Pulse Modulation MCQ Quiz in বাংলা - Objective Question with Answer for Pulse Modulation - বিনামূল্যে ডাউনলোড করুন [PDF]

Last updated on Mar 9, 2025

পাওয়া Pulse Modulation उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). এই বিনামূল্যে ডাউনলোড করুন Pulse Modulation MCQ কুইজ পিডিএফ এবং আপনার আসন্ন পরীক্ষার জন্য প্রস্তুত করুন যেমন ব্যাঙ্কিং, এসএসসি, রেলওয়ে, ইউপিএসসি, রাজ্য পিএসসি।

Latest Pulse Modulation MCQ Objective Questions

Top Pulse Modulation MCQ Objective Questions

Pulse Modulation Question 1:

The main advantage of frequency modulation over amplitude modulation is

  1. That there will be no distortion
  2. That the complete information is contained in the side bands
  3. That is uses a wider band of frequencies
  4. The elimination of noises

Answer (Detailed Solution Below)

Option 4 : The elimination of noises

Pulse Modulation Question 1 Detailed Solution

  • The Noise affects the amplitude of the modulated signal
  • In the Amplitude Modulated system, the message is contained in the amplitude modulations of the carrier; hence the introduction of noise distorts the amplitude and hence the message contained in the signal
  • In a frequency modulated system the message is contained in the frequency variations of the carrier, therefore, the introduction of noise does not affect the message contained in the message signal
  • Since the amplitude of FM remains constant, the noise can be eliminated using an amplitude limiter at the demodulator.

26 June 1

Features:

AM

FM

Noise immunity

In AM, the message is stored in the form of

variation in amplitude. Noise affects the amplitude of signal most so AM is less noise immune.

In FM, the message is stored in the form of

variation in frequency so it has better noise immunity.

Bandwidth

B.W. required in AM is = 2fm.

Hence, less bandwidth is required in case

of AM.

B.W. required in FM is = 2(β+1)fm.

Hence, more bandwidth is required in the case

of FM.

Transmitted power

Power transmitted in AM is given by:

\({P_T} = {P_c}\left( {1 + \frac{{{\mu ^2}}}{2}} \right)\)

As the modulation index ‘μ’

increases power in AM increases.

In FM, power transmitted is always equal to the

total power of carrier before modulation.

Hence, FM requires less power than AM.

Pulse Modulation Question 2:

Which of the following systems is analog?

  1. PCM
  2. DM
  3. DPCM
  4. PAM

Answer (Detailed Solution Below)

Option 4 : PAM

Pulse Modulation Question 2 Detailed Solution

  • Analog modulation is a process in which analog low-frequency baseband signals (like an audio or TV signal) are transmitted over a larger distance without getting faded away, by superimposing over a higher frequency carrier signal such as a radio frequency band.
  • Different modulation techniques are explained with the help of the following block diagram:

F1 P.Y Madhu 9.03.20 D6

  • If the amplitude of a pulse or duration of a pulse is varied according to the instantaneous values of the baseband modulating signal, then such a technique is called as Pulse Amplitude Modulation (PAM).

Pulse Modulation Question 3:

A PCM system used a uniform quantizer followed by a 6 bit encoder. The system bit rate is 60 M bits/sec. The Maximum bandwidth of the message signal for which this system operates satisfactorily would be

  1. 2.5 MHz
  2. 4.5 MHz
  3. 5 MHz
  4. 10 MHz

Answer (Detailed Solution Below)

Option 3 : 5 MHz

Pulse Modulation Question 3 Detailed Solution

The bit rate for a PCM system is given by

R= nfs

Where

Rb = Bit rate

n = number of bits in encoder

fs = sampling frequency

From Nyquist criteria

fs = 2fm

Where fm = message signal frequency

R= nfs

60 MHz = 6 × 2 × fm

fm = 5 MHz

Pulse Modulation Question 4:

In an amplitude modulated system, if the total power is 600 W and the power in the carrier is 400 W. then the modulation index is 

  1. 0.50
  2. 0.75
  3. 0.90
  4. 1.00

Answer (Detailed Solution Below)

Option 4 : 1.00

Pulse Modulation Question 4 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

Calculation:

Given Pt = 600 W, Pc = 400 W

Putting these values in the total power of AM wave expression, we get:

\(600 = 400\left( {1 + \frac{{{μ^2}}}{2}} \right)\)

\(\frac{6}{4}=\left( {1 + \frac{{{μ^2}}}{2}} \right)\)

\(\frac{6}{4}=\left( {1 + \frac{{{μ^2}}}{2}} \right)\)

\(\frac{μ^2}{2}=\frac{1}{2}\)

μ = 1.00

Pulse Modulation Question 5:

A 400 W carrier is amplitude modulated and has side-band power of 50 W. The depth of modulation is

  1. 1
  2. 0.45
  3. 0.50
  4. 0.55

Answer (Detailed Solution Below)

Option 3 : 0.50

Pulse Modulation Question 5 Detailed Solution

In Amplitude modulated system

Power in side band is given by

\({P_{SB}} = {P_c}\left( {\frac{{{\mu ^2}}}{2}} \right)\)

Hence

\(\frac{{{P_{SB}}}}{{{P_c}}} = \left( {\frac{{{\mu ^2}}}{2}} \right)\)

\(\frac{{50}}{{400}} = \left( {\frac{{{\mu ^2}}}{2}} \right)\)

μ = 0.5

Pulse Modulation Question 6:

Pulse Shaping in line coding is used for

  1. Using Optimum Power Consumption
  2. Using Optimum Bandwidth
  3. Maximize Inter symbol Interference
  4. Simplicity of Circuits

Answer (Detailed Solution Below)

Option 2 : Using Optimum Bandwidth

Pulse Modulation Question 6 Detailed Solution

Pulse shaping is the process of changing the waveform of transmitted pulses. Its purpose is to make the transmitted signal better suited to its purpose or the communication channel, typically by limiting the effective bandwidth of the transmission.

