Chebyshev order, poles and TF (1 dB, 18 dB)

Derivation of the order

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

Given $\alpha_{max} = 1$ dB, $\alpha_{min} = 18$ dB, $\frac{\omega_s}{\omega_p} = \frac{1400}{1000} = 1.4$.

$$\begin{aligned} 10^{0.1\alpha_{max}} - 1 &= 10^{0.1} - 1 = 0.25893 \\ 10^{0.1\alpha_{min}} - 1 &= 10^{1.8} - 1 = 62.0957 \\ \sqrt{\frac{62.0957}{0.25893}} &= 15.4862 \\ \cosh^{-1}(15.4862) &= 3.4320 \\ \cosh^{-1}(1.4) &= 0.8670 \\ n &\ge \frac{3.4320}{0.8670} = 3.958 \end{aligned}$$

Answer: n = 4 (next integer above 3.958).

Pole locations

Poles of the normalized Chebyshev filter: $s_k = -\sinh a\sin\theta_k + j\cosh a\cos\theta_k$, with $\theta_k = (2k-1)\pi/2n$ and $a = \frac{1}{n}\sinh^{-1}(1/\varepsilon)$.

$$\begin{aligned} \varepsilon &= \sqrt{0.25893} = 0.50885, \quad 1/\varepsilon = 1.96523 \\ a &= \tfrac{1}{4}\sinh^{-1}(1.96523) = 0.35699 \\ \sinh a &= 0.36463, \quad \cosh a = 1.06440 \end{aligned}$$

Transfer function

Each conjugate pair gives $s^2 + 2|\sigma_k|s + (\sigma_k^2+\omega_k^2)$:

$$\begin{aligned} T_N(s) &= \frac{K}{(s^2+0.27907s+0.98650)(s^2+0.67374s+0.27940)} \\ &= \frac{K}{s^4 + 0.95281s^3 + 1.45393s^2 + 0.74262s + 0.27563} \end{aligned}$$

For even $n$, the DC gain is $1/\sqrt{1+\varepsilon^2}$ (passband ripple starts at the bottom), so

$$K = \frac{0.27563}{\sqrt{1.25893}} = 0.24565$$

Denormalizing with $s \to s/1000$:

$$T(s) = \frac{2.4565\times10^{11}}{(s^2+279.07s+986505)(s^2+673.74s+279398)}$$

Answer: $n = 4$; poles at $-139.54 \pm j983.38$ and $-336.87 \pm j407.33$ rad/s; $T(s)$ as above (DC gain $0.891$, i.e. $-1$ dB). If unity DC gain is preferred, use $K = 0.27563$ (i.e. $2.7563\times10^{11}$).