8/26/09
8/23/09
Classless addressing
Classless addressing
Classful addressing has created many problems
Many ISPs and service users need more addresses
Idea is to have variable-length blocks that belong to no class
Three restrictions on classless address blocks;
The addresses in a block must be contiguous, one after another
The number of addresses in a block must be a power of 2
The first address must be evenly divisible by the number of addresses
Mask and Address Blocks
In IPv4 addressing, a block of addresses can be defined as x.y.z.t /n in which x.y.z.t defines one of the addresses and the /n defines the mask.
The first address in the block can be found by setting the rightmost 32 − n bits to 0s
The last address in the block can be found by setting the rightmost 32 − n bits to 1s
The number of addresses in the block can be found by using the formula 232−n
Example: 205.16.37.39/28
The binary representation is 1100110 00010000 00100101 00100111
If we set 32 − 28 rightmost bits to 0, we get 11001101 00010000 00100101 00100000
205.16.37.32 (First address)
If we set 32 − 28 rightmost bits to 1, we get 11001101 00010000 00100101 00101111
205.16.37.47 (Last address)
The value of n is 28, which means that number of addresses is 232−28 or 16
Network Address
The first address in a block is normally not assigned to any device; it is used as the network address that represents the organization to the rest of the world
Two-Level Hierarchy: No Subnetting
Each address in the block can be considered as a two-level hierarchical structure: the leftmost n bits (prefix) define the network; the rightmost 32 − n bits define the host
Address Allocation and Distribution: Example
The first group has 64 customers; each needs 256 addresses.
The second group has 128 customers; each needs 128 addresses.
The third group has 128 customers; each needs 64 addresses.
Network Address Translation: NAT
NAT enables a user to have a large set of addresses internally and one address, or a small set of addresses, externally.
Addresses Translation
Address translation for source address of outgoing packet and for destination address of incoming packet
Translation Table
Using (1) one IP address, (2) a pool of IP address, and (3) both IP addresses and port numbers
Five-Column Translation Table
IPv6 Addresses
Despite all short-term solutions, such as classless addressing, DHCP (Dynamic Host Configuration Protocol), and NAT, still address-hungry
An IPv6 address is 128 bits long
Hexadecimal colon notation:
IPv6 Address Space
IPv6 Addresses
Unicast addresses: define a single computer
Two types: geographically based and provider-based
Prefixes for provider-based unicast address
Type id (3 bits), Registry id (5 bits)
IPv6 Addresses
Anycast addresses: define a group of nodes
Unlike multicast, a packet is delivered to only one of the members of the anycast group, the nearest
Classful addressing has created many problems
Many ISPs and service users need more addresses
Idea is to have variable-length blocks that belong to no class
Three restrictions on classless address blocks;
The addresses in a block must be contiguous, one after another
The number of addresses in a block must be a power of 2
The first address must be evenly divisible by the number of addresses
Mask and Address Blocks
In IPv4 addressing, a block of addresses can be defined as x.y.z.t /n in which x.y.z.t defines one of the addresses and the /n defines the mask.
The first address in the block can be found by setting the rightmost 32 − n bits to 0s
The last address in the block can be found by setting the rightmost 32 − n bits to 1s
The number of addresses in the block can be found by using the formula 232−n
Example: 205.16.37.39/28
The binary representation is 1100110 00010000 00100101 00100111
If we set 32 − 28 rightmost bits to 0, we get 11001101 00010000 00100101 00100000
205.16.37.32 (First address)
If we set 32 − 28 rightmost bits to 1, we get 11001101 00010000 00100101 00101111
205.16.37.47 (Last address)
The value of n is 28, which means that number of addresses is 232−28 or 16
Network Address
The first address in a block is normally not assigned to any device; it is used as the network address that represents the organization to the rest of the world
Two-Level Hierarchy: No Subnetting
Each address in the block can be considered as a two-level hierarchical structure: the leftmost n bits (prefix) define the network; the rightmost 32 − n bits define the host
Address Allocation and Distribution: Example
The first group has 64 customers; each needs 256 addresses.
The second group has 128 customers; each needs 128 addresses.
The third group has 128 customers; each needs 64 addresses.
Network Address Translation: NAT
NAT enables a user to have a large set of addresses internally and one address, or a small set of addresses, externally.
Addresses Translation
Address translation for source address of outgoing packet and for destination address of incoming packet
Translation Table
Using (1) one IP address, (2) a pool of IP address, and (3) both IP addresses and port numbers
Five-Column Translation Table
IPv6 Addresses
Despite all short-term solutions, such as classless addressing, DHCP (Dynamic Host Configuration Protocol), and NAT, still address-hungry
An IPv6 address is 128 bits long
Hexadecimal colon notation:
IPv6 Address Space
IPv6 Addresses
Unicast addresses: define a single computer
Two types: geographically based and provider-based
Prefixes for provider-based unicast address
Type id (3 bits), Registry id (5 bits)
IPv6 Addresses
Anycast addresses: define a group of nodes
Unlike multicast, a packet is delivered to only one of the members of the anycast group, the nearest
Logical Addressing
Logical Addressing
Position of network layer
The network layer is responsible for the delivery of individual packets from the source to the destination host
Network Layer: Logical Addressing
19.1 IPv4 Addresses
19.2 IPv6 Addresses
IPv4 Addresses
An IP address is a 32-bits long
The IP addresses are unique and universal
The address space of IPv4 is 232 or 4,294,967,296
Binary notation: 01110101 10010101 00011101 00000010
Dotted-decimal notation: 117.149.29.2
Example
Change the following IP addresses from binary notation to dotted-decimal notation.
a. 10000001 00001011 00001011 11101111
b. 11111001 10011011 11111011 00001111
We replace each group of 8 bits with its equivalent decimal number and add dots for separation:
a. 129.11.11.239
b. 249.155.251.15
Classful addressing
In classful addressing, the address space is divided into five classes: A, B, C, D, E
A new architecture, called classless addressing was introduced in the mid-1990s
Classful Addressing: Example
Find the class of each address.
a. 00000001 00001011 00001011 11101111
b. 11000001 10000011 00011011 11111111
c. 14.23.120.8
d. 252.5.15.111
Classful Addressing: Example Solution
Solution
a. The first bit is 0. This is a class A address.
b. The first 2 bits are 1; the third bit is 0. This is a class C address.
c. The first byte is 14; the class is A.
d. The first byte is 252; the class is E.
Position of network layer
The network layer is responsible for the delivery of individual packets from the source to the destination host
Network Layer: Logical Addressing
19.1 IPv4 Addresses
19.2 IPv6 Addresses
IPv4 Addresses
An IP address is a 32-bits long
The IP addresses are unique and universal
The address space of IPv4 is 232 or 4,294,967,296
Binary notation: 01110101 10010101 00011101 00000010
Dotted-decimal notation: 117.149.29.2
Example
Change the following IP addresses from binary notation to dotted-decimal notation.
a. 10000001 00001011 00001011 11101111
b. 11111001 10011011 11111011 00001111
We replace each group of 8 bits with its equivalent decimal number and add dots for separation:
a. 129.11.11.239
b. 249.155.251.15
Classful addressing
In classful addressing, the address space is divided into five classes: A, B, C, D, E
A new architecture, called classless addressing was introduced in the mid-1990s
Classful Addressing: Example
Find the class of each address.
a. 00000001 00001011 00001011 11101111
b. 11000001 10000011 00011011 11111111
c. 14.23.120.8
d. 252.5.15.111
Classful Addressing: Example Solution
Solution
a. The first bit is 0. This is a class A address.
b. The first 2 bits are 1; the third bit is 0. This is a class C address.
c. The first byte is 14; the class is A.
d. The first byte is 252; the class is E.
SONET/SDH
SONET/SDH
17.1 Architecture
17.2 SONET Layers
17.3 SONET Frames
17.4 STS Multiplexing
17.5 SONET Networks
17.6 Virtual Tributaries
Digital transmission standards for fiber-optic cable
Independently developed in USA & Europe
SONET(Synchronous Optical Network) by ANSI
SDH(Synchronous Digital Hierarchy) by ITU-T
Synchronous network using synchronous TDM multiplexing
All clocks in the system are locked to a master clock
It contains the standards for fiber-optic equipments
Very flexible to carry other transmission systems (DS-0, DS-1, etc)
SONET/SDH Architecture
Architecture of a SONET system: signals, devices, and connections
Signals: SONET(SDH) defines a hierarchy of electrical signaling levels called STSs(Synchronous Transport Signals, (STMs)). Corresponding optical signals are called OCs(Optical Carriers)
SONET/SDH Architecture
SONET devices: STS multiplexer/demultiplexer, regenerator, add/drop multiplexer, terminals
SONET/SDH Architecture
Connections: SONET devices are connected using sections, lines, and paths
Section: optical link connecting two neighbor devices: mux to mux, mux to regenerator, or regenerator to regenerator
Lines: portion of network between two multiplexers
Paths: end-to-end portion of the network between two STS multiplexers
SONET Layers
SONET defines four layers: path, line, section, and photonic
Path layer is responsible for the movement of a signal from its optical source to its optical destination
Line layers is for the movement of a signal across a physical line
Section layer is for the movement of a signal across a physical section, handling framing, scrambling, and error control
Photonic layer corresponds to the physical layer of OSI model
SONET Layers
SONET Frames
Each synchronous transfer signal STS-n is composed of 8000 frames. Each frame is a two-dimensional matrix of bytes with 9 rows by 90 × n columns.
A SONET STS-n signal is transmitted at 8000 frames per second
Each byte in a SONET frame can carry a digitized voice channel
774 voice channels are supported by STS 1 signaling simultaneously.
Prove that STS-1 has data rate of 51.84 Mbps.
SONET Frames
In SONET, the data rate of an STS-n signal is n times the data rate of an STS-1 signal
In SONET, the duration of any frame is 125 μs
Concatenated Signal
The suffix c (for concatenated) means that the STS-n is not considered as n STS-1 signals. So, it cannot be demultiplexed into n STS-1 signals
An STS-3c signal can carry 44 ATM cells as its SPE
SPE of an STS-3c can carry 9 x 260 = 2340 which can accommodate approximately 44 ATM cells, each of 53 bytes
Add/Drop Multiplexer
Only remove the corresponding bytes and replace them with the new bytes including the bytes in the section and line overhead
17.1 Architecture
17.2 SONET Layers
17.3 SONET Frames
17.4 STS Multiplexing
17.5 SONET Networks
17.6 Virtual Tributaries
Digital transmission standards for fiber-optic cable
Independently developed in USA & Europe
SONET(Synchronous Optical Network) by ANSI
SDH(Synchronous Digital Hierarchy) by ITU-T
Synchronous network using synchronous TDM multiplexing
All clocks in the system are locked to a master clock
It contains the standards for fiber-optic equipments
Very flexible to carry other transmission systems (DS-0, DS-1, etc)
SONET/SDH Architecture
Architecture of a SONET system: signals, devices, and connections
Signals: SONET(SDH) defines a hierarchy of electrical signaling levels called STSs(Synchronous Transport Signals, (STMs)). Corresponding optical signals are called OCs(Optical Carriers)
SONET/SDH Architecture
SONET devices: STS multiplexer/demultiplexer, regenerator, add/drop multiplexer, terminals
SONET/SDH Architecture
Connections: SONET devices are connected using sections, lines, and paths
Section: optical link connecting two neighbor devices: mux to mux, mux to regenerator, or regenerator to regenerator
Lines: portion of network between two multiplexers
Paths: end-to-end portion of the network between two STS multiplexers
SONET Layers
SONET defines four layers: path, line, section, and photonic
Path layer is responsible for the movement of a signal from its optical source to its optical destination
Line layers is for the movement of a signal across a physical line
Section layer is for the movement of a signal across a physical section, handling framing, scrambling, and error control
Photonic layer corresponds to the physical layer of OSI model
SONET Layers
SONET Frames
Each synchronous transfer signal STS-n is composed of 8000 frames. Each frame is a two-dimensional matrix of bytes with 9 rows by 90 × n columns.
A SONET STS-n signal is transmitted at 8000 frames per second
Each byte in a SONET frame can carry a digitized voice channel
774 voice channels are supported by STS 1 signaling simultaneously.
Prove that STS-1 has data rate of 51.84 Mbps.
SONET Frames
In SONET, the data rate of an STS-n signal is n times the data rate of an STS-1 signal
In SONET, the duration of any frame is 125 μs
Concatenated Signal
The suffix c (for concatenated) means that the STS-n is not considered as n STS-1 signals. So, it cannot be demultiplexed into n STS-1 signals
An STS-3c signal can carry 44 ATM cells as its SPE
SPE of an STS-3c can carry 9 x 260 = 2340 which can accommodate approximately 44 ATM cells, each of 53 bytes
Add/Drop Multiplexer
Only remove the corresponding bytes and replace them with the new bytes including the bytes in the section and line overhead
Wired LANs: Ethernet
Wired LANs: Ethernet
IEEE Standards
Standard Ethernet
Changes in the Standard
Fast Ethernet
Gigabit Ethernet
IEEE Standards
In 1985, the Computer Society of the IEEE started a project, called Project 802, to set standards to enable intercommunication among equipment from a variety of manufacturers. Project 802 is a way of specifying functions of the physical layer and the data link layer of major LAN protocols.
