Analog and Digital Communication Prepared According to R 2017

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ANALOG AND DIGITAL

COMMUNICATION

Prepared according to Anna university syllabus R-201

7 (Common to III semester-CSE/IT )

G. E

lumalai

,

M.E.,(Ph.D)

Assistant Professor (Grade I)

Department of Electronics and Communication Engineering

Panimalar Engineering College

Chennai.

Er

. m. J

aiGanEsh

,

M.E.,

Assistant Professor

Department of Electronics and Communication Engineering

Panimalar Engineering College

Chennai.

SREE KAMALAMANI PUBLICATIONS

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Publised by SREE KAMALAMANI PUBLICATIONS. New No. AJ. 21, old No. AJ. 52, Plot No. 2614, 4th_C_ross_{, 9}th_M_ain_R_oad_{, A}_nna_N_agar_{-600 040,}

Chennai, Tamilnadu, India Landline: 91-044-42170813, Mobile: 91-9840795803

EMAilid: skmpulbicationsmumdad@gmail.com 1ST EdiTioN 2014

2Nd REViSEd EdiTioN 2016

No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording or otherwise or stored in a database or retrieval system without the prior written permission of the publishers.

This edition can be exported from India only by the Publishers, Sree Kamalamani Publications.

ISBN (13 digits): 978-93-85449-12-3

Typeset & Coverpage :

Sree Kamalamani PublicationS

New No. AJ. 21, Old No. AJ. 52, Plot No 2614, 9th_{Main, 4}th_{cross, Anna Nagar-600 040}

Chennai, Tamilnadu, India.

lANdliNE: 91-044-42170813, MobilE: 91-9840795803

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G.Elumalai M.E., is working as an Assistant Professor (Grade – I) in the Department of Electronics and Communication Engineering, Panimalar Engineering College, Chennai. He obtained his B.E. in Electronics and Communication Engineering; M.E. in Applied Electronics and Ph.D pursing in Wireless Sensor Network. His areas of interests are Communication System, Digital communication, Digital signal processing and Wireless Sensor Network. He has more than 13 years of experience.

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Dear Students,

We are extremely happy to present the book “Analog and Digital Communication” for you. This book has been written strictly as per the revised syllabus (R2013) of Anna University. We have divided the subject

into five units so that the topics can be arranged and understood proper -ly. The topics within the units have been arranged in a proper sequence

to ensure smooth flow of the subject.

Unit I - Introduce the basic concepts of communication, need of modulation and different types of analog modulation (Amplitude modu-lation, Frequency modulation and Phase modulation).

Unit II - Deals with basic concepts of digital communication which includes ASK, FSK, PSK, QPSK and QAM.

Unit III - Discuss about concept of data communication and various pulse modulation technique.

Unit IV - Concentrate on various techniques for error control cod-ing.

Unit V – Describe about multiuser radio communication.

A large number of solved university examples and university questions have been included in each unit, so we are sure that this book will cater all your needs for this subject.

We have made every possible effort to eliminate all the errors in

this book. However if you find any, please let we know, because that will

help for us to improve further.

G.Elumalai

M.Jaiganesh

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UNIT I ANALOG COMMUNICATION

Noise: Source of Noise - External Noise- Internal Noise- Noise Calculation. Introduction to Communication Systems: Modulation – Types - Need for Modulation. Theory of Amplitude Modulation - Evolution and Description of SSB Techniques - Theory of Frequency and Phase Modulation – Comparison of various Analog Communication System (AM – FM – PM).

UNIT II DIGITAL COMMUNICATION

Amplitude Shift Keying (ASK) – Frequency Shift Keying (FSK) Minimum Shift Keying (MSK) –Phase Shift Keying (PSK) – BPSK – QPSK – 8 PSK – 16 PSK - Quadrature Amplitude Modulation (QAM) – 8 QAM – 16 QAM – Bandwidth Efficiency– Comparison of various Digital Communication System (ASK– FSK – PSK – QAM).

UNIT III DATA AND PULSE COMMUNICATION

Data Communication: History of Data Communication - Standards Organiza-tions for Data Communication- Data Communication Circuits - Data Communication Codes - Error Detection and Correction Techniques - Data communication Hardware - serial and parallel interfaces. Pulse Communication: Pulse Amplitude Modulation (PAM) –Pulse Time Modulation (PTM) – Pulse code Modulation (PCM) - Comparison of various Pulse Communication System (PAM – PTM – PCM).

UNIT IV SOURCE AND ERROR CONTROL CODING

Entropy, Source encoding theorem, Shannon fano coding, Huffman coding, mutual information, channel capacity, channel coding theorem, Error Control Coding, linear block codes, cyclic codes,

convolution codes, viterbi decoding algorithm.

UNIT V MULTI-USER RADIO COMMUNICATION

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Unit

1 ANALOG COMMUNICATION

1.1 INTRODUCTION

Communication is the process of establishing connection (or link) between two points for information exchange.

The science of communication involving long distances is called telecommunication ,the word tele stands for long distance

The information can be of different type such as sound, picture, music computer data etc.,

The basic communication components are

t A Transmitter

t A communication channel or medium and

t A receiver

1.1.1 Elements of communication system:

The block diagram of elements of communication system is as

shown in figure 1.1

Information Source

Transmitter Channel Receiver Destination

Noise and Distortion

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The elements of basic communication system are as follows

1. Information or input signal

2. Input transducer

3. Transmitter

4. Communication channel

5. Noise

6. Receiver

7. Output transducer

Information or input signal

• The communication system has been developing for communicating useful information from one place to the other.

• This information can be in the form of a sound signal like speech or music, or it can be in the form of pictures or it can be data informa-tion coming from a computer.

Input transducer

The information in the form of sound, picture and data signals cannot be transmitted as it is.

Input transducer is used to convert the information signal from source into suitable electrical signal .

The input transducer used usually in the communication systems are microphones, TV camera etc..

Transmitter

The function of transmitter block is to convert the electrical equivalent of the information to a suitable form corresponding to communicate through communication medium (or) channel.

The transmitter consists of the electronic circuits such as modulator,

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In addition to that it increases the power level of the signal. The power level should be increased in order to cover a large range.

