COURSE CODE: 20EC0451 R20

SIDDARTH INSTITUTE OF ENGINEERING & TECHNOLOGY:: PUTTUR

(AUTONOMOUS)
Siddharth Nagar, Narayanavanam Road – 517583
QUESTION BANK (DESCRIPTIVE)

Subject with Code: ICS(20EC0451) Course& Branch: B.Tech & CSE,CSM,CIC

Year & Sem: III-B.Tech.& I-Sem. Regulation: R20

UNIT –I
INTRODUCTION TO COMMUNICATION SYSTEMS

1 a) Define Communication and draw the basic block diagram of communication

system.
Communication is the process of establishing a link/ connection between two points
(Transmitter, Receiver) for information exchange Via a channel.

(or)

Communication is a process of conveying/ exchanging information.

1 b) Explain the function of each block of communication system.

Information Source:

we know that the communication system serves to communicate a message (or) information
source. This message/ information originates is the information source. In general there can be
various messages in the form of words, group of words codes, symbols, sound signal etc.
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In Short form, we can say that the function of information source is to produce required
message which has to be transmitted.

Input Transducer:

A transducer is a device which converts one form of energy or signal into another form of
energy or signal. The message form the source may not be an electrical signal.so there is need
to convert input message signal into electrical signal (Analog (or) Digital).
Eg:- A microphone converts message signal which is in the form of sound waves into electrical
form.

Transmitter:

The main function of the transmitter is to process the electrical signal from different aspects
for example in radio board casting the electrical signal obtained from sound signal is processed
to restrict its range of audio frequencies (upto 5KHz) in amplitude modulation radio board cast
and is often amplified. In wire telephone no real processing is needed. However in long distance
radio communication (or) board cast signal amplification is necessary before modulation.

Modulation is the process the main function of the transmitter. In modulation, the message
signal is super imposed upon the high frequency carrier signal.

Communication Channel:

The communication channel is a medium through which the signal travels.

or
The communication channel is a wired or wireless medium through which the signal
(information) travels from source (transmitter) to destination (receiver).
Communication channels are divided into two categories: wired and wireless. Some examples
of wired channels include co-axial cables, fiber optic cables, and twisted pair telephone lines.
Examples of wireless channels are air, water, and vacuum.

Noise:

Noise is an unwanted signal that enters the communication system via the communication
channel and interferes with the transmitted signal. The noise signal (unwanted signal) degrades
the transmitted signal (signal containing information).

Receiver:

The main function of a receiver is to receive signal from the communication channel. This
received signal is the distorted version with noise included into it. The original signal can
reproduced by a process called Demodulation.

Output Transducer:
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A transducer is a device which converts one form of energy or signal into another form of
energy or signal. An output transducer is used to convert electrical signal into message signal.
Eg: - A loud speaker converts electrical signal back to sound signal

Destination:

Destination is the final stage which is used to convert an electrical message signal into its
original form by using transducer. In radio board casting the destination is a loud speaker. In
destination the actual information will be received.

2 a) Define wired communication and wireless communication.

Wired Communication:

A wired network uses cables to connect devices such as laptop or desktop computers to the
internet or another network. A wired network has some dis advantages when compared to a
wireless network. The biggest advantage is that your device tethered to a router. The most
common wired networks use cables connected at one end to an Ethernet port on the network
router and at the other end to a computer or other device.

Wireless Communication:

A wireless network allows devices to stay connected to the network but room untethered to any
wires. Access points amplify Wi-Fi signals, so a device can be far from a router but still be
connected to the network. When your connection to a Wi-Fi, hot spot at a café, a hotel, an
airport lounge or another public place, you are connecting to that business wireless network.

2 b) Compare Analog and Digital communication.

Analog Communication Digital Communication

1. If the message signal is analog. 1. If the message signal is digital then it
is known as digital Modulation.
2. Get affected by noise and distortion. 2. Immune to Noise and Distortion.
3. Low Bandwidth requirement. 3. High Bandwidth requirement.
4. Hardware is complicated and less 4. Hardware is flexible and less
flexible than digital system. complicated than Analog system.
5. High Power is required. 5. Low Power is required.
6. Error Probability is High. 6. Error Probability is Low.
7. It consists of continuous values 7. It consists of discrete values.
8. Analog signal can be represented by 8. Digital signal can be represented by
sine wave square wave.
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3 a) Define modulation. Classify different types of modulation.

Modulation:

Modulation is defined as changing the characteristics of a carrier signal in accordance with the
instantaneous values of the another signal called message signal/ modulating signal. Signals
containing information are referred as modulating signals. This information bearing signals is
also called base band signal. The high frequency signal is known as carrier signal. The signal
resulting from the process of modulation is called modulated signal.

Types of Modulation:

There are three types of modulation

1. Amplitude Modulation
2. Frequency Modulation
3. Phase Modulation

Amplitude Modulation:

Amplitude Modulation is defined as the modulation in which the amplitude of the carrier wave
is varied in accordance with the instantaneous amplitude of the modulating signal, keeping its
carrier frequency and phase constant.

Frequency Modulation:

Frequency Modulation is defined as the modulation in which the frequency of the carrier wave
is varied in accordance with the instantaneous amplitude of the modulating signal, keeping its
carrier amplitude and phase constant.

Phase Modulation:

Phase Modulation is defined as the modulation in which the Phase of the carrier wave is varied
in accordance with the instantaneous amplitude of the modulating signal, keeping its carrier
amplitude and phase constant.

3 b) Explain the need for Modulation.

Need of Modulation:

1. Reduces the height of Antenna:

Height of antenna is a function of wave length λ. The minimum height of antenna is

given by λ/4.

i.e. Height of the Antenna= λ/4=c/4f

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where λ=c/f

c=3*108, velocity of light.

f= Transmitting Frequency

Eg: - (i) f=15KHz

Height of the Antenna= λ/4=c/4f= 3*108/4*15*103= 5000 meters.

Eg: - (i) f=1 MHz

Height of the Antenna= λ/4=c/4f= 3*108/4*1*106= 7 meters.

For the above two examples it is clear that as the transmitting frequency is increased, height of
the antenna is decreased.

2. Avoid missing of signals:

All audio (message) signals ranges from 20Hz to 20 KHz. The transmission of
message signals from various sources causes that missing of signals and then it is difficult to
separate these signals at the receiver end.

3. Increases the range of communication:

Low frequency signals have poor radiation and they get highly attenuated.
Therefore baseband signals cannot be transmitted directly over long distance. Modulation
increases the frequency of the signal and this they can be transmitted over long distance.

4. Allows adjustments in the Bandwidth:

Bandwidth of a modulated signal may be mode smaller or larger.

5. Improves quality of reception:

Modulation techniques like frequency modulation, pulse code modulation reduces

the effect of noise to great extent. Reduction of noise improves quality of reception.

4 a) Define Amplitude Modulation. Derive expression for AM wave.

Amplitude Modulation:

Amplitude Modulation is defined as the modulation in which the amplitude of the carrier wave
is varied in accordance with the instantaneous amplitude of the modulating signal, keeping its
carrier frequency and phase constant.

Expression for Amplitude Modulation wave:

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Fig (a) Message Signal, Fig (b) AM wave S(t)

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4 b) Determine the modulation index of AM, Percentage Modulation and

Bandwidth of AM.
Modulation Index:

The ratio of change in amplitude of modulating signal to the amplitude of carrier wave is known
as modulation index (or) modulation factor (or) modulation co-efficient (or) depth of
modulation (or) degree of modulation ‘M’.

Percentage Modulation Index:

Note:

1. If Am is greater than Ac then distortion is introduced in to the system.

2. The modulating signal voltage ‘Am’ must be less than carrier signal voltage ‘Ac’ for
proper amplitude modification.

Transmission Bandwidth (BT):

The difference between upper side band and lower side band frequencies defines the
transmission bandwidth ‘BT’.
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Therefore, Bandwidth requires for transmission of an Amplitude Modulation wave is twice

the modulating signal frequency that is 2fm

5 a) Explain shortly about i) Sidebands ii) Justify the reason for selecting the DSB-
SC over DSB FC.

I) Sidebands:

In electronic signal transmission, a sideband is the portion of a modulated carrier wave that
is either above or below the basic ( Baseband ) signal. The portion above the baseband signal
is the upper sideband; the portion below is the lower sideband. In regular amplitude modulation
(AM) transmission, both sidebands are used to carry a message. In some forms of transmission,
one sideband is removed (single-sideband transmission) or a portion of one sideband is
removed.

There are two sidebands in the Amplitude Modulated wave. One is the Upper Side Band which
is termed in-short as USB and another is the Lower Side Band, also called LSB.

The frequency of the Upper Side Band is given by,

fUSB = fc + fm and

The frequency of the Lower Side Band is given by,

fLSB = fc - fm

II) DSB-SC and DSB-SC are the types of Amplitude Modulation Schemes. DSB-SC is an
acronym for Double Sideband Suppressed Carrier and DSB-FC is an acronym for Double
Sideband Full Carrier.

In DSB-FC around 67% or two-thirds of the total power is wasted by the carrier.

So, in DSB-SC the carrier is suppressed, but this suppression won't affect the message signal.

Therefore, the DSB-SC is selected over DSB-FC.

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5 b) A modulating signal 10 cos (2𝜋 × 103𝑡) is used to modulate a carrier signal

20 cos (2𝜋 × 104𝑡). Compute the modulation index, % of modulation index,
frequency of sideband components and their amplitudes. What will be the
bandwidth of modulated signal?

Modulating Signal Vm=10 cos (2𝜋 × 103𝑡)

Carrier Signal Vc= 20 cos (2𝜋 × 104𝑡)

Bandwidth of modulated signal =2fm=2*1KHz=2KHz

6 a) Illustrate the Amplitude modulation for single tone information.

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COURSE CODE: 20EC0451 R20

6 b) Discuss the advantages and disadvantages of DSB-SC.

Advantages of SSB-SC:

1. SSB required half the bandwidth required of Amplitude wave and DSB-SC signals.
2. Due to suppression of carrier and one side band power is saved.
3. Reduced interference of noise. This is due to the reduced bandwidth as the bandwidth
increases the amount of noise added to the signal with increase.
4. Fading does not occur in SSB transmission.
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5. Fading means that a signal alternating increases and decreases in strength as it is picked up
by the receiver.
6. It occurs because the carrier and sideband may reach the receiver shifted in time and phase
with respect to each other.

Disadvantages of SSB-SC:

1. The generation and reception of SSB signal is a complex process.

2. Since carrier is absent the SSB transmitter and receiver need to have an excellent frequency
stability.
3. The SSB modulation is expensive and highly complex to implement.

7 a) What is DSB-SC Modulation? Explain the Time and Frequency domain

expressions of DSB-SC wave.
To overcome the drawback of power voltage is AM wave (DSB-FC) an DSB-SC method is
used.

DSB-SC is a method of transmission where only the two sidebands are transmitted without the
carrier (suppressing carrier)

(or)

The conventional AM wave in which the carrier is suppressed is called DSB-SC Modulation.

Time domain Representation of DSB-SC wave:

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The amplitude spectrum drawn above exhibits the following factor

1. on either sides of +fc or - fc, we have two sidebands designated as lower and upper
sideband.
2. The impulse are absent at +fc or - fc in the amplitude spectrum signifying the fact that the
carrier form is suppressed in the transmitted wave.
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3. The minimum transmission bandwidth required is 2W that is twice the message bandwidth.