IEEE 802 Working Group
Active working groups Inactive or disbanded working groups
802.1 Higher Layer LAN Protocols Working
Group
802.3 Ethernet Working Group
802.11 Wireless LAN Working Group
802.15 Wireless Personal Area Network
(WPAN) Working Group
802.16 Broadband Wireless Access Working
Group
802.17 Resilient Packet Ring Working Group
802.18 Radio Regulatory TAG
802.19 Coexistence TAG
802.20 Mobile Broadband Wireless Access
(MBWA) Working Group
802.21 Media Independent Handoff Working
Group
802.22 Wireless Regional Area Networks 802.2 Logical Link Control Working Group
802.4 Token Bus Working Group
802.5 Token Ring Working Group
802.7 Broadband Area Network Working
Group
802.8 Fiber Optic TAG
802.9 Integrated Service LAN Working
Group
802.10 Security Working Group
802.12 Demand Priority Working Group
802.14 Cable Modem Working Group
Logical Link Control (LLC)
Framing: LLC defines a protocol data unit (PDU)
To provide flow and error control for the upper-layer protocols that actually demand these services
Standard Ethernet
The original Ethernet was created in 1976 at Xerox’s Palo Alto Research Center (PARC). Since then, it has gone through four generations
MAC Sublayer
Preamble: alerting the receiving system to the coming frame and enabling it to synchronize its input timing
CRC: CRC-32
Addressing
The least significant bit of the first byte defines the type of address. If the bit is 0, the address is unicast; otherwise, it is multicast
The broadcast destination address is a special case of the multicast address in which all bits are 1s
Ethernet address in hexadecimal notation
Ethernet
Access method: 1-persistent CSMA/CD
Slot time = rount-trip time + time required to send the jam sequence
512 bits for Ethernet, 51.2 μs for 10 Mbps Ethernet
Slot time and collision
Slot time and maximum network length
MaxLength = PropagationSpeed x SlotTime/2
MaxLength = (2 x 108) x (51.2 x 10-6/2) = 5120 m
MaxLength = 2500 m 48 % of the theoretical calculation by considering delay times in repeaters and interfaces, and the time required to send the jam sequence
Fast Ethernet
Under the name of IEEE 802.3u
Upgrade the data rate to 100 Mbps
Make it compatible with Standard Ethernet
Keep the same 48-bit address and the same frame format
Keep the same min. and max. frame length
MAC Sublayer
CSMA/CD for the half-duplex approach
No need for CSMA/CD for full-duplex Fast Ethernet
Autonegotiation: allow two devices to negotiate the mode or data rate of operation
Gigabit Ethernet
Under the name of IEEE 802.3z
Upgrade the data rate to 1 Gbps
Make it compatible with Standard or Fast Ethernet
Keep the same 48-bit address and the same frame format
Keep the same min. and max. frame length
Support autonegotiation as defined in Fast Ethernet
MAC Sublayer
Most of all implmentations follows full-duplex approach
In the full-duplex mode of Gigabit Ethernet, there is no collision; the maximum length of the cable is determined by the signal attenuation in the cable.
Half-duplex mode (very rare)
Traditional: 0.512 μs (25m)
Carrier Extension: 512 bytes (4096 bits) min. length
Frame bursting to improve the inefficiency of carrier extension
Ten-Gigabit Ethernet
Under the name of IEEE 802.3ae
Upgrade the data rate to 10 G bps
Make it compatible with Standard, Fast, and Giga Ethernet
Keep the same 48-bit address and the same frame format
Keep the same min. and max. frame length
Allow the interconnection of existing LANs into a MAN or WAN
Make Ethernet compatible with Frame Relay and ATM
IEEE Standards
Standard Ethernet
Changes in the Standard
Fast Ethernet
Gigabit Ethernet
IEEE Standards
In 1985, the Computer Society of the IEEE started a project, called Project 802, to set standards to enable intercommunication among equipment from a variety of manufacturers. Project 802 is a way of specifying functions of the physical layer and the data link layer of major LAN protocols.
IEEE 802 Working Group
Active working groups Inactive or disbanded working groups
802.1 Higher Layer LAN Protocols Working
Group
802.3 Ethernet Working Group
802.11 Wireless LAN Working Group
802.15 Wireless Personal Area Network
(WPAN) Working Group
802.16 Broadband Wireless Access Working
Group
802.17 Resilient Packet Ring Working Group
802.18 Radio Regulatory TAG
802.19 Coexistence TAG
802.20 Mobile Broadband Wireless Access
(MBWA) Working Group
802.21 Media Independent Handoff Working
Group
802.22 Wireless Regional Area Networks 802.2 Logical Link Control Working Group
802.4 Token Bus Working Group
802.5 Token Ring Working Group
802.7 Broadband Area Network Working
Group
802.8 Fiber Optic TAG
802.9 Integrated Service LAN Working
Group
802.10 Security Working Group
802.12 Demand Priority Working Group
802.14 Cable Modem Working Group
Logical Link Control (LLC)
Framing: LLC defines a protocol data unit (PDU)
To provide flow and error control for the upper-layer protocols that actually demand these services
Standard Ethernet
The original Ethernet was created in 1976 at Xerox’s Palo Alto Research Center (PARC). Since then, it has gone through four generations
MAC Sublayer
Preamble: alerting the receiving system to the coming frame and enabling it to synchronize its input timing
CRC: CRC-32
Addressing
The least significant bit of the first byte defines the type of address. If the bit is 0, the address is unicast; otherwise, it is multicast
The broadcast destination address is a special case of the multicast address in which all bits are 1s
Ethernet address in hexadecimal notation
Ethernet
Access method: 1-persistent CSMA/CD
Slot time = rount-trip time + time required to send the jam sequence
512 bits for Ethernet, 51.2 μs for 10 Mbps Ethernet
Slot time and collision
Slot time and maximum network length
MaxLength = PropagationSpeed x SlotTime/2
MaxLength = (2 x 108) x (51.2 x 10-6/2) = 5120 m
MaxLength = 2500 m 48 % of the theoretical calculation by considering delay times in repeaters and interfaces, and the time required to send the jam sequence
Fast Ethernet
Under the name of IEEE 802.3u
Upgrade the data rate to 100 Mbps
Make it compatible with Standard Ethernet
Keep the same 48-bit address and the same frame format
Keep the same min. and max. frame length
MAC Sublayer
CSMA/CD for the half-duplex approach
No need for CSMA/CD for full-duplex Fast Ethernet
Autonegotiation: allow two devices to negotiate the mode or data rate of operation
Gigabit Ethernet
Under the name of IEEE 802.3z
Upgrade the data rate to 1 Gbps
Make it compatible with Standard or Fast Ethernet
Keep the same 48-bit address and the same frame format
Keep the same min. and max. frame length
Support autonegotiation as defined in Fast Ethernet
MAC Sublayer
Most of all implmentations follows full-duplex approach
In the full-duplex mode of Gigabit Ethernet, there is no collision; the maximum length of the cable is determined by the signal attenuation in the cable.
Half-duplex mode (very rare)
Traditional: 0.512 μs (25m)
Carrier Extension: 512 bytes (4096 bits) min. length
Frame bursting to improve the inefficiency of carrier extension
Ten-Gigabit Ethernet
Under the name of IEEE 802.3ae
Upgrade the data rate to 10 G bps
Make it compatible with Standard, Fast, and Giga Ethernet
Keep the same 48-bit address and the same frame format
Keep the same min. and max. frame length
Allow the interconnection of existing LANs into a MAN or WAN
Make Ethernet compatible with Frame Relay and ATM
Multiple Access
Multiple Access
Random Access
Controlled Access
Channelization
Data Link Layer: Two sublayers
Data link layer divided into two functionality-oriented sublayers
IEEE made this division for LANs
Medium Access Protocols
Random Access
Each station has the right to the medium without being controlled by any other station
Collision, a access conflict, if more than one station tries to send
ALOHA
The earliest random access method developed at the Univ. of Hawaii in the early 1970s
Designed for a radio (wireless) LAN
Pure ALOHA and Slotted ALOHA
Pure ALOHA Protocol: Procedure
The throughput for pure ALOHA is S = G × e −2G .
The maximum throughput Smax = 0.184 when G= (1/2).
Slotted ALOHA
Pure ALOHA vulnerable time = 2 x Tfr because there is no rule that defines when the station can send
Slotted ALOHA was invented to improve the efficiency of pure ALOHA
throughput for slotted ALOHA is S = G × e−G .
The maximum throughput Smax = 0.368 when G = 1
Slotted ALOHA vulnerable time = Tfr
Carrier Sense Multiple Access (CSMA)
CSMA
“Sense before transmit”
“Listen before talk”
CSMA can reduce the possibility of collision, but it can not eliminate it
CSMA: Vulnerable Time
Vulnerable time for CSMA is the propagation time Tp needed for a signal to propagate from one end of the medium to the other
Persistence Strategy
Nonpersistent strategy
Reduces the chance of collision
Reduces the efficiency of the network
1-persistent
Increases the chance of collision
p-persistent
Reduces the chance of collision and improves the efficiency by combining the other two strategies.
CSMA/CD: Min. Frame Size
Example: A network using CSMA/CD has a bandwidth of 10 Mbps. If the maximum propagation time (including the delays in the devices and ignoring the time needed to send a jamming signal, as we see later) is 25.6 μs, what is the minimum size of the frame?
Solution
The frame transmission time is Tfr = 2 × Tp = 51.2 μs. This means, in the worst case, a station needs to transmit for a period of 51.2 μs to detect the collision. The minimum size of the frame is 10 Mbps × 51.2 μs = 512 bits or 64 bytes. This is actually the minimum size of the frame for Standard Ethernet.
CSMA/CD: Energy Level & Throughput
Throughput of CSMA/CD is greater than that of ALOHA
The max. throughput occurs at a different value of G and is based on the persistent method and the value of p in the p-persistent approach
The max throughput is around 50% when G=1 for 1-persistent, up to 90% when G is between 3 and 8 for non-persistent
CSMA/CA (Collision Avoidance)
Invented for wireless network where we cannot detect collisions
Collision are avoided through the use of CSMA/CA’s three strategies: the interframe space, the contention windows, and acknowledgement
IFS can also be used to define the priority of a station or a frame
If the station finds the channel busy, it does not restart the timer of the contention window; it stops the timer and restarts it when the channel becomes idle
Controlled Access
The stations consult one another to find which station has the right to send
Reservation/Polling/ Token passing
Reservation access method
Polling: Select and Poll Functions
Channelization: FDMA
FDMA
Available bandwidth of the common channel is divided into bands that are separated by guard bands
FDMA is an access method in data link layer protocol. But, FDM is a physical layer technique
Channelization: TDMA
TDMA
The bandwidth is just one channel that is timeshared between different stations
TDMA is an access method. But, TDM is a physical layer technique
One channel carries all transmissions simultaneously
Two properties: If we multiply each code by another, we get 0. If we multiply each code by itself, we get 4
Data = (d1.c1 + d2.c2 + d3.c3 + d4.c4) .c1
= d1.c1.c1 + d2.c2.c1 + d3.c3.c1 + d4.c4.c1 = 4.d1
CDMA: Chips
Orthogonal sequences have the following properties:
Each sequence is made of N elements, where N is the number of stations
If we multiply a sequence by a number, every element in the sequence is multiplied by that element (scalar multiplication)
If we multiply two equal sequence, element by element, and add the results, we get N (inner product)
If we multiply two different sequence, element by element, and add the results, we get 0
Adding two sequence means adding the corresponding elements. The result is another sequence
Data representation in CDMA
Random Access
Controlled Access
Channelization
Data Link Layer: Two sublayers
Data link layer divided into two functionality-oriented sublayers
IEEE made this division for LANs
Medium Access Protocols
Random Access
Each station has the right to the medium without being controlled by any other station
Collision, a access conflict, if more than one station tries to send
ALOHA
The earliest random access method developed at the Univ. of Hawaii in the early 1970s
Designed for a radio (wireless) LAN
Pure ALOHA and Slotted ALOHA
Pure ALOHA Protocol: Procedure
The throughput for pure ALOHA is S = G × e −2G .
The maximum throughput Smax = 0.184 when G= (1/2).
Slotted ALOHA
Pure ALOHA vulnerable time = 2 x Tfr because there is no rule that defines when the station can send
Slotted ALOHA was invented to improve the efficiency of pure ALOHA
throughput for slotted ALOHA is S = G × e−G .
The maximum throughput Smax = 0.368 when G = 1
Slotted ALOHA vulnerable time = Tfr
Carrier Sense Multiple Access (CSMA)
CSMA
“Sense before transmit”
“Listen before talk”
CSMA can reduce the possibility of collision, but it can not eliminate it
CSMA: Vulnerable Time
Vulnerable time for CSMA is the propagation time Tp needed for a signal to propagate from one end of the medium to the other
Persistence Strategy
Nonpersistent strategy
Reduces the chance of collision
Reduces the efficiency of the network
1-persistent
Increases the chance of collision
p-persistent
Reduces the chance of collision and improves the efficiency by combining the other two strategies.
