Chebyshev order (0.25 dB, 18 dB, 1650 rad/s)

Order of a Chebyshev low-pass filter

The Chebyshev response, normalized to $\omega_p$ = 1 rad/s, is

$$|T(j\omega)|^2 = \frac{1}{1 + \varepsilon^2 C_n^2(\omega)}, \qquad C_n(\omega) = \begin{cases} \cos(n\cos^{-1}\omega), & |\omega| \le 1 \\ \cosh(n\cosh^{-1}\omega), & |\omega| > 1 \end{cases}$$

so $\alpha(\omega) = 10\log_{10}[1 + \varepsilon^2 C_n^2(\omega)]$.

Passband: $C_n^2(\omega)$ oscillates between 0 and 1 for $|\omega| \le 1$, so the maximum passband loss (ripple) is

$$\alpha_{max} = 10\log_{10}(1 + \varepsilon^2) \;\Rightarrow\; \varepsilon^2 = 10^{0.1\alpha_{max}} - 1$$

Stopband: at $\omega_s$ (normalized $\omega_s/\omega_p > 1$):

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

rounded up to the next integer. ($\cosh^{-1}x = \ln(x + \sqrt{x^2-1})$.)

Order for the given specifications

$\alpha_{max}$ = 0.25 dB, $\alpha_{min}$ = 18 dB, $\omega_p$ = 1000 rad/s, $\omega_s$ = 1650 rad/s.

$$\begin{aligned} 10^{0.1\alpha_{min}} - 1 &= 10^{1.8} - 1 = 62.0957 \\ 10^{0.1\alpha_{max}} - 1 &= 10^{0.025} - 1 = 0.05925 \\ \sqrt{62.0957/0.05925} &= 32.3723, \qquad \frac{\omega_s}{\omega_p} = \frac{1650}{1000} = 1.65 \\ n &\ge \frac{\cosh^{-1}(32.3723)}{\cosh^{-1}(1.65)} = \frac{4.1702}{1.0860} = 3.840 \end{aligned}$$

So n = 4.

Check: with $n = 4$, attenuation at $\omega_s$ = $10\log_{10}[1 + 0.05925\cosh^2(4\cosh^{-1}1.65)] = 19.49$ dB $\ge 18$ dB.

Answer: a 4th-order Chebyshev filter. (A Butterworth filter would need $n = 7$ for the same specifications.)