SZE 3533SZE 3533Topic VI Topic VI

MULTIPLEXINGMULTIPLEXING

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6.1 Introduction6.1 Introduction

Pemodulatan Digit

• Process of transmitting two or more signals simultaneously is known as multiplexing

• The main objective of this system is to increase the capacity of a transmission channel

• 3 multiplexing methods: frequency (FDM), time (TDM) and space (SDM)

6.2 FREQUENCY DIVISION MULTIPLEXING(FDM)

FDM system can be used in both analog and digital system

FDM uses a different carrier frequency for each user

frequency

time

Space

fc3

fc2

fc1

FDM Concept2

BPF

BPF

BPF

vm1(t)

vFDM(t)

fc1

vm3(t)

vm2(t)

fc3

fc2

An example of FDM system with 3 channels:

• In FDM process, each information signal is multiplied by a different sub-carrier frequency i.e fc1 for

vm1(t), fc2 for vm2(t) and fc3 for

vm3(t).

• Output of the multiplier will be a double sideband supressed carrier signal (DSBSC)

• Each signal will then passed through a BPF with bandwidth fc+fm or fc-fm.

• Ouput of BPF will then be either lower sideband (LSB) or upper sideband (USB)

• Output from each filter will be multiplexed together.

multiplexer

3

fc1 fc2 fc3 f

channel 1

channel 2

channel 3

FDM signal spectrum for 3 sub-carrier

Bandwidth of FDM depends on the number of sub-carrier used. For N sub-carrier, the bandwidth of FDM is:

mFDM NfBW

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Example 6.1 :

4 information signals vm1(t), vm2(t), vm3(t) and vm4(t) with frequency range

10 Hz x 4 kHz were transmitted using FDM system. Center frequency, fo for this FDM system is 900 MHz. Determine the minimum

sub-carrier frequencies for each information signals. Next, calculate the bandwidth for the system. Solution :

Minimum sub-carrier frequency depends on the minimum bandwidth of the sampled signal. Minimum bandwidth for each information signal:

kHzBW

kHzBW

ffBW makN

8

)4(2

2

FDM signal spectrum with center freq., fo

= 900 MHz.

fc1 fc2 fo fc3 fc4 f (MHz) 900

channel 1

fc2+ fm

fc3 - fm

fc3+ fm

fc4 - fm

fc1- fm fc4+ fm fc1+ fm

fc2 - fm

channel 2

channel 3

channel 4

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Sub-carrier frequency is the center frequency for each sampled signal Therefore :

MHzf

kHzMHzf

fff

c

c

moc

988.899

)4(3900

3

1

1

1

MHzf

kHzMHzf

fff

c

c

moc

996.899

4900

2

2

2

MHzf

kHzMHzf

fff

c

c

moc

004.900

4900

3

3

3

MHzf

kHzMHzf

fff

c

c

moc

012.900

)4(3900

3

4

4

4

The number of sub-carrier , N is 4. Therefore, the bandwidth of the system is:

mFDM NfBW

= 4(4kHz)

= 16kHz

fc1 fc2 fo fc3 fc4 f (MHz) 900

channel 1

fc2+ fm

fc3 - fm

fc3+ fm

fc4 - fm

fc1- fm fc4+ fm fc1+ fm

fc2 - fm

channel 2

channel 3

channel 4

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6.3 TIME DIVISION MULTIPLEXING(TDM)6.3 TIME DIVISION MULTIPLEXING(TDM)

• TDM is used in digital system.

• TDM system uses a different time slot for each user, but with the same carrier frequency.

time (t)

f

space

t1 t2 t3

TDM concept

2 types of TDM system:

• TDM PAM – input signal is a PAM signal. This signal is transmitted directly as PAM samples.

• TDM PCM – Input to this system is the output of PCM

• TDM PCM system can also received signal in the form of PAM samples. However, these PAM samples need to be changed to digital form (by using quantizer and encoder or analog to digital converter) after sampled by the commutator before transmitting on the line

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t1 t2 t3 t

channel 1 channel 2 channel 3

Sampled bit vm1

Sampled Bit vm2

SampledBit vm3

channel 1 channel 2 channel 3

Sampled Bit vm1

SampledBit vm2

SampledBit vm3

Frame period, TF

sample vm1

sample vm2

sample vm3

commutator

TDM PCM signal

A/Dconverter

TDM PCM multiplexing process for 3 information signals:

All sampled bits for one complete cycle of the commutator is known as a frame.

