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

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

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2078 Bhadra, 8 marks
- Video: [Steady-State Error #1, using Error Constants](https://www.youtube.com/watch?v=Idk9OkB2fuY) — The Ryder Project
- 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-2-5t
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

The open loop transfer function of 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 + 2t².

## Short answer

The system is type 2, so Kp = infinity, Kv = infinity, Ka = 2.25. The steady-state error for input 2+5t+2t^2 is e_ss = 4/Ka = 1.78.

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

- Step part $2$: $e_1 = \dfrac{2}{1+\infty} = 0$
- Ramp part $5t$: $e_2 = \dfrac{5}{\infty} = 0$
- Parabolic part $2t^2 = 4\cdot\dfrac{t^2}{2}$, so $C = 4$: $e_3 = \dfrac{4}{K_a} = \dfrac{4}{2.25} = 1.78$

By superposition:
$$
e_{ss} = 0 + 0 + \frac{4}{2.25} = \frac{16}{9} = 1.78
$$

**Answer:** $K_p = \infty$, $K_v = \infty$, $K_a = 2.25$, and $e_{ss} = 1.78$ (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 unstable and strictly no steady state exists. The value $1.78$ is the result the error-coefficient method gives, which is the answer normally expected here.

## 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 0
   B. Type 2, because there are two poles at the origin
   C. Type 1
   D. Type 4
   Answer: B. The type equals the number of poles at s = 0, which is 2 here. (Key point: System type)

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

3. How is the parabolic part 2t^2 of the input treated?
   A. As C = 2, so error is 2/Ka
   B. As C = 8, so error is 8/Ka
   C. As 4 x t^2/2, so C = 4 and the error is 4/Ka
   D. As a ramp with B = 2
   Answer: C. The standard parabola is C t^2/2, so 2t^2 = 4(t^2/2) and C = 4. (Key point: Parabolic component)

4. What is ess by superposition?
   A. 0 + 5/Kv + 4/Ka = 1.78 + infinity
   B. 2/(1+Kp) + 5 + 4 = 9
   C. 2.25/4 = 0.5625
   D. 0 + 0 + 4/2.25 = 1.78
   Answer: D. The step and ramp terms vanish for a Type 2 system, leaving 16/9 = 1.78. (Key point: Total steady state 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 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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