Inverse Chebyshev characteristics, order, poles and zeros

Characteristics of the inverse Chebyshev response

• Maximally flat-like, monotonic passband: no ripple in the passband; $|T(0)| = 1$.
• Equiripple stopband: the response bounces between 0 and $1/\sqrt{1+1/\varepsilon^2}$ above $\omega_s$, so the minimum stopband attenuation is exactly $\alpha_{min}$.
• Finite transmission zeros on the $j\omega$ axis at $\omega_s/\cos\frac{(2k-1)\pi}{2n}$, so it is not an all-pole function (numerator has $s^2+\omega_z^2$ factors).
• Transition band as sharp as a Chebyshev filter of the same order (same order formula).
• Better phase and delay in the passband than Chebyshev (poles are farther from the $j\omega$ axis), so less ringing.
• For odd $n$ the response falls to zero at infinity at $-20$ dB/decade; for even $n$ it levels off at the stopband ripple value.

 |T|
 1 |----.__
   |       \       passband flat
   |        \
   |         \  .-.   .-.  equal ripple
   |          \/   \ /   \ stopband
   +-----------+----+-----+---> w
              ws   zeros

Order of the inverse Chebyshev low-pass filter

The inverse Chebyshev (Chebyshev Type II) response, normalized to the stopband edge ($\Omega = \omega/\omega_s$, stopband begins at $\Omega=1$), is

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

It is obtained by taking a Chebyshev response, subtracting it from 1 (turning it into a high-pass with equiripple stopband) and replacing $\Omega$ by $1/\Omega$.

Step 1: stopband edge. At $\Omega = 1$, $C_n(1) = 1$:

$$\alpha_{min} = 10\log_{10}\left(1+\frac{1}{\varepsilon^2}\right) \;\Rightarrow\; \varepsilon^2 = \frac{1}{10^{0.1\alpha_{min}}-1}$$

Step 2: passband edge. At $\Omega_p = \omega_p/\omega_s$, $1/\Omega_p = \omega_s/\omega_p > 1$, and the attenuation must not exceed $\alpha_{max}$:

$$\alpha_{max} \ge 10\log_{10}\left(1+\frac{1}{\varepsilon^2 C_n^2(\omega_s/\omega_p)}\right) \;\Rightarrow\; C_n^2\!\left(\frac{\omega_s}{\omega_p}\right) \ge \frac{1}{\varepsilon^2(10^{0.1\alpha_{max}}-1)} = \frac{10^{0.1\alpha_{min}}-1}{10^{0.1\alpha_{max}}-1}$$

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

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

This is the same expression as for the Chebyshev filter: for the same specification both need the same order.

Poles and zeros for the given specification

Order. $\alpha_{max} = 0.25$ dB, $\alpha_{min} = 18$ dB, $\omega_s/\omega_p = 1.4$:

$$\begin{aligned} 10^{0.025}-1 &= 0.05925, \quad 10^{1.8}-1 = 62.0957 \\ \sqrt{62.0957/0.05925} &= 32.372, \quad \cosh^{-1}(32.372) = 4.1702 \\ \cosh^{-1}(1.4) &= 0.8670 \\ n &\ge 4.1702/0.8670 = 4.81 \;\Rightarrow\; n = 5 \end{aligned}$$

Poles. Compute Chebyshev poles with the stopband $\varepsilon$, invert them, and multiply by $\omega_s$:

$$\begin{aligned} \varepsilon &= \frac{1}{\sqrt{62.0957}} = 0.12690, \quad \frac{1}{\varepsilon} = 7.88008 \\ a &= \tfrac{1}{5}\sinh^{-1}(7.88008) = 0.55230 \\ \sinh a &= 0.58081, \quad \cosh a = 1.15643 \\ p_k &= -\sinh a \sin\theta_k + j\cosh a\cos\theta_k, \quad \theta_k = \tfrac{(2k-1)\pi}{10} \\ s_k &= \omega_s / p_k \end{aligned}$$

Zeros. On the $j\omega$ axis at $\omega_{z,k} = \omega_s/\cos\theta_k$:

$$\begin{aligned} \omega_{z1} &= \frac{1400}{\cos 18^\circ} = \frac{1400}{0.95106} = 1472.05\ \text{rad/s} \\ \omega_{z2} &= \frac{1400}{\cos 54^\circ} = \frac{1400}{0.58779} = 2381.82\ \text{rad/s} \end{aligned}$$

($\theta = 90°$ gives a zero at infinity, since $n$ is odd.)

Answer: $n = 5$; poles at $-202.34 \pm j1239.90$, $-963.40 \pm j1393.66$ and $-2410.44$ rad/s; zeros at $\pm j1472.05$, $\pm j2381.82$ rad/s and one at $\infty$.