---
title: "Butterworth order (0.1 dB to 200 rad/s, 30 dB)"
subject: "Filter Design"
chapter: 2
key: "semester_5/filter_design:butterworth-order-200-2000"
source: https://www.bctengineer.com/q/semester_5/filter_design/butterworth-order-200-2000
site: BCT Engineering
---

# Butterworth order (0.1 dB to 200 rad/s, 30 dB)

Filter Design · Chapter 2: Approximation Methods

- Asked in: 2069 Chaitra, 5+3 marks
- Video: [Design of Analog Butterworth Filter - Problem#1 Solved - IIR Filters - DTSP](https://www.youtube.com/watch?v=WfYFv8IIzz0) — Padmasri Naban
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/filter_design/butterworth-order-200-2000
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-2

## Question

Derive the expression to calculate the order of Butterworth approximation for given lowpass filter specifications. Calculate the order of Butterworth low pass filter having following specification; i) Passband extends from ω = 0 to ω = 200 rad/s and the attenuation in the passband should not exceed 0.1 dB. ii) Stopband extends from ω = 2000 rad/s to ω = ∞ and the attenuation in the stopband should not be less than 30 dB

## Short answer

Butterworth order n ≥ log10((10^0.1αmin−1)/(10^0.1αmax−1)) / (2 log10(ωs/ωp)). With ωp=200 rad/s, αmax=0.1 dB, ωs=2000 rad/s, αmin=30 dB, n≥2.316, so required order is n=3.

## Answer

#### Order of a Butterworth low-pass filter
For the normalized Butterworth response (passband edge $\omega_p$ = 1 rad/s)
$$
|T(j\omega)|^2 = \frac{1}{1+\varepsilon^2\omega^{2n}}, \qquad \alpha(\omega) = 10\log_{10}(1 + \varepsilon^2\omega^{2n})\ \text{dB}
$$
**Passband condition** ($\omega = 1$, $\alpha = \alpha_{max}$):
$$
\alpha_{max} = 10\log_{10}(1+\varepsilon^2) \;\Rightarrow\; \varepsilon^2 = 10^{0.1\alpha_{max}} - 1
$$
**Stopband condition** (at $\omega = \omega_s/\omega_p$ the attenuation must be at least $\alpha_{min}$):
$$
\begin{aligned}
10\log_{10}\left[1 + \varepsilon^2\left(\frac{\omega_s}{\omega_p}\right)^{2n}\right] &\ge \alpha_{min} \\
\left(\frac{\omega_s}{\omega_p}\right)^{2n} &\ge \frac{10^{0.1\alpha_{min}} - 1}{10^{0.1\alpha_{max}} - 1}
\end{aligned}
$$
Taking logarithms:
$$
n \ge \frac{\log_{10}\left[\dfrac{10^{0.1\alpha_{min}} - 1}{10^{0.1\alpha_{max}} - 1}\right]}{2\log_{10}(\omega_s/\omega_p)}
$$
$n$ is rounded **up** to the next integer.

#### Order for the given specifications
Passband 0 to $\omega_p$ = 200 rad/s with $\alpha_{max}$ = 0.1 dB; stopband from $\omega_s$ = 2000 rad/s with $\alpha_{min}$ = 30 dB.
$$
\begin{aligned}
10^{0.1\alpha_{min}} - 1 &= 10^{3} - 1 = 999.0000 \\
10^{0.1\alpha_{max}} - 1 &= 10^{0.01} - 1 = 0.02329 \\
\frac{\omega_s}{\omega_p} = \frac{2000}{200} = 10 \\
n &\ge \frac{\log_{10}(999.0000/0.02329)}{2\log_{10}(10)} = \frac{4.6323}{2.0000} = 2.316
\end{aligned}
$$
The order must be an integer, so **n = 3**.

Check: with $n = 3$, $\varepsilon^2 = 0.02329$, attenuation at $\omega_s$ is $10\log_{10}(1 + 0.02329 \times 10^{6}) = 43.67$ dB $\ge 30$ dB.

**Answer:** a 3rd-order Butterworth low-pass filter.

## Check yourself

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

1. In the specification wp = 200 rad/s and ws = 2000 rad/s, what ratio goes into the order formula?
   A. ws/wp = 10
   B. ws/wp = 100
   C. ws - wp = 1800
   D. wp/ws = 0.1
   Answer: A. The formula uses the ratio ws/wp = 10, so log(ws/wp) = 1. (Key point: Frequency ratio)

2. With amax = 0.1 dB and amin = 30 dB the formula gives n >= 2.316. What order is chosen?
   A. 4
   B. 3
   C. 2
   D. 2.316
   Answer: B. A 2nd-order filter would not meet the stopband spec, so round up to n = 3. (Key point: Result and rounding)

3. Which step gives the order inequality in the derivation?
   A. Requiring loss at w = 0 to equal amax
   B. Requiring the phase to be linear at ws
   C. Requiring poles on the jw axis
   D. Requiring loss at ws to be at least amin: 10log[1 + eps^2 (ws/wp)^(2n)] >= amin
   Answer: D. The stopband condition 10log[1+eps^2(ws/wp)^(2n)] >= amin, with eps^2 from the passband edge, is solved for n. (Key point: Derivation basis)

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