Chebyshev order (0.25 dB, 20 dB, 1500 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, $\omega_p$ = 1000 rad/s, $\alpha_{min}$ = 20 dB, $\omega_s$ = 1500 rad/s.

$$\begin{aligned} 10^{0.1\alpha_{min}} - 1 &= 10^{2} - 1 = 99.0000 \\ 10^{0.1\alpha_{max}} - 1 &= 10^{0.025} - 1 = 0.05925 \\ \sqrt{99.0000/0.05925} &= 40.8752, \qquad \frac{\omega_s}{\omega_p} = \frac{1500}{1000} = 1.5 \\ n &\ge \frac{\cosh^{-1}(40.8752)}{\cosh^{-1}(1.5)} = \frac{4.4035}{0.9624} = 4.575 \end{aligned}$$

So n = 5.

Check: with $n = 5$, attenuation at $\omega_s$ = $10\log_{10}[1 + 0.05925\cosh^2(5\cosh^{-1}1.5)] = 23.52$ dB $\ge 20$ dB.

Answer: a 5th-order Chebyshev filter.