Ec8394 - Analog and Digital Communication unit I

KARPAGAM INSTITUTE OF TECHNOLOGY,
COIMBATORE - 105
Course Code with Name : EC8394 Analog and Digital Communication
Staff Name / Designation : Manoj Kumar T AP
Department : ECE
Year / Semester :II/III
Department of Information Technology

COURSE SYLLABUS
UNIT I ANALOG COMMUNICATION 9
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 Analog Communication Systems (AM – FM – PM).
UNIT II PULSE AND DATA COMMUNICATION 9
Pulse Communication: Pulse Amplitude Modulation (PAM) – Pulse Time Modulation (PTM) – Pulse code
Modulation (PCM) - Comparison of various Pulse Communication System (PAM – PTM – PCM). Data
Communication: History of Data Communication - Standards Organizations for Data Communication- Data
Communication Circuits - Data Communication Codes - Data communication Hardware - serial and
parallel interfaces.
UNIT III DIGITAL COMMUNICATION 9
Amplitude Shift Keying (ASK) – Frequency Shift Keying (FSK)–Phase Shift Keying (PSK) – BPSK –
QPSK – Quadrature Amplitude Modulation (QAM) – 8 QAM – 16 QAM – Bandwidth Efficiency–
Comparison of various Digital Communication System (ASK – FSK – PSK – QAM).
UNIT IV SOURCE AND ERROR CONTROL CODING 9
Entropy, Source encoding theorem, Shannon fano coding, Huffman coding, mutual information, channel
capacity, Error Control Coding, linear block codes, cyclic codes - ARQ Techniques.
UNIT V MULTI-USER RADIO COMMUNICATION 9
Global System for Mobile Communications (GSM) - Code division multiple access (CDMA) – Cellular Concept
and Frequency Reuse - Channel Assignment and Handover Techniques - Overview of Multiple Access
Schemes - Satellite Communication - Bluetooth.
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Karpagam Insitute of Technology

COURSE OBJECTIVE
• Understand analog and digital communication
techniques.
• Learn data and pulse communication
techniques.
• Be familiarized with source and Error control
coding.
• Gain knowledge on multi-user radio
communication.
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Karpagam Insitute of Technology

COURSE OUTCOME
At the end of the course, the student should be
able to:
Apply analog and digital communication
techniques
Use data and pulse communication techniques
Analyze Source and Error control coding
Utilize multi-user radio communication
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PROGRAM OUTCOMES
ENGINEERING GRADUATES WILL BE ABLE TO:
1. Engineering knowledge: Apply the knowledge of mathematics, science, engineering fundamentals, and
an engineering specialization to the solution of complex engineering problems.
2. Problem analysis: Identify, formulate, review research literature, and analyze complex engineering
problems reaching substantiated conclusions using first principles of mathematics, natural sciences, and
engineering sciences.
3. Design/development of solutions: Design solutions for complex engineering problems and design system
components or processes that meet the specified needs with appropriate consideration for the public health
and safety, and the cultural, societal, and environmental considerations.
4. Conduct investigations of complex problems: Use research-based knowledge and research methods
including design of experiments, analysis and interpretation of data, and synthesis of the information to
provide valid conclusions.
5. Modern tool usage: Create, select, and apply appropriate techniques, resources, and modern
engineering and IT tools including prediction and modeling to complex engineering activities with an
understanding of the limitations.

PROGRAM OUTCOMES - CONT
6. The engineer and society: Apply reasoning informed by the contextual knowledge to assess
societal, health, safety, legal and cultural issues and the consequent responsibilities relevant to the
professional engineering practice.
7. Environment and sustainability: Understand the impact of the professional engineering solutions in
societal and environmental contexts, and demonstrate the knowledge of, and need for sustainable
development.
8. Ethics: Apply ethical principles and commit to professional ethics and responsibilities and norms of the
engineering practice.
9. Individual and team work: Function effectively as an individual, and as a member or leader in
diverse teams, and in multidisciplinary settings.
10. Communication: Communicate effectively on complex engineering activities with the engineering
community and with society at large, such as, being able to comprehend and write effective reports and
design documentation, make effective presentations, and give and receive clear instructions.
11. Project management and finance: Demonstrate knowledge and understanding of the engineering and
management principles and apply these to one‘s own work, as a member and leader in a team, to manage
projects and in multidisciplinary environments.
12. Life-long learning: Recognize the need for, and have the preparation and ability to engage in
independent and life-long learning in the broadest context of technological change.

