Q technique is applied to HDSL technology. Among these parameters are the
transmission power, the number and position of quaternary modulation symbols,
sensitivity, and channel attenuation losses.
6.6.2 Quadrature Amplitude Modulation
Below are some aspects of modulation and demodulation in QAM. In this practice,
in addition to phase modulation, amplitude modulation appears. The tribits are
experienced, with special attention to the tribit that is modulated in amplitude.
6.7 WCDMA
205
Fig. 6.18 Wi-Fi communication system using spread spectrum technique
Fig. 6.19 Digital Video Broadcasting-Terrestial modulation scheme
Direct Sequence Spread Spectrum on Wi-Fi. With the use of this modulation, the
bandwidth required for the signals to be transmitted greatly decreases, however
this can also affect as it can create greater intersymbol interference between the
signals. On the other hand, DSSS encoding is recommended to encrypt a message
since it can only be decoded by someone who has its respective pseudorandom
code and due to the importance of keeping the header data secure, without
requiring much quality, it is a good option for the application. Finally, thanks to
the fact that the subcarriers used when generating this modulation are orthogonal,
information can be transmitted in a more optimal way since a channel can be used
to send multiple messages using frequency multiplexing. This system divides the
.2.4 GHz band into channels from 11 to 22 MHz. The channels overlap partially
(Zarifeh et al. 2007). Figure 6.18 illustrates a communication system, which aims
to establish a link between two points for the proper transmission of a signal.
Usually, a 16-QAM modulation scheme is used.
QAM analysis in a digital broadcasting system. Figure 6.19 illustrates a Digital
Video Broadcasting-Terrestial (DVB-T) system, which aims to establish a link
between two points using 16 or 64 QAM, for proper transmission of a signal
(Jadán Elizalde and Paspuel Revelo 2003).
6.7 WCDMA
Most third-generation (3G) mobile telephone standards are based on the application
of WCDMA (Wideband Code Division Multiple Access) technology. Increases the
data transmission rates of GSM systems using the CDMA (Code Division Multiple
Access) air interface (Fajardo 2004). Compared to previous access technologies,
Time Division Multiple Access (TDMA) and Frequency Division Multiple Access
(FDMA) (García et al. 2003), WCDMA provides greater spectral efficiency,
represented by the separation of messages sent through different channels within
a same bandwidth or link (Machado Aristizabal and García Álvarez 2003).
206
6 Time–Frequency Channel Assignment
Fig. 6.20 Comparison between FDMA, TDMA y CDMA technologies
When based on the CDMA scheme, it has the advantage that all users transmit
simultaneously thanks to the fact that there is no separation in time or frequency
due to the use of the same bandwidth. Thus each user can be discriminated by
assigning a code that identifies them independently. To have greater clarity regarding
the differences between WCDMA technology and its predecessors, Fig. 6.20 is
presented where the change from the frequency and time division to a code division
where the time and the frequency remain constant (Brunagirvent 2014).
However, in order to guarantee the effectiveness of this technology, the optimization of the transmission medium must be carried out. For this reason, multiplexing
has been developed, which is a set of techniques that allows the transmission of
multiple signals through a single link and in one bandwidth (Gonzalez Yamil and
Garcia-Alvarez 2002).
Frequency Division Multiplexing (FDM) brings great advantages to the application of WCDMA technology. This multiplexing process allows the bandwidth of a
given channel to be divided into several independent logical channels. A user can be
added to the system simply by connecting to one of these channels (Taghall 2011).
Due to the greater spectral efficiency offered by this technology compared to
its predecessors due to the use of code division and frequency multiplexing, much
higher data transmission speeds can be achieved in mobile and portable wireless
devices (it can reach speeds of up to 2 Mbps for voice, video, data, and image
transmission) (Morfín 2007).
In the work of Morfin (Morfín 2007), WCDMA is described as a standard divided
into layers (Fig. 6.21): the physical layer, the data link layer, and the RRC (Radio
Resource Control) layer (Garcia-Alvarez and Perez-Camacho 2007).
• Physical layer: It is the lowest layer in the WCDMA protocol model, its purpose
being to condition the digital signal from higher layers so that it can be transmitted through the medium. The physical layer is the place where the coding,
modulation, multiplexing, equalization, bit synchronization, carrier detection,
and transmission mode determination functions are performed (Tanenbaum
2002).
• Data link layer: It is located above the physical layer and is in turn divided into
two layers: MAC layer (Medium Access Control, for its acronym in English) and
6.7 WCDMA
207
Fig. 6.21 WCDMA protocol
layers of Universal Terrestrial
Radio Access (UTRA)
the RLC layer (RLC) layer, for its acronym in English (Bermeo 2014). The MAC
layer provides data to the RLC layer through logical channels; these determine
the type of information that is transmitted. The RLC layer is responsible for
ensuring that the information is transmitted.
• RRC Layer: It carries all the parameters required to establish, modify, and
release the entities of layers 1 and 2. RRC messages carry in their payload all
the signaling of the mobility, connection, and session layers. User mobility is
controlled by RRC signaling through measurements, call transfer, and updates.
Given the importance of WCDMA technology for the development of mobile
communications, the constant demand from users for greater capacity to access the
network contributed to the evolution of this technology. To satisfy this demand,
High Speed Downlink Packet Access (HSDPA) was created, also belonging to the
3G network; highlighting that the latter is the optimization of the first and is known
as 3.5G. In HSDPA, the maximum information transfer capacity is significantly
improved, allowing download speeds of up to 14 Mbps. The development of an
application that simulates a communication system with the principles of WCDMA
technology is focused on the operations performed on the physical layer for this
technology. Three message signals are used to carry out the following three stages:
• Encoding
• Multiplexing
• Modulation
The first stage is done using a Manchester encoding scheme. This strategy in
turn allows Multiplexing. In the final stage, the Modulation QFSK (Quadrature
Frequency-Shift Keying) strategy is implemented. In the results section, the separation of messages into different channels within the same bandwidth is evident, as
mentioned previously.
208
6 Time–Frequency Channel Assignment
6.7.1 Mathematical Modeling
Fig. 6.22 represents the physical layer in the WCDMA scheme, there are three
stages (coding, multiplexing and modulation), which make it possible to send the
three messages simultaneously through a single channel, locating each message in
a portion of the total bandwidth. In order to give the reader an opportunity to reply
the WCDMA scheme, so computational cost could be at minimum, the assigned
frequencies of carriers will be significantly shifted from GHz to Hz.
Coding Stage. In this first stage, Manchester is used. The first Manchester code
was published by G. E. Thomas in 1949 and later by numerous authors such as
Tanenbaum (2002) pages. 274–275. In this type of coding each bit period is divided
into two equal intervals. A binary 1 bit is sent with the voltage high during the first
interval and low during the second. A binary 0 is just the opposite: first low and then
high. This scheme ensures that each bit period has a half transition, making it easier
for the receiver to synchronize with the sender (Tanenbaum 2002).
The Manchester technique is used in order to manipulate the bandwidth of each
message and place each spectrum in a specific band as presented in Fig. 6.23.
The Manchester coding process is developed, which is based on the implementation of an “XOR” gate between the clock signal and the message signal as seen in
Fig. 6.24, said gate has the truth table shown in Table 6.5.
The three messages proposed in the mathematical modeling are encoded,
manipulating the frequencies of the clock signals, in such a way that the spectra
of the coded signals do not overlap as seen in Fig. 6.25.
Fig. 6.22 WCDMA scheme
Fig. 6.23 Increase of bandwidth due to Manchester coding
6.7 WCDMA
209
Fig. 6.24 Signal processing using Manchester coding
Table 6.5 Truth table for
logical “XOR” gate
Input
0
0
1
1
Clock
0
1
0
1
Output
1
0
0
1
Fig. 6.25 Spectrum of coded signals
Multiplexing Stage. In this stage, the sum of the three coded signals is made
in order to unify the three messages to be transmitted over a channel as shown
in Fig. 6.26. The results of multiplexing are adequate if the coded signals are
orthogonal to each other; these results can be verified in the spectrum in Fig. 6.27.
If these spectra are not located properly, there may be loss of information in the
transmission.
210
6 Time–Frequency Channel Assignment
Fig. 6.26 Frequency multiplexing of three orthogonal signals
Fig. 6.27 Signal spectrum generated form multiplexing stage
From the signal that results from the multiplexing stage, the spectrum shown in
Fig. 6.27 is obtained. It can be seen that it coincides with the sum of the spectra in
Fig. 6.25, fulfilling the additivity property of the Fourier transform.
Modulation Stage. After the coding and multiplexing stage, we proceed with the
Modulation strategy QFSK. Due to the multiplexing of the signals and also by
using bipolar signals to perform the Coding, the resulting signal will have 4 voltage
levels, For this reason, the QFSK modulation format is used, which adjusts to the
characteristics of the signal.
6.7 WCDMA
211
Fig. 6.28 Signal processing by QFSK modulation
Fig. 6.29 Signals generated by QFSK, according to the signal level
The general Eq. 6.6 follows a frequency modulation such as:
VQFSK (t) = Ac cos(2πfi t)
.
(6.6)
where .Ac is the amplitude of the carrier signal. The QFSK modulation format
consists of 4 frequency values, these are obtained depending on the voltage of
the modulating signal (.V1 , V2 , V3 , V4 ), such as every symbol or value .v = {vi }
corresponds to a given frequency .ω = {ωi }
In Figs. 6.28 and 6.29 a modulated signal using the QFSK technique is presented.
The change in frequencies of the modulated signal is observed as a function of
the signal amplitude. To modulate signals with more than four amplitude levels, a
different technique must be chosen, for example 8FSK or 16FSK.
The simulation was carried out with signals whose parameters are presented in
Tables 6.6 and 6.7. In addition, the simulation time was 40 seconds, 5 bits are sent
for each channel, the sampling frequency is .3.2 kHz, the number of sampling points
is .131.072, and the modulation frequencies are 65, 70, 75, and 80 Hz.
212
6 Time–Frequency Channel Assignment
Table 6.6 Signal levels
Signal type
Level value
Alphabet
0, 1
Carrier information
0, 1
Coded data
Output data
.−1, 1
.−3, .−1, 1, 3
Channel 2
[10010]
6
0.75 Hz
Channel 3
[01101]
12
1.5 Hz
Table 6.7 Signal parameters
Parameter
Data sequence
Bit normalized cycles
Carrier frequency
Channel 1
[01011]
3
0.375 Hz
Fig. 6.30 Time domain and spectra of clock signals
Fig. 6.30 shows the three clock signals used in coding. The selected frequencies
guarantee that the signals to be transmitted do not overlap when multiplexing is
carried out. The signals over time and their spectra are shown. The fundamental
frequency peak of each carrier is observed in the spectrum, in this case 0.375, 0.75,
and 1.5 Hz, respectively.
In Fig. 6.31 the signals are presented in the time domain; it is evident that for
the coded signals of each message the clock signal is inverted when the message is
zero and that it has only two logical levels (1 and .−1), while the total coded signal
has 4 logical levels (3,1,.−1,.−3). The modulated signal presents frequency changes
depending on the value of the encoded signal. For visualization purposes, only 10 s
of the modulated signal are displayed.
In Fig. 6.32 the spectra of the signals are presented; it is observed that for the
clock signals there is a sequence of Dirac delta whose envelope corresponds to a
cardinal sine of the form .sinc(x), for messages there are components up to 0.5 Hz
and for coded signals there is the spectrum of the message signal aligned with the
spectral components of the clock signal.
In Fig. 6.33 the effect of frequency shift is observed, grouping into spectrum in
each Dirac delta located at the modulation frequencies.
Fig. 6.31 Clock, message and encoded signal in time domain
6.7 WCDMA
213
Fig. 6.32 Spectrum of clock, message and encoded signals
214
6 Time–Frequency Channel Assignment
6.8 Multiplexing Techniques in a 4G System
215
Fig. 6.33 Coded and modulated signal comparison
Fig. 6.34 Communication system with channel restricted by noise
6.8 Multiplexing Techniques in a 4G System
Figure 6.34 illustrates a communication system, which aims to establish a link
between two points using 16-QAM, for the proper transmission of a signal.
The Long Term Evolution technology (LTE) is used for the 4G communications
system. This technology is based on the Orthogonal frequency division multiple
access (OFDMA) technique. A 16-QAM modulation scheme is preferred due to
its advantages such as the use of frequency domain equalizers (Al-Kamali et al.
2016). Among these parameters are the transmission power, the number and position
of symbols of the phase constellation, sensitivity, channel attenuation losses, and
antenna gain. Defining an arbitrary scenario, the analysis of the link between 4G
transmitter and receiver takes into account the distance restrictions defined for the
216
6 Time–Frequency Channel Assignment
standard. The values of received power and link losses can be simulated by different
a simulation program, where the Signal-to-Noise Ratio values, using a power value
for the channel noise, as it is shown in the work of Sabbah et al. (2018). For recent
VoLTE, the link between two points uses 32-QAM, for the proper transmission of a
signal.
