Communication Systems MCQ Quiz - Objective Question with Answer for Communication Systems - Download Free PDF

Last updated on May 30, 2025

Latest Communication Systems MCQ Objective Questions

Communication Systems Question 1:

Which of the following components is required to detect (demodulate) Pulse Position Modulation (PPM)?
a) Pulse Generator
b) RS Flip-Flop
c) PWM Demodulator

  1. Only a
  2. All a, b, and c
  3. Only c
  4. Only a and b

Answer (Detailed Solution Below)

Option 3 : Only c

Communication Systems Question 1 Detailed Solution

Explanation:

Pulse Position Modulation (PPM) Demodulation

Pulse Position Modulation (PPM) is a type of signal modulation where the position of a pulse, relative to the position of a reference pulse, is varied according to the sampled value of the modulating signal. In simpler terms, the timing of the pulse is varied to encode the information.

In PPM, the position of each pulse is shifted in time to represent the data being transmitted. The time displacement of each pulse is proportional to the amplitude of the analog signal at the corresponding sampling time. To demodulate this signal, the receiver needs to be able to detect these time shifts accurately.

Components Required for PPM Demodulation:

To demodulate a PPM signal, the following components are typically required:

  • Pulse Generator: Used to create reference pulses that help in determining the position of the incoming pulses.
  • RS Flip-Flop: Helps in maintaining the state information needed for the timing and synchronization of the demodulated signal.
  • PWM Demodulator: While not directly used for PPM demodulation, understanding Pulse Width Modulation (PWM) demodulation can be beneficial as PPM can be derived from PWM signals.

Correct Option Analysis:

The correct option is:

Option 4: Only c

This option correctly identifies that a PWM Demodulator (Option c) is required for the demodulation of PPM. The reasoning is that a PPM signal can be derived from a PWM signal. Therefore, understanding and utilizing a PWM demodulator is crucial for effectively demodulating a PPM signal

Communication Systems Question 2:

Which of the following statements about the AM detection using PLL circuit is/are correct?
S1: It has higher noise immunity than the conventional peak detector type AM detector.
S2: The PLL is locked to the carrier frequency of the AM signal.

  1. Only S1
  2. Only S2
  3. Neither S1 nor S2
  4. Both S1 and S2

Answer (Detailed Solution Below)

Option 4 : Both S1 and S2

Communication Systems Question 2 Detailed Solution

Explanation:

AM Detection Using PLL Circuit

Definition: Phase-Locked Loop (PLL) circuits are used for a variety of applications in communication systems, including frequency synthesis, demodulation, and signal synchronization. In the context of AM (Amplitude Modulation) detection, PLLs are used to demodulate the AM signal by locking onto the carrier frequency and extracting the modulating signal.

Working Principle: A PLL circuit consists of a phase detector, a low-pass filter, and a voltage-controlled oscillator (VCO). The phase detector compares the phase of the incoming AM signal with the phase of the VCO output. The resulting error signal is filtered and used to adjust the VCO frequency. When the PLL is locked, the VCO frequency matches the carrier frequency of the AM signal, and the error signal corresponds to the modulating signal (audio or data).

Advantages:

  • Higher noise immunity compared to conventional peak detector type AM detectors.
  • Precise demodulation of the AM signal even in the presence of noise and signal distortion.
  • Ability to lock onto the carrier frequency, providing stable and accurate demodulation.

Disadvantages:

  • Increased complexity and cost compared to simple peak detector circuits.
  • Requires careful design and tuning to ensure proper locking and demodulation.

Applications: PLL-based AM detectors are used in high-fidelity radio receivers, communication systems, and other applications where accurate and reliable demodulation of AM signals is critical.

Correct Option Analysis:

The correct option is:

Option 4: Both S1 and S2

This option correctly identifies both key aspects of AM detection using PLL circuits:

  • S1: It has higher noise immunity than the conventional peak detector type AM detector.
    This statement is true because PLL circuits can reject noise and maintain lock on the carrier frequency, leading to better demodulation performance in noisy environments.
  • S2: The PLL is locked to the carrier frequency of the AM signal.
    This statement is also true because the fundamental operation of a PLL involves locking onto the carrier frequency of the input signal, ensuring accurate extraction of the modulating signal.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Only S1

While S1 is correct, this option is incomplete because it does not acknowledge the importance of S2, which is also a true statement about PLL-based AM detectors.

