--- title: "Root locus of a sampled system (T = 0.5 s)" subject: "Digital Control System" chapter: 4 key: "semester_6/digital_control_system:draw-root-locus-t-0-5" source: https://www.bctengineer.com/q/semester_6/digital_control_system/draw-root-locus-t-0-5 site: BCT Engineering --- # Root locus of a sampled system (T = 0.5 s) Digital Control System · Chapter 4: Design and Compensation of Discrete Time Control System - Asked in: 2071 Bhadra, 6 marks - Video: [Digital control 24: The digital root locus, Part 2](https://www.youtube.com/watch?v=92getAXrYFc) — Corné van Daalen - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/digital_control_system/draw-root-locus-t-0-5 - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4 ## Question Draw root locus plot for the given system below. [Figure: unity-feedback loop: R(s)=1/s → summing junction → sampler (T = 0.5 sec) → ZOH → 1/s(s+1) → C(s)] ## Short answer Open-loop poles at z=1 and 0.6065, zero at -0.8467. Real-axis locus lies between 0.6065 and 1, and left of -0.8467. Breakaway at 0.7915, break-in at -2.485. Complex branches form a circle centered at -0.8467 with radius 1.638. System is stable for 0 < K < 4.36; at K=4.36 the poles cross the unit circle at 0.571±j0.821. ## Answer The root locus is drawn in the z-plane for the open-loop pulse transfer function with gain K in the forward path. #### Open-loop pulse transfer function With $T = 0.5$ s and $e^{-0.5} = 0.6065$: $$ \begin{aligned} G(z) &= K(1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+1)}\right] = K\,\frac{(T-1+e^{-T})z + (1-e^{-T}-Te^{-T})}{(z-1)(z-e^{-T})}\\ &= \frac{K(0.1065z + 0.0902)}{(z-1)(z-0.6065)} = \frac{0.1065K(z+0.8467)}{(z-1)(z-0.6065)} \end{aligned} $$ - Open-loop poles: $z = 1$, $z = 0.6065$ - Open-loop zero: $z = -0.8467$ (the other branch ends at $-\infty$) #### Construction 1. **Real-axis locus:** between 0.6065 and 1, and to the left of −0.8467. 2. **Breakaway / break-in points** from $dK/dz = 0$, where $K = -\dfrac{(z-1)(z-0.6065)}{0.1065(z+0.8467)}$: $$ z^2 + 1.6935z - 1.9667 = 0 \;\Rightarrow\; z = 0.7915\ \ (K = 0.221),\quad z = -2.4850\ \ (K = 61.7) $$ 3. **Complex part:** for one real zero and two real poles the complex part is a circle centred at the zero: centre $= -0.8467$, radius $= \sqrt{(1+0.8467)(0.6065+0.8467)} = 1.638$. 4. **Crossing of the unit circle:** characteristic equation $$ z^2 + (0.1065K - 1.6065)z + (0.6065 + 0.0902K) = 0 $$ For complex roots $|z|^2 = 0.6065 + 0.0902K = 1$, so $K_c = 4.36$. The roots are then $z = 0.571 \pm j0.821$ (angle 55.2°, $\omega = 1.93$ rad/s). #### Sketch ```text Im | . - - - | # K = 4.36: z = 0.57 + j0.82 (on unit circle) . | . . | . <======*=========o-------+----x==*==x----> Re -2.48 -0.85 0 0.61 0.79 1 break-in zero breakaway . | . . | . . - - - | # K = 4.36: z = 0.57 - j0.82 | x = open-loop pole, o = open-loop zero, * = break point, # = unit-circle crossing, == real-axis locus Complex branches: circle, centre -0.85, radius 1.64 ``` #### Result - For $0 < K < 0.221$: two real closed-loop poles between 0.6065 and 1 (overdamped). - For $0.221 < K < 61.7$: complex poles on the circle. They lie inside the unit circle only while $K < 4.36$. - **The system is stable for $0 < K < 4.36$**; at $K = 4.36$ it oscillates at about 1.93 rad/s. - The input $R(s) = 1/s$ does not change the locus; it only decides the step response. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For 1/(s(s+1)) with ZOH and T = 0.5 s, what are the open-loop poles and finite zero of G(z)? A. Poles at 0 and 1, zero at 0.6065 B. Poles at 1 and 0.6065, zero at -0.8467 C. Poles at 1 and -0.6065, zero at 0.8467 D. Poles at 0.6065 and -0.8467, no zero Answer: B. G(z) = 0.1065K(z + 0.8467)/((z - 1)(z - 0.6065)). (Key point: Open-loop poles and zero) 2. Which parts of the real axis belong to the locus? A. Between -0.8467 and 0.6065 B. Everywhere right of z = 1 C. Between 0.6065 and 1, and to the left of -0.8467 D. Only between 0 and 1 Answer: C. A real-axis point is on the locus if an odd number of poles and zeros lie to its right. (Key point: Real-axis locus) 3. The complex part of the locus is: A. A straight line along the imaginary axis B. A spiral C. A parabola D. A circle centred at the zero -0.8467, radius 1.638 Answer: D. For one real zero and two real poles, the complex locus is a circle centred on the zero. (Key point: Shape of complex locus) 4. How is the critical gain found and what is it? A. Set |z|^2 = constant term = 1, giving Kc = 4.36 B. Set K at the breakaway point, giving 0.221 C. Set the sum of roots to zero, giving 61.7 D. Set the zero at the origin, giving 1 Answer: A. For complex roots |z|^2 = 0.6065 + 0.0902K = 1, so K = 4.36; the system is stable for 0 < K < 4.36. 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