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) 5I 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

What is the order of the normalized low pass Butterworth filter used t

1.	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) 5I 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» Correct CHOICE is (b) 3 The EXPLANATION: Given information is Ω1=2π20=125.663 rad/sec Ω2=2π4510^3=2.82710^5 rad/sec Ωu=2π2010^3=1.25710^5 rad/sec Ωl=2π50=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.

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