EECB3023

Communication Systems
Chapter 6 Multiplexing and Multiple Access
Multiplexing and Multiple Access

 6.1 Introduction
 6.2 Frequency Division Multiplexing (FDM)
 6.3 Time Division Multiplexing (TDM)
 6.4 Code Division Multiple Access (CDMA), Frequency Division
 Multiple Access (FDMA) and Time Division Multiple Access
 (TDMA)
 6.1 Introduction
 Multiplexing – methods of transmitting more than one signal along a single
transmission path/stream i.e many to one.
 Demultiplexing – separate the stream back into its component transmission
i.e one to many.
 Path– refers to the physical link.
 Channel – refers to a portion that carries a transmission between a given
pair of devices. One path can have many channels.
 Two common form of multiplexing are Frequency Division Multiplexing(FDM) and
Time Division Multiplexing(TDM).
 Two variations of these basic methods are frequency division multiple access (FDMA)
and time division multiple access (TDMA)
 Another form of multiple access is known as code-division multiple access (CDMA)
 Advantages:
1. Increase number of channels so that more info can be transmitted
2. Save cost by using one channel to send many info signals
 6.1 Frequency Division Multiplexing (FDM)
 In FDM, multiple sources that originally occupied the same frequency spectrum are
each converted to different frequency band and transmitted simultaneously.
 FDM is an analog technique – the information entering an FDM system must be
analog. If the source is digital, it must be converted to analog before being
frequency-division multiplexed.
 The total channel bandwidth is divided into several smaller channels of different
frequency bands.
 Guard bands keep the modulated signals from overlapping and interfering with one
another.
 Modulation is used to translate the center frequency of the baseband signal up into a
preassigned frequency slot.
 FDM System - Transmitter

 A number of signal, 𝑚𝑖 (𝑡) where 𝑖 = 1, 2, … , 𝑛are to be multiplexed onto the same

transmission medium.
 Each signal 𝑚𝑖 𝑡 is modulated onto a carrier 𝑓𝑖 , refers as subcarrier.
 Modulated signals are then summed to produce a composite signal 𝑚𝑏 𝑡 .
 Spectrum of Composite Baseband Modulating Signal

 This figure shows each signal 𝑚𝑖 𝑡 is centered at its respective sub-carrier 𝑓𝑖 .

 𝑓𝑖 must be chosen so that the bandwidth of the various signals do not overlap i.e channel must be
separated by guard band (unused bandwidth).
 FDM System - Receiver

 At the receiver, the FDM signal is demodulated to retrieved 𝑚𝑏 𝑡 , which is then passed
through n bandpass filters.
 Example:

• This figure shows a simple FDM system where four 5kHz

channels, which are frequency-division multiplexed
into a single 20kHz combined channel.

• With FDM, each narrowband channels are stacked on

top of one another in the frequency domain.
FDM in Telephone System
 The principle of FDM is best illustrated by
looking at the formation of a standard
telephone group which consists of 12
standard telephone channels in FDM.
 Each telephone channel starts off within
the nominal baseband bandwidth of 0 –
4 kHz and then is SSB (LSB) modulated up
on to one of a sequence of subcarriers
spaced at 4 kHz apart.
 It is then placed in a 4 kHz frequency slot in
the group transmission bandwidth from 60
– 180 kHz.
 These subcarriers will be at 64 kHz, 68 kHz
and so on up to 104 kHz and finally 108 kHz.
 All 12 SSB signals are summed in a linear
mixer to produce a single frequency
multiplexed signal to form basic group.

 If more than 12 voice channels are

needed, multiple basic groups are used.
 Group :
 12 voice channels (4 kHz each) = 48 kHz
 Range from 60 kHz to 108 kHz
 Supergroup :
 60 channels
 FDM of 5 group signals on carriers between 420 kHz and 612 kHz.
 Each group is treated as a separated signal with 48 kHZ bandwidth
 Mastergroup:
 10 supergroup
 600 voice channels with a bandwidth of 2.52 MHz
 Q1: A cable TV service uses a single coaxial cable with bandwidth of 860
MHz to transmit multiple TV signals to subscribers. Each TV signal is 6 MHz
wide. How many channels can be carried?

𝐵𝑊 = 2 × 𝑓𝑚 = 2 × 6𝑀 = 12 𝑀𝐻𝑧

𝑇𝑜𝑡𝑎𝑙 𝑏𝑎𝑛𝑑𝑤𝑖𝑑𝑡ℎ 860𝑀

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑇𝑉 𝑐ℎ𝑎𝑛𝑛𝑒𝑙𝑠 = = = 71.67 𝑐ℎ𝑎𝑛𝑛𝑒𝑙𝑠 ≈ 71 𝑐ℎ𝑎𝑛𝑛𝑒𝑙𝑠
𝐵𝑎𝑛𝑑𝑤𝑖𝑑𝑡ℎ 𝑓𝑜𝑟 𝑒𝑎𝑐ℎ 𝑐ℎ𝑎𝑛𝑛𝑒𝑙 12𝑀
 Q2: For a particular telephone company, the first sub carrier frequency is at
60 kHz and the total bandwidth is 96 kHz. Design a FDM system, given a
general rule of 12 channels per basic group and 4 kHz per channel applies
to the design.
i. How many basic groups are required?
ii. Draw the circuit diagram of your design
iii. Draw the frequency spectrum of your multiplexed system

