---
title: "Impulse invariance Butterworth, 200 Hz and 500 Hz"
subject: "Digital Signal Analysis and Processing"
chapter: 6
key: "semester_7/digital_signal_analysis_and_processing:impulse-invariance-butter-200-500hz"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/impulse-invariance-butter-200-500hz
site: BCT Engineering
---

# Impulse invariance Butterworth, 200 Hz and 500 Hz

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

- Asked in: 2072 Chaitra, 12 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-200-500hz
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-6

## Question

Design a digital low pass Butterworth filter using impulse invariant transformation with pass band and stop band frequencies 200 Hz and 500 Hz respectively. The pass band and stop band attenuation are −5 dB and −12 dB respectively. The sampling frequency is 5 kHz. Use IIM method.

## Short answer

Order N=2, analog cut-off Ωc=1036.3 rad/s. The digital filter is H(z)=0.03697z⁻¹/(1−1.7088z⁻¹+0.7459z⁻²).

## Answer

#### Step 1: specifications
Pass-band edge 200 Hz with −5 dB (|H| ≥ 0.5623), stop-band edge 500 Hz with −12 dB (|H| ≤ 0.2512), Fs = 5 kHz, T = 0.2 ms.

#### Step 2: analog edge frequencies (impulse invariance: Ω = ω/T)
Ωp = **1256.6371** rad/s, Ωs = **3141.5927** rad/s  (T = 0.0002 s)

#### Step 3: order
N ≥ log[(1/0.2512² − 1)/(1/0.5623² − 1)] / (2 log(Ωs/Ωp)) = log[14.8489/2.1623] / (2 log 2.5000) = 1.0514

**N = 2**

#### Step 4: cut-off
Ωc = Ωp / (2.1623)^(1/4) = **1036.2916** rad/s

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

- s0 = -732.7689 + j732.7689
- s1 = -732.7689 − j732.7689

H(s) = Ωc^2 / Π(s − s_k) = Σ A_k/(s − s_k), with residues

- A0 = 0.0000 − j732.7689
- A1 = -0.0000 + j732.7689

#### 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:

- (0.00000 + 0.03697z⁻¹) / (1 − 1.70884z⁻¹ + 0.74594z⁻²)

H(z) is the sum of these terms (parallel form). Here the factor T = 0.0002 is included in the numerator so that the DC gain of the digital filter equals that of the analog filter (if T is left out, as some textbooks do, the numerator becomes 1/T = 5000 times larger, i.e. 184.8 z⁻¹).

#### Step 7: check
|H(e^jωp)| = 0.5633 (required ≥ 0.5623); |H(e^jωs)| = 0.1114 (required ≤ 0.2512); |H(1)| = 0.9964.

**Answer: N = 2, Ωc = 1036.3 rad/s, H(s) = Ωc²/(s² + 1465.5 s + Ωc²), H(z) = 0.03697 z⁻¹ / (1 − 1.7088z⁻¹ + 0.7459z⁻²).**

## Check yourself

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

1. How are 200 Hz and 500 Hz with Fs = 5 kHz turned into analog edges?
   A. w = 2 pi f/Fs: 0.08 pi and 0.2 pi used directly in tan
   B. Omega = 2 pi f, i.e. 1256.6 and 3141.6 rad/s
   C. Omega = f: 200 and 500
   D. Omega = 2 pi f x T
   Answer: B. Omega = w/T = 2 pi f for impulse invariance. (Key point: Edge frequencies)

2. What |H| is -5 dB and -12 dB?
   A. 0.5 and 0.12
   B. 0.316 and 0.063
   C. 0.5623 and 0.2512
   D. 0.891 and 0.794
   Answer: C. 10^(-5/20) = 0.5623 and 10^(-12/20) = 0.2512. (Key point: Attenuation to gain)

3. What is the order?
   A. N = 1
   B. N = 3
   C. N = 5
   D. N = 2 (1.05 rounded up)
   Answer: D. 1.05 is rounded up to 2. (Key point: Order)

4. Why does the numerator include the factor T?
   A. So the digital DC gain equals the analog DC gain
   B. To make the filter linear phase
   C. To cancel aliasing
   D. T has no effect
   Answer: A. h[n] = T*h_a(nT), so T scales the numerator. (Key point: Role of T)

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