Sampling MCQ Quiz - Objective Question with Answer for Sampling - Download Free PDF

Last updated on Jun 20, 2025

Latest Sampling MCQ Objective Questions

Sampling Question 1:

Which of the following represent impulse response of an ideal LPF ?

  1. qImage684c26cd095c94ba958e8023
  2. qImage684c26cd095c94ba958e8025
  3. qImage684c26ce095c94ba958e8026
  4. qImage684c26ce095c94ba958e8028

Answer (Detailed Solution Below)

Option 1 : qImage684c26cd095c94ba958e8023

Sampling Question 1 Detailed Solution

Explanation:

Impulse Response of an Ideal Low Pass Filter (LPF)

Definition: The impulse response of a system is the output of the system when an impulse signal is applied as input. For an ideal Low Pass Filter (LPF), the frequency response is a rectangular function in the frequency domain, which allows frequencies within a specified range (passband) to pass through and attenuates frequencies outside this range (stopband).

Mathematical Representation:

An ideal low-pass filter has the following frequency response:

H(f) = 1 for |f| ≤ fc

H(f) = 0 for |f| > fc

Where:

  • H(f): Frequency response of the filter
  • f: Frequency variable
  • fc: Cutoff frequency

The impulse response, h(t), of the filter is obtained by taking the inverse Fourier Transform of its frequency response:

h(t) = ∫-∞ H(f) × ej2πft df

For the ideal LPF, substituting the value of H(f):

h(t) = ∫-fcfc ej2πft df

Solving this integral gives:

h(t) = 2fc × sinc(2fct)

Where the sinc function is defined as:

sinc(x) = sin(πx) / (πx)

Key Characteristics:

  • The impulse response of an ideal LPF is a sinc function, which extends infinitely in both directions in time.
  • The sinc function exhibits oscillatory behavior, with the main lobe centered at t = 0 and side lobes that decrease in amplitude as |t| increases.
  • The main lobe width is inversely proportional to the cutoff frequency fc.

Correct Option Analysis:

The correct option is:

Option 1: The impulse response of an ideal LPF is represented by the sinc function.

This option correctly identifies that the impulse response of an ideal low-pass filter is a sinc function, as derived mathematically above

Sampling Question 2:

For a band limited signal with max frequency fmax the Nyquist rate of signal is _____

  1. \(\rm \frac{fmax}{2}\)
  2. \(\rm \frac{1}{fmax}\)
  3. fmax
  4. 2f max

Answer (Detailed Solution Below)

Option 4 : 2f max

Sampling Question 2 Detailed Solution

Concept:

Nyquist Sampling interval is defined as the maximum time that:

1) is between regularly spaced samples that will permit the signal waveform to be completely determined.

2) and is equal to reciprocal to twice of the highest frequency in the sampling signal.

i.e., if signal frequency is fm then :

Nyquist rate fs ≥ 2fmax

Sampling Question 3:

According to the Nyquist theorem, to produce the original analog signal, one necessary condition is : 

  1. Sampling rate must be at least 2 times the highest frequency contained in the signal.
  2. Sampling rate must be at least 2 times the lowest frequency contained in the signal.
  3. Sampling rate must be at least \(\frac{1}{2}\) times the highest frequency contained in the signal.
  4. Sampling rate must be at least \(\frac{1}{2}\) times the lowest frequency contained in the signal.

Answer (Detailed Solution Below)

Option 1 : Sampling rate must be at least 2 times the highest frequency contained in the signal.

Sampling Question 3 Detailed Solution

The correct answer is Sampling rate must be at least 2 times the highest frequency contained in the signal.

Key Points

  • The Nyquist theorem, also known as the Nyquist-Shannon sampling theorem, states that in order to accurately reconstruct an analog signal from its samples, the sampling rate must be at least twice the highest frequency present in the original signal.
  • This minimum sampling rate is known as the Nyquist rate. If the sampling rate is lower than this, aliasing can occur, which means different signal frequencies become indistinguishable and the original signal cannot be properly reconstructed.

