---
title: "K limiting disturbance effect to 2%"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:disturbance-5k-s10-6-s2"
source: https://www.bctengineer.com/q/semester_3/control_systems/disturbance-5k-s10-6-s2
site: BCT Engineering
---

# K limiting disturbance effect to 2%

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2079 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/disturbance-5k-s10-6-s2
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

The system below shows a potential tracking problem with the reference input r(t) and disturbance d(t). What value of K will limit the steady state component of c(t) due to d(t) to 2% of d(t)? [Figure: r(t) → summing point (+, −) → 5K/(s+10) → summing point (+, + with disturbance d(t)) → 6/(s+2) → c(t); unity feedback from c(t) to the first summing point (−)]

## Short answer

The steady-state output from the disturbance is c_ss = 3d/(1+1.5K). Setting this to 0.02d gives K >= 99.33, so choose K = 100 to limit it to 2%.

## Answer

With $R = 0$, the output due to disturbance is found from the loop with $G_2$ forward and $G_1$ in feedback; then the final value theorem gives its steady state.

Let $G_1 = \dfrac{5K}{s+10}$ and $G_2 = \dfrac{6}{s+2}$.

#### Transfer function from $D$ to $C$ (set $R = 0$)
$$
\frac{C(s)}{D(s)} = \frac{G_2}{1 + G_1G_2} = \frac{\frac{6}{s+2}}{1 + \frac{30K}{(s+10)(s+2)}} = \frac{6(s+10)}{(s+10)(s+2) + 30K}
$$

#### Steady state for a step disturbance $D(s) = d/s$
$$
c_{ss} = \lim_{s\to 0} s\cdot\frac{d}{s}\cdot\frac{6(s+10)}{(s+10)(s+2)+30K} = \frac{60\,d}{20 + 30K} = \frac{3d}{1 + 1.5K}
$$

#### Condition: $c_{ss} \le 2\%$ of $d$
$$
\begin{aligned}
\frac{3}{1 + 1.5K} &\le 0.02 \\
1 + 1.5K &\ge 150 \\
K &\ge \frac{149}{1.5} = 99.33
\end{aligned}
$$

**Stability check:** characteristic equation $s^2 + 12s + 20 + 30K = 0$; all coefficients positive for $K > 0$, so the second order system is stable for this $K$.

**Answer:** $K \ge 99.33$ (about $K = 100$) keeps the steady state output due to the disturbance within 2% of $d(t)$.

## Check yourself

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

1. With R = 0, what is C(s)/D(s)?
   A. G2/(1 + G1G2) = 6(s+10)/((s+10)(s+2) + 30K)
   B. G1G2/(1 + G1G2)
   C. G2(1 + G1G2)
   D. G1/(1 + G1G2)
   Answer: A. The disturbance enters before G2, so the forward path is G2 and the loop is G1G2. (Key point: C/D transfer function)

2. What is c_ss for a step disturbance of size d?
   A. 6d/(1 + K)
   B. 3d/(1 + 1.5K), from the final value theorem
   C. d/(1 + 30K)
   D. 3d K
   Answer: B. s -> 0 gives 60d/(20 + 30K) = 3d/(1 + 1.5K). (Key point: Steady state due to step disturbance)

3. What condition on K limits c_ss to 2% of d?
   A. K >= 9.9
   B. K >= 1.33
   C. 3/(1+1.5K) <= 0.02, so K >= 99.33
   D. K <= 99.33
   Answer: C. 1 + 1.5K >= 150 gives K >= 149/1.5 = 99.33. (Key point: Requirement for K)

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