Digital Signal Processing Questions and Answers – Design of Low Pass Butterworth Filters – 1

This set of Digital Signal Processing Multiple Choice Questions & Answers (MCQs) focuses on “Design of Low Pass Butterworth Filters-1”.

1. Which of the following is a frequency domain specification?
a) 0 ≥ 20 log|H(jΩ)|
b) 20 log|H(jΩ)| ≥ KP
c) 20 log|H(jΩ)| ≤ KS
d) All of the mentioned
View Answer

Answer: d
Explanation: We are required to design a low pass Butterworth filter to meet the following frequency domain specifications.
KP ≤ 20 log|H(jΩ)| ≤ 0
and 20 log|H(jΩ)| ≤ KS.

2. What is the value of gain at the pass band frequency, i.e., what is the value of KP?
a) -10 \(log⁡ [1-(\frac{\Omega_P}{\Omega_C})^{2N}]\)
b) -10 \(log⁡ [1+(\frac{\Omega_P}{\Omega_C})^{2N}]\)
c) 10 \(log⁡ [1-(\frac{\Omega_P}{\Omega_C})^{2N}]\)
d) 10 \(log⁡ [1+(\frac{\Omega_P}{\Omega_C})^{2N}]\)
View Answer

Answer: b
Explanation: We know that the formula for gain is K = 20 log|H(jΩ)|
We know that
\(|H(j\Omega)|=\frac{1}{\sqrt{(1+(\frac{\Omega}{\Omega_C})^{2N}}}\)
By applying 20log on both sides of above equation, we get
K = 20 \(log|H(j \Omega)|=-20 [log⁡[1+(\frac{\Omega}{\Omega_C})^{2N}]]^{1/2}\)
= -10 \(log⁡[1+(\frac{\Omega}{\Omega_C})^{2N}]\)
We know that K= KP at Ω=ΩP
=> KP=-10 \(log⁡[1+(\frac{\Omega_P}{\Omega_C})^{2N}]\).

3. What is the value of gain at the stop band frequency, i.e., what is the value of KS?
a) -10 \(log⁡[1+(\frac{\Omega_S}{\Omega_C})^{2N}]\)
b) -10 \(log⁡[1-(\frac{\Omega_S}{\Omega_C})^{2N}]\)
c) 10 \(log⁡[1-(\frac{\Omega_S}{\Omega_C})^{2N}]\)
d) 10 \(log⁡[1+(\frac{\Omega_S}{\Omega_C})^{2N}]\)
View Answer

Answer: a
Explanation: We know that the formula for gain is
K = 20 log|H(jΩ)|
We know that
\(|H(j \Omega)|=\frac{1}{\sqrt{(1+(\frac{\Omega}{\Omega_C})^{2N}}}\)
By applying 20log on both sides of above equation, we get
K = 20 \(log|H(j\Omega)|=-20 [log⁡[1+(\frac{\Omega}{\Omega_C})^{2N}]]^{1/2}\)
= -10 \(log⁡[1+(\frac{\Omega}{\Omega_C})^{2N}]\)
We know that K= KS at Ω=ΩS
=> KS=-10 \(log⁡[1+(\frac{\Omega_S}{\Omega_C})^{2N}]\).
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4. Which of the following equation is True?
a) \([\frac{\Omega_P}{\Omega_C}]^{2N} = 10^{-K_P/10}+1\)
b) \([\frac{\Omega_P}{\Omega_C}]^{2N} = 10^{K_P/10}+1\)
c) \([\frac{\Omega_P}{\Omega_C}]^{2N} = 10^{-K_P/10}-1\)
d) None of the mentioned
View Answer

Answer: c
Explanation: We know that,
KP=-10 \(log⁡[1+(\frac{\Omega_P}{\Omega_C})^{2N}]\)
=>\([\frac{Ω_P}{Ω_C}]^{2N} = 10^{\frac{-K_P}{10}}-1\)

5. Which of the following equation is True?
a) \([\frac{\Omega_S}{\Omega_C} ]^{2N} = 10^{-K_S/10}+1\)
b) \([\frac{\Omega_S}{\Omega_C} ]^{2N} = 10^{K_S/10}+1\)
c) \([\frac{\Omega_S}{\Omega_C} ]^{2N} = 10^{-K_S/10}-1\)
d) None of the mentioned
View Answer

