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

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

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.

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

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:

Given that,

f_m = 3.1 kHz 

⇒ fs ≥ 2f_m

⇒ f_s ≥ 2 × 3.1 = 6.4 kHz

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:

The maximum frequency of the band-limited signal.

f_m = 5 kHz

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 = 2 × 5 = 10 kHz

So the sampling frequency f_s ≥ f_N

f_s ≥ 10 kHz

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

∴ 5 kHz is not a valid sampling frequency.

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">Ts<12Tm

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

Nyquist criteria says that

Sampling frequency, f_s > 2f _max

f_max  →  maximum frequency of message signal.

Maximum frequency is given signal, ω_max = 14 π

⇒  f_max = 7

∴ sampling frequency, f_s > 2 × 7

f_s > 14

Given,

x(t) = cos(6πt) + sin(8πt)

The bandwidth of the signal will be:

$\frac{8\pi }{2\pi }=4\mathrm{H}\mathrm{z}$

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

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

Taking the Fourier transform, we get:

$\mathrm{Y}\left(\mathrm{j}\omega \right)=\frac{1}{\left|2\right|}\mathrm{X}\left(\frac{\mathrm{j}\omega }{2}\right){\mathrm{e}}^{\frac{\mathrm{j}5\omega }{2}}$

$\mathrm{Y}\left(\mathrm{j}\omega \right)=\frac{1}{2}\mathrm{X}\left(\frac{\mathrm{j}\omega }{2}\right){\mathrm{e}}^{\frac{\mathrm{j}5\omega }{2}}$

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:

f_s = 2 × 8 = 16 samples/sec

Concept:

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

                         $E=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\left|x\left(n\right)\right|}^2$

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

                         $E=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\left(x\left(t\right)\right)}^{2}dt$

Explanation:

$x_1\left(t\right)=\{\begin{array}{cc}\left|t\right|& -1\le t\le 1\\ 0& otherwise\end{array}$

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

Energy of $x_1\left(n\right)=E_1=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\left|x_1\left(n\right)\right|}^2$

= 0² + 2 [0.25² + 0.5² + 0.75² + 1²]

= 3.75

$x_2\left(t\right)=\{\begin{array}{cc}1-\left|t\right|& -1\le t\le 1\\ 0& otherwise\end{array}$

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

Energy of $x_2\left(n\right)=E_2=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\left|x_2\left(n\right)\right|}^2$

= 1² + 2 [0.75² + 0.5² + 0.25²]

= 2.75

⇒ E₁ > E₂

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.

$f_s=2f_m$

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

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

Sampling frequency,

f_s = 2f_m = 2 × 350 = 700 Hz

Sampling time period = 1/700 sec

Concept:

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

ω_s ≥ 2 ω_s sample per sec.

In other word minimum sample rate is f_s = 2f_m Hz

Calculation:

Minimum sampling rate f_s should be

8 ≤ f_s - 5

_fs­ ≥ 8 + 5 = 13 kHz

Period of sampling train, T_s = 20 ms

$\therefore f_s=\frac{1}{20\times {10}^{-3}}=50\mathrm{H}\mathrm{z}$

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

f_s - f_x = 50 - f_x and f_s + f_x = 50 + f_s

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

f_c = 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.

f_s - f_x < 25

and f_s - f_x = 20

50 - f_x­ = 20

f_x = 50 - 20 = 30 Hz

Calculation:

f_s = Sampling Frequency = 56 MHz

The frequencies present at the input are:

f_m1 = 10 MHz, f_m2 = 50 MHz and, f_m3 = 70 MHz

Once sampled, the frequency content at the output will be: f_m ± f_s

For, f_m1 = 10 MHz, the output will contain frequencies of,

= f_m1, f_m1 ±, f_m1 ± 2f_s ….

= 10 MHz, 66 MHz, 122 MHz

For, f_m2 = 50 MHz, the output will contain frequencies of

= f_m2, f_m2 ± f_s, f_m2 ± 2f_s ….

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

For, f_m3 = 70 MHz, the output will contain frequencies of

= f_m3, f_m3 ± f_s, f_m3 ± 2f_s ….

= 70 MHz, 14 MHz, 182 MHz ….

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)

Maximum frequency components of

x1(ω) = B, rad/sec

Fourier transform of x2(t) = x2(ω)

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

∴ x(t) = x1(t). x2(t)

Using multiplication property in Fourier transform,

$F\{x(t)\}=x(\omega )=\frac{1}{2\pi }\{x_1(\omega )\ast x_2(\omega )\}$

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,

$H(\omega )=\frac{4}{4+{\omega }^2}$

Maximum frequency component of H(ω) = ∞

∴ y(t) = x(t) * h(t)

Fourier Transform, Y(ω) = x(ω) H(ω)

$=\frac{1}{2\pi }\{x_1(\omega )\ast x_2(\omega )\}.H(\omega )$

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

Concept:

Duality property,

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

$A\tau \sin C\left(\frac{t\tau }{2\pi }\right)\stackrel{\mathrm{F}.\mathrm{T}.}{\leftrightarrow }2\pi \mathrm{A}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\left(\frac{\omega }{\tau }\right)$

Explanation:

z(t) = x(t) cos (1000 πt)

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

Spectrum of impulse response is,

$\frac{\sin \mathrm{a}\mathrm{t}}{\pi \mathrm{t}}\leftrightarrow \{\begin{array}{cc}1,& \left|\omega \right|<\mathrm{a}\\ 0,& \left|\omega \right|>\mathrm{a}\end{array}$

Spectrum of output is,

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

Sampling rate, f_s = 2f_m = 2 × 750 = 1500 Hz

NOTE:

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