Therefore, the correct condition is: Sampling rate must be at least 2 times the highest frequency contained in the signal.

Sampling Question 4:

Direction: The item consists of two statements, one labelled as 'Statement (I)' and the other as 'Statement (II)'. You are to examine these two statements carefully and select the answers to these items using the code given below.

Statement (I): In the sampling and quantizing operations, errors are introduced into the digital signal. These errors are reversible and it is possible to produce an exact replica of the original analog signal from its digital representation.

Statement (II): The use of digital communication offers flexibility and compatibility in that the adoption of a common digital format makes it possible for a transmission system to sustain many different sources of information in a flexible manner.

  1. Both Statement (I) and Statement are individually true and Statement (II) is the correct explanation of Statement (I)
  2. Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I)
  3. Statement (I) is true but Statement (II) is false
  4. Statement (I) is false but Statement (II) is true

Answer (Detailed Solution Below)

Option 4 : Statement (I) is false but Statement (II) is true

Sampling Question 4 Detailed Solution

Here, statement 1 is false. Suggests that errors introduced in sampling and quantizing operations are reversible, allowing for an exact replica of the original analog signal to be produced from its digital representation. However, this is not entirely true. In the process of quantization, some information is lost, and these errors are not completely reversible.

Statement 2 is true. Discusses the advantages of digital communication, emphasizing flexibility and compatibility due to the adoption of a common digital format. This statement is generally true, as digital communication does provide flexibility and compatibility for handling various sources of information.

Here, option 4 is correct.

Sampling Question 5:

A band limited signal with a maximum frequency of 7 kHz is to be sampled. According to the sampling theorem the sampling frequency in kHz which is not valid is

  1. 12 kHz
  2. 15 kHz
  3. 18 kHz
  4. 21 kHz

Answer (Detailed Solution Below)

Option 1 : 12 kHz

Sampling Question 5 Detailed Solution

The correct answer is option 12 kHz

Concept:

According to the Nyquist sampling theorem, the sampling frequency must be greater than the Nyquist frequency which is given as fN = 2 fm

The sampling frequency fs ≥ fN

Calculation:

fm =  7 kHz 

f= 2 fm

= 14  kHz 

fs ≥ fN

Option 1 is invalid as fs is less than fN

Top Sampling MCQ Objective Questions

If the lower sideband overlaps the basebands, the distortion is called

  1. Cross-Over distortion
  2. Aliasing
  3. Crosstalk
  4. None of these

Answer (Detailed Solution Below)

Option 2 : Aliasing

Sampling Question 6 Detailed Solution

Download Solution PDF

If the lower sideband overlaps the basebands, the distortion is called Aliasing. The main reason for aliasing is undersampling \(i.e.\;{f_s} < 2{f_m}\)

BSNL TTA 25 18Q 27th first shift Part5 Updated Rishi Hindi 2

To satisfy the sampling theorem, a 100 Hz sine wave should be sampled at

  1. 10 Hz
  2. 100 Hz
  3. 200 Hz
  4. 50 Hz

Answer (Detailed Solution Below)

Option 3 : 200 Hz

Sampling Question 7 Detailed Solution

Download Solution PDF

Nyquist Sampling Theorem:

A continuous-time signal can be represented in its samples and can be recovered back when sampling frequency fs is greater than or equal to twice the highest frequency component of the message signal, i.e.

f≥ 2fm

Therefore when we want to convert continuous signals to discrete signals, the sampling must be done at the Nyquist rate.

