2. Unit–I Amplitude Modulation: Introduction to communication system, need for
modulation, classification of modulation techniques. Amplitude Modulation:
Time domain and frequency domain description; Single tone and multi tone AM
modulation; Power relations in AM wave; Generation of AM Waves – Square Law
Modulator, Switching Modulator. Detection of AM wave: Square Law Detector,
Envelope Detector.
Unit – II DSBSC Modulation: Time domain and frequency domain description,
Generation of DSBSC Wave - Balanced Modulators, Ring Modulator. Coherent
detection of DSBSC Modulated wave , COSTAS Loop.
SSB Modulation: Frequency domain description, Frequency discrimination
method for generation of AM SSB Modulated wave, Time domain description,
Phase discrimination method for generating AM SSB Modulated wave.
Demodulation of SSB wave, VSB Modulation, Comparison of AM Techniques.
Unit – III Angle Modulation: Introduction, Spectral Analysis of Sinusoidal FM and
PM signals, Differences between FM and PM signals, Narrow band FM, Wide
band FM, Generation of FM and PM Signals - Direct and indirect methods.
Detection of FM wave - Balanced Frequency discriminator, phase discriminator,
Phase locked loop, Comparison of FM & AM.
3. Unit – IV Radio Transmitters: Classification of Transmitters, AM Transmitter-low
level and high level modulation, FM Transmitters – Variable reactance and phase
modulated types.
Radio Receivers: Tuned radio frequency receiver, super heterodyne receiver,
Image frequencies, AGC Circuit, FM Receiver, Comparison with AM Receiver.
Unit – V Pulse Modulation: Sampling theorem, sampling techniques, sample and
hold circuit, Types of Pulse modulation, Generation and demodulation of PAM;
Generation and demodulation of PWM; Generation and demodulation of PPM.
Unit – VI Multiplexing: Frequency Division Multiplexing, Time Division
Multiplexing, Comparison between TDM and FDM.
Noise in analog modulation: Noise sources, Thermal noise, Noise Figure, Signal-
to-Noise ratios, AM receiver model, Noise in DSB-SC using coherent detection,
noise in AM receivers in using envelope detection, FM receiver model, FM
Threshold effect, Pre-emphasis & de-emphasis
Text Books:
1. An Introduction to Analog and Digital Communications - Simon Haykin, John Wiley,
2/e.
2. 2. Principles of Communication Systems – H Taub and D. Schilling, TMH, 2017, 4/e.
4. UNIT- I
1) Explain the basic elements of a communication system with a block diagram? (CO1)
2) Define modulation? Why do we need modulation? Classify various Analog modulation
schemes? (CO1)
3) Explain amplitude modulation for an arbitrary baseband signal m(t) with necessary
expressions, wave forms and spectrums. (CO1)
4) What is AM? Derive the expression for transmitted power in terms of carrier power and
Modulation index? (CO1)
5) How AM is generated using square law modulator? Derive relevant expressions. (CO1)
6) Explain the principle of operation of Envelope detector used for AM detection, with
necessary equations. (CO1)
UNIT- II
1) Explain the balanced modulator method to generate DSBSC waveform with neat diagram.
(CO2)
2) Explain the demodulation of DSB-SC using Synchronous detection method. (CO2)
3) Explain about COSTAS loop with a neat block diagram for demodulating DSB-SC wave.(CO2)
4) Explain the Frequency discrimination method for generating SSB signal. (CO2)
5) Explain the phase discrimination method of generating SSB modulated wave with neat
diagram. (CO2)
6) With neat diagrams, explain about the VSB modulation system and also explain its
Applications. (CO2)
5. UNIT- III
1) Define modulation index for FM and PM and obtain the relation between FM and PM.
(CO3)
2) Describe the spectral analysis of sinusoidal FM signal. (CO3)
3) Explain the generation of FM wave using Armstrong method. (CO3)
4) Explain the generation of FM wave using Direct method. (CO3)
5) Describe the process of detection of FM wave using phase discriminator method. (CO3)
6) What are the different demodulation techniques of FM? Explain the demodulation of FM
signal with the help of PLL. (CO3)
UNIT- IV
1) Explain about a Low Level AM Transmitter, with a neat block diagram. (CO4)
2) Explicate the working of AM transmitter using high level modulation with a neat block
diagram. (CO4)
3) Explain the working of Phase modulated FM transmitter with a neat block diagram.
(CO4)
4) Draw the block diagram of a tuned radio frequency receiver and elaborate its
limitations. (CO4)
5) Draw the block diagram of Superhetrodyne receiver and explain the function of each
block. (CO4)
6) Draw a block diagram of a FM receiver and explain the function of each stage. (CO4)
6. UNIT- V
1) State and prove Sampling theorem. (CO5)
2) Explain about sample and hold circuit. (CO5)
3) Draw the PAM wave forms for single polarity and double polarity cases. Explain
the generation of PAM signal. (CO5)
4) Explain the generation and detection of PWM. (CO5)
5) Explain, how a PPM signal can be generated from PWM signal? (CO5)
6) Explain about Demodulation of Pulse Position Modulation signals with neat
sketches. (CO5)
UNIT- VI
1) With a neat block diagram, explain the operation of Frequency division
multiplexing technique. (CO6)
2) With a neat block diagram, explain the operation of Time division multiplexing
technique. (CO6)
3) Discuss the noise performance of AM system using envelope detection. (CO6)
4) Prove that the figure of merit of an AM system for single tone modulation with
100% modulation is 1/3. (CO6)
5) Describe about pre emphasis and de-emphasis with neat figure? (CO6)
6) What is threshold effect in FM? Explain how it can be reduced? (CO6)
7. Assignment-1 (Batch-1)
1)a) Derive an expression for AM wave in time and frequency domain
representation with necessary waveforms. (CO1)
b) An amplitude modulated signal represented in time domain as 4Cos (1800πt)
+ 10Cos (2000πt) + 4Cos (2200πt). Sketch the spectrum and calculate the band
width and total power? (CO1)
2) Explain the generation of AM wave using square law modulator. (CO1)
3) a) Describe the generation of DSB-SC wave using Ring modulator. (CO2)
b) A carrier signal c (t) = 10Cos (2π10^6t) is modulated by a message signal
m(t) = 2Cos (8π10^3t) to generate a DSB-SC signal. Sketch the spectrum,
calculate the B.W and power?
