---
title: "System function from all-pole lattice K's"
subject: "Digital Signal Analysis and Processing"
chapter: 4
key: "semester_7/digital_signal_analysis_and_processing:all-pole-lattice-system-function"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/all-pole-lattice-system-function
site: BCT Engineering
---

# System function from all-pole lattice K's

Digital Signal Analysis and Processing · Chapter 4: Discrete Filter Structures

- Asked in: 2071 Shrawan, 5 marks
- Video: [LATTICE Form Representation of Digital Filters | Third-Order and Second-Order Examples | EC Academy](https://www.youtube.com/watch?v=p6nBPNa2OdQ) — EC Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/all-pole-lattice-system-function
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-4

## Question

If a 3 stage lattice filter for all pole polynomial has coefficients K₁ = 1/4, K₂ = 1/2 and K₃ = 1/3, obtain the system function of this filter.

## Short answer

Using the step-up recursion: A₃(z) = 1 + (13/24)z⁻¹ + (5/8)z⁻² + (1/3)z⁻³. Thus H(z) = 1 / (1 + 0.5417z⁻¹ + 0.625z⁻² + 0.3333z⁻³).

## Answer

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).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For an all-pole lattice, what is the system function?
   A. H(z) = A3(z)
   B. H(z) = 1/A3(z)
   C. H(z) = A3(z)/A3(z^-1)
   D. H(z) = z^-3 A3(z^-1)
   Answer: B. An all-pole lattice has H(z) = 1/A_M(z). (Key point: Form of H(z))

2. Which recursion builds A3(z) from the K's?
   A. a_m(m) = K_m, a_m(i) = a_(m-1)(i) + K_m a_(m-1)(m-i)
   B. a_m(m) = 1/K_m, a_m(i) = a_(m-1)(i) K_m
   C. a_m(m) = K_m, a_m(i) = a_(m-1)(i) - K_m a_(m-1)(m-i)
   D. a_m(i) = [a_(m-1)(i) - K_m]/(1 - K_m^2)
   Answer: A. The step-up (Levinson) recursion sets a_m(m) = K_m and adds K_m times the reversed lower-order coefficient. (Key point: Step-up recursion)

3. What is H(z) for K1 = 1/4, K2 = 1/2, K3 = 1/3?
   A. 1/(1 + 0.25z^-1 + 0.5z^-2 + 0.3333z^-3)
   B. 1/(1 + 0.5417z^-1 + 0.625z^-2 + 0.3333z^-3)
   C. 1/(1 + 0.375z^-1 + 0.5z^-2 + 0.3333z^-3)
   D. 1/(1 - 0.5417z^-1 - 0.625z^-2 - 0.3333z^-3)
   Answer: B. a2(1) = 0.375, then a3(1) = 0.375 + 0.5/3 = 0.5417 and a3(2) = 0.5 + 0.375/3 = 0.625. (Key point: Result)

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