chap 4 multiplexing cont

7/31/2019 Chap 4 Multiplexing Cont

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Frequency Division Multiplexing FDM

The SSB filters are the same as in the encoder, i.e. each one centred on f1, f2and f3 to select the appropriate sideband and reject the others. These are thenfollowed by a synchronous demodulator, each fed with a synchronous LO, fc1, fc2and fc3 respectively.

For the 3 channel system shown there is 1 design for the BLF (used 3 times), 3designs for the SSB filters (each used twice) and 1 design for the LPF (used 3times).

A co-axial cable could accommodate several thousand 4 kHz channels, forexample 3600 channels is typical. The bandwidth used is thus 3600 x 4kHz =

14.4Mhz. Potentially therefore there are 3600 different SSB filter designs. Notonly this, but the designs must range from kHz to MHz.

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Frequency Division Multiplexing FDM

For designs around say 60kHz, QkHz

kHz

60

4= 15 which is reasonable.

However, for designs to have a centre frequency at around say 10Mhz,

QkHz

kHz

10 000

4

,gives a Q = 2500 which is difficult to achieve.

To overcome these problems, a hierarchical system for telephony used the FDM

principle to form groups, supergroups, master groups and supermaster groups.

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Basic 12 Channel Group

The diagram below illustrates the FDM principle for 12 channels (similar to 3 channels)

to a form a basic group.

m1(t)

m2(t)

m3(t)

m12(t)

Multiplexer

12kHz 60kHz

freq

i.e. 12 telephone channels are multiplexed in the frequency band 12kHz 60 kHz in

4kHz channels basic group.

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Basic 12 Channel Group

A design for a basic 12 channel group is shown below:

300Hz 3400kHz

4kHz

300Hz 3400kHz

4kHz

300Hz 3400kHz

4kHz

Band Limiting Filters

DSBSC

8.6 15.4kHz

12.6 19.4kHz

52.6 59.4kHz

f1 = 12kHz

f1 = 16kHz

f12 = 56kHz

Increase in 4kHz steps

FDM OUT

1260kHz

12.3 15.4kHz

16.3 19.4kHz

56.3 59.4kHz

CH1

m1(t)

CH2

m2(t)

CH12

m12(t)

SSB Filter

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Super Group

These basic groups may now be multiplexed to form a super group.

BASIC

GROUP

1260kHz

12

Inputs

SSB

FILTER

420kHz

BASIC

GROUP

1260kHz

12

Inputs

SSB

FILTER

468kHz

BASIC

GROUP

1260kHz

12

Inputs

SSB

FILTER

516kHz

BASIC

GROUP

1260kHz12

Inputs

SSB

FILTER

564kHz

BASIC

GROUP

12

60kHz

12

Inputs

SSB

FILTER

612kHz

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Super Group

5 basic groups multiplexed to form a super group, i.e. 60 channels in one super group.

Notethe channel spacing in the super group in the above is 48kHz, i.e. each carrier

frequency is separated by 48kHz. There are 12 designs (low frequency) for one basic

group and 5 designs for the super group.

The Q for the super group SSB filters is QkHz

kHz

612

4812 - which is reasonable

Hence, a total of 17 designs are required for 60 channels. In a similar way, super groups

may be multiplexed to form a master group, and master groups to form super master

groups

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Multiplexing application(contd) Analog Hierarchy

To maximize the efficiency of their infrastructure,telephone companies have traditionally multiplexedsignals from lower bandwidth lines onto higher

bandwidth lines.


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Multiplexing application(contd) Analog hierarchy


10/20McGraw-Hill The McGraw-Hill Companies, Inc., 2004

Example 3

Four data channels (digital), each transmitting at 1 Mbps,

use a satellite channel of 1 MHz. Design an appropriate

configuration using FDM

SolutionThe satellite channel is analog. We divide it into four

channels, each channel having a 250-KHz bandwidth.

