---
title: "Bilinear Butterworth, 0.27π and 0.58π"
subject: "Digital Signal Analysis and Processing"
chapter: 6
key: "semester_7/digital_signal_analysis_and_processing:bilinear-butter-0-27-0-58"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/bilinear-butter-0-27-0-58
site: BCT Engineering
---

# Bilinear Butterworth, 0.27π and 0.58π

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

- Asked in: 2074 Asoj, 15 marks
- Video: [IIR Filter Design using BLT - Butterworth filter design in DSP](https://www.youtube.com/watch?v=7LsjhXjzhZs) — Shrenik Jain
- 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-0-27-0-58
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-6

## Question

Design a low pass discrete IIR filter by Bilinear Transformation method to an approximate Butterworth filter having specifications: pass band edge frequency (ωp) = 0.27π radians, stop band edge frequency (ωs) = 0.58π radians, passband ripple (δp) = 0.11, stopband ripple (δs) = 0.21. Consider sampling frequency 0.5 Hz.

## Short answer

After pre-warping (T=2 s), Ωp=0.4515 rad/s and Ωs=1.2892 rad/s; Butterworth order N=3, Ωc=0.5643 rad/s. Bilinear substitution gives H(z) with numerator [0.06101,0.18303,0.18303,0.06101] and denominator [1,-1.00260,0.60224,-0.11157].

## Answer

#### Step 1: specifications
Low-pass Butterworth, ωp = 0.27π, ωs = 0.58π. δp = 0.11 and δs = 0.21 are read as |H(ωp)| ≥ 0.89 and |H(ωs)| ≤ 0.21. Sampling frequency 0.5 Hz gives T = 2 s.

#### Step 2: pre-warp the edges (bilinear: Ω = (2/T) tan(ω/2), T = 2 s)
- Ωp = (2/2) tan(0.27π/2) = **0.4515** rad/s
- Ωs = (2/2) tan(0.58π/2) = **1.2892** rad/s

#### Step 3: order of the Butterworth filter
N ≥ log[(1/0.21² − 1)/(1/0.89² − 1)] / (2 log(Ωs/Ωp)) = log[21.6757/0.2625] / (2 log 2.8552) = 2.1035

So **N = 3**.

#### Step 4: cut-off frequency
Using the pass-band condition (1 + (Ωp/Ωc)^(2N)) = 1/0.89²:
Ωc = Ωp / (0.2625)^(1/6) = **0.5643** 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) = 0.1797 / [(s + 0.5643) (s² + 0.5643 s + 0.3184)]

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

- H₀(z) = 0.3607(1 + z⁻¹) / (1 − 0.2785 z⁻¹)
- H1(z) = 0.16913(1 + 2z⁻¹ + z⁻²) / (1 − 0.7241 z⁻¹ + 0.4006 z⁻²)

H(z) = H0(z) · H1(z)

Multiplied out:

```text
        0.06101 0.18303 0.18303 0.06101   (numerator coefficients, powers z⁰ … z⁻3)
H(z) = ---------------------------------------------
        1.00000 -1.00260 0.60224 -0.11157   (denominator coefficients)
```

#### Step 7: check
|H(e^jω)| at ωp = 0.8900 (required ≥ 0.89); at ωs = 0.0836 (required ≤ 0.21). Both conditions are satisfied. 

**Answer: N = 3, Ωc = 0.5643 rad/s, H(z) = H₀(z)·H₁(z) as above.**

## Check yourself

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

1. How are delta_p = 0.11 and delta_s = 0.21 used?
   A. |H(wp)| >= 0.11 and |H(ws)| <= 0.79
   B. |H(wp)| <= 0.89 and |H(ws)| >= 0.21
   C. |H(wp)| >= 0.89 and |H(ws)| <= 0.21
   D. |H(wp)| = 1.11 and |H(ws)| = 0.21
   Answer: C. The pass band may drop by 0.11 from 1, and the stop band must stay below 0.21. (Key point: Reading ripples)

2. With T = 2 s, what are the analog edges?
   A. Omega_p = 0.27 pi and Omega_s = 0.58 pi
   B. Omega_p = 0.8481 and Omega_s = 3.1416
   C. Omega_p = 0.2257 and Omega_s = 0.6446
   D. Omega_p = tan(0.135 pi) = 0.4515 and Omega_s = tan(0.29 pi) = 1.2892
   Answer: D. Omega = (2/T) tan(w/2) with 2/T = 1. (Key point: Pre-warping)

3. Which expression gives N?
   A. log[(1/ds^2 - 1)/(1/(1-dp)^2 - 1)] / (2 log(Omega_s/Omega_p)), rounded up to 3
   B. log(Omega_s/Omega_p)/2, giving 1
   C. log(ds/dp), giving 2
   D. Same expression but rounded down to 2
   Answer: A. The ratio of the stop-band and pass-band conditions is used and the result is rounded up. (Key point: Order formula)

4. How is Omega_c found?
   A. As the average of Omega_p and Omega_s
   B. Using the pass-band condition 1 + (Omega_p/Omega_c)^(2N) = 1/0.89^2, giving 0.5643
   C. As Omega_s itself
   D. From N = 3 only, without a condition
   Answer: B. Pass-band condition is met exactly, so the stop band is exceeded. (Key point: Cut-off)

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