Signals and Systems MCQ Quiz - Objective Question with Answer for Signals and Systems - Download Free PDF

Concept: Fourier Transform Symmetry Property

If a time-domain function $f(t)$ is:

• Real: It has no imaginary part

• Even: f(t) = f(-t)" id="MathJax-Element-103-Frame" role="presentation" style="position: relative;" tabindex="0"> f(t) = f(-t)" id="MathJax-Element-1-Frame" role="presentation" style="position: relative;" tabindex="0">$f(t) = f(-t)$ $f(t) = f(-t)$ $f(t) = f(-t)$

Then it's  Fourier Transform " id="MathJax-Element-104-Frame" role="presentation" style="position: relative;" tabindex="0">" id="MathJax-Element-2-Frame" role="presentation" style="position: relative;" tabindex="0">  F(ω) will be:

• Purely real

• Even: F(ω)=F(−ω)" id="MathJax-Element-105-Frame" role="presentation" style="position: relative;" tabindex="0">" id="MathJax-Element-3-Frame" role="presentation" style="position: relative;" tabindex="0">

🔍 Example: Cosine Wave

Let’s take: f(t) = cos(ω0​t)

• It is real.

• It is even, because " id="MathJax-Element-107-Frame" role="presentation" style="position: relative;" tabindex="0">" id="MathJax-Element-4-Frame" role="presentation" style="position: relative;" tabindex="0">  cos(−ω_0​t)=cos(ω_0​t)

Fourier Transform of " id="MathJax-Element-108-Frame" role="presentation" style="position: relative;" tabindex="0">" id="MathJax-Element-5-Frame" role="presentation" style="position: relative;" tabindex="0"> cos(ω_0​t) is: F(ω)=π [δ(ω−ω_0​)+δ(ω+ω_0​)]

This result is:

• Purely real (involves delta functions, no imaginary component)

• Even since it's symmetric around " id="MathJax-Element-109-Frame" role="presentation" style="position: relative;" tabindex="0">" id="MathJax-Element-6-Frame" role="presentation" style="position: relative;" tabindex="0"> ω=0

Concept:

Sinc pulse shaping is commonly used in digital communication systems to achieve ideal Nyquist pulse shaping, minimizing intersymbol interference (ISI).

The sinc function is defined as: \( \text{sinc}(t) = \frac{\sin(\pi t)}{\pi t} \)

Explanation:

The sinc function arises as the inverse Fourier Transform of a rectangular function in the frequency domain.

In other words, if the frequency spectrum is a perfect rectangular shape (ideal low-pass filter), its time-domain representation is a sinc function.

Conclusion:

Correct Answer: Option 4) rectangular

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

Z Transform:

The z transform for a discrete signal x[n] is given by:
           X[z] = 

The set of all values of z where X(z) converges to a finite value is called as Radius of Convergence (ROC).
The ROC does not contain any poles.
If x[n] is a finite duration causal sequence or right-sided sequence, then the ROC is entire z-plane except at z = 0.
If x[n] is a finite duration anti-causal sequence or left-sided sequence, then the ROC is the entire z-plane except at z =∞ .

Time Invariance in Systems

Definition: A system is considered time-invariant if its behavior and characteristics do not change over time. In other words, if an input signal is delayed by a certain amount of time, the output signal will be equally delayed by the same amount. Mathematically, a system is time-invariant if for any input signal x[n] and any time delay n0, the following condition holds true:

Y[n] = T{x[n]} implies Y[n - n₀] = T{x[n - n₀]}

Here, T represents the system's transformation, Y[n] is the output, and x[n] is the input.

Analysis of the Given Options:

Let's analyze each option to determine if the system is time-invariant:

Option 1: Y[n] = nx[n]

To check for time invariance, we need to examine if a time shift in the input signal results in an equivalent time shift in the output signal:

If the input is x[n], the output is Y[n] = nx[n].

If the input is shifted by n₀, the new input is x[n - n₀], and the output should be:

Y[n - n0] = (n - n₀)x[n - n₀]

Since the output is not simply a time-shifted version of the original output (nx[n]), this system is not time-invariant.

Option 2: Y[n] = x[n] - x[n - 1]

For this option, we again check if a time shift in the input results in an equivalent time shift in the output:

If the input is x[n], the output is Y[n] = x[n] - x[n - 1].

If the input is shifted by n₀, the new input is x[n - n₀], and the output should be:

Y[n - n₀] = x[n - n₀] - x[n - n₀ - 1]

This is indeed the time-shifted version of the original output. Therefore, this system is time-invariant.

Option 3: Y[n] = x[-n]

Here, we check the time invariance by examining the time shift:

If the input is x[n], the output is Y[n] = x[-n].

If the input is shifted by n₀, the new input is x[n - n₀], and the output should be:

Y[n - n₀] = x[-(n - n₀)] = x[-n + n₀]

This is not simply a time-shifted version of the original output. Therefore, this system is not time-invariant.

