--- title: "Lattice structure of 1+2z⁻¹−3z⁻²+4z⁻³" subject: "Digital Signal Analysis and Processing" chapter: 4 key: "semester_7/digital_signal_analysis_and_processing:lattice-coeff-1-2-minus3-4" source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/lattice-coeff-1-2-minus3-4 site: BCT Engineering --- # Lattice structure of 1+2z⁻¹−3z⁻²+4z⁻³ Digital Signal Analysis and Processing · Chapter 4: Discrete Filter Structures - Asked in: 2072 Kartik, 6 marks; 2069 Chaitra, 6 marks (asked 2 times) - 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-minus3-4 - 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 − 3z⁻² + 4z⁻³. ## Short answer The reflection coefficients are K1=-0.5385, K2=0.7333, K3=4, obtained by step-down recursion from the polynomial coefficients. The lattice cascades three stages; |K3|>1 indicates zeros outside the unit circle. ## Answer #### Method The FIR lattice of order M uses the stage equations ```text f0(n) = g0(n) = x(n) fm(n) = f(m−1)(n) + Km · g(m−1)(n−1) gm(n) = Km · f(m−1)(n) + g(m−1)(n−1) m = 1 … M y(n) = fM(n) ``` The reflection coefficient of the last stage is Km = am(m) (the last coefficient of the polynomial Am(z)). The lower-order polynomials come from the step-down recursion ```text a(m−1)(i) = [ am(i) − Km · am(m−i) ] / (1 − Km²), i = 1 … m−1 ``` #### Computation A3(z) = 1 + 2z⁻¹ − 3z⁻² + 4z⁻³ - A3: coefficients {1, 2, -3, 4}, so **K3 = a3(3) = 4** - A2: coefficients {1, -0.9333, 0.7333}, so **K2 = a2(2) = 0.7333** - A1: coefficients {1, -0.5385}, so **K1 = a1(1) = -0.5385** | Stage | Km | Size | |---|---|---| | 1 | -0.5385 | magnitude below 1 | | 2 | 0.7333 | magnitude below 1 | | 3 | 4 | magnitude above 1 | **Lattice coefficients:** K1 = -0.5385, K2 = 0.7333, K3 = 4 #### Structure One lattice stage (repeated for m = 1 … 3 with K1 … K3): ```text f(m-1)[n] ──────●────────────────────(+)──► f(m)[n] │ ┌───────────▲ │ Km │ Km └─────────┼───────────┐ │ │ │ ▼ g(m-1)[n] ─[z⁻¹] ─────────●──────────(+)──► g(m)[n] ``` Cascade (f and g pass together from stage to stage): ```text +------+ +------+ +------+ x[n]-+->| |-f1->| |-f2->| |-f3-> y[n] = f3[n] | | K1 | | K2 | | K3 | +->| |-g1->| |-g2->| | +------+ +------+ +------+ ``` (The upper path carries f and the lower path carries g; f0 = g0 = x[n].) **Check:** rebuilding A3 from K1 = −7/13, K2 = 11/15, K3 = 4 with A_m(z) = A_(m−1)(z) + K_m z⁻ᵐ A_(m−1)(z⁻¹) reproduces {1, 2, −3, 4}, and simulating the lattice with a unit impulse gives the impulse response 1, 2, −3, 4. ✓ Since |K3| = 4 > 1, the filter has zeros outside the unit circle (it is not minimum-phase). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the FIR lattice, how is the last reflection coefficient K3 found? A. K3 = a3(1), the first nonunity coefficient B. K3 = a3(3), the last coefficient of A3(z) C. K3 = 1 - a3(3) D. K3 is the sum of all coefficients Answer: B. K_m = a_m(m). (Key point: Last coefficient) 2. Which formula gives the lower-order coefficients? A. a(m-1)(i) = a_m(i) + K_m a_m(m-i) B. a(m-1)(i) = a_m(i)(1 - K_m^2) C. a(m-1)(i) = [a_m(i) - K_m a_m(m-i)]/(1 - K_m^2) D. a(m-1)(i) = a_m(i)/K_m Answer: C. This is the step-down recursion. (Key point: Step-down) 3. What are K1, K2 and K3? A. K1 = 2, K2 = -3, K3 = 4 B. K1 = 0.5385, K2 = -0.7333, K3 = 4 C. K1 = -0.5385, K2 = 0.7333, K3 = 0.25 D. K1 = -0.5385, K2 = 0.7333, K3 = 4 Answer: D. K3 = 4, K2 = 0.7333, K1 = -0.5385 (= -7/13). (Key point: Result) 4. What does |K3| = 4 > 1 tell us? A. The filter has zeros outside the unit circle, so it is not minimum-phase B. The filter is unstable C. The lattice cannot be built D. The filter is all-pass Answer: A. An FIR filter is always stable; |K| > 1 means some zeros are outside the unit circle. 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