Answer:
i.Given BWtotal = 96kHz;12 channels/basic group
1 channel = 4 kHz,
then 12 channels = 12x4k = 48kHz/basic group
Thus 96/48 = 2 basic group
 ii. Circuit Diagram

Ch 12 BPF
fc = 60kHz Linear
Voice Mixer

Ch 1 BPF
Linear
fc = 104kHz Mixer
Ch 24 BPF
fc = 108kHz Linear
Mixer

Ch 13 BPF
fc = 152kHz

iii. Frequency Spectrum

48 kHz 48 kHz

f (kHz)
60 104 108 152
Q3:

Without guard band:

𝑇𝑜𝑡𝑎𝑙 𝑏𝑎𝑛𝑑𝑤𝑖𝑑𝑡ℎ (50𝑀 − 0𝑀)
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑝𝑒𝑒𝑐ℎ 𝑐ℎ𝑎𝑛𝑛𝑒𝑙𝑠 = = = 12,500 𝑐ℎ𝑎𝑛𝑛𝑒𝑙𝑠
𝐵𝑎𝑛𝑑𝑤𝑖𝑑𝑡ℎ 𝑓𝑜𝑟 𝑒𝑎𝑐ℎ 𝑐ℎ𝑎𝑛𝑛𝑒𝑙 4𝑘

With guard band:

𝑇𝑜𝑡𝑎𝑙 𝑏𝑎𝑛𝑑𝑤𝑖𝑑𝑡ℎ
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑝𝑒𝑒𝑐ℎ 𝑐ℎ𝑎𝑛𝑛𝑒𝑙𝑠 =
𝐵𝑎𝑛𝑑𝑤𝑖𝑑𝑡ℎ 𝑓𝑜𝑟 𝑒𝑎𝑐ℎ 𝑐ℎ𝑎𝑛𝑛𝑒𝑙 + 𝐺𝑢𝑎𝑟𝑑 𝑏𝑎𝑛𝑑
(50𝑀 −0𝑀) 50𝑀
= = = 10,000 𝑐ℎ𝑎𝑛𝑛𝑒𝑙𝑠
4𝑘+1𝑘 5𝑘
 Q4. Several 10 kHz baseband signals are amplitude modulated, multiplexed and
transmitted over radio frequency of 930 kHz to 978 kHz. The modulated signal are
separated by a 2 kHz guard band.
a. Determine the maximum number of baseband signal that can be multiplexed.
State your assumption(s) if any.
b. Draw complete block diagram of the multiplexing system, showing all the
important
parameter.
a. Number of baseband signal

Given 𝑓𝑚 = 10 𝑘𝐻𝑧
Available 𝐵𝑊 = 978 𝑘𝐻𝑧 − 930 𝑘𝐻𝑧 = 48 𝑘𝐻𝑧
48 𝑘𝐻𝑧
Assumption : SSBSC 𝑁𝑜. 𝑜𝑓 𝑠𝑖𝑔𝑛𝑎𝑙𝑠 =
10 𝑘𝐻𝑧 + 2 𝑘𝐻𝑧
𝐵𝑊 = 𝑓𝑚 = 10 𝑘𝐻𝑧
𝐺𝑢𝑎𝑟𝑑 𝐵𝑎𝑛𝑑 = 2 𝑘𝐻𝑧 𝑁𝑜. 𝑜𝑓 𝑠𝑖𝑔𝑛𝑎𝑙𝑠 = 𝟒 𝒔𝒊𝒈𝒏𝒂𝒍𝒔/𝒖𝒔𝒆𝒓𝒔
 b. Block diagram Frequency Planning

User Frequency Subcarrier,

Range 𝒇𝑪𝒊
BM 1 BPF
User 1 930 – 940 𝒇𝑪𝟏 = 930 kHz
𝑓𝑐𝑒𝑛𝑡𝑒𝑟 = 935 kHz
𝒇𝑪𝟏 =930 kHz kHz
BW = 10 kHz
User 2 942 – 952 kHz 𝒇𝑪𝟐 = 942 kHz
BM 1 BPF FDM

Linear Summer
Output User 3 954– 964 kHz 𝒇𝑪𝟑 = 954 kHz
𝑓𝑐𝑒𝑛𝑡𝑒𝑟 = 947 kHz
𝒇𝑪𝟐 =942 kHz BW = 10 kHz
930 – 976 kHz
User 4 966 – 976 kHz 𝒇𝑪𝟒 = 966 kHz
BM 1 BPF
𝑓𝑐𝑒𝑛𝑡𝑒𝑟 = 959 kHz
𝒇𝑪𝟑 =954 kHz BW = 10 kHz
Assumptions
• SSBSC
BM 1 BPF
• USB is used
𝒇𝑪𝟒 =966 kHz 𝑓𝑐𝑒𝑛𝑡𝑒𝑟 = 971 kHz
BW = 10 kHz
 Q5. THREE (3) FM signal are multiplexed and share the frequency spectrum of 88 MHz to 91
MHz. The specifications of each FM signal are as follows:

• Modulating signal frequency : 90 kHz

• Peak frequency deviation : 216 kHz
• Guard band : 100kHz
• Equal bandwidth
Design a transmitter system for the above scenario. Present the block diagram indicating all
the important parameters and the proposed frequency planning.

