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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?(a) 4(b) 5(c) 6(d) 3I had been asked this question in my homework.I'm obligated to ask this question of Design of Low Pass Butterworth Filters in portion Digital Filters Design of Digital Signal Processing

Answer»

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.



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