--- title: "Lattice coefficients of 1+2z⁻¹+z⁻²" subject: "Digital Signal Analysis and Processing" chapter: 4 key: "semester_7/digital_signal_analysis_and_processing:lattice-coeff-1-2-1" source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/lattice-coeff-1-2-1 site: BCT Engineering --- # Lattice coefficients of 1+2z⁻¹+z⁻² Digital Signal Analysis and Processing · Chapter 4: Discrete Filter Structures - Asked in: 2072 Chaitra, 5 marks - Video: [Q6. a. Lattice Structure of FIR Filters | DSP | EnggClasses](https://www.youtube.com/watch?v=AcMcntjNpM8) — EnggClasses - 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-coeff-1-2-1 - Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-4 ## Question Compute the lattice coefficients and draw the lattice structure of the following FIR system H(z) = 1 + 2z⁻¹ + z⁻². ## Short answer K₂ = a₂(2) = 1; using the build-up relation gives K₁ = a₂(1)/(1 + a₂(2)) = 1. The lattice coefficients are K₁ = 1, K₂ = 1. The structure is a cascade of two identical lattice stages, each computing forward and backward signals with these coefficients. ## Answer A₂(z) = 1 + 2z⁻¹ + z⁻² (M = 2, already monic). #### Computation - Last stage: **K₂ = a₂(2) = 1**. - The step-down division 1 − K₂² = 0 is 0/0 here, because a₂(1) − K₂a₂(1) = 0 as well. Use the build-up relation instead: A₂(z) = A₁(z) + K₂z⁻²A₁(z⁻¹) with A₁ = 1 + K₁z⁻¹ gives 1 + K₁(1 + K₂)z⁻¹ + K₂z⁻² = 1 + 2K₁z⁻¹ + z⁻². Matching with 1 + 2z⁻¹ + z⁻²: 2K₁ = 2, so **K₁ = a₂(1)/(1 + a₂(2)) = 2/2 = 1**. **Lattice coefficients:** K₁ = 1, K₂ = 1 (both have magnitude 1, consistent with the double zero at z = −1 on the unit circle). #### Structure Stage equations (m = 1, 2): fm(n) = f(m−1)(n) + Km g(m−1)(n−1); gm(n) = Km f(m−1)(n) + g(m−1)(n−1); f0 = g0 = x(n), y(n) = f2(n). ```text f(m-1)[n] ──────●────────────────────(+)──► f(m)[n] │ ┌───────────▲ │ Km │ Km └─────────┼───────────┐ │ │ │ ▼ g(m-1)[n] ─[z⁻¹] ─────────●──────────(+)──► g(m)[n] ``` Two such stages in cascade, with K1 = 1 in the first and K2 = 1 in the second: ```text +------+ +------+ x[n]--->| K1=1 |---->| K2=1 |---> y[n] = f2[n] g0=f0 +------+ +------+ ``` **Check:** f1 = x(n) + x(n−1); f2 = f1(n) + g1(n−1), g1 = x(n) + x(n−1) = f1, so f2 = x(n) + 2x(n−1) + x(n−2). ✓ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is K2 for A2(z) = 1 + 2z^-1 + z^-2? A. 2 B. 0.5 C. 1 D. -1 Answer: C. K2 = a2(2) = 1. (Key point: Last stage) 2. Why does the step-down formula fail here? A. K2 is negative B. The filter is IIR C. a2(1) is zero D. 1 - K2^2 = 0 gives 0/0 Answer: D. Both numerator and denominator vanish. (Key point: Step-down failure) 3. How is K1 found instead? A. Match A2 = 1 + K1(1 + K2)z^-1 + K2 z^-2, so K1 = a2(1)/(1 + a2(2)) = 1 B. K1 = a2(1) = 2 C. K1 = 0 D. K1 = a2(1)/(1 - a2(2)) Answer: A. 2K1 = 2 so K1 = 1. (Key point: Build-up) 4. What does |K| = 1 for both stages mean? A. The filter is unstable B. Consistent with the double zero at z = -1 on the unit circle C. The zeros are at the origin D. The zeros are inside the unit circle Answer: B. H(z) = (1 + z^-1)^2 has a double zero at z = -1. 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