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

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

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## 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.
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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.

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