Communication channel

The communication channel is the medium used for trans-mission of electronic signal from one place to the other. The

communication medium can be conducting wires, cables, optical fibre or

free space. Depending on the type of communication medium, two types of communication systems will exist. They are:

• Wire communication (or) line communication

• Wireless communication (or) radio communication

Noise

Noise is an unwanted electrical signal which gets added to the transmitted signal when it is travelling towards the receiver

Due to noise, quality of the transmitted information degrades. Once added, the noise cannot be separated out from the information.

Hence noise is a big problem in the communication systems.

Even though noise cannot be completely eliminated, its effect can be reduced by using various techniques.

Receiver

The reception is exactly the opposite process of transmission. That is extract original signal form transmitted signal.

The receiver consists of the electronic circuits such as

demodula-tor, amplifier, mixer, oscillator and power amplifier.

Output transducer

The output transducer converts the electrical signal at the output of the receiver back to the original form (i.e) Sound, picture and data signals.

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1.2 NOISE

Noise is an unwanted signal that interferes with the desired message signal.

In audio and video systems electrical disturbances are appearing as interference is called as noise.

In general noise may be predictable or unpredictable (random) in nature.

Predictable noise

The predictable noise can be estimated and eliminated by proper engineering design.

The predictable noise is generally man made noise and it can be eliminated easily.

Examples: power supply hum, ignition radiation pickup,

spurious oscillations in feedback amplifiers, fluorescent lightening.

Unpredictable noise

This type of noise varies randomly with time, and we have no control over this noise.

The term noise is generally used to represent random noise.

Presents of random noise ,complicate the communication system Sources of noise

1. Internal noise

2. External noise

Internal noise may be classified as

1. Shot noise

2. Thermal Noise

3. Partition Noise

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External Noise may be classified as

1. Natural Noise

2. Manmade Noise

1.2.1 Natural noise

This type of noise randomly occurs in atmosphere due to lightning, electrical storms and other atmospheric disturbances. This noise is unpredictable in nature.

This noise is also called as atmospheric noise (or) static noise 1.2.2 Manmade Noise

Manmade noise results from undesired pickups from electrical appliances, such as motors, automobiles and aircraft ignition etc.,

This type of noise can be eliminated by removing the source of

the noise. This noise is effective in frequency range of 1 MHz - 500 MHz

1.2.3 Internal noise

Internal noise is created by active and passive components present within the communication system

1.2.3.1 Shot noise

Shot noise present in active devices due to random fluctuation of

charge carriers crossing the potential barriers. In electron tubes, shot noise is generated due to random emission from cathodes.

In semi-conductor devices, it is caused due to random diffusion of minority carriers (or) random generation of recombination of electron hole.

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_{Shot noise has a flat response spectrum. The mean squared}

noise component is proportional to the DC-flowing and for most

of the devices the mean square shot noise current is given by,

I_n2 _{= 2I}

oqeBn amperes ...(1)

Where

I₀ = DC in amperes

q_e = Magnitude of electron charge (1.6 x 10-19_C)

B_n = Equivalent noise Bandwidth 1.2.3.2 Thermal noise

? This type of noise arises due to random motion of electrons in a conducting medium such as a resistor, and this motion in

turn is randomized through collisions caused by imperfection

in the structure of conductors. The net effect of motion of all

electrons constitutes an electric current flowing through the

resistor, causing the noise

This noise is also known as resistor noise (or) Johnson noise.

? The power density spectrum of the current contributing the thermal noise is given by

S_i ω KTG

ω α

( )

= + 

  _ 2

1

2 ...(2)

Where,

T- Ambient temperature in degree kelvin

G- Conductance of the resistor in mhos

K - Boltzman constant

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1.2.3.3 Partition noise

? This noise is generated whenever a current has to divide between two (or) more electrodes and results from random

fluctuation in the division.

? It would be expected therefore that a diode would be less noisy than a transistor, if third electrode draws current.

? For this reason, the input stage of microwave receiver is often a

diode circuit. The spectrum of the partition is flat

1.2.3.4 Flicker noise (or) low frequency noise

Flicker noise occurs due to imperfection in cathode surface of electron tubes and surface around the junctions of semiconductor

devices. In the semiconductor, flicker noise arise from fluctuation in the carrier density, which in turn give rise to fluctuation in the conductivity of the material. The power density of the flicker noise is inversely

proportional to frequency (ie) S (w) a 1

f

.

Hence, this noise becomes significant at very low frequencies (below a few KHz)

1.2.4 Calculation of noise

i. Signal to noise Ratio (SNR)

It is defined as the ratio of signal power to noise power either

input side (or) at output side of the circuit (or) device

SNR_i = Signal power at the input Noise power at the input

Output Signal power Output Noise power

SNR₀ =

ii. Noise Figure

Noise figure is defined as, the ratio of the signal to noise power

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to noise power ratio supplied to the output terminal (or) load resistor (SNR₀)

Therefore,

Noise figure (F) = _(SNR)(SNR)i 0

Calculation of Noise Figure

Voltage gain

A

R_i R_L

V₀

Amplifier (receiver) Generator (Antenna)

Figure 1.1 (a) Block Diagram of calculation of noise figure

Calculate noise figure consider a network shown in figure 1.1(a).

The network has the following

1. Input impedance R_t 2. Output impedance R_L 3. An Overall voltage gain

It is led from a source that is antenna of internal resistance R_a. The internal resistance R_a, may or may not be equal to R_t. The figure

1.1(a) shows the block diagram of such 4 terminals network.