Option 2: Only S2

Similar to Option 1, this option is incomplete as it only considers S2 and ignores the validity of S1, which is an important aspect of the advantages of PLL-based AM detectors.

Option 3: Neither S1 nor S2

This option is incorrect because both S1 and S2 are true statements about PLL-based AM detectors. Ignoring both statements would lead to a misunderstanding of the advantages and operation of PLL circuits in AM detection.

Conclusion:

Understanding the role of PLL circuits in AM detection is crucial for recognizing their advantages over conventional peak detector circuits. PLLs offer higher noise immunity and precise locking to the carrier frequency, ensuring accurate demodulation of AM signals. Both S1 and S2 are correct statements that highlight these key benefits, making Option 4 the accurate choice.

Communication Systems Question 3:

In a 100% modulated AM signal with a carrier power of 100 W, what is the power in the lower sideband?

  1. 50 W
  2. 25 W
  3. 15 W
  4. 150 W

Answer (Detailed Solution Below)

Option 2 : 25 W

Communication Systems Question 3 Detailed Solution

Concept:

The total transmitted power for an AM system is given by:

\({P_t} = {P_c}\left( {1 + \frac{{{μ^2}}}{2}} \right)\)

Pc = Carrier Power

μ = Modulation Index

The above expression can be expanded to get:

\({P_t} = {P_c} + P_c\frac{{{μ^2}}}{2}\)

The total power is the sum of the carrier power and the sideband power, i.e.

\({P_s} = P_c\frac{{{μ^2}}}{2}\)

Calculation:

Given: Pc = 100 W and μ = 1

We can write:

\({P_s} = P_c\frac{{{μ^2}}}{2}\)

\(= 100 \times \frac{1^2}{2}\)

\(P_s=50~W\)

The total sideband power = 50 W

power in upper sideband + power in lower side band = 50 W

power in upper sideband = power in lower sideband = 25 W

Communication Systems Question 4:

Which of the following is NOT a commonly used pulse-shaping filter in communication systems?

  1. Raised cosine filter
  2. Sinc filter
  3. Gaussian filter
  4. High-pass filter

Answer (Detailed Solution Below)

Option 4 : High-pass filter

Communication Systems Question 4 Detailed Solution

Concept:

In digital communication systems, pulse-shaping filters are used to control bandwidth and reduce intersymbol interference (ISI). Commonly used pulse-shaping filters include Raised Cosine, Gaussian, and Sinc filters. These are specifically designed to shape pulses for transmission.

High-pass filters, on the other hand, are general-purpose frequency filters used to block low frequencies, and they are not typically used for pulse shaping in baseband communication.

Evaluation of Options:

Option 1: Raised cosine filter –  Incorrect
Widely used in pulse shaping due to excellent ISI performance.

Option 2: Sinc filter –  Incorrect
Ideal pulse-shaping filter (though impractical due to infinite length).

Option 3: Gaussian filter –  Incorrect
Used in GMSK and other modulation schemes.

Option 4: High-pass filter –  Correct
Not used for pulse shaping; used for removing low-frequency components.

 

Communication Systems Question 5:

In which modulation technique is the carrier modulated using phase shifts of 0°, 90°, 180°, and 270°?

  1. QPSK (Quadrature Phase Shift Keying)
  2. ASK (Amplitude Shift Keying)
  3. BPSK (Binary Phase Shift Keying)
  4. FSK (Frequency Shift Keying)

Answer (Detailed Solution Below)

Option 1 : QPSK (Quadrature Phase Shift Keying)

Communication Systems Question 5 Detailed Solution

QPSK modulation represents symbols by a constellation of four-phase angles of the carrier signal, orthogonal to each other.

There are two bits per symbol. So for example 00 = 0°, 01 = 90° , 10 = 180° and 11= 270°. 

QPSK transmits twice the data rate in a given bandwidth compared to BPSK, at the same BER. 

The constellation diagram for 4 different symbols is as shown:

 F1 Shubham.B 03-12-20 Savita D1

Top Communication Systems MCQ Objective Questions

On which principle optical fibre works?