7 b) Define demodulation. Explain any one amplitude demodulation technique.

Demodulation:

Demodulation or detector is the process of recovering the original message signal from the
modulated wave at the receiver. Demodulation is the reverse of the modulation process.

There are two types of detectors

1. Square law detector

2. Envelope detector

Square Law Detector:

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COURSE CODE: 20EC0451 R20

8 a) Explain single tone modulation for transmitting only upper side band (USB)
frequency of SSB modulation.

Upper side band (USB) frequency of SSB Modulation:

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8 b) Explain briefly about the various applications of SSB-SC.

Application of SSB-SC:

1. SSB is used in the applications where the power saving is required in mobile systems.
2. SSB is also used in applications in which bandwidth requirements are low.
Eg: - Point to Point communication, Land, Air and Maritime Mobile Communications,
Television, Telemetry, Military communications, Radio navigation and Amateur radio are the
greatest users of SSB in one form or another.

9 a) Explain single tone modulation for transmitting only lower side band (LSB)
frequency of SSB modulation.

Lower side band (LSB) frequency of SSB Modulation:

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9 b) What are the advantages and disadvantages of SSB-SC signal

Advantages of SSB-SC:

1. SSB required half the bandwidth required of Amplitude wave and DSB-SC signals.
2. Due to suppression of carrier and one side band power is saved.
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3. Reduced interference of noise. This is due to the reduced bandwidth as the bandwidth
increases the amount of noise added to the signal with increase.
4. Fading does not occur in SSB transmission.
5. Fading means that a signal alternating increases and decreases in strength as it is picked up
by the receiver.
6. It occurs because the carrier and sideband may reach the receiver shifted in time and phase
with respect to each other.

Disadvantages of SSB-SC:

1. The generation and reception of SSB signal is a complex process.

2. Since carrier is absent the SSB transmitter and receiver need to have an excellent frequency
stability.
3. The SSB modulation is expensive and highly complex to implement.

10 a) Comparison of Amplitude modulation techniques.

10 b) List the advantages and disadvantages of Double side-band Full carrier.

Advantages of DSB-FC:
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Disadvantages of DSB-FC:
COURSE CODE: 20EC0451 R20
SIDDARTH INSTITUTE OF ENGINEERING & TECHNOLOGY: PUTTUR
(AUTONOMOUS)
Siddharth Nagar, Narayanavanam Road – 517583
QUESTION BANK (DESCRIPTIVE)

Subject with Code: ICS(20EC0451) Course& Branch: B.Tech & CSE,CSM.CIC

Year & Sem: III-B.Tech.& I-Sem. Regulation: R20

UNIT- II
Angle Modulation & Demodulation

1 a) Define angle modulation. Classify different types of angle modulation and

advantages of Angle modulation.

Angle Modulation:

Angle Modulation is the process in which either the frequency or phase of the carrier wave is
varied in accordance with the instantaneous amplitude of the message signal is termed as Angle
Modulation.

Types of Angle Modulation:

Angle Modulation is classified into two types. They are

1. Frequency Modulation
2. Phase Modulation

Frequency Modulation:

The modulation in which frequency of carrier wave is varied in accordance with the instantaneous
amplitude of the message signal is known as frequency modulation.

Phase Modulation:

The modulation in which Phase of carrier wave is varied in accordance with the instantaneous
amplitude of the message signal is known as phase modulation.

Advantages of Angle Modulation:

1. It is used in Noise Reduction.

2. It is used in Improved system fidelity.
3. It is used in efficient use of power.
4. It is used in wide bandwidth.
5. It is used in high system performance.
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1 b) Analyze the expression of single tone NBFM.

Narrow Band FM:

1. A Narrow Band FM is the FM with a small bandwidth. The Modulation Index ‘β’ of NBFM is small as
compared to one radian.
2. The NBFM has a Narrow Bandwidth which is equal to twice the message bandwidth.
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COURSE CODE: 20EC0451 R20

2 a) Define Frequency Modulation with necessary waveforms.

Frequency Modulation:

The modulation in which frequency of carrier wave is varied in accordance with the instantaneous
amplitude of the message signal is known as frequency modulation.

Mathematical Representation
Let the carrier frequency be fc
The frequency at maximum amplitude of the message signal = fc + Δf
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The frequency at minimum amplitude of the message signal = fc − Δf
The difference between FM modulated frequency and normal frequency is termed as Frequency
Deviation and is denoted by Δf.
The deviation of the frequency of the carrier signal from high to low or low to high can be termed
as the Carrier Swing.
Carrier Swing = 2 × frequency deviation
= 2 × Δf

Equation for FM WAVE

The equation for FM wave is
s(t) = Ac cos [2πfct + βsin(2πfmt]
Where,

Ac = the amplitude of the carrier

fc = frequency of the carrier

m(t) = message signal

FM can be divided into Narrowband FM and Wideband FM.

2 b) Derive the expression of Frequency modulation.

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COURSE CODE: 20EC0451 R20

3 a) Explain the generation of Narrowband FM and Wideband FM.

Generation of NBFM:
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The block diagram of NBFM modulator as shown in the above figure. Here the integrator is used to
integrate the modulating signal m(t)m(t). The Carrier signal A ccos(2πfct) is the phase shifted by -90o to
get Acsin(2πfct) with the help of -90o phase shifter. The product modulator has two inputs
∫ 𝑚(𝑡)𝑑𝑡 ∫ 𝑚(𝑡)𝑑𝑡 and Acsin(2πfct). It produces an output which is the product of these two points.

Positive and Negative signs are assigned for the carrier signal and other form at the input of the summer
block. Finally, the summer block produces NBFM wave.

Generation of WBFM:

Consider the following block diagram

COURSE CODE: 20EC0451 R20

A narrowband FM signal can be generated easily using the block diagram of the
narrowband FM modulator that was described in a previous lecture. The narrowband FM
modulator generates a narrowband FM signal using simple components such as an
integrator (an Op-Amp), oscillators, multipliers, and adders. The generated narrowband FM
signal can be converted to a wideband FMsignal by simply passing it through a non–linear
device with power P. Both the carrier frequency and the frequency deviation Δf of the
narrowband signal are increased by a factor P. Sometimes, the desired increase in the carrier
frequency and the desired increase in Δf are different. In this case, we increase Δf to the
desired value and use a frequency shifter (multiplication by a sinusoid followed by a BPF)
to change the carrier frequency to the desired value.

3 b) What are the advantages, disadvantages, and applications of FM.

Advantages and Disadvantages of Frequency Modulation:

Advantages Disadvantages
Less interference and noise. Equipment cost is higher. Has a large bandwidth.
Power Consumption is less as More complicated receiver and transmitter
compared to AM.
Adjacent FM channels are separated The antennas for FM systems should be kept close for
by guard bands. better communication.

Applications of Frequency Modulation

If we talk about the applications of frequency modulation, it is mostly used in radio

broadcasting. It offers a great advantage in radio transmission as it has a larger signal-to-
noise ratio. Meaning, it results in low radio frequency interference. This is the main reason
that many radio stations use FM to broadcast music over the radio.
Additionally, some of its uses are also found in radar, telemetry, seismic prospecting and
in EEG, different radio systems, music synthesis as well as in video-transmission
instruments. In radio transmission, frequency modulation has a good advantage over other
modulation. It has a larger signal-to-noise ratio meaning it will reject radio frequency
interferences much better than an equalpower amplitude modulation (AM) signal.
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4 a) Explain the generation of FM using direct method.

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4 b) What are the differences between NBFM &WBFM?

Sr No. Narrow Band FM Wide Band FM

1 Modulation index is less than 1 Modulation index is greater than 1

2 Frequency Deviation = 5KHz Frequency Deviation = 75KHz

3 Modulating Frequency = 3KHz Modulating frequency range from 30 Hz to

15KHz

Bandwidth 15 times NBFM, Bandwidth

4 Bandwidth = 2fm = 2(δ+fmmax)

5 Maximum modulation index is Maximum modulation index between 5 to 2500

slightly greater than 1
6 It is used for mobile It is used for FM broad casting.
communication.

5 a) Classify Frequency modulation techniques.

FM can be divided into Narrowband FM and Wideband FM.

Narrowband FM:

Narrow band FM, NBFM, is used for signals where the deviation is small enough that the terms
in the Bessel function is small and the main sidebands are those appearing at ± modulation
frequency. The sidebands further out are negligible. For NBFM, the FM modulation index must
be less than 0.5, although a figure of 0.2 is often used. For NBFM the audio or data bandwidth
is small, but this is acceptable for this type of communication. Narrowband FM is widely used
for two-way radio communications. Although digital technologies are taking over, NBFM is
still widely used and very effective. Many two way radios or walkie talkies use NBFM,
especially those which conform to the license-free standards like PMR446 and FRS radio
communications systems. NBFM is ideal for the low cost radio communication systems,
especially those that use small walkie talkies because it can be implemented with a minimum
of amount of circuitry, most of which is low cost. Although digital technology is becoming
much cheaper, narrow band FM is still very cost effective.

Wideband FM:
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Wideband FM is typical used for signals where the FM modulation index is above about 0.5.
For these signals the sidebands beyond the first two terms are not insignificant. Broadcast FM
stations use wide-band FM which enables them to transmit high quality audio, as well as other
facilities like stereo, and other facilities like RDS, etc. The wide bandwidth of wide band FM
is enables high quality broadcast transmissions to be made, combining a wide frequency
response with low noise levels. Once the signal is sufficiently strong, the audio signal to noise
ratio is very good. Sometimes high fidelity FM tuners may use a wide-band filter for strong
signals to ensure the optimum fidelity and performance. Here the quieting effect of the strong
signal will allow for wide-band reception and the full audio bandwidth. For lower strength
signals they may switch to a narrower filter to reduce the noise level, although this will result
in the audio bandwidth being reduced. However, on balance the narrower bandwidth will give
a more pleasing sound when the received signal is low.

5 b) Explain the generation of Narrowband FM wave.

Generation of NBFM:
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The block diagram of NBFM modulator as shown in the above figure. Here the integrator is
used to integrate the modulating signal m(t)m(t). The Carrier signal A ccos(2πfct) is the phase
shifted by -90o to get Acsin(2πfct) with the help of -90o phase shifter. The product modulator
has two inputs ∫ 𝑚(𝑡)𝑑𝑡 ∫ 𝑚(𝑡)𝑑𝑡 and Acsin(2πfct). It produces an output which is the product
of these two points.

Positive and Negative signs are assigned for the carrier signal and other form at the input of
the summer block. Finally, the summer block produces NBFM wave.

6 a) Discuss about transmission bandwidth and Carson’s rule of FM signal.

Theoretically FM has infinite number of side bands. So, the bandwidth required for
transmission is also infinite.

Carson generalized the bandwidth formula for an FM wave. According to him the
approximate formula for approximate formula for computing the bandwidth of an FM signal
generated by a single tone modulating signal frequency ‘fm, is

The above Formula holds good for all values of β

The transmission bandwidth ‘BT’ can also be expressed in terms of frequency deviation ‘Δf’
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6 b) A 20 MHz carrier is frequency modulated by a sinusoidal signal such that the

peak frequency deviation is 100 kHz. Determine the modulation index and the
approximate bandwidth of the FM signal if the frequency of the modulating signal
is: (i) 1kHz (ii) 15 kHz.

