Example H8.4

MATLAB code for example 8.4 from the book "Regeltechniek voor het HBO"

First order process with tamed PD controller

Date : 02/03/2022
Revision : 1.0

Table of Contents

First order process with tamed PD controller Root locus equation for K Root locus equation for tau_d Starting and end points of the -locus Analyzing the -locus Conclusion MATLAB code for this example (under construction)

In example 6.13 we met already a tamed PD- controller. Now, in figure 8.20, the first order process is controlled by a tamed PD-action, which causes the addition of one pole and one zero to the system.

Root locus equation for K

The root locus equation for

to detect the poles of the system is:

→

(To find the zeros we would have to solve the equation:

, but that aside for now).

Root locus equation for tau_d

The root locus equation for

is also:

→

If we choose

, the root locus equation for

will be:

Starting and end points of the τ-locus

For

and still

the solutions of the root locus equation are:

and

The latter is true because:

and:

too.

and

are starting points of the (sometimes so called) τ-locus (instead of root locus).

Still for

and

the solutions are:

and

. So 0 and

are ending points of the τ-locus.

In combination with what we know of the parts of the real axis that belong to the root locus (alternating and starting to the right with 'not' as we discussed before): to the right of 0 no, between 0 and

yes, between

and

no, to the left of

again yes ), it is immediately clear that the root locus proceeds as shown in figure 8.21 for

Analyzing the τ-locus

At first sight, it could be concluded from figure 8.21, that by adding the D-action the system becomes slower due to the dominant (!) pole moving from

to the right, so closer to the imaginary axis which causes a longer transient as a function of time in a step response (see figure 3.11). But in addition to the behaviour as a function of time, the magnitude of each transient also is important. Generally we don't pay much attention to it, but a very fast but large transient can still surpass a very small but slow transient (both e-powers actually last infinitely long), and then the fast transient determines the duration of the transient. That appears to be true in this case. For example for

and still

, the transfer function of the total system is:

with

and

Figure 8.22a shows the pole and zero plot in this case. So, the transient in a step response contains the components

and

In section 3.4.4 we discussed how to determine

and

graphically:

and

, almost

times bigger than

. In this case, because of the large difference between

and

, the response is almost entirely determined by the pole in

and the system has become much faster due to the D-action (in fact the pole in

has become dominant).

If we take

and still

, then the poles and zero of

appear to be in the s-plane as is shown in the pole and zero plot of figure 8.22b. The pole and zero in

compensate each other. Now the system is a pure first order system with one pole in

and much faster due to the D-action.

Even for larger values of

than

the influence of the rightmost pole in the total step response remains small. This is easy to check by simulation in MATLAB.

Conclusion

From this example we can draw the general conclusion that a zero close to a pole will reduce the contribution (in magnitude!) of that pole in the transient. In section 3.4.4 we clearly see, that the numerator of

(belonging to

) contains a very small coefficient because of the very small distance from that zero to the pole. So

is very small. The opposite holds for two poles close to each other: from section 3.4.4 again we can see, that the denominator of the component

of each pole contains a very small coefficient because of the very small distance to the other pole. So the contribution

of each pole to the transient (in magnitude) is very big. A simulation in MATLAB you can do will confirm this.

MATLAB code for this example (under construction)

% clear all variables from Workspace and close all figures.
clear variables;
close all;
 
% Define 's' variable
s=tf('s');

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