Computer Network & Security

Course Learning Outcome (CLOs): After Completing this course successfully, the student will be able to…

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Summary of Course Content:

Recommended Books:

1. Kurose and Ross, “Computer Networking: A Top-Down Approach”, Fifth Edition, Pearson(ISBN: 9788131790540)
2. Trivedi, “Computer Networks”, Oxford University Press(ISBN: 9780198066774)
3. Black, “Computer Networks: Protocols, Standards and Interface”, PHI (ISBN: 9788120310414)
4. Gupta, “Data Communications and Computer Networks”, PHI (ISBN: 9788120328464)
5. Halsall and Kulkarni, “Computer Networking and The Internet” Fifth Edition, Pearson (ISBN: 97881775847

Assessment Pattern

CIE- Continuous Internal Evaluation ( 90 Marks)

SEE- Semester End Examination (60 Marks)

Course Plan

Course Plan

1.7

Week: 01

Chapter 1

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Introduction

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

1.8

1-1 DATA COMMUNICATIONS

The term telecommunication means communication at a distance. The word data refers to information presented in whatever form is agreed upon by the parties creating and using the data. Data communications are the exchange of data between two devices via some form of transmission medium such as a wire cable.

• Components of a data communications system
• Data Flow

Topics discussed in this section:

1.9

Figure 1.1 Components of a data communication system

1.10

Figure 1.2 Data flow (simplex, half-duplex, and full-duplex)

1.11

1-2 NETWORKS

A network is a set of devices (often referred to as nodes) connected by communication links. A node can be a computer, printer, or any other device capable of sending and/or receiving data generated by other nodes on the network. A link can be a cable, air, optical fiber, or any medium which can transport a signal carrying information.

• Network Criteria
• Physical Structures
• Categories of Networks

Topics discussed in this section:

• Performance
• Depends on Network Elements
• Measured in terms of Delay and Throughput
• Reliability
• Failure rate of network components
• Measured in terms of availability/robustness
• Security
• Data protection against corruption/loss of data due to:
• Errors
• Malicious users

1.12

Network Criteria

• Type of Connection
• Point to Point - single transmitter and receiver
• Multipoint - multiple recipients of single transmission
• Physical Topology
• Connection of devices
• Type of transmission - unicast, mulitcast, broadcast

1.13

Physical Structures

1.14

Figure 1.3 Types of connections: point-to-point and multipoint

1.15

Figure 1.4 Categories of topology

1.16

Figure 1.5 A fully connected mesh topology (five devices)

1.17

Figure 1.6 A star topology connecting four stations

1.18

Figure 1.7 A bus topology connecting three stations

1.19

Figure 1.8 A ring topology connecting six stations

1.20

Figure 1.9 A hybrid topology: a star backbone with three bus networks

• Local Area Networks (LANs)
• Short distances
• Designed to provide local interconnectivity
• Wide Area Networks (WANs)
• Long distances
• Provide connectivity over large areas
• Metropolitan Area Networks (MANs)
• Provide connectivity over areas such as a city, a campus

1.21

Categories of Networks

1.22

Figure 1.10 An isolated LAN connecting 12 computers to a hub in a closet

1.23

Figure 1.11 WANs: a switched WAN and a point-to-point WAN

1.24

Figure 1.12 A heterogeneous network made of four WANs and two LANs

1.25

1-3 THE INTERNET

The Internet has revolutionized many aspects of our daily lives. It has affected the way we do business as well as the way we spend our leisure time. The Internet is a communication system that has brought a wealth of information to our fingertips and organized it for our use.

Organization of the Internet

Internet Service Providers (ISPs)

Topics discussed in this section:

1.26

Figure 1.13 Hierarchical organization of the Internet

1.27

1-4 PROTOCOLS

A protocol is synonymous with rule. It consists of a set of rules that govern data communications. It determines what is communicated, how it is communicated and when it is communicated. The key elements of a protocol are syntax, semantics and timing

• Syntax
• Semantics
• Timing

Topics discussed in this section:

• Syntax
• Structure or format of the data
• Indicates how to read the bits - field delineation
• Semantics
• Interprets the meaning of the bits
• Knows which fields define what action
• Timing
• When data should be sent and what
• Speed at which data should be sent or speed at which it is being received.

1.28

Elements of a Protocol

2.29

Week: 02

Chapter 2

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Network Models

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2.30

2-1 LAYERED TASKS

We use the concept of layers in our daily life. As an example, let us consider two friends who communicate through postal mail. The process of sending a letter to a friend would be complex if there were no services available from the post office.

Sender, Receiver, and Carrier�Hierarchy

Topics discussed in this section:

2.31

Figure 2.1 Tasks involved in sending a letter

2.32

2-2 THE OSI MODEL

Established in 1947, the International Standards Organization (ISO) is a multinational body dedicated to worldwide agreement on international standards. An ISO standard that covers all aspects of network communications is the Open Systems Interconnection (OSI) model. It was first introduced in the late 1970s.

Layered Architecture�Peer-to-Peer Processes

Encapsulation

Topics discussed in this section:

2.33

ISO is the organization.�OSI is the model.

2.34

Figure 2.2 Seven layers of the OSI model

2.35

Figure 2.3 The interaction between layers in the OSI model

2.36

Figure 2.4 An exchange using the OSI model

2.37

2-3 LAYERS IN THE OSI MODEL

In this section we briefly describe the functions of each layer in the OSI model.

Physical Layer�Data Link Layer

Network Layer

Transport Layer

Session Layer

Presentation Layer

Application Layer

Topics discussed in this section:

2.38

Figure 2.5 Physical layer

2.39

The physical layer is responsible for movements of

individual bits from one hop (node) to the next.

2.40

Figure 2.6 Data link layer

2.41

The data link layer is responsible for moving �frames from one hop (node) to the next.

2.42

Figure 2.7 Hop-to-hop delivery

2.43

Figure 2.8 Network layer

2.44

The network layer is responsible for the �delivery of individual packets from

the source host to the destination host.

2.45

Figure 2.9 Source-to-destination delivery

2.46

Figure 2.10 Transport layer

2.47

The transport layer is responsible for the delivery �of a message from one process to another.

2.48

Figure 2.11 Reliable process-to-process delivery of a message

2.49

Figure 2.12 Session layer

2.50

The session layer is responsible for dialog �control and synchronization.

2.51

Figure 2.13 Presentation layer

2.52

The presentation layer is responsible for translation, compression, and encryption.

2.53

Figure 2.14 Application layer

2.54

The application layer is responsible for �providing services to the user.

2.55

Figure 2.15 Summary of layers

2.56

2-4 TCP/IP PROTOCOL SUITE

The layers in the TCP/IP protocol suite do not exactly match those in the OSI model. The original TCP/IP protocol suite was defined as having four layers: host-to-network, internet, transport, and application. However, when TCP/IP is compared to OSI, we can say that the TCP/IP protocol suite is made of five layers: physical, data link, network, transport, and application.

Physical and Data Link Layers�Network Layer�Transport Layer

Application Layer

Topics discussed in this section:

2.57

Figure 2.16 TCP/IP and OSI model

2.58

2-5 ADDRESSING

Four levels of addresses are used in an internet employing the TCP/IP protocols: physical, logical, port, and specific.

Physical Addresses�Logical Addresses�Port Addresses�Specific Addresses

Topics discussed in this section:

2.59

Figure 2.17 Addresses in TCP/IP

2.60

Figure 2.18 Relationship of layers and addresses in TCP/IP

2.61

In Figure 2.19 a node with physical address 10 sends a frame to a node with physical address 87. The two nodes are connected by a link (bus topology LAN). As the figure shows, the computer with physical address 10 is the sender, and the computer with physical address 87 is the receiver.

