Upload
hafiz-azman
View
272
Download
1
Embed Size (px)
Citation preview
SZE 3533SZE 3533Topic VI Topic VI
MULTIPLEXINGMULTIPLEXING
1
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
4
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
5
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
6
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
7
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
10
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.
11
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
12
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.
13
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
14
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
15
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.
16
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.
17
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.
18
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.
19
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
20