---
title: "Derivative feedback for ramp input"
subject: "Control Systems"
chapter: 7
key: "semester_3/control_systems:derivative-feedback-ramp"
source: https://www.bctengineer.com/q/semester_3/control_systems/derivative-feedback-ramp
site: BCT Engineering
---

# Derivative feedback for ramp input

Control Systems · Chapter 7: Performance Specifications and Compensation Design

- Asked in: 2078 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-ramp
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7

## Question

In response to unit ramp input, discuss role of derivative feedback controller for a second order system.

## Short answer

Derivative feedback feeds back the output derivative, increasing the effective damping ratio to ζ'=ζ+Kt ωn/2. This reduces overshoot and improves stability without altering ωn. For a unit ramp input, however, the steady-state error increases by Kt, so the forward gain must be raised to recover accuracy.

## Answer

**Derivative (rate) feedback** feeds back $K_t\,\dfrac{dc}{dt}$ in addition to the output, for example with a tachogenerator on a position servo. It increases damping but, for a ramp input, it **increases the steady-state error**.

Take the standard second-order plant with derivative feedback in a minor loop:

$$
G(s) = \frac{\omega_n^2}{s(s + 2\zeta\omega_n)}, \qquad H(s) = 1 + K_t s
$$

Closed-loop transfer function:

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

#### Effect on damping
Comparing with $s^2 + 2\zeta'\omega_n s + \omega_n^2$:

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

The natural frequency is unchanged, but damping rises, so overshoot and oscillation fall and the system becomes more stable.

#### Effect on the unit ramp response
Error $E(s) = R(s) - C(s)$ with $R(s) = 1/s^2$:

$$
\begin{aligned}
E(s) &= \frac{s^2 + (2\zeta\omega_n + K_t\omega_n^2)s}{s^2 + (2\zeta\omega_n + K_t\omega_n^2)s + \omega_n^2}\cdot\frac{1}{s^2} \\
e_{ss} &= \lim_{s\to0} sE(s) = \frac{2\zeta\omega_n + K_t\omega_n^2}{\omega_n^2} = \frac{2\zeta}{\omega_n} + K_t
\end{aligned}
$$

Without rate feedback $e_{ss} = 2\zeta/\omega_n$. Rate feedback adds $K_t$ to the ramp error.

**Example:** $\omega_n = 4$, $\zeta = 0.25$, $K_t = 0.15$: $\zeta' = 0.25 + 0.3 = 0.55$ (overshoot drops from 44% to 12.6%) but $e_{ss}$ rises from 0.125 to 0.275.

**Remedy:** increase the forward gain (amplifier) to bring the ramp error back down; the extra damping from $K_t$ keeps the response well damped. In contrast, a PD controller in the forward path adds similar damping while leaving $K_v$, and so the ramp error $2\zeta/\omega_n$, unchanged. Derivative feedback therefore trades ramp accuracy for damping.

## Check yourself

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

1. With H(s) = 1 + Kt*s feedback, what happens to damping and natural frequency?
   A. Damping rises to zeta + Kt*wn/2; wn is unchanged
   B. Both rise
   C. Damping falls; wn rises
   D. Both unchanged
   Answer: A. The characteristic equation becomes s^2 + (2*zeta*wn + Kt*wn^2)s + wn^2, so only the s coefficient grows. (Key point: Effect on damping)

2. What is the steady-state ramp error with derivative feedback?
   A. 2*zeta/wn + Kt
   B. Kt only
   C. 2*zeta/wn - Kt
   D. 2*zeta/wn
   Answer: A. ess = (2*zeta*wn + Kt*wn^2)/wn^2 = 2*zeta/wn + Kt, so derivative feedback increases the error. (Key point: Ramp error)

3. For wn = 4, zeta = 0.25 and Kt = 0.15, what is the new damping?
   A. 0.25
   B. 0.55
   C. 0.40
   D. 0.85
   Answer: B. zeta' = 0.25 + 0.15*4/2 = 0.55. (Key point: Example)

4. How is the larger ramp error reduced, and how does PD in the forward path differ?
   A. Increase forward gain; PD leaves Kv and the ramp error unchanged
   B. Remove the tachogenerator; PD raises the type
   C. Decrease forward gain; PD doubles the error
   D. Increase Kt further; PD increases error too
   Answer: A. A higher amplifier gain restores accuracy. PD in the forward path adds damping while leaving Kv unchanged. (Key point: Remedy and comparison)

## Related in Control Systems

- [Derivative feedback in second order time response](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Aderivative-feedback-second-order) — Control Systems, chapter 7
- [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×
- [Underdamped second order unit step response derivation](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Asecond-order-step-derivation) — Control Systems, chapter 3
- [Impulse and ramp response from first order step response](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Afirst-order-impulse-ramp) — Control Systems, chapter 3
- [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

## Related in other subjects

- [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
- [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×
- [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
- [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