Example 2.1

2.62

Figure 2.19 Physical addresses

2.63

Most local-area networks use a 48-bit (6-byte) physical address written as 12 hexadecimal digits; every byte (2 hexadecimal digits) is separated by a colon, as shown below:

Example 2.2

07:01:02:01:2C:4B�

A 6-byte (12 hexadecimal digits) physical address.

2.64

Figure 2.20 shows a part of an internet with two routers connecting three LANs. Each device (computer or router) has a pair of addresses (logical and physical) for each connection. In this case, each computer is connected to only one link and therefore has only one pair of addresses. Each router, however, is connected to three networks (only two are shown in the figure). So each router has three pairs of addresses, one for each connection.

Example 2.3

2.65

Figure 2.20 IP addresses

2.66

Figure 2.21 shows two computers communicating via the Internet. The sending computer is running three processes at this time with port addresses a, b, and c. The receiving computer is running two processes at this time with port addresses j and k. Process a in the sending computer needs to communicate with process j in the receiving computer. Note that although physical addresses change from hop to hop, logical and port addresses remain the same from the source to destination.

Example 2.4

2.67

Figure 2.21 Port addresses

2.68

The physical addresses will change from hop to hop,

but the logical addresses usually remain the same.

2.69

Example 2.5

A port address is a 16-bit address represented by one decimal number as shown.

753�

A 16-bit port address represented �as one single number.

3.70

Week: 03

Chapter 3

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Data and Signals

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

3.71

To be transmitted, data must be transformed to electromagnetic signals.

3.72

3-1 ANALOG AND DIGITAL

Data can be analog or digital. The term analog data refers to information that is continuous; digital data refers to information that has discrete states. Analog data take on continuous values. Digital data take on discrete values.

• Analog and Digital Data
• Analog and Digital Signals
• Periodic and Nonperiodic Signals

Topics discussed in this section:

Analog and Digital Data

• Data can be analog or digital.
• Analog data are continuous and take continuous values.
• Digital data have discrete states and take discrete values.

3.73

Analog and Digital Signals

• Signals can be analog or digital.
• Analog signals can have an infinite number of values in a range.
• Digital signals can have only a limited �number of values.

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3.74

3.75

Figure 3.1 Comparison of analog and digital signals

3.76

3-2 PERIODIC ANALOG SIGNALS

In data communications, we commonly use periodic analog signals and nonperiodic digital signals.

Periodic analog signals can be classified as simple or composite. A simple periodic analog signal, a sine wave, cannot be decomposed into simpler signals. A composite

periodic analog signal is composed of multiple sine waves.

• Sine Wave
• Wavelength
• Time and Frequency Domain
• Composite Signals
• Bandwidth

Topics discussed in this section:

3.77

Figure 3.2 A sine wave

3.78

Figure 3.3 Two signals with the same phase and frequency, � but different amplitudes

3.79

Frequency and period are the inverse of each other.

3.80

Figure 3.4 Two signals with the same amplitude and phase,� but different frequencies

3.81

Table 3.1 Units of period and frequency

3.82

The power we use at home has a frequency of 60 Hz. The period of this sine wave can be determined as follows:

Example 3.1

3.83

The period of a signal is 100 ms. What is its frequency in kilohertz?

Example 3.2

Solution

First we change 100 ms to seconds, and then we calculate the frequency from the period (1 Hz = 10⁻³ kHz).

Frequency

• Frequency is the rate of change with respect to time.
• Change in a short span of time means high frequency.
• Change over a long span of �time means low frequency.

3.84

3.85

If a signal does not change at all, its frequency is zero.

If a signal changes instantaneously, its frequency is infinite.

3.86

Phase describes the position of the waveform relative to time 0.

3.87

Figure 3.5 Three sine waves with the same amplitude and frequency,� but different phases

3.88

A sine wave is offset 1/6 cycle with respect to time 0. What is its phase in degrees and radians?

Example 3.3

Solution

We know that 1 complete cycle is 360°. Therefore, 1/6 cycle is

3.89

Figure 3.6 Wavelength and period

3.90

Figure 3.7 The time-domain and frequency-domain plots of a sine wave

3.91

A complete sine wave in the time domain can be represented by one single spike in the frequency domain.

3.92

The frequency domain is more compact and useful when we are dealing with more than one sine wave. For example, Figure 3.8 shows three sine waves, each with different amplitude and frequency. All can be represented by three spikes in the frequency domain.

Example 3.7

3.93

Figure 3.8 The time domain and frequency domain of three sine waves

Signals and Communication

• A single-frequency sine wave is not useful in data communications
• We need to send a composite signal, a signal made of many simple sine waves.
• According to Fourier analysis, any composite signal is a combination of simple sine waves with different frequencies, amplitudes, and phases.

3.94

Composite Signals and Periodicity

• If the composite signal is periodic, the decomposition gives a series of signals with discrete frequencies.
• If the composite signal is nonperiodic, the decomposition gives a combination of sine waves with continuous frequencies.

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3.95

3.96

Figure 3.9 shows a periodic composite signal with frequency f. This type of signal is not typical of those found in data communications. We can consider it to be three alarm systems, each with a different frequency. The analysis of this signal can give us a good understanding of how to decompose signals.

Example 3.4

3.97

Figure 3.9 A composite periodic signal

3.98

Figure 3.10 Decomposition of a composite periodic signal in the time and� frequency domains

3.99

Figure 3.11 shows a nonperiodic composite signal. It can be the signal created by a microphone or a telephone set when a word or two is pronounced. In this case, the composite signal cannot be periodic, because that implies that we are repeating the same word or words with exactly the same tone.

Example 3.5

3.100

Figure 3.11 The time and frequency domains of a nonperiodic signal

Bandwidth and Signal Frequency

• The bandwidth of a composite signal is the difference between the highest and the lowest frequencies contained in that signal.

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3.101

3.102

Figure 3.12 The bandwidth of periodic and nonperiodic composite signals

3.103

If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700, and 900 Hz, what is its bandwidth? Draw the spectrum, assuming all components have a maximum amplitude of 10 V.

Solution

Let f_h be the highest frequency, f_l the lowest frequency, and B the bandwidth. Then

Example 3.6

The spectrum has only five spikes, at 100, 300, 500, 700, and 900 Hz (see Figure 3.13).

3.104

Figure 3.13 The bandwidth for Example 3.6

3.105

A periodic signal has a bandwidth of 20 Hz. The highest frequency is 60 Hz. What is the lowest frequency? Draw the spectrum if the signal contains all frequencies of the same amplitude.

Solution

Let f_h be the highest frequency, f_l the lowest frequency, and B the bandwidth. Then

Example 3.7

The spectrum contains all integer frequencies. We show this by a series of spikes (see Figure 3.14).

3.106

Figure 3.14 The bandwidth for Example 3.7

3.107

A nonperiodic composite signal has a bandwidth of 200 kHz, with a middle frequency of 140 kHz and peak amplitude of 20 V. The two extreme frequencies have an amplitude of 0. Draw the frequency domain of the signal.

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Solution

The lowest frequency must be at 40 kHz and the highest at 240 kHz. Figure 3.15 shows the frequency domain and the bandwidth.

Example 3.8

3.108

Figure 3.15 The bandwidth for Example 3.8

3.109

An example of a nonperiodic composite signal is the signal propagated by an AM radio station. In the United States, each AM radio station is assigned a 10-kHz bandwidth. The total bandwidth dedicated to AM radio ranges from 530 to 1700 kHz. We will show the rationale behind this 10-kHz bandwidth in Chapter 5.

Example 3.9

3.110

Another example of a nonperiodic composite signal is the signal propagated by an FM radio station. In the United States, each FM radio station is assigned a 200-kHz bandwidth. The total bandwidth dedicated to FM radio ranges from 88 to 108 MHz. We will show the rationale behind this 200-kHz bandwidth in Chapter 5.

