---
title: "Find a and b for 25% overshoot"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:a-b-j-overshoot-25-tp2"
source: https://www.bctengineer.com/q/semester_3/control_systems/a-b-j-overshoot-25-tp2
site: BCT Engineering
---

# Find a and b for 25% overshoot

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2076 Asoj, 6 marks
- Video: [Numerical 3: Determine K1 & K2 (2nd Order Control System)](https://www.youtube.com/watch?v=f7pp-7VMnl4) — Learning Electronics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/a-b-j-overshoot-25-tp2
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Determine values of a and b of the closed loop control system shown below, so that maximum overshoot for unit step input is 25% and the peak time is 2 sec. Assume that J = 1 kg-m². [Figure: R(s) → summing point (+, −) → summing point (+, −) → a/(Js) → node → 1/s → C(s); minor loop: the node after a/(Js) fed back through b to the second summing point (−); major loop: C(s) fed back with unity gain to the first summing point (−)]

## Short answer

Reduce inner loop to a/[s(Js+ab)], with J=1 closed-loop denominator s²+ab s + a. From 25% overshoot, ζ ≈ 0.4037; from peak time 2 s, ωn ≈ 1.717 rad/s. Thus a = ωn² ≈ 2.95, b = 2ζωn/a ≈ 0.47.

## Answer

Reduce the minor (velocity) feedback loop first, then compare the closed-loop characteristic equation with $s^2 + 2\zeta\omega_n s + \omega_n^2$.

#### Step 1: Reduce the minor loop
The block $\dfrac{a}{Js}$ has feedback $b$:
$$
\frac{a/(Js)}{1 + ab/(Js)} = \frac{a}{Js + ab}
$$
Open-loop transfer function (in series with $1/s$):
$$
G(s) = \frac{a}{s(Js + ab)}
$$

#### Step 2: Closed-loop transfer function (unity feedback, $J = 1$)
$$
\frac{C(s)}{R(s)} = \frac{a}{s^2 + ab\,s + a}
\;\Rightarrow\;
\omega_n^2 = a,\quad 2\zeta\omega_n = ab
$$

#### Step 3: $\zeta$ from $M_p = 25\%$
$$
\zeta = \frac{-\ln 0.25}{\sqrt{\pi^2 + (\ln 0.25)^2}} = \frac{1.3863}{\sqrt{9.8696 + 1.9218}} = \frac{1.3863}{3.4339} = 0.4037
$$

#### Step 4: $\omega_n$ from $t_p = 2$ s
$$
\omega_n = \frac{\pi}{t_p\sqrt{1-\zeta^2}} = \frac{3.1416}{2\sqrt{1 - 0.4037^2}} = \frac{3.1416}{2(0.9149)} = 1.717\ \text{rad/s}
$$

#### Step 5: Find $a$ and $b$
$$
\begin{aligned}
a &= \omega_n^2 = 1.717^2 = 2.948 \\
b &= \frac{2\zeta\omega_n}{a} = \frac{2(0.4037)(1.717)}{2.948} = \frac{1.386}{2.948} = 0.470
\end{aligned}
$$

**Answer:** $a \approx 2.95$ and $b \approx 0.47$ (with $J = 1$ kg-m²).

## Check yourself

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

1. What is the minor loop result for a/(Js) with feedback b?
   A. a/(Js - ab)
   B. ab/(Js + a)
   C. a/(Js + ab)
   D. a/(Js)(1 + b)
   Answer: C. G/(1+GH) with G = a/(Js), H = b gives a/(Js + ab). (Key point: Minor loop reduction)

2. With J = 1, what is C/R?
   A. a/(s^2 + ab s + a), so wn^2 = a and 2 zeta wn = ab
   B. a/(s^2 + a s + b)
   C. ab/(s^2 + ab s + a)
   D. a/(s^2 + b s + a)
   Answer: A. Unity feedback around a/[s(s + ab)] gives a/(s^2 + ab s + a). (Key point: Closed loop transfer function)

3. What zeta gives 25% overshoot?
   A. 0.25
   B. 0.4037
   C. 0.707
   D. 0.5
   Answer: B. zeta = -ln 0.25/sqrt(pi^2 + ln^2 0.25) = 1.3863/3.4339 = 0.4037. (Key point: Damping ratio)

4. With wn = 1.717 rad/s, how are a and b found?
   A. a = wn^2 = 2.95 and b = 2 zeta wn/a = 0.47
   B. a = 2 zeta wn and b = wn^2
   C. a = wn = 1.717 and b = 2 zeta
   D. a = wn^2 = 2.95 and b = 2 zeta wn = 1.39
   Answer: A. wn^2 = a and 2 zeta wn = ab, so b = 1.386/2.948 = 0.47. (Key point: a and b)

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