Bandwidth of each channel

𝐵𝑊𝐵𝑒𝑠𝑠𝑒𝑙 = 2𝑛𝑓𝑚

Δ𝑓 216 𝑘 𝒏=𝟓
𝑚= = = 2.4
𝑓𝑚 90 𝑘

𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝐵𝑊𝐵𝑒𝑠𝑠𝑒𝑙 = 2 5 90𝑘 = 𝟗𝟎𝟎 𝒌𝑯𝒛

b. Block diagram Frequency Planning

User Frequency Subcarrier, 𝒇𝑪𝒊

Range
FM Modulator
User 1 88.0 – 88.9 MHz 𝒇𝑪𝟏 = 88.45
𝒇𝑪𝟏 = 88.45 MHz

Linear Summer
MHz
FDM
Output User 2 89.0 – 89.9 MHz 𝒇𝑪𝟐 = 89.45 MHz
FM Modulator
88.0 MHz User 3 90.0 – 90.9 MHz 𝒇𝑪𝟑 = 90.45 MHz
𝒇𝑪𝟐 = 89.45 MHz –
90.9 MHz

FM Modulator

𝒇𝑪𝟑 = 90.45 MHz

Introduction: TDM
 It is the transmission of information from multiple occur on the same
facility but NOT at the same time.
 Transmissions from various sources are interleaved in the time
domain.
 PCM is the most prevalent encoding technique used for TDM digital
signals.
 In PCM-TDM system, two or more voice channels are sampled,
converted to PCM codes, and then time-division multiplexed onto a
single metallic or optical fiber cable.
 For example: For an 8 kHz sample rate and an 8 bit PCM code, it
produces PCM line speed:
 Time Division Multiplexing
 Definition – TDM is the time interleaving of samples from several
sources so that the info from these sources can be transmitted
serially over a single communication channel.
 In brief, TDM is a digital multiplexing technique for combining
several low-rate channels into one high-rate one.
 Can be used for analog & digital information signal.

Frame

Figure gives a conceptual view of TDM. Note that the same link is used as in FDM;
here the link is sectioned by time rather than frequency
Synchronous vs Asynchronous TDM
 TDM can be implemented in two ways, synchronous &
asynchronous TDM.

Synchronous TDM
 In synchronous TDM, the data rate of the link is n times faster, and
the unit duration is n times shorter.
Synchronous TDM
 Both transmitter and receiver must be synchronized.
 Also called fixed cycle operation i.e data carry in a repetitive
frames
 Each frame consists of a set of time slots and each source is
assigned one or more time slots per frame.
 The channel is divided into time slots and each user is allocated
a slot – whether it is empty or not.
 Synchronous TDM

❑ TDM can be visualized as two fast rotating switches, one on the

multiplexer side and the other on the demultiplexer side.
❑ The switches are synchronized and rotate at the same speed,
but in opposite directions.
Asynchronous TDM (Statistical TDM)

 Time slots are allocated on demand instead of a fixed cycle

 Allows unused slots to be allocated to active users
 Many more devices can be connected and provides more
efficient use of the available BW.
 Time slots are not preassigned to particular data sources.
 Rather, user data are buffered and transmitted as rapidly as
possible using available time slots
Synchronous vs Asynchronous TDM
 Figure 8.12 shows four data sources produced in four time
periods (t0, t1, t2, t3, t4).

 In case of synchronous TDM, during each period, data are

collected from all 4 sources. Resulted sources C and D produce
no data. Thus 2 of 4 time slots transmitted by mux are empty.

 For statistical TDM, the multiplexer does not send empty slots if
there are data to send. During the first period, only slot A and B
are sent. Address info is required to assure proper delivery. Thus,
more overhead per slot is needed because each slot carries an
address as well as data.
TDM PCM System
TDM PCM System
 Figure 3.35 illustrates the TDM concept applied to 3 analog sources
multiplexed over a PCM system.
 An electronic switch is used for the commutation (sampler)
 Pulse width for TDM PAM is Ts/3 = 1/3fs
 Pulse width for TDM PCM is Ts/3n where n is the number of bits used in
PCM word and fs = 1/Ts i.e the frequency of rotation
 fs should satisfies the Nyquist rate for the analog source with the
largest bandwidth.
 Larger bandwidth sources may be connected to several switch
position on the sampler.
 At the receiver, the sampler has to be synchronized with the
incoming waveform.
 LPF are used to reconstruct the analog signals from the PAM signals
 TDM of Digital Signal (Identical bit rate)
 Multiplexing can be done on a bit-by-bit basis (bit or digit
interleaving) or on a word-by-word basis (byte or word interleaving)
TDM of Digital Signal (Different bit rate)

