--- title: "Parallel and ladder realization of a filter" subject: "Digital Control System" chapter: 4 key: "semester_6/digital_control_system:parallel-and-ladder-realization" source: https://www.bctengineer.com/q/semester_6/digital_control_system/parallel-and-ladder-realization site: BCT Engineering --- # Parallel and ladder realization of a filter Digital Control System · Chapter 4: Design and Compensation of Discrete Time Control System - Asked in: 2072 Asoj, 8 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-and-ladder-realization - 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 and ladder scheme. ## Short answer Parallel: G(z) = -1.6667 + 4.5/(1-0.2z^-1) - 0.8333/(1+0.6z^-1) with two first-order branches. Ladder: continued fraction expansion yields A0=2, B1=0.7143, A1=16.333, B2=0.0357, A2=-20; the ladder diagram alternates constant gains and delay blocks 1/(Bi z). ## Answer Given: $$ G(z) = \frac{2 + 2.2z^{-1} + 0.2z^{-2}}{1 + 0.4z^{-1} - 0.12z^{-2}} $$ #### (a) Parallel scheme 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. #### (b) Ladder scheme Write the filter in positive powers of $z$: $$ G(z) = \frac{2 + 2.2z^{-1} + 0.2z^{-2}}{1 + 0.4z^{-1} - 0.12z^{-2}} = \frac{2z^2 + 2.2z + 0.2}{z^2 + 0.4z - 0.12} $$ In ladder programming, $G(z)$ is expanded as a continued fraction: $$ G(z) = A_0 + \cfrac{1}{B_1z + \cfrac{1}{A_1 + \cfrac{1}{B_2z + \cfrac{1}{A_2}}}} $$ Each $1/(B_iz)$ is realized by a delay $z^{-1}/B_i$, and each $A_i$ by a constant. **Step 1: $A_0$.** $$ G(z) = 2 + \frac{1.4z + 0.44}{z^2 + 0.4z - 0.12} $$ So $A_0 = 2$. **Step 2: $B_1$.** $$ \frac{z^2 + 0.4z - 0.12}{1.4z + 0.44} = 0.7143z + \frac{0.0857z - 0.12}{1.4z + 0.44} $$ So $B_1 = 1/1.4 = 0.7143$. **Step 3: $A_1$.** $$ \frac{1.4z + 0.44}{0.0857z - 0.12} = 16.333 + \frac{2.4}{0.0857z - 0.12} $$ So $A_1 = 49/3 = 16.333$. **Step 4: $B_2$ and $A_2$.** $$ \frac{0.0857z - 0.12}{2.4} = 0.0357z - 0.05 $$ So $B_2 = 1/28 = 0.0357$ and $A_2 = -1/0.05 = -20$. $$ G(z) = 2 + \cfrac{1}{0.7143z + \cfrac{1}{16.333 + \cfrac{1}{0.0357z - 0.05}}} $$ | $A_0$ | $B_1$ | $A_1$ | $B_2$ | $A_2$ | |---|---|---|---|---| | 2 | 0.7143 | 16.333 | 0.0357 | −20 | **Ladder diagram:** ```text X(z)--o-----------------[A0]--------------->(S)--> Y(z) | ^ v + | Y1 (S1)-->[z^-1/B1]-------------o----------+ ^ - | + | v o<--E1--[1/A1]<----------- (S2) | ^ - v + | (S3)-->[z^-1/B2]-------------o Y2 ^ - | | | +<--E2--[1/A2]<-------------+ ``` **Signal equations** of the ladder ($o$ = take-off point, (S) = summer): $$ \begin{aligned} Y_1 &= \tfrac{z^{-1}}{B_1}(X - E_1), & E_1 &= \tfrac{1}{A_1}(Y_1 - Y_2) \\ Y_2 &= \tfrac{z^{-1}}{B_2}(E_1 - E_2), & E_2 &= \tfrac{1}{A_2}Y_2 \\ Y &= A_0X + Y_1 \end{aligned} $$ With the values: $1/B_1 = 1.4$, $1/A_1 = 0.06122$, $1/B_2 = 28$, $1/A_2 = -0.05$. Only two delays are needed. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does the partial-fraction expansion of G(z) contain a constant term C0 = -1.6667? A. Because the numerator and denominator have the same degree in z^-1, so a direct feed-through remains B. Because the denominator has a repeated pole C. Because the poles are complex D. Because the filter is unstable Answer: A. Equal degrees of numerator and denominator leave a constant after division. C0 is the ratio of the highest-power coefficients, 0.2/(-0.12). (Key point: Constant term C0) 2. For parallel realization, the denominator 1 + 0.4z^-1 - 0.12z^-2 is factored as: A. (1 + 0.2z^-1)(1 - 0.6z^-1) B. (1 + 0.6z^-1)(1 - 0.2z^-1) C. (1 - 0.6z^-1)(1 - 0.2z^-1) D. (1 + 0.6z^-1)(1 + 0.2z^-1) Answer: B. Expanding (1+0.6z^-1)(1-0.2z^-1) gives 1 + 0.4z^-1 - 0.12z^-2. The poles are z = -0.6 and z = 0.2. (Key point: Pole factorisation) 3. In the parallel scheme, the branch for the pole at z = 0.2 uses which difference equation? A. w1(k) = x(k) - 0.2w1(k-1) B. w1(k) = x(k) + 0.2w1(k-1) C. w1(k) = x(k) + 0.2w1(k+1) D. w1(k) = 4.5x(k) + 0.2x(k-1) Answer: B. The branch 1/(1 - 0.2z^-1) gives w1(k) = x(k) + 0.2w1(k-1), and the output is scaled by 4.5. (Key point: Parallel branch equations) 4. In the ladder scheme, G(z) is expanded as a continued fraction. What does A0 = 2 represent here? A. The first pole of G(z) B. The constant obtained by dividing the numerator by the denominator (the direct term) C. The gain of the first delay D. The DC gain of the filter Answer: B. Dividing 2z^2+2.2z+0.2 by z^2+0.4z-0.12 leaves quotient 2 and remainder (1.4z+0.44). A0 is that quotient. 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