The calculation procedures are as follows

Step 1: Determination of input signal power ‘P_st’

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power P_si as

Substituting equation (1) in (2) we get,

P_si = Vsi

Step 2: Determination of input noise power ‘P_ni’

Similarly the noise input voltage V_ni and power P_nican be

Substitute V_ni value here, we get

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Step 3 Calculation of input SNR

SNR_i = Psi

P_ni

Using equation (3) and (5), we get

SNR_i = (Vsi

2_R

t/(Ra + Rt) 2₎

4KTB (R_a/ R_a + R_t)

= Vsi

2_.R t

4KTB. R_a (R_a + R_t) ...(6) Step 4: Determination of signal output power P_so

The output signal power will be given as,

P_so = Vso

2

R_L

(AV_si)2

R_L =

P_so =

A2_.V si

2

R_L ...(7)

Substitute equation (1) in (7), we get

P_so =

A2_(V

siRt/ RaRt) 2

R_L

P_so = A

2_V si

2_.R t

2

R_L(R_a + R_t)2 ...(8)

Step 5 Determination of noise output power P_no

The noise output power may be quite difficult to calculate for

instance, it can be simply written as,

P_no = output noise power ...(9)

Step 6 Calculation of the output SNR

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SNR₀ = P_so P_no

Using equation (8) and (9) we get

SNR₀ =

Step 7 Calculation of Noise figure (F)

The general expression for noise figure is

F = SNRi

SNR₀

Using equation (6) and (10), we get

F V R

This is the necessary equation.

1.3 INTRODUCTION TO COMMUNICATION SYSTEMS

Electronics communication system can be classified into various

categories based on the following parameters

1. Whether the system is unidirectional (or) bidirectional

2. Whether it uses an analog (or) digital information signal

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Electronics communication systems

Unidirectional/ Bidirectional communication

Nature of Information

signal

Technique of transmission

Simplex system

Half duplex

Analog Digital Baseband transmission

Communication using modulation Full

Duplex

Figure 1.2 Classification of communication system

1.3.1 Classifications based on directions of Communication

Based on whether the system communicates only in one direction

(or) otherwise, the communication systems are classified as,

1. Simplex system

2. Half duplex systems

3. Full duplex systems

Communication System

Unidirectional

(Simplex)

Bidirectional

(Duplex)

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Simplex system

In these systems the information is communicated in only one direction , they cannot receive.

For example, the radio, TV-broadcasting and telemetry System of a satellite to earth.

Half duplex system

These systems are bidirectional they can transmit as well as receive but not simultaneously.

At a time these systems can either transmit (or) receive, for example a trans-receiver (or)walky talky set.

Full duplex System

These are truly bidirectional systems as they allow the communication to take place in both the direction simultaneously.

These systems can transmit as well as receive simultaneously , for example the telephone Systems.

Transmitter + Receiver 1

Transmitter + Receiver 2 Bidirectional flow

of information

Communication

link

Figure 1.3 Basic Block diagram of full duplex system

1.3.2 Classifications based on the nature of Information signal

Based on nature of information signal, Communication system

classified into two categories namely,

1. Analog Communication system.

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Analog Communication

In this communication technique, the transmitted signal is in the form of analog (or) continuous in nature through the communication channel (or) media.

Digital communication

In this communication technique, the transmitted signal is in the form of digital pulses of constant amplitude, frequency and phase.

1.3.3 Classification based on the technique of transmission

Based on the technique used for the signal transmission. we can categories into two namely,

1. Base band transmission.

2. Communication systems using modulation.

Base-band transmission

In this technique, the baseband signal (original information signals) are directly transmitted.

Examples of these type of systems are telephone networks where the sound signal converted into electrical signal is placed directly on the telephone lines for transmission (local calls).

Another example of baseband transmission is computer data transmission over a Co-axial Cables in the computer networks (eg. RS 232 cables).

Thus , the base band transmission is the transmission of the original information signal as it is.

Limitations of Baseband transmission

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free space.

This is because the Voice signal (in the electrical form) cannot travel long distance in air.

It gets suppressed after a short distance. Therefore for the radio Communication of baseband signals a technique called “Modulation” is used.

Drawbacks of baseband transmission (without modulation)

1. Excessively large antenna heights.

2. Signals get mixed up.

3. Short range of communication.

4. Multiplexing is not possible and

5. Poor quality of reception.

Why modulation

The baseband transmission has many limitations which can be overcome using modulation.

In radio communication, signals from various sources are transmitted through a common medium that is in open (free) space .This causes interference among various signals, and no useful message is received by the receiver.

The problem of interference is solved by translating the message signals to different radio frequency spectra. This is done by the transmitter by a process known as ”Modulation”.

1.4 MODULATION

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Modulating signal Baseband signal Low frequency signal

Carrier signal High frequency signal (or)

(or)

Modulating signal Modulator Modulated signal

Carrier signal

Modulation is the process of changing the characteristics of carrier signal (such as amplitude, frequency and phase) in accordance with the instantaneous value of modulating signal.

In simple, modulation is the process of mixing of modulating signal and carrier signal together.

In the process of modulation , the baseband signal is translated (i.e) shifted from low frequency to high frequency.

1.5 NEED FOR MODULATION (OR) ADVANTAGES OF MODULATION

The advantages of modulation are,

(1) Easy of radiation.

(2) Adjustment of bandwidth.

(3) Reduction in height of antenna.

(4) Avoids mixing of signals.

(5) Increases the range of information.

(6) Multiplexing and

(7) Improves quality of reception.

1.5.1 Easy of radiation

As the signals are translated to higher frequencies,

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antenna systems at these increased frequencies.

1.5.2 Adjustment of bandwidth

Bandwidth of a modulated signal may be made smaller (or) larger than the original signal.

Signal to noise ratio (SNR) in the receiver which is a function of the signal Bandwidth can thus be improved by proper control of bandwidth at the modulating stage.

1.5.3 Reduction in antenna height

When free space is used as a communication media, messages are transmitted with the help of antennas.

If the signals are transmitted without modulation, the size of

antenna needed for an effective radiation would be of the order of the half of the wavelength, given as,

= λ

2

c

2f ...(1) In broadcast systems, the maximum audio frequency transmitted from a radio station is 5 KHZ. Therefore,the antenna height required is,

= λ 2

c 2f

c 2 x 5 x 103

3 x 108 10 x 103

= = = = 30 km

The antenna of this height is practically impossible to install.

Now consider a modulated signal f=10 MHZ. The minimum

antenna height is given by,

Antenna height is = λ 2

c 2f =

= 3 x 10

8 2 x 10 x 106 = 15 metre

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1.5.4 Avoid mixing of signals

Each modulating signal (message signal) is modulated with different carrier then they will occupy different slot in the frequency domain (different channels).Thus modulation avoids mixing of signals.

