Sampling Theorem and Nyquist Rate MCQ Quiz - Objective Question with Answer for Sampling Theorem and Nyquist Rate - Download Free PDF

Last updated on May 30, 2025

Latest Sampling Theorem and Nyquist Rate MCQ Objective Questions

Sampling Theorem and Nyquist Rate Question 1:

What type of distortion occurs if a signal that contains frequency components up to 15 kHz is sampled using 20 kHz?

  1. Quantization error
  2. Aliasing
  3. Slope Overload
  4. No distortion

Answer (Detailed Solution Below)

Option 2 : Aliasing

Sampling Theorem and Nyquist Rate Question 1 Detailed Solution

Explanation:

Understanding Aliasing in Signal Processing

Definition: Aliasing is a phenomenon that occurs when a signal is sampled at a rate that is insufficient to capture the changes in the signal accurately. Specifically, aliasing happens when the sampling rate is below the Nyquist rate, which is twice the maximum frequency present in the signal. When aliasing occurs, different signals become indistinguishable from each other after sampling, leading to distortion.

Detailed Explanation: In the context of the given problem, a signal contains frequency components up to 15 kHz, and it is sampled using a 20 kHz sampling rate. According to the Nyquist theorem, the sampling rate must be at least twice the maximum frequency present in the signal to avoid aliasing. Therefore, the Nyquist rate for a signal with a maximum frequency of 15 kHz is 30 kHz. Since the given sampling rate of 20 kHz is below this Nyquist rate, aliasing will occur.

When aliasing occurs, the higher frequency components of the signal are "folded" back into the lower frequencies, causing distortion that makes the original signal indistinguishable from its aliased counterpart. This effect results in a loss of information and can significantly degrade the quality of the reconstructed signal.

Illustrative Example: Consider a simple sine wave with a frequency of 15 kHz. If we sample this wave at 20 kHz, the sampling points will not capture the wave's true form accurately. Instead, the samples will represent a different, lower frequency signal, leading to a distorted representation. This misrepresentation can be visualized by plotting the sampled points and observing that the reconstructed signal does not match the original 15 kHz sine wave.

Importance of Proper Sampling: To avoid aliasing, it is crucial to sample signals at a rate that is at least twice the maximum frequency present in the signal. This requirement ensures that the original signal can be accurately reconstructed from the sampled data. In practical applications, engineers often use anti-aliasing filters to remove high-frequency components before sampling, ensuring that the sampled signal adheres to the Nyquist criterion.

Conclusion: In the given problem, the correct answer is option 2 (Aliasing) because the signal with frequency components up to 15 kHz is sampled at a rate of 20 kHz, which is below the Nyquist rate of 30 kHz, leading to aliasing and resulting in distortion.

Important Information:

Let's analyze the other options to understand why they are incorrect:

Quantization error occurs during the process of converting a continuous signal into a discrete digital signal by approximating the signal's amplitude to the nearest value within a finite set of levels. This type of error is related to the precision of the digital representation and is not specifically related to the sampling rate. In the given problem, the primary issue is the sampling rate being too low, leading to aliasing rather than quantization error.

Slope overload is a phenomenon that occurs in delta modulation when the rate of change of the input signal exceeds the ability of the modulator to track it. This results in distortion due to the inability to follow steep slopes in the signal. Slope overload is not related to the sampling rate but to the modulation technique and its parameters. Therefore, it is not applicable to the given problem.

Option 4 suggests that no distortion occurs, which is incorrect. Given that the sampling rate of 20 kHz is below the Nyquist rate for a signal with frequency components up to 15 kHz, aliasing will occur, causing distortion. Thus, this option does not apply to the scenario described in the problem.

  • Option 1: Quantization Error
  • Option 3: Slope Overload
  • Option 4: No Distortion

Sampling Theorem and Nyquist Rate Question 2:

Which of the following is not related to frequency band of Television transmission?   

