Inverse Chebyshev characteristics and order (0.4 dB, 16 dB)

Characteristics of the inverse Chebyshev response

• Maximally flat-like, monotonic passband: no ripple in the passband; $|T(0)| = 1$.
• Equiripple stopband: the response bounces between 0 and $1/\sqrt{1+1/\varepsilon^2}$ above $\omega_s$, so the minimum stopband attenuation is exactly $\alpha_{min}$.
• Finite transmission zeros on the $j\omega$ axis at $\omega_s/\cos\frac{(2k-1)\pi}{2n}$, so it is not an all-pole function (numerator has $s^2+\omega_z^2$ factors).
• Transition band as sharp as a Chebyshev filter of the same order (same order formula).
• Better phase and delay in the passband than Chebyshev (poles are farther from the $j\omega$ axis), so less ringing.
• For odd $n$ the response falls to zero at infinity at $-20$ dB/decade; for even $n$ it levels off at the stopband ripple value.

Order of the inverse Chebyshev low-pass filter

The inverse Chebyshev (Chebyshev Type II) response, normalized to the stopband edge ($\Omega = \omega/\omega_s$, stopband begins at $\Omega=1$), is

$$|T(j\Omega)|^2 = \frac{\varepsilon^2 C_n^2(1/\Omega)}{1+\varepsilon^2 C_n^2(1/\Omega)}$$

It is obtained by taking a Chebyshev response, subtracting it from 1 (turning it into a high-pass with equiripple stopband) and replacing $\Omega$ by $1/\Omega$.

Step 1: stopband edge. At $\Omega = 1$, $C_n(1) = 1$:

$$\alpha_{min} = 10\log_{10}\left(1+\frac{1}{\varepsilon^2}\right) \;\Rightarrow\; \varepsilon^2 = \frac{1}{10^{0.1\alpha_{min}}-1}$$

Step 2: passband edge. At $\Omega_p = \omega_p/\omega_s$, $1/\Omega_p = \omega_s/\omega_p > 1$, and the attenuation must not exceed $\alpha_{max}$:

$$\alpha_{max} \ge 10\log_{10}\left(1+\frac{1}{\varepsilon^2 C_n^2(\omega_s/\omega_p)}\right) \;\Rightarrow\; C_n^2\!\left(\frac{\omega_s}{\omega_p}\right) \ge \frac{1}{\varepsilon^2(10^{0.1\alpha_{max}}-1)} = \frac{10^{0.1\alpha_{min}}-1}{10^{0.1\alpha_{max}}-1}$$

Step 3: solve. With $C_n(x) = \cosh(n\cosh^{-1}x)$ for $x>1$:

$$n \ge \frac{\cosh^{-1}\sqrt{\dfrac{10^{0.1\alpha_{min}}-1}{10^{0.1\alpha_{max}}-1}}}{\cosh^{-1}\left(\dfrac{\omega_s}{\omega_p}\right)}$$

This is the same expression as for the Chebyshev filter: for the same specification both need the same order.

Order for the given specification

$\omega_p = 10000$ rad/s, $\omega_s = 20000$ rad/s, $\alpha_{max} = 0.4$ dB, $\alpha_{min} = 16$ dB.

$$\begin{aligned} 10^{0.04}-1 &= 0.09648 \\ 10^{1.6}-1 &= 38.8107 \\ \sqrt{38.8107/0.09648} &= 20.057, \quad \cosh^{-1}(20.057) = 3.6911 \\ \cosh^{-1}(2) &= 1.3170 \\ n &\ge \frac{3.6911}{1.3170} = 2.803 \end{aligned}$$

Answer: n = 3.