1.5.5 Increases the range of communication

The modulation process increases the frequency of the signal to be transmitted. Hence, increases the range of communication.

1.5.6 Multiplexing

If different message signals are transmitted without modulation through a single channel may causes interference with one another. (i.e) overlap with one another.

To overcome this interference means, we need n-number of channels for n-message signals separately.

But different message signals can be transmitted over a same channel (single channel) without interference using the techniques “Multiplexing”.

Simultaneous transmission of multiple message (more than one message) over a channel is known as “multiplexing”.

Due to multiplexing, the number of channels needed are less. This reduces the cost of installation and maintenance of more channels.

1.5.7 Improves quality of reception

Due to modulation, the effect of noise is reduced to great extent. This improves quality of reception.

The two basic types of communications systems are analog and digital.

Analog communication Message - continuous signal

Carrier - continuous signal

Message - Digital (or) analog signal

Carrier - continuous signal (analog)

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1.6 CLASSIFICATIONS OF MODULATION

Modulation

Analog modulation Digital modulation

Amplitude- modulation(AM)

Angle modulation Continuous

modulation

Analog pulse

modulation DPCM DM ADM PCM

PAM PWM PPM

Phase modulation

(PM)

Frequency modulation

(FM)

Figire 1.4 Classifications of Modulation

Where,

PAM - Pulse amplitude modulation.

PWM - Pulse width modulation.

PPM - Pulse Position modulation.

PCM – Pulse code modulation.

DM – Delta modulation.

ADM – Adaptive delta modulation.

DPCM – Differential Pulse code modulation.

Linear modulation

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theorem of spectra is known as linear modulation system.

Non-linear modulation

The modulation system which does not follow the superposition theorem of spectra is known as non-linear modulation system.

1.7 SOME IMPORTANT DEFINITIONS RELATED TO

COMMUNICATION

1.7.1 Frequency(f)

The frequency is defined as the number of cycles of a waveform per second. It is expressed in hertz (Hz).

Frequency is simply the number of times a periodic motion, such as a sine wave of voltage (or) current, occurs in a given period of time.

Time

1 Cycle

Amp

Figure 1.5 One cycle

1.7.2 Wave length (λ)

Wave length (λ ) is defined as the distance between two points of

similar cycles of a periodic wave.

Wavelength

Figure 1.6 Wavelength

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electromagnetic wave during the time of one cycle.

1.7.3 Bandwidth

Bandwidth is defined as the frequency range over which an

information signals is transmitted .

Bandwidth is the difference between the upper and lower frequency limits of the signal.

Bandwidth (BW) = f₂f₁

Where, f₂ – upper frequency f₁ – lower frequency

f₁ f₂ Frequency BW

1.7.4 Transmission frequencies

The total usable radio frequency (RF) spectrum is divided into narrower frequency bands, which are descriptive names and several of these band are further broken down into various types of services.

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Frequency deSignation

Frequencyrange

Wavelength range Extremely High

frequency (EFH) 30 - 300 GHZ 1mm - 1cm super High frequency

(SHF) 3 - 30 GHZ 1 - 10 cm

Ultra High

Frequency (UHF) 300MHZ -3GHZ 10cm - 1m Very High frequency

(VHF) 30 - 300MHZ 1 -10m

High frequency(HF) 3 - 30MHZ 10-100m Medium frequency (MF) 300KHZ - 3MHZ 100m-1km

Low frequency (LF) 30 KHZ - 300KHZ 1km - 10km

Table 1.1 The radio frequency spectrum

Solved Problem

1. Find the wavelength of a signal at each of the following frequencies. (1) 850 MHZ (2) 1.9 GHZ (3) 28 GHZ.

Solution

Given data

(1)f = 850 MHZ

(2)f =1.9 GHZ and (3) f = 28 GHZ.

Wavelength ‘ λ ‘ = Velocity of light

Frequency

= c

f

(i) l = 3 x 10

8

850 x 106 = 0.35 M

(ii) l = 3 x 10

8

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(iii) l = 3 x 10

8

28 x 109 = 0.0107 m

1.7.5 Frequency spectrum

Frequency spectrum is the representation of a signal in the frequency domain . It can be obtained by using either fourier series (or) fourier transform.

It consists of the amplitude and phase spectrums of the signal. The frequency spectrum indicates the amplitude and phase of various frequency components present in the given signal.

The frequency spectrum enables us to analyze and synthesize a

signal.

1.7.6 Demodulation (or) Detection

The process of extracting a modulating (or) baseband signal from the modulated signal is called “demodulation”.

In other words , Demodulation (or) detection is the process by which the message signal is recovered from the modulated signal at receiver.

1.8 THEORY OF AMPLITUDE MODULATION

Definition

Amplitude modulation (AM) is the process by which amplitude of the carrier signal is varied in accordance with the instantaneous value (amplitude) of the modulating signal, but frequency and phase remains constant.

1.8.1 Mathematical Representation of an AM wave

Let us consider,

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V_c __ _{Amplitude of the carrier signal (volts)} V_m_ Amplitude of the modulating signal (volts)

ωm _ Frequency of the modulating signal (HZ) ωc _ Frequency of the carrier signal (HZ)

According to the definition of amplitude modulation, the

amplitude of the carrier signal is changed after modulation with respect to message signal,

V_AM(t) = (V_c+V_m(t))sinω_ct ...(3) Substitute the value of V_m(t) in equation (3) we get,

V_AM(t) = (V_c+V_msinω_mt)sinω_ct

= V V V

t t

c

m

c

m c

1+  

 

  sinω sinω

Where, Vm

V_c = ma = modulation index.

Modulation index is defined as the ratio of amplitude of message

signal to amplitude of carrier signal.

Modulation index ‘m_a’= Vm V_c

V_AM(t) = V_c[1+ m_a sinω_mt] sinω_ct ...(4) V_AM(t) = V_c[1+m_asin (2f_m)t] sin (2f_c)t ...(4)a

The equation (4)a represents the time domain representation of an AM-signal.

