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. V_O= V_in/(1+0.0314jf)
B. V_O= V_in×(1+0.0314jf)
C. V_O= V_in+0.0314jf/(1+jf)
D. None of the mentioned

Answer: A

The voltage across the capacitor,

V_O= V_in/(1+j2πfR_C)

=> V_O= V_in/(1+j2π×5k×1µF×f)

=> V_O= V_in/(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. A_F/[1+j(f/f_H)].
B. A_F/√ [1+j(f/f_H)²].
C. A_F×[1+j(f/f_H)].
D. None of the mentioned

Answer: A

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

V_O/ V_in=A _F/(1+j(f/f_H)).

3. Compute the passband gain and high cut-off frequency for the first order high pass filter.

A. A_F=11, f_H=796.18Hz
B. A_F=10, f_H=796.18Hz
C. A_F=2, f_H=796.18Hz
D. A_F=3, f_H=796.18Hz

Answer: C

The pass band gain of the filter,

A_F =1+(R_F/R₁)

=>A_F=1+(10kΩ/10kΩ)=2.

The high cut-off frequency of the filter

f_H=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. V_O/V_in ≅ A_F/√2
2. f=f_H	ii. V_O/V_in ≤ A_F
3. f>f_H	iii. V_O/V_in ≅ A_F

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

Answer: D

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

|(V_o/V_in)|=A_F/√(1+(f/f_H)²).

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

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

Answer: A

Given the phase angle

φ =-tan^-1(f/f_H)

=> f/f_H=- φtan(φ)

= -tan(59.77^o)

=> f/f_H= -1.716.

Substituting the above value in gain of the filter

|(V_O/V_in)|

= A_F/√ (1+(f/f_H)²)

=7/√[1+(-1.716)²)] =7/1.986

=>|(V_O/V_in)|=3.5.

6. In a low pass Butterworth filter, the condition at which f=f_H is called

A. Cut-off frequency
B. Break frequency
C. Corner frequency
D. All of the mentioned

Answer: D

The frequency, f=f_H 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

Answer: C

High cut-off frequency of a filter

f_H= 0.707×A_F

= 0.707×10

=>f_H=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

Answer: B

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 f_H 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

Answer: B

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: A_F=2 and f_H=1kHz.

A. 4.28 dB
B. 5.85 dB
C. 1.56 dB
D. None of the mentioned

Answer: C

When f₁=200Hz, V_O(1)/V_in =A_F/√ [1+(f/f_H)²]

=2/√ [1+(200/1kHz) ²] =2/1.0198.

=> V_O(1)/V_in =1.96

=>20log|(V_O/V_in)|=5.85dB.

When f=700Hz

V_O(2)/V_in= 2/√ [1+(700/1kHz) ²]

=2/1.22=1.638.

=> V_O(2)/V_in =20log|(V_O/V_in|

=20log(1.638) = 4.28.

Therefore, the difference in the gain magnitude is given as

V_O(1)/V_in-V_O(2)/V_in

=5.85-4.28 =1.56 dB.

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