System function from all-pole lattice K's

For an all-pole lattice, H(z) = 1/A₃(z). A₃(z) is found from the reflection coefficients by the step-up (Levinson) recursion:

a_m(m) = K_m, a_m(i) = a_{m−1}(i) + K_m · a_{m−1}(m−i), i = 1, …, m−1

Given K₁ = 1/4, K₂ = 1/2, K₃ = 1/3.

Stage 1

a₁(1) = K₁ = 0.25

Stage 2

• a₂(2) = K₂ = 0.5
• a₂(1) = a₁(1) + K₂·a₁(1) = 0.25 + 0.5×0.25 = 0.375

Stage 3

• a₃(3) = K₃ = 1/3
• a₃(1) = a₂(1) + K₃·a₂(2) = 0.375 + (1/3)(0.5) = 0.5417 (= 13/24)
• a₃(2) = a₂(2) + K₃·a₂(1) = 0.5 + (1/3)(0.375) = 0.625 (= 5/8)

So

A₃(z) = 1 + (13/24)z⁻¹ + (5/8)z⁻² + (1/3)z⁻³

System function

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

The difference equation is y[n] = x[n] − 0.5417y[n−1] − 0.625y[n−2] − 0.3333y[n−3]. Since |K₁|, |K₂|, |K₃| < 1, all poles lie inside the unit circle and the filter is stable.

Answer: H(z) = 1 / (1 + 0.5417z⁻¹ + 0.625z⁻² + 0.3333z⁻³).