Chebyshev Filter

Question

This question is about the formula for Chebyshev Filter. (Low Pass Filter is considered throughout the question). According to Wikipedia, the formula for type-I Chebyshev Filter is given by: \begin{equation} |H_n(s)|^2 = \frac{1}{1+\varepsilon^2T_n^2(\frac{\Omega}{\Omega_c})} \end{equation} where, \$\Omega_c\$ is the cut-off frequency (not the pass-band frequency)

But according to [Proakis] the Type-I Chebyshev Filter transfer function is given by: \begin{equation} |H_n(s)|^2=\frac{1}{1+\varepsilon^2T_n^2(\frac{\Omega}{\Omega_p})} \end{equation} where, \$\Omega_p\$ is the pass-band frequecy. Taking an analogy with Butterworth Filter, its Transfer function is given by \begin{equation} |H(s)|^2=\frac{1}{1+\big(\frac{s}{j\Omega_c}\big)^{2N}} = \frac{1}{1+\varepsilon^2\big(\frac{s}{j\Omega_p}\big)^{2N}} \end{equation} So, here the ripple factor \$\varepsilon\$ appears only when the denominator is the pass-band frequency and also by definition, \begin{equation} \varepsilon=\sqrt{10^{0.1\alpha_{\text{max}}}-1} \end{equation}

So, according to this the second one must be correct for Chebyshev filter isn't it?

According to Science Direct, most of them use the second one too. But some other references such as [Proakis] use the first one.

Which one is correct? I really need a help with this. Thanks a lot in advance.

Reference: [Proakis] Proakis, J. G., & Manolakis, D. G. (2014). Digital signal processing (4th ed.). Upper Saddle River, NJ: Pearson.

Answer 1

John B., the cut-off frequency is defined as the end of the passband - hence, both are identical. But note that for Chebyshev responses the end of the passband is defined by the ripple width (epsilon) within the passband. Typical values are (0.1, 0.2, 0.5, 1.0) dB....

Furthermore, in your first formula the quantity T must appear to be be squared (Tn²).

Answer 2

The only difference between your first and second equations seems to be the choice of Ω_p and Ω_c. They are both the same frequency, just different terms for it, at which the filter is the ripple amplitude down, not 3dB down as it is for most other filter types. Perhaps that's the reason for the different names, people attempting to use something other than something that sounds like a 3dB frequency.

I don't have Proakis, but I do have A. B. Williams. He, your Science Direct link, and Wikipedia share a more significant difference to your equations. Their denominator is 1+ε²T_n²() rather than the 1+ε²T_n() you've shown. The notation is variously T() or C() for the Chebyshev polynomial, which needs to be squared.

Answer 3

Just a small addition to the already existent answers: while the passband is considered up to the end of the ripples, it doesn't mean that you can't have a custom frequency scaling. This is valid for all filters. It is possible to have a 1dB ripple and a 2dB cutoff, it's just a bit non-standard. In the case of Chebyshev I, you simply impose \$\epsilon_pT_n(\frac{\omega_{scale}}{\omega_p})=\epsilon_{scale}\$, and solve for \$\omega_{scale}\$, which gives:

$$\omega_{scale}=\omega_p\textrm{cosh}\left[{\frac{\textrm{acosh}\left(\frac{\epsilon_{scale}}{\epsilon_p}\right)}{N}}\right] \\ \epsilon_p=\sqrt{10^{\frac{A_p}{10}-1}} \\ \epsilon_{scale}=\sqrt{10^{\frac{A_{scale}}{10}}-1}$$

Note that this considers the scaling falling outside the passband, where \$T_n(x)=\textrm{cosh}(N\textrm{acosh}x)\$. While you could, in theory, scale to be inside the passband, it has no practical value. Scaling outside the passband is a valid choice when you are interested in, for example, an acceptable ripple in the passband (whatever that requirement happens to be), but you need half-power bandwidth, and you accept this trade-off for the benefits of a steeper rolloff that Butterworth doesn't have.