CSMA/CD: Min. Frame Size
Example: A network using CSMA/CD has a bandwidth of 10 Mbps. If the maximum propagation time (including the delays in the devices and ignoring the time needed to send a jamming signal, as we see later) is 25.6 μs, what is the minimum size of the frame?
Solution
The frame transmission time is Tfr = 2 × Tp = 51.2 μs. This means, in the worst case, a station needs to transmit for a period of 51.2 μs to detect the collision. The minimum size of the frame is 10 Mbps × 51.2 μs = 512 bits or 64 bytes. This is actually the minimum size of the frame for Standard Ethernet.
CSMA/CD: Energy Level & Throughput
Throughput of CSMA/CD is greater than that of ALOHA
The max. throughput occurs at a different value of G and is based on the persistent method and the value of p in the p-persistent approach
The max throughput is around 50% when G=1 for 1-persistent, up to 90% when G is between 3 and 8 for non-persistent
CSMA/CA (Collision Avoidance)
Invented for wireless network where we cannot detect collisions
Collision are avoided through the use of CSMA/CA’s three strategies: the interframe space, the contention windows, and acknowledgement
IFS can also be used to define the priority of a station or a frame
If the station finds the channel busy, it does not restart the timer of the contention window; it stops the timer and restarts it when the channel becomes idle
Controlled Access
The stations consult one another to find which station has the right to send
Reservation/Polling/ Token passing
Reservation access method
Polling: Select and Poll Functions
Channelization: FDMA
FDMA
Available bandwidth of the common channel is divided into bands that are separated by guard bands
FDMA is an access method in data link layer protocol. But, FDM is a physical layer technique
Channelization: TDMA
TDMA
The bandwidth is just one channel that is timeshared between different stations
TDMA is an access method. But, TDM is a physical layer technique
One channel carries all transmissions simultaneously
Two properties: If we multiply each code by another, we get 0. If we multiply each code by itself, we get 4
Data = (d1.c1 + d2.c2 + d3.c3 + d4.c4) .c1
= d1.c1.c1 + d2.c2.c1 + d3.c3.c1 + d4.c4.c1 = 4.d1
CDMA: Chips
Orthogonal sequences have the following properties:
Each sequence is made of N elements, where N is the number of stations
If we multiply a sequence by a number, every element in the sequence is multiplied by that element (scalar multiplication)
If we multiply two equal sequence, element by element, and add the results, we get N (inner product)
If we multiply two different sequence, element by element, and add the results, we get 0
Adding two sequence means adding the corresponding elements. The result is another sequence
Data representation in CDMA
Data Link Control
Data Link Control
Line Discipline
Flow Control
Error Control
Data Link Layer
FLOW CONTROL
ERROR CONTROL
Automatic Repeat Request (ARQ)
Stop and Wait
Go Back N
Selective Repeat
Flow and Error Control
Flow control refers to a set of procedures used to restrict the amount of data that the sender can send before waiting for acknowledgment.
Error control is based on automatic repeat request (ARQ), which is the retransmission of data.
Example
In a Stop-and-Wait ARQ system, the bandwidth of the line is 1 Mbps, and 1 bit takes 20 ms to make a round trip. What is the bandwidth-delay product? If the system data frames are 1000 bits in length, what is the utilization percentage of the link?
The bandwidth-delay product is 1 106 20 10-3 = 20,000 bits The system can send 20,000 bits during the time it takes for the data to go from the sender to the receiver and then back again. However, the system sends only 1000 bits. We can say that the link utilization is only 1000/20,000, or 5%. For this reason, for a link with high bandwidth or long delay, use of Stop-and-Wait ARQ wastes the capacity of the link.
Line Discipline
Flow Control
Error Control
Data Link Layer
FLOW CONTROL
ERROR CONTROL
Automatic Repeat Request (ARQ)
Stop and Wait
Go Back N
Selective Repeat
Flow and Error Control
Flow control refers to a set of procedures used to restrict the amount of data that the sender can send before waiting for acknowledgment.
Error control is based on automatic repeat request (ARQ), which is the retransmission of data.
Example
In a Stop-and-Wait ARQ system, the bandwidth of the line is 1 Mbps, and 1 bit takes 20 ms to make a round trip. What is the bandwidth-delay product? If the system data frames are 1000 bits in length, what is the utilization percentage of the link?
The bandwidth-delay product is 1 106 20 10-3 = 20,000 bits The system can send 20,000 bits during the time it takes for the data to go from the sender to the receiver and then back again. However, the system sends only 1000 bits. We can say that the link utilization is only 1000/20,000, or 5%. For this reason, for a link with high bandwidth or long delay, use of Stop-and-Wait ARQ wastes the capacity of the link.
Error Detection and Correction
Introduction
Block Coding
Linear Block Codes
Cyclic Codes
Checksum
Hamming Codes
Type of Errors
An electromagnetic signal is subject to interference from heat, magnetism, and other forms of electricity
Single-bit error: 0 1 or 1 0
Burst error: 2 or more bits have changed
Sender site:
1. The message is divided into 16-bit words.
2. The value of the checksum word is set to 0.
3. All words including the checksum are added using one’s complement
addition.
4. The sum is complemented and becomes the checksum.
5. The checksum is sent with the data.
Receiver site:
1. The message (including checksum) is divided into 16-bit words.
2. All words are added using one’s complement addition.
3. The sum is complemented and becomes the new checksum.
4. If the value of checksum is 0, the message is accepted; otherwise, it is
rejected.
Forward Error Correction
Purpose: An FEC (n, m) encoder
Take m-bit original data as input
Add r=n-k check bits to the original data to produce a n-bit codeword
The receiver can fix any error
Hamming Distance
Code: set of codewords
Hamming distance between 2 codewords is the number of bit positions where the 2 codewords differ
HammingDist(10001001, 10110001) = 3
Hamming distance of a code is the minimum Hamming distance between any two codewords in the code
HammingDist({0000000000,0000011111,1111100000,1111111111}) = 5
Interesting Findings
To detect d single-bit errors
Need a distance-(d+1) code
Appling d single-bit errors to a codeword must result in an invalid codeword. (WHY?)
To correct d single-bit errors
Need a distance-(2d+1) code
Given an incorrect codeword, the corresponding correct one must be the codeword closest in Hamming distance (WHY?)
Linear Block Code: Hamming Code
All Hamming codes discussed in our textbook have dmin = 3.
The relationship between k and n in these codes is n = 2k − 1
Single-Bit Error
Only one bit of a given data unit is changed
The least likely type of error in serial transmission
Single-bit error can happen in parallel transmission
Burst Error
Two or more bits in the data unit have changed
Burst error does not necessarily mean that the errors occur in consecutive bits
Most likely to happen in a serial transmission
Number of bits affected depends on the data rate and duration of noise
Single Bit Error vs. Burst Error
Error Detection
Error detection uses the concept of redundancy, which means adding extra bits for detecting errors at the destination.
Redundancy for Error Detection
Parity Check
Modular Arithmetic
In modulo-N arithmetic, we use only the integers in the range 0 to N-1, inclusive.
Adding: 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 0
Subtracting: 0 – 0 = 0 0 – 1 = 1 1 – 0 = 1 1 – 1 = 0
XORing of two single bits or two words
Block Coding
Divide the message into blocks, each of k bits, called datawords.
Add r redundant bits to each block to make the length n = k + r. The resulting n-bit blocks are called codewords
Example: 4B/5B block coding
k = 4 and n = 5.
2k = 16 datawords and 2n = 32 codewords.
Error Detection in Block Coding
Assume the sender encodes the dataword 01 as 011 and sends it to the receiver. Consider the following cases:
1. The receiver receives 011 which is a valid codeword. The receiver extracts the dataword 01 from it.
2. The codeword is corrupted during transmission, and 111 is received. This is not a valid codeword and is discarded.
3. The codeword is corrupted during transmission, and 000 is received. This is a valid codeword. The receiver incorrectly extracts the dataword 00. Two corrupted bits have made the error undetectable.
An error-detecting code can detect only the types of errors for which it is designed; other types of errors may remain undetected
Linear Block Code: Parity-Check Code
A simple parity-check code is a single-bit error-detecting code in which n = k + 1 with dmin = 2.
Encoder and Decoder for Parity-Check Code
The result of addition over all 5 bits: syndrome
2-d Parity Check
2 Dimensional Parity-Check Code
2 Dimensional Parity-Check Code
Cyclic Code: CRC
Cyclic codes are special linear block codes with one extra property.
Cyclic Redundancy Check (CRC)
Checksum
Tendency is to replace the checksum with a CRC
Not as strong as CRC in error-checking capability
One’s complement arithmetic
We can represent unsigned numbers between 0 and 2n – 1 using only n bits
If the number has more than n bits, the extra leftmost bits need to be added to the n rightmost bits (wrapping)
A negative number can be represented by inverting all bits. It is the same as subtracting the number from 2n – 1
Internet Checksum
Block Coding
Linear Block Codes
Cyclic Codes
Checksum
Hamming Codes
Type of Errors
An electromagnetic signal is subject to interference from heat, magnetism, and other forms of electricity
Single-bit error: 0 1 or 1 0
Burst error: 2 or more bits have changed
Sender site:
1. The message is divided into 16-bit words.
2. The value of the checksum word is set to 0.
3. All words including the checksum are added using one’s complement
addition.
4. The sum is complemented and becomes the checksum.
5. The checksum is sent with the data.
Receiver site:
1. The message (including checksum) is divided into 16-bit words.
2. All words are added using one’s complement addition.
3. The sum is complemented and becomes the new checksum.
4. If the value of checksum is 0, the message is accepted; otherwise, it is
rejected.
Forward Error Correction
Purpose: An FEC (n, m) encoder
Take m-bit original data as input
Add r=n-k check bits to the original data to produce a n-bit codeword
The receiver can fix any error
Hamming Distance
Code: set of codewords
Hamming distance between 2 codewords is the number of bit positions where the 2 codewords differ
HammingDist(10001001, 10110001) = 3
Hamming distance of a code is the minimum Hamming distance between any two codewords in the code
HammingDist({0000000000,0000011111,1111100000,1111111111}) = 5
Interesting Findings
To detect d single-bit errors
Need a distance-(d+1) code
Appling d single-bit errors to a codeword must result in an invalid codeword. (WHY?)
To correct d single-bit errors
Need a distance-(2d+1) code
Given an incorrect codeword, the corresponding correct one must be the codeword closest in Hamming distance (WHY?)
Linear Block Code: Hamming Code
All Hamming codes discussed in our textbook have dmin = 3.
The relationship between k and n in these codes is n = 2k − 1
Single-Bit Error
Only one bit of a given data unit is changed
The least likely type of error in serial transmission
Single-bit error can happen in parallel transmission
Burst Error
Two or more bits in the data unit have changed
Burst error does not necessarily mean that the errors occur in consecutive bits
Most likely to happen in a serial transmission
Number of bits affected depends on the data rate and duration of noise
Single Bit Error vs. Burst Error
Error Detection
Error detection uses the concept of redundancy, which means adding extra bits for detecting errors at the destination.
Redundancy for Error Detection
Parity Check
Modular Arithmetic
In modulo-N arithmetic, we use only the integers in the range 0 to N-1, inclusive.
Adding: 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 0
Subtracting: 0 – 0 = 0 0 – 1 = 1 1 – 0 = 1 1 – 1 = 0
XORing of two single bits or two words
Block Coding
Divide the message into blocks, each of k bits, called datawords.
Add r redundant bits to each block to make the length n = k + r. The resulting n-bit blocks are called codewords
Example: 4B/5B block coding
k = 4 and n = 5.
2k = 16 datawords and 2n = 32 codewords.
Error Detection in Block Coding
Assume the sender encodes the dataword 01 as 011 and sends it to the receiver. Consider the following cases:
1. The receiver receives 011 which is a valid codeword. The receiver extracts the dataword 01 from it.
2. The codeword is corrupted during transmission, and 111 is received. This is not a valid codeword and is discarded.
3. The codeword is corrupted during transmission, and 000 is received. This is a valid codeword. The receiver incorrectly extracts the dataword 00. Two corrupted bits have made the error undetectable.
An error-detecting code can detect only the types of errors for which it is designed; other types of errors may remain undetected
Linear Block Code: Parity-Check Code
A simple parity-check code is a single-bit error-detecting code in which n = k + 1 with dmin = 2.
Encoder and Decoder for Parity-Check Code
The result of addition over all 5 bits: syndrome
2-d Parity Check
2 Dimensional Parity-Check Code
2 Dimensional Parity-Check Code
Cyclic Code: CRC
Cyclic codes are special linear block codes with one extra property.
Cyclic Redundancy Check (CRC)
Checksum
Tendency is to replace the checksum with a CRC
Not as strong as CRC in error-checking capability
One’s complement arithmetic
We can represent unsigned numbers between 0 and 2n – 1 using only n bits
If the number has more than n bits, the extra leftmost bits need to be added to the n rightmost bits (wrapping)
A negative number can be represented by inverting all bits. It is the same as subtracting the number from 2n – 1
Internet Checksum
Circuit Switching
Circuit Switching
Switched network
A switched network consists of a series of inter-linked nodes, called switches.
Switches are hardware and/or software devices capable of creating temporary connections between two or more devices linked to the switch but not to each other.
