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

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| ≤ f_c

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

Where:

• H(f): Frequency response of the filter
• f: Frequency variable
• f_c: 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) × e^j2πft df

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

h(t) = ∫_-fc^f_c e^j2πft df

Solving this integral gives:

h(t) = 2f_c × sinc(2f_ct)

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 f_c.

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

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

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.