---
title: "Find K and Kt for ζ=0.6, ts=0.1 s"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:k-kt-zeta-ts"
source: https://www.bctengineer.com/q/semester_3/control_systems/k-kt-zeta-ts
site: BCT Engineering
---

# Find K and Kt for ζ=0.6, ts=0.1 s

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2082 Baisakh, 8 marks; 2080 Bhadra, 8 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/k-kt-zeta-ts
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

For a control system shown in figure below, find the value of K & Kt so that the damping ratio (ζ) of system is 0.6 & settling time (ts) is 0.1 sec for the unit step response. [Figure: R(s) → summing point (+, −) → K → summing point (+, −) → 100/(1+0.2s) → 1/(20s) → C(s). Minor loop: output of 100/(1+0.2s) block fed back through Kt to the second summing point (−). Major loop: C(s) fed back with unity gain to the first summing point (−).]

## Short answer

Using 2% settling criterion: ts = 4/(ζ ωn) = 0.1 s → ζ ωn = 40, ωn = 66.67 rad/s. Then K = ωn^2/25 ≈ 177.8, Kt = (2ζ ωn/5 − 1)/100 = 0.15.

## Answer

#### Step 1: Reduce the minor (tachometer) loop
The minor loop has forward block $\dfrac{100}{1 + 0.2s}$ and feedback $K_t$ taken from its own output:
$$
G_{in}(s) = \frac{\frac{100}{1 + 0.2s}}{1 + \frac{100K_t}{1 + 0.2s}} = \frac{100}{0.2s + 1 + 100K_t}
$$

#### Step 2: Open-loop transfer function
$$
G(s) = K\cdot\frac{100}{0.2s + 1 + 100K_t}\cdot\frac{1}{20s} = \frac{5K}{s(0.2s + 1 + 100K_t)} = \frac{25K}{s\left[s + 5(1 + 100K_t)\right]}
$$

#### Step 3: Closed-loop characteristic equation (unity feedback)
$$
s^2 + 5(1 + 100K_t)s + 25K = 0
$$
Comparing with $s^2 + 2\zeta\omega_ns + \omega_n^2 = 0$:
$$
\omega_n^2 = 25K, \qquad 2\zeta\omega_n = 5(1 + 100K_t)
$$

#### Step 4: Use the specifications
Settling time (2% criterion):
$$
t_s = \frac{4}{\zeta\omega_n} = 0.1 \;\Rightarrow\; \zeta\omega_n = 40
$$
$$
\omega_n = \frac{40}{0.6} = 66.67\ \text{rad/s}
$$

#### Step 5: Solve for $K$ and $K_t$
$$
\begin{aligned}
25K &= \omega_n^2 = (66.67)^2 = 4444.4 \;\Rightarrow\; K = 177.8 \\
5(1 + 100K_t) &= 2\zeta\omega_n = 80 \;\Rightarrow\; 1 + 100K_t = 16 \;\Rightarrow\; K_t = 0.15
\end{aligned}
$$

**Answer:** $K \approx 177.8$ and $K_t = 0.15$ (using the 2% settling criterion).

If the 5% criterion $t_s = 3/(\zeta\omega_n)$ is used instead, $\zeta\omega_n = 30$, $\omega_n = 50$ rad/s, giving $K = 100$ and $K_t = 0.11$.

Note how the tachometer feedback $K_t$ sets the damping, while $K$ sets the natural frequency, so both specifications can be met independently.

## Check yourself

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

1. What does the minor loop 100/(1+0.2s) with feedback Kt reduce to?
   A. 100/(0.2s + 1 + 100Kt)
   B. 100/(0.2s + 1 - 100Kt)
   C. 100Kt/(0.2s + 1)
   D. 100/(1 + 0.2s + Kt)
   Answer: A. G/(1+GKt) with G = 100/(1+0.2s) gives 100/(1+0.2s+100Kt). (Key point: Minor loop reduction)

2. What is the closed-loop characteristic equation?
   A. s^2 + 5(1+100Kt)s + 25K = 0
   B. s^2 + 5s + 25K = 0
   C. s^2 + (1+100Kt)s + K = 0
   D. s^2 + 5(1+100Kt)s + 5K = 0
   Answer: A. Open loop is 25K/(s[s+5(1+100Kt)]); adding 1 gives this equation. (Key point: Characteristic equation)

3. With the 2% criterion ts = 0.1 s, what is ζωn?
   A. 40 rad/s
   B. 30 rad/s
   C. 0.4 rad/s
   D. 4 rad/s
   Answer: A. ts = 4/(ζωn) = 0.1 gives ζωn = 40. (Key point: Settling time)

4. What values meet the specifications (2% criterion)?
   A. K ≈ 177.8, Kt = 0.15
   B. K = 100, Kt = 0.11
   C. K ≈ 177.8, Kt = 0.11
   D. K = 100, Kt = 0.15
   Answer: A. ωn = 66.67 gives 25K = 4444 so K = 177.8; 5(1+100Kt) = 80 gives Kt = 0.15. (Key point: Results)

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