---
title: "Compensator for 10/[s(s+1)], ζ=0.5, ωn=3"
subject: "Control Systems"
chapter: 7
key: "semester_3/control_systems:comp-10-s-s1-zeta05-wn3"
source: https://www.bctengineer.com/q/semester_3/control_systems/comp-10-s-s1-zeta05-wn3
site: BCT Engineering
---

# Compensator for 10/[s(s+1)], ζ=0.5, ωn=3

Control Systems · Chapter 7: Performance Specifications and Compensation Design

- Asked in: 2081 Bhadra, 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/comp-10-s-s1-zeta05-wn3
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7

## Question

Design a compensator for unity feedback system with open loop transfer function G(s) = 10/[s(s+1)] such that the damping ratio would become 0.5 and natural frequency of oscillation 3 rad/sec.

## Short answer

The desired dominant poles are −1.5 ± j2.598. The plant angle deficiency is 40.9°; place the compensator zero at −1 to cancel the plant pole, putting the pole at −3. With Kc=0.9, Gc(s)=0.9(s+1)/(s+3) gives closed-loop ζ=0.5 and ωn=3 rad/s.

## Answer

The specification is on the closed-loop pole positions, so a **lead compensator is designed by the root-locus method**.

#### Step 1: Present system

$$
\frac{C}{R}=\frac{10}{s^2+s+10}\Rightarrow\omega_n=\sqrt{10}=3.16\ \text{rad/s},\ \zeta=\frac{1}{2\times3.16}=0.158
$$

The response is very oscillatory ($M_p\approx60\%$).

#### Step 2: Desired dominant poles

$$
s_d=-\zeta\omega_n\pm j\omega_n\sqrt{1-\zeta^2}=-1.5\pm j2.598
$$

#### Step 3: Angle deficiency

Angle of $G(s)=\dfrac{10}{s(s+1)}$ at $s_d=-1.5+j2.598$:

$$
\begin{aligned}
\angle s_d&=180^\circ-\tan^{-1}\frac{2.598}{1.5}=120^\circ\\
\angle(s_d+1)&=180^\circ-\tan^{-1}\frac{2.598}{0.5}=100.9^\circ\\
\angle G(s_d)&=-(120^\circ+100.9^\circ)=-220.9^\circ
\end{aligned}
$$

To satisfy the angle condition ($-180^\circ$), the compensator must add

$$
\phi=-180^\circ-(-220.9^\circ)=40.9^\circ\ \text{(lead)}
$$

#### Step 4: Locate zero and pole

Choose the compensator zero at $s=-1$ to cancel the plant pole at $-1$. Then the pole $-p$ must satisfy

$$
-\angle s_d-\angle(s_d+p)=-180^\circ\Rightarrow-120^\circ-\angle(s_d+p)=-180^\circ\Rightarrow\angle(s_d+p)=60^\circ
$$

$$
\tan^{-1}\frac{2.598}{p-1.5}=60^\circ\Rightarrow p-1.5=\frac{2.598}{1.732}=1.5\Rightarrow p=3
$$

Check of angle added: zero $+100.9^\circ$, pole $-60^\circ$, net $+40.9^\circ$ ✓.

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

$$
G_c(s)G(s)=K_c\frac{s+1}{s+3}\cdot\frac{10}{s(s+1)}=\frac{10K_c}{s(s+3)}
$$

$$
\left|\frac{10K_c}{s_d(s_d+3)}\right|=1\Rightarrow10K_c=|s_d|\,|s_d+3|=3\times3=9\Rightarrow K_c=0.9
$$

$$
G_c(s)=0.9\,\frac{s+1}{s+3}=0.3\,\frac{1+s}{1+0.333s}
$$

#### Step 6: Verification

$$
\frac{C(s)}{R(s)}=\frac{9}{s^2+3s+9}
$$

$\omega_n=3$ rad/s, $2\zeta\omega_n=3\Rightarrow\zeta=0.5$ ✓. Closed-loop poles: $-1.5\pm j2.598$.

```text
         jw
   x sd  |  j2.598
    \    |
 ---x--o-x------- sigma
   -3 -1 0   (zero cancels pole at -1)
```

| Item | Before | After |
|---|---|---|
| $\zeta$ | 0.158 | 0.5 |
| $\omega_n$ | 3.16 rad/s | 3 rad/s |
| $M_p$ | 60% | 16.3% |
| $t_s$ (2%) | 8 s | 2.67 s |
| $K_v$ | 10 | 3 |

Note: $K_v$ falls to $10\times0.9/3=3$; if a higher $K_v$ is needed, the bisector method or an extra lag section can be used.

**Answer: $G_c(s)=0.9\,\dfrac{s+1}{s+3}$, giving closed-loop poles at −1.5 ± j2.6 (ζ = 0.5, ωn = 3 rad/s).**

## Check yourself

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

1. The specification is zeta = 0.5 and wn = 3 rad/s. Which design method fits?
   A. Bode plot lag design
   B. Root-locus lead design placing dominant poles at -1.5 +/- j2.598
   C. Nyquist plot design
   D. Routh-Hurwitz design
   Answer: B. Specs on closed-loop pole positions point to root-locus design with s_d = -zeta wn +/- j wn sqrt(1-zeta^2). (Key point: Method)

2. The plant angle at s_d = -1.5 + j2.598 is -220.9 degrees. What must the compensator contribute?
   A. +40.9 degrees (lead)
   B. -40.9 degrees (lag)
   C. +220.9 degrees
   D. 0 degrees
   Answer: A. Angle condition requires -180 degrees, so compensator must add -180 - (-220.9) = +40.9 degrees. (Key point: Angle deficiency)

3. With the zero chosen at s = -1 to cancel the plant pole, where is the compensator pole?
   A. s = -1
   B. s = -3
   C. s = -5
   D. s = -0.3
   Answer: B. Angle of (s_d + p) must be 60 degrees, so p - 1.5 = 2.598/tan60 = 1.5 and p = 3. (Key point: Pole placement)

4. Gc G = 10 Kc/[s(s+3)]. The gain Kc is found from
   A. The angle condition, Kc = 1
   B. |10 Kc/(s_d(s_d+3))| = 1, so 10 Kc = 3 x 3 = 9, Kc = 0.9
   C. Kv = 10, so Kc = 10
   D. Kc = 3/10 x 10 = 3
   Answer: B. Magnitude condition at s_d gives |s_d||s_d+3| = 9, so Kc = 0.9 and C/R = 9/(s^2 + 3s + 9). (Key point: Gain by magnitude condition)

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