250+ TOP MCQs on First & Second Order High Pass Butterworth Filter and Higher Order Filters & Answers

Linear Integrated Circuit Multiple Choice Questions on “First & Second Order High Pass Butterworth Filter and Higher Order Filters”.

1. How is the higher order filters formed?
A. By increasing resistors and capacitors in low pass filter
B. By decreasing resistors and capacitors in low pass filter
C. By inter changing resistors and capacitors in low pass filter
D. All of the mentioned

Answer: C
Clarification: High pass filter are often formed by interchanging frequency determining resistors and capacitors in low pass filters. For example, a first order high pass filter is formed from a first order low pass filter by inter changing components Rand C.

2. In a first order high pass filter, frequencies higher than low cut-off frequencies are called
A. Stop band frequency
B. Pass band frequency
C. Centre band frequency
D. None of the mentioned

Answer: B
Clarification: Low cut-off frequency, f_L is 0.707 times the pass band gain voltage. Therefore, frequencies above f_L are pass band frequencies.

3. Compute the voltage gain for the following circuit with input frequency 1.5kHz.
A. 4dB
B. 15dB
C. 6dB
D. 12dB

Answer: D
Clarification: |V_O/V_in|= [A_F×(f/f_L)]/ [√1+(f/f_L)²] = [4×(1.5kHz/225.86)] / √[1+(1.5kHz/225.86)²] =26.56/6.716=3.955 =20log(3.955)=11.9.
|V_O/V_in|≅12 dB
A_F= 1+(R_F /R₁)= 1+(12kΩ/4kΩ) =4.
f_L= 1/(2πRC. = 1/2π×15kΩ×0.047µF= 1/4.427×10^-3 =225.86Hz.

4. Determine the expression for output voltage of first order high pass filter?
A. V_O = [1+(R_F /R₁)]× [(j2πfRC/(1+j2πfRC.] × V_in
B. V_O = [-(R_F /R₁)]× [(j2πfRC/(1+j2πfRC.] × V_in
C. V_O = {[1+(R_F /R₁)]× /[1+j2πfRC] }× V_in
D. None of the mentioned

Answer: A
Clarification: The first order high pass filter uses non-inverting amplifier. So, A_F= 1+(R_F /R₁) and the output voltage, V_O = [1+(R_F /R₁)]× [(j2πfRC/(1+j2πfRC.]× V_in.

5. The internal resistor of the second order high pass filter is equal to 10kΩ. Find the value of feedback resistor?
A. 6.9kΩ
B. 5.86kΩ
C. 10kΩ
D. 12.56kΩ

Answer: B
Clarification: Pass band gain for second order butterworth response, A_F =1.586.
=> A_F= [1+(R_F/R₁)]
=> R_F= (A_F-1)×R₁ =(1.586-1)×10kΩ =5860 =5.86kΩ.

6. Consider the following circuit and calculate the low cut-off frequency value?
A. 178.7Hz
B. 89.3Hz
C. 127.65Hz
D. 255.38Hz

Answer: A
Clarification: The low cut-off frequency for the given filter is f_L =1/√[2π√(R₂×R₃×C₂×C₃)]=178.7Hz.

7. Determine voltage gain of second order high pass butterworth filter.
Specifications R₃ =R₂=33Ω, f=250hz and f_L=1khz.
A. -11.78dB
B. -26.51dB
C. -44.19dB
D. None of the mentioned

Answer: C
Clarification: Since R₃ =R₂
=> C₂ = 1/(2π ×f_L×R₂) = 1/(2π ×1kHz×33Ω)
=> C₃ =C₂= 4.82µF.
Voltage gain of filter |V_O/V_in|=A_F / [√ 1+(f_L/f)⁴] = 1.586/[1+(1kHz/250kz)⁴] =1.586/252=6.17×10^-3 =20log(6.17×10^-3)= -44.19dB.

8. From the given specifications, determine the value of voltage gain magnitude of first order and second order high pass butterworth filter?
Pass band voltage gain=2;
Low cut-off frequency= 1kHz;
Input frequency=500Hz.
A. First order high pass filter =-4.22dB , Second order high pass filter=-0.011dB
B. First order high pass filter =-0.9688dB , Second order high pass filter=-6.28dB
C. First order high pass filter =-11.3194dB , Second order high pass filter=-9.3257dB
D. First order high pass filter =-7.511dB , Second order high pass filter=-5.8999dB

9. How is the higher order filters formed?
A. Using first order filter
B. Using second order filter
C. Connecting first and second order filter in series
D. Connecting first and second order filter in parallel

Answer: C
Clarification: Higher filters are formed by using the first and second order filters. For example, a third order low pass filter is formed by cascading first and second order low pass filter.

10. State the disadvantage of using higher order filters?
A. Complexity
B. Requires more space
C. Expensive
D. All of the mentioned

Answer: D
Clarification: Although higher order filter than necessary gives a better stop band response, the higher order type is more complex, occupies more space and is more expensive.

11. The overall gain of higher order filter is
A. Varying
B. Fixed
C. Random
D. None of the mentioned

Answer: B
Clarification: The overall gain of higher order filter is fixed because all the frequency determining resistor and capacitors are equal.

12. Find the roll-off rate for 8^th order filter
A. -160dB/decade
B. -320dB/decade
C. -480dB/decade
D. -200dB/decade

Answer: A
Clarification: For n^th order filter the roll-off rate will be -n×20dB/decade.
=>∴ for 8^th order filter= 8×20=160dB/decade.