---
title: "Open loop TF and ramp error from (Ks+b)/(s²+as+b)"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:cltf-ks-b-ramp-ess"
source: https://www.bctengineer.com/q/semester_3/control_systems/cltf-ks-b-ramp-ess
site: BCT Engineering
---

# Open loop TF and ramp error from (Ks+b)/(s²+as+b)

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2074 Chaitra, 5 marks; 2069 Chaitra, 6 marks (asked 2 times)
- Video: [Problem to Calculate Steady State Error and Open Loop Transfer Function](https://www.youtube.com/watch?v=96Ep5bGxr70) — TutorialsPoint
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/cltf-ks-b-ramp-ess
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Consider a unity feedback control system with the closed loop transfer function C(s)/R(s) = (Ks + b)/(s² + as + b). Determine the open loop transfer function. Show that the steady state error in the unit ramp input response is given by ess = (a − K)/b.

## Short answer

Open-loop transfer function G(s) = (Ks + b) / [s(s + a - K)]. The system is type 1, so velocity error constant Kv = b/(a - K), giving steady-state error for unit ramp ess = (a - K)/b.

## Answer

#### Open-loop transfer function
For unity feedback, $T(s) = \dfrac{G(s)}{1 + G(s)}$, so
$$
G(s) = \frac{T(s)}{1 - T(s)}
$$
$$
1 - T(s) = \frac{s^2 + as + b - Ks - b}{s^2 + as + b} = \frac{s^2 + (a - K)s}{s^2 + as + b}
$$
$$
G(s) = \frac{Ks + b}{s^2 + (a - K)s} = \frac{Ks + b}{s\left[s + (a - K)\right]}
$$
The open-loop system is **type 1** (one pole at the origin).

#### Steady-state error for unit ramp
For a unit ramp $R(s) = 1/s^2$, $e_{ss} = \dfrac{1}{K_v}$ where the velocity error constant is
$$
K_v = \lim_{s\to 0} sG(s) = \lim_{s\to 0}\frac{Ks + b}{s + (a - K)} = \frac{b}{a - K}
$$
Therefore
$$
e_{ss} = \frac{1}{K_v} = \frac{a - K}{b}
$$

#### Direct check from the error transfer function
$$
E(s) = R(s)\left[1 - T(s)\right] = \frac{1}{s^2}\cdot\frac{s\left[s + (a - K)\right]}{s^2 + as + b}
$$
$$
e_{ss} = \lim_{s\to 0}sE(s) = \lim_{s\to 0}\frac{s + (a - K)}{s^2 + as + b} = \frac{a - K}{b}
$$
Hence proved (valid when the closed loop is stable, i.e. $a > 0$, $b > 0$). Choosing $K = a$ makes the ramp error zero, because the open loop then becomes type 2.

## Check yourself

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

1. For unity feedback, how is G(s) obtained from T(s)?
   A. G = T/(1 - T)
   B. G = T/(1 + T)
   C. G = 1/(1 - T)
   D. G = T(1 - T)
   Answer: A. T = G/(1+G) rearranges to G = T/(1-T). (Key point: Open-loop from closed-loop)

2. What is G(s)?
   A. (Ks + b)/(s[s + (a - K)])
   B. (Ks + b)/(s^2 + as + b)
   C. (Ks + b)/(s[s + a + K])
   D. b/(s[s + a])
   Answer: A. 1 - T = (s^2 + (a-K)s)/(s^2+as+b), so G = (Ks+b)/(s(s+a-K)). (Key point: Result for G(s))

3. What is Kv?
   A. b/(a - K)
   B. (a - K)/b
   C. b/a
   D. K/b
   Answer: A. Kv = lim s G(s) = b/(a - K) since G is type 1. (Key point: Velocity constant)

4. What happens to the ramp error when K = a?
   A. It becomes zero as the system becomes type 2
   B. It becomes infinite
   C. It stays (a)/b
   D. It doubles
   Answer: A. With K = a the denominator becomes s^2, a type 2 system, and ess = (a-K)/b = 0. (Key point: Special case)

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## Related in other subjects

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