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

The SNR in a binary PCM system can be expressed as:

SNR = (6 × n) dB

where n is the number of bits used for quantization. This indicates that the SNR increases linearly with the number of bits used in the PCM system.

Effect of Increasing Number of Bits:

When the number of bits in the PCM system is increased from n to n + 2, the SNR also increases. The improvement in SNR can be calculated using the difference in SNR values before and after increasing the number of bits:

Improvement in SNR = SNR (n + 2) - SNR (n)

Substituting the values:

Improvement in SNR = (6 × (n + 2)) - (6 × n)

Improvement in SNR = 6 × n + 12 - 6 × n

Improvement in SNR = 12 dB

Hence the correct answer is 12 dB

Explanation:

 According to Carson's rule, the bandwidth of an FM signal that occupies 98% of its power can be expressed as:

Bandwidth (B) ≈ 2 × (Δf + fm)

Where:

• Δf: The frequency deviation, which represents the maximum shift in the carrier frequency due to the modulation signal.
• fm: The maximum frequency of the modulating signal.

In this case, the FM signal is given as:

10 sin(2π × 10⁶t + 4 cos(2π × 10⁴t))

From the given FM signal, we can extract the following parameters:

• Carrier Frequency (fc): 10⁶ Hz (1 MHz)
• Frequency of the Modulating Signal (fm): 10⁴ Hz (10 kHz)
• Frequency Deviation (Δf): The frequency deviation is determined by the modulation index multiplied by the frequency of the modulating signal. In this case, the modulation index is 4, so:

Δf = Modulation Index × fm

Δf = 4 × 10⁴

Δf = 40 kHz

Now, applying Carson's rule to calculate the bandwidth:

Bandwidth (B) ≈ 2 × (Δf + fm)

B ≈ 2 × (40 kHz + 10 kHz)

B ≈ 2 × 50 kHz

B ≈ 100 kHz

Amplitude Modulation and Power Calculation

Problem Statement: A carrier signal is amplitude modulated with a modulation index of 60%, resulting in a transmitted power of 472 W. We are tasked with determining how much power can be saved by suppressing the carrier and one sideband of the modulated signal.

Solution:

To solve this problem, let us analyze the power distribution in an amplitude-modulated signal.

Step 1: Power Components in an Amplitude-Modulated Signal

In amplitude modulation (AM), the total transmitted power (P_t) consists of:

• Carrier power (P_c).
• Power in the two sidebands (P_USB and P_LSB), which are equal in magnitude.

The total transmitted power is given by:

P_t = P_c + P_USB + P_LSB

The power in each sideband is determined by the modulation index (m), and is given as:

P_USB = P_LSB = (m²/4) × P_c

Thus, the total transmitted power can be expressed as:

P_t = P_c (1 + m²/2)

Step 2: Calculate the Carrier Power (P_c)

From the problem, the modulation index (m) is 60%, or m = 0.6. The total transmitted power (P_t) is 472 W. Substituting these values into the formula for total transmitted power:

472 = P_c (1 + (0.6)²/2)

First, calculate (0.6)²/2:

(0.6)²/2 = 0.36/2 = 0.18

Now substitute this value into the equation:

472 = P_c (1 + 0.18)

472 = P_c × 1.18

P_c = 472 / 1.18

P_c ≈ 400 W

Thus, the carrier power is 400 W.

Step 3: Calculate the Sideband Power

The power in the sidebands is given by:

P_sidebands = P_USB + P_LSB = (m²/2) × P_c

Substitute m = 0.6 and P_c = 400 W:

P_sidebands = (0.6)²/2 × 400

P_sidebands = 0.36/2 × 400

P_sidebands = 0.18 × 400

P_sidebands = 72 W

The total power in the sidebands is 72 W, with each sideband contributing 36 W.

Step 4: Power Saved by Suppressing the Carrier and One Sideband

When the carrier and one sideband are suppressed, the remaining power is the power in the other sideband. Thus, the power saved is:

Power saved = P_t - P_remaining

The remaining power is the power in one sideband:

P_remaining = P_LSB or P_USB = 36 W

Substitute the values:

Power saved = 472 - 36

Power saved = 436 W

Final Answer: The power saved by suppressing the carrier and one sideband is 436 W.

Minimum Permissible Bit Rate Calculation:

Given Data:

• Bandwidth (BW) of the audio signal = 15 kHz
• Number of quantization levels = 1024
• No overheads in PCM encoding

Step 1: Determine the number of bits per sample

The number of quantization levels (L) is given as 1024. The number of bits required to represent each quantization level can be calculated using the formula:

Number of bits per sample = log₂(L)

Here, log₂(1024) = 10. Therefore, each sample requires 10 bits to represent the quantized value.

Step 2: Apply Nyquist Theorem to determine the sampling rate

According to Nyquist theorem, the minimum sampling rate (f_s) should be at least twice the bandwidth of the signal:

f_s = 2 × BW

Given BW = 15 kHz:

f_s = 2 × 15 kHz = 30 kHz

Step 3: Calculate the minimum permissible bit rate

The bit rate (R) is given by:

R = Sampling Rate × Number of bits per sample

Substituting the values:

R = 30 kHz × 10 bits/sample = 300 kbps

Therefore, the minimum permissible bit rate for the given communication system is 300 kbps.

Correct Option Analysis:

The correct option is:

Option 2) 300 kbps

This option correctly represents the minimum permissible bit rate for the given communication system using PCM encoding, considering the bandwidth, quantization levels, and sampling rate.

Bandwidth of the Main Lobe in BPSK Signal Spectrum:

Definition: In Binary Phase Shift Keying (BPSK), the spectrum of the modulated signal consists of a main lobe and side lobes. The bandwidth of the main lobe is a critical parameter that determines the frequency range over which most of the signal's power is concentrated. This bandwidth is influenced by the bit rate of the signal.

Formula for Bandwidth: The bandwidth of the main lobe for the spectrum of a BPSK signal can be approximated using the following relation:

Bandwidth of the main lobe = 2 × Bit Rate

Here:

• Bit Rate: The rate at which bits are transmitted, typically measured in bits per second (bps).
• 2: A factor that accounts for the symmetric nature of the main lobe around the carrier frequency.

Given Data:

• Carrier Frequency = 1 MHz (This value does not affect the bandwidth of the main lobe.)
• Bit Rate = 1 kbps = 1000 bps

Calculation:

The bandwidth of the main lobe is calculated as:

Bandwidth of the main lobe = 2 × Bit Rate

Bandwidth of the main lobe = 2 × 1000 bps

Bandwidth of the main lobe = 2000 Hz =  2 kHz

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.

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 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.

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.

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

$V=\frac{KT}{q}\mathrm{l}\mathrm{n}\left(\frac{N_a{N}_d}{n_i^2}\right)$

​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 

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

$\therefore {\left(BW\right)}_{\mathrm{m}\mathrm{i}\mathrm{n}}=\frac{2R_b}{{\log }_{2}N}Hz=\frac{2M}{{\log }_{2}16}kHz$

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

$=\frac{2M}{4\times {\log }_{2}2}kHz$

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

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

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 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 nf_s
• 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​

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

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.