Fig. 6.5: The position of the sampled bit information signal

in time domain8

• Period for 1 time slot:

Analysis for Fig. 6.5 : The sampled information signal uses sampling frequency 8 kHz and coded at a rate of 8 kbps.

samplechannel TT

sTkHz

T

channel

channel

1258

1

• No of bits for each channel:

samplechannel nn

bitnchannel 8• Bit rate for each channel:

fb(channel) = fs = 2fm

fb = 8 kbps

• Period for 1 frame:

sT

sT

TT

frame

frame

sframe

375

)125(3

*)channel of no(

• No of bits for 1 frame:

bitn

n

nchanneln channel

24

)8(3

*

frame

frame

frame

• Bit rate for each frame, fb(frame) :fb(frame = no of channels x fb(channel

fb(frame = 3 x 8 kbps

fb(frame) = 24 kbps

• BWTDM = fb(frame) = 24 kbps9

6.4 SPACE DIVISION MULTIPLEXING(SDM)6.4 SPACE DIVISION MULTIPLEXING(SDM)

Pemodulatan Digit

Antena Tx1

Antena Tx2

Antena Tx3

Antena Rx1

Antena Rx2

Antena Rx3

Transmitter 1

Transmitter 2

Transmitter 3

DSP

Receiver 1

Receiver 2

Receiver 3

DSP

transmission medium

SDM uses smart antenna

Transmitter 1

Transmitter 2

Transmitter 3

receiver 1

receiver 2

receiver 3

transmission cable

SDM uses multiple cables

• The channel can still operates eventhough there is a fault in one of the cables.

• Easier maintenance works.

• Increase cable cost and the size of the cables becomes bigger and entangled

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6.5 FDM TELECOMMUNICATION SYSTEM6.5 FDM TELECOMMUNICATION SYSTEM An example of an application using FDM system is the L-carrier Line telephone system by AT&T in US and ITU-T.

MUX1

MUX2

channel 1Basic group

Super group 1 MUX

3

Super group

Super master group

MUX4

Master group

Master group 1

channel 12

Basic group 5

Basic group 1

Super group 5

Master group 3

.

.

.

.

.

.

.

.

.

.

.

.

FDM hierarchy in telephone system (ITU-T standard) used in MalaysiaChannel – telephone

signal with BW 4 kHz

Basic Group consists of 12 telephone channels. BW = 48 kHz (12x4 kHz). Carrier frequency used is in the range of 60-108 kHz.

Super Group = carries 60 telephone channels. BW = 240 kHz (5x48 kHz). Carrier frequency range 312-552 kHz.

Master Group = 300 telephone channels with BW 1.23 MHz. Carrier frequency range 812-2044 kHz.

Master group = 900 telephone chanels with BW = 3.87 MHz. The carrier frequency range is 8.516-12.388 MHz.

900 telephone channels were transmitted using one coaxial cable with BW 4 MHz.

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6.6 TDM TELECOMMUNICATION SYSTEM6.6 TDM TELECOMMUNICATION SYSTEM

Pemodulatan Digit

• 2 standards in telephone system that uses PCM TDM, i.e T Line by AT&T in US and E Line by CEPT in European countries.

• Telephone system in Malaysia uses the standard fixed by CEPT.

• Transmission process of TDM signals were done by using 32 time channels for each E1 Line.

30 channels carrying voice signals

2 channels used for synchronization and signalling

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

Frame period = 125second

Synchronization SignalingVoice channels Voice channels

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Pemodulatan Digit

In a CEPT system, a frame contains 256 bits (8 bit x 32 time slots). Therefore, bit rate is:

bittotal

Tperiod,frame,periodBit kerangkabT

256

second125,periodBit

bT

second1028.488,periodBit 9bT

Mbpsf

Tf

b

b

b

048.2,rateBit

1,rateBit

Bit rate 2.048 Mbps represents the bit rate for one E1 Line in CEPT system.

This bit rate increases through multiplexing process of E1 Line to E2 Line and so on.

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Each frame is transmitted in a group of multi-frame that contains 16 frames (frame 0-15). In this system, 500 multi-frames is transmitted every second. Fig. 6.11 shows the frame arrangement for a multi frame.

