Example H5.7

MATLAB code for example 5.7 from the book "Regeltechniek voor het HBO"

Root locus of a second order process

Date : 02/03/2022
Revision : 1.0

Table of Contents

Root locus of a second order process Root locus of a second order process MATLAB code for this example Rootlocus plot

Root locus of a second order process

In the block diagram of figure 5.11 we now suppose

is a pure second order process:

For the closed sytem we find:

So the root locus is determined by the expression:

, or:

For

we find the two starting points of the locus:

and

(the poles of the open system).

For

we find with the abc-formula:

; poles on the real axis!

So for

we see:

; two equal poles on the real axis!

For

we get:

The real value of

remains constant to

and the imaginary values are getting bigger and bigger (in absolute sense) with increasing

The root locus is shown in figure 5.13. We see that the system remains stable for all

. In terms of stability everything goes well. Other consequences ared discussed later in this book.

Usually we look at the course of a root locus for only positive K. Here we therefore omit the course for negative K. But in that case we are already sure that the startingpoints for

must be

and

. From there the so called branches will depart.

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;

p1 = 1;

p2 = 4;

% Create the 2nd order proces

H=K/((s+p1)*(s+p2));

% 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);