Given Data

Fc =20MHz,
Δf=100 kHz
i) If the frequency of the modulating signal is 1KHz then,
Modulation Index of FM signal is,
β (or) mf = Δf/fm
=100KHz/1
KHzβ = 100
According to Carson’s rule, Bandwidth of FM signal is
B=2(Δf+fm) =2(100KHz+1KHz)
=2(101KHz)
B=202KHz.
ii) If the frequency of the modulating signal is 15KHz then,
Modulation Index of FM signal is,
β (or) mf = Δf/fm
=100KHz/15
KHzβ = 6.66
According to Carson’s rule, Bandwidth of FM signal is
B=2(Δf+fm) =2(100KHz+15KHz)
=2(115KHz)
B=230KHz.

7 a) Differentiate between the Amplitude Modulation and Frequency

Modulation.

S.N
FM AM
o

Amplitude of AM wave willchange with

Amplitude of FM wave is constant.
1. the modulatingvoltage.
It is independent of the modulation
index.
Hence, transmitted power remains Transmitted power is dependent on the
2.
constant.It is independent of mf . modulation index.

Carrier power and one sidebandpower

3. All the transmitted power is useful.
are useless.
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AM receivers are not immune to noise.

4. FM receivers are immune to noise.

It is possible to decrease noise

5. This feature is absent in AM.
further byincreasing deviation.

Bandwidth = 2fm. It is not dependent on

Bandwidth = 2[Δf+fm]. The
6. the modulation index.
bandwidthdepends on modulation
index.
BW is large. Hence, wide channel is
7. BW is much less than FM.
required.

Space wave is used for propagation. Ground wave and sky wave propagation
8. So, radius of transmission is limited to is used. Therefore, large area is covered
line of sight. than FM.

Hence, it is possible to operate Not possible to operate morechannels

9.
severaltransmitters on same on same frequency.
frequency.
FM transmission and reception
10. AM equipment’s are less complex.
equipment are more complex.

The number of sidebands having

Number of sidebands in AM willbe
11. significant amplitudes depends on
constant and equal to 2.
modulation index mf .
The information is contained in the
The information is contained
12. amplitude variation of the carrier.
in thefrequency
variation of the carrier.
13. FM Wave AM Wave

14. Applications: Radio, TV

Applications: Radio and TV
broadcasting, police wireless, point to
broadcasting.
point communications

7 b) Describe the construction and functionality of balanced slope detector.

Balanced Slope Detector:

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8 a) Describe the functionality of each block of phase shift discriminator.

Phase shift discriminator:

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Phase shift discriminator is also call it as a foster seeley discriminator and it converts
frequency variations to amplitude variations.

The basic operation of the circuit can be explained by looking at the instances when the
instantaneous input equals the carrier frequency, the two halves of the tuned transformer circuit
produce the same rectified voltage and the output is zero.

If the frequency of the input changes, the balance between the two halves of the transformer
secondary changes, and the result is a voltage proportional to the frequency deviation of the
carrier.

Looking in more detail at the circuit,the Foster-Seeley circuit operates using a phase difference
between signals. To obtain the different phased signals a connection is made to the primary side
of the transformer using a capacitor, and this is taken to the center tap of the transformer. This
gives a signal that is 90° out of phase. Whenan un-modulated carrier is applied at the center
frequency, both diodes conduct, to produce equaland opposite voltages across their respective
load resistors.

The capacitors C1 and C2 provide a similar filtering function. The voltage of diode D1 is not
equal to voltage at diode D2.

The amplitude variations are rectified and filtered to produce a DC output voltage.

8 b) Explain the block diagram of indirect method in FM generation.

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Figure shows the block diagram of a indirect FM signal system.

In indirect method the message signal m(t) is first period through an integrator before applying
it to phase modulator as shown in figure 1

The carrier signal is generated by using crystal oscillator.

Carrier signal is generated by using crystal oscillator because it provides very high frequency
stability.

The operation of indirect method is divided into two parts as follows

1. Generate a NBFM wave using a phase modulator.

2. using the frequency multiplier of and mixer to obtain the required values of frequency
deviation and modulator modulation index (that is WBFM)

In order to minimize the distortion in the phase modulator, the maximum phase deviation or
modulation index ‘β’ is kept small there by resulting a NBFM signal.

The output of the narrow band phase modulator is then multiplied by a frequency multiplier,
producing the desired WBFM wave as shown in figure 2
COURSE CODE: 20EC0451 R20

A frequency multiplier consists of a memoryless nonlinear device followed by a BPF as shown

in Figure 3

9 a) Explain briefly about Phase Modulation with necessary waveforms.

Phase Modulation:

The modulation in which Phase of carrier wave is varied in accordance with the
instantaneous amplitude of the message signal is known as phase modulation.

The amplitude of the carrier signal remains constant analysis

Message signal is expressed as

m(t)= Amcos(2πfmt)
where
Am is the amplitude of the message signal
fm is the frequency of the message signal

Carrier signal is expressed as

c(t)=Ac cos (fct+∅)

c(t)= Ac cos( ɵ(𝑡))
(or)

c(t)=Ac cos (fct)

For phase modulation the instantaneous phase deviation is given as

ɵi(t)= (fct+kpm(t))

kp is known as phase sensitivity

The phase modulation signal is given as

Spm(t)= Ac cos (2𝜋fct+kpm(t))

Substituting m(t)

Spm(t)= Ac cos (2𝜋fct+kpAmcos(2πfmt))

Modulation Index of PM
COURSE CODE: 20EC0451 R20

mp=kpAm

In phase modulation, the modulation index depends only on the amplitude of the modulating
signal.

Modulation index in phase modulation is directly proportional to amplitude of message

signal.

9 b) Derive the expression of modulation index of Phase modulation

COURSE CODE: 20EC0451 R20

Mathematical Representation

The equation for instantaneous phase ϕi in phase modulation is

ϕi=kpm(t)
Where,
 kp is the phase sensitivity
 m(t)m(t) is the message signal
The standard equation of angle modulated wave is
s(t)=Accos(2πfct+ϕi)
Substitute, ϕi value in the above equation.
s(t)=Accos(2πfct+kpm(t))
This is the equation of PM wave.
If the modulating signal, m(t)=Amcos(2πfmt), then the equation of PM wave
will be
s(t)=Accos(2πfct+βcos(2πfmt))
Where,
 ββ = modulation index = Δϕ=kpAm
 Δϕ is phase deviation
Phase modulation is used in mobile communication systems, while frequency
modulation is used mainly for FM broadcasting.

10 a) Compare Phase Modulation and Frequency Modulation.

S.N
FM PM
o
The frequency deviation is linearly The phase shift of the carrier is linearly
1. proportional to instantaneous amplitude proportional to instantaneous amplitude of
of the modulating signal. the modulating signal.

Frequency modulation is direct method Phase modulation is In direct method of

2.
of producing FM signal Producing FM
 COURSE CODE: 20EC0451 R20
The modulation index of an FM signal is The modulation index is proportional to
3. the ratio of the frequency deviation to the the maximum amplitude of the
modulating system. modulating signal.
Amplitude of the FM wave is constant. Amplitude of the PM wave is constant.
4.

Noise is better suppressed in FM systems Noise immunity is interior to that of FM

5.
as compared to PM system
FM is mainly used for FM broad casting PM is used in mobile communication
6. used for entertainment purposes system.

10 b) Differentiate between the Frequency Modulation and Phase Modulation

with its modulated waveforms.
FM PM
S.N
The frequency deviation is linearly The phase shift of the carrier is linearly
1. proportional to instantaneous amplitude proportional to instantaneous amplitude of
of the modulating signal. the modulating signal.

Frequency modulation is direct method Phase modulation is In direct method of

2.
of producing FM signal Producing FM
The modulation index of an FM signal is The modulation index is proportional to
3. the ratio of the frequency deviation to the the maximum amplitude of the
modulating system. modulating signal.
Amplitude of the FM wave is constant. Amplitude of the PM wave is constant.
4.

Noise is better suppressed in FM Noise immunity is interior to that of FM

5.
systems as compared to PM system
FM is mainly used for FM broad PM is used in mobile communication
6. casting used for entertainment purposes system.

7. FM Wave: PM Wave:
COURSE CODE: 20EC0451 R20
COURSE CODE: 20EC0451 R20

SIDDARTH INSTITUTE OF ENGINEERING & TECHNOLOGY: PUTTUR

(AUTONOMOUS)
Siddharth Nagar, Narayanavanam Road – 517583
QUESTION BANK (DESCRIPTIVE)

Subject with Code: ICS(20EC0451) Course& Branch: B.Tech & CSE,CSM.CIC

Year & Sem: III-B.Tech.& I-Sem. Regulation: R20

UNIT III
Noise in Communication Systems

1 a) Define Noise and list the different types of noises.

Noise:

Noise is an unwanted signal, which interferes with the original message signal and corrupts the
parameters of the message signal. This alteration in the communication process, leads to the message
getting altered. It most likely enters at the channel or the receiver.
The noise signal can be understood by taking a look at the following figure.

Hence, it is understood that the noise is some signal which has no pattern and no constant frequency
or amplitude. It is quite random and unpredictable. It can’t be completely eliminated through Measures
are usually taken to reduce it.
Most common examples of noise are −
 Hiss sound in radio receivers
COURSE CODE: 20EC0451 R20
 Buzz sound amidst of telephone conversations
 Flicker in television receivers, etc
Types of Noise
The classification of noise is done depending on the type of the source, the effect it shows or the
relation it has with the receiver, etc.
There are two main ways in which noise is produced. One is through some external source while the
other is created by an internal source, within the receiver section.
Internal Noise:
 Thermal Noise
 Shot Noise
 Excess Noise
External Noise:
 Atmospheric Noise
 Industrial Noise
 Space or Extra-terrestrial Noise
Thermal Noise:
The noise which is produced by random motion of electrons in a conductor due to thermal agitation is
termed as ‘Thermal Noise’ or ‘white noise’ or ‘Johnson noise’.
Generally, the thermal noise power (PN) is directly proportional to the absolute temperature (T). This
noise also has a direct relation with the noise power bandwidth B (i.e. the bandwidth over which the
noise is measured)
Mathematically PN∝ TB
K is Proportional constant (or) Boltzmann’s constant (1.38*10 -23 joules/kelvin)
T is Absolute temperature (in kelvins)
B is Noise power bandwidth (in Hertz)
Shot Noise:
Short noise is caused due to random variation in current flow in active components.
Main source of short noise are
 Semiconductor diodes
 Transistors
 Tubes
Excess Noise:
Excess noise may be produced due to variations in the carrier density. It usually occurs in tubes,
semiconductors and carbon resistors.
COURSE CODE: 20EC0451 R20
Atmospheric Noise:
The noise originated from the natural sources which cause disturbances in atmosphere is known as
Atmospheric noise. This noise is generally caused due to spurious radio waves that include voltage in
the antennas.
Lighting dis charge in thunderstorms is the best example of atmospheric noise. This noise is also termed
as static because of a static electricity discharge cause during the lightening.
Industrial Noise:
Industrial noise is the noise produced due to the equipment’s that produce sparks.
The common source of equipment noise are
 Automatic and aircraft ignition.
 Electric motors
 Switching equipment
 Leakage from high voltage lines
 Heavy electric machines
 Computers
Space Noise:
Space noise is further classified into two groups
 Solar Noise
 Cosmic Noise
Solar Noise:
Solar noise which is produced from the sun generates significant amount of noise.
Cosmic Noise:
Cosmic noise is the noise produced due to star or group of stars
The impact of space noise is more at very high frequency range (VHF).