PROGRAM SPECIFIC OUTCOMES
1. To create, select, and apply appropriate techniques, resources, modern engineering and IT tools
including prediction and modelling to complex engineering activities with an understanding
of the limitations.
2. To manage complex IT projects with consideration of the human, financial, ethical and
environmental factors and an understanding of risk management processes, and operational
and policy implications.

COURSE OUTCOME Vs. PROGRAM OUTCOMES AND
PROGRAM SPECIFIC OUTCOMES MAPPING
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Course Outcomes PO1 PO2 PO3 PO4 PO5 PO6 PO7 PO8 PO9 PO10 PO11 PO12 PSO2 PSO2 PSO3
CO1
CO2
CO3
CO4

UNIT I
ANALOG COMMUNICATION

UNIT I : SYLLABUS
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 Analog Communication
Systems (AM – FM – PM).
UNIT I ANALOG COMMUNICATION

OBJECTIVES:
• Understand analog and digital communication techniques.
• Learn data and pulse communication techniques.
• Be familiarized with source and Error control coding.
• Gain knowledge on multi-user radio communication.

Introduction to Communication Systems
 Communication is the process of establishing connection (or link) between
two points for information exchange
 The Two basic types of communication systems are
(i) Analog – Transmitted signals are analog in nature
(ii) Digital - Transmitted signals are 0’s and 1’s or discrete levels in nature
Analog
Digital

Advantages of Analog communications
 Transmitters and Receivers are simple
 Low bandwidth requirement
 Frequency Division Multiplexing can be used
Disadvantages of Analog communication
 Noise affects the signal quality
 It is not possible to separate noise and signal
 Repeaters can‘t be used between transmitter
 Coding is not possible
 It is not suitable for the transmission of secret information

General Communication Systems:

Transmitter:
A collection of one or more electronic devices or circuits that converts the original
source information to a form more suitable for transmission over a particular
transmission medium.
Transmission Medium:
Provides a means of transporting signals between a transmitter and a receiver.
Receiver:
A collection of electronic devices and circuits that accepts the transmitted signals fro the
transmission medium and then converts those signals back to their original form.
System Noise:
Is any unwanted electrical signals that interfere with the information signal.

Drawbacks of Baseband Transmission
 Excessively large antenna heights.
 Signals get mixed up.
 Short range of communication.
 Multiplexing is not possible.
 Poor quality of reception.

Modulation
Process of changing characteristics of the carrier signal with respect to the
instantaneous change in message signal
In the modulation process some properties of the high frequency carrier wave is
varied in accordance with the modulating or message signal
Message Signal - Original message Carrier Signal - A high frequency
analog signal
Frequency = 1/time

Types of Modulation

Types of Modulation
(i) Amplitude (ii) Frequency (iii) Phase

Types of Modulation
Amplitude modulation Frequency modulation Phase modulation

Need for Modulation
 Reduction in antenna height
 Long distance communication
 Avoid mixing up of other signals
 Multiplexing
 Ease of radiation
 Improve the quality of reception

Theory of Amplitude Modulation
Amplitude of the carrier signal is changed with respect to the instantaneous
change in message signal.
Message Signal : em = Em sin ωmt
Carrier Signal : ec = Ec sin ωct
Em – Maximum amplitude of message signal
Ec - Maximum amplitude of Carrier signal
ωm - Frequency of message signal 2πfm
ωc - Frequency of carrier signal 2πfc
eAM = (Em sin ωmt + Ec )sin ωct

Modulation Index
The ratio of maximum amplitude of modulating signal to maximum
amplitude carrier signal is called modulation index.
Value of Em must be less than value of Ec to avoid any distortion in the
modulated signal . 0<=m=>1
Maximum value of modulation index will be equal to 1 when Em= Ec
When modulation index is expressed in percentage , it is called as
Percentage Modulation or Depth of modulation.

