In Chebyshev filter the critical freq occurs with gain Gp But in butterworth filter it is always at -3dB; why does this happen? Can anyone explain intuitively on this?


If I understood your question correctly, Butterworth filter is maximally flat (no ripples) in the passband frequencies, and at the cutoff frequency, the gain is down by 3dB (3.01 to be exact). But disadvantage is slow roll off from passband. See https://en.wikipedia.org/wiki/Butterworth_filter especially the figure showing magnitude roll off.

For Chebyshev filter type 1 filter, the roll off is more sharp but result is ripples in pass band. You can control this ripple, which will in turn vary the Gp that you mentioned. See https://en.wikipedia.org/wiki/Chebyshev_filter. If you specify $\delta$ as the ripple factor in dB scale (like 1dB), $\epsilon = \sqrt{10^{\delta /10}-1}$ in linear scale. The gain Gp = $\frac{1}{\sqrt{1 + \epsilon^2}}$ in linear scale. The sharpness of your passband and passband ripple specification will affect $\epsilon$ which will inturn affect Gp.

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  • $\begingroup$ Thank you for your response sir Actually my doubt is say if we are about to design a Band Pass Filter from Normalized LPF of a Butterworth filter We will be calculating the critical frequency and the order of the NLPF then convert the NLPF to a LPF with that critical frequency and then substitute s by (s^2+Wp1*Wp2)/(s*(Wp2-Wp1)). But this step is not required in Chebyshev design. In case of Chebyshev once we get the order of NLPF we directly substitue s by (s^2+Wp1*Wp2)/(s*(Wp2-Wp1)) without calculating Wc. Why is it so? $\endgroup$ – Sangeerth Prabakar Mar 31 at 3:55

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