250+ TOP MCQs on Design of Low Pass Butterworth Filters – 1 & Answers

Digital Signal Processing Multiple Choice Questions 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
Answer: D
Clarification: We are required to design a low pass Butterworth filter to meet the following frequency domain specifications.
K_P ≤ 20 log|H(jΩ)| ≤ 0
and 20 log|H(jΩ)| ≤ K_S.

2. What is the value of gain at the pass band frequency, i.e., what is the value of K_P?
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}])
Answer: B
Clarification: We know that the formula for gain is K = 20 log|H(jΩ)|
We know that
(|H(jOmegA.|=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= K_P at Ω=Ω_P
=> K_P=-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 K_S?
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}])
Answer: A
Clarification: 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(jOmegA.|=-20 [log⁡[1+(frac{Omega}{Omega_C})^{2N}]]^{1/2})
= -10 (log⁡[1+(frac{Omega}{Omega_C})^{2N}])
We know that K= K_S at Ω=ΩS
=> K_S=-10 (log⁡[1+(frac{Omega_S}{Omega_C})^{2N}]).

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
Answer: C
Clarification: We know that,
K_P=-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
Answer: B
Clarification: We know that,
K_P=-10 (log⁡[1+(frac{Omega_S}{Omega_C})^{2N}])
=>([frac{Ω_S}{Ω_C}]^{2N} = 10^{frac{-K_S}{10}}-1)

6. What is the order N of the low pass Butterworth filter in terms of K_P and K_S?
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})})
Answer: D
Clarification:
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
Answer: A
Clarification: 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}}).

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
Answer: C
Clarification: 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
Answer: A
Clarification: The arithmetic mean of the two cutoff frequencies as found above is the final cutoff frequency of the low pass Butterworth filter.

10. What is the lowest order of the Butterworth filter with a pass band gain K_P=-1 dB at Ω_P=4 rad/sec and stop band attenuation greater than or equal to 20dB at Ω_S = 8 rad/sec?
A. 4
B. 5
C. 6
D. 3