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QUESTION: 1

Which of the following is a frequency domain specification?

Solution:

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.

QUESTION: 2

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

Solution:

Explanation: We know that the formula for gain is

QUESTION: 3

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

Solution:

Explanation: We know that the formula for gain is

QUESTION: 4

Which of the following equation is True?

Solution:

Explanation: We know that,

QUESTION: 5

Which of the following equation is True?

Solution:

Explanation: We know that,

QUESTION: 6

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

Solution:

Explanation:

QUESTION: 7

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

Solution:

Explanation: We know that,

QUESTION: 8

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

Solution:

Explanation: We know that,

QUESTION: 9

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

Solution:

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

QUESTION: 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?

Solution:

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.

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