1.8.2 AM – voltage distribution

The time domain representation of an AM – signal is given by,

V_AM(t)= V_c[1+ m_a sinω_mt]sinω_ct ...(1)

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sin A . sin B = cos(A-B) - cos(A+B) 2

Equation (2) becomes,

V_AM(t) = V_csinω_ct +

m_aV_c

2 cos (ωc -ωm)t- m_aV_c

2 cos(ωc+ ωm)t ...(3)

In equation (3),

• First term represents carrier signal (volts).

• Second term represents lower side band signal (volts)

• Third term represents upper sideband signal (volts).

The Figure 1.7 shows the voltage spectrum for an AM – DSBFC

wave (or) AM – signal.

m_aV_c 2

m_aV_c 2 V_c

V_c

f_LSB f_c f_USB

frequency

Voltage (V)

Figure 1.7 Voltage spectrum for an AM – DSBFC wave

1.8.3 Frequency spectrum of AM

The AM wave is given by,

V_AM(t) =V_c(1+m_asin ω_mt) sin ω_ct ...(1)

= V_c(1+ m_a sin 2f_mt) sin 2f_ct

=V_csin 2f_ct + m_aV_c sin 2pf_ct sin 2f_mt

= maVc

2

m_aV_c

2

carrier _USB

LSB

{

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The (-) sign associated with the USB – represents a phase shift of

180 .The figure 1.8 shows the frequency domain representation.

m_aV_c

2

m_aV_c

2

V_c

f_c+ f_m f_c-f_m fc

f_m _f

m

Frequency

Amplitude

BW=2f_m

Figure 1.8 Frequency domain representation of AM-wave.

The equation (2) shows the frequency domain representation of AM- signal.

• First term represents the unmodulated carrier signal with the fre-quency of f_c

• Second term represents the lower sideband signal with the frequency of ( f_c- f_m ).

• Third term represents the upper sideband signal with the frequency of ( f_c + f_m ).

1.8.4 Bandwidth of AM

The bandwidth of the AM-signal is given by the subtraction of the highest frequency component and the lowest frequency component in the frequency spectrum.

BW = Bandwidth = f_USB- f_LSB

= ( f_c + f_m ) - ( f_c - f_m ) = f_c+ f_m- f_c+f_m

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Where, BW = Bandwidth is hertz.

f_m = Highest modulation frequency in Hertz.

f_c = Highest carrier frequency in Hertz.

Thus, the Bandwidth of the AM-signal is the twice that of the maximum frequency of modulating signal.

1.8.5 AM- Envelope (or) Graphical representation of AM-wave

AM- DSBFC is sometimes called conventional AM (or) simply AM.

AM is simply called as Double sideband Full carrier (DSBFC) is

probably the most commonly used. The figure 1.9 shows the graphical

representation of AM – signal.

• The shape of the modulated wave (AM) is called AM –envelope which contains all the frequencies and is used to transfer the information through the systems.

• An increase in the modulating signal amplitude causes the amplitude of the carrier to increase.

• Without signal, the AM output waveform is simply the carrier signal.

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{

No message signal _{Modulation signal}

time

Modulated signal AM-DSBFC No modulation

Carrier signal V_m

V_m

-V_m

V_m(t)

V_Am(t)

V_c(t) -V_m

V_c

-V_c -V_c

0

0 Amplitude

0

Figure 1.9 AM envelope

(40)

1.8.6 Phasor Representation of an AM-wave

The amplitude variation in an AM-system can be explained with

the help of a phasor diagram as shown in figure 1.10

o Carrier

Resultant AM phasor

m_aV_c 2

LSB USB

ωm

-ωm

V_AM(t) V_c

Figure 1.10 Phasor representation of AM-wave

• The phasor for the upper sideband rotate anticlockwise at an angular frequency of w_m, faster than the carrier frequency ω_c (i.e) (ω_m>ω_c).

• The phasor for the lower sideband rotates clockwise at an angular frequency of w_m, slower than the carrier frequency (ω_c) (i.e)(ω_m<ω_c).

• The resulting amplitude of the modulated wave at any instant is the vector sum of the two- sideband phasors.

• V_c is carrier wave phasor, taken as reference phasor and the resulting phasor is V_AM(t).

• The phasors for carrier and the upper and lower side frequencies combine, sometimes in phase (adding) and sometimes out of phase (subtracting).

1.8.7 Modulation index and percentage modulation

Modulation index

(41)

Modulation index ‘m_a’ = V_m V_c

Modulation index is used to describe the amount of amplitude change (modulation) present in an AM – Waveform .

Percentage Modulation

When modulator index is express in percentage, it is called percent modulation.

Percentage modulation gives the percentage changes in the amplitude of the output wave when the carrier is acted on by a modulating signal.

Percentage modulation = Peak amplitude of modulating signal

Peak amplitude of carrier signal x 100

= V_m

V_c x 100 = m_a x 100

1.8.8 Calculation of modulation index from AM-Waveform

V_max = V_c +V_m

V_max = V_c -V_m V_c

V_m _V

min = Vc -Vm

Amplitude

0

Figure 1.11 AM-Envelope for Calculation of modulation index

(42)

From the figure. 1.11 the maximum and minimum amplitude of

AM-signal is represented by,

V_max = V_c+ V_m ...(1)

V_min = V_c - V_m ...(2)

From equation (1) - (2) we obtained

V_max- V_min = V_c+ V_m -V_c + V_m V_max - V_min = 2V_m

V_max = 2V_m + V_min ...(3)

(or)

V_m = Vmax- Vmin

2 ...(4)

From equation (1) and (2) , the Vc can be calculated as, Equation (1) becomes,

V_c = V_max - V_m ...(5)

Equation (2) becomes,

V_c = V_max + V_m...(6)

From equation (5) v_c can be calculated by substitute V_m value,

\ V_c = V_max - Vmax −Vmin 

 

  2

=

2 V_max -V_max + V_min 2

\ V_c =

V_max + V_min

2 ...(7)

According to the modulation index definition, the modulation

(43)

According to the modulation index definition, the modulation index for AM is given by,

M_a =

V_m

V_c ...(8) Substitute V_m, V_c values in this equation (8)

m_a =

V_max- V_min/2 V_max+ V_min/2

Vmax- Vmin

V_max+ V_min

\m_a = ...(9)

Percentage of modulation index in terms of V_max & V_min can be

ex-pressed as,

V_max- V_min V_max+ V_min

% m_a = x 100

...(10)

The Peak changes in the amplitude of the output wave v_m is the sum of the voltages from the upper & lower side frequencies.

