Sampling Theorem MCQ Quiz - Objective Question with Answer for Sampling Theorem - Download Free PDF
Last updated on Jun 26, 2025
Latest Sampling Theorem MCQ Objective Questions
Sampling Theorem Question 1:
A signal Vm sin 100πt + 2Vm sin 200πt must be sampled and stored in a data acquisition system. To extract the signal effectively, the original sampling frequency should be:
X 1.
Answer (Detailed Solution Below)
Sampling Theorem Question 1 Detailed Solution
Explanation:
Sampling Frequency Analysis
Definition: Sampling frequency, also known as the sampling rate, is the number of samples per second taken from a continuous signal to convert it into a discrete signal. According to the Nyquist theorem, the sampling frequency must be at least twice the highest frequency present in the signal to accurately reconstruct the original signal without aliasing.
Given Signal:
The signal provided is: Vm × sin(100πt) + 2Vm × sin(200πt)
Here, the frequencies of the components are:
- First Component: sin(100πt), where f = 50 Hz (since ω = 2πf and ω = 100π).
- Second Component: sin(200πt), where f = 100 Hz (since ω = 2πf and ω = 200π).
Highest Frequency in the Signal:
The highest frequency present in the signal is 100 Hz (from the second component).
Nyquist Criterion:
To sample and reconstruct the signal effectively without aliasing, the sampling frequency must be at least twice the highest frequency in the signal. Mathematically:
Sampling Frequency (fs) ≥ 2 × Highest Frequency (fmax)
Substituting the values:
fs ≥ 2 × 100 Hz
fs ≥ 200 Hz
Correct Option:
The correct sampling frequency is 200 Hz, as per the Nyquist criterion. This ensures that the signal can be sampled and reconstructed accurately without any loss of information or occurrence of aliasing.
Additional Information
To analyze the given options:
- Option 1 (50 Hz): This sampling frequency is insufficient as it is less than the highest frequency in the signal (100 Hz). Sampling at 50 Hz would lead to aliasing, making it impossible to reconstruct the original signal accurately.
- Option 2 (200 Hz): This is the correct choice as it satisfies the Nyquist criterion by being at least twice the highest frequency in the signal (100 Hz). Sampling at this frequency ensures accurate signal extraction and reconstruction.
- Option 3 (100 Hz): Although 100 Hz matches the highest frequency in the signal, it does not satisfy the Nyquist criterion, which requires the sampling frequency to be at least twice the highest frequency. Sampling at 100 Hz would result in aliasing and loss of information.
- Option 4 (150 Hz): This sampling frequency is greater than the highest frequency (100 Hz) but still does not meet the Nyquist criterion. Sampling at 150 Hz could lead to partial aliasing and inaccuracies in signal reconstruction.
Conclusion:
To avoid aliasing and ensure accurate signal extraction, the sampling frequency must adhere to the Nyquist criterion. For the given signal Vm × sin(100πt) + 2Vm × sin(200πt), the correct sampling frequency is 200 Hz, as it is twice the highest frequency (100 Hz). Sampling at this rate guarantees effective signal processing without loss of information or distortion.
```Sampling Theorem Question 2:
For a band limited signal with max frequency fmax the Nyquist rate of signal is _____
Answer (Detailed Solution Below)
Sampling Theorem 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 Theorem Question 3:
According to the Nyquist theorem, to produce the original analog signal, one necessary condition is :
Answer (Detailed Solution Below)
Sampling Theorem 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 Theorem Question 4:
Answer (Detailed Solution Below) 0.7
Sampling Theorem Question 4 Detailed Solution
Concept:
For raised cosine pulses
Minimum bandwidth for M-ary PSK (B) =
Where,
α → Roll of factor
Rb → Bit rate
Rb = nfs
n → number of bits used for a symbol
fs → Sampling frequency
Calculation:
Given;
fmax = 40 kHz
Number of levels : L = 256
Sampling frequency;
fs = 1.5 (2× fm) [∵ sampled at 50% higher than the Nyquist rate ]
=120 kHz
L = 2n
⇒256 = 2n
∴ n = 8
Bit rate ;
Rb = nfs = 8 × 120 = 960 kHz
The channel has 408 kHz bandwidth available for transmission
Then
Sampling Theorem Question 5:
Which of the following is TRUE when the sampling time becomes less than the Nyquist interval?