6.9 System Assembly
Necessary Elements:
•
•
•
•
Two BNC cables (coaxial)
Two oscilloscope probes
Plastic screwdriver
Bifilar cable
Instrumental Required:
• Oscilloscope
• Signal Generator
6.10 Modulator and Demodulator Operation
In the following experiment, two data terminals are connected with a twisted pair
cable. Once the system is operating, the following variables are displayed:
Transmitter
•
•
•
•
•
Signal input (next)
Antialiasing filter: ON. compressor. OFF(OFF: canceled, led off)
Modulation: QAM
Interference channel (Direct)
Bifilar output
Receiver:
•
•
•
•
Two-wire input
Demodulation: QAM
Signal output
Reconstruction filter: ON. Expander:OFF
Make sure all SW3 microswitches are OFF. Leave the carrier recovery PLL as it
was at the end of the previous lab.
Example 6.8 Generate the carrier signal SAudio, with .Vmax = 2 V at .1 kHz.
Make a graph of the transmitted signal using the 16-QAM modulation scheme.
6.10 Modulator and Demodulator Operation
217
Connect channel 1 of the oscilloscope, with the generator off, to TPE4 (UART
output). Measure with channel 2, the tribits generated, at points TPE17, TPE18, and
TPE19. If any of the tribits do not change level, it is because the input combination
with the generator off produces a constant value of that tribit. Turn on the generator
and verify that the tribit now changes. (NOTE: If you turn off the antialiasing
filter to improve oscilloscope synchronization, remember to turn it back on.)
Measure the time of a tribit. To do this, with the generator on, connect channel 2
to the tribit of TPE17, synchronizing with itself. Taking advantage of the brightness
variations of the oscilloscope beam, and adjusting the intensity, measure the time
elapsed between two consecutive edges of TPE17. Note that the trigger time
comprises the tribit’s elapsed transmission time, allowing the clock generator to
provide the timing required to measure the minimum affordable tribit time window.
6.10.1 Demodulator
The circuit used to demodulate the two-phase tribits of QAM is the same as that
of QPSK, already seen in a previous practice. Therefore we will focus on the
demodulation of the amplitude tribit. Put the generator in TTL mode, and connect
it through the corresponding input (BNC2), selecting TTL with the button on the
transmitter. This way we will ensure that all the tribits are generated and the
visualization will be simple. Before viewing the tribits, make sure that the received
phase is correct: to do this, display TPE4 and TPR36, where you should see the
TTL signals from the generator and the recovered one. Manually search for the
phase with the SW3 microswitches until achieved. When this is done, display points
TPE17 (tribit 1 emitted) and TPR23 (tribit 1 demodulated) and verify that the tribit
demodulation is correct. Do the same with the other phase tribit (tribit 2) measuring
at TPE18 and at TPR24. Points TPR18 and TPR19 are equivalent, respectively, to
TPR23 and TPR24 after passing through a D flip-flop to synchronize the data.
Measure the emitted (TPE19) and received (TPR17) amplitude tribit, and verify
correct demodulation.
To demodulate the QAM amplitude, the signal is passed through an envelope
detector (TPR26) and compared to the average level (TPR25). Visualize the two
points above, setting the mass of the two channels to the same point and setting the
same scale.
Adjusting the channel impairment block without interference or noise gives a
perspective about the signal integrity on noisless conditions. There are two level
modes: the average and the fixed. In this experiment, reselect direct mode on the
transmitter (no channel degradations) and reconnect the function generator so that
it outputs a sinusoidal signal. It is done by nulling the pilot signal.
Note that phase variation affects QAM in the same way as QPSK because in both
cases, four phases are transmitted, requiring a reference symbol. For this extent, by
activating the adequete modulation combination, it would provide the reference with
correct phase.
218
6 Time–Frequency Channel Assignment
Carrier Recovery Malfunction
Visualize points TPE1 and TPR41, with the generator on and making sure that
all SW3 microswitches (automatic phase lock control) are OFF. The recovering of
carrier syncronization leaves to QAM system working correctly.
6.11 Spread Spectrum
This system uses FSK as modulation scheme, thus this is known as FSK-Spread
Spectrum (FSK-SS). This modulation represents the ideal transmission medium for
the energy of the wireless load because this modulation is immune to noise, since all
the energy is transmitted in information through the frequency. An special case is the
Hopping Spread Spectrum (FH-SS), which transmits a part of the data at a certain
frequency during a time interval, the transmission is carried out by jumping from
one frequency to another, in a certain order (pseudorandom sequence), stored in a
table. This is how each piece of information is transmitted at a different frequency
for a very short interval of time. The number of hops per second is regulated by
each country. The United States sets a minimum hop rate of .2.5 per second. This
system changes around 1600 times per second. It is adequate, except when the
transmitter is close to the stations or electrical devices. In this case, transmitter
frequency should not be the same. Spread spectrum technology offers following
fundamental advantages:
• They are highly resistant to noise and interference.
• They are difficult to be intercepted.
• They can share a frequency band with different conventional transmitters with
almost no interference.
The frequency of a conventional wireless signal remains constant over time due
to its high frequency; it is easy to locate to recover information since its bandwidth
is in a known range.
The frequency of the transmitted signal is varied by comparing with a segment
of the spectrum; the frequencies are tuned and the main purpose is to intersect the
signal. Most signals use the digital frequency hopping scheme, where the transmission frequency changes suddenly every second between hops; the transmitter
frequency is stable. The length of time the transmitter remains on a given frequency
between “hops” is known as the dwell time. Some spread spectrum circuits employ
continuous frequency variation, which is an analogous scheme. As advantages, this
technique has the following:
•
•
•
•
It resists either intentional or unintentional interference.
It has the ability to eliminate or mitigate the effect of multipath interference.
You can share the same frequency band (overlay) with other users.
Privacy due to random pseudo code sequence (code division multiplexing).
6.11 Spread Spectrum
219
One of its disadvantages is the non-optimal bandwidth use, because it is necessary
to a wide variation of frequency.
6.11.1 Frequency Hoping Spread Spectrum
Hedy Lamarr (1914–2000) discovered that a set of frequencies could be defined for
each signal or message, just as each set of piano keys produced a distinct tone when
played. She therefore created the frequency hoping (FH) technology, which was put
to use after 1962. The structure of Fig. 6.35 represents the communications system
that is developed for three channels of FHSS modulation, performed for energy
transfer.
Since FSK modulation is used, a digital signal for three different channels is
arriving at the transmitter. In the Matlab program, a function is created, which
requests as input variable the two frequencies of the carrier signal .f0 and .f1 and the
digital signal. Resulting in modulation in each channel, this modulated signal passes
through the first high-pass filter, and white Gaussian noise is added to simulate the
real system of the thermal effect, then it is passed through a second high-pass filter.
The filters are defined in the wireless power transfer system itself. The modulation
power and bandwidth are taken from the modulated signal plus the noise, from
which the power losses in the energy transmission of the wireless charge system
are subtracted, to find the power of the receiver. With this power, the sensitivity and
capacity parameters of the channel are determined. By the superposition principle,
the spectrum of the FHSS modulation behavior is shown, which consists of adding
the spectra of the modulated signals in each channel into a single graph, where the
modulation behavior is shown using an eye diagram.
Thus, for channel one, a digital modulating signal of .Sm = [10] is used since
a FSK-SS modulation is proposed, an amplitude of the carrier signal of 2 with
Fig. 6.35 Communication scheme by each channel
220
6 Time–Frequency Channel Assignment
frequencies of .ω0 /2π = 105 kHz and .f1 = 108 kHz. The performance and the
eye-diagram of the channel are illustrated in Figs. 6.36 and 6.37, respectively.
For channel two, a digital modulating signal of .Sm = [01] is used, a carrier
signal amplitude of 4 with frequencies of .f0 = 110 kHz and .f1 = 114 kHz. The
performance and eye diagram of the channel 2 are illustrated in Figs. 6.38 and 6.39,
respectively.
For channel three, a digital modulating signal of .Sm = [11] is used, a carrier
signal amplitude of 2 with frequencies of .f0 = 116 kHz and .f1 = 120 kHz is
illustrated in Fig. 6.40.
In Fig. 6.41 the modulated signal is a sinusoidal signal that jumps from frequencies .f0 to .f1 , around the center frequency of the carrier at .f 1 − f 0, for which
results in a greater bandwidth (Table 6.1) and therefore a better channel capacity
(Table 6.2). In each channel, a modulated signal is seen that does not present abrupt
changes in its phase or that presents changes less than .90◦ , which proves a narrow
band coherent frequency modulation. Then, because the modulation does not handle
phase or amplitude information but rather the frequency, these figures also show how
this modulation is immune to noise.
Fig. 6.36 Modulation performance of channel 1
Fig. 6.37 Eye diagram of channel 1
6.11 Spread Spectrum
221
Fig. 6.38 Modulation performance of channel 2
Fig. 6.39 Eye diagram of channel 2
Fig. 6.40 Modulation performance of channel 3
It can be seen in Fig. 6.41, how in FHSS modulation the synchronization is
maintained in the frequency jumps, even if the physical channel of a logical level is
changed over time, it is maintained. A single channel through which communication
takes place. From the abrupt jump in the carrier frequencies of the FSK modulation,
222
6 Time–Frequency Channel Assignment
Fig. 6.41 Spectra of signals processed by FHSS
the main objective is to increase the capacity of the system, improve its quality, and
reduce the capacity of missed bits (Sung 2009).
Review Questions
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
Explain the use of guard bands.
Freely, draw how a QAM modulated signal looks like.
How does demultiplexing work in FDM?
What are some FDM Applications?
How does demultiplexing work in CDM?
What are some CDM Applications?
What is channel crosstalk?
What advantages does the TDM method offer and how is it different from
FDM?
Images of the (MUX) input and (DMUX) output signals (Channel1,
Channel2).
Image of the Multiplexed signal.
What causes the noise in DMUX?
What role does the clock signal play for TDM?
What can be observed to vary .At?
What is the difference between TDM standards?
The speed for each input connection is 1 kbps, if the Multiplexing unit is 1
bit, calculate:
1. Duration of each input slot
2. Duration of each output slot
3. Duration of each frame
4. Transmission rate
(continued)
6.12 Problems
223
Example 6.9 Draw the dibits you would get from the following sequence of
bits:
00100011100100
.
Example 6.10 How many bits can be grouped to generate a single symbol?
What advantage does it suppose?
Example 6.11 How many different phases are transmitted for some implemented modulation?
Example 6.12 Determine the parameters required for the adequate recovering of a signal.
6.12 Problems
Problem 6.1 24 speech signals are uniformly sampled and then time division
multiplexed. The sampling operation uses flat peak samples with duration of 1.5 μs.
The multiplexing operation includes the provision of synchronization to add an extra
pulse of sufficient amplitude and also of 1 μs in duration. The highest frequency
component of each voice signal is 4 kHz. Assuming a sampling rate of 10 kHz,
determine the spacing value between successive pulses of the multiplexed signal.
Problem 6.2 For the FDM and TDM communication systems, define:
a. The modulation hierarchy, as the relationship between the modulating and carrier
signal
b. The type of modulator/carrier signal used
c. Applications
Problem 6.3 Determine the transmission parameters of the 16-QAM system,
applied to Wi-Fi technology. Among these parameters are the transmission power,
the number and position of symbols of the phase constellation, sensitivity, channel
attenuation losses, and antenna gain. Set the spectral efficiency.
Problem 6.4 Defining an arbitrary scenario, perform the analysis of the link
between Wi-Fi transmitter and receiver, taking into account the distance restrictions
defined for the Wi-Fi standard. Determine the values of received power and link
losses using a simulation program. Determine the Signal-to-Noise Ratio values,
using a power value for the channel noise.
Problem 6.5 Using a simulation program, add noise to the channel of the LTE
system shown in Fig. 6.34, using a Weibull model with 10 different variance values.
Set the Bit Error Rate and Channel Capacity.
224
6 Time–Frequency Channel Assignment
Problem 6.6 Defining an arbitrary scenario, perform the analysis of the link
between HDSL transmitter and receiver, taking into account the distance restrictions
defined for the standard. Determine the values of power received and losses of the
transmission medium using a simulation program. Determine the Signal-to-Noise
Ratio values, using a power value for the channel noise.
Problem 6.7 Determine the Power Spectral Density (PDS) of the 2B1Q modulation. Find the respective bandwidth efficiency η.
Problem 6.8 Analyze the changes produced by the added noise in the channel of
the DSL system shown in Fig. 6.17, using 10 uniform noise variance measurements.
Calculate the reception error.
Problem 6.9 Describe the algorithm for generating SDigital and the corresponding
transmission using 2B1Q modulation.
Problem 6.10 Plot the waveform of a 2B1Q modulated signal, and its respective
spectral density, using a pseudorandom sequence for the message (modulator
signal).
Problem 6.11 Determine the transmission parameters of the 32-QAM system,
applied to VoLTE system of Fig. 6.34. Among these parameters are the transmission
power, the number and position of symbols of the phase constellation, sensitivity,
channel attenuation losses, and antenna gain. Set the spectral efficiency.
Problem 6.12 Defining an arbitrary scenario, perform the analysis of the link
between VoLTE transmitter and receiver, taking into account the distance restrictions defined for the standard. Determine the values of received power and link
losses using a simulation program. Determine the Signal-to-Noise Ratio values,
using a power value for the channel noise.
Problem 6.13 Add noise to the constellation diagram of the 32-QAM system,
using a Weibull model with 10 different variance values. Set the Bit Error Rate
and Channel Capacity.
