Sampling Theorem and Nyquist Rate MCQ Quiz in தமிழ் - Objective Question with Answer for Sampling Theorem and Nyquist Rate - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

Convolution in time domain is multiplication in frequency domain

y(t) = x₁(t) × x₂(t)

Y(ω) = X₁(ω) X₂(ω)

Calculations:

X₁(ω) = 0 for |ω| > 1000 π

X₂(ω) = 0 for |ω| > 2000 π

Output Y(ω) = X₁(ω) X₂(ω)

Maximum frequency of Y(ω) = 1000 π radians

Nyquist rate = 2 × 1000 π = 2000 π

Sampling period T

\(T = \frac{{2\pi }}{{2000\;\pi }} = {10^{ - 3}}\;sec\)

Sampling period T = 1 m sec

Y(jω) = (jω)X (jω)

The message y(t) has same maximum frequency as x(t) with a multiplication factor of jω in the amplitude. 

The bandwidth the signal y(t) will be same as x(t) hence the Nyquist rate is same i.e ω₀

Concept:

Final value theorem,

\(Y(∞) = \displaystyle\lim _{Z \rightarrow 1} ( 1 - Z^{-1}) Y(Z)\)

Calculation:

We have:

\(\frac{Y(z)}{X(z)} = \frac{1}{N} \left( \frac{1 - Z^{-N}}{1 - Z^{-1}} \right)\)....(1)

and x(t) = 2 + 5sin(100πt)

sampling frequency, f_s = 400 Hz

put t = nT_s, the output of the sampling process is,

x(nT_s) = 2 + sin(100πnT_s)

x(nT_s) = 2 + 5 sin\(\left( 100 π n \times \frac{1}{400} \right)\)

x(nT_s) = 2 + 5\(\sin \left(\dfrac{n π}{4} \right)\)

where,

ω₀ = π/4

\(N = \frac{2\pi}{\omega_0} = \frac{2\pi}{\frac{\pi}{4}} =8\)

The z-transform of x(n) is,

\(\rm ZT[x(n)] = ZT \left[ 2 + 5 \sin \left( \frac{\pi n}{4} \right) \right]\)

\(\rm X(Z) = 2 + \frac{5Z \sin \left( \frac{\pi} {4} \right)}{Z^2 - 2Z \cos \left( \frac{\pi}{4} \right) + 1}\)

\(\rm X(Z) = 2 + \frac{\frac{5Z}{\sqrt 2}}{Z^2 - \frac{2z}{\sqrt Z } + 1}\)

\(X(Z) = 2+ \frac{2.5 \sqrt 2 Z}{Z^2 - \sqrt 2 Z + 1}\)

From equation (i)

\(\frac{Y(z)}{X(z)} = \frac{1}{N} \left( \frac{1 - Z^{-N}}{1 - Z^{-1}} \right)\)

\(\rm Y(Z) = \frac{1}{8} \left( \frac{1 - Z^{-8}}{1 - Z^{-1}} \right)X(Z)\)

\(\rm Y(Z) = \frac{1}{8} \left( \frac{1 - Z^{-8}}{1 - Z^{-1}} \right) \left[ 2+ \frac{2.5 \sqrt 2 Z}{Z^2 - \sqrt 2 Z + 1} \right]\)

using final value theorem,

\(Y(∞) = \displaystyle\lim _{Z \rightarrow 1} ( 1 - Z^{-1}) Y(Z)\)

\(Y(∞) = \displaystyle\lim _{Z \rightarrow 1} ( 1 - Z^{-1}) \frac{1}{8} \left( \frac{1 - Z^{-N}}{1 - Z^{-1}} \right) \left[ 2+ \frac{2.5 \sqrt 2 Z}{Z^2 - \sqrt 2 Z + 1} \right]\)

