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All questions of Fourier Transform in Signals & Systems for Electrical Engineering (EE) Exam

F(t) and G(t) are the one-sided z-transforms of discrete time functions f(nt) and g(nt), the z-transform of ∑f(kt)g(nt-kt) is given by _____________
  • a)
    ∑f(nt)g(nt)z-n
  • b)
    ∑f(nt)g(nt)zn
  • c)
    ∑f(kt)g(nt-kt) z-n
  • d)
    ∑f(nt-kt)g(nt)z-n
Correct answer is option 'A'. Can you explain this answer?

Sarita Yadav answered
Given that F (t) and G (t) are the one-sided z-transforms.
Also, f (nt) and g (nt) are discrete time functions, which means that property of Linearity, time shifting and time scaling will be similar to that of continuous Fourier transform. Since, for a continuous Fourier transform, the value of ∑f(kt)g(nt-kt) is given by∑f(nt)g(nt)z-n.
∴ z-transform of ∑f(kt)g(nt-kt) is given by∑f(nt)g(nt)z-n.

A discrete time signal is as given below 

The period of the signal X [n] is _____________
  • a)
    16
  • b)
    4
  • c)
    2
  • d)
    Non-periodic
Correct answer is option 'A'. Can you explain this answer?

Sarita Yadav answered
Given that, N1 = 4, N2 = 16, N3 = 8
We know that period of X [n] (say N) = LCM (N1, N2, N3)
∴ Period of X [n] = LCM (4, 16, 8) = 16.

The time system which operates with a continuous time signal and produces a continuous time output signal is _________
  • a)
    CTF system
  • b)
    DTF System
  • c)
    Time invariant System
  • d)
    Time variant System
Correct answer is option 'A'. Can you explain this answer?

Charvi Reddy answered
Continuous Time System (CTF System)

A continuous time system is a system that operates with a continuous time signal as its input and produces a continuous time output signal. The continuous time signal is a function of time that is defined for all values of time in a given interval. Continuous time systems are commonly used in various fields such as electronics, control systems, and signal processing.

Key Points:

1. Definition of a Continuous Time System:
- A continuous time system is a mathematical representation of a physical system that processes continuous time signals.
- It can be described by a continuous time input-output relationship, where the input and output signals are functions of time.

2. Characteristics of a Continuous Time System:
- Continuous time systems are characterized by their response to continuous time signals.
- They operate on signals that are continuous in both amplitude and time.
- The output of a continuous time system is also a continuous time signal.

3. Examples of Continuous Time Systems:
- Analog filters: Analog filters are commonly used in communication systems to remove unwanted noise or interference from signals. These filters operate on continuous time signals to attenuate or amplify specific frequency components.
- Continuous time control systems: Control systems in industrial applications often operate with continuous time signals. These systems control the behavior of physical processes using continuous time feedback signals.
- Analog audio systems: Analog audio systems, such as amplifiers and equalizers, process continuous time audio signals to amplify, filter, or modify the sound.

4. Advantages of Continuous Time Systems:
- Continuous time systems provide a high level of accuracy and precision in signal processing applications.
- They allow for smooth and continuous variations in signals, resulting in high-quality output.
- Continuous time systems can handle a wide range of frequencies and amplitudes, making them suitable for various applications.

In conclusion, a continuous time system is a system that operates with continuous time signals and produces continuous time output signals. It is widely used in electronics, control systems, and signal processing applications.

Let x(t) and y(t) (with Fourier transform X(ω) and Y(ω) be related as shown in figure below

Then Y(ω) in terms of X(ω) is
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

Ravi Singh answered
From the given pictures of x(t) and y(t)
We get,
y(t) = - x(2t + 2)
y(t) is time scaled and time shifted version of x(t)
Step 1:
If x(t) ↔ X(ω)
Then, x(t + 2) ↔ ej.2.ω X(ω)
Step 2:
Using time shifting property
x(2t + 2) ↔ ½ e X(ω/2)
Step 3:
Using time scaling property:

The given mathematical representation belongs to:
y(t) = x(t - T)
  • a)
    time multiplication
  • b)
    time division
  • c)
    time scaling
  • d)
    time reversal
  • e)
    time shifting
Correct answer is option 'E'. Can you explain this answer?

Ravi Singh answered
Time-shifting property: When a signal is shifted in time domain it is said to be delayed or advanced based on whether the signal is shifted to the right or left.
For example:

Given a discrete time signal x[k] defined by x[k] = 1, for -2 ≤ k ≤ 2 and 0, for |k| > 2. Then, y[k] = x[3k - 2] is ______________
  • a)
    y[k] = 1, for k = 0, 1 and 0 otherwise
  • b)
    y[k] = 1, for k = 1 and -1 for  k= -1
  • c)
    y[k] = 1, for k = 0, 1 and -1 otherwise
  • d)
    None of these
Correct answer is option 'A'. Can you explain this answer?

