---
title: "Kaiser low-pass FIR, passband 0.9π, stop 0.14π"
subject: "Digital Signal Analysis and Processing"
chapter: 5
key: "semester_7/digital_signal_analysis_and_processing:kaiser-lowpass-0-9-0-14"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/kaiser-lowpass-0-9-0-14
site: BCT Engineering
---

# Kaiser low-pass FIR, passband 0.9π, stop 0.14π

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

- Asked in: 2073 Chaitra, 8 marks
- Video: [Kaiser Windowing and Quality-Measuring a Lowpass FIR Digital Filter](https://www.youtube.com/watch?v=mW5VWORoLPU) — DSP Creations
- 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-0-9-0-14
- 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: 0.98 ≤ |H(e^jw)| ≤ 1.02 for 0 ≤ w ≤ 0.9π; |H(e^jw)| ≤ 0.01 for 0.14π ≤ w ≤ π.

## Short answer

The printed passband to 0.9π overlaps the stopband; the intended passband edge is 0.09π. Kaiser design gives δ=0.01, A=40 dB, β=3.395, Δω=0.05π, M=90 (91 taps), cutoff 0.115π. Use M=94 (95 taps) to meet the 0.01 stopband strictly.

## Answer

#### How the specification is read
As printed, the pass band (up to 0.9π) overlaps the stop band (from 0.14π), which is impossible for a low-pass filter. The pass-band edge must be **0.09π** (the 0 was misplaced). I design for ωp = 0.09π, ωs = 0.14π, with pass-band tolerance δ₁ = 0.02 (0.98 to 1.02) and stop-band tolerance δ₂ = 0.01.

#### Kaiser parameters
δ = min(δ₁, δ₂) = 0.01, so A = −20 log₁₀ 0.01 = **40 dB**.

Δω = ωs − ωp = 0.14π − 0.09π = 0.05π

β = 0.5842(40 − 21)^0.4 + 0.07886(40 − 21) = **3.395**

M = (A − 8)/(2.285 Δω) = 32/(2.285 × 0.05π) = 89.15, so **M = 90** (α = M/2 = 45, length 91).

#### Cut-off
ωc = (0.09π + 0.14π)/2 = **0.115π**

#### Ideal low-pass response
h_d(n) = sin(0.115π(n − 45)) / (π(n − 45)),  n ≠ 45;  h_d(45) = 0.115

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

#### Impulse response
h(n) = h_d(n) · w(n), h(n) = h(90 − n):

| n | h_d(n) | w(n) | h(n) |
|---|---|---|---|
| 0 | −0.00370 | 0.1480 | −0.00055 |
| 1 | −0.00136 | 0.1673 | −0.00023 |
| 2 | 0.00127 | 0.1874 | 0.00024 |
| 3 | 0.00386 | 0.2082 | 0.00080 |
| 4 | 0.00606 | 0.2298 | 0.00139 |
| 5 | 0.00757 | 0.2519 | 0.00191 |
| 45 | 0.11500 | 1.0000 | 0.11500 |

H(z) = Σ (n = 0 to 90) h(n) z⁻ⁿ, with linear phase −45ω.

Numerical check: the pass band lies between 0.992 and 1.011, so the 0.02 tolerance is met, but the stop-band peak is 0.0105 (−39.6 dB), slightly above the 0.01 limit. This is because the Kaiser length formula is only an estimate. Increasing the length to M = 94 (95 taps) lowers the stop-band peak to 0.0099 (pass band 0.994 to 1.010), and both tolerances are then met.

**Answer: A = 40 dB, β = 3.395, M = 90 (91 taps) from the Kaiser formula (M = 94 to meet the 0.01 stop-band limit strictly), ωc = 0.115π, 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. Why is the pass-band edge read as 0.09 pi instead of 0.9 pi?
   A. A pass band to 0.9 pi would overlap the stop band starting at 0.14 pi
   B. Kaiser windows only work below 0.1 pi
   C. 0.9 pi gives an unstable filter
   D. The stop band must start at 0.9 pi
   Answer: A. A low-pass pass-band edge must be below the stop-band edge, so 0.09 pi is the sensible reading. (Key point: Interpreting the spec)

2. How is the Kaiser delta chosen when delta1 = 0.02 and delta2 = 0.01?
   A. The larger one, 0.02, giving A = 34 dB
   B. The smaller one, 0.01, giving A = 40 dB
   C. Their sum, 0.03
   D. Their product, 0.0002
   Answer: B. The same ripple applies in both bands, so the stricter tolerance is used. (Key point: Delta and A)

3. With A = 40 dB and delta-omega = 0.05 pi, which gives the filter order?
   A. M = 40/0.05 = 800
   B. M = 89.15 kept as non-integer
   C. M = (A-8)/(2.285 d_omega) = 89.15, take M = 90 so that alpha = 45 is an integer
   D. M = 45
   Answer: C. Round up and use even M so the delay M/2 is an integer; length is 91. (Key point: Length M)

4. What cut-off is used for the ideal low-pass h_d(n)?
   A. 0.09 pi, the pass-band edge
   B. 0.14 pi, the stop-band edge
   C. 0.05 pi, the transition width
   D. 0.115 pi, the mid-point of 0.09 pi and 0.14 pi
   Answer: D. The ideal filter cut-off is the middle of the transition band. (Key point: Cut-off)

## Close variants set in other exams

- [Kaiser low-pass FIR, 0.19π and 0.21π](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Akaiser-lowpass-0-19-0-21) — Digital Signal Analysis and Processing, chapter 5, asked 2×

## Related in Digital Signal Analysis and Processing

- [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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