Chebyshev minimum order and TF (ωs = 2.33 rad/s)

For a Chebyshev low-pass filter, $|T(j\omega)|^2 = 1/[1+\varepsilon^2 C_n^2(\omega)]$ with $\varepsilon^2 = 10^{0.1\alpha_{max}}-1$, and the minimum order is

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

Minimum order

$$\begin{aligned} \varepsilon^2 &= 10^{0.05}-1 = 0.12202, \quad \varepsilon = 0.34931 \\ 10^{2.2}-1 &= 157.4893 \\ \sqrt{157.4893/0.12202} &= 35.9263, \quad \cosh^{-1}(35.9263) = 4.2744 \\ \cosh^{-1}(2.33) &= 1.4894 \\ n &\ge \frac{4.2744}{1.4894} = 2.870 \;\Rightarrow\; n = 3 \end{aligned}$$

Pole locations ($\omega_p = 1$ rad/s, so no denormalization)

$$\begin{aligned} a &= \frac{1}{n}\sinh^{-1}\frac{1}{\varepsilon} = \frac{1}{3}\sinh^{-1}(2.86278) = 0.59138 \\ \sinh a &= 0.62646, \quad \cosh a = 1.18002 \\ s_k &= -\sinh a\,\sin\theta_k + j\cosh a\,\cos\theta_k, \quad \theta_k = \frac{(2k-1)\pi}{6} \end{aligned}$$

Transfer function

The complex pair gives $s^2 + 2(0.31323)s + (0.31323^2 + 1.02193^2) = s^2 + 0.62646s + 1.14245$.

$$D(s) = (s+0.62646)(s^2+0.62646s+1.14245) = s^3 + 1.25291s^2 + 1.53490s + 0.71569$$

For odd $n$, $|T(0)| = 1$, so the numerator equals the constant term of $D(s)$:

$$T(s) = \frac{0.71569}{(s+0.62646)(s^2+0.62646s+1.14245)} = \frac{0.71569}{s^3 + 1.25291s^2 + 1.53490s + 0.71569}$$

Answer: minimum order $n = 3$, with $T(s)$ as above (passband edge 1 rad/s, 0.5 dB ripple).