Problem 6.14 Determine the transmission parameters of the 64-QAM system,
applied to DVB-T technology. Among these parameters are the transmission power,
the number and position of symbols of the phase constellation, sensitivity, channel
attenuation losses, and antenna gain. Set the spectral efficiency.
Problem 6.15 Defining an arbitrary scenario, perform the analysis of the link
between the DVB-T transmitter and receiver, taking into account the distance
restrictions defined for the standard. Determine the values of received power and
link losses using a simulation program. Determine the Signal-to-Noise Ratio values,
using a power value for the channel noise.
Problem 6.16 Add noise to the 16-QAM constellation diagram, using a Weibull
model with 10 different variance values. Set the Bit Error Rate and Channel
Capacity.
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Chapter 7
Multiple Input Multiple Output Systems
7.1 Multipath
In the 5G scheme, Fig. 7.1 illustrates a communication system, which pretends to
establish a link using a 32-QAM scheme, in order to provide a proper transmission
of the signal.
The system supports S users using the same carrier frequency. The receiver is
equipped with a linear antenna array consisting of L uniformly spaced elements, in
order to achieve user separation in the angular domain (Rappaport 2002), as shown
in Fig. 7.1. Assuming that the channel is non-dispersive, the frequency variatons do
not induce intersymbol interference. The received signal vector becomes .x(k) =
[x1 (k)x2 (k) · · · xL(k)]T at receiver.
7.2 Wireless Networks
The cellular system uses efficient use of available channels with low-power
transmitters by frequency reuse technique. Each base station is assigned a group
of channels to use within a small geographic area. Cell coverage is called footprint
and is determined from field measurements or propagation models. A regular shape
is chosen for design purposes and to avoid gaps or overlaps. It can be quadrangular,
triangular, or hexagonal. Hexagonal is chosen to allow a greater coverage area
and minimize co-channel interference, describing a coverage area similar to omnidirectional antennas. However, omnidirectional antennas are used in the center of
the cell and high-directional antennas at the corners. Let us consider the hexagonal
cell configuration of Fig. 7.2, with a rotated coordinate system .(u, y). The distance
.d12 between a cell with base station coordinate position .(u1 , y1 ) and another cell
with .(u2 , y2 ) is given geometrically by
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024
J. C. García-Álvarez, Digital Electronic Communications,
https://doi.org/10.1007/978-3-031-53118-7_7
227
228
7 Multiple Input Multiple Output Systems
Fig. 7.1 Communication system with channel restricted by noise
Fig. 7.2 Hexagonal cell
wireless network
d12 =
.
x 2 = (u2 − u1 )2 (cos 30◦ )2
.
√
X2 + Y 2
[
]2
Y 2 = (y2 − y1 ) + (u2 − u1 )(sin 30◦ )
When a base station is referenced to the origin of coordinates .(u1 , y1 ) = (0, 0),
d12 =
.
/
u22 + y22 + u2 y2
The inradius (minimal radius) r of each hexagonal cell is the distance between
the center of the hexagon and the middle point of any side of the hexagon. Therefore,
the distance between adjacent cells is equal to the distance between the centers of
adjacent cells. In Fig. 7.2, it is equal to 2r, leading to
√
2r = 2 (cos 30◦ ) R = R 3
.
7.3 Wireless Fidelity
229
7.3 Wireless Fidelity
IEEE 802.11 is a family of standards that lays the foundation for wireless
communication. This standard has been revised since it was introduced in 1997.
The IEEE LAN/MAN Standards Committee (IEEE 802) develops and maintains the
Wi-Fi brand, which is used to produce wireless network equipment. Each version is
largely marketed (Puentes Vargas 2005).
Among them, the 802.11b standard appeared on the market in early 2000, and
it has a maximum payload speed of 11 Mbit/s. The 802.11b standard has higher
performance than the original, which caused its acceptance as the definitive wireless
LAN technology. On the negative side, they suffer interference with products that
operate in the 2.4 GHz (Puentes Vargas 2005) band.
Another important standard is 802.11g, which appears in 2003, and it uses an
OFDM modulation scheme in the payload, but for the header it can use either OFDM
or DSSS (Direct Sequence Spread Spectrum). It is also important to highlight its
hardware compatibility with 802.11b. Accepted in 2003 due to its higher data rates
as well as reduced manufacturing costs, in an 802.11g network the activity of an
802.11b participant will reduce the data rate of the overall 802.11g network. And
also on the negative side they suffer interference with devices that operate at 2.4 GHz
(Puentes Vargas 2005).
There is a growing allocation of standards, as is the case of 802.11n/ac/ad/
af/ah/ai/aj/aq/ax/ay. Then some improvements continued to be made,
such as in 2009 802.11n to which they added the MIMO system. Then in 2013
with 802.11ac, they increased the data speed and channel width and introduced
multiple MIMO users. 802.11ad defines a new physical layer to operate at 60 GHz;
the world’s first router with 802.11ad was announced in January 2016. For 802.11af,
OFDM is used for the physical layer and is based on 802.11ac with improvements
in the data speed. Afterwards, 802.11ah was approved in September 2016 and has
advantageous low-frequency spectrum propagation characteristics.
Finally, there are the .802.11ai/aj/aq/ax/ay that will introduce improvements
in the initial establishment time, allow the detection of devices for the discovery
of pre-association of services, and increase performance by up to four times with
respect to the previous ones.
7.3.1 Mathematical Modeling
This section shows the results obtained through the simulation of a system that
communicates using the IEEE 802.11g standard; for this occasion it was chosen
to use the option in which the DSSS modulation is used for the message header,
which will be discussed. Later and for the payload part, OFDM modulation is used,
which will also be talked about later. Figure 7.3 shows the option to be implemented
(Puentes Vargas 2005).
230
7 Multiple Input Multiple Output Systems
Fig. 7.3 Implementation alternative for the standard
Fig. 7.4 Power density spectrum of different subchannels multiplexed by ODFM (Hall 2006)
7.3.2 Orthogonal Frequency Division Modulation
Frequency division multiplexing is used to transmit several signals over the same
channel. The IEEE 802.11g standard uses orthogonal frequency division multiplexing (OFDM), which is a form of multicarrier modulation. An OFDM signal consists
of a number of closely spaced modulated carriers. For this reason it is necessary
that the receiver be capable of receiving the entire signal so that it can correctly
demodulate the data. As a result, when signals are transmitted close together, they
must be spaced so that the receiver can separate them using a filter and there must
be a guard band between them. This is not the case with OFDM. Although the
sidebands of each carrier overlap, they can still be received without the interference
that might be expected because they are orthogonal to each other. This is achieved
by making the carrier spacing equal to the inverse of the symbol period. In Fig. 7.4
you can see the power spectral density of several carriers transmitted on the same
channel, providing a subcarrier space of .1/TF F T . It gives a several number of null
values, so there is no intersymbol interference.
In time we have a signal with the configuration shown in Fig. 7.5 in which we
can see that there must be a space between symbols to avoid interference.
In an OFDM symbol sequence, the kth OFDM symbol is
gk (t) =
N/2
Σ
.
n=−N/2
(
an,k exp j 2π
n
TF F T
)
t (k − 1)T < t < kT
(7.1)
n/=0
where N is the number of subcarriers, .T = TG − TF F T is the symbol period, and
an,k is the complex value of the nth subcarrier at the kth instant of time.
.
7.3 Wireless Fidelity
231
Fig. 7.5 Time domain values for OFDM signal (Hall 2006)
Fig. 7.6 Wide and narrow
PSD (Poole 2014)
Unbroaded
Broaded
–ƒc
–ƒs
ƒs
ƒc
7.3.3 Direct Sequence Spread Spectrum (DSSS)
Direct sequence spread spectrum is a coding technique that uses a pseudonoise
code to digitally “modulate” a carrier in a way that increases the transmission
bandwidth and reduces the spectral power density (i.e., the level of power at any
given frequency). The resulting signal has a spectrum very similar to that of noise,
so that it will appear to be noise to all radio receivers except the one to which the
signal is directed. This is because if two signals are multiplied in time, a convolution
occurs in frequency, and in this way the resulting signal will have approximately the
same spectrum as the noise. To better illustrate it can be seen in Fig. 7.6.
In part a of Fig. 7.7, it is seen how the message signal .b(t) is multiplied by the
pseudorandom signal .c(t) producing the modulated signal .m(t) = b(t)c(t). Then,
in the transmission channel, an interference associated with RBG denoted as .i(t) is
added so that a signal equivalent to .r(t) = m(t)+i(t) arrives at the receiver as shown
(part b of Fig. 7.7). Finally, in part c of this figure, it is shown that the signal after
it reaches the receiver stops through a multiplier that must be synchronized with
the sender and fed by the same pseudorandom signal of this; in this way we obtain
2
2
.z(t) = c (t)b(t) + c(t)i(t). Taking into account that .c(t) = ±1, then .c (t) = 1;
therefore .z(t) = b(t) + c(t)i(t). Because of the broadband spectral density of the
term .c(t)i(t), signal .b(t) can be obtained once more with the help of a low-pass
filter. When the received signal is multiplied by the pilot signal .c(t), the narrow
band signal can be reconstructed.
232
7 Multiple Input Multiple Output Systems
b(t)
+
m(t)
X
m(t)
r(t)
+
c(t)
i(t)
a)
b)
z(t)
r(t)
Tb
X
v
dt
Decision
0
1 v>0
0 v<0
Threshold (commonly 0)
c(t)
c)
Fig. 7.7 Approximated model of direct-sequence spread spectrum system. (a) Transmitter. (b)
Channel. (c) Receiver (Poole 2014)
Value
Ts
Tc
1
Digital signal (A)
–1
1
Pseudo-randomly Sequence (B)
–1
1
Modulated signal (A)* (B)* (–1)
–1
1
Copy of pseudo-randomly sequence
for reconstruction
–1
1
Recovered signal
–1
t
Fig. 7.8 Signal diagram alongside transmission and receiving process (Poole 2014)
Figure 7.8 shows the behavior of the signal over time for each of the previous
processes.
Since the payload is the main reason for communication through the IEEE
802.11g standard, a multiplexing scheme is attributed to it that allows a high
transmission speed for this data and at the same time offers a high quality of service;
that is, for this reason, the use of OFDM for the work has been determined. Below
is shown, through the use of Matlab®, when transmitting signals with QPSK
modulation through 4 different channels; as a first step, four random signals are
generated that will be transmitted on the channels assigned for multiplexing, as seen
in Fig. 7.9.
7.3 Wireless Fidelity
233
Fig. 7.9 Test signals used for the experiment
Fig. 7.10 Constellation
diagram for the transmitted
signals
Likewise, each signal has the constellation diagram shown in Fig. 7.10, where
the QPSK distribution for the modulation is shown in Fig. 7.11.
In this way, although each channel may have a coincident constellation, each is
transmitted at different frequencies, using 200 Hz as the main one. This guarantees
collisions and intersymbol interference. Having a QPSK modulation with an initial
phase shift of .π/4, the resulting signals are shown in Fig. 7.11.
To have better details of the spectral components for each of the modulated
signals in Fig. 7.11, illustrations of their spectrum are included in Fig. 7.12.
234
7 Multiple Input Multiple Output Systems
Fig. 7.11 Output signals
from QPSK modulation
Fig. 7.12 Spectrum of
signals generated by QPSK
technique
By showing each signal within the same bandwidth, where the signals are
superimposed, we can easily see how this interaction between their spectra appears
in the image in Fig. 7.13.
Once the spectra for each signal are known, they are superimposed on the same
signal, which will be transmitted by the established communication line, in our
specific case, a wireless line. The spectrum of the signal ready to be transmitted
is seen in Fig. 7.13. Figure 7.14a illustrates the pseudorandom behavior of the
transmitted signal graphically, and it serves as adequate for optimizing the allocated
bandwidth usage. Figure 7.14b illustrates the loss of information that occurs during
multiplexing because of the combination of spectral components between the
signals.
Next, the respective demultiplexing, demodulation, and detection of the message
sent by wireless are carried out with the IEEE 802.11g standard. For this, the
7.3 Wireless Fidelity
235
0.9
0.9
Channel 1
Channel 2
Channel 3
Channel 4
0.8
0.7
Normalized Power
Normalized Power
0.7
0.8
0.6
0.5
0.4
0.3
0.6
0.5
0.4
0.3
0.2
0.2
0.1
0.1
0
0
100
200
300
400
500
0
0
600
100
200
300
400
Scaled Frequency
Scaled Frequency
a.
b.
500
600
Fig. 7.13 (a) Spectrum of a QPSK signal, discriminated by channel. (b) Spectrum of the channel
with concatenated QPSK signals
0.45
8
0.4
6
0.35
Scaled Power
Amplitude
4
2
0.3
0.25
0
-2
0.2
0.15
-4
0.1
-6
0.05
-8
0
0.05 0.1
0.15 0.2
0.25 0.3
0.35 0.4
0.45 0.5
0
0
0.2
0.4
0.6
0.8
1
1.2
Scaled Time
Frequency, f, Hz
a.
b.
