Digital Signal Processing Questions and Answers – Frequency Transformations

This set of Digital Signal Processing Multiple Choice Questions & Answers (MCQs) focuses on “Frequency Transformations”.

1. What is the pass band edge frequency of an analog low pass normalized filter?
a) 0 rad/sec
b) 0.5 rad/sec
c) 1 rad/sec
d) 1.5 rad/sec
View Answer

Answer: c
Explanation: Let H(s) denote the transfer function of a low pass analog filter with a pass band edge frequency ΩP equal to 1 rad/sec. This filter is known as analog low pass normalized prototype.

2. If H(s) is the transfer function of a analog low pass normalized filter and Ωu is the desired pass band edge frequency of new low pass filter, then which of the following transformation has to be performed?
a) s → s/Ωu
b) s → s.Ωu
c) s → Ωu/s
d) none of the mentioned
View Answer

Answer: a
Explanation: If Ωu is the desired pass band edge frequency of new low pass filter, then the transfer function of this new low pass filter is obtained by using the transformation s → s/Ωu.

3. Which of the following is a low pass-to-high pass transformation?
a) s → s / Ωu
b) s → Ωu / s
c) s → Ωu.s
d) none of the mentioned
View Answer

Answer: b
Explanation: The low pass-to-high pass transformation is simply achieved by replacing s by 1/s. If the desired high pass filter has the pass band edge frequency Ωu, then the transformation is
s → Ωu / s.
advertisement
advertisement

4. Which of the following is the backward design equation for a low pass-to-low pass transformation?
a) \(\Omega_S=\frac{\Omega_S}{\Omega_u}\)
b) \(\Omega_S=\frac{\Omega_u}{\Omega’_S}\)
c) \(\Omega’_S=\frac{\Omega_S}{\Omega_u}\)
d) \(\Omega_S=\frac{\Omega’_S}{\Omega_u}\)
View Answer

Answer: d
Explanation: If Ωu is the desired pass band edge frequency of new low pass filter, then the transfer function of this new low pass filter is obtained by using the transformation s → s/Ωu. If ΩS and Ω’S are the stop band frequencies of prototype and transformed filters respectively, then the backward design equation is given by
\(\Omega_S=\frac{\Omega’_S}{\Omega_u}\).

5. Which of the following is a low pass-to-band pass transformation?
a) s→\(\frac{s^2+Ω_u Ω_l}{s(Ω_u+Ω_l)}\)
b) s→\(\frac{s^2-Ω_u Ω_l}{s(Ω_u-Ω_l)}\)
c) s→\(\frac{s^2+Ω_u Ω_l}{s(Ω_u-Ω_l)}\)
d) s→\(\frac{s^2-Ω_u Ω_l}{s(Ω_u+Ω_l)}\)
View Answer

Answer: c
Explanation: If Ωu and Ωl are the upper and lower cutoff pass band frequencies of the desired band pass filter, then the transformation to be performed on the normalized low pass filter is
s→\(\frac{s^2+Ω_u Ω_l}{s(Ω_u-Ω_l)}\)

6. Which of the following is the backward design equation for a low pass-to-high pass transformation?
a) \(\Omega_S=\frac{\Omega_S}{\Omega_u}\)
b) \(\Omega_S=\frac{\Omega_u}{\Omega’_S}\)
c) \(\Omega’_S=\frac{\Omega_S}{\Omega_u}\)
d) \(\Omega_S=\frac{\Omega’_S}{\Omega_u}\)
View Answer

Answer: b
Explanation: If Ωu is the desired pass band edge frequency of new high pass filter, then the transfer function of this new high pass filter is obtained by using the transformation s → Ωu /s. If ΩS and Ω’S are the stop band frequencies of prototype and transformed filters respectively, then the backward design equation is given by
\(\Omega_S=\frac{\Omega_u}{\Omega’_S}\).

7. Which of the following is a low pass-to-band stop transformation?
a) s→\(\frac{s(Ω_u-Ω_l)}{s^2+Ω_u Ω_l}\)
b) s→\(\frac{s(Ω_u+Ω_l)}{s^2+Ω_u Ω_l}\)
c) s→\(\frac{s(Ω_u-Ω_l)}{s^2-Ω_u Ω_l}\)
d) none of the mentioned
View Answer

Answer: c
Explanation: If Ωu and Ωl are the upper and lower cutoff pass band frequencies of the desired band stop filter, then the transformation to be performed on the normalized low pass filter is
s→\(\frac{s(Ω_u-Ω_l)}{s^2-Ω_u Ω_l}\)
advertisement

8. If A=\(\frac{-Ω_1^2+Ω_u Ω_l}{Ω_1 (Ω_u-Ω_l)}\) and B=\(\frac{Ω_2^2-Ω_u Ω_l}{Ω_2 (Ω_u-Ω_l)}\), then which of the following is the backward design equation for a low pass-to-band pass transformation?
a) ΩS=|B|
b) ΩS=|A|
c) ΩS=Max{|A|,|B|}
d) ΩS=Min{|A|,|B|}
View Answer

Answer: d
Explanation: If Ωu and Ωl are the upper and lower cutoff pass band frequencies of the desired band pass filter and Ω1 and Ω2 are the lower and upper cutoff stop band frequencies of the desired band pass filter, then the backward design equation is
ΩS=Min{|A|,|B|}
where, A=\(\frac{-Ω_1^2+Ω_u Ω_l}{Ω_1 (Ω_u-Ω_l)}\) and B=\(\frac{Ω_2^2-Ω_u Ω_l}{Ω_2 (Ω_u-Ω_l)}\).

9. If A=\(\frac{Ω_1 (Ω_u-Ω_l)}{-Ω_1^2+Ω_u Ω_l}\) and B=\(\frac{Ω_2 (Ω_u-Ω_l)}{Ω_2^2-Ω_u Ω_l}\), then which of the following is the backward design equation for a low pass-to-band stop transformation?
a) ΩS=Max{|A|,|B|}
b) ΩS=Min{|A|,|B|}
c) ΩS=|B|
d) ΩS=|A|
View Answer

Answer: b
Explanation: If Ωu and Ωl are the upper and lower cutoff pass band frequencies of the desired band stop filter and Ω1 and Ω2 are the lower and upper cutoff stop band frequencies of the desired band stop filter, then the backward design equation is
ΩS=Min{|A|,|B|}
where, A=\(\frac{Ω_1 (Ω_u-Ω_l)}{-Ω_1^2+Ω_u Ω_l}\) and B=\(\frac{Ω_2 (Ω_u-Ω_l)}{Ω_2^2-Ω_u Ω_l}\).
advertisement

10. Which of the following is a low pass-to-high pass transformation?
a) s → s / Ωu
b) s → Ωu/s
c) s → Ωu.s
d) none of the mentioned
View Answer

Answer: b
Explanation: The low pass-to-high pass transformation is simply achieved by replacing s by 1/s. If the desired high pass filter has the pass band edge frequency Ωu, then the transformation is
s → Ωu/s.

Sanfoundry Global Education & Learning Series – Digital Signal Processing.

To practice all areas of Digital Signal Processing, 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.