Signal and System Frequency Analysis MCQ Quiz – Objective Question with Answer for Signal and System Frequency Analysis

11. Which of the following relation is correct between Fourier transform X(F) and Fourier series coefficient ck?

A. ck = X(F0/k)
B. ck = 1/TP (X(F0/k))
C. ck = 1/TP(X(kF0))
D. none of the mentioned

Answer: C

Let us consider a signal x(t) whose Fourier transform X(F) is given as

X(F) = \(\int_{-∞}^∞ x(t)e^{-j2πF_0 t}dt\)

and the Fourier series coefficient is given as

ck = \(\frac{1}{T_p} \int_{-∞}^∞ x(t)e^{-j2πkF_0 t}dt\)

By comparing the above two equations, we get

ck = \(\frac{1}{T_p} X(kF_0)\)

 

12. According to Parseval’s Theorem for non-periodic signal, \(\int_{-∞}^∞|x(t)|^2 dt\).

A. \(\int_{-∞}^∞|X(F)|^2 dt \)

B. \(\int_{-∞}^∞|X^* (F)|^2 dt \)

C. \(\int_{-∞}^∞ X(F).X^*(F) dt \)

D. All of the mentioned

Answer: D

Let x(t) be any finite energy signal with Fourier transform X(F). Its energy is

Ex = \(\int_{-∞}^∞|x(t)|^2 dt\)

which in turn, can be expressed in terms of X(F) as follows

Ex = \(\int_{-∞}^∞ x^* (t).x(t)\) dt

= \(\int_{-∞}^∞ x(t) dt[\int_{-∞}^∞X^* (F)e^{-j2πF_0 t} dt]\)

= \(\int_{-∞}^∞ X^* (F) dt[\int_{-∞}^∞ x(t)e^{-j2πF_0 t} dt] \)

\( = \int_{-∞}^∞ |X(F)|^2 dt = \int_{-∞}^∞|X^* (F)|^2 dt = \int_{-∞}^∞X(F).X^* (F) dt\)

 

13. What is the Fourier series representation of a signal x(n) whose period is N?

A. \(\sum_{k = 0}^{N+1}c_k e^{j2πkn/N}\)

B. \(\sum_{k = 0}^{N-1}c_k e^{j2πkn/N}\)

C. \(\sum_{k = 0}^Nc_k e^{j2πkn/N}\)

D. \(\sum_{k = 0}^{N-1}c_k e^{-j2πkn/N}\)

Answer: B

Here, the frequency F0 of a continuous-time signal is divided into 2π/N intervals.
So, the Fourier series representation of a discrete-time signal with period N is given as

x(n) = \(\sum_{k = 0}^{N-1}c_k e^{j2πkn/N}\)

where ck is the Fourier series coefficient

 

14. What is the expression for Fourier series coefficient ck in terms of the discrete signal x(n)?

A. \(\frac{1}{N} \sum_{n = 0}^{N-1}x(n)e^{j2πkn/N}\)

B. \(N\sum_{n = 0}^{N-1}x(n)e^{-j2πkn/N}\)

C. \(\frac{1}{N} \sum_{n = 0}^{N+1}x(n)e^{-j2πkn/N}\)

D. \(\frac{1}{N} \sum_{n = 0}^{N-1}x(n)e^{-j2πkn/N}\)

Answer: D

We know that, the Fourier series representation of a discrete signal x(n) is given as

x(n) = \(\sum_{n = 0}^{N-1}c_k e^{j2πkn/N}\)

Now multiply both sides by the exponential e-j2πln/N and summing the product from n = 0 to n = N-1. Thus,

\(\sum_{n = 0}^{N-1} x(n)e^{-j2πln/N} = \sum_{n = 0}^{N-1}\sum_{k = 0}^{N-1}c_k e^{j2π(k-l)n/N}\)

If we perform summation over n first in the right hand side of above equation, we get

\(\sum_{n = 0}^{N-1} e^{-j2πkn/N}\) = N, for k-l = 0,±N,±2N…
= 0, otherwise

Therefore, the right hand side reduces to Nck

So, we obtain ck = \(\frac{1}{N} \sum_{n = 0}^{N-1}x(n)e^{-j2πkn/N}\)

 

15. Which of the following represents the phase associated with the frequency component of discrete-time Fourier series(DTFS)?

