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

Last updated on Jun 10, 2025

Latest Signals and Systems MCQ Objective Questions

Signals and Systems Question 1:

The Fourier Transform of an real and even function results in:

  1. a purely real and odd function
  2. a purely imaginary and even function
  3. an imaginary and odd function
  4. a purely real and even function

Answer (Detailed Solution Below)

Option 4 : a purely real and even function

Signals and Systems Question 1 Detailed Solution

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)

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(ω0t)

  • 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(ω0t)=cos(ω0t)

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(ω0t) is: F(ω)=π [δ(ωω0)+δ(ω+ω0)]

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

Signals and Systems Question 2:

Sinc pulse shaping is derived from the Fourier Transform of a _______ function.

  1. sine
  2. sawtooth
  3. triangular
  4. rectangular

Answer (Detailed Solution Below)

Option 4 : rectangular

Signals and Systems Question 2 Detailed Solution

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:

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

Signals and Systems Question 3:

What type of distortion occurs if a signal that contains frequency components up to 15 kHz is sampled using 20 kHz?

  1. Quantization error
  2. Aliasing
  3. Slope Overload
  4. No distortion

Answer (Detailed Solution Below)

Option 2 : Aliasing

Signals and Systems Question 3 Detailed Solution

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

Signals and Systems Question 4:

In order to recover the original signal from the Sampled one, what is the condition to be satisfied for sampling frequency ωs and highest frequency component ωm

  1. ωm < ωs ≤ 2 ωm
  2. ωs ≥ 2 ωm
  3. ωs < ωm
  4. ωs = ωm

Answer (Detailed Solution Below)

Option 1 : ωm < ωs ≤ 2 ωm

Signals and Systems Question 4 Detailed Solution

Explanation:

Condition for Sampling Theorem

Definition: The sampling theorem, also known as the Nyquist-Shannon sampling theorem, is a fundamental principle in signal processing. It states that a continuous-time signal can be completely represented and reconstructed from its samples if the sampling frequency satisfies a specific condition in relation to the highest frequency component of the signal.

Statement: To accurately recover the original signal from its sampled version, the sampling frequency (ωs) must be at least twice the highest frequency component (ωm) of the signal. Mathematically, this condition is expressed as:

ωs ≥ 2ωm

This condition ensures that no aliasing occurs during the sampling process, allowing the original signal to be reconstructed without distortion.

Correct Option Analysis:

The correct option is:

Option 2: ωs ≥ 2ωm

This option correctly represents the Nyquist criterion. According to the theorem, the sampling frequency must be at least twice the highest frequency component of the signal (ωm) to prevent aliasing and ensure complete reconstruction of the original signal. If this condition is satisfied, the sampled signal contains all the necessary information to recover the original continuous-time signal.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: ωm s ≤ 2ωm

This option is incorrect because it suggests that the sampling frequency can be less than twice the highest frequency component (ωm) and still allow accurate signal reconstruction. However, if ωs is less than 2ωm, aliasing occurs, making it impossible to recover the original signal without distortion. The condition ωs ≤ 2ωm does not guarantee the Nyquist criterion is met.

Option 3: ωs m

This option is incorrect because it violates the Nyquist criterion. If the sampling frequency is less than the highest frequency component of the signal, severe aliasing occurs, and the original signal cannot be reconstructed. This condition leads to significant distortion in the sampled signal.

Option 4: ωs = ωm

This option is also incorrect. If the sampling frequency is equal to the highest frequency component of the signal, the Nyquist criterion is not satisfied. In such a case, aliasing still occurs, and the original signal cannot be accurately reconstructed. The sampling frequency must be at least twice the highest frequency component to meet the Nyquist criterion.

Conclusion:

The correct condition for accurately recovering the original signal from its sampled version is ωs ≥ 2ωm. This ensures that the sampling theorem is satisfied, preventing aliasing and enabling complete reconstruction of the original signal. Other options either violate this condition or fail to guarantee the accurate recovery of the signal, leading to distortion or loss of information.

Signals and Systems Question 5:

The ROC of a system is the

  1. range in which the signal is free of noise
  2. range of frequency for which the z-transform exists
  3. range of frequency for which the signal gets transmitted
  4. range of z for which the z-transform converges
  5. None of the above.