4) Explain the principle of Coherent Detector to demodulate SSB modulated
waveform? (CO2)
5) a) Explain the generation of FM wave using Armstrong method. (CO3)
b) For an FM modulator with a modulating signal m(t)= Am sin(300×10^3t), the
carrier Signal c(t) = 8 sin(6.5×10^6t) and the modulator index = 2. Find out the
significant side band frequencies and their amplitudes.
6) Explain demodulation of FM signal with the help of PLL. (CO3)
8. Assignment-1 (Batch-2)
1) a) Derive the expression for transmitted power in terms of carrier power and
Modulation index? (CO1)
b) A carrier signal c(t)=20cos(2π10^6t) is modulated by a message signal having three
frequencies 5 kHz,10 kHz and 20 kHz. The corresponding modulation indices are 0.4,
0.5 and 0.6. Sketch the spectrum and calculate bandwidth, power and modulation
efficiency.
2) Explain the detection of AM wave using envelope detector. (CO1)
3) Explain the generation of SSB wave using Frequency Discrimination method. (CO2)
4) a) Explain about COSTAS loop with a neat block diagram? (CO2)
b) In an DSB-SC system, modulating signal is a single tone sinusoid 6cos(2π10^3t)
which modulates a carrier signal 8cos(2π10^6t). Write the equation of modulated
wave. Plot the two sided spectrum of the modulated wave. Calculate the amount of
power transmitted.
5) Describe the spectral analysis of sinusoidal FM signal. (CO3)
6) a) Explain the detection of FM wave using balanced frequency discriminator. (CO3)
b) An angle modulated signal is given by s(t) =100Cos[2πfct + 4Sin2000πt] where, fc
= 10MHz, find (i) the average transmitted power (ii) Find the peak frequency deviation.
9. Assignment-2 (Batch-1)
1) a) Explain about a High Level AM Transmitter, with a neat block diagram.
(CO4)
b) A super heterodyne receiver having R.F amplifier is tuned to 555kHz .The
local oscillator is adjusted to 1010kHz. Then calculate the I.F and image
frequency.
2) Draw a block diagram of a FM receiver and explain the function of each stage.
(CO4)
3) State and prove Sampling theorem. (CO5)
4) Explain the generation and detection of PAM. (CO5)
5) a) Calculate the figure of merit for a DSB-SC system. (CO6)
b) Prove that the figure of merit of an AM system for single tone modulation
with 100% modulation is 1/3.
6) a) Distinguish between pre emphasis and de-emphasis with neat figures.
(CO6)
b) Three signals m1, m2 and m3 are to be multiplexed, m1 and m2 have a 50
kHz bandwidth, and m3 has a 10 KHz bandwidth. Design a commutator
switching system so that each signal is sampled at its Nyquist rate.
10. Assignment-2 (Batch-2)
1) Explain the working of Reactance type FM transmitter with a neat block diagram.
(CO4)
2) a) Draw a block diagram of an AM receiver and explain the function of each stage.
(CO4)
b) An AM superhetrodyne receiver is tuned to 600 kHz, if the ‘Q’ of the RF amplifier
tank circuit is 60 and the IF is 455 kHz. Find image frequency and its rejection ratio.
3) a) Explain various Analog Pulse Modulations with suitable diagrams. (CO5)
b) For a PAM transmission of voice signal having maximum frequency equal to
fm=3KHz calculate the transmission bandwidth. It is given that the sampling
frequency fs=8KHz and the pulse duration T =0.1 TS.
4) Explain how a PPM signal can be generated from PWM signal. (CO5)
5) a) Derive the expression for figure of merit of AM system using envelope
detector.(CO6)
b) An AM receiver operating with a sinusoidal modulating wave and 80% modulation
has an output signal to noise ratio of 30 dB, find the corresponding carrier to noise
ratio.
6) What is multiplexing? Write short notes on Time division multiplexing. (CO6)
11. Frequency Bands
BAND Frequency Range
VLF 3 kHz - 30 kHz
LF 30 kHz - 300 kHz
MF 300 kHz - 3 MHz
HF 3 MHz - 30 MHz
VHF 30 MHz -300 MHz
UHF 300 MHz - 3 GHz
SHF 3 GHz - 30 GHz
EHF 30 GHz - 300GHz
Wavelength, λ= c/f
Frequency is the rate of change with respect to time.
Change in a short span of time means high frequency.
Change over a long span of time means low frequency.
13. 1844 - Telegraph:
1876 - Telephony:
1904 - Radio:
1923-1938 -Television:
1936 - Armstrong’s case of FM radio
1938-1945 - World War II Radar and microwave systems
1948-1950 - Information Theory and coding. C. E. Shannon
1962 - Satellite communications begins with Telstar I.
1962-1966 - High Speed digital communication
1972 - Motorola develops cellular telephone.
Brief Chronology of Communication Systems
14. Signal: any physical phenomenon that conveys or carries some information is
called a signal. (or) Any physical quantity that varies with time or space or any
other independent variable called a signal.
For examples: Speech, audio, light, radio, TV, radar, supersonic, temperature,
ECG, EEG, etc .
1) Analog (or) continuous time signal.
2) Discrete time signal.
3) Digital signal.
Continuous time signal:
If the amplitude of signal continuously varies with respect to time or if the
signal contains infinite number of amplitudes, it is called Analog or continuous
signal.
When a signal is defined continuous signal for any value of independent
variable is called analog or continuous time signal.
Here the amplitude of the signal and the time both are continuous.
15. Discrete time signal:
The signals that are defined at discrete instants of time are known as discrete
time signal.
The discrete time signals are discrete in time and continuous in amplitude.
The discrete time signal is obtained by sampling the continuous time signal.
Digital signal:
A digital signal is a special form of discrete time signal which discrete in both
time and amplitude.
A digital signal is obtained by quantizing each value of discrete time signal.
These signals are digital because their samples represented by numbers or
digits.
