lecture9 &10_tdm

8/7/2019 Lecture9 &10_TDM

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Time Division Multiplexing

Data Communication and

Networks

Prof. Dr. Abdul Qadeer Khan Rajput

By



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In the previous figure , the sampled amplitude of 5.3 is



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TDM



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,and f10 (t) represent 10 different voice channels. At the



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TDM Definition

Time Division Multiplex (TDM) is theprocess of combining signals together

in the time domain.



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TDM Basics

The basic principle of TDM is SamplingTheorem.

The basic unit inTDM is Time Slot.

The data in TDM isgrouped into Frames,which consists of onecycle of time slots.

T

3T



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TDM

TDM is a digitalmultiplexingtechnique to combinedata

Instead of sharing aportion of thebandwidth as in FDM,time is shared.

Each connectionoccupies a portion of time in the link.

In a TDM, the datarate of the link is ntimes faster, and theunit duration is ntimes shorter



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Interleaving

Interleaving is the process of multiplexing.

In TDM, synchronization between the sender and receiver isvery important.



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Interleaving

TDM can be visualizedas two fast rotatingswitches, one on themultiplexing side andthe other on the de-

multiplexing side. Theswitches aresynchronized androtate at the samespeed, but in oppositedirections. On themultiplexing side, asthe switch opens infront of a connection,that connection hasthe opportunity tosend a unit onto thepath. This process iscalled interleaving.



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Four channels are multiplexed using TDM. If each channel

sends 100 bytes/s and we multiplex 1 byte per channel, show

the frame traveling on the link, the size of the frame, the

duration of a frame, the frame rate, and the bit rate for the link.

SolutionSolution

The multiplexer is shown in Figure.

Example 6



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T-1 line for multiplexingtelephone lines

T lines are digitallines designed for thetransmission of digitaldata, audio, or video.

T lines also can beused for analogtransmission (regulartelephoneconnections),provided the analog

signals are sampledfirst, then time-division multiplexed.



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A multiplexer combines four 100-Kbps channels using a time slot

of 2 bits. Show the output with four arbitrary inputs. What is the

frame rate? What is the frame duration? What is the bit rate? What

is the bit duration?

SolutionSolution

Figure shows the output for four arbitrary inputs.

Example 7



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Framing bits

For synchronization betweenmultiplexer and demultiplexer,one or more synchronization

bits are usually added to thebeginning of each frame. Thesebits, called framing bits, followa pattern, frame to frame, thatallows the demultiplexer tosynchronize with the incoming

stream so that it can separatethe time slots accurately.In bit padding, the multiplexeradds extra bits to a device’ssource stream to force thespeed relationships among the

various devices into integermultiples of each other.



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We have four sources, each creating 250 characters per

second. If the interleaved unit is a character and 1

synchronizing bit is added to each frame, find (1) the data

rate of each source, (2) the duration of each character in

each source, (3) the frame rate, (4) the duration of eachframe, (5) the number of bits in each frame, and (6) the

data rate of the link.

Example 8



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We can answer the questions as follows:

1. The data rate of each source is 2000 bps = 2 Kbps.2. The duration of a character is 1/250 s, or 4 ms.

3. The link needs to send 250 frames per second.

4. The duration of each frame is 1/250 s, or 4 ms.

5. Each frame is 4 x 8 + 1 = 33 bits.

6. The data rate of the link is 250 x 33, or 8250 bps.

SolutionSolution



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Two channels, one with a bit rate of 100 Kbps and another

with a bit rate of 200 Kbps, are to be multiplexed. How this

can be achieved? What is the frame rate? What is the frame

duration? What is the bit rate of the link?

SolutionSolution

We can allocate one slot to the first channel and two slots

to the second channel. Each frame carries 3 bits. The

frame rate is 100,000 frames per second because it

carries 1 bit from the first channel. The frame duration is

1/100,000 s, or 10 ms. The bit rate is 100,000 frames/s x

3 bits/frame, or 300 Kbps.

Example 9



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hierarchy

DS-0 service is a singledigital channel of 64Kbps.

DS-1 is a 1.544-Mbps

service [24 times 64Kbps plus 8Kbps of overhead]

And so on. DS-0, DS-1, and so on are

the names of services. To

implement thoseservices, the telephonecompanies use T lines (T-1 to T-4).



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Table DS and T lines ratesTable DS and T lines rates

Service Line Rate

(Mbps)

Voice Channels

DS-1DS-1 T-1T-1 1.5441.544 2424

DS-2DS-2 T-2T-2 6.3126.312 9696

DS-3DS-3 T-3T-3 44.73644.736 672672

DS-4DS-4 T-4T-4 274.176274.176 40324032



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Table .Table .

E lineE line

ratesrates

E Line Rate(Mbps)

VoiceChannels

E-1E-1 2.0482.048 3030

E-2E-2 8.4488.448 120120

E-3E-3 34.36834.368 480480

E-4E-4 139.264139.264 19201920

Europeans use a version of T lines called E lines

The two systems are conceptually identical, but their capacities differ.



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Synchronous TDM

In synchronous TDM, each source is pre-assigned a fixed location of time slot.

Each source can and only can send informationat the time slot given to it. If a source has nodata to send, its time slot remains empty. Thiscan lead to inefficiency.

If n sources are grouped together, the total datarate of the path is n times the original data rate

of each source.

Synchronous TDM



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Synchronous TDMExample

S h TDM



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Synchronous TDM -Multiplexing

S h TDM



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Synchronous TDM -Demultiplexing



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We have four sources, each creating 250 characters per

second. If the interleaved unit is a character and 1

synchronizing bit is added to each frame, find (1) the datarate of each source, (2) the duration of each character in

each source, (3) the frame rate, (4) the duration of each

frame, (5) the number of bits in each frame, and (6) the

data rate of the link.

TDM Example



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We can answer the questions as follows:

1. The data rate of each source is (250 chars x 8 bits) =

2000 bps = 2 kbps.

2. The duration of a character is 1/250 s, or 4 ms.3. The link needs to send 250 frames per second.

4. The duration of each frame is 1/250 s, or 4 ms.

5. Each frame is (4sources x 8bits) + 1sync = 33 bits.

6. The data rate of the link is 250 x 33, or 8250 bps.(strictly, ‘bit rate’ of the link is 8250bps – the actual

data rate is only 8000 data bits/s)

SolutionSolution



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Asynchronous TDM

In asynchronous TDM or statistical TDM,only sources containing data will be sentwith time slot. Therefore, asynchronous

TDM can avoid bandwidth waste insynchronous TDM.

But, in order to distinguish data fromdifferent sources, address should be added

into the frame structure, increasing theoverhead of the transmission.

Asynchronous TDM



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Asynchronous TDMExample



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Address and Overhead

As shown below, address is added before the datafrom each source.

It is practical only when the data size for each time

slot is relatively larger than the address.

In this example, the addressing

information takes up as much space as

the actual data - inefficient



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TDM in Telephone System

FDM was used in the original telephonesystem, but now the telephone lines (exceptthe subscriber line) are all in digital form, soFDM is not in use now.

There are two types of TDM used intelephone system. In US, T-1 line with basicrate of 1.544 Mbps is used, while in Europeand China, E-1 Line of 2.048 Mbps is in use.

DS is a ‘service’, T-n is a ‘line’

Digital Signal (DS)



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Digital Signal (DS)Hierarchy in US

8k sample/sec x 8bit samplesgives 64kbps basic line

DS1 = 24 x DS0 = 24x64k=1536k

1536k + 8k(sync) =1544k



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T-1 Frame Structure

Data sync



One frame = 24 * 8 + 1 sync bit

Sync bit

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