Lattice coefficients for fractional FIR

Simplify: 52/96 = 13/24 and 25/40 = 5/8, so A₃(z) = 1 + (13/24)z⁻¹ + (5/8)z⁻² + (1/3)z⁻³.

Method

The FIR lattice of order M uses the stage equations

f0(n) = g0(n) = x(n)
fm(n) = f(m−1)(n) + Km · g(m−1)(n−1)
gm(n) = Km · f(m−1)(n) + g(m−1)(n−1)     m = 1 … M
y(n)  = fM(n)

The reflection coefficient of the last stage is Km = am(m) (the last coefficient of the polynomial Am(z)). The lower-order polynomials come from the step-down recursion

a(m−1)(i) = [ am(i) − Km · am(m−i) ] / (1 − Km²),   i = 1 … m−1

Computation

A3(z) = 1 + 0.5417z⁻¹ + 0.625z⁻² + 0.3333z⁻³

• A3: coefficients {1, 0.5417, 0.625, 0.3333}, so K3 = a3(3) = 0.3333
• A2: coefficients {1, 0.375, 0.5}, so K2 = a2(2) = 0.5
• A1: coefficients {1, 0.25}, so K1 = a1(1) = 0.25

Lattice coefficients: K1 = 0.25, K2 = 0.5, K3 = 0.3333

Structure

One lattice stage (repeated for m = 1 … 3 with K1 … K3):

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

Cascade (f and g pass together from stage to stage):

        +------+     +------+     +------+
x[n]-+->|      |-f1->|      |-f2->|      |-f3-> y[n] = f3[n]
     |  |  K1  |     |  K2  |     |  K3  |
     +->|      |-g1->|      |-g2->|      |
        +------+     +------+     +------+

(The upper path carries f and the lower path carries g; f0 = g0 = x[n].)

Exact values: K₁ = 1/4, K₂ = 1/2, K₃ = 1/3. All |K| < 1, so all the zeros are inside the unit circle (minimum-phase FIR).

Check: build-up: A₁ = 1 + ¼z⁻¹; A₂ = A₁ + ½z⁻²A₁(z⁻¹) = 1 + 0.375z⁻¹ + 0.5z⁻²; A₃ = A₂ + ⅓z⁻³A₂(z⁻¹) = 1 + 0.5417z⁻¹ + 0.625z⁻² + 0.3333z⁻³. ✓