V_m 2

V_USB = = V_LSB ...(11)

Substituting V_m value in equation (11) we get,

\ V_USB= Vmax- Vmin/2

2 = VLSB

V_USB= V_LSB=

V_max- V_min

4 ...(12)

Where,

(44)

1.8.9 Modulation index for multiple modulating signal frequency

When more than one modulating signals are modulated by a

single carrier, then the modulation index is given by,

m_a(t) (or) m_a_{= m}

1 2_+m

2 2_+m

3 2_... Where,

m_a(or) m_a(t) → Total modulation index.

m₁,m₂... → Modulation index due to individual modulating components.

1.8.10 Degree of modulation

In AM, there are three types of modulations are available. It

depends upon the amplitude of the modulating signal relative to carrier amplitude.

(1) Under modulation.

(2) Critical modulation.

(3) Over modulation.

(1) Under modulation

If V_m<V_cthen, the modulation index m_a<1

• In this type of modulation , the envelope of amplitude modulated

signal does not reach the zero axis. Hence the message signal is fully preserved in the envelope of the AM wave.

(45)

Amplitude

Envelope

time

Figure 1.12 AM wave for M_a <1 (2) Critical modulation

If V_m = V_c , then the modulation index m_a=1

In this type of modulation , the envelope of amplitude modulated signal just reaches the zero amplitude axis. The message signal remains preserved.

V_m

V_c

Amplitude

Time

Figure 1.13 AM wave for m_a=1

(46)

(3) Over modulation

If V_m > V_c ,then the modulation index m_a > 1

• Here , both positive and negative extensions of the modulating signals are cancelled (or) clipped out.

Amplitude

time

Distortion due to Over Modulation

Figure 1.14 AM- wave for m_a>1 V_m

V_c 0

• The envelope of message signal are not same. Due to this enve-lope detector provides distorted message signal.

• According to the three types of modulation,m_a<1, m_a=1 ,m_a>1. The modulating signals are fully preserved without any distorted from the modulated signal if and only if V_m <V_c then M_a<1

1.8.11 AM-Power distribution

According to the voltage distribution of an AM wave, it consist of three components namely carrier, lower side band and upper side band.

(47)

Total transmitted power for AM- signal is given by

P_t = P_c+ P_LSB+ P_USB ...(1) Carrier Power (P_c)

The average power dissipated in a load by an unmodulated carrier is equal to the rms carrier voltage squared divided by the load resistance.

P_c = V

2 carrier

R ...(2)

P_c = (Vc/√2)

2

R

P_c = Vc

2

2R ...(3)

Where,

P_c = Carrier power (watts). V_c = Peak carrier voltage (volts).

R = Load resistance (ohms).

Power in the sidebands

The upper and lower sideband power are expressed mathematically is,

P_USB = P_LSB =

(V_sub/ 2)2

R ...(4)

The peak voltage of the upper and lower side band frequencies,

V_SUB = maVc

2 ...(5)

Substitute equation (5) in (4)

P_USB = P_LSB = [maVc/2] 2) R

(

)

2

P_USB = P_LSB = m_a2_V

c 2

(48)

Where,

Total power in AM -wave

Substitute equation (3) and (8) into equation (1), we get

P_t = P_c+P_LSB + P_USB

The equation (9) shows the total power required for transmission of AM-signal or DSBFC. The figure 1.15 shows the power spectrum for

(49)

V_c2

Figure 1.15 Power spectrum for an AM-wave

For critical modulation m_a =1; If m_a=1 equation (9) becomes

Total power is 1.5 times of carrier power is for critical modulation.

(50)

2 P 1

I_t = Total transmit current (ampere).

I_C = Carrier current (ampere)

R = Antenna resistance (ohms).

We know that, the total transmit power is given by,

P_t =P_c 1

(51)

1.8.14 Modulation index in terms of current

Transmission efficiency of an AM-wave is defined as the ratio of

the transmitted power which contains the information (i.e side band power) to the total transmitted power.

(52)

=

m_a2_/2 m_a2

2

1 +

x 100

₌ ma 2_/2

(2+m_a2_)/2

x 100 = ma 2

2+m_a2 x 100

m_a2

2 +m_a2 x 100 % η =

For critical modulation m_a =1, then the transmission efficiency becomes,

if m_a=1, % η = 12

2+12 x 100 = 1

3 x 100

% η = 33.3 %

The maximum transmission efficiency of the AM-DSBFC is

33.3%.This means , that only one-third of the total power is carried by the sidebands and the rest two-third is a wasted power.

1.8.16 Total transmitted power for AM-DSBSC

Total power required for transmission of AM- DSBSC is depends only on sideband power, because carrier is suppressed.

\P_t= P_LSB +P_USB ...(1)

(\Carrier is suppressed)

Sideband powers

The upper and lower sidebands are having equal power which is equal to,

P_USB = P_LSB= (Vsub)

2 R

(V_sub/ 2)2 _R

= ...(1)a

(53)

V_sub = maVc 2

\ Equation (1) becomes

P_USB = P_LSB = m Va c/ 2

Substitute equation (2) in (1)

P_t = P_LSB+P_USB

For critical modulation, m_a=1 P_t = P_C 1

2

( )

P_t = 0.5 P_c ...(4)

1.8.17 Transmission efficiency (η) (for AM-DSBSC)

(54)

=

The maximum transmission efficiency of AM-DSBSC is 100%. This means the total transmitted power is fully utilized by the AM-

DSBSC. There is no power is wasted.

1.8.18

Comparison of different AM techniques

S.No Parameters DSBFC DSBSC SSB

1. Carrier suppression

Not applicable Fully Fully

2. Sideband suppression

Not applicable Not applicable

6 Complexity Simple Simple Complex

7 Power

requirement to cover are

High Medium Very small

8 Application Radio

1.8.19 Advantages,disadvantages and applications of AM Advantages

• AM is used for long distance communications.