  1. 76-88 MHz
  2. 420-890 MHz
  3. 174-216 MHz
  4. 896-901 MHz

Answer (Detailed Solution Below)

Option 4 : 896-901 MHz

Sampling Theorem and Nyquist Rate Question 2 Detailed Solution

The correct answer is 896-901 MHz

Key Points

  • The frequency band is not related to television transmission is 896-901 MHz
  • The VHF (Very High Frequency) bands for Television transmission are around 54-216 MHz (including 76-88 MHz and 174-216 MHz), and the UHF (Ultra High Frequency) bands are around 470-890 MHz.
  • Hence, the frequency band 896-901 MHz does not fall into any typical TV transmission band.

Sampling Theorem and Nyquist Rate Question 3:

Let x1(t) and x2(t) be two band-limited signals having bandwidth 𝐵 = 4𝜋 × 103 rad/s each. In the figure below, the Nyquist sampling frequency, in rad/s, required to sample y(t), is

F1 Engineering Arbaz 27-12-23 D13

  1. 20𝜋 × 103
  2. 40𝜋 × 103 
  3. 8𝜋 × 103
  4. 32𝜋 × 103

Answer (Detailed Solution Below)

Option 4 : 32𝜋 × 103

Sampling Theorem and Nyquist Rate Question 3 Detailed Solution

Given that, x1(t) and x2(t) are two bandlimited signals having bandwidth B = 4π × 103 rad/sec

F1 Engineering Arbaz 27-12-23 D14

and y(t) = x2(t)cos(12π × 103t) + x1(t) cos(4π × 103t)

F1 Engineering Arbaz 27-12-23 D15

So, Nyquist rate = 2ωmax

= 2[16π × 103]

= 32π × 103 rad/sec

Sampling Theorem and Nyquist Rate Question 4:

The highest frequency component of a speech signal needed for telephonic communications is about 3.1 kHz. What is the suitable value for the sampling rate ?

  1. 1 kHz
  2. 2 kHz
  3. 4 kHz
  4. 8 kHz

Answer (Detailed Solution Below)

Option 4 : 8 kHz

Sampling Theorem and Nyquist Rate Question 4 Detailed Solution

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:

Given that,

fm = 3.1 kHz 

⇒ f≥ 2fm

⇒ fs ≥ 2 × 3.1 = 6.4 kHz

Sampling Theorem and Nyquist Rate Question 5:

If a 100 Hz sinusoidal signal is sampled at the rates of 140 Hz, 90 Hz, and 30 Hz, then the aliased frequencies correspond to each sampling rate will be respectively:

  1. 180 Hz, 40 Hz, 50 Hz
  2. 30 Hz, 60 Hz, 50 Hz
  3. 40 Hz, 10 Hz, 70 Hz
  4. 60 Hz, 80 Hz, 10 Hz
  5. 40 Hz, 190 Hz, 130 Hz

Answer (Detailed Solution Below)

Option 3 : 40 Hz, 10 Hz, 70 Hz

Sampling Theorem and Nyquist Rate Question 5 Detailed Solution

Concept:

Aliased frequency component is given by:

fa = |fs – fm|

fs = sampling frequency

fm = message signal frequency

Calculation:

For a sampling rate of 140 Hz, since fs > fm:

Aliased Frequency = fs - fm,               

= 140 – 100 = 40 Hz

For a sampling rate of 90 Hz, since fs < fm:

Aliased Frequency = fm – fs,       

= 100 – 90 = 10 Hz

For a sampling rate of 30 Hz, since fs < fm:

Aliased Frequency = fm – fs    

= 100 – 30 = 70 Hz

26 June 1

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

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

Top Sampling Theorem and Nyquist Rate MCQ Objective Questions

A band-limited signal with a maximum frequency of 5 kHz is to be sampled. According to the sampling theorem, the sampling frequency which is not valid is

  1. 5 kHz
  2. 12 kHz
  3. 15 kHz
  4. 20 kHz

Answer (Detailed Solution Below)

Option 1 : 5 kHz

Sampling Theorem and Nyquist Rate Question 6 Detailed Solution

Download Solution PDF

The maximum frequency of the band-limited signal.

fm = 5 kHz

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

fN = 2 fm = 2 × 5 = 10 kHz

So the sampling frequency fs ≥ fN

fs ≥ 10 kHz

Only the option (A) does not satisfy the condition.