Example 3.10

3.111

Another example of a nonperiodic composite signal is the signal received by an old-fashioned analog black-and-white TV. A TV screen is made up of pixels. If we assume a resolution of 525 × 700, we have 367,500 pixels per screen. If we scan the screen 30 times per second, this is 367,500 × 30 = 11,025,000 pixels per second. The worst-case scenario is alternating black and white pixels. We can send 2 pixels per cycle. Therefore, we need 11,025,000 / 2 = 5,512,500 cycles per second, or Hz. The bandwidth needed is 5.5125 MHz.

Example 3.11

Fourier Analysis

3.112

Fourier analysis is a tool that changes a time domain signal to a frequency domain signal and vice versa.

Fourier Series

• Every composite periodic signal can be represented with a series of sine and cosine functions.
• The functions are integral harmonics of the fundamental frequency “f” of the composite signal.
• Using the series we can decompose any periodic signal into its harmonics.

3.113

Fourier Series

3.114

Examples of Signals and the Fourier Series Representation

3.115

Sawtooth Signal

3.116

Fourier Transform

• Fourier Transform gives the frequency domain of a nonperiodic time domain signal.

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3.117

Example of a Fourier Transform

3.118

Inverse Fourier Transform

3.119

Time limited and Band limited Signals

• A time limited signal is a signal for which the amplitude s(t) = 0 for t > T₁ and t < T₂
• A band limited signal is a signal for which the amplitude S(f) = 0 for f > F₁ and f < F₂

3.120

3.121

Week: 04

Chapter 3

​

Data and Signals

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

3.122

3-3 DIGITAL SIGNALS

In addition to being represented by an analog signal, information can also be represented by a digital signal. For example, a 1 can be encoded as a positive voltage and a 0 as zero voltage. A digital signal can have more than two levels. In this case, we can send more than 1 bit for each level.

• Bit Rate
• Bit Length
• Digital Signal as a Composite Analog Signal
• Application Layer

Topics discussed in this section:

3.123

Figure 3.16 Two digital signals: one with two signal levels and the other� with four signal levels

3.124

A digital signal has eight levels. How many bits are needed per level? We calculate the number of bits from the formula

Example 3.16

Each signal level is represented by 3 bits.

3.125

A digital signal has nine levels. How many bits are needed per level? We calculate the number of bits by using the formula. Each signal level is represented by 3.17 bits. However, this answer is not realistic. The number of bits sent per level needs to be an integer as well as a power of 2. For this example, 4 bits can represent one level.

Example 3.17

3.126

Assume we need to download text documents at the rate of 100 pages per sec. What is the required bit rate of the channel?

Solution

A page is an average of 24 lines with 80 characters in each line. If we assume that one character requires 8 bits (ascii), the bit rate is

Example 3.18

3.127

A digitized voice channel, as we will see in Chapter 4, is made by digitizing a 4-kHz bandwidth analog voice signal. We need to sample the signal at twice the highest frequency (two samples per hertz). We assume that each sample requires 8 bits. What is the required bit rate?�

Solution

The bit rate can be calculated as

Example 3.19

3.128

What is the bit rate for high-definition TV (HDTV)?�

Solution

HDTV uses digital signals to broadcast high quality video signals. The HDTV screen is normally a ratio of 16 : 9. There are 1920 by 1080 pixels per screen, and the screen is renewed 30 times per second. Twenty-four bits represents one color pixel.

Example 3.20

The TV stations reduce this rate to 20 to 40 Mbps through compression.

3.129

Figure 3.17 The time and frequency domains of periodic and nonperiodic� digital signals

3.130

Figure 3.18 Baseband transmission

3.131

A digital signal is a composite analog signal with an infinite bandwidth.

3.132

Figure 3.19 Bandwidths of two low-pass channels

3.133

Figure 3.20 Baseband transmission using a dedicated medium

3.134

Baseband transmission of a digital signal that preserves the shape of the digital signal is possible only if we have a low-pass channel with an infinite or very wide bandwidth.

3.135

An example of a dedicated channel where the entire bandwidth of the medium is used as one single channel is a LAN. Almost every wired LAN today uses a dedicated channel for two stations communicating with each other. In a bus topology LAN with multipoint connections, only two stations can communicate with each other at each moment in time (timesharing); the other stations need to refrain from sending data. In a star topology LAN, the entire channel between each station and the hub is used for communication between these two entities.

Example 3.21

3.136

Figure 3.21 Rough approximation of a digital signal using the first harmonic � for worst case

3.137

Figure 3.22 Simulating a digital signal with first three harmonics

3.138

^{In baseband transmission, the required bandwidth is proportional to the bit rate;}

^{if we need to send bits faster, we need more bandwidth.}

In baseband transmission, the required bandwidth is proportional to the bit rate;

if we need to send bits faster, we need more bandwidth.

3.139

Table 3.2 Bandwidth requirements

3.140

What is the required bandwidth of a low-pass channel if we need to send 1 Mbps by using baseband transmission?�

Solution

The answer depends on the accuracy desired.

a. The minimum bandwidth, is B = bit rate /2, or 500 kHz.�

b. A better solution is to use the first and the third� harmonics with B = 3 × 500 kHz = 1.5 MHz.�

c. Still a better solution is to use the first, third, and fifth� harmonics with B = 5 × 500 kHz = 2.5 MHz.

Example 3.22

3.141

We have a low-pass channel with bandwidth 100 kHz. What is the maximum bit rate of this

channel?�

Solution

The maximum bit rate can be achieved if we use the first harmonic. The bit rate is 2 times the available bandwidth, or 200 kbps.

Example 3.22

3.142

Figure 3.23 Bandwidth of a bandpass channel

3.143

If the available channel is a bandpass channel, we cannot send the digital signal directly to the channel; �we need to convert the digital signal to an analog signal before transmission.

3.144

Figure 3.24 Modulation of a digital signal for transmission on a bandpass � channel

3.145

An example of broadband transmission using modulation is the sending of computer data through a telephone subscriber line, the line connecting a resident to the central telephone office. These lines are designed to carry voice with a limited bandwidth. The channel is considered a bandpass channel. We convert the digital signal from the computer to an analog signal, and send the analog signal. We can install two converters to change the digital signal to analog and vice versa at the receiving end. The converter, in this case, is called a modem which we discuss in detail in Chapter 5.

Example 3.24

3.146

A second example is the digital cellular telephone. For better reception, digital cellular phones convert the analog voice signal to a digital signal (see Chapter 16). Although the bandwidth allocated to a company providing digital cellular phone service is very wide, we still cannot send the digital signal without conversion. The reason is that we only have a bandpass channel available between caller and callee. We need to convert the digitized voice to a composite analog signal before sending.

Example 3.25

3.147

Week: 05

Chapter 3

​

Data and Signals

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

3.148

3-4 TRANSMISSION IMPAIRMENT

Signals travel through transmission media, which are not perfect. The imperfection causes signal impairment. This means that the signal at the beginning of the medium is not the same as the signal at the end of the medium. What is sent is not what is received. Three causes of impairment are attenuation, distortion, and noise.

• Attenuation
• Distortion
• Noise

Topics discussed in this section:

3.149

Figure 3.25 Causes of impairment

3.150

Attenuation

• Means loss of energy -> weaker signal
• When a signal travels through a medium it loses energy overcoming the resistance of the medium
• Amplifiers are used to compensate for this loss of energy by amplifying the signal.

3.151

Measurement of Attenuation

• To show the loss or gain of energy the unit “decibel” is used.

​

dB = 10log₁₀P₂/P₁

P₁ - input signal

P₂ - output signal

3.152

3.153

Figure 3.26 Attenuation

3.154

Suppose a signal travels through a transmission medium and its power is reduced to one-half. This means that P2 is (1/2)P1. In this case, the attenuation (loss of power) can be calculated as

Example 3.26

A loss of 3 dB (–3 dB) is equivalent to losing one-half the power.

