Inverse Chebyshev order (0.5 dB, 25 dB, 1800 rad/s)

Derivation

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

$\omega_p = 1000$ rad/s, $\omega_s = 1800$ rad/s, $\alpha_{max} = 0.5$ dB, $\alpha_{min} = 25$ dB.

$$\begin{aligned} 10^{0.05}-1 &= 0.12202 \\ 10^{2.5}-1 &= 315.2278 \\ \sqrt{315.2278/0.12202} &= 50.828, \quad \cosh^{-1}(50.828) = 4.6215 \\ \cosh^{-1}(1.8) &= 1.1929 \\ n &\ge \frac{4.6215}{1.1929} = 3.874 \end{aligned}$$

Answer: n = 4. The resulting filter has a flat passband up to 1000 rad/s (at most 0.5 dB loss) and at least 25 dB attenuation everywhere above 1800 rad/s, with transmission zeros at $1800/\cos 22.5^\circ = 1948.3$ and $1800/\cos 67.5^\circ = 4703.6$ rad/s.