∴ 5 kHz is not a valid sampling frequency.

For the signal f(t) = 3 sin 8πt + 6 sin 12πt + sin 14πt, the minimum sampling frequency (in Hz) satisfying the Nyquist criterion is -----------.

Answer (Detailed Solution Below) 14

Sampling Theorem and Nyquist Rate Question 7 Detailed Solution

Download Solution PDF

Concept:

Nyquist rate is the minimum rate at which a signal can be sampled without introducing errors, which is twice the highest frequency present in the signal.

Nyquist rate = 2 × highest frequency present in the signal

fs = 2fm

The condition on the sampling interval is: fs > 2fm

Calculation:Ts<12Tm" id="MathJax-Element-19-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0">

f(t) = 3 sin 8πt + 6 sin 12πt + sin 14πt

Nyquist criteria says that

Sampling frequency, fs > 2f max

fmax  →  maximum frequency of message signal.

Maximum frequency is given signal, ωmax = 14 π

⇒  fmax = 7

∴ sampling frequency, fs > 2 × 7

fs > 14

∴ minimum sampling frequency is 14 Hz

Consider the signal x(t) = cos(6πt) + sin(8πt), where t is in seconds. The Nyquist sampling rate (in samples/second) for the signal y(t) = x(2t + 5) is

  1. 8
  2. 12
  3. 16
  4. 32

Answer (Detailed Solution Below)

Option 3 : 16

Sampling Theorem and Nyquist Rate Question 8 Detailed Solution

Download Solution PDF

Given,

x(t) = cos(6πt) + sin(8πt)

The bandwidth of the signal will be:

\(\frac{{8{\rm{π }}}}{{2{\rm{π }}}} = 4~{\rm{Hz}}\)

Now, y(t) = x(2t + 5) can be written as:

\(y(t) = {\rm{x}}\left( {2\left( {{\rm{t}} + \frac{5}{2}} \right)} \right)\)

Taking the Fourier transform, we get:

\({\rm{Y}}\left( {{\rm{jω }}} \right) = \frac{1}{{\left| 2 \right|}}{\rm{X}}\left( {\frac{{{\rm{jω }}}}{2}} \right){{\rm{e}}^{\frac{{{\rm{j}}5{\rm{ω }}}}{2}}}\)

\( {\rm{Y}}\left( {{\rm{jω }}} \right) = \frac{1}{2}{\rm{X}}\left( {\frac{{{\rm{jω }}}}{2}} \right){{\rm{e}}^{\frac{{{\rm{j}}5{\rm{ω }}}}{2}}} \)

Thus, the frequency spectrum of X(jω)  expands by 2.

This will make the highest frequency component of Y(jω) as:

2× 4 = 8 Hz 

Hence, the Nyquist rate will be:

fs = 2 × 8 = 16 samples/sec

Consider the two continuous-time signals defined below:

\({x_1}\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {\left| t \right|, - 1 \le t \le 1\;}\\ {0,\;otherwise\;} \end{array},} \right.{x_2}\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {1 - \left| t \right|, - 1 \le t \le 1\;}\\ {0,\;otherwise\;} \end{array}} \right.\)

These signals are sampled with a sampling period of T = 0.25 seconds to obtain discrete time signals x1[n] and x2[n], respectively. Which one of the following statements is true?

  1. The energy of x1[n] is greater than the energy of x2[n].
  2. The energy of x2[n] is greater than the energy of x1[n].
  3. x1[n] and x2 have equal energies.
  4. Neither x1[n] nor x2[n] is a finite-energy signal.