Classification of Switched Networks
Circuit-Switching
Circuit switching creates a direct physical connection between two devices such as phones or computers.
Circuit switch connects input links to output links
We can use switches to reduce the number and length of links.
Circuit switch
A circuit switch is a device with n inputs and m outputs that creates a temporary connection between an input link and an output link.
The number of inputs does not have to match the number of outputs.
Circuit Switch Types
Evolution of Switching System:
Manual operators
Step-by-step switch (Strowger switch)
Crossbar switch
Space-Division switches
Multistage switches
Time-Division switches
Time-slot interchange technique
Time-space-time switches
Hybrids combine Time & Space switching
Manual Switching
Patch-cord panel switch invented in 1877
Operators connect users on demand
Establish circuit to allow electrical current to flow from inlet to outlet
Only N connections required to central office
Step-by-step (Strowger) Switch
Electromagnetic switch
Sequence of dialed numbers selects output line
Crossbar Space Switch
N x N array of crosspoints
Connect an input to an output by closing a crosspoint
Non blocking: Any input can connect to idle output
Complexity: N2 crosspoints
Crossbar Switches: Problem
The number of switches is huge.
connect n inputs by m output -- require n * m crosspoint.
Ex : 1000 input, 1000 output → 1,000,000 crosspoint
Inefficient
fewer than 25% of the crosspoints are in use at a given time.
Folded switch
An n-by-n folded switch can connect n lines in full-duplex mode. For example, it can connect n telephones in such a way that each phone can be connected to every other phone.
Circuit switching uses space-division switch [paths in the circuit are separated from each other spatially] or time-division switch.
Circuit Switching
Uses a dedicated path between two stations
Has three phases
Establish (Call Setup)
Transfer
Disconnect
Inefficient
Channel capacity dedicated for duration of connection
if no data, capacity wasted
Set up (connection) takes time
Once connected, transfer is transparent
Blocking or Non-blocking
blocking network
may be unable to connect stations because all paths are in use
used on voice systems
non-blocking network
permits all stations to connect at once
used for some data connections
Space-Division Switch
Paths in the circuit are separated from each other spatially.
Crossbar Switch
Crossbar switch connects n inputs to m outputs in a grid, using electronic micro-switches (transistors) at each cross-point.
Limitation is the number of cross-points required.
Multistage switch
Multistage switch combines crossbar switches in several stages.
Design of a multistage switch depends on the number of stages and the number of switches required (or desired) in each stage.
Normally, the middle stages have fewer switches than do the first and last stages.
MULTIPLE Switching paths
Multiple paths are available in multistage switches.
Blocking refers to times when two inputs are looking for the same output. The output port is blocked.
Design of a Three-Stage Switch
Three steps
Divide the N input lines into groups, each of n lines. For each group, use one crossbar of size n x k, where k is the number of crossbars in the middle stage
Use k crossbars, each of size (N/n) x (N/n) in the middle stage
Use N/n crossbars, each of size k x n at the third stage
N/n(n x k) + k(N/n x N/n) + N/n(k x n) = 2kN + k(N/n)2
In a three-stage switch, the total number of crosspoints is 2kN + k(N/n)2 which is much smaller than the number of crosspoints in a single-stage switch (N2).
Multistage Switch: Example
Design a three-stage, 200 × 200 switch (N = 200) with k = 4 and n = 20.
In the first stage we have N/n or 10 crossbars, each of size 20 × 4. In the second stage, we have 4 crossbars, each of size 10 × 10. In the third stage, we have 10 crossbars, each of size 4 × 20. The total number of crosspoints is 2kN + k(N/n)2, or 2000 crosspoints. This is 5 percent of the number of crosspoints in a single-stage switch (200 × 200 = 40,000).
Time Division Switching
Modern digital systems use intelligent control of space & time division elements
Use digital time division techniques to set up and maintain virtual circuits
Partition low speed bit stream into pieces that share higher speed stream
Individual pieces manipulated by control logic to flow from input to output
Time-slot interchange
TSI consists of random access memory (RAM) with several memory locations. The size of each location is the same as the size of a single time slot.
The number of locations is the same as the number of inputs.
The RAM fills up with incoming data from time slots in the order received. Slots are then sent out in an order based on the decisions of a control unit.
TDM bus
Input and output lines are connected to a high-speed bus through input and output gates (microswitches)
Each input gate is closed during one of the four slots.
During the same time slot, only one output gate is also closed. This pair of gates allows a burst of data to be transferred from one specific input line to one specific output line using the bus.
The control unit opens and closes the gates according to switching need.
Comparison of SDM and TDM
SDM
Advantage:
Instantaneous.
Disadvantage:
Number of cross points required.
TDM
Advantage:
No cross points.
Disadvantage:
Processing delay.
Time- and Space-Division Switch Combination
Space division: instantaneous, crosspoints
Time division: no crosspoint, processing delay
Space-and time-division switching combinations take advantage of the best of both
TST(time-space-time)
TSST(time-space-space-time)
STTS(space-time-time-space)
TST switch
Combine Space division and time division switching.
This results in switches that are optimized both physically (the number of crosspoints) and temporally (the amount of delay).
Switched network
A switched network consists of a series of inter-linked nodes, called switches.
Switches are hardware and/or software devices capable of creating temporary connections between two or more devices linked to the switch but not to each other.
Classification of Switched Networks
Circuit-Switching
Circuit switching creates a direct physical connection between two devices such as phones or computers.
Circuit switch connects input links to output links
We can use switches to reduce the number and length of links.
Circuit switch
A circuit switch is a device with n inputs and m outputs that creates a temporary connection between an input link and an output link.
The number of inputs does not have to match the number of outputs.
Circuit Switch Types
Evolution of Switching System:
Manual operators
Step-by-step switch (Strowger switch)
Crossbar switch
Space-Division switches
Multistage switches
Time-Division switches
Time-slot interchange technique
Time-space-time switches
Hybrids combine Time & Space switching
Manual Switching
Patch-cord panel switch invented in 1877
Operators connect users on demand
Establish circuit to allow electrical current to flow from inlet to outlet
Only N connections required to central office
Step-by-step (Strowger) Switch
Electromagnetic switch
Sequence of dialed numbers selects output line
Crossbar Space Switch
N x N array of crosspoints
Connect an input to an output by closing a crosspoint
Non blocking: Any input can connect to idle output
Complexity: N2 crosspoints
Crossbar Switches: Problem
The number of switches is huge.
connect n inputs by m output -- require n * m crosspoint.
Ex : 1000 input, 1000 output → 1,000,000 crosspoint
Inefficient
fewer than 25% of the crosspoints are in use at a given time.
Folded switch
An n-by-n folded switch can connect n lines in full-duplex mode. For example, it can connect n telephones in such a way that each phone can be connected to every other phone.
Circuit switching uses space-division switch [paths in the circuit are separated from each other spatially] or time-division switch.
Circuit Switching
Uses a dedicated path between two stations
Has three phases
Establish (Call Setup)
Transfer
Disconnect
Inefficient
Channel capacity dedicated for duration of connection
if no data, capacity wasted
Set up (connection) takes time
Once connected, transfer is transparent
Blocking or Non-blocking
blocking network
may be unable to connect stations because all paths are in use
used on voice systems
non-blocking network
permits all stations to connect at once
used for some data connections
Space-Division Switch
Paths in the circuit are separated from each other spatially.
Crossbar Switch
Crossbar switch connects n inputs to m outputs in a grid, using electronic micro-switches (transistors) at each cross-point.
Limitation is the number of cross-points required.
Multistage switch
Multistage switch combines crossbar switches in several stages.
Design of a multistage switch depends on the number of stages and the number of switches required (or desired) in each stage.
Normally, the middle stages have fewer switches than do the first and last stages.
MULTIPLE Switching paths
Multiple paths are available in multistage switches.
Blocking refers to times when two inputs are looking for the same output. The output port is blocked.
Design of a Three-Stage Switch
Three steps
Divide the N input lines into groups, each of n lines. For each group, use one crossbar of size n x k, where k is the number of crossbars in the middle stage
Use k crossbars, each of size (N/n) x (N/n) in the middle stage
Use N/n crossbars, each of size k x n at the third stage
N/n(n x k) + k(N/n x N/n) + N/n(k x n) = 2kN + k(N/n)2
In a three-stage switch, the total number of crosspoints is 2kN + k(N/n)2 which is much smaller than the number of crosspoints in a single-stage switch (N2).
Multistage Switch: Example
Design a three-stage, 200 × 200 switch (N = 200) with k = 4 and n = 20.
In the first stage we have N/n or 10 crossbars, each of size 20 × 4. In the second stage, we have 4 crossbars, each of size 10 × 10. In the third stage, we have 10 crossbars, each of size 4 × 20. The total number of crosspoints is 2kN + k(N/n)2, or 2000 crosspoints. This is 5 percent of the number of crosspoints in a single-stage switch (200 × 200 = 40,000).
Time Division Switching
Modern digital systems use intelligent control of space & time division elements
Use digital time division techniques to set up and maintain virtual circuits
Partition low speed bit stream into pieces that share higher speed stream
Individual pieces manipulated by control logic to flow from input to output
Time-slot interchange
TSI consists of random access memory (RAM) with several memory locations. The size of each location is the same as the size of a single time slot.
The number of locations is the same as the number of inputs.
The RAM fills up with incoming data from time slots in the order received. Slots are then sent out in an order based on the decisions of a control unit.
TDM bus
Input and output lines are connected to a high-speed bus through input and output gates (microswitches)
Each input gate is closed during one of the four slots.
During the same time slot, only one output gate is also closed. This pair of gates allows a burst of data to be transferred from one specific input line to one specific output line using the bus.
The control unit opens and closes the gates according to switching need.
Comparison of SDM and TDM
SDM
Advantage:
Instantaneous.
Disadvantage:
Number of cross points required.
TDM
Advantage:
No cross points.
Disadvantage:
Processing delay.
Time- and Space-Division Switch Combination
Space division: instantaneous, crosspoints
Time division: no crosspoint, processing delay
Space-and time-division switching combinations take advantage of the best of both
TST(time-space-time)
TSST(time-space-space-time)
STTS(space-time-time-space)
TST switch
Combine Space division and time division switching.
This results in switches that are optimized both physically (the number of crosspoints) and temporally (the amount of delay).
Transmission Media
Guided Media
Unguided Media: Wireless
Transmission Medium & Physical Layer
Classes of Transmission Media
Twisted-pair Cable
UTP and STP Cables
UTP Connector
UTP Performance
Parallel Flat Wire
Coaxial Cable
Optical Fiber
Noise Effect on Twisted-Pair
Cumulative effect of noise is equal on both sides
Twisting does not always eliminate the noise, but does significantly reduce it
Optical Fiber
Optical fiber is made of glass or plastic
It transmits signals in the form of light
The Nature of Light
The speed of light
300,000 Km/sec in a vacuum
Depends on the density of the medium through which it is traveling
Other properties of light
Refraction, Critical angle, Reflection
Bending of Light Ray
Optical Fiber
Propagation Modes
Modes
Fiber Types
Fiber Construction
Fiber-optic Cable Connectors
SC (subscriber channel) connector used for cable TV. It uses push/pull locking system
ST (straight-tip) connector used for connecting cable to networking device. It uses a reliable bayonet locking system
MT-RJ is the same size as RJ45
Advantages/Disadvantages of Optical Fiber
Advantages
Higher bandwidth
Less signal attenuation
Immunity to electromagnetic interference
Resistance to corrosive materials
Light weight
Greater immunity to tapping
Disadvantages
Installation and maintenance
Unidirectional light propagation
Cost
Electromagnetic Spectrum for wireless
Propagation Methods
Bands
Wireless Transmission Waves
Wireless Transmission Waves
Radio waves are used for multicast communications, such as radio and television, and paging systems
Microwaves are used for unicast communication such as cellular telephones, satellite networks, and wireless LANs
Infrared signals can be used for short-range communication in a closed area using line-of-sight propagation
Omnidirectional & Unidirectional Antennas
Infrared
Infrared waves with frequencies from 300 GHz to 400 THz for short-range communication in a closed area using line-of-sight propagation
Having high frequencies, it cannot penetrate walls
IrDA (Infrared Data Association) for standards
Example: IrDA port for wireless keyboard
Originally defined a data rate of 75 kbps for a distance up to 8 m
Recent standard for a data rate of 4 Mbps
Unguided Media: Wireless
Transmission Medium & Physical Layer
Classes of Transmission Media
Twisted-pair Cable
UTP and STP Cables
UTP Connector
UTP Performance
Parallel Flat Wire
Coaxial Cable
Optical Fiber
Noise Effect on Twisted-Pair
Cumulative effect of noise is equal on both sides
Twisting does not always eliminate the noise, but does significantly reduce it
Optical Fiber
Optical fiber is made of glass or plastic
It transmits signals in the form of light
The Nature of Light
The speed of light
300,000 Km/sec in a vacuum
Depends on the density of the medium through which it is traveling
Other properties of light
Refraction, Critical angle, Reflection
Bending of Light Ray
Optical Fiber
Propagation Modes
Modes
Fiber Types
Fiber Construction
Fiber-optic Cable Connectors
SC (subscriber channel) connector used for cable TV. It uses push/pull locking system
ST (straight-tip) connector used for connecting cable to networking device. It uses a reliable bayonet locking system
MT-RJ is the same size as RJ45
Advantages/Disadvantages of Optical Fiber
Advantages
Higher bandwidth
Less signal attenuation
Immunity to electromagnetic interference
Resistance to corrosive materials
Light weight
Greater immunity to tapping
Disadvantages
Installation and maintenance
Unidirectional light propagation
Cost
Electromagnetic Spectrum for wireless
Propagation Methods
Bands
Wireless Transmission Waves
Wireless Transmission Waves
Radio waves are used for multicast communications, such as radio and television, and paging systems
Microwaves are used for unicast communication such as cellular telephones, satellite networks, and wireless LANs
Infrared signals can be used for short-range communication in a closed area using line-of-sight propagation
Omnidirectional & Unidirectional Antennas
Infrared
Infrared waves with frequencies from 300 GHz to 400 THz for short-range communication in a closed area using line-of-sight propagation
Having high frequencies, it cannot penetrate walls
IrDA (Infrared Data Association) for standards
Example: IrDA port for wireless keyboard
Originally defined a data rate of 75 kbps for a distance up to 8 m
Recent standard for a data rate of 4 Mbps
Multiplexing and Spreading Circuit Switching and Telephone Network
In this lecture we will cover the following topics:
6. Multiplexing and Spreading
6.1 Multiplexing
6.2 Spread spectrum
6.3 Summary (part 6)
7. Circuit Switching and Telephone Network
7.1 Circuit-switched networks
7.2 Datagram networks
7.3 Virtual circuit networks
7.4 Structure of a switch
7.5 Telephone network
7.6 Dial-up modem
7.7 Digital subscriber line
7.8 Summary (part 7)
NOTE
Bandwidth utilization is the wise use of
available bandwidth to achieve
specific goals.
