Question

An autonomous robot to pick coffee is capable of following planting rows has an orientation system with the transfer function below. Make a sketch of (t) in response to Orer(t)=3u(t). Indicate the % of overshoot, Peak time and settling time. Also identify the type of system. - 53.176 Oref 4.6s2 + 31.281s+ 53. 176

Answer 1

e(s) Orer(s) 53.176 4.6s2 + 31.281s+ 53.176

(5) 53.176 Orop (s) 4.6 (52 +31.2815 + 53.176)

(s) Orer(s) 11.56 52 + 6.80022s + 11.56

The denominator can be factorized as

() Orer(s) 11.56 (s +3.3728) (s +3.4275)

O(S) 11.56 T5 + 3.3728) (s +3.4275 3728)(s +3.4275) Cres (s)

The input is given as

Orer(t) = 3u(t)

Taking Laplace Transform, we get

r(s) =

So

11.56 X 3 0(s) ss +3.3728) (s +3.4275)

34.68 s(s+ 3.3728) (s + 3.4275)

Taking partial fraction expansion, we get

в 34.68 s(s+3.3728) (s+ 3.4275) А s s+ 3.3728 s+ 3.4275

34.68 34.68 A = 5(8+3.3728)(8 + 3.4275)*sx 3.3728 * 3.4275 = 2.999929 * 3

34.68 s(s + 3.3728) (s + 3.4275) = 34.68 20) -187.975 's=-3.3728 -3.3728 x 0.0547

34.68 34.68 s's + 3.3728)(s + 3.4275) 3.4275) 's=-3.4275 -3.4275 -0.0547 – 184.976

So

3 As) =-- 187.975 184.976 s +3.3728's+3.4275 s

Taking Inverse Laplace Transform we get

e(t) = 3u(t) – 187.975e-3.3728t u(t) + 184.976e -3.4275* u(t)

This can be plotted to obtain the response.

Using MATLAB:

The transfer function is given as

e(s) Orer(s) 53.176 4.6s2 + 31.281s+ 53.176

So the numerator coefficient is

num = [53.176]

Denominator coefficient is

den = [4.6, 31.28, 53.176]

The MATLAB Code

clc;
clear all;
close all;

num = [53.176];
den = [4.6, 31.281, 53.176];

sys = tf(3*num, den);

step(sys);
grid;
title('Response of the system');

After executing, we get

Response of the system 2.5 Amplitude 0.5 NL 0.5 2.5 1.5 Time (seconds)

The specifications are marked as follows

Response of the system -.-.- System: sys Settling time (seconds): 1.72 System: sys Peak amplitude: >= 3 Overshoot (%): 0 At time (seconds): > 3 2.5 Amplitude 0.5 NL 0.5 2.5 1.5 Time (seconds)

There is no overshoot. The setting time is 1.72 seconds