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Chebyshev order (0.1 dB, 20 dB, 2500 rad/s)

Short answer

The order n must satisfy n ≥ arcosh(√[(10^(0.1αmin)-1)/(10^(0.1αmax)-1)]) / arcosh(ωs/ωp). For αmax=0.1 dB, αmin=20 dB, ωs/ωp=2.5: ε²=0.02329, arcosh(65.1936)=4.8704, arcosh(2.5)=1.5668, so n ≥ 3.109 → n=4.

Derive an expression to calculate the order of Chebyshev low pass filter. Use this formula to estimate the order of Chebyshev low pass filter having following specification: αmax = 0.1 dB, ωp = 1000 rad/s, αmin = 20 dB, ωs = 2500 rad/s

Derivation of the order

The Chebyshev low-pass magnitude response (normalized so that the passband edge is Ω=ω/ωp=1\Omega = \omega/\omega_p = 1) is

∣T(jΩ)∣2=11+ε2Cn2(Ω)|T(j\Omega)|^2 = \frac{1}{1+\varepsilon^2 C_n^2(\Omega)}

where Cn(Ω)C_n(\Omega) is the Chebyshev polynomial of order nn:

Cn(Ω)={cos⁡(ncos⁡−1Ω),∣Ω∣≤1cosh⁡(ncosh⁡−1Ω),∣Ω∣>1C_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 α(Ω)=10log⁡10[1+ε2Cn2(Ω)]\alpha(\Omega) = 10\log_{10}\left[1+\varepsilon^2 C_n^2(\Omega)\right].

Step 1: ripple factor from the passband edge. At Ω=1\Omega = 1, Cn(1)=1C_n(1) = 1 for every nn, so the attenuation is exactly αmax\alpha_{max}:

αmax=10log⁡10(1+ε2)  ⇒  ε2=100.1αmax−1\alpha_{max} = 10\log_{10}(1+\varepsilon^2) \;\Rightarrow\; \varepsilon^2 = 10^{0.1\alpha_{max}} - 1

Step 2: stopband condition. At Ωs=ωs/ωp (>1)\Omega_s = \omega_s/\omega_p\ (>1) the attenuation must be at least αmin\alpha_{min}:

10log⁡10[1+ε2Cn2(Ωs)]≥αmin  ⇒  Cn2(Ωs)≥100.1αmin−1ε210\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 Ωs>1\Omega_s > 1, Cn(Ωs)=cosh⁡(ncosh⁡−1Ωs)C_n(\Omega_s) = \cosh(n\cosh^{-1}\Omega_s):

cosh⁡(ncosh⁡−1Ωs)≥100.1αmin−1100.1αmax−1\cosh(n\cosh^{-1}\Omega_s) \ge \sqrt{\frac{10^{0.1\alpha_{min}}-1}{10^{0.1\alpha_{max}}-1}} n≥cosh⁡−1100.1αmin−1100.1αmax−1cosh⁡−1(ωsωp)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⁡−1x=ln⁡(x+x2−1)\cosh^{-1}x = \ln\left(x+\sqrt{x^2-1}\right).)

Estimate of the order

Given αmax=0.1\alpha_{max} = 0.1 dB, αmin=20\alpha_{min} = 20 dB, ωsωp=25001000=2.5\frac{\omega_s}{\omega_p} = \frac{2500}{1000} = 2.5.

100.1αmax−1=100.01−1=0.02329100.1αmin−1=102−1=99.000099.00000.02329=65.1936cosh⁡−1(65.1936)=4.8704cosh⁡−1(2.5)=1.5668n≥4.87041.5668=3.109\begin{aligned} 10^{0.1\alpha_{max}} - 1 &= 10^{0.01} - 1 = 0.02329 \\ 10^{0.1\alpha_{min}} - 1 &= 10^{2} - 1 = 99.0000 \\ \sqrt{\frac{99.0000}{0.02329}} &= 65.1936 \\ \cosh^{-1}(65.1936) &= 4.8704 \\ \cosh^{-1}(2.5) &= 1.5668 \\ n &\ge \frac{4.8704}{1.5668} = 3.109 \end{aligned}

Answer: n = 4 (next integer above 3.109).

A small ripple (0.1 dB) makes ε\varepsilon small, so a higher order is needed than with 0.5 dB or 1 dB ripple for the same stopband.

✓ Answer checked by Claude Sonnet 5.5

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