250+ TOP MCQs on Butterworth Filters & Answers

Digital Signal Processing Multiple Choice Questions 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
Answer: D
Clarification: 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 Ω_C?
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
Answer: A
Clarification: 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 Ω_C 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
Answer: C
Clarification: 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=Ω_C, |H(jΩ_C)|=1/√2=1/√2(|H(jΩ)|)
Thus the filter magnitude at the cutoff frequency is 1/√2 times the dc gain.

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
Answer: B
Clarification: 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

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Answer: A
Clarification: 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.

6. |H(jΩ)| is a monotonically increasing function of frequency.
A. True
B. False
Answer: B
Clarification: |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+Ω^2N
D. (frac{1}{1+Ω^{2N}})
Answer: D
Clarification: 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, Ω_C =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}}).

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}})
Answer: A
Clarification: 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}}) => H_N(jΩ).H_N(-jΩ)=(frac{1}{1+Ω^{2N}})
Replacing jΩ by ‘s’ and hence Ω by s/j in the above equation, we get
H_N(s).H_N(-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

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Answer: C
Clarification: The transfer function of normalized low pass Butterworth filter is given as
H_N(s).H_N(-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
=> s_k=(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.

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
Answer: D
Clarification: 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
=> s_k=(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)
Answer: D
Clarification: 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, s_k=(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
Answer: B
Clarification: 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, s_k=(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)
Answer: A
Clarification: We know that the Butterworth polynomial of a 2^nd 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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