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Electronics and Communication Engineering (ECE) Exam  >  Electronics and Communication Engineering (ECE) Tests  >  Test: Fourier - 3 - Electronics and Communication Engineering (ECE) MCQ

Test: Fourier - 3 - Electronics and Communication Engineering (ECE) MCQ


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10 Questions MCQ Test - Test: Fourier - 3

Test: Fourier - 3 for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Test: Fourier - 3 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Test: Fourier - 3 MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Fourier - 3 below.
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Test: Fourier - 3 - Question 1
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Consider a periodic signal x[n] with period N and FS coefficients X [k]. Determine the FS coefficients Y [k] of the signal y[n] given in question.
Q. y[n] = (-1)n x[n], (assume that N is even)

Detailed Solution for Test: Fourier - 3 - Question 1

With N even

Test: Fourier - 3 - Question 2
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Consider a discrete-time periodic signal

Also a function y[n] is defined as y[n] = x[n] - x[n - 1] 
Q. The fundamental period of y[n] is

Detailed Solution for Test: Fourier - 3 - Question 2

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Test: Fourier - 3 - Question 3
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Consider a discrete-time signal with Fourier representation.

In question the FS coefficient Y [k] is given. Determine the corresponding signal y[n] and choose correct option.
Q. Y [k] = X[k - 5] + X[k + 5]

Detailed Solution for Test: Fourier - 3 - Question 3


Test: Fourier - 3 - Question 4
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Consider a discrete-time signal with Fourier representation.

In question the FS coefficient Y [k] is given. Determine the corresponding signal y[n] and choose correct option.
Q. Y [k] = cos (πk/5)X[k]
Detailed Solution for Test: Fourier - 3 - Question 4


Test: Fourier - 3 - Question 5
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Consider a discrete-time signal with Fourier representation.

In question the FS coefficient Y [k] is given. Determine the corresponding signal y[n] and choose correct option.
Q. Y [ k] = X [k] * X [k]
Detailed Solution for Test: Fourier - 3 - Question 5

Y [ k] = X [k] * X [k] ⇒ y[n] = x[n] x[n] = ( x[n])2

Test: Fourier - 3 - Question 6
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Consider a discrete-time signal with Fourier representation.

In question the FS coefficient Y [k] is given. Determine the corresponding signal y[n] and choose correct option.
Q. Y [k] = Re{ X [ k]}
Detailed Solution for Test: Fourier - 3 - Question 6

Y [k] = Re{ X[k]} ⇒ y[n] = Ev{ x[n]} = 

Test: Fourier - 3 - Question 7
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Consider a sequence x[n] with following facts :
1. x[n] is periodic with N = 6

4. x[n] has the minimum power per period among the set of signals satisfying the preceding three condition.
The sequence would be.
Detailed Solution for Test: Fourier - 3 - Question 7


From fact 

By Parseval’s relation, the average power in x[n] is

The value of P is minimized by choosing
X [1] = X [2] = X [ 4 ] = X [5] = 0
Therefore

Test: Fourier - 3 - Question 8
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A real and odd periodic signal x[n] has fundamental period N = 7 and FS coefficients X [k]. Given that X [15] = j, X [16] = 2j, X [17] = 3j. The values of X [0],X [ -1], X [-2 ], and X [-3] will be
Detailed Solution for Test: Fourier - 3 - Question 8

Since the FS coefficient repeat every N . Thus
X [1] = X [15], X [2 ] = X [16], X [ 3] = X [17]
The signal real and odd, the FS coefficient X [k] will be purely imaginary and odd. Therefore X [0] = 0
X [ -1] = -X [1], X [ -2 ] = - X [2 ], X [-3] = -X [ 3]
Therefore (D) is correct option.

Test: Fourier - 3 - Question 9
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In the Fourier transform, if the time domain signal x(t) is real and even, then the frequency domain signal X(jΩ) will be:
Detailed Solution for Test: Fourier - 3 - Question 9

Let F(ω) is the Fourier transform of f(t).

Test: Fourier - 3 - Question 10
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If X(ω) = δ(ω - ω0) then x(t) is
Detailed Solution for Test: Fourier - 3 - Question 10

Concept:

Fourier Transform:

It is used for frequency analysis of any Bounded Input and Bounded Output (BIBO) signal.

Fourier Transform for any function x(t) is given by

Inverse Fourier Transform for any function X(ω) is given by

Frequency shifting

If X(ω) has the inverse Fourier transform as x(t).

Calculation:

δ (ω) is impulse function that exists only at t = 0

so its inverse Fourier transform will be

Given signal is δ(ω – ω0)

Inverse Fourier transform is

Impulse function exists at ω0.
We know that the area of the impulse is 1.

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