--- title: "Series programming of a third-order system" subject: "Digital Control System" chapter: 4 key: "semester_6/digital_control_system:series-programming-z-plus-3" source: https://www.bctengineer.com/q/semester_6/digital_control_system/series-programming-z-plus-3 site: BCT Engineering --- # Series programming of a third-order system Digital Control System ยท Chapter 4: Design and Compensation of Discrete Time Control System - Asked in: 2068 Magh, 8 marks - Video: [Cascade Digital Controller Implementation ๐ŸŒŸ Third-Order System Flowchart ๐Ÿ’ก Example 4](https://www.youtube.com/watch?v=4flqGLW8KrU) โ€” CAN Education - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/digital_control_system/series-programming-z-plus-3 - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4 ## Question Realize system having pulse transfer function G(z)=(z+3)(zยฒ-2z+10)/((z-5)(zยฒ-z+8)) by series programming. ## Short answer Series programming splits G(z) into cascaded sections: G1(z)=(1+3z^-1)/(1-5z^-1) and G2(z)=(1-2z^-1+10z^-2)/(1-z^-1+8z^-2). Each section is realized by standard programming, with one delay for first order and two delays for second order, total three delays. ## Answer In **series programming**, the pulse transfer function is split into a cascade of first- and second-order sections. Each section is realized by standard programming. #### Step 1: Split into sections (in powers of $z^{-1}$) $$ G(z) = \underbrace{\frac{1 + 3z^{-1}}{1 - 5z^{-1}}}_{G_1(z)}\cdot\underbrace{\frac{1 - 2z^{-1} + 10z^{-2}}{1 - z^{-1} + 8z^{-2}}}_{G_2(z)} $$ The quadratics have complex roots, so they stay as one second-order section with real coefficients. #### Step 2: Difference equations **Section 1** (input $x$, output $y_1$): - $w_1(k) = x(k) + 5w_1(k-1)$ - $y_1(k) = w_1(k) + 3w_1(k-1)$ **Section 2** (input $y_1$, output $y$): - $w_2(k) = y_1(k) + w_2(k-1) - 8w_2(k-2)$ - $y(k) = w_2(k) - 2w_2(k-1) + 10w_2(k-2)$ #### Step 3: Block diagram Section 1: ```text x(k)-->(S)------------o--[1]-------->(S)--> y1(k) ^ | ^ | [z^-1] | | | w1(k-1) | +<[5]---------o--[3]----------+ ``` Section 2: ```text y1(k)-->(S)------------o--[1]-------->(S)--> y(k) ^ | ^ | [z^-1] | | | w2(k-1) | (S)<[1]--------o--[-2]------->(S) ^ | ^ | [z^-1] | | | w2(k-2) | +<[-8]--------o--[10]---------+ ``` ```text x(k) --> [ G1(z) ] --y1(k)--> [ G2(z) ] --> y(k) ``` A total of 3 delays are used. The order of the sections can be swapped. **Note:** The poles are $z = 5$ and $z = 0.5 \pm j2.78$ (with $|z| = 2.83$), all outside the unit circle. So this system is unstable, although it can still be realized. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does series programming do to the pulse transfer function? A. Splits it into a sum of partial fractions B. Expands it as a continued fraction C. Splits it into a cascade of first- and second-order sections, each realized by standard programming D. Realizes it with one chain of delays for all terms Answer: C. Series programming multiplies low-order sections in cascade; this is different from the parallel (sum) form. (Key point: Series splitting) 2. For G1(z) = (1+3z^-1)/(1-5z^-1), the standard-programming equations are: A. w1(k) = x(k) - 5w1(k-1); y1(k) = w1(k) + 3w1(k-1) B. w1(k) = x(k) + 5w1(k-1); y1(k) = w1(k) + 3w1(k-1) C. w1(k) = x(k) + 3w1(k-1); y1(k) = w1(k) + 5w1(k-1) D. w1(k) = x(k) + 5w1(k-1); y1(k) = w1(k) - 3w1(k-1) Answer: B. Denominator 1-5z^-1 gives +5 feedback; numerator gives +3 feed-forward. (Key point: Section 1 equations) 3. Why is z^2 - z + 8 not split into first-order factors? A. It has a repeated root B. It has a zero at the origin C. Splitting would increase the number of delays D. Its roots are complex, so it stays as a real-coefficient second-order section Answer: D. Complex conjugate roots would need complex coefficients if split; one real second-order block avoids this. (Key point: Quadratic kept whole) 4. How many delays are used, and is the system stable? A. 3 delays; stable B. 3 delays; unstable, since the poles lie outside the unit circle C. 2 delays; unstable D. 4 delays; stable Answer: B. One delay for the first-order section plus two for the second-order section = 3. Poles at 5 and 0.5 +/- j2.78 are outside the unit circle. 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