Example H5.8

MATLAB code for example 5.8 from the book "Regeltechniek voor het HBO"

Root locus of a first order proces with a zero to the right of the imaginary axis

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Root locus of a first order proces with a zero to the right of the imaginary axis

In a first order process we want to examine the influence of a zero to the right of the imaginary axis. In figure 5.11 we take:
.
The transfer function of the closed system equals:
The equation for the root locus is:
.
For we get: , the pole of the open system; stable because left of the imaginary axis and starting point of the root locus (because ).
For ; unstable or marginally stable.
For ; unstable and the so called end point of the root locus (because ).
Figure 5.14 shows the complete course of the root locus for . We see for that the pole is in the same place as the zero of the open system. Note: despite the fact that for the pole position should be represented by a cross, we still take the circle symbol of the zero of the open system. Actually not correct, but usual.
From figure 5.14 we can conclude that the system remains a first order system (always only one pole). The system behaviour is determined by a variable pole and a constant zero (see formula 5.17). In terms of stability we see, that the system becomes unstable for . See figure 3.11 for the transient in a step response for or simulate a step response with the live script (limit the time to avoid a huge output signal)

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=3;
p1 = 4;
z1 = 1;
 
% Create the 2nd order proces
H=K*(s-z1)/(s+p1);
 
% Create feedback
Hcl=feedback(H,1);
 
 
% Show stepresonse of system with feedback
figure(101);
Tsim=10; % Simulation time in seconds
step(Hcl, Tsim);
grid on;

Rootlocus plot

figure(102);
 
% Show the root locus plot of H
rlocus(H);
grid on;