6. Multiplexing

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It is the set of techniques that allows the simultaneoustransmission of multiple signals across a single datalink.

Whenever the bandwidth of a medium linking twodevices is greater than the bandwidth needs of thedevices, the link can be shared.

Efficiency can be achieved by multiplexing ; Privacyand anti-jamming can be achieved by spreading.


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Dividing a link into channels MUX combines the input lines into a single stream (Many-to-

One)

DEMUX separates the stream back into its componenttransmission (One-to-Many)

CHANNEL the portion of the link that carries a transmission

between a given pair of lines


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MULTIPLEXING

ANALOG

FDM WDM

DIGITAL

TDM

SynchronousTDM Statistical TDM


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It is an analog technique that can be applied when the bandwidthof the link is greater than the combined bandwidth of the signalto be transmitted.


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Guard Bands are used to prevent signals from overlapping.

FDM is an analog multiplexing technique that combines analog

signal


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Multiplexing Process


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De-multiplexing Process


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Example - 1Assume that a voice channel occupies a bandwidth of 4KHz. 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 domainwithout the use of guard bands.

Solution

Shift (modulate) each of the three voice channels to adifferent bandwidth, as shown in Figure.


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Example 2 Five channels, each with a 100-KHz bandwidth, are to bemultiplexed together. What is the minimum bandwidth of the link if there is a need for a guard band of 10 KHz

between the channels to prevent interference?

Solution

For five channels, we need at least four guard bands.This means that the required bandwidth is at least

5 x 100 + 4 x 10 = 540 KHz,

as shown in Figure .


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Example 3Four data channels (digital), each transmitting at 1 Mbps,use a satellite channel of 1 MHz. Design an appropriateconfiguration using FDM

Solution

The satellite channel is analog. We divide it into fourchannels, each channel having a 250 -KHz bandwidth.Each digital channel of 1 Mbps is modulated such thateach 4 bits are modulated to 1 Hz. One solution is 16 -QAM modulation. Next Figure shows one possibleconfiguration.


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Analog Hierarchy12 x 4 = 48

60 x 4 = 240600 x 4 = 2.4 MHzGuard Bands Used

2.52 x 6 = 15.12 MHz

Guard Bands Used


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Designed to use the high Data rate capability of

fiber-optic cable.

WDM is same as FDM; except that themultiplexing and de-multiplexing involve opticalsignals transmitted through Fiber-optic channel.


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BASIC IDEA :

We want to combine multiple light sources into one single light at the multiplexerand do the reverse at the de-multiplexer.

The combining and splitting of light sources are easily handled by a prism.


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APPLICATION :

In SONET network where multiple optical fiber lines are multiplexed and de-multiplexed.

Dense WDM (DWDM), can multiplex a very large number of channels by spacingchannels very close to one another. It achieves even greater efficiency.


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TDM is a process that allows several connections to share the high bandwidth of a

link.Here time is shared. Each connection occupies a portion of time in the link.

TDM is a digital multiplexing technique for combining several low-ate channels intoone high-rate one.


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SYNCHRONOUS TDM

Data flow of each input connection is divided into units.

Each input occupies one input time slot.

The duration of an o/p time slot is n times shorter than the duration of an i/p time slot.

If an i/p time slot is T sec, the o/p time slot is T/n sec, where n is the no. of connections.


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SYNCHRONOUS TDM

If we have n connections, a frame is divided into n time slots andone slot is allocated for each unit.

The data rate of the link is n times faster, and the unit duration isn times shorter.


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Example 5

Four 1-Kbps connections are multiplexed together. A unit is 1 bit.Find (1) the duration of 1 bit before multiplexing, (2) thetransmission rate of the link, (3) the duration of a time slot, and (4)the duration of a frame?

Solution

We can answer the questions as follows: 1. The duration of 1 bit is 1/1 Kbps, or 0.001 s ( 1 ms).2. The rate of the link is 4 Kbps.3. The duration of each time slot ( 1/4) ms or 250 m s.4. The duration of a frame 1 ms.


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INTERLEAVING :

TDM can be visualized as two fast rotating switches , one on the multiplexing side andother is in the de-multiplexing side.

The switches are synchronized and rotate in same speed, but opposite in directions.


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INTERLEAVING :

On multiplexing side, as the switch opens in front of a connection, that connection hasthe opportunity to send a unit onto the path.

On de-multiplexing side, as the switch opens in front of a connection, that connectionhas the opportunity to receive a unit onto the path.

This process is called interleaving.


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Example 6

Four channels are multiplexed using TDM. If each channel sends 100bytes/s and we multiplex 1 byte per channel, show the frame traveling onthe link, the size of the frame, the duration of a frame, the frame rate, andthe bit rate for the link.

Solution


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Example 7

A multiplexer combines four 100-Kbps channels using a time slot of 2bits. 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?

Solution


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Data RateManagement

MultilevelMultiplexing

Multiple-slotAllocation Pulse Stuffing


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Multilevel Multiplexing

Used when the data rate of an input line is a multiple of others.


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Multiple-slot Multiplexing

Sometimes it is more efficient to allot more than one slot in a frame to a single input line.


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Pulse Stuffing

Sometimes the bit rates of sources are not multiple integers of each other.


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Frame Synchronizing

If the multiplexer & de-multiplexer are not synchronized, a bit belonging to one channel may be received by the wrong channel.

So one or more synchronization bits are usually added to thebeginning of each frame (Framing bits).


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Example 8 We have four sources, each creating 250 characters per second. If theinterleaved unit is a character and 1 synchronizing bit is added to eachframe, find (1) the data rate of each source, (2) the duration of eachcharacter 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 thelink.

Solution1. The data rate of each source is 250 x 8 = 2000 bps = 2 Kbps (1 char = 8 bits).

2. The duration of a character is 1/250 s, or 4 ms.3. The link needs to send 250 frames per second. (Each frame has one character fromeach source)

4. The duration of each frame is 1/250 s, or 4 ms.5. Each frame is 4 x 8 + 1 = 33 bits. (4 characters & 1 sync. bit)6. The data rate of the link is 250 x 33, or 8250 bps. (250 frames & each frame has 33 bits)


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Example 9

Two channels, one with a bit rate of 100 Kbps and another with a bit rateof 200 Kbps, are to be multiplexed. How this can be achieved? What isthe frame rate? What is the frame duration? What is the bit rate of the

link?

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

the second channel. Each frame carries 3 bits. The framerate is 100,000 frames per second because it carries 1 bitfrom 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.


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Statistical TDM

Slots are dynamically allocated to improve bandwidth efficiency.

Only when an input line has a slots worth of data to send is it given a slot in the output frame.

Number of slots in each frame is less than the number of input lines.

There is no fixed relationship between the inputs and outputsbecause there is no pre-assigned or reserved slots.


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Statistical TDM


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Statistical TDM

We need to include the address of the receiver inside each slot toshow where it is to be delivered.

Slot carries both data and address. A block of data is usually many bytes while the address is just few bytes.

There is no need of synchronizing bit in frame level.


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6. Multiplexing

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