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

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).