Calculation of modulation index from AM waveform:
𝐸𝑚 =
𝐸𝑚𝑎𝑥 − 𝐸𝑚𝑖𝑛
2
𝐸𝑐 = 𝐸𝑚𝑎𝑥 − 𝐸𝑚
= 𝐸𝑚𝑎𝑥 − (
2
)
=
2𝐸𝑚𝑎𝑥 – (𝐸𝑚𝑎𝑥 − 𝐸min ⁡
⁡
)
2
=
2𝐸𝑚𝑎𝑥 −𝐸𝑚𝑎𝑥 + 𝐸𝑚𝑖𝑛
2
=
𝐸𝑚𝑎𝑥 + 𝐸𝑚𝑖𝑛
2
𝑚 =
𝐸𝑚
𝐸𝑐
=
2
2
𝑚 =

𝑒𝐴𝑀 = (𝐸𝑐 + 𝐸𝑚 sin𝜔𝑚 𝑡) sin𝜔𝑐 𝑡
𝑒𝐴𝑀 = (𝐸𝑐 + 𝑚𝐸𝑐 sin𝜔𝑚 𝑡) sin 𝜔𝑐 𝑡 𝑚 =
𝐸𝑚
𝐸𝑐
𝑜𝑟 𝑚𝐸𝑐 = 𝐸𝑚
𝑒𝐴𝑀 = (𝐸𝑐 sin𝜔𝑐 𝑡 + 𝑚𝐸𝑐 sin 𝜔𝑚 𝑡 sin𝜔𝑐 𝑡)
𝑒𝐴𝑀 = 𝐸𝑐 sin 𝜔𝑐 𝑡 +
𝑚𝐸𝑐
2
cos 𝜔𝑚 𝑡 − 𝜔𝑐𝑡 −
𝑚𝐸𝑐
2
cos(𝜔𝑚 𝑡 + 𝜔𝑐𝑡)
sin𝐴 sin 𝐵 = 1/2 cos 𝐴 − 𝐵 − 1/2 cos(𝐴 + 𝐵)
𝑚𝐸𝑐
2
cos 𝜔𝑐𝑡 − 𝜔𝑚 𝑡 −
𝑚𝐸𝑐
2
cos(𝜔𝑐𝑡 + 𝜔𝑚 𝑡) cos 𝐴 − 𝐵 = cos(𝐵 − 𝐴)
𝑚𝐸𝑐
2
cos 𝜔𝑐 − 𝜔𝑚 𝑡 −
𝑚𝐸𝑐
2
cos(𝜔𝑐 + 𝜔𝑚 )𝑡
𝜔𝑚 = 2𝛱𝑓𝑚 𝑎𝑛𝑑 𝜔𝑐 = 2𝛱𝑓𝑐
𝑒𝐴𝑀 = 𝐸𝑐 sin 2𝛱𝑓𝑐𝑡 +
𝑚𝐸𝑐
2
cos 2𝛱𝑓𝑐 − 2𝛱𝑓𝑚 𝑡 −
𝑚𝐸𝑐
2
cos(2𝛱𝑓𝑐 + 2𝛱𝑓𝑚 ) 𝑡
𝑚𝐸𝑐
2
cos 2𝛱 𝑓𝑐 − 𝑓𝑚 𝑡 −
𝑚𝐸𝑐
2
cos2𝛱(𝑓𝑐 + 𝑓𝑚 ) 𝑡
𝑚𝐸𝑐
2
cos 2𝛱𝑓𝑙𝑠𝑏 𝑡 −
𝑚𝐸𝑐
2
cos2𝛱 𝑓𝑢𝑠𝑏 𝑡

(i) Amplitude of the carrier signal remains unaltered after modulation
(ii) Amplitude of upper side and lower side frequencies depends on both amplitude
of carrier and modulation index
(iii) At the beginning of each cycle of envelope, carrier is out of phase with the upper
and lower side frequencies
(iv) Upper and lower side frequencies are 1800 out of phase with each other
𝑉
𝑚𝑎𝑥 = 𝐸𝑐 +
𝐸𝑐
2
+
𝐸𝑐
2
=
2𝐸𝑐+𝐸𝑐+𝐸𝑐
2
=
4𝐸𝑐
2
= 2𝐸𝑐
𝑉𝑚𝑖𝑛 = 𝐸𝑐 −
𝐸𝑐
2
−
𝐸𝑐
2
=
2𝐸𝑐−𝐸𝑐−𝐸𝑐
2
= 0

Bandwidth of the
signal can be obtained
by taking the difference
between highest and
lowest frequencies
Frequency domain Representation of AM Wave