3.155

A signal travels through an amplifier, and its power is increased 10 times. This means that P₂ = 10P₁ . In this case, the amplification (gain of power) can be calculated as

Example 3.27

3.156

One reason that engineers use the decibel to measure the changes in the strength of a signal is that decibel numbers can be added (or subtracted) when we are measuring several points (cascading) instead of just two. In Figure 3.27 a signal travels from point 1 to point 4. In this case, the decibel value can be calculated as

Example 3.28

3.157

Figure 3.27 Decibels for Example 3.28

3.158

Sometimes the decibel is used to measure signal power in milliwatts. In this case, it is referred to as dB_m and is calculated as dB_m = 10 log10 P_m , where P_m is the power in milliwatts. Calculate the power of a signal with dB_m = −30.

​

Solution

We can calculate the power in the signal as

Example 3.29

3.159

The loss in a cable is usually defined in decibels per kilometer (dB/km). If the signal at the beginning of a cable with −0.3 dB/km has a power of 2 mW, what is the power of the signal at 5 km?

Solution

The loss in the cable in decibels is 5 × (−0.3) = −1.5 dB. We can calculate the power as

Example 3.30

3.160

Distortion

• Means that the signal changes its form or shape
• Distortion occurs in composite signals
• Each frequency component has its own propagation speed traveling through a medium.
• The different components therefore arrive with different delays at the receiver.
• That means that the signals have different phases at the receiver than they did at the source.

3.161

3.162

Figure 3.28 Distortion

3.163

Noise

• There are different types of noise
• Thermal - random noise of electrons in the wire creates an extra signal
• Induced - from motors and appliances, devices act are transmitter antenna and medium as receiving antenna.
• Crosstalk - same as above but between two wires.
• Impulse - Spikes that result from power lines, lighning, etc.

3.164

3.165

Figure 3.29 Noise

Signal to Noise Ratio (SNR)

• To measure the quality of a system the SNR is often used. It indicates the strength of the signal wrt the noise power in the system.
• It is the ratio between two powers.
• It is usually given in dB and referred to as SNR_dB.

3.166

3.167

The power of a signal is 10 mW and the power of the noise is 1 μW; what are the values of SNR and SNR_dB ?

​

Solution

The values of SNR and SNRdB can be calculated as follows:

Example 3.31

3.168

The values of SNR and SNRdB for a noiseless channel are

Example 3.32

We can never achieve this ratio in real life; it is an ideal.

3.169

Figure 3.30 Two cases of SNR: a high SNR and a low SNR

3.170

3-5 DATA RATE LIMITS

A very important consideration in data communications is how fast we can send data, in bits per second, over a channel. Data rate depends on three factors:

1. The bandwidth available

2. The level of the signals we use

3. The quality of the channel (the level of noise)

• Noiseless Channel: Nyquist Bit Rate
• Noisy Channel: Shannon Capacity
• Using Both Limits

Topics discussed in this section:

3.171

Increasing the levels of a signal increases the probability of an error occurring, in other words it reduces the reliability of the system. Why??

Capacity of a System

• The bit rate of a system increases with an increase in the number of signal levels we use to denote a symbol.
• A symbol can consist of a single bit or “n” bits.
• The number of signal levels = 2ⁿ.
• As the number of levels goes up, the spacing between level decreases -> increasing the probability of an error occurring in the presence of transmission impairments.

3.172

Nyquist Theorem

• Nyquist gives the upper bound for the bit rate of a transmission system by calculating the bit rate directly from the number of bits in a symbol (or signal levels) and the bandwidth of the system (assuming 2 symbols/per cycle and first harmonic).
• Nyquist theorem states that for a noiseless channel:

C = 2 B log₂2ⁿ

C= capacity in bps

B = bandwidth in Hz

3.173

3.174

Does the Nyquist theorem bit rate agree with the intuitive bit rate described in baseband transmission?

​

Solution

They match when we have only two levels. We said, in baseband transmission, the bit rate is 2 times the bandwidth if we use only the first harmonic in the worst case. However, the Nyquist formula is more general than what we derived intuitively; it can be applied to baseband transmission and modulation. Also, it can be applied when we have two or more levels of signals.

Example 3.33

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Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with two signal levels. The maximum bit rate can be calculated as

Example 3.34

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Consider the same noiseless channel transmitting a signal with four signal levels (for each level, we send 2 bits). The maximum bit rate can be calculated as

Example 3.35

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We need to send 265 kbps over a noiseless channel with a bandwidth of 20 kHz. How many signal levels do we need?

Solution

We can use the Nyquist formula as shown:

Example 3.36

Since this result is not a power of 2, we need to either increase the number of levels or reduce the bit rate. If we have 128 levels, the bit rate is 280 kbps. If we have 64 levels, the bit rate is 240 kbps.

Shannon’s Theorem

• Shannon’s theorem gives the capacity of a system in the presence of noise.

​

C = B log₂(1 + SNR)

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3.179

Consider an extremely noisy channel in which the value of the signal-to-noise ratio is almost zero. In other words, the noise is so strong that the signal is faint. For this channel the capacity C is calculated as

Example 3.37

This means that the capacity of this channel is zero regardless of the bandwidth. In other words, we cannot receive any data through this channel.

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We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of 3000. The signal-to-noise ratio is usually 3162. For this channel the capacity is calculated as

Example 3.38

This means that the highest bit rate for a telephone line is 34.860 kbps. If we want to send data faster than this, we can either increase the bandwidth of the line or improve the signal-to-noise ratio.

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The signal-to-noise ratio is often given in decibels. Assume that SNR_dB = 36 and the channel bandwidth is 2 MHz. The theoretical channel capacity can be calculated as

Example 3.39

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For practical purposes, when the SNR is very high, we can assume that SNR + 1 is almost the same as SNR. In these cases, the theoretical channel capacity can be simplified to

Example 3.40

For example, we can calculate the theoretical capacity of the previous example as

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We have a channel with a 1-MHz bandwidth. The SNR for this channel is 63. What are the appropriate bit rate and signal level?

​

Solution

First, we use the Shannon formula to find the upper limit.

Example 3.41

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The Shannon formula gives us 6 Mbps, the upper limit. For better performance we choose something lower, 4 Mbps, for example. Then we use the Nyquist formula to find the number of signal levels.

Example 3.41 (continued)

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The Shannon capacity gives us the upper limit; the Nyquist formula tells us how many signal levels we need.

3.186

3-6 PERFORMANCE

One important issue in networking is the performance of the network—how good is it? We discuss quality of service, an overall measurement of network performance, in greater detail in Chapter 24. In this section, we introduce terms that we need for future chapters.

• Bandwidth - capacity of the system
• Throughput - no. of bits that can be pushed through
• Latency (Delay) - delay incurred by a bit from start to finish
• Bandwidth-Delay Product

Topics discussed in this section:

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In networking, we use the term bandwidth in two contexts.

• The first, bandwidth in hertz, refers to the range of frequencies in a composite signal or the range of frequencies that a channel can pass.
• The second, bandwidth in bits per second, refers to the speed of bit transmission in a channel or link. Often referred to as Capacity.

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The bandwidth of a subscriber line is 4 kHz for voice or data. The bandwidth of this line for data transmission

can be up to 56,000 bps using a sophisticated modem to change the digital signal to analog.

Example 3.42

3.189

If the telephone company improves the quality of the line and increases the bandwidth to 8 kHz, we can send 112,000 bps by using the same technology as mentioned in Example 3.42.

Example 3.43

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A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network?

​

Solution

We can calculate the throughput as

Example 3.44

The throughput is almost one-fifth of the bandwidth in this case.

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Propagation & Transmission delay

• Propagation speed - speed at which a bit travels though the medium from source to destination.
• Transmission speed - the speed at which all the bits in a message arrive at the destination. (difference in arrival time of first and last bit)

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Propagation and Transmission Delay

• Propagation Delay = Distance/Propagation speed

​

• Transmission Delay = Message size/bandwidth bps

​

• Latency = Propagation delay + Transmission delay + Queueing time + Processing time

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3.194

What is the propagation time if the distance between the two points is 12,000 km? Assume the propagation speed to be 2.4 × 108 m/s in cable.

