---
title: "Controller for damping ratio of 1"
subject: "Control Systems"
chapter: 7
key: "semester_3/control_systems:controller-zeta-1"
source: https://www.bctengineer.com/q/semester_3/control_systems/controller-zeta-1
site: BCT Engineering
---

# Controller for damping ratio of 1

Control Systems · Chapter 7: Performance Specifications and Compensation Design

- Asked in: 2072 Chaitra, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/controller-zeta-1
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7

## Question

If desired damping ratio is '1', which controller do you suggest? Explain.

## Short answer

To achieve damping ratio of 1 (critical damping), use a PD controller or derivative output feedback. The derivative term increases effective damping without lowering natural frequency, yielding fastest non-overshooting response. Rate feedback gain Kt = 2(1-ζ)/ω_n.

## Answer

To obtain a damping ratio of **1 (critical damping: fastest response with no overshoot)**, a controller that **adds damping** is needed: a **PD (proportional–derivative) controller**, or equivalently **derivative (tachometer) output feedback**. Gain adjustment alone cannot usually do it without making the response very slow.

#### Why PD

For a typical underdamped plant $G(s)=\dfrac{\omega_n^2}{s(s+2\zeta\omega_n)}$, PD control $G_c(s)=1+T_ds$ gives

$$
\frac{C(s)}{R(s)}=\frac{\omega_n^2(1+T_ds)}{s^2+(2\zeta\omega_n+T_d\omega_n^2)s+\omega_n^2}
$$

$$
\zeta'=\zeta+\frac{T_d\omega_n}{2}
$$

The derivative term raises ζ while $\omega_n$ stays the same. For $\zeta'=1$:

$$
T_d=\frac{2(1-\zeta)}{\omega_n}
$$

With **derivative output feedback** ($K_ts$ in the minor loop), $\zeta'=\zeta+\frac{K_t\omega_n}{2}$, so $K_t=\frac{2(1-\zeta)}{\omega_n}$; this has no added zero, so the response is truly non-overshooting.

#### Why not others

| Controller | Effect on ζ |
|---|---|
| P (lower gain) | ζ rises but $\omega_n$ and speed fall; steady-state error rises |
| PI | adds lag: ζ falls |
| PD / rate feedback | ζ rises, speed kept |

**Example:** $G(s)=\dfrac{16}{s(s+2)}$: $\omega_n=4$, $\zeta=0.25$. Rate feedback with $K_t=\dfrac{2(1-0.25)}{4}=0.375$ gives $s^2+(2+6)s+16=(s+4)^2$, i.e. $\zeta=1$, a critically damped response with $t_s\approx\frac{5.8}{\omega_n}=1.46$ s and no overshoot.

## Check yourself

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

1. Which controller is suggested to get zeta = 1?
   A. PD controller, or derivative (tachometer) feedback
   B. PI controller
   C. Lower P gain only
   D. Lag compensator
   Answer: A. It raises damping while keeping wn, so speed is kept. (Key point: Suggested controller)

2. With PD control on wn^2/[s(s+2 zeta wn)], what Td gives zeta' = 1?
   A. Td = (1 - zeta)/(2 wn)
   B. Td = 2(1 - zeta)/wn
   C. Td = 2 zeta/wn
   D. Td = 2(1 + zeta)/wn
   Answer: B. Setting zeta + Td wn/2 = 1 gives Td = 2(1 - zeta)/wn. (Key point: Required Td)

3. Why can derivative output feedback be better than PD for non-overshooting response?
   A. It removes the pole at the origin
   B. It increases wn
   C. It adds no zero, so the response is truly non-overshooting
   D. It adds an integrator
   Answer: C. PD adds a zero which can add some overshoot; rate feedback does not. (Key point: Rate feedback advantage)

4. For G = 16/[s(s+2)], what Kt gives critical damping?
   A. Kt = 0.75
   B. Kt = 0.25
   C. Kt = 1.5
   D. Kt = 0.375
   Answer: D. wn = 4, zeta = 0.25, so Kt = 2(0.75)/4 = 0.375, giving s^2 + 8s + 16 = (s+4)^2. (Key point: Example)

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