Linear Integrated Circuit Questions and Answers – First Order Low Pass Butterworth Filter

This set of Linear Integrated Circuit Multiple Choice Questions & Answers (MCQs) focuses on “First Order Low Pass Butterworth Filter”.

1. Find the voltage across the capacitor in the given circuit
Find the voltage across the capacitor in the given circuit
a) VO= Vin/(1+0.0314jf)
b) VO= Vin×(1+0.0314jf)
c) VO= Vin+0.0314jf/(1+jf)
d) None of the mentioned
View Answer

Answer: a
Explanation: The voltage across the capacitor, VO= Vin/(1+j2πfRC)
=> VO= Vin/(1+j2π×5k×1µF×f)
=> VO= Vin/(1+0.0314jf).

2. Find the complex equation for the gain of the first order low pass butterworth filter as a function of frequency.
a) AF/[1+j(f/fH)].
b) AF/√ [1+j(f/fH)2].
c) AF×[1+j(f/fH)].
d) None of the mentioned
View Answer

Answer: a
Explanation: Gain of the filter, as a function of frequency is given as VO/ Vin=A F/(1+j(f/fH)).

3. Compute the pass band gain and high cut-off frequency for the first order high pass filter.
Compute the pass band gain and high cut-off frequency for the first order high pass filter
a) AF=11, fH=796.18Hz
b) AF=10, fH=796.18Hz
c) AF=2, fH=796.18Hz
d) AF=3, fH=796.18Hz
View Answer

Answer: c
Explanation: The pass band gain of the filter, AF =1+(RF/R1)
=>AF=1+(10kΩ/10kΩ)=2. The high cut-off frequency of the filter, fH=1/2πRC =1/(2π×20kΩ×0.01µF) =1/1.256×10-3 =796.18Hz.
advertisement
advertisement

4. Match the gain of the filter with the frequencies in the low pass filter

Frequency Gain of the filter
1. f < fH i. VO/Vin ≅ AF/√2
2. f=fH ii. VO/Vin ≤ AF
3. f>fH iii. VO/Vin ≅ AF

a)1-i,2-ii,3-iii
b)1-ii,2-iii,3-i
c)1-iii,2-ii,3-i
d)1-iii,2-i,3-ii
View Answer

Answer: d
Explanation: The mentioned answer can be obtained, if the value of frequencies are substituted in the gain magnitude equation |(Vo/Vin)|=AF/√(1+(f/fH)2).
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

5. Determine the gain of the first order low pass filter if the phase angle is 59.77o and the pass band gain is 7.
a) 3.5
b) 7
c) 12
d) 1.71
View Answer

Answer: a
Explanation: Given the phase angle, φ =-tan-1(f/fH)
=> f/fH=- φtan(φ) = -tan(59.77o)
=> f/fH= -1.716.
Substituting the above value in gain of the filter, |(VO/Vin)| = AF/√ (1+(f/fH)2) =7/√[1+(-1.716)2)] =7/1.986
=>|(VO/Vin)|=3.5.

6. In a low pass butterworth filter, the condition at which f=fH is called
a) Cut-off frequency
b) Break frequency
c) Corner frequency
d) All of the mentioned
View Answer

Answer: d
Explanation: The frequency, f=fH is called cut-off frequency, because the gain of the filter at this frequency is down by 3dB from 0Hz. Cut-off frequency is also called as break frequency, corner frequency or 3dB frequency.
advertisement

7. Find the High cut-off frequency if the pass band gain of a filter is 10.
a) 70.7Hz
b) 7.07kHz
c) 7.07Hz
d) 707Hz
View Answer

Answer: c
Explanation: High cut-off frequency of a filter, fH=0.707×AF =0.707×10
=>fH=7.07Hz.

8. To change the high cutoff frequency of a filter. It is multiplied by R or C by a ratio of original cut-off frequency known as
a) Gain scaling
b) Frequency scaling
c) Magnitude scaling
d) Phase scaling
View Answer

Answer: b
Explanation: Once a filter is designed, it may sometimes be a need to change it’s cut-off frequency. The procedure used to convert an original cut-off frequency fH to a new cut-off frequency is called frequency scaling.
advertisement

9. Using the frequency scaling technique, convert 10kHz cut-off frequency of the low pass filter to a cutoff frequency of 16kHz.(Take C=0.01µF and R=15.9kΩ)
a) 6.25kΩ
b) 9.94kΩ
c) 16kΩ
d) 1.59kΩ
View Answer

Answer: b
Explanation: To change a cut-off frequency from 10kHz to 16kHz,multiply 15.9kΩ resistor.
[Original cut-off frequency/New cut-off frequency] =10kHz/16kHz =0.625.
∴ R =0.625×15.9kΩ =9.94kΩ. However 9.94kΩ is not a standard value. So, a potentiometer of 10kΩ is taken and adjusted to 9.94kΩ.

10. Find the difference in gain magnitude for a filter ,if it is the response obtained for frequencies f1=200Hz and f2=3kHz. Specification: AF=2 and fH=1kHz.
a) 4.28 dB
b) 5.85 dB
c) 1.56 dB
d) None of the mentioned
View Answer

Answer: c
Explanation: When f1=200Hz, VO(1)/Vin =AF/√ [1+(f/fH)2] =2/√ [1+(200/1kHz) 2] =2/1.0198.
=> VO(1)/Vin =1.96
=>20log|(VO/Vin)|=5.85dB.
When f=700Hz, VO(2)/Vin= 2/√ [1+(700/1kHz) 2] =2/1.22=1.638.
=> VO(2)/Vin =20log|(VO/Vin|=20log(1.638) = 4.28.
Therefore, the difference in the gain magnitude is given as VO(1)/Vin-VO(2)/Vin =5.85-4.28 =1.56 dB.

11. Design a low pass filter at a cut-off frequency 1.6Hz with a pass band gain of 2.
a)
Find the cut-off frequency 1.6Hz from the given diagram
b)
Find the pass band gain of 2 from the given diagram
c)
Find the RF and R1 from the given diagram
d)
Find the difference in gain magnitude from the given diagram
View Answer

Answer: a
Explanation: From the answer, it is clear that all the C values are the same . Therefore, c= 0.01µF
Given, fH = 1kHz,
=> R= 1/(2πCfm) = 1/2π×0.01µF×1kHz
R= 9.9kΩ ≅ 10kΩ. Since the pass band gain is 2.
=> 2=1+ (RF/R1). Therefore, RF and R1 must be equal.

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.