Example H5.6

MATLAB code for example 5.6 from the book "Regeltechniek voor het HBO"

The effect of feedback on stability

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Table of Contents

Feedback on a first order system without delay

In the block diagram of figure 5.11 is a pure first order process without time delay:
with .
The open system thus has one pole in .
After feedback the transfer function becomes:
with .
We now see that the pole of the closed process depends on K. For the pole equals the pole of the open system. With increasing (and thus positive) K the pole in figure 5.12 shifts further and further to the left and the transient part in a response will therefore pass faster and faster.
(the transient part equals )
The route of the pole in the s-plane as a function of K is the so called root locus (figure 5.12). As can be seen it is not represented by a continuous row of crosses, but by a solid line. For K-values the pole would move from to the right (about the real axis) and pass the imaginary axis in , making the systen unstable.
From we immediatly see that this is the case for .

Note aside

Take care of the signs/polarities in expressions. As you know the general expression of a transfer function is:
.
So for a certain we have got two zeros in and , and three poles in , and . Don't make mistakes here.

MATLAB code for this example

Notice that if you increase K, the output of the system with feedback start to look more and more like the input of the system (a unit step).
% clear all variables from Workspace and close all figures.
clear variables;
close all;
 
% Define 's' variable
s=tf('s');
K=95;
Tau = 0.1;
H=K/(s*Tau+1);
Hcl=feedback(H,1);
% calculate poles and zeros
zpk(Hcl)
ans = 950 ------- (s+960) Continuous-time zero/pole/gain model.
 
% Show stepresonse of system with feedback
figure(101);
Tsim=0.5; % Simulation time in seconds
step(Hcl, Tsim);
grid on;

Rootlocus plot

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