 When the bit rates of incoming

channels are not identical, the
high-bit-rate channel is
allocated proportionately more
slots –(Figure c and d)

 Figure show 4-channel

multiplexing consisting of 3
channels (B, C and D) of
identical bit rate R and 1
channel (A) with a bit rate of 3R
Handling Different Data Rates
 It is possible for a synchronous TDM device to handle sources of
different data rates. For example, the slowest input device could
be assigned one slot per cycle, while faster devices are assigned
multiple slots per cycle.
 The most difficult problem in the design of a synchronous TDM
system is the synchronization of various data sources.
 3 types of handling different data rates:
 Multilevel multiplexing
 Multiple-slot multiplexing
 Pulse stuffing
  Multilevel multiplexing

 Multiple Slot Multiplexing

 Pulse Stuffing
 With pulse stuffing, the outgoing data rate of
the multiplexing, is higher than the sum of the
maximum instantaneous incoming rates.
 The extra capacity is used by stuffing extra
dummy bits or pulses into each incoming
signal until its rate is raised to that of a locally
generally clock signal.
 The stuffed pulses (bits) are inserted at fixed
locations in the multiplexing frame format so
that they may be identified and removed at
the demultiplexer.
Digital Carrier Signal (North American)
 Digital Carrier Signal (North American)

 A DS-0 service is a single digital channel of 64Kbps

 DS-1 is a 1.544Mbps service – 1.544Mbps is 24 times 64kbps plus 8kbps of
overhead.
 It can be used to mux 24 DS-0 channels or to carry any other combination
desired by the user that can fit within its 1.544Mbps capacity.
 DS-2 is a 6.312Mbps service – 6.312Mbps is 96 times 64kbps plus 168kbps
of overhead.
 It can be used as a single service for 6.132Mbps tx or to mux 4 DS-1 channels,
96 DS-0 channels or to carry any other combination of these service type.
 Etc..
Digital Carrier Signal (CCITT Standard)
Example
Design a TDM that will accommodate 11 sources with this
specification:
 Source 1: Analog, 2 kHz bandwidth
 Source 2: Analog, 4 kHz bandwidth
 Source 3: Analog, 2 kHz bandwidth
 Source 4-11: Digital, 7200 bps synchronous
Suppose the analog sources are converted to digital using 4-bit
PCM words.
Solution
Exercises:
1. In a certain telemetry system, there are four analog signals u(t), v(t),
w(t) and x(t). The bandwidth of u(t) is 3.6kHz but those of the
remaining signals are 1.4kHz each. These signals are to be sampled
at rates no less than respective Nyquist rates and are to be analog
TDM multiplexed. This can be achieved by multiplexing the PAM
samples of the four signals and then binary coding the multiplexed
samples.
i. Design a suitable multiplexing scheme for this purpose
ii. What is the commutator frequency in rotation per second?
iii. What is the total bit rate?

2. Repeat question 1 if there are four analog signals u(t), v(t), w(t) and
x(t) with bandwidth of 1200Hz, 700Hz, 300Hz and 200Hz, respectively.
Signal Max. freq, 𝒇𝒂 Sampling rate Bit rate, 𝒇𝒃
𝑢(𝑡) 3.6 kHz 𝑓𝑠 = 2 × 𝑓𝑎 = 2 × 3.6𝑘 = 7.2𝑘𝑠𝑎𝑚𝑝𝑙𝑒𝑠/𝑠𝑒𝑐 𝑓𝑏 = 𝑛 × 𝑓𝑠 = 9 × 7.2𝑘 = 64.8𝑘𝑏𝑝𝑠
𝑣(𝑡) 1.4 kHz 𝑓𝑠 = 2 × 𝑓𝑎 = 2 × 1.4𝑘 = 2.8𝑘𝑠𝑎𝑚𝑝𝑙𝑒𝑠/𝑠𝑒𝑐 𝑓𝑏 = 𝑛 × 𝑓𝑠 = 9 × 2.8𝑘 = 25.6𝑘𝑏𝑝𝑠
𝑤(𝑡) 1.4 kHz 𝑓𝑠 = 2 × 𝑓𝑎 = 2 × 1.4𝑘 = 2.8𝑘𝑠𝑎𝑚𝑝𝑙𝑒𝑠/𝑠𝑒𝑐 𝑓𝑏 = 𝑛 × 𝑓𝑠 = 9 × 2.8𝑘 = 25.6𝑘𝑏𝑝𝑠
𝑥(𝑡) 1.4 kHz 𝑓𝑠 = 2 × 𝑓𝑎 = 2 × 1.4𝑘 = 2.8𝑘𝑠𝑎𝑚𝑝𝑙𝑒𝑠/𝑠𝑒𝑐 𝑓𝑏 = 𝑛 × 𝑓𝑠 = 9 × 2.8𝑘 = 25.6𝑘𝑏𝑝𝑠
Total bit rate 141.6 kbps