Calculation:

With fm = 100 Hz, the minimum sampling frequency will be:

fs = 2fm = 2 × 100

fs = 200 Hz

A system has a sampling rate of 50,000 samples per second. The maximum frequency of the signal it can acquire to reconstruct is

  1. 25 kHz
  2. 50 kHz
  3. 100 kHz
  4. 10 kHz

Answer (Detailed Solution Below)

Option 1 : 25 kHz

Sampling Question 8 Detailed Solution

Download Solution PDF

Nyquist Sampling Theorem: 

A continuous-time signal can be represented in its samples and can be recovered back when sampling frequency fs is greater than or equal to twice the highest frequency component of the message signal, i.e.

f≥ 2fm

Therefore when we want to convert continuous signals to discrete signals, the sampling must be done at the Nyquist rate.

Application:

Given fs = 50 k

We can write:

\(50k \ge 2 \times f_m\)

\({f_m} \leq 25k\)

∴ The maximum frequency of the signal it can acquire to reconstruct is 25 kHz

When aliasing takes place

  1. Sampling signals less than Nyquist rate
  2. Sampling signals more than Nyquist rate
  3. Sampling signals equal to Nyquist rate
  4. Sampling signals at a rate which is twice of Nyquist rate

Answer (Detailed Solution Below)

Option 1 : Sampling signals less than Nyquist rate

Sampling Question 9 Detailed Solution

Download Solution PDF
  • The sampling theorem states that “a signal can be exactly reproduced if it is sampled at the rate fs which is greater than twice the maximum frequency W in the modulating signal."
  • The minimum rate at which a signal can be sampled and still be reconstructed from its samples is known as Nyquist rate.
  • If the sampled frequency is less than the Nyquist frequency, overlapping of lower and upper sidebands known as aliasing takes place.
  • The main reason for aliasing is undersampling \(i.e.\;{f_s} < 2{f_m}\)

           fs = Sampling frequency and fm = Modulating frequency

Aliasing is explained with the help of the spectrum as shown:

BSNL TTA 25 18Q 27th first shift Part5 Updated Rishi Hindi 2

In sampling theorem the Nyquist interval is given by

  1. \(T_s=\frac{1}{f_m}\)
  2. \(T_s=\frac{1}{2f_m}\)
  3. \(T_s=\frac{1}{\pi f_m}\)
  4. \(T_s=\frac{\pi}{f_m}\)

Answer (Detailed Solution Below)

Option 2 : \(T_s=\frac{1}{2f_m}\)

Sampling Question 10 Detailed Solution

Download Solution PDF

Nyquist sampling theorem states that “a signal can be exactly reproduced if it is sampled at the rate fs which is greater than or equal to twice the maximum frequency fm of the modulating signal."​, i.e.

The minimum sampling frequency is:

fs = 2 fm

In terms of the time period:

\(\frac{1}{T_s}=2f_m\)

\(T_s=\frac{1}{2f_m}\)

Ts = Nyquist sampling interval

Calculate the Nyquist rate for sampling when a continuous time signal is given by:

x(t) = 5 cos 100πt +10 cos 200πt - 15 cos 300πt

  1. 100 Hz
  2. 150 Hz
  3. 300 Hz
  4. 600 Hz

Answer (Detailed Solution Below)

Option 3 : 300 Hz

Sampling Question 11 Detailed Solution

Download Solution PDF

Minimum sampling rate to avoid aliasing fs = 2fm= (Nyquist rate)

ωm = max {100π, 200π, 300π} = 300π

fm = 150 Hz = maximum frequency of signal

Sampling frequency or Nyquist rate; fs = 2 × 150 = 300 Hz

A signal of maximum frequency of 10 kHz is sampled at Nyquist rate. The time interval between two successive samples is

  1. 50 μs
  2. 100 μs
  3. 500 μs
  4. 1000 μs

Answer (Detailed Solution Below)

Option 1 : 50 μs

Sampling Question 12 Detailed Solution

Download Solution PDF

Concept:

Nyquist Sampling interval is defined as the maximum time that:

1) is between regularly spaced samples that will permit the signal waveform to be completely determined.

