Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Tests  >  Signals and Systems  >  Test: Fourier - 3 - Electrical Engineering (EE) MCQ

Test: Fourier - 3 - Electrical Engineering (EE) MCQ


Test Description

10 Questions MCQ Test Signals and Systems - Test: Fourier - 3

Test: Fourier - 3 for Electrical Engineering (EE) 2024 is part of Signals and Systems preparation. The Test: Fourier - 3 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Fourier - 3 MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Fourier - 3 below.
Solutions of Test: Fourier - 3 questions in English are available as part of our Signals and Systems for Electrical Engineering (EE) & Test: Fourier - 3 solutions in Hindi for Signals and Systems course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt Test: Fourier - 3 | 10 questions in 30 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study Signals and Systems for Electrical Engineering (EE) Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Fourier - 3 - Question 1

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

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

Test: Fourier - 3 - Question 3

Consider a discrete-time periodic signal

with period N = 10. Also y[n] = x[n] - x[n - 1]
Q. The FS coefficients of y[n] are

Detailed Solution for Test: Fourier - 3 - Question 3


Test: Fourier - 3 - Question 4

Consider a discrete-time periodic signal

with period N = 10. Also y[n] = x[n] - x[n - 1]
Q. The FS coefficients of x[n] are

Detailed Solution for Test: Fourier - 3 - Question 4

y [n] = x [n] - x [n - 1]

Test: Fourier - 3 - Question 5

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 5


Test: Fourier - 3 - Question 6

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 6


Test: Fourier - 3 - Question 7

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 7

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

Test: Fourier - 3 - Question 8

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 8

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

Test: Fourier - 3 - Question 9

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 9


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 10

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 10

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.

44 videos|44 docs|33 tests
Information about Test: Fourier - 3 Page
In this test you can find the Exam questions for Test: Fourier - 3 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Fourier - 3, EduRev gives you an ample number of Online tests for practice

Up next

44 videos|44 docs|33 tests
Download as PDF

Up next

Download the FREE EduRev App
Track your progress, build streaks, highlight & save important lessons and more!