Chebyshev order and elliptical pole locus

Order of a Chebyshev low-pass filter

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: $\cosh^{-1}x = \ln\left(x+\sqrt{x^2-1}\right)$.)

Locus of the poles is an ellipse

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).

Now eliminate $\theta_k = (2k-1)\pi/2n$. From the pole expressions:

$$\frac{\sigma_k}{\sinh a} = -\sin\theta_k, \qquad \frac{\omega_k}{\cosh a} = \cos\theta_k$$

Squaring and adding, and using $\sin^2\theta_k + \cos^2\theta_k = 1$:

$$\frac{\sigma_k^2}{\sinh^2 a} + \frac{\omega_k^2}{\cosh^2 a} = 1$$

This is the equation of an ellipse centred at the origin of the s-plane. Since $\cosh a > \sinh a$:

• semi-major axis $= \cosh a$, along the $j\omega$ axis;
• semi-minor axis $= \sinh a$, along the $\sigma$ axis;
• foci at $\pm j1$ (because $\cosh^2 a - \sinh^2 a = 1$), i.e. at the passband edge.

            jw
      .-----+-----.   cosh a
     /  x   |   .  \
    |       |       |
 ---x-------+-------+--- sigma
    |  sinh a       |
     \  x   |   .  /
      '-----+-----'
  (only left-half poles x are used)

As $\varepsilon \to 0$ (smaller ripple), $a$ grows and $\sinh a \approx \cosh a$, so the ellipse approaches a circle. Relative to its size the response then approaches the Butterworth response.