Efficiency can be achieved by multiplexing; privacy
and anti-jamming can be achieved by spreading
6.1 MULTIPLEXING
Whenever the bandwidth of a medium linking two devices is greater than the bandwidth needs of the devices, the link can be shared. Multiplexing is the set of techniques that allows the simultaneous transmission of multiple signals across a single data link. As data and telecommunications use increases, so does traffic.
Topics discussed in this section:
Frequency Division Multiplexing (FDM)
Wavelength Division Multiplexing (WDM)
Time Division Multiplexing (TDM)
Statistical Time Division Multiplexing (Stat TDM)
Dividing a link into channels
Categories of multiplexing
Frequency Division Multiplexing (FDM)
FDM can be used when the BW of a link is greater than the combined BW of signals to be transmitted.
Signals generated by each sending device modulate different carrier frequencies, which are then combined into a single composite signal
Guard bands are used to prevent signals from overlapping
Example
Assume that a voice channel occupies a bandwidth of 4 kHz. We need to combine three voice channels into a link with a bandwidth of 12 kHz, from 20 to 32 kHz. Show the configuration, using the frequency domain. Assume there are no guard bands.
Solution
We shift (modulate) each of the three voice channels to a different bandwidth, as shown in figure in next slide. We use the 20-24 kHz bandwidth for the first channel, the 24-28 kHz bandwidth for the second channel, and the 28-32 kHz bandwidth for the third one. Then we combine them as shown in the figure.
Wavelength Division Multiplexing (WDM)
WDM is designed to use the high data rate capability of fiber optic
cable.
• Using a fiber-optic cable for one single line wastes the available
bandwidth. Multiplexing allows us to connect several lines into one.
• WDM is conceptually the same as FDM, except that the multiplexing
and demultiplexing involve optical signals
Time Division Multiplexing (TDM)
Note:
TDM is a digital multiplexing technique for combining several low-rate channels into one high-rate one.
Synchronous Time Division Multiplexing
Note: In synchronous TDM, the data rate of the link is n times faster, and the unit duration is n times shorter
Example
Figure in next slide shows synchronous TDM with a data stream for each
input and one data stream for the output. The unit of data is 1 bit. Find (a)
the input bit duration, (b) the output bit duration, (c) the output bit rate, and
(d) the output frame rate.
Solution
a. The input bit duration is the inverse of the bit rate: 1/1 Mbps = 1 μs.
b. The output bit duration is one-fourth of the input bit duration, or 0.25 μs.
c. The output bit rate is the inverse of the output bit duration or 1/(4μs) or
4 Mbps. This can also be deduced from the fact that the output rate is 4
times as fast as any input rate; so the output rate = 4 × 1 Mbps = 4
Mbps.
d. The frame rate is always the same as any input rate. So the frame rate is
1,000,000 frames per second. Because we are sending 4 bits in each
frame, we can verify the result of the previous question by multiplying
the frame rate by the number of bits per frame.
Interleaving
TDM can be visualized as two fast rotating switches, one on the MUX
side and the other on the DEMUX side. The switches are synchronized
and rotate at the same speed but in opposite directions. On the MUX
side, as the switch opens in front of a connection, that connection has
the opportunity to send a unit onto the path. This process is called
inteleaving
Digital hierarchy
Telephone companies implement TDM through a hierarchy of digital signals, called Digital Signal (DS) Service.
The following figure shows the data rates supported by each level:
T-1 line for multiplexing telephone lines
T-1 frame structure
DS-1 requires 8 kbps overhead due to the synchronization bit:
T1 line = 24 slots x 8 bits + 1 bit for synchronization = 193 bits x 8kbps
= 1.544Mbps
E line rates
6.2 SPREAD SPECTRUM
In spread spectrum (SS),, we combine signals from different
sources tto fit into a larger bandwidth,, butt our goals are to
prevent eavesdropping and jamming.. To achieve these
goals,, spread spectrum techniques add redundancy.
Topics discussed in this section:
Frequency Hopping Spread Spectrum (FHSS)
Direct Sequence Spread Spectrum (DSSS)
Spread spectrum
Input is fed into a channel encoder that produces an analog signal
with a relatively narrow BW around a center frequency.
• This signal is further modulated using a sequence of digits known as a
spreading code or spreading sequence.
• The effect of this modulation is to increase significantly the BW
(spread the spectrum) of the signal to be transmitted.
• On the receiving end, the same digit sequence is used to demodulate
the spread spectrum signal.
Frequency hopping spread spectrum (FHSS)
With FHSS the signal is broadcast over a random series of
radio frequencies, hopping from frequency to frequency at
fixed intervals.
A receiver, hopping between frequencies in synchronization with the transmitter picks up the message.
For transmission, binary data are fed into a modulator using FSK or
PSK. The resulting signal is entered on a base frequency. A
pseudorandom code generator serves as an index into a table of
frequencies (spreading code). Each k bits of the spreading sequence
specifies one of the 2k carrier frequencies. At each k-bit interval a new
carrier frequency is selected.
This frequency is then
modulated to produce a
new signal with the same
shape but centered on the
selected carrier frequency.
Direct Sequence Spread Spectrum (DSSS)
With DSSS each bit in the original signal is represented by multiple bits in the transmitted signal using a spreading code.
The spreading code spreads the signal across a wider frequency band in direct proportion to the number of bits used. eg. a 10-bit spreading code spreads the signal across a frequency band that is 10 times greater than a 1-bit spreading code.
6. Multiplexing and Spreading
6.1 Multiplexing
6.2 Spread spectrum
6.3 Summary (part 6)
7. Circuit Switching and Telephone Network
7.1 Circuit-switched networks
7.2 Datagram networks
7.3 Virtual circuit networks
7.4 Structure of a switch
7.5 Telephone network
7.6 Dial-up modem
7.7 Digital subscriber line
7.8 Summary (part 7)
NOTE
Bandwidth utilization is the wise use of
available bandwidth to achieve
specific goals.
Efficiency can be achieved by multiplexing; privacy
and anti-jamming can be achieved by spreading
6.1 MULTIPLEXING
Whenever the bandwidth of a medium linking two devices is greater than the bandwidth needs of the devices, the link can be shared. Multiplexing is the set of techniques that allows the simultaneous transmission of multiple signals across a single data link. As data and telecommunications use increases, so does traffic.
Topics discussed in this section:
Frequency Division Multiplexing (FDM)
Wavelength Division Multiplexing (WDM)
Time Division Multiplexing (TDM)
Statistical Time Division Multiplexing (Stat TDM)
Dividing a link into channels
Categories of multiplexing
Frequency Division Multiplexing (FDM)
FDM can be used when the BW of a link is greater than the combined BW of signals to be transmitted.
Signals generated by each sending device modulate different carrier frequencies, which are then combined into a single composite signal
Guard bands are used to prevent signals from overlapping
Example
Assume that a voice channel occupies a bandwidth of 4 kHz. We need to combine three voice channels into a link with a bandwidth of 12 kHz, from 20 to 32 kHz. Show the configuration, using the frequency domain. Assume there are no guard bands.
Solution
We shift (modulate) each of the three voice channels to a different bandwidth, as shown in figure in next slide. We use the 20-24 kHz bandwidth for the first channel, the 24-28 kHz bandwidth for the second channel, and the 28-32 kHz bandwidth for the third one. Then we combine them as shown in the figure.
Wavelength Division Multiplexing (WDM)
WDM is designed to use the high data rate capability of fiber optic
cable.
• Using a fiber-optic cable for one single line wastes the available
bandwidth. Multiplexing allows us to connect several lines into one.
• WDM is conceptually the same as FDM, except that the multiplexing
and demultiplexing involve optical signals
Time Division Multiplexing (TDM)
Note:
TDM is a digital multiplexing technique for combining several low-rate channels into one high-rate one.
Synchronous Time Division Multiplexing
Note: In synchronous TDM, the data rate of the link is n times faster, and the unit duration is n times shorter
Example
Figure in next slide shows synchronous TDM with a data stream for each
input and one data stream for the output. The unit of data is 1 bit. Find (a)
the input bit duration, (b) the output bit duration, (c) the output bit rate, and
(d) the output frame rate.
Solution
a. The input bit duration is the inverse of the bit rate: 1/1 Mbps = 1 μs.
b. The output bit duration is one-fourth of the input bit duration, or 0.25 μs.
c. The output bit rate is the inverse of the output bit duration or 1/(4μs) or
4 Mbps. This can also be deduced from the fact that the output rate is 4
times as fast as any input rate; so the output rate = 4 × 1 Mbps = 4
Mbps.
d. The frame rate is always the same as any input rate. So the frame rate is
1,000,000 frames per second. Because we are sending 4 bits in each
frame, we can verify the result of the previous question by multiplying
the frame rate by the number of bits per frame.
Interleaving
TDM can be visualized as two fast rotating switches, one on the MUX
side and the other on the DEMUX side. The switches are synchronized
and rotate at the same speed but in opposite directions. On the MUX
side, as the switch opens in front of a connection, that connection has
the opportunity to send a unit onto the path. This process is called
inteleaving
Digital hierarchy
Telephone companies implement TDM through a hierarchy of digital signals, called Digital Signal (DS) Service.
The following figure shows the data rates supported by each level:
T-1 line for multiplexing telephone lines
T-1 frame structure
DS-1 requires 8 kbps overhead due to the synchronization bit:
T1 line = 24 slots x 8 bits + 1 bit for synchronization = 193 bits x 8kbps
= 1.544Mbps
E line rates
6.2 SPREAD SPECTRUM
In spread spectrum (SS),, we combine signals from different
sources tto fit into a larger bandwidth,, butt our goals are to
prevent eavesdropping and jamming.. To achieve these
goals,, spread spectrum techniques add redundancy.
Topics discussed in this section:
Frequency Hopping Spread Spectrum (FHSS)
Direct Sequence Spread Spectrum (DSSS)
Spread spectrum
Input is fed into a channel encoder that produces an analog signal
with a relatively narrow BW around a center frequency.
• This signal is further modulated using a sequence of digits known as a
spreading code or spreading sequence.
• The effect of this modulation is to increase significantly the BW
(spread the spectrum) of the signal to be transmitted.
• On the receiving end, the same digit sequence is used to demodulate
the spread spectrum signal.
Frequency hopping spread spectrum (FHSS)
With FHSS the signal is broadcast over a random series of
radio frequencies, hopping from frequency to frequency at
fixed intervals.
A receiver, hopping between frequencies in synchronization with the transmitter picks up the message.
For transmission, binary data are fed into a modulator using FSK or
PSK. The resulting signal is entered on a base frequency. A
pseudorandom code generator serves as an index into a table of
frequencies (spreading code). Each k bits of the spreading sequence
specifies one of the 2k carrier frequencies. At each k-bit interval a new
carrier frequency is selected.
This frequency is then
modulated to produce a
new signal with the same
shape but centered on the
selected carrier frequency.
Direct Sequence Spread Spectrum (DSSS)
With DSSS each bit in the original signal is represented by multiple bits in the transmitted signal using a spreading code.
The spreading code spreads the signal across a wider frequency band in direct proportion to the number of bits used. eg. a 10-bit spreading code spreads the signal across a frequency band that is 10 times greater than a 1-bit spreading code.
Analog Transmission
Analog Transmission
Binary amplitude shift keying
Binary phase shift keying
QPSK
Example
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.