𝑇𝑜𝑡𝑎𝑙 𝑏𝑖𝑡 𝑟𝑎𝑡𝑒 141.6 𝑘

Rotation speed= = =35.4 kHz
𝑇𝑜𝑡𝑎𝑙 𝑛𝑜.𝑜𝑓 𝑡𝑖𝑚𝑒 𝑠𝑙𝑜𝑡 4

Signal Max. freq, 𝒇𝒂 Sampling rate Bit rate, 𝒇𝒃

𝑢(𝑡) 1.2 kHz 𝑓𝑠 = 2 × 𝑓𝑎 = 2 × 1.2𝑘 = 2.4𝑘𝑠𝑎𝑚𝑝𝑙𝑒𝑠/𝑠𝑒𝑐 𝑓𝑏 = 𝑛 × 𝑓𝑠 = 9 × 2.4𝑘 = 21.6𝑘𝑏𝑝𝑠
𝑣(𝑡) 700 Hz 𝑓𝑠 = 2 × 𝑓𝑎 = 2 × 700 = 1.4𝑘𝑠𝑎𝑚𝑝𝑙𝑒𝑠/𝑠𝑒𝑐 𝑓𝑏 = 𝑛 × 𝑓𝑠 = 9 × 1.4𝑘 = 12.6𝑘𝑏𝑝𝑠
𝑤(𝑡) 300 Hz 𝑓𝑠 = 2 × 𝑓𝑎 = 2 × 300 = 600 𝑠𝑎𝑚𝑝𝑙𝑒𝑠/𝑠𝑒𝑐 𝑓𝑏 = 𝑛 × 𝑓𝑠 = 9 × 600 = 5.5𝑘𝑏𝑝𝑠
𝑥(𝑡) 200 Hz 𝑓𝑠 = 2 × 𝑓𝑎 = 2 × 200 = 400 𝑠𝑎𝑚𝑝𝑙𝑒𝑠/𝑠𝑒𝑐 𝑓𝑏 = 𝑛 × 𝑓𝑠 = 9 × 400 = 3.6𝑘𝑏𝑝𝑠
Total bit rate 43.2 kbps

𝑇𝑜𝑡𝑎𝑙 𝑏𝑖𝑡 𝑟𝑎𝑡𝑒 43.2 𝑘

Rotation speed= = = 10.8 kHz
𝑇𝑜𝑡𝑎𝑙 𝑛𝑜.𝑜𝑓 𝑡𝑖𝑚𝑒 𝑠𝑙𝑜𝑡 4

Assume that 1 channel is given 1 time slot and n = 9 bits

 3. A signal u(t) is band-limited to 3.6kHz and the three other signals v(t),
w(t) and x(t) are band-limited to 1.2kHz each. These signals are
sampled at their Nyquist rate and binary coded using 512 levels
(L=512).
i. Determine the individual Nyquist sampling rate
ii. Find the individual bit rate required for each of the signal
iii. Design a suitable multiplexing scheme for this purpose
iv. What is the commutator frequency in rotation per second?
v. What is the total bit rate?
Signal Max. freq, 𝒇𝒂 Sampling rate Bit rate, 𝒇𝒃
𝑢(𝑡) 3.6 kHz 𝑓𝑠 = 2 × 𝑓𝑎 = 2 × 3.6𝑘 = 7.2𝑘𝑠𝑎𝑚𝑝𝑙𝑒𝑠/𝑠𝑒𝑐 𝑓𝑏 = 𝑛 × 𝑓𝑠 = 9 × 7.2𝑘 = 64.8𝑘𝑏𝑝𝑠
𝑣(𝑡) 1.2 kHz 𝑓𝑠 = 2 × 𝑓𝑎 = 2 × 1.2𝑘 = 2.4𝑘𝑠𝑎𝑚𝑝𝑙𝑒𝑠/𝑠𝑒𝑐 𝑓𝑏 = 𝑛 × 𝑓𝑠 = 9 × 2.4𝑘 = 21.6𝑘𝑏𝑝𝑠
𝑤(𝑡) 1.2 kHz 𝑓𝑠 = 2 × 𝑓𝑎 = 2 × 1.2𝑘 = 2.4𝑘𝑠𝑎𝑚𝑝𝑙𝑒𝑠/𝑠𝑒𝑐 𝑓𝑏 = 𝑛 × 𝑓𝑠 = 9 × 2.4𝑘 = 21.6𝑘𝑏𝑝𝑠
𝑥(𝑡) 1.2 kHz 𝑓𝑠 = 2 × 𝑓𝑎 = 2 × 1.2𝑘 = 2.4𝑘𝑠𝑎𝑚𝑝𝑙𝑒𝑠/𝑠𝑒𝑐 𝑓𝑏 = 𝑛 × 𝑓𝑠 = 9 × 2.4𝑘 = 21.6𝑘𝑏𝑝𝑠
Total bit rate 129.6 kbps
The analog signal is quantized using 512 levels
2n=512. Hence, number of bits per sample, 𝑛 = log 2(512) = 9 𝑏𝑖𝑡𝑠
 A TDM system is designed to transmit signals from three users. The signals have the following
specifications.
Signal 1: Analog. 187.5 kHz bandwidth.
Signal 2: Analog. 375 kHz bandwidth.
Signal 3: Digital synchronous at 6300 kbit/s
Assume that the analog signals are sampled at 1.2 times Nyquist rate, and quantized and PCM
encoded using 128 levels. Draw a block diagram for your design and label the output of all stages.
Justify your answer with calculations.

