--- title: "Root-locus design: 8 samples per cycle, T = 0.2 s" subject: "Digital Control System" chapter: 4 key: "semester_6/digital_control_system:root-locus-8-samples-t02" source: https://www.bctengineer.com/q/semester_6/digital_control_system/root-locus-8-samples-t02 site: BCT Engineering --- # Root-locus design: 8 samples per cycle, T = 0.2 s Digital Control System · Chapter 4: Design and Compensation of Discrete Time Control System - Asked in: 2073 Magh, 12 marks - 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-8-samples-t02 - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4 ## Question Using the root locus method in z-plane determine a digital controller for a system shown below so that the dominant closed loop poles of the compensated system will have damping ratio of 0.5 and number of samples per cycle of damped sinusoidal oscillation to be 8. Assume that the sampling period T as 0.2 sec. [Figure: unity-feedback loop: R(z) → summing junction → sampler → digital controller → zero order hold → 1/s(s+1) → C(z)] ## Short answer Digital controller: GD(z)=13.93(z-0.8187)/(z-0.1596), a lead compensator. Static velocity error constant Kv=3.0 s^-1. ## Answer #### Step 1: Desired dominant poles With 8 samples per cycle, $\omega_dT = 2\pi/8 = 45^\circ$, and $$ \begin{aligned} |z| &= \exp\left(-\frac{2\pi\zeta}{\sqrt{1-\zeta^2}}\cdot\frac{\omega_d}{\omega_s}\right) = \exp\left(-\frac{2\pi(0.5)}{0.866}\cdot\frac18\right) = 0.6354 \\ z_{1,2} &= 0.4493 \pm j0.4493 \end{aligned} $$ With $T = 0.2$ s, this means $\omega_d = \omega_s/8 = 3.927$ rad/s, $\omega_n = 4.534$ rad/s, and $t_s \approx 4/(\zeta\omega_n) = 1.76$ s. #### Step 2: Plant with ZOH ($T = 0.2$ s) $$ \begin{aligned} G(z) &= \frac{(T-1+e^{-T})z + (1-e^{-T}-Te^{-T})}{(z-1)(z-e^{-T})} = \frac{0.01873z + 0.01752}{(z-1)(z-0.8187)} \\ &= \frac{0.01873(z + 0.9355)}{(z-1)(z-0.8187)} \end{aligned} $$ #### Step 3: Angle deficiency at $z_1$ | Factor | Angle | |---|---| | zero at $-0.9355$ | $+17.98^\circ$ | | pole at $1$ | $-140.79^\circ$ | | pole at $0.8187$ | $-129.43^\circ$ | | **Total** | $-252.24^\circ$ | The deficiency is $72.24^\circ$, so a lead compensator is needed. #### Step 4: Controller $G_D(z) = K\dfrac{z-0.8187}{z-\beta}$ The zero cancels the pole at 0.8187 and adds $+129.43^\circ$. The controller pole must then contribute $129.43^\circ - 72.24^\circ = 57.19^\circ$: $$ \beta = 0.4493 - \frac{0.4493}{\tan 57.19^\circ} = 0.1596 $$ #### Step 5: Gain $$ K = \frac{|z_1-1|\,|z_1-0.1596|}{0.01873\,|z_1+0.9355|} = \frac{0.7107 \times 0.5346}{0.01873 \times 1.4559} = 13.93 $$ $$ G_D(z) = 13.93\,\frac{z - 0.8187}{z - 0.1596} $$ #### Step 6: Check $$ G_DG = \frac{0.2610(z + 0.9355)}{(z-1)(z-0.1596)} $$ The characteristic equation is $z^2 - 0.8986z + 0.4038 = 0$, with roots $0.4493 \pm j0.4493$. ✔ $$ K_v = \frac{1}{T}\cdot\frac{0.2610 \times 1.9355}{1 - 0.1596} = 3.01\ \text{s}^{-1} $$ **Answer:** $G_D(z) = 13.93\dfrac{z-0.8187}{z-0.1596}$, a lead compensator, with $K_v \approx 3.0\ \text{s}^{-1}$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With 8 samples per cycle and zeta = 0.5, why are the desired z-plane poles the same (0.4493 +/- j0.4493) regardless of T? A. They depend on T through e^-T B. Because T is always 0.2 C. Because the plant is type 1 D. They depend only on zeta and samples per cycle; T only fixes the real-time values like wd Answer: D. Angle = 360/8 = 45 degrees and magnitude come from zeta and samples per cycle; T then sets wd = 3.927 rad/s. (Key point: Dominant pole independent of T) 2. The plant angles at z1 total -252.24 degrees. What does the controller need to add? A. 72.24 degrees B. -72.24 degrees C. 180 degrees D. 252.24 degrees Answer: A. The plant gives -252.24 degrees; to reach -180 the controller must add 72.24 degrees of lead. (Key point: Angle deficiency) 3. After the zero at 0.8187 cancels the plant pole (+129.43 degrees), the controller pole must contribute: A. 72.24 degrees B. 57.19 degrees, giving beta = 0.1596 C. 129.43 degrees D. 0 degrees Answer: B. 129.43 - 72.24 = 57.19 degrees, so beta = 0.4493 - 0.4493/tan 57.19 degrees = 0.1596. (Key point: Controller pole) 4. How is the gain K found after the angle condition is satisfied? A. From the final value theorem B. By setting K = 1 C. From the magnitude condition: |GD G| = 1 at z1 D. From the angle condition Answer: C. K = product of distances from poles / (plant gain x distance from zero) so the magnitude at z1 equals 1, giving 13.93. 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