​

Solution

We can calculate the propagation time as

Example 3.45

The example shows that a bit can go over the Atlantic Ocean in only 50 ms if there is a direct cable between the source and the destination.

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What are the propagation time and the transmission time for a 2.5-kbyte message (an e-mail) if the bandwidth of the network is 1 Gbps? Assume that the distance between the sender and the receiver is 12,000 km and that light travels at 2.4 × 108 m/s.

​

Solution

We can calculate the propagation and transmission time as shown on the next slide:

Example 3.46

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Note that in this case, because the message is short and the bandwidth is high, the dominant factor is the propagation time, not the transmission time. The transmission time can be ignored.

Example 3.46 (continued)

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What are the propagation time and the transmission time for a 5-Mbyte message (an image) if the bandwidth of the network is 1 Mbps? Assume that the distance between the sender and the receiver is 12,000 km and that light travels at 2.4 × 10⁸ m/s.

​

Solution

We can calculate the propagation and transmission times as shown on the next slide.

Example 3.47

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Note that in this case, because the message is very long and the bandwidth is not very high, the dominant factor is the transmission time, not the propagation time. The propagation time can be ignored.

Example 3.47 (continued)

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Figure 3.31 Filling the link with bits for case 1

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We can think about the link between two points as a pipe. The cross section of the pipe represents the bandwidth, and the length of the pipe represents the delay. We can say the volume of the pipe defines the bandwidth-delay product, as shown in Figure 3.33.

Example 3.48

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Figure 3.32 Filling the link with bits in case 2

3.202

The bandwidth-delay product defines the number of bits that can fill the link.

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Figure 3.33 Concept of bandwidth-delay product

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Week: 06

Chapter 4

​

Digital Transmission

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

4.205

4-1 DIGITAL-TO-DIGITAL CONVERSION

In this section, we see how we can represent digital data by using digital signals. The conversion involves three techniques: line coding, block coding, and scrambling. Line coding is always needed; block coding and scrambling may or may not be needed.

• Line Coding
• Line Coding Schemes
• Block Coding
• Scrambling

Topics discussed in this section:

Line Coding

• Converting a string of 1’s and 0’s (digital data) into a sequence of signals that denote the 1’s and 0’s.
• For example a high voltage level (+V) could represent a “1” and a low voltage level (0 or -V) could represent a “0”.

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Figure 4.1 Line coding and decoding

Mapping Data symbols onto Signal levels

• A data symbol (or element) can consist of a number of data bits:
• 1 , 0 or
• 11, 10, 01, ……
• A data symbol can be coded into a single signal element or multiple signal elements
• 1 -> +V, 0 -> -V
• 1 -> +V and -V, 0 -> -V and +V
• The ratio ‘r’ is the number of data elements carried by a signal element.

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Relationship between data rate and signal rate

• The data rate defines the number of bits sent per sec - bps. It is often referred to the bit rate.
• The signal rate is the number of signal elements sent in a second and is measured in bauds. It is also referred to as the modulation rate.
• Goal is to increase the data rate whilst reducing the baud rate.

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Figure 4.2 Signal element versus data element

Data rate and Baud rate

• The baud or signal rate can be expressed as:

S = c x N x 1/r bauds

where N is data rate

c is the case factor (worst, best & avg.)

r is the ratio between data element & signal element

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4.212

A signal is carrying data in which one data element is encoded as one signal element ( r = 1). If the bit rate is 100 kbps, what is the average value of the baud rate if c is between 0 and 1?

Solution

We assume that the average value of c is 1/2 . The baud rate is then

Example 4.1

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Although the actual bandwidth of a digital signal is infinite, the effective bandwidth is finite.

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The maximum data rate of a channel (see Chapter 3) is Nmax = 2 × B × log₂ L (defined by the Nyquist formula). Does this agree with the previous formula for N_max?

Solution

A signal with L levels actually can carry log₂L bits per level. If each level corresponds to one signal element and we assume the average case (c = 1/2), then we have

Example 4.2

Considerations for choosing a good signal element referred to as line encoding

• Baseline wandering - a receiver will evaluate the average power of the received signal (called the baseline) and use that to determine the value of the incoming data elements. If the incoming signal does not vary over a long period of time, the baseline will drift and thus cause errors in detection of incoming data elements.
• A good line encoding scheme will prevent long runs of fixed amplitude.

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Line encoding C/Cs

• DC components - when the voltage level remains constant for long periods of time, there is an increase in the low frequencies of the signal. Most channels are bandpass and may not support the low frequencies.
• This will require the removal of the dc component of a transmitted signal.

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Line encoding C/Cs

• Self synchronization - the clocks at the sender and the receiver must have the same bit interval.
• If the receiver clock is faster or slower it will misinterpret the incoming bit stream.

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Figure 4.3 Effect of lack of synchronization

4.219

In a digital transmission, the receiver clock is 0.1 percent faster than the sender clock. How many extra bits per second does the receiver receive if the data rate is �1 kbps? How many if the data rate is 1 Mbps?

Solution

At 1 kbps, the receiver receives 1001 bps instead of 1000 bps.

Example 4.3

At 1 Mbps, the receiver receives 1,001,000 bps instead of 1,000,000 bps.

Line encoding C/Cs

• Error detection - errors occur during transmission due to line impairments.
• Some codes are constructed such that when an error occurs it can be detected. For example: a particular signal transition is not part of the code. When it occurs, the receiver will know that a symbol error has occurred.

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Line encoding C/Cs

• Noise and interference - there are line encoding techniques that make the transmitted signal “immune” to noise and interference.
• This means that the signal cannot be corrupted, it is stronger than error detection.

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Line encoding C/Cs

• Complexity - the more robust and resilient the code, the more complex it is to implement and the price is often paid in baud rate or required bandwidth.

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Figure 4.4 Line coding schemes

Unipolar

• All signal levels are on one side of the time axis - either above or below
• NRZ - Non Return to Zero scheme is an example of this code. The signal level does not return to zero during a symbol transmission.
• Scheme is prone to baseline wandering and DC components. It has no synchronization or any error detection. It is simple but costly in power consumption.

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Figure 4.5 Unipolar NRZ scheme

Polar - NRZ

• The voltages are on both sides of the time axis.
• Polar NRZ scheme can be implemented with two voltages. E.g. +V for 1 and -V for 0.
• There are two versions:
• NZR - Level (NRZ-L) - positive voltage for one symbol and negative for the other
• NRZ - Inversion (NRZ-I) - the change or lack of change in polarity determines the value of a symbol. E.g. a “1” symbol inverts the polarity a “0” does not.

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Figure 4.6 Polar NRZ-L and NRZ-I schemes

4.228

In NRZ-L the level of the voltage determines the value of the bit. �In NRZ-I the inversion �or the lack of inversion �determines the value of the bit.

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NRZ-L and NRZ-I both have an average signal rate of N/2 Bd.

4.230

NRZ-L and NRZ-I both have a DC component problem and baseline wandering, it is worse for NRZ-L. Both have no self synchronization &no error detection. Both are relatively simple to implement.

4.231

A system is using NRZ-I to transfer 1-Mbps data. What are the average signal rate and minimum bandwidth?

Solution

The average signal rate is S= c x N x R = 1/2 x N x 1 = 500 kbaud. The minimum bandwidth for this average baud rate is Bmin = S = 500 kHz.

​

Note c = 1/2 for the avg. case as worst case is 1 and best case is 0

Example 4.4

Polar - RZ

• The Return to Zero (RZ) scheme uses three voltage values. +, 0, -.
• Each symbol has a transition in the middle. Either from high to zero or from low to zero.
• This scheme has more signal transitions (two per symbol) and therefore requires a wider bandwidth.
• No DC components or baseline wandering.
• Self synchronization - transition indicates symbol value.
• More complex as it uses three voltage level. It has no error detection capability.