  1. Scattering of light
  2. Total internal absorption
  3. Total internal reflection
  4. Optical rotation

Answer (Detailed Solution Below)

Option 3 : Total internal reflection

Communication Systems Question 6 Detailed Solution

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The correct answer is total internal reflection.

Key Points

  • Optical fiber works on the principle of total internal reflection.
  • An optical fiber is a flexible transparent fiber.
  • It is made up of Glass (silica) or plastic.
  • It is slightly thicker than human hair.
  • It transmits higher bandwidth data over a longer distance than other forms of communication.
  • It is usually used in endoscopy and fiber optic communications.

Additional Information

  •  Scattering of light:
    • Scattering of light occurs when light rays collide with an obstruction such as dust, gas molecules, or water vapors and divert from their straight path.
    • Examples of Scattering of light:
      • The Tyndall effect.
      • The red hues of sunrise and sunset 
      • The white color of the sky at noon.
      • The blue color of the sky.
  • Optical rotation:
    • The optical rotation is the angle through which the plane of polarization rotates when a polarized light ray flows through a layer of a liquid.
    • The effect of optical rotation in a substance is governed by the concentration of chiral molecules and their chemical structure.
    • It is used to test the purity of the material.

Where was the First experimental satellite telecommunication earth station set up in 1967 in India?

  1. Ahmednagar
  2. Allahabad
  3. Ahmedabad
  4. Aurangabad

Answer (Detailed Solution Below)

Option 3 : Ahmedabad

Communication Systems Question 7 Detailed Solution

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The correct answer is Ahmedabad.

Key Points

  • The first 'Experimental Satellite Communication Earth Station (ESCES)' was operationalized in Ahmedabad in 1967, and it also served as a training facility for Indian and international scientists and engineers.
  • ISRO was clear that it did not need to wait for its own satellites to begin application development, and that foreign satellites may be utilized in the early phases to demonstrate that a satellite system can contribute to national development.
  • The Satellite Telecommunication Experiments Project (STEP), a cooperative project of ISRO and the Post and Telegraphs Department (P&T) in 1977-79, used the Franco-German Symphonie satellite.

Important Points

  • The ‘Kheda Communications Project (KCP)' followed SITE, which served as a field laboratory for need-based and locale-specific programme transmission in Gujarat's Kheda area.
  • In 1984, KCP received the UNESCO-IPDC (International Programme for the Development of Communication) award for efficiency in rural communication.
  • The first Indian spacecraft, 'Aryabhata,' was developed during this time and launched using a Soviet Launcher.
  • Another significant milestone was the creation of the SLV-3, the first launch vehicle capable of placing 40 kg in Low Earth Orbit (LEO), which flew for the first time in 1980.

Fiber optics generally work on the principle of ________.

  1. Reflection
  2. Refraction
  3. Dispersion
  4. Total internal reflection

Answer (Detailed Solution Below)

Option 4 : Total internal reflection

Communication Systems Question 8 Detailed Solution

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  • Fiber optics generally work on the principle of Total internal reflection.
  • A fiber optic cable consists of a bundle of glass threads, each of which is capable of transmitting messages modulated onto light waves.

Which of the following is fastest memory cell?

  1. core memory
  2. semiconductor memory
  3. bubble memory
  4. superconductor memory

Answer (Detailed Solution Below)

Option 2 : semiconductor memory

Communication Systems Question 9 Detailed Solution

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The correct answer is semiconductor memory.

Concept:

Core Memory:

  • Core memory was a common form of random-access memory (RAM).
  • The memory made use of magnetic rings called cores that had wires passing through them for selecting and detecting the contents of the cores

 

Semiconductor Memory:

  • In a semiconductor memory chip, each bit of binary data is stored in a tiny circuit called a memory cell consisting of one to several transistors
  • The two major categories of semiconductor memories are the RAM and the ROM.
  • These are the fastest memory.

 

Bubble Memory:

  • Bubble memory is a type of non-volatile memory that makes use of a thin layer of magnetic material that holds small magnetized areas known as bubbles or domains, which are able to store one bit of data each.

