--- title: "Lattice coefficients of all-pole IIR; stability" subject: "Digital Signal Analysis and Processing" chapter: 4 key: "semester_7/digital_signal_analysis_and_processing:lattice-iir-all-pole-stability" source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/lattice-iir-all-pole-stability site: BCT Engineering --- # Lattice coefficients of all-pole IIR; stability Digital Signal Analysis and Processing · Chapter 4: Discrete Filter Structures - Asked in: 2071 Chaitra, 4+2+1 marks - Video: [Numerical Lattice Realization of IIR filter #DTSP #DSP #IIRfilter #latticerealization](https://www.youtube.com/watch?v=ETIWCb6ILqs) — We The Engineers - 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-all-pole-stability - Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-4 ## Question Compute lattice coefficients and draw lattice structure for given IIR system H(z) = 1/(1 − 0.01z⁻¹ − 0.23z⁻² + 0.5z⁻³). Also check the stability of given system. ## Short answer Step-down recursion gives K₁ = 0.2, K₂ = -0.3, K₃ = 0.5. The all-pole lattice consists of three cascaded stages with these reflection coefficients. All |K_m| < 1, so the filter is stable. ## Answer #### Method For the all-pole system H(z) = 1/A₃(z), with A₃(z) = 1 − 0.01z⁻¹ − 0.23z⁻² + 0.5z⁻³, the reflection coefficients are those of the polynomial A₃(z), found by the step-down recursion K_m = a_m(m), a_(m−1)(i) = [a_m(i) − K_m a_m(m−i)] / (1 − K_m²) #### Computation - A₃ = {1, −0.01, −0.23, 0.5}: **K₃ = a₃(3) = 0.5** - 1 − K₃² = 0.75. a₂(1) = (−0.01 − 0.5(−0.23))/0.75 = 0.105/0.75 = 0.14; a₂(2) = (−0.23 − 0.5(−0.01))/0.75 = −0.225/0.75 = −0.3. A₂ = {1, 0.14, −0.3}: **K₂ = −0.3** - 1 − K₂² = 0.91. a₁(1) = (0.14 − (−0.3)(0.14))/0.91 = 0.182/0.91 = 0.2. A₁ = {1, 0.2}: **K₁ = 0.2** **Lattice coefficients:** K₁ = 0.2, K₂ = −0.3, K₃ = 0.5. #### Lattice structure (all-pole) Stage equations, from m = 3 down to 1 (input enters at f₃): ```text f3(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) y(n) = f0(n) = g0(n) ``` ```text f(m)[n] ─────(+)────────●──► f(m-1)[n] ▲────────┼─────────┐ │ │ -Km │ Km │ ┌──────────┘ │ ▼ │ g(m)[n] ◄────(+)◄─────────────────●◄─[z⁻¹]◄─ g(m-1)[n] ``` Three stages in cascade: ```text +------+ +------+ +------+ x[n]=f3-->| K3 |---->| K2 |---->| K1 |--> y[n]=f0=g0 | =0.5 | |=-0.3 | | =0.2 | +------+ +------+ +------+ ``` (the f path moves left to right, the g path returns right to left; g0 = f0 at the end). #### Stability An all-pole lattice is stable if and only if all |K_m| < 1. Here |0.2|, |−0.3|, |0.5| are all less than 1, so the system is **stable**. Check with the poles: the roots of z³ − 0.01z² − 0.23z + 0.5 are −0.886 and 0.448 ± 0.603j (radius 0.751). All have magnitude < 1. ✓ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For H(z) = 1/(1 - 0.01z^-1 - 0.23z^-2 + 0.5z^-3), what is K3? A. -0.23 B. -0.01 C. 0.5 D. 1 Answer: C. K3 = a3(3), the last coefficient of A3(z), which is +0.5. (Key point: Coefficients of A3) 2. Which set gives the lattice coefficients K1, K2, K3? A. 0.5, -0.3, 0.2 B. 0.2, -0.3, 0.5 C. 0.2, 0.3, 0.5 D. 0.14, -0.3, 0.5 Answer: B. Step-down gives A2 = {1, 0.14, -0.3} so K2 = -0.3, then A1 = {1, 0.2} so K1 = 0.2, with K3 = 0.5. (Key point: Lattice coefficient set) 3. What is the stability condition for an all-pole lattice, and the verdict here? A. Sum of K's is below 1; unstable B. All |Km| < 1; stable C. All Km positive; unstable because K2 < 0 D. Any |Km| > 0.5 is unstable; unstable Answer: B. An all-pole lattice is stable iff all |Km| < 1; here 0.2, 0.3 and 0.5 all satisfy this. (Key point: Stability test) 4. In the all-pole lattice, how is f(m-1)(n) computed? A. f(m-1)(n) = fm(n) - Km g(m-1)(n-1) B. f(m-1)(n) = fm(n) + Km g(m-1)(n-1) C. f(m-1)(n) = Km fm(n) D. f(m-1)(n) = fm(n) - Km f(m-1)(n-1) Answer: A. The all-pole lattice runs from f3 = x down to f0 = y, subtracting Km times the delayed backward signal. 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