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4.233

Figure 4.7 Polar RZ scheme

Polar - Biphase: Manchester and Differential Manchester

• Manchester coding consists of combining the NRZ-L and RZ schemes.
• Every symbol has a level transition in the middle: from high to low or low to high. Uses only two voltage levels.
• Differential Manchester coding consists of combining the NRZ-I and RZ schemes.
• Every symbol has a level transition in the middle. But the level at the beginning of the symbol is determined by the symbol value. One symbol causes a level change the other does not.

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Figure 4.8 Polar biphase: Manchester and differential Manchester schemes

4.236

In Manchester and differential Manchester encoding, the transition

at the middle of the bit is used for synchronization.

4.237

The minimum bandwidth of Manchester and differential Manchester is 2 times that of NRZ. The is no DC component and no baseline wandering. None of these codes has error detection.

Bipolar - AMI and Pseudoternary

• Code uses 3 voltage levels: - +, 0, -, to represent the symbols (note not transitions to zero as in RZ).
• Voltage level for one symbol is at “0” and the other alternates between + & -.
• Bipolar Alternate Mark Inversion (AMI) - the “0” symbol is represented by zero voltage and the “1” symbol alternates between +V and -V.
• Pseudoternary is the reverse of AMI.

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Figure 4.9 Bipolar schemes: AMI and pseudoternary

Bipolar C/Cs

• It is a better alternative to NRZ.
• Has no DC component or baseline wandering.
• Has no self synchronization because long runs of “0”s results in no signal transitions.
• No error detection.

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Multilevel Schemes

• In these schemes we increase the number of data bits per symbol thereby increasing the bit rate.
• Since we are dealing with binary data we only have 2 types of data element a 1 or a 0.
• We can combine the 2 data elements into a pattern of “m” elements to create “2^m” symbols.
• If we have L signal levels, we can use “n” signal elements to create Lⁿ signal elements.

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Code C/Cs

• Now we have 2^m symbols and Lⁿ signals.
• If 2^m > Lⁿ then we cannot represent the data elements, we don’t have enough signals.
• If 2^m = Lⁿ then we have an exact mapping of one symbol on one signal.
• If 2^m < Lⁿ then we have more signals than symbols and we can choose the signals that are more distinct to represent the symbols and therefore have better noise immunity and error detection as some signals are not valid.

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4.243

In mBnL schemes, a pattern of m data elements is encoded as a pattern of n signal elements in which 2^m ≤ Lⁿ.

Representing Multilevel Codes

• We use the notation mBnL, where m is the length of the binary pattern, B represents binary data, n represents the length of the signal pattern and L the number of levels.
• L = B binary, L = T for 3 ternary, L = Q for 4 quaternary.

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Figure 4.10 Multilevel: 2B1Q scheme

Redundancy

• In the 2B1Q scheme we have no redundancy and we see that a DC component is present.
• If we use a code with redundancy we can decide to use only “0” or “+” weighted codes (more +’s than -’s in the signal element) and invert any code that would create a DC component. E.g. ‘+00++-’ -> ‘-00--+’
• Receiver will know when it receives a “-” weighted code that it should invert it as it doesn’t represent any valid symbol.

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Figure 4.11 Multilevel: 8B6T scheme

Multilevel using multiple channels

• In some cases, we split the signal transmission up and distribute it over several links.
• The separate segments are transmitted simultaneously. This reduces the signalling rate per link -> lower bandwidth.
• This requires all bits for a code to be stored.
• xD: means that we use ‘x’ links
• YYYz: We use ‘z’ levels of modulation where YYY represents the type of modulation (e.g. pulse ampl. mod. PAM).
• Codes are represented as: xD-YYYz

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Figure 4.12 Multilevel: 4D-PAM5 scheme

Multitransition Coding

• Because of synchronization requirements we force transitions. This can result in very high bandwidth requirements -> more transitions than are bits (e.g. mid bit transition with inversion).
• Codes can be created that are differential at the bit level forcing transitions at bit boundaries. This results in a bandwidth requirement that is equivalent to the bit rate.
• In some instances, the bandwidth requirement may even be lower, due to repetitive patterns resulting in a periodic signal.

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Figure 4.13 Multitransition: MLT-3 scheme

MLT-3

• Signal rate is same as NRZ-I
• But because of the resulting bit pattern, we have a periodic signal for worst case bit pattern: 1111
• This can be approximated as an analog signal a frequency 1/4 the bit rate!

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Table 4.1 Summary of line coding schemes

Block Coding

• For a code to be capable of error detection, we need to add redundancy, i.e., extra bits to the data bits.
• Synchronization also requires redundancy - transitions are important in the signal flow and must occur frequently.
• Block coding is done in three steps: division, substitution and combination.
• It is distinguished from multilevel coding by use of the slash - xB/yB.
• The resulting bit stream prevents certain bit combinations that when used with line encoding would result in DC components or poor sync. quality.

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4.255

Block coding is normally referred to as mB/nB coding;

it replaces each m-bit group with an �n-bit group.

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Figure 4.14 Block coding concept

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Figure 4.15 Using block coding 4B/5B with NRZ-I line coding scheme

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Table 4.2 4B/5B mapping codes

4.259

Figure 4.16 Substitution in 4B/5B block coding

Redundancy

• A 4 bit data word can have 24 combinations.
• A 5 bit word can have 25=32 combinations.
• We therefore have 32 - 26 = 16 extra words.
• Some of the extra words are used for control/signalling purposes.

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4.261

We need to send data at a 1-Mbps rate. What is the minimum required bandwidth, using a combination of 4B/5B and NRZ-I or Manchester coding?

Solution

First 4B/5B block coding increases the bit rate to 1.25 Mbps. The minimum bandwidth using NRZ-I is N/2 or 625 kHz. The Manchester scheme needs a minimum bandwidth of 1.25 MHz. The first choice needs a lower bandwidth, but has a DC component problem; the second choice needs a higher bandwidth, but does not have a DC component problem.

Example 4.5

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Figure 4.17 8B/10B block encoding

More bits - better error detection

• The 8B10B block code adds more redundant bits and can thereby choose code words that would prevent a long run of a voltage level that would cause DC components.

​

4.263

Scrambling

• The best code is one that does not increase the bandwidth for synchronization and has no DC components.
• Scrambling is a technique used to create a sequence of bits that has the required c/c’s for transmission - self clocking, no low frequencies, no wide bandwidth.
• It is implemented at the same time as encoding, the bit stream is created on the fly.
• It replaces ‘unfriendly’ runs of bits with a violation code that is easy to recognize and removes the unfriendly c/c.

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Figure 4.18 AMI used with scrambling

4.266

For example: B8ZS substitutes eight consecutive zeros with 000VB0VB.

The V stands for violation, it violates the line encoding rule

B stands for bipolar, it implements the bipolar line encoding rule

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Figure 4.19 Two cases of B8ZS scrambling technique

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HDB3 substitutes four consecutive zeros with 000V or B00V depending

on the number of nonzero pulses after the last substitution.

If # of non zero pulses is even the substitution is B00V to make total # of non zero pulse even.

If # of non zero pulses is odd the substitution is 000V to make total # of non zero pulses even.

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Figure 4.20 Different situations in HDB3 scrambling technique

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Week: 07

Chapter 4

​

Digital Transmission

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

4.271

4-2 ANALOG-TO-DIGITAL CONVERSION

A digital signal is superior to an analog signal because it is more robust to noise and can easily be recovered, corrected and amplified. For this reason, the tendency today is to change an analog signal to digital data. In this section we describe two techniques, pulse code modulation and delta modulation.

