---
title: "Linear phase FIR condition; check h(n)={−1,0,1}"
subject: "Digital Signal Analysis and Processing"
chapter: 5
key: "semester_7/digital_signal_analysis_and_processing:linear-phase-fir-condition"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/linear-phase-fir-condition
site: BCT Engineering
---

# Linear phase FIR condition; check h(n)={−1,0,1}

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

- Asked in: 2070 Asar, 2+4 marks
- 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/linear-phase-fir-condition
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-5

## Question

What is the condition satisfied by linear phase FIR filter? Show that the filter with h(n) = {−1, 0, 1} is a linear phase filter.

## Short answer

An FIR filter has linear phase if its impulse response is symmetric (h(n)=h(M−1−n)) or anti-symmetric (h(n)=−h(M−1−n)). The sequence {−1,0,1} is anti-symmetric with h(0)=−h(2) and h(1)=0, meeting the condition, so it is a linear-phase filter.

## Answer

#### Condition for linear phase
An FIR filter of length M has exactly linear phase if its impulse response is either

- **symmetric:** h(n) = h(M − 1 − n), or
- **anti-symmetric:** h(n) = −h(M − 1 − n),  for 0 ≤ n ≤ M − 1.

The phase is then θ(ω) = −αω (symmetric) or θ(ω) = β − αω with β = ±π/2 (anti-symmetric), with a constant delay α = (M − 1)/2. Anti-symmetric filters need h(α) = 0 when M is odd.

#### Check h(n) = {−1, 0, 1}
Here M = 3, so M − 1 = 2.

- h(0) = −1 and h(2) = 1, so h(0) = −h(2).
- h(1) = 0, and −h(1) = 0.

Hence h(n) = −h(2 − n): the sequence is **anti-symmetric**, and the centre sample is zero as required (M odd). The delay is α = 1.

Frequency response:

H(e^jω) = −1 + 0·e^{−jω} + e^{−j2ω} = e^{−jω}(e^{−jω} − e^{jω}) = e^{−jω} · (−2j sin ω)

H(e^jω) = 2 sin ω · e^{−j(ω + π/2)}

So the amplitude is 2 sin ω and the phase is θ(ω) = −ω − π/2 (taking the sign of sin ω into account, it is −ω − π/2 for 0 < ω < π). This is a straight line in ω with slope −1 (constant group delay of 1 sample) and intercept −π/2.

**Hence h(n) = {−1, 0, 1} is a linear-phase filter (Type III, a differentiator-like response).**

## Check yourself

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

1. Which condition on h(n) of length M gives an exactly linear-phase FIR filter?
   A. h(n) = h(M-1-n) or h(n) = -h(M-1-n)
   B. h(n) = h(n+M) for all n
   C. h(n) = 0 for n < 0 and h(n) real only
   D. h(n) = h(M-n) with no sign change allowed
   Answer: A. Linear phase needs the impulse response to be symmetric or anti-symmetric about its centre (M-1)/2. (Key point: Linear phase condition)

2. For a linear-phase FIR filter of length M, what is the constant delay alpha?
   A. M/2 + 1
   B. (M-1)/2
   C. M-1
   D. M
   Answer: B. The centre of symmetry is at (M-1)/2, which is the delay in samples. (Key point: Constant delay)

3. For an anti-symmetric h(n) with M odd, what must hold for the centre sample?
   A. It must equal 1
   B. It must be the largest sample
   C. h((M-1)/2) = 0
   D. It can have any value
   Answer: C. Anti-symmetry h(a) = -h(a) at the centre forces it to be zero. (Key point: Anti-symmetric centre sample)

4. Why is h(n) = {-1, 0, 1} a linear-phase filter?
   A. h(n) = h(2-n), so it is symmetric
   B. It has only three samples, so any 3-tap filter is linear phase
   C. The sum of its samples is zero, which gives linear phase
   D. h(n) = -h(2-n), so it is anti-symmetric with a zero centre sample
   Answer: D. h(0) = -1 = -h(2) and h(1) = 0, which satisfies the anti-symmetric condition. (Key point: Checking {-1,0,1})

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