--- title: "Root-locus design: ζ = 0.5, ts = 2 s, plant 1/(s(s+1))" subject: "Digital Control System" chapter: 4 key: "semester_6/digital_control_system:root-locus-design-plant-s1" source: https://www.bctengineer.com/q/semester_6/digital_control_system/root-locus-design-plant-s1 site: BCT Engineering --- # Root-locus design: ζ = 0.5, ts = 2 s, plant 1/(s(s+1)) Digital Control System · Chapter 4: Design and Compensation of Discrete Time Control System - Asked in: 2071 Magh, 10 marks; 2070 Magh, 12 marks (asked 2 times) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/digital_control_system/root-locus-design-plant-s1 - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4 ## Question Design a digital controller such that the compensated system should have damping ratio ζ=0.5 and settling time = 2 seconds of the following system using root locus approach. [Figure: unity-feedback loop: R(s) → summing junction → E(s) → sampler (period T) → digital controller G₁*(s) → ZOH → 1/s(s+1) → C(s)] ## Short answer With T=0.2s, desired z-domain poles are 0.5158±j0.4281. Plant G(z)=0.01873(z+0.9355)/[(z-1)(z-0.8187)]. A lead compensator cancels the plant pole at 0.8187, adds a pole at 0.2474, and gain 11.52: G_D(z)=11.52(z-0.8187)/(z-0.2474). ## Answer The sampling period is not given, so we choose it first. #### Step 1: Desired poles and choice of $T$ From $\zeta = 0.5$ and $t_s = 4/(\zeta\omega_n) = 2$ s: $$ \zeta\omega_n = 2, \quad \omega_n = 4\ \text{rad/s}, \quad \omega_d = 4\sqrt{1-0.25} = 3.464\ \text{rad/s} $$ Choose **$T = 0.2$ s**. Then $\omega_s = 2\pi/T = 31.4$ rad/s, giving about 9 samples per damped cycle. This is fast enough (the usual rule is at least 8–10), and the ZOH adds little phase lag. $$ \begin{aligned} |z| &= e^{-\zeta\omega_nT} = e^{-0.4} = 0.6703 \\ \angle z &= \omega_dT = 0.6928\ \text{rad} = 39.69^\circ \\ z_{1,2} &= 0.5158 \pm j0.4281 \end{aligned} $$ #### Step 2: Plant with ZOH $$ \begin{aligned} G(z) &= (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+1)}\right] = \frac{(T-1+e^{-T})z + (1-e^{-T}-Te^{-T})}{(z-1)(z-e^{-T})} \\ &= \frac{0.01873(z + 0.9355)}{(z-1)(z-0.8187)} \end{aligned} $$ #### Step 3: Angle condition at $z_1$ | Factor | Angle | |---|---| | zero at $-0.9355$ | $+16.44^\circ$ | | pole at $1$ | $-138.52^\circ$ | | pole at $0.8187$ | $-125.28^\circ$ | | **Total** | $-247.36^\circ$ | The angle deficiency is $247.36^\circ - 180^\circ = 67.36^\circ$, so a **lead** controller is needed. #### Step 4: Lead controller $G_D(z) = K\dfrac{z-\alpha}{z-\beta}$ Cancel the slow plant pole with the zero: $\alpha = 0.8187$, which adds $+125.28^\circ$. The controller pole must give $125.28^\circ - 67.36^\circ = 57.92^\circ$: $$ \beta = 0.5158 - \frac{0.4281}{\tan 57.92^\circ} = 0.2474 $$ #### Step 5: Gain $$ K = \frac{|z_1-1|\,|z_1-0.2474|}{0.01873\,|z_1+0.9355|} = \frac{0.6464 \times 0.5053}{0.01873 \times 1.5131} = 11.52 $$ $$ G_D(z) = 11.52\,\frac{z - 0.8187}{z - 0.2474} $$ #### Step 6: Check The open loop is $\dfrac{0.2158(z+0.9355)}{(z-1)(z-0.2474)}$. The characteristic equation is $z^2 - 1.0316z + 0.4493 = 0$, whose roots are $0.5158 \pm j0.4281$. ✔ $$ K_v = \frac{1}{T}\lim_{z\to1}(1-z^{-1})G_DG = \frac{0.2158 \times 1.9355}{0.2 \times 0.7526} = 2.78\ \text{s}^{-1} $$ **Answer** (for $T = 0.2$ s): $G_D(z) = 11.52\dfrac{z-0.8187}{z-0.2474}$. The closed-loop poles are at $0.5158 \pm j0.4281$ ($\zeta = 0.5$, $t_s = 2$ s), and $K_v = 2.78\ \text{s}^{-1}$. A different $T$ changes the numbers but not the method. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The sampling period is not given. Why is T = 0.2 s chosen? A. It gives about 9 samples per damped cycle, within the rule of at least 8-10, so the ZOH adds little lag B. It makes the plant gain unity C. It is the only value giving zeta = 0.5 D. It cancels the plant pole Answer: A. ws = 31.4 rad/s versus wd = 3.464 gives about 9 samples per cycle. (Key point: Choice of T) 2. The plant angle at the desired pole is -247.36 degrees, so the controller must A. Add 67.36 degrees of lead B. Subtract 67.36 degrees C. Add 180 degrees D. Not be used Answer: A. The deficiency from -180 degrees is 67.36 degrees, requiring a lead controller. (Key point: Angle deficiency) 3. The controller zero is placed at 0.8187. What is this? A. The slow plant pole e^-T, cancelled B. The desired pole's real part C. The plant zero D. The origin Answer: A. The zero cancels the plant pole at e^-0.2 = 0.8187, contributing +125.28 degrees. (Key point: Pole-zero choice) 4. The controller pole beta is found as A. 0.5158 - 0.4281/tan(57.92 degrees) = 0.2474 B. 0.8187 - 0.4281/tan(67.36 degrees) C. 1 - 0.2474 D. 0.5158 + 0.4281 Answer: A. The pole must contribute 125.28 - 67.36 = 57.92 degrees, giving beta = 0.2474. (Key point: Controller pole location) 5. The designed controller is A. 11.52 (z - 0.8187)/(z - 0.2474) B. 0.2158 (z - 0.8187)/(z - 0.2474) C. 11.52 (z - 0.2474)/(z - 0.8187) D. 2.78 (z - 0.8187)/(z - 1) Answer: A. K = 0.6464*0.5053/(0.01873*1.5131) = 11.52, giving the stated GD(z). 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