• Pulse Code Modulation (PCM)
• Delta Modulation (DM)

Topics discussed in this section:

PCM

• PCM consists of three steps to digitize an analog signal:
1. Sampling
2. Quantization
3. Binary encoding
• Before we sample, we have to filter the signal to limit the maximum frequency of the signal as it affects the sampling rate.
• Filtering should ensure that we do not distort the signal, ie remove high frequency components that affect the signal shape.

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Figure 4.21 Components of PCM encoder

Sampling

• Analog signal is sampled every T_S secs.
• T_s is referred to as the sampling interval.
• f_s = 1/T_s is called the sampling rate or sampling frequency.
• There are 3 sampling methods:
• Ideal - an impulse at each sampling instant
• Natural - a pulse of short width with varying amplitude
• Flattop - sample and hold, like natural but with single amplitude value
• The process is referred to as pulse amplitude modulation PAM and the outcome is a signal with analog (non integer) values

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Figure 4.22 Three different sampling methods for PCM

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According to the Nyquist theorem, the sampling rate must be

at least 2 times the highest frequency contained in the signal.

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Figure 4.23 Nyquist sampling rate for low-pass and bandpass signals

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For an intuitive example of the Nyquist theorem, let us sample a simple sine wave at three sampling rates: f_s = 4f (2 times the Nyquist rate), f_s = 2f (Nyquist rate), and �f_s = f (one-half the Nyquist rate). Figure 4.24 shows the sampling and the subsequent recovery of the signal.

​

It can be seen that sampling at the Nyquist rate can create a good approximation of the original sine wave (part a). Oversampling in part b can also create the same approximation, but it is redundant and unnecessary. Sampling below the Nyquist rate (part c) does not produce a signal that looks like the original sine wave.

Example 4.6

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Figure 4.24 Recovery of a sampled sine wave for different sampling rates

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Consider the revolution of a hand of a clock. The second hand of a clock has a period of 60 s. According to the Nyquist theorem, we need to sample the hand every 30 s (T_s = T or f_s = 2f ). In Figure 4.25a, the sample points, in order, are 12, 6, 12, 6, 12, and 6. The receiver of the samples cannot tell if the clock is moving forward or backward. In part b, we sample at double the Nyquist rate (every 15 s). The sample points are 12, 3, 6, 9, and 12. The clock is moving forward. In part c, we sample below the Nyquist rate (T_s = T or f_s = f ). The sample points are 12, 9, 6, 3, and 12. Although the clock is moving forward, the receiver thinks that the clock is moving backward.

Example 4.7

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Figure 4.25 Sampling of a clock with only one hand

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An example related to Example 4.7 is the seemingly backward rotation of the wheels of a forward-moving car in a movie. This can be explained by under-sampling. A movie is filmed at 24 frames per second. If a wheel is rotating more than 12 times per second, the under-sampling creates the impression of a backward rotation.

Example 4.8

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Telephone companies digitize voice by assuming a maximum frequency of 4000 Hz. The sampling rate therefore is 8000 samples per second.

Example 4.9

4.284

A complex low-pass signal has a bandwidth of 200 kHz. What is the minimum sampling rate for this signal?

Solution

The bandwidth of a low-pass signal is between 0 and f, where f is the maximum frequency in the signal. Therefore, we can sample this signal at 2 times the highest frequency (200 kHz). The sampling rate is therefore 400,000 samples per second.

Example 4.10

4.285

A complex bandpass signal has a bandwidth of 200 kHz. What is the minimum sampling rate for this signal?

Solution

We cannot find the minimum sampling rate in this case because we do not know where the bandwidth starts or ends. We do not know the maximum frequency in the signal.

Example 4.11

Quantization

• Sampling results in a series of pulses of varying amplitude values ranging between two limits: a min and a max.
• The amplitude values are infinite between the two limits.
• We need to map the infinite amplitude values onto a finite set of known values.
• This is achieved by dividing the distance between min and max into L zones, each of height Δ.

Δ = (max - min)/L

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Quantization Levels

• The midpoint of each zone is assigned a value from 0 to L-1 (resulting in L values)
• Each sample falling in a zone is then approximated to the value of the midpoint.

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Quantization Zones

• Assume we have a voltage signal with amplitutes V_min=-20V and V_max=+20V.
• We want to use L=8 quantization levels.
• Zone width Δ = (20 - -20)/8 = 5
• The 8 zones are: -20 to -15, -15 to -10, -10 to -5, -5 to 0, 0 to +5, +5 to +10, +10 to +15, +15 to +20
• The midpoints are: -17.5, -12.5, -7.5, -2.5, 2.5, 7.5, 12.5, 17.5

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Assigning Codes to Zones

• Each zone is then assigned a binary code.
• The number of bits required to encode the zones, or the number of bits per sample as it is commonly referred to, is obtained as follows:

n_b = log₂ L

• Given our example, n_b = 3
• The 8 zone (or level) codes are therefore: 000, 001, 010, 011, 100, 101, 110, and 111
• Assigning codes to zones:
• 000 will refer to zone -20 to -15
• 001 to zone -15 to -10, etc.

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Figure 4.26 Quantization and encoding of a sampled signal

Quantization Error

• When a signal is quantized, we introduce an error - the coded signal is an approximation of the actual amplitude value.
• The difference between actual and coded value (midpoint) is referred to as the quantization error.
• The more zones, the smaller Δ which results in smaller errors.
• BUT, the more zones the more bits required to encode the samples -> higher bit rate

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Quantization Error and SN_QR

• Signals with lower amplitude values will suffer more from quantization error as the error range: Δ/2, is fixed for all signal levels.
• Non linear quantization is used to alleviate this problem. Goal is to keep SN_QR fixed for all sample values.
• Two approaches:
• The quantization levels follow a logarithmic curve. Smaller Δ’s at lower amplitudes and larger Δ’s at higher amplitudes.
• Companding: The sample values are compressed at the sender into logarithmic zones, and then expanded at the receiver. The zones are fixed in height.

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Bit rate and bandwidth requirements of PCM

• The bit rate of a PCM signal can be calculated form the number of bits per sample x the sampling rate

Bit rate = n_b x f_s

• The bandwidth required to transmit this signal depends on the type of line encoding used. Refer to previous section for discussion and formulas.
• A digitized signal will always need more bandwidth than the original analog signal. Price we pay for robustness and other features of digital transmission.

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4.294

We want to digitize the human voice. What is the bit rate, assuming 8 bits per sample?

Solution

The human voice normally contains frequencies from 0 to 4000 Hz. So the sampling rate and bit rate are calculated as follows:

Example 4.14

PCM Decoder

• To recover an analog signal from a digitized signal we follow the following steps:
• We use a hold circuit that holds the amplitude value of a pulse till the next pulse arrives.
• We pass this signal through a low pass filter with a cutoff frequency that is equal to the highest frequency in the pre-sampled signal.
• The higher the value of L, the less distorted a signal is recovered.

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Figure 4.27 Components of a PCM decoder

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We have a low-pass analog signal of 4 kHz. If we send the analog signal, we need a channel with a minimum bandwidth of 4 kHz. If we digitize the signal and send 8 bits per sample, we need a channel with a minimum bandwidth of 8 × 4 kHz = 32 kHz.

Example 4.15

Delta Modulation

• This scheme sends only the difference between pulses, if the pulse at time t_n+1 is higher in amplitude value than the pulse at time t_n, then a single bit, say a “1”, is used to indicate the positive value.
• If the pulse is lower in value, resulting in a negative value, a “0” is used.
• This scheme works well for small changes in signal values between samples.
• If changes in amplitude are large, this will result in large errors.

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Figure 4.28 The process of delta modulation

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Figure 4.29 Delta modulation components

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Figure 4.30 Delta demodulation components

Delta PCM (DPCM)

• Instead of using one bit to indicate positive and negative differences, we can use more bits -> quantization of the difference.
• Each bit code is used to represent the value of the difference.
• The more bits the more levels -> the higher the accuracy.

