DCN A05 Multiplexing

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DATA COMMUNICATION

AND

NETWORKING

( 07CS51 )



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Multiplexing : set of techniques that

allows simultaneous transmission ofmultiple signals across a single data link

an efficient method that allows maximization

of utilization of bandwidth

Multiplexing

n lines share thebandwidth of one link



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Multiplexing techniques :

• FDM• WDM• TDM

Multiplexing



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Frequency - Division Multiplexing (FDM)

analog techniqueused when the available bandwidth of a linkis greater than the combined bandwidths ofthe signals to be transmitted

method :• signals generated by each sending device

modulate different carrier frequencies

• modulated signals are combined into asingle composite signal that can betransported on the link

Multiplexing



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Frequency - Division Multiplexing (FDM)Multiplexing

one link - many channels



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sometimes, channels are separated byslots of unused bandwidth…..

to ensure carrier frequencies do notinterfere i.e. to prevent signals fromoverlapping these slots are called guard bands

carrier frequencies must also not interferewith the original data frequencies

Multiplexing



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Example BF4e6.1

Multiplexing

Multiplexing of 3 voice channels of 4 KHz bandwidth each ontoa link with bandwidth range 24 - 32 KHz without guard band



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Example BF4e6.2

Five channels each with 100 KHz

bandwidth are to me multiplexed together.

What is the minimum bandwidth of thelink if there is a need for a guard band of

10 KHz between the channels to preventinterference ?

Multiplexing



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Example BF4e6.6.13

Ten voice channels of bandwidth 4 KHzeach are required to be multiplexed using

FDM.Guard bands of width 500 Hz each areused.

Calculate the required bandwidth.

Multiplexing



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Example BF4e6.6.13

Bandwidth= (4 x 10 KHz) + (9 x 500 Hz)

= 44.5 KHz

Multiplexing



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The Analog Carrier System

Multiplexing

Analog hierarchy

Telephone companiesmultiplex signals from lower-bandwidth lines onto higher-bandwidth lines to maximizethe efficiency of their infrastructure



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The Analog Carrier Systemmaximization of efficiency of infrastructuremultiplexing of signals from lowerbandwidth lines onto higher bandwidth lines

Analog hierarchy :12 voice channels are multiplexed onto ahigher bandwidth line to create a group

5 groups → super group10 super groups → master group6 master groups → jumbo group

Multiplexing



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Example BF4e6.6.15

In the analog hierarchy shown, find theoverhead in each hierarchy level.

Multiplexing



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Example BF4e6.6.15

(a) Group level: 0 Hz

(b) Supergroup level: 0 Hz(c) Master group: overhead = 2520 KHz − (10 × 240 KHz) =

= 120 KHz

(d) Jumbo Group:overhead = 16.984 MHz − (6 × 2.52 MHz)

= 1.864 MHz

Multiplexing



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Wavelength - Division Multiplexing (WDM)

is designed to use the high - data - ratecapability of fiber-optic (OFC) cableconceptually same as FDM

uses optical signals(very high frequencies)

transmitted through fibre-optic channels

Multiplexing

combination ofoptic signals



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Wavelength - Division Multiplexing (WDM)

combining of multiple light sources intoone single light at the multiplexer ….. and splitting at the demultiplexer isaccomplished using prismsbending of light depends on angle ofincidence and wavelength

Multiplexing



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End of Class - 15 10.10.2011

Data Communication & Networking



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Time - Division Multiplexing (TDM)

digital methodallows several connections (low - rate) toshare a high bandwidth (high - rate) of a link

sharing is based on each connectionoccupying a portion of time in the link

Multiplexing



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two schemes :• synchronous TDM• statistical TDM

Synchronous TDMin synchronous TDM, each input connection

is allotted an output even if it is not sendingdata

Multiplexing



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Synchronous TDM : time slots and framesunit : can be one bit, one character or oneblock of datadata flow of each input connection isdivided into unitseach input occupies one input time slotafter multiplexing :•

each input unit becomes one output unit• one output unit occupies one output time slot• duration of an output slot is n times shorterthan the duration of an input time slot

Multiplexing

( n denotes the number of connections)



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Synchronous TDM : time slots and framesa round of data units from each inputconnection is collected into a frame

for n connections, a frame is divided inton time slots ...one slot is allocated for each unit i.e. onefor each input line

if the duration of the input unit is T, theduration of each slot is (T ÷ n )duration of each frame is T

Multiplexing



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Synchronous TDM : time slots and frameslet the input time slot be T secondsthe output time slot will be (T ÷ n) seconds

Multiplexing

a combination of data unitsfrom each input connection

is called a frame



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Synchronous TDM : Example BF4e6.6

Multiplexing

given : unit of data in the above example = 1 bit

input bit duration ? = 1 / 1 Mbps = 1 μsecoutput bit duration ? = 1 / 4 x 1 = 0.25 μsec

output bit rate ? = 1 / 0.25 μsec = 4 Mbpsframe rate? = input rate = 106 frames / sec



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Synchronous TDM : empty slots

Multiplexing

if the source does not have data to send,the corresponding slot in the output isempty → reduced efficiency