AM Time Domain Analysis
If 𝑓
𝑚 = 5 𝐻𝑧 , 𝑓
𝑐 = 25 𝐻𝑧 , 𝑚 = 1 𝑎𝑛𝑑 𝐸𝑐 = 1 𝑉
𝑒𝐴𝑀 = 𝐸𝑐 sin 2𝛱𝑓
𝑐𝑡 +
𝑚𝐸𝑐
2
cos 2𝛱 𝑓
𝑐 − 𝑓
𝑚 𝑡 −
𝑚𝐸𝑐
2
cos2𝛱(𝑓
𝑐 + 𝑓
𝑚 ) 𝑡
𝑒𝐴𝑀 = sin 2𝛱25𝑡 +
1
2
cos 2𝛱 20 𝑡 −
1
2
cos2𝛱(30) 𝑡

AM Power Distribution:
AM signal has three components : Unmodulated carrier, lower sideband and upper sideband. Hence total
power of AM wave is the sum of carrier power and powers in the two sidebands .
𝑃𝑡𝑜𝑡𝑎𝑙 = 𝑃𝑐 + 𝑃𝑙𝑠𝑏 + 𝑃𝑢𝑠𝑏 𝑃𝑜𝑤𝑒𝑟 =
𝑉𝑜𝑙𝑡𝑎𝑔𝑒 2
𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑉𝑜𝑙𝑡𝑎𝑔𝑒 =
𝐸
2
𝑃𝑐 =
(
𝐸𝑐
2
)2
𝑅
𝑃𝑙𝑠𝑏 =
(
𝐸𝑚
2 2
)2
𝑅
𝑃𝑢𝑠𝑏 =
(
𝐸𝑚
2 2
)2
𝑅
𝐸𝑚 = 𝑚𝐸𝑐
𝑃𝑡𝑜𝑡𝑎𝑙 =
𝐸𝑐
2
2𝑅
+
(
𝑚 𝐸𝑐
2 2
)2
𝑅
+
(
𝑚 𝐸𝑐
2 2
)2
𝑅
𝐸𝑐
2
2𝑅
+
𝑚2𝐸𝑐
2
8𝑅
+
𝑚2𝐸𝑐
2
8𝑅
𝐸𝑐
2
2𝑅
+ 2(
𝑚2𝐸𝑐
2
8𝑅
)
𝐸𝑐
2
2𝑅
+
𝑚2𝐸𝑐
2
4𝑅
𝐸𝑐
2
2𝑅
(1 +
𝑚2
2
)
𝑃𝑡𝑜𝑡𝑎𝑙 = 𝑃𝑐(1 +
𝑚2
2
)
𝑃𝑡𝑜𝑡𝑎𝑙
𝑃𝑐
= (1 +
𝑚2
2
) ;
𝑃𝑐
− 1 =
𝑚2
2
; 2
𝑃𝑐
− 1 = 𝑚2
𝑚 = 2(
𝑃𝑐
− 1)

Evolution and Description of SSB Techniques
Single sideband, SSB modulation is basically a derivative of amplitude
modulation, AM. By removing some of the components of the ordinary AM
signal it is possible.
Advantages of SSB
(i) Carrier is removed - it can be re-introduced in the receiver
(ii) Secondly one sideband is removed - both sidebands are mirror images
of one another and the carry the same information.

SSB – Single Side Band
SSBFC - Single Side Band Full Carrier
Half of the side band is transmitted
Requires less bandwidth than DSBFC
Power in the AM modulated wave is also halved
Repetition rate is equal to the frequency of the message signal

SSBSC - Single Side Band Suppressed Carrier
One sideband and carrier is removed
Requires only half of the bandwidth
Requires even less power

SSBRC - Single Side Band Reduced Carrier or Single Side Band reinserted Carrier
One side band is completely removed
Carrier voltage is reduced by 10 % of its unmodulated wave ie 0.1Ec
Carrier is totally suppressed and then reinserted at reduced voltage or
amplitude
Main purpose is for demodulation purpose
Carrier amplitude must be elevated to its original amplitude at receiver
Requires only half of the bandwidth
Also called as Exalted carrier transmission

ISB- Independent Side Band
Single carrier frequency is independently modulated by two different
message signal
Consists of two independent side band signals
Conserves both power and bandwidth

VSB – Vestigial Side Band
Part of side band is suppressed
One side band, Full carrier, part of side band is transmitted

Advantages
Power Conservation
Bandwidth Conservation
Selective Fading
Noise Reduction
Disadvantages
Complex Receivers
Tuning difficulties