Answer:
Channel 1 bandwidth
= 1.2 × Nyquist rate
= 1.2 × 2 × Baseband bandwidth
= 1.2 × 2 × 187.5 kHz = 450 kHz

Channel 2 bandwidth
= 1.2 × Nyquist rate
= 1.2 × 2 × Baseband bandwidth
= 1.2 × 2 × 375 kHz = 900 kHz

The analog signal is quantized using 128 levels

2n=128. Hence, number of bits per sample, n = 7
 6.4 Multiple Access Techniques
 Multiple access means multiple users can access the channel or link.
 Difference between multiple access and multiplexing:
 Sources of multiple access may be geographically dispersed, while
 Sources of multiplexing are confined within a local site (or point).
 For multiple users to be able to share a common resource in a managed and
effective way requires some form of access protocol that defines how or when the
sharing is to take place and the means for identifying individual messages.
 Process is known as multiplexing in wired networks and multiple access in wireless
digital communications.
 Three classes of multiple access techniques will be
considered:
 techniques where individual users are identified by assigning different
frequency slots (FDMA)
 techniques where individual users are given different time slots (TDMA)
and
 techniques where individual users are given the same time and
frequency slots but are identified by a different code (CDMA).
 CDMA
 CDMA is a multiplexing technique used with spread spectrum
 Narrowband message signal is multiplied by very large bandwidth spreading
signal which is referred to as pseudo-noise (PN).
 All users can use same carrier frequency and may transmit simultaneously
 Each user has own PN code which is approximately orthogonal to other PN
codes.
 Receiver performs time correlation operation to detect only specific PN code,
other PN codes appear as noise due to decorrelation

 Advantages  Disadvantages
 No timing coordination unlike  Implementation complexity of
TDMA spread spectrum
 CDMA uses spread spectrum,  Power control is essential for
resistant to interference (multipath practical operation
fading) CDMA can provide more
users per cell
 No hard limit on number of users
 In Code Division Multiple Access (CDMA), one channel carries all transmissions
simultaneously.
 Assume we have 4 different stations 1,2,3 and 4 connected to the same channel.
 The data from station 1 are d1, from station 2 are d2 and so on.
 The code assigned to station 1 is c1, to station 2 is c2 and so on.
 CDMA – Chip Sequences (PN codes)

 The sequence were not chosen randomly, they were carefully selected.
They are called orthogonal sequence and have the following properties:
1. C2 * C2 = [ +1 -1 +1 -1] * [ +1 -1 +1 -1] = 4 (inner product of two equal
sequence equal to number of element in each sequence)
2. C2 * C4 = [+1 -1 +1 -1] * [+1 -1 -1 +1] = 0 (inner product of two different
sequence equal to 0)
3. C2+C4 = [+1 -1 +1 -1] + [+1 -1 -1 +1] = [+2 -2 0 0] (adding two sequence will
result another sequence)
4. A * C4 = [+A –A –A +A] (scalar multiplication)
 Suppose station 1 and 2 are talking to each other.
 Station 2 wants to hear what station 1 is saying.
 It multiplies the data on the channel by c1 (the code from station 1)
 Since (c1.c1) is 4, but (c2.c1), (c3.c1) and (c4.c1) are all 0s, station 2 divides the
result by 4 to get the data from station 1.
 Note: Data sent via common channel = d1.c1 + d2.c2 + d3.c3 + d4.c4
 Data at station 2 receiver = (d1.c1 + d2.c2 + d3.c3 + d4.c4). c1
= d1.c1.c1 + d2.c2 .c1+ d3.c3 .c1+ d4.c4.c1 = 4.d1
 CDMA – Data Representation

Silence == idle

 Example – Four stations share the link during a 1-bit interval.

 Station 1 and 2 are sending a 0 bit and station 4 is sending 1 bit while station 3 is
silent.
 The data at the sender site are translated to be [ -1, -1, 0, +1].
 Note:
 Each station multiplies the corresponding number by its chip (its orthogonal sequence),
which is unique for each station.
 The sequence on the channel is the sum of all four sequences.
CDMA – Sharing Channel
CDMA – Digital signal created by four stations

Station 1: d1.c1 = [-1 -1 -1 -1]

Station 2: d2.c2 = [-1 +1 -1 +1]
Station 3: d3.c3 = [0 0 0 0]
Station 4: d4.c4 = [+1 -1 -1 +1]
Sum : [-1 -1 -3 +1]
CDMA - Decoding of the composite signal
• Suppose a station wants to decipher the data
being transmitted by station 2.
• The station must know the Station 2’s PN code.
• The transmitted composite data must be
multiplied with Station 2’s PN code.