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Synchronous TDM : data rate managementso far, it has been seen that the data rateof all input lines were samehow to handle the situation if the inputdata rates are different on different lines ?

three strategies :• multilevel multiplexing• multiple-slot allocation• pulse stuffing

Multiplexing



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Synchronous TDM : data rate managementMultilevel multiplexingused when data rate of an input line is afraction of others

Multiplexing



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Synchronous TDM : data rate managementMulti-slot allocationused when data rate of an input line is amultiple of others → allot more than oneslot in a frame to a single input line

Multiplexing

serial to parallelconverter



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Synchronous TDM : data rate managementPulse stuffing (or bit padding or bit stuffing)

used when data rate of sources are notmultiple integers of each otherhighest rate = dominant data rate (50 Kbps)

Multiplexing

add dummy bits to the input lines with lower data rates



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Synchronous TDM : frame synchronization

lack of synchronization between multiplexerand demultiplexer can result in ……

a bit belonging to one channel beingreceived by a wrong channel

to overcome this, one or more sync bits,called framing bits , are added at thebeginning of each frame

Multiplexing



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Synchronous TDM : frame synchronization

Multiplexing

framing bits are added according to a pattern

usually 1 bit per frame, alternatingbetween 0 and 1



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End of Class - 16 10.10.2011




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Synchronous TDM : frame synchronizationExample BF4e6.10

Given :• 4 sources, each creating 250 charactersper second

• Multiplexing done at character-level• 1 sync bit is added to each frame

Multiplexing



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Synchronous TDM : frame synchronizationExample BF4e6.10

Given :(a) data rate of each source = 250 x 8 = 2 Kbps

(b) duration of character = (1 / 250) = 4 msec

(c) each frame comprises one character from

each source

→ frame rate = 250 frames per sec

(d) duration of each frame = 4 msec

(e) size of each frame = (4 x 8) + 1 = 33 bits

Multiplexing



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Synchronous TDM : frame synchronizationExample BF4e6.6.17

20 digital sources each of 100 Kbps arecombined using synchronous TDM.

Each output slot carries 2 bits from eachdigital source.

One extra sync bit is added to each frame.

Multiplexing



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(a) What is the size of an output frame in bits ?

(b) What is the output frame rate ?

(c) What is the duration of an output frame ?

(d) What is the output data rate ?

(e) What is the efficiency (ratio of useful

bits to total bits) of the system ?

Multiplexing



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(a) Frame size = (20 × 2) + 1 = 41 bits.

(b) Frame rate = 20 x 100 x 1000 / 2 = 106

fr / sec (c) Frame duration = 1 / (frame rate)

= 1 /106 = 1 μs

(d) Data rate = (106 frames /s) × (41 bits/frame)

= 41 Mbps

(e) Efficiency = 40 / 41= 97.5%

Multiplexing



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2 channels, one with a bit rate of 190 Kbpsand the other with a bit rate of 180 Kbpsare to be multiplexed with no sync bits.

Answer the following questions :

Multiplexing



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(a) What is the size of frame in bits ?

(b) What is the frame rate ?

(c) What is the duration of frame ?

(d) What is the data rate ?

Multiplexing



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(a) 2 bits

(b) 190,000 frames per second

(c) 5.26 micro seconds

(d) 380 Kbps

Multiplexing



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Digital Signal (DS) Service / Hierarchy

Multiplexing

T- 1

T- 2

T- 3

T- 4(24 x 64 Kbps)+ 8 Kbps O/H

(96 x 64 Kbps)+ 168 Kbps O/H

(672 x 64 Kbps)+ 1368 Kbps O/H

(4032 x 64 Kbps)+ 16.128 Mbps O/H

DS-0 to DS-4 are names of services;

T-1 to T- 4 are T lines used for implementation



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Digital Signal (DS) Service / Hierarchy

Multiplexing



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T Lines used for Analog Transmissionanalog signals are first sampled and thenTDM is used

Multiplexing



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T - 1 frame

Multiplexing

frame size = (24 slots x 8 bits) + 1 sync bitT-1 line carries 8000 framesdata rate = 8000 x 193 = 1.544 Mbps



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E linesEuropean version of T linesconceptually same; capacities differ

Multiplexing

M l i l i



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Synchronous TDM features :

• each input connection is allotted an outputeven if it is not sending data

• synchronization and pre - assignedrelationships exist between inputs andoutputs →

• destination addressing is not required

Multiplexing

M l i l i



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Statistical TDM features :• transmission capacity of the link will beshared by only those users who have

data to be sent• a communication channel is divided intoan arbitrary number of variable bit-ratedigital channels or data streams

• link sharing is adapted to the instantaneous

traffic demands of the data streams that aretransferred over each channel

Multiplexing

M l i l i



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Statistical TDM features :

• there are no pre assigned or reservedslots →

• there is no fixed relationship betweeninputs and outputs

• address of the receiver needs to be

included in each slot to indicate where ithas to be delivered

• no need to have synchronization bits

Multiplexing

M lti l i



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Statistical TDM :

Multiplexing



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End of Class - 17 13.10.2011


DCN A05 Multiplexing

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