6/10/2015 unit-1 : data communications 1 cs 1302 computer networks — unit - 1 — — data...
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04/18/23 Unit-1 : Data Communications 1
CS 1302Computer Networks
— Unit - 1 —
— Data Communications —
Text Book Behrouz .A. Forouzan, “Data communication and Networking”, Tata McGrawHill, 2004
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Overview of Overview of Data Communications Data Communications
and and NetworkingNetworking
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Overview
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Introduction
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1.1 Data Communication
Components
Data Representation
Direction of Data Flow
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Figure 1.1 Five components of data communication
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Figure 1.2 Simplex
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Figure 1.3 Half-duplex
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Figure 1.4 Full-duplex
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1.2 Networks
Distributed Processing
Network Criteria
Physical Structures
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Figure 1.5 Point-to-point connection
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Figure 1.6 Multipoint connection
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Figure 1.7 Categories of topology
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Figure 1.8 Fully connected mesh topology (for five devices)
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Figure 1.9 Star topology
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Figure 1.10 Bus topology
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Figure 1.11 Ring topology
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Figure 1.12 Categories of networks
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Figure 1.13 LAN
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Figure 1.13 LAN (Continued)
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Figure 1.14 MAN
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Figure 1.15 WAN
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1.3 The Internet1.3 The Internet
A Brief History
The Internet Today
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Figure 1.16 Internet today
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1.4 Protocols and Standards1.4 Protocols and Standards
Protocols
Standards
Standards Organizations
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NetworkModels
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2.1 Layered Tasks
Sender, Receiver, and Carrier
Hierarchy
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Figure 2.1 Sending a letter
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2.2 Internet Model
Peer-to-Peer Processes
Functions of Layers
Summary of Layers
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Figure 2.2 Internet layers
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Figure 2.3 Peer-to-peer processes
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Figure 2.4 An exchange using the Internet model
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Figure 2.5 Physical layer
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The physical layer is responsible for transmitting individual bits from one
node to the next.
Note:Note:
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Figure 2.6 Data link layer
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The data link layer is responsible for transmitting frames from
one node to the next.
Note:Note:
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Figure 2.7 Node-to-node delivery
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Example 1Example 1
In Figure 2.8 a node with physical address 10 sends a frame to a node with physical address 87. The two nodes are connected by a link. At the data link level this frame contains physical addresses in the header. These are the only addresses needed. The rest of the header contains other information needed at this level. The trailer usually contains extra bits needed for error detection
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Figure 2.8 Example 1
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Figure 2.9 Network layer
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The network layer is responsible for the delivery of packets from the
original source to the final destination.
Note:Note:
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Figure 2.10 Source-to-destination delivery
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Example 2Example 2
In Figure 2.11 we want to send data from a node with network address A and physical address 10, located on one LAN, to a node with a network address P and physical address 95, located on another LAN. Because the two devices are located on different networks, we cannot use physical addresses only; the physical addresses only have local jurisdiction. What we need here are universal addresses that can pass through the LAN boundaries. The network (logical) addresses have this characteristic.
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Figure 2.11 Example 2
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Figure 2.12 Transport layer
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The transport layer is responsible for delivery of a message from one process
to another.
Note:Note:
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Figure 2.12 Reliable process-to-process delivery of a message
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Example 3Example 3
Figure 2.14 shows an example of transport layer communication. Data coming from the upper layers have port addresses j and k (j is the address of the sending process, and k is the address of the receiving process). Since the data size is larger than the network layer can handle, the data are split into two packets, each packet retaining the port addresses (j and k). Then in the network layer, network addresses (A and P) are added to each packet.
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Figure 2.14 Example 3
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Figure 2.15 Application layer
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The application layer is responsible for providing services to the user.
Note:Note:
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Figure 2.16 Summary of duties
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2.3 OSI Model2.3 OSI Model
A comparison
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Figure 2.17 OSI model
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DigitalTransmission
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4.1 Line Coding
Some Characteristics
Line Coding Schemes
Some Other Schemes
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Figure 4.1 Line coding
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Figure 4.2 Signal level versus data level
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Figure 4.3 DC component
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Example 1Example 1
A signal has two data levels with a pulse duration of 1 ms. We calculate the pulse rate and bit rate as follows:
Pulse Rate = 1/ 10Pulse Rate = 1/ 10-3-3= 1000 pulses/s= 1000 pulses/s
Bit Rate = Pulse Rate x logBit Rate = Pulse Rate x log22 L = 1000 x log L = 1000 x log22 2 = 1000 bps 2 = 1000 bps
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Example 2Example 2
A signal has four data levels with a pulse duration of 1 ms. We calculate the pulse rate and bit rate as follows:
Pulse Rate = = 1000 pulses/sPulse Rate = = 1000 pulses/s
Bit Rate = PulseRate x logBit Rate = PulseRate x log22 L = 1000 x log L = 1000 x log22 4 = 2000 bps 4 = 2000 bps
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Figure 4.4 Lack of synchronization
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Example 3Example 3
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?