2) and is equal to reciprocal to twice of the highest frequency in the sampling signal.

i.e., if signal frequency is fm then :

Nyquist rate fs ≥ 2fm

And Nyquist interval:

\({T_s} \le \frac{1}{{2{f_m}}}\)

Calculation:

fm = 10 kHz

fs = 20 kHz

\({T_s} \le \frac{1}{{2\times{10}}}\)

Ts = 50 μs

The spectrum of a band pass signal spans from 20 kHz to 30 kHz. The signal can be recovered ideally from the sampled values when the sampling rate is at least:

  1. 20 kHz
  2. 60 kHz
  3. 50 kHz
  4. 40 kHz

Answer (Detailed Solution Below)

Option 1 : 20 kHz

Sampling Question 13 Detailed Solution

Download Solution PDF

Concept:

Sampling frequency of bandpass signal is given by:

\({f_S} \ge \frac{{2{f_H}}}{k}\)

Where

\(k = \left[ {\frac{{{f_H}}}{{{f_H} - {f_L}}}} \right]\)  (Where [ .] denotes the floor value)

Analysis:

fH = 30 kHz

fL = 20 kHz

fH - fL = 10 kHz

\(k = \left[ {\frac{{30}}{10}} \right] = 3\) (floor value is 3)

\({f_S} \ge \frac{{2{f_H}}}{k} \ge 2\frac{{30}}{3} \ge 20~ kHz\)

so minimum sampling frequency is 20 kHz

How many minimum number of samples are required to exactly describe the following signal?

x(t) = 10 cos(6πt) + 4 sin(8πt)

  1. 4 samples per second
  2. 6 samples per second
  3. 8 samples per second
  4. 2 samples per second

Answer (Detailed Solution Below)

Option 3 : 8 samples per second

Sampling Question 14 Detailed Solution

Download Solution PDF

Sampling Theorem:

 A continuous-time signal can be represented in its samples and can be recovered back when sampling frequency (fs)is greater than or equal to twice the highest frequency component of the message signal (fm).

The sampling frequency is given by:

\(f_s=2\times max(f_{m1},f_{m2})\)

Calculation:

Given, x(t) = 10 cos(6πt) + 4 sin(8πt)

6π = 2πfm1

fm1 = 3 Hz

8π = 2πfm2

fm2 = 4 Hz

\(f_s=2\times max(3,4)\)

fs = 8 Hz

fs = 8 samples per second

A band limited signal is sampled at Nyguist rate, The signal can be recovered by passing the samples through

  1. RC filter
  2. envelope detector
  3. PLL
  4. ideal low pass filter with appropriate band width

Answer (Detailed Solution Below)

Option 4 : ideal low pass filter with appropriate band width

Sampling Question 15 Detailed Solution

Download Solution PDF

Concept:

Nyquist Sampling Theorem:

A continuous-time signal can be represented in its samples and can be recovered back when sampling frequency fs is greater than or equal to twice the highest frequency component of the message signal, i.e.

f≥ 2fm

Therefore when we want to convert continuous signals to discrete signals, the sampling must be done at the Nyquist rate.

If the low pass filter is not ideal, just sampling at the Nyquist frequency will not work we need to sample above the Nyquist frequency. 

For example:

We have assumed spectrum x(t) is band limited to ωm.

F1 Tapesh 24.9.20 Pallavi D7

Let, ωs > 2ωm

Let ωs = 3ωm

The spectrum of the sampled signal will be:

F1 Tapesh 24.9.20 Pallavi D8

We see for ωs > 2ωm, there is no overlap between the shifted spectrum of X(ω). Thus, as long as the sampling frequency ωs is greater than the twice the signal bandwidth (2 ω­m), x(t) can be recovered by passing the sampled signal xs(t) through an ideal or practical low pass filter having bandwidth between ωm and (ωs - ωm) rad/sec.

Get Free Access Now
Hot Links: teen patti pro teen patti 100 bonus teen patti master gold teen patti all games