Answer (Detailed Solution Below)
Sampling Theorem Question 5 Detailed Solution
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:
Ts = Nyquist sampling interval
Top Sampling Theorem MCQ Objective Questions
To satisfy the sampling theorem, a 100 Hz sine wave should be sampled at
Answer (Detailed Solution Below)
Sampling Theorem Question 6 Detailed Solution
Download Solution PDFNyquist 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.
fs ≥ 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
When aliasing takes place
Answer (Detailed Solution Below)
Sampling Theorem Question 7 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
fs = Sampling frequency and fm = Modulating frequency
Aliasing is explained with the help of the spectrum as shown:
A system has a sampling rate of 50,000 samples per second. The maximum frequency of the signal it can acquire to reconstruct is
Answer (Detailed Solution Below)
Sampling Theorem Question 8 Detailed Solution
Download Solution PDFNyquist 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.
fs ≥ 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:
∴ The maximum frequency of the signal it can acquire to reconstruct is 25 kHz
In sampling theorem the Nyquist interval is given by
Answer (Detailed Solution Below)
Sampling Theorem Question 9 Detailed Solution
Download Solution PDFNyquist 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:
Ts = Nyquist sampling interval
A signal of maximum frequency of 10 kHz is sampled at Nyquist rate. The time interval between two successive samples is
Answer (Detailed Solution Below)
Sampling Theorem Question 10 Detailed Solution
Download Solution PDFConcept:
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:
Calculation:
fm = 10 kHz
fs = 20 kHz
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:
Answer (Detailed Solution Below)
Sampling Theorem Question 11 Detailed Solution
Download Solution PDFConcept:
Sampling frequency of bandpass signal is given by:
Where
Analysis:
fH = 30 kHz
fL = 20 kHz
fH - fL = 10 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)
Answer (Detailed Solution Below)
Sampling Theorem Question 12 Detailed Solution
Download Solution PDFSampling 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:
Calculation:
Given, x(t) = 10 cos(6πt) + 4 sin(8πt)
6π = 2πfm1
fm1 = 3 Hz
8π = 2πfm2
fm2 = 4 Hz
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
Answer (Detailed Solution Below)
Sampling Theorem Question 13 Detailed Solution
Download Solution PDFConcept:
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.
fs ≥ 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.
Let, ωs > 2ωm
Let ωs = 3ωm
The spectrum of the sampled signal will be:
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.
The phenomenon of some of the higher frequencies in the spectrum of the signal g(t) appearing as lower frequencies in the spectrum of its sampled version is called
Answer (Detailed Solution Below)
Sampling Theorem Question 14 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."
- 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
fs = Sampling frequency and fm = Modulating frequency
Aliasing is explained with the help of the spectrum as shown:
Consider a real-value based-band signal x(t), band limited to 10 kHz. The Nyquist rate for the signal y(t) = x(t)
Answer (Detailed Solution Below)
Sampling Theorem Question 15 Detailed Solution
Download Solution PDFConcept:
Consider x(t) → f(kHz) signal
Then:
Since multiplication in one domain is the convolution in another domain, and in convolution, due to the time-invariant system, the duration of the signal gets added.
Hence the frequency of both signals gets added in the frequency domain.
More clearly, we can understand from the following specturem diagram:
Application:
Given x(t) = 10 kHz signal
Then,
∴ y(t) will be (10 + 5) = 15 kHz signal
Nyquist rate of y(t) = 2 [frequency of y(t)]
= 2 × 15 kHz
= 30 kHz