---
title: "Radar servo velocity error coefficient"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:radar-velocity-error"
source: https://www.bctengineer.com/q/semester_3/control_systems/radar-velocity-error
site: BCT Engineering
---

# Radar servo velocity error coefficient

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2078 Bhadra, 4 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/radar-velocity-error
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

A servomechanism as shown in block diagram below is designed to keep a radar antenna pointed at a flying aeroplane. If the aeroplane is flying with a velocity of 600 km/hr, at a range of 2 km and the maximum tracking error is to be within 0.1°, determine the required velocity error coefficient. [Figure: R(s) → summing point (+, −) → Kv/[s(1+sT)] → C(s), unity negative feedback]

## Short answer

Angular velocity is 4.775 °/s. Required velocity error coefficient Kv ≥ 47.75 s⁻¹.

## Answer

For a type-1 system with a ramp input of slope $\omega$ (constant angular velocity), the steady-state error is $e_{ss} = \frac{\omega}{K_v}$. So the required $K_v = \frac{\omega}{e_{ss,\max}}$.

#### Angular velocity of the line of sight

The antenna must turn at the angular rate of the aeroplane as seen from the radar:

$$
\begin{aligned}
v &= 600\ \text{km/h} = \frac{600\times1000}{3600} = 166.67\ \text{m/s} \\
\omega &= \frac{v}{r} = \frac{166.67}{2000} = 0.08333\ \text{rad/s} \\
&= 0.08333\times\frac{180}{\pi} = 4.775^\circ/\text{s}
\end{aligned}
$$

So the input to the servo is a ramp $\theta_r(t) = 4.775\,t$ degrees.

#### Velocity error coefficient

For $G(s)=\frac{K_v}{s(1+sT)}$ (type 1):

$$
\lim_{s\to0}sG(s) = K_v
$$

$$
e_{ss} = \frac{\omega}{K_v} \le 0.1^\circ
\quad\Rightarrow\quad
K_v \ge \frac{4.775}{0.1} = 47.75\ \text{s}^{-1}
$$

**Answer:** Required velocity error coefficient $K_v \ge 47.75\ \text{s}^{-1}$ (about $48\ \text{s}^{-1}$).

## Check yourself

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

1. Why is the input to the antenna servo treated as a ramp?
   A. The aeroplane's line of sight turns at a constant angular rate
   B. The aeroplane accelerates uniformly
   C. The range changes at constant speed
   D. The antenna has a type-0 plant
   Answer: A. For constant velocity at fixed range, the angle to the aeroplane grows at a constant rate, which is a ramp input. (Key point: Input type)

2. What is the angular velocity of the line of sight for v = 600 km/h at range 2 km?
   A. 0.3 rad/s
   B. 0.0833 rad/s (4.775 deg/s)
   C. 83.3 rad/s
   D. 4.775 rad/s
   Answer: B. v = 166.67 m/s and r = 2000 m, so omega = v/r = 0.0833 rad/s = 4.775 deg/s. (Key point: Angular velocity)

3. For a type-1 system with ramp slope omega, the steady-state error is
   A. omega*Kv
   B. omega/Kv
   C. 1/(1+Kv)
   D. Kv/omega
   Answer: B. For a ramp input, ess = slope/Kv, with Kv = lim s G(s). (Key point: Steady-state error formula)

4. With a maximum error of 0.1 degree, the required Kv is
   A. 0.4775 per s
   B. 4.775 per s
   C. 47.75 per s
   D. 477.5 per s
   Answer: C. Kv >= 4.775 / 0.1 = 47.75 per second. (Key point: Required Kv)

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