 

Superconducting Memory:

  • This is just a concept of memory and has not been developed into practice.

The diffusion potential across a p-n junction __________.

  1. decreases with increasing doping concentration
  2. increases with decreasing band gap
  3. does not depend on doping concentrations
  4. increases with increase in doping concentrations

Answer (Detailed Solution Below)

Option 4 : increases with increase in doping concentrations

Communication Systems Question 10 Detailed Solution

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In a pn junction, if the doping concentration increases, the recombination of electrons and holes increases, thereby increases the voltage across the barrier.

\(V = \frac{{KT}}{q}{\rm{ln}}\left( {\frac{{{N_a}{N_d}}}{{n_i^2}}} \right)\)

ISRO has established the _______, at Bharati station, Antarctica, for receiving IRS data. 

  1. NCAOR
  2. NRSC
  3. AGEOS
  4. IMGEOS

Answer (Detailed Solution Below)

Option 3 : AGEOS

Communication Systems Question 11 Detailed Solution

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​The correct answer is AGEOS.

Key Points

  • ISRO has established the AGEOS, at Bharati Station, Antarctica, for receiving IRS data.
  • ISRO has established the Antarctica Ground Station for Earth Observation Satellites (AGEOS), at Bharati Station, Larsemann Hills, Antarctica, for receiving Indian Remote sensing Satellites (IRS) data.
  • This state-of-the-art advanced Ground station was commissioned during August 2013 and is receiving data from IRS satellites (like CARTOSAT-2 Series, SCATSAT-1, RESOURCESAT-2/2A, CARTOSAT-1) and transferring the same to NRSC, Shadnagar near Hyderabad.
  • This Satcom station is providing vital communication support to the Indian scientific community for pursuing their research work at Maitri throughout the year. With the commissioning of the Earth station at NCAOR, Goa, the Indian station, Maitri has been brought into the ambit of the World Wide Web.
  • The AGEOS is continuously operated and maintained by the Engineers of ISRO who are under deputation to Bharati Station, Antarctica on a regular basis.

Additional Information 

Organization Description
NCAOR
  • The National Centre for Antarctic & Ocean Research (NCAOR), an Autonomous Society under the Ministry of Earth Sciences, Govt. of India, New Delhi, is the nodal agency responsible for planning, coordinating and executing the Indian Polar Programme.
  • In addition, the Centre has a well-focused scientific mandate of basic and applied research in various disciplines and themes of polar and ocean sciences.
NRSC
  • National Remote Sensing Centre (NRSC) is one of the primary centres of the Indian Space Research Organisation (ISRO), Department of Space (DOS).
  • NRSC operates through multiple campuses to meet the national and regional remote sensing data and applications needs of the country.
IMGEOS
  • Integrated Multi-Mission Ground Segment for Earth Observation Satellites (IMGEOS) facility is established in Shadnagar campus.
  • This facility is equipped with state of the art data acquisition systems which receive data from various satellites. 

 

The bit rate of the digital communication system is M kbps. The modulation used is 16 QAM. The minimum bandwidth required for ideal transmission is _________.

  1. M/2 kHz
  2. M/16 kHz
  3. M kHz
  4. M/8 kHz

Answer (Detailed Solution Below)

Option 1 : M/2 kHz

Communication Systems Question 12 Detailed Solution

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Concept:

In the M-Array modulation scheme, the minimum bandwidth required for ideal transmission is given by:

\({\left( {BW} \right)_{min}} = \frac{{{2R_b}}}{{{{\log }_2}N}}Hz\)

Where,

Rb = bit rate in bps

N = number of levels in M-Array scheme

Calculation:

Given that,

Bit rate = M kbps

Number of levels = N = 16

\(\therefore {\left( {BW} \right)_{{\rm{min}}}} = \frac{{{2R_b}}}{{{{\log }_2}N}}Hz = \frac{2M}{{{{\log }_2}16}}kHz\)

\( = \frac{2M}{{{{\log }_2}{2^4}}}kHz\)

\( = \frac{2M}{{4\; \times\; {{\log }_2}2}}kHz\)

\( (BW)_{min}= \frac{M}{{2}}kHz\)

So. The minimum bandwidth will be M/2 kHz, for ideal transmission.