Based on the given information, the discrete time signal x[k] is defined as follows:

x[k] = 1 for -2 <= k=""><=>
x[k] = 0 otherwise

This means that the signal x[k] has a value of 1 for values of k between -2 and 2 (inclusive), and has a value of 0 for all other values of k.

To visualize this signal, you can plot its values on a graph. The x-axis represents the values of k, and the y-axis represents the values of x[k].

Here is a plot of the signal x[k]:

| 1 |
1 ---| |---
| |
0 ---| 1 |---
| |
-1---| |---
| |
-2---|-------|---
-2 2

In this plot, the signal x[k] has a value of 1 for k = -2, -1, 0, 1, 2, and has a value of 0 for all other values of k.

The system described by the difference equation y(n) – 2y(n-1) + y(n-2) = X(n) – X(n-1) has y(n) = 0 and n<0. If x (n) = δ(n), then y (z) will be?
  • a)
    2
  • b)
    1
  • c)
    0
  • d)
    -1
Correct answer is option 'C'. Can you explain this answer?

Kanika Nair answered
= a * y(n-1) + b * x(n) is a recursive system, where y(n) is the output at time n, y(n-1) is the previous output at time n-1, x(n) is the input at time n, and a and b are constants.

 What is the Fourier transform of the signal x(n)=a|n|, |a|<1? 
  • a)
    (1+a2)/(1-2acosω+a2)
  • b)
    (1-a2)/(1-2acosω+a2)
  • c)
    2a/(1-2acosω+a2 )
  • d)
    None of the mentioned
Correct answer is option 'B'. Can you explain this answer?

Arindam Malik answered
Explanation: First we observe x(n) can be expressed as
x(n)=x1(n)+x2(n)
where x1(n)= an, n>0
=0, elsewhere
 
x2(n)=a-n, n<0 =0, elsewhere Now applying Fourier transform for the above two signals, we get X1(ω)= 1/(1-ae)/((1-ae(-jω) )(1-ae)) = (1-acosω-jasinω)/(1-2acosω+a2 )
Now, X(ω)= X1(ω)+ X2(ω)= 1/(1-ae^(-jω) )+(ae^jω)/(1-ae^jω ) = (1-a2)/(1-2acosω+a2).

A discrete time signal is as given below X [n] = cos (n/8) cos (πn/8)
The period of the signal X [n] is _____________
  • a)
    16 π
  • b)
    16(π+1)
  • c)
    8
  • d)
    Non-periodic
Correct answer is option 'D'. Can you explain this answer?

Sarita Yadav answered
We know that for X [n] = X1 [n] × X2 [n] to be periodic, both X1 [n] and X2 [n] should be periodic with finite periods.
Here X2 [n] = cos (πn/8), is periodic with fundamental period as 8/n
But X1 [n] = cos (n/8) is non periodic.
∴ X [n] is a non-periodic signal.

 What is the convolution of the sequences of x1(n)=x2(n)={1,1,1}?
  • a)
    {1,2,3,2,1}
  • b)
    {1,2,3,2,2}
  • c)
    {1,1,1,1,1}
  • d)
    {1,1,1,1,2}
Correct answer is option 'A'. Can you explain this answer?

Explanation: Given x1(n)=x2(n)={1,1,1}
By calculating the Fourier transform of the above two signals, we get
X1(ω)= X2(ω)=1+ ejω + e -jω = 1+2cosω
From the convolution property of Fourier transform we have,
X(ω)= X1(ω). X2(ω)=(1+2cosω)2=3+4cosω+2cos2ω
By applying the inverse Fourier transform of the above signal, we get
x1(n)*x2(n)={1,2,3,2,1}

What is the value of XI(ω) given X(ω)=1/(1-ae-jω ) ,|a|<1? 
  • a)
    asinω/(1-2acosω+a2 )
  • b)
    (1+acosω)/(1-2acosω+a2 )
  • c)
    (1-acosω)/(1-2acosω+a2 )
  • d)
    (-asinω)/(1-2acosω+a2 )
Correct answer is option 'D'. Can you explain this answer?

Explanation: Given, X(ω)= 1/(1-ae-jω ) ,|a|<1
By multiplying both the numerator and denominator of the above equation by the complex conjugate of the denominator, we obtain
X(ω)= (1-ae)/((1-ae(-jω) )(1-ae)) = (1-acosω-jasinω)/(1-2acosω+a2 )
This expression can be subdivided into real and imaginary parts, thus we obtain
XI(ω)= (-asinω)/(1-2acosω+a2 ).