Quantizing: Rounding of the sample values to the nearest integer.
16. Communication: It is the process of conveying or transferring information from
one point to another. (or) It is the process of establishing connection or link
between two points for information exchange.
Communication refers to sending, receiving and processing of information by
electronic means. The components of a communication system are
the transmitter, communication channel, and receiver
Types of Communication : Broadcast vs Point-to-point
Broadcast: A method of sending a signal where multiple parties may hear a
single sender.
Radio stations are a good example of everyday life "Broadcast Network".
Point-to-point: A method of communication where one "point" (person or entity)
speaks to another entity.
Classification Based on Direction of Communication
Based on whether the system communicates only in one direction or otherwise,
1. Simplex System
2. Half duplex System
3. Full duplex System
17.
18. Modulation:Modulation is nothing but a process of changing any one of the
characteristics (Amplitude or frequency or phase angle) of (high frequency)
carrier signal in accordance with the instantaneous values (amplitude) of the
message signal.
Demodulation: is the reverse process of modulation, which is used to get back
the original message signal.
The process of extracting the message signal from modulated (received) signal
is called demodulation or detection.
Continuous wave modulation: When the carrier wave is continuous in nature the
modulation process is known as continuous wave modulation.
Amplitude modulation: Amplitude Modulation is a process of varying amplitude
of high frequency carrier signal in accordance with the instantaneous amplitude
of the message signal and also the frequency and phase are kept constant.
Frequency modulation: Frequency Modulation is a process of varying
Frequency of high frequency carrier signal in accordance with the instantaneous
amplitude of the message signal and also the amplitude and phase are kept
constant.
19. Phase modulation (PM): Phase Modulation is a process of varying Phase of high
frequency carrier signal in accordance with the instantaneous amplitude of the
message signal and also the amplitude and frequency are kept constant.
Pulse modulation: When the carrier wave is a pulse in nature the modulation
process is known as Pulse modulation
Pulse Amplitude Modulation: The amplitude of the pulse carrier varies in
accordance with the instantaneous amplitude of the message signal is called
Pulse Amplitude Modulation. But the duration or width and position of the carrier
pulse remain constant.
Pulse width Modulation: The width of the pulse carrier varies in accordance with
the instantaneous amplitude of the message signal is called Pulse width
Modulation. But the amplitude and position of the carrier pulse remain constant .
Pulse Position Modulation: The Position of the pulse carrier varies in
accordance with the instantaneous amplitude of the message signal is called
Pulse Position Modulation. But the amplitude and duration of the carrier pulse
remain constant.
20. In modulation process, three types of signals are used to transmit information
from source to destination.
They are:
1) Message signal
2) Carrier signal
3) Modulated signal
Message signal: The signal which contains a message to be transmitted to the
destination is called a message signal. The message signal is also known as a
modulating signal or baseband signal.
Carrier signal: The high energy or high frequency signal which has
characteristics like amplitude, frequency, and phase but contains no information
is called a carrier signal.
Modulated signal: When the message signal is mixed with the carrier signal, a
new signal is produced. This new signal is known as a modulated signal.
The modulated signal is the combination of the carrier signal and modulating
signal.
21. Elements of a Communication System
Communication is the process of transformation of information from source
to destination or from transmitter to receiver.
The main objective of a communication system is to transmit an information
signal from a source located at one point to a destination located at another
point with minimum distortion and noise so that to produce accurate output at
the receiver end.
The communication system consists of three basic components. (Transmitter,
Channel and Receiver)
Voice – 300Hz to 3.4KHz
Audio – 20Hz to 20 KHz
Video – 0 to 4.5 KHz
22. Input transducer: is the equipment which converts physical message, such as
sound, words, pictures etc., into corresponding electrical signal. Ex. Microphone.
Transmitter: The transmitter is a collection of electronic components and circuits
that converts the electrical signal into a signal suitable for transmission over a
given medium.
(Transducer, amplifier, modulator, oscillator, power amplifier, antenna)
23. Channel : Channel is the medium trough which signal is transmitted from one
place to another. which provides transmission path between transmitter and
receiver.
Wired communication is preferred for short distance communication, where
the channel will be Twisted pair wires, Coaxial cable, optical fiber etc
For long distance communication wireless communication is proffered
where the Channel will be free space.
Noise: Noise is random, undesirable electronic energy that enters the
communication system via the communicating medium and interferes with the
transmitted message.
Receiver : A receiver is a collection of electronic components and circuits that
accepts the transmitted message from the channel and converts it back into a
form understandable by humans.
(antenna, amplifier, demodulator, oscillator, power amplifier, transducer)
Output Transducer: which converts the electric signal as physical signal.
Example: Loudspeaker, personal computer.
24. Need for modulation. Modulation is extremely necessary in communication
system due to the following reasons :
1. Reduce the height of the antenna
2. Avoiding mixing of Signals ,
3. Increase the range of communication.
4. Allows multiplexing of signals
5. Improves Quality of reception.
6. Narrow banding.
1. Reduce the height of the antenna
Height of the antenna is a function of wavelength ‘λ’. The minimum height of
antenna is given by λ/4.
Wavelength, λ= c/f
As the audio frequencies range from 20 Hz to 20 kHz, therefore, if they are
transmitted directly into space, the length of the transmitting antenna required
would be extremely large.
For Example: f = 20 kHz, λ= 3 × 108 / 20 × 103 = 15,000m, antenna length = λ/4 =
15,000/4 = 3,750 meters.
25. This is too long antenna to be constructed practically.
For this reason, it is impracticable to radiate audio signal directly into space.
If f = 1000 kHz, λ= 300 meters, antenna length = λ/4 = 75 m. This antenna can
be easily installed practically.
Thus, modulation reduces the height of the antenna.
2. Avoiding mixing of Signals
If the baseband sound signals are transmitted without using the modulation
by more than one transmitter, then all the signals will be in the same
frequency range i.e. 20Hz to 20KHz.
Therefore, all the signals get mixed together and a receiver can not separate
them from each other .
Hence, if each baseband sound signal is used to modulate a different carrier
then they will occupy different slots in the frequency domain (different
channels).
Thus, modulation avoids mixing of signals .