Concept of a constellation diagram
Constellation diagrams for ASK (OOK), BPSK, and QPSK
Constellation diagrams for some QAMs
Modem Standards
Modem = Modulator/Demodulator
Telephone modem:
Telephone Line Bandwidth
V.32 Modem
Use 32-QAM
Data is divided into 4-bit sections, each adding a redundant bit to form a 5-bit => reduce value density => reduce noise interference (how?)
baud rate = 2400 (why?)
V.32bis Modem
1st modem standard to support 14,400bps
Can adjust upstream or downstream speed depending on line or signal quality
Use 128-QAM => 6-bit data => Bit rate = 14,400 bps
V.34bis Modem
960-point constellation => bit rate = 28,800 bps
1664-point constellation => bit rate = 33,600 bps
33.6Kbps: max bit rate of traditional modems
Traditional Modems
56K Modem: V.90
56K Modem: V.92
Similar to V.90
Modem can adjust speed
If noise allows => upload max 48 Kbps, download still 56 Kbps
V.92: can interrupt the Internet connection when there is an incoming call (if call-waiting service is installed)
Binary amplitude shift keying
Binary phase shift keying
QPSK
Example
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.
Concept of a constellation diagram
Constellation diagrams for ASK (OOK), BPSK, and QPSK
Constellation diagrams for some QAMs
Modem Standards
Modem = Modulator/Demodulator
Telephone modem:
Telephone Line Bandwidth
V.32 Modem
Use 32-QAM
Data is divided into 4-bit sections, each adding a redundant bit to form a 5-bit => reduce value density => reduce noise interference (how?)
baud rate = 2400 (why?)
V.32bis Modem
1st modem standard to support 14,400bps
Can adjust upstream or downstream speed depending on line or signal quality
Use 128-QAM => 6-bit data => Bit rate = 14,400 bps
V.34bis Modem
960-point constellation => bit rate = 28,800 bps
1664-point constellation => bit rate = 33,600 bps
33.6Kbps: max bit rate of traditional modems
Traditional Modems
56K Modem: V.90
56K Modem: V.92
Similar to V.90
Modem can adjust speed
If noise allows => upload max 48 Kbps, download still 56 Kbps
V.92: can interrupt the Internet connection when there is an incoming call (if call-waiting service is installed)
Encoding and Modulating
Encoding and Modulating
Different Conversion Schemes
Digital to Digital Encoding
Types of Digital to Digital Encoding
Unipolar Encoding
Unipolar --->DC component (Average Amplitude is non zero)
--->Synchronization ( Due to unvarying signal)
Bipolar Encoding Scheme
Types of Bipolar Encoding
Bipolar AMI Encoding
B8ZS Encoding
HDB3 Encoding
Analog to Digital Encoding
PAM
Digital to Analog Encoding
Digital to Analog Modulation
Digital-to-analog modulation: the process of changing one of the characteristics of an analog signal based on the information in a digital signal
“Don’t forget”: Characteristics of a sine wave are amplitude, frequency, phase.
Carrier Signal
Sender
Produce a high-frequency signal that acts as a basis for the information signal => carrier signal
Modulate the carrier signal to reflect the digital information. The information signal is called the modulating signal
Receiver
Tune in the carrier frequency to receive
Digital-Analog Modulation Schemes
ASK
FSK
PSK
Now we will study
ASK
FSK
PSK
QAM
Amplitude Shift Keying (ASK)
Peak amplitude during each bit duration is constant
Bit Rate vs. Baud Rate
Bit rate is the number of bits per second
- More important in speaking of computer efficiency
Baud rate is the number of signal units per second that are required to represent those bits
More important in speaking of data transmission
Determine the bandwidth required to send the signal
Analogy in transportation: a baud is analogous to a car while a bit is analogous to a passenger (1: male, 0: female). The number of cars determines the traffic; that of passengers does not
Baud Rate Example
An analog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second, find the baud rate and the bit rate
Baud rate = 1000 bauds per second (baud/s)
Bit rate = 1000 x 4 = 4000 bps
The bit rate of a signal is 3000. If each signal unit carries 6 bits, what is the baud rate?
Baud rate = 3000 / 6 = 500 baud/s
ASK Disadvantage(2)
Disadvantage
Highly susceptible to noise interference because ASK relies on amplitude to differentiate between 1 and 0
Need a great gap between amplitude values so that noise can be detected and removed
OOK (on/off keying)
A popular ASK technique
Zero voltage represent a bit value (e.g., 0)
Save energy in transmitting information
Used to transmit digital data over optical fiber
ASK: Bandwidth Requirement
Question: What is the min bw for an ASK signal transmitting at 2000 bps? The transmission mode is half-duplex.
Frequency Shift Keying (FSK)
FSK: Baud Rate and Bandwidth
Question: Find the maximum bit rates for an FSK signal if the bandwidth of the medium is 12,000 Hz and the difference between the two carriers is 2000 Hz. Transmission is in full-duplex mode.
FSK vs. ASK
FSK
Less susceptible to error
Commonly used for high-freq (3-30 MHz) radio
Also used at even high freq on LANs that use coaxial cable
Phase Shift Keying (PSK)
2-PSK: only 2 phase values are used, each for 1 or 0
4-PSK
PSK: Baud Rate and Bandwidth
Same as in ASK, but can transmit more bps given same bandwidth
PSK: Questions
Given a bandwidth of 5000 Hz for an 8-PSK signal, what are the baud rate and bit rate? Transmission is in half-duplex mode.
For PSK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But in 8-PSK the bit rate is 3 times the baud rate, so the bit rate is 15,000 bps.
PSK: Drawback
Modulation device is not able to distinguish small differences in phase => limit BitRate
Why not combine PSK and ASK: x variations in phase with y variations in amplitude result in xy variations => increase bit rate
Quadrature Amplitude Modulation
A combination of ASK and PSK: both phase and amplitude varied
#amplitude shifts << #phase shifts
Lower susceptible to noise than ASK, higher bit rate than PSK
8-QAM
16-QAM
QAM Modulator
QAM: Bandwidth
Bandwidth requirement is the same as in ASK and PSK
QAM: BitRate vs. Baud Rate
ASK, FSK, 2-PSK Bit 1 N N
4-PSK, 4-QAM Dibit 2 N 2N
8-PSK, 8-QAM Tribit 3 N 3N
16-QAM Quadbit 4 N 4N
32-QAM Pentabit 5 N 5N
64-QAM Hexabit 6 N 6N
128-QAM Septabit 7 N 7N
256-QAM Octabit 8 N 8N
Analog to Analog Modulation
Amplitude Modulation: AM
Carrier signal is modulated so that its amplitude varies with the changing amplitudes of the modulating signal
Freq, phase remain same
Example: AM
Let x(t) = cos(2fmt)
Derive an express for s(t)
Answer:
s(t) = cos(2fct) + (na/2)cos(2(fc-fm)t) + (na/2)cos(2(fc+fm)t)
AM: Bandwidth
Audio signal bandwidth is ~5Khz => an AM radio station needs a minimum 10Khz
Frequency Modulation: FM
Freq. of carrier signal is modified to reflect the changing amplitudes of the modulating signal
Amp., phase remain same
FM: Bandwidth
The bandwidth of a stereo audio signal is usually 15 KHz. Therefore, an FM station needs at least a bandwidth of 150 KHz. Let PTA requires the minimum bandwidth to be at least 200 KHz (0.2 MHz).
Phase Modulation: PM
Only phase is varied to reflect the change of amplitude in modulating signal
Require simpler hardware than FM
Use in some systems as an alternative to FM
Analog to Analog Modulation
Representation of analog information by an analog signal
Why do we need it? Analog is already analog!!!
Because we may have to use a band-pass channel
Think about radio…
Schemes
Amplitude modulation (AM)
Frequency modulation (FM)
Phase modulation (PM)
Different Conversion Schemes
Digital to Digital Encoding
Types of Digital to Digital Encoding
Unipolar Encoding
Unipolar --->DC component (Average Amplitude is non zero)
--->Synchronization ( Due to unvarying signal)
Bipolar Encoding Scheme
Types of Bipolar Encoding
Bipolar AMI Encoding
B8ZS Encoding
HDB3 Encoding
Analog to Digital Encoding
PAM
Digital to Analog Encoding
Digital to Analog Modulation
Digital-to-analog modulation: the process of changing one of the characteristics of an analog signal based on the information in a digital signal
“Don’t forget”: Characteristics of a sine wave are amplitude, frequency, phase.
Carrier Signal
Sender
Produce a high-frequency signal that acts as a basis for the information signal => carrier signal
Modulate the carrier signal to reflect the digital information. The information signal is called the modulating signal
Receiver
Tune in the carrier frequency to receive
Digital-Analog Modulation Schemes
ASK
FSK
PSK
Now we will study
ASK
FSK
PSK
QAM
Amplitude Shift Keying (ASK)
Peak amplitude during each bit duration is constant
Bit Rate vs. Baud Rate
Bit rate is the number of bits per second
- More important in speaking of computer efficiency
Baud rate is the number of signal units per second that are required to represent those bits
More important in speaking of data transmission
Determine the bandwidth required to send the signal
Analogy in transportation: a baud is analogous to a car while a bit is analogous to a passenger (1: male, 0: female). The number of cars determines the traffic; that of passengers does not
Baud Rate Example
An analog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second, find the baud rate and the bit rate
Baud rate = 1000 bauds per second (baud/s)
Bit rate = 1000 x 4 = 4000 bps
The bit rate of a signal is 3000. If each signal unit carries 6 bits, what is the baud rate?
Baud rate = 3000 / 6 = 500 baud/s
ASK Disadvantage(2)
Disadvantage
Highly susceptible to noise interference because ASK relies on amplitude to differentiate between 1 and 0
Need a great gap between amplitude values so that noise can be detected and removed
OOK (on/off keying)
A popular ASK technique
Zero voltage represent a bit value (e.g., 0)
Save energy in transmitting information
Used to transmit digital data over optical fiber
ASK: Bandwidth Requirement
Question: What is the min bw for an ASK signal transmitting at 2000 bps? The transmission mode is half-duplex.
Frequency Shift Keying (FSK)
FSK: Baud Rate and Bandwidth
Question: Find the maximum bit rates for an FSK signal if the bandwidth of the medium is 12,000 Hz and the difference between the two carriers is 2000 Hz. Transmission is in full-duplex mode.
FSK vs. ASK
FSK
Less susceptible to error
Commonly used for high-freq (3-30 MHz) radio
Also used at even high freq on LANs that use coaxial cable
Phase Shift Keying (PSK)
2-PSK: only 2 phase values are used, each for 1 or 0
4-PSK
PSK: Baud Rate and Bandwidth
Same as in ASK, but can transmit more bps given same bandwidth
PSK: Questions
Given a bandwidth of 5000 Hz for an 8-PSK signal, what are the baud rate and bit rate? Transmission is in half-duplex mode.
For PSK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But in 8-PSK the bit rate is 3 times the baud rate, so the bit rate is 15,000 bps.
PSK: Drawback
Modulation device is not able to distinguish small differences in phase => limit BitRate
Why not combine PSK and ASK: x variations in phase with y variations in amplitude result in xy variations => increase bit rate
Quadrature Amplitude Modulation
A combination of ASK and PSK: both phase and amplitude varied
#amplitude shifts << #phase shifts
Lower susceptible to noise than ASK, higher bit rate than PSK
8-QAM
16-QAM
QAM Modulator
QAM: Bandwidth
Bandwidth requirement is the same as in ASK and PSK
QAM: BitRate vs. Baud Rate
ASK, FSK, 2-PSK Bit 1 N N
4-PSK, 4-QAM Dibit 2 N 2N
8-PSK, 8-QAM Tribit 3 N 3N
16-QAM Quadbit 4 N 4N
32-QAM Pentabit 5 N 5N
64-QAM Hexabit 6 N 6N
128-QAM Septabit 7 N 7N
256-QAM Octabit 8 N 8N
Analog to Analog Modulation
Amplitude Modulation: AM
Carrier signal is modulated so that its amplitude varies with the changing amplitudes of the modulating signal
Freq, phase remain same
Example: AM
Let x(t) = cos(2fmt)
Derive an express for s(t)
Answer:
s(t) = cos(2fct) + (na/2)cos(2(fc-fm)t) + (na/2)cos(2(fc+fm)t)
AM: Bandwidth
Audio signal bandwidth is ~5Khz => an AM radio station needs a minimum 10Khz
Frequency Modulation: FM
Freq. of carrier signal is modified to reflect the changing amplitudes of the modulating signal
Amp., phase remain same
FM: Bandwidth
The bandwidth of a stereo audio signal is usually 15 KHz. Therefore, an FM station needs at least a bandwidth of 150 KHz. Let PTA requires the minimum bandwidth to be at least 200 KHz (0.2 MHz).
Phase Modulation: PM
Only phase is varied to reflect the change of amplitude in modulating signal
Require simpler hardware than FM
Use in some systems as an alternative to FM
Analog to Analog Modulation
Representation of analog information by an analog signal
Why do we need it? Analog is already analog!!!
Because we may have to use a band-pass channel
Think about radio…
Schemes
Amplitude modulation (AM)
Frequency modulation (FM)
Phase modulation (PM)
TRANSMISSION MODES
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.
Data transmission and modes
Parallel transmission
Serial transmission
Asynchronous transmission
Synchronous transmission
for more detail
cool_pkjanu@yahoo.com
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.
