This set of Digital Signal Processing Multiple Choice Questions & Answers (MCQs) focuses on “Design of Low Pass Butterworth Filters-1”.

1. Which of the following is a frequency domain specification?

a) 0≥ 20 log|H(jΩ)|

b) 20 log|H(jΩ)| ≥ KP

c) 20 log|H(jΩ)| ≤ KS

d) All of the mentioned

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Explanation: 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

2. What is the value of gain at the pass band frequency, i.e., what is the value of KP?

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3. What is the value of gain at the stop band frequency, i.e., what is the value of KS?

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4. Which of the following equation is True?

d) None of the mentioned

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5. Which of the following equation is True?

d) None of the mentioned

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6. What is the order N of the low pass Butterworth filter in terms of KP and KS?

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7. What is the expression for cutoff frequency in terms of pass band gain?

d) None of the mentioned

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8. What is the expression for cutoff frequency in terms of stop band gain?

d) None of the mentioned

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

a) True

b) False

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

10. 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?

a) 4

b) 5

c) 6

d) 3

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Explanation: 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.

**Sanfoundry Global Education & Learning Series – Digital Signal Processing.**

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