Lattice structure of fourth order IIR

Approach

The system is H(z) = B(z)/A(z) with

• B(z) = 1 + (1/3)z⁻¹ + (9/8)z⁻² + (4/3)z⁻³ + z⁻⁴
• A(z) = 1 + (2/3)z⁻¹ + (5/8)z⁻² + (2/3)z⁻³ + z⁻⁴

A pole-zero IIR system is realised by a lattice-ladder: the lattice built from A(z) gives the reflection coefficients K₁ … K₄, and the numerator is formed by tapping the g-nodes with ladder coefficients v₀ … v₄ and adding: y(n) = Σ v_m g_m(n).

Step 1. Reflection coefficients of A(z)

Step-down: K_m = a_m(m), a_(m−1)(i) = [a_m(i) − K_m a_m(m−i)]/(1 − K_m²).

• A₄ = {1, 2/3, 5/8, 2/3, 1}: K₄ = a₄(4) = 1.

K₄ has magnitude 1 because A(z) is symmetric (a₄(i) = a₄(4−i)), so all four poles lie exactly on the unit circle (computed radius = 1.000, angles about ±63.7° and ±140.9°). For such a polynomial the step-down division is 0/0, so A₃ is not unique: any A₃ with a₃(i) + a₃(4−i) = a₄(i) steps back up to A₄ when K₄ = 1. Taking the symmetric split a₃(i) = a₄(i)/2 for i = 1 … 3 gives:

• A₃ = {1, 1/3, 5/16, 1/3}: K₃ = 1/3
• 1 − K₃² = 8/9: a₂(1) = (1/3 − (1/3)(5/16))·(9/8) = 33/128, a₂(2) = (5/16 − 1/9)·(9/8) = 29/128. A₂ = {1, 33/128, 29/128}: K₂ = 29/128 ≈ 0.2266
• a₁(1) = a₂(1)/(1 + K₂) = (33/128)/(157/128) = 33/157. K₁ = 33/157 ≈ 0.2102

Check: building up A₁ → A₂ → A₃ → A₄ with these K's gives back {1, 2/3, 5/8, 2/3, 1}. ✓

Step 2. Ladder coefficients from the numerator

B(z) = Σ v_m B̃_m(z), where B̃_m(z) = z⁻ᵐA_m(z⁻¹) is the reversed polynomial. Starting from the top: v_m = b_m − Σ(i = m+1 to 4) v_i a_i(i − m).

• v₄ = b₄ = 1
• v₃ = b₃ − v₄a₄(1) = 4/3 − 2/3 = 2/3
• v₂ = b₂ − v₃a₃(1) − v₄a₄(2) = 9/8 − (2/3)(1/3) − 5/8 = 5/18 ≈ 0.2778
• v₁ = b₁ − v₂a₂(1) − v₃a₃(2) − v₄a₄(3) = 1/3 − (5/18)(33/128) − (2/3)(5/16) − 2/3 = −157/256 ≈ −0.6133
• v₀ = b₀ − v₁a₁(1) − v₂a₂(2) − v₃a₃(3) − v₄a₄(4) = 1 + (157/256)(33/157) − (5/18)(29/128) − (2/3)(1/3) − 1 = −5/32 ≈ −0.1563

Structure (lattice-ladder)

Stage equations: f₄(n) = x(n); f(m−1)(n) = fm(n) − Km g(m−1)(n−1); gm(n) = Km f(m−1)(n) + g(m−1)(n−1); g₀ = f₀; output y(n) = Σ(m = 0 to 4) vm gm(n).

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

        +------+   +------+   +------+   +------+
x[n]--->| K4=1 |-->|K3=1/3|-->|K2=.23|-->|K1=.21|--> f0=g0
 =f4    +------+   +------+   +------+   +------+
          g4         g3         g2         g1      g0
          |v4        |v3        |v2        |v1     |v0
          +----------+----(+)---+----------+-------+--> y[n]

(Each g_m node is multiplied by v_m, and all the products are added to form y(n).)

Verification: simulating this lattice-ladder with a unit impulse reproduces the impulse response of B(z)/A(z) exactly (first 30 samples). ✓

Stability: |K₄| = 1, so the system is not strictly stable: all poles are on the unit circle (marginally stable, undamped oscillation), consistent with the pole radii found from A(z).