Lattice structure of second order pole-zero system

A pole-zero IIR filter is realised as a lattice-ladder structure: the denominator gives the all-pole lattice (reflection coefficients K₁, K₂) and the numerator is built by tapping the lattice with ladder coefficients c₀, c₁, c₂.

Step 1: reflection coefficients from the denominator

A₂(z) = 1 + a₂(1)z⁻¹ + a₂(2)z⁻² = 1 − 0.3z⁻¹ + 0.4z⁻², so a₂(1) = −0.3, a₂(2) = 0.4.

• K₂ = a₂(2) = 0.4
• Step-down: a₁(1) = [a₂(1) − K₂·a₂(1)] / (1 − K₂²) = (−0.3 − 0.4×(−0.3)) / (1 − 0.16) = −0.18/0.84 = −0.2143
• K₁ = a₁(1) = −0.2143 (= −3/14)

Both |K| < 1, so the filter is stable.

Step 2: ladder coefficients from the numerator

B(z) = 0.5 + 2z⁻¹ + 0.6z⁻², so b₀ = 0.5, b₁ = 2, b₂ = 0.6. Use c_m = b_m − Σ (i = m+1 to 2) c_i · a_i(i−m):

• c₂ = b₂ = 0.6
• c₁ = b₁ − c₂·a₂(1) = 2 − 0.6×(−0.3) = 2.18
• c₀ = b₀ − c₁·a₁(1) − c₂·a₂(2) = 0.5 − 2.18×(−0.2143) − 0.6×0.4 = 0.7271

Step 3: structure

Equations of each stage (f = forward, g = backward signal):

• f₂[n] = x[n]
• f₁[n] = f₂[n] − K₂ g₁[n−1], f₀[n] = f₁[n] − K₁ g₀[n−1]
• g₀[n] = f₀[n]
• g₁[n] = K₁ f₀[n] + g₀[n−1], g₂[n] = K₂ f₁[n] + g₁[n−1]
• y[n] = c₀ g₀[n] + c₁ g₁[n] + c₂ g₂[n]

One lattice stage m (the filter has two such stages in cascade, stage 2 then stage 1):

 f_m[n] ──►(+)────────────────┬──► f_(m-1)[n]
            ▲  ╲             │
        −K_m│   ╲            │
            │    ╲           │
            │     ╳          │
            │    ╱           │
            │   ╱  K_m       │
            │  ╱             ▼
 g_m[n] ◄──(+)◄──────[z⁻¹]◄──┴── g_(m-1)[n]

Overall: x[n] = f₂ enters stage 2, f₁ enters stage 1, and f₀ = g₀ is the end of the lattice. Taps g₀, g₁, g₂ are multiplied by c₀, c₁, c₂ and summed to form y[n].

(Each stage has a forward adder, a backward adder, two cross multipliers −K and K, and one delay z⁻¹ in the backward path.)

I checked the equations by simulating the lattice against the direct difference equation y[n] = 0.3y[n−1] − 0.4y[n−2] + 0.5x[n] + 2x[n−1] + 0.6x[n−2]; the outputs are identical.

Answer: K₁ = −0.2143, K₂ = 0.4; c₀ = 0.7271, c₁ = 2.18, c₂ = 0.6.