---
title: "Root-locus design: ζ = 0.5, ts = 2 s, plant 1/(s(s+2))"
subject: "Digital Control System"
chapter: 4
key: "semester_6/digital_control_system:root-locus-design-plant-s2"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/root-locus-design-plant-s2
site: BCT Engineering
---

# Root-locus design: ζ = 0.5, ts = 2 s, plant 1/(s(s+2))

Digital Control System · Chapter 4: Design and Compensation of Discrete Time Control System

- Asked in: 2082 Chaitra, 12 marks; 2076 Bhadra, 12 marks; 2073 Bhadra, 12 marks (asked 3 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-s2
- 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 tₛ=2 s using root locus approach. [Figure: unity-feedback loop: R(s) → summing junction → E(s) → sampler (T = 0.2) → digital controller → ZOH → 1/s(s+2) → C(s)]

## Short answer

Desired closed-loop poles: |z| = e^{−ζωnT} = 0.6703, angle ωdT = 39.69°. Plant G(z) = 0.01758(z+0.8753)/[(z−1)(z−0.6703)] has angle deficiency +51.26°. The lead controller zero cancels the plant pole at 0.6703; pole placed at 0.2543. Gain K = 12.67 gives G_D(z) = 12.67(z−0.6703)/(z−0.2543) with static velocity error constant K_v = 2.80 s⁻¹.

## Answer

#### Step 1: Desired closed-loop poles
From the specifications $\zeta = 0.5$ and $t_s = 2$ s (2% criterion, $t_s = 4/\zeta\omega_n$):

$$
\begin{aligned}
\zeta\omega_n &= \frac{4}{t_s} = 2, \qquad \omega_n = 4\ \text{rad/s} \\
\omega_d &= \omega_n\sqrt{1-\zeta^2} = 3.464\ \text{rad/s}
\end{aligned}
$$

With $T = 0.2$ s, the samples per cycle are $\omega_s/\omega_d = 31.42/3.464 \approx 9$.

$$
\begin{aligned}
|z| &= e^{-\zeta\omega_nT} = e^{-0.4} = 0.6703 \\
\angle z &= \omega_dT = 0.6928\ \text{rad} = 39.69^\circ \\
z_{1,2} &= 0.6703\angle \pm 39.69^\circ = 0.5158 \pm j0.4281
\end{aligned}
$$

#### Step 2: Plant pulse transfer function
$$
\begin{aligned}
G(z) &= (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+2)}\right] = \frac{(2T-1+e^{-2T})z + (1-e^{-2T}-2Te^{-2T})}{4(z-1)(z-e^{-2T})} \\
&= \frac{0.01758(z + 0.8753)}{(z-1)(z-0.6703)}
\end{aligned}
$$

#### Step 3: Angle deficiency at $z_1 = 0.5158 + j0.4281$
| Factor | Angle at $z_1$ |
|---|---|
| zero at $-0.8753$ | $+17.11^\circ$ |
| pole at $1$ | $-138.52^\circ$ |
| pole at $0.6703$ | $-109.85^\circ$ |
| **Total** $\angle G(z_1)$ | $-231.26^\circ$ |

The angle condition requires $-180^\circ$, so the controller must add **$+51.26^\circ$** of phase lead.

#### Step 4: Lead controller
Take $G_D(z) = K\dfrac{z-\alpha}{z-\beta}$. Put the zero on the plant pole, $\alpha = 0.6703$, to cancel it. This zero adds $+109.85^\circ$. The 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 from the magnitude condition
After cancellation, the open-loop function is $\dfrac{0.01758K(z+0.8753)}{(z-1)(z-0.2543)}$, and $|G_DG| = 1$ at $z_1$:

$$
K = \frac{|z_1-1|\,|z_1-0.2543|}{0.01758\,|z_1+0.8753|} = \frac{0.6464 \times 0.5017}{0.01758 \times 1.4555} = 12.67
$$

$$
G_D(z) = 12.67\,\frac{z - 0.6703}{z - 0.2543}
$$

#### Step 6: Check
$$
\begin{aligned}
G_D(z)G(z) &= \frac{0.2228(z + 0.8753)}{(z-1)(z-0.2543)} \\
\text{Characteristic eq.:}\quad z^2 &- 1.0316z + 0.4493 = 0
\end{aligned}
$$

The roots are $z = 0.5158 \pm j0.4281$, the desired poles. ✔ This is a second-order closed loop, so the poles are exactly dominant.

**Static velocity error constant:**

$$
K_v = \lim_{z\to1}\frac{(1-z^{-1})G_DG}{T} = \frac{1}{0.2}\cdot\frac{0.2228 \times 1.8753}{0.7457} = 2.80\ \text{s}^{-1}
$$

**Answer:** $G_D(z) = 12.67\dfrac{z-0.6703}{z-0.2543}$, a lead compensator, with $K_v = 2.80\ \text{s}^{-1}$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For zeta = 0.5, ts = 2 s (ts = 4/(zeta wn)) and T = 0.2 s, the desired z-plane poles are
   A. 0.5158 +/- j0.4281
   B. 0.8187 +/- j0.2780
   C. 0.6703 +/- j0.6703
   D. 0.4 +/- j0.3464
   Answer: A. zeta*wn = 2, wd = 3.464, |z| = e^-0.4 = 0.6703 and angle = wd*T = 39.69 degrees. (Key point: Desired poles)

2. The plant angle at the desired pole is -231.26 degrees. What does the controller need to add?
   A. -51.26 degrees
   B. +51.26 degrees of phase lead
   C. +231.26 degrees
   D. 0 degrees
   Answer: B. The total must be -180 degrees, so the controller must supply +51.26 degrees. (Key point: Angle deficiency)

3. In the lead controller, the zero alpha is placed at 0.6703. Why?
   A. To cancel the plant pole at 0.6703
   B. To cancel the plant zero
   C. To place it at the origin
   D. To make the controller unstable
   Answer: A. The zero cancels the slow plant pole at e^-2T = 0.6703 and gives +109.85 degrees. (Key point: Zero placement)

4. With the zero giving 109.85 degrees, the controller pole must contribute
   A. 109.85 + 51.26 = 161.11 degrees
   B. 109.85 - 51.26 = 58.59 degrees
   C. 51.26 degrees
   D. 180 degrees
   Answer: B. The pole's angle is subtracted, so its angle is 109.85 - 51.26 = 58.59 degrees, giving beta = 0.2543. (Key point: Controller pole angle)

5. The gain from the magnitude condition gives the controller
   A. 12.67 (z - 0.6703)/(z - 0.2543)
   B. 0.2228 (z - 0.6703)/(z - 0.2543)
   C. 12.67 (z - 0.2543)/(z - 0.6703)
   D. 1.267 (z - 0.6703)/(z - 1)
   Answer: A. K = |z1-1||z1-0.2543| / (0.01758|z1+0.8753|) = 12.67. (Key point: Gain K and final controller)

## Close variants set in other exams

- [Root-locus design: ζ = 0.5, ts = 2 s, plant 1/(s(s+1))](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aroot-locus-design-plant-s1) — Digital Control System, chapter 4, asked 5×

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