Each digital channel of 1 Mbps is modulated such that

each 4 bits are modulated to 1 Hz. One solution is 16-

QAM modulation. Figure 6.8 shows one possible

configuration.



Figure 6.8 Example 3



Figure 6.9 Analog hierarchy



Example 4

The Advanced Mobile Phone System (AMPS) uses two

bands. The first band, 824 to 849 MHz, is used forsending; and 869 to 894 MHz is used for receiving. Each

user has a bandwidth of 30 KHz in each direction. The 3-

KHz voice is modulated using FM, creating 30 KHz of

modulated signal. How many people can use their cellular

phones simultaneously?

Solution

Each band is 25 MHz. If we divide 25 MHz into 30 KHz,

we get 833.33. In reality, the band is divided into 832

channels.


14/20McGraw-Hill The McGraw-Hill Companies, Inc., 200411 May 201214



TDM

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1611 May 2012

TDM

Example :If the system consists of 24 PCM voice channel,multiplexed by using TDM where each channel issampled at 8 kHz with 8 bit per sample.

Find the total bit rate transmitted:Solution : 24 x 8 + 1 (framing bit) 193 bits.

193 x 8 kbps = 1.544 Mbps

The example of the system that use TDMA is digital cellular radio

system where several signals from mobile units are combined on

one channel by assigning each a time slot.



Example 7.2TDM with sources having different data rates Consider the case of three streams with bit rates of8

kbit/sec,16 kbit/sec, and 24 kbit/sec, respectively. We want tocombine these streams into a single high-speed stream usingTDM. The high-speed stream in this case must have atransmission rate of 48 kbit/sec, which is the sum of the bit

rates of the three sources. To determine the number of timeslots to be assigned to each source in the multiplexing process.we must reduce the ratio of the rates, 8:16:24, to the lowestpossible form, which in this case is 1:2:3. The sum of thereduced ratio is 6, which will then represent the minimumlength of the repetitive cycle of slot assignments in themultiplexing process. The solution is now readily obtained: In

each cycle of six time slots we assign one slot to Source A (8kbit/sec), two slots to Source B (16 kbit/sec), and three slots toSource: C (24 kbit/sec). Figure 7-4 illustrates this assignment,using a to indicate data from Source A, b to indicate datafrom Source B, and c to indicate data from Source C.

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8.5 Multiplexing application :Telephone system

Telephone Network



Prior to transmission, we divide each stream of bits coming from asource into fixed-size blocks. We then add a small group of bitscalled a headerto each block, with the header containing theaddresses of the source and intended user for that block. The blockand the header are then transmitted together across the channel.Combined, the block and header are called a packet.

Actually, the header may contain other information besides thesource and user addresses, such as extra bits for error control (seeChapter 10) or additional bits for link control (used, for example, to

indicate the position of a particular block in a sequence of blockscoming from the same user, or to indicate priority level for aparticular message). Extra bits can also be added to the beginningand end of a block for synchronization; a particular pattern of bits,called a start flag, can be used in the header to mark the start of ablock, and another particular pattern of bits, called an end flag, canbe used to conclude the block. Each block transmitted across thechannel thus contains a group of information bits that the userwants, plus additional bits needed by the system to ensure proper

transmission. These additional bits, while necessary to systemoperation, reduce the effective transmission rate on the channel.Figures 7-5 and 7-6 present the statistical TDM technique and thestructure of a typical packet.

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20/20McGraw Hill The McGraw Hill Companies Inc 2004

Review questions Why is multiplexing so cost effective? How is interference avoided by using FDM What is echo-cancellation?

Define upstream and downstream with respect to subscriber lines. Explain how synchronous TDM works. Why is statistical TDM more efficient than a synchronous TDM multiplexer?

Draw the block diagrams of an FDM and TDM communication system Why is ADSL best suited for residential customers for accessing the Internet?

chap 4 multiplexing cont

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