---
title: "Impulse invariance Butterworth, 0.7 dB and 14 dB"
subject: "Digital Signal Analysis and Processing"
chapter: 6
key: "semester_7/digital_signal_analysis_and_processing:impulse-invariance-butter-0-7db-14db"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/impulse-invariance-butter-0-7db-14db
site: BCT Engineering
---

# Impulse invariance Butterworth, 0.7 dB and 14 dB

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

- Asked in: 2074 Chaitra, 12 marks
- Video: [Q9a Design a digital lowpass Butterworth filter with the following specifications](https://www.youtube.com/watch?v=30rrdEfKBkQ) — EnggClasses
- 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/impulse-invariance-butter-0-7db-14db
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-6

## Question

Design a digital low-pass filter with the following specification: i) Pass-band magnitude constant to 0.7 dB below the frequency of 0.15π. ii) Stop-band attenuation at least 14 dB for the frequencies between 0.6π to π. Use Butterworth approximation as a prototype and use impulse invariance method to obtain the digital filter.

## Short answer

Order N=3, Ωc=0.6301 rad/s. H(z)=0.63015/(1−0.53251z⁻¹) + (−0.63015+0.53085z⁻¹)/(1−1.24748z⁻¹+0.53251z⁻²); parallel realization.

## Answer

#### Step 1: specifications
Pass band within 0.7 dB of 0 dB up to 0.15π (|H| ≥ 10^(−0.7/20) = 0.9226); stop-band attenuation at least 14 dB (|H| ≤ 0.1995) from 0.6π to π. T is not given; take T = 1 s.

#### Step 2: analog edge frequencies (impulse invariance: Ω = ω/T)
Ωp = **0.4712** rad/s, Ωs = **1.8850** rad/s  (T = 1 s)

#### Step 3: order
N ≥ log[(1/0.1995² − 1)/(1/0.9226² − 1)] / (2 log(Ωs/Ωp)) = log[24.1189/0.1749] / (2 log 4.0000) = 1.7769

The analog formula gives N = 2. However, impulse invariance causes aliasing: the usual N = 2 design (Ωc = 0.7287 rad/s, H(z) = 0.3033z⁻¹/(1 − 1.0396z⁻¹ + 0.3568z⁻²)) gives only 0.899 at 0.15π, below the required 0.9226, and a numerical search shows that no choice of Ωc lets a second-order design meet both edges. So take **N = 3**.

#### Step 4: cut-off
Ωc = Ωp / (0.1749)^(1/6) = **0.6301** rad/s

#### Step 5: analog poles and H(s)
Poles: s_k = Ωc exp[jπ(2k + N + 1)/(2N)], k = 0 … 2  (left half plane):

- s0 = −0.3151 + j0.5457
- s1 = −0.6301
- s2 = −0.3151 − j0.5457

H(s) = Ωc³ / [(s + Ωc)(s² + Ωc s + Ωc²)] = 0.2502 / [(s + 0.6301)(s² + 0.6301s + 0.3971)] = Σ A_k/(s − s_k), with residues

- A0 = −0.3151 − j0.1819
- A1 = 0.6301
- A2 = −0.3151 + j0.1819

#### Step 6: impulse invariant transformation
Each term A/(s − s_k) becomes T·A/(1 − e^(s_k T) z⁻¹), so h[n] = T·h_a(nT). Combining the complex-conjugate pair into a real second-order section:

- 0.63015 / (1 − 0.53251z⁻¹)
- (−0.63015 + 0.53085z⁻¹) / (1 − 1.24748z⁻¹ + 0.53251z⁻²)

H(z) is the sum of these two terms (parallel form).

#### Step 7: check
|H(e^jωp)| = 0.9231 (required ≥ 0.9226); |H(e^jωs)| = 0.0360 (required ≤ 0.1995); |H(1)| = 0.9996. Both conditions are satisfied.

**Answer: N = 3, Ωc = 0.6301 rad/s, H(s) = 0.2502/[(s + 0.6301)(s² + 0.6301s + 0.3971)], H(z) = 0.63015/(1 − 0.53251z⁻¹) + (−0.63015 + 0.53085z⁻¹)/(1 − 1.24748z⁻¹ + 0.53251z⁻²).**

## Check yourself

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

1. What |H| are 0.7 dB loss and 14 dB attenuation?
   A. 0.7 and 0.14
   B. 0.9226 and 0.1995
   C. 0.5 and 0.2
   D. 0.99 and 0.05
   Answer: B. 10^(-0.7/20) = 0.9226 and 10^(-14/20) = 0.1995. (Key point: Specs to gains)

2. What N is found?
   A. N = 1
   B. N = 3
   C. N = 2 (1.78 rounded up)
   D. N = 4
   Answer: C. Rounded up from 1.78. (Key point: Order)

3. Why is the pass-band value 0.8986 and below target?
   A. Wrong T
   B. Butterworth has ripples
   C. Round-off in Omega_c
   D. Impulse invariance causes aliasing, so the response drops slightly
   Answer: D. Overlap of shifted spectra changes the response. (Key point: Aliasing effect)

4. What remedy does the answer suggest if the pass band must be met strictly?
   A. Use N = 3 with Omega_c >= 0.64
   B. Use N = 1
   C. Use bilinear with T=2
   D. Reduce Omega_s
   Answer: A. A higher order gives more margin against aliasing. (Key point: Fix)

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