Inverse Chebyshev order (0.25 dB, 18 dB, 1400 rad/s)

Order of an inverse Chebyshev low-pass filter

The inverse Chebyshev response is flat (monotonic) in the passband and equiripple in the stopband. Normalized to the stopband edge $\omega_s$ = 1 rad/s:

$$|T(j\omega)|^2 = \frac{\varepsilon^2 C_n^2(1/\omega)}{1 + \varepsilon^2 C_n^2(1/\omega)}, \qquad \alpha(\omega) = 10\log_{10}\left[1 + \frac{1}{\varepsilon^2 C_n^2(1/\omega)}\right]$$

Stopband ($\omega \ge 1$): $1/\omega \le 1$, so $C_n^2(1/\omega)$ swings between 0 and 1. The least attenuation (where $C_n^2 = 1$) must be $\alpha_{min}$:

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

Passband: the normalized passband edge is $\omega_p/\omega_s < 1$, where the loss must not exceed $\alpha_{max}$:

$$\begin{aligned} 10\log_{10}\left[1 + \frac{1}{\varepsilon^2 C_n^2(\omega_s/\omega_p)}\right] &\le \alpha_{max} \\ C_n^2\left(\frac{\omega_s}{\omega_p}\right) &\ge \frac{1}{\varepsilon^2(10^{0.1\alpha_{max}} - 1)} = \frac{10^{0.1\alpha_{min}} - 1}{10^{0.1\alpha_{max}} - 1} \end{aligned}$$

Since $\omega_s/\omega_p > 1$, $C_n = \cosh(n\cosh^{-1}x)$, so

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

This is the same expression as for the Chebyshev filter.

Order for the given specifications

$\alpha_{max}$ = 0.25 dB, $\alpha_{min}$ = 18 dB, $\omega_p$ = 1000 rad/s, $\omega_s$ = 1400 rad/s.

$$\begin{aligned} 10^{0.1\alpha_{min}} - 1 &= 10^{1.8} - 1 = 62.0957 \\ 10^{0.1\alpha_{max}} - 1 &= 10^{0.025} - 1 = 0.05925 \\ \sqrt{62.0957/0.05925} &= 32.3723, \qquad \frac{\omega_s}{\omega_p} = \frac{1400}{1000} = 1.4 \\ n &\ge \frac{\cosh^{-1}(32.3723)}{\cosh^{-1}(1.4)} = \frac{4.1702}{0.8670} = 4.810 \end{aligned}$$

So n = 5.

Answer: a 5th-order inverse Chebyshev filter is required, with $\varepsilon^2 = 1/62.096 = 0.01610$.