​

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4-3 TRANSMISSION MODES

The transmission of binary data across a link can be accomplished in either parallel or serial mode. In parallel mode, multiple bits are sent with each clock tick. In serial mode, 1 bit is sent with each clock tick. While there is only one way to send parallel data, there are three subclasses of serial transmission: asynchronous, synchronous, and isochronous.

• Parallel Transmission
• Serial Transmission

Topics discussed in this section:

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Figure 4.31 Data transmission and modes

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Figure 4.32 Parallel transmission

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Figure 4.33 Serial transmission

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In asynchronous transmission, we send 1 start bit (0) at the beginning and 1 or more stop bits (1s) at the end of each byte. There may be a gap between �each byte.

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Asynchronous here means “asynchronous at the byte level,”

but the bits are still synchronized; �their durations are the same.

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Figure 4.34 Asynchronous transmission

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In synchronous transmission, we send bits one after another without start or stop bits or gaps. It is the responsibility of the receiver to group the bits. The bits are usually sent as bytes and many bytes are grouped in a frame. A frame is identified with a start and an end byte.

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Figure 4.35 Synchronous transmission

Isochronous

• In isochronous transmission we cannot have uneven gaps between frames.
• Transmission of bits is fixed with equal gaps.

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Chapter 5

​

Analog Transmission

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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5-1 DIGITAL-TO-ANALOG CONVERSION

Digital-to-analog conversion is the process of changing one of the characteristics of an analog signal based on the information in digital data.

• Aspects of Digital-to-Analog Conversion
• Amplitude Shift Keying
• Frequency Shift Keying
• Phase Shift Keying
• Quadrature Amplitude Modulation

Topics discussed in this section:

Digital to Analog Conversion

• Digital data needs to be carried on an analog signal.
• A carrier signal (frequency f_c) performs the function of transporting the digital data in an analog waveform.
• The analog carrier signal is manipulated to uniquely identify the digital data being carried.

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Figure 5.1 Digital-to-analog conversion

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Figure 5.2 Types of digital-to-analog conversion

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Bit rate, N, is the number of bits per second (bps). Baud rate is the number of signal

elements per second (bauds).

In the analog transmission of digital data, the signal or baud rate is less than �or equal to the bit rate.

S=Nx1/r bauds

Where r is the number of data bits per signal element.

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An analog signal carries 4 bits per signal element. If 1000 signal elements are sent per second, find the bit rate.

Solution

In this case, r = 4, S = 1000, and N is unknown. We can find the value of N from

Example 5.1

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Example 5.2

An analog signal has a bit rate of 8000 bps and a baud rate of 1000 baud. How many data elements are carried by each signal element? How many signal elements do we need?

Solution

In this example, S = 1000, N = 8000, and r and L are unknown. We find first the value of r and then the value of L.

Amplitude Shift Keying (ASK)

• ASK is implemented by changing the amplitude of a carrier signal to reflect amplitude levels in the digital signal.
• For example: a digital “1” could not affect the signal, whereas a digital “0” would, by making it zero.
• The line encoding will determine the values of the analog waveform to reflect the digital data being carried.

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Bandwidth of ASK

• The bandwidth B of ASK is proportional to the signal rate S.

B = (1+d)S

• “d” is due to modulation and filtering, lies between 0 and 1.

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Figure 5.3 Binary amplitude shift keying

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Figure 5.4 Implementation of binary ASK

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Example 5.3

We have an available bandwidth of 100 kHz which spans from 200 to 300 kHz. What are the carrier frequency and the bit rate if we modulated our data by using ASK with d = 1?

Solution

The middle of the bandwidth is located at 250 kHz. This means that our carrier frequency can be at f_c = 250 kHz. We can use the formula for bandwidth to find the bit rate (with d = 1 and r = 1).

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Example 5.4

In data communications, we normally use full-duplex links with communication in both directions. We need to divide the bandwidth into two with two carrier frequencies, as shown in Figure 5.5. The figure shows the positions of two carrier frequencies and the bandwidths. The available bandwidth for each direction is now 50 kHz, which leaves us with a data rate of 25 kbps in each direction.

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Figure 5.5 Bandwidth of full-duplex ASK used in Example 5.4

Frequency Shift Keying

• The digital data stream changes the frequency of the carrier signal, f_c.
• For example, a “1” could be represented by f₁=f_c +Δf, and a “0” could be represented by f₂=f_c-Δf.

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Figure 5.6 Binary frequency shift keying

Bandwidth of FSK

• If the difference between the two frequencies (f₁ and f₂) is 2Δf, then the required BW B will be:

B = (1+d)xS +2Δf

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Example 5.5

We have an available bandwidth of 100 kHz which spans from 200 to 300 kHz. What should be the carrier frequency and the bit rate if we modulated our data by using FSK with d = 1?

Solution

This problem is similar to Example 5.3, but we are modulating by using FSK. The midpoint of the band is at 250 kHz. We choose 2Δf to be 50 kHz; this means

Coherent and Non Coherent

• In a non-coherent FSK scheme, when we change from one frequency to the other, we do not adhere to the current phase of the signal.
• In coherent FSK, the switch from one frequency signal to the other only occurs at the same phase in the signal.

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Multi level FSK

• Similarly to ASK, FSK can use multiple bits per signal element.
• That means we need to provision for multiple frequencies, each one to represent a group of data bits.
• The bandwidth for FSK can be higher

B = (1+d)xS + (L-1)/2Δf = LxS

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Figure 5.7 Bandwidth of MFSK used in Example 5.6

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Example 5.6

We need to send data 3 bits at a time at a bit rate of 3 Mbps. The carrier frequency is 10 MHz. Calculate the number of levels (different frequencies), the baud rate, and the bandwidth.

Solution

We can have L = 2³ = 8. The baud rate is S = 3 Mbps/3 = 1 Mbaud. This means that the carrier frequencies must be 1 MHz apart (2Δf = 1 MHz). The bandwidth is B = 8 × 1M = 8M. Figure 5.8 shows the allocation of frequencies and bandwidth.

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Figure 5.8 Bandwidth of MFSK used in Example 5.6

Phase Shift Keyeing

• We vary the phase shift of the carrier signal to represent digital data.
• The bandwidth requirement, B is:

B = (1+d)xS

• PSK is much more robust than ASK as it is not that vulnerable to noise, which changes amplitude of the signal.

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Figure 5.9 Binary phase shift keying

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Figure 5.10 Implementation of BASK

Quadrature PSK

• To increase the bit rate, we can code 2 or more bits onto one signal element.
• In QPSK, we parallelize the bit stream so that every two incoming bits are split up and PSK a carrier frequency. One carrier frequency is phase shifted 90^o from the other - in quadrature.
• The two PSKed signals are then added to produce one of 4 signal elements. L = 4 here.

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Figure 5.11 QPSK and its implementation

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Example 5.7

Find the bandwidth for a signal transmitting at 12 Mbps for QPSK. The value of d = 0.

Solution

For QPSK, 2 bits is carried by one signal element. This means that r = 2. So the signal rate (baud rate) is S = N × (1/r) = 6 Mbaud. With a value of d = 0, we have B = S = 6 MHz.

Constellation Diagrams

• A constellation diagram helps us to define the amplitude and phase of a signal when we are using two carriers, one in quadrature of the other.
• The X-axis represents the in-phase carrier and the Y-axis represents quadrature carrier.

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Figure 5.12 Concept of a constellation diagram

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Example 5.8

Show the constellation diagrams for an ASK (OOK), BPSK, and QPSK signals.

Solution

Figure 5.13 shows the three constellation diagrams.

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Figure 5.13 Three constellation diagrams

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Quadrature amplitude modulation is a combination of ASK and PSK.

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Figure 5.14 Constellation diagrams for some QAMs

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