Chebyshev response and order (0.5 dB, 22 dB)

Chebyshev response

The Chebyshev approximation chooses

$$|T(j\Omega)|^2 = \frac{1}{1+\varepsilon^2 C_n^2(\Omega)},\qquad \Omega = \frac{\omega}{\omega_p}$$

where $C_n$ is the Chebyshev polynomial, defined by $C_n(\Omega) = \cos(n\cos^{-1}\Omega)$ for $|\Omega|\le1$ and $\cosh(n\cosh^{-1}\Omega)$ for $|\Omega|>1$. It obeys the recursion

$$C_0 = 1,\quad C_1 = \Omega,\quad C_{n+1}(\Omega) = 2\Omega C_n(\Omega) - C_{n-1}(\Omega)$$

so $C_2 = 2\Omega^2-1$, $C_3 = 4\Omega^3-3\Omega$, $C_4 = 8\Omega^4-8\Omega^2+1$.

Behaviour of the response:

• In the passband ($|\Omega|\le1$), $C_n$ swings between $-1$ and $+1$, so $C_n^2$ is between 0 and 1 and $|T|$ ripples between $1$ and $1/\sqrt{1+\varepsilon^2}$ (equal ripple).
• In the stopband ($|\Omega|>1$), $C_n$ grows like $2^{n-1}\Omega^n$, so $|T|$ falls monotonically and quickly.
• At $\Omega=0$: $|T|=1$ for odd $n$, $1/\sqrt{1+\varepsilon^2}$ for even $n$. At $\Omega=1$: $|T| = 1/\sqrt{1+\varepsilon^2}$ for all $n$.

 |T|
 1   |\  /\  /\
     | \/  \/  \
     |          \
     |           \__
     +-------------+--------> W
                   1

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

Order for the given specification

As printed, $\omega_p = \omega_s = 2000$ rad/s. With $\omega_s/\omega_p = 1$, $\cosh^{-1}(1) = 0$ and $n \to \infty$: a zero-width transition band cannot be met by any finite filter, so the value of $\omega_s$ is clearly a misprint. The companion Butterworth question with the same $\omega_p$, $\alpha_{max}$ and $\alpha_{min}$ uses $\omega_s = 3000$ rad/s, so assume $\omega_s = 3000$ rad/s ($\omega_s/\omega_p = 1.5$).

$$\begin{aligned} \varepsilon^2 &= 10^{0.05}-1 = 0.12202 \\ 10^{2.2}-1 &= 157.4893 \\ \sqrt{157.4893/0.12202} &= 35.9263, \quad \cosh^{-1}(35.9263) = 4.2744 \\ \cosh^{-1}(1.5) &= 0.9624 \\ n &\ge \frac{4.2744}{0.9624} = 4.441 \end{aligned}$$

Answer: n = 5 (next integer above 4.441).

Check: with $n = 5$ the attenuation at $\omega_s$ is $10\log_{10}[1+0.12202\cosh^2(5\cosh^{-1}1.5)] = 26.65$ dB $\ge 22$ dB, while $n = 4$ gives only 18.35 dB, which is not enough.

The same steps apply to whatever $\omega_s$ is intended; only $\cosh^{-1}(\omega_s/\omega_p)$ changes. For example, $\omega_s = 4000$ rad/s ($\omega_s/\omega_p = 2$) gives $n \ge 4.2744/1.3170 = 3.246$, so $n = 4$.