1.4
1.6
1.8 2
× 104
Fig. 7.14 (a) Signal burst of ODFM wireless transmission. (b) Spectrum of ODFM wireless
transmission
spectrum of the sent signal can be seen, shown in Fig. 7.14, in the image in Fig. 7.14
where the signal is added to a 15 MHz carrier, to be transmitted. The signal is
demodulated to recover the carriers; thus using a low-pass filter and then bandpass with a bandwidth of 100 Hz, with a central frequency located in each of the
harmonics taken for the OFDM, the decoded signals and spectra are shown in
Figs. 7.15 and 7.16, respectively.
Since the demultiplexed signals are now available, all that remains is to find the
phase angles that each of the signals has at the different symbol times established
by the message. The constellation diagram obtained for the signals received through
communication is shown in Fig. 7.17.
Finally, as the angles of the signals shown in Fig. 7.15 were detected, and by
approximating the symbols that each of these represents in the signal, it is possible
to recover the message shown in Fig. 7.18.
236
Fig. 7.15 Recovered signals
from OFDM demultiplexer
Fig. 7.16 Spectrum of the
recovered signal after OFDM
process
Fig. 7.17 Constellation
diagram for received signals
7 Multiple Input Multiple Output Systems
7.3 Wireless Fidelity
237
Fig. 7.18 Recovered message after wireless transmission according to standard IEEE 801.22g
There is evidence of spectral component loss and the introduction of minor
interference by neighboring signals throughout the transmission process due to the
attenuation provided by the wireless environment and the noise that is unavoidably
supplied to the channel. Fig. 7.17 illustrates the efficient message recovery despite
a slight variation from the constellation diagram depicted in Fig. 7.10 due to
transmitted errors.
7.3.4 Message Header
For the header part, a direct sequence spread spectrum (DSSS) technique is used.
In this instance, addressing, encryption, or protocol information are sent using this
particular element. For this reason, a random message created by random is used to
symbolize the head message of the code. For this case we will use a bit time of .0.25
s, we will have a 20-bit message, and for each bit of these, we will use 10 encryption
bits for the pseudorandom sequence.
At the same time, this pseudorandom sequence that is used as a carrier or
encryption can be generated by clock, but the particularity is emphasized that for
each bit of the original message it must have at least 10 change bits to expand the
spectrum according to the IEEE 801.22g standard, as can be seen in Fig. 7.19
The following stage of this procedure involves multiplying the initial transmission signal by another signal that possesses pseudo-random behavior, enabling the
transmitted signal’s spectrum to be broader. The strategy also pertains to security,
238
7 Multiple Input Multiple Output Systems
Fig. 7.19 Data and
pseudorandom DSSS
sequence
1.5
Message
Pseudorandom
Sequence
1
Amplitude
0.5
0
-0.5
-1
Fig. 7.20 (a) Multiplied
output sequence. (b) DSSS
passband signal
Normalized amplitude
-1.5
0
0.5
1
1.5
2
0.5
1
1.5
2
2.5
Scaled time
3
3.5
4
4.5
5
3
3.5
4
4.5
5
1
0.5
0
-0.5
-1
0
2.5
Scaled time
Normalized amplitude
a
0.5
0
-0.5
-1
0.05
0.1
0.15
0.2
0.25
Scaled time
0.3
0.35
0.4
0.45
0.5
b
as anyone who intercepts the signal using unauthorized equipment will receive a
scrambled signal lacking of information. The authorizer receiver is the only one
who can determine the frequencies of the sinusoidal components, even if the sine
wave is always utilized as the carrier, because the encrypted header provides the
time domain multiplication sequence. An illustration of this procedure is shown in
second graph of Fig. 7.20.
Now we proceed to observe the spectrum of the original signal like this in
Fig. 7.21. However, we can also observe the way in which the spectrum of the
original message is expanded or spread since its amplitude can clearly be seen to
decrease almost by half, going from .0.5 to .0.2, and it spreads on the frequency axis
with greater stake. In this way, for the rest of the people it can be perceived as noise
in the baseband, and it should be noted that the more encryption bits used for each
message bit sent, the more the spectrum will widen and the more it will resemble
noise. This can be contrasted in Fig. 7.21.
It is then sent to a carrier frequency of the IEEE Wi-Fi standard in the range
of 2500 MHz using a cosine; such a transfer can be seen in Fig. 7.21b. For the
demodulation process, the first step is to multiply the received signal by the same
cosine of the sending carrier in order to obtain the original spectrum and another
base at twice the carrier frequency, in this case at 5 GHz, as can be seen in Fig. 7.21b.
In this way we use a low-pass filter with a cutoff frequency of 1.25 times the
carrier frequency in order to obtain the baseband signal with pseudorandom code
7.3 Wireless Fidelity
239
Fig. 7.21 (a) Spectrum of baseband digital signal. (b) Spectrum of the signal multiplied by the
pseudorandom sequence. (c) Spectrum of the DSSS sequence through the channel. (d) Spectrum
of the demodulated signal using same carrier
Fig. 7.22 Spectrum of the
signal after the low-pass filter
0.05
0.045
0.04
Scaled power
0.035
0.03
0.025
0.02
0.015
0.01
0.005
0
0
100
200
300
400
500
Scaled frequency
600
700
800
as analyzed in Fig. 7.22. From there we use that received signal and multiply it
by the coding of the received pseudorandom signal to obtain the original signal,
because the squared pseudorandom generates a unitary signal in this case to obtain
the original message. From there we can compare time of the received signal with
the original, and we realize that it is not received in the best way and with changes
240
Scaled amplitude
1.5
Received
Source
1
0.5
0
-0.5
-1
-1.5
0
0.5
1
1.5
2
2.5
Scaled time
3
3.5
4
5
4.5
a
0.5
Scaled power
Fig. 7.23 (a) The transmitted
passband sequence, drawn
over the recovered signal at
the receiver. (b) Spectrum of
recovered signal, drawn over
the spectrum of the passband
sequence
7 Multiple Input Multiple Output Systems
Received
Source
0.4
0.3
0.2
0.1
0
0
0.5
1
2.5
2
1.5
Scaled frequency
b
Recovered
Source
1
0.8
0.6
Normalized amplitude
Fig. 7.24 Recovered signal
using threshold detection.
Baseband digital signal is
drawn over
0.4
0.2
0
–0.2
–0.4
–0.6
–0.8
–1
0
0.5
1
1.5
2
2.5
Scaled time
3
3.5
4
4.5
5
in amplitude per bit, but it is maintained with the required level changes as seen in
Fig. 7.23.
The last step of the demodulation is used in the threshold detector; in this case
for sign detection, for each bit since because we know the bit time, we can use
the average per bit to recover the original sequence. Such recovered signal can be
compared with the original one in Fig. 7.24 and its respective spectrum in the second
graph in Fig. 7.23.
7.4 Visible Light Communication Modulation
The demand for wireless connectivity has continuously grown in the last decade.
Recent market forecasts announce that more than one zettabyte (.1021 bytes) of data
will be transmitted by air each year from 2015 [1]. However, in some countries the
network infrastructure is somehow limited and often unable to satisfy the mobile
data traffic. The widening of the wireless channels via Visible Light Communication
(VLC) leads to a promising potential for next generation networks (Garcia-Alvarez
et al. 2006), being a suitable solution for short-range indoor wireless links, as
in IEEE 802.11 (Wi-Fi). These technologies become able to meet high traffic
demands in applications like data downstream. Acting as a high-efficient energy
device with fast data switching times, the Solid State Lighting (SSL) devices are the
7.4 Visible Light Communication Modulation
241
core component in VLC (Henao et al. 2019). Nonetheless, the following challenges
are derived from the use of this device on VLC:
• Flicker must be minimized into acceptable perceptual visual levels.
• Brightness control (Dimming) should be affordable for energy efficiency.
• A compromise between a constant rate of data transmission and dimming of the
light source has to be achieved.
The general term, visible light communication (VLC), includes any use of the
visible light portion of the electromagnetic spectrum to transmit information. Harald
Haas, a scientist from the University of Edinburgh, United Kingdom, is the one who
gives an idea for working with the visible light spectrum, in addition to accessing
the Internet wirelessly through LED bulbs, calling this technology Li-Fi (Fidelity
Light). Li-Fi, also known as VLC, stands for Visible Light Communications, and it
is the term used to label fast, low-cost wireless communications systems, the optical
equivalent of Wi-Fi.
Harald Haas is known as the father of Li-Fi technology, but research begins in
the countries of Germany, Korea, and Japan where LED technology is developed
in the year 1990, presenting LEDs in three categories: Infrared, Visible light, and
Ultraviolet (Mazumder and Hossain 2014).
More intense research on visible light communication began in 2003 in the
United Kingdom, the United States, and Japan, focusing on LEDs that work in
the light spectrum visible to the human eye, with the purpose of a duality in their
operation, which not only serves as lighting but can also send wireless information
through these devices.
In 2007, with the help of his assistant, Hass sent information through light for the
first time. With small variations in the intensity of the LEDs that are not perceived
by the human eye, Hass sends information within the visible light, modulating,
encoding the information, and sending it at a frequency of 100 MHz (Mazumder
and Hossain 2014).
In October 2011, a group of companies and industry groups formed the LiFi Consortium to promote high-speed wireless optical systems and overcome the
limitations of the radio spectrum by exploiting a completely different part of the
electromagnetic spectrum. This consortium believes that it is possible to reach
speeds of more than 10 Gbps, theoretically allowing a high-definition movie to be
downloaded in 30 seconds.
In 2013, important advances in VLC systems were made. In August, a singlecolor LED transmitter was used to achieve data speeds exceeding 1.6 Gbps.
According to a press release from September, line-of-sight conditions are typically
not necessary for Light Fidelity (Li-Fi) technology, which is a particular use of
VLC systems. It was reported in October that Li-Fi development kits were being
developed by Chinese producers.
VLC communication is the model of IEEE 802.15.7 communication protocols for this working group, which defines the following fundamental characteristics: the physical layer (known by the acronym PHY) and the media access control
242
7 Multiple Input Multiple Output Systems
Fig. 7.25 OFDM channel
(known by the acronym MAC for Media Access Control). This standard is capable
of providing sufficient data speed to transmit audio, video, and multimedia services.
The techniques used in wireless optical communications and that are widely
used in Li-Fi are mainly based on the principle of orthogonal frequency division
multiplexing (OFDM); this technique allows the division of a frequency channel into
several equidistant frequency bands, as shown in Fig. 7.25. Each of them carries a
certain amount of information while maintaining orthogonality in frequency, which
is an important point since it allows there to be no interference in the information.
Starting from the ODFM scheme of Fig. 7.32b, an OFDM signal is expressed
as the sum of pulses exchanging in time and frequency and multiplying by each
symbol. The mathematical notation for how to write the kth symbol is described as
v(t) =
N−1
Σ
.
Xk ej 2kπ t/T ,
0≤t <T
(7.2)
k=0
Donde:
• .Xk symbols
• N is the number of subcarriers.
• T is the time interval given for each symbol.
• . T1 is the spacing between subcarriers.
The orthogonality property is expressed in the following way:
.
1
T
{
f T(
)(
)
1,
ej 2k1 π t/T ej 2k2 π t/T dt =
0,
0
k1 = k2
k1 /= k2
(7.3)
To avoid intersymbol interference (ISI), a guard interval (GI) or a cyclic prefix (CP)
is inserted of length .Tg before the OFDM block. The OFDM signal together with
this GI (Fig. 7.26) is expressed as
7.4 Visible Light Communication Modulation
243
Fig. 7.26 Guard interval (GI)
v(t) =
N
−1
Σ
.
Xk ej 2kπ t/T ,
−Tg ≤ t < T
(7.4)
k=0
Still, intercarrier interference (ICI) can cause orthogonal loss between subcarriers. To solve this problem, the final part of the OFDM symbol is cyclically
extended in the guard interval; in this way, any direct or delayed replica of the signal
will continue to have an integer number of cycles. This ensures the orthogonality
of the different subchannels as long as the delay is less than the chosen guard
interval. Since the reverse process of transmission was carried out and based on the
signal modulated with QPSK in reception, the constellation diagram of each of the
phases of the information was obtained. With these data, the constellation diagram
performs the maximum probability demodulation of the message. Now returning
to the modulated signal and taking into account Orthogonal Frequency Division
Multiplexing (OFDM), it is necessary that we identify the three main channels that
will be multiplexed, we will denote these channels by R (Channel 1), G (Channel
2), and B (Channel 3), and the constellation diagram can be seen for an RGB LED
receiver in Fig. 7.27 (Henao et al. 2019).
244
7 Multiple Input Multiple Output Systems
Fig. 7.27 VLC constellation diagram
7.4.1 Modeling and Simulation of an Adaptive Spatial
Modulation Scheme
For the transmission of data on VLC, most of coding and modulation techniques
focus on increasing symbol transmission rate, leaving out brightness control,
sometimes without guaranteeing the minimum acceptable perceptual human eye
level. Specifically, losing brightness control by increasing transmission speed affects
Quality of Service (QoS) parameters. Various authors proposed the use of channel
multiplexing schemes that maximize light control and transmission rates, and the
optimal scheme model is still an open issue. Since the illumination-level parameters
(such as the perceptual contrast with the daylight or the Quality of Experience
(QoE) of the user) depend on data transmission variations, the double purpose of
lighting and transmission is a challenging task. Therefore, the problem relies on
finding a modulation scheme that supports brightness control, allowing us to keep
transmission rates without degrading the perceptual light levels. Thus, the available
bandwidth/perceptual brightness optimization is possible for VLC systems, using an
adaptive multi-transmitter modulation model, independent of the used SSL (Oweiss
2006; Ojaroudiparchin et al. 2015). A possible approach for a spatial adaptive
distributed coding scheme (such as a spatial and geometric source optimization)
could be multi-channel signal processing, which is employed in multiplexing
techniques including Multiple Input Single Output (MISO). This set of techniques
could achieve the following features: (1) increase of the spectral efficiency through
wavelength modulation by adding number of carriers and (2) a flexible brightness
control by giving bi-orthogonal modulation format control.
