# Chebyshev Filter

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.

• Did you mean $\omega$ or $\Omega$? If you want the lower case Greek letter, you should use a lower case in the "code" too $\omega$ or to &omega; Mar 27, 2020 at 10:47
• $\Omega$ is used in [Proakis]. Hence I follow the same notation here. Mar 27, 2020 at 18:20
• John.....what do you mean with "first one" and "second one"?
– LvW
Mar 28, 2020 at 9:29
• John - just a question: When speaking of "cut-off frequency" (index "c") , do you mean "3dB frequency" - in contrast to "end of passband" (index "p") which can be defined differently?
– LvW
Mar 28, 2020 at 9:48
• Yes, right @LvW Mar 29, 2020 at 8:04

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).

• Thanks a lot @LvW. Yeah, sorry. I forgot to square $$T_n$$ Mar 27, 2020 at 18:19

The only difference between your first and second equations seems to be the choice of Ωp and Ωc. 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+ε2Tn2() rather than the 1+ε2Tn() you've shown. The notation is variously T() or C() for the Chebyshev polynomial, which needs to be squared.

• Thanks a lot @Neil_UK. Sorry, I forgot to square $$T_n$$ Mar 27, 2020 at 18:18
• And. $$\Omega_s = 2F_s\tan\Bigg(\frac{\omega_s}{2}\Bigg)$$. And hence, due to transfer function of Butterworth filter given above, $$\Omega_c = \Omega_s \big(\alpha_s-1\big)^{-\frac{1}{2N}}$$ and here, the prewarped pass-band frequency $$\Omega_p = 54.83135021$$ and the cut-off frequency (pre-warped) $$\Omega_c = 5072.373022$$ which are not the same though. And, $$\alpha_s = 10^{|A_s|/10}$$ Mar 27, 2020 at 18:43
• But look at this example (designing IIR filter using BLT method): Stop-band frequency $f_s: 2 \text{kHz}$; Pass-band frequency $f_p: 500 \text{Hz}$; Pass-band ripple $R_p: 3 \text{dB}$ and stop-band attenuation $A_s: 20 \text{dB}$ and sampling frequency $F_s: 8 \text{kHz}$. For this, the normalized frequencies will be: $$\omega_x = 2\pi\frac{f_x}{F_s}$$ $x = p \text{ or } s$. And hence the pre-warped frequencies will be: $$\Omega_p = 2F_s\tan\Bigg(\frac{\omega_p}{2}\Bigg)$$. [contd. in the below comment] Mar 27, 2020 at 18:51
• Due to more number of edits I had to rewrite an entire comment again. So, the order change. Sorry for this. The order for the example is 3rd comment and then the 2nd comment. BLT is Bilinear Transformation Method. Mar 27, 2020 at 18:56

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.

• Thanks a lot @a concerned citizen Mar 27, 2020 at 18:20