---
title: "Bilinear Butterworth, 0.89125 and 0.17783 limits"
subject: "Digital Signal Analysis and Processing"
chapter: 6
key: "semester_7/digital_signal_analysis_and_processing:bilinear-butter-0-89125"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/bilinear-butter-0-89125
site: BCT Engineering
---

# Bilinear Butterworth, 0.89125 and 0.17783 limits

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

- Asked in: 2073 Shrawan, 12 marks
- Video: [Design Butterworth LPF using Bilinear Transformation | DSAP | IOE BCT Exam Preparation](https://www.youtube.com/watch?v=xo7cleaoouQ) — AFTERIMAGE
- 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-0-89125
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-6

## Question

Using bilinear transformation, design a Butterworth low pass filter which satisfies the following magnitude response: 0.89125 ≤ |H(e^jw)| ≤ 1 for 0 ≤ ω ≤ 0.2π; |H(e^jw)| ≤ 0.17783 for 0.3π ≤ ω ≤ π.

## Short answer

After pre-warping (T=1 s), Ωp=0.6498 rad/s and Ωs=1.0191 rad/s; Butterworth order N=6, Ωc=0.7273 rad/s. Bilinear substitution gives H(z) as a cascade of three second-order sections; numerator coefficients [0.00058,0.00348,0.00870,0.01159,0.00870,0.00348,0.00058] over denominator [1,-3.31430,4.95011,-4.14326,2.02754,-0.54583,0.06284].

## Answer

#### Step 1: specifications
Butterworth low-pass: 0.89125 ≤ |H| ≤ 1 for 0 ≤ ω ≤ 0.2π (−1 dB) and |H| ≤ 0.17783 for 0.3π ≤ ω ≤ π (−15 dB). T is not given; take T = 1 s (the result does not depend on T because 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(0.3π/2) = **1.0191** rad/s

#### Step 3: order of the Butterworth filter
N ≥ log[(1/0.17783² − 1)/(1/0.89125² − 1)] / (2 log(Ωs/Ωp)) = log[30.6220/0.2589] / (2 log 1.5682) = 5.3044

So **N = 6**.

#### Step 4: cut-off frequency
Using the pass-band condition (1 + (Ωp/Ωc)^(2N)) = 1/0.89125²:
Ωc = Ωp / (0.2589)^(1/12) = **0.7273** 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) = 0.1480 / [(s² + 0.3765 s + 0.5290) (s² + 1.0285 s + 0.5290) (s² + 1.4050 s + 0.5290)]

#### 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) = 0.10014(1 + 2z⁻¹ + z⁻²) / (1 − 1.3143 z⁻¹ + 0.7149 z⁻²)
- H2(z) = 0.08031(1 + 2z⁻¹ + z⁻²) / (1 − 1.0541 z⁻¹ + 0.3753 z⁻²)
- H3(z) = 0.07207(1 + 2z⁻¹ + z⁻²) / (1 − 0.9459 z⁻¹ + 0.2342 z⁻²)

H(z) = H1(z) · H2(z) · H3(z)

Multiplied out:

```text
        0.00058 0.00348 0.00870 0.01159 0.00870 0.00348 0.00058   (numerator coefficients, powers z⁰ … z⁻6)
H(z) = ---------------------------------------------
        1.00000 -3.31430 4.95011 -4.14326 2.02754 -0.54583 0.06284   (denominator coefficients)
```

#### Step 7: check
|H(e^jω)| at ωp = 0.8913 (required ≥ 0.89125); at ωs = 0.1310 (required ≤ 0.17783). Both conditions are satisfied. 

**Answer: N = 6, Ωc = 0.7273 rad/s, H(z) = H₁(z)·H₂(z)·H₃(z).** Realise it as a cascade of three second-order sections (Direct Form II).

## Check yourself

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

1. What are 0.89125 and 0.17783 in dB?
   A. -0.89 dB and -0.18 dB
   B. -3 dB and -15 dB
   C. -1 dB and -15 dB
   D. -10 dB and -20 dB
   Answer: C. 20 log10(0.89125) = -1 dB and 20 log10(0.17783) = -15 dB. (Key point: Meaning of specs)

2. Why can T be taken as 1 s when it is not given?
   A. It does not affect the order only
   B. All filters use T = 1
   C. Butterworth filters do not depend on frequency
   D. T cancels between pre-warping and the bilinear substitution
   Answer: D. The 2/T factors cancel, so H(z) is independent of T. (Key point: T cancels)

3. Order from Omega_p = 0.6498 and Omega_s = 1.0191 gives
   A. N = 6 (5.30 rounded up)
   B. N = 5
   C. N = 3
   D. N = 12
   Answer: A. The order formula gives 5.30; round up to 6. (Key point: Order)

4. How is a 6th-order result best realized?
   A. As a single direct-form section with 6 delays only
   B. As a cascade of three second-order sections
   C. As three first-order sections
   D. As a lattice with 12 coefficients
   Answer: B. Cascade second-order sections reduce coefficient sensitivity. (Key point: Realization)

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