Linear Integrated Circuit Questions and Answers – First & Second Order High Pass Butterworth Filter and Higher Order Filters

This set of Linear Integrated Circuit Multiple Choice Questions & Answers (MCQs) focuses 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
View Answer

Answer: c
Explanation: 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
View Answer

Answer: b
Explanation: Low cut-off frequency, fL is 0.707 times the pass band gain voltage. Therefore, frequencies above fL are pass band frequencies.

3. Compute the voltage gain for the following circuit with input frequency 1.5kHz.
Compute the voltage gain for the following circuit with input frequency 1.5kHz
a) 4dB
b) 15dB
c) 6dB
d) 12dB
View Answer

Answer: d
Explanation: |VO/Vin|= [AF×(f/fL)]/ [√1+(f/fL)2] = [4×(1.5kHz/225.86)] / √[1+(1.5kHz/225.86)2] =26.56/6.716=3.955 =20log(3.955)=11.9.
|VO/Vin|≅12 dB
AF= 1+(RF /R1)= 1+(12kΩ/4kΩ) =4.
fL= 1/(2πRC) = 1/2π×15kΩ×0.047µF= 1/4.427×10-3 =225.86Hz.
advertisement
advertisement

4. Determine the expression for output voltage of first order high pass filter?
a) VO = [1+(RF /R1)]× [(j2πfRC/(1+j2πfRC)] × Vin
b) VO = [-(RF /R1)]× [(j2πfRC/(1+j2πfRC)] × Vin
c) VO = {[1+(RF /R1)]× /[1+j2πfRC] }× Vin
d) None of the mentioned
View Answer

Answer: a
Explanation: The first order high pass filter uses non-inverting amplifier. So, AF= 1+(RF /R1) and the output voltage, VO = [1+(RF /R1)]× [(j2πfRC/(1+j2πfRC)]× Vin.

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
View Answer

Answer: b
Explanation: Pass band gain for second order butterworth response, AF =1.586.
=> AF= [1+(RF/R1)] => RF= (AF-1)×R1 =(1.586-1)×10kΩ =5860 =5.86kΩ.

6. Consider the following circuit and calculate the low cut-off frequency value?
Calculate the low cut-off frequency value for the given circuit
a) 178.7Hz
b) 89.3Hz
c) 127.65Hz
d) 255.38Hz
View Answer

Answer: a
Explanation: The low cut-off frequency for the given filter is fL =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 fL=1khz.
a) -11.78dB
b) -26.51dB
c) -44.19dB
d) None of the mentioned
View Answer

Answer: c
Explanation: Since R3 =R2
=> C2 = 1/(2π ×fL×R2) = 1/(2π ×1kHz×33Ω)
=> C3 =C2= 4.82µF.
Voltage gain of filter |VO/Vin|=AF / [√ 1+(fL/f)4] = 1.586/[1+(1kHz/250kz)4] =1.586/252=6.17×10-3 =20log(6.17×10-3)= -44.19dB.
advertisement

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
View Answer

Answer: b
Explanation: For first order high pass filter,
|VO/Vin|=AF ×(f/fL) / [ √1+(f/fL)2] =(2×(500Hz/1kHz)) /√[1+(500Hz/1kHz)2] => |VO/Vin| = 1/1.118= 0.8944 =20log(0.8944) =-0.9686dB.
For second order high pass filter,
|VO/Vin|=AF / [ √ 1 +(fL/f)4] =2/√[1+ (1kHz/500Hz)2] =>|VO/Vin|=2/4.123= =0.4851 = 20log(0.4851) = -6.28dB.

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
View Answer

Answer: c
Explanation: 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.
advertisement

10. State the disadvantage of using higher order filters?
a) Complexity
b) Requires more space
c) Expensive
d) All of the mentioned
View Answer

Answer: d
Explanation: 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
View Answer

Answer: b
Explanation: 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 8th order filter
a) -160dB/decade
b) -320dB/decade
c) -480dB/decade
d) -200dB/decade
View Answer

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

Sanfoundry Global Education & Learning Series – Linear Integrated Circuits.

To practice all areas of Linear Integrated Circuits, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & discussions at Telegram SanfoundryClasses.