COMMUNICATION SYSTEM EEEB453
Chapter 6
DIGITAL MODULATION
Intan Shafinaz Mustafa
Dept of Electrical Engineering
Universiti Tenaga Nasional
http://metalab.uniten.edu.my/~shafinaz
Digital-to-Analog Conversion
 Modulating a digital signal
 Eg. Sending computer data through public telephone line
2
Digital-to-Analog Conversion

Process of changing one of the characteristics of an analog signal based on
the information in digital signals (0’s and 1’s).

Modulation involved switching (known as keying) between short bursts of
different signals to transmit the encoded message.

A general carrier wave may be written:
C t   A sin  2  ft   

Modulation methods based on varying the amplitude, A, frequency, f and
phase, to transmit digital data is known as Amplitude Shift Keying (ASK),
Frequency Shift Keying (FSK) and Phase Shift Keying (PSK) respectively.

Example – Tx of digital data over telephone wire (modem)
B Mbps
B MHz
3
Information Capacity, Bits, Bit Rate, Baud
and M-ary Encoding.

Information Capacity – is a measure of how much information can
be propagated through a communications system and is a function
of BW and Tx line.
It represents the number of independent symbols that can be carried
through a system in a given unit of time.
 Hartley’s law is

I  Bxt
Where
I = information capacity (bps)
B = bandwidth (hertz)
t = transmission time (sec)
M-ary Encoding





M-ary is a term derived from the word binary.
M simply represents a digit that corresponds to the number of condition, levels
or combinations possible for a given number of binary variables.
For example, a digital signal with four possible conditions is an M-ary system
where M = 4.
The number of bits necessary to produce a given number of condition is
expressed mathematically as
N = log 2 M
Where N = number of bits necessary
M = Number of conditions, levels or combination possible with N bits
Rearranged the equation to express the number of conditions possible with N
bits as
2N = M
Example, with one bit, only 21 = 2 conditions are possible, with two bits, 22 = 4
conditions are possible etc..
Bit Rate, Baud Rate and Minimum BW

Two basic aspects of digital-to-analog modulation; bit and baud rate.

Bit rate - number of bits per second (rate at which bit changes, bps).
 Computer Efficiency – how long it takes to process each piece of
information (time to send)

Baud rate – number of signal units per second (rate at which signal element
changes). Also called modulation rate or symbol rate.
 Data Transmission Efficiency – how efficient we can move those data
from place to place.

Analogy – In transportation, baud ≈ car and bit ≈ passenger
 A car can carry one or more passengers. If 1000 cars go from one point to another, carrying
only 1 passenger (i.e the driver), then 1000 passengers are transported.
 However, if each car carries four passengers (carpooling), then 4000 passengers are
transported.
 Note that the number of cars not the number of passengers, determine the traffic, and
therefore, the need for wider highways.
 Similarly, the number of bauds determines the required bandwidth, not the number of bits.
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Bit Rate, Baud Rate and Minimum BW

According to H. Nyquist, binary digital signals can be propagated through an
ideal noiseless transmission medium at a rate equal to two times the
bandwidth of the medium.

The minimum theoretical bandwidth necessary to propagate a signal is
called the minimum Nyquist bandwith or the minimum Nyquist frequency.

Thus,
Where

fb = 2B
fb = bit rate (bps)
B = ideal Nyquist bandwidth (hertz)
Mathematically,
Baud = 1  f b (baud per second)
ts
N
where
ts = time of one signaling element (second)
N = number of bits per signal element
fb = bit rate (bps)
In analog Tx of digital data, the baud rate is less than or equal to the bit rate.
- In binary system such as binary FSK and binary PSK, baud and bits per second are equal.
However, in higher-level system such as QPSK and 8-PSK, bps is always greater than baud.
Bit Rate, Baud rate and Minimum BW
Summary of the Terms:
Term
Units
Definition
Data element
Bits
A single binary one or zero
Data rate
Bits per second (bps)
The rate at which data
elements are transmitted.
Signal element
Digital: a voltage pulse of constant
amplitude
Analog: a pulse of constant frequency,
phase and amplitude
That part of a signal that
occupies the shortest
interval of a signaling code
Signaling rate or
modulation rate
Signal elements per second (baud)
The rate at which signal
elements are transmitted.
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Example 1 – An analog signal carries four bits in each
signal element. If 1000 signal elements are sent per
second, find the baud rate and bit rate.
Example 2 – The bit rate of a signal is 3000. If each signal
element carries 6 bits, what is the baud rate?
9
Types of digital-to-analog conversion
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Amplitude Shift Keying (ASK) or On-Off
Keying (OOK)



