Digital Signal Processing Questions and Answers – Frequency Analysis of Discrete Time Signal – 2

This set of Digital Signal Processing Interview Questions & Answers focuses on “Frequency Analysis of Discrete Time Signal-2 “.

1. Which of the following relation is true if the signal x(n) is real?
a) X*(ω)=X(ω)
b) X*(ω)=X(-ω)
c) X*(ω)=-X(ω)
d) None of the mentioned
View Answer

Answer: b
Explanation: We know that,
X(ω)=\(\sum_{n=-∞}^∞ x(n)e^{-jωn}\)
=> X*(ω)=\([\sum_{n=-∞}^∞ x(n)e^{-jωn}]^*\)
Given the signal x(n) is real. Therefore,
X*(ω)=\(\sum_{n=-∞}^∞ x(n)e^{jωn}\)
=> X*(ω)=X(-ω).

2. For a signal x(n) to exhibit even symmetry, it should satisfy the condition |X(-ω)|=| X(ω)|.
a) True
b) False
View Answer

Answer: a
Explanation: We know that, if a signal x(n) is real, then
X*(ω)=X(-ω)
If the signal is even symmetric, then the magnitude on both the sides should be equal.
So, |X*(ω)|=|X(-ω)| => |X(-ω)|=|X(ω)|.

3. What is the energy density spectrum Sxx(ω) of the signal x(n)=anu(n), |a|<1?
a) \(\frac{1}{1+2acosω+a^2}\)
b) \(\frac{1}{1+2asinω+a^2}\)
c) \(\frac{1}{1-2asinω+a^2}\)
d) \(\frac{1}{1-2acosω+a^2}\)
View Answer

Answer: d
Explanation: Since |a|<1, the sequence x(n) is absolutely summable, as can be verified by applying the geometric summation formula.
\(\sum_{n=-∞}^∞|x(n)| = \sum_{n=-∞}^∞ |a|^n = \frac{1}{1-|a|} \lt ∞\)
Hence the Fourier transform of x(n) exists and is obtained as
X(ω) = \(\sum_{n=-∞}^∞ a^n e^{-jωn}=\sum_{n=-∞}^∞ (ae^{-jω})^n\)
Since |ae-jω|=|a|<1, use of the geometric summation formula again yields
X(ω)=\(\frac{1}{1-ae^{-jω}}\)
The energy density spectrum is given by
Sxx(ω)=|X(ω)|2= X(ω).X*(ω)=\(\frac{1}{(1-ae^{-jω})(1-ae^{jω})} = \frac{1}{1-2acosω+a^2}\).
advertisement
advertisement

4. What is the Fourier transform of the signal x(n) which is defined as shown in the graph below?
a) Ae-j(ω/2)(L)\(\frac{sin⁡(\frac{ωL}{2})}{sin⁡(\frac{ω}{2})}\)
b) Aej(ω/2)(L-1)\(\frac{sin⁡(\frac{ωL}{2})}{sin⁡(\frac{ω}{2})}\)
c) Ae-j(ω/2)(L-1)\(\frac{sin⁡(\frac{ωL}{2})}{sin⁡(\frac{ω}{2})}\)
d) None of the mentioned
View Answer

Answer: c
Explanation: The Fourier transform of this signal is
X(ω)=\(\sum_{n=0}^{L-1} Ae^{-jωn}\)
=A.\(\frac{1-e^{-jωL}}{1-e^{-jω}}\)
=\(Ae^{-j(ω/2)(L-1)}\frac{sin⁡(\frac{ωL}{2})}{sin⁡(\frac{ω}{2})}\)

5. Which of the following condition is to be satisfied for the Fourier transform of a sequence to be equal as the Z-transform of the same sequence?
a) |z|=1
b) |z|<1
c) |z|>1
d) Can never be equal
View Answer

Answer: a
Explanation: Let us consider the signal to be x(n)
Z{x(n)}=\(\sum_{n=-∞}^∞ x(n)z^{-n} and X(ω)=\sum_{n=-∞}^∞ x(n)e^{-jωn}\)
Now, represent the ‘z’ in the polar form
=> z=r.e
=>Z{x(n)}=\(\sum_{n=-∞}^∞ x(n)r^{-n} e^{-jωn}\)
Now Z{x(n)}= X(ω) only when r=1=>|z|=1.
Note: Join free Sanfoundry classes at Telegram or Youtube

6. The sequence x(n)=\(\frac{sin⁡ ω_c n}{πn}\) does not have both z-transform and Fourier transform.
a) True
b) False
View Answer

Answer: b
Explanation: The given x(n) do not have Z-transform. But the sequence have finite energy. So, the given sequence x(n) has a Fourier transform.

