---
title: "Error constants of 1000/[s(s+10)(s+100)], r = 2+t"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:sse-1000-s-s10-s100"
source: https://www.bctengineer.com/q/semester_3/control_systems/sse-1000-s-s10-s100
site: BCT Engineering
---

# Error constants of 1000/[s(s+10)(s+100)], r = 2+t

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2070 Chaitra, 6 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-1000-s-s10-s100
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Find all static error constants for a unity feedback system with feedforward transfer function G(s) = 1000/[s(s+10)(s+100)]. Evaluate steady state error if system is excited with r(t) = 2 + t.

## Short answer

K_p = ∞, K_v = 1 s⁻¹, K_a = 0. For r(t) = 2 + t, steady‑state error is e_ss = 2/(1+K_p) + 1/K_v = 0 + 1 = 1.

## Answer

The steady state error of a unity feedback system follows from its static error constants. The system is **Type 1** (one pole at the origin).

#### Stability check
$1 + G = 0$ gives $s^3 + 110s^2 + 1000s + 1000 = 0$. Routh: $110 \times 1000 > 1000$, so all first-column terms are positive and the system is stable.

#### Static error constants
$$
\begin{aligned}
K_p &= \lim_{s\to 0} G(s) = \lim_{s\to 0}\frac{1000}{s(s+10)(s+100)} = \infty \\
K_v &= \lim_{s\to 0} sG(s) = \frac{1000}{(10)(100)} = 1 \\
K_a &= \lim_{s\to 0} s^2G(s) = 0
\end{aligned}
$$

#### Steady state error for $r(t) = 2 + t$
By superposition:
$$
\begin{aligned}
e_{ss} &= \frac{2}{1 + K_p} + \frac{1}{K_v} \\
&= \frac{2}{1 + \infty} + \frac{1}{1} = 0 + 1 = 1
\end{aligned}
$$

| Input part | Error |
|---|---|
| Step 2 | 0 |
| Ramp t | 1 |

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

## Check yourself

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

1. G(s)=1000/[s(s+10)(s+100)] is of type
   A. 0
   B. 2
   C. 3
   D. 1
   Answer: D. One pole at the origin means Type 1. (Key point: System type)

2. Kv = lim sG(s) equals
   A. 1000
   B. 1
   C. 10
   D. infinity
   Answer: B. Kv = 1000/(10*100) = 1 per second. (Key point: Kv)

3. Kp and Ka for a Type 1 system are
   A. Kp = 0, Ka = infinity
   B. Kp = infinity, Ka = 0
   C. both infinity
   D. both 0
   Answer: B. Type 1 has Kp = infinity and Ka = 0. (Key point: Error constants)

4. For r(t)=2+t, ess is
   A. 0
   B. 2
   C. 3
   D. 1
   Answer: D. Step part gives 2/(1+Kp)=0, ramp part gives 1/Kv=1, so ess = 1. (Key point: Total error)

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