---
title: "Error constants of 4/[s(s²+4s+4)], r = 4t+1"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:sse-4-s-s2-4s-4"
source: https://www.bctengineer.com/q/semester_3/control_systems/sse-4-s-s2-4s-4
site: BCT Engineering
---

# Error constants of 4/[s(s²+4s+4)], r = 4t+1

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2065 Shrawan (old course), 6 marks
- Video: [#3 Error Coefficients & Steady State Error Calculation When Input is Given | Control System](https://www.youtube.com/watch?v=kZhCDjtXmlE) — ELECTRICAL HUB
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/sse-4-s-s2-4s-4
- 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)H(s) = 4/[s(s²+4s+4)]. Find the static error constants and calculate error due to input r(t) = 4t + 1.

## Short answer

Kp = ∞, Kv = 1 s⁻¹, Ka = 0. For r(t)=4t+1, steady-state error is ess = 4.

## Answer

Static error constants are the limits $K_p=\lim_{s\to0}G(s)H(s)$, $K_v=\lim_{s\to0}sG(s)H(s)$ and $K_a=\lim_{s\to0}s^2G(s)H(s)$. They give the steady-state error for step, ramp and parabolic inputs.

$$
G(s)H(s) = \frac{4}{s(s^2+4s+4)} = \frac{4}{s(s+2)^2}
$$

There is one pole at the origin, so the system is **type 1**.

**Stability check** (needed before using final value theorem): characteristic equation $s^3+4s^2+4s+4=0$. All coefficients are positive and $4\times4=16>1\times4=4$, so by Routh's condition for a cubic the closed loop is stable.

#### Static error constants

$$
\begin{aligned}
K_p &= \lim_{s\to0}\frac{4}{s(s+2)^2} = \infty \\
K_v &= \lim_{s\to0}s\cdot\frac{4}{s(s+2)^2} = \frac{4}{4} = 1\ \text{s}^{-1} \\
K_a &= \lim_{s\to0}s^2\cdot\frac{4}{s(s+2)^2} = 0
\end{aligned}
$$

#### Steady-state error for $r(t)=4t+1$

By superposition, split the input into a step and a ramp:

- Step part: $r_1(t)=1\cdot u(t)$, $A=1$: $e_{ss1} = \frac{A}{1+K_p} = \frac{1}{1+\infty} = 0$
- Ramp part: $r_2(t)=4t$, $A=4$: $e_{ss2} = \frac{A}{K_v} = \frac{4}{1} = 4$

$$
e_{ss} = e_{ss1} + e_{ss2} = 0 + 4 = 4
$$

| Input | Error constant | Error |
|---|---|---|
| Step $1$ | $K_p=\infty$ | 0 |
| Ramp $4t$ | $K_v=1$ | 4 |
| Total | | 4 |

**Answer:** $K_p=\infty$, $K_v=1\ \text{s}^{-1}$, $K_a=0$; steady-state error for $r(t)=4t+1$ is $e_{ss}=4$ (units of the output).

## Check yourself

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

1. For G=4/[s(s+2)^2], Kv equals
   A. 4
   B. 0
   C. 1
   D. infinity
   Answer: C. Kv = 4/4 = 1. (Key point: Kv)

2. For r = 4t + 1, the step part gives error
   A. 1
   B. 0
   C. 4
   D. 0.5
   Answer: B. Kp = infinity so 1/(1+Kp)=0. (Key point: Superposition)

3. The ramp 4t gives error A/Kv =
   A. 1
   B. 0
   C. 4
   D. 0.25
   Answer: C. A = 4, Kv = 1, so 4. (Key point: Ramp part)

4. Why check stability first?
   A. Because Kv needs it
   B. Because the final value theorem is valid only for stable systems
   C. To find Type
   D. To find Ka
   Answer: B. For an unstable loop the error does not settle. (Key point: Stability check)

## Close variants set in other exams

- [Error constants of 20/[s(0.5s+1)(s+2)], r = 10+5t](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Asse-20-s-05s1-s2) — Control Systems, chapter 3, asked 2×
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