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

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

View Answer

Explanation: 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

View Answer

Explanation: 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 pass band 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

View Answer

Explanation: The pass band gain of the filter, A

_{F}=1+(R

_{F}/R

_{1})

=>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 < 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

View Answer

Explanation: The mentioned answer can be obtained, if the value of frequencies are 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 pass band gain is 7.

a) 3.5

b) 7

c) 12

d) 1.71

View Answer

Explanation: 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

View Answer

Explanation: 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. Cut-off frequency is also called as break frequency, corner frequency or 3dB frequency.

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

Explanation: 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 original cut-off frequency known as

a) Gain scaling

b) Frequency scaling

c) Magnitude scaling

d) Phase scaling

View Answer

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 f

_{H}to a new cut-off frequency is called frequency scaling.

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

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 f_{1}=200Hz and f_{2}=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

View Answer

Explanation: When f

_{1}=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.

11. Design a low pass filter at a cut-off frequency 1.6Hz with a pass band gain of 2.

View Answer

Explanation: From the answer, it is clear that all the C values are the same . Therefore, c= 0.01µF

Given, f

_{H}= 1kHz,

=> R= 1/(2πCf

_{m}) = 1/2π×0.01µF×1kHz

R= 9.9kΩ ≅ 10kΩ. Since the pass band gain is 2.

=> 2=1+ (R

_{F}/R

_{1}). Therefore, R

_{F}and R

_{1}must be equal.

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