7.4 Visible Light Communication Modulation
245
7.4.2 Definition of VLC
Visible Light Communications (VLCs) are in the field of optical communications,
located at the .[400−800] THz band (visible spectra). This topic has been researched
since the Alexander Graham Bell’s photo-phone in 1880. The discovery of SolidState Lighting (SSL) devices as data transmitters, like the Light Emitted Diodes
(LEDs), leaded to the problem of maintaining the illumination levels to an enclosure, without any additional infrastructure or complex devices. A modulator at the
transmitter part performs the switching of the SLL device so that it is imperceptible
to the human eye. Thus, the photodetector picks up the unperceptive variations
in the light energy and transforms it into an electric signal. Early SSL devices as
fluorescent lamps achieved few transmission speeds. They were overcome by the
LED devices, which are able to give faster switching speeds. From different studies
on VLC issues, different standards and innovations come to light, like the case of
Professor Harald Haas at the University of Edinburgh, which in 2011 developed
the Li-Fi standard, sending a 10-Mbps video signal with a single LED (Henao-Ríos
et al. 2021). VLC technology has the following advantages among others: (1) the
use of channel frequencies different from those that are saturated, such as Wi-Fi,
and (2) these channel frequencies share the light-beam of LEDs, so we can create
either a sparse bundle that provides broad coverage or a very thin beam, illuminating
small regions by increasing the beam directive gain (Henao et al. 2019).
As part of the proposal, we consider the transmission of an Orthogonal Frequency
Division Multiplexing (OFDM) signal over some of the different wavelength bands,
using an LED as SSL device. In this wavelength–frequency hybrid modulation
scheme, each symbol .X̂ is first transmitted at time interval .Tx , through the
OFDM scheme .X̂ = xI φˆI + xQ φˆQ , where .φI and .φQ are the} In Phase and}
Quadrature orthogonal basis, respectively. Every nth combination . xI (n), xQ (n)
is called a subcarrier .X̂(n). OFDM standard establishes a limited number of
subcarriers N according to the optimal available bandwidth for the error-free
transmission of symbol. These subcarriers can be graphically measured via phase
constellation, where each subcarrier represents a point of the I–Q plane. Thereafter,
the OFDM-modulated symbol is decomposed into the wavelength bands of the
LED. {In this case, }
an SSL-LED device works as a three-channel signal transmitter
.Ŝ =
φˆR , φˆG , φˆB , corresponding to the chromatic wavelength bands Red (R),
Green (G), and Blue (B), respectively. The weighted combination of these three
components forms a signal that the human eye interprets as White (W) light. So the
frequency/wavelength modulated symbol .Ŷ is transmitted as
[
.
Ŷ =
y(I,R) y(I,G) y(I,B)
y(Q,R) y(Q,G) y(Q,B)
]
⎡
⎤
φˆR
⎣ φˆG ⎦
φˆB
(7.5)
246
7 Multiple Input Multiple Output Systems
Fig. 7.28 Communication system with channel restricted by noise
So, Eq. (7.5) represents each symbol transmitted per subcarrier n at interval .Tx .
Since it creates a hyperspace (or hybrid OFDM) signal, this technology was initially
referred to as orthogonal hybrid division multiplexing, or OHDM. When used with
VLC systems, this methodology maximizes the effective bandwidth on a group of
light transmitters, which are often solid-state (SSL) devices such as LEDs, once
the brightness level necessary to overcome the Signal to Noise Ratio has been
tuned. This methodology is based on an adaptive spatial encoding scheme that
combines multiple input single output (MISO) wavelength division multiplexing
with orthogonal frequency division multiplexing.
7.4.3 Acousto-Optic Modulators (AOMs)
These devices allow the frequency, intensity, and direction of a laser beam to be
modulated. In these devices, incident light is diffracted by the Bragg effect using
acoustic wave fronts that propagate through a crystal. Modulation of incident light
can be achieved by varying the amplitude and frequency of the acoustic waves
traveling through the glass.
Sound waves traveling through glass can be modeled as crests of increasing
refractive index, alternating with valleys of decreasing refractive index. The light
incident in the form of a gradient for a given refractive index is dispersed, so the
light is diffracted from the acoustic wavefronts. In an AOM, light scattered from
successive wave fronts interferes constructively. Figure 7.1 illustrates two light rays
passing over two consecutive wave fronts in the crystal. Note that only part of the
light is scattered from these wave fronts. The optical and acoustic wavelengths are
denoted, respectively, as .λ and .A, while .θi and .θd are the respective angles of the
incident and scattered rays, modulated by acoustic wave fronts. The condition for
constructive interference of scattered light is
nλ = A(sin θi + sin θd )
.
where the modulation index n is an integer. Figure 7.28 illustrates a communication
system, which aims to establish a link between two points using AOM, for the proper
transmission of a signal.
7.4 Visible Light Communication Modulation
247
7.4.4 Implementation of a VLC
For the implementation of the VLC system, the mathematical tool Matlab(R) is
used. The functions that we can perform in the Interface include:
• Generation of pseudorandom binary messages
• Cyclic prefix variation between subcarriers
• Expansion of the figures
We can see the interface in Fig. 7.29.
This section shows step by step the transmission and reception of data in VLC.
First, we proceed to generate a 96-bit pseudorandom binary signal, said signal
includes the information that will be sent, and we can see it in Fig. 7.30.
By having the message signal, QPSK modulation is necessary to be able to
transmit in the medium. Modulating with a carrier frequency .F c = 400 THz, the
message information is at the desired frequencies, as it is shown via the spectrum of
the modulated signal in Fig. 7.31.
By having all subcarriers in the frequency domain, frequency multiplexing is
performed. This happens by adding a cyclic prefix to each of the subcarriers, which
guarantees that they do not overlap each other and also that they are within the
spectrum: visible light (.400 THz − 800 THz).
Figure 7.32 shows the R channel multiplexed in the initial components of
the visible light spectrum. Figure 7.32 shows the multiplexed G channel in the
central components of the visible light spectrum. Figure 7.32b shows the B channel
multiplexed in the final components of the visible light spectrum. Having the three
multiplexed channels, we proceed to add them to obtain the OFDM signal, the signal
said to be between 400 THz and 800 THz; this signal is shown in Fig. 7.32b.
Fig. 7.29 Graphical interface elaborated for simulate a VLC signal tracking. (Source: Henao JL,
Guerrero Gonzalez N, Garcia-Alvarez JC (2019))
248
7 Multiple Input Multiple Output Systems
Fig. 7.30 Pseudorandom sequence of 96 bit lenght, transmitted through the VLC channel
Fig. 7.31 Spectrum of QPSK modulated signal
7.5 Problems
249
Fig. 7.32 Cyclic prefix between subcarriers for (a) channel R, (b) channel G, and (c) channel B.
(d) OFDM signal for all channels
7.5 Problems
The experiment consists of comparing the performance of the three wireless links
considered. To do this, the following measurements are used: signal-to-noise ratio,
eye diagram, link calculation (powers, gain, and noise factor), and constellation
diagram.
Problem 7.1 Determine the transmission parameters of the 64-QAM system,
applied to 5G technology. Among these parameters are the transmission power,
the number and position of symbols of the phase constellation, sensitivity, channel
attenuation losses, and antenna gain. Set the spectral efficiency.
Problem 7.2 Defining an arbitrary scenario, perform the analysis of the link
between transmitter and receiver 5G, taking into account the distance restrictions
defined for the standard. Determine the values of received power and link losses
250
7 Multiple Input Multiple Output Systems
Fig. 7.33 Logic connection
of a TDM network from
station A to station D Wi-Fi
Full–Duplex
Full–Duplex
Full–Duplex
Table 7.1 Budget of link
.A − B
Element
Antenna
Parameter
Link
Power loss
Used bandwidth
Assigned frequency
Sensitivity
Receiver
Value (units)
using a simulation program. Determine the signal-to-noise ratio values, using a
power value for the channel noise.
Problem 7.3 Add noise to the constellation diagram, using a Weibull model with
ten different variance values. Set the Bit Error Rate and Channel Capacity.
Problem 7.4 Wireless Link. Figure 7.33 shows the wireless link .A → D, for a
single frequency of a MHz. The distance between sections is .b km on average. In
addition, according to the relief plane, the reach areas of the link are allowed to
be located. Explain the need to use repeaters in this type of links, relating it to
the sensitivity of the receivers. Fill in Table 7.1 with the measured values of the
proposed link.
Problem 7.5 Using the IEEE 802.11a protocol and a simulation software, draw
the map where you can observe the power analysis on link .A → D and the network
architecture. The report must contain the location map of the stations, regardless of
the city where they are located.
Problem 7.6 Using the quote data, determine the consumption bandwidth efficiency and the relationship .Eb /N0 that can occur in the system IEEE802.11a,
7.5 Problems
251
according to the scheme-related modulation. With a bit error rate constraint of
BER.= 10−5 , determine whether the system meets these requirements.
Problem 7.7 A Pulse-Width Modulated (PWM) signal is used to maintain the
luminosity levels in a switched dimmer-controlled lamp. Simulate the PWM signal,
giving a graphic of the modulating index vs. the spectral efficiency. Use two cases:
(1) a unipolar pulse is sent at the beginning of its period .Tb , so the symbol 1 is
sent in a period .τ1 and the 0 is sent in another period .τ2 < τ1 , and (2) a positionsymmetric pulse is sent, the symbol 1 at the interval .0 ≤ t < (T b/2) and the 0 at
.(T b/2) ≤ t ≤ Tb .
Problem 7.8 An OFDM signal is transmitted, using a PWM signal as carrier.
Simulate the OFDM/PWM constellation and signal spectrum, using as PWM carrier
a unipolar pulse, sent at the beginning of its period .Tb
Problem 7.9 A Light-Emitting Diode (LED) uses three channels called Red,
Green, and Blue, for lighting purposes. Determine the wavelength values of these
channels. Simulate the transmission of a PWM pulse, and draw its spectrum
per channel using Wavelength Division Multiplexing (WDM) at the following
scenarios:
• Each pulse is transmitted for one of the channels, and the other channels remain
idle while the pulse duration is .Tb .
• A pulse triad is transmitted, one pulse for channel, at the same time. Use a
uniform probability function for the generation of pulse widths.
Problem 7.10 A special type of LED can transmit OFDM signals on each light
channel, called Red, Green, and Blue. The transmission is based on Power-Line
Communications (PLCs), where the OFDM digital (Pulsed Modulating/Analog
Carrier) signal is mounted on a pulse of duration .Tb . Simulate the OFDM/PLC
constellation, and evaluate the spectral efficiency, by varying the .Tb parameter.
Problem 7.11 The RGB map is used to determine the power levels for each
color carrier, called Red, Green, and Blue, to obtain certain color. By using the
information of the RGB map, build a program that outputs the amplitude values of
two color carriers to construct a white light, by giving the amplitude value of one of
the carriers.
Problem 7.12 Determine the transmission parameters of the AOM system. Among
these parameters are the transmission power, the number and position of symbols of
the phase constellation, sensitivity, channel attenuation losses, and the gain of the
laser transmitters. Set the spectral efficiency.
Problem 7.13 Defining an arbitrary scenario, perform the analysis of the link
between the AOM transmitter and receiver, taking into account the distance
restrictions defined for the standard. Determine the values of received power and
link losses using a simulation program. Determine the signal-to-noise ratio values,
using a power value for the channel noise.
252
7 Multiple Input Multiple Output Systems
Problem 7.14 Simulate an OFDM constellation, with added Gaussian noise.
Change noise variance in the intervals between .0.1 and .0.9 of the mean energy
of each OFDM symbol.
Problem 7.15 Determine by simulation the modulation index per symbol .ηb of an
OFDM signal, based on the general equation of modulation index:
ηb =
.
max Eb
Ec
Problem 7.16 Diversity vs. Spatial Multiplexing. The matrix .H ∈ R 2 represents
the attenuation of a multipath channel, where every path .hij ∈ H is the attenuation
value of some path between the ith light transmitter and the j th opto-coupling
receiver.
} }
• At the Diversity model, a single symbol .ŝn is transmitted by different paths . hij
at symbol duration .Tn . As advantage, this model increases system reliability.
• At
} the
} Spatial model, different symbols .{sn } are transmitted by different paths
. hij , independent of each symbol duration .Tn , increasing transmission rate.
By simulation, calculate the spectral efficiency of the received signal for each of
the mentioned models. The values of .H are generated using a Rayleigh probability
density function, with the mean values going around the center positions of the
matrix. Compare the results with a matrix .H' generated by a uniform probability
density function.
