4123702
Data Communications System
By
Ajarn Preecha Pangsuban
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Chapter 5 – Analog Transmission
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Analog Signals
 Infinite number of values in a range
 More susceptible to noise and is less precise than a
digital signal
 Since they are more variable than digital signals, they can
convey greater subtleties
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Bit Rate and Baud Rate
 Bit rate – number of bits transmitted in 1 second
Computer efficiency

 Baud rate – number of signal units per second
required to represent those bits; equal to or less
than bit rate


Network transmission efficiency
Determines bandwidth required
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Example 1
An analog signal carries 4 bits in each signal unit. If
1000 signal units are sent per second, find the baud
rate and the bit rate.
Solution
Baud rate = 1000 bauds per second (baud/s)
Bit rate = 1000 x 4 = 4000 bps
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Example 2
The bit rate of a signal is 3000. If each signal unit
carries 6 bits, what is the baud rate?
Solution
Baud rate = bit rate/bits per signal
= 3000 / 6
= 500 baud/s
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Modulation of Digital Data
 Carrier Signal - High-frequency signal used for
digital-to-analog or analog-to-digital modulation
 May change any aspect of an analog signal to
represent digital data: amplitude, frequency, and
phase
 Modulation - process of changing one of the
characteristics of an analog signals based on
information in digital signal
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Digital-to-Analog Conversion
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Methods
 Three mechanisms
Amplitude shift keying (ASK)
 Frequency shift keying (FSK)
 Phase shift keying (PSK)
 Combined approach: quadrature amplitude
modulation (QAM)

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Amplitude Shift Keying (ASK)
 Strength of carrier signal is varied to represent a
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1 or 0
Frequency and phase remain constant while
amplitude is changed
Amplitude remains constant during bit duration
Highly susceptible to noise interference since
noise usually affects amplitude
Minimum bandwidth required is equal to baud
rate
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Amplitude Shift Keying (ASK)
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Relationship between baud rate and
bandwidth in ASK
BW = (1+d)Nbaud
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Example 3
Find the minimum bandwidth for an ASK signal
transmitting at 2000 bps. The transmission mode is halfduplex.
Solution
In ASK the baud rate and bit rate are the same. The baud
rate is therefore 2000. An ASK signal requires a
minimum bandwidth equal to its baud rate. Therefore,
the minimum bandwidth is 2000 Hz.
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Example 4
Given a bandwidth of 5000 Hz for an ASK signal, what
are the baud rate and bit rate?
Solution
In ASK the baud rate is the same as the bandwidth,
which means the baud rate is 5000. But because the baud
rate and the bit rate are also the same for ASK, the bit
rate is 5000 bps.
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Example 5
Given a bandwidth of 10,000 Hz (1000 to 11,000 Hz), draw
the full-duplex ASK diagram of the system. Find the carriers
and the bandwidths in each direction. Assume there is no gap
between the bands in the two directions.
Solution
For full-duplex ASK, the bandwidth for each direction is
BW = 10000 / 2 = 5000 Hz
The carrier frequencies can be chosen at the middle of each
band (see Fig. 5.5).
fc (forward) = 1000 + 5000/2 = 3500 Hz
fc (backward) = 11000 – 5000/2 = 8500 Hz
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Solution to Example 5
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Frequency Shift Keying (FSK)
 Frequency is varied to represent a 1 or 0
 Frequency during bit duration is constant
 Amplitude and phase remain constant
 Avoids most of noise problems of ASK; can
ignore voltage spikes
 Limited by physical capabilities of medium
 Bandwidth required is equal to baud rate plus
frequency shift
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Frequency Shift Keying (FSK)
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Relationship between baud rate and
bandwidth in FSK
BW = fc1-fc0+Nbaud
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Example 6
Find the minimum bandwidth for an FSK signal
transmitting at 2000 bps. Transmission is in half-duplex
mode, and the carriers are separated by 3000 Hz.
Solution
For FSK
BW = baud rate + fc1 - fc0
BW = bit rate + fc1 - fc0 = 2000 + 3000 = 5000 Hz
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Example 7
Find the maximum bit rates for an FSK signal if the bandwidth
of the medium is 12,000 Hz and the difference between the two
carriers is 2000 Hz. Transmission is in full-duplex mode.
Solution
Because the transmission is full duplex, only 6000 Hz is
allocated for each direction.
BW = baud rate + fc1 - fc0
Baud rate = BW - (fc1 - fc0 ) = 6000 - 2000 = 4000
But because the baud rate is the same as the bit rate, the bit rate
is 4000 bps.
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Phase Shift Keying (PSK)
 Phase of carrier is varied to represent 1 or 0
 Peak amplitude and frequency remain constant
 Phase remains constant during each bit duration
 Not as susceptible to noise degradation as ASK
or to bandwidth limitations of FSK
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Phase Shift Keying (PSK)
2-PSK or Binary PSK
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PSK constellation (Phase state diagram)
Constellation is a graphical representation of the phase
and amplitude of different bit combinations in digital-toanalog modulation.
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4-PSK; 8-PSK
 4-PSK
May utilize four variations of phase shift by 90 degrees
Each phase shift represents 2 bits (dibit); technique is
referred to as 4-PSK
Allows data transmission two times as fast as 2-PSK