SolutionSolution
At 1 Kbps:1000 bits sent 1001 bits received1 extra bpsAt 1 Mbps: 1,000,000 bits sent 1,001,000 bits received1000 extra bps
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Figure 4.5 Line coding schemes
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Unipolar encoding uses only one voltage level.
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Figure 4.6 Unipolar encoding
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Polar encoding uses two voltage levels Polar encoding uses two voltage levels (positive and negative).(positive and negative).
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Figure 4.7 Types of polar encoding
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In NRZ-L the level of the signal is In NRZ-L the level of the signal is dependent upon the state of the bit.dependent upon the state of the bit.
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In NRZ-I the signal is inverted if a 1 is In NRZ-I the signal is inverted if a 1 is encountered.encountered.
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Figure 4.8 NRZ-L and NRZ-I encoding
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Figure 4.9 RZ encoding
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A good encoded digital signal must A good encoded digital signal must contain a provision for contain a provision for
synchronization.synchronization.
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Figure 4.10 Manchester encoding
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In Manchester encoding, the In Manchester encoding, the transition at the middle of the bit is transition at the middle of the bit is
used for both synchronization and bit used for both synchronization and bit representation.representation.
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Figure 4.11 Differential Manchester encoding
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In differential Manchester encoding, In differential Manchester encoding, the transition at the middle of the bit is the transition at the middle of the bit is
used only for synchronization. used only for synchronization. The bit representation is defined by the The bit representation is defined by the
inversion or noninversion at the inversion or noninversion at the beginning of the bit.beginning of the bit.
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In bipolar encoding, we use three In bipolar encoding, we use three levels: positive, zero, levels: positive, zero,
and negative.and negative.
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Figure 4.12 Bipolar AMI encoding
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Figure 4.13 2B1Q
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Figure 4.14 MLT-3 signal
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4.2 Block Coding
Steps in Transformation
Some Common Block Codes
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Figure 4.15 Block coding
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Figure 4.16 Substitution in block coding
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Table 4.1 4B/5B encodingTable 4.1 4B/5B encoding
Data Code Data Code
0000 1111011110 1000 1001010010
0001 0100101001 1001 1001110011
0010 1010010100 1010 1011010110
0011 1010110101 1011 1011110111
0100 0101001010 1100 1101011010
0101 0101101011 1101 1101111011
0110 0111001110 1110 1110011100
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Table 4.1 4B/5B encoding (Continued)Table 4.1 4B/5B encoding (Continued)
Data Code
Q (Quiet) 0000000000
I (Idle) 1111111111
H (Halt) 0010000100
J (start delimiter) 1100011000
K (start delimiter) 1000110001
T (end delimiter) 0110101101
S (Set) 1100111001
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Figure 4.17 Example of 8B/6T encoding
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4.3 Sampling4.3 Sampling
Pulse Amplitude ModulationPulse Code ModulationSampling Rate: Nyquist TheoremHow Many Bits per Sample?Bit Rate
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Figure 4.18 PAM
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Pulse amplitude modulation has some Pulse amplitude modulation has some applications, but it is not used by itself applications, but it is not used by itself in data communication. However, it is in data communication. However, it is the first step in another very popular the first step in another very popular
conversion method called conversion method called pulse code modulation.pulse code modulation.
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Figure 4.19 Quantized PAM signal
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Figure 4.20 Quantizing by using sign and magnitude
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Figure 4.21 PCM
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Figure 4.22 From analog signal to PCM digital code
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According to the Nyquist theorem, the According to the Nyquist theorem, the sampling rate must be at least 2 times sampling rate must be at least 2 times
the highest frequency.the highest frequency.
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Figure 4.23 Nyquist theorem
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Example 4Example 4
What sampling rate is needed for a signal with a bandwidth of 10,000 Hz (1000 to 11,000 Hz)?
SolutionSolution
The sampling rate must be twice the highest frequency in the signal:
Sampling rate = 2 x (11,000) = 22,000 samples/sSampling rate = 2 x (11,000) = 22,000 samples/s
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Example 5Example 5
A signal is sampled. Each sample requires at least 12 levels of precision (+0 to +5 and -0 to -5). How many bits should be sent for each sample?
SolutionSolution
We need 4 bits; 1 bit for the sign and 3 bits for the value. A 3-bit value can represent 23 = 8 levels (000 to 111), which is more than what we need. A 2-bit value is not enough since 22 = 4. A 4-bit value is too much because 24 = 16.
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Example 6Example 6
We want to digitize the human voice. What is the bit rate, assuming 8 bits per sample?
SolutionSolution
The human voice normally contains frequencies from 0 to 4000 Hz. Sampling rate = 4000 x 2 = 8000 samples/sSampling rate = 4000 x 2 = 8000 samples/s
Bit rate = sampling rate x number of bits per sample Bit rate = sampling rate x number of bits per sample = 8000 x 8 = 64,000 bps = 64 Kbps= 8000 x 8 = 64,000 bps = 64 Kbps
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Note that we can always change a Note that we can always change a band-pass signal to a low-pass signal band-pass signal to a low-pass signal
before sampling. In this case, the before sampling. In this case, the sampling rate is twice the bandwidth.sampling rate is twice the bandwidth.