A simple version of amplitude modulation used for digital
modulation.
Both freq and phase remain constant while the amplitude changes.
Uses logic levels in the data to control the amplitude of the carrier
wave.
‘1’ for high amplitude (switch ON)
‘0’ for low amplitude (switch OFF).
11
Amplitude Shift Keying (ASK) or On-Off
Keying (OOK)
Basic implementation of Binary ASK
12
Amplitude Shift Keying (ASK) or On-Off
Keying (OOK)

ASK Modulator
- The modulator cct has 2 inputs:
1. data to be transmitted
2. high freq carrier sinewave
At the Tx, let the input of a
data stream is 0110001011
13
Amplitude Shift Keying (ASK) or On-Off
Keying (OOK)

At the Rx, the data stream need to extracted:
Step 1 – Rectify the input ASK
waveform to contain only +ve
signal but it will still contain
unwanted carrier wave
component.
Step 2 – Pass through a LPF
to remove the carrier
component.
Step 3 – Pass through a
voltage comparator to get a
true copy of the original
data stream
14
Amplitude Shift Keying (ASK)

Example 3 – Determine the baud and minimum bandwidth
necessary to pass a 10 kbps binary signal using amplitude shift
keying.
Solution
For ASK, N=1
Bmin = fb/N
Bmin = 10k/1 = 10kHz
Baud = fb/N = 10kbaud/sec
15
Amplitude Shift Keying (ASK)

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.
The baud rate and the bit rate are also the same for ASK,
the bit rate is 5000bps.
16
Frequency Shift Keying (FSK)

Use logic levels in the data to control the frequency of the carrier
wave.
 Data = ‘1’ for high frequency
 Data = ‘0’ for low frequency
17
FSK Bit Rate, Baud and Bandwidth



It can be seen that the time of one bit (tb) is the same as the time the FSK
output is a mark or a space frequency (ts).
Thus the bit time equals the time of an FSK signaling element and the bit
rate equals the baud.
The baud for binary FSK can also be determined by substituting N = 1,
baud =
fb
N

fb
1
 fb
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FSK Bit Rate, Baud and Bandwidth

The minimum bandwidth for FSK is given as
B  ( f s  fb )  ( fm  fb )
 f s  fm  2 fb
f s  fm  2f
And since
Then the minimum bandwidth can be approximated as
(**)
B  2(f  f )
b
Where B = minimum Nyquist bandwidth (hertz)
f = frequency deviation (|fm - fs|) (hertz)
fb = input bit rate (bps)
Note that equation (**) resembles Carson’s rule for determining the approximate
bandwidth for an FM wave. The only difference in the two equations is that, for FSK,
the bit rate, fb is substituted for the modulation signal freq fm.
FSK Bit Rate, Baud and Bandwidth

Example 5 – A binary FSK with a mark frequency of 49kHz, a space
frequency of 51kHz and an input bit rate of 2kbps, determine
a. The peak frequency deviation
b. The minimum bandwidth
c. Baud
Solution
a.
From 2f =
|fm - fs|,
and f = |49kHz – 51kHz|/2 = 1kHz
b. Min BW, B = 2(f + fb) = 2(1000+2000) = 6kHz
c. For FSK, N = 1, the baud is fb/N =2000/1 = 2000
20
FSK Bit Rate, Baud and Bandwidth

Bessel function can also be used to determine the approximate
bandwidth for an FSK wave.
The fastest rate of change
i.e highest fundamental freq
occurs when alternating 1s
and 0s are occuring.
Therefore,
Where
fa 
fb
2
fa = highest fund freq (hz)
fb = input bit rate (bps)
The formula used for modulation index in FM is also valid for FSK, thus
h
f
(unitless)
fa
Where
h = FM modulation index called h-factor in FSK
f = peak freq deviation (Hz)
FSK Bit Rate, Baud and Bandwidth