26 June 1

For Baseband

For Passband

Binary:

1) B.W. = Rb

Binary:

1) BW = 2 Rb

Raised cosine (α) :

2)  \(BW = \frac{{{R_b}}}{2}\left( {1 + \alpha } \right)\)

Raised cosine (α) :

\(2)\;BW = \frac{{2{R_b}}}{2}\left( {1 + \alpha } \right)\)

= Rb (1 + α) 

M-ary:

1)  \(B.W. = \frac{{{R_b}}}{{{{\log }_2}M}}\)

M-ary:

1)  \(B.W = \frac{{2{R_b}}}{{{{\log }_2}M}}\)

Raised cosine (α):

2)  \(B.W. = \frac{{{R_b}\left( {1 + \alpha } \right)}}{{2{{\log }_2}M}}\)

Raised cosine (α) :

2)  \(B.W = \frac{{{R_b}\left( {1 + \alpha } \right)}}{{{{\log }_2}M}}\)

GPRS stands for

  1. General Packet Radio Service
  2. Global Positioning Radio Service
  3. Geological Packet Radio Service
  4. Geological Positioning Radio Service

Answer (Detailed Solution Below)

Option 1 : General Packet Radio Service

Communication Systems Question 13 Detailed Solution

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General Packet Radio Service

  • GPRS, or General Packet Radio Service, is a best-effort packet-switching communications protocol for cellular networks.  
  • GPRS was one of the first widely used data transfer protocols on cellular networks.
  • GPRS is a third-generation step toward internet access.
  • GPRS is also known as GSM-IP that is a Global-System Mobile Communications Internet Protocol as it keeps the users of this system online, allows them to make voice calls, and access the internet on-the-go.

In digital transmission, the modulation technique that requires the minimum bandwidth is:

  1. PCM
  2. PAM
  3. DPCM
  4. Delta modulation

Answer (Detailed Solution Below)

Option 4 : Delta modulation

Communication Systems Question 14 Detailed Solution

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  • In PCM an analog signal is sampled and encoded into different levels before transmission
  • The bandwidth of PCM depends on the number of levels
  • If each sample is encoded into n bits, then the bandwidth of PCM is nfs
  • The bandwidth of DPCM is almost the same as that of PCM signal, the only difference between PCM and DPCM is that the dynamic range is reduced in the DPCM signal
  • However, in the case of Delta modulation, each sample is sent using only 1 bit which is +Δ or -Δ
  • Hence there is bandwidth saving in Delta modulation

26 June 1

A comparison of different modulation schemes is as shown in the table below:

Parameter

PCM

DM

 DPCM

Number of bits

It can use 4, 8

or 16 bits per sample

It uses only

one bit for one sample

Bits can be more than one but are less than PCM

Level/Step size

Step size

is fixed

Step size is fixed

and

cannot be varied

A fixed number of levels are used.

Quantiz-ation error or Distortion

Quanti-zation error depends

on the number

of levels used

Slope

overload distortion and granular

noise is

present

Slope overload distortion

and quantization noise is

present

Bandwidth

of the transmi-ssion channel

 Highest bandwidth is required since the number of bits is high

The lowest band-width is required

The bandwidth required is lower than PCM

Signal to Noise ratio

Good

Poor

Fair

Area of Applic-ation

Audio and

Video

Telephony

Speech and images

Speech and video

Each of the following sentences has been divided into four parts – (a), (b), (c), (d) – one of which has an error. Choose the part which contains the error.

I talked (a) / to my neighbourer (b) / to settle the issue (c) / that had been hanging for a long time.(d)

  1. (a)
  2. (b)
  3. (c)
  4. (d)

Answer (Detailed Solution Below)

Option 2 : (b)

Communication Systems Question 15 Detailed Solution

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The correct answer is '(b)'.

Key Points 

  • There is an error in part (b).
  • 'neighbourer' should be replaced by 'neighbour'.
  • neighbour means a person living next door to or very near to the speaker or person referred to.​
    • Example:- "our garden was the envy of the neighbours"
  • The correct sentence will be - I talked to my neighbour to settle the issue that had been hanging for a long time.

​Hence, the correct answer is Option 2.

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