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}).

^{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. Frequency scaling is done using

a) Standard capacitor

b) Varying capacitor

c) Standard resistance

d) None of the mentioned

View Answer

Explanation: In frequency scaling standard capacitors are chosen, because for non standard value of resistor, a potentiometer is used.

_{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.

12. 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.

13. Arrange the series of step involved in designing a filter for first order low pass filter

Step 1: Select a value of C less than or equal to 1µF

Step 2: Choose a value of high cut-off frequency f_{H}

Step 3: Select a value of R_{1}C and R_{F} depending on the desired pass band gain

Step 4: Calculate the value of R

a) Steps- 2->4->3->1

b) Steps- 4->1->3->2

c) Steps- 2->1->4->3

d) Steps- 1->3->4->2

View Answer

Explanation: The mentioned option is the sequence of steps followed for designing a low pass filter.

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