---
title: "Error constants of 5/[s(s+2)(s²+2s+8)], r = 2+5t+2t²"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:sse-5-s-s2-s2-2s"
source: https://www.bctengineer.com/q/semester_3/control_systems/sse-5-s-s2-s2-2s
site: BCT Engineering
---

# Error constants of 5/[s(s+2)(s²+2s+8)], r = 2+5t+2t²

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2067 Asar (old course), 8 marks
- Video: [Steady state error | Positional, Velocity, Acceleration error | Examples solved |Mathspedia |](https://www.youtube.com/watch?v=2gF-m2maHiE) — Mathspedia
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/sse-5-s-s2-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) = 5/[s(s+2)(s²+2s+8)]. Find the static error coefficients and steady state error of the system when subjected to an input given by r(t) = 2 + 5t + 2t².

## Short answer

The system is type 1. Static error coefficients: Kp=∞, Kv=0.3125, Ka=0. For r(t)=2+5t+2t², the steady-state error is infinite because the parabolic term gives an infinite error component.

## Answer

The steady state error of a unity feedback system follows from its static error coefficients. $G(s)$ has one pole at the origin, so the system is **Type 1**.

#### Stability check
$1 + G = 0$: $s^4 + 4s^3 + 12s^2 + 16s + 5 = 0$.

| Row | | | |
|---|---|---|---|
| $s^4$ | 1 | 12 | 5 |
| $s^3$ | 4 | 16 | |
| $s^2$ | 8 | 5 | |
| $s^1$ | 13.5 | | |
| $s^0$ | 5 | | |

No sign change, so the system is stable.

#### Static error coefficients
$$
\begin{aligned}
K_p &= \lim_{s\to 0} G(s) = \infty \\
K_v &= \lim_{s\to 0} sG(s) = \frac{5}{(2)(8)} = 0.3125 \\
K_a &= \lim_{s\to 0} s^2G(s) = 0
\end{aligned}
$$

#### Steady state error for $r(t) = 2 + 5t + 2t^2$
Write $2t^2 = 4\cdot\dfrac{t^2}{2}$.
$$
\begin{aligned}
e_{ss} &= \frac{2}{1 + K_p} + \frac{5}{K_v} + \frac{4}{K_a} \\
&= 0 + \frac{5}{0.3125} + \frac{4}{0} = 0 + 16 + \infty = \infty
\end{aligned}
$$

| Input part | Error |
|---|---|
| Step 2 | 0 |
| Ramp 5t | 16 |
| Parabola 2t² | ∞ |

**Answer:** $K_p = \infty$, $K_v = 0.3125$, $K_a = 0$; $e_{ss} = \infty$. A Type 1 system cannot follow the parabolic part of the input.

## Check yourself

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

1. G(s)=5/[s(s+2)(s^2+2s+8)] is Type
   A. 1
   B. 0
   C. 2
   D. 4
   Answer: A. Only one pole at the origin. (Key point: System type)

2. Kv = lim sG(s) equals
   A. 5/16 = 0.3125
   B. 5/8
   C. 5/2
   D. 0
   Answer: A. Kv = 5/(2*8) = 0.3125. (Key point: Kv)

3. For r = 2 + 5t + 2t^2, the parabola part is written as A*t^2/2 with A =
   A. 2
   B. 4
   C. 1
   D. 8
   Answer: B. 2t^2 = 4*(t^2/2), so A = 4. (Key point: Parabolic input)

4. The total steady state error is
   A. 16
   B. 0
   C. infinite
   D. 0.3125
   Answer: C. The parabolic part gives 4/Ka = infinity for a Type 1 system, so the total is infinite. (Key point: Total error)

## Close variants set in other exams

- [Error constants of 108/[s²(s+4)(s²+3s+12)], r = 2+5t+8t²](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Asse-108-s2-input-2-2) — Control Systems, chapter 3, asked 4×
- [Error constants of 108/[s²(s+4)(s²+3s+12)], r = 2+5t+2t²](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Asse-108-s2-input-2-5t) — Control Systems, chapter 3, asked 4×

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