1 b) Explain briefly about Noise in communication system.

Noise:

Noise is an unwanted signal, which interferes with the original message signal and corrupts the
parameters of the message signal. This alteration in the communication process, leads to the message
getting altered. It most likely enters at the channel or the receiver.
The noise signal can be understood by taking a look at the following figure.
COURSE CODE: 20EC0451 R20

Hence, it is understood that the noise is some signal which has no pattern and no constant frequency
or amplitude. It is quite random and unpredictable. It can’t be completely eliminated through Measures
are usually taken to reduce it.
Most common examples of noise are −
 Hiss sound in radio receivers
 Buzz sound amidst of telephone conversations
 Flicker in television receivers, etc
Types of Noise
The classification of noise is done depending on the type of the source, the effect it shows or the
relation it has with the receiver, etc.
There are two main ways in which noise is produced. One is through some external source while the
other is created by an internal source, within the receiver section.
Internal Noise:
 Thermal Noise
 Shot Noise
 Excess Noise
External Noise:
 Atmospheric Noise
 Industrial Noise
 Space or Extra-terrestrial Noise
COURSE CODE: 20EC0451 R20
Thermal Noise:
The noise which is produced by random motion of electrons in a conductor due to thermal agitation is
termed as ‘Thermal Noise’ or ‘white noise’ or ‘Johnson noise’.
Generally, the thermal noise power (PN) is directly proportional to the absolute temperature (T). This
noise also has a direct relation with the noise power bandwidth B (i.e. the bandwidth over which the
noise is measured)
Mathematically PN∝ TB
K is Proportional constant (or) Boltzmann’s constant (1.38*10 -23 joules/kelvin)
T is Absolute temperature (in kelvins)
B is Noise power bandwidth (in Hertz)
Shot Noise:
Short noise is caused due to random variation in current flow in active components.
Main source of short noise are
 Semiconductor diodes
 Transistors
 Tubes
Excess Noise:
Excess noise may be produced due to variations in the carrier density. It usually occurs in tubes,
semiconductors and carbon resistors.
Atmospheric Noise:
The noise originated from the natural sources which cause disturbances in atmosphere is known as
Atmospheric noise. This noise is generally caused due to spurious radio waves that include voltage in
the antennas.
Lighting dis charge in thunderstorms is the best example of atmospheric noise. This noise is also termed
as static because of a static electricity discharge cause during the lightening.
Industrial Noise:
Industrial noise is the noise produced due to the equipment’s that produce sparks.
The common source of equipment noise are
 Automatic and aircraft ignition.
 Electric motors
 Switching equipment
 Leakage from high voltage lines
 Heavy electric machines
 Computers
COURSE CODE: 20EC0451 R20
Space Noise:
Space noise is further classified into two groups
 Solar Noise
 Cosmic Noise
Solar Noise:
Solar noise which is produced from the sun generates significant amount of noise.
Cosmic Noise:
Cosmic noise is the noise produced due to star or group of stars
The impact of space noise is more at very high frequency range (VHF).

2 a) Explain noise figure and derive its expression.

Noise Figure:
The ratio of signal to noise at input to the signal to noise at the output at the receiver. It is denoted by
‘NF’ or ‘F’.
NF= Signal to noise (S/N) input/(S/N) output
COURSE CODE: 20EC0451 R20

2 b) A mixer stage has a noise figure of 20 dB and it is preceded by another amplifier with a noise figure of 9
dB and an available power gain of 15 dB. Calculate the overall noise figure referred to the input. A cellular
telephone system provides a wireless connection to the PSTN for any user location within the radio range
of the system.

Solution:
(i) First, we convert dB into equivalent power ratios as under:
F₁ = 9 dB or 9 = 10 log F₁ or F₁ = Antilog (0.9) = 7.94
F₂ = 20 dB or 20 = 10 log F₂ or F₂ = Antilog (2) = 100

G₁ = 15 dB or 15 = 10 log G₁ or G₁ = Antilog (1.5) = 31.62

(ii) Then, we calculate the overall noise factor as under:
𝐹2 − 1
𝐹 = 𝐹1 +
𝐺1
100−1
𝐹 = 7.94 + = 11.07
31.62

(iii) The overall noise figure = 10 log F = 10 log10 (11.07) = 10.44 dB

3 a) Explain briefly about Signal to Noise Ratio.

COURSE CODE: 20EC0451 R20
COURSE CODE: 20EC0451 R20
3 b) Calculate the input signal to noise ratio for an amplifier with an output signal to noise ratio of 16
dB and a noise figure of 5.4 dB.

Given, (S/N) output = 16 dB

Noise Figure, FdB = 5.4 dB

(S/N) input =?

We know that,

FdB = (S/N) output dB – (S/N) input dB

5.4 dB = 16 dB – (S/N) input dB

(S/N) input dB = 21.4 dB

or Input signal to noise ratio, (S/N) input = 138.04

4 a) Explain Pulse Amplitude modulation with its waveforms.

Pulse Amplitude Modulation:

The carrier is in the form of narrow pulses having frequency fc. The uniform sampling
takes place in multiplier to generate PAM signal. Samples are placed Ts sec away from
each other.

Fig. PAM Modulator

Mathematical Analysis
In a flat top PAM, the top of the samples remains constant and is equal to the instantaneous value of
the baseband signal n(t) at the start of sampling.

The duration or width of each sample is τ and sampling rate is equal to,
COURSE CODE: 20EC0451 R20
From fig.1 (b), it may be noted that only starting edge of the pulse represents instantaneous value of
the baseband signal x(t).

Also, the flat top pulse of g(t) is mathematically equivalent to the convolution of instantaneous sample
and a pulse h(t) as depicted in fig.2.

Fig.2 : Convolution of any function with delta function is equal to that function

This means that the width of the pulse in g(t) is determined by the width of h(t) and the sampling instant
is determined by the delta function.

In fig.1 (b), the starting edge of the pulse represents the point where baseband signal is sampled and
width is determined by function h(t).

Therefore, g(t) will be expressed as,

………… (1)
This equation has been explained in fig.3 below.

Fig.3: (a) Baseband signal x(t), (b) Instantaneously sample signal s(t), (c) Constant pulse width function
h(t), (d) Flat top sampled PAM signal g(t) obtained through convolution of h(t) and s(t)
COURSE CODE: 20EC0451 R20
4 b) Explain the process of demodulation of a PAM signals.

PAM Demodulator:
⚫ The PAM demodulator circuit which is just an envelope detector followed by a second order op-amp
low pass filter (to have good filtering characteristics) is as shown below

5 a) What are the advantages and disadvantages of PAM signal.

Advantages of pulse amplitude modulation:

 Transmitter and receiver circuits are simple and easy to construct.

 PAM can generate other pulse modulation signals and can carry the message at the same time.
 Simple process for both modulation and demodulation is used in PAM.
 Data can be transmitted quickly, efficiently and effectively through usual copper wires.

Disadvantages of pulse amplitude modulation:

 Bandwidth should be large for transmission PAM modulation.

 Noise will be great.
 Pulse amplitude signal varies so the power required for transmission will be more.
 For this modulation, noise immunity is low as compared to other types

5 b) Define Pulse Width Modulation and classify it with proper diagram.

PWM is also called Pulse Duration Modulation (PDM), Pulse Length Modulation (PLM) and
Definition: In PWM, Width of the pulses of the carrier pulse train is varied in accordance with the
modulating signal.

Types of Pulse Width Modulation Technique

There are three conventional types of pulse width modulation technique and they are named as follows:
COURSE CODE: 20EC0451 R20
 Trail Edge Modulation – In this technique, the signal’s lead edge is modulated, and the trailing
edge is kept fixed.
 Lead Edge Modulation – In this technique, the signal’s lead edge is fixed, and the trailing edge
is modulated.
 Pulse Center Two Edge Modulation – In this technique, the pulse centre is fixed and both
edges of the pulse are modulated.

6 a) Explain the process involved in generation of PWM wave.

PWM is also called Pulse Duration Modulation (PDM), Pulse Length Modulation (PLM) and
Definition: In PWM, Width of the pulses of the carrier pulse train is varied in accordance with the
modulating signal.

1.The PWM pulses obtained at the comparator output are applied to a mono stable multi vibrator
which is negative edge triggered.
2.Hence for each trailing edge of PWM signal, the monostable output goes high.
3.It remains high for a fixed time decided by its RC components.
COURSE CODE: 20EC0451 R20
4.Thus, as the trailing edges of the PWM signal keeps shifting in proportion with the modulating
signal, the PPM pulses also keep shifting.
5.Therefore, all the PPM pulses have the same amplitude and width. The information is conveyed via
changing position of pulses.

6 b) Describe the demodulation technique of PWM signal.

Demodulation:
COURSE CODE: 20EC0451 R20

7 a) What are the advantages and disadvantages of PWM signal?

Advantages of pulse width modulation:

 The demodulation process employed in PWM system is simple

 PWM has good noise immunity as amplitude of the signal is kept constant.
 PWM system does not require synchronization between transmitter and receiver.

Disadvantages of pulse width modulation:

 Due to pulses varying in width, the signal power is also varied.

 The transmission of PWM signal requires large bandwidth when compared to PAM signal.
 PWM is not suitable for time division multiplexing.

7 b) Differentiate between the Pulse Amplitude Modulation and Pulse Width Modulation with its
modulated waveforms.
COURSE CODE: 20EC0451 R20

8 a) Explain about the generation of PPM signal.

1.The PWM pulses obtained at the comparator output are applied to a mono stable multi vibrator
which is negative edge triggered.
2.Hence for each trailing edge of PWM signal, the monostable output goes high.
3.It remains high for a fixed time decided by its RC components.
4.Thus, as the trailing edges of the PWM signal keeps shifting in proportion with the modulating
signal, the PPM pulses also keep shifting.
5.Therefore, all the PPM pulses have the same amplitude and width. The information is conveyed via
changing position of pulses.
COURSE CODE: 20EC0451 R20

8 b) Elaborate demodulation of PPM signal.

COURSE CODE: 20EC0451 R20

9 a) What are the advantages and disadvantages of PPM signal

Advantages of pulse Position modulation:

 Demodulation process is simple in PPM.

 In PPM, noise interference is less as amplitude is kept constant.
 Because of the constant pulse widths and amplitudes, signal power is also constant in PPM.

Disadvantages of pulse Position modulation:

 PPM signals require synchronization between transmitter and receiver.

 The transmission bandwidth required for a PPM signal is large when compared to that of
PAM signal.
 PPM is not suitable for time division multiplexing.
COURSE CODE: 20EC0451 R20
9 b) Differentiate between the Pulse Position Modulation and Pulse Width Modulation with its
modulated waveforms.