Data transmission and modes
Parallel transmission
Serial transmission
Asynchronous transmission
Synchronous transmission
for more detail
cool_pkjanu@yahoo.com
ANALOG-TO-DIGITAL CONVERSION
ANALOG-TO-DIGITAL CONVERSION
A digital signal is superior to an analog signal.
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
Components of PCM encoder
Three different sampling methods for PCM
Nyquist sampling rate for low-pass and bandpass signals
According to the Nyquist theorem,
the sampling rate must be at least 2 times the highest frequency contained in the signal
Recovery of a sampled sine wave for different sampling rates
Sampling at the Nyquist rate can create a good approximation of the original sine wave.
Oversampling can also create the same approximation, but is redundant and unnecessary.
Sampling below the Nyquist rate does not produce a signal that looks like the original sine wave.
Sampling of a clock with only one hand
The second hand of a clock has a period of 60 s.
According to the Nyquist theorem, we need to sample hand every 30 s
Examples
An example of under-sampling is the seemingly backward rotation of the wheels of a forward-moving car in a movie.
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.
Telephone companies digitize voice by assuming a maximum frequency of 4000 Hz.
The sampling rate therefore is 8000 samples per second
Example
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
Quantization and encoding of a sampled signal
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
A digital signal is superior to an analog signal.
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
Components of PCM encoder
Three different sampling methods for PCM
Nyquist sampling rate for low-pass and bandpass signals
According to the Nyquist theorem,
the sampling rate must be at least 2 times the highest frequency contained in the signal
Recovery of a sampled sine wave for different sampling rates
Sampling at the Nyquist rate can create a good approximation of the original sine wave.
Oversampling can also create the same approximation, but is redundant and unnecessary.
Sampling below the Nyquist rate does not produce a signal that looks like the original sine wave.
Sampling of a clock with only one hand
The second hand of a clock has a period of 60 s.
According to the Nyquist theorem, we need to sample hand every 30 s
Examples
An example of under-sampling is the seemingly backward rotation of the wheels of a forward-moving car in a movie.
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.
Telephone companies digitize voice by assuming a maximum frequency of 4000 Hz.
The sampling rate therefore is 8000 samples per second
Example
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
Quantization and encoding of a sampled signal
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
Digital Transmission
Digital Transmission
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.
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
Topics discussed in this section:
Line Coding
Line Coding Schemes
Block Coding
Scrambling
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”.
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.
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.
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
Example
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?
We assume that the average value of c is 1/2 . The baud rate is then
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
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.
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.
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
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.
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.
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.
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.
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.
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
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
In Manchester and differential Manchester encoding, the transition
at the middle of the bit is used for synchronization
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.
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.
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 “2m” symbols.
If we have L signal levels, we can use “n” signal elements to create Ln signal elements.
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.
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
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.
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!
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.
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.
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.
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.
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.
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
Topics discussed in this section:
Line Coding
Line Coding Schemes
Block Coding
Scrambling
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”.
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.
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.
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
Example
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?
We assume that the average value of c is 1/2 . The baud rate is then
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
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.
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.
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
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.
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.
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.
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.
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.
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
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
In Manchester and differential Manchester encoding, the transition
at the middle of the bit is used for synchronization
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.
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.
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 “2m” symbols.
If we have L signal levels, we can use “n” signal elements to create Ln signal elements.
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.
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
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.
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!
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.
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.
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.
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.
TRANSMISSION IMPAIRMENT
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
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
A loss of 3 dB (–3 dB) is equivalent to losing one-half the power.
Distortion
Noise
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)
Increasing the levels of a signal may reduce the reliability of the system.
Nyquist Theorem
In baseband transmission, we said 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.
Shannon Capacity
In reality, we can not have a noisless channel
For noisy channel,
Capacity = Bandwith x log2(1+SNR)
The Shannon capacity gives us the upper limit;
the Nyquist formula tells us how many signal levels we need
Example
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. What is the channel capacity?
Solution
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.
PERFORMANCE
One important issue in networking is the performance of the network—how good is it?
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.
Examples
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.
If the telephone company improves the quality of the line and increases the bandwidth to 8 kHz,
we can send 112,000 bps.
Concept of bandwidth-delay product
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.
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
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
A loss of 3 dB (–3 dB) is equivalent to losing one-half the power.
Distortion
Noise
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)
Increasing the levels of a signal may reduce the reliability of the system.
Nyquist Theorem
In baseband transmission, we said 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.
Shannon Capacity
In reality, we can not have a noisless channel
For noisy channel,
Capacity = Bandwith x log2(1+SNR)
The Shannon capacity gives us the upper limit;
the Nyquist formula tells us how many signal levels we need
Example
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. What is the channel capacity?
Solution
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.
PERFORMANCE
One important issue in networking is the performance of the network—how good is it?
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.
Examples
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.
If the telephone company improves the quality of the line and increases the bandwidth to 8 kHz,
we can send 112,000 bps.
Concept of bandwidth-delay product
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.
DIGITAL SIGNALS
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.
Two digital signals: one with two signal levels and
the other with four signal levels
Examples
A digital signal has 8 levels. How many bits are needed per level?
We calculate the number of bits from the formula
Each signal level is represented by 3 bits.
A digital signal has 9 levels. How many bits are needed per level?
Each signal level is represented by 3.17 bits.
The number of bits sent per level needs to be an integer as well as a power of 2.
Hence, 4 bits can represent one level.
Example
HDTV uses digital signals to broadcast high quality video signals. There are 1920 by 1080 pixels per screen, and the screen is renewed 30 times per second. Also, 24 bits represents one color pixel.
What is the bit rate for high-definition TV (HDTV)?
The TV stations reduce this rate to 20 to 40 Mbps through compression.
Baseband transmission
A digital signal is a composite analog signal with an infinite bandwidth.
Baseband transmission using a dedicated medium
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.
Example
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
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
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.
Bandwidth of a bandpass channel
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.
Modulation of a digital signal for transmission
on a bandpass channel
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
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.
Two digital signals: one with two signal levels and
the other with four signal levels
Examples
A digital signal has 8 levels. How many bits are needed per level?
We calculate the number of bits from the formula
Each signal level is represented by 3 bits.
A digital signal has 9 levels. How many bits are needed per level?
Each signal level is represented by 3.17 bits.
The number of bits sent per level needs to be an integer as well as a power of 2.
Hence, 4 bits can represent one level.
Example
HDTV uses digital signals to broadcast high quality video signals. There are 1920 by 1080 pixels per screen, and the screen is renewed 30 times per second. Also, 24 bits represents one color pixel.
What is the bit rate for high-definition TV (HDTV)?
The TV stations reduce this rate to 20 to 40 Mbps through compression.
Baseband transmission
A digital signal is a composite analog signal with an infinite bandwidth.
Baseband transmission using a dedicated medium
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.
Example
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
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
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.
Bandwidth of a bandpass channel
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.
Modulation of a digital signal for transmission
on a bandpass channel
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
ANALOG AND DIGITAL
ANALOG AND DIGITAL
Data can be analog or digital
Analog data refers to information that is continuous
Analog data take on continuous values
Analog signals can have an infinite number of values in a range
Digital data refers to information that has discrete states
Digital data take on discrete values
Digital signals can have only a limited number of values
In data communications, we commonly use
periodic analog signals and nonperiodic digital signals.
PERIODIC ANALOG 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.
Signal amplitude and 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.
If a signal does not change at all, its frequency is zero
If a signal changes instantaneously, its frequency is infinite.
Frequency and Period.
Frequency and period are the inverse of each other.
Phase
Phase describes the position of the waveform
relative to time 0
Wavelength and period
Wavelength = Propagation speed x Period
= Propagation speed / Frequency
Time-domain and frequency-domain plots of a sine wave
A complete sine wave in the time domain can be represented by one single spike in the frequency domain.
Frequency Domain
The frequency domain is more compact and useful when we are dealing with more than one sine wave.
A single-frequency sine wave is not useful in data communication
We need to send a composite signal, a signal made of many simple sine waves.
Fourier analysis
According to Fourier analysis,
any composite signal is a combination of simple sine waves with different frequencies, amplitudes, and phases.
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.
A composite periodic signal
Decomposition of the composite periodic signal in the time and frequency domains
Time and frequency domains of a nonperiodic signal
A non-periodic composite signal
It can be a 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.
Bandwidth
The bandwidth of a composite signal is
the difference between the highest and the lowest frequencies contained in that signal.
Data can be analog or digital
Analog data refers to information that is continuous
Analog data take on continuous values
Analog signals can have an infinite number of values in a range
Digital data refers to information that has discrete states
Digital data take on discrete values
Digital signals can have only a limited number of values
In data communications, we commonly use
periodic analog signals and nonperiodic digital signals.
PERIODIC ANALOG 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.
Signal amplitude and 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.
If a signal does not change at all, its frequency is zero
If a signal changes instantaneously, its frequency is infinite.
Frequency and Period.
Frequency and period are the inverse of each other.
Phase
Phase describes the position of the waveform
relative to time 0
Wavelength and period
Wavelength = Propagation speed x Period
= Propagation speed / Frequency
Time-domain and frequency-domain plots of a sine wave
A complete sine wave in the time domain can be represented by one single spike in the frequency domain.
Frequency Domain
The frequency domain is more compact and useful when we are dealing with more than one sine wave.
A single-frequency sine wave is not useful in data communication
We need to send a composite signal, a signal made of many simple sine waves.
Fourier analysis
According to Fourier analysis,
any composite signal is a combination of simple sine waves with different frequencies, amplitudes, and phases.
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.
A composite periodic signal
Decomposition of the composite periodic signal in the time and frequency domains
Time and frequency domains of a nonperiodic signal
A non-periodic composite signal
It can be a 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.
Bandwidth
The bandwidth of a composite signal is
the difference between the highest and the lowest frequencies contained in that signal.
Data & Signals
Analog & digital
Periodic analog signal
Digital signal
Transmission Impairment
Data rate limits
Performance
Physical Layers
Bit-to-Signal Transformation
To be transmitted, data must be transformed to electromagnetic signals.
Bit-Rate Control
The transmission medium provides the upper limit on data rate, the physical layer controls it
Bit Synchronization
The physical layer provides clocking mechanisms to control the timing of bit transfer
Multiplexing
A physical link may need to be divided into logical channels, each used by a transmission, for better efficiency
Circuit Switching
Allowing pairs of nodes to have their dedicated links is a function of the physical layer
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.
In data communications, we commonly use periodic analog signals and aperiodic digital signals.
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.
If a signal does not change at all, its frequency is zero. If a signal changes instantaneously, its frequency is infinite.
Phase
Question: A sine wave is offset one-sixth of a cycle with respect to time zero. What is its phase in degrees and radians???
Sine Wave: One more example
An analog signal is best represented in the frequency domain.
Use of Sine Waves
A power company sends a single sine wave with frequency 60Hz to distribute electric energy to houses and businesses
We can use a single sine wave to send an alarm to a security center
But single sine waves are useless in data communications. WHY?
Think about phone conversation
Composite Analog Signals
In reality, signals are more complicated than sine waves => can be represented as a combination of sine waves with different frequencies, amplitude, and phases
(using Fourier transformation)
Fourier Transformation
s(t)=A1sin(2f1t+ф1) + A2sin(2f2t+ф1) + A3sin(2f3t+ф3) +…
Frequency Spectrum
The description of a signal using the frequency domain and containing all its sine-wave components is called the frequency spectrum of that signal
Signal Over Transmission Medium
Signal corruption: Transmission medium may block or weaken some frequencies. As a result, the output signal may be different from the input signal.
Perfect medium: preserve frequency, amplitude, and phase values
Bandwidth of Transmission Medium
The bandwidth is a property of a medium: It is the difference between the highest and the lowest frequencies that the medium can satisfactorily pass.
Digital vs. Analog
Digital’s advantage:
Digital design less expensive and more reliable
Greater dynamic range and error detection and recovery by the use of coding
Many sources can be combine (voice, video, data) and sent over the same channel
Digital’s disadvantage
More bandwidth needed
Baseband vs. Broadband Signals
Baseband signal: frequency spectrum extends from 0 to some maximum frequency (similar to low-pass channel’s property)
Broadband signal: frequency spectrum belongs to a range between a min freq and a max freq (similar to band-pass channel’s property)
Data Rate Limits
How fast can we send data (in bps) over a link?
The answer depends on
Bandwidth available
Levels of signals we can use
Quality of channel (level of noise)
Noiseless Channel: Nyquist Bit rate
Bit Rate = 2 Bandwidth log2L
L: the number of signal levels used to represent data
Noisy Channel: Shannon Capacity
Channel Capacity = Bandwidth log2(1+SNR)
SNR: signal to noise ratio – statistical ratio of the signal power and noise power
Bandwidth: analog channel bandwidth
Bit Rate = 2 3000 log2 2 = 6000 bps
Periodic analog signal
Digital signal
Transmission Impairment
Data rate limits
Performance
Physical Layers
Bit-to-Signal Transformation
To be transmitted, data must be transformed to electromagnetic signals.
