Example H6.1
Matlab code for example 6.1 from the book "Regeltechniek voor het HBO"
Root locus of a first order system
- Date : 02/03/2022
- Revision : 1.0
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Table of Contents
Root locus of a first order system
Again we take the first order process with feedback of example 5.6 with 
(figure 6.3a).
(figure 6.3a).The open system has one pole in 
With feedback we get the transfer function:
and for the closed system there is a pole in 
. For 
, this pole is of course equal to that of the open system. An increase in K means the pole moves futher and further to the left along the negative real axis. The set of these poles as a function of K is the root locus of the system (figure 6.3b). In theory a large K-value is attractive (short transient; see figure 3.11), but in practice that is the question. A large K means a very powerfull control of the process and this might not be feasable or desirable.
, this pole is of course equal to that of the open system. An increase in K means the pole moves futher and further to the left along the negative real axis. The set of these poles as a function of K is the root locus of the system (figure 6.3b). In theory a large K-value is attractive (short transient; see figure 3.11), but in practice that is the question. A large K means a very powerfull control of the process and this might not be feasable or desirable.Matlab code for this example
% clear all variables from Workspace and close all figures.
clear variables;
close all;
% Define 's' variable
s=tf('s');
% Determine parameters.
K=1;
Tau = 2;
% Create the 2nd order proces
H=K/(Tau*s+1);
% Create feedback
Hcl=feedback(H,1);
% Show stepresonse of system with feedback
figure(101);
Tsim=5*Tau; % Simulation time in seconds
step(Hcl, Tsim);
grid on;
Rootlocus plot
figure(102);
% Show the root locus plot of H
rlocus(H);
grid on;