26. 3. Increase the range of communication
Low frequency signals have poor radiation and they get highly attenuated.
Therefore baseband signals cannot be transmitted directly over long
distances.
The modulation process increases the frequency of the signal to be
transmitted.
Therefore, it increases the range of communication.
4. Allows multiplexing of signals
Multiplexing is a process in which two or more signals can be transmitted
over the same communication channel simultaneously.
Modulation allows the multiplexing to be used.
Multiplexing allows the same channel to be used by many signals.
5. Improves Quality of reception
Modulation techniques like Frequency modulation and the digital
communication techniques such as Pulse code modulation reduces the effect
of noise to great extent.
Reduction of noise improves quality of reception.
27. 6. Narrow banding
Suppose that to transmit audio signal ranging from 50 - 104 Hz using suitable
antenna. The ratio of highest to lowest frequency is 200.
Therefore an antenna suitable for use at one end of the frequency range would
be entirely too short or too long for the other end.
Suppose that the audio spectrum is translated so that it occupies the range
from 50+106 to 104+106 Hz. Then the ratio of highest to lowest frequency
becomes 1.01.
Thus the process of frequency translation is useful to change wideband
signals to narrow band signals.
Continuous-Time Sinusoidal Signal
A sinusoidal signal which is defined for every instant of time is
called continuous-time sinusoidal signal. The continuous time sinusoidal
signal is given as follows − 𝑥(𝑡) = 𝐴 sin(ω𝑡 + φ) = 𝐴 sin(2𝜋𝑓𝑡 + φ)
All the continuous-time sinusoidal signals are periodic signal.
The time period (T) of a continuous-time sinusoidal signal is given by,
T=2π/ω=1/f.
28. Amplitude Modulation
Amplitude modulation is a type of modulation where the amplitude of the
carrier signal is varied (changed) in accordance with the amplitude of the
message signal while the frequency and phase of carrier signal remain
constant.
Time-domain Representation:
Let the modulating signal be
m(t)=Amcos(2πfm t)
and the carrier signal be c(t)=Ac cos(2πfct)
Where,
Am and Ac are the amplitude of the modulating
signal and the carrier signal respectively.
fm and fc are the frequency of the modulating signal
and the carrier signal respectively.
Superimposing modulating signal into a
carrier wave and also varying the amplitude
of the carrier wave in accordance with the
amplitude of the modulating signal, and the
amplitude modulated wave s(t) will be
29. s(t)=[Ac + m(t)] cos(2πfct) ⇒ s(t)=[Ac + Am cos(2πfmt)] cos(2πfct)
This is the general form of amplitude modulated wave. S(t) = is the amplitude
modulated wave
Where, A = Ac + Amcos(2πfmt) = Amplitude of the modulated wave
Rearrange the Equation 1 as below.
s(t)=Ac[1+(Am/Ac)cos(2πfmt)]cos(2πfct) ⇒s(t)=Ac[1+μcos(2πfmt)]cos(2πfct)
Where, μ is Modulation index and it is the ratio of the Amplitude of modulating
signal to the amplitude of the carrier wave. Mathematically, μ = Am/Ac
⇒s(t)=Ac cos(2πfct)+𝑨𝒄𝝁 cos(2πfct) cos(2πfmt)
cosa cosb = (1/2)[cos(a + b) + cos(a - b)]
⇒s(t)=Ac cos(2πfct)+
𝑨𝒄𝝁
𝟐
cos[2π(fc+fm)t]+
𝑨𝒄𝝁
𝟐
cos[2π(fc−fm)t]
Hence, the amplitude modulated wave has three frequencies.
Those are carrier frequency fc, upper sideband frequency fc+fm and lower
sideband frequency fc − fm
Bandwidth: (BW) It is the difference between the highest and lowest frequencies
of the signal.
BW = upper sideband frequency(fc+fm) – lower sideband frequency (fc − fm)
30. or BW= fmax−fmin
Here, fmax = fc+fm and fmin = fc−fm
Substitute, fmax and fmin values in bandwidth formula.
BW = fc+fm−(fc−fm) ⇒BW = 2fm
BW = 2fm = twice the frequency of modulating signal
Let Amax and Amin be the maximum and minimum amplitudes of the modulated
wave.
The maximum amplitude of the modulated wave, when cos(2πfmt) is 1.
⇒Amax = Ac + Am
The minimum amplitude of the modulated wave, when cos (2πfmt) is -1.
⇒Amin = Ac − Am
Amax+Amin=Ac+Am+Ac−Am=2Ac ⇒ Ac = (Amax+Amin)/2
Amax−Amin= Ac + Am− (Ac−Am)=2Am ⇒ Am = (Amax−Amin)/2
31. The ratio of (Amax+Amin)/2 and (Amax−Amin)/2 will be as follows.
Am (Amax−Amin)/2
---- = -------------------
Ac (Amax + Amin)/2
The modulation index or modulation depth is often denoted in percentage
called as Percentage of Modulation.
For a perfect modulation, the value of modulation index should be 1, which
implies the percentage of modulation should be 100%.
For instance, if this value is less than 1, i.e., the modulation index is 0.5, then
the modulated output would look like the following figure. Such a wave is
called as an under-modulated wave.
If the value of the modulation index is greater than 1, i.e., 1.5 or so, then the
wave will be an over-modulated wave.
Amax − Ami
μ = ----------------
Amax + Amin
32. Frequency-Domain Representation or Spectrum of AM Wave
Frequency-domain representation means description of a signal with respect
to its frequency and a spectrum analyzer is a frequency-domain instrument.
Consider the following equation of amplitude modulated wave.
s(t)=Ac cos(2πfct)+Ac𝝁cos(2πfct) cos(2πfmt)
⇒s(t)=Ac cos(2πfct) + Ac kam(t) cos(2πfmt) since ka = 𝝁/Am
s(t)=Ac[1+kam(t)] cos(2πfct)
Let S(f) denote the Fourier transform of s(t), C(f) denotes the Fourier
transform of c(t) and M(f) denotes the Fourier transform of m(t).
