Lattice structure of 1+2z⁻¹−3z⁻²+4z⁻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 + 2z⁻¹ − 3z⁻² + 4z⁻³

• A3: coefficients {1, 2, -3, 4}, so K3 = a3(3) = 4
• A2: coefficients {1, -0.9333, 0.7333}, so K2 = a2(2) = 0.7333
• A1: coefficients {1, -0.5385}, so K1 = a1(1) = -0.5385

Lattice coefficients: K1 = -0.5385, K2 = 0.7333, K3 = 4

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].)

Check: rebuilding A3 from K1 = −7/13, K2 = 11/15, K3 = 4 with A_m(z) = A_(m−1)(z) + K_m z⁻ᵐ A_(m−1)(z⁻¹) reproduces {1, 2, −3, 4}, and simulating the lattice with a unit impulse gives the impulse response 1, 2, −3, 4. ✓ Since |K3| = 4 > 1, the filter has zeros outside the unit circle (it is not minimum-phase).