The worst-case modulation index (deviation ratio) yields the widest BW.
The widest BW occurs when both the freq deviation and the modulating
signal freq are at their maximum values, thus
or
Thus, bandwidth,
B = 2(n x fa)
FSK Bit Rate, Baud and Bandwidth

Example 6 – A binary FSK with a mark frequency of 49kHz, a space
frequency of 51kHz and an input bit rate of 2kbps, using Bessel
table, determine
a. The modulation index, h
b. The bandwidth
Solution
a. h = |49kHz – 51kHz|/2kbps = 1
b. From a Bessel table, for modulation index of 1, n = 3,
then
B = 2(3 x 1000)
= 6kHz
23
Frequency Shift Keying (FSK)
 There are many different ways of
generating an FSK waveform.
 One way is by combining 2 different
ASK waveform/modulator.
24
Frequency Shift Keying (FSK)

Lets assume that the above data stream is applied to an ASK
modulator using the higher freq as the carrier. The resulting
output:
 Inverting the original data stream:
25
Frequency Shift Keying (FSK)

This inverted data stream will be the input to another ASK
modulator using a lower carrier freq - the original data 0 periods
filled with a lower freq carrier.
26
Frequency Shift Keying (FSK)

Summing amplifier is used to add the two ASK waveforms:
Output from
Modulator 1
Output from
Modulator 2
27
Frequency Shift Keying (FSK)

Advantage of FSK over ASK – higher reliability in term of data
accuracy.

Disadvantage – requires higher BW (the actual increase depends on
the 2 freqs used). The higher the freq and the more they differ
from each other, the wider the BW required.
28
29
Multiple FSK (MFSK)

A signal that is more bandwidth efficient, but also more
susceptible to errors is multiple FSK (MFSK), in which more than
two frequencies are used. (i.e each signaling element represents more than
one bit)

The transmitted MFSK signal for one signal element time can be
defined as follows:
s i ( t )  A cos 2 f i t ,
Where
1 i  M
f i  f c  ( 2i  1  M )  f
fc = the carrier frequency
∆f = the difference frequency
M = number of signal element (2n)
n = number of bits per signal element
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
Example 7– With fc = 350kHz, ∆f = 20kHz, and M=8
(n=3), the following frequency assignments for each of
the eight possible 3-bit data combinations:
fi
Frequency
Assignment
Freq (kHz)
f1
f2
f3
f4
f5
f6
f7
f8
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




Example – Figure shows an example of MFSK with M = 4.
An input bit streams of 20 bits is encoded 2 bits at a time, with each of
the four possible 2-bit combinations transmitted as a different frequency.
The display in the figure shows the frequency transmitted (y-axis) as a
function of time (x-axis).
Each column represents a time Ts in which a single 2-bit signal element is
transmitted.
The shaded rectangle in the column indicates the frequency transmitted
during that time unit.
32
Phase Shift Keying (PSK)


In PSK, the phase of the carrier is shifted to represent data.
Two-Level PSK(Binary PSK) - BPSK
 Uses two phases (0 and 180°) to represent the two binary
digits.
 The resulting transmitted signal for one bit time is:
 A cos( 2  f c t )
s (t )  
 A cos( 2  f c t   )
 A cos( 2  f c t )
 
  A cos( 2  f c t )
binary 1
binary 0
33
Phase Shift Keying (PSK)


Example: binary 1 is represent with a phase 0°, while binary 0 is
represented with a phase of 180°.
PSK is equivalent to multiplying the carrier by +1 when the info is
1, and by -1 when the info is 0.
Bipolar
NRZ
binary ‘1’
binary ‘0’
34
Phase Shift Keying (PSK)

