Question

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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Answer 1

Right answer is (b) 5

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

N=\(\frac{log⁡[(10^{-K_P/10}-1)/(10^{-K_s/10}-1)]}{2 log⁡(\frac{Ω_P}{Ω_S})}\)

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.