Example H6.4

Matlab code for example 6.4 from the book "Regeltechniek voor het HBO"

Beneficial use of the symmetry of a root locus

Date : 02/03/2022
Revision : 1.0

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Beneficial use of the symmetry of a root locus Beneficial use of the symmetry of a root locus Matlab code for this example

Beneficial use of the symmetry of a root locus

Suppose

in the system of figure 6.4 is:

and thus the transfer function of the feed back system is:

and thus the root locus equation in "normaalvorm" is:

and thus there is only one starting point in

and one end point in

, and thus there is only one root locus branche.

It has been posted before, that complex poles (and also zeros) in physical systems only occur as complex conjugated pairs. That's because the parameters of the differential equation describing the behaviour in the real time domain are real. See for instance the spring mass damper system of section 2.5.2. The poles are determined by solving

with the abc-formula, and if the solutions (poles!) are complex, than they are complex conjugated because

and

are real.

Because of the symmetry requirement around the real axis, the pole branch that starts in

and ends in

can only run along the real axis. The root locus of this system therefore looks as shown in figure 6.5.

Generally we indicate by means of arrows in the root locus how the poles move with increasing

If we want to know the value of

at some point s (on the root locus!) we can substitute that value for s in the root locus equation.

For instance for

we will find

Sustituting a value of s outside the root locus will yield of course a value of

that is not allowed.

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 = 6;

z1 = 2;

% Create the 2nd order proces

H=K*(s+z1)/(s+p1);

figure(101);

% Show the root locus plot of H

rlocus(H);

grid on;