250+ TOP MCQs on Second Order Low Pass Butterworth Filter & Answers

Linear Integrated Circuit Multiple Choice Questions on “Second Order Low Pass Butterworth Filter”.

1. How can a first order low pass filter can be converted into second order low pass filter
A. By adding LC network
B. By adding RC network
C. By adding RC || LC network
D. None of the mentioned

Answer: B
Clarification: The addition of RC network makes the stop band response having a 40dB/decade.

2. Consider the following specifications and calculate the high cut-off frequency for the circuit given?
A. 95Hz
B. 48Hz
C. 14Hz
D. 33Hz

Answer: D
Clarification: The high cut-off frequency, f_H= 1/[2 π√(R₂ ×R₃× C₂× C₃)]
= 1/[2π√(33kΩ×15kΩ×0.47µF×0.1µF)]= 1/[2π× 4.82×(10^-3)]= 33Hz.

3. Find the gain and phase angle of the second order low pass filter?
Where pass band gain of the filter is 5, frequency and the high cut-off frequency of the filter are 3000Hz and 1kHz.
A. None of the mentioned
B. Gain magnitude = -1.03dB , φ =63.32^o
C. Gain magnitude = -5.19dB , φ =71.56^o
D. Gain magnitude = -4.94dB , φ =90^o

Answer: C
Clarification: The gain of the second order low pass filter, [V_O /V_in] =A_F/ √ [1+(f/f_h)²] =5/ √[1+(3000/1000)⁴] =5/9.055 =0.55.
=> [V_O /V_in] = 20log(0.55) =-.519dB.
Phase angle of second order low pass filter is given as φ= tan^-1(f/f_H)
=> φ =71.56^o.

4. Design a second order low pass butterworth filter at a high cut-off frequency of 2.2kHz. Given R_F=20kΩ and capacitor 0.047µF.

Answer: C
Clarification: Given f_H=21.2kHz, C₂=C₃ =0.047µF
R₃ = 1/(2πf_HC₃ = 1/(2π×2.2kHz×0.047µF) =1/5.9032×10^-4 =1.69kΩ.
=>R₃ =R₂=1.67kΩ
Since, R_F=0.586R₁
=> R₁=R_F/0.586 =20kΩ/0.586
R₁ = 34.13kΩ.

5. A second order low pass filter is given an input frequency of 30kHz and produce a output having phase angle of 79^o. Determine the pass band gain of the filter?
A. 11 dB
B. 89.11 dB
C. 46.78 dB
D. None of the mentioned

Answer: C
Clarification: Phase angle of the filter, φ = tan^-1(f/f_H)
=> f_h =f×tan(φ) =30kHz × tan(79^{)= 154.34kHz.
Therefore, the pass band gain A_F = f_H/0.707 = 154.34kHz/0.707
A_F= 218.3 =20log(218.3)= 46.78dB.}

6. The pass band voltage gain of a second order low pass butterworth filter is
A. 1.586
B. 8.32
C. 0.586
D. 0.707

Answer: A
Clarification: Second order low pass filter has a pass band voltage gain equal to 1.586 because of equal resistor and capacitor values. This gain is necessary to guarantee butterworth response.

7. Arrange the series of step involved in designing a filter for first order low pass filter
Step 1: Select a value of C less than or equal to 1µF
Step 2: Choose a value of high cut-off frequency f_H
Step 3: Select a value of R₁C and R_F depending on the desired pass band gain
Step 4: Calculate the value of R
A. Steps- 2->4->3->1
B. Steps- 4->1->3->2
C. Steps- 2->1->4->3
D. Steps- 1->3->4->2

Answer: B
Clarification: The mentioned option is the sequence of steps followed for designing a low pass filter.

8. Frequency scaling is done using
A. Standard capacitor
B. Varying capacitor
C. Standard resistance
D. None of the mentioned

Answer: A
Clarification: In frequency scaling standard capacitors are chosen, because for non standard value of resistor, a potentiometer is used.