Sampling MCQ Quiz - Objective Question with Answer for Sampling - Download Free 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 ≥ 2fmax

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.

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.

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 f_N = 2 f_m

The sampling frequency f_s ≥ f_N

Calculation:

f_m =  7 kHz 

fN = 2 fm

= 14  kHz 

fs ≥ fN

Option 1 is invalid as f_s is less than f_N

The correct answer is Snowball sampling.

Key Points Snowball Sampling:

• Snowball sampling is a non-probability sampling method where new units are recruited by other units to form part of the sample.
• Snowball sampling can be a useful way to conduct research about people with specific traits who might otherwise be difficult to identify (e.g., people with a rare disease).
• Also known as chain sampling or network sampling, snowball sampling begins with one or more study participants.
• It then continues on the basis of referrals from those participants. This process continues until you reach the desired sample or a saturation point.
• Here, in this example, the survey first tries to sort out a name of the friends that would likely purchase their product. then they are contacted. Hence, it is an example of Snowball sampling.

Additional Information Judgemental Sampling- Judgment sampling, also referred to as judgmental sampling or authoritative sampling, is a non-probability sampling technique where the researcher selects units to be sampled based on his own existing knowledge, or his professional judgment

Expert Sampling- Expert sampling is used when the research requires individuals with a high level of knowledge about a particular subject. The experts are thus selected based on a demonstrable skill set, or level of experience possessed.

Quota Sampling- Quota sampling is a non-probability sampling method that relies on the non-random selection of a predetermined number or proportion of units. This is called a quota sampling.

Hence, the correct answer is Snowball sampling.

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

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.

Calculation:

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

fs = 2fm = 2 × 100

f_s = 200 Hz

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.

Application:

Given f_s = 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

• 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:

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 f_m of the modulating signal."​, i.e.

The minimum sampling frequency is:

f_s = 2 f_m

In terms of the time period:

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

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

T_s = Nyquist sampling interval

Minimum sampling rate to avoid aliasing f_s = 2f_m= (Nyquist rate)

ω_m = max {100π, 200π, 300π} = 300π

f_m = 150 Hz = maximum frequency of signal

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

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:

f_m = 10 kHz

f_s = 20 kHz

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

Ts = 50 μs

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

Sampling Theorem:

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

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 = 2πf_m1

f_m1 = 3 Hz

8π = 2πfm2

fm2 = 4 Hz

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

f_s = 8 Hz

fs = 8 samples per second

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.

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.