Digital Signal Processing Questions and Answers – Butterworth Filters

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

1. Which of the following is true in the case of Butterworth filters?
a) Smooth pass band
b) Wide transition band
c) Not so smooth stop band
d) All of the mentioned
View Answer

Answer: d
Explanation: Butterworth filters have a very smooth pass band, which we pay for with a relatively wide transmission region.

2. What is the magnitude frequency response of a Butterworth filter of order N and cutoff frequency ΩC?
a) \(\frac{1}{\sqrt{1+(\frac{Ω}{Ω_C})^{2N}}}\)
b) \(1+(\frac{Ω}{Ω_C})^{2N}\)
c) \(\sqrt{1+(\frac{Ω}{Ω_C})^{2N}}\)
d) None of the mentioned
View Answer

Answer: a
Explanation: A Butterworth is characterized by the magnitude frequency response
|H(jΩ)|=\(\frac{1}{\sqrt{1+(\frac{Ω}{Ω_C})^{2N}}}\)
where N is the order of the filter and ΩC is defined as the cutoff frequency.

3. What is the factor to be multiplied to the dc gain of the filter to obtain filter magnitude at cutoff frequency?
a) 1
b) √2
c) 1/√2
d) 1/2
View Answer

Answer: c
Explanation: The dc gain of the filter is the filter magnitude at Ω=0.
We know that the filter magnitude is given by the equation
|H(jΩ)|=\(\frac{1}{\sqrt{1+(\frac{Ω}{Ω_C})^{2N}}}\)
At Ω=ΩC, |H(jΩC)|=1/√2=1/√2(|H(jΩ)|)
Thus the filter magnitude at the cutoff frequency is 1/√2 times the dc gain.
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4. What is the value of magnitude frequency response of a Butterworth low pass filter at Ω=0?
a) 0
b) 1
c) 1/√2
d) None of the mentioned
View Answer

Answer: b
Explanation: The magnitude frequency response of a Butterworth low pass filter is given as
|H(jΩ)|=\(\frac{1}{\sqrt{1+(\frac{Ω}{Ω_C})^{2N}}}\)
At Ω=0 => |H(jΩ)|=1 for all N.

5. As the value of the frequency Ω tends to ∞, then |H(jΩ)| tends to ____________
a) 0
b) 1
c) ∞
d) None of the mentioned
View Answer

Answer: a
Explanation: We know that the magnitude frequency response of a Butterworth filter of order N is given by the expression
|H(jΩ)|=\(\frac{1}{\sqrt{1+(\frac{Ω}{Ω_C})^{2N}}}\)
In the above equation, if Ω→∞ then |H(jΩ)|→0.
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6. |H(jΩ)| is a monotonically increasing function of frequency.
a) True
b) False
View Answer

Answer: b
Explanation: |H(jΩ)| is a monotonically decreasing function of frequency, i.e., |H(jΩ2)| < |H(jΩ1)| for any values of Ω1 and Ω2 such that 0 ≤ Ω1 < Ω2.

7. What is the magnitude squared response of the normalized low pass Butterworth filter?
a) \(\frac{1}{1+Ω^{-2N}}\)
b) 1+Ω-2N
c) 1+Ω2N
d) \(\frac{1}{1+Ω^{2N}}\)
View Answer

Answer: d
Explanation: We know that the magnitude response of a low pass Butterworth filter of order N is given as
|H(jΩ)|=\(\frac{1}{\sqrt{1+(\frac{Ω}{Ω_C})^{2N}}}\)
For a normalized filter, ΩC =1
=> |H(jΩ)|=\(\frac{1}{\sqrt{1+(Ω)^{2N}}}\) => |H(jΩ)|2=\(\frac{1}{1+Ω^{2N}}\)
Thus the magnitude squared response of the normalized low pass Butterworth filter of order N is given by the equation,
|H(jΩ)|2=\(\frac{1}{1+Ω^{2N}}\).
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8. What is the transfer function of magnitude squared frequency response of the normalized low pass Butterworth filter?
a) \(\frac{1}{1+(s/j)^{2N}}\)
b) \(1+(\frac{s}{j})^{-2N}\)
c) \(1+(\frac{s}{j})^{2N}\)
d) \(\frac{1}{1+(s/j)^{-2N}}\)
View Answer

Answer: a
Explanation: We know that the magnitude squared frequency response of a normalized low pass Butterworth filter is given as
H(jΩ)|2=\(\frac{1}{1+Ω^{2N}}\) => HN(jΩ).HN(-jΩ)=\(\frac{1}{1+Ω^{2N}}\)
Replacing jΩ by ‘s’ and hence Ω by s/j in the above equation, we get
HN(s).HN(-s)=\(\frac{1}{1+(\frac{s}{j})^{2N}}\) which is called the transfer function.

