Question

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

(a) \(\frac{log⁡[(10^\frac{K_P}{10}-1)/(10^\frac{K_s}{10}-1)]}{2 log⁡(\frac{\Omega_P}{\Omega_S})}\)

(b) \(\frac{log⁡[(10^\frac{K_P}{10}+1)/(10^\frac{K_s}{10}+1)]}{2 log⁡(\frac{\Omega_P}{\Omega_S})}\)

(c) \(\frac{log⁡[(10^\frac{-K_P}{10}+1)/(10^\frac{-K_s}{10}+1)]}{2 log⁡(\frac{\Omega_P}{\Omega_S})}\)

(d) \(\frac{log⁡[(10^\frac{-K_P}{10}-1)/(10^\frac{-K_s}{10}-1)]}{2 log⁡(\frac{\Omega_P}{\Omega_S})}\)

This question was posed to me by my college director while I was bunking the class.

Question is taken from Design of Low Pass Butterworth Filters in portion Digital Filters Design of Digital Signal Processing

Answer 1

The correct choice is (d) \(\frac{log⁡[(10^\frac{-K_P}{10}-1)/(10^\frac{-K_s}{10}-1)]}{2 log⁡(\frac{\Omega_P}{\Omega_S})}\)

To explain I would say: We know that, \([\frac{Ω_P}{Ω_C}]^{2N} = 10^{-K_P/10}-1\) and \([\frac{Ω_P}{Ω_C}]^{2N} = 10^{-K_S/10}-1\).

By dividing the above two equations, we get

=> \([Ω_P/Ω_S]^{2N} = (10^{-K_S/10}-1)(10^{-K_P/10}-1)\)

By taking log in both sides, we get

=> N=\(\frac{log⁡[(10^\frac{-K_P}{10}-1)/(10^\frac{-K_s}{10}-1)]}{2 log⁡(\frac{\Omega_P}{\Omega_S})}\).