\(Y(∞) = \displaystyle\lim _{Z \rightarrow 1} \frac{1}{8}( 1 - Z^{-8}) \left[ 2+ \frac{2.5 \sqrt 2 Z}{Z^2 - \sqrt 2 Z + 1} \right]\)

y(∞) = 0

x(t) = cos (500 πt) * cos (800 πt)

\(\cos {\omega _0}t\mathop \to \limits^{FT} \pi \left[ {\delta \left( {\omega - {\omega _0}} \right) + \delta \left( {\omega + {\omega _0}} \right)} \right]\)

\(\Rightarrow \cos \left( {500\pi t} \right) \to \pi \left[ {\delta \left( {\omega - 500\pi } \right) + \delta \left( {\omega + 500\pi } \right)} \right]\)

\(\Rightarrow x\left( t \right) = \pi \left[ {\left\{ {\delta \left( {\omega - 500\pi } \right) + \delta \left( {\omega + 500\pi } \right)} \right\} \cdot \left\{ {\delta \left( {\omega - 800\pi + \delta \left( {\omega + 800\pi } \right)} \right)} \right\}} \right]\)

Multiplication exist between the intersection range so

So ω_m = 500 π rad / sec

f_m = 250 Hz

Nyquist rate = 2f_m

= 500 Hz

We know that,

Nyquist sampling freq. of \(\rm sinc^2(50t) \ is \ 50 \times 2 = 100Hz\)

Nyquist sampling freq. of \(\rm sinc^3(100t)\ is\ 100 \times 3 = 300Hz\)

Hence, Nyquist sampling freq of \(\rm x(t)\) is \(\rm 300\ Hz\)_.

Given that x(t) is sampled at twice of Nyquist sampling freq. hence sampling freq. of \(\rm x(t) \ is\ 300 \times 2 = 600 Hz\)

Hence sampling interval \(\rm = {1 \over {sampling\ frequency}} = {1 \over {600}} = 1.67\ msec.\)

To avoid aliasing sampling frequency \(f_s\) should be greater that twice the input signal frequency. The sampling frequency \(\omega_s=\frac{2\pi}{T}\). We have time period of sampling signal \(T=10^{-3} sec\). Thus, according to sampling theorem \(\rm{\omega_c>2\omega_m}\). 

\(\rm{\Rightarrow \frac{2\pi}{10^{-3}}>2\omega_m}\)

\(\rm{\Rightarrow 2\omega_m< 2000\pi \Rightarrow \omega_m<1000\pi}\)

Thus the sampled signal frequency spectrum should be such that its spectrum does not exists for \(\rm{\omega>1000\pi}\)

Concept:

Aliased frequency component is given by:

f_a = |f_s – f_m|

f_s = sampling frequency

f_m = message signal frequency

Calculation:

For a sampling rate of 140 Hz, since f_s > f_m:

Aliased Frequency = f_s - f_m,               

= 140 – 100 = 40 Hz

For a sampling rate of 90 Hz, since f_s < f_m:

Aliased Frequency = f_m – f_s,       

= 100 – 90 = 10 Hz

For a sampling rate of 30 Hz, since f_s < f_m:

Aliased Frequency = f_m – f_s    

= 100 – 30 = 70 Hz

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

Concept:

Discrete signal is periodic if

\(\frac{N}{K}=\frac{2\pi }{{{\omega }_{0}}}\) is integer

Application

\(\cos 15~t\underset{Sampling}{\mathop{\to }}\,\cos \left( 15\dot{n}{{T}_{s}} \right)\)

Period:

\(\frac{N}{K}=\frac{2\pi }{15\left( {{T}_{s}} \right)}\)

\({{T}_{s}}=\frac{2\pi }{15}\times \frac{K}{N}\)

\({{T}_{s}}=\frac{2\pi }{15}\times \frac{K}{4}\)

\({{T}_{s}}=\frac{\pi }{30}K\)

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

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

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

So, Nyquist rate = 2ω_max

= 2[16π × 10³]

= 32π × 10³ rad/sec