Answer (Detailed Solution Below)

Option 1 : The energy of x1[n] is greater than the energy of x2[n].

Sampling Theorem and Nyquist Rate Question 9 Detailed Solution

Download Solution PDF

Concept:

Energy of the discrete time signal x(n) is,

                         \({E} = \mathop \sum \limits_{n = - \infty }^\infty {\left| {{x}\left( n \right)} \right|^2}\)

Energy of the continuous time signal x(t) is,

                         \(E = \mathop \smallint \limits_{ - \infty }^\infty {\left( {x\left( t \right)} \right)^2}dt\)

Explanation:

\({x_1}\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {\left| t \right|}&{ - 1 \le t \le 1}\\ 0&{otherwise} \end{array}} \right.\)

 

GATE EE 2018 Techinical 54Q images Q38

After sampling the above signal for T = 0.25 sec, the discrete signal will be

GATE EE 2018 Techinical 54Q images Q38a

Energy of \({x_1}\left( n \right) = {E_1} = \mathop \sum \limits_{n = - \infty }^\infty {\left| {{x_1}\left( n \right)} \right|^2}\)

= 02 + 2 [0.252 + 0.52 + 0.752 + 12]

= 3.75

\({x_2}\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {1 - \left| t \right|}&{ - 1 \le t \le 1}\\ 0&{otherwise} \end{array}} \right.\)

GATE EE 2018 Techinical 54Q images Q38b

After sampling the above signal for T = 0.25 sec, the discrete signal will be

GATE EE 2018 Techinical 54Q images Q38c

Energy of \({x_2}\left( n \right) = {E_2} = \mathop \sum \limits_{n = - \infty }^\infty {\left| {{x_2}\left( n \right)} \right|^2}\)

= 12 + 2 [0.752 + 0.52 + 0.252]

= 2.75

⇒ E1 > E2

The Nyquist sampling interval, for the signal sinc(700t) + sinc(500t) is:

  1. \(\frac{1}{{350}}sec\)
  2. \(\frac{\pi }{{350}}sec\)
  3. \(\frac{1}{{700}}sec\)
  4. \(\frac{\pi }{{175}}sec\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{1}{{700}}sec\)

Sampling Theorem and Nyquist Rate Question 10 Detailed Solution

Download Solution PDF

Concept:

Nyquist rate: The minimum sampling rate is often called the Nyquist rate. The Nyquist sampling rate is two times the highest frequency of the input or message signal.

\({\rm{\;}}{f_s} = 2{f_m}\)

Where,

fs is the minimum sampling frequency or Nyquist rate

fm is the highest frequency of the input or message signal.

Calculation:

sinc(700t) + sinc(500t)

\( = \frac{{\sin 700\pi t}}{{700\pi t}} + \frac{{\sin 500\pi t}}{{500\pi t}}\)

f1 = 700π/2π = 350 Hz

f2 = 500π/2π = 250 Hz

fm = max (f1, f2) = max (350, 250) = 350 Hz

Sampling frequency,

fs = 2fm = 2 × 350 = 700 Hz

Sampling time period = 1/700 sec

A band-limited low-pass signal x(𝑡) of bandwidth 5 kHz is sampled at a sampling rate 𝑓𝑠. The signal x(𝑡) is reconstructed using the reconstruction filter H(𝑓) whose magnitude response is shown below:

1107

The minimum sampling rate fs(in kHz) for perfect reconstruction of x(t) is _______.

Answer (Detailed Solution Below) 13

Sampling Theorem and Nyquist Rate Question 11 Detailed Solution

Download Solution PDF

Concept:

According to sampling theorem "A band-limited signal of finite energy can be completed reconstructed from its samples taken uniformly at a rates 

ωs ≥ 2 ωs sample per sec.

