---
title: "Impulse invariance Butterworth, 0.2π, 0.5 dB"
subject: "Digital Signal Analysis and Processing"
chapter: 6
key: "semester_7/digital_signal_analysis_and_processing:impulse-invariance-butter-0-5db"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/impulse-invariance-butter-0-5db
site: BCT Engineering
---

# Impulse invariance Butterworth, 0.2π, 0.5 dB

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

- Asked in: 2070 Asar, 13 marks
- Video: [Design digital filter using Using Impulse Invariant Transformation.](https://www.youtube.com/watch?v=B92Q-2NwM7U) — Engg-Course-Made-Easy
- 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-0-5db
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-6

## Question

Design a low pass digital filter by impulse invariance method to an approximate Butterworth filter, if passband edge frequency is 0.2π radians and maximum deviation of 0.5 dB below 0 dB gain in the passband. The maximum gain of −15 dB and frequency is 0.35π radian in stopband, consider sampling frequency 1 Hz.

## Short answer

Order N=5, Ωc=0.7754 rad/s. H(z) is realized as sum of three parallel sections: (−0.21432−0.22420z⁻¹)/(1−1.16493z⁻¹+0.61926z⁻²) + (−1.25466+1.00752z⁻¹)/(1−0.95901z⁻¹+0.28517z⁻²) + 1.46898/(1−0.46051z⁻¹).

## Answer

#### Step 1: specifications
Pass-band edge 0.2π with at most 0.5 dB loss (|H| ≥ 0.94406). Stop band: −15 dB (|H| ≤ 0.17783) at 0.35π. Sampling frequency 1 Hz, so T = 1 s.

#### Step 2: analog edge frequencies (impulse invariance: Ω = ω/T)
Ωp = **0.6283** rad/s, Ωs = **1.0996** rad/s  (T = 1 s)

#### Step 3: order
N ≥ log[(1/0.17783² − 1)/(1/0.94406² − 1)] / (2 log(Ωs/Ωp)) = log[30.6228/0.1220] / (2 log 1.7500) = 4.9367

**N = 5**

#### Step 4: cut-off
Ωc = Ωp / (0.1220)^(1/10) = **0.7754** rad/s

#### Step 5: analog poles and H(s)
Poles: s_k = Ωc exp[jπ(2k + N + 1)/(2N)], k = 0 … 4  (left half plane):

- s0 = −0.2396 + j0.7375
- s1 = −0.6273 + j0.4558
- s2 = −0.7754
- s3 = −0.6273 − j0.4558
- s4 = −0.2396 − j0.7375

H(s) = Ωc^5 / Π(s − s_k) = Σ A_k/(s − s_k), with residues

- A0 = −0.1072 + j0.3298
- A1 = −0.6273 − j0.8634
- A2 = 1.4690
- A3 = −0.6273 + j0.8634
- A4 = −0.1072 − j0.3298

#### 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 each complex-conjugate pair into a real second-order section:

- (−0.21432 − 0.22420z⁻¹) / (1 − 1.16493z⁻¹ + 0.61926z⁻²)
- (−1.25466 + 1.00752z⁻¹) / (1 − 0.95901z⁻¹ + 0.28517z⁻²)
- 1.46898 / (1 − 0.46051z⁻¹)

H(z) is the sum of these terms (parallel form). Aliasing makes the digital pass-band gain fall a hair below the target (0.94403 against 0.94406, i.e. a loss of 0.5003 dB instead of 0.5 dB). If the specification must be met strictly, a slightly larger cut-off such as Ωc = 0.776 rad/s gives 0.9444 at ωp and still keeps the stop band below 0.1726.

#### Step 7: check
|H(e^jωp)| = 0.94403 (required ≥ 0.94406, short by 0.00003); |H(e^jωs)| = 0.1719 (required ≤ 0.17783); |H(1)| = 1.0000.

**Answer: N = 5, Ωc = 0.7754 rad/s, H(z) = sum of the three parallel sections above.**

## Check yourself

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

1. How are digital and analog edges related in impulse invariance?
   A. Omega = (2/T) tan(w/2)
   B. Omega = w/T (linear)
   C. Omega = T w
   D. Omega = ln(w)/T
   Answer: B. Impulse invariance preserves the frequency axis linearly, so no pre-warping. (Key point: Frequency mapping)

2. What |H| does a 0.5 dB loss correspond to?
   A. 0.5
   B. 0.891
   C. 0.94406
   D. 0.995
   Answer: C. 10^(-0.5/20) = 0.94406. (Key point: Pass-band spec)

3. What is N?
   A. N = 4
   B. N = 3
   C. N = 6
   D. N = 5 (4.94 rounded up)
   Answer: D. The formula gives 4.94, so round up. (Key point: Order)

4. Under impulse invariance, each pole term A/(s - s_k) becomes
   A. T*A/(1 - e^(s_k T) z^-1)
   B. A/(1 + s_k T z^-1)
   C. A/(1 - s_k z^-1)
   D. T*A/(s_k - z^-1)
   Answer: A. Poles map to z_k = e^(s_k T). (Key point: Transformation)

## 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×

## Related in Digital Signal Analysis and Processing

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- [Bilinear Butterworth, 0.27π and 0.58π](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Abilinear-butter-0-27-0-58) — Digital Signal Analysis and Processing, chapter 6

## Related in other subjects

- [Impulse response of ideal low pass filter](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aideal-lpf-impulse-response) — Signals and Systems, chapter 5, asked 5×
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