Chebyshev order and poles (0.5 dB, 15 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 $\alpha_{max} = 0.5$ dB, $\alpha_{min} = 15$ dB, $\frac{\omega_s}{\omega_p} = \frac{2000}{1000} = 2$.

$$\begin{aligned} 10^{0.1\alpha_{max}} - 1 &= 10^{0.05} - 1 = 0.12202 \\ 10^{0.1\alpha_{min}} - 1 &= 10^{1.5} - 1 = 30.6228 \\ \sqrt{\frac{30.6228}{0.12202}} &= 15.8420 \\ \cosh^{-1}(15.8420) &= 3.4548 \\ \cosh^{-1}(2) &= 1.3170 \\ n &\ge \frac{3.4548}{1.3170} = 2.623 \end{aligned}$$

Answer: n = 3 (next integer above 2.623).

Pole locations

The poles are the left-half-plane roots of $1+\varepsilon^2 C_n^2(s/j) = 0$, i.e. $C_n = \pm j/\varepsilon$. Putting $\cos^{-1}(s/j) = u + jv$:

$$\cos(nu)\cosh(nv) - j\sin(nu)\sinh(nv) = \pm \frac{j}{\varepsilon}$$

Equating real parts: $\cos(nu)=0 \Rightarrow u_k = \dfrac{(2k-1)\pi}{2n}$. Equating imaginary parts: $\sin(nu_k)=\pm1 \Rightarrow v = a = \dfrac{1}{n}\sinh^{-1}\dfrac{1}{\varepsilon}$. Then $s = j\cos(u_k + ja)$ gives (taking the left-half-plane roots)

$$s_k = \sigma_k + j\omega_k,\quad \sigma_k = -\sinh a\,\sin\frac{(2k-1)\pi}{2n},\quad \omega_k = \cosh a\,\cos\frac{(2k-1)\pi}{2n}$$

for $k = 1, 2, \dots, n$ (normalized to $\omega_p = 1$; multiply by $\omega_p$ to denormalize).

With $n = 3$ and $\varepsilon = \sqrt{0.12202} = 0.34931$:

$$\begin{aligned} a &= \frac{1}{3}\sinh^{-1}\left(\frac{1}{0.34931}\right) = \frac{1}{3}\sinh^{-1}(2.86278) = 0.59138 \\ \sinh a &= 0.62646, \qquad \cosh a = 1.18002 \end{aligned}$$

These three poles lie on an ellipse with semi-minor axis $\sinh a = 0.626$ (along $\sigma$) and semi-major axis $\cosh a = 1.180$ (along $j\omega$), scaled by 1000 rad/s. All of them are in the left half of the s-plane:

                    jw
                    |
   -313 + j1022  x  |
                    |
   -626 x ----------+----------> sigma
                    |
   -313 - j1022  x  |
                    |

Answer: $n = 3$; poles at $s = -626.46$ rad/s and $s = -313.23 \pm j1021.93$ rad/s. The corresponding (normalized) denominator is $(s+0.62646)(s^2+0.62646s+1.14245)$.