FT [Ac cos(2πfct)] <--> Ac/2 [δ(f – fc) +δ(f + fc)]
FT [m(t) Ac cos(2πfct)] <--> Ac/2 [M(f – fc) + M(f + fc)]
Therefore S(f) = Ac /2 [δ(f – fc) +δ(f + fc)] + Ac ka/2 [M(f – fc) + M(f + fc)]
Above equation for Amplitude Modulated Wave contains two factors.
The first factor represents the presence of carrier signal i.e., two impulses
each having strength equal to Ac /2.
The second factor represents the spectrum of original or baseband signal
shifted in the positive as well as in the negative direction by the factor fc.
33. Thus the spectrum of modulated signal contains shifted spectrum of
modulating signal and the spectrum of carrier signal as shown in figure.
In the frequency domain, amplitude modulation corresponds to translating the
baseband spectrum to a band surrounding the carrier frequency.
spectrum of modulated signal
spectrum of carrier signal
spectrum of message signal
34. Power Relations in the AM Wave
Let the modulating signal be m(t)=Amcos(2πfm t)
and the carrier signal be ct)=Ac cos(2πfct)
s(t)=[Ac + m(t)] cos(2πfct) ⇒ s(t)=[Ac + Am cos(2πfmt)] cos(2πfct)
s(t)=Ac[1+(Am/Ac)cos(2πfmt)]cos(2πfct) ⇒s(t)=Ac[1+μcos(2πfmt)]cos(2πfct)
⇒s(t)=Ac cos(2πfct)+𝑨𝒄𝝁 cos(2πfct) cos(2πfmt)
cosa cosb = (1/2)[cos(a + b) + cos(a - b)]
s(t)=Ac cos(2πfct)+
𝑨𝒄𝝁
𝟐
cos[2π(fc+fm)t]
𝑨𝒄𝝁
𝟐
cos[2π(fc−fm)t]
The amplitude modulated wave is the sum of three high-frequency
components. They are original carrier wave frequency, lower sideband
frequency, and upper sideband frequency.
Hence, the power of the amplitude modulated wave is equal to the sum of
powers of the carrier, upper sideband, and lower sideband frequency
components. Pt = Pc + PUSB + PLSB
The standard formula for power of signal is Pc =
𝑽𝒓𝒎𝒔
𝟐
𝑹
=
𝒗𝒎/ 𝟐 𝟐
𝑹
Where, Vrms is the rms value of cos signal. vm is the peak value of cos signal.
35. Then, the power of the carrier wave is given by, Pc =
𝑨𝒄/ 𝟐
𝟐
𝑹
=
𝑨𝒄
𝟐
𝟐𝑹
and power of
the side bands is given by PUSB = PLSB =
𝑨𝒄𝝁/𝟐 𝟐
𝟐
𝑹
=
𝑨𝒄
𝟐𝝁𝟐
𝟖𝑹
Now, let us add these three powers in order to get the power of AM wave.
𝑨𝒄
𝟐
𝟐𝑹
+
𝑨𝒄
𝟐𝝁𝟐
𝟖𝑹
+
𝑨𝒄
𝟐𝝁𝟐
𝟖𝑹
It is used to calculate the power of AM wave, when the carrier power and the
modulation index are known. 𝑷𝒕=
𝑨𝒄
𝟐
𝟐𝑹
𝟏 +
𝝁𝟐
𝟒
+
𝝁𝟐
𝟒
⇒𝑷𝒄 𝟏 +
𝝁𝟐
𝟐
If the modulation index μ=1 then the power of AM wave is equal to 1.5 times the
carrier power, 𝑷𝒕 = 𝑷𝒄 𝟏 +
𝝁𝟐
𝟐
⇒𝑷𝒕 = 𝟏. 𝟓𝑷𝒄
Current calculations: Let Ic be the unmodulated current and It the total, or
modulated, current of an AM transmitter, both being rms values. If R is the
resistance in which these currents flow, then 𝑷𝒕 = 𝑰𝒕
𝟐
𝑹 and 𝑷𝒄 = 𝑰𝒄
𝟐
𝑹
𝑷𝒕 = 𝑷𝒄 𝟏 +
𝝁𝟐
𝟐
= 𝑰𝒕
𝟐
𝑹 = 𝑰𝒄
𝟐𝑹 𝟏 +
𝝁𝟐
𝟐
= 𝑰𝒕= 𝑰𝒄 𝟏 +
𝝁𝟐
𝟐
36. Transmission Efficiency of an AM wave
The transmission efficiency of an AM wave is defined as the percentage of total
power contributed by the sidebands.
Transmission efficiency of an AM wave is the ratio of the transmitted power which
contains the information (i.e. the total sideband power) to the total transmitted
power. 𝜼 = efficiency =
𝑷𝑺𝑩
𝑷𝒕
where: 𝑷𝑺𝑩 = the power in all the side-bands
𝑷𝒕 = the total transmitted power (includes carrier and side-bands)
The percent transmission efficiency is given by
The maximum efficiency is obtained for μ = 1, i.e. 33.33 %
The efficiency may be increased by increasing the value of the modulation index,
𝝁, but if you use a value > 1.0 there will be distortion introduced in the receiver.
2
1
8R
A
8R
A
2
2
2
c
2
2
c
c
P
2
1
2R
A
8R
A
8R
A
2
2
c
2
2
c
2
2
c
2
2
2
%
100
2 2
2
37. Carrier wave of frequency f = 1mHz with pack voltage of 20V used to
modulate a signal of frequency 1kHz with pack voltage of 10v. Find out the
following (i) μ (ii) Frequencies of the modulated wave. (iii) Bandwidth.
Solution:(i) μ = Am/Ac =10/20=0.5
(ii) frequencies of modulated wave, f → fc, fc + fm and fc – fm
fc = 1mHz, fm = 1kHz
fc + fm = 1×10^6 + 1×10^3 = 1001 ×10^3 = 1001 kHz
fc – fm = 1×10^6 – 1×10^3 = 999 × 10^3 = 999 kHz
(iii) Band width: = USB frequency – SSB frequency = fc + fm – (fc – fm)= 2fm
= 1001 kHz – 999 kHz = 2 kHz
A modulating signal m(t)=10cos(2π×103t) is amplitude modulated with a
carrier signal c(t)=50cos(2π×105t). Find the modulation index, the carrier
power, and the power required for transmitting AM wave.
