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

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 Ω_{C}?

d) None of the mentioned

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Explanation: A Butterworth is characterized by the magnitude frequency response

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

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

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

Explanation: The magnitude frequency response of a Butterworth low pass filter is given as

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

Explanation: We know that the magnitude frequency response of a Butterworth filter of order N is given by the expression

In the above equation, if Ω→∞ then |H(jΩ)|→0.

6. |H(jΩ)| is a monotonically increasing function of frequency.

a) True

b) False

View Answer

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) 1/(1+Ω^{-2N})

b) 1+Ω^{-2N}

c) 1+Ω^{2N}

d) 1/(1+Ω^^{2N})

View Answer

Explanation: We know that the magnitude response of a low pass Butterworth filter of order N is given as

8. What is the transfer function of magnitude squared frequency response of the normalized low pass Butterworth filter?

View Answer

Explanation: We know that the magnitude squared frequency response of a normalized low pass Butterworth filter is given as

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

Explanation: The transfer function of normalized low pass Butterworth filter is given as

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) π/N k+π/2N k=0,1,2…N-1

b) π/N k+π/2N+π/2 k=0,1,2…2N-1

c) π/N k+π/2N+π/2 k=0,1,2…N-1

d) π/N k+π/2N k=0,1,2…2N-1

View Answer

11. What is the Butterworth polynomial of order 3?

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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)

12. What is the Butterworth polynomial of order 1?

a) s-1

b) s+1

c) s

d) None of the mentioned

View Answer

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((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) 1/(s^{2}+√2 s+1)

b) 1/(s^{2}-√2 s+1)

c) s^{2}-√2 s+1

d) s^{2}+√2 s+1

View Answer

Explanation: We know that the Butterworth polynomial of a 2nd order low pass filter is

B2(s)= s

^{2}+√2 s+1

Thus the transfer function is given as 1/(s

^{2}+√2 s+1).

**Sanfoundry Global Education & Learning Series – Digital Signal Processing.**

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