---
title: "Error constants and K range for K/[s(s+1)(s²+2s+2)]"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:sse-k-s-s1-s2-2s"
source: https://www.bctengineer.com/q/semester_3/control_systems/sse-k-s-s1-s2-2s
site: BCT Engineering
---

# Error constants and K range for K/[s(s+1)(s²+2s+2)]

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2066 Jestha (old course), 8 marks
- Video: [Q5.b Find Kp, Kv, Ka and Steady State Error for a system](https://www.youtube.com/watch?v=KzoFcW82DOo) — EnggClasses
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/sse-k-s-s1-s2-2s
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

The open loop transfer function of a unity feedback system is given by G(s) = K/[s(s+1)(s²+2s+2)]. Calculate the static error constants for this system and find the range of K if static error is less than 0.5 for input r(t) = 10 + 50t.

## Short answer

Type 1 system: Kp=∞, Kv=K/2, Ka=0. For r(t)=10+50t, ess=100/K. ess<0.5 demands K>200, but Routh stability limits K to 0<K<2.22, so no K satisfies the error condition with a stable system.

## Answer

$G(s)$ has one pole at the origin, so the system is **Type 1**: the step error is zero and the ramp error is $1/K_v$.

#### Static error constants
$$
\begin{aligned}
K_p &= \lim_{s\to 0} G(s) = \infty \\
K_v &= \lim_{s\to 0} sG(s) = \frac{K}{(1)(2)} = \frac{K}{2} \\
K_a &= \lim_{s\to 0} s^2G(s) = 0
\end{aligned}
$$

#### Steady state error for $r(t) = 10 + 50t$
$$
e_{ss} = \frac{10}{1 + K_p} + \frac{50}{K_v} = 0 + \frac{50}{K/2} = \frac{100}{K}
$$

#### Condition $e_{ss} < 0.5$
$$
\frac{100}{K} < 0.5 \;\Rightarrow\; K > 200
$$

#### Stability check (needed for the result to be valid)
Characteristic equation: $s^4 + 3s^3 + 4s^2 + 2s + K = 0$.

| Row | | | |
|---|---|---|---|
| $s^4$ | 1 | 4 | $K$ |
| $s^3$ | 3 | 2 | |
| $s^2$ | $10/3$ | $K$ | |
| $s^1$ | $2 - 0.9K$ | | |
| $s^0$ | $K$ | | |

For stability: $2 - 0.9K > 0$ and $K > 0$, so $0 < K < 2.22$.

**Answer:** the error condition needs $K > 200$, but the system is stable only for $0 < K < 2.22$. The two ranges do not overlap, so **no value of K** meets $e_{ss} < 0.5$ with a stable system. A compensator (for example lag compensation) is needed to raise $K_v$ without causing instability.

## Check yourself

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

1. For G=K/[s(s+1)(s^2+2s+2)], Kv equals
   A. K
   B. K/2
   C. 2K
   D. K/4
   Answer: B. Kv = K/(1*2) = K/2. (Key point: Kv)

2. For r = 10 + 50t, ess equals
   A. 50/K
   B. 10/K
   C. 200/K
   D. 100/K
   Answer: D. ess = 50/Kv = 50/(K/2) = 100/K. (Key point: Error expression)

3. ess < 0.5 requires
   A. K > 200
   B. K < 200
   C. K > 50
   D. K < 2.22
   Answer: A. 100/K < 0.5 gives K > 200. (Key point: Error condition)

4. The Routh test shows stability only for
   A. K > 200
   B. 0 < K < 2.22
   C. K > 2.22
   D. all K
   Answer: B. The s^1 row 2-0.9K must be positive. The ranges do not overlap, so no K satisfies both. (Key point: Stability conflict)

## Close variants set in other exams

- [Error constants of 4/[s(s²+4s+4)], r = 4t+1](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Asse-4-s-s2-4s-4) — Control Systems, chapter 3, asked 3×

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