# First Order Low Pass Butterworth Filter MCQ [Free PDF] – Objective Question Answer for First Order Low Pass Butterworth Filter Quiz

1. 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

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

The gain of the filter, as a function of frequency, is given as

VO/ Vin=A F/(1+j(f/fH)).

3. Compute the passband 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

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.

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

 Frequency Gain of the filter 1. f H 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

The mentioned answer can be obtained if the value of frequencies is substituted in the gain magnitude equation

|(Vo/Vin)|=AF/√(1+(f/fH)2).

5. Determine the gain of the first order low pass filter if the phase angle is 59.77o and the passband gain is 7.

A. 3.5
B. 7
C. 12
D. 1.71

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

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

7. Find the High cut-off frequency if the passband gain of a filter is 10.

A. 70.7Hz
B. 7.07kHz
C. 7.07Hz
D. 707Hz

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 the original cut-off frequency known as

A. Gain scaling
B. Frequency scaling
C. Magnitude scaling
D. Phase scaling

Once a filter is designed, it may sometimes be a need to change its cut-off frequency. The procedure used to convert an original cut-off frequency fH to a new cut-off frequency is called frequency scaling.

9. Using the frequency scaling technique, convert the 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

To change a cut-off frequency from 10kHz to 16kHz, multiply the 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

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.

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