A. ej2πkn/N
B. e-j2πkn/N
C. ej2πknN
D. none of the mentioned

Answer: A

We know that,

x(n) = \(\sum_{k = 0}^{N-1}c_k e^{j2πkn/N}\)

In the above equation, ck represents the amplitude and ej2πkn/N represents the phase associated with the frequency component of DTFS.

 

16. The Fourier series for the signal x(n) = cos√2πn exists.

A. True
B. False

Answer: B

For ω0 = √2π, we have f0 = 1/√2.

Since f0 is not a rational number, the signal is not periodic. Consequently, this signal cannot be expanded in a Fourier series.

 

17. What are the Fourier series coefficients for the signal x(n) = cosπn/3?

A. c1 = c2 = c3 = c4 = 0,c1 = c5 = 1/2
B. c0 = c1 = c2 = c3 = c4 = c5 = 0
C. c0 = c1 = c2 = c3 = c4 = c5 = 1/2
D. none of the mentioned

Answer: A

In this case, f0 = 1/6 and hence x(n) is periodic with fundamental period N = 6.

Given signal is x(n) = cosπn/3 = cos2πn/6

= \(\frac{1}{2} e^{j2πn/6}+\frac{1}{2} e^{-j2πn/6}\)

We know that -2π/6 = 2π-2π/6 = 10π/6 = 5(2π/6)

∴ x(n) = \(\frac{1}{2} e^{j2πn/6}+\frac{1}{2} e^{j2π(5)n/6}\)

Compare the above equation with

x(n) = \(\sum_{k = 0}^{N-1}c_k e^{j2πkn/N}\)

So, we get c1 = c2 = c3 = c4 = 0 and c1 = c5 = 1/2.

 

18. What is the Fourier series representation of a signal x(n) whose period is N?

A. \(\sum_{k = 0}^{\infty}|c_k|^2\)

B. \(\sum_{k = -\infty}^{\infty}|c_k|\)

C. \(\sum_{k = -\infty}^0|c_k|^2\)

D. \(\sum_{k = -\infty}^{\infty}|c_k|^2\)

Answer: B

The average power of a periodic signal x(t) is given as

\(\frac{1}{T_p}\int_{t_0}^{t_0+T_p}|x(t)|^2 dt\)

= \(\frac{1}{T_p}\int_{t_0}^{t_0+T_p} x(t).x^* (t) dt\)

= \(\frac{1}{T_p}\int_{t_0}^{t_0+T_p}x(t).\sum_{k = -∞}^∞ c_k^* e^{-j2πkF_0 t} dt\)

By interchanging the positions of integral and summation and by applying the integration, we get

= \(\sum_{k = -∞}^∞|c_k |^2\)

 

19. What is the average power of the discrete-time periodic signal x(n) with period N?

A. \(\frac{1}{N} \sum_{n = 0}^{N}|x(n)|\)

B. \(\frac{1}{N} \sum_{n = 0}^{N-1}|x(n)|\)

C. \(\frac{1}{N} \sum_{n = 0}^{N}|x(n)|^2\)

D. \(\frac{1}{N} \sum_{n = 0}^{N-1}|x(n)|^2 \)

Answer: D

Let us consider a discrete-time periodic signal x(n) with period N.

The average power of that signal is given as

Px = \(\frac{1}{N} \sum_{n = 0}^{N-1}|x(n)|^2\)

 

20. What is the equation for the average power of discrete-time periodic signal x(n) with period N in terms of Fourier series coefficient ck?

A. \(\sum_{k = 0}^{N-1}|c_k|\)

B. \(\sum_{k = 0}^{N-1}|c_k|^2\)

C. \(\sum_{k = 0}^N|c_k|^2\)

D. \(\sum_{k = 0}^N|c_k|\)

Answer: B

We know that

Px = \(\frac{1}{N} \sum_{n = 0}^{N-1}|x(n)|^2\)

= \(\frac{1}{N} \sum_{n = 0}^{N-1}x(n).x^*(n)\)

= \(\frac{1}{N} \sum_{n = 0}^{N-1}x(n) \sum_{k = 0}^{N-1}c_k * e^{-j2πkn/N}\)

= \(\sum_{k = 0}^{N-1}c_k * \frac{1}{N} \sum_{n = 0}^{N-1}x(n)e^{-j2πkn/N}\)

= \(\sum_{k = 0}^{N-1}|c_k |^2\)

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