---
title: "Error constants of 108/[s²(s+4)(s²+3s+12)], r = 2+5t+8t²"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:sse-108-s2-input-2-2"
source: https://www.bctengineer.com/q/semester_3/control_systems/sse-108-s2-input-2-2
site: BCT Engineering
---

# Error constants of 108/[s²(s+4)(s²+3s+12)], r = 2+5t+8t²

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2074 Asoj, 8 marks; 2081 Baisakh, 6 marks (asked 2 times)
- 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-108-s2-input-2-2
- 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) = 108/[s²(s+4)(s²+3s+12)]. Find the static error coefficients and steady state error of the system when subjected to an input given by r(t) = 2 + 5t + 8t².

## Short answer

The system is type 2, so Kp = ∞, Kv = ∞, Ka = 2.25. For r(t)=2+5t+8t², step and ramp errors are zero; the parabolic component (C=16) gives ess = 16/Ka = 7.11.

## Answer

The steady state error of a unity feedback system depends on the **type** of $G(s)$ and on the static error coefficients $K_p$, $K_v$ and $K_a$.

#### Type of the system
$$
G(s) = \frac{108}{s^2(s+4)(s^2+3s+12)}
$$
There are two poles at the origin, so the system is **Type 2**.

#### Static error coefficients
$$
\begin{aligned}
K_p &= \lim_{s\to 0} G(s) = \infty \\
K_v &= \lim_{s\to 0} sG(s) = \lim_{s\to 0}\frac{108}{s(s+4)(s^2+3s+12)} = \infty \\
K_a &= \lim_{s\to 0} s^2G(s) = \frac{108}{(4)(12)} = \frac{108}{48} = 2.25
\end{aligned}
$$

#### Error for each input term
For a unity feedback system:

| Input | Error formula | Value here |
|---|---|---|
| Step $A$ | $A/(1+K_p)$ | $0$ |
| Ramp $Bt$ | $B/K_v$ | $0$ |
| Parabola $Ct^2/2$ | $C/K_a$ | finite |

Input: $r(t) = 2 + 5t + 8t^2$.

- Step part $2$: $e_1 = \dfrac{2}{1+\infty} = 0$
- Ramp part $5t$: $e_2 = \dfrac{5}{\infty} = 0$
- Parabolic part $8t^2 = 16\cdot\dfrac{t^2}{2}$, so $C = 16$: $e_3 = \dfrac{16}{K_a} = \dfrac{16}{2.25} = 7.11$

By superposition:
$$
e_{ss} = 0 + 0 + \frac{16}{2.25} = \frac{64}{9} = 7.11
$$

**Answer:** $K_p = \infty$, $K_v = \infty$, $K_a = 2.25$, and $e_{ss} = 7.11$ (units of the output).

**Note on stability:** the closed-loop characteristic equation is
$$
s^5 + 7s^4 + 24s^3 + 48s^2 + 0\cdot s + 108 = 0
$$
The $s$ term is missing, so by the R-H necessary condition the closed loop is actually unstable, and strictly a steady state does not exist. The value $7.11$ is the result the error-coefficient method gives, which is what is normally expected for this question.

## Check yourself

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

1. What is the type of G(s) = 108/[s^2(s+4)(s^2+3s+12)]?
   A. Type 2
   B. Type 1
   C. Type 0
   D. Type 4
   Answer: A. There are two poles at the origin, so it is type 2. (Key point: System type)

2. Which set of coefficients is correct?
   A. Kp = ∞, Kv = ∞, Ka = 2.25
   B. Kp = 0, Kv = 0, Ka = 2.25
   C. Kp = ∞, Kv = 2.25, Ka = 0
   D. Kp = ∞, Kv = ∞, Ka = ∞
   Answer: A. Ka = lim s^2 G(s) = 108/(4×12) = 2.25; Kp and Kv are infinite for type 2. (Key point: Static constants)

3. For r(t) = 2 + 5t + 8t^2, what is the constant C in C t^2/2?
   A. 16
   B. 8
   C. 4
   D. 2
   Answer: A. 8t^2 = 16 × (t^2/2), so C = 16. (Key point: Parabolic term)

4. What error does the coefficient method give, and what caveat applies?
   A. 7.11, but the closed loop is actually unstable (missing s term)
   B. 0, since the system is type 2
   C. 2.25, and the system is stable
   D. 7.11, and the system is stable
   Answer: A. ess = 16/2.25 = 7.11, but the characteristic equation has no s term, so it is unstable. (Key point: Steady-state error)

## Close variants set in other exams

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

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