Theory of Frequency and Phase Modulation
Deviation ratio: Deviation ratio is the worst case modulation index and is
equal to the maximum peak frequency deviation divided by the maximum
modulating signal frequency.
Carson’s rule : Carson‘s rule states that the bandwidth required to transmit an
angle modulated wave as twice the sum of the peak frequency deviation
and the highest modulating signal frequency.
B=2(δ +fm) Hz

Angle modulation can be expressed as e(t) = 𝐸𝑐 sin( 𝜔𝑐 𝑡 + 𝜃𝑡)
(i) For FM signal, the maximum frequency deviation takes place when
modulating signl is at positive and negative peaks.
(ii) For PM signal the maximum requency deviation takes place near zero
crossings of the modulating signal.
(iii) Both FM and PM waveforms are identical except the phase shit.
(iv) From modulated waveform it is difficult to know, whether the
modulation is FM or PM
Message Signal - 𝑒𝑚 = 𝐸𝑚 sin 𝜔𝑚 𝑡
Carrier Signal - 𝑒𝑐 = 𝐸𝑐 sin 𝜔𝑐 𝑡

PM wave : 𝑒𝑝𝑀 𝑡 = 𝐸𝑐 sin[𝜔𝑐 𝑡 + 𝐾𝑝𝑒𝑚 ]
PM wave : 𝑒𝑝𝑀 𝑡 = 𝐸𝑐 sin[𝜔𝑐 𝑡 + 𝐾𝑝𝐸𝑚 sin 𝜔𝑚 𝑡]
Modulation Index for PM: K𝐸𝑚
PM wave : 𝑒𝑝𝑀 𝑡 = 𝐸𝑐 sin[𝜔𝑐 𝑡 + 𝑚 sin 𝜔𝑚 𝑡]
FM wave : 𝑒𝐹𝑀 𝑡 = 𝐸𝑐 sin[𝜔𝑐 𝑡 + 𝐾𝑓 𝑒𝑚 ]
FM wave : 𝑒𝐹𝑀 𝑡 = 𝐸𝑐 sin[𝜔𝑐 𝑡 + 𝐾𝑓 𝐸𝑚 sin 𝜔𝑚 𝑡]
FM wave : 𝑒𝑓𝑀 𝑡 = 𝐸𝑐 sin[𝜔𝑐 𝑡 + 𝐾𝑓𝐸𝑚
cos 𝜔𝑚 𝑡
𝜔𝑚
.]
Modulation Index for FM:
𝐾𝐸𝑚
𝜔𝑚
or
𝐾𝐸𝑚
2𝛱𝑓𝑚
FM wave : 𝑒𝑝𝑀 𝑡 = 𝐸𝑐 sin[𝜔𝑐 𝑡 + mcos 𝜔𝑚 𝑡 .]
Modulation Index for FM:
Modulation Index for PM
2𝛱𝑓𝑚
Maximum Frequency deviation (δ):
Modulation Index for PM
2𝛱
Modulation Index for FM :
𝛿
𝑓𝑚
Bandwidth : 2(δ+𝑓𝑚 𝑚𝑎𝑥 )

Advantages of FM
Resilience to noise
Easy to apply modulation at a low power stage of the transmitter
It is possible to use efficient RF amplifiers with frequency modulated signals
Disadvantages of FM
Requires more complicated demodulator
Sidebands extend to infinity either side

Comparison of Analog Communication Systems
(AM – FM – PM)
AM FM PM
Amplitude gets
varied
Frequency gets
varied
Phase gets varied
Bandwidth: 2fm Bandwidth :
2(δ+fm)
Bandwidth :
2(δ+fm)
High noise
distortion
Lesser noise
distortion
Less noise
distortion
Multiplexing is not
possible
Multiplexing is
possible
Multiplexing is
possible
High complexity in
design
Low complexity in
design
Low complexity is
design

• Communication
Process of Exchanging Information Between Transmitter and Receiver
• Types of Communication – Analog and Digital
• Components of Communication – Source, Transmitter, Channel, Receiver,
Destination
• Modulation and Demodulation
• Need for modulation
• Amplitude Modulation – Changing amplitude of carrier with respect to the
message signal
• SSB Techniques – SSBFC, SSBSC, SSBRC, ISB, VSB
• Angle Modulation – Frequency and Phase Modulation

