Chebyshev characteristics and minimum order

The Chebyshev (Type I) response is an all-pole approximation that allows equal ripple in the passband to get a much steeper transition into the stopband.

Characteristics of the Chebyshev magnitude response

• Equiripple passband: $|T|$ oscillates between $1$ and $1/\sqrt{1+\varepsilon^2}$ in $0 \le \Omega \le 1$; the ripple height is $\alpha_{max}$ dB and is set by $\varepsilon$.
• Number of ripples: the passband contains $n$ half-cycles of ripple between $\Omega = 0$ and $\Omega = 1$, so a higher order gives more ripples (of the same height).
• DC value: $|T(0)| = 1$ for odd $n$ and $1/\sqrt{1+\varepsilon^2}$ for even $n$.
• Monotonic stopband: beyond $\Omega = 1$ the response falls monotonically; the asymptotic roll-off is $-20n$ dB/decade, but the transition is much sharper than Butterworth of the same order (about $6(n-1)$ dB more stopband attenuation for the same $\varepsilon$).
• All-pole function: no finite transmission zeros; poles lie on an ellipse in the s-plane.
• Phase/delay: phase is more non-linear and group delay has a larger peak near the band edge than Butterworth (poorer transient response, more ringing).
• For a given specification it needs a lower order than Butterworth, so fewer components.

 |T|
 1  |~\  /~\  /~\
    |  \/   \/   \   <- equal ripple (alpha_max)
    |             \
    |              \_
    |                \___  monotonic stopband
    +-----------------+-------> w
                     wp

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

Minimum order for the given specification

Given $\alpha_p = 1$ dB, $\alpha_s = 25$ dB, $\omega_s/\omega_p = 1.5$.

$$\begin{aligned} \varepsilon^2 &= 10^{0.1} - 1 = 0.25893 \\ 10^{2.5} - 1 &= 315.2278 \\ \sqrt{\frac{315.2278}{0.25893}} &= 34.8919 \\ \cosh^{-1}(34.8919) &= 4.2452 \\ \cosh^{-1}(1.5) &= 0.9624 \\ n &\ge \frac{4.2452}{0.9624} = 4.411 \end{aligned}$$

Answer: minimum order n = 5.