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Appendix A
Probability Theory
A.1 Conditional Probability
Let A and B two events with joint probability function .P (A, B).
P (A/B) =
.
P (A.B)
P (B)
Then
P (A, B) = P (A/B)P (B) = P (B/A)P (A)
.
Bayes’ Theorem. For mutually excluded events .Ai :
P (Ai /B) = Σ
.
P (Ai , B)
P (B/Ai
=
P (B)
P
(B/A
)P
(A
)
j
j
j
Statistical Independence. Events A and B are statistically independent if:
P (A, B) = P (A)P (B)
.
Therefore, conditional probability is .P (A/B) = P (A)
A.2 Probability Functions
The cumulative (distribution) probability function (cpf) of event variable X is
defined as .FX (x) = P (X ≤ x).
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024
J. C. García-Álvarez, Digital Electronic Communications,
https://doi.org/10.1007/978-3-031-53118-7
255
256
A Probability Theory
P (x1 < X ≤ x2 ) = FX (x2 ) − FX (x1 )
.
The probability density function (pdf) is defined as
pX (x0 ) =
.
dFX (x)
dx
FX (x) =
f x
−∞
pX (u)du
Note that cumulative functions are bounded: .F (−∞) → 0, .F (∞) → 1. Moreover:
P (x1 < x ≤ x2 ) =
f x2
pX (x)dx
.
x1
For a joint probability function of random variables x, y, and z:
f x f y f z
F (XY Z)(x, y, z) =
pXY Z (u1 , u2 , u3 )du1 du2 du3
.
−∞ −∞ −∞
pXY Z (x, y, z) =
.
∂3
F (XY Z)(x, y, z)
∂x∂y∂z
A.3 Statistical Analysis of the Signal
For this type of analysis, we consider that the source X generates a continuous
signal .X = x(t) with probability distribution .pX (x) = P {X ≤ x}, thus considering
that it contains an amount of information. Considering that the occurrence of
some signal value follows a probability distribution, it is possible to calculate
parameters through statistical moments. These characteristics result from the need
to make measurements of these signals, such that they are relevant to the design
of a communication system. These characteristics are expressed by the statistical
moments of the source .x ∈ X, with probability distribution function .pX (x).
Expected Value. Corresponds to the proportional average or average value of the
occurrence of a signal.
E {X} =
f ∞
.
−∞
xpX (x)dx.
(A.1)
Mean Square Value. Determines the average energy value of the probability
distribution. It is applied directly to electrical signals.
{ } f ∞
E X2 =
x 2 pX (x)dx
.
−∞
(A.2)
A
Probability Theory
Variance.
mean.
257
It is the expected value of the squared deviation with respect to the
f ∞(
{ }
)
x 2 − E {X} pX (x)dx = E 2 {X} − E X2
.V ar {X} = σ =
2
−∞
(A.3)
From this result, the variance is equal to the square of the expected value minus
the root mean square value.
Similarly, for a source that generates discrete signal elements .X = x[k], the
statistical moments can be calculated. It will only be illustrated for the expected
value case, as an application of Eq. (A.1):
E {X} =
Σ
.
xi P {X = xi }
(A.4)
i∈X
Variance is calculated by
.
{ }
Var {X} = E2 {X} − E X2
A.4 Continuous Probability Functions
A.4.1 Exponential
Assigned with the function .expo(β), its density function is
{
pX (x) =
.
1
β exp (−x/β) ,
x≥0
0
other case
Furthermore, its distribution function is
⎧
(
)
⎨1 − exp − x
β
.F (x) =
⎩0
si x ≥ 0
(A.5)
(A.6)
other case
This function has the parameters of Table A.1 and the following characteristics:
1. The distribution .expo(β) is a special case of the Gamma and Weibull distributions (shape parameter .α = 1 and scale parameter .β in both cases).
2. If .X1 , X2 , . . . , Xm are independent random variables .expo(β), then the sum of
these variables is a gamma distribution .X1 + X2 + · · · + Xm ∼ .gamma(m, β),
which is called the Erlang-m distribution.
258
A Probability Theory
Table A.1 Properties of function .expo(β)
Function
Parameter
Range
Mean
Variance
Mode
.Expo
.Gamma
Scale .β > 0
.[0, ∞)
.β
2
.β
0
Shape .α > 0, Scale .β > 0
.[0, ∞)
.αβ
2
.αβ
.β(α − 1) si .α ≥ 1, 0 si .α < 1
.Weibull
3. The exponential distribution is the only continuous distribution with the property
of being memoryless.
A.4.2 Gamma
The Gamma function, which we define here as .gamma(α, β), follows the probability
density function:
pX (x) =
.
⎧
)
(
⎨ β −α x α−1 exp − x ,
r(α)
⎩0,
β
x>0
(A.7)
other case
f∞
where .r(α) is the gamma function, defined by .r(z) = 0 τ z−1 eτ dτ for any real
number .z > 0. The following properties of the Gamma function are: .r(z + 1) =
zr(z) for any value .z > 0, r(k + 1) = k! for any nonnegative integer .k, r(k + 12 ) =
√
√
π · 1 · 3 · 5 · · · (2k − 1)/2k for any positive integer k, and .r(1/2) = π . The
distribution function cannot be expressed in terms of usual mathematical functions
when .α is not an integer. For the case where .α is an integer, the gamma distribution
function is calculated as follows:
⎧
(
)
( )j
x α−1
⎪
⎨1 − e − β Σ 1 x ,
x>0
j! β
(A.8)
.F (x) =
j =0
⎪
⎩
0,
other case
With the parameters of Table A.1, the gamma function has the following characteristics:
1. The distributions .expo (β) and .gamma (1, β) are identical.
2. For a positive number m, the distribution .gamma (1, β) is called the distribution
.m − Erlang (β).
3. The chi-square distribution with k degrees of freedom is equivalent to the
.gamma (k/2, 2) distribution.
A
Probability Theory
259
4. If the random variables .X1 , X2 , . . . , Xm are independent, following functions
.Xi ∼ gamma(α, β), then the sum of the random variables is equivalent to .X1 +
X2 + · · · + Xn ∼ gamma(α1 + α2 + . . . + αm , β).
5. If .X1 and .X2 are independent random variables that follow .Xi ∼ gamma(α, β),
then the random variable .X1 / (X1 + X2 ) ∼ beta(α1 , α2 ).
6. A random variable .X ∼ gamma(α, β) if and only if the random variable .Y =
1/X follows a type five Pearson distribution of shape parameters .α and of scale
.1/β, noted as .PT5(α, 1/β).
7. The limit of the density function .pX (x) for gamma is
⎧
⎪
⎪
⎨∞,
1
. lim pX (x) =
β,
⎪
x→0
⎪
⎩0
α<1
α=1
if α > 1
A.4.3 Weibull
The parameters of the Weibull function are found in Table A.1. The following are
some features of this feature:
1. The distribution functions .expo(β) and .Weibull(1, β) are equivalent.
2. A random variable is .X ∼ Weibull(α, β) if and only if .Xα ∼ expo (β α ).
3. The natural logarithm of a Weibull random variable has a distribution known as
the extreme value distribution or Gumbel distribution.
4. The distribution .Weibull(2, β) is also called a Rayleigh distribution with parameter .β, .Rayleigh(β). If the random variables .Y, Z are normally
with
(√distributed
)
√
2β .
zero mean and variance .β 2 , then .X = Y 2 + Z 2 ∼ Rayleigh
5. The limit of the density function .pX (x) for Weibull is
⎧
⎪
⎪
⎨∞,
1
. lim pX (x) =
β,
⎪
x→0
⎪
⎩0,
α<1
α=1
α>1
A.4.4 Normal
)
(
The Normal function, expressed as .N μ, σ 2 , has as applications the estimation
of various sources of error, which we will later denote as noise, in addition to the
sum of large numbers of other quantities, by virtue of the central limit theorem. The
260
A Probability Theory
)
(
Table A.2 Properties of functions .N μ, σ 2
Function
Parameters
Range
Mean
Variance
Mode
.N
.LN
Location .μ ∈ (−∞, ∞), scale .σ > 0
.(−∞, ∞)
.μ
2
.σ
.μ
Shape .σ > 0, scale .μ ∈ (−∞, ∞)
.(0, ∞)
(
)
μ + σ 2 /2
)
[
( )
]
(
2 exp α 2 − 1
.exp 2μ + σ
)
(
2
.exp μ − σ
.exp
Normal density function is
f (x) = √
.
(
1
2π σ 2
e −(x−μ) /2σ
2
2
)
−∞<x <∞
(A.9)
However, it does not have a closed-form distribution function, called the error
function, so it is calculated numerically. The parameters are found in Table A.2.
The following are normal function characteristics:
1. If two normal random variables with joint distribution are uncorrelated, they are
also independent.
2. Assuming that the joint distribution of the random variables
) .X1 , X2 , . . . , Xm is
(
multivariable and having .μi E {Xi } and .Cij = Cov Xi , Xj the expected value
and covariance, respectively, for any real number .a, b1 , b2 , . . . , bm , the random
variable resulting from the linear combination .a + b1 X1 + b2 X2 + . . . + bm Xm
has a normal distribution with mean and variance, respectively:
μ=a+
m
Σ
.
bi μi ;
σ =
2
m
m Σ
Σ
bi bj Cij
i=1 j =1
i=1
Note that the independence of the random variables X is not assumed. As the
previous paragraph expresses, if the random variables are independent:
σ2 =
m
Σ
.
bi2 Var (Xi )
i=1
3. The distribution .N(0, 1) is usually called the standard or unit normal distribution.
4. If the random variables .X1 , X2 , . . . , Xk are independent standardized normal
functions, then the random variable .X12 + X22 + · · · + Xk2 has a chi-square distribution with k degrees of freedom, which
is)also the distribution .gamma(k/2 , 2).
(
5. If the random variable is .X ∼ N μ, σ 2 , the random variable .eX has a log)
(
normal distribution with parameters .μ and .σ , which is called .LN μ, σ 2 .
6. If the random variable .X ∼ N(0, 1), furthermore if the random variable Y
has a chi-square distribution with k degrees of freedom, these variables being
A
Probability Theory
261
√
independent, then .X/ Y /k has a Student-t distribution with k degrees of
freedom.
7. If the normal distribution is used to represent a nonnegative signal, for example
time, the density function must be truncated at .x = 0.
8. As .σ → 0, the normal distribution degenerates toward the value .μ.
A.4.5 Log Normal
)
(
The function denoted as .LN μ, σ 2 is used to estimate the execution time of
some task, similar to the functions .gamma(α, β) and .Weibull(α, β). However, this
function has a peak value close to .x = 0, which makes it useful for certain
operations. It is also used to estimate quantities that are the product of a large
number of other quantities, due to the central limit theorem. Without closed form of
distribution function, the log-normal density function is
{
f (x) =
.
√1
eg(x,σ,μ) ,
x 2π σ 2
x>0
0,
otherwise
g(x, σ, μ) =
.
−(ln x − μ)2
2σ 2
The parameters of the function are found in Table A.2 and have the following
characteristics:
)
)
(
(
1. A random variable .X ∼ LN μ, σ 2 if and only if .ln X ∼ N μ, σ 2 . Thus, if
we have the samples .X1 , X2 , . . . , X2 . . . Xn supposedly as log-normal functions,
the logarithm calculated on these samples is .ln X1 , ln X2 . . . , ln Xn that can
be treated as normal-distributed in order to throw an hypothesis about some
distribution, parameter estimation, or test of adjustment and confidence.
2. As .σ → 0, the log-normal distribution degenerates to the point .eμ . Then, the
log-normal densities for small values of .σ have a sharp peak in the mode.
3. .limx→0 f (x) = 0, independent of the value of the parameters.
A.4.6 Beta
This function is described as .beta (α1 , α2 ) and is used as a raw model in case of the
absence of data. It is also used for preliminary estimation of the distribution of a
random variable. Its density function is
262
A Probability Theory
{ α1 −1
x
pX (x) =
.
(1−x)α2 −1
,
B(α1 ,α2 )
0,
0<x<1
else case
where .B is the beta function, defined by
B (z1 , z2 ) =
f 1
.
τ z1 −1 (1 − τ )z2 −1 dτ
0
for any real number .z1 > 0 and .z2 > 0. The following are some properties of the
beta feature:
B (z1 , z2 ) = B (z2 , z1 ) ,
.
B (z1 , z2 ) =
The beta function has the distribution function:
⎧
⎪
⎪
⎨0,'
B (x,α1 ,α2 )
.FX (x) =
B(α1 ,α2 ) ,
⎪
⎪
⎩1,
r (z1 ) r (z2 )
r (z1 + z2 )
x<0
0≤x≤1
x>1
where .B' is the incomplete beta function, which is calculated as
B' (x, z1 , z2 ) =
f x
.
τ z1 −1 (1 − τ )z2 −1 dτ
0
This integral is calculated numerically. The properties of this function are given in
Table A.3.
Table A.3 Properties of the function .beta (α1 , α2 )
Parameters
Range
.[0, 1]
Mean
.
α1
α1 +α2
Variance
.