Option 4: Y[n] = x[n] cos(2πf₀n)

For this option, we check the time invariance by examining the time shift:

If the input is x[n], the output is Y[n] = x[n] cos(2πf₀n).

If the input is shifted by n₀, the new input is x[n - n_0], and the output should be:

Y[n - n₀] = x[n - n₀] cos(2πf₀(n - n₀))

This is not simply a time-shifted version of the original output. Therefore, this system is not time-invariant.

Conclusion:

After analyzing all the options, it is clear that:

Option 2: Y[n] = x[n] - x[n - 1] is the correct answer as it satisfies the condition for time invariance.
 

Definition:

Z transform is defined as

\(X\left( z \right) = \mathop \sum \limits_{ - \infty }^\infty x\left[ n \right]{z^{ - n}}\)

Properties:

Differentiation in z domain:

If X(z) is a z transform of x(n), then the z transform of n x(n) is,

\(nx\left( n \right) \leftrightarrow - z\frac{d}{{dz}}\left( {X\left( z \right)} \right)\)

Time-shifting:

If X(z) is a z transform of x(n), then the z transform of x(n – n₀) is,

\(x\left( {n - {n_0}} \right) \leftrightarrow {z^{ - {n_0}}}X\left( z \right)\)

Application:

Let x(n) = 1

\(X\left( z \right) = \mathop \sum \limits_0^\infty {z^{ - n}} = \frac{1}{{1 - {z^{ - 1}}}} = \frac{z}{{z - 1}}\)

Now, by applying the property of differentiation in the z domain,

\(x\left( n \right) = n \leftrightarrow - z\frac{d}{{dz}}\left( {\frac{z}{{z - 1}}} \right) = \frac{z}{{{{\left( {z - 1} \right)}^2}}}\)

Now, by applying the property of differentiation in the z domain,

\(x\left( n \right) = {n^2} \leftrightarrow - z\frac{d}{{dz}}\left( {\frac{z}{{{{\left( {z - 1} \right)}^2}}}} \right) = \frac{{z\left( {z + 1} \right)}}{{{{\left( {z - 1} \right)}^3}}}\)

Now, by applying the property of time-shifting,

\(x\left( n \right) = {\left( {n + 1} \right)^2} \leftrightarrow z.\frac{{z\left( {z + 1} \right)}}{{{{\left( {z - 1} \right)}^3}}} = \frac{{{z^2}\left( {z + 1} \right)}}{{{{\left( {z - 1} \right)}^3}}}\)

Concept:

Shifting property of impulse function

\(\rm\displaystyle\int_{-\infty}^{\infty} x(t) \delta (t -a ) dt = x(a)\)

Scaling property of impulse function

\(\delta (at) = \frac{1}{|a|)} \delta (t)\)

Explanation:

Let:

\(\rm I =\displaystyle\int_{-\infty}^{\infty}e^{-t} \delta(2t - 2) dt\)

\( I = \rm\displaystyle\int_{-\infty}^{\infty}e^{-t} \delta[2(t - 1)] dt\)

Using scaling property of impulse function in the above equation, we'll get:

\( I =\rm\displaystyle\int_{-\infty}^{\infty}e^{-t} \frac{1}{|2|}\delta(t - 1) dt\)

Applying Shifting property of impulse function to the above equation, we'll get:

\( I =\frac{1}{2}\rm\displaystyle\int_{-\infty}^{\infty}e^{-t} \delta(t - 1) dt\)

\(I = \frac{1}{2}. \left. e^{-t} \right|_{t = 1}\)

\(\frac{1}{{2e}}\)

Concept:

Minimum sampling rate to avoid aliasing:

f_s = 2f_m = (Nyquist rate)

Calculation:

Given that, ω_m = 400 π

f_m = 200 Hz = maximum frequency of signal

Sampling frequency f_s = 2 × 200 = 400 Hz

Concept:

Bilateral Laplace transform:

\(L\left[ {x\left( t \right)} \right] = x\left( s \right) = \;\mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - st}}dt\)

Unilateral Laplace transform:

\(L\left[ {x\left( t \right)} \right] = x\left( s \right) = \;\mathop \smallint \limits_0^\infty x\left( t \right){e^{ - st}}dt\)

Some important Laplace transforms:

Calculation:

\(\sin \omega t. u(t)\leftrightarrow \frac{\omega }{{{s^2} + {\omega ^2}}}\)

By applying frequency differentiation property,

\({e^{ - at}}\sin \omega t. u(t) \leftrightarrow \frac{\omega }{{{{\left( {s + a} \right)}^2} + {\omega ^2}}}\)

Modulation:

• The process in which the characteristics of carrier signal is varied in accordance with baseband message signal to make bandpass signal.

Demultiplexing :

• The process or technique of transmitting multiple analog or digital input signals or data streams over a single channel is called multiplexing.
• The reverse of the multiplexing process. Demultiplexing is a process reconverting a signal containing multiple analog or digital signal streams back into the original signals.

