Question

What is the order of the normalized low pass Butterworth filter used to design a analog band pass filter with -3.0103dB upper and lower cutoff frequency of 50Hz and 20KHz and a stop band attenuation 20dB at 20Hz and 45KHz?

(a) 2

(b) 3

(c) 4

(d) 5

I got this question in a job interview.

My doubt stems from Design of Low Pass Butterworth Filters topic in chapter Digital Filters Design of Digital Signal Processing

Answer 1

Correct choice is (b) 3

The explanation: Given information is

Ω1=2π*20=125.663 rad/sec

Ω2=2π*45*10^3=2.827*10^5 rad/sec

Ωu=2π*20*10^3=1.257*10^5 rad/sec

Ωl=2π*50=314.159 rad/sec

We know that

A=\(\frac{-Ω_1^2+Ω_u Ω_l}{Ω_1 (Ω_u-Ω_l)}\) and B=\(\frac{Ω_2^2-Ω_u Ω_l}{Ω_2 (Ω_u-Ω_l)}\)

=> A=2.51 and B=2.25

Hence ΩS=Min{|A|,|B|}=> ΩS=2.25 rad/sec.

The order N of the normalized low pass Butterworth filter is computed as follows

N=\(\frac{log⁡[(10^{-K_P/10}-1)(10^{-K_s/10}-1)]}{2 log⁡(\frac{1}{Ω_S})}\)=2.83

Rounding off to the next large integer, we get, N=3.