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Butterworth order for ωs/ωp = 1.5 with poles

Short answer

Order from stopband condition n ≥ log10((10^0.1αmin−1)/(10^0.1αmax−1)) / (2 log10(ωs/ωp)). For αmax=1 dB, αmin=20 dB, ωs/ωp=1.5, n≥7.333 so n=8. Pole radius ωo=ε^(-1/n)=1.0881 rad/s, angles ±11.25°, ±33.75°, ±56.25°, ±78.75°; transfer function is product of four quadratic factors with numerator scaled to give unity gain at DC.

Derive the expression to calculate the order n of a Butterworth Low pass filter and use it to find the order for given specification: αmax = 1 dB, αmin = 20 dB and ωs/ωp = 1.5. Also determine pole locations and transfer functions.

Order of a Butterworth low-pass filter

For the normalized Butterworth response (passband edge ωp\omega_p = 1 rad/s)

∣T(jω)∣2=11+ε2ω2n,α(ω)=10log⁡10(1+ε2ω2n) dB|T(j\omega)|^2 = \frac{1}{1+\varepsilon^2\omega^{2n}}, \qquad \alpha(\omega) = 10\log_{10}(1 + \varepsilon^2\omega^{2n})\ \text{dB}

Passband condition (ω=1\omega = 1, α=αmax\alpha = \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

Stopband condition (at ω=ωs/ωp\omega = \omega_s/\omega_p the attenuation must be at least αmin\alpha_{min}):

10log⁡10[1+ε2(ωsωp)2n]≥αmin(ωsωp)2n≥100.1αmin−1100.1αmax−1\begin{aligned} 10\log_{10}\left[1 + \varepsilon^2\left(\frac{\omega_s}{\omega_p}\right)^{2n}\right] &\ge \alpha_{min} \\ \left(\frac{\omega_s}{\omega_p}\right)^{2n} &\ge \frac{10^{0.1\alpha_{min}} - 1}{10^{0.1\alpha_{max}} - 1} \end{aligned}

Taking logarithms:

n≥log⁡10[100.1αmin−1100.1αmax−1]2log⁡10(ωs/ωp)n \ge \frac{\log_{10}\left[\dfrac{10^{0.1\alpha_{min}} - 1}{10^{0.1\alpha_{max}} - 1}\right]}{2\log_{10}(\omega_s/\omega_p)}

nn is rounded up to the next integer.

Order for αmax\alpha_{max} = 1 dB, αmin\alpha_{min} = 20 dB, ωs/ωp\omega_s/\omega_p = 1.5

100.1αmin−1=102−1=99.0000100.1αmax−1=100.1−1=0.25893ωsωp=1.5n≥log⁡10(99.0000/0.25893)2log⁡10(1.5)=2.58250.3522=7.333\begin{aligned} 10^{0.1\alpha_{min}} - 1 &= 10^{2} - 1 = 99.0000 \\ 10^{0.1\alpha_{max}} - 1 &= 10^{0.1} - 1 = 0.25893 \\ \frac{\omega_s}{\omega_p} = 1.5 \\ n &\ge \frac{\log_{10}(99.0000/0.25893)}{2\log_{10}(1.5)} = \frac{2.5825}{0.3522} = 7.333 \end{aligned}

The order must be an integer, so n = 8.

Check: with n=8n = 8, ε2=0.25893\varepsilon^2 = 0.25893, attenuation at ωs\omega_s is 10log⁡10(1+0.25893×1.516)=22.3310\log_{10}(1 + 0.25893 \times 1.5^{16}) = 22.33 dB ≥20\ge 20 dB.

Pole locations

Work with the passband edge normalized to ωp\omega_p = 1 rad/s. ε=0.25893=0.50885\varepsilon = \sqrt{0.25893} = 0.50885. With the passband edge met exactly, the poles lie on a circle of radius

ωo=ε−1/n=0.50885−1/8=1.0881 rad/s\omega_o = \varepsilon^{-1/n} = 0.50885^{-1/8} = 1.0881\ \text{rad/s}

at pk=−ωosin⁡θk±jωocos⁡θkp_k = -\omega_o\sin\theta_k \pm j\omega_o\cos\theta_k, θk=(2k−1)π/16\theta_k = (2k-1)\pi/16 (spacing 22.5°):

Polesθk\theta_kLocation
p1,p8p_{1}, p_{8}11.25°11.25°−0.2123±j1.0672-0.2123 \pm j1.0672
p2,p7p_{2}, p_{7}33.75°33.75°−0.6045±j0.9047-0.6045 \pm j0.9047
p3,p6p_{3}, p_{6}56.25°56.25°−0.9047±j0.6045-0.9047 \pm j0.6045
p4,p5p_{4}, p_{5}78.75°78.75°−1.0672±j0.2123-1.0672 \pm j0.2123

Transfer function

Each conjugate pair gives a factor s2+2ωosin⁡θk s+ωo2s^2 + 2\omega_o\sin\theta_k\, s + \omega_o^2 with ωo2=1.1840\omega_o^2 = 1.1840:

T(s)=1.9652(s2+0.4246s+1.1840)(s2+1.2091s+1.1840)(s2+1.8095s+1.1840)(s2+2.1344s+1.1840)T(s) = \frac{1.9652}{(s^2 + 0.4246s + 1.1840)(s^2 + 1.2091s + 1.1840)(s^2 + 1.8095s + 1.1840)(s^2 + 2.1344s + 1.1840)}

where the numerator ωo8=1.9652\omega_o^8 = 1.9652 makes T(0)=1T(0) = 1. For an actual ωp\omega_p, replace ss by s/ωps/\omega_p. (Check: loss at ω\omega = 1 is 1.00 dB and at ω\omega = 1.5 is 22.33 dB.)

✓ Answer checked by Claude Sonnet 5.5

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