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

Last updated on Jun 18, 2025

Latest Communication Systems MCQ Objective Questions

Communication Systems Question 1:

Which of the following is an example of a digital modulation technique?

  1. Single Sideband Modulation (SSB)
  2. Narrow Band FM
  3. Phase Shift Keying (PSK) 
  4. Double side band Amplitude Modulation 

Answer (Detailed Solution Below)

Option 3 : Phase Shift Keying (PSK) 

Communication Systems Question 1 Detailed Solution

Explanation:

Digital Modulation Technique

Definition: Digital modulation refers to the process of modifying a carrier signal to encode digital information. This is achieved by altering the carrier's properties such as amplitude, frequency, or phase. Digital modulation techniques are fundamental in modern communication systems, as they enable efficient and reliable transmission of digital data over various communication channels.

Phase Shift Keying (PSK):

Definition: Phase Shift Keying (PSK) is a digital modulation technique where the phase of the carrier signal is varied to represent digital data. Each unique phase state corresponds to a specific binary or multi-bit symbol, allowing the transmission of information.

Working Principle: In PSK, the carrier signal's phase is shifted to encode digital information. For example:

  • BPSK (Binary Phase Shift Keying): Two distinct phase states (e.g., 0° and 180°) are used to represent binary data (0 and 1).
  • QPSK (Quadrature Phase Shift Keying): Four phase states (e.g., 0°, 90°, 180°, and 270°) are used, with each phase state representing two bits of data.
  • Higher-Order PSK: More phase states (e.g., 8-PSK, 16-PSK) are used to represent more bits per symbol, increasing the data transmission rate.

Advantages of PSK:

  • High spectral efficiency, allowing more data to be transmitted within a given bandwidth.
  • Resilience to noise and interference, especially in low signal-to-noise ratio (SNR) environments.
  • Widely used in modern digital communication systems, including wireless networks, satellite communications, and optical fiber systems.

Disadvantages of PSK:

  • Complexity increases with higher-order PSK, requiring sophisticated demodulation techniques.
  • Performance can degrade in the presence of severe phase noise or synchronization issues.

Applications: PSK is extensively used in various digital communication systems, including Wi-Fi, cellular networks, and satellite communications, due to its efficiency and robustness.

Correct Option Analysis:

The correct option is:

Option 3: Phase Shift Keying (PSK)

This option correctly identifies PSK as an example of a digital modulation technique. PSK is a widely used method for encoding digital information by varying the phase of the carrier signal, making it a fundamental concept in digital communication systems.

Important Information

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

Option 1: Single Sideband Modulation (SSB)

SSB is a type of analog modulation technique used in amplitude modulation (AM) systems. It involves transmitting only one sideband (either the upper or lower sideband) of the modulated signal while suppressing the carrier and the other sideband. SSB is primarily used in analog communication systems, such as amateur radio and long-distance voice communication, and is not an example of a digital modulation technique.

Option 2: Narrow Band FM

Narrow Band FM is another analog modulation technique where the frequency of the carrier signal is varied in proportion to the modulating signal. It is used in applications such as two-way radio communication and public safety communication. Similar to SSB, it is not a digital modulation technique.

Option 4: Double Sideband Amplitude Modulation

This is a classical analog modulation method where both the upper and lower sidebands are transmitted along with the carrier signal. It is commonly used in AM radio broadcasting but is not related to digital modulation techniques.

Conclusion:

Understanding the distinction between analog and digital modulation techniques is crucial for identifying the correct option. Among the options provided, only Phase Shift Keying (PSK) is a digital modulation technique, as it involves encoding digital information by varying the phase of the carrier signal. The other options, including Single Sideband Modulation (SSB), Narrow Band FM, and Double Sideband Amplitude Modulation, are examples of analog modulation techniques, which are fundamentally different from digital modulation.

Communication Systems Question 2:

Return loss of a stand-alone antenna is 10 dB. One 6 dB attenuator with return. loss of 40 dB is connected at the input of that antenna. What will be the approximate return loss of the combined (Antenna + Attenuator) port?

  1. 40 dB
  2. 22 dB
  3. 16 dB
  4. 10 dB

Answer (Detailed Solution Below)

Option 2 : 22 dB

Communication Systems Question 2 Detailed Solution

Concept:

Return loss (RL) measures how well a device or system matches its load. It is defined as the ratio (in dB) of the power sent into a device to the power reflected back due to impedance mismatch.

When an attenuator is placed in front of an antenna, it improves the effective return loss at the input port of the system.