• AM is relatively inexpensive. Disadvantages

(55)

• Receiver cost and complexity.

• 66.67% of transmitted power is wasted.

• Large bandwidth is required. Applications

• Sound and audio broadcasting.

• Point to point link communication

• Aircraft communications in the VHF frequency range. Problems

1. What is the bandwidth needed to transmit 4 KHZ voice signal using AM.

Solution

Bandwidth required for AM = 2f_m BW =2 f_m

BW = 2(4 KHZ)

BW =8 KHZ

2. In an AM –transmitter , the carrier power is 10 KW and the modulation index is 0.5 .calculate the total RF-power delivered.

Solution

Given data

P_c =10 KW

m_a = 0.5

The total RF-power delivered P_t = P_c 1 2

2 +  

 

  m

a

1 0 5

2 2 +  

 

  .

=10 KW

(56)

3. The output of AM-transmitter is given by

U_AM(t)=500(1+0.4 sin 3140 t)sin 6.28 x 107 _t

Calculate

(i) Carrier frequency (ii) Modulating frequency (iii) Modulation index (iv) Carrier power if the load is 600 W (v) Total power.

Solution

The output of AM-DSBFC is given by,

V_AM(t)= V_c(1+ m_a sinω_mt)sinω_ct ...(1) Compare equation (1)with the given output equation.

We get,

V_c =500 volts

m_a =0.4

ωm = 3140 ωc = 6.28 x 10

7 (i) ωm = 2fm = 3140

f_m = 3140

2 499.7 f_m = 500 HZ

2f_c = 6.28 x 107 ₌w c

(ii) w_c = 2pf_c = 6.28 x 107

f_c = 6.28 x10

7 2

f_c = 0.999 x 107_HZ

(57)

(iii) Modulation index m_a = 0.4

(iv) Carrier power if the load is 600 Ω

P_c _{= =}Vc 2

2R

5002

2 x 600

P_c = 208.3 watts

(v)Total Power (P_t) = P 1 2

2 +  

 

  m

a

= 208.3 1 0 4

2 2 +  

 

  .

= 224.9

Pt = 225 watts

The total power required for AM is 225 watts.

4. For an AM-DSBFC modulator with a carrier frequency

of 100 KHZ and maximum modulating signal frequency of 5 KHZ.

Determine upper and lower side band frequency and the Bandwidth.

Solution

Given data:

f_c =100KHZ f_m =5 KHZ

(1)Upper sideband frequency

fUSB =f_c +f_m

=100 KHZ + 5 KHZ

=105 KHZ

(58)

f_lSb =f_c - f_m

=100 KHZ - 5 KHZ

=95 KHZ

(3)Bandwidth ‘BW’ =2 f_m

=2(5 KHZ)

=10 KHZ

5. In an AM-Modulator, 500 KHZ carrier of amplitude 20 V is modulated

by 10 KHZ modulating signal which causes a change in the output

wave of ± 7.5 V. Determine

(i) Upper and lower sideband frequencies

(ii) Modulation index

(iii) Peak amplitude of upper and lower side frequency.

(iv) Maximum and minimum amplitudes of envelope.

Solution

Given data

f_c= 500 KHZ , f_m=10 KHZ, V_c= 20 V,V_m=15 V (i) f_USB = f_c+f_m

= 500 K +10 K

= 510 KHZ

f_LSB = f_c- f_m

= 500 K +10 K

(59)

(ii) Modulation index m_a = Vm

V_c

= 15

20

= 0.75

(iii) Peak amplitude of upper and lower side frequencies

= maVc

2

= 0.75 x 20

2

= ± 7.5 V

(iv). Maximum and minimum amplitude of envelope.

V_max = V_c+ V_m

= 20 + 15

= 35 V

V_min = V_c - V_m = 20 - 15

= 5 V

6. If a 10V carrier is amplitude modulated by two different frequencies

with amplitudes 2 V and 3 V respectively. Find modulation index.

Solution

Given data

V

c = 10 V, Vm1 = 2V, Vm2 = 3V

Modulation index m₁ = Vm1

V_c =

2

(60)

= 0.2

Modulation index m₂ = Vm1

V_c =

3

10 = 0.3

Total modulation index ‘m_a’ = _m

12 +m2 2 0.04 +0.09

=

m_a = 0.66

7. An AM-signal has the equation V(t)=[15 + 4 sin 44 x 103_{t] (sin 46.5 x} 106_{t)V. Find}

(i)carrier frequency (ii)modulating frequency (iii)modulation index (iv)sketch the signal in the time domain, showing voltage and time scales.(v) Peak voltage of unmodulated carrier.

Solution

V(t) =[15+4 sin 44 x 103 t] sin(46.5 x 106 _t) =15[1+ 4

15 sin 44 x 103 t](sin 46.5 x 106 t)

=15[1+ 0.26 sin 44 x 103 t] (sin 46.5 x 106 t) ...(1)

For AM-DSBFC output is given by

V(t) =V_c[1+m_a sin ω_mt]sin ω_ct ...(2)

Compare equation (1) and (2) we get,

V_c =15 V

m_a = 0.26

ωm = 44 x 10 3_HZ ωc =46.5 x 10

6_HZ

(i) Carrier frequency ‘f_c'

2fc = 46.5 x 10

6

_f

c =

46.5 x106 2

_f

c =7.4 x 10

(61)

(ii) Modulating frequency ‘f_m’

2fm =44 x 10

3

_f

m =

44 x 103 2

fm = 7 x 103 _HZ

(iii) Modulation index ‘m_a’=0.26

(iv) Peak voltage of unmodulated carrier is =15 V

(v) AM-Voltage spectrum.

Voltage maVc

2 =1.95 V

m_aV_c

2 =1.95 V

Vc = 15 V

LSB USB

Frequency

f_LSB fC fUSB

Fig 1.16 AM-Voltage Spectrum

8. For a modulation co-efficient = 0.2 and an unmodulated, carrier

power P_c=1000 W. Determine

(i)Total sideband power (ii)Upper and lower sideband power(iii) modulated carrier power (iv)Total transmitted power.