10 a) Define pulse modulation and different types of pulse modulation in analog and digital
communication.
Pulse Modulation: Pulse Modulation can be defined as the variation of amplitude or width or position
of a higher frequency discrete carrier signal in accordance with the amplitude of an analog modulating
signal.
Pulse Modulation can be broadly classified into two major types. They are
 Analog pulse modulation
 Digital pulse modulation
Analog Pulse Modulation: In analog pulse modulation technique amplitude or time of a carrier is
varied in accordance with the instantaneous value of analog modulating signal.
Analog pulse modulation technique is further classified into two types namely
Pulse Amplitude Modulation (PAM): A modulation technique in which the amplitude of the carrier
signal consisting of periodic train or pulses is varied linearly with amplitude of the message signal is
known as pulse amplitude modulation.
COURSE CODE: 20EC0451 R20
Pulse Time Modulation (PTM): The modulation technique in which the timing of the carrier pulse is
changed with respect to the amplitude of the message signal is known as pulse time modulation.
The pulse time modulation is further divided into two types. They are
Pulse Width Modulation (PWM): A modulation technique in which, the width of the carrier signal
consisting of periodic train pulse is varied linearly with the amplitude of the message signal is known
as pulse width modulation. It is also termed as Pulse Duration Modulation(PDM).
Pulse Position Modulation (PPM): A modulation technique in which, the position of the carrier signal
consisting of periodic train pulse is varied linearly with the amplitude of the message signal is called
pulse position modulation.

Digital Pulse Modulation: In digital pulse modulation technique, analog modulation signal is
converted into discrete signal by changing the amplitude of carrier pulse train. These discrete levels are
then represented by digital codes for transmission.
Digital pulse modulation technique is further classified into two types. They are
 Pulse Code Modulation (PCM)
 Delta Modulation (DM)
Pulse Code Modulation (PCM): Pulse code modulation can be defined as a signal encoding technique,
where in an analog information signal is sampled and amplitude of these samples is approximated to
the nearest value among the finite set of discrete levels. This approximation is carried out such that both
amplitude and times is indicated in discrete format.
PCM is further classified into Differential Pulse Code Modulation (DPCM). In this type of modulation,
difference in the amplitude levels of two successive samples is transmitted instead of the absolute value
of the actual sample.
Delta Modulation (DM): Delta modulation is the simplest form of DPCM wherein difference between
successive samples are encoded into data streams of n-bits. It employs single bit DPCM code to digitally
transmit analog signals.
It is further classified into Adaptive Delta Modulation (ADM) technique. ADM can be defined as a
delta modulation technique which varies step size of the signal, based on the amplitude characteristics
of the applied analog signal.

10 b) Compare PAM, PWM and PPM techniques.

COURSE CODE: 20EC0451 R20
COURSE CODE: 20EC0451 R20

SIDDARTH INSTITUTE OF ENGINEERING & TECHNOLOGY: PUTTUR

(AUTONOMOUS)
Siddharth Nagar, Narayanavanam Road – 517583
QUESTION BANK (DESCRIPTIVE)

Subject with Code: ICS(20EC0451) Course& Branch: B.Tech & CSE,CSM.CIC

Year & Sem: III-B.Tech.& I-Sem. Regulation: R20

UNIT IV
Digital Communication

1 a) Define Digital Communication and draw the basic block diagram of Digital

communication system.
Digital Communication: Digital communication is defined as the process of exchanging information
between two or more communicating points using digital signals.

1 b) Explain the function of each block of Digital communication system.

Source of information: There are two types of source information.
 Analog Information Source
COURSE CODE: 20EC0451 R20
 Digital information source
Analog Information source: Microphone actuated by a speech, T.V channel, camera, scanning a sense
continuous amplitude signal
Digital Information source: Analog information source can be transferred from discrete to sampling.
Source Encoder and Decoder: The symbol can’t be transmitted directly and then converted into 0’s
and 1’s (Binary) is called source encoder. It converts binary output of the channel.
Channel of the decoder into a symbol form is called source decoder.
Channel Encoder and Decoder: Channel encoder adds some binary bits to the input signal is called
channel encoder.
Channel decoder at the receiver is able to reconstruct error sequence and reduce the distortion is
called channel decoder.
Modulator: Modulator can be used to minimized the effect of channel noise to match the frequency
of spectrum and capability for many signals is called Modulator.
Demodulator: The extraction of the message from the information waveforms produced by the
modulator and reduce the demodulator output of the demodulation is bit stream.
Channel: Channel provides the connection between sources and destination is called channel.

2 a) Explain the Process of Quantization with suitable example.

Quantization: Quantization of signals can be defined as the conversion of an analog information signal
into discrete form, where in infinite number of levels are transformed into finite number of conditions.
In the process of quantization, the peak to peak range of input sample values is divided into decision
levels or thresholds of finite set (or value). Among the available finite set of representation levels, the
output is allotted with a discrete value in the quantization process.
During the rounding off process of analog sample values, a significant amount of error or noise known
as quantization error is produced in the quantizer. This error is directly proportional to the difference
between consecutive quantization levels and inversely proportional to the number of levels for
amplitude range.
COURSE CODE: 20EC0451 R20

2 b) Discuss the different types of Quantization in detail.

Types of Quantization: Based on the distribution of quantization levels to be spaced uniformly, the
process of quantization is classified into two types. They are
 Uniform Quantization
 Non uniform Quantization
Uniform Quantization: The process of quantization wherein the quantization levels (or step size) are
spaced uniformly over the entire range of input signal is known as uniform quantization.
Based on the position of origin, uniform quantization is again subdivided into two types namely,
 Midtread uniform quantization
 Midriser uniform quantization
Midtread Uniform Quantization: In this type of quantization, origin lies in the middle of a tread of
quantization staircase process as shown in figure (i).
COURSE CODE: 20EC0451 R20

It can be observed from figure(i) that the gap between representation levels and decision thresholds are
designated by a common value known as step size (△).
The origin lies at the middle of a staircase riser when representation levels are at +△/2 and -△/2, +3△/2
and -3△/2, +5△/2 and -5△/2, +7△/2 and -7△/2….. and decision thresholds are maintained at 0, +△ and
-△, +2△ and -2△, +3△ and -3△, ………
The number of quantized levels in the midtread quantization is an odd number and is given as,
Number of quantized level=2 m-1
Where m is number of bits used for encoding a sample.
This type of quantization process finds its major application in voice communications.

Midriser Uniform Quantization: In this type of quantization origin lies in the middle of a rise of
quantization staircase process as shown in figure (ii).
COURSE CODE: 20EC0451 R20

The number of quantized levels in the midriser quantization is an even number and is given as,
Number of quantized level=2m-1
Where m is number of bits used for encoding a sample.
Non Uniform Quantization: The process of quantization where in the quantization levels are not
spaced uniformly and step size varies with respect to the relative amplitude level of sampled value is
known as non-uniform quantization.

3 a) Illustrate the different types of Quantization noise.

Quantization Noise: The difference between the original sampled value are quantized signal is
known as quantization error.
E=xq(nT s)-x(nT s)
Let us consider the input signal α(t) is continuous and linearly varied with the range of -xmax to xmax
The input signal x(t) is sampled to produce x(nT s) and quantized into q quantization levels
The total amplitude= xmax –(-xmax) = xmax +xmax=2 xmax
Let the total amplitude is divided into q quantization levels then the step size is
2 xmax
Step size △= 𝑞
COURSE CODE: 20EC0451 R20
If the input is normalized to the amplitude of unity, then xmax =1 and - xmax =-1 and △=2/q
If the step size is small, then the quantization error e will be an uniformly distributed variable
therefore the maximum quantization error is
Emax=△/2
Therefore, the quantization error depends on step size. If step size is large the quantization error is
high. To reduce the quantization error of PCM, step size △ should be less than or equal to one
△≥1
Types of Quantization: Based on the distribution of quantization levels to be spaced uniformly, the
process of quantization is classified into two types. They are
 Uniform Quantization
 Non uniform Quantization
Uniform Quantization: The process of quantization wherein the quantization levels (or step size) are
spaced uniformly over the entire range of input signal is known as uniform quantization.
Non Uniform Quantization: The process of quantization where in the quantization levels are not
spaced uniformly and step size varies with respect to the relative amplitude level of sampled value is
known as non-uniform quantization.
3 b) State sampling theorem. What is Nyquist rate and Nyquist interval?
Sampling Theorem: According to sampling theorem, a continuous time signal can be completely
represented in its samples and recovered back, if the sampling frequency fs is greater than (or) equal to
the twice of highest frequency component of the message signal fm.
i.e. fs≥ 2fm
Nyquist Rate: The minimum sampling rate at which both sampling and reconstruction of a signal (from
its samples) can be performed without any distortion is called Nyquist rate. It is denoted by f s and is
given as
i.e. fs=2fm
Nyquist Interval: Nyquist interval can be defined as the maximum time interval between the equally
spaced samples of the signal during which sampling rate is equal to Nyquist rate.
Nyquist interval is equal to reciprocal of Nyquist rate and is given by
Ts=1/2fmax
4 a) Illustrate with a neat block diagram explain PCM transmitter and receiver.
Pulse code Modulation (PCM): Pulse code modulation is an analog to digital convertor. The block
diagram of pulse code modulation system is shown in figure
COURSE CODE: 20EC0451 R20

The block diagram of a pulse code modulation system is shown in figure. It consists of
 Transmitter
 Regenerative repeater
 Receiver
PCM Transmitter: In PCM Transmitter consisting Low pass filter, samples, quantizer and encoder
and finally it generated PCM signal.
Transmission Path: In transmission path one major circuitry is there that is Regenerative repeater
circuit.
PCM Receiver: PCM Receiver consisting decoder Low pass filter and destination finally it gives
D/A (Digital to Analog) conversion and get the original signal.

4 b) What are the advantages & disadvantages of PCM?

Advantages of Pulse Code Modulation:
 It has a higher noise immunity.
 It has a higher transmitter efficiency.
 Easily multiplexed.
 Uniform transmission quality.
 Low manufacturing effect.
 Due to its digital nature we can easily store PCM signals.
COURSE CODE: 20EC0451 R20
Disadvantages of Pulse Code Modulation:
 It requires large bandwidth as compared to other method called an analog system.
 Over loaded appears when modulation signal change between sampling by an amount greater
than the size of the step.
 The difference between the original analog signal and the translated digital signal is called
quantization error.
 Encoding, decoding and also have quantizing circuit of PCM is very complex.
 Noise and cross talk leave low but rise attenuation.

5 a) Explain DPCM system with neat diagram.

Differential Pulse Code Modulation: In the use of Pulse Code Modulation for the digization of a voice
or video signal, the signal is sampled of a rate slightly higher than the Nyquist rate. The resulting
sampled signal is then found to exhibit a high correlation between adjacent samples. It is standard PCM
system.
The base band signal x(t) is sampled of the rate fs=1/Ts sequences of correlated samples Ts second a
part.
COURSE CODE: 20EC0451 R20
COURSE CODE: 20EC0451 R20

5 b) What are the advantages & disadvantages of DPCM.

Advantages of DPCM:
 Bandwidth Requirement of DPCM is less compared to PCM.
 Quantization error is reduced because of prediction filter.
 Number of bits used to represent one sample value of are also reduced compared to PCM.
 Reduce the redundance information.
Disadvantages of DPCM:
 High bit rate.
 Needs prediction filter circuit to be used which is very high complex.
 Practical usage is limit.

6 a) Explain DM (delta modulation system) with suitable diagrams.

Delta Modulation (DM): Delta Modulation transmits only one bit per sample that is the present sample
value is compared with the previous sample value and the indication whether the amplitude is increased
or decreased is sent.
COURSE CODE: 20EC0451 R20
The input signal x(t) is approximated to step signal by the data modulator. The difference is between
input signal x(t) and staircase approximated signal is quantized into only two levels that is +△ or -△
COURSE CODE: 20EC0451 R20
COURSE CODE: 20EC0451 R20

6 b) Compare PCM, DPCM, and DM.