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 8-PSK
Vary signal by shifts of 45 degrees; each shift may
then represent three bits (tribit) and send data three
times as fast

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The 4-PSK method (Q-PSK)
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The 4-PSK characteristics
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The 8-PSK characteristics
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Relationship between baud rate and
bandwidth in PSK
BW = (1+d)Nbaud
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Example 8
Find the bandwidth for a 4-PSK signal transmitting at
2000 bps. Transmission is in half-duplex mode.
Solution
For 4-PSK the baud rate is one –half of the bit rate (1 baud :
2 bit). The baud rate is therefore 1000. A PSK signal requires
a bandwidth equal to its baud rate . Therefore, the bandwidth
is 1000 Hz
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Example 9
Given a bandwidth of 5000 Hz for an 8-PSK signal, what
are the baud rate and bit rate?
Solution
For PSK the baud rate is the same as the bandwidth,
which means the baud rate is 5000. But in 8-PSK the bit
rate is 3 times the baud rate, so the bit rate is 15,000 bps.
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Quadrature Amplitude Modulation (QAM)
 Combination of ASK and PSK so that a maximum
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contrast between each signal unit (bit, dibit, tribit, and
so on) is achieved
x number of variations in phase and y variations in
amplitude
Number of phase shifts is always larger than number
of amplitude shifts due to amplitude susceptibility to
noise
QAM is therefore less susceptible to noise than ASK
Same bandwidth is required for ASK and PSK
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The 4-QAM and 8-QAM constellations
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Time domain for an 8-QAM signal
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16-QAM constellations
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Bit/Baud Comparison
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Bit and baud rate comparison
Modulation
Units
Bits/Baud Baud rate Bit Rate
ASK, FSK, 2-PSK
Bit
1
N
N
4-PSK, 4-QAM
Dibit
2
N
2N
8-PSK, 8-QAM
Tribit
3
N
3N
16-QAM
Quadbit
4
N
4N
32-QAM
Pentabit
5
N
5N
64-QAM
Hexabit
6
N
6N
128-QAM
Septabit
7
N
7N
256-QAM
Octabit
8
N
8N
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Example 10
A constellation diagram consists of eight equally spaced
points on a circle. If the bit rate is 4800 bps, what is the
baud rate?
Solution
The constellation indicates 8-PSK with the points 45
degrees apart. Since 23 = 8, 3 bits are transmitted with
each signal unit. Therefore, the baud rate is
4800/3 = 1600 baud
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Example 11
Compute the bit rate for a 1000-baud 16-QAM signal.
Solution
A 16-QAM signal has 4 bits per signal unit since
log216 = 4.
Thus,
(1000)(4) = 4000 bps
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Example 12
Compute the baud rate for a 72,000-bps 64-QAM signal.
Solution
A 64-QAM signal has 6 bits per signal unit since
log2 64 = 6.
Thus,
72000/6 = 12,000 baud
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5.2 Telephone Modems
 Traditional phone lines carry frequencies between 300 and
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3300 Hz; bandwidth of 3000 Hz
All this range is used for transmitting voice
Data signals require higher degree of accuracy to ensure
integrity (the edges of this range are not used)
Effective bandwidth for data transmission is 2400 Hz
(between 600 and 3000 Hz)
MODEM is a composite word refer to signal modulator and
signal demodulator
modulator – creates a band-pass analog signal from binary
data
demodulator – recovers binary data from modulated signal
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Telephone line bandwidth
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Modulation/demodulation
Telco is Telephone Company or Your local telephone service provider
(such as TOT, TT&T)
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Modem Standards
 The most popular MODEMs available are based
on the V-series standards by the ITU-T
 V-series standards that define data transmission
over telephone lines.
(ITU-T : International Telecommunications UnionTelecommunication Standardization Sector)
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V.32 (32-QAM)
 A combined modulation and encoding technique
called Trellis Coded Modulation (TCM)
 Trellis is essentially QAM plus a redundant bit
 A pentabit is transmitted (4 bit data stream and 1
extra bit )
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The V.32 constellation and bandwidth
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V.32 bis (128-QAM)
 The first of the ITU-T standards to support 14,400
bps transmission
 Uses 128-QAM transmission (7 bit/baud with 1 bit
for error control)
 Data rate of 2400 baud is 2400*6 = 14,400 bps
 Automatic speed adjustment depending on quality
of line or signal
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The V.32bis constellation and bandwidth
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V.34 bis
 For bit rate of 28,800 with a 960 point
constellation to a bit rate 33,600 with 1664 point
constellation
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V.90
 V.90 MODEM with a bit rate 56,00bps, called 56 k
MODEMs
 The downloading rate is a maximum of 56kbps,
while the uploading rate can be maximum of 33.6
kbps (asymmetric)
 These MODEM may be used only if one party is
using digital signaling (such as through an ISP)
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Traditional modems (33.6kbps)
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56K modems
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V.92
 These MODEM can adjust their speed , and if the
noise allows, the can upload data at the rate of 48
kbps
 The downloading rate is a maximum of 56kbps
 The MODEM can interrupt the Internet connection
when there is an incoming call if the line has callwaiting service
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MODEM type
 Asynchronous, Synchronous
Low-speed modems are designed to operate
asynchronously
High-speed modems as well as leased-lines modems
use synchonous tranmission
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 Half-Duplex, Full-Duplex
A half-duplex modem must alternately send and
received signals Half-duplex
A full-duplex modem can simultaneously handle two
signals using two carriers to transmit and receive
data

