---
title: "Kaiser bandpass FIR; minimum length and β"
subject: "Digital Signal Analysis and Processing"
chapter: 5
key: "semester_7/digital_signal_analysis_and_processing:kaiser-bandpass-length-beta"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/kaiser-bandpass-length-beta
site: BCT Engineering
---

# Kaiser bandpass FIR; minimum length and β

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

- Asked in: 2074 Chaitra, 9+3 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-bandpass-length-beta
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-5

## Question

Design a FIR linear phase filter using Kaiser window that meets the following specifications: |H(e^jw)| ≤ 0.01, 0 ≤ |w| ≤ 0.25π; 0.95 ≤ |H(e^jw)| ≤ 1.05, 0.35π ≤ |w| ≤ 0.6π; |H(e^jw)| ≤ 0.01, 0.65π ≤ |w| ≤ π. Also determine the minimum length (M+1) of the impulse response and Kaiser window parameter β.

## Short answer

Kaiser bandpass design: δ=0.01, A=40 dB, β=3.395, Δω=0.05π, M=90 (91 taps), cutoffs 0.30π and 0.625π; h(n)=hd(n)w(n). The Kaiser formula is an estimate, so use M=92 (93 taps) to meet the 0.01 stopband limit strictly.

## Answer

#### Specifications
- Stop band 1: 0 ≤ |ω| ≤ 0.25π, δ₁ = 0.01
- Pass band: 0.35π ≤ |ω| ≤ 0.6π, tolerance ±0.05 about 1, δ₂ = 0.05
- Stop band 2: 0.65π ≤ |ω| ≤ π, δ₃ = 0.01

#### Kaiser design parameters
The same δ must serve every band, so δ = min(0.01, 0.05, 0.01) = **0.01**.

A = −20 log₁₀ δ = −20 log₁₀(0.01) = **40 dB**

Transition widths: Δω₁ = 0.35π − 0.25π = 0.10π; Δω₂ = 0.65π − 0.6π = 0.05π. Use the narrower one: **Δω = 0.05π**.

#### β
Since 21 ≤ A ≤ 50:

β = 0.5842(A − 21)^0.4 + 0.07886(A − 21) = 0.5842(19)^0.4 + 0.07886×19 = 1.8946 + 1.4983 = **3.395**

#### Minimum length
M = (A − 8) / (2.285 Δω) = 32 / (2.285 × 0.05π) = 32/0.3589 = 89.15

Round up, keeping M even so that α = M/2 is an integer: **M = 90**, so the impulse response has **M + 1 = 91 samples**, α = 45.

#### Cut-offs (mid-points of the transitions)
ωc1 = (0.25π + 0.35π)/2 = 0.30π,  ωc2 = (0.6π + 0.65π)/2 = 0.625π

#### Ideal band-pass response
h_d(n) = [sin(0.625π(n − 45)) − sin(0.3π(n − 45))] / (π(n − 45)),  n ≠ 45;  h_d(45) = (0.625π − 0.3π)/π = 0.325

#### Kaiser window
w(n) = I₀(3.395 √(1 − ((n − 45)/45)²)) / I₀(3.395),  0 ≤ n ≤ 90

#### Filter
h(n) = h_d(n) · w(n), with h(n) = h(90 − n). First values:

| n | h_d(n) | w(n) | h(n) |
|---|---|---|---|
| 0 | 0.00978 | 0.1480 | 0.00145 |
| 1 | −0.00298 | 0.1673 | −0.00050 |
| 2 | 0.00055 | 0.1874 | 0.00010 |
| 3 | −0.00185 | 0.2082 | −0.00039 |
| 45 | 0.32500 | 1.0000 | 0.32500 |

Numerical check of the full 91-tap filter: the pass band stays between 0.990 and 1.010 (well inside 0.95 to 1.05) and stop band 1 peaks at 0.0059, but stop band 2 peaks at 0.0102, slightly above the 0.01 limit (−39.8 dB instead of −40 dB). The Kaiser length formula is only an estimate; increasing the length to M = 92 (93 taps) brings the stop-band 2 peak down to 0.0094, and then every band meets its specification.

**Answer: δ = 0.01 (A = 40 dB), β = 3.395, M = 90, minimum length from the Kaiser formula M + 1 = 91 samples (M = 92, i.e. 93 taps, if the 0.01 limit must be met strictly); h(n) = h_d(n)·w(n) as above.**

## Check yourself

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

1. Why is delta = 0.01 used for the Kaiser design of this band-pass filter?
   A. One delta must serve all bands, so the smallest tolerance (0.01) is used
   B. The average of 0.01, 0.05 and 0.01 is used
   C. The largest tolerance 0.05 is used to save length
   D. Only the pass-band tolerance matters
   Answer: A. The Kaiser window gives the same ripple in every band, so the strictest tolerance governs. (Key point: Choosing delta)

2. For delta = 0.01, what is A = -20 log10(delta)?
   A. 20 dB
   B. 40 dB
   C. 60 dB
   D. 0.01 dB
   Answer: B. -20 log10(0.01) = 40 dB. (Key point: Attenuation A)

3. Which transition width is used to find the length?
   A. The wider one, 0.10 pi
   B. Their average, 0.075 pi
   C. The narrower one, 0.05 pi
   D. Their sum, 0.15 pi
   Answer: C. The sharper transition needs the longer filter, so the narrowest width is used for both. (Key point: Transition width)

4. Which formula gives beta here, and what value?
   A. 0.1102(A-8.7), beta = 3.45
   B. 0.5842(A-21)^0.4 only, beta = 1.89
   C. beta = A/12, beta = 3.33
   D. 0.5842(A-21)^0.4 + 0.07886(A-21), beta = 3.395
   Answer: D. A = 40 dB lies between 21 and 50, so the middle formula applies, giving 3.395. (Key point: Beta for 21<=A<=50)

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