Solution: Given, the equation of modulating signal as, m(t)=10cos(2π×103t)
The standard equation of modulating signal is, m(t)=Amcos(2πfmt)
By comparing the above two equations, amplitude of modulating signal
is Am=10volts and Frequency of modulating signal is fm=103Hz =1KHz
38. Given, the equation of carrier signal is, c(t)=50cos(2π×105t)
The standard equation of carrier signal is, c(t)=Accos(2πfct)
By comparing these two equations, we will get amplitude of carrier signal
as Ac = 50volts and Frequency of carrier signal as fc= 105Hz =100KHz.
The formula for modulation index is,μ=Am/Ac
Substitute, Am and Ac values in the above formula, μ=10/50=0.2
Therefore, the value of modulation index is 0.2 and percentage of modulation
is 20%.
The formula for Carrier power, Pc =
𝑨𝒄
𝟐
𝟐𝑹
Assume R=1Ω and substitute Ac value in the above formula,
Pc=(50)2/2(1) =1250W
Therefore, the Carrier power, Pc is 1250 watts.
The formula for power required for transmitting AM wave is, 𝑷𝒕 = 𝑷𝒄 𝟏 +
𝝁𝟐
𝟐
Substitute Pc and μ values in the above formula, 𝑷𝒕 = 𝟏𝟐𝟓𝟎 𝟏 +
(𝟎.𝟐)𝟐
𝟐
=1275W
39. Consider an Am signal with a carrier frequency of ′′fc′′ follows:
s(t)=[20+4cos(3000t)+10cos(6000t)]cos(2πfct)
Find the normalized power carried by the sidebands of this AM signal.
Solution
The power carried by the sidebands of an AM signal can be given as,
PUSB = PLSB=
𝑨𝒄
𝟐𝝁𝟐
𝟖𝑹
⇒PSB= PUSB + PLSB=
𝑨𝒄
𝟐𝝁𝟐
𝟒𝑹
=
𝑷𝒄𝝁𝟐
𝟐
𝐬𝐢𝐧𝐜𝐞 Pc =
𝑨𝒄
𝟐
𝟐𝑹
Pc = carrier power =
𝑨𝒄
𝟐
𝟐𝑹
=
𝟐𝟎𝟐
𝟐
=200 W
μ= modulation index, The given AM signal consists of two sinusoidal
message signals.
μ1=4/20=0.2
μ2=10/20=0.5
So, μ= μ𝟏
𝟐+μ𝟐
𝟐√= √(0.2)𝟐
+ (0.5)𝟐
= 0.54
PSB =
𝑷𝒄𝝁𝟐
𝟐
=
(200)(0.54)
𝟐
𝟐
= 29W
40. The equation of amplitude wave is given by s(t) = 20 [1+0.8cos (2π×103t)]
cos(4π×105t) Find the carrier power, the total sideband power, and the band
width of AM wave.
Solution: Given, the equation of Amplitude modulated wave is
s(t)=20[1+0.8cos(2π×103t)]cos(4π×105t)
s(t)=20[1+0.8cos(2π×103t)]cos(2π×2×105t)
The equation of AM wave is, s(t)=Ac[1+μcos(2πfmt)]cos(2πfct)
Amplitude of carrier signal as Ac= 20volts
Modulation index as μ=0.8
Frequency of modulating signal as fm=103Hz =1KHz
Frequency of carrier signal as fc= 2×105Hz =200KHz
Assume R=1Ω and substitute Ac value Pc= (20)2/2(1) = 200W
Therefore, the Carrier power, Pc is 200watts.
We know the formula for total side band power is, PSB=Pcμ2/2
Substitute Pc and μ values in the above formula, PSB=200×(0.8)2/2=64W
Therefore, the total side band power is 64 watts.
BW = 2 fm= 2(1K) =2 KHz Therefore, the bandwidth of AM wave is 2 KHz.
41. Single Tone Amplitude Modulation
Modulating signal is a random signal which contains a large number of
frequency components.
This means that a carrier signal (fixed frequency signal) is modulated by a
large number of frequency components.
If the message signal contains single frequency component and the resulting
modulating signal is called as single tone modulated signal.
Let the modulating signal be m(t)=Amcos(2πfm t)
and the carrier signal be c(t)=Ac cos(2πfct)
Where,
Am and Ac are the amplitude of the modulating signal and the carrier signal
respectively.
fm and fc are the frequency of the modulating signal and the carrier signal
respectively.
Superimposing modulating signal into a carrier wave and also varying the
amplitude of the carrier wave in accordance with the amplitude of the
modulating signal, and the amplitude modulated wave s(t) will be
42. s(t)=[Ac + m(t)] cos(2πfct)
⇒ s(t)=[Ac + Am cos(2πfmt)] cos(2πfct)
This is the general form of amplitude modulated wave. S(t) = is the amplitude
modulated wave
Where, A = Ac + Amcos(2πfmt) = Amplitude of the modulated wave
Rearrange the Equation 1 as below.
s(t)=Ac[1+(Am/Ac)cos(2πfmt)]cos(2πfct)
⇒s(t)=Ac[1+μcos(2πfmt)]cos(2πfct)
Where, μ is Modulation index and it is the ratio of the Amplitude of modulating
signal to the amplitude of the carrier wave. Mathematically, μ = Am/Ac
⇒s(t)=Ac cos(2πfct)+𝑨𝒄𝝁 cos(2πfct) cos(2πfmt)
cosa cosb = (1/2)[cos(a + b) + cos(a - b)]
⇒s(t)=Ac cos(2πfct)+
𝑨𝒄𝝁
𝟐
cos[2π(fc+fm)t]+
𝑨𝒄𝝁
𝟐
cos[2π(fc−fm)t]
The amplitude modulated wave is the sum of three high-frequency
components. They are original carrier wave frequency, lower sideband
frequency, and upper sideband frequency.
