---
title: "Minimum Butterworth order (1 dB, 25 dB, ratio 1.5)"
subject: "Filter Design"
chapter: 2
key: "semester_5/filter_design:butterworth-minimum-order-25db"
source: https://www.bctengineer.com/q/semester_5/filter_design/butterworth-minimum-order-25db
site: BCT Engineering
---

# Minimum Butterworth order (1 dB, 25 dB, ratio 1.5)

Filter Design · Chapter 2: Approximation Methods

- Asked in: 2081 Bhadra, 3 marks
- Video: [Q9a Design a digital lowpass Butterworth filter with the following specifications](https://www.youtube.com/watch?v=30rrdEfKBkQ) — EnggClasses
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/filter_design/butterworth-minimum-order-25db
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-2

## Question

Determine the Butterworth low pass characteristics with the minimum n such that following specifications are satisfied: αp = 1dB, αs = 25 dB, ωs/ωp = 1.5.

## Short answer

Order n ≥ log10((10^2.5-1)/(10^0.1-1)) / (2 log10(1.5)) = 8.761, so n=9. The Butterworth characteristic is |T(jω)|^2 = 1/(1+0.2589 ω^18) with ε=0.5088 and half-power frequency ω3dB=1.0780 rad/s.

## Answer

For a Butterworth response normalized to $\omega_p$ = 1, $\alpha(\omega) = 10\log_{10}(1 + \varepsilon^2\omega^{2n})$, with $\varepsilon^2 = 10^{0.1\alpha_p} - 1$. The stopband condition gives
$$
n \ge \frac{\log_{10}\left[(10^{0.1\alpha_s} - 1)/(10^{0.1\alpha_p} - 1)\right]}{2\log_{10}(\omega_s/\omega_p)}
$$
$$
\begin{aligned}
10^{0.1\alpha_s} - 1 &= 10^{2.5} - 1 = 315.2278 \\
10^{0.1\alpha_p} - 1 &= 10^{0.1} - 1 = 0.25893 \\
\frac{\omega_s}{\omega_p} &= 1.5 \\
n &\ge \frac{\log_{10}(315.2278/0.25893)}{2\log_{10}(1.5)} = \frac{3.0854}{0.3522} = 8.761
\end{aligned}
$$
The order must be an integer, so **n = 9**.

Check: with $n = 9$, $\varepsilon^2 = 0.25893$, attenuation at $\omega_s$ is $10\log_{10}(1 + 0.25893 \times 1.5^{18}) = 25.84$ dB $\ge 25$ dB.

**Butterworth characteristic with minimum order:**
$$
|T(j\omega)|^2 = \frac{1}{1 + 0.2589\,\omega^{18}}, \qquad \varepsilon = 0.5088
$$
($\omega$ normalized to $\omega_p$). It gives exactly 1 dB at $\omega$ = 1 and 25.84 dB at $\omega$ = 1.5. The half-power frequency is $\omega_{3dB} = \varepsilon^{-1/9} = 1.0780$, and the 9 poles lie on a circle of this radius, spaced 20° apart (one real pole at $-1.0780$).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For 1 dB, 25 dB and ws/wp = 1.5 the formula gives 8.761. The minimum n is:
   A. 8.761
   B. 9
   C. 8
   D. 10
   Answer: B. Round up to 9; n = 9 gives 25.84 dB at ws, meeting 25 dB. (Key point: Minimum order)

2. What is the Butterworth characteristic for the answer?
   A. |T|^2 = 1/(1 + 0.2589 w^18)
   B. |T|^2 = 1/(1 + 0.5088 w^18)
   C. |T|^2 = 1/(1 + 0.2589 w^9)
   D. |T|^2 = 1/(1 + w^18)
   Answer: A. |T|^2 = 1/(1+eps^2 w^(2n)) with eps^2 = 0.2589 and 2n = 18. (Key point: Resulting characteristic)

3. What is the 3 dB frequency and where are the poles?
   A. 1.5 rad/s, on an ellipse
   B. epsilon^(-1/9) = 1.078, on a circle of that radius
   C. 1 rad/s, on the unit circle
   D. epsilon^9, on a line
   Answer: B. w_3dB = epsilon^(-1/n) = 1.078; the 9 poles lie on that circle 20 degrees apart, one real at -1.078. (Key point: Half-power frequency)

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