Sampling :

• The process of conversion of continuous-time signals into discrete time signals.

Quantization :

• The process of mapping of continuous-time signals into discrete time signal.
• Hence correct option is "3"

Concept:

Final value theorem:

A final value theorem allows the time domain behavior to be directly calculated by taking a limit of a frequency domain expression

Final value theorem states that the final value of a system can be calculated by

\(x\left( \infty \right) = \mathop {\lim }\limits_{s \to 0} sX\left( s \right)\)

 Where X(s) is the Laplace transform of the function.

For the final value theorem to be applicable system should be stable in steady-state and for that real part of poles should lie in the left side of s plane.

Initial value theorem:

\(x\left( 0 \right) = \mathop {\lim }\limits_{t \to 0} x\left( t \right) = \mathop {\lim }\limits_{s \to \infty } sX\left( s \right)\)

It is applicable only when the number of poles of X(s) is more than the number of zeros of X(s).

Calculation:

Given that, \(X\left( s \right) = \frac{{4s + 1}}{{{s^2} + 6s + 3}}\)

Initial value,

 \(x\left( 0 \right) = \mathop {\lim }\limits_{s \to \infty } s\frac{{4s + 1}}{{{s^2} + 6s + 3}}\\=\mathop {\lim }\limits_{\frac{1}{s} \to 0 } \frac{{4 + \frac{1}{s}}}{{{1} + \frac{6}{s} + \frac{3}{s^2}}} = 4\)

Given, the signal

V (t) = 30 sin 100t + 10 cos 300 t + 6 sin (500t+π/4)

So, we have

ω₁ = 100 rads

ω₂ = 300 rads

ω₃ = 500 rads

∴ The respective time periods are

\(\begin{array}{l} {T_1} = \frac{{2\pi }}{{{\omega _1}}} = \frac{{2\pi }}{{100}}sec\\ {T_1} = \frac{{2\pi }}{{{\omega _2}}} = \frac{{2\pi }}{{300}}sec\\ {T_3} = \frac{{2\pi }}{{500}}sec \end{array}\)

So, the fundamental time period of the signal is

\(LCM\left( {{T_1},{T_2},{T_3}} \right) = \frac{{LCM\left( {2\pi ,2\pi ,2\pi } \right)}}{{HCF\left( {100,\ 300,\ 500} \right)}}\)

as \({T_0} = \frac{{2\pi }}{{100}}\)

∴ The fundamental frequency, \({\omega _0} = \frac{{2\pi }}{{{T_0}}} = 100\ rad/s\)

Concept:

Final value theorem:

It states that:

\(x\left( \infty \right) = \mathop {\lim }\limits_{z \to 1} \left( {1 - {z^1}} \right)X\left( z \right)\)

Conditions:

1. It is valid only for causal systems. 

2. Pole of (1 – z-1) X(z) must lie inside the unit circle.

Calculation:

The final value theorem for z-transform is:

\( = \mathop {\lim }\limits_{z \to 1} \frac{1}{4}\frac{{\left( {1 - {z^{ - 1}}} \right){z^{ - 1}}\left( {1 - {z^{ - 4}}} \right)}}{{{{\left( {1 - {z^{ - 1}}} \right)}^2}}}\)

\(= \mathop {\lim }\limits_{z \to 1} \frac{1}{4}\;\frac{{\left( {{z^2} - 1} \right)\left( {{z^2} + 1} \right)}}{{{z^4}\left( {z - 1} \right)}}\)

\( = \mathop {\lim }\limits_{z \to 1} \frac{1}{4}\;\frac{{\left( {z + 1} \right)\left( {{z^2} + 1} \right)}}{{{z^4}}} \)

= 1/4 × 1 × 2 × 2 = 1

Concept:

The Fourier Transform of a continuous-time signal x(t) is given as:

\(X\left( \omega \right) = \mathop \smallint \limits_{ - \infty}^{\infty} x(t) ~{e^{ - j\omega t}}~dt \)

Analysis:

Given:

x(t) = e^j10t  defined from t = -1 to 1. 

\( X\left( \omega \right) = \mathop \smallint \limits_{ - 1}^1 {e^{j10t}}.{e^{ - j\omega t}}dt = \mathop \smallint \limits_{ - 1}^1 {e^{j\left( {10 - \omega } \right)t}}dt\)

\(X(\omega) = \left. {\frac{{{e^{j\left( {10 - \omega } \right)t}}}}{{j\left( {10 - \omega } \right)}}} \right|_{ - 1}^1 = \frac{{2\sin \left( {\omega - 10} \right)}}{{\left( {\omega - 10} \right)}} \)

Given:

f_m = 5 kHz, 

ω_c = 2πfc = 2000π 

f_c = 1 kHz

Maximum frequency in x(t) is given as:

f_c + f_m = 6 kHz