43. Those are carrier frequency fc, upper sideband frequency fc + fm and lower
sideband frequency fc − fm
Taking Fourier transform on both sides for the above equation
Time domain Frequency domain
)]
(
)
(
[
4
)]
(
)
(
[
4
)]
(
)
(
[
2
)
(
m
c
m
c
c
m
c
m
c
c
c
c
c
f
f
f
f
f
f
A
f
f
f
f
f
f
A
f
f
f
f
A
f
S
44. Multi tone AM modulation
Modulating signal is a random signal which contains a large number of
frequency components.
This means that a carrier signal (fixed frequency signal) is modulated by a
large number of frequency components.
If the message signal contains single frequency component and the resulting
modulating signal is called as single tone modulated signal.
likewise if the message signal contains more than one frequency component
then the resulting modulated signal is called as multi tone modulated signal.
Hence modulation of this type of message signals (which has more than one
frequency component)is called multi tone modulation.
Let us consider there are two modulating signals
m1(t)=Am1 cos(2πfm1t) m2(t)=Am2 cos(2πfm2t)
and the carrier signal be c(t)=Ac cos(2πfct)
Total Modulating signal = m1(t)+ m2(t)
= Am1 cos(2πfm1t) +Am2 cos(2πfm2t)
45. The standard equation of AM wave is s(t)=Ac[1+Kam(t)] cos(2πfct)
s(t)=Ac{1+Ka [m1(t)+ m2(t)]} cos(2πfct)
s(t)=Ac{1+Ka [Am1 cos(2πfm1t) +Am2 cos(2πfm2t)]} cos(2πfct)
s(t)=Ac[1+ KaAm1 cos(2πfm1t) +KaAm2 cos(2πfm2t)} cos(2πfct)
s(t)=Ac[1+μ1cos(2πfmt)+ μ2cos(2πfmt)] cos(2πfct)
cosa cosb = (1/2)[cos(a + b) + cos(a - b)]
Taking Fourier transform on both sides for the above equation
]
)
(
2
cos[
2
]
)
(
2
cos[
2
]
)
(
2
cos[
2
]
)
(
2
cos[
2
)
2
cos(
)
(
2
2
2
2
1
1
1
1
t
f
f
A
t
f
f
A
t
f
f
A
t
f
f
A
t
f
A
t
s
m
c
c
m
c
c
m
c
c
m
c
c
c
c
))
(
))
(
[
4
))
(
))
(
[
4
))
(
(
[
4
))
(
(
))
(
(
[
4
)]
(
)
(
[
2
)
(
2
2
2
2
2
2
1
1
1
1
1
1
m
c
m
c
c
m
c
m
c
c
m
c
m
c
c
m
c
m
c
c
c
c
c
f
f
f
f
f
f
A
f
f
f
f
f
f
A
f
f
f
f
f
f
A
f
f
f
f
f
f
A
f
f
f
f
A
f
S
46. Power Content In Multiple-tone Amplitude Modulation
The total power in the Am wave can be given as
Pt = Pc + PUSB1 + PLSB1 +PUSB2 + PLSB2
8R
A
8R
A
8R
A
8R
A
2R
A
2
2
2
c
2
2
2
c
2
1
2
c
2
1
2
c
2
c
t
P
4
4
4
4
1
2R
A
2
2
2
2
2
1
2
1
2
c
t
P
2
2
1
2R
A
2
2
2
1
2
c
t
P
2
2
1
P
2
2
2
1
c
t
P
2
1
2
t
c
t P
P
2
2
2
2
2
2
1
2
t
where ......
2
2
2
1
t
47. Generation of AM Waves
Let us discuss about the modulators, which generate amplitude modulated
wave. (The device that generates an amplitude modulated wave is called an
amplitude modulator)
The following two modulators generate AM wave.
1.Square law modulator,
2.Switching modulator
Square Law Modulator
A square law modulator has three major parts 1.summer (adder) 2.nonlinear
device 3. BPF.
The block diagram of the square law modulator as shown in figure.
48. Let the modulating and carrier signals be denoted as m(t) and Ac
cos(2πfct) respectively.
These two signals are applied as inputs to the summer (adder) block.
This summer block produces an output, which is the addition of the modulating
and the carrier signal.
Mathematically, V1(t) = m(t)+Ac cos(2πfct). This signal V1(t) is applied as an input
to a nonlinear device like diode.
The characteristics of the diode are closely related to square law.
V2(t)=a1V1(t)+a2V1
2(t) (Eq.1)
Where, a1 and a2 are constants.
Substitute V1(t)) in Eq. 1
V2(t) = a1[m(t)+Accos(2πfct)] + a2[m(t)+Accos(2πfct)]2
⇒V2(t) = a1m(t) + a1Accos(2πfct) + a2m2(t)+ a2Ac
2cos2(2πfct)+ 2a2m(t)Accos(2πfct)
⇒V2(t)=a1m(t)+a2m2(t)+a2Ac
2cos2(2πfct)+a1Ac[1+(2a2/a1) m(t)] cos(2πfct)
The last term of the above equation represents the desired AM wave and the
first three terms of the above equation are unwanted.
So, with the help of band pass filter, it will pass only AM wave and eliminate
the first three terms.
49. Therefore, the output of square law modulator is
s(t)=a1Ac[1+(2a2/a1)m(t)] cos(2πfct)
The standard equation of AM wave is
s(t)=Ac[1+kam(t)]cos(2πfct)
Where, Ka is the amplitude sensitivity
By comparing the output of the square law modulator with the standard
equation of AM wave,
The scaling factor as a1 and the amplitude sensitivity ka as 2a2/a1.
Switching Modulator
Switching modulator is similar to the square law modulator.
The only difference is that in the square law modulator, the diode is operated
in a non-linear mode, whereas, in the switching modulator, the diode has to
operate as an ideal switch. Following is the block diagram of switching
modulator.
Consider a semiconductor diode used as an ideal switch to which the carrier
signal Ac cos(2πfct) and information signal m(t) are applied simultaneously as
shown figure.
50. Let the modulating and carrier signals be denoted as m(t) and c(t) =Ac
cos(2πfct) respectively.
These two signals are applied as inputs to the summer (adder) block.
Summer block produces an output, which is the addition of modulating and
carrier signals.