Pulse Modulation Question 7:

A speech signal is sampled a rate of 20% above the Nyquist rate. The signal has a bandwidth of 10 kHz. The sample is quantized into 1024 levels and then transmitted through 8-level PAM over an AWGN baseband channel. The bandwidth required for transmission is ______

  1. 80 kHz
  2. 44 kHz
  3. 40 kHz
  4. 34 kHz

Answer (Detailed Solution Below)

Option 1 : 80 kHz

Pulse Modulation Question 7 Detailed Solution

Concept:

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

Where,

Rb = nfs

n = no. of bits

M = no. of levels of PAM

Calculation:

fm = 10 kHz

Nyquist sampling rate = 2f­m = 20 kHz.

The signal is sampled at 20% above Nyquist rate:

∴ fs = (1.2 × 20) = 24 kHz.

No. of Quantization levels

= 1024

No. of bits required = log2 (1024)

= log2 210 = 10 bits.

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

\(= \frac{{10\; \times \;24}}{{ \;{{\log }_2}8}}\)

\(= \frac{{240}}{3}\)

= 80 kHz.

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

Pulse Modulation Question 8:

Consider the following statements comparing delta modulation (DM) with PCM system:

DM requires

1. A lower sampling rate

2. A higher sampling rate

3. A lower bandwidth

4. Simple hardware

Which one of the above statements are correct?

  1. 1 and 3 only
  2. 2 and 4 only
  3. 1, 3 and 4
  4. 2, 3 and 4

Answer (Detailed Solution Below)

Option 4 : 2, 3 and 4

Pulse Modulation Question 8 Detailed Solution

  • 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
  • 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
  • DM has a simple hardware requirement in comparison to PCM.

26 June 1

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

Parameter

PCM

Delta Modulation (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

Fixed number of levels are used.

Quantization error or Distortion

Quantization 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 transmission channel

Highest bandwidth is required since the number of bits are high

Lowest bandwidth is required

The bandwidth required is lower than PCM

Signal to Noise ratio

Good

Poor

Fair

Area of Application

Audio and Video Telephony

Speech and images

Speech and video

Pulse Modulation Question 9:

In phase modulation, phase deviation is proportional to:

  1. Carrier phase
  2. Message signal
  3. Carrier amplitude
  4. Message signal frequencies

Answer (Detailed Solution Below)

Option 2 : Message signal

Pulse Modulation Question 9 Detailed Solution

Analysis:

A general expression for a phase-modulated wave is:

xPM (t) = A cos [2πfct + kpm(t)]

The instantaneous angle is given as:

ϕi(t) = 2πfct + kp m(t)

Phase deviation is, therefore:

\(\Delta [ϕ_i(t)] = k_p m(t)\)

Conclusion: The phase deviation is proportional to the message signal.

  

In phase modulation, if the question was asking about the frequency deviation, then it will be proportional to the frequency of the message signal as well. But the phase deviation is proportional to the amplitude of the message signal only as derived above.

Important Points

A wave has 3 parameters Amplitude, Phase, and Frequency. Thus there are 3 types of modulation techniques.

Amplitude Modulation:

The amplitude of the carrier is varied according to the amplitude of the message signal.

Frequency Modulation:

The frequency of the carrier is varied according to the amplitude of the message signal.

Phase Modulation:

The Phase of the carrier is varied according to the amplitude of the message signal.

RRB JE EC 13 6Q 28thAug 2015 Shift3 Hindi images Q1

Pulse Modulation Question 10:

In an ADC, over sampling is a technique used to

  1. reduce the effect of input noise
  2. improve conversion clock stability
  3. reduce quantization noise
  4. improve power supply rejection ratio

Answer (Detailed Solution Below)

Option 3 : reduce quantization noise

Pulse Modulation Question 10 Detailed Solution

To reduce the Quantization noise two things are done.

Oversampling and Noise shaping.

Explanation:

  • An analog signal first undergoes the process of sampling before it is applied to ADC for conversion into a digital signal.
  • Oversampling is a process in which an analog signal is sampled at a sampling frequency that is much greater than the Nyquist rate.
  • Oversampling is:

    fs' = K × fs , ( K > 1)

  • Sigma-Delta ADC is an example of an ADC that employs oversampling.
  • It is the same as “Delta modulation”. That is 1 sample is represented with 1 bit.
  • In oversampling correlation increases and Quantization noise decreases.

26 June 1

 

Delta Modulation: 

A signal is Delta modulated to convert its analog information into a binary sequence, i.e., 1s and 0s. Here only one bit is required to represent the analog input. An over-sampled input is taken to make full use of the signal correlation. A stair-case approximated waveform will be the output of the delta modulator with the step-size as delta (Δ).

Delta Modulator has two main drawbacks:

1. Slope overload distortion: When the rate of rising of the input signal x(t) is so high that the staircase signal can not approximate it. We will see consecutive pulses of the same polarity in the output for such type of distortion. 

2. Granular noise: When the step size is too large compared to a small variation in the input signal (input is almost constant). Here, the output of the Delta modulator will be a sequence of alternate +ve and -ve pulses. 

F1 S.B 30.5.20 Pallavi D6

∴ We can conclude that if the consecutive pulses are of opposite polarity during any time interval then the signal is almost constant during that interval.

Get Free Access Now
Hot Links: teen patti circle teen patti noble teen patti neta