ASSIGNMENT QUESTIONS
1. Calculate the modulation index and percent modulation if the
instantaneous voltage of modulating signal and carrier are 40 sin ωmt and
50 sin ωct respectively
2. An audio frequency signal 10 sin2Π500t is used to amplitude modulate a
carrier of 10 sin2Π105t. Calculate
i. Modulation Index ii. Sideband Frequencies iii. Amplitude of each
side band frequencies iv. Bandwidth v. Total power delivered to the
load of 600 Ω.
3. A 400 W carrier is modulated to a depth of 80 %. Calculate the total
power in modulated wave.
4. A broadcast transmitter radiates 20 kW when the modulation percentage
is 75. Calculate the carrier power and power of each sideband.
5. If a10V carrier is amplitude modulated by two different frequencies with
amplitudes 2V and 3V respectively, find the modulation index.
6. Find the bandwidth of an AM-DSBFC wave for a carrier frequency
fc=100KHz and maximum modulation signal frequency fm=5KHz.

7. Find the carrier power of a AM broadcast radio transmitter that radiates
20KW for which the modulation index is 0.6.
8. For an AM-DSBFC modulator with a carrier frequency of 100KHz and
maximum modulating signal frequency of 5 KHz, determine upper and
lower sideband frequency and the bandwidth.
9. Determine the power of an AM-DSBFC with peak unmodulated carrier
voltage of Vc=20Vp and a load resistance RL =20Ω. Assume the
modulation index as 0.6.
10. What is the bandwidth required for and FM signal in which the
modulating signal is 2 KHz and maximum deviation is 10 KHz.
11. A transmitter supplies 8KW to the antenna when modulated. Determine
the total power radiated when modulated to 70%.
12. In an AM transmitter the carrier power is 10KW and the modulation
index is 0.5. Calculate the total RF power delivered.
13. In an amplitude modulation system the carrier frequency is fc = 100
KHz. The maximum frequency of the signal is 5 KHz. Determine the lower
and upper sidebands and the bandwidth of AM signal.
14. The maximum frequency deviation in an FM is 10 KHz and signal
frequency is 10 KHz. Find the bandwidth using Carson’s rule and
modulation index.

To a conventional AM modulator, one input is a 500KHz carrier with an
amplitude of 20 V (peak). The other input is a 10KHz modulating signal
that is of sufficient amplitude to cause a change in the output wave of
±7.5V (peak). Determine
(i) Upper and lower side frequencies.
(ii) Modulation coefficient and percent modulation.
(iii) Peak amplitude of the modulated carrier and the upper and lower side
frequency voltages.
(iv) Maximum and minimum amplitudes of the envelope.
(v) The expression for the modulated wave.
(vi) Draw the output spectrum.
(vii) Sketch the output envelope. (14)

For an FM modulator with deviation sensitivity K1= 4 KHz/V and a
modulating signal vm(t) = 10Sin(2π2000 t) , détermine,
(i) The peak frequency deviation
(ii) The carrier swing
(iii) and the modulation index.
(iv) What is the peak frequency deviation produced if the modulating signal
were to double in amplitude.
For an FM modulator with a modulation index m=1, a modulating signal
vm(t) = VmSin(2π1000 t) and an unmodulated carrier vc(t) = 10Sin(2π500Kt),
determine,
(i) The number of sets of significant side frequencies,
(ii) Their amplitudes.
(iii) Draw the frequency spectrum showing their relative amplitudes. (6)

The output of an AM transmitter is given by
Vm(t) = 500(1 + 0.4 sin3140t)sin(6.28x107t). Calculate
(1) Carrier frequency
(2) Modulating frequency
(3) Modulation index
(4) Carrier power if load is 600 Ω.
(5) Total power.

Assignment Answers
Calculate the modulation index and percent modulation if the instantaneous
voltage of modulating signal and carrier are 40 sin ωmt and 50 sin ωct
respectively

An audio frequency signal 10 sin2Π500t is used to amplitude modulate a
carrier of 10 sin2Π105t. Calculate
i. Modulation Index ii. Sideband Frequencies iii. Amplitude of each
side band frequencies iv. Bandwidth v. Total power delivered to the
load of 600 Ω.

A 400 W carrier is modulated to a depth of 80 %. Calculate the total power in
modulated wave

A broadcast transmitter radiates 20 kW when the modulation percentage is 75.
Calculate the carrier power and power of each sideband.

If a10V carrier is amplitude modulated by two different frequencies with
amplitudes 2V and 3V respectively, find the modulation index.