α1 α2
(α1 +α2 )2 (α1 +α2 +1)
Mode
Shape .α1 > 0 and .α2 > 0
⎧
α1 −1
⎪
⎪
α1 +α2 −2
⎪
⎪
⎪
⎪
⎪ 0 and1
⎨
.
0,
⎪
⎪
⎪
⎪
1,
⎪
⎪
⎪
⎩ no unique value,
α1 > 1, α2 > 1
if α1 < 1, α2 < 1
(α1 < 1, α2 ≥ 1) ∨ (α1 = 1, α2 > 1)
(α1 ≥ 1, α2 < 1) ∨ (α1 > 1, α2 = 1)
α1 = α2 = 1
A
Probability Theory
263
A.5 Discrete Probability Functions
A.5.1 Binomial
This function has as density (The properties of this function are given in Table A.4)
pX (x) =
.
⎧( )
⎪
⎪
⎨ τ
⎪
⎪
⎩0,
x
px (1 − p)τ −x ,
x ∈ {0, 1, . . . , τ }
other case
( )
τ
where .
is the binomial coefficient, defined as
x
.
( )
τ!
τ
=
x
x!(τ − x)!
Its distribution function is defined as
⎧
⎪
0,
⎪
⎪
( )
⎪
⎨ [x]
Σ τ
.F (x) =
p ' (1 − p)τ −i ,
⎪
i
τ
=0
⎪
⎪
⎪
⎩
1,
x<0
0≤x≤t
t <x
A.5.2 Poisson
The function noted as .Poisson(λ) is used to estimate the number of events that can
occur in a time interval as long as the events occur at a constant rate. The density
function is calculated by
Table A.4 Properties of functions .bin (t, p)
Function
Parameters
Range
Mean
Variance
Mode
.bin
.Poisson
Positive integer n, .p ∈ (0, 1)
.λ > 0
.{0, 1, . . . , n}
.{0, 1, . . .}
.τp
.λ
.τp(1 − p)
.p(τ + 1) − 1 y .p(τ + 1)
.[p(τ + 1)] other case
.λ
si p(t+1) is a integer
.λ − 1 y .λ if .λ is integer
.[λ] other case
264
A Probability Theory
{ x
pX (x) =
.
λ
x! exp (−x)
si x ∈ {0, 1, . . .}
0
otro caso
The distribution function is equal to
F (x) =
.
⎧
⎪
⎨0,
x<0
⎪
⎩exp (−x)
[x]
Σ
i=0
λi
i! ,
0≤x
A.6 Maximum Likelihood Estimation
The following are the parameter estimators according to the observed data:
Exponential Function .β̂ = X̄(n).
Gamma Function It must satisfy the following equation system:
1Σ
ln Xi
n
n
ln β̂ + W(α̂) =
.
(A.10)
i=1
α̂ β̂ = X̄(n)
The bigamma function is .W(α̂) = r ' (α)/ r(α), with .r ' the derivative
of gamma function. The estimation of parameter .α can be calculated by the
Newton–Raphson method, where, for some iteration .k = 1, 2 . . .,
α̂[k] = α̂[k − 1] −
.
ξ ' (α̂[k − 1])
ξ '' (α̂[k − 1])
where
ξ ' (α) = n ln
.
ξ '' (α) =
(
α
X̄
)
+ nW(α) +
n
Σ
ln Xi
i=1
n
− nW ' (α)
α
and .W ' (α) is the derivative of bigamma function. Note that second equation in
the equation system (A.10), .β̂[k] = α̂[k]/X̄(n).
• Weibull function: The first can be solved for .α̂ by Newton’s method, and
the second directly returns the value .β̂. The general recursive iteration for this
method is
A
Probability Theory
265
α̂[k + 1] = α̂[k] +
.
A + 1/α̂[k] − Ck /Bk
2
1/α̂ [k] + (Bk Hk − Ck 2) /Bk2
where the coefficients .A, B, C, H are calculated as follows:
1Σ
ln Xi ,
n
n
A=
.
Bk =
n
Σ
i=1
Hk =
n
Σ
Xiα̂[k] ,
Ck =
i=1
n
Σ
Xiα̂[k] ln Xi ,
i=1
Xiα̂[k] (ln Xi )2
i=1
You can use the work of Thoman, Bain, and Antle (Thoman et al. 1969) to
obtain these formulas, as well as the confidence intervals to find the exact value
of .α and .β. As a starting point for the iterations, the following estimate can be
used:
⎧
⎨
⎡
)2 ⎤⎫−1/2
( n
n
⎬
Σ
Σ
1
6
2
⎣
⎦
.α̂[0] =
−
ln
X
X
(ln
)
i
i
⎩ (n − 1)π 2
⎭
n
i=1
i=1
This estimate is the work of Menon (1963) and suggested in the work of
Thoman, Bain, and Antle (Thoman et al. 1969). With this choice of .α̂[0],
Thoman, Bain, and Antle (Thoman et al. 1969) reported that only an average
of .3.5 Newton iterations are required to achieve a precision of four significant
figures.
• Normal distribution.
/
n−1
Svar(n)
.μ̂ = X̄(n),
σ̇ =
n
where .Svar(n) is the variance of the samples.
Log-Normal Function.
1Σ
ln Xr ,
.μ̂ =
n
n
i=1
r
| n
|1 Σ
σ̂ = ]
(ln X1 − μ̇)2
n
i=1
Beta Function. To estimate the parameters, Gnanadesikan, Pinkham, and
Hughes (Gnanadesikan et al. 1967) established that the following two equations
must be satisfied:
(
)
( )
W α̂1 − W α̂1 + α̂2 = ln G1 ,
.
(
)
( )
W α̂2 − W α̂1 + α̇2 = ln G2
where .W is the digamma function and the coefficients are calculated by
266
A Probability Theory
G1 =
[ n
| |
.
i−1
]1/n
Xi
,
G2 =
[ n
| |
]1/n
(1 − Xi )
i−1
Note that .G1 + G2 ≤ 1. Beckman and Tietjen determined a numerical method to
solve these equations.
• Binomial. If .τ is known, then .p̂ = X̄(n)/τ . If both .τ and p are not known, .τ̂
and .p̂ exist if and only if .X̄(n) > (n − 1) Svar2 (n)/n = V (n).
References
Gnanadesikan R, Pinkham RS, Hughes LP (1967) Maximum likelihood estimation of the
parameters of the beta distribution from smallest order statistics. Technometrics 9(4):607–
620. https://doi.org/10.1080/00401706.1967.10490509, http://www.tandfonline.com/doi/abs/
10.1080/00401706.1967.10490509
Thoman DR, Bain LJ, Antle CE (1969) Inferences on the parameters of the Weibull distribution. Technometrics 11(3):445. https://doi.org/10.2307/1267019, https://www.jstor.org/stable/
1267019?origin=crossref
Appendix B
The Fourier Transform
The Fourier transform is a linear mapping between time space and frequency space
and vice versa.
f (t) ⇔ F (ω)
.
The Fourier transform is a tool for analyzing signals in frequency. In practice, the
Short Fourier Transform (STF) is usually used to estimate frequency components
from a finite sweep of the signal:
• Nonintuitive analysis. The graph of the spectral coefficients is known as the
Fourier spectrum of the signal.
• Allows direct frequency and phase analysis.
• Many complicated calculations in the time domain are often simple in the
frequency domain.
Time
Convolution
Offset
.⇔
.⇔
.⇔
Frequency
Multiplication
Scaling
• Fourier transform is an infinite series that approximates a periodical and continuous function.
• It is used in signal analysis to represent the sinusoidal components of a
nonsinusoidal periodic wave.
• Any periodic function on an interval T can be represented as a linear combination
of harmonically related sines and cosines.
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024
J. C. García-Álvarez, Digital Electronic Communications,
https://doi.org/10.1007/978-3-031-53118-7
267
268
B The Fourier Transform
f (t) ≈
.
) Σ
)
(
(
∞
∞
A0 Σ
2π nt
2π nt
+
+
An cos
Bn sin
2
T
T
n=1
n=1
• Fourier transform:
f+∞
.F (ω) =
f (t) e−j ωt dt
−∞
• Inverse Fourier Transform:
1
.f (t) =
2π
f+∞
F (ω) ej ωt dω
−∞
The following are the Fourier transforms for continuous signals.
B.1 Properties of the Fourier Transform
The following are the properties of a spectral density function:
1. Complex Function:
F (ω) = |F (ω)| ej θ(t)
.
2. Condition of Existence:
f+∞|
|
|
|
.
|f (t) e−j ωt | dt < ∞
−∞
3. Symmetry:
f (t) ⇔ F (ω)
.
4. Linearity:
a.g (t) + b.f (t) ⇔ a.G (ω) + b.F (ω)
.
B
The Fourier Transform
269
5. Scaling:
f (at) =
.
1 (ω)
F
|a|
a
6. Frequency Shift:
f (t) ej ωc t ⇔ F (ω − ωc )
.
7. Displacement in Time:
f (t − t0 ) ⇔ F (ω) .e−j ωt0
.
8. Differentiation:
.
df (t)
⇔ j ωF (ω)
dx
9. Integration:
ft
.
f (τ ) dτ ⇔
−∞
1
F (ω)
jω
10. Convolution in Time:
f (t) ∗ g (t) ⇔ F (ω) G (ω)
.
11. Convolution in Frequency:
f (t) g (t) ⇔
.
1
[F (ω) ∗ G (ω)]
2π
B.1.1 Fourier Functions
F {cos ω0 t} = π [δ(ω + ω0 ) + δ(ω − ω0 )]
.
F {sin ω0 t} = j π [δ(ω + ω0 ) − δ(ω − ω0 )]
F {f (t) cos ω0 t} =
1
[F (ω + ω0 ) + F (ω − ω0 )]
2
270
B The Fourier Transform
F {f (t) sin ω0 t} =
j
[F (ω + ω0 ) − F (ω − ω0 )]
2
B.1.2 Fourier Multipliers
A detailed information can be found in Mattila (2015). Let .m ∈ L∞ (Rd ) and .f ∈
L1 (Rd ). Consider the linear operator
F {Tm f } (ξ ) = m(ξ )fˆ(ξ )
.
The function m is a Fourier multiplier and .Tm is a Fourier multiplier operator. In
particular, the operator .Tm is defined on the Schwartz space .S(Rd ) by .Tm f =
F {mf } (Muscalu and Schlag 2013).
Example B.1 Translation by .τ ∈ Rd with .mτ (ξ ) = e2πj τ ·ξ through
{
}
Tτ f (x) = f (x − τ ) = F −1 e2πj τ ·ξ fˆ(ξ ) (x)
.
Example B.2 Partial differentiation with respect to .xi , .i = 1, . . . through:
.
{
}
∂
f (x) = F −1 (2π iξi ) fˆ(ξ )
∂xi
Example B.3 A low-pass filter through:
{
}
Or f (x) = F −1 χRr (ξ )fˆ(ξ )
.
where .Rr = [−r, r] × . . . × [−r, r] and .χ is the indicator function.
B.1.3 Fourier Tensors
In signal processing, we sometimes encounter the case when the input signal
is multidimensional. A multidimensional signal is a function of more than two
variables. For example, a video signal is a function of three independent variables
which are time and two spatial coordinates .(X, Y ). The high-order (not higher
dimensional) Fourier analysis arises when we consider the Fourier transform of a
multidimensional input signal (Xu 2021). A 2m-order n-dimensional Fourier tensor,
denoted by .Wm,n , is defined by
B
The Fourier Transform
271
2π Σ
(ik − 1)(jk − 1)
n
m
Wm,n = n−m/2 ej W (m,n)
W (m, n) =
.
k=1
where .ik , jk ∈ [n].
B.2 Cauchy–Bunyakovsky–Schwartz Inequality
Particular cases of it include Cauchy’s inequality for the dot product and Holder’s
inequality for square-integrable functions.
Let V a vector space with inner product, .u, v ∈ V . Then,
.
|<u, v>|2 ≤ <u, u> · <v, v>
In .L2 :
|f
|2 f
f
|
|
∗
|
| ≤ |f (x)|2 dx · |g(x)|2 dx
f
(x)g
.
(x)dx
|
|
B.3 Bessel Function
Bessel function of the first kind is a solution of the differential equation:
β2
.
)
d 2y
dy ( 2
2
+
β
y(β) = 0
+
β
−
n
dx
dx 2
Bessel function defined for negative and positive real integers. We assume the
following solutions. .n = λ
y=
∞
Σ
.
Ck x k+n
k=0
Thus, the equation becomes
y(x) =
.
∞
Σ
(−1)k
k=0
x 2k+n
22k+n k!r(k + n + 1)
272
B The Fourier Transform
B.3.1 Properties of Bessel Function
It can be shown that for integer values of n:
j−n (β) = (−1)n jn (β)
.
jn−1 (β) + jn+1 (β) =
.
∞
Σ
.
2n
jn (β)
β
jn2 (β) = 1
n=−∞
B.3.2 Bessel Equation of Order λ and Parameter p
The differential equation
x2
.
)
d 2y
dy ( 2 2
2
+
p
y=0
+
x
x
−
λ
dx
dx 2
has as solution the Bessel function .yp (x) = Ajλ (px).
