---
title: "Kaiser low-pass FIR, 0.19π and 0.21π"
subject: "Digital Signal Analysis and Processing"
chapter: 5
key: "semester_7/digital_signal_analysis_and_processing:kaiser-lowpass-0-19-0-21"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/kaiser-lowpass-0-19-0-21
site: BCT Engineering
---

# Kaiser low-pass FIR, 0.19π and 0.21π

Digital Signal Analysis and Processing · Chapter 5: FIR Filter Design

- Asked in: 2072 Kartik, 8 marks
- Video: [FIR Filter Design NUMERICAL Problem (Kaiser Window) | Step-by-Step in DSP](https://www.youtube.com/watch?v=MdhS91q4Pm0) — EC Academy
- 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/kaiser-lowpass-0-19-0-21
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-5

## Question

Design an FIR linear phase filter using Kaiser window to meet the following specifications: 0.99 ≤ |H(e^jw)| ≤ 1.01 for 0 ≤ w ≤ 0.19π; |H(e^jw)| ≤ 0.01 for 0.21π ≤ w ≤ π.

## Short answer

Kaiser lowpass: δ=0.01, A=40 dB, β=3.395, Δω=0.02π, M=224 (225 taps), cutoff 0.2π; h(n)=hd(n)w(n). The formula minimum is 223; M=226 meets both passband and stopband tolerances strictly.

## Answer

#### Specifications
Pass band: 0.99 ≤ |H| ≤ 1.01 for 0 ≤ ω ≤ 0.19π, so δ₁ = 0.01. Stop band: |H| ≤ 0.01 for 0.21π ≤ ω ≤ π, so δ₂ = 0.01. Hence ωp = 0.19π, ωs = 0.21π.

#### Kaiser parameters
δ = 0.01, A = −20 log₁₀(0.01) = **40 dB**.

Δω = ωs − ωp = 0.02π

β = 0.5842(19)^0.4 + 0.07886(19) = **3.395**

M = (A − 8)/(2.285 Δω) = 32/(2.285 × 0.02π) = 222.9, so M ≥ 223. Take **M = 224** (even, so α = M/2 = 112 is an integer and the filter is Type I); length 225.

#### Cut-off
ωc = (0.19π + 0.21π)/2 = **0.2π**

#### Ideal response
h_d(n) = sin(0.2π(n − 112)) / (π(n − 112)),  n ≠ 112;  h_d(112) = 0.2

#### Kaiser window
w(n) = I₀(3.395 √(1 − ((n − 112)/112)²)) / I₀(3.395),  0 ≤ n ≤ 224

#### Impulse response
h(n) = h_d(n) w(n) = h(224 − n):

| n | h_d(n) | w(n) | h(n) |
|---|---|---|---|
| 0 | 0.00270 | 0.1480 | 0.00040 |
| 1 | 0.00169 | 0.1556 | 0.00026 |
| 2 | 0.00000 | 0.1635 | 0.00000 |
| 3 | −0.00172 | 0.1714 | −0.00029 |
| 4 | −0.00280 | 0.1794 | −0.00050 |
| 5 | −0.00283 | 0.1876 | −0.00053 |
| 112 | 0.20000 | 1.0000 | 0.20000 |

Numerical check of the 225-tap filter: the stop-band peak is 0.0097, so the stop band is met. The pass band stays between 0.9899 and 1.0097; its lowest value, right at the edge 0.19π, is a hair below 0.99, because the Kaiser length formula is only an estimate. Taking M = 226 (227 taps) gives a pass band of 0.9912 to 1.0093 and a stop-band peak of 0.0094, meeting both tolerances strictly.

**Answer: A = 40 dB, β = 3.395, M = 224 (formula minimum 223; M = 226 meets the pass band strictly), ωc = 0.2π, h(n) = h_d(n)·w(n).**

## Check yourself

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

1. Pass-band ripple 0.01 and stop-band 0.01 give which A?
   A. 40 dB
   B. 20 dB
   C. 80 dB
   D. 0.01 dB
   Answer: A. A = -20 log10(0.01) = 40 dB. (Key point: Delta and A)

2. What is delta-omega for this design?
   A. 0.19 pi
   B. 0.02 pi
   C. 0.21 pi
   D. 0.4 pi
   Answer: B. It is the stop-band edge minus the pass-band edge: 0.21 pi - 0.19 pi. (Key point: Transition width)

3. Why is M = 224 chosen when the formula gives about 222.9?
   A. Because 224 is a power of 2
   B. To make the filter anti-symmetric
   C. Round up to 223 and use even M so that alpha = 112 is an integer (Type I)
   D. To increase beta
   Answer: C. An even M gives integer delay and a symmetric odd-length Type I filter. (Key point: Length)

4. What is the cut-off frequency of the ideal low-pass?
   A. 0.19 pi
   B. 0.21 pi
   C. 0.02 pi
   D. 0.2 pi
   Answer: D. It is the mid-point (0.19 pi + 0.21 pi)/2 = 0.2 pi. (Key point: Cut-off)

## Close variants set in other exams

- [Kaiser low-pass FIR, passband 0.9π, stop 0.14π](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Akaiser-lowpass-0-9-0-14) — Digital Signal Analysis and Processing, chapter 5, asked 2×

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- [Low-pass FIR window design, αs = 40 dB](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Afir-lowpass-window-0-3-0-5-40) — Digital Signal Analysis and Processing, chapter 5, asked 2×
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