Test: Fourier Part - 3


10 Questions MCQ Test Signal and System | Test: Fourier Part - 3


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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)

Solution:

With N even

QUESTION: 2

Consider a discrete-time periodic signal

with period N = 10. Also y[n] = x[n] - x[n - 1]
Q. The fundamental period of y[n] is

Solution:

y[n] is shown is fig. S5.8.21. It has fundamental

period of 10.

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

Solution:


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

Solution:

y [n] = x [n] - x [n - 1]

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]

Solution:


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]

Solution:


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]

Solution:

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

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]}

Solution:

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

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.

Solution:


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

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

Solution:

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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