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Period for 1 multi frame = 2 msecond

Frame

Fig. 6.11: Multi frame format for E1 Line telephone system

Frame Time slot

0 0 and 16

1 1 and 17

2 2 and 18

3 3 and 19

4 4 and 20

5 5 and 21

6 6 and 22

7 7 and 23

8 8 and 24

9 9 and 25

10 10 and 26

11 11 and 27

12 12 and 28

13 13 and 29

14 14 and 30

15 15 and 31

Table 6.1: Position of time slot for synchronization and signalling for 1 multi frame

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Pemodulatan Digit(a) bits per time slot (b) time slots per frame (c) frames per multiframe

488 ns

3.9 s

3.9 s

125 s

125 s

2 ms

8 bits per time slot

Bit duration

30 signal + 2 control = 32 channels = 1 frame

Signalling & synchronization

16 frames = 1 multiframeDuration of multiframe

Example : PCM-TDM CEPT SystemExample : PCM-TDM CEPT System

Frame structure and Timing : European standard PCM system : E Line

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Pemodulatan Digit

MUX1

MUX2

MUX3

MUX4

30 Voice channels

.

.

.

.

.

.

E1 Line2.048 Mbps

E2 Line8.448 MbpsE1

E1E1

E2E2E2

E3E3E3

E3 Line34.368 Mbps

E4 Line139.264 Mbps

Line Hierarchy of CEPT telephone system

For E1 Line = 30 voice channels is multiplexed for transmission.

Data rate for E1 Line = 2.048 Mbps.

4 E1 Line were next multiplexed to become E2 Line.

Data rate = 8.448 Mbps.

Next, 4 E2 Line were multiplexed becoming E3 Line.

Data rate = 34.368 Mbps.

Lastly, 4 E3 Line were multiplexed becoming E4 Line.

Data rate = 139.264 Mbps.

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Pemodulatan Digit

Example 6.2 :

A PCM TDM communication system contains 30 voice channels and 2 control channels. The information data rate is 64 kbps for each channel. The audio channel represents information signal that is quantumed to 256 levels. Determine the bit period, Tb and bit rate for each frame, fb. Next, calculate the maximum audio frequency

that can be transmitted.

Solution:

Given 1 frame = 32 time slot, and fb(channel)= 64 kbps

Bit rate for 1 frame, fb(frame) =

32 x 64 kbps= 2.048 Mbps

second488.0

048.2

1

)frame(

1

b

b

b

b

T

MbpsT

fT

Bit period:

Sampling frequency for an audio signal is the same as the sampling frequency used by the commutator of this system. Given sampling period for 1 frame is 125 s. Therefore, the sampling frequency is 8 kHz.

mN ff 2

Maximum audio frequency,

kHzf

kHzf

ff

m

m

Nm

42

82

An example of audio signal that has a maximum frequency is the voice signal.

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Example 6.3 :

Design a TDM communication system that can accomodate 11 information signals with the following characteristics:

Information 1 : Analog signal, bandwidth = 2 kHz

Information 2 : Analog signal, bandwidth = 4 kHz

Information 3 : Analog signal, bandwidth = 2 kHz

Information 4-11 : Digital signal, synchronized at a bit rate of 7200 bps

Solution :

The first step is to change the analog signals to 4 bit PCM. Nyquist rate is 4 kHz, 8kHz and 4 kHz for information signals 1, 2 and 3 respectively.

To satisfy the three Nyquist rates, the commutator is set at 16 kHz. Therefore, information signal 2 will be sampled twice by this commutator.

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Pemodulatan Digit

Sample vm1

Sample vm2

Sample vm3

commutator

16 kHz

4 bit A/D

Ouput of 4 bit A/D is 64 kbps.

A second commutator is needed to sample the ouput of 4 bit A/D and the digital signal (information signal 4 to 11).

This digital signal has a bit rate of 7.2 kbps.

The second commutator is chosen with sampling rate of 8 kHz.

Therefore, pulse stuffing is needed for each information signal 4 to 11 to generate a bit rate of 8 kbps.

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Pemodulatan Digit

PCM TDMsignal

Commutator 116 kHz

4 bit A/DSample vm2

Sample vm1

Sample vm3

Commutator 2

8 kHz

Pulse stuffing

Pulse stuffing

Pulse stuffing

Pulse stuffing

Pulse stuffing

Pulse stuffing

Pulse stuffing

Pulse stuffing

Information 4

Information 5

Information 6

Information 7

Information 8

Information 9

Information 10

Information 11

64 kbps

8 kbps

7.2 kbps

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