---
title: "Bilinear Butterworth, 0.2π passband, −15 dB at 0.4π"
subject: "Digital Signal Analysis and Processing"
chapter: 6
key: "semester_7/digital_signal_analysis_and_processing:bilinear-butter-0-2-1db-0-4"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/bilinear-butter-0-2-1db-0-4
site: BCT Engineering
---

# Bilinear Butterworth, 0.2π passband, −15 dB at 0.4π

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

- Asked in: 2071 Shrawan, 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-0-2-1db-0-4
- 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 frequency is 0.2π radians and maximum deviation of 1 dB below 0 dB gain in the pass band. The maximum gain of −15 dB and frequency is 0.4π radians in stop band, consider sampling frequency 1 Hz.

## Short answer

Order N=3, Ωc=0.8140 rad/s. H(z)=H₀(z)·H₁(z) where H₀(z)=0.2893(1+z⁻¹)/(1−0.4215z⁻¹) and H₁(z)=0.10533(1+2z⁻¹+z⁻²)/(1−1.0611z⁻¹+0.4824z⁻²); cascade of a first-order and second-order section.

## Answer

#### Step 1: specifications
Pass-band edge 0.2π with a maximum drop of 1 dB (|H| ≥ 0.89125); stop-band gain at most −15 dB (|H| ≤ 0.17783) at 0.4π. 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(0.2π/2) = **0.6498** rad/s
- Ωs = (2/1) tan(0.4π/2) = **1.4531** 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 2.2361) = 2.9656

So **N = 3**.

#### Step 4: cut-off frequency
Using the pass-band condition (1 + (Ωp/Ωc)^(2N)) = 1/0.89125²:
Ωc = Ωp / (0.2589)^(1/6) = **0.8140** 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) = 0.5393 / [(s + 0.8140) (s² + 0.8140 s + 0.6626)]

#### 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:

- H₀(z) = 0.2893(1 + z⁻¹) / (1 − 0.4215 z⁻¹)
- H1(z) = 0.10533(1 + 2z⁻¹ + z⁻²) / (1 − 1.0611 z⁻¹ + 0.4824 z⁻²)

H(z) = H0(z) · H1(z)

Multiplied out:

```text
        0.03047 0.09140 0.09140 0.03047   (numerator coefficients, powers z⁰ … z⁻3)
H(z) = ---------------------------------------------
        1.00000 -1.48258 0.92964 -0.20333   (denominator coefficients)
```

#### Step 7: check
|H(e^jω)| at ωp = 0.8913 (required ≥ 0.89125); at ωs = 0.1731 (required ≤ 0.17783). Both conditions are satisfied. 

**Answer: N = 3, Ωc = 0.8140 rad/s, H(z) = H₀(z)·H₁(z).**

## Check yourself

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

1. What are the analog edges for T = 1?
   A. 0.2 pi and 0.4 pi
   B. 0.3249 and 0.7265
   C. Omega_p = 0.6498 and Omega_s = 1.4531
   D. 1.2 and 2.4
   Answer: C. 2 tan(0.1 pi) = 0.6498, 2 tan(0.2 pi) = 1.4531. (Key point: Pre-warping)

2. What is the order?
   A. N = 2
   B. N = 4
   C. N = 6
   D. N = 3 (2.97 rounded up)
   Answer: D. The formula gives 2.97, so N = 3. (Key point: Order)

3. What does the third-order analog H(s) contain?
   A. One first-order factor and one second-order factor
   B. Three first-order factors
   C. One third-order polynomial that cannot be factored
   D. Three second-order factors
   Answer: A. Odd order gives a real pole plus a complex pair. (Key point: Odd order factors)

4. Which condition gives Omega_c?
   A. The stop-band condition at Omega_s, giving 1.45
   B. The pass-band condition at Omega_p, giving 0.8140
   C. The 3 dB condition at 0.2 pi
   D. The average of Omega_p and Omega_s
   Answer: B. Using the pass band condition meets it exactly. (Key point: Cut-off)

## Close variants set in other exams

- [Bilinear Butterworth, 0.25π passband, −15 dB at 0.45π](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Abilinear-butter-0-25-1db-0-45) — Digital Signal Analysis and Processing, chapter 6, asked 3×

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