Station 2 code: c2 = [+1 -1 +1 -1]

Transmitted data: [-1 -1 -3 +1]
Sum of inner product: (+1)(-1)+(-1)(-1)+(+1)(-3)+(-1)(+1)
= -1 + 1 -3 -1 = -4
Station 2 transmitted data = (-4)/4 = -1 →(Bit 0)
 Example 1:
 Sender is sending 2 bits.
 Sending bit 1 in the time slot 0.
 Sending bit 0 in the time slot 1.
 Sender has 8 bits of PN-code
[+1 +1 +1 -1 +1 -1 -1 -1]
 Hence, transmitted output:
Slot 0: d0.c1 = (+1) [+1 +1 +1 -1 +1 -1 -1 -1]
= [+1 +1 +1 -1 +1 -1 -1 -1]
Slot 1: d1.c1 = (-1) [+1 +1 +1 -1 +1 -1 -1 -1]
= [-1 -1 -1 +1 -1 +1 +1 +1]
 At the receiver, sum of inner product:
Slot 0: [+1 +1 +1 -1 +1 -1 -1 -1]
Slot 1: [+1 +1 +1 -1 +1 -1 -1 -1]
[+1 +1 +1 -1 +1 -1 -1 -1]
[-1 -1 -1 +1 -1 +1 +1 +1]
=(+1)(+1)+(+1)(+1)+(+1)(+1)+(-1)(-1)+(+1)(+1)+(-1)(-
1)+ )+(-1)(-1)+ )+(-1)(-1)] =(+1)(-1)+(+1)(-1)+(+1)(-1)+(-1)(+1)+(+1)(-1)+(-1)(+1)+(-

= +1+1+1+1+1+1+1+1=+8 1)(+1)+(-1)(+1)
= -1-1-1-1-1-1-1-1=-8
Transmitted bit in Slot 0 = (+8)/8 = +1 → Bit 1
Transmitted bit in Slot 1 = (-8)/8 = -1 → Bit 0
Example 2: two-sender interference
 Sender 1 is sending 2 bits.
 Sending bit 1 in the time slot 0.
 Sending bit 0 in the time slot 1.
 Sender 1 has 8 bits of PN-code
sender 1
[+1 +1 +1 -1 +1 -1 -1 -1]
 Sender 2 is sending 2 bits.
 Sending bit 1 in the time slot 0.
sender 2
 Sending bit 1 in the time slot 1.
 Sender 2 has 8 bits of PN-code
[+1 -1 +1 +1 +1 -1 +1 +1]
 Hence, transmitted output:
Slot 0: d0.c1 = (+1) [+1 +1 +1 -1 +1 -1 -1 -1]
= [+1 +1 +1 -1 +1 -1 -1 -1]
Slot 0: d0.c2 = (+1) [+1 -1 +1 +1 +1 -1 +1 +1]
= [+1 -1 +1 +1 +1 -1 +1 +1]
 Resultant transmitted data in Slot 0:
[+1 +1 +1 -1 +1 -1 -1 -1]
uses sender 1 code [+1 -1 +1 +1 +1 -1 +1 +1]
to receive sender 1 data [+2 0 +2 0 +2 -2 0 0]
 At the receiver station 1, sum of inner product in Slot 0:
Station 1’s PN code: [+1 +1 +1 -1 +1 -1 -1 -1]
Received data: [+2 0 +2 0 +2 -2 0 0]
=(+1)(+2)+(+1)(0)+(+1)(+2)+(-1)(0)+(+1)(+2)+(-1)(-2)+(-1)(0)+(-1)(0)
= +2+0+2+0+2+2+0+0=+8
Transmitted bit by sender 1 in Slot 0 = (+8)/8 = +1 → Bit 1
Example 3:
 Given a binary chip sequence and a bipolar chip sequence for 4 stations
as below :
A : 00011011 A : (-1 –1 –1 +1 +1 –1 +1 +1 )
B : 00101110 B : ( -1 –1 +1 –1 +1 +1 +1 –1 )
C : 01011100 C : ( -1 +1 –1 +1 +1 +1 –1 –1 )
D : 01000010 D : ( -1 +1 –1 –1 –1 –1 +1 –1 )

What is the resulting chip sequence if :

i. Only C transmit 1
Answer:
Station A: dA.cA = (0) [-1 –1 –1 +1 +1 –1 +1 +1] = [ 0 0 0 0 0 0 0 0]
Station B: dB.cB = (0) [-1 –1 +1 –1 +1 +1 +1 –1] = [ 0 0 0 0 0 0 0 0]
Station C: dC.cC = (+1)[-1 +1 –1 +1 +1 +1 –1 –1] = [-1 +1 –1 +1 +1 +1 –1 –1]
Station D: dD.cD = (0) [-1 +1 –1 –1 –1 –1 +1 –1] = [ 0 0 0 0 0 0 0 0]
Resultant data : [-1 +1 –1 +1 +1 +1 –1 –1]
ii. B and C transmit 1
Answer:
Station A: dA.cA = (0) [-1 –1 –1 +1 +1 –1 +1 +1] = [ 0 0 0 0 0 0 0 0]
Station B: dB.cB = (+1)[-1 –1 +1 –1 +1 +1 +1 –1] = [-1 –1 +1 –1 +1 +1 +1 –1]
Station C: dC.cC = (+1)[-1 +1 –1 +1 +1 +1 –1 –1] = [-1 +1 –1 +1 +1 +1 –1 –1]
Station D: dD.cD = (0) [-1 +1 –1 –1 –1 –1 +1 –1] = [ 0 0 0 0 0 0 0 0]
Resultant data : [-2 0 0 0 +2 +2 0 –2]

