--- title: "Lattice structure of fourth order IIR" subject: "Digital Signal Analysis and Processing" chapter: 4 key: "semester_7/digital_signal_analysis_and_processing:lattice-iir-pole-zero-4th" source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/lattice-iir-pole-zero-4th site: BCT Engineering --- # Lattice structure of fourth order IIR Digital Signal Analysis and Processing · Chapter 4: Discrete Filter Structures - Asked in: 2073 Chaitra, 10 marks - Video: [DSAP5.11 Lattice Structure IIR Filter - Shiva Gyawali](https://www.youtube.com/watch?v=FQzA03FNuX8) — Shiva Gyawali - 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/lattice-iir-pole-zero-4th - Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-4 ## Question Draw the Lattice Structure for the following system function: H(z) = (1 + (1/3)z⁻¹ + (9/8)z⁻² + (4/3)z⁻³ + z⁻⁴)/(1 + (2/3)z⁻¹ + (5/8)z⁻² + (2/3)z⁻³ + z⁻⁴). ## Short answer From denominator polynomial A(z): K₄=1, K₃=1/3, K₂≈0.2266, K₁≈0.2102. Ladder coefficients: v₀≈-0.1563, v₁≈-0.6133, v₂≈0.2778, v₃=2/3, v₄=1. The lattice-ladder structure cascades four stages with these K’s; output is sum of v_m g_m(n). ## Answer #### 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** | m | 0 | 1 | 2 | 3 | 4 | |---|---|---|---|---|---| | Km | – | 0.2102 | 0.2266 | 0.3333 | 1 | | vm | −0.1563 | −0.6133 | 0.2778 | 0.6667 | 1 | #### 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). ```text f(m)[n] ─────(+)────────●──► f(m-1)[n] ▲────────┼─────────┐ │ │ -Km │ Km │ ┌──────────┘ │ ▼ │ g(m)[n] ◄────(+)◄─────────────────●◄─[z⁻¹]◄─ g(m-1)[n] ``` ```text +------+ +------+ +------+ +------+ 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). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is a pole-zero IIR system H(z) = B(z)/A(z) realised in lattice form? A. Lattice from B(z) only, with feedback taps B. A lattice-ladder: lattice K's from A(z), ladder taps v_m on the g nodes for B(z) C. Two separate lattices in cascade, one for B and one for A D. A lattice with all K equal to 1 Answer: B. The denominator gives the reflection coefficients and the numerator is formed by tapping the g-nodes with ladder coefficients v_m and summing. (Key point: Lattice-ladder structure) 2. Why is K4 = 1 here, and what does it mean? A. A(z) is anti-symmetric, so the poles are at the origin B. A(z) is symmetric, so all four poles lie on the unit circle (marginally stable) C. B(z) is symmetric, so the zeros are on the unit circle D. A(z) has real coefficients only, so the system is stable Answer: B. a4(4) = 1 because A(z) is symmetric; |K4| = 1 puts the poles exactly on the unit circle, so the system is not strictly stable. (Key point: K4 = 1 case) 3. Which set gives the reflection coefficients K1, K2, K3, K4 of A(z)? A. 0.2102, 0.2266, 0.3333, 1 B. 1, 0.3333, 0.2266, 0.2102 C. 0.6667, 0.625, 0.6667, 1 D. 0.2102, 0.2266, 0.6667, 1 Answer: A. Step-down gives K4 = 1, K3 = 1/3, K2 = 29/128 = 0.2266 and K1 = 33/157 = 0.2102. (Key point: Step-down result) 4. How are the ladder coefficients v_m found from B(z)? A. v_m = b_m - sum over i>m of v_i a_i(i-m), starting from v4 = b4 B. v_m = b_m for every m C. v_m = b_m / a_m(m) D. v_m = b_m + sum over i