Low Pass Butterworth Filter Design MCQ Quiz – Objective Question with Answer for Low Pass Butterworth Filter Design

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

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 \(log⁡ [1+(\frac{\Omega_P}{\Omega_C})^{2N}]\)

Answer: B

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 \(log⁡[1+(\frac{\Omega_S}{\Omega_C})^{2N}]\)

Answer: A

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 \(log⁡[1+(\frac{\Omega_S}{\Omega_C})^{2N}]\).

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

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

Answer: B

We know that,

KP=-10 \(log⁡[1+(\frac{\Omega_S}{\Omega_C})^{2N}]\)

=>\([\frac{Ω_S}{Ω_C}]^{2N} = 10^{\frac{-K_S}{10}}-1\)

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. \(\frac{log⁡[(10^\frac{-K_P}{10}-1)/(10^\frac{-K_s}{10}-1)]}{2 log⁡(\frac{\Omega_P}{\Omega_S})}\)

Answer: D

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 passband 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

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 stopband 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

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

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 passband gain KP=-1 dB at ΩP=4 rad/sec and stopband attenuation greater than or equal to 20dB at ΩS = 8 rad/sec?

A. 4
B. 5
C. 6
D. 3

Answer: B

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
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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