In other word minimum sample rate is fs = 2fm Hz

Calculation:

1108

Minimum sampling rate fshould be

8 ≤ fs - 5

f ≥ 8 + 5 = 13 kHz

A sinusoid x(t) of unknown frequency is sampled by an impulse train of period 20 ms. The resulting sample train is next applied to an ideal lowpass filter with cutoff at 25 Hz. The filter output is seen to be a sinusoid of frequency 20 Hz. This means that x(t) is

  1. 10 Hz
  2. 60 Hz
  3. 30 Hz
  4. 90 Hz

Answer (Detailed Solution Below)

Option 3 : 30 Hz

Sampling Theorem and Nyquist Rate Question 12 Detailed Solution

Download Solution PDF

Period of sampling train, Ts = 20 ms

\(\therefore {f_s} = \frac{1}{{20 \times {{10}^{ - 3}}}} = 50{\rm{Hz}}\)

If frequency of x(t) is fx, then after sampling the signal, the sampled signal has the frequency,

fs - fx = 50 - fx and fs + fx = 50 + fs

 Now, the sampled signal is applied to and ideal low pass filter with cut off frequency.

fc = 25 Hz

Now, the o/p of filter carried a single frequency component of 20 Hz

∴, only \(\left( {{f_s} - {f_x}} \right)\) component passes through the filter, ie.

fs - fx < 25

and fs - fx = 20

50 - f = 20

fx = 50 - 20 = 30 Hz

A continuous-time signal has frequency content at f = 10 MHz, 50 MHz, and 70 MHz. The signal is sampled at a sampling frequency of 56 MHz and is then passed through a low-pass filter with a cutoff frequency of 15 MHz. The frequency content of the output of the filter will be:

  1. 10 MHz
  2. 10 MHz and 6 MHz
  3. 10 MHz, 6 MHz, and 14MHz
  4. 46 MHz

Answer (Detailed Solution Below)

Option 3 : 10 MHz, 6 MHz, and 14MHz

Sampling Theorem and Nyquist Rate Question 13 Detailed Solution

Download Solution PDF

Calculation:

fs = Sampling Frequency = 56 MHz

The frequencies present at the input are:

fm1 = 10 MHz, fm2 = 50 MHz and, fm3 = 70 MHz

Once sampled, the frequency content at the output will be: fm ± fs

For, fm1 = 10 MHz, the output will contain frequencies of,

= fm1, fm1 ±, fm1 ± 2fs ….

= 10 MHz, 66 MHz, 122 MHz

For, fm2 = 50 MHz, the output will contain frequencies of

= fm2, fm2 ± fs, fm2 ± 2fs ….

= 50 MHz, 6 MHz, 106 MHz, 162 MHz

For, fm3 = 70 MHz, the output will contain frequencies of

= fm3, fm3 ± fs, fm3 ± 2fs ….

= 70 MHz, 14 MHz, 182 MHz ….

When these frequencies are passed through a low pass filter with a cutoff frequency of 15 MHz, the only remaining frequencies at its output will be 6MHz, 10 MHz and 14 MHz.

Let \({x_1}\left( t \right) \leftrightarrow {X_1}\left( \omega \right)\) and \({x_2}\left( t \right) \leftrightarrow {X_2}\left( \omega \right)\) be two signals whose Fourier Transforms are as shown in the figure below. In the figure, \(h\left( t \right) = {e^{ - 2\left| t \right|}}\) denotes the impulse response.

Gate EE 2016 paper 2 Images-Q28

For the system shown above, the minimum sampling rate required to sample y(t), so that y(t) can be uniquely reconstructed from its samples, is

  1. \(2{B_1}\)
  2. \(2\left( {{B_1} + {B_2}} \right)\)
  3. \(4\left( {{B_1} + {B_2}} \right)\)
  4. \(\infty\)

Answer (Detailed Solution Below)

Option 2 : \(2\left( {{B_1} + {B_2}} \right)\)

Sampling Theorem and Nyquist Rate Question 14 Detailed Solution

Download Solution PDF

Concept:

  • Lower limit of y(t) is sum of lower limit of x(t) & h(t).
  • Upper limit of y(t) is sum of upper limit of x(t) & h(t).