iii. A, B and D transmit 1, C transmit 0

Answer:
Station A: dA.cA = (+1)[-1 –1 –1 +1 +1 –1 +1 +1] = [-1 –1 –1 +1 +1 –1 +1 +1]
Station B: dB.cB = (+1)[-1 –1 +1 –1 +1 +1 +1 –1] = [-1 –1 +1 –1 +1 +1 +1 –1]
Station C: dC.cC = (-1)[-1 +1 –1 +1 +1 +1 –1 –1] = [+1 -1 +1 -1 -1 -1 +1 +1]
Station D: dD.cD = (+1)[-1 +1 –1 –1 –1 –1 +1 –1] = [-1 +1 –1 –1 –1 –1 +1 –1]
Resultant data : [-2 -2 0 -2 0 -2 +4 0]
Example 4:
 Two stations (station A and station B) are using the same carrier frequency and
able to transmit simultaneously. The PN code for each station is as follows:
▪ Station A: 010011
▪ Station B: 110101
 Determine the resultant data if Station A is transmitting bit ‘0’ and Station B is
transmitting bit ‘1’.

Answer:
Station A: dA.cA = (-1)[ -1 +1 –1 -1 +1 +1] = [+1 –1 +1 +1 -1 –1]
Station B: dB.cB = (+1)[+1 +1 -1 +1 -1 +1] = [+1 +1 -1 +1 -1 +1]
Resultant data : [+2 0 0 +2 -2 0]
Example 5:
 Figure below depicts a simplified scheme for code division multiple access
(CDMA) encoding with 3 users. All users utilized a PN code of 8 bits. A 1
MHz(a) band has to be shared by 3 users.

User A
1
Encoder
Data bits 1 =0 0 =1

PN codes -1
Slot 1 Slot 0 Slot 1 Slot 0

User B
Encoder Transmitted
Data bits =0 =0 1 signal
1 0
+
PN codes
-1
Slot 1 Slot 0 Slot 1 Slot 0

User C
Encoder
Data bits 1 =1 0 =
Encoded signal
PN codes
Slot 1 Slot 0
a. What are the PN codes for User A and User B? State assumption used.
Answer:
 User A is sending bit ‘1’ → +1 in Slot 0
The encoded Slot 0 for User A: [+1 +1 +1 -1 +1 -1 -1 -1]
Thus, User A’s PN code : [+1 +1 +1 -1 +1 -1 -1 -1]
 User B is sending bit ‘0’ → -1 in Slot 0
The encoded Slot 0 for User B: [+1 -1 +1 +1 +1 -1 +1 +1]
Thus, User B’s PN code is : [-1 +1 -1 -1 -1 +1 -1 -1]
b. Determine the 8 bits PN codes for User C to ensure they can coexist without interference.
Prove that the PN codes are perfectly orthogonal.
Answer:
User C’s PN code must be orthogonal with User A’s and User B’s PN codes.
User A’s PN code: [+1 +1 +1 -1 +1 -1 -1 -1]
User C’s PN code: [+1 +1 +1 -1 -1 +1 +1 +1]
Sum of Inner product: [+1 +1 +1 +1 -1 -1 -1 -1] = 0
User B’s PN code: [-1 +1 -1 -1 -1 +1 -1 -1]
User C’s PN code: [+1 +1 +1 -1 -1 +1 +1 +1]
Sum of Inner product: [-1 +1 -1 +1 +1 +1 -1 -1] = 0
c. Determine the transmitted signals Slot 0.
Answer:
User A: dA.cA = (+1) [+1 +1 +1 -1 +1 -1 -1 -1] = [+1 +1 +1 -1 +1 -1 -1 -1]
User B: dB.cB = (-1) [-1 +1 -1 -1 -1 +1 -1 -1] = [+1 -1 +1 +1 +1 -1 +1 +1]
User C: dC.cC = (0) [+1 +1 +1 -1 -1 +1 +1 +1] = [ 0 0 0 0 0 0 0 0]
Resultant data : [+2 0 +2 0 +2 -2 0 0]

d. What is the bandwidth each user will get?

Answer:
Each user will get 1 MHz bandwidth.
Example 5:
Suppose that during the first time slot, two users in a CDMA system transmit
signals simultaneously. All of the users utilized a PN code of 8 bits. The PN codes
of each users are as follow:
CA: +1 -1 +1 -1 +1 +1 -1 +1
CB: +1 -1 +1 +1 -1 -1 +1 +1
a) If the resultant transmitted signal during the first time slot is given as:
rx: +1 -1 +1 -1 +1 +1 -1 +1
What was the bit transmitted by User A during the first time slot? [2 marks]
b) If an intruder with a PN code of -1 +1 -1 +1 +1 +1 -1 +1 appears in the
communication link, what would happen to station A and B? Justify your
answer with calculations. [5 marks]
Answer: