Digital Modulation

11. Modulation and
Multiplexing
: Analog Signal

Modulation of Digital Data
Digital-to-Analog Conversion
Amplitude Shift Keying (ASK)
Frequency Shift Keying (FSK)
Phase Shift Keying (PSK)
Quadrature Amplitude Modulation
Bit/Baud Comparison

Introduction
 While digital transmission is very desirable , a low
pass channel is needed.
 Analog transmission is the only choice if we
have a bandpass channel
 Converting digital data to a bandpass analog signal
is traditionally called digital-to-analog
conversion.
 Converting a low-pass analog signal to a bandpass
analog signal is traditionally called analog to
analog conversion.

Modulation of Digital
Data
Modulation of binary data or digital-to-analog modulation is the
process of changing one of the characteristics of an analog
signal based on the information in a digital signal (0s and 1s).

5.5
Digital-to-analog conversion

Data
Bit Rate and Baud Rate
As defined earlier, bit rate is the number of bits transmitted
during 1 s.
Baud rate refers to the number of signal units per second that
are required to represent those bits.
Implications:
# For computer efficiency, bit rate is more relevant.
# For data transmission, baud rate is more important. If fewer
signals are required to send a piece of data, it means the
system is more efficient and less bandwidth is required.
#In the analog transmission of digital data, the signal or baud
rate is less than or equal to the bit rate.
S=Nx1/r bauds
Where r is the number of data bits per signal element
Important:
Bit Rate = Baud Rate X number of bits represented by each

Example 1
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
Solution
Baud rate = 1000 bauds per second (baud/s)
Baud rate = 1000 bauds per second (baud/s)
Bit rate = 1000 x 4 = 4000 bps
Bit rate = 1000 x 4 = 4000 bps
Data
Bit Rate and Baud Rate

Data
Very Important:
# In analog transmission, the sending device produces a high-
frequency signal that acts as a basis for the information signal.
This base signal is called the carrier signal or carrier frequency.
# The receiving device is tuned to the frequency of the carrier
signal that it expects from the sender.
# Digital information then modulates the carrier signal by
modifying one or more of its characteristics (amplitude,
frequency, or phase).
# This kind of modification is called Modulation (or shift keying),
and the information signal is called the modulating signal.

5.9
Figure 5.3 Binary amplitude shift keying

5.10
Figure 5.4 Implementation of binary ASK

Data
In ASK, the strength of the carrier signal is varied to represent
binary 1 or 0. Both frequency and phase remain constant while
the amplitude changes. Figure below clears the concept.

Data
When we decompose an ASK-modulated signal, we get a
spectrum of many simple frequencies. Therefore bandwidth is
as shown in following diagram:
Bandwidth for ASK

Example 3
Example 3
Find the minimum bandwidth for an ASK signal
transmitting at 2000 bps. The transmission mode is half
duplex.
Solution
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.
Data
Bandwidth for ASK

Example 5
Example 5
Given a bandwidth of 10,000 Hz (1000 to 11,000 Hz),
draw the fullduplex 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
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
Data
Bandwidth for ASK

Data
Bandwidth for ASK
Example 5
Example 5

5.16
Figure 5.6 Binary frequency shift keying

Data
In FSK, the frequency of the carrier signal is varied to
represent binary 1 or 0. Both peak amplitude and phase remain
constant. Figure below clears the concept.

Data
Bandwidth for FSK
Note that FSK spectrum is a combination of two ASK spectra
centered on fc0 and fc1, where
fc0 : frequency of the signal representing 0.
fc1 : frequency of the signal representing 1.
The following figure clears the concept:

Example 6
Example 6
Find the minimum bandwidth for an FSK signal
transmitting at 2000 bps. Transmission is in halfduplex
mode, and the carriers are separated by 3000 Hz.
Solution
Solution
For FSK
BW = baud rate + f
BW = baud rate + fc1
c1 
 f
fc0
c0
BW = bit rate + fc1
BW = bit rate + fc1 
 fc0 = 2000 + 3000 = 5000 Hz
fc0 = 2000 + 3000 = 5000 Hz
Data
Bandwidth for FSK

Data
In PSK, the phase of the carrier signal is varied to represent
binary 1 or 0. Both peak amplitude and frequency remain
constant. Figure below clears the concept.

Data
Constellation (Phase state) Diagram
The constellation or phase state diagram shows the
relationship by illustrating only the phases. Depicting two
phases on the diagram is known as 2-PSK.

Data
In phase modulation, if we use four phase variations, then we
can send 2 bits per phase as illustrated in the following
diagram.
Depicting four phases on the diagram is known as 4-PSK.

Data
Depicting Eight phases on the diagram is known as 8-PSK.

Data
Bandwidth requirement is the same as in the case of ASK. But
keep in mind that in PSK bit rate for the same bandwidth can
be 2 or more times than that in ASK.

Data Quadrature Amplitude Modulation (QAM)
PSK is limited by the ability of the equipment to distinguish
small differences in phase and thus limiting potential bit rate.
Quadrature amplitude modulation is a combination of ASK and
PSK so that a maximum contrast between each signal unit (bit,
dibit, tribit, and so on) is achieved.
A constellation diagram helps us to define the amplitude and
phase of a signal when we are using two carriers, one in
quadrature of the other.
The X-axis represents the in-phase carrier and the Y-axis
represents quadrature carrier.

Data Quadrature Amplitude Modulation (QAM)

5.28
Figure 5.14 Constellation diagrams for some QAMs

Data
Bit/Baud Comparison

Modulation
Modulation Units
Units Bits/Baud
Bits/Baud Baud rate
Baud rate Bit Rate
ASK, FSK, 2-PSK
ASK, FSK, 2-PSK Bit 1 N N
4-PSK, 4-QAM
4-PSK, 4-QAM Dibit 2 N 2N
8-PSK, 8-QAM
8-PSK, 8-QAM Tribit 3 N 3N
16-QAM
16-QAM Quadbit 4 N 4N
32-QAM
32-QAM Pentabit 5 N 5N
64-QAM
64-QAM Hexabit 6 N 6N
128-QAM
128-QAM Septabit 7 N 7N
256-QAM
256-QAM Octabit 8 N 8N
Data
Bit/Baud Comparison

5.32
ANALOG AND DIGITAL
ANALOG AND DIGITAL
Analog-to-analog conversion is the representation of
Analog-to-analog conversion is the representation of
analog information by an analog signal. One may ask
analog information by an analog signal. One may ask
why we need to modulate an analog signal; it is
why we need to modulate an analog signal; it is
already analog. Modulation is needed if the medium is
already analog. Modulation is needed if the medium is
bandpass in nature or if only a bandpass channel is
bandpass in nature or if only a bandpass channel is
available to us.
available to us.
Amplitude Modulation
Frequency Modulation
Phase Modulation
Topics discussed in this section:
Topics discussed in this section:

5.33
Figure 5.15 Types of analog-to-analog modulation

5.34
Figure 5.16 Amplitude modulation

5.35
Figure 5.18 Frequency modulation

5.36
Figure 5.20 Phase modulation

WCB/McGraw-Hill
No Multiplexing vs. Multiplexing

Multiplexing is the set of techniques that allows the
simultaneous transmission of multiple signals across a
single data link.
In a multiplexed system, n lines share the bandwidth of
one link.
The word link refers to the physical path whereas the
word channel refers to the portion of a link that carries a
transmission between a given pair of lines. One link can
have many channels.

6.1 FDM
Multiplexing Process
Demultiplexing Process
The Analog Hierarchy
Other Applications of FDM
Implementation

FD
MFrequency division multiplexing (FDM) is an analog technique
that can be applied when the bandwidth of a link (in hertz) is
greater than the combined bandwidth of the signals to be
transmitted.
In FDM, signals generated by each sending device modulate
different carrier frequencies. These modulated signals are
then combined into a single composite signal. Carrier
frequencies are separated by sufficient bandwidth to
accommodate the modulated signal. These bandwidth ranges
are the channels. Channels must be separated by strips of
unused bandwidth (guard bands) to prevent signals from
overlapping.

Example 1
Example 1
Assume that a voice channel occupies a bandwidth of 4
KHz. We need to combine three voice channels into a
link with a bandwidth of 12 KHz, from 20 to 32 KHz.
Show the configuration using the frequency domain
without the use of guard bands.
FD
M
Multiplexing/Demulti
plexing

FD
M
plexing

Example 2
Example 2
Five channels, each with a 100KHz bandwidth, are to be
multiplexed together. What is the minimum bandwidth of
the link if there is a need for a guard band of 10 KHz
between the channels to prevent interference?
Solution
Solution
For five channels, we need at least four guard
bands. This means that the required
bandwidth is at least
5 x 100 + 4 x 10 = 540 KHz,
FD
M
plexing

Example 3
Example 3
Four data channels (digital), each transmitting at 1 Mbps,
use a satellite channel of 1 MHz. Design an appropriate
configuration using FDM
Solution
Solution
The 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.
FD
M
plexing

FD
M
Other applications
of FDM
# AM radio stations: Special band from 530 to 1700 KHz is
assigned to AM radio. All radio stations need to share this
band. Each radio station needs 10 KHz of bandwidth, therefore
multiplexing is used.
#FM radio stations: Same operation with frequencies ranging
from 88 to 108 MHz, with each station needing 200Khz.
# Television broadcasting: Same with each television channel
needing 6 MHz.
# First generation cellular phones: Each user is assigned two
30 KHz channels. The voice signal (3KHz) is modulated using
FSK, therefore bandwidth of 30 KHz (3X10 KHz).

Example 4
Example 4
The Advanced Mobile Phone System (AMPS) uses two
bands. The first band, 824 to 849 MHz, is used for
sending; 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
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.
FD
M
Other applications
of FDM

FD
M
Implementa
tion
#No need for special devices to do
multiplexing/Demultiplexing in case of radio and television
broadcasting.
# Why?
# In cellular phone base station needs to assign a carrier
frequency to telephone user.

6.2 WDM
Wave Division Multiplexing

WD
M
Wave Division
Multiplexing
WDM: A narrow bans of light from different sources are
combined to make a wider band of light. At the receiver, the
signals are separated by the demultiplexer.

WD
M
Wave Division
Multiplexing
Basic
Principle

6.3 TDM
6.3 TDM
Time Slots and Frames
Interleaving
Synchronizing
Bit Padding
Digital Signal (DS) Service

TD
MTDM: is a digital process that allows several connections to
share the high bandwidth of a link. Each connection occupies a
portion of time in the link.

TD
M
Time slots and
Frames
The data flow of each connection is divided into units, and link
combines one unit of each connection to make a frame. For n
input connections, a frame is organized into a minimum of n
time slots, each time slot carrying one unit from each
connection.
In TDM, the data rate of the link that carries data from n
connections must be n times the data rate of a connection to
guarantee the flow of data. Therefore, duration of a unit in a
connection is n times the duration of a time slot in a frame.

Example 5
Example 5
Four 1Kbps connections are multiplexed together. A unit
is 1 bit. Find (1) the duration of 1 bit before multiplexing,
(2) the transmission rate of the link, (3) the duration of a
time slot, and (4) the duration of a frame?
Solution
Solution
We can answer the questions as follows:
1. The duration of 1 bit is 1/1 Kbps, or
0.001 s (1 ms).
2. The rate of the link is 4 Kbps.
3. The duration of each time slot 1/4 ms
or 250 s.
TD
M
Time slots and
Frames

TD
M
Synchroniz
ing
If the multiplexer and demultiplexer are out of
synchronization, a bit belonging to one of channel may be
received by the wrong channel. Therefore one or more
synchronization bits are usually added to the beginning of
each frame, called framing bits.

TD
M
Bit
Padding
It is possible to multiplex data from devices of different data
rates. For example, device A could use one time slot, while
faster device B could use two. The number of slots in a frame
and the input lines to which they are assigned remain fixed
throughout a given system, but devices of different data rates
may control different number of those slots
Therefore, the different data rates must be integer multiples
of each other.
When speed are not integer multiples of each other, they can
be made to behave as if they were, by a technique called bit
padding, in which multiplexer adds extra bits to device’s
source stream. The extra bits are, then, discarded by the
demultiplexer.

Example 9
Example 9
Two channels, one with a bit rate of 100 Kbps and
another with a bit rate of 200 Kbps, are to be multiplexed.
How this can be achieved? What is the frame rate? What
is the frame duration? What is the bit rate of the link?
Solution
Solution
We can allocate one slot to the first channel
and two slots to the second channel. Each
frame carries 3 bits. The frame rate is
100,000 frames per second because it carries
1 bit from the first channel. The frame
duration is 1/100,000 s, or 10 ms. The bit
TD
M
Bit
Padding

Digital Modulation

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