 What is the energy density spectrum of the signal x(n)=anu(n), |a|<1? 
  • a)
    1/(1+2acosω+a2 )
  • b)
    1/(1-2acosω+a2 )
  • c)
    1/(1-2acosω-a2 )
  • d)
    1/(1+2acosω-a2 )
Correct answer is option 'B'. Can you explain this answer?

Given x(n)= anu(n), |a|<1 The auto correlation of the above signal is rxx(l)=1/(1-a2 ) a|l|, -8< l <8 According to Wiener-Khintchine Theorem, Sxx(?)=F{ rxx(l)}= [1/(1-a2)].F{a|l|} = 1/(1-2acos?+a2 )

The Fourier series coefficients of signal x(t) shown in Figure (A) are given as:
c0 = 1/π, c1 = −j/4, cn = 1/π(1 − n2), for k even.


Which of the following Fourier series coefficients of y(t) shown in Figure (B) is/are correct?
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Concept:
Time shifting property of Fourier series states that:

Since ω0 = 2π/T0, the above expression can be written as:

Application:
Observing the two figures, we can write:
y(t) = x(t - 1)

Where cn = Fourier series coefficient of x(t)
Since, T0 = 2

Using the above expression, we get:

Also for even values of n, e-jπn = 1

The Nyquist frequency for the signal x (t) = 3 cos 50πt + 10 sin 300πt – cos 100t is ___________
  • a)
    50 Hz
  • b)
    100 Hz
  • c)
    200 Hz
  • d)
    300 Hz
Correct answer is option 'D'. Can you explain this answer?

Qiana Iyer answered
The Nyquist frequency is the highest frequency that can be accurately represented or sampled in a digital signal. It is equal to half the sampling rate.

In this case, we don't have information about the sampling rate, so we cannot determine the exact Nyquist frequency.

What is the value of |X(ω)| given X(ω)=1/(1-ae-jω ) ,|a|<1? 
  • a)
    1/√(1-2acosω+a2 )
  • b)
    1/√(1+2acosω+a2)
  • c)
    1/(1-2acosω+a2 )
  • d)
    1/(1+2acosω+a2 )
Correct answer is option 'A'. Can you explain this answer?

Arka Bajaj answered
Explanation: For the given X(ω)=1/(1-ae-jω ) ,|a|<1 we obtain
XI(ω)= (-asinω)/(1-2acosω+a2 ) and XR(ω)= (1-acosω)/(1-2acosω+a2 )
We know that |X(ω)|=√(〖X_R (ω)〗2+〖X_I (ω)〗2 )
Thus on calculating, we obtain
|X(ω)|= 1/√(1-2acosω+a2 )

What is the value of XR(ω) given X(ω)=1/(1-ae-jω ) ,|a|<1? 
  • a)
    asinω/(1-2acosω+a2 )
  • b)
    (1+acosω)/(1-2acosω+a2 )
  • c)
    (1-acosω)/(1-2acosω+a2 )
  • d)
    (-asinω)/(1-2acosω+a2 )
Correct answer is option 'C'. Can you explain this answer?

Palak Saini answered
Explanation: Given, X(ω)= 1/(1-ae-jω ) ,|a|<1
By multiplying both the numerator and denominator of the above equation by the complex conjugate of the denominator, we obtain
X(ω)= (1-ae)/((1-ae(-jω) )(1-ae)) = (1-acosω-jasinω)/(1-2acosω+a2 )
This expression can be subdivided into real and imaginary parts, thus we obtain
XR(ω)= (1-acosω)/(1-2acosω+a2 ).

What is the steady state value of The DT signal F (t), if it is known that F(s)
  • a)
    1/16
  • b)
    Cannot be determined
  • c)
    0
  • d)
    1/8
Correct answer is option 'C'. Can you explain this answer?

Sarita Yadav answered
The steady state value of the DT signal F(s) exists since all poles of the given Laplace transform have negative real part.
∴F (∞) = lims→0 s F(s)

= 0.

Let X(t) be Wide Sense Stationary random process with power spectral density Sx(f). If Y(t) is a random process defined as Y(t) = X(2t − 1), the power spectral density Sy(f) is:
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

Ravi Singh answered
Power density has no effect of shifting. It is affected only by scaling
We know,
E[X(t)X(t + τ)] = Rx(τ)
and
E[X(2t + 1)X(2(t + τ) + 1)] = E[X(2t + 1)X(2t + 2τ + 1)] = Rx(2τ)

then using the scaling property of Fourier transforms we have,

Thus, the power spectral density of 

 If x(n) is a real signal, then 
  • a)
    True
  • b)
    False
Correct answer is option 'A'. Can you explain this answer?

Shivani Saha answered
We know that if x(n) is a real signal, then xI(n)=0 and xR(n)=x(n)
We know that,

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