---
title: "Error constants of 20/[s(0.5s+1)(s+2)], r = 10+5t"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:sse-20-s-05s1-s2"
source: https://www.bctengineer.com/q/semester_3/control_systems/sse-20-s-05s1-s2
site: BCT Engineering
---

# Error constants of 20/[s(0.5s+1)(s+2)], r = 10+5t

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2066 Bhadra (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-20-s-05s1-s2
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

The open loop transfer function of a unity negative feedback system is given by G(s) = 20/[s(0.5s+1)(s+2)]. Calculate the static error constants for this system. Also calculate the steady state error due to input r(t) = 10 + 5t.

## Short answer

The system is type 1: Kp=∞, Kv=10 s⁻¹, Ka=0. For r(t)=10+5t, steady-state error ess = 0 + 5/10 = 0.5.

## Answer

The steady state error follows from the static error constants. $G(s)$ has one pole at the origin, so the system is **Type 1**.

#### Static error constants
$$
G(s) = \frac{20}{s(0.5s + 1)(s + 2)}
$$
$$
\begin{aligned}
K_p &= \lim_{s\to 0} G(s) = \infty \\
K_v &= \lim_{s\to 0} sG(s) = \frac{20}{(1)(2)} = 10\ \text{s}^{-1} \\
K_a &= \lim_{s\to 0} s^2G(s) = 0
\end{aligned}
$$

#### Steady state error for $r(t) = 10 + 5t$
$$
\begin{aligned}
e_{ss} &= \frac{10}{1 + K_p} + \frac{5}{K_v} \\
&= \frac{10}{1 + \infty} + \frac{5}{10} = 0 + 0.5 = 0.5
\end{aligned}
$$

**Answer:** $K_p = \infty$, $K_v = 10\ \text{s}^{-1}$, $K_a = 0$; $e_{ss} = 0.5$.

**Note on stability:** $G(s) = \dfrac{40}{s(s+2)^2}$, so the characteristic equation is $s^3 + 4s^2 + 4s + 40 = 0$. Routh's condition for a cubic needs $4 \times 4 > 40$, which fails. So the closed loop as given is unstable, and the error value above (the usual expected answer) holds only in the formal sense. The gain would have to be below 16 (i.e. $K < 8$ in $K/[s(0.5s+1)(s+2)]$) for the result to apply.

## Check yourself

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

1. For G=20/[s(0.5s+1)(s+2)], Kv equals
   A. 20
   B. 5
   C. 10
   D. 40
   Answer: C. Kv = 20/(1*2) = 10. (Key point: Kv)

2. The steady state error for r(t)=10+5t is
   A. 0.5
   B. 1
   C. 5
   D. 0
   Answer: A. Step part is 0 and ramp part is 5/Kv = 0.5. (Key point: Error for 10+5t)

3. The system type is 1, so Ka is
   A. infinity
   B. 10
   C. 20
   D. 0
   Answer: D. A Type 1 system has Ka = 0. (Key point: Type)

4. Why is the quoted error only formal?
   A. Kv is negative
   B. The closed loop is unstable (cubic Routh fails)
   C. The system is Type 0
   D. The input is a parabola
   Answer: B. s^3+4s^2+4s+40 needs 4*4 > 40, which fails. (Key point: Stability caveat)

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