Digital Signal Processing Questions and Answers – Frequency Analysis of Discrete Time Signal – 1

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

Answer: b
Explanation: 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

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

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

3. Which of the following represents the phase associated with the frequency component of discrete-time Fourier series(DTFS)?
a) ej2πkn/N
b) e-j2πkn/N
c) ej2πknN
d) none of the mentioned
View Answer

Answer: a
Explanation: 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.
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4. The Fourier series for the signal x(n)=cos√2πn exists.
a) True
b) False
View Answer

Answer: b
Explanation: 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.

5. 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
View Answer

Answer: a
Explanation: 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)
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 c1=c2=c3=c4=0 and c1=c5=1/2.
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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

Answer: b
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

Answer: d
Explanation: Let us consider a discrete time periodic signal x(n) with period N.
The average power of that signal is given as
Px=\(\frac{1}{N} \sum_{n=0}^{N-1}|x(n)|^2\)
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8. What is the equation for 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) \(\sum_{k=0}^N|c_k|\)
View Answer

Answer: b
Explanation: We know that Px=\(\frac{1}{N} \sum_{n=0}^{N-1}|x(n)|^2\)
=\(\frac{1}{N} \sum_{n=0}^{N-1}x(n).x^*(n)\)
=\(\frac{1}{N} \sum_{n=0}^{N-1}x(n) \sum_{k=0}^{N-1}c_k * e^{-j2πkn/N}\)
=\(\sum_{k=0}^{N-1}c_k * \frac{1}{N} \sum_{n=0}^{N-1}x(n)e^{-j2πkn/N}\)
=\(\sum_{k=0}^{N-1}|c_k |^2\)

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

Answer: a
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)=\(\sum_{n=-∞}^∞x(n)e^{-jωn}\)
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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

Answer: d
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

Answer: c
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) ωc
b) -ωc
c) ωc
d) none of the mentioned
View Answer

Answer: c
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 ωc/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

Answer: a
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

Answer: a
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

Answer: b
Explanation: We know that, Ex=\(\sum_{n=-∞}^∞ |x(n)|^2\)
=\(\sum_{n=-∞}^∞ x(n).x^*(n)\)
=\(\sum_{n=-∞}^∞ x(n)\frac{1}{2π} \int_{-π}^π X^*(ω) e^{-jωn} dω\)
=\(\frac{1}{2π} \int_{-π}^π|X(ω)|^2 dω\)

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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