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Electronics and Communication Engineering (ECE) Exam  >  Electronics and Communication Engineering (ECE) Tests  >  Digital Signal Processing  >  Test: Butterworth Filters Design - 1 - Electronics and Communication Engineering (ECE) MCQ

Test: Butterworth Filters Design - 1 - Electronics and Communication Engineering (ECE) MCQ


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10 Questions MCQ Test Digital Signal Processing - Test: Butterworth Filters Design - 1

Test: Butterworth Filters Design - 1 for Electronics and Communication Engineering (ECE) 2024 is part of Digital Signal Processing preparation. The Test: Butterworth Filters Design - 1 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Test: Butterworth Filters Design - 1 MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Butterworth Filters Design - 1 below.
Solutions of Test: Butterworth Filters Design - 1 questions in English are available as part of our Digital Signal Processing for Electronics and Communication Engineering (ECE) & Test: Butterworth Filters Design - 1 solutions in Hindi for Digital Signal Processing course. Download more important topics, notes, lectures and mock test series for Electronics and Communication Engineering (ECE) Exam by signing up for free. Attempt Test: Butterworth Filters Design - 1 | 10 questions in 10 minutes | Mock test for Electronics and Communication Engineering (ECE) preparation | Free important questions MCQ to study Digital Signal Processing for Electronics and Communication Engineering (ECE) Exam | Download free PDF with solutions
Test: Butterworth Filters Design - 1 - Question 1
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 The cutoff frequency of the low pass Butterworth filter is the arithmetic mean of the two cutoff frequencies as found above.

Detailed Solution for Test: Butterworth Filters Design - 1 - Question 1

The arithmetic mean of the two cutoff frequencies as found above is the final cutoff frequency of the low pass Butterworth filter.

Test: Butterworth Filters Design - 1 - Question 2
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 Which of the following is a frequency domain specification?

Detailed Solution for Test: Butterworth Filters Design - 1 - Question 2
  • We are required to design a low pass Butterworth filter to meet the following frequency domain specifications.
  • KP ≤ 20 log|H(jΩ)| ≤ 0 and 20 log|H(jΩ)| ≤ KS.
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Test: Butterworth Filters Design - 1 - Question 3
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What is the value of gain at the pass band frequency, i.e., what is the value of KP?

Detailed Solution for Test: Butterworth Filters Design - 1 - Question 3

We know that the formula for gain is:

Test: Butterworth Filters Design - 1 - Question 4
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 What is the value of gain at the stop band frequency, i.e., what is the value of KS?
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We know that the formula for gain is:

Test: Butterworth Filters Design - 1 - Question 5
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 Which of the following equation is true?
Detailed Solution for Test: Butterworth Filters Design - 1 - Question 5

We know that,

Test: Butterworth Filters Design - 1 - Question 6
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 Which of the following equation is true? 
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We know that:

Test: Butterworth Filters Design - 1 - Question 7
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What is the order N of the low pass Butterworth filter in terms of KP and KS?
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Explanation:

Test: Butterworth Filters Design - 1 - Question 8
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What is the expression for cutoff frequency in terms of pass band gain?
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We know that:

Test: Butterworth Filters Design - 1 - Question 9
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 What is the expression for cutoff frequency in terms of stop band gain?
Detailed Solution for Test: Butterworth Filters Design - 1 - Question 9

We know that:

Test: Butterworth Filters Design - 1 - Question 10
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What is the lowest order of the Butterworth filter with a pass band gain KP= -1 dB at ΩP= 4 rad/sec and stop band attenuation greater than or equal to 20dB at ΩS= 8 rad/sec?
Detailed Solution for Test: Butterworth Filters Design - 1 - Question 10

We know that the equation for the order of the Butterworth filter is given as:

From the given question:
KP= -1 dB, ΩP= 4 rad/sec, KS= -20 dB and ΩS= 8 rad/sec
Upon substituting the values in the above equation, we get N = 4.289
Rounding off to the next largest integer, we get N=5.

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