---
title: "Derivative feedback in second order time response"
subject: "Control Systems"
chapter: 7
key: "semester_3/control_systems:derivative-feedback-second-order"
source: https://www.bctengineer.com/q/semester_3/control_systems/derivative-feedback-second-order
site: BCT Engineering
---

# Derivative feedback in second order time response

Control Systems · Chapter 7: Performance Specifications and Compensation Design

- Asked in: 2080 Bhadra, 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/derivative-feedback-second-order
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7

## Question

Describe working of derivative feedback controller in time response of second order system.

## Short answer

Derivative feedback feeds back a signal proportional to the output rate, increasing the effective damping ratio to ζ + (Kt ωn)/2 without changing ωn. This reduces overshoot and settling time, but it increases the steady-state error to ramp inputs.

## Answer

In **derivative (rate/tachometer) feedback control**, a signal proportional to the rate of change of the output, $K_t\,\dot c(t)$, is fed back negatively in an inner loop, besides the main unity feedback. In a motor position system, a tachogenerator on the shaft provides this velocity signal.

```text
R --(+)--->(+)--> wn^2/(s(s+2z wn)) --+--> C
     ^-     ^-                        |
     |      +------[ Kt s ]<----------+
     +--------------------------------+
```

#### Analysis

Inner loop:

$$
G_1(s)=\frac{\omega_n^2}{s(s+2\zeta\omega_n+K_t\omega_n^2)}
$$

Closed loop:

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

$$
\zeta_{eff}=\zeta+\frac{K_t\omega_n}{2},\qquad\omega_n\ \text{unchanged}
$$

#### Working / effect on time response

1. When the output moves fast towards the reference, the feedback signal $K_t\dot c$ is large and subtracts from the actuating signal, **braking the system early**. This is extra damping.
2. **Overshoot falls** as $\zeta_{eff}$ rises; $t_s=4/(\zeta_{eff}\omega_n)$ falls.
3. $\omega_n$ is unchanged and **no zero is added**, so the response is smooth (unlike PD, whose zero can add overshoot).
4. **Ramp error increases**: $K_v=\dfrac{\omega_n}{2\zeta+K_t\omega_n}$; this is offset by raising the amplifier gain.

| Property | Effect |
|---|---|
| Damping | increases |
| $M_p$, $t_s$ | decrease |
| $\omega_n$ | same |
| Ramp error | increases |

**Example:** $\omega_n=5$ rad/s, $\zeta=0.2$ ($M_p=52.7\%$). With $K_t=0.12$: $\zeta_{eff}=0.2+0.3=0.5$, $M_p=16.3\%$.

## Check yourself

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

1. With derivative (tachometer) feedback Kt, what is the effective damping ratio?
   A. zeta + Kt*wn/2
   B. zeta*Kt*wn
   C. zeta - Kt*wn/2
   D. zeta + Kt*wn
   Answer: A. The characteristic equation becomes s^2 + (2 zeta wn + Kt wn^2)s + wn^2, so zeta_eff = zeta + Kt wn/2. (Key point: Effective damping)

2. What does derivative feedback do to wn and the closed-loop zeros?
   A. wn increases and a zero is added
   B. wn is unchanged and no zero is added
   C. wn decreases and no zero is added
   D. wn is unchanged but a zero is added, as in PD
   Answer: B. Only the s-term coefficient changes; the numerator stays wn^2, so no zero appears, unlike PD control. (Key point: Natural frequency and zero)

3. How does the feedback signal Kt*dc/dt improve the response?
   A. It adds to the error so the output reaches the target sooner
   B. It removes steady-state error for a step input
   C. When the output moves fast, it subtracts from the actuating signal and brakes the system early, adding damping
   D. It raises wn so the output rises faster
   Answer: C. The velocity signal opposes fast motion, giving extra damping and lower overshoot. (Key point: Physical working)

4. What is the drawback of derivative feedback for ramp input?
   A. Ramp error decreases because damping rises
   B. Settling time increases
   C. The system becomes unstable for any Kt
   D. Kv = wn/(2 zeta + Kt wn) falls, so ramp error increases
   Answer: D. Kv decreases as Kt increases, so the ramp error grows; it is offset by raising the amplifier gain. (Key point: Price paid)

## Related in Control Systems

- [Derivative feedback controller: block diagram and effect](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Aderivative-feedback-bd-effect) — Control Systems, chapter 7, asked 2×
- [Derivative feedback for ramp input](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Aderivative-feedback-ramp) — Control Systems, chapter 7
- [Derive maximum overshoot of second order system](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Amp-expression-derivation) — Control Systems, chapter 3
- [Second order model from step response (peak 1.25)](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Asecond-order-from-step-125) — Control Systems, chapter 3
- [PI controller and its effect on time response](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Api-controller-role) — Control Systems, chapter 7
- [Control system and why feedback makes response faster](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Awhat-is-cs-feedback-faster) — Control Systems, chapter 1

## Related in other subjects

- [Magnitude response of second order system](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Amagnitude-response-0-5-0-3) — Digital Signal Analysis and Processing, chapter 3, asked 2×
- [Second-order step response and damping](https://www.bctengineer.com/question.md?key=semester_4%2Finstrumentation_and_measurement%3Asecond-order-step-response-damping) — Instrumentation and Measurement, chapter 4
- [Second-order transducer frequency and step response](https://www.bctengineer.com/question.md?key=semester_4%2Finstrumentation_and_measurement%3Asecond-order-transducer-numerical) — Instrumentation and Measurement, chapter 4
- [Controller design with 8 samples per cycle, step response and Kv](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Adesign-8-samples-step-kv) — Digital Control System, chapter 4