Bit-Rate Control
The transmission medium provides the upper limit on data rate, the physical layer controls it
Bit Synchronization
The physical layer provides clocking mechanisms to control the timing of bit transfer
Multiplexing
A physical link may need to be divided into logical channels, each used by a transmission, for better efficiency
Circuit Switching
Allowing pairs of nodes to have their dedicated links is a function of the physical layer
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.
In data communications, we commonly use periodic analog signals and aperiodic digital signals.
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.
If a signal does not change at all, its frequency is zero. If a signal changes instantaneously, its frequency is infinite.
Phase
Question: A sine wave is offset one-sixth of a cycle with respect to time zero. What is its phase in degrees and radians???
Sine Wave: One more example
An analog signal is best represented in the frequency domain.
Use of Sine Waves
A power company sends a single sine wave with frequency 60Hz to distribute electric energy to houses and businesses
We can use a single sine wave to send an alarm to a security center
But single sine waves are useless in data communications. WHY?
Think about phone conversation
Composite Analog Signals
In reality, signals are more complicated than sine waves => can be represented as a combination of sine waves with different frequencies, amplitude, and phases
(using Fourier transformation)
Fourier Transformation
s(t)=A1sin(2f1t+ф1) + A2sin(2f2t+ф1) + A3sin(2f3t+ф3) +…
Frequency Spectrum
The description of a signal using the frequency domain and containing all its sine-wave components is called the frequency spectrum of that signal
Signal Over Transmission Medium
Signal corruption: Transmission medium may block or weaken some frequencies. As a result, the output signal may be different from the input signal.
Perfect medium: preserve frequency, amplitude, and phase values
Bandwidth of Transmission Medium
The bandwidth is a property of a medium: It is the difference between the highest and the lowest frequencies that the medium can satisfactorily pass.
Digital vs. Analog
Digital’s advantage:
Digital design less expensive and more reliable
Greater dynamic range and error detection and recovery by the use of coding
Many sources can be combine (voice, video, data) and sent over the same channel
Digital’s disadvantage
More bandwidth needed
Baseband vs. Broadband Signals
Baseband signal: frequency spectrum extends from 0 to some maximum frequency (similar to low-pass channel’s property)
Broadband signal: frequency spectrum belongs to a range between a min freq and a max freq (similar to band-pass channel’s property)
Data Rate Limits
How fast can we send data (in bps) over a link?
The answer depends on
Bandwidth available
Levels of signals we can use
Quality of channel (level of noise)
Noiseless Channel: Nyquist Bit rate
Bit Rate = 2 Bandwidth log2L
L: the number of signal levels used to represent data
Noisy Channel: Shannon Capacity
Channel Capacity = Bandwidth log2(1+SNR)
SNR: signal to noise ratio – statistical ratio of the signal power and noise power
Bandwidth: analog channel bandwidth
Bit Rate = 2 3000 log2 2 = 6000 bps
OSI Model
Internet Model
OSI Model
In the late 1970’s ISO formulated a Reference (Open Systems Interconnection) Model:
To provide a common basis for the coordination of standards development
To allow existing and evolving standards to be placed into perspective with one another
has become the stand model for classifying communications functions
The Model is concerned with the structuring of Communication to provide a reliable, open communication service which is independent of any specific manufacturer’s equipment or conventions
ISO - OSI Model
Consists of seven layers:
Application
Presentation User
Session
Transport
Network
Data Link Network
Physical
Basic Concepts
Line Configuration
Topology
Transmission Mode
Categories of Networks
Internetworks
OSI Model (physiccal layer)
Note: The physical layer is responsible for transmitting individual bits from one node to the next
Define the characteristics of the interface between the devices and the transmission media
Encode bits into signals and decode signals to get bits
Define transmission rate, which must be the same for both sender and receiver
Synchronize clocks
The data link layer is responsible for transmitting frames from one node to the next
Framing: divide the data stream into manageable data units called “frames”
Physical addressing: insert the physical address of the next node into frame’s header
Flow control: prevent overflow at receiver
Error control: make sure that frames are correctly received
Access control: make sure that there is no link access conflict
The network layer is responsible for the delivery of packets from the original source to the final destination over multiple networks
Logical addressing: e.g., IP addresses
Routing: how to get to the destination
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.
The transport layer is responsible for delivery of a message from one process to another
OSI Model
In the late 1970’s ISO formulated a Reference (Open Systems Interconnection) Model:
To provide a common basis for the coordination of standards development
To allow existing and evolving standards to be placed into perspective with one another
has become the stand model for classifying communications functions
The Model is concerned with the structuring of Communication to provide a reliable, open communication service which is independent of any specific manufacturer’s equipment or conventions
ISO - OSI Model
Consists of seven layers:
Application
Presentation User
Session
Transport
Network
Data Link Network
Physical
Basic Concepts
Line Configuration
Topology
Transmission Mode
Categories of Networks
Internetworks
OSI Model (physiccal layer)
Note: The physical layer is responsible for transmitting individual bits from one node to the next
Define the characteristics of the interface between the devices and the transmission media
Encode bits into signals and decode signals to get bits
Define transmission rate, which must be the same for both sender and receiver
Synchronize clocks
The data link layer is responsible for transmitting frames from one node to the next
Framing: divide the data stream into manageable data units called “frames”
Physical addressing: insert the physical address of the next node into frame’s header
Flow control: prevent overflow at receiver
Error control: make sure that frames are correctly received
Access control: make sure that there is no link access conflict
The network layer is responsible for the delivery of packets from the original source to the final destination over multiple networks
Logical addressing: e.g., IP addresses
Routing: how to get to the destination
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.
The transport layer is responsible for delivery of a message from one process to another
Basic information of networking
Chapter 2
Basic Concepts
• Line Configuration
• Topology
• Transmission Mode
• Categories of Networks
• Internetworks
Local Area Network
A LAN (local area network) supplies networking capability to a group of computers in close proximity to each other such as in an office building
Metropolitan Area Network
Metropolitan area networks, or MANs, are large computer networks usually spanning a city. They typically use wireless infrastructure
Wide Area Network
A wide area network (WAN) spans a large geographic area. A WAN often contains multiple local area networks (LANs) and/or metropolitan area networks (MANs).
Basic Concepts
• Line Configuration
• Topology
• Transmission Mode
• Categories of Networks
• Internetworks
Local Area Network
A LAN (local area network) supplies networking capability to a group of computers in close proximity to each other such as in an office building
Metropolitan Area Network
Metropolitan area networks, or MANs, are large computer networks usually spanning a city. They typically use wireless infrastructure
Wide Area Network
A wide area network (WAN) spans a large geographic area. A WAN often contains multiple local area networks (LANs) and/or metropolitan area networks (MANs).
Fundamental Parameters of Data
The effectiveness of data communication system depends upon following four fundamental characteristics:
1. Delivering Data Packets
2. Accuracy
3. Timelessness
4. Jitter
Variation in packet arrival time
Data Communication System Components
1. Message
2. Sender
3. Receiver
4. Medium
5. Protocol
Data Communication System
Data Representation
• It has different forms
Text, number, image, audio, video
Mode of Transmission
• Mode of transmission or data flow between two devices can be
1. Simplex
2. Half Duplex
3. Full Duplex
Networks
• Define Network
– Set of devices connected through communication link.
• Node
– It can be computer, printer or any other device capable of sending/ receiving data.
• Distributed Processing
1. Security
2. Distributed Databases
3. Faster problem solving
4. Security through redundancy
5. Collaborative processing
Network Criteria
Performance
Performance can be measured in many ways including transit and response time.
– Transit time
– Response time
• The performance of network depends upon
– Number of Users
– Type of Transmission Medium
– Hardware( buffer, speed etc)
– Software ( raw data conversion and reassembly)
Reliability ( robustness)
• Frequency of Failure
• Recovery time of a network after a failure
• Catastrophe
Security
• Unauthorized Access
• Viruses
• Hacking etc..
Protocols & Standards
• Define Protocol? ( what, how and when)
• Entity ( data base management, browse, FTP)
– Any thing capable of sending and receiving the information
• Key elements of Protocol
1. Syntax ( format of data)
2. Semantics ( meaning of each section of bits)
3. Timing ( when to send and how fast to send)
Standards
• Standards are essential in creating and maintaining an open and competitive market for the manufacturer
• Standards provide guidelines
• Data communication standards fall into two catagories
• De Jure ( By Law)
– Those standards that has been legislated by an officially recognized body , forums and government agencies.
• De Facto ( By Fact)
– Those standards established by manufacturer, who seek functionality of new product.
Standard Organizations
• Developed by Cooperation among standards creation committees, forums and government regulatory agencies
• Standards Creation Committees
1. ISO
2. ITU
3. ANSI
4. IEEE
5. EIA
6. Telcordia
Forums
• Frame relay Forum
• ATM forum
• Internet Society( ISOC) and Internet Engg Task force
( IETF)
Regulatory Agency
• Purpose is to protect the public interest
• PTA
• FCC
• BTC
Regulatory Agencies
Following are the responsibilities of regulatory agencies
• To review rate and service charge applications made by telegraph and telephone providers.
• To review the technical specifications of communications hardware.
• To establish reasonable common carrier rates of return.
• To divide and allocate radio frequencies.
• To assign carrier frequencies for radio and TV broadcast.
Mesh Topology
Mesh
• Dedicated link
• Robust
• Security
• Fault identification is easier
Disadvantage
• Amount of cabling and I/O ports
• Installation is difficult and reconfiguration too
• Bulk of wiring may be more than spacing
Star Topology
Star
• Less expensive
• Simple due to one I/O port
• Installation and conf is easier
• Robustness
• Self fault identification
Disadvantages
• Hub is down then…..
Tree Topology
Tree
• Same as Star
Bus Topology
Bus
• Ease of installation
• Less cabling
• In same room 4 cables for star where as 1 cable in case of bus
Disadvantage
• Difficult in reconfig and fault isolation
• Low scalability
• When bus is down all network will be down.
Ring
• Fault isolation is simple
• Alarming
Disadvantages
• Ring breaks then…
• Station is dead
The effectiveness of data communication system depends upon following four fundamental characteristics:
1. Delivering Data Packets
2. Accuracy
3. Timelessness
4. Jitter
Variation in packet arrival time
Data Communication System Components
1. Message
2. Sender
3. Receiver
4. Medium
5. Protocol
Data Communication System
Data Representation
• It has different forms
Text, number, image, audio, video
Mode of Transmission
• Mode of transmission or data flow between two devices can be
1. Simplex
2. Half Duplex
3. Full Duplex
Networks
• Define Network
– Set of devices connected through communication link.
• Node
– It can be computer, printer or any other device capable of sending/ receiving data.
• Distributed Processing
1. Security
2. Distributed Databases
3. Faster problem solving
4. Security through redundancy
5. Collaborative processing
Network Criteria
Performance
Performance can be measured in many ways including transit and response time.
– Transit time
– Response time
• The performance of network depends upon
– Number of Users
– Type of Transmission Medium
– Hardware( buffer, speed etc)
– Software ( raw data conversion and reassembly)
Reliability ( robustness)
• Frequency of Failure
• Recovery time of a network after a failure
• Catastrophe
Security
• Unauthorized Access
• Viruses
• Hacking etc..
Protocols & Standards
• Define Protocol? ( what, how and when)
• Entity ( data base management, browse, FTP)
– Any thing capable of sending and receiving the information
• Key elements of Protocol
1. Syntax ( format of data)
2. Semantics ( meaning of each section of bits)
3. Timing ( when to send and how fast to send)
Standards
• Standards are essential in creating and maintaining an open and competitive market for the manufacturer
• Standards provide guidelines
• Data communication standards fall into two catagories
• De Jure ( By Law)
– Those standards that has been legislated by an officially recognized body , forums and government agencies.
• De Facto ( By Fact)
– Those standards established by manufacturer, who seek functionality of new product.
Standard Organizations
• Developed by Cooperation among standards creation committees, forums and government regulatory agencies
• Standards Creation Committees
1. ISO
2. ITU
3. ANSI
4. IEEE
5. EIA
6. Telcordia
Forums
• Frame relay Forum
• ATM forum
• Internet Society( ISOC) and Internet Engg Task force
( IETF)
Regulatory Agency
• Purpose is to protect the public interest
• PTA
• FCC
• BTC
Regulatory Agencies
Following are the responsibilities of regulatory agencies
• To review rate and service charge applications made by telegraph and telephone providers.
• To review the technical specifications of communications hardware.
• To establish reasonable common carrier rates of return.
• To divide and allocate radio frequencies.
• To assign carrier frequencies for radio and TV broadcast.
Mesh Topology
Mesh
• Dedicated link
• Robust
• Security
• Fault identification is easier
Disadvantage
• Amount of cabling and I/O ports
• Installation is difficult and reconfiguration too
• Bulk of wiring may be more than spacing
Star Topology
Star
• Less expensive
• Simple due to one I/O port
• Installation and conf is easier
• Robustness
• Self fault identification
Disadvantages
• Hub is down then…..
Tree Topology
Tree
• Same as Star
Bus Topology
Bus
• Ease of installation
• Less cabling
• In same room 4 cables for star where as 1 cable in case of bus
Disadvantage
• Difficult in reconfig and fault isolation
• Low scalability
• When bus is down all network will be down.
Ring
• Fault isolation is simple
• Alarming
Disadvantages
• Ring breaks then…
• Station is dead
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