---
title: "Bilinear Butterworth, 0.25π passband, −15 dB at 0.45π"
subject: "Digital Signal Analysis and Processing"
chapter: 6
key: "semester_7/digital_signal_analysis_and_processing:bilinear-butter-0-25-1db-0-45"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/bilinear-butter-0-25-1db-0-45
site: BCT Engineering
---

# Bilinear Butterworth, 0.25π passband, −15 dB at 0.45π

Digital Signal Analysis and Processing · Chapter 6: IIR Filter Design

- Asked in: 2072 Kartik, 15 marks
- Video: [Digital Butterworth Filter using Bilinear Transformation - Problem Solved - DTSP - DSP](https://www.youtube.com/watch?v=4yNIBnYwLuQ) — Padmasri Naban
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/bilinear-butter-0-25-1db-0-45
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-6

## Question

Design a low pass digital filter by Bilinear Transformation method to an approximate Butterworth filter, if passband edge frequency is 0.25π radians and maximum deviation of 1 dB below 0 dB gain in the passband. The maximum gain of −15 dB and frequency is 0.45π radians in stopband. Consider sampling frequency 1 Hz.

## Short answer

Order N=4, analog cut-off Ωc=0.9809 rad/s. The digital filter is H(z)=H₁(z)·H₂(z) with H₁(z)=0.14885(1+2z⁻¹+z⁻²)/(1−0.9400z⁻¹+0.5354z⁻²) and H₂(z)=0.11204(1+2z⁻¹+z⁻²)/(1−0.7076z⁻¹+0.1557z⁻²); cascade of two second-order sections.

## Answer

#### Step 1: specifications
Pass-band edge 0.25π with a maximum drop of 1 dB, so 20 log|H(0.25π)| ≥ −1 dB (|H| ≥ 0.89125). Stop-band: gain at most −15 dB (|H| ≤ 0.17783) at 0.45π. Sampling frequency 1 Hz gives T = 1 s.

#### Step 2: pre-warp the edges (bilinear: Ω = (2/T) tan(ω/2), T = 1 s)
- Ωp = (2/1) tan(0.25π/2) = **0.8284** rad/s
- Ωs = (2/1) tan(0.45π/2) = **1.7082** rad/s

#### Step 3: order of the Butterworth filter
N ≥ log[(1/0.17783² − 1)/(1/0.89125² − 1)] / (2 log(Ωs/Ωp)) = log[30.6228/0.2589] / (2 log 2.0619) = 3.2979

So **N = 4**.

#### Step 4: cut-off frequency
Using the pass-band condition (1 + (Ωp/Ωc)^(2N)) = 1/0.89125²:
Ωc = Ωp / (0.2589)^(1/8) = **0.9809** rad/s
(This meets the pass-band exactly and exceeds the stop-band requirement.)

#### Step 5: analog transfer function
Poles of the normalised filter lie on the left half of the circle of radius Ωc. Pairing complex poles:

H(s) = 0.9256 / [(s² + 0.7507 s + 0.9621) (s² + 1.8124 s + 0.9621)]

#### Step 6: bilinear transformation
Replace s by (2/T)(1 − z⁻¹)/(1 + z⁻¹) = 2(1 − z⁻¹)/(1 + z⁻¹) in each factor:

- H1(z) = 0.14885(1 + 2z⁻¹ + z⁻²) / (1 − 0.9400 z⁻¹ + 0.5354 z⁻²)
- H2(z) = 0.11204(1 + 2z⁻¹ + z⁻²) / (1 − 0.7076 z⁻¹ + 0.1557 z⁻²)

H(z) = H1(z) · H2(z)

Multiplied out:

```text
        0.01668 0.06671 0.10006 0.06671 0.01668   (numerator coefficients, powers z⁰ … z⁻4)
H(z) = ---------------------------------------------
        1.00000 -1.64759 1.35628 -0.52524 0.08338   (denominator coefficients)
```

#### Step 7: check
|H(e^jω)| at ωp = 0.8913 (required ≥ 0.89125); at ωs = 0.1081 (required ≤ 0.17783). Both conditions are satisfied. 

**Answer: N = 4, Ωc = 0.9809 rad/s, H(z) = H₁(z)·H₂(z).** The cascade of two second-order sections is the usual realisation.

## Check yourself

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

1. What |H| does a 1 dB maximum drop correspond to?
   A. 0.1
   B. 0.5
   C. 0.89125
   D. 0.99
   Answer: C. 10^(-1/20) = 0.89125. (Key point: Spec in ratio)

2. What T is used?
   A. T = 2 s
   B. T = 0.5 s
   C. T = 2 pi s
   D. T = 1 s, from Fs = 1 Hz
   Answer: D. T = 1/Fs. (Key point: Sampling frequency)

3. What order meets the specs?
   A. N = 4 (3.30 rounded up)
   B. N = 3
   C. N = 5
   D. N = 2
   Answer: A. Computing the formula gives 3.30, so N = 4. (Key point: Order)

4. After bilinear transformation of each analog section, what form results?
   A. First-order sections only
   B. Second-order sections with numerator (1 + 2z^-1 + z^-2)
   C. FIR-like polynomial with no denominator
   D. Sections with numerator 1 - z^-2
   Answer: B. Each analog section with no zeros maps to a numerator with a double zero at z = -1. (Key point: Digital realization)

## Close variants set in other exams

- [Bilinear Butterworth, 0.2π passband, −15 dB at 0.4π](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Abilinear-butter-0-2-1db-0-4) — Digital Signal Analysis and Processing, chapter 6, asked 2×
- [Impulse invariance Butterworth, 0.2π, 0.5 dB](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Aimpulse-invariance-butter-0-5db) — Digital Signal Analysis and Processing, chapter 6, asked 2×

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