--- title: "z-plane design with Kv, plant 1/(s(s+2))" subject: "Digital Control System" chapter: 4 key: "semester_6/digital_control_system:z-plane-design-kv-s2" source: https://www.bctengineer.com/q/semester_6/digital_control_system/z-plane-design-kv-s2 site: BCT Engineering --- # z-plane design with Kv, plant 1/(s(s+2)) Digital Control System · Chapter 4: Design and Compensation of Discrete Time Control System - Asked in: 2079 Chaitra, 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/z-plane-design-kv-s2 - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4 ## Question Consider the digital control system shown in figure below. In the z plane, design a digital controller such that the dominant closed-loop poles have a damping ratio of 0.5 and a settling time of 2 seconds. The sampling period is assumed to be 0.2 sec. Also, obtain the static velocity error constant of the system. [Figure: unity-feedback loop: r(t) → summing junction → e(t) → sampler → e(kT) → digital controller → ZOH → 1/s(s+2) → C(t)] ## Short answer Desired closed-loop poles are z=0.5158±j0.4281 with ζ=0.5 and ts=2 s. Plant with ZOH: G(z)=0.01758(z+0.8753)/[(z-1)(z-0.6703)]. Controller: G_D(z)=12.67(z-0.6703)/(z-0.2543). Static velocity error constant Kv=2.80 s^-1. ## Answer #### Step 1: Desired closed-loop poles With $\zeta = 0.5$ and $t_s = 4/(\zeta\omega_n) = 2$ s: $$ \zeta\omega_n = 2, \quad \omega_n = 4, \quad \omega_d = 3.464\ \text{rad/s} $$ With $T = 0.2$ s: $$ \begin{aligned} |z| &= e^{-0.4} = 0.6703, \qquad \angle z = \omega_dT = 39.69^\circ \\ z_{1,2} &= 0.5158 \pm j0.4281 \end{aligned} $$ #### Step 2: ZOH + plant $1/[s(s+2)]$ $$ G(z) = \frac{(2T-1+e^{-2T})z + (1-e^{-2T}-2Te^{-2T})}{4(z-1)(z-e^{-2T})} = \frac{0.01758(z + 0.8753)}{(z-1)(z-0.6703)} $$ #### Step 3: Angle deficiency At $z_1$: the zero gives $+17.11^\circ$, the pole at 1 gives $-138.52^\circ$, and the pole at 0.6703 gives $-109.85^\circ$. The total is $-231.26^\circ$. The controller must supply $+51.26^\circ$ (lead). #### Step 4: Lead controller $G_D(z) = K\dfrac{z-\alpha}{z-\beta}$ Choose $\alpha = 0.6703$ to cancel the plant pole. This contributes $+109.85^\circ$. The controller pole must then contribute $109.85^\circ - 51.26^\circ = 58.59^\circ$: $$ \beta = 0.5158 - \frac{0.4281}{\tan 58.59^\circ} = 0.2543 $$ #### Step 5: Gain $$ K = \frac{|z_1-1|\,|z_1-0.2543|}{0.01758\,|z_1+0.8753|} = \frac{0.6464 \times 0.5017}{0.01758 \times 1.4555} = 12.67 $$ $$ G_D(z) = 12.67\,\frac{z - 0.6703}{z - 0.2543} $$ #### Step 6: Check $$ G_DG = \frac{0.2228(z + 0.8753)}{(z-1)(z-0.2543)} $$ The characteristic equation $z^2 - 1.0316z + 0.4493 = 0$ gives $z = 0.5158 \pm j0.4281$. ✔ #### Step 7: Static velocity error constant $$ \begin{aligned} K_v &= \lim_{z\to1}\frac{(1-z^{-1})G_D(z)G(z)}{T} = \frac{1}{0.2}\cdot\frac{0.2228(1 + 0.8753)}{1 - 0.2543} \\ &= 5 \times \frac{0.4178}{0.7457} = 2.80\ \text{s}^{-1} \end{aligned} $$ **Answer:** $G_D(z) = 12.67\dfrac{z-0.6703}{z-0.2543}$ and $K_v = 2.80\ \text{s}^{-1}$. The steady-state error for a unit ramp is $1/K_v = 0.357$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With zeta = 0.5 and ts = 2 s (ts = 4/(zeta wn)), and T = 0.2 s, what are the desired poles? A. |z| = e^-2, angle = 30 degrees B. |z| = 0.5, angle = 60 degrees C. |z| = e^-4, angle = 39.69 degrees D. |z| = e^-0.4, angle = wd T = 39.69 degrees, giving 0.5158 +/- j0.4281 Answer: D. zeta wn = 2, wn = 4, wd = 3.464; |z| = e^(-zeta wn T) = e^-0.4 and angle = wd T. (Key point: Desired pole location) 2. The plant angles at z1 sum to -231.26 degrees. What must the controller supply? A. +51.26 degrees (lead), to reach -180 B. -51.26 degrees (lag) C. +231.26 degrees D. -180 degrees Answer: A. The root-locus angle condition needs a total of -180 degrees; the plant gives -231.26, so the controller must add +51.26 degrees (lead). (Key point: Angle deficiency) 3. Why is the controller zero placed at alpha = 0.6703? A. To cancel the integrator B. To cancel the plant pole at 0.6703 C. To make Kv infinite D. To place a zero at z = 1 Answer: B. Cancelling the plant pole removes its angle (+109.85 degrees contributed) and leaves a first-order controller pole to set. (Key point: Pole-zero cancellation) 4. How is Kv computed here? A. Kv = lim GD(z) G(z) as z -> 0 B. Kv = lim (z - 1) GD G as z -> infinity C. Kv = lim (1 - z^-1) GD(z) G(z) / T as z -> 1 D. Kv = 1/T times the gain K Answer: C. For a type-1 sampled system, Kv = (1/T) lim (1 - z^-1) GD G, which gives 2.80 per second. 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