--- title: "Lattice structure of second order pole-zero system" subject: "Digital Signal Analysis and Processing" chapter: 4 key: "semester_7/digital_signal_analysis_and_processing:lattice-iir-pole-zero-2nd" source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/lattice-iir-pole-zero-2nd site: BCT Engineering --- # Lattice structure of second order pole-zero system Digital Signal Analysis and Processing · Chapter 4: Discrete Filter Structures - Asked in: 2070 Asar, 6 marks - Video: [CHAP4 lattice ladder polezero DSAP IOE](https://www.youtube.com/watch?v=DOlcz6FOWAs) — Suyan Shrestha - 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-2nd - Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-4 ## Question Draw lattice structure for given pole-zero system H(z) = (0.5 + 2z⁻¹ + 0.6z⁻²)/(1 − 0.3z⁻¹ + 0.4z⁻²). ## Short answer Reflection coefficients from denominator: K₂ = 0.4, K₁ = -0.2143. Ladder coefficients: c₀ = 0.7271, c₁ = 2.18, c₂ = 0.6. The lattice-ladder structure cascades two stages with these K's and taps g₀, g₁, g₂ weighted by c's to form output. ## Answer 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): ```text 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For A(z) = 1 - 0.3z^-1 + 0.4z^-2, which values are K2 and K1? A. K2 = 0.4, K1 = -0.2143 B. K2 = -0.3, K1 = 0.4 C. K2 = 0.4, K1 = -0.3 D. K2 = -0.4, K1 = 0.2143 Answer: A. K2 = a2(2) = 0.4, and a1(1) = (-0.3 + 0.12)/(1 - 0.16) = -0.2143 = K1. (Key point: Reflection coefficients) 2. What is c2, and how does c1 follow from it? A. c2 = b2 = 0.6; c1 = b1 - c2 a2(1) = 2.18 B. c2 = b0 = 0.5; c1 = 2 C. c2 = 0.4; c1 = 2 - 0.4 D. c2 = 0.6; c1 = b1 + c2 a2(1) = 1.82 Answer: A. c2 = b2 = 0.6, and c1 = 2 - 0.6(-0.3) = 2.18. (Key point: Ladder coefficients) 3. How is the output y[n] formed in this lattice-ladder? A. y[n] = f0[n] B. y[n] = c0 g0[n] + c1 g1[n] + c2 g2[n] C. y[n] = c0 f0 + c1 f1 + c2 f2 D. y[n] = g2[n] only Answer: B. The g taps are multiplied by the ladder coefficients c0, c1, c2 and summed to create the zeros. (Key point: Output formation) 4. How can stability of this pole-zero system be judged from the lattice? A. From the ladder coefficients c_m being below 1 B. From the numerator zeros C. Both |K1| and |K2| are below 1, so it is stable D. It cannot be judged without poles Answer: C. Stability depends on the denominator's reflection coefficients; |-0.2143| and |0.4| are both less than 1. 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