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4.4 Transmission Mode4.4 Transmission Mode
Parallel Transmission
Serial Transmission
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Figure 4.24 Data transmission
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Figure 4.25 Parallel transmission
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Figure 4.26 Serial transmission
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In asynchronous transmission, we In asynchronous transmission, we send 1 start bit (0) at the beginning send 1 start bit (0) at the beginning
and 1 or more stop bits (1s) at the end and 1 or more stop bits (1s) at the end of each byte. There may be a gap of each byte. There may be a gap
between each byte.between each byte.
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Asynchronous here means Asynchronous here means “asynchronous at the byte level,” but “asynchronous at the byte level,” but the bits are still synchronized; their the bits are still synchronized; their
durations are the same.durations are the same.
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Figure 4.27 Asynchronous transmission
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In synchronous transmission, In synchronous transmission, we send bits one after another without we send bits one after another without
start/stop bits or gaps. start/stop bits or gaps. It is the responsibility of the receiver to It is the responsibility of the receiver to
group the bits.group the bits.
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Figure 4.28 Synchronous transmission
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5.2 Telephone Modems
Modem Standards
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A telephone line has a bandwidth of almost 2400 Hz for data transmission.
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Figure 5.18 Telephone line bandwidth
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Modem stands for modulator/demodulator.
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Figure 5.19 Modulation/demodulation
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Figure 5.20 The V.32 constellation and bandwidth
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Figure 5.21 The V.32bis constellation and bandwidth
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Figure 5.22 Traditional modems
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Figure 5.23 56K modems
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TransmissionMedia
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Figure 7.1 Transmission medium and physical layer
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Figure 7.2 Classes of transmission media
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7.1 Guided Media7.1 Guided Media
Twisted-Pair Cable
Coaxial Cable
Fiber-Optic Cable
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Figure 7.3 Twisted-pair cable
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Figure 7.4 UTP and STP
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Table 7.1 Categories of unshielded twisted-pair cablesTable 7.1 Categories of unshielded twisted-pair cables
Category Bandwidth Data Rate Digital/Analog Use
1 very low < 100 kbps Analog Telephone
2 < 2 MHz 2 Mbps Analog/digital T-1 lines
3 16 MHz 10 Mbps Digital LANs
4 20 MHz 20 Mbps Digital LANs
5 100 MHz 100 Mbps Digital LANs
6 (draft) 200 MHz 200 Mbps Digital LANs
7 (draft) 600 MHz 600 Mbps Digital LANs
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Figure 7.5 UTP connector
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Figure 7.6 UTP performance
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Figure 7.7 Coaxial cable
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Table 7.2 Categories of coaxial cablesTable 7.2 Categories of coaxial cables
Category Impedance Use
RG-59RG-59 75 Cable TV
RG-58RG-58 50 Thin Ethernet
RG-11RG-11 50 Thick Ethernet
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Figure 7.8 BNC connectors
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Figure 7.9 Coaxial cable performance
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Figure 7.10 Bending of light ray
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Figure 7.11 Optical fiber
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Figure 7.12 Propagation modes
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Figure 7.13 Modes
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Table 7.3 Fiber typesTable 7.3 Fiber types
Type Core Cladding Mode
50/12550/125 50 125 Multimode, graded-index
62.5/12562.5/125 62.5 125 Multimode, graded-index
100/125100/125 100 125 Multimode, graded-index
7/1257/125 7 125 Single-mode
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Figure 7.14 Fiber construction
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Figure 7.15 Fiber-optic cable connectors
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Figure 7.16 Optical fiber performance
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7.2 Unguided Media: Wireless7.2 Unguided Media: Wireless
Radio Waves
Microwaves
Infrared
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Figure 7.17 Electromagnetic spectrum for wireless communication
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Figure 7.18 Propagation methods
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Table 7.4 BandsTable 7.4 Bands
BandBand RangeRange PropagationPropagation ApplicationApplication
VLFVLF 3–30 KHz Ground Long-range radio navigation
LFLF 30–300 KHz GroundRadio beacons and
navigational locators
MFMF 300 KHz–3 MHz Sky AM radio
HF HF 3–30 MHz SkyCitizens band (CB),
ship/aircraft communication
VHF VHF 30–300 MHzSky and
line-of-sightVHF TV, FM radio
UHF UHF 300 MHz–3 GHz Line-of-sightUHF TV, cellular phones,
paging, satellite
SHF SHF 3–30 GHz Line-of-sight Satellite communication
EHFEHF 30–300 GHz Line-of-sight Long-range radio navigation
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Figure 7.19 Wireless transmission waves
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Figure 7.20 Omnidirectional antennas
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Radio waves are used for multicast communications, such as radio and
television, and paging systems.
NoteNote::
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Figure 7.21 Unidirectional antennas
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Microwaves are used for unicast communication such as cellular
telephones, satellite networks, and wireless LANs.
NoteNote::
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Infrared signals can be used for short-range communication in a closed area
using line-of-sight propagation.
NoteNote::
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