PSK Transmitter – same modulator as in ASK system
 Since a sinewave is symmetrical, it is impossible for the Rx to know
whether signal is inverted form or not.
 Thus need to apply some data conditioning to the incoming stream to
convert it to a form which recognizes logic levels by changes that occur and
not by the absolute levels.
 One such code is bipolar NRZ.
 The amplitude of the carrier is controlled by the bipolar signal on the
modulation input. When the signal goes negative, the sinewave inverts. 35
PSK Bit Rate, Baud and Bandwidth

Mathematically, the output of a BPSK modulator is proportional to
BPSK output  [sin( 2  f a t )]  [sin( 2  f c t )]

Solving for the trig identity for the product of two sine function,
1
2

cos[ 2  ( f c  f a ) t 
1
2
cos[ 2  ( f c  f a ) t
Thus, the minimum double-side Nyquist BW, B is
B  f usf  f lsf

But f a 
then
fb
 ( fc  fa )  ( fc  fa )  2 fa
,
2
 fb 
B  2
  fb
 2 
PSK Bit Rate, Baud and Bandwidth

Example 8 – For a BPSK modulator with a carrier frequency of 70
MHz and an input bit rate of 10Mbps,
determine the maximum and minimum upper and lower side frequencies
b. Draw the output spectrum
c. Determine the minimum bandwidth
d. Calculate the baud.
a.
37
Solution
a.
b. Output spectrum
c.
d.
39
Quadrature PSK (QPSK)




The term “quadrature” implies that there are four possible phases
(4-PSK) which the carrier can have at a given time.
The pair of bits represented by each phase is called dibit.
The rate of change (baud) in this signal determines the signal
bandwidth.
BUT the throughput or bit rate for QPSK is twice the baud rate.
40
QPSK = 4-PSK
Assumption
41
Modulation for Data Communication: QPSK
 One
way to increase the binary data rate while not
increasing the bandwidth required for the signal
transmission is to encode more than 1 bit per phase
change.
 In the system known as quadrature, quarternary, or
quadra phase PSK (QPSK or 4-PSK), more bits per
baud are encoded, the bit rate of data transfer can
be higher than the baud rate, yet the signal will not
take up additional bandwidth.
 In QPSK, each pair of successive digital bits in the
transmitted word is assigned a particular phase.
 Each pair of serial bits, called a dibit, is represented
by a specific phase.
Quadrature PSK modulation.
(a) Phase angle of carrier for different pairs of bits. (b) Phasor
representation of carrier sine wave. (c) Constellation diagram
of QPSK.
QPSK Bit Rate, Baud and Bandwidth

Example 9 – The CCITT V.22 (like Bell 212A) modem uses QPSK to
send data at 1200 bits per second; What is the baud rate.
The CCITT V.22 (like Bell 212A) modem uses QPSK to send data
at 1200 bits per second; however, the phases change only 600
times per second, conveying two bits per change - this is a 600
baud modem.
44
QPSK Bit Rate, Baud and Bandwidth

Example 10 – For a QPSK system, with the following
input bit sequence 100010101101, an input bit rate
equal to 20Mbps,
a. Draw
the QPSK modulated waveform, state assumption
used.
b. Determine
the minimum bandwidth required.
45
QPSK Bit Rate, Baud and Bandwidth
Solution eg. 9 – 100010101101

