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

Question

TuteeHUB · Accepted Answer

NA

TuteeHUB · Answer

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

TuteeHUB · Answer

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

TuteeHUB · Answer

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

TuteeHUB · Answer

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

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

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