7. If x(n) is a stable sequence so that X(z) converges on to a unit circle, then the complex cepstrum signal is defined as ____________
a) X(ln X(z))
b) ln X(z)
c) X-1(ln X(z))
d) None of the mentioned
View Answer

Answer: c
Explanation: Let us consider a sequence x(n) having a z-transform X(z). We assume that x(n) is a stable sequence so that X(z) converges on to the unit circle. The complex cepstrum of the signal x(n) is defined as the sequence cx(n), which is the inverse z-transform of Cx(z), where Cx(z)=ln X(z)
=> cx(z)= X-1(ln X(z))
advertisement

8. If cx(n) is the complex cepstrum sequence obtained from the inverse Fourier transform of ln X(ω), then what is the expression for cθ(n)?
a) \(\frac{1}{2π} \int_0^π \theta(ω) e^{jωn} dω\)
b) \(\frac{1}{2π} \int_{-π}^π \theta(ω) e^{-jωn} dω\)
c) \(\frac{1}{2π} \int_0^π \theta(ω) e^{jωn} dω\)
d) \(\frac{1}{2π} \int_{-π}^π \theta(ω) e^{jωn} dω\)
View Answer

Answer: d
Explanation: We know that,
cx(n)=\(\frac{1}{2π} \int_{-π}^π ln⁡(X(ω))e^{jωn} dω\)
If we express X(ω) in terms of its magnitude and phase, say
X(ω)=|X(ω)|ejθ(ω)
Then ln X(ω)=ln |X(ω)|+jθ(ω)
=> cx(n)=\(\frac{1}{2π} \int_{-π}^π[ln|X(ω)|+jθ(ω)]e^{jωn} dω\) => cx(n)=cm(n)+jcθ(n)(say)
=> cθ(n)=\(\frac{1}{2π} \int_{-π}^πθ(ω) e^{jωn} dω\)

9. What is the Fourier transform of the signal x(n)=u(n)?
a) \(\frac{1}{2sin⁡(ω/2)} e^{j(ω+π)}\)
b) \(\frac{1}{2sin⁡(ω/2)} e^{j(ω-π)}\)
c) \(\frac{1}{2sin⁡(ω/2)} e^{j(ω+π)/2}\)
d) \(\frac{1}{2sin⁡(ω/2)} e^{j(ω-π)/2}\)
View Answer

Answer: d
Explanation: Given x(n)=u(n)
We know that the z-transform of the given signal is X(z)=\(\frac{1}{1-z^{-1}}\) ROC:|z|>1
X(z) has a pole p=1 on the unit circle, but converges for |z|>1.
If we evaluate X(z) on the unit circle except at z=1, we obtain
X(ω) = \(\frac{e^{jω/2}}{2jsin(ω/2)} = \frac{1}{2sin⁡(ω/2)} e^{j(ω-π)/2}\)
advertisement

10. If a power signal has its power density spectrum concentrated about zero frequency, the signal is known as ______________
a) Low frequency signal
b) Middle frequency signal
c) High frequency signal
d) None of the mentioned
View Answer

Answer: a
Explanation: We know that, for a low frequency signal, the power signal has its power density spectrum concentrated about zero frequency.

11. What are the main characteristics of Anti aliasing filter?
a) Ensures that bandwidth of signal to be sampled is limited to frequency range
b) To limit the additive noise spectrum and other interference, which corrupts the signal
c) All of the mentioned
d) None of the mentioned
View Answer

Answer: c
Explanation: The anti aliasing filter is an analog filter which has a twofold purpose. First, it ensures that the bandwidth of the signal to be sampled is limited to the desired frequency range. Using an anti aliasing filter is to limit the additive noise spectrum and other interference, which often corrupts the desired signal. Usually, additive noise is wide band and exceeds the bandwidth of the desired signal.

12. In general, a digital system designer has better control of tolerances in a digital signal processing system than an analog system designer who is designing an equivalent analog system.
a) True
b) False
View Answer

Answer: a
Explanation: Analog signal processing operations cannot be done very precisely either, since electronic components in analog systems have tolerances and they introduce noise during their operation. In general, a digital system designer has better control of tolerances in a digital signal processing system than an analog system designer who is designing an equivalent analog system.

13. The term ‘bandwidth’ represents the quantitative measure of a signal.
a) True
b) False
View Answer

Answer: a
Explanation: In addition to the relatively broad frequency domain classification of signals, it is often desirable to express quantitatively the range of frequencies over which the power or energy density spectrum is concentrated. This quantitative measure is called the ‘bandwidth’ of a signal.

14. If F1 and F2 are the lower and upper cutoff frequencies of a band pass signal, then what is the condition to be satisfied to call such a band pass signal as narrow band signal?
a) (F1-F2)>\(\frac{F_1+F_2}{2}\)(factor of 3 or less)
b) (F1-F2)⋙\(\frac{F_1+F_2}{2}\)(factor of 10 or more)
c) (F1-F2)<\(\frac{F_1+F_2}{2}\)(factor of 3 or less)
d) (F1-F2)⋘\(\frac{F_1+F_2}{2}\)(factor of 10 or more)
View Answer

Answer: d
Explanation: If the difference in the cutoff frequencies is much less than the mean frequency, the such a band pass signal is known as narrow band signal.

15. What is the frequency range(in Hz) of Electroencephalogram(EEG)?
a) 10-40
b) 1000-2000
c) 0-100
d) None of the mentioned
View Answer

Answer: c
Explanation: Electroencephalogram(EEG) signal has a frequency range of 0-100 Hz.

16. Which of the following electromagnetic signals has a frequency range of 30kHz-3MHz?
a) Radio broadcast
b) Shortwave radio signal
c) RADAR
d) Infrared signal
View Answer

Answer: a
Explanation: Radio broadcast signal is an electromagnetic signal which has a frequency range of 30kHz-3MHz.

Sanfoundry Global Education & Learning Series – Digital Signal Processing.

To practice all areas of Digital Signal Processing for Interviews, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
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.

Subscribe to his free Masterclasses at Youtube & discussions at Telegram SanfoundryClasses.