Answer (Detailed Solution Below)

Option 4 : range of z for which the z-transform converges

Signals and Systems Question 5 Detailed Solution

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 =∞ .

Top Signals and Systems MCQ Objective Questions

Answer (Detailed Solution Below)

Option 1 :

Signals and Systems Question 6 Detailed Solution

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Definition:

Z transform is defined as

Properties:

Differentiation in z domain:

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

Time-shifting:

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

Application:

Let x(n) = 1

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

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

Now, by applying the property of time-shifting,

The value of , where  is the Dirac delta function, is

Answer (Detailed Solution Below)

Option 1 :

Signals and Systems Question 7 Detailed Solution

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Concept:

Shifting property of impulse function

Scaling property of impulse function

Explanation:

Let:

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

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

 

Calculate the minimum sampling rate to avoid aliasing when a continuous-time signal is given by x(t) = 5 cos 400πt

  1. 100 Hz
  2. 250 Hz
  3. 400 Hz
  4. 20 Hz

Answer (Detailed Solution Below)

Option 3 : 400 Hz

Signals and Systems Question 8 Detailed Solution

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Concept:

Minimum sampling rate to avoid aliasing:

fs = 2f= (Nyquist rate)

Calculation:

Given that, ωm = 400 π

fm = 200 Hz = maximum frequency of signal

Sampling frequency fs = 2 × 200 = 400 Hz

The Laplace transform of e-at sin ωt u(t) is:

Answer (Detailed Solution Below)

Option 1 :

Signals and Systems Question 9 Detailed Solution

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Concept:

Bilateral Laplace transform:

Unilateral Laplace transform:

Some important Laplace transforms:

 

f(t)

f(s)

ROC

1.

δ(t)

1

Entire s-plane

2.

e-at u(t)

s > - a

3.

e-at u(-t)

s

4.

cos ω0 t u(t)

s > 0

5.

te-at u(t)

s > - a

6.

sin ω0t u(t)

s > 0

7.

u(t)

1/s

s > 0

 

Calculation:

By applying frequency differentiation property,

The process of converting the analog sample into discrete form is called ______.

  1. modulation
  2. demultiplexing
  3. sampling
  4. quatization

Answer (Detailed Solution Below)

Option 3 : sampling

Signals and Systems Question 10 Detailed Solution

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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"

The Laplace transform of a signal x(t) is Determine the initial value x(0).

  1. 1/3
  2. 4
  3. 1/6
  4. 4/3

Answer (Detailed Solution Below)

Option 2 : 4

Signals and Systems Question 11 Detailed Solution

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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

 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:

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, 

Initial value,

 

For a periodic signal v(t) = 30 sin100t + 10 cos300t + 6 sin(500t + π/4), the fundamental frequency in rad/s is _____.

  1. 100
  2. 300
  3. 500
  4. 1500

Answer (Detailed Solution Below)

Option 1 : 100

Signals and Systems Question 12 Detailed Solution

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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

So, the fundamental time period of the signal is

as 

∴ The fundamental frequency, 

The z-transform of a signal is given by , its final value is

  1. 1/4
  2. Zero
  3. 1
  4. Infinity

Answer (Detailed Solution Below)

Option 3 : 1

Signals and Systems Question 13 Detailed Solution

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Concept:

Final value theorem:

It states that:

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:

= 1/4 × 1 × 2 × 2 = 1

Consider a signal defined by

1} \end{array}} \right.\)

Its Fourier Transform is

Answer (Detailed Solution Below)

Option 1 :

Signals and Systems Question 14 Detailed Solution

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Concept:

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

Analysis:

Given:

x(t) = ej10t  defined from t = -1 to 1. 

Suppose the maximum frequency in a band-limited signal  is . Then, the maximum frequency in , in , is ________.

Answer (Detailed Solution Below) 6

Signals and Systems Question 15 Detailed Solution

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Given:

fm = 5 kHz, 

ωc = 2πfc = 2000π 

fc = 1 kHz

Maximum frequency in x(t) is given as:

fc + fm = 6 kHz

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