Explanation:

Fourier transform of x1(t) = x1(ω)

F1 Gaurav EE  10-08-21 Savita D10

Maximum frequency components of

x1(ω) = B, rad/sec

Fourier transform of x2(t) = x2(ω)

F1 Gaurav EE  10-08-21 Savita D11

maximum frequency component of x2(ω) = B2 rad/sec

∴ x(t) = x1(t). x2(t)

Using multiplication property in Fourier transform,

\(F\{x(t)\}=x(ω)=\frac{1}{2\pi}\{x_1(ω)*x_2(ω)\}\)

So, maximum frequency component of x(ω) = B1 + B2 rad/sec

Impulse response  \(h(t)=e^{-2|t|}\)

Taking Fourier transform of h(t) is,

\(H(ω)=\frac{4}{4+ω^2}\)

Maximum frequency component of H(ω) = ∞

∴ y(t) = x(t) * h(t)

Fourier Transform, Y(ω) = x(ω) H(ω)

\(=\frac{1}{2\pi}\{x_1(ω) *x_2(ω)\}.H(ω)\)

Highest frequency component of

Y(ω) = min {x(ω), H(ω)}

Y(ω) = min {B1 + B2, ∞}

Y(ω) = B1 + B2

Minimum sampling rate = Nyquist rate

= 2 × maximum frequency component of Y(ω)

= 2 (B1 + B2) rad/sec

The output y(t) of the following system is to be sampled, so as to reconstruct it from its samples uniquely. The required minimum sampling rate is

F1 U.B. N.J 3-08-2019 D 13

  1. 1000 samples/s
  2. 1500 samples/s
  3. 2000 samples/s
  4. 3000 samples/s

Answer (Detailed Solution Below)

Option 2 : 1500 samples/s

Sampling Theorem and Nyquist Rate Question 15 Detailed Solution

Download Solution PDF

Concept:

Duality property,

A rect \(\left(\frac{t}{\tau}\right)\rm\overset{F.T.}{\leftrightarrow}A\tau \sin C(f\tau)=A\tau \sin C \left(\frac{\omega \tau}{2\pi}\right)\)

\(A\tau \sin C \left(\frac{t \tau}{2\pi}\right)\rm\overset{F.T.}{\leftrightarrow}2\pi A\ rect\left(\frac{\omega}{\tau}\right)\)

Explanation:

 

F1 Gaurav EE  10-08-21 Savita D18

F1 U.B M.P 28.08.19 D 1

z(t) = x(t) cos (1000 πt)

\({\rm{Z}}\left( {\rm{\omega }} \right) = \frac{1}{2}\left[ {{\rm{x}}\left( {{\rm{\omega }} - 1000{\rm{\;\pi }}} \right) + {\rm{x}}\left( {{\rm{\omega }} + 1000{\rm{\pi }}} \right)} \right]\)

F1 U.B M.P 28.08.19 D 2

Spectrum of impulse response is,

F1 U.B M.P 28.08.19 D 3

\(\frac{{\sin {\rm{at}}}}{{{\rm{\pi t}}}} \leftrightarrow \left\{ {\begin{array}{*{20}{c}} {1,}&{\left| {\rm{\omega }} \right| < {\rm{a}}}\\ {0,}&{\left| {\rm{\omega }} \right| > {\rm{a}}} \end{array}} \right.\)

Spectrum of output is,

F1 U.B M.P 28.08.19 D 4

Maximum frequency of signal y(t), fm = 750 Hz

Sampling rate, fs = 2fm = 2 × 750 = 1500 Hz

 

NOTE:

\(\frac{\sin xt}{\pi t}\rm \overset{F.T.}{\leftrightarrow}\left\{\begin{matrix}1;&|\omega|x\end{matrix}\right.\)

Get Free Access Now
Hot Links: teen patti master 51 bonus teen patti gold new version teen patti star apk