COURSE CODE: 20EC0451 R20

7 a) Draw the block diagram of ASK modulator and demodulator and explain the

operation.
Amplitude Shift Keying: Figure (i) illustrates the block diagram of a ASK waveform generator that
employs aa product modulator with a unipolar binary wave m(t) and carrier wave A c cos(2πfct) as its
inputs.

In the above modulation process, amplitude of the carrier signal switches between the binary values 1
and 0. During the binary value 1 a sinusoidal carrier of amplitude A c and frequency f is transmitted for
the bit duration of Tb secs. On the other hand, the binary value 0, the carrier wave is switched off for
the bit duration of Tb secs.
Figure(ii) illustrates the generation of ASK signal waveform for digital input signal and sinusoidal
analog carrier signal.
COURSE CODE: 20EC0451 R20

It can be observed from figure(ii) that

 ASK signal waveform is undergoes one change for every variation in the binary stream from
logic 1 to logic 0 or from logic 0 to logic 1.
 Logic 1 of the input binary data produces constant amplitude ASK and constant frequency
carrier signals.
 Logic 0 of input binary data produces zero ASK signal and no carrier signal at the ASK
system.
The mathematical expression of ASK signal is given as,

ASK Demodulator:
COURSE CODE: 20EC0451 R20
It can be observed from figure that the ASK demodulator consists of three elements namely balance
modulator, integrator and a decision device.
 Balance modulator is provided with a binary ASK signal along with a high frequency sinusoidal
carrier signal. The output of the balance modulator is operated by the integrator circuit for
successive bit intervals (Tb). this integrator basically performs the function of a low pass filter.
 The output of the integrator circuit is fed to the decision making device. This device compares
the integrator output with a preset threshold level. It produces symbol 1 when the threshold
value is exceeded and produced symbol 0 when the threshold value does not exceed.
Thus, the output of the ASK demodulator produces ASK demodulated output or original digital data.

7 b) Explain with suitable waveforms Amplitude Shift Keying.

Amplitude Shift Keying: Figure (i) illustrates the block diagram of a ASK waveform generator that
employs aa product modulator with a unipolar binary wave m(t) and carrier wave A c cos(2πfct) as its
inputs.

In the above modulation process, amplitude of the carrier signal switches between the binary values 1
and 0. During the binary value 1 a sinusoidal carrier of amplitude A c and frequency f is transmitted for
the bit duration of Tb secs. On the other hand, the binary value 0, the carrier wave is switched off for
the bit duration of Tb secs.

Figure(ii) illustrates the generation of ASK signal waveform for digital input signal and sinusoidal
analog carrier signal.
COURSE CODE: 20EC0451 R20

It can be observed from figure(ii) that

 ASK signal waveform is undergoes one change for every variation in the binary stream from
logic 1 to logic 0 or from logic 0 to logic 1.
 Logic 1 of the input binary data produces constant amplitude ASK and constant frequency
carrier signals.
 Logic 0 of input binary data produces zero ASK signal and no carrier signal at the ASK
system.
The mathematical expression of ASK signal is given as,

8 a) Explain the Binary Frequency shift keying in detail.

BFSK Modulator: Figure illustrates the functional block diagram of a BFSK Modulator.
COURSE CODE: 20EC0451 R20

It can be observed from figure (a) that the digital input data is processed through a bipolar NRZ encoder.
This encoded signal is fed to two independent balance modulator (M1 and M2) which multiply the signal
with a high frequency carrier signal. A linear adder circuit is employed to add the outputs of the two
balance modulators. Thus, a BFSK signal is obtained is obtained at the adder circuit which has a
frequency shift from f1 to f2.

BFSK Demodulator: A BFSK signal can be recovered at the receiver by employing BFSK
demodulator figure illustrates the block diagram of a BFSK demodulator.
COURSE CODE: 20EC0451 R20
It can be observed from figure that a BFSK demodulator consists of two correlators a subtraction and
a decision making device. Each correlator has one balance modulator and one integrator circuit. A
subtractor is employed to perform subtraction operation of two correlator outputs. The output of the
subtractor is then compared with a preset threshold level (usually a zero volt) if using decision value.
The output detected is equal to binary 1 if the compared signal greater than 0 volt. On the other hand,
the output is equal to binary 0 if the signal is less than 0 volt.

8 b) Explain with suitable waveforms Binary Frequency Shift Keying.

BFSK Modulator: Figure illustrates the functional block diagram of a BFSK Modulator.

It can be observed from figure (a) that the digital input data is processed through a bipolar NRZ encoder.
This encoded signal is fed to two independent balance modulator (M1 and M2) which multiply the signal
with a high frequency carrier signal. A linear adder circuit is employed to add the outputs of the two
balance modulators. Thus, a BFSK signal is obtained is obtained at the adder circuit which has a
frequency shift from f1 to f2.
COURSE CODE: 20EC0451 R20
Illustrates the waveforms of a BFSK signal for input digital data and carrier sinusoidal signal.

9 a) Explain the Binary Phase Shift Keying modulator and demodulator.

BPSK Modulator: Figure illustrates the functional block diagram of a BPSK modulator.

It can be observed from figure that the binary input data is converted into its corresponding bipolar
NRZ signals using NRZ encoder. This output is fed to the balance modulator along with a higher
frequency sinusoidal carrier signal. The balance modulator produces BPSK output signal depending on
the phase relationship with the carrier oscillator.
 For binary data 1- Modulator output is in phase with the reference carrier oscillator.
 For binary data 0- Modulator output is 180o out of phase with reference carrier oscillator.
In a Binary Shift Keying(BPSK) technique, the phase of a carrier sinusoidal signal is varied in
accordance with the applied digital data input.
The phase of the sinusoidal carrier signal is usually varied from 0o to 180o. BPSK is known as biphase
modulation or phase reversal keying technique.
COURSE CODE: 20EC0451 R20
BPSK signal undergoes a phase shift from 0o to 180o whenever a transition occurs at the digital binary
data input.
BPSK Demodulator: figure illustrates the functional block diagram of a BPSK demodulator

It can be observed from figure that the BPSK is fed to the correlator circuit which comprises one
balance modulator and one integrator.
A carrier signal generated from carrier oscillator is also provided as input to the balance modulator.
The correlator output is then compared with a preset threshold level (usually zero volt) using decision
making device.
 The detected output is equal to 1 when the input of decision making device is greater than 0
volt.
 The detected output is equal to 0 when the input of decision making is less than 0 volt.

9 b) Explain with suitable waveforms Binary Phase Shift Keying.

BPSK Modulator: Figure illustrates the functional block diagram of a BPSK modulator.

It can be observed from figure that the binary input data is converted into its corresponding bipolar
NRZ signals using NRZ encoder. This output is fed to the balance modulator along with a higher
frequency sinusoidal carrier signal. The balance modulator produces BPSK output signal depending on
the phase relationship with the carrier oscillator.
 For binary data 1- Modulator output is in phase with the reference carrier oscillator.
 For binary data 0- Modulator output is 180o out of phase with reference carrier oscillator.
COURSE CODE: 20EC0451 R20
In a Binary Shift Keying(BPSK) technique, the phase of a carrier sinusoidal signal is varied in
accordance with the applied digital data input.
The phase of the sinusoidal carrier signal is usually varied from 0 o to 180o. BPSK is known as biphase
modulation or phase reversal keying technique.

BPSK signal undergoes a phase shift from 0o to 180o whenever a transition occurs at the digital binary
data input.

10 a) Explain Slope overload distortion & Granular Noise.

Slope overload distortion:

 Slope overload distortion arises because of the large dynamic range of the input signal.
COURSE CODE: 20EC0451 R20
 In figure it can be seen that the rate of rise of input signal x(t) is so high that the staircase signal
cannot approximate it the step size δ becomes the small stair case signal x(t) to follow the stop
segment of x(t). thus large error between the stair case approximated signal and the original
input signal x(t). this error is called slope overloaded distortion.
 To reduce this error, the step size should be increased when slope of the signal x(t) is high.
That is, slope of the staircase u(t)≥ slope of the message signal
𝛿⁄
𝑇𝑠 ≥ max [ d/dt x(t)]

Granular Noise:
 This noise occurs when the step size is too large compared to small variations in the input signal
i.e. for very small variations in the input signal, the staircase signal is changed by large amount
because of large step size δ.
 The error between the input and approximated signal is called Granular noise. The solution of
this problem is to make step size small.

10 b) Compare ASK, FSK, and PSK.

COURSE CODE: 20EC0451 R20
COURSE CODE: 20EC0451 R20

SIDDARTH INSTITUTE OF ENGINEERING & TECHNOLOGY: PUTTUR

(AUTONOMOUS)
Siddharth Nagar, Narayanavanam Road – 517583
QUESTION BANK (DESCRIPTIVE)

Subject with Code: ICS(20EC0451) Course& Branch: B.Tech & CSE,CSM.CIC

Year & Sem: III-B.Tech.& I-Sem. Regulation: R20

UNIT-V
Introduction to Wireless Communication Systems

1 a) Discuss briefly about the evolution of Mobile radio communication.

 Wireless communications is enjoying its fastest growth period in history, due to enabling
technologies which permit widespread deployment.
 The ability to provide wireless communications to an entire population was not evenconceived
until Bell Laboratories developed the cellular concept in the 1960s and 1970s.
 With the development of highly reliable, miniature, solid-state radio frequency hardware in
the 1970s, the wireless communications era was born.
 Following figure illustrates how mobile telephony has penetrated our daily lives compared
with other popular inventions of the 20th century.

Fig: The growth of mobile telephony as compared with other popular inventions of the20th
century.
COURSE CODE: 20EC0451 R20
 In 1935, Edwin Armstrong demonstrated frequency modulation (FM) for the firsttime.
 Since the late 1930s, FM has been the primary modulation technique used for mobile
communication systems throughout the world.
 The vast majority of mobile users in the 1960s were not connected to the public switched
telephone network (PSTN), and thus were not able to directly dial telephone numbers from their
vehicles.
 With the boom in CB radio and cordless appliances such as garage door openers and
telephones, the number of users of mobile and portable radio in 1995 was about 100 million.
 In the first few years of the 21st century, it is clear there will be an equal number of wireless
and conventional wireline customers throughout the world.
 At the beginning of the 21st century, over 1% of the worldwide wireless subscriber population
had already abandoned wired telephone service for home use, and had begun to rely solely on
their cellular service provider for telephone access.

1 b) Explain second generation (2G) cellular networks.