The formula for combined return loss is:

\(RL_{\text{combined}} = RL_{\text{antenna}} + 2 \times \text{attenuation}\)

Given:

Return loss of antenna = 10 dB

Attenuator = 6 dB (with own return loss = 40 dB, but only antenna return is affected here)

Calculation:

\(RL_{\text{combined}} = 10 + 2 \times 6 = 10 + 12 = 22~dB\)

 

Communication Systems Question 3:

8 signals, each limited to 2 kHz and sampled at Nyquist rate, are converted into binary PCM signal using 256 quantization levels. The data rate for TDM signal will be:

  1. 256 kbps
  2. 32 kbps
  3. 128 kbps
  4. 512 kbps

Answer (Detailed Solution Below)

Option 1 : 256 kbps

Communication Systems Question 3 Detailed Solution

Explanation:

1. Nyquist Sampling Theorem:

According to the Nyquist sampling theorem, the sampling rate for a signal must be at least twice the highest frequency present in the signal to avoid aliasing. For each signal:

  • Signal bandwidth = 2 kHz
  • Nyquist sampling rate = 2 × 2 kHz = 4 kHz

2. Bits per Sample:

The number of quantization levels is given as 256. The number of bits required to represent each sample is calculated as:

  • Bits per sample = log2(Number of quantization levels)
  • Bits per sample = log2(256) = 8 bits

3. Data Rate for a Single Signal:

The data rate for a single signal is the product of the sampling rate and the number of bits per sample:

  • Data rate for one signal = Sampling rate × Bits per sample
  • Data rate for one signal = 4 kHz × 8 bits
  • Data rate for one signal = 32 kbps

4. Data Rate for 8 Signals:

Since there are 8 signals, the total data rate for the TDM signal is:

  • Total data rate = Data rate for one signal × Number of signals
  • Total data rate = 32 kbps × 8
  • Total data rate = 256 kbps

Final Answer:

Based on the calculations, the data rate for the TDM signal is 256 kbps.

Communication Systems Question 4:

When a carrier is AM modulated by three single tone signals of modulation percentages 50%, 50% and 20%, then the effective modulation index of the resulting signal is:

  1. \(\sqrt{0.54}\)
  2. \(\sqrt{1.2}\)
  3. 1.2
  4. \(\sqrt{0.9}\)

Answer (Detailed Solution Below)

Option 1 : \(\sqrt{0.54}\)

Communication Systems Question 4 Detailed Solution

Concept:

When a carrier is amplitude modulated by multiple independent modulating signals, the effective modulation index is calculated using the square root of the sum of the squares of individual modulation indices.

Effective modulation index, \( m_{eff} = \sqrt{m_1^2 + m_2^2 + m_3^2} \)

Given:

Modulation percentages are 50%, 50%, and 20%.

So, \( m_1 = 0.5, \; m_2 = 0.5, \; m_3 = 0.2 \)

Calculation:

\( m_{eff} = \sqrt{(0.5)^2 + (0.5)^2 + (0.2)^2} = \sqrt{0.25 + 0.25 + 0.04} = \sqrt{0.54} \)

 

Communication Systems Question 5:

The noise phenomena arranged in the order of their dominance when plotted as a function of increasing frequency are:

  1. Transit time noise, Flicker noise, White noise
  2. Flicker noise, White noise, Transit time noise
  3. White noise, Flicker noise, Transit time noise
  4. Transit time noise, White noise, Flicker noise

Answer (Detailed Solution Below)

Option 2 : Flicker noise, White noise, Transit time noise

Communication Systems Question 5 Detailed Solution

Explanation:

Noise Phenomena in Electronics

Noise phenomena in electronics refer to the unwanted disturbances or random fluctuations that interfere with the desired signal in electronic circuits and systems. These phenomena are categorized into different types based on their frequency characteristics and sources, such as flicker noise, white noise, and transit time noise.

Correct Option: The noise phenomena arranged in the order of their dominance when plotted as a function of increasing frequency are:

Option 2: Flicker noise, White noise, Transit time noise

This sequence correctly explains how different noise phenomena dominate over specific frequency ranges. Let's delve into the explanation for each type of noise to understand why Option 2 is correct:

1. Flicker Noise (1/f Noise):

  • Flicker noise is most dominant at low frequencies. It arises due to imperfections in the electronic devices, such as fluctuations in the mobility of charge carriers, or irregularities in material properties.
  • The power spectral density (PSD) of flicker noise is inversely proportional to frequency (1/f). As frequency increases, the dominance of flicker noise diminishes.
  • Flicker noise is a key concern in low-frequency applications like precision amplifiers and audio systems.

2. White Noise:

  • White noise is frequency-independent and has a constant power spectral density over a wide range of frequencies.
  • It originates from thermal agitation of charge carriers (thermal noise) or quantum effects (shot noise).
  • At intermediate frequencies, white noise dominates as flicker noise decreases with increasing frequency, and transit time noise has not yet become significant.

3. Transit Time Noise:

  • Transit time noise becomes significant at very high frequencies, often in the gigahertz range, and is associated with the finite time it takes for charge carriers to traverse a semiconductor device.
  • It is proportional to the frequency and becomes a limiting factor in the performance of high-frequency devices like microwave transistors and high-speed amplifiers.

Conclusion: When plotted as a function of increasing frequency, flicker noise dominates at low frequencies, white noise dominates at intermediate frequencies, and transit time noise becomes significant at very high frequencies. This sequence is captured accurately in Option 2: Flicker noise, White noise, Transit time noise.

Top Communication Systems MCQ Objective Questions

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 6 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.

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 7 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.

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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