Mathematically, we can write it as
V1(t) = m(t)+c(t) = m(t)+Ac cos(2πfct)
This signal V1(t) is applied as an input of diode.
Assume, the magnitude of the modulating signal is very small when
compared to the amplitude of carrier signal Ac.
51. So, the diode’s ON and OFF action is controlled by carrier signal c(t).
This means, the diode will be forward biased when c(t)>0 and it will be reverse
biased when c(t)<0.
Therefore, the output of the diode is
Approximate this as V2(t) = V1(t) x(t)
Where, x(t) is a periodic pulse train with time period T=1/fc
Substitute, V1(t) and x(t) values in Eq. 2.
0
)
(
0
)
(
0
2
1
t
c
if
t
V
t
c
if
t
V
)
)
1
2
(
2
cos(
1
2
)
1
(
2
2
1
)
(
1
1
t
f
n
n
t
x c
n
n
....
)
6
cos(
3
2
)
2
cos(
2
2
1
)
(
t
f
t
f
t
x c
c
....
)
6
cos(
3
2
)
2
cos(
2
2
1
)]
2
cos(
)
(
[
)
(
2 t
f
t
f
t
f
A
t
m
t
V c
c
c
c
53. Demodulation
The process of extracting an original message signal from the modulated
wave is known detection or demodulation.
The circuit, which demodulates the modulated wave is known as
the demodulator.
The following demodulators (detectors) are used for demodulating AM wave.
1. Square Law Demodulator, 2. Envelope Detector
Square Law Demodulator
Square law demodulator is used to demodulate low level AM wave.
Following is the block diagram of the square law demodulator.
This demodulator contains a square law device and low pass filter.
The AM wave V1(t) is applied as an input to this demodulator.
The standard form of AM wave is
V1(t) = Ac[1+kam(t)] cos (2πfct)
54. We know that the mathematical relationship between the input and the output
of square law device is
V2(t) = a1V1(t) + a2V1
2(t) (Eq.1)
Where, V1(t) is the input of the square law device, which is nothing but the AM
wave
V2(t) is the output of the square law device, a1 and a2 are constants
Substitute V1(t) in Eq.1
V2(t)=a1 [Ac[1+kam(t)] cos(2πfct) ]+a2 [Ac[1+kam(t)] cos(2πfct)]2
2
t)
f
cos(4
1
m(t)]
2k
(t)
m
k
[1
A
t)
f
m(t)cos(2
k
A
a
t)
f
cos(2
A
a
(t)
V
c
a
2
2
2
c
2
c
a
c
1
c
c
1
2
a
a
t)
f
m(t)cos(4
k
A
m(t)
k
A
t)
f
cos(4
2
(t)
m
k
A
a
2
(t)
m
k
A
a
t)
f
cos(4
2
A
a
2
A
a
t)
f
m(t)cos(2
k
A
a
t)
f
cos(2
A
a
(t)
V
c
a
2
c
2
a
2
c
2
c
2
2
2
c
2
2
2
2
c
2
c
2
c
2
2
c
2
c
a
c
1
c
c
1
2
a
a
a
a
55. Envelope detector consists of a diode
and low pass filter. Here, the diode is the
main detecting element.
Hence, the envelope detector is also
called as the diode detector.
In the above equation, the term is the scaled version of the
message signal.
It can be extracted by passing the above signal through a low pass filter and
the
DC component can be eliminated with the help of a coupling capacitor.
Envelope Detector
Envelope detector is used to detect (demodulate) high level AM wave.
Following is the block diagram of the envelope detector.
m(t)
k
A
a a
2
c
2
2
A
a 2
c
2
56. The low pass filter contains a parallel combination of the resistor and the
capacitor.
It is a simple and highly effective system. This method is used in most of the
commercial AM radio receivers.
The AM wave s(t) is applied as an input to this detector.
V1(t) = Ac[1+kam(t)] cos (2πfct)
e(t) = Ac|1 + kam(t)| is called the envelope of the AM signal.
During the positive half cycles of the input signals, the diode D is forward
biased and the capacitor C charges up rapidly to the peak of the input signal.
When the input signal falls below this value, the diode becomes reverse
biased and the capacitor C discharges through the load resistor RL.
The discharge process continues until the next positive half cycle.
When the input signal becomes greater than the voltage across the capacitor,
the diode conducts again and the process is repeated.
The charge time constant RsC must be short compared with the carrier period,
the capacitor charges rapidly and there by follows the applied voltage up to
the positive peak when the diode is conducting.
57. That is the charging time constant shall satisfy the condition,
On the other hand, the discharging time-constant RLC must be long enough to
ensure that the capacitor discharges slowly through the load resistor RL
between the positive peaks of the carrier wave,
but not so long that the capacitor voltage will not discharge at the maximum
rate of change of the modulating wave.
That is the discharge time constant shall satisfy the condition,
The result is that the capacitor voltage or detector output is nearly the same as
the envelope of AM wave. (As a result, we will get the capacitor voltage
waveform same as that of the envelope of AM wave, which is almost similar to
the modulating signal.)
The detector output usually has a small ripple the carrier frequency, this ripple
is easily removed by low pass filtering.
c
s
f
C
R
1
W
C
R
f
L
c
1
1
58. Advantages of Amplitude Modulation
1) Few components needed: At the receiver side, the original signal is extracted
(demodulated) using a circuit consisting of very few components.
2) Low cost: The components used in AM is very cheap. So the AM transmitter and AM
receiver build at low cost.
3) It is simple to implement.
4) Long distance communication: Amplitude modulated waves can travel a longer
distance.
Disadvantages of Amplitude Modulation
1) Amplitude modulation is inefficient in terms of its power usage: As we know that the
message signal contains information whereas the carrier signal does not contain any
information.
In amplitude modulation, most of the power is concentrated in the carrier signal which
contains no information.
At the receiver side, the power consumed by the carrier wave is wasted.
2) It requires high bandwidth: The AM is not efficient in terms of its use of bandwidth. It
requires a bandwidth equal to twice that of the highest audio signal frequency.
3) This type of transmission can be easily affected by the external radiation.
4) This type of transmission is also affected by the man-made noises or radiations like
waves from other antennas or channels.
