--- title: "Parallel realization of a second-order filter" subject: "Digital Control System" chapter: 4 key: "semester_6/digital_control_system:parallel-realization-filter" source: https://www.bctengineer.com/q/semester_6/digital_control_system/parallel-realization-filter site: BCT Engineering --- # Parallel realization of a second-order filter Digital Control System · Chapter 4: Design and Compensation of Discrete Time Control System - Asked in: 2073 Magh, 6 marks - Video: [Parallel realization for the system described by 𝐻(𝑧)](https://www.youtube.com/watch?v=ujAbWGj8xQI) — Engg-Course-Made-Easy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/digital_control_system/parallel-realization-filter - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4 ## Question Consider the digital filter defined by G(z)=(2+2.2z⁻¹+0.2z⁻²)/(1+0.4z⁻¹-0.12z⁻²). Realize this filter in the parallel scheme. ## Short answer Partial fraction expansion gives G(z) = -1.6667 + 4.5/(1-0.2z^-1) - 0.8333/(1+0.6z^-1). The parallel form consists of two first-order branches: w1(k)=x(k)+0.2w1(k-1), w2(k)=x(k)-0.6w2(k-1), and output y(k)=-1.6667x(k)+4.5w1(k)-0.8333w2(k) using two delays. ## Answer In **parallel programming**, $G(z)$ is expanded into partial fractions in $z^{-1}$, and each term is realized as a separate branch. #### Step 1: Factor the denominator $$ 1 + 0.4z^{-1} - 0.12z^{-2} = (1 + 0.6z^{-1})(1 - 0.2z^{-1}) $$ The poles are $z = -0.6$ and $z = 0.2$. #### Step 2: Partial fractions The numerator and denominator have the same degree in $z^{-1}$, so $$ G(z) = C_0 + \frac{C_1}{1 - 0.2z^{-1}} + \frac{C_2}{1 + 0.6z^{-1}} $$ Let $w = z^{-1}$. $$ \begin{aligned} C_0 &= \lim_{w\to\infty}\frac{2 + 2.2w + 0.2w^2}{1 + 0.4w - 0.12w^2} = \frac{0.2}{-0.12} = -1.6667 \\ C_1 &= \frac{2 + 2.2w + 0.2w^2}{1 + 0.6w}\Big|_{w=5} = \frac{2 + 11 + 5}{4} = 4.5 \\ C_2 &= \frac{2 + 2.2w + 0.2w^2}{1 - 0.2w}\Big|_{w=-5/3} = \frac{2 - 3.6667 + 0.5556}{1.3333} = -0.8333 \end{aligned} $$ $$ G(z) = -1.6667 + \frac{4.5}{1 - 0.2z^{-1}} - \frac{0.8333}{1 + 0.6z^{-1}} $$ **Check** at $z^{-1} = 0$: $-1.6667 + 4.5 - 0.8333 = 2$. ✔ #### Step 3: Difference equations - $w_1(k) = x(k) + 0.2w_1(k-1)$ - $w_2(k) = x(k) - 0.6w_2(k-1)$ - $y(k) = -1.6667x(k) + 4.5w_1(k) - 0.8333w_2(k)$ #### Step 4: Block diagram Overall structure (three branches added): ```text +-------------[-1.6667]-------------+ | v x(k)-o--+-->[ 4.5/(1-0.2z^-1) ]----------->(S)--> y(k) | ^ +-->[ -0.8333/(1+0.6z^-1) ]---------+ ``` Branch 1: ```text x(k)-->(S)------------o--[1]-------->(S)->[4.5]-> y1(k) ^ | | [z^-1] | | w1(k-1) +<[0.2]-------o ``` Branch 2: ```text x(k)-->(S)------------o--[1]-------->(S)->[-0.8333]-> y2(k) ^ | | [z^-1] | | w2(k-1) +<[-0.6]------o ``` Two delays are used, one per pole. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The denominator 1 + 0.4z^-1 - 0.12z^-2 factors as A. (1 + 0.6z^-1)(1 - 0.2z^-1) B. (1 - 0.6z^-1)(1 + 0.2z^-1) C. (1 + 0.4z^-1)(1 - 0.12z^-1) D. (1 + 0.3z^-1)(1 - 0.4z^-1) Answer: A. Expanding gives 1 + 0.4z^-1 - 0.12z^-2, with poles at z = -0.6 and 0.2. (Key point: Denominator factors) 2. The constant term C0 in the partial fractions is A. -1.6667 B. 2 C. 4.5 D. -0.8333 Answer: A. C0 = limit as z^-1 goes to infinity = 0.2/(-0.12) = -1.6667. (Key point: Constant term) 3. The branch coefficients C1 (for pole 0.2) and C2 (for pole -0.6) are A. 4.5 and -0.8333 B. -0.8333 and 4.5 C. 2 and 1 D. -1.6667 and 4.5 Answer: A. C1 = evaluate at w = 5 gives 4.5; C2 at w = -5/3 gives -0.8333. (Key point: Branch coefficients) 4. The check at z^-1 = 0 requires the terms to sum to A. 0 B. 1 C. 2 D. 4.5 Answer: C. -1.6667 + 4.5 - 0.8333 = 2, the numerator constant. (Key point: Check) 5. The output equation is A. y(k) = -1.6667x(k) + 4.5w1(k) - 0.8333w2(k) B. y(k) = 4.5w1(k) + 0.8333w2(k) C. y(k) = w1(k) w2(k) D. y(k) = -1.6667x(k) - 4.5w1(k) + 0.8333w2(k) Answer: A. The three branch outputs are added, with w1(k) = x(k) + 0.2w1(k-1) and w2(k) = x(k) - 0.6w2(k-1). 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