Assumption:
a.
b.
B = fb/N
B = 20M/2 = 10MHz.
46
QUADRATURE AMPLITUDE MODULATION
-Uses both amplitude and phase modulation of the carrier;
not only are different phase shifts produced but also the
amplitude of the carrier is varied
8 – QAM
Is an M–ary encoding technique where M=8
47
8-QAM
Figure : 8-QAM Modulator a) Truth Table b)phasor diagram c)
Constellation diagram
48
8-QAM
Figure : Output phase and amplitude versus-time realtionship
for 8-QAM
Min. bandwidth required = fb/3
49
Quadrature Amplitude Modulation
A combination of ASK and PSK: both phase and amplitude varied
#amplitude shifts << #phase shifts
Lower susceptible to noise than ASK, higher bit rate than PSK
50
8-QAM
51
16-QAM
Better susceptible to noise because
-Not all possibilities are used
-Sometimes, Amp and Phase have a relationship
52
Example 12 - Calculate the bit rate and baud rate of BPSK
and 8-PSK if the bit duration, Tb and signaling time, Ts
are both 0.01msec.
BPSK
Bit rate = baud rate
= 1/0.01
= 100kbps
Or 100kbaud/sec
8-PSK
baud rate = 1/0.01
= 100kbaud/s
Then bit rate, fb
= 3x100k
=300kbps
53
MORE EXAMPLE :
Example 13: An analog signal has a bit rate of 8000 bps
and a baud rate of 1000 baud. How many data elements
are carried by each signal element? How many number
of conditions ?
Example 14: We need to send data 3 bits at a time at a
bit rate of 3 Mbps. The carrier frequency is 10 MHz.
Calculate the number of levels (different frequencies)
and the baud rate
 Example 15 - For a 16-QAM modulator with input data rate
fb =10 Mbps, and a carrier frequency of 70 MHz, determine
the minimum bandwidth and the baud rate
 Example 16 – For the following modulation schemes,
construct a table showing the number of bits encoded,
number of output conditions, min. bandwidth, and baud for
an information data rate of 12 kbps; QPSK, 8-PSK, 8QAM,16-PSK and 16 QAM
Modulation
n
M
B(Hz)
baud
QPSK
8-PSK
8-QAM
16-PSK
16 QAM
55
Bandwidth Efficiency
Used to compare the performance of one digital modulation
technique to another
Ratio of the transmission bit rate to the minimum
bandwidth required for a particular modulation scheme.
B 
transmissi onbitrate ( bps )
min .bandwidth ( Hz )
Example 17 – For an 8-PSK system, operating with an
information bit rate of 24kbps, determine:
a) baud rate
b) min. bandwidth
c) bandwidth efficiency
56
Summary

ASK
 demodulation: only the presence or absence of a sinusoid in a
given time interval needs to be determined
 advantage: simplicity
 disadvantage: ASK is very susceptible to noise interference–
noise usually (only) affects the amplitude, therefore ASK is the
modulation technique most affected by noise
 application: ASK is used to transmit digital data over optical
fiber
57
Summary

FSK
 demodulation: demodulator must be able to determine which
of two possible frequencies is present at a given time
 advantage: FSK is less susceptible to errors than ASK –receiver
is looking for specific frequency changes over a number of
intervals, so voltage (noise) spikes can be ignored
 disadvantage: FSK spectrum is 2x ASK spectrum
 application: over voice lines, in high-frequency radio
transmission, etc.
58
Summary

PSK
 demodulation: demodulator must be able to determine the
phase of received sinusoid with respect to some reference
phase
 advantage: (i) PSK is less susceptible to errors than ASK, while
it requires/occupies the same bandwidth as ASK; (ii) more
efficient use of bandwidth (higher data-rate) are possible.
 disadvantage: more complex signal detection / recovery
process, than in ASK and FSK.
59
Exercise – past year Qs
1. There are quite a number of signaling techniques used in digital pulse modulation. Explain
the difference between unipolar NRZ signaling and unipolar RZ signaling. [2 marks]
2. Given a binary data stream of 11101010, draw the pulse train for unipolar NRZ signaling
and unipolar RZ signaling. [3 marks]
3. What is the major disadvantage in using NRZ encoding? How does RZ encoding attempt to
solve the problem? [2 marks]
4. For a data stream 10011101, draw the output waveform for Manchester signaling
technique. Assume the output level is initially LOW. [2 marks]
5. An analog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second,
find the baud and the bit rate. [2 marks]
6. Frequency Shift Keying (FSK) is a modulation technique where the frequency of the
carrier signal is modulated with respect to the input data stream. There are several methods
in generating FSK waveforms. Given a data bit stream of 1011000110, explain fully with
the help of diagrams, how we can generate the FSK waveform for the data stream given. [9
marks]
7. What is baud rate? [1 mark]
8. Calculate the bit rate and baud rate of BPSK and 8-PSK if the bit duration, Tb and
signaling time, Ts are both 0.01msec. [2 marks]
9. For a data stream 1010 1110, draw the output waveform for Manchester signaling
technique. Assume the output level is initially LOW. [2 marks]