---
title: "Lead compensator for 4/[s(s+2)], ts=2 s"
subject: "Control Systems"
chapter: 7
key: "semester_3/control_systems:lead-4-s-s2-ts2"
source: https://www.bctengineer.com/q/semester_3/control_systems/lead-4-s-s2-ts2
site: BCT Engineering
---

# Lead compensator for 4/[s(s+2)], ts=2 s

Control Systems · Chapter 7: Performance Specifications and Compensation Design

- Asked in: 2080 Baisakh, 12 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/lead-4-s-s2-ts2
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7

## Question

For a unity feedback system with open loop transfer function G(s) = 4/[s(s+2)], design a Lead compensator such that settling time would become 2 seconds without change in maximum overshoot of the system.

## Short answer

A lead compensator designed via root locus method gives Gc(s) = 4.73 (s+2.93)/(s+5.46). The dominant closed-loop poles become -2 ± j3.464, achieving settling time 2 s with unchanged damping ratio 0.5 and approximately the same 16.3% overshoot.

## Answer

The specification is in the time domain, so the lead compensator is designed by the **root-locus method**.

#### Step 1: Present system

$$
\frac{C(s)}{R(s)}=\frac{4}{s^2+2s+4}\Rightarrow\omega_n=2\ \text{rad/s},\ \zeta=\frac{2}{2\times2}=0.5
$$

$$
t_s=\frac{4}{\zeta\omega_n}=\frac{4}{1}=4\ \text{s (2\% criterion)},\qquad M_p=e^{-\pi\zeta/\sqrt{1-\zeta^2}}=16.3\%
$$

#### Step 2: Desired poles

Same overshoot means same $\zeta=0.5$. New settling time 2 s:

$$
\zeta\omega_n=\frac{4}{2}=2\Rightarrow\omega_n=4\ \text{rad/s},\quad\omega_d=4\sqrt{1-0.25}=3.464
$$

$$
s_d=-2\pm j3.464
$$

#### Step 3: Angle deficiency

$$
\begin{aligned}
\angle s_d&=180^\circ-\tan^{-1}\frac{3.464}{2}=120^\circ\\
\angle(s_d+2)&=90^\circ\\
\angle G(s_d)&=-(120^\circ+90^\circ)=-210^\circ\\
\phi&=-180^\circ-(-210^\circ)=30^\circ
\end{aligned}
$$

#### Step 4: Pole and zero by the bisector method

1. Draw a horizontal line PA from $P=s_d$ to the left, and line PO to the origin. Angle APO $=180^\circ-60^\circ=120^\circ$.
2. Bisect it: the bisector PB makes $60^\circ$ with PA (direction $240^\circ$).
3. Draw PC and PD at $\pm\phi/2=\pm15^\circ$ from PB; they meet the real axis at the zero and pole.

$$
\begin{aligned}
\text{zero: }z&=-2-\frac{3.464}{\tan75^\circ}=-2-0.928=-2.93\\
\text{pole: }p&=-2-\frac{3.464}{\tan45^\circ}=-2-3.464=-5.46
\end{aligned}
$$

Check: $\angle(s_d+2.93)=75^\circ$, $\angle(s_d+5.46)=45^\circ$, net lead $=30^\circ$ ✓.

#### Step 5: Gain (magnitude condition)

$$
K_c=\frac{|s_d|\,|s_d+2|\,|s_d+5.46|}{4\,|s_d+2.93|}=\frac{4\times3.464\times4.899}{4\times3.586}=4.73
$$

$$
G_c(s)=4.73\,\frac{s+2.93}{s+5.46}
$$

#### Step 6: Compensated system

$$
G_c(s)G(s)=\frac{18.93(s+2.93)}{s(s+2)(s+5.46)}
$$

Closed-loop poles: $-2\pm j3.464$ (desired) and a third pole at $-3.46$, which lies close to the zero at $-2.93$, so its effect is small and the response is dominated by the desired pair.

```text
          jw
    sd x  | j3.46
       \  |
 --x---o--x---x--- sigma
 -5.46 -2.93 -2  0
```

| Item | Before | After |
|---|---|---|
| $\zeta$ | 0.5 | 0.5 |
| $\omega_n$ | 2 rad/s | 4 rad/s |
| $t_s$ | 4 s | 2 s |
| $M_p$ | 16.3% | ≈16.3% |
| $K_v$ | 2 | 5.07 |

Alternative: placing the zero at $-2$ (cancelling the plant pole) needs a pole at $-4$ and gives $G_c=4\frac{s+2}{s+4}$ with closed loop exactly $\frac{16}{s^2+4s+16}$.

**Answer: $G_c(s)=4.73\,\dfrac{s+2.93}{s+5.46}$; dominant poles at −2 ± j3.46 give ts = 2 s with the same ζ = 0.5 (Mp ≈ 16%).**

## Check yourself

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

1. Why is the root-locus method used for this lead design?
   A. The specifications (settling time, overshoot) are in the time domain
   B. The specifications are phase margin and gain margin
   C. The plant is type 0
   D. Root locus is the only way to design a lead network
   Answer: A. A time-domain spec fixes the desired dominant pole location, which is naturally handled on the root locus. (Key point: Design method)

2. Same overshoot (zeta = 0.5) and ts = 2 s: what are the desired poles?
   A. -1 +/- j1.732
   B. -2 +/- j3.464
   C. -4 +/- j6.93
   D. -2 +/- j2
   Answer: B. zeta*wn = 4/2 = 2, so wn = 4 and wd = 4 sqrt(0.75) = 3.464. (Key point: Desired poles)

3. What is the angle the compensator must add at s_d?
   A. 90 deg
   B. 210 deg
   C. 30 deg, since angle of G(s_d) = -210 deg and -180 deg is required
   D. 60 deg
   Answer: C. Angle of s_d is 120 deg and of (s_d+2) is 90 deg, so G has -210 deg; the deficiency is 30 deg. (Key point: Angle deficiency)

4. What do the compensated dominant poles give?
   A. Less overshoot and ts = 4 s
   B. Mp = 0% with ts = 2 s
   C. Higher overshoot with ts = 2 s
   D. The same zeta = 0.5 (Mp about 16.3%) with ts = 2 s
   Answer: D. Poles at -2 +/- j3.46 keep the damping ratio at 0.5 while doubling zeta*wn, which halves ts. (Key point: Result)

## Close variants set in other exams

- [Compensator for 4/[s(s+2)], ts=2 s, Kv=2](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Acomp-4-s-s2-ts2-kv2) — Control Systems, chapter 7, asked 2×

## Related in Control Systems

- [Lead compensator for 10/[s(s+1)], ts=2 s, Mp=5%](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-10-s-s1-ts2-mp5) — Control Systems, chapter 7
- [Lead compensator for 4/[s(s+2)], Mp≤16.3%, ts≤2 s](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-4-s-s2-mp163-ts2) — Control Systems, chapter 7
- [Transfer function from overshoot and settling time](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-from-mp25-ts6) — Control Systems, chapter 3
- [Overshoot, settling time and error for 50/[s(s+2)]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Amp-ts-50-s-s2-sse) — Control Systems, chapter 3
- [Overshoot and settling time from pole plot](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Amp-ts-from-pole-plot) — Control Systems, chapter 3
- [Compensator for 4/[s(s+2)]: Mp 16.3%, ts, Kv](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Acomp-4-s-s2-mp163-ts2) — Control Systems, chapter 7

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