---
title: "Tachometer feedback for 16/[s(s+1)] to get ζ = 0.6"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:tacho-16-s-s1-zeta06"
source: https://www.bctengineer.com/q/semester_3/control_systems/tacho-16-s-s1-zeta06
site: BCT Engineering
---

# Tachometer feedback for 16/[s(s+1)] to get ζ = 0.6

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2081 Bhadra, 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/tacho-16-s-s1-zeta06
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

A unity feedback system having feed forward transfer function G(s) = 16/[s(s+1)], determine the value of undamped natural frequency, damping ratio. If tachometer feedback is introduced, the feedback transfer function becomes (1+ks). What should be the value of 'k' to obtain damping ratio 0.6. Also calculate the percentage peak overshoot for unit step response before and after introduction of feedback.

## Short answer

Without tachometer: ωn = 4 rad/s, ζ = 0.125, peak overshoot 67.31%. With feedback H(s)=1+ks, setting ζ=0.6 requires k = 0.2375 s; then overshoot reduces to 9.48%.

## Answer

A second order system $\omega_n^2/(s^2 + 2\zeta\omega_n s + \omega_n^2)$ has peak overshoot $M_p = e^{-\pi\zeta/\sqrt{1-\zeta^2}}$. Tachometer (derivative output) feedback increases the $s$ coefficient, so it increases damping without changing $\omega_n$.

#### Without tachometer feedback
$$
\frac{C(s)}{R(s)} = \frac{G}{1+G} = \frac{16}{s^2 + s + 16}
$$
Comparing with $s^2 + 2\zeta\omega_n s + \omega_n^2$:
$$
\begin{aligned}
\omega_n^2 &= 16 \;\Rightarrow\; \omega_n = 4\ \text{rad/s} \\
2\zeta\omega_n &= 1 \;\Rightarrow\; \zeta = \frac{1}{2\times 4} = 0.125
\end{aligned}
$$
$$
M_p = e^{-\pi(0.125)/\sqrt{1-0.125^2}} = e^{-0.3958} = 0.6731 = 67.31\%
$$

#### With tachometer feedback $H(s) = 1 + ks$
$$
\frac{C(s)}{R(s)} = \frac{G}{1+GH} = \frac{16}{s(s+1) + 16(1+ks)} = \frac{16}{s^2 + (1+16k)s + 16}
$$
$\omega_n$ is still $4$ rad/s. For $\zeta = 0.6$:
$$
\begin{aligned}
2\zeta\omega_n &= 1 + 16k \\
2(0.6)(4) = 4.8 &= 1 + 16k \\
k &= \frac{3.8}{16} = 0.2375
\end{aligned}
$$

#### Peak overshoot after adding feedback
$$
M_p = e^{-\pi(0.6)/\sqrt{1-0.36}} = e^{-0.75\pi} = e^{-2.356} = 0.0948 = 9.48\%
$$

#### Summary

| Quantity | Before | After |
|---|---|---|
| $\omega_n$ | 4 rad/s | 4 rad/s |
| $\zeta$ | 0.125 | 0.6 |
| $M_p$ | 67.31% | 9.48% |

**Answer:** $\omega_n = 4$ rad/s, $\zeta = 0.125$; $k = 0.2375$ s; $M_p$ falls from **67.31%** to **9.48%**. Tachometer feedback greatly reduces overshoot without changing $\omega_n$.

## Check yourself

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

1. For G = 16/[s(s+1)] with unity feedback, what are ωn and ζ?
   A. ωn = 4 rad/s, ζ = 0.125
   B. ωn = 16 rad/s, ζ = 0.5
   C. ωn = 4 rad/s, ζ = 0.5
   D. ωn = 2 rad/s, ζ = 0.25
   Answer: A. s^2 + s + 16: ωn^2 = 16 and 2ζωn = 1 so ζ = 0.125. (Key point: Original parameters)

2. What does feedback H = 1 + ks do to the characteristic equation?
   A. Changes it to s^2 + (1+16k)s + 16, raising damping but keeping ωn
   B. Changes ωn to 4k
   C. Reduces the damping
   D. Adds a pole at the origin
   Answer: A. The 16ks term adds to the s coefficient only. (Key point: Effect of tachometer)

3. What k gives ζ = 0.6?
   A. 0.2375
   B. 0.3
   C. 0.6
   D. 0.15
   Answer: A. 2(0.6)(4) = 4.8 = 1 + 16k, so k = 3.8/16 = 0.2375. (Key point: Value of k)

4. How does Mp change?
   A. From 67.31% to 9.48%
   B. From 9.48% to 67.31%
   C. From 67.31% to 0%
   D. From 25% to 9.48%
   Answer: A. Mp = exp(-πζ/sqrt(1-ζ^2)) gives 67.31% at ζ = 0.125 and 9.48% at ζ = 0.6. (Key point: Overshoot)

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