---
title: "Velocity feedback increases ζ and steady-state error"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:velocity-feedback-zeta-sse"
source: https://www.bctengineer.com/q/semester_3/control_systems/velocity-feedback-zeta-sse
site: BCT Engineering
---

# Velocity feedback increases ζ and steady-state error

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2071 Chaitra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/velocity-feedback-zeta-sse
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Show that using the velocity feedback technique shown in figure below damping ratio and steady state error are both increased. [Figure: R(s) → summing point (+, −) → summing point (+, −) → ωn²/[s(s + 2ζωn)] → C(s); minor loop: output fed back through skd to the second summing point (−); major loop: unity feedback from output to the first summing point (−)]

## Short answer

Velocity feedback adds k_d ω_n/2 to the damping ratio, so ζ' = ζ + k_d ω_n/2, which is larger. The ramp steady‑state error becomes e_ss' = 2ζ/ω_n + k_d, increasing by k_d; therefore both damping ratio and ramp error increase.

## Answer

Velocity (derivative output) feedback adds a term proportional to $\dot{c}$ to the feedback signal. This increases the coefficient of $s$ in the characteristic equation.

#### Without velocity feedback ($k_d = 0$)
$$
G(s) = \frac{\omega_n^2}{s(s + 2\zeta\omega_n)}, \qquad
\frac{C(s)}{R(s)} = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}
$$
Damping ratio $= \zeta$. Velocity error constant and ramp error:
$$
K_v = \lim_{s\to 0}sG(s) = \frac{\omega_n}{2\zeta}, \qquad e_{ss} = \frac{1}{K_v} = \frac{2\zeta}{\omega_n}
$$

#### With velocity feedback
The minor loop has forward gain $\dfrac{\omega_n^2}{s(s+2\zeta\omega_n)}$ and feedback $sk_d$:
$$
G'(s) = \frac{\frac{\omega_n^2}{s(s+2\zeta\omega_n)}}{1 + \frac{\omega_n^2 sk_d}{s(s+2\zeta\omega_n)}} = \frac{\omega_n^2}{s(s + 2\zeta\omega_n + k_d\omega_n^2)}
$$
Closed loop (unity outer feedback):
$$
\frac{C(s)}{R(s)} = \frac{\omega_n^2}{s^2 + (2\zeta\omega_n + k_d\omega_n^2)s + \omega_n^2}
$$

#### New damping ratio
Comparing with $s^2 + 2\zeta'\omega_n s + \omega_n^2$ ($\omega_n$ unchanged):
$$
2\zeta'\omega_n = 2\zeta\omega_n + k_d\omega_n^2
\;\Rightarrow\;
\zeta' = \zeta + \frac{k_d\omega_n}{2}
$$
Since $k_d > 0$, $\zeta' > \zeta$. **Damping increases**, so overshoot falls.

#### New steady state error (unit ramp)
$$
K_v' = \lim_{s\to 0}sG'(s) = \frac{\omega_n^2}{2\zeta\omega_n + k_d\omega_n^2} = \frac{\omega_n}{2\zeta + k_d\omega_n}
$$
$$
e_{ss}' = \frac{1}{K_v'} = \frac{2\zeta + k_d\omega_n}{\omega_n} = \frac{2\zeta}{\omega_n} + k_d
$$
Since $k_d > 0$, $e_{ss}' > e_{ss}$. **The steady state error to a ramp increases** by $k_d$. (The step error stays zero, as the system is still Type 1.)

| Quantity | Without $k_d$ | With $k_d$ |
|---|---|---|
| $\omega_n$ | $\omega_n$ | $\omega_n$ |
| Damping ratio | $\zeta$ | $\zeta + k_d\omega_n/2$ |
| $K_v$ | $\omega_n/2\zeta$ | $\omega_n/(2\zeta + k_d\omega_n)$ |
| Ramp $e_{ss}$ | $2\zeta/\omega_n$ | $2\zeta/\omega_n + k_d$ |

So both the damping ratio and the steady state error increase. This error can be reduced again by raising the forward gain.

## Check yourself

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

1. What does velocity feedback do to the characteristic equation?
   A. Increases the s^2 coefficient
   B. Increases the s coefficient to 2*zeta*wn + kd*wn^2
   C. Changes wn^2
   D. Adds a pole at the origin
   Answer: B. The denominator becomes s^2+(2*zeta*wn+kd*wn^2)s+wn^2, so only the s term grows. (Key point: Effect on characteristic equation)

2. The new damping ratio with velocity feedback is
   A. zeta - kd*wn/2
   B. zeta*kd
   C. zeta + kd*wn/2
   D. zeta/(1+kd)
   Answer: C. Equating 2*zeta'*wn = 2*zeta*wn + kd*wn^2 gives zeta' = zeta + kd*wn/2, which is larger. (Key point: New damping ratio)

3. With velocity feedback, Kv' is
   A. wn/(2*zeta)
   B. wn/(2*zeta + kd*wn)
   C. wn*kd
   D. infinite
   Answer: B. G' = wn^2/[s(s+2*zeta*wn+kd*wn^2)], so Kv' = wn/(2*zeta+kd*wn), which is smaller than before. (Key point: Velocity error constant)

4. What happens to the ramp steady state error and why?
   A. It decreases because damping increases
   B. It is unchanged
   C. It becomes zero
   D. It increases by kd because Kv falls
   Answer: D. e_ss' = 2*zeta/wn + kd, larger than 2*zeta/wn. Better damping costs ramp accuracy. (Key point: Ramp error)

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