9. Where does the poles of the transfer function of normalized low pass Butterworth filter exists?
a) Inside unit circle
b) Outside unit circle
c) On unit circle
d) None of the mentioned
View Answer

Answer: c
Explanation: The transfer function of normalized low pass Butterworth filter is given as
HN(s).HN(-s)=\(\frac{1}{1+(\frac{s}{j})^{2N}}\)
The poles of the above equation is obtained by equating the denominator to zero.
=> \(1+(\frac{s}{j})^{2N}\)=0
=> s=(-1)1/2N.j
=> sk=\(e^{jπ(\frac{2k+1}{2N})} e^{jπ/2}\), k=0,1,2…2N-1
The poles are therefore on a circle with radius unity.
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10. What is the general formula that represent the phase of the poles of transfer function of normalized low pass Butterworth filter of order N?
a) \(\frac{π}{N} k+\frac{π}{2N}\) k=0,1,2…N-1
b) \(\frac{π}{N} k+\frac{π}{2N}+\frac{π}{2}\) k=0,1,2…2N-1
c) \(\frac{π}{N} k+\frac{π}{2N}+\frac{π}{2}\) k=0,1,2…N-1
d) \(\frac{π}{N} k+\frac{π}{2N}\) k=0,1,2…2N-1
View Answer

Answer: d
Explanation: The transfer function of normalized low pass Butterworth filter is given as
HN(s).HN(-s)=\(\frac{1}{1+(\frac{s}{j})^{2N}}\)
The poles of the above equation is obtained by equating the denominator to zero.
=> \(1+(\frac{s}{j})^{2N}\)=0
=> s=(-1)1/2N.j
=> sk=\(e^{jπ(\frac{2k+1}{2N})} e^{jπ/2}\), k=0,1,2…2N-1
The poles are therefore on a circle with radius unity and are placed at angles,
θk=\(\frac{π}{N} k+\frac{π}{2N}\) k=0,1,2…2N-1

11. What is the Butterworth polynomial of order 3?
a) (s2+s+1)(s-1)
b) (s2-s+1)(s-1)
c) (s2-s+1)(s+1)
d) (s2+s+1)(s+1)
View Answer

Answer: d
Explanation: Given that the order of the Butterworth low pass filter is 3.
Therefore, for N=3 Butterworth polynomial is given as B3(s)=(s-s0) (s-s1) (s-s2)
We know that, sk=\(e^{jπ(\frac{2k+1}{2N})} e^{jπ/2}\)
=> s0=(-1/2)+j(√3/2), s1= -1, s2=(-1/2)-j(√3/2)
=> B3(s)= (s2+s+1)(s+1).

12. What is the Butterworth polynomial of order 1?
a) s-1
b) s+1
c) s
d) none of the mentioned
View Answer

Answer: b
Explanation: Given that the order of the Butterworth low pass filter is 1.
Therefore, for N=1 Butterworth polynomial is given as B3(s)=(s-s0).
We know that, sk=\(e^{jπ(\frac{2k+1}{2N})} e^{jπ/2}\)
=> s0=-1
=> B1(s)=s-(-1)=s+1.

13. What is the transfer function of Butterworth low pass filter of order 2?
a) \(\frac{1}{s^2+\sqrt{2} s+1}\)
b) \(\frac{1}{s^2-\sqrt{2} s+1}\)
c) \(s^2-\sqrt{2} s+1\)
d) \(s^2+\sqrt{2} s+1\)
View Answer

Answer: a
Explanation: We know that the Butterworth polynomial of a 2nd order low pass filter is
B2(s)=s2+√2 s+1
Thus the transfer function is given as \(\frac{1}{s^2+\sqrt{2} s+1}\).

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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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