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5.3 Modulation of Analog Signals
 Analog to analog conversion is the presentation of
analog information by an analog signal
 Example is radio
 Can accomplished in three ways :



Amplitude Modulation (AM)
Frequency Modulation (FM)
Phase Modulation (PM)
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Analog-to-analog modulation
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Amplitude modulation (AM)
 Amplitude varies with
changing amplitudes of
the modulating signal
 Frequency and phase
remain the same
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AM bandwidth
 Carrier frequency between 530 and 1700 kHz
 AM radio station needs a minimum BW of 10 kHz (By
FCC)
 The total bandwidth required for AM can be determined
from the bandwidth of the audio signal
 Equal to twice the bandwidth of the audio signal
 Carrier frequencies must be separated on either side to
avoid interference
 Total bandwidth determined by:
BWt = 2 * BWm
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AM bandwidth (cont)
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AM band allocation
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Example 13
We have an audio signal with a bandwidth of 4 KHz.
What is the bandwidth needed if we modulate the signal
using AM? Ignore FCC regulations.
Solution
An AM signal requires twice the bandwidth of the
original signal:
BW = 2 x 4 KHz = 8 KHz
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Frequency modulation (FM)
 Frequency of carrier
signal is modulated to
follow changing voltage
level (amplitude) of the
modulating signal
 Peak amplitude and
phase remain constant
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FM bandwidth
 Carrier frequency between 88MHz and 108 MHz
 The bandwidth of a stereo audio signal is usually 15 KHz.
 Therefore, an FM station needs at least a bandwidth of 150
KHz.
 The FCC requires the minimum bandwidth to be at least 200
KHz (0.2 MHz).
 The total bandwidth required for FM can be determined from
the bandwidth of the audio signal:
BWt = 10 * BWm
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FM bandwidth (cont)
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FM band allocation
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Example 14
We have an audio signal with a bandwidth of 4 MHz.
What is the bandwidth needed if we modulate the signal
using FM? Ignore FCC regulations.
Solution
An FM signal requires 10 times the bandwidth of the
original signal:
BW = 10 x 4 MHz = 40 MHz
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Phase Modulation (PM)
 Sometimes used as an alternative to FM
 Simpler hardware requirements
 Phase is modulated to follow changing voltage
level (amplitude) of modulating signal
 Peak amplitude and frequency remains constant
 The analysis and the final result are similar FM
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Credits
 All figures obtained from publisher-provided
instructor downloads
Data Communications and Networking, 3rd edition by
Behrouz A. Forouzan. McGraw Hill Publishing, 2004
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