Example H6.13
MATLAB code for example 6.13 from the book "Regeltechniek voor het HBO"
Effect of a PD-controller on the course of the root locus
- Date : 02/03/2022
- Revision : 1.0
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Table of Contents
Given problem
In figure 6.16 a second order process is controlled by a so called tamed PD-controller. In chapter 8 this controller is discussed in more detail.
- Determine the value of
such that the absolute damping λ is 
and independent of K. - Then determine the value of K for which
.
Solution assignment 1
From the block diagram we can see that the root locus equation is:
.In "normaal vorm" we get:
. If the absolute damping λ is to be independent of the gain 
the root locus will have to move at a fixed distance (parallel) to the left of the imaginary axis. Figures 6.13d to 6.13f show, this is only possible if there are only 2 closed loop poles. In figure 6.16 this means the value of must be 1/4, because this creates a zero in the numerator of the root locus equation and the pole in 
of the denominator is compensated. So with 
the root locus equation becomes:
the root locus will have to move at a fixed distance (parallel) to the left of the imaginary axis. Figures 6.13d to 6.13f show, this is only possible if there are only 2 closed loop poles. In figure 6.16 this means the value of must be 1/4, because this creates a zero in the numerator of the root locus equation and the pole in of the denominator is compensated. So with the root locus equation becomes:
and the root locus takes the desired course according to figure 6.17. When the poles are complex conjugated the absolute damping factor λ is constant and equal to 
Solution assignment 2
For the second requirement 
, we must demand 
(see example 6.10). If we substitute one of these values in the root locus equation we find the desired value of K: 
.
, we must demand (see example 6.10). If we substitute one of these values in the root locus equation we find the desired value of K: .A simulation in MATLAB (see example 6.12) with of course produces the same root locus. Other fixed values of and variable K can be examined too, but fixed K and a variable for 
is also possible. In the latter case the root locus equation must be reduced to the "normaalvorm" 
, because in that form the expression to the right of the "=" character must be entered in MATLAB. For example for 
, after some calculations, for the root locus equation for , it can be found:
of course produces the same root locus. Other fixed values of and variable K can be examined too, but fixed K and a variable for is also possible. In the latter case the root locus equation must be reduced to the "normaalvorm" , because in that form the expression to the right of the "=" character must be entered in MATLAB. For example for , after some calculations, for the root locus equation for , it can be found:
. We now have the special situation that the degree of the numerator is higher than that of the denominator, so there are more end points (zeros) than start point (poles). So there must be one starting point in infinite whose brache ends in a finite end point (so that branch only can be the real axis, right?). We won't go into that here.