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4.1. Digital Modulation (APM)

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0% found this document useful (0 votes)
24 views21 pages

4.1. Digital Modulation (APM)

Uploaded by

Amin Najm
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PPT, PDF, TXT or read online on Scribd
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COMMUNICATION SYSTEMS

BTB35203

Digital Modulation
Analogue Pulse Modulation

01/21/25 1
Introduction
oIn the early 90’s, telecommunication networks is changing
towards digital world. With the rapid advancement in the
fields of VLSI and microprocessor, several telecommunication
components (e.g. transmission line channel) has been using
digital signals in their operation.
oTherefore, information signals must be changed to digital
form so that it can be transmitted through this network.

January 21, 2025 2


Why digital modulation???
 The RF spectrum must be shared, yet every day there are
more users for that spectrum as demand for
communications services increases.
 Digital modulation schemes have greater capacity to
convey large amounts of information than analog
modulation schemes

01/21/25 3
21/01/25 4
Why digital modulation???
 Advantages :
 Immunity to noise (due to its finite process)
 Easy storage and processing
 Regeneration
 Easy to measure
 Enables encryption
 Error correction detection can be utilized

 Disadvantages :
 Requires a bigger bandwidth
 Analog signal need to be changed to digital first
 Not compatible to analog system
 Need synchronization

01/21/25 5
Digital Modulation Chart
Communication
System

Continuous Wave Digital Wave

Amplitude Angle Analogue Pulse Digital Pulse


Modulation Modulation Modulation Modulation
(AM)

Frequency Pulse
Modulation Modulation
(FM) (PM)

DSBFC DSBSC Vestigial SSB

01/21/25 6
Pulse Modulation

 Pulse modulation includes many difference methods of


converting information into pulse form for transferring
pulses from a source to a destination.

 Pulse modulation
 Analog Pulse Modulation (APM)
 Digital Pulse Modulation (DPM)

 Pulse modulation can be used to transmit analogue


information, it is first converted into pulses by the
process of sampling.

01/21/25 7
Sampling
 Sampling is the process of taking a periodic sample of the
waveform to be transmitted.

 The sampling theorem (Nyquist theorem) is used to


determine minimum sampling rate for any signal so that
the signal will be correctly restored at the receiver.

 Nyquist-Shahnon Sampling theorem:

f s 2 f m
Where fs = sampling frequency
fm(max) = maximum frequency of the modulating signal

01/21/25 8
Sampling Theorem

Message signal Modulated signal

Sample pulse

A process of periodically sampling the continually


changing analog input voltage and convert it to a series of
constant amplitude pulses

01/21/25 9
Sampling

Three basic condition of sampling process:

1. Sampling at fs=2fm(max)

V (volt)

f (Hz)
fs 2fs 3fs
fm(max) fs+fm(max)
fs-fm(max)

01/21/25 10
Sampling
2. Sampling at fs>2fm(max)
Shannon sampling
V (volt)
theorem=> fs  2fm
Guard band

Nyquist frequency
 fs = 2fm= fN
f (Hz)
fs 2fs
fm(max) fs-fm(max) fs+fm(max)
This sampling rate creates a guard band between fm(max) and the lowest
frequency component fs-fm(max) of the sampling harmonics. Guard band is
an unused part of the radio spectrum between radio bands, for the
purpose of preventing interference.
a narrow frequency range used to separate two wider frequency ranges
to ensure that both can transmit simultaneously without interfering each
other.
used in Frequency division multiplexing
01/21/25 11
Sampling
3. Sampling at fs<2fm(max)

V (volt)
Aliasing distortion

f (Hz)
fs 2fs 3fs
fs-fm(max) fs+fm(max)
fm(max)
Aliasing: the distortion produced by the overlapping
components from adjacent bands
Is an effect that causes different signals to become
indistinguishable (or aliases of one another) when sampled
Aliasing occurs when a signal is sampled below its Nyquist rate

01/21/25 12
Analogue Pulse Modulation Chart

Communication
System

Continuous Wave Digital Wave

Analogue Pulse Digital Pulse


Modulation Modulation

PAM PWM PPM

01/21/25 13
Analog Pulse Modulation (APM)
 In APM, the carrier signal is in the form of pulse
form, and the modulated signal is where one of
the characteristics either (amplitude, width,
or position) is changed according to the
modulating/audio signal.

 Three common techniques of APM:


 Pulse Amplitude modulation (PAM)
 Pulse Width Modulation (PWM)
 Pulse Position Modulation (PPM)

01/21/25 14
Analogue Pulse Modulation
1) pulse amplitude modulation (PAM)
 the amplitude of a constant width, constant-position pulse is
varied according to the amplitude of the sample of the analog
signal.

2) pulse width modulation (PWM)


 sometimes called pulse duration modulation (PDM) or pulse
length modulation (PLM), as the width (active portion of the
duty cycle) of a constant amplitude pulse is varied proportional
to the amplitude of the analog signal at the time the signal is
sampled.

3) pulse position modulation (PPM)


 the position of a constant-width pulse and constant amplitude
within a prescribed time slot is varied according to the
amplitude of the sample of the analog signal.

01/21/25 15
Waveforms for PAM, PWM and PPM

Modulating signal

carrier signal

PAM
(dual polarity)

PWM

PPM

01/21/25 16
Pulse Position Modulation

01/21/25 17
Pulse Amplitude Modulation

2 types of sampling:

1. Natural Sampling

tops of the sample pulses retain their natural shape,


making it difficult for ADC to convert to PCM codes

2. Flat-top Sampling

input voltage is sampled with narrow pulses and then held


relatively constant until next sampling

01/21/25 18
m(t)

Information signal t

s(t)

Pulse signal t
Ts
Sampled signal (PAM)
ms(t)  ms(t)
   Ts 
t t
Ts Ts

Natural Sampling Flat-top


Sampling
Example 1: Example 2:

 A television signal  Ifthe sample have


has a message 1024 levels,
frequency of 4.5 determine the
MHz. Determine the number of bit
sampling rate if the required to encode
signal is to be each sample
sampled at a rate
20% above the
Nyquist rate

January 21, 2025 20


Solution 1: Solution 2:

2N =1024
fs = 2fm= 9MHz
N= log 1024 / log 2 = 10 bit
20% ((M) = 1.8 MHz
Therefore,
Sampling rate = 9 MHz + 1.8 MHz

=10.8 MHz

January 21, 2025 21

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