--- title: "Direct form and lattice of FIR, 2+1.8z⁻¹…" subject: "Digital Signal Analysis and Processing" chapter: 4 key: "semester_7/digital_signal_analysis_and_processing:lattice-fir-2-1-8" source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/lattice-fir-2-1-8 site: BCT Engineering --- # Direct form and lattice of FIR, 2+1.8z⁻¹… Digital Signal Analysis and Processing · Chapter 4: Discrete Filter Structures - Asked in: 2073 Shrawan, 3+7 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-fir-2-1-8 - Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-4 ## Question The system function of a filter is H(z) = 2 + 1.8z⁻¹ − 1.6z⁻² + z⁻³. Draw the Direct Form and Lattice Structure implementation of the above filter. ## Short answer Direct form: y[n]=2x[n]+1.8x[n-1]-1.6x[n-2]+x[n-3]. Lattice: normalize to A(z)=1+0.9z^{-1}-0.8z^{-2}+0.5z^{-3}; reflection coefficients K1=-2.6, K2=-1.6667, K3=0.5. Cascade three stages, then multiply output by 2. ## Answer #### Direct form H(z) = 2 + 1.8z⁻¹ − 1.6z⁻² + z⁻³, i.e. y[n] = 2x[n] + 1.8x[n−1] − 1.6x[n−2] + x[n−3]. ```text x[n] ─┬──►[z⁻¹]──┬──►[z⁻¹]──┬──►[z⁻¹]──┐ │ │ │ │ ×b0 ×b1 ×b2 ×b3 │ │ │ │ └──►(+)◄───┘ │ │ └───►(+)◄────────┘ │ └────►(+)◄─────────┘ │ ▼ y[n] ``` with b0 = 2, b1 = 1.8, b2 = -1.6, b3 = 1. Output: y[n] = b0x[n] + b1x[n−1] + b2x[n−2] + b3x[n−3]; 3 delays, 4 multipliers, 3 adders. #### Lattice structure The lattice needs a monic polynomial, so normalise by b₀ = 2: H(z) = 2 · A₃(z), with A₃(z) = 1 + 0.9z⁻¹ − 0.8z⁻² + 0.5z⁻³. #### 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 + 0.9z⁻¹ − 0.8z⁻² + 0.5z⁻³ - A3: coefficients {1, 0.9, -0.8, 0.5}, so **K3 = a3(3) = 0.5** - A2: coefficients {1, 1.7333, -1.6667}, so **K2 = a2(2) = -1.6667** - A1: coefficients {1, -2.6}, so **K1 = a1(1) = -2.6** | Stage | Km | Size | |---|---|---| | 1 | -2.6 | magnitude above 1 | | 2 | -1.6667 | magnitude above 1 | | 3 | 0.5 | magnitude below 1 | **Lattice coefficients:** K1 = -2.6, K2 = -1.6667, K3 = 0.5 #### 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-> ×2 -> y[n] | | K1 | | K2 | | K3 | +->| |-g1->| |-g2->| | +------+ +------+ +------+ ``` (The upper path carries f and the lower path carries g; f0 = g0 = x[n].) The filter is 2 times the monic polynomial above, so the lattice output is multiplied by 2: y[n] = 2 · f3[n]. **Check:** the lattice impulse response is {1, 0.9, −0.8, 0.5}; multiplied by 2 it gives {2, 1.8, −1.6, 1}. ✓ (|K1| and |K2| > 1, so some zeros lie outside the unit circle.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the filter divided by b0 = 2 first? A. The lattice needs a monic polynomial A3(z) with leading coefficient 1 B. To make the filter stable C. To remove the delays D. To make K3 equal to 1 Answer: A. H(z) = 2*A3(z), and the lattice realises A3. (Key point: Normalisation) 2. What is A3(z)? A. 2 + 1.8z^-1 - 1.6z^-2 + z^-3 B. 1 + 0.9z^-1 - 0.8z^-2 + 0.5z^-3 C. 1 + 1.8z^-1 - 1.6z^-2 + z^-3 D. 0.5 + 0.9z^-1 - 0.8z^-2 + z^-3 Answer: B. Divide every coefficient by 2. (Key point: Monic polynomial) 3. What are K1, K2, K3? A. 0.9, -0.8, 0.5 B. -2.6, -1.6667, 2 C. -2.6, -1.6667, 0.5 D. 0.5, -1.6667, -2.6 Answer: C. K3 = 0.5, then step-down gives K2 = -1.6667 and K1 = -2.6. (Key point: Lattice coefficients) 4. How is the final output obtained? A. y[n] = f3[n]/2 B. y[n] = f3[n] + g3[n] C. y[n] = g3[n] D. y[n] = 2*f3[n] Answer: D. The lattice gives A3; multiply by 2. 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