 Unlike first generation cellular systems that relied exclusively on FDMA/FDD and analog
FM, second generation standards use digital modulation formats and TDMA/FDD and
CDMA/FDD multiple access techniques.
 The most popular second generation standards include the following three TDMA standards
and one CDMA standard:
 (a) Global System Mobile (GSM), which supports eight time slotted users for each 200 kHz
radio channel and has been deployed widely by service providers in Europe, Asia, Australia,
South America, and some parts of the US.
 (b) Interim Standard 136 (IS-136), also known as North American Digital Cellular (NADC),
which supports three time slotted users for each 30 kHz radio channel and is a popular choice
for carriers in North America, South America, and Australia.
 (c) Pacific Digital Cellular (PDC), a Japanese TDMA standard that is similar to IS- 136 with
more than 50 million users and
 (d) the popular 2G CDMA standard Interim Standard 95 Code Division Multiple Access (15-
95), also known as cdma-One, which supports up to 64 users that are orthogonally coded and
simultaneously transmitted on each 1.25MHz. channel.
 CDMA is widely deployed by carriers in North America, as well as in Korea, Japan, China,
South America, and Australia.
 Second generation systems were first introduced in the early 1990s, and evolved fromthe first
generation of analog mobile phone systems (e.g., AMPS, ETACS, and JTACS).
 Following figure illustrates how the world subscriber base was divided between the 1G and
2G technologies of late 2001.
COURSE CODE: 20EC0451 R20

Fig: Worldwide subscriber base as a function of cellular technology (2001)

 In many countries, 2G wireless networks are designed and deployed for conventionalmobile
telephone service, as a high capacity replacement for, or in competition with, existing older
first generation cellular telephone systems.
 Modern cellular systems are also being installed to provide fixed (non-mobile) telephone
service to residences and businesses in developing nations—this is particularly cost effective
for providing plain old telephone service (POTS).
 The 2G technologies offer at least a three-times increase in spectrum efficiency as compared
to first generation analog technologies, the need to meet a rapidly growing customer base
justifies the gradual, ongoing change out of analog to digital 2G technologies in any growing
wireless network.

2 a) Explain cordless telephone systems.

Cordless Telephone Systems:

 Cordless telephone systems are full duplex communication systems that use radio to connect
a portable handset to a dedicated base station, which is then connected to a dedicated
telephone line with a specific telephone number on the public switched telephone network
(PSTN).
 In first generation cordless telephone systems, the portable unit communicates only to the
dedicated base unit and only over distances of a few tens of meters.
 Early cordless telephones operate solely as extension telephones to a transceiver connected
to a subscriber line on the PSTN and are primarily for in-home use.
 Second generation cordless telephones allow subscribers to use their handsets at many
outdoor locations within urban centers.
 Modern cordless telephones are sometimes combined with paging receivers so that a
subscriber may first be paged and then respond to the page using the cordless telephone.
 Cordless telephone systems provide the user with limited range and mobility, as it is usually
not possible to maintain a call if the user travels outside the range of the basestation.
 Typical second generation base stations provide coverage ranges up to a few hundred meters.
COURSE CODE: 20EC0451 R20

2 b) Explain paging systems.

Paging Systems:
 Paging systems are communication systems that send brief messages to a subscriber.
 Depending on the type of service, the message may be either a numeric message, an
alphanumeric message, or a voice message.
 Paging systems are typically used to notify a subscriber of the need to call a particular
telephone number or travel to a known location to receive further instructions.
 In modern paging systems, news headlines, stock quotations, and faxes may be sent.
 A message is sent to a paging subscriber via the paging system access number (usually a toll-
free telephone number) with a telephone keypad or modem. The issued message is called a
page.
 The paging system then transmits the page throughout the service area using base stations
which broadcast the page on a radio carrier.
 Paging systems vary widely in their complexity and coverage area.
 While simple paging systems may cover a limited range of 2 to 5 km.
 Wide area paging systems can provide worldwide coverage.
 Wide area paging systems consist of a network of telephone lines, many base station
transmitters, and large radio towers that simultaneously broadcast a page from each base
station (this is called simulcasting).
 Paging systems are designed to provide reliable communication to subscriberswherever they
are; whether inside a building, driving on a highway, or flying in an airplane.
 This necessitates large transmitter powers (on the order of kilowatts) and low data rates (a
couple of thousand bits per second) for maximum coverage from each base station.

3 a) Explain cellular telephone system.

 A cellular telephone system provides a wireless connection to the PSTN for any user location
within the radio range of the system.
COURSE CODE: 20EC0451 R20

 Cellular systems accommodate a large number of users over a large geographic area, within a
limited frequency spectrum.
 Cellular radio systems provide high quality service that is often comparable to thatof
landline telephone systems.
 High capacity is achieved by limiting the coverage of each base station transmitter to a small
geographic area called a cell.
 A sophisticated switching technique called a handoff enables a call to proceed
uninterrupted when the user moves from one cell to another.
 The following figure shows a basic cellular system that consists of mobile stations, base
stations and a mobile switching center (MSC).

 The mobile switching center is sometimes called a mobile telephone switching office(MTSO),
since it is responsible for connecting all mobiles to the PSTN in a cellular system.
 Each mobile communicates via radio with one of the base stations and may be
handed-off to any number of base stations throughout the duration of a call.
 The mobile station contains a transceiver, an antenna, and control circuitry, and may be
mounted in a vehicle or used as a portable hand-held unit.
 The base stations consist of several transmitters and receivers which simultaneously handle
full duplex communications.
 The base station serves as a bridge between all mobile users in the cell and connects the
simultaneous mobile calls via telephone lines or microwave links to the MSC.
 The MSC coordinates the activities of all of the base stations and connects the entirecellular
system to the PSTN.
 A typical MSC handles 100,000 cellular subscribers and 5,000 simultaneous conversations at
a time, and accommodates all billing and system maintenance functions, as well.
 In large cities, several MSCs are used by a single carrier.

3 b) Discuss about frequency division duplexing in wireless communication.

Frequency division duplex (FDD):
 Full-duplex systems, allow simultaneous radio transmission and reception between a
subscriber and a base station, by providing two simultaneous but separate channels (frequency
COURSE CODE: 20EC0451 R20
division duplex, or FDD) or adjacent time slots on a single radio channel (time division
duplex, or TDD) for communication to and from the user.
 FDD provides simultaneous radio transmission channels for the subscriber and the base
station, so that they both may constantly transmit while simultaneously receiving signals from
one another.
 At the base station, separate transmit and receive antennas are used to accommodate the two
separate channels.
 At the subscriber unit, a single antenna is used for both transmissions to and reception from
the base station.
 A device called a duplexer is used inside the subscriber unit to enable the same antenna to be
used for simultaneous transmission and reception.
 To facilitate FDD, it is necessary to separate the transmit and receive frequencies by about
5% of the nominal RF frequency.
 FDD is used exclusively in analog mobile radio systems.

4 a) Explain third generation (3G) wireless networks.

Third Generation (3G) Wireless Networks
 3G systems promise unparalleled wireless access in ways that have never been possible
before.
 Multi-megabit Internet access, communications using Voice over Internet Protocol (VoIP),
voice-activated calls, unparalleled network capacity, and ubiquitous “always-on” access are
just some of the advantages being touted by 3G developers.
 Companies developing 3G equipment cenvision users having the ability to receive live music,
conduct interactive web sessions, and have simultaneous voice and data access with multiple
parties at the same time using a single mobile handset.
 The International Telecommunications Union (ITU) formulated a plan to implement a global
frequency band in the 2000 MHz range that would support a single, ubiquitous wireless
communication standard for all countries throughout the world.
 This plan, called International Mobile Telephone 2000 (IMT-2000), has been successful in
helping to cultivate active debate and technical analysis for new high speed mobile telephone
solutions when compared to 2G.
 However, as can be seen in following figure, the hope for a single worldwide standard has not
materialized, as the world-wide user community remains spit between two camps: GSM/IS-
136/PDC and CDMA.
COURSE CODE: 20EC0451 R20
Fig: Various upgrade paths for 2G-3G technologies
 The eventual 3G evolution for CDMA systems leads to cdma2000. Several variants of CDMA
2000 are currently being developed, but they all are based on the fundamentals of 1-95 and
IS-95B technologies.
 The eventual 3G evolution for GSM, IS-136, and PDC systems leads to Wideband CDMA
(W-CDMA), also called Universal Mobile Telecommunications Service (UMTS).
 W-CDMA is based on the network fundamentals of GSM, as well as the merged versions of
GSM and IS-136 through EDGE.
 The ITU IMT-2000 standards organizations are currently separated into two major
organizations reflecting the two 3G camps:
 (i) 3GPP (3G Partnership Project for Wideband CDMA standards based on backward
compatibility with GSM and IS-136/PDC) and
 (ii) 3GPP2 (3G Partnership Project for cdma2000 standards based on backward compatibility
with IS-95).
 Countries throughout the world are currently determining new radio spectrum bands to
accommodate the 3G networks that will likely be deployed in the 2004-2005 timeframe.
 ITU's 2000 World Radio Conference established the 2500-2690 MHz, 1710-1885 MHz, and
806-960 MIHZ bands as candidates for 3G.
 In the US, additional spectrum in the upper UHF television bands near 700 MHz is also being
considered for 3G.

4 b) A spectrum of 30 MHz of bandwidth is allocated to a particular FDD cellular

telephone system which uses two 25 kHz simplex channels to provide full duplex voice
and control channels, compute the number of channels available per cell if a system uses
(i) four-cell reuse, (ii) seven-cell reuse, and (iii) 12-cell reuse. If 1 MHz of the allocated
spectrum is dedicated to control channels, determine an equitable distribution of control
channels and voice channels in each cell for each of the three systems.

Given, Total bandwidth=30 MHz

Channel bandwidth = 25 kHz × 2 simplex channels = 50 kHz/duplex channelTotal
available channels = 30,000/50 = 600 channels
(a) For N = 4,
Total number of channels available per cell = 600/4 = 150 channels.
(b) For N = 7,
Total number of channels available per cell = 600/7 = 85 channels.
(c) For N = 12,
Total number of channels available per cell = 600/12 = 50 channels.

1 MHz spectrum for control channels implies that there are 1000/50 = 20 control channels out of the
600 channels available. To evenly distribute the control and voice channels, simplyallocate the same
number of voice channels in each cell wherever possible.

(a) For N = 4, we can have five control channels and 145 voice channels per cell. In practice,
however, each cell only needs a single control channel (the control channels have a greater
reuse distance than the voice channels). Thus, one control channel and 145 voice channels
would be assigned to each cell.

(b) For N = 7, four cells with three control channels and 82 voice channels [(600- 20)/7=82], two
COURSE CODE: 20EC0451 R20
cells with three control channels and 90 voice channels, and one cellwith two control channels
and 92 voice channels could be allocated.

(c) For N = 12, we can have eight cells with two control channels and 48 voice channels, and four
cells with one control channel and 49 voice channels each. In an actual system, each cell
would have one control channel, eight cells would have 48 voicechannels, and four cells
would have 49 voice channels.

5 a) Explain the multiple access schemes for narrowband systems.

Multiple access techniques (FDMA, TDMA, CDMA & SDMA) can be grouped as narrowband and
wideband systems, depending on how the available bandwidth is allocated to the users.

Narrowband Systems:
 The term narrowband is used to relate the bandwidth of a single channel to the expected
coherence bandwidth of the channel.
 In a narrowband multiple access system, the available radio spectrum is divided into a
large number of narrowband channels.
 The channels are usually operated using FDD.
 To minimize interference between forward and reverse links on each channel, the frequency
separation is made as great as possible within the frequency spectrum.
 In narrowband FDMA, a user is assigned a particular channel which is not shared by otherusers
in the vicinity, and if FDD is used (that is, each duplex channel has a forward and reverse
simplex channel), then the system is called FDMA/FDD.
 Narrowband TDMA, on the other hand, allows users to share the same radio channel but
allocates a unique time slot to each user in a cyclical fashion on the channel, thus separating
a small number of users in time on a single channel.
 For narrowband TDMA systems, there generally are a large number of radio channels
allocated using either FDD or TDD, and each channel is shared using TDMA. Such systems
are called TDMA/FDD or TDMA/TDD access systems.