To a conventional AM modulator, one input is a 500KHz carrier with an amplitude of 20 V (peak).
The other input is a 10KHz modulating signal that is of sufficient amplitude to cause a change
in the output wave of ±7.5V (peak). Determine
(i) Upper and lower side frequencies.
(ii) Modulation coefficient and percent modulation.
(iii) Peak amplitude of the modulated carrier and the upper and lower side frequency voltages.
(iv) Maximum and minimum amplitudes of the envelope.
(v) The expression for the modulated wave.
(vi) Draw the output spectrum.
(vii) Sketch the output envelope.

What is the bandwidth required for PM and FM signal in which the
modulating signal is 2 KHz and maximum deviation is 10 KHz.
Given : δ = 10000 Hz fm = 2000 Hz
B=2(δ +fm) Hz
B= 2(10000+2000)
=2(12000)
= 24000Hz

For an FM modulator with deviation sensitivity K1= 4 KHz/V and a modulating signal vm(t) =
10Sin(2π2000 t) , détermine,
(iv) What is the peak frequency deviation produced if the modulating signal were to double in amplitude.

How fast the cycle is completed is carrier swing

(1) Carrier frequency = 107 Hz
(2) Modulating frequency = 50 Hz
(3) Modulation index m= 0.4
(4) Carrier power if load is 600 Ω.
(5) Total power.
Vm(t) = 500(1 + 0.4 sin314t)sin(6.28x107t)
= 500(sin6.28*107t+0.4sin314t*sin(6.28*107t)
= 500sin(6.28*107t)+500*0.4sin314t*sin(6.28*107t)
= 500sin(6.28*107t)+200sin314t8sin(6.28*107t)
= 500sin(6.28*107t)+100[cos(314-6.28*107)t-cos(314+6.28*107)t
= 500sin(6.28*107t)+100cos(6.28*50-6.28*107)t-100cos(6.28*50+6.28*107)t
=500sin(6.28*107t)+100cos6.28(50-107)t-100cos6.28(50+107)t
Ec = 500 V Fc= 107 Hz mEc/2 = 100 Fc-Fm = 107-50 Fc+fm = 107+50 fm = 50 Hz

mEC/2 = 100
M*500/2 = 100
m*250=100
M= 100/250
M=0.4
Pc = Ec*Ec/(2R)
= 500*500/(2*600)
= 250000/1200 = 208.33 W
Ptotal = Pc(1+m*m/2)
= 208.33(1+0.16/2)
=208.33(1+0.08)
=208.33(1.08)
= 225 W
Ptotal = Pc+2Plsb

(1) Carrier frequency = 1*107 Hz
(2) Modulating frequency = 50 hz
(3) Modulation index m = 0.4
(4) Carrier power if load is 600 Ω. = 208.33 W
(5) Total power
(6) Side band power
eAM = (Em sin ωmt + Ec )sin ωct
Vm(t) = (500+200sin314t) sin(6.28x107t)
Vm(t) = (200sin314t+500) sin(6.28x107t)
Em = 200 Ec = 500
ωm = 314 ωc=6.28x107
Fm = ωm /2*3.14 Fc = ωc/2*3.14
Fm = 314/2*3.14 Fc = 6.28x107/6.28 = 1*107
= 50
M = Em/Ec = 200/500 = 0.4
Pc = Ec*Ec/2R = 500*500/2*600
= 208.33
Pt= Pc(1+m*m/2)
= 208.33(1+0.4*0.4/2)
= 208.33(1+0.8)
= 208.33*1.8
= 374.994

Pt = Pc + Plsb+Pusb
= Pc + 2Plsb Plsb = Pusb
374.994 = 208.33 + 2*plsb
374.994 – 208.33 = 2*Plsb
166.664=2*Plsb
166.664/2 = Plsb
Plsb = Pusb = 83.332 W
M*m*ec*ec/8R = Em*Em/8R = Plsb

For an FM modulator with deviation sensitivity K1= 4 KHz/V and a modulating signal vm(t) = 10Sin(2π2000 t) , détermine,
(iv) What is the peak frequency deviation produced if the modulating signal were to double in amplitude
K= 4 KHz
Em = 10
ωm = 2 π2000
Fm = ωm /2 π
Fm = 2000
δ = K Em/2 π
= 4000*2*10/2 * 3.14
= 6369.4*2
=
Carrier swing = m fm = 3.18*2000 = 6369.4
m = K Em/(2 π fm)
= 6369.4/2000
= 3.18

Ec8394 - Analog and Digital Communication unit I

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Ec8394 - Analog and Digital Communication unit I