This set of Digital Signal Processing Multiple Choice Questions & Answers (MCQs) focuses on “Frequency Analysis of Discrete Time Signals-1”.

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

View Answer

Explanation: Here, the frequency F

_{0}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 c

_{k}is the Fourier series coefficient

2. What is the expression for Fourier series coefficient c_{k} 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) \(\frac{1}{N} \sum_{n=0}^{N-1}x(n)e^{-j2πkn/N}\)

View Answer

Explanation: 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 Nc

_{k}

So, we obtain c

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

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

a) e^{j2πkn/N}

b) e^{-j2πkn/N}

c) e^{j2πknN}

d) none of the mentioned

View Answer

Explanation: We know that,

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

In the above equation, c

_{k}represents the amplitude and e

^{j2πkn/N}represents the phase associated with the frequency component of DTFS.

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

a) True

b) False

View Answer

Explanation: For ω

_{0}=√2π, we have f

_{0}=1/√2. Since f

_{0}is not a rational number, the signal is not periodic. Consequently, this signal cannot be expanded in a Fourier series.

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

a) c_{1}=c_{2}=c_{3}=c_{4}=0,c_{1}=c_{5}=1/2

b) c_{0}=c_{1}=c_{2}=c_{3}=c_{4}=c_{5}=0

c) c_{0}=c_{1}=c_{2}=c_{3}=c_{4}=c_{5}=1/2

d) none of the mentioned

View Answer

Explanation: In this case, f

_{0}=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)

Therefore, 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 c

_{1}=c

_{2}=c

_{3}=c

_{4}=0 and c

_{1}=c

_{5}=1/2.

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

View Answer

Explanation: 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

=\(\sum_{k=-∞}^∞|c_k |^2\)

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

View Answer

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

The average power of that signal is given as

P

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

8. What is the equation for average power of discrete time periodic signal x(n) with period N in terms of Fourier series coefficient c_{k}?

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

View Answer

Explanation: We know that P

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

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

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

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

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

9. What is the Fourier transform X(ω) of a finite energy discrete time signal x(n)?

a) \(\sum_{n=-∞}^∞x(n)e^{-jωn}\)

b) \(\sum_{n=0}^∞x(n)e^{-jωn}\)

c) \(\sum_{n=0}^{N-1}x(n)e^{-jωn}\)

d) None of the mentioned

View Answer

Explanation: If we consider a signal x(n) which is discrete in nature and has finite energy, then the Fourier transform of that signal is given as

X(ω)=\(\sum_{n=-∞}^∞x(n)e^{-jωn}\)

10. What is the period of the Fourier transform X(ω) of the signal x(n)?

a) π

b) 1

c) Non-periodic

d) 2π

View Answer

Explanation: Let X(ω) be the Fourier transform of a discrete time signal x(n) which is given as

X(ω)=\(\sum_{n=-∞}^∞x(n)e^{-jωn}\)

Now X(ω+2πk)=\(\sum_{n=-∞}^∞ x(n)e^{-j(ω+2πk)n}\)

=\(\sum_{n=-∞}^∞ x(n)e^{-jωn}e^{-j2πkn}\)

=\(\sum_{n=-∞}^∞ x(n)e^{-jωn}= X(ω)\)

So, the Fourier transform of a discrete time finite energy signal is periodic with period 2π.

11. What is the synthesis equation of the discrete time signal x(n), whose Fourier transform is X(ω)?

a) \(2π\int_0^2π X(ω) e^jωn dω\)

b) \(\frac{1}{π} \int_0^{2π} X(ω) e^jωn dω\)

c) \(\frac{1}{2π} \int_0^{2π} X(ω) e^jωn dω\)

d) None of the mentioned

View Answer

Explanation: We know that the Fourier transform of the discrete time signal x(n) is

X(ω)=\(\sum_{n=-∞}^∞ x(n)e^{-jωn}\)

By calculating the inverse Fourier transform of the above equation, we get

x(n)=\(\frac{1}{2π} \int_0^{2π} X(ω) e^{jωn} dω\)

The above equation is known as synthesis equation or inverse transform equation.

12. What is the value of discrete time signal x(n) at n=0 whose Fourier transform is represented as below?

a) ω_{c}.π

b) -ω_{c}/π

c) ω_{c}/π

d) none of the mentioned

View Answer

Explanation: We know that, x(n)=\(\frac{1}{2\pi} \int_{-\pi}^{\pi}X(\omega)e^{j\omega n} dω\)

=\(\frac{1}{2\pi} \int_{-ω_c}^{ω_c}1.e^{j\omega n} dω\)

At n=0,

x(n)=x(0)=\(\int_{-ω_c}^{ω_c}1 dω=\frac{1}{2\pi}(2 ω_c)=\frac{ω_c}{\pi_ω}\)

Therefore, the value of the signal x(n) at n=0 is ω

_{c}/π.

13. What is the value of discrete time signal x(n) at n≠0 whose Fourier transform is represented as below?

a) \(\frac{ω_c}{\pi}.\frac{sin ω_c.n}{ω_c.n}\)

b) \(\frac{-ω_c}{\pi}.\frac{sin ω_c.n}{ω_c.n}\)

c) \(ω_c.\pi \frac{sin ω_c.n}{ω_c.n}\)

d) None of the mentioned

View Answer

Explanation: We know that, x(n)=\(\frac{1}{2\pi} \int_{-\pi}^{\pi}X(\omega)e^{j\omega n} dω\)

=\(\frac{1}{2\pi} \int_{-ω_c}^{ω_c}1.e^{j\omega n} dω=\frac{sin ω_c.n}{ω_c.n}\) =\(\frac{ω_c}{\pi}.\frac{sin ω_c.n}{ω_c.n}\)

14. The oscillatory behavior of the approximation of XN(ω) to the function X(ω) at a point of discontinuity of X(ω) is known as Gibbs phenomenon.

a) True

b) False

View Answer

Explanation: We note that there is a significant oscillatory overshoot at ω=ωc, independent of the value of N. As N increases, the oscillations become more rapid, but the size of the ripple remains the same. One can show that as N→∞, the oscillations converge to the point of the discontinuity at ω=ωc. The oscillatory behavior of the approximation of XN(ω) to the function X(ω) at a point of discontinuity of X(ω) is known as Gibbs phenomenon.

15. What is the energy of a discrete time signal in terms of X(ω)?

a) \(2π\int_{-π}^π |X(ω)|^2 dω\)

b) \(\frac{1}{2π} \int_{-π}^π |X(ω)|^2 dω\)

c) \(\frac{1}{2π} \int_0^π |X(ω)|^2 dω\)

d) None of the mentioned

View Answer

Explanation: We know that, E

_{x}=\(\sum_{n=-∞}^∞ |x(n)|^2\)

=\(\sum_{n=-∞}^∞ x(n).x^*(n)\)

=\(\sum_{n=-∞}^∞ x(n)\frac{1}{2π} \int_{-π}^π X^*(ω) e^{-jωn} dω\)

=\(\frac{1}{2π} \int_{-π}^π|X(ω)|^2 dω\)

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