Lattice coefficients of 1+2z⁻¹+z⁻²

A₂(z) = 1 + 2z⁻¹ + z⁻² (M = 2, already monic).

Computation

• Last stage: K₂ = a₂(2) = 1.
• The step-down division 1 − K₂² = 0 is 0/0 here, because a₂(1) − K₂a₂(1) = 0 as well. Use the build-up relation instead: A₂(z) = A₁(z) + K₂z⁻²A₁(z⁻¹) with A₁ = 1 + K₁z⁻¹ gives 1 + K₁(1 + K₂)z⁻¹ + K₂z⁻² = 1 + 2K₁z⁻¹ + z⁻². Matching with 1 + 2z⁻¹ + z⁻²: 2K₁ = 2, so K₁ = a₂(1)/(1 + a₂(2)) = 2/2 = 1.

Lattice coefficients: K₁ = 1, K₂ = 1 (both have magnitude 1, consistent with the double zero at z = −1 on the unit circle).

Structure

Stage equations (m = 1, 2): fm(n) = f(m−1)(n) + Km g(m−1)(n−1); gm(n) = Km f(m−1)(n) + g(m−1)(n−1); f0 = g0 = x(n), y(n) = f2(n).

f(m-1)[n] ──────●────────────────────(+)──► f(m)[n]
                │         ┌───────────▲
                │ Km      │ Km
                └─────────┼───────────┐
                          │           │
                          │           ▼
g(m-1)[n] ─[z⁻¹] ─────────●──────────(+)──► g(m)[n]

Two such stages in cascade, with K1 = 1 in the first and K2 = 1 in the second:

        +------+     +------+
x[n]--->| K1=1 |---->| K2=1 |---> y[n] = f2[n]
 g0=f0  +------+     +------+

Check: f1 = x(n) + x(n−1); f2 = f1(n) + g1(n−1), g1 = x(n) + x(n−1) = f1, so f2 = x(n) + 2x(n−1) + x(n−2). ✓