Solution

Given Data

(62)

Un modulated carrier power =1000 W

(i) Upper & lower sideband power,

P_USB =P_LSB= ma (ii) The total sideband power is,

P_TSB = P_USB + P_LSB

(iii)The total power required for transmission of AM-wave,

P_c =P_C 1

(iv) Modulated Carrier Power,

(63)

1000W = P_c

Pc =1000 watts

9. A 200 W carrier is modulated to a depth of 75%. Calculate the total

power in the modulated wave.

Solution

Given data

P_c =200 W

% m_a =75% \m_a=0.75 Total power in the modulated wave,

P_t = P_c 1 2

2 +  

 

  m

a

= 200 1 0 75

2

+  

 

  .

P_t = 256.25 watts

10. For an AM-DSBFC wave with an unmodulated carrier voltage of 18V

and a load resistance of 72 W. Determine the following.

(i) Unmodulated carrier power (ii) Modulated carrier power (iii) Total sideband power (iv) Upper & lower sideband power (v) Total transmitted power.

Solution

Given

V_c =18V

(64)

(i) Unmodulated carrier power, (ii) Total sideband power,

P_TSB =

(iii) upper &lower sideband power,

P_USB =P_LSB =

20 KW, for the modulation index is 0.6.

(65)

12. For an AM-DSBSC modulator with a carrier frequency fc = 100 KHz and a maximum modulating signal frequency of 5 KHz. Determine.

(i) frequency limits for the upper and lower sidebands.

(ii) Bandwidth

(iii) sketch the output frequency spectrum.

Solution

(i) Frequency limits for upper & lower sidebands,

f_USB = f_c+ f_m=100 KHz+5 KHz = 105 KHz f_LSB = f_c- f_m=100KHz-5KHz =95KHz

(ii) Bandwidth ‘BW = 2f_m = 2 x 5 KHz

=10 KHz

Voltage

BW = 10 KHz

LSB USB

f_LSB = 95 KHz fUSB = 105 KHz

(66)

13. For an AM-DSB-FC with a Peak unmodulted carrier voltage Vc=12V,

(iii) Total modulated power,

(67)

14. A modulating signal 20 sin (2 x 103 _{t) is used to modulate a carrier} signal 40 sin (2 x 104 _{t).Find out,}

(i) Modulation index

(ii) Percentage modulation

(iii) Frequencies of the sideband components and their amplitudes.

(iv) Bandwidth of the modulating signal

(v) Also draw the spectrum of the AM-wave.

Solution

Given, modulating signal,

V_m(t) =20 sin(2 x 103 t) ...(1)

Carrier signal,

V_c(t) = 40 sin (2 x 104 _t₎_...(2)

(i) Modulating signal, is represented by

V_c(t) = V_m sin (2f_mt) ...(3)

Compare , equation (1) & (3),

V_m= 20 V, f_m=103 HZ =1 KHZ

(ii) Carrier signal is represented by,

V_c(t) = V_c. sin (2f_ct) ...(4) Compare equation (2) and (4)

V_c= 40V, f_c=104 HZ =10 KHZ

(iii) Modulation index,m_a= = Vm

V_c =

20

(68)

(iv) Frequency of sideband components,

Upper sideband frequency f_USB= f_c+ f_m

f_USB = f_c+ f_m= 10+1 = 11 KHz

Lower sideband frequency,

f_USB = f_c-f_m= 10-1= 9 KHz

(v) Amplitudes of sidebands

The amplitudes of upper as well as the lower sideband is given by,

= maVc

2

= 0.5 x 40

2

=10 volts

(vi) Bandwidth BW =2f_m

=2 x 1 KHZ =2 KHz

(vii) Spectrum

Amplitude

10 V 10 V

40 V

9 KHz 10 KHz 11 KHz

Frequency in KHz

(69)

1.9 GENERATION OF SSB

To generate the single side band suppressed carrier (SSBSC), we have to suppress the carrier as well as one of the side bands. The various techniques to suppress one of the side bands are

1. Filter method (or) Frequency discrimination method

2. Phase shift method (or) Hartley Method

3. Third Method (or) Weaver’s Method

1.9.1 Frequency Discrimination method

In Frequency discrimination method, first a DSBSC signal is

generated by using an ordinary product modulator or balanced modulator. Then, this DSBSC signal is passed through

suitable band pass filter to obtain SSBSC signal, where, one of the sidebands is filtered out.

The design of band pass filter is very critical and there are some

limitations on the modulating and carrier frequencies. They are

1. The base band is approximately related to carrier frequency. it is

very difficult to design a band pass filter, If the carrier frequency

is much greater than the bandwidth of the baseband or modu-lating signal.

2. The frequency discrimination method is useful only when the baseband is restricted at its lower edge, so that the upper side bands and lower side bands are non-overlapping.

The filter method is used in speech communication where lowest spectral component is 70Hz and it may not be taken as 300 Hz, without

(70)

Product

Figure 1.20 Frequency discrimination method Applications

Since the modulation and demodulation is complex, costlier, this system is not used for commercial broadcasting.

It is mainly used in wireless system for ultra-high frequency and very high frequency communication process.

1.9.2 Phase Discrimination (or) Phase shift method

The block diagram of the phase discriminator is shown in the

figure 1.21

In this method, there are two balanced modulators and two phase shifters used.

_{One modulator accepts carrier with 90° phase shift from carrier}

oscillator and modulating signal directly.

_{Another modulator accepts modulating signal with phase shift of 90}0

and the carrier signal directly.

_{Balanced modulator 1 accepts direct modulating signal}

V_m(t) = V_m sinwmt and 90° phase shifted carrier signal

V_c(t) = V_c sin (w_ct+ 90).

_{Balanced modulator 2 accepts 90° phase shifted modulating signal}

(71)

_{The output of balanced modulator 1 is also, output of balanced}

also, output of balanced modulator 2 is

==

(

+

)

Therefore, output of the sum will be

0 1 2

1. Each balanced modulator need to be carefully balanced in order to suppress the carrier.

2. Each modulator should have equal sensitivity to baseband signal. 3. It is difficult to design wide band phase shifting network.

Analog and Digital Communication Prepared According to R 2017