---
title: "Bilinear Butterworth, 0.8 to 1 and 0.2 limits"
subject: "Digital Signal Analysis and Processing"
chapter: 6
key: "semester_7/digital_signal_analysis_and_processing:bilinear-butter-0-8-0-2"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/bilinear-butter-0-8-0-2
site: BCT Engineering
---

# Bilinear Butterworth, 0.8 to 1 and 0.2 limits

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

- Asked in: 2070 Chaitra, 10 marks
- 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-0-8-0-2
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-6

## Question

Using bilinear transformation method, design a digital filter using Butterworth approximation which satisfies the following conditions: 0.8 ≤ |H(e^jw)| ≤ 1 for 0 ≤ w ≤ 0.2π; |H(e^jw)| ≤ 0.2 for 0.6π ≤ w ≤ π.

## Short answer

Order N=2, Ωc=0.7504 rad/s. H(z)=0.08422(1+2z⁻¹+z⁻²)/(1−1.02819z⁻¹+0.36508z⁻²).

## Answer

#### Step 1: specifications
Butterworth low-pass: 0.8 ≤ |H| ≤ 1 for 0 ≤ ω ≤ 0.2π, and |H| ≤ 0.2 for 0.6π ≤ ω ≤ π. T is not given; take T = 1 s (it cancels in the bilinear transformation).

#### 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(0.6π/2) = **2.7528** rad/s

#### Step 3: order of the Butterworth filter
N ≥ log[(1/0.2² − 1)/(1/0.8² − 1)] / (2 log(Ωs/Ωp)) = log[24.0000/0.5625] / (2 log 4.2361) = 1.3000

So **N = 2**.

#### Step 4: cut-off frequency
Using the pass-band condition (1 + (Ωp/Ωc)^(2N)) = 1/0.8²:
Ωc = Ωp / (0.5625)^(1/4) = **0.7504** 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) = 0.5631 / [(s² + 1.0612 s + 0.5631)]

#### 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) = 0.08422(1 + 2z⁻¹ + z⁻²) / (1 − 1.0282 z⁻¹ + 0.3651 z⁻²)

H(z) = H1(z)

Multiplied out:

```text
        0.08422 0.16844 0.08422   (numerator coefficients, powers z⁰ … z⁻2)
H(z) = ---------------------------------------------
        1.00000 -1.02819 0.36508   (denominator coefficients)
```

#### Step 7: check
|H(e^jω)| at ωp = 0.8000 (required ≥ 0.8); at ωs = 0.0741 (required ≤ 0.2). Both conditions are satisfied. 

**Answer: N = 2, Ωc = 0.7504 rad/s, H(z) = 0.08422(1 + 2z⁻¹ + z⁻²)/(1 − 1.02819z⁻¹ + 0.36508z⁻²).**

## Check yourself

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

1. What are Omega_p and Omega_s for T = 1?
   A. 0.2 pi and 0.6 pi
   B. 0.3249 and 1.3764
   C. 0.6498 and 2.7528 rad/s
   D. 0.8 and 0.2
   Answer: C. Omega = 2 tan(w/2). (Key point: Pre-warped edges)

2. What order is needed?
   A. N = 1
   B. N = 3
   C. N = 4
   D. N = 2 (1.30 rounded up)
   Answer: D. The formula gives 1.30 so use N = 2. (Key point: Order)

3. How is Omega_c found?
   A. From the pass-band condition 1 + (Omega_p/Omega_c)^4 = 1/0.64
   B. From Omega_c = Omega_s
   C. From the stop band at 0.2
   D. From N alone
   Answer: A. Pass-band equality fixes Omega_c at 0.7504. (Key point: Cut-off)

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