Chebyshev order (3.2 kHz, 0.4 dB; 9.8 kHz, 52 dB)

Derivation

The Chebyshev low-pass magnitude response (normalized so that the passband edge is $\Omega = \omega/\omega_p = 1$) is

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

where $C_n(\Omega)$ is the Chebyshev polynomial of order $n$:

$$C_n(\Omega) = \begin{cases} \cos(n\cos^{-1}\Omega), & |\Omega| \le 1 \\ \cosh(n\cosh^{-1}\Omega), & |\Omega| > 1 \end{cases}$$

The attenuation in dB is $\alpha(\Omega) = 10\log_{10}\left[1+\varepsilon^2 C_n^2(\Omega)\right]$.

Step 1: ripple factor from the passband edge. At $\Omega = 1$, $C_n(1) = 1$ for every $n$, so the attenuation is exactly $\alpha_{max}$:

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

Step 2: stopband condition. At $\Omega_s = \omega_s/\omega_p\ (>1)$ the attenuation must be at least $\alpha_{min}$:

$$10\log_{10}\left[1+\varepsilon^2 C_n^2(\Omega_s)\right] \ge \alpha_{min} \;\Rightarrow\; C_n^2(\Omega_s) \ge \frac{10^{0.1\alpha_{min}}-1}{\varepsilon^2}$$

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

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

The order is the next whole number above this value. (Useful identity: $\cosh^{-1}x = \ln\left(x+\sqrt{x^2-1}\right)$.)

Order for the given specification

Passband edge $f_p = 3.2$ kHz with $\alpha_{max} = 0.4$ dB; stopband edge $f_s = 9.8$ kHz with $\alpha_{min} = 52$ dB. Only the ratio matters, so Hz can be used directly: $\omega_s/\omega_p = 9.8/3.2 = 3.0625$.

$$\begin{aligned} \varepsilon^2 &= 10^{0.04} - 1 = 0.09648 \\ 10^{5.2} - 1 &= 158488.32 \\ \sqrt{\frac{158488.32}{0.09648}} &= 1281.69 \\ \cosh^{-1}(1281.69) &= 7.8491 \\ \cosh^{-1}(3.0625) &= 1.7846 \\ n &\ge \frac{7.8491}{1.7846} = 4.398 \end{aligned}$$

Answer: n = 5.