B.4 Bernoulli Memristors
From the first order ordinary differential equation:
.
dy
+ P (x)y = Q(x)y n
dx
with .n /= 1 the change of variable .u = y 1−n turns the equation into a linear
equation in u. This equation is known as Bernoulli’s equation. For .n = 3, the
equation comes up in modeling frictional forces. The Bernoulli memristors are
certain class of memristors, whose dynamics comply with Bernoulli’s differential
equation (Georgiou et al. 2011).
References
Georgiou PS, Yaliraki SN, Drakakis EM, Barahona M (2011) Quantitative measure of hysteresis
for memristors through explicit dynamics. Technical report, https://arxiv.org/abs/1011.0060
B
The Fourier Transform
273
Mattila P (2015) Fourier analysis and Hausdorff dimension. Cambridge University Press, Cambridge
Muscalu C, Schlag W (2013) Classical and multilinear harmonic analysis: volume 1. Cambridge
University Press, Cambridge
Xu C (2021) Fourier matrices and Fourier tensors. Front Math China 16(4):1099–1115. https://doi.
org/10.1007/s11464-021-0904-y, https://link.springer.com/10.1007/s11464-021-0904-y
Appendix C
Python Routines
The following routines are used for the generation of most figures in this book.
C.1 Chapter 2
Generate and plot the signal of Fig. 2.1
import matplotlib.pyplot as plt
import numpy as np
## Parameters used
StopTime = 1 # End of signal
Fs = 1024 # Sampling rate
f = 4
# Frequency of simulated signal
## Generate sample times
t = np.linspace(0,StopTime, StopTime*Fs)
## Generate signal
x = np.sin(2*np.pi*t*f)-np.sin(2*np.pi*t*3*f)-\
np.sin(2*np.pi*t*10*f)
plt.figure(1)
plt.plot(t[80:110],x[80:110])
(continued)
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024
J. C. García-Álvarez, Digital Electronic Communications,
https://doi.org/10.1007/978-3-031-53118-7
275
276
C
Python Routines
plt.figure(2)
plt.plot(t[80:110],x[80:110],’bo’)
plt.show()
Generate and plot the signal of Figs. 2.10,~2.11, and~2.12
from scipy import signal
import matplotlib.pyplot as plt
import numpy as np
#Create an arbitrary noise signal
def fftnoise(f):
f = np.array(f, dtype=’complex’)
Np = (len(f) - 1) // 2
phases = np.random.rand(Np) * 2 * np.pi
phases = np.cos(phases) + 1j * np.sin(phases)
f[1:Np+1] *= phases
f[-1:-1-Np:-1] = np.conj(f[1:Np+1])
return np.fft.ifft(f).real
def band_limited_noise(min_freq, max_freq, samples=1024,
samplerate=1):
freqs = np.abs(np.fft.fftfreq(samples, 1/samplerate))
f = np.zeros(samples)
idx = np.where(np.logical_and(freqs>=min_freq,
freqs<=max_freq))[0]
f[idx] = 1
return fftnoise(f)
## Parameters used
StopTime = 1 # End of signal
Fs = 1024 # Sampling rate
f = 50
# Frequency of simulated signal
## Generate sample times
t = np.linspace(0,StopTime, StopTime*Fs)
## Generate signal
x = np.sin(2*np.pi*t*f)
## Add noise to signal
## Generate white Gaussian noise
xawgn = np.random.randn(len(x))
#print(’rms noise value:’, sqrt(mean(randn(10000)**2)))
## Generate band-limited (pink) noise
xn = band_limited_noise(30, 300, Fs, Fs)
## Figure(1)
fig,(ax1, ax2) = plt.subplots(1, 2)
(continued)
C
Python Routines
277
y = x
ax1.plot(t[0:100],y[0:100])
fn, Pxx_spec = signal.welch(y, Fs, ’flattop’, 1024,
scaling=’spectrum’)
ax2.semilogy(fn, np.sqrt(Pxx_spec))
## Figure(2)
fig,(ax1, ax2) = plt.subplots(1, 2)
y = x+xawgn
ax1.plot(t[0:100],y[0:100])
fn, Pxx_spec = signal.welch(y, Fs, ’flattop’, 1024,
scaling=’spectrum’)
ax2.semilogy(fn, np.sqrt(Pxx_spec))
## Figure(3)
fig,(ax1, ax2) = plt.subplots(1, 2)
y = x+xn
ax1.plot(t[0:100],y[0:100])
fn, Pxx_spec = signal.welch(y, Fs, ’flattop’, 1024,
scaling=’spectrum’)
ax2.semilogy(fn, np.sqrt(Pxx_spec))
plt.show()
C.2 Chapter 5
Generate and plot the signal of Fig. 5.7
from scipy import signal
import matplotlib.pyplot as plt
import numpy as np
## Parameters used
StopTime = 1 # End of signal
Fs = 1024 # Sampling rate
fc = 200 # Frequency of carrier signal
fm = 15 # Frequency of modulating signal
b = 0.2 # Modulation Index
## Generate sample times
t = np.linspace(0,StopTime, StopTime*Fs)
## Generate signal
xm = np.sin(2*np.pi*t*fm) #Modulating signal
xmod = np.cos((2*np.pi*t*fc)+(b*xm)) #Modulated signal
(continued)
278
C
Python Routines
fn, Pxx_spec = signal.welch(xmod, Fs, ’flattop’, 1024,
scaling=’spectrum’)
plt.semilogy(fn, np.sqrt(Pxx_spec))
plt.show()
Glossary
Antenna Interface between electromagnetic waves propagating through space and
electric currents moving in metal conductors, connected to a transmitter or receiver.
Bandwidth (1) Range between lowest and highest attainable frequencies of a
signal. (2) Maximum rate of data transfer across a given path.
Channel A range of frequencies used to broadcast modulated signals, especially
those given by a transmitter.
Circuit A set of elements connected in such a way to establish a closed path. In
electrical extent, if one of the elements of the circuit is an electrical source, an
electrical current passes by all elements of the circuit.
Downlink To receive or send signals from a device of higher hierarchy, assigned
carrier frequency or height.
Frequency Number of waves or cycles produced within a particular period,
especially one second.
Message Limited quantity of information transported from one point to another.
Modulation Process of varying one or more properties of a periodic waveform,
called the carrier signal, with a separate signal called the modulation signal that
typically contains information to be transmitted.
Oscillator Element, commonly an electronic circuit, that produces a periodic,
oscillating or alternating signal, usually a sine wave, square wave, or a triangle wave.
Spread Spectrum A set of techniques by which the spectral components of a
signal, located in a particular bandwidth, are deliberately spread in the frequency
domain over a wider frequency band.
Uplink To receive or send signals from a device of lower hierarchy, assigned carrier
frequency or height.
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024
J. C. García-Álvarez, Digital Electronic Communications,
https://doi.org/10.1007/978-3-031-53118-7
279
Index
A
Analog electronics, 11, 143
Attenuation, xi, 7, 14, 71, 73, 78, 99, 129,
131–133, 136, 141, 163, 164, 169, 176,
192, 200, 201, 204, 215, 223, 224, 237,
249, 251, 252
C
Carrier, viii, 4, 7, 8, 12–14, 16, 31, 40, 41, 52,
56, 74–76, 96, 102, 104, 105, 107–115,
117, 118, 121, 122, 127–130, 134–136,
139, 140, 143, 144, 147, 151, 153–155,
157, 164–168, 174, 175, 177, 180, 190,
193–197, 202, 206, 208, 211, 212, 216,
218–221, 223, 227, 230, 231, 235,
237–239, 244, 247, 251
Channel, x, xi, 7, 13, 16–20, 39–41, 43, 45,
53, 55, 57, 59–71, 74, 76–78, 82, 90,
91, 95, 96, 102–105, 107, 109, 110,
112–114, 123–125, 127–134, 136, 137,
141, 142, 147, 156, 157, 159–161, 164,
167, 168, 171, 172, 176, 178, 179, 183,
185–199, 201, 202, 204–209, 211, 212,
215–217, 219–221, 223, 224, 227–233,
235, 237, 239, 240, 242–252
Channel assignment, 185–224
Channel capacity, x, 1, 7, 62–64, 66–67, 70,
71, 103, 105, 127, 147, 157, 159, 175,
220, 223, 224, 250
Citation, 196
Coding, 4, 43, 59, 60, 63–66, 68, 77, 84,
88–94, 102, 104, 105, 107, 196, 197,
206, 208–211, 231, 239, 244
Coding stage, 208
Communication systems, viii–xi, 1, 3–10,
26, 29, 36, 39, 40, 57, 64, 70, 73, 74,
107–112, 114, 131, 140, 141, 147, 153,
164, 171, 183, 192, 204, 205, 207, 215,
223, 227, 228, 246, 256
Continuous wave (CW), 7, 74, 113, 135, 190
D
Dedication, v
Delay, 4, 88, 101, 146, 170, 187, 243
Derivative amplifier, 143
Diode, 111, 112, 134–137, 164–166
Diversity, 252
E
Electromagnetic fields, 12
Entropy, 7, 61, 62, 65, 69–71, 84, 90, 103, 104,
156, 157, 159
F
Fading, 39, 183
Filter, 7, 29, 55, 76, 109, 121, 127, 129,
133–136, 140–151, 153, 155, 161–164,
168, 171, 174, 175, 191, 195, 197, 216,
217, 219, 230, 231, 235, 238, 239, 270
Full-wave, 135, 165, 166
G
Gamma function, 258, 264
© The Editor(s) (if applicable) and The Author(s), under exclusive license to
Springer Nature Switzerland AG 2024
J. C. García-Álvarez, Digital Electronic Communications,
https://doi.org/10.1007/978-3-031-53118-7
281
282
H
Half-wave, 135–137, 165, 166
I
Information measure, 59, 87
Integrating amplifier, 146, 169
Interference, x, 6, 12–15, 23, 24, 39, 78, 96,
110, 112–114, 137, 141, 150, 183, 188,
193, 195, 197, 205, 216–218, 227,
229–231, 233, 237, 242, 243, 246
M
Mathematical modeling, 195, 202–203,
208–215, 229–230
Miller integrator, 170
Modulation, x, 1, 7, 8, 11–13, 28, 38, 43, 65,
73–105, 108–110, 112–117, 119–128,
134–141, 143, 147, 153–157, 159, 161,
165, 166, 168, 171, 174, 177–179, 186,
190, 192–196, 202, 204–208, 210–212,
215–221, 223, 224, 229–234, 240–249,
251, 252
Modulation stage, 108, 210
Multipath, 39, 183, 218, 227, 252
Multiplexing, 12, 74, 183, 185–210, 212,
215–216, 223, 230, 232, 234, 242–247,
251, 252
Multiplexing stage, 209, 210
N
Noise, x, xi, 6, 13–15, 23, 24, 31, 39–52,
54–57, 59, 60, 62, 64–66, 70, 71, 73, 74,
76–78, 96, 99, 100, 102, 104, 105, 110,
112–114, 122, 127, 136, 137, 139–141,
144, 147, 154, 156, 157, 159, 162, 164,
165, 167, 168, 171, 172, 174, 175, 178,
192, 196, 204, 215–220, 223, 224, 228,
231, 237, 238, 246, 249–252
Index
Path loss (PL), 132, 133
Phase transitions, 85
Power, 7, 8, 13, 16–28, 38, 41–46, 48, 51, 52,
54–57, 60, 65, 66, 71, 76, 78, 79, 81,
85, 87, 89, 90, 100, 102–104, 107–109,
114, 117–119, 130, 132–135, 137,
139–141, 143, 151, 156, 157, 159, 161,
162, 165–169, 171, 172, 176, 178, 179,
200, 201, 204, 215, 216, 219, 223, 224,
227, 230, 231, 249–251
Probability density function, 56, 103, 163, 252,
256, 258
R
Receiver, xi, 4, 5, 7, 8, 17, 39, 40, 43–46, 57,
59, 65, 66, 69, 70, 90, 91, 99–101, 105,
109, 113, 122, 126, 127, 129, 133–143,
145–149, 151, 153, 165, 167, 168, 171,
174, 177, 179, 188, 189, 191, 203, 208,
215, 216, 219, 223, 224, 227, 230–232,
238, 240, 243, 249–252
Rectifier, 134–136, 165, 166
S
Serial communication, 62
Signal, viii, x, 3–7, 11–60, 62, 63, 65–67, 70,
71, 73–92, 94–100, 102–105, 107–130,
133–141, 144–157, 160, 164–171,
173–180, 183, 185, 186, 188–199,
202–220, 223, 224, 227, 230–240,
242–249, 251, 252, 256–257, 261, 267,
268, 270
Slot, 185–188, 222
Source, x, xi, 5, 7, 21, 23, 24, 31, 39, 41, 42,
47, 59–61, 63, 65, 67–69, 71, 84, 91,
96, 102–104, 107–109, 130, 132, 135,
141, 156, 157, 179, 186, 187, 190, 241,
244, 256, 257, 259
O
Orthogonal signals, 36, 98, 210
P
Passband, 12, 37, 38, 66, 75, 110, 112–113,
127–128, 140, 141, 156, 238, 240
W
Wave, 7, 11–13, 41, 52, 63, 74, 75, 82, 86, 110,
112–114, 116, 124, 126, 128, 129, 135,
140, 154, 155, 166–169, 174, 177, 238,
246, 267