Answer: b
Explanation: We know that,
KP=-10 \(log⁡[1+(\frac{\Omega_S}{\Omega_C})^{2N}]\)
=>\([\frac{Ω_S}{Ω_C}]^{2N} = 10^{\frac{-K_S}{10}}-1\)
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6. What is the order N of the low pass Butterworth filter in terms of KP and KS?
a) \(\frac{log⁡[(10^\frac{K_P}{10}-1)/(10^\frac{K_s}{10}-1)]}{2 log⁡(\frac{\Omega_P}{\Omega_S})}\)
b) \(\frac{log⁡[(10^\frac{K_P}{10}+1)/(10^\frac{K_s}{10}+1)]}{2 log⁡(\frac{\Omega_P}{\Omega_S})}\)
c) \(\frac{log⁡[(10^\frac{-K_P}{10}+1)/(10^\frac{-K_s}{10}+1)]}{2 log⁡(\frac{\Omega_P}{\Omega_S})}\)
d) \(\frac{log⁡[(10^\frac{-K_P}{10}-1)/(10^\frac{-K_s}{10}-1)]}{2 log⁡(\frac{\Omega_P}{\Omega_S})}\)
View Answer

Answer: d
Explanation:
We know that, \([\frac{Ω_P}{Ω_C}]^{2N} = 10^{-K_P/10}-1\) and \([\frac{Ω_P}{Ω_C}]^{2N} = 10^{-K_S/10}-1\).
By dividing the above two equations, we get
=> \([Ω_P/Ω_S]^{2N} = (10^{-K_S/10}-1)(10^{-K_P/10}-1)\)
By taking log in both sides, we get
=> N=\(\frac{log⁡[(10^\frac{-K_P}{10}-1)/(10^\frac{-K_s}{10}-1)]}{2 log⁡(\frac{\Omega_P}{\Omega_S})}\).

7. What is the expression for cutoff frequency in terms of pass band gain?
a) \(\frac{\Omega_P}{(10^{-K_P/10}-1)^{1/2N}}\)
b) \(\frac{\Omega_P}{(10^{-K_P/10}+1)^{1/2N}}\)
c) \(\frac{\Omega_P}{(10^{K_P/10}-1)^{1/2N}}\)
d) None of the mentioned
View Answer

Answer: a
Explanation: We know that,
\([\frac{Ω_P}{Ω_C}]^{2N} = 10^{-K_P/10}-1\)
=> \(Ω_C = \frac{Ω_P}{(10^{-K_P/10}-1)^{1/2N}}\).
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8. What is the expression for cutoff frequency in terms of stop band gain?
a) \(\frac{\Omega_S}{(10^{-K_S/10}-1)^{1/2N}}\)
b) \(\frac{\Omega_S}{(10^{-K_S/10}+1)^{1/2N}}\)
c) \(\frac{\Omega_S}{(10^{K_S/10}-1)^{1/2N}}\)
d) None of the mentioned
View Answer

Answer: c
Explanation: We know that,
\([\frac{Ω_S}{Ω_C}]^{2N} = 10^{-K_S/10}-1\)
=> \(Ω_C = \frac{Ω_S}{(10^{-K_S/10}-1)^{1/2N}}\).

9. The cutoff frequency of the low pass Butterworth filter is the arithmetic mean of the two cutoff frequencies as found above.
a) True
b) False
View Answer

Answer: a
Explanation: The arithmetic mean of the two cutoff frequencies as found above is the final cutoff frequency of the low pass Butterworth filter.
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10. What is the lowest order of the Butterworth filter with a pass band gain KP=-1 dB at ΩP=4 rad/sec and stop band attenuation greater than or equal to 20dB at ΩS = 8 rad/sec?
a) 4
b) 5
c) 6
d) 3
View Answer

Answer: b
Explanation: We know that the equation for the order of the Butterworth filter is given as
N=\(\frac{log⁡[(10^{-K_P/10}-1)/(10^{-K_s/10}-1)]}{2 log⁡(\frac{Ω_P}{Ω_S})}\)
From the given question,
KP=-1 dB, ΩP= 4 rad/sec, KS=-20 dB and ΩS= 8 rad/sec
Upon substituting the values in the above equation, we get
N=4.289
Rounding off to the next largest integer, we get N=5.

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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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