5 b) Discuss about time division duplexing in wireless communication.

 Time division multiple access (TDMA) systems divide the radio spectrum into time slots, and
in each slot only one user is allowed to either transmit or receive.
 It can be seen from figure that each user occupies a cyclically repeating time slot, so a channel
may be thought of as a particular time slot that reoccurs every frame, where N time slots
comprise a frame.
COURSE CODE: 20EC0451 R20
Figure 3 TDMA scheme where each channel occupies a cyclically repeating time slot.
 TDMA systems transmit data in a buffer-and-burst method, thus the transmission for anyuser
is non-continuous.
 The transmission from various users is interlaced into a repeating frame structure as shownin
Figure 4. It can be seen that a frame consists of a number of slots. Each frame is made up of a
preamble, an information message, and tail bits.
 Guard times are utilized to allow synchronization of the receivers between different slots
and frames.

6 a) Explain the multiple access schemes for wideband systems.

Wideband system:
 In wideband systems, the transmission bandwidth of a single channel is much larger than the
coherence bandwidth of the channel.
 Thus, multipath fading does not greatly vary the received signal power within a wideband
channel, and frequency selective fades occur in only a small fraction of the signal bandwidth
at any instance of time.
 In wideband multiple access systems, a large number of transmitters are allowed to transmit
on the same channel.
 TDMA allocates time slots to the many transmitters on the same channel and allows onlyone
transmitter to access the channel at any instant of time, whereas spread spectrum CDMA
allows all of the transmitters to access the channel at the same time.
 TDMA and CDMA systems may use either FDD or TDD multiplexing techniques.

6 b) Draw the TDMA frame structure and briefly explain the fields.
COURSE CODE: 20EC0451 R20
 Preamble consists of Address and synchronization bits used by base stations and mobile
stations to identity each other.

 Information message consists of slots(users)-user information is located.

 Each slot will have trail bits, sync bits, Information data and guard bits.

 Trail bits helps in power control mechanism which help in transmission of the signal.

 Sync bits helps in synchronizing the transmitting and receiver at the time of communications.

 Information data for each slot have the data about that user.

 Guard bits are guard bands between the user to avoid interference between the users.

7 a) Describe the features of the frequency division multiple access (FDMA) scheme.
 Frequency division multiple access (FDMA) assigns individual channels to individual
users.
 It can be seen from figure that each user is allocated a unique frequency band or channel.
These channels are assigned on demand to users who request service.

Figure: FDMA where different channels are assigned different frequency bands
 During the period of the call, no other user can share the same channel.
 In FDD systems, the users are assigned a channel as a pair of frequencies; one frequency is
used for the forward channel, while the other frequency is used for the reverse channel.
The features of FDMA are as follows:
 The FDMA channel carries only one phone circuit at a time.
 If an FDMA channel is not in use, then it sits idle and cannot be used by other users to increase
or share capacity. It is essentially a wasted resource.
 After the assignment of a voice channel, the base station, and the mobile transmit
simultaneously and continuously.
 The bandwidths of FDMA channels are relatively narrow (30 kHz in AMPS) as each channel
supports only one circuit per carrier. That is, FDMA is usually implemented in narrowband
systems.
 The complexity of FDMA mobile systems is lower when compared to TDMA systems,
though this is changing as digital signal processing methods improve for TDMA.
 Since FDMA is a continuous transmission scheme, fewer bits are needed for overhead
purposes (such as synchronization and framing bits) as compared to TDMA.
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 FDMA systems have higher cell site system costs as compared to TDMA systems, becauseof
the single channel per carrier design.
 The FDMA mobile unit uses duplexers since both the transmitter and receiver operate at the
same time.
 This results in an increase in the cost of FDMA subscriber units and base stations.
 FDMA requires tight RF filtering to minimize adjacent channel interference.

7 b) Describe the features of code division multiple access (CDMA) scheme.

 In code division multiple access (CDMA) systems, the narrowband message signal is
multiplied by a very large bandwidth signal called the spreading signal.
 The spreading signal is a pseudo-noise code sequence that has a chip rate which is ordersof
magnitudes greater than the data rate of the message.
 All users in a CDMA system, as seen from Figure, use the same carrier frequency and may
transmit simultaneously.

Figure: Spread spectrum multiple access in which each channel is assigned a unique
PNcode which is orthogonal or approximately orthogonal to PN codes used by other
users.

 Each user has its own pseudorandom code word which is approximately orthogonal to all
other code words.
 The receiver performs a time correlation operation to detect only the specific desired
code word. All other code words appear as noise due to de-correlation.
The features of CDMA including the following:
 Many users of a CDMA system share the same frequency. Either TDD or FDD may be
used.
 Unlike TDMA or FDMA, CDMA has a soft capacity limit. Increasing the number of users in
a CDMA system raises the noise floor in a linear manner.
 Thus, there is no absolute limit on the number of users in CDMA.
 Rather, the system performance gradually degrades for all users as the number of users is
increased, and improves as the number of users is decreased.
 Multipath fading may be substantially reduced because the signal is spread over a large
spectrum.
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 If the spread spectrum bandwidth is greater than the coherence bandwidth of the channel, the
inherent frequency diversity will mitigate the effects of small-scale fading.
 Channel data rates are very high in CDMA systems.
 A RAKE receiver can be used to improve reception by collecting time delayed versionsof
the required signal.
 Self-jamming is a problem in the CDMA system. Self-jamming arises from the fact thatthe
spreading sequences of different users are not exactly orthogonal.
 The near-far problem occurs at a CDMA receiver if an undesired user has a high detected
power as compared to the desired user.

8 a) Describe the features of time division multiple access (TDMA) scheme.

 Time division multiple access (TDMA) systems divide the radio spectrum into time slots, and
in each slot only one user is allowed to either transmit or receive.
 It can be seen from figure that each user occupies a cyclically repeating time slot, so a channel
may be thought of as a particular time slot that reoccurs every frame, where N time slots
comprise a frame.

Figure 3 TDMA scheme where each channel occupies a cyclically repeating time slot.
 TDMA systems transmit data in a buffer-and-burst method, thus the transmission for anyuser
is non-continuous.
 The transmission from various users is interlaced into a repeating frame structure as shown
Figure 4. It can be seen that a frame consists of a number of slots. Each frame is madeup of a
preamble, an information message, and tail bits.
 TDMA systems transmit data in a buffer-and-burst method, thus the transmission for anyuser
is non-continuous.
 The transmission from various users is interlaced into a repeating frame structure as shownin
Figure 4. It can be seen that a frame consists of a number of slots. Each frame is made up of a
preamble, an information message, and tail bits.
 Guard times are utilized to allow synchronization of the receivers between different slots
and frames.
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The features of TDMA include the following:

 TDMA shares a single carrier frequency with several users, where each user makes use of
non-overlapping time slots.
 Data transmission for users of a TDMA system is not continuous but occurs in bursts.
 This results in low battery consumption, since the subscriber transmitter can be turned off
when not in use (which is most of the time).
 Because of discontinuous transmissions in TDMA, the handoff process is much simpler for
a subscriber unit, since it is able to listen for other base stations during idle time slots.
 TDMA uses different time slots for transmission and reception, thus duplexers are not
required.

8 b) Evaluate the efficiency of time division multiple access (TDMA) scheme.

Efficiency of TDMA:
 The efficiency of a TDMA system is a measure of the percentage of transmitted data that
contains information as opposed to providing overhead for the access scheme.
 The frame efficiency ηf, is the percentage of bits per frame which contain transmitted data.
 Note that the transmitted data may include source and channel coding bits, so the raw end-
user efficiency of a system is generally less than ηf.
 The frame efficiency can be found as follows.
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9 a) Differentiate FDMA, TDMA and CDMA.

FDMA TDMA CDMA

FDMA stands for Frequency TDMA stands for Time CDMA stands for Code
Division Multiple Access. Division Multiple Access. Division Multiple Access.

In this, sharing of bandwidth In this, only the sharing of In this, there is sharing of both
among different stations takes time of satellite i.e. bandwidth and time among
place. transponder takes place. different stations takes place.

There is no need of any code There is no need of any

word. code word. Code word is necessary.

In this, there is only need of

guard bands between the In this, guard time of the
adjacent channels are adjacent slots are In this, both guard bands and
necessary. necessary. guard time are necessary.

Synchronization is not Synchronization is

required. required. Synchronization is not required.
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FDMA TDMA CDMA

The rate of data is

The rate of data is low. medium. The rate of data is high.

Mode of data transfer is Mode of data transfer is Mode of data transfer is digital
continuous signal. signal in bursts. signal.

It is little flexible. It is moderate flexible. It is highly flexible.

9 b) Illustrate with a timing diagram how call initiated by a mobile user is established.
When a cellular phone is turned ON, but not yet engaged in a call, it first scans the group of forward
control channels to determine the one with the strongest signal, and then monitors that control
channel until the signal level drops below a usable level.
Call initiation by a landline (PSTN) subscriber to mobile user:

 The mobile switching centre (MSC) dispatches the request to all base station in a cellular
system.
 The Mobile Identification Number (MIN) which is subscriber telephone number is then
broadcast as a paging message over all of the forward control channels throughout the cellular
system.
 The mobile receives the paging message sent by BS which s monitors, and responds by
identifying itself over the RCC.
 The BS relays the acknowledgement sent by the mobile and informs the MSC of handshake.
 The MSC instructs the BS to move the call to an unused voice channel pair within the cell.
 The BS signals the mobile to change frequencies to an unused forward and reverse voice
channel pair.
 Another data message is transmitted on forward channel to instruct the mobile telephone to
ring and mobile user to answer the phone.
 Figure below shows sequence of events involved in call connection.
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10 a) Explain various hybrid spread spectrum techniques in CDMA.

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10 b) Describe space division multiple access (SDMA) scheme.

 Space division multiple access (SDMA) controls the radiated energy for each user in
space.
 It can be seen from Figure that SDMA serves different users by using spot beam antennas.

Figure: A spatially filtered base station antenna serving different users by using spot
beams.
 These different areas covered by the antenna beam may be served by the same frequency(in
a TDMA or CDMA system) or different frequencies (in an FDMA system).
 Sectorized antennas may be thought of as a primitive application of SDMA.
 The reverse link presents the most difficulty in cellular systems for several reasons.
 First, the base station has complete control over the power of all the transmitted signals on
the forward link.
 Second, transmit power is limited by battery consumption at the subscriber unit, thereforethere
are limits on the degree to which power may be controlled on the reverse link.
 If the base station antenna is made to spatially filter each desired user so that more energyis
detected from each subscriber, then the reverse link for each user is improved and less power
is required.
 Adaptive antennas used at the base station (and eventually at the subscriber units) promiseto
mitigate some of the problems on the reverse link.
 In the limiting case of infinitesimal beam-width and infinitely fast tracking ability, adaptive
antennas implement optimal SDMA, thereby providing a unique channel that is free from the
interference of all other users in the cell.
 With SDMA, all users within the system would be able to communicate at the same time using
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the same channel.
 In addition, a perfect adaptive antenna system would be able to track individual multipath
components for each user and combine them in an optimal manner to collect all of the
available signal energy from each user.
 The perfect adaptive antenna system is not feasible since it requires infinitely large
antennas.